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https://oeis.org/A336837 | 1,726,602,487,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651829.57/warc/CC-MAIN-20240917172631-20240917202631-00762.warc.gz | 406,747,414 | 4,303 | The OEIS is supported by the many generous donors to the OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A336837 Numerator of ratio A336841(n)/A000005(n). 4
0, 1, 2, 14, 3, 9, 5, 17, 44, 13, 6, 32, 8, 21, 23, 284, 9, 163, 11, 137, 37, 25, 14, 105, 30, 33, 86, 73, 15, 81, 18, 547, 44, 37, 53, 1622, 20, 45, 58, 149, 21, 129, 23, 260, 401, 57, 26, 1662, 230, 109, 65, 114, 29, 297, 63, 237, 79, 61, 30, 263, 33, 73, 213, 4010, 83, 153, 35, 383, 100, 183, 36, 1715, 39, 81 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,3 LINKS Antti Karttunen, Table of n, a(n) for n = 1..16384 Antti Karttunen, Data supplement: n, a(n) computed for n = 1..65537 Index entries for sequences computed from indices in prime factorization FORMULA a(n) = A336841(n) / A336856(n) = A336841(n) / gcd(A000005(n), A336841(n)). PROG (PARI) A003961(n) = { my(f = factor(n)); for(i=1, #f~, f[i, 1] = nextprime(f[i, 1]+1)); factorback(f); }; A336841(n) = ((numdiv(n)*A003961(n)) - sigma(A003961(n))); A336837(n) = numerator(A336841(n)/numdiv(n)); CROSSREFS Cf. A000005, A003961, A003973, A336838, A336841, A336856. Cf. A336839 (denominators). Sequence in context: A227628 A298800 A138907 * A276189 A281944 A306724 Adjacent sequences: A336834 A336835 A336836 * A336838 A336839 A336840 KEYWORD nonn,frac AUTHOR Antti Karttunen, Aug 07 2020 STATUS approved
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Last modified September 17 15:47 EDT 2024. Contains 375987 sequences. (Running on oeis4.) | 677 | 1,725 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2024-38 | latest | en | 0.545112 |
http://learninglover.com/blog/tag/compliment/ | 1,669,769,080,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710712.51/warc/CC-MAIN-20221129232448-20221130022448-00254.warc.gz | 30,021,262 | 12,919 | # Probability: Sample Spaces
I’ve been doing a few games lately (can be seen here, here and here) and, while I think those are very good ways to become interested in some of the avenues of math research, I also have had a few people come to me with questions regarding help with their classes. So I decided to write a script to try to help understand some elementary probability theory, focusing on discrete sample spaces.
In statistics, any process of observation is referred to as an experiment.
The set of all possible outcomes of an experiment is called the sample space and it is usually denoted by S. Each outcome in a sample space is called an element of the sample space. An event is a subset of the sample space or which the event occurs. Two events are said to be mutually exclusive if they have no elements in common.
Similar to set theory, we can form new events by performing operations like unions, intersections and compliments on other events. If A and B are any two subsets of a sample space S, then their union A ∪ B is the subset of S that contains all the elements that are in either A, in B, or in both; their intersection A ∩ B is the subset of S that contains all the elements that are in both A and B; the compliment A’ of A is the subset of S that contains all the elements of S that are not in A.
A probability is a function that assigns real numbers to events of a sample space. The following are the axioms of probability that apply when the sample space is discrete (finite or countable).
Axiom 1: The probability of an event is a non-negative real number; that is P(A) ≥ 0 for any subset A of S.
Axiom 2: The probability of the entire sample space is 1; that is P(S) = 1.
Axiom 3: If A1, A2, A3, … , is a finite or infinite sequence of mutually exclusive events of S, then
P(A1 ∪ A2 ∪ A3 ∪ …) = P(A1) + P(A2) + P(A3) + …
If A and B are any two events in a sample space S and P(A) ≠ 0, the conditional probability of B given A is
P(B | A) =
P(A ∩ B)P(A)
Two events A and B are independent if and only if P(A | B) = P(A) ∙ P(B).
# Learning Math through Set Theory
In grade school, we’re taught that math is about numbers. When we get to college (the ones of us who are still interested in math), we’re taught that mathematics is about sets, operations on sets and properties of those sets.
Understanding Set Theory is fundamental to understanding advanced mathematics. Iv wrote these scripts so that users could begin to play with the different set operations that are taught in a basic set theory course. Here, the sets are limited to positive integers and we’re only looking at a few operations, in particular the union, intersection, difference, symmetric difference, and cross product of two sets. I will explain what each of these is below.
The union of the sets S1 and S2 is the set S1 [union] S2, which contains the elements that are in S1 or S2 (or in both).
Note: S1 [union] S2 is the same as S2 [union] S1.
The intersection of the sets S1 and S2 is the set S1 [intersect] S2, which contains the elements that are in BOTH S1 and S2.
Note: S1 [intersect] S2 is the same as S2 [intersect] S1.
The difference between the sets S1 and S2 is the set S1 / S2, which contains the elements that are in S1 and not in S2.
. Note. S1 / S2 IS NOT the same as S2 / S1.
Note. S1 / S2 is the same as S1 [intersect] [not]S2.
The symmetric difference between the sets S1 and S2 is the set S1 [symm diff] S2, which contains the elements that are in S1 and not in S2, or the elements that are in S2 and not in S1.
Note. S1 [symm diff] S2 is the same as S2 [symm diff] S1.
Note. S1 [symm diff] S2 is the same as (S1 [intersect] [not] S2) [union] (S2 [intersect] [not] S1).
The cartesian product of the two sets S1 and S2 is the set of all ordered pairs (a, b), where a [in] S1 and b [in] S2. | 1,006 | 3,823 | {"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-2022-49 | longest | en | 0.965503 |
https://community.oracle.com/thread/1205434?tstart=73545&messageID=4965963 | 1,394,254,597,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1393999653202/warc/CC-MAIN-20140305060733-00080-ip-10-183-142-35.ec2.internal.warc.gz | 626,810,368 | 36,026 | This discussion is archived
5 Replies Latest reply: Nov 18, 2007 6:06 AM by 807600
# Could someone help me with arrays & recursion??
Currently Being Moderated
Hi, I'm new here and I'm having trouble.
I have to write a java application which uses recursion to find the maximal contiguous sum in a list of integers.
The input is read from a text file where each line represents a list containing the following integers:
For example, it might contain
7 -2 -4 30 15 -7 -5 1000
2 -50 100
6 1000 -2000 900 2800 -2900 2801
4 100 -10 5 8
4 100 -50 5 8
And for the output, for each line of data read, I have to display the largest sum of consecutive integers in the list followed by the list itself. For example for the lists above, my output would be:
(on the left side is the largest sum of consecutive integers and on the right is the list of integers)
1033
• ###### 1. Re: Could someone help me with arrays & recursion??
Currently Being Moderated
Please don't double post. Since this is the smaller post, let's consider this one dead.
• ###### 2. Re: Could someone help me with arrays & recursion??
Currently Being Moderated
I'm sorry about that. I didn't mean to double post. I accidently clicked "post" before I finished typing.
• ###### 3. Re: Could someone help me with arrays & recursion??
Currently Being Moderated
R.S. wrote:
Hi, I'm new here and I'm having trouble.
I have to write a java application which uses recursion to find the maximal contiguous sum in a list of integers.
The input is read from a text file where each line represents a list containing the following integers:
For example, it might contain
7 -2 -4 30 15 -7 -5 1000
2 -50 100
6 1000 -2000 900 2800 -2900 2801
4 100 -10 5 8
4 100 -50 5 8
And for the output, for each line of data read, I have to display the largest sum of consecutive integers in the list followed by the list itself. For example for the lists above, my output would be:
(on the left side is the largest sum of consecutive integers and on the right is the list of integers)
1033
Where did this 1033 come from? Could you explain your problem in a bit more detail and also post what you have done so far and explain where you are stuck: only posting the problem description sound a bit like "can some one do it for me", it's better to ask a specific question here.
• ###### 4. Re: Could someone help me with arrays & recursion??
Currently Being Moderated | 626 | 2,411 | {"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-2014-10 | latest | en | 0.905966 |
https://fixanswer.com/will-your-shadow-be-long-or-short-in-the-early-morning/ | 1,686,253,284,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224655092.36/warc/CC-MAIN-20230608172023-20230608202023-00378.warc.gz | 293,933,787 | 22,452 | # Will Your Shadow Be Long Or Short In The Early Morning?
Shadows are longest in the early morning and late afternoon/early evening when the sun appears low in the sky. As the Earth rotates on its axis, the sun hits each location in the morning at an angle. This becomes more vertical as the sun appears to pass more directly overhead around noon.
The shortest shadow occurs when the sun reaches its highest point, at local noon. By tracking the length of the shadow cast by a stationary gnomon throughout the day, it is easy to determine the point at which the shadow is shortest.
The Sun’s position in the sky affects the length of the shadow. When the Sun is low on the horizon, the shadows are long. When the Sun is high in the sky, the shadows are much shorter.
## What is the length of shadow in early morning?
Shadows are longer in the morning and evening and short in the noon because the sun rays are slanting in the morning and evening whereas at noon it is just above our head. The size of the shadow depends on the position of the object from the source of light.
