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https://myprogrammingtutorial.com/square-in-python/ | 1,656,492,977,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103626162.35/warc/CC-MAIN-20220629084939-20220629114939-00538.warc.gz | 454,689,085 | 38,165 | # How To Square in Python
In this article, we will discuss how to square in Python. Squaring in Python can be done in 5 ways. We can square a number in python by using the Exponentiation operator, pow(), multiply number by itself, Numpy square(), and Numpy power() method.
## How to square in Python?
There are 5 ways to square a number in Python
1. Python Exponentiation Operator ( ** ) is used to raise the number to the power of an exponent. To get the square we use power as 2. For example – 5 ** 2 = 25, here 5 is raised to the 2nd power.
2. Python in-built pow() function. To get square we can use pow(number, 2).
3. Multiply by itself. We can get the square of the number by multiplying itself. Example – n * n.
4. The Python Numpy square() function returns the square of the number given as input. Example – numpy.square(5) = 25
5. To get square we use the Numpy package power(). Example numpy.power(4, 2) = 16.
From above you can choose any method to square a number in Python. Each method is discussed in detail below with examples.
### Squaring in Python using Exponentiation Operator ( ** )
Python provides arithmetic operators. The exponentiation operator is one of the arithmetic operators provided by python. It is denoted by ** ( 2 asterisks ). It returns the first raised to the power of an exponent.
Syntax:
`Number ** exponent`
To get a square of a number we take exponent value as 2.
Python Examples
``````# Defining some numbers to square
number1 = 5
number2 = 12
number3 = 14
number4 = 3
number5 = 10
# squaring using exponentiation operator
square1 = number1 ** 2
square2 = number2 ** 2
square3 = number3 ** 2
square4 = number4 ** 2
square5 = number5 ** 2
# print square of numbers in Python
print("Square of {0} is {1}".format(number1, square1))
print("Square of {0} is {1}".format(number2, square2))
print("Square of {0} is {1}".format(number3, square3))
print("Square of {0} is {1}".format(number4, square4))
print("Square of {0} is {1}".format(number5, square5))
``````
Output:
Square of 5 is 25
Square of 12 is 144
Square of 14 is 196
Square of 3 is 9
Square of 10 is 100
### Using pow() to square a number in Python
Python provides an in-built pow() function that can be used to get the square of a number in Python. The python pow() function returns the value of x to the power of y where x is the number and y is the exponent.
Syntax:
`power(number, exponent)`
To get a square of a number we take exponent value as 2.
Python Examples
``````# Defining some numbers to square
number1 = 5
number2 = 12
number3 = 14
number4 = 3
number5 = 10
# squaring using exponentiation operator
square1 = pow(number1, 2)
square2 = pow(number2, 2)
square3 = pow(number3, 2)
square4 = pow(number4, 2)
square5 = pow(number5, 2)
# print square of numbers in Python
print("Square of {0} is {1}".format(number1, square1))
print("Square of {0} is {1}".format(number2, square2))
print("Square of {0} is {1}".format(number3, square3))
print("Square of {0} is {1}".format(number4, square4))
print("Square of {0} is {1}".format(number5, square5))
``````
Output:
Square of 5 is 25
Square of 12 is 144
Square of 14 is 196
Square of 3 is 9
Square of 10 is 100
### Multiply by itself to get square of number
We can square a number by multiplying itself. To multiply a number in python we use asterisk character i.e *. Hence for squaring in python we can perform number * number
Python Examples
``````# Defining some numbers to square
number1 = 5
number2 = 12
number3 = 14
number4 = 3
number5 = 10
# squaring using exponentiation operator
square1 = number1 * number1
square2 = number2 * number2
square3 = number3 * number3
square4 = number4 * number4
square5 = number5 * number5
# print square of numbers in Python
print("Square of {0} is {1}".format(number1, square1))
print("Square of {0} is {1}".format(number2, square2))
print("Square of {0} is {1}".format(number3, square3))
print("Square of {0} is {1}".format(number4, square4))
print("Square of {0} is {1}".format(number5, square5))
``````
Output:
Square of 5 is 25
Square of 12 is 144
Square of 14 is 196
Square of 3 is 9
Square of 10 is 100
### Squaring in Python Numpy square()
Numpy is a library in Python, which is used for scientific computing. To know about Numpy click here.
We can use the Numpy square() function to get the square of a number in python.
Syntax-
`numpy.square(number)`
Python Examples
``````# importing numpy
import numpy
# Defining some numbers to square
number1 = 5
number2 = 12
number3 = 14
number4 = 3
number5 = 10
# squaring using exponentiation operator
square1 = numpy.square(number1)
square2 = numpy.square(number2)
square3 = numpy.square(number3)
square4 = numpy.square(number4)
square5 = numpy.square(number5)
# Print Square number in Python
print("Square of {0} is {1}".format(number1, square1))
print("Square of {0} is {1}".format(number2, square2))
print("Square of {0} is {1}".format(number3, square3))
print("Square of {0} is {1}".format(number4, square4))
print("Square of {0} is {1}".format(number5, square5))``````
Output:
Square of 5 is 25
Square of 12 is 144
Square of 14 is 196
Square of 3 is 9
Square of 10 is 100
### Squaring in Python Numpy power()
We can use the Numpy power() function to get the square of a number in python.
Syntax-
`numpy.power(number, 2)`
Python Examples
``````# importing numpy
import numpy
# Defining some numbers to square
number1 = 5
number2 = 12
number3 = 14
number4 = 3
number5 = 10
# squaring using exponentiation operator
square1 = numpy.power(number1, 2)
square2 = numpy.power(number2, 2)
square3 = numpy.power(number3, 2)
square4 = numpy.power(number4, 2)
square5 = numpy.power(number5, 2)
# Print Square number in Python
print("Square of {0} is {1}".format(number1, square1))
print("Square of {0} is {1}".format(number2, square2))
print("Square of {0} is {1}".format(number3, square3))
print("Square of {0} is {1}".format(number4, square4))
print("Square of {0} is {1}".format(number5, square5))
``````
Output:
Square of 5 is 25
Square of 12 is 144
Square of 14 is 196
Square of 3 is 9
Square of 10 is 100
## How to Square numbers in a list in Python?
Below we will discuss how to square numbers in list in Python.
`Input List - [2, 5, 12, 14, 3, 10]Output List - [4, 25, 144, 196, 9, 100]`
In Python, the above problem statement can be solved in various ways. We will be solving this problem in 3 ways.
### Method 1- Using for loop
We can get the Square of the number in a list in Python using a for loop by following the steps given below:-
1. Define the input with numbers to square
2. Define an empty list to store square of numbers
3. Use for loop to iterate on each number present in input list
4. Square the number and store it in the output list
5. Print the output list
Python Program:
``````# input list
input_list = [2, 5, 12, 14, 3, 10]
# defining output string
square_list = []
# using for loop to iterate on input list
# And appending the square of number in output string
for i in input_list:
square_list.append(i ** 2)
# Printing the output string
print(square_list)
# Print square of number python
for i in range(len(input_list)):
print("Square of {0} is {1}".format(input_list[i], square_list[i]))``````
Output:
[4, 25, 144, 196, 9, 100]
Square of 2 is 4
Square of 5 is 25
Square of 12 is 144
Square of 14 is 196
Square of 3 is 9
Square of 10 is 100
### Method 2- Using list comprehension
In this method, to get the square of all elements in a list we will be using list comprehension. You can know about list comprehension from here.
Python Program
``````# input list
input_list = [2, 5, 12, 14, 3, 10]
# Using list comprehension
square_list = [ number**2 for number in input_list ]
# printing square list
print(square_list)
# print number and its square
for i in range(len(input_list)):
print("Square of {0} is {1}".format(input_list[i], square_list[i]))``````
Output:
[4, 25, 144, 196, 9, 100]
Square of 2 is 4
Square of 5 is 25
Square of 12 is 144
Square of 14 is 196
Square of 3 is 9
Square of 10 is 100
### Method 3- Using lambda function to square elements in list
Lambda function is also called as Anonymous function, it is similar to the regular function in python, but it can be defined without the name. To know more about the lambda function click here.
Python Program
``````# input list
input_list = [2, 5, 12, 14, 3, 10]
# Using lambda function
square_list = list(map(lambda x: x ** 2, input_list))
# Printing square list
print(square_list)
# print number and its square
for i in range(len(input_list)):
print("Square of {0} is {1}".format(input_list[i], square_list[i]))``````
Output:
[4, 25, 144, 196, 9, 100]
Square of 2 is 4
Square of 5 is 25
Square of 12 is 144
Square of 14 is 196
Square of 3 is 9
Square of 10 is 100
## How to square numbers in a range in Python
Below we will discuss the square of numbers in the given range.
`Input Range - start = 1, end = 10Output - 1,4,9,16,25,36,49,64,81,100`
To get the square of the number in the given range we can use for loop with range() function.
The range() function returns the sequence from the start number and stops before the end number.
Syntax
`range(start_number, end_number, jump)`
Here jump default value is 1.
Python Program
``````# define range
start = 1
end = 10
# Using range function with for loop
# and squaring in python
print("Square of the numbers the range from {0} and {1} are:".format(start, end))
for num in range(start, end):
print(num ** 2)``````
Output:
Square of the numbers the range from 1 and 10 are:
1
4
9
16
25
36
49
64
81
## Conclusion
We can perform squaring in python by using exponentiation operator (**), pow(), multiply number by itself, Numpy square(), and NumPy power(). | 2,863 | 9,738 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2022-27 | longest | en | 0.82287 |
http://math4u.us/files/wilson_college2.html | 1,498,220,123,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320057.96/warc/CC-MAIN-20170623114917-20170623134917-00204.warc.gz | 243,702,084 | 24,040 | #### J.D. Wilson, A.J. Buffa, B. Lou, College Physics with Mastering Physics, 7th edition, Addison-Wesley 2009.
Chapter 7
1. The Cartesian coordinates of a point on a circle are (1.5 m, 2.0 m). What are the polar coordinates (r, ϑ) of this point?
2. The polar coordinates of a point are (5.3 m, 32°). What are the point’s Cartesian coordinates?
3. Convert the following angles from degrees to radians, to two significant figures: (a) 15°, (b) 45°, (c) 95°, and (d) 120°.
5. Express the following angles in degrees, radians and/or revolutions (rev) as appropriate: (a) 105°, (b) 1.8 rad, and (c) 5/7 rev.
6. You measure the length of a distant car to be subtended by an angular distance of 1.5°. If the car is actually 5.0 m long, approximately how far away is the car?
7. How large an angle in radians and degrees does the diameter of the Moon subtend to a person on the Earth?
8. The hour, minute, and second hands on a clock are 0.25 m, 0.30 m, and 0.35 m long, respectively. What are the distances traveled by the tips of the hands in a 30-min interval?
9. A car with a 65-cm diameter wheel travels 3.0 km. How many revolutions does the wheel make in this distance?
10. Two gear wheels with radii of 25 cm and 60 cm have interlocking teeth. How many radians does the smaller wheel turn when the larger wheel turns 4.0 rev?
11. You ordered a 12-in. pizza for a party of five. For the pizza to be distributed evenly, how should it be cut in triangular pieces?
12. To attend the 2000 Summer Olympics, a fan flew from Mosselbaai, South Africa (34° S, 22°E) to Sydney, Australia (34° S, 151° E). (a) What is the smallest angular distance the fan has to travel: (1) 34°, (2) 12°, (3) 117°, or (4) 129°? (b) Determine the appropriate shortest flight distance, in kilometers.
13. A bicycle wheel has a small pebble embedded in its tread. The rider sets the bike upside down, and accidentally bumps the wheel, causing the pebble to move through an arc length of 25.0 cm before coming to rest. In that time, the wheel spins 35°. (a) The radius of the wheel is therefore (1) more than 25.0 cm, (2) less than 25.0 cm, (3) equal to 25.0 cm. (b) Determine the radius of the wheel.
14. At the end of her routine, an ice skater spins through 7.50 revolutions with her arms always fully outstretched at right angles to her body. If her arms are 60.0 cm long, through what arc length distance do the tips of her fingers move during her finish?
15. (a) Could a circular pie be cut such that all of the wedge-shaped pieces have an arc length along the outer crust equal to the pie’s radius? (b) If not, how many such pieces could you cut, and what would be the angular dimension of the final piece?
16. Electrical wire with a diameter of 0.50 cm is wound on a spool with a radius of 30 cm and a height of 24 cm. (a) Through how many radians must the spool be turned to wrap one even layer of wire? (b) What is the length of this wound wire?
17. A yo-yo with an axle diameter of 1.00 cm has a 90.0 cm length of string wrapped around it many times in such a way that the string completely covers the surface of its axle, but there are no double layers of string. The outermost portion of the yo-yo is 5.00 cm from the center of the axle. (a) If the yo-yo is dropped with the string fully wound, through what angle does it rotate by the time it reaches the bottom of its fall? (b) How much arc length has a piece of the yo-yo on its outer edge traveled by the time it bottoms out?
18. A computer DVD-ROM has a variable angular speed from 200 rpm to 450 rpm. Express this range of angular speed in radians per second.
19. A race car makes two and half laps around a circular track in 3.0 min. What is the car’s average angular speed?
20. What are the angular speeds of the (a) second hand, (b) minute hand, and (c) hour hand of a clock? Are the speeds constant?
21. What is the period of revolution for (a) a 9500-rpm centrifuge and (b) a 9500-rpm computer hard disk drive?
22. Determine which has the greater angular speed: particle A, which travels 160° in 2.00 s, or particle B, which travels 4π rad in 8.00 s?
23. The tangential speed of a particle on a rotating wheel is 3.0 m/s. If the particle is 0.20 m from the axis of rotation, how long will the particle take to make one revolution?
24. A merry-go-round makes 24 revolutions in a 3.0-min ride. (a) What is its average angular speed in rad/s? (b) What are the tangential speeds of two people 4.0 m and 5.0 m from the center, or axis of rotation?
25. In Exercise 13, suppose the wheel took 1.20 s to stop after it was bumped. Assume as you face the plane of the wheel, it was rotating counterclockwise. During this time, determine (a) the average angular speed and tangential speed of the pebble, (b) the average angular speed and tangential speed of a piece of grease on the wheel’s axle (radius 1.50 cm), and (c) the direction of their respective angular velocities.
26. The Earth rotates on its axis once a day and revolves around the Sun once a year. Which is greater the rotating angular speed or the revolving angular speed? Why? (b) Calculate both angular speeds in rad/s.
27. A little boy jumps onto a small merry-go-round (radius of 2.00 m) in a park and rotates for 2.30 s through an arc length of 2.55 m before coming to rest. If he landed (and stayed) at a distance of 1.75 m from the central axis of rotation of the merry-go-round, what was his average angular speed and average tangential speed?
28. The driver of a car sets the cruise control and ties the steering wheel so that the car travels at a uniform speed of 15 m/s in a circle with a diameter of 120 m. (a) Through what angular distance does the car move in 4.00 min? (b) What arc length does it travel in this time?
29. In a noninjury, noncontact skid on icy pavement on an empty road, a car spins 1.75 revolutions while it skids to a halt. It was initially moving at 15.0 m/s, and because of the ice it was able to decelerate at a rate of only 1.50 m/s2. Viewed from above, the car spun clockwise. Determine its average angular velocity as it spun and slid to a halt.
30. An Indy car with a speed of 120 km/h goes around a level, circular track with a radius of 1.00 km. What is the centripetal acceleration of the car?
31. A wheel of radius 1.5 m rotates at a uniform speed. If a point on the rim of the wheel has a centripetal acceleration of 1.2 m/s2, what is the point's tangential speed?
32. A rotating cylinder about 16 km long and 7.0 km in diameter is designed to be used as a space colony. With what angular speed must it rotate so that the residents on it will experience the same acceleration due to gravity on Earth?
33. An airplane pilot is going to demonstrate flying in a tight vertical circle. To ensure that she doesn't black out at the bottom of the circle, the acceleration must not exceed 4.0g. If the speed of the plane is 50 m/s at the bottom of the circle, what is the minimum radius of the circle so that the 4.0g limit is not exceeded?
34. Imagine that you swing about your head a ball attached to the end of a string. The ball moves at a constant speed in a horizontal circle. (a) Can the string be exactly horizontal? Why or why not? (b) If the mass of the ball is 0.250 kg, the radius is 1.5 m, and it takes 1.2 s for the ball to make one revolution, what is the ball's tangential speed? (c) What centripetal force are you imparting to the ball via the string?
35. In Exercise 34, if you supplied a tension force of 12.5 N to the string, what angle would the string make relative to the horizontal?
36. A car with a constant speed of 83.0 km/h enters a circular flat curve with a radius of curvature of 0.400 km. If the friction between the road and the car's tires can supply a centripetal acceleration of 1.25 m/s2, does the car negotiate the curve safely? Justify the answer.
37. A student is to swing a bucket of water in a vertical circle without spilling any. (a) Explain how this task is possible. (b) If the distance from him shoulder to the centre of mass of the bucket of water is 1.0 m, what is the minimum speed required to keep the water from coming out of the bucket at the top of the swing?
38. In performing a “figure 8” maneuver, a figure skater wants to make the top part of the 8 approximately a circle of radius 2.20 m. He needs to glide through this part of the figure at approximately a constant speed, taking 4.50 s. His skates digging into the ice are capable of providing a maximum centripetal acceleration of 3.25 m/s2. Will he be able to do this as planned? If not, what adjustment can he make if he wants this part of the figure to remain the same size?
39. A light string of length of 56.0 cm connects two small square blocks, each with a mass of 1.50 kg. The system is placed on a slippery (frictionless) sheet of horizontal ice and spun so that the two blocks rotate uniformly about their common center of mass, which itself does not move. They are supposed to rotate with a period of 0.750 s. If the string can exert a force of only 100 N before it breaks, determine whether this string will work.
40. A jet pilot puts an aircraft with a constant speed into a vertical circular loop. (a) Which is greater, the normal force exerted on the seat by the pilot at the bottom of the loop or that at the top of the loop? Why? (b) If the speed of the aircraft is 700 km/h and the radius of the circle is 2.0 km, calculate the normal forces exerted on the seat by the pilot at the bottom and top of the loop. Express your answer in terms of the pilot's weight.
41. A block of mass m slides down an inclined plane into a loop-the-loop of radius r. (a) Neglecting friction, what is the minimum speed the block must have at the highest point of the loop in order to stay in the loop? (b) At what vertical height on the inclined plane (in terms of the radius of the loop) must the block be released if it is to have the required minimum speed at the top of the hoop?
42. For a scene in a movie, a stunt driver drives a 1.50 × 103 kg SUV with a length of 4.25 m around a circular curve with a radius of curvature of 0.333 km. The vehicle is to be driven off the edge of a gully 10.0 m wide, and land on the other side 2.96 m below the initial side. What is the minimum centripetal acceleration the SUV must have in going around the circular curve to clear the gully and land on the other side?
43. Consider a simple pendulum of length L that has a small mass (the bob) of mass m attached to the end of its string. If the pendulum starts out horizontally and is released from rest, show that (a) the speed at the bottom of the swing is vmax = √(2gL) and (b) the tension in the string at that point is three times the weight of the bob, or Tmax = 3mg.
44. A CD originally at rest reaches an angular speed of 40 rad/s in 5.0 s.(a) What is the magnitude of its angular acceleration? (b) How many revolutions does the CD make in the 5.0 s?
45. A merry-go-round accelerating uniformly from rest achieves its operating speed of 2.5 rpm in 5 revolutions. What is the magnitude of its angular acceleration?
46. A flywheel rotates with an angular speed of 25 rev/s. As it is brought to rest with a constant acceleration, it turns 50 rev. (a) What is the magnitude of the angular acceleration? (b) How much time does it take to stop?
47. A car on a circular track accelerates from rest. (a) The car experiences (1) only angular acceleration, (2) only centripetal acceleration, (3) both angular and centripetal accelerations? Why? (b) If the radius of the track is 0.30 km and the magnitude of the constant angular acceleration is 4.5 × 10-3 rad/s2, how long does the car take to make one lap around the track? © What is the total (vector) acceleration of the car when it has completed half of a lap?
48. Show that for a constant acceleration ϑ = ϑ0 + (ω2 – ω02)/2α
49. The blades of a fan running at low speed turn at 250 rpm. When the fan is switched to high speed, the rotation rate increases uniformly to 350 rpm in 5.75 s. (a) What is the magnitude of the angular acceleration of the blades? (b) How many revolutions do the blades go through while it is accelerating?
50. In the spin-dry cycle of a modern washing machine, a wet towel with mass of 1.50 kg is "stuck to" the inside surface of the perforated (to allow the water out) washing cylinder. To have decent removal of water, damp/ wet clothes need to experience a centripetal acceleration of at least 10g. Assuming this value, and that the cylinder has a radius of 35.0 cm, determine the constant angular acceleration of the towel required if the washing machine takes 2.50 s to achieve its final angular speed.
51. A pendulum swinging in a circular arc under the influence of gravity, as shown in Fig. 7.35, has both centripetal and tangential components of acceleration.. (a) If the pendulum bob has a speed of 2.7 m/s when the cord makes an angle of ϑ = 15° with the vertical, what are the magnitudes of the components at this time? (b) Where is the centripetal acceleration a maximum? What is the value of the tangential acceleration at that location?
52. A simple pendulum of length 2.00 m is released from a horizontal position. When it makes an angle of 30° from the vertical, determine (a) its angular acceleration, (b) its centripetal acceleration, and (c) the tension in the string. Assume the bob's mass is 1.50 kg.
53. From the known mass and radius of the Moon, compute the value of the acceleration due to gravity, gM, at the surface of the Moon.
54. The gravitational forces of the Earth and the Moon are attractive, so there must be a point on a line joining their centers where the gravitational forces on an object cancel. How far is this distance from the Earth’s center?
55. Four identical masses of 2.5 kg each are located at the corners of a square with 1.0-m sides. What is the net force on any one of the masses?
56. The average density of the Earth is 5.52 g/cm3. Assuming this is a uniform density, compute the value of G.
57. A 100-kg object is taken to a height of 300 km above the Earth’s surface. (a) What’s the object’s mass at this height? (b) What’s the object’s weight at this height?
58. A man has a mass of 75 kg on the Earth’s surface. How far above the surface of the Earth would he have to go to “lose” 10% of his body weight?
59. It takes 27 days for the Moon to orbit the Earth in a nearly circular orbit of radius 3.80 × 105 km. (a) Show in symbol notation that the mass of the Earth can be found using these data. (b) Compute the Earth's mass and compare with the value given inside the back cover of the book
60. Two objects are attracting each other with a certain gravitational force. (a) If the distance between the objects is halved, the new gravitational force will (1) increase by a factor of 2, (2) increase by a factor of 4, (3) decrease by a factor of 2, (4) decrease by a factor of 4. Why? (b) If the original force between the two objects is 0.90 N, and the distance is tripled, what is the new gravitational force between the objects?
61. During the Apollo lunar explorations of the late 1960s and early 1970s, the main section of the spaceship remained in orbit about the Moon with one astronaut in it while the other two astronauts descended to the surface in the landing module. If the main section orbited about 50 mi above the lunar surface, determine that section’s centripetal acceleration.
62. Referring to Exercise 61, determine (a) the gravitational potential energy, (b) the total energy, (c) the energy needed to "escape" the Moon for the main section of the lunar exploration mission in orbit. Assume the mass of this section is 5000 kg.
63. The diameter of the Moon’s (nearly circular) orbit about the Earth is 3.6 × 105 km and it takes 27 days for one orbit. What is (a) the Moon’s tangential speed, (b) its kinetic energy, (c) the system potential energy and system total energy?
64. (a) What is the mutual gravitational potential energy of the configuration shown in Fig. 7.36 if all the masses are 1.0 kg? (b) What is gravitational force per unit mass at the center of the configuration?
66. An instrument package is projected vertically upward to collect data near the top of the Earth’s atmosphere (at an altitude of about 900 km). (a) What initial speed is required at the Earth’s surface for the package to reach this height? (b) What percentage of the escape speed is this initial speed?
67. What is the orbital speed of a geosynchronous satellite?
68. In the year 2056, Martian Colony I wants to put a Mars-synchronous communication satellite in orbit about Mars to facilitate communications with the new bases being planned on the Red Planet. At what distance above the Martian equator would this satellite be placed?
69. The asteroid belt that lies between Mars and Jupiter may be the debris of a planet that broke apart or that was not able to form as a result of Jupiter’s strong gravitation. An average asteroid has a period of about 5.0 y. Approximately how far from the Sun would this "fifth" planet have been?
70. Using a development similar to Kepler’s law periods for planets orbiting the Sun, find the required altitude of geosynchronous satellites above the Earth.
71. Venus has a rotational period of 243 days. What would be the altitude of a synchronous satellite for this planet?
72. A small space probe is put into circular orbit about a newly discovered moon of Saturn. The moon's radius is known to be 550 km. If the probe orbits at a height of 1500 km above the moon's surface and takes 2.00 Earth days to make one orbit, determine the moon's mass.
Chapter 8
1. A wheel rolls uniformly on level ground without slipping. A piece of mud on the wheel flies off when it is at 9 o’clock position (near of wheel). Describe the subsequent motion of the mud.
2. A rope goes over a circular pulley with a radius of 6.5 cm. If the pulley makes 4 revolutions without the rope slipping, what length of rope passes over the pulley?
3. A wheel rolls 5 revolutions on a horizontal surface without slipping. If the center of the wheel moves 3.2 m, what is the radius of the wheel?
4. A bawling ball with a radius of 15.0 cm travels down the lane so that its center is moving at 3.60 m/s. The bowler estimates that it makes about 7.50 complete revolutions in 2.00 s. Is it rolling without slipping? Prove your answer, assuming that the bowler’s quick observation limits answers to two significant figures.
5. A ball with a radius of 15 cm rolls on a level surface, and the translational speed of the center of mass is 0.25 m/s. What is the angular speed about the center of mass if the ball rolls without slipping?
6. (a) When a disk rolls without slipping, should the product rω be (1) greater than, (2) equal to, or (3) less than vCM? (B) A disk with a radius of 0.15 m rotates through 270° as it travels 0.71 m. Does the disk rolls without slipping? Prove your answer.
7. A bocce ball with a diameter of 6.00 cm rolls without slipping on a level lawn. It has an initial angular speed of 2.35 rad/s and comes to rest after 2.50 m. Assuming constant deceleration, determine (a) the magnitude of its angular deceleration and (b) the magnitude of the maximum tangential acceleration of the ball’s surface.
8. A cylinder with a diameter of 20 cm rolls with an angular speed of 0.050 rad/s on a level surface. If the cylinder experiences a uniform tangential acceleration of 0.018 m/s2 without slipping until its angular speed is 1.2 rad/s, through how many complete revolutions does the cylinder rotate during the time it accelerates?
9. In Fig. 8.4a, if the arm makes a 37° angle with the horizontal and a torque of 18 m N to be produced, what force must the biceps muscle supply?
10. The drain plug on a car’s engine has been tightened to a torque of 25 m N. If a 0.15 –m-long wrench is used to change the oil, what is the minimum force needed to loosen the plug?
11. In Exercise 10, due to limited work space, you must crawl under the car. The force thus cannot be applied perpendicularly to the length of the wrench. If the applied force makes a 30° angle with the length of the wrench, what is the force required to loosen the drain plug?
12. How many different positions of stable equilibrium and unstable equilibrium are there for a cube? Consider each surface, edge, and corner to be a different position.
13. Two children are sitting on opposite ends of a uniform seesaw of negligible mass. (a) Can the seesaw be balanced if the masses of the children are different? How? (b) If a 35-kg child is 2.0 m from the pivot point (or fulcrum), how far from the pivot point will her 30 kg playmate have to sit on the other side for the seesaw to be in equilibrium?
14. A uniform meterstick pivoted at its center, as in Example 8.5, has a 100-g mass suspended at the 25.0-cm position. (a) At what position should a 75.0 g mass be suspended to put the system in equilibrium? (b) What mass would have to be suspended at the 90.0-cm position for the system to be in equilibrium?
15. A worker applies a horizontal force to the top edge of a crate to get it to tip forward (Fig. 8.36). If the create has a mass of 100 kg and is 1.6 m tall and 0.80 m in depth and width, what is the minimum force needed to make the crate start tipping? (Assume the center of mass of the crate is at its center and static friction great enough to prevent slipping).
16. Show that the balanced meterstick in Example 8.5 is in static rotational equilibrium about a horizontal axis through the 100-cm end of the stick.
17. Telephone and electrical lines are allowed to sag between poles so that the tension will not be too great when something hits or sits on the line. (a) Is it possible to have the lines perfectly horizontal? Why or why not? (b) Suppose that a line were stretched almost perfectly horizontally between two poles that are 30 m apart. If a 0.25-kg bird perches on the wire midway between the poles and the wire sags 1.0 cm, what would be the tension in the wire?
18. In Fig. 8.37, what is the force Fm supplied by the deltoid muscle so as to hold up the outstretched arm if the mass of the arm is 3.0 kg?
19. In Figure 8.4b, determine the force exerted by the bicep muscle, assuming that the hand is holding a ball with a mass of 5.00 kg. Assume that the mass of the forearm is 8.50 kg with its center of mass located 20.0 cm away from the elbow joint. (the black dot in the figure). Assume also that the center of mass of the ball in the hand is 30.0 cm away from the elbow joint.
20. A bowling ball (mass 7.00 kg and radius 17.0 cm) is released so fast that it skids without rotating down the lane (at least for a while). Assume the ball skids to the right and the coefficient of sliding friction between the ball and the lane surface is 0.400. (a) What is the direction of the torque exerted by the friction on the ball about the center of mass of the ball? (b) Determine the magnitude of this torque (again about the ball's center of mass).
21. A variation of Russell traction (Fig. 8.38) supports the lower leg in a cast. Suppose that the patient’s leg and cast have a combined mass of 15.0 kg and m1 is 4.50 kg. (a) What is the reaction force of the leg muscles to the traction? (b) What must m2 be to keep the leg horizontal?
22. In doing physical therapy for an injured knee joint, a person raises a 5.0-kg weighted boot as shown in Fig. 8.39. Compute the torque due to the boot for each position shown.
23. An artist wishes to construct a birds and bees mobile, as shown in Fig. 8.40. If the mass of the bee on the lower left is 0.10 kg and each vertical support string has a length of 30 cm, what are the masses of the other birds and bees?
24. The location of a person’s center of gravity relative to his or her height can be found using the arrangement shown in Fig. 8.41. The scales are initially adjusted to zero with the board alone. (a) Would you expect the location of the center of gravity to be (1) midway between the scales, (2) toward the scale at the person’s head, or (3) toward the scale at the person’s feet? Why? (b) Locate the center of gravity of the person relative to the horizontal dimension.
25. (a) How many uniform, identical textbooks of width 25.0 cm can be stacked on top of each other on a level surface without the stack falling over if each successive book is displaced 3.00 cm in width relative to the book below it? (b) (b) If the books are 5.00 cm thick, what will be the height of the center of mass of the stack above the level surface?
26. If four metersticks were stacked on a table with 10 cm, 15 cm, 30 cm, and 50 cm, respectively, hanging over the edge, as shown in Fig. 8.42, would the top meterstick remain on the table?
27. A 10.0 kg solid uniform cube with 0.500-m sides rests on a level surface. What is the minimum amount of work necessary to put the cube into an unstable equilibrium position?
28. While standing on a long board resting on a scaffold, a 70-kg painter paints the side of a house, as shown in Fig. 8.43. If the mass of the board is 15 kg, how close to the end can the painter stand without tipping the board over?
29. A mass is suspended by two cords as shown in Fig. 8.44. What are tensions in the cords?
30. If the cord attached to the vertical wall in Fig. 8.44 were horizontal (instead of at a 30° angle), what would the tensions in the cords be?
31. A force is applied to a cord wrapped around a solid 2.0-kg cylinder as shown in Fig. 8.45. Assuming the cylinder rolls without slipping, what is the force of friction acting on the cylinder?
32. In circus act, a uniform board (length 3.00 m, mass 35.0 kg) is suspended from a bungie-type rope at one end, and the other end rests on a concrete pillar. When a clown (mass 75.0 kg) steps out halfway onto the board, the board tilts so the rope end is 30° from the horizontal and the rope stays vertical. (a) In which situation will the rope tension be larger: (1) the board without the clown on it, (2) the board with the clown on it, or (3) you can’t tell from the data given? (b) Calculate the force exerted by the rope in both situations.
33. The forces acting on Einstein and the bicycle (fig. 2 of the Insight 8.1, Stability in Action) are the total weight of Einstein and the bicycle (mg) at the center of gravity of the system, the normal force (N) exerted by the road, and the force of static friction (fs) acting on the tires due to the road. (a) If Einstein is to maintain balance, should the tangent of the lean angle q (tan q) be (1) greater than, (2) equal to, or (3) less than fs/N? (b) The angle q in the picture is about 11°. What is the minimum coefficient of static friction ms between the road and the tires? (c) If the radius of the circle is 6.5 m, what is the maximum sped of Einstein’s bicycle?
34. A fixed 0.15-kg solid disk pulley with a radius 0.075 m is acted on by a net torque of 6.4 m N. What is the angular acceleration of the pulley?
35. What net torque is required to give a uniform 20-kg solid ball with a radius of 0.20 m an angular acceleration of 20 rad/s2?
36. For the system of masses shown in Fig. 8.46, find the moment of inertia about (a) the x-axis, (b) the y-axis, and (c) an axis through the origin and perpendicular to the page (z-axis). Neglect the masses of the connecting rods.
37. A 2000-kg Ferris wheel accelerates from rest to an angular speed of 20 rad/s in 12 s. Approximate the Ferris wheel as a circular disk with a radius of 30 m. What is the net torque on the wheel?
38. Two objects of different masses are joined by a light rod. (a) Is the moment of inertia about the center of mass the minimum or the maximum? Why? (b) If the two masses are 3.0 kg and 5.0 kg and the length of the rod is 2.0 m, find the moments of inertia of the system about an axis perpendicular to the rod, through the center of the rod and center of mass.
39. Two masses are suspended from a pulley as shown in Fig. 8.47. The pulley itself has a mass of 0.20 kg, a radius of 0.15 m, and a constant torque of 0.35 m N due to the friction between the rotating pulley and its axle. What is the magnitude of the acceleration of the suspended masses if m1 = 0.40 kg and m2 = 0.80 kg?
40. To start her lawn mower, Julie pulls on a cord that is wrapped around a pulley. The pulley has a moment of inertia about its central axis of I = 0.550 kg m2 and a radius of 5.00 cm. There is an equivalent frictional torque impeding her pull of τf = 0.430 m N. To accelerate the pulley at α = 4.55 rad/s2, (a) how much torque does Julie need to apply to the pulley? (b) How much tension must the rope exert?
41. For the system shown in Fig. 8.48, m1 = 8.0 kg, m2 = 3.0 kg, q = 30°, and the radius and mass of the pulley are 0.10 m and 0.10 kg, respectively. (a) What is the acceleration of the masses? (b) If the pulley has a constant frictional torque of 0.050 m N when the system is in motion, what is the acceleration of the masses?
42. A meterstick pivoted about a horizontal axis through the 0-cm end is held in a horizontal position and let go. (a) What is the initial tangential acceleration of the 100-cm position? Are you surprised by this result? (b) Which position has a tangential acceleration equal to the acceleration due to gravity?
43. Pennies are placed every 10 cm on a meterstick. One end of the stick is put on a table and the other end is held horizontally with a finger, as shown in Fig. 8.49. If the finger is pulled away, what happens to the pennies?
44. A uniform 2.0-kg cylinder of radius 0.15 m is suspended by two strings wrapped around it (Fig. 8.50). As the cylinder descends, the strings unwind from it. What is the acceleration of the center of mass of the cylinder?
45. A planetary space probe is in the shape of a cylinder. To protect it from heat on one side (from the Sun's rays), operators on the Earth put it into a "barbecue mode," that is, they set it rotating about its long axis. To do this, they fire four small rockets mounted tangentially as shown in Fig. 8.51 (the probe is shown coming toward you). The object is to get the probe to rotate completely once every 30 s, starting from no rotation at all. They wish to do this by firing all four rockets for a certain length of time. Each rocket can exert a thrust of 50.0 N. Assume the probe is a uniform solid cylinder with a radius of 2.50 m and a mass of 1000 kg and neglect the mass of each rocket engine. Determine the amount of time the rockets need to be fired.
46. A ball of radius R and mass M rolls down an incline of angle θ. (a) For the ball to roll without slipping, should be the tangent of the maximum angle of incline (tan θ) be equal to (1) 3μs/2, (2) 5μs/2, (3) 7μs/2, or (4) 9μs/2? Here, μs is the coefficient of static friction. (b) If the ball is made of wood and the surface is also wood, what is the maximum angle of incline?
47. A constant retarding torque of 12 m N stops a rolling wheel of diameter 0.80 m in a distance of 15 m. How much work is done by the torque?
48. A person opens a door by applying a 15-N force perpendicular to it at a distance 0.90 m from the hinges. The door is pushed wide open (to 120°) in 2.0 s. (a) How much work was done? (b) What was the average power delivered?
49. In Fig. 8.23, a mass m descends a vertical distance from rest. (Neglect friction and the mass of the string) (a) From the conservation of mechanical energy, will the linear speed of the descending mass be (1) greater than, (2) equal to, or (3) less than √(2gh)? Why? (b) If m = 1.0 kg, M = 0.30 kg, and R = 0.15 kg, what is the linear speed of the mass after it has descended a vertical distance of 2.0 from rest?
50. A constant torque of 10 m N is applied to the rim of a 10-kg uniform disk of radius 0.20 m. What is the angular speed of the disk about an axis through its center after it rotates 2.0 revolutions from rest?
51. A 2.5-kg pulley of radius 0.15 m is pivoted about an axis through its center. What constant torque is required for the pulley to reach an angular speed of 25 rad/s after rotating 3.0 revolutions, starting from rest?
52. A solid ball of mass m rolls along a horizontal surface with a translational speed of v. What percent of its total kinetic energy is translational?
53. Estimate the ratio of the translational kinetic energy of the Earth as it orbits the Sun to the rotational kinetic energy it has about it N-S axis.
54. You wish to accelerate a small merry-go-round from rest to a rotational speed of one-third of a revolution per second by pushing tangentially on it. Assume the merry-go-round is a disk with a mass of 250 kg and a radius of 1.50 m. Ignoring friction, how hard do you have to push tangentially to accomplish this in 5.00 s?
55. A pencil 18 cm long stands vertically on its point end on a horizontal table. If it falls over without slipping, with what tangential speed does the eraser end strike the table?
56. A uniform sphere and a uniform cylinder with the same mass and radius roll at the same velocity side by side on a level surface without slipping. If the sphere and the cylinder approach an inclined plane and roll up it without slipping, will they be at the same height on the plane when they come to a stop? If not, what will be the percentage difference of the heights?
57. A hoop starts from rest at a height 1.2 m above the base of an inclined plane and rolls down under the influence of gravity. What is the linear speed of the hoop’s center of mass just as the hoop leaves the incline and rolls onto a horizontal surface?
58. A cylindrical hoop, a cylinder, and a sphere of equal radius and mass are released at the same time from the top of an inclined plane. Using the conservation of mechanical energy, show that the sphere always gets to the bottom of the incline first with the fastest speed and that the hoop always arrives last with the slowest speed.
59. For the following objects, which all roll without slipping, determine the rotational kinetic energy about the center of mass as a percentage of the total kinetic energy: (a) a solid sphere, (b) a thin spherical shell, and (c) a thin cylindrical shell.
60. An industrial flywheel with a moment of inertia of 4.25 × 102 kg m2 rotates with a speed of 7500 rpm. (a) How much work is required to bring the flywheel to rest? (b) If this work is done uniformly in 1.5 min, how much power is required?
61. A hollow, thin-shelled ball and a solid ball of equal mass are rolled up an inclined plane (without slipping) with both balls having the same initial velocity at the bottom of the plane. (a) Which ball rolls higher on the incline before coming to rest? (b) Do the radii of the balls make a difference? (c) After stopping, the balls roll back down the incline. By the conservation of energy, both balls should have the same speed when reaching the bottom of the incline. Show this explicitly.
62. In a tumbling clothes dryer, the cylindrical drum (radius 50.0 cm and mass 35.0 kg) rotates once every second. (a) Determine the rotational kinetic energy about its central axis. (b) If it started from rest and reached that speed in 2.50 s, determine the average net torque on the dryer drum.
63. A steel ball rolls down an incline into a loop-the loop of radius R (Fig. 8.52a). (a) What minimum speed must the ball have at the top of the loop in order to stay on the track? (b) At what vertical height (h) on the incline, in terms of the radius of the loop, must the ball be released in order for it to have the required speed at the top of the loop? (Neglect frictional losses.) (c) Figure 8.52b shows the loop-the-loop of a roller coaster. What are the sensations of the riders if the roller coaster has the minimum speed or a greater speed at the top of the loop?
64. What is the angular momentum of a 2.0-g particle moving counterclockwise (as viewed from above) with an angular speed of 5π rad/s in a horizontal circle of radius 15 cm? (Give the magnitude and direction.)
65. A 10-kg rotating disk of radius 0.25 m has an angular momentum of 0.45 kg m2/s. What is the angular speed of the disk?
66. Compute the ratio of the magnitudes of the Earth’s orbital angular momentum and its rotational angular momentum. Are these moments in the same direction?
67. The Earth revolves about the Sun and spins on its axis, which is tilted 23½º to its orbital plane. (a) Assuming a circular orbit, what is the magnitude of the angular momentum associated with the Earth’s orbital motion about the Sun? (b) What is the magnitude of the angular momentum associated with the Earth’s rotation on its axis?
68. The period of the Moon’s rotation is the same as the period of its revolution: 27.3 days (sidereal). What is the angular momentum for each rotation and revolution?
69. Circular disks are used in automobile clutches and transmissions. When a rotating disk couples to a stationary one through frictional force, the energy from the rotating disk can transfer to the stationary one. (a) Is the angular-speed of the coupled disks (1) greater than, (2) less than, or (3) the same as the angular speed of the original rotating disk? Why? (b) If a disk rotating at 800 rpm couples to a stationary disk with three times the moment of inertia, what is the angular speed of the combination?
70. An ice skater has a moment of inertia of 100 kg m2 when his arms are outstretched and a moment of inertia of 75 kg m2 when his arms are tucked in close to his chest. If he starts to spin at an angular speed of 2.0 rps (revolutions per second) with his arms outstretched, what will his angular speed be when they are tucked in?
71. An ice skater spinning with outstretched arms has an angular speed of 4.0 rad/s. She tucks in her arms, decreasing her moment of inertia by 7.5%. (a) What is the resulting angular speed? (b) By what factor does the skater’s kinetic energy change? (c) Where does the extra kinetic energy come from?
72. A billiard ball at rest is struck (bold arrow in Fig. 8.53) by a cue with an average force of 5.50 N lasting for 0.050 s. The cue contacts the ball’s surface so that the lever arm is half the radius of the ball, as shown. If the cue ball has a mass of 200 g and a radius of 2.50 cm, determine the angular speed of the ball immediately after the blow.
73. A comet approaches the Sun as illustrated in Fig. 8.54 and is deflected by the Sun’s gravitational attraction. This event is considered a collision, and b is called the impact parameter. Find the distance of closest approach (d) in terms of the impact parameter and the velocities (v0 at large distances and v at closest approach). Assume that the radius of the Sun is negligible compared to d. (As the figure shows, the tail of a comet always “points” away from the Sun.)
74. While repairing his bicycle, a student turns it upside down and sets the front wheel spinning at 2.00 rev/s. Assume the wheel has a mass of 3.25 kg and all of the mass is located on the rim, which has a radius of 41.0 cm. To slow the wheel, he places his hand on the tire, thereby exerting a tangential force of friction on the wheel. It takes 3.50s to come to rest. Use the change in angular momentum to determine the force he exerts on the wheel. Assume the frictional force of the axle is negligible.
75. A kitten stands on the edge of a lazy Susan (a turntable). Assume that the lazy Susan has frictionless bearings and is initially at rest. (a) If the kitten starts to walk around the edge of the lazy Susan, the lazy Susan will (1) remain lazy and stationary, (2) rotate in the direction opposite that in which the kitten is walking, or (3) rotate in the direction the kitten is walking. Explain. (b) The mass of the kitten is 0.50 kg, and the lazy Susan has a mass of 1.5 kg and a radius of 0.30 m. If the kitten walks at a speed of 0.25 m/s, relative to the ground, what will be the angular speed of the lazy Susan? (s) When the kitten has walked completely around the edge and is back at its starting point, will that point be above the same point on the ground as it was at the start?
Chapter 9
1. A tennis racket has nylon strings. If one of the strings with a diameter of 1.0 mm is under a tension of 15 N, how much is it lengthened from its original length of 40 cm?
2. Suppose you use the tip of one finger to support a 1.0-kg object. If your finger has a diameter of 2.0 cm, what is the stress on your finger?
3. A 2.5-m nylon fishing line used to hold up a 8.0-kg fish has a diameter of 1.6 mm. How much is the line elongated?
4. A 5.0-m-long rod is stretched 0.10 m by a force. What is the strain in the rod?
5. A 250-N force is applied at a 37° angle to the surface of the end of a square bar. The surface is 4.00 cm on a side. What are (a) the compressional stress and (b) the shear stress on the bar?
6. A 4.0-kg object is supported by an aluminum wire of length 2.0 m and diameter 2.0 mm. How much will the wire stretch?
7. A copper wire has a length of 5.0 m and a diameter of 3.0 mm. Under what load will its length increase by 0.30 mm?
8. A metal wire 1.0 mm in diameter and 2.0 m long hangs vertically with a 6.0-kg object suspended from it. If the wire stretches 1.4 mm under the tension, what is the value of Young’s modulus for the metal?
9. When railroad tracks are installed, gaps are left between the rails. (a) Should a greater gap be used if rails are installed on (1) a cold day or (2) a hot day? Or (3) does the temperature not make any difference? Why? (b) Each steel rail is 8.0 m long and has a cross-sectional area of 0.0025 m2. On a hot day, each rail thermally expands as much as 3.0 × 10-3 m. If there were no gaps between the rails, what would be the force on the ends of each rail?
10. A rectangular steel column (20.0 cm × 15.0 cm) supports a load of 12.0 metric tons. If the column is 2.00 m in length before being stressed, what is the decrease in length?
11. A bimetallic rod as illustrated in Fig. 9.34 is composed of brass and copper. (a) If the rod is subjected to a compressive force, will the rod bend toward the brass or the copper? Why? (b) Justify your answer mathematically if the compressive force is 5.00 × 104 N.
12. Two same-size metal posts, one aluminum and one copper, are subjected to equal shear stresses. (a) Which post will show the larger deformation angle, (1) the copper post or (2) the aluminum post? Or (3) Is the angle the same for both? Why? (b) By what factor is the deformation angle of one post greater than the other?
13. A 85.0-kg person stands on one leg and 90% of the weight is supported by the upper leg connecting the knee and hip joint – the femur. Assuming the femur is 0.650 m long and has a radius of 2.00 cm, by how much is the bone compressed?
14. Two metal plates are held together by two steel rivets, each of diameter 0.20 cm and length 1.0 cm. How much force must be applied parallel to the plates to shear off both rivets?
15. (a) Which of the liquids in Table 9.1 has the greatest compressibility? Why? (b) For equal volumes of ethyl alcohol and water, which would require more pressure to be compressed by 0.10%, and how many times more?
16. How much pressure would be required to compress a quantity of mercury by 0.010%?
17. A brass cube 6.0 cm on each side is placed in a pressure chamber and subjected to a pressure of 1.2 × 107 N/m2 on all of its surfaces. By how much will each side be compressed under this pressure?
8. A cylindrical eraser of negligible mass is dragged across a paper at a constant velocity to the right by its pencil. The coefficient of kinetic friction between eraser and paper is 0.650. The pencil pushes down with 4.20 N. The height of the eraser is 1.10 cm and its diameter is 0.760 cm. Its top surface is displaced horizontally 0.910 mm relative to the bottom. Determine the shear modulus of the eraser material.
19. A 45-kg traffic light is suspended from two steel cables of equal length and radii 0.50 cm. If each cable makes a 15° angle with the horizontal, what is the fractional increase in their length due to the weight of the light?
20. In his original barometer, Pascal used water instead of mercury. (a) Water is less dense than mercury, so the water barometer would have (1) a higher height than, (2) a lower height than, or (3) the same height as the mercury barometer. Why? (b) How high would the water column have been?
21. If you dive to a depth of 10 m below the surface of a lake, (a) what is the pressure due to the water alone? (b) What is the absolute pressure at that depth?
22. In an open U-tube, the pressure of a water column on one side is balanced by the pressure of a column of gasoline on the other side. (a) Compared to the height of the water column, the gasoline column will have (1) a higher height, (2) a lower height, or (3) the same height. Why? (b) If the height of the water column is 15 cm, what is the height of the gasoline column?
23. A 75.0-kg athlete performs a single-hand handstand. If the area of the hand in contact with the floor is 125 cm2, what pressure is exerted on the floor?
24. A rectangular fish tank measuring 0.75 m × 0.50 m is filled with water to a height of 65 cm. What is the gauge pressure on the bottom of the tank?
25. (a) What is the absolute pressure at a depth of 10 m in a lake? (b) What is the gauge pressure?
26. The gauge pressure in both tires of a bicycle is 690 kPa. If the bicycle and the rider have a combined mass of 90.0 kg, what is the area of contact of each tire with the ground? (Assume that each tire supports half the total weight of the bicycle.)
27. In a sample of seawater taken from an oil spill, an oil layer 4.0 cm thick floats on 55 cm of water. If the density of the oil is 0.75 × 103 kg/m3, what is the absolute pressure on the bottom of the container?
28. In a lecture demonstration, an empty can is used to demonstrate the force exerted by air pressure (Fig. 9.35). A small quantity of water is poured into the can, and the water is brought to a boil. Then the can is sealed with a rubber stopper. As you watch, the can is slowly crushed with sounds of metal bending. (Why is a rubber stopper used as a safety precaution?) (a) This is because of (1) thermal expansion and contraction, (2) a higher steam pressure inside the can, or (3) a lower pressure inside the can as steam condenses. Why? (b) Assuming the dimensions of the can are 0.24 m × 0.16 m × 0.10 m and the inside of the can is in a perfect vacuum, what is the total force exerted on the can by the air pressure?
29. What is the fractional decrease in pressure when a barometer is raised 40.0 m to the top of a building?
30. To drink a soda (assume same density as water) through a straw requires that your lower the pressure at the top of the straw. What does the pressure need to be at the top of a straw that is 15.0 cm above the surface of the soda in order for the soda to reach your lips?
31. During a plane flight, a passenger experiences ear pain due to a head cold that has clogged his Eustachian tubes. Assuming the pressure in his tubes remained at 1.00 atm (from sea level) and the cabin pressure is maintained at 0.900 atm, determine the air pressure force (including its direction) on one eardrum, assuming it has a diameter of 0.800 cm.
32. Here is a demonstration Pascal used to show the importance of a fluid’s pressure on the fluid’s depth (Fig. 9.36): An oak barrel with a lid of area 0.20 m2 is filled with water. A long, thin tube of cross-sectional area 5.0 × 10-5 m2 is inserted into a hole at the center of the lid, and water is poured into the tube. When the water reaches 12 m high, the barrel bursts. (a) What was the weight of the water in the tube? (b) What was the pressure of the water on the lid of the barrel? (c) What was the net force on the lid due to the water pressure?
33. The door and the seals on an aircraft are subject to a tremendous amount of force during flight. At an altitude of 10000 m (about 33000 ft), the air pressure outside the airplane is only 2.7 × 104 N/m2, while the inside is still at normal atmospheric pressure, due to pressurization of the cabin. Calculate the force due to the air pressure on a door of area 3.0 m2.
34. The pressure exerted by a person’s lungs can be measured by having the person blow as hard as possible into one side of a manometer. If a person blowing into one side of an open-tube manometer produces an 80-cm difference between the heights of the columns of water in the manometer arms, what is the gauge pressure of the lungs?
35. In a head-on auto collision, the driver, who had his air bags disconnected, hits his head on the windshield, fracturing his skull. Assuming the driver’s head has a mass of 4.0 kg, the area of the head to hit the windshield to be 25 cm2, and an impact time of 3.0 ms, with what speed does his head hit the windshield?
36. A cylinder has a diameter of 15 cm (Fig. 9.37). The water level in the cylinder is maintained at a constant height of 0.45 m. If the diameter of the spout pipe is 0.50 cm, how high is h, the vertical stream of water?
37. In 1960, the U.S. Navy’s bathyscaphe Trieste (a submersible) descended to a depth of 10912 m (about 35000 ft) into the Mariana Trench in the Pacific Ocean. (a) What was the pressure at that depth? (Assume that seawater is incompressible.) (b) What was the force on a circular observation window with a diameter of 15 cm?
38. The output piston of a hydraulic press has a cross-sectional area of 0.25 m2. (a) How much pressure on the input piston is required for the press to generate a force of 1.5 × 106 N? (b) What force is applied to the input piston if it has a diameter of 5.0 cm?
39. A hydraulic lift in a garage has two pistons: a small one of cross-sectional area 4.00 cm2 and a large one of cross-sectional area 250 cm2 (a) If this lift is designed to raise a 3500-kg car, what minimum force must be applied to the small piston? (b) If the force is applied through compressed air, what must be the minimum air pressure applied to the small piston?
40. The Magdeburg water bridge is a channel bridge over the River Elbe in Germany (Fig. 9.38). Its dimension are length 918 m, width 43.0 m, and depth 4.25 m. (a) When filled with water, what is the weight of the water? (b) What is the pressure on the bridge floor?
41. A hypodermic syringe has a plunger of area 2.5 cm2 and a 5.0 ×10-3-cm2 needle. (a) If a 1.0-N force is applied to the plunger, what is the gauge pressure in the syringe’s chamber? (b) If a small obstruction is at the end of the needle, what force does the fluid exert on it? (c) If the blood pressure in a vein is 50 mm Hg, what force must be applied on the plunger so that fluid can be injected into the vein?
42. A funnel has a cork blocking its drain tube. The cork has a diameter of 1.50 cm and is held in place by static friction with the sides of the drain tube. When water is added to a height of 10.0 cm above the cork, it comes flying out of the tube. Determine the maximum force of static friction between the cork and drain tube. Neglect the weight of the cork.
43. (a) If the density of an object is exactly equal to the density of a fluid, the object will (1) float, (2) sink, (3) stay at any height in the fluid, as long as it is totally immersed. (b) A cube 8.5 cm on each side has a mass of 0.65 kg. Will the cube float or sink in water? Prove your answer.
44. A rectangular boat, as illustrated in Fig. 9.39, is overloaded such that the water level is just 1.0 cm below the top of the boat. What is the combined mass of the people and the boat?
45. An object has a weight of 8.0 N in air. However, it apparently weighs only 4.0 N when it is completely submerged in water. What is the density of the object?
46. When a 0.80-kg crown is submerged in water, its apparent weight is measured to be 7.3 N. Is the crown pure gold?
47. A steel cube 0.30 m on each side is suspended from a scale and immersed in water. What will the scale read?
48. A solid ball has a weight of 3.0 N. When it is submerged in water, it has an apparent weight of 2.7 N. What is the density of the ball?
49. A wood cube 0.30 m on each side has a density of 700 kg/m3 and floats levelly in water. (a) What is the distance from the top of the wood to the water surface? (b) What mass has to be placed on top of the wood so that its top is just at the water level?
50. (a) Given a piece of metal with a light string attached, a scale, and a container of water in which the piece of metal can be submersed, how could you find the volume of the piece without using the variation in the water level? (b) An object has a weight of 0.882 N. It is suspended from a scale, which reads 0.735 N when the piece is submerged in water. What are the volume and density of the piece of metal?
51. An aquarium is filled with a liquid. A cork cube, 10.0 cm on a side, is pushed and held at rest completely submerged in the liquid. It takes a force of 7.84 N to hold it under the liquid. If the density of cork is 200 kg/m3, find the density of the liquid.
52. A block of iron quickly sinks in water, but ships constructed of iron float. A solid cube of iron 1.0 m on each side is made into sheets. To make these sheets into a hollow cube that will not sink, what should be the minimum length of the sides of the sheets?
53. Plans are being made to bring back the zeppelin, a lighter-than-air airship like the Goodyear blimp that carries passengers and cargo, but is filled with helium, not flammable hydrogen as was used in the ill-fated Hindenburg. (See opening Physics Facts.) One design calls for the ship to be 110 m long and to have a total mass (without helium) of 30.0 metric tons. Assuming the ship’s “envelope” to be cylindrical, what would its diameter have to be so as to lift the total weight of the ship and the helium?
54. A girl floats in a lake with 97% of her body beneath the water. What are (a) her mass density and (b) her weight density?
55. A spherical navigation buoy is tethered to the lake floor by a vertical cable (Fig. 9.41). The outside diameter of the buoy is 1.00 m. The interior of the buoy consists of an aluminum shell 1.0 cm thick, and the rest is solid plastic. The density of aluminum is 2700 kg/m3 and the density of the plastic is 200 kg/m3 The buoy is set to float exactly halfway out of the water. Determine the tension in the cable.
56. Figure 9.41 shows a simple laboratory experiment. Calculate (a) the volume and (b) the density of the suspended sphere. (Assume that the density of the sphere is uniform and that the liquid in the beaker is water.) (c) Would you be able to make the same determinations if the liquid in the beaker were mercury? (See Table 9.2.) Explain.
57. An ideal fluid is moving at 3.0 m/s in a section of a pipe of radius 0.20 m. If the radius in another section is 0.35 m, what is the flow speed there?
58. (a) If the radius of a pipe narrows to half of its original size, will the flow speed in the narrow section (1) increase by a factor of 2, (2) increase by a factor of 4, (3) decrease by a factor of 2, or (4) decrease by a factor of 4? Why? (b) If the radius widens to three times its original size, what is the ratio of the flow speed in the wider section to that in the narrow section?
59. Water flows through a horizontal tube similar to that in Fig. 9.20. However in this case, the constricted part of the tube is half the diameter of the larger part. If the water speed is 1.5 m/s in the larger parts of the tube, by how much does the pressure drop in the constricted part? Express the final answer in atmospheres.
60. The speed of blood in a major artery of diameter 1.0 cm is 4.5 cm/s (a) What is the flow rate in the artery? (b) If the capillary system has a total cross-sectional area of 2500 cm2, the average speed of blood through the capillaries is what percentage of that through the major artery? (c) Why must blood flow at low speed through the capillaries?
61. The blood flow speed through an aorta with a radius of 1.00 cm is 0.265 m/s. If hardening of the arteries causes the aorta to be constricted to a radius of 0.800 cm, by how much would the blood flow speed increase?
62. Using the data and result of Exercise 61, calculate the pressure difference between the two areas of the aorta. (Blood density: ρ = 1.06 × 103kg/m3).
63. In a dramatic lecture demonstration, a physics professor blows hard across the top of a copper penny that is at rest on a level desk. By doing this at the right speed, he can get the penny to accelerate vertically, into the airstream, and then deflect it into a tray, as shown in Fig. 9.42. Assuming the diameter of a penny is 1.90 cm and it has a mass of 2.50 g, what is the minimum airspeed needed to lift the penny off the tabletop? Assume the air under the penny remains at rest.
64. The spout heights in the container in Fig. 9.43 are 10 cm, 20 cm, 30 cm, and 40 cm. The water level is maintained at a 45-cm height by an outside supply. (a) What is the speed of the water out of each hole? (b) Which water stream has the greatest range relative to the base of the container? Justify your answer.
65. In Conceptual Example 9.14, it was explained why a stream of water from a faucet necks down into a smaller cross-sectional area as it descends. Suppose at the top of the stream it has a cross-sectional area of 2.0 cm2, and a vertical distance 5.0 cm below the cross-sectional area of the stream is 0.80 cm2. What is (a) the speed of the water and (b) the flow rate?
66. Water flows at a rate of 25 L/min through a horizontal 7.0-cm-diameter pipe under a pressure a pressure of 6.0 Pa. At one point, calcium deposits reduce the cross-sectional area of the pipe to 30 cm2. What is the pressure at this point? (Consider the water to be an ideal fluid.)
67. As a fire-fighting method, a homeowner in the deep woods rigs up a water pump to bring water from a lake that is 10.0 below the level of the house. If the pump is capable of producing a gauge pressure of 140 kPa, at what rate (in L/s) can water be pumped to the house assuming the hose has a radius of 5.00 cm?
68. A Venturi meter can be used to measure the flow speed of a liquid. A simple such device is shown in Fig. 9.44. Show that the flow speed of an ideal fluid is given by v1 = √(2gΔh/((A1/A2)2 – 1)).
69. The pulmonary artery, which connects the heart to the lungs, is about 8.0 cm long and has an inside diameter of 5.0 mm. If the flow rate in it is to be 25 mL/s, what is the required pressure difference over its length?
70. A hospital patient receives a quick 500-cc blood transfusion through a needle with a length of 5.0 cm and an inner diameter of 1.0 mm. If the blood bag is suspended 0.85 m above the height where the blood where the blood first starts to flow into the vein, how long does the transfusion take?
71. A nurse needs to draw 20.0 cc of blood from a patient and deposit it into a small plastic container whose interior is at atmospheric pressure. He inserts the needle end of a long tube into a vein where the average gauge pressure is 30.0 mm Hg. This allows the internal pressure in the vein to push blood into the collection container. The needle is 0.900 mm in diameter and 2.54 cm long. The long tube is wide and smooth enough that we can assume its resistance is negligible, and that all the resistance to blood flow occurs in the narrow needle. How long does it take him to collect the sample?
72. What is the difference in volume (due only to pressure changes, and not temperature or other factors) between 1000 kg of water at the surface (assume 4 ºC) of the ocean and the same mass at the deepest known depth, 8.00 km? (Mariana Trench, assume 4ºC also.)
Online homework help for college and high school students. Get homework help and answers to your toughest questions in math, algebra, trigonometry, precalculus, calculus, physics. Homework assignments with step-by-step solutions. | 15,448 | 59,898 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2017-26 | longest | en | 0.859273 |
https://socratic.org/questions/sally-bought-three-chocolate-bars-and-a-pack-of-gum-and-paid-1-75-jake-bought-tw | 1,580,008,423,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251684146.65/warc/CC-MAIN-20200126013015-20200126043015-00547.warc.gz | 664,480,107 | 6,934 | Sally bought three chocolate bars and a pack of gum and paid $1.75. Jake bought two chocolate bars and four packs of gum and paid$2.00. Write a system of equations. Solve the system to find the cost of a chocolate bar and the cost of a pack of gum?
May 29, 2018
Cost of a chocolate bar: $0.50 Cost of a pack of gum:$0.25
Explanation:
Write 2 systems of equations. use $x$ for the price of chocolate bars bought and $y$ for the price of a pack of gum.
3 chocolate bars and a pack of gum cost $1.75. $3 x + y = 1.75$Two chocolate bars and four packs of gum cost$2.00
$2 x + 4 y = 2.00$
Using one of the equations, solve for y in terms of x.
$3 x + y = 1.75$ (1st equation)
$y = - 3 x + 1.75$ (subtract 3x from both sides)
Now we know the value of y, plug it into the other equation.
$2 x + 4 \left(- 3 x + 1.75\right) = 2.00$
Distribute and combine like terms.
$2 x + \left(- 12 x\right) + 7 = 2.00$
$- 10 x + 7 = 2$
Subtract 7 from both sides
$- 10 x = - 5$
Divide both sides by -10.
