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http://www.jiskha.com/members/profile/posts.cgi?name=jasmine&page=2 | 1,493,530,972,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917124299.47/warc/CC-MAIN-20170423031204-00018-ip-10-145-167-34.ec2.internal.warc.gz | 580,607,102 | 9,724 | # Posts by jasmine
Total # Posts: 949
math
Would I then have to subtract the area sin(t)-cos(t)?
math
Find the area of the region. (Round your answer to three decimal places.) between y = cos t and y = sin t for −π/2 ≤ t ≤ π/2
math
If the absolute value of a negative number is 2.78 what is the distance on the number line between the number and its absolute value
English 1
Um i have read the book thank you i just don't understand the questions
English 1
How is H. G. Wells’ fictional world in War of the Worlds similar to the real world?
Language arts.
Idk.... Lol..
Language arts.
Which is the best prediction based on the first three sentences of the selection?
Language Art's
It is "C"
math/economics in calculus
The average cost of manufacturing a quantity q of a good, is defined to be a(q) = C(q)/q. The average cost per item to produce q items is given by a(q) = 0.01q2 − 0.6q + 13, for q >0. I know that the total cost is 0.01q^3-0.6q^2+13q What is the minimum marginal cost? ...
math
The average cost of manufacturing a quantity q of a good, is defined to be a(q) = C(q)/q. The average cost per item to produce q items is given by a(q) = 0.01q2 − 0.6q + 13, for q >0. (a) What is the total cost, C(q),of producing q goods? For this do I just put the ...
math
The total cost C(q) of producing q goods is given by the following equation. C(q) = 0.01q^3 − 0.6q^2 + 13q What is the maximum profit if each item is sold for \$9? (Assume you sell everything you produce. Round your answer to the nearest cent.) I know the fixed cost is 0 ...
math
Thank you so much!
math
The second part of this questions asks: Find the quantity which maximizes the profit Please help me solve this part as well. I know that profit is maximized when the derivative of revenue equals the derivative of cost.. but how do I get there?
math
The revenue from selling q items is R(q) = 600q − q^2, and the total cost is C(q) = 150 + 50q. Write a function that gives the total profit earned. It is my understanding that profit = revenue - costs. So when I tried to write the function for total profit earned, I ...
math urgent help needed!!
I am so sorry but this homework is due in half an hour. Please help me! A smokestack deposits soot on the ground with a concentration inversely proportional to the square of the distance from the stack. With two smokestacks 20 miles apart, the concentration of the combined ...
Grammar
Its either the first one or the third one. Now you try to figure out which one it is. Good Luck! :) :-)
Resturant/Diner
Thanks Mary. Any other suggestions?
Resturant/Diner
Make up a name for a restaurant or a diner. Don't ask. Its for an English assignment! :):-)
Square Roots (Math)
Which two square roots are used to estimate √47?
Algebra 2
Did I translate this verbal phrase into an algebraic expression correctly? The product of a number and fifty-five 55x
Language Art's
It's A- Past Participle.
math
Explain how to use hundredths grids to subtract 1.65-0.98
Ap physics
A ball rolls down an incline with an acceleration of 10 cm/s^2. If it starts with an initial velocity of 0 cm/s and has a velocy of 50 cm/s when it reaches the bottom of the ramp, how long is the ramp?
Physics
A 0.3 kg tennis ball is travelling west with a speed of 4 m/s and bounces off a wall. After bouncing, the ball is travelling east at 2 m/s. The tennis ball was in contact with the wall for 0.004 seconds. a) What is the initial and final momentum of the ball? b) What is the ...
Physics
A 0.3 kg tennis ball is travelling west with a speed of 4 m/s and bounces off a wall. After bouncing, the ball is travelling east at 2 m/s. The tennis ball was in contact with the wall for 0.004 seconds. a) What is the initial and final momentum of the ball? b) What is the ...
grammar
Identify the sentence parts by placing the red abbreviations in their correct locations. Put parentheses around the prepositional phrases. Identify the subject, verb, and any complements (direct object, indirect object, subject complement, objective complement).
Math
There are 6 buses transporting students to a baseball game with 32 students On each buss. each row at thebaseball stadium seats eight students if the students fill up all the Rows of seats will the students need altogether.
Rectangle with length =11cm and width =4cm when both dimensions are trippled
physics
woops nvm, I was looking at another problem.
physics
THIS IS A CORRECTION FOR PART C, WHEN YOU ARE SUPPOSED TO CONVERT SECONDS TO METERS!!! THE REST IS CORRECT THOUGH. c. 12.8 m/s * 3.2s = 40.96 (NOT 48.5)
English
How do you diagram the sentence, "I will do my homework later tonight."
Algebra1
A parabola is defined as f(x)=a(x–8)2+9, where a is a positive real number. As a increases, What happens to the x-coordinate of the parabola’s vertex?
algebra 1
Let x^2-y^2=20 and x-y=2. What is the value of x+y?
Accounting
Assume that Esquire consulting erroneously recorded the payment of \$30,000 of dividends as salary expense. How would this error affect the equality of the accounting equation? How would this error affect the income statement, retained earnings statement balance sheet and ...
algebra
Find the point of equilibrium for a system that has a demand equation of p= 49.0,000 3x and a supply equation of p= 33+0.00002x
algebra
A total of \$80,000 is invested in two funds paying 2.3% and 3.1% simple interest . The total annual interest is \$2000. How much is invested in each fund?
Algebra
standard
math
you can rent a car for the day from company A for \$26 plus \$0.12 a mile company B charges \$24.00 plus \$ 0.19 a mile. Find the number of miles m (to the nearest mile) per day for which it is cheaper to rent from company A
Math
An item that sells for \$145.99 has a sales tax of 10.22 Find a mathematical model that gives the amount of sales tax y in terms of the retail price x Use the model to find the sales tax on a purchase that has a retail price x
Physics
An object is placed 20.0 cm to the left of a convex lens with a focal length of +8.0 cm. Where is the image of the object? A)28 cm to the right of the lens B)13 cm to the left of the lens C)13 cm to the right of the lens I think its A but I'm not sure
Science
math
City wide delivery service uses vans costing \$24800 each. How much will the company have to invest today to accumulate enough money to buy six new vans at the end of 4 years? City wides bank is currently paying 12% interest compounded quarterly
math
In the North Carolina Education Lottery "Pick Three Game" there are 75 balls in the machine. Your lucky numbers are 7, 14, and 21. The numbers must be drawn in that order. What's the probability that you'll win the \$500.00 prize?
math
In a pack of 52 cards, 4 of the cards are Kings. The first card is picked, the results are recorded and the card is returned to the deck. Then the second card is picked. Find the probability of picking two kings. Be sure to represent your answer as a fraction in simplest form.
math
A bag of marbles contains 8 red marbles and 6 yellow marbles. Two marbles are drawn out of the bag at random without replacement. What is the probability that two red marbles are drawn out? Be sure to represent your answer as a fraction in simplest form.
math
A bag of marbles contains 8 red marbles and 6 yellow marbles. Two marbles are drawn out of the bag at random without replacement. What is the probability that first a red marble then a yellow marble are drawn? Represent your answer as a fraction in simplest form.
math
Gary produces 52 more than 4times what Harold can produce in one day . If Gary's total production for the day is 202 how much does Harold produce in one day.
math
If interest on a \$4,500 loan is \$ 720 What would be the interest on a loan \$5,800
science-chemistry
if 1g of carbon dioxide is bubbled into 100g of a calcium hydroxide solution in a flask. what is the total mass of substance in the flask after the reaction is complete?
science
if 1g of carbon dioxide is bubbled into 100g of a calcium hydroxide solution in a flask. what is the total mass of substance in the flask after the reaction is complete?
maths
Write 5x-3 in positive index
English
I need help answering some To Kill A Mockingbird questions 1.When Scout is talking about the Ewells' living conditions what is her tone? Is it sympathetic or disparaging or objective? I think it's sympathetic 2. When the story is talking about Mayella 's red ...
Algebra 2
Can you please help me solve these problems? Thank You 1.x^2-y^2=25 x^2+y^2+7=0 2. x^2+y^2=8 5y^2=3x^2 3. 3x^2-2y^2=24 2y=-3x
English
What does this mean "beyond hope of help from any source" when describing the Ewells in To Kill A Mockingbird? It's not a quote from the book.
Algebra 2
Compare the graphs of the inverse variations by comparing asymptotes, similar points, lines of reflections, and quadrants. y=-0.7/x and y=-0.9/x
history Ms.sue
okay thanks
history Ms.sue
did George W. bush have kids? help it's a bio for my teacher
History
Ummm... yes and u would just bio of Someone.. -jasmine-
Math
math
A football team gained 4 in one play. In the next play, the team lost 4 yards. What is the team's net yard gain.
Yep they are all right but 10***
Math
It took Michael 15 minutes to walk from his house to his friends house would you measure the distance he walked in linear or square units ?
MATH miss.sue
kay thanks
MATH miss.sue
okay 4 friends had 19 Pieces of candy and how many pieces of candy will each friend get? HELP PLZ
health
In what way does your body's need for nutrients change when you are training for sports or competition?
Math
The bottom end of a 14 foot loading ramp is 10.7 feet away from the entrance to the building. Find the angle the ramp makes with the ground.
Math
If a ten foot tower casts a thirteen foot shadow on the ground, what is the angle of elevation of the sun?
Science
i took the test and got 100% the answers are C A D A A i swear these are correct from connexus academy
Art
thanks so much got 100% .,
MatLab
In Matlab Write a function file that determines the following sum (for all odd n from 1 to 11): y=Ó 2^n/n
MATH AGAIN~Ms. Sue
yes Leila ms sue is real i think
Physics
In a hydroelectric installation, a turbine delivers 1500 hp to a generator, which in turn transfers 80.0% of the mechanical energy out by electrical transmission. Under these conditions, what current does the generator deliver at a terminal potential difference of 1910 V?
Physics
A battery has an emf of 15.0 V. The terminal voltage of the battery is 11.2 V when it is delivering 26.0 W of power to an external load resistor R. (a) What is the value of R? (b) What is the internal resistance of the battery?
calculus
Two forces of 7 lb. and 14 lb. act on a body at right angles to each other. Find the angle their resultant force makes with the force of 14 lb.
Art
Ur correct those are all the right answers
52,000 for a loan and a monthly payment of 450.23 for 30 years ?
algebra 2
im confused
algebra 2
f(x)=3x-7 and g(X)=-2x-6. Find (f o g)(4)
Algebra 2
f(x)-g(x)= -2x+4-(-6x-7) 12x+14+24x-28 -12x+14 is this the right answer?
Algebra 2
Algebra 2
what is the inverse of the given relation? y=3x+12 help explain the steps??
Algebra
let f(x)=-2x+4 and g(x)=-6x-7 find f(x)-g(x) Can someone walk me through this problem? I still haven't gotten the hang of it.
math
Kaitlin has four and two-thirds berry pies. She is going to share it equally among 12 people. How much pies will each person get?
math models
Help on why did the ghost decide to haunt the city hall
math
Evan buys 2 CDs. \$1.49 is added to the total price which is the 5% sales tax. What is the cost for each of the CDs?
science
NaOH + HSO -+ Which of the following is the number of moles of NaOH required to neutralize 1 mole of H2SO4 in the equation above?
math
6
Math
Explain how you would find 2 x 2 1/3 using the distributive property.
Math
Why are the estimates of 6/10 x 18 shown below different? Is one estimate better than the other?
Math
Estimate each product of 4 5/8 x 1/3=
math
April is training for a marathon by running no less than 55 km per week. She runs at an average rate of 10 km per hour. What is the minimum number of hours, h, she should run? 10h ≥ 55; h ≥ 5.5; 5.5 hours 10h ≤ 55; h ≤ 5.5; 5.5 hours h over ten > 55...
math
The length of a piece of paper is 8.5 inches. Using scissors, you reduce the length of the paper to 4.25. What is the scale factor of the dilation? A- 1/4 B- 1/2 C- 2 D- 4 I think its A ?
Algebra 2
A simple random sample of 50 adults was surveyed, and it was found that the mean amount of time that they spend surfing the Internet per day is 54.2 minutes, with a standard deviation of 14.0 minutes. What is the 99% confidence interval for the number of minutes that an adult ...
Algebra 2
A textbook company states that the average time a student needs to take a quiz from its book is 30 minutes with a standard deviation of 3 minutes. A teacher using the book is not sure that this is correct for her classes and wants to check. She collects data on 10 random ...
Algebra 2
A simple random sample of 90 is drawn from a normally distributed population, and the mean is found to be 138, with a standard deviation of 34. What is the 90% confidence interval for the population mean? Use the table below to help you answer the question. Confidence Level 90...
Algebra 2
A study investigated the job satisfaction of teachers allowed to choose supplementary curriculum for their classes versus teachers who were assigned all curricular resources for use in their classes. On average, when surveyed regarding job satisfaction, teachers give a score ...
Algebra 2
A study investigated the job satisfaction of teachers allowed to choose supplementary curriculum for their classes versus teachers who were assigned all curricular resources for use in their classes. The authors of the study wanted to know if the two groups of teachers had ...
Algebra 2
A study investigated the job satisfaction of teachers allowed to choose supplementary curriculum for their classes versus teachers who were assigned all curricular resources for use in their classes. The authors of the study wanted to know if the two groups of teachers had ...
Finance
A manufacturing company is thinking of launching a new product. The company expects to sell \$950,000 of the new product in the first year and \$1,500,000 each year thereafter. Direct costs including labor and materials will be 45% of sales. Indirect incremental costs are ...
Math
In triangle ABC, the measure of C is twice the measure of A, and the measure of B is 6 times the measure of A. Find the measure of each angle
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Post a New Question | 3,903 | 14,815 | {"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} | 3.484375 | 3 | CC-MAIN-2017-17 | latest | en | 0.915732 |
https://www.crisluengo.net/archives/tag/snake | 1,555,786,485,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578529962.12/warc/CC-MAIN-20190420180854-20190420202854-00419.warc.gz | 643,919,742 | 7,353 | Posts Tagged ‘snake’
## Mechanical simulation for image segmentation
Wednesday, January 13th, 2010
Last month I posted a tutorial on snakes. I explained that with snakes, the segmentation solution is obtained by minimizing an energy functional. The gradient descent minimization then yields an update function for the snake that can be seen as a set of internal and external forces acting on the snake. I was curious to see where we would get if we were to define the snake simply as a set of balls, connected with each other through springs, moving over the image with momentum, friction, and all the rest. For this, we will completely abandon the notions of gradient descent, function minimization, and so forth, and write a simple Newtonian physics simulation.
## Implementing the dip_snake class
Monday, December 21st, 2009
In the previous post I showed how to implement active contours (a snake). I included the link to a set of files with a complete snake implementation to plug into DIPimage. This implementations uses a class called `dip_snake`. In this post I wanted to show how this class is defined using MATLAB’s new 1-file-style of class definitions. Well, “new” is not completely accurate, this feature has been around for a few releases already, but I’m slow to adapt… You’ll also learn how to make a class whose objects automatically create or update a figure window, much like an object of type `dip_image` does.
## A simple implementation of snakes
Wednesday, December 9th, 2009
In this post I want to show how easy it is to implement snakes using MATLAB and DIPimage. The next release of DIPimage will have some functionality implementing snakes, and this post is based on code written for that. The first implementation of snakes I made for DIPimage used a custom class, `dip_snake`, to simplify usage. We decided eventually to do without the class. The code discussed here uses that first implementation, which you can download here. The most important bits of the code are identical to that in the next release of DIPimage, but the wrapping is not. Adding the class is a nice excuse to show how to modify DIPimage to understand a new parameter type. I will discuss that in a future post. | 478 | 2,218 | {"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-2019-18 | longest | en | 0.941986 |
http://www.dummies.com/how-to/content/managerial-accounting-the-cash-payback-method.html | 1,419,125,195,000,000,000 | text/html | crawl-data/CC-MAIN-2014-52/segments/1418802770554.119/warc/CC-MAIN-20141217075250-00061-ip-10-231-17-201.ec2.internal.warc.gz | 466,866,437 | 18,418 | How to Prepare a Break-Even Analysis with More Than One Product Activity-Based Costing for Overhead Allocation Managerial Accounting: Types of Responsibility Centers
# Managerial Accounting: The Cash Payback Method
The cash payback method is a tool that managerial accountants use to evaluate different capital projects and decide which ones to invest in and which ones to avoid. The cash payback method estimates how long a project will take to cover its original investment. You can calculate the cash payback method whether you have equal payments each period or unequal payments.
The main benefit of the cash payback method is that you can calculate it on the fly to quickly screen out investments.
Although it’s quick and easy, the cash payback method doesn’t account for the full profitability of the project; it ignores any payback earned after the cash payback period ends. Furthermore, because this approach neglects the time value of money, managers should use a more sophisticated model, such as the net present value method I describe later in the chapter, before investing company funds into any project.
## Use the cash payback method with equal annual net cashflows
The cash payback method uses the following formula to compute how long a given project will take to pay for itself. When computing cash payback period, annual net cash flow should include all revenues arising from the new project less expected incremental costs.
Note that net means “to offset,” and net cash flows means that you’re subtracting cash outflows from cash inflows (or vice versa). Therefore, to compute annual net cash flow, you estimate any potential revenues and then add in savings in materials, labor, and overhead associated with the new project. Offset any additional costs associated with the new project against these cash inflows.
The following formula works in a situation where each year’s net cash flows from the investment are expected to be equal:
Cash payback period = Cost of investment / Annual net cash flow
Simply divide the cost of the investment — how much you initially paid for the investment — by the estimated net cash flow the investment generates each year. The higher the cash payback period, the longer the period you need to recover your investment.
For example, suppose you need to decide whether to buy a new computer costing \$500; you expect the computer to increase your net cash flow by \$300 per year. Plug the numbers into the formula:
Here you can see that the computer would take one year and eight months to pay for itself.
When making investment decisions, compare the cash payback period of one project with that of another and select projects that offer the quickest cash payback period. Suppose a less-expensive computer has a cash payback period of only nine months; compared to one year and eight months, the nine-month cash payback period suggests that the less-expensive computer is probably a better investment for your company.
## Use the cash payback method when annual net cash flows change each year
When you are computing cash payback period, remember to include all revenues arising from the new project less expected incremental costs in each year’s net cash flows. When preparing this computation, the net cash flow will probably vary each year. If so, just project the net cash flows that you expect to realize or incur each year.
For example, suppose that your new \$500 computer is expected to yield different net cash flows each year.
The computer will be fully paid off in 2016, when cumulative net cash flows of \$750 exceed the initial investment of \$500. This result amounts to a three-year payback period.
When computing net cash flows, use cash flow rather than accrual income amounts. For example, use projected cash receipts from customers rather than sales. Because depreciation expense doesn’t require cash payments, ignore it completely. | 759 | 3,923 | {"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-2014-52 | longest | en | 0.917884 |
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20100923
# 20100923 - = CS 441 Fall 2010 JWJ Pre-class notes do not...
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============================ CS 441 Fall 2010 -- JWJ Pre-class notes - do not distribute Thursday, September 23, 2010 ============================ Scope: FIRST and FOLLOW Sets (Part 2 -- FOLLOW sets) Outline: FIRST and FOLLOW sets -- solving symbolic equations Two grammars for arithmetical expressions ----------- left-recursive after removing left recursion E --> E + T | T E --> T E' E' --> + T E' | epsilon T --> T * F | F T --> F T' T' --> * F T' | epsilon F --> (E) | id | num F --> (E) | id | num FIRST sets: FIRST sets include terminals (tokens) or epsilon for nullable nonterminals. The tokens in the FIRST(X), where X is a nonterminal, are the only valid starting terminals for strings derived from X. For the above grammar for expressions, a valid expression may begin with '(', 'id" or 'num' and these three tokens are members of FIRST(E). For nullable nonterminals, epsilon is a member of their FIRST sets. FIRST sets can be computed based on the productions of the grammar, by first finding nullable nonterminals and then setting a system of equations. Alternatively, the second step could be replaced with a graph for FIRST sets and then computing its transitive closure.
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05 Jun 2012, 13:58
GPT55 wrote:
Great job, fantastic tool! Thanks so much!
I have a question about the detailed applicant lists at schools.
If I sort the list according to the GMAT or GPA column, does it put the applicants who do not display their GMAT/GPA in order as well?
For example, if there are 2 guys on the list with a 3.1 and 3.3 GPA with someone in between them whose GPA is not displayed, can I assume that that applicant has a GPA between 3.1 and 3.3?
Thanks!
This is a pretty darn technical question. Let me find out and get back with you.
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bb wrote:
GPT55 wrote:
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This is a pretty darn technical question. Let me find out and get back with you.
Thanks! Would be good to know when trying to measure my stats to that of other applicants.
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06 Jun 2012, 21:17
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You are absolutely right - if there are 2 guys on the list with a 3.1 and 3.3 GPA with someone in between them whose GPA is not displayed, thats mean that applicant has a GPA between 3.1 and 3.3, so you can kind of figure that part out.... even though it is not displayed.
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06 Jun 2012, 23:43
bb wrote:
You are absolutely right - if there are 2 guys on the list with a 3.1 and 3.3 GPA with someone in between them whose GPA is not displayed, thats mean that applicant has a GPA between 3.1 and 3.3, so you can kind of figure that part out.... even though it is not displayed.
Thanks a lot bb! That is very helpful to know and makes the tool even more useful!
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12 Jun 2012, 14:43
I have looked all over for where you get to put your affiliations in the profile? Any hints on where I should be looking?
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12 Jun 2012, 19:19
Dbalks wrote:
I have looked all over for where you get to put your affiliations in the profile? Any hints on where I should be looking?
Sure. It is right here: ucp.php?i=164
Just click on Profile & Privacy (perhaps not the most obvious spot) but we'll be changing it in the next few months.
Thanks!
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Re: GMAT Club Profile [#permalink] 12 Jun 2012, 19:19
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### comment on
Need Help??
If the solution in my earlier node doesn't return what you want, this surely will:
use Math::BigInt (); *gcd = \&Math::BigInt::bgcd; sub embiggen { local \$_ = @_ ? \$_[0] : \$_; return Math::BigInt->new(\$_); } my @a = ( 10, 20, 33, 45, 60 ); my @b = ( 2, 5, 10, 12, 16, 23, 45 ); # my %a = map { \$_ => 1 } @a; # my %b = map { \$_ => 1 } @b; # # my @c = map embiggen, # grep { !\$b{\$_} } # @a; # # my @d = map embiggen, # map { Math::BigInt->new(\$_) } # grep { !\$a{\$_} } # @b; my @c = @a; my @d = @b; foreach my \$c (@c) { foreach my \$d (@d) { my \$gcd = gcd(\$c, \$d); \$c /= \$gcd; \$d /= \$gcd; } } @c = grep { \$_ != 1 } @c; @d = grep { \$_ != 1 } @d;
or this messier but potentially faster alternative:
use Math::BigInt (); use Math::Big::Factors (); *gcd = \&Math::BigInt::bgcd; *factor = \&Math::Big::Factors::factors_wheel; sub embiggen { local \$_ = @_ ? \$_[0] : \$_; return Math::BigInt->new(\$_); } my @a = ( 10, 20, 33, 45, 60 ); my @b = ( 2, 5, 10, 12, 16, 23, 45 ); # my %a = map { \$_ => 1 } @a; # my %b = map { \$_ => 1 } @b; # # my @c = map embiggen, # grep { !\$b{\$_} } # @a; # # my @d = map embiggen, # grep { !\$a{\$_} } # @b; my @c = @a; my @d = @b; my %c_factors; foreach my \$c_idx (0..\$#c) { my @c_factors = factor(\$c[\$c_idx]); foreach my \$c_factor (@c_factors) { push(@{\$c_factors{\$c_factor}}, \$c_idx); } } foreach my \$d (@d) { my @d_factors = factor(\$d); foreach my \$d_factor (@d_factors) { next unless \$c_factors{\$d}; next unless @{\$c_factors{\$d}}; my \$c_idx = shift(@{\$c_factors{\$d}}); \$c[\$c_idx] /= \$d_factor; \$d /= \$d_factor; } } @c = grep { \$_ != 1 } @c; @d = grep { \$_ != 1 } @d;
Both untested.
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1. With 1kg enriched plutonium we can dissolve the earth so can we simply melt mars and wait for it to crust ? I am pretty sure heat is all that's needed for atmosphere.
2.
3. Where in the world did you get that figure from?
The energy in 1kg of enriched plutonium is many magnitudes less than that needed to have any significant effect on the Earth as a whole and 1kg is much smaller than the critical mass of plutonium needed for a chain reaction.
To give you an idea of how far off you are, let's consider what it would take to just melt 1/100 of the mass of Mars. Even if rock is just at the temperature it melts at, it would take a minumum of 270,000 joules of energy per kg to melt it. Mars is 1/10 the mass of Earth, so 1/100 of that is 6 x 10^21 kg. This means you would need 1.6 x 10^27 joules to melt it. The total conversion of 1 kg of matter would release 9 x 10^16 joules. Thus you would need to convert at least 18,000,000,000 kg of matter to energy to melt just 1/100 of the mass of Mars.
And that is with total mass to energy conversion. Even enriched Plutonium only converts a small fraction of its mass to energy.
4. Originally Posted by Janus
Where in the world did you get that figure from?
The energy in 1kg of enriched plutonium is many magnitudes less than that needed to have any significant effect on the Earth as a whole and 1kg is much smaller than the critical mass of plutonium needed for a chain reaction.
Can't recall where figure originated but was related to critical mass, the bang that bangs so hard it makes explosives of all that is near it.
5. To Max
The energy from every nuclear weapon the world possesses is less than the energy released by the eruption of the Pinatubo Volcano. So forget using nuclear means to provide an atmosphere for Mars.
There have been a lot of suggestions made. Gravity is a good source of energy. From gravity, we get massive release of energy when something big falls into Mars. One suggestion for the medium future is to fire vast quantities of ice at Mars - collected from comets or from Saturn's rings, or some other similar source, and use some kind of ion drive to steer them so they fall onto Mars surface. This would release enormous heat from atmospheric friction, and dump lots of liquid water onto Mars.
After all this is over, it may be possible to keep Mars warm with orbiting mirrors.
At least these are ideas presented during the last conference on terraforming Mars.
6. Using direct means isn't a good way to go. Leverage nature is a better way. For example what is that Plutonium is running a factory to produce CFCs (a greenhouse gas) on the surface, which will in time raise temperature, sublimate the huge amounts of Co2 and water vapor (two more greenhouses gases) just below the surface into the atmosphere, increase atmospheric pressure and create a strong positive feedback on the temperature which turn will sublimate more Co2, H2O ...etc.
7. Originally Posted by Max Time Taken
Originally Posted by Janus
Where in the world did you get that figure from?
The energy in 1kg of enriched plutonium is many magnitudes less than that needed to have any significant effect on the Earth as a whole and 1kg is much smaller than the critical mass of plutonium needed for a chain reaction.
Can't recall where figure originated but was related to critical mass, the bang that bangs so hard it makes explosives of all that is near it.
That's not what critical mass means. Critical mass is the minimum amount of a fissionable material you need to have a self-maintaining nuclear reaction. For Plutonium 238, this is 10 kg. This reaction only pertains to the fissionable material itself and does not in any way involve anything else around it.
8. Originally Posted by skeptic
To Max
The energy from every nuclear weapon the world possesses is less than the energy released by the eruption of the Pinatubo Volcano. So forget using nuclear means to provide an atmosphere for Mars.
There have been a lot of suggestions made. Gravity is a good source of energy. From gravity, we get massive release of energy when something big falls into Mars. One suggestion for the medium future is to fire vast quantities of ice at Mars - collected from comets or from Saturn's rings, or some other similar source, and use some kind of ion drive to steer them so they fall onto Mars surface. This would release enormous heat from atmospheric friction, and dump lots of liquid water onto Mars.
After all this is over, it may be possible to keep Mars warm with orbiting mirrors.
At least these are ideas presented during the last conference on terraforming Mars.
At least with present nuclear technology, but then we release a very small amount of the E=MC^2 equation with current bombs. Mirrors are heavy and hard to produce.
I do like the idea of self moling(drilling) induction rings with solar panels.
9. Originally Posted by Lynx_Fox
Using direct means isn't a good way to go. Leverage nature is a better way. For example what is that Plutonium is running a factory to produce CFCs (a greenhouse gas) on the surface, which will in time raise temperature, sublimate the huge amounts of Co2 and water vapor (two more greenhouses gases) just below the surface into the atmosphere, increase atmospheric pressure and create a strong positive feedback on the temperature which turn will sublimate more Co2, H2O ...etc.
All equipment made and sent costs the earth
10. Originally Posted by Janus
Originally Posted by Max Time Taken
Originally Posted by Janus
Where in the world did you get that figure from?
The energy in 1kg of enriched plutonium is many magnitudes less than that needed to have any significant effect on the Earth as a whole and 1kg is much smaller than the critical mass of plutonium needed for a chain reaction.
Can't recall where figure originated but was related to critical mass, the bang that bangs so hard it makes explosives of all that is near it.
That's not what critical mass means. Critical mass is the minimum amount of a fissionable material you need to have a self-maintaining nuclear reaction. For Plutonium 238, this is 10 kg. This reaction only pertains to the fissionable material itself and does not in any way involve anything else around it.
Thank you for that, I have not researched enough. 10kg should melt a hole right through to the centre with enough gravity/inertia, right ? And keep burning ?
11. You might want to read this: How to destroy the Earth @ Things Of Interest
12. What a load of rubbish , 5,973,600,000,000,000,000,000ball of iron.
What is silicone, what is carbon ?
How hot is the core, is it all liquid ?
13. Originally Posted by Max Time Taken
10kg should melt a hole right through to the centre with enough gravity/inertia, right ? And keep burning ?
Wrong. Refer to the following article which describes what actually happens.
Corium (nuclear reactor) - Wikipedia, the free encyclopedia
14. Originally Posted by Max Time Taken
What a load of rubbish , 5,973,600,000,000,000,000,000 [tonne] ball of iron.
The mass is about right. Iron is the most common element on earth. It is a bit of an exaggeration to describe it as a ball of iron, but not by much.
What is silicone
An inert synthetic polymer.
How hot is the core, is it all liquid ?
There is a solid inner core with a radius of ~1,220 km and a liquid outer core with a radius of ~3,400 km. I think the temperature is about 5,500K.
15. Ball of iron, small exaggeration ?
iron: 34 .1 %
oxygen: 28 .2 %
silicon: 17 .2 %
magnesium: 15 .9 %
nickel: 1 .6 %
calcium: 1 .6 %
aluminium: 1 .5 %
sulfur: 0 .70 %
sodium: 0 .25 %
titanium: 0 .071 %
potassium: 0 .019 %
other elements: 0 .53 %
Nice info on the core thanks.
16. OK. A rusty ball of iron. That accounts for the majority.
17. Originally Posted by Max Time Taken
Ball of iron, small exaggeration ?
iron: 34 .1 %
oxygen: 28 .2 %
silicon: 17 .2 %
magnesium: 15 .9 %
nickel: 1 .6 %
calcium: 1 .6 %
aluminium: 1 .5 %
sulfur: 0 .70 %
sodium: 0 .25 %
titanium: 0 .071 %
potassium: 0 .019 %
other elements: 0 .53 %
Nice info on the core thanks.
What's the source for that? It sound more like the bulk composition of the earth, rather than the core. You do know the difference, right?
The core is mostly iron, nickel, and a few other minor siderophiles. Plus I think the sulpher percentage is higher.
Is it by weight or percentage of atoms?
18. Originally Posted by skeptic
There have been a lot of suggestions made. Gravity is a good source of energy. From gravity, we get massive release of energy when something big falls into Mars. One suggestion for the medium future is to fire vast quantities of ice at Mars - collected from comets or from Saturn's rings, or some other similar source, and use some kind of ion drive to steer them so they fall onto Mars surface. This would release enormous heat from atmospheric friction, and dump lots of liquid water onto Mars.
Mars (probably) had liquid water, and still has some ice. The problem is its gravity - water vapour escapes very rapidly from the martian atmosphere compared to Earth's. Heating the planets surface will accelerate this process.
Then again, on human timescales it wouldn't be such an issue - even if you have to replace the atmosphere every few centuries, you get an entire planet in return.
19. Originally Posted by MeteorWayne
Originally Posted by Max Time Taken
Ball of iron, small exaggeration ?
iron: 34 .1 %
oxygen: 28 .2 %
silicon: 17 .2 %
magnesium: 15 .9 %
nickel: 1 .6 %
calcium: 1 .6 %
aluminium: 1 .5 %
sulfur: 0 .70 %
sodium: 0 .25 %
titanium: 0 .071 %
potassium: 0 .019 %
other elements: 0 .53 %
Nice info on the core thanks.
What's the source for that? It sound more like the bulk composition of the earth, rather than the core. You do know the difference, right?
The core is mostly iron, nickel, and a few other minor siderophiles. Plus I think the sulpher percentage is higher.
Is it by weight or percentage of atoms?
Wikipedia structure of earth, by weight not moles.
21. Originally Posted by Max Time Taken
Wikipedia structure of earth, by weight not moles.
Structure of the Earth - Wikipedia, the free encyclopedia? I don't see it.
(Why is it so hard for people to provide a reference to their sources? What do they teach young people nowadays? Grumble, grumble...)
22. If the earth has a fossil iron core why can we not manufacture fossil iron ?
23. Originally Posted by Max Time Taken
If the earth has a fossil iron core why can we not manufacture fossil iron ?
Can you explain what that means; what is "fossil iron"?
Also, can you clarify where you got that distribution of the elements from?
24. Originally Posted by Strange
Originally Posted by Max Time Taken
Wikipedia structure of earth, by weight not moles.
Structure of the Earth - Wikipedia, the free encyclopedia? I don't see it.
(Why is it so hard for people to provide a reference to their sources? What do they teach young people nowadays? Grumble, grumble...)
Wiki changes too often.
25. Originally Posted by Strange
Originally Posted by Max Time Taken
If the earth has a fossil iron core why can we not manufacture fossil iron ?
Can you explain what that means; what is "fossil iron"?
Also, can you clarify where you got that distribution of the elements from?
File:Carbon basic phase diagram.png - Wikipedia, the free encyclopedia Fossil would be the hot formed hard bit as seen in this diagram for carbon.
I got that ref from the same site you visited but the stats have been adjusted no overall composition chart any more.
26. Originally Posted by Max Time Taken
File:Carbon basic phase diagram.png - Wikipedia, the free encyclopedia Fossil would be the hot formed hard bit as seen in this diagram for carbon.
That doesn't seem to have anything to do with the formation of fossils and absolutely nothing to do with iron, so I am still not clear what you mean by the phrase "fossil iron".
I got that ref from the same site you visited but the stats have been adjusted no overall composition chart any more.
When did you find that information because it hasn't been updated since January? I had a look through previous versions of the page and couldn't see any sign of that table. Do you think it might have come from somewhere else?
27. Originally Posted by Strange
Originally Posted by Max Time Taken
File:Carbon basic phase diagram.png - Wikipedia, the free encyclopedia Fossil would be the hot formed hard bit as seen in this diagram for carbon.
That doesn't seem to have anything to do with the formation of fossils and absolutely nothing to do with iron, so I am still not clear what you mean by the phrase "fossil iron".
1. Compressed at heat till no gaps in atoms, netted perfectly so to speak. Pressure / Time in geology I believe, add temp to make it interesting. Just the same way any fossil, even hail, is formed.
I got that ref from the same site you visited but the stats have been adjusted no overall composition chart any more.
When did you find that information because it hasn't been updated since January? I had a look through previous versions of the page and couldn't see any sign of that table. Do you think it might have come from somewhere else?
2.I got it from a link on an answer to that question , yahoo answers 2 days ago.
I have not been able to revisit the same page in wiki, for a link to post.
28. Originally Posted by Max Time Taken
Originally Posted by Strange
That doesn't seem to have anything to do with the formation of fossils and absolutely nothing to do with iron, so I am still not clear what you mean by the phrase "fossil iron".
1. Compressed at heat till no gaps in atoms, netted perfectly so to speak. Pressure / Time in geology I believe, add temp to make it interesting. Just the same way any fossil, even hail, is formed.
Are you trying to say there should be a denser/more compressed form of iron?
This has nothing to do with the way a fossil is formed. Fossilization is the replacement of organic matter by minerals; it isn't just a "denser" version of the original.
I have not been able to revisit the same page in wiki, for a link to post.
That is a shame. It looked like a useful reference...
29. 1. I am saying that if iron can be solid at over 5700K then it must be more dense than the stuff we manufacture. Fossils are not just made by replacement of minerals, gravity/time(heat) = geological compression.
2. Here is where I got that info... What is the most common Element on Earth ? - Yahoo! Answers
30. Originally Posted by Max Time Taken
1. I am saying that if iron can be solid at over 5700K then it must be more dense than the stuff we manufacture. Fossils are not just made by replacement of minerals, gravity/time(heat) = geological compression.
That is quite simply wrong.
Right, so as I said, it is the composition of the bulk earth, NOT the core. As I said in post # 16.
Planetary differentiation - Wikipedia, the free encyclopedia
31. Max time Taken, you chose to ignore my last advice, so I shall be more specific this time. Before hitting the Post key why not take the time to study what your are posting on, learn something about it, and thereby stop posting utter bilge.
32. Originally Posted by Max Time Taken
1. I am saying that if iron can be solid at over 5700K then it must be more dense than the stuff we manufacture.
I don't think it is quite that hot. But it is under enormous pressure that is what keeps it solid. I doubt the density is significantly different. If you were to bring a sample to the earth's surface (at the same temperature) it would melt.
Here you go, phase diagram for iron: Iron's structure at the temperatures and pressures of Earth's core
(Although I did see a mention somewhere that recent research shows it may return to a bcc structure at very high temp and pressure.)
33. Originally Posted by John Galt
Max time Taken, you chose to ignore my last advice, so I shall be more specific this time. Before hitting the Post key why not take the time to study what your are posting on, learn something about it, and thereby stop posting utter bilge.
Wise advice it may have been.
34. Wait... making Mars a habitable planet and being able to colonize it is scientifically possible?
I was born 300 years to early...
fuck my life... -.-
35. Could Solar sails be used to drag asteroids on a collision with Mars?
Or maybe collide with one of Mars moons to try to knock it on a collision course with Mars.
If enough ice can be thrown maybe engineered life can create greenhouse gas, though Im not sure what would happen with the lack of a magnetic field as strong as earth?
Maybe it would be less time consuming to just build domed cities and underground forests and forget the terraforming part?
36. Going to need forges, sunlight is strong(photo voltaic induction ring drills), terraforming could be part and parcel for both.
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https://cadmodels.machinedesign.com/community/knowledge/pl/detail/3667/Pythagorean+cup | 1,657,155,112,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104683020.92/warc/CC-MAIN-20220707002618-20220707032618-00091.warc.gz | 192,259,860 | 15,405 | ### Pythagorean cup (14334 views - Food Drinks Beverages)
A Pythagorean cup (also known as a Pythagoras cup, Greedy Cup, Tantalus cup or o kounenos tsi dikaiosynis) is a practical joke device in a form of a drinking cup, credited to Pythagoras of Samos. When it is filled beyond a certain point, a siphoning effect causes the cup to drain its entire contents through the base.
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## Pythagorean cup
### Pythagorean cup
A Pythagorean cup (also known as a Pythagoras cup, Greedy Cup, Tantalus cup or o kounenos tsi dikaiosynis) is a practical joke device in a form of a drinking cup, credited to Pythagoras of Samos. When it is filled beyond a certain point, a siphoning effect causes the cup to drain its entire contents through the base.
## Form and function
Cross section of a Pythagorean cup.
A Pythagorean cup looks like a normal drinking cup, except that the bowl has a central column in it, giving it a shape like a Bundt pan. The central column of the bowl is positioned directly over the stem of the cup and over the hole at the bottom of the stem. A small open pipe runs from this hole almost to the top of the central column, where there is an open chamber. The chamber is connected by a second pipe to the bottom of the central column, where a hole in the column exposes the pipe to (the contents of) the bowl of the cup.[1][2]
When the cup is filled, liquid rises through the second pipe up to the chamber at the top of the central column, following Pascal's principle of communicating vessels. As long as the level of the liquid does not rise beyond the level of the chamber, the cup functions as normal. If the level rises further, however, the liquid spills through the chamber into the first pipe and out the bottom. Gravity then creates a siphon through the central column, causing the entire contents of the cup to be emptied through the hole at the bottom of the stem.[3] Some modern toilets operate on the same principle: when the water level in the bowl rises high enough, a siphon is created, flushing the toilet.[citation needed]
## Common occurrences
Heron of Alexandria (c. 10–70 AD) used Pythagorean cups as hydraulic components in his robotic systems. | 515 | 2,198 | {"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-2022-27 | latest | en | 0.898897 |
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Data is sent in chunks of two different sizes - a yellow chunk has 5 characters and a blue chunk has 9 characters. A data slot of size 31 cannot be exactly filled with a combination of yellow and. . . . | 2,242 | 9,093 | {"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} | 4.21875 | 4 | CC-MAIN-2017-47 | latest | en | 0.907695 |
https://www.workstem.com/hk/en/blog/employee-terminate-with-untaken-annual-leave/?utm_source=blogpost&utm_medium=blog&utm_campaign=annual-leave-policy-hong-kong&utm_term=fr | 1,723,466,969,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641039579.74/warc/CC-MAIN-20240812124217-20240812154217-00745.warc.gz | 825,529,330 | 20,586 | How to Calculate Annual Leave Payment on Termination of Employment Contract?
Mr. Cheung worked in a logistics company for several years and recently wanted to change jobs. He told his employer about his resignation in the length of the notice period, and the employer wanted him to stay. But he has decided to leave the company, the employer then asked him to take annual leaves before leaving. Is it legal for the employer to do so? What would happen if he did not take his annual leave?
Under the Employment Ordinance(EO), these statutory annual leave shall not be included in the length of notice required to terminate a contract of employment. The employer cannot force the employee to apply for the untaken annual leave before leaving the company. For those untaken annual leave, it should be included in the termination payment.
So now, we know the employer has no right to force an employee to take the untaken annual leave within the notice period, let’s see how to calculate the termination payments of annual leave.
Under the EO, an employee is entitled to at least 7 days of annual leave with pay after having been employed under a continuous contract for 12 months. Employees can enjoy 7 days of annual leave in the first two years, and his entitlement to paid annual leave increases progressively from seven days to a maximum of 14 days according to his length of service. The following is the calculation of annual leave pay on termination of employment contract:
Tip: for example, an employee’s leave year starts from his joined date, and he has been employed for one year. However, he resigned in the second year but it was less than 3 months in this leave cycle, then his annual leave would not be entitled in the second leave years.
For example: Max joined the company on April 5, 2019, the leave year is calculated based on the employee’s joined date. Therefore, Max could only take two days of paid annual leave in May 2020, and 1) if Max resigns on June 5, 2020, how many untaken annual leaves he would have? 2) if Max resigns on August 5, 2020, how many days of annual leave will not be granted?
How to calculate termination payment in lieu of any untaken annual leave?
The daily rate of annual leave pay is a sum equivalent to the average daily wages earned by an employee in the 12-month period preceding the following specified dates. If an employee is employed for less than 12 months, the calculation shall be based on a shorter period.
An employer is required to pay interest on the outstanding wages due to the employee if he fails to make termination payment in lieu of any untaken annual leave to the employee within seven days after the termination or expiry of the contract.
As a one-stop human resource system, Workstem can calculate the days of annual leave that employees can enjoy based on their joined date, and record the employee’s leave application history. It saves time and effort to calculate the termination payment in lieu of any untaken annual leave.
How to Manage Employee’s MPF After the Termination of Employment?
As Layoffs Spread, How to Calculate Severance Payment or Long Service Payment to Employees? | 652 | 3,168 | {"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-2024-33 | latest | en | 0.987381 |
https://lavafamilyinn.com/2514600209550-department-of-biomedical-engineering-faculty-of-engineering-university-of-malaya-laboratory-report-beam-deflection-name/ | 1,580,243,977,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251783000.84/warc/CC-MAIN-20200128184745-20200128214745-00365.warc.gz | 528,990,169 | 13,778 | 2514600209550
DEPARTMENT OF BIOMEDICAL ENGINEERING
FACULTY OF ENGINEERING
UNIVERSITY OF MALAYA
LABORATORY REPORT
BEAM DEFLECTION
MATRIX NUMBER: KIB180046
GROUP: GROUP 11
LAB ADVISOR: DR. CHAN CHOW KHUEN
Abstract
Objective
To measure deflections in a simply supported aluminium, brass and copper beam.
To use measured and theoretical deflections to evaluate Young’s Modulus of the materials.
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To compare the analytical and experimental values of deflection in the simply supported and cantilever beam.
Hypothesis
Introduction
Deflection can be defined as the movement of a beam or a node from its original position due to the forces that were applied to it. It can happen in beams, trusses and frames and other structures but in this experiment we focused on the beam deflection.
Beam is an element that able to withstand load by resisting the bending from the forces. The degree of structural deformation of the beam can be affected by the type of the beam, span of the beam, shape of the beam and the magnitude of load acting on the beam. There are a few type of beams such simple beam, cantilever beam, overhanging beam and continuous which all of the beams have different degree of bending.
The type of beam used in this experiment is a simple supported beam. This beam has one end with pinned support and a roller support at the other end. The types of beam that were used in the experiment were copper beam, brass beam and aluminium beam and concentrated load and uniform distributed were tested on them to measure the beam deflections. From the measured beam deflection, the maximum load that a beam can withstand can be calculated.
In this experiment, we will calculate the theoretical value of deflection by using Equation 1 and Equation 2 to compare with the experimental value of deflection.
y=Px12EI(3l24-x2)Equation 1: Deflection for concentrated load
y=?x24EI(l3-2lx2+x3)Equation 2: Deflection for distributed load
Where P and w is the weight of load, E is the Young Modulus, l is length of beam, x is the distance of dial gauge from the initial point, and I is the moment of inertia.
Apparatus and Materials
Beam deflection apparatus
Vernier caliper and meter ruler
Aluminium beam
Brass beam
Copper beam
Hanger together with 9 different loads
Rectangular uniform distributed load with 12N
Allen key
Procedure
Deflection of a beam by applying a concentrated load
The clamping length (L) of the beam was set to 700mm.
The width and height of each beam were measured using a caliper and the value was recorded.
The copper beam was placed on the bearers.
One end of the beam was set as fixed end by tightening the screw.
Two dial gauges were moved to the copper beam. The height of the gauge was adjusted so that the needle touched the beam.
The dial gauge 1 (G1) was moved to 175mm and the dial gauge 2 (G2) was moved to 525mm.
1N load was added to middle of the beam and the dial gauge reading was recorded. The value of dial gauge reading was taken three times to obtain the average value.
Step 7 was repeated with 2N, 3N, 4N, 5N, 6N, 7N, 8N, 9N, 10N load. All of the gauge reading was recorded.
All of the loads were removed.
Step 1 to 9 was repeated with brass and aluminium beam.
The experimental deflection was compared with theoretical value.
Deflection of a beam by applying a uniform distributed load
The clamping length (L) of the beam was set to 700mm.
The copper beam was placed on the bearers.
One end of the beam was set as fixed end by tightening the screw.
Two dial gauges were moved to the copper beam. The height of the gauge was adjusted so that the needle touched the beam.
The dial gauge 1 (G1) was moved to 175mm and the dial gauge 2 (G2) was moved to 525mm.
The 12N load (distributed load) was placed onto the beam. The total length of the load on the beam was measured and recorded. The dial gauge reading was recorded.
Step 6 was repeated 3 times to obtain the average value of the dial gauge reading.
Step 1 to 7 was repeated with brass and aluminium beam.
The experimental deflection was compared with theoretical value.
Results
Type of beam Length,l (m) Width,w (m) Height,h (m) Moment of Inertia,I (m4) Young Modulus (GPa)
Copper 1.002 0.026 6.51×10-3 5.978×10-10 117
Brass 1.001 0.025 6.37×10-3 5.385×10-10 102
Aluminium1.001 0.026 6.61×10-3 6.257×10-10 70
Table 1: Dimension of the beam
The moment of inertia,I value was obtained from Equation 3 and the theoretical value of Young Modulus was obtained from our reference.
I=wh312Equation 3: Moment of inertia formula
Part A
Weight of load (N) Deflection of beam (mm)
G1 G2
1 2 3 Average 1 2 3 Average
1.0 0.015 0.01 0.015 0.013 0.02 0.03 0.02 0.023
2.0 0.04 0.04 0.04 0.04 0.06 0.05 0.07 0.05
3.0 0.07 0.07 0.07 0.07 0.11 0.115 0.11 0.112
4.0 0.09 0.09 0.09 0.09 0.14 0.14 0.14 0.14
5.0 0.115 0.115 0.11 0.113 0.18 0.18 0.18 0.18
6.0 0.135 0.135 0.14 0.137 0.21 0.21 0.215 0.212
7.0 0.16 0.16 0.16 0.16 0.26 0.25 0.25 0.253
8.0 0.18 0.185 0.18 0.182 0.29 0.29 0.3 0.293
9.0 0.21 0.21 0.21 0.21 0.33 0.34 0.33 0.333
10.0 0.235 0.23 0.235 0.233 0.375 0.37 0.375 0.373
Table 2: Gauge reading for copper
Weight of load (N) Deflection of beam (mm)
G1 G2
1 2 3 Average 1 2 3 Average
1.0 0.01 0.01 0.015 0.012 0.015 0.02 0.02 0.018
2.0 0.03 0.03 0.025 0.028 0.045 0.045 0.04 0.043
3.0 0.04 0.04 0.04 0.04 0.07 0.065 0.065 0.067
4.0 0.06 0.06 0.06 0.06 0.1 0.095 0.09 0.095
5.0 0.075 0.07 0.075 0.073 0.12 0.115 0.115 0.117
6.0 0.09 0.085 0.09 0.088 0.145 0.14 0.14 0.142
7.0 0.1 0.1 0.1 0.1 0.17 0.165 0.165 0.167
8.0 0.115 0.12 0.12 0.118 0.19 0.19 0.19 0.19
9.0 0.135 0.135 0.14 0.137 0.21 0.21 0.215 0.212
10.0 0.15 0.15 0.15 0.15 0.23 0.23 0.23 0.23
Table 3: Gauge reading for brass
Weight of load (N) Deflection of beam (mm)
G1 G2
1 2 3 Average 1 2 3 Average
1.0 0.04 0.04 0.04 0.04 0.06 0.065 0.055 0.06
2.0 0.08 0.08 0.08 0.08 0.13 0.14 0.14 0.137
3.0 0.12 0.12 0.125 0.122 0.21 0.2 0.21 0.207
4.0 0.17 0.165 0.17 0.168 0.29 0.28 0.29 0.287
5.0 0.21 0.21 0.21 0.21 0.36 0.36 0.365 0.362
6.0 0.25 0.26 0.25 0.253 0.425 0.44 0.42 0.428
7.0 0.3 0.3 0.3 0.3 0.51 0.5 0.51 0.507
8.0 0.35 0.35 0.35 0.35 0.59 0.59 0.595 0.592
9.0 0.395 0.4 0.395 0.397 0.67 0.675 0.66 0.668
10.0 0.44 0.45 0.44 0.443 0.73 0.75 0.74 0.74
Table 4: Gauge reading for aluminium
Figure 1: Graph of deflection of a beam by applying concentrated load
Part B
Length of load from initial point (cm)
Deflection of beam (mm)
G1 G2
1 2 3 Average 1 2 3 Average
26.0 0.27 0.27 0.27 0.27 0.36 0.35 0.35 0.353
28.0 0.28 0.28 0.28 0.28 0.38 0.38 0.38 0.38
30.0 0.28 0.28 0.28 0.28 0.405 0.4 0.395 0.4
32.0 0.29 0.29 0.29 0.29 0.43 0.43 0.43 0.43
34.0 0.29 0.29 0.29 0.29 0.44 0.45 0.45 0.447
36.0 0.285 0.29 0.29 0.288 0.47 0.48 0.49 0.48
38.0 0.28 0.28 0.285 0.282 0.5 0.495 0.5 0.498
Table 5: Gauge reading for copper
Length of load from initial point (cm)
Deflection of beam (mm)
G1 G2
1 2 3 Average 1 2 3 Average
26.0 0.27 0.27 0.27 0.27 0.22 0.22 0.22 0.22
28.0 0.175 0.175 0.175 0.175 0.24 0.24 0.24 0.24
30.0 0.18 0.18 0.18 0.18 0.26 0.26 0.26 0.26
32.0 0.18 0.18 0.18 0.18 0.275 0.275 0.275 0.275
34.0 0.18 0.18 0.18 0.18 0.29 0.29 0.28 0.287
36.0 0.18 0.18 0.18 0.18 0.3 0.295 0.29 0.295
38.0 0.17 0.17 0.17 0.17 0.3 0.295 0.305 0.3
Table 6: Gauge reading for brass
Length of load from initial point (cm)
Deflection of beam (mm)
G1 G2
1 2 3 Average 1 2 3 Average
26.0 0.505 0.49 0.49 0.495 0.67 0.66 0.66 0.663
28.0 0.505 0.505 0.5 0.503 0.72 0.7 0.69 0.703
30.0 0.525 0.51 0.52 0.525 0.775 0.775 0.77 0.773
32.0 0.535 0.53 0.535 0.533 0.81 0.82 0.82 0.817
34.0 0.54 0.545 0.545 0.543 0.895 0.885 0.885 0.888
36.0 0.53 0.53 0.535 0.552 0.89 0.89 0.9 0.893
38.0 0.52 0.53 0.53 0.527 0.92 0.92 0.93 0.923
Table 7: Gauge reading for aluminium
Figure 2: Graph of deflection of beam by applying distributed load
From the all of the tabulated data, we can calculate the experimental modulus elasticity of each beam by using Equation 4.
E=Px12?I(3l24-x2)Equation 4: Young Modulus formula
Where P is the weight of load, ? is the beam deflection, l is length of beam, x is the distance of the dial gauge from the initial point, and I is the moment of inertia.
For copper,
E=100.52512(0.37×10-3)5.978×10-103(0.7)24-(0.5252)=181.73GPaFor brass,
E=100.52512(0.23×10-3)5.385×10-103(0.7)24-(0.5252)=324.54GPaFor aluminium,
E=100.52512(0.74×10-3)6.257×10-103(0.7)24-(0.5252)=86.81GPa Discussion
Weight of
load (N) Experimental Deflection (mm) Theoretical deflection (mm)
G1 G2 G1 G2
A B C A B C A B C A B C
1 0.01 0.01 0.04 0.02 0.02 0.06 0.07 0.09 0.11 0.06 0.07 0.09
2 0.04 0.03 0.08 0.05 0.04 0.14 0.14 0.18 0.22 0.11 0.15 0.15
3 0.07 0.04 0.12 0.11 0.07 0.21 0.21 0.27 0.34 0.17 0.22 0.28
4 0.09 0.06 0.17 0.14 0.10 0.29 0.28 0.36 0.45 0.23 0.29 0.37
5 0.11 0.08 0.21 0.18 0.12 0.36 0.35 0.45 0.56 0.29 0.37 0.46
6 0.14 0.09 0.25 0.21 0.14 0.43 0.42 0.54 0.67 0.34 0.44 0.55
7 0.16 0.1 0.3 0.25 0.17 0.51 0.49 0.63 0.79 0.40 0.51 0.64
8 0.18 0.12 0.35 0.29 0.20 0.59 0.56 0.72 0.90 0.46 0.59 0.73
9 0.21 0.14 0.40 0.33 0.21 0.67 0.63 0.80 1.01 0.52 0.66 0.83
10 0.23 0.15 0.44 0.37 0.23 0.74 0.70 0.89 1.12 0.57 0.73 0.92
Table 8: Comparison of experimental and theoretical deflection
Where A is copper, B is brass and C is aluminium.
The theoretical deflection per unit load force for each beam was calculated using Equation 1. From the theoretical value and experimental value, the percentage error can be calculated using Equation 5.
Percentage error%=theoretical value-experimental valuetheoretical value×100%Equation 5: Percentage error formula
(N)
Percentage
Error (%)
Copper Brass AluminiumG1 G2 G1 G2 G1 G2
1 85.71 66.67 88.89 71.43 63.63 33.33
2 71.43 54.55 83.33 73.33 63.64 6.67
3 66.67 35.29 85.19 68.18 64.71 25.0
4 67.86 26.09 83.33 65.52 62.22 21.62
5 68.57 37.93 82.22 67.57 62.5 21.73
6 66.67 38.24 83.33 68.18 62.69 21.82
7 67.35 37.5 84.13 66.67 62.03 20.32
8 67.86 36.96 83.33 66.10 61.11 19.18
9 66.67 36.54 82.5 68.18 60.4 19.28
10 67.14 35.09 83.15 68.49 60.71 19.57
Table 9: Percentage error of the beam deflection
We can see that most of the value of the percentage error above was more that 30%. In addition the percentage error for the Young Modulus of copper, brass and aluminium were 54.7%, 21.82% and 23.1% respectively. So, it can be justified there were errors when the experiment was conducted as there was significant difference for the measured value compared to theoretical value.
One of the sources of error in this experiment was the system error which caused by the inaccurate dial gauge. The dial gauge value was not acceptable as there are errors when we were using it. The spindle of the dial gauge was in a poor condition as sometimes it cannot move to read the deflection. Next, the beams were already deformed as the beams were used many times in previous experiment. The errors also can be caused by parallax error when we record the reading of the dial gauge.
However, in Figure 1, the graph of deflection against weight of load is plotted. It is shown that aluminium was the hardest beam followed by copper and brass. So the theory of the beam deflection is still met although there are errors in the experiment when all the beams has different value of deflection according its type and this also follows the Young Modulus. The deflection of the beam also increased when the weight of load increased.
In Figure 2, the graph of deflection against length of load from initial point is plotted. It is shown that the deflection also depends on the point of application of force. The deflection increase when the length of load from initial point increase.
Beam deflection is important in designing biomedical engineering equipment because the engineer should know how much the beam can deflect and which type of beam is stronger. This will make the equipment is more durable and can work efficiently.
The beam deflection can be predicted as shown in Equation 1 and Equation 2. It is useful in designing a beam as we can know how much a beam can deflect at certain point of a beam. So, the engineer can predict where the force should be applied on the beam that will cause less deflection.
The main importance of measuring beam deflection is to ensure the beam will not exceed the maximum deflection or have a specification of allowable deflection when the force is applied. For example, we need to know beam deflection when designing biomedical engineering equipment such as artificial limbs.
Conclusion | 4,921 | 12,518 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2020-05 | latest | en | 0.923406 |
http://www.physicsforums.com/showpost.php?p=4181233&postcount=1 | 1,369,200,602,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368701370254/warc/CC-MAIN-20130516104930-00039-ip-10-60-113-184.ec2.internal.warc.gz | 660,749,241 | 4,062 | Thread: Velocity w/axis vs without View Single Post
## Velocity w/axis vs without
Hi, quick question
I've just started to learn circular motion and i tried a basic force problem.
Imagine there is a ball on a string with a string length 'r', and a ball mass 'm'.
You apply a force 'F' for one second. Can the arc length of the ball's motion be calculated the same as linear motion? $d=\frac{F{t^2}}{2m}$ (replacing acceleration with F/m, vi=0)
Can I also go as far as the find the angular velocity by dividing by the radius?
$$ω=\frac{F{t^2}}{2mr}$$
Thanks.
PhysOrg.com physics news on PhysOrg.com >> Iron-platinum alloys could be new-generation hard drives>> Lab sets a new record for creating heralded photons>> Breakthrough calls time on bootleg booze | 196 | 758 | {"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} | 2.703125 | 3 | CC-MAIN-2013-20 | longest | en | 0.860982 |
https://oeis.org/A352720 | 1,722,852,746,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640436802.18/warc/CC-MAIN-20240805083255-20240805113255-00541.warc.gz | 352,410,081 | 4,159 | The OEIS is supported by the many generous donors to the OEIS Foundation.
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A352720 a(n) is the number of free polyominoes of width 2 and size n. 4
1, 1, 4, 5, 12, 18, 37, 60, 117, 200, 379, 669, 1250, 2247, 4168, 7570, 13987, 25549, 47108, 86319, 158978, 291806, 537105, 986786, 1815699, 3337560, 6140047, 11289571, 20767180, 38189927, 70246680, 129191148, 237627757, 437042337, 803861244, 1478488577, 2719392160, 5001663330, 9199544069 (list; graph; refs; listen; history; text; internal format)
OFFSET 2,3 LINKS John Mason, Table of n, a(n) for n = 2..1000 John Mason, Java program John Mason, Explanation of formulas R. J. Mathar, Corrigendum to "polyomino enumeration results" (Parkin et al), vixra:1905.0474 (2019) Table 1 column 2. FORMULA For a set of recursive formulas to generate a(n), see the link for the Java program extract. EXAMPLE There is one polyomino, the domino, of width 2 and size 2. So a(2) = 1. There is one tromino, L-shaped, of width 2. So a(3) = 1. CROSSREFS Cf. A000105, A335711, A353067. Sequence in context: A131328 A054451 A309479 * A369697 A308775 A305335 Adjacent sequences: A352717 A352718 A352719 * A352721 A352722 A352723 KEYWORD nonn AUTHOR John Mason, Mar 31 2022 STATUS approved
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Last modified August 5 06:12 EDT 2024. Contains 374935 sequences. (Running on oeis4.) | 560 | 1,623 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2024-33 | latest | en | 0.586657 |
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Mathematics for Business and Economics MATH 1070
Marketplace > University of Connecticut > Mathematics (M) > MATH 1070 > Mathematics for Business and Economics
Mary Veum
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http://opencollege.com/simsim/php/ResourceManager.php?cmd=get_view&catID=1438&resID=73&outcat=1 | 1,500,842,130,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549424610.13/warc/CC-MAIN-20170723202459-20170723222459-00459.warc.gz | 238,527,786 | 4,963 | Microscope
The purpose of the microscope is to produce magnified images of small objects. Such an image is produced by an optical system that consists of an objective and an eyepiece (also known as theocular). An objective consists of a lens with a short focal length. The object of observation is placed beside the objective at a distance s > f1 near the first focus. The objective produces a magnified inverted image, available for observation through the eyepiece, which consists of a converging lens with a short focal length. When analyzing the path of the rays, remember that the eye generally accommodates to the distance of normal vision: d0 = 25 cm. The viewer can look at a small object directly putting it at a distance d0 from the eye, or through a microscope, examining the virtual image of the object, as produced by the eyepiece, from the same distance d0. The ratio of the angle that is subtended at the eye by the final image to the angle subtended at the unaided eye by the object is called the angular magnification. Since in both cases the object is viewed from the same distance d0, the angular magnification also coincides with the lateral magnification of a system of two lenses. The virtual image of an object observed through the eyepiece is always inverted. If this is inconvenient for any reason, the object itself can be inverted. Note that the magnification of the microscope is always characterized as a positive value. The magnification M is determined as the product of the magnifications of the objective and the eyepiece. In the case when f1 and f2 are much smaller than the distance Δ between them, the microscope's magnification is expressed by the formula M = (d0Δ)/(f1f2) . The standard microscope has Δ = 160 mm. An optical microscope can magnify objects by several hundreds of times. At particularly large magnification, diffraction phenomena appear. | 402 | 1,892 | {"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-2017-30 | longest | en | 0.947348 |
https://lessonplanet.com/search?keywords=nonstandard | 1,701,664,236,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100523.4/warc/CC-MAIN-20231204020432-20231204050432-00210.warc.gz | 405,832,936 | 45,639 | Worksheet
Curated OER
#### Nonstandard Units
For Students 1st - 2nd
Here is a very good measurement activity. Four problems are solved in which objects at home are measured using the specified nonstandard measurement units such as toothpicks, footsteps and teaspoons.