## At what time shadow is longest?
Shadows are longest in the early morning and late afternoon/early evening when the sun appears low in the sky. As the Earth rotates on its axis, the sun hits each location in the morning at an angle. This becomes more vertical as the sun appears to pass more directly overhead around noon.
## How long is my shadow at noon?
on June 21, 2011, at noon the sun’s altitude is 70.9 degrees. Rewrite the following formula with the numerical equivalents: Object Height / tan θ = Shadow Length. For example, for the 790-foot high Prudential Tower in Boston, the formula would be 790 / 2.89 = Shadow Length.
## At what point during the year is the person’s shadow the shortest?
One of those aspects is the varying length of shadows over the span of the year. As the Sun moves through the zodiac constellations over the annual cycle of the seasons, the Sun moves to its highest point overhead in the sky on the summer solstice. On this day, the noon shadows are the shortest.
## How long is a person’s shadow?
The length of shadow to your height is proportional to 1/Tangent (sun’s altitude). If the sun is low in the sky (10 degrees), your shadow would be 5.67 times as long as your height. The corresponding ratio at 5 degrees is 11.43 . (So an average height person (5.8 feet) would have a 66 foot long shadow).
## Do you have a shadow at noon?
At exactly noon, the stick’s shadow (or your shadow) will point due north because the sun is due south. So even with a digital watch, you can find north at noon. At any other time of day, a shadow is at some angle away from north. The size of the angle changes with the motion of the sun.
## Where will the shadow be in the morning?
In morning, sun rises in the east and so any shadow falls towards the west.
## In which direction will the shadow of the stick be in the morning?
Explanation: n the Northern Hemisphere, they will move from west to east, and will point north at noon. In the Southern Hemisphere, shadows will indicate south at noon. With practice, you can use shadows to determine both direction and time of day.
## Why our shadow is tallest in the morning?
Answer: Our shadow is the tallest in the morning. It is because the sun rays are slanting at the time of sunrise. | 751 | 3,356 | {"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} | 4.03125 | 4 | CC-MAIN-2023-23 | longest | en | 0.937228 |
http://reu.dimacs.rutgers.edu/~claryka/ | 1,516,482,143,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084889733.57/warc/CC-MAIN-20180120201828-20180120221828-00281.warc.gz | 276,199,836 | 2,641 | ### General Information
Student: Kaleigh Clary
Office: CoRE 434
School: Hendrix College
E-mail: claryka@hendrix.edu
Project: Forbidden Subgraphs of Competition Graphs
Mentor: Dr. Gene Fiorini
### Project Description
Competition graphs arose from analysis of food webs and have applications in biology and network systems. The competition graph C(D) of a digraph D = (V,E) has the same vertex set as D, and two distinct vertices x, y are adjacent in C(D) if there is some vertex zV (possibly x or y) such that xz,yzE.
We define a digraph of a relation on ℝn where each element in the set has a vertex there is an edge from
(a1, a2, ..., an) to (b1, b2, ..., bn) if and only if ai > bii.
We are interested in finding forbidden subgraphs of competition graphs of this relation on ℝ3.
### Weekly Log
Week 1:
I selected a problem and reviewed the relevant literature. At the end of the week, I gave a presentation which can be found at the bottom of this page.
Week 2:
I continued a literature review and wrote a Maple script to generate competition graphs of our relation. I also began working on a proof showing that some claw graphs are forbidden in competition graphs of our relation.
Week 3:
I completed a proof showing subgraphs of maximum degree n-1 are forbidden in competition graphs of our relation and started examining subgraphs of competition graphs of this relation with maximum degree n-2. I also looked at the conditions necessary for claw subgraphs of competition graphs of this relation.
Week 4:
I completed a proof showing there are no forbidden subgraphs of maximum degree n-2 in competition graphs of our relation. I then looked at possible restrictions on the digraphs of our relation or the competition graphs thereof to examine forbidden subgraphs of the competition graphs under those restrictions.
Week 5:
I investigated the conditions necessary for split subgraphs to exist in competition graphs our our relation, and I started a writeup of my review of the literature in weeks 1 and 2. I also modified the Maple script to allow restrictions of D.
Week 6:
After consulting with Dr. Brian Nakamura, I looked at a specific restriction of the digraphs of our relation. Suppose you have two different lines through the origin, and suppose you distribute n points along these two lines. We want to determine properties of competition graphs that exist under these constraints. I wrote a second Maple script to examine the digraphs and competition graphs in this restriction and identified specific cases where we can describe the competition graphs.
Week 7:
I showed that there are no forbidden subgraphs of maximum degree n-2 in competition graphs of our relation in the restricted domain with two lines in ℝ2. Now we are looking at extending this to three dimensions. I also gave a presentation over my work this summer, which can be found at the bottom of this page. | 629 | 2,893 | {"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-2018-05 | latest | en | 0.945542 |
https://techcommunity.microsoft.com/t5/excel/excel-formula-help/td-p/352351 | 1,582,079,696,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875143963.79/warc/CC-MAIN-20200219000604-20200219030604-00263.warc.gz | 584,074,510 | 61,220 | • 543K Members
• 3,242 Online
• 647K Conversations
## Excel formula help
Highlighted
Occasional Visitor
# Excel formula help
Hello, I am learning as I go a Excel. I hope you all can help me, as this is out of my range of understanding and not sure if this is possible or where to start. Here is an example of my data in Excel.
I am checking for accuracy here. Column C contains receipt numbers. E:8 for example has a numeric value of 10 and is highlighted in green and is associated with receipt 337068. Basically how this works is I have to take each receipt number and add the values associated with it then add \$17.22. This must equal the new act plan payment also highlighted in green. In this case we have 10 + 17.22 must equal the new activation plan payment of \$27.22. In this case it does and this receipt is good. Moving down to 337290. As you can see this one has an add on of \$6. So we have E16 + E18 + \$17.22 must equal G19 (6+50+\$17.22 must equal \$73.22. And again in this case it does, so I need nothing. What I am wondering is if there is a way to have the program return to me the receipt numbers of the ones that DO NOT equal. This is something I do for the company and can be very time consuming. Any help would be greatly appreciated. Thanks
Highlighted
# Re: Excel formula help
I suggest that you modify the layout of your data so that each column contains only one type of data. For example, your column labels might be Receipt Number, Plan Type, Plan Description, Handset Price, Activation Price, Markup, and Total.
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IFS with AND and multiple falses | 448 | 1,767 | {"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-2020-10 | latest | en | 0.901276 |
https://metanumbers.com/1522290000000 | 1,601,397,773,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400202418.22/warc/CC-MAIN-20200929154729-20200929184729-00478.warc.gz | 505,156,772 | 8,171 | ## 1522290000000
1,522,290,000,000 (one trillion five hundred twenty-two billion two hundred ninety million) is an even thirteen-digits composite number following 1522289999999 and preceding 1522290000001. In scientific notation, it is written as 1.52229 × 1012. The sum of its digits is 21. It has a total of 18 prime factors and 1024 positive divisors. There are 315,840,000,000 positive integers (up to 1522290000000) that are relatively prime to 1522290000000.
## Basic properties
• Is Prime? No
• Number parity Even
• Number length 13
• Sum of Digits 21
• Digital Root 3
## Name
Short name 1 trillion 522 billion 290 million one trillion five hundred twenty-two billion two hundred ninety million
## Notation
Scientific notation 1.52229 × 1012 1.52229 × 1012
## Prime Factorization of 1522290000000
Prime Factorization 27 × 3 × 57 × 7 × 11 × 659
Composite number
Distinct Factors Total Factors Radical ω(n) 6 Total number of distinct prime factors Ω(n) 18 Total number of prime factors rad(n) 1522290 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 0 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 1,522,290,000,000 is 27 × 3 × 57 × 7 × 11 × 659. Since it has a total of 18 prime factors, 1,522,290,000,000 is a composite number.
## Divisors of 1522290000000
1024 divisors
Even divisors 896 128 64 64
Total Divisors Sum of Divisors Aliquot Sum τ(n) 1024 Total number of the positive divisors of n σ(n) 6.31123e+12 Sum of all the positive divisors of n s(n) 4.78894e+12 Sum of the proper positive divisors of n A(n) 6.16331e+09 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 1.23381e+06 Returns the nth root of the product of n divisors H(n) 246.992 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 1,522,290,000,000 can be divided by 1024 positive divisors (out of which 896 are even, and 128 are odd). The sum of these divisors (counting 1,522,290,000,000) is 6,311,233,843,200, the average is 6,163,314,300.
## Other Arithmetic Functions (n = 1522290000000)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 315840000000 Total number of positive integers not greater than n that are coprime to n λ(n) 987000000 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 56334265281 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 315,840,000,000 positive integers (less than 1,522,290,000,000) that are coprime with 1,522,290,000,000. And there are approximately 56,334,265,281 prime numbers less than or equal to 1,522,290,000,000.
## Divisibility of 1522290000000
m n mod m 2 3 4 5 6 7 8 9 0 0 0 0 0 0 0 3
The number 1,522,290,000,000 is divisible by 2, 3, 4, 5, 6, 7 and 8.
