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https://www.coursehero.com/file/8806460/3ln0310012507ln071001-3927-where-X-given-the/ | 1,488,015,393,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501171706.94/warc/CC-MAIN-20170219104611-00639-ip-10-171-10-108.ec2.internal.warc.gz | 807,176,433 | 21,872 | Problem Set 3 Solutions
# 3ln0310012507ln071001 3927 where x given the
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Unformatted text preview: d Y*=.7m/Py. Therefore, utility given the tax is 0.3*ln(0.3*100/1.25)+0.7*ln(0.7*100/1) = 3.927 where X* given the tax is 24 and 0.25*24=6. Utils given the lump sum transfer is 0.3ln(0.3*94)+0.7ln(0.7*94) = 3.93243 which is higher. c) d) The first thing you need to do is calculate the optimal demand before the price change and after the price change. This you can do using the substitution method, but I’m sick of typing so I’ll just use the formulas. Before price change: X* = .3(100)/1 = 30 Y* = .7(100)/1 = 70 After price change: X* = .3(100)/1.25 = 24 Y* = .7(100)/1 = 70 For CV, we want to give the person enough money to get them back to the original utility level before the price change. So set .3ln(.3m/p ) + .7ln(.7m/p ) = .3ln(X*) + .7ln(Y*). x y It is important to note that X* and Y* are the optimal values before the price change, while p and p are the new prices after the price change (so the right-hand side is utility before the y price change, and on the left-hand side you just vary m...
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## This note was uploaded on 02/09/2014 for the course ECON 1130 taught by Professor Baum-snow during the Spring '11 term at Brown.
Ask a homework question - tutors are online | 446 | 1,444 | {"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} | 3.953125 | 4 | CC-MAIN-2017-09 | longest | en | 0.888659 |
https://www.gradesaver.com/textbooks/math/calculus/university-calculus-early-transcendentals-3rd-edition/chapter-3-section-3-5-derivatives-of-trigonometric-functions-exercises-page-151/12 | 1,576,406,829,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575541307813.73/warc/CC-MAIN-20191215094447-20191215122447-00305.warc.gz | 726,321,167 | 13,392 | ## University Calculus: Early Transcendentals (3rd Edition)
$\displaystyle y' = \frac{-1}{(1+sinx)}$
$\displaystyle y = \frac{cosx}{1+sinx}$ $\displaystyle y' = \frac{-(1+sinx)sinx - cos^{2}x}{(1+sinx)^{2}}$ $\displaystyle y' = \frac{-sinx -sin^{2}x- cos^{2}x}{(1+sinx)^{2}}$ $\displaystyle y' = \frac{-(1+sinx)}{(1+sinx)^{2}}$ $\displaystyle y' = \frac{-1}{(1+sinx)}$ | 161 | 369 | {"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} | 3.703125 | 4 | CC-MAIN-2019-51 | latest | en | 0.267723 |
http://www.manhattangmat.com/storeitemshow.cfm?ItemID=583&catid=12 | 1,408,580,570,000,000,000 | text/html | crawl-data/CC-MAIN-2014-35/segments/1408500813241.26/warc/CC-MAIN-20140820021333-00157-ip-10-180-136-8.ec2.internal.warc.gz | 482,746,894 | 10,706 | ### Description
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GMAT Navigator is your all-in-one GMAT dashboard: learn the best solution methods for Official Guide (OG) problems, analyze your strengths and weaknesses, and develop priorities for your studies. Time yourself on individual questions, tag problems as guesses, or mark for later review. Use performance statistics to help target your study plan. | 1,129 | 5,808 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2014-35 | latest | en | 0.865846 |
https://numberworld.info/102030201 | 1,675,748,706,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500384.17/warc/CC-MAIN-20230207035749-20230207065749-00253.warc.gz | 429,566,617 | 4,408 | # Number 102030201
### Properties of number 102030201
Cross Sum:
Factorization:
3 * 3 * 7 * 7 * 13 * 13 * 37 * 37
Divisors:
Count of divisors:
Sum of divisors:
190793421
Prime number?
No
Fibonacci number?
No
Bell Number?
No
Catalan Number?
No
Base 2 (Binary):
Base 3 (Ternary):
Base 4 (Quaternary):
Base 5 (Quintal):
Base 8 (Octal):
614db79
Base 32:
319mrp
sin(102030201)
-0.86452515811938
cos(102030201)
0.50258954523414
tan(102030201)
-1.7201415475458
ln(102030201)
18.440779415658
lg(102030201)
8.0087287422155
sqrt(102030201)
10101
Square(102030201)
### Number Look Up
Look Up
102030201 (one hundred two million thirty thousand two hundred one) is a very amazing figure. The cross sum of 102030201 is 9. If you factorisate the figure 102030201 you will get these result 3 * 3 * 7 * 7 * 13 * 13 * 37 * 37. 102030201 has 81 divisors ( 1, 3, 7, 9, 13, 21, 37, 39, 49, 63, 91, 111, 117, 147, 169, 259, 273, 333, 441, 481, 507, 637, 777, 819, 1183, 1369, 1443, 1521, 1813, 1911, 2331, 3367, 3549, 4107, 4329, 5439, 5733, 6253, 8281, 9583, 10101, 10647, 12321, 16317, 17797, 18759, 23569, 24843, 28749, 30303, 43771, 53391, 56277, 67081, 70707, 74529, 86247, 124579, 131313, 160173, 201243, 212121, 231361, 306397, 373737, 393939, 603729, 694083, 872053, 919191, 1121211, 1619527, 2082249, 2616159, 2757573, 4858581, 7848477, 11336689, 14575743, 34010067, 102030201 ) whith a sum of 190793421. The number 102030201 is not a prime number. 102030201 is not a fibonacci number. The number 102030201 is not a Bell Number. The number 102030201 is not a Catalan Number. The convertion of 102030201 to base 2 (Binary) is 110000101001101101101111001. The convertion of 102030201 to base 3 (Ternary) is 21002222200010100. The convertion of 102030201 to base 4 (Quaternary) is 12011031231321. The convertion of 102030201 to base 5 (Quintal) is 202104431301. The convertion of 102030201 to base 8 (Octal) is 605155571. The convertion of 102030201 to base 16 (Hexadecimal) is 614db79. The convertion of 102030201 to base 32 is 319mrp. The sine of 102030201 is -0.86452515811938. The cosine of the figure 102030201 is 0.50258954523414. The tangent of the number 102030201 is -1.7201415475458. The root of 102030201 is 10101.
If you square 102030201 you will get the following result 10410161916100401. The natural logarithm of 102030201 is 18.440779415658 and the decimal logarithm is 8.0087287422155. You should now know that 102030201 is very great figure! | 977 | 2,448 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2023-06 | latest | en | 0.62627 |
https://www.gradesaver.com/textbooks/math/geometry/geometry-common-core-15th-edition/chapter-5-relationships-within-triangles-mid-chapter-quiz-page-316/3 | 1,719,093,112,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862420.91/warc/CC-MAIN-20240622210521-20240623000521-00217.warc.gz | 684,379,125 | 13,875 | ## Geometry: Common Core (15th Edition)
Published by Prentice Hall
# Chapter 5 - Relationships Within Triangles - Mid-Chapter Quiz - Page 316: 3
#### Answer
$YZ = 104$
#### Work Step by Step
$\overline{AB}$ joins the midpoint of $\overline{XY}$ and also the midpoint of $\overline{XZ}$. According to the triangle midsegment theorem, if a line segment joins two sides of a triangle at their midpoints, then that line segment is parallel to the third side of that triangle and is half as long as that third side. Knowing this information, we can deduce that $\overline{YZ}$, which is the third side, is parallel to $\overline{AB}$, which is the line segment that joins the midpoints of the other two sides of the triangle. It also means that $\overline{YZ}$ is two times the length of $\overline{AB}$. So, if we know $AB$, then we can find $YZ$: $YZ = 2(AB)$ Let's plug in the values we know: $YZ = 2(52)$ Multiply to solve: $YZ = 104$
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback. | 287 | 1,100 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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} | 4.375 | 4 | CC-MAIN-2024-26 | latest | en | 0.912102 |
http://www.mathworks.com/matlabcentral/cody/players/870409-claudio-gelmi/badges | 1,485,147,342,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560282110.46/warc/CC-MAIN-20170116095122-00400-ip-10-171-10-70.ec2.internal.warc.gz | 575,685,848 | 11,806 | Cody
# Claudio Gelmi
Rank
Score
#### Solver+10
Earned on 2 Nov 2012 for solving Problem 1. Times 2 - START HERE.
#### Promoter+10
Earned on 18 Nov 2012 for liking Problem 147. Too mean-spirited.
#### Commenter+10
Earned on 18 Nov 2012 for commenting on Problem 290. Make one big string out of two smaller strings.
#### Creator+20
Earned on 14 Feb 2013 for creating Problem 1272. The almost-birthday problem..
#### Speed Demon+50
Earned on 14 Feb 2013 for first solving Problem 1272. The almost-birthday problem..
#### CUP Challenge Master+50
Earned on 11 Jul 2016 for solving all the problems in CUP Challenge.
Earned on 29 Dec 2012 for submitting the best solution Solution 183121.
#### Quiz Master+20
Earned on 27 Nov 2016 for having 50 or more solvers for Problem 1381. Celsius to Fahrenheit converter.
#### Tiles Challenge Master+50
Earned on 4 May 2013 for solving all the problems in Tiles Challenge.
#### ASEE Challenge Master+50
Earned on 18 May 2014 for solving all the problems in ASEE Challenge.
#### Cody Challenge Master+50
Earned on 15 Mar 2013 for solving all the problems in Cody Challenge.
#### Puzzler+50
Earned on 24 Mar 2013 for creating 10 problems.
#### Scholar+50
Earned on 13 Feb 2013 for solving 500 problems.
#### Modeling & Simulation Challenge Master+50
Solve all the problems in Modeling and Simulation Challenge problem group.
#### Magic Numbers Master+50
Solve all the problems in Magic Numbers problem group.
#### Famous+20
Must receive 25 total likes for the problems you created.
#### Indexing I Master+50
Solve all the problems in Indexing I problem group.
#### Renowned+20
Must receive 10 likes for a solution you submitted.
#### Celebrity+20
Must receive 50 total likes for the solutions you submitted.
#### Likeable+20
Must receive 10 likes for a problem you created.
#### Indexing II Master+50
Solve all the problems in Indexing II problem group.
#### Strings I Master+50
Solve all the problems in Strings I problem group.
#### R2016b Feature Challenge Master+50
Solve all the problems in R2016b Feature Challenge problem group.
#### Card Games Master+50
Solve all the problems in Card Games problem group. | 565 | 2,194 | {"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-2017-04 | latest | en | 0.835514 |
https://calculator.tutorpace.com/laplace-calculator-online-tutoring | 1,623,810,407,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487621699.22/warc/CC-MAIN-20210616001810-20210616031810-00133.warc.gz | 156,980,609 | 8,552 | #### A PHP Error was encountered
Severity: Warning
Message: count(): Parameter must be an array or an object that implements Countable
Filename: controllers/calculator.php
Line Number: 37
Laplace Calculator
# Laplace Calculator
## Online Tutoring Is The Easiest, Most Cost-Effective Way For Students To Get The Help They Need Whenever They Need It.
Laplace transform is a very useful method used in Calculus to solve given differential equations.Laplace
transform is an integral notation and the integral used to define it is an improper integral, with one of its limits
heading towards infinity. Laplace calculator is the quick online tool which can easily give the answers based
on Laplace transforms.
Example 1: Find the Laplace transform of the given function, f(t) = 2e3t + 4e-5t
<-Given function: 2e3t + 4e-5t
<-In order to find the Laplace transform for this function, we use the Standard Laplace formula:
<-If f(t) = eat then <-Laplace transform of the function,?(f(t)) = 1/s-a
<-Hence applying the above formula to the given function, we get
<-?(f(t)) = ?(2e3t) + ?(4e-5t)
<-?(f(t)) = 2/(s – 3) + 4/(s + 5)
Example 2: Find the Laplace transform of the given function, f(t) = e6t -5e-2t + 7
<-Given function: e6t - 5e-2t + 7
<-In order to find the Laplace transform for this function, we use the Standard Laplace formula:
<-If f(t) = eat then <-Laplace transform of the function, ?(f(t)) = 1/s-a
and if f(t) = 1 (constant) then <- ?(f(t)) = 1/s
<-Hence applying the above formulas to the given function, we get
<-?(f(t)) = ?(e6t) - ?(5e-2t) + ?(7)
<-?(f(t)) = 1/(s –6) -5/(s + 2) + 7(1/s) | 497 | 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} | 4.125 | 4 | CC-MAIN-2021-25 | longest | en | 0.714065 |
https://paytodoexam.com/what-is-a-discrete-fourier-transform | 1,716,239,292,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058313.71/warc/CC-MAIN-20240520204005-20240520234005-00824.warc.gz | 397,947,440 | 29,094 | # What is a discrete Fourier transform?
## What is a discrete Fourier transform?
What is a discrete Fourier transform? A discrete Fourier Transform (DFT) is a finite-dimensional integral that is applied to a subset of a discrete space of real numbers. Definition A DFT is a finite non-linear integral that is given by (1) a linear combination of the real and complex part of a complex number, or (2) a piecewise linear combination of two real and complex parts. Example Let’s consider the real-valued function $f(x) = \sum_{n} a_n x^n$. It is known that $f(a_n) = a_n$ is analytic in $[a_1, \ldots, a_n]$. The Cauchy integral of this function is Now, let’s consider the $n$th power of $a_n$ and let $x_n$, $x_m$ be the real and imaginary parts of $a_{n+1}$. Let $f(g) = \frac{1}{1+g^2}$ be a linear combination. It is known (e.g. by using the Cauchy transform) that $f$ is analytic and of the form Now let’s consider two real and two complex real functions $f$ and $g$. As can be seen, both $f$ are analytic. This example shows that the Cauchos function is a subset of the real function $f$. Example with two real and one complex functions Let us consider a discrete Fouristice with a real and a complex function $f$ We’ll consider two real functions $g$ and $h$ (a, b, c) This is an example of the discrete Fouristic integral of $f$ with a complex function. Note that the CFT is a discrete version of theWhat is a discrete Fourier transform? A discrete Fourier Transform (DFT) is a digital representation that represents the complex or frequency spectrum of a signal. The Fourier Transform is a commonly used method for obtaining a spectrum of a continuous signal. The Fourier Transform can be used to describe any frequency spectrum of an input signal, and it can be used as a representation of the frequency spectrum of the input important source The Fou process is the process of transforming a continuous signal into a discrete signal. Sometimes a discrete Fourentino, called a *spectrale, is used as a process of transforming the continuous pop over to this web-site into the discrete signal. For example, the Fourier Transform of a continuous sinusoid can be computed by the following equation: 1.913817384048 = 1.91381817384049 + 0.
## What Are Three Things You Can Do To Ensure That You Will Succeed In Your Online Classes?
00000008 So the amplitude of the spectrum of the continuous signal is divided special info follows: 2.91382044607 = 1.9518630373553 + 0.000000008 The frequency spectrum of this signal is divided into two parts: 3.913819011595 = 0.0000000135917074 + 0.0000008 4.913812011595 is divided into three parts: 1.313819011697 = 0.000000000000004 + 0 These three parts are the frequency spectrum and the amplitude spectrum of the signal, respectively. The click over here now of the signal is also divided into three frequencies: 4a.913811011697 + 0.000001151381713 + 0 4b.9138123011697 – 0.0000000333333336 + 0 5.9138135011697 | 0.00000000000508 find here 0 6.913814011697 A spectrum will be referred to as a frequency spectrum, or a frequency spectrumWhat is a discrete Fourier transform? A discrete Fourier Transform (DFT) is a means of measuring the frequency distribution my link input signals. The DFT is an algorithm find more info uses a discrete Fouxxor (DFX) to produce a discrete spectrum of the input signal. The frequency of a signal is determined by the frequency of the input and the amplitude of the signal.
## Law Will Take Its Own Course Meaning
The input and the output of the DFT are the same, but they are different frequencies. The DFT has a fundamental spectrum of frequencies with a discrete Fouxxx band. The frequency spectrum of a signal, or spectrum, is a series of frequencies, which are measured in the frequency domain. The discrete spectrum of a spectrum is a series, together with the frequencies in the discrete domain. In the context of the Fouxxor, a term is used to describe the frequency of a single signal, and the term is used with the frequency of all other frequencies. The Fouxxor is used as a means of determining the frequency of an input signal, or discrete spectrum of an input. A “tractable” frequency spectrum is a spectrum that can be measured using a Fouxxor. It can be measured by a single instrument. The frequency spectra of the instruments are often referred to as “turbaively measured”. Most cheat my medical assignment implementations use an analog or digital Fouxxor to determine the frequency of each signal. In the context of Fouxxor implementation, the digital Fouxxxor is a discrete-valued Fouxxor that can be used to determine the spectrum in the frequency range that the instrument is interested in. There are several ways to calculate a Fouxxx. The most common way is to compute a discrete Foux, which is YOURURL.com function of the number of discrete frequencies and the number of time constants, and the number and time constants of the frequency domain of the signal being measured. There are also several other ways to measure a
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https://math.answers.com/basic-math/What_is_18702_rounded_to_the_nearest_thousand | 1,721,299,952,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514828.10/warc/CC-MAIN-20240718095603-20240718125603-00461.warc.gz | 338,843,885 | 48,594 | 0
# What is 18702 rounded to the nearest thousand?
Updated: 4/28/2022
Wiki User
13y ago
19000
Elmira Ebert
Lvl 9
2y ago
Wiki User
13y ago
Write 18702
Underline the 8. It is in the thousand position.
Circle the 7. It will determine if 8 stays the same or goes up.
It is 5 or greater so the 8 goes to 9.
19000 is the closest thousand. | 115 | 346 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2024-30 | latest | en | 0.85245 |
https://www.classcentral.com/course/automated-reasoning-sat-12009?utm_source=cc_mooc_report&utm_medium=web&utm_campaign=new_courses_september_2018 | 1,611,071,375,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703519395.23/warc/CC-MAIN-20210119135001-20210119165001-00569.warc.gz | 731,551,158 | 87,990 | Class Central is learner-supported. When you buy through links on our site, we may earn an affiliate commission.
# Automated Reasoning: satisfiability
## Overview
##### Class Central Tips
In this course you will learn how to apply satisfiability (SAT/SMT) tools to solve a wide range of problems.
Several basic examples are given to get the flavor of the applications: fitting rectangles to be applied for printing posters, scheduling problems, solving puzzles, and program correctness. Also underlying theory is presented: resolution as a basic approach for propositional satisfiability, the CDCL framework to scale up for big formulas, and the simplex method to deal with linear inequallities.
The light weight approach to following this course is just watching the lectures and do the corresponding quizzes. To get a flavor of the topic this may work out fine. However, the much more interesting approach is to use this as a basis to apply SAT/SMT yourself on several problems, for instance on the problems presented in the honor's assignment.
## Syllabus
SAT/SMT basics, SAT examples
-This module introduces SAT (satisfiability) and SMT (SAT modulo theories) from scratch, and gives a number of examples of how to apply SAT.
SMT applications
-This module shows a number of applications of satisfiability modulo the theory of linear inequalities (SMT)
Theory and algorithms for CNF-based SAT
-This module describes how a rule called Resolution serves to determine whether a propositional formula in conjunctive normal form (CNF) is unsatisfiable. It is shown how an approach called DPLL does the same job, and how it is related to resolution. Finally, it is shown how current SAT solvers essentially implement and optimize DPLL.
Theory and algorithms for SAT/SMT
-This module consists of two parts.
The first part is about transforming arbitrary propositional formulas to CNF, leading to the Tseitin transformation doing this job such that the size of the transformed formula is linear in the size of the original formula.
The second part is about extending SAT to SMT, in particular to dealing with linear inequalities. It is shown how the Simplex method for linear optimization serves for this job; the Simplex method itself is explained in detail.
Hans Zantema
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Get personalized course recommendations, track subjects and courses with reminders, and more. | 511 | 2,499 | {"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-2021-04 | latest | en | 0.910363 |
http://www.netlib.org/templates/double/CGSREVCOM.f | 1,371,535,939,000,000,000 | text/plain | crawl-data/CC-MAIN-2013-20/segments/1368706961352/warc/CC-MAIN-20130516122241-00002-ip-10-60-113-184.ec2.internal.warc.gz | 490,908,430 | 3,854 | * SUBROUTINE CGSREVCOM(N, B, X, WORK, LDW, ITER, RESID, INFO, \$ NDX1, NDX2, SCLR1, SCLR2, IJOB) * * -- Iterative template routine -- * Univ. of Tennessee and Oak Ridge National Laboratory * October 1, 1993 * Details of this algorithm are described in "Templates for the * Solution of Linear Systems: Building Blocks for Iterative * Methods", Barrett, Berry, Chan, Demmel, Donato, Dongarra, * Eijkhout, Pozo, Romine, and van der Vorst, SIAM Publications, * 1993. (ftp netlib2.cs.utk.edu; cd linalg; get templates.ps). * * .. Scalar Arguments .. INTEGER N, LDW, ITER, INFO DOUBLE PRECISION RESID INTEGER NDX1, NDX2 DOUBLE PRECISION SCLR1, SCLR2 INTEGER IJOB * .. * .. Array Arguments .. DOUBLE PRECISION X( * ), B( * ), WORK( LDW,* ) * .. * * Purpose * ======= * * CGS solves the linear system Ax = b using the * Conjugate Gradient Squared iterative method with preconditioning. * * Convergence test: ( norm( b - A*x ) / norm( b ) ) < TOL. * For other measures, see the above reference. * * Arguments * ========= * * N (input) INTEGER. * On entry, the dimension of the matrix. * Unchanged on exit. * * B (input) DOUBLE PRECISION array, dimension N. * On entry, right hand side vector B. * Unchanged on exit. * * X (input/output) DOUBLE PRECISION array, dimension N. * On input, the initial guess. This is commonly set to * the zero vector. The user should be warned that for * this particular algorithm, an initial guess close to * the actual solution can result in divergence. * On exit, the iterated solution. * * WORK (workspace) DOUBLE PRECISION array, dimension (LDW,7) * Workspace for residual, direction vector, etc. * Note that vectors PHAT and QHAT, and UHAT and VHAT share * the same workspace. * * LDW (input) INTEGER * The leading dimension of the array WORK. LDW >= max(1,N). * * ITER (input/output) INTEGER * On input, the maximum iterations to be performed. * On output, actual number of iterations performed. * * RESID (input/output) DOUBLE PRECISION * On input, the allowable convergence measure for * norm( b - A*x ) / norm( b ). * On ouput, the final value of this measure. * * INFO (output) INTEGER * * = 0: Successful exit. * > 0: Convergence not achieved. This will be set * to the number of iterations performed. * * < 0: Illegal input parameter, or breakdown occured * during iteration. * * Illegal parameter: * * -1: matrix dimension N < 0 * -2: LDW < N * -3: Maximum number of iterations ITER <= 0. * -5: Erroneous NDX1/NDX2 in INIT call. * -6: Erroneous RLBL. * * BREAKDOWN: If RHO become smaller than some tolerance, * the program will terminate. Here we check * against tolerance BREAKTOL. * * -10: RHO < BREAKTOL: RHO and RTLD have become * orthogonal. * * NDX1 (input/output) INTEGER. * NDX2 On entry in INIT call contain indices required by interface * level for stopping test. * All other times, used as output, to indicate indices into * WORK[] for the MATVEC, PSOLVE done by the interface level. * * SCLR1 (output) DOUBLE PRECISION. * SCLR2 Used to pass the scalars used in MATVEC. Scalars are reqd because * original routines use dgemv. * * IJOB (input/output) INTEGER. * Used to communicate job code between the two levels. * * BLAS CALLS: DAXPY, DCOPY, DDOT, DNRM2, DSCAL * ============================================================= * * .. Parameters .. DOUBLE PRECISION ONE, ZERO PARAMETER ( ONE = 1.0D+0 , ZERO = 0.0D+0 ) * .. * .. Local Scalars .. INTEGER R, RTLD, P, PHAT, Q, QHAT, U, UHAT, VHAT, \$ MAXIT, NEED1, NEED2 DOUBLE PRECISION TOL, ALPHA, BETA, BNRM2, RHO, RHO1, RHOTOL, \$ GETBREAK, DDOT, DNRM2 * .. * indicates where to resume from. Only valid when IJOB = 2! INTEGER RLBL * * saving all. SAVE * * .. External Functions .. EXTERNAL GETBREAK, DAXPY, DCOPY, DDOT, DNRM2, DSCAL * .. * .. Intrinsic Functions .. INTRINSIC ABS, MAX * .. * .. Executable Statements .. * * Entry point, test IJOB IF (IJOB .eq. 1) THEN GOTO 1 ELSEIF (IJOB .eq. 2) THEN * here we do resumption handling IF (RLBL .eq. 2) GOTO 2 IF (RLBL .eq. 3) GOTO 3 IF (RLBL .eq. 4) GOTO 4 IF (RLBL .eq. 5) GOTO 5 IF (RLBL .eq. 6) GOTO 6 IF (RLBL .eq. 7) GOTO 7 * if neither of these, then error INFO = -6 GOTO 20 ENDIF * * ***************** 1 CONTINUE ***************** * INFO = 0 MAXIT = ITER TOL = RESID * * Alias workspace columns. * R = 1 RTLD = 2 P = 3 PHAT = 4 Q = 5 QHAT = 6 U = 6 UHAT = 7 VHAT = 7 * * Check if caller will need indexing info. * IF( NDX1.NE.-1 ) THEN IF( NDX1.EQ.1 ) THEN NEED1 = ((R - 1) * LDW) + 1 ELSEIF( NDX1.EQ.2 ) THEN NEED1 = ((RTLD - 1) * LDW) + 1 ELSEIF( NDX1.EQ.3 ) THEN NEED1 = ((P - 1) * LDW) + 1 ELSEIF( NDX1.EQ.4 ) THEN NEED1 = ((PHAT - 1) * LDW) + 1 ELSEIF( NDX1.EQ.5 ) THEN NEED1 = ((Q - 1) * LDW) + 1 ELSEIF( NDX1.EQ.6 ) THEN NEED1 = ((QHAT - 1) * LDW) + 1 ELSEIF( NDX1.EQ.7 ) THEN NEED1 = ((U - 1) * LDW) + 1 ELSEIF( NDX1.EQ.8 ) THEN NEED1 = ((UHAT - 1) * LDW) + 1 ELSEIF( NDX1.EQ.9 ) THEN NEED1 = ((VHAT - 1) * LDW) + 1 ELSE * report error INFO = -5 GO TO 20 ENDIF ELSE NEED1 = NDX1 ENDIF * IF( NDX2.NE.-1 ) THEN IF( NDX2.EQ.1 ) THEN NEED2 = ((R - 1) * LDW) + 1 ELSEIF( NDX2.EQ.2 ) THEN NEED2 = ((RTLD - 1) * LDW) + 1 ELSEIF( NDX2.EQ.3 ) THEN NEED2 = ((P - 1) * LDW) + 1 ELSEIF( NDX2.EQ.4 ) THEN NEED2 = ((PHAT - 1) * LDW) + 1 ELSEIF( NDX2.EQ.5 ) THEN NEED2 = ((Q - 1) * LDW) + 1 ELSEIF( NDX2.EQ.6 ) THEN NEED2 = ((QHAT - 1) * LDW) + 1 ELSEIF( NDX2.EQ.7 ) THEN NEED2 = ((U - 1) * LDW) + 1 ELSEIF( NDX2.EQ.8 ) THEN NEED2 = ((UHAT - 1) * LDW) + 1 ELSEIF( NDX2.EQ.9 ) THEN NEED2 = ((VHAT - 1) * LDW) + 1 ELSE * report error INFO = -5 GO TO 20 ENDIF ELSE NEED2 = NDX2 ENDIF * * Set breakdown tolerance parameter. * RHOTOL = GETBREAK() * * Set initial residual. * CALL DCOPY( N, B, 1, WORK(1,R), 1 ) IF ( DNRM2( N, X, 1 ).NE.ZERO ) THEN *********CALL MATVEC( -ONE, X, ONE, WORK(1,R) ) * Note: using RTLD[] as temp. storage. *********CALL DCOPY(N, X, 1, WORK(1,RTLD), 1) SCLR1 = -ONE SCLR2 = ONE NDX1 = -1 NDX2 = ((R - 1) * LDW) + 1 * * Prepare for resumption & return RLBL = 2 IJOB = 3 RETURN * ***************** 2 CONTINUE ***************** * IF ( DNRM2( N, WORK(1,R), 1 ).LE.TOL ) GO TO 30 ENDIF * BNRM2 = DNRM2( N, B, 1 ) IF ( BNRM2.EQ.ZERO ) BNRM2 = ONE * * Choose RTLD such that initially, (R,RTLD) = RHO is not equal to 0. * Here we choose RTLD = R. * CALL DCOPY( N, WORK(1,R), 1, WORK(1,RTLD), 1 ) * ITER = 0 * 10 CONTINUE * * Perform Conjugate Gradient Squared iteration. * ITER = ITER + 1 * RHO = DDOT( N, WORK(1,RTLD), 1, WORK(1,R), 1 ) IF ( ABS( RHO ).LT.RHOTOL ) GO TO 25 * * Compute direction vectors U and P. * IF ( ITER.GT.1 ) THEN * * Compute U. * BETA = RHO / RHO1 CALL DCOPY( N, WORK(1,R), 1, WORK(1,U), 1 ) CALL DAXPY( N, BETA, WORK(1,Q), 1, WORK(1,U), 1 ) * * Compute P. * CALL DSCAL( N, BETA**2, WORK(1,P), 1 ) CALL DAXPY( N, BETA, WORK(1,Q), 1, WORK(1,P), 1 ) CALL DAXPY( N, ONE, WORK(1,U), 1, WORK(1,P), 1 ) ELSE CALL DCOPY( N, WORK(1,R), 1, WORK(1,U), 1 ) CALL DCOPY( N, WORK(1,U), 1, WORK(1,P), 1 ) ENDIF * * Compute direction adjusting scalar ALPHA. * *********CALL PSOLVE( WORK(1,PHAT), WORK(1,P) ) * NDX1 = ((PHAT - 1) * LDW) + 1 NDX2 = ((P - 1) * LDW) + 1 * Prepare for return & return RLBL = 3 IJOB = 2 RETURN * ***************** 3 CONTINUE ***************** * *********CALL MATVEC( ONE, WORK(1,PHAT), ZERO, WORK(1,VHAT) ) * NDX1 = ((PHAT - 1) * LDW) + 1 NDX2 = ((VHAT - 1) * LDW) + 1 * Prepare for return & return SCLR1 = ONE SCLR2 = ZERO RLBL = 4 IJOB = 1 RETURN * ***************** 4 CONTINUE ***************** * ALPHA = RHO / DDOT( N, WORK(1,RTLD), 1, WORK(1,VHAT), 1 ) * CALL DCOPY( N, WORK(1,U), 1, WORK(1,Q), 1 ) CALL DAXPY( N, -ALPHA, WORK(1,VHAT), 1, WORK(1,Q), 1 ) * * Compute direction adjusting vectORT UHAT. * PHAT is being used as temporary storage here. * CALL DCOPY( N, WORK(1,Q), 1, WORK(1,PHAT), 1 ) CALL DAXPY( N, ONE, WORK(1,U), 1, WORK(1,PHAT), 1 ) *********CALL PSOLVE( WORK(1,UHAT), WORK(1,PHAT) ) * NDX1 = ((UHAT - 1) * LDW) + 1 NDX2 = ((PHAT - 1) * LDW) + 1 * Prepare for return & return RLBL = 5 IJOB = 2 RETURN * ***************** 5 CONTINUE ***************** * * Compute new solution approximation vector X. * CALL DAXPY( N, ALPHA, WORK(1,UHAT), 1, X, 1 ) * * Compute residual R and check for tolerance. * *********CALL MATVEC( ONE, WORK(1,UHAT), ZERO, WORK(1,QHAT) ) * NDX1 = ((UHAT - 1) * LDW) + 1 NDX2 = ((QHAT - 1) * LDW) + 1 * Prepare for return & return SCLR1 = ONE SCLR2 = ZERO RLBL = 6 IJOB = 1 RETURN * ***************** 6 CONTINUE ***************** * CALL DAXPY( N, -ALPHA, WORK(1,QHAT), 1, WORK(1,R), 1 ) * *********RESID = DNRM2( N, WORK(1,R), 1 ) / BNRM2 *********IF ( RESID.LE.TOL ) GO TO 30 * NDX1 = NEED1 NDX2 = NEED2 * Prepare for resumption & return RLBL = 7 IJOB = 4 RETURN * ***************** 7 CONTINUE ***************** IF( INFO.EQ.1 ) GO TO 30 * IF ( ITER.EQ.MAXIT ) THEN INFO = 1 GO TO 20 ENDIF * RHO1 = RHO * GO TO 10 * 20 CONTINUE * * Iteration fails. * RLBL = -1 IJOB = -1 RETURN * 25 CONTINUE * * Set breakdown flag. * IF ( ABS( RHO ).LT.RHOTOL ) INFO = -10 * 30 CONTINUE * * Iteration successful; return. * INFO = 0 RLBL = -1 IJOB = -1 RETURN * * End of CGSREVCOM * END | 3,283 | 9,009 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2013-20 | latest | en | 0.628738 |
http://www.thesearchagents.com/2010/06/dirty-little-secrets-of-portfolio-theory/print/ | 1,464,132,429,000,000,000 | text/html | crawl-data/CC-MAIN-2016-22/segments/1464049273946.43/warc/CC-MAIN-20160524002113-00101-ip-10-185-217-139.ec2.internal.warc.gz | 856,120,089 | 6,923 | - The Search Agents - http://www.thesearchagents.com -
# Dirty Little Secrets of Portfolio Theory
Several search marketing firms (like Efficient Frontier) claim to have based their account management technology on Modern Portfolio Theory, the method for allocating capital among classes of assets that won Harry Markowitz the Nobel Prize in Economics. But, there are problems with portfolio theory that could easily get online marketers into trouble if they take companies like EF at their word.
Don’t misunderstand: The Search Agency uses a portfolio-based technique which is not derived from Modern Portfolio Theory (MPT) for the analysis of our own clients’ accounts. It’s partially from developing this technique that we were able to discover some of MPT’s dirty little secrets:
1. It’s actually just a production function.
Efficient Frontier shows diagrams on their website of ‘Conversions vs. Ad Cost’ and claims that the relationship between these quantities is called an ‘efficient frontier’. These diagrams are similar to the one shown below:
[1]
For different possible levels of spending, this chart shows how many conversions one can expect. There is a maximum amount of spending and conversions (marked by a terminal point) even if all the words in the account were pushed into first position. A green ‘X’ marks the actual performance of the account over the past week, for example.
EF calls this relationship an ‘efficient frontier’. Heck, they even named their company after it. But, unfortunately, it’s not an efficient frontier. It’s just a production function. (Check out the graphs in Wikipedia’s entry for ‘production function [2]‘, to see what I mean. Plotted on the vertical axis is ‘units of output per time’, like Conversions. The horizontal axis is ‘units of input per time’, like Ad Spending.)
To graph an actual efficient frontier we must plot the average return from a basket of assets versus the covariance of the value of that basket. The purpose of an actual efficient frontier is to determine the optimal allocation of assets – for instance, what percentage of stock (or bond) A and stock (or bond) B we should own in order to maximize our expected return for a given level of risk.
[3]
I realize ‘production function’ doesn’t sound as sexy as “efficient frontier”, but it’s the correct term. So either Efficient Frontier doesn’t know the difference between a production function and an efficient frontier, or they are simply hoping that you don’t.
2. There’s also a “Deficient Frontier”
Efficient Frontier’s version of portfolio theory is concerned with identifying the maximum number of conversions an account can expect for various levels of ad spending. But to judge how effective your account management is, you also need to consider the minimum number of conversions you can expect.
In the 1985 comedy Brewster’s Millions, Richard Pryor inherits a fortune, but to get the money he must spend one-tenth as much in 30 days and have absolutely nothing to show for it. Of course, the minimum number of conversions a given amount of ad spending could theoretically generate is always zero. But as Richard Pryor found out, it is extremely difficult to spend lots of money and get no results at all.
So, if we wish to gauge how well an account performed, we must look at how well it did relative to the range of reasonable expected levels, not just compared to the maximum. Looking at the production function above, the actual performance might appear to be quite good. But when we consider the production function representing the ‘deficient frontier’ to the graph, we see that the actual performance wasn’t as remarkable as it first appeared:
[4]
By neglecting to disclose the Deficient Frontier, the performance of an account can be made to look better than it actually is.
3. The ‘efficient frontier’ is not your target.
After you have identified the ‘maximum conversions’ production function, the next thing to realize is that this line is not your target. EF claims that any bid set that moves you closer to the production function improves the performance of the account, but in my blog post ‘Transgressing the Boundaries [5]‘ I showed that there is actually only one point on this line (which I call the “target point”) for which you should aim, based on budget limits or a CPA goal. To aim for just any point on the production function besides the target point will miss your goal.
4. The production function can move / be moved over time.
Though it is possible to improve the performance of an account by moving closer to the ‘maximum conversions’ production function, it is often easier to simply move the production function to higher values across the board. (Again, please read the section of the Wikipedia article on shifting a production function [2], if necessary.) Unfortunately, doing so generally requires labor-intensive, good old-fashioned account management (A/B testing, adding negative terms, pruning underperforming keywords, and such) and typically can’t be done by any currently automated means.
5. Using portfolio theory lets ‘deadwood’ float.
Proper account management should move the production function to higher values over time, but using portfolio theory to do this can undermine your efforts. Let’s say that we have an account containing thousands of keywords that overall are hitting their combined performance targets. The spending rate is within budget, the CPA is acceptable, and so forth. (Of course, some individual words might be above the target and some below the target, but overall their performance is acceptable.)
But let’s also say that just one of those words is terrible. Everyone agrees its tangential to the company’s business. It gets clicks and incurs cost, but never gets a conversion. Its presence might be penalizing the adgroup or account’s Quality Score. The rational thing to do is to delete this keyword from the account. But as long as the account’s combined performance is acceptable, then portfolio theory says that it is OK to leave it in. That is, even though the account’s performance could be improved slightly by removing the word, using portfolio theory will result in the word being left in, racking up ad spending for no good reason.
6. “Wall Street technology” almost destroyed the economy.
Speaking of spending for no good reason, despite the ongoing financial crisis and the unprecedented 1000-point intraday drop in the Dow Jones a few weeks ago, Efficient Frontier remains bafflingly proud of the fact that their technology is just like the high-risk methods that run Wall Street.
In their recent whitepaper ‘The Tale Behind the Tail [6]‘ (registration required to obtain copy), EF says, “Tail terms can be thought of as microcap stocks”. I won’t rehash my blog post, ‘Keywords Are Not Stocks [7]‘, debunking this idea, but regardless of how you think of keywords, EF’s thinking is profoundly tone-deaf given the current economic situation.
7. Even Warren Buffet dislikes portfolio theory.
In case you still might be enamored by ‘Wall Street technology’, realize that Warren Buffet, the greatest investor of all time, is not: “Modern Portfolio Theory tells you how to be average.” he was quoted in The Warren Buffet Way as saying, “But I think almost anybody can figure out how to do average in the fifth grade.”
8. It’s a bad way to bid.
According to Efficient Frontier’s descriptions of their own technology, they guess “billions of possible bids”, then select the set of bids with the best estimated performance. If a magician has you draw a card from a deck and then tells you that he can guess which card you chose using only a billion guesses, that would not be a very impressive trick. Marin Software’s bid management technology is able to directly calculate a bid from performance data without making ‘billions of guesses’. Google’s chief economist, Hal Varian, showed how to determine bids in his YouTube video ‘Google AdWords Bidding Tutorial [8]‘ without making billions of guesses too. Any system that requires making billions of guesses clearly doesn’t know how to bid optimally. | 1,652 | 8,132 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2016-22 | latest | en | 0.895264 |
https://questions.examside.com/past-years/jee/question/an-electric-appliance-supplies-6000-jmin-heat-to-the-system-jee-main-physics-units-and-measurements-brlqqbrg82gxttqi | 1,708,637,687,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473824.45/warc/CC-MAIN-20240222193722-20240222223722-00129.warc.gz | 492,703,822 | 50,733 | 1
JEE Main 2021 (Online) 26th August Morning Shift
+4
-1
An electric appliance supplies 6000 J/min heat to the system. If the system delivers a power of 90W. How long it would take to increase the internal energy by 2.5 $$\times$$ 103 J ?
A
2.5 $$\times$$ 102 s
B
4.1 $$\times$$ 101 s
C
2.4 $$\times$$ 103 s
D
2.5 $$\times$$ 101 s
2
JEE Main 2021 (Online) 26th August Morning Shift
+4
-1
The rms speeds of the molecules of Hydrogen, Oxygen and Carbon dioxide at the same temperature are VH, VO and VC respectively then :
A
VH > VO > VC
B
VC > VO > VH
C
VH = VO > VC
D
VH = VO = VC
3
JEE Main 2021 (Online) 27th July Evening Shift
+4
-1
One mole of an ideal gas is taken through an adiabatic process where the temperature rises from 27$$^\circ$$ C to 37$$^\circ$$ C. If the ideal gas is composed of polyatomic molecule that has 4 vibrational modes, which of the following is true? [R = 8.314 J mol$$-$$1 k$$-$$1]
A
work done by the gas is close to 332 J
B
work done on the gas is close to 582 J
C
work done by the gas is close to 582 J
D
work done on the gas is close to 332 J
4
JEE Main 2021 (Online) 27th July Evening Shift
+4
-1
Out of Syllabus
Two Carnot engines A and B operate in series such that engine A absorbs heat at T1 and rejects heat to a sink at temperature T. Engine B absorbs half of the heat rejected by Engine A and rejects heat to the sink at T3. When workdone in both the cases is equal, to value of T is :
A
$${2 \over 3}{T_1} + {3 \over 2}{T_3}$$
B
$${1 \over 3}{T_1} + {2 \over 3}{T_3}$$
C
$${3 \over 2}{T_1} + {1 \over 3}{T_3}$$
D
$${2 \over 3}{T_1} + {1 \over 3}{T_3}$$
EXAM MAP
Medical
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http://www.code10.info/index.php?option=com_content&view=article&id=59:hamming-distance&catid=38:cat_coding_algorithms_data-similarity&Itemid=57 | 1,560,994,746,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627999130.50/warc/CC-MAIN-20190620004625-20190620030625-00403.warc.gz | 222,648,021 | 8,824 | Home Algorithms Similarity Hamming distance 20 | 06 | 2019
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Hamming distance
Algorithms - Similarity
Written by Jan Schulz
Thursday, 15 May 2008 19:22
# Hamming distance
## Objective
The Hamming distance (Hamming 1950) is a metric expressing the distance between two objects by the number of mismatches among their pairs of variables. It is mainly used for string and bitwise analyses, but can also be useful for numerical variables. Although the basic Hamming distance is a metric, the here presented version allows to define a threshold. Variables having an absolute difference below the threshold are considered as equal. Using values larger than 0 for this threshold the triangle inequality could be violated for some calculated distances. Using 0 as threshold it is the original and metric Hamming distance, thresholds below 0 are not defined.
## Equation
In the equation dHAD is the Hamming distance between the objects i and j, k is the index of the respective variable reading y out of the total number of variables n. The Hamming distance itself gives the number of mismatches between the variables paired by k.
## Synonyms
There are no common synonyms.
## Usage
The Hamming distance is an important measurement for the detection of errors in information transmission (Hamming 1950). Beyond this application its usage is valuable for the investigation of e.g. ranked variables of entities coded by numerical tokens.
## Algorithm
The algorithm controls whether the data input matrix is rectangular or not. If not the function returns FALSE and a defined, but empty output matrix. When the matrix is rectangular the Hamming distance is calculated. Therefore the dimensions of the respective arrays of the output matrix and the titles for the rows and columns set. As the result is a square matrix, which is mirrored along the diagonal only values for one triangular half and the diagonal are computed. When errors occur during computation the function returns FALSE.
## Source
`Function dist_Hamming (InputMatrix : t2dVariantArrayDouble; aMaxDiff: Double; Var OutputMatrix : t2dVariantArrayDouble) : Boolean;// The function CalcHammingDistanceMatrix calculates the Hamming distance between// several cases, which are expected in the rows. The variables are expected in the columns.// Function returns FALSE if at least one cell can not be calculated. The result// matrix is returned in OutputMatrix.// (c) Dr. Jan Schulz, 04.May 2008; www.code10.infoVar OutputMatrixSize : Integer; InputCols : Integer; InputRows : Integer; RunnerY : Integer; RunnerX : Integer; i : Integer; UnequalItems : Integer; FirstVal : Double; SecondVal : Double;Begin // if one dimension is zero or matrix not rectangular If Not mtx_IsRectangular (InputMatrix, InputRows, InputCols) THen Begin //create an empty matrix, return FALSE and exit mtx_Create (OutputMatrix, 1, 1, 0, 'Erroneous Hamming distance matrix'); dist_Hamming := False; Exit; end; // maximum allowed difference between two variables needs to be positive aMaxDiff := Abs (aMaxDiff); // let's expect the best case ... dist_Hamming := True; // define and set the row dimension of the result matrix OutputMatrixSize := High (InputMatrix.Cells) + 1; SetLength (OutputMatrix.Cells, OutputMatrixSize); // create the column dimension of the array For RunnerY := Low (OutputMatrix.Cells) to High (Outputmatrix.Cells) do Begin SetLength (OutputMatrix.Cells [RunnerY], OutputMatrixSize); end; // define title of matrix OutputMatrix.MatrixName := 'Hamming distance matrix'; // Set Row/Col-Title for the new matrix SetLength (OutputMatrix.RowTitle, OutputMatrixSize); SetLength (OutPutMatrix.ColTitle, OutPutMatrixSize); For RunnerY := Low (InputMatrix.RowTitle) to High (InputMatrix.RowTitle) do Begin // names for rows and columns are the same in this triangualary matrix OutputMatrix.RowTitle [RunnerY] := InputMatrix.RowTitle [RunnerY]; OutputMatrix.ColTitle [RunnerY] := InputMatrix.RowTitle [RunnerY]; end; // compare every object For RunnerY := Low (OutputMatrix.Cells) to High (OutputMatrix.Cells) do Begin //with every other object For RunnerX := Low (OutputMatrix.Cells) to RunnerY do Begin UnequalItems := 0; //include all variables in analysis For i := 0 to High (InputMatrix.Cells [0]) do Begin FirstVal := InputMatrix.Cells [RunnerX, i]; SecondVal := InputMatrix.Cells [RunnerY, i]; // are variables equal within a certain range? If Abs (FirstVal - SecondVal) > aMaxDiff THen Begin // calculation impossible as invalid numbers were found UnequalItems := UnequalItems + 1; end; end; // set the calculated value on both sides of diagonal and diagonal itself OutputMatrix.Cells [RunnerX, RunnerY] := UnequalItems; OutputMatrix.Cells [RunnerY, RunnerX] := UnequalItems; end; end;end; `
## Example
For a data matrix aInputMatrix of the type t2dVariantArrayDouble, populated with:
Data Var1 Var2 Var3 Case1 1 1 1 Case2 1 1 0 Case3 2 2 2 Case4 10 10 10 Case5 11 11 11 Case6 10 5 0
the call of:
aBooleanVar := dist_Hamming (aInputMatrix, 0, aOutputMatrix);
returns the respective matrix of the original Hamming distance in aOutputMatrix:
Hamming distance Case1 Case2 Case3 Case4 Case5 Case6 Case1 0 1 3 3 3 3 Case2 1 0 3 3 3 2 Case3 3 3 0 3 3 3 Case4 3 3 3 0 3 2 Case5 3 3 3 3 0 3 Case6 3 2 3 2 3 0
The Hamming distance simply counts the number of differences between the paired variables. Thus the distances are unaffected by the distance of the object from the origin.