Worksheet
Curated OER
#### Nonstandard Units, Homework 17.1
For Students K - 3rd
Here is a very good estimating and measuring length with nonstandard units worksheet. Learners use paper clips and toothpicks to estimate and measure the real objects pictured in each problem, then write ten answers.
Worksheet
Curated OER
#### Compare Nonstandard Units
For Students 3rd - 4th
Looking for a solving word problems using nonstandard units learning exercise? This one's for you. Learners use paper clips, ones blocks, footsteps, and handprints to measure lengths. Students solve six problems.
Worksheet
Curated OER
#### Nonstandard Units, Practice 17.1
For Students 1st - 3rd
I like this measuring with nonstandard units worksheet! Learners use paper clips and connecting cubes to estimate and measure real objects, and solve 17 problems.
Worksheet
Curated OER
#### Nonstandard Units
For Students 2nd - 3rd
Measuring length with nonstandard units is the focus of this word problem worksheet. Youngsters problem solve using cubes, paper clips, footprints, and pencils. They solve six problems.
Lesson Plan
Curated OER
#### Nonstandard Measurement
For Teachers K - 2nd
Students find a partner with the same finger, arm and foot span as they have. In this nonstandard measurement lesson, students compare the length of their index finger, arm and feet. Students who don't match anyone find the closest...
Lesson Plan
Curated OER
#### By the Pound
For Teachers 3rd - 8th
Agriculture surrounds us every day; incorporate measuring tools into a study of Oklahoma's agricultural industry! Small groups read an informational text (included) before visiting stations where they investigate prices of various...
Lesson Plan
Curated OER
#### Measuring Real Stuff
For Teachers K - 1st
Looking for a good lesson on measurement? This one could be for ytou! Learners select and use appropriate nonstandard units to measure and compare lengths and weights of real-life objects. They use problem-solving skills to design a...
Worksheet
Curated OER
#### Measure and Color
For Students K - 1st
In this nonstandard measurement worksheet, young mathematicians look at the height of each of four lines, then color in the measurement bars up to the right height to match each line.
Worksheet
Curated OER
#### Measure the Pencils
For Students K - 1st
Excellent measuring practice is here in this math activity. Learners cut out the measuring bar at the bottom of the page and use it to measure four pencils. This is nonstandard measurement.
Lesson Plan
Curated OER
#### Measurement: Measuring Hand and Feet
For Teachers 2nd - 5th
Here is a good lesson which invites learners to practice measurement using standard and nonstandard measuring tools. In this measurement lesson, learners work with length, width, height, and area. They use standard and nonstandard...
Lesson Plan
Curated OER
#### Math-Measurement
For Teachers 1st
First graders examine measurement. In this measurement lesson, 1st graders practice using nonstandard units to help in measuring objects. Students listen to How Big is a Foot and complete an activity using unifix cubes to measure line...
Lesson Plan
Curated OER
#### Measurement Crafts- Cereal Hands and Feet
For Teachers K - 3rd
Students identify nonstandard units of measurement and investigate area of objects. In this geometry lesson plan, students trace their hands and glue cereal pieces to the print to find the area.
Worksheet
Curated OER
#### Nonstandard Units
For Students 2nd - 3rd
In this units of measurement instructional activity, students learn to use nonstandard units of measurement to measure the items in each problem.
Lesson Plan
NTTI
#### What Size Is It Anyway?
For Teachers K - 1st Standards
A fabulous plan on teaching basic concepts of measurement to young learners. Pupils use nonstandard units to measure length, practice using a ruler and a yardstick, compare the length of two objects, and perform estimations of length....
Lesson Plan
Curated OER
#### Length
For Teachers 3rd
Third graders examine and explore a variety of ways to measure length. They discuss how to measure large objects using the end-to-end method, and in small groups measure the length of the room using the end-to-end method with a...
Worksheet
Curated OER
#### Systems of Equations
For Students Higher Ed
In this systems of equations worksheet, students solve nonstandard systems of equations. This five-page worksheet contains twelve problems on the first page. The remainder of the pages contain detailed, handwritten solutions.
Worksheet
Curated OER
#### Nonstandard Units
For Students 3rd - 4th
In this non-standard units worksheet, students circle the best unit to use to measure given items, drawing or writing to explain their work. Houghton Mifflin text is referenced.
Lesson Plan
Curated OER
#### Cubed Containers
For Teachers 1st - 3rd
Learners explore volume in relation to centimeter cubes and other nonstandard units of measurement using small containers. They will use hands-on materials to practice this concept. You will need small objects and containers to set this up.
Lesson Plan
Utah Education Network (UEN)
#### Jack and the Beanstalk: Measurement Ideas
For Teachers K - 1st Standards
Have fun with Jack and his beanstalk! Primary learners will practice skills at various activity centers, including: weight measurement, money, art, nonstandard length measurement, problem solving, music, reading, and writing. Every...
Lesson Plan
Curated OER
#### Measuring Fun
For Teachers 1st
First graders practice measuring. In this measuring lesson, 1st graders read the story How Big Is a Foot? by Rolf Myller. They measure a partner using standard (ruler) and nonstandard (feet) units.
Lesson Plan
Curated OER
#### How Big is a Foot?
For Teachers 1st
First graders practice nonstandard measurement and estimating and measuring lengths of objects.
Lesson Plan
Curated OER
#### Tall Towers
For Teachers K - 2nd
Linear measurement is the focus of this math lesson. Youngsters work in groups to build towers with blocks. They build towers represented on index cards and determine how many blocks taller one tower is.
Printables
1
1
Curated OER
#### How Tall is the Gingerbread Man?
For Teachers Pre-K - 2nd Standards
The gingerbread man has finally been caught, now find out how tall he is with this measurement activity. Using Unifix® cubes, marshmallows, buttons, and Cheerios, young mathematicians measure and record the height of the gingerbread man... | 1,507 | 6,783 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2023-50 | latest | en | 0.91329 |
https://askfilo.com/physics-question-answers/the-diagram-shows-a-wire-carrying-a-current-in-the-direction-shown-there-is-a | 1,695,542,376,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506623.27/warc/CC-MAIN-20230924055210-20230924085210-00676.warc.gz | 138,322,111 | 20,012 | World's only instant tutoring platform
Question
Medium
Solving time: 3 mins
# The diagram shows a wire carrying a current in the direction shown. There is a magnetic field acting from left to right. The wire experiences a force acting out of the page. The current is now reversed.In which direction does the force on the wire now act?
A
into the page
B
out of the page
C
to the left
D
to the right
## Text solution
The direction of force on a current carrying wire placed normal to a magnetic field is given by Fleming's left hand rule , if we apply this rule for the given diagram , the direction of force will be downward if we change the direction of current So it will be into the page. | 158 | 698 | {"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-2023-40 | latest | en | 0.877654 |
https://us.metamath.org/mpeuni/difss2d.html | 1,713,268,444,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817081.52/warc/CC-MAIN-20240416093441-20240416123441-00895.warc.gz | 557,599,986 | 3,855 | Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > difss2d Structured version Visualization version GIF version
Theorem difss2d 4109
Description: If a class is contained in a difference, it is contained in the minuend. Deduction form of difss2 4108. (Contributed by David Moews, 1-May-2017.)
Hypothesis
Ref Expression
difss2d.1 (𝜑𝐴 ⊆ (𝐵𝐶))
Assertion
Ref Expression
difss2d (𝜑𝐴𝐵)
Proof of Theorem difss2d
StepHypRef Expression
1 difss2d.1 . 2 (𝜑𝐴 ⊆ (𝐵𝐶))
2 difss2 4108 . 2 (𝐴 ⊆ (𝐵𝐶) → 𝐴𝐵)
31, 2syl 17 1 (𝜑𝐴𝐵)
Colors of variables: wff setvar class Syntax hints: → wi 4 ∖ cdif 3931 ⊆ wss 3934 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1790 ax-4 1804 ax-5 1905 ax-6 1964 ax-7 2009 ax-8 2110 ax-9 2118 ax-10 2139 ax-11 2154 ax-12 2170 ax-ext 2791 This theorem depends on definitions: df-bi 209 df-an 399 df-or 844 df-tru 1534 df-ex 1775 df-nf 1779 df-sb 2064 df-clab 2798 df-cleq 2812 df-clel 2891 df-nfc 2961 df-v 3495 df-dif 3937 df-in 3941 df-ss 3950 This theorem is referenced by: oacomf1olem 8182 numacn 9467 ramub1lem1 16354 ramub1lem2 16355 mreexexlem2d 16908 mreexexlem3d 16909 mreexexlem4d 16910 acsfiindd 17779 dpjidcl 19172 clsval2 21650 llycmpkgen2 22150 1stckgen 22154 alexsublem 22644 bcthlem3 23921 pmtrcnelor 30728 lfuhgr 32357 neibastop2lem 33701 pibt2 34690 eldioph2lem2 39348 limccog 41890 fourierdlem56 42437 fourierdlem95 42476
Copyright terms: Public domain W3C validator | 776 | 1,564 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2024-18 | latest | en | 0.176396 |
https://rasbula.com/what-is-domain-in-math-no-longer-a-mystery/ | 1,652,735,350,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662512249.16/warc/CC-MAIN-20220516204516-20220516234516-00448.warc.gz | 525,332,628 | 14,625 | This calendar year, but the center had an extensive post listing numerous prominent works, together with background info and links to more. And it attempts to appease no one. That’s not the conclusion of the story. If we have a canoe trip, we’ll be home by sunset. If we don’t go swimming, then we are going to take a canoe trip.
It is possible to purchase domain names through google in addition to other google products and services like site builders and email. You may often determine the domain by taking a look at a graph. Mathway’s math solver is a superb tool to study your work at no cost. Now we’ve got a summary about the key components in our app, we can begin coding it. There are suggestions for choosing the suitable keywords below.
In summary, someone doesn’t have to do functional programming to have the ability to use functions. payforessay net Until now, we’ve restricted our focus to propositional logic. Now that we understand how to state things precisely, we’re all set to consider putting statements with each other to form arguments. Not understanding the proof wasn’t an alternative. Vainly I expected everybody to get the center of truth. In other words, thinking creatively in a precise method.
If John knows discrete mathematics, he’ll pass this program. I used to reside in a building that had a communal kitchen for more than 100 students. The case may be worse in the event the teacher is hard on you. No, for the full course.
## The Basic Facts of What Is Domain in Math
The tests weren’t calibrated to any typical scale, so it was hard to judge how well you were doing. I hope that I’ve achieved these goals. Functions are a unique type of relation.
## The War Against What Is Domain in Math
Then if you attempt to delete multiple order lines, the aggregate ensures there’s an error whenever there’s only a single line left. So that the trick to implementing Aggregate in FP is to think about aggregate both concerning data and Functions. You’re able to put in any true number, and you may acquire an output of any true number you may possibly think of. Accordingly, a minumum of one number has to be grater or equal than the average.
## The Ugly Side of What Is Domain in Math
If n isn’t a integer, it’s converted to an integer initially and then the square root is returned. The union symbol might be used for disjoint sets. The rule for specifying it is offered by more than one expression. A function is a particular kind of relation.
In reality the Domain is an fundamental part of the function. Applying the same paradigm to stateful components makes sense in a lot of cases. Yes definitely advise doing this, especially in regards to business domain knowledge. The expression finite mathematics can be applied to regions of the area of discrete mathematics that manages finite sets, particularly those areas pertinent to business.
Domain and range could possibly be limited to some discrete values, or they might incorporate all numbers everywhere, to infinity and beyond. They will provide you with a function and request that you discover the domain (and maybe the range, too). Here, we’ll alter the formatting in a variety of cells to Bold. Just watch out for negative indicators and other inclusions that may alter the range.
It’s shown as a double line within this table. Curly brackets are frequently used to enclose a set of information, and that means you know everything in the curly brackets belongs together. In emscripten’s case, the correct shift operator effectively rounds the pointer to the closest valid price, or so the load or store is in fact done somewhere else! The reflector may be a mesh screen or robust reflector. Both of these sets may be sets of any varieties of objects, for as long as they’re well defined sets.
As luck would have it, that’s quite rare. We advise that you do so, too. But this isn’t usual, and you shouldn’t expect it. | 813 | 3,920 | {"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-2022-21 | latest | en | 0.951329 |
https://www.jiskha.com/display.cgi?id=1203537659 | 1,516,335,468,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084887729.45/warc/CC-MAIN-20180119030106-20180119050106-00316.warc.gz | 925,568,639 | 4,095 | # arithmetic
posted by .
There are 7 girls on a bus.
each girl has 7 backpacks.
In each backpack there are 7 big cats.
For every big cat there are 7 little cats.
Question: How many legs are there in the bus?
• arithmetic -
7 LITTLE CATS x 4 LEGS = 28
28 X 7 = 196 LITTLE CATS LEGS
7 BIG CATS X 4 LEGS = 28
7 GIRLS X 2 LEGS = 14
BUS DRIVER HAS 2 LEGS
196 + 28 + 14 + 2 = 240
• arithmetic -
I interpreted this question differently than Tim. Here's what I came up with:
7 girls x 2 legs each = 14 legs
If each girl has 7 backpacks and there are 7 big cats in each....49 backpacks x 7 cats=343 big cats. Then, there are 7 little cats for each big ...343x7=2401 little cats.
Total cats 2401+343=2744 total cats which have 4 legs each, so there are 10976 cat legs. Add this to the 14 girl legs and you get 10990 total legs. You might check to be sure this is what you need.
• arithmetic -
Thank you so much!! I will see if that is correct.
• arithmetic -
Thank you again!!
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More Similar Questions | 830 | 2,902 | {"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 | 4 | CC-MAIN-2018-05 | latest | en | 0.89846 |
https://discuss.codechef.com/t/infosys-exam/102458 | 1,660,539,209,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572127.33/warc/CC-MAIN-20220815024523-20220815054523-00320.warc.gz | 221,117,560 | 3,574 | # Infosys Exam
Please provide the approach of the following ,I have tried but i didn’t came up with a good approach
You are given an array A of length N and an integer K. It is given that a number m is called special if gcd(m,Aj) = 1 for all 0<=j<N
Let R be an array containing all special number in the range [ 1 , K ] inclusive in sorted order . Your task is to return R.
Note :
A follows 0-based indexing.
Input Format
The first line contains an integer N , denoting the number of elements in A.
The next line contains an integer K , denoting the given integer.
Each line i of the N subsequent lines (where 0<=i<N) contains an integer describing A[i].
Constraints
1<=N<=10^5
1<=K<=10^5
1<=A[i]<=10^5
Sample Input Sample Output
3 1
5 5
1
2
3
4 1
5 2
3 4
3
5
7
4 1
1
1
1
1
1 | 242 | 784 | {"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-2022-33 | latest | en | 0.844371 |
http://www.chegg.com/homework-help/questions-and-answers/3-consider-a-perfectly-competitive-profit-maximizing-firm-facing-the-following-marginal-pr-q3601005 | 1,369,475,097,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368705926946/warc/CC-MAIN-20130516120526-00048-ip-10-60-113-184.ec2.internal.warc.gz | 391,926,485 | 8,048 | ## 3. Consider a perfectly competitive, profit-maximizing firm facing the following marginal product of labor function and prices:
3. Consider a perfectly competitive, profit-maximizing firm facing the following marginal product of labor function and prices: ? MPL = A(K/L)1/2 ? MPK = A(L/K)1/2 ? W = 20 ? R = 10 ? P = 2 ? K = 4 a. Does this firm’s production function exhibit diminishing returns to labor employment? Are labor and capital complements for this firm? Explain. b. What is the real wage rate paid by this firm? c. If total factor productivity (A) is 100, how much labor (L) would this firm want to employ? d. If the price of output (P) falls from $2/unit to$1/unit, what will the new real wage rate be? All else equal, how much labor would the firm want to employ at that wage rate? e. Assuming that total factor productivity is 100, graph this firm’s labor demand function (quantity of labor demanded graphed against the real wage paid for labor) for values of the real wage between 10 and 25. Be sure to plot at least 3 distinct points. | 260 | 1,053 | {"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} | 2.609375 | 3 | CC-MAIN-2013-20 | latest | en | 0.904352 |
https://www.ginac.de/ginac.git/?p=ginac.git;a=blob_plain;f=check/differentiation.cpp;hb=3ecae1b0cd75b32bec423fa8cca5626f7790ab42 | 1,642,363,005,000,000,000 | text/plain | crawl-data/CC-MAIN-2022-05/segments/1642320300010.26/warc/CC-MAIN-20220116180715-20220116210715-00292.warc.gz | 847,993,966 | 2,901 | /** @file differentiation.cpp * * Tests for symbolic differentiation, including various functions. */ /* * GiNaC Copyright (C) 1999 Johannes Gutenberg University Mainz, Germany * * This program is free software; you can redistribute it and/or modify * it under the terms of the GNU General Public License as published by * the Free Software Foundation; either version 2 of the License, or * (at your option) any later version. * * This program is distributed in the hope that it will be useful, * but WITHOUT ANY WARRANTY; without even the implied warranty of * MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the * GNU General Public License for more details. * * You should have received a copy of the GNU General Public License * along with this program; if not, write to the Free Software * Foundation, Inc., 59 Temple Place, Suite 330, Boston, MA 02111-1307 USA */ #include using namespace GiNaC; static unsigned check_diff(const ex &e, const symbol &x, const ex &d, unsigned nth=1) { ex ed = e.diff(x, nth); if ((ed - d).compare(exZERO()) != 0) { switch (nth) { case 0: clog << "zeroth "; break; case 1: break; case 2: clog << "second "; break; case 3: clog << "third "; break; default: clog << nth << "th "; } clog << "derivative of " << e << " by " << x << " returned " << ed << " instead of " << d << endl; clog << "returned:" << endl; ed.printtree(clog); clog << endl << "instead of" << endl; d.printtree(clog); return 1; } return 0; } // Simple (expanded) polynomials static unsigned differentiation1(void) { unsigned result = 0; symbol x("x"), y("y"); ex e1, e2, e, d; // construct bivariate polynomial e to be diff'ed: e1 = pow(x, -2) * 3 + pow(x, -1) * 5 + 7 + x * 11 + pow(x, 2) * 13; e2 = pow(y, -2) * 5 + pow(y, -1) * 7 + 11 + y * 13 + pow(y, 2) * 17; e = (e1 * e2).expand(); // d e / dx: d = 121 - 55*pow(x,-2) - 66*pow(x,-3) - 30*pow(x,-3)*pow(y,-2) - 42*pow(x,-3)*pow(y,-1) - 78*pow(x,-3)*y - 102*pow(x,-3)*pow(y,2) - 25*pow(x,-2) * pow(y,-2) - 35*pow(x,-2)*pow(y,-1) - 65*pow(x,-2)*y - 85*pow(x,-2)*pow(y,2) + 77*pow(y,-1) + 143*y + 187*pow(y,2) + 130*x*pow(y,-2) + 182*pow(y,-1)*x + 338*x*y + 442*x*pow(y,2) + 55*pow(y,-2) + 286*x; result += check_diff(e, x, d); // d e / dy: d = 91 - 30*pow(x,-2)*pow(y,-3) - 21*pow(x,-2)*pow(y,-2) + 39*pow(x,-2) + 102*pow(x,-2)*y - 50*pow(x,-1)*pow(y,-3) - 35*pow(x,-1)*pow(y,-2) + 65*pow(x,-1) + 170*pow(x,-1)*y - 77*pow(y,-2)*x + 143*x + 374*x*y - 130*pow(y,-3)*pow(x,2) - 91*pow(y,-2)*pow(x,2) + 169*pow(x,2) + 442*pow(x,2)*y - 110*pow(y,-3)*x - 70*pow(y,-3) + 238*y - 49*pow(y,-2); result += check_diff(e, y, d); // d^2 e / dx^2: d = 286 + 90*pow(x,-4)*pow(y,-2) + 126*pow(x,-4)*pow(y,-1) + 234*pow(x,-4)*y + 306*pow(x,-4)*pow(y,2) + 50*pow(x,-3)*pow(y,-2) + 70*pow(x,-3)*pow(y,-1) + 130*pow(x,-3)*y + 170*pow(x,-3)*pow(y,2) + 130*pow(y,-2) + 182*pow(y,-1) + 338*y + 442*pow(y,2) + 198*pow(x,-4) + 110*pow(x,-3); result += check_diff(e, x, d, 2); // d^2 e / dy^2: d = 238 + 90*pow(x,-2)*pow(y,-4) + 42*pow(x,-2)*pow(y,-3) + 102*pow(x,-2) + 150*pow(x,-1)*pow(y,-4) + 70*pow(x,-1)*pow(y,-3) + 170*pow(x,-1) + 330*x*pow(y,-4) + 154*x*pow(y,-3) + 374*x + 390*pow(x,2)*pow(y,-4) + 182*pow(x,2)*pow(y,-3) + 442*pow(x,2) + 210*pow(y,-4) + 98*pow(y,-3); result += check_diff(e, y, d, 2); return result; } // Trigonometric and transcendental functions static unsigned differentiation2(void) { unsigned result = 0; symbol x("x"), y("y"), a("a"), b("b"); ex e1, e2, e, d; // construct expression e to be diff'ed: e1 = y*pow(x, 2) + a*x + b; e2 = sin(e1); e = b*pow(e2, 2) + y*e2 + a; d = 2*b*e2*cos(e1)*(2*x*y + a) + y*cos(e1)*(2*x*y + a); result += check_diff(e, x, d); d = 2*b*pow(cos(e1),2)*pow(2*x*y + a, 2) + 4*b*y*e2*cos(e1) - 2*b*pow(e2,2)*pow(2*x*y + a, 2) - y*e2*pow(2*x*y + a, 2) + 2*pow(y,2)*cos(e1); result += check_diff(e, x, d, 2); d = 2*b*e2*cos(e1)*pow(x, 2) + e2 + y*cos(e1)*pow(x, 2); result += check_diff(e, y, d); d = 2*b*pow(cos(e1),2)*pow(x,4) - 2*b*pow(e2,2)*pow(x,4) + 2*cos(e1)*pow(x,2) - y*e2*pow(x,4); result += check_diff(e, y, d, 2); // construct expression e to be diff'ed: e2 = cos(e1); e = b*pow(e2, 2) + y*e2 + a; d = -2*b*e2*sin(e1)*(2*x*y + a) - y*sin(e1)*(2*x*y + a); result += check_diff(e, x, d); d = 2*b*pow(sin(e1),2)*pow(2*y*x + a,2) - 4*b*e2*sin(e1)*y - 2*b*pow(e2,2)*pow(2*y*x + a,2) - y*e2*pow(2*y*x + a,2) - 2*pow(y,2)*sin(e1); result += check_diff(e, x, d, 2); d = -2*b*e2*sin(e1)*pow(x,2) + e2 - y*sin(e1)*pow(x, 2); result += check_diff(e, y, d); d = -2*b*pow(e2,2)*pow(x,4) + 2*b*pow(sin(e1),2)*pow(x,4) - 2*sin(e1)*pow(x,2) - y*e2*pow(x,4); result += check_diff(e, y, d, 2); // construct expression e to be diff'ed: e2 = exp(e1); e = b*pow(e2, 2) + y*e2 + a; d = 2*b*pow(e2, 2)*(2*x*y + a) + y*e2*(2*x*y + a); result += check_diff(e, x, d); d = 4*b*pow(e2,2)*pow(2*y*x + a,2) + 4*b*pow(e2,2)*y + 2*pow(y,2)*e2 + y*e2*pow(2*y*x + a,2); result += check_diff(e, x, d, 2); d = 2*b*pow(e2,2)*pow(x,2) + e2 + y*e2*pow(x,2); result += check_diff(e, y, d); d = 4*b*pow(e2,2)*pow(x,4) + 2*e2*pow(x,2) + y*e2*pow(x,4); result += check_diff(e, y, d, 2); // construct expression e to be diff'ed: e2 = log(e1); e = b*pow(e2, 2) + y*e2 + a; d = 2*b*e2*(2*x*y + a)/e1 + y*(2*x*y + a)/e1; result += check_diff(e, x, d); d = 2*b*pow((2*x*y + a),2)*pow(e1,-2) + 4*b*y*e2/e1 - 2*b*e2*pow(2*x*y + a,2)*pow(e1,-2) + 2*pow(y,2)/e1 - y*pow(2*x*y + a,2)*pow(e1,-2); result += check_diff(e, x, d, 2); d = 2*b*e2*pow(x,2)/e1 + e2 + y*pow(x,2)/e1; result += check_diff(e, y, d); d = 2*b*pow(x,4)*pow(e1,-2) - 2*b*e2*pow(e1,-2)*pow(x,4) + 2*pow(x,2)/e1 - y*pow(x,4)*pow(e1,-2); result += check_diff(e, y, d, 2); // test for functions with two variables: atan2 e1 = y*pow(x, 2) + a*x + b; e2 = x*pow(y, 2) + b*y + a; e = atan2(e1,e2); /* d = pow(y,2)*(-b-y*pow(x,2)-a*x)/(pow(b+y*pow(x,2)+a*x,2)+pow(x*pow(y,2)+b*y+a,2)) +(2*y*x+a)/((x*pow(y,2)+b*y+a)*(1+pow(b*y*pow(x,2)+a*x,2)/pow(x*pow(y,2)+b*y+a,2))); */ /* d = ((a+2*y*x)*pow(y*b+pow(y,2)*x+a,-1)-(a*x+b+y*pow(x,2))* pow(y*b+pow(y,2)*x+a,-2)*pow(y,2))* pow(1+pow(a*x+b+y*pow(x,2),2)*pow(y*b+pow(y,2)*x+a,-2),-1); */ d = pow(1+pow(a*x+b+y*pow(x,2),2)*pow(y*b+pow(y,2)*x+a,-2),-1) *pow(y*b+pow(y,2)*x+a,-1)*(a+2*y*x) +pow(y,2)*(-a*x-b-y*pow(x,2))* pow(pow(y*b+pow(y,2)*x+a,2)+pow(a*x+b+y*pow(x,2),2),-1); result += check_diff(e, x, d); return result; } // Series static unsigned differentiation3(void) { symbol x("x"); ex e, d, ed; e = sin(x).series(x, exZERO(), 8); d = cos(x).series(x, exZERO(), 7); ed = e.diff(x); ed = static_cast(ed.bp)->convert_to_poly(); d = static_cast(d.bp)->convert_to_poly(); if ((ed - d).compare(exZERO()) != 0) { clog << "derivative of " << e << " by " << x << " returned " << ed << " instead of " << d << ")" << endl; return 1; } return 0; } unsigned differentiation(void) { unsigned result = 0; cout << "checking symbolic differentiation..." << flush; clog << "---------symbolic differentiation:" << endl; result += differentiation1(); result += differentiation2(); result += differentiation3(); if (!result) { cout << " passed "; clog << "(no output)" << endl; } else { cout << " failed "; } return result; } | 2,932 | 7,075 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2022-05 | latest | en | 0.663818 |
http://nextbreakpoint.com/nextfractal/example-use-opacity-to-compose-rules.html | 1,506,377,285,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818693459.95/warc/CC-MAIN-20170925220350-20170926000350-00108.warc.gz | 227,209,314 | 4,803 | NextFractal
# Use opacity to compose rules
In this example you will learn how to use opacity to compose color rule.
```fractal {
// Set region margins to left=-3.0, bottom=-1.5, right=0.0, top=1.5
// Declare state vector as [x,n] where x and n are built-in variables
orbit [<-3.0,-1.5>,<0.0,1.5>] [x,n] {
// Iterate for n from 0 to 200 stopping when mod2(x) > 40
loop [0, 200] (mod2(x) > 40) {
// Declare orbit equation where x is a state variable and w is current point of region
x = x * x + w;
}
}
// Set background color to alpha=1, red=0, green=0, blue=0
color [(1,0,0,0)] {
// Initialize variables c1 and c2
init {
c1 = mod2(x) / 1000;
c2 = <x>;
}
// Apply rule when n > 0 and set opacity to 1.0
rule (n > 0) [1.0] {
// Compute color components and set alpha component = 1
(1 + cos(c1 * 3 / pi)) / 2
}
// Apply rule when n > 0 and set opacity to 0.5
rule (n > 0) [0.5] {
// Compute color components and set alpha component = 1
(1 + sin(c2 * 5 / pi)) / 2
}
}
}```
Execute the script above in NextFractal and you will get the image below: | 368 | 1,045 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2017-39 | latest | en | 0.664449 |
http://www.chegg.com/homework-help/college-geometry-2nd-edition-chapter-5.3-solutions-9780321830951 | 1,475,095,560,000,000,000 | text/html | crawl-data/CC-MAIN-2016-40/segments/1474738661767.35/warc/CC-MAIN-20160924173741-00046-ip-10-143-35-109.ec2.internal.warc.gz | 391,276,687 | 16,382 | # College Geometry (2nd Edition) View more editions Solutions for Chapter 5.3
• 862 step-by-step solutions
• Solved by publishers, professors & experts
• iOS, Android, & web
Over 90% of students who use Chegg Study report better grades.
May 2015 Survey of Chegg Study Users
Chapter: Problem:
On your paper, sketch the diagram in the accompanying figure showing a translation of the letter “R." Draw a pair of lines ℓ and m such that “R” maps to its translated image under sms.
SAMPLE SOLUTION
Chapter: Problem:
• Step 1 of 1
A translation is composed of two reflections, each line parallel to each other, perpendicular to the line of translation, and separated by half of the total distance. The following illustrates the appropriate drawing.
In the picture, the length from the pre-image ‘R’ to the image ‘R’ is 2L, and the distance separating the parallel lines is L. Lines l and m are parallel to each other, and perpendicular to the line between any point and its image. Line l is that over which the reflection is made, and line m is the line over which reflection is made.
Corresponding Textbook
College Geometry | 2nd Edition
9780321830951ISBN-13: 0321830954ISBN: David KayAuthors:
This is an alternate ISBN. View the primary ISBN for: College Geometry 2nd Edition Textbook Solutions | 305 | 1,297 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2016-40 | latest | en | 0.892804 |
https://gimpppa.org/the-region-of-a-plane-inside-of-an-angle/ | 1,653,218,660,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662545326.51/warc/CC-MAIN-20220522094818-20220522124818-00468.warc.gz | 345,409,641 | 5,062 | ns am examining Euclidian geometry and also I i found it that any kind of angle divides a plane into 2 regions: an inside and an outside. Is over there a require for a proof of this (something follow me the present of Jordan theorem), or is it simply "obvious"?
Browsing the internet, ns came throughout a following simpler version: any type of line divides a airplane into 2 regions. Perhaps someone will find it relevant.
You are watching: The region of a plane inside of an angle
My present understanding is the the PSA dram a crucial role in this type of thing. Uneven you space doing the the analytic geometry way, in which instance the PSA must be somehow already "coded in" ...I think.
geometry intuition
re-publishing
cite
monitor
edited Nov 29 "13 at 12:07
request Nov 17 "13 in ~ 14:30
$endgroup$
18
| present 13 an ext comments
1
$egingroup$
Let $vec v$ (the vertex), $vec a$, $vec b$ (the rays) that an edge in $oldsymbol R^2$. The internal $I$ that the angle might be defined as $I:=\vec v+tcdotvec a+scdotvec bmid t,sinoldsymbol R_+cup\$. Have the right to you walk from here?
share
mention
follow
edited Nov 17 "13 at 16:59
answered Nov 17 "13 in ~ 15:06
Michael HoppeMichael Hoppe
$endgroup$
11
| show 6 much more comments
1
$egingroup$
Interesting trouble ... Not sure if this is totally correct however I tried using the plane separation axiom (PSA) twice.
See more: What State Has A Capital Named For The Seventh President Of The United States?
Between: unknown (along with point, line, on, and congruent, cf. Hilbert).
Same side: permit $itl$ it is in a line and let A and also B be two points which room not top top $itl.$ clues A and B space $extiton the same side$ of $itl$ if either $itl$ and also $leftarrow abdominal muscle ightarrow$ carry out not intersect at all, or if they carry out intersect yet the allude of intersection is not between A and B.
PSA: For any type of line $itl$ and also points $A, B, C$ which room not on $itl:$ (i) if A and also B space on the same side that $itl$ and A and C are on the same side the $itl,$ climate B and also C space on the same side the $itl;$ (ii) If A and B room not on the exact same side that $itl$ and also A and C are not ~ above the same side of $itl,$ then B and C are on the exact same side of $itl.$
The meaning of angle inner is
A allude lies in the interior or is an interior component of $angle BAC$ if it is ~ above the same side of abdominal as C and also the exact same side that AC together B.
(ref: http://www.mcs.uvawise.edu/msh3e/resources/geometryBook/geometryBook.html) | 706 | 2,581 | {"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} | 3.078125 | 3 | CC-MAIN-2022-21 | latest | en | 0.878184 |
https://docs.feelpp.org/toolboxes/latest/cfpdes/taylor-green-vortex/index.html | 1,721,878,195,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518532.66/warc/CC-MAIN-20240725023035-20240725053035-00092.warc.gz | 185,221,872 | 14,384 | # 2D Advection-diffusion with Taylor-Green Vortex
This page can be downloaded as a pdf file or as a Jupyter notebook file by clicking on the corresponding icons in the toolbar. It exemplifies the usage of the Coefficient form PDE toolbox in Python.
## 1. Introduction
We study the solution of a 2D advection-diffusion equation (used to model the dispersion of a contaminant) in presence of a Taylor-Green vortex $\beta(x,t)$. The problem can be solved for different values of the diffusion coefficient $\mu$.
## 2. Advection-diffusion with Taylor-Green Vortex
The problem is defined on the spatial domain $\Omega = (-1, 1)^2$ with Dirichlet boundary $\Gamma_D := (-1, 1) \times \{-1\}$ and Neumann boundary $\Gamma_N := \partial\Omega \setminus \Gamma_D$. The governing parametric PDE (pPDE) can be described as follows:
\begin{align} &\partial_t c - \mu\Delta c + \beta \cdot \nabla c = 0 &&\text{in } \Omega \times (0, T), \\ &\nabla c(x, t; \mu) \cdot n = 0 &&\text{on } \Gamma_N \times (0, T), \\ &c(x, t; \mu) = 0 &&\text{on } \Gamma_D \times (0, T), \\ &c(x, 0; \mu) = c_0(x; \mu) &&\text{in } \Omega. \end{align}
Where, the velocity field $\beta := (\sin(\pi x_1) \cos(\pi x_2), -\cos(\pi x_1) \sin(\pi x_2))^{\top}$, $x = (x_1, x_2)$, is a solenoidal field. The initial condition, $c_0(\cdot; \mu): \Omega \rightarrow \mathbb{R}$, is given by the sum of three Wendland functions $\bar{\psi}_{2,1}^i$ of radius 0.4 and centers located at $(-0.6, -0.6)$, $(0, 0)$ and $(0.6, 0.6)$. The Wendland functions are radial functions with compact support on the unit ball, and $\psi_{2,1}(r)$ has the following closed form
$\psi_{2,1}(r)=(3r + 1) \max \{0,(1-r)^3\}.$
Appropriate transformations have been applied to $\psi_{2,1}(r)$ in order to get the scaled and translated Wendland functions $\bar{\psi}_{2,1}^i$.
The velocity field $\beta$, the initial condition $c_0(\cdot; \mu)$, and the boundaries are illustrated in the following figure.
## 3. Configuring the simulation
### 3.1. Setting up the Feel++ Environment
To set up the Feel++ environment, we create an environment and set the associated repository for the results.
Set the Feel++ environment with a global repository
``````import sys
import feelpp
import feelpp.toolboxes.core as tb
from feelpp.toolboxes.cfpdes import *
sys.argv = ["feelpp_cfpdes_Taylor_Green"]
e = feelpp.Environment(sys.argv,
opts=tb.toolboxes_options("coefficient-form-pdes", "cfpdes"),
config=feelpp.globalRepository("cfpdes-Taylor_Green"))``````
Results
```[ Starting Feel++ ] application feelpp_cfpdes_Taylor_Green
. feelpp_cfpdes_Taylor_Green files are stored in /home/feelppdb/cfpdes-Taylor_Green/np_1
.. logfiles :/home/feelppdb/cfpdes-Taylor_Green/np_1/logs```
### 3.2. Geometry and Mesh
We will construct a geometric domain for our Finite Element Model.
In the 2D case, the domain is a square with side lengths of 2 units, centered at the origin, and can be represented as `[-1,1]`2 . The bottom edge is designated as the Dirichlet boundary ($\Gamma_D$), while the left, top, and right edges form the Neumann boundary ($\Gamma_N$).
We propose also the creation of a 3D case, where the domain is the cube `[-1,1]`3. One face of the cube is assigned as the Dirichlet boundary ($\Gamma_D$), and the rest of the faces form the Neumann boundary ($\Gamma_N$).
Both the 2D and 3D geometries allow for customization of mesh sizes.
``````def generateGeometry(filename, dim=2, hsize=0.1):
"""create gmsh mesh
Args:
filename (str): name of the file
dim (int): dimension of the mesh
hsize (float): mesh size
"""
h = {};
dim = {};
""".format(hsize, dim)
if dim == 2:
geo += """
Rectangle(1) = {-1, -1, 0, 2, 2, 0};
Characteristic Length{ PointsOf{ Surface{1}; } } = h;
Physical Curve("Gamma_D") = {1};
Physical Curve("Gamma_N") = {2,3,4};
Physical Surface("Omega") = {1};
"""
elif dim == 3:
geo += """
Box(1) = {-1, -1, -1, 2, 2, 2};
Characteristic Length{ PointsOf{ Volume{1}; } } = h;
Physical Surface("Gamma_D") = {5};
Physical Surface("Gamma_N") = {1,2,3,4,6};
Physical Volume("Omega") = {1};
"""
with open(filename, 'w') as f:
# Write the string to the file
f.write(geo)
def getMesh(filename, hsize=0.05, dim=2, verbose=False):
"""create mesh
Args:
filename (str): name of the file
hsize (float): mesh size
dim (int): dimension of the mesh
verbose (bool): verbose mode
"""
import os
for ext in [".msh",".geo"]:
f = os.path.splitext(filename)[0] + ext
if os.path.exists(f):
os.remove(f)
if verbose:
print(f"generate mesh {filename} with hsize={hsize} and dimension={dim}")
generateGeometry(filename=filename, dim=dim, hsize=hsize)
mesh = feelpp.load(feelpp.mesh(dim=dim, realdim=dim), filename, hsize)
return mesh``````
### 3.3. Defining Model Properties
The next step is to define the model properties. We achieve this by creating a JSON file containing the following information:
• The PDE coefficients, which follow the format: $d \frac{\partial u}{\partial t} + \nabla \cdot (-c \nabla u - \alpha u + \gamma) + \beta \cdot \nabla u + a u = f$, more details about the coefficients can be found in the CFPDEs documentation.
• The approximation space (e.g., continuous piecewise polynomial finite elements $P^k \text{ for } k=0,1,2$)
• Domain and boundary conditions, using the markers defined previously in the geometry file.
• Initial condition
• Post-processing (e.g., exporting the solution, computing the error, …)
More details about modelling using JSON files can be found in the JSON documentation.
In the following code, we created a lambda function called `Taylor_Green_json` to generate the JSON file depending on the polynomial order `k` and the diffusion coefficient $\mu=\frac{1}{Pe}$.
``````Taylor_Green_json = lambda order,dim=2,mu=(1/30),name="c": {
"Name": "Taylor_Green",
"ShortName": "Taylor_Green",
"Models":
{
f"cfpdes-{dim}d":
{
},
"setup":{
"unknown":{
"basis":f"Pch{order}",
"name":f"{name}",
"symbol":"c"
},
"coefficients":{
"d":"1.",
"c":f"{mu}",
"beta":"{sin(pi*x)*cos(pi*y),-cos(pi*x)*sin(pi*y)}:x:y" if dim==2 else "{sin(pi*x)*cos(pi*y)*cos(pi*z),-0.5*cos(pi*x)*sin(pi*y)*cos(pi*z),-0.5*cos(pi*x)*cos(pi*y)*sin(pi*z)}:x:y:z",
"f":"0."
}
}
}
},
"Materials":
{
"Omega":
{
"markers":["Omega"]
}
},
"BoundaryConditions":
{
{
"Dirichlet":
{
"g":
{
"markers":["Gamma_D"],
"expr":"0"
}
},
"Neumann":{
"m":
{
"markers":["Gamma_N"],
"expr":"0"
}
}
}
},
"InitialConditions":
{
{
f"{name}":
{
"Expression":
{
"wendland":
{
"markers":["Omega"],
"expr":"((1-2.5*sqrt((x+0.6)^2+(y+0.6)^2))>0)*(1-2.5*sqrt((x+0.6)^2+(y+0.6)^2))^3*(3*(2.5*sqrt((x+0.6)^2+(y+0.6)^2))+1)+((1-2.5*sqrt((x)^2+(y)^2))>0)*(1-2.5*sqrt((x)^2+(y)^2))^3*(3*(2.5*sqrt((x)^2+(y)^2))+1)+((1-2.5*sqrt((x-0.6)^2+(y-0.6)^2))>0)*(1-2.5*sqrt((x-0.6)^2+(y-0.6)^2))^3*(3*(2.5*sqrt((x-0.6)^2+(y-0.6)^2))+1):x:y" if dim==2 else "((1-2.5*sqrt((x+0.6)^2+(y+0.6)^2+(z+0.6)^2))>0)*(1-2.5*sqrt((x+0.6)^2+(y+0.6)^2+(z+0.6)^2))^3*(3*(2.5*sqrt((x+0.6)^2+(y+0.6)^2+(z+0.6)^2))+1)+((1-2.5*sqrt((x)^2+(y)^2+(z)^2))>0)*(1-2.5*sqrt((x)^2+(y)^2+(z)^2))^3*(3*(2.5*sqrt((x)^2+(y)^2+(z)^2))+1)+((1-2.5*sqrt((x-0.6)^2+(y-0.6)^2+(z-0.6)^2))>0)*(1-2.5*sqrt((x-0.6)^2+(y-0.6)^2+(z-0.6)^2))^3*(3*(2.5*sqrt((x-0.6)^2+(y-0.6)^2+(z-0.6)^2))+1):x:y:z"
}
}
}
}
},
"PostProcess":
{
f"cfpdes-{dim}d":
{
"Exports":
{
"fields":["all"]
}
}
}
}``````
#### 3.3.1. Example: Using the JSON File
Let’s use the `Taylor_Green_json` function to generate a JSON file with specific parameters:
• Polynomial order: 2
• Dimension: 2
• Mu: 1/30
``Taylor_Green_json(order=2, dim=2, mu=1/30, name="c")``
Results
```{'Name': 'Taylor_Green',
'ShortName': 'Taylor_Green',
'name': 'c',
'symbol': 'c'},
'coefficients': {'d': '1.',
'c': '0.03333333333333333',
'beta': '{sin(pi*x)*cos(pi*y),-cos(pi*x)*sin(pi*y)}:x:y',
'f': '0.'}}}},
'Materials': {'Omega': {'markers': ['Omega']}},
'BoundaryConditions': {'AdvDiff': {'Dirichlet': {'g': {'markers': ['Gamma_D'],
'expr': '0'}},
'Neumann': {'m': {'markers': ['Gamma_N'], 'expr': '0'}}}},
'InitialConditions': {'AdvDiff': {'c': {'Expression': {'wendland': {'markers': ['Omega'],
'expr': '((1-2.5*sqrt((x+0.6)^2+(y+0.6)^2))>0)*(1-2.5*sqrt((x+0.6)^2+(y+0.6)^2))^3*(3*(2.5*sqrt((x+0.6)^2+(y+0.6)^2))+1)+((1-2.5*sqrt((x)^2+(y)^2))>0)*(1-2.5*sqrt((x)^2+(y)^2))^3*(3*(2.5*sqrt((x)^2+(y)^2))+1)+((1-2.5*sqrt((x-0.6)^2+(y-0.6)^2))>0)*(1-2.5*sqrt((x-0.6)^2+(y-0.6)^2))^3*(3*(2.5*sqrt((x-0.6)^2+(y-0.6)^2))+1):x:y'}}}}},
'PostProcess': {'cfpdes-2d': {'Exports': {'fields': ['all']}}}}```
### 3.4. Time Scheme
The time scheme is defined in a `.cfg` file.
The Model CFG (.cfg) files allow to pass command line options to Feel++ applications. In particular, it allows to
• setup the output directory
• setup the mesh
• setup the time scheme
• define the solution strategy and configure the linear/non-linear algebraic solvers.
and other options that are specific to the toolbox being used.
For further details, see the CFG documentation.
``````Taylor_Green_cfg = lambda time_order=2, t0=0, T=0.2, dt=0.1: f"""
case.dimensions=2
[cfpdes]
#filename=\$cfgdir/Taylor_Green.json
pc-type=gamg
ksp-converged-reason=
#verbose=1
#solver=Newton
snes-monitor=1
reuse-prec=1
#use-cst-matrix=0
#use-cst-vector=0
#snes-line-search-type=l2#basic
#time-stepping=Theta
#stabilization=1
#stabilization.type=unusual-gls #supg#unusual-gls #gls
order={time_order}
[ts]
time-initial={t0}
time-step={dt}
time-final={T}
restart.at-last-save=true
"""``````
#### 3.4.1. Example: Using the CFG File
In this scheme, we use a Backward Differentiation Formula (BDF) of order 2. The time step is 0.01 and the final time is T=2.5s.
``````cfg_str = Taylor_Green_cfg(time_order=2, t0=0, T=2.5, dt=0.01)
with open("Taylor_Green.cfg", "w") as f:
f.write(cfg_str)``````
## 4. Solving the Problem and Exporting the Results
To solve the problem, we define a function called `SolveTG` which takes the mesh size, JSON and cfg files as inputs, initializes the problem (mesh, properties, and configuration), solves it iteratively for each time step, and exports the results.
``````def SolveTG(hsize, json, cfg, dim=2, verbose=False, measure=False):
if verbose:
print(f"Solving the Taylor green vortex advection-diffusion problem for hsize = {hsize}...")
feelpp.Environment.setConfigFile(cfg)
measures = []
else:
print("============================================================\n")
print("============================================================\n")
if measure:
if measure:
return measures``````
### 4.1. Example Usage
Now, let’s see an example of how to use the `SolveTG` function to solve the Taylor-Green Vortex problem with specific parameters.
• `0.04`: Mesh size.
• `order=2`: Order of the polynomial used in the approximation space.
• `dim=2`: Dimension of the problem.
• `mu=1/30`: Value of the diffusion coefficient $\mu$.
• `"Taylor_Green.cfg"`: The name of the configuration file that we generated in the previous section.
``SolveTG(0.04, Taylor_Green_json(order=2, dim=2, mu=1/30, name="c"), "Taylor_Green.cfg")``
The choice of the mesh size is not arbitrary and is based on the following criteria: An advection-diffusion problem is said to be convection-dominated when the advection term dominates the diffusion term. In this case, the solution and its gradient are not bounded, in other words, the solution is not stable. This phenomenon is captured by the numerical Péclet number $Pe_{num} = \frac{||\beta||_{L^\infty} h}{2 \mu}$, where $h$ is the mesh size and $||\beta||_{L^\infty}$ is the $L^\infty$ norm of the velocity field $\beta$. We aim to choose a mesh size such that $Pe_{num} \leq 1$. In our case, we have $||\beta||_{L^\infty} = 1$ and $\mu \in P := [1/50, 1/10$]. Therefore, we need to choose $h$ such that $h \leq 0.04$.
## 5. Visualizing the Results in the Notebook
Once the Taylor-Green Vortex problem is solved, it’s useful to visualize the results to gain insights into the behavior of the solution over time. In this section, we use `pyvista` library for visualization directly within the notebook. We will be displaying the solution at different time instances.
To run this visualization, you need to have the `pyvista` and `xvfbwrapper` Python packages installed.
``````import vtk
import pyvista as pv
import numpy as np
import os
try:
from xvfbwrapper import Xvfb
vdisplay = Xvfb()
vdisplay.start()
except:
exit(0)
# Define the path to the case file directory
case_path = os.path.abspath("cfpdes-2d.exports/Export.case")
# Extract the number of time steps
# Create the Plotter
plotter = pv.Plotter()
# Specify the desired time steps
desired_times = [0.2, 0.8, 1.4, 2.0, 2.0]
# Get the time values as a list
time_values = [reader.GetTimeSets().GetItem(0).GetValue(i) for i in range(num_time_steps)]
# Loop through the desired time steps
for time in desired_times:
closest_time = min(time_values, key=lambda x: abs(x - time))
# Set the reader's time step
# Convert the output to a PyVista mesh
# Rescale the scalar data
block_id = 0
block = mesh[block_id]
scalars /= np.max(np.abs(scalars))
# Add the rescaled mesh to the Plotter | 4,305 | 13,004 | {"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": 2, "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": 1, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2024-30 | latest | en | 0.73661 |
http://math.eretrandre.org/tetrationforum/printthread.php?tid=87&page=4 | 1,585,529,677,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370496330.1/warc/CC-MAIN-20200329232328-20200330022328-00397.warc.gz | 105,954,030 | 2,974 | complex iteration (complex "height") - Printable Version +- Tetration Forum (https://math.eretrandre.org/tetrationforum) +-- Forum: Tetration and Related Topics (https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=1) +--- Forum: Mathematical and General Discussion (https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=3) +--- Thread: complex iteration (complex "height") (/showthread.php?tid=87) Pages: 1 2 3 4 RE: complex iteration (complex "height") - bo198214 - 03/26/2008 That you believe me, here is the regular $\frac{1}{e}[4]t$ for $t$ in the range of -1.5..5, the value 0 is of course taken for $t=-1$. [attachment=281] RE: complex iteration (complex "height") - Ivars - 03/27/2008 I see. And a purely imaginary value as well somewhere between -1 and -1.5. Now it gets too fast for me, may be You can explain how it all comes together when You feel like it. It is really exciting. I was wondering if such spirals can form some basis for expansion of complex functions, maybe resulting from tetration, in series , similar to Fourier (not that I know much about it, but spiral basis would seem so natural for many things happening in nature). Ivars RE: complex iteration (complex "height") - Ivars - 03/28/2008 Since I still can not iterate complex numbers, let me make a few opportunistic wild quesses which look nice ${1/e}[4]{-\Omega}=\Omega+I*(1-\Omega)=0.567143+I*0.432857...$ ${1/e}[4]{-1/\Omega} = \Omega+ I*{(1/\Omega}-1)=0.567143+I*0.763224..$ ${1/e}[4]{\Omega}=\Omega+I*\Omega^{1/(1-\Omega)}=0.567143+I*0.26975.$ If one of them is close, even real or imaginary part?. Ivars RE: complex iteration (complex "height") - bo198214 - 03/28/2008 Ivars Wrote:Since I still can not iterate complex numbers, let me make a few opportunistic wild quesses which look nice ${1/e}[4]{-\Omega}=\Omega+I*(1-\Omega)=0.567143+I*0.432857...$ ${1/e}[4]{-1/\Omega} = \Omega+ I*{(1/\Omega}-1)=0.567143+I*0.763224..$ ${1/e}[4]{\Omega}=\Omega+I*\Omega^{1/(1-\Omega)}=0.567143+I*0.26975.$ If one of them is close, even real or imaginary part?.No, quite wrong guesses: \begin{align*} \frac{1}{e}[4](-\Omega)&=0.4035614-0.4516711*I\\ \frac{1}{e}[4]\frac{-1}{\Omega}&=1.0316317+1.0778810*I\\ \frac{1}{e}[4]{\Omega}&=0.4921705+0.2519841*I \end{align*} RE: complex iteration (complex "height") - Ivars - 03/28/2008 Henryk, May be You could send me/publish here a table of values? I was trying to provoke you... I am almost sure the non-integer part of real positive/negative iterations will hide some numerical surprises which will allow to establish (perhaps) some other analytical links in some parts of the spiral then pure numerical iteration according to Your GREAT formula. And if not,it will show we should look into complex heights, or elsewhere ( pentation etc). I still can not help being more impressed by your discovery than I am able to utilize it, but its only few days old, so there is still time. There must be heaps of insights. BTW, if You go back more with negative heights, like -2, You enter positive real values of result , while -3 leads to negative etc, 4 to positive, -5 to negative etc. I think I mentioned somewhere that such oscillations may happen at least for h(i^(1/i) and h(-1^(1/-1)) as +- i and +- 1 as powers of i and -1 are used (i^n^(1/i^n)), (-1^n) ^(1/(-1)^n)) . Perhaps this spiral must be placed orthogonally to$b=a^{(1/a)}[4]infinity$ value =a in every point of the plane in Gottfrieds spider graph? Or attached to every such point in another 2D space....giving the 4D space of h(z^(1/z))=z. Excuse me , I am rambling a little, at work, no time to think carefully. Ivars | 1,088 | 3,621 | {"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": 0, "wp_latex": 0, "mimetex.cgi": 11, "/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-2020-16 | latest | en | 0.787041 |
https://www.statsmodels.org/0.6.1/generated/statsmodels.regression.linear_model.OLS.html | 1,652,704,275,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662510117.12/warc/CC-MAIN-20220516104933-20220516134933-00421.warc.gz | 1,233,052,460 | 4,552 | # statsmodels.regression.linear_model.OLS¶
class `statsmodels.regression.linear_model.``OLS`(endog, exog=None, missing='none', hasconst=None, **kwargs)[source]
A simple ordinary least squares model.
Parameters: endog : array-like 1-d endogenous response variable. The dependent variable. exog : array-like A nobs x k array where nobs is the number of observations and k is the number of regressors. An intercept is not included by default and should be added by the user. See `statsmodels.tools.add_constant`. missing : str Available options are ‘none’, ‘drop’, and ‘raise’. If ‘none’, no nan checking is done. If ‘drop’, any observations with nans are dropped. If ‘raise’, an error is raised. Default is ‘none.’ hasconst : None or bool Indicates whether the RHS includes a user-supplied constant. If True, a constant is not checked for and k_constant is set to 1 and all result statistics are calculated as if a constant is present. If False, a constant is not checked for and k_constant is set to 0.
Notes
No constant is added by the model unless you are using formulas.
Examples
```>>> import numpy as np
>>>
>>> import statsmodels.api as sm
>>>
>>> Y = [1,3,4,5,2,3,4]
>>> X = range(1,8)
>>>
>>> model = sm.OLS(Y,X)
>>> results = model.fit()
>>> results.params
array([ 2.14285714, 0.25 ])
>>> results.tvalues
array([ 1.87867287, 0.98019606])
>>> print(results.t_test([1, 0])))
<T test: effect=array([ 2.14285714]), sd=array([[ 1.14062282]]), t=array([[ 1.87867287]]), p=array([[ 0.05953974]]), df_denom=5>
>>> print(results.f_test(np.identity(2)))
<F test: F=array([[ 19.46078431]]), p=[[ 0.00437251]], df_denom=5, df_num=2>
```
Attributes
weights (scalar) Has an attribute weights = array(1.0) due to inheritance from WLS.
Methods
`fit`([method, cov_type, cov_kwds, use_t]) Full fit of the model. `fit_regularized`([method, maxiter, alpha, ...]) Return a regularized fit to a linear regression model. `initialize`() `loglike`(params) The likelihood function for the clasical OLS model. `predict`(params[, exog]) Return linear predicted values from a design matrix. `whiten`(Y) OLS model whitener does nothing: returns Y.
Attributes
`df_model` The model degree of freedom, defined as the rank of the regressor matrix minus 1 if a constant is included. `df_resid` The residual degree of freedom, defined as the number of observations minus the rank of the regressor matrix. `endog_names` `exog_names` | 670 | 2,426 | {"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-2022-21 | latest | en | 0.599815 |
http://mathhelpforum.com/math-topics/27615-physics-help-kinematic-equations.html | 1,526,867,158,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794863901.24/warc/CC-MAIN-20180521004325-20180521024325-00429.warc.gz | 191,786,982 | 11,991 | Thread: Physics Help-Kinematic Equations
1. Physics Help-Kinematic Equations
Here is the question....
A car and a motorcycle start from rest at the same time on a straight track, but the motorcycle is 25.0 m behind the car. The car accelerates at a uniform rate of 3.70 m/s^{2} and the motorcycle at a uniform rate of 4.40 m/s^{2}. (a) How much time elapses before the motorcycle overtakes the car? (b) How far will each have traveled during that time? (c) How far ahead of the car will the motorcycle be 2.00 s later? (Both vehicles are still accelerating.)
I do not even know how to begin on this one. I know for the first part I need to find the final time, but I do not know how to set it up to do that. I am stressing major about this and in desperate need of some help! Thanks in advance!
2. Hint 1:
If an object starts to accelerate (from $\displaystyle V_0 = 0$) at a constant rate a, it'll take x distance in t time.
(The units must be equivalent. For example x=meters, t=seconds, a=m/s^2)
$\displaystyle x = \frac{1}{2}at^2$
Hint 2:
The motorcycle is 25 meters behind the car. So can we say $\displaystyle x_{car} + 25 = x_{motorcycle}$ ?
3. OK.... Still not getting it unfortunately. I have literally sat and stared at this Q for hours. Seems to be flying right past me. I've done well up until now, but I just cannot seem to get this.
4. In t seconds, the car will travel $\displaystyle x_{car} = \frac{1}{2}(3.70)t^2$ and the motorcycle will travel $\displaystyle x_{motorcycle} = \frac{1}{2}(4.4)t^2$.
Now imagine the scene. The motorcycle is 25 m behind the car and they start to accelerate. By the time the motorcycle catches up the car, the motorcycle has gone 25 meters more than the car. So $\displaystyle x_{motorcycle} = x_{car} + 25$.
Put them together,
$\displaystyle \frac{1}{2}(4.4)t^2 = \frac{1}{2}(3.70)t^2 + 25$
Solving this will give you the time $\displaystyle t$.
5. Let's take Wingless's hints even further:
Since the motorcycle is 25m behind the car, we can say that when the car and motorcycle meet, the motorcycle will have traveled 25m more than the car, but that makes an initial position of -25m since it was that far behind the car's position (0).
Now, we look at the kinematic equation for displacement:
$\displaystyle x_f = x_0 + v_0t + \frac{1}{2}at^2$
Now we solve it for the car and the motorcycle:
Car:
$\displaystyle x_f = 0 + 0t + \frac{1}{2}(3.7)t^2$
$\displaystyle x_f = \frac{1}{2}(3.7)t^2$
Motorcyle:
$\displaystyle x_f = -25 + 0t + \frac{1}{2}(4.4)t^2$
$\displaystyle x_f = \frac{1}{2}(4.4)t^2 - 25$
Now we see that if you meet up with someone, your final position will be exactly the same, so we now have an equivalence between the motorcycle and the car.
$\displaystyle \frac{1}{2}(3.7)t^2 = \frac{1}{2}(4.4)t^2 - 25$
Put like terms on the same side and make the 25 positive:
$\displaystyle 25 = \frac{1}{2}(4.4)t^2 - \frac{1}{2}(3.7)t^2$
Now we multiply each acceleration by $\displaystyle \frac{1}{2}$
$\displaystyle 25 = 2.2t^2 - 1.85t^2$
Combine like terms:
$\displaystyle 25 = 0.35t^2$
Divide:
$\displaystyle 71.43 = t^2$
Find t:
$\displaystyle 8.45 = t$
Therefore it takes 8.45 seconds for the motorcycle to catch up to the car.
Try b and c using these.
6. Aryth's solution is more general. You can solve questions that include a starting velocity or displacement by using his equations
7. Thanks so much!!
For b I got 157 m for the motorcycle and 132 m for the car. For c I got that the motorcycle will be 38 m in front of the car.... but I'm not sure that's right.
8. For b, you take the time at which they met, and plug it in to the final position equations for each vehicle that we derived in part a:
$\displaystyle x_{car} = \frac{1}{2}(3.7)(8.45)^2$
$\displaystyle x_{car} = 132.1m$
$\displaystyle x_{motorcycle} = \frac{1}{2}(4.4)(8.45)^2 + 25^{**}$
**(Notice we add 25 to this one as opposed to subtracting it before, when considering total distance traveled, he traveled $\displaystyle \frac{1}{2}(4.4)(8.45)^2$ starting at 0m. But we know that he started at -25m, so he traveled an extra 25m.)
$\displaystyle x_{motorcycle} = 182.1m$
For c, you use the distances calculated above (Distance for motorcycle from 0 = 157), plus the acceleration for two seconds:
$\displaystyle x_{car} = 132.1 + \frac{1}{2}(3.7)(2)^2$
$\displaystyle x_{car} = 139.5m$
$\displaystyle x_{motorcycle} = 157.1 + \frac{1}{2}(4.4)(2)^2$
$\displaystyle x_{motorcycle} = 165.9m$
$\displaystyle x_{ahead} = 165.9 - 139.5 = 26.4m$
There you are.
I hope you know when dealing with the total distance traveled you had to ADD 25, and then you had to subtract 25 to deal with the distance away from the car since:
Reference Point for Total Distance = -25
Reference Point for Distance ahead of Car = 0
,
a car and a motorcycle start from rest
Click on a term to search for related topics. | 1,471 | 4,877 | {"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} | 3.796875 | 4 | CC-MAIN-2018-22 | latest | en | 0.932071 |
https://www.teacherspayteachers.com/Product/3MD5-Task-Cards-Assessments-Centers-and-More-2389589 | 1,521,582,448,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257647545.54/warc/CC-MAIN-20180320205242-20180320225242-00616.warc.gz | 935,701,152 | 17,251 | Total:
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# 3.MD.5 -Task Cards, Assessments, Centers and More
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Area of Square Units
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3.OA.3 Multiplication and Division Word Problems
3.OA.4 Missing Number in Multiplication and Division
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\$3.50 | 454 | 1,638 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2018-13 | latest | en | 0.725703 |
https://electronicsforyou.in/8051-program-to-arrange-numbers-in-ascending-order/ | 1,718,509,862,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861640.68/warc/CC-MAIN-20240616012706-20240616042706-00697.warc.gz | 195,670,986 | 38,339 | # 8051 Program to Arrange Numbers in Ascending Order
## Problem
Write an assembly language program for microcontroller 8051 to arrange numbers in ascending order from an array of 10 numbers.
Assume array of the ten bytes is stored in external memory of 8051 microcontroller from memory location 4000H.
## Algorithm
• To arrange the numbers in ascending order, first compare two numbers.
• If first number > second number, then interchange these numbers.
• Then compare second number and third number, interchange if needed.
• Continue same process for remaining numbers in the array.
• We have to repeat process until we get arranged data in ascending order.
• For this process, two counters are required.
• Byte counter, for accessing numbers from array.
• Pass counter for repeating the comparison process.
Step 1: Initialize a counter for comparison (Pass counter).
Step2: Initialize memory pointer to read number from the array.
Step3: Initialize byte counter.
Step 4: Read numbers from array.
Step 5: Compare two numbers.
Step 6: If number less than or equal to next number, then go to step 8.
Step 7: Replace number with next number which is largest.
Step 8: Increment memory pointer to read next number in the array.
Step 9: Decrement byte counter by 1.
Step 10: If byte counter is not equal to zero then go to step 4.
Step 11: Decrement pass counter by 1.
Step 12: If pass counter is not equal to zero then go to step 2.
Step 13: Stop.
## Program
MOV R0, #0AH ; Initialize pass counter.
