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# Solve and graph inequality calculator
This can help the student to understand the problem and how to Solve and graph inequality calculator. Keep reading to learn more!
## Solving and graph inequality calculator
As a student, there are times when you need to Solve and graph inequality calculator. Perform all algebraic or geometric operations you can think of in detail. Use formal reasoning or intuitive insight, or, if possible, use both methods to determine the correctness of each step. If your topic is very complex, you can distinguish between big steps and small steps, and each big step contains several small steps. Check the big steps first, and then go deep into some small steps in turn.
It is closely related to algebra, number theory, topology, differential geometry and mathematical physics. Many modern developments in this field are deeply influenced by these related fields, and in turn affect them. There are many tools needed in this field, ranging from complex analysis to finite field and P-progression. Some of the basic ideas in this topic are profound, such as dynamic form, modulus, or methods from complex numbers to finite fields and back.
The personal income tax calculator can be viewed through Baidu app. Users can download Baidu app, search for the personal income tax calculator, find the personal income tax wizard calculator applet, and enter information such as pre tax income to calculate after tax income. The steps are as follows: 1 find out the important function keys. Scientific calculators have several function keys, which are very important for learning algebra, trigonometry, geometry, calculus, etc.
Do simple math questions first, and then complex math questions; According to your own actual situation, skip the college entrance examination math questions that are really without ideas, from easy to difficult. And the students also said that the math questions in the college entrance examination were very simple, and they believed that all students could get full marks. The students said this lightly, but we in front of the screen are really trembling. Is this something we can say casually? Don't make mathematics seem very simple.
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## Does a hexagon have 6 or 12 lines of symmetry?
The regular hexagon has Dih6 symmetry, order 12.
### Does a hexagon have 4 lines of symmetry?
For all regular polygons, the number of lines of symmetry is equal to the number of sides. That is an equilateral triangle has 3 lines of symmetry, a square has 4 lines of symmetry, similarly a regular hexagon has 6 lines of symmetry. The answer is 6.
#### Does a hexagon have 12 lines of symmetry?
question_answer Answers(12) A hexagon has 6 sides. We cannot tell exactly how many lines of symmetry a hexagon has. But for a regular hexagon (all sides are equal in length) there are 6 lines of symmetry. A regular hexagon with six equal sides has six lines of symmetry.
How do I find the line of symmetry?
To find the line of symmetry algebraically, you need to identify if the equation is written in standard form or vertex form. Standard form is y = ax^2 + bx + c, where a, b, and c equal all real numbers. You can use the formula x = -b / 2a to find the line of symmetry.
What is the angle of a hexagon?
120 degrees
A hexagon has six sides, and we can use the formula degrees = (# of sides – 2) * 180. Then degrees = (6 – 2) * 180 = 720 degrees. Each angle is 720/6 = 120 degrees.
## Is a hexagon the strongest shape?
The hexagon is the strongest shape known. It’s also one of the only shapes which tessellates perfectly (think tiles, if you tiled a wall with hexagons then there wouldn’t be any gaps.
### Which figure has only one line of symmetry?
Kite Rhombus (all sides equal length) 1 Line of Symmetry 2 Lines of Symmetry
#### Which quadrilateral has only one line of symmetry?
Kite
Kite. A kite has one line of symmetry. It has rotational symmetry of order one.
What is a line of symmetry in math?
A line of symmetry is a line that cuts a shape exactly in half. This means that if you were to fold the shape along the line, both halves would match exactly. Equally, if you were to place a mirror along the line, the shape would remain unchanged. A square has 4 lines of symmetry, as shown below.
What is the exterior angle of a hexagon?
60 degrees
d=180(n−1)n, “d” represents interior angle and n is the number of sides in the polygon. So, the value of d or interior angle is 120 degrees. Therefore, the exterior angle of a regular hexagon will be = 180 – 120 = 60 degrees.
## Can a hexagon be any shape?
A hexagon is an example of a polygon, or a shape with many sides. ‘ A regular hexagon has six sides that are all congruent, or equal in measurement. A regular hexagon is convex, meaning that the points of the hexagon all point outward. All of the angles of a regular hexagon are congruent and measure 120 degrees.
### How many axis of symmetry does a hexagon have?
A regular hexagon has 6 axes of symmetry. A circle has an infinite number of axes of symmetry. The figure below has no axis of symmetry.
#### How many intersecting lines does a hexagon have?
The Lemoine hexagon is a cyclic hexagon (one inscribed in a circle) with vertices given by the six intersections of the edges of a triangle and the three lines that are parallel to the edges that pass through its symmedian point.
How do you calculate the length of a hexagon?
The simplest, and by far most common, way of finding the length of a regular hexagon’s sides is using the following formula: s = P ÷ 6, where P is the perimeter of the hexagon, and s is the length of any one of its sides.
Does a hexagon have point symmetry?
This means that a regular hexagon has 6 sides, 6 lines of symmetry and an order of rotational symmetry of 6. Following from this, then a square, which is a regular polygon, has 4 sides, 4 lines of symmetry and an order of rotational symmetry of 4. If a shape has rotational symmetry, it must have either line symmetry or point symmetry or both.
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# Factors of 216 | Find the Factors of 216 by Factoring Calculator
Factoring Calculator calculates the factors and factor pairs of positive integers.
Factors of 216 can be calculated quickly with the help of Factoring Calculator i.e. 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 27, 36, 54, 72, 108, 216 positive integers that divide 216 without a remainder.
Factors of 216 are 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 27, 36, 54, 72, 108, 216. There are 16 integers that are factors of 216. The biggest factor of 216 is 216.
Factors of:
### Factor Tree of 216 to Calculate the Factors
216 2 108 2 54 2 27 3 9 3 3
Factors of 216 are 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 27, 36, 54, 72, 108, 216. There are 16 integers that are factors of 216. The biggest factor of 216 is 216.
Positive integers that divides 216 without a remainder are listed below.
• 1
• 2
• 3
• 4
• 6
• 8
• 9
• 12
• 18
• 24
• 27
• 36
• 54
• 72
• 108
• 216
### Factors of 216 in pairs
• 1 × 216 = 216
• 2 × 108 = 216
• 3 × 72 = 216
• 4 × 54 = 216
• 6 × 36 = 216
• 8 × 27 = 216
• 9 × 24 = 216
• 12 × 18 = 216
• 18 × 12 = 216
• 24 × 9 = 216
• 27 × 8 = 216
• 36 × 6 = 216
• 54 × 4 = 216
• 72 × 3 = 216
• 108 × 2 = 216
• 216 × 1 = 216
### Factors of 216 Table
FactorFactor Number
1one
2two
3three
4four
6six
8eight
9nine
12twelve
18eighteen
24twenty four
27twenty seven
36thirty six
54fifty four
72seventy two
108one hundred eight
216two hundred sixteen
### How to find Factors of 216?
As we know factors of 216 are all the numbers that can exactly divide the number 216 simply divide 216 by all the numbers up to 216 to see the ones that result in zero remainders. Numbers that divide without remainder are factors and in this case below are the factors
1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 27, 36, 54, 72, 108, 216 are the factors and all of them can exactly divide number 216.
### Frequently Asked Questions on Factors of 216
1. What are the factors of 216?
Answer: Factors of 216 are the numbers that leave a remainder zero. The ones that can divide 216 exactly i.e. factors are 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 27, 36, 54, 72, 108, 216.
2.What are Factor Pairs of 216?
• 1 × 216 = 216
• 2 × 108 = 216
• 3 × 72 = 216
• 4 × 54 = 216
• 6 × 36 = 216
• 8 × 27 = 216
• 9 × 24 = 216
• 12 × 18 = 216
• 18 × 12 = 216
• 24 × 9 = 216
• 27 × 8 = 216
• 36 × 6 = 216
• 54 × 4 = 216
• 72 × 3 = 216
• 108 × 2 = 216
• 216 × 1 = 216
3. What is meant by Factor Pairs?
Answer:Factor Pairs are numbers that when multiplied together will result in a given product.
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# Algebra- Check Please
posted by .
A farmer is tracking two wild honey bees in his field. He maps the first bee's path back to the hive on the line y=9/7x. the second bee's bee path follows the line y=-3x+12. Their paths cross at the hive. At what coordinate will the farmer find the hive?
I got (-6/7, 6)
• Algebra- Check Please -
that isn't the answer, use substitution replacing y to the left of the equation 9/7x=-3x+12 solve this for x. then multiply the x you find by 9/7 to get the y
• Algebra- Check Please -
At x=0,y=12.at y=0,x=4 therefore x=4,y=12.
• Algebra- Check Please -
At codinate -3x+12
• Algebra- Check Please -
pq(p-10)+(p-10)
### Answer This Question
First Name: School Subject: Answer:
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# NCERT Solutions for Class 9th Maths: Chapter 3 Coordinate Geometry
Notes Assignments NCERT Solutions Revision sheet
In this page we have NCERT Solutions for Class 9th Maths: Chapter 3 Coordinate Geometry for All exercise Hope you like them and do not forget to like , social share and comment at the end of the page.
Question 1
How will you describe the position of a table lamp on your study table to another person?
We can use the concept of Coordinate Geometry to describe the position of a table lamp on the study table. we have to take two lines, a perpendicular and horizontal. Considering the table as a plane and taking perpendicular line as Y axis and horizontal as X axis. Take one corner of table as origin where both X and Y axes intersect each other. Now, the length of table is Y axis and breadth is X axis. From The origin, join the line to the lamp and mark a point. Calculate the distance of this point from both X and Y axes and then write it in terms of coordinates.
Let the distance of point from X axis is x and from Y axis is y then the position of the table lamp in terms of coordinates is (x, y).
Question 2
(Street Plan): A city has two main roads which cross each other at the centre of the city. These two roads are along the North-South direction and East-West direction.
All the other streets of the city run parallel to these roads and are 200 m apart. There are 5 streets in each direction. Using 1cm = 200 m, draw a model of the city on your notebook. Represent the roads/streets by single lines. There are many cross- streets in your model. A particular cross-street is made by two streets, one running in the North - South direction and another in the East - West direction. Each cross street is referred to in the following manner: If the 2nd street running in the North - South direction and 5th in the East - West direction meet at some crossing, then we will call this cross-street (2, 5). Using this convention, find:
(i) how many cross - streets can be referred to as (4, 3).
(ii) how many cross - streets can be referred to as (3, 4)
Let us taken X axis along S-N direction and Y axis along W-E direction. Now drawing the 5 lines representing roads parallel to X and Y axis each separated equally.
(i) Only one street can be referred to as (4, 3) as we see from the figure.
(ii) Only one street can be referred to as (3, 4) as we see from the figure.
Question 3
Write the answer of each of the following questions:
(i) What is the name of horizontal and the vertical lines drawn to determine the position of any point in the Cartesian plane?
(ii) What is the name of each part of the plane formed by these two lines?
(iii) Write the name of the point where these two lines intersect.
What is the name of horizontal and the vertical lines drawn to determine the position of any point in the Cartesian plane? The name of horizontal lines and vertical lines drawn to determine the position of any point in the Cartesian plane is x-axis and y-axis respectively. What is the name of each part of the plane formed by these two lines? The name of each part of the plane formed by these two lines x-axis and y-axis is quadrants. Write the name of the point where these two lines intersect The point where these two lines intersect is called origin.
Question 4
Refer the below figure and write the following:
(i) The coordinates of B.
(ii) The coordinates of C.
(iii) The point identified by the coordinates (-3, -5).
(iv) The point identified by the coordinates (2, -4).
(v) The abscissa of the point D.
(vi) The ordinate of the point H.
(vii)The coordinates of the point L.
(viii) The coordinates of the point M.
We can answer the above question by recalling these statements
1) The distance of a point from y axis is called x –coordinate or abscissa and the distance of the point from x –axis is called y – coordinate or Ordinate
2) The x-coordinate and y –coordinate of the point in the plane is written as (x, y) for point and is called the coordinates of the point
3) A point on the x –axis has zero distance from x-axis so coordinate of any point on the x-axis will be (x, 0)
4) A point on the y –axis has zero distance from y-axis so coordinate of any point on the y-axis will be (0, y)
i) The coordinates of B is (-5, 2) ii) The coordinates of C is (5, -5) iii) The point identified by the coordinates (-3, -5) is E. iv) The point identified by the coordinates (2, -4) is G. v) Abscissa means x coordinate of point D. So, abscissa of the point D is 6. vi) Ordinate means y coordinate of point H. So, ordinate of point H is -3. vii) The coordinates of the point L is (0, 5). viii) The coordinates of the point M is (- 3, 0).
Question 5
In which quadrant or on which axis do each of the points (-2, 4), (3, -1), (-1, 0), (1, 2) and (-3, -5) lie? Verify your answer by locating them on the Cartesian plane.
We know the quadrant signs are given below
Quadrant x-coordinate y-coordinate Ist Quadrant + + IInd quadrant - + IIIrd quadrant - - IVth quadrant + -
So on that basis we can derive the answer as
(-2, 4) Second quadrant (3, -1) Fourth quadrant (-1, 0) Second Quadrant and X axis (1, 2) First quadrant (-3, -5) Third quadrant
Now Let us draw the Cartesian plane as given below to verify the answer
Question 6
Plot the points (x, y) given in the following table on the plane, choosing suitable units of distance on the axes.
x -2 -1 0 1 3 y 8 7 -1.25 3 -1
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Get Full Access to Physics: Principles With Applications - 6 Edition - Chapter 7 - Problem 81gp
Get Full Access to Physics: Principles With Applications - 6 Edition - Chapter 7 - Problem 81gp
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# The gravitational slingshot effect. Figure 7-51 shows the
ISBN: 9780130606204 3
## Solution for problem 81GP Chapter 7
Physics: Principles with Applications | 6th Edition
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Problem 81GP
The gravitational slingshot effect. Figure shows the planet Saturn moving in the negative direction at its orbital speed (with respect to the Sun) of $$9.6 \mathrm{~km} / \mathrm{s}$$. The mass of Saturn is $$5.69 \times 10^{26} \mathrm{~kg}$$. A spacecraft with mass approaches Saturn. When far from Saturn, it moves in the direction at $$10.4 \mathrm{~km} / \mathrm{s}$$. The gravitational attraction of Saturn (a conservative force) acting on the spacecraft causes it to swing around the planet (orbit shown as dashed line) and head off in the opposite direction. Estimate the final speed of the spacecraft after it is far enough away to be considered free of Saturn's gravitational pull.
FIGURE 7-47 Problem 81
Equation Transcription:
Text Transcription:
9.6 km/s
5.69 x 10^26 kg
10.4 km/s
Step-by-Step Solution:
Solution 81GP:
We have to determine the final speed with which spacecraft was moving, as its direction of motion was reversed due to gravitational attraction between the spacecraft and Saturn.
Step 1 of 5
Concept:
The interaction between the Saturn and the spacecraft is elastic, because the force of gravity is conservative. Thus kinetic energy is conserved in this interaction.
Relative velocity of body A with respect to body B having velocity respectively, is given as,
Step 2 of 5
Step 3 of 5
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# Search by Topic
#### Resources tagged with Mathematical reasoning & proof similar to Why 8?:
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### Is it Magic or Is it Maths?
##### Stage: 3 Challenge Level:
Here are three 'tricks' to amaze your friends. But the really clever trick is explaining to them why these 'tricks' are maths not magic. Like all good magicians, you should practice by trying. . . .
### 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. . . .
### Happy Numbers
##### Stage: 3 Challenge Level:
Take any whole number between 1 and 999, add the squares of the digits to get a new number. Make some conjectures about what happens in general.
### Children at Large
##### Stage: 3 Challenge Level:
There are four children in a family, two girls, Kate and Sally, and two boys, Tom and Ben. How old are the children?
##### Stage: 3 Challenge Level:
A little bit of algebra explains this 'magic'. Ask a friend to pick 3 consecutive numbers and to tell you a multiple of 3. Then ask them to add the four numbers and multiply by 67, and to tell you. . . .
### More Mathematical Mysteries
##### Stage: 3 Challenge Level:
Write down a three-digit number Change the order of the digits to get a different number Find the difference between the two three digit numbers Follow the rest of the instructions then try. . . .
### Eleven
##### Stage: 3 Challenge Level:
Replace each letter with a digit to make this addition correct.
### Largest Product
##### Stage: 3 Challenge Level:
Which set of numbers that add to 10 have the largest product?
### The Genie in the Jar
##### Stage: 3 Challenge Level:
This jar used to hold perfumed oil. It contained enough oil to fill granid silver bottles. Each bottle held enough to fill ozvik golden goblets and each goblet held enough to fill vaswik crystal. . . .
### Calendar Capers
##### Stage: 3 Challenge Level:
Choose any three by three square of dates on a calendar page...
### Aba
##### Stage: 3 Challenge Level:
In the following sum the letters A, B, C, D, E and F stand for six distinct digits. Find all the ways of replacing the letters with digits so that the arithmetic is correct.
### One O Five
##### Stage: 3 Challenge Level:
You can work out the number someone else is thinking of as follows. Ask a friend to think of any natural number less than 100. Then ask them to tell you the remainders when this number is divided by. . . .
### Even So
##### Stage: 3 Challenge Level:
Find some triples of whole numbers a, b and c such that a^2 + b^2 + c^2 is a multiple of 4. Is it necessarily the case that a, b and c must all be even? If so, can you explain why?
### Dicing with Numbers
##### Stage: 3 Challenge Level:
In how many ways can you arrange three dice side by side on a surface so that the sum of the numbers on each of the four faces (top, bottom, front and back) is equal?
### 1 Step 2 Step
##### Stage: 3 Challenge Level:
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##### Stage: 3 Challenge Level:
Make a set of numbers that use all the digits from 1 to 9, once and once only. Add them up. The result is divisible by 9. Add each of the digits in the new number. What is their sum? Now try some. . . .
### Seven Squares - Group-worthy Task
##### Stage: 3 Challenge Level:
Choose a couple of the sequences. Try to picture how to make the next, and the next, and the next... Can you describe your reasoning?
### Tourism
##### Stage: 3 Challenge Level:
If you can copy a network without lifting your pen off the paper and without drawing any line twice, then it is traversable. Decide which of these diagrams are traversable.
### Sticky Numbers
##### Stage: 3 Challenge Level:
Can you arrange the numbers 1 to 17 in a row so that each adjacent pair adds up to a square number?
### What Numbers Can We Make?
##### Stage: 3 Challenge Level:
Imagine we have four bags containing a large number of 1s, 4s, 7s and 10s. What numbers can we make?
### What Numbers Can We Make Now?
##### Stage: 3 Challenge Level:
Imagine we have four bags containing numbers from a sequence. What numbers can we make now?
### Konigsberg Plus
##### Stage: 3 Challenge Level:
Euler discussed whether or not it was possible to stroll around Koenigsberg crossing each of its seven bridges exactly once. Experiment with different numbers of islands and bridges.
### Go Forth and Generalise
##### Stage: 3
Spotting patterns can be an important first step - explaining why it is appropriate to generalise is the next step, and often the most interesting and important.
### Leonardo's Problem
##### Stage: 4 and 5 Challenge Level:
A, B & C own a half, a third and a sixth of a coin collection. Each grab some coins, return some, then share equally what they had put back, finishing with their own share. How rich are they?
### 9 Weights
##### Stage: 3 Challenge Level:
You have been given nine weights, one of which is slightly heavier than the rest. Can you work out which weight is heavier in just two weighings of the balance?
### Always Perfect
##### Stage: 4 Challenge Level:
Show that if you add 1 to the product of four consecutive numbers the answer is ALWAYS a perfect square.
### The Pillar of Chios
##### Stage: 3 Challenge Level:
Semicircles are drawn on the sides of a rectangle ABCD. A circle passing through points ABCD carves out four crescent-shaped regions. Prove that the sum of the areas of the four crescents is equal to. . . .
### More Number Pyramids
##### Stage: 3 Challenge Level:
When number pyramids have a sequence on the bottom layer, some interesting patterns emerge...
### Cross-country Race
##### Stage: 3 Challenge Level:
Eight children enter the autumn cross-country race at school. How many possible ways could they come in at first, second and third places?
### Our Ages
##### Stage: 4 Challenge Level:
I am exactly n times my daughter's age. In m years I shall be ... How old am I?
### Diophantine N-tuples
##### Stage: 4 Challenge Level:
Can you explain why a sequence of operations always gives you perfect squares?
### Euler's Squares
##### Stage: 4 Challenge Level:
Euler found four whole numbers such that the sum of any two of the numbers is a perfect square...
### Pyramids
##### Stage: 3 Challenge Level:
What are the missing numbers in the pyramids?
### Always the Same
##### Stage: 3 Challenge Level:
Arrange the numbers 1 to 16 into a 4 by 4 array. Choose a number. Cross out the numbers on the same row and column. Repeat this process. Add up you four numbers. Why do they always add up to 34?
### Logic
##### Stage: 2 and 3
What does logic mean to us and is that different to mathematical logic? We will explore these questions in this article.
### Square Mean
##### Stage: 4 Challenge Level:
Is the mean of the squares of two numbers greater than, or less than, the square of their means?
### Top-heavy Pyramids
##### Stage: 3 Challenge Level:
Use the numbers in the box below to make the base of a top-heavy pyramid whose top number is 200.
### 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}.
### Online
##### Stage: 2 and 3 Challenge Level:
A game for 2 players that can be played online. Players take it in turns to select a word from the 9 words given. The aim is to select all the occurrences of the same letter.
### Con Tricks
##### Stage: 3
Here are some examples of 'cons', and see if you can figure out where the trick is.
### Cycle It
##### Stage: 3 Challenge Level:
Carry out cyclic permutations of nine digit numbers containing the digits from 1 to 9 (until you get back to the first number). Prove that whatever number you choose, they will add to the same total.
### Shuffle Shriek
##### Stage: 3 Challenge Level:
Can you find all the 4-ball shuffles?
### Multiplication Square
##### Stage: 4 Challenge Level:
Pick a square within a multiplication square and add the numbers on each diagonal. What do you notice?
### Clocked
##### Stage: 3 Challenge Level:
Is it possible to rearrange the numbers 1,2......12 around a clock face in such a way that every two numbers in adjacent positions differ by any of 3, 4 or 5 hours?
### Concrete Wheel
##### Stage: 3 Challenge Level:
A huge wheel is rolling past your window. What do you see?
### Perfectly Square
##### Stage: 4 Challenge Level:
The sums of the squares of three related numbers is also a perfect square - can you explain why?
### Tessellating Hexagons
##### Stage: 3 Challenge Level:
Which hexagons tessellate?
##### Stage: 4 Challenge Level:
Four jewellers share their stock. Can you work out the relative values of their gems?
### Converse
##### Stage: 4 Challenge Level:
Clearly if a, b and c are the lengths of the sides of an equilateral triangle then a^2 + b^2 + c^2 = ab + bc + ca. Is the converse true?
### Hockey
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After some matches were played, most of the information in the table containing the results of the games was accidentally deleted. What was the score in each match played?
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# computing the digist of Pi
I tried to code the Salamin-Brent iteration. I calculated with exact numbers and the procedure returned the numerical value with the .n() function. The iteration was very very slow, for 20 iterations it is approximately 900 secs. (In Maple the same code is 0.06 secs). Can I modify the numerical precision in every step with some method? (Because in every step the number of correct digits is doubling )
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Sort by ยป oldest newest most voted
sage: def BS(N):
sage: a0 = 1
sage: b0 = 1/sqrt(2)
sage: s0 = 1/2
sage: B = 2
sage: a1 = a0
sage: b1 = b0
sage: s = s0
sage: for k in range(N+1):
sage: a = (a1+b1)/2
sage: b2 = a1 * b1
sage: b = sqrt(b2)
sage: c = a^2-b2
sage: s -= B * c
sage: p = 2* a^2/s
sage: a1 = a
sage: b1 = b
sage: s1 = s
sage: B = 2*B
sage: return p.n(digits=2^N)
When you write b0 = 1/sqrt(2), you define an element of the Symbolic Ring:
sage: b0 = 1/sqrt(2)
sage: b0.parent()
Symbolic Ring
Which means that all compurations are done in a symboloc way, there is no numerical approximation at all there. So, with square roots, products and sums, Sage will deal with bigger and bigger formulas, which explains your timings.
For example, if i replace return p.n(digits=2^N) by return p in your code, i got:
sage: BS(5)
-1/32*(4*sqrt(1/2)*2^(1/4) + 4*sqrt((sqrt(2) + 2)*sqrt(1/2)*2^(1/4)) + 4*sqrt((4*sqrt(1/2)*2^(1/4) + sqrt(2) + 2)*sqrt((sqrt(2) + 2)*sqrt(1/2)*2^(1/4))) + 4*sqrt((4*sqrt(1/2)*2^(1/4) + 4*sqrt((sqrt(2) + 2)*sqrt(1/2)*2^(1/4)) + sqrt(2) + 2)*sqrt((4*sqrt(1/2)*2^(1/4) + sqrt(2) + 2)*sqrt((sqrt(2) + 2)*sqrt(1/2)*2^(1/4)))) + 4*sqrt((4*sqrt(1/2)*2^(1/4) + 4*sqrt((sqrt(2) + 2)*sqrt(1/2)*2^(1/4)) + 4*sqrt((4*sqrt(1/2)*2^(1/4) + sqrt(2) + 2)*sqrt((sqrt(2) + 2)*sqrt(1/2)*2^(1/4))) + sqrt(2) + 2)*sqrt((4*sqrt(1/2)*2^(1/4) + 4*sqrt((sqrt(2) + 2)*sqrt(1/2)*2^(1/4)) + sqrt(2) + 2)*sqrt((4*sqrt(1/2)*2^(1/4) + sqrt(2) + 2)*sqrt((sqrt(2) + 2)*sqrt(1/2)*2^(1/4))))) + sqrt(2) + 2)^2/(32*(sqrt(2) + 2)^2 + 16*(4*sqrt(1/2)*2^(1/4) + sqrt(2) + 2)^2 + 8*(4*sqrt(1/2)*2^(1/4) + 4*sqrt((sqrt(2) + 2)*sqrt(1/2)*2^(1/4)) + sqrt(2) + 2)^2 + 4*(4*sqrt(1/2)*2^(1/4) + 4*sqrt((sqrt(2) + 2)*sqrt(1/2)*2^(1/4)) + 4*sqrt((4*sqrt(1/2)*2^(1/4) + sqrt(2) + 2)*sqrt((sqrt(2) + 2)*sqrt(1/2)*2^(1/4))) + sqrt(2) + 2)^2 + 2*(4*sqrt(1/2)*2^(1/4) + 4*sqrt((sqrt(2) + 2)*sqrt(1/2)*2^(1/4)) + 4*sqrt((4*sqrt(1/2)*2^(1/4) + sqrt(2) + 2)*sqrt((sqrt(2) + 2)*sqrt(1/2)*2^(1/4))) + 4*sqrt((4*sqrt(1/2)*2^(1/4) + 4*sqrt((sqrt(2) + 2)*sqrt(1/2)*2^(1/4)) + sqrt(2) + 2)*sqrt((4 ...
more
Thank's, it's very nice! Mainly I'm in trouble with variables after defining a ring.
( 2013-06-25 07:46:08 -0500 )edit
Looking forward to the real number tutorial!
( 2013-06-25 09:40:08 -0500 )edit
Could you provide your code ? My guess is that you compute your iteration in the symbolic ring, and then use .n() method to get your numbers at the end, but i am not sure.
Note that, if you want to get X digits of precision of an exact number a, you can do:
sage: a.n(digits=X)
more
def BS(N): a0 = 1; b0 = 1/sqrt(2); s0 = 1/2; B = 2 a1 = a0; b1 = b0; s = s0 for k in range(N+1): a = (a1+b1)/2; b2 = a1*b1; b = sqrt(b2) c = a^2-b2 s -= B*c p = 2*a^2/s a1 = a; b1 = b; s1 = s; B = 2*B return p.n(digits=2^n)
( 2013-06-25 02:35:22 -0500 )edit
def BS(N): a0 = 1; b0 = 1/sqrt(2); s0 = 1/2; B = 2 a1 = a0; b1 = b0; s = s0 for k in range(N+1): a = (a1+b1)/2; b2 = a1*b1; b = sqrt(b2) c = a^2-b2 s -= B*c p = 2*a^2/s a1 = a; b1 = b; s1 = s; B = 2*B return p.n(digits=2^n)
( 2013-06-25 02:35:22 -0500 )edit
it would be enough once... the last n is N, naturally so, i would like to compute in every iteration with 2^k digits
( 2013-06-25 02:37:48 -0500 )edit
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math
posted by isaac
The sum of
8
and
y
is less than or equal to
−22
.
1. Steve
8+y <= -22
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# Given planes P1: cy+bz=xP2 : az+cx =yP3: bx+ay=zP1, P2 and P3 pass through one line, if(A) a2+b2+c2=ab+bc+ca(B) a2+b2+c2+2abc=1(C)a2+b2+c2=1(D) a2+b2+c2+2ab+2bc+2ca+2abc=1
Yash Jain
55 Points
9 years ago
I’ve got it using a bit long method so try to solve it with any better and easier one. I used the concept of family of Planes.What I did is as follows:
The equation of family of planes from P1 and P2 is given as,
-→ -x+cy+bz + L(cx-y+az) = 0
-→ (-1+Lc)x + (c-L)y + (b+La)z = 0 ---(i)
Since P3 also passes through the line of intersection, i.e.P1, P2 and P3 pass through one line,Thus the coeff. of P3 and that of eq (i) are proportional. So,
(-1+Lc)/b = (c-L)/a = (b+La)/-1 = k (say) ---(ii)
Now, lets start eliminating the assumed constants.
From eq (ii) we get L = (bk+1)/c {= c-ka = -(1+b)/a}--(iii)
From (iii), conclude that k = -(bc+a)/(ab+c)
Putting this value of k in the equation (bk+1)/c = c-ka, we’ll get to our answer which is...
a2+b2+c2+2abc=1
I am sure you’ll be able to find more convenient method for the same. Best wishes.
Vivekanandan
13 Points
5 years ago
Try it it with coplanarity conditions
So
Determinant of |-1 c b|
|c -1 a| =0
|b a -1|
So. -1(1-a2) -c(-c-ab) +b(ac+b)
= -1+a2 + c2+abc +abc+b2 = 0
= a2+b2+c2+2abc = 1
Hence proved
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Interest Formula Online - Tutorpace
# Interest Formula
## Online Tutoring Is The Easiest, Most Cost-Effective Way For Students To Get The Help They Need Whenever They Need It.
If the Principal amount, rate of interest and time period is known, then the simple interest can be calculated using the interest formula,
I = P * T * R / 100
Here P is the principal amount over which interest is to be calculated
T is the time period in years
R is the annual rate at which interest is calculated
This formula is used for calculating the interest on a loan taken.
Example -1: Find the total amount of simple interest that is paid over a period of 5 years on a principal of \$ 30,000 at a simple interest rate of 6%.
Solution: We are given P = \$ 30,000
R = 6 %
T = 5 yrs
We can calculate the simple interest using the interest formula.
I = P * T * R / 100
Now we put the values in the formula we get,
I = 30,000 * 5 * 6 / 100
I = 900,000 / 100
I = 9000
Thus the interest on the given amount is \$ 9000.
Example 2: Find the simple interest on \$ 10,000 at the rate of 5% for 3 years.
Solution: It is given that the Principal = \$ 10,000 Rate = 5 % Time = 3 year
The interest formula is I = P * T * R / 100
On putting the values in the interest formula and solving,
I = 10,000 * 5 * 3 / 100
I = 150, 000 / 100
I = 1500
Thus we get an interest of \$ 1500 on the given principal amount.
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# Search by Topic
#### Resources tagged with Area similar to In a Box:
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### There are 83 results
Broad Topics > Measures and Mensuration > Area
### Can They Be Equal?
##### Stage: 3 Challenge Level:
Can you find rectangles where the value of the area is the same as the value of the perimeter?
### Perimeter Possibilities
##### Stage: 3 Challenge Level:
I'm thinking of a rectangle with an area of 24. What could its perimeter be?
### Blue and White
##### Stage: 3 Challenge Level:
Identical squares of side one unit contain some circles shaded blue. In which of the four examples is the shaded area greatest?
### Pick's Theorem
##### Stage: 3 Challenge Level:
Polygons drawn on square dotty paper have dots on their perimeter (p) and often internal (i) ones as well. Find a relationship between p, i and the area of the polygons.
### The Pillar of Chios
##### Stage: 3 Challenge Level:
Semicircles are drawn on the sides of a rectangle ABCD. A circle passing through points ABCD carves out four crescent-shaped regions. Prove that the sum of the areas of the four crescents is equal to. . . .
### Curvy Areas
##### Stage: 4 Challenge Level:
Have a go at creating these images based on circles. What do you notice about the areas of the different sections?
### Squaring the Circle
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Bluey-green, white and transparent squares with a few odd bits of shapes around the perimeter. But, how many squares are there of each type in the complete circle? Study the picture and make. . . .
### An Unusual Shape
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Can you maximise the area available to a grazing goat?
### Isosceles Triangles
##### Stage: 3 Challenge Level:
Draw some isosceles triangles with an area of $9$cm$^2$ and a vertex at (20,20). If all the vertices must have whole number coordinates, how many is it possible to draw?
### Inscribed in a Circle
##### Stage: 3 Challenge Level:
The area of a square inscribed in a circle with a unit radius is, satisfyingly, 2. What is the area of a regular hexagon inscribed in a circle with a unit radius?
### Fence It
##### Stage: 3 Challenge Level:
If you have only 40 metres of fencing available, what is the maximum area of land you can fence off?
### The Pi Are Square
##### Stage: 3 Challenge Level:
A circle with the radius of 2.2 centimetres is drawn touching the sides of a square. What area of the square is NOT covered by the circle?
### Hallway Borders
##### Stage: 3 Challenge Level:
A hallway floor is tiled and each tile is one foot square. Given that the number of tiles around the perimeter is EXACTLY half the total number of tiles, find the possible dimensions of the hallway.
### Shear Magic
##### Stage: 3 Challenge Level:
What are the areas of these triangles? What do you notice? Can you generalise to other "families" of triangles?
### Salinon
##### Stage: 4 Challenge Level:
This shape comprises four semi-circles. What is the relationship between the area of the shaded region and the area of the circle on AB as diameter?
### Trapezium Four
##### Stage: 4 Challenge Level:
The diagonals of a trapezium divide it into four parts. Can you create a trapezium where three of those parts are equal in area?
### Areas of Parallelograms
##### Stage: 4 Challenge Level:
Can you find the area of a parallelogram defined by two vectors?
### Square Areas
##### Stage: 3 Challenge Level:
Can you work out the area of the inner square and give an explanation of how you did it?
### Tilted Squares
##### Stage: 3 Challenge Level:
It's easy to work out the areas of most squares that we meet, but what if they were tilted?