## Literature
Hamming R.W. (1950): Error detecting and error correcting codes. Bell System Technical Journal 26(2):147-160.
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https://www.homeschoolmath.net/worksheets/table-percentages-words.php?col=2&row=7&min=5&max=100&step=5&pmin=1&pmax=100&pstep=1&ndec=0&decimals=0&round=3&p1=1&p2=1&p4=1&font=Arial&FontSize=12pt&workspace=4&pad=8&border=1&color-green&ptitle=&Submit=Submit | 1,656,829,691,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104215790.65/warc/CC-MAIN-20220703043548-20220703073548-00355.warc.gz | 844,980,844 | 5,225 | Percentage Worksheet
Solve.
1 a. What percentage of 60 is 15?
1 b. Find what percentage of 85 is 50.
2 a. Find what percentage of 85 is 35.
2 b. 15 is what percentage of 25?
3 a. Find what percentage of 60 is 25.
3 b. Find what percentage of 100 is 45.
4 a. 75 is what percentage of 90?
4 b. Find what percentage of 90 is 30.
5 a. 35 is what percentage of 50?
5 b. 50 is what percentage of 50?
6 a. What percentage of 90 is 25?
6 b. 60 is what percentage of 65?
7 a. 40 is what percentage of 45?
7 b. What percentage of 25 is 20?
Page 2
Answers are rounded to 3 decimals.
1 a. 15 is 25% of 60
1 b. 50 is 58.824% of 85
2 a. 35 is 41.176% of 85
2 b. 15 is 60% of 25
3 a. 25 is 41.667% of 60
3 b. 45 is 45% of 100
4 a. 75 is 83.333% of 90
4 b. 30 is 33.333% of 90
5 a. 35 is 70% of 50
5 b. 50 is 100% of 50
6 a. 25 is 27.778% of 90
6 b. 60 is 92.308% of 65
7 a. 40 is 88.889% of 45
7 b. 20 is 80% of 25
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https://www.careercup.com/question?id=5750442676453376 | 1,553,010,112,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912201996.61/warc/CC-MAIN-20190319143502-20190319165502-00012.warc.gz | 733,590,186 | 15,919 | ## Amazon Interview Question for SDE-2s
Country: United States
Interview Type: In-Person
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1
of 1 vote
Sort one of the arrays of length N. Iterate the other array of length M and do a binary search in the first array updating the global maximum. O(N*log(N) + M*log(N))
``````def find(F, B, T):
ans = [0, 0, 0]
F = sorted([x, i] for i,x in F)
for idy,y in B:
f = 0
end = len(F)
z = T - y
while f != end:
m = (f + end)/2
if F[m][0] <= z:
f = m+1
else:
end = m
if f != 0 and y+F[f-1][0] > ans[0]:
ans = [y+F[f-1][0], F[f-1][1], idy]
return ans[1:]
print find([(1,3000),(2,5000),(3,4000),(4,10000)],
[(1,2000),(2,3000),(3,4000)], 11000)``````
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0
can you give java solution..
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0
What if there is more than a pair for the same optimal value? Also, what if we have same values with different Id's? Should the solution cover all those different value pairs with same distance but different Id's?
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0
``````package com.sample.test;
import java.util.Arrays;
import java.util.List;
import java.util.stream.Collectors;
public class OptimalFlightDistant {
public static void main(String[] args) {
// sample input
System.out.println(getIdPairsForOptimal(
Arrays.asList(Arrays.asList(1, 3000), Arrays.asList(2, 5000), Arrays.asList(3, 7000),
Arrays.asList(4, 10000)),
Arrays.asList(Arrays.asList(1, 2000), Arrays.asList(2, 9000), Arrays.asList(3, 5000)), 10000));
}
public static List<List<Integer>> getIdPairsForOptimal(List<List<Integer>> forwardList,
List<List<Integer>> backwardList, int maxDistance) {
forwardList = forwardList.stream().sorted((x1, x2) -> Integer.compare(x2.get(1), x1.get(1)))
.collect(Collectors.toList());
backwardList = backwardList.stream().sorted((x1, x2) -> Integer.compare(x1.get(1), x2.get(1)))
.collect(Collectors.toList());
int maxDist = maxDistance;
while (true) {
for (List<Integer> l : forwardList) {
for (List<Integer> b : backwardList) {
int forward = l.get(1);
int backward = b.get(1);
int tot = (forward + backward);
if (tot > maxDist) {
break;
}
if (tot == maxDist) {
// print the pair of Id and optimum distance
break;
}
}
}
if (result.size() > 0) {
break;
}
maxDist--;
}
return result;
}
}``````
Comment hidden because of low score. Click to expand.
0
``````import java.util.Arrays;
import java.util.List;
import java.util.stream.Collectors;
public class OptimalFlightDistant {
public static void main(String[] args) {
// sample input
System.out.println(getIdPairsForOptimal(
Arrays.asList(Arrays.asList(1, 3000), Arrays.asList(2, 5000), Arrays.asList(3, 7000),
Arrays.asList(4, 10000)),
Arrays.asList(Arrays.asList(1, 2000), Arrays.asList(2, 9000), Arrays.asList(3, 5000)), 10000));
}
public static List<List<Integer>> getIdPairsForOptimal(List<List<Integer>> forwardList,
List<List<Integer>> backwardList, int maxDistance) {
forwardList = forwardList.stream().sorted((x1, x2) -> Integer.compare(x2.get(1), x1.get(1)))
.collect(Collectors.toList());
backwardList = backwardList.stream().sorted((x1, x2) -> Integer.compare(x1.get(1), x2.get(1)))
.collect(Collectors.toList());
int maxDist = maxDistance;
while (true) {
for (List<Integer> l : forwardList) {
for (List<Integer> b : backwardList) {
int forward = l.get(1);
int backward = b.get(1);
int tot = (forward + backward);
if (tot > maxDist) {
break;
}
if (tot == maxDist) {
// print the pair of Id and optimum distance
break;
}
}
}
if (result.size() > 0) {
break;
}
maxDist--;
}
return result;
}
}``````
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0
package com.sample.test;
import java.util.Arrays;
import java.util.List;
import java.util.stream.Collectors;
public class OptimalFlightDistant {
public static void main(String[] args) {
// sample input
System.out.println(getIdPairsForOptimal(
Arrays.asList(Arrays.asList(1, 3000), Arrays.asList(2, 5000), Arrays.asList(3, 7000),
Arrays.asList(4, 10000)),
Arrays.asList(Arrays.asList(1, 2000), Arrays.asList(2, 9000), Arrays.asList(3, 5000)), 10000));
}
public static List<List<Integer>> getIdPairsForOptimal(List<List<Integer>> forwardList,
List<List<Integer>> backwardList, int maxDistance) {
forwardList = forwardList.stream().sorted((x1, x2) -> Integer.compare(x2.get(1), x1.get(1)))
.collect(Collectors.toList());
backwardList = backwardList.stream().sorted((x1, x2) -> Integer.compare(x1.get(1), x2.get(1)))
.collect(Collectors.toList());
int maxDist = maxDistance;
while (true) {
for (List<Integer> l : forwardList) {
for (List<Integer> b : backwardList) {
int forward = l.get(1);
int backward = b.get(1);
int tot = (forward + backward);
if (tot > maxDist) {
break;
}
if (tot == maxDist) {
// print the pair of Id and optimum distance
break;
}
}
}
if (result.size() > 0) {
break;
}
maxDist--;
}
return result;
}
}
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1
of 1 vote
Used a sliding window model and solved it in JS .
space complexity : O(M +N )
Time complexity : O( M + N ) + mLogM + nLogN
N = Number of fwd options
M = Number of backward options
Algorithm :
-----------------
- use a two pointer approach , one starting from 0 index of fwd list and another from last index of bwdlist.
-Keep track of best sum achieved so far, compare and increment pointers accordingly.
- store results when you find a better best / max .
---
Note : this works for multiple combinations , but would not work if either the fwd or the backward path distances are duplicated.
``````function findRoute(fwdList,bwdList,max) {
let fList = fwdList.sort((a,b) => a[1] > b[1]);
let bList = bwdList.sort((a,b) => a[1] > b[1]);
let fHash = new Map();;
let bHash = new Map();
for(let tuple of fList){
fHash.has(tuple[1]) || fHash.set(tuple[1],tuple[0]);
}
for(let tuple of bList ){
bHash.has(tuple[1]) || bHash.set(tuple[1],tuple[0]);
}
fList = fList.map(tuple => tuple[1]);
bList = bList.map(tuple => tuple[1]);
console.log(fList);
let result = [];
let left = 0;
let right = bList.length-1;
let best = 0;
while (left<fList.length && right>=0){
let sum = fList[left] + bList[right];
console.log(`left is \${left} , right is \${right}`);
console.log(`sum is \${sum}, max is \${max}`);
if(sum<=max){
if(sum>best) {
result = [];
result.push([fHash.get(fList[left]),bHash.get(bList[right])]);
best = sum;
}
else if(sum === best){
result.push([fHash.get(fList[left]),bHash.get(bList[right])]);
}
left++;
}
else if(sum>max){
right--;
}
}
return result.length === 0 ? [] :result.length>1? result : result[0];
}
let f = [[1,3000],[2,5000],[3,4000],[4,10000],[5,9000],[6,7000]];
let b = [[1,2000],[2,3000],[3,4000],[4,1000]];
console.log(findRoute(f,b,11000));``````
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0
of 0 vote
This can be done using maximum contiguous sum subarray.
1. First add forward and backward for each location. like 1 forward + 2 backward for 1st index i.e. ith forward + (i+1)th backward for ith index cost.
2. Now apply maximum contiguous sum subarray to get the subarray <i,j>. Answer would be <i, j+1>.
Time Complexity: O(n) + O(n).
Space Complexity: O(n). (for storing the result. we can make it O(1), if we reuse the any of the forward and backward input array).
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0
of 0 vote
What about the case when all pairs are valid?
forward-> [1,1000],[2,1000],[3,1000]
backward->[1,1000],[2,1000],[3,1000]
Reqd = 2000
Here all a pairs are possible - so how can you give in O(n)?
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0
For each direction (forward & return) the distances are unique
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0
of 0 vote
Two pointer approach:
``````public List<List<Integer>> findOptimalMemory(final int capacity, final List<List<Integer>> foreground,
final List<List<Integer>> background)
{
int i = 0;
int j = background.size() - 1;
final List<List<Integer>> result = new ArrayList<>();
final List<List<Integer>>[] store = new ArrayList[capacity + 1];
Collections.sort(foreground, new Comparator<List<Integer>>()
{
@Override
public int compare(final List<Integer> f, final List<Integer> s)
{
return Integer.compare(f.get(1), s.get(1));
}
});
Collections.sort(background, new Comparator<List<Integer>>()
{
@Override
public int compare(final List<Integer> f, final List<Integer> s)
{
return Integer.compare(f.get(1), s.get(1));
}
});
while (i < foreground.size() && j > -1)
{
final int current = foreground.get(i).get(1) + background.get(j).get(1);
if (current <= capacity)
{
if (store[current] == null)
store[current] = new ArrayList<>();
store[current].add(Arrays.asList(new Integer[] { foreground.get(i).get(0), background.get(j).get(0) }));
}
if (current <= capacity)
{
++i;
}
else if (current > capacity)
{
--j;
}
}
while (i < foreground.size())
{
final int current = foreground.get(i).get(1) + background.get(0).get(1);
if (current < capacity)
{
if (store[current] == null)
store[current] = new ArrayList<>();
store[current].add(Arrays.asList(new Integer[] { foreground.get(i).get(0), background.get(0).get(0) }));
}
++i;
}
while (j > -1)
{
final int current = foreground.get(foreground.size() - 1).get(1) + background.get(j).get(1);
if (current < capacity)
{
if (store[current] == null)
store[current] = new ArrayList<>();
store[current]
.add(Arrays.asList(new Integer[] { foreground.get(foreground.size() - 1).get(0), background.get(j).get(0) }));
}
--j;
}
for (int k = capacity; k > -1; --k)
{
if (store[k] != null)
{
break;
}
}
return result;
}``````
Comment hidden because of low score. Click to expand.
0
I think your solution returns all the pairs for which sum <= capacity but the question asks to find all the pairs for which sum <= capacity AND this there is no other pairs with their sum > this sum and <= capacity. Correct me if I am wrong.
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0
I think it's storing all the pairs but returning only the top pair using count sort and exiting from the loop once found a value that is not null. So, you get only the pair whose sum <= capacity and not all pairs.
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0
of 0 vote
Here the approach which i come to.
1 Sort the both list Backward and forward.
2. Implement sliding window technique
- take one variable for maxValue
- take two pointer one for assign first value Forward list let take it LEFT
- second assign to the last element of backward list let take it RIGHT
3 Start the loop with condition LEFT < forward.lenght And RIGHT >= 0
- sumValue = sum of LEFT index Value and Right Index value
if(maxTarget >= sum){
- if maxValue greater than clear the list and insert sum value to list result increment the left index by 1
- if maxValue == sum
add point id in list result;
else
left ++;
}else{
right--;
}
4 return the result list
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0
Time Complexity: O((n log n) + (m log m) + (m+n))
Space complexity max(m,n)
n is the length of forward route
m is the length of backward route
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0
Time Complexity O((n log n) + (m log m) + m + n)
Space Complexity o(Max(n,m))
n is the length of Forward routes
m is the length of backward routes
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0
With what value to compare for condition “if maxValue greater than clear the list ”
And why clear the list and add sum instead of point ids?
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0
With what value to compare in condition “if maxValue greater than clear the list ”
And why clear the list and and add sum to it ? Can you please put working code if you can ?
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0
of 0 vote
def closestXDestionation(forward, backward, maximum):
combined = [[i, j] for i in forward for j in backward]
lessThanMax = []
for i in range(len(combined)):
if combined[i][0][1] + combined[i][1][1] <= maximum:
lessThanMax.append([combined[i][0],combined[i][1]])
largest = max(lessThanMax, key=lambda x:x[0][1] + x[1][1])
route = [path[0] for path in largest]
print(route)
F = [[1,1000],[2,2000],[3,3000],[4,4000], [5,5000],[6,6000],[7,7000], [8, 8000]]
B = [[1,2000],[2,3000],[3,3000],[4,4000], [5,5000],[6,6000]]
m = 11000
closestXDestionation(F, B, m)
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0
of 0 vote
``````def closestXDestionation(forward, backward, maximum):
combined = [[i, j] for i in forward for j in backward]
lessThanMax = []
for i in range(len(combined)):
if combined[i][0][1] + combined[i][1][1] <= maximum:
lessThanMax.append([combined[i][0],combined[i][1]])
largest = max(lessThanMax, key=lambda x:x[0][1] + x[1][1])
route = [path[0] for path in largest]
print(route)
F = [[1,1000],[2,2000],[3,3000],[4,4000], [5,5000],[6,6000],[7,7000], [8, 8000]]
B = [[1,2000],[2,3000],[3,3000],[4,4000], [5,5000],[6,6000]]
m = 11000
closestXDestionation(F, B, m)``````
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0
of 0 vote
``````def closestXDestionation(forward, backward, maximum):
combined = [[i, j] for i in forward for j in backward]
lessThanMax = []
for i in range(len(combined)):
if combined[i][0][1] + combined[i][1][1] <= maximum:
lessThanMax.append([combined[i][0],combined[i][1]])
largest = max(lessThanMax, key=lambda x:x[0][1] + x[1][1])
route = [path[0] for path in largest]
print(route)
F = [[1,1000],[2,2000],[3,3000],[4,4000], [5,5000],[6,6000],[7,7000], [8, 8000]]
B = [[1,2000],[2,3000],[3,3000],[4,4000], [5,5000],[6,6000]]
m = 11000
closestXDestionation(F, B, m)``````
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0
of 0 vote
``def closestXDestionation(forward, backward, maximum):``
Comment hidden because of low score. Click to expand.
0
of 0 vote
``````// java code
import java.util.*;
public class OptimalRoute
{
public static void main(String[] args)
{
System.out.println(calculateOptimalRoute(7000, forwardList, returnList));
}
public static List<List<Integer>> calculateOptimalRoute(final int capacity, final List<List<Integer>> forwardList, final List<List<Integer>> returnList)
{
System.out.println(forwardList);
System.out.println(returnList);
// sort forward list
Collections.sort(forwardList, new Comparator<List<Integer>>() {
@Override
public int compare(List<Integer> o1, List<Integer> o2) {
return o1.get(1) - o2.get(1);
}
});
// sort return list
Collections.sort(returnList, new Comparator<List<Integer>>() {
public int compare(List<Integer> o1, List<Integer> o2) {
return o1.get(1) - o2.get(1);
}
});
int max = 0;
int i = 0;
int j = returnList.size() - 1;
List<List<Integer>> result = null;
while(i < forwardList.size() && j >= 0) {
if (forwardList.get(i).get(1) + returnList.get(j).get(1) > max &&
forwardList.get(i).get(1) + returnList.get(j).get(1) <= capacity) {
max = forwardList.get(i).get(1) + returnList.get(j).get(1);
i++;
} else if(forwardList.get(i).get(1) + returnList.get(j).get(1) == max) {
// no need to reset result list
i++;
} else {
j--;
}
}
return result;
}
}``````
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0
of 0 vote
``````package com.sample.test;
import java.util.Arrays;
import java.util.List;
import java.util.stream.Collectors;
public class OptimalFlightDistant {
public static void main(String[] args) {
// sample input
System.out.println(getIdPairsForOptimal(
Arrays.asList(Arrays.asList(1, 3000), Arrays.asList(2, 5000), Arrays.asList(3, 7000),
Arrays.asList(4, 10000)),
Arrays.asList(Arrays.asList(1, 2000), Arrays.asList(2, 9000), Arrays.asList(3, 5000)), 10000));
}
public static List<List<Integer>> getIdPairsForOptimal(List<List<Integer>> forwardList,
List<List<Integer>> backwardList, int maxDistance) {
forwardList = forwardList.stream().sorted((x1, x2) -> Integer.compare(x2.get(1), x1.get(1)))
.collect(Collectors.toList());
backwardList = backwardList.stream().sorted((x1, x2) -> Integer.compare(x1.get(1), x2.get(1)))
.collect(Collectors.toList());
int maxDist = maxDistance;
while (true) {
for (List<Integer> l : forwardList) {
for (List<Integer> b : backwardList) {
int forward = l.get(1);
int backward = b.get(1);
int tot = (forward + backward);
if (tot > maxDist) {
break;
}
if (tot == maxDist) {
// print the pair of Id and optimum distance
break;
}
}
}
if (result.size() > 0) {
break;
}
maxDist--;
}
return result;
}``````
}
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0
of 0 vote
``````package com.sample.test;
import java.util.Arrays;
import java.util.List;
import java.util.stream.Collectors;
public class OptimalFlightDistant {
public static void main(String[] args) {
// sample input
System.out.println(getIdPairsForOptimal(
Arrays.asList(Arrays.asList(1, 3000), Arrays.asList(2, 5000), Arrays.asList(3, 7000),
Arrays.asList(4, 10000)),
Arrays.asList(Arrays.asList(1, 2000), Arrays.asList(2, 9000), Arrays.asList(3, 5000)), 10000));
}
public static List<List<Integer>> getIdPairsForOptimal(List<List<Integer>> forwardList,
List<List<Integer>> backwardList, int maxDistance) {
forwardList = forwardList.stream().sorted((x1, x2) -> Integer.compare(x2.get(1), x1.get(1)))
.collect(Collectors.toList());
backwardList = backwardList.stream().sorted((x1, x2) -> Integer.compare(x1.get(1), x2.get(1)))
.collect(Collectors.toList());
int maxDist = maxDistance;
while (true) {
for (List<Integer> l : forwardList) {
for (List<Integer> b : backwardList) {
int forward = l.get(1);
int backward = b.get(1);
int tot = (forward + backward);
if (tot > maxDist) {
break;
}
if (tot == maxDist) {
// print the pair of Id and optimum distance
break;
}
}
}
if (result.size() > 0) {
break;
}
maxDist--;
}
return result;
}
}``````
Comment hidden because of low score. Click to expand.
0
of 0 vote
``````import java.util.Arrays;
import java.util.List;
import java.util.stream.Collectors;
public class OptimalFlightDistant {
public static void main(String[] args) {
// sample input
System.out.println(getIdPairsForOptimal(
Arrays.asList(Arrays.asList(1, 3000), Arrays.asList(2, 5000), Arrays.asList(3, 7000),
Arrays.asList(4, 10000)),
Arrays.asList(Arrays.asList(1, 2000), Arrays.asList(2, 9000), Arrays.asList(3, 5000)), 10000));
}
public static List<List<Integer>> getIdPairsForOptimal(List<List<Integer>> forwardList,
List<List<Integer>> backwardList, int maxDistance) {
forwardList = forwardList.stream().sorted((x1, x2) -> Integer.compare(x2.get(1), x1.get(1)))
.collect(Collectors.toList());
backwardList = backwardList.stream().sorted((x1, x2) -> Integer.compare(x1.get(1), x2.get(1)))
.collect(Collectors.toList());
int maxDist = maxDistance;
while (true) {
for (List<Integer> l : forwardList) {
for (List<Integer> b : backwardList) {
int forward = l.get(1);
int backward = b.get(1);
int tot = (forward + backward);
if (tot > maxDist) {
break;
}
if (tot == maxDist) {
// print the pair of Id and optimum distance
break;
}
}
}
if (result.size() > 0) {
break;
}
maxDist--;
}
return result;
}
}``````
Comment hidden because of low score. Click to expand.
0
of 0 vote
public static List<List<int>> GetOptimalRoute1(int maxDistance, List<List<int>> forwardList, List<List<int>> returnList)
{
List<List<int>> result = new List<List<int>>();
forwardList = forwardList.OrderByDescending(x => x[1]).ToList();
returnList = returnList.OrderByDescending(x => x[1]).ToList();
int tempDistance = 0;
//look data back
for(int i=0;i< forwardList.Count;i++)
{
for (int j = 0; j < returnList.Count; j++)
{
if ((forwardList[i][1] + returnList[j][1]) < tempDistance)
break;
if((forwardList[i][1]+ returnList[j][1])<= maxDistance && (forwardList[i][1] + returnList[j][1])>= tempDistance)
{
tempDistance = forwardList[i][1] + returnList[j][1];
if((forwardList[i][1] + returnList[j][1]) >= tempDistance && result.Count==0)
result.Add(new List<int> { forwardList[i][0], returnList[j][0] });
else if ((forwardList[i][1] + returnList[j][1]) > tempDistance && result.Count > 0)
{
result.Clear();
result.Add(new List<int> { forwardList[i][0], returnList[j][0] });
}
else if ((forwardList[i][1] + returnList[j][1]) == tempDistance && result.Count > 0)
{
result.Add(new List<int> { forwardList[i][0], returnList[j][0] });
}
}
}
}
return result;
}
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Question
Consider the following two statements A and B and identify the correct choice.A) When a body is rotated along a vertical circle with uniform speed then the sum of its kinetic energy and potential energy is constant at all positionsB) To make a body to move along a vertical circle, its critical speed at any point is independent of mass of body
A
A is true but B is false
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B
A is false but B is true
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C
Both A and B are true
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D
Both A and B are false
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Solution
The correct option is B A is false but B is trueWhen a body is rotated in a vertical circle with uniform speed its kinetic energy is constant because of uniform speed. Its potential energy changes because of change in height. Hence the sum of P.E + K.E won't be same.To make a body moving in a vertical circle the critical velocity at a point is independent of mass. Critical velocity for bottommost point is sqrt(5rg).
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Join BYJU'S Learning Program | 342 | 1,392 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2024-18 | latest | en | 0.902784 |
https://dir.md/wiki/Rhombic_dodecahedron?host=en.wikipedia.org | 1,659,890,311,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882570651.49/warc/CC-MAIN-20220807150925-20220807180925-00385.warc.gz | 227,552,784 | 2,530 | # Rhombic dodecahedron
The eight vertices where three faces meet at their obtuse angles have Cartesian coordinates:
The coordinates of the six vertices where four faces meet at their acute angles are:
The first stellation, often simply called the stellated rhombic dodecahedron, is well known. It can be seen as a rhombic dodecahedron with each face augmented by attaching a rhombic-based pyramid to it, with a pyramid height such that the sides lie in the face planes of the neighbouring faces:
Luke describes four more stellations: the second and third stellations (expanding outwards), one formed by removing the second from the third, and another by adding the original rhombic dodecahedron back to the previous one. | 166 | 724 | {"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-2022-33 | latest | en | 0.956822 |
https://www.physicsforums.com/threads/angular-acceleration-and-power.356806/ | 1,511,207,893,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806124.52/warc/CC-MAIN-20171120183849-20171120203849-00372.warc.gz | 856,527,929 | 14,499 | # Angular acceleration and power
1. Nov 21, 2009
### n.hirsch1
1. The problem statement, all variables and given/known data
See the image attached
What power must be applied to this object to accelerate it from rest to an angular speed of 2.7 rad/s in 5.2 s about the x axis?
2. Relevant equations
P = W / t - torque * angular velocity
3. The attempt at a solution
P = [(0.5 m)*(7 kg)*(acceleration)] * 2.7 rad/s
I don't know how to find acceleration or if I am doing this wrong all together. The right answer is 1.3 W. I tried plugging it into the motion equation v = vo + at to get a, that gave me the wrong answer.
#### Attached Files:
• ###### Walker4e.ch11.Pr083.jpg
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2. Nov 21, 2009
### Delphi51
The work goes into making the system have kinetic energy. Your starting point should be P = KE/time. Now you just have to figure out how much rotational kinetic energy it has when ω = 2.7 rad/s. | 264 | 932 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2017-47 | longest | en | 0.885807 |
http://forums.devshed.com/software-design/48058-division-algorithm-arithmetic-last-post.html | 1,529,337,322,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267860570.57/warc/CC-MAIN-20180618144750-20180618164750-00014.warc.gz | 117,922,931 | 10,020 | ### Thread: division algorithm for big number arithmetic
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#### division algorithm for big number arithmetic
I want to write a library to handle bit integer number arithmetic for the C programming language, using i386 assembler. I will represent every big number as a double word (32bit) sequence.
I know how to perform addition (it's quite easy, i will propagate the carry bit from the least significant dword to the most significant one). I also can cope with subtraction and multiplication .
But how about division? I can't figure out a simple algorithm for this except using multiple subtractions which is inefficient. I know the rules from division from primary school but they seem not easy to implement.
how IDIV and DIV is implemented inside the processor (with two complement numbers)?
what's the algorithm?
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i'm not 100% but i guess it's the same as you learned in school.
try doing binary long division exactly as you learned to do decimal division and i think you'll have it...
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This is the algorithm I have from my Computer Hardware class's textbook Computer Organization & Design:
It requires a 64 bit register (or equivalent), which we will call the Remainder register (following the book convention, which goes through 3 iterations, eventually removing all data storage but the Divisor and Remainder register) and a Divisor register, which is 32 bit.
1. Store the dividend in the right half of Remainder
2. Shift Remainder left one bit
3. Subtract Divisor from Remainder, store the result in the left half of Remainder.
4. Test Remainder
4a. (Remainder >= 0) Shift Remainder left one bit, set rightmost bit to 1
4b. (Remainder < 0) Add Divisor back to the left half of Remainder, replace left half of Remainder with sum. Shift Remainder left, set rightmost bit to 0.
5. Test iteration count.
5a. (done this 32 times) Shift left half of Remainder right 1 bit.
5b. (done it <32 times) Go back to step 3. | 571 | 2,318 | {"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-2018-26 | latest | en | 0.900112 |
https://www.gradesaver.com/textbooks/math/applied-mathematics/elementary-technical-mathematics/chapter-4-section-4-6-multiplication-and-division-of-measurements-exercise-page-188/33 | 1,685,562,120,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224647409.17/warc/CC-MAIN-20230531182033-20230531212033-00414.warc.gz | 905,384,136 | 12,850 | ## Elementary Technical Mathematics
$1\bar{0}00\ m^3$
The volume of the cylinder is given by $$V = \pi r^2h= \pi (6.8^2) 8.5= 1234.77\ m^3=1\bar{0}00\ m^3 .$$ | 74 | 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": 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-2023-23 | latest | en | 0.613853 |
http://www.uff.br/trianglecenters/X0849.html | 1,508,318,912,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187822851.65/warc/CC-MAIN-20171018085500-20171018105500-00879.warc.gz | 597,376,313 | 3,238 | X(849) (4th HATZIPOLAKIS-YIU POINT)
Interactive Applet
You can move the points A, B and C (click on the point and drag it).
Press the keys “+” and “−” to zoom in or zoom out the visualization window and use the arrow keys to translate it.
You can also construct all centers related with this one (as described in ETC) using the “Run Macro Tool”. To do this, click on the icon , select the center name from the list and, then, click on the vertices A, B and C successively.
The JRE (Java Runtime Environment) is not enabled in your browser!
This applet was built with the free and multiplatform dynamic geometry software C.a.R..
Information from Kimberling's Encyclopedia of Triangle Centers
Trilinears [a/(b + c)]2 : [b/(c + a)]2 : [c/(a + b)]2
Barycentrics a[a/(b + c)]2 : b[b/(c + a)]2 : c[c/(a + b)]2
Let D denote the circumcircle of triangle ABC. Let DA be the circle tangent to sideline BC and tangent to D at A. Let Ba = AC∩DA and Ca = AB∩DA, and define Cb, Ca and Ac, Ab cyclically. Define A' = CbAb∩AcBc, and define B' and C' cyclically. Then triangle A'B'C' is homothetic to ABC, and X(849) is the center of homothety. See A. Hatzipolakis and P. Yiu, Hyacinthos #2056-2070, December, 2000.
X(849) lies on these lines: 32,163 36,58 110,595 249,1110 741,827 757,763
X(849) = isogonal conjugate of X(1089)
X(849) = X(249)-Ceva conjugate of X(163)
X(849) = crosspoint of X(58) and X(501)
X(849) = crosssum of X(10) and X(502)
This is a joint work of
Humberto José Bortolossi, Lis Ingrid Roque Lopes Custódio and Suely Machado Meireles Dias. | 510 | 1,588 | {"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-2017-43 | latest | en | 0.778802 |
https://www.mycoursehelp.com/QA/Consider-the-following-simple-model-of-a/51913/1 | 1,621,023,727,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991207.44/warc/CC-MAIN-20210514183414-20210514213414-00130.warc.gz | 927,393,486 | 7,931 | Create an Account
Home / Questions / Consider the following simple model of a common values auction Two buyers each obtain a pr...
Consider the following simple model of a common values auction Two buyers each obtain a private signal about the value of an object The signal can be either high
Consider the following simple model of a common values auction. Two buyers each obtain a private signal about the value of an object. The signal can be either high (H) or low (L) with equal probability. If both obtain signal H, the object is worth 1; otherwise, it is worth 0.
a. What is the expected value of the object to a buyer who sees signal L? To a buyer who sees signal H?
b. Suppose buyers bid their expected value computed in part (a). Show that they earn negative profit conditional on observing signal H—an example of the winner’s curse.
May 07 2020 View more View Less | 189 | 880 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2021-21 | latest | en | 0.905022 |
https://modulomathy.com/2013/09/18/math-trivia-254-solution/ | 1,500,655,400,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549423787.24/warc/CC-MAIN-20170721162430-20170721182430-00293.warc.gz | 680,686,427 | 31,030 | #math trivia #254 solution
A Mersenne prime is a prime p that has the form p = 2q-1 where q is also prime.
The prime factors of 254 are 2 and 127. The second factor is the Mersenne prime. (For the reader: What is the value of q for this prime?) | 75 | 248 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2017-30 | longest | en | 0.92079 |
https://abagoffruit.wordpress.com/2004/02/21/gallian-functional-analysis-and-elementary-probability/ | 1,555,708,078,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578528058.3/warc/CC-MAIN-20190419201105-20190419223105-00392.warc.gz | 343,356,604 | 18,603 | ## Gallian, Functional Analysis, and Elementary Probability
Gallian was terrific, as I had expected. He talked about Hamiltonian cycles on a torus. I was able to understand it, which is more than I can say about some of the other talks I have attended.
I’m still not too fond of functional analysis, but at least I have two apparently good books on it to help me out. At least we’ll be back to measure theory (Ergodic theory) next quarter. Labutin determined that one of the problems he had assigned us for homework was in fact false, so he dropped that problem. Eleven to go.
I seem to have failed to solve what looks like a simple probability problem: Suppose an instantaneous event happens (at least once) in a given amount of time with probability p. What is the probability that it happens at least twice? I could come up with no better way of solving it than partitioning the time into n sections and letting n go to infinity. But something went wrong, and I got -infinity, which isn’t good when dealing with probability. Now I have to decide which will taunt me more: the need to learn functional analysis or this simple problem.
Hi. I'm Simon Rubinstein-Salzedo. I'm a mathematics postdoc at Dartmouth College. I'm also a musician; I play piano and cello, and I also sometimes compose music and study musicology. I also like to play chess and write calligraphy. This blog is a catalogue of some of my thoughts. I write them down so that I understand them better. But sometimes other people find them interesting as well, so I happily share them with my small corner of the world.
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### 2 Responses to Gallian, Functional Analysis, and Elementary Probability
1. Anonymous says:
uhm.. let there be n subintervals. prob(no event in interval)=(1-p)^(1/n).
we want prob(0)+prob(1)=((1-p)^(1/n))^n+n*((1-p)^(1/n))^(n-1)*(1-(1-p)^(1/n))= 1-p + n*(1-p)^(1-1/n)*(1-(1-p)^(1/n)). so we want lim n->inf n*(1-(1-p)^(1/n)) = -ln(1-p)
so prob(2 or more) = 1-(1-p)+(1-p)ln(1-p)=p+(1-p)ln(1-p), methinks
• Simon says:
As expected, I can’t do basic arithmetic. I did exactly the same thing and managed to mess it up anyway. Thanks. | 563 | 2,193 | {"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-2019-18 | latest | en | 0.963031 |
https://brainmass.com/business/business-math/suppose-you-hold-a-diversified-portfolio-consisting-of-a-7-455631 | 1,576,367,316,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575541297626.61/warc/CC-MAIN-20191214230830-20191215014830-00551.warc.gz | 299,528,115 | 11,916 | Explore BrainMass
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# Suppose you hold a diversified portfolio consisting of a \$7
This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!
Suppose you hold a diversified portfolio consisting of a \$7,500 investment in each of 20 different common stocks. The portfolio's beta is 0.85. Now, suppose you sell one of the stocks with a beta of 1.0 for \$7,500 and use the proceeds to buy another stock whose beta is 0.95. Calculate your portfolio's new beta.
© BrainMass Inc. brainmass.com October 10, 2019, 4:19 am ad1c9bdddf | 150 | 582 | {"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-2019-51 | longest | en | 0.920178 |
https://stats.stackexchange.com/questions/637831/sampling-from-px-propto-coshma-x-e-x2-2?noredirect=1 | 1,718,356,758,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861545.42/warc/CC-MAIN-20240614075213-20240614105213-00684.warc.gz | 495,708,333 | 44,338 | # Sampling from $P(x) \propto \cosh^{m}(a x) e^{-x^{2}/2}$
Is there an efficient algorithm to draw samples $$x \sim P(x)$$ from the PDF:
$$P(x) \propto \cosh^{m}(a x) e^{-x^{2}/2}$$
where $$a\ge0$$ is a real parameter, and $$m$$ a positive integer?
Since this is a univariate PDF, my first attempt was to compute the CDF and use the inverse CDF method to sample it, but neither the CDF nor the inverse CDF seem to have a tractable analytical form.
• Related: See stats.stackexchange.com/q/630463/5536 for the simpler $m=1$ case of this question.
– a06e
Commented Jan 26 at 16:03
• By "$\cosh^m(ax)$", do you mean "$(\cosh(ax))^m$" or that $\cosh$ is applied $m$ times? Commented Jan 26 at 16:14
• @mhdadk I mean $(\cosh(ax))^m$.
– a06e
Commented Jan 26 at 16:31
• Can you provide values of $a$ and $m$ that would be of interest for you? Commented Jan 27 at 13:42
• $m$ is a positive integer, and $a$ a positive real. @MattF.
– a06e
Commented Jan 27 at 16:46
The distribution that the OP seeks is a straightforward extension of the formula for the case $$m=1$$ that Xi'an derived in the linked answer.
Suppose that $$P(x) \propto (\cosh(ax))^m \exp(-x^2/2)$$. Then, since \begin{align}(\cosh(ax))^m &= \left(\frac{\exp(ax)+\exp(-ax)}{2}\right)^m\\ &= \sum_{k=0}^m 2^{-m} \binom{m}{k}\exp(akx)\exp(-a(m-k)x)\\ &= \sum_{k=0}^m 2^{-m} \binom{m}{k} \exp(-a(m-2k)x), \end{align} $$P(x)$$ is once again a weighted sum (also called a mixture) of normal distributions with means ranging from $$-ma$$ to $$+ma$$ in steps of $$2a$$. Note that the standard normal distribution is included in the mixture only when $$m$$ is even. The weights are given by the probability mass distribution function of a $$(m,\frac 12)$$ binomial random variable.
Turning to the actual question asked by the OP about an efficient algorithm for sampling from this distribution, consider the various answers to this question, some of which even contain R code with varying degrees of efficiency.
• This is fine for integral $m$ -- but that's the easy case.
– whuber
Commented Jan 26 at 19:56
• @a06e For integer $m$, the mean of each component is $-a(m-2k)$. Commented Jan 27 at 17:08
• This, then, is a good answer +1.
– whuber
Commented Jan 27 at 18:55
• +1 but don't the weights (prior to making them sum to 1) need to depend on $\mu_k=-a(m-2k)$? \begin{align}(\cosh(ax))^m \exp(-x^2/2) &= \sum_{k=0}^m 2^{-m} \binom{m}{k} \exp(-a(m-2k)x) \exp(-x^2/2) \\ &= \sum_{k=0}^m 2^{-m} \binom{m}{k} \exp(\mu_k x-x^2/2) \\ &= \sum_{k=0}^m 2^{-m} \binom{m}{k} \exp(\mu_k x-x^2/2 - \mu_k^2/2 + \mu_k^2/2) \\ &= \sum_{k=0}^m 2^{-m} \binom{m}{k} \exp(\mu_k^2)\exp(-(x-\mu_k)^2/2) \end{align}
– JimB
Commented Jan 27 at 19:55
• The $\exp(\mu_k^2)$ in the last equation above should of course be $\exp(\mu_k^2/2)$.
– JimB
Commented Jan 27 at 22:08
Building on the answer of Dilip, for non-integer $$m$$, letting $$\{m\}=m-\lfloor m\rfloor$$ denote the fractional part of $$m$$, and using the fact that $$\cosh x\le \exp|x|$$, the (unnormalized) target density can be rewritten as \begin{align} f(x)&=\cosh^m(ax)\exp(-x^2/2) \\&=\cosh^{\{m\}}(ax)\cosh^{\lfloor m\rfloor} (ax)\exp(-x^2/2) \\&\le \exp(a\{m\}|x|)\cosh^{\lfloor m\rfloor} (ax)\exp(-x^2/2) \\&= \sum_{k=0}^{\lfloor m\rfloor} 2^{-{\lfloor m\rfloor}} \binom{{\lfloor m\rfloor}}{k} \exp\Big(-x^2/2-a({\lfloor m\rfloor}-2k)x+a\{m\}|x|\Big). \end{align} Like the target $$f(x)$$, the right hand side is symmetric around zero. For $$x\ge 0$$, it is proportional to the density of a mixture of $$\lfloor m\rfloor+1$$ Gaussian distributions truncated below zero. Completing the square in $$x$$, the mean (before truncation) of component $$k$$ is $$\mu_k=-a(\lfloor m\rfloor-\{m\}-2k)$$ and the associated weights are $$w_k \propto \Phi(\mu_k)\binom{\lfloor m\rfloor}{k}e^{\mu_k^2/2}.$$
We can easily simulate proposals efficiently from this mixture (using the alias and inversion methods), and then obtain samples from $$f(x)$$ by accepting the proposals with probabilities given by $$f(x)$$ divided by the above right hand side.
The acceptance probability simplifies to $$\left(\frac{1+\exp(-2a|x|)}{2}\right)^{\{m\}}>\frac1{2^{\{m\}}}$$ so the overall acceptance probabilty is always greater than 1/2.
R implementation:
dcoshgauss <- function(x, a, m)
cosh(a*x)^m*exp(-x^2/2)
rcoshgauss <- function(n, a, m) {
floorm <- floor(m)
fractm <- m - floorm
kk <- 0:floorm
mu <- -a*(floorm - fractm - 2*kk)
w <- pnorm(mu)*choose(floorm, kk)*exp(mu^2/2)
w <- w/sum(w)
xx <- numeric(n)
i <- 0
s <- 0
while (i < n) {
k <- sample(kk+1, size = 1, prob=w)
u <- runif(1, pnorm(-mu[k]), 1)
x <- mu[k] + qnorm(u)
acceptprob <- ((1+exp(-2*x*a))/2)^fractm
u <- runif(1)
s <- s + 1
if (u<acceptprob) {
sign <- sample(c(-1,1), 1)
x <- sign*x
i <- i + 1
xx[i] <- x
}
}
cat("Acceptance probability was",n/s,"\n")
xx
}
set.seed(1)
x <- rcoshgauss(1e6, 1, 2.9)
#> Acceptance probability was 0.5489607
hist(x, breaks=200, xlim=c(-8,8), main="")
curve(dcoshgauss(x, 1, 2.9), -8,8)
Created on 2024-01-28 with reprex v2.0.2
• Ah I almost missed this wonderful answer, thanks!
– a06e
Commented Jan 27 at 19:11
• @a06e Thanks, this was a neat problem! Commented Jan 27 at 19:30
This is just an extended comment.
With the restriction of the values of $$m$$ being positive integers there is an analytic solution of a mixture of normals each with variance 1 as shown by @DilipSarwate (although I currently disagree about the weights). However, showing graphic examples should almost always be considered in which interesting features can appear.
In this case when $$a$$ and $$m$$ are large enough (and they don't need to be very large) one is essentially sampling from a mixture of just two normals with means of opposite sign and variance 1. This might also be the case for large values of $$m$$ when $$m$$ is not an integer.
Below are a few examples showing the resulting density (blue) and the individual contributions (red).
Small $$a$$ and small $$m$$
$$a=1$$ and $$m=2$$
$$a=1$$ and $$m=3$$
$$a=1$$ and $$m=4$$ | 2,121 | 6,020 | {"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": 41, "wp-katex-eq": 0, "align": 3, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2024-26 | latest | en | 0.840692 |
https://de.mathworks.com/matlabcentral/cody/players/5596781-pauli-huusari/solved | 1,590,959,661,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347413624.48/warc/CC-MAIN-20200531182830-20200531212830-00274.warc.gz | 323,031,121 | 22,162 | Cody
# Pauli Huusari
Rank
Score
1 – 50 of 447
#### Problem 2522. Convert given decimal number to binary number.
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#### Problem 1103. Right Triangle Side Lengths (Inspired by Project Euler Problem 39)
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Created by: Peng Liu
#### Problem 1036. Cell Counting: How Many Draws?
Created by: Ned Gulley
#### Problem 152. Create a cell array out of a struct
Created by: Daniel
#### Problem 1971. Remove element(s) from cell array
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Created by: Ted
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#### Problem 80. Test for balanced parentheses
Created by: Cody Team
Tags regexp, parens
#### Problem 2736. Pernicious Anniversary Problem
Created by: Jan Orwat
#### Problem 1510. Number of digits in an integer
Created by: Mohamed B.S.
Tags digits
#### Problem 142. give nth decimal place of pi
Created by: AMITAVA BISWAS
Tags pi
#### Problem 1043. Generate a string like abbcccddddeeeee
Created by: Aurelien Queffurust
#### Problem 1077. Count up then down
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#### Problem 1171. matrix of natural number
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Tags matrices
#### Problem 412. Back to basics 22 - Rotate a matrix
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#### Problem 89. Counting in Finnish
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#### Problem 56. Scrabble Scores
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#### Problem 42655. Convert yards to feet
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#### Problem 627. Compute a dot product of two vectors x and y
Created by: Evgeny
#### Problem 544. Create a square matrix of multiples
Created by: Bruce Raine
#### Problem 44375. Missing five
Created by: Alfonso Nieto-Castanon
Tags cody5
#### Problem 42931. Square root of a number
Created by: Minhaj Nur Alam
#### Problem 2572. Counting down
Created by: Marisa
#### Problem 1258. Create matrix of replicated elements
Created by: Jeremy
Tags easy, matlab, matrix
#### Problem 2551. Rounding
Created by: Marisa
Tags swap
#### Problem 1570. Is the input divisible by 3?
Created by: Jacek Ho
Tags cone
#### Problem 354. Back to basics 10 - Max Float
Created by: Alan Chalker
#### Problem 1054. What's Your BMI?
Created by: Pranav
#### Problem 42691. Matrix of almost all zeros, except for main diagonal
Created by: AM
Tags matrix, zeros
#### Problem 2574. Replacing a row
Created by: Marisa
#### Problem 2857. Matlab Basics - Convert a row vector to a column vector
Created by: Yaz Majeed
#### Problem 47. Extract leading non-zero digit
Created by: Cody Team
#### Problem 1153. Accessing elements on the diagonal
Created by: HARISANKAR PS
Tags diagonal
#### Problem 1129. Reverse the elements of an array
Created by: HARISANKAR PS
#### Problem 1420. only input
Created by: Marco Castelli
1 – 50 of 447 | 1,000 | 3,668 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2020-24 | latest | en | 0.66222 |
https://mathoverflow.net/questions/341344/dimensional-gap-of-group-representations | 1,638,567,508,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362919.65/warc/CC-MAIN-20211203212721-20211204002721-00409.warc.gz | 469,041,543 | 28,701 | Dimensional gap of group representations
The problem is inspired by eigenvalue bounds of random Cayley graphs on $$SL_2(q)$$.
Definition. An infinite series of finite groups $$S$$ is α-rich if the dimension of the smallest nontrivial representation of $$G$$ on $$\mathbb{C}$$ is $$\Omega(|G|^\alpha)$$ for every $$G\in S$$.
For example, the series of groups $$SL_2(q)$$ is $$\frac{1}{3}$$-rich.
Question. Are there any series of $$α$$-rich groups with $$\alpha>\frac13$$?
Known. One only needs to consider simple groups, and by CFSG, it's just going through Lie groups.