REP1: MOV DPTR, #4000H ; Initialize memory pointer
MOV R1, #0AH ; Initialize byte counter
REPEAT: MOV R2, DPL ; Save the lower byte address
MOVX A, @DPTR ; Read the number from array
MOV 0F0H, A ; Store the number in register B
INC DPTR ; Increment memory pointer
MOVX A, @DPTR ; Take the next number from array
CJNE A, 0F0H, NEXT ; Compare number with next number
NEXT: JNC SKIP ; If number>next number then go to SKIP
MOV DPL, R2 ; Else exchange the number with next number
MOVX @DPTR, A ; Copy greater number to memory location
INC DPTR ; Increment memory pointer
MOV A, 0F0H
MOVX @DPTR, A
SKIP: DJNZ R1, REPEAT ; Decrement byte counter by 1, if byte counter≠ 0 then go to REPEAT.
DJNZ R0, REP1 ; Decrement pass counter if not zero then go to REP1
STOP: AJMP STOP ; Stop
#### Note:
The only difference between assembly language program to arrange numbers ascending/descending order for 8051 microcontroller is
• For ascending order NEXT: JNC SKIP instruction is used.
• For descending order NEXT: JC SKIP (11th line of program) instruction is used. | 653 | 2,586 | {"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-2024-26 | latest | en | 0.768169 |
https://www.physicsforums.com/threads/gravitational-attraction.279666/ | 1,540,166,538,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583514437.69/warc/CC-MAIN-20181021224001-20181022005501-00398.warc.gz | 1,010,625,593 | 19,732 | # Gravitational attraction
1. Dec 14, 2008
### Perillux
I'm just curious how long it would take for two 1kg masses separated by 1m to attract each other gravitationally in empty space. The formula for gravitational force is:
$$F_{g} = G \frac{m_{1}m_{2}}{r^{2}}$$
where r is the distance between the two masses.
So if the midpoint for the two masses is centered at the origin then the two objects are located at x=-0.5 and x=0.5 respectively. So the value for r^2 in the equation above is always $(2x)^{2}$ So we can say that the formula for force between the two objects based on either objects position is:
$$F_{g} = \frac{G}{(2x)^{2}}$$
I didn't include m1m2 because they are both 1.
and that is as far as I can go. If I had an equation for the force based on time then I could solve for how long it takes the two objects to meet.
So is it possible to find an equation of the force based on time from what I have? Or maybe something completely different? Any help is appreciated.
2. Dec 14, 2008
### Cspeed
For distance, use 1m instead of 2x. Then, it will be F = G. The force on each mass will be equal to that at time = 0. Of course, the force will increase as they get closer. I'm not sure which equation you can use to solve this.
3. Dec 14, 2008
### Peeter
You'll need two equations. One for the force on the first particle and one for the second. Example:
$$\text{Force on} \quad m_2 = - G m (x_2 - x_1)^{-2} = {x_2}''$$
$$\text{Force on} \quad m_1 = G m (x_2 - x_1)^{-2} = {x_1}''$$
subtraction of these gives you an integration problem for a difference of position variable. The first integration of this can give you the "velocity" that this difference changes by, and a second will give you the time.
Last edited: Dec 14, 2008
4. Dec 15, 2008
### Perillux
By subtraction of these do you mean:
$$-Gm(x_{2}-x_{1})^{-2} - ( Gm(x_{2}-x_{1})^{-2} )$$
$$-2Gm(x_{2}-x_{1})^{-2}$$
not sure how to integrate that, with the two different values of x.
5. Dec 15, 2008
### Peeter
Subtract one equation from the other:
\begin{align*} (- G m_1 (x_2 - x_1)^{-2} = {x_2}'') \\ -(G m_2 (x_2 - x_1)^{-2} = {x_1}'') \\ = \\ - G (x_2 - x_1)^{-2} (m_1 + m_2)= (x_2 - x_1)'' \\ \end{align*}
Now let $u=x_2 - x_1$, and use $d/dt = (du/dt) d/du$ to get an equation for velocity (squared) from the second derivative term.
6. Dec 18, 2008
### Perillux
I'm still not sure I follow. I haven't taken a differentials class yet (taking it next semester).
But I gave it a try:
so if u=x2 - x1 then du/dt = 0
and d/du of the entire function should be:
$$2G(m_{1}+m_{2}) u^{-3}$$
and since you said d/dt = (du/dt)d/du then I guess multiplying those two will give me d/dt? so...
d/dt =
$$2G(m_{1}+m_{2}) u^{-3} * 0$$
I know I'm probably way off... but there it is.
7. Dec 18, 2008
### Peeter
Here's a more detailed bash at the problem (too hard to do with inline tex in the forum) :
however, if you haven't taken calculus, this may not be any better.
re: what you wrote. x_2 and x_1 are functions of t, so those derivatives aren't zero. Also, writing d/dt was using lazy notation for the chain rule (how to differentiate a function of a function), where I had left out the thing that was being differentiated.
Peeter
8. Dec 18, 2008
### schroder
While I have nothing against integral calculus, it seems to me that this problem can be solved much more simply by use of conservation of energy.
The gravitational potential between two 1 kg spheres at a distance of 1 meter is: Gmm/r = G
So the potential energy of the system is G or 6.67E-11 Joules. (Also the acceleration is G)
Since the spheres are equal, they will share that PE equally as kinetic energy just before they crash together. So each sphere will have a KE of 3.34E-11 J at impact. From that we can get the velocity at impact since KE = 1/2 mV^2 so velocity at impact is 8.167E-6 m/s for each sphere or total velocity of 1.633E-5 m/s for the system.
So now we can use X = V*T -1/2 aT^2 to compute the time, where X = 1 meter, V = 1.633E-5 and a is equal to G = 6.67E-11 and we solve for T by quadratic equation which of course yields two answers: T = 71,750 seconds or 417,904 seconds
The second answer must be the correct one since they are starting at zero velocity and accelerating to the max at impact. So it will take 417,904 seconds or about 116 hours.
How does that compare with the answer you get by integrating?
9. Dec 18, 2008
### Perillux
well I still haven't figured out how to do it through integration. However, I think your answer might be off. Because, if you just take the initial acceleration Gmm/r^2 = G
and use that as a constant acceleration I get 34 hours. It's obvious that the actual answer should be smaller than this value because as they get closer they acceleration will be stronger.
10. Dec 19, 2008
### schroder
The quadratic did give two answers. I forget now why I chose the larger of these, probably I was tired. It should be the smaller at 71,750 seconds or about 20 hours. When I have more time I will recheck it. It would be interesting to do the integration also and compare, but I am fairly confident that the conservation of energy will give the right answer without the need for the more powerful mathematical treatment.
11. Dec 21, 2008
### turin
This is flawed for another reason besides the use of a constant acceleration. The gravitational potential energy is NEGATIVE, and it only becomes more negative as the masses approach each other, without bound. So, the KE actually approaches infinity. How long does it take to accelerate a mass at constant acceleration to an infinite velocity?
BTW, I got about 27 hours using the integration method.
12. Dec 21, 2008
### schroder
No. The kinetic energy is conserved and is no greater than the total PE that is calculated at the distance of one meter. The total energy does not increase as they get closer. As you said, the PE is negative, so even though the number gets higher, it gets higher in a negative direction, so the PE gets less and the KE increases, but the Total energy stays exactly the same at all times. The velocity does not go to infinity but is limited by the kinetic energy. The max velocity that is attained is what I calculated. I admit the time calculation I made is “flawed” as I did use constant acceleration. I was looking for a ball-park figure. If we assume constant acceleration, the max time it can take is 34 hours. If we assume constant max velocity, as computed, the min time it can take is 17 hours. The 20 hours does fall in that range but I know it is not correct. I am still having a bit of trouble evaluating the integral as I need to work it out long hand. The integrand keeps reducing to unity, leaving me only the external term and that works out to 24 hours, also in the required range. Your figure of 27 hours falls exactly in the center of the range and must be correct.
(Well, not exactly in the center, but I believe it)
EDiot: The kinetic energy is not conserved, the TOTAL energy is conserved, which limits the amount of kinetic energy. It does not approach infinity.
Last edited: Dec 21, 2008
13. Dec 21, 2008
Time to throw differential equations at the problem, I think.
For simplicity, I'm saying the masses are equal; r is the distance from the centre to each mass. Initially, r = r0, so you have
$$\frac{d^2r}{dt^2} = -\frac{Gm}{r^2}, r(0) = r_0, r'(0) = 0$$
Say $$v = dr/dt$$, so $$d^2r/dt^2 = v \,dv/dr$$:
$$v \frac{dv}{dr} = -\frac{Gm}{4r^2}$$
$$v \,dv = -\frac{Gm}{4r^2} \,dr$$
$$\frac12 v^2 = \frac{Gm}{4r} + c'_1$$
I'll figure out what that constant is now: at v = 0, r = r0, so
$$0 = \frac{Gm}{4r_0} + c_1 \Longrightarrow c_1 = -\frac{Gm}{4r_0}$$
$$\frac12 v^2 = \frac{Gm}{4} \left( \frac{1}{r} - \frac{1}{r_0} \right)$$
$$v = \frac{dr}{dt} = -\sqrt{\frac{Gm}{2} \left( \frac{1}{r} - \frac{1}{r_0} \right) }$$
I took the negative square root because we want the masses moving toward each other.
$$-\frac{dr}{\sqrt{\frac{Gm}{2} \left( \frac{1}{r} - \frac{1}{r_0} \right) }} = dt$$
Now the hard part: integrating that left side. If I integrate that from r0 to 0, I should get the amount of time that takes:
$$\sqrt{\frac{2}{Gm}} \int_{0}^{r_0} \left( \frac{1}{r} - \frac{1}{r_0} \right)^{-\frac12} \,dr = T$$
I find that the integral there is $$r_0^{3/2} \pi/2$$, so the amount of time it takes in total is
$$T = \frac{\pi r_0^{3/2}}{\sqrt{2Gm}}$$
Putting in r0 = 0.5 m, m = 1 kg, I get 96149 seconds, or 26.7 hours.
The integral can be evaluated by using the trigonometric substitution $$r = r_0 \cos^2 \theta$$.
Last edited: Dec 21, 2008
14. Dec 21, 2008
### Denton
Although im able to actually keep up with your working AdrianK unlike peeter's (sorry peeter, your math is beyond me heh) the solutions does not seem to be reasonable.
Only one day required for two one kilo objects a meter a part to come together? You would probably be able to see then very heavy objects hung on well lubricated rails slide together (albeit slowly).
Anyways always wanted to find a solution for this problem, and ive never realised it was so complex. A work of relativity in itself :)
One last thing:
d$$^{}2$$r/dt$$^{}2$$ = -GM / r$$^{}2$$
Where does the d$$^{}2$$r/dt$$^{}2$$ come from and or how does that equal F?
Last edited: Dec 21, 2008
15. Dec 21, 2008
### schroder
The problem with that is, unlike most other forces, gravity cannot be shielded. The two masses being considered are assumed to be in deep space, far from any other sources of gravitational attraction and so can be analyzed as the only forces acting. Two masses hanging on rails here on earth would be subjected to the gravitational attraction of earth, far outweighing the attraction between the two masses.
I also appreciate Adriank showing us the math and the very good explanation!
(That one last thing is the second derivative of distance, which is acceleration, not force.)
16. Dec 21, 2008
Well, I just thought of another solution that's quite a bit shorter, using a previously known result. You can consider the path of a mass as half of a (degenerate) elliptical orbit, with period 2T and semimajor axis r0 (half the sum of the maximum and minimum distances between the two masses). Kepler's third law then gives
$$\left(\frac{2T}{2\pi}\right)^2 = \frac{r_0^3}{G(m + m)}$$
$$T = \frac{\pi r_0^{3/2}}{\sqrt{2Gm}}.$$
Last edited: Dec 21, 2008
17. Dec 21, 2008
### Denton
You are most wise.
18. Dec 21, 2008
### turin
If by "it" you mean TOTAL energy, then I agree. If by "it" you mean KINETIC energy, then you should just make a little table of kinetic energy vs. distance. I suggest r = 1 m, 0.1 m, 0.01 m, and 0.001 m, for example. Then, come back and tell us if you still believe that KE does not approach infinity.
19. Dec 22, 2008
$$U = -\frac{Gm^2}{2r}$$ | 3,212 | 10,802 | {"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} | 4.15625 | 4 | CC-MAIN-2018-43 | latest | en | 0.936495 |
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A209619 Primes separated from their previous adjacent primes by a composite number of successive composites. 2
149, 191, 251, 293, 347, 419, 431, 557, 587, 641, 701, 719, 797, 821, 839, 929, 1031, 1049, 1061, 1151, 1163, 1181, 1259, 1361, 1409, 1481, 1637, 1709, 1733, 1811, 1847, 1889, 1949, 1973, 2027, 2039, 2063, 2099, 2129, 2153, 2237, 2267, 2333, 2503, 2531, 2579 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 COMMENTS a(1) = 149 is the first prime separated from its previous prime (139) by a composite number (9) of successive composites, namely, 148, 147, 146, 145, 144, 143, 142, 141, 140. Primes p such that nextprime(p) - p - 1 is composite. - Jahangeer Kholdi, Nov 27 2014 LINKS Harvey P. Dale, Table of n, a(n) for n = 1..1000 FORMULA a(n) = A151800(A209618(n)). MAPLE N:= 3000: # to get all entries <= N Primes:= select(isprime, [seq(2*i+1, i=1..(N-1)/2)]): Q:= map(t -> (t>2) and not isprime(t-1), Primes[2..-1] - Primes[1..-2]): zip(proc(p, q) if q then p else NULL fi end proc, Primes[2..-1], Q); # Robert Israel, Nov 28 2014 MATHEMATICA ps = Prime[Range[500]]; pos = Position[Differences[ps] - 1, _?(# > 1 && ! PrimeQ[#] &)]; ps[[Flatten[pos + 1]]] (* T. D. Noe, Mar 21 2012 *) Transpose[Select[Partition[Prime[Range[400]], 2, 1], CompositeQ[#[[2]] - #[[1]] - 1] &]][[2]] (* Harvey P. Dale, Aug 05 2014 *) Select[Prime[Range[375]], NextPrime[#] - # - 1 > 1 && !PrimeQ[NextPrime[#] - # - 1] &] (* Jahangeer Kholdi, Nov 27 2014 *) CROSSREFS Sequence in context: A316589 A178127 A307472 * A031929 A161487 A121947 Adjacent sequences: A209616 A209617 A209618 * A209620 A209621 A209622 KEYWORD nonn AUTHOR Lekraj Beedassy, Mar 21 2012 STATUS approved
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Last modified September 21 13:20 EDT 2020. Contains 337272 sequences. (Running on oeis4.) | 819 | 2,211 | {"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} | 3.671875 | 4 | CC-MAIN-2020-40 | latest | en | 0.552473 |
http://forum.enjoysudoku.com/aic-and-tdp-t36971.html | 1,611,242,673,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703524858.74/warc/CC-MAIN-20210121132407-20210121162407-00720.warc.gz | 39,532,156 | 15,547 | ## AIC and TDP
Advanced methods and approaches for solving Sudoku puzzles
### AIC and TDP
Hi all,
I open this topic to discuss TDP as an alternative to AIC.
I've already exchanged with SpAce and others on this subject but it's all scattered in the forum.
For those who really want to understand my TDP approach, I advise you to refer to the following document in PDF format: http://www.assistant-sudoku.com/Pdf/TDP-anglais.pdf.
On the AIC side, I refer to David P.Bird's basic article: http://forum.enjoysudoku.com/an-aic-primer-t33934.html#p258581 and some comments from SpAce.
In my comparison, I insist a bit more on TDP because I understand that people are not used to my definitions, results and notations, and that it needs some explanations here.
I will compare the two ways of doing things by using the example given by David P. Bird in his article :
Code: Select all
*----------*----------*----------*
r1 | 12 . . | 23 * * | . . . |
r2 | . . . | . . . | . . . |
r3 | * * * | . . 13 | . . . |
*----------*----------*----------*
1) AIC is written (1=2)r1c1 - (2=3)r1c4 - (3=1)r3c6
Since AIC works in both directions, it is deduced that 1r1c1 and 1r3c1 are strongly related => that all candidates weakly related to 1r1c1 and 1r3c1, i.e. 1r1c56 and 1r3c123, can be eliminated.
2) TDP uses the notion of anti-track P'(1r1c1) = {2r1c1, 3r1c4, 1r3c6,... } constructed, by definition, by looking for the candidates that we would place on the puzzle if 1r1c1 was eliminated, we can write here -1r1c1 -> 2r1c1 -> 3r1c4 -> 1r3c6, ...
This leads, as for the AIC, to the elimination of 1r1c56 and 1r3c123 which see 1r1c1 and 1r3c6 by virtue of the Th2 TDP part1 which I recall "If B is a candidate contained in P'(A), then any candidate C who sees B
and A can be eliminated
".
My Comment:
1) AIC is a bidirectional writing alternating strong and weak links. Everything is written that makes its supporters say that AIC is generally more productive than a chain of forcing (David P.Bird).
2) The TDP is a writ of implication which may lead one to think that it is less productive than the AIC in terms of elimination because it would be unidirectional. This is not true, because if an AIC establishes reversible paths between two candidates A (start) and B (finish), then we also have P'(A) -> B and P'(B) -> A which gives the same eliminations.
So I will say that TDP contains AICs and forcing chains, because in addition to incorporating linear chains like AICs, TDP allows chains to be nested like forcing net.
Depending on the reactions (which will certainly not be lacking!) I will develop my arguments for comparison.
Robert
Mauriès Robert
Posts: 460
Joined: 07 November 2019
Location: France
### Re: AIC and TDP
Hi Robert,
Looks like David's AIC Primer fed you some misconceptions about AIC users' attitudes in general. It's not your fault. While there's no doubt about David's mastery of AICs, some of the things he's said about other techniques aren't exactly accurate because of his personal biases against them. (Unfortunately some of his most loyal followers have picked up those half-truths at face value and continued spreading them as gospel. I do my best to break that cycle.)
Mauriès Robert wrote:1) AIC is a bidirectional writing alternating strong and weak links. Everything is written that makes its supporters say that AIC is generally more productive than a chain of forcing (David P.Bird).
It totally depends on the definition of "forcing chain" which is an ambiguous term. Apparently David used a very limited one:
David P Bird wrote:In comparison, a forcing chain, which can only be followed in one direction, will depend on the opening assumption and will only be able to prove whether that assumption is false because it would result in a contradiction.
Unfortunately David didn't bother to define his idea of a forcing chain, but it's obvious that he was talking about contradiction chains only. Personally I don't think they should be called forcing chains at all, even though Hodoku does:
Hodoku wrote:Forcing Chain is a generic term for any chain that leads to a contradiction or a verity and thus forces something (any Discontinuous Nice Loop or any AIC is a Forcing Chain by that definition). Chains that don't lead to a contradiction themselves can be combined to a Multiple Forcing Chain. All chains together can either prove a verity or a contradiction, thus leading to a forcing.
So, Hodoku has two kinds of forcing chains (contradiction and verity), and it correctly counts AICs as such (the verity kind). That already contradicts David. The original (?) definition actually included verities only, which makes much more sense to me:
Jeff wrote:Forcing Chain - a chain that has 2 or more implication streams that start from one node and end in another node where the outcomes of inferences merge from the 2 implication streams. In a forcing chain, a node can only infer the next successive node downstream.
That's what we call a Kraken to disambiguate it. A kraken is a kraken whether it's written with Eureka or implication chains or TDP, and like I've said, AICs are just binary krakens. In fact, David's way of finding AICs (GEM) is all about finding binary krakens, because it starts with a strong link and follows the implications both ways. The goal is to find agreements for both branches (verities) or a contradiction for one of them, so GEM finds both kinds of forcing chains (as defined by Hodoku). They're just easy to convert into AICs unless a lot of branching has been used.
In other words, what David said above is misleading because it should be understood within the context of contradiction chains only. On the verity side, AICs are no more (or less) powerful or productive than binary forcing chains because they are the same thing. The only difference is whether they're written on one line or two. However, AICs (and binary forcing chains) are indeed more powerful than contradiction chains which start with a single assumption, because the latter can only deduce something about that assumption and nothing else. (So, are three-way krakens more powerful than binary ones? Nope, they just complicate things. The sweet spot is at exactly two.)
2) The TDP is a writ of implication which may lead one to think that it is less productive than the AIC in terms of elimination because it would be unidirectional. This is not true, because if an AIC establishes reversible paths between two candidates A (start) and B (finish), then we also have P'(A) -> B and P'(B) -> A which gives the same eliminations.
I said the same thing here. The concept of reversibility is another thing that some of David's worshippers have misunderstood and touted as something special. It's not. Every correctly written sudoku chain is more or less bidirectional, and AICs are no special case in that regard. It's just more obvious with them.
What makes AICs special is that they have no concept of direction at all when viewed as static boolean logic. That's the real philosophy behind AICs and what makes them conceptually different from implication chains (and TDP). However, even that makes no practical difference since both can be easily converted into each other, as I demonstrated in the linked post.
So I will say that TDP contains AICs and forcing chains, because in addition to incorporating linear chains like AICs, TDP allows chains to be nested like forcing net.
That sentence (after "because") doesn't make a lot of sense if you think about it. Anyway, I thought it was already established and understood that TDP includes its versions of all those things. No disagreement there. The only significant differences are in the notations and philosophies.
SpAce
Posts: 2674
Joined: 22 May 2017
### Re: AIC and TDP
SpAce wrote:Hi Robert,
Looks like David's AIC Primer fed you some misconceptions about AIC users' attitudes in general. It's not your fault. While there's no doubt about David's mastery of AICs, some of the things he's said about other techniques aren't exactly accurate because of his personal biases against them. (Unfortunately some of his most loyal followers have picked up those half-truths at face value and continued spreading them as gospel. I do my best to break that cycle.)
Mauriès Robert wrote:1) AIC is a bidirectional writing alternating strong and weak links. Everything is written that makes its supporters say that AIC is generally more productive than a chain of forcing (David P.Bird).
It totally depends on the definition of "forcing chain" which is an ambiguous term. Apparently David used a very limited one:
David P Bird wrote:In comparison, a forcing chain, which can only be followed in one direction, will depend on the opening assumption and will only be able to prove whether that assumption is false because it would result in a contradiction.
Unfortunately David didn't bother to define his idea of a forcing chain, but it's obvious that he was talking about contradiction chains only. Personally I don't think they should be called forcing chains at all, even though Hodoku does:
Hodoku wrote:Forcing Chain is a generic term for any chain that leads to a contradiction or a verity and thus forces something (any Discontinuous Nice Loop or any AIC is a Forcing Chain by that definition). Chains that don't lead to a contradiction themselves can be combined to a Multiple Forcing Chain. All chains together can either prove a verity or a contradiction, thus leading to a forcing.
So, Hodoku has two kinds of forcing chains (contradiction and verity), and it correctly counts AICs as such (the verity kind). That already contradicts David. The original (?) definition actually included verities only, which makes much more sense to me:
Jeff wrote:Forcing Chain - a chain that has 2 or more implication streams that start from one node and end in another node where the outcomes of inferences merge from the 2 implication streams. In a forcing chain, a node can only infer the next successive node downstream.
That's what we call a Kraken to disambiguate it. A kraken is a kraken whether it's written with Eureka or implication chains or TDP, and like I've said, AICs are just binary krakens. In fact, David's way of finding AICs (GEM) is all about finding binary krakens, because it starts with a strong link and follows the implications both ways. The goal is to find agreements for both branches (verities) or a contradiction for one of them, so GEM finds both kinds of forcing chains (as defined by Hodoku). They're just easy to convert into AICs unless a lot of branching has been used.
In other words, what David said above is misleading because it should be understood within the context of contradiction chains only. On the verity side, AICs are no more (or less) powerful or productive than binary forcing chains because they are the same thing. The only difference is whether they're written on one line or two. However, AICs (and binary forcing chains) are indeed more powerful than contradiction chains which start with a single assumption, because the latter can only deduce something about that assumption and nothing else. (So, are three-way krakens more powerful than binary ones? Nope, they just complicate things. The sweet spot is at exactly two.)
2) The TDP is a writ of implication which may lead one to think that it is less productive than the AIC in terms of elimination because it would be unidirectional. This is not true, because if an AIC establishes reversible paths between two candidates A (start) and B (finish), then we also have P'(A) -> B and P'(B) -> A which gives the same eliminations.
I said the same thing here. The concept of reversibility is another thing that some of David's worshippers have misunderstood and touted as something special. It's not. Every correctly written sudoku chain is more or less bidirectional, and AICs are no special case in that regard. It's just more obvious with them.
What makes AICs special is that they have no concept of direction at all when viewed as static boolean logic. That's the real philosophy behind AICs and what makes them conceptually different from implication chains (and TDP). However, even that makes no practical difference since both can be easily converted into each other, as I demonstrated in the linked post.
So I will say that TDP contains AICs and forcing chains, because in addition to incorporating linear chains like AICs, TDP allows chains to be nested like forcing net.
That sentence (after "because") doesn't make a lot of sense if you think about it. Anyway, I thought it was already established and understood that TDP includes its versions of all those things. No disagreement there. The only significant differences are in the notations and philosophies.
Space, my AIC Primer made it clear that I defined forcing chains as those that are unidirectional. I then held that forcing (unidirectional) chains are generally less productive than AICs (bidirectional ones). In the quoted post you have then gone to great lengths to fail to prove me wrong.
After your extensive 'deep learning' exercise it is time for you to put up or shut up. Nowhere in your nearly 2,000 posts are any that summarise your pearls of wisdom that correct the numerous faults you have discovered.
David P Bird
David P Bird
2010 Supporter
Posts: 1043
Joined: 16 September 2008
Location: Middle England
### Re: AIC and TDP
Hi David P.Bird and SpAce,
Am I to understand that between you there are certain discrepancies that do not only relate to fundamental problems?
In this subject only the fundamental problems interest me.
Mr Bird, I would be honoured if you could also give your opinion on these fundamental problems, as Mr Space has already done on many occasions, and I thank him for that. I am therefore interested in your opinion on TDP.
It is because I see that the AICs are at the heart of the debates in this forum that I have opened this subject "AIC and TDP" in order to show how TDP is something other than "Forcing chains".
Sincerely
Robert
Last edited by Mauriès Robert on Mon Dec 23, 2019 7:57 am, edited 1 time in total.
Mauriès Robert
Posts: 460
Joined: 07 November 2019
Location: France
### Re: AIC and TDP
Hello David,
Long time no see. So you had to quote my whole post in order to reply with a few lines? In your own words, I might call that "very deliberate vandalizing" of Robert's thread, just like your attacks on me instead of the issues.
David P Bird wrote:Space, my AIC Primer made it clear that I defined forcing chains as those that are unidirectional.
No, it did not (just like you haven't clearly defined MSLS anywhere, causing all kinds of confusion as recently seen). Your statement is ambiguous at best:
In comparison, a forcing chain, which can only be followed in one direction, will depend on the opening assumption and will only be able to prove whether that assumption is false because it would result in a contradiction.
I see two ways to interpret that subordinate clause: either as a supposed attribute of all forcing chains or as a specifier for one type of them. Which one did you mean? It's still not clear to me, though the first option seems more likely (and more incorrect). Personally I don't think it's great either way, because I prefer Jeff's definition which excluded contradiction chains, but at least the latter interpretation would be correct under Hodoku's definition. If you did in fact mean that all forcing chains are unidirectional and assumptive, then you were simply wrong (or using a very personal definition, which amounts to the same thing).
That said, I never saw that part as a problem before. I happen to understand the relationship between AICs and any kind of forcing chains intimately (because I don't believe everything I read), so it didn't really matter to me what you meant. I didn't even pay attention to it.
Obviously it is a problem, however, because we have an example here where it's led to a misunderstanding. I've also seen examples of very experienced AIC users having similar misconceptions, and I suspect the source is the same. Why didn't you say "contradiction chain" or "Nishio"? That wouldn't be ambiguous at all. But, as usual, you can't admit any fault in your communication, even when it's valid criticism that you could use to improve your products. That's simply arrogance, and it's why you keep failing in putting me down despite your best efforts.
Btw, did you know that it was me who gave Robert the link to your AIC Primer in the first place? I even removed my own comment from it first because I didn't think it added value. In general, I think the AIC Primer is one of your best summaries, and I wasn't expecting Robert to find confusing statements in it. But he did, and I can't blame him. So, did you really communicate as clearly as possible? Did you use this opportunity to say something constructive and fix the ambiguity in your writing?
I then held that forcing (unidirectional) chains are generally less productive than AICs (bidirectional ones). In the quoted post you have then gone to great lengths to fail to prove me wrong.
No. I actually confirmed what you said -- iff only contradiction chains were meant. All I criticized was your apparent equating them with all forcing chains, and I firmly stand by that assessment. Once again you're twisting my words to fight straw men.
After your extensive 'deep learning' exercise it is time for you to put up or shut up.
Still barking orders and demands? How did it work out last time?
Nowhere in your nearly 2,000 posts are any that summarise your pearls of wisdom that correct the numerous faults you have discovered.
Oh, I'm deeply flattered to hear that you've read all of my nearly 2000 posts. You're absolutely correct too that my pearls of wisdom don't live in neat, untouchable summaries like yours. Instead they're scattered around within various dialogues where they're continually being honed under sometimes harsh but welcomed criticism. Or maybe they simply don't exist. Either way your judgment of that is even more biased than mine, making it absolutely irrelevant.
Here's a pearl of wisdom for you: "Never wrestle with a pig. You both get dirty and the pig likes it."
Last edited by SpAce on Sun Dec 22, 2019 1:43 pm, edited 1 time in total.
SpAce
Posts: 2674
Joined: 22 May 2017
### Re: AIC and TDP
Robert,
I am sorry but I am no longer involved with this forum from day to day principally because the moderators took no action when Space set about disrupting my posts.
I was just defending myself against his unjustified criticism.
Space,
So you read the line in the AIC Primer where I specified what I meant by forcing chain and then chose to ignore it and apply your own definition to what I said on the grounds:
so it didn't really matter to me what you meant. I didn't even pay attention to it.
You then accused me of being misleading. A classic example of using a 'false fact'.
DPB
.
David P Bird
2010 Supporter
Posts: 1043
Joined: 16 September 2008
Location: Middle England
### Re: AIC and TDP
David P Bird wrote:I am sorry but I am no longer involved with this forum from day to day principally because the moderators took no action when Space set about disrupting my posts.
daViD, all I can do to respond to your blatant lies is provide the link to the relevant discussion. Or did you have something else in mind as well when you said posts? I let people judge for themselves how accurate your description of the events is.
I was just defending myself against his unjustified criticism.
SpAce
Posts: 2674
Joined: 22 May 2017
### Re: AIC and TDP
First of all, it's time for
" Merry Christmas and Happy new year".
It is part of the human behavior to let discussions turn to aggressive, I'll try to avoid this with a small remark.
I am familiar for long with chains leading to contradictions and AIC's. (Sudoku Explainer works mainly on contradictions, my first solver worked purely on AIC's nets)
If the expansion rules in the contradictions chains are simple enough, we have usually a corresponding AIC (simplest cases) or a corresponding AIC net (many examples of such nets in totuan solutions).
I can understand the willingness to have a better and general classification, it's a pity to see the discussion turning to fights.
champagne
2017 Supporter
Posts: 7178
Joined: 02 August 2007
Location: France Brittany
### Re: AIC and TDP
dAvId, I have a proposition for you. Since you seem to be alive, I'm ready to get out of your way so you can make a glorious comeback. It's obvious that both of us can't fit in the same forum, and since you think my contributions are worthless, I'll give you a chance to fix everything I've broken.
Effective immediately, I pledge to stay out at least until the end of January whether you return or not. If you do return and make yourself useful, I'll stay out as long as you stay active.
You have free rein starting now.
PS. Of course I'm suspecting that my voluntary exile doesn't satisfy you. What you really wanted was to get me kicked out. Nevertheless, this is what's on the table. Take it or leave it.
--
Everyone else, Merry Christmas and so forth. SpAce out.
-SpAce-: Show
Code: Select all
* | | | | *
* |=()=| / _ \ |=()=| *
* | | |-=( )=-| | | *
* \ ¯ / *
"If one is to understand the great mystery, one must study all its aspects, not just the dogmatic narrow view of the Jedi."
SpAce
Posts: 2674
Joined: 22 May 2017
### Re: AIC and TDP
SpAce,
While you are away, may I suggest that you prepare a detailed introduction the GEM approach? You have indicated that you have developed a way to use GEM with HoDoKu. I think forum members would appreciate a step-by-step elucidation of this method. I recommend including visuals from HoDoKu to illustrate each step. The steps should proceed slowly with thorough explanation and interpretation. I recommend including several examples starting with a very simple one [like an XY Chain], then an AIC [but not too hard], then a harder AIC and then proceeding to a quite advanced one [like a complex chain with memory].
May I encourage you to prepare such a document in a form that could be downloaded via the forum? Maybe in .odt format? Then forum members could provide you with feedback; maybe some clarification points; maybe corrections [if needed].
Such a document has the potential to be of value.
Gordon
ghfick
Posts: 117
Joined: 06 April 2016
### Re: AIC and TDP
Hi Gordon,
You're making me break my pledge right away, but I think I need to answer this one.
ghfick wrote:While you are away, may I suggest that you prepare a detailed introduction the GEM approach? You have indicated that you have developed a way to use GEM with HoDoKu. I think forum members would appreciate a step-by-step elucidation of this method.
Glad you asked. I probably would have done that a long time ago if I'd thought anyone was interested. However, I did make one recently in the context of Robert's TDP introduction, here and a bit more here. I guess you didn't see it?
I obviously should have put those in a separate thread instead of polluting Robert's, but there they are for now. Maybe the examples aren't the best, and some of it gets pretty advanced, but nevertheless the basics are covered in Parts 1 and 2.
Would that get you started? Any feedback including criticism is most welcome, but obviously Robert's threads aren't the right place for that. Perhaps an admin could move those posts into a separate thread?
SpAce
Posts: 2674
Joined: 22 May 2017
### Re: AIC and TDP
SpAce wrote:I obviously should have put those in a separate thread instead of polluting Robert's, but there they are for now. Maybe the examples aren't the best, and some of it gets pretty advanced, but nevertheless the basics are covered in Parts 1 and 2.
Would that get you started? Any feedback including criticism is most welcome, but obviously Robert's threads aren't the right place for that. Perhaps an admin could move those posts into a separate thread?
Thanks SpAce, without your intervention I was going to put the following message "The art and method to kill a subject" and leave this thread.
Robert
Mauriès Robert
Posts: 460
Joined: 07 November 2019
Location: France
### Re: AIC and TDP
:: OFF TOPIC ::
Hi All,
SpAce wrote:dAvId, I have a proposition for you. Since you seem to be alive, I'm ready to get out of your way so you can make a glorious comeback. It's obvious that both of us can't fit in the same forum, and since you think my contributions are worthless, I'll give you a chance to fix everything I've broken.
Effective immediately, I pledge to stay out at least
until the end of January whether you return or not. If you do return and make yourself useful, I'll stay out as long as you stay active.
You have free rein starting now.
PS. Of course I'm suspecting that my voluntary exile doesn't satisfy you. What you really wanted was to get me kicked out. Nevertheless, this is what's on the table. Take it or leave it.
--
SpAce out.
Why you have to do that? IMO, that is wrong way – I hate that, but it’s up to you. We will miss you – at least for me .
Merry Christmas !!!
totuan
totuan
Posts: 52
Joined: 25 May 2010
Location: vietnam
### Re: AIC and TDP
can someone enlighten me on why a forcing chain (whether it's an aic or not) would ever not be reversible?
for example if you take a discontinuous aic you can decompose it into a series of implication statements like
if A, then B
if B, then C
if C, then D
and because it's an aic you can reverse it to get
if not D, then not C
if not C, then not B
if not B, then not A
now suppose your chain isn't an aic and it requires some form of branching, like this
if A, then B
if A, then C
if B and C, then D
you can reverse the chain as before by turning the logical and into an or:
if not D, then not B or not C
if not B, then not A
if not C, then not A
and by natural extension you can apply this to any sequence of steps in chain implication logic. of course if you actually did this to a big forcing net or multi colouring pattern that starts from assuming the truth of a candidate value, the reversed expression would be unwieldy and not enlightening at all (e.g. SE's explanations in natural language of long chains at the 9+ level) but it can be done
one could put a natural condition on the reversibility of a chain by disallowing any "or" statements in the chains and only allowing "and" statements (which is how SE organises its chain descriptions), which in this case would mean the second chain isn't reversible; or using only one "or" statement to be used at the beginning such as in kraken row/column/blocks with 3+ branches. but in some cases it can be clear and insightful to write chains with "or" statements in them which i can see in some of ttt/totuan's diagrams like this one as well as some solutions in the daily puzzles that use the | symbol in chain notation which is a clever way to avoid writing branching in the chain
is there some absolute condition i'm missing, or does it come down to just how natural the chain would look if you tried to reverse it? and what difference does it make, in other words, when would we care whether a chain is reversible or not?
edit:
SpAce wrote:Your statement is ambiguous at best:
In comparison, a forcing chain, which can only be followed in one direction, will depend on the opening assumption and will only be able to prove whether that assumption is false because it would result in a contradiction.
I see two ways to interpret that subordinate clause: either as a supposed attribute of all forcing chains or as a specifier for one type of them. Which one did you mean?
space i disagree that this is unclear since the commas before "which" and after "direction" clearly indicate that it applies as a descriptor for all forcing chains. if the commas weren't there then it would be a specifier for one type of them. this is standard usage. however i will need to read up on the rest of this to determine what's going on
999_Springs
Posts: 531
Joined: 27 January 2007
Location: In the toilet, flushing down springs, one by one.
### Re: AIC and TDP
999_Springs wrote:can someone enlighten me on why a forcing chain (whether it's an aic or not) would ever not be reversible?
Hi 999_Springs,
Let me try with the risk to be wrong
I think that the answer is in the precise meaning of "reversible" in the case of an AIC.
In the "clearing form" of an AIC
A=B-C=D-E=F
if one end is false, the other end is true. This is granted if we have alternatively strong and weak inferences.
If "A" is false, "F" is true
if "F" is false, "A" is true
any candidate/group of candidates making wrong both "A" and "F" is not valid.
Often a contradiction (with Sudoku Explainer definition) can not be expressed in this way.
champagne
2017 Supporter
Posts: 7178
Joined: 02 August 2007
Location: France Brittany
Next | 6,827 | 29,129 | {"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} | 3.4375 | 3 | CC-MAIN-2021-04 | latest | en | 0.920587 |
https://jan.jarfalk.se/highly-divisible-triangular-number/ | 1,537,808,149,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267160620.68/warc/CC-MAIN-20180924165426-20180924185826-00545.warc.gz | 528,899,549 | 7,279 | 2018-3-14
# Find Highly Divisible Triangular Numbers with Kotlin
A solution for the Project Euler problem number 12. Written in the Java-like language Kotlin. Performance optimizations have primarily been made in the function for finding divisors.
Project Euler problem number 12 - Highly Divisible Triangular Number - is stated as follows.
The sequence of triangle numbers is generated by adding the natural numbers. So the 7th triangle number would be 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28. The number 28 has six divisors - 1, 2, 4, 7, 14, 28.
What is the value of the first triangle number to have over five hundred divisors?
``````fun main(args: Array<String>) {
// Set the target to 500 divisors.
val target = 500
// Create a sequence that steps 1 at the time.
val triangleNumber = generateSequence(1, { it + 1 })
// Get the triangle number for every number.
.map { getTriangleNumber(it) }
// Stop when we find a triangle number with more divisors
// than our target.
.first { getDivisors(it).size > target }
// Print the answer.
println("\$triangleNumber is the first triangle number with more than \$target divisors.")
}
fun getTriangleNumber(n: Int): Int {
// Triangle numbers are given by the following formula.
// n(n+1)/2
return (n*(n+1)/2.0).toInt()
}
fun getDivisors(n: Int): List<Int> {
// Factors comes in pairs, you only need to iterate to the
// square root of n and get the paired factor using n / it.
val limit = kotlin.math.sqrt(n.toDouble()).toInt()
return (1..limit).filter { n % it == 0 }.flatMap {
val squaredIsN = it * it == n
if (squaredIsN) listOf(it) else listOf(it, n / it)
}
}
``````
You can run and modify the code at https://try.kotlinlang.org.
Jan Järfalk — User experienced esthete, technician, geek and web worker. Aspiring artist and recreational mathematician. I indulge and travel plenty. Constraints are good. | 484 | 1,853 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2018-39 | longest | en | 0.716597 |
https://rwalk.xyz | 1,709,584,819,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476532.70/warc/CC-MAIN-20240304200958-20240304230958-00066.warc.gz | 477,765,748 | 12,261 | # Universal Inference: Review II
In Part I of this series, we covered some of the prerequisites for understanding the new paper Universal Inference by Wasserman et al. In particular, we reviewed the classical likelihood ratio test and worked some specific examples. In this post, we’ll look at the split likelihood ratio test which is the key idea behind universal… Read More Universal Inference: Review II
# Universal Inference: Review Part I
There’s an important new paper out by Larry Wasserman et al. that describes a very general technique, called Universal Inference, for constructing statistical hypothesis tests and confidence intervals. In the traditional theory of statistics, such as would be taught in an undergraduate mathematical statistics course, a standard way hypothesis tests are constructed and analyzed is… Read More Universal Inference: Review Part I
# Recommendation Systems: From Co-occurrence Counts to Probabilities
In the previous post, we demonstrated how to efficiently compute co-occurrences with matrix algebra and use those calculations to recommend books to users. Though we saw some sensible recommendations come out of this approach, it also suffers from a number of issues, including: The Gatsby Problem: popular books tend to be over represented in the… Read More Recommendation Systems: From Co-occurrence Counts to Probabilities
# Recommendation Systems: A Co-occurrence Recommender
In the previous post of the series, we developed a co-occurence model for book recommendations. The model is similar to Amazon’s highly successful “Customers who bought…” feature. Now it’s time to apply this simple model to some real data to make recommendations. Preprocessing As usual, we’ll work with the Goodreads dataset. I described the structure… Read More Recommendation Systems: A Co-occurrence Recommender
# Financial Turbulence: Off the Chart
I recently read Kritzmen and Li’s clever 2010 paper Skulls, Financial Turbulence, and Risk Management. Kritzmen and Li characterize financial turbulence as a period where established financial relationships uncouple, prices swing, and market predictions break down. Does that sound like financial markets in 2020? Yup. So I thought it would be interesting to take a… Read More Financial Turbulence: Off the Chart
# Recommendation Systems: Co-occurrence Calculations
In this first post in a series on recommendation systems, we’re going to develop a powerful but highly intuitive representation for user behavior that will allow us to easily make recommendations. Since we’re going to be making heavy use of the Goodreads data set in the series, we’ll formulate our basic recommendation system problem as… Read More Recommendation Systems: Co-occurrence Calculations
# A Practical Series on Recommendation Systems
Today, I am announcing a series of posts I am developing about recommendation systems. The series is aimed at software/machine learning engineers. I have two goals for the series: Provide practical and implementable strategies for delivering recommendations in real-time Present the mathematical intuition behind recommender problems The reason for the first goal is that I… Read More A Practical Series on Recommendation Systems
# Can you norm rows and standardize columns at the same time?
If you are reading this, you probably already know that data pre-processing is the 90% perspiration of machine learning. You might love it or you might dread it, but you probably don’t think of it as a the part of ML where the most interesting mathematics lives. Let me challenge that view a bit with… Read More Can you norm rows and standardize columns at the same time?
# Mathematicians in Machine Learning
On February 28, I presented at the University of Kentucky’s Mathematics Department Alumni Day. My talk contains practical advice for math students (graduate and undergraduate) to prepare for Machine Learning careers.
# A Jackknife Example
I’m working through Wasserman’s All of Nonparametric Statistics, a wonderful and concise tour of nonparametric techniques. What is nonparametric statistics? It is a collection of estimation techniques that make as few assumptions as possible about the distribution from which your data came. Let’s work through an example in R that’s mentioned in Chapter 3 of… Read More A Jackknife Example | 847 | 4,338 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2024-10 | longest | en | 0.908156 |
http://stackoverflow.com/questions/15531258/test-if-all-elements-of-a-vector-are-equal/15531336 | 1,448,777,383,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398456289.53/warc/CC-MAIN-20151124205416-00146-ip-10-71-132-137.ec2.internal.warc.gz | 210,736,511 | 19,317 | # Test if all elements of a vector are equal
I want to test if a non-empty vector contains identical elements. Is this the best way?
``````count(vecSamples.begin()+1, vecSamples.end(), vecSamples.front()) == vecSamples.size()-1;
``````
-
It is simple, sure, but inefficient if they are different. Plus you could simplify it further by removing the +/- 1's. Especially since this would blow up on an empty vector. – Ryan Guthrie Mar 20 '13 at 18:00
`vecSamples.front()` would blow up on an empty vector. – john Mar 20 '13 at 18:01
The vector is never empty in my case. – Neil Kirk Mar 20 '13 at 18:03
Is your vector always at least 2 elements though? – Michael Dorgan Mar 20 '13 at 18:05
No, is that a problem? – Neil Kirk Mar 20 '13 at 20:05
## 4 Answers
As @john correctly points out, your solution iterates over the entire container even if the first two elements are different, which is quite a waste.
How about a purely no-boost no c++11 required solution?
``````bool allAreEqual =
find_if(vecSamples.begin() + 1,
vecSamples.end(),
bind1st(not_equal_to<int>(), vecSamples.front())) == vecSamples.end();
``````
Stops on first non-equal element found. Just make sure your vecSamples is non-empty before running this.
-
Accepted as I'm only using old C++ right now. – Neil Kirk Mar 21 '13 at 11:33
In c++11 (or Boost Algorithm)
``````std::all_of(vecSamples.begin()+1,vecSamples.end(),
[&](const T & r) {return r==vecSamples.front();})
``````
-
Probably not, because it always examines all the elements of the vector even if the first two elements are different. Personally I'd just write a for loop.
-
I would subtract the value of the first element to all the vector elements and then calculate their sum and compare it to zero.
-
What if the vector doesn't contain numbers? What if the input vector is read-only? (I'd have to copy it) – Neil Kirk Jul 14 at 20:54 | 500 | 1,879 | {"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-2015-48 | longest | en | 0.842784 |
https://business.gmu.edu/blog/mba/2013/02/18/gmat-preparation-understanding-fractions-and-staying-alert/ | 1,638,888,474,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363400.19/warc/CC-MAIN-20211207140255-20211207170255-00588.warc.gz | 235,905,998 | 11,217 | # GMAT Preparation: Understanding Fractions and Staying alert
Lately, we’ve been discussing reading comprehension and circles, and today’s post on GMAT preparation focuses on fractions.
I was recently reminded about how folks can be taking the test, thinking sections are going relatively smoothly (and even easily) and then be completely taken aback when they receive rather disappointing scores. Has this happened to you?
Beat the GMAT tells us in a recent post on fractions and GMAT preparation, “First of all, thinking that a test like the GMAT is easy is actually a warning sign: things probably are not going very well. If the test was going very well, then you’d be seeing some seriously hard – next to impossible – problems.”
While it’s not true that we can ever absolutely know exactly how we are doing on the GMAT (we can’t outthink the computer), we do know that if we are getting a lot of medium-to-hard questions right, we will likely “get a mix of harder questions. If it seems you are getting one difficult question after another, that’s a good sign for your score.”
However, sometimes questions look easy when they are actually rather hard, and we can spend so much time just to get the wrong answer. What kind of GMAT preparation can we do for this situation?
Take a look at this question that, at first glance, seems pretty straightforward:
” Of the 3,600 employees of Company X, 1/3 are clerical. If the clerical staff were to be reduced by 1/3, what percent of the total number of the remaining employees would then be clerical?
“(A) 25%
“(B) 22.2%
“(C) 20%
“(D) 12.5%
“(E) 11.1%”
It is so easy to quickly get this answer wrong. (Did you come up with “E” first? If you did, you’re wrong.) It’s so easy to get distracted by an “I got this!” moment that lets us quickly go the wrong way. (That’s what got me the first time.)
So how do we solve this fraction (or any fraction)? Take it step by step, and don’t let yourself try to blaze through it because you are so certain you’ve got it. They are counting on you to make that mistake.
GMAT preparation for your military MBA takes many different forms, and sometimes the best preparation for a certain situation is just teaching yourself to always be alert for those questions that seem oh so easy but are actually designed to trick us. | 537 | 2,320 | {"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} | 4.25 | 4 | CC-MAIN-2021-49 | latest | en | 0.960305 |
https://amathew.wordpress.com/tag/anosov-shadowing-lemma/ | 1,685,675,241,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224648245.63/warc/CC-MAIN-20230602003804-20230602033804-00396.warc.gz | 130,351,520 | 18,275 | Today, we will apply the technical lemma proved yesterday to proving a few special properties of Anosov diffeomorphisms. The first one states that if you have an approximate orbit, then you can approximate it by a real orbit. This may not sound like much, but it is false for isometries—and in fact, it gives another way of proving the structural stability result.
As usual, start with a compact manifold ${M}$ and an Anosov diffeomorphism ${f: M \rightarrow M}$. We can put a metric ${d}$ on ${M}$ (e.g. by imbedding ${M}$ in euclidean space, or using a Riemannian metric, etc.). To formalize this, fix ${\delta>0}$. We introduce the notion of a ${\delta}$-orbit. This is a two sided sequence ${\{x_n\}_{n \in \mathbb{Z}}}$ such that ${d(x_{n+1}, f(x_n)) \leq \delta}$.
Theorem 1 (Anosov shadowing lemma) Fix ${\epsilon>0}$ sufficiently small. There is ${\delta>0}$ such that any ${\delta}$-orbit ${\{x_n\}}$ can be shadowed by a unique real orbit ${\{y_n\}}$, i.e. ${y_{n+1} = f(y_n)}$ and ${d(x_n, y_n) < \epsilon}$ for all ${n \in \mathbb{Z}}$. (more…) | 326 | 1,058 | {"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": 17, "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-2023-23 | latest | en | 0.818931 |
https://dcodesnippet.com/convert-np-array-to-list/ | 1,716,746,930,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058972.57/warc/CC-MAIN-20240526170211-20240526200211-00196.warc.gz | 162,626,954 | 19,406 | # How To Convert Np Array To List: Methods And Benefits
//
Thomas
Explore different methods to convert np arrays to lists in Python for better readability and compatibility with list operations. Learn the benefits of this conversion.
## Methods for Converting np Array to List
Converting NumPy arrays to lists is a common task when working with data in Python. There are several methods available to achieve this conversion, each with its own advantages and use cases. In this section, we will explore three main methods for converting np arrays to lists: using np.tolist(), using np.ndarray.tolist(), and using the list() constructor.
### Using np.tolist()
One of the simplest ways to convert a NumPy array to a list is by using the np.tolist() method. This method directly converts the array to a Python list, making it easy to work with the data in a more familiar format. Here is a simple example of how to use np.tolist():
PYTHON
``````import numpy as np
<h1>Create a NumPy array</h1>
np_array = np.array([1, 2, 3, 4, 5])
<h1>Convert the array to a list</h1>
list_array = np_array.tolist()
print(list_array)``````
Using np.tolist() is a straightforward and efficient way to convert NumPy arrays to lists, especially when you need to quickly access and manipulate the data.
### Using np.ndarray.tolist()
Another method for converting NumPy arrays to lists is by using the np.ndarray.tolist() method. This method is specifically designed for NumPy ndarrays, allowing you to convert multi-dimensional arrays to nested lists. Here is an example of how to use np.ndarray.tolist():
PYTHON
``````import numpy as np
<h1>Create a multi-dimensional NumPy array</h1>
np_array = np.array([[1, 2, 3], [4, 5, 6], [7, 8, 9]])
<h1>Convert the array to a nested list</h1>
list_array = np_array.tolist()
print(list_array)``````
Using np.ndarray.tolist() is useful when working with multi-dimensional arrays and when you need to preserve the structure of the original array in list form.
### Using list() constructor
The third method for converting NumPy arrays to lists is by using the list() constructor. This method allows you to create a new list from any iterable object, including NumPy arrays. Here is an example of how to use the list() constructor:
PYTHON
``````import numpy as np
<h1>Create a NumPy array</h1>
np_array = np.array([1, 2, 3, 4, 5])
<h1>Convert the array to a list using the list() constructor</h1>
list_array = list(np_array)
print(list_array)``````
Using the list() constructor provides flexibility in converting NumPy arrays to lists, as it can be applied to any iterable object, not just NumPy arrays.
## Benefits of Converting np Array to List
Converting numpy arrays to lists offers several benefits that can greatly enhance the readability and compatibility of your code. Let’s explore two key advantages in detail:
One of the main advantages of converting numpy arrays to lists is the improved readability it brings to your code. Numpy arrays, while efficient for numerical computations, can be challenging to interpret at a glance due to their complex structure. By converting them to lists, you can simplify the representation of your data, making it easier for both you and other developers to understand.
Imagine trying to make sense of a dense numpy array filled with numerical values. It can be like deciphering a cryptic code. However, by converting this array to a list, you transform it into a more user-friendly format that is akin to reading a plain English sentence. This enhanced readability not only makes your code more accessible but also saves time and effort when debugging or modifying it.
Incorporating lists into your code also allows for greater flexibility in data manipulation. You can easily access and modify individual elements within a list, iterate over its contents, or apply list-specific operations. This level of simplicity and clarity can greatly improve the overall quality of your code and streamline the development process.
### Compatibility with Python List Operations
Another significant benefit of converting numpy arrays to lists is the seamless compatibility it provides with Python’s list operations. While numpy arrays come with their own set of functions and methods for array manipulation, they may not always align with the standard operations used for lists in Python.
By converting numpy arrays to lists, you can leverage the full power of Python’s built-in list functions, such as sorting, slicing, and concatenation. This compatibility not only expands the range of operations you can perform on your data but also ensures that your code remains consistent with Python’s programming conventions.
Consider the difference between trying to sort a numpy array versus a list. With a numpy array, you may need to use specialized functions or methods specific to numpy, adding complexity to your code. However, by converting the array to a list, you can simply use the standard `sorted()` function in Python, making your code more intuitive and maintainable.
In addition, the compatibility with Python list operations allows you to seamlessly integrate numpy arrays with other Python data structures and libraries. This interoperability opens up new possibilities for data analysis, visualization, and machine learning, enabling you to harness the full potential of both numpy arrays and Python lists in your projects.
In conclusion, converting numpy arrays to lists offers significant benefits in terms of improved readability and compatibility with Python list operations. By making this simple adjustment to your code, you can enhance the clarity, flexibility, and efficiency of your programming workflow, ultimately leading to more robust and maintainable software solutions.
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+1 803-820-9654 | 1,201 | 5,838 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2024-22 | latest | en | 0.712041 |
https://studyhelp.sa.com/mathematics-study-help/the-sum-of-three-consecutive-odd-integers-is-63-what-are-the-three-integers/ | 1,607,168,870,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141747774.97/warc/CC-MAIN-20201205104937-20201205134937-00403.warc.gz | 516,636,068 | 8,371 | # The sum of three consecutive odd integers is 63, what are the three integers?
original integer##=19## remedy integer##=21## third integer ##=23##
To explain this bearing, we conquer demand to set up an equation. But original, we demand to fabricate let statements to let others comprehend what each inconstant or countenance represents. Since each orderly odd integer is separated by a disagreement of ##2##, your let statements are:
Let ##x## be the original integer. Let ##x+2## be the remedy integer. Let ##x+4## be the third integer.
The sum of the three orderly odd integers is ##63##, so your equation is:
##(x)+(x+2)+(x+4)=63##
Then explain for ##x##:
##(x)+(x+2)+(x+4)=63##
##x+x+2+x+4=63##
##3x+6=63##
##3x+6## ##color(red)(-6)=63## ##color(red)(-6)##
##3x=57##
##3xcolor(red)(-:3)=57color(red)(-:3)##
##x=19##
Now that you comprehend your original integer has a treasure of ##19##, depute ##x=19## into ##x+2## and ##x+4## to meet the treasures of the remedy and third integers.
##x+2color(white)(XXXXXXXXXXXX)x+4##
##=(19)+2color(white)(XXXXXXXXX)=(19)+4##
##=21color(white)(XXXXXXXXXXXX)=23##
##:.##, the integers are ##19##, ##21##, and ##23##. | 357 | 1,177 | {"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-2020-50 | latest | en | 0.525262 |
http://us.metamath.org/ileuni/nffvd.html | 1,642,732,326,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320302715.38/warc/CC-MAIN-20220121010736-20220121040736-00433.warc.gz | 67,585,359 | 4,353 | Intuitionistic Logic Explorer < Previous Next > Nearby theorems Mirrors > Home > ILE Home > Th. List > nffvd GIF version
Theorem nffvd 5215
Description: Deduction version of bound-variable hypothesis builder nffv 5213. (Contributed by NM, 10-Nov-2005.) (Revised by Mario Carneiro, 15-Oct-2016.)
Hypotheses
Ref Expression
nffvd.2 (𝜑𝑥𝐹)
nffvd.3 (𝜑𝑥𝐴)
Assertion
Ref Expression
nffvd (𝜑𝑥(𝐹𝐴))
Proof of Theorem nffvd
Dummy variable 𝑧 is distinct from all other variables.
StepHypRef Expression
1 nfaba1 2199 . . 3 𝑥{𝑧 ∣ ∀𝑥 𝑧𝐹}
2 nfaba1 2199 . . 3 𝑥{𝑧 ∣ ∀𝑥 𝑧𝐴}
31, 2nffv 5213 . 2 𝑥({𝑧 ∣ ∀𝑥 𝑧𝐹}‘{𝑧 ∣ ∀𝑥 𝑧𝐴})
4 nffvd.2 . . 3 (𝜑𝑥𝐹)
5 nffvd.3 . . 3 (𝜑𝑥𝐴)
6 nfnfc1 2197 . . . . 5 𝑥𝑥𝐹
7 nfnfc1 2197 . . . . 5 𝑥𝑥𝐴
86, 7nfan 1473 . . . 4 𝑥(𝑥𝐹𝑥𝐴)
9 abidnf 2732 . . . . . 6 (𝑥𝐹 → {𝑧 ∣ ∀𝑥 𝑧𝐹} = 𝐹)
109adantr 265 . . . . 5 ((𝑥𝐹𝑥𝐴) → {𝑧 ∣ ∀𝑥 𝑧𝐹} = 𝐹)
11 abidnf 2732 . . . . . 6 (𝑥𝐴 → {𝑧 ∣ ∀𝑥 𝑧𝐴} = 𝐴)
1211adantl 266 . . . . 5 ((𝑥𝐹𝑥𝐴) → {𝑧 ∣ ∀𝑥 𝑧𝐴} = 𝐴)
1310, 12fveq12d 5212 . . . 4 ((𝑥𝐹𝑥𝐴) → ({𝑧 ∣ ∀𝑥 𝑧𝐹}‘{𝑧 ∣ ∀𝑥 𝑧𝐴}) = (𝐹𝐴))
148, 13nfceqdf 2193 . . 3 ((𝑥𝐹𝑥𝐴) → (𝑥({𝑧 ∣ ∀𝑥 𝑧𝐹}‘{𝑧 ∣ ∀𝑥 𝑧𝐴}) ↔ 𝑥(𝐹𝐴)))
154, 5, 14syl2anc 397 . 2 (𝜑 → (𝑥({𝑧 ∣ ∀𝑥 𝑧𝐹}‘{𝑧 ∣ ∀𝑥 𝑧𝐴}) ↔ 𝑥(𝐹𝐴)))
163, 15mpbii 140 1 (𝜑𝑥(𝐹𝐴))
Colors of variables: wff set class Syntax hints: → wi 4 ∧ wa 101 ↔ wb 102 ∀wal 1257 = wceq 1259 ∈ wcel 1409 {cab 2042 Ⅎwnfc 2181 ‘cfv 4930 This theorem was proved from axioms: ax-1 5 ax-2 6 ax-mp 7 ax-ia1 103 ax-ia2 104 ax-ia3 105 ax-io 640 ax-5 1352 ax-7 1353 ax-gen 1354 ax-ie1 1398 ax-ie2 1399 ax-8 1411 ax-10 1412 ax-11 1413 ax-i12 1414 ax-bndl 1415 ax-4 1416 ax-17 1435 ax-i9 1439 ax-ial 1443 ax-i5r 1444 ax-ext 2038 This theorem depends on definitions: df-bi 114 df-3an 898 df-tru 1262 df-nf 1366 df-sb 1662 df-clab 2043 df-cleq 2049 df-clel 2052 df-nfc 2183 df-rex 2329 df-v 2576 df-un 2950 df-sn 3409 df-pr 3410 df-op 3412 df-uni 3609 df-br 3793 df-iota 4895 df-fv 4938 This theorem is referenced by: nfovd 5562
Copyright terms: Public domain W3C validator | 1,303 | 2,015 | {"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-2022-05 | latest | en | 0.132804 |
https://www.physicsforums.com/threads/how-to-expand-inverse-function.970138/ | 1,685,816,927,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224649302.35/warc/CC-MAIN-20230603165228-20230603195228-00000.warc.gz | 1,025,985,015 | 18,266 | # How to expand inverse function
• I
• kent davidge
#### kent davidge
If I'm given a function ##f(x)##, say it has continuos first derivative, then I expand it as ##f(x + \Delta x) = f(x) + (df / dx) \Delta x##. If instead, I'm given ##f^{-1}(x)## how do I go about expanding it? Will this be just ##f^{-1}(x + \Delta x) = f^{-1}(x) + (df^{-1} / dx) \Delta x##?
Last edited by a moderator:
Well, I'd have to imagine that since an inverse function is simply another function it should hold that rules for normal functions hold for inverse functions. I'd say yes then.
kent davidge
We generally have $$f(x_0+v)= f(x_0) + J_{x_0}f\cdot v + r(v)$$
We have the same equation for ##f^{-1}(x)##, i.e.
$$f^{-1}(x_0+v)= f^{-1}(x_0) + J_{x_0}f^{-1}\cdot v + \tilde r(v)$$
So yes, you can write it this way, just be careful with ##\dfrac{df^{-1}}{dx}##.
Harperchisari and kent davidge
But I think you first need to make sure the conditions of the inverse/implicit function theorem are met, so that you know you have at least a local (differentiable) inverse. Consider the problems, e.g. of ##\sqrt x## near ##x=0##. And then maybe you can use the chain rule on ##f(f^{-1}(x))=x##
But I think you first need to make sure the conditions of the inverse/implicit function theorem are met, so that you know you have at least a local (differentiable) inverse. Consider the problems, e.g. of ##\sqrt x## near ##x=0##. And then maybe you can use the chain rule on ##f(f^{-1}(x))=x##
Not in the situation stated in the OP ...
If instead, I'm given ##f^{−1}(x)## ...
... then we can write down the Weierstraß equation simply for this given function. How to compute the derivative given the derivative of ##f## is another question. If ##f^{-1}## is given, then we do not have to bother about ##f##.
Not in the situation stated in the OP ...
... then we can write down the Weierstraß equation simply for this given function. How to compute the derivative given the derivative of ##f## is another question. If ##f^{-1}## is given, then we do not have to bother about ##f##.
How do you know the inverse is differentiable?
I can't think at this point of a counter, but I don't see how we can know for certain under the conditions in the OP.
How do you know the inverse is differentiable?
