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2019-02-28, 23:25 #1 SALMONMILK Feb 2019 1 Posts Applications of factoring besides crypto? Hi all, I am wondering if there are any uses for number factorization besides cryptography. I have been trying to research the answer for a long time and cannot seem to find anything besides cryptography. I am familiar with RSA and the other crypto schemes that are related on Wikipedia, but that is about it when it comes to factoring. I have come across a version of the fast-fourier transform (https://en.wikipedia.org/wiki/Prime-..._FFT_algorithm) that says prime factor in its name, but I do not have the knowledge to be able to understand what it does or means. Does it utilize number factorization? So really the main question: Are there any known applications for number factorization besides cryptography? Any information is greatly appreciated. I am just deeply curious and have been unable to find any information. Thanks so much
2019-03-01, 09:10 #2
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Quote:
Originally Posted by SALMONMILK I have come across a version of the fast-fourier transform (https://en.wikipedia.org/wiki/Prime-..._FFT_algorithm) that says prime factor in its name, but I do not have the knowledge to be able to understand what it does or means. Does it utilize number factorization?
The classical FFT, which applies to data with a length of 2n exploits the (almost trivial) factorization to earn the subriquet "Fast" by dividing a range of data into two sub-ranges each half the size of the original. Similar approaches, though not as efficient, can exploit small prime factors, p, to divide the range into p portions, each 1/p the size of the original.
2019-03-02, 17:26 #3 chris2be8 Sep 2009 96916 Posts AFAIK it's main other use is mathematical research. Read the "Links to factoring projects" https://mersenneforum.org/showthread.php?t=9611 thread in this forum and follow the links to see what is being worked on now. Chris
2021-04-01, 07:36 #4 naturevault 2·3·7·13·17 Posts Technically it is still crypto but a different sort, proof of work for crypto collectibles.
2021-04-01, 11:59 #5 kriesel "TF79LL86GIMPS96gpu17" Mar 2017 US midwest 22×1,777 Posts There are applications in engineering design sometimes for factoring of what mathematicians would consider small to trivial numbers or ratios. Clock drives for equatorial mounts for telescopes or tracking solar arrays when electronic or computer speed controls are impractical, obtaining gear reduction ratios near a given desired ratio with smooth numbers. The selection of prime factors in commercially available power gears or even instrument size gears or toothed-belt sprockets is pretty limited. The method of continued fractions is a way to derive a series of successive approximations of ratio, each of which can be evaluated for practicality. https://en.wikipedia.org/wiki/Continued_fraction
2021-04-03, 15:59 #6 alpertron Aug 2002 Buenos Aires, Argentina 101101110112 Posts Solving Diophantine equations of second degree on two variables requires integer factorization. For example: find the integer solutions (x, y) of a circle represented by x2 + y2 = N where N is a big integer number. You cannot solve this problem without factoring N.
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# Probabilistic Reasoning in AI - A way to deal with Uncertainty
In this tutorial, we will learn about the probability theory probabilistic reasoning while dealing with Uncertainty. We will study what probability theory is, how an agent implements probabilistic reasoning in its decision making and we will also study how this theory solves the problem of uncertainty in the environment of the agent. By Monika Sharma Last updated : April 15, 2023
## Overview
As we know that there are many cases where the answer to the problem is neither completely true nor completely false. For example, the statement- "Student will pass in the board exams". We cannot say anything about a student's result before the results are declared. However, we can draw some predictions based on the student's past performances in academics.
## Probabilistic Reasoning
In these types of situations, probabilistic theory can help us give an estimate of how much an event is likely to occur or happen? In this theory, we find the probabilities of all the alternatives that are possible in any experiment. The sum of all these probabilities for an experiment is always 1 because all these events/alternatives can happen only within this experiment.
## Example
As in the above example, the statement can either be true or false, not anything other than that. That means, the student will either pass in board exams or will fail. So, if we are given the following probability:
P (Student will pass in board exams) = 0.80
Therefore, P (Student will fail in board exams) = 0.20
Then this means that there are 80 percent chances that the student will pass and 20 percent chances that the student will fail. And as we can observe that, the probability that one of these events will occur is 100 percent.
Therefore, in all those cases where there is a fixed number of outcomes possible for any given experiment, the probabilistic theory is applicable.
Another example in this theory can be taken of picking a card from a deck (Excluding the Joker). If the stated events in this experiment are as follows:
``` A: The chosen card is of Spade
B: The chosen card is of Hearts
C: The chosen card is of Clubs
D: The chosen card is of Diamond
Then the probability of each event is:
P(A) = P(B) = P(C) = P(D) = 0.25
```
AS there are 13 cards of each of them in a deck. And the probability of the events to occur when the experiment is taking place successfully is:
``` P(E)= P(A) + P(B) + P(C) + P(D)= 1
``` | 557 | 2,491 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2024-30 | latest | en | 0.939314 |
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1. Which of the following systems is stable?
a) y(t) = log(x(t))
b) y(t) = sin(x(t))
c) y(t) = exp(x(t))
d) y(t) = tx(t) + 1
Answer: b [Reason:] Stability implies that a bounded input should give a bounded output. In a,b,d there are regions of x, for which y reaches infinity/negative infinity. Thus the sin function always stays between -1 and 1, and is hence stable.
2. State whether the integrator system is stable or not.
a) Unstable
b) Stable
c) Partially Stable
d) All of the mentioned
Answer: a [Reason:] The integrator system keep accumulating values and hence may become unbounded even for a bounded input in case of an impulse.
3. For what values of k is the following system stable, y = (k^2 – 3k -4)log(x) + sin(x)?
a) k=1,4
b) k=2,3
c) k=5,4
d) k =4,-1
Answer: d [Reason:] The values of k for which the logarithmic function ceases to exist, gives the condition for a stable system.
4. For a bounded function, is the integral of the odd function from -infinity to +infinity defined and finite?
a) Yes
b) Never
c) Not always
d) None of the mentioned
Answer: a [Reason:] The odd function will have zero area over all real time space.
5. When a system is such that the square sum of its impulse response tends to infinity when summed over all real time space,
a) System is marginally stable
b) System is unstable
c) System is stable
d) None of the mentioned
Answer: b [Reason:] The system turns out to be unstable. Only if it is zero/finite it is stable.
6. Is the system h(t) = exp(-jwt) stable?
a) Yes
b) No
c) Can’t say
d) None of the mentioned
Answer: c [Reason:] If w is a complex number with Im(w) < 0, we could have an unstable situation as well. Hence, we cannot conclude [no constraints on w given].
7. Is the system h(t) = exp(-t) stable?
a) Yes
b) No
c) Can’t say
d) None of the mentioned
Answer: a [Reason:] The integral of the system from -inf to +inf equals to a finite quantity, hence it will be a stable system.
8. Comment on the stability of the following system, y[n] = n*x[n-1].
a) Stable
b) Unstable
c) Partially Stable
d) All of the mentioned
Answer: b [Reason:] Even if we have a bounded input as n tends to inf, we will have an unbounded output. Hence, the system resolves to be an unstable one.
9. Comment on the stability of the following system, y[n] = (x[n-1])n.
a) Stable
b) Unstable
c) Partially Stable
d) All of the mentioned
Answer: a [Reason:] Even if we have a bounded input as n tends to inf, we will have an bounded output. Hence, the system resolves to be a stable one.
10. What is the consequence of marginally stable systems?
a) The system will turn out to be critically damped
b) The system will be an overdamped system
c) It will be a damped system
d) Purely oscillatory system | 822 | 2,981 | {"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} | 3.15625 | 3 | CC-MAIN-2019-22 | latest | en | 0.853234 |
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## The sum of three numbers in AP is 3 and their product is -35. Find the numbers.
• 0
One of the basic question from arithmetic progression chapter in which we are to find the numbers, if their product ...Read more
• 0
## Show that (a−b)^2, (a^2+b^2) and (a+b)^2 are in AP.
• 0
One of the basic question from arithmetic progression chapter in which we have been asked to prove that (a−b)^2, (a^2+b^2) and ...Read more | 129 | 429 | {"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-2023-06 | latest | en | 0.951305 |
https://www.sciencefacts.net/conduction.html | 1,701,693,724,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100529.8/warc/CC-MAIN-20231204115419-20231204145419-00595.warc.gz | 1,112,259,338 | 39,386 | Home / Physics / Conduction
# Conduction
Thermal conduction is the flow of thermal energy (heat) from higher to lower temperatures through molecular vibrations and collisions. Conduction occurs within an object or from a hot object to a cold object in contact with the former. It can occur in solids, liquids, and gases but is primarily observed in solids where molecules are closely packed. Heat will continue to flow until thermal equilibrium is reached.
## Examples
• Warming hands by touching a hot body
• Heating one end of a metal rod
• Heating a frying pan on top of a stove
• Hot air immediately above the Earth’s surface
How is Heat Transferred Through Thermal Conduction
According to kinetic theory, matter is made of particles that are in constant random motion. This motion manifests as thermal energy, which depends on the temperature. The higher the temperature, the higher the thermal energy. The motion of particles, whether translatory or vibrational, leads to collisions. The particles transfer energy among themselves. Consequently, heat travels from a high-temperature region to a low-temperature region.
## Equation
### Fourier’s Law of Thermal Conduction
For thermal conduction to occur, there has to be a temperature gradient. Fourier’s law of thermal conduction states that the time rate of heat transfer through a material is proportional to the negative temperature gradient and the area through which the heat flows at right angles to that gradient. Mathematically, Fourier’s law can be written as
$Q = – K A \frac{\Delta T}{\Delta x}$
Where
Q: Heat transfer rate (Js-1 or W)
K: Thermal conductivity (Wm-1K-1)
A: Cross-sectional area (m2)
$$\frac{\Delta T}{\Delta x}$$: Temperature gradient (Km-1)
Suppose heat flows through a conductor of thickness d whose ends are at temperatures TA (hot) and TB (cold). The heat flowing per second through the conductor is
$Q = KA\frac{T_B – T_A}{d}$
## Thermal Conductivity
Thermal conductivity is a physical property of substances. In the above equation, suppose A = 1, d = 1, and (TA – TB) = 1. Then
K = Q
Thus, thermal conductivity is defined as the amount of heat flowing through a conductor of unit length, whose cross-section has a unit area and whose ends are at a unit temperature difference.
Metals have high thermal conductivity because their valence electrons are delocalized and can efficiently conduct heat. For example, the thermal conductivities of silver and copper are 406 Wm-1K-1 and 385 Wm-1K-1, respectively.
Insulators are poor conductors of heat. They have voids in between the atoms, which interfere with heat transfer. For example, the thermal conductivity of wood ranges from 0.04 to 0.12 Wm-1K-1. Air is a poor conductor of heat. Its thermal conductivity at 0 ˚C is 0.024 Wm-1K-1.
Article was last reviewed on Monday, January 2, 2023 | 660 | 2,848 | {"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} | 3.875 | 4 | CC-MAIN-2023-50 | longest | en | 0.885861 |
https://www.unitsconverters.com/en/Ac-To-Varas-Castellanas-Cuad/Utu-308-339 | 1,620,320,218,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988758.74/warc/CC-MAIN-20210506144716-20210506174716-00190.warc.gz | 1,069,229,923 | 36,232 | Formula Used
1 Acre = 4046.85611888692 Square Meter
1 Square Meter = 1.4311536386 Varas Castellanas Cuad
1 Acre = 5791.67285943569 Varas Castellanas Cuad
## ac to Varas Castellanas Cuad Conversion
The abbreviation for ac and Varas Castellanas Cuad is acre and varas castellanas cuad respectively. 1 ac is 5792 times bigger than a Varas Castellanas Cuad. To measure, units of measurement are needed and converting such units is an important task as well. unitsconverters.com is an online conversion tool to convert all types of measurement units including ac to Varas Castellanas Cuad conversion.
## Acre to Varas Castellanas Cuad
Check our Acre to Varas Castellanas Cuad converter and click on formula to get the conversion factor. When you are converting area from Acre to Varas Castellanas Cuad, you need a converter that is elaborate and still easy to use. All you have to do is select the unit for which you want the conversion and enter the value and finally hit Convert.
## ac to Varas Castellanas Cuad
The formula used to convert ac to Varas Castellanas Cuad is 1 Acre = 5791.67285943569 Varas Castellanas Cuad. Measurement is one of the most fundamental concepts. Note that we have Fahrenheit as the biggest unit for length while Per Degree Celsius is the smallest one.
## Convert ac to Varas Castellanas Cuad
How to convert ac to Varas Castellanas Cuad? Now you can do ac to Varas Castellanas Cuad conversion with the help of this tool. In the length measurement, first choose ac from the left dropdown and Varas Castellanas Cuad from the right dropdown, enter the value you want to convert and click on 'convert'. Want a reverse calculation from Varas Castellanas Cuad to ac? You can check our Varas Castellanas Cuad to ac converter.
How to convert ac to Varas Castellanas Cuad?
The formula to convert ac to Varas Castellanas Cuad is 1 Acre = 5791.67285943569 Varas Castellanas Cuad. ac is 5791.6729 times Bigger than Varas Castellanas Cuad. Enter the value of ac and hit Convert to get value in Varas Castellanas Cuad. Check our ac to Varas Castellanas Cuad converter. Need a reverse calculation from Varas Castellanas Cuad to ac? You can check our Varas Castellanas Cuad to ac Converter.
How many m² is 1 ac?
1 ac is equal to 4046.85611888692 m². 1 ac is 4046.85611888692 times Bigger than 1 m².
How many km² is 1 ac?
1 ac is equal to 0.00404685611888692 km². 1 ac is 247.1054 times Smaller than 1 km².
How many cm² is 1 ac?
1 ac is equal to 40468561.1888692 cm². 1 ac is 40468561.1888692 times Bigger than 1 cm².
How many mm² is 1 ac?
1 ac is equal to 4046856118.88692 mm². 1 ac is 4046856118.88692 times Bigger than 1 mm².
## ac to Varas Castellanas Cuad Converter
Units of measurement use the International System of Units, better known as SI units, which provide a standard for measuring the physical properties of matter. Measurement like area finds its use in a number of places right from education to industrial usage. Be it buying grocery or cooking, units play a vital role in our daily life; and hence their conversions. unitsconverters.com helps in the conversion of different units of measurement like ac to Varas Castellanas Cuad through multiplicative conversion factors. When you are converting area, you need a Acre to Varas Castellanas Cuad converter that is elaborate and still easy to use. Converting ac to Varas Castellanas Cuad is easy, for you only have to select the units first and the value you want to convert. If you encounter any issues to convert Acre to Varas Castellanas Cuad, this tool is the answer that gives you the exact conversion of units. You can also get the formula used in ac to Varas Castellanas Cuad conversion along with a table representing the entire conversion.
Let Others Know | 983 | 3,750 | {"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-2021-21 | latest | en | 0.724706 |
http://www.nag.com/numeric/FL/nagdoc_fl24/examples/source/g01abfe.f90 | 1,386,460,730,000,000,000 | text/plain | crawl-data/CC-MAIN-2013-48/segments/1386163056120/warc/CC-MAIN-20131204131736-00086-ip-10-33-133-15.ec2.internal.warc.gz | 464,929,886 | 1,449 | Program g01abfe ! G01ABF Example Program Text ! Mark 24 Release. NAG Copyright 2012. ! .. Use Statements .. Use nag_library, Only: g01abf, nag_wp ! .. Implicit None Statement .. Implicit None ! .. Parameters .. Integer, Parameter :: nin = 5, nout = 6 ! .. Local Scalars .. Integer :: i, ifail, iwt, n ! .. Local Arrays .. Real (Kind=nag_wp) :: res(13) Real (Kind=nag_wp), Allocatable :: wt(:), wtin(:), x1(:), x2(:) ! .. Executable Statements .. Write (nout,*) 'G01ABF Example Program Results' Write (nout,*) ! Skip heading in data file Read (nin,*) ! Read in the problem size Read (nin,*) n, iwt Allocate (wt(n),wtin(n),x1(n),x2(n)) ! Read in data Read (nin,*)(x1(i),x2(i),i=1,n) If (iwt==1) Then Read (nin,*) wtin(1:n) wt(1:n) = wtin(1:n) End If ! Display data Write (nout,99999) 'Number of cases', n Write (nout,*) 'Data as input -' Write (nout,*) ' Var 1 Var 2 Var 1 & & Var 2 Var 1 Var 2' Write (nout,99995)(x1(i),x2(i),i=1,n) If (iwt==1) Then Write (nout,*) 'Weights as input -' Write (nout,99994) wtin(1:n) End If Write (nout,*) ! Calculate summary statistics ifail = -1 Call g01abf(n,x1,x2,iwt,wt,res,ifail) If (ifail/=0) Then If (ifail/=2) Then Go To 100 End If End If ! Display results Write (nout,99999) 'No. of valid cases', iwt Write (nout,99993) 'Variable 1', 'Variable 2' Write (nout,99998) 'Mean ', res(1), res(2) Write (nout,99997) 'Corr SSP', res(5), res(6), res(7) Write (nout,99998) 'Minimum ', res(9), res(11) Write (nout,99998) 'Maximum ', res(10), res(12) Write (nout,99998) 'Sum of weights ', res(13) If (ifail==0) Then Write (nout,99998) 'Std devn', res(3), res(4) Write (nout,99996) 'Correln ', res(8) Else Write (nout,*) 'Std devn and Correln not defined' End If 100 Continue 99999 Format (1X,A,I5) 99998 Format (1X,A,F15.1,F30.1) 99997 Format (1X,A,3E15.5) 99996 Format (1X,A,F30.4) 99995 Format (5X,6F11.1) 99994 Format (13X,F9.3) 99993 Format (13X,A,20X,A) End Program g01abfe | 714 | 1,906 | {"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} | 2.703125 | 3 | CC-MAIN-2013-48 | longest | en | 0.484292 |
http://nrich.maths.org/public/leg.php?code=5027&cl=2&cldcmpid=7779 | 1,484,942,826,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560280872.69/warc/CC-MAIN-20170116095120-00537-ip-10-171-10-70.ec2.internal.warc.gz | 202,628,133 | 6,401 | # Search by Topic
#### Resources tagged with Comparing data similar to Our Sports:
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Broad Topics > Handling, Processing and Representing Data > Comparing data
### Our Sports
##### Stage: 2 Challenge Level:
This problem explores the range of events in a sports day and which ones are the most popular and attract the most entries.
### Real Statistics
##### Stage: 2 Challenge Level:
Have a look at this table of how children travel to school. How does it compare with children in your class?
### Compare the Squares
##### Stage: 2 Challenge Level:
In this problem you will do your own poll to find out whether your friends think two squares on a board are the same colour or not.
### Terrariums
##### Stage: 3 Challenge Level:
Build a mini eco-system, and collect and interpret data on how well the plants grow under different conditions.
### It's a Scrabble
##### Stage: 2 Challenge Level:
Letters have different values in Scrabble - how are they decided upon? And would the values be the same for other languages?
### Inspector Morse
##### Stage: 3 Challenge Level:
You may like to read the article on Morse code before attempting this question. Morse's letter analysis was done over 150 years ago, so might there be a better allocation of symbols today?
### Which List Is Which?
##### Stage: 3 and 4 Challenge Level:
Six samples were taken from two distributions but they got muddled up. Can you work out which list is which?
### Substitution Transposed
##### Stage: 3 and 4 Challenge Level:
Substitution and Transposition all in one! How fiendish can these codes get?
##### Stage: 3 Challenge Level:
When Charlie retires, he's looking forward to the quiet life, whereas Alison wants a busy and exciting retirement. Can you advise them on where they should go?
### Olympic Records
##### Stage: 3 Challenge Level:
Can you deduce which Olympic athletics events are represented by the graphs?
### Reaction Timer
##### Stage: 3 Challenge Level:
This problem offers you two ways to test reactions - use them to investigate your ideas about speeds of reaction.
### Substitution Cipher
##### Stage: 3 Challenge Level:
Find the frequency distribution for ordinary English, and use it to help you crack the code.
### Secondary Cipher Challenge Part 1
##### Stage: 3 and 4 Challenge Level:
Here is the start of a six-part challenge. Can you get to the end and crack the final message?
### Warmsnug Double Glazing
##### Stage: 3 Challenge Level:
How have "Warmsnug" arrived at the prices shown on their windows? Which window has been given an incorrect price?
### Decathlon: the Art of Scoring Points
##### Stage: 3, 4 and 5
How do decisions about scoring affect who wins a combined event such as the decathlon?
### The Time Is ...
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Can you put these mixed-up times in order? You could arrange them in a circle. | 649 | 2,957 | {"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} | 3.5625 | 4 | CC-MAIN-2017-04 | longest | en | 0.894103 |
https://metric-calculator.com/convert-cup-to-gal.htm | 1,718,231,856,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861261.53/warc/CC-MAIN-20240612203157-20240612233157-00041.warc.gz | 368,905,761 | 5,489 | # Cups to Gallons Converter
Select conversion type:
Rounding options:
Convert Gallons to Cups (gal to cup) ▶
## Conversion Table
cups to gallons cup gal 10 cup 0.625 gal 20 cup 1.25 gal 30 cup 1.875 gal 40 cup 2.5 gal 50 cup 3.125 gal 60 cup 3.75 gal 70 cup 4.375 gal 80 cup 5 gal 90 cup 5.625 gal 100 cup 6.25 gal 110 cup 6.875 gal 120 cup 7.5 gal 130 cup 8.125 gal 140 cup 8.75 gal 150 cup 9.375 gal 160 cup 10 gal 170 cup 10.625 gal 180 cup 11.25 gal 190 cup 11.875 gal 200 cup 12.5 gal
## How to convert
1 cup (cup) = 0.0625 gallon (gal). Cup (cup) is a unit of Volume used in Cooking system. Gallon (gal) is a unit of Volume used in Standard system.
## Cups: A Unit of Volume
Cups are a unit of volume that are used to measure liquids, such as water, milk, oil, vinegar, etc. They are also used to measure some dry ingredients, such as sugar, flour, rice, etc. They are different from tablespoons and teaspoons, which are smaller units of volume. They are also different from quarts and gallons, which are larger units of volume. They are also different from barrel of oil equivalent (BOE), which is a unit of energy based on the approximate energy released by burning one barrel of crude oil.
## How to Convert Cups
To convert cups to other units of volume, one can use the following formulas:
• To convert cups to milliliters: multiply by 250
• To convert cups to fluid ounces: multiply by 8.45
• To convert cups to tablespoons: multiply by 16
• To convert cups to quarts: multiply by 0.25
• To convert cups to gallons: multiply by 0.0625
• To convert cups to BOE: divide by 1,200
US customary cup can be abbreviated as c., = 236.5882365 millilitres = 1/16 U.S. customary gallon = 1/4 U.S. customary quart
## Gallons: A Unit of Volume
Gallons are a unit of volume that are used to measure liquids, such as water, milk, oil, wine, etc. They are also used to measure some dry goods, such as grains, fruits, nuts, etc. They are different from cups, which are a smaller unit of volume. They are also different from liters, which are a larger unit of volume. They are also different from barrel of oil equivalent (BOE), which is a unit of energy based on the approximate energy released by burning one barrel of crude oil.
## How to Convert Gallons
To convert gallons to other units of volume, one can use the following formulas:
• To convert US liquid gallons to liters: multiply by 3.785
• To convert US liquid gallons to cubic inches: multiply by 231
• To convert US liquid gallons to fluid ounces: multiply by 128
• To convert US liquid gallons to UK gallons: multiply by 0.833
• To convert US liquid gallons to BOE: divide by 5
• To convert US dry gallons to liters: multiply by 4.405
• To convert US dry gallons to cubic inches: multiply by 268.8
• To convert US dry gallons to fluid ounces: multiply by 148.9
• To convert US dry gallons to UK gallons: multiply by 0.969
• To convert US dry gallons to BOE: divide by 4.6
• To convert UK gallons to liters: multiply by 4.546
• To convert UK gallons to cubic inches: multiply by 277.4
• To convert UK gallons to fluid ounces: multiply by 160
• To convert UK gallons to US liquid gallons: multiply by 1.2
• To convert UK gallons to BOE: divide by 6.1
The US gallon is equal to 3.785411784 liters and defined as 231 cubic inches.
Español Russian Français | 921 | 3,350 | {"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-2024-26 | latest | en | 0.916095 |
https://us.metamath.org/mpeuni/cevathlem2.html | 1,721,320,693,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514848.78/warc/CC-MAIN-20240718161144-20240718191144-00273.warc.gz | 525,926,499 | 8,991 | Mathbox for Saveliy Skresanov < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > Mathboxes > cevathlem2 Structured version Visualization version GIF version
Theorem cevathlem2 43475
Description: Ceva's theorem second lemma. Relate (doubled) areas of triangles 𝐶𝐴𝑂 and 𝐴𝐵𝑂 with of segments 𝐵𝐷 and 𝐷𝐶. (Contributed by Saveliy Skresanov, 24-Sep-2017.)
Hypotheses
Ref Expression
cevath.sigar 𝐺 = (𝑥 ∈ ℂ, 𝑦 ∈ ℂ ↦ (ℑ‘((∗‘𝑥) · 𝑦)))
cevath.a (𝜑 → (𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ))
cevath.b (𝜑 → (𝐹 ∈ ℂ ∧ 𝐷 ∈ ℂ ∧ 𝐸 ∈ ℂ))
cevath.c (𝜑𝑂 ∈ ℂ)
cevath.d (𝜑 → (((𝐴𝑂)𝐺(𝐷𝑂)) = 0 ∧ ((𝐵𝑂)𝐺(𝐸𝑂)) = 0 ∧ ((𝐶𝑂)𝐺(𝐹𝑂)) = 0))
cevath.e (𝜑 → (((𝐴𝐹)𝐺(𝐵𝐹)) = 0 ∧ ((𝐵𝐷)𝐺(𝐶𝐷)) = 0 ∧ ((𝐶𝐸)𝐺(𝐴𝐸)) = 0))
cevath.f (𝜑 → (((𝐴𝑂)𝐺(𝐵𝑂)) ≠ 0 ∧ ((𝐵𝑂)𝐺(𝐶𝑂)) ≠ 0 ∧ ((𝐶𝑂)𝐺(𝐴𝑂)) ≠ 0))
Assertion
Ref Expression
cevathlem2 (𝜑 → (((𝐶𝑂)𝐺(𝐴𝑂)) · (𝐵𝐷)) = (((𝐴𝑂)𝐺(𝐵𝑂)) · (𝐷𝐶)))
Distinct variable groups: 𝑥,𝑦,𝐴 𝑥,𝐵,𝑦 𝑥,𝐶,𝑦 𝑥,𝐷,𝑦 𝑥,𝑂,𝑦 𝑥,𝐸,𝑦 𝑥,𝐹,𝑦
Allowed substitution hints: 𝜑(𝑥,𝑦) 𝐺(𝑥,𝑦)
Proof of Theorem cevathlem2
StepHypRef Expression
1 cevath.sigar . . . . . . 7 𝐺 = (𝑥 ∈ ℂ, 𝑦 ∈ ℂ ↦ (ℑ‘((∗‘𝑥) · 𝑦)))
2 cevath.b . . . . . . . . 9 (𝜑 → (𝐹 ∈ ℂ ∧ 𝐷 ∈ ℂ ∧ 𝐸 ∈ ℂ))
32simp2d 1140 . . . . . . . 8 (𝜑𝐷 ∈ ℂ)
4 cevath.a . . . . . . . . 9 (𝜑 → (𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ))
54simp1d 1139 . . . . . . . 8 (𝜑𝐴 ∈ ℂ)
64simp2d 1140 . . . . . . . 8 (𝜑𝐵 ∈ ℂ)
73, 5, 63jca 1125 . . . . . . 7 (𝜑 → (𝐷 ∈ ℂ ∧ 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ))
8 cevath.c . . . . . . . 8 (𝜑𝑂 ∈ ℂ)
95, 8subcld 10990 . . . . . . . . . 10 (𝜑 → (𝐴𝑂) ∈ ℂ)
103, 8subcld 10990 . . . . . . . . . 10 (𝜑 → (𝐷𝑂) ∈ ℂ)
119, 10jca 515 . . . . . . . . 9 (𝜑 → ((𝐴𝑂) ∈ ℂ ∧ (𝐷𝑂) ∈ ℂ))
12 cevath.d . . . . . . . . . 10 (𝜑 → (((𝐴𝑂)𝐺(𝐷𝑂)) = 0 ∧ ((𝐵𝑂)𝐺(𝐸𝑂)) = 0 ∧ ((𝐶𝑂)𝐺(𝐹𝑂)) = 0))
1312simp1d 1139 . . . . . . . . 9 (𝜑 → ((𝐴𝑂)𝐺(𝐷𝑂)) = 0)
141, 11, 13sigariz 43470 . . . . . . . 8 (𝜑 → ((𝐷𝑂)𝐺(𝐴𝑂)) = 0)
158, 14jca 515 . . . . . . 7 (𝜑 → (𝑂 ∈ ℂ ∧ ((𝐷𝑂)𝐺(𝐴𝑂)) = 0))
161, 7, 15sigaradd 43473 . . . . . 6 (𝜑 → (((𝐴𝐵)𝐺(𝐷𝐵)) − ((𝑂𝐵)𝐺(𝐷𝐵))) = ((𝐴𝐵)𝐺(𝑂𝐵)))
171sigarperm 43467 . . . . . . 7 ((𝐵 ∈ ℂ ∧ 𝐴 ∈ ℂ ∧ 𝑂 ∈ ℂ) → ((𝐵𝑂)𝐺(𝐴𝑂)) = ((𝐴𝐵)𝐺(𝑂𝐵)))
186, 5, 8, 17syl3anc 1368 . . . . . 6 (𝜑 → ((𝐵𝑂)𝐺(𝐴𝑂)) = ((𝐴𝐵)𝐺(𝑂𝐵)))
1916, 18eqtr4d 2839 . . . . 5 (𝜑 → (((𝐴𝐵)𝐺(𝐷𝐵)) − ((𝑂𝐵)𝐺(𝐷𝐵))) = ((𝐵𝑂)𝐺(𝐴𝑂)))
2019oveq1d 7154 . . . 4 (𝜑 → ((((𝐴𝐵)𝐺(𝐷𝐵)) − ((𝑂𝐵)𝐺(𝐷𝐵))) · (𝐶𝐷)) = (((𝐵𝑂)𝐺(𝐴𝑂)) · (𝐶𝐷)))
215, 6subcld 10990 . . . . . . 7 (𝜑 → (𝐴𝐵) ∈ ℂ)
223, 6subcld 10990 . . . . . . 7 (𝜑 → (𝐷𝐵) ∈ ℂ)
2321, 22jca 515 . . . . . 6 (𝜑 → ((𝐴𝐵) ∈ ℂ ∧ (𝐷𝐵) ∈ ℂ))
241, 23sigarimcd 43469 . . . . 5 (𝜑 → ((𝐴𝐵)𝐺(𝐷𝐵)) ∈ ℂ)
258, 6subcld 10990 . . . . . . 7 (𝜑 → (𝑂𝐵) ∈ ℂ)
2625, 22jca 515 . . . . . 6 (𝜑 → ((𝑂𝐵) ∈ ℂ ∧ (𝐷𝐵) ∈ ℂ))
271, 26sigarimcd 43469 . . . . 5 (𝜑 → ((𝑂𝐵)𝐺(𝐷𝐵)) ∈ ℂ)
284simp3d 1141 . . . . . 6 (𝜑𝐶 ∈ ℂ)
2928, 3subcld 10990 . . . . 5 (𝜑 → (𝐶𝐷) ∈ ℂ)
3024, 27, 29subdird 11090 . . . 4 (𝜑 → ((((𝐴𝐵)𝐺(𝐷𝐵)) − ((𝑂𝐵)𝐺(𝐷𝐵))) · (𝐶𝐷)) = ((((𝐴𝐵)𝐺(𝐷𝐵)) · (𝐶𝐷)) − (((𝑂𝐵)𝐺(𝐷𝐵)) · (𝐶𝐷))))
3120, 30eqtr3d 2838 . . 3 (𝜑 → (((𝐵𝑂)𝐺(𝐴𝑂)) · (𝐶𝐷)) = ((((𝐴𝐵)𝐺(𝐷𝐵)) · (𝐶𝐷)) − (((𝑂𝐵)𝐺(𝐷𝐵)) · (𝐶𝐷))))
326, 28, 53jca 1125 . . . . 5 (𝜑 → (𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ∧ 𝐴 ∈ ℂ))
33 cevath.e . . . . . . 7 (𝜑 → (((𝐴𝐹)𝐺(𝐵𝐹)) = 0 ∧ ((𝐵𝐷)𝐺(𝐶𝐷)) = 0 ∧ ((𝐶𝐸)𝐺(𝐴𝐸)) = 0))
3433simp2d 1140 . . . . . 6 (𝜑 → ((𝐵𝐷)𝐺(𝐶𝐷)) = 0)
353, 34jca 515 . . . . 5 (𝜑 → (𝐷 ∈ ℂ ∧ ((𝐵𝐷)𝐺(𝐶𝐷)) = 0))
361, 32, 35sharhght 43472 . . . 4 (𝜑 → (((𝐴𝐵)𝐺(𝐷𝐵)) · (𝐶𝐷)) = (((𝐴𝐶)𝐺(𝐷𝐶)) · (𝐵𝐷)))
376, 28, 83jca 1125 . . . . 5 (𝜑 → (𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ∧ 𝑂 ∈ ℂ))
381, 37, 35sharhght 43472 . . . 4 (𝜑 → (((𝑂𝐵)𝐺(𝐷𝐵)) · (𝐶𝐷)) = (((𝑂𝐶)𝐺(𝐷𝐶)) · (𝐵𝐷)))
3936, 38oveq12d 7157 . . 3 (𝜑 → ((((𝐴𝐵)𝐺(𝐷𝐵)) · (𝐶𝐷)) − (((𝑂𝐵)𝐺(𝐷𝐵)) · (𝐶𝐷))) = ((((𝐴𝐶)𝐺(𝐷𝐶)) · (𝐵𝐷)) − (((𝑂𝐶)𝐺(𝐷𝐶)) · (𝐵𝐷))))
405, 28subcld 10990 . . . . . . 7 (𝜑 → (𝐴𝐶) ∈ ℂ)
413, 28subcld 10990 . . . . . . 7 (𝜑 → (𝐷𝐶) ∈ ℂ)
421sigarim 43458 . . . . . . 7 (((𝐴𝐶) ∈ ℂ ∧ (𝐷𝐶) ∈ ℂ) → ((𝐴𝐶)𝐺(𝐷𝐶)) ∈ ℝ)
4340, 41, 42syl2anc 587 . . . . . 6 (𝜑 → ((𝐴𝐶)𝐺(𝐷𝐶)) ∈ ℝ)
4443recnd 10662 . . . . 5 (𝜑 → ((𝐴𝐶)𝐺(𝐷𝐶)) ∈ ℂ)
458, 28subcld 10990 . . . . . . 7 (𝜑 → (𝑂𝐶) ∈ ℂ)
4645, 41jca 515 . . . . . 6 (𝜑 → ((𝑂𝐶) ∈ ℂ ∧ (𝐷𝐶) ∈ ℂ))
471, 46sigarimcd 43469 . . . . 5 (𝜑 → ((𝑂𝐶)𝐺(𝐷𝐶)) ∈ ℂ)
486, 3subcld 10990 . . . . 5 (𝜑 → (𝐵𝐷) ∈ ℂ)
4944, 47, 48subdird 11090 . . . 4 (𝜑 → ((((𝐴𝐶)𝐺(𝐷𝐶)) − ((𝑂𝐶)𝐺(𝐷𝐶))) · (𝐵𝐷)) = ((((𝐴𝐶)𝐺(𝐷𝐶)) · (𝐵𝐷)) − (((𝑂𝐶)𝐺(𝐷𝐶)) · (𝐵𝐷))))
503, 5, 283jca 1125 . . . . . . 7 (𝜑 → (𝐷 ∈ ℂ ∧ 𝐴 ∈ ℂ ∧ 𝐶 ∈ ℂ))
511, 50, 15sigaradd 43473 . . . . . 6 (𝜑 → (((𝐴𝐶)𝐺(𝐷𝐶)) − ((𝑂𝐶)𝐺(𝐷𝐶))) = ((𝐴𝐶)𝐺(𝑂𝐶)))
521sigarperm 43467 . . . . . . 7 ((𝐶 ∈ ℂ ∧ 𝐴 ∈ ℂ ∧ 𝑂 ∈ ℂ) → ((𝐶𝑂)𝐺(𝐴𝑂)) = ((𝐴𝐶)𝐺(𝑂𝐶)))
5328, 5, 8, 52syl3anc 1368 . . . . . 6 (𝜑 → ((𝐶𝑂)𝐺(𝐴𝑂)) = ((𝐴𝐶)𝐺(𝑂𝐶)))
5451, 53eqtr4d 2839 . . . . 5 (𝜑 → (((𝐴𝐶)𝐺(𝐷𝐶)) − ((𝑂𝐶)𝐺(𝐷𝐶))) = ((𝐶𝑂)𝐺(𝐴𝑂)))
5554oveq1d 7154 . . . 4 (𝜑 → ((((𝐴𝐶)𝐺(𝐷𝐶)) − ((𝑂𝐶)𝐺(𝐷𝐶))) · (𝐵𝐷)) = (((𝐶𝑂)𝐺(𝐴𝑂)) · (𝐵𝐷)))
5649, 55eqtr3d 2838 . . 3 (𝜑 → ((((𝐴𝐶)𝐺(𝐷𝐶)) · (𝐵𝐷)) − (((𝑂𝐶)𝐺(𝐷𝐶)) · (𝐵𝐷))) = (((𝐶𝑂)𝐺(𝐴𝑂)) · (𝐵𝐷)))
5731, 39, 563eqtrrd 2841 . 2 (𝜑 → (((𝐶𝑂)𝐺(𝐴𝑂)) · (𝐵𝐷)) = (((𝐵𝑂)𝐺(𝐴𝑂)) · (𝐶𝐷)))
586, 8subcld 10990 . . . 4 (𝜑 → (𝐵𝑂) ∈ ℂ)
591sigarac 43459 . . . 4 (((𝐵𝑂) ∈ ℂ ∧ (𝐴𝑂) ∈ ℂ) → ((𝐵𝑂)𝐺(𝐴𝑂)) = -((𝐴𝑂)𝐺(𝐵𝑂)))
6058, 9, 59syl2anc 587 . . 3 (𝜑 → ((𝐵𝑂)𝐺(𝐴𝑂)) = -((𝐴𝑂)𝐺(𝐵𝑂)))
6160oveq1d 7154 . 2 (𝜑 → (((𝐵𝑂)𝐺(𝐴𝑂)) · (𝐶𝐷)) = (-((𝐴𝑂)𝐺(𝐵𝑂)) · (𝐶𝐷)))
629, 58jca 515 . . . . 5 (𝜑 → ((𝐴𝑂) ∈ ℂ ∧ (𝐵𝑂) ∈ ℂ))
631, 62sigarimcd 43469 . . . 4 (𝜑 → ((𝐴𝑂)𝐺(𝐵𝑂)) ∈ ℂ)
64 mulneg12 11071 . . . 4 ((((𝐴𝑂)𝐺(𝐵𝑂)) ∈ ℂ ∧ (𝐶𝐷) ∈ ℂ) → (-((𝐴𝑂)𝐺(𝐵𝑂)) · (𝐶𝐷)) = (((𝐴𝑂)𝐺(𝐵𝑂)) · -(𝐶𝐷)))
6563, 29, 64syl2anc 587 . . 3 (𝜑 → (-((𝐴𝑂)𝐺(𝐵𝑂)) · (𝐶𝐷)) = (((𝐴𝑂)𝐺(𝐵𝑂)) · -(𝐶𝐷)))
6628, 3negsubdi2d 11006 . . . 4 (𝜑 → -(𝐶𝐷) = (𝐷𝐶))
6766oveq2d 7155 . . 3 (𝜑 → (((𝐴𝑂)𝐺(𝐵𝑂)) · -(𝐶𝐷)) = (((𝐴𝑂)𝐺(𝐵𝑂)) · (𝐷𝐶)))
6865, 67eqtrd 2836 . 2 (𝜑 → (-((𝐴𝑂)𝐺(𝐵𝑂)) · (𝐶𝐷)) = (((𝐴𝑂)𝐺(𝐵𝑂)) · (𝐷𝐶)))
6957, 61, 683eqtrd 2840 1 (𝜑 → (((𝐶𝑂)𝐺(𝐴𝑂)) · (𝐵𝐷)) = (((𝐴𝑂)𝐺(𝐵𝑂)) · (𝐷𝐶)))
Colors of variables: wff setvar class Syntax hints: → wi 4 ∧ w3a 1084 = wceq 1538 ∈ wcel 2112 ≠ wne 2990 ‘cfv 6328 (class class class)co 7139 ∈ cmpo 7141 ℂcc 10528 ℝcr 10529 0cc0 10530 · cmul 10535 − cmin 10863 -cneg 10864 ∗ccj 14451 ℑcim 14453 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1797 ax-4 1811 ax-5 1911 ax-6 1970 ax-7 2015 ax-8 2114 ax-9 2122 ax-10 2143 ax-11 2159 ax-12 2176 ax-ext 2773 ax-sep 5170 ax-nul 5177 ax-pow 5234 ax-pr 5298 ax-un 7445 ax-resscn 10587 ax-1cn 10588 ax-icn 10589 ax-addcl 10590 ax-addrcl 10591 ax-mulcl 10592 ax-mulrcl 10593 ax-mulcom 10594 ax-addass 10595 ax-mulass 10596 ax-distr 10597 ax-i2m1 10598 ax-1ne0 10599 ax-1rid 10600 ax-rnegex 10601 ax-rrecex 10602 ax-cnre 10603 ax-pre-lttri 10604 ax-pre-lttrn 10605 ax-pre-ltadd 10606 ax-pre-mulgt0 10607 This theorem depends on definitions: df-bi 210 df-an 400 df-or 845 df-3or 1085 df-3an 1086 df-tru 1541 df-ex 1782 df-nf 1786 df-sb 2070 df-mo 2601 df-eu 2632 df-clab 2780 df-cleq 2794 df-clel 2873 df-nfc 2941 df-ne 2991 df-nel 3095 df-ral 3114 df-rex 3115 df-reu 3116 df-rmo 3117 df-rab 3118 df-v 3446 df-sbc 3724 df-csb 3832 df-dif 3887 df-un 3889 df-in 3891 df-ss 3901 df-nul 4247 df-if 4429 df-pw 4502 df-sn 4529 df-pr 4531 df-op 4535 df-uni 4804 df-br 5034 df-opab 5096 df-mpt 5114 df-id 5428 df-po 5442 df-so 5443 df-xp 5529 df-rel 5530 df-cnv 5531 df-co 5532 df-dm 5533 df-rn 5534 df-res 5535 df-ima 5536 df-iota 6287 df-fun 6330 df-fn 6331 df-f 6332 df-f1 6333 df-fo 6334 df-f1o 6335 df-fv 6336 df-riota 7097 df-ov 7142 df-oprab 7143 df-mpo 7144 df-er 8276 df-en 8497 df-dom 8498 df-sdom 8499 df-pnf 10670 df-mnf 10671 df-xr 10672 df-ltxr 10673 df-le 10674 df-sub 10865 df-neg 10866 df-div 11291 df-2 11692 df-cj 14454 df-re 14455 df-im 14456 This theorem is referenced by: cevath 43476
Copyright terms: Public domain W3C validator | 5,990 | 7,652 | {"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} | 3.203125 | 3 | CC-MAIN-2024-30 | latest | en | 0.193599 |
https://www.media4math.com/NY-5.NBT.3a | 1,685,587,108,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224647525.11/warc/CC-MAIN-20230601010402-20230601040402-00730.warc.gz | 975,571,669 | 11,887 | ## NY-5.NBT.3a: Read and write decimals to thousandths using base-ten numerals, number names, and expanded form, e.g., 347.392 = 3 × 100 + 4 × 10 + 7 × 1 + 3 × (1/10) + 9 × (1/100) + 2 × (1/1000).
There are 14 resources.
Title Description Thumbnail Image Curriculum Topics
## Math Example--Numbers--Writing Four-Digit Numbers in Expanded Form--Example 1
Math Example--Numbers--Writing Four-Digit Numbers in Expanded Form--Example 1
This is part of a collection of math examples that focus on numbers and their properties.
Place Value
## Math Example--Numbers--Writing Four-Digit Numbers in Expanded Form--Example 2
Math Example--Numbers--Writing Four-Digit Numbers in Expanded Form--Example 2
This is part of a collection of math examples that focus on numbers and their properties.
Place Value
## Math Example--Numbers--Writing Four-Digit Numbers in Expanded Form--Example 3
Math Example--Numbers--Writing Four-Digit Numbers in Expanded Form--Example 3
This is part of a collection of math examples that focus on numbers and their properties.
Place Value
## Math Example--Numbers--Writing Four-Digit Numbers in Expanded Form--Example 4
Math Example--Numbers--Writing Four-Digit Numbers in Expanded Form--Example 4
This is part of a collection of math examples that focus on numbers and their properties.
Place Value
## Math Example--Numbers--Writing Four-Digit Numbers in Expanded Form--Example 5
Math Example--Numbers--Writing Four-Digit Numbers in Expanded Form--Example 5
This is part of a collection of math examples that focus on numbers and their properties.
Place Value
## Math Example--Numbers--Writing Four-Digit Numbers in Expanded Form--Example 6
Math Example--Numbers--Writing Four-Digit Numbers in Expanded Form--Example 6
This is part of a collection of math examples that focus on numbers and their properties.
Place Value
## Math Example--Numbers--Writing Four-Digit Numbers in Expanded Form--Example 7
Math Example--Numbers--Writing Four-Digit Numbers in Expanded Form--Example 7
This is part of a collection of math examples that focus on numbers and their properties.
Place Value
## Math Example--Numbers--Writing Four-Digit Numbers in Expanded Form--Example 8
Math Example--Numbers--Writing Four-Digit Numbers in Expanded Form--Example 8
This is part of a collection of math examples that focus on numbers and their properties.
Place Value
## Math Example--Numbers--Writing Four-Digit Numbers in Expanded Form--Example 9
Math Example--Numbers--Writing Four-Digit Numbers in Expanded Form--Example 9
This is part of a collection of math examples that focus on numbers and their properties.
Place Value
## Worksheet: Working with Decimal Powers of 10, Worksheet 1
Worksheet: Working with Decimal Powers of 10, Worksheet 1
This is part of a collection of math worksheets on the topic of decimal powers of ten.
Multiply Decimals
## Worksheet: Working with Decimal Powers of 10, Worksheet 2
Worksheet: Working with Decimal Powers of 10, Worksheet 2
This is part of a collection of math worksheets on the topic of decimal powers of ten.
Multiply Decimals
## Worksheet: Working with Decimal Powers of 10, Worksheet 3
Worksheet: Working with Decimal Powers of 10, Worksheet 3
This is part of a collection of math worksheets on the topic of decimal powers of ten.
Multiply Decimals
## Worksheet: Working with Decimal Powers of 10, Worksheet 4
Worksheet: Working with Decimal Powers of 10, Worksheet 4
This is part of a collection of math worksheets on the topic of decimal powers of ten.
Multiply Decimals
## Worksheet: Working with Decimal Powers of 10, Worksheet 5
Worksheet: Working with Decimal Powers of 10, Worksheet 5
This is part of a collection of math worksheets on the topic of decimal powers of ten.
Multiply Decimals | 865 | 3,798 | {"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} | 3.1875 | 3 | CC-MAIN-2023-23 | longest | en | 0.78167 |
https://www.studypool.com/questions/223984/how-to-do-parabola-equation | 1,480,898,947,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541517.94/warc/CC-MAIN-20161202170901-00125-ip-10-31-129-80.ec2.internal.warc.gz | 1,001,653,274 | 750,442 | # how to do parabola equation
Sigchi4life
Category:
Mathematics
Price: \$5 USD
Question description
1.Find a parabola with equation
y = ax2 + bx + c
that has slope
3
at x = 1, slope
−17
at x = −1, and passes through the point
(1, 1).
2.Where does the normal line to the parabola
y = x − x2
at the point
(1, 0)
intersect the parabola a second time?
(Top Tutor) Daniel C.
(997)
School: UC Berkeley
Studypool has helped 1,244,100 students
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https://nrich.maths.org/public/leg.php?code=-77&cl=2&cldcmpid=7513 | 1,569,140,111,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514575402.81/warc/CC-MAIN-20190922073800-20190922095800-00109.warc.gz | 603,284,148 | 8,247 | # Search by Topic
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Learn how to use the Excel functions LCM and GCD. | 1,674 | 7,764 | {"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} | 3.796875 | 4 | CC-MAIN-2019-39 | latest | en | 0.755793 |
http://www.algebra.com/algebra/homework/Probability-and-statistics/Probability-and-statistics.faq.question.320334.html | 1,369,440,148,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368705300740/warc/CC-MAIN-20130516115500-00057-ip-10-60-113-184.ec2.internal.warc.gz | 313,848,269 | 4,783 | # SOLUTION: A box contains 24 widgets, 4 of which are defective. If 4 are sold at random, find the probability that a) all are defective b) none are defective.
Algebra -> Algebra -> Probability-and-statistics -> SOLUTION: A box contains 24 widgets, 4 of which are defective. If 4 are sold at random, find the probability that a) all are defective b) none are defective. Log On
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Algebra: Probability and statistics Solvers Lessons Answers archive Quiz In Depth
Click here to see ALL problems on Probability-and-statistics Question 320334: A box contains 24 widgets, 4 of which are defective. If 4 are sold at random, find the probability that a) all are defective b) none are defective.Found 2 solutions by nyc_function, jrfrunner:Answer by nyc_function(2733) (Show Source): You can put this solution on YOUR website!Part (a): p(all are defective) = 4/24 = 1/6 ================================ Part (b): p(none of defective) = 1 - 4/24 p(none of defective) = 24/24 - 4/24 p(none of defective) = 20/24 = 5/6 Answer by jrfrunner(365) (Show Source): You can put this solution on YOUR website!a) all are defective. This means selecting four defectives in a row. The first one being defective is 4/24, the second being defective is 3/23 the third one being defective is 2/22 and the fourth one being defective is 1/21, therefore P(all defective) = P(1st defective)*P(2nd defective)*P(3rd defectie)*P(4th defective) = 4/24*3/23*2/22*1/21=24/255024 = 0.000094 b)one being defective This means 1 defective and 3 non-defective =4C1* P(D)*P(N)*P(N)*P(N)*P(N) since there are 4C1 ways to arrange that "one" defective. 4*4/24*20/23*19/22*18/21=109440/255024=0.429 | 530 | 1,848 | {"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-2013-20 | latest | en | 0.918403 |
https://15.fofoneontany.org/breadboard-circuits-schematics/ | 1,566,701,053,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027322170.99/warc/CC-MAIN-20190825021120-20190825043120-00310.warc.gz | 342,978,076 | 32,660 | In Wiring Diagram195 views
4.06 / 5 ( 183votes )
Breadboard Circuits Schematics Functional circuit sculptures have been gaining popularity what the full work expands on is a sizable breadboard mounted Whether a schematic is used to assemble a circuit by hand such as on a breadboard or used to produce a printed circuit Designed to help you easily expand your circuit playground projects offering a bolt on perma proto enabling you to solder.
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Author: Eva Pevasik
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Top | 1,441 | 7,226 | {"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-2019-35 | latest | en | 0.917082 |
https://www.unitconverters.net/length/break-to-hand.htm | 1,669,876,932,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710801.42/warc/CC-MAIN-20221201053355-20221201083355-00431.warc.gz | 1,109,699,673 | 4,067 | Home / Length Conversion / Convert Break to Hand
# Convert Break to Hand
Please provide values below to convert break to hand, or vice versa.
From: break To: hand
### Break to Hand Conversion Table
BreakHand
0.01 break9.8425196850394E+29 hand
0.1 break9.8425196850394E+30 hand
1 break9.8425196850394E+31 hand
2 break1.9685039370079E+32 hand
3 break2.9527559055118E+32 hand
5 break4.9212598425197E+32 hand
10 break9.8425196850394E+32 hand
20 break1.9685039370079E+33 hand
50 break4.9212598425197E+33 hand
100 break9.8425196850394E+33 hand
1000 break9.8425196850394E+34 hand
### How to Convert Break to Hand
1 break = 9.8425196850394E+31 hand
1 hand = 1.016E-32 break
Example: convert 15 break to hand:
15 break = 15 × 9.8425196850394E+31 hand = 1.4763779527559E+33 hand | 283 | 777 | {"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} | 2.96875 | 3 | CC-MAIN-2022-49 | latest | en | 0.450351 |
http://www.gradesaver.com/textbooks/science/physics/physics-principles-with-applications-7th-edition/chapter-3-kinematics-in-two-dimensions-vectors-problems-page-71/50 | 1,524,133,345,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125936833.6/warc/CC-MAIN-20180419091546-20180419111546-00349.warc.gz | 344,717,936 | 12,479 | Physics: Principles with Applications (7th Edition)
44.4$^o$ north of east.
Call east the positive x direction and north the positive y direction. Let P denote the Plane, A the Air, and G the Ground. The pair “PG”, for example, represents the plane’s motion relative to the ground. $$\vec{v_{PG}} = \vec{v_{PA}} + \vec{v_{AG}}$$ See the diagram. The vectors form a triangle. Apply the Law of Sines. $$\frac{ v_{AG}}{sin \alpha} = \frac{ v_{PA}}{sin 128 ^{\circ} }$$ We find the small angle $\alpha$ to be 6.4$^o$. Therefore the heading of the plane should be 38 + 6.4 = 44.4$^o$ north of east. | 187 | 594 | {"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} | 3.671875 | 4 | CC-MAIN-2018-17 | latest | en | 0.732705 |
http://octave.1599824.n4.nabble.com/DAE-daspk-solver-problem-td4680377.html | 1,558,842,518,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232258621.77/warc/CC-MAIN-20190526025014-20190526051014-00346.warc.gz | 149,183,068 | 21,293 | # DAE daspk solver problem
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## DAE daspk solver problem
Hi all, I'd like to use the daspk solver for DAE that comes with Octave but I can't find any official example on how to use it so I decided to give it a go myself. I've tried to solve the pendulum equations in cartesian coordinates but I'm having a hard time working out why the solver it is not working. the variables are as follows x(1)=position x x(2)=velocity x x(3)=position y x(4)=velocity y x(5)=reaction force intensity xdot(1)=velocity x xdot(2)=acceleration x xdot(4)=velocity y xdot(5)=velocity y and the script is t=linspace(0,5,1000)'; m=1; g=9.81; function f=effe(x,xdot, t) m=1; g=9.81; f(1) = xdot(1)-x(2); f(2) = m*xdot(2)+x(5)*x(1); f(3) = xdot(3)-x(4); f(4) = m*xdot(4) + x(5)*x(3) +m*g; f(5) = x(1)^2 + x(3)^2 - 1; endfunction daspk_options('print initial condition info', 1); daspk_options('compute consistent initial condition', 1); daspk_options('algebraic variables', [0,0,0,0,1]); [x, XDOT, ISTATE, MSG] = daspk (@effe, [1,0, 0, 0, m*g], [0,0,0,g,0], t); any help would be appreciated. Regards Marco
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## Re: DAE daspk solver problem
Lazy_Tom wrote f(1) = xdot(1)-x(2); f(2) = m*xdot(2)+x(5)*x(1); f(3) = xdot(3)-x(4); f(4) = m*xdot(4) + x(5)*x(3) +m*g; f(5) = x(1)^2 + x(3)^2 - 1; I think there are a few typos, e.g., it should be -x(5)*x(3) in eq (4) but this is probably not the issue here. You are trying to solve a very difficult problem. This formulation of the pendulum is a DAE of index 3 and afaik daspk can only deal with DAEs of index<3. There are many good books on this topic. If you are looking for a good free ressource, you may want to look at https://www.mathematik.hu-berlin.de/~steffen/pub/introduction_to_daes_497.pdf (page 2 and 13). Seb.