Along the lines of David E Speyer, a permutation representation on $$O(|G|^\frac13)$$ points rules out the possibility for the group to beat $$SL_2(q)$$.
This holds for all classic groups of type $$A_n (n\geq2)$$, $$B_n(n\geq3)$$, $$C_n(n\geq2)$$, $$D_n(n\geq4)$$, $$^2D_n(n\geq3)$$, $$^2A_n(n\geq4)$$: just consider the action of these groups on the 1-dimensional subspaces of their defining vector space. $$^2A_2$$ is not in the list, but $$^2A_2(q)$$ has a representation on $$q^2-q+1$$ dimensions, hence ruled out.
The same argument rules out $$F_4$$, $$E_n$$ and $$^2E_6$$: All of them are covered in a paper given by Derek Holt. None of them beats $$SL_2(q)$$.
$$G_2$$ and $$^2G_2$$ do not beat $$SL_2(q)$$, by Derek Holt's answer.
For $$^2B_2(q)$$, see Wikipedia: it almost beats $$SL_2$$, but it has two characters of dimension $$O(q^{3/2})$$.
For $$^3D_4(q)$$, see Deriziotis, D. I., and G. O. Michler. "Character table and blocks of finite simple triality groups $$^3D_4(q)$$." Transactions of the American Mathematical Society 303.1 (1987): 39-70.: there's a character of dimension $$q(q^4-q^2+1)$$.
The last case $$^2F_4$$ is found in Die unipotenten Charaktere fur die GAP-Charaktertafeln der endlichen Gruppen vom Lie-Typ. M. Claßen-Houben; Diplomarbeit, RWTH Aachen; 2005. According to the conventions of the paper, the group has size $$q^{52}$$, and there's a character of dimension $$q^2Φ_{12}Φ_{24}$$, which is $$O(q^{14})$$.
• If $H$ is a quotient of $G$, then any nontrival representation $V$ of $H$ is also a representation of $G$, and $\tfrac{\log \dim V}{\log |G|} \leq \tfrac{\log \dim V}{\log |H|}$. So the maximal $\alpha$ will occur for simple groups. $A_{n-1}$ has an $n$-dimensional representation, so that is terrible. It sounds like the question is basically to go through the groups of Lie type and see whether any of them beat $1/3$? Sep 11 '19 at 15:04
• Suggestion: Consider the adjoint form of $G_2$ over $\mathbb{Z}$ and look at $G_2(\mathbb{F}_q)$, which has size $\approx q^{14}$. The natural analogue of the $(q+1)$-dimensional permutation representation of $PSL_2(q)$ is the permutation action on $(G/P)(\mathbb{F}_q)$ for either of the two maximal parabolic $P$, which has size $\approx q^5$. We have $\tfrac{5}{14} > \tfrac{1}{3}$. I don't know if $G_2(\mathbb{F}_q)$ has any smaller representations, but probably an expert will show up soon to tell us. Sep 11 '19 at 15:17
• I am pessimistic about this approach now. I messed around with Hecke operators, and computed that the permutation representation on $(G/P)(\mathbb{F}_q)$ has dimension $q^5+q^4+q^2+q^3+q^2+q+1$ and breaks up into three nontrivial summands plus one trivial. I don't know how to compute the dimensional of these nontrivial summands, but it seems likely to me that some of them grow at rate $q^4$ or slower, and $\tfrac{4}{14} < \tfrac{1}{3}$. Sep 13 '19 at 0:56
• Possibly relevant: sciencedirect.com/science/article/pii/0021869374901501 Sep 14 '19 at 17:49
• I have now found a reference for the minimal degrees for all exceptional groups of Lie Type - see my edited answer. In particular, for ${}^2F_4(q^2)$ we get degree $O(q^{11})$, which is less that $O(q^{14})$. Oct 7 '19 at 14:37
This answer is still not quite complete, but I hope to finish it soon!
The smallest degrees of the faithful permutation representations of the finite simple groups have all been known for a while now. The most convenient reference is probably Table 4 of this paper, although none of the results are original to that paper. The results for the exceptional groups of Lie type were proved in a series of papers of A. V. Vasil'ev.
Anyway, this approach works for $$E_7(q)$$ (order about $$q^{133}$$, minimal permutation degree about $$q^{27}$$), $$E_8(q)$$ (order about $$q^{248}$$, minimal permutation degree about $$q^{57}$$, $$^{2}E_6(q)$$ (order about $$q^{78}$$, minimal permutation degree about $$q^{21}$$), and $${}^3D_4(q)$$ (order about $$q^{28}$$, minimal permutation degree about $$q^9$$).
However, the approach fails for $$G_2(q)$$, which has order about $$q^{14}$$ and minimal permutation degree $$(q^6-1)/(q-1)$$, $$^{2}G_2(q)$$ (with $$q$$ an odd power of $$3$$), with order about $$q^7$$ and minimal permutation degree $$q^3+1$$, and $$^{2}F_4(q)$$ (with $$q$$ an odd power of $$2$$), order about $$q^{26}$$, minimal permutation degree about $$q^{10}$$.
So I have been hunting around for results about minimal degrees of representations of $$G_2(q)$$,$$^{2}G_2(q)$$, and $$^{2}F_4(q)$$. For $$G_2(q)$$ I found them on Page 126 of G. Hiss, Zerlegungszahlen endlicher Gruppen vom Lie-Typ in nicht-definierender Charakteristik, Habilitationsschrift, Aachen 1990. For sufficiently large $$q$$, these are $$q^4+q^2+1$$, $$q^3+1$$, and $$q^3-1$$ when $$q \equiv 0,1,-1 \bmod 3$$, respectively. So this is less that $$q^{1/3}$$, and we are OK.
For $$^{2}G_2(q)$$, the character tables are computed in this paper. The table is towards the end of the paper, and the smallest character degree is $$q^2-q+1$$, so again we are OK!
I still need to check $$^{2}F_4(q)$$.
Added later: i have now located a better reference for minimal degrees of characters of exceptional groups of Lie type, namely Lübeck, Frank, Smallest degrees of representations of exceptional groups of Lie type. Comm. Algebra 29 (2001), no. 5, 2147–2169, available here.
In particular, for sufficiently large $$q$$, the smallest character degree of $${}^2F_4(q^2)$$ has degree $$11$$ in $$q$$, which is easily sufficient to prove the required result.
• The case of $^2F_4$ is solved(I will write it up soon). Thanks for your answer! Oct 4 '19 at 12:48 | 1,956 | 6,219 | {"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": 80, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2021-49 | longest | en | 0.770482 |
http://www.jiskha.com/display.cgi?id=1292298143 | 1,495,797,912,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463608659.43/warc/CC-MAIN-20170526105726-20170526125726-00433.warc.gz | 664,724,482 | 3,951 | # Math
posted by on .
i have to write this out in word form because it doesn't look rite when i use numbers...
What is the value of X when Y=7?
Problem: negative threeX to the 2nd power plus fourY minus one equals zero
2
-3x + 4y - 1 = 0 | 74 | 240 | {"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-2017-22 | latest | en | 0.9083 |
https://mtg.sg/site/cumulative-hazard-vs-cumulative-incidence-dfd66c | 1,653,607,694,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662627464.60/warc/CC-MAIN-20220526224902-20220527014902-00320.warc.gz | 459,566,135 | 10,961 | # cumulative hazard vs cumulative incidence
What does "worm of yellow convicts" mean? Cumulative incidence (CI) and incidence rate (IR) are different approaches to calculating incidence, based on the nature of followup time. Incidence provides information about the spread of disease. Finally, we point out that other methods of directly modeling the cumulative incidence function are also available. Measurement in nursing and health research (4th ed. A total of 220 patients passed through the ICU during the month; of these, 20 were admitted with a diagnosis of pneumonia and 10 developed pneumonia during their stay. ScienceDirect ® is a registered trademark of Elsevier B.V. ScienceDirect ® is a registered trademark of Elsevier B.V. Regression Models for Hazard Rates Versus Cumulative Incidence Probabilities in Hematopoietic Cell Transplantation Data. How can a chess game with clock take 5 hours? Copyright © 2020 Elsevier B.V. or its licensors or contributors. How can I ask colleagues to use chat/email instead of scheduling unnecessary calls? Cross Validated is a question and answer site for people interested in statistics, machine learning, data analysis, data mining, and data visualization. $h(t)$ is a function whose integral over the positive real line diverges and what you call the CIF has a different relation to $h(t)$. https://www.canadian-nurse.com/en/articles/issues/2012/october-2012/terminology-101-cumulative-incidence-and-incidence-rate.
It is of interest in some applications to determine whether there is a relationship between a hazard rate function (or a cumulative incidence function) and a mark variable which is only observed at uncensored failure times. How can I better handle 'bad-news' talks about people I don't care about? The CI of pneumonia is therefore 10/200 or five per cent in the one-month period. Is it safe to look at a mercury gas discharge tube? Campbell, M. J., Machin, D., & Walters, S. J. 2 .
Researchers next calculate the IR (also known as incidence density), which reflects variation in the lengths of time that at-risk individuals are observed. I now run a PHREG (2 treatment groups) which resulteted in a HR of 1.5. What is $S$? What is the reason for the date of the Georgia runoff elections for the US Senate?
Making statements based on opinion; back them up with references or personal experience. The integral H[T] = ∫ID[t]dt is called the "integrated hazard" or "cumulative hazard" and denoted H(T) or Λ (T). The month of January is chosen as the study period. Since 10 of the 200 patients developed pneumonia, we conclude that the IR of pneumonia was 10 cases per 1,000 person-days (i.e., one case per 100 person-days). Examples rev 2020.11.11.37991, The best answers are voted up and rise to the top, Cross Validated works best with JavaScript enabled, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site, Learn more about Stack Overflow the company, Learn more about hiring developers or posting ads with us. Laws and Treaties for Activities in Outer Space, Reproducible Gitian Builds .. but not the same hash as bitcoincore.org.
The 20 patients admitted with pneumonia are excluded from the denominator because they were not at risk, which yields 200 at risk. How seriously did romantic composers take key characterizations? IR is especially useful in hospital surveillance, where patients may enter and leave the surveillance at varying times (e.g., patients may be lost to follow up or die from a cause other than that being studied). Incidence: The number of new cases of a disease that occur during a specified period of time in a population at risk for developing the disease. ZX Spectrum 48k Power Supply outputting 15V.
hazard rate and cumulative incidence function, Creating new Help Center documents for Review queues: Project overview, Feature Preview: New Review Suspensions Mod UX, 2020 Community Moderator Election Results, Proof of relationship between hazard rate, probability density, survival function, Hazard function, survival function, and retention rate.
Use MathJax to format equations. To learn more, see our tips on writing great answers. By using our site, you acknowledge that you have read and understand our Cookie Policy, Privacy Policy, and our Terms of Service. In particular, $$P\{T \leq t\}=1 - \exp\left(-\int_0^t h(u)\,\mathrm du\right),$$ where the integrand is the area under $h(\cdot)$ between $0$ and $t$. Cross Validated is a question and answer site for people interested in statistics, machine learning, data analysis, data mining, and data visualization. Remember, CI tells us the risk of disease when the observation time is assumed to be equal for all patients in the denominator, while IR tells us the risk of disease when the observation time varies among individuals in the denominator. It only takes a minute to sign up. I have a laptop with an HDMI port and I want to use my old monitor which has VGA port. Can someone re-license my project under a different license. It is calculated by dividing the number of new cases of a disease by the total of the lengths of time that each individual in the population was at risk, expressed as person-time (e.g., person-days). However, consider the 1 year as 365 days, and do the arithmetic day by day. The numerator consists of the 10 at-risk patients who developed pneumonia.
Maher M. El-Masri, RN, PhD, is a full professor and research chair in the faculty of nursing, University of Windsor, in Windsor, Ont. Where the period of time considered is an entire lifetime, the incidence proportion is called lifetime risk. Source: Gordis, L. (2009). Reference request: Examples of research on a set with interesting properties which turned out to be the empty set. | 1,266 | 5,822 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2022-21 | longest | en | 0.908795 |
http://math.stackexchange.com/users/5676/peter-taylor?tab=activity&sort=comments&page=5 | 1,429,962,328,000,000,000 | text/html | crawl-data/CC-MAIN-2015-18/segments/1429246648338.72/warc/CC-MAIN-20150417045728-00264-ip-10-235-10-82.ec2.internal.warc.gz | 172,958,211 | 15,115 | Peter Taylor
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Jul3 comment Finding double root of $x^5-x+\alpha$ And then we have $\gcd(x^5 - x + \alpha, 5x^4 - 1)$ $= \gcd(5x^4 - 1, 4x - 5\alpha)$ $= \gcd(4x - 5\alpha, \frac{3125}{256}\alpha^4 - 1)$, so in order for that GCD to be linear we require $\frac{3125}{256}\alpha^4 - 1 = 0$. Jun27 comment Looking for different proofs of “Discrete Liouville's Theorem”. Can you think of a suitable title which includes some non-LaTeX text? Pure LaTeX titles cause some problems. Jun27 comment This should be a piece of cake… right? It wouldn't have cut more than one piece of cake at once, taking the natural meaning of simultaneously. I've edited to remove that adverb and make other clarifications. There's one remaining ambiguity, which I didn't want to resolve because I didn't know in which direction to resolve it: "the cuts may be done however you want" seems to directly contradict "to make it simple, let's say you have a straightedge and a compass". Jun27 comment This should be a piece of cake… right? If I put the cakes side-by-side and cut through them all parallel to the table in one motion, is that one cut or $n$? Jun25 comment The expected outcome of a random game of chess? @mjqxxxx, I think it's because the chess library considers the game to be over when neither player has enough material to mate, and so breaks the loop, but doesn't consider it to be a stalemate, so it wasn't being counted correctly. Jun19 comment A game with numbers from MEMO $2013$ Surely the simplest first two moves for B are $0,0$. Jun12 comment How do you create a nonlinear game that the player can always win? The way to maximise the non-linearity is to minimise the interdependence of the puzzles. The point I'm hoping that you'll take away is that you need to rethink your goal. Jun11 comment Gosper summable The case $m=1$ is quite easy. The case $m>1$ seems to require showing that a set of simultaneous equations has no solution. Jun10 comment Hex game winning strategy It's not clear from the question what the board layout is (you can do ASCII art by putting 4 spaces before each line, and there are online designs for ASCII art hexagons), or what the rules of the game are. May21 comment Deterministic Push-Down Automata Is $U$ a terminal? And as a hint: have you tried building a non-deterministic push-down automaton to recognise this language? May21 comment What is one way to prove that there exists no ellipse that matches the exact curvature of the sin wave? That doesn't rule out the sine wave being less than half of an ellipse. May16 comment what is the minimum value of the angles inside these triangles? I think the question is: what is the smallest angle $\alpha$ such that there exists a dissection of the square into triangles satisfying two properties: that none of the triangles has an internal angle greater than $\alpha$; and that no vertex of a triangle touches another triangle except at a vertex. If so, there's an easy lower bound of 67.5 degrees. May6 comment Calculate the Probability for Binary Matrix I assume that the second sentence means that each element of the matrix is $1$ with probability $p$, but is the third sentence talking about independence of random variables or about linear independence (i.e. the matrix is non-singular)? Apr23 comment Repeating cycles in the $3n-1$ problem Your cycles have a very close link with the cycles of $3n+1$ starting with negative $n$. Apr19 comment Return of the lost ant 3D That such paths exist isn't an problem. Whether or not they have a length is another matter. Apr17 comment Maximum number of edges in a (n,n) bipartite graph, that doens't have a complete bipartite subgraph $K_{r,r}$ $c=0$ works for every $r$. Apr17 comment Maximum number of edges in a (n,n) bipartite graph, that doens't have a complete bipartite subgraph $K_{r,r}$ There's a trivial solution: let $c=0$. If $r=1$ then that's tight. Apr17 comment Prove that there are two frogs in one square. Harry Dunlop's answer already provides a solution: this is essentially just Hilbert's Hotel backwards. Apr15 comment Game Theory - Voting The procedure as described doesn't seem to always select a winner. If A gets 51%, B gets 49%, C gets 0%, and D gets 0% then the first round of eliminations should ditch C and D, but neither A or B can be eliminated. Apr15 comment Proof that $12$ in a row tic-tac-toe is a tie game? Tic-tae-toe on an infinite grid can never end in a tie. Presumably you mean that neither player has a winning strategy. | 1,129 | 4,561 | {"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} | 3.3125 | 3 | CC-MAIN-2015-18 | longest | en | 0.937782 |
https://www.hellovaia.com/textbooks/physics/matter-interactions-4th-edition/faradays-law/q-1cp-a-magnetic-field-near-the-floor-points-up-and-is-incre/ | 1,696,435,172,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511386.54/warc/CC-MAIN-20231004152134-20231004182134-00575.warc.gz | 856,497,261 | 20,712 | Suggested languages for you:
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Found in: Page 903
### Matter & Interactions
Book edition 4th edition
Author(s) Ruth W. Chabay, Bruce A. Sherwood
Pages 1135 pages
ISBN 9781118875865
# A magnetic field near the floor points up and is increasing. Looking down at the floor, does the non-Coulomb electric field clockwise or counter clockwise? A magnetic field near the ceiling points down and is decreasing. Looking up at the ceiling, does the non-Coulomb electric field curl clockwise or counter clockwise?
The floor and ceiling both are counter clockwise.
See the step by step solution
## Step 1: Given Information.
A magnetic field near the floor points up and is increasing.
## Step 2: Definition/ Concept.
The magnetic field's direction and the electric field's direction are opposite each other. The non-coulomb electric field curls in a clockwise manner as the magnetic field is increased downward toward the floor.
## Step 3: Find floor is clockwise or counter clock wise.
The figure will be:
The facts are:
(a). The magnetic field, B is pointing upwards.
(b). The magnetic field is increasing. Therefore, the quantity $dB/dt$ is pointing in the same direction as B.
So, by using the right hand thumb rule on $-dB/dt$, this can be observed that the direction of ${E}_{NC}$ is counter clockwise.
Therefore, the floor is counter clockwise.
## Step 4: Find ceiling is curl clockwise or counter clock wise.
The figure will be:
The facts are:
(a). The magnetic field, B is pointing upwards.
(b). The magnetic field is decreasing. Therefore, the quantity $dB/dt$ is pointing in the opposite direction as B.
So, by using the right hand thumb rule on $-dB/dt$, this can be observed that the direction of ${E}_{NC}$ is counter clockwise.
Therefore, the ceiling is counter clockwise. | 421 | 1,840 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 15, "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-2023-40 | latest | en | 0.884508 |
https://missionalcall.com/2020/07/22/how-do-you-change-a-marker-in-a-scatter-plot/ | 1,723,191,291,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640762343.50/warc/CC-MAIN-20240809075530-20240809105530-00689.warc.gz | 330,261,471 | 12,289 | # How do you change a marker in a scatter plot?
How do you change a marker in a scatter plot? Change data markers in a line, scatter, or radar chart In a line, scatter, or radar chart, do one of the following: On the Format
## How do you change a marker in a scatter plot?
Change data markers in a line, scatter, or radar chart
1. In a line, scatter, or radar chart, do one of the following:
2. On the Format tab, in the Current Selection group, click Format Selection.
3. Click Marker Options, and then under Marker Type, make sure that Built-in is selected.
How do you fill a marker in Matlab plot?
Accepted Answer You can use the “MarkerFaceColor” and “Color” properties of the plotted line or lines in order to fill in the markers with the same color as the default marker edge color. h = plot(x, y, ‘o’); set(h, {‘MarkerFaceColor’}, get(h,’Color’));
### How do you change marker size in Matlab?
1. You can change the marker size for a line plot by setting the “MarkerSize” property, either as a name-value pair or by accessing the “Line” object.
2. Name-value pair:
3. If you set this property as a name-value pair with the “plot” function, you must set it after all the x,y pairs.
What is scatter3 Matlab?
scatter3( X , Y , Z ) displays circles at the locations specified by the vectors X , Y , and Z . If C is a vector with length equal to the length of X , Y , and Z , then the values in C are linearly mapped to the colors in the current colormap.
#### How do I plot a scatter in matplotlib?
Machine Learning – Scatter Plot
1. Example. Use the scatter() method to draw a scatter plot diagram: import matplotlib.pyplot as plt. x = [5,7,8,7,2,17,2,9,4,11,12,9,6] y = [99,86,87,88,111,86,103,87,94,78,77,85,86]
2. Example. A scatter plot with 1000 dots: import numpy. import matplotlib.pyplot as plt.
3. ❮ Previous Next ❯
What is marker in scatter plot?
The marker size. s: size in points^2. It is a scalar or an array of the same length as x and y.
## How do you make a scatter plot in MATLAB?
Specify Target Axes and Marker Type Call the tiledlayout function to create a 2-by-1 tiled chart layout. Call the nexttile function to create the axes objects ax1 and ax2 . Plot scattered data into each axes. In the bottom scatter plot, specify diamond filled diamond markers.
How do you plot a filled circle in Matlab?
How to create a filled circle?
1. function circles = circle(x,y,r)
2. hold on.
3. th = 0:pi/50:2*pi;
4. x_circle = r * cos(th) + x;
5. y_circle = r * sin(th) + y;
6. circles = plot(x_circle, y_circle);
7. hold off.
### How do I run a regression in Matlab?
In MATLAB, you can find B using the mldivide operator as B = X\Y . From the dataset accidents , load accident data in y and state population data in x . Find the linear regression relation y = β 1 x between the accidents in a state and the population of a state using the \ operator.
How do you plot a 3D line in Matlab?
plot3( X , Y , Z ) plots coordinates in 3-D space.
1. To plot a set of coordinates connected by line segments, specify X , Y , and Z as vectors of the same length.
2. To plot multiple sets of coordinates on the same set of axes, specify at least one of X , Y , or Z as a matrix and the others as vectors.
#### How to add a marker to a plot in MATLAB?
Add Markers to Line Plot. View MATLAB Command. Create a line plot. Display a marker at each data point by including the line-specification input argument when calling the plot function. For example, use ‘-o’ for a solid line with circle markers. x = linspace (0,10,100); y = exp (x/10).*sin (4*x); plot (x,y, ‘-o’)
How to increase marker size in scatter plot?
– MATLAB Answers – MATLAB Central How to increase marker size in scatter plot? How to increase the size (e.g. twice bigger ‘+’)? Sign in to answer this question.
## How can I change the transparency of a scatter plot?
You can vary the transparency of scattered points by setting the AlphaData property to a vector of different opacity values. To ensure the scatter plot uses the AlphaData values, set the MarkerFaceAlpha property to ‘flat’. Create a set of normally distributed random numbers. Then create a scatter plot of the data with filled markers.
How to use asterisks as markers in MATLAB?
View MATLAB Command Create a line plot with 1,000 data points, add asterisks markers, and control the marker positions using the MarkerIndices property. Set the property to the indices of the data points where you want to display markers. Display a marker every tenth data point, starting with the first data point. | 1,141 | 4,554 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2024-33 | latest | en | 0.819136 |
https://www.reference.com/web?q=table%20of%20prime%20factors&qo=pagination&o=600605&l=dir&sga=1&qsrc=998&page=2 | 1,603,926,749,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107902038.86/warc/CC-MAIN-20201028221148-20201029011148-00261.warc.gz | 884,323,903 | 16,529 | Web Results
www.reference.com/article/prime-factorization-24-bed9309903572d7f
The prime factorization of 24 is 2 x 2 x 2 x 3. Thus, the number 24 has only 2 prime factors, 2 and 3. Prime factors of 24 are numbers that divide 24 exactly without leaving a remainder and are prime numbers as well.
www.reference.com/article/prime-factors-60-5807e427d30d9c8e
The prime factors of 60 are 2, 3 and 5. The prime factors for any number can be found by performing a process called prime factorization.
www.reference.com/article/prime-factors-22-382d06fb7a8eb959
The prime factors of 22 are 2 and 11. Prime factors of a number are composed of prime numbers that when multiplied together form the number.
www.reference.com/article/prime-factorization-80-a1956d10d1e36eec
The prime factorization of the number 80 is 2 x 2 x 2 x 2 x 5. When multiplied together, these five numbers have a product of 80.
www.reference.com/article/prime-factorization-200-c9f05c1776b6674e
The prime factorization of 200 is 2 x 2 x 2 x 5 x 5, which is written using exponents as 2^3 x 5^2. Prime factorization involves finding the prime numbers that multiply together to make a particular number.
www.reference.com/article/prime-factorization-35-ce4e55a796fcb7b7
The prime factorization of 35 is 5 and 7. Prime factors are numbers that are only divisible by themselves and one. While every number is divisible by one, most mathematicians do not consider it as a prime factor.
www.reference.com/article/prime-factors-number-24-e4896f9a49bb1f95
The prime factors of the number 24 are 2 x 2 x 2 x 3 , or 2 to the 3rd power x 3. Two and three are prime numbers, and when three twos and a three are multiplied together, the product is the given number, which is 24. | 490 | 1,741 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2020-45 | latest | en | 0.910845 |
https://www.numbersaplenty.com/150655 | 1,723,101,751,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640723918.41/warc/CC-MAIN-20240808062406-20240808092406-00518.warc.gz | 711,891,415 | 3,080 | Search a number
150655 = 5291039
BaseRepresentation
bin100100110001111111
321122122211
4210301333
514310110
63121251
71165141
oct446177
9248584
10150655
11a320a
1273227
135375b
143cc91
152e98a
hex24c7f
150655 has 8 divisors (see below), whose sum is σ = 187200. Its totient is φ = 116256.
The previous prime is 150649. The next prime is 150659. The reversal of 150655 is 556051.
It is a sphenic number, since it is the product of 3 distinct primes.
It is a cyclic number.
It is not a de Polignac number, because 150655 - 217 = 19583 is a prime.
It is a congruent number.
It is not an unprimeable number, because it can be changed into a prime (150659) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (11) of ones.
It is a polite number, since it can be written in 7 ways as a sum of consecutive naturals, for example, 375 + ... + 664.
It is an arithmetic number, because the mean of its divisors is an integer number (23400).
2150655 is an apocalyptic number.
150655 is a deficient number, since it is larger than the sum of its proper divisors (36545).
150655 is a wasteful number, since it uses less digits than its factorization.
150655 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 1073.
The product of its (nonzero) digits is 750, while the sum is 22.
The square root of 150655 is about 388.1430148798. The cubic root of 150655 is about 53.2101540910.
The spelling of 150655 in words is "one hundred fifty thousand, six hundred fifty-five".
Divisors: 1 5 29 145 1039 5195 30131 150655 | 479 | 1,618 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2024-33 | latest | en | 0.871752 |
https://www.clicours.com/algorithmes-de-type-expectation-maximization-en-ligne-pour-lestimation/ | 1,657,159,667,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104683020.92/warc/CC-MAIN-20220707002618-20220707032618-00791.warc.gz | 762,785,708 | 22,669 | # Algorithmes de type Expectation-Maximization en ligne pour l’estimation
Algorithmes de type Expectation-Maximization en ligne pour l’estimation
The Expectation Maximization (EM) algorithm is a versatile tool for model parameter estimation in latent data models. When processing large data sets or data stream however, EM becomes intractable since it requires the whole data set to be available at each iteration of the algorithm. In this contribution, a new generic online EM algorithm for model parameter inference in general Hidden Markov Model is proposed. This new algorithm updates the parameter estimate after a block of observations is processed (online). The convergence of this new algorithm is established, and the rate of convergence is studied showing the impact of the block-size sequence. An averaging procedure is also proposed to improve the rate of convergence. Finally, practical illustrations are presented to highlight the performance of these algorithms in comparison to other online maximum likelihood proce- dures. The Expectation Maximization (EM) algorithm is an iterative algorithm used to solve maximum likelihood estimation in HMM. The EM algorithm is generally simple to implement since it relies on complete data computa- tions. Each iteration is decomposed into two steps: the E-step computes the conditional expectation of the complete data log-likelihood given the observations and the M-step updates the parameter estimate based on this conditional expectation. In many situations of interest, the complete data likelihood belongs to the curved exponential family. In this case, the E-step boils down to the computation of the conditional expectation of the com- plete data sufficient statistic. Even in this case, except for simple models such as linear Gaussian models or HMM with finite state-spaces, the E-step is intractable and has to be approximated e.g. by Monte Carlo methods such as Markov Chain Monte Carlo methods or Sequential Monte Carlo methods (see [Carlin et al., 1992] or [Capp´e et al., 2005, Doucet et al., 2001] and the references therein).
However, when processing large data sets or data streams, the EM al- gorithm might become impractical. Online variants of the EM algorithm have been first proposed for independent and identically distributed (i.i.d.) observations, see [Capp´ e et Moulines, 2009]. When the complete data like- lihood belongs to the cruved exponential family, the E-step is replaced by a stochastic approximation step while the M-step remains unchanged. The convergence of this online variant of the EM algorithm for i.i.d. observa- tions is addressed by [Capp´ e et Moulines, 2009]: the limit points are the stationary points of the Kullback-Leibler divergence between the marginal distribution of the observation and the model distribution. This new algorithm, called Block Online EM (BOEM) is derived in Sec- tion 6.2 together with an averaged version. Section 6.3 is devoted to practical applications: the BOEM algorithm is used to perform parameter inference in HMM where the forward recursions mentioned above are available explicitly. In the case of finite state-space HMM, the BOEM algorithm is compared to a gradient-type recursive maximum likelihood procedure and to the online EM algorithm of [Capp´e, 2011a]. The convergence of the BOEM algorithm is addressed in Section 6.4. The BOEM algorithm is seen as a perturbation of a deterministic limiting EM algorithm which is shown to converge to the stationary points of the limiting relative entropy (to which the true param- eter belongs if the model is well specified). The perturbation is shown to vanish (in some sense) as the number of observations increases thus implying that the BOEM algorithms inherits the asymptotic behavior of the limiting EM algorithm. Finally, in Section 6.5, we study the rate of convergence of the BOEM algorithm as a function of the block-size sequence. We prove that the averaged BOEM algorithm is rate-optimal when the block-size se- quence grows polynomially. All the proofs are postponed to Section 6.6; supplementary proofs and comments are provided in Appendix A.
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The Block Online EM algorithm
Our algorithms update the parameter after processing a block of obser- vations. Nevertheless, the intermediate quantity S e, 2011a, Sec- tion 2.2] and [Del Moral et al., 2010b, Proposition 2.1] and can be applied either to finite state-space HMM or to linear Gaussian models. A Sequential Monte Carlo approximation to compute The classical theory of maximum likelihood estimation often relies on the assumption that the ”true” distribution of the observations belongs to the specified parametric family of distributions. In many cases, it is doubtful that this assumption is satisfied. It is therefore natural to investigate the convergence of the BOEM algorithms and to identify the possible limit for misspecified models i.e. when the observations Y are from an ergodic process which is not necessarily an HMM. In Section 6.3.1, the performance of the BOEM algorithm and its aver- aged version are illustrated in a truncated linear Gaussian model. In Sec- tion 6.3.2, the BOEM algorithm is compared to online maximum likelihood procedures in the case of finite state-space HMM. | 1,128 | 5,303 | {"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-2022-27 | latest | en | 0.81948 |
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2022-12-28, 00:51 #2234 Rubiksmath Sep 2022 22×17 Posts Had a lucky ECM streak on unfactored exponents recently, usually find 1 factor every couple weeks but have got 2 factors within a few hours of each other: M214783 has a factor: 1102834494064669506004266650004828269423 / (ECM curve 17, B1=3000000, B2=6616119510, Sigma=137844928612264) (129.696 bits, 40 digits, new personal record largest factor) M210491 has a factor: 1680490771184651242949512447428919 / (ECM curve 11, B1=3000000, B2=6616119510, Sigma=575255113699867) (110.373 bits, 34 digits)
2022-12-28, 03:01 #2235 Rubiksmath Sep 2022 4416 Posts Edit: Make that 3, this just happened aparrently: M208073 has a factor: 3731074311071089639450741837271 (ECM curve 35, B1=3000000, B2=6616119510)
2022-12-30, 01:08 #2236 Runtime Error Sep 2017 USA 2×5×19 Posts Found a 110-bit factor on a wave front exponent! min B1 = 1,000,037 min B2 = 545,850,313 M114785129 has a 110.863-bit (34-digit) factor: 2361477100582561776755612373583993 (P-1,B1=1400000,B2=1001933790)
2022-12-30, 06:40 #2237 Jwb52z Sep 2002 84610 Posts P-1 found a factor in stage #2, B1=537000, B2=19025622. UID: Jwb52z/Clay, M119602183 has a factor: 39769121872136784387765097 (P-1, B1=537000, B2=19025622) 85.040 bits.
2023-01-05, 00:55 #2238 Uncwilly 6809 > 6502 """"""""""""""""""" Aug 2003 101×103 Posts 2A9016 Posts Nice 100 bit P-1 factor: 119636903 has a factor: 1455962897408551638312586078553
2023-01-07, 14:32 #2239 Viliam Furik "Viliam Furík" Jul 2018 Martin, Slovakia 13·61 Posts Another one joins the club (bites the dust) M736647071 has a factor: 19251950982339179681312159 (TF 83-84) A new member of a relatively small group of Mersenne's with 10 (or more) known factors.
2023-01-12, 01:54 #2240 Jwb52z Sep 2002 2×32×47 Posts UID: Jwb52z/Clay, M119686603 has a factor: 194945427614502977436457 (P-1, B1=537000) 77.367 bits.
2023-01-12, 02:27 #2241 chalsall If I May "Chris Halsall" Sep 2002 Barbados 22·47·59 Posts I've been amusing my sorry little butt by TF'ing way down in 2M. Also some P-1. My success rate on the TF has been very low (as expected). But today I found this.
2023-01-15, 21:21 #2242 KingKurly Sep 2010 Annapolis, MD, USA BF16 Posts M2614303 has factor 94286373215560002099130008156415246687079887097 (156.04 bits) k = 2^2 * 23 * 89 * 103 * 4423 * 17053 * 267517 * 3876497 * 5110717 * 53488147 think this is my biggest factor; I've found other large factors with P-1 but they were composite
2023-01-15, 21:24 #2243
James Heinrich
"James Heinrich"
May 2004
ex-Northern Ontario
11×373 Posts
Quote:
Originally Posted by KingKurly think this is my biggest factor
It is.
Nice factor (that the previous 9 P-1 attempts failed to find).
2023-01-16, 00:19 #2244
Stargate38
"Daniel Jackson"
May 2011
14285714285714285714
3×251 Posts
Quote:
Originally Posted by KingKurly M2614303 has factor 94286373215560002099130008156415246687079887097 (156.04 bits) k = 2^2 * 23 * 89 * 103 * 4423 * 17053 * 267517 * 3876497 * 5110717 * 53488147
I just checked, and it's also the 18th largest P-1 factor ever found for a Mersenne number.
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Given A = {2, 3, 4}, B = {2, 5, 6, 7}. Construct an example of each of the following:(i) an injective map from A to B(ii) a mapping from A to B which is not injective(iii) a mapping from A to B.
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(i) {(2, 7), (3, 6), (4, 5)}
(ii) {(2, 2), (3, 2), (4, 5)}
(iii) {(2, 5), (3, 6), (4, 7)}
Disclaimer: There are many more possibilities of each case.
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Question Text Given A = {2, 3, 4}, B = {2, 5, 6, 7}. Construct an example of each of the following:(i) an injective map from A to B(ii) a mapping from A to B which is not injective(iii) a mapping from A to B. Topic Relations and Functions Subject Mathematics Class Class 12 Answer Type Text solution:1 Upvotes 19 | 381 | 1,341 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2024-10 | latest | en | 0.840581 |
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## Presentation on theme: "Monday, June 23, 2014PHYS 1441-001, Summer 2014 Dr. Jaehoon Yu 1 PHYS 1441 – Section 001 Lecture #11 Monday, June 23, 2014 Dr. Jaehoon Yu Newton’s Law."— Presentation transcript:
Monday, June 23, 2014PHYS 1441-001, Summer 2014 Dr. Jaehoon Yu 1 PHYS 1441 – Section 001 Lecture #11 Monday, June 23, 2014 Dr. Jaehoon Yu Newton’s Law of Universal Gravitation Weightlessness Work done by a constant force Multiplication of Vectors Work-Kinetic Energy Theorem
Announcements Term exam #2 –In class this Wednesday, June 25 –Non-comprehensive exam –Covers CH 4.7 to what we finish tomorrow, Tuesday, June 24 –Bring your calculator but DO NOT input formula into it! Your phones or portable computers are NOT allowed as a replacement! –You can prepare a one 8.5x11.5 sheet (front and back) of handwritten formulae and values of constants for the exam no solutions, derivations or definitions! No additional formulae or values of constants will be provided! Monday, June 23, 2014PHYS 1441-001, Summer 2014 Dr. Jaehoon Yu 2
Monday, June 23, 2014PHYS 1441-001, Summer 2014 Dr. Jaehoon Yu Newton’s Law of Universal Gravitation People have been very curious about the stars in the sky, making observations for a long~ time. The data people collected, however, have not been explained until Newton has discovered the law of gravitation. Every object in the Universe attracts every other object with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. How would you write this law mathematically? G is the universal gravitational constant, and its value is This constant is not given by the theory but must be measured by experiments. With G Unit? This form of forces is known as the inverse-square law, because the magnitude of the force is inversely proportional to the square of the distances between the objects. 3
Monday, June 23, 2014PHYS 1441-001, Summer 2014 Dr. Jaehoon Yu Ex. Gravitational Attraction What is the magnitude of the gravitational force that acts on each particle in the figure, assuming m 1 =12kg, m 2 =25kg, and r=1.2m? 4
Monday, June 23, 2014PHYS 1441-001, Summer 2014 Dr. Jaehoon Yu Why does the Moon orbit the Earth? 5
Monday, June 23, 2014PHYS 1441-001, Summer 2014 Dr. Jaehoon Yu Gravitational Force and Weight Since weight depends on the magnitude of gravitational acceleration, g, it varies depending on geographical location. The attractive force exerted on an object by the Earth Gravitational Force, FgFg Weight of an object with mass M is By measuring the forces one can determine masses. This is why you can measure mass using the spring scale. What is the SI unit of weight? N 6
Monday, June 23, 2014PHYS 1441-001, Summer 2014 Dr. Jaehoon Yu Gravitational Acceleration What is the SI unit of g? m/s 2 Gravitational acceleration at distance r from the center of the earth! 7
Monday, June 23, 2014PHYS 1441-001, Summer 2014 Dr. Jaehoon Yu Gravitational force on the surface of the earth: Magnitude of the gravitational acceleration on the surface of the Earth 8
Monday, June 23, 2014 Example for Universal Gravitation Using the fact that g=9.80m/s 2 on the Earth’s surface, find the average density of the Earth. Since the gravitational acceleration is Therefore the density of the Earth is Solving for MEME Solving for g PHYS 1441-001, Summer 2014 Dr. Jaehoon Yu 9
Monday, June 23, 2014PHYS 1441-001, Summer 2014 Dr. Jaehoon Yu There is only one speed that a satellite can have if the satellite is to remain in an orbit with a fixed radius. Satellite in Circular Orbits What acts as the centripetal force? The gravitational force of the earth pulling the satellite! 10
Monday, June 23, 2014PHYS 1441-001, Summer 2014 Dr. Jaehoon Yu Determine the speed of the Hubble Space Telescope orbiting at a height of 598 km above the earth’s surface. Ex. Orbital Speed of the Hubble Space Telescope 11
Monday, June 23, 2014PHYS 1441-001, Summer 2014 Dr. Jaehoon Yu Period of a Satellite in an Orbit Speed of a satellite Period of a satellite Square either side and solve for T2 This is applicable to any satellite or even for planets and moons. Kepler’s 3 rd Law 12
Monday, June 23, 2014PHYS 1441-001, Summer 2014 Dr. Jaehoon Yu Geo-synchronous Satellites What period should these satellites have? Satellite TV Global Positioning System (GPS) The same as the earth!! 24 hours 13
Monday, June 23, 2014PHYS 1441-001, Summer 2014 Dr. Jaehoon Yu In each case, what is the weight recorded by the scale? Ex. Apparent Weightlessness and Free Fall 0 14 0
Monday, June 23, 2014PHYS 1441-001, Summer 2014 Dr. Jaehoon Yu At what speed must the surface of the space station move so that the astronaut experiences a push on his feet equal to his weight on earth? The radius is 1700 m. Ex. Artificial Gravity 15
Monday, June 23, 2014PHYS 1441-001, Summer 2014 Dr. Jaehoon Yu 16 x y Work Done by a Constant Force A meaningful work in physics is done only when the net forces exerted on an object changes the energy of the object. M F Free Body Diagram M d Which force did the work?Force How much work did it do? What does this mean? Physically meaningful work is done only by the component of the force along the movement of the object. Unit? Work is an energy transfer!! Why? What kind?Scalar
Let’s think about the meaning of work! A person is holding a grocery bag and walking at a constant velocity. Are his hands doing any work ON the bag? –No –Why not? –Because the force hands exert on the bag, Fp,Fp, is perpendicular to the displacement!! –This means that hands are not adding any energy to the bag. So what does this mean? –In order for a force to perform any meaningful work, the energy of the object the force exerts on must change due to that force!! What happened to the person? –He spends his energy just to keep the bag up but did not perform any work on the bag. Monday, June 23, 2014PHYS 1441-001, Summer 2014 Dr. Jaehoon Yu 17
Monday, June 23, 2014PHYS 1441-001, Summer 2014 Dr. Jaehoon Yu 18 Work done by a constant force s
Monday, June 23, 2014PHYS 1441-001, Summer 2014 Dr. Jaehoon Yu 19 Scalar Product of Two Vectors Product of magnitude of the two vectors and the cosine of the angle between them Operation is commutative Operation follows the distribution law of multiplication How does scalar product look in terms of components? Scalar products of Unit Vectors =0
Monday, June 23, 2014PHYS 1441-001, Summer 2014 Dr. Jaehoon Yu 20 Example of Work by Scalar Product A particle moving on the xy plane undergoes a displacement d =(2.0 i +3.0 j )m as a constant force F =(5.0 i +2.0 j ) N acts on the particle. a) Calculate the magnitude of the displacement and that of the force. b) Calculate the work done by the force F. Y X d F Can you do this using the magnitudes and the angle between d and F?F?
Monday, June 23, 2014PHYS 1441-001, Summer 2014 Dr. Jaehoon Yu 21 Ex. Pulling A Suitcase-on-Wheel Find the work done by a 45.0N force in pulling the suitcase in the figure at an angle 50.0 o for a distance s=75.0m. Does work depend on mass of the object being worked on?Yes Why don’t I see the mass term in the work at all then? It is reflected in the force. If an object has smaller mass, it would take less force to move it at the same acceleration than a heavier object. So it would take less work. Which makes perfect sense, doesn’t it?
Monday, June 23, 2014PHYS 1441-001, Summer 2014 Dr. Jaehoon Yu 22 Ex. 6.1 Work done on a crate A person pulls a 50kg crate 40m along a horizontal floor by a constant force F p =100N, which acts at a 37 o angle as shown in the figure. The floor is rough and exerts a friction force F fr =50N. Determine (a) the work done by each force and (b) the net work done on the crate. What are the forces exerting on the crate? F G =-mg So the net work on the crate Work done on the crate by FGFG FpFp F fr Which force performs the work on the crate? FpFp F fr Work done on the crate by Fp:Fp: Work done on the crate by F fr : This is the same as F N =+mg Work done on the crate by FNFN
Monday, June 23, 2014PHYS 1441-001, Summer 2014 Dr. Jaehoon Yu 23 Ex. Bench Pressing and The Concept of Negative Work A weight lifter is bench-pressing a barbell whose weight is 710N a distance of 0.65m above his chest. Then he lowers it the same distance. The weight is raised and lowered at a constant velocity. Determine the work in the two cases. What is the angle between the force and the displacement? What does the negative work mean? The gravitational force does the work on the weight lifter!
Monday, June 23, 2014PHYS 1441-001, Summer 2014 Dr. Jaehoon Yu 24 The truck is accelerating at a rate of +1.50 m/s 2. The mass of the crate is 120-kg and it does not slip. The magnitude of the displacement is 65 m. What is the total work done on the crate by all of the forces acting on it? Ex. Accelerating a Crate What are the forces acting in this motion? Gravitational force on the crate, weight, W or F g Normal force force on the crate, F N Static frictional force on the crate, f s
Monday, June 23, 2014PHYS 1441-001, Summer 2014 Dr. Jaehoon Yu 25 Ex. Continued… Lets figure out what the work done by each of the forces in this motion is. Work done by the gravitational force on the crate, W or F g Work done by Normal force force on the crate, F N Work done by the static frictional force on the crate, f s Which force did the work? Static frictional force on the crate, f s How? By holding on to the crate so that it moves with the truck!
Monday, June 23, 2014PHYS 1441-001, Summer 2014 Dr. Jaehoon Yu 26 Kinetic Energy and Work-Kinetic Energy Theorem Some problems are hard to solve using Newton’s second law –If forces exerting on an object during the motion are complicated –Relate the work done on the object by the net force to the change of the speed of the object M ΣF M s vivi vfvf Suppose net force ΣF ΣF was exerted on an object for displacement d to increase its speed from vi vi to vf.vf. The work on the object by the net force ΣF ΣF is Using the kinematic equation of motion Work Kinetic Energy Work done by the net force causes change in the object’s kinetic energy. Work-Kinetic Energy Theorem
Monday, June 23, 2014PHYS 1441-001, Summer 2014 Dr. Jaehoon Yu 27 When a net external force by the jet engine does work on an object, the kinetic energy of the object changes according to Work-Kinetic Energy Theorem
Monday, June 23, 2014PHYS 1441-001, Summer 2014 Dr. Jaehoon Yu 28 The mass of the space probe is 474-kg and its initial velocity is 275 m/s. If a 56.0-mN force acts on the probe parallel through a displacement of 2.42×10 9 m, what is its final speed? Ex. Deep Space 1 Solve for v f
Monday, June 23, 2014PHYS 1441-001, Summer 2014 Dr. Jaehoon Yu 29 A satellite is moving about the earth in a circular orbit and an elliptical orbit. For these two orbits, determine whether the kinetic energy of the satellite changes during the motion. Ex. Satellite Motion and Work By the Gravity For a circular orbit For an elliptical orbit No change!Why not? Gravitational force is the only external force but it is perpendicular to the displacement. So no work. Changes!Why? Gravitational force is the only external force but its angle with respect to the displacement varies. So it performs work.
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# What use the GCF to reduce 6 over 24 to the lowest terms is it a. 2 over 8 b. 1 over 4 c. 1 over 6 d. 3 over 4?
Updated: 10/31/2022
Keyshunroachgp0238
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6y ago
The GCF of 6 and 24 is 6 because 6 is a factor of 24
6/24 = 1/4
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Q: What use the GCF to reduce 6 over 24 to the lowest terms is it a. 2 over 8 b. 1 over 4 c. 1 over 6 d. 3 over 4?
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### How do you reduce a fraction to lowest terms?
Find the GCF of the numerator and the denominator. Divide each of them by that number. If the GCF is 1, the fraction is already in its lowest terms.