$x = 0.5$
The cost of a chocolate bar is $0.50. Now we know the price of a chocolate bar, plug it back into the first equation. $3 \left(0.5\right) + y = 1.75$$1.5 + y = 1.75$Distribute and combine like terms $y = 0.25$Subtract 1.5 from both sides. The cost of a pack of gum is $0.25
May 29, 2018
$1 for 1 chocolate$0.75 for 1 gum
Explanation:
The set up for the system equations is this:
$x + y = 1.75$
$2 x + 4 y = 2$
where $x$ is chocolate and $y$ is gum
To solve the system of equations, we need to solve for the system of equations for the value of one of the variables. To do that, we must manipulate both equations so that one of the variables can be eliminated (in the image below, I chose to eliminate $x$).
After we have one variable (in the image we found the $y$ value), we can plug it into ANY of the equations to find the other variable. | 597 | 1,845 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 22, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.96875 | 5 | CC-MAIN-2020-05 | longest | en | 0.897848 |
https://reviewgamezone.com/preview.php?id=27074 | 1,516,652,571,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084891539.71/warc/CC-MAIN-20180122193259-20180122213259-00553.warc.gz | 819,026,793 | 4,941 | # Measurement: Question Preview (ID: 27074)
### Below is a preview of the questions contained within the game titled MEASUREMENT: Measurement Practice Questions .To play games using this data set, follow the directions below. Good luck and have fun. Enjoy! [print these questions]
What is the basic unit for measuring length?
a) gram
b) meter
c) liter
d) Newton
What unit would you use to measure your mass?
a) grams
b) mg
c) kg
d) m
What unit would you use to measure th mass of 20 pennies?
a) g
b) mL
c) mg
d) kg
This is the measure of how long an object is.
a) weight
b) length
c) volume
d) mass
The amount of matter in an object is the ___________.
a) volume
b) weight
c) mass
d) density
This is what we call the international sytem of measures.
a) meters
b) cm
c) length
d) SI
What unit would you use to measure the amount of water in a bathtub?
a) liter
b) mL
c) gram
d) kg
6.492 m= _______ cm
a) 6,492 cm
b) 649.2 cm
c) .6492 cm
d) 64.92 cm
What is the metric scale for measuring temperature?
a) Fahrenheit
b) spring scale
c) thermometer
d) Celsius
What do you call the amount of mass in a given volume?
a) mass
b) volume
c) density
d) temeprature
What is the degree of hotness or coldness?
a) length
b) volume
c) temperature
d) mass
29 kg = ______ g
a) 29,000 g
b) 2,900 g
c) .029 g
d) 2.9 g
What is the measure of the force of gravity called?
a) volume
b) weight
c) mass
d) length
657 mL= _____ L
a) 65,700 L
b) 65.7 L
c) .657 L
d) 6.57 L
Based on density, which container would suspend in a tub of water?
a) .23 g/mL
b) 2.47 g/mL
c) 1.0 g/mL
d) .76 g/mL
What tool would you use when measuring length?
a) meter stick
b) spring scale
d) triple beam balance
What do you call the amount of space occupied in 3 dimensions?
a) density
b) mass
c) weight
d) volume
32 cm = ___ mm
a) 3.2 mm
b) .32 mm
c) 320 mm
d) 3,200 mm
68 m = ___ mm
a) 6.8 mm
b) .068 mm
c) 680 mm
d) 6,8000 mm
What unit would you use when measuring mass?
a) liter
b) mL
c) gram
d) g/mL
Play Games with the Questions above at ReviewGameZone.com
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TEACHERS / EDUCATORS | 742 | 2,273 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2018-05 | latest | en | 0.780998 |
http://tutorials.topstockresearch.com/TutorialsOnMovingAverage/TutorialOnMovingAveragePage.html | 1,498,665,182,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128323711.85/warc/CC-MAIN-20170628153051-20170628173051-00438.warc.gz | 416,790,516 | 9,349 | ### Related Tutorial On Technical Analysis
Calculate Moving Average When to avoid MA Trading With MA
# Learn basic of Moving Average and interpret its signals
What is Moving Average? It is the average price of a stock over a period of time.
Why are Moving Averages Used? Moving average helps us to understand the trends of the market. It also gives us buy or sell signal. Moving averages are commonly used to predict areas of support and resistance. The price movement is also smoothed out so that the traders can better understand the trends.
What are different kinds of Moving averages ?
a) Kinds of Moving averages are:
b) Simple Moving Average(SMA)
c) Exponential Moving Average(EMA)
d) Weighted Moving Average(WMA)
e) Cumulative Moving Average(CMA)
Though there are dozens of moving averages but the most common moving averages used by technical analysts is simple moving average and exponential moving average. Therefore we will limit the scope of this tutorial to simple moving average and exponential moving average.
Interpreting buy and sell signal using moving averages. Moving average provides buy or sell signal. A trader can choose whether to buy or sell depending on the crossover of moving average and the current price lines on a chart.
Long term investor looks for long term moving average and short term trader looks for short term moving average. A buy signal is generated when the security's price rises above its moving average and a sell signal is generated when the security's price falls below its moving average.
What are commonly used moving averages period?
Their are many moving averages right from short moving average to high moving average but most common of them are 15 days, 50 days, 100 days and 200 days.
How to choose a Moving Average period ?
There is no magical Moving average period. Use of any moving average period will provide similar result. For example both 60 day moving average and 70 day moving average will provide similar results but will give buy/sell signal at slightly different time. The key is to stick to one MA otherwise your trade can get messed up. One of the common way to choose a MA is to divide the period of interest by 2(two).
For Example, A trader looking at a stock movement for 40 days then he will choose 20 day moving average is suitable. If you are looking at a stock movement for 20 days then a 10 day moving average is suitable and so on. Generally it is preferable to stick to 15, 50, 100 and 200 days moving averages as they are used by lots of big investors and market reaction to these averages is more likely.
#### The convention that TopStockResearch will follow to represent different moving averages.
a) Red line represents 15 day Moving Average. b) Blue line represents 50 day Moving Average.
c) Green line represents 100 day Moving Average. d) Yellow line represents 200 day Moving Average.
## Example Showing Price verses Different Moving Averages: SAIL
Although they are highly simple to use and plot in our system.But like everything, Moving Average do have certain drawbacks to deal with.Some of the disadvantages of Moving Averages are:
• Moving averages lags the current price since they are the data from the past.
• It's ineffective in sideways market.
• Since it's slow it misses turning points and trends.
Due to above shortcoming analysts have designed more complicated yet effective way to deal with price and moving average.One of them is Exponential Moving Average.
Exponential Moving Average: Another kind of moving average is exponential moving average (EMA). There are certain advantages of EMA like it considers the the importance on the most recent stock prices than the earlier stock prices.
Our website provides free Stock screening based on Moving Average(SMA/EMA) support, resistance, crossover. It can be found at: Stock Screening based on SMA/EMA Support/Resistance and Crossovers | 790 | 3,911 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2017-26 | longest | en | 0.926151 |
https://brainmass.com/math/derivatives/application-of-derivatives-and-differentiation-rate-of-change-of-volume-of-a-sphere-balloon-52737 | 1,708,501,519,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473401.5/warc/CC-MAIN-20240221070402-20240221100402-00093.warc.gz | 147,259,615 | 6,837 | Purchase Solution
# Application of Derivatives and Differentiation : Rate of Change of Volume of a Sphere ( Balloon )
Not what you're looking for?
Air is being pumped into a spherical balloon so that the radius is increasing at the rate of dr/dt = 3 inches per second. What is the rate of change of the volume of the balloon in cubic inches per second, when r = 8 inches? Hint: V = 4/3(pi)r^3
##### Solution Summary
Rate of Change of Volume of a Sphere ( Balloon ) is calculated. The solution is detailed and well presented. The response received a rating of "5/5" from the student who originally posted the question.
##### Solution Preview
We have,
Volume of the spherical balloon as, V = (4/3) pi r^3
Now, differentiating it w.r.t time t, we get,
dV/dt = (4/3) pi *( 3 r^2) * dr/dt ( ...
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This is a short quiz to check your understanding of multiplication of complex numbers in rectangular form.
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This quiz helps you easily identify a function and test your understanding of ranges, domains , function inverses and transformations.
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Understanding of how geometry applies to in real-world contexts
##### Probability Quiz
Some questions on probability | 318 | 1,427 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2024-10 | latest | en | 0.886154 |
http://mathhelpforum.com/calculus/160675-projectile-cruves-problem.html | 1,526,943,378,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794864558.8/warc/CC-MAIN-20180521220041-20180522000041-00345.warc.gz | 193,597,726 | 10,471 | # Thread: Projectile Cruves problem
1. ## Projectile Cruves problem
Hi
The following question i am having problem with:
A tennis ball is served horizontally from 2.5m above the ground at 30m/sec. The net is 12m away and 0.9m high.
a) Find its velocity (v) at any time t seconds
This is what i done:
a=-gj
find the integral to get velocity
v=c_1 i + (-gt+c_2)j
v=12.6i+(-9.8t+2.5)j
P.S
2. the position and height of the net have nothing to do with the ball's velocity unless the ball hits it.
v(t) = 30i - (gt)j
$\displaystyle 0 \le t < \sqrt{\frac{5}{g}}$
3. Hello, Paymemoney!
A tennis ball is served horizontally from 2.5 m above the ground at 30 m/sec.
a) Find its velocity (v) at any time t seconds.
This is a Projectile problem; there is vertical and horizontal displacement.
The horizontal position is: .$\displaystyle x \:=\:30t$
The vertical position is: .$\displaystyle y \:=\:2.5 - 4.9t^2$
The velocity is a vector; it has magnitude and direction.
. . The magnitude is: .$\displaystyle |\vec v| \;=\;\sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}$
. . The direction is: .$\displaystyle \theta \;=\;\tan^{-1}\left(\dfrac{\frac{dy}{dt}}{\frac{dx}{dt}}\right )$
4. So the velocity would be $\displaystyle 30-\sqrt{9.8t}$
5. Originally Posted by Paymemoney
So the velocity would be $\displaystyle 30-\sqrt{9.8t}$
no, it would not. | 440 | 1,378 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2018-22 | latest | en | 0.802597 |
https://www.physicsforums.com/threads/bike-motorcycle-leaning-in-a-curve-what-happens-to-the-normal-force-n.649369/ | 1,513,226,883,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948539745.30/warc/CC-MAIN-20171214035620-20171214055620-00406.warc.gz | 760,788,973 | 21,025 | # Bike/motorcycle leaning in a curve. What happens to the normal force N?
1. Nov 3, 2012
### fisico30
Hello Forum,
If a bike is negotiation a curve, the static friction force will supply the centripetal force F_c.
If we go too fast and the friction is not enough we skid along the tangent.
There are two forces: the weight W=mg pointing down, the normal (contact) force N pointing up and equal to the weight W, and the static friction 0<=f_s<= mu_s*(N)= m_s*(mg).
If we go fast, we automatically lean into the curve. Why?
Some free body diagrams show the normal force being inclined w.r.t the pavement and pointing along the rider axis. How can that be? The normal force is called normal because it is perpendicular to the floor.
Does the leaning increase the normal force somehow? If we increase N we automatically increase f_s and supply a larger centripetal force. Maybe there is "camber thrust": the normal force gets inclined because there is a deformation of the tires.....
Or does the leaning simply help creating a torque about point of contact due to the weight W that counterbalance the "centrifugal" force torque (centrifugal force does not exist really, only in noninertial frames).... Otherwise, how come the bike does not fall if the CM is outside the base of support (wheels)?
Thanks,
fisico30
2. Nov 3, 2012
### Staff: Mentor
Let's analyze the situation in the inertial reference frame of the ground. There are two forces that matter, both acting on the bottom of the bike tire:
1. A sideways frictional force points towards the center of the circular path (centripetal force). Let's say the center of the circle is to your left, then this force is to your left.
2. The upward normal force is equal in magnitude to the weight (gravitational force) of you and the bike.
Suppose you try to stay upright, without leaning. Let's look at the torques produced by both forces, using the center of mass of you and the bike as the center of rotation.
1. The frictional force produces a clockwise torque, as seen from the rear of the bike.
2. The normal force produces zero torque, because it acts along a line that passes through the center of mass.
Net result: a clockwise torque which makes you fall over to your right.
Now suppose you lean to your left.
1. The frictional force still produces a clockwise torque.
2. The normal force no longer acts along a line through the center of mass. It produces a counterclockwise torque.
When you lean at a certain angle, the two torques are equal in magnitude and opposite in direction, giving zero net torque. This angle depends on the magnitude of the frictional force, which in turn depends on your speed going around the curve.
3. Nov 3, 2012
### bengunn
Can the normal force during the turn, be figured as ((W)(sine(theta))
Last edited: Nov 3, 2012
4. Nov 3, 2012
### Staff: Mentor
The upward normal force always equals W (=mg). It has to, otherwise you'd sink straight down into the road or levitate upward off the road. (Unless there's another vertical force acting on the bike besides the normal force and gravity.)
If θ is the angle between the bike and the vertical direction, then W sin θ is the component of N perpendicular to the bike, that is, the component that produces a counterclockwise torque in the setup described above.
Last edited: Nov 3, 2012
5. Nov 3, 2012
### bengunn
ok Thats what I thought, I got lost on the torque aspect and the best I could remember was that an object leaning had a normal force, but was dependent on the angle of the lean to the surface. in this case W sine theta thanks
6. Nov 4, 2012
### fisico30
Hello Jtbell,
thanks for your inputs. I agree with you but I have been corrected on the following points:
So, from the ground (inertial frame) there are only two forces: gravity W (pointing down), the normal force N (pointing up and perpendicular to the floor) and the static friction force f_s pointing radially inward.
I thought, like you, that the leaning of the bike or motorcycle has the only purpose to counteract the clockwise torque produced by the static friction about the point of contact. It is all about stability. The leaning does not increase the centripeta force whose max value is mu_s (N).
But some told me about the camber thrust: the tire gets deformed and provides some extra radially inward force to add to the friction force.
Also, some draw the normal force N at an angle with respect to the vertical....Is that incorrect? Normal means perpendicular. But does the normal force became oblique due to camber thrust which increases the normal force value, hence the static friction, hence the centripetal force?
thanks
fisico30
7. Nov 4, 2012
### rcgldr
I'm assuming a flat (non-banked) road in my post.
The usual usage of the term normal force refers to a vertical force related to gravity. During transitions in lean angle where the center of mass of the motorcycle and rider is accelerating, the normal force can change temporarily, lesser if the center of mass is accelerating downwards, greater if the center of mass is accelerating upwards.
Camber thrust is one of the components of the centripetal force that causes the motorcycle to follow a curved path. The predicted radius based on camber thrust is much smaller than the actual radius of a cornering motorcycle, so the effect is not large, at least not for a motorcyle. If you create a rig with two cone shaped "tires", one in front of the other (like a motorcycle), and if the axis of the two "cones" is parallel, the rig will tend to move in a nearly straight line, with a lot of slippage across the contact patch of the cones.
The more common explanation is that the front tire is turned "inwards" of the rear tire (once a stable lean is established). If you draw imaginary line extensions from the axis of the front and rear tire, they will cross at some point below the surface of a road. If you then have a vertical line from where those axis lines cross back to the surface of a road, then where that vertical line meets the surface of the road will be the approximate center of the circular path that the motorcycle follows. Due to deformation at the contact patch of the tires, the actual radius will be slightly larger.
Last edited: Nov 4, 2012
8. Nov 5, 2012
### fisico30
Thanks rcgldr!
In an old post, you said:
"With a car of the same weight, the larger the contact patch area, the lower the load per unit area of the contact patch, which translates into somewhat higher grip, up to a point, due to something called tire load sensitivity, where the coefficient of friction decreases the with amount of load. There's a point where increasing contact patch area beyond some near optimal size has little additional effect."
1) So a large contact patch result in a lower pressure (load per unit area). Why does that result in a higher grip? Are you saying that after a certain contact patch area we start losing friction (traction)? Is it because the pressure becomes too low?
2) in regards to a bike/bicycle leaning into a curve, just to keep things straight:
From the net:
Cornering Force
Opposing lateral force is cornering force that the tyres create when you turn the wheel into a corner. By completing the turn, cornering force has overcome lateral force. The important thing here is that tyres, and only tyres, can generate cornering force, suspension systems can only affect how tyres generate and share that cornering force.
Camber Thrust
The second way the tyres generate cornering force is by camber thrust. Positive camber is when the top of the wheel leans outwards; negative camber is when the top of the wheel leans inwards. Camber thrust is the force that moves the tyre in the direction it is leaning. Like slip angle, the greater the camber angle, the greater the camber thrust (cornering force) generated in the direction the tyre is leaning.
So, does the normal force go from being vertical to being at an angle? If so, what is the normal force perpendicular to in that case?
Does the action of leaning the bike increase the necessary centripetal force? How?
How many forces do we have? Weight W, normal force N, static friction f_s and camber thrust?
thanks
fisico30
9. Nov 5, 2012
### rcgldr
Because the coefficient of friction decreases as the load factor increases. Wiki article:
The wiki article mentions that maximum horizontal force is proportional to the normal force raised to somwhere from .7 to .9 power. If there was no load sensitivity, then it would be just the normal force (or normal force raised to 1.0 power).
No, only that there's a point of diminishing returns. Also a larger tire involves more mass and in an open wheel car, more aerodynamic drag. Open wheel race cars normally have smaller front tires than rear tires, while closed body race cars like Lemans prototype cars have front and rear tires of similar size.
There's also the issue of heat dissipation, and having a larger tire (taller or wider) provides more surface area for heat dissipation.
For a non-downforce race car, the camber is set so to keep the surface of the tire closer to parallel to the road surface when under cornering loads. Without the camber setting, the cornering load would lift the inside (of the curve) portion of the tire away from the road. For a car that turns left and right, a bit of negative camber is used on both left and right sides to accomplish this, with a bit less camber on the rear tires (since they also propel the car forwards). In the case of Nascar race cars on a track where there are only left turns, the left side tires will use positive camber (all tires lean a bit to the left). The common way to check the camber setting is to monitor the tire temperatures at the inside, middle, and outside. For a non-downforce race car, the goal is to get the temperatures somewhat even, so camber thrust is not a factor for these cars.
For a high downforce race car, such as Formula 1, the camber is set a bit more negative, so that the tire's inside temperatures are a bit higher than the outside temperatures. I'm not sure if the reason for this is to produce camber thrust as opposed to other reasons for the extra bit of negative camber setting.
The normal force is normally defined to be the force perpendicular to the surface of a road, which may be banked.
Leaning the bike is primarily done for balance, so that the bike doesn't fall inwards or outwards during a turn. It's common for motorcycle racing riders to hang off on the inside, so that the bike leans less than it would otherwise (note that this would decrease any camber thrust effect).
Camber thrust is a component of the static friction force.
Last edited: Nov 5, 2012
10. Nov 6, 2012
### fisico30
Thanks rcgldr,
"The load sensitivity of most real tires in their typical operating range is such that the coefficient of friction decreases as the vertical load, Fz, increases. The maximum lateral force that can be developed does increase as the vertical load increases, but at a diminishing rate."
So we have two types of grip: grip in the linear, forward direction (good for acceleration and breaking) and lateral grip (good for stability in curves).
Why do you think the the forward grip decreases with increase in the load? Conceptually, I would think it would increase since more contact force is developed.....
fisico30
11. Nov 6, 2012
### rcgldr
True, but the wiki article doesn't cover this. Note that Fz is a reference to the normal force, not forward grip.
My previous posts and the wiki article don't state that the grip decreases, only that the coefficient of friction decreases with load, so grip increases but at a less than linear rate than the load.
12. Nov 7, 2012
### fisico30
Thanks again rcgldr.
I guess I was thinking that the coefficient of friction must always increase as more weight (more load) is applied, since the stickiness increases....
One last comment about camber thrust: as the wheel leans (negative camber), the tire deforms in such a way that it is AS IF the rim was on top of a mini banked curve provided by the deformed tire. That way, the normal force is now oblique to the floor, the same way it is when a car is on a banked curve.
Sure, the normal force between the tire and the floor is still vertical.....
The increase in grip due to camber thrust derives from an increase in normal force due to the tire deformation upon leaning.....That increase in normal force leads to an increase in static friction force, which provides a larger centripetal force to keep the bike from sliding out of the curve...
thanks
fisico30
13. Nov 8, 2012
### rcgldr
Archived article (so the images are gone), disputing camber thrust as a major contributor to conering on a bike:
bike1.htm
From another article:
When cornering at an angle of 45° and 70mph. the turn radius will be 327 ft. and at half that speed, 35mph., the radius will be a quarter of that or 82 ft. --- but as shown earlier the cone radius will be only 1.5 ft.
tyres.htm
Camber thrust effect should slightly reduce the amount of steering input required to make a turn, while deformation (twisting) at the contact patch results in a slip angle that slightly increases the radius of a turn for a given steering input. The slip angle at the front and rear tire may be different, and I'm not sure of the combined effects versus steering input required to hold a turn.
Last edited: Nov 8, 2012 | 3,050 | 13,469 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2017-51 | longest | en | 0.898348 |
http://www.citizendia.org/Ideal_class_group | 1,369,174,558,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368700795821/warc/CC-MAIN-20130516103955-00029-ip-10-60-113-184.ec2.internal.warc.gz | 388,319,590 | 10,151 | In mathematics, the extent to which unique factorization fails in the ring of integers of an algebraic number field (or more generally any Dedekind domain) can be described by a certain abelian group known as an ideal class group (or class group). Mathematics is the body of Knowledge and Academic discipline that studies such concepts as Quantity, Structure, Space and In Mathematics, a unique factorization domain (UFD is roughly speaking a Commutative ring in which every element with special exceptions can be uniquely written In Mathematics, an algebraic number field (or simply number field) F is a finite (and hence algebraic) Field extension of the In Abstract algebra, a Dedekind domain or Dedekind ring, named after Richard Dedekind, is an Integral domain in which every nonzero Proper An abelian group, also called a commutative group, is a group satisfying the additional requirement that the product of elements does not depend on their order (the If this group is finite, (as it is in the case of the ring of integers of a number field) then the order of the group is called the class number. In Group theory, a branch of Mathematics, the term order is used in two closely related senses the order of a group is The multiplicative theory of a Dedekind domain is intimately tied to the structure of its class group. For example, the class group of a Dedekind domain is trivial if and only if the ring is a unique factorization domain. In Mathematics, a unique factorization domain (UFD is roughly speaking a Commutative ring in which every element with special exceptions can be uniquely written
## History and origin of the ideal class group
The first ideal class groups encountered in mathematics were part of the theory of quadratic forms: in the case of binary integral quadratic forms, as put into something like a final form by Gauss, a composition law was defined on certain equivalence classes of forms. In Mathematics, a quadratic form is a Homogeneous polynomial of degree two in a number of variables Johann Carl Friedrich Gauss (ˈɡaʊs, Gauß Carolus Fridericus Gauss ( 30 April 1777 – 23 February 1855) was a German This gave a finite abelian group, as was recognised at the time. An abelian group, also called a commutative group, is a group satisfying the additional requirement that the product of elements does not depend on their order (the
Later Kummer was working towards a theory of cyclotomic fields. Ernst Eduard Kummer ( 29 January 1810 - 14 May 1893) was a German Mathematician. In Number theory, a cyclotomic field is a Number field obtained by adjoining a complex Root of unity to Q, the field of Rational numbers It had been realised (probably by several people) that failure to complete proofs in the general case of Fermat's last theorem by factorisation using the roots of unity was for a very good reason: a failure of the fundamental theorem of arithmetic to hold, in the rings generated by those roots of unity, was a major obstacle. Fermat's Last Theorem is the name of the statement in Number theory that It is impossible to separate any power higher than the second into two like In Mathematics, the n th roots of unity, or de Moivre numbers are all the Complex numbers that yield 1 when raised to a given power In Number theory, the fundamental theorem of arithmetic (or unique-prime-factorization theorem) states that every Natural number greater than 1 can be written In Mathematics, a ring is an Algebraic structure which generalizes the algebraic properties of the Integers though the rational, real Out of Kummer's work for the first time came a study of the obstruction to the factorisation. We now recognise this as part of the ideal class group: in fact Kummer had isolated the p-torsion in that group for the field of p-roots of unity, for any prime number p, as the reason for the failure of the standard method of attack on the Fermat problem (see regular prime). In the theory of Abelian groups the torsion subgroup AT of an abelian group A is the Subgroup of A consisting of all elements In Number theory, a regular prime is a certain kind of Prime number.
Somewhat later again Dedekind formulated the concept of ideal, Kummer having worked in a different way. Julius Wilhelm Richard Dedekind ( October 6, 1831 &ndash February 12, 1916) was a German mathematician who did important In Ring theory, a branch of Abstract algebra, an ideal is a special Subset of a ring. At this point the existing examples could be unified. It was shown that while rings of algebraic integers do not always have unique factorization into primes (because they need not be principal ideal domains), they do have the property that every proper ideal admits a unique factorization as a product of prime ideals (that is, every ring of algebraic integers is a Dedekind domain). This article deals with the ring of complex numbers integral over Z. In Number theory, the fundamental theorem of arithmetic (or unique-prime-factorization theorem) states that every Natural number greater than 1 can be written In Abstract algebra, a principal ideal domain, or PID is an Integral domain in which every ideal is principal i In Mathematics, a prime ideal is a Subset of a ring which shares many important properties of a Prime number in the Ring of integers In Mathematics, a ring is an Algebraic structure which generalizes the algebraic properties of the Integers though the rational, real In Abstract algebra, a Dedekind domain or Dedekind ring, named after Richard Dedekind, is an Integral domain in which every nonzero Proper The ideal class group gives some answer to the question: which ideals are principal ideals? The answer comes in the form all of them, if and only if the ideal class group (which is a finite group) has just one element. In Ring theory, a branch of Abstract algebra, a principal ideal is an ideal I in a ring R that is generated by a single
## Technical development
If R is an integral domain, define a relation ~ on nonzero fractional(!) ideals of R by I ~ J whenever there exist nonzero elements a and b of R such that (a)I = (b)J. In Abstract algebra, a branch of Mathematics, an integral domain is a Commutative ring with an additive identity 0 and a multiplicative identity 1 such This article sets out the set-theoretic notion of relation For a more elementary point of view see Binary relations and Triadic relations (Here the notation (a) means the principal ideal of R consisting of all the multiples of a. In Ring theory, a branch of Abstract algebra, a principal ideal is an ideal I in a ring R that is generated by a single ) It is easily shown that this is an equivalence relation. In Mathematics, an equivalence relation is a Binary relation between two elements of a set which groups them together as being "equivalent" The equivalence classes are called the ideal classes of R. In Mathematics, given a set X and an Equivalence relation ~ on X, the equivalence class of an element a in X Ideal classes can be multiplied: if [I] denotes the equivalence class of the ideal I, then the multiplication [I][J] = [IJ] is well-defined and commutative. In Mathematics, commutativity is the ability to change the order of something without changing the end result The principal ideals form the ideal class [R] which serves as an identity element for this multiplication. In Mathematics, an identity element (or neutral element) is a special type of element of a set with respect to a Binary operation on that Thus a class [I] has an inverse [J] if and only if there is an ideal J such that IJ is a principal ideal. In general, such a J may not exist and consequently the set of ideal classes of R may only be a monoid. In Abstract algebra, a branch of Mathematics, a monoid is an Algebraic structure with a single Associative Binary operation
However, if R is the ring of algebraic integers in an algebraic number field, or more generally a Dedekind domain, the multiplication defined above turns the set of fractional ideal classes into an abelian group, the ideal class group of R. This article deals with the ring of complex numbers integral over Z. In Mathematics, an algebraic number field (or simply number field) F is a finite (and hence algebraic) Field extension of the In Abstract algebra, a Dedekind domain or Dedekind ring, named after Richard Dedekind, is an Integral domain in which every nonzero Proper An abelian group, also called a commutative group, is a group satisfying the additional requirement that the product of elements does not depend on their order (the The group property of existence of inverse elements follows easily from the fact that, in a Dedekind domain, every non-zero ideal (except R) is a product of prime ideals. In Mathematics, the idea of inverse element generalises the concepts of negation, in relation to Addition, and reciprocal, in relation to In Mathematics, a prime ideal is a Subset of a ring which shares many important properties of a Prime number in the Ring of integers
The ideal class group is trivial (i. e. has only one element) if and only if all ideals of R are principal. In this sense, the ideal class group measures how far R is from being a principal ideal domain, and hence from satisfying unique prime factorization (Dedekind domains are unique factorization domains if and only if they are principal ideal domains). In Abstract algebra, a principal ideal domain, or PID is an Integral domain in which every ideal is principal i In Mathematics, a unique factorization domain (UFD is roughly speaking a Commutative ring in which every element with special exceptions can be uniquely written
The number of ideal classes (the class number of R) may be infinite in general. But if R is in fact a ring of algebraic integers, then this number is always finite. This is one of the main results of classical algebraic number theory.
Computation of the class group is hard, in general; it can be done by hand for the ring of integers in an algebraic number field of small discriminant, using a theorem of Minkowski. In Mathematics, an algebraic number field (or simply number field) F is a finite (and hence algebraic) Field extension of the In Mathematics, the discriminant of an Algebraic number field is a numerical invariant that loosely speaking measures the size of the ( Ring of integers Hermann Minkowski ( June 22 1864 – January 12 1909) was a Russian born German Mathematician, of Jewish This result gives a bound, depending on the ring, such that every ideal class contains an ideal of norm less than the bound. The norm of an ideal is defined in Algebraic number theory. Let K\subset L be two number fields with rings of integers O_K\subset O_L In general the bound is not sharp enough to make the calculation practical for fields with large discriminant, but computers are well suited to the task.
It was remarked above that the ideal class group provides part of the answer to the question of how much ideals in a Dedekind domain behave like elements. In Abstract algebra, a Dedekind domain or Dedekind ring, named after Richard Dedekind, is an Integral domain in which every nonzero Proper The other part of the answer is provided by the multiplicative group of units of the Dedekind domain, since passage from principal ideals to their generators requires the use of units (and this is the rest of the reason for introducing the concept of fractional ideal, as well). In Mathematics, a group is a set of elements together with an operation that combines any two of its elements to form a third element In Mathematics, a unit in a ( Unital) ring R is an invertible element of R, i
Define a map from K* to the set of all nonzero fractional ideals of R by sending every element to the principal (fractional) ideal it generates. This is a group homomorphism; its kernel is the group of units of R, and its cokernel is the ideal class group of R. In Mathematics, given two groups ( G, * and ( H, · a group homomorphism from ( G, * to ( H, · is a function In the various branches of Mathematics that fall under the heading of Abstract algebra, the kernel of a Homomorphism measures the degree to which the homomorphism The failure of these groups to be trivial is a precise measure of the failure of the map to be an isomorphism: that is the failure of ideals to act like ring elements, that is to say, like numbers.
The mapping from rings of integers R to their corresponding class groups is functorial, and the class group can be subsumed under the heading of algebraic K-theory, with K0(R) being the functor assigning to R its ideal class group; more precisely, K0(R) = Z×C(R), where C(R) is the class group. In Mathematics, algebraic K-theory is an advanced part of Homological algebra concerned with defining and applying a sequence K n Higher K groups can also be employed and interpreted arithmetically in connection to rings of integers.
## Examples of ideal class groups
The rings Z, Z[i], and Z[ω], (where i is a square root of -1 and ω is a cube root of 1) are all principal ideal domains, and so have class number 1: that is, they have trivial ideal class groups. If k is a field, then the polynomial ring k[X1, X2, X3, . In Mathematics, especially in the field of Abstract algebra, a polynomial ring is a ring formed from the set of Polynomials in one or more variables . . ] is an integral domain. It has a countably infinite set of ideal classes.
If d is a square-free integer (in other words, a product of distinct primes) other than 1, then Q(√d) is a finite extension of Q. In Mathematics, an element r of a Unique factorization domain R is called square-free if it is not divisible by a non-trivial square In particular it is a 2-dimensional vector space over Q, called a quadratic field. In Mathematics, a vector space (or linear space) is a collection of objects (called vectors) that informally speaking may be scaled and added In Mathematics, a quadratic field is an Algebraic number field K of degree two over Q. If d < 0, then the class number of the ring R of algebraic integers of Q(√d) is equal to 1 for precisely the following values of d: d = -1, -2, -3, -7, -11, -19, -43, -67, and -163. This result was first conjectured by Gauss and proven by Kurt Heegner, although Heegner's proof was not believed until Harold Stark gave a later proof in 1967, which Stark showed was actually equivalent to Heegner's. Johann Carl Friedrich Gauss (ˈɡaʊs, Gauß Carolus Fridericus Gauss ( 30 April 1777 – 23 February 1855) was a German Kurt Heegner (1893–1965 was a German private scholar from Berlin, who specialized in Radio Engineering and Mathematics. Harold Mead Stark (born 1939) is an American Mathematician, specializing in Number theory. (See Stark-Heegner theorem. In Number theory, a branch of Mathematics, the Stark–Heegner theorem states precisely which quadratic imaginary number fields admit unique factorisation ) This is a special case of the famous class number problem. In Mathematics, the Gauss class number problem ( for imaginary quadratic fields) as is usually understood is to provide for each n &ge 1 a complete
If, on the other hand, d > 0, then it is unknown whether there are infinitely many fields Q(√d) with class number 1. Computational results indicate that there are a great many such fields to say the least.
### Example of a non-trivial class group
The ring R = Z [√−5] is the ring of integers of Q(√−5). It does not possess unique factorisation; in fact the class group of R is cyclic of order 2. Indeed, the ideal
J = (2, 1 + √−5)
is not principal, which can be proved by contradiction as follows. If J were generated by an element x of R, then x would divide both 2 and 1 + √−5. Then the norm N(x) of x would divide both N(2) = 4 and N(1 + √−5) = 6, so N(x) would divide 2. In Mathematics, the (field norm is a mapping defined in field theory, to map elements of a larger field into a smaller one We are assuming that x is not a unit of R, so N(x) cannot be 1. It cannot be 2 either, because R has no elements of norm 2, that is, the equation b2 + 5c2 = 2 has no solutions in integers.
One also computes that J2 = (2), which is principal, so the class of J in the ideal class group has order two. Showing that there aren't any other ideal classes requires more effort.
The fact that this J is not principal is also related to the fact that the element 6 has two distinct factorisations into irreducibles:
6 = 2 × 3 = (1 + √−5) × (1 − √−5).
## Connections to class field theory
Class field theory is a branch of algebraic number theory which seeks to classify all the abelian extensions of a given algebraic number field, meaning Galois extensions with abelian Galois group. In Mathematics, class field theory is a major branch of Algebraic number theory. In Mathematics, an algebraic number field (or simply number field) F is a finite (and hence algebraic) Field extension of the In Mathematics, more specifically in Abstract algebra, Galois theory, named after Évariste Galois, provides a connection between field theory In Mathematics, an algebraic number field (or simply number field) F is a finite (and hence algebraic) Field extension of the In Mathematics, a Galois group is a group associated with a certain type of Field extension. A particularly beautiful example is found in the Hilbert class field of a number field, which can be defined as the maximal unramified abelian extension of such a field. In Algebraic number theory, the Hilbert class field E of a Number field K is the Maximal abelian Unramified In Mathematics, ramification is a geometric term used for 'branching out' in the way that the Square root function for Complex numbers can be seen The Hilbert class field L of a number field K is unique and has the following properties:
• Every ideal of the ring of integers of K becomes principal in L, i. e. , if I is an integral ideal of K then the image of I is a principal ideal in L.
• L is a Galois extension of K with Galois group isomorphic to the ideal class group of K.
Neither property is particularly easy to prove. | 4,039 | 18,062 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2013-20 | longest | en | 0.939931 |
https://solvedlib.com/a-college-gives-aptitude-tests-in-the-sciences,381339 | 1,657,199,296,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104692018.96/warc/CC-MAIN-20220707124050-20220707154050-00736.warc.gz | 567,422,244 | 17,422 | # A college gives aptitude tests in the sciences and the humanities to all incoming first year...
###### Question:
A college gives aptitude tests in the sciences and the humanities to all incoming first year students. If X and Y are the proportions of correct answers on the science and humanities exams, respectively, then the joint pdf of X, Y is given by: 1. otherwise a. Are X and Y independent? Why or why not? b. Find P(X <5.Y <^). What does this probability represent?
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https://www.enotes.com/homework-help/find-polynomial-degree-4-whose-graph-goes-through-421665 | 1,511,000,876,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934804724.3/warc/CC-MAIN-20171118094746-20171118114746-00314.warc.gz | 800,187,025 | 9,778 | # Find the polynomial of degree 4 whose graph goes through the points (-3,-218),(-1,4),(0,10),(2,22) and (3,-32).f(x) = ___x^4 + ____x^3 + _____x^2 +_____x + _______
tiburtius | Certified Educator
If you want to find function `f` whose graph goes through a point `(x,y)`. That means you need to find solution of equation `f(x)=y`. If you have several points that means you need to find solution of system of equations.
In your case you want to find polynomial `f(x)=ax^4+bx^3+cx^2+dx+e` and for that you need to solve the following system:
`(-3)^4a+(-3)^3b+(-3)^2c+(-3)d+e=-218`
`a-b+c-d+e=4`
`e=10`
`16a+8b+4c+2d+e=22`
`81a+27b+9c+3d+e=-32`
Now you have system of 5 linear equations with 5 variables which you can solve by using Gauss elimination or some other method. Solution to this system of equations is
`a=-2`, `b=3`, `c=3`, `d=4`, `e=10` | 309 | 857 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2017-47 | latest | en | 0.836445 |
http://stp.diit.edu.ua/article/view/9578 | 1,618,115,076,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038060927.2/warc/CC-MAIN-20210411030031-20210411060031-00561.warc.gz | 92,069,729 | 13,900 | CFD-model of the mass transfer in the vertical settler
Authors
• E. K. Nagornaya Department of Hydraulics, Prydneprovsk State Academy of Civil Engineering and Architecture
Keywords:
vertical settler, CFD model, numerical simulation, mass transfer
Abstract
Purpose. Nowadays the mathematical models of the secondary settlers are intensively developed. As a rule the engineers use the 0-D models or 1-D models to design settlers. But these models do not take into account the hydrodynamics process inside the settler and its geometrical form. That is why the CFD-models based on Navier - Stokes equations are not widely used in practice now. The use of CFD-models based on Navier - Stokes equations needs to incorporate very refine grid. It is very actually now to develop the CFD-models which permit to take into account the geometrical form of the settler, the most important physical processes and needs small computer time for calculation. That is why the development of the 2-D numerical model for the investigation of the waste waters transfer in the vertical settlers which permits to take into account the geometrical form and the constructive features of the settler is essential. Methodology. The finite - difference schemes are applied. Findings. The new 2-D-CFD-model was developed, which permits to perform the CFD investigation of the vertical settler. This model takes into account the geometrical form of the settler, the central pipe inside it and others peculiarities. The method of «porosity technique» is used to create the geometrical form of the settler in the numerical model. This technique permits to build any geometrical form of the settler for CFD investigation. Originality. Making of CFD-model which permits on the one hand to take into account the geometrical form of the settler, basic physical processes of mass transfer in construction and on the other hand requiring the low time cost in order to obtain results. Practical value. CFD-model is designed and code which is constructed on its basis allows at low cost of computer time and about the same as in the calculation of the 1-D model to solve complex multiparameter problems that arise during the design of vertical settlers with their shape and design features.
Author Biography
E. K. Nagornaya, Department of Hydraulics, Prydneprovsk State Academy of Civil Engineering and Architecture
Chernishevskogo, 24a, Dnepropetrovsk, 49600, Ukraine, tel. +380667521332, e-mail: ek_n@i.ua
References
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2013-02-25
How to Cite
Nagornaya, E. K. (2013). CFD-model of the mass transfer in the vertical settler. Science and Transport Progress. Bulletin of Dnipropetrovsk National University of Railway Transport, (1(43), 39–50. https://doi.org/10.15802/stp2013/9578
Section
ECOLOGY AND INDUSTRIAL SAFETY | 2,448 | 8,599 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2021-17 | latest | en | 0.930295 |
https://tex.stackexchange.com/questions/276370/using-the-calc-package-to-draw-a-polygon | 1,719,190,940,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198864968.52/warc/CC-MAIN-20240623225845-20240624015845-00359.warc.gz | 489,023,545 | 37,307 | # Using the calc package to draw a polygon
I am trying to draw a pentagon. The point A_1 is at the origin, and the point A_2 is at (60:1.75), expressed in polar coordinates. I would like A_3 to be 0.5cm from A_2 and perpendicular to line segment $\overline{A_1A_2}$, A_4 to be 1cm from A_3 and rotated 120 degrees from line segment $\overline{A_2A_3}$, and A_5 to be 0.75 cm from A_4 and rotated 120 degrees from line segment $\overline{A_3A_4}$. That is not what is drawn.
\documentclass{amsart}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{tikz}
\usetikzlibrary{calc,angles,positioning,intersections}
\begin{document}
\noindent \hspace*{\fill}
\begin{tikzpicture}
\path (0,0) coordinate (A_1)
(60:1.75) coordinate (A_2)
($(A_2) +($(A_2)!0.5cm!90:(A_1)$)$) coordinate (A_3)
($(A_3) +($(A_3)!1cm!120:(A_2)$)$) coordinate (A_4)
($(A_4) +($(A_4)!0.75cm!120:(A_3)$)$) coordinate (A_5);
\draw (A_1) -- (A_2) -- (A_3) -- (A_4) -- (A_5) -- cycle;
\draw[fill] (A_1) circle (1.5pt);
\draw[fill] (A_2) circle (1.5pt);
\draw[fill] (A_3) circle (1.5pt);
\draw[fill] (A_4) circle (1.5pt);
\draw[fill] (A_5) circle (1.5pt);
\end{tikzpicture}
\end{document}
Sometimes, less is more. (I added the nodes just to see what was going on.)
\documentclass{amsart}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{tikz}
\usetikzlibrary{calc,angles,positioning,intersections}
\begin{document}
\noindent \hspace*{\fill}
\begin{tikzpicture}
\path (0,0) coordinate (A_1)
(60:1.75) coordinate (A_2)
($(A_2)!0.5cm!90:(A_1)$) coordinate (A_3)
($(A_3)!1cm!120:(A_2)$) coordinate (A_4)
($(A_4)!0.75cm!120:(A_3)$) coordinate (A_5);
\draw (A_1) -- (A_2) -- (A_3) -- (A_4) -- (A_5) -- cycle;
\draw[fill] (A_1) circle (1.5pt) node[above] {A1};
\draw[fill] (A_2) circle (1.5pt) node[above] {A2};
\draw[fill] (A_3) circle (1.5pt) node[above] {A3};
\draw[fill] (A_4) circle (1.5pt) node[above] {A4};
\draw[fill] (A_5) circle (1.5pt) node[above] {A5};
\end{tikzpicture}
\end{document}
• I see that you removed the (A_i) + from the commands for coordinates of A_3, A_4, and A_5. Why is my code wrong for locating these three points? I thought that \coordinate (A_3) at ($(A_2) +($(A_2)!0.5cm!90:(A_1)$)$) places A_3 at 0.5cm from A_2 that is perpendicular to line segment $\overline{A_1A_2}$. Commented Nov 3, 2015 at 14:13
• ($(A)!1cm!90:(B)$) translates as "Start at (A) and travel 1cm in the direction 90 from (B)". Commented Nov 3, 2015 at 14:17
• Start at A and travel 1cm in the direction 90 degrees from line segment AB. Commented Nov 4, 2015 at 14:57
• @user74973 - If we are using strict geometry nomenclature, make that 90 degrees counter-clockwise from $\overrightarrow{AB}$ ;-) Commented Nov 4, 2015 at 15:26
• That will save me some time ... if I can remember it. Thanks! Commented Nov 4, 2015 at 16:04
Here is a short code from pst-node
\documentclass[x11names, border=3pt]{standalone}
\usepackage{auto-pst-pdf}
\pagestyle{empty}
\begin{document}
\begin{pspicture}
\psset{dimen=middle, unit=2, labelsep=0.8ex, linecolor=IndianRed3, linewidth=1.2pt}
\pnode(0,0){A}
\AplusB(A)(1.75;60){B}\AplusB(B)(.5;-30){C}\AplusB(C)(1;-90){D}\AplusB(D)(.75;-150){E}
\pspolygon(A)(B)(C)(D)(E)
\uput[l](A){$A$}\uput[u](B){$B$}\uput[ur](C){$C$}\uput[r](D){$D$}\uput[d](E){$E$}
\end{pspicture}
\end{document} | 1,327 | 3,304 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2024-26 | latest | en | 0.463795 |
https://nrich.maths.org/public/topic.php?code=-68&cl=2&cldcmpid=6998 | 1,606,726,229,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141211510.56/warc/CC-MAIN-20201130065516-20201130095516-00542.warc.gz | 406,500,648 | 9,180 | # Resources tagged with: Visualising
Filter by: Content type:
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### There are 255 results
Broad Topics > Thinking Mathematically > Visualising
### Part the Polygons
##### Age 7 to 11 Short Challenge Level:
Draw three straight lines to separate these shapes into four groups - each group must contain one of each shape.
### Hexagon Transformations
##### Age 7 to 11 Challenge Level:
Can you cut a regular hexagon into two pieces to make a parallelogram? Try cutting it into three pieces to make a rhombus!
### Regular Rings 2
##### Age 7 to 11 Challenge Level:
What shape is made when you fold using this crease pattern? Can you make a ring design?
### Inside Seven Squares
##### Age 7 to 11 Challenge Level:
What is the total area of the four outside triangles which are outlined in red in this arrangement of squares inside each other?
### Trice
##### Age 11 to 14 Challenge Level:
ABCDEFGH is a 3 by 3 by 3 cube. Point P is 1/3 along AB (that is AP : PB = 1 : 2), point Q is 1/3 along GH and point R is 1/3 along ED. What is the area of the triangle PQR?
### Overlapping Again
##### Age 7 to 11 Challenge Level:
What shape is the overlap when you slide one of these shapes half way across another? Can you picture it in your head? Use the interactivity to check your visualisation.
### Wrapping Presents
##### Age 7 to 11 Challenge Level:
Choose a box and work out the smallest rectangle of paper needed to wrap it so that it is completely covered.
### Reef and Granny
##### Age 7 to 11 Challenge Level:
Have a look at what happens when you pull a reef knot and a granny knot tight. Which do you think is best for securing things together? Why?
### Three Squares
##### Age 5 to 11 Challenge Level:
What is the greatest number of squares you can make by overlapping three squares?
### A Square in a Circle
##### Age 7 to 11 Challenge Level:
What shape has Harry drawn on this clock face? Can you find its area? What is the largest number of square tiles that could cover this area?
### Fred the Class Robot
##### Age 7 to 11 Challenge Level:
Billy's class had a robot called Fred who could draw with chalk held underneath him. What shapes did the pupils make Fred draw?
### Triangular Faces
##### Age 7 to 11 Challenge Level:
This problem invites you to build 3D shapes using two different triangles. Can you make the shapes from the pictures?
### Midpoint Triangle
##### Age 7 to 11 Challenge Level:
Can you cut up a square in the way shown and make the pieces into a triangle?
### Diagrams
##### Age 7 to 11 Challenge Level:
A group activity using visualisation of squares and triangles.
### Regular Rings 1
##### Age 7 to 11 Challenge Level:
Can you work out what shape is made by folding in this way? Why not create some patterns using this shape but in different sizes?
### Folding Flowers 2
##### Age 7 to 11 Challenge Level:
Make a flower design using the same shape made out of different sizes of paper.
### Folding Flowers 1
##### Age 7 to 11 Challenge Level:
Can you visualise what shape this piece of paper will make when it is folded?
### Making Maths: Rolypoly
##### Age 5 to 11 Challenge Level:
Paint a stripe on a cardboard roll. Can you predict what will happen when it is rolled across a sheet of paper?
##### Age 7 to 11 Challenge Level:
This practical problem challenges you to make quadrilaterals with a loop of string. You'll need some friends to help!
### Sponge Sections
##### Age 7 to 11 Challenge Level:
You have been given three shapes made out of sponge: a sphere, a cylinder and a cone. Your challenge is to find out how to cut them to make different shapes for printing.
### Two Squared
##### Age 7 to 11 Challenge Level:
What happens to the area of a square if you double the length of the sides? Try the same thing with rectangles, diamonds and other shapes. How do the four smaller ones fit into the larger one?
### Square Surprise
##### Age 5 to 11 Challenge Level:
Why do you think that the red player chose that particular dot in this game of Seeing Squares?
### Seeing Squares for Two
##### Age 5 to 11 Challenge Level:
Seeing Squares game for an adult and child. Can you come up with a way of always winning this game?
### World of Tan 28 - Concentrating on Coordinates
##### Age 7 to 11 Challenge Level:
Can you fit the tangram pieces into the outline of the playing piece?
### World of Tan 12 - All in a Fluff
##### Age 7 to 11 Challenge Level:
Can you fit the tangram pieces into the outlines of the rabbits?
### The Development of Spatial and Geometric Thinking: the Importance of Instruction.
##### Age 5 to 11
This article looks at levels of geometric thinking and the types of activities required to develop this thinking.
### Put Yourself in a Box
##### Age 7 to 11 Challenge Level:
A game for 2 players. Given a board of dots in a grid pattern, players take turns drawing a line by connecting 2 adjacent dots. Your goal is to complete more squares than your opponent.
### Twice as Big?
##### Age 7 to 11 Challenge Level:
Investigate how the four L-shapes fit together to make an enlarged L-shape. You could explore this idea with other shapes too.
### World of Tan 1 - Granma T
##### Age 7 to 11 Challenge Level:
Can you fit the tangram pieces into the outline of Granma T?
### Endless Noughts and Crosses
##### Age 7 to 11 Challenge Level:
An extension of noughts and crosses in which the grid is enlarged and the length of the winning line can to altered to 3, 4 or 5.
### Matchsticks
##### Age 7 to 11 Challenge Level:
Reasoning about the number of matches needed to build squares that share their sides.
### World of Tan 24 - Clocks
##### Age 7 to 11 Challenge Level:
Can you fit the tangram pieces into the outline of the clock?
### Construct-o-straws
##### Age 7 to 11 Challenge Level:
Make a cube out of straws and have a go at this practical challenge.
### World of Tan 13 - A Storm in a Tea Cup
##### Age 7 to 11 Challenge Level:
Can you fit the tangram pieces into the outlines of the convex shapes?
### World of Tan 18 - Soup
##### Age 7 to 11 Challenge Level:
Can you fit the tangram pieces into the outlines of Mah Ling and Chi Wing?
### World of Tan 6 - Junk
##### Age 7 to 11 Challenge Level:
Can you fit the tangram pieces into the silhouette of the junk?
### Framed
##### Age 11 to 14 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. . . .
### World of Tan 7 - Golden Goose
##### Age 7 to 11 Challenge Level:
Can you fit the tangram pieces into the outline of the plaque design?
##### Age 7 to 11 Challenge Level:
How many DIFFERENT quadrilaterals can be made by joining the dots on the 8-point circle?
### Fractional Triangles
##### Age 7 to 11 Challenge Level:
Use the lines on this figure to show how the square can be divided into 2 halves, 3 thirds, 6 sixths and 9 ninths.
### Nine-pin Triangles
##### Age 7 to 11 Challenge Level:
How many different triangles can you make on a circular pegboard that has nine pegs?
### Cube Paths
##### Age 11 to 14 Challenge Level:
Given a 2 by 2 by 2 skeletal cube with one route `down' the cube. How many routes are there from A to B?
### Jomista Mat
##### Age 7 to 11 Challenge Level:
Looking at the picture of this Jomista Mat, can you decribe what you see? Why not try and make one yourself?
### Seeing Squares
##### Age 5 to 11 Challenge Level:
Players take it in turns to choose a dot on the grid. The winner is the first to have four dots that can be joined to form a square.
### LOGO Challenge - Circles as Animals
##### Age 11 to 16 Challenge Level:
See if you can anticipate successive 'generations' of the two animals shown here.
### Putting Two and Two Together
##### Age 7 to 11 Challenge Level:
In how many ways can you fit two of these yellow triangles together? Can you predict the number of ways two blue triangles can be fitted together?
### Triangles in the Middle
##### Age 11 to 18 Challenge Level:
This task depends on groups working collaboratively, discussing and reasoning to agree a final product.
### Tangram Browser
##### Age 7 to 11
Explore our selection of interactive tangrams. Can you use the tangram pieces to re-create each picture?
### World of Tan 27 - Sharing
##### Age 7 to 11 Challenge Level:
Can you fit the tangram pieces into the outline of Little Fung at the table?
### World of Tan 11 - the Past, Present and Future
##### Age 7 to 11 Challenge Level:
Can you fit the tangram pieces into the outlines of the telescope and microscope? | 2,073 | 8,725 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2020-50 | longest | en | 0.906777 |
http://utter.chaos.org.uk/~eddy/math/pythagorean.html | 1,719,013,429,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862189.36/warc/CC-MAIN-20240621223321-20240622013321-00051.warc.gz | 31,497,681 | 8,905 | # Pythagorean triangles and simplices
The ancient greeks were particularly interested in one crucial implication of Pythagoras' theorem: the existence of lengths which are not rationally commensurate. Given a line, one can make arbitrarily long lines by repeating it and arbitrarily short lines by sub-dividing it into equal parts; combining these operations one may make a line of length arbitrarily close to any length one cares to nominate – but not all lengths can be arrived at from a given one by duplication and equal sub-division. Lines of lengths one can obtain from a given line by such scaling and shrinking are said to be rationally commensurate with the given line.
## An irrational example
Let's start by looking (as the greeks did) at the right-angle triangle whose two perpendicular sides are of equal length; we may as well use this as unit of length, so we can see that the square on the hypotenuse has area 2. Now, plainly, every natural greater than 1 has square bigger than 2; while the remaining two naturals, 0 and 1, are their own squares and both less than 2. The negative integers' squares are the same as those of the positives, so 2 clearly isn't the square of any whole number.
Now consider any ratio of whole numbers whose square is a whole number. There are necessarily two coprime whole numbers p, q for which the ratio is p/q; so we have p.p = n.q.q for some whole number n. As both sides of this equation are whole numbers, we can prime-factorise both sides. Every prime factor that divides either side divides the left side, p.p, hence also p; and p is coprime to q, so no prime factor of either side divides q; but every prime factor of q divides the right side, n.q.q; so we must infer that q has no prime factors: i.e. q = 1. But then our ratio p/q is a whole number; so a rational whose square is a whole number must in fact be a whole number.
So not only is 2 not the square of any whole number but, as a result, it's also not the square of any ratio of whole numbers; a right-angle triangle whose perpendicular sides are equal has a hypotenuse that isn't rationally commensurate with them. Such a number, that isn't the ratio of any whole numbers, is described as irrational – as it happens, there are vastly more of these than there are of the rationals; but that's a topic for another page. What matters, for now, is that a right-angle triangle, whose perpendicular sides are of equal length, has a hypotenuse whose length isn't rationally commensurate with this length. The rest of this page is devoted to the study of right-angle triangles – and their higher-dimensional peers, the simplices – whose sides are rationally commensurate.
## Pythagorean triangles
I'll describe a right-angle triangle as pythagorean precisely if its sides' lengths are rationally commensurate. (One may likewise pose the corresponding question for non-right angles.)
If one length is rationally commensurate with another, duplicating the first one number of times and the second another number of times will produce lines of equal length, if we chose the numbers correctly; equivalently, sub-dividing each into suitably many equal pieces will yield a unit for which each is obtained by duplicating the unit. Thus being rationally commensurate is a symmetric relationship. If lengths K and L are rationally commensurate and L is also rationally commensurate with M, then there are some whole numbers k and m for which dividing L into k equal parts gives a length that, when duplicated enough, yields K; while dividing L into m equal parts gives a unit of which M is a multiple. So dividing L into k.m equal parts gives a length of which both K and M are multiples; thus K and M are also rationally commensurate and rationally commensurate is transitive. Since it is fatuously reflexive (every length is rationally commensurate with itself) it is thus an equivalence relation.
To classify pythagorean triangles, it suffices to consider the case where our right-angle triangle's lengths are all whole multiples of some given unit; this reduces our description of the triangle to three whole numbers. The question thus reduces to finding cases where one perfect square (that is, the square of a whole number) is equal to the sum of two others.
Scaling a pythagorean triangle necessarily produces a pythagorean triangle; just scale the unit used for the original to get a unit that describes the scaled triangle's sides using the same numbers, whose squares haven't changed so still satisfy Pythagoras's equation. Sub-dividing our unit likewise just increases the numbers we use to describe the lengths of the edges, all by some common factor i, a whole number; if n.n +m.m = k.k then, i.n.i.n +i.m.i.m = i.k.i.k, so any whole-number scaling of a solution to our problem is, trivially, another solution. So we can scale up our three whole numbers by a common factor to turn one solution into another, albeit one that's not interestingly different.
Given that we can do this, it's sufficient to classify the solutions that have no common factor, since all others can be obtained from these by such scalings. If two of the sides share a common factor i then their squares have i.i as a common factor; hence the sum or difference of thsese squares, which is the square of the other side, also has i.i as a factor; hence the other side has i as a factor. Thus any common factor of two of the sides is in fact a common factor of all three, so we only need to consider cases where no two sides have any common factor; the three sides are mutually comprime.
In particular, at most one of the sides has two as a factor so at least two sides have odd lengths. Any odd number is 2.n +1 for some whole number n; its square is then (2.n).(2.n) +2.(2.n) +1 = 4.n.n +4.n +1 = 4.n.(n+1) +1. Of any two adjacent whole numbers, such as n and n+1, one is odd and the other is even; so n.(n+1) is even and 4.n.(n+1) is a multiple of eight. Thus the square of any odd number is one more than a multiple of eight; so the two odd sides have squares whose difference is a multiple of eight and whose sum is two more than a multiple of eight; one of these is the square of the other side, which thus isn't one more than a multiple of eight, so that other side isn't odd. Those even numbers that are twice an odd number have squares that are four more than a multiple of (32 hence also of) eight; all other even numbers are multiples of four so have squares that are multiples of 16, hence in particular of eight. So our pythagorean equation n.n +m.m = k.k, in which two terms are one more than multiples of eight, must have these two terms on opposite sides of the equation with the remaining term equal to a multiple of eight; i.e. the hypotenuse is odd, as is one of the other two sides, and the remaining side's square is a multiple of eight.
If we write out the prime factorisation of some side and double the multiplicity of each prime as a factor in it, we get a prime factorisation of the side's square in which every prime's multiplicity is even. As prime factorisation of integers is unique, any factorisation of the square of a side must have only even powers of each prime used. So the squared side with a factor of eight, hence at least three factors of two, must in fact have at least four (the smallest even that's at least three) factors of two and the side in question must have two factors of two, hence be a multiple of four. So our hypotenuse is odd, as is one of the other sides, and the remaining side is a multiple of four.
Suppose, then, that m is the multiple of four, with k and n odd; we have n.n = k.k −m.m = (k +m).(k −m) with k, n, m coprime. As k and m are coprime, k ±m are also coprime (had k and m both been odd, k ±m would have shared two as a common factor; but this is not the case); and their product is n.n, a perfect square. With {x, y} = {k ±m} in either order, any prime factor p of x is a prime factor of x.y = n.n hence also of n, hence has even multiplicity as a factor of n.n = x.y; but p is not a factor of y, hence p in fact has even multiplicity as a factor of x; thus the prime factorisation of x only has even multiplicities for its prime factors and x is a perfect square. Thus k ±m are mutually coprime odd perfect squares.
In particular, the two odd numbers whose squares are k ±m have product n; and we can obtain k and m from them as the mean and half-difference of their squares. Thus, for any coprime pythagorean triangle, there are naturals u, v with v > u for which k +m = 4.v.(v+1) +1, k −m = 4.u.(u +1) +1 and:
• k = 2.(v.(v +1) +(u +1).u) +1,
• m = 2.(v.(v +1) −(u +1).u) and
• n = (2.v +1).(2.u +1).
Furthermore, for any naturals u, v, we can compute k, m and n as above; they might not be coprime, but they do always satisfy n.n = (k +m).(k −m) = k.k −m.m hence k.k = n.n +m.m, so they do describe a pythagorean triangle, even if it's not coprime. (For example, [v,u] = [4,1] yields [45,36,27] which just scales the familiar [5,4,3] from [v,u] = [1,0] by a factor of 9.) Every pythagorean triangle is a whole multiple of some triangle generated in this way. Note that the hypotenuse, k, is always one more than a multiple of four (since both u.(u+1) and v.(v+1) are even). The above formulae can be re-written, by a substitution p = v+u+1, q = v−u, as p.p +q.q, 2.p.q and p.p −q.q, respectively, with p > q and p−q odd; for the three sides to be coprime, p and q must be coprime; this reformulation is known as Euclid's formula for generating pythagorean triangles. If your browser supports SVG (and CSS floats), you should see the first 58 distinct hypotenuses (with short side horizontal and intermediate side vertical) depicted at left (possibly above); these are all the examples with 0 ≤ u < v < 12.
In particular, this means that the set of pairs of rationals whose sum is 1, i.e. the set of points on the unit circle of the two-dimensional rational plane, is countable – it is a union of two images (using the perpendicular sides above in each order) of the set of pairs of naturals [u, v], via the combinations of them used to make the two perpendicular sides above. These give us (countably many) angles whose Sin and Cos are rational. (Technically, these are all in the first quadrant (from zero through a quarter turn); we can add and subtract them to obtain angles in the full range, still with rational Sin and Cos – but this in fact necessarily just gives the same set of angles as adding multiples of a quarter turn to the ones obtained directly, since each implies a pythagorean triangle, albeit in another quadrant.) While this set of angles is (though I have not shown this) dense in the turn, it leaves plenty of gaps in the unit circle of the rational plane; e.g. the line {[r, r]: r is a positive rational} runs from [0,0] inside the unit circle to outside that circle but has no intersection with the circle.
### Inscribed circle
I learn, via Catriona Shearer and Ben Orlin, that the inscribed circle of a pythagorean triangle has a radius that's a whole multiple of the same unit as the triangle's sides are whole multiples of. We now have the general formula for the pythagorean triangle's sides, at least when they're coprime, so what does that give us as the radius of the incircle ?
Let the radius be r; use our two perpendicular sides of the triangle as co-ordinate axes, with [0, m] and [n, 0] as the other two corners, distance k apart. The centre of the circle is at [r, r]; the radius that meets the hypotenuse does so at [x, y] = [r, r] +r.[m, n]/k = r.[k +m, k +n]/k; the one-form m.x +n.y is constant on the hypotenuse, equal to m.n, so
n.m
= r.(k.m +m.m +k.n +n.n)/k
= r.(m +n +k), so
r = n.m / (m +n +k)
= 2.(v.(v +1) −(u +1).u).(2.v +1).(2.u +1) / (2.(v.(v +1) −(u +1).u) +(2.v +1).(2.u +1) +2.(v.(v +1) +(u +1).u) +1)
= 2.(v.v +v −u.u −u).(2.v +1).(2.u +1) / (4.v.(v +1) +(2.v +1).(2.u +1) +1)
= (v.v −u.u +v −u).(2.v +1).(2.u +1) / (2.v.v +2.v +2.v.u +u +v +1)
= ((v −u).(v +u) +v −u).(2.v +1).(2.u +1) / (2.v.v +v +2.v.u +u +2.v +1)
= (v −u).(2.u +1)
and, sure enough, the radius of the incircle of a pythagorean triangle is a whole number multiple of the highest common factor of its sides.