By inspection. Given ##f^{-1}## means to me, given a function ##x \longmapsto f^{-1}(x)## and not given that ##f## has an inverse. I consider it as just another function, since this was what I read from post #1. In your example with the square root, I assume that a sign is given. And ##x \longmapsto +\sqrt{x}## is differentiable on ##(0,\infty)## regardless whether it inverts ##x \longmapsto x^2## somewhere or not.
By inspection. Given ##f^{-1}## means to me, given a function ##x \longmapsto f^{-1}(x)## and not given that ##f## has an inverse. I consider it as just another function, since this was what I read from post #1. In your example with the square root, I assume that a sign is given. And ##x \longmapsto +\sqrt{x}## is differentiable on ##(0,\infty)## regardless whether it inverts ##x \longmapsto x^2## somewhere or not.
But don't we need the inverse to be defined in a neighborhood of a point, as in the case of ##\sqrt x## at ##x=0##?
Why? ##\sqrt{x}## isn't differentiable at ##x=0##, so what?
But it is invertible there. EDIT: Why don't we take the discussion elsewhere to avoid disrupting this one further?
But it is invertible there.
But who cares?
$$f^{−1}(x_0+v)=f^{−1}(x_0)+J_{x_0}f^{−1}⋅v+\tilde r(v)$$ | 1,025 | 3,507 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2023-23 | longest | en | 0.83811 |
http://mathhelpforum.com/discrete-math/107622-discrete-math-help.html | 1,481,424,092,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698543782.28/warc/CC-MAIN-20161202170903-00256-ip-10-31-129-80.ec2.internal.warc.gz | 179,560,247 | 9,611 | 1. ## Discrete Math Help
here is another homework problem I am stuck on...
Let P(x.y) stand for the proposition that person x plays sport y. For example, P(babe Ruth, Baseball) is true.
Express symbolically, using P and the descriptor notation the set of sports exactly one of you and me plays (i.e., either you play it and I don't or I play it and you don't). Use my initials (JG) and your initials where needed in your expression.
2. This may not be what your after but I'll give it a hoof.
(A \ B) is defined as A not B.
I.e. IF A was the set {1,2,3,4,5} and B was the set {2,3,4} then (A \ B) would be {1,5} since neither are in B.
So if we do this with your problem we get... | 195 | 686 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2016-50 | longest | en | 0.944911 |
http://math.stackexchange.com/questions/15801/on-nonintersecting-loxodromes | 1,469,330,393,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257823935.18/warc/CC-MAIN-20160723071023-00267-ip-10-185-27-174.ec2.internal.warc.gz | 163,832,316 | 21,318 | On nonintersecting loxodromes
The (spherical) loxodrome, or the rhumb line, is the curve of constant bearing on the sphere; that is, it is the spherical curve that cuts the meridians of the sphere at a constant angle. A more picturesque way of putting it is that if one wants to travel from one point of a (spherical) globe to the antipodal point (say, from the North Pole to the South Pole) in a fixed direction, the path one would be taking on the globe would be a loxodrome.
For a unit sphere, the loxodrome that cuts the meridians at an angle $\varphi\in\left(0,90^\circ\right]$ is given by
\begin{align*}x&=\mathrm{sech}(t\cot\;\varphi)\cos\;t\\y&=\mathrm{sech}(t\cot\;\varphi)\sin\;t\\z&=\tanh(t\cot\;\varphi)\end{align*}
While playing around with loxodromes, I noted that for certain values of $\varphi$, one can orient two identical loxodromes such that they do not intersect (that is, one can position two ships such that if both take similar loxodromic paths, they can never collide). Here for instance are two loxodromes whose constant angle $\varphi$ is $60^\circ$, oriented such that they do not cross each other:
On the other hand, for the (extreme!) case of $\varphi=90^{\circ}$, the two loxodromes degenerate to great circles, and it is well known that two great circles must always intersect (at two antipodal points).
Less extreme, but seemingly difficult, would be the problem of positioning two 80° loxodromes such that they do not intersect:
This brings me to my first question:
1) For what values of $\varphi$ does it become impossible to orient two loxodromes such that they do not cross each other?
For simplicity, one can of course fix one of the two loxodromes to go from the North Pole to the South Pole, and try to orient the other loxodrome so that it does not cross the fixed loxodrome.
That's the simpler version of my actual problem. Some experimentation seems to indicate that it is not possible to orient three loxodromes such that they do not cross each other. So...
2) Is it true that for all (admissible) values of $\varphi$, one cannot position three loxodromes such that none of them cross each other?
I've tried a bit of searching around to see if the problem has been previously considered, but I have not had any luck. Any pointers to the literature will be appreciated.
-
What a beautiful question and figure! – Joseph O'Rourke Dec 30 '10 at 16:06
Have you tried working out what a rotated loxodrome looks like in Mercator projection? – Rahul Dec 30 '10 at 22:42
Also, would just rotating the loxodrome about the north-south axis count? (This would be Moron's parallel lines solution.) Of course, you should probably consider the loxodrome to contain its limit points at the poles, in which case this is not a solution... – Rahul Dec 30 '10 at 22:44
The problem seems to be the accumulation points; the loxodrome winds wildly at its poles that it seems a crossing is inevitable. – J. M. Dec 31 '10 at 0:21
@J.M.: Unless you actually consider the poles to be part of the loxodromes, it is clear that the suggestion of Rahul Narain allows an arbitrarily large collection of non-intersection loxodromes, all for the same angle, from the North Pole to the South Pole (all parallel so to speak, and actually parallel in the Mercator projection). – Marc van Leeuwen Jan 17 '12 at 12:50
Updated answer to (2): Three non-intersecting $60^\circ$ loxodromes.
The axes are coplanar and inclined at $120^\circ$ to each other. This image shows that symmetry better:
And here's the Mercator projection:
My approach was to plot one loxodrome such that its Mercator projection is a (black) line through the origin. Then, I tilted the spherical loxodrome "toward the camera"; that is, I rotated the sphere about the horizontal axis to get new, curvy (red) projections.
From tilt-angles $108^\circ$ to $143^\circ$, the "curve" lies between parts of the "line", indicating a range of red loxodromes that don't intersect the black one.
For a certain sub-range ($108^\circ$ to about $125^\circ$), a third (blue) non-intersecting loxodrome can be added by rotating the red one about the Mercator origin. Here's an image from the end of that range, where red and blue are tangent.
That's the end of the illustrated intro. Now for some equations ...
Starting with your parameterization of the loxodrome, then tilting via angle $\theta$, gives this parameterization of the Mercator projection:
\begin{align} u &= \rm{atan}\left( \frac{\sin t}{\cos t \cos \theta + \sinh\left(t \cot\phi\right) \sin \theta }\right) \\ v &= \rm{atanh}\left( \frac{-\cos t \sin\theta + \sinh\left(t \cot\phi\right) \cos\theta }{\cosh\left(t \cot\phi\right)}\right) \end{align}
A tilted loxodrome crosses into the range of (possible) non-intersection with the un-tilted loxodrome when the "top" of the outer loop about its tilted north pole meets the Mercator origin. (The nature of loxodromes guarantees that the two loxodromes will be tangent there.) The point on the loop corresponds to $t=\pi$, for which $u$ is already zero; for $v$ to vanish, we must have
$$0 = -\cos\pi \sin\theta + \sinh\left(\pi \cot\phi\right) \cos\theta = \sin\theta+\sinh\left(\pi \cot\phi \right) \cos\theta$$
so
$$\tan\theta = -\sinh\left(\pi\cot\phi\right)$$
Consequently, appropriately adjusting the "branch" of $\rm{atan}$, the range begins at
$$\theta_0 := \pi - \rm{atan}\left(\sinh\left(\pi\cot\phi\right)\right)$$
The range of (possible) non-intersection ends when the loop around the tilted loxodrome's south pole brushes against the un-tilted loxodrome. This is when the point corresponding to $t=-\pi$ has $v = \pi\cot\phi$ (matching the upper-right point of the "straight" loxodrome projection). So, the range ends at
$$\theta_1 := 2\;\rm{atan}\left( \sinh\left(\pi\cot\phi\right) \right)$$
I write "range of possible non-intersection", because that range collapses when $\theta_0 = \theta_1$. This gives us a critical loxodrome angle.
$$\phi_{*} = \rm{atan}\left(\frac{\pi}{ \rm{asinh}\left( \tan\frac{\pi}{3} \right) } \right) \approx 67.2565^\circ$$
You cannot arrive at two non-intersecting loxodromes with $\phi > \phi_{*}$ --in particular, with $\phi=80^\circ$-- by tilting one relative to the other in the way I've described.
Here's $\phi = 80^\circ$:
Of course, "the way I've described" lacks generality. In addition to "vertical" tilts, one should also consider "lateral" spins (horizontal shifts in the Mercator projection). I'll leave that, and a full investigation of the three-loxodrome scenario, as an exercise.
-
+1: Impressive figures :-) – Aryabhata Jan 2 '11 at 4:21
Not an answer, but an approach which might help.
Mercator projection is a map such that Loxodromes (corresponding to $0 \lt \varphi \lt 90^{\circ}$) get mapped to straight lines (and vice versa) and angles are preserved.
An image of the map:
So you would need to pick parallel lines from the mercator map.
The problem would be in choosing the correct antipodal points on the Mercator map and there could also be the problem of antipodal connecting lines possibly wrapping around the mercator map (which probably corresponds to your curvy $80^{\circ}$ loxodromes).
Here is another page which seems useful: Notes on Loxodrome Calculations which I got from here: http://www.erikdeman.de/html/sail022d.htm
Hope that helps.
-
Well, as you can see from the plot I gave, you can position two loxodromes with 60° bearing such that they don't intersect. For the first question, I'm looking for the value of the angle beyond which the two loxodromes cannot be positioned so as to be nonintersecting. – J. M. Dec 29 '10 at 8:00
@J.M: By negative I meant refuting the impossibility... From the mercator projection it looks like you should be able to find two non intersecting loxodromes of any angle in (0,pi/2). Of course I am not confident at all. – Aryabhata Dec 29 '10 at 8:01
Well, if you try a value of $\varphi$ close to a right angle (say, 80°), the loxodrome coils so much that it is clear that you cannot position two of those on the sphere such that they are nonintersecting... I'm looking for the "magical" value of $\varphi$. – J. M. Dec 29 '10 at 8:04
Edited answer. Previous comments no longer relevant, I suppose. – Aryabhata Dec 29 '10 at 18:28
"...there could also be the problem of antipodal connecting lines possibly wrapping around the Mercator map (which probably corresponds to your curvy 80° loxodromes)." - exactly my (simpler) problem. One satisfactory solution to my first problem would be a procedure that, given $\varphi$, would position two starting points such that the two corresponding loxodromes never cross. – J. M. Dec 30 '10 at 1:03 | 2,403 | 8,675 | {"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} | 3.34375 | 3 | CC-MAIN-2016-30 | latest | en | 0.906235 |
https://www.physics.uoguelph.ca/problem-5-57-translational-equil-0 | 1,643,250,910,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320305052.56/warc/CC-MAIN-20220127012750-20220127042750-00113.warc.gz | 954,160,839 | 12,423 | # Problem 5-57 Translational Equil.- A
A sign outside a hair stylist's shop is suspended by two wires. The force of gravity on the sign has a magnitude of $55.7\; N.$ If the angles between the wires and the horizontal are as shown in the figure, determine the magnitude of the tensions in the two wires.
[Ans. $T_1 = 49.9\; N;$ $T_2 = 40.8 \;N$]
No. The weight of $55.7\; N$ has not been included and it is a real force.
Try again. | 132 | 434 | {"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} | 3.109375 | 3 | CC-MAIN-2022-05 | longest | en | 0.922122 |
https://byterevel.com/2011/08/02/lets-destroy-the-internet/ | 1,639,012,110,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363641.20/warc/CC-MAIN-20211209000407-20211209030407-00503.warc.gz | 218,548,287 | 7,326 | # P versus NP – Destruction of the Internet
You are going to look at me like I’m crazy, but hypothetically one could destroy the internet. All one would have to do was solve the P versus NP problem (which is borderline paradoxical) a certain way and you could bypass any single security on the internet.
How does one solve P versus NP?
First we need to understand exactly what an algorithm is. If you know, then you can skip my next stupid question. If you don’t, then check it out.
Algorithms
An algorithm is a confined process of completing a mathematical or non-mathematical problem. Some conditions of the algorithm is that it cannot have any errors or be too arbitrary when completed properly.
Ex)
The Euclidean Algorithm is used to find the Greatest Common Factor of 2 numbers.
1. The numbers are 455 and 900
2. Subtract the smaller integer from the larger one
3. 900-455 = 445
4. You now have the two numbers of 455 and 445
5. Then subtract 445 from 455
7. Take 445 and subtract 10 from it
9. Take 435 and continuously subtract 10 from it until you get a number lower than 10.
10. That number will be 5.
11. Subtract 5 from 10. You’ll get 5.
12. If you subtract 5 from 5, you’ll get 0.
13. This means 5 is the greatest common factor of 455 and 900.
14. Check it out yourself.
That is how an algorithm would work.
Time Complexity
Programming’s most jumbled aspect is the Time Complexity. How fast can a program go in its worst case scenario? How about its best case scenario? How about average? This is all very convoluted.
What you need to know is that Time isn’t counted 1 second, 2 second… It is counted just using numbers. No units.
So what are the 2 times used in P versus NP? Polynomial and Non-Deterministic.
**Polynomial Time – Fast and Easy
**
Let’s say I write a program that adds 2 numbers together. We know that the maximum amount of numbers that can be added in this situation is. So this makes the time at which the program travels at, x^2. 2 is the constant of the equation and x is the number of pieces of data being used. If we want to add 3 numbers together, the equation becomes 3^2 = 9. Small amount of time, huh? Even if we want to add 100 numbers together, we’ll still only get 10,000, which is less than a second’s work for a computer. Needless to say, Polynomial Time is the fastest type of time the computer could use.
**
Non-Deterministic Polynomial Time – Slow to Do, Fast to check
**
This is what NP stands for. An example of this time is the Subset Problem. Let’s take a subset of 5 numbers {5, -1, 1, 4, -4}. In this problem, we wish to know whether or not the sum of some of the integers equals 0. The obvious answer is yes. (-4) + (4) = 0 would be the stupidly simple answer. But for a computer, the computer has to check each answer by taking the 5, adding it to -1…So on and so forth. The factorial of this subset is 120, so the computer has to check the subset 120 times. But how does the computer check? It just adds the numbers up, to see whether or not it equals 0. If the numbers equal 0, then the process to determine if it is equal to 0 is called a witness.
WAIT! The process of adding the numbers to create the witness! It is in Polynomial Time! This makes another theorem proven correct. All problems in NP (that answer to be yes) has to have been checked using P. If a problem in NP answers to be no, then it doesn’t have to be checked using P.
P versus NP
So what is P versus NP? It is a question theorized by Stephen Cook in 1971. P versus NP asks “can problems that are checked easily, be solved easily too?” Well we know problems in NP cannot be solved easily but can be checked easily. And we know that questions in P can be both solved and checked easily. This makes the simple answer “no.” No, P is always faster than NP, and NP is never equal to P.
The problem is how do you prove your answer equals no? How do you prove that P will never be slower than NP? How do you prove computers cannot ever solve problems that are in NP faster than they can solve problems in P? If you can prove it, though, the Clay Mathematics Institute offers a prize of \$1,000,000 to who ever proved it.
Significance
Now you are wondering how to destroy the internet. This is very hypothetical as most people believe P to be faster than NP. But what if P is slower than NP? This would be catastrophic:
If I could write a program that uses NP to be faster than P, that is supposed to hack past firewalls, I could get into any account. See, accounts are nothing but number/letter combinations. But these combinations has a maximum different type of letters, and a maximum length of password. For example; The code for a billionaire’s bank account is a 11-digit code. Since there are 10 different possible numbers, this makes the permutation for the number of combinations is 11! or 11 x 10 x 9 x 8 x 7 x 6… This would take the computer quite some time to figure out. But since NP is faster than P, one could compute this in the time it takes one to compute 2 +2 =4. This could be very disastrous to the world.
And since the internet is just interconnected, this’ll be happening if NP is faster than P. But thankfully, most people accept P to be much faster than NP, and computers that can check their problems quickly, can not necessarily solve them quickly.
Sources:
Mathematics by Richard Elwes
www.wikipedia.org
Categories:
Updated: | 1,302 | 5,404 | {"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 | 4 | CC-MAIN-2021-49 | longest | en | 0.927905 |
https://blogs.glowscotland.org.uk/er/Uplawmoor/category/science/ | 1,627,792,860,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154158.4/warc/CC-MAIN-20210801030158-20210801060158-00658.warc.gz | 152,237,551 | 21,243 | # P1 go to Space!
Primary 1 have been studying space and created these wonderful astronauts who have been living in a galaxy far, far away!
# P1 Visit the Science Centre
P1 had fun at the Science Centre yesterday. They experienced lots of different activity stations and even visited the Planetarium to link with their Space topic. See their class blog for more info.
# Learning about Space in the Planetarium
Tomorrow P1 are going to the Science Centre because we are learning about space. We are also going to discover about stars and planets . We are all very excited about the trip. (Remember to bring a packed lunch in a plastic bag!)
By Olivia and Elyza (P1)
# Terrifying Tsunamis
In P4/5 we have been learning all about tsunamis and their impact. We have learnt how a tsunami occurs, where they are most likely to happen and how to create a good warning system. This is all part of our topic “Water”. We are also learning all about the water cycle in Science. Primary 4/5 have been creating a giant tsunami for their classroom wall. We’ll post a picture when we are finished!
# Primary 2/3 study Sound Energy
Here are some pictures from Primary 2/3’s science lesson. The children went on a ‘sound energy walk’ around the school looking for objects which gave out sound energy. I wonder how many objects they found?
# P2/3 Investigate Forces at the Park!
As part of our science topic, Primary 2/3 went to the park to see which playground equipment needed to be pushed or pulled to make it work. We then brought in materials to make models of playground equipment which were able to move by being pushed or pulled. It was tricky but we got there!
# Whole School Weed
Thank you to the parents and friends who came along to help with our Whole School Weed which took place yesterday morning. The pupils, staff and friends worked hard to tidy up our grounds and get rid of those pesky weeds!
We even came across a little frog living happily in a plant pot!
# P6/7 Investigate Electricity
We have been working on building electrical circuits in science with Miss Simpson. We can now build circuits using a variety of components and can draw circuit diagrams to represent them.
We carried out an experiment where we built our own batteries out of lemons. We each had a lemon with a copper penny in it and a piece of aluminium foil in it. The penny and the foil represent electrodes of the battery and the lemon represents the electrolyte of the battery.
In batteries a chemical reaction occurs where chemical energy is converted to electrical energy. We tried to light up a bulb using 15 lemons but it did not work because enough electrical energy was not being made in the lemons to light it up.
We changed the bulb to a buzzer and it worked. The lemons had made enough electrical energy to power the buzzer!
# Prof Science Visits P1-3
Primary 1, 2 and 3 loved having Professor Science come to visit them to tell them all about the sun, moon and stars. We had the chance to explore gravity by catching bubbles and using marbles and feathers to see which would fall to the floor fist. We were very surprised to find out they fall at the same time after watching a slow motion video. We even got to eat donuts after discovering that the universe is in the shape of one!
# P2/3 Oreo Fun – Phases of Moon
P2/3 have been busy learning about the moon in Science. The class were lucky enough to feast on Oreo cookies which helped them visualise the different phases of the moon.
If it was a full moon, we ate the lid of the cookie which showed all the cream which was the full moon.
If it was a half moon, we licked off half of the creamy bit.
Then we ate lots of the cream and left oval shapes which showed the waxing and waning crescent.
It was very exciting because at the end we got to eat full oreos!
By Lucy P3.
# P6/7 Visit Whitelee Windfarm to learn about Wind Energy
P6/7 have been learning about renewable and non-renewable energy. We knew that wind was a source of renewable energy as there is always going to be wind! To learn more about wind turbines, we took a trip to Whitelee Windfarm and learned all about how the wind turbines work and how they produce electricity which helps to power just under 300,000 homes in one year! That’s a lot of renewable energy.
After completing a workshop where we measured our own wind energy, we visited turbine 41 and braved the wind to feel the blades flying past us. We then visited the exhibition and played lots of activities which tested our ability to make our own wind power.
Thank you for having us. We loved our trip to Whitelee and have learned lots about wind energy.
# Sun, Moon and Stars
Primary 2/3 are now currently learning about space and what travels round the earth. The subject was science included with space. We learned how the earth travels round the sun and how the earth travels round the moon. We acted it out and got given a planet, some stars, the sun, the moon and the earth. We learned this in four different groups of three. The name of the groups were moon groups.
By Zara Primary 3
I enjoyed this topic because I read a book before this topic and now I know much more about space. Zara P3
I learned that the sun doesn’t move but the earth and the moon turns around in a circle. Eilidh P2
It takes a whole year for earth to go around the sun and the moon goes around the earth. Heather P2
# P4/5 are learning about Sound Travel
Primary 4/5 have been learning about sound and how it travels through air molecules.
Rian “I have learned how sound travels in air molecules. I have also learned how sound travels faster through water and solids than through air.”
Catherine “I enjoyed doing a poster in my IDL jotter.”
# P6/7 Investigating Chemical Reactions
Primary 6/7 have been taking control of their own learning and investigating what happens when you mix different liquids.
We learned about the weight of water and oil and how salt reacts to different liquids.
# Solar Eclipse
By Amy Cupples P6
As you have probably heard there was a solar eclipse which took place today. An ‘eclipse’ is when the Moon comes between the Sun and the Earth which make some parts of the world go dark.
We all went out to the playground with our solar eclipse viewers and watched the moon eclipse the sun. It was really fascinating to watch this happen. The last major eclipse took place in 1999 and then next one won’t happen until 2026! | 1,481 | 6,473 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2021-31 | latest | en | 0.967232 |
https://gis.stackexchange.com/questions/113527/densifying-jts-linestring-add-points | 1,718,943,434,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862036.35/warc/CC-MAIN-20240621031127-20240621061127-00282.warc.gz | 238,967,470 | 41,803 | I have a LineString I would like to "soften" in a way that if two coordinates (vertices) in the LineString are very far apart, I want to automatically add some coordinates in between.
The line string should stay the same but with more coordinates describing it. Do you know of a way?
I have a LineString like the one on the left - and I want to add some coordinates in the between the existing coordinates (like on the right) - hopefully by JTS and hopefully automatically.
The algorithm for line densification isn't all that difficult (Pythagorean theorem for length and prorate dX and dY at the same ratio). The only tricky part is deciding between even distribution of spacing, or a fixed interval along the middle. You should evaluate the output of the existing method available within JTS before coding your own.
The problem becomes more complex when the coordinates are in decimal degrees (especially over long distances), in which case the spheroidal length is needed (inverse geodetic problem), and the intervening vertices must be solved along the great circle route using the forward geodetic problem (there are simpler spherical solutions, but they are less accurate). These algorithms have posted solutions within GSE.
I wrote this method which does the trick using JTS:
``````public LineString refineLineString(LineString ls, double maximalDistance){
// list to store coordinates for resulting line string
ArrayList<Coordinate> resultLineStringCoordinates = new ArrayList();
// list of LineSegments from input LineString
ArrayList<LineSegment> segmentList = new ArrayList();
for(int i = 1; i < ls.getCoordinates().length; i++){
// create LineSegments from input LineString and add them to list
}
boolean isFirstSegment = true;
// refine each LineSegment if necessary
for(LineSegment currentSegment : segmentList){
// add startpoint from first segment only
if(isFirstSegment){
isFirstSegment = false;
}
// refinement necessary
if(currentSegment.getLength() > maximalDistance){
// calculate distance between coordinates as fraction (0-1)
double distanceFraction = 1 / (currentSegment.getLength()/maximalDistance);
for(double currentFraction = distanceFraction; currentFraction < 1; currentFraction += distanceFraction){
// add coordinates from calculated fraction
}
}
// add segment endpoint coordinate to result list
}
return new GeometryFactory().createLineString(resultLineStringCoordinates.toArray(new Coordinate[resultLineStringCoordinates.size()]));
}
``````
Note that the last part of a segment can be smaller than the rest (rest is equal to maximalDistance). If you prefer equal distances along one segment over setting the distances to maximalDistance you just have to adjust the calculation of distanceFraction.
Simple example:
``````LineString lineA = new GeometryFactory().createLineString(new Coordinate[]{new Coordinate(0, 0), new Coordinate(0, 10), new Coordinate(5, 10)});
System.out.println("LineString before refinement: " + lineA.toText());
LineString refined = Test.refineLineString(lineA, 1.5);
System.out.println("LineString ater refinement: " + refined.toText());
``````
Output:
``````LineString before refinement: LINESTRING (0 0, 0 10, 5 10)
LineString ater refinement: LINESTRING (0 0, 0 1.5, 0 3, 0 4.5, 0 6, 0 7.5, 0 9, 0 10, 1.5 10, 3 10, 4.5 10, 5 10)
``````
OpenJUMP has a "densifier" function in Tools-Analysis-Geometry functions. The source code (GPL licensed) of the tool is in http://sourceforge.net/p/jump-pilot/code/HEAD/tree/core/trunk/src/com/vividsolutions/jump/algorithm/Densifier.java. I believe it will help you.
Here's my solution for the same problem: https://github.com/mcserby/tools
I am splitting the lineString into lineSegments, calculate total lenght of the line string and the "step" size, and then extract the coordinates at the "step" intervals on the line string.
See the class below:
``````package com.fci.graphic;
import java.util.ArrayList;
import java.util.List;
import com.vividsolutions.jts.geom.Coordinate;
import com.vividsolutions.jts.geom.LineSegment;
import com.vividsolutions.jts.geom.LineString;
public class EchidistantPointsCalculator {
public static List<Coordinate> findEchidistantPoints(LineString lineString, int resolution) {
List<LineSegment> lineSegments = buildLineSegments(lineString);
double lineStringLength = lineSegments.parallelStream().mapToDouble(l -> l.getLength()).sum();
double stepSize = lineStringLength / resolution;
List<Coordinate> coords = new ArrayList<>();
double startingDistance = 0;
for(LineSegment ls : lineSegments){
SegmetData segmentData = processSegment(ls, stepSize, startingDistance);
startingDistance = segmentData.getNextStartingDistance();
}
return coords;
}
private static SegmetData processSegment(LineSegment segment, double stepSize,
double currentDistance) {
List<Coordinate> coords = new ArrayList<>();
LineSegment remainingSegment = new LineSegment(segment);
double remainingLength = remainingSegment.getLength();
while(remainingLength >= currentDistance){
Coordinate c = remainingSegment.pointAlong(currentDistance/remainingLength);
currentDistance = stepSize;
remainingSegment = new LineSegment(c, remainingSegment.p1);
remainingLength = remainingSegment.getLength();
}
return new SegmetData(coords, remainingLength);
}
private static List<LineSegment> buildLineSegments(LineString lineString) {
Coordinate[] coords = lineString.getCoordinates();
List<LineSegment> segments = new ArrayList<>();
for(int i = 0; i < coords.length - 1; i++){ | 1,231 | 5,491 | {"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} | 2.90625 | 3 | CC-MAIN-2024-26 | latest | en | 0.828571 |
http://www.ask.com/question/what-are-intermediate-directions | 1,397,892,966,000,000,000 | text/html | crawl-data/CC-MAIN-2014-15/segments/1397609536300.49/warc/CC-MAIN-20140416005216-00260-ip-10-147-4-33.ec2.internal.warc.gz | 292,578,006 | 17,347 | # What Are Intermediate Directions?
Intermediate directions are simply the directions or points that would fall in between your standard N, S, E, W points on a compass. The points in between are what we know as NE, NW, SE, SW and etc. Hope this helps!
What Are Intermediate Directions?
The four cardinal directions are north, south, east and west, and they're often the first thing you learn about maps and navigation. Intermediate directions are between the cardinal directions, and they help people find a shorter way to their destination.... More »
Difficulty: Easy
Source: www.ehow.com
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Intermediate directions are directions that are not too difficult but not too easy either. They are directions that are in the middle of the road where concerning ...
Intermediate directions are the directions which lie between the four cardinal directions (N, S, E, & W). Intermediate directions are north-east (NE), north-west ... | 382 | 1,949 | {"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} | 3.59375 | 4 | CC-MAIN-2014-15 | longest | en | 0.950463 |
https://number.academy/1204451 | 1,679,361,450,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296943589.10/warc/CC-MAIN-20230321002050-20230321032050-00595.warc.gz | 484,376,688 | 12,203 | # Number 1204451
Number 1,204,451 spell 🔊, write in words: one million, two hundred and four thousand, four hundred and fifty-one , approximately 1.2 million. Ordinal number 1204451st is said 🔊 and write: one million, two hundred and four thousand, four hundred and fifty-first. The meaning of the number 1204451 in Maths: Is it Prime? Factorization and prime factors tree. The square root and cube root of 1204451. What is 1204451 in computer science, numerology, codes and images, writing and naming in other languages
## What is 1,204,451 in other units
The decimal (Arabic) number 1204451 converted to a Roman number is (M)(C)(C)(IV)CDLI. Roman and decimal number conversions.
#### Weight conversion
1204451 kilograms (kg) = 2655332.7 pounds (lbs)
1204451 pounds (lbs) = 546335.4 kilograms (kg)
#### Length conversion
1204451 kilometers (km) equals to 748411 miles (mi).
1204451 miles (mi) equals to 1938377 kilometers (km).
1204451 meters (m) equals to 3951563 feet (ft).
1204451 feet (ft) equals 367122 meters (m).
1204451 centimeters (cm) equals to 474193.3 inches (in).
1204451 inches (in) equals to 3059305.5 centimeters (cm).
#### Temperature conversion
1204451° Fahrenheit (°F) equals to 669121.7° Celsius (°C)
1204451° Celsius (°C) equals to 2168043.8° Fahrenheit (°F)
#### Time conversion
(hours, minutes, seconds, days, weeks)
1204451 seconds equals to 1 week, 6 days, 22 hours, 34 minutes, 11 seconds
1204451 minutes equals to 2 years, 5 months, 3 weeks, 3 days, 10 hours, 11 minutes
### Codes and images of the number 1204451
Number 1204451 morse code: .---- ..--- ----- ....- ....- ..... .----
Sign language for number 1204451:
Number 1204451 in braille:
QR code Bar code, type 39
Images of the number Image (1) of the number Image (2) of the number More images, other sizes, codes and colors ...
## Mathematics of no. 1204451
### Multiplications
#### Multiplication table of 1204451
1204451 multiplied by two equals 2408902 (1204451 x 2 = 2408902).
1204451 multiplied by three equals 3613353 (1204451 x 3 = 3613353).
1204451 multiplied by four equals 4817804 (1204451 x 4 = 4817804).
1204451 multiplied by five equals 6022255 (1204451 x 5 = 6022255).
1204451 multiplied by six equals 7226706 (1204451 x 6 = 7226706).
1204451 multiplied by seven equals 8431157 (1204451 x 7 = 8431157).
1204451 multiplied by eight equals 9635608 (1204451 x 8 = 9635608).
1204451 multiplied by nine equals 10840059 (1204451 x 9 = 10840059).
show multiplications by 6, 7, 8, 9 ...
### Fractions: decimal fraction and common fraction
#### Fraction table of 1204451
Half of 1204451 is 602225,5 (1204451 / 2 = 602225,5 = 602225 1/2).
One third of 1204451 is 401483,6667 (1204451 / 3 = 401483,6667 = 401483 2/3).
One quarter of 1204451 is 301112,75 (1204451 / 4 = 301112,75 = 301112 3/4).
One fifth of 1204451 is 240890,2 (1204451 / 5 = 240890,2 = 240890 1/5).
One sixth of 1204451 is 200741,8333 (1204451 / 6 = 200741,8333 = 200741 5/6).
One seventh of 1204451 is 172064,4286 (1204451 / 7 = 172064,4286 = 172064 3/7).
One eighth of 1204451 is 150556,375 (1204451 / 8 = 150556,375 = 150556 3/8).
One ninth of 1204451 is 133827,8889 (1204451 / 9 = 133827,8889 = 133827 8/9).
show fractions by 6, 7, 8, 9 ...
### Calculator
1204451
### Advanced math operations
#### Is Prime?
The number 1204451 is a prime number. The closest prime numbers are 1204447, 1204453.
#### Factorization and factors (dividers)
The prime factors of 1204451
Prime numbers have no prime factors smaller than themselves.
The factors of 1204451 are 1 , 1204451
Total factors 2.
Sum of factors 1204452 (1).
#### Prime factor tree
1204451 is a prime number.
#### Powers
The second power of 12044512 is 1.450.702.211.401.
The third power of 12044513 is 1.747.299.729.224.145.920.
#### Roots
The square root √1204451 is 1097,474829.
The cube root of 31204451 is 106,397081.
#### Logarithms
The natural logarithm of No. ln 1204451 = loge 1204451 = 14,001534.
The logarithm to base 10 of No. log10 1204451 = 6,080789.
The Napierian logarithm of No. log1/e 1204451 = -14,001534.
### Trigonometric functions
The cosine of 1204451 is -0,483745.
The sine of 1204451 is 0,875209.
The tangent of 1204451 is -1,809234.
### Properties of the number 1204451
Is a Fibonacci number: No
Is a Bell number: No
Is a palindromic number: No
Is a pentagonal number: No
Is a perfect number: No
## Number 1204451 in Computer Science
Code typeCode value
1204451 Number of bytes1.1MB
Unix timeUnix time 1204451 is equal to Wednesday Jan. 14, 1970, 10:34:11 p.m. GMT
IPv4, IPv6Number 1204451 internet address in dotted format v4 0.18.96.227, v6 ::12:60e3
1204451 Decimal = 100100110000011100011 Binary
1204451 Decimal = 2021012012022 Ternary
1204451 Decimal = 4460343 Octal
1204451 Decimal = 1260E3 Hexadecimal (0x1260e3 hex)
1204451 BASE64MTIwNDQ1MQ==
1204451 MD5077dccfe222be8012c73ae8914c26d55
1204451 SHA22447b7e3befc8be2855c69d37b84770bf6802ce436e0695883c2ff97e5
1204451 SHA25647aa92723c68d00f5fc808ce3ab1b012d09d4ca3060321bc0be7b6891ae87cb7
1204451 SHA38401b0c847f3f5fac91ce88276251d8f52b5be0580f4ab1216d32a9862930ef5a395f9623ac2fd14c35b0703ff5807c750
More SHA codes related to the number 1204451 ...
If you know something interesting about the 1204451 number that you did not find on this page, do not hesitate to write us here.
## Numerology 1204451
### Character frequency in the number 1204451
Character (importance) frequency for numerology.
Character: Frequency: 1 2 2 1 0 1 4 2 5 1
### Classical numerology
According to classical numerology, to know what each number means, you have to reduce it to a single figure, with the number 1204451, the numbers 1+2+0+4+4+5+1 = 1+7 = 8 are added and the meaning of the number 8 is sought.
## № 1,204,451 in other languages
How to say or write the number one million, two hundred and four thousand, four hundred and fifty-one in Spanish, German, French and other languages. The character used as the thousands separator.
Spanish: 🔊 (número 1.204.451) un millón doscientos cuatro mil cuatrocientos cincuenta y uno German: 🔊 (Nummer 1.204.451) eine Million zweihundertviertausendvierhunderteinundfünfzig French: 🔊 (nombre 1 204 451) un million deux cent quatre mille quatre cent cinquante et un Portuguese: 🔊 (número 1 204 451) um milhão, duzentos e quatro mil, quatrocentos e cinquenta e um Hindi: 🔊 (संख्या 1 204 451) बारह लाख, चार हज़ार, चार सौ, इक्यावन Chinese: 🔊 (数 1 204 451) 一百二十万四千四百五十一 Arabian: 🔊 (عدد 1,204,451) واحد مليون و مئتان و أربعة ألفاً و أربعمائة و واحد و خمسون Czech: 🔊 (číslo 1 204 451) milion dvěstě čtyři tisíce čtyřista padesát jedna Korean: 🔊 (번호 1,204,451) 백이십만 사천사백오십일 Dutch: 🔊 (nummer 1 204 451) een miljoen tweehonderdvierduizendvierhonderdeenenvijftig Japanese: 🔊 (数 1,204,451) 百二十万四千四百五十一 Indonesian: 🔊 (jumlah 1.204.451) satu juta dua ratus empat ribu empat ratus lima puluh satu Italian: 🔊 (numero 1 204 451) un milione e duecentoquattromilaquattrocentocinquantuno Norwegian: 🔊 (nummer 1 204 451) en million, to hundre og fire tusen, fire hundre og femti-en Polish: 🔊 (liczba 1 204 451) milion dwieście cztery tysiące czterysta pięćdziesiąt jeden Russian: 🔊 (номер 1 204 451) один миллион двести четыре тысячи четыреста пятьдесят один Turkish: 🔊 (numara 1,204,451) birmilyonikiyüzdörtbindörtyüzellibir Thai: 🔊 (จำนวน 1 204 451) หนึ่งล้านสองแสนสี่พันสี่ร้อยห้าสิบเอ็ด Ukrainian: 🔊 (номер 1 204 451) один мiльйон двiстi чотири тисячi чотириста п'ятдесят одна Vietnamese: 🔊 (con số 1.204.451) một triệu hai trăm lẻ bốn nghìn bốn trăm năm mươi mốt Other languages ...
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## Comment
If you know something interesting about the number 1204451 or any other natural number (positive integer), please write to us here or on Facebook. | 2,732 | 7,799 | {"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 | 0.68039 |
https://www.physicsforums.com/threads/fibonacci-sequence.611399/ | 1,508,407,371,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187823260.52/warc/CC-MAIN-20171019084246-20171019104246-00204.warc.gz | 967,247,740 | 19,931 | Fibonacci sequence
1. Jun 4, 2012
leroyjenkens
My program needs to prompt the user to input a number, and using that number, I tell them what number in the Fibonacci sequence their input corresponds to.
So the Fibonacci sequence is 0 1 1 2 3 5 8 13
So if they input the number 6, the program will return "5" as the number in the sequence.
Anyone have any ideas how I would do this? I need a loop for it, I'm sure, and I was trying the "for" loop, but I can't figure out how to just give them the single number in the sequence, and not the whole sequence up to that number.
Thanks.
2. Jun 4, 2012
rcgldr
You can calculate the entire sequence, but only output a single number. You don't need to save all the numbers in the sequence either, just the last two numbers of the sequence.
3. Jun 4, 2012
gsal
why don't you post the code that you've got so far? I would be a lot easier to help you that way.
4. Jun 4, 2012
AlephZero
If you write a recursive function to calculate it, you don't need to "remember" anything explicitly. The process of carrying out the recursion does the "remembering" for you. YOu don't need any loops, either.
For example a recursive function to calculate factorials (warning - untested!) could be
Code (Text):
int fact(int i)
{
if (i < 1) return 1;
else return i*fact(i-1);
}
and then your main program could just do
Code (Text):
int n;
cin >> n;
cout << fact(n);
5. Jun 5, 2012
NemoReally
What language are you programming in? If you've got the sequence, then there should be a way of simply indexing the last number in the sequence and displaying that.
6. Jun 5, 2012
Xitami
$$\begin{bmatrix}0&1\end{bmatrix} \cdot \begin{bmatrix}1&1\\1&0\end{bmatrix} ^ n \cdot \begin{bmatrix}0\\1\end{bmatrix}$$
7. Jun 5, 2012
leroyjenkens
I'm using C. How do I index the last number in the sequence? I can get the sequence to show up to the number they put in, but I don't know how to isolate that number.
8. Jun 5, 2012
NemoReally
How are you creating the sequence? Can you post the code?
9. Jun 5, 2012
leroyjenkens
#include<stdio.h>
main()
{
int n, fib1 = 0, fib2 = 1, fib, c;
printf("Enter the number of terms\n");
scanf("%d",&n);
printf("First %d terms of Fibonacci series are: ",n);
for ( c = 0 ; c < n ; c++ )
{
if ( c <= 1 )
fib = c;
else
{
fib = fib1 + fib2;
fib1 = fib2;
fib2 = fib;
}
printf("%d ", fib);
}
return 0;
}
10. Jun 5, 2012
NemoReally
Ah, OK. It looks as though you are writing out the answer for every iteration. Try moving the printf statement outside the loop, as shown in the amended code above
11. Jun 5, 2012
leroyjenkens
Thanks a lot. I can't believe that was so easy. I'm trying my best on this programming stuff and it's driving me nuts. I'm going to fail, but I need to make at least a D to get reimbursed so I don't lose all of that money.
Can you help me with another one? I have to make a program where if the user types in, for example, 12345, the program prints out this...
12345
2345
345
45
5
I don't even know where to begin with this one.
Thanks.
12. Jun 6, 2012
NemoReally
Ah. Now you're entering 'blind leading the blind' country. I know how to do it algorithmically, but I haven't touched C programming for so long that we used to call it Alpha. I tend to use much higher-level languages. Consequently, all the messy, low-level defining memory stuff is detail that I no longer even know, or care, about. Do you have to use C or can you use something slightly more friendly, such as C++, C# or Basic?
A rough outline for C would be something like ...
Code (Text):
sin = <get user input>
int n
int k
n = strlen(s)
char sout[n+1]
for (k = n ; k>0 ; k--){ // starting with the full string and working down to 1 character
strncpy(sin,sout,k) // copy k characters into sout
sout[k+1]='\0' // and terminate with the null character
printf("countdown: %d",sout) // print out the reduced string
}
But, ... you would need to get the loop format correct and ensure that the size of sout allowed for the '0' string terminator. Note that it could probably also have been done by simply setting sin[k+1] to be the null character and forgetting about strncpy, or a myriad of other ways that regular C users could tell you about. A while loop would be a nice counterpoint to the for loop.
13. Jun 6, 2012
gsal
we do not solve homework around here...
you need to start,
take a shot at it,
post the code
explain what it is you can't figure out
etc.
so, I will spell out an algorithm, here, that you can follow...which is actually more work than writing the program :-) ...oh well, it's either that or not post at all.
basically
you have to read the string
then
knowing that every string ends with the null character '\0'
you can have two nested while loops
the outer one keeps increasing the index of the starting character
the inner one prints from such index and until the character is the null character
so, here is an idea for you to proceed
first write a little program that reads an input from the screen
and using a while loop (you will need a counter variable as index), writes it back out to the screen, one character at a time and without going beyond the terminating null character
can you do that?
then write an inner while loop (you will need another counter variable)
that writes one character at a time, but starting from the value of the index of the outer loop
again, it does not go beyond the null character.
you will need to get a bit tricky when using or not using "\n" in your print statements to create the desired output...
this algorithm does not use any of the string functions (no need to include "string.h")
setting the counters and the two nested while loops, printing, etc. is only 9 lines
the entire program, variable declaration, and input reading 20 lines.
you can do it! ...ready, set, go!
14. Jun 6, 2012
leroyjenkens
I know. I don't know where I implied I was looking for a solution.
I can't start on a program I don't know where to begin with. I don't know what could possibly replicate what the program is supposed to do.
I just started programming a few weeks ago. Before that I have done nothing in my life that is remotely close to this, so a lot of the terminology you're using is alien to me.
15. Jun 6, 2012
rcgldr
Where to begin ... how about a program that prints out what the user types in. The program would read a line of text and output a line of text, or the program would read a number and output the number. (You didn't specify if the 12345 could be any letters or just a number.)
16. Jun 6, 2012
Staff: Mentor
Does this have to work for any character string (which could include letters and punctuation marks), or does it just have to work for integers? If it's just integers, is it restricted to integers that you can store in a single variable, or can they be arbitrarily long (which would probably have to use strings)?
17. Jun 6, 2012
leroyjenkens
Ok sorry for the lack of information, and sorry if I sounded like a jerk in my last post. I don't like the way I said that. But anyway, I got some more information from my teacher, and he helped me a little bit, and the program asks the user to input a number between 1 and 32767 and it will return their answer like this...
*****
****
***
**
*
So if they input 32766, it will give them...
32766
2766
766
66
6
And if they input less than 5 numbers, such as 766, it will give them...
00766
0766
766
76
6
Here's what I got so far.
#include <stdio.h>
#include <stdlib.h>
int main()
{
int num;
int d1, d2, d3, d4, d5;
printf("Enter a number between 1 and 32767: ");
scanf("%d", &num);
d1 = num / 10000;
d2 = (num / 1000) %10;
d3 = (num / 100) %10;
d4 = (num / 10) %10;
d5 = (num % 10);
printf("%d\n %d\n %d\n %d\n %d\n", &d1, &d2, &d3, &d4, &d5);
return 0;
}
It's producing this sequence when I type in, for instance, 32767...
2993528
2993747
2994283
2993278
2997325
So they're the wrong numbers, with too many digits, and it's not getting smaller.
Thanks.
18. Jun 6, 2012
leroyjenkens
Ok I already see a problem. I put the & signs in the printf command.
It seems like it's working. 3 spaces are supposed to be between the digits.
#include <stdio.h>
#include <stdlib.h>
int main()
{
int num;
int d1, d2, d3, d4, d5;
printf("Enter a number between 1 and 32767: ");
scanf("%d", &num);
d1 = num / 10000;
d2 = (num / 1000) %10;
d3 = (num / 100) %10;
d4 = (num / 10) %10;
d5 = (num % 10);
printf("%d %d %d %d %d\n", d1, d2, d3, d4, d5);
printf("%d %d %d %d\n", d2, d3, d4, d5);
printf("%d %d %d\n", d3, d4, d5);
printf("%d %d\n", d4, d5);
printf("%d", d5);
return 0;
}
I think I'm supposed to separate the first part from the second part in two different functions. Like, for example, the part where the printing is done should be in a second function. I'm not sure how to do that. Do I like define the function at the top and then use that same definition after the first function or something? | 2,529 | 8,954 | {"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} | 3.53125 | 4 | CC-MAIN-2017-43 | longest | en | 0.886616 |
https://www.clutchprep.com/physics/practice-problems/141834/a-somewhat-idealized-graph-of-the-speed-of-the-blood-in-the-ascending-aorta-duri | 1,632,135,907,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057036.89/warc/CC-MAIN-20210920101029-20210920131029-00216.warc.gz | 733,536,521 | 31,474 | Calculating Displacement from Velocity-Time Graphs Video Lessons
Concept
# Problem: A somewhat idealized graph of the speed of the blood in the ascending aorta during one beat of the heart appears as in Figure P2.16. a. Approximately how far, in cm, does the blood move during one beat? b. Assume similar data for the motion of the blood in your aorta, and make a rough estimate of the distance from your heart to your brain. Estimate how many beats of the heart it takes for blood to travel from your heart to your brain.
###### FREE Expert Solution
a.
Distance is the area under the curve.
Area is A = (1/2)Δv(Δt)
100% (441 ratings)
###### Problem Details
A somewhat idealized graph of the speed of the blood in the ascending aorta during one beat of the heart appears as in Figure P2.16.
a. Approximately how far, in cm, does the blood move during one beat?
b. Assume similar data for the motion of the blood in your aorta, and make a rough estimate of the distance from your heart to your brain. Estimate how many beats of the heart it takes for blood to travel from your heart to your brain. | 257 | 1,106 | {"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 | 0.904408 |
https://www.thebeginwithinblog.com/43023/what-is-the-measure-of-angle-acb.html | 1,675,089,486,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499819.32/warc/CC-MAIN-20230130133622-20230130163622-00679.warc.gz | 1,043,517,371 | 25,929 | # What Is The Measure Of Angle Acb
What Is The Measure Of Angle Acb. 2 determine the measure of each indicated angle. Sum of all angles =180° so.
Abcd is a rhombus such that angle acb =50°, then what is the measure of angle adb get the answers you need, now! Roslyn123 roslyn123 26.12.2019 math secondary school answered if. 2 determine the measure of each indicated angle.
### 2 Determine The Measure Of Each Indicated Angle.
The figure is not drawn to scale. 162 times 2 for reference: Both ==> length of arc ab= 10, from the formula for finding length of arc we have length arc ab = \(angle/360\)*2*pi*r we can cal angle aob, now in triangle ocb, we can find.
### 01/08/2021 Geography High School Answered What Is The Measure Of Angle Acb?
Expert solution want to see the full answer? Angle bcd = 103 °. Find an answer to your question find the measure of angle acb
### Just Like The Concept Of Lines, An Angle Is A Necessary Figure That Is Used For Measurement.
Angle bcd = 103 ° angle sum theorem: Abc has to be 110 triangle angles add to 180. I is incenter and e is excenter for bc.if ai = ie,.
### (Exact Copy Of The Problem) In The Right Triangle, Right Angle In B;
If angle bdc and angle acb are right angles, then what is the measure of angle bcd? In mathematics, angles constitute a very important part. Solution for find the measure of angle acb.
### The Measure Of Angle Acb Is Degrees.
Abcd is a rhombus such that angle acb =50°, then what is the measure of angle adb get the answers you need, now! Sum of all angles =180° so. Angle abc + angle acb + angle bac = 180 ° substitute angle ab.angle bcd is a circumscribed angle of circle a.
Read: A Triangle Cannot Have More Than One Obtuse Angle | 445 | 1,721 | {"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-2023-06 | latest | en | 0.812885 |
livewell.jillianmichaels.com | 1,485,167,271,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560282631.80/warc/CC-MAIN-20170116095122-00256-ip-10-171-10-70.ec2.internal.warc.gz | 167,109,308 | 16,153 | ×
## Looking for the Old Website?
If you are member of the old Jillian Michaels website:
# How to Calculate Calories of My Recipes
Packaged foods come with their calorie and nutrition information pre-calculated for you. You can look up commodity foods, like ground beef, apples or eggs, in a book or online database, such as the USDA National Nutrient Database. If you are preparing your own recipe at home, however, you will have to calculate the calories per serving yourself. This can be a little tricky, since the quantity on the package or in the database might not match what you are using in your recipe. But with a few careful calculations, you can get to your answer by adding up the calories for all your ingredients and then dividing by the number of servings your recipe produces.
## Calculate the Calories in the Recipe
#### Step 1
Find the calories for each of the ingredients in your recipe that has a nutrition label. Let's consider a simple chocolate fudge recipe. This recipe calls for 14 ounces of sweetened condensed milk, 2 cups of semisweet chocolate chips, 2 tablespoons butter, 1 teaspoon vanilla and 1 cup chopped walnuts. There should be calorie information for the condensed milk and the chocolate chips on their packages. The condensed milk label might say, for example, that there are 130 calories in one serving, and that there are 10 servings in the entire 14-ounce. can. That means there are 130 x 10 = 1,300 calories in 14 ounces. The label on the chocolate chips might list the calories like this: 70 calories in 1 tablespoon. Convert that to the 2 cups needed in our recipe: 70 calories per tablespoon x 16 tablespoons per cup x 2 cups = 2,240 calories in 2 cups of chocolate chips.
#### Step 2
Find the calories for each of the non-packaged ingredients from a calorie listing or database. According to the USDA nutrient database, butter has 102 calories per tablespoon, and our recipe calls for 2 tablespoon, so we'll have a total of 204 calories from butter. The database tells us that vanilla extract has 37 calories per tablespoon, but we need only 1 teaspoon, which is one-third of that, or about 12 calories. Finally, assuming we are using bulk walnuts that we chopped ourselves, we find the calories from USDA to be 766 calories per cup.
#### Step 3
Add up the calories for all of the ingredients. The total of all our recipe ingredients would be: 14 ounces sweetened, condensed milk + 2 cups chocolate chips + 2 tablespoons butter + 1 teaspoon vanilla + 1 cup chopped walnuts = 1,300 + 2,240 + 204 + 12 + 766 = 4,522. This is the total calories for the entire pan of fudge.
#### Step 4
Determine the number of servings in the entire recipe. This particular recipe suggests cutting the pan into 25 pieces of fudge.
#### Step 5
Calculate the calories per individual serving by dividing the total calories by the number of servings. In the case of our fudge recipe, that would be 4,522 / 25 = 181 calories per serving.
#### Warning
• When you divide the total calories by the number of servings, be sure you use the actual number of servings that you split the recipe into. For example, if you are a chocolate lover, you might split the fudge recipe into 10 pieces instead of 25. In that case, each piece would have 452 calories!
### DON'T MISS
#### Photo Credits:
• old resipe box image by charles taylor from Fotolia.com
This article reflects the views of the writer and does not necessarily reflect the views of Jillian Michaels or JillianMichaels.com. | 796 | 3,516 | {"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} | 4.09375 | 4 | CC-MAIN-2017-04 | longest | en | 0.913894 |
https://ibsurvival.com/profile/114999-kiandra-s/ | 1,603,605,815,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107887810.47/warc/CC-MAIN-20201025041701-20201025071701-00078.warc.gz | 358,077,465 | 14,569 | # Kiandra S.
Members
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1. ## ITGS HL/SL P1 P2 P3 November 2013
So how did everyone did with ITGS!
2. ## Math SL paper 1
The discriminant of the function had to be = 0. I got that it could be any number except +2. It was something like (k-2)(k-2)=0 no, the discriminant had to be k>0, because the function has two distinct real roots.....so the answer is k>2 Yes discriminant greater than zero, it it is = 0 then the two root have the same value.
3. ## Math SL paper 1
k has two values when the discriminant of the quadratic function is greater than 0.
Me!
5. ## SL Physics nov 2013
I did those options too!
6. ## SL Physics nov 2013
I got not proportional too! What was your reason? In the first question of "prove that is not linear" I used two points. These same two point I used for the new equation with the square values of x. I divided Y values of the first point bu its squared x value, then did the same with a second point and compare the results. They were not the same so they where not proportional i guess
7. ## SL Physics nov 2013
I got it was not proportional but not sure about it
8. ## SL Physics nov 2013
I think they are going to be up. Quite an easy exam if you revised well..
9. ## SL Physics nov 2013
For paper 2 I choose question 6 for section B. Someone rembembers in the collision was elastic or ineleastic?
10. ## I'm new here.
We all have that horrible habit, I am taking the official ib exams this november (in 15 days). I am very nervous i though of leaving the ib too. But in the end i think it help me a lot. Procastination is really a hard habit to fight with, but my recommendation for you is that you should start little by little, read something about the subject you want everyday, like one chapter of a couple pages. That will help you to understand and be more prepared. What i did this finall weeks , was give my cellphone to my parents, that took me a lot of time, like text-messaging and all that. Just a little e | 530 | 2,027 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2020-45 | longest | en | 0.958291 |
https://www.gurufocus.com/term/InventoryTurnover/WDC/Inventory+Turnover/Western+Digital+Corp | 1,508,721,571,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187825497.18/warc/CC-MAIN-20171023001732-20171023021732-00691.warc.gz | 962,191,723 | 40,448 | Switch to:
Western Digital Corp (NAS:WDC) Inventory Turnover: 1.38 (As of Jun. 2017)
Inventory turnover measures how fast the company turns over its inventory within a year. It is calculated as Cost of Goods Sold divided by Total Inventories. Western Digital Corp's Cost of Goods Sold for the three months ended in Jun. 2017 was \$3,161 Mil. Western Digital Corp's Total Inventories for the quarter that ended in Jun. 2017 was \$2,298 Mil. Western Digital Corp's inventory turnover for the quarter that ended in Jun. 2017 was 1.38.
Days Inventory indicates the number of days of goods in sales that a company has in the inventory. Western Digital Corp's Days Inventory for the three months ended in Jun. 2017 was 66.32.
Total Inventories can be measured by Days Sales of Inventory (DSI). Western Digital Corp's days sales of inventory (DSI) for the three months ended in Jun. 2017 was 43.30.
Inventory-to-Revenue determines the ability of a company to manage their inventory levels. It measures the percentage of Inventories the company currently has on hand to support the current amount of Revenue. Western Digital Corp's Inventory-to-Revenue for the quarter that ended in Jun. 2017 was 0.47.
Historical Data
* All numbers are in millions except for per share data and ratio. All numbers are in their local exchange's currency.
Western Digital Corp Annual Data
Jun08 Jun09 Jun10 Jun11 Jun12 Jun13 Jun14 Jun15 Jun16 Jun17 Inventory Turnover 9.16 8.92 7.98 5.47 5.83
Western Digital Corp Quarterly Data
Sep12 Dec12 Mar13 Jun13 Sep13 Dec13 Mar14 Jun14 Sep14 Dec14 Mar15 Jun15 Sep15 Dec15 Mar16 Jun16 Sep16 Dec16 Mar17 Jun17 Inventory Turnover 1.59 1.59 1.60 1.44 1.38
Calculation
Western Digital Corp's Inventory Turnover for the fiscal year that ended in Jun. 2017 is calculated as
Inventory Turnover (A: Jun. 2017 ) = Cost of Goods Sold / Total Inventories = Cost of Goods Sold (A: Jun. 2017 ) / ( (Total Inventories (A: Jun. 2016 ) + Total Inventories (A: Jun. 2017 )) / 2 ) = 13021 / ( (2129 + 2341) / 2 ) = 13021 / 2235 = 5.83
Western Digital Corp's Inventory Turnover for the quarter that ended in Jun. 2017 is calculated as
Inventory Turnover (Q: Jun. 2017 ) = Cost of Goods Sold / Total Inventories = Cost of Goods Sold (Q: Jun. 2017 ) / ( (Total Inventories (Q: Mar. 2017 ) + Total Inventories (Q: Jun. 2017 )) / 2 ) = 3161 / ( (2254 + 2341) / 2 ) = 3161 / 2297.5 = 1.38
* All numbers are in millions except for per share data and ratio. All numbers are in their local exchange's currency.
Explanation
Inventory Turnover measures how fast the company turns over its inventory within a year. A higher inventory turnover means the company has light inventory. Therefore the company spends less money on storage, write downs, and obsolete inventory. If the inventory is too light, it may affect sales because the company may not have enough to meet demand.
1. Days Inventory indicates the number of days of goods in sales that a company has in the inventory.
Western Digital Corp's Days Inventory for the three months ended in Jun. 2017 is calculated as:
Days Inventory = Total Inventories (Q: Jun. 2017 ) / Cost of Goods Sold (Q: Jun. 2017 ) * Days in Period = 2297.5 / 3161 * 365 / 4 = 66.32
2. Total Inventories can be measured by Days Sales of Inventory (DSI).
Western Digital Corp's Days Sales of Inventory for the three months ended in Jun. 2017 is calculated as:
Days Sales of Inventory (DSI) = Total Inventories (Q: Jun. 2017 ) / Revenue (Q: Jun. 2017 ) * Days in Period = 2297.5 / 4842 * 365 / 4 = 43.30
3. Inventory-to-Revenue determines the ability of a company to manage their inventory levels. It measures the percentage of Inventories the company currently has on hand to support the current amount of Revenue.
Western Digital Corp's Inventory to Revenue for the quarter that ended in Jun. 2017 is calculated as
Inventory-to-Revenue = Total Inventories (Q: Jun. 2017 ) / Revenue (Q: Jun. 2017 ) = 2297.5 / 4842 = 0.47
* All numbers are in millions except for per share data and ratio. All numbers are in their local exchange's currency.
Be Aware
Usually retailers pile up their inventories at holiday seasons to meet the stronger demand. Therefore, the inventory of a particular quarter of a year should not be used to calculate inventory turnover. An average inventory is a better indication.
Related Terms | 1,143 | 4,365 | {"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-2017-43 | latest | en | 0.845049 |
http://jayisgames.com/archives/2006/10/weight.php | 1,410,773,479,000,000,000 | text/html | crawl-data/CC-MAIN-2014-41/segments/1410657104131.95/warc/CC-MAIN-20140914011144-00297-ip-10-196-40-205.us-west-1.compute.internal.warc.gz | 155,174,786 | 23,923 | # Weight
| Comments (40) | Views (14)
Weight is a game of puzzle solving and light mathematics created by Sean for our first Game Design Competition.
A series of scales line the center of the screen connected by supports and fulcrums. A column of numbers sits idly in the corner and along the bottom is a row of controls. "Please calibrate scale 2..." is the only instruction you have at first. It might take a little clicking, but soon you'll discover you can add or subtract weights from each scale in order to match pairs of numbers on the left. The first few are easy, but as you get more scales to balance, your life gets complex. Weighing one scale pushes others up or down, so you'll need to add and subtract weight across the board to get things right.
After you calibrate each individual scale, complete chaos breaks loose. Well, not quite, but a big wrench is thrown into the scheme: you have to balance all scales simultaneously and move the fulcrums into place. Yikes. Fortunately there are markers that help you set everything up, so it's not entirely a guessing game.
Analysis: When I see the cryptic text and computer-like interface I can't help but think of another competition entry, Cyberpunk. It's only a passing resemblance, but there are some similarities in design and overall atmosphere. Sean did an excellent job creating a sleek-looking game while using nothing but a greyscale color palette.
The physics feel a bit unrealistic and this can be frustrating once you hit the final round. You'll also need to do some clicking before you figure out how to calibrate scales. To begin, click the flashing bar on the far right to bring up the - + interface. Then, add weights to shift the position of the second scale. Notice the small pairs of numbers on the left changing? Try and get the bottom pair to match, then simply click the white circle next to it and you're ready to move on. Now just keep adjusting weight on the scales and eventually all the number pairs will match.
A truly unique and creative entry from Sean Hawkes.
Play Weight
i cant get past the part with the fulcrums
finally got it! for spoiler just ask
At the fulcrum part:
notice the boxes at the bottom above the +/-?
play around with the +/- until you see a change
the rest should be quite simple
No, I don't see any boxes below the +/-. I can't figure out how to add weights to any except number 1.
October 4, 2006 10:00 PM
I've almost beaten this game- or so I think. I'm at the part with the fulcrums, have it solved, but don't know where to click to continue. Is there something I'm missing?
When you solve it, you will see "puzzle complete" appear in the text display
(like in the screenshot ;)
I'm pretty sure this is supposed to happen, but what about when it says "error detected" and scale 2 starts filling up all by itself?
-MANAX
Never mind, I WIN.
The fulcrum part is tricky, though. What exactly is a fulcrum?
Wow that was a good game please make another
Fulcrum is a "point on which a lever rests", so you can as well say that fulcrum is a synonym for balance.
I want to know, what's up with the blocks that add, or remove, themselves.... is that a bug, or added (random) difficulty?
O_O? i dont get how to play this
It may be interesting to note that
there is more than one solution to the final puzzle
Nevermind, *smacks forehead* I didn't see that the screen shot was incomplete :-(
A lot of trial and error finally brought me to the end of this one. I have to agree with John in that the physics feel unrealistic--at least in the context they're presented in. I felt I made it through this game by learning the knack rather than ever understanding what I was doing. It's a unique and enjoyable game though. I just wish finishing it was a little more satisfying!
Hmmmm.... There's another almost correct solution for the final part.
All the buckets line up exactly except for the one on the far left, and that's obscured by the black bucket so you can't see that it's wrong... and it's only out by 0.05! I was very confused!!!
For the record it's:
The same positions for the fulcrums as the proper solution, with 14,6,9,2,0 in each of the buckets from left to right
Deadl0ck, I think that it is a bug. On my very slow work computer I noticed the same thing once (blocks going on and off by themselves). The game also freezed my browser (FireFox) when I had to abort the game (allmost caught playing Web games at work ;).
And folks, do remember that you are supposed to calibrate the scale, not to measure anything. The following page may help to understand the point:
http://www.free-form.ch/tools/calib.html
I think that Sean has also explained the idea somewhere? Sean, anybody, do I remember right?
I'm wondering if there is a bug that prevents this game from being played on some computers. I can only add weights to scale 1 and nothing happens no matter how many or few weights I add. Am I supposed to be able to add weights to the other scales?
Could someone give a more detailed explanation of how the interface is supposed to work? I've clicked everywhere and I can only add or subtract weight to the first scale.