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## Re: DAE daspk solver problem
Sebastian Schöps wrote Lazy_Tom wrote f(1) = xdot(1)-x(2); f(2) = m*xdot(2)+x(5)*x(1); f(3) = xdot(3)-x(4); f(4) = m*xdot(4) + x(5)*x(3) +m*g; f(5) = x(1)^2 + x(3)^2 - 1; I think there are a few typos, e.g., it should be -x(5)*x(3) in eq (4) but this is probably not the issue here. You are trying to solve a very difficult problem. This formulation of the pendulum is a DAE of index 3 and afaik daspk can only deal with DAEs of index<3. There are many good books on this topic. If you are looking for a good free ressource, you may want to look at https://www.mathematik.hu-berlin.de/~steffen/pub/introduction_to_daes_497.pdf (page 2 and 13). Seb. Hi Seb, thanks for pointing out daspk max index, I wasn't aware. it spared a lot of headache. indeed, for some obscure reason I was convinced that the option "maximum order" was actually "maximum index"... -.- I don't know why DAE solvers index capabilities are not reported in Octave manual. Matlab at least tells you that you can go no further than index 1 (at least without doing some "magic"...) about the paper you cited, yup, I'm aware of it and I agree, it is quite nice. Well, at least I'll gave it a try, I think that for DAE problems I'll move to openModelica Cheers, Marco
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## Re: DAE daspk solver problem
Lazy_Tom wrote I don't know why DAE solvers index capabilities are not reported in Octave manual. Matlab at least tells you that you can go no further than index 1 (at least without doing some "magic"...) You are right. Feel free to oben a bug report at savannah, maybe immediately suggesting a string that can be added to the help. However, I assume that dassl and daspk will be deprecated in some future version since we are moving to odeXY compatible solvers. Lazy_Tom wrote about the paper you cited, yup, I'm aware of it and I agree, it is quite nice. Well, at least I'll gave it a try, I think that for DAE problems I'll move to openModelica I think that ode5r from odepkg should be able to solve index3 problems since its based on Radau 5 [1]. We hope to have a new release of odepkg quickly after Octave 4.2. I will keep in mind to include such information in the help. Regarding openModelica: yes, they do some clever tricks for higher index. As far as I recall it's based on dummy derivatives [2]. Such stuff is easier for modelica since it is in some sense a symbolic language. However, to my best knowledge, it is not really well suited for big problems. So, as usual, the best choice depends on what you want to do :) Seb. [1] http://www.unige.ch/~hairer/prog/stiff/radau5.f[2] http://dx.doi.org/10.1137/0914043
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## Re: DAE daspk solver problem
I was able to solve this using ode5r from odepkg. My test code is below:function radau5PendulumDAEvopt = odeset ('InitialStep', 1e-3, 'AbsTol', 1e-8);len=1; theta0=pi/2; m=1; g=9.81;y0=[ len*sin(theta0), -len*cos(theta0), 0, 0, m*g];odef = @(t,x) odeFunc (t, x, m, g, len)sol=ode5r (odef, [0, 5], y0, vopt);figure; plot(sol.x, sol.y(:,1));endfunction y = odeFunc (t, x, m, g, len)lam = x(5);y=[x(2); -lam*x(1); x(4); -lam*x(3)-m*g; x(1)^2 + x(3)^2 - 1];endOn Fri, Oct 28, 2016 at 2:18 AM, Sebastian Schöps wrote:Lazy_Tom wrote > I don't know why DAE solvers index capabilities are not reported in Octave > manual. Matlab at least tells you that you can go no further than index 1 > (at least without doing some "magic"...) You are right. Feel free to oben a bug report at savannah, maybe immediately suggesting a string that can be added to the help. However, I assume that dassl and daspk will be deprecated in some future version since we are moving to odeXY compatible solvers. Lazy_Tom wrote > about the paper you cited, yup, I'm aware of it and I agree, it is quite > nice. > Well, at least I'll gave it a try, I think that for DAE problems I'll move > to openModelica I think that ode5r from odepkg should be able to solve index3 problems since its based on Radau 5 [1]. We hope to have a new release of odepkg quickly after Octave 4.2. I will keep in mind to include such information in the help. Regarding openModelica: yes, they do some clever tricks for higher index. As far as I recall it's based on dummy derivatives [2]. Such stuff is easier for modelica since it is in some sense a symbolic language. However, to my best knowledge, it is not really well suited for big problems. So, as usual, the best choice depends on what you want to do :) Seb. [1] http://www.unige.ch/~hairer/prog/stiff/radau5.f [2] http://dx.doi.org/10.1137/0914043 -- View this message in context: http://octave.1599824.n4.nabble.com/DAE-daspk-solver-problem-tp4680377p4680395.html Sent from the Octave - General mailing list archive at Nabble.com. _______________________________________________ Help-octave mailing list [hidden email] https://lists.gnu.org/mailman/listinfo/help-octave _______________________________________________ Help-octave mailing list [hidden email] https://lists.gnu.org/mailman/listinfo/help-octave
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## Re: DAE daspk solver problem
In reply to this post by Sebastian Schöps As I understand it, Radau5.f demands a structure of DAEs with Linearly implicit form: M Y' = F(Y,T) DASPK admits allows a more general form F(Y,Y',T)=0. There can be problems that can not be transformed to the linearly implicit form. For large scale problems, (> 1e5 variables), retaining sparsity is an important issue, so DASPK structure can offer some advantages. Even when the equations can be symbolically transformed, symbolic manipulation may not scale well (perhaps the problem with scaling in Modelica), or may destroy sparsity and therefore have inefficient discrete propagation. Lastly, the fortran DASPK / DASKR can take advantage of implicit sparse solvers. Do octave implementations allow this?
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## Re: DAE daspk solver problem
I did not try this simple test in daspk but I did try it in ida (with C-api).Even after carefullyinsuring that I had consistent initial conditions, ida could not advance to thefirst time step. So I am very doubtful that daspk will solve this example in itsindex-3 form.Frankly, I was quite surprised that radau5 was able to do so particularly sinceit doesn't appear to have an option to specify an initial y'. On Fri, Oct 28, 2016 at 1:17 PM, SunilShah wrote:As I understand it, Radau5.f demands a structure of DAEs with Linearly implicit form: M Y' = F(Y,T) DASPK admits allows a more general form F(Y,Y',T)=0. There can be problems that can not be transformed to the linearly implicit form. For large scale problems, (> 1e5 variables), retaining sparsity is an important issue, so DASPK structure can offer some advantages. Even when the equations can be symbolically transformed, symbolic manipulation may not scale well (perhaps the problem with scaling in Modelica), or may destroy sparsity and therefore have inefficient discrete propagation. Lastly, the fortran DASPK / DASKR can take advantage of implicit sparse solvers. Do octave implementations allow this? -- View this message in context: http://octave.1599824.n4.nabble.com/DAE-daspk-solver-problem-tp4680377p4680409.html Sent from the Octave - General mailing list archive at Nabble.com. _______________________________________________ Help-octave mailing list [hidden email] https://lists.gnu.org/mailman/listinfo/help-octave _______________________________________________ Help-octave mailing list [hidden email] https://lists.gnu.org/mailman/listinfo/help-octave
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## Re: DAE daspk solver problem
DASPK/DASKR are not expected to solve index 2+ problems like the index-3 pendulum problem. My comment is about problem structure (mass matrix vs fully implicit), sparsity and scaling.
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## Re: DAE daspk solver problem
In reply to this post by sshah On 28 Oct 2016, at 19:17, SunilShah <[hidden email]> wrote: > For large scale problems, (> 1e5 variables), retaining sparsity is an > important issue, so DASPK structure can offer some advantages. Even when > the equations can be symbolically transformed, symbolic manipulation may not > scale well (perhaps the problem with scaling in Modelica), or may destroy > sparsity and therefore have inefficient discrete propagation. Unfortunately the sparsity is not exploited in the Octave interface. > Lastly, the fortran DASPK / DASKR can take advantage of implicit sparse > solvers. Do octave implementations allow this? No, AFAIK there is no option in Octave to invoke the matrix-free iterative solver in DASPK from Octave. c. _______________________________________________ Help-octave mailing list [hidden email] https://lists.gnu.org/mailman/listinfo/help-octave
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## Re: DAE daspk solver problem
In reply to this post by Bill Greene-3 Bill Greene-3 wrote I was able to solve this using ode5r from odepkg. My test code is below: function radau5PendulumDAE vopt = odeset ('InitialStep', 1e-3, 'AbsTol', 1e-8); len=1; theta0=pi/2; m=1; g=9.81; y0=[ len*sin(theta0), -len*cos(theta0), 0, 0, m*g]; odef = @(t,x) odeFunc (t, x, m, g, len) sol=ode5r (odef, [0, 5], y0, vopt); figure; plot(sol.x, sol.y(:,1)); end function y = odeFunc (t, x, m, g, len) lam = x(5); y=[x(2); -lam*x(1); x(4); -lam*x(3)-m*g; x(1)^2 + x(3)^2 - 1]; end Hi Bill, I don't think your description is correct, at the moment you are solving a regular ODE. The solver is not aware that the last equation doesn't contain the derivative of x(5). And this needs to be specified with odeset 'Mass' option and other parameters Out of curiosity, getting odest documentation (command 'doc odepkg' doesn't work on my windows release) and odepkg.pdf document has been a problem. The pdf is mentioned in odeset help, odepkg help and website (http://octave.sourceforge.net/odepkg/overview.html) but to the best of my knowledge the actual location of it doesn't exist. In fact, after installing the odepkg I did a search odepkg.pdf on my octave directory with no results. What I've found though is an odepkg.texi file which is indeed the package manual but it's not a pdf. This is how I finally found out about the 'Mass' option (yes there is a tiny example in odepkg_examples_dae but I don't think this is enough) Do you or anyone else have this problem as well? Marco
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## Re: DAE daspk solver problem
Lazy_Tom wrote I don't think your description is correct, at the moment you are solving a regular ODE. The solver is not aware that the last equation doesn't contain the derivative of x(5). And this needs to be specified with odeset 'Mass' option and other parameters Yes, you need something like M = diag([1 1 1 1 0]); vopt = odeset ('InitialStep', 1e-3, 'AbsTol', 1e-8,'Mass',M); I cannot test it since I do not have a working odepkg due to Octave 4.2 incompatibilities. Lazy_Tom wrote Out of curiosity, getting odest documentation (command 'doc odepkg' doesn't work on my windows release) and odepkg.pdf document has been a problem. Actually "doc" has also never worked for me. Not sure, if this is a bug or an expected fail. Lazy_Tom wrote The pdf is mentioned in odeset help, odepkg help and website (http://octave.sourceforge.net/odepkg/overview.html) but to the best of my knowledge the actual location of it doesn't exist. In fact, after installing the odepkg I did a search odepkg.pdf on my octave directory with no results. What I've found though is an odepkg.texi file which is indeed the package manual but it's not a pdf. The fact that odepkg.pdf is not correctly build, is a bug of the packaging script of odepkg. It seems the variable "TEXI2PDF = texi2pdf --clean" is not set in the Makefile. We will fix it in the next release. Lazy_Tom wrote This is how I finally found out about the 'Mass' option (yes there is a tiny example in odepkg_examples_dae but I don't think this is enough) This is extensively explained in the pdf (e.g. page 14 in http://cosy.informatik.uni-bremen.de/sites/default/files/odepkg.pdf).
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## Re: DAE daspk solver problem
In reply to this post by Sebastian Schöps Sebastian Schöps wrote Lazy_Tom wrote I don't know why DAE solvers index capabilities are not reported in Octave manual. Matlab at least tells you that you can go no further than index 1 (at least without doing some "magic"...) You are right. Feel free to oben a bug report at savannah, maybe immediately suggesting a string that can be added to the help. However, I assume that dassl and daspk will be deprecated in some future version since we are moving to odeXY compatible solvers. Hi Seb. Oh, so there are other solvers that support full implicit ODE form? I see, any chance you can send me some links like you did for Radu 5? About bug report, I don't know how to do it and I don't understand the "suggesting a string" (string = comment?), can you give me more details? Or, considering future deprecation Sebastian Schöps wrote Lazy_Tom wrote about the paper you cited, yup, I'm aware of it and I agree, it is quite nice. Well, at least I'll gave it a try, I think that for DAE problems I'll move to openModelica I think that ode5r from odepkg should be able to solve index3 problems since its based on Radau 5 [1]. We hope to have a new release of odepkg quickly after Octave 4.2. I will keep in mind to include such information in the help. I will probably give Radau a go, assuming I understood how to set up odeset :) Cheers for the index info Sebastian Schöps wrote Regarding openModelica: yes, they do some clever tricks for higher index. As far as I recall it's based on dummy derivatives [2]. Such stuff is easier for modelica since it is in some sense a symbolic language. However, to my best knowledge, it is not really well suited for big problems. So, as usual, the best choice depends on what you want to do :) Indeed. About index reduction, at the moment I can see it has three methods (except for the third, no idea what the others mean): Uode, Dynamic State Selection and Dummy Derivatives BTW, nice link! Cheers, Marco
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## Re: DAE daspk solver problem
In reply to this post by Sebastian Schöps Sebastian Schöps wrote Lazy_Tom wrote I don't think your description is correct, at the moment you are solving a regular ODE. The solver is not aware that the last equation doesn't contain the derivative of x(5). And this needs to be specified with odeset 'Mass' option and other parameters Yes, you need something like M = diag([1 1 1 1 0]); vopt = odeset ('InitialStep', 1e-3, 'AbsTol', 1e-8,'Mass',M); I cannot test it since I do not have a working odepkg due to Octave 4.2 incompatibilities. I can give it a go. This is what I tried, it is based on the odepkg_examples_dae Demonstration 1 (which I know it works) to make things simple m=1, g=10, pole length=1, both ode2r and ode5r has been tried. function [vyd] = frobertson (vt, vy, varargin) vyd(1,1) = vy(2); vyd(2,1) = -vy(1) * vy(5); vyd(3,1) = vy(4); vyd(4,1) = -vy(3) * vy(5) - 10; vyd(5,1) = vy(1)^2 + vy(3)^2 - 1; endfunction function [vmass] = fmass (vt, vy, varargin) vmass = diag([1, 1, 1, 1, 0]); endfunction vopt = odeset ('Mass', @fmass, 'NormControl', 'on'); vsol = ode5r (@frobertson, [0, 17], [1, 0, 0, 0, 0], vopt); plot (vsol.x, vsol.y); the output was : warning: Option "RelTol" not set, new value 1.0e-006 is used warning: called from radau5_test_1_V04 at line 19 column 6 warning: Option "AbsTol" not set, new value 1.0e-006 is used warning: Option "NormControl" will be ignored by this solver warning: Option "InitialStep" not set, new value 1.0e-006 is used warning: Option "MaxStep" not set, new value 1.4e+000 is used warning: Option "Mass" only supports constant mass matrices M() and not M(t,y) warning: Option "NewtonTol" not set, default value is used warning: Option "MaxNewtonIterations" not set, default value 7 is used EXIT OF RADAU5 AT X= 0.2537E-02 STEP SIZE T0O SMALL, H= 1.7937759160684990E-018 error: missing implementation error: called from radau5_test_1_V04 at line 19 column 6 Unless I have made some typos, I suspect this might take more than expected to set up... Sebastian Schöps wrote Lazy_Tom wrote This is how I finally found out about the 'Mass' option (yes there is a tiny example in odepkg_examples_dae but I don't think this is enough) This is extensively explained in the pdf (e.g. page 14 in http://cosy.informatik.uni-bremen.de/sites/default/files/odepkg.pdf). Cheers, I assume there are no changes in odeset from 0.8.2 to 0.8.5 Marco | 5,295 | 18,500 | {"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-2019-22 | latest | en | 0.825326 |
https://nylearns.org/module/Standards/Tools/Browse?linkStandardId=0&standardId=98076 | 1,708,514,652,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473472.21/warc/CC-MAIN-20240221102433-20240221132433-00864.warc.gz | 453,756,757 | 11,765 | Hello, Guest
## Browse Standards
View all PreK-12 NYS Learning Standards in a dropdown list format.
• Standard Area - TECH: Learning Standards for Technology
(see MST standards under Previous Standard Versions)
• Standard - 5.NF.3:
Interpret a fraction as division of the numerator by the denominator (a/b = a / b). Solve word problems involving division of whole numbers leading to answers in the form of fractions or mixed numbers, e.g., by using visual fraction models or equations to represent the problem. For example, interpret 3/4 as the result of dividing 3 by 4, noting that 3/4 multiplied by 4 equals 3, and that when 3 wholes are shared equally among 4 people each person has a share of size 3/4. If 9 people want to share a 50-pound sack of rice equally by weight, how many pounds of rice should each person get? Between what two whole numbers does your answer lie?
Emphasis: M
• Component - 5.NF.4.a:
Interpret the product (a/b) * q as a parts of a partition of q into b equal parts; equivalently, as the result of a sequence of operations a * q / b. For example, use a visual fraction model to show (2/3) * 4 = 8/3, and create a story context for this equation. Do the same with (2/3) * (4/5) = 8/15. (In general, (a/b) * (c/d) = ac/bd.)
• Component - 5.NF.4.b:
Find the area of a rectangle with fractional side lengths by tiling it with unit squares of the appropriate unit fraction side lengths, and show that the area is the same as would be found by multiplying the side lengths. Multiply fractional side lengths to find areas of rectangles, and represent fraction products as rectangular areas. | 406 | 1,615 | {"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} | 3.890625 | 4 | CC-MAIN-2024-10 | latest | en | 0.901215 |
https://www.teacherspayteachers.com/Product/6NS1-Dividing-Fractions-stations-activity-2142521 | 1,531,949,081,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676590329.25/warc/CC-MAIN-20180718193656-20180718213656-00240.warc.gz | 978,340,045 | 16,936 | # 6.NS.1 - Dividing Fractions (stations activity)
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10 station problems (designed to cut into half sheets)
The page with the photo is designed to be folded in half as a folder for the station cards.
Recording sheet provided
Exit ticket provided
Mixed skills review homework (depends on where you are in the curriculum)
Worked out solutions provided for all questions.
10 station problems fashioned after this lesson problem:
Saundra has 12 ! cups of punch and is pouring ! for each guest at her party.
Part A: Does she have enough for 16 guests?
Part B: How much should she pour in each glass in order for each guest to get a serving?
(Hint: Here we were given 16 guests in PART A---Each serving is an equal size)
Part C: If Saundra sticks with 4/5 of a cup for each person, how many cups of lemonade does he need?
Total Pages
19 pages
Included
Teaching Duration
1 hour
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\$2.00 | 256 | 1,056 | {"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-2018-30 | latest | en | 0.941491 |
https://civilengineering.blog/2017/11/18/ | 1,590,423,540,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347389309.17/warc/CC-MAIN-20200525161346-20200525191346-00405.warc.gz | 293,568,922 | 78,087 | Analysis of T beam working stress method
## Analysis of T beam working stress method
Learn : Analysis of T beam working stress method : Neutral Axis is Within the Flange (n < Df), Neutral Axis Lies in the Web of the Beam (n >Df) ANALYSIS OF T BEAM Consider the section of a T-beam shown in Fig. 2.14 (a). The analysis of a T-beam comprises of following two cases : (i) Neutral axis is within the flange. (ii) Neutral axis is in the web. Case 1 : Neutral Axis is Within the Flange (n < Df) Equivalent or Transformed Section The equivalent section of the T-beam in terms of concrete is shown in Fig. 2.13 (b). The concrete below the neutral axis is assumed to be cracked and the area of steel is replaced by an equivalent concrete area which is equal to m.Ast. The compression area is rectangular in shape as n < Df. Thus, this flanged beam can be analyzed exactly as…
Continue Reading Analysis of T beam working stress method
## T beams and terms used in T beams in Reinforced cement concrete
T beams and terms used in T beams : Breadth of Web (bw),Thickness of the Flange (Df),Overall Depth of the Beam (D),Effective Width of the Flange (bf),Effective width of the compression flange of the flanged beam in Reinforced cement concrete T beams and terms used in T beams in Reinforced cement concrete T BEAMS In RCC construction, slabs and beams are cast monolithic-ally. In such construction, a portion of the slab act integrally with the beam and bends along with the beam under the loads. This phenomenon is seen in the beams supported slab system as shown in Fig. 2.11. The portion of the slab which acts integrally with the beam to resist loads is called as Flange of the T-beam or L-beam. The portion of the beam below the flange is called as Web or Rib of the beam. The intermediate beams supporting the slab are called as T-beams and the…
Continue Reading T beams and terms used in T beams in Reinforced cement concrete
Steel beam theory is used to find the approximate value of the moment of resistance of a doubly reinforced beam specially when the area of compression steel is equal to or more than the area of the tensile steel. Steel beam theory moment of resistance of a doubly reinforced beam The moment of resistance of a doubly reinforced beam consists of : (i) Moment of resistance of compression concrete and the corresponding tensile steel (Ast1) i.e., moment of resistance of balanced section (M1). (ii) Moment of Resistance M' of the compression steel (Asc) and the additional tensile steel (Ast2). In the steel beam theory : (i) Concrete is completely neglected. (ii) The moment of resistance is taken equal to the amount of the couple of compressive and tensile steel. (iii) The permissible stress in compressive steel is taken as equal to the permissible stress in tensile steel. $\therefore M _{r}=\sigma… Continue Reading Steel beam theory is used to find the MR of doubly reinforced beam ## Types of problem in doubly reinforced beams working stress method Types of problem in doubly reinforced beams working stress method: Determination of moment of resistance of the given section,Determination of actual stresses in concrete and steel,Design of the section. Types of problem in doubly reinforced beams working stress method Determination of moment of resistance of the given section. Determination of actual stresses in concrete and steel. Design of the section. Determination of Moment of Resistance Given : (i) Dimension of the beam section (b and d) (ii) Area of tensile steel Ast and area of compressive steel Asc (iii) Permissible stress in concrete {σcbc) and permissible stress in steel (σst) Procedure : Calculate \[m=\frac{280}{3\sigma_{cbc}}$ Calculate critical neutral axis (nc) $\frac{n_{c}}{d-n_{c}}=\frac{m.\sigma_{cbc}}{\sigma_{st}}$ Calculate actual neutral axis depth (nc) $\frac{b.n^{2}}{2}+(1.5m-1)A_{sc}(n-d_{c})=m.A_{st}(d-n)$ Compare n and nc (a) If n>nc the section is under reinforced (fully stressed) Maximum tensile stress developed in steel = σst Maximum compressive stress developed in concrete \[\sigma_{cbc}(where \sigma'_{cbc})is less …
Continue Reading Types of problem in doubly reinforced beams working stress method | 982 | 4,243 | {"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} | 2.765625 | 3 | CC-MAIN-2020-24 | latest | en | 0.906309 |
http://whoswhoandnew.blogspot.com/2014/04/how-to-teach-your-students-how-to-check.html?m=1 | 1,511,548,305,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934808742.58/warc/CC-MAIN-20171124180349-20171124200349-00104.warc.gz | 327,115,733 | 20,685 | ## Monday, April 28, 2014
### How to Teach Your Students How to CHECK THEIR WORK
A student walks up with with his finished math test and hands it to me. I glance at the clock. I think, Hmm, 10 minutes ago I gave you this test.
But I say, "Did you check your work?"
"Whoops," he says and takes back the test. He remains standing in front of me, and proceeds to give each page a quick glance, nodding.
10 seconds later, "Okay, I checked it. Here you go."
Oh, thank you, child. Thank you for your thoroughness. Thank you for NOT noticing the simple calculation error you made on number 5, and for forgetting to answer both parts of number 8, and for misreading the question on number 9.
And by the way, you skipped number 10. But don't worry, I will notice these silly mistakes for you tonight as I grade your test, wondering how much of this content you actually know or not.
After many of these interactions with students, or something similar, I decided to have another crack at teaching my students how to check their work. Simply reminding them to check their work was rarely helpful. Did students really know what I meant? Did I really know what I meant?
After some reflection, I broke the concept of "checking your work" into three levels. I modeled how to do each level, and then I taught students when each "level of checking" was best to use. This idea has worked well in math especially, but can be applied to other subjects too.
Level One Checking
I consider a "level one check" to be the lowest level of checking your work (but better than nothing). After completing your test or assignment, you return to the beginning and check to be sure that you have answered every question. Basically, you are checking to see if you skipped anything. It is quick and takes little thought, but might be the right choice in certain situations, like if you are running out of time.
Level Two Checking
When you check your work at level two, you return to the beginning of your test or assignment and you reread the first question. You then look at your answer and see if it makes sense. If it does, you move on and do the same for the rest of the questions. If it doesn't, you work the problem again to try to find your error. With level two, you really keep your brain turned on. It helps eliminate those "silly" mistakes.
Level Three Checking
Checking your work at level three is like a full attack. This is when you read each question again and rework the problem. You then compare your answer to your first attempt to see if you got the same thing. This takes a lot of time and mental stamina, but might be appropriate for certain sections, like a problem with multi-digit addition or subtraction.
Getting students to care about checking their work isn't easy. It takes a change of mindset that completing the last question is not "the end." It also takes the realization that we all are capable of making absent-minded mistakes. I've found that teaching students how to use these different levels of checking has helped give students more ownership and the ability to make wiser choices when it comes to checking their work. I hope the idea can help your students too!
1. Love your levels! This is the bane of my existence lately...for exactly the reasons you mentioned. I even went so far as to add a simple checklist at the bottom of each page of a recent test. Didn't quite cure all our ills, but it was interesting to see who checked the box but forgot to do what it said!
2. Thanks, Suzy! I love that idea of a little checklist at the end of the test as a reminder!
3. I. Love. This.
Tomorrow starts our state test and I am kicking myself for not thinking of something like this sooner! Thanks so much. I already love following your blog, but I'm going to have to stop by more often because you have such awesome ideas :)
Jenny
4. Thanks, Jenny! Yeah, we just had our state testing at the beginning of the week. Yuck. Hope it goes well!
5. As usual, I love your ideas and your thought process on this. It's a great time of year for the reminder too, as we are all beginning to mentally begin summer vacation. Ha ha!
6. Totally agree about the time of the year, Jenny! Thanks for your feedback!
7. I am so glad I found this...our state testing begins on Tuesday, and I am forever asking students if they checked their work only to find that they just glanced over it! I love the explanations given for each level. Thanks for sharing!
Sarah Beth
Miss White's Classroom
8. This has been the bane of my existence! Such a great idea! I have four week so of school left, so it's not too late to at least introduce this, right? ;)
9. Very simple, but effective. Thank you for sharing.
Debbie
Crockett's Classroom, Forever in Third Grade
10. Thank you for sharing this intervention - I can't wait to see the results next year :)
11. Helps eliminate the age old question, "Is this good?"
12. I love this! I look forward to using it with my students this year, thanks so much for sharing the poster as well.
13. Ide bagus! saya senang sekali terima kasih, ini sangat membantu
14. Great ideas! Do you have any ideas how to rework level 3 for Reading? (That's the subject I teach.) I struggle with getting students to "check their work," especially now that our tests are computer based - no paper/pencils allowed.
15. These are wonderful! Thank you for sharing. We just finished high stakes testing. I teach primary- but this is something I would like to share on my blog for other teachers :) I am going to link to it! http://feinmandoramoore.blogspot.com/2015/03/how-to-teach-your-students-to-check.html
16. Definitely using this in my new 5th grade class this year! I have struggled with this exact issue for several years! You rock!
17. I always tell my kids to check their work before they turn in their test. So many leave answers blank because they want to be done. Thanks for sharing.
Beti
18. Thank you so much! I'm printing this out and trying it this week. I love your anecdote and teaching/writing voice - I am totally on the same page with you!!!! | 1,379 | 6,089 | {"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} | 2.515625 | 3 | CC-MAIN-2017-47 | longest | en | 0.96838 |
http://stackoverflow.com/questions/10066518/prolog-addition?answertab=active | 1,406,676,593,000,000,000 | text/html | crawl-data/CC-MAIN-2014-23/segments/1406510267876.49/warc/CC-MAIN-20140728011747-00117-ip-10-146-231-18.ec2.internal.warc.gz | 276,513,584 | 15,518 | I am really stuck up with the addition predicate in prolog for long time. Any help will greatly help me..
This is my predicate :
p(s(U,I), [s(A,B)|P1], T1,H(N, E), P,C) :- m(s(E,F), p(s(U,I), [s(A,B)|P1], +(C,F) ,H(N,E), P,C).
Everything works fine except for this part :
+(C,F) always gives value like 3+2 and not like 5.
I have tried all possible methods like
p(s(U,I), [s(A,B)|P1], T1,H(N, E), P,C) :- m(s(E,F)), C2 is C+F, p(s(U,I), [s(A,B)|P1], C2 ,H(N,E), P,C).
But everytime it returns as someinteger+someinteger like 3+2 instead of 5.
Any help would be greatly helpful.
Thanks!
-
Can you add some of the use cases you've tried with the p predicate? It's not clear to me what all the extra arguments are for. – EMS Apr 8 '12 at 21:10
H(N, E) not is valid Prolog – CapelliC Apr 9 '12 at 6:26
you have to use the is/2 predicate to perform arithmetic, ie X is 3 + 2.. Else it's not "evaluated"!
try:
p(s(U, I),
[s(A, B)|P1],
C1,
H(N, E),
P,
C) :-
m(s(E, F)),
C2 is F + C1,
p(s(U, I),
[s(A, B)|P1],
C2,
H(N, E),
P,
C).
predicates aren't executed if you place them in other predicate argument spots (well if we're speaking about non meta predicates as we're here). That was your problem.
-
I tried that but the same error occured. I used this , p(s(U,I), [s(A,B)|P1], T1,H(N, E), P,C) :- m(s(E,F)), **C2 is C+F**, p(s(U,I), [s(A,B)|P1], **C2** ,H(N,E), P,C). But , I still get it as 3+2 instead of 5 .Any pointers? – stackuser Apr 8 '12 at 22:43
Sorry for it. I used the is/2 in the predicate : p(s(U,I), [s(A,B)|P1],C1,H(N, E), P,C) :- m(s(E,F)), p(s(U,I), [s(A,B)|P1], **C2 is F+C1** ,H(N,E), P,C).' But still I get as,C = _h553 is (_h532 is 0 + 5) + 5;' . It is not evaluating to 10 in arithmetic. Really confused here.. – stackuser Apr 8 '12 at 23:06
Is there any means to send code privately through stackoverflow? – stackuser Apr 8 '12 at 23:12
I tried even that earlier, and got instantiation error as , '++Error[XSB/Runtime/P]: [Instantiation] ++Error[XSB]: [Runtime/C] Uninstantiated argument of evaluable function +/2 Goal: 0 + _Var, probably as 2nd arg of is/2 Forward Continuation... ... m_#303/9 ... p/6' . The instantiation error is pointing to the function m which has no errors. – stackuser Apr 8 '12 at 23:21
@VaniJayram it means that either F or C1 is a free/partially free variable. Arguments of is/2 must be ground. I can't help you without more code. Your problem isn't the is/2 predicate once you use my code, it's the fact that you don't instantiate properly your variables and that could come from any other place in your code. – m09 Apr 8 '12 at 23:23 | 873 | 2,601 | {"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} | 2.828125 | 3 | CC-MAIN-2014-23 | latest | en | 0.848455 |
https://mathoverflow.net/questions/tagged/moments | 1,718,769,797,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861797.58/warc/CC-MAIN-20240619025415-20240619055415-00200.warc.gz | 352,186,524 | 36,554 | # Questions tagged [moments]
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### Convergence in probability of sample covariance for permutation invariant triangular arrays
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1 vote | 684 | 2,484 | {"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": 1, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2024-26 | latest | en | 0.839904 |
http://www.jiskha.com/display.cgi?id=1302673681 | 1,481,123,287,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542213.61/warc/CC-MAIN-20161202170902-00441-ip-10-31-129-80.ec2.internal.warc.gz | 546,263,672 | 3,774 | Wednesday
December 7, 2016
# Homework Help: math
Posted by Anonymous on Wednesday, April 13, 2011 at 1:48am.
Suppose that the population of a certain townis made up of 45% of men and 55% of women. Of the men, 40% wear glasses, abd of the women, 20% wear glasses. Given that a person chosen at random from the town wears glasses, what is the probability that the person is a woman?
• math - Raj, Wednesday, April 13, 2011 at 8:06am
Probability: 11 out of 29;
or 37.9310345%.
Method: Take a sample of 100.
45 are men; 40% or 18 men wear glasses
55 are women; 20% or 11 women wear glasses.
So out of 100, 18+11=29 wear glasses out of which 11 are women.
Probability: 11 out of 29 | 213 | 681 | {"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-2016-50 | longest | en | 0.956292 |
http://mathoverflow.net/questions/73647?sort=newest | 1,369,116,275,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368699730479/warc/CC-MAIN-20130516102210-00036-ip-10-60-113-184.ec2.internal.warc.gz | 167,288,139 | 10,708 | ## Do the converses of [weak law of large numbers / central limit theorem] hold?
Let $\; X_0,X_1,X_2,X_3,...\;$ be independent and identically distributed (real-valued) random variables.
1. Suppose $\frac1n \cdot\sum\limits_{m=0}^n X_m$ converges in probability. Does it follow that $\operatorname{E}(X_0)$ exists?
2. Suppose $\operatorname{E}(X_0) = 0$ and that $\frac1{\sqrt n} \cdot\sum\limits_{m=0}^n X_m$ converges in distribution to a normal random variable. Does it follow that $\operatorname{E}((X_0)^2)$ is finite?
(I already found that the converse of the strong law of large numbers holds.)
-
a remark. The weak law fails for the Cauchy distribution. – Gerald Edgar Aug 25 2011 at 12:01
A classical example for 1'. is a symmetric integer-valued X with P(X=n)=P(X=-n)=c/(n^2log(n)). Then phi is C^1 but X is not integrable. On the other hand, if phi is C^2 then X^2 is integrable. – Didier Piau Aug 25 2011 at 12:44
Didier, this gives a counterexample to 1, right? I think the last line in the question means that if one replaces, in 1, convergence in probability by a.s. convergence, then the answer is yes (by say the converse to Borel-Cantelli). – Mikael de la Salle Aug 25 2011 at 12:59
Necessary and sufficient conditions (in terms close to those you want) for the WLLN and the CLT can be found, e.g., in "Foundations of modern probability" by Kallenberg (Theorems 4.16 and 4.17). – Yvan Velenik Aug 25 2011 at 14:09
@Ricky in the 2nd Edition it's Theorems 5.16 and 5.17 – pgassiat Aug 25 2011 at 23:48 | 490 | 1,522 | {"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} | 3.1875 | 3 | CC-MAIN-2013-20 | latest | en | 0.823632 |
https://se.mathworks.com/matlabcentral/answers/478846-why-do-i-get-nan-as-output | 1,623,803,919,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487621699.22/warc/CC-MAIN-20210616001810-20210616031810-00459.warc.gz | 440,347,046 | 26,433 | # why do i get Nan as output?
39 views (last 30 days)
Sunetra CV on 4 Sep 2019
Commented: Sunetra CV on 16 Sep 2019
im trying to solve four such equations using an input vector containing 4 values.
fn = 1/(rho(1)*Y(1))*(1.113-2.14*Y(1)+2.81*Y(1)^2-1.76*Y(1)^3+0.396*Y(1)^4+0.08694*Y(1)^5-0.0579*Y(1)^6 +0.0077*Y(1)^7);
using ode45.
But the output vector consists of Nan values ; where as its returining numerical values on using the command window.
function file:
function fn = plume5(t,Y)
global rho cp
fn(1) = 1/(rho(1)*cp)*(760.91-1608.08*Y(1)+2449.22*Y(1)^2-2001.61*Y(1)^3+904.51*Y(1)^4-212.087*Y(1)^5+20.095*Y(1)^6);
fn(2) = 1/(rho(2)*cp)*(760.91-1608.08*Y(2)+2449.22*Y(2)^2-2001.61*Y(2)^3+904.51*Y(2)^4-212.087*Y(2)^5+20.095*Y(2)^6);
fn(3) = 1/(rho(3)*cp)*(760.91-1608.08*Y(3)+2449.22*Y(3)^2-2001.61*Y(3)^3+904.51*Y(3)^4-212.087*Y(3)^5+20.095*Y(3)^6);
fn(4) = 1/(rho(4)*cp)*(760.91-1608.08*Y(4)+2449.22*Y(4)^2-2001.61*Y(4)^3+904.51*Y(4)^4-212.087*Y(4)^5+20.095*Y(4)^6);
fn(5) = 1/(rho(5)*Y(5))*(1.113-2.14*Y(5)+2.81*Y(5)^2-1.76*Y(5)^3+0.396*Y(5)^4+0.08694*Y(5)^5-0.0579*Y(5)^6 +0.0077*Y(5)^7);
fn(6) = 1/(rho(6)*Y(6))*(1.113-2.14*Y(6)+2.81*Y(6)^2-1.76*Y(6)^3+0.396*Y(6)^4+0.08694*Y(6)^5-0.0579*Y(6)^6 +0.0077*Y(6)^7);
fn(7) = 1/(rho(7)*Y(7))*(1.113-2.14*Y(7)+2.81*Y(7)^2-1.76*Y(7)^3+0.396*Y(7)^4+0.08694*Y(7)^5-0.0579*Y(7)^6 +0.0077*Y(7)^7);
fn(8) = 1/(rho(8)*Y(8))*(1.113-2.14*Y(8)+2.81*Y(8)^2-1.76*Y(8)^3+0.396*Y(8)^4+0.08694*Y(8)^5-0.0579*Y(8)^6 +0.0077*Y(8)^7);
fn = fn';
main file
clear all
close all
global rho cp
rho = [0.4878;0.7862;1.1449;1.1611;0.4878;0.7862;1.1449;1.1611]
cp = 1.4857
t_span= 0:10
y = [0.0182;0.3105;1.2882;2.7341;0.0182;0.3105;1.2882;2.7341]
[A,P]= ode45('plume5',t_span,y);
##### 2 CommentsShowHide 1 older comment
Sunetra CV on 5 Sep 2019
Thank You
Jan on 4 Sep 2019
Edited: Jan on 4 Sep 2019
Use te debugger to determine the first occurrence of a NaN. Type this in the command window:
dbstop if naninf
Then run the code again. When Matlab stops, check the value of the trajectory. Most likely there is a pole and NaN or Inf is the correct numerical value. You can check this by:
plot(t, Y)
By the way, vectorization can simplify the code:
function fn = plume5(t,Y, rho, cp)
fn = zeros(8, 1);
fn(1:4) = 1 ./ (rho(1:4)*cp) .* (760.91 - 1608.08 * Y(1:4) + ...
2449.22 * Y(1:4) .^ 2 - 2001.61 * Y(1:4) .^ 3 + ...
904.51 * Y(1:4) .^ 4 - 212.087 * Y(1:4) .^ 5 + 20.095 * Y(1:4) .^6 );
fn(5:6) = 1 ./ (rho(5:8) .* Y(5:8)) * (1.113 - 2.14 * Y(5:8) + ...
2.81 * Y(5:8) .^ 2 - 1.76 * Y(5:8) .^ 3 + 0.396 * Y(5:8) .^ 4 + ...
0.08694 * Y(5:8) .^ 5 - 0.0579 * Y(5:8) .^ 6 + 0.0077 * Y(5:8) .^ 7);
Sunetra CV on 16 Sep 2019
As in making the coefficients of y(i) in both the equations of the same range. | 1,443 | 2,728 | {"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} | 2.859375 | 3 | CC-MAIN-2021-25 | latest | en | 0.559744 |
https://metacpan.org/search?q=module%3AMath%3A%3ABaseCalc | 1,679,604,260,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945183.40/warc/CC-MAIN-20230323194025-20230323224025-00060.warc.gz | 453,692,663 | 6,144 | ### Math::BaseCalc - Convert numbers between various bases River stage one • 6 direct dependents • 9 total dependents 1 ++
This module facilitates the conversion of numbers between various number bases. You may define your own digit sets, or use any of several predefined digit sets. The to_base() and from_base() methods convert between Perl numbers and strings which repr...
/Math-BaseCalc-1.019 - 08 Jun 2017 05:28:22 UTC
### Math::Base::Convert::CalcPP - standard methods used by Math::Base::Convert River stage two • 1 direct dependent • 78 total dependents 3 ++
This module contains the standard methods used by Math::Base::Convert to convert from one base number to another base number. * \$carry = addbaseno(\$reg32ptr,\$int) This function adds an integer < 65536 to a long n*32 bit register and returns the carry...
/Math-Base-Convert-0.11 - 22 Oct 2015 22:34:07 UTC
### Units/Convert/Base.pm River stage one • 2 direct dependents • 7 total dependents 1 ++
/Math-Calc-Units-1.07 - 04 Aug 2009 16:55:12 UTC
### Units/Convert/Base2Metric.pm River stage one • 2 direct dependents • 7 total dependents 1 ++
/Math-Calc-Units-1.07 - 04 Aug 2009 16:55:12 UTC
### Math::BaseCnv - basic functions to CoNVert between number Bases River stage two • 6 direct dependents • 25 total dependents ++
BaseCnv provides a few simple functions for converting between arbitrary number bases. You're probably better off using Michael Robinton's Math::Base::Convert benchmarked high-performance Perl modules. Another object syntax for number-base conversion...
/Math-BaseCnv-1.14 - 30 Jul 2016 12:06:39 UTC
### Math::Int2Base - Perl extension for converting decimal (base-10) integers into another number base from base-2 to base-62, and back to decimal. River stage one • 2 direct dependents • 2 total dependents ++
Math::Int2Base provides * "int2base( \$int, \$base, \$minlen )" for converting from decimal to another number base, * "base2int( \$num, \$base )" for converting from another base to decimal, and * "base_chars( \$base )" for retrieving the string of charact...
/Math-Int2Base-1.01 - 30 Jun 2017 18:20:04 UTC
### Math::Fleximal - Integers with flexible representations. River stage zero No dependents 3 ++
This is a package for doing integer arithmetic while using a different base representation than normal. In base n arithmetic you have n symbols which have a representation. I was going to call them "glyphs", but being text strings they are not really...
/Math-Fleximal-0.06 - 10 Apr 2005 06:23:56 UTC
### Math::BaseConvert - fast functions to CoNVert between number Bases River stage zero No dependents 1 ++
BaseConvert provides a few simple functions for converting between arbitrary number bases. It is as fast as I currently know how to make it (of course relying only on the lovely Perl). If you would rather utilize an object syntax for number-base conv...
/Math-BaseConvert-1.8 - 14 Apr 2016 16:32:23 UTC
### Bundle::Math::Base - Bundle related to calculations between different bases River stage zero No dependents ++
This is a bundle of modules related to calculations between different bases. Please have a look at Bundle::Math. If you would like to see a specific module included in a future version of this bundle, please send me an email or use rt.cpan.org....
/Bundle-Math-Base-1.00 - 03 Apr 2004 16:12:23 UTC
9 results (0.032 seconds) | 870 | 3,378 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2023-14 | latest | en | 0.678579 |
https://tw511.com/2/17/416.html | 1,656,884,670,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104249664.70/warc/CC-MAIN-20220703195118-20220703225118-00626.warc.gz | 630,514,466 | 8,738 | # C語言運算子
• 算術運算子
• 關係運算子
• 邏輯運算子
• 按位元運算子
• 賦值運算子
• 其它運算子
## 1.算術運算子
+ 將兩個運算元相加 A + B = 30
- 從第一個運算元減去第二個運算元 A ? B = -10
* 將兩個運算元相乘 A * B = 200
/ 將第一個運算元除以第二個運算元
% 模數運算子和整數除法後的餘數。 B % A = 0
++ 遞增運算子將整數值增加1 A++ = 11
-- 遞減運算子將整數值減1。 A-- = 9
#include <stdio.h>
void main() {
int a = 21;
int b = 10;
int c ;
c = a + b;
printf("Line 1 - Value of c is %d\n", c );
c = a - b;
printf("Line 2 - Value of c is %d\n", c );
c = a * b;
printf("Line 3 - Value of c is %d\n", c );
c = a / b;
printf("Line 4 - Value of c is %d\n", c );
c = a % b;
printf("Line 5 - Value of c is %d\n", c );
c = a++;
printf("Line 6 - Value of c is %d\n", c );
c = a--;
printf("Line 7 - Value of c is %d\n", c );
}
Line 1 - Value of c is 31
Line 2 - Value of c is 11
Line 3 - Value of c is 210
Line 4 - Value of c is 2
Line 5 - Value of c is 1
Line 6 - Value of c is 21
Line 7 - Value of c is 22
## 2.關係運算子
== 檢查兩個運算元的值是否相等。 如果相等，則條件成立。 (A == B)結果為false
!= 檢查兩個運算元的值是否相等。 如果值不相等，則條件成立。 (A != B) 結果為true
> 檢查左運算元的值是否大於右運算元的值。 如果是，則條件成立。 (A > B) 結果為false
< 檢查左運算元的值是否小於右運算元的值。 如果是，則條件成立。 (A < B)結果為true
>= 檢查左運算元的值是否大於等於右運算元的值。 如果是，則條件成立。 (A >= B) 結果為false
<= 檢查左運算元的值是否小於等於右運算元的值。 如果是，則條件成立。 (A <= B)結果為true
#include <stdio.h>
main() {
int a = 21;
int b = 10;
int c ;
if( a == b ) {
printf("Line 1 - a is equal to b\n" );
}
else {
printf("Line 1 - a is not equal to b\n" );
}
if ( a < b ) {
printf("Line 2 - a is less than b\n" );
}
else {
printf("Line 2 - a is not less than b\n" );
}
if ( a > b ) {
printf("Line 3 - a is greater than b\n" );
}
else {
printf("Line 3 - a is not greater than b\n" );
}
/* Lets change value of a and b */
a = 5;
b = 20;
if ( a <= b ) {
printf("Line 4 - a is either less than or equal to b\n" );
}
if ( b >= a ) {
printf("Line 5 - b is either greater than or equal to b\n" );
}
}
Line 1 - a is not equal to b
Line 2 - a is not less than b
Line 3 - a is greater than b
Line 4 - a is either less than or equal to b
Line 5 - b is either greater than or equal to b
## 3.邏輯運算子
&& 邏輯與運算子。 如果兩個運算元都不為零，則條件成立。 (A && B)結果為false
! 稱為邏輯非運算子，它用於反轉其運算元的邏輯狀態。如果條件為真，則邏輯NOT運算子將使其結果為false
#include <stdio.h>
main() {
int a = 5;
int b = 20;
int c ;
if ( a && b ) {
printf("Line 1 - Condition is true\n" );
}
if ( a || b ) {
printf("Line 2 - Condition is true\n" );
}
/* lets change the value of a and b */
a = 0;
b = 10;
if ( a && b ) {
printf("Line 3 - Condition is true\n" );
}
else {
printf("Line 3 - Condition is not true\n" );
}
if ( !(a && b) ) {
printf("Line 4 - Condition is true\n" );
}
}
Line 1 - Condition is true
Line 2 - Condition is true
Line 3 - Condition is not true
Line 4 - Condition is true
## 4.按位元運算子
p q p & q p/q p ^ q
0 0 0 0 0
0 1 0 1 1
1 1 1 1 0
1 0 0 1 1
A = 0011 1100
B = 0000 1101
-----------------
A&B = 0000 1100
A|B = 0011 1101
A^B = 0011 0001
~A = 1100 0011
& 如果二進位制AND運算子存在於兩個運算元中，則二進位制AND運算子將對結果複製一位。 (A&B)= 12，即0000 1100
^ 二進位制XOR操作符複製該位，如果它設定在一個運算元中，而不是兩者。 (A ^ B) = 49, 即, 0011 0001
~ 二進位制二補數運算子是一元的，具有「翻轉」位的作用。 (~A)= -61，即 1100 0011的二補數形式。
<< 二進位制左移操作符，左運算元值左移由右運算元指定的位數。 A << 2 = 240 即, 1111 0000
>> 二進位制右移操作符，左運算元值被右運算元指定的位移動。 A >> 2 = 15 即,0000 1111
#include <stdio.h>
main() {
unsigned int a = 60; /* 60 = 0011 1100 */
unsigned int b = 13; /* 13 = 0000 1101 */
int c = 0;
c = a & b; /* 12 = 0000 1100 */
printf("Line 1 - Value of c is %d\n", c );
c = a | b; /* 61 = 0011 1101 */
printf("Line 2 - Value of c is %d\n", c );
c = a ^ b; /* 49 = 0011 0001 */
printf("Line 3 - Value of c is %d\n", c );
c = ~a; /*-61 = 1100 0011 */
printf("Line 4 - Value of c is %d\n", c );
c = a << 2; /* 240 = 1111 0000 */
printf("Line 5 - Value of c is %d\n", c );
c = a >> 2; /* 15 = 0000 1111 */
printf("Line 6 - Value of c is %d\n", c );
}
Line 1 - Value of c is 12
Line 2 - Value of c is 61
Line 3 - Value of c is 49
Line 4 - Value of c is -61
Line 5 - Value of c is 240
Line 6 - Value of c is 15
## 5.賦值運算子
= 簡單賦值運算子，將右側運算元的值分配給左側運算元 C = A + B，將A + B的值分配給C
+= 相加與賦值運算子。它將右運算元新增到左運算元，並將結果分配給左運算元。 C + = A等價於C = C + A
-= 相減與賦值運算子。它從左運算元中減去右運算元，並將結果分配給左運算元。 C -= A等價於 C = C - A
*= 乘以與賦值運算子。它將右運算元與左運算元相乘，並將結果分配給左運算元。 C * = A等價於C = C * A
/= 除以與賦值運算子。它將左運算元與右運算元分開，並將結果分配給左運算元。 C /= A等價於C = C / A
%= 模數與賦值運算子。它需要使用兩個運算元的模數，並將結果分配給左運算元。 C %= A等價於C = C % A
<<= 左移與賦值運算子 C <<= 2等價於C = C << 2
>>= 右移與賦值運算子 C >> = 2等價於C = C >> 2
&= 按位元與賦值運算子 C &= 2等價於C = C & 2
^= 按位元互斥或運算子和賦值運算子。 C ^= 2等價於C = C ^ 2
#include <stdio.h>
void main() {
int a = 21;
int c ;
c = a;
printf("Line 1 - = Operator Example, Value of c = %d\n", c );
c += a;
printf("Line 2 - += Operator Example, Value of c = %d\n", c );
c -= a;
printf("Line 3 - -= Operator Example, Value of c = %d\n", c );
c *= a;
printf("Line 4 - *= Operator Example, Value of c = %d\n", c );
c /= a;
printf("Line 5 - /= Operator Example, Value of c = %d\n", c );
c = 200;
c %= a;
printf("Line 6 - %= Operator Example, Value of c = %d\n", c );
c <<= 2;
printf("Line 7 - <<= Operator Example, Value of c = %d\n", c );
c >>= 2;
printf("Line 8 - >>= Operator Example, Value of c = %d\n", c );
c &= 2;
printf("Line 9 - &= Operator Example, Value of c = %d\n", c );
c ^= 2;
printf("Line 10 - ^= Operator Example, Value of c = %d\n", c );
c |= 2;
printf("Line 11 - |= Operator Example, Value of c = %d\n", c );
}
Line 1 - = Operator Example, Value of c = 21
Line 2 - += Operator Example, Value of c = 42
Line 3 - -= Operator Example, Value of c = 21
Line 4 - *= Operator Example, Value of c = 441
Line 5 - /= Operator Example, Value of c = 21
Line 6 - = Operator Example, Value of c = 11
Line 7 - <<= Operator Example, Value of c = 44
Line 8 - >>= Operator Example, Value of c = 11
Line 9 - &= Operator Example, Value of c = 2
Line 10 - ^= Operator Example, Value of c = 0
Line 11 - |= Operator Example, Value of c = 2
## 6.其他操作符：sizeof和三元運算子
sizeof() 返回變數的大小 sizeof(a)，其中a為整數，將返回4
& 返回變數的地址 &a; 返回變數的實際地址。
* 指向變數的指標 *a;
? : 條件表示式 如果條件是真的？ 那麼返回值X：否則返回Y
#include <stdio.h>
void main() {
int a = 4;
short b;
double c;
int* ptr;
/* example of sizeof operator */
printf("Line 1 - Size of variable a = %d\n", sizeof(a));
printf("Line 2 - Size of variable b = %d\n", sizeof(b));
printf("Line 3 - Size of variable c= %d\n", sizeof(c));
/* example of & and * operators */
ptr = &a; /* 'ptr' now contains the address of 'a'*/
printf("value of a is %d\n", a);
printf("*ptr is %d.\n", *ptr);
/* example of ternary operator */
a = 10;
b = (a == 1) ? 20 : 30;
printf("Value of b is %d\n", b);
b = (a == 10) ? 20 : 30;
printf("Value of b is %d\n", b);
}
Line 1 - Size of variable a = 4
Line 2 - Size of variable b = 2
Line 3 - Size of variable c= 8
value of a is 4
*ptr is 4.
Value of b is 30
Value of b is 20
## 7.運算子優先順序
int value = 10 + 20 * 10;
value變數計算結果為：210，因為*(乘法運算子)的優先順序比+(加法運算子)高，所以在+(加法運算子)之前進行求值。
C語言運算子的優先順序和關聯性如下：
#include <stdio.h>
void main() {
int a = 20;
int b = 10;
int c = 15;
int d = 5;
int e;
e = (a + b) * c / d; // ( 30 * 15 ) / 5
printf("Value of (a + b) * c / d is : %d\n", e);
e = ((a + b) * c) / d; // (30 * 15 ) / 5
printf("Value of ((a + b) * c) / d is : %d\n", e);
e = (a + b) * (c / d); // (30) * (15/5)
printf("Value of (a + b) * (c / d) is : %d\n", e);
e = a + (b * c) / d; // 20 + (150/5)
printf("Value of a + (b * c) / d is : %d\n", e);
return 0;
}
Value of (a + b) * c / d is : 90
Value of ((a + b) * c) / d is : 90
Value of (a + b) * (c / d) is : 90
Value of a + (b * c) / d is : 50 | 3,525 | 7,428 | {"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} | 3.78125 | 4 | CC-MAIN-2022-27 | longest | en | 0.348297 |
https://datasciencewiki.net/gabor-regression/ | 1,701,708,866,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100531.77/warc/CC-MAIN-20231204151108-20231204181108-00074.warc.gz | 230,469,569 | 10,324 | # Gabor Regression
## Gabor Regression :
Gabor regression is a type of machine learning algorithm that is used for modeling and predicting continuous values. It is based on the concept of Gabor functions, which are mathematical functions used to represent signals in a way that is similar to the way that the human auditory system processes sound.
Gabor regression is similar to other regression algorithms, such as linear regression or support vector regression, in that it is used to find the best fit line or curve for a given set of data points. However, unlike these other algorithms, Gabor regression uses Gabor functions to model the data instead of using a simple linear or polynomial function.
One advantage of using Gabor functions in regression is that they are able to capture both the frequency and the time domain characteristics of a signal. This means that Gabor regression can be used to model signals that have complex, nonlinear patterns, such as speech or other audio signals.
Here are two examples of how Gabor regression can be used:
Predicting the price of a stock: In this example, Gabor regression could be used to model the historical price data of a stock. By using Gabor functions to capture the complex, nonlinear patterns in the data, the algorithm can then be used to make predictions about the future price of the stock.
Modeling the relationship between weather and crop yield: In this example, Gabor regression could be used to model the relationship between weather variables, such as temperature and precipitation, and the yield of a crop. By using Gabor functions to capture the complex, nonlinear patterns in the data, the algorithm can then be used to make predictions about the yield of a crop given certain weather conditions.
Overall, Gabor regression is a powerful tool for modeling and predicting continuous values, especially for signals that have complex, nonlinear patterns. By using Gabor functions to represent the data, Gabor regression is able to capture the frequency and time domain characteristics of a signal, which allows it to model and predict a wide variety of phenomena. | 407 | 2,132 | {"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} | 2.84375 | 3 | CC-MAIN-2023-50 | longest | en | 0.907614 |
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```STATISTICS 200
Lecture #25
Textbook: 14.1 and 14.3
Tuesday, November 15, 2016
Objectives:
• Formulate null and alternative hypotheses involving
regression coefficients.
• Calculate T-statistics; determine correct degrees of freedom.
Refresher
• Chapter 3 covered regression equations
and the relationship between two
quantitative
__________ variables.
Example:
Q. What is your height in
inches?
Q. What is your father’s
height in inches?
Which variable should
be the explanatory
variable?
A.Father’s height
B.Student’s height
response on the y-axis
__________
Regression equation
R-squared
explanatory
___________ on the x-axis
Regression terminology
• The linear regression equation looks like this:
slope
Estimated
(mean)
response
explanatory
y-intercept
• Recall that “y-hat” is the estimated mean of the
response (y). It can also refer to a predicted value of y.
In our example:
If the dad was 0 inches tall,
the predicted student height
is 30.34 inches. CAREFUL!
y-hat = 30.34 + 0.53 x (dadheight)
For every one inch increase in
_______, the predicted _______
dadheight
height
increases by ______
.53 inches.
___________
Refresher
• An important point to remember is that if the
slope (b1) is different from zero, then the two
quantitative variables are linearly related.
Parameters for Regression
• When regression was discussed before, we only talked
b0 (sample
about it in terms of the sample statistics: _____
b1 (sample slope).
intercept) and ___
parameters for the slope and
• However, there are also ____________
intercept in a linear regression equation.
• These parameters represent the intercept and slope that
would be found if the whole population for both variables
was used to create a regression equation.
Parameters for Regression
β0 is the population intercept. It is estimated by the
• ____
sample intercept (b0).
β1 is the population slope. It is estimated by the
• ____
sample slope (b1).
Parameters for Regression
E(y) is the population mean response (i.e. expected
• _____
value of y for all individuals in the population who have
the particular value of x.)
y-hat is an estimate of E(y)
• Note that ______
____.
• The value epsilon is called the error or the deviation.
It has a mean of zero. (And we assume it is normally
distributed.)
Parameters for Regression
• If two variables have a linear relationship, then β1
(the population slope) would be different from
0
_____.
Hypothesis Testing About the Slope
• Statistical significance of a linear relationship can be
evaluated by testing whether the population slope is
0
______
or not.
• This test is done in a similar way to tests with
proportions and means.
• First, the null and alternative hypotheses need
to be determined.
Null and Alternative Hypotheses
• Null hypothesis
β1 = 0
H0: ______
• This would mean that our two variables, x and y,
are
not linearly related
_______
• Alternative hypothesis
β1 ≠ 0
Ha: _________
are linearly related
• The variables x and y ____
Null and Alternative Hypotheses
• The alternative hypothesis can be 1-sided
_______ as well
(β1 > 0 or β1 < 0), but most software use the 2-sided
_______
alternative hypothesis (β1 ≠ 0)
2-sided alternative hypothesis
• We will only use the ________
The Test Statistic
t-statistic is
• For the hypothesis tests for slope, the __________
used.