### Changing Areas, Changing Perimeters
##### Stage: 3 Challenge Level:
How can you change the area of a shape but keep its perimeter the same? How can you change the perimeter but keep the area the same?
### Areas and Ratios
##### Stage: 4 Challenge Level:
What is the area of the quadrilateral APOQ? Working on the building blocks will give you some insights that may help you to work it out.
### Great Squares
##### Stage: 2 and 3 Challenge Level:
Investigate how this pattern of squares continues. You could measure lengths, areas and angles.
### Disappearing Square
##### Stage: 3 Challenge Level:
Do you know how to find the area of a triangle? You can count the squares. What happens if we turn the triangle on end? Press the button and see. Try counting the number of units in the triangle now. . . .
### Appearing Square
##### Stage: 3 Challenge Level:
Make an eight by eight square, the layout is the same as a chessboard. You can print out and use the square below. What is the area of the square? Divide the square in the way shown by the red dashed. . . .
### Partly Circles
##### Stage: 4 Challenge Level:
What is the same and what is different about these circle questions? What connections can you make?
### Gutter
##### Stage: 4 Challenge Level:
Manufacturers need to minimise the amount of material used to make their product. What is the best cross-section for a gutter?
### Lying and Cheating
##### Stage: 3 Challenge Level:
Follow the instructions and you can take a rectangle, cut it into 4 pieces, discard two small triangles, put together the remaining two pieces and end up with a rectangle the same size. Try it!
### Overlap
##### Stage: 3 Challenge Level:
A red square and a blue square overlap so that the corner of the red square rests on the centre of the blue square. Show that, whatever the orientation of the red square, it covers a quarter of the. . . .
### Towers
##### Stage: 3 Challenge Level:
A tower of squares is built inside a right angled isosceles triangle. The largest square stands on the hypotenuse. What fraction of the area of the triangle is covered by the series of squares?
### Isosceles
##### Stage: 3 Challenge Level:
Prove that a triangle with sides of length 5, 5 and 6 has the same area as a triangle with sides of length 5, 5 and 8. Find other pairs of non-congruent isosceles triangles which have equal areas.
### Semi-square
##### Stage: 4 Challenge Level:
What is the ratio of the area of a square inscribed in a semicircle to the area of the square inscribed in the entire circle?
### Equilateral Areas
##### Stage: 4 Challenge Level:
ABC and DEF are equilateral triangles of side 3 and 4 respectively. Construct an equilateral triangle whose area is the sum of the area of ABC and DEF.
### Rati-o
##### Stage: 3 Challenge Level:
Points P, Q, R and S each divide the sides AB, BC, CD and DA respectively in the ratio of 2 : 1. Join the points. What is the area of the parallelogram PQRS in relation to the original rectangle?
### Exploration Versus Calculation
##### Stage: 1, 2 and 3
This article, written for teachers, discusses the merits of different kinds of resources: those which involve exploration and those which centre on calculation.
### Semi-detached
##### Stage: 4 Challenge Level:
A square of area 40 square cms is inscribed in a semicircle. Find the area of the square that could be inscribed in a circle of the same radius.
### Square Pizza
##### Stage: 4 Challenge Level:
Can you show that you can share a square pizza equally between two people by cutting it four times using vertical, horizontal and diagonal cuts through any point inside the square?
### Framed
##### Stage: 3 Challenge Level:
Seven small rectangular pictures have one inch wide frames. The frames are removed and the pictures are fitted together like a jigsaw to make a rectangle of length 12 inches. Find the dimensions of. . . .
### Square Pegs
##### Stage: 3 Challenge Level:
Which is a better fit, a square peg in a round hole or a round peg in a square hole?
### Making Rectangles
##### Stage: 2 and 3 Challenge Level:
A task which depends on members of the group noticing the needs of others and responding.
### Covering Cups
##### Stage: 3 Challenge Level:
What is the shape and dimensions of a box that will contain six cups and have as small a surface area as possible.
### Carpet Cuts
##### Stage: 3 Challenge Level:
You have a 12 by 9 foot carpet with an 8 by 1 foot hole exactly in the middle. Cut the carpet into two pieces to make a 10 by 10 foot square carpet.
### Compare Areas
##### Stage: 4 Challenge Level:
Which has the greatest area, a circle or a square inscribed in an isosceles, right angle triangle?
### Dissect
##### Stage: 3 Challenge Level:
It is possible to dissect any square into smaller squares. What is the minimum number of squares a 13 by 13 square can be dissected into?
### Warmsnug Double Glazing
##### Stage: 3 Challenge Level:
How have "Warmsnug" arrived at the prices shown on their windows? Which window has been given an incorrect price?
### Tiling Into Slanted Rectangles
##### Stage: 2 and 3 Challenge Level:
A follow-up activity to Tiles in the Garden.
### Extending Great Squares
##### Stage: 2 and 3 Challenge Level:
Explore one of these five pictures.
### Growing Rectangles
##### Stage: 3 Challenge Level:
What happens to the area and volume of 2D and 3D shapes when you enlarge them?
### Pie Cuts
##### Stage: 3 Challenge Level:
Investigate the different ways of cutting a perfectly circular pie into equal pieces using exactly 3 cuts. The cuts have to be along chords of the circle (which might be diameters).
### Of All the Areas
##### Stage: 4 Challenge Level:
Can you find a general rule for finding the areas of equilateral triangles drawn on an isometric grid?
### Cylinder Cutting
##### Stage: 2 and 3 Challenge Level:
An activity for high-attaining learners which involves making a new cylinder from a cardboard tube.
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Second Order ODEs - Maple Help
ODE Steps for Second Order ODEs
Overview
• This help page gives a few examples of using the command ODESteps to solve second order ordinary differential equations.
• See Student[ODEs][ODESteps] for a general description of the command ODESteps and its calling sequence.
Examples
> $\mathrm{with}\left(\mathrm{Student}:-\mathrm{ODEs}\right):$
> $\mathrm{ode1}≔2x\left(\frac{ⅆ}{ⅆx}y\left(x\right)\right)-9{x}^{2}+\left(2\left(\frac{ⅆ}{ⅆx}y\left(x\right)\right)+{x}^{2}+1\right)\left(\frac{{ⅆ}^{2}}{ⅆ{x}^{2}}y\left(x\right)\right)=0$
${\mathrm{ode1}}{≔}{2}{}{x}{}\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right){-}{9}{}{{x}}^{{2}}{+}\left({2}{}\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right){+}{{x}}^{{2}}{+}{1}\right){}\left(\frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right){=}{0}$ (1)
> $\mathrm{ODESteps}\left(\mathrm{ode1}\right)$
$\begin{array}{lll}{}& {}& \text{Let's solve}\\ {}& {}& {2}{}{x}{}\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right){-}{9}{}{{x}}^{{2}}{+}\left({2}{}\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right){+}{{x}}^{{2}}{+}{1}\right){}\left(\frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right){=}{0}\\ \text{•}& {}& \text{Highest derivative means the order of the ODE is}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}2\\ {}& {}& \frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\\ \text{•}& {}& \text{Make substitution}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}u=\frac{ⅆ}{ⅆx}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}y{}\left(x\right)\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{to reduce order of ODE}\\ {}& {}& {2}{}{x}{}{u}{}\left({x}\right){-}{9}{}{{x}}^{{2}}{+}\left({2}{}{u}{}\left({x}\right){+}{{x}}^{{2}}{+}{1}\right){}\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{u}{}\left({x}\right)\right){=}{0}\\ \text{▫}& {}& \text{Check if ODE is exact}\\ {}& \text{◦}& \text{ODE is exact if the lhs is the total derivative of a}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{C}^{2}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{function}\\ {}& {}& \left[{}\right]{=}{0}\\ {}& \text{◦}& \text{Compute derivative of lhs}\\ {}& {}& \frac{{\partial }}{{\partial }{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{F}{}\left({x}{,}{u}\right){+}\left(\frac{{\partial }}{{\partial }{u}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{F}{}\left({x}{,}{u}\right)\right){}\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{u}{}\left({x}\right)\right){=}{0}\\ {}& \text{◦}& \text{Evaluate derivatives}\\ {}& {}& {2}{}{x}{=}{2}{}{x}\\ {}& \text{◦}& \text{Condition met, ODE is exact}\\ \text{•}& {}& \text{Exact ODE implies solution will be of this form}\\ {}& {}& \left[{F}{}\left({x}{,}{u}\right){=}{\mathrm{C1}}{,}{M}{}\left({x}{,}{u}\right){=}\frac{{\partial }}{{\partial }{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{F}{}\left({x}{,}{u}\right){,}{N}{}\left({x}{,}{u}\right){=}\frac{{\partial }}{{\partial }{u}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{F}{}\left({x}{,}{u}\right)\right]\\ \text{•}& {}& \text{Solve for}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}F{}\left(x,u\right)\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{by integrating}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}M{}\left(x,u\right)\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{with respect to}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}x\\ {}& {}& {F}{}\left({x}{,}{u}\right){=}\left[{}\right]{+}{\mathrm{_F1}}{}\left({u}\right)\\ \text{•}& {}& \text{Evaluate integral}\\ {}& {}& {F}{}\left({x}{,}{u}\right){=}{{x}}^{{2}}{}{u}{-}{3}{}{{x}}^{{3}}{+}{\mathrm{_F1}}{}\left({u}\right)\\ \text{•}& {}& \text{Take derivative of}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}F{}\left(x,u\right)\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{with respect to}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}u\\ {}& {}& {N}{}\left({x}{,}{u}\right){=}\frac{{\partial }}{{\partial }{u}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{F}{}\left({x}{,}{u}\right)\\ \text{•}& {}& \text{Compute derivative}\\ {}& {}& {{x}}^{{2}}{+}{2}{}{u}{+}{1}{=}{{x}}^{{2}}{+}\frac{{ⅆ}}{{ⅆ}{u}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{\mathrm{_F1}}{}\left({u}\right)\\ \text{•}& {}& \text{Isolate for}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\frac{ⅆ}{ⅆu}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}\mathrm{_F1}{}\left(u\right)\\ {}& {}& \frac{{ⅆ}}{{ⅆ}{u}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{\mathrm{_F1}}{}\left({u}\right){=}{2}{}{u}{+}{1}\\ \text{•}& {}& \text{Solve for}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\mathrm{_F1}{}\left(u\right)\\ {}& {}& {\mathrm{_F1}}{}\left({u}\right){=}{{u}}^{{2}}{+}{u}\\ \text{•}& {}& \text{Substitute}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\mathrm{_F1}{}\left(u\right)\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{into equation for}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}F{}\left(x,u\right)\\ {}& {}& {F}{}\left({x}{,}{u}\right){=}{{x}}^{{2}}{}{u}{-}{3}{}{{x}}^{{3}}{+}{{u}}^{{2}}{+}{u}\\ \text{•}& {}& \text{Substitute}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}F{}\left(x,u\right)\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{into the solution of the ODE}\\ {}& {}& {{x}}^{{2}}{}{u}{-}{3}{}{{x}}^{{3}}{+}{{u}}^{{2}}{+}{u}{=}{\mathrm{C1}}\\ \text{•}& {}& \text{Solve for}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}u{}\left(x\right)\\ {}& {}& \left\{{u}{}\left({x}\right){=}{-}\frac{{{x}}^{{2}}}{{2}}{-}\frac{{1}}{{2}}{-}\frac{\sqrt{{{x}}^{{4}}{+}{12}{}{{x}}^{{3}}{+}{2}{}{{x}}^{{2}}{+}{4}{}{\mathrm{C1}}{+}{1}}}{{2}}{,}{u}{}\left({x}\right){=}{-}\frac{{{x}}^{{2}}}{{2}}{-}\frac{{1}}{{2}}{+}\frac{\sqrt{{{x}}^{{4}}{+}{12}{}{{x}}^{{3}}{+}{2}{}{{x}}^{{2}}{+}{4}{}{\mathrm{C1}}{+}{1}}}{{2}}\right\}\\ \text{•}& {}& \text{Solve 1st ODE for}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}u{}\left(x\right)\\ {}& {}& {u}{}\left({x}\right){=}{-}\frac{{{x}}^{{2}}}{{2}}{-}\frac{{1}}{{2}}{-}\frac{\sqrt{{{x}}^{{4}}{+}{12}{}{{x}}^{{3}}{+}{2}{}{{x}}^{{2}}{+}{4}{}{\mathrm{C1}}{+}{1}}}{{2}}\\ \text{•}& {}& \text{Make substitution}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}u=\frac{ⅆ}{ⅆx}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}y{}\left(x\right)\\ {}& {}& \frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right){=}{-}\frac{{{x}}^{{2}}}{{2}}{-}\frac{{1}}{{2}}{-}\frac{\sqrt{{{x}}^{{4}}{+}{12}{}{{x}}^{{3}}{+}{2}{}{{x}}^{{2}}{+}{4}{}{\mathrm{C1}}{+}{1}}}{{2}}\\ \text{•}& {}& \text{Integrate both sides to solve for}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}y{}\left(x\right)\\ {}& {}& {\int }\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}{=}{\int }\left({-}\frac{{{x}}^{{2}}}{{2}}{-}\frac{{1}}{{2}}{-}\frac{\sqrt{{{x}}^{{4}}{+}{12}{}{{x}}^{{3}}{+}{2}{}{{x}}^{{2}}{+}{4}{}{\mathrm{C1}}{+}{1}}}{{2}}\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}{+}{\mathrm{C2}}\\ \text{•}& {}& \text{Compute lhs}\\ {}& {}& {y}{}\left({x}\right){=}{\int }\left({-}\frac{{{x}}^{{2}}}{{2}}{-}\frac{{1}}{{2}}{-}\frac{\sqrt{{{x}}^{{4}}{+}{12}{}{{x}}^{{3}}{+}{2}{}{{x}}^{{2}}{+}{4}{}{\mathrm{C1}}{+}{1}}}{{2}}\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}{+}{\mathrm{C2}}\\ \text{•}& {}& \text{Solve 2nd ODE for}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}u{}\left(x\right)\\ {}& {}& {u}{}\left({x}\right){=}{-}\frac{{{x}}^{{2}}}{{2}}{-}\frac{{1}}{{2}}{+}\frac{\sqrt{{{x}}^{{4}}{+}{12}{}{{x}}^{{3}}{+}{2}{}{{x}}^{{2}}{+}{4}{}{\mathrm{C1}}{+}{1}}}{{2}}\\ \text{•}& {}& \text{Make substitution}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}u=\frac{ⅆ}{ⅆx}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}y{}\left(x\right)\\ {}& {}& \frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right){=}{-}\frac{{{x}}^{{2}}}{{2}}{-}\frac{{1}}{{2}}{+}\frac{\sqrt{{{x}}^{{4}}{+}{12}{}{{x}}^{{3}}{+}{2}{}{{x}}^{{2}}{+}{4}{}{\mathrm{C1}}{+}{1}}}{{2}}\\ \text{•}& {}& \text{Integrate both sides to solve for}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}y{}\left(x\right)\\ {}& {}& {\int }\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}{=}{\int }\left({-}\frac{{{x}}^{{2}}}{{2}}{-}\frac{{1}}{{2}}{+}\frac{\sqrt{{{x}}^{{4}}{+}{12}{}{{x}}^{{3}}{+}{2}{}{{x}}^{{2}}{+}{4}{}{\mathrm{C1}}{+}{1}}}{{2}}\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}{+}{\mathrm{C2}}\\ \text{•}& {}& \text{Compute lhs}\\ {}& {}& {y}{}\left({x}\right){=}{\int }\left({-}\frac{{{x}}^{{2}}}{{2}}{-}\frac{{1}}{{2}}{+}\frac{\sqrt{{{x}}^{{4}}{+}{12}{}{{x}}^{{3}}{+}{2}{}{{x}}^{{2}}{+}{4}{}{\mathrm{C1}}{+}{1}}}{{2}}\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}{+}{\mathrm{C2}}\end{array}$ (2)
> $\mathrm{ode2}≔\frac{{ⅆ}^{2}}{ⅆ{x}^{2}}y\left(x\right)-\left(\frac{ⅆ}{ⅆx}y\left(x\right)\right)-x{ⅇ}^{x}=0$
${\mathrm{ode2}}{≔}\frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right){-}\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right){-}{x}{}{{ⅇ}}^{{x}}{=}{0}$ (3)
> $\mathrm{ODESteps}\left(\mathrm{ode2}\right)$
$\begin{array}{lll}{}& {}& \text{Let's solve}\\ {}& {}& \frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right){-}\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right){-}{x}{}{{ⅇ}}^{{x}}{=}{0}\\ \text{•}& {}& \text{Highest derivative means the order of the ODE is}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}2\\ {}& {}& \frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\\ \text{•}& {}& \text{Isolate 2nd derivative}\\ {}& {}& \frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right){=}\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right){+}{x}{}{{ⅇ}}^{{x}}\\ \text{•}& {}& \text{Group terms with}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}y{}\left(x\right)\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear}\\ {}& {}& \frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right){-}\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right){=}{x}{}{{ⅇ}}^{{x}}\\ \text{•}& {}& \text{Characteristic polynomial of homogeneous ODE}\\ {}& {}& {{r}}^{{2}}{-}{r}{=}{0}\\ \text{•}& {}& \text{Factor the characteristic polynomial}\\ {}& {}& {r}{}\left({r}{-}{1}\right){=}{0}\\ \text{•}& {}& \text{Roots of the characteristic polynomial}\\ {}& {}& {r}{=}\left({0}{,}{1}\right)\\ \text{•}& {}& \text{1st solution of the homogeneous ODE}\\ {}& {}& {{y}}_{{1}}{}\left({x}\right){=}{1}\\ \text{•}& {}& \text{2nd solution of the homogeneous ODE}\\ {}& {}& {{y}}_{{2}}{}\left({x}\right){=}{{ⅇ}}^{{x}}\\ \text{•}& {}& \text{General solution of the ODE}\\ {}& {}& {y}{}\left({x}\right){=}{\mathrm{C1}}{}{{y}}_{{1}}{}\left({x}\right){+}{\mathrm{C2}}{}{{y}}_{{2}}{}\left({x}\right){+}{{y}}_{{p}}{}\left({x}\right)\\ \text{•}& {}& \text{Substitute in solutions of the homogeneous ODE}\\ {}& {}& {y}{}\left({x}\right){=}{\mathrm{C1}}{+}{\mathrm{C2}}{}{{ⅇ}}^{{x}}{+}{{y}}_{{p}}{}\left({x}\right)\\ \text{▫}& {}& \text{Find a particular solution}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{y}_{p}{}\left(x\right)\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{of the ODE}\\ {}& \text{◦}& \text{Use variation of parameters to find}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{y}_{p}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{here}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}f{}\left(x\right)\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{is the forcing function}\\ {}& {}& \left[{{y}}_{{p}}{}\left({x}\right){=}{-}{{y}}_{{1}}{}\left({x}\right){}\left({\int }\frac{{{y}}_{{2}}{}\left({x}\right){}{f}{}\left({x}\right)}{{W}{}\left({{y}}_{{1}}{}\left({x}\right){,}{{y}}_{{2}}{}\left({x}\right)\right)}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}\right){+}{{y}}_{{2}}{}\left({x}\right){}\left({\int }\frac{{{y}}_{{1}}{}\left({x}\right){}{f}{}\left({x}\right)}{{W}{}\left({{y}}_{{1}}{}\left({x}\right){,}{{y}}_{{2}}{}\left({x}\right)\right)}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}\right){,}{f}{}\left({x}\right){=}{x}{}{{ⅇ}}^{{x}}\right]\\ {}& \text{◦}& \text{Wronskian of solutions of the homogeneous equation}\\ {}& {}& {W}{}\left({{y}}_{{1}}{}\left({x}\right){,}{{y}}_{{2}}{}\left({x}\right)\right){=}\left[\begin{array}{cc}{1}& {{ⅇ}}^{{x}}\\ {0}& {{ⅇ}}^{{x}}\end{array}\right]\\ {}& \text{◦}& \text{Compute Wronskian}\\ {}& {}& {W}{}\left({{y}}_{{1}}{}\left({x}\right){,}{{y}}_{{2}}{}\left({x}\right)\right){=}{{ⅇ}}^{{x}}\\ {}& \text{◦}& \text{Substitute functions into equation for}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{y}_{p}{}\left(x\right)\\ {}& {}& {{y}}_{{p}}{}\left({x}\right){=}{-}\left({\int }{x}{}{{ⅇ}}^{{x}}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}\right){+}{{ⅇ}}^{{x}}{}\left({\int }{x}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}\right)\\ {}& \text{◦}& \text{Compute integrals}\\ {}& {}& {{y}}_{{p}}{}\left({x}\right){=}{{ⅇ}}^{{x}}{}\left({1}{-}{x}{+}\frac{{1}}{{2}}{}{{x}}^{{2}}\right)\\ \text{•}& {}& \text{Substitute particular solution into general solution to ODE}\\ {}& {}& {y}{}\left({x}\right){=}{\mathrm{C1}}{+}{\mathrm{C2}}{}{{ⅇ}}^{{x}}{+}{{ⅇ}}^{{x}}{}\left({1}{-}{x}{+}\frac{{1}}{{2}}{}{{x}}^{{2}}\right)\end{array}$ (4)
> $\mathrm{ode3}≔\frac{{ⅆ}^{2}}{ⅆ{x}^{2}}y\left(x\right)+\frac{5{\left(\frac{ⅆ}{ⅆx}y\left(x\right)\right)}^{2}}{y\left(x\right)}=0$
${\mathrm{ode3}}{≔}\frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right){+}\frac{{5}{}{\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right)}^{{2}}}{{y}{}\left({x}\right)}{=}{0}$ (5)
> $\mathrm{ODESteps}\left(\mathrm{ode3}\right)$
$\begin{array}{lll}{}& {}& \text{Let's solve}\\ {}& {}& \frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right){+}\frac{{5}{}{\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right)}^{{2}}}{{y}{}\left({x}\right)}{=}{0}\\ \text{•}& {}& \text{Highest derivative means the order of the ODE is}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}2\\ {}& {}& \frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\\ \text{•}& {}& \text{Define new dependent variable}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}u\\ {}& {}& {u}{}\left({x}\right){=}\left[{}\right]\\ \text{•}& {}& \text{Compute}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\frac{{ⅆ}^{2}}{ⅆ{x}^{2}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}y{}\left(x\right)\\ {}& {}& \left[{}\right]{=}\left[{}\right]\\ \text{•}& {}& \text{Use chain rule on the lhs}\\ {}& {}& \left(\left[{}\right]\right){}\left(\left[{}\right]\right){=}\left[{}\right]\\ \text{•}& {}& \text{Substitute in the definition of}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}u\\ {}& {}& {u}{}\left({y}\right){}\left(\left[{}\right]\right){=}\left[{}\right]\\ \text{•}& {}& \text{Make substitutions}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\frac{ⅆ}{ⅆx}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}y{}\left(x\right)=u{}\left(y\right)\text{,}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\frac{{ⅆ}^{2}}{ⅆ{x}^{2}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}y{}\left(x\right)=u{}\left(y\right){}\left(\frac{ⅆ}{ⅆy}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}u{}\left(y\right)\right)\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{to reduce order of ODE}\\ {}& {}& {u}{}\left({y}\right){}\left(\frac{{ⅆ}}{{ⅆ}{y}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{u}{}\left({y}\right)\right){+}\frac{{5}{}{{u}{}\left({y}\right)}^{{2}}}{{y}}{=}{0}\\ \text{•}& {}& \text{Separate variables}\\ {}& {}& \frac{\frac{{ⅆ}}{{ⅆ}{y}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{u}{}\left({y}\right)}{{u}{}\left({y}\right)}{=}{-}\frac{{5}}{{y}}\\ \text{•}& {}& \text{Integrate both sides with respect to}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}y\\ {}& {}& {\int }\frac{\frac{{ⅆ}}{{ⅆ}{y}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{u}{}\left({y}\right)}{{u}{}\left({y}\right)}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{y}{=}{\int }{-}\frac{{5}}{{y}}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{y}{+}{\mathrm{C1}}\\ \text{•}& {}& \text{Evaluate integral}\\ {}& {}& {\mathrm{ln}}{}\left({u}{}\left({y}\right)\right){=}{-}{5}{}{\mathrm{ln}}{}\left({y}\right){+}{\mathrm{C1}}\\ \text{•}& {}& \text{Solve for}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}u{}\left(y\right)\\ {}& {}& {u}{}\left({y}\right){=}\frac{{{ⅇ}}^{{\mathrm{C1}}}}{{{y}}^{{5}}}\\ \text{•}& {}& \text{Solve 1st ODE for}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}u{}\left(y\right)\\ {}& {}& {u}{}\left({y}\right){=}\frac{{{ⅇ}}^{{\mathrm{C1}}}}{{{y}}^{{5}}}\\ \text{•}& {}& \text{Revert to original variables with substitution}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}u{}\left(y\right)=\frac{ⅆ}{ⅆx}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}y{}\left(x\right)\text{,}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}y=y{}\left(x\right)\\ {}& {}& \frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right){=}\frac{{{ⅇ}}^{{\mathrm{C1}}}}{{{y}{}\left({x}\right)}^{{5}}}\\ \text{•}& {}& \text{Separate variables}\\ {}& {}& \left(\frac{{ⅆ}}{{ⅆ}}\right)\end{array}$
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# What Is Graph Data Structure in Computer Science?
//
Heather Bennett
In computer science, the graph data structure is a fundamental concept used to represent relationships between objects. It is widely used in various applications such as social networks, routing algorithms, and recommendation systems. A graph consists of a set of vertices or nodes connected by edges or arcs.
## Key Terminology
Before diving deeper into the details of graphs, let’s familiarize ourselves with some key terminology:
• Vertex: Also known as a node, it represents an object or entity in a graph.
• Edge: It represents the relationship between two vertices. An edge can be directed (one-way) or undirected (two-way).
• Weight: In some graphs, an edge may have a weight or cost associated with it. This weight can represent factors like distance, time, or importance.
## Types of Graphs
In computer science, there are several types of graphs that serve different purposes:
### 1. Directed Graph (Digraph)
A directed graph is a graph in which all edges have a direction. It means that if there is an edge from vertex A to vertex B, it doesn’t imply that there exists an edge from B to A.
### 2. Undirected Graph
An undirected graph is a graph in which all edges are bidirectional. If there is an edge from vertex A to vertex B, it implies that there exists an edge from B to A as well.
### 3. Weighted Graph
A weighted graph is a graph in which each edge has a weight associated with it. These weights can represent various factors like distance, cost, or importance.
### 4. Cyclic Graph
A cyclic graph is a graph that contains at least one cycle, which is a path that starts and ends at the same vertex.
### 5. Acyclic Graph
An acyclic graph is a graph that does not contain any cycles.
## Graph Representation
There are multiple ways to represent a graph in computer science:
An adjacency matrix is a 2D matrix where the rows and columns represent the vertices of the graph.
Each cell in the matrix represents an edge between two vertices. If an edge exists between vertex A and vertex B, the corresponding cell of the matrix will be marked as 1 or contain a weight value.
An adjacency list is a collection of linked lists or arrays where each list/array represents a vertex in the graph. Each element in these lists/arrays represents an edge connected to that vertex.
## Graph Traversal
The process of visiting all the vertices and edges of a graph is known as graph traversal. There are two commonly used algorithms for this purpose:
### 1. Depth-First Search (DFS)
In DFS, we start from an initial vertex and explore as far as possible along each branch before backtracking. It uses a stack data structure to keep track of visited vertices. Breadth-First Search (BFS)
In BFS, we start from an initial vertex and explore all its neighboring vertices before moving on to their neighbors. It uses a queue data structure to keep track of visited vertices.
## Applications of Graphs
Graphs have a wide range of applications in computer science:
• Social Networks: Graphs are used to represent connections between users in social media platforms.
• Routing Algorithms: Graphs help find the shortest path between two nodes in network routing algorithms.
• Recommendation Systems: Graphs are used to analyze user preferences and make personalized recommendations.
• Dependency Management: Graphs help manage dependencies between software components.
In conclusion, the graph data structure is a powerful tool for representing relationships between objects. Understanding graphs and their various types can greatly enhance your problem-solving skills as a computer scientist or programmer. So, dive into the fascinating world of graphs and explore their endless possibilities!
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# The Rational Zero Theorem
## Presentation on theme: "The Rational Zero Theorem"— Presentation transcript:
The Rational Zero Theorem
3.4: Zeros of Polynomial Functions The Rational Zero Theorem The Rational Zero Theorem gives a list of possible rational zeros of a polynomial function. Equivalently, the theorem gives all possible rational roots of a polynomial equation. Not every number in the list will be a zero of the function, but every rational zero of the polynomial function will appear somewhere in the list. The Rational Zero Theorem If f (x) = anxn + an-1xn-1 +…+ a1x + a0 has integer coefficients and (where is reduced) is a rational zero, then p is a factor of the constant term a0 and q is a factor of the leading coefficient an.
EXAMPLE: Using the Rational Zero Theorem
3.4: Zeros of Polynomial Functions EXAMPLE: Using the Rational Zero Theorem List all possible rational zeros of f (x) = 15x3 + 14x2 - 3x – 2. Solution The constant term is –2 and the leading coefficient is 15. Divide 1 and 2 by 1. by 3. by 5. by 15. There are 16 possible rational zeros. The actual solution set to f (x) = 15x3 + 14x2 - 3x – 2 = 0 is {-1, -1/3, 2/5}, which contains 3 of the 16 possible solutions.
EXAMPLE: Solving a Polynomial Equation
3.4: Zeros of Polynomial Functions EXAMPLE: Solving a Polynomial Equation Solve: x4 - 6x2 - 8x + 24 = 0. Solution Recall that we refer to the zeros of a polynomial function and the roots of a polynomial equation. Because we are given an equation, we will use the word "roots," rather than "zeros," in the solution process. We begin by listing all possible rational roots.
EXAMPLE: Solving a Polynomial Equation
3.4: Zeros of Polynomial Functions EXAMPLE: Solving a Polynomial Equation Solve: x4 - 6x2 - 8x + 24 = 0. Solution The graph of f (x) = x4 - 6x2 - 8x + 24 is shown the figure below. Because the x-intercept is 2, we will test 2 by synthetic division and show that it is a root of the given equation. x-intercept: 2 2 The zero remainder indicates that 2 is a root of x4 - 6x2 - 8x + 24 = 0.
EXAMPLE: Solving a Polynomial Equation
3.4: Zeros of Polynomial Functions EXAMPLE: Solving a Polynomial Equation Solve: x4 - 6x2 - 8x + 24 = 0. Solution Now we can rewrite the given equation in factored form. x4 - 6x2 + 8x + 24 = This is the given equation. (x – 2)(x3 + 2x2 - 2x - 12) = 0 This is the result obtained from the synthetic division. x – 2 = or x3 + 2x2 - 2x - 12 = Set each factor equal to zero.
EXAMPLE: Solving a Polynomial Equation
3.4: Zeros of Polynomial Functions EXAMPLE: Solving a Polynomial Equation Solve: x4 - 6x2 - 8x + 24 = 0. Solution We can use the same approach to look for rational roots of the polynomial equation x3 + 2x2 - 2x - 12 = 0, listing all possible rational roots. However, take a second look at the figure of the graph of x4 - 6x2 - 8x + 24 = 0. Because the graph turns around at 2, this means that 2 is a root of even multiplicity. Thus, 2 must also be a root of x3 + 2x2 - 2x - 12 = 0, confirmed by the following synthetic division. x-intercept: 2 These are the coefficients of x3 + 2x2 - 2x - 12 = 0. The zero remainder indicates that 2 is a root of x3 + 2x2 - 2x - 12 = 0.
EXAMPLE: Solving a Polynomial Equation
3.4: Zeros of Polynomial Functions EXAMPLE: Solving a Polynomial Equation Solve: x4 - 6x2 - 8x + 24 = 0. Solution Now we can solve the original equation as follows. x4 - 6x2 + 8x + 24 = This is the given equation. (x – 2)(x3 + 2x2 - 2x - 12) = 0 This was obtained from the first synthetic division. (x – 2)(x – 2)(x2 + 4x + 6) = 0 This was obtained from the second synthetic division. x – 2 = 0 or x – 2 = 0 or x2 + 4x + 6 = Set each factor equal to zero. x = x = x2 + 4x + 6 = Solve.
EXAMPLE: Solving a Polynomial Equation
3.4: Zeros of Polynomial Functions EXAMPLE: Solving a Polynomial Equation Solve: x4 - 6x2 - 8x + 24 = 0. Solution We can use the quadratic formula to solve x2 + 4x + 6 = 0. We use the quadratic formula because x2 + 4x + 6 = 0 cannot be factored. Let a = 1, b = 4, and c = 6. Multiply and subtract under the radical. Simplify. The solution set of the original equation is {2, -2 - i i }.
Properties of Polynomial Equations
3.4: Zeros of Polynomial Functions Properties of Polynomial Equations 1. If a polynomial equation is of degree n, then counting multiple roots separately, the equation has n roots. 2. If a + bi is a root of a polynomial equation (b 0), then the nonreal complex number a - bi is also a root. Nonreal complex roots, if they exist, occur in conjugate pairs.
Descartes' Rule of Signs
3.4: Zeros of Polynomial Functions Descartes' Rule of Signs If f (x) = anxn + an-1xn-1 + … + a2x2 + a1x + a0 be a polynomial with real coefficients. 1. The number of positive real zeros of f is either equal to the number of sign changes of f (x) or is less than that number by an even integer. If there is only one variation in sign, there is exactly one positive real zero. 2. The number of negative real zeros of f is either equal to the number of sign changes of f (-x) or is less than that number by an even integer. If f (-x) has only one variation in sign, then f has exactly one negative real zero.
EXAMPLE: Using Descartes’ Rule of Signs
3.4: Zeros of Polynomial Functions EXAMPLE: Using Descartes’ Rule of Signs Determine the possible number of positive and negative real zeros of f (x) = x3 + 2x2 + 5x + 4. Solution 1. To find possibilities for positive real zeros, count the number of sign changes in the equation for f (x). Because all the terms are positive, there are no variations in sign. Thus, there are no positive real zeros. 2. To find possibilities for negative real zeros, count the number of sign changes in the equation for f (-x). We obtain this equation by replacing x with -x in the given function. f (-x) = (-x)3 + 2(-x)2 + 5(-x) + 4 f (x) = x x x This is the given polynomial function. Replace x with -x. = -x3 + 2x2 - 5x + 4
EXAMPLE: Using Descartes’ Rule of Signs
3.4: Zeros of Polynomial Functions EXAMPLE: Using Descartes’ Rule of Signs Determine the possible number of positive and negative real zeros of f (x) = x3 + 2x2 + 5x + 4. Solution Now count the sign changes. f (-x) = -x3 + 2x2 - 5x + 4 There are three variations in sign. The number of negative real zeros of f is either equal to the number of sign changes, 3, or is less than this number by an even integer. This means that there are either 3 negative real zeros or = 1 negative real zero.