• Arithmetic
• Abundant
• Polite
• Practical
• Frugal
## Base conversion (1522290000000)
Base System Value
2 Binary 10110001001101111100011100000000010000000
3 Ternary 12101112022000012110102010
4 Quaternary 112021233203200002000
5 Quinary 144420122220000000
6 Senary 3123155205452520
8 Octal 26115743400200
10 Decimal 1522290000000
12 Duodecimal 207043a11140
20 Vigesimal 2j95fca000
36 Base36 jfbw5thc
## Basic calculations (n = 1522290000000)
### Multiplication
n×i
n×2 3044580000000 4566870000000 6089160000000 7611450000000
### Division
ni
n⁄2 7.61145e+11 5.0743e+11 3.80572e+11 3.04458e+11
### Exponentiation
ni
n2 2317366844100000000000000 3527704373104989000000000000000000000 5370189090133993704810000000000000000000000000000 8174985150020077276895214900000000000000000000000000000000000
### Nth Root
i√n
2√n 1.23381e+06 11503.6 1110.77 273.212
## 1522290000000 as geometric shapes
### Circle
Diameter 3.04458e+12 9.56483e+12 7.28022e+24
### Sphere
Volume 1.47768e+37 2.91209e+25 9.56483e+12
### Square
Length = n
Perimeter 6.08916e+12 2.31737e+24 2.15284e+12
### Cube
Length = n
Surface area 1.39042e+25 3.5277e+36 2.63668e+12
### Equilateral Triangle
Length = n
Perimeter 4.56687e+12 1.00345e+24 1.31834e+12
### Triangular Pyramid
Length = n
Surface area 4.0138e+24 4.15744e+35 1.24294e+12
## Cryptographic Hash Functions
md5 6de9a33b192ced3b28a362e3bbef48b1 0cab284c1b01ef021f74da906258254beb1b6cfc c55b5dab9f1454cd0950c830cf8263668c6b715c2a5483c49f801ab1fe72be2d 15db830a5f6aed98486da06a44199150d9711ff258df3bb5c722cd0f4d159dbca52d84bbc55d45deccb726660f1a59367a245fdf54cd098c0d9288db90f2188a bd68bbc44a396ed1084153b18f63a80e1d74871f | 1,751 | 4,799 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2020-40 | latest | en | 0.768507 |
https://www.physicsforums.com/threads/rise-in-temperature-after-a-collision.680712/ | 1,508,227,947,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187820930.11/warc/CC-MAIN-20171017072323-20171017092323-00199.warc.gz | 961,194,667 | 16,998 | # Rise in temperature after a collision
1. Mar 24, 2013
### runningninja
1. The problem statement, all variables and given/known data
"A body of mass m, moving with velocity v, collides with a body of mass 2m at rest, in a head-on collision. The coefficient of restitution is 1/3. If the 2m body has a specific heat c, and if it is assumed that the two bodies share the heat generated in the collision equally (not a very reasonable assumption), and that no heat is lost (a ridiculous assumption), how much does the temperature of the 2m body rise? (Keep fractions throughout in solution.)"
2. Relevant equations
$$\text{Coefficient of restitution}\ =~e~= \frac{v_{2F} - v_{1F}}{v_{1I} - v_{2I}}$$
$$p_I = p_F$$
$$\Delta K_2 = Q_2$$
3. The attempt at a solution
Since it is head on, this is a one dimensional problem, with $v_{2I} = 0$ and $v_{1I} = v$. I started off by resolving my momentum equation into
$$mv = -mv_{1F} + 2mv_{2F}$$
(1) $v_{1F} = 2v_{2f} - v$
I then set my restitution equation equal to 1/3 and substituted (1) in.
$$\frac{1}{3}\ = \frac{v_{2f} - 2v_{2f} + v}{v}$$
Which simplifies to
(2) $$v_{2F} = \frac{2v}{3}$$
I then expand my Q equation to
(3) $$\frac{1}{2}\ 2 m (v_{2f})^2 = 2 m c \Delta T$$
I then plug (2) into (3) and solve for $\Delta T$ :
$$\Delta T\ = \frac{2v^2}{9c}$$
However, the answer has a fraction of 2/27 rather than 2/9. I'm off by a factor of a third, and I don't know why.
Last edited: Mar 24, 2013
2. Mar 25, 2013
### voko
Why the minus sign?
3. Mar 25, 2013
### runningninja
Because it is a one dimensional collision. The first object has to bounce backwards because there is no other place for it to go. Even if I omit the minus sign, I get 8/81 for the fraction.
4. Mar 25, 2013
### voko
It does not have to go backward. It can continue forward at a reduced speed. You do not have to guess what is going to happen. Use the equations.
Then you must have made another error somewhere. Show your new derivation. | 640 | 1,976 | {"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": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2017-43 | longest | en | 0.90756 |
https://www.markedbyteachers.com/as-and-a-level/science/heating-effects-of-a-current.html | 1,701,426,632,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100286.10/warc/CC-MAIN-20231201084429-20231201114429-00886.warc.gz | 969,065,291 | 11,714 | # Heating effects of a current
Heating effects of a current
Aim- My aim is to see how the increase and decrease of the current effects the temperature of the water
Equipment List-
• Wires
• Beaker
• Water
• Variable Resistor
• Power Pack
• Ammeter
• Crocodile Clips
• Thermometer
• Goggles
Fair Test- I will use an equal quantity of water each time. I will make sure the equipment is working properly. I will start the experiment at the same temperature every time.
Pre Test- I am going to do a pre test to decide what voltage to use in the experiment. I will try 4 volts, 6 volts, 8 volts and 10 volts. The voltage was most effective at 10 volts because that’s were the temperature rose the most.
Method- I will gather the equipment on my list. Connect the equipment together making sure it is set up in parallel. I will fill the beaker with a 100ml of water and put the wire into it. I will then switch on the experiment. I will note down the starting temperature and the amp reading. I will then finish after 10 minutes, noting down the temperature. I will then figure out the temperature change. Then I will strengthen the variable resistor and do the procedure. I will do the experiment with the variable resistor at 5 different levels. I will then repeat this whole experiment again so I know my results are accurate.
Prediction- I predict that when the electric current runs through the wire, it will heat up. I think this because I know that the electrons will collide creating friction and therefore heat. The heat is then transferred through convection currents. Convection happens because the water that is heated becomes less dense, this is because the particles are moving a lot more. Then the water that is less dense spreads by making the rest of the water particles vibrate. I predict that the larger the current the larger the temperature will be. This is because as I said above the current will make the electrons collide, and the larger the current the faster they will move so the more they will collide so the larger the heat. I predict that the higher the resistant the more the temperature of the water will rise. This is because if the resistant is low the current will pass through the top part of the series circuit because it is a shorter distance round there and electricity always takes the shortest route. But if the resistance is high it will be easier for the current to go through the bottom part of the circuit because there is less resistance there. This will therefore mean that the wire will heat up because it will have more current running through it. If the wire heats up this will mean the water will heat up because it will conduct the heat from the wire. This is like a smaller version of an immersion heater. The wire in the experiment compares to the pipe, and when the pipe heats up the water around it heats up. I predict the graph will go toward upwards diagonally right because as the current rises so will the change in temperature.
Results-
Average results
Analysing
The evidence shows that as I predicted that the larger the current the more the temperature will rise. It also shows that the higher the resistance the more the temperature rose. I no this because the ammeter is set up in parallel to the resistor so if the current is higher on the ammeter then it must mean there is less current flowing through the resistor. This also means the current is flowing into the water.
My graph shows that the more the current increases the more ... | 728 | 3,510 | {"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-2023-50 | latest | en | 0.937238 |
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# mbvoyager
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https://de.mathworks.com/matlabcentral/answers/1815595-to-write-a-for-loop-in-which-a-variable-is-updated-with-each-run | 1,670,554,517,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711376.47/warc/CC-MAIN-20221209011720-20221209041720-00798.warc.gz | 218,789,935 | 31,529 | # To write a For loop in which a variable is updated with each run.
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Andrei_43214 on 1 Oct 2022
Answered: Stephen23 on 1 Oct 2022
I want to write a script file that generates the Cartesian coordinates of the vertices of a regular Octadecagon (18-sided regular polygon) centered at the origin and with a radius, r = 1. The coordinates can be obtained by rotating repeatedly by the angle . This can be done by matrix multiplication of
by the coordinates of a vertex of the polygon as a column vector.The first vertex is the following:
Matrix multiplication ... until will give the coordinates of all 18 vertices. I've been trying to do this in a for loop which would ideally give me the 18 coordinates as an 18x2 array but I am still very new to coding and I'm not managing to do it. I have been trying to do this so far:
R = [cos(pi/9) -sin(pi/9); sin(pi/9) cos(pi/9)];
v_1 = [sin(pi/18); -cos(pi/18)];
A = [v_1'];
for r = 1:18
column = num2str(r);
v_column = R*v_column;
v_column+('1') = v_column';
A = [A;v_column+('1')];
end
I am unable to let Matlab know that I want to update v_column to the next integer with each run. Any help/tips would be greatly appreciated.
Basically, all I want is to make the following piece of code more efficient by using a for loop (if it can be done).