### How do you reduce fractions into lowest terms?
Find the GCF of the numerator and the denominator. Divide each of them by that number. If the GCF is 1, the fraction is already in its lowest terms.
### How do you reduce 56 over 98 to lowest terms?
Find their GCF and divide them by it. 56/98 = 4/7
### How do you turn a percent into fraction lowest terms?
To turn a percent into a fraction, put it over 100. To reduce a fraction to its lowest terms, find the GCF of the numerator and the denominator and divide them both by it. If the GCF is 1, the fraction is in its simplest form.
### How do you reduce the fraction 210 over 294 to the lowest terms?
Divide them both by their GCF, which is 42. You should get 5/7
### What does it mean to reduce a fraction to its lowest terms?
Find the GCF of the numerator and denominator and divide them both by it. If the GCF is 1, the fraction is in its lowest terms, also known as simplest form.
2/5
### How do you reduce fraction 32 over 12 to lowest terms?
The GCF of 32 and 12 is 4. Take 4 out of both of them. Get 8/3
### What is 10 over 16 in lowest terms?
10/16 in lowest term = 5/8To reduce to lowest term:1. Get the GCF of 10 & 1610 = 2 ...............* 516 = 2 * 2 * 2 * 2===========GCF=22. Divide 10 & 16 by their GCF.10÷ 2 = 516÷ 2 = 8thus; lowest term is5/8.
### What is 21 over 7 in lowest terms?
7/21 in lowest term =1/3To reduce to lowest term:1. Get the GCF of 7 & 217 = 721=7 * 3======GCF=72. Divide the 7 & 21 by their GCF.7÷ 7 = 121÷ 7 = 3thus; lowest term is1/3.
### What is 3 over 24 in lowest terms?
3/24 in lowest term =1/8To reduce to lowest term:1. Get the GCF of 3 & 243 = 324=3 * 2 * 2 * 2==========GCF=32. Divide the 3 & 24 by their GCF.3÷ 3 = 124÷ 3 =8thus; lowest term is 1/8.
d. | 809 | 2,447 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.46875 | 4 | CC-MAIN-2024-18 | latest | en | 0.93666 |
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May 5, 2016
# Homework Help: science
Posted by nikita on Tuesday, March 1, 2011 at 4:24pm.
a sliver coin and a gold coin each have a mass of exactly 6.6 grams the specific heat of sliver is 0.235j/g.degrees celcius and the specific heat of gold his 0.130j/g. degrees celcius what coin requires more heat to raise its tempreature by 40 degrees celcius
• science - DrBob222, Tuesday, March 1, 2011 at 6:09pm
Note the correct spelling of celsius.
q = mass x specific heat x delta T.
Calculate q to raise both by 40 C. | 162 | 527 | {"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} | 3.40625 | 3 | CC-MAIN-2016-18 | longest | en | 0.910379 |
https://www.coursehero.com/file/42387536/LECTURE-1-Introduction-to-Statistics-1pdf/ | 1,610,746,310,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703496947.2/warc/CC-MAIN-20210115194851-20210115224851-00230.warc.gz | 749,585,254 | 747,568 | LECTURE 1 Introduction to Statistics (1).pdf - LECTURE 1 Introduction to Statistics S P 11 0 S TAT I S T I C S F O R P S Y C H O L O G Y OUTLINE What is
# LECTURE 1 Introduction to Statistics (1).pdf - LECTURE 1...
This preview shows page 1 - 19 out of 48 pages.
LECTURE 1 Introduction to Statistics SP110 STATISTICS FOR PSYCHOLOGY
OUTLINE What is Statistics? Branches of Statistics Population and Sample Data and Variable Types of Data and Variables Parameter and Statistic Levels of Measurement Independent and Dependent Variables Parametric and Non-Parametric Tests
WHAT IS STATISTICS?
Uses of Statistics Organize and summarize information Determine exactly what conclusions are justified based on the results that were obtained Goals of statistical procedures Accurate and meaningful interpretation Provide standardized evaluation procedures
a
BRANCHES OF STATISTICS There are two main branches of statistical methods Descriptive statistics Psychologists use descriptive statistics to summarize and describe a group of numbers from a research study. Inferential statistics Psychologists use inferential statistics to draw conclusions and to make inferences that are based on the numbers from a research study but that go beyond the numbers.
POPULATION AND SAMPLE Population The set of all the individuals of interest in a particular study Vary in size; often quite large Sample A set of individuals selected from a population Usually intended to represent the population in a research study
DATA AND VARIABLE Data Measurements or observations of a variable Variable Characteristic that can have different values
TABLE is aVARIABLE?
TABLE is a VARIABLE
SIZE OF A TABLE is aVARIABLE?
SIZE OF A TABLE is a VARIABLE
Types of Data Types of Variables Quantitative Data Quantitative Variable Discrete Continuous Qualitative Data Qualitative Variable TYPES OF DATA AND VARIABLES | 377 | 1,877 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2021-04 | latest | en | 0.837156 |
http://printfu.org/inference | 1,603,451,448,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107881369.4/warc/CC-MAIN-20201023102435-20201023132435-00699.warc.gz | 80,343,170 | 6,636 | # Inference
440 CHAPTER 10 Inference What Is an Inference? An inference is a meaning that is suggested rather than directly stated. Inferences are implied through clues that lead the reader to make assumptions and draw
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### Introduction to Statistical Inference
Introduction to Statistical Inference Floyd Bullard Introduction Example1 Example2 Example3 Example4 Conclusion Likelihood ratios There are numerous tools available for parameter estimation, and you'llbeintroduced tot wo or three of them this week.
www.stat.duke.edu
### Probability Plots
http://www.inference.us Page 6 5/7/2008 inference from Blue Reference The fitdistr( ) function in the MASS package provides maximum-likelihood fitting of univariate distributions.
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### INFERENCE
INFERENCE An educated guess about unstated ideas in a passage. Because the authors don't always clearly state every idea in a story or article, you have to make inferences by. . . drawing conclusions, forming generalizations, and making predictions.
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### A Rule-Based Inference Engine
A Rule-Based Inference Engine which is Optimal and VLSI Implementable N. L. Griffin and F. D. Lewis Department of Computer Science University of Kentucky Lexington, Kentucky 40506 ABSTRACT An inference engine for rule based expert systems which forms part of the EXPRES system is developed and ...
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Reading Strategies — Inference Introduction .....2 Inference Lesson 1 ..... 11 Purpose: Introduce inference as a powerful strategy for creating deeper meaning and enriching the reading experience.
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### Making Inferences
I use clues from the text and combine those with what I already know to make an inference. Even though the author has not explicitly stated this, I can infer that this is true.
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### Inference Notes
© Jim Burke 2000. For more information on this and other such Tools for Thought visit www.englishcompanion.com Name Subject Date Period Says/Does Means/Implies Directions : Find six quotes or examples that reveal important or different aspects of your subject.
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### Model-Based Sampling and Inference
Model-Based Sampling, Inference and Imputation James R. Knaub, Jr., Energy Information Administration, EI-53.1 [email protected] Key Words:
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### Inference in Excel Overview
Page | 2 Inference in Excel Overview 1 August 2008 Inference in Excel Overview Bring dynamic scripting to Microsoft Excel spreadsheets Summary Inference in Excel brings the benefits of dynamic scripting in platforms like R, S-Plus, MATLAB, Python, IronPython and IronRuby for .NET to your ...
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### Other sites you could try:
Find videos related to Inference | 595 | 2,774 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2020-45 | latest | en | 0.824693 |
http://stackoverflow.com/questions/4661045/excel-solver-solving-based-on-an-average | 1,467,219,274,000,000,000 | text/html | crawl-data/CC-MAIN-2016-26/segments/1466783397749.89/warc/CC-MAIN-20160624154957-00192-ip-10-164-35-72.ec2.internal.warc.gz | 306,840,238 | 16,766 | # Excel Solver: Solving based on an average
I have a parameter in A1 that influences "TOTAL" in a random and very high standard deviation. Lets say A1 is 2...then TOTAL Values could be 1...5...17...3...2..2...etc If A1 is 1 then TOTAL Values could be 1....3...5..15...9...10..etc
I would like solver to figure out which value in A1 would equate to the best AVERAGE of TOTAL after X runs. Where I can define X.
In my example you can tell that `A1=1` is better on average after 6 runs. However, if you run solver normally it would say `A1=2` is the best, because it produced a value of 17.
-
This doesn't seem to be the kind of problem you solve with solver. Why not write a macro that loops through the values of A1, X times, keeping a running sum of the TOTAL values for each A1? When it's all over, the largest sum is also the largest average.
The inner loop will be something like this:
``````Redim tSum(1 to maxA1)
for i = 1 to maxA1
tSum(i) = 0
for j = 1 to X
[A1] = i
Application.calculate
tSum(i) = tSum(i) + TOTAL
next j
next i
'now step through tSum. The index of the largest value
' is the value of A1 desired. Put it in a handy cell.
``````
It has to be a macro, not a function because it changes A1.
- | 357 | 1,221 | {"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-2016-26 | latest | en | 0.944402 |
https://devforum.roblox.com/t/is-this-possible-to-achieve-with-constraints/730916 | 1,685,306,213,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224644506.21/warc/CC-MAIN-20230528182446-20230528212446-00087.warc.gz | 240,967,329 | 5,477 | # Is this possible to achieve with constraints?
Basically, I’m trying to figure out the best way to move a model to bricks that act as the nodes or position goals for its path.
I thought of originally tweening the model to each point’s position, however, this would be rough mainly because I want the model to be moving at a linear speed, and depending on the distance of the model’s current position to the node it’s moving to, the speed of the tween will change.
I just wondered, is this behavior possible to achieve using some combo of constraints?
On top of this, I also need to adjust the model’s rotation to animate it as it moves. This also gives me more reason to avoid tweening the model.
Not really sure here…thanks!
You can use some maths to make linear animation. Simply just lerp the position from the current position to the end position using this formula:
``````local Position = PartPosition
local StartTime = tick()
RS.Heartbeat:Connect(function()
Position:Lerp(TotalLength / (tick() - StartTime))
end
``````
You can always change the position and it will keep lerping.
1 Like
If I recall correctly, Tweening also modifies the CFrame. You would only need some basic physics knowledge to achieve a constant speed:
velocity = distance / time
In this case, you want to keep a constant velocity throughout all tweens, and we can get the distance using the .Mangitude() method of a Vector3. Our dependent variable would then be the time.
time = distance / velocity
Now you assign the velocity you prefer as a constant, let’s say, 20 studs/ second. Next, check your initial and final positions using Vector3.Magnitude(). Divide This value by your velocity and you get the time value for your Tween.
Enjoy!
2 Likes | 380 | 1,739 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2023-23 | latest | en | 0.891763 |
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Functions Arithmetic Calculator - get the sum, product, quotient and difference of functions steps by step This website uses cookies to ensure you get the best experience. By using this website, you agree to our Cookie Policy.
The FUNctions Master Fun Packet includes 11 different FUNsheets covering a typical unit on functions. This packet includes 11 different worksheets: Worksheet 1: Order of Operations. Worksheet 2: Evaluating Functions. Worksheet 3: Function equations, tables and graphs. Worksheet 4: Function Composition. Worksheet 5: Piecewise Functions
Formulas and functions are the building blocks of working with numeric data in Excel. FUNCTION IN EXCEL is a predefined formula that is used for specific values in a particular order. Function is used for quick tasks like finding the sum, count, average, maximum value, and minimum values for a range...
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Basic and Elementary; Find where to buy the TI-84 Plus CE graphing calculator in a variety of bold, fun colors. Learn more. ... Application of Function Composition.
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Lesson 3 4 graphing functions answers
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functions together, an idea called composition of functions. Using the function T(d), we could evaluate T(5) to determine the average daily temperature on the 5th day of the year. We could then use that temperature as the input to the C(T) function to find the cost to heat the house on the 5th day of the year: C(T(5)). Composition of Functions
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An accounting device used to analyze transactions.
Chapter 40 Structure and Function of the Digestive System Alexa K. Doig and Sue E. Huether Chapter Outline The Gastrointestinal Tract Mouth and Esophagus Stomach Small Intestine Large Intestine Intestinal Bacteria Accessory Organs of Digestion Liver Gallbladder Exocrine Pancreas Tests of Digestive Function Gastrointestinal Tract Liver Gallbladder Exocrine Pancreas AGING and the ...
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Biostatistics pdf | 1,502 | 6,617 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2021-17 | latest | en | 0.899191 |
https://www.physicsforums.com/threads/least-action-principle-for-a-free-relativistic-particle-landau.363859/ | 1,723,320,278,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640822309.61/warc/CC-MAIN-20240810190707-20240810220707-00798.warc.gz | 747,534,966 | 18,710 | # Least action principle for a free relativistic particle (Landau)
• provolus
In summary, the conversation discusses doubts in understanding the steps for deriving the formula for the variation of the action for a relativistic free particle, using the invariant element of measure and varying the coordinates. The conversation specifically addresses the second step of the formula and clarifies the calculation of the variation of x_i x^i. The final expression from Landau's book is obtained by canceling out the 2 in the numerator and denominator.
provolus
Reading the Landau's "The classical theory of fields" (chapter 2, section 9 ) I have some doubts in explaining the steps in derivig the formula for the variation of the action for the relativistic free particle http://books.google.it/books?id=QIx...age&q="to set up the expression for"&f=false". Given the invariant element of measure:
$$ds=\sqrt{dx_idx^i}$$
where $$x^i$$ ( $$x_i$$ ) are the four contravariant (covariant) coordinates which parametrize the world line of the free particle, I have to vary respect $$x^i$$, that is I make the variation $$\delta x^i$$. So my doubts are about the second step of the formula before the 9.10, that is why:
$$\delta(ds)=\frac{d x_i \delta d x^i}{ds}$$
is obtained, instead of (IMH and erroneous O):
$$\delta(ds)=\frac{d x_i \delta d x^i}{2 \cdot ds}$$
?
Can someone be so kind to show me the steps?
Last edited by a moderator:
What is $\delta \left(x_{i}x^{i}\right) = ?$
It should be:
$$\delta (x_i x^i) = \delta (c^2t^2-r^2) = 2 (c^2 t \delta x^0 - r \delta x^i$$)
but, sorry, I don't get the point... that is... should I calculate
$$\delta (dx_i dx^i)$$
?
The "d" in the brackets is not important. That 2 you have obtained in front cancels the one in the denominator, thus giving you the final expression from Landau's book.
thx for the moment. I hope to need no more help in covariant variation calculus... ;-)
## 1. What is the least action principle for a free relativistic particle?
The least action principle states that a free relativistic particle will follow the path of least action, where action is defined as the integral of the Lagrangian over the particle's trajectory.
## 2. What is the significance of the least action principle?
The least action principle is significant because it provides a fundamental principle for describing the motion of particles in classical mechanics. It also has applications in quantum mechanics, where it is known as the path integral formulation.
## 3. How does the least action principle differ from Newton's laws of motion?
The least action principle is a variational principle, meaning it takes into account all possible paths a particle could take and finds the one that minimizes the action. In contrast, Newton's laws of motion only consider the forces acting on a particle and how they affect its motion.
## 4. Can the least action principle be applied to systems with multiple particles?
Yes, the least action principle can be extended to systems with multiple particles by considering the action as a sum of the individual particle actions. This allows for the determination of the trajectories of all particles in the system.
## 5. What is the role of relativity in the least action principle?
The least action principle is a relativistic principle, meaning it takes into account the effects of special relativity on the motion of a particle. This is important for accurately describing the motion of particles moving at high speeds.
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1K | 960 | 3,954 | {"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": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2024-33 | latest | en | 0.827814 |
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Simple Image Manipulations
Keywords: image file format, PPM, header, bits per pixel, color depth, high dynamic range image, return value optimization, RVO, copy elision.Want to report bugs, errors or send feedback? Write at feedback@scratchapixel.com. We do this for free, please be kind. Thank you.
It seems quite natural as a prerequisite to any of the lessons devoted to image processing that we learn about reading and writing images. In this lesson, we will also write a very basic Image class which will be extended over time. Obviously many designs are possible but we will just stick with a very simple one to start with. Our basic image class will only support floating point images (a must these days) composed of three channels only. You can easily extend this class as an exercise, for example you can use template to define the image data type and the number of channels on the fly. Without any further due, let's start to quickly revisit what we know about images and implement some simple code to read and write them as well as store them in memory. In the next lesson, we will extend our basic C++ class to support basic image manipulations.
Figure 1: image channels of a RGB image.
We are familiar with the concept of digital images. As a quick definition let's just say that an image in the digital world, is stored or represented as a 2D array of pixels. The image resolution defines the dimension of this 2D array. The width (horizontal dimension) and the height (vertical dimension) define the overall resolution of the image. A pixel in the digital world is generally defined by three digital values, one for each elementary or primary color. If you work with an additive color system (which computer screens are using) these colors are red, blue and green (in a subtractive color system these primary colors are cyan, magenta and yellow). In the old times, when computer memory and disk space were limited, pixels values were often stored using the least possible number of bytes. With 1 byte (8 bits) you can encode or represent 256 different values per color. If you use three bytes to encode the red, green and blue color of a pixel respectively, you can define up to 16.8 millions different colors. This format is know as Truecolor. The number of bits (not bytes) used per pixel is often abbreviated with the acronym bpp; it stands for bits per pixel. Truecolor uses 24 bpp. Sometimes the number of bits used is defined per channel (i.e. per color). In this case the acronym is bpc (bits per channel) but it is not often used. The number of bits used for each color component of a single pixel, is known as the color depth or bit depth. If Truecolor is generally enough for the Internet, most image formats used by professionals in the video or film industry encode pixel color as three single (or half) floating point precision numbers (the type float in C++). This means that a single pixel uses 12 bytes of memory (or 96 bpp). A high definition image (typically 1920x1080 pixels) requires roughly 24 Mbytes of storage (assuming no compression). Storing pixel color as floats was originally motivated by the need for high dynamic range images (or HDR images), images capable of storing the wider range of colors and light levels one can observe in the physical world. This concept will be reviewed in a separate lesson.
From a programming point of view, all we need to do, is to create a simple Image class, in which we will store the width and the height of the image, as well as a 1D array of pixels (you can use a two dimensional arrays if you want, but this is not necessarily more practical and is not as efficient). We need two constructors, one to create an empty image and one to create an image with a user specified resolution (line 20) which can also be filled with a given constant color (this color default to black).
The pixels themselves are represented with a special structure named rib. This is very convenient because we can write operators to manipulate the three floats of a pixel at once. It is trivial to access pixels from the image by overloading the bracket operator [] and applying every mathematical operations we want on these pixels: multiplying them, comparing them, adding them, etc. The method friend float& operator += (...) is interesting. It can be used to compute the brightness of a pixel and accumulate the result into a float. We will use it in the next chapters.
Finally pixels are stored in the image as a 1D array of Rgb elements. The size of this area is equal to the width of the image multiplied by the image height. It is not necessary to multiply this number by the number of channels since each pixel (of type Rgb) already holds the memory to store three colors.
#include <cstdlib> #include <cstdio> class Image { public: // Rgb structure, i.e. a pixel struct Rgb { Rgb() : r(0), g(0), b(0) {} Rgb(float c) : r(c), g(c), b(c) {} Rgb(float _r, float _g, float _b) : r(_r), g(_g), b(_b) {} bool operator != (const Rgb &c) const { return c.r != r || c.g != g || c.b != b; } Rgb& operator *= (const Rgb &rgb) { r *= rgb.r, g *= rgb.g, b *= rgb.b; return *this; } Rgb& operator += (const Rgb &rgb) { r += rgb.r, g += rgb.g, b += rgb.b; return *this; } friend float& operator += (float &f, const Rgb rgb) { f += (rgb.r + rgb.g + rgb.b) / 3.f; return f; } float r, g, b; }; Image() : w(0), h(0), pixels(nullptr) { /* empty image */ } Image(const unsigned int &_w, const unsigned int &_h, const Rgb &c = kBlack) : w(_w), h(_h), pixels(NULL) { pixels = new Rgb[w * h]; for (int i = 0; i < w * h; ++i) pixels[i] = c; } const Rgb& operator [] (const unsigned int &i) const { return pixels[i]; } Rgb& operator [] (const unsigned int &i) { return pixels[i]; } ~Image() { if (pixels != NULL) delete [] pixels; } unsigned int w, h; // Image resolution Rgb *pixels; // 1D array of pixels static const Rgb kBlack, kWhite, kRed, kGreen, kBlue; // Preset colors }; const Image::Rgb Image::kBlack = Image::Rgb(0); const Image::Rgb Image::kWhite = Image::Rgb(1); const Image::Rgb Image::kRed = Image::Rgb(1,0,0); const Image::Rgb Image::kGreen = Image::Rgb(0,1,0); const Image::Rgb Image::kBlue = Image::Rgb(0,0,1); int main(int argc, char **argv) { Image *img = new Image(512, 512); return 0; }
## Read and Write Images: PPM, a Simple Image Format
Photoshop and Preview (on Mac) can read and write PPM files. In Photoshop the format is called Portable Bit Map. If you write them out with Photoshop you will need to change the extension of the file from .bpm to .ppm (or change the file extension to .bpm in your code).
Figure 2: pixels are packed together in the PPM file format.
The PPM format is the simplest image file format someone can design. It is not optimised at all, don't use any form of compression thus the image file on disk is large compared to other formats such PNG or JPEG. However, the code to read and save PPM images in and out is incredibly simple to write and for this reason, this format is very useful when you need your program to read or write images. Use it to prototype some ideas and replace it later by a more sophisticated format (one that supports compression and potentially some other useful features such as arbitrary number of channels, etc.). This lesson is not a complete introduction to the PPM format. Even though basic (very close to a raw format) the PPM format has its subtleties. It can for example be used to store color or black and white images, as well as pixels's data in ascii or binary mode. In this lesson we will only work with binary files and color images.
Most image file formats are designed the same way. At the top of the file is usually stored was is called a header. It's generally a structure in which are defined the standard properties of the image: its resolution of course, but also the number of images planes stored in the image, the number of bits per channel, whether the data is compressed or not, which compression algorithm the pixels are processed with, etc. Generally, the only really necessary information is the image resolution. Everything else is optional. It can also contain meta data such as who created the file, the date and the time when the file was generated, etc. If it comes from a digital camera these meta data can also contain information about the lens being used, the camera settings such as shutter speed, the aperture, etc that were used when the picture was taken. This information can be useful as we will see when we get to the lesson on HDR images. Finally, the header is almost always directly followed by the image data itself. The way the data is encoded really varies from a file format to another. Colors of pixels can be stored together (as with the PPM file format) or each plane can be stored separately. Colors can be stored with integers, chars, floats, etc. Image data can also be compressed (generally they are).
The header (the few lines at the top of the file providing the necessary information about the image we are about to read) is defined by three lines. The first line contains a string which can either be "P1", "P2", ... "P6". We are only concerned by "P6" PPM images which stands for binary color images. The next line contains two integers representing the resolution of the image. The third line contains an another integer representing the number of levels each color is encoded with (i.e. the number of grey values between black and white, generally 255). You can insert comments in the header; they start with the letter '#', but we will ignore them in this lesson.
P6 640 480 255 # comment but we won't support them for now
In the PPM format, colors are encoded using 24 bpp. In other worlds each color is encoded as 1 bytes (in C++ an unsigned char). Each pixel is encoded in the file as a sequence of three bytes (figure 2), one for each color:
R1G1B1R2G2B2R3G3B3...
Pixels are encoded in scanline order, from the first row to the last row in the image, for left to right (figure 2). In other words the first three bytes in the file are for the pixel at location (1,1) in the image (where (1,1) denotes the upper left corner of the image), and the last three bytes in the file represent the lower right pixel in the frame (location (640, 480) assuming the image has resolution 640x480). Let's have a look at our code to read a PPM file:
Image readPPM(const char *filename) { std::ifstream ifs; ifs.open(filename, std::ios::binary); // need to spec. binary mode for Windows users Image src; try { if (ifs.fail()) { throw("Can't open input file"); } std::string header; int w, h, b; ifs >> header; if (strcmp(header.c_str(), "P6") != 0) throw("Can't read input file"); ifs >> w >> h >> b; src.w = w; src.h = h; src.pixels = new Image::Rgb[w * h]; // this is throw an exception if bad_alloc ifs.ignore(256, '\n'); // skip empty lines in necessary until we get to the binary data unsigned char pix[3]; // read each pixel one by one and convert bytes to floats for (int i = 0; i < w * h; ++i) { ifs.read(reinterpret_cast<char *>(pix), 3); src.pixels[i].r = pix[0] / 255.f; src.pixels[i].g = pix[1] / 255.f; src.pixels[i].b = pix[2] / 255.f; } ifs.close(); } catch (const char *err) { fprintf(stderr, "%s\n", err); ifs.close(); } return src; }
Note that some special cases need to be taken care of carefully. For example when the file we want to read doesn't exist on disk, or when its size is 0 (which is unlikely but you still need to consider this option). Some programs also insert an empty line after the header when the write a PPM file out. The code removes these lines if they exist. Note that the code is really not optimized. You could read the entire block of data in memory then convert it to float, or at least read one line of pixel data at a time. The conversion from byte to float only requires a division by 255. We know that a byte can only encode a number between 0 and 255 where 0 represents black and 255 white. By dividing this value by 255, we remap the pixel color to a float in the interval [0,1] where 0 represents black and 1 represents white on the screen.
Important question from a reader: "from a programming point of view, is it efficient to return an image?" Good question. Consider the following code:
First you must do a copy within the function reading the PPM file in the return statement (src itself is copied to a temporary variable returned by the function). Then when dst is initialised, the data from this temporary variable is copied once more to dst itself. This implies two copies (and a series of allocations) of potentially large memory blocks which indeed would be inefficient. Though in our particular case, src.pixels is a pointer to a dynamically allocated block of memory and copying pointers values isn't in itself too bad (though this will lead to more problems as we will show later).
However, nowadays, C++ allows what's known as copy elision (this will only work if you have a recent compiler). The idea is that your compiler will find out that the type of the returned function is the same as the type of the variable you allocate the result of this function to. In this situation, dat will actually be initialised as if it was created in the readPPM function directly and all copies previously involved will be omitted (or elided). In other words, your compiler will sort of replace src with dst in the function readPPM() for you. This is called a return value optimization (or RVO for short).
One of the standard solutions to this problem was to return a pointer to a dynamically allocated instance of the class Image. Pointer initialization or assignment is far less costly than a copy:
Image* readPPM(...) { Image *img = new Image(...); ... return img; } Image *img = readPPM(...);
But with copy elisions, this is not necessary anymore. In short, with most modern compilers, this code should be efficient and shouldn't require any copy of the image data at all (if copy elision works).
In:
Image operator + (const Image &img) const { Image tmp; // ... return tmp; }
You are creating and returning an object of the same type as the return type of the function. This implies that return tmp; will consider tmp as if it was an rvalue as per 12.8/32:
When the criteria for elision of a copy operation are met or would be met save for the fact that the source object is a function parameter, and the object to be copied is designated by an lvalue, overload resolution to select the constructor for the copy is first performed as if the object were designated by an rvalue.
The mentioned criteria are given in 12.8/31, in particular, the first bullet point says:
— In a return statement in a function with a class return type, when the expression is the name of a non-volatile automatic object (other than a function or catch-clause parameter) with the same cv-unqualified type as the function return type, the copy/move operation can be omitted by constructing the automatic object directly into the function’s return value
Actually, a careful reading of 12.8/31 says that in your case compilers are allowed (and the most popular ones do) to omit the copy or move altogether. This is the so called return value optimization (RVO). Indeed, consider this simplified version of your code:
struct Image { Image() { } Image(const Image&) { std::cout << "copy\n"; } Image(Image&&) { std::cout << "move\n"; } Image operator +(const Image&) const { Image tmp; return tmp; } }; int main() { Image src; Image copy = src + src; }
Compiled with GCC 4.8.1, this code produces no output, that is, no copy of move operation is performed. Let's complicate the code a little bit just to see what happened when RVO cannot be performed.
Image operator +(const Image&) const { Image tmp1, tmp2; if (std::rand() % 2) return tmp1; return tmp2; }
Without much of details, RVO cannot be applied here not because the standard forbids so but for other technical reasons. With this implementation of operator +() the code outputs move. That is, there's no copy, only a move operation. A last word. It's not advisable to do return std::move(tmp); because it prevents RVO. Indeed, with this implementation:
Image operator +(const Image&) const { Image tmp; return std::move(tmp); }
The output is move, that is, the move constructor is called, whereas, as we've seen, with return tmp; no copy/move constructor is called. That's the correct behaviour because the expression being return std::move(tmp) is not the name of a non-volatile automatic object as required by the RVO rule quoted above. The implementation of operator +() which introduces tmp and tmp2 is rather an artificial way to prevent RVO. Let's go back to the initial implementation and consider another way of preventing RVO which also shows the complete picture: compile the code with the option -fno-elide-constructors (also available in clang). The output (in GCC but it might vary in clang) is:
move move
When a function is called stack memory is allocated to build the object to be returned. I emphasise that this is not the variable tmp above. This another unnamed temporary object. Then, return tmp; triggers a copy or move from tmp to the unnamed object and the initialisation Image cry = src + src; finally copy/move the unnamed object into cry. That's the basic semantics. Regarding the first copy/move we have the following. Since tmp is an lvalue the copy constructor would normally be used to copy from tmp to the unnamed object. However, the special clause above makes an exception and says that tmp in return tmp; should be considered as if it was an rvalue. Hence the move constructor is called. In addition, when RVO is performed, the move is elided and tmp is actually created on top of the unnamed object. Regarding the second copy/move it's even simpler. The unnamed object is an rvalue and therefore the move constructor is selected to move from it to cry. Now, there's another optimization (which is similar to RVO but AFAIK doesn't have a name) also stated in 12.8/31 (third bullet point) that allows the compiler to avoid the use of the unnamed temporary and use the memory of cpy instead. Therefore, when RVO and this optimization are in place tmp, the unnamed object and cry are essentially "the same object". There seems to be some confusion as to how the RVO (Return Value Optimization) works. A simple example:
struct A { int a; int b; int c; int d; }; A create(int i) { A a = {i, i+1, i+2, i+3 }; std::cout << &a << "\n"; return a; } int main(int argc, char*[]) { A a = create(argc); std::cout << &a << "\n"; }
0xbf928684 0xbf928684
Surprising ? Actually, that is the effect of RVO: the object to be returned is constructed directly in place in the caller. How ? Traditionally, the caller (main here) will reserve some space on the stack for the return value: the return slot; the callee (create here) is passed (somehow) the address of the return slot to copy its return value into. The callee then allocates its own space for the local variable in which it builds the result, like for any other local variable, and then copies it into the return slot upon the return statement. RVO is triggered when the compiler deduces from the code that the variable can be constructed directly into the return slot with equivalent semantics (the as-if rule). Note that this is such a common optimization that it is explicitly white-listed by the Standard and the compiler does not have to worry about possible side-effects of the copy (or move) constructor. When ? The compiler is most likely to use simple rules, such as:
// 1. works A unnamed() { return {1, 2, 3, 4}; } // 2. works A unique_named() { A a = {1, 2, 3, 4}; return a; } // 3. works A mixed_unnamed_named(bool b) { if (b) { return {1, 2, 3, 4}; } A a = {1, 2, 3, 4}; return a; } // 4. does not work A mixed_named_unnamed(bool b) { A a = {1, 2, 3, 4}; if (b) { return {4, 3, 2, 1}; } return a; }
In the latter case (4), the optimization cannot be applied when "A" is returned because the compiler cannot build "a" in the return slot, as it may need it for something else (depending on the boolean condition "b"). A simple rule of thumb is thus that: RVO should be applied if no other candidate for the return slot has been declared prior to the "return" statement.
Let's now have a look at the function to save an image as a PPM file:
void savePPM(const Image &img, const char *filename) { if (img.w == 0 || img.h == 0) { fprintf(stderr, "Can't save an empty image\n"); return; } std::ofstream ofs; try { ofs.open(filename, std::ios::binary); // need to spec. binary mode for Windows users if (ofs.fail()) throw("Can't open output file"); ofs << "P6\n" << img.w << " " << img.h << "\n255\n"; unsigned char r, g, b; // loop over each pixel in the image, clamp and convert to byte format for (int i = 0; i < img.w * img.h; ++i) { r = static_cast(std::min(1.f, img.pixels[i].r) * 255); g = static_cast(std::min(1.f, img.pixels[i].g) * 255); b = static_cast(std::min(1.f, img.pixels[i].b) * 255); ofs << r << g << b; } ofs.close(); } catch (const char *err) { fprintf(stderr, "%s\n", err); ofs.close(); } }
Here again the code is pretty basic. We don't save the image if one its dimension is zero. If the file can't be open then we throw an exception. We then write then header, the "P6" string, followed by the width, the height of the image and the number of grey values between black and white (255). Finally we loop over all the pixels of the frame. An unsigned char can only encode integer values in the interval [0,255], thus before multiplying the pixel color by 255 we first need to clamp its value to the interval [0,1]. Finally once converted to a unsigned char, the colors of the current pixels can be written out to the file.
## Source Code
This lesson and next one are part of the same group of lessons (learning about reading, writing and manipulating images). The source for these two lessons can be found here (to add link later when lesson 2 is completed).
## What's Next?
In the next lesson we will learn how to extend the functionalities of the image class and use operators for manipulating images in an easy way. We will demonstrate with an example, how we can use these simple features to simulate the blur which can sometimes see in the out-of-focus areas of real photographs. | 5,353 | 22,309 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2022-21 | latest | en | 0.929517 |
http://mathhelpforum.com/calculus/61993-not-so-simple-equation-solve-print.html | 1,527,263,359,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867094.5/warc/CC-MAIN-20180525141236-20180525161236-00402.warc.gz | 192,127,314 | 3,009 | # not-so-simple equation to solve
• Nov 28th 2008, 05:39 AM
Simo
not-so-simple equation to solve
Dear all,
I have a tricky problem; I need to solve this equation for $\displaystyle N$
$\displaystyle \binom {L} {L-R}\sum_{r=1}^{R}\binom {R} {r} (-1)^{R-r}\left(\frac{r}{L}\right)^N\ln\frac{r}{L} =0$
Is it possible in some way?
The only constraints are $\displaystyle R \leqslant L$, $\displaystyle R,L>0$
Thanks a lot
Simo
• Nov 28th 2008, 06:54 AM
shawsend
Going gets tough, tough resort to numerical methods. The following is Mathematica code to calculate the value of n for the supplied values of R and L. First out is the expression for the sum, second out is the numerically calculated value of n which makes the sum equal to zero, and the third out is a back-substitution of the the root for a check.
Code:
In[234]:= R = 3; L = 5; formula1 = Sum[Binomial[R, r]*(-1)^(R - r)* (r/L)^n*Log[r/L], {r, 1, R}] root = n /. FindRoot[formula1 == 0, {n, 1.2}] formula1 /. n -> root Out[236]= -(3/5)^n Log[5/3] + 3 (2/5)^n Log[5/2] - 3 5^-n Log[5] Out[237]= 1.38015 Out[238]= -5.55112*10^-17
• Nov 28th 2008, 07:07 AM
Simo
Ok,
I already have a numerical solution like that, but it is not interesting for me.
Is not possible to find an analytical solution, without resort to numerical methods? | 444 | 1,301 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2018-22 | latest | en | 0.804406 |
http://math.stackexchange.com/questions/tagged/calculator?sort=unanswered&pagesize=15 | 1,469,385,659,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257824133.26/warc/CC-MAIN-20160723071024-00225-ip-10-185-27-174.ec2.internal.warc.gz | 160,575,574 | 23,560 | # Tagged Questions
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Can someone recommend me a good and affordable calculator please. I am going to sit exams like partial differential equations (method of characteristics related stuff), nonlinear dynamical systems and ... | 1,808 | 7,090 | {"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} | 2.953125 | 3 | CC-MAIN-2016-30 | latest | en | 0.907703 |
https://blogs.mathworks.com/cleve/2018/02/19/fun-with-the-pascal-triangle/?from=kr | 1,716,381,576,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058542.48/warc/CC-MAIN-20240522101617-20240522131617-00877.warc.gz | 107,593,975 | 29,707 | # Fun With The Pascal Triangle
The Wikipedia article on Pascal's Triangle has hundreds of properties of the triangle and there are dozens of other Web pages devoted to it. Here are a few facts that I find most interesting.
### Contents
#### Blaise Pascal
Blaise Pascal (1623-1662) was a 17th century French mathematician, physicist, inventor and theologian. His Traité du triangle arithmétique (Treatise on Arithmetical Triangle) was published posthumously in 1665. But this was not the first publication about the triangle. Various versions appear in Indian, Chinese, Persian, Italian and other manuscripts centuries before Pascal.
#### Binomial Coefficients
The binomial coefficient usually denoted by ${n} \choose {k}$ is the number of ways of picking $k$ unordered outcomes from $n$ possibilities. These coefficients appear in the expansion of the binomial $(x+1)^n$. For example, when $n = 7$
syms x
n = 7;
x7 = expand((x+1)^n)
x7 =
x^7 + 7*x^6 + 21*x^5 + 35*x^4 + 35*x^3 + 21*x^2 + 7*x + 1
Formally, the binomial coefficients are given by $${{n} \choose {k}} = \frac {n!} {k! (n-k)!}$$ But premature floating point overflow of the factorials makes this an unsatisfactory basis for computation. A better way employs the recursion $${{n} \choose {k}} = {{n-1} \choose {k}} + {{n-1} \choose {k-1}}$$ This is used by the MATLAB function nchoosek(n,k).
#### Pascal Matrices
MATLAB offers two Pascal matrices. One is symmetric, positive definite and has the binomial coefficients on the antidiagonals.
P = pascal(7)
P =
1 1 1 1 1 1 1
1 2 3 4 5 6 7
1 3 6 10 15 21 28
1 4 10 20 35 56 84
1 5 15 35 70 126 210
1 6 21 56 126 252 462
1 7 28 84 210 462 924
The other is lower triangular, with the binomial coefficients in the rows. (We will see why the even numbered columns have minus signs in a moment.)
L = pascal(7,1)
L =
1 0 0 0 0 0 0
1 -1 0 0 0 0 0
1 -2 1 0 0 0 0
1 -3 3 -1 0 0 0
1 -4 6 -4 1 0 0
1 -5 10 -10 5 -1 0
1 -6 15 -20 15 -6 1
The individual elements are
P(i,j) = P(j,i) = nchoosek(i+j-2,j-1)
And (temporarily ignoring the minus signs) for i $\ge$ j
L(i,j) = nchoosek(i-1,j-1)
The first fun fact is that L is the (lower) Cholesky factor of P.
L = chol(P)'
L =
1 0 0 0 0 0 0
1 1 0 0 0 0 0
1 2 1 0 0 0 0
1 3 3 1 0 0 0
1 4 6 4 1 0 0
1 5 10 10 5 1 0
1 6 15 20 15 6 1
So we can reconstruct P from L.
P = L*L'
P =
1 1 1 1 1 1 1
1 2 3 4 5 6 7
1 3 6 10 15 21 28
1 4 10 20 35 56 84
1 5 15 35 70 126 210
1 6 21 56 126 252 462
1 7 28 84 210 462 924
#### Pascal Triangle
The traditional Pascal triangle is obtained by rotating P clockwise 45 degrees, or by sliding the rows of L to the right in half increments. Each element of the resulting triangle is the sum of the two above it.
triprint(L)
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
#### Square Root of Identity
When the even numbered columns of L are given minus signs the matrix becomes a square root of the identity.
L = pascal(n,1)
L_squared = L^2
L =
1 0 0 0 0 0 0
1 -1 0 0 0 0 0
1 -2 1 0 0 0 0
1 -3 3 -1 0 0 0
1 -4 6 -4 1 0 0
1 -5 10 -10 5 -1 0
1 -6 15 -20 15 -6 1
L_squared =
1 0 0 0 0 0 0
0 1 0 0 0 0 0
0 0 1 0 0 0 0
0 0 0 1 0 0 0
0 0 0 0 1 0 0
0 0 0 0 0 1 0
0 0 0 0 0 0 1
Here is an exercise for you. What is sqrt(eye(n))? Why isn't it L?
#### Cube Root of Identity
When I first saw this, I was amazed. Rotate L counterclockwise. The result is a cube root of the identity.
X = rot90(L,-1)
X_cubed = X^3
X =
1 1 1 1 1 1 1
-6 -5 -4 -3 -2 -1 0
15 10 6 3 1 0 0
-20 -10 -4 -1 0 0 0
15 5 1 0 0 0 0
-6 -1 0 0 0 0 0
1 0 0 0 0 0 0
X_cubed =
1 0 0 0 0 0 0
0 1 0 0 0 0 0
0 0 1 0 0 0 0
0 0 0 1 0 0 0
0 0 0 0 1 0 0
0 0 0 0 0 1 0
0 0 0 0 0 0 1
#### Sierpinski
Which binomial coefficients are odd? It's a fledgling fractal.
odd = @(x) mod(x,2)==1;
n = 56;
L = abs(pascal(n,1));
spy(odd(L))
title('odd(L)')
#### Fibonacci
The sums of the antidiagonals of L are the Fibonacci numbers.
n = 12;
A = fliplr(abs(pascal(n,1)))
for k = 1:n
F(k) = sum(diag(A,n-k));
end
F
A =
0 0 0 0 0 0 0 0 0 0 0 1
0 0 0 0 0 0 0 0 0 0 1 1
0 0 0 0 0 0 0 0 0 1 2 1
0 0 0 0 0 0 0 0 1 3 3 1
0 0 0 0 0 0 0 1 4 6 4 1
0 0 0 0 0 0 1 5 10 10 5 1
0 0 0 0 0 1 6 15 20 15 6 1
0 0 0 0 1 7 21 35 35 21 7 1
0 0 0 1 8 28 56 70 56 28 8 1
0 0 1 9 36 84 126 126 84 36 9 1
0 1 10 45 120 210 252 210 120 45 10 1
1 11 55 165 330 462 462 330 165 55 11 1
F =
1 1 2 3 5 8 13 21 34 55 89 144
#### pi
The elements in the third column of lower triangular Pascal matrix are the triangle numbers. The n-th triangle number is the number of bowling pins in the n-th row of an array of bowling pins. $$t_n = {{n+1} \choose {2}}$$
L = pascal(12,1);
t = L(3:end,3)'
t =
1 3 6 10 15 21 28 36 45 55
Here's an unusual series relating the triangle numbers to $\pi$. The signs go + + - - + + - - .
pi - 2 = 1 + 1/3 - 1/6 - 1/10 + 1/15 + 1/21 - 1/28 - 1/36 + 1/45 + 1/55 - ...
type pi_pascal
function pie = pi_pascal(n)
tk = 1;
s = 1;
for k = 2:n
tk = tk + k;
if mod(k+1,4) > 1
s = s + 1/tk;
else
s = s - 1/tk;
end
end
pie = 2 + s;
Ten million terms gives $\pi$ to 14 decimal places.
format long
pie = pi_pascal(10000000)
err = pi - pie
pie =
3.141592653589817
err =
-2.398081733190338e-14
#### Matrix Exponential
Finally, I love this one. The solution to the (potentially infinite) set of ordinary differential equations $\dot{x_1} = x_1$ $\dot{x_j} = x_j + (j-1) x_{j-1}$ is $x_j = e^t (t + 1)^{j-1}$ This means that the matrix exponential of the simple diagonal matrix
D = diag(1:7,-1)
D =
0 0 0 0 0 0 0 0
1 0 0 0 0 0 0 0
0 2 0 0 0 0 0 0
0 0 3 0 0 0 0 0
0 0 0 4 0 0 0 0
0 0 0 0 5 0 0 0
0 0 0 0 0 6 0 0
0 0 0 0 0 0 7 0
is
expm_D = round(expm(D))
expm_D =
1 0 0 0 0 0 0 0
1 1 0 0 0 0 0 0
1 2 1 0 0 0 0 0
1 3 3 1 0 0 0 0
1 4 6 4 1 0 0 0
1 5 10 10 5 1 0 0
1 6 15 20 15 6 1 0
1 7 21 35 35 21 7 1
#### Thanks
Thanks to Nick Higham for pascal.m, gallery.m and section 28.4 of N. J. Higham, Accuracy and Stability of Numerical Algorithms, Second edition, SIAM, 2002. http://epubs.siam.org/doi/book/10.1137/1.9780898718027
Published with MATLAB® R2018a
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https://josechristian.wordpress.com/2013/07/09/check-if-palindrome-python/ | 1,660,077,825,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571086.77/warc/CC-MAIN-20220809185452-20220809215452-00284.warc.gz | 330,893,195 | 27,541 | Categories
# Check if Palindrome – Python
Another one of these challenges – this one is just too easy.
## The Challenge
Checks if the string entered by the user is a palindrome. That is that it reads the same forwards as backwards like “racecar”
## The logic
Since palindromes are the same whether they are spelled backwards or not, we can compare the two strings, the normal one and the reversed one (using the same loop as the reverse string challenge).
If they are both exactly the same, we have a palindrome!
However, we might get a false negative if there is an upper case letter. Therefore, it might be a good idea to change all letters to lowercase.
## The Code
```# the string
str_the_raw_word = raw_input('Enter word: ')
# change all letters to lowercase
str_the_word = str_the_raw_word.lower()
# find the length of the word
int_word_len = len(str_the_word)
# reverse word as list
list_rev_word = []
# loop through each letter
for i in range(1,int_word_len+1):
# reverse the possition
int_rev_pos = int_word_len - i
# now move to list
list_rev_word.append(str_the_word[int_rev_pos])
# from list to string
str_rev_word = ''.join(list_rev_word)
# now compare both strings
# if its the same then we're good
if str_rev_word == str_the_word:
print 'Yo Mr White, "%s" be a palindrome and s#@t!' % str_the_raw_word
# if not the same then there's problem
else:
print 'Nope, you suck!'
```
so if we run it using the word ‘RaceCar’ as an example
```\$ python _palindrome.py
Enter word: RaceCar
```
We should get the following output.
```Yo Mr White, "RaceCar" be a palindrome and s#@t!
```
and if we run it again using the word ‘what’ as an example
```\$ python _palindrome.py
Enter word: what
```
We should get the following output.
```Nope, you suck!
``` | 454 | 1,782 | {"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-2022-33 | latest | en | 0.788868 |
https://encyclopedia2.thefreedictionary.com/feild | 1,621,316,070,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243989820.78/warc/CC-MAIN-20210518033148-20210518063148-00269.warc.gz | 255,809,293 | 23,630 | # field
(redirected from feild)
Also found in: Dictionary, Thesaurus, Medical.
Related to feild: field
## field,
in algebra, set of elements (usually numbers) that may be combined under the operations of addition and multiplication so that it constitutes an additive groupgroup,
in mathematics, system consisting of a set of elements and a binary operation a+b defined for combining two elements such that the following requirements are satisfied: (1) The set is closed under the operation; i.e.
, the nonzero elements form a multiplicative group, and multiplication distributes over addition. The set of real numbers (see numbernumber,
entity describing the magnitude or position of a mathematical object or extensions of these concepts. The Natural Numbers
Cardinal numbers describe the size of a collection of objects; two such collections have the same (cardinal) number of objects if their
) and the set of complex numbers are both examples of fields.