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# Find the sequence of partial sums for the series $a_n = (-1)^n$ Does this series converge?
Find the sequence of partial sums for the series $$\sum_{n=0}^\infty (-1)^n = 1 -1 + 1 -1 + 1 - \cdots$$
Does this series converge ?
My answer is that the sequence $= 0.5 + 0.5(-1)^n$. This makes a sequence that alternates between $1$ and $0$.
I know that the sequence does not converge since it is not monotone. But how can I prove this?
• Do there exist subsequences which converge to disparate limits? – abiessu May 29 '15 at 0:00
• – Victor May 29 '15 at 0:02
• A series does not have to be monotone to converge, e.g., $(-1)^n/n$. What theorems/tests do you have for convergence? Cauchy? For instance, if $S_n$ is the partial sum, then $|S_n - S_{n+1}| = 1$ and hence is not Cauchy. – Simon S May 29 '15 at 0:02
hint: $$s_{2n} = 0, s_{2n+1} = 1$$.
If it converges to $\ell$ then $|s_n-\ell|<\varepsilon$, once $\varepsilon>0$ is fixed, and $n$ is sufficiently large. Here $s_n$ is exactly the $n$-th partial sum.
But if you set, for example, $\ell=1/10$, then the above inequality cannot be satisfied even for $n$ enough large.
All you need to show is that the the sequence $x_{n}=(-1)^{n}$ does not converge. Many ways to do it : First, using the theorem saying that if a sequence converges , then any subsequence converges to the same limit. $x_{2n}$ and $x_{2n+1}$ are subsequences converging obviously to different limits. Not familiar with this? No problem! Assume that $x_{n}$ converges, say to $l$. Then, given $\epsilon >0$, we can find an integer $N(\epsilon)$ such that $n \geq N(\epsilon)$ implies $|x_{n}-l| <\epsilon.$ Now, $|x_{n+1}-x_{n}| = |x_{n+1}-l+l-x_{n}| \leq |x_{n+1}-l|+|l-x_{n}| <2\epsilon$, provided $n \geq N(\epsilon)$. Now, we chose $\epsilon=1$. And I will leave it to you to see the obvious contradiction!.
$$\sum\limits_{n=0}^\infty (-1)^n = \lim\limits_{m\to\infty}\sum\limits_{n=0}^m (-1)^n$$ $$= \lim\limits_{m\to\infty} \frac12\left((-1)^m + 1\right)$$ Note that the sequence $$a_m=\frac12\left((-1)^m + 1\right)$$ is convergent if $$\lim\limits_{m\to\infty} a_{2m} = \lim\limits_{m\to\infty} a_{2m+1} = L$$ However, in this case we have $$\lim\limits_{m\to\infty} \frac12\left((-1)^{2m}+1\right) =\frac22= 1$$ And $$\lim\limits_{m\to\infty} \frac12\left((-1)^{2m+1}+1\right) =\frac02= 0$$ Therefore $a_m$ is a divergent sequence.
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Understanding r^2: A Key Measure in Statistics
What is r^2?
When it comes to analyzing data and conducting statistical analyses, one important concept to understand is the coefficient of determination, commonly referred to as r^2. R^2 is a numerical measure that quantifies the proportion of variation in one variable, explained by another variable or a regression model. In simpler terms, it tells us how well a regression model fits the observed data.
Digging Deeper into r^2
In statistical terms, r^2 ranges from 0 to 1, where 0 indicates that the regression model explains none of the variability in the data, and 1 signals that the model accounts for all the variability. The closer the value is to 1, the better the model fits the data.
To put it into perspective, let's consider an example: suppose we have a regression model that predicts the housing prices based on the size of the house. If the r^2 for this model is 0.80, it tells us that 80% of the variation in the housing prices can be explained by the size of the house. In this case, the model is considered reasonably good at predicting housing prices.
Furthermore, r^2 can also be interpreted as the square of the correlation coefficient (r) between two variables. The correlation coefficient measures the strength and direction of the linear relationship between two variables, while the squared value provides the proportion of shared variance between them.
The Importance of Assessing Knowledge in r^2 for Hiring Process
When it comes to making informed hiring decisions, assessing a candidate's understanding of r^2 is crucial. Proficiency in this statistical measure allows organizations to identify individuals who possess the necessary analytical skills to interpret and analyze data accurately.
A strong grasp of r^2 enables candidates to evaluate the strength of relationships between variables, gauge the effectiveness of regression models, and make data-driven decisions. By assessing a candidate's knowledge in r^2, companies can ensure they hire individuals who can contribute to successful data analysis and decision-making processes.
Candidates who excel in r^2 possess essential skills that benefit a wide range of industries, including finance, marketing, research, and more. Whether it's predicting sales trends, optimizing marketing strategies, or analyzing experimental data, proficiency in r^2 is a valuable asset for professionals working with quantitative data.
By incorporating assessments that evaluate a candidate's understanding of r^2, organizations can streamline their hiring processes, identify top talent, and make data-driven decisions that drive success in today's data-centric business world. At Alooba, we provide comprehensive assessments to help companies gauge a candidate's aptitude in r^2 and ensure they hire the right individuals for data-related roles.
Assessing Candidates on r^2 with Alooba
To evaluate a candidate's understanding of r^2 effectively, Alooba offers tailored assessments designed to assess this specific skillset. Through our platform, organizations can evaluate candidates' knowledge and application of r^2 through relevant test types.
1. Concepts & Knowledge Test: Our Concepts & Knowledge test provides a comprehensive assessment of a candidate's theoretical understanding of r^2. With carefully crafted multiple-choice questions, this test evaluates candidates' knowledge of the concept, its interpretation, and its significance in statistical analysis.
2. Written Response Test: For a more in-depth evaluation, our Written Response test allows candidates to demonstrate their understanding of r^2 through written explanations. This test provides an opportunity for candidates to showcase their ability to articulate the concept, its applications, and its limitations effectively.
By utilizing these assessments, organizations can accurately gauge candidates' proficiency in r^2 and make informed hiring decisions. Alooba's user-friendly platform facilitates the administration and evaluation of these tests, providing organizations with valuable insights into a candidate's understanding of r^2 and their ability to apply it in practical situations.
Partner with Alooba and leverage our assessment expertise to identify candidates with strong knowledge in r^2, ensuring that your organization can build a data-savvy team capable of driving informed decision-making processes.
Exploring Topics Covered in r^2
When delving into the concept of r^2, there are several subtopics that one might encounter. These topics provide a deeper understanding of how r^2 is calculated, interpreted, and utilized in statistical analysis. Here are some key areas that form the foundation of r^2:
1. Calculation and Interpretation: Understanding how to calculate r^2 is essential. This involves computing the sum of squared residuals, also known as the sum of squares of errors or SSE. Interpreting r^2 involves knowing how the value relates to the goodness-of-fit of a regression model.
2. Comparison and Evaluation: Comparing different models based on their respective r^2 values allows researchers and analysts to assess the effectiveness of each model. Additionally, interpreting the magnitude of r^2 values helps in evaluating the strength and significance of relationships between variables.
3. Limitations and Considerations: It is important to recognize the limitations and assumptions associated with r^2. For example, r^2 only quantifies the proportion of variation explained by the independent variable(s), and it may not capture all the complexities of a relationship. Understanding these limitations assists in conducting comprehensive analyses and making appropriate conclusions.
4. Extensions and Variations: Beyond the standard r^2, specialized variations and extensions of the measure exist. These include adjusted r^2, which accounts for the number of predictors in the model, and hierarchical r^2, which examines the unique contribution of each predictor in a nested regression model.
By exploring these various topics, individuals can gain a comprehensive understanding of r^2 and its applications in statistical analysis. Alooba's assessment platform provides a space for individuals to demonstrate their knowledge in these areas and showcase their ability to utilize r^2 in practical data analysis scenarios.
Practical Applications of r^2
The concept of r^2 finds application in various fields where statistical analysis plays a crucial role. Here are some practical ways in which r^2 is used:
1. Model Evaluation and Selection: In regression analysis, r^2 serves as a valuable tool for evaluating the goodness-of-fit of different regression models. By comparing the r^2 values of alternative models, researchers can determine which model best captures the relationship between variables, helping them make informed decisions and predictions.
2. Predictive Analytics: r^2 is commonly used to assess the predictive power of a regression model. By examining the r^2 value, one can estimate the proportion of variability in the dependent variable that can be explained by the independent variable(s). This aids in forecasting outcomes, such as predicting sales based on advertising expenditure or estimating future customer behavior.
3. Performance Evaluation: r^2 is utilized as a performance metric in various fields. For example, in finance, it is employed to evaluate investment portfolios' performance by measuring their ability to explain the variability in returns. In marketing, it helps assess the effectiveness of advertising campaigns and their impact on consumer behavior.
4. Scientific Research: r^2 is widely applied in scientific research across disciplines. Whether studying the impact of environmental factors on biological processes or investigating the relationship between variables in social sciences, r^2 helps researchers quantify the strength and significance of relationships, enabling them to draw meaningful conclusions.
Understanding how r^2 is used in these real-world scenarios allows professionals to harness its power to make data-driven decisions. At Alooba, we recognize the importance of assessing candidates' proficiency in r^2, ensuring organizations can confidently evaluate and hire individuals who possess the necessary skills to excel in fields where statistical analysis is essential.
Roles that Require Strong r^2 Skills
Proficiency in r^2 is highly beneficial for individuals working in various data-centric roles. The following roles particularly require good r^2 skills:
• Data Analyst: Data analysts rely on r^2 to evaluate the strength of relationships between variables and to make data-driven decisions. They utilize r^2 to assess the effectiveness of regression models and interpret their findings accurately.
• Data Scientist: Data scientists heavily employ r^2 to assess model performance, compare different models, and determine the adequacy of variables in prediction tasks. Strong r^2 skills are essential for extracting meaningful insights and creating robust predictive models.
• Data Engineer: Data engineers benefit from good r^2 skills as they are responsible for processing and transforming data efficiently. Understanding r^2 helps them ensure the accuracy of their data pipelines and validate the output of their processes effectively.
• Insights Analyst: Insights analysts leverage r^2 to evaluate the performance of marketing campaigns, measure customer behavior, and provide valuable insights to improve business strategies. They rely on r^2 to assess the impact of various factors on business outcomes.
• Financial Analyst: Financial analysts use r^2 to assess the performance of investment portfolios, measure risk, and evaluate the effectiveness of financial models. They rely on r^2 to make informed decisions and provide accurate financial forecasts.
• Research Data Analyst: Research data analysts utilize r^2 to analyze data, test hypotheses, and evaluate the strength of relationships in scientific research. They rely on r^2 to draw meaningful conclusions and support research findings.
These roles, among others, require individuals to possess strong r^2 skills to excel in the field. At Alooba, we provide assessments tailored to evaluate candidates' proficiency in r^2, helping organizations identify top talent for these data-focused positions.
Associated Roles
Data Analyst
Data Analysts draw meaningful insights from complex datasets with the goal of making better decisions. Data Analysts work wherever an organization has data - these days that could be in any function, such as product, sales, marketing, HR, operations, and more.
Data Engineer
Data Engineers are responsible for moving data from A to B, ensuring data is always quickly accessible, correct and in the hands of those who need it. Data Engineers are the data pipeline builders and maintainers.
Data Scientist
Data Scientists are experts in statistical analysis and use their skills to interpret and extract meaning from data. They operate across various domains, including finance, healthcare, and technology, developing models to predict future trends, identify patterns, and provide actionable insights. Data Scientists typically have proficiency in programming languages like Python or R and are skilled in using machine learning techniques, statistical modeling, and data visualization tools such as Tableau or PowerBI.
Financial Analyst
Financial Analysts are experts in assessing financial data to aid in decision-making within various sectors. These professionals analyze market trends, investment opportunities, and the financial performance of companies, providing critical insights for investment decisions, business strategy, and economic policy development. They utilize financial modeling, statistical tools, and forecasting techniques, often leveraging software like Excel, and programming languages such as Python or R for their analyses.
GIS Data Analyst
GIS Data Analysts specialize in analyzing spatial data and creating insights to inform decision-making. These professionals work with geographic information system (GIS) technology to collect, analyze, and interpret spatial data. They support a variety of sectors such as urban planning, environmental conservation, and public health. Their skills include proficiency in GIS software, spatial analysis, and cartography, and they often have a strong background in geography or environmental science.
HR Analyst
HR Analysts are integral in managing HR data across multiple systems throughout the employee lifecycle. This role involves designing and launching impactful reports, ensuring data integrity, and providing key insights to support strategic decision-making within the HR function. They work closely with various stakeholders, offering training and enhancing HR data reporting capabilities.
Insights Analyst
Insights Analysts play a pivotal role in transforming complex data sets into actionable insights, driving business growth and efficiency. They specialize in analyzing customer behavior, market trends, and operational data, utilizing advanced tools such as SQL, Python, and BI platforms like Tableau and Power BI. Their expertise aids in decision-making across multiple channels, ensuring data-driven strategies align with business objectives.
Marketing Analyst
Marketing Analysts specialize in interpreting data to enhance marketing efforts. They analyze market trends, consumer behavior, and campaign performance to inform marketing strategies. Proficient in data analysis tools and techniques, they bridge the gap between data and marketing decision-making. Their role is crucial in tailoring marketing efforts to target audiences effectively and efficiently.
Product Analyst
Product Analysts utilize data to optimize product strategies and enhance user experiences. They work closely with product teams, leveraging skills in SQL, data visualization (e.g., Tableau), and data analysis to drive product development. Their role includes translating business requirements into technical specifications, conducting A/B testing, and presenting data-driven insights to inform product decisions. Product Analysts are key in understanding customer needs and driving product innovation.
Other names for R^2 include Coefficient of determination, and R-squared.
Discover How Alooba Can Help You Assess r^2 Skills and More
Schedule a Call to Explore the Benefits of Alooba's Assessment Platform
Unlock the power of data analysis and make confident hiring decisions with Alooba. Our platform offers tailored assessments to evaluate candidates' proficiency in r^2 and other essential skills. Experience the benefits of streamlined candidate assessment, improved hiring outcomes, and a data-driven approach to talent acquisition.
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0
Science
Linear Algebra
What is pressure force?
Wiki User
Pressure is defined as the following Pressure = force / area. So what this means is pressure is how much force is exerted over a area of given size. For examle a concrete slab is on the floor this slab has a width of 2m and a length of 2 m its area is 2 x 2 which gives 2m Squared. Its force it exerts is 1000 newtons ok what the hell does this mean? if you place this in the formula you will get pressure = 1000/4 Pressure = 250 newtons per square metre. I hope this answers you question on what pressure is.
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## Intermediate Algebra (6th Edition)
$4c+3=7$
"Three more than": addition, +3 "The product of 4 and c": multiplication (can use $\times$ or a coefficient, e.g. $4c$) "Is": $=$
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## 【学习笔记】[AGC063E] Child to Parent
news/2024/2/25 14:07:20/文章来源:https://blog.csdn.net/cqbzlydd/article/details/135633089
f u , i = ∑ f u , i − k r × g v . k f_{u,i}=\sum f_{u,i-kr}\times g_{v.k}
F u ( x ) = ∑ f u , i x i G u ( x ) = ∑ g u , i x i F_u(x)=\sum f_{u,i}x^i\\G_u(x)=\sum g_{u,i}x^i
F u ( x ) ← ∏ G v ( x ) F u ( x ) ← x a i F u ( x r ) G u ( x ) ← F u ( x ) + F u ( 1 ) − F u ( x ) 1 − x F_u(x)\gets \prod G_v(x)\\F_u(x)\gets x^{a_i}F_u(x^r)\\ G_u(x)\gets F_u(x)+\frac{F_u(1)-F_u(x)}{1-x}
F u ′ ( x ) ← ∏ G v ′ ( x ) F'_u(x)\gets \prod G'_v(x)\\
F u ′ ( x ) = F u ( x + 1 ) = ∏ G v ( x + 1 ) = ∏ G v ′ ( x ) F'_u(x)=F_u(x+1)=\prod G_v(x+1)=\prod G'_v(x)
F u ′ ( x ) ← ( x + 1 ) a i F u ′ ( ( x + 1 ) r − 1 ) F_u'(x)\gets (x+1)^{a_i}F_u'((x+1)^r-1)
F u ′ ( x ) = F u ( x + 1 ) = ( x + 1 ) a i F u ( ( x + 1 ) r ) = ( x + 1 ) a i F u ′ ( ( x + 1 ) r − 1 ) F'_u(x)=F_u(x+1)=(x+1)^{a_i}F_u((x+1)^r)=(x+1)^{a_i}F_u'((x+1)^r-1)
G u ′ ( x ) = F u ′ ( x ) + F u ′ ( 0 ) − F u ′ ( x ) − x G'_u(x)=F'_u(x)+\frac{F'_u(0)-F'_u(x)}{-x}
#include<bits/stdc++.h>
#define fi first
#define se second
#define pb push_back
#define inf 0x3f3f3f3f
#define ll long long
using namespace std;
const int mod=998244353;
int n,fa[305],dep[305];
ll r,fac[305],inv[305],ifac[305],to[305][305],f[305][305],g[305][305],a[305],res;
vector<int>G[305];
ll fpow(ll x,ll y=mod-2){ll z(1);for(;y;y>>=1){if(y&1)z=z*x%mod;x=x*x%mod;}return z;
}
ll binom(int x,int y){if(x<0||y<0||x<y)return 0;return fac[x]*inv[y]%mod*inv[x-y]%mod;
}
void init(int n){fac[0]=1;for(int i=1;i<=n;i++)fac[i]=fac[i-1]*i%mod;inv[n]=fpow(fac[n]);for(int i=n;i>=1;i--)inv[i-1]=inv[i]*i%mod;ifac[1]=1;for(int i=2;i<=n;i++)ifac[i]=mod-ifac[mod%i]*(mod/i)%mod;
}
void dfs(int u){f[u][0]=1;for(auto v:G[u]){dep[v]=dep[u]+1,dfs(v);memset(f[0],0,sizeof f[0]);for(int i=0;i<=dep[v];i++)for(int j=0;i+j<=dep[v];j++)(f[0][i+j]+=f[u][i]*g[v][j])%=mod;memcpy(f[u],f[0],sizeof f[0]);}memset(f[0],0,sizeof f[0]);for(int i=0;i<=dep[u]+1;i++)for(int j=0;j<=dep[u]+1;j++)(f[0][j]+=f[u][i]*to[i][j])%=mod;memset(f[u],0,sizeof f[u]);ll mul=1;for(int i=0;i<=dep[u]+1;i++){for(int j=0;i+j<=dep[u]+1;j++)(f[u][i+j]+=mul*f[0][j])%=mod;mul=mul*(a[u]-i)%mod*ifac[i+1]%mod;}for(int i=0;i<=dep[u];i++)g[u][i]=(f[u][i]+f[u][i+1])%mod;
}
signed main(){//freopen("data.in","r",stdin);ios::sync_with_stdio(false);cin.tie(0),cout.tie(0);cin>>n;for(int i=2;i<=n;i++)cin>>fa[i],G[fa[i]].pb(i);cin>>r;init(n);to[0][0]=1;for(int i=1;i<=n;i++){for(int j=0;j<=i;j++){ll mul=1;int sgn=(i-j&1)?-1:1;for(int k=0;k<=n;k++){(to[i][k]+=sgn*binom(i,j)*mul)%=mod;mul=mul*((j*r-k)%mod)%mod*ifac[k+1]%mod;}}}for(int i=1;i<=n;i++)cin>>a[i];dfs(1);cout<<(f[1][0]+mod)%mod;
}
### Java NIO （二）NIO Buffer类的重要方法
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∘ Φ \circ_Φ ∘Φ denotes a convolution Let b l o c k d i a g blockdiag blockdiag(A) be a n1n3-by-n2n3 block diagonal matrix， f o l d fold fold indicate its inverse operator diagonal degree tensor D \mathcal{D} D 作者未提供代码
### ASP.NET Core 的 Web Api 实现限流 中间件
Microsoft.AspNetCore.RateLimiting 中间件提供速率限制（限流）中间件。 它是.NET 7 以上版本才支持的中间件，刚看了一下，确实挺好用，下面给大家简单介绍一下： RateLimiterOptionsExtensions 类提供下列用…
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Sedimentation II
Analytical
Ultracentrifugation
by: Andrew Rouff and Andrew Gioe
Partial Specific Volume (v)
•
•
Partial Specific Volume is defined as the
specific volume of the solute, “which is
related to volume increase of of the solution
You can calculate by using different
densities of solution, and creating a graph
Partial Specific Volume (v)
ρ= density of solution
ρo= density of solvent
w= weight
concentration of
solute
ρ= ρo+w(1-ρov)
How we find v
V is found by
calculating the slope
of a graph of p as a
function of w
ρ= ρo+w(1-ρov)
ρ-ρo= w(1-ρov)
ρ-ρo/w= (1-ρov)
ρ/w=v
Molecular Mass Sedimentation and
Diffusion Equation
s/D= M(1-vρo)/ RT
M=sRT/D(1-vρo)
D4.19 found by
combining previous
equations
First Svedberg
Equation
Assumption
The First Svedberg Equation works if diffusion
and sedimentation friction coefficients are the
same (s and D)
In reality, this is not the actual case, causes
slight error in formula
Sedimentation Equilibrium
•
•
As a solution is
centrifuged for a
long time,
eventually the
diffusion and
sedimentation stop
changing over time
This is when it is at
equilibrium
Lamm Equation
(dC/dt) = -1/r[d/dr(w2r2sCDr(dc/dr])
+ diffusion
coefficient
Sedimentation Equilibrium
Rearrange Svedberg equation, D= sRT/m(1-vρo)
Plug into Lamm Equation, flux equal to zero at equilibrium
[-sRT/m(1-vρo)](dC/dx)+ sw2r2C(x)=0
rearrange- dln(C)/d(r2/2)= M(1-vρo)w2/ RT
How to use slope
Slope of Ln[C(r)] vs
r2/2 graph is M(1vρo)w2/ RT
As shown by graph,
line is linear,
meaning term is
constant
Slope= Molecular
weight
Number average molecular mass
Second graph is not linear because there is
more than one macromolecule present
ΣNimi/ΣNi is number average molecular mass
eg. 10 molecules of particle A which is 2Da and 5
molecules of particle B which is 3Da
[(10)(2) + (5)(3)] / 15 = 2.3Da = Mn
Weight Average Molecular Mass
Σ(Nimi)mi/ΣNimi= Weight average molecular mass
eg. 5 molecules of A which is 2Da and 10 molecules of B
which is 3Da
[(5)(2)(2)+(10)(3)(3)](15)(5)= 1.47Da
Concentration dependence of
average molecular mass
C(r)= C(a)exp[w2M(1-vρ)(r2-a2)/2RT]= second svedberg
equation
exp[w2M(1-vρ)(r2-a2)/2RT] known as * for now
C(r) = Ca*+ Cb* + Cab*
Cab= CaCb/Kab
Can find equilibrium constant from sedimentation data
A Closer Look at the
Forces in Centrifugation
Centrifugation results in a
center of Rotor
The Two Techniques
1. Analytical Zonal Sedimentation Velocity
a.
High velocity, low spin time
a.
Low velocity, High spin time
Analytical Zonal Sedimentation Velocity
● Upon Centrifugation, the analyte particles
sediment through the gradient to separate
zones based on their sedimentation velocity
● Linear 5-20% sucrose gradients are a
material
● Separates the molecules in mixtures
according to their sedimentation coefficients
(S’s)
The Process of Sedimentation in a
Centrifugal Field
● Velocity zonal sedimentation separates
molecules in the mixture according to their
sedimentation coefficients.
● Analyte particles when exposed to the
centrifugal field settle down through the sucrose
solution until their density is equal to the density
of the sucrose solution .
● A density gradient of the analyte particles
results with the the densest particles migrating
the farthest through the sucrose solution.
A sample containing mixtures of particles of varying size, shape and density is added
on the top of a preformed density gradient. The gradient is higher in density toward
the bottom of the tube. Centrifugation results in separation of the particles
depending on thier size,shape and buoyant density. Fractions of defined volume
Example of an Automated Volume
Fraction Collector ● Since Density
upon cessation of
Centrifugation, the
sample tube may
transferred to a
fraction collector.
● A hole is poked in the
bottom of the sample
tube and then the
density fraction are
dripped into collection
tubes one level at a
time
Equilibrium
● Pre sample injection, a solution containing a
heavy such as CsCl or RbCl is spun until a
small solute density gradient forms within the
cell from the force of the Centrifugal field.
● Three components are in tube/cell: solvent
molecules, solutes molecules (salts) and
analyte molecules.
● The small solute becomes distributed in the
cell in just the same way as a large molecule
A Mathematical Description of the
equilibrium sedimentation distribution
● An equation describing the equilibrium sedimentation
distribution is obtained by setting the total flux equal in the
cell to zero in since at equilibrium, there are no changes in
concentration with time
Because of the small molecular mass of the solute
molecules we can expand the Svedberg eqation and
Thus the Svedberg form of the macromolecular
concentration distribution between meniscus a and
point r in the cell reduces to:
Where:
C is the concentration distribution of the analyte particles.
M is the mass of the analyte particles
v is the partial specific volume of the analyte particles
p is the density
w is the angular velocity
R is the Universal Gas Constant and T is the Temperature in the
cell
Applying Equilibrium Sedimentation to prove the Semi Conservative Nature of DNA
Messelson and Stahl grew E.coli cells in a medium in which the sole nitrogen source was 15 Nlabelled ammonium chloride.
The 15 N-containing E.coli cell culture was then transferred to a light 14 N medium and allowed to
continue growing. Samples were harvested at regular intervals.
The DNA was extracted and its buoyant density determined by centrifugation in CsCl density
gradients. The isolated DNA showed a single band in the density gradient, midway between the
light 14 N-DNA and the heavy 15 N-DNA bands (Fig. D4.24(c)).
After two generations in the 14 N medium the isolated DNA exhibited two bands, one with a density
equal to light DNA and the other with a density equal to that of the hybrid DNA observed after one
generation (Fig. D4.24(d)).
After three generations in the 14 N medium the DNA still has two bands, similar to those observed
after two generations (Fig. D4.24(e)). The results were exactly those expected from the
semiconservative replication hypothesis.
Macromolecular Shape from
Sedimentation Data
● Molecules of the same shape but different molecular
mass are called homologous series.
● The relationship between mass M and sedimentation
coefficient s is as follows
Homologous series of quasi-spherical
particles: globular proteins in water
● The frictional coefficient of a sphere of radius R0
under slip boundary conditions in a solvent of
viscosity η0 is given by Stokes Law:
● A good straight line fit of log s* versus log M
is obtained according to the single equation:
● This establishes that globular proteins
actually form a homologous series.
● Small deviations from perfect spherical
shapes and the existence of hydration shelly
modify the relation between the Stokes
radius and the partial specific volume without
changing the power law.
● Thus we conclude that the fact that their
sedimentation behavior can be described by
the single previous equation means that
globular proteins are very close to spherical
in shape and hydrated to abou the same
extent.
● Three proteins that do not obey the
equation:
Can you identify them?
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# Showing that the maximum likelihood estimator (MLE) exists but is not unique
I have a few questions with regards to a solution to the problem below:
1. How is it possible to have $max_{1\leq{i}\leq{n}}x_i-1<\theta<min_{1\leq{i}\leq{n}}x_i$? How can a value of $\theta$ be both greater than a larger value and less than a smaller value? Isn't that contradictory?
2. How is it that $max_{1\leq{i}\leq{n}}x_i-1 < min_{1\leq{i}\leq{n}}x_i$? For example, what if my $max_{1\leq{i}\leq{n}}x_i = 10$ and $min_{1\leq{i}\leq{n}}x_i = 1$? Would this then mean that $10<1$?
3. Finally, if $x_1,...,x_n= 1,0,1,0,1,0,...,1,0$, where the maximum is $1$ and the minimum is $0$, wouldn't it mean that $0<\theta<0$ and so implies that $\theta = 0$ and thus is unique?
Thanks everyone!!!
• Did you notice that $\text{max}(x_i) < \text{min}(x_i)+1$? What happens when you subtract 1 from both sides? This relates to both (1) and (2). It looks like you need to pay careful attention to how the density has been defined. Pick $\theta=1.5$, say, and draw the density. Now try $\theta=2.7$. If you do both of those, the answers to several of your questions should be more obvious. In particular, you should see why the question just before (3) makes no sense. – Glen_b Feb 2 '14 at 11:29
• If you actually have a sample where its min value is 1 and its max value is 10, then simply, this sample cannot be a realization from the assumed distribution, and you have a case of misspecification. – Alecos Papadopoulos Feb 2 '14 at 11:30
1. The value of θ is not greater than a larger value and less than a smaller value. Note that you subtract 1 from the maximum value, and the difference between the maximum and minimum can never be greater than 1.
2. Your example is not consistent with the probability distribution. The probability distribution says that all possible values of X are between θ and θ + 1 (for some fixed number θ). Note that you actually have a uniform distribution on this interval.
3. Yes. In this case θ itself (not just its estimate) must be 0. The same thing will happen whenever the largest value in the samples is 1 greater than the smallest value. This has of course probability zero of occurring, but is theoretically possible.
• It's worth noting that the pdf is a uniform distribution on the open interval between θ and θ + 1. Both you and the solution seem to have a slight error here. – Roland Feb 2 '14 at 11:00
• @Roland I think it’s an unfortunate (though valid) definition in the textbook. Whether you define the support of X as an open, half-open or closed interval, the distribution of X is the same; that is, the cumulative distribution functions are identical. – Karl Ove Hufthammer Feb 2 '14 at 11:10
• I see that the cdf are identical, and this makes sense; I haven't thought that far. However, it seems to make a difference for the likelihood function, as only an open interval would allow strict inequalities - cf. my answer to question 2. – Roland Feb 2 '14 at 11:16
1. You might be missing the $-1$ after the max: $\theta$ lies above the maximum value of the $x_i$s $-1$ and the minimum value of the $x_i$s. Another way is two write $\max x_i \leq \theta + 1 \leq \min x_i +1$.
2. The likelihood function of $\theta$ given $n$ independent observations $x_1, \dots, x_n$ is given by the product of their probabilities - in this case $\Pi_{i=1}^n f(x_i\vert \theta)$, which is the the product of the indicator functions of the interval $(\theta, \theta+1)$. This means the following: If one of the $x_i$s doesn't lie in this interval, the likelihood is zero. The case where the likelihood is not equal to zero can be translated to: The largest value of the $x_i$s is less than $\theta +1$, and the smallest is more than $\theta$, i.e. $$\max_{i=1, \dots n}x_i< \theta, \quad \min_{i=1, \dots n}x_i> \theta + 1.$$ Subtracting $1$ from both sides gives us the desired formula.
3. Taking the inquality from the solution for the case where the likelihood is equal to $1$ is, as you correctly observed, $0<\theta <0$. Since no number is smaller than itself, this case simply can't happpen. Thus, for your example, all $\theta$s have the same likelihood, namely $0$, and all values of $\theta$ maximize the likelihood, albeit in a rather unsatisfactory way. The argument from the solution is concerning the case where there is some $\theta$ such that the likelihood is equal to one.
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BIDMAS or BODMAS Questions, Revision and Worksheets
GCSE 1 - 3KS3AQAEdexcelOCRWJECAQA 2022Edexcel 2022OCR 2022WJEC 2022
BIDMAS or BODMAS
BIDMAS (sometimes BODMAS) is an acronym that helps us remember which order to perform operations in. We start from left to right. The letters stand for:
• Brackets
• Indices (or as they can be called, Orders)
• Division
• Multiplication
• Subtraction
Level 1-3 GCSE KS3
Using BIDMAS
When performing calculations, always follow the BIDMAS order of operations.
Example: Work out the value of $3 \times(3^2 + 4) - 8$
Step 1: The first letter of BIDMAS is B, meaning the first thing we should do is look to what’s inside the brackets. (If there are not brackets, move onto indices, then divide and so on..)
Here, we have two operations happening: a power/index, and an addition. The letter I comes before the letter A in BIDMAS which means we first work out the result of $3^2$ and then add $4$ to it.
$(3^2 + 4) = (9 + 4) = 13$
Step 2: We are left with a multiplication and a subtraction, so because M comes before S, we do the multiplication first and the subtraction second,
$3 \times 13 - 8 = 39 - 8 = 31$
Level 1-3 GCSE KS3
BIDMAS and Fractions
For fractions, we work out what the values of the top (numerator) and bottom (denominator) are separately (using the rules of BIDMAS), and then lastly, we look at the fraction we have and see if it can be simplified.
Example: Simplify the fraction $\dfrac{3 \times 4 - 5}{11 + (9 \div 3)}$
Step 1: First, considering the numerator. There’s a multiplication and a subtraction, so we do the multiplication first and the subtraction second.
$3 \times 4 - 5 = 12 - 5 = 7$
Step 2: Now, the denominator. That contains a division inside brackets, so that will be the first bit of the calculation, and then the addition will be second.
$11 + (9 \div 3) = 11 + (3) = 14$
Step 3: Therefore, our fraction is $\dfrac{7}{14}$. Both top and bottom have a factor of $7$, so the simplified answer is
$\dfrac{7}{14} = \dfrac{1}{2}$
Level 1-3 GCSE KS3
Note:
• For division and multiplication, work them out in the order that they appear (from left to right).
• For addition and subtraction, calculate them in the order that they appear (from left to right) when they are the only two operations left in the sum.
Example: BIDMAS and Algebra
Write the expression $4xy \times 9y - 13 \times xy^2$ in its simplest form.
[3 marks]
Step 1: There are two multiplications in this expression, so it doesn’t matter which order we do them in providing we do them both before the subtraction. The first one becomes:
$4xy \times 9y = 4 \times 9 \times x \times y \times y = 36xy^2$
Step 2: The second multiplication becomes:
$13 \times xy^2 = 13xy^2$
Step 3: So, now we subtract the second from the first one, to get the expression in its simplest form.
$36xy^2 - 13xy^2 = 23xy^2$
Level 1-3 GCSE KS3
Example Questions
Mathematical operators must be carried out in the correct order. The acronym BIDMAS (or BODMAS) is a helpful way to remember this order.
There are two brackets (B) to first calculate,
$(2 \times 3^3)$ and $(15-9)$
Inside the first bracket, there is a power or index number (I or O),
$2 \times 3^3 = 2 \times 27$
Carry out any divisions or multiplications (DM) then additions or subtractions (AM) inside the brackets,
$(2 \times 27 = 54)$ and $(15-9 = 6)$
Complete the calculation,
\begin{aligned} 54 \div 6 &= 9 \\ (2 \times 3^3) \div (15 - 9) &= 9 \end{aligned}
The first operation to consider following BIDMAS is the calculation inside the brackets (B),
$12 \div 4 = 3$
As this does not simplify we can move onto the indices (I),
$(3)^2 = 9$
Again as this does not simplify, the last operation of the expression is multiplication (M), and we get
$16\times 9= 144$
The first operation to consider following BIDMAS is the calculation inside the brackets (B) dealing with the numerator and denominator separately for the moment.