• The t-statistic is calculated in the same way as before:
The Test Statistic
• When we are using the t-test for the test of the slope, the
degrees of freedom are equal to the sample size minus
two.
n–2
• df = ________
The Test Statistic
• The calculations for the sample slope and its
standard error are complicated
• Luckily, Minitab can do this for us:
Coefficients
Term
Constant
dadheight
b1
Coef
30.34
0.5280
SE Coef
5.08
0.0732
T-Value
5.98
7.21
s.e.(b1)
P-Value
0.000
0.000
VIF
1.00
p-value
t-stat
Example: Age and Reading Distance
• A sample was taken in which subjects were asked their
age, and then they were measured to see how far away
they could read a road sign.
• Age was treated as the explanatory variable, and reading
distance was the response variable.
• There are n = 30 observations
Example: Age and Reading Distance
• The sample slope was –3.0068, which means that for
each additional year of age, the estimated reading
decreased by about 3 feet.
distance ____________
• The standard error for the slope was 0.4243.
Example: Age and Reading Distance
• The t-statistic is calculated like this:
• The Minitab output would look like this:
Example: Age and Reading Distance
• The correct conclusion is that, since the p-value is
< 0.05 , the null hypothesis should be rejected
______
________.
• This would mean the slope is significantly
0
__________
different from ______.
linearly related
• So age and reading distance are
___________________.
Confidence Intervals for Slope
• Just like with means and proportions, confidence intervals
can be made for slopes.
• These intervals are used to estimate the true
value for the population slope.
Confidence Intervals for Slope
• Just like with hypothesis testing, the value for degrees of
Two fewer than the sample size.
freedom is ___________
n–2
• df = ______
Example: Age and Reading Distance
• The 95% confidence interval for the slope from
the reading distance example is
Example: Age and Reading Distance
• The correct interpretation for this confidence
interval is that we are 95% confident that the true
population slope
____________________for
the linear
relationship between reading distance and age is
-3.88 _______
-2.14and ________.
between
• Does this agree with our conclusion from the
YES!
hypothesis test? __________.
Correlation
• Remember, correlation (r) is a measure of
direction and _________
strength for a linear
_________
relationship
• As a note, if you find a significant hypothesis test
for the population slope (so β1 ≠ 0), then the
correlation will also be significantly different from
zero.
Sample Size and Significance
• An important concept to keep in mind is that the
larger the sample size, the more likely it is that
significance would be found for a hypothesis
test
n
• ___increases
p-value decreases
• _________
n increases
• _____
p-value decreases
significance increases
significance increases
If you understand today’s lecture…
14.1, 14.2, 14.4, 14.5, 14.7, 14.9, 14.21,
14.22, 14.24, 14.25, 14.27, 14.28
Objectives:
• Formulate null and alternative hypotheses involving
regression coefficients.
• Calculate T-statistics; determine correct degrees of freedom.
```
Document related concepts
Student's t-test wikipedia, lookup
Taylor's law wikipedia, lookup
Bootstrapping (statistics) wikipedia, lookup
Resampling (statistics) wikipedia, lookup
Misuse of statistics wikipedia, lookup
Degrees of freedom (statistics) wikipedia, lookup
Psychometrics wikipedia, lookup
Foundations of statistics wikipedia, lookup
Omnibus test wikipedia, lookup
Similar | 1,776 | 7,129 | {"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-2018-43 | latest | en | 0.855864 |
http://slidegur.com/doc/343072/rounding | 1,534,409,890,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221210559.6/warc/CC-MAIN-20180816074040-20180816094040-00075.warc.gz | 388,564,301 | 10,479 | ### Rounding
```Rounding
3.7-3.10 (combined)
How we round numbers to any place.
How rounding is used in everyday life.
Questions to ponder today:
What is rounding?
Why do we need to know how to round?
When will we use rounding in real life?
Jack’s parents bought a used car, a
new motorcycle, and a used fourwheeler. The car cost \$8,999. The
motorcycle cost \$9,690. The fourwheeler cost \$4,419. About how much
money did they spend on the three
items?
Let’s talk number lines!
What is a number line?
Why do we use them?
10
15
midpoint
20
WHAT?!?!
20
15
10
20
15
What’s halfway between 10 and 20?
10
200
What’s halfway between 100 and 200?
150
100
Let’s Try!
5,000
How many thousands are in 4,100?
4,500
1 more thousand would be?
4,000
What’s halfway between
4 thousands and 5 thousands?
Round 14,500 to the nearest thousand. How many
thousands are there in 14,500?
What’s 1 more thousand?
15,000
What is halfway between
14,000 and 15,000?
The halfway point is nearer to 15,000
14,500
so 14,500 rounded to the nearest thousand
is ________.
14,000
On your dry erase board – You try!
Round 214,569 the nearest thousand.
Round to
6,700~__________
the nearest thousand. Use
the number
line
16,400~
_________
Practice Problems
The 2012 Super Bowl had an attendance of just 68,658
people. If the headline in the newspaper the next day
did the newspaper round to estimate the total
number of people in attendance?
Application Problem
34,123 people attended a basketball game. 28,310
people attended a football game. About how many
more people attended the basketball game than the
football game? Round to the nearest ten thousands to
might be a better way to compare attendance?
Does it make any difference to round to different place
value spots?
Let’s round 942, 234 to the ten thousands.
Round to the thousands
What changes about the numbers when rounding to
smaller and smaller units? -Discuss with your shoulder
partner.
How can you determine the best estimate
in a situation?
We can notice patterns or data that might
inform my estimate.
Rounding to the largest unit will effect different results for different
numbers.
We must choose the unit based on the situation and what is most
reasonable.
When you round to the largest unit, every other place will have a zero.
Rounding to the largest unit gives you the easiest number to add,
subtract, multiply, or divide.
As you round to smaller units, there are not as many zeros in the number.
Rounding to smaller units gives an estimate that is closer to the actual
value of the number.
STATION TIME!
Look at station chart and decide where you need
to go. Wiggle fingers when you know where you
need to be once we start.
When you see all classmate’s fingers wiggling in
the air- tiptoe to your station and get started.
REMEMBER OUR STATION EXPECTATIONS.
You will need a pencil only.
Now let’s try…
Jack’s parents bought a used car, a
new motorcycle, and a used fourwheeler. The car cost \$8,999. The
motorcycle cost \$9,690. The fourwheeler cost \$4,419. About how much
money did they spend on the three
items?
The cost of tuition at Cornell University is \$43,000 per
year when rounded to the nearest thousand. What is
the greatest possible amount the tuition could be?
What is the least possible amount the tuition could
be?
STATION PRODUCT
Solve the real-life word problem at your desk that matches your symbol.
(pumpkin, ghost, or bat), or you may create a word problem of your own that
includes a time when you had to, or you would have to, use rounding. Be sure
Make sure that you have all of the required components that you need to
have, based on the rubric we made.
**Remember when rounding: Complete the rounding checklist (song)!**
Find the number,
look next door,
5 or higher – add one more,
4 or less- let it rest.
**Be sure to show on your rounded problem(s) that you used this strategy.
Early finisher: Halloween rounding worksheet at desk.
Early FinisherWhat would you do?
You were shopping for Halloween candy, for the
class. You have \$15 to spend. Each bag of candy is
\$5.99. Determine how much money you will have to
spend to satisfy each classmate. (You decide what
they get.)
Think about cost of candy
Think about distribution of candy
SHOW WHAT YOU KNOW!
Exit ticket
On the back of your paper tell me:
3-Things that helped me, or that I
enjoyed
2-New things that I learned
1-Thing I still need clarification
on, or a question I still have.
Just for fun!
History of Halloween!
1000 B.C. – 100 B.C.
The Celts were the first to celebrate a Halloweenish-type of party known as the festival of the dead. This pre-Christian holiday was called Samhain,
pronounced Sah-ween. Sound familiar? Samhain was a time when ghosts walked among the living. Muhahaha!
43 A.D. - 600 A.D.
Christian missionaries were not fans of Samhain and attempted to eradicate the pagan holiday. When that didn’t work, they tried to revamp it and make it
more Christian-like.
800 A.D. – 1000 A.D.
Still unable to shake the Pagan celebration, Pope Gregory III rules that All Saints’ Day, also known as All Hallows’ Day (Nov. 1st) be honored with Samhain.
In celebration, villagers are told to dress up as saints and young men are encouraged to go door-to-door begging for food for the poor. It’s starting to
sound a little more like the Halloween we know, right?
1500s – early 1800s
Fast forward a few hundred years and the night before All Hallows’ Day is now called All Hallows’ Evening, Hallowe’en for short. Protestants make another
attempt to get rid of the pagan practice, but the Celts continue the celebrations with bonfires and children are allowed to beg for money. In America, the
Puritans ban Halloween, Christmas and Easter because they consider them to be Catholic. Only Catholics and Episcopalians take part in Halloween
activities.
Around 1850
Millions of Irish immigrants bring their Halloween traditions to America. They dress up in costumes and visit neighborhood homes asking for food or
money, very similar to today’s trick-or-treating.
1900
Candy Corn hits the market. It’s not yet the quintessential Halloween sweet, but it’s a big hit with farmers because of its “corn” shape.
1921
Anoka, Minnesota is the first American city to officially sanction a citywide Halloween celebration. New York follows in 1923 and L.A. in 1925. By this time,
kids are trick-or-treating and throwing Halloween parties.
1966 & 1978
“It’s the Great Pumpkin, Charlie Brown” makes its TV debut in 1966. And the horror flick “Halloween” hits theaters. Are there any two better Halloween
classics?
2012
Halloween is celebrated by millions around the world and by people of all ages. It’s also a huge money-maker. Costumes, candy, party supplies – the year’s
most spooktacular day is big business.
The modern Halloween celebrates imagination and creativity – but maintains the eerie-edge of its Pagan origin, Samhein.
``` | 1,825 | 6,955 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2018-34 | longest | en | 0.930606 |
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Measurement of the W Boson Mass Ashutosh Kotwal Duke University High Energy Physics Seminar DESY July 14-15, 2009
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Measurement of the W Boson MassAshutosh KotwalDuke University
High Energy Physics SeminarDESY
July 14-15, 2009
� The electroweak gauge sector of the standard model is constrained by three precisely known parameters
� αEM (MZ) = 1 / 127.918(18)
� GF = 1.16637 (1) x 10-5 GeV-2
� MZ = 91.1876 (21) GeV
� At tree-level, these parameters are related to MW by
� MW2 = παΕΜ / 2G√ F sin2θW
� Where θW is the weak mixing angle, defined by
cosθW = MW/MZ
Motivation
� Radiative corrections due to heavy quark and Higgs loops and exotica
Motivation
Motivate the introduction of the ρ parameter: MW2 = ρ [MW(tree)]2
with the predictions (ρ−1) ∼ Μtop2 and (ρ−1) ∼ ln MH
� In conjunction with Mtop, the W boson mass constrains the mass of the Higgs boson, and possibly new particles beyond the standard model
Progress on Mtop at the Tevatron
� From the Tevatron, δMtop = 1.3 GeV => δMH / MH = 11%
� equivalent δMW = 8 MeV for the same Higgs mass constraint
� Current world average δMW = 23 MeV
� progress on δMW now has the biggest impact on Higgs constraint!
� SM Higgs fit: MH = 83+30-23 GeV (gfitter.desy.de)
� LEPII direct searches: MH > 114.4 GeV @ 95% CL (PLB 565, 61)
Motivation
In addition to the Higgs, is there another missing piece in this puzzle?
( AFBb vs ALR: 3.2σ )
Must continue improvingprecision of MW , Mtop ...
other precision measurementsconstrain Higgs, equivalent to δMW ~ 15 MeV
Motivate direct measurement of MW at the 15 MeV level
� SM Higgs fit: MH = 83+30-23 GeV (gfitter.desy.de)
� LEPII direct searches: MH > 114.4 GeV @ 95% CL (PLB 565, 61)
Motivation
?MW
GF
Sin2θW
Mtop MZ
In addition to the Higgs, is there another missing piece in this puzzle?
( AFBb vs ALR: 3.2σ )
Must continue improvingprecision of MW , Mtop ...
other precision measurementsconstrain Higgs, equivalent to δMW ~ 15 MeV
Motivate direct measurement of MW at the 15 MeV level
νN
Mtop vs MHiggs( LEP & SLD Collaborations and LEPEWWG, SLD EW & HF Groups, Physics Reports, Vol. 427 Nos. 5-6, 257 (May 2006) )
MW and sin2θeff provide complementary constraints on MHiggs
Analysis Strategy
W Boson Production at the Tevatron
Neutrino
LeptonW
GluonQuark
Antiquark
Quark-antiquark annihilationdominates (80%)
Lepton pT carries most of W mass information, can be measured precisely (achieved 0.03%)
Initial state QCD radiation is O(10 GeV), measure as soft 'hadronic recoil' incalorimeter (calibrated to ~1%)Pollutes W mass information, fortunately pT(W) << MW
W Boson Production at the Tevatron
Neutrino
LeptonW
GluonQuark
Antiquark
Quark-antiquark annihilationdominates (80%)
Lepton pT carries most of W mass information, can be measured precisely (achieved 0.03%)
Initial state QCD radiation is O(10 GeV), measure as soft 'hadronic recoil' incalorimeter (calibrate to ~1%)Pollutes W mass information, fortunately pT(W) << MW
e
Quadrant of Collider Detector at Fermilab (CDF)
.η = 1Central electromagnetic calorimeter
Central hadronic calorimeter
Select W and Z bosons with central ( | η | < 1 ) leptons
COT providesprecise lepton track momentummeasurement
EM calorimeter provides preciseelectron energymeasurement
Calorimeters measure hadronic recoil particles
Collider Detector at Fermilab (CDF)
Centralhadroniccalorimeter
Muondetector
Centraloutertracker(COT)
CDF W & Z Data Samples
� W, Z, J/ψ and Upsilon decays triggered in the dilepton channel
Analysis of 2.3 fb-1 data in progress
CDF's analysis published in 2007, based on integrated luminosity (collected between February 2002 – September 2003):
� Electron channel: L = 218 pb-1
� Muon channel: L = 191 pb-1
Event selection gives fairly clean samples
� W boson samples' mis-identification backgrounds ~ 0.5%
Outline of AnalysisEnergy scale measurements drive the W mass measurement
� Tracker Calibration
� alignment of the COT (~2400 cells) using cosmic rays
� COT momentum scale and tracker non-linearity constrained using J/ψ µµ and Υ µµ mass fits
� Confirmed using Z µµ mass fit
� EM Calorimeter Calibration
� COT momentum scale transferred to EM calorimeter using a fit to the peak of the E/p spectrum, around E/p ~ 1
� Calorimeter energy scale confirmed using Z ee mass fit
� Tracker and EM Calorimeter resolutions
� Hadronic recoil modelling
� Characterized using pT-balance in Z ll events
Drift Chamber (COT) Alignment
COT endplategeometry
Internal Alignment of COT
� Use a clean sample of ~200k cosmic rays for cell-by-cell internal alignment
� Fit COT hits on both sides simultaneously to a single helix (AK, H. Gerberich and C. Hays, NIMA 506, 110 (2003))
� Time of incidence is a floated parameter in this 'dicosmic fit'
Residuals of COT cells after alignment
Final relative alignment of cells ~5 µm (initial alignment ~50 µm)
Res
idua
l (m
icro
ns)
Cell number (φ)
Cell number (φ)
Before alignment
after alignment
CDFII
Cross-check of COT alignment
� Final cross-check and correction to beam-constrained track curvature based on difference of <E/p> for positrons vs electrons
� Smooth ad-hoc curvature corrections fitted and applied as a function of polar and azimuthal angle: statistical errors => δMW = 6 MeV
CDFII L = 200 pb-1
Signal Simulation and Fitting
Signal Simulation and Template Fitting
� All signals simulated using a fast Monte Carlo
� Generate finely-spaced templates as a function of the fit variable
perform binned maximum-likelihood fits to the data
! Custom fast Monte Carlo makes smooth, high statistics templates
" And provides analysis control over key components of the simulation
MW = 80 GeV
MW = 81 GeVMonte Carlo template
# We will extract the W mass from six kinematic distributions: Transverse mass, charged lepton pT and neutrino pT using both electron and muon channels
Generator-level Signal Simulation
\$ Generator-level input for W & Z simulation provided by RESBOS (C. Balazs & C.-P. Yuan, PRD56, 5558 (1997) and references therein), which
% Calculates triple-differential production cross section, and pT-dependent double-differential decay angular distribution
& calculates boson pT spectrum reliably over the relevant pT range: includes tunable parameters in the non-perturbative regime at low pT
' Radiative photons generated according to energy vs angle lookup table from WGRAD (U. Baur, S. Keller & D. Wackeroth, PRD59, 013002 (1998))
RESBOS
WGRAD
Constraining Boson pT Spectrum
( Fit the non-perturbative parameter g2 in RESBOS to pT(ll) spectra: find g2 = 0.685 ± 0.048
) Consistent with global fits (Landry et al, PRD67, 073016 (2003))
* Negligible effect of second non-perturbative parameter g3
DataSimulation
DataSimulation
∆MW = 3 MeV
Position of peak in boson pT spectrum depends on g2
Fast Monte Carlo Detector Simulation
+ A complete detector simulation of all quantities measured in the data
, First-principles simulation of tracking
- Tracks and photons propagated through a high-resolution 3-D lookup table of material properties for silicon detector and COT
. At each material interaction, calculate
/ Ionization energy loss according to complete Bethe-Bloch formula
0 Generate bremsstrahlung photons down to 4 MeV, using detailed cross section and spectrum calculations
1 Simulate photon conversion and compton scattering
2 Propagate bremsstrahlung photons and conversion electrons
3 Simulate multiple Coulomb scattering, including non-Gaussian tail
4 Deposit and smear hits on COT wires, perform full helix fit including optional beam-constraint
Fast Monte Carlo Detector Simulation
5 A complete detector simulation of all quantities measured in the data
6 First-principles simulation of tracking
7 Tracks and photons propagated through a high-resolution 3-D lookup table of material properties for silicon detector and COT
8 At each material interaction, calculate
9 Ionization energy loss according to complete Bethe-Bloch formula
: Generate bremsstrahlung photons down to 4 MeV, using detailed cross section and spectrum calculations
; Simulate photon conversion and compton scattering
< Propagate bremsstrahlung photons and conversion electrons
= Simulate multiple Coulomb scattering, including non-Gaussian tail
> Deposit and smear hits on COT wires, perform full helix fit including optional beam-constraint
e-
e-
e+Calorim
eter
e-
γ
Tracking Momentum Scale
Tracking Momentum Scale
? Set using J/ψ µµ and Υ µµ resonance and Z µµ masses
@ All are individually consistent with each other
A J/ψ:
B Extracted by fitting J/ψ mass in bins of <1/pT(µ)>, and extrapolating momentum scale to zero curvature
∆p/p = ( -1.64 ± 0.06stat ± 0.24sys ) x 10 -3
<1/pT(µ)> (GeV-1)
∆p/pJ/ψ µµ mass independent of pT(µ)
Default energy loss * 0.94 J/ψ µµ mass fit
DataSimulation
CDFII preliminary L ~ 200 pb-1
Tracking Momentum Scale
C Υ µµ resonance provides
D Momentum scale measurement at higher pT
E Validation of beam-constaining procedure (upsilons are promptly produced)
F Non-beam-constrained and beam-constrained (BC) fits statistically consistent
BC Υ µµ mass fit
DataSimulation
Tracking Momentum Scale Systematics
Systematic uncertainties on momentum scale
Uncertainty dominated by QED radiative corrections and magnetic fieldnon-uniformity
Z µµ Mass Cross-check & Combination
G Using the J/ψ and Υ momentum scale, measured Z mass is consistent with PDG value
H Final combined:� ∆p/p = ( -1.50 ± 0.15independent ± 0.13QED ± 0.07align ) x 10 -3
M(µµ) (GeV)
DataSimulation
CDF II preliminary L ~ 200/pb
∆MW = 17 MeV
Eve
nts
/ 0.5
GeV
EM Calorimeter Response
EM Calorimeter Scale
I E/p peak from W eν decays provides measurements of EM calorimeter scale and its (ET-dependent) non-linearity
J SE = 1 ± 0.00025stat ± 0.00011X0 ± 0.00021Tracker
K Setting SE to 1 using E/p calibration
DataSimulation
ECAL / ptrack
Tail region of E/p spectrumused for tuning model ofradiative material
Consistency of Radiative Material Model
L Excellent description of E/p spectrum tail
M radiative material tune factor: SX0 = 1.004 ± 0.009stat ± 0.002background
achieves consistency with E/p spectrum tail
N CDFSim geometry confirmed as a function of pseudorapidity: SMAT
independent of | η |
Calorimeter tower |iη|
SX0 vs |iη|
ECAL / ptrack
DataSimulation
Default energy loss * 1.004
Measurement of EM Calorimeter Non-linearity
O Perform E/p fit-based calibration in bins of electron ET
P Pameterize non-linear response as: SE = 1 + ξ (ET/GeV – 39)
Q Tune on W and Z data: ξ = (6 ± 7stat) x 10-5
R => ∆MW = 23 MeV
Z dataW data
ET (e) (GeV)ET (e) (GeV)
CDF II L ~ 200/pbCDF II L ~ 200/pbSE
SE
Z ee Mass Cross-check and Combination
S Z mass consistent with E/p-based measurements
T Combining E/p-derived scale & non-linearity measurement with Z ee mass yields the most precise calorimeter energy scale:
U SE = 1.00001 ± 0.00037
M(ee) ( GeV)
DataSimulation
∆MW = 30 MeV
Hadronic Recoil Model
Constraining the Hadronic Recoil Model
Exploit similarity in productionand decay of W and Z bosons
Detector response model forhadronic recoil tuned usingpT-balance in Z ll events
Transverse momentum of Hadronic recoil (u) calculated as 2-vector-sum over calorimeter towers
Tuning Recoil Response Model with Z events
Project the vector sum of pT(ll) and u on a set of orthogonal axes definedby lepton directions
Mean and rms of projections as a function of pT(ll) provideinformation hadronic model parameters
DataSimulation
mea
n of
pT-
bala
nce
(GeV
)
µµ
η
u
Hadronic model parameterstuned by minimizing χ2
between data and simulation
∆MW = 9 MeV
Tuning Recoil Resolution Model with Z events
At low pT(Z), pT-balance constrains hadronic resolution due tounderlying event
At high pT(Z), pT-balance constrains jet resolution
DataSimulation
Res
olut
ion
of p
T-ba
lanc
e (G
eV)
µµ
η
u
∆MW = 7 MeV
Testing Hadronic Recoil Model with W events
lu (recoil)
Recoil projection (GeV) on lepton direction
Compare recoil distributions between simulation and data
DataSimulation
pT(W) comparison
DataSimulation
W Mass Fits
Blind Analysis Technique
V All W mass fit results were blinded with a random [-100,100] MeV offset hidden in the likelihood fitter
W Blinding offset removed after the analysis was declared frozen
X Technique allows to study all aspects of data while keeping W mass result unknown within 100 MeV
W Transverse Mass Fits
Muons DataSimulation
W Lepton pT Fits
Electrons
DataSimulation
Transverse Mass Fit Uncertainties (MeV)
electrons commonW statistics 48 54 0Lepton energy scale 30 17 17Lepton resolution 9 3 -3Recoil energy scale 9 9 9Recoil energy resolution 7 7 7Selection bias 3 1 0Lepton removal 8 5 5Backgrounds 8 9 0pT(W) model 3 3 3Parton dist. Functions 11 11 11QED rad. Corrections 11 12 11
Y[Z \] ^ _` _ \a b ] \ c[d ef gh gi
Y[Z \] ^ i g i j
muons
Systematic uncertainties shown in green: statistics-limited by control data samples
W charge asymmetryfrom Tevatronhelps with PDFs
(CDF, PRL 99:151801, 2007; Phys. Rev. D 77:112001, 2008)
Tevatron Run 1 (100 pb-1) W Mass Systematic Uncertainties (MeV)
W statistics 100 65 60Lepton energy scale 85 75 56Lepton resolution 20 25 19Recoil model 35 37 35pT(W) 20 15 15Selection bias 18 - 12Backgrounds 25 5 9Parton dist. Functions 15 15 8QED rad. Corrections 11 11 12
10 10 10
kml npo q rs s r rt us
CDF µ CDF v D0 e
Γ(W)
For comparison to run 2 analysis
Comparisons
(CDF Run II: PRL 99:151801, 2007; PRD 77:112001, 2008)
Updated MW vs MtopMW vs Mtop
Standard Model Higgs Constraints
Improvement of MW Uncertainty with Sample Statistics
Preliminary Studies of 2.3 fb-1 Data
CDF has started the analysis of 2.3 fb-1 of data, with the goal of measuringMW with precision better than 25 MeV
Tracker alignment with cosmic rays has been completed for this dataset
Lepton resolutions as good as they were in 200 pb-1 sample
J/ψ->µµ Υ->µµ
Preliminary Studies of 2.3 fb-1 DataStatistical errors on all leptoncalibration fits have scaled with statistics
Detector and data qualitymaintained over time
detailed calibrations in progress
W->eν
Ζ->ee
Ζ->µµ
Preliminary Studies of 2.3 fb-1 Data
Recoil resolution not significantly degradedat higher instantaneousluminosity
W->eν
statistical errors on transverse mass fits are scaling with statistics
W->µν
Summary
w The W boson mass is a very interesting parameter to measure with increasing precision
x CDF Run 2 W mass result with 200 pb-1 data:
y MW = 80413 ± 34stat ± 34syst MeV = 80413 ± 48 MeV
z D0 Run 2 W mass result with 1 fb-1 data:
{ MW = 80401 ± 21stat ± 38syst MeV = 80401 ± 43 MeV
| Most systematics limited by statistics of control samples
} CDF and D0 are both working on δMW < 25 MeV measurements from ~ 2 fb-1 (CDF) and ~ 4 fb-1 (D0)
E/p Calibration vs Z ee mass consistency
~ Inclusion of hadronic calorimeter leakage distribution has a ~150 MeV effect on the fitted EM calorimeter scale from the E/p distribution
� Modelling the bremsstrahlung spectrum down to 4 MeV (from 40 MeV cutoff) has a ~60 MeV effect on the E/p calibration
� Modelling the calorimeter non-linearity as a property of individual particles has a ~30 MeV effect
� Collectively, these simulated effects in the Run 2 analysis affect the consistency of the Z mass by ~240 MeV
Tracking Momentum Scale
� Υ µµ resonance provides
� Momentum scale measurement at higher pT
� Validation of beam-constaining procedure (upsilons are promptly produced)
� Non-beam-constrained and beam-constrained fits statistically consistent
BC Υ µµ mass fitNon-BC Υ µµ mass fit
DataSimulation
DataSimulation
W Mass Measurement at the Tevatron
pT(W)=0
pT(W) 0≠
measured
(figures from Abbott et. al. (D0 Collaboration), PRD 58, 092003 (1998))
MT = (2 p√ Tl pT
ν (1 – cos φlν)Insensitive to pT(W) to first order
Reconstruction of pTν sensitive to
hadronic response and multiple interactions
pT(l) fit: provides cross-check of production model:
Needs theoretical model of pT(W)
PT(ν) fit provides cross-check of hadronic modelling
W mass information contained in location of transverse Jacobian edge
Lepton Resolutions
� Tracking resolution parameterized in the fast Monte Carlo by
� Drift chamber hit resolution σh= 150 ± 3stat µm
� Beamspot size σb= 39 ± 3stat µm
� Tuned on the widths of the Z µµ (beam constrained) and Y µµ (both beam constrained and non-beam constrained) mass peaks
. => ∆MW = 3 MeV (muons)
� Electron cluster resolution parameterized in the fast Monte Carlo by
� 13.5% / E√ T (sampling term)
� Primary constant term κ = 0.89 ± 0.15stat %
� Secondary photon resolution κγ = 8.3 ± 2.2stat %
� Tuned on the widths of the E/p peak and the Z ee peak (selecting radiative electrons) => ∆MW = 9 MeV (electrons)
Lepton Tower Removal
� We remove the calorimeter towers containing lepton energy from the hadronic recoil calculation
� Lost underlying event energy is measured in φ-rotated windows
Electron channel W data
Muon channel W data
∆MW = 8 MeV
Calorimeter Simulation for Electrons and Photons
� Distributions of energy loss calculated based on expected shower profiles as a function of ET
� Leakage into hadronic calorimeter
� Absorption in the coil
� Relevant for E/p lineshape
��
� Energy-dependent gain (non-linearity) parameterized and fit from data
� Energy resolution parameterized as fixed sampling term and two tunable constant terms
� Constant terms are fit from the width of E/p peak and Z ee mass peak
Combined Results
� Combined electrons (3 fits): MW = 80477 ± 62 MeV, P(χ2) = 49%
� Combined muons (3 fits): MW = 80352 ± 60 MeV, P(χ2) = 69%
� All combined (6 fits): MW = 80413 ± 48 MeV, P(χ2) = 44%
Lepton pT and Missing ET Fit Uncertainties
Backgrounds in the W sample
Source
Mis-identified QCD jetsDecays-in-flightCosmic rays
Fraction (electrons) Fraction (muons)Z -> ll 0.24 ± 0.04 % 6.6 ± 0.3 %W -> τν 0.93 ± 0.03 % 0.89 ± 0.02 %
0.25 ± 0.15 % 0.1 ± 0.1 %0.3 ± 0.2 %
0.05 ± 0.05 %
Backgrounds are small (except Z µµ with a forward muon)
backgrounds contribute systematic uncertainty of 9 MeV on transverse mass fit | 5,173 | 18,392 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2024-18 | latest | en | 0.799246 |
https://www.nag.com/numeric/nl/nagdoc_27cpp/flhtml/g03/g03gaf.html | 1,627,955,289,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154408.7/warc/CC-MAIN-20210802234539-20210803024539-00128.warc.gz | 934,496,215 | 8,586 | # NAG FL Interfaceg03gaf (gaussian_mixture)
## 1Purpose
g03gaf performs a mixture of Normals (Gaussians) for a given (co)variance structure.
## 2Specification
Fortran Interface
Subroutine g03gaf ( n, m, x, ldx, isx, nvar, ng, popt, prob, w, g, sopt, s, lds, sds, f, tol,
Integer, Intent (In) :: n, m, ldx, isx(m), nvar, ng, popt, lprob, riter, sopt, lds, sds Integer, Intent (Inout) :: niter, ifail Real (Kind=nag_wp), Intent (In) :: x(ldx,m), tol Real (Kind=nag_wp), Intent (Inout) :: prob(lprob,ng), g(nvar,ng), s(lds,sds,*), f(n,ng) Real (Kind=nag_wp), Intent (Out) :: w(ng), loglik
#include <nag.h>
void g03gaf_ (const Integer *n, const Integer *m, const double x[], const Integer *ldx, const Integer isx[], const Integer *nvar, const Integer *ng, const Integer *popt, double prob[], const Integer *lprob, Integer *niter, const Integer *riter, double w[], double g[], const Integer *sopt, double s[], const Integer *lds, const Integer *sds, double f[], const double *tol, double *loglik, Integer *ifail)
The routine may be called by the names g03gaf or nagf_mv_gaussian_mixture.
## 3Description
A Normal (Gaussian) mixture model is a weighted sum of $k$ group Normal densities given by,
$p x∣w,μ,Σ = ∑ j=1 k wj g x∣μj,Σj , x∈ℝp$
where:
• $x$ is a $p$-dimensional object of interest;
• ${w}_{j}$ is the mixture weight for the $j$th group and $\sum _{\mathit{j}=1}^{k}{w}_{j}=1$;
• ${\mu }_{j}$ is a $p$-dimensional vector of means for the $j$th group;
• ${\Sigma }_{j}$ is the covariance structure for the $j$th group;
• $g\left(·\right)$ is the $p$-variate Normal density:
$g x∣μj,Σj = 1 2π p/2 Σj 1/2 exp - 12 x-μj Σ j -1 x-μj T .$
Optionally, the (co)variance structure may be pooled (common to all groups) or calculated for each group, and may be full or diagonal.
## 4References
Hartigan J A (1975) Clustering Algorithms Wiley
## 5Arguments
1: $\mathbf{n}$Integer Input
On entry: $n$, the number of objects. There must be more objects than parameters in the model.
Constraints:
• if ${\mathbf{sopt}}=1$, ${\mathbf{n}}>{\mathbf{ng}}×\left({\mathbf{nvar}}×{\mathbf{nvar}}+{\mathbf{nvar}}\right)$;
• if ${\mathbf{sopt}}=2$, ${\mathbf{n}}>{\mathbf{nvar}}×\left({\mathbf{ng}}+{\mathbf{nvar}}\right)$;
• if ${\mathbf{sopt}}=3$, ${\mathbf{n}}>2×{\mathbf{ng}}×{\mathbf{nvar}}$;
• if ${\mathbf{sopt}}=4$, ${\mathbf{n}}>{\mathbf{nvar}}×\left({\mathbf{ng}}+1\right)$;
• if ${\mathbf{sopt}}=5$, ${\mathbf{n}}>{\mathbf{nvar}}×{\mathbf{ng}}+1$.
2: $\mathbf{m}$Integer Input
On entry: the total number of variables in array x.
Constraint: ${\mathbf{m}}\ge 1$.
3: $\mathbf{x}\left({\mathbf{ldx}},{\mathbf{m}}\right)$Real (Kind=nag_wp) array Input
On entry: ${\mathbf{x}}\left(\mathit{i},\mathit{j}\right)$ must contain the value of the $\mathit{j}$th variable for the $\mathit{i}$th object, for $\mathit{i}=1,2,\dots ,n$ and $\mathit{j}=1,2,\dots ,{\mathbf{m}}$.
4: $\mathbf{ldx}$Integer Input
On entry: the first dimension of the array x as declared in the (sub)program from which g03gaf is called.
Constraint: ${\mathbf{ldx}}\ge {\mathbf{n}}$.
5: $\mathbf{isx}\left({\mathbf{m}}\right)$Integer array Input
On entry: if ${\mathbf{nvar}}={\mathbf{m}}$ all available variables are included in the model and isx is not referenced; otherwise the $j$th variable will be included in the analysis if ${\mathbf{isx}}\left(\mathit{j}\right)=1$ and excluded if ${\mathbf{isx}}\left(\mathit{j}\right)=0$, for $\mathit{j}=1,2,\dots ,{\mathbf{m}}$.
Constraint: if ${\mathbf{nvar}}\ne {\mathbf{m}}$, ${\mathbf{isx}}\left(\mathit{j}\right)=1$ for nvar values of $\mathit{j}$ and ${\mathbf{isx}}\left(\mathit{j}\right)=0$ for the remaining ${\mathbf{m}}-{\mathbf{nvar}}$ values of $\mathit{j}$, for $\mathit{j}=1,2,\dots ,{\mathbf{m}}$.
6: $\mathbf{nvar}$Integer Input
On entry: $p$, the number of variables included in the calculations.
Constraint: $1\le {\mathbf{nvar}}\le {\mathbf{m}}$.
7: $\mathbf{ng}$Integer Input
On entry: $k$, the number of groups in the mixture model.
Constraint: ${\mathbf{ng}}\ge 1$.
8: $\mathbf{popt}$Integer Input
On entry: if ${\mathbf{popt}}=1$, the initial membership probabilities in prob are set internally; otherwise these probabilities must be supplied.
9: $\mathbf{prob}\left({\mathbf{lprob}},{\mathbf{ng}}\right)$Real (Kind=nag_wp) array Input/Output
On entry: if ${\mathbf{popt}}\ne 1$, ${\mathbf{prob}}\left(i,j\right)$ is the probability that the $i$th object belongs to the $j$th group. (These probabilities are normalised internally.)
On exit: ${\mathbf{prob}}\left(i,j\right)$ is the probability of membership of the $i$th object to the $j$th group for the fitted model.
10: $\mathbf{lprob}$Integer Input
On entry: the first dimension of the array prob as declared in the (sub)program from which g03gaf is called.
Constraint: ${\mathbf{lprob}}\ge {\mathbf{n}}$.
11: $\mathbf{niter}$Integer Input/Output
On entry: the maximum number of iterations.
Suggested value: $15$
On exit: the number of completed iterations.
Constraint: ${\mathbf{niter}}\ge 1$.
12: $\mathbf{riter}$Integer Input
On entry: if ${\mathbf{riter}}>0$, membership probabilities are rounded to $0.0$ or $1.0$ after the completion of every riter iterations.
Suggested value: $5$
13: $\mathbf{w}\left({\mathbf{ng}}\right)$Real (Kind=nag_wp) array Output
On exit: ${w}_{j}$, the mixing probability for the $j$th group.
14: $\mathbf{g}\left({\mathbf{nvar}},{\mathbf{ng}}\right)$Real (Kind=nag_wp) array Output
On exit: ${\mathbf{g}}\left(i,j\right)$ gives the estimated mean of the $i$th variable in the $j$th group.
15: $\mathbf{sopt}$Integer Input
On entry: determines the (co)variance structure:
${\mathbf{sopt}}=1$
Groupwise covariance matrices.
${\mathbf{sopt}}=2$
Pooled covariance matrix.
${\mathbf{sopt}}=3$
Groupwise variances.
${\mathbf{sopt}}=4$
Pooled variances.
${\mathbf{sopt}}=5$
Overall variance.
Constraint: ${\mathbf{sopt}}=1$, $2$, $3$, $4$ or $5$.
16: $\mathbf{s}\left({\mathbf{lds}},{\mathbf{sds}},*\right)$Real (Kind=nag_wp) array Output
Note: the last dimension of the array s must be at least ${\mathbf{ng}}$ if ${\mathbf{sopt}}=1$, and at least $1$ otherwise.
On exit: if ${\mathbf{sopt}}=1$, ${\mathbf{s}}\left(i,j,k\right)$ gives the $\left(i,j\right)$th element of the $k$th group.
If ${\mathbf{sopt}}=2$, ${\mathbf{s}}\left(i,j,1\right)$ gives the $\left(i,j\right)$th element of the pooled covariance.
If ${\mathbf{sopt}}=3$, ${\mathbf{s}}\left(j,k,1\right)$ gives the $j$th variance in the $k$th group.
If ${\mathbf{sopt}}=4$, ${\mathbf{s}}\left(j,1,1\right)$ gives the $j$th pooled variance.
If ${\mathbf{sopt}}=5$, ${\mathbf{s}}\left(1,1,1\right)$ gives the overall variance.
17: $\mathbf{lds}$Integer Input
On entry: the first dimension of the (co)variance structure s.
Constraints:
• if ${\mathbf{sopt}}=5$, ${\mathbf{lds}}=1$;
• otherwise ${\mathbf{lds}}={\mathbf{nvar}}$.
18: $\mathbf{sds}$Integer Input
On entry: the second dimension of the (co)variance structure s.
Constraints:
• if ${\mathbf{sopt}}=1$ or $2$, ${\mathbf{sds}}\ge {\mathbf{nvar}}$;
• if ${\mathbf{sopt}}=3$, ${\mathbf{sds}}\ge {\mathbf{ng}}$;
• if ${\mathbf{sopt}}=4$ or $5$, ${\mathbf{sds}}\ge 1$.
19: $\mathbf{f}\left({\mathbf{n}},{\mathbf{ng}}\right)$Real (Kind=nag_wp) array Output
On exit: ${\mathbf{f}}\left(i,j\right)$ gives the $p$-variate Normal (Gaussian) density of the $i$th object in the $j$th group.
20: $\mathbf{tol}$Real (Kind=nag_wp) Input
On entry: iterations cease the first time an improvement in log-likelihood is less than tol. If ${\mathbf{tol}}\le 0$ a value of ${10}^{-3}$ is used.
21: $\mathbf{loglik}$Real (Kind=nag_wp) Output
On exit: the log-likelihood for the fitted mixture model.
22: $\mathbf{ifail}$Integer Input/Output
On entry: ifail must be set to $0$, $-1$ or $1$ to set behaviour on detection of an error; these values have no effect when no error is detected.
A value of $0$ causes the printing of an error message and program execution will be halted; otherwise program execution continues. A value of $-1$ means that an error message is printed while a value of $1$ means that it is not.
If halting is not appropriate, the value $-1$ or $1$ is recommended. If message printing is undesirable, then the value $1$ is recommended. Otherwise, the value $0$ is recommended. When the value $-\mathbf{1}$ or $\mathbf{1}$ is used it is essential to test the value of ifail on exit.
On exit: ${\mathbf{ifail}}={\mathbf{0}}$ unless the routine detects an error or a warning has been flagged (see Section 6).
## 6Error Indicators and Warnings
If on entry ${\mathbf{ifail}}=0$ or $-1$, explanatory error messages are output on the current error message unit (as defined by x04aaf).
Errors or warnings detected by the routine:
${\mathbf{ifail}}=1$
On entry, ${\mathbf{n}}=〈\mathit{\text{value}}〉$ and $p=〈\mathit{\text{value}}〉$.
Constraint: ${\mathbf{n}}>p$, the number of parameters, i.e., too few objects have been supplied for the model.
${\mathbf{ifail}}=2$
On entry, ${\mathbf{m}}=〈\mathit{\text{value}}〉$.
Constraint: ${\mathbf{m}}\ge 1$.
${\mathbf{ifail}}=4$
On entry, ${\mathbf{ldx}}=〈\mathit{\text{value}}〉$ and ${\mathbf{n}}=〈\mathit{\text{value}}〉$.
Constraint: ${\mathbf{ldx}}\ge {\mathbf{n}}$.
${\mathbf{ifail}}=5$
On entry, ${\mathbf{nvar}}=〈\mathit{\text{value}}〉$ and ${\mathbf{m}}=〈\mathit{\text{value}}〉$.
Constraint: $1\le {\mathbf{nvar}}\le {\mathbf{m}}$.
${\mathbf{ifail}}=6$
On entry, ${\mathbf{nvar}}\ne {\mathbf{m}}$ and isx is invalid.
${\mathbf{ifail}}=7$
On entry, ${\mathbf{ng}}=〈\mathit{\text{value}}〉$.
Constraint: ${\mathbf{ng}}\ge 1$.
${\mathbf{ifail}}=8$
On entry, ${\mathbf{popt}}\ne 1$ or $2$.
${\mathbf{ifail}}=9$
On entry, row $〈\mathit{\text{value}}〉$ of supplied prob does not sum to $1$.
${\mathbf{ifail}}=10$
On entry, ${\mathbf{lprob}}=〈\mathit{\text{value}}〉$ and ${\mathbf{n}}=〈\mathit{\text{value}}〉$.
Constraint: ${\mathbf{lprob}}\ge {\mathbf{n}}$.
${\mathbf{ifail}}=11$
On entry, ${\mathbf{niter}}=〈\mathit{\text{value}}〉$.
Constraint: ${\mathbf{niter}}\ge 1$.
${\mathbf{ifail}}=16$
On entry, ${\mathbf{sopt}}\ne 1$, $2$, $3$, $4$ or $5$.
${\mathbf{ifail}}=18$
On entry, ${\mathbf{lds}}=〈\mathit{\text{value}}〉$ was invalid.
${\mathbf{ifail}}=19$
On entry, ${\mathbf{sds}}=〈\mathit{\text{value}}〉$ was invalid.
${\mathbf{ifail}}=44$
A covariance matrix is not positive definite, try a different initial allocation.
${\mathbf{ifail}}=45$
An iteration cannot continue due to an empty group, try a different initial allocation.
${\mathbf{ifail}}=-99$
See Section 7 in the Introduction to the NAG Library FL Interface for further information.
${\mathbf{ifail}}=-399$
Your licence key may have expired or may not have been installed correctly.
See Section 8 in the Introduction to the NAG Library FL Interface for further information.
${\mathbf{ifail}}=-999$
Dynamic memory allocation failed.
See Section 9 in the Introduction to the NAG Library FL Interface for further information.
Not applicable.
## 8Parallelism and Performance
g03gaf is threaded by NAG for parallel execution in multithreaded implementations of the NAG Library.
g03gaf makes calls to BLAS and/or LAPACK routines, which may be threaded within the vendor library used by this implementation. Consult the documentation for the vendor library for further information.
Please consult the X06 Chapter Introduction for information on how to control and interrogate the OpenMP environment used within this routine. Please also consult the Users' Note for your implementation for any additional implementation-specific information.
None.
## 10Example
This example fits a Gaussian mixture model with pooled covariance structure to New Haven schools test data, see Table 5.1 (p. 118) in Hartigan (1975).
### 10.1Program Text
Program Text (g03gafe.f90)
### 10.2Program Data
Program Data (g03gafe.d)
### 10.3Program Results
Program Results (g03gafe.r) | 4,082 | 11,801 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 200, "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-2021-31 | latest | en | 0.591488 |
https://www.jiskha.com/display.cgi?id=1301947417 | 1,501,137,839,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549427749.61/warc/CC-MAIN-20170727062229-20170727082229-00184.warc.gz | 787,026,794 | 3,961 | # math
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the perimeter of a rectangle is twice the sum of its length and its width. the perimeter is 40 meters and its lenth is 2 meters more then twice its width. what is the length?
• math -
p=2.(w+l) p=40 l=2+w 40=2(w+(2=w)divide by 2 20=2w=2 subtract by 2 18=2w divide by 2 l=2+w substitude w for 9 l=11 answer =11
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Post a New Question | 650 | 2,500 | {"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-2017-30 | longest | en | 0.918492 |
http://adriandorn.com/complex.htm | 1,544,479,749,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376823445.39/warc/CC-MAIN-20181210212544-20181210234044-00294.warc.gz | 12,167,567 | 6,326 | The Elegant Way To Generate The Complex Number Field
Definition: A ring <R, +, •> is a set R together with two binary operations + and • of addition and multiplication defined on R such that the following axioms are satisfied:
1) <R, +> is an abelian group.
2) Multiplication is closed and associative.
3) For all a, b, c contained in R, the left distributive law, a(b + c) = (ab) + (ac), and the right distributive law, (a + b) = (ac) + (bc), hold.
Definition: A ring in which the multiplication is commutative is a commutative ring. A ring R with a multiplicative identity 1 such that 1x = x1 = x for all x contained in R is a ring with unity. A multiplicative identity in a ring is unity, and this unity is unique.
Definition: If a and b are two nonzero elements of a ring R such that a•b = 0, then a and b are divisors of zero.
Definition: An integral domain is a commutative ring with unity containing no divisors of zero.
If an integral domain is such that every nonzero element has a multiplicative inverse, then it is a field. Although many integral domains, such as the integers Z, do not form a field, every integral domain can be regarded as being contained in a certain field, a field of quotients of the integral domain. For example, the integers are contained in the field Q, the rational numbers, whose elements can all be expressed as quotients of integers.
* A note on notation. Notation is very important in mathematics, witness the notation differences between Newton's and Leibnitz's calculus. Who won out? I've tried to avoid confusion by trying to be as careful as I can with the relevant notation. In particular, [.....] stands for equivalence class or coset, and <.....> for an ideal of a ring. However, in the act of factoring, the ideal takes on the role of zero equivalence class -- the additive identity element -- in the quotient ring thus generated. So, it can be a little confusing. *
Definition: If R is a commutative ring with unity and a is in R, the ideal {ra | r is in R} of all multiples of a is the principle ideal generated by a and is denoted by "<a>".
An ideal is to a ring as a normal subgroup is to a group.
Definition: Let R be a ring. A polynomial f(x) with coefficients in R is an infinite sum
∑ aixi = a0 + a1x + ..... + anxn + .....,
*[the index i ranges from 0 to ∞]
where ai is in R and ai = 0 for all but a finite number of values of i. The ai are coefficients of f(x).
If F is a field, then F[x] -- the set of all polynomials in an indeterminate x with coefficients in F -- is an integral domain. Note that F[x] is not a field, for x is not a unit in F[x]. An element of a field is a unit of that field if it has a multiplicative inverse; and there is no polynomial f(x) in F[x] such that x•f(x) = 1. For that to be the case, we would have to have the possibility of negative exponents for the x's, which we don't by the definition above.
Definition: Let F be a subfield of a field E, and let α be an element of E. Let f(x) = a0 + a1x + ... + anxn be in F[x], and let φα: F[x} → E be the basic homomorphism taking f(x) to f(α). Let "f(α)" denote
(f(x))φα = a0 + a1α + ... + anαn.
If f(α) = 0, then α is a zero of f(x).
[A homomorphism can be thought of as a map that preserves structure and computational integrity.]
If f(x) has no zero in F, we have to manufacture in some fashion a field E containing F such that there is an α in E with f(α) = (f(x))φα = 0. How can we build E? The key is to realize that E is to contain the image (F[x])φα of F[x] under our basic homomorphism φα.
* It's almost like we're going in the back door to develop F[x] into a field. Find a polynomial that has no zeros or roots in the field F, use it to generate a maximal (principle) ideal, then use this to factor the ring of polynomials thereby constructing a field of polynomials, and by so accomplishing, ensure that all polynomials over the extended field have zeros or solutions. The existence of multiplicative inverses insures the absence of zero divisors, an important feature when factoring. *
Theorem: Let R be a commutative ring with unity. Then M is a maximal ideal if and only if R/M is a field.
Theorem: (Fundamental Homomorphism Theorem). Let φ be a homomorphism of a ring R into a ring R′ with kernel K. Then is a ring and there is a canonical isomorphism of Rφ with R/K.
[An isomorphism is a homomorphism that is one-to-one and onto. Intuitively, the domain and its image under an isomorphism contain the same number of elements [cardinality] in their sets and respect the same operations of addition and multiplication. Canonical just means natural in the sense of what one would expect.]
This theorem suggests that we try to form E by forming a factor ring F[x]/N for some suitable ideal N of F[x].
Theorem: F[x]/(p(x)) is a field if and only if p(x) is irreducible.
* Outline of Proof: Given: F[x] is a commutative ring with unity [the multiplicative identity]. We need to show F[x]/(p(x)), for p(x) irreducible, is a field, and in order to do that we need to show that nonzero elements are invertible, that is to say, have inverses. The proof of inverses for non-zero polynomials is too lengthy to include here. It can be found on the page on Quotient Rings below. In the proof, as it turns out, if p(x) is not irreducible, then F[x]/(p(x)) has zero divisors, in which case it fails to even be an integral domain.*
The Complex Number Field: Let F = R, and let f(x) = (x2 + 1), which is well known to have no zeros in R, and thus is irreducible over R. Then <x2 + 1> is a maximal ideal in R[x], so R[x]/<x2 + 1> is a field. Identifying r in R with r + <x2 + 1>, we can view R as a subfield of E = R[x]/<x2 + 1>. Let
α = x + <x2 + 1>.
Computing in R[x]/<x2 + 1>, we find
α2 + 1 = (x + <x2 + 1>)2 + (1 + <x2 + 1>)
= (x2 + 1) + <x2 + 1> = 0.
Thus α is a zero of (x2 + 1) in R[x]/<x2 + 1>. We can view R[x]/<x2 + 1> as an extension of R. Then R(α) = R[x]/<x2 + 1> and consists of all elements of the form a + bα for a, b in R. But since α2 + 1 = 0, we see that α plays the role of i in C, and a + bα plays the role of (a + bi) in C. Therefore R(α) is isomorphic to C.
We're identifying <x2 + 1> with zero through the factor ring where the kernel is the zero element, analogous to a factor group, the role of normal subgroup being played by the ideal. The polynomial (x2 + 1) has no solution in R, the Real Numbers, but it does in C, the Complex Numbers -- the solution being the square root of minus one, i = √-1.
(x2 + 1)φα = i2 + 1 = 0
We're basically extending the field R to C; that is, in the act of deflating, we expand. The external appearance of R[x] shrinks and at the same instant its internal configuration embeds itself in the newly defined algebraic structure, infusing its integral domain structure into that of the field -- R[x]/<x2 + 1>.
This is very similar to how the field of rational numbers, Q, includes the integral domain of integers, Z. Z before the transition is all of the universe, after, it resides inside the more complex Q. Imagine the integers spread out on a line, suddenly the formerly blank space between pairs is filled with an infinite number of fractions -- quotients of integers. By virtue of these fractions, the non-zero integers take on the new role of units. Z shrinks in relative size, yet grows, through sub-divisions, to a vast sea of multi-dimensional relationships and interconnections.
Here's what's happening: The axiom we needed to jump from an integral domain to a field was the existence of a multiplicative inverse for each non-zero polynomial -- f(x) + <p(x)> -- in our quotient ring. Those are our units. A proof that they exist can be found at the link on Quotient Rings below. That gives us the added internal structure of an abelian multiplicative group and hence, a field. The symmetry of connections after collapse increases by a whole new order. Topologically, a refinement has been made.
If you're interested in understanding how Quotient Rings work, what constitutes a coset, etcetera, check out Quotient Rings of Polynomial Rings in HTML format, or Quotient Rings of Polynomial Rings in PDF format, by Professor Bruce Ikenaga of Millersville University.
Here's the beauty and magic of this. Regardless of whether or not you understand the theory, the math, what is going on is quite remarkable. We're creating an extension of an integral domain to a field for purposes of solving an irreducible polynomial, one that's unsolvable in the environment restricted to the real numbers. In the world of algebraic hierarchy, a field is one level above an integral domain in structural identity and therefore in complexity, similar to the difference in atom arrangements between liquid water and ice crystals. So, we're generating a more complex structure by superimposing a conditional symmetry on the parent structure through the process of factoring. A transformation to a new species occurs, one whose interconnections are far more intricate.
By identifying the irreducble polynomial with the zero element of the factor ring, we map all polynomials of the parent ring that contain the irreducble as a factor to the kernel [the maximal ideal], the zero element of the factor ring, our manufactured field. All other non-zero polynomials [units] are considered congruent if they differ by a multiple of the zero-class -- [x2 + 1] -- and are thereby grouped into cosets of equivalence classes.
By placing boundaries, setting initial conditions, on something infinite and incomprehensible, we make it comprehensible. To reiterate, by partitioning the ring of polynomials with real number coefficients -- R[x] -- into distinct, disjoint cosets modulo <x2 + 1> we are essentially collapsing R[x] by a subring containing all polynomials that share (x2 + 1) as a factor -- (x2 + 1)•f(x) for all polynomials f(x) in R[x]. Simultaneously we are increasing the complexity of R[x]. We're going from a one-dimensional system -- the Real Numbers -- to a two-dimensional system -- the Complex Numbers.
To paraphrase from above: Then R(i) = R[x]/<x2 + 1> and consists of all elements of the form a + bi for a, b in R. The extension creates the vector space of two-dimensions. From this we see that the Complex Number Field is isomorphic to the Euclidean Plane. One axis for the Reals and the other for "i", the imaginary number -- √-1. Any point in the plane-space can thus be mapped by a complex number.
By factoring the parent ring with the stamp of the irreducible polynomial, we force (x2 + 1) to equal zero and thereby introduce the number √-1. By increasing the complexity of the integral domain structure, we generate another dimension in the number field. R thus becomes a subfield component of C; that is to say, <R, √-1> is an extension over the fixed field R.
The mathematics of evolution.
For p(x) = (x2 + 1):
The real line is the function line, each point, a function or polynomial, f(x). To it is added a function-multiple of i = p(x). The linear combination, f(x) + <p(x)>, maps isomorphically to a + bi on the complex plane.
The field of polynomials, R[x]/<p(x)> equal to R(i) ≈ C can be mapped to the complex plane by cosines and sines.
The exponential function e equals the sum of the power series expansions for cos(θ) and i sin(θ).
From: e = cos(θ) + i sin(θ),
We get: ei f(x) = cos(f(x)) + i sin(f(x))
Replacing θ with f(x) in R[x] maps them isomorphically to the complex plane. Each non-zero number on the Real line must equal, or is isomorphic to, a function in R[x].
The zero equivalence class, [p(x)], is orthogonal to the function line, and is therefore at 90 degrees, or π/2 radians. Therefore, for x = 0, the cosine of f(x) + 0[p(x)] equals one, and the sine equals zero. And for a nonzero multiple of [p(x)], the sine [p(x)], or the i axis in the complex plane, is equal to one (1) and the cosine, zero.
ei[p(x)] = cos([p(x)]) + i sin([p(x)]) = i
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black & white version
************************** | 3,092 | 11,993 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2018-51 | latest | en | 0.941755 |
https://support.nag.com/numeric/nl/nagdoc_28.6/flhtml/s/sconts.html | 1,701,657,678,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100523.4/warc/CC-MAIN-20231204020432-20231204050432-00695.warc.gz | 631,137,961 | 9,907 | # NAG FL InterfaceS (Specfun)Approximations of Special Functions
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S (Specfun) Chapter Introduction – A description of the Chapter and an overview of the algorithms available.