## Right-angle simplices
A simplex is the generalisation of a triangle to other than two dimensions; the triangle is the two-simplex, the tetrahedron is the three-simplex, a simple line segment is the one-simplex and a point is the zero-simplex. For any natural n, {mappings ({positives}:f:1+n): sum(f) = 1} provides a canonical (open) n-simplex. However, I'm here more interested in a less symmetric n-simplex – namely, one in which n of the edges are (of non-zero length and) mutually perpendicular.
### Simplices
One important feature of a simplex is that each vertex is connected to every other vertex by an edge (and, in fact, for every set of i vertices, there is an i-simplex face of the simplex which has these vertices as its corners). Given a set of edges, a path among those edges is a list (:f:1+i) of vertices among which, for each j in i, f(j) and f(1+j) are connected by one of the edges in our set; f(0) and f(i) are described as the ends of the path and the path is described as a path between these two vertices, or from f(0) to f(i). A set of edges is connected if, between any two vertices of edges in the set, there is some path within the set.
A path is closed if its ends are the same vertex; i.e. f(0) is f(i). This is easy to achieve by traversing some sequence of edges and then simply reversing the sequence of edges to come back to where you started; but this is the boring case. If you can get back to where you started without doing that, then your path can be reduced to one which never re-uses an edge; a path is described as a closed loop if it is closed and uses each edge at most once. Among a set of mutually perpendicular edges there is, necessarily, no closed loop. A path among some edges, among which there is no closed loop, can always be reduced to a path that uses each edge at most once.
A set of i edges (for positive natural i), with as few vertices as possible subject to containing no closed loop, always has i+1 vertices and is connected: when i is 1, this is trivial. Thereafter, each added edge must add at least one vertex – because the prior edges are connected, so connecting two of their vertices would form a loop. If all vertices are reached by prior edges, it is no longer possible to add an edge without creating a loop; otherwise, connecting any vertex of an existing edge to any vertex not reached by existing edges does indeed add only the one vertex it must, to the set of edges, without adding a closed loop. If we have n edges of an n-simplex and no closed loop, this requires us to use all 1+n vertices that our n-simplex has; and (since this means our set of edges uses as few vertices as possible, while having no closed loop) ensures that they are all connected. So any n mutually perpendicular edges of an n-simplex are necessarily all connected and reach every vertex.
### Hypotenuses
Every vertex of our n-simplex is an end of at least one of our mutually perpendicular edges; and the mutually perpendicular edges are all connected, so there is some path among these edges from any given vertex to any other. As there are no closed loops, this path need only use each edge at most once; and there is only one such path, for each given pair of vertices. Thus, for every edge of our n-simplex, there is such a path connecting its end-points; if this path traverses i edges (once each), I'll refer to the edge whose end-points it connects as an i-hypotenuse of our n-simplex. Each of our perpendicular edges is a 1-hypotenuse (which is boring).
If the n mutually perpendicular edges form a path, that uses each edge exactly once, then we have an n-hypotenuse and, for each positive i in n, each n−i sub-path of length i provide us with an i-hypotenuse. However, if three (or more) of the perpendicular edges meet in one vertex, no path among the perpendicular edges can use each exactly once; at most two of the edges at any given vertex can appear in any such path, as we have no closed loops.
In two dimensions, we had two perpendicular sides to a triangle and they couldn't avoid forming a connected path, so life was simple: we only had one other side and it was a 2-hypotenuse. In higher dimensions, we have more things to consider:
• We can restrict attention to the case with a single n-hypotenuse,
• We can consider all the i-hypotenuses, for each i in n, or
• We can restrict attention to the case where all the perpendicular edges meet in one vertex; all other edges are 2-hypotenuses.
This leads to various possible analogues of pythagorean that we could investigate for n-simplices with n perpendicular sides. Describe an edge as tidy, for present purposes, precisely if its length is rationally commensurate with those of the perpendicular sides. No edge can be tidy unless all the mutually perpendicular edges have rationally commensurate lengths, of course. We could study n-simplices with n perpendicular edges and all edges tidy. A little less generally, we could restrict attention to the case where the perpendicular edges form a path. As a simplified case of this, we could ignore all edges except the n-hypotenuse; this amounts to looking for lists ({naturals}:|n) whose sum of squares is a perfect square.
### Incrementalism
Refining the simplest notion slightly, we can ask for lists of naturals where each initial sub-list has a perfect square as its sum of squares; thus, for example, [3, 4, 12, 84, 132] has 3×3 +4×4 = 5×5; adding 12×12 to that we get 13×13; adding 84×84 to that we get 85×85; and adding 132×132 to that we get 157×157. Each of [3, 4], [3, 4, 12], [3, 4, 12, 84] is an initial prefix of the list and has a perfect square as its sum of squares. This means we can have an n-simplex with these as the lengths of successive perpendicular edges, such that each i-hypotenuse associated with a path starting from the start of the first edge is tidy. However, notice that 4×4+12×12 = 160 isn't a perfect square, so at least one of the other other 2-hypotenuses isn't tidy.
From the above, it should be clear that we can build such incrementally pythagorean lists quite straightforwardly; indeed, in the example above, [3, 4, 12, 84] follow a pattern that can be continued indefinitely. As 3 is odd, its square is the sum of adjacent naturals, 4 and 5, hence equal to the difference of their squares, so [3, 4] has length 5. Then 5×5 is odd and likewise 12+13, so [5, 12] has length 13, whose odd square is 84+85, so [13, 84] has length 85; whose square is 7225 = 3612 +3613, so [85, 3612] has length 3613, which is (inevitably) odd again, so we can keep doing this for ever. Such a list grows rapidly, approximating (: 2.power(power(2, 1+i), x/2) ←i :) for some x, but can clearly be done for any odd start-value: for example, [7, 24, 312, 48984]. None the less, as illustrated above, they're not the only game in town; we could use 132 after 84 instead of 3612.
Even so, all cases of this kind can be built using the two-dimensional pythagorean triangles, incrementally: the sum of squares of the first n entries in such a list is the square of some natural, p, so any pythagorean triangle of which p is a perpendicular side can be used to obtain the other perpendicular side as n-th entry in the list, making the hypotenuse take the place of p for the thus-extended list.
Written by Eddy. | 5,000 | 19,526 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5 | 4 | CC-MAIN-2024-26 | latest | en | 0.941915 |
https://blog.earthandmoondesign.com/en/multistep-inequalities-worksheet.html | 1,713,068,851,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816864.66/warc/CC-MAIN-20240414033458-20240414063458-00041.warc.gz | 124,228,640 | 7,593 | # Multistep Inequalities Worksheet
Multistep Inequalities Worksheet - First subtract [math processing error] 6 from both sides: Web algebra multistep inequalities practice riddle worksheet this is an 15 question riddle practice worksheet designed to. Solve equations with the distributive property. Web multi step inequalities worksheets. Web ©u f2m0 p1e2b xkouwtza2 ysbo7fut bwja erje z olql bco.n t ha dlply grwigohqt8sr hrleis pewr. Graphing an inequality in 2 dimensional. Web test and worksheet generator for algebra 1. Web 25 carefully thought out problems on solving multistep inequalities. Solve each inequality and graph the solution. 6 4 5 7 8.
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First subtract [math processing error] 6 from both sides: Solve equations with the distributive property. Web solution p < 6 to graph this inequality, you draw an open circle at the end point, 6, on the number line. 6 4 5 7 8. The problems start off with the basic 2 step equations and slowly.
## Web Solution P < 6 To Graph This Inequality, You Draw An Open Circle At The End Point, 6, On The Number Line.
First subtract [math processing error] 6 from both sides: Algebra 1 > unit 2 lesson 5: Solve equations with the distributive property. [math processing error] 2 x + 6 ≤ 10 → [math processing error] 2 x + 6 − 6 ≤ 10 − 6.
## Web Algebra Multistep Inequalities Practice Riddle Worksheet This Is An 15 Question Riddle Practice Worksheet Designed To.
Printable math worksheets @ www.mathworksheets4kids.com solve. Infinite algebra 1 covers all typical algebra material, over 90 topics in all, from adding. Web 25 carefully thought out problems on solving multistep inequalities. Graphing an inequality in 2 dimensional.
## Web ©U F2M0 P1E2B Xkouwtza2 Ysbo7Fut Bwja Erje Z Olql Bco.n T Ha Dlply Grwigohqt8Sr Hrleis Pewr.
___________ so much more online! Web multi step inequalities worksheets. The problems start off with the basic 2 step equations and slowly. Solve each inequality and graph the solution.
## Web Test And Worksheet Generator For Algebra 1.
Web multi step inequalities worksheets. Web these inequality notes and worksheets cover:graphing one variable inequality on a number linesolving one, two, and multi. 6 4 5 7 8. | 844 | 3,292 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2024-18 | latest | en | 0.801217 |
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# lecture_12 (dragged) 2 - MA 36600 LECTURE NOTES MONDAY...
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MA 36600 LECTURE NOTES: MONDAY, FEBRUARY 9 3 We explain the relationship with solutions to these equations. Recall how to solve the diferential equation when the equation is “homogeneous” i.e., when g ( t ) is the zero Function: dy dt + p ( t ) y =0 dy dt = p ( t ) y 1 y dy dt = p ( t ) d dt ln | y | = p ( t ) ln | y | = ° t p ( τ ) = y ( t )= y 0 exp ± ° t t 0 p ( τ ) ² . We can solve the diference equation by considering its iterates: y 1 = ρ 0 y 0 = y 0 ³ 1 p ( t 0 ) h ´ y 2 = ρ 1 y 1 = y 0 ³ 1 p ( t 0 ) h ´³ 1 p ( t 1 ) h ´ y 3 = ρ 2 y 2 = y 0 ³ 1 p ( t 0 ) h ´³ 1 p ( t 1 ) h ´³ 1 p ( t 2 ) h ´ = y n = y 0 n 1 µ k =0 ± 1 p ( t k )∆ t k ² where we have set ∆ t k = h . Note the similarity with the exponential and the product. Autonomous Equations. Say that we have a linear autonomous equation in the Form dy dt + py = g where p and g are constants. The associated diference equation is in the Form y n +1 = ρy n + b where ρ =1 ph, b = gh.
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Ask a homework question - tutors are online | 448 | 1,242 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2017-51 | latest | en | 0.814563 |
https://kr.mathworks.com/matlabcentral/profile/authors/21287581?page=2 | 1,638,899,745,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363405.77/warc/CC-MAIN-20211207170825-20211207200825-00250.warc.gz | 419,827,236 | 2,695 | 해결됨
Determine whether a vector is monotonically increasing
Return true if the elements of the input vector increase monotonically (i.e. each element is larger than the previous). Return f...
4달 전
해결됨
Check if number exists in vector
Return 1 if number _a_ exists in vector _b_ otherwise return 0. a = 3; b = [1,2,4]; Returns 0. a = 3; b = [1,...
4달 전
해결됨
Select every other element of a vector
Write a function which returns every other element of the vector passed in. That is, it returns the all odd-numbered elements, s...
4달 전
해결됨
Swap the first and last columns
Flip the outermost columns of matrix A, so that the first column becomes the last and the last column becomes the first. All oth...
4달 전
해결됨
Swap the input arguments
Write a two-input, two-output function that swaps its two input arguments. For example: [q,r] = swap(5,10) returns q = ...
4달 전
해결됨
Column Removal
Remove the nth column from input matrix A and return the resulting matrix in output B. So if A = [1 2 3; 4 5 6]; ...
4달 전
해결됨
Reverse the vector
Reverse the vector elements. Example: Input x = [1,2,3,4,5,6,7,8,9] Output y = [9,8,7,6,5,4,3,2,1]
4달 전
해결됨
Magic is simple (for beginners)
Determine for a magic square of order n, the magic sum m. For example m=15 for a magic square of order 3.
4달 전
해결됨
Length of the hypotenuse
Given short sides of lengths a and b, calculate the length c of the hypotenuse of the right-angled triangle. <<https://i.imgu...
8달 전
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Triangle Numbers
Triangle numbers are the sums of successive integers. So 6 is a triangle number because 6 = 1 + 2 + 3 which can be displa...
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해결됨
Generate a vector like 1,2,2,3,3,3,4,4,4,4
Generate a vector like 1,2,2,3,3,3,4,4,4,4 So if n = 3, then return [1 2 2 3 3 3] And if n = 5, then return [1 2 2...
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해결됨
Make the vector [1 2 3 4 5 6 7 8 9 10]
In MATLAB, you create a vector by enclosing the elements in square brackets like so: x = [1 2 3 4] Commas are optional, s...
10달 전
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Return area of square
Side of square=input=a Area=output=b
10달 전
해결됨
Maximum value in a matrix
Find the maximum value in the given matrix. For example, if A = [1 2 3; 4 7 8; 0 9 1]; then the answer is 9.
10달 전
해결됨
Find the sum of all the numbers of the input vector
Find the sum of all the numbers of the input vector x. Examples: Input x = [1 2 3 5] Output y is 11 Input x ...
10달 전
해결됨 | 789 | 2,370 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2021-49 | latest | en | 0.622818 |
https://stats.stackexchange.com/questions/47227/drawing-conclusion-from-fixed-significance-level-or-p-value-in-a-two-sample-test | 1,642,362,262,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300010.26/warc/CC-MAIN-20220116180715-20220116210715-00266.warc.gz | 597,363,238 | 35,382 | # Drawing conclusion from fixed significance level or p-value in a two-sample test
This was an example done in class, However I was sick
An experiment was performed to determine whether the average nicotine content of brand A cigarette exceeds that of brand B cigarette by 0.20 milligram. If 50 cigarettes of brand A had a sample mean of 2.61 milligrams whereas 40 brand B cigarettes had an average nicotine of 2.38 milligrams. The population standard deviations of the nicotine content for the two brands of cigarettes are known to be 0.12 and 0.14 for brand A and B, respectively.
(a) Based on a significance level of 5%, what can you conclude about the difference between the two brands of cigarettes?
(b) Base on a p−value, what can you conclude about the difference between the two brands of cigarettes?
My Attempt:
(a)
$H_{0} :\mu_{A}-\mu_{B} =0.2$
$H_{1} :\mu_{A}-\mu_{B} \ne 0.2$
Significance Level : $\alpha = 0.05$
Rejection Region : $|z| >1.96$
Test Statistic : $z = \frac{2.61-2.38 -0.2}{\sqrt{\frac{0.12^2}{50}+\frac{0.14^2}{40}}} =1.08$
Conclusion : Since $1.08 <1.96$ I fail to reject $H_{0}$ at 5%
I really need Help with B
• What did you try?
– ThiS
Jan 8 '13 at 13:09
• I added the homework tag; this reads very much like a HW problem. Jan 8 '13 at 13:24
• Welcome to the site, @Jason. Please don't remove the HW tag, even if this was an in-class question, & not technically homework. The tag doesn't exist just to label questions that come from someone's actual HW, but to identify any "routine question from a textbook, course, or test used for a class or self-study". Your Q does come from a course, & it seems you are using this for self-study, in a sense, now. You can read more about this here: should-we-tag-questions-that-smell-like-homework & on the FAQ. Jan 8 '13 at 14:06
• Is your issue with part (b) one of not knowing how to compute the p-value, or not knowing how p-values relate to the conclusion of a statistical test at a given significance level, or both? Jan 8 '13 at 15:38
• Your statement of the hypothesis is incorrect according to "n experiment was performed to determine whether the average nicotine content of brand A cigarette exceeds that of brand B cigarette by 0.20 mg" May 13 '14 at 18:19
## 2 Answers
The area of the standard normal curve corresponding to a z-score of 1.08 is 0.1251. Because this test is two-tailed, that figure is doubled to yield a probability of 0.2502 (25%) that the population means are the same.
• @Glen_b not knowing how p-values relate to the conclusion of a statistical test at a given significance level Jan 8 '13 at 17:20
• @Jason Given the definition of the p-value, if the p-value is larger than the significance level, what would that imply? What would it imply if it was less than or equal to the significance level? Try the first few sentences of en.wikipedia.org/wiki/P-value for further explanation. Jan 9 '13 at 2:31
The essential question seems to be "how to interpret a p-value". This is my favorite paper on the subject, which explains the evolution (and misuse) of null-hypothesis significance testing.
Edit Thanks @Ben Bolker, corrected. Well spotted.
The short answer is it's the probability of data equal to or more extreme than the observed values given the null hypothesis (which, of course, was stated before performing the experiment). In your case, the probability of the data in your experiment, given the means are in fact the same, is <5% (actually approx. <14% based on the distribution function for the standard normal distribution).
• this is false! it's the probability of data equal to or more extreme than the observed values, under the null. $P(D|H)$ is the likelihood, not the p-value. Jan 4 '14 at 15:14 | 968 | 3,735 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2022-05 | latest | en | 0.920968 |
https://docs.cupy.dev/en/v9.6.0/reference/generated/cupy.random.Generator.html | 1,660,379,390,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571911.5/warc/CC-MAIN-20220813081639-20220813111639-00072.warc.gz | 223,732,021 | 16,645 | # cupy.random.Generator¶
class cupy.random.Generator(bit_generator)
Container for the BitGenerators.
Generator exposes a number of methods for generating random numbers drawn from a variety of probability distributions. In addition to the distribution-specific arguments, each method takes a keyword argument size that defaults to None. If size is None, then a single value is generated and returned. If size is an integer, then a 1-D array filled with generated values is returned. If size is a tuple, then an array with that shape is filled and returned. The function numpy.random.default_rng() will instantiate a Generator with numpy’s default BitGenerator. No Compatibility Guarantee Generator does not provide a version compatibility guarantee. In particular, as better algorithms evolve the bit stream may change.
Parameters
bit_generator – (cupy.random.BitGenerator): BitGenerator to use as the core generator.
Methods
beta(self, a, b, size=None, dtype=numpy.float64)
Beta distribution.
Returns an array of samples drawn from the beta distribution. Its probability density function is defined as
$f(x) = \frac{x^{\alpha-1}(1-x)^{\beta-1}}{B(\alpha,\beta)}.$
Parameters
• a (float) – Parameter of the beta distribution $$\alpha$$.
• b (float) – Parameter of the beta distribution $$\beta$$.
• size (int or tuple of ints) – The shape of the array. If None, a zero-dimensional array is generated.
• dtype – Data type specifier. Only numpy.float32 and numpy.float64 types are allowed.
Returns
Samples drawn from the beta distribution.
Return type
cupy.ndarray
exponential(self, scale=1.0, size=None)
Exponential distribution.
Returns an array of samples drawn from the exponential distribution. Its probability density function is defined as
$f(x) = \frac{1}{\beta}\exp (-\frac{x}{\beta}).$
Parameters
• scale (float or array_like of floats) – The scale parameter $$\beta$$.
• size (int or tuple of ints) – The shape of the array. If None, a zero-dimensional array is generated.
Returns
Samples drawn from the exponential distribution.
Return type
cupy.ndarray
gamma(self, shape, scale=1.0, size=None)
Gamma distribution.
Returns an array of samples drawn from the gamma distribution. Its probability density function is defined as
$f(x) = \frac{1}{\Gamma(k)\theta^k}x^{k-1}e^{-x/\theta}.$
Parameters
• shape (float or array_like of float) – The shape of the gamma distribution. Must be non-negative.
• scale (float or array_like of float) – The scale of the gamma distribution. Must be non-negative. Default equals to 1
• size (int or tuple of ints) – The shape of the array. If None, a zero-dimensional array is generated.
integers(self, low, high=None, size=None, dtype=numpy.int64, endpoint=False)
Returns a scalar or an array of integer values over an interval.
Each element of returned values are independently sampled from uniform distribution over the [low, high) or [low, high] intervals.
Parameters
• low (int) – If high is not None, it is the lower bound of the interval. Otherwise, it is the upper bound of the interval and lower bound of the interval is set to 0.
• high (int) – Upper bound of the interval.
• size (None or int or tuple of ints) – The shape of returned value.
• dtype – Data type specifier.
• endpoint (bool) – If True, sample from [low, high]. Defaults to False
Returns
If size is None, it is single integer sampled. If size is integer, it is the 1D-array of length size element. Otherwise, it is the array whose shape specified by size.
Return type
int or cupy.ndarray of ints
poisson(self, lam=1.0, size=None)
Poisson distribution.
Returns an array of samples drawn from the poisson distribution. Its probability mass function is defined as
$f(x) = \frac{\lambda^xe^{-\lambda}}{x!}.$
Parameters
• lam (array_like of floats) – Parameter of the poisson distribution $$\lambda$$.
• size (int or tuple of ints) – The shape of the array. If None, this function generate an array whose shape is lam.shape.
Returns
Samples drawn from the poisson distribution.
Return type
cupy.ndarray
random(self, size=None, dtype=numpy.float64, out=None)
Return random floats in the half-open interval [0.0, 1.0).
Results are from the “continuous uniform” distribution over the stated interval. To sample $$Unif[a, b), b > a$$ multiply the output of random by (b-a) and add a:
(b - a) * random() + a
Parameters
• size (None or int or tuple of ints) – The shape of returned value.
• dtype – Data type specifier.
• out (cupy.ndarray, optional) – If specified, values will be written to this array
Returns
Samples uniformly drawn from the [0, 1) interval
Return type
cupy.ndarray
standard_exponential(self, size=None, dtype=numpy.float64, method='inv', out=None)
Standard exponential distribution.
Returns an array of samples drawn from the standard exponential distribution. Its probability density function is defined as
$f(x) = e^{-x}.$
Parameters
• size (int or tuple of ints) – The shape of the array. If None, a zero-dimensional array is generated.
• dtype – Data type specifier. Only numpy.float32 and numpy.float64 types are allowed.
• method (str) – Method to sample. Currently only 'inv', sampling from the default inverse CDF, is supported.
• out (cupy.ndarray, optional) – If specified, values will be written to this array
Returns
Samples drawn from the standard exponential distribution.
Return type
cupy.ndarray
standard_gamma(self, shape, size=None, dtype=numpy.float64, out=None)
Standard gamma distribution.
Returns an array of samples drawn from the standard gamma distribution. Its probability density function is defined as
$f(x) = \frac{1}{\Gamma(k)}x^{k-1}e^{-x}.$
Parameters
• shape (float or array_like of float) – The shape of the gamma distribution. Must be non-negative.
• size (int or tuple of ints) – The shape of the array. If None, a zero-dimensional array is generated.
• dtype – Data type specifier.
• out (cupy.ndarray, optional) – If specified, values will be written to this array
standard_normal(self, size=None, dtype=numpy.float64, out=None)
Standard normal distribution.
Returns an array of samples drawn from the standard normal distribution.
Parameters
• size (int or tuple of ints) – The shape of the array. If None, a zero-dimensional array is generated.
• dtype – Data type specifier.
• out (cupy.ndarray, optional) – If specified, values will be written to this array
Returns
Samples drawn from the standard normal distribution.
Return type
cupy.ndarray
__eq__(value, /)
Return self==value.
__ne__(value, /)
Return self!=value.
__lt__(value, /)
Return self<value.
__le__(value, /)
Return self<=value.
__gt__(value, /)
Return self>value.
__ge__(value, /)
Return self>=value. | 1,595 | 6,735 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2022-33 | latest | en | 0.710302 |
https://math.libretexts.org/TextMaps/Applied_Mathematics/Book%3A_Introduction_to_Social_Network_Methods_(Hanneman)/8%3A_Embedding/8.3%3A_Reciprocity | 1,542,079,220,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039741192.34/warc/CC-MAIN-20181113020909-20181113042909-00193.warc.gz | 720,782,813 | 15,334 | $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$
# 8.3: Reciprocity
[ "article:topic", "authorname:rhanneman" ]
$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$
$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$
With symmetric dyadic data, two actors are either connected, or they are not. Density tells us pretty much all there is to know.
With directed data, there are four possible dyadic relationships: A and B are not connected, A sends to B, B sends to A, or A and B send to each other. A common interest in looking at directed dyadic relationships is the extent to which ties are reciprocated. Some theorists fell that there is an equilibrium tendency toward dyadic relationships to be either null or reciprocated, and that asymmetric ties may be unstable. A network that has a predominance of null or reciprocated ties over asymmetric connections may be a more "equal" or "stable" network than one with a predominance of asymmetric connections (which might be more of a hierarchy).
There are (at least) two different approaches to indexing the degree of reciprocity in a population. Consider the very simple network shown in Figure 8.3. Actors A and B have reciprocated ties, actors B and C have a non-reciprocated tie, and actors A and C have no tie.
Figure 8.3: Definitions of reciprocity
What is the prevalence of reciprocity in this network? One approach is to focus on the dyads, and ask what proportion of pairs have a reciprocated tie between them? This would yield one such tie for three possible pairs (AB, AC, BC), or a reciprocity rate of 0.333. More commonly, analysts are concerned with the ratio of the number of pairs with a reciprocated tie relative to the number of pairs with any tie. In large populations, usually most actors have no direct ties to most other actors, and it may be more sensible to focus on the degree of reciprocity among pairs that have any ties. In our simple example, this would yield one reciprocated pair divided by two tied pairs, or a reciprocity rate of 0.500. The method just described is called the dyad method in Network>Cohesion>Reciprocity.
Rather than focusing on actors, we could focus on relations. We could ask, what percentage of all possible ties (or "arcs" of the direct graph) are parts of reciprocated structures? Here, two such ties (A to B and B to A) are a reciprocated structure among the six possible ties (AB, BA, AC, CA, BC, CB) or a reciprocity of 0.333. Analysts usually focus, instead, on the number of ties that are involved in reciprocal relations relative to the total number of actual ties (not possible ties). Here, this definition would give us 2/3 or 0.667. This approach is called the arc method in Network>Cohesion>Reciprocity. Here's a typical dialog for using this tool.
Figure 8.4: Dialog for Network>Cohesion>Reciprocity
We've specified the "hybrid" method (the default) which is the same as the dyad approach. Note that it is possible to block or partition the data by some pre-defined attribute (like in the density example previously) to examine the degree of reciprocity within and between sub-populations. Figure 8.5 shows the results for the Knoke information network.
Figure 8.5: Reciprocity in the Knoke information network
We see that, of all pairs of actors that have any connection, $$53\%$$ of the pairs have a reciprocated connection. This is neither "high" nor "low" in itself, but does seem to suggest a considerable degree of institutionalized horizontal connection within this organizational population.
The alternative method of "arc" reciprocity (not shown here) yields a result of 0.6939. That is, of all the relations in the graph, $$69\%$$ are parts of reciprocated ties. | 1,058 | 4,124 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2018-47 | latest | en | 0.867389 |
https://dogfuranddandelions.com/2022/10/26/ | 1,716,264,064,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058383.61/warc/CC-MAIN-20240521025434-20240521055434-00809.warc.gz | 182,128,718 | 14,112 | # How to Simplify the Basic Blackjack Strategy
Blackjack is a game that requires strategy to win. The winning hand is a combination of an Ace and a 10 value card. This combination is known as Blackjack, and it beats all other hands. If both the player and the dealer get Blackjack, then it results in a tie, and the bet is returned to the player.
## Basic strategy
A basic blackjack strategy is an important part of any game of blackjack. The strategy dictates when to stand or risk another card. This is based on the probability of getting closer to 21 than the dealer. It is also important to keep a specific order while making decisions. There are several ways to simplify the basic blackjack strategy.
One way to simplify basic blackjack strategy is to use flashcards. These can be used to practice the decisions that are most profitable.
## Probability of busting
Probability of busting blackjack is a mathematical formula used to calculate the chances of busting a blackjack hand. It is the product of the probability that the dealer will receive an ace and your hand will not be higher than twenty-two. This mathematical formula is not possible to calculate any other way.
The probability of busting is higher the higher your hand total is. For example, the probability of busting if you get an ace and a ten-valued card is 92%. However, if you have a 21-point hand, your chances of busting are only 4.83%.
## House edge
The house edge of blackjack is not set in stone, but it is an important concept for any blackjack player. A higher house edge means you will lose more money than you make playing the game. This is because the casino keeps every penny you wager. If you make bad decisions, you could end up with a bad house edge.
The house edge is based on several factors. For instance, the number of hands you play in one session varies greatly. This explains why many simulations show a variable house edge, such as when a player plays a ploppy strategy and only plays the first thousand hands. The longer you play, the more likely you are to make a profit.
## Payouts
Blackjack is a game of chance that has different rules and bonus amounts that can increase your winnings or decrease them. The traditional payout is 3 to 2 when you have a blackjack, but more casinos are offering payouts that are lower. For example, you can now find blackjack tables that pay six to five, one to one, and so on. However, these games also raise the house edge.
The house edge on side bets is even higher than in the blackjack game itself, and this makes them particularly vulnerable to card counting. There are, however, certain side bets that have a sufficient win rate to make advantage play worthwhile. These side bets are most often played by teams. This means that a team of people can be dedicated to counting a single side bet in order to give each player a fair shot.
## Basic strategy table
A basic blackjack strategy table helps a player find the correct play in a game. It consists of two parts, the left column represents the hand of the player and the top column represents the hand of the dealer. The goal is to line up your hands with the value of the dealer’s cards. There are many different basic blackjack strategies, and one of the easiest to use is to follow the card combinations in the left column with the cards on the dealer’s right.
The Hit or Stand decision is the most important and frequent decision in the game of blackjack. Blackjack Basic Strategy is often taught in one of two ways: a traditional togel hari ini way in which each decision is listed in a table, containing the dealer’s up-card and the player’s first two cards. While this method can be helpful, it may be difficult to remember all of the decisions in the game.
Categories: Gambling | 786 | 3,794 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2024-22 | latest | en | 0.953416 |
https://math.stackexchange.com/questions/3731180/sagemath-defining-class-of-functions-on-elliptic-curves | 1,701,914,042,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100626.1/warc/CC-MAIN-20231206230347-20231207020347-00887.warc.gz | 439,465,461 | 40,069 | # SageMath: defining class of functions on Elliptic Curves
In SageMath,
I would like to manipulate rational functions on elliptic curves (defined on finite fields). For example, for $$P = (x,y)$$ on some curve $$E$$
$$f = x+y-12$$ $$g = \frac{x+y-3}{(x-3)^2}$$ etc.
Is there a natural class ?
I am looking to make a toy example with pairings, so I need to define stuff like $$P \rightarrow f_P$$ where $$f_P:Q \rightarrow f_P(Q)$$ is a function
I can't see how to do that, and I'm able to make computations if I define
def f (P,Q):
....
but I can only compute the values taken by the function $$f_P$$, I cannot "see" the function $$f_P$$. Basically I'm trying as an exercise to re-write the following Magma code to SageMath:
http://www.craigcostello.com.au/pairings/beginners/5-3-1-TateWeilMiller.txt
EDIT: my question hasn't attracted much interest so i'm going to give a more concrete example:
# this is the beginning of the code related to the example in Costello
q=47
F = GF(q)
R.<x> = F[] ; R
F4.<u>= F.extension(x^4-4*x^2+5)
a = 21 ; b= 15
E = EllipticCurve(F4,[a,b])
r=17
k=4
(q^4-1) % r # r=17 is a divisor of q^4 - 1 = 47^4 - 1
P = E([45,23])
P.order()
h = E.cardinality() / r^2
O = E(0)
Q = E([5*u^3 + 37*u + 13,7*u^3 + 45*u^2 + 10*u + 7])
lamb_da=(Q[1]-P[1])/(Q[0]-P[0])
c =P[1]-lamb_da*P[0]
l =(y-(lamb_da*x+c))
v =(x-(lamb_da^2-P[0]-P[1]))
return (l/v)
fADD_(P,Q,x,y) will return an error while i would have liked it to return a rational function in x,y
--------------------------------------------------------------------------- KeyError Traceback (most recent call last) /Applications/SageMath-9.1.app/Contents/Resources/sage/local/lib/python3.7/site-packages/sage/structure/coerce.pyx in sage.structure.coerce.CoercionModel.bin_op (build/cythonized/sage/structure/coerce.c:9946)() 1195 try: -> 1196 action = self._action_maps.get(xp, yp, op) 1197 except KeyError:
/Applications/SageMath-9.1.app/Contents/Resources/sage/local/lib/python3.7/site-packages/sage/structure/coerce_dict.pyx in sage.structure.coerce_dict.TripleDict.get (build/cythonized/sage/structure/coerce_dict.c:7917)() 1327
if not valid(cursor.key_id1): -> 1328 raise KeyError((k1, k2, k3)) 1329 value = cursor.value
KeyError: (Finite Field in u of size 47^4, Symbolic Ring, )
During handling of the above exception, another exception occurred:
TypeError Traceback (most recent call last) in () ----> 1 fADD_(P,Q,x,y)
in fADD_(P, Q, x, y) 2 lamb_da=(Q[Integer(1)]-P[Integer(1)])/(Q[Integer(0)]-P[Integer(0)]) 3 c =P[Integer(1)]-lamb_daP[Integer(0)] ----> 4 l =(y-(lamb_dax+c)) 5 v =(x-(lamb_da**Integer(2)-P[Integer(0)]-P[Integer(1)])) 6 return (l/v)
/Applications/SageMath-9.1.app/Contents/Resources/sage/local/lib/python3.7/site-packages/sage/structure/element.pyx in sage.structure.element.Element.mul (build/cythonized/sage/structure/element.c:12034)() 1515
return (left).mul(right) 1516 if BOTH_ARE_ELEMENT(cl): -> 1517 return coercion_model.bin_op(left, right, mul) 1518 1519 cdef long value
/Applications/SageMath-9.1.app/Contents/Resources/sage/local/lib/python3.7/site-packages/sage/structure/coerce.pyx in sage.structure.coerce.CoercionModel.bin_op (build/cythonized/sage/structure/coerce.c:9996)() 1196
action = self._action_maps.get(xp, yp, op) 1197 except KeyError: -> 1198 action = self.get_action(xp, yp, op, x, y) 1199 if action is not None: 1200 if (action)._is_left:
/Applications/SageMath-9.1.app/Contents/Resources/sage/local/lib/python3.7/site-packages/sage/structure/coerce.pyx in sage.structure.coerce.CoercionModel.get_action (build/cythonized/sage/structure/coerce.c:16783)() 1725
except KeyError: 1726 pass -> 1727 action = self.discover_action(R, S, op, r, s) 1728 action = self.verify_action(action, R, S, op) 1729
self._action_maps.set(R, S, op, action)
/Applications/SageMath-9.1.app/Contents/Resources/sage/local/lib/python3.7/site-packages/sage/structure/coerce.pyx in sage.structure.coerce.CoercionModel.discover_action (build/cythonized/sage/structure/coerce.c:18201)() 1856 """ 1857 if isinstance(R, Parent): -> 1858 action = (R).get_action(S, op, True, r, s) 1859 if action is not None: 1860 return action
/Applications/SageMath-9.1.app/Contents/Resources/sage/local/lib/python3.7/site-packages/sage/structure/parent.pyx in sage.structure.parent.Parent.get_action (build/cythonized/sage/structure/parent.c:19901)() 2475
action = self.get_action(S, op, self_on_left) 2476 if action is None: -> 2477 action = self.discover_action(S, op, self_on_left, self_el, S_el) 2478 2479 if action is not None:
/Applications/SageMath-9.1.app/Contents/Resources/sage/local/lib/python3.7/site-packages/sage/structure/parent.pyx in sage.structure.parent.Parent.discover_action (build/cythonized/sage/structure/parent.c:20878)() 2554
# detect actions defined by rmul, lmul, act_on, and acted_upon methods 2555 from .coerce_actions import
detect_element_action -> 2556 action = detect_element_action(self, S, self_on_left, self_el, S_el) 2557 if action is not None: 2558 return action
/Applications/SageMath-9.1.app/Contents/Resources/sage/local/lib/python3.7/site-packages/sage/structure/coerce_actions.pyx in sage.structure.coerce_actions.detect_element_action (build/cythonized/sage/structure/coerce_actions.c:5026)() 215 if isinstance(x, ModuleElement) and isinstance(y, Element): 216 try: --> 217 return (RightModuleAction if X_on_left else LeftModuleAction)(Y, X, y, x) 218 except CoercionException as msg: 219 _record_exception()
/Applications/SageMath-9.1.app/Contents/Resources/sage/local/lib/python3.7/site-packages/sage/structure/coerce_actions.pyx in sage.structure.coerce_actions.ModuleAction.init (build/cythonized/sage/structure/coerce_actions.c:6778)() 361 if not isinstance(g, Element) or not isinstance(a, ModuleElement): 362 raise CoercionException("not an Element acting on a ModuleElement") --> 363 res = self.act(g, a) 364 if parent(res) is not the_set: 365 # In particular we will raise an error if res is None
/Applications/SageMath-9.1.app/Contents/Resources/sage/local/lib/python3.7/site-packages/sage/categories/action.pyx in sage.categories.action.Action.act (build/cythonized/sage/categories/action.c:4115)() 213 5*x 214 """ --> 215 return self._act_convert(g, x) 216 217 def invert(self):
/Applications/SageMath-9.1.app/Contents/Resources/sage/local/lib/python3.7/site-packages/sage/categories/action.pyx in sage.categories.action.Action._act_convert (build/cythonized/sage/categories/action.c:3759)() 169 if parent(x) is not U: 170 x = U(x) --> 171 return self.act(g, x) 172 173 cpdef act(self, g, x):
/Applications/SageMath-9.1.app/Contents/Resources/sage/local/lib/python3.7/site-packages/sage/structure/coerce_actions.pyx in sage.structure.coerce_actions.RightModuleAction.act (build/cythonized/sage/structure/coerce_actions.c:8600)() 629 g = <Element?>self.connecting.call(g) 630 if self.extended_base is not None: --> 631 a = <ModuleElement?>self.extended_base(a) 632 return (a).lmul(g) # a * g 633
/Applications/SageMath-9.1.app/Contents/Resources/sage/local/lib/python3.7/site-packages/sage/structure/parent.pyx in sage.structure.parent.Parent.call (build/cythonized/sage/structure/parent.c:9218)() 898 if mor is not None: 899 if no_extra_args: --> 900 return mor.call(x) 901 else: 902 return mor._call_with_args(x, args, kwds)
/Applications/SageMath-9.1.app/Contents/Resources/sage/local/lib/python3.7/site-packages/sage/structure/coerce_maps.pyx in sage.structure.coerce_maps.DefaultConvertMap_unique.call (build/cythonized/sage/structure/coerce_maps.c:4556)() 159 print(type(C), C) 160 print(type(C._element_constructor), C._element_constructor) --> 161 raise 162 163 cpdef Element _call_with_args(self, x, args=(), kwds={}):
/Applications/SageMath-9.1.app/Contents/Resources/sage/local/lib/python3.7/site-packages/sage/structure/coerce_maps.pyx in sage.structure.coerce_maps.DefaultConvertMap_unique.call (build/cythonized/sage/structure/coerce_maps.c:4448)() 154 cdef Parent C = self._codomain 155 try: --> 156 return C._element_constructor(x) 157 except Exception: 158 if print_warnings:
/Applications/SageMath-9.1.app/Contents/Resources/sage/local/lib/python3.7/site-packages/sage/symbolic/ring.pyx in sage.symbolic.ring.SymbolicRing.element_constructor (build/cythonized/sage/symbolic/ring.cpp:6648)() 377 elif isinstance(x, (RingElement, Matrix)): 378 if x.parent().characteristic(): --> 379 raise TypeError('positive characteristic not allowed in symbolic computations') 380 exp = x 381 elif isinstance(x, Factorization):
TypeError: positive characteristic not allowed in symbolic computations
• Certainly the keyword here is probably "function field" since your rational functions live inside $\bar{\mathbb{F}_p}(E)$, although I am not sure how they are implemented in Sage, or how to evaluate them at points. Jun 23, 2020 at 10:45
• I don't understand that piece of Magma, so I can't tell exactly what you want. If you just want to take the function $f(P,Q)$ and fix $P$ to get a function $f_P(Q)$ then in Sage you can do something like f_P = lambda Q : f(P,Q).
– user208649
Jun 24, 2020 at 1:11
• stupid example $f_P$ could be $Q \rightarrow P+Q$ Jun 24, 2020 at 1:45
• it's like doing a partial evaluation of a function of 2 variables F(X,Y). if you fix $X=X_0$ you are left with a function $F_{X_0}(Y)$ i am interested in seeing the details of $F_{X_0}$ because i want to manipulate the coefficients to create new functions Jun 24, 2020 at 1:53
• I don't know what you mean by "see the details". If you have $f(P,Q) = f(P_1,P_2,Q_1,Q_2)$ defined as a rational function in sage, then to figure out $f_P$ as a rational function, you can just evaluate $f(P_1,P_2,x,y)$ where $x,y$ are two variables. It will then print out what you want, I think.
– user208649
Jun 24, 2020 at 4:48
It was hard (for me) to find the starting point of the question. You are working i a concrete situation and the task is clear, but i had to guess which is the point of the question. (We want to "translate" some magma code to sage? Or we want to correct the sage code snippet only? Or the question is the reason for the long traceback information?)
This answer is "making the sage code work". (If something else is the subject of the question, please edit.) Note first that the given code is not complete, well y is not defined. This may be also the reason for the error. (If it is a variable, or it is not "married" structurally with x, then i would expect such errors.)
The following piece of code worked for me:
p = 47
F = GF(p)
R.<x> = PolynomialRing(F)
F4.<u> = GF(p^4, modulus=x^4 - 4*x^2 + 5)
a, b = 21, 15
E = EllipticCurve(F4, [a,b])
print(f"Working with the elliptic curve E:\n{E}")
r = 17
k = 4
print("Is r = {} dividing p^4 - 1 = {}? {}"
.format(r, p^4 - 1, r.divides(p^4 - 1)))
P = E.point([45, 23])
print(f"The order of the point {P.xy()} is {P.order()}")
h = E.cardinality() / r^2
O = E(0)
Q = E.point([5*u^3 + 37*u + 13, 7*u^3 + 45*u^2 + 10*u + 7])
R.<x,y> = PolynomialRing(F)
# or F4 instead of F, to avoid structural coerce
def myadd(P, Q, x, y):
m = (Q[1] - P[1]) / (Q[0] - P[0])
c = P[1] - m*P[0]
return (y - (m*x+c)) / (x - (m^2 - P[0] - P[1]))
myadd(P, Q, x, y)
(Please try to give a readable shape to the code, i also break pep8 rules, but spaces make reading (and error searching) easier.)
The above gave me:
Working with the elliptic curve E:
Elliptic Curve defined by y^2 = x^3 + 21*x + 15
over Finite Field in u of size 47^4
Is r = 17 dividing p^4 - 1 = 4879680? True
The order of the point (45, 23) is 17
((5*u^3 + 22*u^2 + 2*u - 4)*x
+ y + (10*u^3 - 3*u^2 + 4*u + 16))
/
(x + (12*u^3 - 4*u^2 - 12*u + 17))
(... manually rearrange to better fit the stackexchange format.)
And indeed, the result of myadd(P, Q, x, y) is a rational function in x and y, defined over F4.
(I will try to see what happens in the other code, but i do not have magma on this own linux machine.) | 3,644 | 11,886 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 8, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2023-50 | longest | en | 0.767963 |
https://physics.stackexchange.com/questions/96304/understanding-watts-available-in-miniature-batteries/96324 | 1,571,441,424,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986685915.43/warc/CC-MAIN-20191018231153-20191019014653-00013.warc.gz | 663,979,912 | 33,233 | # Understanding watts available in miniature batteries?
I am doing a project with some small miniature incandescent light bulbs (like a CM7333).
Sorry for not providing more links, the system will only allow 2.
The power source is a E11A battery.
The issue I am having is the bulb is not bright enough. We need to basically double or triple its current light out put or lumens. Theses bulbs are rated in MSCP. something Candle Power I presume.
These little bulbs are available with different amp, volt, and MSCP ratings, as precribed in this [chart][3].
Ok, so it would seem a simple matter of getting a bulb with an increased amp rating or filament design (I want to use the same battery in my design, which is 6v) so the total watts is higher and hence the bulb will burn brighter. From the [chart][4]
I could say grab a CM3150 which indicates a MSCP rating 3 times higher than the current bulb I am working with, for the same Volts and amps. I assume it brighter because the filament is a lighter duty design, which burns brighter. At least that's my way of wrapping my mind around it.
This were I run into my question or were I need some education. These little batteries seem to have some kind of current limiting capability or attribute. I have reviewed the technical data but its not clear to me how many amps the battery can supply.
I don't know how to properly determine how many amps my little battery will provide. I know it says "38 mAh to 3.0 volts", but I dont know how to properly apply that. The data sheet also states the drain as ".5 mA continuous", if I ma reading it correctly. Does that mean the battery can provide .5 mAs. or .0005 of an amp? Is it saying HALF a miliamp? or half a amp? half an amp sounds like a lot and half a milliamp sounds tiny.
So, in closing I hope I asked my question in a way that can be understood. Basically, I need to understand the maximum out put of that battery in even divisions of the amp. Like my bench top power supply. .01 .357, etc. Do these little guys have a current limit over time. I don't think they can discharge all their energy in a second... I think its the current limiting that is preventing my bulb from burning brighter... I dont know
Thanks, Robert | 508 | 2,232 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2019-43 | latest | en | 0.955263 |
http://experiment-ufa.ru/x=25*tan(41) | 1,524,338,730,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125945317.36/warc/CC-MAIN-20180421184116-20180421204116-00338.warc.gz | 107,206,162 | 6,457 | # x=25*tan(41)
## Simple and best practice solution for x=25*tan(41) equation. Check how easy it is, and learn it for the future. Our solution is simple, and easy to understand, so dont hesitate to use it as a solution of your homework.
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## Solution for x=25*tan(41) equation:
Simplifying
x = 25tan(41)
Reorder the terms for easier multiplication:
x = 25 * 41ant
Multiply 25 * 41
x = 1025ant
Solving
x = 1025ant
Solving for variable 'x'.
Move all terms containing x to the left, all other terms to the right.
Simplifying
x = 1025ant` | 186 | 640 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2018-17 | longest | en | 0.886236 |
https://stat.ethz.ch/pipermail/r-help/2020-September/468758.html | 1,701,818,076,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100568.68/warc/CC-MAIN-20231205204654-20231205234654-00370.warc.gz | 613,104,676 | 2,192 | # [R] How can we get a predicted value that are used to plot the figure using a plot_model function of sjPlot?
Peter Wagey peter@w@g|ey09 @end|ng |rom gm@||@com
Sat Sep 19 10:28:56 CEST 2020
```Hi R users,
I was trying to create a figure of three-way-interactions. There is a
function "plot-model" but I was wondering whether we can extract the
predicted value before we run the "plot-model" function.
For example:
in this example,
plot_model(fit, type = "pred", terms = c("c12hour", "barthtot [30,50,70]",
"c161sex"))
How can we see the predicted values that are used to plot the figure? If we
can see the data (predicted values), we could use other functions to create
another type of figures.
Thank you very much for your suggestions.
Thanks,
Peter
#############
library(sjPlot)
library(sjmisc)
library(ggplot2)
data(efc)
theme_set(theme_sjplot())
# make categorical
efc\$c161sex <- to_factor(efc\$c161sex)
# fit model with 3-way-interaction
fit <- lm(neg_c_7 ~ c12hour * barthtot * c161sex, data = efc)
# select only levels 30, 50 and 70 from continuous variable Barthel-Index
plot_model(fit, type = "pred", terms = c("c12hour", "barthtot [30,50,70]",
"c161sex"))
How can we get the predicted value that is used to plot the graph? we would
like to see the predicted value using three groups of barthtot
[30,50,70].Is there any way we can extract the data (predicted value) so
that we can use other graphic functions to create figures?
[[alternative HTML version deleted]]
``` | 418 | 1,489 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2023-50 | latest | en | 0.793536 |
https://betselection.cc/bally's-blog/bricklayers'-wall-meets-vdw-ap/?prev_next=prev | 1,591,138,471,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347426956.82/warc/CC-MAIN-20200602224517-20200603014517-00185.warc.gz | 270,107,144 | 8,352 | 08 Most recent two
### Topic: Most recent two (Read 4949 times)
0 Members and 1 Guest are viewing this topic.
#### Bally6354
• Moderator
• Posts: 1123
##### Most recent two
« on: August 04, 2013, 08:40:14 pm »
• Here is a pretty simple idea that may be worth testing a bit to see how it goes.
I will give a quick example using the sixlines and the quads.
Sixlines...
1-6
7-12
13-18
19-24
25-30
31-36
1-9
10-18
19-27
28-36
I got the following numbers from Spielbank Wiesbaden.
24
13
31
19
27
29
24
4
18
29
30
14
22
29
17
I will convert the above numbers to the Quads first.
3
2
4
3
3
4
3
1
2
4
4
2
3
4
2
Now it's pretty simple. All you do is look for the most recent two of the same and bet for that until there is a change and you have a new most recent two.
So here is how it would go for the Quads.
3
2
4
3 **The 3 here is the most recent two**
3 win (+3)
4 loss (+2)
3 win (+5)
1 loss (+4)
2 loss (+3)
4 loss (+2) **The 4 is now the most recent two**
4 win (+5)
2 loss (+4)
3 loss (+3)
4 win (+6)
2 loss (+5) **The 2 is now the most recent two**
Now here are the numbers converted to the Sixlines.
4
3
6
4
5
5
4
1
3
5
5
3
4
5
3
Here is how it would go for the Sixlines.
4
3
6
4 **The 4 is the most recent two**
5 loss (-1)
5 loss (-2) **The 5 is the most recent two**
4 loss (-3)
1 loss (-4)
3 loss (-5)
5 win (level)
5 win (+5)
3 loss (+4)
4 loss (+3)
5 win (+8)
3 loss (+7) **The 3 is the most recent two**
That's it basically.
I was thinking about progressions for this.
Maybe something like 111 222 333
For the Sixlines...
1111 2222 3333
or something along those lines.
cheers
Sometimes it is the people who no one imagines anything of who do the things that no one can imagine. | 619 | 1,740 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2020-24 | longest | en | 0.855546 |
http://modea.mobi/a-geometric-introduction-to-topology-wall-pdf/ | 1,511,407,763,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806720.32/warc/CC-MAIN-20171123031247-20171123051247-00414.warc.gz | 204,972,926 | 11,821 | # A geometric introduction to topology wall pdf
0
3
In the geometry of higher dimensions – the problem of deciding whether a Wang domino set can tile the plane is also undecidable. A a geometric introduction to topology wall pdf set of Wang dominoes can tile the plane, homeomorphisms of the Möbius band described in the previous paragraph. Like the Klein bottle – voronoi tilings with randomly placed points can be used to construct random tilings of the plane.
By using this site; Decorative mosaic tilings made of small squared blocks called tesserae were widely employed in classical antiquity, ueber diejenigen Fälle in welchen die Gaussichen hypergeometrische Reihe eine algebraische Function ihres vierten Elementes darstellt”. Any polyhedron that fits this criterion is known as a plesiohedron — Generated as Wythoff constructions, which has eight tetrahedra and six octahedra at each polyhedron vertex. which has eight cubes at each polyhedron vertex.
Escher explained that “No single component of all the series – half of which was on each side of the scissors. The result is sometimes called the “Sudanese Möbius Band”, edge because the long side of each rectangular brick is shared with two bordering bricks. But becomes impractical after sufficiently many folds, giving it extra twists and reconnecting the ends produces figures called paradromic rings. Such as can be used to generate some Penrose patterns using assemblies of tiles called rhombs, having the same angle between adjacent edges for every tile.
the Russian crystallographer Yevgraf Fyodorov proved that every periodic tiling of the plane features one of seventeen different groups of isometries. Tilings exists with convex N, Tile and glass, to avoid ambiguity one needs to specify whether the colours are part of the tiling or just part of its illustration. Such foams present a problem in how to pack cells as tightly as possible: in 1887, the underlying topological spaces within the Möbius strip are homeomorphic in each case. The Gilbert tessellation is a mathematical model for the formation of mudcracks – which use tiles that cannot tessellate periodically.
whose fibres are great semicircles. Möbius strips are common in the manufacture of fabric computer printer and typewriter ribbons, one of the three regular tilings of the plane. Later civilisations also used larger tiles, escher’s Legacy: A Centennial Celebration. | 522 | 2,414 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2017-47 | longest | en | 0.918925 |
https://healthyliving.azcentral.com/figure-bowling-handicap-4943.html | 1,600,590,947,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400196999.30/warc/CC-MAIN-20200920062737-20200920092737-00766.warc.gz | 402,150,571 | 28,697 | Handicapping allows competitors of different skill levels to compete equitably in a bowling league or a tournament. The United States Bowling Congress, the sports sanctioning body, recommends that league or tournament officials set a standard score higher than the average posted by the best player or team. Handicaps are then based on the difference between each player or team’s average and the league standard.
## Individual Handicapping
#### Step 1
Establish a standard score that’s higher than any individual’s average within the league or tournament.
#### Step 2
Choose a handicap percentage, typically 70 to 100 percent.
#### Step 3
Establish each individual’s average. Averages are typically based on the previous season’s scores.
#### Step 4
Subtract each individual’s average from the league or tournament standard selected in Step 1.
#### Step 5
Multiply the difference between each individual’s average and the standard score by the percentage figure selected in Step 2. The result is the individual’s handicap. For example, if the standard score is 195 and a bowler’s average is 175, the difference is 20. In an 80 percent handicap league, multiply 20 times 0.8, giving the bowler a handicap of 16 pins. In a 100 percent league, the multiplication step is not necessary, because the difference between a bowler’s average and the standard score becomes the bowler’s handicap.
## Standard Team Handicapping
#### Step 1
Establish a standard score that’s higher than any team’s average within the league or tournament.
#### Step 2
Choose a handicap percentage, typically 70 to 100 percent.
#### Step 3
Establish each individual’s average. Averages are typically based on the previous season’s scores.
#### Step 4
Add each team’s individual averages to determine the team average.
#### Step 5
Subtract each team’s average from the league or tournament standard selected in Step 1.
#### Step 6
Multiply the difference between a team’s average and the standard score by the percentage figure selected in Step 2. The result is the team’s handicap. For example, if the standard score is 790 and a team’s average is 760, the difference is 30. In a 70 percent handicap league, multiply 30 by 0.7, giving the team a handicap of 21 pins, rounded to the nearest whole number. In a 100 percent league, the multiplication step is not necessary, because the difference between a team’s average and the standard score becomes the team’s handicap.
## Team Difference Handicapping
#### Step 1
Choose a handicap percentage, typically 70 to 100 percent.
#### Step 2
Establish each individual’s average. Averages are typically based on the previous season’s scores.
#### Step 3
Add each team’s individual averages to determine the team average.
#### Step 4
Subtract the lower team’s average from its opponent’s team average.
#### Step 5
Multiply the difference between the teams’ averages by the percentage selected in Step 1. For example, if one team’s average is 800 and its opponent’s average is 750, the difference is 50 pins. In a 90 percent league, multiply 50 by 0.9, giving the lesser team a handicap of 45 pins. In a 100 percent league, the multiplication step is not necessary, because the difference between the teams’ averages becomes the lesser team’s handicap.
SHARE | 720 | 3,306 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2020-40 | longest | en | 0.872858 |
http://pakturkmaths.com/finding-slope-from-a-graph-worksheet/ | 1,566,693,375,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027322160.92/warc/CC-MAIN-20190825000550-20190825022550-00341.warc.gz | 156,749,360 | 28,733 | # Finding Slope From A Graph Worksheet
Finding Slope From A Graph Worksheet – When it comes to attaining a goal you may just put down on paper what it’s you want. Nevertheless many of us discover that we regularly lose these pieces of paper, or we do not use a very good format like the SMART objectives format. Subsequently it might really be significantly better should you were to make use of one of many many goal setting Finding Slope From A Graph Worksheet on supply right now.
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## Using Microsoft Excel to Create a Financial Worksheet Template
Excel users can make the most of texts, formulas, and double click on changes to create a template Finding Slope From A Graph Worksheet for any home, enterprise, or church. We will outline here the right way to set up a worksheet template in Microsoft Excel. This basic template can then be used for fundamental record retaining or modified for a lot of different uses.
The subsequent step is to create a formulation to calculate your complete stability of all columns. Within the H13 textbox enter the components =sum(f11:h11), what it will do is total the unfavourable bills and the optimistic deposits, creating a grand total amount. Additionally, you will wish to create a beginning stability (begin of the month balance) at J2. If you are using this template for a brand new undertaking, then your starting balance will likely be zero.
## A Budgeting Worksheet Gets You Started With a Price range
You need to start someplace and a budgeting Finding Slope From A Graph Worksheet can make the dreaded task of budgeting much simpler. Whether you utilize pc based software or plan to maintain your funds on paper, a worksheet may help you to brainstorm the categories you will have to finances for. Whereas most of them will likely be slightly different than the price range you ultimately provide you with, they can function a helpful tool that may make your entire process a lot simpler.
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## Using Math Worksheets
What are math worksheets and what are they used for? These are math forms which can be used by mother and father and academics alike to help the young children study fundamental math such as subtraction, addition, multiplication and division. This software is very important and you probably have a small kid and you don’t have a worksheet, then its time you bought yourself one or created one in your child. There are a number of web sites over the internet that offer free worksheets that are downloadable and printable to be used by mother and father and lecturers at dwelling or at college.
If you happen to can not buy a math work sheet since you suppose you could not have time to, then you’ll be able to create on using your property computer and customise it on your child. Doing that is straightforward. All you need is Microsoft phrase software in your computer to achieve this. Simply open the phrase utility in your laptop and begin a new document. Make sure that the new doc you might be about to create is based on a template. Then, be sure that your web connection is on earlier than you possibly can search the term “math worksheet” from the internet. You will get templates of all kinds in your worksheet. Select the one you need after which obtain.
Once downloaded, you may customize the math worksheet to suit your child. The level of the kid at school will decide the look and content material of the worksheet. Use the school textbook that your youngster makes use of at school as a reference guide that can assist you in the creation of the maths worksheet. This may be certain that the worksheet is completely related to the child and will assist the child enhance his or her grades in school.
Thank you for reading my article about Finding Slope From A Graph Worksheet, i hope you enjoy it. Please come back to read my one other article
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1. Sep 29, 2011
hermano
Hi,
I want to compare (statistically) different models which predict the values y at several x-values. Therefore I want to calculate the 'total error' between the exact (measured) y-values and the calculated y-values using different models. My problem is that I'm not sure which method to use to calculate the 'total error' for each model. Should I use the sum of squared errors, the sum of the absolute errors or some other technique?
Thanks
2. Sep 29, 2011
Bacle
Bro:
It will depend on what aspect of the error you are interested in.
3. Sep 30, 2011
hermano
What do you mean whit 'the aspect'? I want to quantify the accuracy of my models by calculating the difference between the predicted values of the model and the measured values for the whole set of points (thus for each x-value). I can easily calculate the error for each x-value, but I want to add all these errors together on a way like the sum of absolute errors or something for the whole set to get a total error which is a number that quantities the total accuracy of my model. The question is: Which method should I use to add all these 'separate' errors together?
4. Sep 30, 2011
Bacle
Well, modeling of processes can be done either from the perspective/approach of
Least Squares, or from Maximum-Likelihood estimation. Ijust wondered what
perspective you are using to get some insight.
5. Sep 30, 2011
Stephen Tashi
The question is what YOU mean by aspect.
This expresses an intuitive desire but it is not a well posed mathematical problem. For example, Suppose you have a model F(x) for x values in the range 0 to 100, Is it more or less important to fit the values of x from 0 to 50 than the values from 90 to 100? Do you care about errors as measured by the arithmetic difference between measured and predicted values or do you care about the percentage error? Is an over prediction by 10 as bad as an un-prediction by 10? Is the data that you have equally spaced over all the x values, or do I have a lot of data for one particular subset of those values?
Most importantly, what are you trying to accomplish? Are you looking for a number that "quantifies the total accuracy of your model" to publish in a paper, or in an advertising flyer? Are you trying to do a statistical hypothesis tests that accepts or rejects the model?
6. Sep 30, 2011
Bacle
You put it much more nicely and precisely than I did, Stephen. Many people seem
not to realize the need for specific details of what they want when they make
a request. Nice job!.
7. Oct 1, 2011
hermano
Hi Stephen and Bacle,
Indeed, I want a number (which reflects the total error) that quantifies the total accuracy of my model so I can compare different models with each other.
I will try to explain my problem:
Lets say that the data I have measured is a rough sine wave in function of the angular position (0 to 2*pi, which is the independent variable x) which I measured with three sensors under three different angular positions. The sample frequency determines the number of data points, lets say that for one revolution this is 1000 equidistant points. I add all these three measurements together (three vectors of 1000 points) and this is my input for my model. With my model I want to separate the data again for each sensor. In order to quantify each model, I want to compute the difference between the measured data of each sensor and the separated data of my model for each sensor. This gives me again three vectors of 1000 points which is the ABSOLUTE error on each angular position for the three sensors. My question is: How can I define/calculate one number for each of these vectors that quantifies the total error of my model?
At the end I want to compare these numbers for each model in order to select the model which gives me the lowest error between the measured and calculated data based on the total error!
I hope it is more clear now to help me with my problem!
Thanks
8. Oct 1, 2011
Stephen Tashi
One of the first things to determine is if there is imprecision in the data. In a simplistic view of the world, the model would be $y = f(x)$ and the data would be perfectly accurate. In a slightly more complicated view, data of the form $(x_i, y_i)$ has $x_i$ measured perfectly but the $y_i$ have measurement errors. In an even more complicated view, the $x_i$ are not be perfectly accurate either.
For example, models are often fit by defining "best fit" to mean a fit f(x) that minimizes the average of the quantities $(y_i - f(x_i))^2$ which doesn't account for any error in the $x_i$. The different approach of "total least squares" assumes that there are also errors in the $x_i$ measurements. http://en.wikipedia.org/wiki/Total_least_squares
I'm guessing that there are complications in your problem that haven't been explained yet because you speak of "adding" the data from the 3 sensors and then separating it again. If by "adding", you simply mean putting the 3 sets of data into one file, then separating it again seems a trivial operation, so I don't know why you would bother to mention it. It would be best if you explained the actual nature of the sensors and what they measure. Do you process the raw sensor measurements by assuming the sensors are at some known angle relative to the thing they measure when $x_i = 0$. Is the placing of the 1000 equally spaced angles done by taking measurements equally spaced in time and assuming a constant rate of rotation of something? | 1,262 | 5,528 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2018-22 | latest | en | 0.917287 |
https://www.in2013dollars.com/1940-AUD-in-1937?amount=1 | 1,579,510,224,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250598217.23/warc/CC-MAIN-20200120081337-20200120105337-00332.warc.gz | 907,878,854 | 12,193 | \$
# \$1 in 1940 → \$0.93 in 1937
### Australian Inflation Rate, \$1 in 1940 to 1937
According to the Bureau of Statistics consumer price index, prices in 1937 are 7.41% lower than average prices since 1940. The Australian dollar experienced an average inflation rate of 2.60% per year during this period, meaning the real value of a dollar decreased.
In other words, \$1 in 1940 is equivalent in purchasing power to about \$0.93 in 1937, a difference of \$-0.07 over 3 years.
The 1937 inflation rate was 4.17%. The inflation rate in 1940 was 3.85%. The 1940 inflation rate is lower compared to the average inflation rate of 4.82% per year between 1940 and 2020.
Cumulative price change -7.41% Average inflation rate 2.60% Converted amount (\$1 base) \$0.93 Price difference (\$1 base) \$-0.07 CPI in 1940 2.700 CPI in 1937 2.500 Inflation in 1937 4.17% Inflation in 1940 3.85%
AUD Inflation since 1922
Annual Rate, the Bureau of Statistics CPI
### Buying power of \$1 in 1937
This chart shows a calculation of buying power equivalence for \$1 in 1937 (price index tracking began in 1922).
For example, if you started with \$1, you would need to end with \$0.93 in order to "adjust" for inflation (sometimes refered to as "beating inflation").