R = [cos(pi/9) -sin(pi/9); sin(pi/9) cos(pi/9)];
% For i = 1,...,18: v_i = (x_i,y_i) %
v_1 = [sin(pi/18); -cos(pi/18)];
v_2 = R*v_1;
v_3 = R*v_2;
v_4 = R*v_3;
v_5 = R*v_4;
v_6 = R*v_5;
v_7 = R*v_6;
v_8 = R*v_7;
v_9 = R*v_8;
v_10 = R*v_9;
v_11 = R*v_10;
v_12 = R*v_11;
v_13 = R*v_12;
v_14 = R*v_13;
v_15 = R*v_14;
v_16 = R*v_15;
v_17 = R*v_16;
v_18 = R*v_17;
A = [v_1';v_2';v_3';v_4';v_5';v_6';v_7';v_8';v_9';v_10';v_11';v_12';v_13';v_14';v_15';v_16';v_17';v_18']
##### 2 CommentsShowHide 1 older comment
Andrei_43214 on 1 Oct 2022
I don't want to plot the octadecagon; I just want Matlab to generate the 18 (x,y) coordinate points of the 18 vertices of the regular octadecagon as an 18x2 array where the first column denotes the x-coordinates and the second column denotes the y-coordinates, respectively. Consider the following the piece of code. I want to make this more efficient by using a for loop (if it can be done).
R = [cos(pi/9) -sin(pi/9); sin(pi/9) cos(pi/9)];
% For i = 1,...,18: v_i = (x_i,y_i) %
v_1 = [sin(pi/18); -cos(pi/18)];
v_2 = R*v_1;
v_3 = R*v_2;
v_4 = R*v_3;
v_5 = R*v_4;
v_6 = R*v_5;
v_7 = R*v_6;
v_8 = R*v_7;
v_9 = R*v_8;
v_10 = R*v_9;
v_11 = R*v_10;
v_12 = R*v_11;
v_13 = R*v_12;
v_14 = R*v_13;
v_15 = R*v_14;
v_16 = R*v_15;
v_17 = R*v_16;
v_18 = R*v_17;
A = [v_1';v_2';v_3';v_4';v_5';v_6';v_7';v_8';v_9';v_10';v_11';v_12';v_13';v_14';v_15';v_16';v_17';v_18']
A = 18×2
0.1736 -0.9848 0.5000 -0.8660 0.7660 -0.6428 0.9397 -0.3420 1.0000 -0.0000 0.9397 0.3420 0.7660 0.6428 0.5000 0.8660 0.1736 0.9848 -0.1736 0.9848
Stephen23 on 1 Oct 2022
"I want to make this more efficient by using a for loop (if it can be done)."
Of course it can be done, you just need to avoid numbering variable names like that.
Here is a basic MATLAB approach using one matrix and indexing:
R = [cos(pi/9) -sin(pi/9); sin(pi/9) cos(pi/9)];
A = [sin(pi/18), -cos(pi/18); nan(17,2)];
for k = 2:18
A(k,:) = R * A(k-1,:).';
end
display(A)
A = 18×2
0.1736 -0.9848 0.5000 -0.8660 0.7660 -0.6428 0.9397 -0.3420 1.0000 -0.0000 0.9397 0.3420 0.7660 0.6428 0.5000 0.8660 0.1736 0.9848 -0.1736 0.9848
### More Answers (1)
Matt J on 1 Oct 2022
Edited: Matt J on 1 Oct 2022
Why not just do,
pgon=nsidedpoly(18);
pgon.Vertices
ans = 18×2
-0.1736 -0.9848 -0.5000 -0.8660 -0.7660 -0.6428 -0.9397 -0.3420 -1.0000 0 -0.9397 0.3420 -0.7660 0.6428 -0.5000 0.8660 -0.1736 0.9848 0.1736 0.9848
plot(pgon); axis equal
##### 1 CommentShowHide None
Andrei_43214 on 1 Oct 2022
Hmm, that does indeed help. Now, the actual problem is to generate those coordinates for 5 different octadecagons (having radii varying by 1cm). I have 5 octadecagons, first having radius = 1.75, second having radius = 2.75, ... fifth having radius = 5.75. I want to generate those 18 coordinates for each octadecagon and then append them into one final array of size 90x2.
for r=1.75:1:5.75
pgon_r.Vertices
end
ans = 18×2
-0.3039 -1.7234 -0.8750 -1.5155 -1.3406 -1.1249 -1.6445 -0.5985 -1.7500 0 -1.6445 0.5985 -1.3406 1.1249 -0.8750 1.5155 -0.3039 1.7234 0.3039 1.7234
ans = 18×2
-0.4775 -2.7082 -1.3750 -2.3816 -2.1066 -1.7677 -2.5842 -0.9406 -2.7500 0 -2.5842 0.9406 -2.1066 1.7677 -1.3750 2.3816 -0.4775 2.7082 0.4775 2.7082
ans = 18×2
-0.6512 -3.6930 -1.8750 -3.2476 -2.8727 -2.4105 -3.5238 -1.2826 -3.7500 0 -3.5238 1.2826 -2.8727 2.4105 -1.8750 3.2476 -0.6512 3.6930 0.6512 3.6930
ans = 18×2
-0.8248 -4.6778 -2.3750 -4.1136 -3.6387 -3.0532 -4.4635 -1.6246 -4.7500 0 -4.4635 1.6246 -3.6387 3.0532 -2.3750 4.1136 -0.8248 4.6778 0.8248 4.6778
ans = 18×2
-0.9985 -5.6626 -2.8750 -4.9796 -4.4048 -3.6960 -5.4032 -1.9666 -5.7500 0 -5.4032 1.9666 -4.4048 3.6960 -2.8750 4.9796 -0.9985 5.6626 0.9985 5.6626
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Translated by | 2,267 | 5,199 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2022-49 | latest | en | 0.845654 |
https://www.mathworksheets4kids.com/activities/ | 1,508,660,430,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187825154.68/warc/CC-MAIN-20171022075310-20171022095310-00563.warc.gz | 974,395,555 | 8,926 | # Common Core Math Worksheets
The printable worksheets in this page are designed according to the Common Core State Standards. Detailed view of worksheets classified according to grades (preschool through 8th grade) are here. Math quizzes with answer keys are also available for online practice.
Preschool math worksheets have numerous colorful charts to recognize numbers and skills on counting objects up to 5. Activities to learn the basic shapes and also the measurable attributes of an object are described here. Amusing coloring activities are available to interest the kid.
Preschool Math Worksheets
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Kindergarten Math Worksheets
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Grade eight math worksheets are carefully prepared for students to find square root and cube root of numbers; expressing numbers in scientific notations; simplifying and evaluating algebraic expressions, finding the volume of geometric shapes, statistics and probability. | 561 | 3,127 | {"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-2017-43 | latest | en | 0.925448 |
https://airhostsforum.com/t/i-just-discovered-something-about-the-rating-system/38457 | 1,680,211,746,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949387.98/warc/CC-MAIN-20230330194843-20230330224843-00419.warc.gz | 110,626,017 | 11,509 | # I just discovered something about the Rating System
One of my favourite past times is to check other hosts ratings on AirBnB. I spend a lot of time reading reviews and analyzing the guests’ experience. It helps me to become a better host. I used to think that anything above 4.8 Overall Experience is an exceptionally good host and therefore is a direct competitor. Today, for the first time, I sat down with a calculator and realised that someone with an average of 4.8 is actually a pretty average host in my books: every fifth stay that they host is a FOUR STAR. If I was hosting with every fifth rating a four star I wouldn’t be happy at all. Even a 4.9 means that every tenth visit is a four star. Even then I would think to myself “what am I doing wrong to get four stars so frequently?”
The reason why I bring this up because it totally changed my perception of who my competitors are:
4.9-5 Direct Competition
4.8-4.9 Good hosts with occasional slip ups
4.7-4.8 Alright hosts
4.6-4.7 Mediocre hosts
Anything below 4.6 I would regard as poor
Anything below 4 is comically bad
I know the Superhost minimum 4.8 sounds high at first but if you think about it in purely mathematical terms (e.g. 4.8 is the equivalent of 20% of your guests not being entirely happy with their stay) then it is much more achievable in my mind.
What do people think about these numbers? Am I on the right track?
Also, does anyone know of a listing with a high number of ratings where the Overall Experience is 5*? I know one host who managed to get 50 reviews and she is still on 5* overall average which I find very impressive.
I’ve thought that a long time and have gotten heat for saying so here. It’s been pointed out to me that there are good reasons why someone good might not be a superhost but I still use that filter when searching for a place to stay. And I’ve never stayed (and never will) with hosts below 4.7+.
There are tons of them. You must mean a 5.00 rating, not a 4.98? I have 578 reviews and a 4.98 average. I think there are multiple members here with similar ratings. I think we established just this week that @RebeccaF has over 300 ratings across two rooms and has a 5.0.
I look forward to what others have to say.
2 Likes
Thanks for getting back to me on this one.
Yes, I agree, after making this calculation I also wouldn’t want to stay anywhere below 4.7. What 4.7 indicates is that 30% of guests weren’t entirely happy with their stay. Too high a ratio for me.