Routine
Mark of
Introduction
Purpose
s01baf 14 nagf_specfun_log_shifted
$\mathrm{ln}\left(1+x\right)$
s01eaf 14 nagf_specfun_exp_complex
Complex exponential, ${e}^{z}$
s07aaf 1 nagf_specfun_tan
$\mathrm{tan}x$
s09aaf 1 nagf_specfun_arcsin
$\mathrm{arcsin}x$
s09abf 3 nagf_specfun_arccos
$\mathrm{arccos}x$
s10aaf 3 nagf_specfun_tanh
$\mathrm{tanh}x$
s10abf 4 nagf_specfun_sinh
$\mathrm{sinh}x$
s10acf 4 nagf_specfun_cosh
$\mathrm{cosh}x$
s11aaf 4 nagf_specfun_arctanh
$\mathrm{arctanh}x$
s11abf 4 nagf_specfun_arcsinh
$\mathrm{arcsinh}x$
s11acf 4 nagf_specfun_arccosh
$\mathrm{arccosh}x$
s13aaf 1 nagf_specfun_integral_exp
Exponential integral ${E}_{1}\left(x\right)$
s13acf 2 nagf_specfun_integral_cos
Cosine integral $\mathrm{Ci}\left(x\right)$
Sine integral $\mathrm{Si}\left(x\right)$
s14aaf 1 nagf_specfun_gamma
Gamma function
s14abf 8 nagf_specfun_gamma_log_real
Log gamma function, real argument
s14acf 14 nagf_specfun_polygamma
$\psi \left(x\right)-\mathrm{ln}x$
Scaled derivatives of $\psi \left(x\right)$
s14aef 20 nagf_specfun_psi_deriv_real
Polygamma function ${\psi }^{\left(n\right)}\left(x\right)$ for real $x$
s14aff 20 nagf_specfun_psi_deriv_complex
Polygamma function ${\psi }^{\left(n\right)}\left(z\right)$ for complex $z$
s14agf 21 nagf_specfun_gamma_log_complex
Logarithm of the gamma function $\mathrm{ln}\Gamma \left(z\right)$, complex argument
s14ahf 23 nagf_specfun_gamma_log_scaled_real
Scaled log gamma function
s14anf 27 nagf_specfun_gamma_vector
Gamma function, vectorized $\Gamma \left(x\right)$
s14apf 27 nagf_specfun_gamma_log_real_vector
Log gamma function, vectorized $\mathrm{ln}\left(\Gamma \left(x\right)\right)$
s14baf 14 nagf_specfun_gamma_incomplete
Incomplete gamma functions $P\left(a,x\right)$ and $Q\left(a,x\right)$
s14bnf 27 nagf_specfun_gamma_incomplete_vector
Incomplete gamma functions, vectorized $P\left(a,x\right)$ and $Q\left(a,x\right)$
s14cbf 24 nagf_specfun_beta_log_real
Logarithm of the beta function $\mathrm{ln}B\left(a,b\right)$
s14ccf 24 nagf_specfun_beta_incomplete
Regularized incomplete beta function ${I}_{x}\left(a,b\right)$ and its complement $1-{I}_{x}$
s14cpf 27 nagf_specfun_beta_log_real_vector
Logarithm of the beta function, vectorized $\mathrm{ln}B\left(a,b\right)$
s14cqf 27 nagf_specfun_beta_incomplete_vector
Regularized incomplete beta function, vectorized ${I}_{x}\left(a,b\right)$ and its complement $1-{I}_{x}$
s15abf 3 nagf_specfun_cdf_normal
Cumulative Normal distribution function $P\left(x\right)$
s15acf 4 nagf_specfun_compcdf_normal
Complement of cumulative Normal distribution function $Q\left(x\right)$
Complement of error function $\mathrm{erfc}\left(x\right)$
s15aef 4 nagf_specfun_erf_real
Error function $\mathrm{erf}\left(x\right)$
s15aff 7 nagf_specfun_dawson
Dawson's integral
s15agf 22 nagf_specfun_erfcx_real
Scaled complement of error function, $\mathrm{erfcx}\left(x\right)$
s15apf 27 nagf_specfun_cdf_normal_vector
Cumulative Normal distribution function, vectorized $P\left(x\right)$
s15aqf 27 nagf_specfun_compcdf_normal_vector
Complement of cumulative Normal distribution function, vectorized $Q\left(x\right)$
s15arf 27 nagf_specfun_erfc_real_vector
Complement of error function, vectorized $\mathrm{erfc}\left(x\right)$
s15asf 27 nagf_specfun_erf_real_vector
Error function, vectorized $\mathrm{erf}\left(x\right)$
s15atf 27 nagf_specfun_dawson_vector
Dawson's integral, vectorized
s15auf 27 nagf_specfun_erfcx_real_vector
Scaled complement of error function, vectorized $\mathrm{erfcx}\left(x\right)$
s15ddf 14 nagf_specfun_erfc_complex
Scaled complex complement of error function, $\mathrm{exp}\left(-{z}^{2}\right)\mathrm{erfc}\left(-iz\right)$
s15drf 27 nagf_specfun_erfc_complex_vector
Scaled complex complement of error function, vectorized $\mathrm{exp}\left(-{z}^{2}\right)\mathrm{erfc}\left(-iz\right)$
s17acf 1 nagf_specfun_bessel_y0_real
Bessel function ${Y}_{0}\left(x\right)$
Bessel function ${Y}_{1}\left(x\right)$
s17aef 5 nagf_specfun_bessel_j0_real
Bessel function ${J}_{0}\left(x\right)$
s17aff 5 nagf_specfun_bessel_j1_real
Bessel function ${J}_{1}\left(x\right)$
s17agf 8 nagf_specfun_airy_ai_real
Airy function $\mathrm{Ai}\left(x\right)$
s17ahf 8 nagf_specfun_airy_bi_real
Airy function $\mathrm{Bi}\left(x\right)$
s17ajf 8 nagf_specfun_airy_ai_deriv
Airy function ${\mathrm{Ai}}^{\prime }\left(x\right)$
s17akf 8 nagf_specfun_airy_bi_deriv
Airy function ${\mathrm{Bi}}^{\prime }\left(x\right)$
s17alf 20 nagf_specfun_bessel_zeros
Zeros of Bessel functions ${J}_{\alpha }\left(x\right)$, ${J}_{\alpha }^{\prime }\left(x\right)$, ${Y}_{\alpha }\left(x\right)$ or ${Y}_{\alpha }^{\prime }\left(x\right)$
s17aqf 24 nagf_specfun_bessel_y0_real_vector
Bessel function vectorized ${Y}_{0}\left(x\right)$
s17arf 24 nagf_specfun_bessel_y1_real_vector
Bessel function vectorized ${Y}_{1}\left(x\right)$
s17asf 24 nagf_specfun_bessel_j0_real_vector
Bessel function vectorized ${J}_{0}\left(x\right)$
s17atf 24 nagf_specfun_bessel_j1_real_vector
Bessel function vectorized ${J}_{1}\left(x\right)$
s17auf 24 nagf_specfun_airy_ai_real_vector
Airy function vectorized $\mathrm{Ai}\left(x\right)$
s17avf 24 nagf_specfun_airy_bi_real_vector
Airy function vectorized $\mathrm{Bi}\left(x\right)$
s17awf 24 nagf_specfun_airy_ai_deriv_vector
Derivatives of the Airy function, vectorized ${\mathrm{Ai}}^{\prime }\left(x\right)$
s17axf 24 nagf_specfun_airy_bi_deriv_vector
Derivatives of the Airy function, vectorized ${\mathrm{Bi}}^{\prime }\left(x\right)$
s17dcf 13 nagf_specfun_bessel_y_complex
Bessel functions ${Y}_{\nu +a}\left(z\right)$, real $a\ge 0$, complex $z$, $\nu =0,1,2,\dots$
s17def 13 nagf_specfun_bessel_j_complex
Bessel functions ${J}_{\nu +a}\left(z\right)$, real $a\ge 0$, complex $z$, $\nu =0,1,2,\dots$
s17dgf 13 nagf_specfun_airy_ai_complex
Airy functions $\mathrm{Ai}\left(z\right)$ and ${\mathrm{Ai}}^{\prime }\left(z\right)$, complex $z$
s17dhf 13 nagf_specfun_airy_bi_complex
Airy functions $\mathrm{Bi}\left(z\right)$ and ${\mathrm{Bi}}^{\prime }\left(z\right)$, complex $z$
s17dlf 13 nagf_specfun_hankel_complex
Hankel functions ${H}_{\nu +a}^{\left(j\right)}\left(z\right)$, $j=1,2$, real $a\ge 0$, complex $z$, $\nu =0,1,2,\dots$
s17gaf 26.1 nagf_specfun_struve_h0
Struve function of order $0$, ${H}_{0}\left(x\right)$
s17gbf 26.1 nagf_specfun_struve_h1
Struve function of order $1$, ${H}_{1}\left(x\right)$
s18acf 1 nagf_specfun_bessel_k0_real
Modified Bessel function ${K}_{0}\left(x\right)$
Modified Bessel function ${K}_{1}\left(x\right)$
s18aef 5 nagf_specfun_bessel_i0_real
Modified Bessel function ${I}_{0}\left(x\right)$
s18aff 5 nagf_specfun_bessel_i1_real
Modified Bessel function ${I}_{1}\left(x\right)$
s18aqf 24 nagf_specfun_bessel_k0_real_vector
Modified Bessel function vectorized ${K}_{0}\left(x\right)$
s18arf 24 nagf_specfun_bessel_k1_real_vector
Modified Bessel function vectorized ${K}_{1}\left(x\right)$
s18asf 24 nagf_specfun_bessel_i0_real_vector
Modified Bessel function vectorized ${I}_{0}\left(x\right)$
s18atf 24 nagf_specfun_bessel_i1_real_vector
Modified Bessel function vectorized ${I}_{1}\left(x\right)$
s18ccf 10 nagf_specfun_bessel_k0_scaled
Scaled modified Bessel function ${e}^{x}{K}_{0}\left(x\right)$
s18cdf 10 nagf_specfun_bessel_k1_scaled
Scaled modified Bessel function ${e}^{x}{K}_{1}\left(x\right)$
s18cef 10 nagf_specfun_bessel_i0_scaled
Scaled modified Bessel function ${e}^{-|x|}{I}_{0}\left(x\right)$
s18cff 10 nagf_specfun_bessel_i1_scaled
Scaled modified Bessel function ${e}^{-|x|}{I}_{1}\left(x\right)$
s18cqf 24 nagf_specfun_bessel_k0_scaled_vector
Scaled modified Bessel function vectorized ${e}^{x}{K}_{0}\left(x\right)$
s18crf 24 nagf_specfun_bessel_k1_scaled_vector
Scaled modified Bessel function vectorized ${e}^{x}{K}_{1}\left(x\right)$
s18csf 24 nagf_specfun_bessel_i0_scaled_vector
Scaled modified Bessel function vectorized ${e}^{-|x|}{I}_{0}\left(x\right)$
s18ctf 24 nagf_specfun_bessel_i1_scaled_vector
Scaled modified Bessel function vectorized ${e}^{-|x|}{I}_{1}\left(x\right)$
s18dcf 13 nagf_specfun_bessel_k_complex
Modified Bessel functions ${K}_{\nu +a}\left(z\right)$, real $a\ge 0$, complex $z$, $\nu =0,1,2,\dots$
s18def 13 nagf_specfun_bessel_i_complex
Modified Bessel functions ${I}_{\nu +a}\left(z\right)$, real $a\ge 0$, complex $z$, $\nu =0,1,2,\dots$
s18gaf 26.1 nagf_specfun_struve_l0
Modified Struve function of order $0$, ${L}_{0}\left(x\right)$
s18gbf 26.1 nagf_specfun_struve_l1
Modified Struve function of order $1$, ${L}_{1}\left(x\right)$
s18gcf 26.1 nagf_specfun_struve_i0ml0
The function ${I}_{0}\left(x\right)-{L}_{0}\left(x\right)$, where ${I}_{0}\left(x\right)$ is a modified Bessel function and ${L}_{0}\left(x\right)$ is a Struve function
s18gdf 26.1 nagf_specfun_struve_i1ml1
The function ${I}_{1}\left(x\right)-{L}_{1}\left(x\right)$, where ${I}_{1}\left(x\right)$ is a modified Bessel function and ${L}_{1}\left(x\right)$ is a Struve function
s18gkf 21 nagf_specfun_bessel_j_seq_complex
Bessel function of the 1st kind ${J}_{\alpha ±n}\left(z\right)$
s19aaf 11 nagf_specfun_kelvin_ber
Kelvin function $\mathrm{ber}x$
s19abf 11 nagf_specfun_kelvin_bei
Kelvin function $\mathrm{bei}x$
s19acf 11 nagf_specfun_kelvin_ker
Kelvin function $\mathrm{ker}x$
Kelvin function $\mathrm{kei}x$
s19anf 24 nagf_specfun_kelvin_ber_vector
Kelvin function vectorized $\mathrm{ber}x$
s19apf 24 nagf_specfun_kelvin_bei_vector
Kelvin function vectorized $\mathrm{bei}x$
s19aqf 24 nagf_specfun_kelvin_ker_vector
Kelvin function vectorized $\mathrm{ker}x$
s19arf 24 nagf_specfun_kelvin_kei_vector
Kelvin function vectorized $\mathrm{kei}x$
s20acf 5 nagf_specfun_fresnel_s
Fresnel integral $S\left(x\right)$
Fresnel integral $C\left(x\right)$
s20aqf 24 nagf_specfun_fresnel_s_vector
Fresnel integral vectorized $S\left(x\right)$
s20arf 24 nagf_specfun_fresnel_c_vector
Fresnel integral vectorized $C\left(x\right)$
s21baf 8 nagf_specfun_ellipint_symm_1_degen
Degenerate symmetrised elliptic integral of 1st kind ${R}_{C}\left(x,y\right)$
s21bbf 8 nagf_specfun_ellipint_symm_1
Symmetrised elliptic integral of 1st kind ${R}_{F}\left(x,y,z\right)$
s21bcf 8 nagf_specfun_ellipint_symm_2
Symmetrised elliptic integral of 2nd kind ${R}_{D}\left(x,y,z\right)$
s21bdf 8 nagf_specfun_ellipint_symm_3
Symmetrised elliptic integral of 3rd kind ${R}_{J}\left(x,y,z,r\right)$
s21bef 22 nagf_specfun_ellipint_legendre_1
Elliptic integral of 1st kind, Legendre form, $F\left(\varphi \mid m\right)$
s21bff 22 nagf_specfun_ellipint_legendre_2
Elliptic integral of 2nd kind, Legendre form, $E\left(\varphi \mid m\right)$
s21bgf 22 nagf_specfun_ellipint_legendre_3
Elliptic integral of 3rd kind, Legendre form, $\Pi \left(n;\varphi \mid m\right)$
s21bhf 22 nagf_specfun_ellipint_complete_1
Complete elliptic integral of 1st kind, Legendre form, $K\left(m\right)$
s21bjf 22 nagf_specfun_ellipint_complete_2
Complete elliptic integral of 2nd kind, Legendre form, $E\left(m\right)$
s21caf 15 nagf_specfun_jacellip_real
Jacobian elliptic functions sn, cn and dn of real argument
s21cbf 20 nagf_specfun_jacellip_complex
Jacobian elliptic functions sn, cn and dn of complex argument
s21ccf 20 nagf_specfun_jactheta_real
Jacobian theta functions ${\theta }_{k}\left(x,q\right)$ of real argument
s21daf 20 nagf_specfun_ellipint_general_2
General elliptic integral of 2nd kind $F\left(z,{k}^{\prime },a,b\right)$ of complex argument
s22aaf 20 nagf_specfun_legendre_p
Legendre functions of 1st kind ${P}_{n}^{m}\left(x\right)$ or $\overline{{P}_{n}^{m}}\left(x\right)$
s22baf 24 nagf_specfun_hyperg_confl_real
Real confluent hypergeometric function ${}_{1}F_{1}\left(a;b;x\right)$
s22bbf 24 nagf_specfun_hyperg_confl_real_scaled
Real confluent hypergeometric function ${}_{1}F_{1}\left(a;b;x\right)$ in scaled form
s22bef 25 nagf_specfun_hyperg_gauss_real
Real Gauss hypergeometric function ${}_{2}F_{1}\left(a,b;c;x\right)$
s22bff 25 nagf_specfun_hyperg_gauss_real_scaled
Real Gauss hypergeometric function ${}_{2}F_{1}\left(a,b;c;x\right)$ in scaled form
s22caf 27 nagf_specfun_mathieu_ang_periodic_real
Calculates values of real periodic angular Mathieu functions
s30aaf 22 nagf_specfun_opt_bsm_price
Black–Scholes–Merton option pricing formula
s30abf 22 nagf_specfun_opt_bsm_greeks
Black–Scholes–Merton option pricing formula with Greeks
s30acf 27.1 nagf_specfun_opt_imp_vol
Black–Scholes–Merton implied volatility
s30baf 22 nagf_specfun_opt_lookback_fls_price
Floating-strike lookback option pricing formula in the Black-Scholes-Merton model
s30bbf 22 nagf_specfun_opt_lookback_fls_greeks
Floating-strike lookback option pricing formula with Greeks in the Black-Scholes-Merton model
s30caf 22 nagf_specfun_opt_binary_con_price
Binary option, cash-or-nothing pricing formula
s30cbf 22 nagf_specfun_opt_binary_con_greeks
Binary option, cash-or-nothing pricing formula with Greeks
s30ccf 22 nagf_specfun_opt_binary_aon_price
Binary option, asset-or-nothing pricing formula
s30cdf 22 nagf_specfun_opt_binary_aon_greeks
Binary option, asset-or-nothing pricing formula with Greeks
s30faf 22 nagf_specfun_opt_barrier_std_price
Standard barrier option pricing formula
s30jaf 22 nagf_specfun_opt_jumpdiff_merton_price
Jump-diffusion, Merton's model, option pricing formula
s30jbf 22 nagf_specfun_opt_jumpdiff_merton_greeks
Jump-diffusion, Merton's model, option pricing formula with Greeks
s30naf 22 nagf_specfun_opt_heston_price
Heston's model option pricing formula
s30nbf 23 nagf_specfun_opt_heston_greeks
Heston's model option pricing formula with Greeks
s30ncf 25 nagf_specfun_opt_heston_term
Heston's model option pricing with term structure
s30ndf 28.5 (Experimental) nagf_specfun_opt_heston_more_greeks
Heston's model option pricing formula with Greeks, sensitivities of model parameters and negative rates
s30qcf 22 nagf_specfun_opt_amer_bs_price
American option, Bjerksund and Stensland pricing formula
s30saf 22 nagf_specfun_opt_asian_geom_price
Asian option, geometric continuous average rate pricing formula
s30sbf 22 nagf_specfun_opt_asian_geom_greeks
Asian option, geometric continuous average rate pricing formula with Greeks | 4,840 | 14,257 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 152, "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 | 0.437853 |
https://mathematica.stackexchange.com/questions/136561/evaluation-of-autocorrelation-function | 1,568,833,886,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514573323.60/warc/CC-MAIN-20190918172932-20190918194932-00141.warc.gz | 584,726,775 | 28,880 | # Evaluation of autocorrelation function [closed]
I am attempting to evaluate this double integral:
Integrate[
Integrate[
Exp[-R (t1 - t2)] s[t1] s[t2], {t2, 0, t1}
], {t1, 0, t}
]
where s(t) is a piecewise-constant function. The final t value will be within a portion of the domain of s(t) is constant. The s(t) function is determined first by generating an array, lArray, that contains the domain for the piecewise-function, and ModArray, that contains the function values for those domains. I then define the following identifying function that returns the value of ModArray when fed an arbitrary l value:
IdenS[lArray_, ModArray_, Nlen_, l_] :=
For[i = 1, i < Nlen + 1, i++,
TempSum = Sum[lArray[[1, j]], {j, 1, i}];
TotalSum = Sum[lArray[[1, j]], {j, 1, Nlen}];
Which[
((TempSum > l) && (l < TotalSum) && (l > 0)),
(Return[ModArray[[1, i]]];
Break;),
l < 0,
(Return[0];
Break;),
l > TotalSum,
(Return[0];
Break;
)
]
]
When trying to evaluate the integral using the IdenS function defined above, the integral returns something in terms of Null values. The IdenS function is having a problem with accepting the t1 and t2 values before evaluation. Is there some workaround that will result in an analytic result in terms of t and R for t values greater than the sum of all the t values in lArray? Perhaps some usage of Hold, Evaluate, and Assumptions?
Example result:
Nlen = 5;
lArray = ConstantArray[1, {1, Nlen}];
ModArray = ConstantArray[0, {1, Nlen}];
For[j = 1, j < Nlen + 1, j++,
ModArray[[1, j]] = (-1)^(j + 1);
]
Integrate[
Integrate[
Exp[-R (t1 - t2)] IdenS[lArray, ModArray, Nlen, t1] IdenS[lArray,
ModArray, Nlen, t2], {t2, 0, t1}
], {t1, 0, 1}
]
(* Result: *)
(E^-R Null^2 (1 + E^R (-1 + R)))/R^2
## closed as off-topic by MarcoB, m_goldberg, Sascha, Feyre, corey979Jan 31 '17 at 13:10
This question appears to be off-topic. The users who voted to close gave this specific reason:
• "This question arises due to a simple mistake such as a trivial syntax error, incorrect capitalization, spelling mistake, or other typographical error and is unlikely to help any future visitors, or else it is easily found in the documentation." – m_goldberg, Sascha, Feyre
If this question can be reworded to fit the rules in the help center, please edit the question.
• Shouldn't s(t1) and s(t2) actually be s[t1] and s[t2] instead? Also, you have shown how you are trying to implement your solution. But could you comment on what exactly you are trying to accomplish? It is not that clear to me what the underlying problem is. – MarcoB Jan 30 '17 at 16:43
• Sorry, it should be s[t1] and s[t2], but I never actually execute that code, so that shouldn't be a problem. Thanks for pointing that out. My goal is generally to calculate something that looks like an autocorrelation of the s(t) function convolved with an exponential decay. The general problem is that the IdenS function is having a difficult time dealing with the symbolic t1 and t2 values. – Aaron Ross Jan 30 '17 at 16:45
• Aaron, I wonder if IdenS is correct. That function returns 1 throughout the $[0,1[$ domain, except at 1, where its value is -1. Is that correct? See Plot[IdenS[lArray, ModArray, Nlen, t], {t, 0, 1}]. – MarcoB Jan 30 '17 at 16:55
• Hi MarcoB: this is correct! The domain of the s(t) function here is intended to be [0,5], oscillating back and forth between 1 and -1, and 0 outside of those bounds. – Aaron Ross Jan 30 '17 at 17:02
• In your example, the IdenS function generates a square wave. Is that always the case, or is it peculiar to the example? If it is always the case, then you can replace it with the built-in function SquareWave. – m_goldberg Jan 31 '17 at 8:13
I believe your IdenS can be replaced with a function defined by much simpler code.
f[lArray_, ModArray_, l_] :=
Which[
l <= 0, ModArray[[1]],
l >= Total[lArray], ModArray[[-1]],
True,
Module[{val = Pick[ModArray, l <= # & /@ Accumulate[lArray]][[1]]},
If[NumericQ[val], val, 1, 1]]]
Using the example parameters
n = 5;
lArray = ConstantArray[1, n];
modArray = Table[(-1)^(j + 1), {j, n}];
f generates a square wave.
Plot[f[lArray, modArray, t] , {t, 0, 5}]
Evaluating your double integral with f replacing IdesS
Integrate[
Exp[-R (t1 - t2)] f[lArray, modArray, t1] f[lArray, modArray, t2],
{t1, 0, 1}, {t2, 0, t1}]
gives
The above integral is equivalent to
Integrate[
Exp[-R (t1 - t2)] SquareWave[t1/2] SquareWave[t2/2], {t1, 0, 1}, {t2, 0, t1}]
• this is perfect! Thanks for working through this. I am not entirely sure that I understand the syntax in the Module portion of the function though. I need to dig into the # % /@ usage a bit. The f function is necessary for me to be able to input more complicated modulation functions other than the square wave. – Aaron Ross Jan 31 '17 at 15:36 | 1,442 | 4,786 | {"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} | 3.234375 | 3 | CC-MAIN-2019-39 | latest | en | 0.690399 |
https://en.x-mol.com/paper/article/1414521883245875200 | 1,638,542,713,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362879.45/warc/CC-MAIN-20211203121459-20211203151459-00082.warc.gz | 311,494,397 | 5,620 | Find Paper, Faster
Example：10.1021/acsami.1c06204 or Chem. Rev., 2007, 107, 2411-2502
Additive properties of numbers with restricted digits
Algebra & Number Theory (IF0.938), Pub Date : 2021-06-30, DOI: 10.2140/ant.2021.15.1283
Han Yu
We consider some additive properties of integers with restricted digit expansions. Let $b\ge 3$ be an integer and ${B}_{b}$ be the set of integers whose base $b$ expansions have only digits $\left\{0,1\right\}$. Let $a,b,c$ be three integers greater than $2$. We give some estimates on the size of $\left({B}_{a}+{B}_{b}\right)\cap {B}_{c}$. In particular, under mild conditions, $\left({B}_{a}+{B}_{b}\right)\cap {B}_{c}$ is a very thin set in the sense that for each $𝜖>0$, as $N\to \infty$,
$#\left(\left({B}_{a}+{B}_{b}\right)\cap {B}_{c}\cap \left[1,N\right]\right)=O\left({N}^{𝜖}\right).$ | 299 | 833 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 11, "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-2021-49 | latest | en | 0.593005 |
https://nrich.maths.org/public/leg.php?code=-334&cl=3&cldcmpid=6158 | 1,474,812,169,000,000,000 | text/html | crawl-data/CC-MAIN-2016-40/segments/1474738660214.50/warc/CC-MAIN-20160924173740-00296-ip-10-143-35-109.ec2.internal.warc.gz | 877,450,803 | 9,823 | # Search by Topic
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https://www.gmpuzzles.com/blog/tag/thermo-sudoku-2/page/4/ | 1,660,721,667,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572870.85/warc/CC-MAIN-20220817062258-20220817092258-00420.warc.gz | 674,604,601 | 11,633 | ### Thermo-Sudoku by Kishore Kumar Sridharan
PDF
or solve online (using our beta test of Penpa-Edit tools)
Theme: Sun
Author/Opus: This is the 2nd puzzle from guest contributor Kishore Kumar Sridharan.
Rules: Standard Thermo-Sudoku rules.
Answer String: Enter the 1st column from top to bottom, followed by a comma, followed by the 4th column from top to bottom.
Time Standards (highlight to view): Grandmaster = 5:00, Master = 7:30, Expert = 15:00
Solution: PDF
Note: Follow this link for more Thermo-Sudoku puzzles. If you are new to this puzzle type, here are our easiest Thermo-Sudoku to get started on. More Thermo-Sudoku puzzles can be found in The Art of Sudoku 2.
### Thermo-Sudoku by John Bulten
PDF
or solve online (using our beta test of Penpa-Edit tools)
Theme: Candlepower
Author/Opus: This is the 53rd puzzle from our contributing puzzlemaster John Bulten.
Rules: Standard Thermo-Sudoku rules.
Answer String: Enter the 1st row from left to right, followed by a comma, followed by the 9th row from left to right.
Time Standards (highlight to view): Grandmaster = 5:30, Master = 11:00, Expert = 22:00
Solution: PDF; a solution video is also available here.
Note: Follow this link for more Thermo-Sudoku puzzles. If you are new to this puzzle type, here are our easiest Thermo-Sudoku to get started on. More Thermo-Sudoku puzzles can be found in The Art of Sudoku 2.
### Thermo-Sudoku by Serkan Yürekli
PDF
or solve online (using our beta test of Penpa-Edit tools)
Theme: Pi Circle in Order
Author/Opus: This is the 182nd puzzle from our contributing puzzlemaster Serkan Yürekli.
Rules: Standard Thermo-Sudoku rules.
Answer String: Enter the 2nd row from left to right, followed by a comma, followed by the 4th row from left to right.
Time Standards (highlight to view): Grandmaster = 4:00, Master = 5:00, Expert = 10:00
Solution: PDF
Note: Follow this link for more Thermo-Sudoku puzzles. If you are new to this puzzle type, here are our easiest Thermo-Sudoku to get started on. More Thermo-Sudoku puzzles can be found in The Art of Sudoku 2.
PDF
or solve online (using our beta test of Penpa-Edit tools)
Theme: Diamonds (for Chris Green)
Author/Opus: This is the 150th puzzle from our contributing puzzlemaster Prasanna Seshadri.
Rules: Classic Sudoku Rules. Also, some arrow shapes are in the grid; the sum of the digits along the path of each arrow must equal the digit in the circled cell. Some thermometer shapes are in the grid; digits must be strictly increasing from the round bulb to each flat end. Finally, some diamond shapes are in the grid; the labeled diamonds below the grid must be placed in some order onto the shapes in the grid. The labeled diamonds can be rotated but cannot be reflected.
Answer String: Enter the 1st row from left to right, followed by a comma, followed by the 9th row from left to right.
Time Standards (highlight to view): Grandmaster = 5:15, Master = 10:45, Expert = 21:30
Solution: PDF
Note: Follow this link for other less common variations of Sudoku and this link for classic Sudoku. If you are new to this puzzle type, here are our easiest Sudoku to get started on. More Arrow and Thermo-Sudoku puzzles can be found in The Art of Sudoku 2.
### Thermo-Sudoku by Swaroop Guggilam
PDF
or solve online (using our beta test of Penpa-Edit tools)
Theme: X-Box
Author/Opus: This is the 8th puzzle from guest contributor Swaroop Guggilam.
Rules: Standard Thermo-Sudoku rules.
Answer String: Enter the 2nd row from left to right, followed by a comma, followed by the 8th row from left to right.
Time Standards (highlight to view): Grandmaster = 6:30, Master = 13:00, Expert = 26:00
Solution: PDF; a solution video is available here.
Note: Follow this link for more Thermo-Sudoku puzzles. If you are new to this puzzle type, here are our easiest Thermo-Sudoku to get started on. More Thermo-Sudoku puzzles can be found in The Art of Sudoku 2.
### Thermo-Sudoku by Serkan Yürekli
PDF
or solve online (using our beta test of Penpa-Edit tools)
Theme: Weightlifter
Author/Opus: This is the 115th puzzle from our contributing puzzlemaster Serkan Yürekli.
Rules: Standard Thermo-Sudoku rules.
Answer String: Enter the 2nd row from left to right, followed by a comma, followed by the 6th column from top to bottom.
Time Standards (highlight to view): Grandmaster = 5:30, Master = 8:30, Expert = 17:00
Solution: PDF
Note: Follow this link for more Thermo-Sudoku puzzles. If you are new to this puzzle type, here are our easiest Thermo-Sudoku to get started on. More Thermo-Sudoku puzzles can be found in The Art of Sudoku 2.
### Thermo-Sudoku (Missing Bulbs) by Hans van Stippent
PDF
or solve online (using our beta test of Penpa-Edit tools)
Theme: Clown Face
Author/Opus: This is the 12th puzzle from guest contributor Hans van Stippent.
Rules: Standard Thermo-Sudoku rules. Also, the bulbs are missing from three of the thermometer shapes. These bulbs were positioned at one of the open ends of those thermometers.
Answer String: Enter the 2nd row from left to right, followed by a comma, followed by the 5th row from left to right.
Time Standards (highlight to view): Grandmaster = 7:00, Master = 9:00, Expert = 18:00
Solution: PDF
Note: Follow this link for more Thermo-Sudoku puzzles. If you are new to this puzzle type, here are our easiest Thermo-Sudoku to get started on. More Thermo-Sudoku puzzles can be found in The Art of Sudoku 2. | 1,424 | 5,449 | {"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-2022-33 | latest | en | 0.849334 |
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Interesting infinite product $\sqrt{2}-1=\dfrac{1\cdot7\cdot9\cdot15\cdot17\cdot23\cdots}{3\cdot5\cdot11\cdot13\cdot19\cdot21\cdots}$
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asked by Lithium Score of 21
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<3 | 1,700 | 6,552 | {"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} | 2.984375 | 3 | CC-MAIN-2023-40 | latest | en | 0.935316 |
http://www.davidespataro.it/fold-operator-brief-detour/ | 1,566,231,078,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027314852.37/warc/CC-MAIN-20190819160107-20190819182107-00491.warc.gz | 241,120,639 | 25,787 | This article aims to be a brief practical introduction to new haskell programmer about the fold high order function (which roughly speaking is a function that takes as input and/or produces as output a function) and its usage. It will also outline a bit of the theory that lies behind this operator (the term operator comes from the field of recursion theory (Kleene, 1952), even if, striclty speaking fold is a function in haskell).
## Haskell Fold - Introduction and formal definition
Many problem in function programming are solved by means of tail recursive functions (one that do not perform any other operations that need some values from the recursive call). For instance let's say we want to get the product of integers stored in a list.
What we would do is to write down a tail recursive function (more or less) like this
```tailmultiply :: [Int] -> Int -> Int
tailmultiply [] acc = acc
tailmultiply (x:xs) acc = tailmultiply xs (acc * x)
```
Note, to stress the tail recursion, a definition like the one that follows would not be tail recursive.
```--No tail recursive version
notailmultiply :: [Int] -> Int
notailmultiply [] acc = 1 --neutral element for multiplication
notailmultiply (x:xs) acc = x * notailmultiply xs
```
as the recursive call is used for the multiplication (this implies that we need a sort of stack representation of recursive call, but this is another story).
Let's say now that we want to compute the sum of the numbers, we cannot somehow reuse the function for the multiplication and we should write down another function definition that look more or less the same of the previous one.
```tailsum :: [Int] -> Int -> Int
tailsum [] acc = acc
tailsum (x:xs) acc = tailsum xs (acc + x)
```
As before the same recursion patter appear, we have an edge case concerning the empty list the a tail recursive step over a non-empty list. It turn out that this patter is effective and natural when addressing a great number of problems. Here that fold operator comes to help, because it encapsulate this pattern and allow us to concentrate and limitate our writing efforts just to the core operation we want to perform on the list. To be precise, two family of fold exists, left and right fold(foldl and foldr respectively left and right associative ). The former straightfowardly extend the recursive patter seen before, but as it will turs out later on, it is less powerful (in terms of problem can solve) than his right version (that we can use to write the left version).
```foldl :: (b -> a -> b) -> b -> [a] -> b
foldl _ acc [] = acc
foldl f acc (x:xs) = foldl f (f acc x) xs
```
It is a function that takes as input a binary function an initial value for the accumulator and the list to operate on. The edge case here use the wildcard to stress the fact that the binary function is not applied. It consumes the list and store the partial result at each recursive call in the accumulator function (exactly as in the multiplication and sum examples). It allow use to easily rewrite out sum and product function as follow:
```foldproduct :: [Int] -> Int
foldproduct = foldl (*) 1
foldsum :: [Int] -> Int
foldsum = foldl (+) 0
```
which looks now more coincise and elengantly written, even if at first sight a bit cryptic.
Just to show a bit more sophisticated example we are gonna redifine the map function in terms of foldl. Remaind that the map takes a function and applies it to all the elements of the list:
```map :: (a -> b) -> [a] -> [b]
map f = foldl (\acc x -> f acc ++ [f x]) []
```
Note that the accumulator value for fold can be anything, in this case is a list, but could even be functions (it would be the case when writing foldl in terms of foldr) or whichever type comes in your mind. Note also that map would implemented more efficenltly using a right fold (and you should always use foldr when producing list out of a fold call) as we will see later (++ operator is more costly compared to the cons operator : we will use) but is works for the purpose of showing this concept.
## Right fold
Righ fold is right associative, and roughly speaking it means that consumes the list from the right. (see the images , from wikipedia, for a graphical explanation of the differences between foldr e foldl)
The foldr operator is defined as follow (this patter should look familiar now):
```fold :: (a -> b -> b ) -> b -> [a] -> b
fold _ v [] = v
fold f v (x:xs) = f x (fold f v xs)
```
it takes a binary function as foldl does but in reverse order (i.e. takes as first parameter an element from the list and an accumulator as second), an initial value and the list that has to consumes. There is a nice and intuitive explaination of how foldr works i.e., it traverse the list to the end and substitute the empty list constructor at the end of the input list with the parameter v, and each cons operator with an application of f.
Suppose we want to write down our version of standard prelude filter function which takes a functions (a predicate test) of type (a -> Bool) and returns a sublist of the input list made of all the elements that satisfy the predicate:
```filter :: (a -> Bool) -> [a] -> [a]
filter _ [] = []
filter f (x:xs)
| f x = x : filter xs
| otherwise = filter xs
```
Our foldr filter function would looks like this:
```filterwithfoldr :: (a -> Bool) -> [a] -> [a]
filterwithfoldr p = foldr f []
where f x acc = if p x then x:acc else acc
```
Let's say we want to filter out of a list of integers all the elements that are even. Obviously we should invoche something like filter (even) [1,2,3,4,5]. We know that foldr will start from the rightmost element (5) testing the p predicates on it (in out case if it is even) and if it is satisfied simply puts the value into the to be returned accumulator value and so on until we consume the whole list.
Let's track an execution of the filterwithfoldr function on the list [1..5]
filter (even) [] 1:2:3:4:(5:[])
(p 5 fails) filter (even) [] 1:2:3:4 do not push anything into the accumulator
(p 4 ok) filter (even) [4] 1:2:3:4 , 4 should be returned as result of filter
(p 3 fails) filter (even) [4] 1:2:3 again 3 fails, so discard it
(p 2 ok) filter (even) [2,4] 1:2 obsiously 2 is even, so keep it
(p 1 fails) filter (even) [4] 1 one is odd
(empty list) filter (even) [2,4] => finally we return the accumulator
Here a list some of few simple functions and their definition with foldr/foldl:
```--prelude any
myAny :: (a->Bool) ->[a] -> Bool-> Bool
myAny f [] acc = acc
myAny f (x:xs) acc = myAny f xs (f x || acc)
myAnyFold :: (a -> Bool) -> [a] -> Bool
myAnyFold p = foldr step False
where step x acc = (p x) || acc
--prelude takewhile
myTakeWhile :: (a->Bool) ->[a]-> [a] ->[a]
myTakeWhile f [] acc = acc
myTakeWhile f (x:xs) acc
|f x = myTakeWhile f xs (acc++[x])
|otherwise = acc
myTakeWhileFoldr :: (a->Bool) -> [a] -> [a]-> [a]
myTakeWhileFoldr p = foldr f id
where f = (\ x y -> (\ acc -> if p x then y (acc++[x]) else acc))
--prelude cycle
myCycle :: [a]->[a]
myCycle l = l ++ myCycle l
myCycleFold :: [a]->[a]
myCycleFold l = foldr step n l
where
n = l ++ myCycleFold l
step x acc= x:acc
```
## Use fold to produce tuples
Suppose you have a list of elements you want to calculate k functions on them. It could be the case of a list of integers for which you need to compute the product, sum, average, max, min ... Each of these function requires you to loop over all the element of the list, with a total cost of O(kn). Using fold you can consume the list only once and produce a tuple that store at each location i the result of the function on the list. How can we do it? We will fist see an example and then try to generalize this method.
Let's so write a function for computing both sum and products of a list of integers. It has to return a tuple, hence not surprisingly the initial value is a tuple which component are the initial value of the corrensponding fold definition (in this caso 0 for the sum and 1 for the product).
```sumproductfold :: [Int] -> (Int,Int)
sumproductfold = foldr f (0,1)
where f x (ss,pp) = (ss+x,pp*x)
```
More in general suppose you have functions (that can be defined with fold obviously)
One can always write a function the computes a tuple s.t.
using just one application of fold (hence traversing the list only once) where,
and
## Conclusion on haskell fold functions
Writing function and programs that use fold is a good habit that every haskell programmer should start to take, even if it requires a bit of practice and experience expecially for haskell novices(at least it was so for me). Fold is very well understood by the haskell community, and its behaviour is somehow more predictable than a recursive function that force us to step into the details of the function we are trying to read and undestand (general recursion can do anything). Moreover as a side effect your code will be less error prone (you know how fold operates).
In one of the next article we will address, a techniques (from Hutton's paper, a very well written paper about fold) that help in converting "canonical" tail recursive function into ones that use fold using the so called universality property that uncoinsciously used in some of the example provided in this article. | 2,368 | 9,287 | {"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-2019-35 | latest | en | 0.896737 |
https://gradeup.co/barc-2020-fluid-mechanics-nuclear-quiz-2-i-8ef42fd0-52d9-11ea-8f7a-0b355e5d455a | 1,623,592,024,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487608856.6/warc/CC-MAIN-20210613131257-20210613161257-00049.warc.gz | 279,311,268 | 38,238 | Time Left - 12:00 mins
# BARC 2020: Fluid Mechanics Nuclear quiz 2
Attempt now to get your rank among 774 students!
Question 1
The velocity of fluid particle in a 2-D flow field is given by, V = . Calculate the slope of streamline at point (4,2)
Question 2
Match the following
Question 3
Couette flow is characterized by
Question 4
A fluid has an absolute viscosity of 0.05 Pa s and a specific gravity of 0.68. When such a fluid flows over a flat plate, the velocity at a point 75 mm from the surface is 1.25m/s. Assuming parabolic velocity distribution with the vertex at the above said point, the shear stress (in N/m2) at 50 mm from the boundary surface will be
Question 5
Match List-I (Type of fluid with List-II (Variation of shear stress) and select the correct answer:
Question 6
A pipe of 0.4 m diameter and 1500 m length is used to carry an oil of specific gravity 0.75. Take coefficient of friction = 0.0025. The amount of power required to maintain the oil at a rate of 0.1 m3/s will be
Question 7
A geometrically similar hydraulic model of a spillway has been constructed on a scale of 1: 16. If the prototype discharge is 2048 m3/s, what would be the discharge in the model_____? (in m3/s)
Question 8
Determine the temperature at the sea level, if the pressure at sea-level was found to be 101.5 kN/m2. The density of air is taken constant = 1.28 kg/m3. Take R = 274
Question 9
A container with two circular vertical tubes of diameter d1 = 39.5 mm and d2 = 12.7 mm is partially filled with mercury. The equilibrium level of the liquid is shown on the left diagram. A cylindrical object made from solid brass is placed in the larger tube so that it floats, as shown in right diagram. Object is D = 37.5 mm in diameter and H = 76.2 mm high. What will be the new equilibrium level, h of mercury with the brass cylinder in place?
[Given, ]
Question 10
A piston of 7.95 cm diameter and 30 cm long works in a cylinder of 8 cm diameter. The annular space of the piston is filled with an oil of viscosity 3 poise. If an axial load of 18 N is applied to piston, the speed of piston will be
• 774 attempts | 569 | 2,130 | {"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} | 3.3125 | 3 | CC-MAIN-2021-25 | latest | en | 0.898084 |
http://quant.stackexchange.com/questions/3323/why-isnt-the-nelson-siegel-model-arbitrage-free/3496 | 1,469,688,127,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257828009.82/warc/CC-MAIN-20160723071028-00249-ip-10-185-27-174.ec2.internal.warc.gz | 196,161,535 | 18,992 | # Why isn't the Nelson-Siegel model arbitrage-free?
Assume $X_t$ is a multivariate Ornstein-Uhlenbeck process, i.e. $$dX_t=\sigma dB_t-AX_tdt$$ and the spot interest rate evolves by the following equation: $$r_t=a+b\cdot X_t.$$ After solving for $X_t$ using $e^{tA}X_t$ and Ito and looking at $\int_0^T{r_s\;ds}$, it turns out that $$\int_0^T{r_s\;ds} \sim \mathcal{N}(aT+b^{T}(I-e^{-TA})A^{-1}X_0,b^{T}V_Tb)$$ where $V_t$ is the covariance matrix of $\int_0^T(I-e^{-(T-u)A})A^{-1}\sigma dB_u$.
This gives us the yield curve $$y(t)=a+\frac{b^{T}(I-e^{-tA})A^{-1}X_0}{t}+\frac{b^{T}V_tb}{2t}$$ and by plugging in $A= \begin{pmatrix} \lambda & 1 \\ 0 & \lambda \\ \end{pmatrix}$ we finally arrive at $$y(t)=a+\frac{1-e^{-\lambda t}}{\lambda t}C_0+e^{-\lambda t}C_1+\frac{b^{T}V_tb}{2t}.$$ The formula above without $\frac{b^{T}V_tb}{2t}$ is known as the Nelson-Siegel yield curve model. Could somebody clarify why neglecting $\frac{b^{T}V_tb}{2t}$ leads to arbitrage opportunities?
So I am essentially asking the following question:
Why is the above model (with $\frac{b^{T}V_tb}{2t}$) arbitrage free?
-
Where did you get this derivation of the NS model? I'm only familiar with Filipovic (1999) (onlinelibrary.wiley.com/doi/10.1111/1467-9965.00073/abstract) showing that arbitrage is possible in the standard NS formulation and the solution of Christensen et al. (2009) (frbsf.org/publications/economics/papers/2007/wp07-20bk.pdf) who add a yield adjustment term. – Bob Jansen May 6 '12 at 11:53
This derivation was given by Chris Rogers, who is teaching Advanced Financial Models at the University of Cambridge this year. Here's a scan of the notes, taken by a student of his class: statslab.cam.ac.uk/~chris/AFM/AFM_L21.pdf – Tom Artiom Fiodorov May 6 '12 at 12:07
The original Nelson Siegel paper describes a parsimonious model of the term structure using only four or three (if $\lambda_t$ is fixed). Filipovic (1999) proves that this model can never be used in a arbitrage free context, paraphrasing the abstract:
We introduce the class of consistent state space processes, which have the property to provide an arbitrage-free interest rate model when representing the parameters of the Nelson–Siegel (NS) family. (We show that) there exists no nontrivial interest rate model driven by a consistent state space Itō process.
This problem is solved by Christensen et al. (2009). They provide some ODE's which must hold for an AFNS and write that the "key difference between Dynamic NS and AFNS is the maturity dependent yield-adjustment term" and show how to solve for this term.
They show that the yield adjustment term is empirically small and that their model
fares well in out-of-sample prediction, consistently outperforming, for example, the canonical $A_0(3)$ model (of Duffee 2002).
-
Let $P(t,T)$ be the time-$t$ price of the zero-coupon bond expiring at $T$.
The no-arbitrage condition forces: $$e^{-\int_0^tr_sds}P(t,T)=\mathbb{E}[e^{-\int_0^Tr_sds}|\mathcal{F_t}],$$ where $\mathcal{F_t}$ is the filtration of the Brownian motion up to time $t$. Note that the expression on the right is a martingale by the tower property of expectations, so by the First Theorem of Asset pricing, there is no arbitrage. It immediately follows that $$P(t,T)=\mathbb{E}[e^{-\int_t^Tr_sds}],$$ which will further result in the yield curve specified as above (with $V_t$ term). Therefore neglecting the covariance term could result in arbitrage.
In fact, I was told that there is a proof showing it is not indeed arbitrage-free, but I am not going to go into that.
- | 1,051 | 3,574 | {"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} | 3.125 | 3 | CC-MAIN-2016-30 | latest | en | 0.766476 |
https://www.12000.org/my_notes/Murphy_ODE/reports/maple_2021_1_and_mma_12_3_1/KERNELsubsubsection828.htm | 1,716,228,045,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058293.53/warc/CC-MAIN-20240520173148-20240520203148-00280.warc.gz | 538,012,388 | 3,345 | ##### 4.17.28 $$x y(x)^2 y'(x)+y'(x)^2+y(x)^3=0$$
ODE
$x y(x)^2 y'(x)+y'(x)^2+y(x)^3=0$ ODE Classification
[[_homogeneous, class G]]
Book solution method
No Missing Variables ODE, Solve for $$x$$
Mathematica
cpu = 0.389853 (sec), leaf count = 57
$\left \{\left \{y(x)\to \frac {\cosh (c_1)-\sinh (c_1)}{-i x+\cosh (c_1)+\sinh (c_1)}\right \},\left \{y(x)\to \frac {\cosh (c_1)-\sinh (c_1)}{i x+\cosh (c_1)+\sinh (c_1)}\right \}\right \}$
Maple
cpu = 0.728 (sec), leaf count = 123
$\left [y \left (x \right ) = \frac {4}{x^{2}}, y \left (x \right ) = \frac {2 \sqrt {2}\, x \textit {\_C1} -2 \textit {\_C1}^{2}}{\textit {\_C1}^{2} \left (\textit {\_C1}^{2}-2 x^{2}\right )}, y \left (x \right ) = -\frac {2 \left (\sqrt {2}\, x \textit {\_C1} +\textit {\_C1}^{2}\right )}{\textit {\_C1}^{2} \left (\textit {\_C1}^{2}-2 x^{2}\right )}, y \left (x \right ) = -\frac {\left (\sqrt {2}\, x \textit {\_C1} -2\right ) \textit {\_C1}^{2}}{2 \left (\textit {\_C1}^{2} x^{2}-2\right )}, y \left (x \right ) = \frac {\left (\sqrt {2}\, x \textit {\_C1} +2\right ) \textit {\_C1}^{2}}{2 \textit {\_C1}^{2} x^{2}-4}\right ]$ Mathematica raw input
DSolve[y[x]^3 + x*y[x]^2*y'[x] + y'[x]^2 == 0,y[x],x]
Mathematica raw output
{{y[x] -> (Cosh[C[1]] - Sinh[C[1]])/((-I)*x + Cosh[C[1]] + Sinh[C[1]])}, {y[x] -
> (Cosh[C[1]] - Sinh[C[1]])/(I*x + Cosh[C[1]] + Sinh[C[1]])}}
Maple raw input
dsolve(diff(y(x),x)^2+x*y(x)^2*diff(y(x),x)+y(x)^3 = 0, y(x))
Maple raw output
[y(x) = 4/x^2, y(x) = 2*(2^(1/2)*x*_C1-_C1^2)/_C1^2/(_C1^2-2*x^2), y(x) = -2*(2^
(1/2)*x*_C1+_C1^2)/_C1^2/(_C1^2-2*x^2), y(x) = -1/2*(2^(1/2)*x*_C1-2)/(_C1^2*x^2
-2)*_C1^2, y(x) = 1/2*(2^(1/2)*x*_C1+2)/(_C1^2*x^2-2)*_C1^2] | 894 | 1,684 | {"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} | 3.5 | 4 | CC-MAIN-2024-22 | latest | en | 0.262179 |
http://www.physicsforums.com/showthread.php?t=369562 | 1,410,903,318,000,000,000 | text/html | crawl-data/CC-MAIN-2014-41/segments/1410657119965.46/warc/CC-MAIN-20140914011159-00176-ip-10-196-40-205.us-west-1.compute.internal.warc.gz | 739,395,176 | 11,456 | # Killing Vector Equations
by TerryW
Tags: equations, killing, vector
PF Gold P: 65 I'm working my way through D'Inverno's Understanding GR and have reached Chapter 7 with no real problems. I'm stuck though on p101 (and the corresponding problem 7.7). D'inverno sets out two forms of the geodesic equations, one of which is the Lagrangian with the Killing Vectors and then has a throwaway line (paraphrased as) '..it is possible ...using these two equations to read off the components of the Christoffel symbols..' I have managed all the problems so far but I just cannot see this one. Am I missing something blindingly obvious?
Mentor
P: 6,248
Quote by TerryW I'm working my way through D'Inverno's Understanding GR and have reached Chapter 7 with no real problems. I'm stuck though on p101 (and the corresponding problem 7.7). D'inverno sets out two forms of the geodesic equations, one of which is the Lagrangian with the Killing Vectors and then has a throwaway line (paraphrased as) '..it is possible ...using these two equations to read off the components of the Christoffel symbols..' I have managed all the problems so far but I just cannot see this one. Am I missing something blindingly obvious?
Welcome to Physics Forums!
Let's work through this. Can you find
$$\frac{\partial K}{\partial x^a}?$$
for $x^a$ equal to $t$, $r$, $\theta$, and $\phi$
If you can, post what you get; if you can't, ask more questions.
PF Gold P: 65 Hi George, and thanks for helping. For a=0, I reckon I'll get 1/2(g00,0 + g11,0 +g22,0 + g33,0). I worked out that da (dx/dsbdx/dsc = 0 when I was verifying the workings on the previous page. I already know what the gααs are. How am I doing? Terry
Mentor
P: 6,248
Killing Vector Equations
Quote by TerryW For a=0, I reckon I'll get 1/2(g00,0 + g11,0 +g22,0 + g33,0).
For a diagonal metric, this is almost, but not quite, correct.
Since $K = g_{ab} \dot{x}^a \dot{x}^b /2$,
$$\frac{\partial K}{\partial x^0} = \frac{1}{2} \left( \frac{\partial g_{ab}}{\partial x^0} \right) \dot{x}^a \dot{x}^b = \frac{1}{2} \left[ \left( \frac{\partial g_{00}}{\partial x^0} \right) \left( \dot{x}^0 \right)^2 + \left( \frac{\partial g_{11}}{\partial x^0} \right) \left( \dot{x}^1 \right)^2 + \left( \frac{\partial g_{22}}{\partial x^0} \right) \left( \dot{x}^2 \right)^2 + \left( \frac{\partial g_{33}}{\partial x^0} \right) \left( \dot{x}^3 \right)^2 \right]$$
for a diagonal metric.
Quote by TerryW I already know what the gααs are. How am I doing?
Very well.
$$\frac{\partial K}{\partial x^0}?$$
If you want to see the code for a mathematical expression, click on the expression, and the code will be displayed in a code window. As you click on different expressions (in any post), new code windows don't appear, the contents of a single code window change.
This is a time consuming process that is sometimes (and sometimes not) worth the effort. For now, don't worry too much about learning how to do this.
P: 665
I'm working my way through D'Inverno's Understanding GR and have reached Chapter 7 with no real problems.
Sorry for asking an irrelevant question: Is this book really great for a self-study, George, or are there any other books which can be more helpful and interesting than D'Inverno's book?
PF Gold P: 65 Thanks George, Oh yes, I forgot about the (dxa/ds)2 terms. It's the next part of the Lagrangian which is the problem though and it seems to be a bit of a circular argument which brings you back to the first equation for the geodesics with d2x0/ds2 and the Christoffel symbols. Do you use Latex and then cut/paste the result into your posts? I'm off to bed now. Hope to hear from you tomorrow. Regards Terry
PF Gold P: 65 Hi Altabeh, I'm finding it OK but I did work my way all the way through Schaum's Tensor Calculus (doing all the problems!) before I started. I think I might have struggled with D'Inverno's chapters on Tensor Calculus if I hadn't done this. Regards Terry
Mentor
P: 6,248
Quote by George Jones Now, what about $$\frac{\partial K}{\partial x^0}?$$
Oops, I forgot the latex \dot. This should read
$$\frac{\partial K}{\partial \dot{x}^0}?$$
Quote by TerryW Do you use Latex and then cut/paste the result into your posts?
I usually type the latex commands as I type my posts.
P: 665
Quote by TerryW Hi Altabeh, I'm finding it OK but I did work my way all the way through Schaum's Tensor Calculus (doing all the problems!) before I started. I think I might have struggled with D'Inverno's chapters on Tensor Calculus if I hadn't done this. Regards Terry
And was Schaum's Tensor Calculus useful and easygoing?
PF Gold P: 65 Hi George, My first venture into Latex: $$\frac{\partial K}{\partial \dot{x}^a} = g_{ab}\dot{x}^b \ \ \ \ (2K = g_{ab}\dot{x}^a\dot{x}^b\ \ \ \frac{\partial g_{ab}}{\partial \dot{x}^a}\ =\ \frac{\partial g_{ab}}{\partial {x}^a}\frac{\partial{x}}{\partial \dot{x}^a} \rightarrow \0\ as\ \ x\rightarrow0)$$ Now you are going to ask me what happens when I take the derivative of $$\frac{\partial K}{\partial \dot{x}^a}$$ with respect to the affine parameter represented by the dot over the x
PF Gold P: 65 Schaum's Tensor Calculus was very useful, but you have to do the problems. There are quite a few printing errors however which leave you pondering for a while. Whether or not it is easy going depends on you!
P: 665
Quote by TerryW Hi George, My first venture into Latex: $$\frac{\partial K}{\partial \dot{x}^a} = g_{ab}\dot{x}^b \ \ \ \ (2K = g_{ab}\dot{x}^a\dot{x}^b\ \ \ \frac{\partial g_{ab}}{\partial \dot{x}^a}\ =\ \frac{\partial g_{ab}}{\partial {x}^a}\frac{\partial{x}}{\partial \dot{x}^a} \rightarrow \0\ as\ \ x\rightarrow0)$$ Now you are going to ask me what happens when I take the derivative of $$\frac{\partial K}{\partial \dot{x}^a}$$ with respect to the affine parameter represented by the dot over the x
See that you are getting professional in using Latex...
$$\frac{\partial K}{\partial \dot{x}^a} = \frac{\partial}{\partial \dot{x}^a}(\frac{1}{2}g_{bc}\dot{x}^b\dot{x^c}) = ...$$
Now you continue this calculation and use the Kronecker's delta to manage combining two terms appearing in the operation. (Remember that g_ab is symmetric).
AB
PF Gold P: 65 Hi Altabeh, I thought I'd done that already $$\frac{\partial K}{\partial \dot{x}^a} = g_{ab}\dot{x}^b}$$
Mentor
P: 6,248
Quote by TerryW Hi George, My first venture into Latex: $$\frac{\partial K}{\partial \dot{x}^a} = g_{ab}\dot{x}^b$$
Yes, this is correct. Did you arrive at this by using Kronecker deltas, as Altabeh suggested?
Quote by TerryW $$(2K = g_{ab}\dot{x}^a\dot{x}^b\ \ \ \frac{\partial g_{ab}}{\partial \dot{x}^a}\ =\ \frac{\partial g_{ab}}{\partial {x}^a}\frac{\partial{x}}{\partial \dot{x}^a} \rightarrow \0\ as\ \ x\rightarrow0)$$
I'm not sure what this means.
Quote by TerryW Now you are going to ask me what happens when I take the derivative of $$\frac{\partial K}{\partial \dot{x}^a}$$ with respect to the affine parameter represented by the dot over the x
Right!
PF Gold P: 65 Hi George, Yes I did arrive at my answer by Kronecker deltas. The other bit was just an explanation of why the partial derivative of gab disappears. So for the next bit; $$\frac{d}{du}(g_{ab}\dot{x}^b) = \partial_{c}g_{ab}\dot{x}^b\dot{x}^c+g_{ab}\ddot{x}^b$$ Where now?