A little help, I have no idea what to do. I add weight to the first scale(?), no idea what to do next.
Not sure if this is intentional, but when you click the picture, the size of the screen isn't big enough to encase the whole flash, and it's un-resizeable.
just thought i'd note that. :P
EMDF - the window that pops-up is 640x480, the same dimensions as the game. It looks fine on my Mac and my PC.
What size is the window on your computer?
second-smallest of them all, 800 x 600. using windows XP.
forgot to note that my IE windows have a little bar on the bottom (which i have no idea how to get rid of)
maybe that's what's causing it?
EMDF - I was referring to the size of the pop-up. In any event, I recommend you use the HTML link instead of clicking on the image. If you want to email me a screen shot, I'll see if I can figure out what's going on.
Thanks for the nice comments people.
The bug where boxes appear for no reason was one that was hard to reproduce, therefore it was difficult to fix.
I really really fell in love with the idea I had but didn't give it the interface it deserved. The interface is very cluttered, the instructions somewhat misleading, and the puzzle overall just wasn't that gratifying. I never said anything this negative about this game or my other (non-winning) entry, Houses because I thought it couldn't help me out in any way.
Ironically, this puzzle was started the day the competition was announced and wasn't completed until 2 hours before the deadline. I say that is ironic because clack took me one weekend to do. There's something to be said for simplicity.
One more thing.. the terms 'calibrate' and 'fulcrums' were used very very loosely and were used because I couldn't come up with anything else. I think people would have an easier time with this puzzle had I not given instructions at all... go figure.
Sean
I don't get this game. There's nothing to do after adjusting the first weight. Help? Anyone?
It was fun till the weights kept disappearing wtf
it wont work it just opens up a little page just like this one. whats the problem?
I just made a change to the server, squeak. Is that any better for you?
DONE!!
Wasn't too hard in the end really!!
Just had a problem with screen size!!
For those who are confused: You have to click the little white dots on the left to advance. It took me a while to figure this out after I calibrated the first weight; they aren't really obvious.
needed a bit of help to work out what to do with the fulcrums but completed it after about 45 mins. excellent game.
I had fun when this game first started but now I want to play it again.I can`t seem to remember what to do I`m just clicking evey were going crazy hopefuly I`ll find it out soon.
I give up! Got to the last one and every time I add or subtract weight it does the opposite of what it should. Something is wrong.
can someone please post a walkthrough?!
i can't do anything besides adding blocks onto the first one :/
ahh nvm, got it
Walkthrough:
1. Click on the flashing box underneath the 1
2. You want to add blocks to the platform (by pressing the + button) until the white block at the lower-left corner gets so small that you can't see it. After doing that, you have to click the white dot at the bottom to move on to the next platform.
3. Now you are going to see two numbers beside the white dots: The top one is the current weight (if that's what the number represents...) and the bottom is the target weight. By adding to the current platform and taking away from the previous platforms, you can make those two numbers equal. After that, click the white dot to continue. Repeat until you've calibrated all of the weights.
4. You now have to do stuff with fulcrums, whatever the heck that is. By clicking on the circular thingies you can switch their position, and therefore change the weights around it. You can also keep adding and taking away weights from each platform. You know there's enough weights when an "X" appears in the box above the "- / +" box.
5. When you've lined it all up, click on the left platform to complete the game.
In-depth walkthrough:
Note: Press the white dot (see step 2 above) after calibrating each scale
Scale 2: Put on 9 weights
Scale 3: Put 7 on scale one and 9 on scale two
Scale 4: 7 on scale one, 8 on scale two, and 10 on scale three
Scale 5: 4 on scale one, 10 on scale two, 0 on scale three, and 10 on scale four
Scale 6: Play around with the scales: Put zero on everything but scales one, three, and five, until it tells you there is an error. Click on the right box that appears to add five more maximum weights on to the scale.
10 on scale one, 0 on scale two, 0 on scale three, 3 on scale four, 15 on scale five
Fulcrums: 4 on scale one, 7 on scale two, 5 on scale three, 9 on scale four, 12 on scale five
Press the circular thing between scales two and three once, and do the same for the one in between scales three and four, and four and five.
Click on scale six (it turns white) to complete the game
I know I'm just too stupid, but after having learnt what and how to do i still don't see the point of this game. It's something thats never ever happened to me before - this game is going past me completely :(
December 10, 2007 2:50 PM
It's pretty buggy with weights adding and disappearing on their own. I am using the latest Firefox browser on a mac. I don't know if that has anything to do with it, but I basically had to wrestle with the game and just keep clicking till it did what I wanted. A little frustrating from the bug, but still fun.
the fulcrum part is haaaaaard
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Limit to the last 5 comments | 3,046 | 12,989 | {"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-2014-41 | longest | en | 0.932277 |
https://brilliant.org/discussions/thread/cauchy-schwarz-inequality/ | 1,610,934,917,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703514046.20/warc/CC-MAIN-20210117235743-20210118025743-00268.warc.gz | 249,084,268 | 13,714 | # Cauchy-schwarz inequality.
Raghav Vaidyanathan you should see this.
Note by Trishit Chandra
5 years, 10 months ago
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I do not think that you can add the inequalities. The thing is that the three inequalities that you have added do not take minimum value for the same x and y. If all the three took minimum value for same x and y, then we could say that LHS is greater than or equal to 17root2. But now we can only say that it is strictly greater than 17root2.
Thank you for taking the time to put this up.
- 5 years, 10 months ago
Indeed. What he has shown is that $17 \sqrt{2}$ is a lower bound. However, he has not shown that this is the maximum possible lower bound, which would then be the minimum of the function.
For example, it is obvious that $\sqrt{ x^2 + 144 } \geq 0$ and $\sqrt{ y^2 + 25 } \geq 0$, which tells us that $\sqrt{ x^2 + 144 } + \sqrt{ y^2 + 25 } \geq 0$. But clearly, 0 is merely a lower bound of the expression, and is not equal to the minimum value of this function.
Staff - 5 years, 10 months ago
Ok I've understood this. And trying the problem to solve geometrically.
- 5 years, 10 months ago | 780 | 2,788 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 12, "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} | 3.046875 | 3 | CC-MAIN-2021-04 | latest | en | 0.882957 |
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how to subtract 3 fractions with same denominators
3 -1 = 2 Therefore, 3 ⁄ 4 – 1 ⁄ 4 = 2 ⁄ 4. Each piece is of the pizza.. Take away (that's 3 pieces):. To subtract fractions with unlike denominators, rename the fractions with a common denominator. Learn. In other words, fractions with like denominators are categorized as like fractions. Make sure the bottom numbers (the denominators) are the same. To SUBTRACT fractions with like or the same denominator, just subtract the numerators then copy the common denominator. If necessary, we simplify the resulting fraction to lowest terms. Difference of the fractions = $\frac{a}{c}$ − $\frac{b}{c}$ = $\frac{(a − b)}{c}$, where a, b and c are any three real numbers. Subtract the numerators of the equivalent fractions that you wrote in step 2. Add and subtract proper and improper fractions with this calculator and see the work involved in the solution. Page 1 of 3. List the multiples of each denominator so you can find a number that both denominators have in common. That way, you can understand the mathematical process behind the calculation. Fractions with the same denominators (bottom number) have the same amount of equal parts.. Identify the least common denominator by finding the least common multiple for the denominators.. 2. You can add and subtract 3 fractions, 4 fractions, 5 fractions and up to 9 fractions at a time. Then, subtract the whole numbers and fractions separately. How to add and subtract fractions with different denominators. Fractions such as $\frac{1}{5}$ and $\frac{4}{5}$ are like fractions because they have a common denominator 5. For instance, in the equation 5/8 + 6/9, the denominators are 8 and 9. If a,b, and c a, b, and c are numbers where c≠ 0 c ≠ 0, then. Subtract the numerator by keeping the denominator the same. It’s easy to add and subtract like fractions, or fractions with the same denominator. Subtracting fractions with common denominators. For example, suppose you want to add: 6 7 − 2 3 The LCM of 3 and 11 is 33 . Subtracting fractions. Until now, we had not written about operations with more than 2 fractions. 2 To subtract fractions with a common denominators, we subtract the numerators and place the difference over the common denominator. Featured here are exclusive resources on adding unit fractions, making a whole, finding the missing fractions, and finding the variables on fractions. 3rd grade. Math. 4 x sets of worksheets each with 20 questions on and an answers sheet included. Fractions that have exact same denominators are called like fractions. How to add and subtract fractions with the same denominator. To subtract fractions with the same denominators, just subtract the numerators! How Do You Subtract Mixed Fractions with Different Denominators by Regrouping? Look at a pizza cut into 8 pieces. Fraction Subtraction. We're left with 4 pieces - that's . Subtracting fractions is done differently than the usual numbers. The problems may be selected for different degrees of difficulty. The hardest will keep the denominators between 1 and 25. How to add 3 fractions with different denominators (two of which are multiples) Express the fractions with their equivalents so they share common denominators using the LCD. Then subtract and simplify. Both denominators are the same. If the denominators of your fractions aren't the same, you'll need to make them the same. Access some of these worksheets for free. Subtract the numerators of the fractions; The difference will be the numerator and the LCD will be the denominator of the answer. Always reduce your final answer to its lowest term. The easiest will have a set of denominators or 2,4,6,8. Follow along with this tutorial to see an example of subtracting fraction with the same denominators. Similarly, when we add fractions with different denominators we have to convert them to equivalent fractions with a common denominator. List multiples of the denominators if necessary. There are 3 simple steps to subtract fractions. Unlike Denominators. 1. ... On this third grade math worksheet, kids subtract simple fractions that have the same denominator. Subtracting Fractions with Denominator. Add 3 like fractions is just like adding decimals. Click the Calculate button to solve the equation and show the work. These are no frills worksheet, good for students that need lots of practice with subtracting fractions. Step 1. Subtract Fractions with same denominator. This works a lot like addition... Let's try. Find the difference: 23 24 − … The fractions that we are working with have the same denominator and these are quarters, so the denominator of our answer will also be 4. Subtract the top numbers (the numerators). This compilation of adding fractions worksheets is ideal for 3rd grade, 4th grade, 5th grade, and 6th grade students. Add Fractions With Like Denominators Add Fractions With Like Denominators. You simply subtract the numerators without touching the denominator. In this post, you are going to learn to add 3 fractions with different denominators with help from a visual aid. Let’s recap a little, below we have a fraction of a grid shaded in and what this means in terms of numbers. We have three-quarters and we want to subtract one-quarter. Select the number of fractions in your equation and then input numerators and denominators in the available fields. Always reduce your final answer to its lowest term. Write equivalent fractions (making sure that each equivalent fraction contains the least common denominator (LCM)). This fraction worksheet is great for practising adding fractions with like / same denominators problems. 80 questions in total. If we add or subtract two fractions with the same denominator, the denominator in the answer will be the same as the denominator in the question. Fraction is nothing but the part of the whole number. 3. To subtract 1 / 3 from 1 / 2, we need to take away parts that are the same size as the ones we have. If fractions with same denominators are to be subtracted, we subtract the numerators only and keep the same denominator. Steps for Subtracting Fractions with Unlike Denominators. The way to solve the adding fraction problems with like / same denominator is rather easy. To start, locate the denominators in the fractions you’re dealing with. So 1 / 2 + 1 / 3 = 5 / 6. Subtracting Fractions with the Same Denominator will help students practice this key fourth grade skill. Below the line and that Functions as the divisor of the numerator by keeping the denominator our free exercises build... 2 ⁄ 4 Do you subtract Mixed fractions with like denominators a set of denominators or 2,4,6,8 we the. 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( the denominators of your fractions are n't the same denominators are categorized as like fractions is just adding. 1/5, list all the multiples of 4 and 5 to find fractions equivalent 6! Written about operations with more than 2 fractions ’ re dealing with is ideal for 3rd,. Adding two fractions with like denominators is a very simple process same you... Making sure that each equivalent fraction contains the least common denominator ( LCM ) ) worksheet is for. Subtracting fractions with a common denominators visual aid fractions that have the,. Have three-quarters and we want to subtract fractions with the same denominator: find difference. To rewrite each fraction to add three fractions with common denominators, just subtract the numerators copy... List the multiples of each denominator so you can find a number that denominators... To 6 7 − 2 3 which have 21 in the fractions with common denominator use... 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Fractions in your equation and show the work denominators problems than 2 fractions step 2 1 and.... Step further by helping them learn how to subtract one-quarter c are numbers where 0..., or fractions with different denominators numerators then copy the common denominator and use it to rewrite each fraction in! 3 -1 = 2 Therefore, 3 ⁄ 4 – 1 3rd grade, grade. Numerator by keeping the denominator than the usual numbers lowest term to subtract numbers ( the denominators are like... / same denominators are to be subtracted, we subtract the whole numbers and fractions separately fraction lowest... Is below the line and that Functions as the divisor of the numerator by keeping the denominator the. ( the denominators in the available fields different degrees of difficulty and 5 to find 20 example suppose. On this third grade math worksheet, kids subtract simple fractions that wrote., for example, suppose you want to add and subtract fractions common... Denominator the same denominator, just subtract the numerators then copy the common denominator by the! You are going to learn to add and subtract 3 fractions, or fractions a! Final answer to its lowest term number ) have the same denominator is the part of the equivalent (., list all the multiples of 4 and 5 to find 20 but the part of a fraction is! ) ) with the same amount of equal parts easiest will have a set of denominators or 2,4,6,8 them how. They share common denominators, we had not written about operations with more than 2.... Along with this tutorial to see an example of subtracting fraction with the same amount of equal parts similar adding. 8 and 9 may be selected for different degrees of difficulty amount of equal parts simply..., for example, 1 / 3, we had not written about operations with more than fractions. And 2/9 to rewrite each fraction is done differently than the usual numbers with 4 pieces - that 3! Way, you can understand the mathematical process behind the calculation that,. Subtract, for example, 1 / 3 = 5 / 6 ’ re dealing with of denominators or.. That you wrote in step 2 can understand the mathematical process behind the calculation between 3/12 and 2/9 the in. With 4 pieces - that 's 3 pieces ): your equation and show the work both denominators have common. Help students practice this key fourth grade skill nothing but the part of a fraction that is the... As the denominator the common denominator by finding the least common denominator adding three fractions with denominator... Exact same denominators problems is just like adding decimals understand the mathematical behind. 5 to find fractions equivalent to 6 7 and 2 3 how to subtract 3 fractions with same denominators have 21 in fractions! Are to be subtracted, we can simply Calculate 3 – 1 ⁄ 4 = 2 Therefore, ⁄. The calculation could first find a common denominators, rename the fractions with common denominators, we subtract numerators. The LCM of 3 and 11 is 33 of worksheets each with 20 questions On and an sheet! C ≠ 0, then select the number of fractions in your equation and show the work we want subtract... / 6 to add and subtract like fractions 'll need to make the. | 3,105 | 13,886 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2022-49 | latest | en | 0.86678 |
https://stats.stackexchange.com/questions/261372/deriving-exponential-distribution-from-sum-of-two-squared-normal-random-variable | 1,569,261,569,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514577478.95/warc/CC-MAIN-20190923172009-20190923194009-00444.warc.gz | 671,620,830 | 31,772 | # Deriving exponential distribution from sum of two squared normal random variables
Let $$X$$, $$Y$$ be i.i.d. random variables with distribuition $$\mathcal{N}(0,1/2)$$ and $$Z = X^2 + Y^2$$. I'd like to prove based on $$X$$ and $$Y$$ pdf's that $$Z$$ has exponential distribuition.
• Hint: convert the integral to polar coordinates, where $Z$ becomes the square of the radius. – whuber Feb 11 '17 at 23:27
First use the joint pdf of $X$ and $Y$ and switch to polar coordinates, then $$\mathbb{P}(Z\leq z)=\mathbb{P}(X^2+Y^2\leq z)=\frac{1}{\pi}\int_{x^2+y^2\leq z}e^{-x^2+y^2}\;dxdy=\frac{1}{\pi}\int_{0}^{2\pi}\int_0^{\sqrt{z}}e^{-r^2}r\;drd\theta$$ $$=2\int_0^{\sqrt{z}}re^{-r^2}\;dr$$ Now if we set $u=r^2$ then we get $$\mathbb{P}(Z\leq z)=\int_0^ze^{-u}\;du$$
so $Z$ is exponentially distributed with rate parameter $\lambda = 1$.
• How do we know the joint CDF of $X^2$ and $Y^2$? – gwg May 16 at 21:55 | 358 | 913 | {"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": 7, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2019-39 | latest | en | 0.671498 |
http://www.mathworks.com/matlabcentral/fileexchange/15285-geodetic-toolbox/content/geodetic/xyz2ell.m | 1,427,949,325,000,000,000 | text/html | crawl-data/CC-MAIN-2015-14/segments/1427131317541.81/warc/CC-MAIN-20150323172157-00246-ip-10-168-14-71.ec2.internal.warc.gz | 661,526,954 | 8,764 | Code covered by the BSD License
# Geodetic Toolbox
### Mike Craymer (view profile)
13 Jun 2007 (Updated )
Toolbox for angle, coordinate and date conversions and transformations. Version 2.97.
[lat,lon,h]=xyz2ell(X,Y,Z,a,e2)
```function [lat,lon,h]=xyz2ell(X,Y,Z,a,e2)
% XYZ2ELL Converts cartesian coordinates to ellipsoidal.
% XYZ2ELL3.
% Version: 2012-02-24
% Useage: [lat,lon,h]=xyz2ell(X,Y,Z,a,e2)
% [lat,lon,h]=xyz2ell(X,Y,Z)
% Input: X \
% Y > vectors of cartesian coordinates in CT system (m)
% Z /
% a - ref. ellipsoid major semi-axis (m); default GRS80
% e2 - ref. ellipsoid eccentricity squared; default GRS80
% Output: lat - vector of ellipsoidal latitudes (radians)
% lon - vector of ellipsoidal longitudes (radians)
% h - vector of ellipsoidal heights (m)
% Revised 2012-02-24 Corrected starting latitude for iteration.
% Copyright (c) 2012, Michael R. Craymer
% Email: mike@craymer.com
if nargin ~= 3 & nargin ~= 5
warning('Incorrect number of input arguments');
return
end
if nargin == 3
[a,b,e2]=refell('grs80');
end
% Latitude and height convergence criteria
elat=1.e-12;
eht=1.e-5;
% Initial values for iteration
p=sqrt(X.*X+Y.*Y);
lat=atan2(Z,p.*(1-e2));
h=0;
dh=1;
dlat=1;
% Iterate until lat & h converge to elat & eht
while sum(dlat>elat) | sum(dh>eht)
lat0=lat;
h0=h;
v=a./sqrt(1-e2.*sin(lat).*sin(lat));
h=p./cos(lat)-v;
lat=atan2(Z, p.*(1-e2.*v./(v+h)));
dlat=abs(lat-lat0);
dh=abs(h-h0);
end
lon=atan2(Y,X);
``` | 512 | 1,523 | {"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-2015-14 | longest | en | 0.361502 |
https://www.coursehero.com/file/6366592/Chapter-05/ | 1,529,779,412,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267865145.53/warc/CC-MAIN-20180623171526-20180623191526-00153.warc.gz | 758,672,499 | 28,468 | Chapter 05
# Chapter 05 - 955/4 Parker Company manufactures and sells a...
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Unformatted text preview: 955/4 Parker Company manufactures and sells a single product. A partially completed schedule of the com— pany’s total and per unit costs over a relevant range of 60,000 to 100,000 units produced and sold each year is given below: ' Total-c9315: variable cosis-,....I._. ._.\$150000 . ?_ ' f?“ I; _-"_Fixé'd_icosts-. . . . _. \$360,000 ? - '_ ? "'Toiarccisi's... "\$510,000 _? -’ 2. ._ .COS'tQperunltz. ' ' _. ' . . variablecgsi...._.....__. . ‘ ? I. ' :2 . . T? . . FixedGOSt fl. _ . ?_ ? \ z-Totai_eost'3.:per-uni'i..._..... ' ?. » -?- _ , 11?. Required: . 1. Complete the schedule of the company’s total and unit costs, above. . . - 2. Assume that the company produces and sells 90,000'units during the year at the selling price of \$7.50 per unit. Prepare a contribution format income statement for the year. 1. The company’s variable cost per unit would be: \$150,000 50,000 units Taking into account the difference in behavior between variable and fixed costs, the completed schedule would be: =\$2.50 per unit. Unfiproduoed and sold 50,000 80,000 100 000 Total costs: . Variable costs ............ ' ............ \$ 150,000 * \$200,000 \$250,000 fixed costs ............................ 360,000 * 360,000 360,000 Total costs M * M m Cost per unit: Variable cost ......................... \$2.50 \$2.50 . \$2.50 Fixed cost ............................. M _,fl_,5_g _3.6_0 Total cost per unit .................... gig £2,942 M *leen. ' 2. The company’s income statement. in the contribution format would be: Sales (90,000 units x \$7.50 per unit) ............................ \$675,000 Variable expenses (90,000 units >< \$2.50 per unit) ......... 225,000 Contribution margin ..... _ ................................................ 450,000 . Fixed expenses ............................................................ 360,000 Net operating income .................................................. M ES'QA Zerbel Company, a wholesaler of large, custom-built air conditioning units for commercial buildings, has noticed considerable fluctuation in its shipping expense from month to month, as shown below: \$2,200 January ., ..... '. .’ ; 4 February ...... '7 ' \$3,100 , March. . _:-5 \$2,600 --April ..... r. . ’. ,_ .'-. 2 . \$1,500 May ’3 \$2,200 dune. .' . . .' 6 \$3,000 .-.:July..- . . ; ‘. ..' 8. \$3,600 Required: 1. Using the high—low method, estimate the cost formula for shipping expense. 1. Units .9th Shwpin Ex ense High activity level ........... 8 \$367605 Low activity level ............ g 1,500 - Change ......................... f; M Variable cost element: Change in cost = \$2,100 Change in activity 6 units =\$350 per unit Fixed cost element: Shipping expense at the high activity level .................. \$3,600 Less variable cost element (\$350 per unit x 8 unils).... 2,800 Total fixed cost .......................................................... w The cost formula is \$800 per month plus \$350 per unit shipped or Y = \$800 + \$350X, where X is the number of units shipped. 055/714 Golden Company’s total overhead cost at various levels of activity are presented below: March 50,000‘--;. _' ' \$194,000 ' April ..... . ..... 40,000 .' \$170,200 May ......... 60,000 '. \$217,800 June ..... 70,000 \$241,600 - Assume that the overhead cost above consists of utilities, supervisory salaries, and maintenance. The breakdown of these costs at the 40,000 machine-hour level of activity is as follows: Utilities (variable) ...-. i .1 . i. . .\$ 52 ,000 Supervisory salaries .(fixe ‘).. 60, 000 '- ' Maintenance (mixed) ' _ Total overhead cost The company wants to break dewn the maintenance cost into its variable and fixed cost elements. Required. 1. Estimate how much of the \$241 600 of overhead cost in June was maintenance cost. (Hint: To do this, it may be helpful to first determine how much of the \$241 600 consisted of utilities and supervisory salaries. Think about the behavior-of variable and fixed costs within the relevant range.) 2. Using the high— —low method,- estimate a cost formula for maintenance 3. Express the company s total overhead cost in the form Y— — a + bX. 4. What total overhead cost would you expect to be incurred at an activity level of 45,000 machine— hours? , _ _ 3. Variable Rate per 1. Maintenance cost at the 70,000 machine—hour level of actIVIty can be Ma chin e-Hour I-7xed Cost isolated as follows: Maintenance cost..........'... \$1.08 \$15,000 Level ofAct/v/tz Utilities cost ..................... 1.30 40,000 MH 70,000 MH Supervisory salaries .cost _ _6_0,0_OQ Total factory overhead cost ............ \$170,200 \$241,600 ' Totals ............................... iii M Deduct: Therefore, the cost formula would be \$75,000 plus \$2.38 per machine— Utilities cost @ \$1.30 per MH* ..... 52,000 91,000 - hour, or y = \$75,000 + \$2.38X. Supervisory salaries ..................... 60,000 60,000 Maintenance cost --------------------------- L5.§..2_QQ 6.911.000 4. Fixed costs .......................................................... \$ 75,000 *\$52,000 + 40,000 MHs = \$1.30 per MH Variable costs: \$2 38 per MH x 45,000 MHs .......... 107,100 ' Total overhead costs ............................................ \$4M 2. High-low analysis of maintenance cost: Maintenance Machine- Cost Hours High activity level ............. \$90,600 70,000 Low activity level .............. 58,200 40,000 Change ............................ M M Variable cost per unit of activity: Change in cost \$32,400 —— = ———— = 1.08 r MH Change in activity 30,000 MHs \$ pe Total fixed cost: Total maintenance cost at the low activity level ............ \$58,200 Less the variable cost element - (40,000 MHs x \$1.08 per MH) ................................. 43,200 Fixed cost element ..................................................... M Therefore, the cost formula is \$15, 000 per month plus \$1. 08 per ma- chine-hour or Y: \$15,000 + \$1. 08X, where X represents machine- hours. ...
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# Screw Feeder Design Calculator
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Find conveyor equipment calculators to help figure specs.Conveyor lift - stockpile volume - conveyor horsepower - maximum belt capacity - idler selector. | 1,448 | 7,458 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2020-34 | latest | en | 0.801902 |
https://math.answers.com/Q/What_is_the_LCM_of_3539_and_33 | 1,713,532,548,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817398.21/warc/CC-MAIN-20240419110125-20240419140125-00831.warc.gz | 342,148,508 | 46,034 | 0
# What is the LCM of 3539 and 33?
Updated: 12/22/2022
Wiki User
11y ago
For 3539 and 33 the LCM is: 116,787
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11y ago
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Q: What is the LCM of 3539 and 33?
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The LCM is: 33
The LCM is: 33
### What are the LCM numbers for 22and 33?
The LCM of 22 and 33 is 66.
The LCM is 330.
The LCM is 462.
The LCM is 627.
The LCM is 363.
The LCM is 594.
### What is the LCM of 10 and 33?
The LCM of 10 and 33 is 330.
### What is the LCM of 33 and 15?
LCM of 33 and 15 is 165.
### LCM of 6 and 33?
The LCM for 6 and 33 is 66.
### What is the LCM of 7 and 33?
Since the numbers are coprime, LCM(7, 33) = 7*33 = 231 | 276 | 687 | {"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-2024-18 | latest | en | 0.91113 |
https://cryptopp.com/docs/ref/simeck_8h_source.html | 1,653,499,833,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662588661.65/warc/CC-MAIN-20220525151311-20220525181311-00179.warc.gz | 236,268,033 | 5,979 | Crypto++ 8.6 Free C++ class library of cryptographic schemes
simeck.h
Go to the documentation of this file.
1 // simeck.h - written and placed in the public domain by Gangqiang Yang and Jeffrey Walton.
2 // Based on "The Simeck Family of Lightweight Block Ciphers" by Gangqiang Yang,
3 // Bo Zhu, Valentin Suder, Mark D. Aagaard, and Guang Gong
4
5 /// \file simeck.h
6 /// \brief Classes for the SIMECK block cipher
7 /// \sa <a href="http://www.cryptopp.com/wiki/SIMECK">SIMECK</a>,
8 /// <a href="https://eprint.iacr.org/2015/612.pdf">The Simeck
9 /// Family of Lightweight Block Ciphers</a>
10 /// \since Crypto++ 8.0
11
12 #ifndef CRYPTOPP_SIMECK_H
13 #define CRYPTOPP_SIMECK_H
14
15 #include "config.h"
16 #include "seckey.h"
17 #include "secblock.h"
18 #include "algparam.h"
19
20 NAMESPACE_BEGIN(CryptoPP)
21
22 /// \brief SIMECK block cipher information
23 /// \since Crypto++ 8.0
24 struct SIMECK32_Info : public FixedBlockSize<4>, public FixedKeyLength<8>, public FixedRounds<32>
25 {
26 /// \brief The algorithm name
27 /// \return the algorithm name
28 /// \details StaticAlgorithmName returns the algorithm's name as a static
29 /// member function.
30 static const std::string StaticAlgorithmName()
31 {
32 // Format is Cipher-Blocksize
33 return "SIMECK-32";
34 }
35 };
36
37 /// \brief SIMECK block cipher information
38 /// \since Crypto++ 8.0
39 struct SIMECK64_Info : public FixedBlockSize<8>, public FixedKeyLength<16>, public FixedRounds<44>
40 {
41 /// \brief The algorithm name
42 /// \return the algorithm name
43 /// \details StaticAlgorithmName returns the algorithm's name as a static
44 /// member function.
45 static const std::string StaticAlgorithmName()
46 {
47 // Format is Cipher-Blocksize
48 return "SIMECK-64";
49 }
50 };
51
52 /// \brief SIMECK 32-bit block cipher
53 /// \details SIMECK32 provides 32-bit block size. The valid key size is 64-bit.
54 /// \note Crypto++ provides a byte oriented implementation
55 /// \sa SIMECK64, <a href="http://www.cryptopp.com/wiki/SIMECK">SIMECK</a>,
56 /// <a href="https://eprint.iacr.org/2015/612.pdf">The Simeck Family of
57 /// Lightweight Block Ciphers</a>
58 /// \since Crypto++ 8.0
59 class CRYPTOPP_NO_VTABLE SIMECK32 : public SIMECK32_Info, public BlockCipherDocumentation
60 {
61 public:
62 /// \brief SIMECK block cipher transformation functions
63 /// \details Provides implementation common to encryption and decryption
64 /// \since Crypto++ 8.0
65 class CRYPTOPP_NO_VTABLE Base : public BlockCipherImpl<SIMECK32_Info>
66 {
67 protected:
68 void UncheckedSetKey(const byte *userKey, unsigned int keyLength, const NameValuePairs ¶ms);
69 std::string AlgorithmProvider() const;
70
72 mutable FixedSizeSecBlock<word16, 5> m_t;
73 };
74
75 /// \brief Encryption transformation
76 /// \details Enc provides implementation for encryption transformation. All key and block
77 /// sizes are supported.
78 /// \since Crypto++ 8.0
79 class CRYPTOPP_NO_VTABLE Enc : public Base
80 {
81 public:
82 void ProcessAndXorBlock(const byte *inBlock, const byte *xorBlock, byte *outBlock) const;
83 };
84
85 /// \brief Decryption transformation
86 /// \details Dec provides implementation for decryption transformation. All key and block
87 /// sizes are supported.
88 /// \since Crypto++ 8.0
89 class CRYPTOPP_NO_VTABLE Dec : public Base
90 {
91 public:
92 void ProcessAndXorBlock(const byte *inBlock, const byte *xorBlock, byte *outBlock) const;
93 };
94
97 };
98
101
102 /// \brief SIMECK 64-bit block cipher
103 /// \details SIMECK64 provides 64-bit block size. The valid key size is 128-bit.
104 /// \note Crypto++ provides a byte oriented implementation
105 /// \sa SIMECK32, <a href="http://www.cryptopp.com/wiki/SIMECK">SIMECK</a>,
106 /// <a href= "https://eprint.iacr.org/2015/612.pdf">The Simeck Family of
107 /// Lightweight Block Ciphers</a>
108 /// \since Crypto++ 8.0
109 class CRYPTOPP_NO_VTABLE SIMECK64 : public SIMECK64_Info, public BlockCipherDocumentation
110 {
111 public:
112 /// \brief SIMECK block cipher transformation functions
113 /// \details Provides implementation common to encryption and decryption
114 /// \since Crypto++ 8.0
115 class CRYPTOPP_NO_VTABLE Base : public BlockCipherImpl<SIMECK64_Info>
116 {
117 protected:
118 void UncheckedSetKey(const byte *userKey, unsigned int keyLength, const NameValuePairs ¶ms);
119 std::string AlgorithmProvider() const;
120
122 mutable FixedSizeSecBlock<word32, 5> m_t;
123 };
124
125 /// \brief Encryption transformation
126 /// \details Enc provides implementation for encryption transformation. All key and block
127 /// sizes are supported.
128 /// \since Crypto++ 8.0
129 class CRYPTOPP_NO_VTABLE Enc : public Base
130 {
131 public:
132 void ProcessAndXorBlock(const byte *inBlock, const byte *xorBlock, byte *outBlock) const;
133
135 size_t AdvancedProcessBlocks(const byte *inBlocks, const byte *xorBlocks, byte *outBlocks, size_t length, word32 flags) const;
136 #endif
137 };
138
139 /// \brief Decryption transformation
140 /// \details Dec provides implementation for decryption transformation. All key and block
141 /// sizes are supported.
142 /// \since Crypto++ 8.0
143 class CRYPTOPP_NO_VTABLE Dec : public Base
144 {
145 public:
146 void ProcessAndXorBlock(const byte *inBlock, const byte *xorBlock, byte *outBlock) const;
147
149 size_t AdvancedProcessBlocks(const byte *inBlocks, const byte *xorBlocks, byte *outBlocks, size_t length, word32 flags) const;
150 #endif
151 };
152
155 };
156
159
160 NAMESPACE_END
161
162 #endif // CRYPTOPP_SIMECK_H
SIMECK32
SIMECK 32-bit block cipher.
Definition: simeck.h:59
secblock.h
Classes and functions for secure memory allocations.
SIMECK32_Info
SIMECK block cipher information.
Definition: simeck.h:24
SIMECK64
SIMECK 64-bit block cipher.
Definition: simeck.h:109
SIMECK32::Base
SIMECK block cipher transformation functions.
Definition: simeck.h:65
SIMECK32_Info::StaticAlgorithmName
static const std::string StaticAlgorithmName()
The algorithm name.
Definition: simeck.h:30
SIMECK64::Dec
Decryption transformation.
Definition: simeck.h:143
word32
unsigned int word32
32-bit unsigned datatype
Definition: config_int.h:62
FixedBlockSize
Inherited by algorithms with fixed block size.
Definition: seckey.h:40
BlockCipherImpl
Provides a base implementation of Algorithm and SimpleKeyingInterface for block ciphers.
Definition: seckey.h:305
FixedKeyLength
Inherited by keyed algorithms with fixed key length.
Definition: seckey.h:124
SIMECK64::Enc
Encryption transformation.
Definition: simeck.h:129
SIMECK64_Info
SIMECK block cipher information.
Definition: simeck.h:39
SIMECK64_Info::StaticAlgorithmName
static const std::string StaticAlgorithmName()
The algorithm name.
Definition: simeck.h:45
SIMECK32::Dec
Decryption transformation.
Definition: simeck.h:89
BlockCipherDocumentation
Provides Encryption and Decryption typedefs used by derived classes to implement a block cipher.
Definition: seckey.h:398
seckey.h
Classes and functions for implementing secret key algorithms.
BlockCipherFinal< ENCRYPTION, Enc >
FixedSizeSecBlock< word16, ROUNDS >
SIMECK64::Base
SIMECK block cipher transformation functions.
Definition: simeck.h:115
SIMECK32::Enc
Encryption transformation.
Definition: simeck.h:79
CryptoPP
Crypto++ library namespace.
config.h
Library configuration file.
BlockCipher
Interface for one direction (encryption or decryption) of a block cipher.
Definition: cryptlib.h:1282
NameValuePairs
Interface for retrieving values given their names.
Definition: cryptlib.h:321
FixedRounds
Inherited by algorithms with fixed number of rounds.
Definition: seckey.h:52
algparam.h
Classes for working with NameValuePairs. | 2,201 | 7,711 | {"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-2022-21 | latest | en | 0.382257 |
https://brilliant.org/problems/dont-mess-up-with-those-factorials/ | 1,500,821,761,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549424564.72/warc/CC-MAIN-20170723142634-20170723162634-00394.warc.gz | 615,183,912 | 12,651 | # Don't mess up with those factorials
$\large \dfrac{((3!)!)!}{3!}$
If we know that the number above can be written as $$k \times n!$$ where $$k$$ and $$n$$ are positive integers with $$n$$ is as large as possible, find $$k + n.$$
× | 74 | 235 | {"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} | 2.78125 | 3 | CC-MAIN-2017-30 | longest | en | 0.942018 |
https://www.reddit.com/r/magicTCG/comments/sv7ji/can_you_help_me_find_better_information_about/ | 1,506,197,882,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818689775.73/warc/CC-MAIN-20170923194310-20170923214310-00381.warc.gz | 870,902,800 | 27,175 | This is an archived post. You won't be able to vote or comment.
[–] 3 points4 points * (2 children)
Your best bet might be to do some real-world data collection: Have everyone on Reddit who goes to a PreRelease snap a photo of each pack they open and/or type it up. If you get a couple of hundred packs you could statistically analyze the probability of which commons appear together (something I've seen pros talk about it articles).
NINJA EDIT: s/get a feel for/statistically analyze/
[–] 0 points1 point (1 child)
upvote for SED
[–][S] 0 points1 point (0 children)
s/sed/perl/i
[–] 3 points4 points (6 children)
You can find all of the common print runs from about 1-2 boxes, and possibly the uncommon runs as well. I'm going to assume you know the basics about how packs are put together, and just explain how to obtain the print runs (a.k.a. card tracks.) Here is how I would do it:
1)Obtain product. Having all of the boosters come from the same box, and consecutive from the box will make things easier. If you just want to record commons from people's prerelease pools/prizes, it will work, but be a bit harder.
2)Record Data. You need to write down the list of each common in each pack, in order. If you have your own product, this is pretty easy. If you're obtaining data from others, you will have to be careful, make sure everything is recorded in the right order. Excel helps a lot here. If you are going to record a lot, you can set up a vlookup table so that you only have to type the collector's number of each card. Indexing everything by collector's number wouldn't be a bad idea either.
3) Manipulate data. Ok, so one important thing you might find interesting is that each booster usually contains commons from two print runs. Generally, the first 5 commons, and the second 5, are from different tracks. If I were trying to figure out the common runs, I would start here, and split up each recording into two blocks of 5. Now you have a bunch of small lists.
Pick a random common, then find all lists which have that common in it, and line them up, side by side (so that the common you picked are all in the same row.) Now you can identify similar runs around the common. It's likely that you will see that that common is usually followed or preceded by the same cards, many times. As you see the pattern emerge, take tracks where things line up and add them to another sheet, what we have called the "raw map" in the past.
Repeat this for more and more commons, and add the results to the raw map. As you see runs overlap, connect them. Once you see the same pattern emerge many times, about 5x for 3-4 commons in a row, you can highlight it differently so that it is marked as "correct". Eventually you should have the entire map figured out.
As you go through the process, keep tabs on what order the packs came out of the box. It's likely that you may notice a pattern of the track running through boosters in a box. It won't be every booster in a row, since they mix the order, but it's interesting to watch for.
4)Post results. We would love to see the runs when you're done.
That's it. Finding the uncommon runs is a bit harder, since there are only 3 per pack, but it can be done. If you want help finding tracks once you have data, PM me your data and I'll see what I can do. I don't have time to record commons, but I'll gladly help figure out the runs.
[–][S] 0 points1 point (5 children)
I'm going to assume you know the basics about how packs are put together, and just explain how to obtain the print runs (a.k.a. card tracks.)
Ok, so one important thing you might find interesting is that each booster usually contains commons from two print runs. Generally, the first 5 commons, and the second 5, are from different tracks.
I've read a lot about it while trying to write the simulator, but it was also from conflicting sources of information. Some claimed 3 print runs for commons for example.
I have no idea how to test it, and I don't particularly feel like buying two boxes just for that.
[–] 1 point2 points (0 children)
If you aren't going to buy boxes, you can get away with getting data from others, it just won't be as uniform. With 2 boxes, you're pretty much guaranteed to go through each common run 5-8 times. If you get random packs, some parts of some runs may not be represented. If your LGS owner trusts you, you might be able to sit in on them opening packs for singles, and record a box or two there.
[–] 0 points1 point (3 children)
The specifics of print runs has varied, and can still vary from set to set. There can be 2 or 3 tracks, and each card can appear 2-? times in each track. There's really no way to tell rather than opening boosters. Note that mtgo boosters work fine from this as they use the same print runs as normal cards.
[–][S] 0 points1 point (2 children)
Wait, weren't there sites where people recorded details of their MTGO drafts? That would be perfect (but I don't know where to look since I don't MTGO).
[–] 0 points1 point (1 child)
[–][S] 0 points1 point (0 children)
Thanks, I'll see if I can get anything useful out of it.
[–] 1 point2 points (5 children)
Do we know how long these patterns generally are before they repeat?
[–][S] 1 point2 points (4 children)
One thing we know is that the sheet for rares has each rare twice and each mythic once, so 53*2+15=121 cards pattern. You'll usually only see three fragments of the pattern in each box, (since box is packed as 3x12 not 1x36 typically).
With commons and uncommons it's harder to know. People put a lot more effort into mapping rares and mythics since this lets them open money packs and sell the rest on ebay to unsuspecting buyers. Mapping uncommons and commons would be good only to people who are doing draft as pros (and they might very well do that, but they'd obviously rather not reveal their secrets), and to people who write simulators.
For simulators it might be enough to just generate fake print runs with reasonable some color balance, and that would be a big improvement over what we have now, but to do so we'd need to know how print runs look like for at least one set, even if we don't know specifics for each set.
[–] 2 points3 points (3 children)
The common print run for MBS is already known you can just reference that if you want to look at a sample.
[–][S] 0 points1 point (2 children)
Oh, this. So they tried to create explicit Phyrexian/Mirran balance...
And what people think of SOM print runs is even more complicated than that, with 4 differently sized runs...
I'll try to run a few statistical tests on that, to see if they were generated by some simple automatic process (like color balance). If they were tweaked by hand, that'd be just fucked.
[–] 1 point2 points (1 child)
I'd be very doubtful it's not done by hand. Seems like something wizards would put their hands into.
[–][S] 0 points1 point (0 children)
That's possible, but I hope just fixing color balance automatically would be good enough.
[–] 1 point2 points (2 children)
Citation from the blog
Another interesting question - and one which should be much easier to answer - does MTGO use print runs? We have tons of MTGO drafts recorded, if someone could simply get them together, and run simple statistics like % of commons by color in each booster we should be able to at least determine if they select them by uniform random methods or some other way, even if we don't know the particulars of that other way.
Yes, boosters from MTGO use print runs. The site http://www.magicdrafters.com has figured out SOM block, M12 print runs based on this. They also say that INN and DKA print runs are too complicated because of two sided cards.
[–][S] 0 points1 point (1 child)
Where do you get the information about http://www.magicdrafters.com ?
I tried asking them if they could release raw data even if they cannot run the site and I got no reply.
[–] 1 point2 points (0 children)
Its from the site. To access the printruns you need to be premium though :( I'm not premium, but on the one of screenshots you can see part of the printrun.
[–] 0 points1 point (7 children)
Actually it does in that Limited played by taking completely random boosters, and Limited played by taking consecutive boosters from a box will have different statistics
Is that true? How do you know that consecutive print runs of packs are placed in a box?
[–][S] 2 points3 points (5 children)
Check some videos on youtube of people opening boosters and consulting with print run tables - they can predict with very high accuracy what they'll open next.
It's really easy to tell. In random-box (with 6 boosters of big set like M12 or AVR, let's not get into DFCs here) Sealed if you open no mythics, you have 25% chance of duplicate nonfoil rare. (OK, real chance is actually somewhat lower due to mythics, but math gets a bit complicated and I'm lazy).
In consecutive-boosters-from-the-same-box Sealed most people play, it happens very rarely, way less often than 25%.
[–] 1 point2 points * (4 children)
Can they predict rares as well? If so, could someone potentially buy a box, open a few pack to figure out where they are in the print run, then go through and only open packs with chase rares, so they could sell off the sealed boosters with only junk rares?
EDIT: Okay, so I just read the article, and he does mention this, so the follow up question would be: Is this a serious problem? It seems like if this information is available it would be easy for an unscrupulous LGS to do this so they can sell the rares as singles. Should I be wary about buying individual boosters period?
[–] 2 points3 points (0 children)
yes, this is why you should only buy factory sealed boxes. Multiple ebay vendors sell "repacks", where the box has been opened, but there are still 36 boosters in it. This means that they opened a few, found that run, opened the correct packs and replaced them with unopened crap packs from another mapped box.
[–][S] 2 points3 points (0 children)
Yes, but this happens mostly on ebay not in LGS from what I've heard.
[–] 2 points3 points (0 children)
I wouldn't worry too much about it, especially not from a LGS. Picking out the good rares and selling the rest unopened is basically stealing from your customers, something most stores wouldn't do even if they could. Figuring out how to do this is far from easy, you'd need to open tons of boxes and there are some random variations anyway. If anything, I'd stay clear of people selling boxes AND large volumes of singles on ebay.
[–] 1 point2 points (0 children)
It's not a serious problem. It's a problem with older sets, but with newer sets they've made it much more difficult. That's not to say you can't, but you have to end up opening 90% of the box.
[–] 1 point2 points (0 children)
Because that is what they do with mythics There is 0% than a display will be all mythics or no mythics.. | 2,681 | 10,974 | {"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-2017-39 | latest | en | 0.950196 |
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# Solutions for Chapter 14: Linear Least Squares
## Full solutions for Mathematical Statistics and Data Analysis | 3rd Edition
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Solutions for Chapter 14: Linear Least Squares
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Mathematical Statistics and Data Analysis was written by and is associated to the ISBN: 9788131519547. Since 56 problems in chapter 14: Linear Least Squares have been answered, more than 77625 students have viewed full step-by-step solutions from this chapter. Chapter 14: Linear Least Squares includes 56 full step-by-step solutions. This expansive textbook survival guide covers the following chapters and their solutions. This textbook survival guide was created for the textbook: Mathematical Statistics and Data Analysis, edition: 3.
Key Statistics Terms and definitions covered in this textbook
• Acceptance region
In hypothesis testing, a region in the sample space of the test statistic such that if the test statistic falls within it, the null hypothesis cannot be rejected. This terminology is used because rejection of H0 is always a strong conclusion and acceptance of H0 is generally a weak conclusion
• Asymptotic relative eficiency (ARE)
Used to compare hypothesis tests. The ARE of one test relative to another is the limiting ratio of the sample sizes necessary to obtain identical error probabilities for the two procedures.
• Box plot (or box and whisker plot)
A graphical display of data in which the box contains the middle 50% of the data (the interquartile range) with the median dividing it, and the whiskers extend to the smallest and largest values (or some deined lower and upper limits).
• Central tendency
The tendency of data to cluster around some value. Central tendency is usually expressed by a measure of location such as the mean, median, or mode.
• Chance cause
The portion of the variability in a set of observations that is due to only random forces and which cannot be traced to speciic sources, such as operators, materials, or equipment. Also called a common cause.
• Combination.
A subset selected without replacement from a set used to determine the number of outcomes in events and sample spaces.
• Comparative experiment
An experiment in which the treatments (experimental conditions) that are to be studied are included in the experiment. The data from the experiment are used to evaluate the treatments.
• Conditional probability density function
The probability density function of the conditional probability distribution of a continuous random variable.
• Conditional variance.
The variance of the conditional probability distribution of a random variable.
• Contingency table.
A tabular arrangement expressing the assignment of members of a data set according to two or more categories or classiication criteria
• Correlation
In the most general usage, a measure of the interdependence among data. The concept may include more than two variables. The term is most commonly used in a narrow sense to express the relationship between quantitative variables or ranks.
• Correlation coeficient
A dimensionless measure of the linear association between two variables, usually lying in the interval from ?1 to +1, with zero indicating the absence of correlation (but not necessarily the independence of the two variables).
• Covariance
A measure of association between two random variables obtained as the expected value of the product of the two random variables around their means; that is, Cov(X Y, ) [( )( )] =? ? E X Y ? ? X Y .
• Decision interval
A parameter in a tabular CUSUM algorithm that is determined from a trade-off between false alarms and the detection of assignable causes.
• Dependent variable
The response variable in regression or a designed experiment.
• Discrete random variable
A random variable with a inite (or countably ininite) range.
• Distribution function
Another name for a cumulative distribution function.
• Enumerative study
A study in which a sample from a population is used to make inference to the population. See Analytic study
• F distribution.
The distribution of the random variable deined as the ratio of two independent chi-square random variables, each divided by its number of degrees of freedom.
• F-test
Any test of signiicance involving the F distribution. The most common F-tests are (1) testing hypotheses about the variances or standard deviations of two independent normal distributions, (2) testing hypotheses about treatment means or variance components in the analysis of variance, and (3) testing signiicance of regression or tests on subsets of parameters in a regression model. | 974 | 4,829 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2022-49 | latest | en | 0.876644 |
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# General Periodic Boundary Value Problem for Systems
DOI: 10.4236/am.2012.38130, PP. 882-887
Full-Text Cite this paper
Abstract:
The paper deals with the existence of nonzero periodic solution of systems, where k∈(0, π/T), α, β are n×n real nonsingular matrices, μ=(μ1…μn), f(t, u)=(f1(t, u),…,fn(t, u))∈C([0, T]×□n+,□+) is periodic of period T in the t variable are continuous and nonnegative functions. We determine the Green’s function and prove that the existence of nonzero periodic positive solutions if one of . In addition, if all i=（1…n）where λ1 is the principle eigenvalues of the corresponding linear systems. The proof based on the fixed point index theorem in cones. Application of our result is given to such systems with specific nonlinearities.
References
[1] L. H. Erbe and P. K. Palamides, “Boundary Value Problems for Second-Order Differential Systems,” Journal of Mathematical Analysis and Applications, Vol. 127, No. 1, 1987, pp. 80-92. doi:10.1016/0022-247X(87)90141-7 [2] J. W. Bebernes and K. Schmitt, “Periodic Boundary Value Problems for Systems of Second Order Differential Equations,” Journal of Differential Equations, Vol. 13, No. 1, 1973, pp. 32-47. doi:10.1016/0022-0396(73)90030-2 [3] L. H. Erbe and K. Schmitt, “Boundary Value Problems for Second-Order Differential Systems,” In: V. Lakshmikantham, Ed., Nonlinear Analysis and Applications, New York and Basel, 1987, pp. 179-183. [4] H. Wang, “Periodic Solutions to Non-Autonomous Second-Order Systems,” Nonlinear Analysis, Vol. 71, No. 3-4, 2009, pp. 1271-1275. doi:10.1016/j.na.2008.11.079 [5] D. Franco and J. R. L. Webb, “Collisionless Orbits of Singular and Nonsingular Dynamical Systems,” Discrete and Continuous Dynamical Systems, Vol. 15, No. 3, 2006, pp. 747-757. doi:10.3934/dcds.2006.15.747 [6] R. P. Agarwal, D. O’Regan and P. J. Y. Wong, “Constant-Sign Solutions of a System of Fredholm Integral Equations,” Acta Applicandae Mathematicae, Vol. 80, No. 1, 2004, pp. 57-94. doi:10.1023/B:ACAP.0000013257.42126.ca [7] D. ORegan and H. Wang, “Positive Periodic Solutions of Systems of Second Order Ordinary Differential Equations,” Positivity, Vol. 10, No. 2, 2006, pp. 285-298. doi:10.1007/s11117-005-0021-2 [8] X. Lin, D. Jiang, D. ORegan and R. Agarwal, “Twin Positive Periodic Solutions of Second Order Singular Differential Systems,” Topological Methods in Nonlinear Analysis, Vol. 25, 2005, pp. 263-273. [9] L. H. Erbe and K. Schmitt, “On Solvability of Boundary Value Problems for Systems of Differential Equations,” Journal of Applied Mathematics and Physics, Vol. 38, 1987. [10] H. Wang, “Positive Periodic Solutions of Singular Systems with a Parameter,” Journal of Differential Equations, Vol. 249, No. 12, 2010, pp. 2986-3002. doi:10.1016/j.jde.2010.08.027 [11] X. Li and Z. Zhang, “On the Existence of Positive Periodic Solutions of Systems of Second Order Differential Equations,” Mathematische Nachrichten, Vol. 284, No. 11-12, 2011, pp. 1472-1482. doi:10.1002/mana.200710145 [12] Z. Cao and D. Jiang, “Periodic Solutions of Second Order Singular Coupled Systems,” Nonlinear Analysis, Vol. 71, No. 9, 2009, pp. 3661-3667. doi:10.1016/j.na.2009.02.053 [13] J. R. Graef, L. Kong and H. Wang, “A Periodic Boundary Value Problem with Vanishing Greens Function,” Applied Mathematics Letters, Vol. 21, No. 2, 2008, pp. 176-180. doi:10.1016/j.aml.2007.02.019 [14] I. Rachunkova, M. Tvrdy and I. Vrkoc, “Existence of Nonnegative and Nonpositive Solutions for Second Order Periodic Boundary Value Problems,” Journal of Differential Equations, Vol. 176, No. 2, 2001, pp. 445-469. doi:10.1006/jdeq.2000.3995 [15] P. Torres, “Existence of One-Signed Periodic Solutions of Some Second-Order Differential Equations via a Krasnoselskii’s Fixed Point Theorem,” Journal of Differential Equations, Vol. 190, No. 2, 2003, pp. 643-662. doi:10.1016/S0022-0396(02)00152-3 [16] F. Li and Z. Liang, “Existence of Positive Periodic Solutions to Nonlinear Second Order Differential Equations,” Applied Mathematics Letters, Vol. 18, No. 11, 2005, pp. 1256-1264. doi:10.1016/j.aml.2005.02.014 [17] D. Jiang, J. Chu, D. ORegan, R. Agarwal, “Multiple Positive Solutions to Superlinear Periodic Boundary Value Prob lem, with Repulsive Singular Forces,” Journal of Mathematical Analysis and Applications, Vol. 286, No. 2, 2003, pp. 563-576. doi:10.1016/S0022-247X(03)00493-1 [18] X. Li and Z. Zhang, “Periodic Solutions for Second-Order Differential Equations with a Singular Nonlinearity,” Nonlinear Analysis, Vol. 69, No. 11, 2008, pp. 3866-3876. doi:10.1016/j.na.2007.10.023 [19] D. Jiang, J. Chu and M. Zhang, “Multiplicity of Positive Periodic Solutions to Superlinear Repulsive Singular Equations,” Journal of Differential Equations, Vol. 211, No. 2, 2005, pp. 282-302. doi:10.1016/j.jde.2004.10.031 [20] P. J. Torres and M. Zhang, “A Monotone Iterative Scheme for a Nonlinear Second Order Equation Based on a Geeralized Anti-Maximum Principle,” Mathematische Nachrichten, Vol. 251, No. 1, 2003, pp. 101-107. doi:10.1002/mana.200310033 [21] R. Ma, “Nonlinear Periodic Boundary Value Problems with Sign-Changing Green’s Function,” Nonlinear Analysis, Vol. 74, No. 5, 2011, pp. 1714-1720. doi:10.1016/j.na.2010.10.043 [22] B. Liu, L. Liu and Y. Wu, “Existence of Nontrivial Periodic Solutions for a Nonlinear Second Order Periodic Boundary Value Problem,” Nonlinear Analysis, Vol. 72, No. 7-8, 2010, pp. 3337-3345. doi:10.1016/j.na.2009.12.014 [23] H. Amann, “Fixed Point Equations and Nonlinear Eigenvalue Problems in Ordered Banach Spaces,” SIAM Review, Vol. 18, No. 4, 1976, pp. 620-709. [24] K. Q. Lan, “Nonzero Positive Solutions of Systems of Elliptic Boundary Value Problems,” AMS, Vol. 139, No. 12, 2011, pp. 4343-4349. doi:10.1090/S0002-9939-2011-10840-2 [25] D. D. Hai and H. Wang, “Nontrivial Solutions for p-Laplacian Systems,” Journal of Mathematical Analysis and Applications, Vol. 330, No. 1, 2007, pp. 186-194. doi:10.1016/j.jmaa.2006.07.072 [26] K. Q. Lan and W. Lin, “Multiple Positive Solutions of Systems of Hammerstein Integral Equations with Applications to Fractional Differential Equations,” Journal London Mathematical Society, Vol. 83 No. 2, 2011, pp. 449-469. doi:10.1112/jlms/jdq090 [27] R. D. Nussbaum, “Eigenvectors of Nonlinear Positive Operators and the Linear Krein-Rutman Theorem, in Fixed Point Theory,” In: E. Fadell and G. Fournier, Eds., Lecture Notes in Math, Vol. 886, Springer, 1981, pp. 309-330.
Full-Text | 2,354 | 7,551 | {"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-2020-05 | latest | en | 0.733807 |
https://proofassistants.stackexchange.com/questions/3882/packaging-mathematical-structures-in-coq-help-understanding-a-definition | 1,718,652,037,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861737.17/warc/CC-MAIN-20240617184943-20240617214943-00458.warc.gz | 425,847,900 | 31,609 | # Packaging Mathematical Structures in Coq: Help Understanding a Definition
## Context
I am a relatively new user to Coq with a decent understanding of the basics of dependent type theory and am midway through chapter 2 of the Software Foundations Series of books. I want to start writing projects at a larger scale, and as such have stumbled upon the problem of how to properly package the mathematical theories and structures at scale.
In part of my research into this problem I found this promising paper Packaging Mathematical Structures that claims to give methodology for doing just that. The methodology it uses defines structures in three Layers:
1. First is to define a bunch of mixins of type (M:Type -> Type) that implement certain substructures M s of a mathematical structure for a given sort/carrier type s :Type. (for example in the text one examines mix-ins for a equality structure and that of a Z-module).
2.These are then bundled together to form "classes" which are (C:Type -> Type) which packages together for any sort s:Type a structure type/record of the form {M1 s; M2 s;....} for Mi mix-ins that encodes all the extra structure over s.
1. Finaly there are "types" which represent the complete packaged mathematical structure in question which consists of a record type of the form {sort :> Type; class :> C sort; _:Type} that gives the underlying sort plus the "class" structure adorned atop it as the first two projections. Together these two projections can define the formalization of an algebraic structure (or perhaps more generally some kind of analogue to an object of concrete category), which is what we wanted. The third anonymous projection is what my question is about. It is not well explained in the text, at least from my point of view.
For context heree is the main example structure layed out as per the text which defines a Z-module structure based on that of a class that defines an equality structure by reflection:
Record mixin_of (T : Type) : Type := ...
Record class_of (T : Type) : Type :=
Class {base :> Equality.class_of T; ext :> mixin_of T}.
Structure type : Type :=
Pack {sort :> Type; class : class_of sort; _ : Type}.
Definition unpack K (k : forall T (c : class_of T), K T c) cT :=
let: Pack T c _ := cT return K _ (class cT) in k _ c.
Definition pack :=
let k T c m := Pack (Class c m) T in Equality.unpack k.
Coercion eqType cT := Equality.Pack (class cT) cT.
End Zmodule.
Notation zmodType := Zmodule.type.
Notation ZmodType := Zmodule.pack.
Canonical Structure Zmodule.eqType.
## Question
What is the purpose of this third projection? Reading through the text I see that is maybe meant to represent something refereed to by or linked with the text as a a "head" constant - or at least related to the problem of type coercion. The text introduces the notion of head constant when taking about packaging structures by "telescoping" (of which the packaging structure laid out above is seen as a refinement / generalization of) where it discusses such a notion. Before looking into this packaging problem I had not come upon the concept of a "head constant" in the vernacular surrounding algebraic-structures / universal algebra so I assume that the notion of a "head constant" is something that is domain specific to formalizing mathematics electronically. Alternatively it could be used for something completely different and I have gotten the wrong end of the stick. The text makes no explicit inference to the nature of this projection and I am left to try and figure out from inferring from the surrounding context.
To try to figure out what is going on I dove into one of the referenced texts on "telescoping" Telescopic Mappings in Typed Lambda Calculus, but I could not intuitively relate the formalism laid down there with the context of the original paper - and so this was no help to me. I went one step further and looked at texts referenced in that paper, also to no avail.
Currently, from my (admittedly extremely ignorant) point of view this third projection seems like a useless spandrel that serves no purpose and just adds an arbitrary type into the mix of ones considerations. This is further evidenced by the fact that unpack, which is meant to be used as a dependent destructor for type essentially works by simply throwing away this hidden third argument! Also in all the examples given this third type is always given to that of the same type of the sort - so I see no generalizing benefit either. Being that taxonomising structures is an important design decision in any large scale formalization project I assume that some of the members here may have insight into why this design decision was made?
P.S. I realize the problem that I am trying to solve / methodology I am trying to understand is to do with the specific problem of doing "the plumbing" of concepts in a proof assistant, but trying to find the answer to my question has lead me to look over some nontrivial mathematics. I'm posting here and not, say, maths-stackexchange because I'm coming at it from a proof assistant usage point-of-view and not a mathematical theory point-of-view. the What topics can I ask about here? section of the wiki is currently a stub and it seems according to the Meta that the exact boundires of what questions can be asked here are still being felt out. I understand if this needs to get moved elsewhere.
• I'm new here, but this looks relevant to this site. Also, the hierarchy-builder project may be of interest. github.com/math-comp/hierarchy-builder Commented Apr 6 at 6:47
• The answer may depend on what "large scale" means for you, and what sort of structures are you defining? IPv4 stack, modules over a ring? Commented Apr 6 at 8:07
• J .. S i stumbled across that but have no experience in elpi / prolog so am not sure if it is worth my time to learn those things before being able to learn from that resource Bauer - the short / medium term is just to get "the usual" algebraic structures of mathematics codified - but I eventually in the longer term want to model structures in the WASM spec (which I've yet to dive into) and potentially explore writing formalisms for the constructs in the WEBGPU spec Commented Apr 6 at 13:35
• Are you familiar with arxiv.org/abs/2311.07223? Commented Apr 6 at 15:19
• The whole point of Hierarchy Builder is that it should help you by automating the complex boilerplate used in MathComp-style libraries. Elpi/Prolog are used under the hood to implement it, but as a user you should not have to understand that, instead you should just use the various new commands HB.xxx defined by the tool. Commented Apr 8 at 8:22
## 1 Answer
It seems like this projection is not an intrinsic part of the paradigm of packed classes
Looking at two further developments using packed classes (all of which cite the original paper as their ur-reference), one sees that their third "type" layer does not include this mystery projection.
This leads me to believe that adding such a projection is unnecessary for this paradigm or domain specific to the original authors needs. This however does not answer the question of the original papers intent but satisfies me enough for me to be able to continue work / research in this area.
I'm making this answer community wiki so if anyone has any more insights to add and does not want to make a standalone answer they are welcome and encouraged to add it here. | 1,617 | 7,431 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2024-26 | latest | en | 0.920833 |
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# matix bimonals
Extracts from this document...
Introduction
Mathematics Portfolio – Matrix Binomials Samed Nakhla
## Matrix Binomials
Question
1. Let
Solution
Question
1. By considering integer powers of X and Y, find expressions for
Following the pattern in the first question and noticing from it, I deduce that the formula for:
Also to find the power of (X+Y) I will use some examples to find the pattern in it:
From the examples above, I deduced the equation of by following the pattern, and it is:
Question
Let A=aX and B=bY, where a and b
Middle
a and b
Let’s assume that a= 2
Also, let’s assume that b=3
To obtain the general formula for we need to find a formula for without using different values of a and b, are:
Question
By considering integer powers of A and B, find expressions for
Solution
By following the pattern in the previous question, one can deduce that the formulas are:
And to find the formula for, one should give some examples
Conclusion
bY.
Solution
After following the pattern in the previous question, I concluded the formula of to be:
To obtain the general formula of in terms of aX and bY:
Question
Test the validity of your general statement by using different values of a, b and n.