In the numerator, we have to first, substitute in the given value of $x$ and apply the power (I), before the addition (A).
$((-3)^2+3)=(9+3)=12$
In the denominator, there are no indices nor any multiplications divisions to consider so we can move straight to the subtraction (S),
$(10-6) = 4$
The last operation of the expression is a division (D), so,
$\dfrac{12}{4}=3$
The first operation to consider following BIDMAS is the calculation inside the brackets (B),
$y^2 + 5y^2 = 6y^2$
There are no indices or divisions in this expression, but there is a multiplication (M),
$3y \times 7y = 21y^2$
The last operation of the expression is subtraction (S), and we get,
$6y^2 - 21y^2 = -15y^2$
The first operation to consider following BIDMAS is the calculation inside the brackets (B) dealing with the numerator and denominator separately for the moment.
In the numerator, there is only one operation in the form of multiplication (M), so
$42q^2 \times pq = 42q^2 \times q \times p = 42q^{3}p$
In the denominator, the first calculation is inside the brackets (B), which is a substitution (S),
$9p - 5p = 4p$
Then, the division (D) operator can be applied,
$28p^3 \div 4p = 7p^2$
So, we are left with a fraction with $p$ on the top and bottom, as well as a factor of $7$. Both of these cancel so,
$\dfrac{42q^{3}p}{7p^2}=\dfrac{6q^3}{p}$
As there are no more common factors, we can not simplify the expression any further.
Worksheet and Example Questions
(NEW) BIDMAS /BODMAS - Exam Style Questions - MME
Level 1-3 GCSENewOfficial MME
Level 1-3 GCSE
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# How do you solve sin(2x) - sin(3x) + sin(4x) = 0?
##### 1 Answer
May 11, 2015
The answer is $\left\{x \in \mathbb{R} : x = k \pi \mathmr{and} x = \frac{2 \pi}{3} + 2 k \pi \mathmr{and} x = - \frac{2 \pi}{3} + 2 k \pi , k \in \mathbb{Z}\right\}$
Remember:
$\sin \left(4 x\right) = 2 \sin \left(2 x\right) \cos \left(2 x\right)$
$\sin \left(3 x\right) = \sin \left(x\right) \cos \left(2 x\right) + \cos \left(x\right) \sin \left(2 x\right)$
$\sin \left(2 x\right) \left(1 + 2 \cos \left(2 x\right)\right) = \sin \left(2 x\right) \cos \left(x\right) + \cos \left(2 x\right) \sin \left(x\right)$
Remember: $1 = 2 - 1$
$\sin \left(2 x\right) \left(2 + 2 \cos \left(2 x\right) - 1 - \cos \left(x\right)\right) = \cos \left(2 x\right) \sin \left(x\right)$
Remember: $1 + \cos \left(2 x\right) = 2 {\cos}^{2} \left(x\right)$
$2 \sin \left(x\right) \cos \left(x\right) \left(4 {\cos}^{2} \left(x\right) - 1 - \cos \left(x\right)\right) = \cos \left(2 x\right) \sin \left(x\right)$
We have one set of solutions: $A = \left\{x \in \mathbb{R} : \sin \left(x\right) = 0\right\} = {\left\{k \pi\right\}}_{k \in \mathbb{Z}}$
We now can simplify
$2 \cos \left(x\right) \left(4 {\cos}^{2} \left(x\right) - 1 - \cos \left(x\right)\right) = \cos \left(2 x\right)$
Remember: $\cos \left(2 x\right) = {\cos}^{2} \left(x\right) - {\sin}^{2} \left(x\right) = 2 {\cos}^{2} \left(x\right) - 1$
$8 {\cos}^{3} \left(x\right) - 2 \cos \left(x\right) - 2 {\cos}^{2} \left(x\right) = 2 {\cos}^{2} \left(x\right) - 1$
$y = \cos \left(x\right)$
$8 {y}^{3} - 4 {y}^{2} - 2 y + 1 = 0$
We notice it's
${\left(2 y\right)}^{3} - {\left(2 y\right)}^{2} - \left(2 y\right) + 1 = 0$
So $2 y = - 1$ and it's the sole solution in $\mathbb{R}$ (we can divide with Ruffini's rule, or calculate the local minimum to prove it)
So we have another set of solutions $B = \left\{x \in \mathbb{R} : \cos \left(x\right) = - \frac{1}{2}\right\} = \left\{\frac{2 \pi}{3} + 2 k \pi\right\} \cup \left\{- \frac{2 \pi}{3} + 2 k \pi\right\}$
We know we don't have any other solution so the set is $A \cup B$
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Binary Math Lessons: The Secret Origin
Unschooling? How did my last post have anything to do with unschooling? As soon as I saw the title on the screen, I cringed. The benefits of binary math, check, anything to do with unschooling? Nada.
As it turns out, I’d started in the middle of the story. Our six year-old, No. 1, and I started heading towards binary math—in more proper unschooling form—because she wandered into the room one day and said, “Dad, I want to learn what you do at work.”
All I do at work is test machines whose sole job it is to move ones and zeroes around: microprocessors and other digital devices also known as computer chips in the vernacular. So, since one and zero are pretty simple concepts, and as it turns out, the logic gate building blocks of digital devices are also really simple, off we went!
The first thing we need to nail down were the handful of logic gates I encounter. What’s a logic gate you ask? It’s just an electrical embodiment of a logical construct, (you know like the one’s you had in philosophy 101). Take the ‘and’ gate we started out with for example. The device takes two inputs that can be either a one or a zero, and outputs a single number in return, again either one or zero. If both the inputs are 1, (known as logical true in the vernacular), then the device outputs a one, if not, then the device outputs a 0,(a logical false value). Hence, if one output AND the other are both true, the ‘and’ gate gives a true output aka 1. Otherwise its output is 0, aka false .
No. 1 and I made up some homework sheets for her to play around with. Her homework was to fill in the logic gates on the page with any sets of inputs and outputs from the table.
"Hey, what's 1 AND 1?"
"1"
"What's 1 AND 0?"
I personally think our constant conversations drive things home more than the homework, but who knows? In any event, there are a lot of MUNI riders who know more about logic gates than they used to or maybe wanted to.
In case you wanted to play along:
And as a picture:
Cool Math Tricks: Deriving the Divergence, (Del or Nabla) into New (Cylindrical) Coordinate Systems
The following is a pretty lengthy procedure, but converting the divergence, (nabla, del) operator between coordinate systems comes up pretty often. While there are tables for converting between common coordinate systems, there seem to be fewer explanations of the procedure for deriving the conversion, so here goes!
What do we actually want?
To convert the Cartesian nabla
to the nabla for another coordinate system, say… cylindrical coordinates.
What we’ll need:
1. The Cartesian Nabla:
2. A set of equations relating the Cartesian coordinates to cylindrical coordinates:
3. A set of equations relating the Cartesian basis vectors to the basis vectors of the new coordinate system:
How to do it:
Use the chain rule for differentiation to convert the derivatives with respect to the Cartesian variables to derivatives with respect to the cylindrical variables.
The chain rule can be used to convert a differential operator in terms of one variable into a series of differential operators in terms of othe…
Lab Book 2014_07_10 More NaI Characterization
Summary: Much more plunking around with the NaI detector and sources today. A Pb shield was built to eliminate cosmic ray muons as well as potassium 40 radiation from the concreted building. The spectra are much cleaner, but still don't have the count rates or distinctive peaks that are expected.
New to the experiment? Scroll to the bottom to see background and get caught up.
Lab Book Threshold for the QVT is currently set at -1.49 volts. Remember to divide this by 100 to get the actual threshold voltage. A new spectrum recording the lines of all three sources, Cs 137, Co 60, and Sr 90, was started at approximately 10:55. Took data for about an hour.
Started the Cs 137 only spectrum at about 11:55 AM
Here’s the no-source background from yesterday
In comparison, here’s the 3 source spectrum from this morning.
The three source spectrum shows peak structure not exhibited by the background alone. I forgot to take scope pictures of the Cs137 run. I do however, have the printout, and…
Unschooling Math Jams: Squaring Numbers in their own Base
Some of the most fun I have working on math with seven year-old No. 1 is discovering new things about math myself. Last week, we discovered that square of any number in its own base is 100! Pretty cool! As usual we figured it out by talking rather than by writing things down, and as usual it was sheer happenstance that we figured it out at all. Here’s how it went.
I've really been looking forward to working through multiplication ala binary numbers with seven year-old No. 1. She kind of beat me to the punch though: in the last few weeks she's been learning her multiplication tables in base 10 on her own. This became apparent when five year-old No. 2 decided he wanted to do some 'schoolwork' a few days back.
"I can sing that song... about the letters? all by myself now!" 2 meant the alphabet song. His attitude towards academics is the ultimate in not retaining unnecessary facts, not even the name of the song :)
After 2 had worked his way through the so…
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## Year 2 Lesson Plan 5 - Fraction addition and perimeter
1. (5 min) Mental Math
1. What is 25% of 16? [4]
2. What is 1/3 of 18? [6]
3. What is the decimal equivalent of 1/3? [.333...]
4. What is 2/3 + 2/3? Express your answer as a mixed number [1 1/3]
5. What is 1/2 of 1/4? [1/8]
6. What is 50% of 1/2? [1/4 or .25]
2. (5 min) Review of selected problems from lesson 4 (no more than 3 problems)
3. (5 min) Review of fractions:
1. Mixed number conversion to improper fractions: Examples:
``` 1 1/2 = 3/2 2 1/3 = 7/3 3 1/4 = 13/4
Ask: How many halves How many thirds How many fourths
are in 1 1/2? are in 2 1/3? are in 3 1/4?
[3] [7] [13]
```
2. Common denominators: Examples:
1. 2/3 + 2/5:
Find a number that both denominators divide into: 3 x 5 = 15
Multiply 2/3 top and bottom by 5 = 10/15
[Reason: to make the new denominator 15]
Multiply 2/5 top and bottom by 3 = 6/15
Add: 10/15 + 6/15 = 16/15 or 1 1/15.
2. 1/4 + 5/6
What denominator should we use?
What do we multiply 1/4 by?
what do we multiply 5/6 by?
4. (10 min) Discuss perimeter [length around the edge of a figure].
1. Use this example:
2. Discuss perimeter of various shapes:
1. Perimeter of a square with sides 5 cm [P=20 cm]
2. Perimeter of an equilateral triangle with sides 5 cm [P= 15 cm]
3. Perimeter of an isoceles triangle with sides 3,3,4 cm [P= 10 cm]
4. Circumference of circle with diameter 5 cm [= pi X diameter = 15.7 cm]
5. Perimeter of regular hexagon with sides 5 cm [P= 30 cm]
(hex = six)
6. Perimeter of regular octagon with sides 5 cm [P= 40 cm]
(oct = 8 [think: octopus])
7. Perimeter of regular pentagon with sides 5 cm [P= 25 cm]
(penta = 5 [think Chrysler symbol])
5. (Remainder of class) In-class exercise
6. Hand out homework as students successfully complete the in-class exercise.
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One side
One side is 36 long with a 15° incline. What is the height at the end of that side?
Result
h = 9.646
Solution:
Leave us a comment of example and its solution (i.e. if it is still somewhat unclear...):
Be the first to comment!
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## Chemistry: Atoms First (2nd Edition)
Published by Cengage Learning
# Chapter 13 - Exercises - Page 575d: 86
#### Answer
From the strongest to the weakest: $HNO_3$, $C_5H_5NH^+$, $N{H_4}^+$, $H_2O$
#### Work Step by Step
- $HNO_3$ is the only strong acid (Ka > 1.0) in the list, so it is the stronger: - $H_2O$ has a $K_a = 10^{-14}$ Analyzing table 13-3, we get that: - $Kb (NH_3) = 1.8 \times 10^{-5}$, therefore: - Since $N{H_4}^+$ is the conjugate acid of $NH_3$ , we can calculate its Ka by using this equation: $K_b * K_a = K_w = 10^{-14}$ $1.8\times 10^{- 5} * K_a = 10^{-14}$ $K_a = \frac{10^{-14}}{ 1.8\times 10^{- 5}}$ $K_a = 5.556\times 10^{- 10}$ and - $Kb(C_5H_5N) = 1.7 \times 10^{-9}$ - Since $C_5H_5NH^+$ is the conjugate acid of $C_5H_5N$ , we can calculate its kb by using this equation: $K_b * K_a = K_w = 10^{-14}$ $1.7\times 10^{- 9} * K_a = 10^{-14}$ $K_a = \frac{10^{-14}}{ 1.7\times 10^{- 9}}$ $K_a = 5.882\times 10^{- 6}$ Therefore, ordering from the higher ka value, to the lower: $HNO_3$, $C_5H_5NH^+$, $N{H_4}^+$, $H_2O$
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1. ## Trigo 7
By using the formula which expresses sin a and cos a in terms of t , where t= tan (a/2) , show that
(1+sin a)/(5+4 cos a )=(1+t)^2/(9+t^2)
I can do this part .
Deduce that $\displaystyle 0\leq\frac{1+\sin a}{5+4 \cos a}\leq\frac{10}{9}$ for all values of a .
I cant figure out this part .
2. You just need to study the function $\displaystyle f(t) = \frac{(1+t)^2}{9+t^2}$ defined over IR
3. Originally Posted by running-gag
You just need to study the function $\displaystyle f(t) = \frac{(1+t)^2}{9+t^2}$ defined over IR
Erm , i can see that t can be any real numbers for the function f(t) to be defined over IR . I don get what u are trying to say , sorry .
4. I am just saying that you need to study the function (variations, limits, ...)
Start by calculating the derivative
5. Hello everyone
Originally Posted by running-gag
You just need to study the function $\displaystyle f(t) = \frac{(1+t)^2}{9+t^2}$ defined over IR
Here's a useful technique for dealing with this type of expression.
Let $\displaystyle y =\frac{(1+t)^2}{9+t^2}$
Then $\displaystyle 9y+t^2y = 1 + 2t+t^2$
$\displaystyle \Rightarrow t^2(y-1) - 2t + (9y-1) =0$
This is a quadratic in $\displaystyle t$ with real roots. So using the discriminant:
$\displaystyle (-2)^2 - 4(y-1)(9y-1) \ge 0$
Which simplifies to:
$\displaystyle -9y^2 +10y \ge 0$
$\displaystyle \Rightarrow y(10-9y) \ge 0$
$\displaystyle \Rightarrow 0 \le y \le \frac{10}{9}$
$\displaystyle \Rightarrow 0 \le \frac{(1+t)^2}{9+t^2} \le\frac{10}{9}$
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# Moment of Inertia of swivel chair
1. Sep 3, 2011
### Afphan
Question:
While sitting in a swivel chair, you push against the floor withyour heel to make the chair spin. The 7.1 frictional force is appliedat a point 45 from the chair'srotation axis, in the direction that causes the greatest angularacceleration.
If that angular acceleration is 1.5 , what is thetotal moment of inertia about the axis of you and the chair?
I solved it like :
Its actually torque that is the tendency to rotate an object is equal to the perpendicular force at a distance --- T= F * d (Newton-meter is SI unit) -- i found out the torque and then there z relation between Torque and angular acceleration and Moment of Inertia similar to Newton's second law F=MA only the difference is that its not for linear processes but rotational processes (So here Torque is the force , Mass is Moment of Inertia and Linear acceleration is converted to angular acceleration) so T = I . a (T = moment of a force , I = Moment of Inertia , a = angular acceleration ) so it would be like I = T / a = ****(7.1 N * 0.45 m) / 1.5 radpersec^2 **** I = (Ans) Kg/m^2 >>>>>>>>> do check it !
Then for my personal experience i did following experiment :-
Observation = While sitting on a swivel chair , i pushed against the floor with my heel to make the chair of 8 kg spin and i used stop watch to calculate its angular velocity , after 1 second and 20 centiseconds (approx) as calculated it maintained constant 1 rev/sec velocity for further 5 seconds . The frictional force was applied at a 55 cm from the chair's rotation axis. I weigh 65 kg.
Find moment of inertia about my axis and chair ----Answer = kg * m^2
Solution :-
1. Since constant angular velocity is calculated through experiment, we can apply formula to measure constant angular acceleration (radi/sec)
*Angular acceleration = omega (angular velocity) / time
a (angular acceleration) = 6.92 / 1.2 = 5.76 radi / sec^2
2. To calculate frictional force applied , Newton's of law of inertia can be used :-
F = m * a
F (tangential) = m * a (tangential)
//a (tan) = r * a (angular.a)//
<F (tan) = m * r a (angular.a)>
(where F = frictional force , m = mass , r = moment arm and a = angular mass)
Lets suppose i need to add both masses = 73 Kg
F = 73 * 0.55 * 5.76 = 231.264 N
3. Calculate Torque = T = F x r = 127.20 Nm
4. In order to calculate Moment of inertia , use its relation with torque and angular acceleration = T = I * a (where T= torque, I= moment of inertia , a = angular acceleration)
I = T / a = 127.2 / 5.76
I = 22.0825 Kg.m^2
Note :-
Just check my observations out and if i'm wrong , how actually to find out moment of inertia or an irregular shaped body like swivel chair
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## Intermediate Algebra: Connecting Concepts through Application
$\color{blue}{-2cd^2}$
Factor the radicand so that at least one factor is a perfect fifth power: $=\sqrt[5]{(-2)^5c^5(d^2)^5}$ Bring out the fifth root of the perfect fifth power factors to obtain: $=\color{blue}{-2cd^2}$
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# Kirchh
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### Kirchh
1. 1. The KIRCHHOFF’S RULES<br />
2. 2. Introduction <br />Kirchhoff’s rules <br />Types of Kirchhoff’s rules <br />Junction rule <br />Example of Junction rule <br />Loop rule <br />Example of Loop rule <br />Sign Convention (Kirchhoff”s rule) <br />Uses of Kirchhoff’s rules <br />TOPICS<br />
3. 3. Introduction <br />Most electrical circuit consist not merely a single source and a single external resistor, but comprise of a number of sources, resistors or other elements such as capacitors, motors etc. interconnected in a complicated manner. The general term applied to such a circuit is called a network In this presentation, we will discuss how Kirchhoff’s Rule are based on charge neutrality in a metal which will greatly help in calculating electrical properties. <br />
4. 4. KIRCHHOFF’S RULES <br />We know that :- 1/R=1/R1 + 1/R2 + 1/R3 +………+ 1/Rn<br />
5. 5. This uses the fact that there is no net current at any junction . The potential difference across any resistor is same that is if we complete the circuit XY -> YX via a path involving any two resistors, the total potential change is 0. These facts, are called Kirchhoff’s Rules which are very useful for many circuit problem.These rules are given by Gustav Kirchhoff in 1845. <br />
6. 6. Junction rule<br />It states that “at any junction of several circuit elements, The sum of currents entering the junction must equal the sum of currents leaving it.” This rule is based on the fact that change can’t accumulate at any point in a conductor in a steady situation. Net positive or negetive charge will accumalate at the junction at a rate equal to the net electrical current at the junction.<br />
7. 7. EXAMPLE OF JUNCTION RULE <br />In this figure, the currents directed towards junction are:- <br />I 1 ,I 2 , -I 3 and -I 4 . So, <br />I 1 + I 2 +(-I 3 ) +(-I 4 )=0 <br />Charges pass through point. <br />So,net charge coming towards point should be equal to that going away from it in same time. <br />
8. 8. LOOP RULE <br />It states that “The algebraic sum of changes in potential around any closed resistor loop must be zero.” This rule is based on energy conservation. Otherwise,one can continuously gain energy by circulating charge around a closed loop in a particular direction . The net charge of all potential differences should be zero. <br />
9. 9. EXAMPLE OF LOOP RULE<br />While using this rule, one starts from a point on the loop and goes along the loop, either clockwise or anti- clockwise, to reach the same point again. <br />Any potential drop encountered is taken to be positive and any potential rise is taken to be negative. <br />
10. 10. The sum of all potential differences should be zero. As we start from A and go along the loop clockwise to reach the same point A,we get the following potential differences: <br />VA – VB = -I1 R1 VB – VC = -I2 R2 VC – VD = - E1 VD – VE = I3 R3 VE – VF =-I4 R4 VF – VA = E2 <br />
11. 11. Adding all these, 0= I 1 R 1 + I 2 R 2 – E1 + I 3 R 3 – I 4 R 4 + E2 The loop will follow the fact that the work done by it in any closed path is zero. <br />
12. 12. SIGN CONVENTION IN APPLYING KIRCHHOFF’S RULES <br />The principal difficulty in these rules is not in understanding basic ideas but in keeping track of algebraic signs. <br />
13. 13. GUIDELINES HELPING IN SOLVING <br />PROBLEMS ARE:- <br />1.Choose any closed loop in the network, and designate a direction (clockwise or counter clockwise) to transverse the loop in applying the loop rule. <br />
14. 14. 2 . Go around the loop in the designated direction, adding emf’s and potential differences. An emf is counted as positive when it is traversed from (-) to (+)and negative when transformed from (+) to (-). An IR term is counted negative if the resistor is traversed in same direction of the assumed current, and positive if in opposition direction.<br />
15. 15. 3. Equate the sum of step (2) to zero . <br />4 . If necessary, choose another loop to obtain different relations between the unknowns . <br />
16. 16. continue until there are as many equations and unknowns or until every circuit element has been included in at least one of the chosen loops.<br />
17. 17. USES OF KIRCHHOFF’S RULES<br />To determine currents in different parts of closed loop : <br />JUNCTION RULE: <br />∑l= 0 (at any junction) <br />LOOP RULE: <br />If directions of travel and current are SAME, the sign of “IR” is(-). <br />If directions of travel and current are opposite,then sign of “IR” is(+). <br />
18. 18. To determine potential difference between any 2 points in loop. <br />Travel from any one point in the circuit to the other, the potential difference between them will be :- <br />= ∑ E + ∑ IR, <br />as applied on the cells and resistances between these points. <br />
19. 19. Thank you<br />
20. 20. Done By<br />M.Rama Krishna<br />
21. 21. Guided by<br />G.PrasadaRao<br />PGT PHYSICS<br />
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## Inscribed conic as envelope of tripolars
Let (e) be a conic inscribed in the triangle ABC with perspector at D. Consider line (a') as a variation of side (a), a' being tangent at a point E, different from the traces D1,D2,D3 of D. The following are complementary remarks to the discussion in HyperbolaPropertyParallels2.html , showing that
Every inscribed in a triangle ABC conic (e) is the envelope of the tripolars E1E2 of points A'' on its perspectrix.
[1] Apply Pascal's theorem (for quadrangles) to quadrangle ED1D2D3. Points H, E* are intersections of opposite sides and C, E1 are intersections of opposite tangents. The four points are collinear and make a harmonic division (E1,C,H,E*)=-1.
[2] Apply Brianchon's theorem (for quadrangles) to quadrangle ABGE2. Lines AG, BE2 are diagonals and lines D2E, D1D3 join opposite contact points. The four lines pass through a common point I.
[3] By considering the pole-polar relation for points H and E2 we deduce that line AG passes through H, line BE2 passes through E* and triangle HIE* is self-polar.
[4] Let J be the intersection point of D2H and BE2. Line D2H is the polar of B. Line BE2 is the polar of H. Thus, line BH is the polar of J.
[5] By [1] (E1,C,H,E*)=-1 and this is equal to (B,E2,I2,E*), which from E1 projects onto (A,G,I,H), which from B projects onto (A,K,E*,A'), A' being the intersection of BH and AE*. These relations imply that A' is the tripole of the tangent at E line E1E2.
[6] By the aforementioned discussion A' is on the tripolar of D.
There are several other relations of coincidence and harmonicity of tetrads of points hidden in this figure. For example, points {K,L,E1} are collinear. See the file InconicsTangents.html for an alternative discussion.
### See Also
Brianchon.html
HyperbolaPropertyParallels2.html
InconicsTangents.html
PascalOnQuadrangles.html
TrilinearPolar.html
Trilinears.html
Produced with EucliDraw© http://users.math.uoc.gr/~pamfilos/
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# Is there a simple relationship between angle of attack and lift coefficient?
Is there an equation relating AoA to lift coefficient?
I have been searching for a while: there are plenty of discussions about the relation between AoA and Lift, but few of them give an equation relating them.
I know that for small AoA, the relation is linear, but is there an equation that can model the relation accurately for large AoA as well? (so that we can see at what AoA stall occurs)
I am not looking for a very complicated equation. Can anyone just give me a simple model that is easy to understand?
(Of course, if it has to be complicated, then please give me a complicated equation)
• Are you asking about a 2D airfoil or a full 3D wing? – MikeY May 19 '19 at 1:42
• @MikeY 3D is good, but 2D is OK – Holding Arthur May 19 '19 at 3:56
• For a 3D wing, you can tailor the chord distribution, sweep, dihedral, twist, wing airfoil selection, and other parameters to get any number of different behaviors of lift versus angle of attack. So your question is just too general. – MikeY May 19 '19 at 13:30
No, there's no simple equation for the relationship.
Here's an example lift coefficient graph:
(Image taken from http://www.aerospaceweb.org/question/airfoils/q0150b.shtml.)
This is actually three graphs overlaid on top of each other, for three different Reynolds numbers. I'll describe the graph for a Reynolds number of 360,000.
We see that the coefficient is 0 for an angle of attack of 0, then increases to about 1.05 at about 13 degrees (the stall angle of attack). From here, it quickly decreases to about 0.62 at about 16 degrees. Then it decreases slowly to 0.6 at 20 degrees, then increases slowly to 1.04 at 45 degrees, then all the way down to -0.97 at 140, then...
Well, in short, the behavior is pretty complex. The most accurate and easy-to-understand model is the graph itself.
• "there's no simple equation". Could you give me a complicated equation to model it? – Holding Arthur May 19 '19 at 4:53
• You could take the graph and do an interpolating fit to use in your code. – MikeY May 19 '19 at 13:28
• @Holding Arthur, the relationship of AOA and Coefficient of Lift is generally linear up to stall. So for an air craft wing you are using the range of 0 to about 13 degrees (the stall angle of attack) for normal flight. There is an interesting second maxima at 45 degrees, but here drag is off the charts. This is why coefficient of lift and drag graphs are frequently published together, – Robert DiGiovanni May 19 '19 at 14:33
• @HoldingArthur Perhaps. What are you planning to use the equation for? I don't want to give you an equation that turns out to be useless for what you're planning to use it for. – Terran Swett May 20 '19 at 0:55
• Great graph and source (there's also a graph for drag), but the conclusion seems wrong. Based on those graphs, lift is clearly proportional to sin(2α) in the post-stall region. – Rainer P. May 21 '19 at 20:27
In the post-stall regime, airflow around the wing can be modelled as an elastic collision with the wing's lower surface, like a tennis ball striking a flat plate at an angle. Lift and drag are thus:
$$c_L = sin(2\alpha)$$ $$c_D = 1-cos(2\alpha)$$
I superimposed those (blue line) with measured data for a symmetric NACA-0015 airfoil and it matches fairly well. I don't know how well it works for cambered airfoils.
The lift coefficient is linear under the potential flow assumptions. So just a linear equation can be used where potential flow is reasonable.
Potential flow solvers like XFoil can be used to calculate it for a given 2D section. Or for 3D wings, lifting-line, vortex-lattice or vortex panel methods can be used (e.g. using XFLR5).
When the potential flow assumptions are not valid, more capable solvers are required.
XFoil has a very good boundary layer solver, which you can use to fit your "simple" model to (e.g. a spline approximation). And I believe XFLR5 has a non-linear lifting line solver based on XFoil results.
For 3D wings, you'll need to figure out which methods apply to your flow conditions. Possible candidates are: experimental data, non-linear lifting line, vortex panel methods with boundary layer solver, steady/unsteady RANS solvers, ...
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# the number e
### Paying interest and the number e
January 24, 2011 |
Suppose I borrow a dollar from you and I’ll pay you 100% interest at the end of the year. How much money will you have then? \$1 * (1 + 1) = \$2 What happens if instead the interest is calculated as 50% twice in the year? \$1 * (1.5 * 1.5) = \$2.25 After … Continue reading →
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q7-sequences
# q7-sequences - not equal to any one of r 1 r 2 r 3(a...
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CSE 260 QUIZ-7– Sequences-summations: ANSWERS (25 minutes) NAME: 1. What are the nth terms of the following sequences. (a) 1,3,5,7,9,. . . 2 n - 1 (b) 2,4,6,8,10,. . . 2 n (c) 3,6,12,24,48,96,192,. . . 3 . 2 n - 1 (d) 1,0,2,0,4,0,8,0,16,0,. . . 2 ( n - 1) / 2 for odd terms 0 for even terms 2. Compute the following double sum. Σ 3 i =1 Σ 2 j =1 ( i - j ) 3 i =1 (( i - 1) + ( i - 2)) =(1 - 1) + (1 - 2) + (2 - 1) + (2 - 2) + (3 - 1) + (3 - 2) =3 3. Find the value of Σ 200 k =100 k Given Σ n k =1 k = n ( n + 1) / 2 Σ 200 k =1 k - Σ 99 k =1 k =(200(200 + 1)) / 2 - (99(99 + 1)) / 2 =20100 - 4950 =15150 1
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4. In showing that the set of all real numbers is uncountable , we show that the subset of all real numbers that fall between 0 and 1 is also uncountable . We show this by contradiction assuming that the subset is countable . Under this assumption, the real numbers between 0 and 1 can be listed in some order, say r 1 , r 2 , r 3 , .... Let the decimal representation of these numbers be r 1 = 0 .d 11 d 12 d 13 d 14 ... r 2 = 0 .d 21 d 22 d 23 d 24 ... r 3 = 0 .d 31 d 32 d 33 d 34 ... - - - where d ij ² { 0 , 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 } We contradict by deFning a real number r that is between 0 and 1 but
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Unformatted text preview: not equal to any one of r 1 , r 2 , r 3 , .... (a) Complete the following deFnition of r: d i = 4 if d ii 6 = 4 d i = 5 if d ii = ?4 (b) Indicate why r is not equal to any of r 1 , r 2 , r 3 , ..... ith digit of r is di±erent from the ith digit of r i ∀ i 5. Indicate if the following sets are countable . (a) Set of all odd positive integers. Yes, countable because of the one to one correspondence between Z + and the set of all odd positive integers, as shown below. 1 2 3 4 . . . ^ ^ ^ ^ | | | | v v v v 1 3 5 7. . . (b) Set of all odd integers. 2 Yes, countable because of the one to one correspondence between Z + and the set of all odd positive integers, as shown below. 1 2 3 4 5 6. . . ^ ^ ^ ^ ^ ^ | | | | | | v v v v v v 1-1 2-2 3-3 . . . 3...
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Maths Quiz for Class 3 Multiplication
# Maths Quiz for Class 3 Multiplication
## Maths Quiz for Class 3 Multiplication
Maths quiz for class 3 multiplication is very helpful to sharp the brain of kids. In this post, you will find 20 maths quiz questions for class 3 multiplication.
Question 1: 7 × 9 = ______
A) 56
B) 72
C) 63
D) 54
Explanation: 7 × 9 = 63
Question 2: 15 × 6 = ______
A) 90
B) 75
C) 80
D) 85
Explanation: 15 × 6 = 90
Question 3: 321 × 3 = ______
A) 963
B) 363
C) 923
D) 961
Explanation: 321 × 3 = 963
Question 4: The product of 254 and 12 is _____.
A) 3008
B) 3048
C) 2448
D) 3028
Explanation: 254 × 12 = 3048
Question 5: The product of 326 and 16 is _____.
A) 5206
B) 5216
C) 5116
D) 4216
Explanation: 326 × 16 = 5216
Question 6: 248 × 10 = ______
A) 248000
B) 24800
C) 2480
D) 248
Explanation: 248 × 10 = 2480
Question 7: 421 × 100 = ______
A) 4210
B) 42100
C) 421
D) 421000
Explanation: 421 × 100 = 42100
Question 8: 57 × 1000 = ______
A) 57
B) 570
C) 5700
D) 57000
Explanation: 57 × 1000 = 57000
Question 9: 576 × 0 = _____
A) 0
B) 1
C) 576
D) 575
Explanation: 576 × 0 = 0
Question 10: 384 × 1 = _____
A) 0
B) 1
C) 384
D) 383
Explanation: 384 × 1 = 384
Question 11: There are 342 beads in a box. The number of beads in 6 such boxes are ________.
A) 2042
B) 2052
C) 1852
D) 1952
Explanation: 342 × 6 = 2052
Question 12: Hema has 4 piggy banks with ₹ 265 in each. How much money does she have?
A) ₹ 860
B) ₹ 1060
C) ₹ 960
D) ₹ 1050
Explanation: ₹ 265 × 4 = ₹ 1060
Question 13: Which product from the following options is equal to 42 + 18?
A) 12 × 5
B) 12 × 4
C) 14 × 5
D) 15 × 5
Explanation: 12 × 5 = 42 + 18 = 60
Question 14: Which one of the following products lies between 50 and 60?
A) 7 × 7
B) 7 × 9
C) 6 × 8
D) 7 × 8
Explanation: 7 × 8 = 56
Question 15: Which one of the following is correct?
A) 13 × 10 = 1300
B) 7 × 100 = 70
C) 8 × 1000 = 8000
D) 11 × 1000 = 1100
Explanation: 8 × 1000 = 8000 is correct and others are wrong.
Question 16: 14 × 4 hundreds = _______
A) 560
B) 5600
C) 56
D) 56000
Explanation: 14 × 4 hundreds = 14 × 400 = 14 × 4 × 100 = 5600
Question 17: There are 5 packets of 50 candies each. The total number of candies is _________.
A) 250
B) 200
C) 150
D) 350
Explanation: 50 × 5 = 250
Question 18: If a packet contains 45 candles, then the total number of candles in 8 such packets is ________.
A) 270
B) 315
C) 360
D) 405
Explanation: 45 × 8 = 360
Question 19: There are 48 students in each section of class 3. If total 5 sections are there in class 3, then the total number of students in class 3 is _____.
A) 250
B) 240
C) 260
D) 280
Explanation: 48 × 5 = 240
Question 20: John bought 13 chocolates for his birthday party. If each chocolate costs ₹ 25, how much money did he pay for the chocolates?
A) ₹ 275
B) ₹ 300
C) ₹ 325
D) ₹ 350
Explanation: ₹ 25 × 13 = ₹ 325
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+0
# Inequality
0
149
1
Solve the inequality 2x - 5 <= -x + 12. Give your answer as an interval.
Oct 1, 2021
#1
+13581
+1
Solve the inequality 2x - 5 $$\leq$$ -x + 12.
Hello Guest!
$$2x - 5 \leq -x + 12.\\ x\leq \dfrac{17}{3}$$
$$x\in \mathbb R\ |\ x\leq \frac{17}{3}$$
!