P: 665
Quote by TerryW Hi George, Yes I did arrive at my answer by Kronecker deltas. The other bit was just an explanation of why the partial derivative of gab disappears. So for the next bit; $$\frac{d}{du}(g_{ab}\dot{x}^b) = \partial_{c}g_{ab}\dot{x}^b\dot{x}^c+g_{ab}\ddot{x}^b$$ Where now?
Nice. Now you are required to calculate $$\frac{\partial}{\partial x^{a}}(\frac{1}{2}g_{bc}\dot{x}^b\dot{x}^c)$$. If you've already calculated these, then arrange all the terms obtained so far in the order given by the geodesic equation (7.46). After that just try to get rid of $$g_{ab}$$ in $$g_{ab}\ddot{x }^b$$. What should you do to lead to pure terms like $$\ddot{x }^b$$? (Note: Don't take the index b of $$\ddot{x}^b$$ seriously here. It must be something else if one still uses terms including $$\dot{x}^b\dot{x}^c$$.)
PF Gold P: 65 Hi Altabeh, Is all this leading towards: $$\ddot{x}^a +\Gamma^a_{bc}\dot{x}^b\dot{x}^c = 0$$ ? (7.42 in D'Inverno) If it is, it isn't the point of my original question which was to find out what D'Inverno means by "It is possible, by (7.42) (ie the equation above) to read off directly from $$\frac{\partial{K}}{\partial{x}^a} - \frac{d}{du}\left(\frac{\partial{K}}{\partial\dot{x}^a}\right)=0$$ (7.46 in D'Inverno) the components of the connection $$\Gamma^a_{bc}$$, and this proves to be a very efficient way of calculating $$\Gamma^a_{bc}$$." I know various techniques for working out the $$\Gamma^a_{bc}$$ values for a given metric, I thought that this was alluding to an even smarter way of doing it but I can't see it.
P: 665
Quote by TerryW Hi Altabeh, Is all this leading towards: $$\ddot{x}^a +\Gamma^a_{bc}\dot{x}^b\dot{x}^c = 0$$ ? (7.42 in D'Inverno) If it is, it isn't the point of my original question which was to find out what D'Inverno means by "It is possible, by (7.42) (ie the equation above) to read off directly from $$\frac{\partial{K}}{\partial{x}^a} - \frac{d}{du}\left(\frac{\partial{K}}{\partial\dot{x}^a}\right)=0$$ (7.46 in D'Inverno) the components of the connection $$\Gamma^a_{bc}$$, and this proves to be a very efficient way of calculating $$\Gamma^a_{bc}$$." I know various techniques for working out the $$\Gamma^a_{bc}$$ values for a given metric, I thought that this was alluding to an even smarter way of doing it but I can't see it.
Yes, it is! But the point is that if you wanted to derive the Christoffel symbols for the desired metric here directly from the Euler-Lagrange equation, you would end up leading to what D'Inverno asked for in reality. We just gave a proof of how one would extract (7.42) from (7.46) which is the Euler-Lagrange equation.
So all you need now is to start from the line element in spherical coordinates, ds2, and devide both sides of it by du2, so this would be your $$\frac{1}{2}g_{bc}\dot{x}^b\dot{x}^c$$. Remember that here $$(ds/du)^2=2K$$ and for the sake of convenience, simply ignore that factor 2 which won't make any trouble ahead.
AB
Related Discussions Special & General Relativity 11 Advanced Physics Homework 3 Special & General Relativity 0 Differential Geometry 0 Special & General Relativity 4 | 3,184 | 10,388 | {"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} | 3.078125 | 3 | CC-MAIN-2014-41 | latest | en | 0.917462 |
https://www.math-edu-guide.com/CLASS-8-Intersection-of-Set-Venn-Diagram.html | 1,718,757,833,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861796.49/warc/CC-MAIN-20240618234039-20240619024039-00620.warc.gz | 759,859,633 | 6,443 | # CLASS-8INTERSECTION OF SET- VENN DIAGRAM
Intersection of Sets
The shaded portion of the following Venn Diagram represents the set A B when 1) A & B are overlapping sets, 2) A & B are disjoint sets and, 3) A B. no portion of the diagram is shaded when A & B are disjoint sets because in this case A B = φ
1) A & B are overlapping sets
In the above picture, A (Red Color) is one, and B (Green Color) is another set and the Common portion of A & B sets is marked or described by Yellow Color.
2) A & B are Disjoint Sets
3) A B
Complement of Set –
The shaded portion of the diagram represents the set A’, which is the complement of the set. | 175 | 650 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2024-26 | latest | en | 0.923813 |
http://www.slideserve.com/satya/chapter-9-tree-graphs | 1,493,250,394,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917121752.57/warc/CC-MAIN-20170423031201-00159-ip-10-145-167-34.ec2.internal.warc.gz | 670,164,589 | 16,891 | # Chapter 9 “ Tree graphs ” - PowerPoint PPT Presentation
1 / 13
Chapter 9 “ Tree graphs ”. Tree. it is graphs it have edeges and vertices but has no cycles. Traversal Tree There are 3 standard ways of traversing a tree with root R. In order traversal. Pre order traversal. Post order traversal. example: Apply traversal to print the tree:. a. b.
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Chapter 9 “ Tree graphs ”
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#### Presentation Transcript
Chapter
9
“Tree graphs”
Tree
it is graphs it have edeges and vertices but has no cycles
Traversal Tree
There are 3 standard ways of traversing a tree with root R.
In order traversal
Pre order traversal
Post order traversal
example:
Apply traversal to print the tree:
a
b
c
d
e
f
g
h
solution:
• (1) In order traversal:
Left : h d b e .Root : a .Right : f c s .The result : h d b e a f c g .
(2) Pre order traversal :-
Root : a .
Left : b d h e .
Right : c f g .
The result : h d e b f g c a .
(3) Post order traversal:-
Left : h d e b .
Right : f g c .
Root : a .
The result : h d e b f g c a .
• Building Tree :
• example:
• Building trees with the data set …
• (1) 50 , 60 , 40 , 30 , 20 , 45 , 65.
• . (2) 10 , 20 , 30 , 40 , 50
solution: (1)
Step1
Step2
Step3
Step4
50
50
50
50
40
60
60
60
40
30
step 5
step 6
step 7
50
50
50
40
60
40
60
40
60
30
30
45
65
30
45
20
20
20
solution: (2)
Step 1
Step 2
Step 3
Step 4
10
10
10
10
20
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Step 5
10
20
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عمل الطالبات :
• سارة مبارك
• مها مبارك
• فاطمة مبارك
• هيا فهد
• غزاله ناصر
• ريم عبد الله
• نوره محمد
• وفاء حمد | 702 | 2,220 | {"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-2017-17 | longest | en | 0.705569 |
https://brilliant.org/problems/its-quite-simple-2/ | 1,490,236,426,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218186608.9/warc/CC-MAIN-20170322212946-00104-ip-10-233-31-227.ec2.internal.warc.gz | 790,025,641 | 18,971 | # Ratio Of Different Trigonometric Functions!
Geometry Level 3
If $$\dfrac {\cos {3A}}{\cos {A}}$$ $$= \dfrac {1}{2}$$, then the value of $$\dfrac {\sin {3A}}{\sin {A}}$$ can be expressed as $$\dfrac {a}{b}$$, where $$a$$ and $$b$$ are coprime positive integers, find $$a+b$$.
× | 101 | 281 | {"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} | 2.796875 | 3 | CC-MAIN-2017-13 | longest | en | 0.656406 |
https://thebestpaperwriters.com/sets-complements-and-probability/ | 1,685,971,912,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224652116.60/warc/CC-MAIN-20230605121635-20230605151635-00073.warc.gz | 590,043,947 | 12,539 | # Sets, Complements and Probability
1. Set x= (5,7,11,13,16,19) Set y= (1,2,5,13,19)
a. what is the union of sets x and y
b. what is the intersection of sets x and y
c. create your own set z that is a subset of x
2.Let set 1 be the entire alphabet. let set 2 = (w,x,y,z)
a. what is the complement of set 2 in set 1
b. set 3 =(w,x,z). is set 3 a proper subset of set 2 . please explain in words
3. suppose you are going to flip a coin. what is the set of possible outcomes for this
suppose you flip the coin twice. what is the set of possible outcomes for this
if you flip the coin twice what are the chances that you will get one head and one tail? ( i.e 1 in 3, 1 in 4)
4. consider 2 regular dices- how many possible outcomes (sum of dots) are there if you throw the two dice on the board? what are the chaces the sum of both dice will be equal to 6. how about both dice again but the sum of 7 ?
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In how many ways can the letters of the word “CORPORATION” be arranged so that vowels always come together.
5760
50400
2880
None of above
Explanation:
Vowels in the word “CORPORATION” are O,O,A,I,O
Lets make it as CRPRTN(OOAIO)
This has 7 lettes, where R is twice so value = 7!/2!
= 2520
Vowel O is 3 times, so vowels can be arranged = 5!/3!
= 20
Total number of words = 2520 * 20 = 50400
In a group of 6 boys and 4 girls, four children are to be selected. In how many different ways can they be selected such that at least one boy should be there
109
128
138
209
Explanation:
In a group of 6 boys and 4 girls, four children are to be selected such that
at least one boy should be there.
So we can have
(four boys) or (three boys and one girl) or (two boys and two girls) or (one boy and three gils)
This combination question can be solved as
From a group of 7 men and 6 women, five persons are to be selected to form a committee so that at least 3 men are there on the committee. In how many ways can it be done
456
556
656
756
Explanation:
From a group of 7 men and 6 women, five persons are to be selected with at least 3 men.
So we can have
(5 men) or (4 men and 1 woman) or (3 men and 2 woman)
A bag contains 2 white balls, 3 black balls and 4 red balls. In how many ways can 3 balls be drawn from the bag, if at least one black ball is to be included in the draw
64
128
132
222
Explanation:
From 2 white balls, 3 black balls and 4 red balls, 3 balls are to be selected such that
at least one black ball should be there.
Hence we have 3 choices
All three are black
Two are black and one is non black
One is black and two are non black
Total number of ways
= 3C3 + (3C2 x 6C1) + (3C1 x 6C2) [because 6 are non black]
=1+[3×6]+[3×((6×5)/(2×1))]=1+18+45=64
Ques. A box contains 4 red, 3 white and 2 blue balls. Three balls are drawn at random. Find out the number of ways of selecting the balls of different colours
12
24
48
168
Explanation:
This question seems to be a bit typical, isn’t, but it is simplest.
1 red ball can be selected in 4C1 ways
1 white ball can be selected in 3C1 ways
1 blue ball can be selected in 2C1 ways
Total number of ways
= 4C1 x 3C1 x 2C1
= 4 x 3 x 2
= 24
Please note that we have multiplied the combination results, we use to add when their is OR condition, and we use to multiply when there is AND condition, In this question it is AND as
1 red AND 1 White AND 1 Blue, so we multiplied.
Three unbiased coins are tossed, what is the probability of getting at least 2 tails ?
1/3
1/6
1/2
1/8
Explanation:
Total cases are = 2*2*2 = 8, which are as follows
[TTT, HHH, TTH, THT, HTT, THH, HTH, HHT]
Favoured cases are = [TTH, THT, HTT, TTT] = 4
So required probability = 4/8 = ½
Ques. In a throw of dice what is the probability of getting number greater than 5
1/2
1/3
1/5
1/6 | 903 | 2,881 | {"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} | 4.34375 | 4 | CC-MAIN-2020-40 | latest | en | 0.939128 |
https://tutorme.com/tutors/5779/interview/ | 1,591,524,542,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590348526471.98/warc/CC-MAIN-20200607075929-20200607105929-00597.warc.gz | 569,088,115 | 50,081 | Enable contrast version
# Tutor profile: Akash S.
Inactive
Akash S.
Student pursuing M.Sc. Physics
Tutor Satisfaction Guarantee
## Questions
### Subject:Physics (Newtonian Mechanics)
TutorMe
Question:
Why is it better to keep the wheels of a car rolling rather than locking up while applying the brakes?
Inactive
Akash S.
This is because when the wheels are rolling, the friction that acts between the wheels and surface is static friction. When the wheels are locked kinetic friction acts between the two surfaces. Since the value of static friction is higher than kinetic friction it is better to keep the wheels of the car rolling.
### Subject:Physics (Electricity and Magnetism)
TutorMe
Question:
Magnetic fields do no work. Then how is an electromagnetic crane able to lift iron pieces, since work is done in lifting the pieces.
Inactive
Akash S.
In this case work is definitely being done but not by the magnetic forces but by the electric forces. The magnetic force does the job of redirecting this electric force which can do work.
### Subject:Physics
TutorMe
Question:
Will the Center of Mass and Center of Gravity be at the exact same point for an object lying on the surface of earth?
Inactive
Akash S.
No. The center of mass is related to mass and the center of gravity is related to weight. Center of mass is a point at which the distribution of weight is equal in all directions. It can be thought of as a point where all the mass of an object is concentrated. Center of Gravity is the point where the the gravitational force acts. It is slightly lower than the center of mass for an object lying on the surface of earth because the the force of gravity is different for different points on the object. This is due to the non uniform gravitational field of earth which decreases as we move away from the surface of earth. In a uniform gravitational field these two points coincide.
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Contact Akash | 421 | 2,010 | {"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-2020-24 | latest | en | 0.930379 |
https://www.folkstalk.com/2022/10/get-all-combinations-from-two-lists-python-with-code-examples.html | 1,709,633,676,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707948234904.99/warc/CC-MAIN-20240305092259-20240305122259-00799.warc.gz | 774,140,269 | 14,787 | # Get All Combinations From Two Lists Python With Code Examples
Get All Combinations From Two Lists Python With Code Examples
In this session, we will try our hand at solving the Get All Combinations From Two Lists Python puzzle by using the computer language. The following piece of code will demonstrate this point.
```a = ["foo", "melon"]
b = [True, False]
c = list(itertools.product(a, b))
>> [("foo", True), ("foo", False), ("melon", True), ("melon", False)]```
We were able to fix the Get All Combinations From Two Lists Python problemcode by looking at a number of different examples.
## How do you generate all possible combinations of two lists?
Add a Custom Column to and name it List1. Enter the formula =List1. Expand out the new List1 column and then Close & Load the query to a table. The table will have all the combinations of items from both lists and we saved on making a custom column in List1 and avoided using a merge query altogether!
## How do you get all the combinations of a string in Python?
Find all permutations of a string in Python
• import itertools.
• if __name__ == '__main__':
• s = 'ABC'
• nums = list(s)
• permutations = list(itertools. permutations(nums))
• # Output: ['ABC', 'ACB', 'BAC', 'BCA', 'CAB', 'CBA']
• print([''. join(permutation) for permutation in permutations])
## How do you get the number of combinations in Python?
Python: Find the number of combinations of a,b,c and d
• Input:
• Sample Solution:
• Python Code: import itertools print("Input the number(n):") n=int(input()) result=0 for (i,j,k) in itertools.product(range(10),range(10),range(10)): result+=(0<=n-(i+j+k)<=9) print("Number of combinations:",result)
• Flowchart:
## What does Itertools combinations return?
combinations(iterable, r) : It return r-length tuples in sorted order with no repeated elements. For Example, combinations('ABCD', 2) ==> [AB, AC, AD, BC, BD, CD].23-Nov-2020
## How do you figure out possible combinations?
To calculate combinations, we will use the formula nCr = n! / r! * (n – r)!, where n represents the total number of items, and r represents the number of items being chosen at a time. To calculate a combination, you will need to calculate a factorial.08-Apr-2022
## How do you calculate combinations?
The number of combinations of n objects taken r at a time is determined by the following formula: C(n,r)=n! (n−r)!
## How do you find all possible combination of a string?
Algorithm is copied below. void combine(String instr, StringBuffer outstr, int index) { for (int i = index; i < instr. length(); i++) { outstr.Each loop iteration proceeds as follows:
• append a character.
• print the result.
• perform a recursive invocation at the level i+1.
• remove the character we added at step 1.
## How do you do permutations in Python without Itertools?
A. To create combinations without using itertools, iterate the list one by one and fix the first element of the list and make combinations with the remaining list. Similarly, iterate with all the list elements one by one by recursion of the remaining list.21-Jun-2022
## How do you write permutations in Python?
To calculate permutations in Python, use the itertools. permutation() method. The itertools. permutations() method takes a list, dictionary, tuple, or other iterators as a parameter and returns the permutations of that list.16-Mar-2022
## What is Permute in Python?
A permutation, also called an “arrangement number” or “order”, is a rearrangement of the elements of an ordered list S into a one-to-one correspondence with S itself. A string of length n has n! permutation. Examples: Input : str = 'ABC' Output : ABC ACB BAC BCA CAB CBA.11-Jul-2022 | 880 | 3,689 | {"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} | 3.140625 | 3 | CC-MAIN-2024-10 | latest | en | 0.71224 |
https://math.stackexchange.com/questions/2754053/let-x-y-be-independent-rvs-if-there-exists-c-in-mathbb-r-s-t-pxy-c | 1,656,285,137,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103322581.16/warc/CC-MAIN-20220626222503-20220627012503-00552.warc.gz | 450,324,428 | 67,548 | # Let $X,Y$ be independent RVs. If there exists $c \in \mathbb R$ s.t. $P(X+Y=c)=1$, then $X,Y$ are constants
I would like to show:
Let $$X,Y$$ be independent RVs. If there exists $$c \in \mathbb R$$ s.t. $$P(X+Y=c)=1$$, then $$X,Y$$ are constants a.s..
What I tried:
Since X,Y are indep,$$1=P(X+Y=c)=\int 1_{x+y=c} \, d \mu _x \, d \mu_y = \int P(X=c-y) \, d \mu_y$$ but it gets nowhere from here.
• 1. It's better to write text without special formatting. 2. My prof told me not to write quantifiers in words in a sentence. Apr 26, 2018 at 1:02
• @GNUSupporter Can u teach me how to highlight? Apr 26, 2018 at 1:04
• $\rm\LaTeX$ provides \emph{...} to emphasize text (default in italics). To render italics in Markdown, you surround the text *like this*; to render boldface in Markdown, you surround the text **like this** Apr 26, 2018 at 1:07
• This is false. $X,Y$ constant a.s. is the best you can get. Think about $X$, $Y$ taking other values on a measure-zero set. Apr 26, 2018 at 1:16
• @GNUSupporter I edited the question Apr 26, 2018 at 1:52
If $$X$$ and $$Y$$ are square integrable (*) then we may consider $$Var(X+Y)$$ to say:
$$0 = Var(X+Y) =Var(X)+Var(Y) \to 0 = Var(X)=Var(Y)$$
Otherwise:
Two integrals for two random variables!
$$1=P(X+Y=c)=\int\int 1_{x+y=c} \, d \mu _x \, d \mu_y$$
$$=\int\int 1_{d+y=c,x=d} \, d \mu _x \, d \mu_y (d \in \operatorname{Range}(X))$$
$$=\int\int 1_{d+y=c}1_{x=d} \, d \mu _x \, d \mu_y$$
$$=\int 1_{d+y=c}\int1_{x=d} \, d \mu _x \, d \mu_y$$
$$=\int 1_{d+y=c}\mu_x(x=d) \, d \mu_y$$
$$=\mu_x(x=d) \int 1_{d+y=c} \, d \mu_y$$
$$=\mu_x(x=d) \mu_y(d+y=c)$$
$$=\mu_x(x=d) \mu_y(y=c-d)$$
$$\to 1 =\mu_x(x=d) = \mu_y(y=c-d)$$
(*) Hmmm...I guess if $$Z=c$$ a.s. then $$E[Z], E[|Z|], E[Z^2], Var(Z) < \infty$$.
But if $$\exists$$ independent $$X, Y$$ s.t. $$Z=X+Y$$, then does that mean that $$X$$ and $$Y$$ are square integrable? I was thinking $$\infty - \infty$$, but I guess that's undefined. Thus, $$X,Y < \infty$$ a.s.
$$\to c=X+Y$$
$$\to c^2=(X+Y)^2$$
$$\to E[c^2]=E[(X+Y)^2]$$
$$\to c^2=E[X^2+2XY+Y^2]$$
$$\to c^2=E[X^2]+2E[XY]+E[Y^2]$$
$$\to c^2=E[X^2]+2E[X]E[Y]+E[Y^2]$$
$$\to E[X^2], E[XY], E[X]E[Y], E[X], E[Y], E[Y^2] < \infty$$
• Is $d$ arbitrary? May 1, 2018 at 12:08
• @izimath Good question. Edited
– BCLC
May 1, 2018 at 12:43
• Fubini implies that $E(X+y)^2 <\infty$ for some real $y$, which implies $X$ is square integrable. Sep 3, 2020 at 5:56
Thanks to @Daniel Schepler, here's my answer:
$$Var(X) + Var(Y) = Var(X+Y)=E[(X+Y)^2]-{E[(X+Y)]}^2 = c^2 -c^2 =0 \\ \Rightarrow Var(X)=0=Var(Y)$$
• If X and Y are independent, then X and Y are square integrable?
– BCLC
May 1, 2018 at 11:38
• @BCLC You are right... May 1, 2018 at 11:40
• izimath, why?
– BCLC
May 1, 2018 at 11:40
• @BCLC It is possible to show that they are square integrable May 1, 2018 at 11:56
$$P(Y=c-X)= 1$$
Thus $$\sigma(Y) \subseteq \sigma(X)$$
But $$\sigma(Y)$$ and $$\sigma(X)$$ are independent.
Thus $$\sigma(Y)$$ is independent of itself!
$$\to P(A) \in \{0,1\} \ \forall A \in \sigma(Y)$$. Choose $$A=\{Y=\inf\{y \mid F_Y(y)=1\}\}.$$
Convince yourself $$P(A) > 0$$.
Thus $$P(A) = 1 \to P(\{X=\inf\{x \mid F_X(x)=1\}\})=1=P(X=c-\inf\{y \mid F_Y(y)=1\})$$ | 1,367 | 3,227 | {"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": 43, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2022-27 | latest | en | 0.637584 |
https://www.coursehero.com/file/1674698/note1/ | 1,529,371,704,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267861641.66/warc/CC-MAIN-20180619002120-20180619022120-00223.warc.gz | 791,265,746 | 95,579 | note1 - Microeconomic Theory Econ 101A Fall 2008 GSI Eva...
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Microeconomic Theory Econ 101A Fall 2008 GSI: Eva Vivalt Section Notes 1: Calculus and Optimization 1 Multi-Variable Calculus 1.1 Partial Differentiation The partial derivative of a multi-variable function f(x 1 , x 2 ) is the incremental change in the function caused by an incremental change in one of the variables while all other variables are held constant. The formal definition of the partial derivative is: ∂f ( x 1 , x 2 ) ∂x 1 = lim h →∞ f ( x 1 + h, x 2 ) - f ( x 1 , x 2 ) h For convenience we sometimes write partial derivatives using the following alternative notations: ∂f ( x 1 , x 2 ) ∂x 1 = f 1 ( x 1 , x 2 ) = f x 1 ( x 1 , x 2 ) Example 1. Say our function of x and y is 2x 3 y. ∂x 2 x 3 y = 2 y ∂x x 3 = 6 x 2 y ∂y 2 x 3 y = 2 x 3 ∂y y = 2 x 3 1.2 Total Differentiation For small changes in the function f(x 1 , x 2 ), a first order Taylor approximation holds exactly, thus the total change in f is just the sum of the partials for each variable (slopes in each direction) times the incremental change in that particular variable. df ( x 1 , x 2 ) = ∂f ( x 1 , x 2 ) ∂x 1 dx 1 + ∂f ( x 1 , x 2 ) ∂x 2 dx 2 = f 1 ( x 1 , x 2 ) dx 1 + f 2 ( x 1 , x 2 ) dx 2 1
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Now let’s assume that y is a known function of x, y=h(x). We could find the total derivative of f(x,y) with respect to x using the formula above so that: df ( x, y ) dx = df ( x, h ( x )) dx = f 1 ( x, h ( x )) dx dx + f 2 ( x, h ( x )) dh ( x ) dx = f 1 ( x, h ( x )) + f 2 ( x, h ( x )) dh ( x ) dx Example 2. Again, take our function of x and y . By our previous work, df ( x, y ) = ∂f ( x, y ) ∂x dx + ∂f ( x, y ) ∂y dy = 6 x 2 ydx + 2 x 3 dy df ( x, y ) dx = ∂f ( x, y ) ∂x + ∂f ( x, y ) ∂y dy dx And if we know the relationship between x and y, e.g. f ( x ) = y = x 2 then we can note: df ( x ) dx = 2 x And consequently: df ( x, y ) dx = 6 x 2 y + 4 x 4 1.3 Implicit Differentiation Sometimes we may be in a situation where y depends on x but there is no ’explicit’ formula that can be solved to express y in terms of x, yet it is known that x and y satisfy some equation such as: f ( x, y ) = k where k is a constant. Even though we can’t explicitly solve for a function h s.t. y = h(x), it may still be possible to find the derivative dh ( x ) dx 2
To see this, start by taking the total derivative of the equation above with respect to x. The derivative of the LHS is: df ( x, y ) dx = f 1 ( x, h ( x )) + f 2 ( x, h ( x )) dh ( x ) dx The derivative of the RHS is: dk dx = 0 thus we can rearrange and solve: f 1 ( x, h ( x )) + f 2 ( x, h ( x )) dh ( x ) dx = 0 dh ( x ) dx = - f 1 ( x, h ( x )) f 2 ( x, h ( x )) which is defined as long as f 2 ( x, h ( x )) 6 = 0 .
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4.Determine whether the series converges or diverges. For convergent series, find the sumof the series.
8.Determine whether the series converges or diverges. For convergent series, find the sumof the series.
10.Determine whether the series converges or diverges. For convergent series, find the sumof the series.
20.Determine whether the series converges or diverges. For convergent series, find the sumof the series.
28.Determine all values ofcsuch that the series converges.021kckC=kSection 8.34.Determine convergence or divergence of the series.(a)8424kk
42241 211111limlim122limlim(ln]lnln1RRxxRRRuRRdxuxdudxduudiverges 4241144241414 ( )lim() ( )1024 ( )...kkkkkkakkbkkkabkxkbothdiverge(b)264(24 )kk22222244 161214411624 161614()lim() ()1/ 4041616()...kkkkkkkakkbkkkkabkxkkbothconverge10. Determine convergence or divergence of the series.
(b)25011kkk2225/21/251/2121/25/21/25/2555/25/2111111/lim10111 /11/kkkkkkakkkbkkkkkkxkkkkdiverges
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WANG | 460 | 1,447 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2022-21 | latest | en | 0.653632 |
https://discourse.mcneel.com/t/prime-factors-suggestions/94016 | 1,638,155,756,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358685.55/warc/CC-MAIN-20211129014336-20211129044336-00373.warc.gz | 297,436,187 | 4,501 | # Prime Factors -suggestions?
Anyone have suggestions on how to find all prime factors of a given number in GH?
Using Python?
``````def prime_factors(n):
i = 2
factors = []
while i * i <= n:
if n % i:
i += 1
else:
n //= i
factors.append(i)
if n > 1:
factors.append(n)
return factors
a = prime_factors(x)
``````
wow great! This is perfect. Thanks a million!
(Also this is my first time using python script component --worked like a charm!)
Great! Glad to help there are doubtless more efficient algorithms out there, depending on the size of numbers you are looking at and how many times you are repeating the operation. | 160 | 625 | {"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} | 2.734375 | 3 | CC-MAIN-2021-49 | latest | en | 0.850922 |
https://www.geeksforgeeks.org/computer-graphics-3d-scaling-transformation/?ref=rp | 1,675,856,950,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500758.20/warc/CC-MAIN-20230208092053-20230208122053-00680.warc.gz | 769,667,119 | 23,869 | # Computer Graphics – 3D Scaling Transformation
• Difficulty Level : Hard
• Last Updated : 02 Jun, 2022
Prerequisite: Computer Graphics – 3D Translation Transformation Scaling Transformation : It is performed to resize the 3D-object that is the dimension of the object can be scaled(alter) in any of the x, y, z direction through Sx, Sy, Sz scaling factors. Matrix representation of Scaling transformation Condition : The following kind of sequences occur while performing the scaling transformations on a fixed point –
• The fixed point is translated to the origin.
• The object is scaled.
• The fixed point is translated to its original position.
Let a point in 3D space is P(x, y, z) over which we want to apply Scaling Transformation operation and we are given with Scaling factor [Sx, Sy, Sz] So, the new position of the point after applying Scaling operation would be –
Note : If Scaling factor (Sx, Sy, Sz), then, in this case, the 3D object will be Scaled up uniformly in all X, Y, Z direction. Problem : Consider the above problem where a cube” OABCDEFG” is given O(0, 0, 0, ), A(0, 4, 0), B(0, 4, 4), C(4, 4, 0), D(4, 4, 4), E(4, 0, 0), F(0, 0, 4), G (4, 0, 4) and we are given with Scaling factor Sx, Sy, Sz. Perform Scaling operation over the cube. Solution : We are asked to perform the Scaling transformation over the given below 3D object Fig.1:
Fig.1
Now, applying the Matrix Scaling transformation condition we get – After performing the Scaling Transformation successfully the Fig.1 will look like as below Fig.2 –
Fig.2
My Personal Notes arrow_drop_up | 417 | 1,579 | {"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} | 3.09375 | 3 | CC-MAIN-2023-06 | latest | en | 0.822004 |
https://macaulay2.com/doc/Macaulay2-1.22/share/doc/Macaulay2/OldPolyhedra/html/_hypercube.html | 1,701,507,915,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100381.14/warc/CC-MAIN-20231202073445-20231202103445-00494.warc.gz | 427,159,749 | 2,642 | next | previous | forward | backward | up | index | toc
# hypercube -- computes the d-dimensional hypercube with edge length 2*s
## Synopsis
• Usage:
P = hypercube(d,s)
• Inputs:
• d, an integer, , strictly positive
• s, ZZ or QQ, positive (optional)
• Outputs:
## Description
The d-dimensional hypercube with edge length 2*s is the convex hull of all points in {+/- s}^d in QQ^d. If s is omitted it is set to 1.
i1 : P = hypercube 3 o1 = {ambient dimension => 3 } dimension of lineality space => 0 dimension of polyhedron => 3 number of facets => 6 number of rays => 0 number of vertices => 8 o1 : Polyhedron i2 : vertices P o2 = | -1 1 -1 1 -1 1 -1 1 | | -1 -1 1 1 -1 -1 1 1 | | -1 -1 -1 -1 1 1 1 1 | 3 8 o2 : Matrix QQ <--- QQ
## Ways to use hypercube :
• "hypercube(ZZ)"
• "hypercube(ZZ,QQ)"
• "hypercube(ZZ,ZZ)"
## For the programmer
The object hypercube is . | 294 | 874 | {"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} | 2.515625 | 3 | CC-MAIN-2023-50 | latest | en | 0.596209 |
https://chrischona2015.org/which-system-of-linear-inequalities-has-the-point-2-1-in-its-solution-set/ | 1,653,544,262,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662601401.72/warc/CC-MAIN-20220526035036-20220526065036-00520.warc.gz | 221,238,058 | 4,436 | A mechanism of straight inequalities in two variables is composed of at least two direct inequalities in the same variables. The systems of a linear inequality is the ordered pair that is a systems to every inequalities in the system and the graph of the linear inequality is the graph of all services of the system.
Example
Graph the system of inequalities
$$\left{eginmatrix yleq 2x-3: : : : : : : : \ ygeq -3: : : : : : : : : : : : : : : \ yleq -1.25x+2.5 endmatrix ight.\$$
Graph one heat at the time in the exact same coordinate aircraft and shade the half-plane the satisfies the inequality.
The solution region which is the intersection of the half-planes is presented in a darker shade
Usually only the solution region is shaded which provides it less complicated to view which an ar is the solution region
## Video lesson
Graph the system of inequalities
$$\left{eginmatrix x > 3 \ y leq -x + 2 endmatrix ight.\$$
You are watching: Which system of linear inequalities has the point (2, 1) in its solution set?
## Next Chapter:
EXPONENTS and also EXPONENTIAL attributes – properties of exponents
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All courses | 459 | 1,997 | {"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} | 3.453125 | 3 | CC-MAIN-2022-21 | latest | en | 0.879114 |
https://www.edaboard.com/threads/how-much-will-be-the-current.106401/ | 1,611,709,194,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610704804187.81/warc/CC-MAIN-20210126233034-20210127023034-00709.warc.gz | 758,659,982 | 17,214 | # How much will be the current
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#### dmmahesh
##### Junior Member level 2
If we invert 12Vdc into 220Vac, 50Hz what extend of current (Amps) we can use from the output. Say it is 500W means we may use upto 2Amps., is it right, after getting that output is it possible to stepup up into higher current rate upto 8Amp with same 230Vac? If so what can we do with the transformer or circuit and what would be the forumla we've to use? Please kindly give me in detail
#### swingbyte
##### Member level 2
P=V*I
500W/220=~2A. You cannot get power from no where so this is the maximum current you can draw from this unit. You will need a more powerful unit if you want 240V at 8A
P=240*8=1.92KW!! That would mean
1.92KW/12 = 160A on the input.
Thats a lot of Amps!
#### Johnny Hu
##### Newbie level 4
Hello everybody, this is my first participation here, i hope to learn from you and help with my little skills.
About the calculation above, since the efficiency of any conversor is never 100%, you actually will need more input power than that, say some 12% or more (It depends on the tecnology used to built the DC-AC conversor). PWM/Power-MOSFETs circuits are the most efficient of all.
Status
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# Real-Life Knowledge You Need to Know for Critical Reading:
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Magoosh GMAT Instructor
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Real-Life Knowledge You Need to Know for Critical Reading: Economics, Business, Statistics, and Law
For the GMAT Verbal section, on CR and RC, you are not expected to have specialized expertise in the topics they discuss. Sometimes, though, it is particularly helpful to have general real world knowledge. If you make a habit of reading regularly either a good newspaper (the New York Times, the Wall Street Journal, etc.) or the Economist magazine, then probably you will have a very good sense of the "push and pull" of real world situations.
If your GMAT is soon, and you don't have time to get up to speed with everything about the practicalities of business, economics, and politics in the real world, this is the first post in a series of five real-life areas you should be familiar with: knowing these can only help you interpret these verbal questions.
Each post will have hyperlinks to articles with more information, if you find the content new or unfamiliar. Again, no question will automatically expect you to know any of these inside out, but insofar as you understand these, they will provide germane background, help you interpret the question, and help you to spot unrealistic incorrect answer choices.
The full list:
Economics: Supply and Demand (below)
Economics: Labor and Wages
Economics: Inflation, unemployment, and interest rates
Law: "beyond any reasonable doubt"
Statistics: Statistical significance
The Law of Supply and Demand
This is a must-know idea if you are heading toward business school. If you studied business or economics in your undergrad years, supply and demand should definitely be familiar ideas. When a product or service is for sale in the free market, as a general rule, there are two forces determining how much it will cost. Demand concerns how much the potential consumers want the product or service --- how willing they would be to pay a certain price for it. High demand means: people would simply line up to buy this product, even at a high price; low demand means that the product is hard to sell, that even lowering the price does little to spur sales of the product. Supply concerns how easy it is to bring a certain product or service to market: for a given price, how willing/able would producers be to bring this product or service to the market for sale. High supply means: the producers easily can bring tons of it to the market. Low supply, or short supply means: it's harder to produce the product, and producers can't make as many or as much as consumers would like to buy.
The Law of Supply and Demand says that, in a free competitive marketplace, the price will come to equilibrium at that level at which supply equals demand. In other words, at this equilibrium price, the producers are willing to make N units, and the consumers are willing to buy N units. The Law of Supply and Demand implies --- as a general rule, if a product or service becomes scarcer, harder to get or provide or make --- i.e. supply drops --- then usually the price will rise; and if a product or service becomes more plentiful, easier to get or provide or make --- i.e. supply rises ---- then the price will tend to fall. For example, when world events threaten the supply of crude oil to the US, then the price of gas in the US rises. Similarly, when demand is very high, the price typically will be high, and when demand drops, usually the price will as well. Think of a new CD or DVD --- when it's first released, everyone wants it --- it's the new "hot" thing --- and since demand is high, the price it typically high; then wait a while --- a few months, or certainly a few years later --- inevitably interest in the item drops, and so does the price.
The Law of Supply and Demand is not a universal rule. There are some rare example of items that do not follow the Law of Supply and Demand, for instance, a Veblan good or a Giffen good. You do not need to know the technical details about these exceptions, but it's good to know that there are exceptions.
Summary
Knowledge of any of this is not "required" for the GMAT, but the more familiar you are with these facts, the easier it will be to place arguments in context and identify unrealistic incorrect answer choices. Furthermore, if you want to demonstrate your competence in business school and beyond, it certainly would be a good idea to understand all of these things well. Here are a couple CR practice questions that touch tangentially on some of these ideas.
1) http://gmat.magoosh.com/questions/1277
2) http://gmat.magoosh.com/questions/1310
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Re: Real-Life Knowledge You Need to Know for Critical Reading: [#permalink] 21 Jul 2017, 18:35
Display posts from previous: Sort by | 1,681 | 7,133 | {"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} | 2.546875 | 3 | CC-MAIN-2017-47 | latest | en | 0.900583 |
https://socratic.org/questions/how-do-solve-the-following-linear-system-2x-3y-1-x-2y-8 | 1,719,325,712,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198866143.18/warc/CC-MAIN-20240625135622-20240625165622-00009.warc.gz | 459,529,976 | 6,181 | # How do solve the following linear system?: -2x+3y=-1 , x-2y=8 ?
Nov 24, 2015
X = -22
Y = -15
#### Explanation:
-2x+3y= -1 Equation 1
x - 2y = 8
Equation 2
MAKE X THE SUBJECT IN EQUATION 2
x = 8 + 2y ____ Equation 3
FIT THE NEW EQUATION 3 INTO EQUATION, SUBSTITUTING THE VALUE OF X
-2 (8 + 2y) + 3y = -1
-16 - 4y + 3y = -1
COLLECT THE LIKE TERMS
-4y + 3y = -1 +16
-y = 15
y = -15
SUBSTITUTE THE NEW VALUE OF Y IN EITHER EQUATION 1 AND EQUATION 2
(Using Equation 2)
x - 2(-15) = 8
x + 30 = 8
x = 8 - 30
x = -22 | 239 | 523 | {"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} | 4.40625 | 4 | CC-MAIN-2024-26 | latest | en | 0.703906 |
https://bashtage.github.io/linearmodels/devel/iv/iv/linearmodels.iv.results.IVResults.wu_hausman.html | 1,720,884,871,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514510.7/warc/CC-MAIN-20240713150314-20240713180314-00670.warc.gz | 104,372,862 | 10,146 | # linearmodels.iv.results.IVResults.wu_hausman¶
IVResults.wu_hausman(variables: str | list[str] | None = None) [source]
Wu-Hausman test of exogeneity
Parameters:
variables: str | list[str] | None = None
List of variables to test for exogeneity. If None, all variables are jointly tested.
Returns:
Object containing test statistic, p-value, distribution and null
Return type:
linearmodels.shared.hypotheses.WaldTestStatistic
Notes
Test statistic is difference between sum of squared OLS and sum of squared IV residuals where each set of residuals has been projected onto the set of instruments in the IV model.
Start by defining
$\delta & = \hat{\epsilon}'_e P_{[z,w]} \hat{\epsilon}_e - \hat{\epsilon}'_c P_{z} \hat{\epsilon}_c$
where $$\hat{\epsilon}_e$$ are the regression residuals from a model where vars are treated as exogenous, $$\hat{\epsilon}_c$$ are the regression residuals from the model leaving vars as endogenous, $$P_{[z,w]}$$ is a projection matrix onto the exogenous variables and instruments (z) as well as vars, and $$P_{z}$$ is a projection matrix only onto z.
The test statistic is then
$\frac{\delta / q}{(\hat{\epsilon}'_e\hat{\epsilon}_e - \delta) / v}$
where $$q$$ is the number of variables iv, $$v = n - n_{endog} - n_{exog} - q$$. The test statistic has a $$F_{q, v}$$ distribution. | 362 | 1,326 | {"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} | 2.890625 | 3 | CC-MAIN-2024-30 | latest | en | 0.799562 |
https://www.nag.com/numeric/nl/nagdoc_26.1/nagdoc_fl26.1/html/d02/d02bhf.html | 1,627,642,761,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153966.52/warc/CC-MAIN-20210730091645-20210730121645-00038.warc.gz | 962,962,181 | 10,066 | # NAG Library Routine Document
## 1Purpose
d02bhf integrates a system of first-order ordinary differential equations over an interval with suitable initial conditions, using a Runge–Kutta–Merson method, until a user-specified function of the solution is zero.
## 2Specification
Fortran Interface
Subroutine d02bhf ( x, xend, n, y, tol, hmax, fcn, g, w,
Integer, Intent (In) :: n, irelab Integer, Intent (Inout) :: ifail Real (Kind=nag_wp), External :: g Real (Kind=nag_wp), Intent (In) :: xend, hmax Real (Kind=nag_wp), Intent (Inout) :: x, y(n), tol Real (Kind=nag_wp), Intent (Out) :: w(n,7) External :: fcn
#include nagmk26.h
void d02bhf_ (double *x, const double *xend, const Integer *n, double y[], double *tol, const Integer *irelab, const double *hmax, void (NAG_CALL *fcn)(const double *x, const double y[], double f[]),double (NAG_CALL *g)(const double *x, const double y[]),double w[], Integer *ifail)
## 3Description
d02bhf advances the solution of a system of ordinary differential equations
$yi′=fix,y1,y2,…,yn, i=1,2,…,n,$
from $x={\mathbf{x}}$ towards $x={\mathbf{xend}}$ using a Merson form of the Runge–Kutta method. The system is defined by fcn, which evaluates ${f}_{i}$ in terms of $x$ and ${y}_{1},{y}_{2},\dots ,{y}_{\mathit{n}}$ (see Section 5), and the values of ${y}_{1},{y}_{2},\dots ,{y}_{\mathit{n}}$ must be given at $x={\mathbf{x}}$.
As the integration proceeds, a check is made on the function $g\left(x,y\right)$ specified by you, to determine an interval where it changes sign. The position of this sign change is then determined accurately by interpolating for the solution and its derivative. It is assumed that $g\left(x,y\right)$ is a continuous function of the variables, so that a solution of $g\left(x,y\right)=0$ can be determined by searching for a change in sign in $g\left(x,y\right)$.
The accuracy of the integration and, indirectly, of the determination of the position where $g\left(x,y\right)=0$, is controlled by tol.
For a description of Runge–Kutta methods and their practical implementation see Hall and Watt (1976).
## 4References
Hall G and Watt J M (ed.) (1976) Modern Numerical Methods for Ordinary Differential Equations Clarendon Press, Oxford
## 5Arguments
1: $\mathbf{x}$ – Real (Kind=nag_wp)Input/Output
On entry: must be set to the initial value of the independent variable $x$.
On exit: the point where $g\left(x,y\right)=0.0$ unless an error has occurred, when it contains the value of $x$ at the error. In particular, if $g\left(x,y\right)\ne 0.0$ anywhere on the range x to xend, it will contain xend on exit.
2: $\mathbf{xend}$ – Real (Kind=nag_wp)Input
On entry: the final value of the independent variable $x$.
If ${\mathbf{xend}}<{\mathbf{x}}$ on entry, integration proceeds in a negative direction.
3: $\mathbf{n}$ – IntegerInput
On entry: $\mathit{n}$, the number of differential equations.
Constraint: ${\mathbf{n}}>0$.
4: $\mathbf{y}\left({\mathbf{n}}\right)$ – Real (Kind=nag_wp) arrayInput/Output
On entry: the initial values of the solution ${y}_{1},{y}_{2},\dots ,{y}_{\mathit{n}}$.
On exit: the computed values of the solution at the final point $x={\mathbf{x}}$.
5: $\mathbf{tol}$ – Real (Kind=nag_wp)Input/Output
On entry: must be set to a positive tolerance for controlling the error in the integration and in the determination of the position where $g\left(x,y\right)=0.0$.
d02bhf has been designed so that, for most problems, a reduction in tol leads to an approximately proportional reduction in the error in the solution obtained in the integration. The relation between changes in tol and the error in the determination of the position where $g\left(x,y\right)=0.0$ is less clear, but for tol small enough the error should be approximately proportional to tol. However, the actual relation between tol and the accuracy cannot be guaranteed. You are strongly recommended to call d02bhf with more than one value for tol and to compare the results obtained to estimate their accuracy. In the absence of any prior knowledge you might compare results obtained by calling d02bhf with ${\mathbf{tol}}={10.0}^{-p}$ and ${\mathbf{tol}}={10.0}^{-p-1}$ if $p$ correct decimal digits in the solution are required.
Constraint: ${\mathbf{tol}}>0.0$.
On exit: normally unchanged. However if the range from $x={\mathbf{x}}$ to the position where $g\left(x,y\right)=0.0$ (or to the final value of $x$ if an error occurs) is so short that a small change in tol is unlikely to make any change in the computed solution, tol is returned with its sign changed. To check results returned with ${\mathbf{tol}}<0.0$, d02bhf should be called again with a positive value of tol whose magnitude is considerably smaller than that of the previous call.
6: $\mathbf{irelab}$ – IntegerInput
On entry: determines the type of error control. At each step in the numerical solution an estimate of the local error, $\mathit{est}$, is made. For the current step to be accepted the following condition must be satisfied:
${\mathbf{irelab}}=0$
$\mathit{est}\le {\mathbf{tol}}×\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left\{1.0,\left|{y}_{1}\right|,\left|{y}_{2}\right|,\dots ,\left|{y}_{\mathit{n}}\right|\right\}$;
${\mathbf{irelab}}=1$
$\mathit{est}\le {\mathbf{tol}}$;
${\mathbf{irelab}}=2$
$\mathit{est}\le {\mathbf{tol}}×\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left\{\epsilon ,\left|{y}_{1}\right|,\left|{y}_{2}\right|,\dots ,\left|{y}_{\mathit{n}}\right|\right\}$, where $\epsilon$ is machine precision.
If the appropriate condition is not satisfied, the step size is reduced and the solution recomputed on the current step.
If you wish to measure the error in the computed solution in terms of the number of correct decimal places, set ${\mathbf{irelab}}=1$ on entry, whereas if the error requirement is in terms of the number of correct significant digits, set ${\mathbf{irelab}}=2$. Where there is no preference in the choice of error test, ${\mathbf{irelab}}=0$ will result in a mixed error test. It should be borne in mind that the computed solution will be used in evaluating $g\left(x,y\right)$.
Constraint: ${\mathbf{irelab}}=0$, $1$ or $2$.
7: $\mathbf{hmax}$ – Real (Kind=nag_wp)Input
On entry: if ${\mathbf{hmax}}=0.0$, no special action is taken.
If ${\mathbf{hmax}}\ne 0.0$, a check is made for a change in sign of $g\left(x,y\right)$ at steps not greater than $\left|{\mathbf{hmax}}\right|$. This facility should be used if there is any chance of ‘missing’ the change in sign by checking too infrequently. For example, if two changes of sign of $g\left(x,y\right)$ are expected within a distance $h$, say, of each other, a suitable value for hmax might be ${\mathbf{hmax}}=h/2$. If only one change of sign in $g\left(x,y\right)$ is expected on the range x to xend, the choice ${\mathbf{hmax}}=0.0$ is most appropriate.
8: $\mathbf{fcn}$ – Subroutine, supplied by the user.External Procedure
fcn must evaluate the functions ${f}_{i}$ (i.e., the derivatives ${y}_{i}^{\prime }$) for given values of its arguments $x,{y}_{1},\dots ,{y}_{\mathit{n}}$.
The specification of fcn is:
Fortran Interface
Subroutine fcn ( x, y, f)
Real (Kind=nag_wp), Intent (In) :: x, y(*) Real (Kind=nag_wp), Intent (Out) :: f(*)
#include nagmk26.h
void fcn (const double *x, const double y[], double f[])
In the description of the arguments of d02bhf below, $\mathit{n}$ denotes the value of n in the call of d02bhf.
1: $\mathbf{x}$ – Real (Kind=nag_wp)Input
On entry: $x$, the value of the argument.
2: $\mathbf{y}\left(*\right)$ – Real (Kind=nag_wp) arrayInput
On entry: ${y}_{\mathit{i}}$, for $\mathit{i}=1,2,\dots ,\mathit{n}$, the value of the argument.
3: $\mathbf{f}\left(*\right)$ – Real (Kind=nag_wp) arrayOutput
On exit: the value of ${f}_{\mathit{i}}$, for $\mathit{i}=1,2,\dots ,\mathit{n}$.
fcn must either be a module subprogram USEd by, or declared as EXTERNAL in, the (sub)program from which d02bhf is called. Arguments denoted as Input must not be changed by this procedure.
Note: fcn should not return floating-point NaN (Not a Number) or infinity values, since these are not handled by d02bhf. If your code inadvertently does return any NaNs or infinities, d02bhf is likely to produce unexpected results.
9: $\mathbf{g}$ – real (Kind=nag_wp) Function, supplied by the user.External Procedure
g must evaluate the function $g\left(x,y\right)$ at a specified point.
The specification of g is:
Fortran Interface
Function g ( x, y)
Real (Kind=nag_wp) :: g Real (Kind=nag_wp), Intent (In) :: x, y(*)
#include nagmk26.h
double g (const double *x, const double y[])
In the description of the arguments of d02bhf below, $\mathit{n}$ denotes the value of n in the call of d02bhf.
1: $\mathbf{x}$ – Real (Kind=nag_wp)Input
On entry: $x$, the value of the independent variable.
2: $\mathbf{y}\left(*\right)$ – Real (Kind=nag_wp) arrayInput
On entry: the value of ${y}_{\mathit{i}}$, for $\mathit{i}=1,2,\dots ,\mathit{n}$.
g must either be a module subprogram USEd by, or declared as EXTERNAL in, the (sub)program from which d02bhf is called. Arguments denoted as Input must not be changed by this procedure.
Note: g should not return floating-point NaN (Not a Number) or infinity values, since these are not handled by d02bhf. If your code inadvertently does return any NaNs or infinities, d02bhf is likely to produce unexpected results.
10: $\mathbf{w}\left({\mathbf{n}},7\right)$ – Real (Kind=nag_wp) arrayWorkspace
11: $\mathbf{ifail}$ – IntegerInput/Output
On entry: ifail must be set to $0$, $-1\text{ or }1$. If you are unfamiliar with this argument you should refer to Section 3.4 in How to Use the NAG Library and its Documentation for details.
For environments where it might be inappropriate to halt program execution when an error is detected, the value $-1\text{ or }1$ is recommended. If the output of error messages is undesirable, then the value $1$ is recommended. Otherwise, if you are not familiar with this argument, the recommended value is $0$. When the value $-\mathbf{1}\text{ or }\mathbf{1}$ is used it is essential to test the value of ifail on exit.
On exit: ${\mathbf{ifail}}={\mathbf{0}}$ unless the routine detects an error or a warning has been flagged (see Section 6).
## 6Error Indicators and Warnings
If on entry ${\mathbf{ifail}}=0$ or $-1$, explanatory error messages are output on the current error message unit (as defined by x04aaf).
Errors or warnings detected by the routine:
${\mathbf{ifail}}=1$
On entry, ${\mathbf{tol}}\le 0.0$, or ${\mathbf{n}}\le 0$, or ${\mathbf{irelab}}\ne 0$, $1$ or $2$.
${\mathbf{ifail}}=2$
With the given value of tol, no further progress can be made across the integration range from the current point $x={\mathbf{x}}$, or dependence of the error on tol would be lost if further progress across the integration range were attempted (see Section 9 for a discussion of this error exit). The components ${\mathbf{y}}\left(1\right),{\mathbf{y}}\left(2\right),\dots ,{\mathbf{y}}\left(\mathit{n}\right)$ contain the computed values of the solution at the current point $x={\mathbf{x}}$. No point at which $g\left(x,y\right)$ changes sign has been located up to the point $x={\mathbf{x}}$.
${\mathbf{ifail}}=3$
tol is too small for d02bhf to take an initial step (see Section 9). x and ${\mathbf{y}}\left(1\right),{\mathbf{y}}\left(2\right),\dots ,{\mathbf{y}}\left(\mathit{n}\right)$ retain their initial values.
${\mathbf{ifail}}=4$
At no point in the range x to xend did the function $g\left(x,y\right)$ change sign. It is assumed that $g\left(x,y\right)=0.0$ has no solution.
${\mathbf{ifail}}=5$ (c05azf)
A serious error has occurred in an internal call to the specified routine. Check all subroutine calls and array dimensions. Seek expert help.
${\mathbf{ifail}}=6$
A serious error has occurred in an internal call to an integration routine. Check all subroutine calls and array dimensions. Seek expert help.
${\mathbf{ifail}}=7$
A serious error has occurred in an internal call to an interpolation routine. Check all (sub)program calls and array dimensions. Seek expert help.
${\mathbf{ifail}}=-99$
See Section 3.9 in How to Use the NAG Library and its Documentation for further information.
${\mathbf{ifail}}=-399$
Your licence key may have expired or may not have been installed correctly.
See Section 3.8 in How to Use the NAG Library and its Documentation for further information.
${\mathbf{ifail}}=-999$
Dynamic memory allocation failed.
See Section 3.7 in How to Use the NAG Library and its Documentation for further information.
## 7Accuracy
The accuracy depends on tol, on the mathematical properties of the differential system, on the position where $g\left(x,y\right)=0.0$ and on the method. It can be controlled by varying tol but the approximate proportionality of the error to tol holds only for a restricted range of values of tol. For tol too large, the underlying theory may break down and the result of varying tol may be unpredictable. For tol too small, rounding error may affect the solution significantly and an error exit with ${\mathbf{ifail}}={\mathbf{2}}$ or ${\mathbf{3}}$ is possible.
The accuracy may also be restricted by the properties of $g\left(x,y\right)$. You should try to code g without introducing any unnecessary cancellation errors.
## 8Parallelism and Performance
d02bhf is not threaded in any implementation.
The time taken by d02bhf depends on the complexity and mathematical properties of the system of differential equations defined by fcn, the complexity of g, on the range, the position of the solution and the tolerance. There is also an overhead of the form $a+b×\mathit{n}$ where $a$ and $b$ are machine-dependent computing times.
For some problems it is possible that d02bhf will return ${\mathbf{ifail}}={\mathbf{4}}$ because of inaccuracy of the computed values y, leading to inaccuracy in the computed values of $g\left(x,y\right)$ used in the search for the solution of $g\left(x,y\right)=0.0$. This difficulty can be overcome by reducing tol sufficiently, and if necessary, by choosing hmax sufficiently small. If possible, you should choose xend well beyond the expected point where $g\left(x,y\right)=0.0$; for example make $\left|{\mathbf{xend}}-{\mathbf{x}}\right|$ about $50%$ larger than the expected range. As a simple check, if, with xend fixed, a change in tol does not lead to a significant change in y at xend, then inaccuracy is not a likely source of error.
If d02bhf fails with ${\mathbf{ifail}}={\mathbf{3}}$, then it could be called again with a larger value of tol if this has not already been tried. If the accuracy requested is really needed and cannot be obtained with this routine, the system may be very stiff (see below) or so badly scaled that it cannot be solved to the required accuracy.
If d02bhf fails with ${\mathbf{ifail}}={\mathbf{2}}$, it is likely that it has been called with a value of tol which is so small that a solution cannot be obtained on the range x to xend. This can happen for well-behaved systems and very small values of tol. You should, however, consider whether there is a more fundamental difficulty. For example:
(a) in the region of a singularity (infinite value) of the solution, the routine will usually stop with ${\mathbf{ifail}}={\mathbf{2}}$, unless overflow occurs first. If overflow occurs using d02bhf, d02pff can be used instead to detect the increasing solution, before overflow occurs. In any case, numerical integration cannot be continued through a singularity, and analytical treatment should be considered; (b) for ‘stiff’ equations, where the solution contains rapidly decaying components, the routine will compute in very small steps in $x$ (internally to d02bhf) to preserve stability. This will usually exhibit itself by making the computing time excessively long, or occasionally by an exit with ${\mathbf{ifail}}={\mathbf{2}}$. Merson's method is not efficient in such cases, and you should try d02ejf which uses a Backward Differentiation Formula method. To determine whether a problem is stiff, d02pef may be used.
For well-behaved systems with no difficulties such as stiffness or singularities, the Merson method should work well for low accuracy calculations (three or four figures). For high accuracy calculations or where fcn is costly to evaluate, Merson's method may not be appropriate and a computationally less expensive method may be d02cjf which uses an Adams' method.
For problems for which d02bhf is not sufficiently general, you should consider d02pff. d02pff is a more general routine with many facilities including a more general error control criterion. d02pff can be combined with the rootfinder c05azf and the interpolation routine d02psf to solve equations involving ${y}_{1},{y}_{2},\dots ,{y}_{\mathit{n}}$ and their derivatives.
d02bhf can also be used to solve an equation involving $x$, ${y}_{1},{y}_{2},\dots ,{y}_{\mathit{n}}$ and the derivatives of ${y}_{1},{y}_{2},\dots ,{y}_{\mathit{n}}$. For example in Section 10, d02bhf is used to find a value of ${\mathbf{x}}>0.0$ where ${\mathbf{y}}\left(1\right)=0.0$. It could instead be used to find a turning-point of ${y}_{1}$ by replacing the function $g\left(x,y\right)$ in the program by:
```Real (kind=nag_wp) Function g(x,y)
Real (kind=nag_wp) x,y(3),f(3)
Call fcn(x,y,f)
g = f(1)
Return
End```
This routine is only intended to locate the first zero of $g\left(x,y\right)$. If later zeros are required, you are strongly advised to construct your own more general root-finding routines as discussed above.