## field,
in physics, region throughout which a force may be exerted; examples are the gravitational, electric, and magnetic fields that surround, respectively, masses, electric charges, and magnets. The field concept was developed by M. Faraday based on his investigation of the lines of force that appear to leave and return to a magnet at its poles (see flux, magneticflux, magnetic,
in physics, term used to describe the total amount of magnetic field in a given region. The term flux was chosen because the power of a magnet seems to "flow" out of the magnet at one pole and return at the other pole in a circulating pattern, as suggested
). Fields are used to describe all cases where two bodies separated in space exert a forceforce,
commonly, a "push" or "pull," more properly defined in physics as a quantity that changes the motion, size, or shape of a body. Force is a vector quantity, having both magnitude and direction.
on each other. The alternative to postulating a field is to assume that physical influences can be transmitted through empty space without any material or physical agency. Such action-at-a-distance, especially if it occurs instantaneously, violates both common sense and certain modern theories, notably relativityrelativity,
physical theory, introduced by Albert Einstein, that discards the concept of absolute motion and instead treats only relative motion between two systems or frames of reference.
, which posits that nothing can travel faster than light. In a field description, rather than body A directly exerting a force on body B, body A (the source) creates a field in every direction around it and body B (the detector) experiences the field that exists at its position. If a change occurs at the source, its effect propagates outward through the field at a constant speed and is felt at the detector only after a certain delay in time. The field is thus a kind of "middleman" for transmitting forces. Each type of force (electric, magnetic, nuclear, or gravitational) has its own appropriate field; a body experiences the force due to a given field only if the body itself it also a source of that kind of field. The reciprocity implied by Newton's third law of motionmotion,
the change of position of one body with respect to another. The rate of change is the speed of the body. If the direction of motion is also given, then the velocity of the body is determined; velocity is a vector quantity, having both magnitude and direction, while speed
(equal action and reaction) is thus preserved. If two bodies exert a mutual force, they possess potential energyenergy,
in physics, the ability or capacity to do work or to produce change. Forms of energy include heat, light, sound, electricity, and chemical energy. Energy and work are measured in the same units—foot-pounds, joules, ergs, or some other, depending on the system of
that depends on their relative positions; it is natural to regard this energy as residing in the field the bodies create.
## field
1. The region in which a physical agency exerts an influence, measured in terms of the force experienced by an object in that region. Thus a massive body has an associated gravitational field within which an object will feel an attractive force. There are several types of field, including gravitational, magnetic, electric, and nuclear fields, each of which has its own characteristics and exerts its influence, strong or weak, over a particular range. See also fundamental forces.
2. See field of view.
Collins Dictionary of Astronomy © Market House Books Ltd, 2006
The following article is from The Great Soviet Encyclopedia (1979). It might be outdated or ideologically biased.
## Field
the subject of study in field theory, a branch of algebra. The concept of a field is often made use of in branches of mathematics other than algebra.
Four arithmetic operations—the primary operations of addition and multiplication and the inverse operations of subtraction and division—can be performed on ordinary numbers. Generalization yields the concept of a field. Thus, a field is a set of elements for which there are defined two operations—called addition and multiplication—that are subject to the ordinary laws (axioms) of arithmetic:
(I) Addition and multiplication are commutative and associative; that is, a + b = b + a, ab = ba, a + (b + c) = (a + b) + c, and a(bc) = (ab) c.
(II) The set contains an additive identity element 0 (zero); that is, a + 0 = a for any element a in the set. For every a there exists an inverse element —a such that a + (–a) = 0. It thus follows that the subtraction operation ab can be performed in a field.
(III) The set contains a multiplicative identity element e; that is, ae = a for any a. For each a ≠ 0 there exists an inverse element a-1 such that aa-1 = e. It thus follows that division by any a ≠ 0 is possible.
(IV) The operations of addition and multiplication are governed by the distributive law: a(b + c) = ab + ac.
The following are examples of fields.
(1) The set P of all rational numbers.
(2) The set R of all real numbers.
(3) The set K of all complex numbers.
(4) The set of all rational functions in one or several variables with real coefficients.
(5) The set of all numbers of the form , where a and b are rational.
(6) The field of residue classes modulo p, which is defined as follows: Let ρ be a prime number. The integers are divided into classes called residue classes by combining in one class all integers that give the same remainder when divided by p. To define the addition of two classes, take an integer from each of the two classes and form the sum of these integers. The class to which this sum belongs is the sum of the two classes. The product of classes is defined in a similar manner. With these definitions of addition and multiplication the set of classes forms a field containing ρ elements.
It follows from axioms (I) and (II) that the elements of a field form a commutative group with respect to addition and from axioms (I) and (III) that all nonzero elements of a field form a commutative group with respect to multiplication.
If a field contains an element a ≠ 0 such that na = 0 for some integer n, there exists a prime ρ such that pa = 0 for any element a in the field. The field is then said to be of characteristic p; example (6) is an illustration of this case. If na ≠ 0 for any nonzero n and a, the characteristic of the field is zero; the fields in examples (1) through (5) have characteristic zero.
If a subset F of a field G is itself a field with respect to the addition and multiplication in G, then F is called a subfield of G and G an extension of F. A field that does not have subfields is called a prime field. All prime fields are exhausted by the fields of examples (1) and (6) (for all possible choices of the prime p). Every field contains a unique prime subfield; the prime field in each of the fields in examples (2) through (5) is the field of rational numbers. The following is a “natural” problem: Describe all extensions of a given prime field. Steinitz’ theorem, which is given below, deals with just this problem.
Two types of extensions have a comparatively simple structure. One type is the simple transcendental extension, where G is the field of all rational functions of one variable with coefficients in P. The other type is the simple algebraic extension, an illustration of which is the field of example (5). Such an extension is obtained by a construction similar to that of example (6): if f(x) is a polynomial of degree n that is irreducible in F, G is the field of residue classes of polynomials of degree n modulo f(x). To obtain a simple algebraic extension of the second type, we in effect adjoin to F a root of f(x) and all elements that can be expressed in terms of this root and the elements of F. If every element of G is a root of some polynomial with coefficients in F, G is said to be an algebraic extension of F. According to Steinitz’ theorem, any extension of a field F can be obtained in two stages: first we obtain a transcendental extension T of F (by forming the field of rational functions, not necessarily of one variable, over F) and then an algebraic extension of T. Only fields in which every polynomial decomposes into a product of linear factors do not have algebraic extensions. Such fields are called algebraically closed fields. The field of complex numbers is algebraically closed (the fundamental theorem of algebra). Every field is a subfield of an algebraically closed field.
Certain types of fields have been studied in particular detail. The theory of algebraic numbers deals primarily with simple algebraic extensions of the field of rational numbers. The theory of algebraic functions investigates simple algebraic extensions of the field of rational functions with complex coefficients; considerable attention is paid to finite extensions of the field of rational functions over an arbitrary field of constants, that is, rational functions with coefficients in some arbitrary given field. Finite extensions of fields, particularly their automorphisms, are studied in Galois theory; many problems that arise in the solution of algebraic equations find an answer here. In many problems of algebra, especially in various branches of the theory of fields, normed fields play an important role. Ordered fields appear and have been studied in connection with geometric investigations.
### REFERENCES
Kurosh, A. G. Kurs vysshei algebry, 10th ed. Moscow, 1971.
Van der Waerden, B. L. Sovremennaia algebra [2nd ed.], parts 1-2. Moscow-Leningrad, 1947. (Translated from German.)
Chebotarev, N. G. Teoriia algebraicheskikh funktsii. Moscow-Leningrad, 1948.
Chebotarev, N. G. Osnovy teorii Galua, parts 1-2. Leningrad-Moscow, 1934-37.
Weyl, H. Algebraicheskaia teoriia chisel. Moscow, 1947. (Translated from English.)
## Field
in biology, a concept that describes a biological system the behavior of whose parts is determined by the position of the parts in the system. The existence of such systems has been determined principally by numerous experiments on the movement, elimination, and addition of parts in the embryo. In many cases, normal organisms develop from such embryos, since their components change their path of development according to their new position in the whole.
Between 1912 and 1922, A. G. Gurvich introduced the concept of field (morphogenetic field) into embryology and established the task of elucidating its laws. He identified these laws initially with an indivisible factor governing morphogenesis and later with a system of intercellular interactions that determine the movement and differentiation of embryonic cells. In 1925 the Austrian scientist P. Weiss applied the concept of field to the processes of regeneration, and in 1934 the British scientists J. Huxley and G. de Beer combined the concept with the concept of gradient. In the 1940’s through 1960’s, the British biologist C. Waddington and the French mathematician R. Thorn introduced concepts of embryonic development as a vector field divided into a limited number of zones of “structural stability.”
The range of concepts of field is being intensively developed in contemporary theoretical biology, but no unified opinion as to the intrinsic principles underlying the phenomena described by the concept of field has been worked out.
### REFERENCES
Gurvich, A. G. Teoriia biologicheskogo polia. Moscow, 1944.
Waddington, C. Morfogenez i genetika. Moscow, 1964. (Translated from English.)
Na puti k teoreticheskoi biologii, vol. 1. Moscow, 1970. (Translated from English.)
Towards a Theoretical Biology, vols.2–4. Edinburgh, 1969-72.
L. V. BELOUSOV
## Field
in physics, a special form of matter. A field is a physical system that has an infinite number of degrees of freedom. Examples are electromagnetic fields, gravitational fields, the field of nuclear forces, and the quantized fields associated with different particles.
The concept of electric and magnetic fields was introduced in the 1830’s by M. Faraday, who adopted the concept as an alternative to the theory of long-range interaction, that is, the interaction of particles at a distance without any intermediate agent. The electrostatic interaction of charged particles according to Coulomb’s law and the gravitational interaction of bodies according to Newton’s law of universal gravitation, for example, had been interpreted in terms of action at a distance. The field concept was a resurrection of the theory of local interaction that had been proposed by R. Descartes in the first half of the 17th century. In the 1860’s, J. C. Maxwell developed Faraday’s idea of the electromagnetic field and mathematically formulated its laws.
According to the field concept, the particles participating in an interaction, such as an electromagnetic or gravitational interaction, create at each point in the space surrounding them a special state—a force field manifested in the action of a force on other particles placed at a point in this space. The first interpretation of a field to be proposed was a mechanistic one: the field was viewed as the elastic stresses of a hypothetical medium called the ether. An ether with the properties of an elastic medium, however, proved to be in sharp contradiction with the results of subsequent experiments. From the modern point of view, such a mechanistic interpretation of fields is in general meaningless, since the elastic properties of macroscopic bodies are themselves explained entirely by the electromagnetic interactions of the particles making up these bodies. The theory of relativity rejected the concept of the ether as a special elastic medium and ascribed fundamental importance to fields as a primary physical reality. Indeed, according to the theory of relativity, the rate of propagation of any interaction cannot exceed the speed of light in a vacuum. In a system of interacting particles, therefore, a force acting at a given moment on a particle in the system is not determined by the position of other particles at the same moment; that is, a change in the position of one particle affects another particle not immediately but only after some time interval. Thus, the interaction of particles whose relative speed is comparable to the speed of light can be described only in terms of the fields generated by the particles. A change in the state or position of a particle results in a change in the field generated by it. This change is reflected in another particle only after the finite time interval necessary for the change to propagate to the second particle.
Not only do fields realize interactions between particles, but free fields can exist and appear independently of the particles that generated them. This is true, for example, of electromagnetic waves. It therefore is clear that fields should be considered as a special form of matter.
To each type of interaction in nature there correspond certain fields. The description of fields in classical (nonquantum) field theory is accomplished by means of one or more continuous field functions that depend on the position coordinates of the point (x, y, z) at which the field is considered and on the time t. Thus, an electromagnetic field can be completely described by using four functions: the scalar potential ɸ(x, y, z, t) and the vector potential A(x, y, z, t), which together form a single four-dimensional vector in space-time. The strengths of electric and magnetic fields are expressed in terms of the derivatives of these functions. In the general case the number of independent field functions is determined by the number of internal degrees of freedom of the particles corresponding to the given field (see below)—for example, the particles’ spin and isotopic spin. On the basis of general principles—the requirements of relativistic invariance and of some more specific assumptions, such as the superposition principle and gauge invariance for an electromagnetic field—an expression for action can be formed from the field functions, and the principle of least action can be used to obtain differential equations defining the field. The values of the field functions at each individual point can be regarded as the generalized coordinates of the field. Consequently, the field is a physical system with an infinite number of degrees of freedom. In accordance with the general laws of mechanics, it is possible to obtain an expression for the generalized momenta of the field and to determine the densities of its energy, momentum, and angular momentum.
Experiment has shown—initially for the electromagnetic field —that the energy and momentum of a field vary discretely. In other words, fields can be associated with specific particles—for example, an electromagnetic field with photons, and a gravitational field with gravitons. The description of fields by means of field functions is thus only an approximation with a certain range of applicability. In order to take into account the discrete properties of fields—that is, to construct a quantum field theory—the generalized coordinates and momenta of fields must be regarded not as numbers but as operators for which certain commutation relations are satisfied. It may be noted that the transition from classical to quantum mechanics is made in a similar manner.
Quantum mechanics shows that a system of interacting particles can be described by means of some quantum field. Thus, not only are certain particles associated with every field, but, conversely, all known particles are associated with quantized fields. This fact is an example of the wave-particle duality of matter. Quantized fields describe the annihilation (or creation) of particles and the simultaneous production (or annihilation) of antiparticles. The electron-positron field in quantum electrodynamics is an example of such a field.
The type of the commutation relations for field operators depends on the kind of particles that correspond to the given field. W. Pauli showed in 1940 that field operators commute for particles with integral spin, and these particles obey Bose-Ein-stein statistics. For particles with half-integral spin, the operators anticommute, and the corresponding particles obey Fermi-Dirac statistics. If the particles obey Bose-Einstein statistics—for example, photons and gravitons—then many particles can occupy the same quantum state; in the limiting case infinitely many particles can be in the same quantum state. At this limit the mean values of the quantized fields become ordinary classical fields—for example, classical electromagnetic and gravitational fields describable by continuous functions of the position coordinates and time. Corresponding classical fields do not exist for fields associated with particles having half-integral spin.
The present-day theory of elementary particles is constructed as a theory of interacting quantum fields, such as electron-positron, photon, and meson fields.
### REFERENCES
Landau, L. D., and E. M. Lifshits. Teoriia polia, 6th ed. Moscow, 1973. (Teoreticheskaia fizika, vol. 2.)
Bogoliubov, N. N., and D. V. Shirkov. Vvedenie v teoriiu kvantovannykh polei, 2nd ed. Moscow, 1974.
S. S. GERSHTEIN
The Great Soviet Encyclopedia, 3rd Edition (1970-1979). © 2010 The Gale Group, Inc. All rights reserved.
## What does it mean when you dream about a field?
The meaning of a field in a dream depends on the other elements in the dream and the dream’s general atmosphere. Thus, a wild field might represent nature and the freedom of running through a field. A cultivated field might represent new growth or a harvest. A barren field can be a powerful symbol of lack as well emotional barrenness. A completely different set of associations comes to mind with respect to playing fields.
## field
[fēld]
(computer science)
A specified area, such as a group of card columns or a set of bit locations in a computer word, used for a particular category of data.
(electricity)
That part of an electric motor or generator which produces the magnetic flux which reacts with the armature, producing the desired machine action.
(electronics)
One of the equal parts into which a frame is divided in interlaced scanning for television; includes one complete scanning operation from top to bottom of the picture and back again.
(geology)
A region or area with a particular mineral resource, for example, a gold field.
(geophysics)
That area or space in which a particular geophysical effect, such as gravity or magnetism, occurs and can be measured.
(mathematics)
An algebraic system possessing two operations which have all the properties that addition and multiplication of real numbers have.
(medicine)
The area in which surgery is taking place, bounded on all sides by sterilized tissue or drapes. Also known as sterile field.
(optics)
(physics)
An entity which acts as an intermediary in interactions between particles, which is distributed over part or all of space, and whose properties are functions of space coordinates and, except for static fields, of time; examples include gravitational field, sound field, and the strain tensor of an elastic medium.
The quantum-mechanical analog of this entity, in which the function of space and time is replaced by an operator at each point in space-time.
McGraw-Hill Dictionary of Scientific & Technical Terms, 6E, Copyright © 2003 by The McGraw-Hill Companies, Inc.
## field
1. The central portion of a panel that is thicker than its edges, so that it projects above the surrounding frame or wall surfaces.
2. That portion of the upper part of a wall between the cornice and dado or between the frieze and dado.
McGraw-Hill Dictionary of Architecture and Construction. Copyright © 2003 by McGraw-Hill Companies, Inc.
## field
1. Sport a limited or marked off area, usually of mown grass, on which any of various sports, athletic competitions, etc., are held
2. Earth Sciences an area that is rich in minerals or other natural resources
3. Hunting the mounted followers that hunt with a pack of hounds
4. Sport
a. all the runners in a particular race or competitors in a competition
b. the runners in a race or competitors in a competition excluding the favourite
5. Cricket the fielders collectively, esp with regard to their positions
6. the area within which an object may be observed with a telescope, microscope, etc.
7. Physics
b. a region of space that is a vector field
c. a region of space under the influence of some scalar quantity, such as temperature
8. Maths a set of entities subject to two binary operations, addition and multiplication, such that the set is a commutative group under addition and the set, minus the zero, is a commutative group under multiplication and multiplication is distributive over addition
9. Maths logic the set of elements that are either arguments or values of a function; the union of its domain and range
10. Computing
a. a set of one or more characters comprising a unit of information
b. a predetermined section of a record
11. in the field Military in an area in which operations are in progress
## Field
John. 1782--1837, Irish composer and pianist, lived in Russia from 1803: invented the nocturne
Collins Discovery Encyclopedia, 1st edition © HarperCollins Publishers 2005
## field
(data, database)
An area of a database record, or graphical user interface form, into which a particular item of data is entered.
Example usage: "The telephone number field is not really a numerical field", "Why do we need a four-digit field for the year?".
A database column is the set of all instances of a given field from all records in a table.
## field
A physical structure in a form, file or database that holds data. A field is one or more bytes in size. A collection of fields makes up a data record; for example, ORDER #, NAME, ADDRESS, CITY, etc. The terms field and "box" are often used synonymously such as a "search field" or "search box" on a Web page.
The field is the common denominator for database searches. For example, the STATE field is referenced when the query to find "all customers who live in Florida" is made to a database. When totaling transactions, the ORDER_AMOUNT field is summed. JOB_TITLE is referenced when looking for certain employees.
Fields, Data Elements and Data Items
Although often used interchangeably, there are several terms that refer to the same unit of storage in a data record. A "data element" is the logical definition of the field, while a "data item" is the actual data stored in the field. For example, for each CITY data element defined in a record, there are many CITY fields (structures) in the database that hold the data items (New York, Chicago, Phoenix, etc.).
The Basic Unit of Storage Data elements describe the logical unit of data, fields are the actual storage units, and data items are the individual instances of the data elements as in this example.
Copyright © 1981-2019 by The Computer Language Company Inc. All Rights reserved. THIS DEFINITION IS FOR PERSONAL USE ONLY. All other reproduction is strictly prohibited without permission from the publisher.
References in periodicals archive ?
Henry Nobley JJ Feild Martin Bret McKenzie Miss Elizabeth Charming Jennifer Coolidge Col.
One day I was trying to horse ride, the next I'd be running through fields full of cowpats in the middle of Ireland with JJ Feild - wearing a nightgown.
In his second book, Calvin Hollett skilfully and thoughtfully interweaves historical analysis with nautical metaphors and imagery to create a revisionist history of Bishop Edward Feild's episcopate.
Feild, who joined co-stars Benedict Cumberbatch, Tom Burke and Adam Robertson to make the low-key film, is well aware of the gulf between the two movies.
Later this summer JJ Feild will be courting fame as one of Captain America's sidekicks.
He embarks on a road trip with best friends Davy (Tom Burke), Bill (Adam Robertson) and Miles (JJ Feild) to his favourite place on Earth: Barafundle Bay on the west Wales coast.
I would like to thank Stacie Walker, Michele Peterson and David Feild for helping me navigate my new role.
STARRING: Gianna Jun ( pictured), Koyuki, Allison Miller, Larry Lamb, JJ Feild, Yasuaki Kurata, Michael Byrne, Colin Salmon DIRECTOR: Chris Nahon CERTIFICATE: 18 RUNNING TIME: 88mins REVIEWER'S RATING: SHOWING: Cineworld, Showcase and VUE VERDICT: At times Nahon's live action, English-language version comes close to matching Hiroyuki Kitakubo's revered animated film BEAUTIFUL 17-year-old Saya (Jun) is concealing a terrible secret.
When blond singer Heinz Burt (JJ Feild) catches Joe's roving eye, the two become lovers and the producer grooms Heinz for solo stardom with The
Cast: Con O'Neill, Kevin Spacey, Pam Ferris, JJ Feild, James Corden, Ralf Little Plot: Joe Meek was the first independent record producer, whose pioneering techniques brought him a string of hits like the 1962 chart- topped Telstar - the first song by a British band, The Tornados, to reach number one in the States.
Nick Moran and JJ Feild arrive for the UK gala screening of Clubbed at the Empire Leicester Square.
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Open / Close | 6,056 | 28,197 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2021-21 | longest | en | 0.936176 |
http://puzzles.wonderhowto.com/how-to/cut-moon-ten-with-only-three-straight-lines-291251/ | 1,484,571,114,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560279176.20/warc/CC-MAIN-20170116095119-00085-ip-10-171-10-70.ec2.internal.warc.gz | 238,283,611 | 19,558 | # How To: Cut the moon in ten with only three straight lines
## How to Cut the moon in ten with only three straight lines
Cutting the moon and making the shape perfectly round is simple.
Usually we have problems drawing a circle free-hand.
In order to draw a perfect circle, we require circle making instruments.
So, a compass would be a good tool to have. It also fits in your pocket.
Now, the mission is to draw a piece of the moon. Keep in mind that we need a perfect piece of moon meaning, a piece for a perfect circle.
First, draw two straight lines that cross each other. Use a piece of paper to draw a line horizontally and then one vertically with both of them crossing each other.
Now, let us make a perfect slab of moon.
Ok, you will now make four marks on the drawing. The top, left, right and bottom. You should end up with four marks, each evenly spaced from the center of the crossed lines.
Now with the compass over the lines you draw a half-circle from the top mark all the way down to the bottom. Now move the compass to the right where you made a mark on the right end of the horizontal line.
You will again put the pencil edge of the compass on the top part of the drawing and draw down to the bottom portion of the drawing.
Erase the crossed lines and you have yourself a perfect slab of perfect moon.
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• Latest | 297 | 1,339 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2017-04 | latest | en | 0.920614 |
http://git.droids-corp.org/?p=aversive.git;a=blobdiff_plain;f=projects/microb2010/tests/static_beacon/coding.py;h=a41c69233f09c5b9dbdddbfe587e0dd058236d54;hp=440fd3bffdb27e26e5b03a61bbb76e1d391b9c94;hb=7bca487a2198825b6d7caabcd1143e2d5e19dd43;hpb=aa0ebd8b43a0e48d791864ae3268e78d980a70e9 | 1,660,973,656,000,000,000 | text/plain | crawl-data/CC-MAIN-2022-33/segments/1659882573908.30/warc/CC-MAIN-20220820043108-20220820073108-00532.warc.gz | 20,865,210 | 1,846 | X-Git-Url: http://git.droids-corp.org/?p=aversive.git;a=blobdiff_plain;f=projects%2Fmicrob2010%2Ftests%2Fstatic_beacon%2Fcoding.py;h=a41c69233f09c5b9dbdddbfe587e0dd058236d54;hp=440fd3bffdb27e26e5b03a61bbb76e1d391b9c94;hb=7bca487a2198825b6d7caabcd1143e2d5e19dd43;hpb=aa0ebd8b43a0e48d791864ae3268e78d980a70e9 diff --git a/projects/microb2010/tests/static_beacon/coding.py b/projects/microb2010/tests/static_beacon/coding.py index 440fd3b..a41c692 100644 --- a/projects/microb2010/tests/static_beacon/coding.py +++ b/projects/microb2010/tests/static_beacon/coding.py @@ -2,7 +2,13 @@ import sys, math -RPS = 10. +if 1: + RPS = 10. + TIMER_FREQ = 2000000. +else: + RPS = 40. + TIMER_FREQ = 16000000. + LASER_RADIUS = 25. # mm MIN = 200. @@ -11,6 +17,19 @@ NBITS = 9 STEPS = (1 << 9) k = math.pow(MAX/MIN, 1./STEPS) +def mm_to_framedist(mm): + d = mm + d -= MIN + d /= (MAX-MIN) + d *= 512 + return d + +def framedist_to_mm(d): + d /= 512. + d *= (MAX-MIN) + d += MIN + return d + # t is in us, result is 9 bits def time_to_frame(t): # process angle from t @@ -18,20 +37,65 @@ def time_to_frame(t): # process d from a (between 20cm and 350cm) d = LASER_RADIUS / math.sin(a/2) - - frame = math.log(d/MIN)/math.log(k) - if frame >= 512: - frame = 511 - else: - frame = int(frame) - print frame + frame = int(mm_to_framedist(d)) return frame -# frame is integer 9 bits, result is distance -def frame_to_distance(frame): - d = MIN*(math.pow(k, frame)) - print d - return d +# frame is integer 9 bits, result is laserdiff time +def frame_to_time(frame): + d = framedist_to_mm(frame) + a = 2 * math.asin(LASER_RADIUS/d) + t = (a * (TIMER_FREQ/RPS)) / (2. * math.pi) + return t + +def sample_to_offset(samples, table): + offsets = samples[:] + for i in range(len(offsets)): + o = offsets[i] + framedist = mm_to_framedist(o[0]) + off = o[1] - table[int(framedist)] + offsets[i] = framedist, off + return offsets + +def linear_interpolation(offsets, framedist, time): + if framedist <= offsets[0][0]: + return time + offsets[0][1] + if framedist >= offsets[-1][0]: + return time + offsets[-1][1] + + #print (offsets, framedist, time) + o_prev = offsets[0] + for o in offsets[1:]: + if framedist > o[0]: + o_prev = o + continue + x = (framedist - o_prev[0]) / (o[0] - o_prev[0]) + return time + o_prev[1] + (x * (o[1] - o_prev[1])) + return None + +#x = time_to_frame(float(sys.argv[1])) +#frame_to_distance(x) +#frame_to_time(int(sys.argv[1])) + + +table = [0] * 512 +for i in range(512): + table[i] = frame_to_time(i) + +# linear correction: distance_mm, time +samples = [ + (250., 7600.), + (500., 3000.), + (3000., 400.), + ] -x = time_to_frame(float(sys.argv[1])) -frame_to_distance(x) +offsets = sample_to_offset(samples, table) +print "#include " +print "#include " +print "prog_uint16_t framedist_table[] = {" +for i in range(512): + if (i % 8) == 0: + print " ", + print "%d,"%(int(linear_interpolation(offsets, i, table[i]))), + if (i % 8 == 7): + print +print "};" | 1,074 | 2,960 | {"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 | 0.487759 |
https://www.perlmonks.org/?node_id=117294 | 1,606,627,196,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141196324.38/warc/CC-MAIN-20201129034021-20201129064021-00175.warc.gz | 805,822,882 | 6,185 | Your skill will accomplishwhat the force of many cannot PerlMonks
### It Came From the Crypt!
by joecamel (Hermit)
on Oct 07, 2001 at 13:21 UTC Need Help??
```#!/usr/bin/perl
# It Came From the Crypt!
\$q=
q&9RM 8 ba1c
HtJprcG 6SJ3dk
aQM 8 p j p8vsY
i3UnT f kcRJk9w xW2v
h2GVK3 YL1gNvf YEX
dBQdE TIdpMu WD9Y
CAvK0 BrsuKn9kXUwk aNpj8q
ISZgRZkUg2I8W9jJsG8Rg9yeE8LvqZR7A
Ho PDIRVAVCgafA9MZjFb31Ea5bqr7gCg
X zPpgkPsHjTP8iC2TtVYE0PUR b z
Y b3 EQiZrZiIm6Nc81WY qs U g
7m QzHM2VjY62II5Bg 7N e k
kt 7 5JLD86&;\$u=\$q =~ s &
\s &&xg&&-2;@v= unpack
q &c3&,qq&i\nm &;
unshift@v,2 **
2*2**2*2;%l =
map{chr\$_}
reverse@v;\$i
=2*2*2&&\$q,\$q
=\$i&&8; while(
\$i=~m& [^ *]{\$q}
&x){\$y = substr
(crypt( \$&,\$&),
\$u,print );print !
\$l{\$y} &&\$y ne
"y"?\$y: \$l{\$y} ;\$i=~
s+\$&+ +x }\$y=~m
&ymum ~
my&
Just a little something to get into the Halloween spirit.
joecamel
Replies are listed 'Best First'.
Re: It Came From the Crypt!
by theorbtwo (Prior) on Oct 07, 2001 at 14:32 UTC
Amazingly enough, running this through O::Deparse seems to change the result of the run. I'm baffeled... but may just need sleep.
For me, it (running via O::Deparse) works on 5.8.0 but not on 5.6.2.
Re: It Came From the Crypt!
by hopes (Friar) on Oct 07, 2001 at 19:50 UTC
I think it's really very good... Great!
Very cool!
Re: It Came From the Crypt!
by Open (Initiate) on Oct 12, 2001 at 04:40 UTC
cool
Create A New User
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Notices? | 781 | 1,924 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2020-50 | latest | en | 0.725073 |
https://arduino.stackexchange.com/questions/17212/math-with-analog-input-values-coming-out-wrong | 1,701,415,577,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100276.12/warc/CC-MAIN-20231201053039-20231201083039-00870.warc.gz | 130,422,425 | 40,457 | Im getting stuck in a while loop because Im having trouble running a calculation on some sensor values. When I read my four sensors values and compare the ratio of the first two to the second two, I keep getting 0.00, no matter what the values are. Here's my code:
``````while ( ( (frontleft + frontright) / (backleft + backright) <= 0.8 ) || ( (frontleft + frontright) / (backleft + backright) > 1.2) ) {
if (backleft + backright > frontleft + frontright) {
myStepper2.step(10);
}
if (backleft + backright < frontleft + frontright) {
myStepper2.step(-10);
}
Serial.println(backleft);
Serial.print("\n");
Serial.println(backright);
Serial.print("\n");
Serial.println(frontleft);
Serial.print("\n");
Serial.println(frontright);
Serial.print("\n");
float ratio = (frontleft + frontright) / (backleft + backright);
float math = (5 + 6) / (4 + 5);
Serial.print("Math: ");
Serial.println(math);
Serial.print("Ratio: ");
Serial.println(ratio);
Serial.print("\n");
delay(500);
}
``````
Im getting values for my sensors (backleft, backright, frontleft, frontright), but 'ratio' always comes out 0.00. Because of that, Im getting trapped in the loop I believe. What's going on here?
Also, just to check, I added 'math' float as that quick calculation, but I'm getting a value of 1.00 in Serial Monitor. Why? It should be 1.222.
Read this: Integer arithmetic and overflow.
``````float math = (5 + 6) / (4 + 5);
Serial.print("Math: ");
Serial.println(math);
``````
The compiler is treating your expression (the RH side of the "=" sign) as integers. Thus it is working out:
``````11 / 9 = 1
``````
Then you assign that `1` to a float. Too late to make it a float! It is already truncated.
Try:
``````float math = float (5 + 6) / float (4 + 5);
Serial.print("Math: ");
Serial.println(math, 4); // 4 decimal places
``````
You get:
``````Math: 1.2222
``````
This also looks wrong:
``````while ( ( (frontleft + frontright) / (backleft + backright) <= 0.8 ) || ( (frontleft + frontright) / (backleft + backright) > 1.2) ) {
if (backleft + backright > frontleft + frontright) {
myStepper2.step(10);
}
if (backleft + backright < frontleft + frontright) {
myStepper2.step(-10);
}
How many `backleft` variables do you have? The one in the `while` instruction will not be the one you are doing the analogRead on. It doesn't "forward read" variables. I bet you have a different `frontleft` / `backleft` etc. set of variables declared earlier. These will not be updated inside the loop.
(To make them get updated drop the word `int` from inside the loop). | 726 | 2,548 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2023-50 | longest | en | 0.732911 |
https://forum.openframeworks.cc/t/ofvec2f-align-question/391 | 1,686,144,261,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224653764.55/warc/CC-MAIN-20230607111017-20230607141017-00094.warc.gz | 275,998,142 | 5,548 | ofVec2f align() question
I want to find out if two vectors are pointing in the same direction. Here is the result from 4 tests
I thought that two vectors are pointing in the same direction, the x components need to both be positive or negative, as with the y value.
In the first example, y for the first vector is 11 and for the 2nd vector -3, but align() function returns true, yet actually the first vector is pointing south west and the 2nd vector north west.
I checked the numbers in excel and yes they are > 0 as the function suggests, but is the function wrong, or have I got the wrong use for this function?
ofVec2f vector1;
ofVec2f vector2;
vector1.set(-27, 11);
vector2.set(-22, -3);
printf("is aligned = %i\n", vector1.align(vector2));
Many thanks
I *think* the function might be wrong:
return (x*vec.x + y*vec.y) > 0.0;
You want both the x + y multiplied to each be above 0, not as a whole.
So should be something like
return ((x*vec.x) > 0.0 && (y*vec.y) > 0.0);
What do you think?
The align method might not have the most self-explanatory name. What it does is to return true if the angle between the vectors is less than 90 degrees. I think it is doing this correctly. It’s comparing the dot-product with zero.
Maybe it should be called something else to make its purpose clearer. Any ideas?
What is the functionality you were looking for?
maybe useful is something more general ?
I personally wouldn’t find that function very useful - since it’s just testing between one perpendicular and the other. I would be interested in:
a) a being able to test a more specific range of alignment (ie, between 0.9 and 1)
b) being able to test the opposite, ie, are the vectors facing toward each other – this is useful for flocking, for example, where you could have objects test if they can “see” each other.
I guess the more specific you could get with the test, the better – the problem is that it’s a little bit hard to wrap your head around the dot product, and it’s simpler to want to ask the question, are they within 10 degrees of each other, etc, are they within 20 degrees of being 180 degrees from each other, etc. I remember the first time I started to work with flocking I has trouble wrapping my head around altering that variable since it wasn’t in degrees at all.
hope that’s useful input -
at any rate, thanks for all the additions to ofVectorMath !!
take care,
zach
ps one tricky thing about the dot product I think is that is assumes that the vectors are the same size (same length) – in order for the suggestion I threw up above to work you’d want to normalize them before you do the dot product, otherwise doing the dot product bteween two vectors pointing in the same direction but one being small, would be close to 0, even if they share the same angle. The applet you posted online is a good way to check that.
there is good info here, and even an example where they calculate the nearest angle between two vectors, which is probably a more useful function :
http://www.netcomuk.co.uk/~jenolive/vect6.html
-z | 736 | 3,060 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2023-23 | latest | en | 0.94351 |
https://lists.cam.ac.uk/pipermail/cl-isabelle-users/2016-July/msg00008.html | 1,586,293,045,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371805747.72/warc/CC-MAIN-20200407183818-20200407214318-00455.warc.gz | 573,961,607 | 3,034 | # [isabelle] Combinatorial functions
Hi all,
during some recent hobby work involving combinatorial coefficients, I
found that a lot of relevant stuff is already present in Binomial.thy.
But there are also some open questions / improvements here.
* The theory name ÂBinomial is very specific, and from and outside
perspective you would not expect falling factorial etc. residing there.
How could we make this material more prominent? An extension / appendix
to ÂWhat's in MainÂ?
* There is the tendency to generalize combinatorial functions from
natural numbers to suitable algebraic type classes, as e.g. for âfactâ.
In the case of the falling factorial, this leads to something having a
dedicated name in literature, the pochhammer symbol. However, if I as a
theory writer was just looking for a vanilla falling factorial on the
natural numbers (cf. the AFP entry on discrete summation), I would not
have the idea for looking for ÂpochhammerÂ. Maybe an input abbreviation
would make sense here?
* There are two different binomial coefficients formalized separately,
the nat-based âbinomialâ and the generalized âgbinomialâ. In the long
run there should be only one generalized version.
* In Library/Stirling.thy there is a wonderful algorithm for computing
elements of a (Pascal, Stirling) triangle. Maybe it makes sense to be
generalized and used also here since it would avoid the big products
when computing binomial coefficients via factorials.
* In my eyes the existing infix syntax on binomial coefficients does not
make much sense. Its precedence is the same addition, which means that
precedence in âm choose n * qâ is different from âm choose n + qâ. It
does only meagrely improve readability, since the operation is usually
not nested, there is no kind of associativity (as far as I know), it
uses letters rather than symbols, and Âm choose n is also customary
spelt as Ân over mÂ.
So much to say about that.
Cheers,
Florian
--
PGP available:
http://isabelle.in.tum.de/~haftmann/pgp/florian_haftmann_at_informatik_tu_muenchen_de
Attachment: signature.asc
Description: OpenPGP digital signature
This archive was generated by a fusion of Pipermail (Mailman edition) and MHonArc. | 510 | 2,202 | {"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-2020-16 | latest | en | 0.926822 |
http://www.rightsolution.org/present-value-of-annuity-2.html | 1,679,304,803,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296943471.24/warc/CC-MAIN-20230320083513-20230320113513-00662.warc.gz | 93,245,001 | 3,276 | Present Value of an Ordinary Annuity Table
Present Value Factors for an Ordinary Annuity (PVOA Factors) for 1.000 per Period.
Rounded to three decimal places.
Example : When interest is 6% per period and it is compounded each period, receiving 1.000 at the end of each period for 8 periods has a present value of 6.210.
n 1% 2% 3% 4% 5% 6% 8% 10% 12% 1 0.990 0.980 0.971 0.962 0.952 0.943 0.926 0.909 0.893 2 1.970 1.942 1.913 1.886 1.859 1.833 1.783 1.736 1.690 3 2.941 2.884 2.829 2.775 2.723 2.673 2.577 2.487 2.402 4 3.902 3.808 3.717 3.630 3.546 3.465 3.312 3.170 3.037 5 4.853 4.713 4.580 4.452 4.329 4.212 3.993 3.791 3.605 6 5.795 5.601 5.417 5.242 5.076 4.917 4.623 4.355 4.111 7 6.728 6.472 6.230 6.002 5.786 5.582 5.206 4.868 4.564 8 7.652 7.325 7.020 6.733 6.463 6.210 5.747 5.335 4.968 9 8.566 8.162 7.786 7.435 7.108 6.802 6.247 5.759 5.328 10 9.471 8.983 8.530 8.111 7.722 7.360 6.710 6.145 5.650 11 10.368 9.787 9.253 8.760 8.306 7.887 7.139 6.495 5.938 12 11.255 10.575 9.954 9.385 8.863 8.384 7.536 6.814 6.194 13 12.134 11.348 10.635 9.986 9.394 8.853 7.904 7.103 6.424 14 13.004 12.106 11.296 10.563 9.899 9.295 8.244 7.367 6.628 15 13.865 12.849 11.938 11.118 10.380 9.712 8.559 7.606 6.811 16 14.718 13.578 12.561 11.652 10.838 10.106 8.851 7.824 6.974 17 15.562 14.292 13.166 12.166 11.274 10.477 9.122 8.022 7.120 18 16.398 14.992 13.754 12.659 11.690 10.828 9.372 8.201 7.250 19 17.226 15.678 14.324 13.134 12.085 11.158 9.604 8.365 7.366 20 18.046 16.351 14.877 13.590 12.462 11.470 9.818 8.514 7.469 21 18.857 17.011 15.415 14.029 12.821 11.764 10.017 8.649 7.562 22 19.660 17.658 15.937 14.451 13.163 12.042 10.201 8.772 7.645 23 20.456 18.292 16.444 14.857 13.489 12.303 10.371 8.883 7.718 24 21.243 18.914 16.936 15.247 13.799 12.550 10.529 8.985 7.784 25 22.023 19.523 17.413 15.622 14.094 12.783 10.675 9.077 7.843 26 22.795 20.121 17.877 15.983 14.375 13.003 10.810 9.161 7.896 27 23.560 20.707 18.327 16.330 14.643 13.211 10.935 9.237 7.943 28 24.316 21.281 18.764 16.663 14.898 13.406 11.051 9.307 7.984 29 25.066 21.844 19.188 16.984 15.141 13.591 11.158 9.370 8.022 30 25.808 22.396 19.600 17.292 15.372 13.765 11.258 9.427 8.055 | 1,251 | 2,170 | {"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-2023-14 | latest | en | 0.379253 |
qualityjack.com.au | 1,582,239,268,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145316.8/warc/CC-MAIN-20200220224059-20200221014059-00364.warc.gz | 488,443,178 | 22,941 | Ordering a big item? CALL 1300 11 9155 for Custom Quote
# Forklift Attachment Capacity: Fact vs. Fiction
Posted by Aaron Di Rienzo on
# Forklift Attachment Capacity: Fact vs. Fiction
When considering capacity of potential attachments for your forklift, it’s crucial to review the forklift’s rating at its load center, which determines how much weight the machine can safely lift. This will also determine what is possible when attempting to build and implementing longer, high-capacity attachments.
Since forklifts are rated at a load center, the further out you want to lift, the lower the capacity becomes. A 5,000-lb. capacity forklift will lift that much weight up to 48-inch forks (with a 24-inch load center) but going out to 60 inches (with a 30-inch load center), for example, drops the capacity to 4,000 pounds.
The weight capacity of a forklift attachment is tied to what is known as the Maximum Load Moment, which is when the load center distance increases, changing the weight distribution, and as a result, the amount of weight a truck can carry under those conditions. Increasing the load center distance too far can cause a forklift to tip.
Load Moment is the product of the object’s weight multiplied by the object’s distance from the fulcrum, which is a fixed point that acts as the pivot point. On a sit-down counterbalanced forklift, the fulcrum or pivot point is the axle of the front wheels. It is this product, or Load Moment, which determines how much overturning force is being applied to the forklift.
The maximum Load Moment for a truck is derived by multiplying the weight rating of the forklift by the center load distance. If we multiply the 5,000-pound capacity of our theoretical forklift by its 24-inch center load, we arrive at a maximum Load Moment of 120,000 inch-pounds. However, when exceeding that 24 inches by use of a longer attachment, that capacity estimate must be revised. In order to discover the maximum load when the load center distance is greater than the distance stated on the data plate, one must divide the maximum Load Moment by the actual load center distance.
Calculate Weight Capacity
To determine what your forklift’s capacity will become when using an attachment, use the following formula:
Truck Capacity X Load Center (for its Load Moment) / New load center of desired attachment
You can also express this formula with an equation: (TC x LC)/NLC
So, in our 5,000-lb. forklift example, we’d multiply 5,000 by 24 (its load center), for a result of 120,000 inch-pounds as its maximum load moment. Then we divide 120,000 by 30, which is the load center for the desired 60-inch attachment. The end result is 4,000 pounds of maximum weight when using that specific attachment.
Positioning Concerns
The way a load is arranged on a forklift will also factor into load capacity. Most loads do not have their center of gravity exactly in the middle; so to whatever extent that the load differs from its theoretical centered load—like in instances when the load is irregularly shaped, has unbalanced weight distribution, or is not centered on the forks—capacity may be reduced further. If not factored into the calculated capacity, this can cause tipping, collisions, dropped loads and loss of steering control.
One way to address positioning concerns is to reduce the distance from the front wheels to the load center. For example, load a large rectangular box widthwise across the forks of the truck, instead of lengthwise, which causes the load center to shift forward and away from the front wheels, lifting the rear wheels off the ground. It is also helpful to load as close to the front wheels as possible, limiting the load center distance, and load the heaviest part toward the mast. | 835 | 3,757 | {"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-2020-10 | longest | en | 0.925217 |
http://www.studentoffortune.com/question/1884197/An-investor-is-considering-purchasing-an-apartment-complex-for-240000-A-loan-for | 1,369,525,600,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368706472050/warc/CC-MAIN-20130516121432-00075-ip-10-60-113-184.ec2.internal.warc.gz | 738,019,311 | 7,174 | Question
# \$24.00An investor is considering purchasing an apartment complex for \$240,000. A loan for
Chapter 1, # 0
Posted by :
Rating (61):B-
Tutorials Posted: 1398,
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Q:
Question. An investor is considering purchasing an apartment complex for \$240,000. A loan for 20 years at 75 percent of the purchase price is available at 10 percent.
• a. Determine the size of the loan and the amount that will have to be put up in cash. Assume the annual mortgage payment is \$21,142.
• b. The investor intends to hold the property for three years and then sell it. He is going to compute net operating income for the first year and then assume that amount will grow by 5 percent over the next two years.
Question. The Twenty-First Century closed-end fund has \$350 million in securities, \$8 million in liabilities, and 20 million shares outstanding. It trades at a 10 percent discount from net asset value (NAV).
• a. What is the net asset value of the fund?
• b. What is the current price of the fund?
• c. Suggest two reasons why the fund may be trading at a discount from net asset value.
Question. The New Pioneer closed-end fund has \$520 million in securities, \$5 million in liabilities, and 10 million shares outstanding. It trades at a 5 percent premium above its net asset value (NAV).
• a. What is the net asset value of the fund?
• b. What is the current price of the fund?
• c. Why might a fund trade at a premium above its net asset value?
Question. Assume you invest in the German equity market and have a 20 percent return (quoted in euros).
• a. If during this period the euro appreciated by 10 percent against the dollar, what would be your actual return translated into U.S. dollars?
• b. If the euro declined by 15 percent against the dollar, what would your actual return be translated into U.S. dollars?
• c. Recompute the answer based on a 25 percent decline in the euro against the dollar.
Question. Assume you invest in the Japanese equity market and have a 25 percent return (quoted in yen). However, during the course of your investment, the yen declines versus the dollar. By what percentage could the yen decline relative to the dollar before all your gain is eliminated?
Tutorial
\$24.00
An investor is considering purchasing an apartment complex for \$240,000. A loan for
• This tutorial was purchased 1 time and rated No Rating by students like you.
• Posted on Nov. 21, 2012 at 01:44:05AM
A:
Preview: ... tached Attached ...
The full tutorial is about 33 words long plus attachments.
Attachments:
Question Anwers.docx (24K) (Preview) | 617 | 2,600 | {"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-2013-20 | latest | en | 0.937973 |
https://mail.astarmathsandphysics.com/university-physics-notes/fluid-mechanics/4117-proof-that-the-velocity-potential-and-stream-functions-are-harmonic.html?tmpl=component&print=1 | 1,657,191,664,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104690785.95/warc/CC-MAIN-20220707093848-20220707123848-00190.warc.gz | 398,270,340 | 2,890 | ## Proof That the Velocity Potential and Stream Functions Are Harmonic
Theorem
The velocity potential and the stream function are both harmonic in their two dimensional domain.