Oct 1, 2021
edited by asinus Oct 2, 2021
edited by asinus Oct 2, 2021
edited by asinus Oct 2, 2021
edited by asinus Oct 2, 2021
edited by asinus Oct 2, 2021
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# Enter math problem
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# ellipse
• May 31st 2009, 04:34 PM
dan123
ellipse
an ellipse has the following points
(-2,3) left
(2,5) top
(6,3) right
(2,1) bottom
write the equation in general and standard form of the ellipse
• May 31st 2009, 05:09 PM
Chris L T521
Quote:
Originally Posted by dan123
an ellipse has the following points
(-2,3) left
(2,5) top
(6,3) right
(2,1) bottom
write the equation in general and standard form of the ellipse
The length of the axis are as follows:
Between $\displaystyle \left(-2,3\right),\left(6,3\right)$: length is 6-(-2)=8
Between $\displaystyle \left(2,5\right),\left(2,1\right)$: length is 5-1=4
So it follows that the major axis is parallel to the x axis and $\displaystyle a=4$, and the minor axis is parallel to the y axis and $\displaystyle b=2$.
The ellipse will take the form $\displaystyle \frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1$.
I leave it for you to verify that the center of the ellipse is (2,3).
Therefore, the equation of the ellipse is $\displaystyle \frac{(x-2)^2}{16}+\frac{(y-3)^2}{4}=1$
Does this make sense?
• May 31st 2009, 05:52 PM
dan123
yeah,it does,I knew the formulas and everything,but I just find it difficult to put it all together
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A328359 Numbers k such that Omega(k - 2) = Omega(k) = Omega(k + 2) where Omega = A001222. 2
5, 68, 93, 121, 143, 172, 185, 188, 203, 215, 217, 219, 244, 284, 289, 301, 303, 321, 342, 393, 404, 413, 415, 428, 436, 471, 490, 517, 535, 570, 581, 604, 669, 687, 697, 788, 791, 815, 858, 870, 892, 1014, 1057, 1079, 1135, 1137, 1139, 1147, 1167, 1205, 1206, 1208, 1210, 1255, 1268, 1276 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 LINKS Harvey P. Dale, Table of n, a(n) for n = 1..1000 EXAMPLE 5 is a term because A001222(3) = A001222(5) = A001222(7) = 1; 68 is a term because A001222(66) = A001222(68) = A001222(70) = 3; 93 is a term because A001222(91) = A001222(93) = A001222(95) = 2. MATHEMATICA Select[Range[10^4], PrimeOmega[#-2]==PrimeOmega[#]==PrimeOmega[#+2]&] (* Metin Sariyar, Oct 14 2019 *) Flatten[Position[Partition[PrimeOmega[Range[2000]], 5, 1], _?(#[[1]]== #[[3]] == #[[5]]&), 1, Heads->False]]+2 (* Harvey P. Dale, Nov 02 2021 *) PROG (MAGMA) [k: k in [4..1300]| forall{m:m in [-2, 2]| &+[p[2]: p in Factorization(k+m)] eq &+[p[2]: p in Factorization(k)] }]; // Marius A. Burtea, Oct 15 2019 CROSSREFS Cf. A001222, A280382, A278311. Sequence in context: A129963 A115764 A252794 * A003361 A336433 A099334 Adjacent sequences: A328356 A328357 A328358 * A328360 A328361 A328362 KEYWORD nonn AUTHOR Juri-Stepan Gerasimov, Oct 14 2019 STATUS approved
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Last modified December 3 11:58 EST 2021. Contains 349462 sequences. (Running on oeis4.)
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# Abelian groups and the order of the group
How can I prove that the groups with order less than 6 are Abelian.?
And the first non Abelian group is of the order 6.
• Depends on what facts you are allowed to use. Do you know that groups of prime order are cyclic, and cyclic groups are abelian, and groups of order square of a prime are abelian? Commented Mar 20, 2015 at 6:13
• So that means 2,3,5 are out and 4 is Abelian.. Yeaa gotit.... Commented Mar 20, 2015 at 6:14
To add to @Yval Filmus' answer, every group of prime order must be abelian, leaving all groups of order 4. Since $S_3$ is of order $3! = 6$ and is not abelian, a group of order $6$ is the smallest order for which a group is abelian.
Since $2$, $3$, and $5$ are prime, this leaves groups of order $1$ and $4$. A group of order $1$ is trivial, and there are only two groups of order $4$. Every group of order $4$ is isomorphic to either $\mathbb{Z}_{4}$ or $\mathbb{Z}_{2}\oplus\mathbb{Z}_{2}$.
$\mathbb{Z}_{4}$ is abelian, and $G \oplus H$ is abelian if and only if $G$ and $H$ are abelian, thus $\mathbb{Z}_{2}\oplus\mathbb{Z}_{2}$ must be abelian since $\mathbb{Z}_{2}$ is abelian.
• In fact we don't need to prove that any group of order $4$ is isomorphic to either $\mathbb Z_4$ or $\mathbb Z_2 \oplus \mathbb Z_2.$ If a group of order $4$ has an element of order $4,$ then it is cyclic. If it has no element of order $4,$ then every element $g$ in that group satisfies $g^2=e.$ Any group with this condition is abelian. Commented Mar 20, 2015 at 7:08
• Right! Only one element of a group can have order $1$ (the identity). This means that all elements must have either order $2$ or $4$ by Lagrange's theorem (all subgroups must divide the order of the group, which is $4$). If there is an element of order $4$, then the group is cyclic, and cyclic groups are necessarily abelian. If there is no element of order $4$, then all elements must be of order $2$, which means the group is necessarily abelian. Thanks for pointing that out :) Commented Mar 20, 2015 at 7:18
• Yes! Your answer is absolutely correct. I merely pointed out that one doesn't need the isomorphisms for degree $4$ as it may create some unnecessary complication. Since we need only to prove that it is abelian, we can avoid the isomorphisms. +1 Commented Mar 20, 2015 at 7:25
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# Wilston Definite Integral Problems And Solutions Pdf
## Unit 4. Applications of integration MIT OpenCourseWare
### CHAPTER 32 Improper Integrals alexnegrescu
PROBLEM SET 7 SOLUTIONS. ArsDigita University. definite integrals ncert problems and solutions PDF may not make exciting reading, but definite integrals ncert problems and solutions is packed with valuable instructions, information and warnings., integral sign. This leaflet explains how to evaluate definite integrals. 1. Definite integrals The quantity Z b a f(x)dx is called the definite integral of f(x) from a to b. The numbers a and b are known as the lower and upper limits of the integral. To see how to evaluate a definite integral consider the following example. Example Find Z 4 1 x2dx. Solution First of all the integration of.
### Step-by-step Solutions for Definite Integrals in Wolfram|Alpha
Unit 4. Applications of integration MIT OpenCourseWare. 2 PROBLEM SET 7 SOLUTIONS (a) R ln(x) x dx ANSWER: You can do this integral by integration by parts (see below), but its much easier to just substitute u = ln(x), because then du = 1 x, Test how well you understand the definition of definite integrals with the mathematics problems found in this interactive quiz. Continue your....
Integration by substitution There are occasions when it is possible to perform an apparently difficult piece of integration by first making a substitution. This has the effect of changing the variable and the integrand. When dealing with definite integrals, the limits of integration can also change. In this unit we Basic Methods of Learning the art of inlegration requires practice. In this chapter, we first collect in a more systematic way some of the integration formulas derived in Chapters 4-6. We then present the two most important general techniques: integration by substitution and integration by parts. As the techniques for evaluating integrals are developed, you will see that integration is a more
integral sign. This leaflet explains how to evaluate definite integrals. 1. Definite integrals The quantity Z b a f(x)dx is called the definite integral of f(x) from a to b. The numbers a and b are known as the lower and upper limits of the integral. To see how to evaluate a definite integral consider the following example. Example Find Z 4 1 x2dx. Solution First of all the integration of Integrals - Exercises. Here you will find problems for practicing. Each problem has hints coming with it that can help you if you get stuck. The main topic is integrals. The ones from Basic methods are for initial practicing of techniques; the aim is not to solve the integrals, but just do the specified step. The Simple problems are genuine
This section contains problem set questions and solutions on the definite integral and its applications. Math exercises on integral of a function. Practice the basic formulas for integrals and the substitution method to find the indefinite integral of a function.
Definite Integral Using U-Substitution •When evaluating a definite integral using u-substitution, one has to deal with the limits of integration . •So by substitution, the limits of integration also change, giving us new Integral in new Variable as well as new limits in the same variable. •The following example shows this. Practice Problems: Integration by Parts (Solutions) Written by Victoria Kala [email protected] November 25, 2014 The following are solutions to the Integration by Parts practice problems …
Practice Problems: Integration by Parts (Solutions) Written by Victoria Kala [email protected] November 25, 2014 The following are solutions to the Integration by Parts practice problems … 26/02/2018 · Here is a set of practice problems to accompany the Computing Definite Integrals section of the Integrals chapter of the notes for Paul Dawkins Calculus I course at Lamar University.
Master the concepts of Definite Integral including properties of definite integral and geometrical interpretation with the help of study material for IIT JEE by askIITians. integral sign. This leaflet explains how to evaluate definite integrals. 1. Definite integrals The quantity Z b a f(x)dx is called the definite integral of f(x) from a to b. The numbers a and b are known as the lower and upper limits of the integral. To see how to evaluate a definite integral consider the following example. Example Find Z 4 1 x2dx. Solution First of all the integration of
Solution of Jee Mains Maths problems Sample Paper 04 (Download Pdf.) Question Paper-05 : Definite & Indefinite Integration The Science Stream students who are preparing for the JEE Advanced exam already know the benefits of having the JEE Mains sample question papers. Practice Problems: Improper Integrals Written by Victoria Kala [email protected] December 6, 2014 Solutions to the practice problems posted on November 30. For each of the following problems: (a) Explain why the integrals are improper. (b) Decide if the integral is convergent or divergent. If it is convergent, nd which value it converges to
Problem: Evaluate the integral Solution: We started to solve this problem in this note as an example of substitution, we prepared it like this: Why did we chose to do so? The root was clearly troublesome, so getting rid of it by substitution seemed like a good idea. Whether it will be possible or not depended on us being able to express dx solely in terms of y. Write a definite integral, here b a f x dx, to express the limit of these sums as the norms of the partitions go to zero. 3. Evaluate the integral numerically or with an antiderivative. EXAMPLE 4 Modeling the Effects of Acceleration A car moving with initial velocity of 5 mph accelerates at the rate of a t 2.4t mph per second for 8 seconds.
Problem: Evaluate the integral Solution: We started to solve this problem in this note as an example of substitution, we prepared it like this: Why did we chose to do so? The root was clearly troublesome, so getting rid of it by substitution seemed like a good idea. Whether it will be possible or not depended on us being able to express dx solely in terms of y. Master the concepts of Definite Integral including properties of definite integral and geometrical interpretation with the help of study material for IIT JEE by askIITians.
2012 Integration Bee Qualifying Test January 13, 2012 Name: Email: This is the qualifying test for the 2012 Integration Bee, held on Friday, January 13th at 4PM–6PM in room 4-149. Finalists will be notified by email by midnight tonight (12:00am, Saturday, January 14th). You have 20 minutes to solve these 25 integrals. Each integral is worth 1 point. In order to receive full credit you must THE CALCULUS PAGE PROBLEMS LIST Problems and Solutions Developed by : D. A. Kouba And brought to you by : Beginning Integral Calculus : Problems using summation notation ; Problems on the limit definition of a definite integral Problems on u-substitution ; Problems on integrating exponential functions ; Problems on integrating trigonometric functions ; Problems on integration by parts
improper integral. divergent if the limit does not exist. RyanBlair (UPenn) Math104: ImproperIntegrals TuesdayMarch12,2013 4/15 . ImproperIntegrals Infinite limits of integration Definition Improper integrals are said to be convergent if the limit is finite and that limit is the value of the improper integral. divergent if the limit does not exist. Each integral on the previous page is Basic Methods of Learning the art of inlegration requires practice. In this chapter, we first collect in a more systematic way some of the integration formulas derived in Chapters 4-6. We then present the two most important general techniques: integration by substitution and integration by parts. As the techniques for evaluating integrals are developed, you will see that integration is a more
Get acquainted with the concepts of Solved Examples on Definite Inetgral with the help of study material for IIT JEE by askIITians. 26/02/2018 · Here is a set of practice problems to accompany the Computing Definite Integrals section of the Integrals chapter of the notes for Paul Dawkins Calculus I course at Lamar University.
Math 114Q Integration Practice Problems 12.! √ π 0 xsin(x2)dx Let u = x2.Then du =2x dx, so π 0 xsin(x2)dx = 1 2! √ π 0 sin(x2)·2x dx1 2! x= π x=0 sin(u)du1 2 " −cos(x2) π 0 =1 13.! x √ 4−x dx [Hint: If u =4−x, what does that make x in terms of u?] If u =4−x, then x =4−u and so dx = −1du.Now just substitute all of this into the integral: INTEGRAL CALCULUS - EXERCISES 43 Homework In problems 1 through 13, find the indicated integral. Check your answers by differentiation. 1. R x5dx 2. R x3 4 dx 3.
19/12/2016 · This calculus video tutorial explains how to calculate the definite integral of function. It provides a basic introduction into the concept of integration. It provides plenty of examples and integral sign. This leaflet explains how to evaluate definite integrals. 1. Definite integrals The quantity Z b a f(x)dx is called the definite integral of f(x) from a to b. The numbers a and b are known as the lower and upper limits of the integral. To see how to evaluate a definite integral consider the following example. Example Find Z 4 1 x2dx. Solution First of all the integration of
2 PROBLEM SET 7 SOLUTIONS (a) R ln(x) x dx ANSWER: You can do this integral by integration by parts (see below), but its much easier to just substitute u = ln(x), because then du = 1 x Master the concepts of Definite Integral including properties of definite integral and geometrical interpretation with the help of study material for IIT JEE by askIITians.
26/02/2018 · Here is a set of practice problems to accompany the Computing Definite Integrals section of the Integrals chapter of the notes for Paul Dawkins Calculus I course at Lamar University. Practice Problems: Improper Integrals Written by Victoria Kala [email protected] December 6, 2014 Solutions to the practice problems posted on November 30. For each of the following problems: (a) Explain why the integrals are improper. (b) Decide if the integral is convergent or divergent. If it is convergent, nd which value it converges to
MATH 105 921 Solutions to Integration Exercises Therefore, Z sintcos(2t)dt= 2 3 cos3 t+ cost+ C 7) Z x+ 1 4 + x2 dx Solution: Observe that we may split the integral as follows: Definite Integral Using U-Substitution •When evaluating a definite integral using u-substitution, one has to deal with the limits of integration . •So by substitution, the limits of integration also change, giving us new Integral in new Variable as well as new limits in the same variable. •The following example shows this.
The connection between the definite integral and indefinite integral is given by the second part of the Fundamental Theorem of Calculus. If f is continuous on [a, b] then . Take note that a definite integral is a number, whereas an indefinite integral is a function. Example: Evaluate. Solution… The connection between the definite integral and indefinite integral is given by the second part of the Fundamental Theorem of Calculus. If f is continuous on [a, b] then . Take note that a definite integral is a number, whereas an indefinite integral is a function. Example: Evaluate. Solution…
164 Chapter 8 Techniques of Integration Z cosxdx = sinx+C Z sec2 xdx = tanx+ C Z secxtanxdx = secx+C Z 1 1+ x2 dx = arctanx+ C Z 1 √ 1− x2 dx = arcsinx+ C 8.1 Substitution Needless to say, most problems we encounter will not be so simple. THE CALCULUS PAGE PROBLEMS LIST Problems and Solutions Developed by : D. A. Kouba And brought to you by : Beginning Integral Calculus : Problems using summation notation ; Problems on the limit definition of a definite integral Problems on u-substitution ; Problems on integrating exponential functions ; Problems on integrating trigonometric functions ; Problems on integration by parts
Math 114Q Integration Practice Problems 12.! √ π 0 xsin(x2)dx Let u = x2.Then du =2x dx, so π 0 xsin(x2)dx = 1 2! √ π 0 sin(x2)·2x dx1 2! x= π x=0 sin(u)du1 2 " −cos(x2) π 0 =1 13.! x √ 4−x dx [Hint: If u =4−x, what does that make x in terms of u?] If u =4−x, then x =4−u and so dx = −1du.Now just substitute all of this into the integral: Test how well you understand the definition of definite integrals with the mathematics problems found in this interactive quiz. Continue your...
26/02/2018 · Here is a set of practice problems to accompany the Computing Definite Integrals section of the Integrals chapter of the notes for Paul Dawkins Calculus I course at Lamar University. In this lesson, we will define and interpret definite integrals geometrically, evaluate definite integrals using properties and apply definite integrals to find area of a bounded region. OBJECTIVES After studying this lesson, you will be able to : • define and interpret geometrically the definite integral as a limit of sum;
### Lecture Notes on Integral Calculus Undergrad Mathematics
Motion problems (with definite integrals) (article) Khan. Basic Methods of Learning the art of inlegration requires practice. In this chapter, we first collect in a more systematic way some of the integration formulas derived in Chapters 4-6. We then present the two most important general techniques: integration by substitution and integration by parts. As the techniques for evaluating integrals are developed, you will see that integration is a more, Math exercises on integral of a function. Practice the basic formulas for integrals and the substitution method to find the indefinite integral of a function..
### PROBLEM SET 7 SOLUTIONS. ArsDigita University
1.1 Integrals as solutions Mathematics LibreTexts. 2012 Integration Bee Qualifying Test January 13, 2012 Name: Email: This is the qualifying test for the 2012 Integration Bee, held on Friday, January 13th at 4PM–6PM in room 4-149. Finalists will be notified by email by midnight tonight (12:00am, Saturday, January 14th). You have 20 minutes to solve these 25 integrals. Each integral is worth 1 point. In order to receive full credit you must https://en.wikipedia.org/wiki/Integral_approximation integral sign. This leaflet explains how to evaluate definite integrals. 1. Definite integrals The quantity Z b a f(x)dx is called the definite integral of f(x) from a to b. The numbers a and b are known as the lower and upper limits of the integral. To see how to evaluate a definite integral consider the following example. Example Find Z 4 1 x2dx. Solution First of all the integration of.
• Math 104 Improper Integrals (With Solutions)
• Understanding Calculus Problems Solutions and Tips
• Get acquainted with the concepts of Solved Examples on Definite Inetgral with the help of study material for IIT JEE by askIITians. definite integrals ncert problems and solutions PDF may not make exciting reading, but definite integrals ncert problems and solutions is packed with valuable instructions, information and warnings.
indefinite integral and definite integral which makes the definite integral as a practical tool for science and engineering. The definite integral is also used to solve many interesting problems from various disciplines like economic s, finance and probability . In this Chapter, we shall confine ourselves to the study of indefinite and definite Integral Challenge Problems 1. Z sin 1 x 2 dx 2. Z xsin 1 xdx 3. Z sin 1 p xdx 4. Z 1 1 tan2 x dx 5. Z ln p. Created Date: 1/6/2010 6:51:29 PM
problems in the workbook; and the supporting materials in the back of the workbook, such as the solutions to all problems, glossary, list of formulas, list of theorems, trigonometry review sheet, and composite study sheet, which can be torn out and used for quick and easy reference. 2 19/12/2016 · This calculus video tutorial explains how to calculate the definite integral of function. It provides a basic introduction into the concept of integration. It provides plenty of examples and
Integration by substitution There are occasions when it is possible to perform an apparently difficult piece of integration by first making a substitution. This has the effect of changing the variable and the integrand. When dealing with definite integrals, the limits of integration can also change. In this unit we INTEGRAL CALCULUS - EXERCISES 43 Homework In problems 1 through 13, find the indicated integral. Check your answers by differentiation. 1. R x5dx 2. R x3 4 dx 3.
Practice Problems: Improper Integrals Written by Victoria Kala [email protected] December 6, 2014 Solutions to the practice problems posted on November 30. For each of the following problems: (a) Explain why the integrals are improper. (b) Decide if the integral is convergent or divergent. If it is convergent, nd which value it converges to Basic Integration Problems I. Find the following integrals. 1. (5 8 5)x x dx2 2. ( 6 9 4 3)x x x dx32 3 3. ( 2 3)x x dx 2 23 8 5 6 4. dx x xx 1 5. ( ) 3 x dx
Math exercises on integral of a function. Practice the basic formulas for integrals and the substitution method to find the indefinite integral of a function. Often in practice an integral can be simplified by using an appropriate transformation or substitution and formula 14.6. The following list gives some transformations and their effects. The following list gives some transformations and their effects.
integral sign. This leaflet explains how to evaluate definite integrals. 1. Definite integrals The quantity Z b a f(x)dx is called the definite integral of f(x) from a to b. The numbers a and b are known as the lower and upper limits of the integral. To see how to evaluate a definite integral consider the following example. Example Find Z 4 1 x2dx. Solution First of all the integration of Basic Integration Problems I. Find the following integrals. 1. (5 8 5)x x dx2 2. ( 6 9 4 3)x x x dx32 3 3. ( 2 3)x x dx 2 23 8 5 6 4. dx x xx 1 5. ( ) 3 x dx
Practice problems on double integrals The problems below illustrate the kind of double integrals that frequently arise in probability applications. The first group of questions asks to set up a double integral of a general function f(x,y) over a giving region in the xy-plane. This means writing the integral as an iterated integral of the form Mathematics Learning Centre, University of Sydney 1 1Introduction This unit deals with the definite integral.Itexplains how it is defined, how it is calculated and some of the ways in which it is used. We shall assume that you are already familiar with the process of finding indefinite inte-
Definite Integral Using U-Substitution •When evaluating a definite integral using u-substitution, one has to deal with the limits of integration . •So by substitution, the limits of integration also change, giving us new Integral in new Variable as well as new limits in the same variable. •The following example shows this. 164 Chapter 8 Techniques of Integration Z cosxdx = sinx+C Z sec2 xdx = tanx+ C Z secxtanxdx = secx+C Z 1 1+ x2 dx = arctanx+ C Z 1 √ 1− x2 dx = arcsinx+ C 8.1 Substitution Needless to say, most problems we encounter will not be so simple.
The connection between the definite integral and indefinite integral is given by the second part of the Fundamental Theorem of Calculus. If f is continuous on [a, b] then . Take note that a definite integral is a number, whereas an indefinite integral is a function. Example: Evaluate. Solution… Math 114Q Integration Practice Problems 12.! √ π 0 xsin(x2)dx Let u = x2.Then du =2x dx, so π 0 xsin(x2)dx = 1 2! √ π 0 sin(x2)·2x dx1 2! x= π x=0 sin(u)du1 2 " −cos(x2) π 0 =1 13.! x √ 4−x dx [Hint: If u =4−x, what does that make x in terms of u?] If u =4−x, then x =4−u and so dx = −1du.Now just substitute all of this into the integral:
Math exercises on integral of a function. Practice the basic formulas for integrals and the substitution method to find the indefinite integral of a function. 06/06/2018 · Chapter 1 : Integration Techniques. Here are a set of practice problems for the Integration Techniques chapter of the Calculus II notes. If you’d like a pdf document containing the solutions the download tab above contains links to pdf’s containing the solutions for the full book, chapter and section.
## 8.9 Evaluating definite integrals mathcentre.ac.uk
DEFINITE INTEGRALS NCERT PROBLEMS AND SOLUTIONS PDF. 164 Chapter 8 Techniques of Integration Z cosxdx = sinx+C Z sec2 xdx = tanx+ C Z secxtanxdx = secx+C Z 1 1+ x2 dx = arctanx+ C Z 1 √ 1− x2 dx = arcsinx+ C 8.1 Substitution Needless to say, most problems we encounter will not be so simple., E. Solutions to 18.01 Exercises 4. Applications of integration a/2 y = 3x 4B-6 If the hypotenuse of an isoceles right triangle has length h, then its area.
### DEFINITE INTEGRALS National Institute of Open Schooling
Math Exercises & Math Problems Indefinite Integral of a. problems in the workbook; and the supporting materials in the back of the workbook, such as the solutions to all problems, glossary, list of formulas, list of theorems, trigonometry review sheet, and composite study sheet, which can be torn out and used for quick and easy reference. 2, indefinite integral and definite integral which makes the definite integral as a practical tool for science and engineering. The definite integral is also used to solve many interesting problems from various disciplines like economic s, finance and probability . In this Chapter, we shall confine ourselves to the study of indefinite and definite.
19/12/2016 · This calculus video tutorial explains how to calculate the definite integral of function. It provides a basic introduction into the concept of integration. It provides plenty of examples and In this lesson, we will define and interpret definite integrals geometrically, evaluate definite integrals using properties and apply definite integrals to find area of a bounded region. OBJECTIVES After studying this lesson, you will be able to : • define and interpret geometrically the definite integral as a limit of sum;
improper integral. divergent if the limit does not exist. RyanBlair (UPenn) Math104: ImproperIntegrals TuesdayMarch12,2013 4/15 . ImproperIntegrals Infinite limits of integration Definition Improper integrals are said to be convergent if the limit is finite and that limit is the value of the improper integral. divergent if the limit does not exist. Each integral on the previous page is In this lesson, we will define and interpret definite integrals geometrically, evaluate definite integrals using properties and apply definite integrals to find area of a bounded region. OBJECTIVES After studying this lesson, you will be able to : • define and interpret geometrically the definite integral as a limit of sum;
Solution of Jee Mains Maths problems Sample Paper 04 (Download Pdf.) Question Paper-05 : Definite & Indefinite Integration The Science Stream students who are preparing for the JEE Advanced exam already know the benefits of having the JEE Mains sample question papers. Definite integrals are commonly used to solve motion problems, for example, by reasoning about a moving object's position given information about its velocity. Learn how this is done and about the crucial difference of velocity and speed.
26/02/2018 · Here is a set of practice problems to accompany the Computing Definite Integrals section of the Integrals chapter of the notes for Paul Dawkins Calculus I course at Lamar University. definite integrals ncert problems and solutions PDF may not make exciting reading, but definite integrals ncert problems and solutions is packed with valuable instructions, information and warnings.
Integration by substitution There are occasions when it is possible to perform an apparently difficult piece of integration by first making a substitution. This has the effect of changing the variable and the integrand. When dealing with definite integrals, the limits of integration can also change. In this unit we Definite integrals are commonly used to solve motion problems, for example, by reasoning about a moving object's position given information about its velocity. Learn how this is done and about the crucial difference of velocity and speed.
Practice problems on double integrals The problems below illustrate the kind of double integrals that frequently arise in probability applications. The first group of questions asks to set up a double integral of a general function f(x,y) over a giving region in the xy-plane. This means writing the integral as an iterated integral of the form The connection between the definite integral and indefinite integral is given by the second part of the Fundamental Theorem of Calculus. If f is continuous on [a, b] then . Take note that a definite integral is a number, whereas an indefinite integral is a function. Example: Evaluate. Solution…
Mathematics Learning Centre, University of Sydney 1 1Introduction This unit deals with the definite integral.Itexplains how it is defined, how it is calculated and some of the ways in which it is used. We shall assume that you are already familiar with the process of finding indefinite inte- E. Solutions to 18.01 Exercises 4. Applications of integration a/2 y = 3x 4B-6 If the hypotenuse of an isoceles right triangle has length h, then its area
Here R.H.S. of the equation means integral of f(x) with respect to x. f(x)is called the integrand. dx is called the integrating agent. a is the upper limit of the integral and b is the lower limit of the integral. Evaluating Definite Integrals – Properties. Let us now discuss important properties of definite … Math exercises on integral of a function. Practice the basic formulas for integrals and the substitution method to find the indefinite integral of a function.
Basic Methods of Learning the art of inlegration requires practice. In this chapter, we first collect in a more systematic way some of the integration formulas derived in Chapters 4-6. We then present the two most important general techniques: integration by substitution and integration by parts. As the techniques for evaluating integrals are developed, you will see that integration is a more 19/12/2016 · This calculus video tutorial explains how to calculate the definite integral of function. It provides a basic introduction into the concept of integration. It provides plenty of examples and
Practice Problems: Improper Integrals Written by Victoria Kala [email protected] December 6, 2014 Solutions to the practice problems posted on November 30. For each of the following problems: (a) Explain why the integrals are improper. (b) Decide if the integral is convergent or divergent. If it is convergent, nd which value it converges to 2 PROBLEM SET 7 SOLUTIONS (a) R ln(x) x dx ANSWER: You can do this integral by integration by parts (see below), but its much easier to just substitute u = ln(x), because then du = 1 x
Practice Problems: Improper Integrals Written by Victoria Kala [email protected] December 6, 2014 Solutions to the practice problems posted on November 30. For each of the following problems: (a) Explain why the integrals are improper. (b) Decide if the integral is convergent or divergent. If it is convergent, nd which value it converges to Integral Challenge Problems 1. Z sin 1 x 2 dx 2. Z xsin 1 xdx 3. Z sin 1 p xdx 4. Z 1 1 tan2 x dx 5. Z ln p. Created Date: 1/6/2010 6:51:29 PM
Here R.H.S. of the equation means integral of f(x) with respect to x. f(x)is called the integrand. dx is called the integrating agent. a is the upper limit of the integral and b is the lower limit of the integral. Evaluating Definite Integrals – Properties. Let us now discuss important properties of definite … Definite integrals are commonly used to solve motion problems, for example, by reasoning about a moving object's position given information about its velocity. Learn how this is done and about the crucial difference of velocity and speed.
Practice problems on double integrals The problems below illustrate the kind of double integrals that frequently arise in probability applications. The first group of questions asks to set up a double integral of a general function f(x,y) over a giving region in the xy-plane. This means writing the integral as an iterated integral of the form Definite Integral Using U-Substitution •When evaluating a definite integral using u-substitution, one has to deal with the limits of integration . •So by substitution, the limits of integration also change, giving us new Integral in new Variable as well as new limits in the same variable. •The following example shows this.
Get acquainted with the concepts of Solved Examples on Definite Inetgral with the help of study material for IIT JEE by askIITians. Here R.H.S. of the equation means integral of f(x) with respect to x. f(x)is called the integrand. dx is called the integrating agent. a is the upper limit of the integral and b is the lower limit of the integral. Evaluating Definite Integrals – Properties. Let us now discuss important properties of definite …
Often in practice an integral can be simplified by using an appropriate transformation or substitution and formula 14.6. The following list gives some transformations and their effects. The following list gives some transformations and their effects. Definite integrals are commonly used to solve motion problems, for example, by reasoning about a moving object's position given information about its velocity. Learn how this is done and about the crucial difference of velocity and speed.
Basic Methods of Learning the art of inlegration requires practice. In this chapter, we first collect in a more systematic way some of the integration formulas derived in Chapters 4-6. We then present the two most important general techniques: integration by substitution and integration by parts. As the techniques for evaluating integrals are developed, you will see that integration is a more indefinite integral and definite integral which makes the definite integral as a practical tool for science and engineering. The definite integral is also used to solve many interesting problems from various disciplines like economic s, finance and probability . In this Chapter, we shall confine ourselves to the study of indefinite and definite
definite integrals ncert problems and solutions PDF may not make exciting reading, but definite integrals ncert problems and solutions is packed with valuable instructions, information and warnings. Do note that the definite integral and the indefinite integral (antidifferentiation) are completely different beasts. The definite integral always evaluates to a number. Therefore, Equation $$\ref{1.1.2}$$ is a formula we can plug into the calculator or a computer, and it will be happy to calculate specific values for us. We will easily be able to plot the solution and work with it just like
Solution of Jee Mains Maths problems Sample Paper 04 (Download Pdf.) Question Paper-05 : Definite & Indefinite Integration The Science Stream students who are preparing for the JEE Advanced exam already know the benefits of having the JEE Mains sample question papers. Do note that the definite integral and the indefinite integral (antidifferentiation) are completely different beasts. The definite integral always evaluates to a number. Therefore, Equation $$\ref{1.1.2}$$ is a formula we can plug into the calculator or a computer, and it will be happy to calculate specific values for us. We will easily be able to plot the solution and work with it just like
Write a definite integral, here b a f x dx, to express the limit of these sums as the norms of the partitions go to zero. 3. Evaluate the integral numerically or with an antiderivative. EXAMPLE 4 Modeling the Effects of Acceleration A car moving with initial velocity of 5 mph accelerates at the rate of a t 2.4t mph per second for 8 seconds. The connection between the definite integral and indefinite integral is given by the second part of the Fundamental Theorem of Calculus. If f is continuous on [a, b] then . Take note that a definite integral is a number, whereas an indefinite integral is a function. Example: Evaluate. Solution…
164 Chapter 8 Techniques of Integration Z cosxdx = sinx+C Z sec2 xdx = tanx+ C Z secxtanxdx = secx+C Z 1 1+ x2 dx = arctanx+ C Z 1 √ 1− x2 dx = arcsinx+ C 8.1 Substitution Needless to say, most problems we encounter will not be so simple. 06/06/2018 · Chapter 1 : Integration Techniques. Here are a set of practice problems for the Integration Techniques chapter of the Calculus II notes. If you’d like a pdf document containing the solutions the download tab above contains links to pdf’s containing the solutions for the full book, chapter and section.
### Calculus I Computing Definite Integrals (Practice Problems)
Calculus II Integration Techniques (Practice Problems). Practice problems on double integrals The problems below illustrate the kind of double integrals that frequently arise in probability applications. The first group of questions asks to set up a double integral of a general function f(x,y) over a giving region in the xy-plane. This means writing the integral as an iterated integral of the form, Do note that the definite integral and the indefinite integral (antidifferentiation) are completely different beasts. The definite integral always evaluates to a number. Therefore, Equation $$\ref{1.1.2}$$ is a formula we can plug into the calculator or a computer, and it will be happy to calculate specific values for us. We will easily be able to plot the solution and work with it just like.
### THE CALCULUS PAGE PROBLEMS LIST
Basic Integration Problems hollandcsd.org. 26/02/2018 · Here is a set of practice problems to accompany the Computing Definite Integrals section of the Integrals chapter of the notes for Paul Dawkins Calculus I course at Lamar University. https://en.wikipedia.org/wiki/Integral_approximation 164 Chapter 8 Techniques of Integration Z cosxdx = sinx+C Z sec2 xdx = tanx+ C Z secxtanxdx = secx+C Z 1 1+ x2 dx = arctanx+ C Z 1 √ 1− x2 dx = arcsinx+ C 8.1 Substitution Needless to say, most problems we encounter will not be so simple..