And yes, I meant a 5.00 rating. More than 300 ratings with an average of 5.00 is mind boggling. Even statistically (never mind the hosting skills you need to have) this seems so beyond reach. The highest I know is 50 ratings at 5.00.
An overall rating of 4.98 is extremely good and that is the type of average I would aspire to myself. Anything above 4.9 is impressive in my books.
1 Like
I have seen a few. I should have bookmarked them.
1 Like
Yes, Rebecca has a listing with 5.0 and 113 reviews. The other room in their home is 4.99 with 155 reviews. A single knucklehead is responsible for that subtraction.
And I had a mistake one star review. If not for that I’d be at 4.99. I have one 1 star, 1 3 star (completely earned and justified) 8 4 star. Only one of the 4 star ratings seems justified in my mind. In one case I even invited the guy to my super bowl party. Another was an Italian who didn’t understand that he hadn’t rented an entire apartment (for \$40 a night). I let him and his wife use my kitchen and dining room. If I hadn’t I guess he would have given an even lower rating. It definitely happens to most of us.
2 Likes
I believe tourists and people on “vacation” are much harder to please. I get few of those types and that works to my advantage.
1 Like
So 5.00 Overall has been achieved by 113 reviews. 113 seems like the highest number. That chimes with my impression to. It gets really difficult to maintain that consistent level after 50 reviews. Anyways, very impressive.
What did you earn the 1 star for?
I didn’t earn the one star. The guest hit the wrong button. Here is the review that went with it:
I am happy to share our experience at K9 Karma Casa. In a word…perfect! Dusty was there to greet us. She is warm and friendly. As a host, she is terrific! The room was very clean, nicely decorated, and one of the best European style showers I have seen. The shower is a large walk-in. Super cool! We also thought the bed was super comfortable. The thoughtful, practical amenities were impressive. From toiletries, multi-outlets, nice lighting, clear instruction on how to operate the lights and fan was helpful. There are so many lovely details. It’s an amazing space. My husband and I highly recommend it. We liked our stay so much that we are staying at K9 Karma Casa on our way back. Thank you, Dusty!
And they did stay a second time. I made sure she understood the importance of not giving me another 1 star.
The 3 star was earned. It was during my first year hosting and at that time I worked full time and guests had to enter my part of the home to get to their room. Their review said this:
The house was beautiful and clean. It was in a nice neighborhood. The room was a little small but clean and comfortable. We knew that there were two dogs in the home, and they were friendly and quiet. There were additional dogs boarded and they were noisy at times.
The mistake I made with them was letting them check in early (2pm) while I wasn’t there. They had trouble with the lock box and tried to call. I can’t answer because I’m teaching. They were barked at by the dogs who were behind a baby gate in my room, but still a cacophony. My listing made it clear that I boarded dogs in my home. I made it clear it was a small room (described as 100 sf). Like the Italians, this middle aged couple trying out Airbnb didn’t know how to read a listing and that not all Airbnbs are the same. However, I take responsibility for not understanding that a host has to guest proof their listing and communication. So that’s why I say I earned it. I believe it was their stay that inspired me to find this forum.
2 Likes
I think I’ve seen higher. But I understand that the more stays you host the harder it is to maintain a perfect score.
1 Like
No.
You would be on the right track if there were only 4 and 5 star ratings.
But your missing the 3,2 and 1 star ratings.
A host with a 4.8 average could have 20 5* reviews and 1 single 1* because the guest did not get his late check-out.
And sooner or later every host gets one of those.
11 Likes
Airbnb’s rating system is annoying in that regard a bit. We have 75 reviews. Out of 75 total, 74 reviews are 5 stars in every category. We have one guest that gave us 5 stars in every category except location (he didn’t like the nearby train - despite being in the listing). In location we got 4 stars and now Airbnb’s system shows his entire review as a “Four star” review. WTF is up with that?
We have 46 reviews and all 5*. One guest left several 4’s on the categories but they don’t count on the overall 5* for us. We would have more on Air, but we are listed elsewhere and several guests did not review.
Do hosts think that a non- review might connote a bad experience by guest who “won’t mention” it??
2 Likes
Are you sure he didn’t give you four stars overall? Have you checked under your progress tab. Click 4 star reviews and see if there’s anything there.
@KKC Yes, thats how I found it. His shows up with a four star review:
But if you look at the granular:
Am I missing something?
The overall and categories are separate. The overall is not an average of the individuals. He gave you a 4 star overall. If you think that sucks try getting all 5s in each category and then a 4 star overall.
2 Likes
When I’ve looked at listings in my area I’ve found that the listings with ratings below 4.7 tend to have reviews that mention cleanliness or old or stained furnishings, or maintenance issues.
One thing I see really often are reviews that point out a ton of cleanliness issues that make it obvious that the house wasn’t cleaned at all after the previous guest. Basically, the cleaners never showed up and the host didn’t ensure the listing was prepped. These hosts are the most deserving of 1-star reviews.
There are frequently outlier reviews, too and they can tank the rating of a listing with only 20 reviews total. You always have to wonder if those are retaliatory. Sometimes you can tell.
This may be outside of the point you are trying to make, but there is nothing purely mathematical about those statements.
In pure mathematical terms, it is possible to get a 4.7 with 92% of guests leaving a 5-star rating; and it is possible to get a 4.8 with 95% of guests leaving 5-star ratings.
3 Likes
Here is my least favorite review ever. I hate this one more than the 3 star or 4 star. This was this woman’s second stay. The first stay she didn’t leave a review at all and if I’d known she would leave a 4 star on her return trip I would have blocked her. On top of that, she is one of those who left nit picky reviews for other hosts and asked for extras like do I have a bowl she can use (I don’t have a kitchen or provide any utensils as stated in listing).
Her review says “need I say more?” I thought, why yes, yes you do. After this review I blocked her so she can’t stay with me again.
4 Likes
That is a very stingy review. The one compliment is just the icing on the cake.
Genuine question, how do you block someone from staying with you again? I thought you could only indicate this when you write the review (e.g. you haven’t seen their review yet) when you respond to the question “Would you host this guest again?”
Yes, you are absolutely right. At the point when I made these calculations I only considered 4 and 5 star reviews. Point taken.
2 Likes
Another genuine question to people on this thread:
Would you block someone (didn’t even know you could do that) who gave you 4 out of 5 for value? Overall was 5 but dinged down on value.
My instinct is to also block this type of guest … Any thoughts on this welcome. | 2,422 | 10,139 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2023-14 | latest | en | 0.975098 |
https://lernsys.com/en/math-3rd-grade | 1,632,729,919,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780058373.45/warc/CC-MAIN-20210927060117-20210927090117-00620.warc.gz | 391,449,396 | 24,794 | Exploring 3rd Grade Math With Jori
Sale ends in
Teacher: Jori
Customers Who Have Viewed This Course: 1112
\$149.00
\$70.00
0 Introduction 07:50
This course dives into the mathematics of the 3rd grade. The teacher of the course finds reasons to study each different area, making it easier for the students to motivate themselves to study. The different themes are gone through in a clear manner and many themes are given concrete examples easy to relate to.
The course goes through the basics of multiplication and division, and the student will also understand regrouping when adding and subtracting numbers until 1000. This, of course, gives also the basis for more complicated addition and subtraction calculations. The student will also learn how to compare numbers with each other. A first peak into probability as well as patterns is also included.
In addition, the student learns the basics of fractions and decimals, will understand about time and temperature, length and distance, perimeter and area, capacity and mass in both the customary units used in North America as well as the International, metric system. The course includes also basic calculations with money (U.S. dollars).
In geometry, the course goes through different 2-D and 3-D shapes, as well as the coordinate grid and graphs, transformations and locations of objects.
All in all, this is a good course for 3rd grade mathematics, giving a good basis to continue onto the 4th grade.
0 Lesson 1: Numbers to 100 16:11
This lesson goes through what the numbers from 1 to 100, what they mean, as well as some basic calculations. Mostly it is math that is needed already in the second grade, but this works as a good repetition of the second grade math before plunging into the content of the 3rd grade.
0 Lesson 1: Worksheet review 07:04
The solutions for the tasks from this lesson.
1 Lesson 2: Addition and subtraction with two digit numbers 12:40
This lesson contains the basics of addition and subtraction with two digit numbers. The teacher goes through basic calculations with math blocks, number line and also goes through the basics of regrouping with two digit calculations.
1 Lesson 2: Worksheet review 10:15
The solutions for the tasks from this lesson.
2 Lesson 3: Numbers to 1000 12:44
This lesson goes through what the numbers from 100 to 1000, what they mean, as well as some calculations. The student gets acquainted with three digits and understands the concept of hundreds, and how they relate to tens and ones.
2 Lesson 3: Worksheet review 07:09
The solutions for the tasks from this lesson.
3 Lesson 4: Comparing numbers from 1 to 1000 12:13
In this lesson, the teacher goes through the basic of comparison of numbers: the greater numbers, the lesser numbers, equal. The idea is to help students understand the values of numbers as well as how to compare them.
3 Lesson 4: Worksheet review 11:35
The solutions for the tasks from this lesson.
4 Lesson 5: Addition and subtraction with three digit numbers 15:31
This lesson contains the basics of addition and subtraction with three digit numbers. The teacher goes through basic calculations with math blocks, number line and also goes through the basics of regrouping with two digit calculations.