## 10Example
This example finds the value ${\mathbf{x}}>0.0$ at which $y=0.0$, where $y$, $v$, $\varphi$ are defined by
$y′ = tanϕ v′ = -0.032tanϕv- 0.02v cosϕ ϕ′ = -0.032v2$
and where at ${\mathbf{x}}=0.0$ we are given $y=0.5$, $v=0.5$ and $\varphi =\pi /5$. We write $y={\mathbf{y}}\left(1\right)$, $v={\mathbf{y}}\left(2\right)$ and $\varphi ={\mathbf{y}}\left(3\right)$ and we set ${\mathbf{tol}}=\text{1.0E−4}$ and ${\mathbf{tol}}=\text{1.0E−5}$ in turn so that we can compare the solutions. We expect the solution ${\mathbf{x}}\simeq 7.3$ and so we set ${\mathbf{xend}}=10.0$ to avoid determining the solution of $y=0.0$ too near the end of the range of integration. The initial values and range are read from a data file.
### 10.1Program Text
Program Text (d02bhfe.f90)
### 10.2Program Data
Program Data (d02bhfe.d)
### 10.3Program Results
Program Results (d02bhfe.r)
© The Numerical Algorithms Group Ltd, Oxford, UK. 2017 | 5,385 | 18,479 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 164, "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} | 2.859375 | 3 | CC-MAIN-2021-31 | longest | en | 0.712372 |
https://web2.0calc.com/questions/system_114 | 1,713,554,457,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817442.65/warc/CC-MAIN-20240419172411-20240419202411-00385.warc.gz | 560,895,813 | 5,796 | +0
# System
0
235
2
Suppose a, b, c, and d are real numbers which satisfy the system of equations
a + 2b + 3c + 4d = 10
4a + b + 2c + 3d = 4
3a + 4b + c + 2d = -10
2a + 3b + 4c + d = −4.
Find a + b + c + d.
May 2, 2022
#1
+2666
0
Adding all 4 equations gives us: \(10(a+b+c+d)=0\)
Dividing both sides by 10, we find \(a+b+c+d=\color{brown}\boxed{0}\)
May 2, 2022
#2
+118608
+2
Hi Builderboi,
I do have a suggestion though.
Try experimenting with giving answers that encourage askers to learn and that make it more difficult for them to just copy.
Like here you could have said.
"Try adding all the left sides and putting the answer equal to the sum of all the right sides. See what you get ;)"
People are much more likely to learn from you if you do not do all the work for them.
Teaching is a learned art.
May 2, 2022 | 292 | 840 | {"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-2024-18 | latest | en | 0.908848 |
https://magicorelearning.com/shop/product-line/escape-rooms/partitions-and-fractions-geometry-escape-room-2nd-grade | 1,670,190,693,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710980.82/warc/CC-MAIN-20221204204504-20221204234504-00433.warc.gz | 419,206,608 | 50,995 | # Fractions 2nd Grade Math Candy Factory Escape Room – Digital & Printable
\$5.00
## Description
### Fractions 2nd Grade Math Escape Room
⭐ This 2nd grade partitions and fractions geometry math escape room is a truly immersive experience and themed perfectly for a class of candy enthusiasts! Students navigate their way through a candy factory by completing challenges to help create delicious creations and become a candy connoisseur. To earn the key to the factory, students complete four unique math standards-based challenges. The challenges require students to practice partitioning rectangles and identifying factions. Students add their answers to the Factory Decoding Tool to help them unlock the secret codes and complete each challenge. Students travel to various floors of their factory while making these delicious treats and mastering key mathematics skills! This resource is a great engaging and low prep activity with printable and digital versions are included!
⭐Comes in both digital Google Slides and printable versions, so it’s perfect for classroom and distance learning. In the digital version, the Factory Decoding Tool is a Google Sheet that self corrects and tells each student when they can continue to the next challenge.
⭐ Integrated videos, thematic items, and a candy factory storyline come together to engage students while developing and enforcing several standards-based mathematics skills.
Key Features:
✏️ Digital & Printable Versions – Google Slides/Forms and print versions make this resource perfect for classroom and distance learning. The digital version can be used dynamically in the classroom as a presentation while students complete each passage and question set in groups or individually.
✏️ Immersive Experience: Candy Factory themed, dynamic challenges, and engaging videos provide positive reinforcement and keep students engaged and focused.
✏️Skill Focused: Each question set was written to focus on the key mathematics skills of partitioning rectangles and identifying fractions.
✏️Scaffolded Challenges: Challenges increase in difficulty as students progress through the escape room.
✏️No Prep/ Low Prep: Digital version requires NO PREP. The printable version is low prep.: Just print the pages, place them in folders, and go. The printable version has the option to incorporate technology and videos through QR codes.
What’s Included:
2nd Grade Partitions and Fractions Candy Factory Escape Room
• Print and Digital Directions
• Candy Man Invitation Letter
• Challenge #1: Partition rectangles to make the marshmallows.
• Challenge #2: Partition rectangles to make the cotton candy.
• Challenge #3: Identify fractions to make the fun-size chocolate.
• Challenge #4: Identify fractions to make the sprinkles.
• Keys to the Factory Certificate
• Ooops Cards
• Recording Brochure
Take a closer look at what this resource looks like in this video preview.
Please Note: This escape room is similar to the 3rd grade version but with problems and skills at a 2nd grade level.
***************************************************************************
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Virtual Field Trip – Growing Mega Bundle
## Reviews
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Only logged in customers who have purchased this product may leave a review. | 729 | 3,617 | {"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} | 2.546875 | 3 | CC-MAIN-2022-49 | latest | en | 0.883943 |
https://labuladong.gitbook.io/algo-en/ii.-data-structure/reverse_part_of_a_linked_list_via_recursion | 1,726,890,603,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725701427996.97/warc/CC-MAIN-20240921015054-20240921045054-00876.warc.gz | 307,862,493 | 41,064 | # Reverse Part of Linked List via Recursion
Translator: CarrieOn
It's easy to reverse a single linked list using iteration, however it's kind of difficult to come up with a recursive solution. Furthermore, if only part of a linked list needs reversed, can you nail it with recursion?
If you haven't known how to recursively reverse a single linked list, no worry, we will start right here and guide you step by step to a deeper level.
``````// node structure for a single linked list
public class ListNode {
int val;
ListNode next;
ListNode(int x) { val = x; }
}``````
To reverse part of a linked list means we only reverse elements in a specific interval and leave others untouched.
Note: Index starts from 1. Two loops needed if solve via iteration: use one for-loop to find the mth element, and then use another for-loop to reverse elements between m and n. While in recursive solution, no loop at all.
Though iterative solution looks simple, you have to be careful with the details. On the contrary, recursive solution is quite elegant. Let's start reversing a whole single linked list in the recursive way.
## 1. Recursively reverse a whole single Linked List
You may have already known the solution below.
``````ListNode reverse(ListNode head) {
return last;
}``````
Do you feel lost in trying to understand code above? Well, you are not the only one. This algorithm is often used to show how clever and elegant recursion can be. Let's dig into the code together.
For recursion, the most important thing is to clarify the definition of the recursive function. Specifically, we define `reverse` as follows:
Input a node `head`, we will reverse the list starting from `head`, and return the new head node.
After clarifying the definition, we look back at the problem. For example, we want to reverse the list below:
So after calling `reverse(head)`, recursion happens:
``ListNode last = reverse(head.next);``
Did you just step into the messy details in recursion? Oops, it's a wrong way, step back now! Focus on the recursion definition (which tells you what it does) to understand how recursive code works the wonder.
After executing `reverse(head.next)`, the whole linked list becomes this:
According to the definition of the recursive function, `reverse` needs to return the new head node, so we use variable `last` to mark it.
Let's continue cracking the next piece of code:
``head.next.next = head;``
Last work to do：
``````head.next = null;
return last;``````
The whole linked list is successfully reversed now. Amazing, isn't it?
Last but not the least, there are two things in recursion you need to pay attention to:
1. Recursion needs a base case.
`` if(head.next == null) return head;``
which means when there is only one node, after reversion, the head is still itself.
2. After reversion, the new head is `last`, and the former `head` becomes the last node, don't forget to point its tail to null.
`` head.next = null;``
After understanding above, now we can proceed further, the problem below is actually an extend to the above solution.
## 2. Reverse first N nodes
This time we will implement a funtion below:
``````// reverse first n nodes in a linked list (n <= length of the list)
Take below as an example, call `reverseN(head, 3)`:
The idea is similar to reversing the whole linked list, only a few modifications needed:
``````ListNode successor = null; // successor node
ListNode reverseN(ListNode head, int n) {
if (n == 1) {
// mark the (n + 1)th node
}
// starts from head.next, revers the first n - 1 nodes
ListNode last = reverseN(head.next, n - 1);
return last;
}``````
Main differences:
1. Base case `n == 1`, if reverse only one element, then new head is itself, meanwhile remember to mark the successor node.
2. In previouse solution, we set `head.next` directly to null, because after reversing the whole list, head becoms the last node. But now `head` may not be the last node after reversion, so we need mark `successor` (the (n+1)th node), and link it to `head` after reversion.
OK, now we are pretty close to reversing part of the linked list.
## 3. Reverse part of a linked list
Given an interval `[m,n]` (index starts from 1), only reverse elements in this section.
``ListNode reverseBetween(ListNode head, int m, int n)``
First, if `m == 1`, it is equal to reversing the first `n` elements as we discussed just now.
``````ListNode reverseBetween(ListNode head, int m, int n) {
// base case
if (m == 1) {
// equals to reversing the first n nodes
}
// ...
}``````
What if `m != 1`? If we take the index of the `head` as 1, then we need to reverse from the `mth` element. And what if we take the index of the `head.next` as 1? Then compared to `head.next`, the reverse section should start from `(m-1)th` element. And what about `head.next.next` ...
Different from iteration, this is how we think in the recursive way, so our code should be:
``````ListNode reverseBetween(ListNode head, int m, int n) {
// base case
if (m == 1) {
}
Mission: Stick to original high quality articles, and make algorithms easy to understand. Welcome to subscribe my Wechat public account `ID:labuladong` for latest articles. | 1,247 | 5,192 | {"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-2024-38 | latest | en | 0.895775 |
1stladysaloon.com | 1,657,057,725,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104628307.87/warc/CC-MAIN-20220705205356-20220705235356-00163.warc.gz | 141,706 | 16,184 | # Concomitantly, I emphasize types of algorithmic ineffability, and showcase how specific procedures are more difficult to mathematically catch
Concomitantly, I emphasize types of algorithmic ineffability, and showcase how specific procedures are more difficult to mathematically catch
I next stroll customers through Bayes’ formula, a numerical formula that sits in the centre of not only spam filter systems, but numerous additional strong computational engineering. 1 we showcase the limits of numerical formulations through the formulations on their own by foregrounding many of the aporia of sieves. As you go along, I theorize different kinds of ontological inertia, showing just how specific presumptions tend to be a€?deepera€? and therefore more difficult to over the years change. Above all else, plus combination using the other areas, this query tries to show how equations and algorithms can concurrently be susceptible to and play a role in anthropological review.
At long last, it usually is useful to remember light polarizers: while two polarizers, at right aspects together, may quit all light from acquiring through, if you place a 3rd polarizer in the middle them, by itself 45 levels away from skew about local hookup near me Kalgoorlie Australia others two, some light will get through. Notice, then, that in sieving for an element, the substances sieved might be suffering from the sieving and thereby visited take on services they did not originally have-in certain, services that allow this type of substances to slip through these sieves. Presume, including, of Sigmund Freud’s ( 1999) strategies in regards to the dream-work. And, much more generally, think about the probability of recoding and rechanneling any information so as to fall past a censor (Kockelman 2010a). We’re going to return to this time below.
One-to-one mapping between feedback and result versus interpretant of register relation to appeal of representative featuring of item.
As a next example, and somewhat much more decisively, we could always just blend- which can be, in some good sense, the alternative of sieving: simply shake, aggregate, amass, pour, muddle, muddy, plus normally smartly discombobulate.
## Recommendations
As to what comes after, after getting sometime in order to make these information considerably clearly relevant to anthropological problems, we explore these methods in significantly more details along with a lot more generality. Audience will identify a Peircean direction in what uses, but it’s the definitions of those words that matter, maybe not the labels. 9
The indices integrate particular actions (variations of throwing and catching activities, plus threading needles, and techniques with the body a lot more normally). Plus the types in issues are kid and girl-though they can were any sociocultural identities under the sun (age.g., Huck’s dad could have eliminated into community trying to go themselves down as wealthy, sober, or sophisticated). Ultimately, observe that tag Twain, due to the fact author of this scenario, keeps a somewhat implicit ontology that also includes in it assumptions concerning the ontologies of men and women like Mrs. Loftus. Particularly, what types of values does she has when it comes to particular types, like female and boy? In doing this, many ontologies are naturally metaontologies-one may have presumptions about other individuals’ presumptions (about a person’s assumptions about others’ assumptions …), an such like, and so forth.
To return to your prior focus, if looks are some sort of type, junk e-mail is a kind of design. Specifically, and prefiguring most issues in the soon after section, filters made to end spam from achieving the email embody an ontology regarding the tendency for someone spam message to evince specific indices (in comparison to a nonspam information). Read Figure 3. Figure 3:
In a few sense, the individual-kind regards (can it be a dog or a wolf) transforms by regard to the individual-index relation (they bayed on moonlight), as the kindindex relationship continues to be constant (wolfs bay from the moon, but dogs do not). 12
Notice, then, that sieves-such as junk e-mail filters-have needs built into all of them (insofar while they selectively permit certain matters and prohibit rest); and they have values constructed into all of them (insofar while they display ontological presumptions). 15 and not just carry out sieves have actually philosophy and desires constructed into all of them (thereby, in a few feeling, embody standards that are fairly derivative regarding producers and users); they may be considered has emergent opinions and desires (and therefore embody unique relatively originary beliefs, but involuntary they and their producers and users include of those). Specifically, the standards of this variables are usually measures ahead of the consciousness associated with the developers (and certainly of users)-and thus constitute a type of prosthetic unconsciousness with very rich and wily temporal dynamics. Mention, next, that after we generate algorithms then ready those algorithms free, there can be typically absolutely no way to understand whatwill result next (Bill Maurer, private communication).
To resolve this matter, and understand the reason behind the solution, truly helpful to diagram the difficulty in a specific means. Figure 4a shows a square with a product region add up to 1. This is actually the space of all possible effects (so that the likelihood of some result is completely). Figure 4b reveals this exact same location split into two elements, among device location 2/3 (revealing the portion of urns which can be of type 1), as well as the other of unit region 1/3 (revealing that portion of urns that are of means 2). They are your a priori probabilities: loosely speaking, the chance the urn are of sort 1 or means 2 if your wanting to pull-out the copper coin. They’ve been designated P(U1) and P(U2), correspondingly. Mention, then, that when you has actually reached into the urn, just by method of how issue was set up, possible claim that the possibility that urn try of type 1 concerns 66 per cent.
## Summation
This equation might interpreted below. Throughout the left-hand area, there is PIndex(sorts), or the possibility that someone try of a certain kinds, in the context of their creating evinced some index. Throughout the right-hand area we possess the product of a probability (that folks of specific sort exhibit indices of certain kinds, or PKind(list)) and an a priori chances (or perhaps the chance, nonetheless subjective or tentative, the individual had been of this sorts earlier evinced the directory, or P(sorts)). And that goods was by itself divided by general likelihood that individual evinces the list aside from the sort, or P(Index). Crucially, although we derived this picture in the context of a global that had best two types of types with two kinds of indicator, it is totally common: one simply needs to sum during the items of likelihoods and a priori possibilities for each and every feasible kind considering the index concerned. 17 | 1,479 | 7,161 | {"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-2022-27 | longest | en | 0.919405 |
https://wiki.physics.wisc.edu/facultywiki/Rotating_Gizmo?action=recall&rev=2 | 1,529,362,787,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267861456.51/warc/CC-MAIN-20180618222556-20180619002556-00361.warc.gz | 734,185,998 | 4,183 | [:PiraScheme#Mechanics: Table of Mechanics Demonstration] [:MEEquipmentList: List of Mechanics Equipment & Supplies] [:Demonstrations:Lecture Demonstrations]
# Rotating Gizmo, 3A40.18
Topic and Concept:
• Newton's First Law, [:Newtons1STLaw#InertiaofRest: 1F20. Inertia of Rest]
Location:
attachment:Gizmo09-400.jpg
Abstract:
A device operated with a hand crank causes two small, circular discs to move in simple harmonic motion. One disc moves in a circular path while the other moves in a vertical line showing the physical correspondence of SHM with the "unit" circle (sinusoidal functions).
Equipment Location ID Number Rotating Gizmo [:MechanicsCabinetBayA11: ME, Bay A11, Shelf T]
Important Setup Notes:
• N/A
Setup and Procedure:
1. Place device on tabletop and operate with the hand crank.
Cautions, Warnings, or Safety Concerns:
• N/A
Discussion:
When the crank is turned, both discs move. One disc moves in a circle while the other moves in a vertical line. The period of oscillation for each disc depends on the speed at which the crank is turned. Regardless of the crank speed, the period of oscillation for each disc will always be the same. This shows how a 1D SHO, or more generally any SHO, can be represented with a sinusoidal function. Such functions have their domain on the unit circle, which is represented by the circularly moving disc.
attachment:Gizmo01-250.jpg attachment:Gizmo02-250.jpg attachment:Gizmo03-250.jpg attachment:Gizmo04-250.jpg attachment:Gizmo05-250.jpg attachment:Gizmo06-250.jpg attachment:Gizmo07-250.jpg attachment:Gizmo08-250.jpg
Videos:
References:
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### Author Topic: Could there be a minimum speed of matter? (Read 9972 times)
#### another_someone
• Guest
##### Could there be a minimum speed of matter?
« on: 07/07/2006 19:04:27 »
Just out of idle speculation: could there be a minimum speed of matter?
I was thinking that, since we have relativity telling us that nothing can exceed the speed of light, could we also have a converse law that says that things must travel with a minimum speed?
If such a minimum speed limit were to exists, its effects could only exist at the quantum level.
Would not a minimum speed limit would create an ambiguity in the speed of a particle that would be not by unlike the uncertainty principle?
George
#### Razak
• Jr. Member
• Posts: 17
##### Re: Could there be a minimum speed of matter?
« Reply #1 on: 08/07/2006 06:22:05 »
I think the uncertanity principal will create problems to mesure such a speed.
RazaK
#### gecko
• Sr. Member
• Posts: 196
##### Re: Could there be a minimum speed of matter?
« Reply #2 on: 08/07/2006 07:09:47 »
no. there is no minimum speed of matter. i have no evidence, so dont ask for any.
#### heikki
• Full Member
• Posts: 56
##### Re: Could there be a minimum speed of matter?
« Reply #3 on: 10/07/2006 08:36:28 »
quote:
Originally posted by another_someone
Just out of idle speculation: could there be a minimum speed of matter?
I was thinking that, since we have relativity telling us that nothing can exceed the speed of light, could we also have a converse law that says that things must travel with a minimum speed?
If such a minimum speed limit were to exists, its effects could only exist at the quantum level.
Would not a minimum speed limit would create an ambiguity in the speed of a particle that would be not by unlike the uncertainty principle?
George
Hi.
If light-particles can go that measured light speed and light speed is different in different matter-stuff(water.air.space,glass. etc.) then if exist smaller matter-particles which can go faster than measured light-speed then it can and speed therefore can be 500000km/s or ever faster. But anypath, space-nature matter-contitions make limits also of this speed (distance/time). And then there is maximun speed-limit.
So other ways, minimun speed? If one matter-particle stay on it's place without moving any direction, then it's speed is 0m/s.
Distance is 0 and speed is 0m/1s.
But, like you make question, can it be 0?
I think that if one matter-particle is alone, then is dont goes any direction and speed is 0m/1s.
But, where is the location or place to our space-nature where some matter-particle is alone without contacts some others particles?
Hmm. a_s your question is quite fine and my thought of this question is that there is minimun speed limit at todays space-nature because it move(live) all time everywhere.
But is it important to know these speed limits on these small matter-particle sizes? I'm sitting and wroting computer and my speed is now 0m/s on to my sittingplace, but my speed when i round sun is much more. What or where technics area or scient to i do or need data of these speed limit knowing? Electricity, maybe? Other areas?
#### another_someone
• Guest
##### Re: Could there be a minimum speed of matter?
« Reply #4 on: 10/07/2006 14:49:46 »
quote:
Originally posted by heikki
If light-particles can go that measured light speed and light speed is different in different matter-stuff(water.air.space,glass. etc.) then if exist smaller matter-particles which can go faster than measured light-speed then it can and speed therefore can be 500000km/s or ever faster. But anypath, space-nature matter-contitions make limits also of this speed (distance/time). And then there is maximun speed-limit.
Since the reason why light can travel at the nominal speed of light (~3x10^8 m/s) is because it has zero mass, then you are correct that if something has less mass than that, it may well travel faster than light, but the only way that matter can have less mass than zero mass is if it can have negative mass. We know of no way to create negative mass, but if such a particle could be created, it would probably also exhibit anti-gravity.
Could not exactly work out what you meant when you referring to the slowing down of light in the vicinity of matter. That slowing down happens because light is interacting with the electrical charge of the atoms (because light is an electromagnetic wave). In a simplistic way, I would imagine you can think of it as light having to hop from atom to atom, so while it is travelling between the atoms, it is still travelling at its normal speed, but then rests a while as it reaches an atom, so its average speed slows down. I accept that this is grossly simplistic.
quote:
So other ways, minimun speed? If one matter-particle stay on it's place without moving any direction, then it's speed is 0m/s.
Distance is 0 and speed is 0m/1s.
But speed is always relative. To say that something is travelling at 3 m/s, one has to say that it is travelling at 3/s relative to something.
If one says that something is travelling at 0 m/s, then it is to say that it is stationary with respect to something else.
To say that there is a minimum speed limit would be to imply that nothing can be stationary with respect to anything else, which is to say that any two objects that are anywhere in the universe, if they are capable of being aware of the other object, must be travelling at a different speed to the other object (i.e. no two objects in the universe can have exactly the same speed – a little like the Pauli exclusion principle applied to fermions of similar spin, but applied to all things in the universe).
A minimum speed limit would not only mean that two objects cannot be stationary with respect to each other, but that there must be a minimum difference in speed between them, would would imply (although not necessitate) that velocity is quantised (i.e. that velocity can inly increase in discrete steps, rather than as a continuous value).
George
#### another_someone
• Guest
##### Re: Could there be a minimum speed of matter?
« Reply #5 on: 11/07/2006 03:20:19 »
quote:
Originally posted by Razak
I think the uncertanity principal will create problems to mesure such a speed.
It would create problems in directly measuring such a speed, but the effects of there being a minimum speed limit would probably be visible even at scales that we can observe, even though those effects might be indirect.
Suppose that the minimum speed limit is a speed that we designate V(0). Suppose, for the sake of argument, that the smallest measurable speed is actually 100 time V(0), thus it is clear that we can never actually measure the difference between V(0) and zero velocity.
Suppose we have 4 objects that we can observe, and each of them are subject to the same minimum speed limit. We will label these objects A through D, and their respective speeds as V(A) through V(D).
What are the minimum speeds that these objects can move at?
If we arbitrarily use object A as a baseline to work from, then we will arbitrarily assume that V(A) is actually zero speed relative to our reference, because it is actually designated as the reference for everything else.
Since object B cannot be travelling less than V(0) relative to the speed of V(A), then
Code: [Select]
` V(B) >= V(A) + V(0).`
Since object C cannot be travelling slower than V(0) relative to either A or B, therefore either
Code: [Select]
` V(C) >= V(B) + V(0) >= V(A) + 2 * V(0)`or
Code: [Select]
` V(C) <= V(A) - V(0)`
Similarly, if V(C) >= V(A) + 2 * V(0) then either
Code: [Select]
` V(D) >= V(C) + V(0) >= V(A) + 3 * V(0)`or
Code: [Select]
` V(D) <= V(A) - V(0)`
but if V(C) <= V(A) – V(0) then
Code: [Select]
` V(D) >= V(B) + V(0) >= V(A) + 2 * V(0)`or
Code: [Select]
` V(D) <= V(C) - V(0) <= V(A) - 2 * V(0)`
Thus, with 4 objects, each object must occupy a unique speed that is and integral multiple of V(0) (this must be so, because if any of the two objects are travelling at the same speed, then there will be another object within the system with which they are at rest with, and thus violating the notion of a minimum speed limit); and the fastest of the objects must be travelling at a minimum speed (for a 4 object system) of twice the minimum speed limit (i.e. it must be travelling at least at V(A) +/- 2 * V(0))..
If one now has a system composed of 1000 objects, then the fastest of these objects must be travelling at a speed of at least V(A) +/- 500 * V(0). Since we can measure any speed of at least 100 * V(0), so it is clear that the fastest of these objects must be travelling at a clearly measurable speed (in fact 80% of the objects must be travelling at a clearly measurable speed); and we can say that in this system, it would be impossible for these object to travel any slower than these speeds, no matter what one does to the system.
George
« Last Edit: 11/07/2006 03:29:33 by another_someone »
• Neilep Level Member
• Posts: 2175
• Scallywag
##### Re: Could there be a minimum speed of matter?
« Reply #6 on: 11/07/2006 14:23:18 »
Is it the object that has the speed or the force that is acting on it?
What you do speaks so loudly that I cannot hear what you say.
#### another_someone
• Guest
##### Re: Could there be a minimum speed of matter?
« Reply #7 on: 11/07/2006 15:27:11 »
quote:
Is it the object that has the speed or the force that is acting on it?
Force does not have speed.
If you are saying that the space in which the objects lie has speed, that is something else, but even then, I am not sure what that means. General relativity allows space to posses acceleration (which manifests itself as an apparent gravitational force), but not speed as such.
Yes, it is very likely that such a minimum speed limit, if it exists, may be associated with some curvature of space, which would imply that local space is accelerating; but the actual observed speed would have to be manifest in the objects themselves.
George
#### heikki
• Full Member
• Posts: 56
##### Re: Could there be a minimum speed of matter?
« Reply #8 on: 13/07/2006 19:50:31 »
quote:
Is it the object that has the speed or the force that is acting on it?
You make the question which is very good basic question to study all objects motion.
My self.
I'm a object and i can make that speed-force when i example running. But, but, can i make it alone? Without else? No i cant. I need ground where i can push my legs to run. If i dont have ground under my legs my mind can control me to run and try to acting me to make that speed-force but nothing dont happend.
I'm a object who has speed of force and i can acting it, but, but, not alone.
Other object, example.
Moon. Moon is a object who has speed of force and something acting and control it.
Basic question.
I have mind to control that speed force. I can acting and control it. But i need ground under my leg, otherwice i cannot do anything.
Moon have mind to control and acting that speed force or not?
Second basic question.
I am object, construction is matter.
Moon is object, construction is matter also.
If moon dont has mind to control it's speed force then something round of moon control it. Earth-ball, sun-ball, space-matter. Philosophically comes clearly next question. What different has moon than sun or earth-ball or space-matter?
Hadrian make fine question. It open many new thinking ways to my mind. Thanks.
#### another_someone
• Guest
##### Re: Could there be a minimum speed of matter?
« Reply #9 on: 14/07/2006 02:20:14 »
quote:
Originally posted by heikki
Basic question.
I have mind to control that speed force. I can acting and control it. But i need ground under my leg, otherwice i cannot do anything.
Moon have mind to control and acting that speed force or not?
Second basic question.
I am object, construction is matter.
Moon is object, construction is matter also.
If moon dont has mind to control it's speed force then something round of moon control it. Earth-ball, sun-ball, space-matter. Philosophically comes clearly next question. What different has moon than sun or earth-ball or space-matter?
What kind of control were you thinking for the moon?
The moon has a fairly constant speed (although, because of its orbit, it also has constant acceleration) – so why would it need active control of its speed?
George
#### heikki
• Full Member
• Posts: 56
##### Re: Could there be a minimum speed of matter?
« Reply #10 on: 14/07/2006 18:49:41 »
quote:
What kind of control were you thinking for the moon?
The moon has a fairly constant speed (although, because of its orbit, it also has constant acceleration) – so why would it need active control of its speed?
Moon, hmm. i dont exactly know.
Moon is matter-ball and goes forward. It dont has actually constant speed and constant acceleration if we speak long time period. Also i thing that if we measure accurace it's daily speed at one year we notice that speed vibrate little.
What control moon speed and place?
We know that sun and earth is part of that control-process. Also moon itself has weight and also therefore speed-force. Then space-matter where moon "swim" or "fly" has also one part of that control-process.
We can see that example flower has purpose, it's a life and be exist.
What is purpose of this nature planet process?
And how planets control that motion process?
I mean that if there is not any kind of join-control-process, then moon must fly away like hammer fly away from sportman hands. But, moon dont fly away, it "want" to round earth.
I think that space-matter has some kind of big part of this join-control-system.
But, why this planets-process exist? Why space-nature planets want to round and grow, bigger and bigger, grow like trees on the forest.
#### neilep
• Withdrawnmist
• Naked Science Forum GOD!
• Posts: 20592
• Thanked: 8 times
##### Re: Could there be a minimum speed of matter?
« Reply #11 on: 14/07/2006 19:35:03 »
There simply must be a minimum speed of matter !!
..Try to get wifey to pay for something and a hundred thousand frame per second camera would not detect any movement !
Conversely, when I'm paying for something, the speed at which she grabs for my wallet puts the speed of light to shame !!
Men are the same as women, just inside out !
#### nexus
• Jr. Member
• Posts: 19
##### Re: Could there be a minimum speed of matter?
« Reply #12 on: 21/07/2006 05:24:31 »
Speed has direction, a vector, if you go lower than zero you are going in reverse.
But if you simply think that there is minimum unit of time above zero I don't think there is.
Take some billiard balls(on a frictionless pool table), the cue ball hits another, say the 8 ball, straight on, all the speed gets transfered to the 8 ball.
Now have the cue ball hit 8 ball at an angle from the direction you want it to travel, only a fraction of the speed is transfered. Now increase the angle a smaller amount of energy is transferred. And so on, approaching the speed of zero but never getting there.
#### heikki
• Full Member
• Posts: 56
##### Re: Could there be a minimum speed of matter?
« Reply #13 on: 21/07/2006 07:32:31 »
quote:
Originally posted by nexus
Speed has direction, a vector, if you go lower than zero you are going in reverse.
I think that not reverse.
Time goes forward always. Speed unit is m/s.
Then all objects which have speed goes forward relation to time.
But relation to place (point a-b) of cource can go reverse direction.
Object speed m/s is always connection to time s.
#### another_someone
• Guest
##### Re: Could there be a minimum speed of matter?
« Reply #14 on: 21/07/2006 13:06:36 »
quote:
Originally posted by nexus
Speed has direction, a vector,
True.
quote:
if you go lower than zero you are going in reverse.
If speed were a scalar, this would be so; but you have yourself said it is a vector – in other words, speed has direction and magnitude, but magnitude cannot be negative..
But this is in some ways semantics – whether or not you regard the speed of an object travelling 180 degrees to the current vector as being arithmetically less than the speed of the reference object, or just a positive speed rotated through 180 degrees, what the real intent of the question was to ask whether there can be a minimum magnitude of speed, whatever its direction or sign.
quote:
But if you simply think that there is minimum unit of time above zero I don't think there is.
In some ways this is a related but not identical question.
There is I believe what is known as the plank unit of time, which is the smallest measurable unit of time, although whether this is the smallest physical unit of time is another question (but this in some ways also relates to all quantum measure, and the general evidence at present seems to argue that the uncertainties involved in quantum measurement are not merely limitations of measurement itself, but limitations in the underlying physical processes).
quote:
Take some billiard balls(on a frictionless pool table), the cue ball hits another, say the 8 ball, straight on, all the speed gets transfered to the 8 ball.
Now have the cue ball hit 8 ball at an angle from the direction you want it to travel, only a fraction of the speed is transfered. Now increase the angle a smaller amount of energy is transferred. And so on, approaching the speed of zero but never getting there.
You cannot infinitely divide up the angle of impact between two billiard balls because you cannot infinitely divide up the balls themselves – at the minimum level, you start getting into the interaction of two atoms on the surface of the billiard balls just grazing each other, and beyond that, you even get into quantum uncertainty (how accurately can you measure the angle of one quantum particle hitting another quantum particle?).
George
« Last Edit: 21/07/2006 13:10:36 by another_someone »
#### nexus
• Jr. Member
• Posts: 19
##### Re: Could there be a minimum speed of matter?
« Reply #15 on: 21/07/2006 13:40:28 »
yes i agree largely with the responeses to my response, in my haste to type it out I left out portions I was thinking (they never made it to my fingertips)
but just because something is uncertain does that mean that it can't happen?
#### another_someone
• Guest
##### Re: Could there be a minimum speed of matter?
« Reply #16 on: 21/07/2006 15:40:10 »
quote:
Originally posted by nexus
but just because something is uncertain does that mean that it can't happen?
This is true, and this is where much of the debate about quantum uncertainty lies – whether there is a precise reality that exists but is really unmeasurable, or whether reality itself is that vague.
The present consensus tends towards the latter interpretation (and some of the arguments about quantum entanglement actually are based upon the fact that there cannot be an underlying but immeasurable precision to the universe).
George
#### pete1024
• First timers
• Posts: 1
##### Could there be a minimum speed of matter?
« Reply #17 on: 05/01/2008 14:58:12 »
Ah, you guys are missing a point here. A stationary object contains atoms, these atoms are moving, but are being bounced around and held in place by the interactive forces around them.
'Speed' as we measure it is the average resultant vector of those atoms.
But in actual fact a stationary object still has 'speed' so long as the atoms are vibrating.
So, would the minimum speed of an atom be zero, if the atoms temperature reached absolute zero?
Can it reach absolute zero, or is there some retained energy as a result of the atoms' first energy level.
Could the minimum speed be 1/c?
#### Soul Surfer
• Neilep Level Member
• Posts: 3345
• keep banging the rocks together
##### Could there be a minimum speed of matter?
« Reply #18 on: 06/01/2008 10:18:21 »
An interesting topic pete and worth reviving, but I dont think you are right. Whilst I do not think there is a minimum speed because it should be possible for two pbjects to be stationary with respect to each other. There may however be a mininmum change of speed so two objects that are moving must move at specific velocities or zero by quantum theory but these changes will be extremely small and not observable.
It is interesting to think that in a universe that had a beginning (big bang) there is a lower limit on the frequency of electromagnetic and gravitiational waves to those with a period less than the time since the big bang. The electromagnetic limit is probably not significant (now ) but the gravitiational waves might be and could possibly have an effect like dark matter dark energy or MOND. I always wonder if that possibility has been factored into the equations by the cosmologists.
Consider the Casimir effect. This produces an attractive force between two plates by the exclusion of some of the states of the quantum mechanical vacuum. Could the whole universe exhibit such an effect particularly when it was very small?
#### The Naked Scientists Forum
##### Could there be a minimum speed of matter?
« Reply #18 on: 06/01/2008 10:18:21 » | 5,082 | 21,228 | {"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-2016-44 | longest | en | 0.928422 |
https://it.mathworks.com/matlabcentral/cody/problems/579-spiral-in/solutions/1507305 | 1,586,479,105,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371880945.85/warc/CC-MAIN-20200409220932-20200410011432-00347.warc.gz | 509,561,910 | 15,785 | Cody
# Problem 579. Spiral In
Solution 1507305
Submitted on 28 Apr 2018 by bainhome
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
m = 3; n = 5; s_correct = [1 12 11 10 9; 2 13 14 15 8; 3 4 5 6 7]; assert(isequal(spiralIn(m,n),s_correct))
2 Pass
m = 5; n = 3; s_correct = [1 12 11; 2 13 10; 3 14 9; 4 15 8; 5 6 7]; assert(isequal(spiralIn(m,n),s_correct))
3 Pass
m = 1; n = 1; s_correct = 1; assert(isequal(spiralIn(m,n),s_correct))
4 Pass
m = 5; n = 0; s_correct = zeros(5,0); assert(isequal(spiralIn(m,n),s_correct))
5 Pass
m = 2; n = 2; s_correct = [1 4; 2 3]; assert(isequal(spiralIn(m,n),s_correct)) | 307 | 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} | 2.75 | 3 | CC-MAIN-2020-16 | latest | en | 0.521417 |
https://www.doubtnut.com/qa-hindi/gunnn-cos-12-cot-14-cot-46-cos-48-cot-48-cos-72-cot-74-sin-162-kaa-maan-m-n-hai-jhaan-m-evn-n-shabha-642920362 | 1,631,952,141,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056348.59/warc/CC-MAIN-20210918062845-20210918092845-00535.warc.gz | 772,306,838 | 66,310 | Home
>
Hindi
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कक्षा 12
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Maths
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Chapter
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Compound Angle (Trigonometry Phase - I)
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गुणन cos 12^(@) cot 14^(@) cot...
# गुणन cos 12^(@) cot 14^(@) cot 46^(@) cos 48^(@) cot 48^(@) cos 72^(@) cot 74^(@) sin 162^(@) का मान (m)/(n) है, जहाँ m एवं n सहअभाज्य है, तो (n - 12m) का मान ज्ञात कीजिए |
उत्तर
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Updated On: 31-3-2021
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1:02 | 476 | 1,062 | {"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} | 3.0625 | 3 | CC-MAIN-2021-39 | latest | en | 0.23471 |
https://downhouse.software/tag/ruby/ | 1,686,034,756,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224652235.2/warc/CC-MAIN-20230606045924-20230606075924-00792.warc.gz | 251,133,742 | 26,803 | # Tag Archives: ruby
Coding Club today.
There’s a chance I may be getting more takers for my ‘Codecademy’ – based coding club at FSCC soon. Several students have shown interest and I look forward to opening up the class towards becoming a more open space with students (including myself) pursuing a number of projects simultaneously.
If anyone (local, at least) is interested in joining our group, please feel free. We take all comers and look forward to building our numbers with anyone interested in learning, teaching or challenging themselves.
If you’re not local, I’d still be interested in hearing from you if you’d like to start an online learning community tied to codecademy, code school, or any other online resource.
Posted by on August 30, 2013 in Uncategorized
## New Coding Challenge – The Quincunx!
The Quincunx – A triangular pegboard that will create a nice normal distribution as balls are dropped from the top and bounce down randomly over the triangular array of pegs.
Society’s greatest achievement, The Price Is Right, demonstrates the use of a plinko board in this video with the most excited player ever.
The coding challenge is to design a quincunx that demonstrates each of the following four points… No animation is required, simply (1)show the board as an array of X’s with a (2) user-determined number of rows (1-20) and the (3)resulting bins filling with integers as a (4)user-determined number of balls (1-100,000) is dropped. This time, I’m awarding prizes to the cleanest, most clearly documented entries in each language represented on Codecademy (Ruby, JS, Python).
As always, the prizes are bragging rights, presentation of your code on my blog with full attribution to you and a promo copy of any of my eBooks on iPad for you to share with the youngster in your life (or keep yourself). Each of my books presents educational material in the form of a story (Heracles and the Gas Laws, Sisyphus and the Laws of Motion, Zombies and Fractions).
Happy Coding!
Posted by on July 10, 2013 in Uncategorized
## RubyMonk
I mention codecademy all the time here and rant about how great a way it is to learn a variety of languages and markups online, on your own time and free. I’ve been following both the web programming track and the Ruby tracks aggressively lately (I’m on a 20 day streak presently). However, I have to also mention another free site that does much the same thing. RubyMonk offers free online courses in Ruby (and Python, under PythonMonk). The Monk websites are clean, well structured and provide an element of atmosphere as well.
Unfortunately, RubyMonk does not provide a forum where I can pitch my project challenges – er, I mean Koans. But if you are learning Ruby or Python with the Monk, please feel free to come here from time to time to see if there are any simple programming challenges open.
Posted by on June 28, 2013 in Uncategorized
## Winding down Coding Challenge I
I’ve received several entries answering my coding challenge to demonstrate / test Goldbach’s Conjecture that all even numbers > 4 are the sum of two primes. So far python has been the language of choice for entries.
I will be closing down this challenge as of June 30 at 11:59pm.
Once I take a look at the entries, I’ll award the prize, a copy of my iBook, In Parts to the winner and post the code here with a walkthrough to show how the problem was tackled as well as any interesting comparisons between entrants.
Posted by on June 26, 2013 in Uncategorized
## Coding Challenge II: Make Mine a MASTERmind
Don’t worry, Coding Challenge I is still open, but someone was writing about games to develop in the codecademy discussion groups.
A while ago, when I was first following the JavaScript pathway, there, I decided to write a MasterMind program. Many of the versions I saw prohibited players from using the same number/color more than once, but I felt that was a cop-out. My solution works, but as usual for me, had some tortuous logic.
So, here’s the challenge:
In any language, write a MasterMind game where the computer chooses the numbers and the user deduces them.
1. use numbers (four of them,#s1-4, randomly chosen by the computer), rather than colors
2. that allows multiple uses of the same numbers, i.e. ‘1122’
3. provides appropriate feedback to the user to help them close in on the correct sequence.
4. keeps track of the number of turns taken
5. (optional) can also be played with 2 users -or- user sets the code and computer guesses
6. (optional) allow user to select # of positions and range of numbers used.
simple code trumps tangly code. I prefer languages I can read (C++, javascript, python) but all are welcome.
Posted by on June 22, 2013 in Uncategorized
## Coding Challenge
Recently, there have been a couple new revelations about number theory published in Science Within the article was a pair of theories about prime numbers that I had never heard before, one of which was:
Goldbach’s conjecture, [which] makes two assertions: that every even number greater than 2 is the sum of two primes, and that every odd number greater than 5 is the sum of three primes.
I thought it would be fun to start with the first part of this problem and write a program to accept user input in the form of an even integer > 2 and then look for the two primes whose sum is equal to the user provided:
prime1 + prime2 = user input
where prime1 and prime2 may be any prime number (even the same number twice)
I could easily see this escaping the processing power of my machine if the numbers get high, but I think it shouldn’t be too hard to at least write a code that could look for them and demonstrate whether this worked with known input.
Are you up for a quick challenge?
Learn Fractions with Zombies
If so, submit your documented answer here as a comment. Feel free to use any language you would like (I just did it in C++, but I’m eager to see better answers than my own). My favorite submissions will win a free copy of my iBook, In Parts, Tales of Fractional Zombies, which you can enjoy yourself or regift to a youngster in your life who wants a fun way to learn the concept of fractions.
You can use these links as resources to help check your work:
If you are new to coding and are looking for a coding environment to work in, check out this posting for help setting up a C++ coding environment using Xcode (on your mac)
Posted by on June 19, 2013 in Codecademy, Coding
## Code School Hall Pass
Code school is currently offering 48 hours of free access to their site. Code School has video classes covering Ruby, JavaScript, iOS and HTML. I just started mine tonight and I’m hoping to get a lot out of it and see if its something I can get more value from. If you’re interested in coding, I highly recommend looking into this site and also Codecademy
1 Comment
Posted by on March 18, 2013 in Uncategorized
## Two minutes into my latest coursera lecture….
Two minutes into my latest coursera lecture (introduction to interactive programming in python) the instructor indicated his frustration in javascript programming saying that it’s a terrible language. This may be the case… so far I don’t have a lot to compare against, but I have been enjoying learning JS in codecademy and I’m dying to know why he thinks so.
If you have experience programming in javascript and python (or other languages…ruby?) let me know if you agree with the above statement and what makes you think so. As a new programmer I am interested in learning as much as possible – if I can understand what faults people see in these languages I think that would be very instructive.
Thanks | 1,722 | 7,667 | {"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-2023-23 | latest | en | 0.928307 |
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Scene not validated
Fruit medley
Input text:
A small banana is inside a bisque bowl. A pear is zero inches to the left of the banana. A small apple is zero inches to the right of the banana. The apple is matte. A strawberry and a small cherry are on the banana. A small grape is zero inches in front of the apple. A tiny pineapple is in the small mango. A tiny plum "plum" sphere is in front of the mango. A small orange "orange" sphere is next to it. The ground is bisque and matte.
Views:
717
3D models:
(legacy) The bisque color is lovely!*br**br*I was playing around with the positioning and stacking of objects and made a few observations.*br**br*Proximity of alignment of objects seems to depend a bit on the object type that's the 'ground'. (It tried 'zero inches' but didn't seem to have an effect on aligning things right next to a 'pear' for example. It would be neat to be able to specify 'X is exactly aligned to Y' or something like that to get this effect.)*br**br*Plurals (for example two cherries) didn't apply when starting from another object and then positioning them relative to this other object. (I also noticed that at a certain point I couldn't add more singular fruits, but perhaps there's a natural limit to the number of objects which fit into the scene or into another object, here a 'bowl')?*br**br*I also noticed that square defaults doesn't apply to things which are also colors (plum, orange).*br**br*
(legacy) If you just say "X is in front of Y" then it will be at 0 inches in front of. The reason it might not have seemed that way is that it uses the bounding box of the object to place objects next to each other. So if the stem of the banana is curved away from the body, that will determine the bounds of the box.*br**br**br*The way to exactly position the objects in cases like this is to say things like "the left side of the banana is .6 inches to the left of the right side of the pear."
(legacy) Ok, that explains it. Is "0 inches" okay? Is "zero inches" (instead of the number) allowed too?
(legacy) "0 inches" and "zero inches" should both work and do the same thing. But they won't really change anything since the default is to place objects directly next to each other anyway.
(legacy) I don't think "-3 inches" works though, right? Is that a problem at the language processing phase or at your end? If the former, I will fix it.*br*
(legacy) That would be great to fix. Looks like it's an nlp problem.*br**br*WE* (trace nlp::paint-world)*br*(NLP::PAINT-WORLD)*br*WE* (pw "the cube is -2 inches to the left of the sphere.")*br* 0: (NLP::PAINT-WORLD ("the cube is -2 inches to the left of the sphere"))*br* 0: NLP::PAINT-WORLD returned [ENTITY: "Main World" #:|world12633|*br**br*WORLD: "Main World" #:|world12633|*br* Entity: "cube" #:|ent12733|*br* New Property: :ATTRIBUTE "inch"*br* Property-of-property: :ATTRIBUTE "-2"*br* Property-of-property: :ATTRIBUTE "to the left of"*br* *br*[ENTITY: "Main World" #:|world12633|*br*WE* (pw "the cube is 2 inches to the left of the sphere.")*br* 0: (NLP::PAINT-WORLD ("the cube is 2 inches to the left of the sphere"))*br* 0: NLP::PAINT-WORLD returned [ENTITY: "Main World" #:|world12741|*br**br*WORLD: "Main World" #:|world12741|*br* Entity: "cube" #:|ent12844|*br* Entity: "sphere" #:|ent12845|*br* Relation: "to the left of" "STATIVE"*br* Property: :MEASURE (2 :INCHES)*br* FIGURE: "cube" #:|ent12844|*br* GROUND: "sphere" #:|ent12845|*br* *br*[ENTITY: "Main World" #:|world12741|*br*WE*
(legacy) Could you run the "-2" example through "tag-sentence" and show me the result.*br*
(legacy) Could you run the "-2" example through "tag-sentence" and show me the result.*br*
(legacy) Anticipating that the problem is with the tagger, change the definition of DEFAULT-TAGS in en-tagger.lisp to:*br**br*(defun DEFAULT-TAGS (word)*br* "Set default tags for some classes of tokens"*br* (cond*br* ;; 10% and things like that which should not be*br* ;; classed as CD*br* ((and (digit-char-p (aref word 0))*br* (eq (aref word (1- (length word))) #\%))*br* `(("RB" 0)))*br* ((or (digit-char-p (aref word 0))*br* (and (* (length word) 1)*br* (eq (aref word 0) #\-)*br* (digit-char-p (aref word 1)))*br* (and*br* (eq (aref word 0) #\.)*br* (digit-char-p (aref word 1))))*br* `(("CD" 0)))*br* ((isquote (aref word 0))*br* `(("NN" 0.693) ("JJ" 0.693)))*br* (t*br* `(("NN" 0.240687) ("VB" 1.54219)))))*br**br*
(legacy) Let's not reveal any proprietary software code in Forum exchanges.
(legacy) I don't think this is too much of a revelation. Good luck to anyone who wants to recreate the system from this fragment. And shame on them for stealing the idea given that this particular fragment is a grotesque hack.*br*
(legacy) right...fragments like that will be of no use to anyone other than us.
(legacy) Let me know if that fixes the problem. If it does not I'll have to poke around further.*br*
(legacy) It works for "-1 feet" but not "-3 feet"*br**br*WE* (pw "the sphere is -3 feet to the left of the cube.")*br**br*WORLD: "Main World" #:|world10723|*br* Collection: "foot" #:|coll10822|*br* *Members:*br* **br* Entity: "cube" #:|ent10824|*br* Relation: "to the left of" "STATIVE"*br* FIGURE: "foot" #:|coll10822|*br* GROUND: "cube" #:|ent10824|*br* *br**br*WE* (pw "the sphere is -1 feet to the left of the cube.")*br**br*WORLD: "Main World" #:|world10831|*br* Entity: "sphere" #:|ent10934|*br* Entity: "cube" #:|ent10935|*br* Relation: "to the left of" "STATIVE"*br* Property: :MEASURE (-1 :FEET)*br* FIGURE: "sphere" #:|ent10934|*br* GROUND: "cube" #:|ent10935|*br*
(legacy) hmm. that makes little sense. okay well i'll have to get set up and try this out. impossible to debug from 30,000 feet.*br*
(legacy) Looking back at your -3 output of paintworld, it isn't clear that the NL component is even getting the "-3". Notice that no number shows up in the world description. Are you sure some preprocessing phase isn't filtering out the -3?*br*
(legacy) I'm definitely passing it through.*br**br*WE* (trace nlp::paint-world)*br**br*(NLP::PAINT-WORLD)*br*WE* (pw "the sphere is -3 feet to the left of the cube.")*br* 0: (NLP::PAINT-WORLD ("the sphere is -3 feet to the left of the cube"))*br* 0: NLP::PAINT-WORLD returned [ENTITY: "Main World" #:|world10567|*br**br*WORLD: "Main World" #:|world10567|*br* Collection: "foot" #:|coll10666|*br* *Members:*br* **br* Entity: "cube" #:|ent10668|*br* Relation: "to the left of" "STATIVE"*br* FIGURE: "foot" #:|coll10666|*br* GROUND: "cube" #:|ent10668|*br* *br*[ENTITY: "Main World" #:|world10567|*br*WE*
(legacy) That makes no sense. I can't imagine what I am doing that would only affect the literal string "-3"
(legacy) Never mind. I found it.*br**br*Comment out the following line in en-tags.lisp:*br**br* ("-3" (("SYM" 0.0)))*br*
(legacy) ok, i'll do that.
Share to | 2,169 | 7,029 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2020-34 | latest | en | 0.951108 |
http://achievethecore.org/peersandpedagogy/tackling-slope/ | 1,725,991,696,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651303.70/warc/CC-MAIN-20240910161250-20240910191250-00002.warc.gz | 845,997 | 23,932 | # Tackling Slope
Digging into 8th-grade standards 8.EE.B.5 and 8.EE.B.6
Welcome back to the Unfinished Learning in Middle School Math series, in which math educators Chrissy Allison and Becca Varon illustrate how to make some of the trickiest standards in grades 6-8 accessible for all students. Over the course of six blog posts, we’ve provided concrete examples of how math educators can address unfinished learning within the context of grade-level lessons, which in the long term will help prevent an entrenched pattern of over-remediation and below-grade-level teaching. You can read our introductory post here. In this final post, we will explore ways to “bridge the gap” with the 8th-grade standards 8.EE.B.5 and 8.EE.B.6.
Let’s take a look at these two impactful standards:
For our final post in this series, we knew we had to tackle slope. This topic pulls together countless threads of prior learning from grades K-8. The intention is that the study of linear relationships in 8th grade should weave those threads into a parachute that will allow them to jump fearlessly into high school, where they’ll apply that knowledge to other types of functions. However, in practice, this unit is often a major source of struggle. Some of those threads of prior knowledge are inevitably missing or frayed, and it can be tough to know how to meet students’ needs in the moment. As a result, data shows that students often depart 8th grade missing this key foundation for Algebra I.
Let’s take a moment to study and understand two of the standards that are essential to building students’ knowledge of how to use slope:
In order to start to see all of those threads of prior knowledge — and to wrap our heads around the “slope triangles” in 8.EE.B.6 — let’s solve this task from Illustrative Mathematics.
As you look to diagnose students’ prerequisite understanding, here are a couple of things we noticed as we solved this task:
• There’s a lot of geometry involved. And not just the 8th-grade standards addressing similarity, translations, and dilations (8.G.A), which are typically taught soon before this learning and have several of their own prerequisites. Students will also need some comfort with finding side lengths of polygons graphed on coordinate planes (6.G.A.3), especially in future tasks where a visual isn’t provided for them. Unfortunately, we know from experience that geometry standards are under-taught before 8th grade. Since they aren’t in the K-7 Major Work of the Grade, they’re often the first to go when teachers make hard decisions about pacing.
• Proportional reasoning is at the heart of this task. Students need to see that the line rises 2 units for every horizontal increase of 3 units, no matter how large the triangle gets. This builds on an array of skills that stretch back to equivalent fractions and multiplicative reasoning in elementary school.
In a prior blog post for ANet, we wrote about some general teaching methods that will boost student understanding of linear relationships. All of those suggestions, like using visual patterns and encouraging students to discuss the connections between representations, still apply.
Here are a couple of additional ideas for ways to take action in supporting students with specific threads of unfinished learning:
1. Use slope triangle manipulatives. There are some excellent tools out there, like this Desmos activity, that allow students to experiment with translating and dilating slope triangles. Supplement these grade-level activities with questions that probe students to think deeply about concepts that may still be unclear to them. For example, if your students have unfinished learning about polygons graphed in the coordinate plane, ask them to predict the coordinates for triangles with different side lengths.
2. Bring in real-world contexts. Tangible scenarios make things click sometimes. For example, check out this lesson in the IM curriculum that investigates proportionality between the shadows cast by different objects. We can totally imagine students having an aha moment about how shadows “match” their objects in predictable (proportional) ways. Or, your students might enjoy experimenting with ways to safely land a plane using equations. Instead of waiting until the culmination of your unit to decide whether students are “ready” for real-world activities, try using them at the beginning to generate curiosity and stories that you can refer back to.
We hope you found this blog series helpful, and we’re cheering you on as you tackle the hard and critical work of supporting your students. We’d love to keep the conversation going in the comments by hearing from you! Please share:
• What approaches have you used to “bridge the gap” to support students with slope?
• How have you helped students find success with linear relationships?
Finally, don’t forget to check out other posts in the Unfinished Learning in Middle School Math series for additional strategies, examples, and ideas.
## One thought on “Tackling Slope”
1. Barbie Herring says:
Thank you for helping me understand slope!!
About the Author: Chrissy Allison is the founder and CEO of Mindful Math Coach whose core purpose is to support secondary math educators in providing positive and equitable learning experience for students of color. A former middle school math teacher, Chrissy became an instructional leader where she led the school's math team to increase summative assessment scores by 30% within three years. Her 5+ years experience serving as Director of Math Professional Learning & Content Design at educational nonprofits gives Chrissy unique insight and expertise about what it takes to shift teacher practice and move the needle with student learning. As host of The Mindful Math Podcast, Chrissy interviews experienced educators and shares her own advice and lessons learned to help teachers reach every learner while finding balance in their own lives. Chrissy lives in Chicago with her husband, Dan, and two children, Liviana and Otto. Learn more about Chrissy, access the podcast, and download free resources at www.mindfulmathcoach.com.
About the Author: Becca Varon is the Director of Math Content Learning at ANet. Prior to joining ANet, Becca taught 6th grade math and science in Glendale, Arizona and 5th grade math in Boston, Massachusetts. Becca now lives in Oakland, CA, where she continues to tutor local students in math. | 1,300 | 6,453 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2024-38 | latest | en | 0.938651 |
https://www.mathplanet.com/education/algebra-2/polynomials-and-radical-expressions/polynomials | 1,723,524,996,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641063659.79/warc/CC-MAIN-20240813043946-20240813073946-00894.warc.gz | 673,048,289 | 8,303 | # Polynomials
When polynomials are multiplied, each term of one expression is multiplied by every term of the other expression. Then the terms in the product are added.
Choose one polynomial, preferably the longest, and then:
• multiply it by the first term of the other polynomial
• then multiply it by the second term of the other polynomial then multiply it by the third term of the other polynomial (if any) etc ...
• lastly, add up the products.
Example
Multiply the following polynomials:
$(a+b+c)\cdot (a+b)$
We first multiply our first polynomial by the first term of the second polynomial then we multiply it by the second term of the other polynomial:
$(a+b+c)\cdot (a+b)=$
$=a\cdot a+b\cdot a+c\cdot a+a\cdot b+b\cdot b+c\cdot b$
Lastly we simplify and add up the products
$a\cdot a+b\cdot a+c\cdot a+a\cdot b+b\cdot b+c\cdot b=$ $a^{2}+2ab+ac+b^{2}+cb$
When we are dividing a polynomial by something more complicated than just a simple monomial, then we will need to use a different method for the simplification. This method is called "long (polynomial) division", and it works just like the long (numerical) division you did back in elementary school, except that now you're dividing with variables.
We will show the procedure in an example.