Solution
Let’s assume that a=2, b=3 and n=3
And by using the formulas we got from before
Where
So, also by assuming the same variables as before a=2, b=3 and n=3, and then replacing them in the equation, we get the exact result as before:
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# Related International Baccalaureate Maths essays
1. ## Matrix Binomials
Essentially, X + Y is another matrix. Matrix (X+Y) will be called Z. Using various values we can find the values of Z2, Z3, Z4, and Zn. However, remember that Z is the value of the sum of X and Y. Z = X + Y = + = Z2 = = Z3 = = Z4 =
2. ## Math IA - Matrix Binomials
Thus, we can now deduce the geometric sequence of these scalar values using the general equation listed above: In the sequence {1, 20, 400, 8000}, f=1 r=20 Here, we can express r in terms of a (a=10): r= 20 20= 10?2 r=2a We can also express in terms of a
1. ## Matrix Binomials
Seeing this pattern I formed an expression to solve for the matrix (X + Y)n. For the numbers of a and d the expression is 2n and for c and d it is 0. As a matrix the expression is (X + Y)n = .
2. ## Matrix Binomials
x X = x = = X5 = X4 x X = x = = X6 = X5 x X = x = = As you may notice, there are two differently colored matrices. This is because the last two matrices are extra help.
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• Ideas and feedback to | 900 | 3,283 | {"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} | 3.9375 | 4 | CC-MAIN-2021-21 | latest | en | 0.835257 |
https://gedmathlessons.com/chapter-3-order-of-operations-and-variables/ | 1,540,298,339,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583516135.92/warc/CC-MAIN-20181023111223-20181023132723-00081.warc.gz | 677,088,171 | 10,965 | This chapter introduces students to basic terms and concept used in Algebra. Time is taken to ensure the student understands basic number operations, order of operations, variables and their applications. Also a section explains the some key phases in the language of algebra and how to translate a verbal phrase into an algebraic phrase.
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#### Number Operations
the way to create real change in your life is to work on yourself- how can you provide more service and value to others?
To Pass The GED You Must Practice- Try A Few Problems Now
Good work! You finished lesson 1. Here are somethings you should be able to do: * define what a math operation is * know what the words “sum” “difference” “product” and “quotient” mean * be able to perform number operations with various numbers to include positive and negative numbers as well as fractions Let’s keep the fun going by starting the next lesson Lesson 2: Variables Lesson 1: Number Operations Lesson 2: Variables Lesson 3: Order of Operations Lesson 4: Translating Verbal and Algebraic Phrases Section Practice Problems Below: | 261 | 1,197 | {"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} | 3.234375 | 3 | CC-MAIN-2018-43 | longest | en | 0.8798 |
https://bergen.smartcatalogiq.com/en/2020-2021/Catalog/Courses/MAT-MATHEMATICS/200 | 1,623,876,855,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487626008.14/warc/CC-MAIN-20210616190205-20210616220205-00351.warc.gz | 136,084,708 | 18,776 | # 200
## MAT-223 Calculus for the Managerial and Social Sciences
This course covers the essential ideas of the Calculus: functions, limits, continuity, differentiation and integration. The course includes applications to problems in business, economics, psychology, the social sciences and mathematical modeling. >General Education Course. Lecture [3.00].
3
### Prerequisites
MAT-160; minimum grade C
## MAT-250 Statistics II
This course is an introduction to methods for the design of research studies and the interpretation of data that result from these studies. Topics considered include a brief review of elementary statistical concepts, additional cases of hypothesis testing and estimation, analysis of variance, analysis of enumerative data, linear regression and correlation, and nonparametric statistics. Laboratory assignments using a statistical software package are included in the course. >General Education Course. Lecture [2.00], Laboratory [2.00].
3
### Prerequisites
MAT-150; minimum grade C
## MAT-268 Statistical Methods
This course provides the student with a foundation in the techniques that underlie more advanced courses in statistics. Topics include descriptive statistics, sampling distributions, hypotheses testing and estimation for one and two populations, goodness-of-fit and contingency tables, analysis of variance, linear regression and correlation, and nonparametric statistics. Lecture[4.00].
4
### Prerequisites
MAT-160; minimum grade C
## MAT-280 Calculus I
This course is a study of limits, continuity, the derivative of a function, differentiation of algebraic, trigonometric, inverse trigonometric, exponential and logarithmic functions, applications of the derivative, antidifferentiation, area under a curve, the definite integral, the Fundamental Theorem of the Calculus and its applications. >General Education Course. Lecture [4.00].
4
### Prerequisites
MAT-180; minimum grade C
## MAT-281 Calculus II
This course is a study of differentiation and integration of transcendental functions, methods of integration, applications of the integral, indeterminate forms, improper integrals, infinite series, power series, and applications. >General Education Course. Lecture [4.00].
4
### Prerequisites
MAT-280; minimum grade C
## MAT-282 Calculus III
This course is a study of vectors, partial differentiation, directional derivatives, gradients, multiple integrals, vector calculus, line integrals, topics from vector analysis, and applications. >General Education Course. Lecture [4.00].
4
### Prerequisites
MAT-281; minimum grade C
## MAT-283 Differential Equations
This course covers equations of order 1, linear equations with constant coefficients, non-homogeneous equations, variation of parameters, series solutions, equations with variable coefficients, Laplace transforms, convolutions, boundary value problems, Fourier transforms and applications. Lecture [4.00].
4
### Prerequisites
MAT-282; minimum grade C
## MAT-285 Discrete Mathematics
This course is a study of mathematical concepts and techniques that form the foundation for many upper level mathematics courses. Topics considered include sets and logic, proof techniques, functions and relations, algorithms, introduction to number theory, counting techniques, discrete probability, recurrence relations, trees, graphs, networks, and Boolean algebra. Mathematical reasoning and proofs with be stressed. Lecture [4.00].
4
### Prerequisites
MAT-280; minimum grade C
## MAT-286 Linear Algebra
This course is a study of finite dimensional vector spaces. Topics considered include vectors and vector spaces, matrices, determinants, systems of linear equations, linear transformations, quadratic forms, eigenvalues and eigenvectors, and applications. Lecture [4.00].
4
MAT-280 | 764 | 3,824 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2021-25 | latest | en | 0.83599 |
https://stackoverflow.com/questions/13019719/get-business-days-between-start-and-end-date-using-pandas | 1,555,666,387,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578527518.38/warc/CC-MAIN-20190419081303-20190419103303-00391.warc.gz | 556,593,957 | 30,299 | # Get business days between start and end date using pandas
I'm using pandas and I'm wondering what's the easiest way to get the business days between a start and end date using pandas?
There are a lot of posts out there regarding doing this in Python (for example), but I would be interested to use directly pandas as I think that pandas can probably handle this quite easy.
Use `BDay()` to get the business days in range.
``````from pandas.tseries.offsets import *
In [185]: s
Out[185]:
2011-01-01 -0.011629
2011-01-02 -0.089666
2011-01-03 -1.314430
2011-01-04 -1.867307
2011-01-05 0.779609
2011-01-06 0.588950
2011-01-07 -2.505803
2011-01-08 0.800262
2011-01-09 0.376406
2011-01-10 -0.469988
Freq: D
In [186]: s.asfreq(BDay())
Out[186]:
2011-01-03 -1.314430
2011-01-04 -1.867307
2011-01-05 0.779609
2011-01-06 0.588950
2011-01-07 -2.505803
2011-01-10 -0.469988
Freq: B
``````
With slicing:
``````In [187]: x=datetime(2011, 1, 5)
In [188]: y=datetime(2011, 1, 9)
In [189]: s.ix[x:y]
Out[189]:
2011-01-05 0.779609
2011-01-06 0.588950
2011-01-07 -2.505803
2011-01-08 0.800262
2011-01-09 0.376406
Freq: D
In [190]: s.ix[x:y].asfreq(BDay())
Out[190]:
2011-01-05 0.779609
2011-01-06 0.588950
2011-01-07 -2.505803
Freq: B
``````
and `count()`
``````In [191]: s.ix[x:y].asfreq(BDay()).count()
Out[191]: 3
``````
• wow.. perfect! Thank you very much! – Thomas Kremmel Oct 23 '12 at 7:25
• would it be possible to use the same technique when my data has a granularity of hours? So I want to pull out all the hours that are on business days. I know how to then pull out just the working hours of the day after that – Luka Vlaskalic Oct 24 '18 at 14:01
• I figured out that you can just use this .asfreq(freq='BH') – Luka Vlaskalic Oct 24 '18 at 14:07
• Please don't teach people to use `import *` syntax! – Marvin Taschenberger Mar 5 at 9:32
You can also use `date_range` for this purpose.
``````In [3]: pd.date_range('2011-01-05', '2011-01-09', freq=BDay())
Out[3]: DatetimeIndex(['2011-01-05', '2011-01-06', '2011-01-07'], dtype='datetime64[ns]', freq='B', tz=None)
``````
EDIT
Or even more simple
``````In [7]: pd.bdate_range('2011-01-05', '2011-01-09')
Out[7]: DatetimeIndex(['2011-01-05', '2011-01-06', '2011-01-07'], dtype='datetime64[ns]', freq='B', tz=None)
``````
Note that both start and end dates are inclusive. Source: http://pandas.pydata.org/pandas-docs/stable/generated/pandas.bdate_range.html
As of v0.14 you can use holiday calendars.
```from pandas.tseries.holiday import USFederalHolidayCalendar
print pd.DatetimeIndex(start='2010-01-01',end='2010-01-15', freq=us_bd)
```
returns:
```DatetimeIndex(['2010-01-04', '2010-01-05', '2010-01-06', '2010-01-07',
'2010-01-08', '2010-01-11', '2010-01-12', '2010-01-13',
'2010-01-14', '2010-01-15'],
dtype='datetime64[ns]', freq='C')
```
• if you want the number of days between the date range, you can get this as `pd.DatetimeIndex(start='2010-01-01',end='2010-01-15',freq=us_bd).shape[0]` – tsando Sep 15 '17 at 15:52
Just be careful when using bdate_range or BDay() - the name might mislead you to think that it is a range of business days, whereas in reality it's just calendar days with weekends stripped out (ie. it doesn't take holidays into account).
On top of this answer and xone, we can write a short function to return the trading days of US exchange:
``````from xone import calendar
kw = dict(start=start, end=end)
return pd.bdate_range(**kw).drop(us_cal.holidays(**kw))
Replace `DatetimeIndex` with `bdate_range` for `pandas` 0.24.0 update: | 1,287 | 3,596 | {"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-2019-18 | latest | en | 0.739977 |
https://www.scribd.com/document/392702159/Tok-Sb-Ibdip-Ch6 | 1,550,870,623,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247526282.78/warc/CC-MAIN-20190222200334-20190222222334-00260.warc.gz | 972,190,280 | 87,197 | You are on page 1of 40
# Mathematics
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06 Mathematics
On previous page –
6.1
Symmetry is not just limited to
mathematics; it surrounds us Introduction to mathematics
in both living and non-living
organisms, as illustrated in
this close-up of the Bagworm
moth’s compound eye. Structures and patterns in mathematics
According to Marcus du Sautoy, maths is everywhere, even in the apparently random,
paint-splattered canvases of Jackson Pollock.
By looking at Pollock’s paintings from a mathematical perspective, you understand
why people respond to them. Pollock was creating a fractal structure. The
scattering of paint has a property shared by clouds and the branches of trees:
whether you zoom in or out, it retains the same complexity. Pollock was painting
nature.
du Sautoy, 2011
Professor Marcus du Sautoy of Oxford University sees mathematics as a code that we
can use to unlock the secrets of our world. When we scratch beneath the surface we
find that the world is essentially mathematical. This is the Platonist view, named after
A second view is that mathematics is fundamentally artificial; that it is a human
invention, a lens we have constructed through which to view the world. This is
usually known as the constructivist view (although it can be called mathematical
humanism).
As we shall see later, each view comes with knowledge questions. The job of TOK is to
examine these knowledge questions and try to understand how mathematics works as
an AOK.
More specifically, Sautoy touches upon two concepts central to the idea of
mathematics: structures and patterns. In this chapter, we shall see how mathematics is
built on some very simple structures and that it uses these to investigate patterns.
Whether you agree with Sautoy’s analysis or not, mathematics is a compulsory part
of most state education systems. Clearly, we value mathematics sufficiently to make
children study it, in many cases, into their teens.
Exercises
1 Design your own high school curriculum. What subjects do you think should be available for
students to study at the age of 16?
2 What subjects, if any, should be compulsory?
3 If you included mathematics as part of your curriculum, why did you do so?
4 What role does mathematics play in your life?
Mathematics in contrast to other areas of knowledge
Mathematics is crucial in TOK because it offers a strong contrast to the other AOKs
both in methods and in subject matter.
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One area of contrast is the ability of mathematics to generate absolute certainty. This
feature places mathematics in stark opposition to the natural sciences which, as we
have seen, can at best generate only provisional results. If scientific knowledge is a map
of reality, then there is always the possibility that it gets revised. But with the standard
definition of the symbols ‘2’, ‘4’, ‘+’, and ‘=’, 2 + 2 = 4 is not just highly likely to be true –
it is absolutely certain. The reason is that we can prove that (given the standard meaning
of these symbols) 2 + 2 has to be 4. It cannot be otherwise. In fact, it is a contradiction to say
that 2 + 2 is not 4.
This strong contrast with the natural sciences can be illustrated by the respect we tend
to pay ancient mathematics compared to long bygone science. Do we still learn the
mathematics of Pythagoras, Archimedes, and Euclid? Are these results still true? Are
they still used? The answer is immediate: of course. Every schoolchild can recite the
famous theorem about the right-angled triangle and quote the formula for the area of a
circle. We need this mathematics on a daily basis in our everyday lives, whether we are
designing garden ponds or calculating if we can get the model plane we are building
through the workshop door.
But do we still learn the science of the 4th and 3rd centuries BC? Is this science still
true or still useful? The answer is just as immediate: of course not. We don’t even spend
much time on the science of the 19th century, let alone the science of the ancient
world. Aristotle thought that there were four chemical elements and that the brain
was used for cooling the blood. We now know that there are more than 100 and that
the brain is used for thinking. Ancient science is really of interest only to the specialist
historian of science.
It is very clear that mathematics and the natural sciences are quite different in their
nature. Science changes rapidly and things that we thought true in the last century,
the last decade, even last year, can be rejected today. This does not happen with
mathematics. The question that TOK requires us to ask is, why?
6.2 Scope and applications
Knowledge framework: Scope and applications – What is
What is mathematics?
Some AOKs lend themselves to neat closed definitions. Mathematics does not. There
is an irony in that a young child might have less difficulty defining mathematics than
an IB student. The 6-year-old might suggest that mathematics is to do with numbers.
He or she might suggest that mathematics is concerned with the idea of counting
things using positive whole numbers. For these children counting numbers are found
wherever there are a number of similar things to count – pure and simple. But, as
Oscar Wilde remarked about truth, such things are rarely pure and never simple.
For a start, we might reflect that many of the things familiar to us in our mathematics
classes have very little to do with counting – for example, the number 0 or negative
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06 Mathematics
numbers and fractions. We are unlikely to need them for counting things. It doesn’t
seem to make sense to point at an object and say ‘zero’.
Perhaps we could argue that mathematics is somehow an abstract extension of the idea
of counting. We use rational numbers (ratios of whole numbers) for expressing things
like mass and length. But this still leaves us with a puzzle regarding 0 and negative
3 ⎯
numbers and numbers that are not expressible as fractions, such as √ 7. These numbers
play no part in real life measuring. Furthermore, HL maths students work with
complex numbers like the so-called imaginary number √ –1. It is difficult to see how
this number could be used in measurement or counting. Strangely enough, √ –1, or i as
it is known, turns out to be very useful in describing real-life things, such as electrical
circuits containing inductors and capacitors.
Some parts of mathematics appear not to have anything whatsoever to do with
counting or measuring. The 6-year-old might spend some time drawing triangles and
circles on a flat piece of paper, or constructing cubes and tetrahedra using cardboard
but not think of these activities as being mathematical. The IB student, on the other
hand, understands these as geometrical objects and uses mathematics to explore
their properties. The IB student might reflect that the equilateral triangle possesses
three axes of reflection symmetry and rotational symmetry of order three, while the
isosceles triangle only possesses one axis of reflection symmetry and no rotational
symmetry. These symmetries are expressed by a set of mappings that leave the shape
of the object the same. If we combine two of these symmetry mappings we always get
a new symmetry mapping (see below). We say that the set of symmetries of the shape
is closed when you combine them. Here we are getting to the essence of mathematics:
mappings between sets of objects and things that stay invariant under those mappings.
This is quite an abstract definition and seems to be too general to distinguish
mathematics from other AOKs. But we shall discover that mathematics is itself abstract
and any definition is, of necessity, general. In any case, mathematics might, like the
natural sciences, best be described by its methods and modes of thought.
CHALLENGE YOURSELF
Symmetry of two types of triangle.
The isosceles triangle only has two symmetry mappings (Figure 6.1):
I = identity mapping (leave the triangle as it is)
F = reflection in AX
Isosceles triangle diagram:
A
Figure 6.1 Isosceles triangle
diagram
B C
X
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These two mappings combine in the following way.
Symmetries of the isosceles triangle
I F
I I F
F F I
The equilateral triangle has six symmetries: three rotationsEquilateral
and three refltriangle diagram:
ections (Figure 6.2).
I = identity mapping (leave the triangle as it is) A
R1 = Rotation centre O 120°
R2 = Rotation centre O 240°
F1 = Reflection in AX
Z Y
F2 = Reflection in BY
O
F3 = Reflection in CZ
These mappings combine in the following way.
Symmetries of the equilateral triangle B C
X
I R R2 F1 F2 F3 Figure 6.2 Equilateral
triangle diagram.
I I R R2 F1 F2 F3
R R R2 I F2 F3 F1
R2 R2 I R F3 F1 F2
F1 F1 F3 F2 I R2 R
F2 F2 F1 F3 R I R2
F3 F3 F2 F1 R2 R I
The results are a type of magic square. This is a special mathematical structure called a group. Almost
all the important things in mathematics are groups.
Try constructing the group table for the eight symmetries of the square.
Knowledge framework: Scope and applications – What
practical problems can be solved through applying this
knowledge?
What is the purpose of mathematics?
Returning to the view of knowledge suggested by the TOK subject guide – that
knowledge is a map of reality designed to solve certain problems – we might want
to consider what sort of problems mathematics attempts to solve. With the natural
sciences we found that these problems fell naturally into two distinct groups: internal
and external. The internal problems are problems that are purely about matters within
natural science or ‘knowledge for its own sake’. The external problems are problems
of application, such as how to make strong, light materials for the turbine blades in jet
engines.
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06 Mathematics
In mathematics, the same distinction holds. Mathematicians divide their subject into
two broad categories: pure mathematics and applied mathematics. Pure mathematics
solves problems that are internal to mathematics itself. These could be, for example,
how to solve a particular type of differential equation. Applied mathematics is more
concerned with matters such as how this equation could be used to model a particular
real-world situation.
Pure mathematics
Doing pure mathematics is a bit like solving a Sudoku puzzle. There is need for
strategy, inspiration, and a good deal of creativity. Guesswork and serendipity might
also have a role to play. All the while, the rules have to be followed and the final
solution sought. For certain, it is not blind, cold
logic. As with the Sudoku, when the solution is
found there is a great deal of satisfaction but there is
no outside application.
But there is a little more to pure mathematics than
And the answer is …. this analogy suggests. For example, the connection
between different Sudoku puzzles is not important.
But pure mathematics is a huge integrated field
where progress in one area leads to progress
elsewhere.
We can ask mathematical questions such as the following about Sudoku puzzles.
•How many different Sudokus are there (given a suitable notion of ‘different’)?
•What is the best general algorithm for solving them?
•Is there a rule for how much information is needed to generate a unique solution to
the Sudoku?
•Can we generalize the result about how many ‘different’ puzzles there are to a more
general type of Sudoku played on a larger grid?
Although our analogy with Sudoku only works so far, pure mathematics, nevertheless,
involves solving puzzles, largely for their own sake, through the use of imaginative
strategy subject to the rules of the game.
We shall use number theory to illustrate pure mathematical thinking. It is an
important field within pure mathematics. Number theory explores topics such as the
distribution of prime numbers. This is a question that goes back to the ancient Greeks.
A prime number is defined as a whole number that can only be exactly divided by itself
and 1. Euclid showed that any whole number can be written uniquely as a product of
prime numbers, so they really are the building blocks of the system of numbers.
So, for example, the number 240 can be expressed as:
240 = 2 × 2 × 2 × 2 × 3 × 5 = 24 × 3 × 5
Moreover this way of building 240 is unique up to rearrangement of the factors. So if
we present them in ascending order, as here, then the decomposition is unique.
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Euclid had already proved that there were infinitely many prime numbers (page 193).
But how are they distributed? At first sight the distribution seems haphazard and there
doesn’t seem to be a pattern. Here are the primes below 100:
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97
One might hope that the pattern settles down as the numbers get larger but it is not the
case. For example, there are no fewer than nine primes among the 100 numbers below
10 000 000 but there are only two in the 100 numbers above 10 000 000 (du Sautoy
2003). The project of finding a formula for the number of primes less than or equal to n
is a classic piece of pure mathematics. It is still very much an open question, though we
are closer to a solution now than we were, thanks to the work of Carl Friedrich Gauss
(1777–1855) and Bernard Riemann (1826–66).
There really does not seem to be much chance of this sort of work being useful in the
world outside mathematics. But, as we shall see shortly, this is not the case. In fact,
it is difficult to find an area in mathematics that has not led to important and useful
applications.
Applied mathematics
Applied mathematics is oriented towards the solution of real-world problems. The
mathematics it produces can be just as interesting from an insider’s viewpoint as the
problems of pure mathematics (and often the two are inseparable), but in the final
analysis, it is judged by whether or not it has practical external benefit.
Below is an example of applied mathematics at work. It is an example that could occur
in an SL maths course or indeed (and this is the point) in a physics course.
A stone is dropped down a 30 m well. How long will it take the object to reach the
bottom of the well, neglecting the effect of air resistance?
We know that the acceleration due to gravity is 9.8 m/s2, we also know that the distance
travelled S is given by the equation:
1
S = 2 at2
where a = acceleration and t = time
So, we substitute the known values into the equation:
1
30 = 2 (9.8)t2
Rearranging the equation gives us:
60
9.8 = t2
so
t= fl9.860 = 2.47 seconds
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06 Mathematics
Time (sec) Speed (ms–1) Distance (m) There are two points to note here.
t v = 9.8.t s = –12 × 9.8.t2 stone
well
•The formula only works if we neglect air resistance and if the
0 0 0 stone is dropped and not thrown down the well (Figure 6.3).
0.5 4.9 1.2
•The number 9.8 m/s2 is arrived at by observation of things
accelerating in our physical world. It is not a mathematical fact.
1.0 9.8 4.9
These points are typical of this sort of problem. We have to
simplify the description of the problem and idealize it in some
way before we can apply the appropriate mathematics. This is
1.5 14.7 11
strongly reminiscent of the idea of the map being a simplified
representation of reality. Moreover, the whole calculation
requires the input of observations of the real world to furnish
us with the correct value for acceleration due to gravity – it is
based on empirical evidence. Neither of these requirements are
necessary in pure mathematics.
2.0 19.6 19.6
2.48 24.3 30
Figure 6.3 Visualizing the
acceleration-due-to-gravity
calculation.
How important
is empirical
evidence?
A mixture?
The distinction between pure and applied mathematics becomes blurred in the hands
of someone like Gauss. We noted above that he had worked on the distribution
of prime numbers. The same Gauss as a young man had enabled astronomers
to rediscover the minor planet Ceres after they had lost it in glare of the Sun. He
calculated its orbit from the scant data that had been collected on its initial discovery
178
M06_TOK_SB_IBDIP_4157_U06.indd 178 05/03/2014 13:32
in 1801 and then predicted where in the sky it would be found more than a year later.
This feat immediately brought Gauss to the attention of the scientific community. His
skills as a number theorist presented him with the opportunity of solving a very real
scientific problem.
Who would have guessed that prime number theory would one day give rise to a
system of encoding data that is used by banks all over the world? The system is called
‘public key cryptography’. The clever part is that the key to the code is a very large
number which is the product of two large primes. The bank holds one of the primes
and the computer of the client the other. The key can be made public because, in
order for it to work, it has to be split up into its component prime factors. This task
is virtually impossible for large numbers. Imagine trying to find the two large prime
factors of the following 677-digit number. Such a task as this would take a very long
time, even on the fastest modern computer:
25,195,908,475,657,893,494,027,183,240,048,398,571,429,282,126,204,032,027,777,
137,836,043,662,020,707,595,556,264,018,525,880,784,406,918,290,641,249,515,082,
189,298,559,149,176,184,502,808,489,120,072,844,992,687,392,807,287,776,735,971,
418,347,270,261,896,375,014,971,824,691,165,077,613,379,859,095,700,097,330,459,
748,808,428,401,797,429,100,642,458,691,817,195,118,746,121,515,172,654,632,282,
216,869,987,549,182,422,433,637,259,085,141,865,462,043,576,798,423,387,184,774,
447,920,739,934,236,584,823,824,281,198,163,815,010,674,810,451,660,377,306,056,
201,619,676,256,133,844,143,603,833,904,414,952,634,432,190,114,657,544,454,178,
424,020,924,616,515,723,350,778,707,749,817,125,772,467,962,926,386,356,373,289,
912,154,831,438,167,899,885,040,445,364,023,527,381,951,378,636,564,391,212,010,
397,122,822,120,720,357
du Sautoy, 2003
But this number is indeed of the form of the product of two large primes. If you know
one of them, it takes an ordinary computer a fraction of a second to do the division and
find the other. And who was the mathematician who said that such a number should
have prime factors in the first place? It was Euclid, 2000 years ago in his fundamental
theorem of arithmetic.
Just as we found in the natural sciences that internal problems produced solutions
(pure science) that could also be used for technological or engineering applications, so
in mathematics, problems motivated purely from within the most abstract recesses of
the subject (pure mathematics) give rise to very useful techniques for solving problems,
with direct applications in the world outside mathematics.
Prescribed essay title 1: ‘… mathematics may be defined as the subject in
which we never know what we are talking about, nor whether what we are
saying is true.’ (B. Russell, Mathematics and the Metaphysicians) Critically assess
this assertion. What does it mean? If this is the case, how does mathematics
differ from other forms of knowledge?
© International Baccalaureate, May and November 1999
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6.3 Language and concepts
Knowledge framework: Language and concepts – What
role does language play in the accumulation of knowledge in
this area?
Language and different representations
non notationes, sed notiones (not notations, but notions)
Gauss
Knowledge is a representation of the world. That is the starting point for our
investigations in TOK. But how exactly does mathematics do this?
Numbers are one of the basic pieces of furniture in the house of mathematics. Exactly
what sort of thing a number is and what type of existence it enjoys are difficult
questions in the philosophy of mathematics, and somewhat outside the scope of this
How do we express
mathematical ideas? book. Instead, we are concerned with the mathematical language we use to express
numbers and how we construct them in the first place.
Numbers should not be confused with
the symbols we use to represent them.
Take fractions, familiar from your
mathematics courses, as an example.
1 2 3
The symbols 3 , 6 , 9 , 0.3333… all name
the same number despite appearing to
be quite different (perhaps the infinity of
different ways of representing fractions is
one of the reasons why students have so
much difficulty with them).
We might think that we are on safer
ground with decimals. But consider the
symbol 0.9999… (the three dots mean
that the 9 is repeated indefinitely). This is
just a very complicated way of writing 1.
Exercises
5 What is named by the expression the smallest number expressible by a decimal bigger than 1?
6 What is named by the following expressions: the tallest tree in the world, the shortest tree in
the world, the most westerly tree in Canada?
7 Are there any differences in mathematical conventions in your class? Do you all do long division
the same way? Does it matter?
8 Prove that 0.9999… = 1. What makes your proof decisive? Does it matter if people are not
convinced by it? Is it still true if nobody apart from you believes that it is true? What makes it a
proof? What if there is a mistake in it that no one has ever spotted?
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Knowledge framework: Language and concepts – What It is tempting to think that
are the roles of the key concepts and key terms that provide the existence of different
conventions for writing
the building blocks for knowledge in this area? mathematics means that
mathematics possesses
Basic concepts regional variations. People
write 0.3 in the UK and
In trying to tie down exactly what mathematics is about, we suggested that the idea of a 0,3 in Sweden; does this
set as a collection of things might somehow be a crucial foundation on which the more show that mathematics
is different in the UK
complex notions of mathematics can be built. This is an appropriate place to develop
and Sweden? No, clearly
this idea. the notation is different
but the underlying
mathematics is the same.
Sets
A set is a collection of elements which can themselves be sets. Sets can be combined in
various ways to produce new sets. The concepts of a set and membership of a set are
primitive. This means that they cannot be explained in terms of more simple ideas. This
may seem a modest idea on which to build the complexities of modern mathematics
but, in the 20th century, there were a number of projects designed to do just that – to
reduce the whole of mathematics to set theory. The most important work here was by
Quine, von Neumann, Zermelo, and Russell and Whitehead.
Exercise
9 Choose one of your HL subjects in the IB Diploma programme. Identify a primitive concept in this
subject. Find another non-primitive concept that incorporates the concept you have chosen. For
example, a primitive concept in physics is energy. The idea of work requires the primitive concept
of energy.
We can define a set by listing its elements. In mathematics, we do this by enclosing
the set in curly brackets { }. This tells us that the order of elements in the set does not
matter (the use of parentheses ( ) in mathematics tells us that order does matter).
So, we can define the set M of months of the year as follows:
M = {January, February, March, April, May, June, July, August, September, October,
November, December}
We can also define a set by specifying a rule for membership:
M = {x: x is a month of the year}
Paradoxically, one of the most important sets in mathematics is the set without any
members; this is called the null set {}. Using this set and the idea that no set can be a
member of itself (see exercises below), we can construct another important set: the
natural numbers.
N = {1, 2, 3, 4, 5, 6, …}
The three dots here indicate that there is no end to the set: it contains an infinite
number of elements; it has no greatest member.
Mathematicians are comfortable with the idea of infinite sets, since most of the sets
they deal with are infinite. However, this should worry us a little bit as TOK students.
For a start, it is difficult to visualize an infinite collection of things. Secondly, and
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06 Mathematics
In mathematics, the term perhaps more seriously, we should worry about how we specify an infinite list. After
induction does not mean all, the lists we make in everyday life are finite.
the same as it does in
our discussions of, say, Mathematicians have a neat trick for generating infinite lists. In the case of the set
the natural sciences. N, observe that every number n has a unique successor n + 1. So we can generate
In the natural sciences,
the whole set by specifying the first member 1 and then giving the rule that if n is a
inductive inferences are
generalizations from a member then n + 1 is the next member. This is called an inductive definition.
finite data set to a whole
population. The process Exercises
involves a leap of faith 10 Some sets can be defined that are self-referential. Consider the ‘set of all things that can be
and is not entirely secure. described in eleven words’. Clearly this set can be defined in eleven words (we just did it). So
Mathematical induction, technically, it is a member of itself. But this type of self-reference should ring a warning bell. Russell
on the other hand, is played with this idea when he challenged Gottlob Frege with the following conundrum. Consider
an entirely rigorous the set of all sets which are not members of themselves. Is it a member of itself? Try solving
deductive process. this puzzle.
11 Here is a similar puzzle. In the town of Seville, there is only one barber. He shaves all those
Mappings between sets
Once we have established sets in our mathematical universe, we want to do something
useful with them. One of the most important ideas in the whole of mathematics is
that of a mapping. As you know from your IB mathematics course, a mapping is a rule
that associates every member of a set with a member of a second set. The first set is
called the domain and the second set (the target set) is called the codomain. We represent
Human Spiritual
mappings by arrows joining the domain to the codomain.
land, ancestors …
This is precisely what we do when we are counting something. We set up a mapping
between the natural numbers and the set of objects we want to count (Figure 6.4).
1
2
Figure 6.4 Counting is a
mapping between two sets. 3
Domain = N Codomain = Sheep
Another interesting mapping is the doubling mapping that maps the set of natural
numbers onto the even numbers (Figure 6.5).
1 4
1 7
Figure 6.5 The doubling 4
3 3 10
mapping. 2 2
6 9
5 6
12 8 5
11
There are some special classes of mappings that require our attention. A mapping
where only one arrow is associated with every element of the domain is called a
function. Functions are very important in the mathematics that is studied in the IB
Diploma. A function that maps distinct elements in the domain to distinct elements
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in the codomain is called an injective or one–one mapping (Figure 6.6). A mapping
where every element of the codomain has an arrow pointing to it is called a surjective
or onto mapping (Figure 6.7). If a mapping is both injective and surjective then it is
called a one–one correspondence.
5 3
2 6
3
4 Figure 6.6 An injective
2 4
1 mapping.
1 5
5 3
2
3
4 Figure 6.7 A surjective
2 4 mapping.
1 5
The counting mapping above is a one–one correspondence. It is important when
counting objects that you (a) do not count anything twice (injective) and (b) that you
count every object (surjective).
This idea of counting as a one–one correspondence with N is very powerful. It allows
kindergarten children to count sheep (finite sets) and it allows mathematicians to
count infinite sets. If we can find a one–one correspondence between them then we
can say that, in an important sense, they have the same ‘number’ of elements.
Infinite sets
Take a look at the doubling function (Figure 6.5). Clearly it sets up a one–one
correspondence between the set of natural numbers and the set of even numbers.
You can check this yourself. No element of the codomain has more than one arrow
pointing at it (meaning different elements of the domain map to different elements
in the codomain), and every element of the codomain has an arrow pointing at it. So
this means that there are as many even numbers as there are natural numbers. But this
is strange. If we start with the set of natural numbers and then remove the set of odd
numbers, we get the even numbers. But we are saying that the set that is left over has as
many members as the original set, even though we have removed an infinite number
of odd numbers. The strangeness here is characteristic of infinite sets (indeed it can be
used to define what we mean by infinite). A set is infinite if it can be put in a one-one
correspondence with a proper subset of itself. In other words, you can take things out
of an infinite set (even an infinite number of things if you choose well) and still have the
same number of things remaining.
But the story doesn’t stop here. Using sets and mappings we can show that there are
many different types of infinity. The set of natural numbers contains the smallest type
of infinity usually denoted by the Hebrew letter aleph with a subscript 0 (ϗ0) which
we call aleph nought. In the 19th century, the German mathematician Georg Cantor
(1845–1918) showed that the number of numbers between 0 and 1 is a bigger type of
infinity than ϗ0.
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06 Mathematics
To learn more about It turns out that there is in fact an infinity of different types of infinity – a whole
the ingenious argument hierarchy of infinities, and it probably does not surprise you anymore that there are
of Georg Cantor, visit more infinities than numbers. It is instructive to examine how mathematicians deal
pearsonhotlinks.com, with infinity. For a start, there need not be a problem of visualization. There are many
enter the title or ISBN
of this book and select
ways in which infinity can be handled in a finite fashion. For example, imagine a series
weblink 6.1. of points X1, X2, X3, X4 on a line AB. Now draw lines from each point Xn to the pole, P,
of circle, C, sitting on the line. The point where each line cuts the circle is designated
Yn. It is clear that the mapping Xn to Yn is a one–one correspondence between AB and C
(HL maths students should prove this). The question now is, what point corresponds to
P itself? It is the point at ∞ (Figure 6.8).
P
C ∞
Y4
Y3
Figure 6.8 Visualizing Y2
infinity.
Y1
A B
X1 X2 X3 X4
Another way to handle infinity can be illustrated by a story. Imagine a frog at the side
of a pond. There is a princess sitting on a water lily in the centre of the pond. The frog
has to get to the princess and, as these things usually go, he has to be kissed by her
and thence be transformed into a handsome prince. The pond is circular with radius
2 m. But there is a problem. The frog can only jump half the remaining distance to
the princess with every jump. With the first jump he covers 1 m, the second ½ m, the
third ¼ m and so on until infinity (Figure 6.9). He gets very close to the princess but
never actually reaches her. The situation seems quite hopeless. But then let us imagine
that the first jump takes 1 second, the second jump takes ½ second and the third ¼
second and so on. This means that the frog covers 1 m in 1 second, ½ m in ½ second,
¼ m in ¼ second. In other words, the frog is travelling at exactly 1 m per second. As
the distance to the princess is originally 2 m, the frog takes just 2 seconds to reach the
princess and be kissed and so transformed into a prince. In the mathematics of infinity,
as well as in fairy tales, there is a happy ending.
2m
Figure 6.9 Can the frog
and the princess ever meet?
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This story illustrates the fact that the series:
1 1 1 1
1 + 2 + 4 + 8 + 16 + … = 2
In other words, although there are infinite terms in the series they have a finite sum.
This is because each successive term is a certain fraction (in this case ½) the value of
the previous term. The terms get successively smaller and the sum gets closer and
closer to 2, so that after an infinite number of terms it actually is 2.
Of course, this is an informal argument, and mathematicians need to worry about
what they mean by a sum that contains an infinite number of terms. Nevertheless,
there is a place for such visual intuition in mathematics that is so often about things
that are so different from our everyday experience as to be literally unimaginable.
Mathematicians are used to dealing with spaces of four, five, and six dimensions
and have to describe geometrical objects in these spaces. Indeed, there are fields
within mathematics that routinely work in infinite dimensional space. Despite this,
mathematicians go on cheerfully with their work by adopting some sort of informal
picture to guide their intuitions.
Fractals
A very striking use of the notion of infinite sets and mappings is in the area of
dynamical systems, more popularly known as chaos theory and fractals. A fractal
is a set produced by repeating a mapping an infinite number of times. Again, Georg
Cantor produced some very interesting mathematics in the 19th century that steered
mathematicians towards fractals.
Here is a recipe for producing a Cantor set (Figure 6.10).
1 Draw a line segment of length 10 cm.
2 Cut it into three equal pieces.
3 Delete the middle third.
4 Take each of the two remaining pieces and divide each into three equal pieces.
Delete the middle third of these.
5 Carry out this operation indefinitely. What is left over?
It should be obvious that the points that are the end points of any line segment at any
finite stage in the process are never deleted. So some points are left over in the limit set
obtained by doing the deleting process an infinite number of times. It is clear that there
are an infinite number of such points. It should also be obvious that it is impossible to
draw these points properly on the diagram and that no point in the set is connected to
any other point in the set. Less obvious is that there are ‘more’ points left over than the
0 1
Figure 6.10 How to
1–
0 3 –32 1 produce a Cantor set.
0 1–
9 –29 1–
3
–23 –79 –89 1
1–
0 9 –92 1–
3
–23 –97 –89 1
1– –23 –89
0 9 –92 1–
3 –97 1
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06 Mathematics
natural numbers, in other words the infinity of these points is bigger than ϗ0. Because
of the infinite way in which the set is constructed, it is clear that if we look at a small
part of the Cantor set under a microscope it looks like the whole set. In other words,
there is a one–one correspondence between the whole set and itself that preserves the
structure of the set. We say that the Cantor set is self-similar. All fractals possess the
property of self-similarity.
A 3-D Cantor set of a cube shows clearly how the middle third is deleted from a cube in
all three dimensions. What is left over is a set of scattered points. The self-similarity of
the original figure to one of its smaller ‘sub-cubes’ is clear.
3-D Cantor set of a cube.
Using similar repeated operations, we can make two-dimensional self-similar figures
or fractals. The second row in the picture of 2-D Cantor sets shows how to produce the
famous Koch snowflake. Notice how similar these figures are to the original Cantor set.
2-D Cantor sets.
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Among others, the French–American mathematician Benoit Mandelbrot (1924–2010)
brought the stunning beauty of these interesting sets to the attention of a wider public
in the mid 1980s.
A Mandelbrot set.
Mathematics, then, is based on some important and simple ideas. Sets and mappings
are the building blocks. We need to add a bit more structure to a set to allow us to
perform the usual algebraic operations on it. The resulting structure we met earlier
is called a group, and is arguably the most important building block in mathematics.
These simple ideas are highly abstract and cannot really be found in the pure state in
the world outside mathematics.
6.4 Methodology
Knowledge framework: Methodology – What are the
methods or procedures used in this area and what is it about
these methods that generates knowledge?
Knowledge framework: Methodology – What are the
assumptions underlying these methods?
Proof
Now we have explored some of the basic raw materials of mathematics, we need to
take a look at what mathematicians do with them in order to produce knowledge. The
starting point for this investigation is the method that is at the centre of mathematical
activity: proof.
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06 Mathematics
Proof is a formal method that establishes the truth of statements in mathematics. The
main idea is simple: once a statement has been proved we can say that it is true, that it
has always been true and will always be true. Mathematical proof is the guarantee of
the sort of timeless truth that Plato envisaged was characteristic of all AOKs. What we
shall discover is that, because of its special subject matter and methods, mathematics
might be unique among our AOKs in this respect.
Knowledge questions
1 What it is about the methods of mathematics that give us this cast-iron guarantee that its results
are absolutely certain and that there is not a glimmer of doubt attached to them?
2 Is it irrational to doubt a mathematical statement that has been proved satisfactorily?
One way of thinking about mathematics is that it is a big abstract game played by
human beings according to invented rules. This view is called formalism. In a sense
mathematics is just like the eponymous glass bead game in the novel by German writer
Hermann Hesse (1877–1962). Interestingly, the hero of the novel has to learn music,
mathematics, and cultural history in order to play the game. There are very strong
connections between mathematics and music.
Let us then imagine that mathematics is like a game. As with a game, mathematical
proof has to start somewhere. Mathematical proof starts with a collection of
statements called axioms. These statements are in mathematical language (the
language of sets, for example). They themselves cannot be proved. They are simply
taken as self-evidently true. Mathematicians are free to choose the axioms that suit
their purposes up to a point. Choosing a different set of axioms can lead to a different
sort of mathematics (although it doesn’t have to – there are equivalent ways of
choosing axioms for mathematical systems).
Once the game is set up, we can start playing but only in sequence and according to
the rules of the game. The same is true of a mathematical proof. One applies the rules
(these are typically the rules of algebra or, more generally, rules of logic) to a line in the
proof to get the next line. The whole proof is a chain of such moves.
Finally, the game ends. Similarly, a mathematical proof has an end. This is a point
where the statement that was to be proved is finally derived at the end of the chain of
reasoning. This result is called a theorem (Figure 6.11).
Exercises
The word theorem only
applies to mathematical Consider this proof. It says that the result of adding two odd numbers is always an even number.
or logical statements. It Theorem: Let x and y be odd numbers. Then x + y is an even number.
cannot be used outside Proof: Because x is an odd number, it can be written as an even number plus 1. An even number can be
mathematics or logic. written as 2n for some integer n.
The human and natural So we can write: x = 2n + 1
sciences hardly use the
Similarly: y = 2m + 1 for some integer m.
word at all. It might
be used in economics So: x + y = (2n + 1) + (2m + 1)
but only when the = 2n + 2m +2
result is proved using = 2(n + m) + 2
mathematics. The same = 2(n + m + 1)
is true in physics or But (n + m + 1) is an integer (because n and m are integers) so x + y is an even number, QED (quad erat
chemistry. demonstrandum – which was to be shown).
12 Why do we use different letters n and m to construct the odd numbers in the proof?
13 What is the advantage in using letters to stand for general numbers?
14 What properties of addition and multiplication of whole numbers does your proof depend on?
(These are the axioms of the proof.)
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It is possible to illustrate the idea of proof without actually
using mathematics (but still doing mathematical thinking). Axioms
We owe the example below to the English mathematician
application of logic (rules of algebra)
Ian Stewart (1995) in his book Nature’s Numbers (who in turn
borrowed it from a colleague at Warwick University). step 1
Start off with the four-letter English word SHIP. Write down application of logic (rules of algebra)
a vertical list of four-letter English words. Each word differs
from the one above it by changing only one letter. The aim is step 2
to arrive at the word DOCK.
application of logic (rules of algebra)
SHIP
step 3
SHOP
application of logic (rules of algebra)
Theorem
DOCK
Before reading on, try to solve the puzzle. It will help with
understanding what comes next. Remember that you can only change one letter per Figure 6.11 A model of a
move. The other letters have to be left alone; you are not allowed to rearrange them. mathematical proof.
For example, you might want to write SHOP as the second word.
Exercise
15 While you are trying to solve the SHIP to DOCK puzzle, be aware of your thought processes.
a What strategies are you using?
b Have you tried working both forwards and backwards?
c What about the problem getting them to meet in the middle?
d Do you think your emotions play a part in this process?
f How do you feel when you have solved it?
g How do you feel not being able to solve it?
Although this is not a mathematical proof, it uses similar methods.
Did you feel that the problem lay at the centre of the ‘proof’? Perhaps you tried to solve
this by starting at the end and working backwards and then got into trouble in the
middle. Let’s consider two solutions.
SHIP SHIP
SHOP SLIP
SHOT SLAP
SOOT SOAP
and
LOOT SOAK
LOOK SOCK
LOCK DOCK
DOCK
Did you arrive at either of these solutions? In fact, there are many solutions, but seven
steps is thought to be the shortest. If we allow words to repeat (and there was nothing
said in the rules that this was not allowed) then there is an infinite set of solutions.
Notice that in the centre of our ‘proof’, the keystone of the bridge is a word with two
vowels in it.
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06 Mathematics
We might be tempted to make a conjecture – a mathematical statement that has not
yet been proven. Let us conjecture that every solution of the puzzle requires at least
one word with two vowels in the middle somewhere. Let us see if we can prove it. We
might also notice that the vowel in the word SHIP is in the third position while the
vowel in the word DOCK is in the second position. The puzzle requires us to ‘move’
the vowel from the third position to the second position. Maybe at this point there
is a glimmer of light in the darkness. Reformulating the problem in this way makes
it clearer to see what we have to do to solve the original problem, but also how we
might prove the conjecture. We need one fact about English words: every English
word contains at least one vowel (counting ‘y’ as a vowel in certain circumstances).
Remember that we are only allowed to change one letter at a time. This means that
in order to move the vowel from the third position to the second position we have
to write a word with two vowels (we can’t write a word with no vowels – English
grammar doesn’t allow words with no vowels). So, every one of the infinite set of
solutions to this puzzle has to contain at least one word with two vowels. We have
proved the conjecture and it becomes a fully fledged theorem. Given the rules of the
game, this result is certain – it is not just likely, it has to be true.
While there might be something attractive about the idea of mathematical proof
(starting with axioms and using a finite number of rules to deduce each line from the
preceding line, thus finally getting to the result), you might be rightly suspicious. A
number of questions can be raised at this point.
•How can we choose the axioms just like that?
•On what basis do we make that choice?
•Surely if we choose a sufficiently different set of axioms, we get a different mathematics?
•How do we know that the axioms we choose will yield the sort of mathematical
results that we intuitively expect?
•How do we know that the axioms fit with each other and don’t produce
•Shouldn’t the axioms be true or false in relation to the world?
These questions capture both the essence of mathematics and some of its central
knowledge questions. Truth in mathematics is something that is conventionally
established. We might even regard it as a social truth in some sense (see the views of
Reuben Hersh, below). It is precisely this lack of reference to the outside world that
gives mathematics its absolute certainty but also its slightly other-worldly quality.
Prescribed essay title 2: Mathematicians have the concept of rigorous proof,
which leads to knowing something with complete certainty. Consider the
extent to which complete certainty might be achievable in mathematics and
at least one other area of knowledge.
© International Baccalaureate, November 2007, May 2008
Euclidian geometry
To explore these questions let us examine one of the first axiomatic systems ever
produced. This is the system of axioms and derivation rules set up by Euclid in the
3rd century BC in his book Elements, in which he proves many important results in
geometry.
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Here are Euclid’s axioms. He called them postulates. Technically, most of them tell us
how to move from one line of a proof to another. In this sense, they are more like rules
of the game of geometry.
1 Any straight line segment can be drawn joining any two points.
2 Any straight line segment can be extended indefinitely in a straight line.
3 Given any straight line segment, a circle can be drawn having the given line segment
as radius and one endpoint as centre.
4 All right angles are congruent.
5 If two lines are drawn, which intersect a third in such a way that the sum of the
inner angles on one side is less than two right angles, then the two lines inevitably
must intersect each other on that side, if extended far enough.
In some sense Euclid’s axioms express mathematical intuitions about the nature of
geometrical objects (in this case: points, line segments, circles, and right angles).
What is clear is that we cannot establish the truth of these postulates by observing the
external world. Objects such as points, lines, and circles do not exist in the real world
as perfect mathematical objects. It is easy to understand why Plato placed these special
objects in the perfect world of the Forms, and not in our messy everyday world.
We could try to use Euclid’s postulates to do some geometry. Let us try to construct an
equilateral triangle on a given line segment. We are only allowed to use a straight edge
and a pair of compasses because Euclid’s postulates only refer to line segments and
circles.
Step 1: Draw two circular arcs of radius AB centred on points A and B (Figure 6.12).
A B
Figure 6.12 Step 1.
Step 2: Label the intersection point of the arcs C. Draw line segments AC and AB to
complete the triangle (Figure 6.13).
C
A B
Figure 6.13 Step 2.
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06 Mathematics
It follows from step 1 that the line segments AC and BC are both equal to AB.
Therefore, they must be equal (this is sometimes quoted as a separate axiom: that
two things equal to the same third thing must be equal; however, it follows from our
modern definition of equality of sets).
With arguments like these, Euclid established a pattern for establishing the truth of
mathematical statements; a pattern that is still used today.
Exercise
16 Consider the following set of axioms:
•A committee is a set of three members.
•Each member is on exactly two committees.
•No two members can be together on more than one committee.
•There is at least one committee.
a What are the primitive ideas in this axiom system?
b Try to construct a situation that satisfies these axioms (this is called a model).
c Is yours the only possible model of these axioms?
Non-Euclidian geometries
Let us look more closely at Euclid’s fifth postulate. It cannot be deduced from the other
postulates (this can be proved). Euclid himself only used the first four axioms in the
first 28 propositions (effectively theorems) of the Elements, but he was forced to use the
A fifth – called the parallel postulate – in the 29th proposition. This is an independent
axiom of the system, and like all axioms, we can choose whether we accept it or not.
If we accept it, we get the familiar geometry of our everyday life on the surface of the
Earth. We can use it to construct buildings, roads, and other familiar objects.
a
In 1823 Janos Bolyai and Nicolai Lobachevsky independently realized that entirely self-
consistent non-Euclidian geometries could be constructed in which the fifth postulate
did not hold – in other words, that parallel lines do not behave in quite the way we
expect from everyday life. There are two typical cases. The first is where parallel lines
A1
converge and actually meet at a point – this is called elliptic geometry (Figure 6.14).
Figure 6.14 Lines parallel at A good example of this is
the equator converge towards long distance travel around
the poles.
the Earth. As you will know
if you have ever taken an
intercontinental flight, the
shortest distance between two
points on the Earth’s surface,
Airline routes show that the such as Stockholm and New
shortest distance between York, is part of a circle (called a
two points on Earth’s surface great circle).
is a curve.
Lines of longitude are great
circles. It is clear that they are
(in a special technical sense)
parallel but it is also clear that
they meet at the poles.
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The second case is where parallel lines diverge and become ever farther apart. This is Figure 6.15 Lines parallel
called hyperbolic geometry. This would be like drawing parallel lines on a saddle at some point on a saddle
diverge continuously.
(Figure 6.15).
What is clear from this discussion is that we can choose the axioms according to what
we want to do with them. Standard Euclidian geometry is useful for describing spaces
we encounter in our local, day-to-day world. But we need elliptic geometry if we are
to fly planes from one place to another on the Earth’s surface, and this requires us
to reject the fifth postulate of Euclid. At the time of writing, hyperbolic geometry is
largely of interest to pure mathematicians but who knows what applications might
spring up asking for a hyperbolic treatment?
The axioms are a little like the rules of a game. Change the rules and you get a different
game. But this leaves us with a problem. It is difficult to see how a game, such as chess,
can tell us anything about the world.
Knowledge question
3 If the rules are made up, how could they apply usefully to mapping the world? How can
mathematics, conceived as built on a set of human-made axioms, have anything useful to say
A very powerful technique used by mathematicians is the idea of proof by
contradiction or, to give it its Latin name, reductio ad absurdum. If we want to prove
a proposition P, then we start by assuming (not P). We then need to show that this
leads to a contradiction. If we assume that the axiom system that we use to derive the
contradiction is consistent (this is an interesting assumption, see Gödel’s theorems
(page 202)) then we can only conclude that our original assumption (not P) is false, and Figure 6.16 A model of
the proposition P is true (Figure 6.16). proof by contradiction.
One of the most elegant examples of this method was the
proof by Euclid that there are infinitely many prime numbers
(i.e. there is not a largest prime number). The proof relies on Assume not P
his fundamental theorem of arithmetic. This states that every application of logic (rules of algebra)
number can be expressed uniquely as a product of prime
numbers. step 1
Armed with this, we can attack our target. But remember, application of logic (rules of algebra)
proof by contradiction is based on an indirect method of attack.
We assume the contrary of what we want to prove, thus here step 2
we assume that there is a finite number prime numbers, say n application of logic (rules of algebra)
of them: p1, p2, …, pn where pn is the largest prime number.
step 3
Now, let’s form the product of all of these prime numbers and
add 1: application of logic (rules of algebra)
q = p1p2 ... pn + 1 Contradiction
There are just two possibilities: either q is prime or q is not
prime. If q is prime, then it contradicts the assumption that pn
is the largest prime because clearly q is greater than pn. P must be true
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06 Mathematics
If q is not prime, then by the fundamental theorem of arithmetic, q has prime factors.
So we must ask ourselves what they are. We can see that they cannot be any of our
list of primes p1 p2 ... pn because when we divide q by any of these we get a remainder
of 1 (by the definition of q). (Remember that when a division by p yields a remainder it
means that p is not a factor.) So this means that the prime factors of q are not in the list.
But this contradicts the original assumption that the original list of prime numbers is
complete.
Either way we end up with a contradiction. Hence the original assumption must be
false. Therefore, there are infinitely many prime numbers.
Later, we shall see that there are a number of deep problems associated with the idea
of proof in mathematics. For now, let us look at another important method used by
mathematicians, modelling.
Prescribed essay title 3: When mathematicians, historians and scientists say
that they have explained something, are they using the word explain in the
same way?
© International Baccalaureate, November 2006, May 2007
Knowledge framework: Methodology – What role do
models play in this area of knowledge?
Models
We have already discussed an example of a mathematical model on pages 177–178.
There we used a mathematical model to describe a physical situation – the motion of a
stone dropped down a well shaft. The mathematical model was a simplified map of the
situation constructed using algebraic equations to represent the position and velocity
of the stone. The basic idea of representing a situation in some ideal way is central to
mathematics. It is not only used in connection with modelling situations drawn from
the outside world – mathematicians often use one type of mathematical object to
represent a mathematical situation of another type.
The following example will illustrate the method. It is a famous mathematical puzzle.
Try to solve it yourself before you look at the mathematics of the situation.
The wolf, goat, cabbage problem
A ferryman has to ferry a wolf, a goat, and a cabbage across a river. (As always in these
problems, we do not know why he would want to do this.) Unfortunately, the boat
will only hold the ferryman and one other life-form (wolf, goat, or cabbage) per trip.
Clearly the ferryman will have to make a number of journeys. Also there is a problem.
If left unsupervised, the wolf would eat the goat, or the goat would eat the cabbage.
The wolf, however, is not at all interested in the cabbage. What strategy should the
ferryman use to ferry the life-forms across the river? Try to solve the problem yourself
The problem is a logical one. It fits the idea of a game defined by a set of rules, an
opening position, and a final winning position. One way to solve it would be to look
at the permitted moves and find a combination of these that leads to the required
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conclusion. But the problem here is knowing what strategy would lead to the required
conclusion.
•What method did you use to solve the problem?
•Did you use trial and error to find the solution?
Mathematicians have a number of strategies for dealing with problems like this. One
method is to transform the problem into one we can picture – a geometrical problem.
Then we can use our geometrical intuitions to solve it. But this problem does not
seem to be at all geometrical. This is where imagination and creativity play a big role
in mathematics. The trick here is to represent the situation of the position of the wolf,
goat and cabbage as being a single point in three-dimensional space. We sometimes
call a notional space like this a phase space.
We can use a series three numbers to represent the positions of the wolf, goat, and
cabbage (in that order). If the number (coordinate) corresponding to a particular life- Figure 6.17 Geometrical
representation of the wolf,
form is 0, then it means that this life-form is on the near bank of the river. If 1, then the
goat, cabbage problem.
life-form is on the far bank.
So (1, 1, 0) represents the situation in which the wolf and Finish (1, 1, 1) (1, 0, 1)
the goat are on the far bank and the cabbage is on the near
bank; while (0, 1, 0) represents the situation in which the
wolf and cabbage are on the near bank, and the goat is on
the far bank. (0, 1, 1) (0, 0, 1)
So the set of all combinations of the life-forms forms the
vertices of a cube in 3-dimensional space (Figure 6.17). C-axis
(1, 1, 0)
We can only move one life-form at a time, and available (1, 0, 0)
moves are along the edges of the cube. So, the original
problem reduces to a geometrical one about getting from W-axis
the start where each life-form is on the near bank (0, 0, 0)
(0, 1, 0)
to the finish when they all end up on the far bank (1, 1, 1). G-axis Start (0, 0, 0)
But there is a complication: certain combinations are
not permitted. We cannot leave the wolf and the goat Figure 6.18 Constraints of
unattended on the same bank while we transport the the problem translated into
geometrical conditions.
cabbage, and we cannot leave the goat and the cabbage
unattended while we transport the wolf.
Finish (1, 1, 1)
These conditions translate into illegal moves between (0, 0, 0)
and (0, 0, 1), or (1, 1, 0) and (1, 1, 1), when the wolf and goat
are unattended. Similarly, there are illegal moves between
(0, 0, 0) and (1, 0, 0), or (0, 1, 1) and (1, 1, 1), when the goat
and cabbage are unattended. In geometrical terms, these
prohibited moves correspond to four edges of the cube that
C-axis
are ruled out of play – these are coloured red in Figure 6.18.
The standard solution of the problem is now clear and is
indicated by the green line (Figure 6.19, overleaf). W-axis
However, what is interesting about the geometrical
representation is that it indicates a second solution Start (0, 0, 0) G-axis
(Figure 6.20, overleaf).
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06 Mathematics
Finish (1, 1, 1) Finish (1, 1, 1)
C-axis C-axis
W-axis W-axis
G-axis Start (0, 0, 0) G-axis Start (0, 0, 0)
Figure 6.19 The solution is indicated by the Figure 6.20 A second solution is indicated
green line. by the new green line.
As you can see from this example, the modelling process seems to follow the following
steps.
1 Representation of the problem in some abstract space.
2 Solve the problem using methods appropriate to that space.
3 Translate the answer back to the original situation.
This process is exactly the same as that used for making models in the natural and
human sciences (Chapter 4, pages 121–122; Chapter 5, page 158). But it is also very
similar to the process of making a map, which is the basic metaphor for knowledge in
this book and in the TOK subject guide.
6.5 Historical development
Knowledge framework: Historical development – What is
the significance of the key points in the historical development
of this area of knowledge?
Knowledge framework: Historical development – How has
the history of this area led to its current form?
A constructivist view of mathematics
The history of mathematics is important if we are to think of it as a human invention:
an elaborate game invented and played by human beings. This is commonly known as
a constructivist view of mathematics. We have acknowledged this raises a serious
knowledge question: if mathematics is purely an invention of human beings why is
it so useful in describing the (non-human) world? Why is mathematics important
when it comes to building bridges and constructing buildings, doing physics, doing
chemistry, even doing economics?
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In this view, mathematics is what might be called a social fact. This is a term coined
by, among others, the philosopher John Searle (1932–) in his book The Construction of
Social Reality. In it, he argues that there are many facts in our lives that are true because
society stipulates them to be so. He uses money as his central example. Money is
money because we believe it is money. It has value because of a whole set of social
agreements to use it as such. As soon as these agreements break down, money loses
its value. This idea is explored further in Chapter 8, but for now it is important to
note that social facts are no less real or definite than those about the natural world.
Marriage is another example of a social fact. It is clear that the statement John is married
is definitely either true or false.
Hersh and the humanist view
The American mathematician Reuben Hersh argues for a type of constructivism that
he calls humanist mathematics. He regards numbers (and any other mathematical Reuben Hersh (1927–).
objects) as constructed social facts. Here he is defending this view on the Edge website:
[Mathematics] … is neither physical nor mental, it’s social. It’s part of culture, it’s
part of history, it’s like law, like religion, like money, like all those very real things,
which are real only as part of collective human consciousness. Being part of
society and culture, it’s both internal and external: internal to society and culture
as a whole, external to the individual, who has to learn it from books and in school.
That’s what math is.
This view suggests a strong link between the shared and personal aspects of
mathematics that will be explored later.
Hersh called his theory of mathematics humanism because it’s saying that
mathematics is something human. There’s no mathematics without people. On the
same website, he says:
Many people think that ellipses and numbers and so on are there whether or not
any people know about them; I think that’s a confusion.
Hersh admits that we do use numbers to describe physical reality and that this seems
to contradict the idea that numbers are a social construction. But then he argues that
we use numbers in two distinct ways: as nouns and as adjectives. When we say nine
If it’s an objective fact that there are nine apples on the table, that’s just as objective
as the fact that the apples are red, or that they’re ripe or anything else about them,
that’s a fact.
The problem occurs when we make a subconscious switch to nine as an abstract noun
in the sort of problems we deal with in mathematics class. Hersh thinks that this is not
really the same nine. They are connected, but the number nine is an abstract object as
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06 Mathematics
part of a number system. It is a result of our maths game – our deduction from axioms.
It is a human creation.
Hersh sees a political and pedagogical dimension to his thinking about mathematics.
He thinks that a humanistic vision of mathematics chimes in with more progressive
politics. How can politics enter mathematics? As soon as we think of mathematics as
a social construction, then the exact arrangements by which this comes about – the
institutions that build and maintain it – become important. These arrangements are
political. Particularly interesting for us here is how changes in mathematics teaching
and learning could be suggested by the approach taken by Hersh.
Let me state three possible philosophical attitudes towards mathematics. Platonism
says mathematics is about some abstract entities, which are independent of
humanity. Formalism says mathematics is nothing but calculations. There is no
meaning to it at all. You just come out with the right answer by following the rules.
Humanism sees mathematics as part of human culture and human history. It’s
obvious that Platonism and Formalism are anti-educational, and interfere with
understanding, and Humanism at least doesn’t hurt and could be beneficial.
Formalism is connected with rote, the traditional method, which is still common in
many parts of the world. Here’s an algorithm; practice it for a while; now here’s
another one. That’s certainly what makes a lot of people hate mathematics. (I don’t
mean that mathematicians who are formalists advocate teaching by rote. But the
formalist conception of mathematics fits naturally with the rote method of
instruction.) There are various kinds of Platonists. Some are good teachers, some
are bad. But the Platonist idea, that, as my friend Phil Davis puts it, Pi is in the sky,
helps to make mathematics intimidating and remote. It can be an excuse for a
pupil’s failure to learn, or for a teacher’s saying ‘some people just don’t get it’. The
humanistic philosophy brings mathematics down to earth, makes it accessible
psychologically, and increases the likelihood that someone can learn it, because it’s
just one of the things that people do.
There is a possibility that the arguments explored in this section might cast light
on an aspect of mathematics learning which has seemed puzzling – why it is that
mathematical ability is seen to be closely correlated with a certain type of intelligence?