4 Lesson 5: Worksheet review 09:32
The solutions for the tasks from this lesson.
5 Lesson 6: Addition of 3-Digit Numbers with regrouping 16:43
This lesson handles how to regroup when doing additions with numbers from 100 to 1000. The rules of regrouping are the same however great the numbers are, and that is why this is one of the most important lessons in this course.
5 Lesson 6: Worksheet review 15:09
The solutions for the tasks from this lesson.
6 Lesson 7: Subtraction of 3-Digit Numbers with regrouping 18:17
This lesson teaches how to regroup when doing subtractions with numbers from 100 to 1000. The rules of regrouping are the same however great the numbers are, and that is why this is one of the most important lessons in this course.
6 Lesson 7: Worksheet review 18:17
The solutions for the tasks from this lesson.
7 Lesson 8: Multiplication 16:14
This lesson goes through the basics of multiplication. Multiplication is related it to addition, and the student understands fundamentally and in a concrete way, what multiplication means.
7 Lesson 8: Worksheet review 10:07
The solutions for the tasks from this lesson.
8 Lesson 9: Multiplication II 17:34
This lesson continues work with multiplication. It also introduces some rules and hints how to calculate some multiplication tables faster in one’s head.
8 Lesson 9: Worksheet review 06:56
The solutions for the tasks from this lesson.
9 Lesson 10: Division 14:02
This lesson goes through the basics of division. Division is related it to multiplication, and the student understands fundamentally and in a concrete way, what division means.
9 Lesson 10: Worksheet review 07:14
The solutions for the tasks from this lesson.
10 Lesson 11: Division II 18:43
This lesson continues to dive into the secrets of division. It relates it further to multiplication as well as introduces the vocabulary of division.
10 Lesson 11: Worksheet review 08:19
The solutions for the tasks from this lesson.
11 Lesson 12: Order of operations 13:23
In mathematics, the calculations are done in a certain order. The order is very important and fundamental, since it affects the outcome of each calculation. In this lesson, we go through the order of operations and the rules needed in 3rd grade math.
11 Lesson 12: Worksheet review 15:11
The solutions for the tasks from this lesson.
12 Lesson 13: Length and Distance 16:42
This lesson introduces the concepts of length and distance. It includes lots of measuring and estimating, as well as the basic units used to measure length and distance.
12 Lesson 13: Worksheet review 07:22
The solutions for the tasks from this lesson.
13 Lesson 14: Perimeter and area 13:22
This lesson introduces the concepts of perimeter and area. The student will understand, how different shapes affect the perimeter, and how perimeter and area relate. Measuring perimeter happens in this lesson in “units” and measuring area in “unit squares”.
13 Lesson 14: Worksheet review 10:47
The solutions for the tasks from this lesson.
14 Lesson 15: Capacity and mass 15:04
This lesson introduces the concepts of capacity and mass. It includes lots of measuring and estimating, as well as the basic units used to measure capacity and mass.
14 Lesson 15: Worksheet review 13:29
The solutions for the tasks from this lesson.
15 Lesson 16: Time and Temperature 14:48
This lesson introduces the concepts of time and temperature. The student learns how to measure time with an analog clock that has minutes and hours. Temperature is introduced through a thermometer.
15 Lesson 16: Worksheet review 10:27
The solutions for the tasks from this lesson.
16 Lesson 17: International measurement system 17:43
The international measurement system is something that is very good to be aware of. This lesson introduces the metric system to students and compares it with the customary system used in North America.
16 Lesson 17: Worksheet review 06:40
The solutions for the tasks from this lesson.
19 Lesson 20: Money 14:06
This lesson goes through U.S. bills and coins and helps the student to get familiar with money, and recognize the value of different bills and coins.
19 Lesson 20: Worksheet review 04:16
The solutions for the tasks from this lesson.
20 Lesson 21: Addition and subtraction with money 18:52
This lesson concentrates on doing calculations with money. It is an important skill in every day life. It also introduces calculations with cents. NOTE: Before this lesson it is important to go through the Decimal lesson.
20 Lesson 21: Worksheet review 09:22
The solutions for the tasks from this lesson.
21 Lesson 22: 2-D Shapes Quadrilaterals 15:27
This lesson introduces different quadrilateral shapes and what their properties are. After this lesson, the student can recognize several quadrilateral shapes, as well as define them.
21 Lesson 22: Worksheet review 06:53
The solutions for the tasks from this lesson.
22 Lesson 23: 2-D Shapes II 17:20
This lesson continues work with two dimensional shapes. It introduces triangles and their properties, as well as other two dimensional shapes.
22 Lesson 23: Worksheet review 07:42
The solutions for the tasks from this lesson.
23 Lesson 24: 3-D shapes 21:56
This lesson introduces three dimensional shapes, the vocabulary related to them, as well as their properties. Some of the shapes are also drawn in the lesson.
23 Lesson 24: Worksheet review 09:25
The solutions for the tasks from this lesson.
24 Lesson 25: Graphs 15:45
This lesson introduces graphs. Graphs are a visual way to compare numbers, and it helps especially when there are more numbers to compare. The student will learn how to interpret as well as draw bar graphs.
24 Lesson 25: Worksheet review 08:42
The solutions for the tasks from this lesson.
25 Lesson 26: The Coordinate grid and locations of shapes 14:34
This lesson introduces the coordinate grid. The coordinate grid is an important concept that is needed often later in math. The student learns about the x- and y-axis, as well as the recognize locations of shapes and objects.
25 Lesson 26: Worksheet review 05:16
The solutions for the tasks from this lesson.
26 Lesson 27: Transformations 23:13
This lesson introduces the transformations of different shapes. Slides, flips and rotations in mathematics are different ways of working with different shapes, and the student becomes familiar with all of them.
26 Lesson 27: Worksheet review 04:51
The solutions for the tasks from this lesson.
17 Lesson 18: Fractions 23:40
This lesson introduces fractions in many ways, to give the student a concrete idea what fractions mean and how to recognize fractions, as well as how to compare them.
17 Lesson 18: Worksheet review 08:43
The solutions for the tasks from this lesson.
18 Lesson 19: Decimals 18:08
This lesson introduces decimals. Decimals are needed in many things, from interpreting sport results to calculating with money. In this lesson, decimals are also related to fractions and the place values of decimals are explained.
18 Lesson 19: Worksheet review 07:58
The solutions for the tasks from this lesson.
27 Lesson 28: Patterns in shapes 19:25
This lesson introduces patterns in shapes to students. The student learns to recognize continuous patterns that are presented as shapes. This lesson is fun detective work!
27 Lesson 28: Worksheet review 14:23
The solutions for the tasks from this lesson.
28 Lesson 29: Patterns in numbers and even and odd numbers 19:45
This lesson continues the work with patterns. The student becomes familiar with patterns in numbers and gets to do lot of fun detective work yet again! The student also becomes familiar with the concept of even and odd numbers.
28 Lesson 29: Worksheet review 11:02
The solutions for the tasks from this lesson.
29 Lesson 30: Probability 18:38
Probability is an important concept in math as well as science. This lesson introduces the basics – how to describe probabilities as well as tell probabilities as fractions.
29 Lesson 30: Worksheet review 11:21
The solutions for the tasks from this lesson.
This course dives into the mathematics of the 3rd grade. The teacher of the course finds reasons to study each different area, making it easier for the students to motivate themselves to study. The different themes are gone through in a clear manner and many themes are given concrete examples easy to relate to.
This course includes:
• 30 WORKSHEETS WITH EXERCISES TO EACH LESSON
• 30 VIDEO-LESSONS
• 30 EXERCISE REVIEW VIDEOS (I explain step by step how to reach the correct answer)
• 30 ONLINE QUIZZES AND EXAMS
• OVER 10 HOURS OF VIDEO-LESSONS
Course Goals
The course goes through the basics of multiplication and division, and the student will also understand regrouping when adding and subtracting numbers until 1000. This, of course, gives also the basis for more complicated addition and subtraction calculations. The student will also learn how to compare numbers with each other. A first peak into probability as well as patterns is also included.
In addition, the student learns the basics of fractions and decimals, will understand about time and temperature, length and distance, perimeter and area, capacity and mass in both the customary units used in North America as well as the International, metric system. The course includes also basic calculations with money (U.S. dollars).
In geometry, the course goes through different 2-D and 3-D shapes, as well as the coordinate grid and graphs, transformations and locations of objects.
Target Audience
This video-course is primarily intended for third-graders.
Course Requirements
Students taking this course will need to have completed 2nd grade mathematics.
All in all, this is a good course for 3rd grade mathematics, giving a good basis to continue onto the 4th grade.