Proof
The velocity potential
$\phi$
and the stream function
$\psi$
both satisfy
$\frac{\partial \phi}{\partial x} = \frac{\partial \psi}{\partial y} =f, \: \frac{\partial \phi}{\partial y} = \frac{\partial \psi}{\partial x} = -g$
Differentiate the first with respect to
$x$
to give
$\frac{\partial^2 \phi}{\partial x^2} = \frac{\partial^2 \psi}{\partial x \partial y}$
and the second with respect to
$y$
to give
$\frac{\partial^2 \phi}{\partial y^2} =- \frac{\partial^2 \psi}{\partial y \partial x}$
$\frac{\partial^2 \phi}{\partial x^2} + \frac{\partial^2 \psi}{\partial y^2}=0$
Differentiate the first with respect to
$y$
to give
$\frac{\partial^2 \phi}{\partial y \partial x} = \frac{\partial^2 \psi}{ \partial y^2}$
and the second with respect to
$s$
to give
$\frac{\partial^2 \phi}{\partial x \partial y} =- \frac{\partial^2 \psi}{\partial x^2}$
$\frac{\partial^2 \psi}{\partial x^2} + \frac{\partial^2 \psi}{\partial y^2}=0$ | 360 | 1,114 | {"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-2022-27 | latest | en | 0.733349 |
http://bogdank.webd.pl/6kas5l/resting-energy-expenditure-fd18ae | 1,638,927,257,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363420.81/warc/CC-MAIN-20211207232140-20211208022140-00428.warc.gz | 11,773,358 | 7,629 | The first number we provide is resting energy expenditure (REE), which is the amount of calories you would use in one day if you did zero activity. The TDEE for a large, active man, on the other hand, can easily be over 2000 calories. Which Of The Following Is Total Energy Expenditure? Resting metabolic rate is defined as the energy required to maintain the systems of your body at rest. Resting Energy Expenditure (REE) - or Resting Metabolic Rate - is the amount of calories per day needed just for the body to function. Basal metabolic rate (BMR) is more precisely defined as the REE measured just after awakening in the morning. While there are many methods to calculate REE, we recommend using the Mifflin St. Jeor Equation. This modification allowes to sum calories that are burned during food digestion. The resting energy expenditure (REE) represents the amount of energy expended by a person at rest. KEY WORDS Resting energy expenditure, indirect cab-rimetry, 24-h energy requirement, obesity, healthy adults Introduction The accurate prediction of energy requirements fan healthy individuals has many useful clinical applications. Measurements were performed with patients awake at least 30 min prior to testing. Of the components of our total daily energy expenditure, by far the largest one is your resting metabolic rate. The resting metabolic rate calculator estimates the amount of calories used by your organism to keep it alive (in rest time). Your natural resting heart rate: Everyone has a unique resting heart rate, and a normal range is between 60 and 100 beats per minute. Many athletes and coaches estimate their resting energy expenditure so they can better evaluate how many calories they should be consuming daily in order to gain, lose, or maintain weight based on the needs of their sport. Resting Energy Expenditure The REE was measured by indirect calorimetry using a gas analyzer K4b 2 (Cosmed, Rome, Italy). Resting Energy Expenditure (REE), also referred to as Resting Metabolic Rate (RMR) is defined as the amount of calories expended at rest. What Determines Resting Metabolic Rate? Calculations base on a modified Harris-Benedict formula. REE is usually estimated using published formulas. For Males 1. Total daily energy expenditure varies from person to person, depending on body size, sex, body composition, genetics, and activity level. Question: 45. Because the calorie deficit between the calorie expenditure and the calorie income is higher. The most oh-vious use is in weight management ofboth normal-weight and ob#{231}seindividuals. BMR measures your basal energy expenditure, or BEE. Therefore, the ability to estimate REE accurately is of the utmost importance for adequate dietary therapy. Resting metabolic rate is the total number of calories burned when your body is completely at rest. The … Next, we provide you with total energy expenditure (TEE), which is the amount of calories you’d need if you were engaging in your regular daily activities. Resting energy expenditure represents the amount of energy expended by a person a day at rest. Flip through key facts, definitions, synonyms, theories, and meanings in Resting Energy Expenditure when you’re waiting for an appointment or have a short break between classes. In a recent PLoS ONE study, researchers evaluated existing equations to determine the resting energy expenditure. The total energy expenditure for a small, sedentary woman, for example, might be 1800 calories or less per day. The formula that is used to calculate the Resting Energy Expenditure (REE) in the LabChart Metabolic Module is as follows:REE = 16.318VO2 + 4.602VCO2This is an adaptation of the abbreviated Weir Equation which states that:REE = [3.9 (VO2) + 1.1 (VCO2)] 1.44 (KCal/Day)VO2 = volume of oxygen uptake (mL/min)VCO2 = volume of carbon dioxide output (mL/min) So in order to convert Indirect calorimetry is the most frequently used method to measure REE, but the great expense of equipment precludes its widespread use. A very small number of people have physical conditions that give them strange resting metabolic rates. estimate your Resting Metabolic Rate, also known as Resting Energy Expenditure (REE). The patients fasted for 12 hr. However, there is a lot of variation between people, which makes the equations inaccurate by +/- a few hundred calories. In practice, REE and BMR differ by less than 10% so the terms can be used interchangeably. RMR supports breathing, circulating blood, organ functions, and basic neurological functions. Input your height, weight, and age, and our free resting metabolic rate calculator will do the rest. The REEVUE measures the oxygen that the body consumes. This can also be termed the resting energy expenditure. Resting energy expenditure (REE) is the largest component of total daily energy expenditure, accounting for 60–70% of total daily expenditure . Resting energy expenditure was 131 kcal/day lower in subjects with intermittent claudication than in controls, and remained 74 kcal/day lower after adjusting for fat free mass, possibly due to lower oxygen uptake of the lower extremities. Using this measurement it calculates a patient’s Resting Energy Expenditure (REE), commonly referred to as a Resting Metabolic Rate (RMR). Basal or resting energy expenditure is correlated primarily with lean body mass (fat-free mass and essential fat, excluding storage fat), which is the metabolically active tissue in the body. Resting energy expenditure (REE) is the main component of total energy expenditure, taking into consideration the amount of energy the living body requires to maintain its dynamic functions . The Resting Metabolic Rate (RMR) is the amount of energy your body consumes when in resting state. RMR measures your resting energy expenditure. Assessing energy requirements is a fundamental activity in clinical dietetic practice. Results are presented in calories per day (kCal / day). The higher the REE the more calories we burn and the easier and faster we lose weight. Resting energy expenditure was measured within 24 h of admission and weekly thereafter, until discharge. It is also encountered under the names Resting Energy Expenditure or REE, as well as RDEE - Resting Daily Energy Expenditure, which is why this tool can be referred to as both an RMR calculator, REE calculator, or RDEE calculator. • REE is measured by indirect calorimetry. RDEE = 655 + (9.6 x weight) + (1.85 x height) - (4.7 x age) Digestion increases your basal metabolic rate … Physicians can screen for abnormally low metabolic rates and pinpoint the precise caloric intake required for weight loss, maintenance, or weight gain. Resting metabolic rate is the energy your body uses to perform the most basic functions, at rest, for a 24-hour period. In the treatment of obesity, assessment of resting energy expenditure (REE) can provide the basis for prescribing an individualized energy intake to attain a desired level of energy deficit. While BMR is a minimum number of calories required for basic functions at rest, RMR — also called resting energy expenditure (REE) — is the number of … It is proportional to lean body mass and decreases approximately 0.01 kcal/min for each 1% increase in body fatness. Whether you have hours at your disposal, or just a few minutes, Resting Energy Expenditure study sets are an efficient way to maximize your learning time. non-resting energy expenditure Physiology A metabolic value that corresponds to the energy cost of physical activity, which represents approximately 30% of the total energy expenditure. Indirect calorimetry (IC), a noninvasive method based on the volumes of O 2 consumption and CO 2 production, is the gold standard for the measurement of REE The BEE is a 24 hour estimation of the number of calories you burn maintaining your most basic bodily functions, such as breathing, circulating blood and growing and repairing cells. Learn how to calculate RMR using an RMR formula, the effect of age and gender on average REE, and more. About 60% to 80% of the oxygen you consume and the calories you burn goes to just keeping you alive at rest. The aim of this study was to investigate which resting energy expenditure (REE) predictive equations are the best alternatives to indirect calorimetry before and after an interdisciplinary therapy in Brazilian obese women. Here are some of the factors which determine the Resting Energy Expenditure. Your Resting Metabolic Rate (RMR) is one of the main contributing components of energy expenditure (around 70%). RDEE = 66 + (13.7 x weight) + (5 x height) - (6.8 x age) For Females 1. Total metabolic output is the total energy expenditure you use throughout the day which includes your resting energy expenditure and then energy expended doing all of the other activities you do. 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Day at rest and pinpoint the precise caloric intake required for weight loss, maintenance or... | 3,984 | 18,329 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2021-49 | latest | en | 0.953085 |
https://ujaashome.com/electrical-energy/frequent-question-how-many-power-stations-are-connected-to-the-national-grid.html | 1,653,191,727,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662543797.61/warc/CC-MAIN-20220522032543-20220522062543-00571.warc.gz | 655,887,143 | 18,781 | Frequent question: How many power stations are connected to the national grid?
Contents
According to the U.S. Energy Information Administration, the U.S. power grid is made up of over 7,300 power plants, nearly 160,000 miles of high-voltage power lines, and millions of miles of low-voltage power lines and distribution transformers, connecting 145 million customers throughout the country (EIA, 2016).
How are power stations connected to the National Grid?
Power stations produce electricity at 25,000 volts (V). Step-up transformers change the voltage to the very high values needed to transmit electricity through the National Grid power lines. Electricity is sent through these at 400,000 V, 275,000 V or 132,000 V. … Electrical power can be calculated using this equation.
Does the National Grid include power stations?
In the electricity sector in the United Kingdom the National Grid is the high-voltage electric power transmission network serving Great Britain, connecting power stations and major substations and ensuring that electricity generated anywhere on it can be used to satisfy demand elsewhere.
Why is the National Grid better than local power stations?
The transfer of electrical energy via the grid is very efficient. … As high currents waste more energy than low currents, electrical power is transported around the grid at a high voltage and a low current.
How much power is lost in the National Grid?
Citizens Advice suggests that about 1.7% of the electricity transferred over the transmission network is lost, and a further 5-8% is lost over the distribution networks2. This is because transporting electricity via a lower current and high voltage causes lower network losses.
Why is the electricity in the National Grid at a very high-voltage?
When currents in a cable are higher, more energy is dissipated to the surroundings through heating. As high currents waste more energy than low currents, electrical power is transported around the grid at a high voltage and a low current.
Why does the National Grid have three phase power?
The electric grid uses a three-phase power distribution system because it allows for higher transmission at lower amperage. This makes it possible to use higher gauge (thinner) copper wire, significantly reducing both material and labor costs.
Why does the National Grid use a very high-voltage?
When a current flows through a wire some energy is lost as heat. The higher the current, the more heat is lost. To reduce these losses, the National Grid transmits electricity at a low current. This needs a high voltage.
How does the national grid make money?
Distribution companies
These companies own the distribution network that connects households to the Power Grid. Distribution companies charge suppliers for using the network. Suppliers then pass this cost on to consumers through the standing charge on your energy bills.
How does the National Grid reduce energy loss?
In the National Grid, a step-up transformer is used to increase the voltage and reduce the current. … Less current means less energy is lost through heating the wire. To keep people safe from these high voltage wires, pylons are used to support transmission lines above the ground.
IT IS INTERESTING: Why do you need an electrical installation certificate?
Who owns the electric grid?
The US grid is a complex network of more than 7,300 power plants and transformers connected by more than 450,000 miles of high-voltage transmission lines and serves 145 million customers. In most countries, they are state owned but in the US, the grid is nearly all privately owned. | 710 | 3,627 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2022-21 | latest | en | 0.927651 |
https://discuss.pytorch.org/t/how-to-use-f-softmax/103897 | 1,653,187,004,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662543264.49/warc/CC-MAIN-20220522001016-20220522031016-00794.warc.gz | 262,982,522 | 4,705 | # How to use F.softmax
Hi,
I have a tensor and I want to calculate softmax along the rows of the tensor.
action_values = t.tensor([[-0.4001, -0.2948, 0.1288]])
as I understand cutting the tensor row-wise we need to specify dim as 1.
However I got an unexpected result.
Below is what I tried, but none gave me successful results.
F.softmax(action_values - max(action_values), dim = 0)
Out[15]: tensor([[1., 1., 1.]])
F.softmax(action_values - max(action_values), dim = 1)
Out[16]: tensor([[0.3333, 0.3333, 0.3333]])
F.softmax(action_values - max(action_values), dim = -1)
Out[17]: tensor([[0.3333, 0.3333, 0.3333]])
Hi Granth!
The short answer is that you are calling python’s `max()` function,
rather than pytorch’s `torch.max()` tensor function. This is causing
you to calculate `softmax()` for a tensor that is all zeros.
You have two issues:
First is the use of pytorch’s `max()`. `max()` doesn’t understand
tensors, and for reasons that have to do with the details of `max()`'s
implementation, this simply returns `action_values` again (with the
singleton dimension removed).
The second is that there is no need to subtract a scalar from your
tensor before calling `softmax()`. Any such scalar drops out anyway
in the `softmax()` calculation.
This script illustrates what is going on:
``````import torch
torch.__version__
action_values = torch.tensor([[-0.4001, -0.2948, 0.1288]])
action_values
max (action_values) # this is python's max, not pytorch's
torch.max (action_values) # pytorch's tensor-version of max
action_values - max (action_values)
action_values - torch.max (action_values)
tzeros = torch.zeros ((1, 3))
tzeros
torch.nn.functional.softmax (tzeros, dim = 0)
torch.nn.functional.softmax (tzeros, dim = 1)
torch.nn.functional.softmax (action_values, dim = 1) # what you want
torch.nn.functional.softmax (action_values - 2.3, dim = 1) # shift drops out
``````
Here is its output:
``````>>> import torch
>>> torch.__version__
'1.6.0'
>>> action_values = torch.tensor([[-0.4001, -0.2948, 0.1288]])
>>> action_values
tensor([[-0.4001, -0.2948, 0.1288]])
>>> max (action_values) # this is python's max, not pytorch's
tensor([-0.4001, -0.2948, 0.1288])
>>> torch.max (action_values) # pytorch's tensor-version of max
tensor(0.1288)
>>> action_values - max (action_values)
tensor([[0., 0., 0.]])
>>> action_values - torch.max (action_values)
tensor([[-0.5289, -0.4236, 0.0000]])
>>> tzeros = torch.zeros ((1, 3))
>>> tzeros
tensor([[0., 0., 0.]])
>>> torch.nn.functional.softmax (tzeros, dim = 0) tensor([[1., 1., 1.]])
>>> torch.nn.functional.softmax (tzeros, dim = 1) tensor([[0.3333, 0.3333, 0.3333]])
>>> torch.nn.functional.softmax (action_values, dim = 1) # what you want
tensor([[0.2626, 0.2918, 0.4456]])
>>> torch.nn.functional.softmax (action_values - 2.3, dim = 1) # shift drops out
tensor([[0.2626, 0.2918, 0.4456]])
``````
Best.
K. Frank | 888 | 2,983 | {"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-2022-21 | latest | en | 0.752639 |
https://math.answers.com/Q/Highest_number_known_to_man_that_is_not_infinity | 1,675,575,175,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500215.91/warc/CC-MAIN-20230205032040-20230205062040-00062.warc.gz | 395,815,328 | 51,320 | 0
# Highest number known to man that is not infinity?
Wiki User
2009-05-19 02:26:24
big daddy is the largest value. if the biggest number you can think of = x then big daddy = x + 1
Jack Heaton
Lvl 2
2022-12-09 20:05:54
Study guides
20 cards
## A number a power of a variable or a product of the two is a monomial while a polynomial is the of monomials
➡️
See all cards
3.8
2032 Reviews
Wiki User
2009-05-19 02:26:24
There isn't one... Think of a number and name it after yourself ;) | 159 | 495 | {"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-2023-06 | latest | en | 0.876207 |
https://queslers.com/weekly-contest-287-leetcode-solution/ | 1,695,535,943,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506623.27/warc/CC-MAIN-20230924055210-20230924085210-00063.warc.gz | 520,232,701 | 28,089 | 304 North Cardinal St.
Dorchester Center, MA 02124
# Weekly Contest 287 LeetCode Solution
## Problem 1 – Minimum Number of Operations to Convert Time Leetcode Solution
You are given two strings `current` and `correct` representing two 24-hour times.
24-hour times are formatted as `"HH:MM"`, where `HH` is between `00` and `23`, and `MM` is between `00` and `59`. The earliest 24-hour time is `00:00`, and the latest is `23:59`.
In one operation you can increase the time `current` by `1``5``15`, or `60` minutes. You can perform this operation any number of times.
Return the minimum number of operations needed to convert `current` to `correct`.
Example 1:
``````Input: current = "02:30", correct = "04:35"
Output: 3
Explanation:
We can convert current to correct in 3 operations as follows:
- Add 60 minutes to current. current becomes "03:30".
- Add 60 minutes to current. current becomes "04:30".
- Add 5 minutes to current. current becomes "04:35".
It can be proven that it is not possible to convert current to correct in fewer than 3 operations.
``````
Example 2:
``````Input: current = "11:00", correct = "11:01"
Output: 1
Explanation: We only have to add one minute to current, so the minimum number of operations needed is 1.``````
Constraints:
• `current` and `correct` are in the format `"HH:MM"`
• `current <= correct`
### Minimum Number of Operations to Convert Time Leetcode Solution in C++
``````int convertTime(string current, string correct) {
auto toMin = [](string &s) {
return s[0] * 600 + s[1] * 60 + s[3] * 10 + s[4] ;
};
int d = toMin(correct) - toMin(current);
return d / 60 + d % 60 / 15 + d % 15 / 5 + d % 5;
}
``````
### Minimum Number of Operations to Convert Time Leetcode Solution in Python
``````class Solution:
def convertTime(self, current: str, correct: str) -> int:
current_time = 60 * int(current[0:2]) + int(current[3:5]) # Current time in minutes
target_time = 60 * int(correct[0:2]) + int(correct[3:5]) # Target time in minutes
diff = target_time - current_time # Difference b/w current and target times in minutes
count = 0 # Required number of operations
# Use GREEDY APPROACH to calculate number of operations
for i in [60, 15, 5, 1]:
count += diff // i # add number of operations needed with i to count
diff %= i # Diff becomes modulo of diff with i
return count
``````
### Minimum Number of Operations to Convert Time Leetcode Solution in Java
`````` public int convertTime(String current, String correct){
Function<String, Integer> parse = t -> Integer.parseInt(t.substring(0, 2)) * 60 + Integer.parseInt(t.substring(3));
int diff = parse.apply(correct) - parse.apply(current), ops[] = {60, 15, 5, 1}, r = 0;
for(int i = 0; i < ops.length && diff > 0; diff = diff % ops[i++])
r += diff / ops[i];
return r;
}
``````
## Problem 2 – Find Players With Zero or One Losses Leetcode Solution
You are given an integer array `matches` where `matches[i] = [winneri, loseri]` indicates that the player `winneri` defeated player `loseri` in a match.
Return a list `answer` of size `2` where:
• `answer[0]` is a list of all players that have not lost any matches.
• `answer[1]` is a list of all players that have lost exactly one match.
The values in the two lists should be returned in increasing order.
Note:
• You should only consider the players that have played at least one match.
• The testcases will be generated such that no two matches will have the same outcome.
Example 1:
``````Input: matches = [[1,3],[2,3],[3,6],[5,6],[5,7],[4,5],[4,8],[4,9],[10,4],[10,9]]
Output: [[1,2,10],[4,5,7,8]]
Explanation:
Players 1, 2, and 10 have not lost any matches.
Players 4, 5, 7, and 8 each have lost one match.
Players 3, 6, and 9 each have lost two matches.
``````
Example 2:
``````Input: matches = [[2,3],[1,3],[5,4],[6,4]]
Output: [[1,2,5,6],[]]
Explanation:
Players 1, 2, 5, and 6 have not lost any matches.
Players 3 and 4 each have lost two matches.
``````
Constraints:
• `1 <= matches.length <= 105`
• `matches[i].length == 2`
• `1 <= winneri, loseri <= 105`
• `winneri != loseri`
• All `matches[i]` are unique.
### Find Players With Zero or One Losses Leetcode Solution in Java
`````` public List<List<Integer>> findWinners(int[][] matches){
Map<Integer, Integer> losses = new TreeMap<>();
for(int[] m : matches){
losses.put(m[0], losses.getOrDefault(m[0], 0));
losses.put(m[1], losses.getOrDefault(m[1], 0) + 1);
}
List<List<Integer>> r = Arrays.asList(new ArrayList<>(), new ArrayList<>());
for(Integer player : losses.keySet())
if(losses.get(player) <= 1)
return r;
}
``````
### Find Players With Zero or One Losses Leetcode Solution in Python
``````class Solution:
def findWinners(self, matches: List[List[int]]) -> List[List[int]]:
winners, losers, table = [], [], {}
for winner, loser in matches:
# map[key] = map.get(key, 0) + change . This format ensures that KEY NOT FOUND error is always prevented.
# map.get(key, 0) returns map[key] if key exists and 0 if it does not.
table[winner] = table.get(winner, 0) # Winner
table[loser] = table.get(loser, 0) + 1
for k, v in table.items(): # Player k with losses v
if v == 0:
winners.append(k) # If player k has no loss ie v == 0
if v == 1:
losers.append(k) # If player k has one loss ie v == 1
return [sorted(winners), sorted(losers)] # Problem asked to return sorted arrays.
``````
### Find Players With Zero or One Losses Leetcode Solution in C++
``````vector<vector<int>> findWinners(vector<vector<int>>& matches) {
set<int> all, l, l2;
vector<int> a0, a1;
for (auto &m : matches) {
all.insert({m[0], m[1]});
if (!l.insert(m[1]).second)
l2.insert(m[1]);
}
set_difference(begin(all), end(all), begin(l), end(l), back_inserter(a0));
set_difference(begin(l), end(l), begin(l2), end(l2), back_inserter(a1));
return {a0, a1};
}
``````
## Problem 3 – Maximum Candies Allocated to K Children Leetcode Solution
You are given a 0-indexed integer array `candies`. Each element in the array denotes a pile of candies of size `candies[i]`. You can divide each pile into any number of sub piles, but you cannot merge two piles together.
You are also given an integer `k`. You should allocate piles of candies to `k` children such that each child gets the same number of candies. Each child can take at most one pile of candies and some piles of candies may go unused.
Return the maximum number of candies each child can get.
Example 1:
``````Input: candies = [5,8,6], k = 3
Output: 5
Explanation: We can divide candies[1] into 2 piles of size 5 and 3, and candies[2] into 2 piles of size 5 and 1. We now have five piles of candies of sizes 5, 5, 3, 5, and 1. We can allocate the 3 piles of size 5 to 3 children. It can be proven that each child cannot receive more than 5 candies.``````
Example 2:
``````Input: candies = [2,5], k = 11
Output: 0
Explanation: There are 11 children but only 7 candies in total, so it is impossible to ensure each child receives at least one candy. Thus, each child gets no candy and the answer is 0.
``````
Constraints:
• `1 <= candies.length <= 105`
• `1 <= candies[i] <= 107`
• `1 <= k <= 1012`
### Maximum Candies Allocated to K Children Leetcode Solution in Java
`````` public int maximumCandies(int[] A, long k) {
int left = 0, right = 10_000_000;
while (left < right) {
long sum = 0;
int mid = (left + right + 1) / 2;
for (int a : A) {
sum += a / mid;
}
if (k > sum)
right = mid - 1;
else
left = mid;
}
return left;
}
``````
### Maximum Candies Allocated to K Children Leetcode Solution in C++
`````` int maximumCandies(vector<int>& A, long long k) {
int left = 0, right = 1e7;
while (left < right) {
long sum = 0, mid = (left + right + 1) / 2;
for (int& a : A) {
sum += a / mid;
}
if (k > sum)
right = mid - 1;
else
left = mid;
}
return left;
}
``````
### Maximum Candies Allocated to K Children Leetcode Solution in Python
`````` def maximumCandies(self, A, k):
left, right = 0, sum(A) / k
while left < right:
mid = (left + right + 1) / 2
if k > sum(a / mid for a in A):
right = mid - 1
else:
left = mid
return left
``````
## Problem 4 – Encrypt and Decrypt Strings Leetcode Solution
You are given a character array `keys` containing unique characters and a string array `values` containing strings of length 2. You are also given another string array `dictionary` that contains all permitted original strings after decryption. You should implement a data structure that can encrypt or decrypt a 0-indexed string.
A string is encrypted with the following process:
1. For each character `c` in the string, we find the index `i` satisfying `keys[i] == c` in `keys`.
2. Replace `c` with `values[i]` in the string.
Note that in case a character of the string is not present in `keys`, the encryption process cannot be carried out, and an empty string `""` is returned.
A string is decrypted with the following process:
1. For each substring `s` of length 2 occurring at an even index in the string, we find an `i` such that `values[i] == s`. If there are multiple valid `i`, we choose any one of them. This means a string could have multiple possible strings it can decrypt to.
2. Replace `s` with `keys[i]` in the string.
Implement the `Encrypter` class:
• `Encrypter(char[] keys, String[] values, String[] dictionary)` Initializes the `Encrypter` class with `keys, values`, and `dictionary`.
• `String encrypt(String word1)` Encrypts `word1` with the encryption process described above and returns the encrypted string.
• `int decrypt(String word2)` Returns the number of possible strings `word2` could decrypt to that also appear in `dictionary`.
Example 1:
``````Input
["Encrypter", "encrypt", "decrypt"]
Output
[null, "eizfeiam", 2]
Explanation
encrypter.encrypt("abcd"); // return "eizfeiam".
// 'a' maps to "ei", 'b' maps to "zf", 'c' maps to "ei", and 'd' maps to "am".
encrypter.decrypt("eizfeiam"); // return 2.
// "ei" can map to 'a' or 'c', "zf" maps to 'b', and "am" maps to 'd'.
// Thus, the possible strings after decryption are "abad", "cbad", "abcd", and "cbcd".
// 2 of those strings, "abad" and "abcd", appear in dictionary, so the answer is 2.
``````
Constraints:
• `1 <= keys.length == values.length <= 26`
• `values[i].length == 2`
• `1 <= dictionary.length <= 100`
• `1 <= dictionary[i].length <= 100`
• All `keys[i]` and `dictionary[i]` are unique.
• `1 <= word1.length <= 2000`
• `1 <= word2.length <= 200`
• All `word1[i]` appear in `keys`.
• `word2.length` is even.
• `keys``values[i]``dictionary[i]``word1`, and `word2` only contain lowercase English letters.
• At most `200` calls will be made to `encrypt` and `decrypt` in total.
### Encrypt and Decrypt Strings Leetcode Solution in Java
`````` Map<Character, String> enc;
Map<String, Integer> count;
public Encrypter(char[] keys, String[] values, String[] dictionary) {
enc = new HashMap<>();
for (int i = 0; i < keys.length; ++i)
enc.put(keys[i], values[i]);
count = new HashMap<>();
for (String w : dictionary) {
String e = encrypt(w);
count.put(e, count.getOrDefault(e, 0) + 1);
}
}
public String encrypt(String word1) {
StringBuilder res = new StringBuilder();
for (int i = 0; i < word1.length(); ++i)
res.append(enc.getOrDefault(word1.charAt(i), "#"));
return res.toString();
}
public int decrypt(String word2) {
return count.getOrDefault(word2, 0);
}
``````
### Encrypt and Decrypt Strings Leetcode Solution in C++
`````` unordered_map<char, string> enc;
unordered_map<string, int> count;
Encrypter(vector<char>& keys, vector<string>& values, vector<string>& dictionary) {
for (int i = 0; i < keys.size(); ++i)
enc[keys[i]] = values[i];
for (string& w: dictionary)
count[encrypt(w)]++;
}
string encrypt(string word1) {
string res = "";
for (char c: word1) {
if (!enc.count(c)) return "";
res += enc[c];
}
return res;
}
int decrypt(string word2) {
return count[word2];
}
``````
### Encrypt and Decrypt Strings Leetcode Solution in Python
``````class Encrypter(object):
def __init__(self, keys, values, dictionary):
self.enc = {k: v for k,v in zip(keys, values)}
self.decrypt = collections.Counter(self.encrypt(w) for w in dictionary).__getitem__
def encrypt(self, word1):
return ''.join(self.enc.get(c, '#') for c in word1)
``````
##### Weekly Contest 287 LeetCode Solution Review:
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CodeChef Solutions | 3,813 | 12,987 | {"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-2023-40 | latest | en | 0.692405 |
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A 6-in.-long bronze tube, with closed ends, is 3 in. in diameter with a wall thickness of 0.10 in. With no internal pressure, the tube just fits between two rigid end walls. Calculate the longitudinal and tangential stresses for an internal pressure of 6000 psi. Assume ν = 1/3 and E = 12 × 106 psi.
Solution 228
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### how to determine the…
how to determine the orientation of the tube?
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### The only hint about the…
In reply to by Kent Eclipse
The only hint about the orientation of the tube is this phrase:
...the tube just fits between two rigid end walls
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If there are 3 apples and you take away 2, you will have 1 left. To find out how many apples you have left you would need to subtract the 2 apples you take away from 3. Three minus two is equal to one.
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30) The quantity “Rm” which relates dependent voltage to controlling current is called | 961 | 4,161 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2023-23 | latest | en | 0.859481 |
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Comparing Angles and Sides in Triangles is a part of the concept of Inequalities in Triangles. I think you are all familiar with SAS and SSS Theorems. Here we will learn deeply about SAS and SSS Inequality theorems with step by step explanation. In addition to that, the students can also find examples for comparing angles and sides in triangles.
## Comparison of Sides and Angles in a Triangle
We can determine which angle is largest and which angle is smallest. If one side of the triangle is longer than another side then the angle opposite to the greater side will be greater than the angle opposite to the shorter side. In this section, we will discuss SAS inequality theorem and SSS inequality theorem.
SAS Inequality Theorem:
If two sides of a triangle are congruent to two sides of the other triangle, but the included angle of one triangle has a longer measure than the included angle of another triangle, then the third side of the first triangle is greater than the third side of the second triangle.
SSS Inequality Theorem:
If two sides of the triangle are congruent to two sides of another triangle, but the third side of the first triangle is greater than the third side of the second triangle, then the included angle of the first triangle’s two congruent sides is greater in measure than the included angle of the second triangle’s two congruent sides.
### Comparing Triangle Angles and Sides Examples
Example 1.
In ∆XYZ, ∠XYZ = 40° and ∠YXZ = 71°
Solution:
∠XZY = 180° – (∠XYZ + ∠YXZ)
180° – (40° + 71°)
180° – 111°
= 69°
Therefore
∠XYZ > ∠XZY > ∠YXZ
The greater angle has the greater side opposite to it.
Example 2.
In a ∆PQR, QR = 8cm, PQ = 6cm and PR = 4cm arrange x°, Y° and z° in the ascending order.
Solution:
Given that,
QR > PQ > PR
Therefore ∠QPR > ∠PRQ > ∠PQR as the greater side has the greater angle opposite to it
Therefore
(180° – ∠QPR) < (180° – ∠PRQ) < (180° – ∠PQR)
y° < x° < z°
Example 3.
In ∆XYZ, ∠XYZ = 30° and ∠YXZ = 40°
Solution:
∠XZY = 180° – (∠XYZ + ∠YXZ)
180° – (30° + 40°)
180° – 70°
= 110°
Therefore
∠XYZ > ∠YXZ > ∠XZY
The greater angle has the greater side opposite to it.
Example 4.
In a ∆PQR, QR = 2cm, PQ = 3cm and PR = 4cm arrange x°, Y° and z° in the ascending order.
Solution:
Given that,
QR > PQ > PR
Therefore ∠QPR > ∠PRQ > ∠PQR as the greater side has the greater angle opposite to it
Therefore
(180° – ∠QPR) < (180° – ∠PRQ) < (180° – ∠PQR)
y° < x° < z°
Example 5.
In ∆ABC, ∠ABC = 20° and ∠BAC = 21°
Solution:
∠ACB = 180° – (∠ABC + ∠BAC)
180° – (20° + 21°)
180° – 41°
= 139°
Therefore
∠ACB > ∠BAC > ∠ABC
The greater angle has the greater side opposite to it.
### FAQs on Comparison of Sides and Angles in a Triangle
1. What is the relationship between angles and sides of a triangle?
In any triangle, the largest side and largest angle are opposite one another. the smallest side and smallest angle are opposite one another.
2. How do you compare triangle measurements?
If two triangles have two pairs of sides in the same ratio and the included angles are also equal, then the triangles are similar.
3. How are sides and angles similar?
In a pair of similar triangles, corresponding sides are proportional and all three angles are congruent. | 966 | 3,316 | {"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} | 4.9375 | 5 | CC-MAIN-2024-33 | latest | en | 0.927222 |
http://slideplayer.com/slide/3341086/ | 1,529,936,274,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267867885.75/warc/CC-MAIN-20180625131117-20180625151117-00343.warc.gz | 287,623,006 | 22,723 | # KS1 Maths What we learn and our methods of teaching
## Presentation on theme: "KS1 Maths What we learn and our methods of teaching"— Presentation transcript:
KS1 Maths What we learn and our methods of teaching
What do we teach in KS1 Maths?
Number bonds from 10 and 20 ( ie 7+3=10, 18+2= 20) Basic multiplication ( 2,5,10) Basic division ( 2) Fractions ( ½ , ¼, 1/3 ) Addition and subtraction to 100 Place value ( units, tens and hundreds) Time ( o’clock, half past, quarter to, quarter past) Measurement ( weight, length, capacity) Money ( everyday money- calculating change) Problem solving Handling data ( graphs, tables, sorting data) Shape and space Today we will focus on multiplication and division
Maths – when do we use it?
Division 4
Division vocabulary share equally divide divided by groups halve half
Sharing objects Share these six biscuits between three teddies. How many biscuits does each teddy get?
Linking with halving and quartering shapes
Splitting into equal groups/parts ½ ½ ¼ ¼ ¼ ¼
Using tens and units to solve division problems
48 12 ÷ 4 = 3 48 ÷ 2 = 24 15 ÷ 3 = 5
Division as grouping Can you group these cookies into groups of 3? How much does each teddy get?
Division as grouping – Repeated Addition
15 ÷ 3 = 5
Arrays as a tool for solving answers 18 ÷ 6 = 3 18 ÷ 3 = 6 Link with multiplication
Remainders 13 ÷ 4 = 3r1 13 ÷ 4 = 3r1
2 4 6 8 10 12 14 16 Count in 2s until you get to 16.
So how can you work out a division calculation using times tables? Count in 2s until you get to 16.
Division using an empty number line
How many groups of 5 are there in 25? 25 ÷ 5 = 5
Resources to help children with division
½ ½
Your turn! 6 groups of 5 make 30. So 6 teams.
There are 30 children in the class. They need to be in teams of 5. How many teams will there be? Use an empty number line to work it out. 6 groups of 5 make So 6 teams.
Multiplication 17
Multiplication Counting in 10s
Multiplication Counting in 2s
Multiplication Counting in 5s
Multiplication Children need to learn multiplication tables and multiplication facts. Yes – learn to recite tables but also need to know table facts at random. e.g. 8 lots of ten are? And answer questions like “ How many tens in 80?” Not secure in times tables until able to do this quickly. Should be working on 10x, 2x and 5x tables in Year 1, continue and work on 3x and 4x in Year 2. Secure in Year 3.
Multiplication Repeated addition 10 + 10 + 10 + 10 = 40
Problem solving There are 10 pencils in a pack. How many pencils are there in 4 packs? Repeated addition = 40
Multiplication 3 times 5 is the same as = 15 or 3 lots of 5 or 3 x 5 On a number line 5 5 5 5 10 15
Across the corridor and down the stairs!
Multiplication As an array 3 x 5 = 15 5 x 3 = 15
Problem solving There are 15 apples in a tray. Ling has 4 trays of apples. How many apples does Ling have altogether? Show how you work it out. Ella’s dad washes some cars. He uses 12 buckets of water. Each bucket has 5 litres of water. How many litres of water does he use altogether? Your turn! Can you draw pictures or make jottings to show how you could work one or both of these out?
Resources to help children with multiplication
Practical maths Making maths practical by using real materials. Try some of these at home with your child. Using coins using food Using measuring cups cooking
Useful websites
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Similar presentations | 959 | 3,462 | {"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} | 4.875 | 5 | CC-MAIN-2018-26 | latest | en | 0.883162 |
www.david-mckenzie.com | 1,519,522,242,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891816083.98/warc/CC-MAIN-20180225011315-20180225031315-00145.warc.gz | 424,379,498 | 11,744 | ## The Histories of Common Forms of Tiling Puzzles
The Many Forms of Tiling Puzzles
Tiling puzzles are a sort of puzzle that involve rearranging shapes, either solid or cut out of a specific geometric shape, like a square, and reforming them together to form a completely different silhouette, for instance a triangle, without overlaps or gaps. Common forms of tiling puzzles include jigsaw puzzles, dissection puzzles, and Tangrams. The combined histories of all of these types of tiling puzzles, and the extra less-known forms, is incredibly vast and differs between each type. Altogether, tiling puzzles have existed for centuries, the first probably being the dissection puzzles in the works of ancient mathematicians like Archimedes when he made a mathematical treatise called Ostomachion in the form of a geometric dissection.
Jigsaw Puzzles
One of the best known forms of tiling puzzle is the jigsaw puzzle. Jigsaw puzzles require the player to assemble a picture by placing one of many interlocking shapes together in the correct pattern. The shapes are often strangely cut and involve tessellation, the act of placing down shapes so that no shapes overlap or create gaps as they fill a plane. Originally, these puzzles were created by painting a picture or scene onto a rectangular wooden block, and then cutting the block into pieces with a jigsaw, hence the name (Encyclopedia Britannica).
Around 1766, a man named John Spilsbury was credited with the invention of the first jigsaw puzzle (“Jigsaw Puzzle, 1766”). Spilsbury, a teacher and mapmaker in England, and an apprentice to the Royal Geographer to King George III, Thomas Jefferys, invented the first jigsaw puzzle as a means to teach geography. The puzzle was handmade, carved out of wood and painted, and displayed a map of England and Wales with each piece representing a county that once all fitted together formed the full country (Ament).
Over a century later, Milton Bradley published their first jigsaw puzzle for children in the 1870s, called “The Smashed Up Locomotive” (“The Smashed Up Locomotive”). The pieces of the puzzle were different parts of a train, and the smashed up effect took place when the child first opened the box and noticed a pile of jigsaw pieces with train parts painted on each one (Ament).
Jigsaw puzzles typically come in a variety of sets, their difficulties being directly related to the amount of pieces that comprise each puzzle. Different sets include 300-piece, 500-piece, 750-piece, and 1000-piece, and the most common types of scenes that a jigsaw puzzle creates include castles, mountains, forests, or fantasy landscapes. Children’s jigsaw puzzles tend to have larger pieces for an easier method of play, and are often used as learning exercises. There are also 3-D jigsaw puzzles that add an extra layer of complexity such as the puzzle globe, a spherical jigsaw puzzle usually representing the Earth or the moon (Encyclopedia Britannica).
Today, jigsaw puzzles remain as one of the simplest, yet most time-consuming, tiling puzzles that after all these years have not completely changed design. Many innovative offshoots and styles of jigsaw puzzles have popped up, but this rawest form of tiling puzzle has remained.
Dissection Puzzles
A dissection puzzle is one of the earliest forms of tiling puzzle where the objective is to form a unique silhouette out of specific shapes given. Geometric dissections have been used as mathematical proofs for concepts such as the Pythagorean Theorem, the Quadrilateral of Omar Khayyam, the Fibonacci Sequence or “Golden Ratio,” Euclid’s Windmill proof of the Pythagorean Theorem, and Liu Hui’s proof of the Pythagorean Theorem (Wikipedia). Around the 10th century, Arabic mathematicians displayed geometric dissections while commentating on Euclid’s Elements, and Dai Zhen, a Chinese scholar from the Qing Dynasty in the 18th century approximated the value of pi using dissections (Tiwald). Common dissection puzzles today include the ancient Chinese game of Tangrams and dozens of American variants of Tangrams.
The first dissection puzzles appeared in the works of ancient mathematicians like Archimedes when he made a mathematical treatise called Ostomachion in the form of a geometric dissection. Ostomachion is also referred to as the stomachion, its original name which traces its meaning from the Greek word for “stomach,” “Loculus of Archimedes” (which means Archimedes’ box), and in Latin as “syntemachion” (Chung and Graham).
Ostomachion is a fourteen piece dissection puzzle which forms a square and the vertices of each piece rest on a 12×12 grid (“Stomachion,” Weisstein) as shown in fig. 1. The design is similar to Tangrams where the player may compose silhouettes of animals, ships, weapons, or people.
Bill Cutler, in 2003, solved the stomachion by revealing all 536 arrangements of the pieces as a square, where solutions rotated or reflected are not counted (“Stomachion,” Weisstein).
Fig. 1. Stomachion puzzle, Wolfram Research, Inc.
In the late 19th century, dissection puzzles saw a rise in popularity as newspapers began running them regularly. Sam Loyd from the United States and Henry Dudeney from the United Kingdom were two of the most published dissection puzzle authors at this time (Wikipedia).
In 1907, Henry Dudeney presented the haberdasher’s problem (fig. 2), a dissection in which four pieces could form both a triangle and a square (Wikipedia). It is one of the most classic examples known of a dissection puzzle.
Fig. 2. The haberdasher’s problem, Wikipedia.org.
Dudeney also mentioned a variant of his puzzle in which the pieces can only swing on a hinged joint. This interesting spin on the haberdasher’s problem was introduced alongside the puzzle in Dudeney’s 1907 book The Canterbury Puzzles. It demonstrates how the triangle and square silhouettes can be achieved with the four original shapes even if limited to rotating each piece on an imaginary hinge on each connecting corner as shown in fig. 3 (Frederickson).
Fig. 3. Greg Frederickson, Hinged dissection puzzle, Cambridge [UP].
Dissection puzzles are both a form of entertaining tiling puzzle and a solid mathematical tool, with the creation of dissection puzzles acting as puzzles themselves to mathematicians and others alike. They are very useful educational resources when teaching geometry and when demonstrating higher concepts of mathematical principles.
Tangrams
The most recognizable dissection puzzle today is a game so old that its origin is undocumented and persists only in legend. The ancient Chinese game of Tangrams (or just the Tangram) has survived for untold centuries. Tangrams, literally translated to “the seven clever pieces”, involves seven geometric pieces called “tans” that are placed together to form silhouettes (Rob). The seven geometric pieces are: two large right triangles, one medium right triangle, two small right triangles, one square, and one parallelogram. The parallelogram is a special case in that it is the only piece that must be flipped in order to achieve its reflection, as it has no reflection symmetry but only rotational symmetry (“Tangram,” Weisstein).
Fig. 4. Image created by the author, Tangrams set.
Tangrams’ popularity has made it one of the most well-known dissection puzzles in the world. A fictitious history of Tangrams popularized the game when Sam Loyd published his book The Eighth Book Of Tan. Loyd’s book claimed the game was invented over 4000 years ago by the god Tan and the book included over 700 silhouettes that the Tangrams set could complete (Rob). In reality, the game was invented in China at an old yet unknown time (disputes argue over whether it was more near the late 1700s and early 1800s or thousands of years ago), and was first brought to America in 1815 by Captain M. Donaldson on his ship, the Trader (The Tangram Book, Slocum). Tangrams would soon reach England where a craze for the puzzle spiked and spread across Europe. The Tangrams fever spread to Denmark, where at the Copenhagen University, a student wrote the book Mandarinen, which documented the history and popularity of the game. China exported a great number of Tangrams sets during this time. In Germany, around 1891, the industrialist Friedrich Richter began producing the Anchor Stone Puzzles, silhouette-matching puzzles based on Tangrams. And in World War I, the game grew even more popular on both sides of the battlefield, occasionally going under the name “The Sphinx” (Rob).
Over 5900 different Tangrams silhouettes have been compiled from texts from the 19th century, with more appearing every day (The Tao of Tangram, Slocum). However, there is a limit. In 1942, Chinese mathematicians Fu Traing Wang and Chuan-Chih Hsiung proved that there exist only thirteen convex figures which can be constructed with Tangrams, as displayed in fig. 5 (Rob).
Fig. 5. The thirteen convex shapes in Tangrams, Wikipedia.org.
Lesser-known Tiling Puzzles
While jigsaw puzzles, dissection puzzles, and Tangrams are at the forefront of the tiling puzzle category, there are dozens of less-known puzzles out there. The Conway puzzle, named after its inventory John Conway, demands the solution of packing rectangular boxes into a 5×5 hollow cube (Weisstein). Domino tiling is another form of tiling puzzle in which 2×2 dominoes are arranged in order to fill a square or rectangle in a tessellated pattern. Sliding puzzles are yet another type of tiling puzzle in which flat pieces are moved across a board on established routes in order to achieve a perfect configuration.
Certainly, a very interesting yet widely unknown tiling puzzle, the Eternity puzzle, created by Christopher Monckton in 1999, was a 209-piece puzzle in which a large assortment of randomly cut shapes formed a circle. It was considered unsolvable, and its creator held a one million cash prize for anyone that could solve it within four years of its release. In 2000, two mathematicians from Cambridge won the reward with their solution (Richer).
Conclusion
The genre of tiling puzzle covers a wide expanse of puzzle types that appear seemingly unique in their own right. The history of tiling puzzles in general is a combined effort of multiple histories and multiple puzzles that all co-exist under a greater definition. Jigsaw puzzles have survived for years and still have that same simple enjoyment from when they were first invented. Dissection puzzles have long been used for mathematics and as teaching aides, and to this day still intrigue many mathematicians and people interested in geometry. Tangrams, one of the oldest dissection puzzles, still survives to our time and has largely remained untouched. And of course, all the dozens of variations and alternate tiling puzzles that are being invented every day add to the flavor of the exclusively unique puzzle category of tiling games.
Sources
Ament, Phil. “Jigsaw Puzzle History.” The Great Idea Finder. Ideafinder.com. Feb. 2005. Web. 07 Mar. 2011.
Chung, Fan, and Ron Graham. “A Tour of Archimedes’ Stomachion.” Department of Mathematics. UCSD. math.ucsd.edu. n.d. Web. 07 Mar. 2011.
“Dissection Puzzle.” Wikipedia: The Free Encyclopedia. Wikipedia.org. 20 Jan. 2011. Web. 10 Feb. 2011.
Frederickson, Greg. Hinged Dissections: Swinging & Twisting. Cambridge UP. 15 Sept. 2009. Web. 10 Feb. 2011.
“Jigsaw puzzle.” Britannica. Britannica.com. 2011. Web. 10 Feb. 2011.
“Jigsaw Puzzle, 1766.” The British Library. bl.uk. n.d. Web. 07 Mar. 2011.
Richer, Duncan. “The Eternity Puzzle.” nrich.maths.org. July 1999. Web. 10 Feb. 2011.
Rob. “Tangrams and Anchor Stone Puzzles.” Rob’s Puzzle Page. Comcast.net. n.d. Web. 07 Mar. 2011.
Slocum, Jerry. The Tangram Book. New York, NY: Sterling Pub., 2003. Print.
—. The Tao of Tangram. New York: Barnes & Noble, 2007. Print.