When \$1 is equivalent to \$0.93 over time, that means that the "real value" of a single Australian dollar decreases over time. In other words, a dollar will pay for fewer items at the store.
This effect explains how inflation erodes the value of a dollar over time. By calculating the value in 1937 dollars, the chart below shows how \$1 buys less over the past 3 years.
According to the Bureau of Statistics, each of these AUD amounts below is equal in terms of what it could buy at the time:
Year Dollar Value Inflation Rate
1937 \$1.00 4.17%
1938 \$1.04 4.00%
1939 \$1.04 0.00%
1940 \$1.08 3.85%
1941 \$1.12 3.70%
1942 \$1.24 10.71%
1943 \$1.28 3.23%
1944 \$1.28 0.00%
1945 \$1.28 0.00%
1946 \$1.28 0.00%
1947 \$1.36 6.25%
1948 \$1.48 8.82%
1949 \$1.60 8.11%
1950 \$1.76 10.00%
1951 \$2.08 18.18%
1952 \$2.44 17.31%
1953 \$2.56 4.92%
1954 \$2.60 1.56%
1955 \$2.64 1.54%
1956 \$2.80 6.06%
1957 \$2.88 2.86%
1958 \$2.88 0.00%
1959 \$2.96 2.78%
1960 \$3.08 4.05%
1961 \$3.12 1.30%
1962 \$3.12 0.00%
1963 \$3.16 1.28%
1964 \$3.24 2.53%
1965 \$3.36 3.70%
1966 \$3.44 2.38%
1967 \$3.56 3.49%
1968 \$3.68 3.37%
1969 \$3.80 3.26%
1970 \$3.92 3.16%
1971 \$4.16 6.12%
1972 \$4.40 5.77%
1973 \$4.80 9.09%
1974 \$5.56 15.83%
1975 \$6.40 15.11%
1976 \$7.24 13.13%
1977 \$8.12 12.15%
1978 \$8.76 7.88%
1979 \$9.56 9.13%
1980 \$10.56 10.46%
1981 \$11.56 9.47%
1982 \$12.84 11.07%
1983 \$14.16 10.28%
1984 \$14.72 3.95%
1985 \$15.68 6.52%
1986 \$17.12 9.18%
1987 \$18.56 8.41%
1988 \$19.92 7.33%
1989 \$21.40 7.43%
1990 \$23.00 7.48%
1991 \$23.72 3.13%
1992 \$23.96 1.01%
1993 \$24.36 1.67%
1994 \$24.84 1.97%
1995 \$26.00 4.67%
1996 \$26.68 2.62%
1997 \$26.76 0.30%
1998 \$26.96 0.75%
1999 \$27.36 1.48%
2000 \$28.60 4.53%
2001 \$29.84 4.34%
2002 \$30.76 3.08%
2003 \$31.60 2.73%
2004 \$32.32 2.28%
2005 \$33.20 2.72%
2006 \$34.36 3.49%
2007 \$35.16 2.33%
2008 \$36.72 4.44%
2009 \$37.36 1.74%
2010 \$38.44 2.89%
2011 \$39.72 3.33%
2012 \$40.40 1.71%
2013 \$41.40 2.48%
2014 \$42.44 2.51%
2015 \$43.08 1.51%
2016 \$43.64 1.30%
2017 \$44.48 1.92%
2018 \$45.04 1.26%
2019 \$45.90 1.90%
2020 \$46.77 1.90%*
* Compared to previous annual rate. Not final. See inflation summary for latest 12-month trailing value.
### How to Calculate Inflation Rate for \$1, 1937 to 1940
This inflation calculator uses the following inflation rate formula:
CPI in 1937CPI in 1940
×
1940 AUD value
=
1937 AUD value
Then plug in historical CPI values. The Australian CPI was 2.7 in the year 1940 and 2.5 in 1937:
2.52.7
×
\$1
=
\$0.93
\$1 in 1940 has the same "purchasing power" or "buying power" as \$0.93 in 1937.
To get the total inflation rate for the 3 years between 1937 and 1940, we use the following formula:
CPI in 1937 - CPI in 1940CPI in 1940
×
100
=
Cumulative inflation rate (3 years)
Plugging in the values to this equation, we get:
2.5 - 2.72.7
×
100
=
-7%
### Data Source & Citation
Raw data for these calculations comes from the government of Australia's annual (CPI) as provided by the Reserve Bank of Australia. The consumer price index was established in 1922 and is tracked by Australian Bureau of Statistics (ABS).
You may use the following MLA citation for this page: “\$1 in 1940 → 1937 | Australia Inflation Calculator.” Official Inflation Data, Alioth Finance, 20 Jan. 2020, https://www.officialdata.org/1940-AUD-in-1937?amount=1.
Special thanks to QuickChart for providing downloadable chart images.
in2013dollars.com is a reference website maintained by the Official Data Foundation.
#### About the author
Ian Webster is an engineer and data expert based in San Mateo, California. He has worked for Google, NASA, and consulted for governments around the world on data pipelines and data analysis. Disappointed by the lack of clear resources on the impacts of inflation on economic indicators, Ian believes this website serves as a valuable public tool. Ian earned his degree in Computer Science from Dartmouth College.
Cumulative price change -7.41% Average inflation rate 2.60% Converted amount (\$1 base) \$0.93 Price difference (\$1 base) \$-0.07 CPI in 1940 2.700 CPI in 1937 2.500 Inflation in 1937 4.17% Inflation in 1940 3.85% | 1,962 | 5,405 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2020-05 | longest | en | 0.891395 |
http://www.informatyczna-pomoc.eu/download-pdf-ccss-hsg-gmda1-2-3-circumference-volume-1-book-by-lorenz-educational-press.pdf | 1,477,411,037,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988720154.20/warc/CC-MAIN-20161020183840-00462-ip-10-171-6-4.ec2.internal.warc.gz | 520,107,468 | 2,710 | CCSS HSG-GMD.A.1, 2, 3 Circumference & Volume 1
4.11 - 1251 ratings - Source
Fill in the gaps of your Common Core curriculum! Each ePacket has reproducible worksheets with questions, problems, or activities that correspond to the packetas Common Core standard. Download and print the worksheets for your students to complete. Then, use the answer key at the end of the document to evaluate their progress. Look at the product code on each worksheet to discover which of our many books it came from and build your teaching library! This ePacket has 7 activities that you can use to reinforce the standard CCSS HSG-GMD.A.1, 2, 3: Circumference a Volume. To view the ePacket, you must have Adobe Reader installed. You can install it by going to http://get.adobe.com/reader/.A.1, 2, 3: Give an informal argument for the formulas for the circumference of a circle, area of a circle, volume of a cylinder, ... of a sphere and other solid figures; Use volume formulas for cylinders, pyramids, cones, and spheres to solve problems. ... Find circumference by multiplying the diameter (or twice the length of the radius) by p: C = pd OR C = 2pr Area of a ... B E N F R A N K L I N M A R K B R U G G E M A N P N E CIRCUMFERENCE RADIUS AREA 18p cm 7 cm 9p in.2 36p m 18anbsp;...
Title : CCSS HSG-GMD.A.1, 2, 3 Circumference & Volume 1 Author : Publisher : Lorenz Educational Press - 2014-01-01 | 379 | 1,387 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2016-44 | latest | en | 0.861028 |
http://mathcentral.uregina.ca/QandQ/topics/interval | 1,590,396,741,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347388012.14/warc/CC-MAIN-20200525063708-20200525093708-00094.warc.gz | 80,198,246 | 7,425 | Math Central - mathcentral.uregina.ca
Quandaries & Queries
Q & Q
Topic: interval
start over
17 items are filed under this topic.
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Positive and negative values of a function 2018-01-30 From Grayson:f(x)=x^6-x^4 Interval: ( negative infinity, negative one ) Test Value: negative two Function Value f(x): positive forty eight Interval: ( negative one, zero ) Test Value: negative one Function Value f(x): zero Interval: ( zero, positive one ) Test Value: positive one Function Value f(x): zero Interval: ( positive one, positive infinity ) Test Value: positive two Function Value f(x): positive forty eight What is the sign of f(x) for each Interval?Answered by Penny Nom. Continuity on a closed interval 2014-09-21 From Pragya:The trouble I'm having is as follows : a continuous function is most of the times defined on a closed interval, but how is it possible to define it on a closed interval ,because to be continuous at endpoints of the interval the function's limit must exist at that endpoint,for which it has to be defined in its neighborhood,but we don't know anything about whether the function is always defined in the neighborhood. Please help...Answered by Penny Nom. Differentiable on an interval 2010-08-12 From Dave:Hi I was wondering if a function can be differentiable at its endpoint. For example if I have Y = X^2 and it is bounded on closed interval [1,4], then is the derivative of the function differentiable on the closed interval [1,4] or open interval (1,4). They always say in many theorems that function is continuous on closed interval [a,b] and differentiable on open interval (a,b) and an example of this is Rolle's theorem. Thank you for your help.Answered by Robert Dawson. Sample size 2010-03-29 From Rae:What sample size was needed to obtain an error range of 2% if the following statement was made? "75% of the workers support the proposed benefit package. These results are considered accurate to within + or - 2%, 18 out of 20 times. This seems like a straight forward question but I'm getting it wrong. Could you please help me out even just the set up would be appreciated so I can see if that's where I'm going wrong. ThanksAnswered by Harley Weston. The intervals where the function is positive and negative 2010-01-10 From Ron:Hello I'm trying to find out the intervals where the function is positive and negative. It's for a polynomial function y= -(x+2)^2 (x-2) and y= (x+1)(x+4)(x-3) I have tried the right and left side of each x-intercepts, but I still don't understand the results thank you for your helpAnswered by Penny Nom. Percent change between two value ranges 2007-11-28 From Joe:How do you calculate a percent change between tow value ranges - for instance if I project a range for 2007 to be between 100 and 120 and a range for 2008 to be between 120 and 140, how do I calculate the estimated increase between the range? Is it 0% to 40% (taking the two inside values rto calculate the minimum and the two outside values rto calculate the maximum?)Answered by Harley Weston. Find the sample size needed 2007-05-13 From Mini:Find the sample size needed to be 98% confident thata marketing survey on the proportion of shoppers who use the internet for holiday shopping is accurate within a margin of error of 0.02. Assume that the conditions for a binomial distribution are met, and that a current estimate for a sample proportion does not exist.Answered by Penny Nom. Interval of the domain 2007-05-13 From Gale:What does the term interval of the domain mean?Answered by Penny Nom and Stephen La Rocque. Write the interval in absolute value notation 2007-03-20 From Timothy:1. Write interval in absolute value notation i) xE[0,9] ii) xE[-2,20]Answered by Penny Nom. A confidence interval 2006-01-21 From Jonathan:I am attempting to calculate how my confidence interval will widen at the 95% confidence level if my response universe increases from 100 to 150 or to 200. There is a universe of 54,000. I take a 5% sample for a test universe of 2,700 If my "yes" universe is 100, at the 95% confidence level, what is my +/- range? (i.e +/- 3? +/-5?) Historically, 6.6% of the 2,700 you say "yes". I am trying to determine how the confidence interval would change if the number of "yes" responders increased to 150 or to 200. Answered by Penny Nom. Computing confidence intervals 2004-11-26 From Christie:I was given a question with N=100, sample proportion is 0.1- compute the 95% confidence interval for P? I have tried this several ways but do not know how to do without means, standard deviations, standard error of the mean? I asked my teacher and she said I have all the info I need. Can you help????Answered by Penny Nom. Sampling distributions 2002-02-18 From A student: given: n = 40, standard deviation is not known, population of individual observations not normal. does the central limit theorem apply in this case? why or why not? for an estimation problem, list two ways of reducing the magnitude of sampling error? What will happen to the magnitude of sampling error if the confidence level is raised all other things remaining the same? justify your answer? Answered by Harley Weston. A sample size problem 2001-10-28 From Charles:The U.S Transportation Dept. will randomly sample traffic reports to estimate the proportion of accidents involving people over the age of 70. The Dept. has no advance estimate of this proportion. how many reports should the dept select to be atleast 97% confident that the estimate is within .01 of the true proportion? Answered by Harley Weston. A confidence interval 2001-06-28 From Murray:An investigator wants to find out of there are any difference in "skills" between full and part time students. Records show the following: ```Student Mean Score Std Dev Number ---------- ----------------- ---------- ----------- Full time 83 12 45 Part time 70 15 55 ``` Compute a 95% confidence interval for the difference in mean scores.Answered by Andrei Volodin. A confidence interval 2001-04-26 From Kim:A poll asked 1528 adults if they were in favor of the death penalty, 1238 said yes, find 99% confidence level for percent of all adult who are in favor of the death penalty.Answered by Andrei Volodin. Estimating the population mean 1999-11-13 From John Barekman:Statitistics: Estimating the population mean when the standard deviation is known: I am not sure which n to use in the formula for the confidence interval equation: x +/- z*(standard deviation/sqrt(n)) If we have data of ten people, and if we have the data of ten sets of ten people each, what is the difference in the n that we use? What is the difference between the standard deviation and the standard error? Are we using the number of sampling means or just the number of samples?Answered by Harley Weston. La somme de deux fonctions 2007-11-19 From maud:Consigne : Ecrire la fonction f comme somme de deux fonctions u et v définies sur I. Donner le sens de variation de u et de v sur I. En déduire le sens de variation de la fonction f sur l'intervalle I indiqué. f(x)=-2x+(1sur x) I=]0;+infini[ Correction : Sens de variation de f sur I=]0;+infini[ On a f(x)=u(x) + v(x), avec {u(x) = -2x et v(x) = 1sur x La fonction u est strictement décroissante sur R, donc sur I ( droite avec coefficient directeur -2 négatif). La fonction v qui est la fonction inverse est stricyement décroissante sur [0;+infini[. Donc, la fonction f = u+v est strictement décroissante sur [0;+infini[. Ma question : Pourquoi la fonction v et la fonction f ne sont pas définies sur le même intervalle que la fonction u c'est-à-dire sur l'intervalle I indiqué ?Answered by Claude Tardif.
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about math central :: site map :: links :: notre site français | 1,929 | 7,824 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2020-24 | latest | en | 0.92391 |
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INFOSYS PAPER ON 15th APRIL AT HYDERABAD
MARKS: 50
TIME: 1Hr
This paper was for experienced people ( with 12 Months to 24 Months of Experience for the post
of Entry Level Software Engineers).
As I have seen previous years papers of 2006 & 2007 the pattern was from RS Agarwal (Verbal
& Non Verbal) so did't prepare puzzles.
The paper which I have written consist of only 10 puzzles of 50 Marks, Time was 1 Hr. No
English.
***Never take a Chance***
So Friends be prepared with
Previous Year Papers- Puzzles(important)
Shakuntala Devi Puzzles
RS Agarwal - Verbal & Non-Verbal
1)
A jeweller prepared a window display each displaying 3 of the 7 jems at a time . They were
methyst, opal, sapphire, emerald, ruby and garnet. Displayed according to the following
conditions:-
1 A sholud always be displayed on the left window and D on the right.
2 Ruby should never come with any of D or G.
3.E should always be with S.
then some 4 questions were asked on this.
#1 which combination is appropriate?
#2 which condition is correct in the right window?
#3 Ruby can be displayed with following other two?
#4 S can be displayed with the following other two on left side window?
(8 Marks)
2)
Yesterday my mother asked me to buy some stamps. Stamps are available in 2
paise,7paise,10paise,15paise and 20paise denominations. For three types of stamps I was
each. Unfortunately I forgot which I was supposed to buy five of and which to buy six of
Luckly my mother had given me the exact money required to buy the stamps , Rs. 3.00 and
the shopkeeper was able to give me the correct stamps. Which stamps did I buy?
(4 Marks)
3.
A cyclist covered 2/3 of his distance by cycle ,then his type was punctured. after that he
covered the remaining distance by walk. He felt that he had walked twice the time of he
cycle. how fast he cycled than he walked. Ans:4 times
(3 Marks)
4)
Four persons A, B, C and D are playing cards. They have one card in their hand. Each card
has two different colours on each of its side. In total there are 2 red, 2 green, and 3 blue
colours. They made following statements about their hidden colour of the card. and exactly
two among them are lying.
A -> Blue or Green
B -> Neither Blue nor Green
C -> Blue or Yellow
D -> Blue or Yellow
The visible colours are Red, Green, Red and Blue respectively in that order. Tell the hidden
colour of card for each of the person.
(8 Marks)
5)
M is sister of P, O`s husband is L`s brother, N is father of K & grandfather of P There are
two fathers and one mother and 3 brothers.
Find
1) Group of brothers
2) How many Male candidates?
3) who is husband of O ?
(6 Marks)
6)
There are two workers ,each takes 10 hrs & other for 9 hrs.Both can work together with 5
bricks laid for 1 hr.. The owner knows this but in a hurry he assigned a work to
both..........How many Bricks they constructed for 1 hr.......(Some thing like that)
(3 Marks)
7)
There are seven friends. First one goes to his friends house on first day. Second one goes to
his friends house on second day. Third one goes to his friends house on third day. Fourth
one goes to his friends house on fourth day. Fifth one goes to his friends house on fifth day.
Sixth one goes to his friends house on sixth day. Seventh one goes to his friends house on
seventh day. Altogether all people can meet on which day?
(4 Marks)
8)
Some puzzle question on ages (like S is older than T, but younger E.F is older than S, but
greater than T, etc)
(6Marks)
9)
Cricket Scores: Find the Number of runs altogether by the cricket team....As Sachin scored
74 less than Dravid. Dravid greater than Jedeja...............Rahul and Azar togther
76................(Something like that)
(4 Marks)
10)
There are 3 Men, in a party...... Each men dancing with three women..
if one women has two pairs of men how many attended the party? (Not Remember the
exact question)
(4 Marks) | 1,024 | 3,888 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2020-10 | latest | en | 0.971667 |
https://www.r-bloggers.com/2014/02/vertical-histogram/ | 1,726,108,412,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651420.25/warc/CC-MAIN-20240912011254-20240912041254-00230.warc.gz | 873,997,464 | 21,452 | Want to share your content on R-bloggers? click here if you have a blog, or here if you don't.
In the process of munging data for my current project I came across the need to compare (visually) the difference between two modes within the same dataset. I was using a simple scatterplot and setting the alpha in the hopes that the over-plotting would indicate which was the major mode. Unfortunately, the size of the data overwhelmed this approach.
I only wanted to use a single image and it was important that I keep the scatterplot to show other features of the data. I started looking for a way to combine a histogram (rotated 90 degrees) with the scatterplot to help describe the density within the plot. A quick search for how to do this in R turned up empty so I decided to implement my own version of such a plot.
Certainly, there are other ways to describe the features that I am trying to present here but in this particular case the following code worked out nicely. Hopefully it proves useful to others as well.
```plot.vertical.hist <- function(data,breaks=500) {
agg <- aggregate(data\$Y, by=list(xs=data\$X), FUN=mean)
hs <- hist(agg\$x / 10000, breaks=breaks, plot=FALSE)
mar.default <- par('mar')
mar.left <- mar.default
mar.right <- mar.default
mar.left[4] <- 0
mar.right[2] <- 0
# Main plot
par (fig=c(0,0.8,0,1.0), mar=mar.left)
plot (agg\$xs, agg\$x / 10000,
xlab="X", ylab="Y",
main="Vertical Histogram Side Plot",
pch=19, col=rgb(0.5,0.5,0.5,alpha=0.5))
grid ()
# Vertical histogram of the same data
par (fig=c(0.8,1.0,0.0,1.0), mar=mar.right, new=TRUE)
plot (NA, type='n', axes=FALSE, yaxt='n',
xlab='Frequency', ylab=NA, main=NA,
xlim=c(0,max(hs\$counts)),
ylim=c(1,length(hs\$counts)))
axis (1)
arrows(rep(0,length(hs\$counts)),1:length(hs\$counts),
hs\$counts,1:length(hs\$counts),
length=0,angle=0)
par(old.par)
invisible ()
}
```
Results look similar to the following:
Initially, I experimented with rug or barplot(..., horiz=TRUE). Unfortunately, rug isn't available on the left or right side and would suffer from the same problem that the alpha settings did and I was unable to get the alignment worked out when using barplot. | 594 | 2,166 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2024-38 | latest | en | 0.822671 |
https://oeis.org/A115980 | 1,721,435,265,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514972.64/warc/CC-MAIN-20240719231533-20240720021533-00336.warc.gz | 375,179,146 | 4,448 | The OEIS is supported by the many generous donors to the OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A115980 Array read by rows distributing the values of A000712 (vertically) and A001519 (horizontally). 0
1, 2, 2, 3, 2, 4, 4, 3, 2, 4, 6, 5, 4, 6, 2, 4, 6, 8, 6, 3, 4, 4, 8, 9, 2, 4, 6, 8, 10, 7, 4, 12, 8, 4, 8, 12, 12, 2, 4, 6, 8, 10, 12, 8, 3, 6, 5, 4, 14, 21, 12, 4, 8, 12, 16, 15, 2, 4, 6, 8, 10, 12, 14, 9 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,2 COMMENTS A001906 records the partial sums of the column sequence A001519 and is also the row sum of A078812 and of A085643; sequences linking a(n) to compositions of n having k parts when there are q kinds of part q. - Alford Arnold, Apr 30 2006 LINKS Table of n, a(n) for n=0..63. EXAMPLE The array begins: 1 ..2 ....2 ....3 ......2 ......4..3 ......4 .........2 .........4..4..3 .........6..6..4 .........5 with column sums beginning 1 2 5 10 20 ...A000712 related to A000041 and sums over each template beginning 1 2 5 13 34 ...A001519 related to A000045 CROSSREFS Cf. A000041, A000045. Cf. A001906, A078812, A085643. Sequence in context: A266470 A371745 A209700 * A088936 A328405 A049822 Adjacent sequences: A115977 A115978 A115979 * A115981 A115982 A115983 KEYWORD easy,nonn,tabf AUTHOR Alford Arnold, Feb 11 2006 STATUS approved
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Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recents
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Last modified July 19 19:50 EDT 2024. Contains 374436 sequences. (Running on oeis4.) | 626 | 1,696 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2024-30 | latest | en | 0.665315 |
https://math.stackexchange.com/questions/194130/need-matlab-help-related-with-the-iteration-method | 1,563,852,695,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195528687.63/warc/CC-MAIN-20190723022935-20190723044935-00275.warc.gz | 459,219,973 | 36,918 | Need matlab help related with the iteration method
I am reading an iteration method for computing the Moore- Penrose genralized inverse of a given matrix $A$, which is given as follows:
$X_{k+1} = (1+\beta)X_{k} - \beta X_{k} A X_{k}$
where $X_{k}$, k = 0,1,... is a sequence of approximations for computing Moore- Penrose genralized inverse
$X_{0} = \beta A'$ is the initial approximation , $0<\beta\leq 1$ and $A'$ is the transpose of matrix $A$
$d_{k} = \|X_{k+1} - X_{k}\|_{fro}$ is the error matrix norm (frobenius norm)
I have made following matlab program for computing Moore- Penrose genralized inverse by above mentioned method. But i am unable to make code for stopping criterion which says that.
perform the iteration untill
$|d_{k+1}/d_{k} - \beta -1|> 10^{-4}$
• You can post the MATLAB code on your post. Check the "code sample" button, the one with the "{}". – Rod Carvalho Sep 11 '12 at 14:02
• Why are you unable to make code for the stopping criterion? Do you mean do you don't know how to do it, or that when you try, it doesn't work? Can't you use a while loop? – Christopher A. Wong Sep 12 '12 at 3:54
• @ChristopherA.Wong Dear sir, i m beginner in matlab. I don't know how to do that? I need help. – srijan Sep 12 '12 at 4:51
• @srijan, you should change your for loop to a while loop that checks the value of the error. – in_mathematica_we_trust Sep 14 '12 at 8:34
The prep before your loop should stay the same. The appropriate script is
A = ...; % as you have given
beta = ...; % whatever you want
X0 = beta*A'; % calculate initial estimate
% (these initial values may need to be changed, I don't have a copy of
% matlab in front of me)
dklast = NaN; dk = NaN; % initialise to begin loop
iter = 0;
maxiter = 100;
while (abs(dk/dklast - beta - 1) > 1e-4) && (iter < maxiter) % loop until tolerance met
iter = iter + 1; % keep count of iteration
X1 = (1+beta)*X0 -beta*X0*A*X0; % calculate new iterate
dklast = dk; % move old difference "new estimate to previous iterate"
dk = norm(X1-X0,'fro'); % determine new difference
X0 = X1; % copy current iterate to "old" iterate for next iteration
end
I am wondering why you are using this convergence test at all. I would recommend using
dk = norm(X1*X0-I,'fro');
which measures how close X1 is to the left inverse of $A$. Your termination criteria would then be
while dk > (some_tolerance) && iter < maxiter
....
end
As you currently have, you are measuring how much X1 changes from X0, which may be small, but still not an approximate inverse (or pseudoinverse) for $A$.
• First of all thanks for replying me. I took beta = 1/norm (A,2)^2 , but program is not running. – srijan Sep 14 '12 at 8:58
• What error are you receiving? I have seen this algorithm with beta=1, not anything else. As I said, those two lines may need to be changes. You could initialise the process with dklast=-30 and dk=-20 and it should work then. – Daryl Sep 14 '12 at 10:05
• Programm is not running..just Matrix A is coming...not getting x1.. – srijan Sep 14 '12 at 10:07
• If X1 is not defined at the end of the iteration, then the loop is not running. As I said in my above comment, try with the values I have given there. The NaN terms may be invalidating the condition. – Daryl Sep 14 '12 at 10:10
• Dear i am not getting what you mean by x1 is not defined at the end of iteration bro..x1 is the sequence of approximation..nd you have defined that – srijan Sep 14 '12 at 10:13 | 1,014 | 3,458 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2019-30 | latest | en | 0.836887 |
https://la.mathworks.com/help/fixedpoint/ref/embedded.fi.html | 1,643,384,527,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320306301.52/warc/CC-MAIN-20220128152530-20220128182530-00270.warc.gz | 404,067,991 | 30,145 | # fi
Construct fixed-point numeric object
## Description
To assign a fixed-point data type to a number or variable, create a `fi` object using the `fi` constructor. You can specify numeric attributes and math rules in the constructor or using the `numerictype` and `fimath` objects.
## Creation
### Syntax
``a = fi``
``a = fi(v)``
``a = fi(v,s)``
``a = fi(v,s,w)``
``a = fi(v,s,w,f)``
``a = fi(v,s, w,slope,bias)``
``a = fi(v,s, w,slopeadjustmentfactor,fixedexponent,bias)``
``a = fi(v,T)``
``a = fi(___,F)``
``a = fi(___,Name,Value)``
### Description
example
````a = fi` returns a `fi` object with no value, 16-bit word length, and 15-bit fraction length.```
example
````a = fi(v)` returns a fixed-point object with value `v` and default property values.```
example
````a = fi(v,s)` returns a fixed-point object with signedness (signed or unsigned) `s`.```
example
````a = fi(v,s,w)` creates a fixed-point object with word length specified by `w`.```
example
````a = fi(v,s,w,f)` creates a fixed-point object with fraction length specified by `f`.```
example
````a = fi(v,s, w,slope,bias)` creates a fixed-point object using slope and bias scaling.```
example
````a = fi(v,s, w,slopeadjustmentfactor,fixedexponent,bias)` creates a fixed-point object using slope and bias scaling.```
example
````a = fi(v,T)` creates a fixed-point object with value `v`, and numeric type properties, `T`.```
example
````a = fi(___,F)` creates a fixed-point object with math settings specified by `fimath` object `F`.```
example
````a = fi(___,Name,Value)` creates a fixed-point object with property values specified by one or more `Name,Value` pair arguments. `Name` must appear inside single quotes (`''`). You can specify several name-value pair arguments in any order as `Name1,Value1,...,NameN,ValueN`.```
### Input Arguments
expand all
Value of the `fi` object, specified as a scalar, vector, matrix, or multidimensional array.
The value of the output `fi` object is the value of the input quantized to the data type specified in the `fi` constructor.
You can specify the non-finite values `-Inf`, `Inf`, and `NaN` as the value only if you fully specify the numeric type of the `fi` object. When `fi` is specified as a fixed-point numeric type,
• `NaN` maps to `0`.
• When the `'OverflowAction'` property of the `fi` object is set to `'Wrap'`, `-Inf`, and `Inf` map to `0`.
• When the `'OverflowAction'` property of the `fi` object is set to `'Saturate'`, `Inf` maps to the largest representable value, and `-Inf` maps to the smallest representable value.
Data Types: `single` | `double` | `int8` | `int16` | `int32` | `int64` | `uint8` | `uint16` | `uint32` | `uint64` | `logical` | `fi`
Signedness of the `fi` object, specified as a boolean. A value of `1`, or `true`, indicates a signed data type. A value of `0`, or `false`, indicates an unsigned data type.
Data Types: `logical`
Word length, in bits, of the `fi` object, specified as a scalar integer.
The `fi` object has a word length limit of 65535 bits.
Data Types: `single` | `double` | `int8` | `int16` | `int32` | `int64` | `uint8` | `uint16` | `uint32` | `uint64` | `logical`
Fraction length, in bits, of the `fi` object, specified as a scalar integer. If you do not specify a fraction length, the `fi` object automatically uses the fraction length that gives the best precision while avoiding overflow for the specified value, word length, and signedness.
Data Types: `single` | `double` | `int8` | `int16` | `int32` | `int64` | `uint8` | `uint16` | `uint32` | `uint64` | `logical`
Slope of the scaling, specified as a scalar integer. The following equation represents the real-world value of a slope bias scaled number.
`$real-worldvalue=\left(slope×integer\right)+bias$`
Data Types: `single` | `double` | `int8` | `int16` | `int32` | `int64` | `uint8` | `uint16` | `uint32` | `uint64` | `logical`
Bias of the scaling, specified as a scalar. The following equation represents the real-world value of a slope bias scaled number.
`$real-worldvalue=\left(slope×integer\right)+bias$`
Data Types: `single` | `double` | `int8` | `int16` | `int32` | `int64` | `uint8` | `uint16` | `uint32` | `uint64` | `logical`
The slope adjustment factor of a slope bias scaled number. The following equation demonstrates the relationship between the slope, fixed exponent, and slope adjustment factor.
`$slope=slopeadjustmentfactor×{2}^{fixedexponent}$`
Data Types: `single` | `double` | `int8` | `int16` | `int32` | `int64` | `uint8` | `uint16` | `uint32` | `uint64` | `logical`
The fixed exponent of a slope bias scaled number. The following equation demonstrates the relationship between the slope, fixed exponent, and slope adjustment factor.
`$slope=slopeadjustmentfactor×{2}^{fixedexponent}$`
Data Types: `single` | `double` | `int8` | `int16` | `int32` | `int64` | `uint8` | `uint16` | `uint32` | `uint64` | `logical`
Numeric type properties of the `fi` object, specified as a `numerictype` object. For more information, see `numerictype`.
Fixed-point math properties of the `fi` object, specified as a `fimath` object. For more information, see `fimath`.
## Examples
collapse all
Create a signed `fi` object with a value of `pi`, a word length of eight bits, and a fraction length of 3 bits.
`a = fi(pi,1,8,3)`
```a = 3.1250 DataTypeMode: Fixed-point: binary point scaling Signedness: Signed WordLength: 8 FractionLength: 3 ```
Create an array of `fi` objects with 16-bit word length and 12-bit fraction length.
`a = fi((magic(3)/10), 1, 16, 12)`
```a = 0.8000 0.1001 0.6001 0.3000 0.5000 0.7000 0.3999 0.8999 0.2000 DataTypeMode: Fixed-point: binary point scaling Signedness: Signed WordLength: 16 FractionLength: 12 ```
When you specify only the value and the signedness of the `fi` object, the word length defaults to 16 bits, and the fraction length is set to achieve the best precision possible without overflow.
`a = fi(pi, 1)`
```a = 3.1416 DataTypeMode: Fixed-point: binary point scaling Signedness: Signed WordLength: 16 FractionLength: 13 ```
If you do not specify a fraction length, input argument `f`, the fraction length of the `fi` object defaults to the fraction length that offers the best precision.
`a = fi(pi,1,8)`
```a = 3.1562 DataTypeMode: Fixed-point: binary point scaling Signedness: Signed WordLength: 8 FractionLength: 5 ```
The fraction length of `fi` object `a` is five because three bits are required to represent the integer portion of the value when the data type is signed. If the `fi` object uses an unsigned data type, only two bits are needed to represent the integer portion, leaving six fractional bits.
`b = fi(pi,0,8)`
```b = 3.1406 DataTypeMode: Fixed-point: binary point scaling Signedness: Unsigned WordLength: 8 FractionLength: 6 ```
The real-world value of a slope bias scaled number is represented by:
`$\mathrm{real}\text{\hspace{0.17em}}\mathrm{world}\text{\hspace{0.17em}}\mathrm{value}=\left(\mathrm{slope}×\mathrm{integer}\right)+\mathrm{bias}$`
To create a `fi` object that uses slope and bias scaling, include the `slope` and `bias` arguments after the word length in the constructor.
`a = fi(pi, 1, 16, 3, 2)`
```a = 2 DataTypeMode: Fixed-point: slope and bias scaling Signedness: Signed WordLength: 16 Slope: 3 Bias: 2 ```
The `DataTypeMode` property of the `fi` object, `a`, is `slope and bias scaling`.
When the value input argument, `v`, of a `fi` object is a non-double, and you do not specify the word length or fraction length properties, the resulting `fi` object retains the numeric type of the input, `v`.
Create a `fi` object from a built-in integer
When the input is a built-in integer, the fixed-point attributes match the attributes of the integer type.
```v1 = uint32(5); a1 = fi(v1)```
```a1 = 5 DataTypeMode: Fixed-point: binary point scaling Signedness: Unsigned WordLength: 32 FractionLength: 0 ```
```v2 = int8(5); a2 = fi(v2)```
```a2 = 5 DataTypeMode: Fixed-point: binary point scaling Signedness: Signed WordLength: 8 FractionLength: 0 ```
Create a `fi` object from a `fi` object
When the input value is a `fi` object, the output uses the same word length, fraction length, and signedness of the input `fi` object.
```v = fi(pi, 1, 24, 12); a = fi(v)```
```a = 3.1416 DataTypeMode: Fixed-point: binary point scaling Signedness: Signed WordLength: 24 FractionLength: 12 ```
Create a `fi` object from a logical
When the input `v` is logical, the `DataTypeMode` property of the output `fi` object is `Boolean`.
```v = true; a = fi(v)```
```a = 1 DataTypeMode: Boolean ```
Create a `fi` object from a single
When the input is single, the `DataTypeMode` property of the output is `Single`.
```v = single(pi); a = fi(v)```
```a = 3.1416 DataTypeMode: Single ```
The arithmetic attributes of a `fi` object are defined by a `fimath` object which is attached to that `fi` object.
Create a `fimath` object and specify the `OverflowAction`, `RoundingMethod`, and `ProductMode` properties.
`F = fimath('OverflowAction', 'Wrap', 'RoundingMethod','Floor', 'ProductMode','KeepMSB')`
```F = RoundingMethod: Floor OverflowAction: Wrap ProductMode: KeepMSB ProductWordLength: 32 SumMode: FullPrecision ```
Create a `fi` object and specify the `fimath` object, `F`, in the constructor.
`a = fi(pi, F)`
```a = 3.1415 DataTypeMode: Fixed-point: binary point scaling Signedness: Signed WordLength: 16 FractionLength: 13 RoundingMethod: Floor OverflowAction: Wrap ProductMode: KeepMSB ProductWordLength: 32 SumMode: FullPrecision ```
Use the `removefimath` function to remove the associated `fimath` object and restore the math settings to their default values.
`a = removefimath(a)`
```a = 3.1415 DataTypeMode: Fixed-point: binary point scaling Signedness: Signed WordLength: 16 FractionLength: 13 ```
A `numerictype` object contains all of the data type information of a `fi` object. By transitivity, `numerictype` properties are also properties of `fi` objects.
You can create a `fi` object that uses all of the properties of an existing `numerictype` object by specifying the `numerictype` object in the `fi` constructor.
`T = numerictype(0,24,16)`
```T = DataTypeMode: Fixed-point: binary point scaling Signedness: Unsigned WordLength: 24 FractionLength: 16 ```
`a = fi(pi, T)`
```a = 3.1416 DataTypeMode: Fixed-point: binary point scaling Signedness: Unsigned WordLength: 24 FractionLength: 16 ```
When you use binary-point representation for a fixed-point number, the fraction length can be greater than the word length. In this case, there are implicit leading zeros (for positive numbers) or ones (for negative numbers) between the binary point and the first significant binary digit.
Consider a signed value with a word length of 8, fraction length of 10, and a stored integer value of 5. Calculate the real-world value using the following equation.
`$\mathrm{real}\text{\hspace{0.17em}}\mathrm{world}\text{\hspace{0.17em}}\mathrm{value}=\mathrm{stored}\text{\hspace{0.17em}}\mathrm{integer}×{2}^{-\mathrm{fraction}\text{\hspace{0.17em}}\mathrm{length}}$`
`realWorldValue = 5*2^(-10)`
```realWorldValue = 0.0049 ```
Create a signed `fi` object with value `realWorldValue`, a word length of 8 bits, and a fraction length of 10 bits.
`a = fi(realWorldValue, 1, 8, 10)`
```a = 0.0049 DataTypeMode: Fixed-point: binary point scaling Signedness: Signed WordLength: 8 FractionLength: 10 ```
Get the stored integer value of `a` using the `int` function.
`int(a)`
```ans = int8 5 ```
Use the `bin` function to view the stored integer value in binary.
`bin(a)`
```ans = '00000101' ```
Because the fraction length is two bits longer than the word length, the binary value of the stored integer is `X.XX00000101`, where `X` is a placeholder for implicit zeroes. 0.0000000101 (binary) is equivalent to 0.0049 (decimal).
When you use binary-point representation for a fixed-point number, the fraction length can be negative. In this case, there are implicit trailing zeros (for positive numbers) or ones (for negative numbers) between the binary point and the first significant binary digit.
Consider a signed data type with a word length of 8, fraction length of -2 and a stored integer value of 5. Calculate the stored integer value using the following equation.
`$\mathrm{real}\text{\hspace{0.17em}}\mathrm{world}\text{\hspace{0.17em}}\mathrm{value}=\mathrm{stored}\text{\hspace{0.17em}}\mathrm{integer}×{2}^{-\mathrm{fraction}\text{\hspace{0.17em}}\mathrm{length}}$`
`realWorldValue = 5*2^(2)`
```realWorldValue = 20 ```
Create a signed `fi` object with value `realWorldValue`, a word length of 8 bits, and a fraction length of -2 bits.
`a = fi(realWorldValue, 1, 8, -2)`
```a = 20 DataTypeMode: Fixed-point: binary point scaling Signedness: Signed WordLength: 8 FractionLength: -2 ```
Get the stored integer value of `a` using the `int` function.
`int(a)`
```ans = int8 5 ```
Get the binary value of `a` using the `bin` function.
`bin(a)`
```ans = '00000101' ```
Because the fraction length is negative, the binary value of the stored integer is `00000101XX`, where `X` is a placeholder for implicit zeros. 0000010100 (binary) is equivalent to 20 (decimal).
You can set math properties, such as rounding and overflow modes during the creation of the `fi` object.
`a = fi(pi, 'RoundingMethod', 'Floor', 'OverflowAction', 'Wrap')`
```a = 3.1415 DataTypeMode: Fixed-point: binary point scaling Signedness: Signed WordLength: 16 FractionLength: 13 RoundingMethod: Floor OverflowAction: Wrap ProductMode: FullPrecision SumMode: FullPrecision ```
The `RoundingMethod` and `OverflowAction` properties are properties of the `fimath` object. Specifying these properties in the `fi` constructor associates a local `fimath` object with the `fi` object.
Use the `removefimath` function to remove the local `fimath` and set the math properties back to their default values.
`a = removefimath(a)`
```a = 3.1415 DataTypeMode: Fixed-point: binary point scaling Signedness: Signed WordLength: 16 FractionLength: 13 ```
When using a `fi` object as an index, the value of the `fi` object must be an integer.
Set up an array to index into.
`x = 10:-1:1;`
Create an integer valued `fi` object and use it to index into `x`.
```a = fi(3); y = x(a)```
```y = 8 ```
Use `fi` as the index in a `for` loop
Create `fi` objects to use as the index of a for loop. The values of the indices must be integers.
```a = fi(1, 0, 8, 0); b = fi(2, 0, 8, 0); c = fi(10, 0, 8, 0); for x = a:b:c x end```
```x = 1 DataTypeMode: Fixed-point: binary point scaling Signedness: Unsigned WordLength: 8 FractionLength: 0 ```
```x = 3 DataTypeMode: Fixed-point: binary point scaling Signedness: Unsigned WordLength: 8 FractionLength: 0 ```
```x = 5 DataTypeMode: Fixed-point: binary point scaling Signedness: Unsigned WordLength: 8 FractionLength: 0 ```
```x = 7 DataTypeMode: Fixed-point: binary point scaling Signedness: Unsigned WordLength: 8 FractionLength: 0 ```
```x = 9 DataTypeMode: Fixed-point: binary point scaling Signedness: Unsigned WordLength: 8 FractionLength: 0 ```
The `fipref` object defines the display and logging attributes for all `fi` objects. Use the `DataTypeOverride` setting of the `fipref` object to override `fi` objects with doubles, singles, or scaled doubles.
Save the current `fipref` settings to restore later.
```fp = fipref; initialDTO = fp.DataTypeOverride;```
Create a `fi` object with the default settings and original `fipref` settings.
`a = fi(pi)`
```a = 3.1416 DataTypeMode: Fixed-point: binary point scaling Signedness: Signed WordLength: 16 FractionLength: 13 ```
Turn on data type override to doubles and create a new `fi` object without specifying its `DataTypeOverride` property so that it uses the data type override settings specified using `fipref`.
`fipref('DataTypeOVerride', 'TrueDoubles')`
```ans = NumberDisplay: 'RealWorldValue' NumericTypeDisplay: 'full' FimathDisplay: 'full' LoggingMode: 'Off' DataTypeOverride: 'TrueDoubles' DataTypeOverrideAppliesTo: 'AllNumericTypes' ```
`a = fi(pi)`
```a = 3.1416 DataTypeMode: Double ```
Now create a `fi` object and set its `DataTypeOverride` setting to `off` so that it ignores the data type override settings of the `fipref` object.
`b = fi(pi, 'DataTypeOverride', 'Off')`
```b = 3.1416 DataTypeMode: Fixed-point: binary point scaling Signedness: Signed WordLength: 16 FractionLength: 13 ```
Restore the fipref settings saved at the start of the example.
`fp.DataTypeOverride = initialDTO;`
To use the non-numeric values `-Inf`, `Inf`, and `NaN` as fixed-point values with `fi`, you must fully specify the numeric type of the fixed-point object. Automatic best-precision scaling is not supported for these values.
Saturate on Overflow
When the numeric type of the `fi` object is specified to saturate on overflow, then `Inf` maps to the largest representable value of the specified numeric type, and `-Inf` maps to the smallest representable value. `NaN` maps to zero.
```x = [-inf nan inf]; a = fi(x,1,8,0,'OverflowAction','Saturate') b = fi(x,0,8,0,'OverflowAction','Saturate') ```
```a = -128 0 127 DataTypeMode: Fixed-point: binary point scaling Signedness: Signed WordLength: 8 FractionLength: 0 RoundingMethod: Nearest OverflowAction: Saturate ProductMode: FullPrecision SumMode: FullPrecision b = 0 0 255 DataTypeMode: Fixed-point: binary point scaling Signedness: Unsigned WordLength: 8 FractionLength: 0 RoundingMethod: Nearest OverflowAction: Saturate ProductMode: FullPrecision SumMode: FullPrecision ```
Wrap on Overflow
When the numeric type of the `fi` object is specified to wrap on overflow, then `-Inf`, `Inf`, and `NaN` map to zero.
```x = [-inf nan inf]; a = fi(x,1,8,0,'OverflowAction','Wrap') b = fi(x,0,8,0,'OverflowAction','Wrap') ```
```a = 0 0 0 DataTypeMode: Fixed-point: binary point scaling Signedness: Signed WordLength: 8 FractionLength: 0 RoundingMethod: Nearest OverflowAction: Wrap ProductMode: FullPrecision SumMode: FullPrecision b = 0 0 0 DataTypeMode: Fixed-point: binary point scaling Signedness: Unsigned WordLength: 8 FractionLength: 0 RoundingMethod: Nearest OverflowAction: Wrap ProductMode: FullPrecision SumMode: FullPrecision ```
## Compatibility Considerations
expand all
Behavior changed in R2020b | 5,097 | 18,247 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 7, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2022-05 | longest | en | 0.638776 |
https://en.m.wikipedia.org/wiki/Ampheck | 1,555,663,372,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578527518.38/warc/CC-MAIN-20190419081303-20190419102244-00034.warc.gz | 398,234,404 | 15,074 | # Logical NOR
(Redirected from Ampheck)
In boolean logic, logical nor or joint denial is a truth-functional operator which produces a result that is the negation of logical or. That is, a sentence of the form (p NOR q) is true precisely when neither p nor q is true—i.e. when both of p and q are false. In grammar, nor is a coordinating conjunction.
Logical NOR
NOR
Definition${\displaystyle {\overline {x+y}}}$
Truth table${\displaystyle (1000)}$
Logic gate
Normal forms
Disjunctive${\displaystyle {\overline {x}}\cdot {\overline {y}}}$
Conjunctive${\displaystyle {\overline {x}}\cdot {\overline {y}}}$
Zhegalkin polynomial${\displaystyle 1\oplus x\oplus y\oplus xy}$
Post's lattices
0-preservingno
1-preservingno
Monotoneno
Affineno
Self-dualno
The NOR operator is also known as Peirce's arrowCharles Sanders Peirce introduced the symbol ↓ for it,[1] and demonstrated that the logical NOR is completely expressible: by combining uses of the logical NOR it is possible to express any logical operation on two variables. Thus, as with its dual, the NAND operator (a.k.a. the Sheffer stroke—symbolized as either ↑, | or /), NOR can be used by itself, without any other logical operator, to constitute a logical formal system (making NOR functionally complete). It is also known as Quine's dagger (his symbol was †), the ampheck (from Ancient Greek ἀμφήκης, amphēkēs, "cutting both ways") by Peirce,[2] or neither-nor.
Other ways of notating ${\displaystyle P\downarrow Q}$ include, P NOR Q, and "Xpq" (in Bocheński notation). It is logically equivalent to ${\displaystyle \neg (P\lor Q)}$, where the symbol ${\displaystyle \lor }$ signifies OR and ${\displaystyle \neg }$ signifies the negation.
The computer used in the spacecraft that first carried humans to the moon, the Apollo Guidance Computer, was constructed entirely using NOR gates with three inputs.[3]
## Definition
The NOR operation is a logical operation on two logical values, typically the values of two propositions, that produces a value of true if and only if both operands are false. In other words, it produces a value of false if and only if at least one operand is true.
### Truth table
The truth table of ${\displaystyle P\downarrow Q}$ (also written as P NOR Q) is as follows:
${\displaystyle P}$ ${\displaystyle Q}$ ${\displaystyle P\downarrow Q}$ T T F T F F F T F F F T
### Logical Equivalences
The logical NOR ${\displaystyle \downarrow }$ is the negation of the disjunction:
${\displaystyle P\downarrow Q}$ ${\displaystyle \Leftrightarrow }$ ${\displaystyle \neg (P\lor Q)}$ ${\displaystyle \Leftrightarrow }$ ${\displaystyle \neg }$
## Properties
Logical NOR does not possess any of the five qualities (truth-preserving, false-preserving, linear, monotonic, self-dual) required to be absent from at least one member of a set of functionally complete operators. Thus, the set containing only NOR suffices as a complete set.
## Other Boolean Operations in terms of the Logical NOR
NOR has the interesting feature that all other logical operators can be expressed by interlaced NOR operations. The logical NAND operator also has this ability.
Expressed in terms of NOR ${\displaystyle \downarrow }$ , the usual operators of propositional logic are:
${\displaystyle \neg P}$ ${\displaystyle \Leftrightarrow }$ ${\displaystyle P\downarrow P}$ ${\displaystyle \neg }$ ${\displaystyle \Leftrightarrow }$
${\displaystyle P\rightarrow Q}$ ${\displaystyle \Leftrightarrow }$ ${\displaystyle {\Big (}(P\downarrow P)\downarrow Q{\Big )}}$ ${\displaystyle \downarrow }$ ${\displaystyle {\Big (}(P\downarrow P)\downarrow Q{\Big )}}$ ${\displaystyle \Leftrightarrow }$ ${\displaystyle \downarrow }$
${\displaystyle P\land Q}$ ${\displaystyle \Leftrightarrow }$ ${\displaystyle (P\downarrow P)}$ ${\displaystyle \downarrow }$ ${\displaystyle (Q\downarrow Q)}$ ${\displaystyle \Leftrightarrow }$ ${\displaystyle \downarrow }$
${\displaystyle P\lor Q}$ ${\displaystyle \Leftrightarrow }$ ${\displaystyle (P\downarrow Q)}$ ${\displaystyle \downarrow }$ ${\displaystyle (P\downarrow Q)}$ ${\displaystyle \Leftrightarrow }$ ${\displaystyle \downarrow }$ | 1,116 | 4,143 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 46, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2019-18 | longest | en | 0.8829 |
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# Allocate Books InterviewBit Solution
Problem Description:
Given an array of integers A of size N and an integer B.
College library has N bags, the ith book has A[i] number of pages.
You have to allocate books to B number of students so that the maximum number of pages allotted to a student is minimum.
• A book will be allocated to exactly one student.
• Each student has to be allocated at least one book.
• Allotment should be in contiguous order, For example: A student cannot be allocated book 1 and book 3, skipping book 2.
Calculate and return that minimum possible number.
Note: Return -1 if a valid assignment is not possible.
Input Format
`The first argument given is the integer array A. The second argument given is the integer B. `
Output Format
`Return that minimum possible number `
Constraints
```1 <= N <= 10^5
1 <= A[i] <= 10^5```
For Example
```Input 1:
A = [12, 34, 67, 90]
B = 2
Output 1:
113
Explanation 1:
There are 2 number of students. Books can be distributed in following fashion :
1) [12] and [34, 67, 90]
Max number of pages is allocated to student 2 with 34 + 67 + 90 = 191 pages
2) [12, 34] and [67, 90]
Max number of pages is allocated to student 2 with 67 + 90 = 157 pages
3) [12, 34, 67] and [90]
Max number of pages is allocated to student 1 with 12 + 34 + 67 = 113 pages
Of the 3 cases, Option 3 has the minimum pages = 113.
Input 2:
A = [5, 17, 100, 11]
B = 4
Output 2:
100```
## Approach
This problem is an unique type of binary search problem, where you don't do binary search on index, but on the range of answer.
Let's talk about the situation when there will be no valid assignment and for that answer is -1. This is possible only when, the number of students exceeds numbers of books, only then you can not allot minimum of one book to some student(s).
So now the second case, when valid answer exists for sure.
In this case the maximum possible answer is the sum of pages of all the books, given only to one student. And Let's take the minimum to be zero, even though according to the given constraints it's not possible to have 0 as an answer, but if there exists a valid answer then binary search will eventually find it.
So we have a range, from 0 to the sum of pages, we can do binary search on this range, if for a given threshold value, it is possible to assign books to students, then we shift towards left, to find minimum, otherwise to the right.
In the below solution, "canBePlaced" function tells us whether the given threshold number of pages can be distributed among the students or not. Here, threshold means, we can not allocate a student more than threshold value of pages.
## Time & Space Complexity
`Time Complexity: O(logN), N being the sum of array P. `
`Space Complexity: O(1), Ignoring the space taken by the given input array.`
Code in C++
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http://hyperphysics.phy-astr.gsu.edu/hbase/ph4060/p406i.html | 1,369,365,274,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368704133142/warc/CC-MAIN-20130516113533-00019-ip-10-60-113-184.ec2.internal.warc.gz | 133,181,336 | 3,265 | ## Physics 4060: Acoustics Laboratory
I. Decibels
Because of the large range of power associated with audible sound (1014 or larger) it is customary to use the logarithmic decibel scale for sound intensity measurement. The decibel level is defined by
where I = sound intensity = sound power per unit area (watts/cm2 is a typical unit).
The reference level for such measurements is
I0 = the intensity for the standard threshold of hearing at 1000 Hz = 10-16 watts/cm2.
By examining the definition when I0 is inserted for I, giving a 1 in the logarithm, you can see that the level I0 is zero decibels. The threshold of pain is about 1013 I0 or 130 decibels. Normal music is 40dB-100dB and conversational speech about 75dB.
The just noticeable change in sound intensity is about 1 decibel. Doubling the power of a source increases the sound level by 10 log10(2) = 3 decibels. This is not as dramatic a change in loudness as you might expect from power doubling - the ear's loudness response is approximately logarithmic. This might be borne in mind if you consider paying a lot more money for a 70 watt stereo rather than a 50 watt - the perceived loudness difference is slight. Subjective experiments have indicated that for a given sound the intensity must be increased by about 10 decibels to be perceived as twice as loud - a tenfold power increase!
In an open area with no reverberation, the sound intensity from a point source will drop off according to the inverse square law:
If the distance from the sound source is doubled, the intensity is down to one fourth, so
A doubling of distance from the source should give you a 6 dB drop if the inverse square law holds. (Note that decibels can be used to express any sound ratio in addition to the use for comparing to hearing threshold.) If the sound intensity in an auditorium followed the inverse square law, there would be a 20 decibel drop from the nearest listener to the most distant one if the nearest were at 10 ft and the most distant at 100 ft:
In any real environment - at least in any enclosed room - reverberant sound and background sound are always present and the inverse square law does not apply. If the intensity of the sound source is much greater than the background, then the sound near the source may approximate inverse square drop off- this region is often called the "direct sound field". At larger distances from the source in an acoustically "live" environment, the sound level may have little variation with distance because the room may be filled with the "reverberent sound field".
II. Use of Contour Filters
Sound intensity and sound "loudness" are not the same - there is not a one-to-one correlation. Two sounds of the same intensity will not in general be perceived by the human ear to have the same loudness unless the two sounds are the same pitch and of identical frequency content. The origin of this difficulty is the fact that the ear's sensitivity is strongly frequency dependent, as indicated by example contours from the human hearing curves sketched below.
Sounds below 100 Hz frequency are strongly discriminated against by the human ear, yet are very important for musical sound. If there is a lot of low frequency content in a sound, then a straight decibel measurement will be quite misleading if you are trying to assess loudness. A 50 dB, 1000 Hz sound will be significantly louder than a 50 dB, 100 Hz sound -- the 100 Hz sound will be barely audible. The relative loudness perception of different frequency sounds is further complicated by the fact that the response curves of the ear change with loudness level. Note that the equal loudness curve for 100 phons in the diagram above is considerably flatter that the curve for 40 phons. The human ear discriminates more and more against very low and very high frequencies as the sound gets softer. The "phon" is a unit of loudness so that a 40 phon curve is an equal-loudness curve for sounds which sound the same loudness as a 40 dB, 1000 Hz tone.
Since the human hearing response is non-uniform, standard contour filters have been developed which approximate the human hearing response. The three international standard contours are denoted A,B,C and measurements using them are denoted dBA, dBB, and dBC. These filter contours are roughly sketched below to illustrate their relative discrimination against very low and high audible frequencies.
The A contour is the most commonly used contour for the measurement of musical sound loudness, since it most accurately approximates the human hearing response for sounds of medium loudness. Note from the equal loudness curves that the hearing curve is flatter at very high sound intensity levels, so that contours B and C might be more appropriate for monitoring loud sounds. C is sometimes used for traffic noise surveys. For most purposes a survey of flat dB and dBA measurements constitutes a practical assessment of the sound field.
### Physics 4060 Laboratory
Index
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Assignment Help Physics
##### Reference no: EM1372426
Q. a. Management consulting firms did very well on a per-employee basis, partly because they are mostly comprised of managers (as opposed to blue-collar or entry-level workers). How big a factor do you think composition of workforce is in likelihood of producing a CEO?
b. Do you think so called leadership factories are also better places for non-leaders to work? Why or why not?
c. presume you had job offers from two companies that differed only in how often they produced CEOs. Would these differences affect your decision?
d. Do these data give any credence to the value of leader selection and leader development? Why or why not?
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GATE CSE 2014 Set 1
Numerical
+1
-0
Consider the following system of equations:
3x + 2y = 1
4x + 7z = 1
x + y + z =3
x - 2y + 7z = 0
The number of solutions for this system is ______________________
2
GATE CSE 2014 Set 3
+1
-0.3
Which one of the following statements is TRUE about every $$n\,\, \times \,n$$ matrix with only real eigen values?
A
If the trace of the matrix is positive and the determinant of the negative, at least one of its eigen values is negative.
B
If the trace of the matrix is positive, all its eigen values are positive.
C
If the determinanant of the matrix is positive, all its eigen values are positive.
D
If the product of the trace and determination of the matrix is positive, all its eigen values are positive.
3
GATE CSE 2014 Set 3
Numerical
+1
-0
If $${V_1}$$ and $${V_2}$$ are 4-dimensional subspaces of a 6-dimensional vector space V, then the smallest possible dimension of $${V_1}\, \cap \,\,{V_2}$$ is _________________.
4
GATE CSE 2014 Set 2
Numerical
+1
-0
If the matrix A is such that $$A = \left[ {\matrix{ 2 \cr { - 4} \cr 7 \cr } } \right]\,\,\left[ {\matrix{ 1 & 9 & 5 \cr } } \right]$$\$ then the determinant of A is equal to _________. | 393 | 1,182 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2024-30 | latest | en | 0.800689 |
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Let $\mathcal{D}$ be a full subcategory of a category $\mathcal{C}$. Let $X, Y$ be objects of $\mathcal{D}$. Suppose a product $X\times Y$ exists in $\mathcal{C}$. Suppose $X\times Y$ is not isomorphic to any object of $\mathcal{D}$. Can we conclude that a product of $X$ and $Y$ does not exist in $\mathcal{D}$? If not, any counter-example?
-
Taking C to be the category of finite abelian groups and D the subcat. of cyclic groups of prime order you can find an example, I think. – Mariano Suárez-Alvarez Dec 1 '12 at 23:44
Another quick example is for instance ${\cal D}=\textbf{Fields}$ and ${\cal C}=\textbf{CommRings}$. The product of two rings can never be a field, but some products exist in the category of fields. – Marc Olschok Dec 4 '12 at 18:13
2 Answers
Two counter-examples: one quite well-known and important, and one more elementary.
The important example: compactly generated (Hausdorff) topological spaces (briefly, k-spaces). The full subcategory of Top of these, kHaus, is a very nice category of spaces — better-behaved in many ways than Top itself.
Anyhow, the ordinary topological product of two k-spaces may not itself be a k-space; but it has an associated space (sometimes called its “k-ification”) which, while not a product in Top, is a product in kHaus.
The more elementary example is with coproducts not products, but sticking “op” on the categories involved makes it an example with products. Here, consider Abelian groups, AbGp, as a subcategory of all groups, Grp. The coproduct of two groups $A, B$ in Gp is their free product $A \ast B$. For Abelian $A$, $B$, the free product will usually not be Abelian: for instance, $\mathbb{Z} \ast \mathbb{Z}$ is the free group on two generators. However, they do also have a coproduct in AbGp, given by $A \oplus B$.
There’s a nice explanation for both these examples, and indeed for all the natural examples of this situation that I can think of.
Suppose D is a reflective subcategory of C, i.e. the inclusion functor has a left adjoint. (The Abelianisation functor Gp $\to$ AbGp is a typical example.) Then coproducts in D — more generally, any colimits — can be computed by taking the colimit in C and applying the reflector to the result. So, for instance, $A \oplus B$ is the Abelianisation of $A \ast B$. Checking that this recipe works whenever the colimits exist in C is a short diagram-chase — it’s essentially the fact that left adjoints preserve colimits.
Dually, if D is co-reflective — the inclusion has a right adjoint — then products (and arbitrary limits) can be computed in a similar fashion.
Reflective and co-reflective categories arise very often in nature; so they give lots of good examples where the (co)product in the ambient category doesn’t lie in the subcategory, but the subcategory has its own product nevertheless.
-
No. There's no reason for the inclusion of $D$ into $C$ to preserve products. Peter LeFanu Lumsdain has given some nice examples.
However, one situation in which the inclusion of $D$ into $C$ is guaranteed to preserve products is when it has a left adjoint (in which case the inclusion in fact preserves all limits). In this case, $D$ is said to be a reflective subcategory, and then the desired conclusion holds. For example:
• Compact Hausdorff spaces are a reflective subcategory of topological spaces (the adjoint is Stone-Čech compactification).
• Abelian groups are a reflective subcategory of groups (the adjoint is abelianization).
• Sheaves are a reflective subcategory of presheaves (the adjoint is sheafification).
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# Problem 46908. Product of elements in a column
Solution 3343543
Submitted on 23 Oct 2020 by KATTA PRATYUSHA
• Size: 10
• This is the leading solution.
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
x =[1 2 3;3 4 5;1 2 1]; y_correct = [3 16 15]; assert(isequal(your_fcn_name(x),y_correct))
2 Pass
x =[1 8 9 4;1 26 7 5]; y_correct = [1 208 63 20]; assert(isequal(your_fcn_name(x),y_correct))
3 Pass
x =[10 20;30 40]; y_correct = [300 800]; assert(isequal(your_fcn_name(x),y_correct))
4 Pass
x =[10 30;20 40]; y_correct = [200 1200]; assert(isequal(your_fcn_name(x),y_correct))
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Prove that the set U={f ∈ C([0,1]): f(1/2)=f(1)} is a subspace of C([0,1]).
I am not sure how to do this. I know what the necessary conditions are for a subspace but I can't quite figure out how to show them given a continuous function. Help please?
2. ## Re: Finding subspace/continuous function
Originally Posted by steph3824
Prove that the set U={f ∈ C([0,1]): f(1/2)=f(1)} is a subspace of C([0,1]).
I am not sure how to do this. I know what the necessary conditions are for a subspace but I can't quite figure out how to show them given a continuous function. Help please?
You have to prove the subspace axioms.
U is nonempty since $f\in U$
Let $f,g\in U$
$(f+g)(1/2)=f(1/2)+g(1/2)=f(1)+g(1)=(f+g)(1)$
How about the last axiom now.
3. ## Re: Finding subspace/continuous function
Would it be (af)(1/2) = a·f(1/2) = a·f(1)?
4. ## Re: Finding subspace/continuous function
Originally Posted by steph3824
Prove that the set U={f ∈ C([0,1]): f(1/2)=f(1)} is a subspace of C([0,1]).
I am not sure how to do this. I know what the necessary conditions are for a subspace but I can't quite figure out how to show them given a continuous function. Help please?
I don't suppose you know the fact that if $T:V\to W$ is a linear equation, then $\ker T$ is a subspace of $V$. Well, I'm sure you can note that $C[0,1]\to\mathbb{R}:f\mapsto f(1)-f(\tfrac{1}{2})$ is a linear transformation and $\ker f=U$.
5. ## Re: Finding subspace/continuous function
Originally Posted by steph3824
Would it be (af)(1/2) = a·f(1/2) = a·f(1)?
Good. | 526 | 1,579 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 8, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2016-50 | longest | en | 0.911445 |
https://dba.stackexchange.com/questions/183616/how-to-count-rows-after-a-specific-row | 1,627,747,190,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154089.68/warc/CC-MAIN-20210731141123-20210731171123-00431.warc.gz | 209,303,269 | 38,464 | # How to count rows after a specific row?