• Teacher: Jori
• Areas of expertise: Math, teaching, drama
• Education: Master of Education Bachelor of Applied Theatre
• Interests: Playback Theatre, Learning, Gamified Education
• Skills: Trainer of teachers, professional perfomer
• Associations: math, classroom teacher
• Issues I care about: building bridges between people
Jori Pitkänen is a Finnish teacher who has a Master’s degree in Education. He specializes in teaching the 1st to 6th grade classes. He is enthusiastic in teaching mathematics and tries to find the reason to study each area in mathematics instead of just teaching. He is interested in his students’ motivation and explains things in an understandable way. In his professional life, apart from teaching, he teaches teaching methods (drama in education and gamified education) to teachers, coaches the unemployed and does theatre professionally, especially Playback Theatre. He is an enthusiastic student of learning and is interested in how learning happens and what makes it easier for students to learn. | 3,081 | 14,163 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2021-39 | latest | en | 0.956268 |
https://diveintopython.org/learn/variables/array/convert | 1,721,793,875,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518154.91/warc/CC-MAIN-20240724014956-20240724044956-00721.warc.gz | 186,353,064 | 9,183 | # How to Convert an Array to Other Types
In Python, converting between lists and arrays is a common task when dealing with numerical data. The process of converting from a list to an array allows us to take advantage of efficient numerical computations and operations provided by libraries like NumPy. On the other hand, in some cases we need to make 2d array from 1d array. So let's review most common topics on conversions related to array.
## Conversion Between Lists and NumPy Arrays
Conversions between lists and NumPy arrays provide flexibility in data manipulation and facilitate seamless integration with other numerical computing libraries in Python.
### How to Convert a NumPy Array to Python List
To convert a NumPy array to a Python list, you can use the `tolist()` method provided by the NumPy library. This method converts a NumPy array into a nested Python list. Here's an example:
``````import numpy as np
numpy_array = np.array([1, 2, 3, 4, 5])
python_list = numpy_array.tolist()
print(python_list) # Output: [1, 2, 3, 4, 5]
``````
Similarly, if you have a multidimensional NumPy array, the `tolist()` method will convert it to a nested Python list structure:
``````import numpy as np
numpy_array = np.array([[1, 2, 3], [4, 5, 6]])
python_list = numpy_array.tolist()
print(python_list) # Output: [[1, 2, 3], [4, 5, 6]]
``````
In this example, the numpy_array is a 2D NumPy array with two rows and three columns. The `tolist()` method converts it into a nested Python list with the same structure.
### List to Array Conversion in Python
To convert a Python list to an array, you can use the `array()` function provided by the NumPy library. The `array()` function creates a new NumPy array from the elements of the given list. Here is an example of converting list to array:
``````import numpy as np
my_list = [1, 2, 3, 4, 5]
my_array = np.array(my_list)
print(my_array) # Output: [1 2 3 4 5]
``````
NumPy arrays are homogeneous, meaning they can only hold elements of the same data type. If the elements of the list are of different types, NumPy will automatically upcast the elements to a common data type.
For example:
``````import numpy as np
my_list = [1, 2.5, 'three', True]
my_array = np.array(my_list)
print(my_array) # Output: ['1' '2.5' 'three' 'True']
``````
In this case, the elements of the list have different data types (int, float, str, and bool). NumPy upcasts all the elements to the common data type, which in this case is str. The resulting array contains strings representing the elements of the list.
## Conversion Between String and Array
Converting between strings and arrays in Python can be useful when working with textual data or when manipulating individual characters.
### Python String into Array Conversion
To convert a Python string into an array of individual characters, you can iterate over the string and create a list of characters. Here's an example:
``````string = "Hello, world!"
array = [char for char in string]
print(array) # Output: ['H', 'e', 'l', 'l', 'o', ',', ' ', 'w', 'o', 'r', 'l', 'd', '!']
``````
In the above example, we iterate over each character in the string using a list comprehension. Each character is added to the list, resulting in an array where each element represents an individual character from the string.
If you want to split the string into an array of words instead of characters, you can use the `split()` method. By default, the `split()` method splits the string based on whitespace characters (spaces, tabs, newlines). Here's an example:
``````string = "Hello, world! How are you?"
array = string.split()
print(array) # Output: ['Hello,', 'world!', 'How', 'are', 'you?']
``````
In this example, the `split()` method splits the string into a list of substrings based on whitespace characters, resulting in an array where each element represents a word from the string.
### Array to String Conversion
To convert an array (or list) to a string in Python, you can use the `join()` method. The `join()` method concatenates the elements of an iterable into a single string, using a specified delimiter. Here's an example:
``````array = ['Hello', 'world', 'How', 'are', 'you?']
string = ' '.join(array)
print(string) # Output: Hello world How are you?
``````
By changing the delimiter passed to the `join()` method, you can modify the way the elements are separated in the resulting string. For example, using a comma (',') as the delimiter:
``````array = ['apple', 'banana', 'orange']
string = ', '.join(array)
print(string) # Output: apple, banana, orange
``````
### Contribute with us!
Do not hesitate to contribute to Python tutorials on GitHub: create a fork, update content and issue a pull request.
Python developer, content manager.
Updated: 05/03/2024 - 21:53
Reviewed and approved | 1,188 | 4,813 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2024-30 | latest | en | 0.73246 |
https://gmatclub.com/forum/charles-lindbergh-for-his-attempt-at-a-solo-transatlantic-48581-20.html | 1,487,725,111,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501170864.16/warc/CC-MAIN-20170219104610-00136-ip-10-171-10-108.ec2.internal.warc.gz | 721,245,904 | 58,811 | Charles Lindbergh, for his attempt at a solo transatlantic : GMAT Sentence Correction (SC) - Page 2
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Charles Lindbergh, for his attempt at a solo transatlantic
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Re: Charles Lindbergh, for his attempt at a solo transatlantic [#permalink]
Show Tags
19 Mar 2013, 01:42
debayan222 wrote:
egmat wrote:
Hi All,
Charles Lindbergh, for his attempt at a solo transatlantic flight, was very reluctant to have any extra weight on his plane, he therefore refused to carry even a pound of mail, despite being offered $1,000 to do so. 1. The function prepositional phrase “for his attempt at a solo transatlantic flight” is obscure. It is not clear what is it modifying. It is not placed correctly. 2. Only comma has been used to join two independent clauses. This structure is completely ungrammatical. POE: Choice A: Charles Lindbergh, for his attempt at a solo transatlantic flight, was very reluctant to have any extra weight on his plane, he therefore: Incorrect for the reasons discussed above. Choice B: When Charles Lindbergh was attempting his solo transatlantic flight, being very reluctant to have any extra weight on his plane, he: Incorrect. The syntax of this answer choice is not correct the clauses have not been placed at proper places to convey the meaning clearly. Choice C: Since he was very reluctant to carry any extra weight on his plane when he was attempting his solo transatlantic flight, so Charles Lindbergh: Incorrect. There is no independent clause in the sentence with this choice. Use of “so” is incorrect here. Choice D: Being very reluctant to carry any extra weight on his plane when he attempted his solo transatlantic flight was the reason that Charles Lindbergh: Incorrect. This structure is ungrammatical. If there were a comma instead of “was the reason that”, then the construction would be correct. Choice E: Very reluctant to have any extra weight on his plane when he attempted his solo transatlantic flight, Charles Lindbergh: Correct. “very reluctant” correctly modifies the subject of the Independent clause Charles Lindbergh. Pronoun “he” and “his” also correctly refer to the same subject. Hope this helps. Thanks. Shraddha Hi Shraddha, I'm not able to understand clearly why B is incorrect here and E is correct ? Could you please come up with a bit detail analysis of this question by digging deep further ? Much appreciate your feedback. P.S : I guess,E doesn't have a sense of the fact that Charles Lindbergh was very reluctant to have any extra weight on his plane hence refused to carry even a pound of mail.... Hi Shradhha, Any update on this? Could you please come up with a detail meaning analysis and help me understand it... Really appreciate your feedback. _________________ e-GMAT Discount Codes Kaplan GMAT Prep Discount Codes Veritas Prep GMAT Discount Codes e-GMAT Representative Joined: 02 Nov 2011 Posts: 2029 Followers: 2125 Kudos [?]: 7348 [0], given: 277 Re: Charles Lindbergh, for his attempt at a solo transatlantic [#permalink] Show Tags 19 Mar 2013, 07:46 debayan222 wrote: Hi Shraddha, I'm not able to understand clearly why B is incorrect here and E is correct ? Could you please come up with a bit detail analysis of this question by digging deep further ? Much appreciate your feedback. P.S : I guess,E doesn't have a sense of the fact that Charles Lindbergh was very reluctant to have any extra weight on his plane hence refused to carry even a pound of mail.... Hi Debayan, Choice B: When Charles Lindbergh was attempting his solo transatlantic flight, being very reluctant to have any extra weight on his plane, he The problem with this choice is that the modifier "being very reluctant..." is placed between two commas. Such placement makes it undlear whether this modifier modifies the preceding noun or the following noun. Here is the scope of ambiguity in modification. Choice E: Very reluctant to have any extra weight on his plane when he attempted his solo transatlantic flight, Charles Lindbergh Here, the modification is absolutely clear. Who is very reluctant? The subject of the following clause Chales Lindbergh. There is no acope of any ambiguity whatsoever in this choice. Hence, this choice is precise and conveys the logical menaing absolutely clearly. Hope this helps. Thanks. Shraddha _________________ | '4 out of Top 5' Instructors on gmatclub | 70 point improvement guarantee | www.e-gmat.com BSchool Forum Moderator Joined: 27 Aug 2012 Posts: 1196 Followers: 131 Kudos [?]: 1523 [0], given: 143 Re: Charles Lindbergh, for his attempt at a solo transatlantic [#permalink] Show Tags 19 Mar 2013, 09:01 egmat wrote: debayan222 wrote: Hi Shraddha, I'm not able to understand clearly why B is incorrect here and E is correct ? Could you please come up with a bit detail analysis of this question by digging deep further ? Much appreciate your feedback. P.S : I guess,E doesn't have a sense of the fact that Charles Lindbergh was very reluctant to have any extra weight on his plane hence refused to carry even a pound of mail.... Hi Debayan, Choice B: When Charles Lindbergh was attempting his solo transatlantic flight, being very reluctant to have any extra weight on his plane, he The problem with this choice is that the modifier "being very reluctant..." is placed between two commas. Such placement makes it undlear whether this modifier modifies the preceding noun or the following noun. Here is the scope of ambiguity in modification. Choice E: Very reluctant to have any extra weight on his plane when he attempted his solo transatlantic flight, Charles Lindbergh Here, the modification is absolutely clear. Who is very reluctant? The subject of the following clause Chales Lindbergh. There is no acope of any ambiguity whatsoever in this choice. Hence, this choice is precise and conveys the logical menaing absolutely clearly. Hope this helps. Thanks. Shraddha Hi Shraddha, Thanks for your reply but I'm afraid that it's still not clear to me unfortunately Option E-this option I think, doesn't convey the intended meaning that as the Charles was very reluctant to carry any extra weight...(hence) he refused to carry even a pound of mail.... Option B-in this option "being very reluctant to have any extra weight on his plane" irrespective of the fact whether it modifies the preceding noun or the following noun, I think clearly conveys the intended meaning that Charles was reluctant to carry any extra weight so he refused to carry even a pound of mail.... Please help! _________________ Manager Joined: 10 Jan 2010 Posts: 81 Schools: IIM Followers: 0 Kudos [?]: 23 [0], given: 11 Re: Charles Lindbergh, for his attempt at a solo transatlantic [#permalink] Show Tags 10 Apr 2013, 02:37 egmat wrote: Hi All, Charles Lindbergh, for his attempt at a solo transatlantic flight, was very reluctant to have any extra weight on his plane, he therefore refused to carry even a pound of mail, despite being offered$1,000 to do so.