Example
Divide x2-2x-8 by x+2:
First, we set up the division, writing x+2 to the left and x2-2x-8 to the right. Our result comes out on top of the polynomials. In order to see exactly how this is done watch our video lesson below.
$x+2{\overline{\smash{\big)}\,x^2-2x-8\phantom{)}}}$
## Video lesson
Divide x2-2x-8 by x+2 (same as the example above). | 434 | 1,625 | {"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} | 4.84375 | 5 | CC-MAIN-2024-33 | latest | en | 0.869286 |
http://hitchhikersgui.de/Cosmological_horizon | 1,542,600,323,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039745015.71/warc/CC-MAIN-20181119023120-20181119045120-00180.warc.gz | 157,737,189 | 10,227 | Cosmological horizon
A cosmological horizon is a measure of the distance from which one could possibly retrieve information.[1] This observable constraint is due to various properties of general relativity, the expanding universe, and the physics of Big Bang cosmology. Cosmological horizons set the size and scale of the observable universe. This article explains a number of these horizons.
Particle horizon
The particle horizon (also called the cosmological horizon, the comoving horizon, or the cosmic light horizon) is the maximum distance from which particles could have traveled to the observer in the age of the universe. It represents the boundary between the observable and the unobservable regions of the universe, so its distance at the present epoch defines the size of the observable universe. Due to the expansion of the universe it is not simply the age of the universe times the speed of light, as in the Hubble horizon, but rather the speed of light multiplied by the conformal time. The existence, properties, and significance of a cosmological horizon depend on the particular cosmological model.
In terms of comoving distance, the particle horizon is equal to the conformal time that has passed since the Big Bang, times the speed of light. In general, the conformal time at a certain time is given in terms of the scale factor ${\displaystyle a}$ by,
${\displaystyle \eta (t)=\int _{0}^{t}{\frac {dt'}{a(t')}}}$
or
${\displaystyle \eta (a)=\int _{0}^{a}{\frac {1}{a'H(a')}}{\frac {da'}{a'}}}$.
The particle horizon is the boundary between two regions at a point at a given time: one region defined by events that have already been observed by an observer, and the other by events which cannot be observed at that time. It represents the furthest distance from which we can retrieve information from the past, and so defines the observable universe.[1]
Hubble horizon
Hubble radius, Hubble sphere, Hubble volume, or Hubble horizon is a conceptual horizon defining the boundary between particles that are moving slower and faster than the speed of light relative to an observer at one given time. Note that this does not mean the particle is unobservable, the light from the past is reaching and will continue to reach the observer for a while. Also, more importantly, in the current expansion model e.g., light emitted from the Hubble radius will reach us in a finite amount of time. It is a common misconception that light from the Hubble radius can never reach us. It is true that particles on the Hubble radius recede from us with the speed of light, but the Hubble radius gets larger over time (because the Hubble parameter H gets smaller over time), so light emitted towards us from a particle on the Hubble radius will be inside the Hubble radius some time later. Only light emitted from the cosmic event horizon or further will never reach us in a finite amount of time.
The Hubble velocity of an object is given by Hubble's law,
${\displaystyle v=xH}$.
Replacing ${\displaystyle v}$ with speed of light ${\displaystyle c}$ and solving for proper distance ${\textstyle x}$ we obtain the radius of Hubble sphere as
${\displaystyle r_{HS}(t)={\frac {c}{H(t)}}}$.
In an ever-accelerating universe, if two particles are separated by a distance greater than the Hubble radius, they cannot talk to each other from now on (as they are now, not as they have been in the past), However, if they are outside of each other's particle horizon, they could have never communicated.[2] Depending on the form of expansion of the universe, they may be able to exchange information in the future. Today,
${\displaystyle r_{HS}(t_{0})={\frac {c}{H_{0}}}}$,
yielding a Hubble horizon of some 4.1 Gpc. This horizon is not really a physical size, but it is often used as useful length scale as most physical sizes in cosmology can be written in terms of those factors.
One can also define comoving Hubble horizon by simply dividing Hubble radius by the scale factor
${\displaystyle r_{HS,comoving}(t)={\frac {c}{a(t)H(t)}}}$.
Event horizon
The particle horizon differs from the cosmic event horizon, in that the particle horizon represents the largest comoving distance from which light could have reached the observer by a specific time, while the event horizon is the largest comoving distance from which light emitted now can ever reach the observer in the future.[3] The current distance to our cosmic event horizon is about 5 Gpc (16 billion light years), well within our observable range given by the particle horizon.[4]
In general, the proper distance to the event horizon at time ${\displaystyle t}$ is given by[5]
${\displaystyle d_{e}(t)=a(t)\int _{t}^{t_{max}}{\frac {cdt'}{a(t')}}}$
where ${\displaystyle t_{max}}$ is the time-coordinate of the end of the universe, which would be infinite in the case of a universe that expands forever.
For our case, assuming that dark energy is due to a cosmological constant, ${\displaystyle d_{e}(t_{0})<\infty }$.
Future horizon
In an accelerating universe, there are events which will be unobservable as ${\displaystyle t\rightarrow \infty }$ as signals from future events become redshifted to arbitrarily long wavelengths in the exponentially expanding de Sitter space. This sets a limit on the farthest distance that we can possibly see as measured in units of proper distance today. Or, more precisely, there are events that are spatially separated for a certain frame of reference happening simultaneously with the event occurring right now for which no signal will ever reach us, even though we can observe events that occurred at the same location in space that happened in the distant past. While we will continue to receive signals from this location in space, even if we wait an infinite amount of time, a signal that left from that location today will never reach us. Additionally, the signals coming from that location will have less and less energy and be less and less frequent until the location, for all practical purposes, becomes unobservable. In a universe that is dominated by dark energy which is undergoing an exponential expansion of the scale factor, all objects that are gravitationally unbound with respect to the Milky Way will become unobservable, in a futuristic version of Kapteyn's universe.[6]
Practical horizons
While not technically "horizons" in the sense of an impossibility for observations due to relativity or cosmological solutions, there are practical horizons which include the optical horizon, set at the surface of last scattering. This is the farthest distance that any photon can freely stream. Similarly, there is a "neutrino horizon" set for the farthest distance a neutrino can freely stream and a gravitational wave horizon at the farthest distance that gravitational waves can freely stream. The latter is predicted to be a direct probe of the end of cosmic inflation.
References
1. ^ a b Margalef-Bentabol, Berta; Margalef-Bentabol, Juan; Cepa, Jordi (8 February 2013). "Evolution of the cosmological horizons in a universe with countably infinitely many state equations". Journal of Cosmology and Astroparticle Physics. 015. 2013 (02). arXiv:1302.2186. Bibcode:2013JCAP...02..015M. doi:10.1088/1475-7516/2013/02/015.
2. ^ Dodelson, Scott (2003). Modern Cosmology. Academic Press. p. 146.
3. ^ Lars Bergström and Ariel Goobar: "Cosmology and Particle Physics", WILEY (1999), page 65. ISBN 0-471-97041-7
4. ^ Lineweaver, Charles; Tamara M. Davis (2005). "Misconceptions about the Big Bang" (PDF). Scientific American. Retrieved 2008-11-06.
5. ^ Massimo Giovannini (2008). A primer on the physics of the cosmic microwave background. World Scientific. pp. 70–. ISBN 978-981-279-142-9. Retrieved 1 May 2011.
6. ^ https://arxiv.org/abs/0704.0221 | 1,810 | 7,781 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 15, "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} | 3.15625 | 3 | CC-MAIN-2018-47 | latest | en | 0.907489 |
https://www.teacherspayteachers.com/Product/Number-Sense-and-Number-Writing-Practice-0-20-946181 | 1,495,500,927,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463607242.32/warc/CC-MAIN-20170522230356-20170523010356-00622.warc.gz | 951,522,503 | 23,967 | # Main Categories
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# Number Sense and Number Writing Practice 0-20
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Math Number Sense 0-20.Number sense develops gradually and varies as a result of exploring numbers, visualizing them in a variety of contexts, and relating to them in different ways. When children have daily opportunities to work (and play) with numbers, you will be continually amazed by the growth in their mathematical thinking, confidence, and enthusiasm about mathematics.
This packet will provide students with meaningful activities to help them understand early number concepts and develop number sense!
On the first day of school I introduce a number a day as a math warm up. I enlarge the number papers to use as a model. I also use concrete manipulatives of each example on the page. I display my so the students can check their work when they are finished.
The printing space is nice and big for children learning to print numbers!
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Included in this packet:
2 Number Sense books 0-20 (Download the preview for a closer look)
10 Frame flash cards to help students learn numbers in a flash!
10 Frame Number Mats (laminate or put in sheet protectors to use as play doh mats or as dry erase boards) or use as coloring pages and rainbow writing.
Key Words: printing practice, number sense, number sense pack, number practice, number sense 0-20
This resource is great for morning work, homework, math warm up, centers, introducing a number a day, creating number posters, introducing ten frame, or intervention blocks. (RTI)
Number Printing 0-20
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\$3.00 | 515 | 2,270 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2017-22 | latest | en | 0.878485 |
https://enigmaticcode.wordpress.com/2013/02/16/enigma-1377-susan-denham/ | 1,586,313,724,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371807538.83/warc/CC-MAIN-20200408010207-20200408040707-00373.warc.gz | 448,614,709 | 23,575 | # Enigmatic Code
Programming Enigma Puzzles
## Enigma 1377: Susan Denham
From New Scientist #2537, 4th February 2006
Regular enigmatists will have encountered many puzzles by Susan Denham but many may not know that this name is a pseudonym, Sue Denham and pseudonym being approximate homophones. So it is appropriate that I can set you the following sum, in which digits have been consistently replaced by letters, different letters representing different digits.
But there is no solution to it in base 10. You must tackle the puzzle in the base that allows just one possible value for PSEUD. Your task is to find that value of PSEUD, convert it into its equivalent in base 10 and find the base 10 equivalent of PSEUD as your answer.
According to this article in the Independent, Susan Denham is a pseudonym of Dr Victor Bryant.
[enigma1377]
### One response to “Enigma 1377: Susan Denham”
1. Jim Randell 16 February 2013 at 9:11 am
My original Python program ran in 2.1s (under PyPy), but using the [[ `SubstitutedSum()` ]] solver from the enigma.py module I can get a much neater program that runs in only 335ms.
```from itertools import count
from collections import defaultdict
from enigma import SubstitutedSum, printf
# there are 9 different letters
for b in count(9):
# make a sum in the required base
p = SubstitutedSum(['PSEUD', 'SUSAN'], 'DENHAM', base=b)
# accumulate solutions by the value of PSEUD
r = defaultdict(list)
for s in p.solve():
printf("[base={b}: {t}]", t=p.substitute(s, p.text))
r[p.substitute(s, 'PSEUD')].append(s)
# check for unique values of PSEUD
n = len(r.keys())
if n != 1: continue
for (k, v) in r.items():
# convert the value to decimal
d = int(k, b)
printf("base={b} PSEUD={k} => {d} decimal [{m} solutions]", m=len(v))
# and we're done
break
```
Solution: PSEUD is 156454 (in decimal).
The sum is in base 11, and PSEUD = A7601 (base 11).
This site uses Akismet to reduce spam. Learn how your comment data is processed. | 563 | 1,971 | {"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} | 3.53125 | 4 | CC-MAIN-2020-16 | latest | en | 0.742242 |
https://www.ques10.com/p/47013/prove-that-for-finding-the-natural-frequency-of--1/ | 1,723,229,430,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640768597.52/warc/CC-MAIN-20240809173246-20240809203246-00056.warc.gz | 734,940,653 | 6,652 | 0
1.9kviews
Prove that for finding the natural frequency of spring mass system, the mass of the spring can be taken in to account by adding one third of its mass of the main mass.
1 Answer
0
76views
Prove that for finding the natural frequency of spring mass system, the mass of the spring can be taken in to account by adding one third of its mass of the main mass.
OR
P.T. $w_n = \sqrt{ \frac{k}{m + \frac{ms}{3}}}$
$\frac{\delta}{y} = \frac{x}{L}$
$\delta = (\frac{y}{L}) x$
$\delta' = (\frac{y}{L}) x'$
$Mass = \rho \times (volume)$
$= \rho \times [ A \times L]$
$M = \rho \times L$
$dM = \rho dy$
KE = (KE)m + (KE)dm
$\frac{1}{2} m x^2 + \int^L_0 \frac{1}{2} dM. \delta '^2$
$= \frac{1}{2} mx^2 + \frac{1}{2} \int^L_0 \rho . dy (\frac{y}{L} x)^2$
$= \frac{1}{2} mx^2 + \frac{1}{2} .\rho. \frac{x^2}{L^2} \int^L_0 y^2.dy$
$KE = \frac{1}{2} mx^2 + \frac{1}{2} \rho . \frac{x^2}{L^2} [ \frac{y^3}{3} ] ^L_0$
$= \frac{1}{2} m x^2 + \frac{1}{2} . \rho . \frac{x^2}{L^2} [ \frac{L^3}{3} ]$
$= \frac{1}{2} m x^2 + v [ [ \frac{\rho L}{3}] x^2$
$= \frac{1}{2} m. x^2 + \frac{1}{2} [ \frac{M}{3}] x^2$
$KE = \frac{1}{2} [ m + \frac{m}{3} ] x^2$
$M_{eg} = m + \frac{M}{3}$
$PE = \frac{1}{2} k. x^2$
$K_{eq} = K$
$w_n = \sqrt{ \frac{k}{m + \frac{m}{3}}}$
Diagram
Spring 1, 2 are in parallel
$\therefore K_{eq} 1,2 = K +K = 2K$
Above spring are in series.
$\therefore$ $\frac{1}{K_{eq}} = \frac{1}{2k} + \frac{1}{k} = \frac{1+2}{2k} = \frac{3}{2k}$
$Keq_{123} = \frac{2k}{3}$
$\therefore Keq = Keq_{123} cos^2_\alpha$
$= \frac{2k}{3} cos^2_\alpha$
$\therefore$ $w_n = \sqrt{ \frac{2k cos^2 \alpha}{3m}}$
Generally, m_{eq} x + k_{eq} x = 0
$m.x + \frac{2}{3} k cos^2 \alpha = 0$ - - - - equation of motion.
$\frac{k}{k_x} = \frac{1}{2k} + \frac{1}{2k} + \frac{1}{2k} + \frac{1}{2k cos^2 45}$
$= 0.5k +0.5k + 0.5k + \frac{0.5k}{cos^2 45}$
$= 0.5k (1+1+1+ \frac{1}{cos^2 45}$
$= 0.5k (3+ \frac{1}{cos^2 45})$
page 32:
$Keq_{5,6} = k + k = 2k$
$\frac{1}{keq_{4,5,6,7}} = \frac{1}{2k} + \frac{1}{2k} + \frac{1}{2k} = \frac{3}{2k}$
$Keq_{4-7} = \frac{2k}{3}$
$Keq_8 = 2k.cos^2 45$
$\frac{1}{Keq_{4,8}} = \frac{1}{\frac{2k}{3}} + \frac{1}{2k.cos^2 45}$
$Keq_{4-8} = \frac{2k}{3} + 2k.cos^2 45$
$Keq_{1-8} = k + k.cos^230 + \frac{2k}{3} + 2k. cos^2 45$
$= K(1 + cos^2 30) + 2k (\frac{1}{3} + cos^2 45)$
= 1.75 k + k = 2.75 k
$K_{eq} = 2.15k$
$w_n = \sqrt{ \frac{k_{eq}}{m_{eq}}}$
$w_n \sqrt{ \frac{2.15k}{m}} rad/sec$
k + 0.75k = 1.75k + 0.4k = 2.15k
page 33:
$(k_1 + k_2) 0.5 + (k_1 + k_2) 0.5$
$0.5k_1 + 0.5k_2 + 0.5k_1 + 0.5k_2$
$K_{eq} = k_q + k_2$
$w_n = \sqrt{ \frac{k_1 + k_2}{m}} rad/s$
$K_{eq} = (k_2 + k_1) cos^2 45$
$= 0.5 (k_1 + k_1)$
$K_{eq} = 0.5 (k_1 + k_2) + 0.5 (k_1 + k_2)$
$= (k_1 + k_2)$
$K_{eq} = (k_1 + k_2) cos^2 45$
$= 0.5 (k_1 + k_2)$
Please log in to add an answer. | 1,474 | 2,839 | {"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": 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 | 4 | CC-MAIN-2024-33 | latest | en | 0.52675 |
https://education.ti.com/en/activity/detail?id=229011B0D0EA4A76B9A91C564B3099CC | 1,627,667,622,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153971.20/warc/CC-MAIN-20210730154005-20210730184005-00133.warc.gz | 245,805,465 | 16,534 | Education Technology
# Activities
• ##### Subject Area
• Math: Algebra I: Equations and Inequalities
9-12
60 Minutes
• ##### Device
• TI-Nspire™ CX/CX II
• TI-Nspire™ CX CAS/CX II CAS
• TI-Nspire™ Navigator™
• TI-Nspire™
• TI-Nspire™ CAS
• ##### Software
TI-Nspire™
TI-Nspire™ CAS
3.2
## The Impossible Task
#### Activity Overview
Students are given a manufacturing situation and asked to write and graph inequalities to represent it and find the solutions.
#### Key Steps
• Students are given a manufacturing situation and asked to write an inequality to represent it. Once they have written the inequality, students examine its solution set by testing values of the variable on a spreadsheet and viewing its graph.
• A second constraint and a second variable are added to the situation. Students graph the second inequality along with the first inequality. They will compare the solution set shown by the graph with the solution found by testing values.
• A third constraint is introduced and explored in a similar fashion. First the solution set of the inequality by itself is discussed, and then students are prompted to search for solutions to the system created by all three inequalities taken together. | 276 | 1,224 | {"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-2021-31 | longest | en | 0.883432 |
https://studylib.net/doc/25362657/d41d8cd9 | 1,590,759,760,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347404857.23/warc/CC-MAIN-20200529121120-20200529151120-00531.warc.gz | 544,178,064 | 14,487 | # d41d8cd9
```Question bank (Part 1)
Power System (3)
Problem 1
For the circuit shown in figure 1, find the following;
1. The bus admittance matrix assuming no mutual coupling between any
of the branches
2. Using the admittance matrix modification procedure and assuming nomutual coupling between branches, modify the admittance bus obtained
in 1 to reflect removal of two branches 1-3 and 2-5 from the circuit
Fig. 1
Problem 2
For the linear graph shown in figure 2 for the circuit shown in figure 1, find
the following disregarding all mutual coupling between branches;
1. The branch-to-node incidence matrix A for the circuit with node 0 as
reference.
2. Find the circuit admittance matrix using the network incidence matrix
Problem 3
For the circuit shown in figure 1, considering that only two branches 1-3 and
2-3 are mutually coupled with mutual impedance is π0.15 per unit, find the
following;
2. Branch-to-node incidence matrix A wit node 0 reference. Then find the
circuit admittance matrix using the network incidence matrix
Problem 4
For the circuit shown in figure 1, solve;
1. The nodal equation to find the voltages at the four buses of Prob. 2.
2. The nodal equation to find the voltages at the four buses of Prob. 3.
2
Using both gaussian elimination and Kron reduction methods.
Problem 5
- Determine the bus admittance matrix (πππ’π ) for the following power
three phase system (note that some of the values have already been
determined for you). Assume a three-phase 100 MVA per unit base.
- Assume that a 75 Mvar shunt capacitance (three phase assuming one
per unit bus voltage) is added at bus 4. Calculate the new value of Y44.
Bus input data
Problem 6
- Form the impedance matrix for the circuit shown in Fig. after removing
node 5 by converting the voltage source to a current source, Determine the
voltages with respect to reference node at each of four other nodes when
π = 1.2∠0° and the load currents are πΌπΏ1 = −π0.1, πΌπΏ2 = −π0.1, πΌπΏ3 =
−π0.2, and πΌπΏ4 = −π0.2, all in per unit.
- Then draw the Thevenin equivalent circuit at bus 4 and use it to determine
the current drawn by a capacitor of reactance 5.4 per unit connected
between bus 4 and reference.
- Calculate the voltage changes at each of the buses due to the capacitor.
3
- Calculate the total reactive power loss in the system.
Problem 7
- Modify the impedance matrix of the Prob. 1 to include a capacitor of
reactance 5.4 per unit connected from bus 4 to reference.
- Calculate the new bus voltages using the modified impedance bus.
Problem 8
For the reactance network of the Fig. find;
- The impedance matrix by direct formulation
- The voltage at each bus
- The current drawn by a capacitor having a reactance of 5.0 per unit
connected from bus 3 to neutral.
- The change in voltage at each bus when the capacitor is connected at
bus 3.
- The voltage at each bus after connecting the capacitor.
The magnitude and angle of each of the generated voltages may be assumed to
remain constant.
4
Problem 9
Find the impedance bus for the three-bus circuit of the Fig. by the impedance
matrix building algorithm.
Problem 10
Consider the simplified electric power system shown in Figure for which the
power flow solution can be obtained without resorting to iterative techniques.
a. Compute the elements of the bus admittance matrix πππ’π .
b. Calculate the phase angle πΏ2 by using the real power equation at bus 2
(voltage-controlled bus).
c. Determine |π3 | and πΏ3 by using both the real and reactive power
equations at bus 3 (load bus).
d. Find the real power generated at bus 1 (swing bus).
e. Evaluate the total real power losses in the system.
Problem 11
- Assume a 0.8 + π0.4 per unit load at bus 2 is being supplied by a
generator at bus 1 through a transmission line with series impedance of
0.05 + π0.1 per unit. Assuming bus 1 is the swing bus with a fixed per
unit voltage of 1.0∠0, use the Gauss-Seidel method to calculate the
voltage at bus 2 after three iterations.
5
- Repeat the above problem with the swing bus voltage changed to 1.0 ∠30°
per unit.
Problem 12
- For the three-bus system whose πππ’π is given below, calculate the second
iteration value of π3 using the Gauss-Seidel method. Assume bus 1 as the
slack (with π1 = 1.0∠0), and buses 2 and 3 are load buses with a per unit
load of π2 = 1 + π0.5 and π3 = 1.5 + π0.75. Use voltage guesses of
1.0∠0 at both buses 2 and 3. The bus admittance matrix for a three-bus
system is
- Repeat the first problem except assume the bus 1 (slack bus) voltage of
π1 = 1.05∠0.
Problem 13
The bus admittance matrix for the power system shown in Figure is given by
With the complex powers on load buses 2, 3, and 4 as shown in Figure,
determine the value for π2 that is produced by the first and second iterations of
the Gauss–Seidel procedure. Choose the initial guess π2 (0) = π3 (0) =
π4 (0) = 1.0 ∠0 per unit.
6
Problem 14
The bus admittance matrix of a three-bus power system is given by
with π1 = 1.0 ∠0 per unit; π2 = 1.0 per unit; π2 = 60 MW; π3 = −80 MW;
π3 = −60 MVAR (lagging) as a part of the power-flow solution of the
system. Find π2 and π3 within a tolerance of 0.01 per unit. Start with πΏ2 = 0,
π3 = 1.0 per unit, and πΏ3 = 0. By using
- Gauss-Seidel iteration method
- Newton-Raphson method use a maximum power flow mismatch of 0.1
MVA.
- Fast decoupled method
Problem 15
A generator bus (with a 1.0 per unit voltage) supplies a 150 MW, 50 Mvar
load through a lossless transmission line with per unit (100 MVA base)
impedance of π0.1 and no line charging. Starting with an initial voltage guess
7
of 1.0 ∠0, iterate until converged using the Newton–Raphson power flow
method and fast decoupled method. For convergence criteria use a maximum
power flow mismatch of 0.1 MVA.
Problem 16
For a three bus power system assume bus 1 is the swing with a per unit
voltage of 1.0 ∠0, bus 2 is a PQ bus with a per unit load of 2.0 + π0.5, and
bus 3 is a PV bus with 1.0 per unit generation and a 1.0 voltage setpoint. The
per unit line impedances are j0.1 between buses 1 and 2, π0.4 between buses 1
and 3, and π0.2 between buses 2 and 3. Using a flat start, use the Newton–
Raphson approach and fast decoupled method to determine the first iteration
phasor voltages at buses 2 and 3.
Problem 17
Figure shows the one-line diagram of a simple power system. Generators are
connected at buses 1 and 4 while loads are indicated at all four buses. Base
values for the transmission system, are 100 MVA, 230 kV. The π values or
load are calculated from the corresponding π values assuming a power factor
or 0.85. The net scheduled values, ππ and ππ are negative at the load buses 2
and 3. Generated ππΊπ is not specified where voltage magnitude is constant. In
the voltage column the values for the load buses are flat start estimates. The
slack bus voltage magnitude |π1 | and angle πΏ1 , and magnitude |π4 | at bus 4,
are to be kept constant at the values listed.
1. Calculate the value of π2 for the first iteration using Gauss-Seidel
method.
2. Calculate the value of π2 for the first iteration using Gauss-Seidel
method with considering acceleration factor equal to 1.6.
3. Calculate the voltage at bus 4 with the originally estimated voltages at
buses 2 and 3 replaced by the accelerated values indicated in (2).
4. Calculate the active and reactive power at the slack bus.
8
Problem 18
The small power system of Ex. 17, power flow study of the system is to be
made by the Newton-Raphson method using the polar form of the equations
for P and Q.
1. Determine the number of rows and columns in the jacobian.
(0)
2. Calculate the initial mismatch βπ3
jacobian elements.
9
and the initial values of the
3. Calculate the state variables required in the power flow solution using
Newton Raphson method.
4. Calculate the active and reactive power at the slack bus.
5. Calculate the state variables required in the power flow solution using
Fast decoupled method.
Problem 19
Two transformers are connected in parallel to supply an impedance to neutral
per phase of 0.8 + π 0.6 per unit at a voltage of π2 = 1 .0∠0° per unit.
Transformer Ta has a voltage ratio equal to the ratio of the base voltages on
the two sides of the transformer. This transformer has an impedance of π0.1
per unit on the appropriate base. The second transformer Tb also has an
impedance of π0.1 per unit on the same base but has a step-up toward the load
of 1.05 times that of Ta (secondary windings on 1.05 tap). Figure 4.10 shows
the equivalent circuit with transformer Tb represented by its impedance and
the insertion of a voltage βπ.
1. Find the complex power transmitted to the load through each
transformer using πππ’π model for each of two parallel transformers.
2. Repeat (1) except that Tb includes both a transformer having the same
turns ratio as Ta and a regulating transformer with a phase shift of 3° (t
= 1 .0∠3° ). The impedance of the two components of Tb is j0.1 per unit
on the base of Ta.
Problem 20
For each bus k, determine which of the variables ππ , πΏπ , ππ , and ππ are input
data and which are unknowns.
10
Table. Bus input data
Table. Line input data
Table. Transformer input data
11
Table. Input data and unknowns
1. Compute the elements of the πππ’π .
2. Determine the DC power-flow solution for the five-bus system.
12
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https://www.resurrectionofgavinstonemovie.com/how-can-you-identify-an-unknown-weak-acid/ | 1,718,520,073,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861643.92/warc/CC-MAIN-20240616043719-20240616073719-00220.warc.gz | 860,286,922 | 9,983 | Resurrectionofgavinstonemovie.com
Live truth instead of professing it
# How can you identify an unknown weak acid?
## How can you identify an unknown weak acid?
Purpose: The identity and concentration of an unknown weak acid is determined by titration with standardized NaOH solution. where Mt is the concentration of the titrant, Vt is the volume of added titrant, Mx is the concentration of the unknown weak acid, and Vx is the volume of the weak acid that is titrated.
### How do you find the equivalence point of an unknown acid?
Calculate the moles of acid originally present in the sample that you titrated. Report this value to three significant figures. At the equivalence point, nOH− = nHA = (molarity of NaOH)(volume of NaOH added).
#### Why is an indicator not reliable for a titration of your unknown acid?
Since a gradual change implies more volume must be added to obtain the same pH change in other acid base titration, the difference between equivalence point and end point(if you use indicator) will be more and therefore indicator will not be suitable for it.
Is a titration an effective way to determine the concentration of an unknown acid the identity of an unknown acid Why?
That is done by knowing the mass you used to make your acid solution, so that you can calculate its starting concentration with a good amount of certainty. If you do NOT know what your acid is, then the titration is done to determine its pKa and thus, possibly its identity.
How do you find the pKa of an unknown acid titration?
Calculate the pKa with the formula pKa = -log(Ka). For example, pKa = -log(1.82 x 10^-4) = 3.74.
## Can I titrate a solution of unknown concentration?
NO. Explanation: You can do such a titration to an end-point, but it will be meaningless because you don’t have 2 unknowns, i.e. the concentration of the both solutions. So, at the end of the titration, you will not be able to conduct any meaningful calculations.
### When performing a titration the substance of unknown concentration is called?
Titrand/Analyte. The titrand/analyte is the solution with unknown molarity. It is the substance whose concentration is to be determined through the process of titration. Titrant/Reagant.
#### What is the difference between C1V1 C2V2 and M1V1 M2V2 formula?
The answer would be the same; the concentration units must be the same. A variation: you may see this C1V1 = C2V2 written as M1V1 = M2V2. Here the M1 and M2 are the molar concentrations specifically. As long as the concentrations are the same, the formula works.
What is the endpoint of titration of the unknown acid?
Titration of the Unknown Acid. Using the same procedure and graphing the titration curve (Figure 2), we found the endpoint to be 42 mL. Within the laboratory instructions, there was a list compiled of potential weak acids our unknown solution could be: acetic, boric, tartaric and phosphoric acids.
How do you titrate sodium hydroxide with hydrochloric acid?
In this experiment, students titrate sodium hydroxide solution with hydrochloric acid. By measuring the temperature change each time a portion of acid is added, students can determine the end-point of the titration, indicated by the highest temperature. They then use this information to calculate the concentration of the hydrochloric acid.
## How do you interpret the pKa of an unknown acid?
Based off the pH’s obtained from the titration of the unknown acid, a pH vs. volume graph was generated in which the pKa of the unknown acid was interpreted. When the pH was being recorded, the electrode usually reached a steady value within several seconds.
### How do you determine the dissociation constant of an unknown acid?
By continuously adding a strong base, sodium hydroxide (NaOH), to a solution of unknown acid and plotting the gathered data, the dissociation constant (pK a) of the unknown acid could be determined. | 878 | 3,904 | {"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} | 3.28125 | 3 | CC-MAIN-2024-26 | latest | en | 0.900072 |
https://math.stackexchange.com/questions/1190052/given-a-drawing-of-a-parabola-is-there-any-geometric-construction-one-can-make-t/1190379 | 1,709,089,757,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474690.22/warc/CC-MAIN-20240228012542-20240228042542-00497.warc.gz | 376,622,008 | 35,638 | # Given a drawing of a parabola is there any geometric construction one can make to find its focus?
This question was inspired by another one I asked myself these days Given a drawing of an ellipse is there any geometric construction we can do to find it's foci?
I think this is harder, I can't even find is axis or vertex.
• Any parabola with any orientation?
– tomi
Mar 15, 2015 at 1:37
• yes the orientation don't matter Mar 15, 2015 at 2:09
Let three tangents to a parabola form a triangle. Then Lambert's theorem states that the focus of the parabola lies on the circumcircle of the triangle.
So draw three tangents. Find the triangle that they form and construct the circumcircle.
Start again with two of the original tangents and a new tangent. Find the triangle they form and construct the circumcircle. The focus must be at one of the two intersections of these circles.
• how to draw perfect tangents without knowing the focus and the axis of simetry ? Mar 15, 2015 at 3:41
• @onlyme Given a parabola $\mathcal{P}$ and any direction not parallel to its axis of symmetry, one can construct a tangent along that direction as follows: Let $\ell_1, \ell_2$ be a pair of lines along that direction which intersect $\mathcal{P}$ at points $a_1, b_1$ and $a_2, b_2$. Let $m_1$ and $m_2$ be mid-points of line segments $\overline{a_1b_1}$ and $\overline{a_2b_2}$. Construct a line through $m_1$ and $m_2$ and let it intersect $\mathcal{P}$ at $c$. The line through $c$ parallet to $\ell_1$ and $\ell_2$ will be a tangent of $\mathcal{P}$ along the desired direction! Mar 15, 2015 at 4:44
As @AchilleHui mentions, the midpoints of two parallel chords lead to the point of tangency ($T$) with a third parallel line. Note that the line of midpoints is parallel to the axis of the parabola. By the reflection property of conics, the line of midpoints and the line $\overleftrightarrow{TF}$ make congruent angles with the tangent line.
So, two sets of parallel chords determine two points of tangency and two lines-of-midpoints, which determine two lines that meet at focus $F$. $\square$
• Wow that solves it even faster! And also the way I found the axis with the other method shown above was way more complex: give the foci I would make two more tangents at some points, draw the perpendicular lines to that tangents through the foci intersect theses lines and get two points on the line parallet to the directrix on the apex of the parabola. Thanks. Mar 15, 2015 at 19:52 | 657 | 2,482 | {"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} | 4.5 | 4 | CC-MAIN-2024-10 | latest | en | 0.910503 |
https://chrisvantienhoven.nl/ql-items/ql-points/ql-p28 | 1,725,979,506,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651255.81/warc/CC-MAIN-20240910125411-20240910155411-00634.warc.gz | 151,503,645 | 18,950 | QL-P28 is the Circumcenter of the circle through the four X(186) points of the Component Triangles of the Reference Quadrilateral also called the QL-CT-versions of X(186).
X(186) is the Inverse of the Orthocenter X(4) in the circumcircle of a triangle.
See Ref-12 for an explanation of ETC-points X(i).
See Ref-33 Anopolis message # 409 for a discussion on this point.
There actually are 3 Triangle-points in ETC in the range X(1)-X(4000) for which their appearances in the Component Triangles of a Quadrilateral are concyclic: X(3), X(186), X(265).
Only for X(4) these appearances are collinear in this range.
It’s remarkable that these points all relate to X(3) and X(4).
Coordinates:
1st CT-coordinate:
a2 (a6 (l - m) (l - n) (2 l - m - n) (m - n)
+ b6 (l3 m - 3 l2 m2 - m n (3 m2 - 3 m n + n2) + l (2 m3 + 3 m2 n - 3 m n2 + n3))
- c6 (l3 n - 3 l2 n2 - m n (m2 - 3 m n + 3 n2) + l (m3 - 3 m2 n + 3 m n2 + 2 n3))
+ b2 c4 ( 9 l m n2 - 3 l2 n (2 m + n) + l3 (m + 2 n) + m n (m2 - 3 m n - n2))
+ b4 c2 (-9 l m2 n - l3 (2 m + n) + 3 l2 m (m + 2 n) + m n (m2 + 3 m n - n2))
+ 3 a2 b4 (l - m) (l - n) (m - n) (m - n)
- 3 a2 c4 (l - m) (l - n) (m - n) (m - n)
+ a4 b2 (l3 (-3 m + 2 n) + m n (m2 + 3 m n - 3 n2) + 3 l2 (m2 + 2 m n - 2 n2) + 3 l n (-3 m2 + m n + n2))
- a4 c2 (l3 (2 m - 3 n) + 3 l m (m2 + m n - 3 n2) + m n (-3 m2 + 3 m n + n2) + l2 (-6 m2 + 6 m n + 3 n2))
+ a2 b2 c2 (m - n) (l3 + 9 l m n - 3 l2 (m + n) - 2 m n (m + n)))
1st DT-coordinate:
+ a8 (l2 - m2) (l2 - n2) (m2 - n2) (5 l4 - l2 m2 - l2 n2 - 3 m2 n2)
+ b8 (l2 - m2) (l2 + 3 m2) (m - n)3 (m + n)3
+ c8 (l2 - n2) (l2 + 3 n2) (m - n)3 (m + n)3
- 4 b6 c2 l2 (l2 - m2) (m - n)3 (m + n)3
- 4 b2 c6 l2 (l2 - n2) (m - n)3 (m + n)3
+ 6 b4 c4 (l2 - m2) (l2 - n2) (m - n)3 (m + n)3
+ 6 a4 b4 (l2 - m2) (l2 - n2) (m2 - n2) (3 l2 m2 + m4 - l2 n2 - 3 m2 n2)
- 6 a4 c4 (l2 - m2) (l2 - n2) (m2 - n2) (l2 m2 - 3 l2 n2 + 3 m2 n2 - n4)
+ 4 a2 b6 (l2 - m2) (-4 l2 m6 + 4 l4 m2 n2 + l2 m4 n2 + 7 m6 n2 - 6 l2 m2 n4 - 6 m4 n4 + l2 n6 + 3 m2 n6)
- 4 a2 c6 (l2 - n2) (l2 m6 + 4 l4 m2 n2 - 6 l2 m4 n2 + 3 m6 n2 + l2 m2 n4 - 6 m4 n4 - 4 l2 n6 + 7 m2 n6)
+ 4 a6 c2 (l2 - n2) (2 l6 m2 +3 l4 m4 - l2 m6 + 2 l6 n2 - 13l4 m2 n2 + 6 l2 m4 n2 - 3 m6 n2 + 2 l4 n4 - l2 m2 n4 + 3 m4 n4)
- 4 a6 b2 (l2 - m2) (2 l6 m2 + 2 l4 m4 + 2 l6 n2 -13 l4 m2 n2 - l2 m4 n2 + 3 l4 n4 + 6 l2 m2 n4 + 3 m4 n4 - l2 n6 - 3 m2 n6)
- 4 a2 b4 c2 (-l4 m6 - 3 l2 m8 + 12 l6 m2 n2 - 18 l4 m4 n2 + 15 l2 m6 n2 + 3 m8 n2 - 21 l4 m2 n4 + 27 l2 m4 n4 - 18 m6 n4 + 4 l4 n6 - 3 l2 m2 n6 + 3 m4 n6)
+ 4 a2 b2 c4 (4 l4 m6 + 12 l6 m2 n2 - 21 l4 m4 n2 - 3 l2 m6 n2 - 18 l4 m2 n4 + 27 l2 m4 n4 + 3 m6 n4 - l4 n6 + 15 l2 m2 n6 - 18 m4 n6 - 3 l2 n8 + 3 m2 n8)
- 4 a4 b2 c2 (m2 - n2) (3 l6 m2 + 5 l4 m4 + 3 l6 n2 - 16 l4 m2 n2 - 3 l2 m4 n2 + 5 l4 n4 - 3 l2 m2 n4 + 6 m4 n4)
Properties:
Vernieuwen | 1,567 | 2,795 | {"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} | 3.140625 | 3 | CC-MAIN-2024-38 | latest | en | 0.454996 |
https://www.lmfdb.org/Character/Dirichlet/2100/127 | 1,601,193,930,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400265461.58/warc/CC-MAIN-20200927054550-20200927084550-00222.warc.gz | 897,048,871 | 5,929 | # Properties
Modulus 2100 Conductor 100 Order 20 Real no Primitive no Minimal yes Parity even Orbit label 2100.cs
# Related objects
Show commands for: Pari/GP / SageMath
sage: from sage.modular.dirichlet import DirichletCharacter
sage: H = DirichletGroup(2100)
sage: M = H._module
sage: chi = DirichletCharacter(H, M([10,0,1,0]))
pari: [g,chi] = znchar(Mod(127,2100))
## Basic properties
sage: chi.conductor() pari: znconreyconductor(g,chi) Modulus = 2100 Conductor = 100 sage: chi.multiplicative_order() pari: charorder(g,chi) Order = 20 Real = no sage: chi.is_primitive() pari: #znconreyconductor(g,chi)==1 \\ if not primitive returns [cond,factorization] Primitive = no Minimal = yes sage: chi.is_odd() pari: zncharisodd(g,chi) Parity = even Orbit label = 2100.cs Orbit index = 71
## Galois orbit
sage: chi.galois_orbit()
pari: order = charorder(g,chi)
pari: [ charpow(g,chi, k % order) | k <-[1..order-1], gcd(k,order)==1 ]
## Values on generators
$$(1051,701,1177,1501)$$ → $$(-1,1,e\left(\frac{1}{20}\right),1)$$
## Values
-1 1 11 13 17 19 23 29 31 37 41 43 $$1$$ $$1$$ $$e\left(\frac{3}{10}\right)$$ $$e\left(\frac{19}{20}\right)$$ $$e\left(\frac{13}{20}\right)$$ $$e\left(\frac{2}{5}\right)$$ $$e\left(\frac{1}{20}\right)$$ $$e\left(\frac{1}{10}\right)$$ $$e\left(\frac{9}{10}\right)$$ $$e\left(\frac{9}{20}\right)$$ $$e\left(\frac{1}{5}\right)$$ $$i$$
value at e.g. 2
## Related number fields
Field of values $$\Q(\zeta_{20})$$ | 549 | 1,464 | {"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} | 3.15625 | 3 | CC-MAIN-2020-40 | latest | en | 0.288932 |
http://au.metamath.org/mpegif/mop.html | 1,513,381,517,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948580416.55/warc/CC-MAIN-20171215231248-20171216013248-00309.warc.gz | 24,190,292 | 5,165 | Mathbox for Frédéric Liné < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > Mathboxes > mop Unicode version
Theorem mop 25288
Description: Meet is a binary internal operation. (Contributed by FL, 12-Dec-2009.)
Hypothesis
Ref Expression
jop1
Assertion
Ref Expression
mop
Proof of Theorem mop
Dummy variables are mutually distinct and distinct from all other variables.
StepHypRef Expression
1 jop1 . . . 4
21islatalg 25286 . . 3
3 simp2 956 . . 3
42, 3syl6bi 219 . 2
543impia 1148 1
Colors of variables: wff set class Syntax hints: wi 4 wa 358 w3a 934 wceq 1632 wcel 1696 wral 2556 cop 3656 cxp 4703 cdm 4705 wf 5267 (class class class)co 5874 clatalg 25284 This theorem is referenced by: clme 25290 This theorem was proved from axioms: ax-1 5 ax-2 6 ax-3 7 ax-mp 8 ax-gen 1536 ax-5 1547 ax-17 1606 ax-9 1644 ax-8 1661 ax-13 1698 ax-14 1700 ax-6 1715 ax-7 1720 ax-11 1727 ax-12 1878 ax-ext 2277 ax-sep 4157 ax-nul 4165 ax-pr 4230 ax-un 4528 This theorem depends on definitions: df-bi 177 df-or 359 df-an 360 df-3an 936 df-tru 1310 df-ex 1532 df-nf 1535 df-sb 1639 df-eu 2160 df-mo 2161 df-clab 2283 df-cleq 2289 df-clel 2292 df-nfc 2421 df-ne 2461 df-ral 2561 df-rex 2562 df-rab 2565 df-v 2803 df-dif 3168 df-un 3170 df-in 3172 df-ss 3179 df-nul 3469 df-if 3579 df-sn 3659 df-pr 3660 df-op 3662 df-uni 3844 df-br 4040 df-opab 4094 df-xp 4711 df-rel 4712 df-cnv 4713 df-co 4714 df-dm 4715 df-rn 4716 df-iota 5235 df-fun 5273 df-fn 5274 df-f 5275 df-fv 5279 df-ov 5877 df-latalg 25285
Copyright terms: Public domain W3C validator | 794 | 1,652 | {"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-2017-51 | latest | en | 0.15301 |
https://www.prepanywhere.com/prep/textbooks/functions-11-mcgraw-hill/chapters/chapter-7-financial-applications/materials/7-1-simple-interest/videos/q8 | 1,624,582,019,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488560777.97/warc/CC-MAIN-20210624233218-20210625023218-00042.warc.gz | 844,954,891 | 7,434 | 8. Q8
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Video Solution
Q1
Q2
Q3
L1
L2
L3
Similar Question 1
<p>Rita borrows <code class='latex inline'>\$500</code> at an annual rate of <code class='latex inline'>8.25\%</code> simple interest to enrol in a driver's education course. She plans to repay the loan in 187 months.</p><p><strong>(a)</strong> What amount must she pay back?</p><p><strong>(b)</strong> How much interest will she pay?</p><p><strong>(c)</strong> How much sooner should Rita repay the loan if she wants t soapy no more than <code class='latex inline'>\$50</code> in interest charges?</p>
Similar Question 2
<p>The graph show the amount of an investment earning simple interest.</p><img src="/qimages/6171" /><p><strong>(a)</strong> Write an equation not relate the interest to time.</p><p><strong>(b)</strong> Use your equation from part (a) to determine how long it will take for the original investment to double.</p>
Similar Question 3
<p>The graph show the amount of an investment earning simple interest.</p><img src="/qimages/6171" /><p><strong>(a)</strong> Write an equation not relate the interest to time.</p><p><strong>(b)</strong> Use your equation from part (a) to determine how long it will take for the original investment to double.</p>
Similar Questions
Learning Path
L1 Quick Intro to Factoring Trinomial with Leading a
L2 Introduction to Factoring ax^2+bx+c
L3 Factoring ax^2+bx+c, ex1
Now You Try
<p>Matthew invests <code class='latex inline'>\$850</code> at <code class='latex inline'>7\%/a</code> simple interest. How long will he have to leave his investment in the bank before earning <code class='latex inline'>\$200</code> in interest?</p>
<img src="/qimages/851" /><p><strong>(a)</strong> Develop a linear model to represent the amount in the GIC versus time.</p><p><strong>(b)</strong> Explain why the model from part (a) is a partial variation. Identify the fixed part and the variable part.</p><p><strong>(c)</strong> How long will it take, to the nearest month of this investment to double from its install value?</p>
<p>The graph show the amount of an investment earning simple interest.</p><img src="/qimages/6171" /><p><strong>(a)</strong> Write an equation not relate the interest to time.</p><p><strong>(b)</strong> Use your equation from part (a) to determine how long it will take for the original investment to double.</p>
<p>The graph show the amount of an investment earning simple interest.</p><img src="/qimages/6171" /><p><strong>(a)</strong> What is the principal?</p><p><strong>(b)</strong> What is the annual interest rate?</p><p><strong>(c)</strong> Write an equation to relate the amount to time.</p><p><strong>(d)</strong> Use your equation from party (c) to determine how long it will take, to the nearest month, for the original investment to double.</p>
<p>To save for a new pair of skis, Sven deposits <code class='latex inline'>\$250</code> into a savings bond that earn <code class='latex inline'>4.5\%</code> per year, simple interest.</p><p><strong>(a)</strong> Write an equation to relate the amount of the investment to time.</p><p><strong>(b)</strong> Graph the function.</p><p><strong>(c)</strong> How long will it take, to the nearest month, for the amount to reach$300?</p><p><strong>(d)</strong> What interest rate is required for the amount to reach $300 in 2 years less than your answer in part (c)?</p> <p>Rita borrows <code class='latex inline'>\$500</code> at an annual rate of <code class='latex inline'>8.25\%</code> simple interest to enrol in a driver's education course. She plans to repay the loan in 187 months.</p><p><strong>(a)</strong> What amount must she pay back?</p><p><strong>(b)</strong> How much interest will she pay?</p><p><strong>(c)</strong> How much sooner should Rita repay the loan if she wants t soapy no more than <code class='latex inline'>\$50</code> in interest charges?</p> <p>Yuri deposits$850 into an account that earns 6.25% per year simple interest. How long will it take for the amount in this account to reach $1000?</p> <p>Lena invests <code class='latex inline'>\$5200</code> at <code class='latex inline'>3\%/a</code> simple interest, while her friend Dan invests <code class='latex inline'>\$3600</code> at <code class='latex inline'>5\%/a</code> simple interest. </p><p>How long will it take for Dan's investment to be worth more than Lena's?</p> <p>Tammie took out a loan for <code class='latex inline'>\$940</code> at an annual rate of <code class='latex inline'>11.5\%</code> simple interest. When she repaid the loan, the amount was <code class='latex inline'>\\$1100</code>. How long did Tammie hold this loan?</p>
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Save videos to My Cheatsheet for later, for easy studying. | 1,315 | 4,823 | {"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} | 3.640625 | 4 | CC-MAIN-2021-25 | latest | en | 0.77833 |
http://mail.scipy.org/pipermail/scipy-user/2013-June/034765.html | 1,394,445,162,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1394010740404/warc/CC-MAIN-20140305091220-00036-ip-10-183-142-35.ec2.internal.warc.gz | 118,963,940 | 3,141 | # [SciPy-User] leastsq
josef.pktd@gmai... josef.pktd@gmai...
Thu Jun 27 08:21:27 CDT 2013
```On Thu, Jun 27, 2013 at 9:04 AM, Frédéric Parrenin
<parrenin.ujf@gmail.com> wrote:
> Dear Josef,
>
> OK to use the curve_fit function with a change of variables to have a
> diagonal covariance matrix.
>
> However, here are two questions/remarks:
> - Curve_fit takes as input both the parameters to fit and a variable x where
> the data are 'located'. This approach seems sub-optimal since in many
> inverse problems, the function is evaluated for all x at a time. Running the
> function independently N times will significantly decrease the computation
> time.
I don't understand this part. We are fitting a curve to N
observations. We need all of them to calculate the residual sum of
squares.
> Maybe in this case the best thing to do is to declare that x is empty, but
> how to do that in practice?
You don't need to use x, you can just write f as a method in a class
and attach whatever attributes you want to reuse in the f method. (I'm
not completely remember how this was implemented, and no time to look
it up right now.)
> - It is not very clear from the scipy doc
> http://docs.scipy.org/doc/scipy/reference/generated/scipy.optimize.curve_fit.html
> what the function f is supposed to return. Is it just a scalar function or
> can it be a ndarray or even something else?
the function should return an array of predicted values, one element
for each observation
>
> Some more complex examples in the doc would really help to better understand
> how it works.
There are several examples on stackoverflow, including the case when f
is a method in a class
Josef
>
> Best regards,
>
> Frédéric Parrenin
>
>
>
>
>
> 2013/6/26 <josef.pktd@gmail.com>
>
>> On Wed, Jun 26, 2013 at 12:26 PM, Frédéric Parrenin
>> <parrenin.ujf@gmail.com> wrote:
>> > Dear all,
>> >
>> > I am experimenting the optimize module of scipy.
>> > My optimization problem is a leastsq problem.
>> > However, the leastsq function seems to be not appropriate for two
>> > reasons:
>> > - there is no possibility to specify a covariance matrix between the
>> > leastsq
>> > terms. They are supposed to be independent, which is a too strong
>> > assumption
>> > in my case.
>> > - the analyzed covariance matrix (i.e. the inverse of the jacobian of
>> > the
>> > cost function) cannot be simply outputed.
>> >
>> > Of course I could use a more generic optimization function, like the
>> > minimize one.
>> > However this seems sub-optimal because the minimisation of a least
>> > squares
>> > problem can dealt more efficiently (the jacobian of the cost function
>> > can be
>> > approximated using the jacobian of the terms to minimize).
>> >
>> > Can anybody help me?
>> > Are there plans to improve the leastsq function?
>>
>> leastsq is a low level function and I think we should not load it up
>> with any options.
>>
>> for weighted least-squares the more highlevel interface with
>> However it doesn't allow for a full covariance matrix for the errors.
>>
>> If you want to use leastsq with a full covariance matrix, then you
>> could transform both sides yourself, similar to what is done in
>> curve_fit, but with the cholesky of the inverse covariance matrix.
>> We use that in statsmodels.GLS, but only for linear models.
>> But, if there a large number of observations, then using the full
>> covariance matrix is inefficient, and in many cases a more direct
>> transformation can be used.
>>
>> nonlinear least squares is still largely missing in statsmodels.
>>
>> I don't know if any of the other packages that are based on leastsq
>> have the option.
>>
>> Josef
>>
>>
>>
>> >
>> > Best regards,
>> >
>> > Frédéric Parrenin
>> >
>> >
>> > _______________________________________________
>> > SciPy-User mailing list
>> > SciPy-User@scipy.org
>> > http://mail.scipy.org/mailman/listinfo/scipy-user
>> >
>> _______________________________________________
>> SciPy-User mailing list
>> SciPy-User@scipy.org
>> http://mail.scipy.org/mailman/listinfo/scipy-user
>
>
>
> _______________________________________________
> SciPy-User mailing list
> SciPy-User@scipy.org
> http://mail.scipy.org/mailman/listinfo/scipy-user
>
``` | 1,067 | 4,211 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2014-10 | longest | en | 0.890516 |
https://www.quesba.com/questions/charisma-top-level-leaders-refer-academy-management-journal-august-2015-stu-631017 | 1,638,900,419,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363405.77/warc/CC-MAIN-20211207170825-20211207200825-00599.warc.gz | 996,484,240 | 30,549 | # Charisma of top-level leaders. Refer to the Academy of Management Journal (August 2015) study of...
Charisma of top-level leaders. Refer to the Academy of Management Journal (August 2015) study of the charisma of top-level leaders, Exercise 11.28 (p. 657). Recall that the researchers used data on 24 U.S. presidential election years to model Democratic vote share (y) as a function of the difference (x) between the Democratic and Republican candidates’ charisma scores. Is there evidence to indicate that the simple linear regression model is statistically useful for predicting Democratic vote share? Test using a = .10.
Exercise 11.28
Charisma of top-level leaders. According to a theory proposed in the Academy of Management Journal (August 2015), top leaders in business are selected based on their organization’s performance as well as the leader’s charisma. To test this theory, the researchers collected data on 24 U.S. presidential elections from 1916 to 2008. The dependent variable of interest was Democratic vote share (y), measured as the percentage of voters who voted for the Democratic candidate in the national election. The charisma of both the Democratic and Republican candidates was measured (on a 150-point scale) based on the candidates’ acceptance speeches at their party’s national convention. One of the independent variables of interest was the difference (x) between the Democratic and Republican charisma values. These data are listed in the accompanying table.
a. Find the least squares line relating Democratic vote share (y) to charisma difference (x).
b. Graph the least squares line on a scatterplot of the data. Is there visual evidence of a linear relationship between the variables? Is the relationship positive or negative?
c. Interpret, practically, the estimated slope of the line.
Jul 22 2021| 12:56 PM | | 382 | 1,855 | {"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} | 2.703125 | 3 | CC-MAIN-2021-49 | latest | en | 0.935433 |
http://mathhelpforum.com/calculus/15782-contour-within-contour.html | 1,524,731,632,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125948119.95/warc/CC-MAIN-20180426070605-20180426090605-00222.warc.gz | 201,549,514 | 14,158 | 1. ## Contour within Contour
Let $\displaystyle f(z)$ be a complex function defined on an open non-empty set $\displaystyle S$. Let $\displaystyle \gamma (t)$ be a contour wholly inside $\displaystyle S$. Does there exist another contour $\displaystyle \zeta (t)$ wholly inside $\displaystyle S$ with the property (abusing notation) that $\displaystyle \partial \gamma \subseteq I(\zeta)$ (meaning such that this contour lie strictly within this new contour).
2. I have an idea how to do this, but I am a noob in topology.
Given a contour $\displaystyle \gamma^*$. Can a circle be "shrunk" to be strictly contained in the interior $\displaystyle \gamma^*$?
3. Originally Posted by ThePerfectHacker
I have an idea how to do this, but I am a noob in topology.
Given a contour $\displaystyle \gamma^*$. Can a circle be "shrunk" to be strictly contained in the interior $\displaystyle \gamma^*$?
Only if the region you are "shrinking" the contour over is connected. (Someone check me on this, I don't think so but it might have to be path connected.)
-Dan
4. Originally Posted by topsquark
Only if the region you are "shrinking" the contour over is connected. (Someone check me on this, I don't think so but it might have to be path connected.)
-Dan
Can you give an example of a contour whose interior is not connected?
5. Originally Posted by ThePerfectHacker
Can you give an example of a contour whose interior is not connected?
By 'contour' do you mean the path is a simple closed curve?
6. Originally Posted by ThePerfectHacker
Can you give an example of a contour whose interior is not connected?
Simple. Consider $\displaystyle \mathbb{R}^2 - \{ (x, y) |x^2 + y^2 = 1 \}$ and the contour which is the circle centered on the origin with radius 2. You can't shrink your contour to, say, a circle centered on the origin with radius 1/2.
-Dan
7. Originally Posted by Plato
By 'contour' do you mean the path is a simple closed curve?
Yes, let $\displaystyle \gamma (t): [a,b] \mapsto \mathbb{C}$ be so that:
1)$\displaystyle \gamma$ is a path, i.e. the components of $\displaystyle \gamma$ (IM and RE) are continous.
2)$\displaystyle \gamma(a) = \gamma(b)$ i.e. it is closed.
3)$\displaystyle 0<|t_1-t_2|<b-a \to f(t_1)\not = f(t_2)$, i.e. it is simple.
Originally Posted by topsquark
Simple.
Is that supposed to be a pun! No, seriously I think you and I have a different understanding of what a contour is. See above.
8. Then look up the Jordan Curve Theorem.
9. Originally Posted by Plato
Then look up the Jordan Curve Theorem.
I am not looking it up because I will not understand it. But I am assuming you are saying "yes", i.e. given any contour we can draw a circle inside of it?
10. The Jordan Curve Theorem states that any simple closed curve partitions the plane into three non-empty disjoint connected sets: the curve itself, a bounded set (the interior) and an unbounded set (the exterior). An arc from an interior point to an exterior point must contain a point of the curve.
Moreover, to answer your particular question: yes about any interior point there is a closed disk (can be centered at the point) that is a subset of the interior.
11. Originally Posted by ThePerfectHacker
No, seriously I think you and I have a different understanding of what a contour is.
I think Plato ended up answering this, but to be clear I am defining a "contour" to be the same as a "curve." I believe the only difference in what we were talking about is that your contour was in $\displaystyle \mathbb{C}$ and mine was in a subset of $\displaystyle \mathbb{R}^2$.