I have a contact mechanism between my website's users (always between two users). User A (sender) can send a message to users B (receiver) based on two rules:
1. User A could send 2 messages at most to user B and he must receive at least one message from user B to be able to send another 2.
2. User A could send 4 messages at most to everybody in daily period, not more.
And here is my table structure:
``````-- contact
+----+-----------+------------+--------------------------+------------+
| id | sender_id | receive_id | message | date_time |
+----+-----------+------------+--------------------------+------------+
| 1 | 123 | 456 | Hi, how are you? | 1492431111 |
| 2 | 123 | 789 | How are you doing? | 1492431112 |
| 3 | 456 | 789 | Why would you say that? | 1492431113 |
| 4 | 123 | 456 | Why don't you answer? | 1492431114 |
| 5 | 789 | 456 | Because the sky is high | 1492431115 |
| 6 | 123 | 789 | Hello? | 1492431116 |
+----+-----------+------------+--------------------------+------------+
``````
And here is my current query:
``````INSERT INTO contact(sender_id, receive_id, message, date_time )
SELECT ?, ?, ?, unix_timestamp()
FROM dual
WHERE NOT EXISTS(
SELECT count(*) AS num_day,
FROM contact
WHERE user_id = ?
AND date_time > unix_timestamp(DATE_SUB(now(), INTERVAL 1 day))
HAVING num_day > 4
)
``````
As you can see, only the second rule is implemented in my query. How can I also implement the first rule to the query?
• you can use `where date_time > (select max(date_time) from ...`. But on a general note: it would be very careless if you would simply silently not insert the messages without a warning. So check that before you insert, and display an error. – Solarflare Aug 16 '17 at 11:27
• Or, to put it differently, no more than two in a row to the same receiver and no more than four in one day to the same receiver, correct? – Andriy M Aug 16 '17 at 22:56
• @AndriyM Nope .. no more than two row to the same receiver (without any reply) and no more than four in one day to everyone (not the same receiver) – stack Aug 16 '17 at 23:08
• I wish tweets had those rules! – Rick James Aug 20 '17 at 15:38
I always find views and subqueries easier to read and develop mainly as you can test and edit one section at a time. Depending on the number of messages and your system it may be a little slower though I doubt it. In this case I am going to use a union query.
I would also check first and if the limits are not exceeded then perform the insert (unless you doing millions of messages). Create a view of the total number of sent messages
`````` CREATE view vwMessagesSent AS
SELECT *
FROM `contact`
WHERE date_time > unix_timestamp(DATE_SUB(now(), INTERVAL 1 day))
``````
This reduces the overall complexity of
``````SELECT Least(4,4 - count(id)) as sendLimit
FROM `vwMessagesSent`
WHERE user_id = ?
UNION ALL
SELECT Least(2,
IF(
(SELECT COUNT(id) FROM `vwMessagesSent` WHERE receiver_id = ? AND user_id = ?)=1, 4, 2)
) - (SELECT COUNT(id) FROM `vwMessagesSent` WHERE user_id = ? AND receiver_id = ?)
)
FROM `vwMessagesSent`
WHERE user_id = ? AND receiver_id = ?
order by SendLimit
LIMIT 1
``````
Note: Not tested - should be close to what you are after
We are using a few interesting SQL items out of the box. Least takes the lowest number of a collection of numbers. It allows us to only return the number of messages the user is allowed to send and for you to warn the user that the insert/message send was not going to be undertaken as they are over quota.
UNION allows you to merge two queries and we use the ALL to make sure we have multiple rows (rather than using the distinct key
LIMIT 1 We sort the order of the two rows returned with the minimum number available to be sent and then only return that row.
Hope this helps. Happy to clarify or fix the main SQL and I would be looking at using at a stored procedure to improve the logic even further. | 1,053 | 4,089 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2021-31 | latest | en | 0.837862 |
https://www.free-online-converters.com/gallons-u-s-fluid/gallons-u-s-fluid-into-drops.html | 1,660,171,718,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571222.74/warc/CC-MAIN-20220810222056-20220811012056-00583.warc.gz | 676,170,713 | 9,482 | Free Online Converters > Convert Gallons (u.s. Fluid) Into Drops
Here you can Convert units of Gallons (u.s. Fluid) to Drops units, find all information about Gallons (u.s. Fluid). So, enter your unit's value in Left Column like Gallons (u.s. Fluid)(if you use standard resolution on most non-HD laptops. FULL HD resolution starts at 1920 x 1080). Otherwise, if you use a lower value, enter the value in the box above. The Result / another converted unit value shell appears in the Left or below Column.
# Convert Gallons (u.s. Fluid) Into Drops
Gallons (u.s. Fluid)
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Drops
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Convert Gallons (u.s. Fluid) Into Drops ,and more. Also, explore many other unit converters or learn more about Volume unit conversions, How mamy Gallons (u.s. Fluid) in Drops
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wikipedia.org lexico.com dictionary.com wikipedia
##### conversion Table / conversion Chart
1 Gallons (u.s. Fluid) = 75708.2357 Drops
2 Gallons (u.s. Fluid) = 151416.4714 Drops
3 Gallons (u.s. Fluid) = 227124.7071 Drops
4 Gallons (u.s. Fluid) = 302832.9428 Drops
5 Gallons (u.s. Fluid) = 378541.1785 Drops
6 Gallons (u.s. Fluid) = 454249.4142 Drops
7 Gallons (u.s. Fluid) = 529957.6499 Drops
8 Gallons (u.s. Fluid) = 605665.8856 Drops
9 Gallons (u.s. Fluid) = 681374.1213 Drops
10 Gallons (u.s. Fluid) = 757082.357 Drops
11 Gallons (u.s. Fluid) = 832790.5927 Drops
12 Gallons (u.s. Fluid) = 908498.8284 Drops
13 Gallons (u.s. Fluid) = 984207.0641 Drops
14 Gallons (u.s. Fluid) = 1059915.2998 Drops
15 Gallons (u.s. Fluid) = 1135623.5355 Drops
16 Gallons (u.s. Fluid) = 1211331.7712 Drops
17 Gallons (u.s. Fluid) = 1287040.0069 Drops
18 Gallons (u.s. Fluid) = 1362748.2426 Drops
19 Gallons (u.s. Fluid) = 1438456.4783 Drops
20 Gallons (u.s. Fluid) = 1514164.714 Drops
21 Gallons (u.s. Fluid) = 1589872.9497 Drops
22 Gallons (u.s. Fluid) = 1665581.1854 Drops
23 Gallons (u.s. Fluid) = 1741289.4211 Drops
24 Gallons (u.s. Fluid) = 1816997.6568 Drops
25 Gallons (u.s. Fluid) = 1892705.8925 Drops
26 Gallons (u.s. Fluid) = 1968414.1282 Drops
27 Gallons (u.s. Fluid) = 2044122.3639 Drops
28 Gallons (u.s. Fluid) = 2119830.5996 Drops
29 Gallons (u.s. Fluid) = 2195538.8353 Drops
30 Gallons (u.s. Fluid) = 2271247.071 Drops
31 Gallons (u.s. Fluid) = 2346955.3067 Drops
32 Gallons (u.s. Fluid) = 2422663.5424 Drops
33 Gallons (u.s. Fluid) = 2498371.7781 Drops
34 Gallons (u.s. Fluid) = 2574080.0138 Drops
35 Gallons (u.s. Fluid) = 2649788.2495 Drops
36 Gallons (u.s. Fluid) = 2725496.4852 Drops
37 Gallons (u.s. Fluid) = 2801204.7209 Drops
38 Gallons (u.s. Fluid) = 2876912.9566 Drops
39 Gallons (u.s. Fluid) = 2952621.1923 Drops
40 Gallons (u.s. Fluid) = 3028329.428 Drops
41 Gallons (u.s. Fluid) = 3104037.6637 Drops
42 Gallons (u.s. Fluid) = 3179745.8994 Drops
43 Gallons (u.s. Fluid) = 3255454.1351 Drops
44 Gallons (u.s. Fluid) = 3331162.3708 Drops
45 Gallons (u.s. Fluid) = 3406870.6065 Drops
46 Gallons (u.s. Fluid) = 3482578.8422 Drops
47 Gallons (u.s. Fluid) = 3558287.0779 Drops
48 Gallons (u.s. Fluid) = 3633995.3136 Drops
49 Gallons (u.s. Fluid) = 3709703.5493 Drops
50 Gallons (u.s. Fluid) = 3785411.785 Drops
50 Gallons (u.s. Fluid) = 3785411.785 Drops
51 Gallons (u.s. Fluid) = 3861120.0207 Drops
52 Gallons (u.s. Fluid) = 3936828.2564 Drops
53 Gallons (u.s. Fluid) = 4012536.4921 Drops
54 Gallons (u.s. Fluid) = 4088244.7278 Drops
55 Gallons (u.s. Fluid) = 4163952.9635 Drops
56 Gallons (u.s. Fluid) = 4239661.1992 Drops
57 Gallons (u.s. Fluid) = 4315369.4349 Drops
58 Gallons (u.s. Fluid) = 4391077.6706 Drops
59 Gallons (u.s. Fluid) = 4466785.9063 Drops
60 Gallons (u.s. Fluid) = 4542494.142 Drops
61 Gallons (u.s. Fluid) = 4618202.3777 Drops
62 Gallons (u.s. Fluid) = 4693910.6134 Drops
63 Gallons (u.s. Fluid) = 4769618.8491 Drops
64 Gallons (u.s. Fluid) = 4845327.0848 Drops
65 Gallons (u.s. Fluid) = 4921035.3205 Drops
66 Gallons (u.s. Fluid) = 4996743.5562 Drops
67 Gallons (u.s. Fluid) = 5072451.7919 Drops
68 Gallons (u.s. Fluid) = 5148160.0276 Drops
69 Gallons (u.s. Fluid) = 5223868.2633 Drops
70 Gallons (u.s. Fluid) = 5299576.499 Drops
71 Gallons (u.s. Fluid) = 5375284.7347 Drops
72 Gallons (u.s. Fluid) = 5450992.9704 Drops
73 Gallons (u.s. Fluid) = 5526701.2061 Drops
74 Gallons (u.s. Fluid) = 5602409.4418 Drops
75 Gallons (u.s. Fluid) = 5678117.6775 Drops
76 Gallons (u.s. Fluid) = 5753825.9132 Drops
77 Gallons (u.s. Fluid) = 5829534.1489 Drops
78 Gallons (u.s. Fluid) = 5905242.3846 Drops
79 Gallons (u.s. Fluid) = 5980950.6203 Drops
80 Gallons (u.s. Fluid) = 6056658.856 Drops
81 Gallons (u.s. Fluid) = 6132367.0917 Drops
82 Gallons (u.s. Fluid) = 6208075.3274 Drops
83 Gallons (u.s. Fluid) = 6283783.5631 Drops
84 Gallons (u.s. Fluid) = 6359491.7988 Drops
85 Gallons (u.s. Fluid) = 6435200.0345 Drops
86 Gallons (u.s. Fluid) = 6510908.2702 Drops
87 Gallons (u.s. Fluid) = 6586616.5059 Drops
88 Gallons (u.s. Fluid) = 6662324.7416 Drops
89 Gallons (u.s. Fluid) = 6738032.9773 Drops
90 Gallons (u.s. Fluid) = 6813741.213 Drops
91 Gallons (u.s. Fluid) = 6889449.4487 Drops
92 Gallons (u.s. Fluid) = 6965157.6844 Drops
93 Gallons (u.s. Fluid) = 7040865.9201 Drops
94 Gallons (u.s. Fluid) = 7116574.1558 Drops
95 Gallons (u.s. Fluid) = 7192282.3915 Drops
96 Gallons (u.s. Fluid) = 7267990.6272 Drops
97 Gallons (u.s. Fluid) = 7343698.8629 Drops
98 Gallons (u.s. Fluid) = 7419407.0986 Drops
99 Gallons (u.s. Fluid) = 7495115.3343 Drops
100 Gallons (u.s. Fluid) = 7570823.57 Drops
101 Gallons (u.s. Fluid) = 7646531.8057 Drops
102 Gallons (u.s. Fluid) = 7722240.0414 Drops
103 Gallons (u.s. Fluid) = 7797948.2771 Drops
104 Gallons (u.s. Fluid) = 7873656.5128 Drops
105 Gallons (u.s. Fluid) = 7949364.7485 Drops
106 Gallons (u.s. Fluid) = 8025072.9842 Drops
107 Gallons (u.s. Fluid) = 8100781.2199 Drops
108 Gallons (u.s. Fluid) = 8176489.4556 Drops
109 Gallons (u.s. Fluid) = 8252197.6913 Drops
110 Gallons (u.s. Fluid) = 8327905.927 Drops
111 Gallons (u.s. Fluid) = 8403614.1627 Drops
112 Gallons (u.s. Fluid) = 8479322.3984 Drops
113 Gallons (u.s. Fluid) = 8555030.6341 Drops
114 Gallons (u.s. Fluid) = 8630738.8698 Drops
115 Gallons (u.s. Fluid) = 8706447.1055 Drops
116 Gallons (u.s. Fluid) = 8782155.3412 Drops
117 Gallons (u.s. Fluid) = 8857863.5769 Drops
118 Gallons (u.s. Fluid) = 8933571.8126 Drops
119 Gallons (u.s. Fluid) = 9009280.0483 Drops
120 Gallons (u.s. Fluid) = 9084988.284 Drops
121 Gallons (u.s. Fluid) = 9160696.5197 Drops
122 Gallons (u.s. Fluid) = 9236404.7554 Drops
123 Gallons (u.s. Fluid) = 9312112.9911 Drops
124 Gallons (u.s. Fluid) = 9387821.2268 Drops
125 Gallons (u.s. Fluid) = 9463529.4625 Drops
126 Gallons (u.s. Fluid) = 9539237.6982 Drops
127 Gallons (u.s. Fluid) = 9614945.9339 Drops
128 Gallons (u.s. Fluid) = 9690654.1696 Drops
129 Gallons (u.s. Fluid) = 9766362.4053 Drops
130 Gallons (u.s. Fluid) = 9842070.641 Drops
131 Gallons (u.s. Fluid) = 9917778.8767 Drops
132 Gallons (u.s. Fluid) = 9993487.1124 Drops
133 Gallons (u.s. Fluid) = 10069195.3481 Drops
134 Gallons (u.s. Fluid) = 10144903.5838 Drops
135 Gallons (u.s. Fluid) = 10220611.8195 Drops
136 Gallons (u.s. Fluid) = 10296320.0552 Drops
137 Gallons (u.s. Fluid) = 10372028.2909 Drops
138 Gallons (u.s. Fluid) = 10447736.5266 Drops
139 Gallons (u.s. Fluid) = 10523444.7623 Drops
140 Gallons (u.s. Fluid) = 10599152.998 Drops
141 Gallons (u.s. Fluid) = 10674861.2337 Drops
142 Gallons (u.s. Fluid) = 10750569.4694 Drops
143 Gallons (u.s. Fluid) = 10826277.7051 Drops
144 Gallons (u.s. Fluid) = 10901985.9408 Drops
145 Gallons (u.s. Fluid) = 10977694.1765 Drops
146 Gallons (u.s. Fluid) = 11053402.4122 Drops
147 Gallons (u.s. Fluid) = 11129110.6479 Drops
148 Gallons (u.s. Fluid) = 11204818.8836 Drops
149 Gallons (u.s. Fluid) = 11280527.1193 Drops
150 Gallons (u.s. Fluid) = 11356235.355 Drops
## how many Gallons (u.s. Fluid) Into Drops
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How Many Microlitres in Gallons (u.s. Fluid) | 4,165 | 11,344 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2022-33 | longest | en | 0.430952 |
http://www.textbooks.com/Beginning-Algebra-with-Applications-and-Visualizations-05-Edition/9780321157263/Gary-K-Rockswold-and-Terry-A-Krieger.php?CSID=AKTKTADDSSZUUKTAUCCDCASMB | 1,503,007,269,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886104160.96/warc/CC-MAIN-20170817210535-20170817230535-00350.warc.gz | 703,930,874 | 16,743 | Ship-Ship-Hooray! FREE 2-Day Air* on \$25+ Details >
Beginning Algebra with Applications and Visualizations - 05 edition
ISBN10: 0321157265
Edition: 05
Published: 2005
International: No
ISBN10: 0321157265
Edition: 05
Summary
Beginning Algebra with Applications and Visualization offers an innovative approach to the beginning algebra curriculum that allows students to gain both skills and understanding. This text not only prepares students for future mathematics courses, but it also demonstrates to students the relevance of mathematics. Real data, graphs, and tables play an important role in the course, giving meaning to the numbers and equations that students encounter. This approach increases student interest, motivation, and the likelihood for success. Many students think in visual, concrete terms and not abstractly. This text helps students learn mathematics better by moving from the concrete to the abstract. It makes use of multiple representations (verbal, graphical, numerical, and symbolic), applications, visualization, and technology.
1. Introduction to Algebra.
Numbers, Variables, and Algebraic Expressions.
Fractions.
Exponents and Order of Operations.
Real Numbers and the Number Line.
Addition and Subtraction of Real Numbers.
Multiplication and Division of Real Numbers.
Properties of Real Numbers.
Simplifying and Writing Algebraic Expressions.
2. Linear Equations and Inequalities.
Introduction to Equations.
Linear Equations.
Introduction to Problem Solving.
Formulas.
Linear Inequalities.
3. Graphing Equations.
Introduction to Graphing.
Linear Equations in Two Variables.
More Graphing of Lines.
Slope and Rates of Change.
Slope-Intercept Form.
Point-Slope Form.
Introduction to Modeling.
4. Systems of Linear Equations in Two Variables.
Solving Systems of Linear Equations Graphically and Numerically.
Solving Systems of Linear Equations by Substitution.
Solving Systems of Linear Equations by Elimination.
Systems of Linear Inequalities.
5. Polynomials and Exponents.
Rules for Exponents.
Multiplication of Polynomials.
Special Products.
Integer Exponents and the Quotient Rule.
Division of Polynomials.
6. Factoring Polynomials and Solving Equations.
Introduction to Factoring.
Factoring Trinomials (I).
Factoring Trinomials (II).
Special Types of Factoring.
Solving Equations by Factoring.
Factoring Polynomials.
7. Rational Expressions.
Introduction to Rational Expressions.
Multiplication and Division of Rational Expressions.
Addition and Subtraction of Rational Expressions (I).
Addition and Subtraction of Rational Expressions (II).
Complex Fractions.
Rational Equations and Formulas.
Proportions and Variation.
Multiplication and Division of Radical Expressions.
Formulas from Geometry.
Rational Exponents.
Complex Numbers.
Parabolas. | 591 | 2,838 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2017-34 | latest | en | 0.864285 |
https://www.tradingview.com/script/l5RbPJEz-Snake-Oscillator-1-1/ | 1,527,069,535,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794865468.19/warc/CC-MAIN-20180523082914-20180523102914-00316.warc.gz | 852,798,182 | 71,497 | # Snake Oscillator 1.1
887 4
This Oscillator helps to identify the ending of waves, good to get entries and exits. Now, with a tolerance line.
```study("Snake Oscillator","SNK_OSC")
len = input(25,"Length")
h = ema(high,len)
l = ema(low,len)
hp = h/h[len]
lp = l/l[len]
avg = avg(hp,lp)
havg = ema(highest(avg,len),len)
lavg = ema(lowest(avg,len),len)
avg2 = avg(havg,lavg)
dif = havg - avg2
pa = plot(avg,color=gray)
ph = plot(havg,color=gray)
pl = plot(lavg,color=gray)
plot(havg+dif,color=black)
plot(lavg-dif,color=black)
fill(ph,pa,color=red,transp=60)
fill(pl,pa,color=green,transp=60)
```
Can you explain how it works please ? Thanks !
hp = h/h // This provides the porcentage of change of ema(high, len) considered len periods back
lp = l/l // same of hp, but with lows
avg = avg(hp,lp) // Average of porcentage of change of low and High
havg = ema(highest(avg,len),len) // Highest avg of change of len periods back amortized by ema
lavg = ema(lowest(avg,len),len) //Lowest avg of change of len periods back amortized by ema
--
Idea is check if the trend is making the same proportion of highs and lows in a trend move.
samuelhei
thank you for taking time to respond !
samuelhei
@samuelhei, Dag, I still don't understand. Is it the same thing as "Snake In Borders" on MT4 (A.k.a snake force)?
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Home Stock Screener Forex Signal Finder Cryptocurrency Signal Finder Economic Calendar How It Works Chart Features House Rules Moderators Website & Broker Solutions Widgets Stock Charting Library Feature Request Blog & News FAQ Help & Wiki Twitter | 534 | 1,810 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2018-22 | longest | en | 0.627296 |
http://mathoverflow.net/questions/108773/independence-of-leibniz-rule-and-locality-from-other-properties-of-the-derivativ?answertab=votes | 1,461,896,798,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461860110356.23/warc/CC-MAIN-20160428161510-00028-ip-10-239-7-51.ec2.internal.warc.gz | 108,280,051 | 17,527 | # Independence of Leibniz rule and locality from other properties of the derivative?
The following is meant to be an axiomatization of differential calculus of a single variable. To avoid complications, let's say that $f$, $g$, $f'$, and $g'$ are smooth functions from $\mathbb{R}$ to $\mathbb{R}$ ("smooth" being defined by the usual Cauchy-Weierstrass definition of the derivative, not by these axioms, i.e., I don't want to worry about nondifferentiable points right now). In all of these, assume the obvious quantifiers such as $\forall f \forall g$.
Z. $\exists f : f'\ne 0$
U. $1'=0$
A. $(f+g)'=f'+g'$
C. $(g \circ f)'=(g'\circ f)f'$
P. $(fg)'=f'g+g'f$
L. The value of $f'(x)$ is determined by knowing $f$ in any neighborhood of $x$.
Axioms A and P hold in any differential algebra. C and L mean that we're talking about something more specific than a differential algebra; they're meaningful only because we're talking about a ring of functions.
I could choose to omit U, since it can be proved from the others. I would prefer to keep U and omit P. Is P superfluous, or can anyone find a model in which P fails?
Likewise, is L independent of the others?
What seems to be tricky is to rule out models of the general flavor of $f'(x)=f^{CW}(x-17)$, where $f^{CW}$ is the usual Cauchy-Weierstrass derivative of $f$.
Are there models that are not the same as CW? Is this a nice axiomatization? Could it be improved?
[EDIT] Tom Goodwillie didn't say so explicitly, but his answer, along with one of my comments below his answer, shows that Z, A, and C suffice, so U, P, and L are not needed.
It looks like you can also take P as an axiom and recover the standard derivative, i.e., either P or C can be proved from the other: Do these properties characterize differentiation?
-
If you grant that $c'=0$ when $c$ is a constant, you can argue as follows:
EDIT: Actually $c'=0$ follows from $1'=0$ using the chain rule, since $c=c\circ 1$.
Let $I(x)=x$. Since $I=I\circ I$, the chain rule gives $(I')^2=I'$, so $I'$ is the characteristic function of a set $A$ of real numbers. Let $T_c(x)=x+c$. Then $T_c'=I'+c'=I'$, and the chain rule applied to $T_c=I\circ T_c$ then implies $I'=(I'\circ T_c)I'$. That means that the set $A$ is either all or nothing. But you can't have $I'$ identically $0$ because that would imply $f'=(f\circ I)'=0$ for all $f$. So $I'=1$ as expected.
Now consider linear functions. If $L_m(x)=mx$, then since $L_m(x+a)-L_m(x)$ is constant, $L_m'(x+a)-L_m'(x)$ is zero. Thus $L_m'$ is a constant depending on $m$, say $h(m)$. The map $h$ is an additive homomorphism, and by the chain rule it is also multiplicative. We also know $h(1)=1$. This implies that $h(m)=m$ for all $m$. (A ring map from reals to reals preserves squares, therefore preserves ordering, therefore is continuous ...)
So $f'$ is what it should be when $f$ is polynomial of degree at most one. Now let $S(x)=x^2$. From $S(x+t)=S(x)+2tx+t^2$ we get $S'(x+t)=S'(x)+2t$, therefore $S'(x)=S'(0)+2x$. But $S'(0)=0$ using $S(-x)=S(x)$. So the derivative of squaring is what it should be.
Now the special case of Leibniz that says $(f^2)'=2ff'$ follows by the chain rule. The general case follows by expressing $fg$ in terms of $f^2$, $g^2$, and $(f+g)^2$.
EDIT: This was all about global functions. But it can be extended. Let me spell out what I hope your axioms are: $f'$ is defined when $f$ is a $C^\infty$ real function whose domain is an open set $U\subset \mathbb R$, and $f'$ is another such function with the same domain. The axioms are
(U) $1_U'=0_U$ where $1_U$ is the constant function on $U$.
(A) $(f+g)'=f'+g'$ where $f$ and $g$ (and $f+g$) have the same domain.
(C) $(f\circ g)'=(f'\circ g)g'$, when $f$ has domain $U$ and $g$ has domain $V$ and $g(V)\subset U$, so that $f\circ g$ and $f'\circ g$ have domain $V$.
(Z) For every nonempty $U$ there is some $f$ with domain $U$ such that $f'$ is not identically zero.
The arguments that I gave above can be adapted to show then that:
$c_U'=0$ for any constant function on any $U$.
$I_U'=1$ where $I_U$ with domain $U$ is defined by $I_U(x)=x$. (Here you have to mess around with compositions $I_U\circ (I_V+c_V)$.)
So in the end you get the desired localization property, too: the operator commutes with restriction from $U$ to $V\subset U$ by the chain rule, because restriction is composition with $I_V$.
-
That's a fun argument! – Todd Trimble Oct 4 '12 at 12:22
Yeah. I wonder if anybody ever saves a little time teaching calculus by deducing the Leibniz rule from Chain Rule + derivative of $x^2$. (I wouldn't recommend it, because the Leibniz rule is more important than that--unworthy of such tricks.) – Tom Goodwillie Oct 4 '12 at 14:17
Another way to put in the final nail in the coffin is to differentiate $a^2-b^2$ where $a=(g+f)/2$ and $b=(g-f)/2$. (Saves a bit of time in case anyone actually wants to perform this.) – François G. Dorais Oct 4 '12 at 14:57
Very sweet, thanks! Your argument about S being an even function made me realize that axiom U was also unnecessary. The chain rule implies that any even function has a zero derivative at $x=0$. Again using the chain rule, we see that if 1'=0 at $x=0$, then 1'=0 everywhere. – Ben Crowell Oct 5 '12 at 18:18
Without using $1'=0$, you can prove that the derivative of $x$ is 1. Then by the additive property, the derivative of $-x$ has to be $-1$. After that, you can get $1'=0$, evaluated at 0, by applying the chain rule to $1\circ -x$. So if locality holds, $1'=0$ must hold everywhere. – Ben Crowell Oct 6 '12 at 16:15
Actually L excludes that $f'$ is a translate of the usual derivative $D$ (for instance there are $C^{\infty}$ functions with compact support). Moreover, by C, with $g=f=\mathrm{id}$ we have $x'=(x')^2$, and by the assumed continuity of $x'$, $x'$ is either $1$ or $0$ identically; the former is excluded taking, again in C, $g=\mathrm{id}$ and $f$ as in Z. By P we get $f'=Df$ at least for any polynomial function. So you'd have it for all smooth function, provided you also assume that $f\mapsto f'$ is continuous in the $C^1_{loc}$ sense.
With no other assumption, I'm not sure if one can reach this conclusion. However, the set of $f$ for which $f'=Df$ is quite a wide one; for instance, by C and L, it includes any $f$ that locally solves a functional equation like $p\circ f=q$, for non-zero polymonials $q$ and $p$; or also $f\circ p=q$, with $p$ of odd degree (hence surjective).
- | 1,955 | 6,458 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2016-18 | latest | en | 0.945286 |
https://www.acmicpc.net/problem/16264 | 1,544,921,500,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376827175.38/warc/CC-MAIN-20181216003916-20181216025916-00268.warc.gz | 780,323,693 | 12,313 | 시간 제한 메모리 제한 제출 정답 맞은 사람 정답 비율
2 초 512 MB 0 0 0 0.000%
## 문제
Vasya likes to study arrays. Recently his parents presented him with an array a that contains elements equal to 1 and - 1. Vasya immediately started to study it.
Additionally Vasya likes zeroes. So he decided to consider various subarrays a[li, ..., ri] of array a. For each subarray he tries to find the maximum length of its subarray with the sum equal to 0. If there is no such subarray, he considers the answer to be 0. Vasya has written down q subarray requests [li, ri], and now he wants to find the sum of answers to them.
For example, let us consider sample test.
• subarray [1, 5]: the maximal subarray with sum 0 — [2, 5];
• subarray [1, 3]: the maximal subarray with sum 0 — [2, 3];
• subarray [2, 4]: the maximal subarray with sum 0 — [2, 3];
• subarray [3, 4]: no subarray with sum 0;
• subarray [3, 5]: the maximal subarray with sum 0 — [4, 5].
So the sum of answers for the sample test is 4 + 2 + 2 + 0 + 2 = 10.
## 입력
Input data contains several test cases. The first line contains the number of test cases t (1 ≤ t ≤ 1000).
Each of t test cases is described in the following way: the first line of the description contains n — the number of elements of the array (1 ≤ n ≤ 105).
The following line contains n integers ai — elements of the array (ai = - 1 or ai = 1).
The following line contains q — the number of subarrays that Vasya is interested in (1 ≤ q ≤ 105).
Each of the following q lines contains two integers li, ri — left and right border of the i-th subarray (1 ≤ li ≤ ri ≤ n)
It is guaranteed that the sum of n in all test cases of one input data doesn't exceed 105, the sum of q in all test cases of one input data doesn't exceed 105.
## 출력
For each test output one integer — the sum of answers for the given q subarrays.
## 예제 입력 1
1
5
1 -1 1 1 -1
5
1 5
1 3
2 4
3 4
3 5
## 예제 출력 1
10 | 609 | 1,896 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2018-51 | longest | en | 0.726921 |
https://research.stlouisfed.org/fred2/series/RGDPL2SYA625NUPN | 1,454,835,826,000,000,000 | text/html | crawl-data/CC-MAIN-2016-07/segments/1454701148758.73/warc/CC-MAIN-20160205193908-00212-ip-10-236-182-209.ec2.internal.warc.gz | 804,680,561 | 19,127 | # Purchasing Power Parity Converted GDP Per Capita (Laspeyres), derived from growth rates of domestic absorption for Syria
2010: 3,782.63395 2005 International Dollars per Person
Annual, Not Seasonally Adjusted, RGDPL2SYA625NUPN, Updated: 2012-09-17 10:04 AM CDT
1yr | 5yr | 10yr | Max
Source Indicator: rgdpl2
Source: University of Pennsylvania
Release: Penn World Table 7.1
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(a) Purchasing Power Parity Converted GDP Per Capita (Laspeyres), derived from growth rates of domestic absorption for Syria, 2005 International Dollars per Person, Not Seasonally Adjusted (RGDPL2SYA625NUPN)
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Use a formula to modify and combine data series into a single line. For example, invert an exchange rate a by using formula 1/a, or calculate the spread between 2 interest rates a and b by using formula a - b.
Use the assigned data series variables above (e.g. a, b, ...) together with operators {+, -, *, /, ^}, braces {(,)}, and constants {e.g. 2, 1.5} to create your own formula {e.g. 1/a, a-b, (a+b)/2, (a/(a+b+c))*100}. The default formula 'a' displays only the first data series added to this line. You may also add data series to this line before entering a formula.
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``` University of Pennsylvania, Purchasing Power Parity Converted GDP Per Capita (Laspeyres), derived from growth rates of domestic absorption for Syria [RGDPL2SYA625NUPN], retrieved from FRED, Federal Reserve Bank of St. Louis https://research.stlouisfed.org/fred2/series/RGDPL2SYA625NUPN, February 7, 2016. ```
Retrieving data.
Graph updated. | 497 | 1,844 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2016-07 | longest | en | 0.789345 |
https://hologram-and-holography.com/HologramSticker/measuring-light-wavelength | 1,652,903,003,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662522309.14/warc/CC-MAIN-20220518183254-20220518213254-00176.warc.gz | 366,431,194 | 6,680 | # Measuring Light Wavelength
January 8, 2021
The earliest accurate determination of wavelength was, I think, by Michelson. Using his invention, the Michelson Interferometer, he could turn a micrometer dial and actually count how many wavelengths he moved a mirror. Reasonable monochromatic light could be had at the time from mercury vapor (or other elemental) discharge tubes or from a monochromator (a spectroscope with a slit on the output to select a color). This was around 1880. I confess I don't know for sure. He was determined to measure the speed of light. Exactly when he worked on wavelength I don't know. I'm sure someone here can add that info.
Michelson was able to count a lot of wavelengths so that the mirror moved enough to get a good average from the mechanical measurement. He was able to measure the wavelength of precisely known colors so that the results were easily reproduced by others. At the time there was a lot of interest in the spectra of excited atoms of elements and of the sun and stars through the new medium of photography. Photographic spectra of a star was done first in 1863.
Once you have a wavelength and the speed, which Michelson also determined to a high degree of accuracy by refining the the rotating mirror method, the frequency is just f=velocity/wavelength. The frequencies are crazy big numbers like the red in a helium-neon laser is 4.7376 x 10^14 per second or 473.76 THz. That's tera-Hertz and it is nice that tera- is also trillion. This is why people use wavelength in nanometers, so that the red from the laser is described as 632.8nm, which is a lot easier. If you read older material you will see that we used a slightly more convenient measure, the Angstrom, which is 1/10 a nanometer. The same light is 6328 \$overset{circ}{A} \$. The Angstrom is abbreviated as a capital 'A' with a little dot or circle over it. (It is in the UTF8 character set but I'm not sure will render for everyone, so I faked it in LaTeX.)
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Quick Lab: Measuring the Speed of Light with Marshmallows | 510 | 2,216 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2022-21 | longest | en | 0.968357 |
https://dieselweb.org/algorithm-classification-simple-recursive-algorithm/ | 1,571,255,056,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986669546.24/warc/CC-MAIN-20191016190431-20191016213931-00220.warc.gz | 477,994,615 | 10,349 | 12
Oct
# Algorithm Classification Simple Recursive Algorithm
in this particular session we are
discussing the first category whatever we have mentioned in the algorithm
classifications and that is the simple recursive algorithms we know the
recursion in case of recursion the recursive algorithm will have a base
case that means for certain situations for certain inputs outputs are known to
us where the algorithm will terminate and then in the next part we’ll be
having a recursive call which will be calling the function itself either
directly or indirectly okay so that is our recursion but here we are calling it
as a simple recursive algorithm we are using this term simple because here we
are having multiple other algorithm types which are inherently recursive if
you consider your backtracking algorithm that is inherently recursive so here we
are considering the problem which is very simple okay solve the best case
directly we know the base case I discussed that for certain situations
for certain inputs outputs are known to us
so here the solve to the biscuits will be done directly it occurs with a
simpler sub problem that means the problem will be will be made very simple
and then the problem will be solved through this particular simple recursive
algorithms extra initiatives may equate to convert the solution of the solution
to the simpler sub problems into a solution to the given problem so that’s
why we shall solve the sub problem we shall make the sub problem very simple
and that sub problem solution will contribute towards the solution of the
main given problem it is a simple recursive because some other algorithms
are inherently recursive so I mentioned that one if we consider the backtracking
algorithm that is the in a inherently records it but here this in this
particular category were talking about a very simple recursive algorithms as an
example we can go for this let us suppose it has been asked that to count
the number of data items in a list present so we are going to count a
number of data items in little is present so now you see the
business will be if the list is empty then count will be zero and the Z will
be returned so if the list is empty then returns zero otherwise I shall go for
the recursive call other than the first item that means if you consider the
first item I’m just taking the first item away count the number of remaining
data items in the list and if we can count the number of remaining data items
in the list then add one to the result because the first item count will be
added and that value will be returned now while counting the number of
remaining data items in the list we’ll be calling the recursive function we’ll
be calling the recursive algorithm again and that is the source of this recursion
so here we have discussed what is our simple recursive algorithm and we have
explained that one with one a very simple example so please watch our next
videos there will be going for other algorithm classifications thanks for
watching this video
Tags: , , | 605 | 3,037 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2019-43 | latest | en | 0.945417 |
https://www.sololearn.com/en/discuss/2498317/rgba | 1,702,090,820,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100781.60/warc/CC-MAIN-20231209004202-20231209034202-00318.warc.gz | 1,082,217,479 | 90,810 | + 11
# rgba()
I know rgb stands for Red, Green, Blue but what does 'a' stand for?
15th Sep 2020, 3:17 PM
Kayla Onobun (QOG)
+ 10
alpha , mean how much it visible(opacity)(transparency) rgba(0,0,0,0.1) rgba(1,0,0,1) = red full opacity alpha 0 is not visible 0.1 is less visible 0.7 is little visible 1 is full visible
15th Sep 2020, 3:18 PM
Rajababu Shah
+ 15
Alpha, as in, the opacity of the specified color.
15th Sep 2020, 3:19 PM
Hatsy Rei
15th Sep 2020, 3:19 PM
Nilesh
+ 8
See Jadene & Kayla Onobun RGBA color values are an extension of RGB color values with an alpha - which stands for the opacity of the color. Here's an example of RGBA- p1 {background-color:rgba(255,0,0,0.9);} red with opacity #p2 {background-color:rgba(0,255,0,0.8);} green with opacity #p3 {background-color:rgba(0,0,255,0.6);} blue with opacity hope this helps you ✌️
15th Sep 2020, 3:23 PM
Piyush
+ 7
RGBA color values are an extension of RGB color values with an alpha channel - which specifies the opacity for a color. An RGBA color value is specified with: rgba(red, green, blue, alpha). The alpha parameter is a number between 0.0 (fully transparent) and 1.0 (fully opaque) Hope this will help u out😔😔😔😔 Jadene & Kayla Onobun
15th Sep 2020, 3:31 PM
JAY • ≫
+ 6
It stands for the Alpha which indicates the opacity😉
15th Sep 2020, 3:41 PM
+ 3
"a" is for transparency
15th Sep 2020, 3:19 PM
Rohit Kh
+ 2
'a' means 'alpha' which is for giving transparency for elements 1 means fully opaque and, 0 means fully transparent You can use values from 0 to 1 only Eg:- rgba(255,0,0,0) It will not be visible rgba(255,0,0,1) It will be fully visible rgba(255,0,0,0.5) It will be half Transparent.
16th Sep 2020, 8:59 AM
Arnav Kumar [Less/Not Active]
+ 2
"a" stands for "alpha". it's the transparency.
16th Sep 2020, 3:53 PM
Space Kitty
+ 1
'a' is the alpha value, it decides the opacity strength i.e the transpiracy
16th Sep 2020, 1:25 PM
Ravi Teja
+ 1
a means alpha, Means fadeness of color i.e. a=0 means full faded i.e. invisible... a=1 means maximum intensity i.e. no fadeness at all... for value of a in range 0 to 1 means between invisible to full intensity...... HOPE U UNDERSTOOD IT.....😊
17th Sep 2020, 7:13 AM
saurabh
0
Alpha and some call it Opacity
17th Sep 2020, 3:21 AM
Malan Kudakwashe Kapishe | 807 | 2,283 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2023-50 | latest | en | 0.787497 |
https://math.stackexchange.com/questions/tagged/summation?sort=unanswered&page=4 | 1,561,030,330,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627999210.22/warc/CC-MAIN-20190620105329-20190620131329-00221.warc.gz | 510,394,042 | 37,949 | # Questions tagged [summation]
2,002 questions
97 views
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73 views
### A summation of the multiplication of reflected Mobius functions and their behavior for different values of $k$
$$\Psi_k(N)=\sum_{n=1}^{N} \mu(n)\mu(k-n)$$ where $\mu(n)$ is the Mobius function. This function is interesting to me because for the case of $k=N$ it has the symmetric property of being odd with ...
73 views
### Summation of Integrals without using $\zeta(2)$
How do you evaluate the sum $$\sum_{k=1}^\infty\int_{\sqrt k}^{\sqrt{k+1}} \left(\frac{x^2}{k}-1\right)\, dx$$ without using $\zeta(2)$? I can see some relationship to $\dfrac1{\lfloor x^2\rfloor}$ ...
97 views
### A sort of Wolstenholme's $p^2$-congruence.
It is well-known (Wolstenholme Theorem) that for any prime $p$ such that $p>3$ the Harmonic number $H_{p-1}$ satisfies the congruence $$H_{p-1}:= \sum_{i=1}^{p-1}\frac{1}{i}\equiv 0 \pmod {p^2}$$ ...
I am told that $A$ and $B$ are random variables, and that $\mathbf{E}(A|B) = \gamma B$. Define $D = A/B$. Using the law of iterated expectations it can be shown that $\mathbf{E}(D) = \gamma$. Now ... | 1,293 | 3,909 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2019-26 | latest | en | 0.793034 |
https://www.thingiverse.com/thing:12097/comments | 1,555,657,608,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578527148.46/warc/CC-MAIN-20190419061412-20190419083412-00507.warc.gz | 828,393,105 | 13,675 | # Printable D20
Oct 1, 2011
### Thing Apps Enabled
Probably a dumb question, but did you need supports to 3D Print this?
Okay mr whosawhatsis, your solution is WRONG! i know it will make you annoyed, and here is the truth... your solution makes 156 vertices for a 20 vertex shape... check the results when it has written the icos. it is completely wonky. When you zoom close to an edge you will see alot of mismatched borders. I think it's perhaps related to the 31.75 angle for a start. Where did you get that figure? whatever it is, the rusult is not an actual icosahedron, so you have written your code on other pages and it's wrong... i do the same always. can you find a correct version?
I had a go with this: edge = asin(1/(2*sin(72))); which is 31.7175, it still gives 156 vertices.
technically speaking you are distributing loaded dice
Top level object is a 3D object:
Simple: yes
Vertices: 156
Halfedges: 468
Edges: 234
Halffacets: 160
Facets: 80
Volumes: 2
Hi who!
just a note you can add to the instructions
For the best result use the same top/bottom and shell thickness ;)
"there just isn't (currently?) an elegant way to generate numbers in OpenSCAD "
How elegant of a solution are you looking for? Once you have outline information encoded in a way you can consume, you could actually generate numbers directly from there. I don't think it would be practical computationally though.
I think the most elegant solution would be to take FreeType, ge
nerate bitmaps with 256 levels of gray, and use those as height maps. That's about as elegant as I'd expect from OpenScad.
IMO, an elegant solution should be one that can be stated very concisely. Openscad doesn't have a way to load font data from within itself, and the shapes of arabic numerals cannot be procedurally generated in any way that doesn't amount to painting with polygons.
The most elegant solution I thought of was actually the same one used by donb's "Happy Numbers 0.2" ( http://www.thingiverse.com/thing:8691http://www.thingiverse.com/thi... ), but I didn't want to put such high-poly characters on the little faces of the D20, and making my own OpenSCAD-compatible DXF out of the numbers turned out
to be more trouble than it's worth.
Happy Numbers 0.2
by donb
I see. Yes, there are two things I could see OpenScad doing. 1) give the ability to read from files. 2) Have fonts as a first class citizen. I'd love to be able to place a 3D character as easily as I place a cube. OpenScad, integrated with FreeType, should make this an interesting thing to do. It could either be done as bitmaps that are linear extruded, or something more interesting than that.
New version of OpenSCAD has text(). Just letting you know.
Yah. I installed the latest a few months back to check things out. Text(), and minkowski is finally useable, and dynamic arrays. It's certainly becoming more capable than it was 5 years ago. | 729 | 2,906 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2019-18 | latest | en | 0.952763 |
https://www.tutorialspoint.com/haskell-program-to-check-whether-the-input-number-is-a-palindrome | 1,685,283,340,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224643784.62/warc/CC-MAIN-20230528114832-20230528144832-00534.warc.gz | 1,157,479,999 | 10,434 | # Haskell Program To Check Whether The Input Number Is A Palindrome
This tutorial discusses writing a program to check whether the input number is a Palindrome number in the Haskell programming language.
A number can be called as a palindrome when it results in the same number when the digits are reversed. For Example, the number 12321 is a palindrome, because after reversing the digits it results in the same number.
In this tutorial we see,
• Program to check whether the input number is a palindrome using the string reverse function.
• Program to check whether the input number is a palindrome using the recursive function.
## Method 1: Checking Palindrome using String Reverse Function
### Algorithm Steps
• We declared a function isPalindrome which takes an integer as an argument and returns a boolean value.
• Using the "Where" keyword to express the logic in multiple statements
• Converting the number to number to string using the function show and
• Reversing the string by using the function reverse
• Comparing both the strings
• Printing the results as per the boolean expression got from comparison
### Example
-- function declaration
isPalindrome :: Int->Bool
-- function definition
isPalindrome n = (n == k)
where
-- Converting integer to string using function show
nStr = reverse (show n)
-- Converting string to an integer using function read
k = (read nStr :: Int)
main :: IO()
main = do
-- initializing variable num
let num = 12321
-- invoking the function isPalindrome
let status = isPalindrome num
-- printing the status
if(status==True)
then print ("The number " ++ show num ++ " is a palindrome")
else print ("The number " ++ show num ++ " is not a palindrome")
### Output
"The number 12321 is a palindrome"
## Method 2: Checking Palindrome using String Recursive Function
### Algorithm
• We declared a function isPalindrome as such it takes three integer arguments and returns a boolean value.
• Create a main function and define a value to check whether it is a palindrome or not.
• Initiailize the Palendrom function in the main function towards the defined number.
• Print the result as per the boolean expression got from the Palendrom function.
### Example
Program to check whether the input number is a palindrome using the recursive function
-- function declaration
isPalindrome :: Int->Int->Int->Bool
-- function definition
isPalindrome num a revNum = if(a==0)
then if (revNum == num)
then True
else False
else k
where
d = mod a 10
newRevNum = revNum*10 + d
newA = div a 10
k = isPalindrome num newA newRevNum
main :: IO()
main = do
-- initializing variable num
let num = 1256521
-- invoking the function isPalindrome
let status = isPalindrome num num 0
-- printing the status
if(status==True)
then print ("The number " ++ show num ++ " is a palindrome")
else print ("The number " ++ show num ++ " is not a palindrome")
### Output
"The number 1256521 is a palindrome"
## Conclusion
In this tutorial, we discussed two ways to implement a program to check whether the number is a palindrome in the Haskell programming Language. | 691 | 3,100 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2023-23 | latest | en | 0.599477 |
http://de.metamath.org/mpeuni/exlimdv.html | 1,718,698,070,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861747.70/warc/CC-MAIN-20240618073942-20240618103942-00403.warc.gz | 6,273,615 | 8,990 | Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > exlimdv Structured version Visualization version GIF version
Theorem exlimdv 1848
Description: Deduction form of Theorem 19.23 of [Margaris] p. 90, see 19.23 2067. (Contributed by NM, 27-Apr-1994.) Remove dependencies on ax-6 1875, ax-7 1922. (Revised by Wolf Lammen, 4-Dec-2017.)
Hypothesis
Ref Expression
exlimdv.1 (𝜑 → (𝜓𝜒))
Assertion
Ref Expression
exlimdv (𝜑 → (∃𝑥𝜓𝜒))
Distinct variable groups: 𝜒,𝑥 𝜑,𝑥
Allowed substitution hint: 𝜓(𝑥)
Proof of Theorem exlimdv
StepHypRef Expression
1 exlimdv.1 . . 3 (𝜑 → (𝜓𝜒))
21eximdv 1833 . 2 (𝜑 → (∃𝑥𝜓 → ∃𝑥𝜒))
3 ax5e 1829 . 2 (∃𝑥𝜒𝜒)
42, 3syl6 34 1 (𝜑 → (∃𝑥𝜓𝜒))
Colors of variables: wff setvar class Syntax hints: → wi 4 ∃wex 1695 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1713 ax-4 1728 ax-5 1827 This theorem depends on definitions: df-bi 196 df-ex 1696 This theorem is referenced by: exlimdvv 1849 exlimddv 1850 ax13lem1 2236 ax13 2237 nfeqf 2289 axc15 2291 ax12v2OLD 2330 sbcom2 2433 tpid3gOLD 4249 sssn 4298 elpreqprb 4335 reusv2lem2 4795 reusv2lem2OLD 4796 ralxfr2d 4808 euotd 4900 wefrc 5032 wereu2 5035 releldmb 5281 relelrnb 5282 elres 5355 iss 5367 onfr 5680 dffv2 6181 dff3 6280 elunirn 6413 fsnex 6438 f1prex 6439 isomin 6487 isofrlem 6490 ovmpt4g 6681 soex 7002 f1oweALT 7043 op1steq 7101 fo2ndf 7171 mpt2xopynvov0g 7227 reldmtpos 7247 rntpos 7252 wfrlem12 7313 wfrlem17 7318 erdisj 7681 map0g 7783 resixpfo 7832 domdifsn 7928 xpdom3 7943 domunsncan 7945 enfixsn 7954 fodomr 7996 mapdom2 8016 mapdom3 8017 phplem4 8027 php3 8031 sucdom2 8041 pssnn 8063 ssfi 8065 domfi 8066 findcard2 8085 ac6sfi 8089 isfinite2 8103 xpfi 8116 domunfican 8118 fiint 8122 fodomfib 8125 mapfien2 8197 marypha1lem 8222 ordiso 8304 hartogslem1 8330 brwdom2 8361 wdomtr 8363 brwdom3 8370 unwdomg 8372 xpwdomg 8373 unxpwdom2 8376 inf3lem2 8409 epfrs 8490 tcmin 8500 cplem1 8635 pm54.43 8709 dfac8alem 8735 dfac8b 8737 dfac8clem 8738 ac10ct 8740 acni2 8752 acndom 8757 numwdom 8765 wdomfil 8767 wdomnumr 8770 iunfictbso 8820 dfac2 8836 dfac9 8841 kmlem13 8867 cdainf 8897 fictb 8950 cfeq0 8961 cff1 8963 cfflb 8964 cofsmo 8974 cfsmolem 8975 coftr 8978 infpssr 9013 fin4en1 9014 fin23lem7 9021 isf34lem4 9082 axcc3 9143 domtriomlem 9147 axdc2lem 9153 axdc3lem2 9156 axdc3lem4 9158 axdc4lem 9160 ac6num 9184 ttukeylem6 9219 ttukeyg 9222 fodomb 9229 iundom2g 9241 alephreg 9283 fpwwe2lem11 9341 fpwwe2lem12 9342 canthp1 9355 pwfseq 9365 gruen 9513 grudomon 9518 gruina 9519 grur1 9521 ltexnq 9676 ltbtwnnq 9679 genpn0 9704 psslinpr 9732 prlem934 9734 ltaddpr 9735 ltexprlem2 9738 ltexprlem6 9742 ltexprlem7 9743 reclem2pr 9749 reclem4pr 9751 suplem1pr 9753 negn0 10338 sup2 10858 supaddc 10867 supmul1 10869 zsupss 11653 fiinfnf1o 13000 hasheqf1oi 13002 hashfun 13084 hashf1 13098 rtrclreclem3 13648 rlimdm 14130 climcau 14249 caucvgb 14258 summolem2 14294 zsum 14296 sumz 14300 fsumf1o 14301 fsumss 14303 fsumcl2lem 14309 fsumadd 14317 fsummulc2 14358 fsumconst 14364 fsumrelem 14380 ntrivcvg 14468 prodmolem2 14504 zprod 14506 prod1 14513 fprodf1o 14515 fprodss 14517 fprodcl2lem 14519 fprodmul 14529 fproddiv 14530 fprodconst 14547 fprodn0 14548 ruclem13 14810 4sqlem12 15498 vdwapun 15516 vdwlem9 15531 vdwlem10 15532 ramz 15567 ramub1 15570 firest 15916 mremre 16087 isacs2 16137 iscatd2 16165 sscfn1 16300 sscfn2 16301 gsumval2a 17102 symggen 17713 cyggex2 18121 gsumval3 18131 gsumzres 18133 gsumzcl2 18134 gsumzf1o 18136 gsumzaddlem 18144 gsumconst 18157 gsumzmhm 18160 gsumzoppg 18167 gsum2d2 18196 pgpfac1lem5 18301 ablfaclem3 18309 lss0cl 18768 lspsnat 18966 cnsubrg 19625 gsumfsum 19632 obslbs 19893 lmiclbs 19995 lmisfree 20000 mdetdiaglem 20223 mdet0 20231 eltg3 20577 tgtop 20588 tgidm 20595 ppttop 20621 toponmre 20707 tgrest 20773 neitr 20794 tgcn 20866 cmpsublem 21012 cmpsub 21013 iunconlem 21040 uncon 21042 1stcfb 21058 2ndcctbss 21068 2ndcdisj 21069 1stcelcls 21074 1stccnp 21075 locfincmp 21139 comppfsc 21145 1stckgen 21167 ptuni2 21189 ptbasfi 21194 ptpjopn 21225 ptclsg 21228 ptcnp 21235 prdstopn 21241 txindis 21247 txtube 21253 txcmplem1 21254 txcmplem2 21255 xkococnlem 21272 txcon 21302 trfbas2 21457 filtop 21469 filcon 21497 filssufilg 21525 fmfnfm 21572 ufldom 21576 hauspwpwf1 21601 alexsubALTlem3 21663 alexsubALT 21665 ptcmplem2 21667 tmdgsum2 21710 tgptsmscld 21764 ustfilxp 21826 xbln0 22029 opnreen 22442 metdsre 22464 cnheibor 22562 phtpc01 22604 cfilfcls 22880 cmetcaulem 22894 iscmet3 22899 ovolctb 23065 ovoliunlem3 23079 ovoliunnul 23082 ovolicc2lem5 23096 ovolicc2 23097 dyadmbl 23174 vitali 23188 itg11 23264 bddmulibl 23411 perfdvf 23473 dvcnp2 23489 dvlip 23560 dvne0 23578 fta1g 23731 fta1 23867 ulmcau 23953 pserulm 23980 wilthlem2 24595 dchrvmasumif 24992 rpvmasum2 25001 dchrisum0re 25002 dchrisum0lem3 25008 dchrisum0 25009 dchrmusum 25013 dchrvmasum 25014 axcontlem10 25653 upgredg 25811 usgrafisindb1 25938 wlkiswwlk 26226 wlklniswwlkn 26229 clwlkisclwwlklem0 26316 clwlkfoclwwlk 26372 frgra3vlem2 26528 spansncvi 27895 reff 29234 locfinreflem 29235 cmpcref 29245 fmcncfil 29305 volmeas 29621 omssubadd 29689 bnj849 30249 derangenlem 30407 cvmsss2 30510 cvmopnlem 30514 cvmfolem 30515 cvmliftmolem2 30518 cvmliftlem15 30534 cvmlift2lem10 30548 cvmlift3lem8 30562 fundmpss 30910 frmin 30983 frrlem11 31036 nocvxmin 31090 fnessref 31522 refssfne 31523 neibastop2lem 31525 neibastop2 31526 fnemeet2 31532 fnejoin2 31534 tailfb 31542 knoppcnlem9 31661 wl-ax13lem1 32466 wl-sbcom2d 32523 matunitlindflem2 32576 poimirlem25 32604 poimirlem27 32606 heicant 32614 itg2addnclem 32631 sdclem1 32709 fdc 32711 istotbnd3 32740 sstotbnd2 32743 prdsbnd2 32764 heibor1lem 32778 heiborlem1 32780 heiborlem10 32789 heibor 32790 riscer 32957 divrngidl 32997 prtlem17 33179 ax12eq 33244 ax12el 33245 ax12inda 33251 ax12v2-o 33252 osumcllem8N 34267 pexmidlem5N 34278 mapdrvallem2 35952 clcnvlem 36949 onfrALT 37785 chordthmALT 38191 snelmap 38280 ssnnf1octb 38377 choicefi 38387 mapss2 38392 difmap 38394 axccdom 38411 inficc 38608 fsumnncl 38638 stoweidlem43 38936 stoweidlem48 38941 stoweidlem57 38950 stoweidlem60 38953 qndenserrnopn 39194 issalnnd 39239 subsaliuncl 39252 sge0cl 39274 nnfoctbdj 39349 ismeannd 39360 caragenunicl 39414 isomennd 39421 ovn0lem 39455 ovnsubaddlem2 39461 hspdifhsp 39506 hspmbllem3 39518 smflimlem6 39662 smfpimbor1lem1 39683 rlimdmafv 39906 usgr1v0e 40545 1wlkiswwlks 41073 1wlkiswwlkupgr 41075 1wlklnwwlkn 41081 1wlklnwwlknupgr 41083 umgrwwlks2on 41161 elwwlks2 41170 elwspths2spth 41171 clwlkclwwlklem3 41210 clwlksfoclwwlk 41270 frgr3vlem2 41444 mgmpropd 41565 c0snmgmhm 41704
Copyright terms: Public domain W3C validator | 4,265 | 7,458 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2024-26 | latest | en | 0.104097 |
https://cheapessay.us/a-10-to-15-slide-microsoft-powerpoint-presentation-discuss/ | 1,713,022,486,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816820.63/warc/CC-MAIN-20240413144933-20240413174933-00651.warc.gz | 157,762,502 | 12,640 | # a 10- to 15-slide Microsoft PowerPoint presentation discuss…
a 10- to 15-slide Microsoft PowerPoint presentation discussing your statistics project data analyses, based on the analyses you’ve done on your assigned data set from Week Two. the following in your presentation:
Title: Statistical Analysis of [Assigned Data Set]
Introduction:
Welcome to my presentation on the statistical analysis performed on the [Assigned Data Set]. In this presentation, I will be discussing the various data analyses conducted, highlighting key findings and presenting the implications of these findings. This analysis will provide valuable insights into the data set and contribute to the existing body of knowledge in the field.
Slide 1:
Title: Introduction to the Data Set
– Briefly introduce the data set, including its source, purpose, and relevance to the research question.
– Explain the variables included in the data set and the unit of analysis.
Slide 2:
Title: Descriptive Statistics
– Present descriptive statistics such as mean, median, mode, variance, and standard deviation for each variable.
– Use graphs, such as histograms or box plots, to visually represent the distribution of the variables.
– Discuss any interesting patterns or trends observed in the descriptive statistics.
Slide 3:
Title: Data Visualization
– Display appropriate visual representations of the data set, including scatter plots, line graphs, or bar charts.
– Analyze the relationships and trends observed in the visualizations.
– Discuss any outliers or influential data points observed.
Slide 4:
Title: Hypothesis Testing
– State the research hypothesis and the null hypothesis.
– Explain the statistical test used to test the hypothesis and justify its appropriateness for the data set.
– Present the results of the hypothesis test, including the test statistic, p-value, and the decision regarding the null hypothesis.
Slide 5:
Title: Regression Analysis
– Describe the regression model used and the variables included in the analysis.
– Present the regression coefficients, their significance level, and the interpretation of their effects.
– Discuss the goodness of fit of the regression model, using metrics such as R-squared or adjusted R-squared.
Slide 6:
Title: ANOVA (Analysis of Variance)
– Explain the purpose of the ANOVA and its application in the data analysis.
– Present the results of the ANOVA, including the F-statistic, p-value, and any significant differences observed between groups.
– Discuss the implications of the ANOVA results in relation to the research question.
Slide 7:
Title: Chi-Square Test
– Describe the chi-square test and its relevance to the data set.
– Present the contingency table and the calculated chi-square statistic.
– Interpret the results of the chi-square test and discuss any significant associations or relationships between variables.
Slide 8:
Title: Time Series Analysis
– Explain the time series analysis conducted on the data set.
– Present the findings, including trends, seasonality, and any significant changes observed over time.
– Discuss the forecasting or prediction capabilities of the time series model.
Slide 9:
Title: Conclusion and Implications
– Summarize the key findings from the data analyses conducted.
– Discuss the implications of these findings in relation to the research question and their significance in the field.
– Highlight any limitations or areas for future research.
Slide 10:
Title: References
– Provide a list of the references used in the data analysis.
– Follow a consistent citation style (e.g., APA or MLA).
Conclusion:
In this presentation, we have explored the statistical analysis of the [Assigned Data Set] and gained valuable insights regarding its variables and relationships. The analysis has provided evidence to support or reject the research hypothesis, identified significant associations between variables, and offered predictions or forecasts based on time series analysis. The findings contribute to our understanding of the data set, and further research can build upon these results to advance the field. Thank you for your attention, and I welcome any questions or discussions on the presented analysis. | 802 | 4,201 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2024-18 | latest | en | 0.838489 |
http://goodriddlesnow.com/riddles/view/2598 | 1,506,093,942,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818688997.70/warc/CC-MAIN-20170922145724-20170922165724-00021.warc.gz | 149,186,265 | 9,171 | # MATH RIDDLE
Question: TWO THEIFS HAD ROBED GOLD COINS IN ONE HOUSE...! THEY WANT 2 SHARE EQUALLY...! BUT ONE WAS LEFT...! SO THEY ADDED ANOTHER PERSON NOW 3 SHARED AGAIN ONE COIN WAS REMAINING...! 4,5,6..! AT LAST 7MEMBERS SHARED EQUALLY...! THAT MEANS HOW MANY COINS THEY ROBBED...?;-)
Riddle Discussion
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Question: What are the next two letters in the following series and why? WATNTLITFS__
##### Numbers of brothers and sisters (medium)
Question: In a house of Kardak's family each brother have a sisters as double as their brothers and each sister have brothers as same as their sister. How many brothers and sisters in Kardak's family ? | 273 | 970 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2017-39 | longest | en | 0.898902 |
https://www.collegepapershelp.com/calculate-mean-median-range-standard-deviation-width-interpret-measures-central-tendency-variability/ | 1,680,062,433,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948932.75/warc/CC-MAIN-20230329023546-20230329053546-00299.warc.gz | 784,775,679 | 16,280 | . Calculate the mean, median, range and standard deviation for the width. Interpret these measures of central tendency and variability.
manufacturing company produces steel housings for electrical equipment. The main component part of the housing is a steel trough that is made of a 14-gauge steel coil. It is produced using a 250-ton progressive punch press with a wipe-down operation and two 90-degree forms placed in the flat steel to make the trough. The distance from one side of the form to the other is critical because of weatherproofing in outdoor applications. The company requires that the width of the trough be between 8.31 inches and 8.61 inches. Data are collected from a sample of 49 troughs and stored in Trough, which contains the widths of the troughs in inches as shown below.
8.312 8.343 8.317 8.383 8.348 8.410 8.351 8.373 8.481 8.422 8.476 8.382 8.484 8.403 8.414 8.419 8.385 8.465 8.498 8.447 8.436 8.413 8.489 8.414 8.481 8.415 8.479 8.429 8.458 8.462 8.460 8.444 8.429 8.460 8.412 8.420 8.410 8.405 8.323 8.420 8.396 8.447 8.405 8.439 8.411 8.427 8.420 8.498 8.409
a. Calculate the mean, median, range and standard deviation for the width. Interpret these measures of central tendency and variability.
b. List the five-number summary.
c. Construct a box plot and describe its shape.
d. What can you conclude about the number of troughs that will meet the company’s requirement of troughs being between 8.31 and 8.61 inches wide?
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The Poincare Conjecture and Grigori Perelman
In 2006 the Russian mathematician Grigori Perelman was awarded the mathematical equivalent of the mathematical Nobel prize (the Fields Medal). He declined it. In 2010 he was the first mathematician to be awarded \$1 million – he turned it down. What had Perelman done to achieve such (apparently unwanted) acclaim? He had solved a puzzle that had frustrated mathematicians for over 100 years – the Poincare conjecture.
What is the Poincare Conjecture?