Practice problems on double integrals The problems below illustrate the kind of double integrals that frequently arise in probability applications. The first group of questions asks to set up a double integral of a general function f(x,y) over a giving region in the xy-plane. This means writing the integral as an iterated integral of the form Often in practice an integral can be simplified by using an appropriate transformation or substitution and formula 14.6. The following list gives some transformations and their effects. The following list gives some transformations and their effects.
integral sign. This leaflet explains how to evaluate definite integrals. 1. Definite integrals The quantity Z b a f(x)dx is called the definite integral of f(x) from a to b. The numbers a and b are known as the lower and upper limits of the integral. To see how to evaluate a definite integral consider the following example. Example Find Z 4 1 x2dx. Solution First of all the integration of Here R.H.S. of the equation means integral of f(x) with respect to x. f(x)is called the integrand. dx is called the integrating agent. a is the upper limit of the integral and b is the lower limit of the integral. Evaluating Definite Integrals – Properties. Let us now discuss important properties of definite …
Practice problems on double integrals The problems below illustrate the kind of double integrals that frequently arise in probability applications. The first group of questions asks to set up a double integral of a general function f(x,y) over a giving region in the xy-plane. This means writing the integral as an iterated integral of the form 19/12/2016 · This calculus video tutorial explains how to calculate the definite integral of function. It provides a basic introduction into the concept of integration. It provides plenty of examples and
19/12/2016 · This calculus video tutorial explains how to calculate the definite integral of function. It provides a basic introduction into the concept of integration. It provides plenty of examples and Contents Preface xvii 1 Areas, volumes and simple sums 1 1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 1.2 Areas of simple shapes
integral sign. This leaflet explains how to evaluate definite integrals. 1. Definite integrals The quantity Z b a f(x)dx is called the definite integral of f(x) from a to b. The numbers a and b are known as the lower and upper limits of the integral. To see how to evaluate a definite integral consider the following example. Example Find Z 4 1 x2dx. Solution First of all the integration of definite integrals ncert problems and solutions PDF may not make exciting reading, but definite integrals ncert problems and solutions is packed with valuable instructions, information and warnings.
Definite Integral Using U-Substitution •When evaluating a definite integral using u-substitution, one has to deal with the limits of integration . •So by substitution, the limits of integration also change, giving us new Integral in new Variable as well as new limits in the same variable. •The following example shows this. Often in practice an integral can be simplified by using an appropriate transformation or substitution and formula 14.6. The following list gives some transformations and their effects. The following list gives some transformations and their effects.
Get acquainted with the concepts of Solved Examples on Definite Inetgral with the help of study material for IIT JEE by askIITians. CHAPTER 32 Improper Integrals 32.2 Determine whether J" (1 Ix2) dx 32.3 For what values of p is J" (1 /x)p dx convergent? By Problem 32.1, we know that the integral is divergent when p = 1. 32.4 For p>l, I In the last step, we used L'Hopital's rule to evaluate
164 Chapter 8 Techniques of Integration Z cosxdx = sinx+C Z sec2 xdx = tanx+ C Z secxtanxdx = secx+C Z 1 1+ x2 dx = arctanx+ C Z 1 √ 1− x2 dx = arcsinx+ C 8.1 Substitution Needless to say, most problems we encounter will not be so simple. This section contains problem set questions and solutions on the definite integral and its applications.
MATH 105 921 Solutions to Integration Exercises Therefore, Z sintcos(2t)dt= 2 3 cos3 t+ cost+ C 7) Z x+ 1 4 + x2 dx Solution: Observe that we may split the integral as follows: indefinite integral and definite integral which makes the definite integral as a practical tool for science and engineering. The definite integral is also used to solve many interesting problems from various disciplines like economic s, finance and probability . In this Chapter, we shall confine ourselves to the study of indefinite and definite
E. Solutions to 18.01 Exercises 4. Applications of integration a/2 y = 3x 4B-6 If the hypotenuse of an isoceles right triangle has length h, then its area Here R.H.S. of the equation means integral of f(x) with respect to x. f(x)is called the integrand. dx is called the integrating agent. a is the upper limit of the integral and b is the lower limit of the integral. Evaluating Definite Integrals – Properties. Let us now discuss important properties of definite …
Integration by substitution There are occasions when it is possible to perform an apparently difficult piece of integration by first making a substitution. This has the effect of changing the variable and the integrand. When dealing with definite integrals, the limits of integration can also change. In this unit we Here R.H.S. of the equation means integral of f(x) with respect to x. f(x)is called the integrand. dx is called the integrating agent. a is the upper limit of the integral and b is the lower limit of the integral. Evaluating Definite Integrals – Properties. Let us now discuss important properties of definite …
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RS Aggarwal Solutions: Volume and Surface Area of Solids- 6
# RS Aggarwal Solutions: Volume and Surface Area of Solids- 6 - RS Aggarwal Solutions for Class 10 Mathematics
## Formative Assessment (Unit Test)
Q.1. Find the number of solid sphere, each of diameter 6 cm, that could be moulded to
Form a solid metallic cylinder of height 45 cm and diameter 4 cm.
Given: Diameter of each solid sphere = 6 cm
∴ Radius of each sphere = 6/2 = rs = 3 cm
Diameter of cylinder = 4 cm
∴ Radius of cylinder = 4/2 = r= 2 cm
Height of cylinder = h = 45 cm
Formula: volume of sphere = (4/3) × π × rs3
Volume of cylinder = π × rc2 × h
Spheres are moulded to form cylinder which means the volume remains the same
Let ‘n’ be the number of spheres required
i.e. volume of n spheres = volume of cylinder
∴ n × (4/3) × π × 33 = π × 22 × 45
n × 4 × 27 = 4 × 3 × 45
n = 45/9 = 5
Number of solid spheres made = 5
Q.2. Two right circular cylinder of equal volume have their height in the ratio 1:2. What is the ratio of their radii?
Let the two cylinders be with volume V1 and V2 with their respective radii and height as r1, r2 and h1, h2
Now given ratio of their heights i.e. h1:h2 = 1:2
∴ h1/h2 = 1/2
Volume of cylinder = πr2h
Given that volumes of both cylinder are equal i.e. V1 = V2
∴ π × r12 × h1 = π × r22 × h2
h1/h2 = r22/ r12
r22/ r12 = 1/2
r2/r1 = 1/√2
r1/r2 = √2/1
Therefore the ratio of their radii is r1:r2 = √2:1
Q.3. A circus tent is cylindrical to a height of 4 m and conical above it. If its diameter is 105 m and its slant height is 40 m, find the total area of the canvas required.
Given: diameter of base of cone and the cylinder = 105 m
∴ Radius of cylinder = rcl = 105/2 = 51 m
Radius of cone = rco = 105/2 = 51 m
Height of cylinder = h = 4 m
Slant height of cone = l = 40 m
Formula: Surface area of cylinder = 2πrclh + 2πrcl2
Surface area of cone = πrco2 + πrcol
Since we don’t require canvas for the top surface and bottom surface of cylinder and also for the base of cone we should subtract those areas from the surface area
Area of upper and lower surfaces of cylinder = 2πrcl2
∴ Area of canvas required for cylinder = 2πrclh + 2πrcl2 - 2πrcl2
= 2πrclh
= 2 × 3.14 × 51 × 4
= 1281.12 m2
Area of base of cone = πrco2
∴ area of canvas required for cone = πrco2 + πrcol - πrco2
= πrcol
= 3.14 × 51 × 40
= 6405.6 m2
Total area of canvas required = Area of canvas required for cylinder + area of canvas required for cone
= 1281.12 + 6405.6
= 7686.72 m2
∴ Total area of canvas required = 7686.72 m2
Q.4. The radii of the top and bottom of a bucket of slant height 45 cm are 28 cm and 7 cm respectively. Find the curved surface area of the bucket.
Given: slant height of bucket = l = 45 cm
Radius of bottom circle = r = 7 cm
Radius of top circle = R = 28 cm
As the bucket is in the form of frustum
Formula: Total surface area of frustum = πr2 + πR2 + π(R + r)l cm2
Now we have asked curved surface area, so we should subtract the top and bottom surface areas which are flat circles.
Surface area of top = πr2
Surface area of bottom = πR2
∴ Curved surface area = total surface area – πr2 - πR2 cm2
= πr2 + πR2 + π(R + r)l – πr2 - πR2 cm2
= π(R + r)l cm2
= 3.14 × (28 + 7) × 45 cm2
= 3.14 × 35 × 45 cm2
= 4945.5 cm2
Therefore, curved surface area of bucket = 4945.5 cm2.
Q.5. A solid metal cone with radius of base 12 cm and height 24 cm is melted to form solid spherical balls of diameter 6 cm each. Find the number of balls formed.
Given: base radius of cone = rc = 12 cm
Height of cone = h = 24 cm
Diameter of spherical ball = 6 cm
Radius of spherical ball = rs = 6/2 = 3 cm
Formula: volume of cone = (1/3)πrc2h
Volume of sphere = (4/3)πrs3
Let n be the number of spherical balls made
As the cone is melted and then the spherical balls are made therefore the volume remains same
i.e. volume of n spherical balls made = volume of cone
∴ n × (4/3) × π × rs3 = (1/3) × π × rc2 × h
n × 4 × 33 = 122 × 24
n × 9 = 12 × 24
n = 32
∴ Number of balls formed = 32
Q.6. A hemisphere bowl of internal diameter 30 cm is full of a liquid. This liquid is filled into cylindrical - shapes bottles each of diameter 5 cm and height 6 cm. How many bottles are required?
Given: diameter of hemisphere = 30 cm
∴ Radius of hemisphere = rh = 30/2 = 15cm
Diameter of cylindrical shaped bottles = 5 cm
∴ radius of cylindrical shaped bottles = 5/2 = rc = 2.5 cm
Height of cylindrical shaped bottle = h = 6 cm
Formula: volume of hemisphere = (volume of sphere/2) = (2/3)πrh3
Volume of cylinder = πrc2h
Let ‘n’ bottles are required
As we are filling the cylindrical bottles with liquid in hemispherical bowl hence we can say that volume of liquid in cylindrical bottles = volume of liquid in hemisphere
∴ n × π × rc2 × h = (2/3) × π × rh3
n × 2.52 × 6 × 3 = 2 × 153
n × 6.25 × 9 = 3375
n = 3375/56.25
n = 60
Therefore 60 cylindrical shaped bottles are required to fill the liquid from hemispherical bowl.
Q.7. A solid metallic sphere of diameter 21 cm is melted and recast into small cones, each of diameter 3.5 cm and height 3 cm. Find the number of cones so formed.
Given: diameter of sphere = 21 cm
∴ radius of sphere = rs = (21/2) cm
Diameter of cone = 3.5 cm
∴ radius of cone = rc = 3.5/2 = 1.75 = (7/4) cm
Height of cone = h = 3 cm
Formula: volume of sphere = (4/3)πrs3
Volume of cone = (1/3)πrc2h
Sphere is melted and then cones are made from molten metal therefore the volume remains same
Let ‘n’ be the number of cones made
i.e. volume of n cones = volume of sphere
∴ n × (1/3) × π × rc2 × h = (4/3) × π × rs3
Therefore number of cones formed = 504
Q.8. The diameter of a sphere is 42 cm. It is melted and drawn into a cylindrical wire of diameter 2.8 cm. Find the length of the wire.
Given: diameter of sphere = 42 cm
∴ radius of sphere = 42/2 = rs = 21 cm
Diameter of cylindrical wire = 2.8 cm
∴ Radius of cylindrical wire = rc = 1.4 = (7/5)cm
Let l be the length of wire
Formula: volume of sphere = (4/3)πrs3
Volume of wire = πrc2l
Sphere is melted and wire is made from it
∴ volume of sphere = volume of wire
(4/3)πrs3 = πrc2l
4 × 213 = 3 × (7/5)2 × l
4 × 21 × 9 × 25 = 3 × l
2100 × 3 = l
l = 6300 cm
Therefore length of wire formed = 6300 cm = 63 meters
Q.9. A drinking glass is in the shape of frustum of a cone of height 21 cm with 6 cm and 4 cm as the diameters of its two circular ends. Find the capacity of the glass.
Given: Height of glass = h = 21 cm
Diameter of lower circular end of glass = 4 cm
Diameter of upper circular end of glass = 6 cm
∴ Radius of lower circular end = r = 4/2 = 2 cm
∴ Radius of upper circular end = R = 6/2 = 3 cm
Capacity of glass = volume of frustum
= 22 × (9 + 4 + 6)
= 22 × 19
= 418 cm3
∴ Capacity of glass = 418 cm3
Q.10. Two cubes, each of volume 64 cm3, are joined end to end. Find the total surface area of the resulting cuboid.
volume of each cube = 64 cm3
Let a be the side length of each cube
∴ a3 = 64
a = 4 cm
The figure shows both the cubes joined together after joining we get a cuboid of length 2a and breadth a and height a
Length of cuboid formed = l = 2a
l = 8 cm
Breadth of cuboid formed = b = 4 cm
Height of cuboid formed = h = 4 cm
In the cuboid so formed there are 4 rectangular surfaces of length l = 8 and breadth b = 4 and 2 square surfaces of length 4
Total surface area of cuboid = 4 × l × b + 2a2
= (4 × 8 × 4) + (2 × 42)
= 128 + 32
= 160 cm2
Total surface area of cuboid = 160 cm2
Q.11. The radius of the base and the height of a solid right circular cylinder are in the ratio 2:3 and its volume is 1617 cm3. Find the total surface area of the cylinder. [Take π = 22/7.]
Given: volume of cylinder = 1617 cm3
Let r be the radius of base and h be the height of cone
r:h = 2:3
∴ r/h = 2/3
3r = 2h
h = 3r/2 …(i)
Formula: volume of cylinder = πr2h
∴ 1617 = (22/7) r2h
r2h = 514.5
Using (i) we have
∴ r2 × (3r/2) = 514.5
3r3 = 1029
r3 = 343
r = 7 cm
h = 21/2 cm
Total surface area of cylinder = 2πr2 + 2πrh
= 2 × (22/7) × 72 + 2 × (22/7) × 7 × (21/2)
= 308 + 462
= 770 cm2
Therefore total surface area of cylinder = 770 cm2
Q.12. A toy is in the form of a cone mounted on a hemisphere on a hemisphere of common base radius 7 cm. The total height of the toy is 31 cm. find the total surface area of the toy.
Given: Height of toy = h = 31 cm
Radius of hemisphere = radius of base of cone = r = 7 cm
From the figure we can calculate height of cone as
Height of cone = hc = h - r
= 31 - 7 = 24 cm
∴ hc = 24 cm
Formula: surface area of hemisphere = surface area of sphere/2 = 2πr2
Curved surface area of cone = πrl
Where l is slant height
l = √(49 + 576) = 25 cm
∴ l = 25 cm
Total surface area of toy = curved surface area of cone + surface area of hemisphere
Surface area of hemisphere = 2 × π × r2
= 2 × (22/7) × 72
= 308 cm2
Curved surface area of cone = π × r × l
= (22/7) × 7 × 25
= 550 cm2
∴ Total surface area of toy = 308 + 550
= 858 cm2
∴ Total surface area of toy = 858 cm2
Q.13. A hemispherical bowl of internal radius 9 cm is full of water. This water is to be filled in cylindrical bottles of diameter 3 cm and height 4 cm. find the number of bottles needed to fill the whole water of the bowl.
Given: Radius of hemisphere = rh = 9 cm
Diameter of cylindrical shaped bottles = 3 cm
∴ radius of cylindrical shaped bottles = rc = 3/2 = 1.5 cm
Height of cylindrical shaped bottle = h = 4 cm
Formula: volume of hemisphere = (volume of sphere/2) = (2/3)πrh3
Volume of cylinder = πrc2h
Let ‘n’ bottles are required
As we are filling the cylindrical bottles with liquid in hemispherical bowl hence we can say that volume of liquid in cylindrical bottles = volume of liquid in hemisphere
∴ n × π × rc2 × h = (2/3) × π × rh3
n × (3/2)2 × 4 × 3 = 2 × 93
n = 33 × 2
n = 27 × 2
n = 54
Therefore 54 cylindrical shaped bottles are required to fill the liquid from hemispherical bowl.
Q.14. The surface areas of a sphere and a cube are equal. Find the ratio of their volumes. [Takes π = 22/7.]
Let r be the radius of sphere and a be the side length of cube.
Let Ss be the surface area of sphere and Sc be the surface area of cube and Vs be volume of sphere and Vc be volume of cube
∴ Ss = 4πr2 and Sc = 4a2
Given that surface area of sphere and cube are equal
∴ Ss = Sc
4πr2 = 6a2
r2/ a2 = 3/2π
Vs = (4/3) πr3
Vc = a3
∴ Vs/Vc = 4πr3/3a3
Using (i)
∴ Vs/Vc = √21/√11
Therefore ratio of their volumes is Vs:Vc = √21:√11.
Q.15. The slant height of the frustum of a cone is 4 cm and the perimeters (i.e. , circumferences) of its circular ends are 18 cm and 6 cm. Find the curved surface area of the frustum.
Given: perimeter of upper circle = 18 cm
Perimeter of lower circle = 6 cm
Slant height of frustum = l = 4 cm
Formula: Total surface area of frustum = πr2 + πR2 + π(R + r)l cm2
Let r be the radius of lower circle and R be the radius of upper circle
Now perimeter of circle = circumference of circle = 2π × radius
∴ Perimeter of upper circle = 2πR
18 = 2 × π × R
R = 9/π cm
Perimeter of lower circle = 2πr
6 = 2 × π × r
r = 3/π cm
Now we have asked curved surface area, so we should subtract the top and bottom surface areas which are flat circles.
Surface area of top = πR2
Surface area of bottom = πr2
∴ Curved surface area = total surface area - πr2 - πR2 cm2
= πr2 + πR2 + π(R + r)l - πr2 - πR2 cm2
= π(R + r)l cm2
= π × [(9/π) + (3/π)] × 4 cm2
= (9 + 3) × 4 cm2
= 48 cm2
∴ curved surface area = 48 cm2
Q.16. A solid is composed of a cylinder with hemisphere ends. If the whole length of the solid is 104 cm and the radius of each hemispherical end is 7 cm, find the surface area of the solid.
Total length of solid = l = 104 cm as shown in figure
The solid consist of a cylinder and two hemispheres
Let the height of cylinder be h
We get the height h by subtracting the radii of left and right hemisphere from the total length l as seen in figure
∴ h = 104 - (7 + 7) cm
∴ h = 90 cm
Let r be the radius of hemisphere and the radius of cylinder
∴ r = 7 cm
There are two hemisphere one at left and one at right both of same radius r and two hemispheres make one sphere
Surface area of sphere = 4πr2
The flat circles i.e. the upper and lower circles of cylinder are not to be considered in the surface area of whole solid as they are covered by the hemispheres therefore for cylinder we will take its curved surface area
Curved surface area of cylinder = 2πrh
Surface area of solid = surface area of sphere + curved surface area of cylinder
= 4πr2 + 2πrh
= 2 × (22/7) × 7 × (2 × 7 + 90)
= 44 × 104
= 4576 cm2
Therefore total surface area of solid = 4576 cm2
Q.17. From a solid cylinder whose height is 15 cm and diameter 16 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid. [Use π = 3.14.]
After removing the conical solid the cylinder would look like this
Given: height of cylinder = height of cone = h = 15 cm
Diameter of cylinder = diameter of base cone = 16 cm
∴ radius of cylinder = radius of base of cone = 16/2 = r = 8 cm
Formula: total surface area of cylinder = 2πr2 + πrh
Total surface area of cone = πrl + πr2
Where l is the slant height
l = √(r2 + h2)
l = √(82 + 152)
l = √289
l = 17 cm
In the solid as seen in figure we have the curved surface of cylinder and the base of cylinder as there is no top circular face of the cylinder we should subtract its area from total surface area of cylinder
Area of top circular surface of cylinder = πr2
∴ surface area of cylinder in solid = 2πr2 + πrh - πr2
= πr2 + πrh
= 3.14 × 8 × (8 + 15)
= 3.14 × 8 × 23
= 577.76 cm2
Now there is a hollow conical part with no base of the cone as seen in the figure therefore we should subtract the surface area of base of cone from the total surface area of cone
Surface area of base of cone = πr2
∴ Surface area of conical part in solid = πrl + πr2 - πr2
= πrl
= 3.14 × 8 × 17
= 427.04 cm2
Therefore total surface area of solid = surface area of cylinder in solid + surface area of conical part in solid
= 577.76 + 427.04
= 1004.8 cm2
Surface area of solid = 1004.8 cm2
Q.18. A solid rectangular block of dimension 4.4 m, 2.6 m, 1 m is cast into a hollow cylindrical pipe of internal radius 30 cm and thickness 5 cm. Find the length of the pipe.
Given: length of the block = l = 4.4 m
Width of the block = w = 2.6 m
Height of the block = h = 1 m
Inner radius of pipe = r = 30 cm = 0.3 m
Thickness of pipe = t = 5 cm = 0.05 m
∴ outer radius of pipe as seen in the cross section of pipe = R = r + t = 30 + 5 = 35 cm = 0.35 m
Let l be the length of the pipe
Formula: volume of block = l × w × h
= 4.4 × 2.6 × 1
= 11.44 m3
Volume of block = 11.44 m3
Volume of pipe = π × (radius)2 × (length)
Volume of pipe material = volume of full pipe(R = 0.35) – volume of hollow cylinder(r = 0.3)
= π × 0.352 × l - π × 0.32 × l
= π × l × [(35/100)2 - (3/10)2]
= π × l × [(35/100) + (3/10)] × [(35/100) - (3/10)]
= (22/7) × l × (13/400) m3
∴ volume of pipe material = (22/7) × l × (13/400) m3
The pipe is made from the block
∴ volume of block = volume of pipe material
∴ 11.44 = (22/7) × l × (13/400)
∴ l = 28 × 4
∴ l = 112 m
Length of the pipe = 112 m
Q.19. An open metal bucket is in the shape of a frustum of a cone, mounted on a hollow cylindrical base made of the same metallic sheet. The diameter of the two circular ends of the bucket are 45 cm and 25 cm, the total vertical height of the bucket is 40 cm and that of the cylindrical base is 6 cm. Find the area of the metallic sheet used to make the bucket. Also, find the volume of water the bucket can hold, in litres.
Given: radius of upper circular end of frustum = R = 45 cm
Radius of lower circular end of frustum = radius of cylindrical base = r = 25 cm
Height of bucket = hb = 40 cm
Height of cylindrical base = hc = 6 cm
From the figure height of frustum = hf = hb - hc
= 40 - 6
∴ hf = 34 cm
Volume of cylinder = πr2hc
Curved surface area of cylinder = 2πrhc
curved surface area of frustum = π(R + r)l cm2
Where l = slant height
= √(400 + 1156)
= 39.44 cm
Area of metallic sheet used = curved surface area of frustum + curved surface area of base cylinder + area of base circle of cylinder
Now, curved surface area of frustum = π × (R + r) × l cm2
= (22/7) × (45 + 25) × 39.44 cm2
= 22 × 10 × 39.44 cm2
= 8676.8 cm2
Curved surface area of base cylinder = 2πrhc
= 2 × (22/7) × 25 × 6
= 942.85 cm2
Surface area of base circle of cylinder = πr2
= (22/7) × 252
= 1964.28 cm2
∴ Area of metallic sheet used = 8676.8 + 942.85 + 1964.28
= 11583.93 cm2
Therefore, area of metallic sheet used to make the bucket is 11583.93 cm2 i.e. 1.158393 m2
Volume of water bucket can hold = volume of bas cylinder + volume of frustum
Volume of base cylinder = πr2hc
= (22/7) × 252 × 6
= 11785.71 cm3
= 35.62 × 3775
= 134465.5 cm3
∴ volume of water bucket can hold = 11785.71 + 134465.5
= 146251.21 cm3
Now 1 litre is 1000 cm3
∴ 146251.21 cm3 = 146251.21/1000 = 146.25121 litres
Volume of water bucket can hold = 146.25121 litres
Q.20. A farmer connect a pipe of internal diameter 20 cm from a canal into a cylindrical tank which is 10 m diameter and 2 m deep. If the water flows through the pipe at the of 4 km/hr, in how much time will the tank be filled completely?
Given: diameter of pipe = 20 cm
∴ radius of pipe = rp = 20/2 = 10 cm = 0.1 m
Diameter of tank = 10 m
∴ radius of cylindrical tank = rc = 10/2 = 5 m
Depth of cylindrical tank = height of cylindrical tank = h = 2 m
Rate of flow of water through pipe = 4 km/hr
1 km = 1000 m
4 km/hr = 4000 m/hr
Volume of water required to completely fill the tank is equal to the volume of cylinder
Time require to fill the tank = volume of cylindrical tank/volume of water flown through pipe per hr volume of cylindrical tank = π × rc2 × h
= 3.14 × 25 × 2
= 157 m3
volume of water flown through pipe per hr = π × rp2 × 4000
= 3.14 × 0.12 × 4000
= 3.14 × 40
= 125.6 m3/hr
Time require to fill the tank = 157/125.6
= 1.25 hrs
Therefore, it will take 1.25 hours to fill the tank completely.
The document RS Aggarwal Solutions: Volume and Surface Area of Solids- 6 | RS Aggarwal Solutions for Class 10 Mathematics is a part of the Class 10 Course RS Aggarwal Solutions for Class 10 Mathematics.
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## Integration by Substitution A.K.A... The Reverse Chain Rule
Integration by substitution is just the reverse chain rule. If you learned your derivatives well, this technique of integration won't be a stretch for you.
Let's say we want to find this integral:
For the sake of clarity, we´ll write it like this:
Now, what is the derivative of sinx? It is cosx. So, what we have inside the integral sign is a composite function and its derivative.
Here we'll use a very clever trick. We'll change variable. Let's invent a variable called u, that would be equal to sinx:
The derivative of u is:
If we substitute these two equations into the integral we get:
At this stage, you may not be very comfortable with differentials, but we can "cancel" them out:
So, now we have an integral we already know how to solve:
Finally, we just need to substitute u with sinx:
And that's all integration by substitution is about. Now, let's derive our answer to check it. This is something you can always do check your answers:
By the chain rule:
And this is the function we wanted to integrate!
What we did with that clever substitution was to use the chain rule in reverse.
We saw that the integral was probably a composite function. This is because, according to the chain rule, the derivative of a composite function is the product of the derivatives of the outer and inner functions.
And that's exactly what is inside our integral sign.
As a rule of thumb, whenever you see a function times its derivative, you may try to use integration by substitution.
Let's do some more examples so you get used to this technique.
### Integration by Substitution Example 2
Let's find this integral:
Now, if you remember your derivatives, you know that the derivative of lnx is 1 over x.
To make clear why we need this fact, we'll write the integral like this:
So, again, what we have is a function and its derivative. How should we choose our u?
Remember that we will derive u. So, we should always choose u as the function whose derivative is inside the integral sign. That means:
So we have:
This is an integral we know:
Substitution back the u:
### Integration by Substitution Example 3
Let's find the integral:
Will our u-substitution work here? Where is the derivative of the function there?
If we look closely, we'll note that we do have a composite function there.
We have the function ax. And then sin(ax), which is a composite function.
In our first example, we had a composite function and the derivative of the "inner" function. This integral is similar:
Can you see it? What is the derivative of x? It is one!
So, let's choose our u:
Its derivative:
If we divide both sides of this equation by a, we get:
Now, we can substitute that 1 in our integral:
And we get:
This integral is easier to solve. What is the function whose derivative is sin(u)? It is -cos(u):
So, finally:
### Example 4
Now we'll do a tricky problem:
We can write this as:
We, again, have a composite function and its derivative. The harder part here is knowing how to choose our u:
So, we have:
If you look at your table of integrals, you'll find that:
This fact is easily proved using trigonometric substitution. So, finally we have:
### Conclusion
• Integration by substitution can be considered the reverse chain rule.
• You'll need to know your derivatives well.
• Whenever you see a function times its derivative, you might try to use integration by substitution.
• With practice it'll become easy to know how to choose your u.
If you have just a general doubt about a concept, I'll try to help you. If you have a problem, or set of problems you can't solve, please send me your attempt of a solution along with your question. These will appear on a new page on the site, along with my answer, so everyone can benefit from it.
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piteraufqvw
2023-02-20
How to integrate $\int \frac{1}{{x}^{2}\left(2x-1\right)}$ using partial fractions?
Peruvianoe4p
We need to find $A,B,C$ such that
$\frac{1}{{x}^{2}\left(2x-1\right)}=\frac{A}{x}+\frac{B}{{x}^{2}}+\frac{C}{2x-1}$
for all $x$.
Multiply both sides by ${x}^{2}\left(2x-1\right)$ to get
$1=Ax\left(2x-1\right)+B\left(2x-1\right)+C{x}^{2}$
$1=2A{x}^{2}-Ax+2Bx-B+C{x}^{2}$
$1=\left(2A+C\right){x}^{2}+\left(2B-A\right)x-B$
Equating coefficients give us
$\left\{\begin{array}{l}2A+C=0\\ 2B-A=0\\ -B=1\end{array}$
And thus we have $A=-2,B=-1,C=4$. When this is substituted in the initial equation, we obtain
$\frac{1}{{x}^{2}\left(2x-1\right)}=\frac{4}{2x-1}-\frac{2}{x}-\frac{1}{{x}^{2}}$
Now, integrate it term by term
to get
$2\mathrm{ln}|2x-1|-2\mathrm{ln}|x|+\frac{1}{x}+C$
davz198888za
Perform the decomposition into partial fractions
$\frac{1}{{x}^{2}\left(2x-1\right)}=\frac{A}{{x}^{2}}+\frac{B}{x}+\frac{C}{2x-1}$
$=\frac{A\left(2x-1\right)+Bx\left(2x-1\right)+C\left({x}^{2}\right)}{{x}^{2}\left(2x-1\right)}$
Compare the numerators because the denominators are the same.
$1=A\left(2x-1\right)+Bx\left(2x-1\right)+C\left({x}^{2}\right)$
Let $x=0$, $⇒$, $1=-A$, $⇒$, $A=-1$
Let $x=\frac{1}{2}$, $⇒$, $1=\frac{C}{4}$, $⇒$, $C=4$
Coefficients of ${x}^{2}$
$0=2B+C$
$B=-\frac{C}{2}=-\frac{4}{2}=-2$
Hence,
$\frac{1}{{x}^{2}\left(2x-1\right)}=-\frac{1}{{x}^{2}}-\frac{2}{x}+\frac{4}{2x-1}$
So,
$\int \frac{1dx}{{x}^{2}\left(2x-1\right)}=-\int \frac{1dx}{{x}^{2}}-\int \frac{2dx}{x}+\int \frac{4dx}{2x-1}$
$=\frac{1}{x}-2\mathrm{ln}\left(|x|\right)+2\mathrm{ln}\left(|2x-1|\right)+C$
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## Algebra 1 Tutorial
#### Intro
Pythagorean Theorem is over complicated in most class rooms and I want to simplify it. Originally Pythagoras, the guy who is credited with this theorem, was not doing “math” at all. He needed to find distance’s between two points and he stumbled across this proof. Pythagoras found that if he went 3 steps to the right, and 4 steps forward than he traveled 5 steps from his original position. Let me show you how and why this is.
#### Sample Problem
Try this: Let’s do the problem. Get a piece of paper, a pencil and a ruler.
Mark any point on the paper, this will be the origin (the place we start from). When the point is marked, move three inches in any direction you like. When that line is complete, move either straight up from that line, or straight down, by four inches. You have reached your ending point.
Now measure from the ending point to the origin. You should get 5 inches!
Pythagoras found that 3*3+4*4=5*5 (or 3^2+4^2=5^2)!
#### Solution
Remember the original theorem: a^2+b^2=c^2
Now we know a=3 and b=4
So
a^2=3^2=9
and
b^2=4^2=16
The theorem says to add the two numbers together AFTER they are squared
So
9+16=25
But that is not 5! Why isn’t it 5? Because we did not find c yet, we found c^2.
Lets look again: a^2+b^2=c^2 and we have 9+16=25 thus 25=c^2.
If we take the square root of 25 we get c=5! Sound familiar?
You can now successfully find any distance between two points if you know two of the sides!
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Mechanical Theory of Machines - Set 30 - ObjectiveBooks
Practice Test: Question Set - 30
1. Critical damping is a function of
(A) Mass and stiffness
(B) Mass and damping coefficient
(C) Mass and natural frequency
(D) Damping coefficient and natural frequency
2. Whirling speed of the shaft is the speed at which
(A) Shaft tends to vibrate in longitudinal direction
(B) Torsional vibrations occur
(C) Shaft tends to vibrate vigorously in transverse direction
(D) Combination of transverse and longitudinal vibration occurs
3. In a hydrodynamic journal bearing, there is
(A) A very thin film of lubricant between the journal and the bearing such that there is contact between the journal and the bearing
(B) A thick film of lubricant between the journal and the bearing
(C) No lubricant between the journal and the bearing
(D) A forced lubricant between the journal and the bearing
4. Which of the following statement is correct for gears?
(A) The addendum is less than the dedendum
(B) The pitch circle diameter is the product of module and number of teeth
(C) The contact ratio means the number of pairs of teeth in contact
(D) All of the above
5. The pitching of a ship produces forces on the bearings which act __________ to the motion of the ship.
(A) Vertically and parallel
(B) Vertically and perpendicular
(C) Horizontally and parallel
(D) Horizontally and perpendicular
6. The periodic time of one oscillation for a simple pendulum is
(A) 2π. √(g/l)
(B) (1/2π). √(g/l)
(C) 2π. √(l/g)
(D) (1/2π). √(l/g)
7. The radial distance from the top of a tooth to the bottom of a tooth in a meshing gear, is called
(A) Dedendum
(C) Clearance
(D) Working depth
8. The radial distance of a tooth from the pitch circle to the bottom of the tooth is called
(A) Dedendum
(C) Clearance
(D) Working depth
9. Which of the following is a higher pair?
(A) Belt and pulley
(B) Turning pair
(C) Screw pair
(D) Sliding pair
10. The frictional torque transmitted in a conical pivot bearing with assumption of uniform pressure is __________ as compared to uniform wear.
(A) Less
(B) More
(C) Same
(D) None of these
11. Two pulleys of radii r₁ and r₂ and at distance x apart are connected by means of a cross belt drive. The length of the belt is
(A) π (r₁ + r₂) + [(r₁ + r₂)²/x] + 2x
(B) π (r₁ + r₂) + [(r₁ - r₂)²/x] + 2x
(C) π (r₁ - r₂) + [(r₁ - r₂)²/x] + 2x
(D) π (r₁ - r₂) + [(r₁ + r₂)²/x] + 2x
12. In a radial cam, the follower moves
(A) In a direction perpendicular to the cam axis
(B) In a direction parallel to the cam axis
(C) In any direction irrespective of the cam axis
(D) Along the cam axis
13. In a steam engine, the earlier cut-off with a simple slide valve may be obtained by increasing the steam lap and the angle of advance of the eccentric but keeping constant the travel and lead of the valve, this method will
(A) Cause withdrawing or throttling of steam
(B) Reduce length of effective stroke of piston
(C) Reduce maximum opening of port to steam
(D) All of these
14. A higher pair has__________.
(A) Point contact
(B) Surface contact
(C) No contact
(D) None of the above
15. What is the number of instantaneous centers for an eight link mechanism?
(A) 15
(B) 28
(C) 30
(D) 8
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Miscellaneous
Chapter 14 Class 11 Probability
Serial order wise
### Transcript
Misc 10 The number lock of a suitcase has 4 wheels, each labelled with ten digits i.e., from 0 to 9. The lock opens with a sequence of four digits with no repeats. What is the probability of a person getting the right sequence to open the suitcase? There are 10 digits out of which 4 digits to be chosen with no repeats Hence , n(S) = Total 4 digit numbers out 10 digits = 10P4 = 10! 10 −4! = 10!6! = 10 × 9 × 8 × 7 × 6!6! = 5040 Let A be the event that correct sequence is selected There can be only 1 correct sequence Hence n(A) = 1 P(A) = n(A)n(S) = 𝟏𝟓𝟎𝟒𝟎
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# 5.4 Dividing polynomials (Page 2/6)
Page 2 / 6
Given a polynomial and a binomial, use long division to divide the polynomial by the binomial.