-Dan
12. Originally Posted by topsquark
I think Plato ended up answering this, but to be clear I am defining a "contour" to be the same as a "curve."
Actually I did not answer anything general about contours. I only answer what TPH was considering a contour, for him it is a simple closed curve. That is by no means standard. Most complex variables textbooks would consider a contour as the image of a continuous one-to-one mapping of [0,1]: a Jordan arc. If we do allow only f(0)=f(1) the we have a simple closed curve.
13. Originally Posted by Plato
Most complex variables textbooks would consider a contour as the image of a continuous one-to-one mapping of [0,1]: a Jordan arc. If we do allow only f(0)=f(1) the we have a simple closed curve.
You mean graduate textbooks in complex anaylis. I brought myself an undergraduate one which is why I have many questions about these contours.
This is Mine 61th Post!!!
14. The original set must be open. If that is the case, then you can inscribe a circle inside all contours C contained in the set.
To contruct it, well...
- Jordan's theorem gives us that the interior of C is open, so there certainly are circles everywhere. Here, we might also drop connectedness (though topsq made an important point).
On the other hand, that is some big theorem, and you don't normally call a bulldozer to strike a nail. There is a procedure that can produce a new curve inside this one, and more importantly - maintain the same winding number on the complement of S (as, in the important cases, S has holes where functions have poles). Take the curve to be C(t), t in [0,1]. Let n(t) be the normal to c(t), pointing inwards (since C is simple, this is always well defined). Now there is a s>0, such that sn(t) is wholly inside S (as this is open) and the contour
C*(t)=C(t)+sn(t), t in [0,1]
is a) inscribed in C and b) for a function f, I(C,f)=I(C*,f) on the compliment of S, so the Cauchy theorems apply.
15. Originally Posted by Rebesques
- Jordan's theorem gives us that the interior of C is open, so there certainly are circles everywhere. Here, we might also drop connectedness (though topsq made an important point).
I think I finally understand, thank you!
I was discussing this with my professor by e-mail. He said that the book by Lang makes the following definition:
Definition: Given a contour $\displaystyle \gamma$ a point $\displaystyle a$ is "inside" $\displaystyle \gamma$ iff $\displaystyle \oint_{\gamma} \frac{dz}{z-a} \not = 0$. (I really like it).
So we define $\displaystyle I(\gamma)$ to be all points in $\displaystyle \mathbb{C}$ such that the contour integral over Western Europe is non-zero.
Now I think I understand, "open" means we can find a $\displaystyle \delta >0$ such that (the neighorbhood) $\displaystyle \{z\in \mathbb{C} : |z - z_0|<\delta\} \subset \mathbb{I}(\gamma)$. So the circle $\displaystyle |z-z_0| = \delta$ is what we seek. And this is possible by the Jordan Closed Curve theorem which Plato explains as saying that $\displaystyle I(\gamma)$ is an open set.
Page 1 of 2 12 Last | 1,706 | 6,654 | {"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} | 3.359375 | 3 | CC-MAIN-2018-17 | latest | en | 0.868428 |
https://groups.google.com/g/misc.transport.road/c/hZj2DqNM3EE | 1,638,388,969,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964360881.12/warc/CC-MAIN-20211201173718-20211201203718-00536.warc.gz | 362,817,327 | 142,900 | # Exit 0
6 views
### SPUI
Jan 31, 2002, 4:45:45 AM1/31/02
to
Mile-based numbering often presents a problem: to use or not to use exit 0.
If it is not used, you usually have at least 1.5 miles of exit 1*, or you
modify it so exit 1* is only before mile, exit 2* is before mile 2, etc.
Here is a full list of which states use exit 0.
Y=yes, N=no, NX=no but there are no places with multiple exits at the
beginning that would favor an exit 0, NP=no, takes the number at a terminus
AL Y
AZ N
AR NP
CO N (I-270 is an exception since that was extended)
FL N
GA N
HI N
ID N
IL N
IN Y
IA N
KS N
KY N
LA N
MD N
MI N
MN N
MS N
MO N
MT Y
NE NP
NV N
NJ N
NM Y
NC N
ND NX
OH N
OK N
OR N
PA N
SC N
SD NX
TN N
TX Y
UT N
VA N
WA N
DC N
WV N
WI N
WY NX
--
Dan Moraseski - 14th grade at MIT
http://spui.cjb.net/ - FL NJ MA route logs and exit lists
"Boston really isn't high on the importance scale for numbered routes
because we don't see the need to waste taxpayer money on it. Thru traffic
uses highways, not numbered surface routes. This isn't the 1950s." - some
MassHighway MassHole to Shawn De Cesari
### SP Cook
Jan 31, 2002, 7:18:51 AM1/31/02
to
SPUI <sp...@mit.BUTIDONTLIKESPeduAM> wrote in message
> Mile-based numbering often presents a problem: to use or not to use exit
0.
> If it is not used, you usually have at least 1.5 miles of exit 1*, or you
> modify it so exit 1* is only before mile, exit 2* is before mile 2, etc.
> Here is a full list of which states use exit 0.
> Y=yes, N=no, NX=no but there are no places with multiple exits at the
> beginning that would favor an exit 0, NP=no, takes the number at a
terminus
> junction from the intersecting road
>
> WV N
Umm, the Wheeling Island exit on I-70 is Exit 0.
SP Cook
### Patrick L. Humphrey
Jan 31, 2002, 9:47:57 AM1/31/02
to
Isn't there still an Exit 0 on the Purchase Parkway in Kentucky? (I seem to
remember one when I was up that way 18 months ago.) You have Texas listed
as using Exit 0 -- but where? (It's not on I-30, I-37, or I-45 -- those are
the three Texas interstates I've been to the ends of, so far.)
--PLH, used to the longer exit numbers, like up between 730 and about 800 in
the Houston area
### Jeffrey Coleman Carlyle
Jan 31, 2002, 10:36:37 AM1/31/02
to
"Patrick I. Humphrey" <pat...@io.com> wrote in message
news:szk1yg6...@hagbard.io.com...
> Isn't there still an Exit 0 on the Purchase Parkway in Kentucky? (I seem to
> remember one when I was up that way 18 months ago.)
The beginning of Kentucky's parkways are usually given the "Exit 0"
number.
### SPUI
Jan 31, 2002, 11:22:01 AM1/31/02
to
"Patrick L. Humphrey" <pat...@io.com> wrote in message
news:szk1yg6...@hagbard.io.com...
> Isn't there still an Exit 0 on the Purchase Parkway in Kentucky? (I seem
to
> remember one when I was up that way 18 months ago.) You have Texas listed
> as using Exit 0 -- but where? (It's not on I-30, I-37, or I-45 -- those
are
> the three Texas interstates I've been to the ends of, so far.)
I-10 has an exit 0.
### sycamore
Jan 31, 2002, 11:32:06 AM1/31/02
to
Illinois must have stopped that recently. When I-72 used to end at I-55, I-55
was exits 0A-B.
--
Sycamore--GO RAMS!!!
We beat the Patsies at home...now we'll beat 'em in a dome!
"Road Trippin"--a developing travel diary at Sycamoreland
http://www.sycamoreland.com
ICQ--7810696
### Jason L. Bennett
Jan 31, 2002, 12:14:20 PM1/31/02
to
"Exit 0" is an album by Steve Earle, who has songs such as "Hillbilly
Jason L. Bennett
Oriskany, NY
--
Microsoft - giving IT Support People job security for the past 20 years
-----= Posted via Newsfeeds.Com, Uncensored Usenet News =-----
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### Elkins, H.B.
Jan 31, 2002, 12:28:22 PM1/31/02
to
pat...@io.com (Patrick L. Humphrey) wrote:
>Isn't there still an Exit 0 on the Purchase Parkway in Kentucky? (I seem to
>remember one when I was up that way 18 months ago.) You have Texas listed
>as using Exit 0 -- but where? (It's not on I-30, I-37, or I-45 -- those are
>the three Texas interstates I've been to the ends of, so far.)
Yes. On my long-neglected Purchase Parkway page
(http://www.users.mis.net/~hbelkins/PurchasePkwy.html) You'll see that
there is, indeed, an Exit 0, for KY 116 and KY 166.
+++++++++++++++++++++++++
H.B. Elkins mailto:hbel...@mis.net or mailto:HB...@aol.com
http://www.millenniumhwy.net
http://www.users.mis.net/~hbelkins
"There's no doubt he's the best race driver in the world."
--Dale Jarrett, on Dale Earnhardt (RIP 2/18/01)
Waltrip, Kentucky, Anybody but North Carolina
To reply, you gotta do what NASCAR won't -- remove the restrictor plates!
+++++++++++++++++++++++++
### MLOTT2002
Jan 31, 2002, 12:58:13 PM1/31/02
to
what is exit 0 in alabama?
### Tulsarama
Jan 31, 2002, 1:12:43 PM1/31/02
to
>You have Texas listed as using Exit 0 -- but where?
Isn't the Glenrio exit on I-40 Exit 0?
Rick Mattioni
Tulsa
### Patrick L. Humphrey
Jan 31, 2002, 3:36:57 PM1/31/02
to
...at the south or west end, I'd presume, but the Pennyrile starts at Exit 7
(because of the original plan to extend it south to I-24 from US 41 south of
Hopkinsville), the Cumberland doesn't have an Exit 0 at I-65, and I don't
remember an Exit 0 on the Daniel Boone the last time I traveled it six years
ago...
--PLH, who'd have added the WK, but the last time I traveled that, it was
still a toll road and didn't have exit numbers :)
### Patrick L. Humphrey
Jan 31, 2002, 3:39:42 PM1/31/02
to
"SPUI" <sp...@mit.BUTIDONTLIKESPeduAM> writes:
Just south of Anthony would make sense...of course, that's just about
halfway to Los Angeles from where I get on I-10 headed west. :-)
--PLH, give me two weeks off, and I'll check it out
### Brandon Kraft
Jan 31, 2002, 3:40:37 PM1/31/02
to
"SPUI" <sp...@mit.BUTIDONTLIKESPeduAM> wrote in message
news:3c596f5c\$0\$3940\$b45e...@senator-bedfellow.mit.edu...
I-44 does not have an exit zero, have two miles worth of Exit 1* (1-1D for
five marked exits, each way only have four though. Exit 1 is only SB, exit
1B is only NB)
### BD
Jan 31, 2002, 3:46:04 PM1/31/02
to
What is odd is that IL labels the exits at the end of I-474 as A/B, not
1A/1B or 0A/0B.. just A/B. IIRC, there is an exit 1 on either IL 6 or
I-474.
"sycamore" <syca...@sycamoreland.com> wrote in message
news:3C597186...@sycamoreland.com...
### Aaron O'Donnell
Jan 31, 2002, 4:16:52 PM1/31/02
to
On Thu, 31 Jan 2002 04:45:45 -0500, "SPUI"
>Mile-based numbering often presents a problem: to use or not to use exit 0.
>If it is not used, you usually have at least 1.5 miles of exit 1*, or you
>modify it so exit 1* is only before mile, exit 2* is before mile 2, etc.
>Here is a full list of which states use exit 0.
>Y=yes, N=no, NX=no but there are no places with multiple exits at the
>beginning that would favor an exit 0, NP=no, takes the number at a terminus
>OR N
Oregon could be marked NX, there are a couple places where an exit 0
would work. OR-22 in Salem has Exits 1A and 1B at the spot where the
I-5 interchange/overpass is, which is also the point where the mile
posts start from zero and increment in both the east & west
directions. I think US-26 in Portland may be mileposted like this too.
+-----------------------------+
| Aaron O'Donnell |
| www.aaroncity.com |
+-----------------------------+
### US71
Jan 31, 2002, 4:51:06 PM1/31/02
to
"SPUI" <sp...@mit.BUTIDONTLIKESPeduAM> wrote in message news:<3c591279\$0\$3959\$b45e...@senator-bedfellow.mit.edu>...
> Mile-based numbering often presents a problem: to use or not to use exit 0.
> If it is not used, you usually have at least 1.5 miles of exit 1*, or you
> modify it so exit 1* is only before mile, exit 2* is before mile 2, etc.
> Here is a full list of which states use exit 0.
> Y=yes, N=no, NX=no but there are no places with multiple exits at the
> beginning that would favor an exit 0, NP=no, takes the number at a terminus
> junction from the intersecting road
>
> AR NP
I-540 is kind of a wierd exception to this:
Originally, it ended with Exit 1 A/B at I-40. After it was extended
north towards Fayetteville, the exit was renumbered 15 (though there
are also still signs noting it as Exit 1).
### Neil Alexander Bratney
Jan 31, 2002, 4:53:28 PM1/31/02
to
Iowa actually seems to use both forms. At the southern terminus of
I-380 is Exit 0. See Jason Hancock's page:
http://iowahwypix.tripod.com/ic/80-380b.jpg
I don't know about the northern (Western?) terminus of I-74. Anyone
else?
> If it is not used, you usually have at least 1.5 miles of exit 1*
I-29, I-129 and I-480 all have exits less than 1/2 mile from the border.
They are all named exit 1.
I-35 has an exit at the Missouri border. It is exit 114 (Mo numbering).
I-80 has two exits within 1.5 miles of Nebraska:
Exit 1A = I-29 @ 0.2 miles
Exit 1B = 24th Street @ 1.4 miles
### Neil Alexander Bratney
Jan 31, 2002, 4:56:13 PM1/31/02
to
Neil Alexander Bratney wrote:
>
> Iowa actually seems to use both forms. At the southern terminus of
> I-380 is Exit 0. See Jason Hancock's page:
>
http://iowahwypix.tripod.com/ic/photos.html
You can even see the 0A exit tab!
### SPUI
Jan 31, 2002, 11:39:04 PM1/31/02
to
"Brandon Kraft" <cheeseRE...@cst.REMOVEnet> wrote in message
news:wYh68.79659\$h31.5...@e420r-atl1.usenetserver.com...
>
> I-44 does not have an exit zero, have two miles worth of Exit 1* (1-1D for
> five marked exits, each way only have four though. Exit 1 is only SB, exit
> 1B is only NB)
Hmmm - looks like TX uses exit 0 in rural areas but not urban areas. Or
maybe it's only in the western part of the state.
### SPUI
Jan 31, 2002, 11:39:29 PM1/31/02
to
"MLOTT2002" <mlot...@aol.com> wrote in message
news:20020131125813...@mb-mu.aol.com...
> what is exit 0 in alabama?
I-65 has exit 0 at I-10 and I think I-85 has exit 0 at I-65.
### SPUI
Jan 31, 2002, 11:40:10 PM1/31/02
to
"Aaron O'Donnell" <aa...@nospam.aaroncity.com> wrote in message
news:29cj5u4t1203suehf...@4ax.com...
> On Thu, 31 Jan 2002 04:45:45 -0500, "SPUI"
>
> >Mile-based numbering often presents a problem: to use or not to use exit
0.
> >If it is not used, you usually have at least 1.5 miles of exit 1*, or you
> >modify it so exit 1* is only before mile, exit 2* is before mile 2, etc.
> >Here is a full list of which states use exit 0.
> >Y=yes, N=no, NX=no but there are no places with multiple exits at the
> >beginning that would favor an exit 0, NP=no, takes the number at a
terminus
> >junction from the intersecting road
>
>
> >OR N
>
> Oregon could be marked NX, there are a couple places where an exit 0
> would work. OR-22 in Salem has Exits 1A and 1B at the spot where the
> I-5 interchange/overpass is, which is also the point where the mile
> posts start from zero and increment in both the east & west
> directions. I think US-26 in Portland may be mileposted like this too.
Actually NX would mean that there's no place like that.
### Chris Lawrence
Feb 1, 2002, 12:51:34 AM2/1/02
to
On Thu, 31 Jan 2002 08:47:57 -0600, Patrick L. Humphrey wrote:
> Isn't there still an Exit 0 on the Purchase Parkway in Kentucky?
Yes. Can't remember what it's for, though :)
Chris
--
Chris Lawrence <ch...@lordsutch.com> - http://www.lordsutch.com/chris/
### Patrick L. Humphrey
Feb 1, 2002, 1:21:32 AM2/1/02
to
"SPUI" <sp...@mit.BUTIDONTLIKESPeduAM> writes:
That might be why I-45 doesn't have an Exit 0 on the Island, even though
the 0 mile point is where the on-ramps from 61st Street hit the end of I-45
and the start of Texas 87 (a/k/a Avenue J, or Broadway). For what it's
worth, the only exits on the Island are 1A (61st), 1B (71st), and 1C
(Teichman Road/Harborside Blvd) -- the last of which is just a few hundred
feet before the Mile 2 post. If the exits were numbered correctly, 61st
Street would be Exit 0, 71st would be Exit 1A, and Teichman/Harborside
would be Exit 1B.) Mile 2 occurs just before the approach to the northbound
Causeway back to the mainland, at which point the numbering manages to
follow the mile markers more or less correctly.
--PLH, I-45 has a long Mile 1, but nothing like I-40's Mile 1 in Memphis :-)
Feb 1, 2002, 1:32:55 AM2/1/02
to
>> what is exit 0 in alabama?
>
>I-65 has exit 0 at I-10 and I think I-85 has exit 0 at I-65.
The end of I-85 has no exit tabs.
I-65's South end is indeed signed as Exit 0.
Alex
--
Highway Kick-off Page:
### Patrick L. Humphrey
Feb 1, 2002, 1:45:02 AM2/1/02
to
Chris Lawrence <qua...@watervalley.net> writes:
>On Thu, 31 Jan 2002 08:47:57 -0600, Patrick L. Humphrey wrote:
>> Isn't there still an Exit 0 on the Purchase Parkway in Kentucky?
>Yes. Can't remember what it's for, though :)
I'm even with you on that one -- I've never taken that exit to see what it's
for, myself. :-)
--PLH, who's usually headed towards more important places in the
Commonwealth, as it is
### Alan Hamilton
Feb 1, 2002, 2:46:33 AM2/1/02
to
On 31 Jan 2002 18:12:43 GMT, tuls...@aol.com (Tulsarama) wrote:
>>You have Texas listed as using Exit 0 -- but where?
>
>Isn't the Glenrio exit on I-40 Exit 0?
Yes: http://www.arizonaroads.com/us70/906p.htm (bottom of the page)
--
/
/ * / Alan Hamilton
### John David Galt
Feb 1, 2002, 11:32:32 AM2/1/02
to
ISTR that US 101 in WA has a freeway section near its end in Olympia.
Does it have exit numbers, and if so, which way do they go? (101 is
unique in having two "south ends"; if it were mileposted in WA, one of
the ends would have to be done backwards.)
### SPUI
Feb 1, 2002, 11:46:35 AM2/1/02
to
"John David Galt" <j...@diogenes.sacramento.ca.us> wrote in message
news:3C5AC320...@diogenes.sacramento.ca.us...
No exit numbers, but mileposts go towards I-5.
### Jeffrey Coleman Carlyle
Feb 1, 2002, 9:40:33 PM2/1/02
to
"Patrick L. Humphrey" <pat...@io.com> wrote in message
news:szkit9i...@hagbard.io.com...
> "Jeffrey Coleman Carlyle" <jef...@carlyle.org> writes:
>
> >"Patrick I. Humphrey" <pat...@io.com> wrote in message
> >news:szk1yg6...@hagbard.io.com...
> >> Isn't there still an Exit 0 on the Purchase Parkway in Kentucky? (I seem to
> >> remember one when I was up that way 18 months ago.)
>
> >The beginning of Kentucky's parkways are usually given the "Exit 0"
> >number.
>
> ...at the south or west end, I'd presume, but the Pennyrile starts at Exit 7
> (because of the original plan to extend it south to I-24 from US 41 south of
> Hopkinsville), the Cumberland doesn't have an Exit 0 at I-65, and I don't
> remember an Exit 0 on the Daniel Boone the last time I traveled it six years
> ago...
Looks like my memory failed me. I thought that the beginning of the
Natcher Parkway and Cumberland Parkway were both numbered exit 0. The
Audubon Parkway at the Pennyrile _IS_ numbered 0. As is the afore
mentioned exit on the Purchase. The other parkways appear to use 1 as
their first exit, also with the other mentioned exception of the
Pennyrile. Also as previously mentioned in this newsgroup, planning
appears to be moving ahead on constructing an extension of the Pennyrile
to I-24.
--
// Jeffrey Coleman Carlyle: Graduate Student in Computer Science at the
// University of Illinois at Urbana-Champaign; Creator of StratoSetup,
// Windows Restart, comp.os.msdos.programmer FAQ; THE RULER OF EARTH!
### SPUI
Feb 1, 2002, 9:58:42 PM2/1/02
to
"SPUI" <sp...@mit.BUTIDONTLIKESPeduAM> wrote in message
news:3c591279\$0\$3959\$b45e...@senator-bedfellow.mit.edu...
> PA N
Hmmm - looks like PA 66 has exit 0. That's Turnpike owned though - no
PennDOT roads appear to have one.
--
Dan Moraseski - 14th grade at MIT
http://spui.cjb.net/ - FL NJ MA route logs and exit lists
"Boston really isn't high on the importance scale for numbered routes
because we don't see the need to waste taxpayer money on it. Thru traffic
uses highways, not numbered surface routes. This isn't the 1950s." - some
MassHighway MassHole to Shawn De Cesari
### Patrick L. Humphrey
Feb 2, 2002, 12:21:25 AM2/2/02
to
"Jeffrey Coleman Carlyle" <jef...@carlyle.org> writes:
>"Patrick L. Humphrey" <pat...@io.com> wrote in message
>news:szkit9i...@hagbard.io.com...
>> "Jeffrey Coleman Carlyle" <jef...@carlyle.org> writes:
>>
>> >"Patrick I. Humphrey" <pat...@io.com> wrote in message
>> >news:szk1yg6...@hagbard.io.com...
>> >> Isn't there still an Exit 0 on the Purchase Parkway in Kentucky? (I
>> >> seem to remember one when I was up that way 18 months ago.)
>> >The beginning of Kentucky's parkways are usually given the "Exit 0"
>> >number.
>> ...at the south or west end, I'd presume, but the Pennyrile starts at Exit 7
>> (because of the original plan to extend it south to I-24 from US 41 south of
>> Hopkinsville), the Cumberland doesn't have an Exit 0 at I-65, and I don't
>> remember an Exit 0 on the Daniel Boone the last time I traveled it six years
>> ago...
>Looks like my memory failed me. I thought that the beginning of the
>Natcher Parkway and Cumberland Parkway were both numbered exit 0. The
>Audubon Parkway at the Pennyrile _IS_ numbered 0. As is the afore
>mentioned exit on the Purchase. The other parkways appear to use 1 as
>their first exit, also with the other mentioned exception of the
>Pennyrile. Also as previously mentioned in this newsgroup, planning
>appears to be moving ahead on constructing an extension of the Pennyrile
>to I-24.
The Pennyrile and I-24 finally meeting? That's got to be one of the Seven
Signs. (I spent a lot of my childhood summers a little farther west in Trigg
County, having great-grandparents there, and remember the hoopla when the
Pennyrile opened -- finally, you could get from Hopkinsville to somewhere else
on something *other* than a *&^%\$#@! two-lane road, even if it was only
Madisonville and Henderson. :-) I've never been on the Natcher, but since US
68 is Exit 5 on it, and I-65 is about five miles southeast of there, I'd
wonder just how the exit at I-65 is numbered. The Cumberland, as you
mentioned, uses 1 for its end at I-65 -- and who knows what it'll be numbered
as if it's designated as I-66. (That's assuming I-66 winds up extending west
of I-65.)
--PLH, who'll settle for a routing of I-69 to be decided so I can start
figuring the exit numbers in the Houston area
### Jeffrey Coleman Carlyle
Feb 2, 2002, 12:52:31 AM2/2/02
to
"Patrick L. Humphrey" <pat...@io.com> wrote:
> The Pennyrile and I-24 finally meeting? That's got to be one of the Seven
> Signs. (I spent a lot of my childhood summers a little farther west in Trigg
> County, having great-grandparents there, and remember the hoopla when the
> Pennyrile opened -- finally, you could get from Hopkinsville to somewhere else
> on something *other* than a *&^%\$#@! two-lane road, even if it was only
Yup, it looks like it might happen. Here are some links to information
http://www.kytc.state.ky.us/proserv/bull/Bull99-5/2-100.htm
news.html+20010131
news.html+20010628
And there was \$1 million set aside for it in the 2002 U.S. Transportation
spending bill.
> I've never been on the Natcher, but since US 68 is Exit 5 on it, and I-65
> is about five miles southeast of there, I'd wonder just how the exit at I-65
> is numbered.
According to this map, it is exit 1:
http://www.kytc.state.ky.us/Traffic_Center/counties/gif/warren98.gif
> The Cumberland, as you mentioned, uses 1 for its end at I-65 -- and who
> knows what it'll be numbered as if it's designated as I-66. (That's
> assuming I-66 winds up extending west of I-65.)
Again there is already planning studies for choosing a route west of the
Cumberland Parkway. The Bowling Green Daily News has found that residents
generally like the plans for a new road. There is a log of information
--
// Jeffrey Coleman Carlyle: Graduate Student in Computer Science at the
// University of Illinois at Urbana-Champaign; Creator of StratoSetup,
// Windows Restart, comp.os.msdos.programmer FAQ; THE RULER OF EARTH!
> "Jeffrey Coleman Carlyle" <jef...@carlyle.org> writes:
>
> >"Patrick L. Humphrey" <pat...@io.com> wrote in message
> >news:szkit9i...@hagbard.io.com...
> >> "Jeffrey Coleman Carlyle" <jef...@carlyle.org> writes:
> >>
> >> >"Patrick I. Humphrey" <pat...@io.com> wrote in message
> >> >news:szk1yg6...@hagbard.io.com...
>
> >> >> Isn't there still an Exit 0 on the Purchase Parkway in Kentucky? (I
> >> >> seem to remember one when I was up that way 18 months ago.)
>
> >> >The beginning of Kentucky's parkways are usually given the "Exit 0"
> >> >number.
>
> >> ...at the south or west end, I'd presume, but the Pennyrile starts at Exit 7
> >> (because of the original plan to extend it south to I-24 from US 41 south of
> >> Hopkinsville), the Cumberland doesn't have an Exit 0 at I-65, and I don't
> >> remember an Exit 0 on the Daniel Boone the last time I traveled it six years
> >> ago...
>
> >Looks like my memory failed me. I thought that the beginning of the
> >Natcher Parkway and Cumberland Parkway were both numbered exit 0. The
> >Audubon Parkway at the Pennyrile _IS_ numbered 0. As is the afore
> >mentioned exit on the Purchase. The other parkways appear to use 1 as
> >their first exit, also with the other mentioned exception of the
> >Pennyrile. Also as previously mentioned in this newsgroup, planning onstructing an extension of the Pennyrile
> >to I-24.
>
>
### Patrick L. Humphrey
Feb 2, 2002, 1:17:46 AM2/2/02
to
"Jeffrey Coleman Carlyle" <jef...@carlyle.org> writes:
>"Patrick L. Humphrey" <pat...@io.com> wrote:
>> The Pennyrile and I-24 finally meeting? That's got to be one of the Seven
>> Signs. (I spent a lot of my childhood summers a little farther west in
>> Trigg County, having great-grandparents there, and remember the hoopla when
>> the Pennyrile opened -- finally, you could get from Hopkinsville to
>> somewhere else on something *other* than a *&^%\$#@! two-lane road, even if
>> it was only Madisonville and Henderson. :-)
Thanks for the updates...this is one of those things you've seen planned for
years, but nothing ever happens, and just when you figure it's been relegated
to history's dustbin, it pops back up.
>And there was \$1 million set aside for it in the 2002 U.S. Transportation
>spending bill.
That'll pay for a sliver of the development at 41 that will have to be
relocated.
>> I've never been on the Natcher, but since US 68 is Exit 5 on it, and I-65
>> is about five miles southeast of there, I'd wonder just how the exit at I-65
>> is numbered.
Yep...that's my native Commonwealth, all right -- consistently inconsistent.
>> The Cumberland, as you mentioned, uses 1 for its end at I-65 -- and who
>> knows what it'll be numbered as if it's designated as I-66. (That's
>> assuming I-66 winds up extending west of I-65.)
>Again there is already planning studies for choosing a route west of the
>Cumberland Parkway. The Bowling Green Daily News has found that residents
>generally like the plans for a new road. There is a log of information
I'm sure the people in Bowling Green would, because at present there's no real
access around the area to the north of town, and the improved US 68 doesn't
kick in until you're out past the Natcher. What happens out west of there,
though? I-66 could be justified to I-24 in Trigg County, maybe, but after
that, it'd be a stretch, at best. 68 doesn't have the AADT to justify an
Interstate corridor (of course, that could partly be because it's still two
lanes from Cadiz westward, including across the LBL), but maybe if it could
share I-24 to Paducah and then split off westward into Missouri along the
I-57/US 60 corridor and end up at I-44 in Springfield...that might be doable.
(I'm not expecting to live long enough to see that.)
--PLH, then again, I didn't think I'd live to see US 119 finished from
Pineville to Cumberland, either
### Jason Hancock
Feb 3, 2002, 2:28:34 PM2/3/02
to
Neil Alexander Bratney wrote:
>
> Iowa actually seems to use both forms. At the southern terminus of
> I-380 is Exit 0. See Jason Hancock's page:
>
> http://iowahwypix.tripod.com/ic/80-380b.jpg
>
> I don't know about the northern (Western?) terminus of I-74. Anyone
> else?
The last exits of I-74 and I-280 are unnumbered now, but until about
1988, both of them were Exits 1A-B. (And the US 6 exit for I-280 used
to be Exit 1C; it's now just Exit 1.)
--Jason <http://iowahighways.cjb.net/>
### Jeffrey Coleman Carlyle
Feb 3, 2002, 9:30:59 PM2/3/02
to
There are plans to improve US 60/KY 80 through the Land Between the
Lakes. Technically I-66 through the Land Between the Lakes is still a
possibility.
I've thought that the reconstruction of US 68/KY 80 from Bowling Green to
Cadiz should have been to I-standards, but unfortunately it was not. Now
there is a nice four lane road, but unfortunately most of it probably
wouldn't be useful for extending I-66 along a more southern corridor.
corridor.
There is a little bit about US 68/KY 80 and LBL at:
http://www.kytc.state.ky.us/Features/Land_Between_The_Lakes.htm
--
// Jeffrey Coleman Carlyle: Graduate Student in Computer Science at the
// University of Illinois at Urbana-Champaign; Creator of StratoSetup,
// Windows Restart, comp.os.msdos.programmer FAQ; THE RULER OF EARTH!
> "Jeffrey Coleman Carlyle" <jef...@carlyle.org> writes:
> is map, it is exit 1:
> >http://www.kytc.state.ky.us/Traffic_Center/counties/gif/warren98.gif
>
> Yep...that's my native Commonwealth, all right -- consistently inconsistent.
>
> >> The Cumberland, as you mentioned, uses 1 for its end at I-65 -- and who
> >> knows what it'll be numbered as if it's designated as I-66. (That's
> >> assuming I-66 winds up extending west of I-65.)
>
> >Again there is already planning studies for choosing a route west of the
> >Cumberland Parkway. The Bowling Green Daily News has found that residents
> >generally like the plans for a new road. There is a log of information
> >about I-66 in Kentucky here:
>
>
>
### John David Galt
Feb 3, 2002, 11:48:57 PM2/3/02
to
SPUI wrote:
> Mile-based numbering often presents a problem: to use or not to use exit 0.
> If it is not used, you usually have at least 1.5 miles of exit 1*, or you
> modify it so exit 1* is only before mile, exit 2* is before mile 2, etc.
A somewhat related question: which states will "stretch" the numbering
(for example, number three exits between mileposts 15 and 16 as 15,16,17
rather than 16A,B,C if it won't create conflicts)? NM does this repeatedly
and I thought it was a neat idea, though of course it will backfire if that
area builds up enough to create a conflict later on).
### Jeff Morrison
Feb 4, 2002, 12:38:08 AM2/4/02
to
I-380 has Exit 0A-B at its intersection with I-80. I-80 SHOULD have the I-29
north exit be 0, but it's not. I-29, I-74, and I-280 all have at least one
mile before their first exit; I-35 uses Missouri's 114 because it's on the
line. I do not know about I-680, I-235, and I-129, all of which have an exit
before the first mile marker.
Personally, I'm a fan of it. If the west or south terminus is an interchange,
it should be Exit Zero.
SPUI wrote:
> Mile-based numbering often presents a problem: to use or not to use exit 0.
> If it is not used, you usually have at least 1.5 miles of exit 1*, or you
> modify it so exit 1* is only before mile, exit 2* is before mile 2, etc.
> Here is a full list of which states use exit 0.
> Y=yes, N=no, NX=no but there are no places with multiple exits at the
> beginning that would favor an exit 0, NP=no, takes the number at a terminus
> junction from the intersecting road
>
>
> IA N
### Stanley Cline
Feb 4, 2002, 12:53:21 AM2/4/02
to
On Sun, 03 Feb 2002 20:48:57 -0800, John David Galt
<j...@diogenes.sacramento.ca.us> wrote:
>A somewhat related question: which states will "stretch" the numbering
>(for example, number three exits between mileposts 15 and 16 as 15,16,17
>rather than 16A,B,C if it won't create conflicts)? NM does this repeatedly
GDOT has done it in at least one case I know of -- I-285 between GA
400 and Ashford-Dunwoody Rd. The interchange for A-D Rd is Exit 29
even though it's *west* of mile marker 28! (The eastern end of the
eastern ramps are around mile marker 27.8; the western end of the
western ramps are around MM 27.2 or so.) GA 400 itself is exit 27;
Peachtree-Dunwoody Rd (half-diamond) is exit 28.
-SC
--
Stanley Cline -- sc1 at roamer1 dot org -- http://www.roamer1.org/
...
"Never put off until tomorrow what you can do today. There might
be a law against it by that time." -/usr/games/fortune
### Steve
Feb 4, 2002, 4:07:33 AM2/4/02
to
The GSP does it, but that's independent of NJDOT I believe. I-80 around
Paterson avoids lettered exits except for the same road or the same ramp
complex, so exits 56-61 are all pretty close to each other.
### Patrick L. Humphrey
Feb 4, 2002, 8:22:58 AM2/4/02
to
"Jeffrey Coleman Carlyle" <jef...@carlyle.org> writes:
>"Patrick L. Humphrey" <pat...@io.com> wrote:
>
>> I'm sure the people in Bowling Green would, because at present there's no
>> real access around the area to the north of town, and the improved US 68
>> doesn't kick in until you're out past the Natcher. What happens out west
>> of there, though? I-66 could be justified to I-24 in Trigg County, maybe,
>> but after that, it'd be a stretch, at best. 68 doesn't have the AADT to
>> justify an Interstate corridor (of course, that could partly be because
>> it's still two lanes from Cadiz westward, including across the LBL), but
>> maybe if it could share I-24 to Paducah and then split off westward into
>> Missouri along the I-57/US 60 corridor and end up at I-44 in
>> Springfield...that might be doable. (I'm not expecting to live long enough
>> to see that.)
>There are plans to improve US 60/KY 80 through the Land Between the
>Lakes. Technically I-66 through the Land Between the Lakes is still a
>possibility.
It's a possibility, but I'd hate to see LBL get split by it when the
justification isn't yet there. (Considering where it would be, think of the
environmental paperwork that would have to be done...my four-greats
grandchildren *might* live to see it done.)
>I've thought that the reconstruction of US 68/KY 80 from Bowling Green to
>Cadiz should have been to I-standards, but unfortunately it was not. Now
>there is a nice four lane road, but unfortunately most of it probably
>wouldn't be useful for extending I-66 along a more southern corridor.
>corridor.
At least the only challenges there are 1) the ROW for I-66 between I-65 and the
WK, and 2) getting the WK rebuilt to Interstate standards. Having grown up
making the 68/80 trip from London to Cadiz (or vice versa) once and sometimes
twice a summer, I can remember the nightmare it used to be -- though building
it to Interstate standards around Cadiz would have wiped out where my
great-aunt and uncle lived for years on the east edge of town, right where the
bypass meets the old route into town. (As it is, the new road is almost at
the front door, which means it's about 250 feet north of where the old road
ran.)
>There is a little bit about US 68/KY 80 and LBL at:
>http://www.kytc.state.ky.us/Features/Land_Between_The_Lakes.htm
I'll have to check that out...and if 68/80 get rebuilt between Cadiz and LBL,
I hope they make allowances to at least preserve an exit at Pete Light
Springs. :-)
--PLH, who's come a *long* way
### Jeffrey Coleman Carlyle
Feb 4, 2002, 1:06:40 PM2/4/02
to
"Patrick L. Humphrey" <pat...@io.com> wrote in message
news:szksn8h...@hagbard.io.com...
One of the reasons for rebuilding US 68/KY 80 through LBL is that the
bridges over the Lake Barkley and Kentucky Lake are getting very old and
probably wouldn't survive an earthquake on the New Madrid fault, so there
is interest in replacing the bridges with earthquake resistant
structures.
There was a study a couple of years ago that said the Purchase area would
be cut off completely from the rest of the Kentucky if there was a major
earthquake in the region because none of the bridges (including I-24)
across the lakes and rivers would survive an earthquake.
### Patrick L. Humphrey
Feb 5, 2002, 7:55:34 AM2/5/02
to
"Jeffrey Coleman Carlyle" <jef...@carlyle.org> writes:
>"Patrick L. Humphrey" <pat...@io.com> wrote in message
>news:szksn8h...@hagbard.io.com...
>> "Jeffrey Coleman Carlyle" <jef...@carlyle.org> writes:
>> >There is a little bit about US 68/KY 80 and LBL at:
>> >http://www.kytc.state.ky.us/Features/Land_Between_The_Lakes.htm
>> I'll have to check that out...and if 68/80 get rebuilt between Cadiz and
>> LBL, I hope they make allowances to at least preserve an exit at Pete Light
>> Springs. :-)
>One of the reasons for rebuilding US 68/KY 80 through LBL is that the
>bridges over the Lake Barkley and Kentucky Lake are getting very old and
>probably wouldn't survive an earthquake on the New Madrid fault, so there
>is interest in replacing the bridges with earthquake resistant
>structures.
That's something that's been needed since around 1812, when both of the 68/80
LBL spans appear to have been built. (The last time I went through there in
July of 2000, I was worried mainly about anything larger than a Saturn SL2
trying to cross either span from the opposite direction -- those bridges are
old and narrow.)
>There was a study a couple of years ago that said the Purchase area would
>be cut off completely from the rest of the Kentucky if there was a major
>earthquake in the region because none of the bridges (including I-24)
>across the lakes and rivers would survive an earthquake.
That wouldn't be surprising, since the origins of the New Madrid activity
weren't really known until the last 20 years of the last century, and most of
the infrastructure in the Purchase and adjacent Tennessee and Missouri was
built prior to that. So far, I've been lucky to not be within 50 miles or so
when even a 3.5 occurs...
--PLH, down here on the Gulf Coast in a part of Texas that hasn't been
seismically active in a few million years
### Jeff Morrison
Feb 8, 2002, 3:40:04 PM2/8/02
to
Neil Alexander Bratney wrote:
> I-29, I-129 and I-480 all have exits less than 1/2 mile from the border.
> They are all named exit 1.
<nitpicking>
I-29 Exit 1 (IA 333) is indeed Exit 1, as my copy of Street Atlas has 1.75
miles of I-29 in IA before it. Right on the other two, though.
</nitpicking>
### Sherman Potter
Feb 8, 2002, 3:33:36 PM2/8/02
to
"SPUI" <sp...@mit.BUTIDONTLIKESPeduAM> wrote in message
news:3c591279\$0\$3959\$b45e...@senator-bedfellow.mit.edu...
> Mile-based numbering often presents a problem: to use or not to use exit
0.
> If it is not used, you usually have at least 1.5 miles of exit 1*, or you
> modify it so exit 1* is only before mile, exit 2* is before mile 2, etc.
> Here is a full list of which states use exit 0.
> Y=yes, N=no, NX=no but there are no places with multiple exits at the
> beginning that would favor an exit 0, NP=no, takes the number at a
terminus
> junction from the intersecting road
>
> AL Y
> AZ N
> AR NP
> CO N (I-270 is an exception since that was extended)
> FL N
> GA N
The ramp from I-16W to I-75S in Macon is signed as "Exit 0."
Feb 9, 2002, 5:18:13 AM2/9/02
to
>
>The ramp from I-16W to I-75S in Macon is signed as "Exit 0."
<ARG!!>
No it's not. It is signed as Exit 1.
<ARG!!> | 10,775 | 35,139 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2021-49 | longest | en | 0.937962 |
http://mathoverflow.net/questions/tagged/generating-functions?sort=votes&pagesize=15 | 1,466,825,374,000,000,000 | text/html | crawl-data/CC-MAIN-2016-26/segments/1466783392069.78/warc/CC-MAIN-20160624154952-00120-ip-10-164-35-72.ec2.internal.warc.gz | 189,081,138 | 26,327 | # Tagged Questions
The tag has no usage guidance.
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Inspired by this question, I have been wondering if there are any useful generating functions with all non-zero coefficients equal to one. Obviously, the trivial generating function $\frac{1}{1-x}$ ... | 894 | 3,320 | {"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} | 3.578125 | 4 | CC-MAIN-2016-26 | latest | en | 0.798886 |
https://justaaa.com/statistics-and-probability/263693-the-lengths-of-a-particular-animals-pregnancies | 1,713,400,249,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817184.35/warc/CC-MAIN-20240417235906-20240418025906-00062.warc.gz | 296,014,406 | 9,307 | Question
# The lengths of a particular animal's pregnancies are approximately normally distributed, with mean muequals276 days and...
The lengths of a particular animal's pregnancies are approximately normally distributed, with mean muequals276 days and standard deviation sigmaequals16 days. (a) What proportion of pregnancies lasts more than 300 days? (b) What proportion of pregnancies lasts between 268 and 288 days? (c) What is the probability that a randomly selected pregnancy lasts no more than 248 days? (d) A "very preterm" baby is one whose gestation period is less than 240 days. Are very preterm babies unusual?
The proportion of pregnancies that last more than 300 days is nothing. (Round to four decimal places as needed.)
(b) The proportion of pregnancies that last between 268 and 288 days is nothing. (Round to four decimal places as needed.)
(c) The probability that a randomly selected pregnancy lasts no more than 248 days is nothing. (Round to four decimal places as needed.)
(d) The probability of a "very preterm" baby is nothing. This event ▼ would be would not be unusual because the probability is ▼ greater less than 0.05. (Round to four decimal places as needed.)
#### Earn Coins
Coins can be redeemed for fabulous gifts. | 296 | 1,283 | {"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} | 2.640625 | 3 | CC-MAIN-2024-18 | latest | en | 0.90316 |
http://mathhelpforum.com/calculus/160395-surface-volume-integration-print.html | 1,519,278,489,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891814002.69/warc/CC-MAIN-20180222041853-20180222061853-00287.warc.gz | 233,619,832 | 3,465 | # surface/volume integration
• Oct 20th 2010, 10:01 AM
floater
surface/volume integration
Why is it that the surface of revolution integral contains the arc lengh bit
but the volume of revolution doesnt?
• Oct 22nd 2010, 03:50 AM
mr fantastic
Quote:
Originally Posted by floater
Why is it that the surface of revolution integral contains the arc lengh bit
but the volume of revolution doesnt?
I suggest you review the derivation of both formulae in either an appropriate textbook or at an appropriate website.
• Oct 22nd 2010, 08:02 AM
HallsofIvy
When you calculate volume, you are approximating the area that is rotated around the axis by a thin rectangle and can use the y value at any point to approximate the height of the rectangle. But when you calculate surface area, it is the slant height that has to be rotated around the axis, not the horizontal length of each small section.
If you take the line from (0, 0) to (1, 1), y= x, you can approximate the area under it taking thin rectangles with width $\Delta x$ and height y= x for x any value in that interval. The total area is [tex]\sum x_i \Delta x[tex] that in the limit becomes $\int x dx$ which gives the correct area.
But if you attempt to approximate the length of the line by using little horizontal sections, the length will always total to 1, not the correct length, no matter how many sections you take. To get the correct length, from $x_i$ to $x_i+ \Delta x$, you have to use the actual length of the line, $\sqrt{(x_i+ \Delta x- x_i)^2+ (y_{i+1}- y_i)^2}= \sqrt{x_i^2+ x_i}= x_i\sqrt{2}$. | 412 | 1,568 | {"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": 5, "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-2018-09 | longest | en | 0.87353 |
http://www.jiskha.com/display.cgi?id=1365427440 | 1,462,513,036,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461861727712.42/warc/CC-MAIN-20160428164207-00196-ip-10-239-7-51.ec2.internal.warc.gz | 589,071,223 | 3,716 | Friday
May 6, 2016
# Homework Help: trig
Posted by thapelo on Monday, April 8, 2013 at 9:24am.
if sin 41'=m. write m in terms of cos 41'
• trig - Reiny, Monday, April 8, 2013 at 9:47am
we know
sin^2 41° + cos^2 41° = 1
m^2 = 1 - cos^2 41°
m = ±√(1 - cos^2 41°) | 125 | 264 | {"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} | 3.375 | 3 | CC-MAIN-2016-18 | longest | en | 0.878678 |
https://www.weegy.com/?ConversationId=A0DYGTOE | 1,558,626,420,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232257259.71/warc/CC-MAIN-20190523143923-20190523165923-00193.warc.gz | 1,001,076,789 | 8,573 | Find the product (n 3)2
The product of (n 3)2 is equal to 6n. [smile]
Question
Asked 4/22/2014 10:03:02 PM
Updated 4/22/2014 10:42:08 PM
This answer has been confirmed as correct and helpful.
Confirmed by yumdrea [4/22/2014 10:23:00 PM], Followed by yeswey, Unfollowed by yeswey
Rating
3
(n^3)^2 = n^6
Added 4/22/2014 10:40:44 PM
This answer has been added to the Weegy Knowledgebase
Is this answer correct? I know the answer was got some an online calculator but i dont think the user mean that. Coefficient is normally put before the variables. I believe 2 and 3 are exponents here.
Added 4/22/2014 10:42:08 PM
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Home | Contact | Blog | About | Terms | Privacy | © Purple Inc. | 768 | 2,197 | {"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} | 2.703125 | 3 | CC-MAIN-2019-22 | latest | en | 0.880793 |
http://isabelle.in.tum.de/library/HOL/FunDef.html | 1,369,009,024,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368698150793/warc/CC-MAIN-20130516095550-00060-ip-10-60-113-184.ec2.internal.warc.gz | 148,072,679 | 4,494 | # Theory FunDef
Up to index of Isabelle/HOL
theory FunDef
imports Partial_Function SAT Wellfounded
`(* Title: HOL/FunDef.thy Author: Alexander Krauss, TU Muenchen*)header {* Function Definitions and Termination Proofs *}theory FunDefimports Partial_Function SAT Wellfoundedkeywords "function" "termination" :: thy_goal and "fun" :: thy_declbeginsubsection {* Definitions with default value. *}definition THE_default :: "'a => ('a => bool) => 'a" where "THE_default d P = (if (∃!x. P x) then (THE x. P x) else d)"lemma THE_defaultI': "∃!x. P x ==> P (THE_default d P)" by (simp add: theI' THE_default_def)lemma THE_default1_equality: "[|∃!x. P x; P a|] ==> THE_default d P = a" by (simp add: the1_equality THE_default_def)lemma THE_default_none: "¬(∃!x. P x) ==> THE_default d P = d" by (simp add:THE_default_def)lemma fundef_ex1_existence: assumes f_def: "f == (λx::'a. THE_default (d x) (λy. G x y))" assumes ex1: "∃!y. G x y" shows "G x (f x)" apply (simp only: f_def) apply (rule THE_defaultI') apply (rule ex1) donelemma fundef_ex1_uniqueness: assumes f_def: "f == (λx::'a. THE_default (d x) (λy. G x y))" assumes ex1: "∃!y. G x y" assumes elm: "G x (h x)" shows "h x = f x" apply (simp only: f_def) apply (rule THE_default1_equality [symmetric]) apply (rule ex1) apply (rule elm) donelemma fundef_ex1_iff: assumes f_def: "f == (λx::'a. THE_default (d x) (λy. G x y))" assumes ex1: "∃!y. G x y" shows "(G x y) = (f x = y)" apply (auto simp:ex1 f_def THE_default1_equality) apply (rule THE_defaultI') apply (rule ex1) donelemma fundef_default_value: assumes f_def: "f == (λx::'a. THE_default (d x) (λy. G x y))" assumes graph: "!!x y. G x y ==> D x" assumes "¬ D x" shows "f x = d x"proof - have "¬(∃y. G x y)" proof assume "∃y. G x y" hence "D x" using graph .. with `¬ D x` show False .. qed hence "¬(∃!y. G x y)" by blast thus ?thesis unfolding f_def by (rule THE_default_none)qeddefinition in_rel_def[simp]: "in_rel R x y == (x, y) ∈ R"lemma wf_in_rel: "wf R ==> wfP (in_rel R)" by (simp add: wfP_def)ML_file "Tools/Function/function_common.ML"ML_file "Tools/Function/context_tree.ML"ML_file "Tools/Function/function_core.ML"ML_file "Tools/Function/sum_tree.ML"ML_file "Tools/Function/mutual.ML"ML_file "Tools/Function/pattern_split.ML"ML_file "Tools/Function/relation.ML"method_setup relation = {* Args.term >> (fn t => fn ctxt => SIMPLE_METHOD' (Function_Relation.relation_infer_tac ctxt t))*} "prove termination using a user-specified wellfounded relation"ML_file "Tools/Function/function.ML"ML_file "Tools/Function/pat_completeness.ML"method_setup pat_completeness = {* Scan.succeed (SIMPLE_METHOD' o Pat_Completeness.pat_completeness_tac)*} "prove completeness of datatype patterns"ML_file "Tools/Function/fun.ML"ML_file "Tools/Function/induction_schema.ML"method_setup induction_schema = {* Scan.succeed (RAW_METHOD o Induction_Schema.induction_schema_tac)*} "prove an induction principle"setup {* Function.setup #> Function_Fun.setup*}subsection {* Measure Functions *}inductive is_measure :: "('a => nat) => bool"where is_measure_trivial: "is_measure f"ML_file "Tools/Function/measure_functions.ML"setup MeasureFunctions.setuplemma measure_size[measure_function]: "is_measure size"by (rule is_measure_trivial)lemma measure_fst[measure_function]: "is_measure f ==> is_measure (λp. f (fst p))"by (rule is_measure_trivial)lemma measure_snd[measure_function]: "is_measure f ==> is_measure (λp. f (snd p))"by (rule is_measure_trivial)ML_file "Tools/Function/lexicographic_order.ML"method_setup lexicographic_order = {* Method.sections clasimp_modifiers >> (K (SIMPLE_METHOD o Lexicographic_Order.lexicographic_order_tac false))*} "termination prover for lexicographic orderings"setup Lexicographic_Order.setupsubsection {* Congruence Rules *}lemma let_cong [fundef_cong]: "M = N ==> (!!x. x = N ==> f x = g x) ==> Let M f = Let N g" unfolding Let_def by blastlemmas [fundef_cong] = if_cong image_cong INT_cong UN_cong bex_cong ball_cong imp_cong Option.map_cong Option.bind_conglemma split_cong [fundef_cong]: "(!!x y. (x, y) = q ==> f x y = g x y) ==> p = q ==> split f p = split g q" by (auto simp: split_def)lemma comp_cong [fundef_cong]: "f (g x) = f' (g' x') ==> (f o g) x = (f' o g') x'" unfolding o_apply .subsection {* Simp rules for termination proofs *}lemma termination_basic_simps[termination_simp]: "x < (y::nat) ==> x < y + z" "x < z ==> x < y + z" "x ≤ y ==> x ≤ y + (z::nat)" "x ≤ z ==> x ≤ y + (z::nat)" "x < y ==> x ≤ (y::nat)"by arith+declare le_imp_less_Suc[termination_simp]lemma prod_size_simp[termination_simp]: "prod_size f g p = f (fst p) + g (snd p) + Suc 0"by (induct p) autosubsection {* Decomposition *}lemma less_by_empty: "A = {} ==> A ⊆ B"and union_comp_emptyL: "[| A O C = {}; B O C = {} |] ==> (A ∪ B) O C = {}"and union_comp_emptyR: "[| A O B = {}; A O C = {} |] ==> A O (B ∪ C) = {}"and wf_no_loop: "R O R = {} ==> wf R"by (auto simp add: wf_comp_self[of R])subsection {* Reduction Pairs *}definition "reduction_pair P = (wf (fst P) ∧ fst P O snd P ⊆ fst P)"lemma reduction_pairI[intro]: "wf R ==> R O S ⊆ R ==> reduction_pair (R, S)"unfolding reduction_pair_def by autolemma reduction_pair_lemma: assumes rp: "reduction_pair P" assumes "R ⊆ fst P" assumes "S ⊆ snd P" assumes "wf S" shows "wf (R ∪ S)"proof - from rp `S ⊆ snd P` have "wf (fst P)" "fst P O S ⊆ fst P" unfolding reduction_pair_def by auto with `wf S` have "wf (fst P ∪ S)" by (auto intro: wf_union_compatible) moreover from `R ⊆ fst P` have "R ∪ S ⊆ fst P ∪ S" by auto ultimately show ?thesis by (rule wf_subset)qeddefinition "rp_inv_image = (λ(R,S) f. (inv_image R f, inv_image S f))"lemma rp_inv_image_rp: "reduction_pair P ==> reduction_pair (rp_inv_image P f)" unfolding reduction_pair_def rp_inv_image_def split_def by forcesubsection {* Concrete orders for SCNP termination proofs *}definition "pair_less = less_than <*lex*> less_than"definition "pair_leq = pair_less^="definition "max_strict = max_ext pair_less"definition "max_weak = max_ext pair_leq ∪ {({}, {})}"definition "min_strict = min_ext pair_less"definition "min_weak = min_ext pair_leq ∪ {({}, {})}"lemma wf_pair_less[simp]: "wf pair_less" by (auto simp: pair_less_def)text {* Introduction rules for @{text pair_less}/@{text pair_leq} *}lemma pair_leqI1: "a < b ==> ((a, s), (b, t)) ∈ pair_leq" and pair_leqI2: "a ≤ b ==> s ≤ t ==> ((a, s), (b, t)) ∈ pair_leq" and pair_lessI1: "a < b ==> ((a, s), (b, t)) ∈ pair_less" and pair_lessI2: "a ≤ b ==> s < t ==> ((a, s), (b, t)) ∈ pair_less" unfolding pair_leq_def pair_less_def by autotext {* Introduction rules for max *}lemma smax_emptyI: "finite Y ==> Y ≠ {} ==> ({}, Y) ∈ max_strict" and smax_insertI: "[|y ∈ Y; (x, y) ∈ pair_less; (X, Y) ∈ max_strict|] ==> (insert x X, Y) ∈ max_strict" and wmax_emptyI: "finite X ==> ({}, X) ∈ max_weak" and wmax_insertI: "[|y ∈ YS; (x, y) ∈ pair_leq; (XS, YS) ∈ max_weak|] ==> (insert x XS, YS) ∈ max_weak"unfolding max_strict_def max_weak_def by (auto elim!: max_ext.cases)text {* Introduction rules for min *}lemma smin_emptyI: "X ≠ {} ==> (X, {}) ∈ min_strict" and smin_insertI: "[|x ∈ XS; (x, y) ∈ pair_less; (XS, YS) ∈ min_strict|] ==> (XS, insert y YS) ∈ min_strict" and wmin_emptyI: "(X, {}) ∈ min_weak" and wmin_insertI: "[|x ∈ XS; (x, y) ∈ pair_leq; (XS, YS) ∈ min_weak|] ==> (XS, insert y YS) ∈ min_weak"by (auto simp: min_strict_def min_weak_def min_ext_def)text {* Reduction Pairs *}lemma max_ext_compat: assumes "R O S ⊆ R" shows "max_ext R O (max_ext S ∪ {({},{})}) ⊆ max_ext R"using assmsapply autoapply (elim max_ext.cases)apply ruleapply auto[3]apply (drule_tac x=xa in meta_spec)apply simpapply (erule bexE)apply (drule_tac x=xb in meta_spec)by autolemma max_rpair_set: "reduction_pair (max_strict, max_weak)" unfolding max_strict_def max_weak_defapply (intro reduction_pairI max_ext_wf)apply simpapply (rule max_ext_compat)by (auto simp: pair_less_def pair_leq_def)lemma min_ext_compat: assumes "R O S ⊆ R" shows "min_ext R O (min_ext S ∪ {({},{})}) ⊆ min_ext R"using assmsapply (auto simp: min_ext_def)apply (drule_tac x=ya in bspec, assumption)apply (erule bexE)apply (drule_tac x=xc in bspec)apply assumptionby autolemma min_rpair_set: "reduction_pair (min_strict, min_weak)" unfolding min_strict_def min_weak_defapply (intro reduction_pairI min_ext_wf)apply simpapply (rule min_ext_compat)by (auto simp: pair_less_def pair_leq_def)subsection {* Tool setup *}ML_file "Tools/Function/termination.ML"ML_file "Tools/Function/scnp_solve.ML"ML_file "Tools/Function/scnp_reconstruct.ML"setup {* ScnpReconstruct.setup *}ML_val -- "setup inactive"{* Context.theory_map (Function_Common.set_termination_prover (ScnpReconstruct.decomp_scnp_tac [ScnpSolve.MAX, ScnpSolve.MIN, ScnpSolve.MS]))*}end` | 2,934 | 8,832 | {"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 | 0.527543 |
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