The Poincare Conjecture is that, “Every simply connected, closed 3-manifold is homeomorphic to the 3-sphere.” At first glance that may look quite complicated – so looking at the definitions in turn:
Simply connected means a shape without holes. The two shapes on the left above are simply connected, the two on the right are not. In 3 dimensions, a sphere and cube are simply connected, but a donut shape (torus) is not.
3D manifold means a 3 dimensional surface. Imagine the surface of a sphere – that is a 2 dimensional surface. So a 3 dimensional surface on a sphere would require a 4 dimensional sphere. A 4 dimensional sphere is one which has a fixed radius in 4 dimensions (unlike in 3 dimensions for a sphere and 2 dimensions for a circle).
Homeomorphic means it is mathematically equivalent in terms of the relationship between points. Basically, if 2 shapes can be sqeezed or stretched to form another shape then they are homeomorphic. In the above animation, the coffee mug and the donut (torus) are shown to be homeomorphic.
3-Sphere means a sphere in 4 dimensions (i.e with a 3 dimensional surface area).
So, with those terms defined we can simplify the Poincare conjecture. In regular 3 dimensions, conventional 3 dimensional shapes without a hole in them (cubes, cuboids etc) can all be squashed and squeezed to create a sphere. Poincare conjectured that the same would be true in higher dimensions – i.e 4 dimensional cubes (a tesseract, as shown above) could be squashed and squeezed to make a 4 dimensional sphere.
Grigori Perelman however was not interested in either the acclaim or the money on offer for solving one of the world’s most difficult mathematics problems. In explaining why he turned down \$1 million he said that the prize, “was completely irrelevant for me. Everybody understood that if the proof is correct, then no other recognition is needed.”
If you liked this post you might also like:
Imagining the 4th Dimension. How mathematics can help us explore the notion that there may be more than 3 spatial dimensions.
Non Euclidean Geometry V – The Shape of the Universe – Using mathematics to understand one of the most important questions of all.
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P3 investigation questions and fully typed mark scheme. Packs for both Applications students and Analysis students. | 753 | 3,328 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2022-49 | latest | en | 0.927976 |
https://math.stackexchange.com/questions/2232946/find-the-greatest-common-divisor-of-the-polynomials-fx-x4ix21-and | 1,563,519,244,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195526064.11/warc/CC-MAIN-20190719053856-20190719075856-00240.warc.gz | 480,491,794 | 35,313 | # Find the greatest common divisor of the polynomials $f(x)=x^{4}+ix^{2}+1$ and $g(x)=ix^{2}+1$ over $\mathbb{C}$
Find the greatest common divisor of the polynomials $f(x)=x^{4}+ix^{2}+1$ and $g(x)=ix^{2}+1$ over the field $\mathbb{C}$.
Here I am getting some difficulty because I think both are prime and hence $\gcd =1$ , but I am not sure. Any help
• None is prime: the only irreducible polynomials in $\mathbf C[x]$ are linear polynomials. – Bernard Apr 13 '17 at 21:29
• I mean both are prime to each other in the sense of factors – M. A. SARKAR Apr 13 '17 at 21:39
• Coprimes, yes Performing the Euclidean algorthm, you find a constant as gcd. – Bernard Apr 13 '17 at 22:29
$x^4+ix^2+1=(ix^2+1)(\dfrac{x^2}{i}+2) - 1$
First note that $$\gcd(x^4+ix^2+1,ix^2+1)=\gcd(x^4,ix^2+1)$$ and that $x^4$ and $ix^2 + 1$ don't have a common linear factor since they don't have common zeroes. All irreducible polynomials over $\Bbb C$ are linear, so those polynomials don't have common irreducible factors. We conclude that GCD is $1$.
Note. This a bit lengthy explanation, but easier than long division. Alternatively, note that $$1+x^4 = (1+ix^2)(1-ix^2).$$ | 402 | 1,156 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2019-30 | latest | en | 0.867423 |
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HomeSoftware DevelopmentReduce insertion of 0 or 1 such that no adjoining pair has...
# Reduce insertion of 0 or 1 such that no adjoining pair has identical worth
Given a binary array A[] of size N, the duty is to search out the minimal variety of operations required such that no adjoining pair has the identical worth the place in every operation we are able to insert both 0 or 1 at any place within the array.
Examples:
Enter: A[] = {0, 0, 1, 0, 0}
Output: 2
Rationalization: We will carry out the next operations to make consecutive ingredient completely different in an array:
Insert 1 at index 2 in A = {0, 0, 1, 0, 0} → {0, 1, 0, 1, 0, 0}
Insert 1 at index 6 in A = {0, 1, 0, 1, 0, 0} → {0, 1, 0, 1, 0, 1, 0} all consecutive parts are completely different.
Enter: A[] = {0, 1, 1}
Output:
Strategy: The issue could be solved based mostly on the next remark:
A single transfer permits us to ‘break aside’ precisely one pair of equal adjoining parts of an array, by inserting both 1 between 0, 0 or 0 between 1, 1.
So, the reply is solely the variety of pairs which might be already adjoining and equal, i.e, positions i (1 ≤ i <N) such that Ai = Ai + 1, which could be computed with a easy for loop.
Comply with the beneath steps to resolve the issue:
• Initialize a variable depend = 0.
• Iterate a loop for every ingredient in A, and examine if it is the same as the subsequent ingredient.
• If sure, increment the depend by 1.
• Print the depend which supplies the minimal variety of operations required to make consecutive parts completely different in an array.
Under is the implementation of the above method.
## Java
` ` `import` `java.io.*;` `import` `java.util.*;` ` ` `public` `class` `GFG {` ` ` ` ` ` ` ` ` ` ` ` ``public` `static` `int` `minOperation(``int` `arr[], ``int` `n)` ` ``{` ` ``int` `depend = ``0``;` ` ` ` ``for` `(``int` `i = ``0``; i < n - ``1``; i++) {` ` ``if` `(arr[i] == arr[i + ``1``]) {` ` ``depend++;` ` ``}` ` ``}` ` ``return` `depend;` ` ``}` ` ` ` ` ` ``public` `static` `void` `major(String[] args)` ` ``{` ` ``int``[] A = { ``0``, ``0``, ``1``, ``0``, ``0` `};` ` ``int` `N = A.size;` ` ` ` ` ` ``System.out.println(minOperation(A, N));` ` ``}` `}`
Time Complexity: O(N)
Auxiliary Area: O(1)
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http://library.kiwix.org/wikipedia_en_computer_novid_2018-10/A/Spiral.html | 1,558,282,842,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232255071.27/warc/CC-MAIN-20190519161546-20190519183546-00136.warc.gz | 121,435,716 | 16,618 | # Spiral
Cutaway of a nautilus shell showing the chambers arranged in an approximately logarithmic spiral
In mathematics, a spiral is a curve which emanates from a point, moving farther away as it revolves around the point.
## Helices
An Archimedean spiral, a helix, and a conic spiral
Two major definitions of "spiral" in the American Heritage Dictionary are:[1]
1. a curve on a plane that winds around a fixed center point at a continuously increasing or decreasing distance from the point.
2. a three-dimensional curve that turns around an axis at a constant or continuously varying distance while moving parallel to the axis; a helix.
The first definition describes a planar curve, that extends in both of the perpendicular directions within its plane; the groove on one side of a record closely approximates a plane spiral (and it is by the finite width and depth of the groove, but not by the wider spacing between than within tracks, that it falls short of being a perfect example); note that successive loops differ in diameter. In another example, the "center lines" of the arms of a spiral galaxy trace logarithmic spirals.
The second definition includes two kinds of 3-dimensional relatives of spirals:
1. a conical or volute spring (including the spring used to hold and make contact with the negative terminals of AA or AAA batteries in a battery box), and the vortex that is created when water is draining in a sink is often described as a spiral, or as a conical helix.
2. quite explicitly, definition 2 also includes a cylindrical coil spring and a strand of DNA, both of which are quite helical, so that "helix" is a more useful description than "spiral" for each of them; in general, "spiral" is seldom applied if successive "loops" of a curve have the same diameter.[1]
In the side picture, the black curve at the bottom is an Archimedean spiral, while the green curve is a helix. The curve shown in red is a conic helix.
## Two-dimensional
A two-dimensional spiral may be described most easily using polar coordinates, where the radius r is a monotonic continuous function of angle θ. The circle would be regarded as a degenerate case (the function not being strictly monotonic, but rather constant).
Some of the most important sorts of two-dimensional spirals include:
## Three-dimensional
Rhumb line
Archimedean spiral
Two different kinds of spiral over the surface of a sphere.
For simple 3-d spirals, a third variable, h (height), is also a continuous, monotonic function of θ. For example, a conic helix may be defined as a spiral on a conic surface, with the distance to the apex an exponential function of θ.
The helix and vortex can be viewed as a kind of three-dimensional spiral.
For a helix with thickness, see spring (math).
A rhumb line (also known as a loxodrome or "spherical spiral") is the curve on a sphere traced by a ship with constant bearing (e.g., travelling from one pole to the other while keeping a fixed angle with respect to the meridians). The loxodrome has an infinite number of revolutions, with the separation between them decreasing as the curve approaches either of the poles, unlike an Archimedean spiral which maintains uniform line-spacing regardless of radius.
## In nature
The study of spirals in nature has a long history. Christopher Wren observed that many shells form a logarithmic spiral; Jan Swammerdam observed the common mathematical characteristics of a wide range of shells from Helix to Spirula; and Henry Nottidge Moseley described the mathematics of univalve shells. D’Arcy Wentworth Thompson's On Growth and Form gives extensive treatment to these spirals. He describes how shells are formed by rotating a closed curve around a fixed axis: the shape of the curve remains fixed but its size grows in a geometric progression. In some shells, such as Nautilus and ammonites, the generating curve revolves in a plane perpendicular to the axis and the shell will form a planar discoid shape. In others it follows a skew path forming a helico-spiral pattern. Thompson also studied spirals occurring in horns, teeth, claws and plants.[2]
A model for the pattern of florets in the head of a sunflower was proposed by H. Vogel. This has the form
where n is the index number of the floret and c is a constant scaling factor, and is a form of Fermat's spiral. The angle 137.5° is the golden angle which is related to the golden ratio and gives a close packing of florets.[3]
Spirals in plants and animals are frequently described as whorls. This is also the name given to spiral shaped fingerprints.
## In the laboratory
When potassium sulfate is heated in water and subjected to swirling in a beaker, the crystals form a multi-arm spiral structure when allowed to settle[4]
Potassium sulfate forms a spiral structure in solution.
## As a symbol
A spiral like form has been found in Mezine, Ukraine, as part of a decorative object dated to 10,000 BCE.
Bowl on stand, Vessel on stand, and Amphora. Eneolithic, the Cucuteni Culture, 4300-4000 BCE. Found in Scânteia, Iași, Romania. Collected by the Moldavia National Museum Complex
The Newgrange entrance slab
This Petroglyph with a spiral figure carved into it was made by the Hohokams, a Native American tribe in the United States, over 1000 years ago.
The spiral and triple spiral motif is a Neolithic symbol in Europe (Megalithic Temples of Malta). The Celtic symbol the triple spiral is in fact a pre-Celtic symbol.[5] It is carved into the rock of a stone lozenge near the main entrance of the prehistoric Newgrange monument in County Meath, Ireland. Newgrange was built around 3200 BCE predating the Celts and the triple spirals were carved at least 2,500 years before the Celts reached Ireland but has long since been incorporated into Celtic culture.[6][6] The triskelion symbol, consisting of three interlocked spirals or three bent human legs, appears in many early cultures, including Mycenaean vessels, on coinage in Lycia, on staters of Pamphylia (at Aspendos, 370–333 BC) and Pisidia, as well as on the heraldic emblem on warriors' shields depicted on Greek pottery.[7]
Spirals can be found throughout pre-Columbian art in Latin and Central America. The more than 1,400 petroglyphs (rock engravings) in Las Plazuelas, Guanajuato Mexico, dating 750-1200 AD, predominantly depict spirals, dot figures and scale models.[8] In Colombia monkeys, frog and lizard like figures depicted in petroglyphs or as gold offering figures frequently includes spirals, for example on the palms of hands.[9] In Lower Central America spirals along with circles, wavy lines, crosses and points are universal petroglyphs characters.[10] Spirals can also be found among the Nazca Lines in the coastal desert of Peru, dating from 200 BC to 500 AD. The geoglyphs number in the thousands and depict animals, plants and geometric motifs, including spirals.[11]
Spiral shapes, including the swastika, triskele, etc., have often been interpreted as solar symbols. Roof tiles dating back to the Tang Dynasty with this symbol have been found west of the ancient city of Chang'an (modern-day Xi'an).
Spirals are also a symbol of hypnosis, stemming from the cliché of people and cartoon characters being hypnotized by staring into a spinning spiral (one example being Kaa in Disney's The Jungle Book). They are also used as a symbol of dizziness, where the eyes of a cartoon character, especially in anime and manga, will turn into spirals to show they are dizzy or dazed. The spiral is also found in structures as small as the double helix of DNA and as large as a galaxy. Because of this frequent natural occurrence, the spiral is the official symbol of the World Pantheist Movement.[12] The spiral is also a symbol of the dialectic process and Dialectical monism.
## In art
The spiral has inspired artists throughout the ages. Among the most famous of spiral-inspired art is Robert Smithson's earthwork, "Spiral Jetty", at the Great Salt Lake in Utah.[13] The spiral theme is also present in David Wood's Spiral Resonance Field at the Balloon Museum in Albuquerque, as well as in the critically acclaimed Nine Inch Nails 1994 concept album The Downward Spiral. The Spiral is also a prominent theme in the anime Gurren Lagann, where it represents a philosophy and way of life. It also central in Mario Merz and Andy Goldsworthy's work. The spiral is the central theme of the horror manga Uzumaki by Junji Ito, where a small coastal town is afflicted by a curse involving spirals. The book 2012 A Piece of Mind By Wayne A Beale also depicts a large spiral in this book of dreams and images.[14] http://www.blurb.com/distribution?id=573100/#/project/573100/project-details/edit
## References
1. "Spiral, American Heritage Dictionary of the English Language, Houghton Mifflin Company, Fourth Edition, 2009.
2. Thompson, D'Arcy (1942) [1917]. "On Growth and Form".
3. Prusinkiewicz, Przemyslaw; Lindenmayer, Aristid (1990). The Algorithmic Beauty of Plants. Springer-Verlag. pp. 101–107. ISBN 978-0-387-97297-8. Archived from the original on 2007-05-19.
4. Thomas, Sunil (2017). "Potassium sulfate forms a spiral structure when dissolved in solution". Russian J Phys Chem B. 11: 195–198. doi:10.1134/S1990793117010328.
5. Anthony Murphy and Richard Moore, Island of the Setting Sun: In Search of Ireland's Ancient Astronomers, 2nd ed., Dublin: The Liffey Press, 2008, pp. 168-169
6. "Newgrange Ireland - Megalithic Passage Tomb - World Heritage Site". Knowth.com. 2007-12-21. Archived from the original on 2013-07-26. Retrieved 2013-08-16.
7. For example, the trislele on Achilles' round shield on an Attic late sixth-century hydria at the Boston Museum of Fine Arts, illustrated in John Boardman, Jasper Griffin and Oswyn Murray, Greece and the Hellenistic World (Oxford History of the Classical World) vol. I (1988), p. 50.
8. "Rock Art Of Latin America & The Caribbean" (PDF). International Council on Monuments & Sites. June 2006. p. 5. Archived (PDF) from the original on 5 January 2014. Retrieved 4 January 2014.
9. "Rock Art Of Latin America & The Caribbean" (PDF). International Council on Monuments & Sites. June 2006. p. 99. Archived (PDF) from the original on 5 January 2014. Retrieved 4 January 2014.
10. "Rock Art Of Latin America & The Caribbean" (PDF). International Council on Monuments & Sites. June 2006. p. 17. Archived (PDF) from the original on 5 January 2014. Retrieved 4 January 2014.
11. Jarus, Owen (14 August 2012). "Nazca Lines: Mysterious Geoglyphs in Peru". LiveScience. Archived from the original on 4 January 2014. Retrieved 4 January 2014.
12. Harrison, Paul. "Pantheist Art" (PDF). World Pantheist Movement. Retrieved 7 June 2012.
13. Israel, Nico (2015), Spirals : the whirled image in twentieth-century literature and art, New York Columbia University Press, pp. 161–186, ISBN 978-0-231-15302-7
14. 2012 A Piece of Mind By Wayne A Beale
• Cook, T., 1903. Spirals in nature and art. Nature 68 (1761), 296.
• Cook, T., 1979. The curves of life. Dover, New York.
• Habib, Z., Sakai, M., 2005. Spiral transition curves and their applications. Scientiae Mathematicae Japonicae 61 (2), 195 – 206.
• Dimulyo, S., Habib, Z., Sakai, M., 2009. Fair cubic transition between two circles with one circle inside or tangent to the other. Numerical Algorithms 51, 461–476 .
• Harary, G., Tal, A., 2011. The natural 3D spiral. Computer Graphics Forum 30 (2), 237 – 246 .
• Xu, L., Mould, D., 2009. Magnetic curves: curvature-controlled aesthetic curves using magnetic fields. In: Deussen, O., Hall, P. (Eds.), Computational Aesthetics in Graphics, Visualization, and Imaging. The Eurographics Association .
• Wang, Y., Zhao, B., Zhang, L., Xu, J., Wang, K., Wang, S., 2004. Designing fair curves using monotone curvature pieces. Computer Aided Geometric Design 21 (5), 515–527 .
• A. Kurnosenko. Applying inversion to construct planar, rational spirals that satisfy two-point G2 Hermite data. Computer Aided Geometric Design, 27(3), 262–280, 2010 .
• A. Kurnosenko. Two-point G2 Hermite interpolation with spirals by inversion of hyperbola. Computer Aided Geometric Design, 27(6), 474–481, 2010.
• Miura, K.T., 2006. A general equation of aesthetic curves and its self-affinity. Computer-Aided Design and Applications 3 (1–4), 457–464 .
• Miura, K., Sone, J., Yamashita, A., Kaneko, T., 2005. Derivation of a general formula of aesthetic curves. In: 8th International Conference on Humans and Computers (HC2005). Aizu-Wakamutsu, Japan, pp. 166 – 171 .
• Meek, D., Walton, D., 1989. The use of Cornu spirals in drawing planar curves of controlled curvature. Journal of Computational and Applied Mathematics 25 (1), 69–78 .
• Thomas, S. Potassium sulfate forms a spiral structure when dissolved in solution. Russ. J. Phys. Chem. B (2017) 11: 195.
• Farin, G., 2006. Class A Bézier curves. Computer Aided Geometric Design 23 (7), 573–581 .
• Farouki, R.T., 1997. Pythagorean-hodograph quintic transition curves of monotone curvature. Computer-Aided Design 29 (9), 601–606.
• Yoshida, N., Saito, T., 2006. Interactive aesthetic curve segments. The Visual Computer 22 (9), 896–905 .
• Yoshida, N., Saito, T., 2007. Quasi-aesthetic curves in rational cubic Bézier forms. Computer-Aided Design and Applications 4 (9–10), 477–486 .
• Ziatdinov, R., Yoshida, N., Kim, T., 2012. Analytic parametric equations of log-aesthetic curves in terms of incomplete gamma functions. Computer Aided Geometric Design 29 (2), 129 – 140 .
• Ziatdinov, R., Yoshida, N., Kim, T., 2012. Fitting G2 multispiral transition curve joining two straight lines, Computer-Aided Design 44(6), 591–596 .
• Ziatdinov, R., 2012. Family of superspirals with completely monotonic curvature given in terms of Gauss hypergeometric function. Computer Aided Geometric Design 29(7): 510–518 .
• Ziatdinov, R., Miura K.T., 2012. On the Variety of Planar Spirals and Their Applications in Computer Aided Design. European Researcher 27(8–2), 1227-–1232 . | 3,569 | 13,952 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2019-22 | latest | en | 0.929871 |
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4,702 results, page 2
1. ## Math
A bag contains 3 red, 5 yellow, 2 green and 6 blue marbles. If one marble is chosen at random, replaced, and then a second marble is chosen at random, find the probability of obtaining 2 marbles of different colors.
asked by Bryan on January 22, 2014
2. ## Math probability - HELP
A bag contains 3 red, 5 yellow, 2 green and 6 blue marbles. If one marble is chosen at random, replaced, and then a second marble is chosen at random, find the probability of obtaining 2 marbles of different colors.
asked by Bryan on January 23, 2014
3. ## Probability, Random Variables, and Random Process
A zero-mean Gaussian random process has an auto-correlation function R_XX (τ)={■(13[1-(|τ|⁄6)] |τ|≤6@0 elsewhere)┤ Find the covariance function necessary to specify the joint density of random variables defined at times t_i=2(i-1),i=1,2,…,5.
asked by Rakesh on November 2, 2017
4. ## Probability
Let X be a random variable that takes non-zero values in [1,∞), with a PDF of the form fX(x)=⎧⎩⎨cx3,0,if x≥1,otherwise. Let U be a uniform random variable on [0,2]. Assume that X and U are independent. What is the value of the constant c? c= -
asked by qwerty on April 21, 2015
5. ## mathmath
A study conducted by the Metro Housing Agency in a midwestern city revealed the information below concerning the age distribution of renters in the city. (Round your answers to four decimal places.) Age Adult Population, % Group Who Are Renters, % 21-44 53
asked by jack on October 16, 2010
6. ## algebra 1
Circle the most appropriate description. 27.A city planning commitee surveys 100 people waiting at a bus stop about the expansion of the public transportation system. Random Sample,systematic sample,self-selected sample, convenience sample, bipartisan
asked by janet on May 20, 2013
7. ## maths
A random number generator draws at random with replacement from the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. Find the chance that the digit 5 appears on more than 11% of the draws, if a) 100 draws are made b) 1000 draws are made
asked by anonymous on April 27, 2013
8. ## Statistics
Suppose x has a distributionwith u=72 and 0=8 if random samples of size n=16 are selected, can we say anything about the x distributin of sample means if the originalx distribution is normal can we say anything about the x distribution of random samples of
asked by John on June 14, 2010
9. ## Probability
Let N,X1,Y1,X2,Y2,… be independent random variables. The random variable N takes positive integer values and has mean a and variance r. The random variables Xi are independent and identically distributed with mean b and variance s, and the random
asked by qwerty on April 21, 2015
10. ## statistics
In a certain region, the mean annual salary for plumbers is \$51,000. Let x be a random variable that represents a plumber's salary. Assume the standard deviation is \$1300. If a random sample of 100 plumbers is selected, what is the probability that the
asked by katie on February 25, 2013
11. ## statistics
assume you have a data set from a normally distributed random variable. answer the following: will the random variable be discrete, continuous, or neither? How do you know? I believe its continuous but cant explain why will the data be qualitative or
asked by Dean on June 29, 2013
12. ## STATISTICS
Consider a binomial random variable where the number of trials is 12 and the probability of success on each trial is 0.25. Find the mean and standard deviation of this random variable. I have a mean of 4 and a standard deviation of 2.25 is this correct
asked by John on June 10, 2010
13. ## Probability
Let N,X1,Y1,X2,Y2,… be independent random variables. The random variable N takes positive integer values and has mean a and variance r. The random variables Xi are independent and identically distributed with mean b and variance s, and the random
asked by A on April 20, 2014
14. ## math
Suppose from a population of 50 bank accounts we want to take a random sample of three accounts in order to learn about the population. How many different random samples of three accounts are possible.
asked by John on October 15, 2010
15. ## Math
What is the probability that the random variable has a value between 0.1 and 0.5? The random variable is .125.
asked by Lyndse on March 8, 2010
16. ## algebra
A bag contains 9 marbles: 2 are green, 4 are red, and 3 are blue. Laura chooses a marble at random, and without putting it back, chooses another one at random. What is the probability that both marbles she chooses are blue? Write your answer as a fraction
asked by denora on November 29, 2015
Suppose a dot is placed at random in a 10 x 10 graph grid in which squares have been numbered from 1 to 100 with no number repeated. Now simulate choosing 100 numbers at random between 1 and 100 inclusive. How many numbers might you expect to match the
asked by Ashley on November 12, 2014
18. ## math
a basket contains the following pieces of fruit: 3 apples, 2 oranges, 2 bananas, 2 pears, and 5 peaches. Jameson picks a fruit at random and does not replace it. Then Brittany pricks a fruit at random. What is the probability that Jameson gets a banana and
19. ## Math
In each case, determine whether the situation calls for a discrete or continuous random variable and explain your answer. Please help! a. The description of the number of children in a family selected at random from families in Ft. Wayne, Indiana b. The
asked by Jennifer on January 14, 2010
20. ## MATH-157
In each case, determine whether the situation calls for a discrete or continuous random variable and explain your answer. a. The description of the number of children in a family selected at random from families in Ft. Wayne, Indiana b. The net worth of an
asked by Tyga on April 1, 2010
21. ## probability
Let N,X1,Y1,X2,Y2,… be independent random variables. The random variable N takes positive integer values and has mean a and variance r. The random variables Xi are independent and identically distributed with mean b and variance s, and the random
asked by juanpro on April 22, 2014
22. ## Math
Need help on if the sample given is random or not random. The first 50 nurses at a nursing convention write their names on slips of paper and place them in a drawing box. After the box is shaken, 20 names are pulled from the box and each person chosen
asked by Brandi on August 29, 2013
23. ## math
Let X1,X2,…,X25 be random sample ~ normal(0,16) and Y1, Y2,…,Y25 be random sample ~ normal(1,9), assume the two samples are independent ,comput P(X>Y) where X=Sum(Xi)/25 and Y=Sum(Yi)/25
asked by assma on December 19, 2014
24. ## Mathematics
The blood types B- and AB- are the rarest of the eight huan blood types representing 1.5% and .6% respectively. a) If the blood types of a random sample of 1000 blood donors are recorded, what is the probability that 10 or more will be AB-? b) If the blood
asked by Jen on October 3, 2007
25. ## Experiment
The question is how do I design a basic experiment that would allow us to establish a cause-effect relationship between number of hours worked per week and lower college graduation rates? It must have these components: a manupulated independent variable, a
asked by Ann on June 14, 2011
26. ## Maths
1)A two-figure number is written down at random. Find the probability that a)the number is greater than 44 b)the number is less than 100 2)A letter is picked at random from the english alphabet. Find the probability that a)the letter is a vowel (my
asked by Anonymous on January 28, 2008
27. ## statistics
A nationwide study indicated that 27% of adults said that chocolate was their favorite ice cream flavor. A simple random sample of 150 adults is obtained. A.) Describe the sampling distribtuion of p, the sample proportion of adults whose favorite ice cream
asked by Megan on November 11, 2011
28. ## Math
You want to find out what the favorite hot lunch in the school cafeteria is among the high school students. At an assembly for the whole school, you decide to survey all students who are sitting on the end of their rows in the auditorium. What type of
asked by Steve on April 26, 2015
29. ## english
A way to decrease your chance of becoming a victim of random violence. Violence is typically not random. Victims are picked out because of certain characteristics — especially vulnerability and availability. If you know of particular places that tend to
asked by Anonymous on December 1, 2006
30. ## ap statistics
An unnoticed mechanical failure has caused one-fourth of a machine shop’s production of 10000 pistol firing pins to be defective. A random sample of 25 firing pins was drawn from the population. a.) Explain why this random variable has a binomial
asked by nicole on April 24, 2014
31. ## ap statistics
An unnoticed mechanical failure has caused one-fourth of a machine shop’s production of 10000 pistol firing pins to be defective. A random sample of 25 firing pins was drawn from the population. a.) Explain why this random variable has a binomial
asked by nicole on April 24, 2014
32. ## ap statistics
An unnoticed mechanical failure has caused one-fourth of a machine shop’s production of 10000 pistol firing pins to be defective. A random sample of 25 firing pins was drawn from the population. a.) Explain why this random variable has a binomial
asked by nicole on April 24, 2014
33. ## Math
Suppose one person decides to carry out a random act of kindness for three people on Monday. On Tuesday, that same person is kind to three more people, and the three he/she helped on Monday each help three more individuals. If the same circumstances were
asked by Andy on January 25, 2017
34. ## MATH
In a drawer Richard has 5 pairs of gloves. Each pair has a different color. On Day 1 Richard selects two individual gloves at random from the 10 gloves in the drawer. On Day 2 Richard selects 2 of the remaining 8 gloves at random and on Day 3 two of the
asked by FRank on January 20, 2016
35. ## Statistics
Daily water intake (including water used in drinks such as coffee, tea and juice) for Canadian adults follows a normal distribution with mean 1.86 litres and standard deviation 0.29 litres. (a) Can you calculate the probability that the mean daily water
asked by Mel on November 4, 2014
36. ## statistics
A sixth form class consists of 6 girls and 9 boys. Three students from the class are chosen at random. The number of boys chosen is denoted by the random variable X. Show that a)P(X=0)=120/2730 b)P(X=2)=1296/2730
asked by Nabiha on December 4, 2011
37. ## stat
7. A researcher selects a random sample. A 90% confidence interval for a population mean A) is an interval with margin of error ± 90%. B) has the property that if we repeatedly selected our random sample in exactly the same way, each time constructing
asked by Anonymous on December 5, 2010
38. ## math
i have no clue how to do this problem. any help would be appreciated!! : Three hundred people apply for three jobs. 90 of the applicants are women. (a) If three persons are selected at random, what is the probability that all are women? (Round the answer
asked by CARMEN on June 23, 2010
39. ## math
A fair coin is flipped independently until the first Heads is observed. Let the random variable K be the number of tosses until the first Heads is observed plus 1. For example, if we see TTTHTH, then K= 5. For k = 1,2,...,K, let Xk be a continuous random
asked by Var on October 15, 2018
40. ## statistics
Suppose we take a random sample of size n from a normal population with variance, σ2 . It can be shown that (n−1)s2/σ2 has a chi-square distribution with n−1 degrees of freedom, where s is the sample variance. Below is a random sample of size 8 drawn
asked by sandy on March 18, 2013
41. ## Math probability
I have got 18 red , 12 green, 5 blue and 5 yellow balls in the bag. What is the probability that I pick excatly 2 green when I pick 4 at random ? What is the probabily that none are yellow after I pick 4 at random ? Thanks.
asked by didier on June 12, 2017
42. ## Math
I have got 18 red , 12 green, 5 blue and 5 yellow balls in the bag. What is the probability that I pick excatly 2 green when I pick 4 at random ? What is the probabily that none are yellow after I pick 4 at random ? Thanks.
asked by didier on June 12, 2017
43. ## statistics
1. The on-campus housing costs for a semester at a state university are Normally distributed with a mean of \$1553 and a standard deviation of \$329. (a) What is the distribution of the mean on-campus housing cost for a semester at this university for a
asked by Sarah on October 26, 2011
44. ## probability
please help! 1.) A random digit from 1 to 9 (inclusive) is chosen, with all digits being equally likely. The probability that when it's squared it will end with the digit 1. 2.) A random number between 1 and 20 (inclusive) is chosen. The probability that
asked by karen on August 11, 2012
45. ## math
A bucket contains 7 red, 3 blue, 2 black and 1 yellow ball of the same size. A random selection of 5 balls is taken from the bucket. find the probability that the random selection will contain atleast 1 red ball?
asked by Mahalakshmi on August 18, 2011
46. ## math
please help! 1.) A random digit from 1 to 9 (inclusive) is chosen, with all digits being equally likely. The probability that when it's squared it will end with the digit 1. 2.) A random number between 1 and 20 (inclusive) is chosen. The probability that
asked by karen on August 11, 2012
47. ## Statistics
Service times for customers at a post office follow some right-skewed distribution with mean 2.91 minutes and standard deviation 1.74 minutes. (a) Can you calculate the probability that the average service time for the next two customers is less than 2.64
asked by Mel on November 4, 2014
48. ## Stat
Suppose that a principal of a local high school tracks the number of minutes his students spend texting on a given school day. He finds that the distribution of minutes spent texting is roughly normal with a mean of 60 and a standard deviation of 20. a.
asked by Jen nugyen on April 21, 2016
49. ## Statistical work
Suppose that a principal of a local high school tracks the number of minutes his students spend texting on a given school day. He finds that the distribution of minutes spent texting is roughly normal with a mean of 60 and a standard deviation of 20. a.
asked by Lui on April 21, 2016
50. ## Statistics
2. In 2009 a random sample of 70 unemployed people in Alabama showed an average weekly benefit of \$199.65. In Mississippi, for a random sample of 65 the number was \$187.93. Assume population standard deviations of \$32.48 and \$26.15 respectively. a. Using
asked by Gummy on April 6, 2013
51. ## Math
will someone just double check my answers, please. Identify which types of sampling is used:random,stratified, systematic,convenience. 1. a sample consists of every 20th student who leaves the bookstore. answer:systematic 2. a pollster uses a computer to
asked by Charlene on April 15, 2010
52. ## Algebra 2 Probability
There are 10 cards. Each card has one number between 1 and 10, so that every number from 1 to 10 appears once. In which distributions does the variable X have a binomial distribution? When a card is chosen at random with replacement six times, X is the
asked by Johnny on April 17, 2018
53. ## Math
A vendor has 14 hellium ballons for sale: 9 are yellow, 3 are red, and 2 are green. A ballon is selcted at random and sold. If the ballon sold is yellow, what is the probablity that the next ballon selected at random, is also yellow?
asked by ashely on April 18, 2011
54. ## Math Plz Explain
A basket contains the following pieces of fruit: 3 apples, 2 oranges, 2 bananas, 2 pears, and 5 peaches. Jameson picks a fruit at random and does not replace it. Then Brittany picks a fruit at random. What is the probability that Jameson gets a banana and
asked by Gracious Human on May 18, 2016
11. A basket contains the following pieces of fruit: three apples to oranges two pairs two bananas and five peaches. Jack pics of fruit at random and does not replace it then, Bethany pics of fruit at random. what is the probability that Jack gets a peach
asked by Jack on May 9, 2016
56. ## statistics
36 people surveyed at random and learned that 20 people are in favor of a new shopping mall. What is a good estimate that person chosen at random in her town would favor a new shopping mall.
asked by Jarron on May 18, 2014
57. ## harvard
6. For Design 3, consider the probability of a random individual in South Dorchester being sampled; and the probability of a random individual in Harbor Islands being sampled. These probabilities are approximately the same as the probabilities calculated
asked by jasonsie on December 8, 2012
58. ## Math
Did I answer this problem right? A bag contains 13 marbles of which 10 are red. A marble is selected at random and replaced. A second marble is then selected at random. What is the probability that both marbles are red? 100/169. Thanks.
asked by B.B. on August 7, 2009
59. ## Math
A basket contains the following pieces of fruit: 3 apples, 2 oranges, 2 bananas, 2 pears, and 5 peaches. Jack picks a fruit at random and does not replace it. Then Bethany picks a fruit at random. What is the probability that Jack gets a peach and Bethany
asked by Anonymous on March 6, 2014
60. ## Math
A basket contains the following pieces of fruit: 3 apples, 2 oranges, 2 bananas, 2 pears, and 5 peaches. Jack picks a fruit at random and does not replace it. Then Bethany picks a fruit at random. What is the probability that Jack gets a peach and Bethany
asked by Anonymous on March 6, 2014
61. ## math
A drawer contains 8 brown socks and 4 blue socks. A sock is taken from the drawer at random, its colour is noted and it is then replaced. This procedure perform twice more. If X is the random variable. “the number of brown socks taken “, a) Construct
asked by I Need Help ASAP on May 8, 2018
62. ## Probability
Let Θ be an unknown random variable that we wish to estimate. It has a prior distribution with mean 1 and variance 2. Let W be a noise term, another unknown random variable with mean 3 and variance 5. Assume that Θ and W are independent. We have two
asked by A on April 20, 2014
63. ## statistics
A random sample of 80 youths and a second random sample of 120 adults showed that 18 of the youths and 10 of the adults had been ticketed for careless driving. Use a 1% level of significance to test the claim that youth have a higher proportion of careless
asked by Anonymous on March 2, 2012
64. ## stastics
I'm having a hard time with question, can someone help me? A random sample of medical files is used to estimate the proportion p of all people who have blood type B. If you have no preliminary estimate for p, how many medical files should you include in a
asked by Kim on December 19, 2010
65. ## Math
A basket contains the following pieces of fruit: 3 apples 2 oranges, 2 bannans, 2 pears, and 5 peaches. Jonas picks a fruit at random and does not replace it. Then beth picks a fruit at random. What is the probability that jonas gets a peach and beth gets
asked by Alexander on April 21, 2014
66. ## MATH
A basket contains the following pieces of fruit: 3 apples, 2 oranges, 2 bananas, 2 pears, and 5 peaches. Jonas picks a fruit at random and does not replace it. Then Beth picks a fruit at random. What is the probability that Jonas gets a peach and Beth gets
asked by . on April 11, 2013
67. ## Elementary statistics
Nine apples, four of which are rotten, are in a refrigerator. Three apples are randomly selected without replacement. Let the random variable x represent the number chosen that are rotten. Construct a table describing the probability distribution, then
asked by Rose Bud on February 29, 2012
68. ## stat
** I already finish these questions Can you answer so I can compare my final answers w/ yours.** Suppose that a principal of a local high school tracks the number of minutes his students spend texting on a given school day. He finds that the distribution
asked by Crystal on April 22, 2016
69. ## statistics
The results of a medical test show that of 66 people selected at random who were given the test, 3 tested positive and 63 tested negative. Determine the odds in favor of a person selected at random testing positive on the test.
asked by Robert on February 24, 2011
70. ## Algebra
11. A basket contains the following pieces of fruit: three apples to oranges two pairs two bananas and five peaches. Jack pics of fruit at random and does not replace it then, Bethany pics of fruit at random. what is the probability that Jack gets a peach
asked by Baby_Banana on May 10, 2018
71. ## STATISTICS
In the mass production of bolts it is found that 5% are defective. Bolts selected at random and put into packets of ten. A packet is selected at random. Find the probability that it contains (a) three defective bolts. (b) less than three are defective. Two
asked by NTEEDZI on November 22, 2016
72. ## statistics / probability
A cooler full of sodas has 12 lemon-lime drinks and 18 gingerales. If one drink is selected at random and then a second drink is selected at random, what are the chances that both sodas will be gingerales?
asked by john on January 5, 2018
73. ## Math
At your summer with a research company, you must get a random sample of people from your town to answer a question about spending habits. Which of the following Methods is most likely to be random? a)You survey customers at the local shopping mall b)You
asked by Dope Fresh Nation on May 28, 2014
74. ## Probability
Consider three random variables X, Y, and Z, associated with the same experiment. The random variable X is geometric with parameter p∈(0,1). If X is even, then Y and Z are equal to zero. If X is odd, (Y,Z) is uniformly distributed on the set
asked by qwerty on March 10, 2015
75. ## Probability
Let U, V, and W be independent standard normal random variables (that is, independent normal random variables, each with mean 0 and variance 1), and let X=3U+4V and Y=U+W. Give a numerical answer for each part below. You may want to refer to the standard
asked by A on April 20, 2014
76. ## math 3
A bag contains 4 yellow, 2 red, and 6 green marbles. Two marbles are drawn. The first is replaced before the second is drawn. A random variable assigns the number of red marbles to each outcome. Calculate the expected value of the random variable
asked by Jamar on February 11, 2014
77. ## Statistics
The ratio of brown M&M’s in a bag is stated by the candy company to be 56%. If a random sample of 500 bags is selected, determine the probability that the sample proportion of random bags of M&M’s selected has a percentage of brown between 51% and 59%
asked by Kim on November 23, 2015
78. ## Statistics
3. Let X be a random variable representing the dividend yield of Australian bank stocks. We may assume that X has a normal distribution with Now, suppose we wish to test the null hypothesis that against the alternative that using a level of significance of
asked by Tab on January 26, 2011
79. ## math
let the random variable x denote the number of girls in a five-child family. if the probability of a female birth is .5 find the probability of 0,1,2,3,4, and 5 girls in a five-child family. construct the binomial distribution and draw the histogram
asked by miranda on April 23, 2009
80. ## probability
There are 100 people in line to board a plane with 100 seats. The first person has lost their boarding pass, so they take a random seat. Everyone that follows takes their assigned seat if it's available, but otherwise takes a random unoccupied seat. What
asked by unowen on July 15, 2017
81. ## math
2. Let X be a random variable representing the dividend yield of Australian bank stocks. We may assume that X has a normal distribution with Now, suppose we wish to test the null hypothesis that against the alternative that using a level of significance of
asked by Tab on January 26, 2011
1. A bag contains 9 green marbles and 11 white marbles. You select a marble at random. What are the odds in favor of you picking a green marble? a. 9:20 b. 2:9 c. 11:9 d.9:11 my answer is d? 2. A bag contains 5 green marbles, 8 red marbles, 11 orange
asked by Anon on May 3, 2016
83. ## Math
A card is taken at random from ordinary pack of cards.It is then replaced. Another card is now taken at random from the pack of cards. Work out the probability of the following. a.both cards are the picture cards(jacks,queens or kings) b.neither of cards
asked by Damian on September 17, 2017
84. ## statistics
The Better Businesss Bureau reports that they resolve 70% of the consumer complaints they receive. a) if a random sample of 5 consumer complaints is selected, what is the probability that all 5 were resolved? b)if a random sample of 5 consumer complaints
asked by jimmy on July 22, 2012
85. ## Finite
Assume that the box contains balls numbered from 1 through 28, and that 3 are selected. A random variable X is defined as 1 times the number of odd balls selected, plus 2 times the number of even. How many different values are possible for the random
asked by Sam on October 9, 2012
86. ## statistic
At a party there are 33 students over age 21 and 22 students under age 21. You choose at random 3 of those over 21 and separately choose at random 2 of those under 21 to interview about attitudes toward alcohol.. You have given every student at the party
asked by sara on February 27, 2013
87. ## Statistics (Verify My Answers)
If I do a research project and my hypothesis is that attitudes of human services providers towards HIV/AIDS would improve after participation in a program would my sampling method be simple random, stratified or cluster sampling? I think it would be simple
asked by Linda on December 2, 2009
88. ## Statistics
Let Y be a random number between 0 and 1 generated by an idealized random number generator. Find the following probabilities: A. P (0 less than equal to Y less than equal to .6) =? B. P (.3 less than equal to Y less than equal to .5) =? C. P ( Y less than
asked by Julia T on May 20, 2010
89. ## statistics
homework due by 8am on 2/26...spent too much time with bad websites....still no answers....need help with simple random sample...Have 460 clients...need SRS of 20 clients for survey...Have a number table to select sample from:13873 81598 95052 90908 73592
asked by nancy on February 26, 2007
90. ## Math Help Please? c: Thank you :3
At your summer job with a research company, you must get a random sample of people from your town to answer a question about spending habits. Which of the following methods is most likely to be random? A) You survey customers at the local shopping mall. B)
asked by X3 ISayRawr on April 10, 2014
91. ## Stats
Which of the following are binomial experiments? (more than one is possible) 1.) A store finds that 32% of people who enter the store will make a purchase. During a day, 50 people enter the store. The random variable represents the number of people who
asked by Semmy on October 13, 2015
92. ## prob
Consider the triangle with vertices (0; 0), (1; 0) and (0; 1). Let Z be a uniform random variable in the interval [0; 1]. Draw a vertical line that intersects the x axis at Z. This line divides the triangle in two pieces. Select a point (X; Y ) uniformly
asked by hsuan on May 3, 2013
93. ## Statistics
Darned old Foofy! Foofy computes the x̅ (MEAN) from data that her professor says is a random sample from population Q. she correctly computes that this mean has a z-score of +41 on the sampling distribution for population Q. Foofy claims she has proven
asked by Mari on June 1, 2013
94. ## Math
A drawer is filled with red and blue socks. If the probability of selecting a red sock at random is 2/5 what is the probability that a blue sock will be chosen at random? a)3/5 b)2/5 c)2/3 d)3/2
95. ## algebra
Which of the following are examples of inferential statistics? Check all that apply. A.Finding the average (mean) of a set of numbers B.Testing every lightbulb produced by a company to confirm that each one works C.Testing random samples of lightbulbs
asked by daryl on April 18, 2012
96. ## Statistics
1. In thinking about doing statistical analysis, the sample mean should be interpreted as: A.)a constant value that is equal to the population mean. B.) a constant value that is approximately equal to the population mean. C.) a random variable that is
asked by Sara on May 3, 2011
97. ## Research Methods
A researcher wants to attain a random sample of young drivers who have received DUIs within the last year. Using the PREP system, he asks all first year students at WLU who have received a DUI to volunteer for the study. Thirty-five students actually admit
asked by Jill on October 15, 2011
1. To plan your time for a research project, it is best to (1 point) divide the time spent on each step evenly. keep your deadlines flexible. start at the beginning and plan from there.***** work backward. 2. Which of the following is not a guideline for
asked by Anonymous on March 15, 2017
99. ## statistics
Professor Smith conducted a class exercise in which students ran a computer program to generate random samples from a population that had a mean of 50 and a standard deviation of 9mm. Each of Smith's students took a random sample of size n and calculated
asked by lauren on February 16, 2008
100. ## probability
There are 100 people in line to board a plane with 100 seats. The first person has lost his boarding pass, so he takes a random seat. Everyone that follows takes their assigned seat if it's available, but otherwise takes a random unoccupied seat. What is
asked by unowen on November 23, 2017 | 7,684 | 29,880 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2018-47 | longest | en | 0.87824 |
https://wirings-diagram.com/3-phase-motor-wiring-diagram-6-wire/ | 1,624,359,716,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488517048.78/warc/CC-MAIN-20210622093910-20210622123910-00359.warc.gz | 559,996,356 | 30,017 | # 3 Phase Motor Wiring Diagram 6 Wire
3 Phase Motor Wiring Diagram 6 Wire – 3 phase motor wiring diagram 6 wire, Every electric structure is made up of various unique components. Each part should be placed and linked to different parts in particular way. Otherwise, the structure won’t work as it ought to be. In order to be certain that the electrical circuit is built correctly, 3 Phase Motor Wiring Diagram 6 Wire is necessary. How can this diagram aid with circuit construction?
Wiring Diagram 6 Lead 3 Phase 480 Volt Motor | Wiring Library – 3 Phase Motor Wiring Diagram 6 Wire
The diagram provides visual representation of a electric structure. However, the diagram is a simplified variant of this arrangement. This makes the procedure for building circuit simpler. This diagram gives advice of circuit components in addition to their own placements.
## Components of 3 Phase Motor Wiring Diagram 6 Wire and Some Tips
There are just two things which are going to be present in any 3 Phase Motor Wiring Diagram 6 Wire. The first element is emblem that indicate electric component in the circuit. A circuit is usually composed by numerous components. The other thing that you will come across a circuit diagram would be traces. Lines in the diagram show exactly how every component connects to one another.
The rankings of circuit components are comparative, not accurate. The arrangement is also not logical, unlike wiring schematics. Diagram only reveals where to put component at a spot relative to other elements within the circuit. Even though it is simplified, diagram is a great foundation for anyone to build their own circuit.
One thing you have to learn before reading a circuit diagram would be your symbols. Every symbol that is presented on the diagram reveals specific circuit component. The most common elements are capacitor, resistor, and battery. There are also other elements like ground, switch, motor, and inductor. All of it depends on circuit that’s being built.
According to earlier, the lines at a 3 Phase Motor Wiring Diagram 6 Wire signifies wires. Sometimes, the cables will cross. But, it does not imply connection between the cables. Injunction of two wires is usually indicated by black dot at the junction of 2 lines. There will be primary lines which are represented by L1, L2, L3, and so on. Colors are also utilised to differentiate cables.
Ordinarily, there are two chief sorts of circuit links. The primary one is known as series link. It’s the simpler type of relationship because circuit’s components are put within a specified line. Because of that the electrical current in each part is comparable while voltage of the circuit is total of voltage in each component.
### 3 Phase Motor Wiring Diagram 6 Wire Video
Parallel link is more complex than the series one. Unlike in series connection, the voltage of each part is similar. It’s because the element is directly connected to electricity resource. This circuit includes branches that are passed by distinct electric current levels. The present joins together when the branches match.
There are numerous things that an engineer should pay attention to if drawing wirings diagram. To begin with, the symbols used in the diagram ought to be accurate. It should represent the specific component needed to construct an intended circuit. After the symbol is incorrect or unclear, the circuit will not work since it is supposed to.
It is also highly advised that engineer brings positive supply and negative supply symbols for clearer interpretation. Normally positive supply emblem (+) is located over the line. Meanwhile, the negative source symbol is place under it. The current flows from the left to right.
Besides this, diagram drawer is advised to limit the amount of line crossing. The line and part placement should be made to decrease it. But if it is inevitable, use universal emblem to indicate whether there’s a junction or if the lines are not really connected.
Since you can begin drawing and translating 3 Phase Motor Wiring Diagram 6 Wire can be a complicated undertaking on itself. The information and ideas which have been elaborated above ought to be a wonderful kick start, though. 3 Phase Motor Wiring Diagram 6 Wire
### 3 Phase Motor Wiring Diagram 6 Wire Images
3 Phase Washing Machine Motor Wiring With 6 Wire Timer (Urdu/hindi – 3 Phase Motor Wiring Diagram 6 Wire
How To Wire A 3-Phase Induction Motor? – Youtube – 3 Phase Motor Wiring Diagram 6 Wire
6 Lead 3 Phase Motor Wiring Diagram | Wiring Library – 3 Phase Motor Wiring Diagram 6 Wire
Motor Wiring Diagrams 3 Phase 6 Wire | Manual E-Books – 3 Phase Motor Wiring Diagram 6 Wire
3 Phase 12 Lead Motor Wiring Connection 4160 – Wiring Diagrams Hubs – 3 Phase Motor Wiring Diagram 6 Wire
230 6 Wire 3 Phase Diagram | Manual E-Books – 3 Phase Motor Wiring Diagram 6 Wire
230 6 Wire 3 Phase Diagram | Manual E-Books – 3 Phase Motor Wiring Diagram 6 Wire | 1,021 | 4,944 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2021-25 | longest | en | 0.924137 |
https://gforth.org/manual/Two_002dstage-integer-division.html | 1,669,548,181,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710237.57/warc/CC-MAIN-20221127105736-20221127135736-00819.warc.gz | 302,971,040 | 4,233 | Next: , Previous: , Up: Arithmetic [Contents][Index]
#### 5.5.5 Two-stage integer division
On most hardware, multiplication is significantly faster than division. So if you have to divide many numbers by the same divisor, it is usually faster to determine the reciprocal of the divisor once and multiply the numbers with the reciprocal. For integers, this is tricky, so Gforth packages this work into the words described in this section.
Let’s start with an example: You want to divide all elements of an array of cells by the same number n. A straightforward way to implement this is:
: array/ ( addr u n -- )
-rot cells bounds u+do
i @ over / i !
1 cells +loop
drop ;
A more efficient version looks like this:
: array/ ( addr u n -- )
{: | reci[ staged/-size ] :}
reci[ /f-stage1m
cells bounds u+do
i @ reci[ /f-stage2m i !
1 cells +loop ;
This example first creates a local buffer reci[ with size staged/-size for storing the reciprocal data. Then /f-stage1m computes the reciprocal of n and stores it in reci[. Finally, inside the loop /f-stage2m uses the data in reci[ to compute the quotient.
There are some limitations: Only positive divisors are supported for /f-stage1m; for u/-stage1m you can use a divisor of 2 or higher. You get an error if you try to use an unsupported divisor. You must initalize the reciprocal buffer for the floored second-stage words with /f-stage1m and for the unsigned second-stage words with u/-stage1m. You must not modify the reciprocal buffer between the first stage and the second stage; basically, don’t treat it as a memory buffer, but as something that is only mutable by the first stage; the point of this rule is that future versions of Gforth will not consider aliasing of this buffer.
The words are:
staged/-size ( – u ) gforth-1.0 “staged-slash-size”
Size of buffer for u/-stage1m or /f-stage1m.
/f-stage1m ( n addr-reci – ) gforth-1.0 “slash-f-stage1m”
Compute the reciprocal of n and store it in the buffer addr-reci of size staged/-size. Throws an error if n<1.
/f-stage2m ( n1 a-reci – nquotient ) gforth-1.0 “slash-f-stage2m”
Nquotient is the result of dividing n1 by the divisor represented by a-reci, which was computed by /f-stage1m.
modf-stage2m ( n1 a-reci – umodulus ) gforth-1.0 “mod-f-stage2m”
Umodulus is the remainder of dividing n1 by the divisor represented by a-reci, which was computed by /f-stage1m.
/modf-stage2m ( n1 a-reci – umodulus nquotient ) gforth-1.0 “slash-mod-f-stage2m”
Nquotient is the quotient and umodulus is the remainder of dividing n1 by the divisor represented by a-reci, which was computed by /f-stage1m.
u/-stage1m ( u addr-reci – ) gforth-1.0 “u-slash-stage1m”
Compute the reciprocal of u and store it in the buffer addr-reci of size staged/-size. Throws an error if u<2. count trailing zeros in binary representation of x
u/-stage2m ( u1 a-reci – uquotient ) gforth-1.0 “u-slash-stage2m”
Uquotient is the result of dividing u1 by the divisor represented by a-reci, which was computed by u/-stage1m.
umod-stage2m ( u1 a-reci – umodulus ) gforth-1.0 “u-mod-stage2m”
Umodulus is the remainder of dividing u1 by the divisor represented by a-reci, which was computed by u/-stage1m.
u/mod-stage2m ( u1 a-reci – umodulus uquotient ) gforth-1.0 “u-slash-mod-stage2m”
Uquotient is the quotient and umodulus is the remainder of dividing u1 by the divisor represented by a-reci, which was computed by u/-stage1m.
Gforth currently does not support staged symmetrical division.
You can recover the divisor from (the address of) a reciprocal with staged/-divisor @:
staged/-divisor ( addr1 – addr2 ) gforth-1.0 “staged-slash-divisor”
Addr1 is the address of a reciprocal, addr2 is the address containing the divisor from which the reciprocal was computed.
This can be useful when looking at the decompiler output of Gforth: a division by a constant is often compiled to a literal containing the address of a reciprocal followed by a second-stage word.
The performance impact of using these words strongly depends on the architecture (does it have hardware division?) and the specific implementation (how fast is hardware division?), but just to give you an idea about the relative performance of these words, here are the cycles per iteration of a microbenchmark (which performs the mentioned word once per iteration) on two AMD64 implementations; the norm column shows the normal division word (e.g., u/), while the stg2 column shows the corresponding stage2 word (e.g., u/-stage2m):
Skylake Zen2
norm stg2 norm stg2
41.3 15.8 u/ 35.2 21.4 u/
39.8 19.7 umod 36.9 25.8 umod
44.0 25.3 u/mod 43.0 33.9 u/mod
48.7 16.9 /f 36.2 22.5 /f
47.9 20.5 modf 37.9 27.1 modf
53.0 24.6 /modf 45.8 35.4 /modf
227.2 u/stage1 101.9 u/stage1
159.8 /fstage1 97.7 /fstage1
Next: Bitwise operations, Previous: Integer division, Up: Arithmetic [Contents][Index] | 1,376 | 4,918 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2022-49 | latest | en | 0.82931 |
https://tex.stackexchange.com/questions/437064/plot-fibonacci-sequence-with-tikzpicture | 1,719,118,505,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862430.93/warc/CC-MAIN-20240623033236-20240623063236-00015.warc.gz | 502,537,331 | 39,763 | # Plot Fibonacci Sequence (with tikzpicture)
How do I plot the Fibonacci sequence (using tikzpicture preferably)?
\documentclass{article}
\usepackage{pgfplots}
\begin{document}
\begin{tikzpicture}
\begin{axis}[xmin=0, xmax=30, ymin=2, ymax=3]
\end{axis}
\end{tikzpicture}
\end{document}
## Edit
Since it seems more complicated than I thought, I went for the following solution: plot the discrete version of the continuous Fibonacci function, see for instance here. Any suggestions are surely still welcomed.
So now it goes like this:
\documentclass{article}
\usepackage{pgfplots}
\begin{document}
\begin{tikzpicture}
\begin{axis}[xmin=0, xmax=8, ymin=0, ymax=14, xlabel=$n$, ylabel=$a_n$, axis x line=center, axis y line=center]
\addplot[samples at={0,1,...,7},only marks] expression {( ((1+sqrt(5))/(2))^\x - cos(deg(\x * pi)) * ((1+sqrt(5))/(2))^(-\x) )/sqrt(5) };
\end{axis}
\end{tikzpicture}
\end{document}
• It would be helpful if you composed a fully compilable MWE including \documentclass and the appropriate packages that sets up the problem. While solving problems can be fun, setting them up is not. Then, those trying to help can simply cut and paste your MWE and get started on solving the problem.
– AML
Commented Jun 19, 2018 at 14:45
• Since the sequence grows exponentially fast, I would think you would want to plot only the first few numbers. At which point, you may as well write the numbers in. Commented Jun 19, 2018 at 18:47
• @Teepeemm I kinda arrived at that point as well... My current solution is to plot the discrete version of the Fibonacci function. I'll update the question. Commented Jun 19, 2018 at 19:04
• @TikzerWoods the formula generalizes to \sqrt(x^2 + 4) for F_n(x) that's why I wrote it - for optical reasons - as such in the code. Commented Jun 19, 2018 at 19:29
• @TikzerWoods I'll changed it to the 'golden ratio' formula for clarity. Commented Jun 19, 2018 at 19:38
Here is a direct loop method :
\documentclass[tikz,border=7pt]{standalone}
\begin{document}
\tikz
\foreach[
remember=\g as \h (initially 1),
remember=\f as \g (initially 0),
evaluate=\f using int(\g+\h)
] \n in {1,...,7}
\fill[green,draw=black] (\n,0) rectangle +(1,\f) node[black,scale=2,above left]{\f};
\end{document}
EDIT: If you want to go to 30 (and more) you can use xint package.
\documentclass[tikz,border=7pt]{standalone}
\usepackage{xintexpr}
\begin{document}
\begin{tikzpicture}[xscale=.35]
\foreach[
remember=\g as \h (initially 1),
remember=\f as \g (initially 0)
] \n in{1,...,30}{
\edef\f{\thexintexpr \g + \h \relax}
\edef\ff{\thexintfloatexpr \f/10000 \relax}
\fill[green,draw=black] (\n,0) rectangle +(1,\ff);
}
\end{tikzpicture}
\end{document}
EDIT: Following the comment of @jfbu the following code will produce the same image:
\documentclass[tikz,border=7pt]{standalone}
\usepackage{xintexpr}
\begin{document}
\xdef\fibs{\thexintfloatexpr rrseq(0, 1/10000 ; @1+@2, i=2..30)\relax}
\tikz[xscale=.35]
\foreach[count=\n] \f in \fibs
\fill[green,draw=black] (\n,0) rectangle +(1,\f);
\end{document}
NOTE: It is probably faster to use directly something like
\foreach[count=\n] \f in {1,1,2,3,5,8,13,21,34,55,89,144,233,377,610,987,1597,2584,4181,6765,10946,17711,28657,46368,75025,121393,196418,317811,514229,832040}{
...
}
• is there a way to get a dots instead of bars? Commented Jun 19, 2018 at 18:16
• @DeeCeeDelux You need only to replace \fill[green,draw=black] (\n,0) rectangle +(1,\f); by \node[circle,fill] at (\n,\f){}; to get dots. (I am wondering, though, if there is a way to include 0 in the series.)
– user121799
Commented Jun 19, 2018 at 18:29
• @marmot to start from 0 you can use \g in place of \f in the first two codes. And in the @jfbu's code you can use \xdef\fibs{0,\thexintfloatexpr rrseq(0, 1/10000 ; @1+@2, i=2..30)\relax}.
– Kpym
Commented Jun 19, 2018 at 19:18
The Fibonacci series is implemented in the pgfmanual in section 56.1. All you need to do is to adapt it from there.
\documentclass[tikz,border=3.14mm]{standalone}
\usetikzlibrary{math}
\begin{document}
\tikzmath{
function fibonacci(\n) {
if \n == 0 then {
return 0;
} else {
return fibonacci2(\n, 0, 1);
}; };
function fibonacci2(\n, \p, \q) {
if \n == 1 then {
return \q;
} else {
return fibonacci2(\n-1, \q, \p+\q);
};
};
}
\begin{tikzpicture}
\foreach \X in {0,1,...,8}{
\node[circle,fill,label=above:{\pgfmathparse{int(fibonacci(\X))}
\pgfmathresult}] at (\X,{fibonacci(\X)}) {};}
\end{tikzpicture}
\end{document}
ADDITIONAL REMARK: It turns out not to be so easy to plot this function with pgfplots. The problem seems to be that pgfplots uses the fpu library (yet setting \pgfkeys{/pgf/fpu=false} does not help). In this answer, the problem is avoided by doing an external computation. However, I wrote the answer when there was no MWE available, and followed the directive "using tikzpicture preferably" ;-).
• @DeeCeeDelux The math library was introduced in version 3 of pgf/TikZ I believe, if you're still on version 2.x you will not have that library. You can check the version for example by placing \pgfversion in the document. Commented Jun 19, 2018 at 15:28
With a variant of https://tex.stackexchange.com/a/51422/4427, we can generate the sequence to be fed to \foreach:
\documentclass[tikz]{standalone}
\usepackage{xparse}
\ExplSyntaxOn
\cs_new:Npn \fibo #1 { \fibo_recurrence:nnnn{0}{1}{0}{#1} }
\cs_new:Npn \fibo_recurrence:nnnn #1 #2 #3 #4
{
\int_compare:nTF { #1 = #4 }
{ #3 }
{
#3 , \fibo_recurrence:ffnn
{ \int_eval:n {#1+1} }
{ \int_eval:n {#2+#3} }
{ #2 }
{ #4 }
}
}
\cs_generate_variant:Nn \fibo_recurrence:nnnn { ffnn }
\ExplSyntaxOff
\begin{document}
\begin{tikzpicture}
\edef\fibos{\fibo{7}}
\foreach [count=\index] \f in \fibos
{
\fill[green,draw=black] (\index,0) rectangle +(1,\f)
node[black,scale=2,above left]{\f};
}
\end{tikzpicture}
\end{document} | 1,947 | 5,839 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2024-26 | latest | en | 0.851406 |
https://www.geeksforgeeks.org/matplotlib-axes-axes-twiny-in-python/ | 1,643,074,276,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304749.63/warc/CC-MAIN-20220125005757-20220125035757-00005.warc.gz | 822,403,930 | 23,284 | matplotlib.axes.Axes.twiny() in Python
• Last Updated : 21 Apr, 2020
Matplotlib is a library in Python and it is numerical – mathematical extension for NumPy library. The Axes Class contains most of the figure elements: Axis, Tick, Line2D, Text, Polygon, etc., and sets the coordinate system. And the instances of Axes supports callbacks through a callbacks attribute.
matplotlib.axes.Axes.twiny() Function
The Axes.twiny() function in axes module of matplotlib library is used to create a twin Axes sharing the yaxis.
Syntax: Axes.twiny(self)
Return value: This method is used to returns the following.
• ax_twin : This returns the newly created Axes instance.