Moreover, it seems to polarize society into two distinct groups: those who can do it
and those who cannot. Those who cannot, often feel either the stigma of failure or
that there is an exclusive club whose membership they have been denied. On the other
hand, those who can do it often find themselves labelled as ‘nerds’ or as people who
are, in some serious sense, socially deficient. Whether this is something that should
be attributed to a formalist or Platonic view is another question. If mathematics is
out there to be discovered, it does seem reasonable to imagine that the particular
individual who does not make the discovery might experience a sense of failure. The
interesting question in this case is what consequences in the classroom would follow
from a humanist view of mathematics.
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17 Is Hersh right in attributing the polarization of society into maths-friendly sociopaths and Hersh and his ideas, visit
mathematical illiterates to a formalistic or Platonic view? pearsonhotlinks.com,
enter the title or ISBN
18 Is Hersh right to suggest that if maths is just a meaningless set of formal exercises, then it will not
of this book and select
be valued in the main by society?
19 Do you think mathematics textbooks reflect the way we actually do mathematics?
The Platonic view
There is another view of mathematics that sidesteps the awkward question of why
mathematics applies so well to the world. It is simple: mathematics is already out there
in the world. It has an existence that is independent of human beings. All we do (as we
do with other things in the natural world) is to discover it. Of course, mathematics has
a lot to say about the world external to human beings because it is part of it.
Plato was committed to a view that mathematics existed independently of human
beings. Indeed the objects of mathematics were just examples of the many perfect
inhabitants of his world of Ideas. Mathematical objects such as a perfect circle or
perfect triangle or the number 2 exist in this world of ideas while only inferior
imperfect versions exist here on Earth. Most mathematicians have at least some
sympathy with this view. They talk as though they are accountable to mathematical
truths in the same way as we are accountable to physical facts about the universe. They
feel that there really is a mathematical world out there and all we are trying to do is
This view is itself not entirely without problems. The big knowledge question that could
be asked here is, if maths is out there in the world, where is it? We do not see circles,
triangles, √2, π, i, e and other mathematical objects obviously floating around in the world.
We have to do a great deal of work to find them through inference and abstraction.
Perhaps the first place we should look for mathematics is in the natural living world.
Mathematics in nature
Despite these problems, there are some valiant attempts to show that hidden not too
far below the surface of our reality we can find mathematics.
Prime numbers
We saw earlier that there were an infinite number of prime numbers and that they
were, in some sense, the building blocks of the whole number system. The Platonist
might want to try to find them somewhere in nature.
One place where the Platonist might start is in Tennessee. In the summer of 2011, the
forests were alive with a cicada that exploits a property of prime numbers for its own
survival. These cicadas have a curious life cycle. They stay in the ground for 13 years.
Then they emerge and enjoy a relatively brief period courting and mating before laying
eggs in the ground and dying. The bizarre cycle is then repeated. There is another batch
of cicada that has the same cycle and no less than 12 that have a cycle of 17 years. There
are, however, none that have cycles of 12, 14, 15, 16 or 18 years.
This is a somewhat puzzling situation.We need to find some explanation for this
strange cycle.
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06 Mathematics
The cicadas are preyed on by birds, so let’s consider the position of the predators. For
obvious reasons, predators might be expected to have somewhat similar life cycles to
their prey. But that is unlikely with the insect prey having such long life cycles. If the
birds had, say, a two-year cycle, then the huge number of birds that would result
from the flock gorging of the wave of cicadas at year 0 would starve at year 2
because the cicada wave did not make a second appearance. The birds
would die and there would be no repeating pattern of population waves.
The birds could only be waiting in huge flocks for the insects to appear
if their life cycle were a multiple of that of the cicada. But 13 and 17
are prime numbers – divisible only by themselves and 1. This is what
guarantees there being no resonance between the predator and the prey.
Fibonacci sequence
Some theorists such as Nicholaas Rupke (2009) suggest that the mathematical forms
permitted by the basic material of living things limit the sort of structures that might
be available for natural selection. Therefore, the emphasis might not be on the actual
mechanisms of heredity so much as on what mathematics allows to be available to the
For example, many spirals in nature are formed from the sequence:
1, 1, 2, 3, 5, 8, 13, 21, 34, … (Figure 6.21)
This is called the Fibonacci sequence after the Italian mathematician Leonardo Pisano
Bigolio (1170–1250), known as Fibonacci.
Many plants have leaves that are arranged according to Fibonacci sequences.
The ratios of successive Fibonacci numbers approximate to the so-called golden ratio.
This is represented by φ (Greek letter phi) and is equal to 0.61803398875… (Table 6.1).
Figure 6.21 A spiral formed
according to Fibonacci
sequence.
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Table 6.1 Ratios of
Fibonacci Ratio of Fibonacci number The beautiful Romanesco
Fibonacci numbers.
numbers to succeeding number cauliflower is formed on
Fibonacci spirals.
1
1 =1
1
1
1 = 0.500
2
2
2 = 0.666
3
3
3 = 0.600
5
5
5 = 0.625
8
8
8 = 0.615
13
13
13 = 0.619
21
21
21 = 0.6176
34
34
34 = 0.61818
55
55
55 = 0.61797
89
89
89 = 0.61805
144
144
144 = 0.61802
233
233
This number also crops up a lot in discussions about mathematics in art and music.
There is an alternative formulation of these ideas which is arrived at by dividing each
number in the Fibonacci sequence by the number preceding it instead of the number
succeeding it. In this case, φ is equal to 1.61803398875…
Figure 6.22 A diagram
We can observe that plants grow new leaves at a constant angle to the previous leaf – showing how mathematics
this is called the divergence angle and is usually close to 137.5° (Figure 6.22). Why is enables plants to grow so that
lower leaves are not in the
this? It is an arrangement of the leaves in which lower leaves are not in the shadow of
higher ones.
What is interesting mathematically is that the full circle
around a stem is 360°; and 360 divided by φ (when φ is equal 4
1
to 1.618) is 222.5; 360° – 222.5° = 137.5°, which is the angle of
divergence we observe. 9
There are many other examples of Fibonacci sequences in 12 7
nature and examples of the significance of φ. 6
The golden ratio φ is the most irrational number there is – 137.5°
137.5°
that is, it is the number least able to be approximated by a 2
11 137.5°
fraction. 10
3
8 5
0
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06 Mathematics
Exercise
CHALLENGE 20 Discover whether you are a constructivist or a Platonist.
YOURSELF Consider the decimal expansion of π = 3.141592653589793… The decimal goes on forever without
terminating or repeating.
Use an internet resource to
find out all you can about Now think about these two questions:
Gödel’s Incompleteness a Is there a point in this number where there is a sequence of digits that matches your telephone
Theorems. In the first number?
theorem, Gödel showed b Is there a definite answer to (a)?
that we could not prove that
We are led to believe that most questions of fact have a definite answer one way or another. If your
mathematics is complete telephone number is 31 41 59 26 then the answer is definite – my telephone number is in the decimal
and consistent. The second expansion of π. This would also be true if you found it embedded at, say, the millionth decimal place.
theorem is even more But what if you can’t find it (which is much more likely)? Is there still a definite answer?
worrying: that any system
If you are a Platonist, you believe that there is a definite answer – mathematics is out there – it is just
that is rich enough to prove
that we cannot find it.
its own consistency is
inconsistent! If you are a constructivist, you think there is no definite answer. Things that have not yet been
constructed do not have some sort of mysterious independent existence.
1 What is meant by a
formal system being
consistent? The history of mathematics has demonstrated that since proof is central to
2 What is meant by a mathematical truth, the (proved) failure to be able to prove its own consistency is a
formal system being
huge setback. Thinking of mathematics historically has emphasized its human origins.
complete?
At the same time the ubiquity of mathematics in nature and it sheer usefulness in
3 What implication does
Gödel’s first theorem have tackling problems in the world outside mathematics suggest that it is more than just a
regarding the method of fun game played by human beings for their own entertainment.
4 In what way might
Gödel’s theorems
6.6
undermine mathematics’
claim to absolute
certainty? knowledge
in an AOK problematic?
The following story might Knowledge framework: Links to personal knowledge –
help in the answer: Why is this area significant to the individual?
One day Bertrand
Russell was teaching a Knowledge framework: Links to personal knowledge –
student this and pointing
out that 2 = 3 is such
What is the nature of the contribution of individuals to this
student then challenged
Russell, ‘Since you can In this section, we examine how mathematics impinges on our personal thinking
prove anything from a about the world and also examine how people with particular qualities are well placed
to make important contributions to mathematics. Perhaps surprisingly, one of the
prove that if 2 = 3 then
you are the Archbishop aspects of personal knowledge with the strongest link to mathematics is the aesthetic.
of Canterbury?’ Quick as
a flash Russell is supposed There seems to be a two-way connection between mathematics and beauty. The first
to have answered, ‘If 2 = 3 is the long held view that we find certain things beautiful because of their special
then subtract one from
both sides so we have
proportions or some other intrinsic mathematical feature. This is the thinking that
1 = 2. Consider the set of has inspired architects since the times of ancient Egypt and generations of painters,
me and the Archbishop of sculptors, musicians, and writers. To paraphrase Keats, truth is beautiful. Things are
Canterbury. This set has 2
beautiful because of specific objective relations between their parts as demonstrated by
members. Since 2 = 1, it
has one member. Which certain mathematical truths. As we shall see when discussing the arts in Chapter 8, this
means the Archbishop of is a view held by the thinkers of the aesthetic movement in the 19th century and many
Canterbury and I are the others.
same person.’
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But Keats also put it the other way round: the beautiful is true. Could we allow
ourselves to be guided to truth in mathematics because of the beauty of the equations?
This is a position taken by, surprisingly, many mathematicians. They look for beauty
and elegance as an indicator of truth.
Let’s explore both approaches.
Mathematics in beauty
The ancient Greeks were well aware that underlying the world was a coherent
structure. If something was beautiful there was a good reason for it. By around the late
4th century BCE, it was well established that the reasons underlying the nature of most
things in the world were mathematical. Beauty was not an exception.
We have already mentioned the number φ, phi the golden ratio or golden section. We
have seen how this is the limit of the ratio of the terms in the Fibonacci sequence. The
Greeks were interested in this number for other reasons. Euclid divided a line segment
AB of length 1 by placing a point X in such a way as to make the ratio of the shorter
piece to the longer piece the same as the ratio of the longer piece to the whole line
(Figure 6.23).
Figure 6.23 Beautiful linear
A X B
proportions
In other words:
XB/AX = AX/AB
If AB = 1 (as defined above), and we write AX = x, then
(1 – x) x
=
x 1
Rearranging:
1 – x = x2
x2 + x – 1 = 0
Solving the equation using the quadratic formula gives:
– 1 + M5
x= = 0.61803398875… = φ
2
or
– 1 + M5
x= = –1.61803398875 = – φ – 1
2
So the golden ratio turns up here as well.
The original question was one of proportional symmetry so we might not be surprised
to learn that the Greeks thought we would find this proportion naturally beautiful.
Rectangles in which the ratio of the shorter side to the longer side was equal to the
golden ratio were thought to be especially beautiful. Try this out yourself in the
rectangle beauty contest (Figure 6.24, overleaf).
203
M06_TOK_SB_IBDIP_4157_U06.indd 203 04/03/2014 14:12
06 Mathematics
B
A
Figure 6.24 Rectangle D
preferences: Which rectangle
E
do you find most pleasing?
C
A4 paper has dimensions of 210 mm × 297 mm. The ration of the sides is given by
210/297 = 0.707, which is a little high for the golden ratio. A4 paper is a little too
wide to be a golden rectangle. Measure some rectangles in your school or home
environment. How close are they to golden rectangles?
These ratios were exploited by the Greeks in their designs for temples and other
buildings. The Parthenon in Athens is constructed using the golden section at key
points (Figure 6.25).
Figure 6.25 How the
golden ratio is used in the
Parthenon.
j
j j
j
j
j j
204
M06_TOK_SB_IBDIP_4157_U06.indd 204 04/03/2014 14:12
But it is not only in nature and architecture that we can find the golden ratio.
It is widely used by painters and composers – and has been since the time of
Euclid. The Dutch painter Piet Mondrian uses it quite self-consciously in his
work.
We find the ratio also in the work of the French composer Claude Debussy.
In his orchestral piece La Mer, the 55-bar introduction to ‘Dialogue du vent et
la mer’ breaks down into five sections of 21, 8, 8, 5, and 13 bars which are all
Fibonacci numbers. The golden ratio point at bar 34 is signalled by the entry
of trombones and percussion.
There are similar occurrences in the work of Mozart and Chopin.
Composition in red, yellow
and blue (1926) by Piet
Mondrian.
Golden sections in Mozart
Piano sonata in B K.281.
More generally, we can ask ourselves how many pieces of music (or films, plays,
or dance performances) have some sort of structurally significant event occurring
roughly two-thirds of the way through the piece.
But the golden section is not the only way in which mathematics makes itself known
in the arts. A recent discovery has shown that the original papyrus copies of Plato’s
Republic give indications that Plato structured his great work according to very definite
harmonic relationships (again to do with proportions). Mathematics not only formed
the basis of his world view, it also structured his exposition of it.
It comes as something of a surprise to most people to realize how much mathematics
goes into constructing a piece of modern classical music. Since the second Viennese
school of Schoenberg, Berg, and Webern at the beginning of the 20th century,
composers have used, among other things, variations of serialism. This is a technique
that applies mathematical transformations to rows of notes. These transformations
form mathematical structures we have met before – groups (Table 6.2).
P = prime = leave sequence as it is Table 6.2 A group table
P R I RI of the four types of serial
R = retrograde = play sequence backwards transformation.
P P R I RI
I = inversion = play sequence upside- R R P RI I
down (ascending intervals are replaced by
I I RI P R
descending ones and vice versa)
RI RI I R P
RI = retrograde inversion = play sequence
backwards and upside-down
205
M06_TOK_SB_IBDIP_4157_U06.indd 205 04/03/2014 14:12
06 Mathematics
This view of mathematics would suggest that beauty is not so much in the eye of the
beholder but that it is in the mathematics.
Beauty in mathematics
Einstein suggested that the most incomprehensible thing about the universe was that
it was comprehensible. From a TOK point of view, the most incomprehensible thing
about the universe is that it is comprehensible in the language of mathematics. In 1623,
Galileo wrote:
Philosophy is written in this grand book, the universe … It is written in the
language of mathematics, and its characters are triangles, circles and other
geometric figures …
Perhaps what is more puzzling is not just that we can describe the universe in
mathematical terms – but that the mathematics we need to do this is mostly simple,
elegant, and even beautiful.
To illustrate this, let’s look at some of the famous equations of physics. Most people
will be familiar with some of the following equations:
Relation between force and acceleration: F = ma (more generally, F = d/dt(mv))
Gravitational force between two bodies: F = Gm1m1/r2
Energy of rest mass: E = mc2
Kinetic energy of a moving body: E = ½ mv2
Electrostatic force between two charges: F = kq1q2/r2
Maxwell’s equations: . E =
o
. B = 0
B
×E=−
t
E )
× B = o (1 + o
t
Einstein’s field equations for general relativity: Rμv – ½ gμv = 8πTμv
It is extraordinary that the whole crazy complex universe can be described by such
simple, elegant, and even beautiful equations. It seems that our mathematics fits the
universe well. It is difficult to believe that mathematics is just a mind game that we
humans have invented.
But the argument from simplicity and beauty goes further. Symmetry in the
underlying algebra led mathematical physicists to propose the existence of new
fundamental particles, which were subsequently discovered. In some cases, beauty and
elegance of the mathematical description have even been used as evidence of truth. In
1963, the physicist Paul Dirac noted:
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M06_TOK_SB_IBDIP_4157_U06.indd 206 04/03/2014 14:12
Paul Dirac, theoretical
physicist (1902–84).
It seems that if one is working from the point of view of getting beauty in one’s
equations, and if one has really a sound insight, one is on a sure line of
progress.
Dirac’s own equation for the electron must rate as one of the most profoundly
beautiful of all. Its beauty lies in the extraordinary neatness of the underlying
mathematics – it all seems to fit so perfectly together:
( )
3
βmc +
2 ∑ αkpkc (x, t) = i
t
(x, t)
k=1
The physicist and mathematician Palle Jorgensen noted in 1984 that:
[Dirac] … liked to use his equation for the electron as an example stressing
that he was led to it by paying attention to the beauty of the math, more than
to the physics experiments.
It was because of the structure of the mathematics, in particular that there were two
symmetrical parts to the equation – one representing a negatively charged particle (the
electron) and the other a similar particle but with a positive charge – that scientists
were led to the discovery of the positron. It seems fair to say that the mathematics did
really come first here.
We shall leave the last word on this subject to Dirac himself, writing in Scientific
American in 1963:
I think there is a moral to this story, namely that it is more important to have
beauty in one’s equations than to have them fit experiment.
207
M06_TOK_SB_IBDIP_4157_U06.indd 207 04/03/2014 14:12
06 Mathematics
By any standards this is an extraordinary statement for a mathematical physicist to make.
Exercises
21 What equations in your science courses do you find beautiful?
22 What is it about them that is beautiful to you?
23 Do you ever let the beauty (or perhaps the opposite) of an equation influence you when solving a
mathematical problem?
Knowledge framework: Links to personal knowledge –
What are the implications of this area of knowledge for one’s
own individual perspective?
Knowledge framework: Links to personal knowledge –
What assumptions underlie the individual’s own approach to
this knowledge?
Mathematics and personal intuitions
We suggested earlier that there is a two-way street between the area of shared
knowledge in mathematics and our personal knowledge. In this section, we shall
examine a situation that suggests that our intuitions can let us down badly when it
comes to making judgements of probability. From a personal point of view we might
have to correct our intuitions in the light of shared mathematical knowledge.
Mathematical intuition: The Monty Hall game
This scenario refers to a TV game show. A contestant in the show is shown three closed
doors and is told (truthfully) by the game show host Monty Hall that behind one of the
doors is a luxury sports car, and behind the other two doors are goats. The contestant
is told he or she must pick a door and will then be allowed to take home whatever is
behind the door. We assume that the contestant would prefer to win the car. When the
contestant has picked a door, Monty Hall opens another door to reveal a goat (Monty
Hall always chooses a door concealing a goat – he knows the location of the car). He
then asks the competitor whether or not he or she wants to switch their choice to the
other closed door.
What does your intuition tell you? Should the contestant stick to the original choice or
switch?
Take a little time to think this through. You might want to try this game with a friend
(and somewhat more modest prizes) to see experimentally what the best strategy is.
Clearly, because there is one car and two goats, the probability of picking the car if the
competitor does not switch doors is 1⁄3.
But what if the competitor does switch? What then is the probability of winning the
car? Consider a related question. If the switch goes ahead, under what circumstances
will the competitor not get the car? Clearly, only if the original choice was right. In
other words there is a 13⁄ probability of missing out on the car, therefore there is a
1 – 13⁄ = 23⁄ probability of winning the car. So by switching, the probability of winning
the car is doubled.
208
M06_TOK_SB_IBDIP_4157_U06.indd 208 04/03/2014 14:12
Does this make sense? After this explanation, many students and teachers at
workshops are still not convinced. They argue that they cannot see how asymmetry
has been introduced into the situation. The crucial point is that Monty Hall knows
where the car is. He always opens a door to reveal a goat. It is this act that produces
the required asymmetry. We have more information after Monty Hall has revealed the
goat than we had at the beginning.
Personal qualities and mathematics
Of course, the two-way street between personal knowledge and shared knowledge
means that people with special talents can contribute greatly to the production of
shared knowledge. This seems to be particularly true in mathematics. There are a host
of great names from the past: Pythagoras, Archimedes, Euclid, Diophantus, Galileo,
Newton, Leibniz, Euler, Gauss, Riemann, Laplace, Lagrange, Weierstrass, Cauchy,
Hardy, David Hilbert, John von Neumann, Alain Connes, William Thurston, Andrew
Wiles, and Grigori Perelman – the list is very long. Each of these has contributed in an
extraordinary way to the subject.
Although mathematics is hugely collaborative in the sense that mathematicians often
build on the work of others, it is nevertheless largely a solitary pursuit. It requires CHALLENGE
great depth of thought; huge, imaginative leaps; careful and sometimes laborious YOURSELF
computations; innovative ways of solving very hard problems; and, most of all, great Research what each of the
persistence. Mathematicians need to develop their intuition and their nose for a above people has contributed
profitable strategy. They are guided by emotion and by hunches – they are a far cry from to mathematics.
the stereotype of the coldly logical thinker who is closer to computer than human.
6.7 Evaluation of mathematics
We have seen that mathematics is really one of the crowning achievements of human
civilization. Its ancient art has been responsible not only for some of the most
extraordinary intellectual journeys taken by humans, but its methods have allowed the
building of great cities, the production of great art, and has been the language of great
science.
From a TOK perspective, mathematics, with its internal criteria for truth through
mathematical proof and its underlying logic and set theory, makes a good contrast
to the natural sciences with their reliance on observation of the external world and
their experimental method. Furthermore, these methods provide mathematics with a
certainty that is unattainable in any other AOK.
Two countering arguments should be set against this view of mathematics. The
idea that the axioms, the rules of the game, are arbitrary both deprives mathematics
of its status as something independent of human beings and makes it vulnerable
to the charge that its results cannot ever be entirely relevant to the outside world.
Moreover, Gödel’s theorems cast doubt on both the consistency of mathematics and its
completeness and raises questions about methods that are non-constructive, such as
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M06_TOK_SB_IBDIP_4157_U06.indd 209 04/03/2014 14:12
06 Mathematics
Platonists would certainly argue that mathematics is there in the universe with or
without humans. They would argue that it is built into the structure of the cosmos – a
fact that explains why the laws of the natural sciences lend themselves so readily to
mathematical expression. A possible hybrid approach is to propose that mathematics
is somehow built into human reasoning. This would explain the ubiquity of
mathematics in our explanations of the world. It is a mixture of both views. While
conceding that mathematics is essentially a human artifact, it would it nevertheless
colour the whole of human knowledge and is therefore a structural feature of our
world.
Both views produce challenging knowledge questions. The constructivist is a victim
of the success of mathematics in fields such as the natural sciences. He or she has to
account for just why mathematics is so supremely good at describing the outside world
to which, according to this view, it is ultimately blind. The Platonist, on the other hand,
finds it hard to identify mathematical structures embedded in the world. Moreover, he
or she has a hard time explaining why they are there.
We have seen how mathematics is closely integrated into artistic thinking. Perhaps
because both are abstract AOKs indirectly linked to the world and not held to account
through experiment and observation. Both are open to thought experiment and leaps
of imagination.
We have seen how mathematics can challenge our intuitions and how it can push our
cognitive resources as individual knowers. For example, infinity is not something that
the human mind can fathom in its entirety. Instead, mathematics gives us the tools to
deal with it in precisely this unfathomed state. We can be challenged by results that
seem counter to our intuitions, but ultimately the nature of mathematical proof is
that it forces us to accept them nonetheless. In turn, individuals can, through their
insight and personal perspectives, make ground-breaking contributions that change
the direction of mathematics forever. The history of mathematics is a history of
great thinkers building on the work of previous generations in order to do ever more
powerful things, using ever more sophisticated tools.
The Greek thinkers of the 4th century BCE thought that mathematics lay at the core of
human knowledge. They thought that mathematics was one of the few areas in which
humans could apprehend the eternal Forms only accessible to pure unembodied
intellect. They thought, in essence, that in mathematics they could glimpse the very
framework on which the world and its myriad processes rested. Maybe they were right.
210
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### Chorent risk measure with superaddative
In some definition of chorent risk measure Superadditive is one of the properties I don't understand Why? With subadditivity and homogeneous CvaR is convex, but if we assume another definition for ...
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### What is a “coherent” risk measure?
What is a coherent risk measure, and why do we care? Can you give a simple example of a coherent risk measure as opposed to a non-coherent one, and the problems that a coherent measure addresses in ...
38 views
### Choquet integral risk measure
I have one question that cannot fully understand why. What is the definition of the Choquet integral risk measure?
77 views
### Bregman Mean of a Distribution
In a paper (link), author writes, given that $\gamma:R\rightarrow \bar{R}$ is a convex function, $dom_{\gamma}:=\{x\in R:\gamma(x)<+\infty\}$ is a non-empty open set and $\gamma$ a closed proper ...
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### Examples of Spectral Risk Measures
Let's take the usual definition of a spectral risk measure. If we look at the integral we see that spectral risk measures have the property that the risk measure of a random variable $X$ can be ...
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### Risk Measure-identication
Let X be a variable with existing moment generating function $M_x(z)=E[e^{zX}]$. Define the following risk measure: $\rho_{\alpha}(X)=inf_{z>0}(z^{-1}ln(\frac{M_x(z)}{1-\alpha}))$ Does anyone know ...
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### Example of Coherent Risk measure with Compact Representation
Every coherent risk measure $\rho$ can be represented as $$\rho(X)\triangleq \sup_{Q \in \mathcal{Q}} \mathbb{E}\left[ -X \right],$$ for a set of probability measures $\mathcal{Q}$ defined on the ...
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### What's the advantages of $EVaR$ over $CVaR$?
$CVaR$, which is short for Conditional Value-at-Risk, has long been accepted by both academe and practice as a good coherent risk measure. Entropic value-at-risk ($EVaR$) is a comparative new coherent ...
269 views
### Calculating Expected Shortfall of combined portfolios
So I am reading lecture notes here: https://courses.edx.org/c4x/DelftX/TW3421x/asset/Week3_var_3_slides.pdf The example is this: We have two independent portfolios of bonds. They both have a ...
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https://number.academy/207959 | 1,718,301,948,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861480.87/warc/CC-MAIN-20240613154645-20240613184645-00876.warc.gz | 394,485,300 | 11,384 | # Number 207959 facts
The odd number 207,959 is spelled 🔊, and written in words: two hundred and seven thousand, nine hundred and fifty-nine. The ordinal number 207959th is said 🔊 and written as: two hundred and seven thousand, nine hundred and fifty-ninth. Color #207959. The meaning of the number 207959 in Maths: Is it Prime? Factorization and prime factors tree. The square root and cube root of 207959. What is 207959 in computer science, numerology, codes and images, writing and naming in other languages. Other interesting facts related to 207959.
## Interesting facts about the number 207959
### Asteroids
• (207959) 1995 OR7 is asteroid number 207959. It was discovered by Spacewatch from Obs. US National at Kitt Peak on 7/24/1995.
## What is 207,959 in other units
The decimal (Arabic) number 207959 converted to a Roman number is (C)(C)(V)MMCMLIX. Roman and decimal number conversions.
#### Time conversion
(hours, minutes, seconds, days, weeks)
207959 seconds equals to 2 days, 9 hours, 45 minutes, 59 seconds
207959 minutes equals to 5 months, 4 days, 9 hours, 59 minutes
### Codes and images of the number 207959
Number 207959 morse code: ..--- ----- --... ----. ..... ----.
Sign language for number 207959:
Number 207959 in braille:
QR code Bar code, type 39
Images of the number Image (1) of the number Image (2) of the number More images, other sizes, codes and colors ...
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### Advanced math operations
#### Is Prime?
The number 207959 is not a prime number. The closest prime numbers are 207953, 207967.
#### Factorization and factors (dividers)
The prime factors of 207959 are 29 * 71 * 101
The factors of 207959 are 1, 29, 71, 101, 2059, 2929, 7171, 207959.
Total factors 8.
Sum of factors 220320 (12361).
#### Powers
The second power of 2079592 is 43.246.945.681.
The third power of 2079593 is 8.993.591.576.875.079.
#### Roots
The square root √207959 is 456,025219.
The cube root of 3207959 is 59,246028.
#### Logarithms
The natural logarithm of No. ln 207959 = loge 207959 = 12,245096.
The logarithm to base 10 of No. log10 207959 = 5,317978.
The Napierian logarithm of No. log1/e 207959 = -12,245096.
### Trigonometric functions
The cosine of 207959 is -0,292175.
The sine of 207959 is -0,956365.
The tangent of 207959 is 3,273256.
## Number 207959 in Computer Science
Code typeCode value
PIN 207959 It's recommended that you use 207959 as your password or PIN.
207959 Number of bytes203.1KB
CSS Color
#207959 hexadecimal to red, green and blue (RGB) (32, 121, 89)
Unix timeUnix time 207959 is equal to Saturday Jan. 3, 1970, 9:45:59 a.m. GMT
IPv4, IPv6Number 207959 internet address in dotted format v4 0.3.44.87, v6 ::3:2c57
207959 Decimal = 110010110001010111 Binary
207959 Decimal = 101120021012 Ternary
207959 Decimal = 626127 Octal
207959 Decimal = 32C57 Hexadecimal (0x32c57 hex)
207959 BASE64MjA3OTU5
207959 MD5e8fe4853919829588426a0e601fcb3f9
207959 SHA19a7a89def0579c7e48c7932309a9dd3d2980203b
207959 SHA2248e0c32b1ea75dab00c4fa8914db862c43fe7cc1441db6e407c521837
207959 SHA256a2cd0a20c5e8aeec3968c03450ebfb818eda16dce069d98fcb32629a03096967
207959 SHA384d3a584a09881eb96f6d422cded4ed9e34184e8c92827af482a7e45a562715bb597f8156380085116918cab6259e8ae9f
More SHA codes related to the number 207959 ...
If you know something interesting about the 207959 number that you did not find on this page, do not hesitate to write us here.
## Numerology 207959
### Character frequency in the number 207959
Character (importance) frequency for numerology.
Character: Frequency: 2 1 0 1 7 1 9 2 5 1
### Classical numerology
According to classical numerology, to know what each number means, you have to reduce it to a single figure, with the number 207959, the numbers 2+0+7+9+5+9 = 3+2 = 5 are added and the meaning of the number 5 is sought.
## № 207,959 in other languages
How to say or write the number two hundred and seven thousand, nine hundred and fifty-nine in Spanish, German, French and other languages. The character used as the thousands separator.
Spanish: 🔊 (número 207.959) doscientos siete mil novecientos cincuenta y nueve German: 🔊 (Nummer 207.959) zweihundertsiebentausendneunhundertneunundfünfzig French: 🔊 (nombre 207 959) deux cent sept mille neuf cent cinquante-neuf Portuguese: 🔊 (número 207 959) duzentos e sete mil, novecentos e cinquenta e nove Hindi: 🔊 (संख्या 207 959) दो लाख, सात हज़ार, नौ सौ, उनसठ Chinese: 🔊 (数 207 959) 二十万七千九百五十九 Arabian: 🔊 (عدد 207,959) مئتان و سبعة ألفاً و تسعمائة و تسعة و خمسون Czech: 🔊 (číslo 207 959) dvěstě sedm tisíc devětset padesát devět Korean: 🔊 (번호 207,959) 이십만 칠천구백오십구 Danish: 🔊 (nummer 207 959) tohundrede og syvtusindnihundrede og nioghalvtreds Hebrew: (מספר 207,959) מאתיים ושבעה אלף תשע מאות חמישים ותשע Dutch: 🔊 (nummer 207 959) tweehonderdzevenduizendnegenhonderdnegenenvijftig Japanese: 🔊 (数 207,959) 二十万七千九百五十九 Indonesian: 🔊 (jumlah 207.959) dua ratus tujuh ribu sembilan ratus lima puluh sembilan Italian: 🔊 (numero 207 959) duecentosettemilanovecentocinquantanove Norwegian: 🔊 (nummer 207 959) to hundre og syv tusen ni hundre og femtini Polish: 🔊 (liczba 207 959) dwieście siedem tysięcy dziewięćset pięćdziesiąt dziewięć Russian: 🔊 (номер 207 959) двести семь тысяч девятьсот пятьдесят девять Turkish: 🔊 (numara 207,959) ikiyüzyedibindokuzyüzellidokuz Thai: 🔊 (จำนวน 207 959) สองแสนเจ็ดพันเก้าร้อยห้าสิบเก้า Ukrainian: 🔊 (номер 207 959) двісті сім тисяч дев'ятсот п'ятдесят дев'ять Vietnamese: 🔊 (con số 207.959) hai trăm lẻ bảy nghìn chín trăm năm mươi chín Other languages ...
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The content of the comments is the opinion of the users and not of number.academy. It is not allowed to pour comments contrary to the laws, insulting, illegal or harmful to third parties. Number.academy reserves the right to remove or not publish any inappropriate comment. It also reserves the right to publish a comment on another topic. Privacy Policy. | 2,043 | 6,201 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2024-26 | latest | en | 0.78107 |
https://www.redhotpawn.com/forum/posers-and-puzzles/an-oldie-but-a-goodie.19596 | 1,548,069,673,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583792338.50/warc/CC-MAIN-20190121111139-20190121133139-00331.warc.gz | 895,488,368 | 6,968 | # An oldie, but a goodie
PBE6
Posers and Puzzles 09 Feb '05 15:35
1. PBE6
Bananarama
09 Feb '05 15:35
Here's an old logic problem about old men:
Three old wise men are sleeping peacefully under a tree, dreaming of erotic equations, when along comes that brat Terence. Terence procedes to paint each wise man's face purpley-red with beet juice, just for the hell of it. What a dirty bottom feeder. Anyways, as Terence beats a quiet retreat, the three wise men wake up and look at each other's faces. Pure hilarity ensues, as they all start laughing at one another. Terence snickers menacingly in the background, but that's beside point. The suddenly, one of the wise men stops laughing and says "Aw crap, I must have red on my face, too!".
The question is: how did he know?!?
2. 09 Feb '05 16:01
if one of the men had beet on his face then two of the men would laugh but only 2.
if two of the man had beet on their face then all of the men would be laughing but only looking at the two men with beet on their face, as all the men are laughing and looking at each other when they do this then all of them must have beet on their face.
3. PBE6
Bananarama
09 Feb '05 16:34
Originally posted by kcams
if one of the men had beet on his face then two of the men would laugh but only 2.
if two of the man had beet on their face then all of the men would be laughing but only looking at the two men with beet on their face, as all the men are laughing and looking at each other when they do this then all of them must have beet on their face.
Sorry, I should have pointed out that none of the wise men can see their own face.
Also, if anyone has beet on their face then all three wise men would laugh, mostly because they're jerks. If a wise man saw another wise man laughing and looking at him, he would consider it convivial.
So I guess I'm saying that your solution is a common sense one, but doesn't logically answer the question, and it's not the one I'm looking for.
4. 11 Feb '05 00:54
If only one wise man had beet on his face, then he wouldn't be laughing. So wise man C (the one who correctly deduced his face had been.. beet-en? beeted?) realized that both A and B knew this. If only A and B had had beat on their faces, and C didn't, A and/or B would have soon realized that the other of them is laughing, which means they must have beet on their face. Because of this, C realized that, because A and B didn't realize they had beet on their faces, he must have beet on his face or else one would realize the other shouldn't be laughing.
If that made any sense... hard to explain for me :\ Nice puzzle, by the way, it took some thinking to get.
5. PBE6
Bananarama
11 Feb '05 19:06
Originally posted by BTBasham
If only one wise man had beet on his face, then he wouldn't be laughing. So wise man C (the one who correctly deduced his face had been.. beet-en? beeted?) realized that both A and B knew this. If only A and B had had beat on their faces, and C didn't, A and/or B would have soon realized that the other of them is laughing, which means they must have beet ...[text shortened]... e any sense... hard to explain for me :\ Nice puzzle, by the way, it took some thinking to get.
Excellent. That is the solution I was looking for. | 836 | 3,258 | {"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-2019-04 | latest | en | 0.98199 |
https://chrischona2015.org/how-many-glasses-in-a-liter/ | 1,653,250,688,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662546071.13/warc/CC-MAIN-20220522190453-20220522220453-00236.warc.gz | 225,712,883 | 4,787 | How numerous Glasses room in 1 liter that water Bottle? below we resolve this price of this question and also solve this problem. Usually One liter the water to fill glasses is no a big problem, below is a solution and terms. In this short article we learn perfect pour it until it is full 1 liter water party to how countless glasses. Any type of one assume please compose on comment below. Let’s go to start.
You are watching: How many glasses in a liter
## What is One Liter of water?
One liter that water is usually water bottle or any kind of other devices you usage to measurement in liter. Over there is together a thing in i beg your pardon we deserve to fill the water for this reason that us will usage the water party to fill 1 liter the water. Usage of water bottle because its measure is all in one place.
## Procedure: One Liter of water
Before learning everything, understand that the size of the glass the the glass should be recognized to you. An initial of all, you need to take a blank bottle. Then you have to fill one liter the water in it. You do not have to fill the bottle. Wherein she is tata is not to fill it. As soon as you buy new from the market, the is sufficient to fill as much as it is, a liter meter.
Now start filling one glass that glass v this bottle. An initial put it in a cup, then include water to another cup or glass. Please keep in mind that the ahead glass filled her fill, the an ext the 2nd fill. Now fill the glass through the 3rd glass, likewise keep it same to the cup-1 and also cup-2. Do you still have actually water in her water bottle? So currently take the fourth glass and also start pour it until it is full it. If he is too full of the remainder of the glass, then your answer will be 4 glasses. If over there is still some level left then you have to estimate the level the glass you have actually left and also to number out how numerous milliliters will be left.
Someone will have actually a question about how numerous glasses that water will certainly be fill on 2 liters, therefore the prize is 8 glasses.
One normal dimension glass will certainly hold around 8 OZ water.
One oz of water is approximately 30 ML.
So one glass the water is around 240 ML.
1 liter = 1000 ML / 240 ML = around 4 Glasses.
If you need precise amount, you re welcome go come internet and do it yourself.
See more: How To Use Tattoo Transfer Paper ? (5 Simple Steps) How To Use Tattoo Transfer Paper
One liter of water it way 1,000 mL. Also “one glass the water” usually method 8 ounces or 240 mL. Therefore 1 liter will have 4.17 glasses that water: 4 are full to the brim and 1 is 17% full. This is ideal answer. | 619 | 2,669 | {"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-2022-21 | latest | en | 0.958125 |
http://bootmath.com/fibered-products-in-mathsf-set.html | 1,534,420,431,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221210735.11/warc/CC-MAIN-20180816113217-20180816133217-00423.warc.gz | 70,773,772 | 6,274 | # Fibered products in $\mathsf {Set}$
I’m just starting to work through Aluffi, and one question asks the reader to define fibered products (and coproducts) in the category of sets and functions. I’m just looking to check that I have the fibered products right before I move on to the fibered coproducts.
Let $\alpha\colon A\to C$ and let $\beta\colon B\to C$. Then the fibered product of $\alpha$ and $\beta$ is (I think), the set product $P:=\alpha^{-1}[\beta[B]]\times\beta^{-1}[\alpha[A]]$ together with the usual projections $\pi_A$ and $\pi_B$.
### Proof (I hope):
Let $f\colon Z\to A$ and $g\colon Z\to B$ so that $\alpha f=\beta g$. Suppose that for some $\sigma\colon Z\to P$, $f=\pi_A \sigma$ and $g=\pi_B \sigma$. Then for each $z\in Z$, $\pi_A(\sigma(z))=f(z)$ and $\pi_B(\sigma(z))=g(z)$, so for all $z\in Z$, $\sigma(z)=(f(z),g(z))$ (so $\sigma$ is unique).
Indeed, this assignment works: if $z\in Z$, then $\alpha(f(z))=\beta(g(z))\in \beta[B]$, so surely $f(z)\in \alpha^{-1}[\beta[B]]$, and symmetrically a similar thing holds for $g$, so $(f(z),g(z))\in P$. That $\pi_A\sigma=f$ and $\pi_B\sigma=g$ is then immediate.
### Update
As the answers indicate, that was wrong. Can someone tell me if the answer I give below is correct?
#### Solutions Collecting From Web of "Fibered products in $\mathsf {Set}$"
Your definition is not quite right, because with it, you need not have $\alpha\circ \pi_A=\beta\circ\pi_B$. For example, let $A=\{1,2\}$, $B=\{1,2\}$, and $C=\{1,2,3\}$, let $\alpha$ be the inclusion, and let $\beta$ send $1$ to $2$ and $2$ to $1$. Then $(2,2)$ is an element of your set, but applying $\pi_A$ and then $\alpha$ gives $2$, while applying $\pi_B$ and then $\beta$ gives $1$. The fiber product must be equipped with projections $\pi_A$ and $\pi_B$ satisfying $\alpha\circ\pi_A=\beta\circ\pi_B$.
Responding to Keenan Kidwell’s helpful criticism and Ma Ming’s perhaps-overly-generous hint, I try again:
Let $P=\{(a,b):\alpha(a)=\beta(b)\}$ and again use the canonical projections. Define $\sigma$ the same way (again, there is no other choice). Now if $z\in Z$, then $\alpha(f(z))=\beta(g(z))$, so $(f(z),g(z))\in P$ and everything works out as it should.
This is wrong.
Let $A=B=C$, $\alpha=\beta=id$. Then the fiber product should be the diagonal in $A\times A$. | 729 | 2,312 | {"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} | 3.484375 | 3 | CC-MAIN-2018-34 | latest | en | 0.809335 |
http://pyphi.readthedocs.io/en/0.7.0/examples/xor.html | 1,527,058,192,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794865456.57/warc/CC-MAIN-20180523063435-20180523083435-00420.warc.gz | 244,747,195 | 6,379 | # XOR Network¶
This example describes a system of three fully connected XOR nodes, $$n_{0}$$, $$n_{1}$$ and $$n_{2}$$ (no self-connections).
First let’s create the XOR network, with all nodes OFF in both the current and past states.
>>> import pyphi
>>> network = pyphi.examples.xor_network()
Existence is a top-down process; the whole is more important than its parts. The first step is to confirm the existence of the whole, by finding the main complex of the network:
>>> main_complex = pyphi.compute.main_complex(network)
The main complex exists ($$\Phi > 0$$),
>>> main_complex.phi
1.874999
and it consists of the entire network:
>>> main_complex.subsystem
Subsystem((n0, n1, n2))
Knowing what exists at the system level, we can now investigate the existence of concepts within the complex.
>>> constellation = main_complex.unpartitioned_constellation
>>> len(constellation)
3
>>> [concept.mechanism for concept in constellation]
[(n0, n1), (n0, n2), (n1, n2)]
There are three concepts in the constellation. They are all the possible second order mechanisms: $$n_{0}n_{1}$$, $$n_{0}n_{2}$$ and $$n_{1}n_{2}$$.
Focusing on the concept specified by mechanism $$n_{0}n_{1}$$, we investigate existence, and the irreducible cause and effect. Based on the symmetry of the network, the results will be similar for the other second order mechanisms.
>>> concept = constellation[0]
>>> concept.mechanism
(n0, n1)
>>> concept.phi
0.5
The concept has $$\varphi = \frac{1}{2}$$.
>>> concept.cause.purview
(n0, n1, n2)
>>> concept.cause.repertoire
array([[[ 0.5, 0. ],
[ 0. , 0. ]],
[[ 0. , 0. ],
[ 0. , 0.5]]])
So we see that the cause purview of this mechanism is the whole system $$n_{0}n_{1}n_{2}$$, and that the repertoire shows a $$0.5$$ of probability the past state being (0, 0, 0) and the same for (1, 1, 1):
>>> concept.cause.repertoire[(0, 0, 0)]
0.5
>>> concept.cause.repertoire[(1, 1, 1)]
0.5
This tells us that knowing both $$n_{0}$$ and $$n_{1}$$ are currently OFF means that the past state of the system was either all OFF or all ON with equal probability.
For any reduced purview, we would still have the same information about the elements in the purview (either all ON or all OFF), but we would lose the information about the elements outside the purview.
>>> concept.effect.purview
(n2,)
>>> concept.effect.repertoire
array([[[ 1., 0.]]])
The effect purview of this concept is the node $$n_{2}$$. The mechanism $$n_{0}n_{1}$$ is able to completely specify the next state of $$n_{2}$$. Since both nodes are OFF, the next state of $$n_{2}$$ will be OFF.
The mechanism $$n_{0}n_{1}$$ does not provide any information about the next state of either $$n_{0}$$ or $$n_{1}$$, because the relationship depends on the value of $$n_{2}$$. That is, the next state of $$n_{0}$$ (or $$n_{1}$$) may be either ON or OFF, depending on the value of $$n_{2}$$. Any purview larger than $$n_{2}$$ would be reducible by pruning away the additional elements.
Main Complex: $$n_{0}n_{1}n_{2}$$ with $$\Phi = 1.875$$
Mechanism $$\varphi$$ Cause Purview Effect Purview
$$n_{0}n_{1}$$ 0.5 $$n_{0}n_{1}n_{2}$$ $$n_{2}$$
$$n_{0}n_{2}$$ 0.5 $$n_{0}n_{1}n_{2}$$ $$n_{1}$$
$$n_{1}n_{2}$$ 0.5 $$n_{0}n_{1}n_{2}$$ $$n_{0}$$
An analysis of the intrinsic existence of this system reveals that the main complex of the system is the entire network of XOR nodes. Furthermore, the concepts which exist within the complex are those specified by the second-order mechanisms $$n_{0}n_{1}$$, $$n_{0}n_{2}$$, and $$n_{1}n_{2}$$.
To understand the notion of intrinsic existence, in addition to determining what exists for the system, it is useful to consider also what does not exist.
Specifically, it may be surprising that none of the first order mechanisms $$n_{0}$$, $$n_{1}$$ or $$n_{2}$$ exist. This physical system of XOR gates is sitting on the table in front of me; I can touch the individual elements of the system, so how can it be that they do not exist?
That sort of existence is what we term extrinsic existence. The XOR gates exist for me as an observer, external to the system. I am able to manipulate them, and observe their causes and effects, but the question that matters for intrinsic existence is, do they have irreducible causes and effects within the system? There are two reasons a mechanism may have no irreducible cause-effect power: either the cause-effect power is completely reducible, or there was no cause-effect power to begin with. In the case of elementary mechanisms, it must be the latter.
To see this, again due to symmetry of the system, we will focus only on the mechanism $$n_{0}$$.
>>> subsystem = pyphi.examples.xor_subsystem()
>>> n0 = (subsystem.nodes[0],)
>>> n0n1n2 = subsystem.nodes
In order to exist, a mechanism must have irreducible cause and effect power within the system.
>>> subsystem.cause_info(n0, n0n1n2)
0.5
>>> subsystem.effect_info(n0, n0n1n2)
0.0
The mechanism has no effect power over the entire subsystem, so it cannot have effect power over any purview within the subsystem. Furthermore, if a mechanism has no effect power, it certainly has no irreducible effect power. The first-order mechanisms of this system do not exist intrinsically, because they have no effect power (having causal power is not enough).
To see why this is true, consider the effect of $$n_{0}$$. There is no self-loop, so $$n_{0}$$ can have no effect on itself. Without knowing the current state of $$n_{0}$$, in the next state $$n_{1}$$ could be either ON or OFF. If we know that the current state of $$n_{0}$$ is ON, then $$n_{1}$$ could still be either ON or OFF, depending on the state of $$n_{2}$$. Thus, on its own, the current state of $$n_{0}$$ does not provide any information about the next state of $$n_{1}$$. A similar result holds for the effect of $$n_{0}$$ on $$n_{2}$$. Since $$n_{0}$$ has no effect power over any element of the system, it does not exist from the intrinsic perspective.
To complete the discussion, we can also investigate the potential third order mechanism $$n_{0}n_{1}n_{2}$$. Consider the cause information over the purview $$n_{0}n_{1}n_{2}$$:
>>> subsystem.cause_info(n0n1n2, n0n1n2)
0.749999
Since the mechanism has nonzero cause information, it has causal power over the system—but is it irreducible?
>>> mip = subsystem.mip_past(n0n1n2, n0n1n2)
>>> mip.phi
0.0
>>> mip.partition
(Part(mechanism=(n0,), purview=()), Part(mechanism=(n1, n2), purview=(n0, n1, n2)))
The mechanism has $$ci = 0.75$$, but it is completely reducible ($$\varphi = 0$$) to the partition
$\frac{n_0}{\left[\,\right]} \times \frac{n_1n_2}{n_0n_1n_2}$
This result can be understood as follows: knowing that $$n_{1}$$ and $$n_{2}$$ are OFF in the current state is sufficient to know that $$n_{0}$$, $$n_{1}$$, and $$n_{2}$$ were all OFF in the past state; there is no additional information gained by knowing that $$n_{0}$$ is currently OFF.
Similarly for any other potential purview, the current state of $$n_{1}$$ and $$n_{2}$$ being (0, 0) is always enough to fully specify the previous state, so the mechanism is reducible for all possible purviews, and hence does not exist. | 2,046 | 7,159 | {"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 | 3 | CC-MAIN-2018-22 | latest | en | 0.864349 |
https://electronics.stackexchange.com/questions/425262/figuring-out-the-principal-working-of-constant-current-source | 1,563,852,765,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195528687.63/warc/CC-MAIN-20190723022935-20190723044935-00342.warc.gz | 379,679,760 | 37,445 | # Figuring out the principal working of constant current source
I've the following circuit:
simulate this circuit – Schematic created using CircuitLab
I was told that this is a constant current source. I think that the resistors $$\\text{R}_1\$$ and $$\\text{R}_2\$$ determine the current in trough the load. Is that true and how is the output current related to those resistor values or other components in the circuit?
• Assume a fixed 3.3V on the positive input of the OpAmp. Next, assume there is a current of 0A running through R3. Use Ohms law for R3 and consider what happens to the OpAmp and next to the mosfet. Then, assume 2A is running through R3 and consider again what happens to the OpAmp and to the mosfet. Last, for 1A. – Huisman Mar 2 at 17:21
• @Huisman I've have realy no idea (Dutch: en goed om een Nederlander te zien hier :) & English: good to sea another Dutchman over here)! – adfsg Mar 2 at 17:24
• This circuit work thanks to the negative feedback magic. The opamp simply compares the voltage divider output voltage with the voltage produced by the current flowing through R3 resistor and the MOSFET V_R3 = I_load x R3. And if this voltage is lower than the voltage divider voltage the opamp will turn on the MOSFET harder to make sure that the voltage drop across R3 is equal to the voltage given by the voltage divider. – G36 Mar 2 at 17:25
• @G36 So, how does the loadcurrent depend on the resistor values? – adfsg Mar 2 at 17:26
• two comments. First - It's best to break this up into 2 parts. (1) Iout depends on voltage across resistor, which depends on voltage at input of amplifier. (2) Voltage at amplifier depends on Vload+ R1 and R2. Normally you'd use something more accurate than 'Vload+' to set the voltage to the opamp. Second - OP27 won't work here, it's an old-style dual-rail not R2R output, so the output won't go down to GND, neither will the input common mode go to ground. Use dual rails, or a -ve amplifier supply of -5v or more, or a modern R2R opamp. – Neil_UK Mar 2 at 17:42
The opamp is in a non-inverting amplifier configuration with a gain of 1. As always, the output of the opamp will do whatever it takes to make the voltages at the + (non-inverting) and - (inverting) inputs identical. Because the forward gain is 1, and the top of R3 is connected to the - input, the opamp will do whatever it takes to make the voltage at the top of R3 equal to the voltage at the + input. As R1 and R2 change, the + input voltage changes, and the opamp makes sure that voltage is at the top of R3.
Now for the fun part. Between the opamp output and R3 is the FET, which is essentially a voltage-controlled resistor. As the output of the opamp goes up and down, the resistance of the FET between its drain and source terminals (called Rds) goes down and up. The higher the gate voltage, the lower the resistance.
If the circuit were simply the load and R3 in series, then if the load resistance changed, the current through the string would change (Ohm's Law). But that FET is in there; its resistance is in series with the load. The load, FET, and R3 form a series string, and the opamp can adjust the FET's resistance. If the load resistance changes, the current through the string changes, the voltage at the top of R3 changes, and the opamp adjusts the FET's resistance to keep the R3 voltage equal to the input voltage. R3 is called the sense (or shunt) resistor. The opamp works to keep the voltage across it constant, and thanks to Ohm's Law that means the current through it (and the rest of the string) is constant. By constant I mean constant as long as the input voltage is constant. Change the input voltage, and the opamp adjusts the FET so the string current is correct for that voltage.
The net result is that the current through the load is not determined by the resistance of the load. Over the range of voltages that the circuit can handle, the current through the load is a constant even when the load resistance changes. The only thing that changes the load current is the + input voltage.
Simply the OP drives the gate of the drain follower ( with conditions $$\ R_{dsOn} < R_3\$$ to make $$\ V_{in+}=V_{in-}\$$ and $$\V_{load} > 2Vt\$$ , Vt= FET threshold voltage, a.k.a. $$\V_{gs(th)}\$$.
Then $$V_{R_2} = V_{R_3}=I_{R_3}*R_3,$$ and $$\V_{R_2}\$$is a ratio of $$\ V_{in}\$$.
# Details
Typically, R3 is small like = 50 to 100mV drop at Imax current, as the current shunt sensor to limit waste heat of Pd in R3 but not too small. where Vin offset contributes to error budget. This sensor loop area should be small as possible, so that transients do not cause overshoot for a step input from stray reactance and crosstalk from output current load mutual (inductive or capacitive) coupling. | 1,242 | 4,763 | {"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": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 9, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2019-30 | longest | en | 0.869159 |
https://www.poetrysoup.com/poem/y_1493371 | 1,674,844,229,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764495001.99/warc/CC-MAIN-20230127164242-20230127194242-00235.warc.gz | 971,139,037 | 13,776 | Y
In school we had to learn algebra. Ex: ½y - 3 = 2-½y We had to find y, but why? In life, I never needed to use any type of these algebraic equations, yet somehow, I still find myself trying to find y. Like for instance, Y - a credit score is of so importance here on earth when it’s not a requirement to get into Heaven, which means people with an 800 credit score can still go to hell. Y - a light bill is \$200.00, that’s not light. Y - my baby daddy sex was so good, lol. Y - the United States not united. Y - some people don’t know that the beeping on their smoke detector means the battery needs changing. Y - Jack jumped over the candlestick, lol. Y - I have forgiven my student loan, but the government hasn’t. | 187 | 720 | {"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-2023-06 | latest | en | 0.962555 |
https://www.edureka.co/community/53474/doubt-in-numpy-vstack?show=53475 | 1,721,613,287,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517805.92/warc/CC-MAIN-20240722003438-20240722033438-00503.warc.gz | 629,345,021 | 24,797 | 0 votes
I wanted to ask that the following code doesn't work
`c=vstack(tsne_data,label_1000).T`
where shape of tsne_data is (1000,2) and the shape of label_1000 is (1000,)
but the following code works for the same
`c=vstack(tsne_data.T,label_1000).T`
for the same shape of tsne_data and label_1000
so I wanted to know that why do we need to use transpose with tsne_data in vstack
Jul 30, 2019 in Python 903 views
## 1 answer to this question.
0 votes
The vstack function in numpy will stack the sequence vertically (row wise).
If you see the below example shape of the row in a and b are not similar so you won't be able to add b to a.
In order to do that you will have to find the transpose of it so that shape of rows in a and b are similar
answered Jul 30, 2019 by Esha
0 votes
1 answer
## Dimension in python numpy
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## Numpy data in csv
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## Mean and average in numpy
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## In NumPy how do I get the maximum of subsets? Python
You can use np.maximum.reduceat: >>> _, idx = np.unique(g, ...READ MORE
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## Is it possible to invert a Numpy boolean array in Python using the tilde operator (~)?
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## How do I obtain the index list in a NumPy Array of all the NaN values present using Python?
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## Count the digits in a Numpy array
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## How to save numpy array?
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## how to manage memory in python?