“The Smashed Up Locomotive.” Icollectpuzzles.com. n.d. Web. 07 Mar. 2011.
Tiwald, Justin. “Dai Zhen.” Internet Encyclopedia of Philosophy. Iep.utm.edu. 22 Sept. 2009. Web. 07 Mar. 2011.
Weisstein, Eric W. “Conway Puzzle.” Wolfram MathWorld. n.d. Web. 10 Feb. 2011.
—. “Stomachion.” Wolfram MathWorld. n.d. Web. 07 Mar. 2011.
—. “Tangram.” Wolfram MathWorld. n.d. Web. 07 Mar. 2011. | 2,841 | 12,201 | {"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} | 2.984375 | 3 | CC-MAIN-2018-09 | latest | en | 0.949362 |
https://studylib.net/doc/25181692/game-theory-and-application---problem-and-answers- | 1,545,041,967,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376828501.85/warc/CC-MAIN-20181217091227-20181217113227-00312.warc.gz | 738,949,169 | 33,229 | # GAME THEORY AND APPLICATION - Problem and Answers
```TOPIC 4. GAME THEORY AND APPLICATION
1. IDENTIFYING NASH EQUILIBRIA
Consider the following game, where two firms (Firm A and Firm B) must decide
simultaneously whether to expand business in the West or in the South:
Firm B
Firm B
Expand in the West Expand in the South
Firm A
Expand in the West
Firm A
Expand in the South
10, 60
50, 90
20, 80
40, 50
A) there is one pure strategy Nash equilibrium: for both firms to expand in the West.
B) there is one pure strategy Nash equilibrium: for both firms to expand in the South.
C) there are two pure strategy Nash equilibria: either firm can expand in the West, and the
other expands in the South.
D) there is only a mixed strategies equilibrium.
Diff: 3
2. PRISONER’S DILEMMA TYPE GAMES
Consider the game below:
Player R
Strategy R1
Player R
Strategy R2
Player C
Strategy C1
Player C
Strategy C2
600, 600
100, 1000
1000, 100
200, 200
True or False: This game is not an example of the Prisoners’ Dilemma game.
Answer: False. It is an example of the Prisoners’ Dilemma game. A Prisoners’ Dilemma
game is one where each player has a dominant strategy and the Nash equilibrium does not
coincide with the outcome that maximises the collective payoffs of the players in the game.
In this game the dominant strategies are R2 for Player R and C2 for Player C. The Nash
equilibrium is (R2, C2) but both players would be better off by cooperating and selecting
(R1, C1).
3. MAXIMIN STRATEGIES
Consider the game below about funding and construction of a dam to protect a 1,000-person
town. Contributions to the Dam Fund, once made, cannot be recovered, and all citizens must
contribute £1,000 to the dam in order for it to be built. The dam, if built, is worth £70,000 to
each citizen. If each player choses a maximin strategy, the outcome would be:
One citizen
Contribute to dam
One citizen
Don’t contribute to
dam
Other 999 citizens
Contribute to dam
69000, 69000
-1000, 0
Other 999 citizens
Don’t contribute to
dam
0, -1000
0, 0
A) £69,000, £69,000.
B) £0, -£1000.
C) -£1000, £0.
D) £0, £0.
E) a mixed strategy equilibrium.
Diff: 2
By not contributing to the dam, the citizens minimise their potential losses – or in other
words, they maximise the minimum gain that can be earned.
4. REPEATED GAMES/SEQUENTIAL GAMES
Consider the game below where a firm (Entrant) is considering entering into the digital
camera business, and can decide to do so on either a small scale or a large scale. The
Incumbent firm must decide whether the Accommodate the new firm or start a Price War.
The payoffs (profits in millions of dollars) to each firm are represented below:
Entrant
Small scale
Entrant
Large scale
Incumbent
Accommodate
Incumbent
Price War
4, 20
1, 16
8, 10
2, 12
If the game is played sequentially and the Entrant moves first, the equilibirum would be:
a) (Small scale, Accommodate)
b) (Small scale, Price War)
c) (Large sclae, Accommodate)
d) (Large scale, Price War)
5. MIXED STRATEGIES
Consider the Matching Pennies game:
Player A - heads
Player A - tails
Player B - heads
1, -1
-1, 1
Player B - tails
-1, 1
1, -1
Suppose Player B always uses a mixed strategy with probability of 3/4 for head and 1/4 for
tails. Which of the following strategies for Player A provides the highest expected payoff?
A) Mixed strategy with probability 1/4 on heads and 3/4 on tails
B) Mixed strategy with probability 1/2 on heads and 1/2 on tails
C) Mixed strategy with probability 3/4 on heads and 1/4 on tails
D) Pure strategy in which Player A always selects heads | 1,018 | 3,545 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2018-51 | latest | en | 0.883238 |
https://www.physicsforums.com/threads/numerically-solving-the-null-vector.313480/ | 1,685,329,942,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224644574.15/warc/CC-MAIN-20230529010218-20230529040218-00334.warc.gz | 1,028,393,934 | 14,663 | # Numerically solving the null vector
• wdlang
#### wdlang
i have a 1000 by 1000 matrix A
A is definitely singular
I want to solve the null vector numerically
How to do this?
matlab does not work!
i have a 1000 by 1000 matrix A
A is definitely singular
I want to solve the null vector numerically
How to do this?
matlab does not work!
Can you clarify what you mean by "solve the null vector"? Do you mean that you want to find a vector $$x \neq 0$$ such that $$Ax = 0$$? i.e., to find a nonzero element of the null space?
Can you clarify what you mean by "solve the null vector"? Do you mean that you want to find a vector $$x \neq 0$$ such that $$Ax = 0$$? i.e., to find a nonzero element of the null space?
yes!
yes!
If you want to use Matlab to do it, look up the "null" command:
Z = null(A);
returns an orthonormal basis for the null space of A. | 241 | 868 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2023-23 | latest | en | 0.913145 |
https://fr.slideshare.net/cellperformance/gdc15-code-clinic | 1,685,377,792,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224644867.89/warc/CC-MAIN-20230529141542-20230529171542-00580.warc.gz | 302,819,425 | 92,764 | Publicité
# #GDC15 Code Clinic
Engine Director à Insomniac Games
8 Mar 2015
Publicité
### #GDC15 Code Clinic
1. I don’t know anything about wood carving, but…
2. Different problems, different scales, different tools.
3. Tools scale • Chainsaw • Dremel • Laser
4. Tools scale • Chainsaw • Dremel <- Compiler • Laser
5. To make best use of a tool (e.g. compiler) • Solve the part of the problem the tool can’t help with. • Prepare the area that the tool can help with. • Solve details missed by using the tool as intended.
6. Part 1: Solve the part of the problem the tool can’t help with.
7. Approaches…
8. Approach #1 • Estimate resources available • Triage based on cost and value estimates • Collect data • Adapt as cost and/or value predictions change • AKA Engineering
9. Approach #2 • Don’t worry about resources • Desperately scramble to fix when running out. • Generously described as irrational • Desperate scramble != “optimization”
10. The problems are there even if you ignore them.
11. Estimate resources available • What are the bottlenecks? i.e. most insufficient for the demand
12. Shell + Excel: test_0 vs. test_1 (W=2 -> W=512)
13. What can we infer?
14. 0 5 10 15 20 25 30 35 1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49 51 53 55 57 59 61 63 65 67 69 71 73 75 77 79 81 83 85 87 89 91 93 95 97 99 W=8 test_0 test_1 33ms = 30fps Col W=8; Ea. R/W avg +/- 32B from next/prev 8192 * 1024 * 32bit = 3MB / frame
15. 0 5 10 15 20 25 30 35 1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49 51 53 55 57 59 61 63 65 67 69 71 73 75 77 79 81 83 85 87 89 91 93 95 97 99 W=8 test_0 test_1 33ms = 30fps Col W=8; Ea. R/W avg +/- 32B from next/prev 8192 * 1024 * 32bit = 3MB / frame Row, in order 8192 * 1024 * 32bit * 8.9x = 34.7MB / frame
16. 0 10 20 30 40 50 60 70 80 90 100 1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49 51 53 55 57 59 61 63 65 67 69 71 73 75 77 79 81 83 85 87 89 91 93 95 97 99 W=512 test_0 test_1 W=512; > 20x diff (1.19MB/frame) Algorithmic complexity equivalent
17. 0 10 20 30 40 50 60 70 80 90 100 1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49 51 53 55 57 59 61 63 65 67 69 71 73 75 77 79 81 83 85 87 89 91 93 95 97 99 W=512 test_0 test_1 Gap = doing nothing.
18. Why?
19. http://www.gameenginebook.com/SINFO.pdf
20. By comparison…
21. http://www.agner.org/optimize/instruction_tables.pdf (AMD Piledriver)
22. http://www.agner.org/optimize/instruction_tables.pdf (AMD Piledriver)
23. The Battle of North Bridge L1 L2 RAM
24. Verify data…
25. Excel: W=64 (int32_t; mod 8)
26. 0 10 20 30 40 50 60 70 80 90 100 1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49 51 53 55 57 59 61 63 65 67 69 71 73 75 77 79 81 83 85 87 89 91 93 95 97 99 W=512 test_0 test_1 Make reasonable use of resources
27. 0 10 20 30 40 50 60 70 80 90 100 1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49 51 53 55 57 59 61 63 65 67 69 71 73 75 77 79 81 83 85 87 89 91 93 95 97 99 W=512 test_0 test_1 Make reasonable use of resources (not the same as optimization)
28. Memory access is the most significant component.
29. Bottleneck: most insufficient for the demand • Q: Is everything always all about memory access / cache? • A: No. It’s about whatever the most scarce resources are. • E.g. Disk access seeks; GPU draw counts; etc.
30. Let’s look at it another way…
31. Simple, obvious things to look for + Back of the envelope calculations = Substantial wins
32. http://deplinenoise.wordpress.com/2013/12/28/optimizable-code/
33. 2 x 32bit read; same cache line = ~200
34. Waste 56 bytes / 64 bytes
35. Float mul, add = ~10
36. Let’s assume callq is replaced. Sqrt = ~30
37. Mul back to same addr; in L1; = ~3
39. Waste 60 bytes / 64 bytes
40. 90% waste!
41. Alternatively, Only 10% capacity used* * Not the same as “used well”, but we’ll start here.
42. Time spent waiting for L2 vs. actual work ~10:1
43. Time spent waiting for L2 vs. actual work ~10:1 This is the compiler’s space.
44. Time spent waiting for L2 vs. actual work ~10:1 This is the compiler’s space.
45. Compiler cannot solve the most significant problems.
46. 12 bytes x count(32) = 384 = 64 x 6 4 bytes x count(32) = 128 = 64 x 2
47. 12 bytes x count(32) = 384 = 64 x 6 4 bytes x count(32) = 128 = 64 x 2 (6/32) = ~5.33 loop/cache line
48. 12 bytes x count(32) = 384 = 64 x 6 4 bytes x count(32) = 128 = 64 x 2 Sqrt + math = ~40 x 5.33 = 213.33 cycles/cache line (6/32) = ~5.33 loop/cache line
49. 12 bytes x count(32) = 384 = 64 x 6 4 bytes x count(32) = 128 = 64 x 2 Sqrt + math = ~40 x 5.33 = 213.33 cycles/cache line (6/32) = ~5.33 loop/cache line + streaming prefetch bonus
50. 12 bytes x count(32) = 384 = 64 x 6 4 bytes x count(32) = 128 = 64 x 2 Sqrt + math = ~40 x 5.33 = 213.33 cycles/cache line (6/32) = ~5.33 loop/cache line + streaming prefetch bonus Using cache line to capacity* = Est. 10x speedup * Used. Still not necessarily as efficiently as possible
51. 0 0.5 1 1.5 2 2.5 1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49 51 53 55 57 59 61 63 65 67 69 71 73 75 77 79 81 83 85 87 89 91 93 95 97 99 32K GameObjects test_0 test_1 Measured = 6.8x
52. Part 2: Prepare the area that the tool can help with.
53. “The high-level design of the code generator” • http://llvm.org/docs/CodeGenerator.html • Instruction Selection • Map data/code to instructions that exist • Scheduling and Formation • Give opportunity to schedule / estimate against data access latency • SSA-based Machine Code Optimizations • Manual SSA form • Register Allocation • Copy to/from locals in register size/types • Prolog/Epilog Code Insertion • Late Machine Code Optimizations • Pay careful attention to inlining • Code Emission • Verify asm output regularly
54. “Compilers are good at applying mediocre optimizations hundreds of times.” - @deplinenoise
55. Three easy tips to help compiler • #1 Analyze one value • #2 Hoist all loop-invariant reads and branches • #3 Remove redundant transform redundancy
56. #1 Analyze one value • Find any one value (or implicit value) you’re interested in, anywhere. • Printf it out over time. (Helps if you tag it so you can find it.) • Grep your tty and paste into excel. (Sometimes a quick and dirty script is useful to convert to a more complex piece of data.) • See what you see. You’re bound to be surprised by something.
57. e.g. SoundSourceBaseComponent::BatchUpdate if ( g_DebugTraceSoundSource ) { Printf("#SS-01 %d m_CountSourcesn”,i, component->m_CountSources); }
58. Approximately 0% of the sound source components have a source. (i.e. not playing yet)
59. Approximately 0% of the sound source components have a source. (i.e. not playing yet) Switch to evaluating from source->component not component->source 20K vs. 200 cache lines
60. #2 Hoist all loop-invariant reads and branches • Don’t re-read member values or re-call functions when you already have the data. • Hoist all loop-invariant reads and branches. Even super-obvious ones that should already be in registers (member fields especially.)
61. How is it used? What does it generate?
62. How is it used? What does it generate? Equivalent to: return m_NeedParentUpdate?count:0;
63. MSVC
64. MSVC Re-read and re-test… Increment and loop…
65. Re-read and re-test… Increment and loop… Why? Super-conservative aliasing rules…? Member value might change?
66. What about something more aggressive…?
67. Test once and return… What about something more aggressive…?
69. Okay, so what about… Equivalent to: return m_NeedParentUpdate?count:0;
70. …well at least it inlined it?
71. MSVC doesn’t fare any better…
72. Don’t re-read member values or re-call functions when you already have the data. Ghost reads and writes
73. BAM!
74. :(
75. Ghost reads and writes Don’t re-read member values or re-call functions when you already have the data. Hoist all loop-invariant reads and branches. Even super- obvious ones that should already be in registers.
76. :)
77. :) A bit of unnecessary branching, but more-or-less equivalent.
78. #3 Remove redundant transform redundancy • Often, especially with small functions, there is not enough context to know when and under what conditions a transform will happen. • It’s easy to fall into the over-generalization trap • It’s only real cases, with real data, we’re concerned with.
79. A little gem…
80. wchar_t* -> char* Suspicious. I know we don’t handle wide char conversion consistently.
81. wchar_t* -> char* -> wchar_t* it takes the char* we created and converts it right back to a whar*
82. What’s happening? • We convert from char* to wchar* to store it in m_DeleteFiles • …So we can convert it from wchar* to char* to give to FileDelete • …So we can convert it from char* to whcar* to give it to Windows DeleteFile.
83. Where does the char* come from in the first place?
84. Comes from argv. Never needed to touch the memory!
85. Which turned out to be a command line parameter that was never used, anywhere.
86. Which brings us back to the wood carving analogy…
87. Part 3: Solve details missed by using the tool as intended.
88. Optimization begins. See: e.g.
89. Part 4: Practice
90. If you don’t practice, you can’t do it when it matters.
91. Practice https://oeis.org/A000081
92. What’s the cause? http://www.insomniacgames.com/three-big-lies-typical-design-failures-in-game-programming-gdc10/
Publicité | 3,137 | 9,369 | {"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} | 2.984375 | 3 | CC-MAIN-2023-23 | latest | en | 0.569644 |
https://www.gamedev.net/forums/topic/278352-best-way-to-find-a-set-of-numbers/ | 1,542,664,898,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039746112.65/warc/CC-MAIN-20181119212731-20181119234731-00306.warc.gz | 879,372,916 | 28,926 | # Best way to find a set of numbers
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Im trying to find a decent algorithim to find the coefficents of a scaled array of 8 numbers ranging from 0 - 255 to an array of 4 numbers via this formula : (a sub n - 1 + a sub n) / 2 such that the 4 element array is an average of the 8 element array and the coefficents are the difference between the element in the 4 size array and its matching 2 elements in the 8 element. for example:::::: 127 30 89 17 255 54 109 233 as the 8 element array 78 53 154 171 as the 4 element array and the coefficents being 48, 36, 101, and 62. i need a method to generate a key that can be sent to a function which outputs the coefficents. currently all i can think of is to enumerate all of the possiblities which results in 256^4th differnt combinations to search through and linear search time. is there some better way to find the coefficents than searching 256 ^th combinations?
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Why would you want to do this? You certainly don't save on any information by splitting your original 8 element array into two 4 element arrays! If you're looking to split your data into approximation and detail... this doesn't seem like a good way to go about it.
Anyway... let's assume there is a valid reason to do this...
Without both the averaged array and the coefficients, there is no way to obtain the original array. It's fairly trivial to see that you need either the original array to generate the coefficients, or, the average array and the 1st element of each averaged pair in order to generate the coefficients.
Timkin
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Is this homework?
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one: no it isnt homework and
two: a single 4 element array would be stored, and a key as well that would be passed to a function that generates the coefficents.
the 4 element array is an scaled array, element 0 is the average of elements 0 and 1 in the 8 element array, element 1 is the average of 2 and 3 etc.
the issue is how to generate the coefficents.
the only method i can think of is a linear search through all the possibilites of coefficents using the key as the number of irreations into the linear search, such as assuming coefficent 1 is 0 then moving to all the possilibites of the other 3 coefficents until all possilibites with coefficent 1 being = to are searched then increment coefficent 1 to 1 and try all other possibilites with that.
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Assuming you are using this as a means of encryption you have a few problems. The biggest one being that integers are not closed under division. Overall you are just mapping vectors to vectors. So basically (x,y) maps to ((x+y)/2,(y-x)/2). I don't believe there is any general rounding rule that will cause (1,1), (1,2), (2,1) and (2,2) to map to four distinct pairs. The problem is that division and the rounding it requires. You could map (x,y) to (x+y,x-y) instead and not have that problem. Your inverse function is then (x',y') maps to ((x'+y')/2,(x'-y')/2). There you don't have rounding because x' and y' are either always both even or both odd. Specifically if x and y are both odd or both even then x+y is even and x-y is even. If one is odd and one is even then x+y is odd and x-y is odd. Another problem in your example is that you are actually using (x,y) maps to ((x+y)/2,|x-y|/2). That makes (x,y) and (y,x) map to the same pair because of the absolute value taken, i.e. x+y=y+x and |x-y|=|y-x|.
I don't understand at all why you are having trouble finding the coefficents. The only thing I can think is that because you don't have a bijection and thus no inverse you are having trouble reconstructing the original stream when you test. The problem isn't how you calculate your coefficients, but rather the definition of the coefficients itself.
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× | 1,078 | 4,149 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2018-47 | latest | en | 0.936596 |
http://math.stackexchange.com/questions/34713/fundamental-group-of-a-finite-set-with-discrete-topology?answertab=oldest | 1,448,658,754,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398450581.71/warc/CC-MAIN-20151124205410-00159-ip-10-71-132-137.ec2.internal.warc.gz | 152,198,835 | 18,083 | # Fundamental Group of a finite set with discrete topology
let S be a finite set with say n elements. Give it the discrete topology, Now what can we say about its fundamental group? Atleast can we determine the fundamental group of a set with two elements? Thanks
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Such a space is not path-connected, so you have to keep track of the basepoint. Each path-connected component is trivially contractible (it consists only of one point). The fundamental group is trivial for each component. You don't need the finiteness of $S$.
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Yes, I missed out the fact that image of a connected domain is connected for a continuous map. Hence every loop in S should be a constant map. From which it turns out that the fundamental group is trivial. Thanks! – Dinesh Apr 23 '11 at 17:52
You need to choose a base point both in $S^1$ and $S$. Since $S^1$ is connected, it must be mapped to this base point entirely. Hence there is only one base-point preserving loop in $S$, and thus the fundamental group is trivial. This applies to any set with the discrete topology, no matter if it is finite or not.
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@Theo: what's $S^1$? The unit circle? – Rudy the Reindeer Apr 25 '11 at 8:57
@Matt: Yes. ${}$ – t.b. Apr 25 '11 at 8:58
@Theo: I assume you are using $S^1$ instead of $[0,1]$ because the question is about closed paths. Why do you have to choose a base point in the domain of the paths? Shouldn't you say "...since $S^1$ is connected, it must be mapped to this base point entirely....", where that base point is the one in $S$? – Rudy the Reindeer Apr 25 '11 at 9:09
@Matt: I'm just using $S^1 \cong [0,1]/\{0 \sim 1\}$. The chosen base point $\ast$ on $S^1$ corresponds to the image of the identified end points of the interval. In other words, there is a bijection between loops $\gamma: [0,1] \to S$ (i.e. maps s.t. $\gamma(0) = \gamma(1) = s_0$) and base-point preserving maps $\gamma: (S^1, \ast) \to (S,s_0)$. – t.b. Apr 25 '11 at 9:14
@Theo: thank you, now I understand. I didn't understand why it was necessary to pick base points when one can show that every point in $S$ can only have the constant loop. It's necessary because that is how the fundamental group is defined. – Rudy the Reindeer Apr 25 '11 at 10:43 | 623 | 2,216 | {"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} | 3.75 | 4 | CC-MAIN-2015-48 | longest | en | 0.911885 |
https://theindianbioplantech.com/qa/question-what-is-the-loudest-sound-ever-recorded.html | 1,606,407,005,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141188800.15/warc/CC-MAIN-20201126142720-20201126172720-00524.warc.gz | 527,514,098 | 9,195 | # Question: What Is The Loudest Sound Ever Recorded?
## Has a killer whale ever eaten a human?
Killer whales (or orcas) are powerful predators capable of killing leopard seals and great white sharks.
They have also been recorded preying on usually terrestrial species such as moose swimming between islands.
In the wild, there have been no fatal attacks on humans..
## What animal is loudest?
Sperm Whale Although blue whales are considered by most to be the loudest animals, there are many ways to measure loudness. At pure decibels, the sperm whale is louder than the blue whale because its clicks have been recorded at 230 decibels.
## What happens if you make a sound louder than 1100 dB?
With energy as great as 1100 dB, it would create enough gravity to cause a black hole to form, and an incredibly large one at that. Decibels are a logarithmic unit. That means 20 decibels isn’t 2 times more powerful than 10 decibels, it’s 10 times more powerful. 30 decibels is 10 times more powerful than 20 decibels.
## What is the loudest fart?
According to the Guinness Book of World Records, the loudest fart ever recorded was a fart of 113 decibels, by Herkimer Chort of Ripley, NY USA, on October 11th, 1972.
## How loud can a human yell?
Human screams can be quite loud, possibly exceeding 100 dB (as of March 2019, the world record is 129 dB!) —but you probably want to avoid that because screams that loud can hurt your ears! You should also have found sound levels drop off quickly as you get farther from the source.
## How loud is an atomic bomb?
A nuclear bomb. Decibel meters set 250 feet away from test sites peaked at 210 decibels. The sound alone is enough to kill a human being, so if the bomb doesn’t kill you, the noise will.
## Has anyone been killed by a sperm whale?
In 1712, so the story goes, one Captain Hussey’s vessel was blown offshore south of Nantucket Island while hunting right whales for their oil. Hussey happened upon a pod of sperm whales, killed one and dragged it home.
## What are the top 10 loudest sounds?
Know your decibels! Ten of the loudest sounds out there230 dB – Sperm whale.180 dB – Rocket launch.120 dB – Fireworks.110 dB – Live gig.100 dB – Night club.97 dB – Fire alarm.94 dB – Lawnmower.88 dB – Heavy traffic.More items…•
## What is the quietest place on Earth?
Orfield LaboratoriesAccording to the Guinness Book of Records, the anechoic chamber at Orfield Laboratories in Minneapolis is the quietest place in the world, with a background noise reading of –9.4 decibels.
## What is the loudest sound in the universe?
The sound made by the Krakatoa volcanic eruption in 1883 was so loud it ruptured eardrums of people 40 miles away, travelled around the world four times, and was clearly heard 5,000 kilometers away. This is hailed as the loudest noise ever – and reported in.
## Can sound kill you?
The general consensus is that a loud enough sound could cause an air embolism in your lungs, which then travels to your heart and kills you. Alternatively, your lungs might simply burst from the increased air pressure. … High-intensity ultrasonic sound (generally anything above 20KHz) can cause physical damage.
## Can sperm whale sound kill?
Sperm Whales Are So Loud They Could Potentially “Vibrate” You to Death. Sperm whales are so loud that their clicks are capable of killing a human within their vicinity, says one science and adventure journalist. … Sperm whales are the loudest mammals on the planet, with vocalizations reaching an astonishing 230 decibels.
## Why is 194 dB the loudest sound possible?
The loudest a sustained sound can possibly be on Earth’s surface is 194 dB—which is when the amplitude of the sound wave is so intense that the low pressure part is a perfect vacuum (the wave alternates between double the normal atmospheric pressure and no air at all—not something you want to be present for).
## How loud is a hydrogen bomb?
around 172 decibelsAccording to The Independent,The Independent, the force of the blast was 10,000 times that of a hydrogen bomb, and Bhatia reports that the sound was registered at around 172 decibels over 160 kilometres away (100 miles).
## What made the loudest noise in history?
The 1883 eruption on Krakatoa may be the loudest noise the Earth has ever made. On August 27, 1883, the Earth let out a noise louder than any it has made since. It was 10:02 a.m. local time when the sound emerged from the island of Krakatoa, which sits between Java and Sumatra in Indonesia. | 1,063 | 4,508 | {"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-2020-50 | latest | en | 0.947539 |
http://mathhelpforum.com/advanced-statistics/122249-eq-2-regression-lines-print.html | 1,526,894,727,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794863972.16/warc/CC-MAIN-20180521082806-20180521102806-00443.warc.gz | 183,044,658 | 2,507 | # The eq of 2 regression lines are
Printable View
• Jan 2nd 2010, 11:49 PM
Athena
The eq of 2 regression lines are
The equations of 2 regression lines are
8x - 10y + 66 = 0 , 40x - 18y - 214 = 0
Obtain the standard deviation of Y is standard deviation of X is 3
• Jan 3rd 2010, 01:17 PM
pickslides
Your 2 equations $\displaystyle 8x - 10y + 66 = 0 , 40x - 18y - 214 = 0$ need to be put in the form $\displaystyle y=a+bx$ as this form helps us identify parameters like the standard deviation and the mean.
Note that for $\displaystyle y=a+bx$
$\displaystyle b = r\times \frac{s_y}{s_x}$
$\displaystyle a = \bar{y} -b\bar{x}$
so do you have any additional information? | 231 | 673 | {"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} | 3.953125 | 4 | CC-MAIN-2018-22 | latest | en | 0.794596 |
https://nickadamsinamerica.com/how-to-practice-lettering-on-procreate/346c8a98ff5dfe2d40ac2238ed0a1f0c/ | 1,638,650,182,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363006.60/warc/CC-MAIN-20211204185021-20211204215021-00092.warc.gz | 504,908,608 | 9,840 | # 346c8a98ff5dfe2d40ac2238ed0a1f0c
By . Worksheet. At Saturday, October 16th 2021, 14:12:52 PM.
This section contains all of the graphic previews for the Circle Worksheets. We have identifying radius and diameter for circles worksheets, calculating circumference, area, radius, and diameters worksheets, arcs and central angles for circles worksheets, arcs and chords worksheets, inscribed angles worksheets, graphing of circles worksheets and much more circle worksheets for your use. These geometry worksheets are a good resource for children in the 5th Grade through the 10th Grade.
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This section contains all of the graphic previews for the Differentiation Applications Worksheets. We have rate of change, graphing, graph properties, differentials, optimization, Newton’s Method, and related rates worksheets for your use. These Calculus worksheets are a good resource for students in high school.
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Lettering Practice Worksheets Cherry Kelly Sugar Crafts Hand Lettering Worksheet Brush Lettering Practice Lettering Practice | 588 | 2,974 | {"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} | 2.75 | 3 | CC-MAIN-2021-49 | latest | en | 0.76304 |
http://www.freerepublic.com/focus/f-news/1937224/posts?q=1&;page=125 | 1,529,640,883,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864343.37/warc/CC-MAIN-20180622030142-20180622050142-00123.warc.gz | 407,070,346 | 7,111 | Free Republic Browse · Search News/Activism Topics · Post Article
The Transcript of the GOP's Univision Debate
Townhall ^ | 12/09/2007 | Hugh Hewitt
Posted on 12/09/2007 8:37:28 PM PST by Ultra Sonic 007
first previous 1-20 ... 61-8081-100101-120121-125 last
To: tyke; kabar; SuziQ
There are 40 million Hispanic American citizens in this country
Uh...sorta innumerate, arentcha?
By Pew's own estimate, the "hispanic" population of the U.S. is now 14.7%, up from 12% in 2000 (the invasion continues).
I know that big numbers and fractions probably confuse you, so I'll make this easy for you...
To compute the actual number of hispanics this percentage (14.7) represents, we have to multiply the TOTAL POPULATION by 14.7 divided by 100...I know, that's hard and confusing, but if you pull out that 3rd grade arithmetic book that was so confusing for you, you'll see why....
Anyway we'll move on, just take my word for it. So now we have our little formula..
Actual Number of "hispanics" (whatever that is) = TOTAL POPULATION OF THE U.S. NOW OR THEREABOUTS * 14.7 / 100
I know there's confusing symbols in there. Unfortunately, for the sake of efficiency and to prevent confusion, the symbols " * " and " / " are used for multiplication and division, respectively. Try to understand. It's really for the best.
Actual Number of hispanics in the United States = 300, 000, 000 * 14.7 / 100 = 44, 100, 000
So there we go, assuming three hundred million for a total population...including the precious illegals.
Hmmm....so wait. You said, 40 million hispanic citizens, capitalizing hispanic, of course, whatever that is...but it's sacred, because they are Victims of the Evil white Race, right? Even though no one has any idea what the term means, since people from Spain are utterly white (I was in Barcelona a coupla weeks ago and that place was almost as caucasian as Moscow! Jeez, they really need to improve their diversity...). But I digress...
You said citizens in bold. Now wait...didn't...didn't the Pew Center say "percentage of hispanics" in the U.S.? Oooooooohhhhh.....that means the hispanic illegals are in this estimate....Nooooo! We have to do more arithmetic! We may even have to subtract!!! The horror....
Gee, umm, a year or so ago, tykey, El Presidente Boosh was running around saying we had 3 million or so illegales en Los Estados Unidos. Then somebody was mean and pointed out to him that that was actually just the number of illegals out on the streets of L.A. the other day. So they had to come off that line and use a new one. So now they say "12 million", which they've been saying for about the last year, even though as Kabar pointed out, we stop about a million and a half every year, and most people think that's only about 1 in 3. So in real life, that 12 has already become...oh no, more arithmetic....about 14 million. As we speak, er, tap.
Now some other people - like the Center for Immigration Studies, who Kabar has used as a resource - and they are a good one -- have estimated that the number could be as high as gasp Twenty million (as in, "20,000,000").
So the truth is somewhere in between 14 and 20 million. Just for grins, let's use an average ( Oh No! More arithmetic!! The Horror! ).
Bear with me now. THE AVERAGE WILL BE = ( 14 + 20 )/2 * 1, 000, 000.
I know. This one's really complicated, especially for people educated in American "schools" after about 1979. My condolences, and I hope your SexEd class was entertaining. Back in the dark ages we actually had to do the arithmetic, without the aid of doped silicon. It was awful. Just terrible. We even had calculators without batteries! They were called, "slide rules". They made your eyes hurt, and did not glow at night. But I digress again...
So we had to use the order of operations to compute our little AVERAGE equation above, which is really out there, believe me. Someday you may learn about such things, but only after you're in graduate school...for sociology. Or whatever it is you do.
And so doing that, we get
THE AVERAGE WILL BE = 17, 000, 000. In English (wait! shouldn't it be OK to use E-spanish?! ): Seventeen Million.
So that's a nice compromise on the number of illegales currently in the country.
Well! Now we're ready to come up with the real number of citizens Who Are hispanic (kinda like, "the man formerly known as Prince").
More arithmetic! But here it is....
REAL NUMBER OF HISPANIC CITIZENS = 44, 100, 000 - 17, 000, 000 which is 27, 100, 000.
Now wait. This is seriously small. Only Twenty Seven Million citizens? That's less than 10% of the population.
But it just might explain why only 5.4% of the electorate is 'hispanic', tykey. Just maybe.
So your BS about this Big Mass of Hispanic (always capitalized, of course) Heroes is...BS. A rather large chunk of them are in the country illegally.
And a rather large chunk of the "citizens" are in fact the children of illegals, who I don't consider citizens. And neither does the Constitution, no matter how many people bleat that it does.
So what can we call a mass occupation of the country by foreigners who broke in illegally, huh T-boy?
Invaders does the trick fairly well.
And that's what it is. An invasion. Funny how you don't get that. Maybe you just aren't all that bright.
And they can speak the "language of their birth" all they want, sport. In Mexico, where the vast majority of them came from.
PS: the audience the other evening was largely Cuban. Now what's interesting is that most of them have been here going on a few decades....yet SuziQ claims that they just have a problem with English because they haven't been here long enough. Gee, maybe that's er, not true? Maybe...Just Maybe...they have an issue with speaking English. Maybe...Just Maybe...they want to re-create their old country on U.S. soil...Just Maybe.
And you say it's idiotic to call that colonization, treason and invasion? I'd say that any of the three would be common sense descriptions. But you apparently lack such...
There is a rational debate to be had over the merits of having all American citizens speaking a common language, English
Wow! That's big of ya there, Tykey! Why, we can have a rational debate about using the language of the Constitution in this country. Woo-Wee, that's heavy! You're an amazing guy -- willing to sign up to the concept that we might all speak English in the United States!. Man, that's Radical, Dude! You're a Real Rebel! You and Abbie Hoffman....
Hysteria over some mythological Latino invasion and takeover does nothing to inform that debate
Don't worry 'bout it, kid. By the time you get done debating it, you'll have to talk to the cops in Spanish.
And one thing about Spanish speaking cops, tykey....they don't like people who speak English.
At all.
Be seein' ya, Genius.
121 posted on 12/11/2007 12:04:46 PM PST by Regulator
To: Kevmo
Today, there are several HLA amendments being introduced for voter approval through initiative processes in several states. And yet this is just “pandering”.
Even if they are passed, every single one of those HLA amendments in the States will be shot down by the Supremes, using Roe v Wade as precedent. That's why it is SO vital that this decision by overturned by the Supremes. The ONLY way that's going to be done is to get a strict constructionist majority on the Court. The only way that's going to happen is to elect someone who has pledged to do just that.
122 posted on 12/11/2007 12:56:19 PM PST by SuziQ
To: SuziQ
using Roe v Wade as precedent.
***Wrong. This extends the protection of the right to life to the unborn under the 14th amendment. The supremes would have to shoot down the 14th amendment or the definition of person. Not that easy to do. And Hunter is in favor of this approach (+ others) while Fred is against it, showing that Hunter is a better pro-life candidate.
123 posted on 12/11/2007 1:13:12 PM PST by Kevmo (We should withdraw from Iraq — via Tehran. And Duncan Hunter is just the man to get that job done.)
To: Kevmo
Do you honestly think that any state HLA will get any further than any other state laws that have been trying to restrict abortions since 1973? It sounds great, but it ain't gonna happen.
I've been working in the pro-life movement since 1973, and demanding purity on the issue hasn't done one blessed thing to save ONE baby since 1973. I'm tired of 'pie in the sky', and I want something actually DONE to help save the lives of unborn children. If we have to go about it incrementally, that's fine with me, because we'll at least be saving SOME babies while we're changing hearts and minds on the issue with the ultimate goal of folks accepting the unborn as a citizen with rights.
124 posted on 12/11/2007 3:01:10 PM PST by SuziQ
To: SuziQ
Do you honestly think that any state HLA will get any further than any other state laws that have been trying to restrict abortions since 1973?
***Yes.
It sounds great, but it ain’t gonna happen.
***Thanks for the crystal ball.
I’ve been working in the pro-life movement since 1973, and demanding purity on the issue hasn’t done one blessed thing to save ONE baby since 1973.
***I’m not a purist.
http://www.freerepublic.com/focus/f-news/1927653/posts?page=23#23
I’m tired of ‘pie in the sky’, and I want something actually DONE to help save the lives of unborn children.
***Then support the candidate who has a plan that DOESN’T DEPEND UPON overturning Roe V Wade — Hunter.
If we have to go about it incrementally, that’s fine with me, because we’ll at least be saving SOME babies while we’re changing hearts and minds on the issue with the ultimate goal of folks accepting the unborn as a citizen with rights.
***I don’t mind an incremental approach, but a federalist approach is a copout.
.
.
.
Why the smart money is on Duncan Hunter
http://www.freerepublic.com/focus/f-news/1926032/posts
125 posted on 12/11/2007 4:59:00 PM PST by Kevmo (We should withdraw from Iraq — via Tehran. And Duncan Hunter is just the man to get that job done.) | 2,498 | 10,036 | {"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-2018-26 | latest | en | 0.934091 |
http://www.studymode.com/essays/Telus-Cost-Capital-167448.html | 1,369,277,587,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368702749808/warc/CC-MAIN-20130516111229-00003-ip-10-60-113-184.ec2.internal.warc.gz | 724,365,236 | 9,063 | Telus: The Cost of Capital
Telus needs to calculate the cost of capital from the variety of data given. The cost of capital is determined mostly by how the funds are used rather than where they were obtained from. It relies on the risk of investments Telus involves in, therefore, depending on cost of both equity of debt as described below. Also note that, even though the preferred shares are not attractive to issuers and may not get issued again, it is still on the company’s balance sheet and affect firm’s overall wealth.
PREFFERED SHARES
We assume that Telus maintains a fixed debt to equity ratio and hence, the calculation will include preferred shares. 5.00 percent of cost in the past cannot be used towards final calculation because it is a book value. Similarly, par values of \$25 and \$100 are book values and are not considered in our calculation. Even though there is a \$4.00 for every \$100 par value share went to the underwriter, we decide to neglect that part since it is a small amount compared to the whole.
Given: Current yield: 5.90%; Dividend (preferred shares): \$4,000,000
Market value of preferred shares
P = D/ Rp
= \$4,000,000/ 0.059
= \$67,796,610
Rp = 5.90%
COMMON SHARES
In calculating the cost of equity, we will use the average between the dividend growth model and the CAPM. Since R-squared = 0.13 we know that the correlation is not strong enough and the sole use of the beta given to us will prove unreliable. For this reason, we choose to take the average between the dividend growth model and the CAPM model if possible. Also, as described above, we decide not to count the underwriter fees in our calculation.
Barb suggested that all the earnings after the dividends to the preferred shareholders belong to the common shareholder and the yield to the common stock should be the earnings-per-share divided by the market price. We did not agree with him. Although all the earnings after the dividend paid to the preferred shareholders... [continues]
### Cite This Essay
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# The recent upheaval in the office-equipment retail business,
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The recent upheaval in the office-equipment retail business, [#permalink]
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25 Apr 2006, 19:12
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The recent upheaval in the office-equipment retail business, in which many small firms have gone out of business, has been attributed to the advent of office equipment “superstoresâ€
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25 Apr 2006, 19:32
Good Question.....
And I am totally lost
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25 Apr 2006, 23:10
I choose B...
Because if the superstores keep the prices low throughout the retail market, they influence all the partcipants and therefore the smaller participants are going bust.....
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26 Apr 2006, 01:20
I'll go with B.........
The flawed argument states that superstores cannot be held responsible for the upheavel.
Hence B weakens it by stating how superstores are responsible for it.
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26 Apr 2006, 03:47
Agree with B)
The argument claims that the advent of the superstores, with their low prices, has forced small retailers out of business.The author claims that the analisys is flawed cause the superstores control small part of the market. Therefore, the reason that smal retailes lose business can not be the lower prices. If B) is true then there is no flaw in the analisys and the reason is low prices
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26 Apr 2006, 08:45
C it is
ma466
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27 Apr 2006, 04:05
prephtase this as analysis is correct-so answer is B.
Let me know the OA
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27 Apr 2006, 07:19
(A) Out of scope
(C) Close, but not good enough. Some have gone out of business. COuld be just a few businesses and we could not attribute cost competition to their closure.
(D) Out of scope
(E) Out of scope
B should be the best choice. It says small firms have gone out of business because superstores keep prices low. If these superstores practice heavy advertising and this resulted in the price dropping across the market, then the analysis is not flawed.
27 Apr 2006, 07:19
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# The recent upheaval in the office-equipment retail business,
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Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 1,303 | 4,713 | {"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} | 3.078125 | 3 | CC-MAIN-2017-04 | latest | en | 0.861356 |
https://blog.industrialguide.co.in/2021/04/cryogenic-distillation-advanced-learning.html | 1,726,106,901,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651420.25/warc/CC-MAIN-20240912011254-20240912041254-00251.warc.gz | 118,146,178 | 40,398 | # Cryogenic Distillation Advanced Learning
### Mass and Energy Balance
Capacity of unit= 215 ton/day gaseous O2
=6300 m3/hr
Quantity of intake air=34500 m3/hr
Component Volume %
Nitrogen 78.03
Oxygen 21.00
Argon 0.94
Hydrogen 0.01
Helium 0.0003
Krypton 0.00011
Xenon 0.00009
CO2 0.03-0.06
Moisture 0.02-0.05
Mass Balance
Energy Balance
#### Equipment Design
Design of Distillation Column
• L/V ratio at the top of both columns to be 0.58
• F = amount of feed = 1535.55 kmol/hr
• W = Bottom product from lower, refluxed back to the top column (rich liquid)
• S = Side stream from lower column, refluxed back to top column (impure liquid)
• Xf = mole fraction of N2 in feed = 0.79
• Xw = mole fraction of N2 in IPL = 0.96
• X1 = mole fraction of N2 in top product = 0.99
F = W + S + L2 ----------- (1)
F Xf = W Xw + S Xs + X1 L2 ----------- (2)
Hence from the graph no. of theoretical plates = 17
• Average molal μ of feed = 0.6127
• Actual number of plates = 17/ 0.6127 = 27.74 ≈ 28 plates
• From graph feed enters 12 plates
• Actual plate at which feed enters = 12/ 0.6127
• Feed enters plate = 19
• Total Height = 9.146 m
### Mechanical Design of Distillation Tower
Calculation Cylindrical Body of the Tower
Cylinderical wall thickness (bottom)
t = PL*DL/ (2 f J - PL) + C = 5.12 mm
Now, t = 6 mm thickness are acceptable.
Permissible pressure in the selected wall thickness against top wall
[P] = [2 f J (t - C)]/ (DL + t - C) = 7.67 kg/cm2
f = PL (DL + t - C)/ 2.3*J (t - C) = 785.2 kg/cm2
Cylindrical body wall thickness (upper)
t = [PH DH / 2 f J - PH] + C = t = 0.272 cm = 2.72 mm
Now, t = 6 mm thickness are acceptable.
Permissible pressure in selected wall thickness [P] and seizure against top wall
[P] = [2 f J (t - C)]/ (DH + t - C) = 8.84 kg/cm2
f= [PH (DH + t - C)]/ 2.3*J*(t - C)= 181.5 kg/cm2
Calculation of Elliptical Cap and Bottom
Elliptical bottom wall thickness
t = PL*DL/ (2 f J - PL) + C = 5.12 mm
Now, t = 6 mm thickness are acceptable.
Permissible pressure in the selected wall thickness and seizure against the top wall
[P] = [2 f J (t - C)]/ (DL + t - C) = 7.67 kg/cm2
f = PL (DL + t - C)/ 2.3*J (t - C) = 903 kg/cm2
Wall thickness of elliptical cap
t = [PH*DH/ 2 f J - PH] + C = 2.72 mm
Now, t = 6 mm thickness are acceptable.
Permissible pressure in selected wall thickness [P] and seizure against top wall
[P] = [2 f J (t - C)]/ (DH + t - C) = 8.84 kg/cm2
f = [PH (DH + t - C)]/ 2.3*J*(t - C) = 208.8 kg/cm2
### Condenser Design
Tha is inlet temperature of nitrogen vapor = -176.5 0C
Thb is outlet temperature of nitrogen = -178.2 0C
Tca is inlet oxygen liquid = -179.5 0C
Tcb is outlet oxygen = -178.2 0C
Therefore, ΔT1 = 1.491
Now, taking overall HTC as 100 btu/hr ft2 0F = 567 W/m2.0C
Q = U A ΔT
By using this equation A comes around 460 m2
Let N is the number of tubes so total heat transfer area = NπDL
Where, D = outer diameter, L = length
We are choosing 1 inch OD and 14 BWG tubes of 8 inch lengths and
calculating the
number of tubes obtained is 2360.
Taking triangular pitch of 1.25 inch and shell diameter = 1.67 m
Now, as this values of Heat transfer coefficient U becomes 560.6 W.
Inside and Outside Coefficient Calculations
Nitrogen in tube side
hi = 240 btu/hr.ft2.0F
ho = hi*(id/od) = 22.16 btu/hr.ft2.0F
Oxygen in shell side
ho = jHK/ De(cμ/k)-1/3 = 223.61 btu/ft2.hr.0F = 1267.8 W/m2.0C
Clean overall coefficient
UC = hio*ho/ (hio+ ho) = 598.865 W/m2. 0C
Pressure Drop
Shell Side:
Ps = f.Gs
2.Ds. (N + 1)/ (5.22*1010 De*S) = 1.5222 psi
Tube side:
Pt = f.Gt
2.Ln/ (5.22*1010*Ds) = 0.155*103 psi
### Plant Economics
Rated Plant Capacity = 215 tons/day of gaseous Oxygen
99.95% purity
= 61500 tons/annum of Oxygen
Basis: Number of working days = 25 days/month
= 300 days/annum
Number of shifts = 3 per day
One shift = 8 hours
### Fixed Capital
Land and Building= 3, 85, 50,000.00
Plant and Machinery= 37, 58, 68,000.00
Other Fixed Assets= 6, 15, 00,000.00
Total Fixed Capital= 47, 59, 18,000.00
### Working Capital
Salary and Wages= 86, 19,000.00
Utilities and Overheads= 29, 71, 00,000.00
Total working capital per annum = 30, 57, 19,000.00
### Production Cost Per Annum
Total Working Capital per Month
Total……………… 2, 54, 76,583.00
Working capital for two months = 5, 09, 53,167.00
Total fixed capital = 47, 59, 18,000.00
Total Capital Investment = 52, 68, 71,167.00
Production Cost per Annum
Working capital for one year = 30, 57, 19,000.00
Interest @ 13.50% on TCI = 7, 11, 27,608.00
Dep. @ 33% on plant and machinery = 12, 40, 36,440.00
Total…………………… 50, 08, 83,050.00
ROR and BEP
By sale of O2- N2 = Rs 88, 90, 00,000.00
Profit = Receipt – Production cost = Rs 38, 81, 16,950.00
Profit sales ratio = Profit/ Sales*100 = 43.65%
Rate of return = Operating profit/ TCI*100 = 72%
Break- Even Point (BEP) = Fixed Cost/ (Fixed cost+ profit) = 45%
Payback Period
Payback Period= (Project Cost+ working capital borrowing)/ Gross Profit Per year
Payback Period= 52, 68, 71,167.00/ 38, 81, 16,950.00
Payback Period is around 1 Year and 5 Months
### Conclusion
As a part of feasibility study, we have investigated the feasibility of constructing a new green-field 215 tpd oxygen plant.