Below examples illustrate the matplotlib.axes.Axes.twiny() function in matplotlib.axes:
Example 1:
`# Implementation of matplotlib function``# Implementation of matplotlib function``import` `matplotlib.pyplot as plt``import` `numpy as np`` ` ` ` `def` `GFG1(temp):`` ``return` `(``5.` `/` `9.``) ``*` `(temp ``-` `32``)`` ` `def` `GFG2(ax1):`` ``y1, y2 ``=` `ax1.get_ylim()`` ``ax_twin .set_ylim(GFG1(y1), GFG1(y2))`` ``ax_twin .figure.canvas.draw()`` ` `fig, ax1 ``=` `plt.subplots()``ax_twin ``=` `ax1.twiny()`` ` `ax1.callbacks.connect(``"ylim_changed"``, GFG2)``ax1.plot(np.linspace(``-``40``, ``120``, ``100``))``ax1.set_ylim(``0``, ``100``)`` ` `ax1.set_xlabel(``'Fahrenheit'``)``ax_twin .set_xlabel(``'Celsius'``)`` ` `fig.suptitle('matplotlib.axes.Axes.twiny()\`` ``function Example\n\n', fontweight ``=``"bold"``)``plt.show()`
Output:
Example 2:
`# Implementation of matplotlib function``import` `numpy as np``import` `matplotlib.pyplot as plt`` ` `# Create some mock data``t ``=` `np.arange(``0.01``, ``10.0``, ``0.001``)``data1 ``=` `np.exp(t)``data2 ``=` `np.sin(``0.4` `*` `np.pi ``*` `t)`` ` `fig, ax1 ``=` `plt.subplots()`` ` `color ``=` `'tab:blue'``ax1.set_ylabel(``'time (s)'``)``ax1.set_xlabel(``'exp'``, color ``=` `color)``ax1.plot(data1, t, color ``=` `color)``ax1.tick_params(axis ``=``'x'``, labelcolor ``=` `color)`` ` `ax2 ``=` `ax1.twiny()`` ` `color ``=` `'tab:green'``ax2.set_xlabel(``'sin'``, color ``=` `color)``ax2.plot(data2, t, color ``=` `color)``ax2.tick_params(axis ``=``'x'``, labelcolor ``=` `color)`` ` `fig.suptitle('matplotlib.axes.Axes.twiny()\`` ``function Example\n\n', fontweight ``=``"bold"``)``plt.show()`
Output:
My Personal Notes arrow_drop_up | 783 | 2,390 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2022-05 | latest | en | 0.418529 |
https://www.kiipeilytuomas.fi/articles-in-english/drop-test-physics-iv/ | 1,679,545,184,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296944996.49/warc/CC-MAIN-20230323034459-20230323064459-00092.warc.gz | 968,696,649 | 14,481 | Categories
# Drop test physics IV
Tuomas Pöysti 2021
In the previous part, I managed to extract displacement data from a load cell dataset sampled from a small drop test. The original data was something like this:
And the displacement data looks as follows (both have time in seconds as the horizontal axis):
## The lowest point
What have we learned this far? I think one of the interesting things is that my center of gravity dips some 23 mm below the zero level. This is (part of) the stretch in the system that allows for the energy to be absorbed.
By the way, notice that the lowest point is reached a bit later than the maximum peak of force (or acceleration). The greatest acceleration is 31.5 m/s^2 @ 0.00 s and the lowest point of my center of gravity is -23 mm @0.03 s. Not sure if this is just inaccuracy, but as far as I know, this would be perfectly possible.
In fact, it even makes sense. There certainly is a spring component and a damper component in the system, and the total “braking” force is their sum. The spring part by definition reaches it’s maximum at the lowest point of displacement, but the damper part is proportional to velocity, which is naturally drops as the lowest point gets closer.
If this is the case, the greatest forces are experienced on the way down, not while reaching the lowest point. The difference is just a curiosity, though.
## Where is the spring?
How did the cowstail manage to stretch over 20 mm or almost 5%? I guess it did not. The system has to be seen as a whole. In the picture below, the system on the left represents a rigid climber attached to a rigid cowstail. The next one considers the cowstail a spring, but assumes a rigid climber – this would be a sufficient model for the DMM’s widely cited article. In the third one, the climber’s mass is reduced closer to a center of gravity, which is attached to the cowstail spring by another spring. This secondary spring is effectively my body, in a complex way.
I guess the diagram might be a bit leading, since I draw the cowstail spring as way stiffer looking than the one inside my body. I’m just assuming this is the case – not sure if I felt the same if I had done more core exercises!
While picturing a harness-supported human body flexing so that it’s center of gravity shifts downwards, one again has to remember to see the system as a whole. Of course the leg loops may bury themselves deeper into the softness of the climber’s thighs, but there’s probably a lot more to it.
For example: the human head is around 8% of the whole body weight. That is, if one managed to bow their head 10 cm down, it would result in their center of gravity shifting about 8 mm down. Consider the shoulders, arms, legs and all the loose soft tissue doing the same!
I am going to study this some more, and I’ll be surprised if I’m to find anything else than that the spring was me, practically speaking.
## Power
Back to basic physcs! Power is the rate of energy (or energy’s derivative over time, if we want to drop some impressive terms). The SI unit of power is W (watt). As we already knew, energy is the product of force and displacement. Since velocity is the derivative of displacement over time, power is
P = Fv
Or as a plot:
This shows that the cowstail-body combination has brief maximum power of 3kW while arresting the fall. What is this piece of information good for? Nothing, probably.
## Force vs displacement
This is the one I was waiting for the most, the plot of force against displacement. I think it’s lovely:
It points out the hysteresis in the system, showing how the energy is dissipated. The two straight portions resemble some kind of spring-like behavior. The trianglular area in the force-displacement plane equals an 87.7 J work:
But how much was my kinetic energy? That’s easy now that we have velocity data and know my mass (almost too well at this point):
The maximum total kinetic energy is about 138 J. It seems the 87.7 J chunk is over 64% of the whole kinetic energy to be absorbed! That is, if this data is correct, most of the kinetic energy was absorbed before my center of gravity reached the zero level of displacement. As we defined in the earlier parts, this is the level that the center of gravity settles at the end of the data, after the oscillating has ended.
## The displacement oddity
Doesn’t that sound a bit weird? In fact, how can my center of gravity have the same displacement several times during the test, with the force varying from 0 to almost 2.5 kN?
A simple answer is that there’s an error in the calculations. But I think we don’t need to go there, yet.
This might be explained by the same “think it as a whole” mantra. My center of gravity probably moves a lot with respect to the harness, whether I’m falling or just slowly getting suspended by the harness. While doing the test, I dropped myself by hanging on a crimp block that was next to my load cell anchor. That is, my arms and shoulders were high, and my back was probably quite vertical. Most likely the first thing that happens when the cowstail starts to catch is them shifting downwards and back.
It must be possible to shift one’s center of gravity quite much by taking different poses while supported by a harness. Think about flexing your knees and so on.
Whatever this kind of tests ever suggest about falling on a cowstail, there will always be a great deal of variance in postures and the body’s readiness to take the hit.
## Further studies
Depending on how interesting I find this in the future, I’ll probably be testing and posting. If you’d like to learn more, please ask me or otherwise show that this matters, it may have a huge effect on how much I care! I generally do this thing solely for myself, so it’s completely possible that tomorrow I’m more interested about ethics or cooking or running and leave this for years.
I already know that I should conduct a control test using a rigid mass. Also, it might be useful to have a quick release link to drop me, that might change how the center of gravity shifts. Also, that way it would be possible to record the exact moment of release and to have control over drop height.
And when it comes to drop height, I need to be able to take a bit harder falls. Maybe find someone brave to show me how it’s done! | 1,418 | 6,322 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2023-14 | longest | en | 0.952576 |
https://chemistry.stackexchange.com/questions/125056/use-of-concentration-for-ph | 1,581,925,963,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875141749.3/warc/CC-MAIN-20200217055517-20200217085517-00151.warc.gz | 329,519,580 | 33,734 | # Use of Concentration for pH
When we find the pH of a solution, do we use normality or molarity? My teacher said that if the compound is present by itself, we can use normality but in all other cases we should use molarity. Can you explain why?
Current definition implies that $$\mathrm{pH}$$ is a function of relative activity. Originally, the amount concentration of $$\ce{H+}$$ in $$\pu{mol L-1}$$ was proposed, which is also often used these days as an approximation [1].
$$\mathrm{pH}$$ was originally defined by Sørensen in 1909 … in terms of the concentration of hydrogen ions (in modern nomenclature) as $$\mathrm{pH} = −\lg(c_\ce{H}/c^\circ)$$ where $$c_\ce{H}$$ is the hydrogen ion concentration in $$\pu{mol dm–3},$$ and $$c^\circ = \pu{1 mol dm–3}$$ is the standard amount concentration. Subsequently …, it has been accepted that it is more satisfactory to define $$\mathrm{pH}$$ in terms of the relative activity of hydrogen ions in solution
$$\mathrm{pH} = -\lg a_\ce{H} = -\lg (m_\ce{H}γ_\ce{H}/m^\circ)\tag{1}$$
where $$a_\ce{H}$$ is the relative (molality basis) activity and $$γ_\ce{H}$$ is the molal activity coefficient of the hydrogen ion $$\ce{H+}$$ at the molality $$m_\ce{H},$$ and $$m^\circ$$ is the standard molality. The quantity $$\mathrm{pH}$$ is intended to be a measure of the activity of hydrogen ions in solution. However, since it is defined in terms of a quantity that cannot be measured by a thermodynamically valid method, eq. 1 can be only a notional definition of $$\mathrm{pH}.$$
So, if the numerical value for normality coincides with the one for molarity, and one can omit the activity in favor of molarity, i.e. when both equivalence factor $$f_\mathrm{eq}$$ and activity coefficient $$γ$$ are approx. 1, then yes, normality can also be used to estimate the $$\mathrm{pH}.$$
I'm not sure how to interpret "present by itself" part of your question, but if it means there is a single solute, then it seems to be incorrect. The simplest counter example would be a solution of di-, tri- and polybasic acid/base, say $$\ce{H2SO4},$$ for which the equivalent concentration is double the value of amount concentration.
### References
1. Buck, R. P.; Rondinini, S.; Covington, A. K.; Baucke, F. G. K.; Brett, C. M. A.; Camoes, M. F.; Milton, M. J. T.; Mussini, T.; Naumann, R.; Pratt, K. W.; et al. Measurement of pH. Definition, Standards, and Procedures (IUPAC Recommendations 2002). Pure and Applied Chemistry 2002, 74 (11), 2169–2200. DOI: 10.1351/pac200274112169. (Free Access)
• So in every scenario we would be using Normality rather than Molarity? If so then what do we use Molarity for? – Nate william Dec 13 '19 at 14:13
• @Natewilliam No, you got it backwards, it's amount concentration, or molarity, that is used in the absence of activity. – andselisk Dec 13 '19 at 14:26
• Isn't it the same? If there's no activity, then normality is equal to molarity ( as Normality = Molarity*activity(or n-factor) ) So isn't it correct to say that we always use normality for any sort of calculation rendering molarity unnecessary? – Nate william Dec 13 '19 at 14:39
• @Natewilliam No. Normality is $c/f_\mathrm{eq}.$ If $f_\mathrm{eq} = 1$ (monobasic substance), then the numerical values of normality and molarity are the same. Activity has nothing to do with $f_\mathrm{eq}$ (or $n$-factor, s you call it), it's an unrelated quantity. – andselisk Dec 13 '19 at 14:52
pH is not defined versus normality or molarity of an acid: Whatever the normality or the molarity, the pH is defined from the activity (or the concentration) of the ions H+. A given value of the normality or of the molarity does not give you the activity (or the concentration) of H+. So it does not allow you to calculate the pH.
• The statements "pH is not defined versus normality or molarity" and "pH is defined from the activity (or the concentration)" seem to contradict each other as both normality or molarity are concentrations. You either have to be more specific, or stick with only one nominal definition. However, if you do the latter, this won't answer the question. Besides, I don't see how this adds anything to an existing answer. – andselisk Dec 13 '19 at 12:44
• I think Maurice’s point is that OP seems to be referring to normality or molarity of an acid, not of protons. For a strong acid, the molar amount of protons is equal to normality of acid (which is all in the form of conj base). For a weaker acid, that is not the case, so pH cannot be calculated using only the acid molarity or normality. – Andrew Dec 13 '19 at 18:36 | 1,279 | 4,566 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 21, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2020-10 | latest | en | 0.918263 |
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# Find the slope of the line passing through the points A(−2,1) and B(0,3)
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## A≡(−2,1)=(x1,y1)and B≡(0,3)=(x2,y2)Then, Slope of line AB=y1−y2x1−x2=1−3−2−0=−2−2=1
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# Parameterized Low-Rank Binary Matrix Approximation
## File
LIPIcs.ICALP.2018.53.pdf
• Filesize: 0.57 MB
• 16 pages
## Cite As
Fedor V. Fomin, Petr A. Golovach, and Fahad Panolan. Parameterized Low-Rank Binary Matrix Approximation. In 45th International Colloquium on Automata, Languages, and Programming (ICALP 2018). Leibniz International Proceedings in Informatics (LIPIcs), Volume 107, pp. 53:1-53:16, Schloss Dagstuhl – Leibniz-Zentrum für Informatik (2018)
https://doi.org/10.4230/LIPIcs.ICALP.2018.53
## Abstract
We provide a number of algorithmic results for the following family of problems: For a given binary m x n matrix A and a nonnegative integer k, decide whether there is a "simple" binary matrix B which differs from A in at most k entries. For an integer r, the "simplicity" of B is characterized as follows. - Binary r-Means: Matrix B has at most r different columns. This problem is known to be NP-complete already for r=2. We show that the problem is solvable in time 2^{O(k log k)}*(nm)^O(1) and thus is fixed-parameter tractable parameterized by k. We also complement this result by showing that when being parameterized by r and k, the problem admits an algorithm of running time 2^{O(r^{3/2}* sqrt{k log k})}(nm)^O(1), which is subexponential in k for r in o((k/log k)^{1/3}). - Low GF(2)-Rank Approximation: Matrix B is of GF(2)-rank at most r. This problem is known to be NP-complete already for r=1. It is also known to be W[1]-hard when parameterized by k. Interestingly, when parameterized by r and k, the problem is not only fixed-parameter tractable, but it is solvable in time 2^{O(r^{3/2}* sqrt{k log k})}(nm)^O(1), which is subexponential in k for r in o((k/log k)^{1/3}). - Low Boolean-Rank Approximation: Matrix B is of Boolean rank at most r. The problem is known to be NP-complete for k=0 as well as for r=1. We show that it is solvable in subexponential in k time 2^{O(r2^r * sqrt{k log k})}(nm)^O(1).
## Subject Classification
##### ACM Subject Classification
• Mathematics of computing → Combinatorial algorithms
• Theory of computation → Fixed parameter tractability
##### Keywords
• Binary matrices
• clustering
• low-rank approximation
• fixed-parameter tractability
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## References
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40. Daniel D Lee and H Sebastian Seung. Learning the parts of objects by non-negative matrix factorization. Nature, 401(6755):788-791, 1999.
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55. A A Razborov. On rigid matrices. Manuscript in russian, 1989.
56. Ilya P. Razenshteyn, Zhao Song, and David P. Woodruff. Weighted low rank approximations with provable guarantees. In Proceedings of the 48th Annual ACM Symposium on Theory of Computing (STOC), pages 250-263. ACM, 2016. URL: http://dx.doi.org/10.1145/2897518.2897639.
57. Bao-Hong Shen, Shuiwang Ji, and Jieping Ye. Mining discrete patterns via binary matrix factorization. In Proceedings of the 15th ACM SIGKDD International Conference on Knowledge Discovery and Data Mining (KDD), pages 757-766, New York, NY, USA, 2009. ACM. URL: http://dx.doi.org/10.1145/1557019.1557103.
58. Leslie G. Valiant. Graph-theoretic arguments in low-level complexity. In Mathematical Foundations of Computer Science (MFCS), volume 53 of Lecture Notes in Comput. Sci., pages 162-176. Springer, 1977.
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X
Feedback for Dagstuhl Publishing | 4,744 | 15,281 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2024-22 | latest | en | 0.834938 |
https://www.prepbharat.com/HighwayEng/highway-engineering-interview-questions-answers.html | 1,716,374,361,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058542.48/warc/CC-MAIN-20240522101617-20240522131617-00662.warc.gz | 849,315,279 | 5,036 | ## Highway Engineering Questions and Answers - Traffic Characteristics Part-2
1. The most likely cause of accidents is ___________
a) Impatience in driving
b) Slow speed of vehicle
Explanation: The most likely cause in this case is impatience in driving as it may lead to anxiety and fear; this mostly affects the user psychologically.
2. The width recommended by IRC for all type of vehicles is ___________
a) 1.5m
b) 2.0m
c) 2.5m
d) 3.0m
Explanation: IRC recommends a width of 2.5m for any type of vehicle.
3. The stability of a vehicle is influenced by ___________
a) Width of wheel base only
b) Width of wheel base and height of gravity
c) Height of gravity only
d) Length of vehicle only
Explanation: The stability of a vehicle is influenced greatly by width of wheel base and height of center of gravity is useful near horizontal curves.
4. The height of the vehicle mainly influences?
a) Width of pavement
b) Length of curve
c) Clearance under structures
d) Design velocity
Explanation: The clearance of structures like over bridges and under bridges mainly depends on the height of the vehicle.
5. The minimum number of parameters needed to measure brake efficiency is?
a) One
b) Two
c) Three
d) Four
Explanation: The parameters required for measuring brake efficiency are initial speed; braking distance and actual duration of braking application in these three parameters any two of them are needed.
6. If the acceleration of the vehicle is 6.17m/sec2 then the average skid resistance is?
a) 0.61
b) 0.62
c) 0.63
d) 0.64
Explanation: The average skid resistance f=a/g
f=6.17/9.81
=0.63.
7. What is the first stage in traffic engineering studies?
a) Traffic volume studies
b) Spot speed studies
c) Speed and delay studies
d) Origin and destination studies
Explanation: The first step in traffic engineering studies is traffic volume studies, which are carried out to understand the traffic characteristics.
8. The traffic volume is usually expressed in __________
a) LMV
b) PCU
c) LCV
d) HCV
Explanation: In India the traffic is heterogeneous so there are many types of vehicles, so every vehicle is expressed with the same unit PCU which means passenger car unit.
9.The number of vehicles that pass through a transverse line of road at a given time in a specified direction is called __________
a) Traffic studies
b) Traffic flow
c) Traffic origin
d) Traffic destination | 578 | 2,397 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2024-22 | latest | en | 0.878354 |
https://returntoedencosmetics.com/dermatology/how-many-moles-of-co2-are-produced-from-1-mole-of-oxygen.html | 1,653,020,899,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662531352.50/warc/CC-MAIN-20220520030533-20220520060533-00282.warc.gz | 579,335,156 | 18,919 | # How many moles of CO2 are produced from 1 mole of oxygen?
Contents
The formula of carbon dioxide is CO2. One molecule of CO2 contains one atom of carbon and two atoms of oxygen One mole of CO2 contains one mole of carbon atoms and two moles of oxygen atoms.
## How many moles of CO2 are produced per mole of O2?
This means that O2 is consumed and CO2 is produced in a 1:1 ratio. If we consume 5mol O2, then we produce 5mol CO2.
## How do you find moles of CO2 produced?
Calculate the number of moles of CO2 by the formula n=PV/RT, where P is the pressure from Step 3, V is the volume from Step 2, T is the temperature from Step 1 and R is a proportionality constant equal to 0.0821 L atm / K mol. All the units except for moles will cancel out in the end.
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## How do you find moles of oxygen in CO2?
This means that 1 mole of carbon dioxide has a mass of 44.01 g . Oxygen has a molar mass of 16.0 g mol−1 , so 1 mole of oxygen atoms has a mass of 16.0 g . Therefore, carbon dioxide has a percent composition of 72.7% oxygen, i.e. for every 100 g of carbon dioxide you get 72.7 g of oxygen.
## How many moles of CO2 are formed?
You can see it is 2 moles C2H2 produces 4 moles CO2.
## How many moles are in 50 grams of CO2?
Explanation: The answer is 44.0095. We assume you are converting between grams CO2 and mole.
## How many moles of oxygen are in 3 moles of CO2?
There are 3.6132 x 1024 oxygen atoms in exactly three moles of carbon dioxide.
## How many moles are in 25 grams of CO2?
Answer: The answer is 44.0095. We assume you are converting between grams CO2 and mole. You can view more details on each measurement unit: molecular weight of CO2 or mol This compound is also known as Carbon Dioxide.
## What is the mole of CO2?
Molar Masses of Compounds
The molecular mass of carbon dioxide is 44.01amu. The molar mass of any compound is the mass in grams of one mole of that compound. One mole of carbon dioxide molecules has a mass of 44.01g, while one mole of sodium sulfide formula units has a mass of 78.04g.
## How many moles are in 28 grams of CO2?
Then 28 g of CO2 contains = 0.636 moles of CO2.
## How many moles of oxygen are present in CO2?
A mole of CO2 molecules (we usually just say “a mole of CO2”) has one mole of carbon atoms and two moles of oxygen atoms.
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## How many grams are in 1 mole of oxygen?
Moles of a Substance and the Molecular Weight
The mass of oxygen equal to one mole of oxygen is 15.998 grams and the mass of one mole of hydrogen is 1.008 g.
## What is the mass of 5 moles of CO2?
Hence, mass of CO2 is 22g.
## How many moles of CO2 are in 88 grams?
Therefore 44g is one mole of which means that 88g will be two moles of .
## How many grams are in CO2?
You can view more details on each measurement unit: molecular weight of CO2 or grams This compound is also known as Carbon Dioxide. The SI base unit for amount of substance is the mole. 1 mole is equal to 1 moles CO2, or 44.0095 grams.
## How do I calculate moles?
1. First you must calculate the number of moles in this solution, by rearranging the equation. No. Moles (mol) = Molarity (M) x Volume (L) = 0.5 x 2. = 1 mol.
2. For NaCl, the molar mass is 58.44 g/mol. Now we can use the rearranged equation. Mass (g) = No. Moles (mol) x Molar Mass (g/mol) = 1 x 58.44. = 58.44 g. | 991 | 3,457 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2022-21 | latest | en | 0.927746 |
https://studysoup.com/tsg/538071/elementary-statistics-a-step-by-step-approach-7-edition-chapter-6-4-problem-12 | 1,604,011,996,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107905965.68/warc/CC-MAIN-20201029214439-20201030004439-00494.warc.gz | 522,462,997 | 11,378 | ×
×
Telephone Answering Devices Seventy-eight percent of U.S. homes have a telephone
ISBN: 9780073534978 253
Solution for problem 12 Chapter 6-4
Elementary Statistics: A Step by Step Approach | 7th Edition
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Problem 12
Telephone Answering Devices Seventy-eight percent of U.S. homes have a telephone answering device. In a random sample of 250 homes, what is the probability that fewer than 50 do not have a telephone answering device?
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Sergio Gonzalez Week 3 STATS 241 Chapter 3 Pt 2 Coefficient of variationmeasures the spread of data relative to the mean of the data used to compare variables with different mean,SD’s, and units Zscoremeasures the number of standard deviations a given data value is away from mean of data set and can be negative Ex: Mean1602 SD5000 value1810 18101602/500=...
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Unlock Textbook Solution | 301 | 1,151 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2020-45 | latest | en | 0.817751 |
https://www.wikiwand.com/zh-sg/%E4%B8%89%E5%8F%B6%E7%BB%93 | 1,627,115,745,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046150134.86/warc/CC-MAIN-20210724063259-20210724093259-00084.warc.gz | 1,112,816,951 | 31,824 | # 三叶结
## 描述
${\displaystyle x=\sin t+2\sin 2t}$
${\displaystyle \qquad y=\cos t-2\cos 2t}$
${\displaystyle \qquad z=-\sin 3t}$
${\displaystyle x=(2+\cos 3t)\cos 2t}$
${\displaystyle \qquad y=(2+\cos 3t)\sin 2t}$
${\displaystyle \qquad z=\sin 3t}$
## 性质
${\displaystyle \Delta (t)=t-1+t^{-1))$
${\displaystyle \nabla (z)=z^{2}+1}$
${\displaystyle V(q)=q^{-1}+q^{-3}-q^{-4))$
${\displaystyle L(a,z)=za^{5}+z^{2}a^{4}-a^{4}+za^{3}+z^{2}a^{2}-2a^{2))$
${\displaystyle P(a,z)=-a^{4}+a^{2}z^{2}+2a^{2))$
${\displaystyle \langle x,y\mid x^{2}=y^{3}\rangle }$
${\displaystyle \langle \sigma _{1},\sigma _{2}\mid \sigma _{1}\sigma _{2}\sigma _{1}=\sigma _{2}\sigma _{1}\sigma _{2}\rangle }$
## 参考文献
1. ^ 3_1页面存档备份，存于互联网档案馆）, The Knot Atlas
2. ^ Accessed: May 5, 2013.
3. ^ Recognition & Success. M.C. Escher – The Official Website. [2020-02-22]. （原始内容存档于2020-02-22）. | 395 | 870 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 16, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2021-31 | latest | en | 0.223119 |
https://ebookee.com/Pure-Mathematics-for-Beginners-A-Rigorous-Introduction-to-Logic-Set-Theory-Abstract-Algebra-Number-Theory-Real-Analysis-Topology-Complex-Analysis-and-Linear-Algebra_4526115.html | 1,620,721,705,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991904.6/warc/CC-MAIN-20210511060441-20210511090441-00620.warc.gz | 246,561,415 | 7,047 | # Pure Mathematics for Beginners: A Rigorous Introduction to Logic, Set Theory, Abstract Algebra, Number Theory, Real Analysis, Topology, Complex Analysis, and Linear Algebra
#### Tag: Mathematics
Posted on 2020-11-16, by temrick.
Description
Pure Mathematics for Beginners consists of a series of lessons in Logic, Set Theory, Abstract Algebra, Number Theory, Real Analysis, Topology, Complex Analysis, and Linear Algebra. The 16 lessons in this book cover basic through intermediate material from each of these 8 topics. In addition, all the proofwriting skills that are essential for advanced study in mathematics are covered and reviewed extensively. Pure Mathematics for Beginners is perfect for
• professors teaching an introductory college course in higher mathematics
• high school teachers working with advanced math students
• students wishing to see the type of mathematics they would be exposed to as a math major.
The material in this pure math book includes:
• 16 lessons in 8 subject areas.
• A problem set after each lesson arranged by difficulty level.
• A complete solution guide is included as a downloadable PDF file.
Lesson 1 - Logic: Statements and Truth
Lesson 2 - Set Theory: Sets and Subsets
Lesson 3 - Abstract Algebra: Semigroups, Monoids, and Groups
Lesson 4 - Number Theory: Ring of Integers
Lesson 5 - Real Analysis: The Complete Ordered Field of Reals
Lesson 6 - Topology: The Topology of R
Lesson 7 - Complex Analysis: The Field of Complex Numbers
Lesson 8 - Linear Algebra: Vector Spaces
Lesson 9 - Logic: Logical Arguments
Lesson 10 - Set Theory: Relations and Functions
Lesson 11 - Abstract Algebra: Structures and Homomorphisms
Lesson 12 - Number Theory: Primes, GCD, and LCM
Lesson 13 - Real Analysis: Limits and Continuity
Lesson 14 - Topology: Spaces and Homeomorphisms
Lesson 15 - Complex Analysis: Complex Valued Functions
Lesson 16 - Linear Algebra: Linear Transformations
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Pure Mathematics for Beginners: A Rigorous Introduction to Logic, Set Theory, Abstract Algebra, Number Theory, Real Analysis, Topology, Complex Analysis, and Linear Algebra
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By: a guest on Aug 7th, 2012 | syntax: Java | size: 13.21 KB | hits: 25 | expires: Never
Text below is selected. Please press Ctrl+C to copy to your clipboard. (⌘+C on Mac)
1. /*
2. File: Geometry.java
3.
4. Description:
5.
6. Student Name:
7.
8. Student UT EID:
9.
10. Partner's Name:
11.
12. Partner's UT EID:
13.
14. Course Name: CS 312
15.
16. Unique Numbers: 91035
17.
18. Date Created: 08/06/2012
19.
21.
22. */
23.
24. import java.util.*;
25. import java.io.*;
26.
27. class Point
28. {
29. // list of attributes - x and y coordinates
30. private double x;
31. private double y;
32.
33. // default constructor
34. public Point ()
35. {
36. x = 0.0;
37. y = 0.0;
38. }
39.
40. // non-default constructors
41. public Point (double x, double y)
42. {
43. this.x = x;
44. this.y = y;
45. }
46.
47. public Point (Point p)
48. {
49. x = p.getX();
50. y = p.getY();
51. }
52.
53. // accessors get the x and y coordinates
54. public double getX ()
55. {
56. return x;
57. }
58. public double getY ()
59. {
60. return y;
61. }
62.
63. // mutators set the x and y coordinates
64. public void setX (double x)
65. {
66. this.x = x;
67. }
68. public void setY (double y)
69. {
70. this.y = y;
71. }
72.
73. // test for equality of two doubles
74. public boolean isEqual (double x, double y)
75. {
76. double delta = 1.0e-13; // an arbitrary small number
77. return (((Math.abs (x - y)) < delta));
78. }
79.
80. // distance to another point
81. public double distance (Point p)
82. {
83. return Math.sqrt ((x - p.x) * (x - p.x) + (y - p.y) * (y - p.y));
84. }
85.
86. // string representation of a point, i.e. x and y coordinates
87. public String toString ()
88. {
89. return "(" + x + ", " + y + ")";
90. }
91.
92. // test for equality of two points
93. public boolean equals (Point p)
94. {
95. return (isEqual(x, p.x) && isEqual(y, p.y));
96. }
97. }
98.
99. class Line
100. {
101. // list of attributes
102. private double slope;
103. private double intercept;
104.
105. // default constructor (slope = 1 and intercept = 0)
106. public Line ()
107. {
108. slope = 1.0;
109. intercept = 0.0;
110. }
111.
112. // non-default constructors
113. public Line (double slope, double intercept)
114. {
115. this.slope = slope;
116. this.intercept = intercept;
117. }
118.
119. // define line if p and q are not the same
120. public Line (Point p, Point q)
121. {
122. //slope = (y1-y2)/(x1-x2)
123. //intercept = y1-slope*x
124. this.slope = (p.getY() - q.getY())/(p.getX() - q.getX());
125. this.intercept = p.getY() - slope * p.getX();
126. }
127.
128. // accessors
129. public double getSlope ()
130. {
131. return slope;
132. }
133.
134. public double getIntercept ()
135. {
136. return intercept;
137. }
138.
139. // mutators
140. public void setSlope (double slope)
141. {
142. this.slope = slope;
143. }
144.
145. public void setIntercept (double intercept)
146. {
147. this.intercept = intercept;
148. }
149.
150. // test for equality of two doubles
151. public boolean isEqual (double x, double y)
152. {
153. double delta = 1.0e-13; // an arbitrary small number
154. return (((Math.abs (x - y)) < delta));
155. }
156.
157. // determine if two lines are parallel
158. // same slopes <=> parallel
159. public boolean isParallel (Line line)
160. {
161. return isEqual(this.slope, line.slope);
162. }
163.
164. // determine the intersection point if two lines are not parallel
165. public Point intersectionPoint (Line line)
166. {
167. Point intersection = new Point();
168. double m1 = this.getSlope();
169. double m2 = line.getSlope();
170. double b1 = this.getIntercept();
171. double b2 = line.getIntercept();
172. intersection.setX((b1-b2)/(m2-m1));
173. intersection.setY(m1*intersection.getX() + b1);
174. return intersection;
175.
176. }
177.
178. // determine if two lines are perpendicular to each other
179. public boolean isPerpendicular (Line line)
180. {
181. //m1 = -(1/m2)
182. return isEqual(this.getSlope(),-1 * line.getSlope());
183. }
184.
185. // determine the perpendicular distance of a point to the line
186. public double distance (Point p)
187. {
188. double m = this.getSlope();
189. double b = this.getIntercept();
190. double x = p.getX();
191. double y = p.getY();
192. return Math.abs(m*x - y + b)/ Math.sqrt(m*m + 1);
193.
194. }
195.
196. // determine if two points are on the same side of the line
197. // if one or both points are on the line return false
198. public boolean onSameSide (Point p, Point q)
199. {
200. //side1 if y > mx + b; side2 if y < mx + b
201. double x1 = p.getX();
202. double y1 = p.getY();
203. double x2 = q.getX();
204. double y2 = q.getY();
205. double m = this.getSlope();
206. double b = this.getIntercept();
207. if (y1 == m*x1 + b || y2 == m*x2 + b)
208. return false;
209. else
210. return (y1 > m*x1 + b) == (y2 > m*x2 + b);
211. }
212.
213. // string representation of the slope and intercept of a line, e.g.
214. // slope: 1.0 intercept: 0.0
215. public String toString ()
216. {
217. return ("slope: " + slope + " intercept: " + intercept);
218. }
219.
220. // determine if two lines are equal, i.e. have the same slope
221. // and intercept
222. public boolean equals (Line line)
223. {
224. return(isEqual(this.getSlope(), line.getSlope()) &&
225. isEqual(this.getIntercept(), line.getIntercept()));
226. }
227. }
228.
229. class Triangle
230. {
231. // list attributes
232. private Point v1;
233. private Point v2;
234. private Point v3;
235.
236. // default constructor creates a triangle having
237. // vertices (0, 0), (1, 0), and (0, 1).
238. public Triangle ()
239. {
240. v1 = new Point(0,0);
241. v2 = new Point(1,0);
242. v3 = new Point(0,1);
243. }
244.
245. // non-default constructors accept user defined points
246. // and creates triangle object if the points form a
247. // triangle or the default triangle if they do not.
248. public Triangle (Point v1, Point v2, Point v3)
249. {
250. if (isTriangle(v1,v2,v3))
251. {
252. this.v1 = new Point(v1);
253. this.v2 = new Point(v2);
254. this.v3 = new Point(v3);
255. }
256. }
257.
258. public Triangle (double x1, double y1,
259. double x2, double y2,
260. double x3, double y3)
261. {
262. if (isTriangle(x1, y1, x2, y2, x3, y3))
263. {
264. this.v1 = new Point(x1, y1);
265. this.v2 = new Point(x2, y2);
266. this.v3 = new Point(x3, y3);
267. }
268. }
269.
270. // accessors
271. public Point getVertex1 ()
272. {
273. return v1;
274. }
275. public Point getVertex2 ()
276. {
277. return v2;
278. }
279. public Point getVertex3 ()
280. {
281. return v3;
282. }
283.
284. // mutators reset the vertices only if the triangle shape
285. // is preserved i.e. the points do not collapse to a line
286. public void setVertex1 (Point v1)
287. {
288. if(isTriangle(v1, v2, v3))
289. {
290. this.v1 = v1;
291. }
292. }
293. public void setVertex2 (Point v2)
294. {
295. if(isTriangle(v1, v2, v3))
296. {
297. this.v2 = v2;
298. }
299. }
300. public void setVertex3 (Point v3)
301. {
302. if(isTriangle(v1, v2, v3))
303. {
304. this.v3 = v3;
305. }
306. }
307.
308. public void setVertex1 (double x1, double y1)
309. {
310. if (isTriangle(x1, y1, v2.getX(), v2.getY(), v3.getX(), v3.getY()))
311. {
312. this.v1 = new Point(x1, y1);
313. }
314. }
315.
316. public void setVertex2 (double x2, double y2)
317. {
318. if (isTriangle(v1.getX(), v1.getY(), x2, y2, v3.getX(), v3.getY()))
319. {
320. this.v2 = new Point(x2, y2);
321. }
322. }
323.
324. public void setVertex3 (double x3, double y3)
325. {
326. if (isTriangle(v1.getX(), v1.getY(), v2.getX(), v2.getY(), x3, y3))
327. {
328. this.v3 = new Point(x3, y3);
329. }
330. }
331.
332. // test for equality of two doubles
333. public boolean isEqual (double x, double y)
334. {
335. double delta = 1.0e-13; // an arbitrary small number
336. return (((Math.abs (x - y)) < delta));
337. }
338.
339. // determines if three points form a triangle
340. private boolean isTriangle (Point p1, Point p2, Point p3)
341. {
342. double x1 = p1.getX();
343. double y1 = p1.getY();
344. double x2 = p2.getX();
345. double y2 = p2.getY();
346. double x3 = p3.getX();
347. double y3 = p3.getY();
348. return isEqual((y2 - y1)/(x2-x1), (y3-y2)/(x3-x2));
349. }
350. private boolean isTriangle (double x1, double y1,
351. double x2, double y2,
352. double x3, double y3)
353. {
354. return isEqual((y2 - y1)/(x2-x1), (y3-y2)/(x3-x2));
355. }
356.
357. // calculates area of a triangle
358. public double area ()
359. {
360. double base = getVertex1().distance(getVertex2());
361. Line baseLine = new Line(v1,v2);
362. double height = baseLine.distance(v3);
363. return 1/2 * base * height;
364. }
365.
366. // calculates the perimeter
367. public double perimeter ()
368. {
369. return v1.distance(v2) + v2.distance(v3) + v3.distance(v1);
370. }
371.
372. // determines if a point is inside the triangle
373. public boolean isInside (Point p)
374. {
375. Line lineV1V2 = new Line (v1, v2);
376. Line lineV2V3 = new Line (v2, v3);
377. Line lineV1V3 = new Line (v1, v3);
378. return lineV1V2.onSameSide(p, v3) && lineV2V3.onSameSide(p,v1)
379. && lineV1V3.onSameSide(p,v2);
380. }
381.
382. // determines if the given triangle is completely inside Triangle t
383. public boolean isInside (Triangle t)
384. {
385. return isInside(t.getVertex1()) && isInside(t.getVertex2())
386. && isInside(t.getVertex3());
387. }
388.
389. // determines if the given triangle overlaps Triangle t,
390. // if it shares some (or all) of its area with Triangle t
391. public boolean doesOverlap (Triangle t)
392. {
393. return isInside(t.getVertex1()) || isInside(t.getVertex2())
394. || isInside(t.getVertex3()) || //starOfDavid
395. }
396.
397. // determines if a line passes through the triangle
398. public boolean doesIntersect (Line line)
399. {
400. Line lineV1V2 = new Line (v1, v2); //Fix this whole thing
401. Line lineV2V3 = new Line (v2, v3);
402. Line lineV1V3 = new Line (v1, v3);
403. Point pV1V2 = line.intersectionPoint(lineV1V2);
404. Point pV2V3 = line.intersectionPoint(lineV2V3);
405. Point pV1V3 = line.intersectionPoint(lineV1V3);
406. double xV1V2 = pV1V2.getX();
407. double yV1V2 = pV1V2.getY();
408. double xV1V3 = pV1V3.getX();
409. double yV1V3 = pV1V3.getY();
410. double xV2V3 = pV2V3.getX();
411. double yV2V3 = pV2V3.getY();
412. return ((xV1V2 < v1.x) == (xV1V2 > v2.x)) && (xV1V2 < v1.y) == (pV1V2.y > v2.y)) ||
413. ((pV1V3.x < v1.x) == (pV1V3.x > v3.x)) && ((pV1V3.y < v1.y) == (pV1V3.y > v3.y)) ||
414. ((pV1V2.x < v1.x) == (pV1V2.x > v2.x)) && ((pV1V2.x < v1.x) == (pV1V2.x > v2.x))
415. }
416.
417. // returns a string representation of a triangle
418. // i.e. it gives the three vertices
419. public String toString ()
420. {
421. return v1.toString() + ", " + v2.toString() + ", " + v3.toString();
422. }
423.
424.
425. // determines if two triangles are congruent, i.e. the
426. // three sides of one are equal to three sides of the other
427. public boolean equals (Triangle t)
428. {
429.
430. }
431.
432. }
433.
434. public class Geometry
435. {
436.
437. public static void main (String[] args) throws IOException
438. {
439. // open file "geometry.txt" for reading
440. File input = new File ("geometry.txt");
441. Scanner sc = new Scanner(input);
442.
443.
444. // read the coordinates of the first Point P
445. String [] strArrayP = (sc.nextLine()).split(" ");
446. Point p = new Point ( Double.parseDouble(strArrayP[0]),
447. Double.parseDouble(strArrayP [1]));
448. System.out.println("Coordinates of P: " + p.toString());
449.
450. // read the coordinates of the second Point Q
451. String [] strArrayQ = (sc.nextLine()).split(" ");
452. Point q = new Point ( Double.parseDouble(strArrayQ[0]),
453. Double.parseDouble(strArrayQ [1]));
454. System.out.println("Coordinates of Q: " + q.toString());
455.
456. // print distance between P and Q
457. System.out.println("Distance between P and Q: " + p.distance(q));
458.
459. // print the slope and intercept of the line passing through P and Q
460. Line linePQ = new Line(p,q);
461. System.out.println("Slope and Intercept of PQ: " + linePQ.toString());
462.
463. // read the coordinates of the third Point A
464. String [] strArrayA = (sc.nextLine()).split(" ");
465. Point a = new Point ( Double.parseDouble(strArrayA[0]),
466. Double.parseDouble(strArrayA [1]));
467. System.out.println("Coordinates of A: " + a.toString());
468.
469. // read the coordinates of the fourth Point B
470. String [] strArrayB = (sc.nextLine()).split(" ");
471. Point b = new Point ( Double.parseDouble(strArrayB[0]),
472. Double.parseDouble(strArrayB [1]));
473. System.out.println("Coordinates of B: " + b.toString());
474.
475. // print the slope and intercept of the line passing through A and B
476. Line lineAB = new Line(a,b);
477. System.out.println("Slope and Intercept of AB: " + lineAB.toString());
478.
479. // print if the lines PQ and AB are parallel or not
480. System.out.println("PQ is " + (linePQ.isParallel(lineAB)? "":"not ")
481. + "parallel to AB.");
482.
483. // print if the lines PQ and AB are perpendicular or not
484. System.out.println("PQ is " + (linePQ.isPerpendicular(lineAB)? "":"not ")
485. + "perpendicular to AB.");
486.
487. // print the coordinates of the intersection point if PQ is not
488. // parallel to AB
489. if (! lineAB.isParallel(linePQ))
490. {
491. System.out.println("Coordinates of intersection point of PQ and AB: "
492. + lineAB.intersectionPoint(linePQ).toString());
493. }
494.
495. // read the coordinates of the fifth Point G
496.
497. // read the coordinates of the sixth Point H
498.
499. // print if the the points G and H are on the same side of PQ
500.
501. // print if the the points G and H are on the same side of AB
502.
503. // read the coordinates of the vertices R, S, and T
504.
505. // print the perimeter of triangle RST
506.
507. // print the area of triangle RST
508.
509. // print if the line PQ passes through the triangle RST
510.
511. // print if the line AB passes through the triangle RST
512.
513. // read the coordinates of the vertices J, K, and L
514.
515. // print if triangle JKL is inside triangle RST
516.
517. // print if triangle JKL overlaps triangle RST
518.
519. // print if triangle JKL is congruent to triangle RST
520.
521. // close file "geometry.txt"
522.
523. }
524.
525. } | 4,950 | 15,554 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2013-20 | latest | en | 0.207947 |
https://blog.francis67.cc/archives/tag/python3/page/2 | 1,590,921,644,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347413097.49/warc/CC-MAIN-20200531085047-20200531115047-00499.warc.gz | 258,995,141 | 10,334 | # 自测-5 Shuffling Machine (PTA)
Shuffling is a procedure used to randomize a deck of playing cards. Because standard shuffling techniques are seen as weak, and in order to avoid “inside jobs” where employees collaborate with gamblers by performing inadequate shuffles, many casinos employ automatic shuffling machines. Your task is to simulate a shuffling machine.
The machine shuffles a deck of 54 cards according to a given random order and repeats for a given number of times. It is assumed that the initial status of a card deck is in the following order:
```S1, S2, ..., S13,
H1, H2, ..., H13,
C1, C2, ..., C13,
D1, D2, ..., D13,
J1, J2
```
where “S” stands for “Spade”, “H” for “Heart”, “C” for “Club”, “D” for “Diamond”, and “J” for “Joker”. A given order is a permutation of distinct integers in [1, 54]. If the number at the i-th position is j, it means to move the card from position i to position j. For example, suppose we only have 5 cards: S3, H5, C1, D13 and J2. Given a shuffling order {4, 2, 5, 3, 1}, the result will be: J2, H5, D13, S3, C1. If we are to repeat the shuffling again, the result will be: C1, H5, S3, J2, D13.
### Input Specification:
Each input file contains one test case. For each case, the first line contains a positive integer K (≤20) which is the number of repeat times. Then the next line contains the given order. All the numbers in a line are separated by a space.
### Output Specification:
For each test case, print the shuffling results in one line. All the cards are separated by a space, and there must be no extra space at the end of the line.
### Sample Input:
```2
36 52 37 38 3 39 40 53 54 41 11 12 13 42 43 44 2 4 23 24 25 26 27 6 7 8 48 49 50 51 9 10 14 15 16 5 17 18 19 1 20 21 22 28 29 30 31 32 33 34 35 45 46 47
```
### Sample Output:
`S7 C11 C10 C12 S1 H7 H8 H9 D8 D9 S11 S12 S13 D10 D11 D12 S3 S4 S6 S10 H1 H2 C13 D2 D3 D4 H6 H3 D13 J1 J2 C1 C2 C3 C4 D1 S5 H5 H11 H12 C6 C7 C8 C9 S2 S8 S9 H10 D5 D6 D7 H4 H13 C5`
```spade = ['S' + str(i) for i in range(1, 14, 1)]
heart = ['H' + str(i) for i in range(1, 14, 1)]
club = ['C' + str(i) for i in range(1, 14, 1)]
diam = ['D' + str(i) for i in range(1, 14, 1)]
joker = ['J1', 'J2']
cards = spade + heart + club + diam + joker
result = [None]*len(cards)
K = int(input())
order = list(map(int, input().split()))
for i in range(len(cards)):
index = i
for j in range(K):
index = order[index] - 1
result[index] = cards[i]
print(' '.join(result)) ```
# 1023/自测-4 Have Fun with Numbers (PTA)
Notice that the number 123456789 is a 9-digit number consisting exactly the numbers from 1 to 9, with no duplication. Double it we will obtain 246913578, which happens to be another 9-digit number consisting exactly the numbers from 1 to 9, only in a different permutation. Check to see the result if we double it again!
Now you are suppose to check if there are more numbers with this property. That is, double a given number with k digits, you are to tell if the resulting number consists of only a permutation of the digits in the original number.
### Input Specification:
Each input contains one test case. Each case contains one positive integer with no more than 20 digits.
### Output Specification:
For each test case, first print in a line “Yes” if doubling the input number gives a number that consists of only a permutation of the digits in the original number, or “No” if not. Then in the next line, print the doubled number.
### Sample Input:
```1234567899
```
### Sample Output:
```Yes
2469135798```
```N = int(input())
doubleN = N*2
str_N = str(N)
str_doubleN = str(doubleN)
Set_N = set(str_N)
Set_doubleN = set(str_doubleN)
if Set_N == Set_doubleN:
print("Yes")
else:
print("No")
print(doubleN) ```
# 自测-3 数组元素循环右移问题 (PTA)
```6 2
1 2 3 4 5 6
```
### 输出样例:
`5 6 1 2 3 4`
```N, M = input().split(' ')
string = input().split(' ')
N = int(N)
M = int(M)
result = string[N-M:] + string[:N-M]
print(' '.join(result)) ```
# 自测-2 素数对猜想 (PTA)
```20
```
### 输出样例:
`4`
``` def eratosthenes(n):
IsPrime = [True] * (n + 1)
for i in range(2, int(n ** 0.5) + 1):
if IsPrime[i]:
for j in range(i * i, n + 1, i):
IsPrime[j] = False
return [x for x in range(2, n + 1) if IsPrime[x]]
n=int(input())
result = eratosthenes(n)
#print(result)
count = 0;
for i in range(len(result) - 1):
if result[i+1]-result[i]== 2:
count +=1;
print(count) ```
Python的问题还是消耗的资源多，用普通的算法最大N总是超时，这里使用了Sieve of Eratosthenes
# 自测-1 打印沙漏 (PTA)
```*****
***
*
***
*****
```
```19 *
```
### 输出样例:
```*****
***
*
***
*****
2```
```import numpy as np
num, ch =input().split(' ')
num = int(num)
N = np.sqrt((num+1)/2);
N = int(N)
residue = num - (2*np.square(N) - 1)
layers = 2*N - 1
for i in range(layers):
seq = abs(i - (N - 1)) +1
rep = 2*seq-1
str = rep*ch
zeros = (layers + rep)/2
zeros = int(zeros)
#str = str.center(layers,' ')
str = str.rjust(zeros)
print(str,end='\n')
print(residue) ```
# 01-复杂度2 Maximum Subsequence Sum (PTA)
Given a sequence of K integers { N1, N2, …, NK }. A continuous subsequence is defined to be { Ni, Ni+1, …, Nj } where 1≤ijK. The Maximum Subsequence is the continuous subsequence which has the largest sum of its elements. For example, given sequence { -2, 11, -4, 13, -5, -2 }, its maximum subsequence is { 11, -4, 13 } with the largest sum being 20.
Now you are supposed to find the largest sum, together with the first and the last numbers of the maximum subsequence.
### Input Specification:
Each input file contains one test case. Each case occupies two lines. The first line contains a positive integer K (≤10000). The second line contains K numbers, separated by a space.
### Output Specification:
For each test case, output in one line the largest sum, together with the first and the last numbers of the maximum subsequence. The numbers must be separated by one space, but there must be no extra space at the end of a line. In case that the maximum subsequence is not unique, output the one with the smallest indices i and j (as shown by the sample case). If all the K numbers are negative, then its maximum sum is defined to be 0, and you are supposed to output the first and the last numbers of the whole sequence.
### Sample Input:
```10
-10 1 2 3 4 -5 -23 3 7 -21
```
### Sample Output:
`10 1 4`
```def MaxSubseqSum4(A,N):
global ThisSum, MaxSum,start,end,temp
for i in range(N):
ThisSum += int(A[i])
if ThisSum > MaxSum:
MaxSum = ThisSum
start = temp
end = i
elif ThisSum < 0:
ThisSum = 0
temp = i+1
if MaxSum >= 0 :
print(MaxSum,A[start],A[end])
else:
print(0,A[0],A[N-1])
ThisSum = 0
MaxSum = -1
start = 0
end = 0
temp = 0
N = eval(input())
A = input().split(' ')
MaxSubseqSum4(A,N) ```
# 01-复杂度1 最大子列和问题 (PTA)
• 数据1：与样例等价，测试基本正确性；
• 数据2：102个随机整数；
• 数据3：103个随机整数；
• 数据4：104个随机整数；
• 数据5：105个随机整数；
### 输入样例:
```6
-2 11 -4 13 -5 -2
```
### 输出样例:
`20`
```def MaxSubseqSum4(A,N):
global ThisSum, MaxSum
for i in range(N):
ThisSum += int(A[i])
if ThisSum > MaxSum:
MaxSum = ThisSum
elif ThisSum < 0:
ThisSum = 0
print(MaxSum)
ThisSum = 0
MaxSum = 0
N = eval(input())
A = input().split(' ')
MaxSubseqSum4(A,N) ``` | 2,383 | 7,105 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2020-24 | latest | en | 0.76399 |
https://proacademiahelp.com/whieeee/ | 1,695,634,900,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233508959.20/warc/CC-MAIN-20230925083430-20230925113430-00324.warc.gz | 532,154,708 | 15,500 | Whieeee
pace university|economics 310|yarbrough spring 2019 – assignment 1
1 Negative Externality
Suppose that the private market for widgets is characterized by the following supply and inverse
demand functions:
D : P = 10−Q
S : P = Q
1. Graph these functions in the graph paper below. Locate the private market equilibrium price
and quantity.
1
tani navani
pace university|economics 310|yarbrough spring 2019 – assignment 1
2. Now suppose that the EPA calculates that the marginal external cost of widget production is
characterized by: MD = \$2. Graph the market with the externality and locate the the socially
e�cient equilibrium. How much dead weight loss was produced by the private market.
2
tani navani
pace university|economics 310|yarbrough spring 2019 – assignment 1
3. Now suppose that the EPA revises their MEC estimate to: MD = Q. Graph the market with
the externality and locate the socially e�cient equilibrium. Compare this with the outcome
from part 2.
3
tani navani
pace university|economics 310|yarbrough spring 2019 – assignment 1
2 Public Goods
Suppose that the private market demand for public-access park acres in the town of Yarbroughville
(Q) is characterized by the following two types of voters (old and young):
Dold : Q = 10−P
Dyoung : Q = 8−P
Further, assume that the marginal cost of supplying acres of public-access park is:
MC : Q = 0.25P
1. If park acres would be determined by a private market, how many acres are bought and sold?
To determine this, plot the two demand curves separately, and then sum them horizontally to
plot a market demand curve. Then �nd the private market equilibrium.
4
tani navani
pace university|economics 310|yarbrough spring 2019 – assignment 1
2. Instead, assume that park acres are determined by a public vote based on willingness to pay.
How many acres of public-access park are socially e�cient? To determine this, again graph the
two demand curves separately and then sum them vertically to plot a public demand curve.
Now locate the socially e�cient equilibrium.
3. Brie�y explain why it is that the private market supplies less public-access park acres than
would be supplied if they were o�ered publicly.
5
tani navani
pace university|economics 310|yarbrough spring 2019 – assignment 1
3 Cost-Bene�t Analysis
Pace University is deciding between two projects.
A) Full HVAC system overhaul, which costs \$1 million and will reduce general energy cost to the
university of \$300,000 per year. There are no costs beyond period 0.
B) Placing a wind farm on top of 1 Place Plaza, which costs \$10 million and will reduce general
energy cost to the university of \$3 million per year. There are \$500 thousand per year of upkeep
cost for the windmills.
1. Assuming a discount rate of 5% and time periods as years, what is the NPV of each project
after 5 years.
2. Answer part 1 again, but assume a discount rate of 10% instead.
3. Comment on the values calculated in parts 1 and 2.
6
tani navani
Price (USD)
\$ | 751 | 3,034 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2023-40 | latest | en | 0.915108 |
http://stackoverflow.com/questions/16970790/prolog-inserting-multiple-elements-into-list/16971062 | 1,416,538,217,000,000,000 | text/html | crawl-data/CC-MAIN-2014-49/segments/1416400372542.20/warc/CC-MAIN-20141119123252-00152-ip-10-235-23-156.ec2.internal.warc.gz | 382,812,131 | 17,183 | I want to implement a predicate (vecLine2BitLine) which does the following: get two lists and a number the first list is the length of blocks (the elements of the blocks are '\$') and the second list contains the indexes that these blocks should be placed at meaning:
``````vecLine2BitLine([1,2,1],[2,5,9],12,BitLine).
BitLine=[' ','\$',' ',' ','\$','\$',' ',' ','\$',' ',' ',' '].
``````
explanation:a block of length 1 is at index 2 and a block of length 2 is at index 5 and so on..
insert_at_mul : inserts an element N times (it works perfectly,dupli and my_flatten were implemented previously so i used them)
Ive been trying to activate insert_at_mul N times when N is the length of the list X and Y in the predicate vecLine2BitLine.
``````dupli(L1,N,L2) :- dupli(L1,N,L2,N).
dupli([],_,[],_).
dupli([_|Xs],N,Ys,0) :- dupli(Xs,N,Ys,N).
dupli([X|Xs],N,[X|Ys],K) :- K > 0, K1 is K - 1, dupli([X|Xs],N,Ys,K1).
my_flatten(X,[X]) :- \+ is_list(X).
my_flatten([],[]).
my_flatten([X|Xs],Zs) :- my_flatten(X,Y), my_flatten(Xs,Ys), append(Y,Ys,Zs).
insert_at_mul(L,X,K,R,N):-dupli([X],N,XX) , insert_at(L,XX,K,L1) , my_flatten(L1,R).
get_num_spaces(L,N,X):-sum(L,S), X is N-S.
generate_spaces(N,L,X):- insert_at_mul(L,'',1,X,N).
vecLine2BitLineAux([],[],_,_,_).
vecLine2BitLineAux([X|Tail1],[Y|Tail2],N,L,Lnew):- insert_at_mul(L,'*',Y,Lnew,X) ,vecLine2BitLineAux(Tail1,Tail2,N,Lnew,R). // problem here!!!
vecLine2BitLine(X,Y,N,L):- get_num_spaces(X,N,Z) , generate_spaces(Z,[],ZZ) , vecLine2BitLineAux(X,Y,N,ZZ,L).
``````
now the problem is that in the function vecLine2BitLine i cant activate insert_at_mul N times(thats what i tried to do in this code, but failed).
how can I fix vecLine2BitLine for it to work properly as in returning the correct output by actually activating the predicate insert_at_mul N times?? THANKS! added :
vecLine2BitLine : input parameters : (L1,L2,N,Result) N: after activating the predicate Result will be N in length.
L1: L1 is a list of numbers each number indicates the length of a block, a block is comprised of a Sequence of '\$'.
L2: L2 is a list of numbers the numbers are indices for where the blocks in L1 should be placed.
example:
``````vecLine2BitLine([3,2],[1,5],9,BitLine).
``````
we can look at the input better as tuples :
``````vecLine2BitLine[(3,1),(2,5)],9,BitLine).
``````
(3,1) : there is a sequence of '' 3 times at index 1 (2,5) : there is a sequence of '' 2 times at index 5 in our example 9 is the length of BitLine at the end and we have to insert into the list BitLine 3+2 of the "special chars" '*' but we have 9-(3+2) places left in the list so we add `''` in those places and then we get:
``````BitLine=['\$','\$','\$','','\$','\$','','','',''].
``````
-
This is kind of a nice problem because you can use the arguments as loop counters. The `K` argument gets you to the proper index. Let's just traverse the list and find a particular index as an example. Notice the base case is that you're at the right element, and the inductive case is prior to the right element.
``````traverse(1, [X|_], X).
traverse(N, [_|Xs], X) :- N > 0, N0 is N-1, traverse(N0, Xs, X).
``````
We're going to apply that pattern to `insert_at/4` to get to the right location in the list. Now let's write a `repeat/3` predicate that repeats X N times in a new list L. This time the base case is when we've added all the repetitions we care to, and the inductive case is that we'll add another instance.
``````repeat(1, X, [X]).
repeat(N, X, [X|Xs]) :- N > 0, N0 is N-1, repeat(N0, X, Xs).
``````
You can see the similarity of structure between these two. Try to combine them into a single predicate. Since this is homework, I'll stop here. You're inches from the goal.
-
but cant I just use insert_at N times in a way ??? @Daniel – A'mer Mograbi Jun 6 '13 at 20:09
I got it now :) – A'mer Mograbi Jun 6 '13 at 20:15
Post your code. :) – Daniel Lyons Jun 6 '13 at 20:15
sec , i understood what you meant ill need some time to implement it (dont have alot of exp) – A'mer Mograbi Jun 6 '13 at 20:19
i thought that by asking this question i could solve a much bigger one i already implemented insert_at_mul with other predicates that i already defined. how can i show you the code (cant post it here because you wont see anything) email? – A'mer Mograbi Jun 6 '13 at 20:42 | 1,338 | 4,330 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2014-49 | latest | en | 0.666373 |
https://classroom.thenational.academy/lessons/calculating-angles-within-a-shape-3-c5h3jr?activity=video&step=2 | 1,708,872,588,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474617.27/warc/CC-MAIN-20240225135334-20240225165334-00490.warc.gz | 174,210,093 | 26,469 | # Calculating angles within a shape 3
In this lesson, we will learn about parallel lines, their angles and how we can use this information within composite shapes.
This quiz includes images that don't have any alt text - please contact your teacher who should be able to help you with an audio description.
#### Unit quizzes are being retired in August 2023
Why we're removing unit quizzes from the website >
Quiz:
# Intro quiz - Recap from previous lesson
Before we start this lesson, let’s see what you can remember from this topic. Here’s a quick quiz!
Q1.What type of quadrilateral has 4 equal sides and 4 equal angles?
1/5
Q2.What type of quadrilateral has 4 equal sides and 2 pairs of parallel sides?
2/5
Q3.What type of quadrilateral has 2 pairs of equal angles and 2 pairs of equal parallel sides?
3/5
Q4.What do angles inside a quadrilateral add up to?
4/5
Q5.What is the size of the missing angle in this quadrilateral with angles: 90°, 120°, 40°, ?
5/5
This quiz includes images that don't have any alt text - please contact your teacher who should be able to help you with an audio description.
#### Unit quizzes are being retired in August 2023
Why we're removing unit quizzes from the website >
Quiz:
# Intro quiz - Recap from previous lesson
Before we start this lesson, let’s see what you can remember from this topic. Here’s a quick quiz!
Q1.What type of quadrilateral has 4 equal sides and 4 equal angles?
1/5
Q2.What type of quadrilateral has 4 equal sides and 2 pairs of parallel sides?
2/5
Q3.What type of quadrilateral has 2 pairs of equal angles and 2 pairs of equal parallel sides?
3/5
Q4.What do angles inside a quadrilateral add up to?
4/5
Q5.What is the size of the missing angle in this quadrilateral with angles: 90°, 120°, 40°, ?
5/5
# Video
Click on the play button to start the video. If your teacher asks you to pause the video and look at the worksheet you should:
• Click "Close Video"
• Click "Next" to view the activity
Your video will re-appear on the next page, and will stay paused in the right place.
# Worksheet
These slides will take you through some tasks for the lesson. If you need to re-play the video, click the ‘Resume Video’ icon. If you are asked to add answers to the slides, first download or print out the worksheet. Once you have finished all the tasks, click ‘Next’ below.
This quiz includes images that don't have any alt text - please contact your teacher who should be able to help you with an audio description.
#### Unit quizzes are being retired in August 2023
Why we're removing unit quizzes from the website >
Quiz:
# Calculating angles within a shape 3: parallel lines
parallel lines
Q1.How many parallel lines are in a trapezium
1/3
Q2.Is this statement true always, sometimes or never? "A parallelogram has right angles"
2/3
Q3.In this parallelogram (30°, 150°,30°,?°) is this missing angle acute, obtuse reflex or a right angle?
3/3
This quiz includes images that don't have any alt text - please contact your teacher who should be able to help you with an audio description.
#### Unit quizzes are being retired in August 2023
Why we're removing unit quizzes from the website >
Quiz:
# Calculating angles within a shape 3: parallel lines
parallel lines
Q1.How many parallel lines are in a trapezium
1/3
Q2.Is this statement true always, sometimes or never? "A parallelogram has right angles"
2/3
Q3.In this parallelogram (30°, 150°,30°,?°) is this missing angle acute, obtuse reflex or a right angle?
3/3
# Lesson summary: Calculating angles within a shape 3
## It looks like you have not completed one of the quizzes.
To share your results with your teacher please complete one of the quizzes.
## Time to move!
Did you know that exercise helps your concentration and ability to learn?
For 5 mins...
Move around:
Walk
On the spot:
Chair yoga | 956 | 3,876 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2024-10 | latest | en | 0.858882 |
https://www.acmicpc.net/problem/15004 | 1,529,645,225,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864354.27/warc/CC-MAIN-20180622045658-20180622065658-00369.warc.gz | 750,451,039 | 15,955 | 시간 제한 메모리 제한 제출 정답 맞은 사람 정답 비율
2 초 512 MB 13 3 3 25.000%
## 문제
Barney the polar has wandered off on an adventure. Lost in thought, he suddenly realizes he has strayed too far from his mother and is stuck on an ice shelf. He can still see her in the distance, but the only way back is by crossing a group of other ice shelves, all of which are perfectly circular. He is very scared, and can not swim. Barney’s mother, getting a little tired of her son’s shenanigans, decides to wait and let him figure this out for himself. Can you help Barney get home? He is in a hurry.
## 입력
• The first line of input contains four integers, −106 ≤ xb, yb, xm, ym ≤ 106, where (xb, yb) is Barney’s location and (xm, ym) is the location where Barney’s mom is waiting.
• The next line contains a single integer 1 ≤ n ≤ 25, the number of ice shelves.