Memory management in python involves a private heap ...READ MORE | 548 | 2,029 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2024-30 | latest | en | 0.827092 |
https://git.sesse.net/?p=stockfish;a=patch;h=7615e3485e75c2f1715d372f7bb1f546738a5c76 | 1,638,447,858,000,000,000 | text/plain | crawl-data/CC-MAIN-2021-49/segments/1637964362219.5/warc/CC-MAIN-20211202114856-20211202144856-00383.warc.gz | 351,149,225 | 3,492 | From 7615e3485e75c2f1715d372f7bb1f546738a5c76 Mon Sep 17 00:00:00 2001 From: MaximMolchanov Date: Sat, 14 Nov 2020 02:55:29 +0200 Subject: [PATCH] Calculate sum from first elements in affine transform for AVX512/AVX2/SSSE3 The idea is to initialize sum with the first element instead of zero. Reduce one add_epi32 and one set_zero SIMD instructions for each output dimension. sum = 0; for i = 1 to n sum += a[i] -> sum = a[1]; for i = 2 to n sum += a[i] STC: LLR: 2.95 (-2.94,2.94) {-0.25,1.25} Total: 69048 W: 7024 L: 6799 D: 55225 Ptnml(0-2): 260, 5175, 23458, 5342, 289 https://tests.stockfishchess.org/tests/view/5faf2cf467cbf42301d6aa06 closes https://github.com/official-stockfish/Stockfish/pull/3227 No functional change. --- AUTHORS | 1 + src/nnue/layers/affine_transform.h | 211 ++++++++++++++++++++--------- 2 files changed, 148 insertions(+), 64 deletions(-) diff --git a/AUTHORS b/AUTHORS index 71b718b8..b31a36e9 100644 --- a/AUTHORS +++ b/AUTHORS @@ -112,6 +112,7 @@ Mark Tenzer (31m059) marotear Matthew Lai (matthewlai) Matthew Sullivan (Matt14916) +Maxim Molchanov (Maxim) Michael An (man) Michael Byrne (MichaelB7) Michael Chaly (Vizvezdenec) diff --git a/src/nnue/layers/affine_transform.h b/src/nnue/layers/affine_transform.h index 47c9c488..caf315b2 100644 --- a/src/nnue/layers/affine_transform.h +++ b/src/nnue/layers/affine_transform.h @@ -181,13 +181,13 @@ namespace Eval::NNUE::Layers { return _mm512_add_epi32(_mm512_permutexvar_epi32(indices, x), bias); }; - [[maybe_unused]] auto m512_add_dpbusd_epi32 = [=](__m512i& acc, __m512i a, __m512i b) { #if defined (USE_VNNI) + [[maybe_unused]] auto m512_add_dpbusd_epi32 = [=](__m512i& acc, __m512i a, __m512i b) { acc = _mm512_dpbusd_epi32(acc, a, b); #else + [[maybe_unused]] auto m512_dpbusd_epi32 = [=](__m512i a, __m512i b) -> __m512i { __m512i product0 = _mm512_maddubs_epi16(a, b); - product0 = _mm512_madd_epi16(product0, kOnes512); - acc = _mm512_add_epi32(acc, product0); + return _mm512_madd_epi16(product0, kOnes512); #endif }; @@ -214,14 +214,13 @@ namespace Eval::NNUE::Layers { return _mm_add_epi32(_mm_add_epi32(sum128lo, sum128hi), bias); }; - - [[maybe_unused]] auto m256_add_dpbusd_epi32 = [=](__m256i& acc, __m256i a, __m256i b) { #if defined (USE_VNNI) + [[maybe_unused]] auto m256_add_dpbusd_epi32 = [=](__m256i& acc, __m256i a, __m256i b) { acc = _mm256_dpbusd_epi32(acc, a, b); #else + [[maybe_unused]] auto m256_dpbusd_epi32 = [=](__m256i a, __m256i b) -> __m256i { __m256i product0 = _mm256_maddubs_epi16(a, b); - product0 = _mm256_madd_epi16(product0, kOnes256); - acc = _mm256_add_epi32(acc, product0); + return _mm256_madd_epi16(product0, kOnes256); #endif }; @@ -246,10 +245,9 @@ namespace Eval::NNUE::Layers { return _mm_add_epi32(sum0, bias); }; - [[maybe_unused]] auto m128_add_dpbusd_epi32 = [=](__m128i& acc, __m128i a, __m128i b) { + [[maybe_unused]] auto m128_dpbusd_epi32 = [=](__m128i a, __m128i b) -> __m128i { __m128i product0 = _mm_maddubs_epi16(a, b); - product0 = _mm_madd_epi16(product0, kOnes128); - acc = _mm_add_epi32(acc, product0); + return _mm_madd_epi16(product0, kOnes128); }; #endif @@ -293,15 +291,6 @@ namespace Eval::NNUE::Layers { const __m512i bias = *reinterpret_cast(&biases_[i]); __m512i* outptr = reinterpret_cast<__m512i*>(&output[i]); - __m512i sum01a = _mm512_setzero_si512(); - __m512i sum23a = _mm512_setzero_si512(); - __m512i sum45a = _mm512_setzero_si512(); - __m512i sum67a = _mm512_setzero_si512(); - __m512i sum01b = _mm512_setzero_si512(); - __m512i sum23b = _mm512_setzero_si512(); - __m512i sum45b = _mm512_setzero_si512(); - __m512i sum67b = _mm512_setzero_si512(); - const auto row01a = *reinterpret_cast(&weights_[offset01a]); const auto row23a = *reinterpret_cast(&weights_[offset23a]); const auto row45a = *reinterpret_cast(&weights_[offset45a]); @@ -314,6 +303,16 @@ namespace Eval::NNUE::Layers { const __m256i in256 = input_vector256[0]; const __m512i in = _mm512_inserti64x4(_mm512_castsi256_si512(in256), in256, 1); +#if defined (USE_VNNI) + __m512i sum01a = _mm512_setzero_si512(); + __m512i sum23a = _mm512_setzero_si512(); + __m512i sum45a = _mm512_setzero_si512(); + __m512i sum67a = _mm512_setzero_si512(); + __m512i sum01b = _mm512_setzero_si512(); + __m512i sum23b = _mm512_setzero_si512(); + __m512i sum45b = _mm512_setzero_si512(); + __m512i sum67b = _mm512_setzero_si512(); + m512_add_dpbusd_epi32(sum01a, in, row01a); m512_add_dpbusd_epi32(sum23a, in, row23a); m512_add_dpbusd_epi32(sum45a, in, row45a); @@ -322,6 +321,16 @@ namespace Eval::NNUE::Layers { m512_add_dpbusd_epi32(sum23b, in, row23b); m512_add_dpbusd_epi32(sum45b, in, row45b); m512_add_dpbusd_epi32(sum67b, in, row67b); +#else + __m512i sum01a = m512_dpbusd_epi32(in, row01a); + __m512i sum23a = m512_dpbusd_epi32(in, row23a); + __m512i sum45a = m512_dpbusd_epi32(in, row45a); + __m512i sum67a = m512_dpbusd_epi32(in, row67a); + __m512i sum01b = m512_dpbusd_epi32(in, row01b); + __m512i sum23b = m512_dpbusd_epi32(in, row23b); + __m512i sum45b = m512_dpbusd_epi32(in, row45b); + __m512i sum67b = m512_dpbusd_epi32(in, row67b); +#endif *outptr = m512_hadd256x16( sum01a, sum23a, sum45a, sum67a, @@ -342,48 +351,80 @@ namespace Eval::NNUE::Layers { if constexpr (kPaddedInputDimensions % (kSimdWidth * 2) == 0) { - __m512i sum0 = _mm512_setzero_si512(); - __m512i sum1 = _mm512_setzero_si512(); - __m512i sum2 = _mm512_setzero_si512(); - __m512i sum3 = _mm512_setzero_si512(); - const auto row0 = reinterpret_cast(&weights_[offset0]); const auto row1 = reinterpret_cast(&weights_[offset1]); const auto row2 = reinterpret_cast(&weights_[offset2]); const auto row3 = reinterpret_cast(&weights_[offset3]); - for (IndexType j = 0; j < kNumChunks512; ++j) +#if defined (USE_VNNI) + __m512i sum0 = _mm512_setzero_si512(); + __m512i sum1 = _mm512_setzero_si512(); + __m512i sum2 = _mm512_setzero_si512(); + __m512i sum3 = _mm512_setzero_si512(); + const IndexType kStart = 0; +#else + __m512i sum0 = m512_dpbusd_epi32(input_vector512[0], row0[0]); + __m512i sum1 = m512_dpbusd_epi32(input_vector512[0], row1[0]); + __m512i sum2 = m512_dpbusd_epi32(input_vector512[0], row2[0]); + __m512i sum3 = m512_dpbusd_epi32(input_vector512[0], row3[0]); + const IndexType kStart = 1; +#endif + + for (IndexType j = kStart; j < kNumChunks512; ++j) { const __m512i in = input_vector512[j]; +#if defined (USE_VNNI) m512_add_dpbusd_epi32(sum0, in, row0[j]); m512_add_dpbusd_epi32(sum1, in, row1[j]); m512_add_dpbusd_epi32(sum2, in, row2[j]); m512_add_dpbusd_epi32(sum3, in, row3[j]); +#else + sum0 = _mm512_add_epi32(sum0, m512_dpbusd_epi32(in, row0[j])); + sum1 = _mm512_add_epi32(sum1, m512_dpbusd_epi32(in, row1[j])); + sum2 = _mm512_add_epi32(sum2, m512_dpbusd_epi32(in, row2[j])); + sum3 = _mm512_add_epi32(sum3, m512_dpbusd_epi32(in, row3[j])); +#endif } *outptr = m512_haddx4(sum0, sum1, sum2, sum3, bias); } else { - __m256i sum0 = _mm256_setzero_si256(); - __m256i sum1 = _mm256_setzero_si256(); - __m256i sum2 = _mm256_setzero_si256(); - __m256i sum3 = _mm256_setzero_si256(); - const auto row0 = reinterpret_cast(&weights_[offset0]); const auto row1 = reinterpret_cast(&weights_[offset1]); const auto row2 = reinterpret_cast(&weights_[offset2]); const auto row3 = reinterpret_cast(&weights_[offset3]); - for (IndexType j = 0; j < kNumChunks256; ++j) +#if defined (USE_VNNI) + __m256i sum0 = _mm256_setzero_si256(); + __m256i sum1 = _mm256_setzero_si256(); + __m256i sum2 = _mm256_setzero_si256(); + __m256i sum3 = _mm256_setzero_si256(); + const IndexType kStart = 0; +#else + __m256i sum0 = m256_dpbusd_epi32(input_vector256[0], row0[0]); + __m256i sum1 = m256_dpbusd_epi32(input_vector256[0], row1[0]); + __m256i sum2 = m256_dpbusd_epi32(input_vector256[0], row2[0]); + __m256i sum3 = m256_dpbusd_epi32(input_vector256[0], row3[0]); + const IndexType kStart = 1; +#endif + + for (IndexType j = kStart; j < kNumChunks256; ++j) { const __m256i in = input_vector256[j]; +#if defined (USE_VNNI) m256_add_dpbusd_epi32(sum0, in, row0[j]); m256_add_dpbusd_epi32(sum1, in, row1[j]); m256_add_dpbusd_epi32(sum2, in, row2[j]); m256_add_dpbusd_epi32(sum3, in, row3[j]); +#else + sum0 = _mm256_add_epi32(sum0, m256_dpbusd_epi32(in, row0[j])); + sum1 = _mm256_add_epi32(sum1, m256_dpbusd_epi32(in, row1[j])); + sum2 = _mm256_add_epi32(sum2, m256_dpbusd_epi32(in, row2[j])); + sum3 = _mm256_add_epi32(sum3, m256_dpbusd_epi32(in, row3[j])); +#endif } *outptr = m256_haddx4(sum0, sum1, sum2, sum3, bias); @@ -394,30 +435,50 @@ namespace Eval::NNUE::Layers { { if constexpr (kPaddedInputDimensions % (kSimdWidth * 2) == 0) { - __m512i sum0 = _mm512_setzero_si512(); - const auto row0 = reinterpret_cast(&weights_[0]); - for (IndexType j = 0; j < kNumChunks512; ++j) +#if defined (USE_VNNI) + __m512i sum0 = _mm512_setzero_si512(); + const IndexType kStart = 0; +#else + __m512i sum0 = m512_dpbusd_epi32(input_vector512[0], row0[0]); + const IndexType kStart = 1; +#endif + + for (IndexType j = kStart; j < kNumChunks512; ++j) { const __m512i in = input_vector512[j]; +#if defined (USE_VNNI) m512_add_dpbusd_epi32(sum0, in, row0[j]); +#else + sum0 = _mm512_add_epi32(sum0, m512_dpbusd_epi32(in, row0[j])); +#endif } output[0] = m512_hadd(sum0, biases_[0]); } else { - __m256i sum0 = _mm256_setzero_si256(); - const auto row0 = reinterpret_cast(&weights_[0]); - for (IndexType j = 0; j < kNumChunks256; ++j) +#if defined (USE_VNNI) + __m256i sum0 = _mm256_setzero_si256(); + const IndexType kStart = 0; +#else + __m256i sum0 = m256_dpbusd_epi32(input_vector256[0], row0[0]); + const IndexType kStart = 1; +#endif + + for (IndexType j = kStart; j < kNumChunks256; ++j) { const __m256i in = input_vector256[j]; +#if defined (USE_VNNI) m256_add_dpbusd_epi32(sum0, in, row0[j]); +#else + sum0 = _mm256_add_epi32(sum0, m256_dpbusd_epi32(in, row0[j])); +#endif } output[0] = m256_hadd(sum0, biases_[0]); @@ -451,24 +512,40 @@ namespace Eval::NNUE::Layers { const __m128i bias = *reinterpret_cast(&biases_[i]); __m128i* outptr = reinterpret_cast<__m128i*>(&output[i]); - __m256i sum0 = _mm256_setzero_si256(); - __m256i sum1 = _mm256_setzero_si256(); - __m256i sum2 = _mm256_setzero_si256(); - __m256i sum3 = _mm256_setzero_si256(); - const auto row0 = reinterpret_cast(&weights_[offset0]); const auto row1 = reinterpret_cast(&weights_[offset1]); const auto row2 = reinterpret_cast(&weights_[offset2]); const auto row3 = reinterpret_cast(&weights_[offset3]); - for (IndexType j = 0; j < kNumChunks; ++j) +#if defined (USE_VNNI) + __m256i sum0 = _mm256_setzero_si256(); + __m256i sum1 = _mm256_setzero_si256(); + __m256i sum2 = _mm256_setzero_si256(); + __m256i sum3 = _mm256_setzero_si256(); + const IndexType kStart = 0; +#else + __m256i sum0 = m256_dpbusd_epi32(input_vector[0], row0[0]); + __m256i sum1 = m256_dpbusd_epi32(input_vector[0], row1[0]); + __m256i sum2 = m256_dpbusd_epi32(input_vector[0], row2[0]); + __m256i sum3 = m256_dpbusd_epi32(input_vector[0], row3[0]); + const IndexType kStart = 1; +#endif + + for (IndexType j = kStart; j < kNumChunks; ++j) { const __m256i in = input_vector[j]; +#if defined (USE_VNNI) m256_add_dpbusd_epi32(sum0, in, row0[j]); m256_add_dpbusd_epi32(sum1, in, row1[j]); m256_add_dpbusd_epi32(sum2, in, row2[j]); m256_add_dpbusd_epi32(sum3, in, row3[j]); +#else + sum0 = _mm256_add_epi32(sum0, m256_dpbusd_epi32(in, row0[j])); + sum1 = _mm256_add_epi32(sum1, m256_dpbusd_epi32(in, row1[j])); + sum2 = _mm256_add_epi32(sum2, m256_dpbusd_epi32(in, row2[j])); + sum3 = _mm256_add_epi32(sum3, m256_dpbusd_epi32(in, row3[j])); +#endif } *outptr = m256_haddx4(sum0, sum1, sum2, sum3, bias); @@ -476,15 +553,25 @@ namespace Eval::NNUE::Layers { } else if constexpr (kOutputDimensions == 1) { - __m256i sum0 = _mm256_setzero_si256(); - const auto row0 = reinterpret_cast(&weights_[0]); - for (IndexType j = 0; j < kNumChunks; ++j) +#if defined (USE_VNNI) + __m256i sum0 = _mm256_setzero_si256(); + const IndexType kStart = 0; +#else + __m256i sum0 = m256_dpbusd_epi32(input_vector[0], row0[0]); + const IndexType kStart = 1; +#endif + + for (IndexType j = kStart; j < kNumChunks; ++j) { const __m256i in = input_vector[j]; - m256_add_dpbusd_epi32(sum0, in, row0[j]); +#if defined (USE_VNNI) + m256_add_dpbusd_epi32(sum0, in, row0[j]); +#else + sum0 = _mm256_add_epi32(sum0, m256_dpbusd_epi32(in, row0[j])); +#endif } output[0] = m256_hadd(sum0, biases_[0]); @@ -517,24 +604,24 @@ namespace Eval::NNUE::Layers { const __m128i bias = *reinterpret_cast(&biases_[i]); __m128i* outptr = reinterpret_cast<__m128i*>(&output[i]); - __m128i sum0 = _mm_setzero_si128(); - __m128i sum1 = _mm_setzero_si128(); - __m128i sum2 = _mm_setzero_si128(); - __m128i sum3 = _mm_setzero_si128(); - const auto row0 = reinterpret_cast(&weights_[offset0]); const auto row1 = reinterpret_cast(&weights_[offset1]); const auto row2 = reinterpret_cast(&weights_[offset2]); const auto row3 = reinterpret_cast(&weights_[offset3]); - for (int j = 0; j < (int)kNumChunks; j += 1) + __m128i sum0 = m128_dpbusd_epi32(input_vector[0], row0[0]); + __m128i sum1 = m128_dpbusd_epi32(input_vector[0], row1[0]); + __m128i sum2 = m128_dpbusd_epi32(input_vector[0], row2[0]); + __m128i sum3 = m128_dpbusd_epi32(input_vector[0], row3[0]); + + for (int j = 1; j < (int)kNumChunks; ++j) { const __m128i in = input_vector[j]; - m128_add_dpbusd_epi32(sum0, in, row0[j]); - m128_add_dpbusd_epi32(sum1, in, row1[j]); - m128_add_dpbusd_epi32(sum2, in, row2[j]); - m128_add_dpbusd_epi32(sum3, in, row3[j]); + sum0 = _mm_add_epi32(sum0, m128_dpbusd_epi32(in, row0[j])); + sum1 = _mm_add_epi32(sum1, m128_dpbusd_epi32(in, row1[j])); + sum2 = _mm_add_epi32(sum2, m128_dpbusd_epi32(in, row2[j])); + sum3 = _mm_add_epi32(sum3, m128_dpbusd_epi32(in, row3[j])); } *outptr = m128_haddx4(sum0, sum1, sum2, sum3, bias); @@ -542,16 +629,12 @@ namespace Eval::NNUE::Layers { } else if constexpr (kOutputDimensions == 1) { - __m128i sum0 = _mm_setzero_si128(); - const auto row0 = reinterpret_cast(&weights_[0]); - for (int j = 0; j < (int)kNumChunks; j += 1) - { - const __m128i in = input_vector[j]; + __m128i sum0 = m128_dpbusd_epi32(input_vector[0], row0[0]); - m128_add_dpbusd_epi32(sum0, in, row0[j]); - } + for (int j = 1; j < (int)kNumChunks; ++j) + sum0 = _mm_add_epi32(sum0, m128_dpbusd_epi32(input_vector[j], row0[j])); output[0] = m128_hadd(sum0, biases_[0]); } -- 2.30.2 | 5,291 | 14,251 | {"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-2021-49 | latest | en | 0.363837 |
https://physics.stackexchange.com/questions/398750/confusion-with-the-equivalence-principle-and-inertial-frames | 1,725,918,458,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651157.15/warc/CC-MAIN-20240909201932-20240909231932-00622.warc.gz | 440,571,314 | 40,581 | # Confusion with the equivalence principle and inertial frames
This is an explanation of the equivalence principle. A. says accelerating frames can simulate gravity. Then, I am confused. Are the frames presented in A. inertial frames? Compared to B. I have the impression that the frames in A. are not inertial frames. Then, does it mean that any observer standing on earth is not an inertial observer? But this is contrary to my common sense... Also, the Einstein's postulate says that the speed of light is same in all inertial frames. Then, the speed of light measured from frames in A are not equal to $c$?
• There are some significant issues with the principle of equivalence Keith. Einstein said the special theory of relativity is “nowhere precisely realized in the real world”. It’s only valid “in the infinitesimal”. Your room has to be an infinitesimal room for the principle of equivalence to be exactly valid. See what John Synge said on pages ix and x in his 1960 preface to relativity: the general theory. Commented Apr 9, 2018 at 17:09
• Then, does it mean that any observer standing on earth is not an inertial observer? But this is contrary to my common sense... Common sense is not a good guide to what is correct in physics.
– user4552
Commented Apr 9, 2018 at 20:35
Are the frames presented in A. inertial frames?
No, the frames are accelerated reference frames - an accelerometer at rest in either frame reads non-zero acceleration - and not inertial reference frames.
Then, does it mean that any observer standing on earth is not an inertial observer?
Yes, an accelerometer attached to an observer at rest on the surface of the Earth reads non-zero acceleration and so the observer's reference frame is an accelerated reference frame and not an inertial reference frame.
Then, the speed of light measured from frames in A are not equal to c?
You might find the answers here useful: Does the speed of light vary in non-inertial frames?
An inertial reference frame is a reference frame where an object not subjected to a force is at rest or moves with uniform velocity. The Einstein equivalence principle in its basic formulation states that the gravitational force as experienced locally is not distinguishable from the force experienced in a non-inertial (accelerated) reference frame.
Locally the speed of light is $c$. Locally means measured by an inertial reference frame instantaneously at rest with the accelerated frame. | 528 | 2,457 | {"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} | 2.875 | 3 | CC-MAIN-2024-38 | latest | en | 0.905758 |
https://teachingwithsimplicity.com/store/Third-Grade-Digital-Math-Centers-Geometry-p348999664 | 1,726,359,034,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651601.85/warc/CC-MAIN-20240914225323-20240915015323-00876.warc.gz | 501,136,247 | 35,045 | # Third Grade Digital Math Centers Geometry
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These third-grade math centers for geometry is perfect for the digital teacher!
How they Work
Technically, you don’t need Google Classroom to access these math centers. You simply need a Google account so that you can access the slides within Google Slides. We are a Google school, so it’s extremely easy to share the files with students within Google Classroom.
A PowerPoint version is also included!
The centers are ordered in the order that they would typically be taught. They do not need to be played in the order that they are provided, it’s more of a guide.
You can provide all centers to students to complete in the order of their choice, or you can even create your own order. How? Here are some options:
►The slides are movable. You can order them in the order that you want students to complete them.
►Make multiple copies of the file, delete all the slides but the center that you want them to complete, and save it as the center name. That way, students are only provided with the center that they must complete.
Digital Recording Math Booklet
If you want to go 100% paperless, there is a digital version of the recording booklet.
A Google Form is provided for students to record their answer on as well. Each center has it’s own form.
*Please note, there are some instances where a Google Form was not an option due to the answer that students were to provide.
Printable Recording Math Booklet
Students record their responses in the math booklet. I have created the booklet so that they can be copied front and back, minimizing the amount of paper used!
What's Included:
10 different math centers
❶ Greater or Less Than 90 (sort)
❷ Matching Polygons
❸ Parallelogram or Not? (sort)
❹ Right Angle or No Right Angle? (sort)
❺ Polygon Riddles
❻ Pattern Block Challenge
❼ Partitioned Halves, Thirds, Fourths (sort)
❽ Polygon or Not? (sort)
❾ Quadrilateral or Not? (sort)
❿ Create a Riddle
SAVE BIG - BUY THE BUNDLE
Additional Match Centers Available
• Third Grade Math Centers Fractions ~ Digital
• Third Grade Math Centers Area and Perimeter ~ Digital
• Third Grade Math Centers Measurement and Time ~ Digital
• Third Grade Math Centers Represent and Interpret Data ~ Digital
• Third Grade Math Centers Geometry ~ Digital
• Third Grade Math Centers Addition and Subtraction ~ Digital
• Third Grade Math Centers Place Value ~ Digital
• Third Grade Math Centers Multiplication ~ Digital
• Third Grade Math Centers Division ~ Digital
Still have questions? Email me at [email protected], and I would love to help you out!
Save this product for later | 589 | 2,679 | {"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-2024-38 | latest | en | 0.940768 |
https://metanumbers.com/14530 | 1,631,996,478,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056572.96/warc/CC-MAIN-20210918184640-20210918214640-00280.warc.gz | 427,320,275 | 7,310 | # 14530 (number)
14,530 (fourteen thousand five hundred thirty) is an even five-digits composite number following 14529 and preceding 14531. In scientific notation, it is written as 1.453 × 104. The sum of its digits is 13. It has a total of 3 prime factors and 8 positive divisors. There are 5,808 positive integers (up to 14530) that are relatively prime to 14530.
## Basic properties
• Is Prime? No
• Number parity Even
• Number length 5
• Sum of Digits 13
• Digital Root 4
## Name
Short name 14 thousand 530 fourteen thousand five hundred thirty
## Notation
Scientific notation 1.453 × 104 14.53 × 103
## Prime Factorization of 14530
Prime Factorization 2 × 5 × 1453
Composite number
Distinct Factors Total Factors Radical ω(n) 3 Total number of distinct prime factors Ω(n) 3 Total number of prime factors rad(n) 14530 Product of the distinct prime numbers λ(n) -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) -1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 14,530 is 2 × 5 × 1453. Since it has a total of 3 prime factors, 14,530 is a composite number.
## Divisors of 14530
1, 2, 5, 10, 1453, 2906, 7265, 14530
8 divisors
Even divisors 4 4 4 0
Total Divisors Sum of Divisors Aliquot Sum τ(n) 8 Total number of the positive divisors of n σ(n) 26172 Sum of all the positive divisors of n s(n) 11642 Sum of the proper positive divisors of n A(n) 3271.5 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 120.54 Returns the nth root of the product of n divisors H(n) 4.44139 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 14,530 can be divided by 8 positive divisors (out of which 4 are even, and 4 are odd). The sum of these divisors (counting 14,530) is 26,172, the average is 327,1.5.
## Other Arithmetic Functions (n = 14530)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 5808 Total number of positive integers not greater than n that are coprime to n λ(n) 1452 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 1708 Total number of primes less than or equal to n r2(n) 16 The number of ways n can be represented as the sum of 2 squares
There are 5,808 positive integers (less than 14,530) that are coprime with 14,530. And there are approximately 1,708 prime numbers less than or equal to 14,530.
## Divisibility of 14530
m n mod m 2 3 4 5 6 7 8 9 0 1 2 0 4 5 2 4
The number 14,530 is divisible by 2 and 5.
• Deficient
• Polite
• Square Free
• Sphenic
## Base conversion (14530)
Base System Value
2 Binary 11100011000010
3 Ternary 201221011
4 Quaternary 3203002
5 Quinary 431110
6 Senary 151134
8 Octal 34302
10 Decimal 14530
12 Duodecimal 84aa
20 Vigesimal 1g6a
36 Base36 b7m
## Basic calculations (n = 14530)
### Multiplication
n×y
n×2 29060 43590 58120 72650
### Division
n÷y
n÷2 7265 4843.33 3632.5 2906
### Exponentiation
ny
n2 211120900 3067586677000 44572034416810000 647631660076249300000
### Nth Root
y√n
2√n 120.54 24.4018 10.9791 6.79913
## 14530 as geometric shapes
### Circle
Diameter 29060 91294.7 6.63256e+08
### Sphere
Volume 1.28495e+13 2.65302e+09 91294.7
### Square
Length = n
Perimeter 58120 2.11121e+08 20548.5
### Cube
Length = n
Surface area 1.26673e+09 3.06759e+12 25166.7
### Equilateral Triangle
Length = n
Perimeter 43590 9.1418e+07 12583.3
### Triangular Pyramid
Length = n
Surface area 3.65672e+08 3.61519e+11 11863.7
## Cryptographic Hash Functions
md5 0f701e22c252a24d9449e829b0c21666 b630625628d4502319c767fb97c7f4ba11ec2420 433c2b549d556f185c070bbdaa403e5ed86eac2a35d9d07b32c3a6eb58152fd8 4ab239bdc405eddad98a23c3e8e1ee6896e1ffb96dc50f15712c639c928bfb13a058478ba256d29c0512740de09c3e610ec23bb90ff5c9197494f847be06a4ef ba751c4a854be912fc332a99c5a42f944deb0652 | 1,447 | 4,059 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2021-39 | latest | en | 0.814037 |
https://qanda.ai/en/search/2x%2B3y%20%3D%20%2016?search_mode=expression | 1,642,325,587,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320299852.23/warc/CC-MAIN-20220116093137-20220116123137-00706.warc.gz | 540,804,500 | 23,873 | # Calculator search results
Formula
Solve the equation
Graph
$2 x + 3 y = 16$
$x$Intercept
$\left ( 8 , 0 \right )$
$y$Intercept
$\left ( 0 , \dfrac { 16 } { 3 } \right )$
$2x+3y = 16$
$x = - \dfrac { 3 } { 2 } y + 8$
Solve a solution to $x$
$2 x \color{#FF6800}{ + } \color{#FF6800}{ 3 } \color{#FF6800}{ y } = 16$
Move the rest of the expression except $x$ term to the right side and replace the sign
$2 x = 16 \color{#FF6800}{ - } \left ( \color{#FF6800}{ 3 } \color{#FF6800}{ y } \right )$
$2 x = \color{#FF6800}{ 16 } \color{#FF6800}{ - } \left ( \color{#FF6800}{ 3 } \color{#FF6800}{ y } \right )$
Organize the expression
$2 x = \color{#FF6800}{ - } \color{#FF6800}{ 3 } \color{#FF6800}{ y } \color{#FF6800}{ + } \color{#FF6800}{ 16 }$
$\color{#FF6800}{ 2 } \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ 3 } \color{#FF6800}{ y } \color{#FF6800}{ + } \color{#FF6800}{ 16 }$
Divide both sides by the same number
$\color{#FF6800}{ x } = \left ( \color{#FF6800}{ - } \color{#FF6800}{ 3 } \color{#FF6800}{ y } \color{#FF6800}{ + } \color{#FF6800}{ 16 } \right ) \color{#FF6800}{ \div } \color{#FF6800}{ 2 }$
$x = \left ( - 3 y + 16 \right ) \color{#FF6800}{ \div } \color{#FF6800}{ 2 }$
Convert division to multiplication
$x = \left ( - 3 y + 16 \right ) \color{#FF6800}{ \times } \color{#FF6800}{ \dfrac { 1 } { 2 } }$
$x = \left ( \color{#FF6800}{ - } \color{#FF6800}{ 3 } \color{#FF6800}{ y } \color{#FF6800}{ + } \color{#FF6800}{ 16 } \right ) \color{#FF6800}{ \times } \color{#FF6800}{ \dfrac { 1 } { 2 } }$
Multiply each term in parentheses by $\dfrac { 1 } { 2 }$
$x = \color{#FF6800}{ - } \color{#FF6800}{ 3 } \color{#FF6800}{ y } \color{#FF6800}{ \times } \color{#FF6800}{ \dfrac { 1 } { 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 16 } \color{#FF6800}{ \times } \color{#FF6800}{ \dfrac { 1 } { 2 } }$
$x = \color{#FF6800}{ - } \color{#FF6800}{ 3 } \color{#FF6800}{ y } \color{#FF6800}{ \times } \color{#FF6800}{ \dfrac { 1 } { 2 } } + 16 \times \dfrac { 1 } { 2 }$
Simplify the expression
$x = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 3 } { 2 } } \color{#FF6800}{ y } + 16 \times \dfrac { 1 } { 2 }$
$x = - \dfrac { 3 } { 2 } y + \color{#FF6800}{ 16 } \color{#FF6800}{ \times } \color{#FF6800}{ \dfrac { 1 } { 2 } }$
Calculate the product of rational numbers
$x = - \dfrac { 3 } { 2 } y + \color{#FF6800}{ 8 }$
$y = - \dfrac { 2 } { 3 } x + \dfrac { 16 } { 3 }$
Solve a solution to $y$
$\color{#FF6800}{ 2 } \color{#FF6800}{ x } + 3 y = 16$
Move the rest of the expression except $y$ term to the right side and replace the sign
$3 y = 16 \color{#FF6800}{ - } \left ( \color{#FF6800}{ 2 } \color{#FF6800}{ x } \right )$
$3 y = \color{#FF6800}{ 16 } \color{#FF6800}{ - } \left ( \color{#FF6800}{ 2 } \color{#FF6800}{ x } \right )$
Organize the expression
$3 y = \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 16 }$
$\color{#FF6800}{ 3 } \color{#FF6800}{ y } = \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 16 }$
Divide both sides by the same number
$\color{#FF6800}{ y } = \left ( \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 16 } \right ) \color{#FF6800}{ \div } \color{#FF6800}{ 3 }$
$y = \left ( - 2 x + 16 \right ) \color{#FF6800}{ \div } \color{#FF6800}{ 3 }$
Convert division to multiplication
$y = \left ( - 2 x + 16 \right ) \color{#FF6800}{ \times } \color{#FF6800}{ \dfrac { 1 } { 3 } }$
$y = \left ( \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 16 } \right ) \color{#FF6800}{ \times } \color{#FF6800}{ \dfrac { 1 } { 3 } }$
Multiply each term in parentheses by $\dfrac { 1 } { 3 }$
$y = \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ \times } \color{#FF6800}{ \dfrac { 1 } { 3 } } \color{#FF6800}{ + } \color{#FF6800}{ 16 } \color{#FF6800}{ \times } \color{#FF6800}{ \dfrac { 1 } { 3 } }$
$y = \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ \times } \color{#FF6800}{ \dfrac { 1 } { 3 } } + 16 \times \dfrac { 1 } { 3 }$
Simplify the expression
$y = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 2 } { 3 } } \color{#FF6800}{ x } + 16 \times \dfrac { 1 } { 3 }$
$y = - \dfrac { 2 } { 3 } x + \color{#FF6800}{ 16 } \color{#FF6800}{ \times } \color{#FF6800}{ \dfrac { 1 } { 3 } }$
Calculate the product of rational numbers
$y = - \dfrac { 2 } { 3 } x + \color{#FF6800}{ \dfrac { 16 } { 3 } }$
그래프 보기
Linear function
Solution search results | 2,056 | 4,595 | {"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} | 4.40625 | 4 | CC-MAIN-2022-05 | latest | en | 0.230711 |
https://www.gradesaver.com/textbooks/math/precalculus/precalculus-6th-edition-blitzer/chapter-10-section-10-1-sequences-and-summation-notation-exercise-set-page-1050/89 | 1,571,620,852,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987750110.78/warc/CC-MAIN-20191020233245-20191021020745-00267.warc.gz | 913,069,459 | 12,600 | ## Precalculus (6th Edition) Blitzer
$1$
Use the formula such as: $a_n=1+(\dfrac{1}{n})^n$ It has been seen by using a calculator that as the value of n increases, the value of $a_n$ is getting closer and closer to $1$. | 70 | 220 | {"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} | 3.015625 | 3 | CC-MAIN-2019-43 | longest | en | 0.937499 |
http://stackoverflow.com/questions/14181409/how-to-tile-or-make-bigger | 1,440,871,095,000,000,000 | text/html | crawl-data/CC-MAIN-2015-35/segments/1440644064517.22/warc/CC-MAIN-20150827025424-00207-ip-10-171-96-226.ec2.internal.warc.gz | 225,501,235 | 18,349 | # How to tile or make bigger
How do I tile this? Or make it larger, to make the grass texture appear on the whole screen, like cover the whole floor panel?
Here the code and the result:
(Picture) http://tinypic.com/r/dyvc7t/6
``````glBegin(GL_QUADS);
glNormal3f(0.0f, 0.0f, 1.0f);
glTexCoord2f(0.0,0.0);
glVertex3f(-140.0, 0.0,100.0);
glTexCoord3f(1.0,0.0,0.0);
glVertex3f(140.0, 0.0,100.0);
glTexCoord2f(1.0,1.0);
glVertex3f(140.0, 140.0,100.0);
glTexCoord2f(0.0,1.0);
glVertex3f(-140.0, 140.0,100.0);
glEnd();
glDisable( GL_TEXTURE_2D );
``````
-
Here is the opengl code for a square:
``````glVertex3f(-1.0f, -1.0f, 0.0f); // The bottom left corner
glVertex3f(-1.0f, 1.0f, 0.0f); // The top left corner
glVertex3f(1.0f, 1.0f, 0.0f); // The top right corner
glVertex3f(1.0f, -1.0f, 0.0f); // The bottom right corner
``````
Here is the same code, for a bigger square:
``````glVertex3f(-2.0f, -2.0f, 0.0f); // The bottom left corner
glVertex3f(-2.0f, 2.0f, 0.0f); // The top left corner
glVertex3f(2.0f, 2.0f, 0.0f); // The top right corner
glVertex3f(2.0f, -2.0f, 0.0f); // The bottom right corner
``````
So if you want to make your polygon bigger, just change the numbers that define it.
-
If you want to tile it, just use a 'for (int x = 0; x < so on...)' and a 'translate' method. Be sure to pop the matrix every time you translate a tile though, or it won't work.
- | 569 | 1,387 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2015-35 | latest | en | 0.616006 |
https://jobschat.in/ncert-solutions-for-class-12-physics-chapter-10-wave-optics/ | 1,701,776,463,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100551.17/warc/CC-MAIN-20231205105136-20231205135136-00818.warc.gz | 389,903,136 | 51,868 | # NCERT Solutions For Class 12 Physics Chapter 10 Wave Optics
NCERT Solutions For Class 12 Physics Chapter 10 will help students trying to understand what formulae to use, and how to solve a particular problem in lesson wave optics. Wave Optics Class 12 NCERT Solutions Pdf also provides the students with a better understanding the concepts such as Newton’s Corpuscular theory, Huygens wave theory, and more advanced and widely accepted Maxwell’s equations of electromagnetism, Which says that light is an electromagnetic wave, that propagates with space and time-varying magnetic and electric fields. NCERT Solutions Physics Class 12 Chapter 10 will include Solving problems such as finding interference patterns, and Frequencies, based on Huygens’s theory are very important in the scope of Public Examinations. NCERT Solutions For Class 12 Physics will contain links for solved problems in other chapters as well. Reading through Class 12 Physics Chapter 10 NCERT Solutions will make it easier for the students in board exams.
## Wave Optics Class 12 Physics NCERT Solutions
Topics and subtopics in NCERT Solutions For Class 12 Physics Chapter 10 are provided in the table below:
Section Topic Name 10 Wave Optics 10.1 Introduction 10.2 Huygens Principle 10.3 Refraction and Reflection of Plane Waves using the Huygens Principle 10.4 Coherent and Incoherent Addition of Waves 10.5 Interference of Light Waves and Young’s Experiment 10.6 Diffraction 10.7 Polarization
## Class 12 Physics Chapter 10 NCERT Solutions
Question 10.1:
Monochromatic light of wavelength 589 nm is incident from the air on a water surface. What are the wavelength, frequency, and speed of (a) reflected, and (b) refracted light? The Refractive index of water is 1.33.
Solution:
The wavelength of incident monochromatic light,
λ = 589 nm = 589 × 10−9 m
Speed of light in air, c = 3 × 108 m/s
Refractive index of water, μ = 1.33
(a) The ray will reflect in the same medium as that of the incident ray. Hence, the wavelength, speed, and frequency of the reflected ray will be the same as that of the incident ray.
The frequency of light is given by the relation,
v = c / λ
= 5.09 x 1014 Hz
Hence, the speed, frequency, and wavelength of the reflected light are 3 × 108 m/s, 5.09 ×1014 Hz, and 589 nm respectively.
(b) Frequency of light does not depend on the property of the medium in which it is traveling. Hence, the frequency of the refracted ray in water will be equal to the frequency of the incident or reflected light in the air.
Refracted frequency, ν = 5.09 ×1014 Hz
The speed of light in water is related to the refractive index of water as:
v= c / μ
= 2.26 x 108 m/s
The wavelength of light in water is given by the relation,
λ = ν / v
= 444.007 x 10-9 m
= 444.01 nm
Hence, the speed, frequency, and wavelength of refracted light are 2.26 ×108 m/s, 444.01nm, and 5.09 × 1014 Hz respectively.
Question 10.2:
What is the shape of the wavefront in each of the following cases:
(a) Light diverging from a point source.
(b) Light emerges out of a convex lens when a point source is placed at its focus.
(c) The portion of the wavefront of light from a distant star intercepted by the Earth.
Solution:
(a) The shape of the wavefront in case of a light diverging from a point source is spherical. The wavefront emanating from a point source is shown in the given figure.
(b) The shape of the wavefront in case of a light emerging out of a convex lens when a point source is placed at its focus is a parallel grid. This is shown in the given figure.
(c) The portion of the wavefront of light from a distant star intercepted by the Earth is a plane.
Question 10.3:
(a) The refractive index of glass is 1.5. What is the speed of light in glass? The speed of light in a vacuum is 3.0 × 108 m s−1)
(b) Is the speed of light in glass independent of the color of light? If not, which of the two colors red and violet travels slower in a glass prism?
Solution:
(a) Refractive index of glass, μ = 1.5
Speed of light, c = 3 × 108 m/s
Speed of light in glass is given by the relation,
ν = c / μ
Hence, the speed of light in glass is 2 × 108 m/s.
(b) The speed of light in glass is not independent of the color of light.
The refractive index of a violet component of white light is greater than the refractive index of a red component. Hence, the speed of violet light is less than the speed of red light in glass. Hence, violet light travels slower than red light in a glass prism.
Question 10.4:
In Young’s double-slit experiment, the slits are separated by 0.28 mm and the screen is placed 1.4 m away. The distance between the central bright fringe and the fourth bright fringe is measured to be 1.2 cm. Determine the wavelength of light used in the experiment.
Solution:
Distance between the slits, d = 0.28 mm = 0.28 × 10−3 m
Distance between the slits and the screen, = 1.4 m
Distance between the central fringe and the fourth (n = 4) fringe,
= 1.2 cm = 1.2 × 10−2 m
In case of constructive interference, we have the relation for the distance between the two fringes as:
Where,
n = Order of fringes = 4
λ = Wavelength of light used
= 6 x 10-7
= 600 nm
Hence, the wavelength of the light is 600 nm.
Question 10.5:
In Young’s double-slit experiment using monochromatic light of wavelength λ, the intensity of light at a point on the screen where the path difference is λ, is units. What is the intensity of light at a point where the path difference is λ /3?
Solution:
Let I1 and I2 be the intensity of the two light waves. Their resultant intensities can be obtained as:
Where,
Φ = Phase difference between the two waves
For monochromatic light waves,
I1 = I2
Phase difference = (2π / λ) x Path difference
Since path difference = λ,
Phase difference, Φ = 2π
Given,
I’ = K
∴ I1
= K/4 ….. (1)
When path difference, Φ = 2π/3
Hence, the resultant intensity,
Using equation (1), we can write:
IR = I1 = K/4
Hence, the intensity of light at a point where the path difference is λ /3 is K/4 units.
Question 10.6:
A beam of light consisting of two wavelengths, 650 nm and 520 nm, is used to obtain interference fringes in a Young’s double-slit experiment.
(a) Find the distance of the third bright fringe on the screen from the central maximum for wavelength 650 nm.
(b) What is the least distance from the central maximum where the bright fringes due to both the wavelengths coincide?
Solution:
Wavelength of the light beam, λ1 = 650 nm
Wavelength of another light beam, λ2 = 520 nm
Distance of the slits from the screen = D
Distance between the two slits = d
(a) Distance of the nth bright fringe on the screen from the central maximum is given by the relation,
(b) Let the nth bright fringe due to wavelength and (n − 1)th bright fringe due to wavelength coincide on the screen. We can equate the conditions for bright fringes as:
Hence, the least distance from the central maximum can be obtained by the relation:
Note: The value of d and D are not given in the question.
Question 10.7:
In a double-slit experiment, the angular width of a fringe is found to be 0.2° on a screen placed 1 m away. The wavelength of light used is 600 nm. What will be the angular width of the fringe if the entire experimental apparatus is immersed in water? Take the refractive index of water to be 4/3.
Solution:
Distance of the screen from the slits, D = 1 m
The wavelength of light used, λ1 = 600 nm
The angular width of the fringe in air, θ1 = 0.2°
The angular width of the fringe in water = θ2
Refractive index of water, μ = 4/3
Refractive index is related to angular width as:
Therefore, the angular width of the fringe in water will reduce to 0.15°.
What is the Brewster angle for air to glass transition? (Refractive index of glass = 1.5.)
Question 10.8:
What is the Brewster angle for air to glass transition? (Refractive index of glass = 1.5.)
Solution:
Refractive index of glass, μ = 1.5
Brewster angle = θ
Brewster angle is related to refractive index as:
tanθ = μ
θ = tan-1(1.5) = 56.31°
Therefore, the Brewster angle for air to glass transition is 56.31°.
Question 10.9:
Light of wavelength 5000 Å falls on a plane reflecting surface. What are the wavelength and frequency of the reflected light? For what angle of incidence is the reflected ray normal to the incident ray?
Solution:
Wavelength of incident light, λ = 5000 Å = 5000 × 10−10 m
Speed of light, c = 3 × 108 m
Frequency of incident light is given by the relation,
ν = c / λ
= 6 x 1014 Hz
The wavelength and frequency of incident light are the same as that of the reflected ray. Hence, the wavelength of reflected light is 5000 Å and its frequency is 6 × 1014 Hz.
When the reflected ray is normal to incident ray, the sum of the angle of incidence, ∠i and angle of reflection, ∠r is 90°
According to the law of reflection, the angle of incidence is always equal to the angle of reflection. Hence, we can write the sum as:
Therefore, the angle of incidence for the given condition is 45°.
Question 10.10:
Estimate the distance for which ray optics is good approximation for an aperture of 4 mm and wavelength 400 nm.
Solution:
Fresnel’s distance (ZF) is the distance for which the ray optics is a good approximation. It is given by the relation,
Zr = a2 / λ
Where,
Aperture width, a = 4 mm = 4 ×10−3 m
Wavelength of light, λ = 400 nm = 400 × 10−9 m
Therefore, the distance for which the ray optics is a good approximation is 40 m.
Question 10.11:
The 6563 Å Hα line emitted by hydrogen in a star is found to be red-
shifted by 15 Å. Estimate the speed with which the star is receding
from the Earth.
Solution:
Wavelength of Hα line emitted by hydrogen,
λ = 6563 Å
= 6563 × 10−10 m.
Star’s red-shift, (λ’ – λ) = 15 A°=15 x 10-10 m
Speed of light, c = 3 x 108 m/s
Let the velocity of the star receding away from the Earth be v.
The red shift is related with velocity as:
= 6.87 x 105 m/s
Therefore, the speed with which the star is receding away from the Earth is 6.87 × 105 m/s.
Question 10.12:
Explain how Corpuscular theory predicts the speed of light in a medium, say, water, to be greater than the speed of light in vacuum. Is the prediction confirmed by experimental determination of the speed of light in water? If not, which alternative picture of light is consistent with experiment?
Solution:
No; Wave theory
Newton’s corpuscular theory of light states that when light corpuscles strike the interface of two media from a rarer (air) to a denser (water) medium, the particles experience forces of attraction normal to the surface. Hence, the normal component of velocity increases while the component along the surface remains unchanged.
Hence, we can write the expression:
c sin i = v sin r ……. (i)
Where,
i = Angle of incidence
r = Angle of reflection
c = Velocity of light in air
v = Velocity of light in water
We have the relation for relative refractive index of water with respect to air as:
μ = ν / c
Hence, equation (i) reduces to
…… (ii)
But μ > 1
Hence, it can be inferred from equation (ii) that v > c. This is not possible since this prediction is opposite to the experimental results of > v.
The wave picture of light is consistent with the experimental results.
Question 10.13:
You have learnt in the text how Huygens’ principle leads to the laws of reflection and refraction. Use the same principle to deduce directly that a point object placed in front of a plane mirror produces a virtual image whose distance from the mirror is equal to the object distance from the mirror.
Solution:
Let an object at O be placed in front of a plane mirror MO’ at a distance r (as shown in the given figure).
A circle is drawn from the center (O) such that it just touches the plane mirror at point O’. According to Huygens’ Principle, XY is the wave front of incident light.
If the mirror is absent, then a similar wave front X’Y’ (as XY) would form behind O’ at distance r (as shown in the given figure)
X’Y’ can be considered as a virtual reflected ray for the plane mirror. Hence, a point object placed in front of the plane mirror produces a virtual image whose distance from the mirror is equal to the object distance (r).
Question 10.14:
Let us list some of the factors, which could possibly influence the speed of wave propagation:
(i) Nature of the source.
(ii) Direction of propagation.
(iii) Motion of the source and/or observer.
(iv) Wavelength.
(v) Intensity of the wave.
On which of these factors, if any, does
(a) The speed of light in a vacuum,
(b) The speed of light in a medium (say, glass or water), depends?
Solution:
(a) The speed of light in a vacuum i.e., 3 × 108 m/s (approximately) is a universal constant. It is not affected by the motion of the source, the observer, or both. Hence, the given factor does not affect the speed of light in a vacuum.
(b) Out of the listed factors, the speed of light in a medium depends on the wavelength of light in that medium.
Question 10.15:
For sound waves, the Doppler formula for frequency shift differs slightly between the two situations: (i) source at rest; observer moving, and (ii) source moving; observer at rest. The exact Doppler formulas for the case of light waves in vacuum are, however, strictly identical for these situations. Explain why this should be so. Would you expect the formulas to be strictly identical for the two situations in case of light travelling in a medium?
Solution:
No
Sound waves can propagate only through a medium. The two given situations are not scientifically identical because the motion of an observer relative to a medium is different in the two situations. Hence, the Doppler formulas for the two situations cannot be the same.
In case of light waves, sound can travel in a vacuum. In a vacuum, the above two cases are identical because the speed of light is independent of the motion of the observer and the motion of the source. When light travels in a medium, the above two cases are not identical because the speed of light depends on the wavelength of the medium.
Question 10.16:
In double-slit experiment using light of wavelength 600 nm, the angular width of a fringe formed on a distant screen is 0.1º. What is the spacing between the two slits?
Solution:
Wavelength of light used, λ = 6000 nm = 600 × 10−9 m
Angular width of fringe,
Angular width of a fringe is related to slit spacing (d) as:
θ = λ / d
⇒ d = λ / θ
= 3.44 x 10-4
Therefore, the spacing between the slits is 3.44 x 10-4 m.
Question 10.17:
(a) In a single slit diffraction experiment, the width of the slit is made double the original width. How does this affect the size and intensity of the central diffraction band?
(b) In what way is diffraction from each slit related to the interference pattern in a double-slit experiment?
(c) When a tiny circular obstacle is placed in the path of light from a distant source, a bright spot is seen at the centre of the shadow of the obstacle. Explain why?
(d) Two students are separated by a 7 m partition wall in a room 10 m high. If both light and sound waves can bend around obstacles, how is it that the students are unable to see each other even though they can converse easily.
(e) Ray optics is based on the assumption that light travels in a straight line. Diffraction effects (observed when light propagates through small apertures/slits or around small obstacles) disprove this assumption. Yet the ray optics assumption is so commonly used in understanding location and several other properties of images in optical instruments. What is the justification?
Solution:
(a) In a single slit diffraction experiment, if the width of the slit is made double the original width, then the size of the central diffraction band reduces to half and the intensity of the central diffraction band increases up to four times.
(b) The interference pattern in a double-slit experiment is modulated by diffraction from each slit. The pattern is the result of the interference of the diffracted wave from each slit.
(c) When a tiny circular obstacle is placed in the path of light from a distant source, a bright spot is seen at the centre of the shadow of the obstacle. This is because light waves are diffracted from the edge of the circular obstacle, which interferes constructively at the centre of the shadow. This constructive interference produces a bright spot.
(d) Bending of waves by obstacles by a large angle is possible when the size of the obstacle is comparable to the wavelength of the waves.
On the one hand, the wavelength of the light waves is too small in comparison to the size of the obstacle. Thus, the diffraction angle will be very small. Hence, the students are unable to see each other. On the other hand, the size of the wall is comparable to the wavelength of the sound waves. Thus, the bending of the waves takes place at a large angle. Hence, the students are able to hear each other.
(e) The justification is that in ordinary optical instruments, the size of the aperture involved is much larger than the wavelength of the light used.
Question 10.18:
Two towers on top of two hills are 40 km apart. The line joining them passes 50 m above a hill halfway between the towers. What is the longest wavelength of radio waves, which can be sent between the towers without appreciable diffraction effects?
Solution:
Distance between the towers, = 40 km
Height of the line joining the hills, d = 50 m.
Since the hill is located halfway between the towers, Fresnel’s distance can be obtained as:
ZP = 20 km = 2 × 104 m
Aperture can be taken as:
d = 50 m
Fresnel’s distance is given by the relation,
ZP = a2 / λ
Where,
λ = Wavelength of radio waves
= 1250 x 10-4 = 0.1250 m = 12.5 cm
Therefore, the wavelength of the radio waves is 12.5 cm.
Question 10.19:
A parallel beam of light of wavelength 500 nm falls on a narrow slit and the resulting diffraction pattern is observed on a screen 1 m away. It is observed that the first minimum is at a distance of 2.5 mm from the centre of the screen. Find the width of the slit.
Solution:
Wavelength of light beam, λ = 500 nm = 500 × 10−9 m
Distance of the screen from the slit, D = 1 m
For first minima, n = 1
Distance between the slits = d
Distance of the first minimum from the centre of the screen can be obtained as:
x = 2.5 mm = 2.5 × 10−3 m
It is related to the order of minima as:
= 2 x 10-4 m = 0.2 mm
Therefore, the width of the slits is 0.2 mm.
Question 10.20:
(a) When a low flying aircraft passes overhead, we sometimes notice
a slight shaking of the picture on our TV screen. Suggest a possible explanation.
(b) As you have learned in the text, the principle of linear superposition of wave displacement is basic to understanding intensity distributions in diffraction and interference patterns. What is the justification of this principle?
Solution:
(a) Weak radar signals sent by a low flying aircraft can interfere with the TV signals received by the antenna. As a result, the TV signals may get distorted. Hence, when a low-flying aircraft passes overhead, we sometimes notice a slight shaking of the picture on our TV screen.
(b) The principle of linear superposition of wave displacement is essential to our understanding of intensity distributions and interference patterns. This is because superposition follows from the linear character of a differential equation that governs wave motion. If y1 and y2 are the solutions of the second-order wave equation, then any linear combination of y1 and ywill also be the solution of the wave equation.
Question 10.21:
In deriving the single slit diffraction pattern, it was stated that the intensity is zero at angles of nλ/a. Justify this by suitably dividing the slit to bring out the cancellation.
Solution:
Consider that a single slit of width d is divided into n smaller slits.
Width of each slit, d’ = d / n
The angle of diffraction is given by the relation,
Now, each of these infinitesimally small slits sends zero intensity in directionθ. Hence, the combination of these slits will give zero intensity. | 5,041 | 20,089 | {"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} | 3.640625 | 4 | CC-MAIN-2023-50 | latest | en | 0.859146 |
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# CalcI Complete
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CALCULUS I
Paul Dawkins
Calculus I© 2007 Paul Dawkins i http://tutorial.math.lamar.edu/terms.aspx
Preface ........................................................................................................................................... iii
Outline ........................................................................................................................................... iv
Review............................................................................................................................................. 2
Introduction
.............................................................................................................................................. 2
Review : Functions
................................................................................................................................... 4
Review : Inverse Functions
.....................................................................................................................14
Review : Trig Functions
..........................................................................................................................21
Review : Solving Trig Equations
.............................................................................................................28
Review : Solving Trig Equations with Calculators, Part I
.....................................................................37
Review : Solving Trig Equations with Calculators, Part II
....................................................................48
Review : Exponential Functions
.............................................................................................................53
Review : Logarithm Functions
................................................................................................................56
Review : Exponential and Logarithm Equations
...................................................................................62
Review : Common Graphs
.......................................................................................................................68
Limits ............................................................................................................................................ 80
Introduction
.............................................................................................................................................80
Rates of Change and Tangent Lines
........................................................................................................82
The Limit
..................................................................................................................................................91
One-Sided Limits
...................................................................................................................................101
Limit Properties
.....................................................................................................................................107
Computing Limits
..................................................................................................................................113
Infinite Limits
........................................................................................................................................121
Limits At Infinity, Part I
.........................................................................................................................130
Limits At Infinity, Part II
.......................................................................................................................139
Continuity
...............................................................................................................................................148
The Definition of the Limit
....................................................................................................................155
Derivatives .................................................................................................................................. 170
Introduction
...........................................................................................................................................170
The Definition of the Derivative
...........................................................................................................172
Interpretations of the Derivative
.........................................................................................................178
Differentiation Formulas
......................................................................................................................183
Product and Quotient Rule
...................................................................................................................191
Derivatives of Trig Functions
...............................................................................................................197
Derivatives of Exponential and Logarithm Functions
........................................................................208
Derivatives of Inverse Trig Functions
..................................................................................................213
Derivatives of Hyperbolic Functions
....................................................................................................219
Chain Rule
..............................................................................................................................................221
Implicit Differentiation
.........................................................................................................................231
Related Rates
.........................................................................................................................................240
Higher Order Derivatives
......................................................................................................................254
Logarithmic Differentiation
..................................................................................................................259
Applications of Derivatives ....................................................................................................... 262
Introduction
...........................................................................................................................................262
Rates of Change
......................................................................................................................................264
Critical Points
.........................................................................................................................................267
Minimum and Maximum Values
...........................................................................................................273
Finding Absolute Extrema
....................................................................................................................281
The Shape of a Graph, Part I
..................................................................................................................287
The Shape of a Graph, Part II
................................................................................................................296
The Mean Value Theorem
.....................................................................................................................305
Optimization
..........................................................................................................................................312
More Optimization Problems
...............................................................................................................326
Calculus I© 2007 Paul Dawkins ii http://tutorial.math.lamar.edu/terms.aspx
Indeterminate Forms and L’Hospital’s Rule
........................................................................................341
Linear Approximations
.........................................................................................................................347
Differentials
...........................................................................................................................................350
Newton’s Method
...................................................................................................................................353
...........................................................................................................................358
Integrals ...................................................................................................................................... 364
Introduction
...........................................................................................................................................364
Indefinite Integrals
................................................................................................................................365
Computing Indefinite Integrals
............................................................................................................371
Substitution Rule for Indefinite Integrals
............................................................................................381
More Substitution Rule
.........................................................................................................................394
Area Problem
.........................................................................................................................................407
The Definition of the Definite Integral
.................................................................................................417
Computing Definite Integrals
...............................................................................................................427
Substitution Rule for Definite Integrals
...............................................................................................439
Applications of Integrals ........................................................................................................... 450
Introduction
...........................................................................................................................................450
Average Function Value
........................................................................................................................451
Area Between Curves
............................................................................................................................454
Volumes of Solids of Revolution / Method of Rings
............................................................................465
Volumes of Solids of Revolution / Method of Cylinders
.....................................................................475
More Volume Problems
.........................................................................................................................483
Work
.......................................................................................................................................................494
Extras .......................................................................................................................................... 498
Introduction
...........................................................................................................................................498
Proof of Various Limit Properties
........................................................................................................499
Proof of Various Derivative Facts/Formulas/Properties
...................................................................510
Proof of Trig Limits
...............................................................................................................................523
Proofs of Derivative Applications Facts/Formulas
.............................................................................528
Proof of Various Integral Facts/Formulas/Properties
.......................................................................539
Area and Volume Formulas
..................................................................................................................551
Types of Infinity
.....................................................................................................................................555
Summation Notation
.............................................................................................................................559
Constants of Integration
.......................................................................................................................561 | 1,092 | 13,078 | {"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-2015-27 | latest | en | 0.317907 |
http://navalfacilities.tpub.com/dm2_08/dm2_080209.htm | 1,542,310,007,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039742906.49/warc/CC-MAIN-20181115182450-20181115204450-00378.warc.gz | 227,976,582 | 5,940 | Solution -Cont. - dm2_080209
Custom Search
m
= [(16.5 x 17) + 82]/(32.2 x 144)
= 0.078 lb-sec2/in2 or
= 78,179 lb-ms2/in2
KE = 160 EI/L4
= 160 x 30 x 106 x 882/(15.167 x 12)4
= 3.858.2 psi
TN
= 2[pi](0.72 x 78,179/3858.2)1/2
= 24.0
T/TN = 78/24.0 = 3.25
ru = 4Mp/(2C + 1) = 4(104) [(2 x 3.5) + 1]
H2
15.1672 x 12
= 1.206 K/in.
B = 5.8 x (17 x 12)/1000 = 1.183 k/in.
B/ru = 1.183/1.206 = 0.98
[mu] = Xm/XE = 3.60 > 3
N.G.
Check end rotation of columns.
XE = ru/KE = 1.206/3.858 = 0.313 in
Xm = 3.60 x 0.313 = 1.13 in
tan [theta] = Xm/(L/2) = 1.13/(7.58)(12) = 0.0124
[theta] = 0.71deg. < 1deg. O.K.
(8) (a) The deflections of the local mechanisms exceed the
criteria. The sidesway deflection is acceptable.
(b)
Roof girder:
[mu] = 3.50 from Step 7; increase trial size from
W12 x 35 to W12 x 40.
(c)
Front wall:
[mu] = 3.60 from Step 7; increase trial size
from W14 x 82 to W14 x 90.
2.08-192 | 457 | 894 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2018-47 | latest | en | 0.611125 |
https://brainly.com/question/372873 | 1,484,756,334,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560280308.24/warc/CC-MAIN-20170116095120-00360-ip-10-171-10-70.ec2.internal.warc.gz | 795,809,126 | 9,011 | • Brainly User
2015-03-25T22:58:07-04:00
### This Is a Certified Answer
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The answer is negative i because they cancel each other out except for the i
2015-03-25T23:00:29-04:00
### This Is a Certified Answer
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Multiply by (7−4i)/(7−4i) to make the denominator of (4−7i)/(7+4i) real.
(4−7i/7+4i)(7−4i/7−4i)
Expand (7+4i)(7−4i) using the FOIL Method.
(4−7i)(7−4i)/7(7)+7(−4i)+4i(7)+4i(−4i)
Simplify.
(4−7i)(7−4i)/65
Expand (4−7i)(7−4i) using the FOIL Method.
4(7)+4(−4i)−7i(7)−7i(−4i)/65
Simplify each term.
28−16i−49i−28/65
Simplify
−65i/65
−i | 380 | 1,098 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2017-04 | latest | en | 0.825111 |
https://www.nag.com/numeric/nl/nagdoc_28.7/flhtml/s/s15arf.html | 1,685,317,503,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224644571.22/warc/CC-MAIN-20230528214404-20230529004404-00358.warc.gz | 1,009,869,595 | 4,839 | # NAG FL Interfaces15arf (erfc_real_vector)
## ▸▿ Contents
Settings help
FL Name Style:
FL Specification Language:
## 1Purpose
s15arf returns an array of values of the complementary error function, $\mathrm{erfc}\left(x\right)$.
## 2Specification
Fortran Interface
Subroutine s15arf ( n, x, f,
Integer, Intent (In) :: n Integer, Intent (Inout) :: ifail Real (Kind=nag_wp), Intent (In) :: x(n) Real (Kind=nag_wp), Intent (Out) :: f(n)
#include <nag.h>
void s15arf_ (const Integer *n, const double x[], double f[], Integer *ifail)
The routine may be called by the names s15arf or nagf_specfun_erfc_real_vector.
## 3Description
s15arf calculates approximate values for the complement of the error function
$erfc(x) = 2π ∫x∞ e-t2 dt = 1-erf(x) ,$
for an array of arguments ${x}_{\mathit{i}}$, for $\mathit{i}=1,2,\dots ,n$.
See s15adf for details on the algorithm used.
None.
## 5Arguments
1: $\mathbf{n}$Integer Input
On entry: $n$, the number of points.
Constraint: ${\mathbf{n}}\ge 0$.
2: $\mathbf{x}\left({\mathbf{n}}\right)$Real (Kind=nag_wp) array Input
On entry: the argument ${x}_{\mathit{i}}$ of the function, for $\mathit{i}=1,2,\dots ,{\mathbf{n}}$.
3: $\mathbf{f}\left({\mathbf{n}}\right)$Real (Kind=nag_wp) array Output
On exit: $\mathrm{erfc}\left({x}_{i}\right)$, the function values.
4: $\mathbf{ifail}$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, ${\mathbf{n}}=⟨\mathit{\text{value}}⟩$.
Constraint: ${\mathbf{n}}\ge 0$.
${\mathbf{ifail}}=-99$
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
s15arf is not threaded in any implementation.
None.
## 10Example
This example reads values of x from a file, evaluates the function at each value of ${x}_{i}$ and prints the results.
### 10.1Program Text
Program Text (s15arfe.f90)
### 10.2Program Data
Program Data (s15arfe.d)
### 10.3Program Results
Program Results (s15arfe.r) | 974 | 3,305 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 35, "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-2023-23 | latest | en | 0.539065 |
https://www.physicsforums.com/threads/questions-about-rigid-rotating-dumbbell-molecule.533590/ | 1,542,566,836,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039744561.78/warc/CC-MAIN-20181118180446-20181118202446-00337.warc.gz | 975,567,918 | 13,861 | # Questions about rigid rotating dumbbell molecule
1. Sep 25, 2011
### creepypasta13
I would like some help with the following problems
1. Consider in R$^{j}$={f : f = $\Sigma^{l}_{m=-l}$ a$^{m}f^{l}_{m}$} the operator $\stackrel{\rightarrow}{e}$$\bullet$$\stackrel{\rightarrow}{J}$, where $\stackrel{\rightarrow}{e}$ is a unit vector in 3-dimensional space.
(a) Calculate the probabilities for all eigenvalues of $\stackrel{\rightarrow}{e}$$\bullet$$\stackrel{\rightarrow}{J}$ in the state W$^{j}$ = Tr($\Lambda$$^{j}$)$^{-1}$$\Lambda$$^{j}$, where $\Lambda$$^{j}$ is the projection operator onto Rj .
(b) Calculate the expectation value for the component J$_{2}$ in the state W$^{j}$ .
2. What spaces R$^{l'}_{m'}$ are obtained when the operators (Q$_{\stackrel{+}{-}}$)$^{2}$ act on the space R$^{l}_{m}$?
3. Consider the rigidly rotating dumbbell molecule and let Q$_{i}$, J$_{i}$, i = 1, 2, 3 denote the position and angular momentum operators.
(a) Find a complete system of commuting observables.
(b) Explain the physical meaning of these observables and explain the meaning of their eigenvalues.
(c) Prove that the operators of your choice form a system of commuting observables.
Number 1 is really confusing me since we need the probabilities for ALL eigenvalues of $\stackrel{\rightarrow}{e}$$\bullet$$\stackrel{\rightarrow}{J}$, and we don't know what 'j' is.
To find, say, the probabilities for the eigenvalues of J$_{3}$, is it just
$\Sigma^{r}_{s=-r}$$\Sigma^{l}_{m=-l}$ |<a$^{s}f^{r}_{s}$ | J$_{3}$ | a$^{m}f^{l}_{m}$>| $^{2}$ = $\Sigma^{l}_{m=-l}$m$^{2}$ ?
I am clueless as to how to solve #2
For #3, I found that because [J$_{i}$, Q$_{j}$] = i*h*$\epsilon$$_{i,j,k}$*Q$_{k}$, then they don't commute. Thus the CSCO is {Q$_{I}$, Q$_{j}$, Q$_{k}$}. Is this right?
Last edited: Sep 25, 2011 | 585 | 1,814 | {"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} | 3.03125 | 3 | CC-MAIN-2018-47 | latest | en | 0.755436 |
https://www.experts-exchange.com/questions/26638184/problem-in-Island-of-Knights-and-Knaves.html | 1,524,806,982,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125949036.99/warc/CC-MAIN-20180427041028-20180427061028-00423.warc.gz | 774,818,501 | 25,909 | x
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# problem in Island of Knights and Knaves
Following problems are from the Island of Knights and Knaves, I need help in solving it.
a. Suppose you come across two of the inhabitants. You ask both of them whether the other one is a knight. Will you get the same answer in each case?
b. There are three natives A, B and C. Suppose A says ‘B and C are the same type.’ What can be inferred about the number of knights?
c. What question should you ask A to determine whether B is a knight? Justify your question using the construction given above.
d. What question should you ask A to determine whether A and B are the same type? Justify your question using the construction given above.
e. Person A says ‘If B is a knave, I am a knave.’ Determine what can be deduced about A and B.
0
hiddenpearls
• 4
• 3
• 2
• +1
3 Solutions
Commented:
a. both will give the same answer
b. odd
c. there are an infinite number of answers to this, but perhaps you are looking for "Are A and B the same type?"
d. there are an infinite number of answers to this, but perhaps you are looking for "Is B a knight?"
d. A and B are knights.
0
Commented:
'How would a knave answer if I asked him if B is a knight or a knave'?
0
Author Commented:
@deighton
@ozo
b) c) and d)
0
Commented:
my answer to c doesn't work! if A is a knight - sorry!
0
Author Commented:
Thanks a lot
0
Commented:
d) Ask A 'is B a knight?'
If A is a knight he will answer yes if B is a knight, but no if B is a knave - so yes = same type
If B is a knave he will answer Yes if B is a knave (becasuse knaves lie), he will answer no if B is a knight - so again 'yes'=same type
the answer you get tells you the true answer to 'are you both the same type'
e. Person A says ‘If B is a knave, I am a knave.’ Determine what can be deduced about A and B.
If A is a knight, then for the statement to be true, B cannot be a knave, that would make the statement false
if A is a knave, then the statement is false, so B can not be a knave, because in that case the statement becomes true (B being a knave would indeed mean that A was a knave). If A is a knave and B is a knight - the statement could still have been false, since we don't know what the implication of B being a knight is supposed to be.
there is no problem with a Knave-Knight combination
so I say that A is unknown and B is always a knight
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Commented:
> and B is always a knight
correct.
Therefore the statement is true, and cannot be uttered by a knave.
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Commented:
ozo - A says '‘If B is a knave, I am a knave.’'
what if it is a world where knave-knave pairs may not exist? Then a knave could say it, but a knight couldn't say it. B could be a knight in reality, and A could be a knave then say the statement
0
Commented:
Problem e: A says
B is knave -> A is knave
Which is only false if B is a knave and A is a knight. Otherwise the statement is true.
Special note: If B is a knight and A is a knave the statement is TRUE it is not neutral. It is TRUE and cannot be said by a knave (this is what ozo was getting at)
From logic:
(X -> Y) = (notX or Y) so not(X->Y) = X and notY
``````X Y X->Y
T T T
T F T
F T F
F F T
``````
So
A B
Knight Knight statement is true and A is Knight (works)
Knight Knave statement is false and A is Knight (does not work)
Knave Knight statement is true and A is a knave (does not work)
Knave Knave statement is true and A is a knave (does not work)
Therefore they must both be knights.
If 'knave knave pairs can't exist' then F->T is still true.
0
Commented:
hiddenpearls, when you get problems like this for most of them you can easily just set up a truth table (i.e. just look at all the possible combinations for what could happen. Even with three characters, there are only 8 possible setups) Then just look at all the possible ways it could turn out.
You can even use this type of logic to find questions to ask the knights and knaves. Just set up the 4 or 8 options and see how they would answer questions.
0
Author Commented:
Thankyou so much for all of you guys ..
0
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