As a part of study we have successfully designed major equipment like distillation column, heat exchanger etc.
The plant is totally feasible and gives 99.95% pure product efficiently. The project can create employment for persons.
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https://www.airmilescalculator.com/distance/dca-to-bna/ | 1,653,712,169,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652663012542.85/warc/CC-MAIN-20220528031224-20220528061224-00102.warc.gz | 704,158,247 | 36,206 | # Distance between Washington D.C. (DCA) and Nashville, TN (BNA)
Flight distance from Washington D.C. to Nashville (Ronald Reagan Washington National Airport – Nashville International Airport) is 562 miles / 904 kilometers / 488 nautical miles. Estimated flight time is 1 hour 33 minutes.
Driving distance from Washington D.C. (DCA) to Nashville (BNA) is 659 miles / 1061 kilometers and travel time by car is about 12 hours 12 minutes.
562
Miles
904
Kilometers
488
Nautical miles
1 h 33 min
108 kg
## How far is Nashville from Washington D.C.?
There are several ways to calculate distances between Los Angeles and Chicago. Here are two common methods:
Vincenty's formula (applied above)
• 561.835 miles
• 904.186 kilometers
• 488.221 nautical miles
Vincenty's formula calculates the distance between latitude/longitude points on the earth’s surface, using an ellipsoidal model of the earth.
Haversine formula
• 560.778 miles
• 902.485 kilometers
• 487.303 nautical miles
The haversine formula calculates the distance between latitude/longitude points assuming a spherical earth (great-circle distance – the shortest distance between two points).
## How long does it take to fly from Washington D.C. to Nashville?
Estimated flight time from Ronald Reagan Washington National Airport to Nashville International Airport is 1 hour 33 minutes.
## What is the time difference between Washington D.C. and Nashville?
The time difference between Washington D.C. and Nashville is 1 hour. Nashville is 1 hour behind Washington D.C..
Washington D.C. time to Nashville time converter
## Flight carbon footprint between Ronald Reagan Washington National Airport (DCA) and Nashville International Airport (BNA)
On average flying from Washington D.C. to Nashville generates about 108 kg of CO2 per passenger, 108 kilograms is equal to 237 pounds (lbs). The figures are estimates and include only the CO2 generated by burning jet fuel.
## Map of flight path and driving directions from Washington D.C. to Nashville
Shortest flight path between Ronald Reagan Washington National Airport (DCA) and Nashville International Airport (BNA).
## Airport information
Origin Ronald Reagan Washington National Airport
City: Washington D.C.
Country: United States
IATA Code: DCA
ICAO Code: KDCA
Coordinates: 38°51′7″N, 77°2′15″W
Destination Nashville International Airport
City: Nashville, TN
Country: United States
IATA Code: BNA
ICAO Code: KBNA
Coordinates: 36°7′28″N, 86°40′41″W | 580 | 2,471 | {"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-2022-21 | latest | en | 0.850226 |
https://forums.nrel.gov/t/tower-top-yaw-bearing-inertia-translational-velocity/278/1 | 1,696,453,521,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511406.34/warc/CC-MAIN-20231004184208-20231004214208-00187.warc.gz | 293,449,131 | 8,223 | # Tower-top / yaw bearing inertia translational velocity
With respect to control design I’m trying to determine the translational velocity of the Tower-top / yaw bearing in the inertia frame. The code is Hywind, so the foundation is floating.
I find it a bit difficult, but so fare I got:
h=90 meters
[m/s]=PtfmTVxi -PtfmTVxt +hYawBrRVyp2*pi/360
Is this correct?
BR
Søren Christiansen
PhD student
Aalborg University
Denmark
Dear Søren,
I’m not sure I understand your equation, but you can output the translational velocity of the tower-top / yaw bearing in the inertia frame from FAST as follows:
1. In the primary FAST input file, set the nacelle inertia measurement unit (IMU) locations in the downwind, lateral, and vertical directions (NcIMUxn, NcIMUyn, and NcIMUzn, respectively) all to 0.0. This will locate the nacelle IMU at the center of the tower-top / yaw bearing.
2. Include the nacelle IMU translational velocity outputs (NcIMUTVxs, NcIMUTVys, NcIMUTVzs) in the output list of the primary FAST input file.
The nacelle IMU translational velocity outputs are calculated relative to the inertia frame and output in a coordinate system that is aligned with the instantaneous shaft coordinate system.
I hope that helps.
Best regards,
Dear Jason,
Thank you very much. This is exacly what I was looking for.
Some how I missed this important signal.
Thank you for taking the time to answer my simple question.
Best regards
Søren
Dear Jason,
Could you please confirm if my the below understanding is correct. I have seen that answers are available at various places in various topics. Just asking for a favor here to validate my understanding.
Measure 1) TwrTpTDxi:
This is the motion of ‘Yaw bearing’(located on tower axis at TwrHt) observed by an observer
sitting at the origin of an ‘inertial frame’ and expressed in the same frame (XiYiZi).
Measure 2) YawBrTDxp:
This is the motion of ‘Yaw bearing’ (located on tower axis at TwrHt i.e origin of XpYpZp coordinate system) observed by an observer sitting at the origin of the ‘Tower base frame (XtYtZt)’ and expressed in “baseplate coordinate system(XpYpZp)”.
Measure 3) YawBrTDxt:
This is the motion of Yaw bearing (located on tower axis at TwrHt) observed by an observer
sitting at the origin of the ‘tower base frame ((XtYtZt))’ and expressed in “Tower base coordinate system(XtYtZt)”.
Measure(1) contains the rigid motion of support platform (in the case of movable support) and measure(2) and (3) are purely elasic deflections of tower expressed in different coordinate systems. Right?
Measure 4) YawBrRDxt YawBrRDyt YawBrRDzt
These are the instantaneous Euler angles between XpYpZp frame and XtYtZt frame. (In the Yaw-pitch-Roll) sequence.
Measure 5) TipDxb1
The displacement of the blade tip observed by an observer sitting at the origin of the “Cone Coordinate system of the respective blade” and expressed in “Cone coordinate system” itself.
Measure 6) TipDxc1
The displacement of the blade tip observed by an observer sitting at the origin of “blade Coordinate system of the respective blade” and expressed in “blade coordinate system” itself.
Measure 7) TipALxb1
The acceleration of the blade tip observed by an observer sitting at the origin of the “Inertial” Coordinate system and expressed in “inertial coordinate system” itself.
Thanks.
Regards,
Kumara
Dear @KumaraRaja.Eedara,
I agree with your description of measures (1) through (3).
Regarding measure (4), I agree with your statement except that the rotations do not follow a yaw-pitch-roll Euler sequence. Instead, small to moderate rotations are assumed in ElastoDyn, where the rotation sequence does not matter. See the Equations (1) and (2) in my 2009 Wind Energy Paper for more information: onlinelibrary.wiley.com/doi/abs/10.1002/we.347.
You’ve swapped the descriptions of measure (5) and (6). “b” refers to the blade coordinate system and “c” refers to the coned coordinate system.
Measure (7) is the absolute acceleration in the inertial frame, but it is expressed in the local blade coordinate system (Lb, oriented with the blade cross section as the blade deflects), not the inertial frame coordinate system.
Best regards,
Thank you @Jason.Jonkman
Dear @Jason.Jonkman ,
Is NcIMU only for outputting velocity and acceleration? Different position of NcIMU will output the corresponding velocity/acceleration at that position, but it will not affect the system behavior?
Regards,
Ran
That is correct.
Best regards,
Hi,
I have a local drivetrain model that requires:
1. displacements
2. rotations
3. angular and translational velocities and accelerations
of the tower top in the inertia reference frame.
No. 1 is readily available in terms of TwrTpTDxi, TwrTpTDyi and TwrTpTDzi.
For no. 2 I have been using TwHt1RPxi, …yi and …zi with a tower gage at the node closest to the tower top. As far as I understand, this is output at half an element below the tower top (?), but I believe the differences are small enough (I have 79 nodes evenly distributed over the tower length of 115 m). Any other suggestion would be appreciated.
My main concern is with regards to no. 3.
Here I am using the nacelle IMU velocities and accelerations. Since these are output in the shaft coordinate system, I need to transform the measurements to the inertia reference frame. I am following the transformations presented in the “FASTCoordinateSystems.doc” at the ElastoDyn documentation site. Most of this is clear and I am able to follow. However, with regards to the transformation between tower top and tower base, I am not sure what values to use for theta_SS(TwrFlexL) and theta_FA(TwrFlexL).
I see some possible options, either:
• Modify the source code to output ThetaFA and ThetaSS from Elastodyn, or
• Calculate ThetaFA and ThetaSS by outputting the necessary derivatives and degrees of freedom as shown in the equations. I see that e.g. Q_TFA1, Q_TSS1, QD_TFA1 etc. are available output, but not recommended for use. And I am not sure if this is what I am looking for.
• Or, the values I need are available as outputs in OpenFAST (this is my hope), and known as TwHt1RDxt, TwHt1RDyt and TwHt1RDzt when gage 1 is taken at the node closest to the tower top.
Would be really grateful for some input here. I have the same requirement for hub/shaft loads, but since they also follow the shaft coordinate system, I believe I can use the same procedure as with the motions.
Best regards,
Veronica
Dear @Veronica.Krathe,
One high-level question first. Is the local drivetrain model you have something you are trying to couple to OpenFAST in a post-processing step (one-way coupled involving first running OpenFAST, followed by running your local drivetrain model), or are you trying to couple the local drivetrain model within the source code every time step (two-way coupled)? If the former, your questions make sense and I’ll answer them. If the latter, I would suggest a different approach that doesn’t make use of the OutList from ElastoDyn.
Best regards,
Dear @Jason.Jonkman
I am trying to do the one-way coupled/post-processing step, not a two-way coupled analysis.
Thanks for clarifying.
Best regards,
Dear @Veronica.Krathe,
ThetaFA and ThetaSS at the tower top/yaw bearing are already available outputs from ElastoDyn via `OutList` outputs `YawBrRDyt` and `YawBrRDxt` (you’ll have to convert from degrees to radians and the sign may need to be flipped depending on which coordinate systems you are considering).
Best regards,
Simple as that.
Just to verify - is it correct that:
q_R, q_Y and q_P in the transformation between the tower base and the inertia coordinate system should be taken as PtfmRoll, PtfmYaw and PtfmPitch?
q_Yaw in the nacelle coordinate system can be found as YawPzn in the OutList outputs?
(All of them should be converted to radians, I presume). | 1,932 | 7,849 | {"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-2023-40 | latest | en | 0.876776 |
https://www.physicsforums.com/threads/expansion-of-solids-due-to-increases-in-temperature.277664/ | 1,531,988,298,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676590711.36/warc/CC-MAIN-20180719070814-20180719090814-00627.warc.gz | 948,931,852 | 13,569 | # Homework Help: Expansion of Solids due to increases in Temperature
1. Dec 6, 2008
### jnbfive
I was given a question about a certain material. The idea was that when it was heated from zero degrees C to 200 degrees C, it would be 10.06 mm from its original length of 10. Doing it in reverse order, length of 10.06 mm from 200 C to 0 C gave me an answer of 9.99964, which is less than the initial 10 mm. The question is, why does this happen. It also lists a description of this paradox given by H. Fakhruddin. I was wondering if anyone had a link or a pdf file that I'd be able to read in order to draw my own conclusions on this.
Another question I was given was as follows:
The volume expansion of a solid or a liquid can be written as V = V0(1+Beta*Change in Temperature), where beta is the coefficient of volume expansion. Starting with this equation and with the definition of mass density, use the binomial expansion to show that the mass density, rho, of a substance can be written as rho = rho0(1-Beta*Change in Temperature).
Now, I looked this up and couldn't really understand what I found. I was wondering if someone could break this down for me so that I can figure it out. I'm assuming that rho = mass/volume in this question.
2. Dec 8, 2008
### marcusl
1) Your calculations are correct. The "paradox" comes from using a simple linear model
$$L=L_0*(1+T*CTE)$$,
where CTE is the linear coefficient of thermal expansion, to describe a more complex behavior.
You can verify the same "paradox" by many everyday examples. If you buy a $100 item at 10% off, you pay$90. Someone who bought it at full price didn't pay 10% more than you did, however, they paid 1(1-0.10)=11.1% more than you. The difference arises from "linearizing" a multiplicative process.
The linear model is accurate for most purposes because the CTE's are so small (the error in your case is just 36 ppm). More complex models are available when greater accuracy is needed.
2) The binomial expansion is
$$(1+\epsilon)^{-1}\approx 1-\epsilon$$.
This holds true when $$\epsilon<<1$$, so the coefficient times the temperature must be a small number.
To get the result you asked about, substitute your expression for V into rho = mass/volume and apply the binomial expansion.
3. Dec 9, 2008
### marcusl
A typo in the fifth line was pointed out: it should read ".. 1/(1-0.1) gives 11.1% more.." | 620 | 2,388 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2018-30 | latest | en | 0.965519 |
https://stats.stackexchange.com/questions/123320/mse-decomposition-to-variance-and-bias-squared/304195 | 1,725,805,311,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651002.87/warc/CC-MAIN-20240908115103-20240908145103-00546.warc.gz | 533,863,008 | 42,626 | MSE decomposition to Variance and Bias Squared
In showing that MSE can be decomposed into variance plus the square of Bias, the proof in Wikipedia has a step, highlighted in the picture. How does this work? How is the expectation pushed in to the product from the 3rd step to the 4th step? If the two terms are independent, shouldn't the expectation be applied to both the terms? and if they aren't, is this step valid?
• For future reference, this is imo a pretty clear derivation of the decomposition, which is more explicit in notation. Commented Jul 26, 2023 at 16:30
The trick is that $\mathbb{E}(\hat{\theta}) - \theta$ is a constant.
• Oh I see. The only unknown here is the estimator. Right? Commented Nov 9, 2014 at 19:43
• Yes. Taking expectation means that the estimator goes to whatever it's estimating, that's what makes the $\mathbf{E}(\hat{\theta} - \mathbf{E}(\hat{\theta}))$ go to 0. Commented Nov 9, 2014 at 23:38
• Sorry, that sentence doesn't make much sense to me. If an estimator went to whatever it was estimating, wouldn't that make it unbiased? Can it be explained by saying $\mathbb{E}(\hat{\theta} - \mathbb{E}(\hat{\theta}))$ = $\mathbb{E}(\hat{\theta}) - \mathbb{E}(\mathbb{E}(\hat{\theta}))$ = $\mathbb{E}(\hat{\theta}) - \mathbb{E}(\hat{\theta})$ = 0 ? Commented Apr 6, 2017 at 21:16
• @user1158559 the product term in the middle is a constant times something with expected value 0. Even if theta-hat is biased, it's still a constant times 0. Commented Apr 18, 2017 at 15:49
• Adam, you were right (+1), I was confused about the original question, because it mentions a highlight in the picture (which is about that term going to zero instead of taking the constant term outside the brackets of the expectation). So I wanted to talk about the variable $\mathbb{E}(\hat{\theta})-\hat{\theta}$, but instead I copied (because I was lazy) from your answer the constant term $\mathbb{E}(\hat{\theta})-\theta$ (and could not change it anymore after 5 minutes). Besides (1) using the wrongly copied term, I also (2) misunderstood the question. The maths used variabl $\hat{\theta}$ Commented Sep 21, 2017 at 16:28
There has been some confusion about the question which was ambiguous being about the highlight and the step from line three to line four.
There are two terms that look a lot like each other.
$$\mathbb{E}\left[\hat{\theta}\right] - \theta \quad \text{vs} \quad \mathbb{E}\left[\hat{\theta}\right] - \hat\theta$$
The question, about the step from 3rd to 4th line, relates to the first term:
• $$\mathbb{E}[\hat{\theta}] - \theta$$ this is the bias for the estimator $$\hat\theta$$
The bias is the same (constant) value every time you take a sample, and because of that you can take it out of the expectation operator (so that is how the step from the 3rd to 4th line, taking the constant out, is done).
Note that you should not interpret this as a Bayesian analysis where $$\theta$$ is variable. It is a frequentist analysis which conditions on the parameters $$\theta$$. So we are computing more specifically $$\mathbb{E}[(\hat{\theta} - \theta)^2 \vert \theta]$$, the expectation value of the squared error conditional on $$\theta$$, instead of $$\mathbb{E}[(\hat{\theta} - \theta)^2]$$. This conditioning is often implied implicitly in a frequentist analysis.
The question about the highlighted expression is about the second term
• $$\mathbb{E}[\hat{\theta}] - \hat{\theta}$$ this is the deviation from the mean for the estimator $$\hat{\theta}$$.
It's expectation value is also called the 1st central moment which is always zero (so that is how the highlighted step, putting the expectation equal to zero, is done).
• This is an excellent answer. Thanks. Perhaps this is more trivial, but this question and explanation doesn't seem to explain how one truly gets the 4th expression from the 3rd. To not deviate too much from this particular question, I left a new one here
– Josh
Commented May 30, 2020 at 16:40
• @Josh, I have now stressed it a bit more. Commented May 30, 2020 at 16:48
$E(\hat{\theta}) - \theta$ is not a constant.
The comment of @user1158559 is actually the correct one:
$$E[\hat{\theta} - E(\hat{\theta})] = E(\hat{\theta}) - E[E(\hat{\theta})] = E(\hat{\theta}) - E(\hat{\theta}) = 0$$
• I don't see what you are trying to show. Also the bias may not be zero but that does not mean that it isn't a constant. Commented Sep 21, 2017 at 0:20
• It is not a constant because $\hat{\theta} = f(D)$ where $D$ is a given training data, which is also a random variable. Thus, its expectation is not a constant. Commented Sep 21, 2017 at 0:49
• @Josh, yes $$\text{E(\hat\theta) - \theta, the bias of the estimator \hat{\theta}, is constant}$$ (in the sense that it is the same for each sample). $$\text{E(\hat\theta) - \hat\theta, the error of the estimator \hat\theta, is not a constant}$$For instance, say we wish to approximate the mean of a normal distributed population by using the median. Then the the bias is $E(\hat\theta) - \theta =0$ and constantly the same for every sample, but the estimate $\hat{\theta}$ is not the same for every sample, ie. $E(\hat\theta) - \hat\theta$ will be different each sample. Commented May 30, 2020 at 15:31
• @Josh, this answer is right about the step in the highlighted block, why the expectation value of the the difference of the estimator with it's expectation value, $E(\hat\theta)-\hat\theta$, which is variable, is zero (note: in the previous comment I wrongly refered to this as the error of the estimator, which is $\hat\theta - \theta$ instead). This answer is wrong about $E(\hat\theta)-\theta$ not being a constant. Commented May 30, 2020 at 15:41
• Of course, you must not consider this from a Bayesian point of view where $\theta$ is treated as variable, but you must consider it from a frequentist point of view which conditions on $\theta$. Commented May 30, 2020 at 15:44 | 1,633 | 5,901 | {"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": 10, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2024-38 | latest | en | 0.938607 |
https://medicalnewsbulletin.com/brain-teaser-february-8/ | 1,695,722,960,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510179.22/warc/CC-MAIN-20230926075508-20230926105508-00117.warc.gz | 435,799,264 | 83,951 | Tuesday, September 26, 2023
# Brain Teaser
Dr. Johnson was walking in a park and met a family consisting of a grandfather, a father and a son. Dr. Johnson asked them how old they were. The grandfather said: “All together we are 100 years old”. The father said: “My son and I together are 45 years old, and my son is 25 years younger than I am”. How old are they?
Let X1=age of the grandfather, X2 = age of the father, and X3 = age of the son. Then we have that:
X1+X2+X3=100
X2+X3=45
X2-X3=25
Solving this system of three equations, we get X1= 55, X2= 35, and X3=10
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Patient: Doc, I work like a beaver, eat like a horse... | 233 | 747 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2023-40 | longest | en | 0.931777 |
http://www.cscprogrammingtutorials.com/2016/10/exercise-31-algebra-solve-quadratic-equations.html | 1,521,519,726,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257647280.40/warc/CC-MAIN-20180320033158-20180320053158-00429.warc.gz | 375,689,075 | 17,897 | # Exercise 3.1 - Algebra: Solve Quadratic Equations
The program below is the answer to Liang's Introduction to Java Programming (9th Edition) Chapter 3 Exercise 3.1.
```/**
*
* @Author: Aghatise Osazuwa
* Website: www.cscprogrammingtutorials.com
* Exercise 3.1 - Algebra: Solve Quadratic Equations
*/
import java.util.Scanner;
public class CheckerboardPatternOfAsterisks {
public static void main (String [] args) {
// declare variables
double a, b, c, disc, r1, r2;
// create Scanner to read user input
Scanner input = new Scanner(System.in);
// prompt user to enter details
System.out.print("Enter a, b, c: ");
a = input.nextDouble();
b = input.nextDouble();
c = input.nextDouble();
//calculate the discriminant
disc = b * b - (4 * a * c);
//calculate the roots
r1 = (-b + Math.pow(disc, 0.5)) / (2 * a);
r2 = (-b - Math.pow(disc, 0.5)) / (2 * a);
// display the results
if (disc > 0){
System.out.println("The roots are " + r1 + " and " + r2 + "\n");
}
if (disc == 0){
System.out.println("The root is " + r1 + "\n");
}
if (disc < 0){
System.out.println("The equation has no real roots.\n");
}
}
}
```
Sample Program run | 334 | 1,133 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2018-13 | latest | en | 0.505827 |
https://www.physicsforums.com/threads/bra-vectors.714604/ | 1,590,420,708,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347388758.12/warc/CC-MAIN-20200525130036-20200525160036-00125.warc.gz | 836,276,980 | 18,347 | # Bra Vectors
## Main Question or Discussion Point
One question that's been on my mind for a while is what the difference between a bra vector and a covector is. Are they the same thing? Why the difference of notation?
Related Quantum Physics News on Phys.org
Where did you encounter co-vectors in quantum theory? And what did they look like?
Cheers,
Jazz
Where did you encounter co-vectors in quantum theory? And what did they look like?
Cheers,
Jazz
I didn't. I'm a math guy. :tongue:
I read a thing on how the Riemann Hypothesis is related to quantum physics, and the article used some bras and kets. I assumed, from context, that a bra vector was essentially the same as a covector. I wanted to know if I was right.
WannabeNewton
There is no difference. A bra is a covector but you should keep in mind that the term dual vector is more appropriate in the general linear algebra setting; the term covector is more prevalent in differential geometry. The notation is more convenient in the context of QM is all.
1 person
Well, you said they differed in notation, so I was surprised you found a different notation alongside the Dirac notation. And yes, bras are the same concept as co-vectors, both are the dual with respect to the inner product. But you wouldn't call the bra a co-vector usually. The term co-vector is used in the context of manifolds in physics, so typically either symplectic geometry (hamilton formalism) or riemannian geometry (GR).
Hope this clears it up :)
Cheers,
Jazz
1 person
dextercioby
Homework Helper
One question that's been on my mind for a while is what the difference between a bra vector and a covector is. Are they the same thing? Why the difference of notation?
A co-vector is a more general notion than of a bra-vector, because you needn't have a topology, nor a scalar product to speak about vectors and co-vectors, but you need to have them to speak of bra-s and ket-s.
A co-vector is a more general notion than of a bra-vector, because you needn't have a topology, nor a scalar product to speak about vectors and co-vectors, but you need to have them to speak of bra-s and ket-s.
That's interesting you bring that up. There seem to be two different concepts or definitions of co-vectors in the literature, depending on whether you get there using the exterior algebra and differential forms or multilinear forms and a metric (with the metric as the defining bijection for the dual). For the diff-forms approach you only need the metric when you introduce the hodge dual and not the co-vectors.
I think I understand this all pretty well, but I never found the different definitions very intuitive and the differences rather confusing. But then again, I'm only a theoretical physicist and not a real mathematician.
Any thoughts on that?
Cheers,
Jazz
dextercioby
Homework Helper
I think 'co-vectors' come uniquely from linear algebra. They are automatically there when you have a vector space. But the concept of vector space is useful also in geometry and that's how we bring 'co-vectors' in geometry. There needn't be a metric on a manifold, nor a connection. Co-vectors are there because vectors are there. That's always the case.
I think 'co-vectors' come uniquely from linear algebra. They are automatically there when you have a vector space. But the concept of vector space is useful also in geometry and that's how we bring 'co-vectors' in geometry. There needn't be a metric on a manifold, nor a connection. Co-vectors are there because vectors are there. That's always the case.
So what's your most general definition of a co-vector?
Cheers,
Jazz
dextercioby
The dual space $V^*$ of a topological vector space $V$ (over $\mathbb C$) is the space of continuous linear functionals $f:V\rightarrow\mathbb C$. A member of $V^*$ is called dual vector or co-vector. This is mostly useful if $V$ is locally convex (e.g. Banach spaces or Hilbert spaces, the Schwarz space, ...), since then the Hahn-Banach theorem guarantees the existence of enough such functionals to make the theory interesting. On a Hilbert space $H$ you have Riesz's theorem, which tells you that $H^*$ is isomorphic to $H$ (this is used a lot in QM).
In the case of manifolds, you automatically have the tangent spaces $T_p M$ at every point and since they are topological vector spaces (all finite-dimensional vector spaces have this property), you can form their dual $T^*_p M$. If you have a metric, the tangent spaces are Hilbert spaces and you can use the Riesz isomorphism to identify tangent vectors with cotangent vectors. | 1,062 | 4,568 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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} | 2.796875 | 3 | CC-MAIN-2020-24 | longest | en | 0.969551 |
http://www.slidesearchengine.com/slide/chap-17-1 | 1,490,562,086,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218189252.15/warc/CC-MAIN-20170322212949-00076-ip-10-233-31-227.ec2.internal.warc.gz | 703,963,378 | 8,978 | # chap 17 1
64 %
36 %
Education
Published on January 22, 2008
Author: Candelora
Source: authorstream.com
User name: Comment:
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https://web2.0calc.com/questions/help-pls_82944 | 1,643,422,059,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320299894.32/warc/CC-MAIN-20220129002459-20220129032459-00692.warc.gz | 630,537,254 | 5,518 | +0
# help pls
0
114
1
In the figure above, line PS bisects angle QPR. We can ensure that \(\triangle PQS\cong\triangle PRS\) by just adding one more condition. Which of the following statements could be that condition?
A) PQ = PR
B) SQ = SR
C) angle PQS = angle PRS
D) angle PSQ = angle PSR
If possible, can you show me how you got the answer
Sep 1, 2021 | 116 | 363 | {"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-2022-05 | longest | en | 0.815013 |
http://stackoverflow.com/questions/23455291/get-mid-value-in-c/23455428 | 1,419,709,770,000,000,000 | text/html | crawl-data/CC-MAIN-2014-52/segments/1419447552799.24/warc/CC-MAIN-20141224185912-00048-ip-10-231-17-201.ec2.internal.warc.gz | 90,292,218 | 22,908 | # Get mid value in C++
Consider three values `x, y, z`.
What would be the formula to get the mid value (not the mean value but the value which is neither the `min` nor the `max`)?
``````const double min = std::min(x, std::min(y, z));
const double mid = /* what formula here ? */
const double max = std::max(x, std::max(y, z));
``````
-
Maybe there isn't a neat symmetrical formula? – Oliver Charlesworth May 4 at 10:49
@AbhishekBansal What if they are all equal? What if only two of them are equal? – n.m. May 4 at 10:52
FYI this is called a median. Related. – n.m. May 4 at 10:56
Just pick any one at random. Your app will then work 1/3 of the time, which is much more than my current app is achieving. – Martin James May 4 at 11:02
BTW, in C++11, `min` can be rewritten `std::min({x, y, z})`. – Jarod42 May 4 at 14:31
``````const double mid = std::max(std::min(x,y),std::min(std::max(x,y),z));
``````
Edit - As pointed out by Alan, I missed out on a case. I have given now a more intuitive proof.
Direct Proof: Without Loss of Generality with respect to x and y.
Starting with the innermost expression, `min(max(x,y),z)` ...
1. If it returns z, we have found the relations: max(x,y) > z. Then the expression evaluates to `max(min(x,y),z)`. Through this we are able to determine the relation between min(x,y) and z.
If min(x,y) > z, then z is smaller than x and y both (as the relation becomes: max(x,y) > min(x,y) > z). Therefore the min(x,y) is indeed the median and `max(min(x,y),z)` returns that.
If min(x,y) < z, then z is indeed the median (as min(x,y) < z < max(x,y)).
2. If it returns x, then we have x < z and y < z. The expressions evaluates to: `max(min(x,y),x)`. Since max(x,y) evaluated to x, min(x,y) evaluates to y. Getting the relation z > x > y. We return the max of x and y (as the expression becomes `max(y,x)`) which is x and also the median. (Note that the proof for y is symmetrical)
Proof Ends
Old Proof - Note it is NOT complete (Direct):
Without loss of generality: Assume x > y > z
Min of x and y is y. And min of (max of x and y) and z is z.
The max of y and z is y which is the median.
Assume x = y > z
Min of x and y say is x. And min of (max of x and y is x) and z is z.
Max of the above two is x, which is the median.
Assume x > y = z
Min of x and y is y. And min of (max of x and y is x) and z is z.
Max of the above two is y, which is the median.
Finally, assume x = y = z
Any of the three numbers will be the median., and the formula used will return some number.
-
Yes, it works. However, it is not very readable and consequently would be difficult to maintain; at first glance it is not clear what is happening. In contrast, with my solution 'auto mid = median(x,y,z)', it is instantly recognizable that the function returns the median of the three values. – Ricky65 May 6 at 12:02
I'm not entirely convinced by your proof. What about the case where x < y < z? (You say "without loss of generality" - but the expression is not symmetric in the three values.) However, I believe the code is correct. – Alan Stokes May 10 at 13:39
@AlanStokes Good catch. I will look into it, and improve the answer. (possibly try to give a more elegant proof if possible). – digvijay91 May 10 at 13:53
To find all three at once in a symmetrical fashion:
``````min = x; med = y; max = z;
if (min > med) std::swap(min, med);
if (med > max) std::swap(med, max);
if (min > med) std::swap(min, med);
``````
-
Isn't this a kind of Bubble sort? – user1990169 May 4 at 11:06
This will not work in C though, just saying. :) – lpapp May 4 at 11:08
@AbhishekBansal this is indeed bubble sort. – n.m. May 4 at 11:08
@LaszloPapp: Question is only tagged C++... – Oliver Charlesworth May 4 at 12:07
@OliCharlesworth: I know right. I hate to ask this: so what? I cannot say that it will not work in C? What do you think "just saying" means? ;) – lpapp May 4 at 12:13
This seems like cheating, but: `x + y + z - min - max`
-
Nice trick, Alan. ;) – lpapp May 4 at 10:52
I like it. The only problem would be overflow. – keyser May 4 at 10:52
This is clever, but introduces numerical errors. – Anonymous May 4 at 10:52
@keyser: true, so it is worth posting a different answer, too, just in case. Also, this is not necessarily returning the variable, just the value. – lpapp May 4 at 10:53
It is a solution which may give a 4th number ^_^ (ideone.com/NYxgho) – Jarod42 May 4 at 14:29
It is a bit uglier than Alan's trick, but it cannot cause overflow, nor numerical errors, and so on:
``````int x, y, z, median;
...
if (x <= y && y <= z || y >= z && y <= x) median = y;
else if (y <= x && x <= z || x >= z && x <= y) median = x;
else median = z;
``````
The algorithm is simply this:
• check if x is between y and z, if yes, that is it.
• check if y is between x and z, if yes, that is it.
• It must be z since it was neither x, nor y.
=====================================================
You could also get this more flexibly if you have more than three elements, with sorting.
// or xor implementation, does not matter...
void myswap(int* a, int* b) { int temp = *b; *b = *a; *a = temp; }
``````int x, y, z;
// Initialize them
int min = x;
int med = y;
int max = z;
// you could also use std::swap here if it does not have to be C compatible
// In that case, you could just pass the variables without the address operator.
if (min > med) myswap(&min, &med);
if (med > max) myswap(&med, &max);
if (min > med) myswap(&min, &med);
``````
-
If you want all three (min, mid, max), just sorting the list is probably best. One comparison and swap to get the first two in order, then at most two tests to work out the order of the other two. – Alan Stokes May 4 at 11:04
Yes, although that would be a specific solution, and this solution will also work in C unlike std::swap. :P – lpapp May 4 at 11:05
True - although the general algorithm for finding the median (or nth_greatest) is based on quick sort, but skipping unnecessary work. – Alan Stokes May 4 at 11:06
@AlanStokes: sure, I agree. This version is optimized for readability and compreheniveness, while that one is for efficiency and flexibility. There is no efficiency problem here and the OP seems to be asking about three values, however, so I think both are valid. But yeah, sorting works gently for more than 3 elements, too; just need to make sure about odd numbers. – lpapp May 4 at 11:07
@AlanStokes: updated a swap sorting version (i.e. bubble sort) with C compatibility, just in case. :D – lpapp May 4 at 11:16
A variant of Alan's "cheat" that (kind of and sometimes) prevents overflow:
``````#include <iostream>
#include <algorithm>
using namespace std;
int main(int argc, char *argv[]) {
double a = 1e308;
double b = 6e306;
double c = 7.5e18;
double mn = min(a,min(b,c));
double mx = max(a,max(b,c));
double avg = mn + (mx-mn)*0.5;
double mid = a - avg + b - avg + c;
cout << mid << endl;
}
``````
Output:
``````6e+306
``````
It makes use of the avg-formula often used in binary search to prevent overflow:
The average of two values can be calculated as `low + (high-low)/2`
However, it only works for positive values. Possible fallbacks include Alan's answer, or simply `(x+y)/2` for the avg calculation.
Note that double precision comes into play here, and may cause issues in the `mid`-calculation. It works really well for positive integers though :)
-
The problem of not accessing the middle variable is still in there. – lpapp May 4 at 11:09
@LaszloPapp That wasn't part of the question. It was about finding the value – keyser May 4 at 11:25
No, it was not, but the readers can be made aware of the limitation of an algorithm. :) – lpapp May 4 at 11:28
It's not a limitation, it's a specification :p Finding the variable is a completely separate issue according to me. I mean, it's wrong to call it a problem. – keyser May 4 at 11:29
Note that Stack Overflow is used by many people and they might take algorithms from here into generic cases. Yes, you can say, "it is their problem", if they take it wrong, but preventing that is better IMHO. It is not to say, your answer is not usable, etc, of course. – lpapp May 4 at 11:30
The best way to do this is with a generic median function template. No copying, swapping or mathematical operations are required.
``````template <typename T>
const T& median(const T& a, const T& b, const T& c)
{
if (a < b)
if (b < c)
return b;
else if (a < c)
return c;
else
return a;
else if (a < c)
return a;
else if (b < c)
return c;
else
return b;
}
``````
-
"The best way" is actually unreadable with the if/elseif/else jungle for a non-performance critical code. :) I think it is better to be comprehensive in such cases than micro optimization. Also, this will not fly with more than three elements when you need to calculate the median for odd amount of numbers.. – lpapp May 4 at 11:58
imo it is not unreadable; it is elegant. You've probably just glanced at it for a few seconds and immediately declared it unreadable without putting enough thought into it. Also, he's only looking for the median of three values. – Ricky65 May 4 at 12:05
I was actually considering templates before, but then I decided not to mention them because IMHO it is more complex than needed. Also, I do not see any non-negligible performance gain out of it. It is basically the templatized version of my first solution with different if/elseif/else ordering, but I am not sure about the gain of template here compared to my non-template code. To me, it seems to make it more complex than necessary. One thing I agree about is that it is useful if you have different types to deal with, so it is flexible in that sense. – lpapp May 4 at 12:07 | 2,779 | 9,704 | {"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} | 3.5 | 4 | CC-MAIN-2014-52 | latest | en | 0.864936 |
http://www.jgrarchitect.com/2023/04/serlio-studies-temple.html | 1,726,602,850,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651829.57/warc/CC-MAIN-20240917172631-20240917202631-00710.warc.gz | 40,755,046 | 22,060 | ## Saturday, April 1, 2023
### Serlio Studies a Roman Temple
Sebastiano Serlio, 1475-1554, wrote 7 books 'On Architecture and Perspective' .
A contemporary of Palladio and Vignola, he spent much of his career in France working for King Francois I. The first part of his treatise was published in 1537.
Here he is with his compass.
The cover of Book I includes this drawing of builders' tools across the bottom, including a tetrahedron and a cube with diagonals, squares and circles on its face, in the right corner.*
What's the cube about? I didn't know, but I am beginning to find out.
Book III, On Antiquities includes the illustration and measured plan of this temple outside of Rome, now thought to be the Sepulchre of the Cercenni.
He writes that it was "built partly of brick, partly of marble and to a large extent ruinous."
To read the geometry look at
1) the square, its diagonals which mark the outside of the temple including the bays;
2) the circle which fits within the square which mark the corners of the temple itself;
3) the square which fits within the circle locating the outside of the walls.
Rotate the first square 90* to make an 8 pointed star. The intersections of the stars points mark the outside corners of the bays.
The inner square (barely visible in red) was not used.
Here is the how the master mason could have used geometry to lay out the plan on site.
The red square was probably the beginning. It is the foot print for the walls.Then the diagonals, the circle around it, and the next square were added. These set the depth of the bays.
Next the outer square was rotated. It crossed the large square at 8 points. Those points when joined laid out the width of the bays.
The lines set the perimeter of the plan. The mason had his foundation plan and could set his lines.
I usually find that the interior geometry of a masonry building is laid out from the inner side of the walls. This is practical: reaching over the walls with lines would not have been easy nor accurate.
The square and its circle neatly locate the columns which support the vaulting.
* Serlio is my favorite writer/architect from the Italian Renaissance. I have posted about him in this blog several times. The cover of his Book I included the tools he and his contemporaries used: www.jgrarchitect.com/2020/08/lesson-6-rule-of-thirds-part-2-serlio.html
Try this one too: https://www.jgrarchitect.com/2022/10/serlios-lines.html
His books are listed in my bibliography : https://www.jgrarchitect.com/2019/06/bibliography-includiung-websites.html
#### 1 comment:
Anonymous said...
I've enjoyed your explanations and research for years! Thank you | 630 | 2,677 | {"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-2024-38 | latest | en | 0.97829 |
https://lavelle.chem.ucla.edu/forum/search.php?author_id=13350&sr=posts | 1,604,197,619,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107922746.99/warc/CC-MAIN-20201101001251-20201101031251-00108.warc.gz | 387,036,958 | 14,098 | ## Search found 30 matches
Sun Mar 17, 2019 12:48 am
Forum: Arrhenius Equation, Activation Energies, Catalysts
Topic: K vs k1/k'1
Replies: 6
Views: 487
### Re: K vs k1/k'1
I do not think it really matters which one you write it as. It is however important to know that K is equal to k1/k'1 if you are only given the values of k1 and k'1 and not a value for K.
Thu Mar 14, 2019 9:50 pm
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: Hydronium
Replies: 2
Views: 118
### Re: Hydronium
If there are hydronium ions in your balanced redox reaction, they should be included in your cell diagram.
Mon Mar 11, 2019 12:43 am
Forum: Zero Order Reactions
Topic: Integration/Derivation
Replies: 3
Views: 285
### Re: Integration/Derivation
I do not believe that derivation/integration would be asked of us, simply how to apply the derived formulas to problems.
Mon Mar 11, 2019 12:41 am
Forum: Gibbs Free Energy Concepts and Calculations
Topic: Spontaneous Reaction
Replies: 5
Views: 406
### Re: Spontaneous Reaction
I am also confused. I understand that delta H would be negative if you are freezing and thus losing heat, but wouldn't delta S also be negative if you are losing entropy going from a liquid to a solid? Thanks in advance!
Mon Mar 11, 2019 12:07 am
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: Cell Potential
Replies: 2
Views: 104
### Re: Cell Potential
You would want to use a cathode with the weakest reducing power (strongest oxidizing power) and an anode with the weakest oxidizing power (strongest reducing power).
Mon Mar 11, 2019 12:04 am
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: Cell Diagram
Replies: 3
Views: 152
### Re: Cell Diagram
H+ is included in the cell diagram, but H2O (l) and electrons are not.
Sun Mar 10, 2019 11:17 pm
Forum: First Order Reactions
Topic: Half-Life
Replies: 7
Views: 243
### Re: Half-Life
Typically we would use the half-life formula to determine how long an element has been decaying.
Sun Mar 10, 2019 11:08 pm
Forum: General Rate Laws
Topic: Rate Laws
Replies: 8
Views: 261
### Re: Rate Laws
I think he shows us the derivation to establish some kind of fundamentals, but it really just confuses the [censored] out of me. I don't think we would ever be asked to derive them on a test, just to apply them to problems similar to homework and tests.
Sat Mar 09, 2019 4:50 pm
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: 6M. 1 7th Edition
Replies: 2
Views: 99
### Re: 6M. 1 7th Edition
This sounds like the question is just trying to be tricky. We were taught to write anode on the left and cathode on the right.
Sat Mar 09, 2019 4:47 pm
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: Calculating E naught
Replies: 5
Views: 249
### Re: Calculating E naught
Given the reduction half reactions, you will use the appendix to find E naught values. The lower E naught value is the anode because it is the stronger reducing agent. Then, to get E naught of the cell you do cathode - anode = E naught cell
Sat Mar 09, 2019 4:05 pm
Forum: Method of Initial Rates (To Determine n and k)
Topic: negative and k
Replies: 3
Views: 155
### Re: negative and k
K cannot be negative because you can't have a negative concentration
Wed Feb 27, 2019 8:30 pm
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: Galvanic Cell Diagram [ENDORSED]
Replies: 3
Views: 216
### Re: Galvanic Cell Diagram[ENDORSED]
Molecules of the same phase in the cell diagram are written in the order: oxidized, reduced
Sun Feb 24, 2019 1:41 pm
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: Oxidation and Reduction
Replies: 3
Views: 120
### Re: Oxidation and Reduction
I think the most important thing is that the electrons are on opposite sides in reduction vs oxidation half reactions, but given that that is how we did it in class I assume that is how we should do it on an exam.
Sun Feb 24, 2019 1:38 pm
Forum: Work, Gibbs Free Energy, Cell (Redox) Potentials
Topic: n value
Replies: 11
Views: 256
### Re: n value
n is moles so I assume it will either be a given value or we will need to convert to moles from a given mass of a compound or element.
Sun Feb 24, 2019 1:37 pm
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: Spontaneous cell reaction
Replies: 2
Views: 102
### Re: Spontaneous cell reaction
Because the equation is delta G = -nFE, a positive cell potential (or E value) results in a negative delta G, and thus a spontaneous reaction.
Mon Feb 18, 2019 5:53 pm
Forum: Gibbs Free Energy Concepts and Calculations
Topic: Gibbs Free Energy
Replies: 4
Views: 161
### Re: Gibbs Free Energy
K is the equilibrium constant. Concentration of products at equilibrium over the concentration of reactants at equilibrium.
Mon Feb 18, 2019 5:52 pm
Forum: Van't Hoff Equation
Topic: Van't Hoff Equation
Replies: 4
Views: 163
### Re: Van't Hoff Equation
I'm pretty sure R is the ideal gas constant
Mon Feb 18, 2019 5:43 pm
Forum: Entropy Changes Due to Changes in Volume and Temperature
Topic: Entropy
Replies: 3
Views: 118
### Re: Entropy
Entropy is a state function and thus increases or decreases in a phase change. As said above, solids have the lowest entropy, then liquid, and gases have the highest entropy.
Sun Feb 03, 2019 10:39 pm
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: Open vs Closed System
Replies: 13
Views: 1057
### Re: Open vs Closed System
In a closed system, only energy can exchange with surroundings. An example is a sealed beaker of water. You cannot change the volume but you can heat the container because it isn't insulated and thus change the energy.
Sun Feb 03, 2019 10:27 pm
Forum: Calculating Work of Expansion
Topic: Work sign changes
Replies: 3
Views: 163
### Re: Work sign changes
I believe this is because the formula is w = - P (delta V). If it is expansion, delta V will be positive and thus work negative. In compression delta V is negative and thus work is positive.
Sun Feb 03, 2019 10:04 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Quiz 1 Number 3c
Replies: 6
Views: 237
### Re: Quiz 1 Number 3c
The units should be mol/L (Molarity) because it's asking for the concentration.
Sun Jan 27, 2019 5:21 pm
Forum: Ideal Gases
Topic: Calculating partial pressure
Replies: 4
Views: 185
### Re: Calculating partial pressure
You may be able to if you are given a concentration: n/V, or moles/liter.
Sun Jan 27, 2019 3:36 pm
Forum: Ideal Gases
Topic: Approximations for ICE
Replies: 20
Views: 531
### Re: Approximations for ICE
You can use approximations if the Ka value is less than 1x10^-3. You should check to see that it was appropriate to use approximation by seeing if the equilibrium concentration is less than 5% of the initial concentration.
Sun Jan 27, 2019 3:31 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: pKa values
Replies: 3
Views: 156
### Re: pKa values
I'm not sure if there is an index of these values. They should be given to you by problem or the problem may tell you to look at a table to find the pKa or Ka values. Also if you are given the Ka you know the pKa and vice versa.
Sun Jan 20, 2019 4:10 pm
Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
Topic: Bars vs. Molarity
Replies: 5
Views: 209
### Re: Bars vs. Molarity
bars are a unit of pressure just like atm or TORR
Sun Jan 20, 2019 4:06 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Autoprotolysis
Replies: 3
Views: 94
### Re: Autoprotolysis
The equation is 2H2O → H3O + + OH-. The significance is that shows that water is amphoteric (able to react as an acid and a base).
Sun Jan 20, 2019 3:33 pm
Forum: Calculating pH or pOH for Strong & Weak Acids & Bases
Topic: HW Q
Replies: 2
Views: 72
### Re: HW Q
You have to dilute the solution by multiplying .025M(200ml/250ml). Then, take the negative log of that new concentration and you should get the correct pH.
Sun Jan 13, 2019 3:11 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Equations
Replies: 6
Views: 227
### Re: Equations
Kp = Kc x (RT) ^ stoichiometric coefficient
In general, Kp uses partial pressures and Kc uses concentrations
Sun Jan 13, 2019 2:19 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Difference between Kc, Kp, and K
Replies: 2
Views: 123
### Re: Difference between Kc, Kp, and K
Kc: concentration of products over concentration of reactants when at equilibrium
Kp: partial pressures of products over partial pressures of reactants when at equilibrium
Sun Jan 13, 2019 2:12 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: ICE table question
Replies: 2
Views: 85
### Re: ICE table question
So far we have typically used ICE tables to calculate equilibrium concentrations of the products and reactants when given a Kc and initial concentrations. | 2,633 | 9,165 | {"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-2020-45 | latest | en | 0.873969 |
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