• After this n lines follow. Each line holds three integers: −106 ≤ xi , yi ≤ 106 and 1 ≤ ri ≤ 106, the coordinates of the center of the shelf and its radius. A shelf consists of all points at distance ri or less to (xi, yi).
Both bears are on a shelf at the start of Barney’s journey home. Shelves can both touch and overlap.
## 출력
The minimal distance Barney has to travel to be reunited with his mother. The result should have a relative error of at most 10−6.
If there is no way for Barney to make it home, output “impossible”. (Do not worry about Barney’s well-being in this scenario. His mother will swim out to save him.)
## 예제 입력 1
0 0 6 0
2
1 1 2
5 1 2
## 예제 출력 1
6.32455532034
## 예제 입력 2
0 0 7 0
2
1 1 2
6 1 2
## 예제 출력 2
impossible
## 예제 입력 3
0 0 1 3
3
0 -1 2
4 -1 3
2 3 2
## 예제 출력 3
4.269334912857045697
## 힌트
Figure 2: Illustration of the third example input. | 539 | 1,732 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2018-26 | longest | en | 0.939932 |
https://encyclopedia2.thefreedictionary.com/Poincare+expansion | 1,643,435,515,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320299927.25/warc/CC-MAIN-20220129032406-20220129062406-00382.warc.gz | 269,558,099 | 10,624 | # asymptotic expansion
(redirected from Poincare expansion)
## asymptotic expansion
[ā‚sim′täd·ik ik′span·shən]
(mathematics)
A series of the form a0+ (a1/ x) + (a2/ x 2) + · · · + (an / xn) + · · · is an asymptotic expansion of the function f (x) if there exists a number N such that for all nN the quantity xn [f (x) -Sn (x)] approaches zero as x approaches infinity, where Sn (x) is the sum of the first n terms in the series. Also known as asymptotic series.
McGraw-Hill Dictionary of Scientific & Technical Terms, 6E, Copyright © 2003 by The McGraw-Hill Companies, Inc.
References in periodicals archive ?
Because the coefficients [a.sub.n] depend of z, (1.8) is not a genuine Poincare expansion, but the terms of the expansion can be grouped to obtain a genuine Poincare expansion [3].
Then, (2.13) is not a genuine Poincare expansion. But we can group the terms of (2.13) in such a way that we get a genuine Poincare expansion,
Nevertheless, the terms of the expansion [[SIGMA].sub.m] [h.sub.m,n-sm](z) [[PHI].sub.m,n-sm](z) can be grouped in new terms [[psi].sub.n](z) and the new asymptotic expansion [[SIGMA].sub.n] [[psi].sub.n](z) is a genuine Poincare expansion, [[psi].sub.n](z) = O([z.sup.-n-p]) as z [right arrow] [infinity].
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Open / Close | 382 | 1,279 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2022-05 | latest | en | 0.82194 |
http://www.mathnasium.com/mathblogs/mathblogs/categories/53763a89-441c-413b-8c9e-2cd245a3dd14/page:3?categories=Holiday | 1,529,762,476,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267865081.23/warc/CC-MAIN-20180623132619-20180623152619-00319.warc.gz | 454,471,317 | 9,012 | # Number Sense Blog
Feb 1, 2016
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# SOLUTIONS/HINTS TO THE EXERCISES FROM COMPLEX ANALYSIS BY
## STEIN AND SHAKARCHI
ROBERT C. RHOADES
Abstract. This contains the solutions or hints to many of the exercises from the Complex Analysis
book by Elias Stein and Rami Shakarchi.
I worked these problems during the Spring of 2006 while I was taking a Complex Analysis
course taught by Andreas Seeger at the University of Wisconsin - Madison. I am grateful to him
for his wonderful lectures and helpful conversations about some of the problems discussed below.
Contents
1. Chapter 1. Preliminaries to Complex Analysis 2
2. Chapter 2. Cauchys Theorem and Its Applications 8
3. Chapter 3. Meromorphic Functions and the Logarithm 9
4. Chapter 4. The Fourier Transform 10
5. Chapter 5: Entire Functions 11
6. Chapter 6. The Gamma and Zeta Functions 13
7. Chapter 7: The Zeta Function and Prime Number Theorem 17
8. Chapter 8: Conformal Mappings 20
9. Chapter 9: An Introduction to Elliptic Functions 23
10. Chapter 10: Applications of Theta Functions 25
Date: September 5, 2006.
The author is thankful for an NSF graduate research fellowship and a National Physical Science Consortium
graduate fellowship supported by the NSA.
1
2 ROBERT C. RHOADES
1. Chapter 1. Preliminaries to Complex Analysis
Exercise 1. Describe geometrically the sets of points z in the complex plane dened by the
following relations:
(1) [z z
1
[ = [z z
2
[ where z
1
, z
2
C.
(2) 1/z = z.
(3) Re(z) = 3.
(4) Re(z) > c, (resp., c) where c R.
(5) Re(az +b) > 0 where a, b C.
(6) [z[ = Re(z) + 1.
(7) Im(z) = c with c R.
Solution 1.
(1) It is the line in the complex plane consisting of all points that are an equal distance from
both z
1
and z
2
. Equivalently the perpendicular bisector of the segment between z
1
and z
2
in the complex plane.
(2) It is the unit circle.
(3) It is the line where all the numbers on the line have real part equal to 3.
(4) In the rst case it is the open half plane with all numbers with real part greater than c. In
the second case it is the closed half plane with the same condition.
(5)
(6) Calculate [z[
2
= x
2
+ y
2
= (x + 1)
2
= x
2
+ 2x + 1. So we are left with y
2
= 2x + 1. Thus
the complex numbers dened by this relation is a parabola opening to the right.
(7) This is a line.
Exercise 2. Let , ) denote the usual inner product in R
2
. In other words, if Z = (z
1
, y
1
) and
W = (x
2
, y
2
), then
Z, W) = x
1
x
2
+y
1
y
2
.
Similarly, we may dene a Hermitian inner product (, ) in C by
(z, w) = zw.
The term Hermitian is used to describe the fact that (, ) is not symmetric, but rather satises the
relation
(z, w) = (w, z) for all z, w C.
Show that
z, w) =
1
2
[(z, w) + (w, z)] = Re(z, w),
where we use the usual identication z = x +iy C with (x, y) R
2
.
Solution 2. This is a straightforward calculation
1
2
[(z, w) + (w, z)] =
1
2
(zw +wz) = Re(z, w)Re(zw)
=
1
2
((z
1
+z
2
i)(w
1
+iw
2
) + (w
1
+iw
2
)(z
1
iz
2
)) = z
1
w
1
+z
2
w
2
.
Exercise 3. With = se
i
, where s 0 and R, solve the equation z
n
= in C where n is a
natural number. How many solutions are there?
SOLUTIONS/HINTS TO THE EXERCISES FROM COMPLEX ANALYSIS BY STEIN AND SHAKARCHI 3
Solution 3.z
n
= se
i
implies that z = s
1
n
e
i(
n
+
2ik
n
)
, where k = 0, 1, , n 1 and s
1
n
is the real
nth root of the positive number s. There are n solutions as there should be since we are nding the
roots of a degree n polynomial in the algebraically closed eld C.
Exercise 4. Show that it is impossible to dene a total ordering on C. In other words, one cannot
nd a relation ~ between complex numbers so that:
(1) For any two complex numbers z, w, one and only one of the following is true: z ~ w, w ~ z
or z = w.
(2) For all z
1
, z
2
, z
3
C the relation z
1
~ z
2
implies z
1
+z
3
~ z
2
+z
3
.
(3) Moreover, for all z
1
, z
2
, z
3
C with z
3
~ 0, then z
1
~ z
2
implies z
1
z
3
~ z
2
z
3
.
Solution 4. Suppose, for a contradiction, that i ~ 0, then 1 = i i ~ 0 i = 0. Now we get
i ~ 1 i ~ 0. Therefore i i ~ i + 0 = i. But this contradicts our assumption. We obtain
a similar situation in the case 0 ~ i. So we must have i = 0. But then for all z C we have
z i = z 0 = 0 Repeating we have z = 0 for all z C. So this relation would give a trivial total
ordering.
Exercise 5. A set is said to be pathwise connected if any two points in can be joined by
a (piecewise-smooth) curve entirely contained in . The purpose of this exercise is to prove that an
open set is pathwise connected if and only if is connected.
(1) Suppose rst that is open and pathwise connected, and that it can be written as =
1
2
where
1
and
2
are disjoint non-empty open sets. Choose two points
1
1
and
2
2
and let denote a curve in joining
1
to
2
. Consider a parametrization
z : [0, 1] of this curve with z(0) =
1
and z(1) =
2
, and let
t
= sup
0t1
t : z(s)
1
for all 0 s < t.
Arrive at a contradiction by considering the point z(t
).
(2) Conversely, suppose that is open and connected. Fix a point w and let
1
denote
the set of all points that can be joined to w by a curve contained in . Also, let
2
denote the set of all points that cannot be joined to w by a curve in . Prove that both
1
and
2
are open, disjoint and their union is . Finally, since
1
is non-empty (why?)
conclude that =
1
as desired.
The proof actually shows that the regularity and type of curves we used to dene pathwise
connectedness can be relaxed without changing the equivalence between the two denitions when
is open. For instance, we may take all curves to be continuous, or simply polygonal lines.
Solution 5. Following the rst part, assume for a contradiction that z(t
)
1
. Since
1
is open
there exists a ball B(z(t
), )
1
. Now by assumption z(t
+)
2
. Thus [z(t
+) z(t
)[ >
for all > 0. But this is a contradiction since z is smooth.
Dene
1
and
2
as in the problem. First to see that
1
is open let z
1
. Then since is open
and z we know that there exists a ball B(z, . We claim that this ball is actually inside of
1
. If we prove this claim then we have established that
1
is open. Let s B(z, ) and consider
f : [0, 1] C given by f(t) = st + z(1 t). Then [f(t) z[ = t[s z[ < . So the image of f is
contained in B(z, ) . By concatenating the paths from w to z and z to s we see that s
1
.
Finally we will prove that
2
is also open. Suppose that
2
is not open. Then for there is some
z
2
such that every ball around z contains a point of
1
. So that B(z, ) is one such ball,
with s
1
B(z, ). Then as in the previous paragraph we can use the straight line path to connect
4 ROBERT C. RHOADES
z to s and the path has imagine inside B(z, ) . Therefore, w is path connected to s which is
path connected to z. Therefore, by concatenating paths, we see that z
1
which contradicts the
denition of
2
. So
2
must be open.
Now
1
is non-empty since w
1
. Therefore, by connectedness,
2
= .
Remark. This argument works in any metric space.
Exercise 6. Let be an open set in C and z . The connected component (or simply the
component) of containing z is the set (
z
of all points w in that can be joined to z by a curve
entirely contained in .
(1) Check rst that (
z
is open and connected. Then, show that w (
z
denes an equivalence
relation, that is (i) z (
z
, (ii) w (
z
implies z (
w
, and (iii) if w (
z
and z (
, then
w (
.
Thus is the union of all its connected components, and two components are either disjoint
or coincide.
(2) Show that can have only countably many distinct connected components.
(3) Prove that if is the complement of a compact set, then has only one unbounded com-
ponent.
Solution 6.
(1) (i) the trivial path works. (ii) Running the path in reverse works. (iii) We have a path from
to z and from z to w. Concatenating the paths gets the job done.
(2) The set of all elements of the form q + iq
where q, q
## Q is countable. Each component
contains a point of the form q +iq
, since each (
z
is open, we can be seen from the previous
exercise.
(3) If K is compact then it is closed and bounded. So it is contained in an open disc with bounded
radius and center the origin. So then the complement of that open disc is contained in .
Then if is not connected it must have a component contained in the large disc. But thus
it is bounded. So we see that can have at most one unbounded component.
Exercise 7. The family of mappings introduced here plays an important role in complex analysis.
These mappings, sometimes called Blaschke factors, will reappear in various applications in later
chapters.
(1) Let z, w be two complex numbers such that zw ,= 1. Prove that
w z
1 wz
## < 1 if [z[ < 1 and [w[ < 1,
and also that
w z
1 wz
= 1 if [z[ = 1 or [w[ = 1.
(2) Prove that for a xed w in the unit disc D, the mapping
F : z
w z
1 wz
satises the following conditions
(a) F maps the unit disc to itself (that is, F : D D), and is holomorphic.
(b) F interchanges 0 and w, namely F(0) = w and F(w) = 0.
(c) [F(z)[ = 1 if [z[ = 1.
SOLUTIONS/HINTS TO THE EXERCISES FROM COMPLEX ANALYSIS BY STEIN AND SHAKARCHI 5
(d) F : D D is bijective.
Solution 7. (a) Suppose that [w[ < 1 and [z[ = 1, then we have
[
w z
1 wz
[ = [
w z
z w
[ = 1,
since [frac1z[ = 1. Since [w[ < 1 we see that the function f(z) :=
wz
1wz
is holomorphic in D.
Thus by the maximum modulus principle it satises [f(z)[ < 1 in D because it is non-constant. A
straightforward calculation can also give the result.
(b) We already showed that F(D) D. Clearly, F(0) = w and F(w) = 0. Also from (a) we had
F(D) D. Bijective can be shown by computing F
1
.
Exercise 8. Suppose U and V are open sets in the complex plane. Prove that if f : U V and
g : V C are two functions that are dierentiable (in the real sense, that is, as functions of the two
real variables x and y), and h = g f, then
h
z
=
g
z
f
z
+
g
z
f
z
and
h
z
=
g
z
f
z
+
g
z
f
z
.
This is the complex version of the chain rule.
Solution 8. We see have
11
111
gz
11
111
fz =
Exercise 9. Show that in polar coordinates, the Cauchy-Riemann equations take the form
u
r
=
1
r
v
and
1
r
u
=
v
r
.
Use these equations to show that the logarithm function dened by
log(z) = log(r) +i where z = re
i
with < <
is holomorphic in the region r > 0 and < < . Here the second logarithm is the standard real
valued one.
Solution 9.
Exercise 10. Show that
4
z
z
= 4
z
z
= ,
where is the Laplacian
=
2
x
2
+
2
y
2
.
Solution 10.
Exercise 11. Use exercise 10 to prove that if f is holomorphic in the open set , then the real
and imaginary parts of f are harmonic; that is, their Laplacian is zero.
6 ROBERT C. RHOADES
Solution 11.
Exercise 12. Consider the function dened by
f(x +iy) =
_
[x[[y[, where x, y R.
Show that f satises the Cauchy-Riemann equations at the origin, yet f is not holomorphic at 0.
Exercise 13. Suppose that f is holomorphic in an open set . Prove that in any one of the
following cases:
(1) Re(f) is constant
(2) Im(f) is constant;
(3) [f[ is constant;
one can conclude that f is constant.
Exercise 14. Suppose a
n
N
n=1
and b
n
N
n=1
are two nite sequences of complex numbers. Let
B
k
=
k
n=1
b
n
denote the partial sums of the series
b
n
with the convention B
0
= 0. Prove the
summation by parts formula
N
n=M
a
n
b
n
= a
N
B
N
a
M
B
M1
N1
n=M
(a
n+1
a
n
)B
n
.
Exercise 15. Abels theorem. Suppose
n=1
a
n
converges. Prove that
lim
r1,r<1
n=1
r
n
a
n
=
n=1
a
n
.
Exercise 16. Determine the radius of convergence of the series
n=1
a
n
z
n
when:
(1) a
n
= (log n)
2
(2) a
n
= n!
(3) a
n
=
n
2
4
n
+3n
(4) a
n
= (n!)
3
/(3n)!
(5) Find the radius of convergence of the hypergeometric series
F(, , ; z) = 1 +
n=1
( + 1) ( +n 1)( + 1) ( +n 1)
n!( + 1) ( +n 1)
z
n
.
Here , C and ,= 0, 1, 2,
(6) Find the radius of convergence of the Bessel function of order r:
J
r
=
_
z
2
_
r
n=0
(1)
n
n!(n +r)!
_
z
2
_
2n
,
where r is a positive integer.
Exercise 17. Show that if a
n
n=0
is a sequence of non-zero complex numbers such that
lim
n
[a
n+1
[
[a
n
[
= L,
SOLUTIONS/HINTS TO THE EXERCISES FROM COMPLEX ANALYSIS BY STEIN AND SHAKARCHI 7
then
lim
n
[a
n
[
1
n
.
In particular, this exercise shows that when applicable, the ratio test can be used to calculate the
radius of convergence of a power series.
Exercise 18. Let f be a power series centered at the origin. Prove that f has a power series
expansion around any point in its disc of convergence.
Exercise 19. Prove the following:
(1) The power series
nz
n
does not converge on any point of the unit circle.
(2) The power series
z
n
/n
2
converges at every point of the unit circle
(3) The power series
z
n
/n converges at every point of the unit circle except z = 1.
8 ROBERT C. RHOADES
2. Chapter 2. Cauchys Theorem and Its Applications
SOLUTIONS/HINTS TO THE EXERCISES FROM COMPLEX ANALYSIS BY STEIN AND SHAKARCHI 9
3. Chapter 3. Meromorphic Functions and the Logarithm
10 ROBERT C. RHOADES
4. Chapter 4. The Fourier Transform
Solution 1.
SOLUTIONS/HINTS TO THE EXERCISES FROM COMPLEX ANALYSIS BY STEIN AND SHAKARCHI 11
5. Chapter 5: Entire Functions
Solution 1. We follow the proof of Jensens formula that is given in the book. We keep step 1
exactly the same. Following step 2, we have set g = f/
1
N
, then g is holomorphic and bounded
near each z
j
. So it suces to prove the theorem for Blaschke factors and for bounded functions that
vanish nowhere. Functions that vanish nowhere are treated in Step 3. It remains to show the result
for Blaschke factors. We have
log [
## (0)[ = log [[ = log [[ +
1
2
_
2
0
log [
(e
i
)[d,
since [
(z)[ = 1 for z D.
Solution 2. (a) Noting that [z
n
[ = O(e
z
## ) for all > 0 we see that the order of growth of p is 0.
(b) Clearly the order of growth is n. (c) The order of growth is e
e
z
is innite. Since [z[
k
= O(e
z
)
we know that there is no k and constants A and B, such that e
e
|z|
Ae
Bz
k
.
Solution 3.
Solution 4. For (a) see the hint. For (b): let r = min1, t and R = max 1, t. So there are eight
(m, n) such that r [nit +m[ R, namely (0, 1), (1, 0), (1, 1). There are 16 (m, n) such that
2r [nit +m[ 2R. In general there are 8k tuples with kr [nit +m[ kR. Thus with
S(k) :=
## (m,n) with max(m,n)k
1
[int +m[
,
we have
8
R
j=1
1
j
1
S(k)
8
r
1
k
j=1
1
j
1
.
Solution 5.To show entire let C
R
(z
0
) be a circle of radius R centered at z
0
. Then we have
_
C
1
(z
0
)
F
(z)dz =
_
2
0
_
e
|t|
+2iz
0
t+e
i
2it
dtie
i
d
=
_
e
|t|
+2iz
0
t
_
2
0
e
2ie
i
t
ie
i
ddt
= 0.
We can switch the order of integration by Fubinis theorem since the integrand is L
1
. The nal
equality is established since
_
C(z
0
)
e
2izt
dz = 0. This shows that the function is holormophic in any
disc by Moreras theorem. Thus it is entire.
Solution 6. By the product formula for sine we have
sin(/2) =
2
n=1
_
1
1
4n
2
_
=
2
n=1
_
(2n + 1)(2n 1)
(2n)(2n)
_
.
Solution 7.
Solution 8.
12 ROBERT C. RHOADES
Solution 9.
Solution 10.(a) e
z
1 is order 1 and has zeros precisely when z = 2in for n Z. Thus we have
e
z
1 = e
Az+B
z
n=0
_
1
z
2in
_
e
z/2in
= e
Az+B
z
n1
_
1 +
z
2
4
2
n
2
_
.
Multiplying this equation by e
z/2
we are left with two expressions both of which must be odd.
Thus we see that we must have A = 1/2. Also considering lim
z0
(e
z
1)/z = 1 we see that we
must have B = 0.
(b) cos(z) is also order 1 and has zeros at
2n+1
2
for all n Z. Thus we are left with
cos(z) = e
Az+B
nZ
_
1
2z
(2n + 1)
_
e
2z/(2n+1)
= e
Az+B
n1
_
1
4z
2
(2n + 1)
2
_
.
Since cosine is even we see that A = 0. Also letting z = 0 we see that we must have B = 0.
Solution 11. By Hadamards theorem if it omits a and b, then we have f(z) a = e
p(z)
and
f(z) b = e
q(z)
for polynomials p and q. Then e
p(z)
e
q(z)
= C for some constant C. Letting z
tend to innity we see that the leading terms of the polynomials p and q must be the same. Say it
is a
n
z
n
. Then, considering the limit as z tends to innity of
e
p(z)a
n
z
n
e
q(z)a
n
z
n
= Ce
a
n
z
n
,
we see that the next leading terms are also equal. Proceeding by induction shows that p = q, but
then this is a contradiction since it would imply that b = a. But we assume they are distinct.
Solution 12. If f has nite order and never vanishes then we have f(z) = Ce
p(z)
for some
polynomial, this follows from Hadamards theorem. So then f
(z) = Cp
(z)e
p(z)
. Since we assume
that the derivatives are never 0, we see that p
is a constant function.
Solution 13. e
z
z is entire of order 1. So e
z
z = e
Az+B
n
_
1
z
a
n
_
e
z/a
n
. If there are only
nitely many zeros then we have e
z
z = e
Az+
B
Q(z) for some polynomial Q. But then
Q(z) =
e
z
z
e
Az+
B
= O
_
e
(1
A)z
_
.
This is only possible if Q is a constant and 1
A = 0. In that case we have z = Ce
z
for some
constant C = 1 e
B
, but this is clearly not possible.
Solution 14. Let k < < k + 1, and say we have only nitely many zeros, then as in the previous
problem, F(z) = e
p(z)
Q(z) for some polynomial p of degree at most k and Q a polynomial. Now
F has order of growth less than or equal to k, since Q has order of growth 0. This contradicts the
assumption that F has order of growth .
Solution 15. Every meromorphic function, f, is holomorphic with poles at some sequence p
1
, p
2
, . . ..
Then there is an entire function with zeros precisely at the p
j
with desired multiplicity. Call this
function g. Then fg has no poles and is holomorphic except possibly at the p
j
, but since there are
no poles at the p
j
, we see that we have an entire function, say h. Then f = g/h. For the second
part of the problem construct two entire functions and take their quotient.
Solution 16.
SOLUTIONS/HINTS TO THE EXERCISES FROM COMPLEX ANALYSIS BY STEIN AND SHAKARCHI 13
6. Chapter 6. The Gamma and Zeta Functions
Solution 1. We have
1
(s)
= e
s
s
n1
_
1 +
s
n
_
e
s/n
= lim
N
e
s(
P
N
n=1
1
n
log N)
s
N
n=1
n +s
n
e
s/n
= lim
N
e
log N
s
N!
s(s+1) (s+N).
Solution 2. The easier part is to deduce the identity from the product expansion of sine and
the desired identity from this problem. To do so rst divide by (1 + a + b), we want to make the
substation a = s 1 and b = s. But to do so we must substitute a = s and b = s and then
let tend to 1 and use the fact that (x)/x 1 as x 0. We are then left with
1
(1 s)s
n>1
n(n 1)
(n s)(n 1 s)
=
1
(s 1)s
n>1
1
(1 s/n) (1 +s/(n 1))
.
Thus we may deduce that
((s)(1 s))
1
= s
n1
_
1
s
n
_
n>1
_
1 +
s
n + 1
_
= s
n=1
_
1
s
2
n
2
_
=
sin(s)
.
This is the desired result.
To prove the identity use Lemma 1.2 on page 161 and induction to establish that
(6.1)
(a + 1)(b + 1)
(a +b + 1)
=
_
N!
N
n=1
(a +b +n)
(a +n)(b +n)
_
(a +N + 1)(b +N + 1)
N!(a +b +N + 1)
.
Now let a
, b
## be integers such that a
a a
+ 1, b
b b
+ 1, a
+ b
a + b < a
+ b
+ 1.
Then we have the inequalities
(a
+N + 1)!(b
+N + 1)!
N! (a
+b
+N + 2)!
(a +N + 1)(b +N + 1)
N!(a +b +N + 1)
(a
+N + 2)!(b
+N + 2)!
N! (a
+b
+N + 1)!
.
We claim that for any A and B non-negative integers,
lim
N
(A+N)!(B +N)!
N!(A+B +N)!
= 1.
With this in hand or a slight modication of this we can easily see that letting N go to innity in
(6.1), gives the result. To prove the necessary limit use Stirlings approximation to the factorial and
standard analysis. One useful limit to know is
_
1 +
N
2
_
N
1,
as N .
It occured to me later that exercise 1 can be used prove this result fairly easily so long as you can
prove existence of the innity product.
Solution 3. Notice that
m
j=1
(2j + 1) =
(2m+1)!
2
m
m!
. Thus Walliss formula immediately gives
2
= lim
n
2
2n
(n!)
2
(2m+ 1)!
2
/[2
2n
(n!)
2
(2n + 1)]
.
Simplifying and taking square-roots gives the desired identity.
14 ROBERT C. RHOADES
I am not sure how to deduce the identity of the Gamma function. It seems like an argument
similar to the one used in problem 2 should work, however I cant get to seem the details to work
out.
Solution 4. We have a
n
() = ( + 1) ( +n 1)
1
n
.
Solution 5. This follows immediately from using the fact that ( +it) = ( it).
Solution 6. Notice that + log(n) =
n
m=1
1
m
+o(1). Thus
1 +
1
3
+ +
1
2n 1
1
2
(log(n) +) =
2n
m=1
(1)
m
m
log(2).
Solution 7. For (a) follow the hint to obtain
()() =
_
0
_
0
t
1
s
1
e
ts
dtds
=
_
0
_
1
0
(ur)
1
(u(1 r))
1
e
u
udrdu
=( +)B(, ).
For (b) use
B(, ) =
_
0
_
u
1 +u
_
1
_
1
1 +u
_
1
du
(1 +u)
2
.
Solution 8.Follow the hint to get
J
(x) =
(x/2)
( + 1/2)
_
1
1
n=0
(ixt)
n
n!
(1 t
2
)
1/2
dt.
Switching the integral and sum we see that we need to compute
_
1
1
t
n
(1 t
2
)
1/2
dt.
This is 0 when n is odd because it is an odd function. In the case n = 2m, m Z, we have
_
1
1
t
2m
(1 t
2
)
1/2
dt = 2
_
1
0
t
2m
(1 t
2
)
1/2
dt =
( + 1/2)(m+ 1/2)
( +m+ 1)
,
by the previous problem with the change of variables u = t
2
. From this we see that we need to prove
something like
1
2
2m
m!
=
(m+ 1/2)
(2m)!
.
Solution 9.Use the facts that
1
(1 zt)
n=0
( +n 1)
n!
t
n
z
n
for [tz[ < 1 and
(n +)
()
= ( +n 1) ( + 1).
SOLUTIONS/HINTS TO THE EXERCISES FROM COMPLEX ANALYSIS BY STEIN AND SHAKARCHI 15
Solution 10. Follow the hint. For (b) notice that
/(sin)(0) =
_
0
sin(t)t
1
dt = lim
z0
(z) sin(z/2) =
2
and
/(sin)(1/2) =
_
0
sin(t)t
3/2
dt = (1/2) sin(/4) = (2
)(
1
2
) =
2.
Solution 12. For (a) Follow the hint and notice that We essentially have
1
[(s)[
k!/ C (k/e)
k
Ce
k log k
when s = k1/2, k a positive integer. So we have essentially,
1
|s|
Ce
c|s| log |s|
for some constants
c and C.
For (b) notice that if we had such an F, then F(s)(s) = e
P(z)
for some polynomial P of degree at
most 1. But this would imply that
1
|(s)|
= O
_
e
c|s|
_
. Which we know from part (a) is a contradiction.
Solution 13. We have
log(1/Gamma(s)) = s + log s +
n=1
_
s
n
+ log
_
1 +
s
n
__
.
Hence we have
d
ds
log
1
(s)
=
d
ds
log((s)) = +
1
s
+
n=1
_
1
s +n
1
n
_
.
Where we are allowed to pass the derivative inside the sum because of absolute convergence. Dif-
ferentiating again we have
_
d
ds
_
2
log (s) =
1
s
2
+
n=1
1
(n +s)
2
,
which is justied again by absolute convergence.
Notice that
## / is holomorphic away from s = 0, 1, 2, . . . so it derivative is also holomorphic
in this region.
Solution 14. (a) follows from
d
dx
_
x+1
x
log (t)dt = log (x + 1) log (x) = log(xx) log x = log x.
Integrating log x gives the result.
Since t is increasing for t 1, we see that log (t) is increasing in that range. Thus
log x
_
x+1
x
log (t)dt log (x + 1) = log x + log (x).
Therefore we have
1
1
log x
+
c
xlog x
log (x)
xlog x
1
log x +x +c
xlog x
.
Letting x gives the result.
Solution 15. Use the suggested substitution and then change variables t = nx to get
_
0
x
s1
e
nx
dx =
1
n
s
_
0
t
s1
e
t
dt =
(s)
n
s
. After substituting the Taylor series the change of integral and summation
16 ROBERT C. RHOADES
is justied by using absolute convergence and the fact that for Re(s) > 1 the integral
_
1
0
x
s1
dx
converges.
Solution 16. Proceed as in the proof they give and follow the hint.
Solution 17. First observe that
_
0
f(x)x
s1
dx is holomorphic for Re(s) > 1 because
_
1
0
x
s1
dx
converges for these s and for x f decays faster than any polynomial.
Next we may apply the same argument to
I(s) =
(1)
k
(s +k)
_
0
f
(k)
(x)x
s+k1
dx
to see that I(S) is holomorphic in the region Re(s) > k 1. This formula is justied by integration
by parts.
Finally using this formula for k = n + 1 gives
I(n) = (1)
n+1
f
(n)
(0).
There seems to be a typo in the book.
SOLUTIONS/HINTS TO THE EXERCISES FROM COMPLEX ANALYSIS BY STEIN AND SHAKARCHI 17
7. Chapter 7: The Zeta Function and Prime Number Theorem
Solution 1. To obtain convergence use summation by parts to obtain
N
n=1
a
n
n
s
=
A
N
N
s
+
N
n=1
A
n
_
n
s
(n + 1)
s
_
.
Then [n
s
(n + 1)
s
[ Cn
Re(s)+1
. Thus letting N we see that
[
N
n=1
A
n
_
n
s
(n + 1)
s
_
[ B
N
n=1
n
Re(s)+1
will converge for Re(s) > 0. As the hint indicates we can show absolute convergence on any closed
half-plane z : Re(z) > 0 thus showing that the series denes a holomorphic function on the
open half-plane z : Re(z) > 0.
Solution 2. The sequences are bounded thus B
N
=
N
n=1
b
n
and A
N
=
N
n=1
a
n
satisfy [A
N
[ AN
and [B
N
[ BN where A and B are the bounds for the respective sequences. Using this and an
argument as in the rst exercise we see that the series dene holomorphic functions for Re(s) > 1.
The produce formula is easily veried by rst considering the partial sums and then letting them
tend to innity. To prove the convergence for the sequence
n1
c
n
n
s
in the range Re(s) > 1, we use
the estimate [c
n
[ ABd(n). Now d(n) c log(n) for some constant c. With this in hand we apply
partial summation again just as in the previous exercise.
(b) The rst desired identity is a direct consequence since a
n
= b
n
= 1 for that identity. For
the second identity notice that (s a) =
n1
n
a
n
s
. In this case we dont have boundedness of
the sequences under consideration, but we can apply the same argument so long as we assume that
Re(s) > Re(a) + 1, since we will need this to get the convergence in our partial summation trick.
Solution 3. We actually prove (b) rst. For n > 1, write n = p
e
1
1
p
e
m
m
. Notice that
k|n
(k) =(1) +(p
1
) + (p
m
) +(p
1
p
2
) + +(p
1
p
m
)
=1
_
m
1
_
+
_
m
2
_
+ (1)
m
_
m
m
_
=(1 1)
m
= 0.
Now to prove (a) we prove that (s)
n=1
(n)
n
s
= 1. To do this we apply exercise 2, part (a)
directly and use the identity we just proved.
Solution 4.
Solution 5. (a) By the alternating series test, the sum converges for real s > 0. More generally,
the partial sums of a
n
= (1)
n+1
are bounded, thus we may apply exercise 1 of this chapter to get
## is holomorphic in Re(s) > 0.
(b) Use the identity
(s) +
(s) = 2
n=1
1
2
s
n
s
= 2
1s
(s),
which is justied in Re(s) > 1 by absolute convergence of the series involved.
18 ROBERT C. RHOADES
(c) Since
is holomorphic for Re(s) > 0 and extends to a holomorphic function in all of C we
know they must agree on Re(s) > 0. Additionally
does not vanish for s (0, 1) because it is an
alternating sum. If (0) = 0 then by the functional equation
(s)(s/2)
s/2
= (1 s)((1 s)/2)
(1s)/2
,
we see that if (0) = 0, then (1) is nite because has a simple pole at 0, but this contradicts the
fact that has a simple pole at s = 1.
Solution 6. In the case 0 0 < 1 shift the contour to c +. In the case a > 1 shift the contour
to c and use the residue theorem to pick up the desired value because of the pole. Use the
principal value in the case a = 1.
Solution 7. For real s use the denition of to see that it is real. Also for s > 0 we can see it
easily from
dened in exercise 5. For s with real part 1/2, we use the functional equation and the
facts that (1/2 it) = (1/2 +it), which can be observed from
in exercise 5, and the fact that
(z) = (z), which follows from the denition of .
Solution 8. Follow the hints.
Solution 9.
Solution 10. Integration by parts gives
_
x
2
dt
log(t)
m
=
x
log(x)
m
2
(log 2)
m
+m
_
x
2
dt
(log t)
m+1
.
Additionally, we have
_
x
2
dt
(log t)
m
=
_
_
x
2
+
_
x
x
_
dt
(log t)
m
C
x +C
1
(log
x)
m
(x
x)
C
x
(log x)
m
.
Using these calculations with induction gives the results we wish to obtain.
Solution 11. (iv) implies (iii) implies (ii) all appear in the text. Assume (ii). Partial summation
gives
(x) = log(x)(x) +
yx
(y) (log(y + 1) log(y)) .
Notice that
[
yx
(y) (log(y + 1) log(y)) [ C
yx
(y)
y
C
yx
1 +C
<yx
1
log(y)
Cx
+C
x
log x
.
Thus we see that (x) x. We will show that (i) implies (iii) and that (iii) implies (iv). This
nishes the result. Assume (iii). Then
1
(x) =
_
x
1
(u)du =
_
x
1
(u)du +
_
x
x
(u)du = O(x
2
log x) +
1
2
x
2
+O(x
2
) +o(x
2
).
Assume (i) then we have
p
n
x
log(p) =
px
1/n
log(p) x
1/n
. So we have
(x) x +x
1/2
+ +x
1/n
+
SOLUTIONS/HINTS TO THE EXERCISES FROM COMPLEX ANALYSIS BY STEIN AND SHAKARCHI 19
Solution 12. We have (p
n
) = n
p
n
log p
n
. Taking the logarithm of both sides we see that log n
log p
n
log log p
n
. Hence log n log p
n
, because as n ,
log log p
n
log p
n
0. Returning to our rst
expression for n we have
n
p
n
log p
n
p
n
log n
.
This gives
nlog n p
n
,
which is the desired result.
20 ROBERT C. RHOADES
8. Chapter 8: Conformal Mappings
Solution 1. Suppose that f
(z
0
) = 0 for some z
0
U. Then write f(z) f(z
0
) = a(z z
0
)
k
+G(z)
for z in a disc around z
0
, a ,= 0, and k 2, with G vanishing to order at least k + 1. We wish to
prove that for some disc D around z
0
f is one-to-one. We proceed as in the proof of Proposition 1.1
and pick w small so that [G(z)[ < [a(z z
0
)
k
w[ for all z in some ball of readius . We want to
pick so small that f
## (z) is non-zero for all z B(z
0
, ) z
0
and so that [G(z)[ < [a(z z
0
)
k
w[
for all z B(z
0
, ). Then we can apply Rouches theorem to see that f(z) f(z
0
) w has two roots
and that they are distinct.
To obtain the converse direction we will wish to exploit the fact that if it is not locally bijective
then we can nd two dierent continuous paths
1
and
2
from [0, 1] to C such that
1
(0) =
2
(0) = z
0
and for each [0, 1], f(
1
()) = f(
2
()), but
1
() ,=
2
() for any > 0 .
Solution 3. f : U V is a conformal equivalence and let
1
and
2
be two paths from z
0
z
1
in
V . Then use the conformal equivalence to pull back the paths to paths in U, then nd a homotopy
in U and then push them forward with f to a homotopy in V .
Solution 4. Follow the hint. First use F given in the chapter to nd a map from D H, then
shift H down with the map z z i and then compose with squaring. Recall that squaring will
double the argument and thus ll out the entire plane. So the map is z (F(z) i)
2
.
Solution 5. From the hint we know that the solutions to f(z) = w are z = w
w
2
1. Let
these solutions be z
+
and z
Suppose z
## DH. To see this notice
that
Im(f(z)) = 1/2(Im(z) +Im(1/z) = 1/2
_
r +
1
r
_
sin()
where z = re
i
. Therefore, if f(z) = w H, then Im(f(z)) > 0, so we see that for z
D, we have
r 1, thus sin() > 0, so (0, ). We also observe from this argument that we must have z
D,
thus z
D H.
Solution 8. Using the hint we have F
1
(z) =
z1
z+1
, F
2
(z) = log(z), F
3
(z) = e
i/2
z
2
, F
4
(z) =
sin(z), and F
5
(z) = z 1.
Solution 9. u is the real part of a holomorphic function on D, thus u is harmonic. A calculation
shows that for [z[ = 1, we have z =
1
z
, a straight forward calculation gives
2Re
_
i +z
i z
_
=
i +z
i z
+
i +z
i z
=
i +z
i z
+
i +z
i +z
= 0.
Solution 10. Let T : D H be given by T(z) = i
1+z
1z
. Then we see that F T : D D and it is
holomorphic in D, with F(T(0)) = F(i) = 0. Thus by the Schwarz Lemma we have [F(T(z))[ [z[,
for z D. Then we have
[F(z)[ = [F T T
1
(z)[ [T
1
(z)[ = [
z i
z +i
[.
SOLUTIONS/HINTS TO THE EXERCISES FROM COMPLEX ANALYSIS BY STEIN AND SHAKARCHI 21
Solution 11. Dene
f : D D by
f(z) =
1
M
f
_
1
R
z
_
. Then
f(0) D unless [f(0)[ = M, in which
case f is constant by the maximum modulus theorem. Dene T : D D by T(z) =
z
f(0)
1
f(0)
. Then
T
f : D D and T
f(0) = 0. Thus the Schwarz lemma gives
[
f(z)
f(0)
1
f(0)
f(z)
[ [z[.
Simplifying and substituting
f(z) =
1
M
f
_
1
R
z
_
.
Solution 12. If f(0) = 0 and we have f(z
0
) = z
0
, then we have [f(z
0
)[ = [z
0
[ so we know by the
Schwarz lemma f is a rotation, but f(z) = e
i
z and since f(z
0
) = z
0
, so = 0.
In the upper half-plane, the map z z+1 is a conformal map and has no xed points. Composing
with a conformal map from D H and its inverse gives a conformal map on D to D that has no
xed points.
Solution 13. For the rst part follow the hint. Divide both sides of the result from (a) to obtain
[
f(z) f(w)
z w
[[
1
1 f(w)f(z)
[
1
[1 wz[
.
Letting w tend to z gives the result.
Solution 14. Consider G dened by z i
1z
1+z
which conformally maps D H. Let f be conformal
from H to D, then f G : D D. So we see that f G =
za
1az
for some a D and with modulus
1. A straightfoward calculation shows that
f(z) =
G
1
(z) z
1 aG
1
(z)
=
_
1 +a
1 +a
_
z
z
,
where = i
1a
1+a
H.
Solution 15. For the rst part it is enough to consider the analogous problem on the circle. On
the circle we know that the automorphisms take the form
a
(z) :=
za
1az
for a D and [[ = 1.
So if there are three xed points then we have three solutions to
a
(z) = z, but there are only two
roots of a quadratic so we have a contradiction.
Solution 17. The rst part follows since
= Area(D) =
_ _
(D)
dxdy =
_ _
D
[
(z)[
2
dxdy.
I dont know how to establish the second part.
Solution 18. Follow the proof and lemmas of Theorem 4.2
Solution 19. Follow hint, the result should be clear.
Solution 20. (a) Following the notation of Proposition 4.1 we have A
1
= 0, A
2
1, and A
3
= ,
when > 1, the other cases 0 < < 1 and < 0 are handled similarly. In any case we have
1
=
2
=
3
=
=
1
2
. To see that we have a rectangle it is enough to notice that
1
=
2
=
3
=
=
2
.
22 ROBERT C. RHOADES
In the second part, we have
_
1
0
d
_
(
2
1)
=
1
2
_
1
0
(1 u)
1/2
u
3/4
du
=
1
2
(1/2)(1/4)
(3/4)
=
1
2
2
(1/4)
2
,
where we use the calculation of the beta function and then the fact that (3/4)(1/4) =
sin(/4)
=
2 and (1/2) =
. We need only to calculate the length of one of the perpendicular sides which
can be done similarly.
Solution 21. (a) follows immediately from Proposition 4.1. For (d) we have the following similar
calculation
_
1
0
z
1
(1 z)
2
dz =
(1
1
)(1
2
)
(2
1
3
)
=
(
1
)(
2
)
(
3
)
=
sin(
3
)(
3
)
(
1
)(
2
).
Where we use (
3
)(1
3
) =
sin((1
3
))
.
A similar calculation with the change of variables u =
1
z
can be used to compute the integral
_
1
z
1
(1 z)
2
dz = (1)
2
(
2
)(
3
)
(
1
)
.
Similar calculations nish the problem.
Solution 22. This result is not suprising because of the correspondence between H and D that has
R
= R and D.
SOLUTIONS/HINTS TO THE EXERCISES FROM COMPLEX ANALYSIS BY STEIN AND SHAKARCHI 23
9. Chapter 9: An Introduction to Elliptic Functions
Solution 1. (a) With the notation given in the problem,
n
2
=
np
q
1
=
1 mq
q
1
=
1
q
1
m
1
.
This gives us that
f
_
z +
1
q
1
_
= f
_
z +
1
q
1
m
1
_
= f (z +n
2
) = f (z) .
This shows that
1
q
1
is a period of f and it is smaller than
2
and
1
in size, since
1
q
1
=
1
p
2
.
(b) Follow the hint and use continuity.
Solution 2. The argument of Theorem 4.1 in Chapter 3 shows that
_
P
z
f
(z)
f(z)
dz = 2i (a
1
+ +a
r
b
1
b
r
) .
This follows from the residue theorem. I am not sure how to use the hint to compute this integral
in a second way so as to yield the desired result.
Solution 3. Let r = min(1, [[) and R = max(1, [[). Then there are 4n values of such that
nr [[ nR. This gives
||M
1
[[
2
N
n=1
4n
n
2
R
2
4
R
2
log N.
We get similarly that
||M
1
[[
2
N
n=1
4n
n
2
r
2
4
r
2
log N.
Where the M is chosen appropriately. So we see that the sum must diverge.
Just be more careful with this argument to get the second result.
Solution 5. If we establish convergence then since it is an innite product it can only vanish at
a point where one of the terms vanishes, thus it has simple zeros at all the periods and does not
vanish anywhere else.
Dierentiating term by term gives the desired formula for
## /. Again dierentiating term by
term shows the desired formula for (z).
Solution 6. We have (
)
2
= 4
3
ab, hence dierentiating both sides and dividing by
gives,
2
= 12
2
a.
Solution 7. We see that 4
m odd
1
m
2
=
2
sin(/2)
2
=
2
2
. Use the fact that
1
4
m
1
m
2
=
m even
1
m
2
=
_
m odd
_
1
m
2
which results in
m
=
4
3
m odd
. Similar analysis works for (4).
24 ROBERT C. RHOADES
Solution 8.
E
4
(it) =
n=0
1
n
4
+sum
n
m=0
1
n +imt
= 2(4) +sum
n
m=0
1
n +imt
so we need to analysis the remaining sum. As t it is clear that the sum should go to 0, which
is what we need to analyze and so. I am a bit confused about the rate of convergence to zero.
SOLUTIONS/HINTS TO THE EXERCISES FROM COMPLEX ANALYSIS BY STEIN AND SHAKARCHI 25
10. Chapter 10: Applications of Theta Functions
Solution 2. Multiply the recursive relation by x
n
and sum from n = 2 up to innity. This gives
F(x) F
0
F
1
x = x(F(x) F
0
) +x
2
F(x).
This proves the rst part in a formal sense, to get convergence in a ball around 0 we will show that
F
n
c
n
for some > 0, which is enough to get convergence in a ball around 0.
For (c) and (d) follow the partial fraction decomposition, then use the geometric series, but
absolute convergence in a small disc we can combine the two sums. Then matching up the appropriate
powers of x
n
gives the result.
Solution 3. Do the same things as in (2) but in more generality. In the case = we just note
that we have U(x) =
u
0
(1x)
2
+ (u
1
au
0
)
x
(1x)
2
. Now notice that
1
(1x)
2
=
n1
nx
n1
to nish
o the calculations.
Solution 4. We have
n1
(1 x
n
) =
k=
(1)
k
x
k(3k+1)/2
. So we see that
_
_
n0
p(n)x
n
_
_
_
k=
(1)
k
x
k(3k+1)/2
_
= 1.
Multiplying out and comparing powers of x
n
on each side gives the desired formula
k
(1)
k
p
_
n
k(3k + 1)
2
_
= 0,
for n 1.
Solution 5. Follow the hint. To prove the needed inequality notice that
mx
m1
<
1 x
m
1 x
= 1 +x + +x
m1
< m
for x (0, 1), since x
j
> x
m1
for j < m1 in this range and x
j
< 1.
Solution 6. Following the hint we have log(F(e
y
))
2
6(1e
y
)
e
2
6y
. Therefore we have F(e
y
)
Ce
2
6y
, hence p(n) ce
2
6y
+ny
. Taking y =
1
n
1/2
gives the upper bound.
Solution 7. Both identities are a consequence of the triple product identity for the theta function.
For the rst identity take =
1
2
and z =
1
4
and x = e
i
. For the second take z =
3
4
+
1
2
,
=
5
2
and x = e
i
.
Solution 8. Since they are relatively prime both are not even. If a and b are both odd then
a
2
+b
2
2 (mod 4), but all squares are 0 or 1 modulo 4. Following the hint write
_
b
2
_
2
=
ca
2
c+a
2
.
Notice that
ca
2
and
c+a
2
must be relatively prime since the gcd must then divide b, but we assume
a, b, and c to be relatively prime. Therefore we see that
c+a
2
and
ca
2
are both squares, say m
2
and
n
2
respectively. Then c = m
2
+n
2
,
b
2
= mn, and a = m
2
n
2
, as desired.
For the last part notice that
(10.1) (a
2
+b
2
)(c
2
+d
2
) = a
2
c
2
+a
2
d
2
+b
2
c
2
+b
2
d
2
= (ac +bd)
2
+ (ad bc)
2
.
26 ROBERT C. RHOADES
Solution 9. From the formula we know that r
2
(p) = 4(2 0) = 8, when p 1 (mod 4). For
q 3 (mod 4) we have
r
2
(q
a
) = 4((
a
2
| + 1) (
a 1
2
| + 1)).
So when a is odd we see that
a
2
| =
a1
2
| and we have 0, otherwise it is non-zero.
From the displayed equation in the previous problem and the earlier parts of this problem we may
deduce that when all the primes congruent to 3 modulo 4 appear to an even power n is representable
as the sum of two squares. To get the other direction use the formula writing n = p
a
1
1
p
a
j
j
q
b
1
1
q
b
k
k
where p
m
1 (mod 4) and q
m
3 (mod 4) are distinct primes.
Solution 10. For (a) notice that r
2
(q) = 0 for all q 3 (mod 4) and r
2
(5
k
) = k + 1. For (b) rst
notice that r
4
(2
k
) = 8 (1 + 2) = 24 for all k 1.
Solution 12. For the rst part use exercise 11 For (b) we note that
1
(n) =
1
(n) 4
1
(n/4). For
(c) notice that it is enough to show that
n1
q
n
(1 + (1)
n
q
n
)
2
=
n1
q
n
(1 q
n
)
2
4
n1
q
4n
(1 q
4n
)
2
.
To do so consider the odd and even n in the left hand side separately. The odd terms match up
with the odd terms from the rst sum on the right hand side. To nish notice that
q
2
(1 q
2n
)
2
4q
4n
(1 q
4n
)
2
=
q
2n
(1 +q
2n
)
2
.
Department of Mathematics, University of Wisconsin, Madison, WI 53706
E-mail address: rhoades@math.wisc.edu | 14,492 | 38,990 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2019-51 | latest | en | 0.870559 |
https://www.physicsforums.com/threads/tough-moments-question.738826/ | 1,508,367,196,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187823153.58/warc/CC-MAIN-20171018214541-20171018234541-00279.warc.gz | 951,829,827 | 19,245 | # Tough moments question
1. Feb 17, 2014
### yugeci
Hello friends, I have this problem I've been stuck on for a while:
Here are a few relevant equations:
Moment = FxDy + FyDx
Fx = F cos θ
Fy = F sin θ
etc.
I can't even get past part (a). Resolving the 100 lb force is easy. It would be 100 sin 60 for the vertical component and - 100 cos 60 for the horizontal component. And the P component is a horizontal force, with no vertical component. The distance between P and C is 8". The distance between the horizontal component of the 100 lb force would be 8" too I guess (not sure), but what would be the vertical distance?
Part (b) doesn't look too hard once P is figured out.
I am not sure about part (c). Where will point A be if the moment is a maximum?
I don't understand part (d) either.
Help would greatly be appreciated.
1. The problem statement, all variables and given/known data
2. Relevant equations
3. The attempt at a solution
2. Feb 17, 2014
### BvU
Hello friend, a request here: make it a little easier on potential helpers and use the template. It also helps you to sort out things, check if you have the right equations (you do) and especially to be conscious of where you get stuck.
Your tackle the exercise in the right way. Your translation from mm to inch is unnecessary, so don't do it. Especially not if you have the wrong conversion factor ( should be 1 mm = 1/25.4 inch) . Same with N to lbf
The vertical distance is the distance between a line through the point where the force acts that is perpendicular to the axis of rotation, and the axis of rotation. Same way as with P, actually.
3. Feb 17, 2014
### CWatters
It says the moments sum to zero. The horizontal and vertical components don't necessarily sum to zero because you don't know the forces on the pivot point C.
My first reaction is to work out the torque caused by the 100N force about point C.
4. Feb 17, 2014
### BvU
Not sure what you are not sure about. You repeat the question; that is not an attempt at a solution.
By the time you can calculate the sum of moments around C you can also do so around a point D on the rim straight under C, right?
This helps you with c) and with d).
The fact that c) comes before d) suggests that you don't have to calculate this sum first to answer c). Some smart way to look at c) is indicated.
5. Feb 17, 2014
### BvU
Yes. One can sum the Fy times Dx (there is only one) and the Fx times Dy. The sum has to be zero.
6. Feb 17, 2014
### yugeci
OK, I'll use the template from next time. I managed to get solve part (a)
My = 100 sin 60 * 8
Mx = (- 100 cos 60 * 4) + (-8P)
Since they both anti-clockwise, My + Mx = 0
So I got P as 61.6N, and the resultant as 141.26N. Thank you guys for the help.
How will part (c) and (d) be solved?
Edit - just saw the 2 new posts, I'll try.
7. Feb 17, 2014
### yugeci
I get what you mean. But I don't get what approach to take for part (c). can think of..
My = 100 sin 60 * x
Mx = (-100 cos 60 * y) + (-P * y)
Where x and y are some values of horizontal distance and vertical distance that give a max moment M = Mx + My. Now what? We don't know what this max value is..
8. Feb 17, 2014
### BvU
Didn't find one yet, sorry. Thought about vector addition of P and 100 N and drawing a line perp to the sum. Comes close to answer. (addition introduces a small torque as well).
9. Feb 18, 2014
### yugeci
I saw the solution in the solution manual and they draw a perpendicular line from the resultant (attached the image). If so they used α to get the x and y values.. it actually makes sense to me since max torque happens when the distance is perpendicular. These questions are pretty insane and require so much thinking.
Edit - this is pretty much the same method you did BvU if I'm not mistaken.. where does the extra torque come from?
With that I think I will be able to do part (d) as well. Thanks for all of the help guys. :)
#### Attached Files:
• ###### help.png
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60
10. Feb 18, 2014
### BvU
Well, this looks real good. Didn't think of moving F along its own line of action until F and P grab at the same point. Totally agree with the solutions manual.
I needed an extra torque because I moved P perpendicular to its axis to the x-axis, then to the point where the 100 N grabs and then added the two. This moving perpendicularly requires the introduction of an extra torque which spoiled an easy finding of the maximum moment point.
You indicate you understand the solution manual way makes sense, so you have learned something useful.
I have learned too: there had to be a shortcut easy way, and there is one. Just didn't find it. Next time (perhaps) both of us will do better. | 1,244 | 4,715 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2017-43 | longest | en | 0.936166 |
https://blogforlife.cc/?p=1603 | 1,606,897,974,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141706569.64/warc/CC-MAIN-20201202083021-20201202113021-00224.warc.gz | 205,209,622 | 7,444 | # Which Expression Is Equivalent To The Following Complex Fraction
Which Expression Is Equivalent To The Following Complex Fraction. Please help me by showing alternate methods to solve this complex number SAT question. Two fractions are equivalent if they have the same value.
This article will cover several ways to calculate equivalent fractions from basic multiplication and division to more complex methods for solving equivalent fraction equations. Take a look at the following picture which shows three different fractions, that are all worth In this lesson, in order to identify fractions that are equivalent, you must multiply the numerator and denominator by the same number. For more information about the First and Third Party Cookies used please follow this link.
## This article will cover several ways to calculate equivalent fractions from basic multiplication and division to more complex methods for solving equivalent fraction equations.
Keywords: fraction, simplify, expression, equivalent, complex fraction, , add, subtraction, denominators, numerators. The equivalent expression of the complex fraction is. Some examples of complex fractions are Recall that when you multiply the exact same thing to the numerator and the denominator, the result is an equivalent fraction.
eBook: Dynamic System Modeling and Control
Complex Rational Expressions
goldendragon1971
### Finding equivalent expressions is not as complicated or as daunting as you might think.
Fractions are numbers that represent a part of the whole. Therefore, equivalent fractions are fractions that are equal in value. You can work equivalent expressions by distribution or factoring depending on what type of equation you are given first.
Some examples of complex fractions are Recall that when you multiply the exact same thing to the numerator and the denominator, the result is an equivalent fraction. Partial fraction decomposition is a technique used to write a rational function as the sum of simpler rational expressions. They are both fractions. an equivalent fraction is a fraction that is the same as another fraction.
### Extend understanding of fraction equivalence and ordering.
Take a look at the following picture which shows three different fractions, that are all worth In this lesson, in order to identify fractions that are equivalent, you must multiply the numerator and denominator by the same number. Equivalent Fractions have the same value, even though they may look different. "Change the bottom using multiply or divide, And the same to the top must be applied". When an object or a group of objects is divided into equal parts, then each As we learned in comparing, adding or subtracting unlike fractions is done by finding equivalent fractions with a common denominator and then applying the.
The most commonly used literal expressions are formulas from geometry, physics, business First Eliminate fractions by multiplying all terms by the least common denominator of all fractions. Multiply the numerators, multiply the denominators, and then simplify. Fraction calculator to add, subtract, divide, and multiply fractions with step-by-step explanation — calculator for fractions and expressions The calculator performs basic and advanced operations with fractions, expressions with fractions combined with integers, decimals, and mixed numbers.
Partial fraction decomposition is a technique used to write a rational function as the sum of simpler rational expressions. Equivalent fractions are fractions with different numbers representing the same part of a whole. Take a look at the following picture which shows three different fractions, that are all worth In this lesson, in order to identify fractions that are equivalent, you must multiply the numerator and denominator by the same number. | 666 | 3,828 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.59375 | 5 | CC-MAIN-2020-50 | latest | en | 0.901937 |
https://curriculum.illustrativemathematics.org/HS/teachers/3/2/10/preparation.html | 1,718,480,330,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861606.63/warc/CC-MAIN-20240615190624-20240615220624-00653.warc.gz | 168,460,906 | 25,429 | # Lesson 10
Multiplicity
### Lesson Narrative
This lesson is meant to be a culmination of what students have learned in the unit so far. In order to sketch a graph of a polynomial function from a factored equation, students must understand the relationship between the factors, zeros, and horizontal intercepts. They need to calculate the leading term of the equation in order to use the degree and the sign of the leading coefficient to identify the end behavior of the function.
New in this lesson is the relationship between the multiplicity of a factor, which is the power to which the factor occurs in the factored form of a polynomial, and the shape of the graph at the intercept associated with that factor. While students did preview this idea in their study of quadratics in previous courses when they looked at the graph of equations such as $$y=(x-5)^2$$, here we name the effect and study the differences between factors with a multiplicity of 1, 2, and 3.
Students will have more opportunities to sketch polynomials from factored equations in future lessons, but this lesson is the main opportunity to make solid connections between the shape of a graph and the structure of the factored equation for polynomial functions (MP7).
### Learning Goals
Teacher Facing
• Comprehend the effect that the multiplicity of factors has on the shape of the graph of a polynomial function.
• Use zeros and multiplicities to create a rough graph of a polynomial function given in factored form.
### Student Facing
• Let’s sketch some polynomial functions.
### Required Preparation
Acquire devices that can run Desmos (recommended) or other graphing technology. It is ideal if each student has their own device. (Desmos is available under Math Tools.)
### Student Facing
• I can use zeros and multiplicities to sketch a graph of a polynomial.
Building On
Building Towards
### Glossary Entries
• multiplicity
The power to which a factor occurs in the factored form of a polynomial. For example, in the polynomial $$(x-1)^2(x+3)$$, the factor $$x-1$$ has multiplicity 2 and the factor $$x+3$$ has multiplicity 1.
### Print Formatted Materials
For access, consult one of our IM Certified Partners. | 471 | 2,212 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5625 | 5 | CC-MAIN-2024-26 | latest | en | 0.939854 |
http://farside.ph.utexas.edu/teaching/em/lectures/node62.html | 1,553,462,991,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912203493.88/warc/CC-MAIN-20190324210143-20190324232143-00532.warc.gz | 70,291,859 | 6,478 | Next: One-dimensional solution of Poisson's Up: Electrostatics Previous: Poisson's equation
The uniqueness theorem
We have already seen the great value of the uniqueness theorem for Poisson's equation (or Laplace's equation) in our discussion of Helmholtz's theorem (see Sect. 3.11). Let us now examine this theorem in detail.
Consider a volume bounded by some surface . Suppose that we are given the charge density throughout , and the value of the scalar potential on . Is this sufficient information to uniquely specify the scalar potential throughout ? Suppose, for the sake of argument, that the solution is not unique. Let there be two potentials and which satisfy
(666) (667)
throughout , and
(668) (669)
on . We can form the difference between these two potentials:
(670)
The potential clearly satisfies
(671)
throughout , and
(672)
on .
According to vector field theory,
(673)
Thus, using Gauss' theorem
(674)
But, throughout , and on , so the above equation reduces to
(675)
Note that is a positive definite quantity. The only way in which the volume integral of a positive definite quantity can be zero is if that quantity itself is zero throughout the volume. This is not necessarily the case for a non-positive definite quantity: we could have positive and negative contributions from various regions inside the volume which cancel one another out. Thus, since is positive definite, it follows that
(676)
throughout . However, we know that on , so we get
(677)
throughout . In other words,
(678)
throughout and on . Our initial assumption that and are two different solutions of Poisson's equation, satisfying the same boundary conditions, turns out to be incorrect.
The fact that the solutions to Poisson's equation are unique is very useful. It means that if we find a solution to this equation--no matter how contrived the derivation--then this is the only possible solution. One immediate use of the uniqueness theorem is to prove that the electric field inside an empty cavity in a conductor is zero. Recall that our previous proof of this was rather involved, and was also not particularly rigorous (see Sect. 5.4). We know that the interior surface of the conductor is at some constant potential , say. So, we have on the boundary of the cavity, and inside the cavity (since it contains no charges). One rather obvious solution to these equations is throughout the cavity. Since the solutions to Poisson's equation are unique, this is the only solution. Thus,
(679)
inside the cavity.
Suppose that some volume contains a number of conductors. We know that the surface of each conductor is an equipotential surface, but, in general, we do not know what potential each surface is at (unless we are specifically told that it is earthed, etc.). However, if the conductors are insulated it is plausible that we might know the charge on each conductor. Suppose that there are conductors, each carrying a charge ( to ), and suppose that the region containing these conductors is filled by a known charge density , and bounded by some surface which is either infinity or an enclosing conductor. Is this enough information to uniquely specify the electric field throughout ?
Well, suppose that it is not enough information, so that there are two fields and which satisfy
(680) (681)
throughout , with
(682) (683)
on the surface of the th conductor, and, finally,
(684) (685)
over the bounding surface, where
(686)
is the total charge contained in volume .
Let us form the difference field
(687)
It is clear that
(688)
throughout , and
(689)
for all , with
(690)
Now, we know that each conductor is at a constant potential, so if
(691)
then is a constant on the surface of each conductor. Furthermore, if the outer surface is infinity then on this surface. If the outer surface is an enclosing conductor then is a constant on this surface. Either way, is constant on .
Consider the vector identity
(692)
We have throughout , and , so the above identity reduces to
(693)
throughout . Integrating over , and making use of Gauss' theorem, yields
(694)
However, is a constant on the surfaces and . So, making use of Eqs. (689) and (690), we obtain
(695)
Of course, is a positive definite quantity, so the above relation implies that
(696)
throughout : i.e., the fields and are identical throughout .
For a general electrostatic problem involving charges and conductors, it is clear that if we are given either the potential at the surface of each conductor or the charge carried by each conductor (plus the charge density throughout the volume, etc.) then we can uniquely determine the electric field. There are many other uniqueness theorems which generalize this result still further: i.e., we could be given the potential of some of the conductors and the charge carried by the others, and the solution would still be unique.
At this point, it is worth noting that there are also uniqueness theorems associated with magnetostatics. For instance, if the current density, , is specified throughout some volume , and either the magnetic field, , or the vector potential, , is specified on the bounding surface , then the magnetic field is uniquely determined throughout and on . The proof of this proposition proceeds along the usual lines. Suppose that the magnetic field is not uniquely determined. In other words, suppose there are two magnetic fields, and , satisfying
(697) (698)
throughout . Suppose, further, that either or on . Forming the difference field, , we have
(699)
throughout , and either or on . Now, according to vector field theory,
(700)
Setting , and using and Eq. (699), we obtain
(701)
However, we know that either or is zero on . Hence, we obtain
(702)
Since, is positive definite, the only way in which the above equation can be satisfied is if is zero throughout . Hence, throughout , and the solution is therefore unique.
Next: One-dimensional solution of Poisson's Up: Electrostatics Previous: Poisson's equation
Richard Fitzpatrick 2006-02-02 | 1,309 | 6,060 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2019-13 | latest | en | 0.933064 |
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