1. Set up the division problem.
2. Determine the first term of the quotient by dividing the leading term of the dividend by the leading term of the divisor.
3. Multiply the answer by the divisor and write it below the like terms of the dividend.
4. Subtract the bottom binomial from the top binomial.
5. Bring down the next term of the dividend.
6. Repeat steps 2–5 until reaching the last term of the dividend.
7. If the remainder is non-zero, express as a fraction using the divisor as the denominator.
## Using long division to divide a second-degree polynomial
Divide $\text{\hspace{0.17em}}5{x}^{2}+3x-2\text{\hspace{0.17em}}$ by $\text{\hspace{0.17em}}x+1.$
The quotient is $\text{\hspace{0.17em}}5x-2.\text{\hspace{0.17em}}$ The remainder is 0. We write the result as
$\frac{5{x}^{2}+3x-2}{x+1}=5x-2$
or
$5{x}^{2}+3x-2=\left(x+1\right)\left(5x-2\right)$
## Using long division to divide a third-degree polynomial
Divide $\text{\hspace{0.17em}}6{x}^{3}+11{x}^{2}-31x+15\text{\hspace{0.17em}}$ by $\text{\hspace{0.17em}}3x-2.\text{\hspace{0.17em}}$
There is a remainder of 1. We can express the result as:
$\frac{6{x}^{3}+11{x}^{2}-31x+15}{3x-2}=2{x}^{2}+5x-7+\frac{1}{3x-2}$
Divide $\text{\hspace{0.17em}}16{x}^{3}-12{x}^{2}+20x-3\text{\hspace{0.17em}}$ by $\text{\hspace{0.17em}}4x+5.\text{\hspace{0.17em}}$
$4{x}^{2}-8x+15-\frac{78}{4x+5}$
## Using synthetic division to divide polynomials
As we’ve seen, long division of polynomials can involve many steps and be quite cumbersome. Synthetic division is a shorthand method of dividing polynomials for the special case of dividing by a linear factor whose leading coefficient is 1.
To illustrate the process, recall the example at the beginning of the section.
Divide $\text{\hspace{0.17em}}2{x}^{3}-3{x}^{2}+4x+5\text{\hspace{0.17em}}$ by $\text{\hspace{0.17em}}x+2\text{\hspace{0.17em}}$ using the long division algorithm.
The final form of the process looked like this:
There is a lot of repetition in the table. If we don’t write the variables but, instead, line up their coefficients in columns under the division sign and also eliminate the partial products, we already have a simpler version of the entire problem.
Synthetic division carries this simplification even a few more steps. Collapse the table by moving each of the rows up to fill any vacant spots. Also, instead of dividing by 2, as we would in division of whole numbers, then multiplying and subtracting the middle product, we change the sign of the “divisor” to –2, multiply and add. The process starts by bringing down the leading coefficient.
We then multiply it by the “divisor” and add, repeating this process column by column, until there are no entries left. The bottom row represents the coefficients of the quotient; the last entry of the bottom row is the remainder. In this case, the quotient is $\text{\hspace{0.17em}}2{x}^{2}–7x+18\text{\hspace{0.17em}}$ and the remainder is $\text{\hspace{0.17em}}–31.\text{\hspace{0.17em}}$ The process will be made more clear in [link] .
## Synthetic division
Synthetic division is a shortcut that can be used when the divisor is a binomial in the form $\text{\hspace{0.17em}}x-k\text{\hspace{0.17em}}$ where $\text{\hspace{0.17em}}k\text{\hspace{0.17em}}$ is a real number. In synthetic division , only the coefficients are used in the division process.
Given two polynomials, use synthetic division to divide.
1. Write $\text{\hspace{0.17em}}k\text{\hspace{0.17em}}$ for the divisor.
2. Write the coefficients of the dividend.
3. Bring the lead coefficient down.
4. Multiply the lead coefficient by $\text{\hspace{0.17em}}k.\text{\hspace{0.17em}}$ Write the product in the next column.
5. Add the terms of the second column.
6. Multiply the result by $\text{\hspace{0.17em}}k.\text{\hspace{0.17em}}$ Write the product in the next column.
7. Repeat steps 5 and 6 for the remaining columns.
8. Use the bottom numbers to write the quotient. The number in the last column is the remainder and has degree 0, the next number from the right has degree 1, the next number from the right has degree 2, and so on.
the third and the seventh terms of a G.P are 81 and 16, find the first and fifth terms.
if a=3, b =4 and c=5 find the six trigonometric value sin
pls how do I factorize x⁴+x³-7x²-x+6=0
in a function the input value is called
how do I test for values on the number line
if a=4 b=4 then a+b=
a+b+2ab
Kin
commulative principle
a+b= 4+4=8
Mimi
If a=4 and b=4 then we add the value of a and b i.e a+b=4+4=8.
Tariq
what are examples of natural number
an equation for the line that goes through the point (-1,12) and has a slope of 2,3
3y=-9x+25
Ishaq
show that the set of natural numberdoes not from agroup with addition or multiplication butit forms aseni group with respect toaaddition as well as multiplication
x^20+x^15+x^10+x^5/x^2+1
evaluate each algebraic expression. 2x+×_2 if ×=5
if the ratio of the root of ax+bx+c =0, show that (m+1)^2 ac =b^2m
By the definition, is such that 0!=1.why?
(1+cosA+IsinA)(1+cosB+isinB)/(cos@+isin@)(cos$+isin$)
hatdog
Mark
jaks
Ryan
how we can draw three triangles of distinctly different shapes. All the angles will be cutt off each triangle and placed side by side with vertices touching
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Question
Please use the y = mx + b form, i have y = 3/2x but i don’t know the y intercept.
through: (4, 3), SLOPE = 3/2
please provide an explanation i am really struggling.
1. okay so the y intercept would be “b” in the y=mx + b formula and since u have slope already we know our “m” is 3/2
and luckily we have 2 points to help us find the y intercept first thing we’d do is plug these numbers into our formula and since we know we already have 2 points we can use “4” for our x and “3” for our y
so y=4 m=3/2 x=4 and b is our unknown
we have to solve 3=3/2(4)+b for b now
and we get b=-3 we can check this as well by plugging in -3 and seeing if we get the same awser which is “3”
2. thugiang
So you have the begining of the equation:
y = 3/2 x + b
To find the b, you substitute in x=4 and y=3 (from 4,3):
3 = 3/2 · 4 + b
Solving that…
3 = 6 + b
-3 = b
So your line is: y = 3/2 x – 3.
It’s also a good idea to double check, but substituting in just the x-value and seeing if it gives you the correct y-value:
y = 3/2 · 4 – 3
y = 6 – 3
y = 3
Yup, that checked.
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程序代写案例-ACS133
February 15, 2021留学咨询
ACS133 – Physical Systems Lecture 8: Electrical and Analogous Systems Simon Pope [email protected] Autumn semester – ACS133 1 Introduction to modelling and analysis of dynamic systems 2 Explore different types of physical systems 2.1 Mechanical Systems 2.2 Electrical Systems 2.3 Thermal Systems 2.4 Flow Systems 3 System simulation using Matlab and Simulink 4 Practical laboratory sessions Electrical Systems Electrical Systems Variables Element Laws Interconnection Laws Developing a system model Analogous Systems Case study: Lung Mechanics – Electrical Analogous System Electrical Systems 1. Electrical Systems 2. Analogous Systems 3. System Linearisation Electrical Systems 1. Variables 2. Element Laws 3. Interconnection Laws 4. Developing a system model Variables Electrical Systems I Electricity is created when electrons travel around a circuit I Each electron carries energy with it and has a negative charge Charge Electric charge (denoted by Q) is the physical property of matter that causes it to experience a force when placed in an electromagnetic field. I The amount of electrical charge that moves in a circuit depends on the current flow and how long it flows for I Electrical charge is measured in coulomb (C) Variables Electrical Systems Current Current (denoted by i) is the flow of charge. I Because charge is measured in C, its flow is in coulombs/second which are also called Amperes (A) Voltage Voltage (denoted by v ) can be thought of as the “force” that pushes the flow of charge. I For the operation of circuit elements it is the voltage difference across those elements that matters I The standard unit is the Volt (V) Element Laws Electrical Systems 1. Resistor Definition An ideal resistor (denoted by R) is a passive two-terminal electrical component that implements electrical resistance as a circuit element. I The standard unit is the Ohm (Ω) Ohm’s law The time domain expression relating voltage and current for the resistor is given by Ohm’s law: vR(t) = iR(t)R (1) Element Laws Electrical Systems 2. Capacitor Definition A capacitor (denoted by C) is a passive two-terminal electrical component that stores electrical energy in an electric field. I The standard unit is the farad (F) I The time domain expression relating voltage and current for the Capacitor is given as: vC(t) = 1 C ∫ iC(t)dt (2) Element Laws Electrical Systems 3. Inductor Definition An inductor (denoted by L) is a passive two-terminal electrical component that stores energy in a magnetic field when electric current flows through it. I The standard unit is the henry (H) I The time domain expression relating voltage and current for the inductor is given as vL(t) = L diL(t) dt (3) Element Laws Electrical Systems Element Laws Electrical Systems 4. Voltage sources Definition A voltage source (denoted by ei ) is a two-terminal device which can maintain a fixed voltage. I An ideal voltage source can maintain the fixed voltage independent of the load resistance or the output current 5. Ground Definition Ground is the reference point in an electrical circuit from which voltages are measured. Interconnection Laws Electrical Systems Kirchhoff’s current law The sum of current flowing into a junction of conductors is zero. ∑ j ij = 0 (4) Kirchhoff’s voltage law The directed sum of voltages around a loop is zero. ∑ j vj = 0 (5) Developing a system model Electrical Systems 1. Express currents/voltages using the element laws 2. Apply Kirchhoff’s laws 3. Derive ODEs Electrical Systems Example: RC circuit Consider a two-port electric circuit as shown in the figure. The input voltage is denoted by vi(t) and the output voltage is denoted by vo(t). Find the transfer function Vo(s)Vi (s) of the circuit. Electrical Systems Example: RC circuit The voltage across the resistor is given by vR(t) = iR(t)R. The voltage across the capacitor is given by vC(t) = 1C ∫ iC(t)dt . Note that the current through the resistor and through the capacitor is the same: iC(t) = iR(t) = i(t). Applying Kirchhoff’s voltage law for the left loop gives vi(t) = vR(t) + vC(t) = i(t)R + 1 C ∫ i(t)dt Applying Kirchhoff’s voltage law for the right loop gives vo(t) = vC(t) = 1 C ∫ i(t)dt Electrical Systems Example: RC circuit vi(t) = i(t)R + 1 C ∫ i(t)dt vo(t) = 1 C ∫ i(t)dt Using the Laplace transform assuming zero initial conditions gives Vi(s) = I(s)R + 1 Cs I(s) Vo(s) = 1 Cs I(s) Substituting I(s) from the second equation into the first one gives Vo(s) Vi(s) = 1 1 + RCs Electrical Systems 1. Electrical Systems 2. Analogous Systems 3. System Linearisation Analogous Systems I We can relate the behaviour of our system’s parameters to an equivalent (analogous) known system that is easier to understand and analyse. I Examples: I The Phillips Hydraulic Computer MONIAC used the flow of water to model economic systems (https: //www.youtube.com/watch?v=rAZavOcEnLg) I Electronic circuits can be used to represent both physiological and ecological systems I A mechanical device can be used to represent mathematical calculations Analogous Systems I We can relate the behaviour of our system’s parameters to an equivalent (analogous) known system that is easier to understand and analyse. I Examples: I The Phillips Hydraulic Computer MONIAC used the flow of water to model economic systems (https: //www.youtube.com/watch?v=rAZavOcEnLg) I Electronic circuits can be used to represent both physiological and ecological systems I A mechanical device can be used to represent mathematical calculations Analogous Systems Lung Mechanics – Electrical Analogous System Analogous Systems Lung Mechanics – Electrical Analogous System Analogous Systems Lung Mechanics – Electrical Analogous System Lung Mechanics – Electrical Analogous System Mathematical model I We can now apply Kirchhoff’s laws to derive the system’s dynamic model I However, the real behaviour of the lung volume/pressure is not linear! I In the analogue electric circuit this corresponds to a Non-linear resistor I dVolume dPressure ≈ dCurrentdVoltage I A good approximation for low-medium lung volume is io = 1 7 e3o Lung Mechanics – Electrical Analogous System Mathematical model I Applying Kirchhoff’s laws gives Lung Mechanics – Electrical Analogous System Mathematical model I Applying Kirchhoff’s laws gives 1 2 e˙o + ( eo − ei(t) ) + 1 7 e3o = 0 I This is equivalent to 1 2 e˙o + 1 7 e3o + eo = ei(t) Lung Mechanics – Electrical Analogous System Mathematical model I Applying Kirchhoff’s laws gives 1 2 e˙o + ( eo − ei(t) ) + 1 7 e3o = 0 I This is equivalent to 1 2 e˙o + 1 7 e3o + eo = ei(t) Non-linear! Lecture 8: Take-home points I Electrical systems I Variables I Elements laws (ideal resistor, capacitor, inductor, voltage source) I Connecting laws – Kirchoff’s current and voltage law I Modelling steps I Analogous systems 欢迎咨询51作业君
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# Equations that Describe Patterns
## Describe numerical sequences by finding a rule.
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When the Price Isn't Right
Credit: Rob DiCaterino
Source: http://www.flickr.com/photos/goodrob13/3454140781/
You've been waiting for months. The latest version of your favorite video game comes out today. You've mowed lawns, done extra chores, and saved up gifts from your grandparents. The new game costs $49.99. You have exactly$50.00 saved. So you're good to go, right? Unless you're lucky enough to live in a state without sales tax, you're going to have to wait for that new game.
#### Figuring Out the Tax
State and local governments all charge different amounts of sales tax when you make purchases. To figure out how much your video game really costs, you'll need to know the tax rate for your area. You can write an equation to help you figure out your total cost. If the price of the game is 49.99, you'll have to pay 49.99+(49.99taxrate)\begin{align*}49.99+\left ( 49.99* \mathrm{tax}\:\mathrm{rate}\right )\end{align*}. So, if you live in a state where sales tax is 10%, you'll need to pay the store54.99 for the game. States usually round sales tax up to the nearest penny.
Credit: Eddie Welker
Source: http://www.flickr.com/photos/ed_welker/4076939458/
States use revenue from sales tax to fund programs such as schools and social services. Most states do not tax groceries, since food is considered a basic necessity; however, restaurant and fast food meals are usually taxed. Items like cigarettes and alcoholic beverages are taxed at higher-than-normal rates. States hope these higher taxes will keep people from smoking and drinking. Some states have no sales tax and use this fact to attract shoppers from other states.
#### Explore More
With the links below, learn about statewide sales tax holidays, potential taxes on online shopping, and the uncertain future of Minnesota's lack of sales tax on clothing.
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
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edHelper subscribers - Create a new printable
Middle School Math
Math
Name _____________________________ Date ___________________
Algebra
Complete.
1 80,000 and 1,100,000 added to a number is 1,619,971. What is the number?
2 Three-fourths of a number equals 999. What is the number?
3 The sum of twenty-seven and forty-seven is twenty more than a number. What is the number?
4 Sixty-six more than a number is three. What is the number?
5 The difference between 330 and half of a number is 114. What is the number?
6 Ten times a number, increased by 5.24, equals 65.24. What is the number?
7 Two hundred nine less than a number is negative one hundred twenty-eight. What is the number?
8 8,252 and 5,306 added to the difference between 565 and half of a number is 13,913. What is the number?
9 The sum of twenty and negative twenty-nine is negative forty-five more than a number. What is the number?
10 Sixty-eight more than a number is ninety-five. What is the number?
11 One-twelfth of a number, increased by 94 is 108. What is the number?
12 * This is a pre-made sheet.Use the link at the top of the page for a printable page.
13 Twelve exceeds one-half of a number by 8. What is the number?
14 If eight is added seventeen times to a number, the result is 188. What is the number?
15 664 exceeds seven times a number by 55. What is the number?
16 A number multiplied by -6 is 62.4. What is the number?
17 Negative nine times a number is negative one hundred ninety-eight. What is the number?
18 A number minus 19.04 is 7.51. What is the number?
19 Fifteen more than 11 times a number is 70. What is the number?
20 Eleven times a number, decreased by forty, equals fifty-nine. What is the number?
21 * This is a pre-made sheet.Use the link at the top of the page for a printable page.
22 One hundred forty less than a number is negative forty-three. What is the number?
23 Thirty-two less than a number is negative ninety-nine. What is the number?
24 If a number is decreased by 40, the result is 11. What is the number?
25 140 exceeds two times a number by 88. What is the number?
26 Five less than 4 times a number is 31. What is the number?
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# Monte Carlo method
Monte Carlo methods, or Monte Carlo experiments, are a broad class of computational algorithms that rely on repeated random sampling to obtain numerical results. The underlying concept is to use randomness to solve problems that might be deterministic in principle. The name comes from the Monte Carlo Casino in Monaco, where the primary developer of the method, physicist Stanislaw Ulam, was inspired by his uncle's gambling habits.
Monte Carlo methods are mainly used in three distinct problem classes: optimization, numerical integration, and generating draws from a probability distribution. They can also be used to model phenomena with significant uncertainty in inputs, such as calculating the risk of a nuclear power plant failure. Monte Carlo methods are often implemented using computer simulations, and they can provide approximate solutions to problems that are otherwise intractable or too complex to analyze mathematically.
Monte Carlo methods are widely used in various fields of science, engineering, and mathematics, such as physics, chemistry, biology, statistics, artificial intelligence, finance, and cryptography. They have also been applied to social sciences, such as sociology, psychology, and political science. Monte Carlo methods have been recognized as one of the most important and influential ideas of the 20th century, and they have enabled many scientific and technological breakthroughs.
Monte Carlo methods also have some limitations and challenges, such as the trade-off between accuracy and computational cost, the curse of dimensionality, the reliability of random number generators, and the verification and validation of the results.
## Overview
Monte Carlo methods vary, but tend to follow a particular pattern:
1. Define a domain of possible inputs
2. Generate inputs randomly from a probability distribution over the domain
3. Perform a deterministic computation of the outputs
4. Aggregate the results
For example, consider a quadrant (circular sector) inscribed in a unit square. Given that the ratio of their areas is π/4, the value of π can be approximated using a Monte Carlo method:[1]
1. Draw a square, then inscribe a quadrant within it
2. Uniformly scatter a given number of points over the square
3. Count the number of points inside the quadrant, i.e. having a distance from the origin of less than 1
4. The ratio of the inside-count and the total-sample-count is an estimate of the ratio of the two areas, π/4. Multiply the result by 4 to estimate π.
In this procedure the domain of inputs is the square that circumscribes the quadrant. One can generate random inputs by scattering grains over the square then perform a computation on each input (test whether it falls within the quadrant). Aggregating the results yields our final result, the approximation of π.
There are two important considerations:
1. If the points are not uniformly distributed, then the approximation will be poor.
2. The approximation is generally poor if only a few points are randomly placed in the whole square. On average, the approximation improves as more points are placed.
Uses of Monte Carlo methods require large amounts of random numbers, and their use benefitted greatly from pseudorandom number generators, which are far quicker to use than the tables of random numbers that had been previously used for statistical sampling.
## Application
Monte Carlo methods are often used in physical and mathematical problems and are most useful when it is difficult or impossible to use other approaches. Monte Carlo methods are mainly used in three problem classes:[2] optimization, numerical integration, and generating draws from a probability distribution.
In physics-related problems, Monte Carlo methods are useful for simulating systems with many coupled degrees of freedom, such as fluids, disordered materials, strongly coupled solids, and cellular structures (see cellular Potts model, interacting particle systems, McKean–Vlasov processes, kinetic models of gases).
Other examples include modeling phenomena with significant uncertainty in inputs such as the calculation of risk in business and, in mathematics, evaluation of multidimensional definite integrals with complicated boundary conditions. In application to systems engineering problems (space, oil exploration, aircraft design, etc.), Monte Carlo–based predictions of failure, cost overruns and schedule overruns are routinely better than human intuition or alternative "soft" methods.[3]
In principle, Monte Carlo methods can be used to solve any problem having a probabilistic interpretation. By the law of large numbers, integrals described by the expected value of some random variable can be approximated by taking the empirical mean (a.k.a. the 'sample mean') of independent samples of the variable. When the probability distribution of the variable is parameterized, mathematicians often use a Markov chain Monte Carlo (MCMC) sampler.[4][5][6] The central idea is to design a judicious Markov chain model with a prescribed stationary probability distribution. That is, in the limit, the samples being generated by the MCMC method will be samples from the desired (target) distribution.[7][8] By the ergodic theorem, the stationary distribution is approximated by the empirical measures of the random states of the MCMC sampler.
In other problems, the objective is generating draws from a sequence of probability distributions satisfying a nonlinear evolution equation. These flows of probability distributions can always be interpreted as the distributions of the random states of a Markov process whose transition probabilities depend on the distributions of the current random states (see McKean–Vlasov processes, nonlinear filtering equation).[9][10] In other instances we are given a flow of probability distributions with an increasing level of sampling complexity (path spaces models with an increasing time horizon, Boltzmann–Gibbs measures associated with decreasing temperature parameters, and many others). These models can also be seen as the evolution of the law of the random states of a nonlinear Markov chain.[10][11] A natural way to simulate these sophisticated nonlinear Markov processes is to sample multiple copies of the process, replacing in the evolution equation the unknown distributions of the random states by the sampled empirical measures. In contrast with traditional Monte Carlo and MCMC methodologies, these mean-field particle techniques rely on sequential interacting samples. The terminology mean field reflects the fact that each of the samples (a.k.a. particles, individuals, walkers, agents, creatures, or phenotypes) interacts with the empirical measures of the process. When the size of the system tends to infinity, these random empirical measures converge to the deterministic distribution of the random states of the nonlinear Markov chain, so that the statistical interaction between particles vanishes.
## Computational costs
Despite its conceptual and algorithmic simplicity, the computational cost associated with a Monte Carlo simulation can be staggeringly high. In general the method requires many samples to get a good approximation, which may incur an arbitrarily large total runtime if the processing time of a single sample is high.[12] Although this is a severe limitation in very complex problems, the embarrassingly parallel nature of the algorithm allows this large cost to be reduced (perhaps to a feasible level) through parallel computing strategies in local processors, clusters, cloud computing, GPU, FPGA, etc.[13][14][15][16]
## History
Before the Monte Carlo method was developed, simulations tested a previously understood deterministic problem, and statistical sampling was used to estimate uncertainties in the simulations. Monte Carlo simulations invert this approach, solving deterministic problems using probabilistic metaheuristics (see simulated annealing).
An early variant of the Monte Carlo method was devised to solve the Buffon's needle problem, in which π can be estimated by dropping needles on a floor made of parallel equidistant strips. In the 1930s, Enrico Fermi first experimented with the Monte Carlo method while studying neutron diffusion, but he did not publish this work.[17]
In the late 1940s, Stanislaw Ulam invented the modern version of the Markov Chain Monte Carlo method while he was working on nuclear weapons projects at the Los Alamos National Laboratory. In 1946, nuclear weapons physicists at Los Alamos were investigating neutron diffusion in the core of a nuclear weapon.[17] Despite having most of the necessary data, such as the average distance a neutron would travel in a substance before it collided with an atomic nucleus and how much energy the neutron was likely to give off following a collision, the Los Alamos physicists were unable to solve the problem using conventional, deterministic mathematical methods. Ulam proposed using random experiments. He recounts his inspiration as follows:
The first thoughts and attempts I made to practice [the Monte Carlo Method] were suggested by a question which occurred to me in 1946 as I was convalescing from an illness and playing solitaires. The question was what are the chances that a Canfield solitaire laid out with 52 cards will come out successfully? After spending a lot of time trying to estimate them by pure combinatorial calculations, I wondered whether a more practical method than "abstract thinking" might not be to lay it out say one hundred times and simply observe and count the number of successful plays. This was already possible to envisage with the beginning of the new era of fast computers, and I immediately thought of problems of neutron diffusion and other questions of mathematical physics, and more generally how to change processes described by certain differential equations into an equivalent form interpretable as a succession of random operations. Later [in 1946], I described the idea to John von Neumann, and we began to plan actual calculations.[18]
Being secret, the work of von Neumann and Ulam required a code name.[19] A colleague of von Neumann and Ulam, Nicholas Metropolis, suggested using the name Monte Carlo, which refers to the Monte Carlo Casino in Monaco where Ulam's uncle would borrow money from relatives to gamble.[17] Monte Carlo methods were central to the simulations required for the Manhattan Project, though severely limited by the computational tools at the time. Von Neumann, Nicholas Metropolis and others programmed the ENIAC computer to perform the first fully automated Monte Carlo calculations, of a fission weapon core, in the spring of 1948.[20] In the 1950s Monte Carlo methods were used at Los Alamos for the development of the hydrogen bomb, and became popularized in the fields of physics, physical chemistry, and operations research. The Rand Corporation and the U.S. Air Force were two of the major organizations responsible for funding and disseminating information on Monte Carlo methods during this time, and they began to find a wide application in many different fields.
The theory of more sophisticated mean-field type particle Monte Carlo methods had certainly started by the mid-1960s, with the work of Henry P. McKean Jr. on Markov interpretations of a class of nonlinear parabolic partial differential equations arising in fluid mechanics.[21][22] We also quote an earlier pioneering article by Theodore E. Harris and Herman Kahn, published in 1951, using mean-field genetic-type Monte Carlo methods for estimating particle transmission energies.[23] Mean-field genetic type Monte Carlo methodologies are also used as heuristic natural search algorithms (a.k.a. metaheuristic) in evolutionary computing. The origins of these mean-field computational techniques can be traced to 1950 and 1954 with the work of Alan Turing on genetic type mutation-selection learning machines[24] and the articles by Nils Aall Barricelli at the Institute for Advanced Study in Princeton, New Jersey.[25][26]
Quantum Monte Carlo, and more specifically diffusion Monte Carlo methods can also be interpreted as a mean-field particle Monte Carlo approximation of FeynmanKac path integrals.[27][28][29][30][31][32][33] The origins of Quantum Monte Carlo methods are often attributed to Enrico Fermi and Robert Richtmyer who developed in 1948 a mean-field particle interpretation of neutron-chain reactions,[34] but the first heuristic-like and genetic type particle algorithm (a.k.a. Resampled or Reconfiguration Monte Carlo methods) for estimating ground state energies of quantum systems (in reduced matrix models) is due to Jack H. Hetherington in 1984.[33] In molecular chemistry, the use of genetic heuristic-like particle methodologies (a.k.a. pruning and enrichment strategies) can be traced back to 1955 with the seminal work of Marshall N. Rosenbluth and Arianna W. Rosenbluth.[35]
The use of Sequential Monte Carlo in advanced signal processing and Bayesian inference is more recent. It was in 1993, that Gordon et al., published in their seminal work[36] the first application of a Monte Carlo resampling algorithm in Bayesian statistical inference. The authors named their algorithm 'the bootstrap filter', and demonstrated that compared to other filtering methods, their bootstrap algorithm does not require any assumption about that state-space or the noise of the system. We also quote another pioneering article in this field of Genshiro Kitagawa on a related "Monte Carlo filter",[37] and the ones by Pierre Del Moral[38] and Himilcon Carvalho, Pierre Del Moral, André Monin and Gérard Salut[39] on particle filters published in the mid-1990s. Particle filters were also developed in signal processing in 1989–1992 by P. Del Moral, J. C. Noyer, G. Rigal, and G. Salut in the LAAS-CNRS in a series of restricted and classified research reports with STCAN (Service Technique des Constructions et Armes Navales), the IT company DIGILOG, and the LAAS-CNRS (the Laboratory for Analysis and Architecture of Systems) on radar/sonar and GPS signal processing problems.[40][41][42][43][44][45] These Sequential Monte Carlo methodologies can be interpreted as an acceptance-rejection sampler equipped with an interacting recycling mechanism.
From 1950 to 1996, all the publications on Sequential Monte Carlo methodologies, including the pruning and resample Monte Carlo methods introduced in computational physics and molecular chemistry, present natural and heuristic-like algorithms applied to different situations without a single proof of their consistency, nor a discussion on the bias of the estimates and on genealogical and ancestral tree based algorithms. The mathematical foundations and the first rigorous analysis of these particle algorithms were written by Pierre Del Moral in 1996.[38][46]
Branching type particle methodologies with varying population sizes were also developed in the end of the 1990s by Dan Crisan, Jessica Gaines and Terry Lyons,[47][48][49] and by Dan Crisan, Pierre Del Moral and Terry Lyons.[50] Further developments in this field were described in 1999 to 2001 by P. Del Moral, A. Guionnet and L. Miclo.[28][51][52]
## Definitions
There is no consensus on how Monte Carlo should be defined. For example, Ripley[53] defines most probabilistic modeling as stochastic simulation, with Monte Carlo being reserved for Monte Carlo integration and Monte Carlo statistical tests. Sawilowsky[54] distinguishes between a simulation, a Monte Carlo method, and a Monte Carlo simulation: a simulation is a fictitious representation of reality, a Monte Carlo method is a technique that can be used to solve a mathematical or statistical problem, and a Monte Carlo simulation uses repeated sampling to obtain the statistical properties of some phenomenon (or behavior).
Here are the examples:
• Simulation: Drawing one pseudo-random uniform variable from the interval [0,1] can be used to simulate the tossing of a coin: If the value is less than or equal to 0.50 designate the outcome as heads, but if the value is greater than 0.50 designate the outcome as tails. This is a simulation, but not a Monte Carlo simulation.
• Monte Carlo method: Pouring out a box of coins on a table, and then computing the ratio of coins that land heads versus tails is a Monte Carlo method of determining the behavior of repeated coin tosses, but it is not a simulation.
• Monte Carlo simulation: Drawing a large number of pseudo-random uniform variables from the interval [0,1] at one time, or once at many different times, and assigning values less than or equal to 0.50 as heads and greater than 0.50 as tails, is a Monte Carlo simulation of the behavior of repeatedly tossing a coin.
Kalos and Whitlock[55] point out that such distinctions are not always easy to maintain. For example, the emission of radiation from atoms is a natural stochastic process. It can be simulated directly, or its average behavior can be described by stochastic equations that can themselves be solved using Monte Carlo methods. "Indeed, the same computer code can be viewed simultaneously as a 'natural simulation' or as a solution of the equations by natural sampling."
Convergence of the Monte Carlo simulation can be checked with the Gelman-Rubin statistic.
### Monte Carlo and random numbers
The main idea behind this method is that the results are computed based on repeated random sampling and statistical analysis. The Monte Carlo simulation is, in fact, random experimentations, in the case that, the results of these experiments are not well known. Monte Carlo simulations are typically characterized by many unknown parameters, many of which are difficult to obtain experimentally.[56] Monte Carlo simulation methods do not always require truly random numbers to be useful (although, for some applications such as primality testing, unpredictability is vital).[57] Many of the most useful techniques use deterministic, pseudorandom sequences, making it easy to test and re-run simulations. The only quality usually necessary to make good simulations is for the pseudo-random sequence to appear "random enough" in a certain sense.
What this means depends on the application, but typically they should pass a series of statistical tests. Testing that the numbers are uniformly distributed or follow another desired distribution when a large enough number of elements of the sequence are considered is one of the simplest and most common ones. Weak correlations between successive samples are also often desirable/necessary.
Sawilowsky lists the characteristics of a high-quality Monte Carlo simulation:[54]
• the (pseudo-random) number generator has certain characteristics (e.g. a long "period" before the sequence repeats)
• the (pseudo-random) number generator produces values that pass tests for randomness
• there are enough samples to ensure accurate results
• the proper sampling technique is used
• the algorithm used is valid for what is being modeled
• it simulates the phenomenon in question.
Pseudo-random number sampling algorithms are used to transform uniformly distributed pseudo-random numbers into numbers that are distributed according to a given probability distribution.
Low-discrepancy sequences are often used instead of random sampling from a space as they ensure even coverage and normally have a faster order of convergence than Monte Carlo simulations using random or pseudorandom sequences. Methods based on their use are called quasi-Monte Carlo methods.
In an effort to assess the impact of random number quality on Monte Carlo simulation outcomes, astrophysical researchers tested cryptographically secure pseudorandom numbers generated via Intel's RDRAND instruction set, as compared to those derived from algorithms, like the Mersenne Twister, in Monte Carlo simulations of radio flares from brown dwarfs. RDRAND is the closest pseudorandom number generator to a true random number generator.[citation needed] No statistically significant difference was found between models generated with typical pseudorandom number generators and RDRAND for trials consisting of the generation of 107 random numbers.[58]
### Monte Carlo simulation versus "what if" scenarios
There are ways of using probabilities that are definitely not Monte Carlo simulations – for example, deterministic modeling using single-point estimates. Each uncertain variable within a model is assigned a "best guess" estimate. Scenarios (such as best, worst, or most likely case) for each input variable are chosen and the results recorded.[59]
By contrast, Monte Carlo simulations sample from a probability distribution for each variable to produce hundreds or thousands of possible outcomes. The results are analyzed to get probabilities of different outcomes occurring.[60] For example, a comparison of a spreadsheet cost construction model run using traditional "what if" scenarios, and then running the comparison again with Monte Carlo simulation and triangular probability distributions shows that the Monte Carlo analysis has a narrower range than the "what if" analysis.[example needed] This is because the "what if" analysis gives equal weight to all scenarios (see quantifying uncertainty in corporate finance), while the Monte Carlo method hardly samples in the very low probability regions. The samples in such regions are called "rare events".
## Applications
Monte Carlo methods are especially useful for simulating phenomena with significant uncertainty in inputs and systems with many coupled degrees of freedom. Areas of application include:
### Physical sciences
Monte Carlo methods are very important in computational physics, physical chemistry, and related applied fields, and have diverse applications from complicated quantum chromodynamics calculations to designing heat shields and aerodynamic forms as well as in modeling radiation transport for radiation dosimetry calculations.[61][62][63] In statistical physics, Monte Carlo molecular modeling is an alternative to computational molecular dynamics, and Monte Carlo methods are used to compute statistical field theories of simple particle and polymer systems.[35][64] Quantum Monte Carlo methods solve the many-body problem for quantum systems.[9][10][27] In radiation materials science, the binary collision approximation for simulating ion implantation is usually based on a Monte Carlo approach to select the next colliding atom.[65] In experimental particle physics, Monte Carlo methods are used for designing detectors, understanding their behavior and comparing experimental data to theory. In astrophysics, they are used in such diverse manners as to model both galaxy evolution[66] and microwave radiation transmission through a rough planetary surface.[67] Monte Carlo methods are also used in the ensemble models that form the basis of modern weather forecasting.
### Engineering
Monte Carlo methods are widely used in engineering for sensitivity analysis and quantitative probabilistic analysis in process design. The need arises from the interactive, co-linear and non-linear behavior of typical process simulations. For example,
### Climate change and radiative forcing
The Intergovernmental Panel on Climate Change relies on Monte Carlo methods in probability density function analysis of radiative forcing.[71]
### Computational biology
Monte Carlo methods are used in various fields of computational biology, for example for Bayesian inference in phylogeny, or for studying biological systems such as genomes, proteins,[72] or membranes.[73] The systems can be studied in the coarse-grained or ab initio frameworks depending on the desired accuracy. Computer simulations allow us to monitor the local environment of a particular molecule to see if some chemical reaction is happening for instance. In cases where it is not feasible to conduct a physical experiment, thought experiments can be conducted (for instance: breaking bonds, introducing impurities at specific sites, changing the local/global structure, or introducing external fields).
### Computer graphics
Path tracing, occasionally referred to as Monte Carlo ray tracing, renders a 3D scene by randomly tracing samples of possible light paths. Repeated sampling of any given pixel will eventually cause the average of the samples to converge on the correct solution of the rendering equation, making it one of the most physically accurate 3D graphics rendering methods in existence.
### Applied statistics
The standards for Monte Carlo experiments in statistics were set by Sawilowsky.[74] In applied statistics, Monte Carlo methods may be used for at least four purposes:
1. To compare competing statistics for small samples under realistic data conditions. Although type I error and power properties of statistics can be calculated for data drawn from classical theoretical distributions (e.g., normal curve, Cauchy distribution) for asymptotic conditions (i. e, infinite sample size and infinitesimally small treatment effect), real data often do not have such distributions.[75]
2. To provide implementations of hypothesis tests that are more efficient than exact tests such as permutation tests (which are often impossible to compute) while being more accurate than critical values for asymptotic distributions.
3. To provide a random sample from the posterior distribution in Bayesian inference. This sample then approximates and summarizes all the essential features of the posterior.
4. To provide efficient random estimates of the Hessian matrix of the negative log-likelihood function that may be averaged to form an estimate of the Fisher information matrix.[76][77]
Monte Carlo methods are also a compromise between approximate randomization and permutation tests. An approximate randomization test is based on a specified subset of all permutations (which entails potentially enormous housekeeping of which permutations have been considered). The Monte Carlo approach is based on a specified number of randomly drawn permutations (exchanging a minor loss in precision if a permutation is drawn twice—or more frequently—for the efficiency of not having to track which permutations have already been selected).
### Artificial intelligence for games
Monte Carlo methods have been developed into a technique called Monte-Carlo tree search that is useful for searching for the best move in a game. Possible moves are organized in a search tree and many random simulations are used to estimate the long-term potential of each move. A black box simulator represents the opponent's moves.[78]
The Monte Carlo tree search (MCTS) method has four steps:[79]
1. Starting at root node of the tree, select optimal child nodes until a leaf node is reached.
2. Expand the leaf node and choose one of its children.
3. Play a simulated game starting with that node.
4. Use the results of that simulated game to update the node and its ancestors.
The net effect, over the course of many simulated games, is that the value of a node representing a move will go up or down, hopefully corresponding to whether or not that node represents a good move.
Monte Carlo Tree Search has been used successfully to play games such as Go,[80] Tantrix,[81] Battleship,[82] Havannah,[83] and Arimaa.[84]
### Design and visuals
Monte Carlo methods are also efficient in solving coupled integral differential equations of radiation fields and energy transport, and thus these methods have been used in global illumination computations that produce photo-realistic images of virtual 3D models, with applications in video games, architecture, design, computer generated films, and cinematic special effects.[85]
### Search and rescue
The US Coast Guard utilizes Monte Carlo methods within its computer modeling software SAROPS in order to calculate the probable locations of vessels during search and rescue operations. Each simulation can generate as many as ten thousand data points that are randomly distributed based upon provided variables.[86] Search patterns are then generated based upon extrapolations of these data in order to optimize the probability of containment (POC) and the probability of detection (POD), which together will equal an overall probability of success (POS). Ultimately this serves as a practical application of probability distribution in order to provide the swiftest and most expedient method of rescue, saving both lives and resources.[87]
Monte Carlo simulation is commonly used to evaluate the risk and uncertainty that would affect the outcome of different decision options. Monte Carlo simulation allows the business risk analyst to incorporate the total effects of uncertainty in variables like sales volume, commodity and labor prices, interest and exchange rates, as well as the effect of distinct risk events like the cancellation of a contract or the change of a tax law.
Monte Carlo methods in finance are often used to evaluate investments in projects at a business unit or corporate level, or other financial valuations. They can be used to model project schedules, where simulations aggregate estimates for worst-case, best-case, and most likely durations for each task to determine outcomes for the overall project.[1] Monte Carlo methods are also used in option pricing, default risk analysis.[88][89] Additionally, they can be used to estimate the financial impact of medical interventions.[90]
### Law
A Monte Carlo approach was used for evaluating the potential value of a proposed program to help female petitioners in Wisconsin be successful in their applications for harassment and domestic abuse restraining orders. It was proposed to help women succeed in their petitions by providing them with greater advocacy thereby potentially reducing the risk of rape and physical assault. However, there were many variables in play that could not be estimated perfectly, including the effectiveness of restraining orders, the success rate of petitioners both with and without advocacy, and many others. The study ran trials that varied these variables to come up with an overall estimate of the success level of the proposed program as a whole.[91]
### Library science
Monte Carlo approach had also been used to simulate the number of book publications based on book genre in Malaysia. The Monte Carlo simulation utilized previous published National Book publication data and book's price according to book genre in the local market. The Monte Carlo results were used to determine what kind of book genre that Malaysians are fond of and was used to compare book publications between Malaysia and Japan.[92]
### Other
Nassim Nicholas Taleb writes about Monte Carlo generators in his 2001 book Fooled by Randomness as a real instance of the reverse Turing test: a human can be declared unintelligent if their writing cannot be told apart from a generated one.
## Use in mathematics
In general, the Monte Carlo methods are used in mathematics to solve various problems by generating suitable random numbers (see also Random number generation) and observing that fraction of the numbers that obeys some property or properties. The method is useful for obtaining numerical solutions to problems too complicated to solve analytically. The most common application of the Monte Carlo method is Monte Carlo integration.
### Integration
Deterministic numerical integration algorithms work well in a small number of dimensions, but encounter two problems when the functions have many variables. First, the number of function evaluations needed increases rapidly with the number of dimensions. For example, if 10 evaluations provide adequate accuracy in one dimension, then 10100 points are needed for 100 dimensions—far too many to be computed. This is called the curse of dimensionality. Second, the boundary of a multidimensional region may be very complicated, so it may not be feasible to reduce the problem to an iterated integral.[93] 100 dimensions is by no means unusual, since in many physical problems, a "dimension" is equivalent to a degree of freedom.
Monte Carlo methods provide a way out of this exponential increase in computation time. As long as the function in question is reasonably well-behaved, it can be estimated by randomly selecting points in 100-dimensional space, and taking some kind of average of the function values at these points. By the central limit theorem, this method displays ${\displaystyle \scriptstyle 1/{\sqrt {N}}}$ convergence—i.e., quadrupling the number of sampled points halves the error, regardless of the number of dimensions.[93]
A refinement of this method, known as importance sampling in statistics, involves sampling the points randomly, but more frequently where the integrand is large. To do this precisely one would have to already know the integral, but one can approximate the integral by an integral of a similar function or use adaptive routines such as stratified sampling, recursive stratified sampling, adaptive umbrella sampling[94][95] or the VEGAS algorithm.
A similar approach, the quasi-Monte Carlo method, uses low-discrepancy sequences. These sequences "fill" the area better and sample the most important points more frequently, so quasi-Monte Carlo methods can often converge on the integral more quickly.
Another class of methods for sampling points in a volume is to simulate random walks over it (Markov chain Monte Carlo). Such methods include the Metropolis–Hastings algorithm, Gibbs sampling, Wang and Landau algorithm, and interacting type MCMC methodologies such as the sequential Monte Carlo samplers.[96]
### Simulation and optimization
Another powerful and very popular application for random numbers in numerical simulation is in numerical optimization. The problem is to minimize (or maximize) functions of some vector that often has many dimensions. Many problems can be phrased in this way: for example, a computer chess program could be seen as trying to find the set of, say, 10 moves that produces the best evaluation function at the end. In the traveling salesman problem the goal is to minimize distance traveled. There are also applications to engineering design, such as multidisciplinary design optimization. It has been applied with quasi-one-dimensional models to solve particle dynamics problems by efficiently exploring large configuration space. Reference[97] is a comprehensive review of many issues related to simulation and optimization.
The traveling salesman problem is what is called a conventional optimization problem. That is, all the facts (distances between each destination point) needed to determine the optimal path to follow are known with certainty and the goal is to run through the possible travel choices to come up with the one with the lowest total distance. However, let's assume that instead of wanting to minimize the total distance traveled to visit each desired destination, we wanted to minimize the total time needed to reach each destination. This goes beyond conventional optimization since travel time is inherently uncertain (traffic jams, time of day, etc.). As a result, to determine our optimal path we would want to use simulation – optimization to first understand the range of potential times it could take to go from one point to another (represented by a probability distribution in this case rather than a specific distance) and then optimize our travel decisions to identify the best path to follow taking that uncertainty into account.
### Inverse problems
Probabilistic formulation of inverse problems leads to the definition of a probability distribution in the model space. This probability distribution combines prior information with new information obtained by measuring some observable parameters (data). As, in the general case, the theory linking data with model parameters is nonlinear, the posterior probability in the model space may not be easy to describe (it may be multimodal, some moments may not be defined, etc.).
When analyzing an inverse problem, obtaining a maximum likelihood model is usually not sufficient, as we normally also wish to have information on the resolution power of the data. In the general case we may have many model parameters, and an inspection of the marginal probability densities of interest may be impractical, or even useless. But it is possible to pseudorandomly generate a large collection of models according to the posterior probability distribution and to analyze and display the models in such a way that information on the relative likelihoods of model properties is conveyed to the spectator. This can be accomplished by means of an efficient Monte Carlo method, even in cases where no explicit formula for the a priori distribution is available.
The best-known importance sampling method, the Metropolis algorithm, can be generalized, and this gives a method that allows analysis of (possibly highly nonlinear) inverse problems with complex a priori information and data with an arbitrary noise distribution.[98][99]
### Philosophy
Popular exposition of the Monte Carlo Method was conducted by McCracken.[100] The method's general philosophy was discussed by Elishakoff[101] and Grüne-Yanoff and Weirich.[102]
## References
### Citations
1. ^
2. ^ Kroese, D. P.; Brereton, T.; Taimre, T.; Botev, Z. I. (2014). "Why the Monte Carlo method is so important today". WIREs Comput Stat. 6 (6): 386–392. doi:10.1002/wics.1314. S2CID 18521840.
3. ^ Hubbard, Douglas; Samuelson, Douglas A. (October 2009). "Modeling Without Measurements". OR/MS Today: 28–33.
4. ^ Metropolis, Nicholas; Rosenbluth, Arianna W.; Rosenbluth, Marshall N.; Teller, Augusta H.; Teller, Edward (June 1, 1953). "Equation of State Calculations by Fast Computing Machines". The Journal of Chemical Physics. 21 (6): 1087–1092. Bibcode:1953JChPh..21.1087M. doi:10.1063/1.1699114. ISSN 0021-9606. OSTI 4390578. S2CID 1046577.
5. ^ Hastings, W. K. (April 1, 1970). "Monte Carlo sampling methods using Markov chains and their applications". Biometrika. 57 (1): 97–109. Bibcode:1970Bimka..57...97H. doi:10.1093/biomet/57.1.97. ISSN 0006-3444. S2CID 21204149.
6. ^ Liu, Jun S.; Liang, Faming; Wong, Wing Hung (March 1, 2000). "The Multiple-Try Method and Local Optimization in Metropolis Sampling". Journal of the American Statistical Association. 95 (449): 121–134. doi:10.1080/01621459.2000.10473908. ISSN 0162-1459. S2CID 123468109.
7. ^ Spall, J. C. (2003). "Estimation via Markov Chain Monte Carlo". IEEE Control Systems Magazine. 23 (2): 34–45. doi:10.1109/MCS.2003.1188770.
8. ^ Hill, Stacy D.; Spall, James C. (2019). "Stationarity and Convergence of the Metropolis-Hastings Algorithm: Insights into Theoretical Aspects". IEEE Control Systems Magazine. 39: 56–67. doi:10.1109/MCS.2018.2876959. S2CID 58672766.
9. ^ a b Kolokoltsov, Vassili (2010). Nonlinear Markov processes. Cambridge University Press. p. 375.
10. ^ a b c Del Moral, Pierre (2013). Mean field simulation for Monte Carlo integration. Chapman & Hall/CRC Press. p. 626. Monographs on Statistics & Applied Probability
11. ^ Del Moral, P.; Doucet, A.; Jasra, A. (2006). "Sequential Monte Carlo samplers". Journal of the Royal Statistical Society, Series B. 68 (3): 411–436. arXiv:cond-mat/0212648. doi:10.1111/j.1467-9868.2006.00553.x. S2CID 12074789.
12. ^ Shonkwiler, R. W.; Mendivil, F. (2009). Explorations in Monte Carlo Methods. Springer.
13. ^ Atanassova, E.; Gurov, T.; Karaivanova, A.; Ivanovska, S.; Durchova, M.; Dimitrov, D. (2016). "On the parallelization approaches for Intel MIC architecture". AIP Conference Proceedings. 1773 (1): 070001. Bibcode:2016AIPC.1773g0001A. doi:10.1063/1.4964983.
14. ^ Cunha Jr, A.; Nasser, R.; Sampaio, R.; Lopes, H.; Breitman, K. (2014). "Uncertainty quantification through the Monte Carlo method in a cloud computing setting". Computer Physics Communications. 185 (5): 1355–1363. arXiv:2105.09512. Bibcode:2014CoPhC.185.1355C. doi:10.1016/j.cpc.2014.01.006. S2CID 32376269.
15. ^ Wei, J.; Kruis, F.E. (2013). "A GPU-based parallelized Monte-Carlo method for particle coagulation using an acceptance–rejection strategy". Chemical Engineering Science. 104: 451–459. Bibcode:2013ChEnS.104..451W. doi:10.1016/j.ces.2013.08.008.
16. ^ Lin, Y.; Wang, F.; Liu, B. (2018). "Random number generators for large-scale parallel Monte Carlo simulations on FPGA". Journal of Computational Physics. 360: 93–103. Bibcode:2018JCoPh.360...93L. doi:10.1016/j.jcp.2018.01.029.
17. ^ a b c
18. ^
19. ^ a b Mazhdrakov, Benov & Valkanov 2018, p. 250.
20. ^ Haigh, Thomas; Priestley, Mark; Rope, Crispin (2014). "Los Alamos Bets on ENIAC: Nuclear Monte Carlo Simulations, 1947-1948". IEEE Annals of the History of Computing. 36 (3): 42–63. doi:10.1109/MAHC.2014.40. S2CID 17470931.
21. ^ McKean, Henry P. (1967). "Propagation of chaos for a class of non-linear parabolic equations". Lecture Series in Differential Equations, Catholic Univ. 7: 41–57.
22. ^ McKean, Henry P. (1966). "A class of Markov processes associated with nonlinear parabolic equations". Proc. Natl. Acad. Sci. USA. 56 (6): 1907–1911. Bibcode:1966PNAS...56.1907M. doi:10.1073/pnas.56.6.1907. PMC 220210. PMID 16591437.
23. ^ Herman, Kahn; Theodore, Harris E. (1951). "Estimation of particle transmission by random sampling" (PDF). Natl. Bur. Stand. Appl. Math. Ser. 12: 27–30.
24. ^ Turing, Alan M. (1950). "Computing machinery and intelligence". Mind. LIX (238): 433–460. doi:10.1093/mind/LIX.236.433.
25. ^ Barricelli, Nils Aall (1954). "Esempi numerici di processi di evoluzione". Methodos: 45–68.
26. ^ Barricelli, Nils Aall (1957). "Symbiogenetic evolution processes realized by artificial methods". Methodos: 143–182.
27. ^ a b Del Moral, Pierre (2004). Feynman–Kac formulae. Genealogical and interacting particle approximations. Probability and Its Applications. Springer. p. 575. ISBN 9780387202686. Series: Probability and Applications
28. ^ a b Del Moral, P.; Miclo, L. (2000). "Branching and interacting particle systems approximations of Feynman–Kac formulae with applications to non-linear filtering". Séminaire de Probabilités XXXIV. Lecture Notes in Mathematics. Vol. 1729. Berlin: Springer. pp. 1–145. doi:10.1007/BFb0103798. ISBN 978-3-540-67314-9. MR 1768060.
29. ^ Del Moral, Pierre; Miclo, Laurent (2000). "A Moran particle system approximation of Feynman–Kac formulae". Stochastic Processes and Their Applications. 86 (2): 193–216. doi:10.1016/S0304-4149(99)00094-0.
30. ^ Del Moral, Pierre (2003). "Particle approximations of Lyapunov exponents connected to Schrödinger operators and Feynman–Kac semigroups". ESAIM Probability & Statistics. 7: 171–208. doi:10.1051/ps:2003001.
31. ^ Assaraf, Roland; Caffarel, Michel; Khelif, Anatole (2000). "Diffusion Monte Carlo Methods with a fixed number of walkers" (PDF). Phys. Rev. E. 61 (4): 4566–4575. Bibcode:2000PhRvE..61.4566A. doi:10.1103/physreve.61.4566. PMID 11088257. Archived from the original (PDF) on November 7, 2014.
32. ^ Caffarel, Michel; Ceperley, David; Kalos, Malvin (1993). "Comment on Feynman–Kac Path-Integral Calculation of the Ground-State Energies of Atoms". Phys. Rev. Lett. 71 (13): 2159. Bibcode:1993PhRvL..71.2159C. doi:10.1103/physrevlett.71.2159. PMID 10054598.
33. ^ a b Hetherington, Jack H. (1984). "Observations on the statistical iteration of matrices". Phys. Rev. A. 30 (2713): 2713–2719. Bibcode:1984PhRvA..30.2713H. doi:10.1103/PhysRevA.30.2713.
34. ^ Fermi, Enrique; Richtmyer, Robert D. (1948). "Note on census-taking in Monte Carlo calculations" (PDF). LAM. 805 (A). Declassified report Los Alamos Archive
35. ^ a b Rosenbluth, Marshall N.; Rosenbluth, Arianna W. (1955). "Monte-Carlo calculations of the average extension of macromolecular chains". J. Chem. Phys. 23 (2): 356–359. Bibcode:1955JChPh..23..356R. doi:10.1063/1.1741967. S2CID 89611599.
36. ^ Gordon, N.J.; Salmond, D.J.; Smith, A.F.M. (April 1993). "Novel approach to nonlinear/non-Gaussian Bayesian state estimation". IEE Proceedings F - Radar and Signal Processing. 140 (2): 107–113. doi:10.1049/ip-f-2.1993.0015. ISSN 0956-375X. S2CID 12644877.
37. ^ Kitagawa, G. (1996). "Monte carlo filter and smoother for non-Gaussian nonlinear state space models". Journal of Computational and Graphical Statistics. 5 (1): 1–25. doi:10.2307/1390750. JSTOR 1390750.
38. ^ a b Del Moral, Pierre (1996). "Non Linear Filtering: Interacting Particle Solution" (PDF). Markov Processes and Related Fields. 2 (4): 555–580. Archived from the original (PDF) on March 4, 2016. Retrieved June 11, 2015.
39. ^ Carvalho, Himilcon; Del Moral, Pierre; Monin, André; Salut, Gérard (July 1997). "Optimal Non-linear Filtering in GPS/INS Integration" (PDF). IEEE Transactions on Aerospace and Electronic Systems. 33 (3): 835–850. Bibcode:1997ITAES..33..835C. doi:10.1109/7.599254. S2CID 27966240. Archived from the original (PDF) on November 10, 2022. Retrieved June 11, 2015.
40. ^ P. Del Moral, G. Rigal, and G. Salut. "Estimation and nonlinear optimal control: An unified framework for particle solutions". LAAS-CNRS, Toulouse, Research Report no. 91137, DRET-DIGILOG- LAAS/CNRS contract, April (1991).
41. ^ P. Del Moral, G. Rigal, and G. Salut. "Nonlinear and non Gaussian particle filters applied to inertial platform repositioning." LAAS-CNRS, Toulouse, Research Report no. 92207, STCAN/DIGILOG-LAAS/CNRS Convention STCAN no. A.91.77.013, (94p.) September (1991).
42. ^ P. Del Moral, G. Rigal, and G. Salut. "Estimation and nonlinear optimal control: Particle resolution in filtering and estimation: Experimental results". Convention DRET no. 89.34.553.00.470.75.01, Research report no.2 (54p.), January (1992).
43. ^ P. Del Moral, G. Rigal, and G. Salut. "Estimation and nonlinear optimal control: Particle resolution in filtering and estimation: Theoretical results". Convention DRET no. 89.34.553.00.470.75.01, Research report no.3 (123p.), October (1992).
44. ^ P. Del Moral, J.-Ch. Noyer, G. Rigal, and G. Salut. "Particle filters in radar signal processing: detection, estimation and air targets recognition". LAAS-CNRS, Toulouse, Research report no. 92495, December (1992).
45. ^ P. Del Moral, G. Rigal, and G. Salut. "Estimation and nonlinear optimal control: Particle resolution in filtering and estimation". Studies on: Filtering, optimal control, and maximum likelihood estimation. Convention DRET no. 89.34.553.00.470.75.01. Research report no.4 (210p.), January (1993).
46. ^ Del Moral, Pierre (1998). "Measure Valued Processes and Interacting Particle Systems. Application to Non Linear Filtering Problems". Annals of Applied Probability. 8 (2) (Publications du Laboratoire de Statistique et Probabilités, 96-15 (1996) ed.): 438–495. CiteSeerX 10.1.1.55.5257. doi:10.1214/aoap/1028903535.
47. ^ Crisan, Dan; Gaines, Jessica; Lyons, Terry (1998). "Convergence of a branching particle method to the solution of the Zakai". SIAM Journal on Applied Mathematics. 58 (5): 1568–1590. doi:10.1137/s0036139996307371. S2CID 39982562.
48. ^ Crisan, Dan; Lyons, Terry (1997). "Nonlinear filtering and measure-valued processes". Probability Theory and Related Fields. 109 (2): 217–244. doi:10.1007/s004400050131. S2CID 119809371.
49. ^ Crisan, Dan; Lyons, Terry (1999). "A particle approximation of the solution of the Kushner–Stratonovitch equation". Probability Theory and Related Fields. 115 (4): 549–578. doi:10.1007/s004400050249. S2CID 117725141.
50. ^ Crisan, Dan; Del Moral, Pierre; Lyons, Terry (1999). "Discrete filtering using branching and interacting particle systems" (PDF). Markov Processes and Related Fields. 5 (3): 293–318.
51. ^ Del Moral, Pierre; Guionnet, Alice (1999). "On the stability of Measure Valued Processes with Applications to filtering". C. R. Acad. Sci. Paris. 39 (1): 429–434.
52. ^ Del Moral, Pierre; Guionnet, Alice (2001). "On the stability of interacting processes with applications to filtering and genetic algorithms". Annales de l'Institut Henri Poincaré. 37 (2): 155–194. Bibcode:2001AIHPB..37..155D. doi:10.1016/s0246-0203(00)01064-5.
53. ^ Ripley 1987
54. ^ a b Sawilowsky 2003
55. ^ Kalos & Whitlock 2008
56. ^ Shojaeefard, M.H.; Khalkhali, A.; Yarmohammadisatri, Sadegh (2017). "An efficient sensitivity analysis method for modified geometry of Macpherson suspension based on Pearson Correlation Coefficient". Vehicle System Dynamics. 55 (6): 827–852. Bibcode:2017VSD....55..827S. doi:10.1080/00423114.2017.1283046. S2CID 114260173.
57. ^ Davenport 1992
58. ^ Route, Matthew (August 10, 2017). "Radio-flaring Ultracool Dwarf Population Synthesis". The Astrophysical Journal. 845 (1): 66. arXiv:1707.02212. Bibcode:2017ApJ...845...66R. doi:10.3847/1538-4357/aa7ede. S2CID 118895524.
59. ^ Vose 2008, p. 13.
60. ^ Vose 2008, p. 16.
61. ^ Jia, Xun; Ziegenhein, Peter; Jiang, Steve B (2014). "GPU-based high-performance computing for radiation therapy". Physics in Medicine and Biology. 59 (4): R151–R182. Bibcode:2014PMB....59R.151J. doi:10.1088/0031-9155/59/4/R151. PMC 4003902. PMID 24486639.
62. ^ Hill, R.; Healy, B.; Holloway, L.; Kuncic, Z.; Thwaites, D.; Baldock, C. (March 2014). "Advances in kilovoltage x-ray beam dosimetry". Physics in Medicine and Biology. 59 (6): R183–R231. Bibcode:2014PMB....59R.183H. doi:10.1088/0031-9155/59/6/R183. PMID 24584183. S2CID 18082594.
63. ^ Rogers, D.W.O. (2006). "Fifty years of Monte Carlo simulations for medical physics". Physics in Medicine and Biology. 51 (13): R287–R301. Bibcode:2006PMB....51R.287R. doi:10.1088/0031-9155/51/13/R17. PMID 16790908. S2CID 12066026.
64. ^ Baeurle 2009
65. ^ Möller, W.; Eckstein, W. (March 1, 1984). "Tridyn — A TRIM simulation code including dynamic composition changes". Nuclear Instruments and Methods in Physics Research Section B: Beam Interactions with Materials and Atoms. 2 (1): 814–818. Bibcode:1984NIMPB...2..814M. doi:10.1016/0168-583X(84)90321-5.
66. ^ MacGillivray & Dodd 1982
67. ^ Golden 1979
68. ^ G. A. Bird, Molecular Gas Dynamics, Clarendon, Oxford (1976)
69. ^ Dietrich, S.; Boyd, I. (1996). "A Scalar optimized parallel implementation of the DSMC technique". Journal of Computational Physics. 126 (2): 328–42. Bibcode:1996JCoPh.126..328D. doi:10.1006/jcph.1996.0141.
70. ^ Chen, Shang-Ying; Hsu, Kuo-Chin; Fan, Chia-Ming (March 15, 2021). "Improvement of generalized finite difference method for stochastic subsurface flow modeling". Journal of Computational Physics. 429: 110002. Bibcode:2021JCoPh.42910002C. doi:10.1016/J.JCP.2020.110002. S2CID 228828681.
71. ^ Climate Change 2013 The Physical Science Basis (PDF). Cambridge University Press. 2013. p. 697. ISBN 978-1-107-66182-0. Retrieved July 6, 2023.
72. ^
73. ^
74. ^ Cassey; Smith (2014). "Simulating confidence for the Ellison-Glaeser Index". Journal of Urban Economics. 81: 93. doi:10.1016/j.jue.2014.02.005.
75. ^ Sawilowsky & Fahoome 2003
76. ^ Spall, James C. (2005). "Monte Carlo Computation of the Fisher Information Matrix in Nonstandard Settings". Journal of Computational and Graphical Statistics. 14 (4): 889–909. CiteSeerX 10.1.1.142.738. doi:10.1198/106186005X78800. S2CID 16090098.
77. ^ Das, Sonjoy; Spall, James C.; Ghanem, Roger (2010). "Efficient Monte Carlo computation of Fisher information matrix using prior information". Computational Statistics & Data Analysis. 54 (2): 272–289. doi:10.1016/j.csda.2009.09.018.
78. ^ Chaslot, Guillaume; Bakkes, Sander; Szita, Istvan; Spronck, Pieter. "Monte-Carlo Tree Search: A New Framework for Game AI" (PDF). Sander.landofsand.com. Retrieved October 28, 2017.
79. ^ "Monte Carlo Tree Search - About". Archived from the original on November 29, 2015. Retrieved May 15, 2013.
80. ^ Chaslot, Guillaume M. J. -B; Winands, Mark H. M.; Van Den Herik, H. Jaap (2008). "Parallel Monte-Carlo Tree Search". Computers and Games. Lecture Notes in Computer Science. Vol. 5131. pp. 60–71. CiteSeerX 10.1.1.159.4373. doi:10.1007/978-3-540-87608-3_6. ISBN 978-3-540-87607-6.
81. ^ Bruns, Pete. Monte-Carlo Tree Search in the game of Tantrix: Cosc490 Final Report (PDF) (Report).
82. ^ Silver, David; Veness, Joel. "Monte-Carlo Planning in Large POMDPs" (PDF). 0.cs.ucl.ac.uk. Archived from the original (PDF) on July 18, 2016. Retrieved October 28, 2017.
83. ^ Lorentz, Richard J. (2011). "Improving Monte–Carlo Tree Search in Havannah". Computers and Games. Lecture Notes in Computer Science. Vol. 6515. pp. 105–115. Bibcode:2011LNCS.6515..105L. doi:10.1007/978-3-642-17928-0_10. ISBN 978-3-642-17927-3.
84. ^ Jakl, Tomas. "Arimaa challenge – comparison study of MCTS versus alpha-beta methods" (PDF). Arimaa.com. Retrieved October 28, 2017.
85. ^
86. ^ "How the Coast Guard Uses Analytics to Search for Those Lost at Sea". Dice Insights. January 3, 2014.
87. ^ Stone, Lawrence D.; Kratzke, Thomas M.; Frost, John R. "Search Modeling and Optimization in USCG's Search and Rescue Optimal Planning System (SAROPS)" (PDF). Ifremer.fr. Retrieved October 28, 2017.
88. ^ Carmona, René; Del Moral, Pierre; Hu, Peng; Oudjane, Nadia (2012). "An Introduction to Particle Methods with Financial Applications". In Carmona, René A.; Moral, Pierre Del; Hu, Peng; et al. (eds.). Numerical Methods in Finance. Springer Proceedings in Mathematics. Vol. 12. Springer Berlin Heidelberg. pp. 3–49. CiteSeerX 10.1.1.359.7957. doi:10.1007/978-3-642-25746-9_1. ISBN 978-3-642-25745-2.
89. ^ Kroese, D. P.; Taimre, T.; Botev, Z. I. (2011). Handbook of Monte Carlo Methods. John Wiley & Sons.
90. ^ Arenas, Daniel J.; Lett, Lanair A.; Klusaritz, Heather; Teitelman, Anne M. (2017). "A Monte Carlo simulation approach for estimating the health and economic impact of interventions provided at a student-run clinic". PLOS ONE. 12 (12): e0189718. Bibcode:2017PLoSO..1289718A. doi:10.1371/journal.pone.0189718. PMC 5746244. PMID 29284026.
91. ^ Elwart, Liz; Emerson, Nina; Enders, Christina; Fumia, Dani; Murphy, Kevin (December 2006). "Increasing Access to Restraining Orders for Low Income Victims of Domestic Violence: A Cost-Benefit Analysis of the Proposed Domestic Abuse Grant Program" (PDF). State Bar of Wisconsin. Archived from the original (PDF) on November 6, 2018. Retrieved December 12, 2016.
92. ^ Dahlan, Hadi Akbar (October 29, 2021). "Perbandingan Penerbitan dan Harga Buku Mengikut Genre di Malaysia dan Jepun Menggunakan Data Akses Terbuka dan Simulasi Monte Carlo" (PDF). Kajian Malaysia. 39 (2): 179–202. doi:10.21315/km2021.39.2.8. S2CID 240435973.
93. ^ a b Press et al. 1996
94. ^ MEZEI, M (December 31, 1986). "Adaptive umbrella sampling: Self-consistent determination of the non-Boltzmann bias". Journal of Computational Physics. 68 (1): 237–248. Bibcode:1987JCoPh..68..237M. doi:10.1016/0021-9991(87)90054-4.
95. ^ Bartels, Christian; Karplus, Martin (December 31, 1997). "Probability Distributions for Complex Systems: Adaptive Umbrella Sampling of the Potential Energy". The Journal of Physical Chemistry B. 102 (5): 865–880. doi:10.1021/jp972280j.
96. ^ Del Moral, Pierre; Doucet, Arnaud; Jasra, Ajay (2006). "Sequential Monte Carlo samplers". Journal of the Royal Statistical Society, Series B. 68 (3): 411–436. arXiv:cond-mat/0212648. doi:10.1111/j.1467-9868.2006.00553.x. S2CID 12074789.
97. ^ Spall, J. C. (2003), Introduction to Stochastic Search and Optimization: Estimation, Simulation, and Control, Wiley, Hoboken, NJ. http://www.jhuapl.edu/ISSO
98. ^ Mosegaard & Tarantola 1995
99. ^ Tarantola 2005
100. ^ McCracken, D. D., (1955) The Monte Carlo Method, Scientific American, 192(5), pp. 90-97
101. ^ Elishakoff, I., (2003) Notes on Philosophy of the Monte Carlo Method, International Applied Mechanics, 39(7), pp.753-762
102. ^ Grüne-Yanoff, T., & Weirich, P. (2010). The philosophy and epistemology of simulation: A review, Simulation & Gaming, 41(1), pp. 20-50
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# Kinetic Energy of a Load Suspended by Spring
A heavy load is suspended on a light spring with upper half rigidity (2k). The spring is slowly pulled down at the midpoint (a certain work (A) is done thereby) and then released. Determine the maximum kinetic energy (W_k) of the load in subsequent motion.
Solution:
If the middle of the spring is stretched out by a distance (x) while doing work (A), the entire spring is stretched out by (x).
Hence, the potential energy of the spring which is equal to the kinetic energy in the subsequent vibrational motion is $$W_k=kx^2/2$$
When the spring is pulled downwards at the midpoint, only its upper half (whose rigidity is (2k)) is stretched, and the work equal to the potential energy of the extension of the upper part of the spring is $$A=2kx^2/2=kx^2$$. Hence, we may conclude that the maximum kinetic energy of the load in the subsequent motion is $$W_k=A/2$$
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