1. The function prepositional phrase “for his attempt at a solo transatlantic flight” is obscure. It is not clear what is it modifying. It is not placed correctly.
2. Only comma has been used to join two independent clauses. This structure is completely ungrammatical.
POE:
Choice A: Charles Lindbergh, for his attempt at a solo transatlantic flight, was very reluctant to have any extra weight on his plane, he therefore: Incorrect for the reasons discussed above.
Choice B: When Charles Lindbergh was attempting his solo transatlantic flight, being very reluctant to have any extra weight on his plane, he: Incorrect. The syntax of this answer choice is not correct the clauses have not been placed at proper places to convey the meaning clearly.
Choice C: Since he was very reluctant to carry any extra weight on his plane when he was attempting his solo transatlantic flight, so Charles Lindbergh: Incorrect. There is no independent clause in the sentence with this choice. Use of “so” is incorrect here.
Choice D: Being very reluctant to carry any extra weight on his plane when he attempted his solo transatlantic flight was the reason that Charles Lindbergh: Incorrect. This structure is ungrammatical. If there were a comma instead of “was the reason that”, then the construction would be correct.
Choice E: Very reluctant to have any extra weight on his plane when he attempted his solo transatlantic flight, Charles Lindbergh: Correct. “very reluctant” correctly modifies the subject of the Independent clause Charles Lindbergh. Pronoun “he” and “his” also correctly refer to the same subject.
Hope this helps.
Thanks.
In option E, "Very reluctant to have any extra weight on his plane when he attempted his solo transatlantic flight" is a dependent clause and "Charles Lindbergh refused to carry even a pound of mail, despite being offered $1,000 to do so" is an independent clause. I think so as in the 1st clause "Very ........flight", the word "when" is a dependent clause marker. Manager Joined: 18 Aug 2006 Posts: 118 Location: United States WE: Consulting (Telecommunications) Followers: 1 Kudos [?]: 51 [3] , given: 44 Re: Charles Lindbergh, for his attempt at a solo transatlantic [#permalink] Show Tags 25 Apr 2013, 22:00 3 This post received KUDOS There are two key things one should understand to answer this questions: 1) The modifier "reluctant to carry even a pound of extra weight" is a necessary modifier, meaning this is essential part of the sentence but not there to provide extra information. 2) The modifier "despite being offered$1000 to do so" should modify Charles not anything else.
Coming to the choices:
Original sentence is wrong. Need not discuss why.
Choice B is wrong because it changes the intended meaning of the sentence by violating 1 above. By placing the modifier 1 (above) after a comma, the choice wrongly implies that there is no correlation between "going on a solo transatlantic flight" and "being reluctant to carry extra weight" which is wrong as it changes the intended meaning.
Choice C has redundancy error Since.....so is a wrong construction.
Choice D is wrong because it violates 2 above. "Being very reluctant....was the reason....., despite being offered $1000 to do so". "Being very reluctant to carry" cannot be offered$1000.
Choice E takes care of both 1 and 2 highlighted above. Right
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24 May 2015, 00:06
egmat wrote:
Hi All,
Charles Lindbergh, for his attempt at a solo transatlantic flight, was very reluctant to have any extra weight on his plane, he therefore refused to carry even a pound of mail, despite being offered \$1,000 to do so.
1. The function prepositional phrase “for his attempt at a solo transatlantic flight” is obscure. It is not clear what is it modifying. It is not placed correctly.
2. Only comma has been used to join two independent clauses. This structure is completely ungrammatical.
POE:
Choice A: Charles Lindbergh, for his attempt at a solo transatlantic flight, was very reluctant to have any extra weight on his plane, he therefore: Incorrect for the reasons discussed above.
Choice B: When Charles Lindbergh was attempting his solo transatlantic flight, being very reluctant to have any extra weight on his plane, he: Incorrect. The syntax of this answer choice is not correct the clauses have not been placed at proper places to convey the meaning clearly.
Choice C: Since he was very reluctant to carry any extra weight on his plane when he was attempting his solo transatlantic flight, so Charles Lindbergh: Incorrect. There is no independent clause in the sentence with this choice. Use of “so” is incorrect here.
Choice D: Being very reluctant to carry any extra weight on his plane when he attempted his solo transatlantic flight was the reason that Charles Lindbergh: Incorrect. This structure is ungrammatical. If there were a comma instead of “was the reason that”, then the construction would be correct.
Choice E: Very reluctant to have any extra weight on his plane when he attempted his solo transatlantic flight, Charles Lindbergh: Correct. “very reluctant” correctly modifies the subject of the Independent clause Charles Lindbergh. Pronoun “he” and “his” also correctly refer to the same subject.
Hope this helps.
Thanks.
According to e-GMAT's SC course, the correct answer has to convey the intended meaning of the original sentence. Here, the original sentence indicates that "being very relectant to...." is the reason why Charles Lindbergh refused to carry... But in choice E, we do not see this relation. In choice E, the "reluctant" thing only becomes a modifier to Charles Lindbergh, and there is no reason for the "refused" thing.
I think choice D, though wordy, is the correct answer.
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25 May 2015, 23:19
tronghieu1987 wrote:
According to e-GMAT's SC course, the correct answer has to convey the intended meaning of the original sentence. Here, the original sentence indicates that "being very relectant to...." is the reason why Charles Lindbergh refused to carry... But in choice E, we do not see this relation. In choice E, the "reluctant" thing only becomes a modifier to Charles Lindbergh, and there is no reason for the "refused" thing.
I think choice D, though wordy, is the correct answer.
Hi tronghieu1987,
Thanks for posting your doubt here.
The context of the sentence is such hat it makes it absolutely clear that Charles Lindbergh refused to carry even a pound of mail because he was reluctant to have any extra weight on his plane. Even though this sentence does not use any "reason" word., the context makes it absolutely clear.
Also, this is an officially correct answer that is always correct. We should not doubt them. rather we should learn new usages from them.
Hope this helps.
Thanks.
SJ
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17 Jul 2016, 21:35
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Re: Charles Lindbergh, for his attempt at a solo transatlantic [#permalink] 17 Jul 2016, 21:35
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May 5, 2016
# Homework Help: Algebra 1
Posted by Matt D. on Wednesday, February 9, 2011 at 5:44am.
Lesson 3 Assignment .Explain the differences between solving these two equations: By steps
log3(x - 1) = 4
AND
log3(x - 1) = log3^4 | 84 | 242 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2016-18 | longest | en | 0.950819 |
https://ask.sagemath.org/question/61964/how-derive-minimization-frequencies-formula/ | 1,679,569,662,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945144.17/warc/CC-MAIN-20230323100829-20230323130829-00142.warc.gz | 145,078,893 | 13,194 | # How derive minimization frequencies formula ?
Fourier DFT transformation of sawtooth wave (0, a, 2 a, 3a..a*(N-1).) has coefficients
$a \frac{N(N+1)}{2}, k=0$
$-a \frac{N}{1-e^{-j2\pi k/N}}$, $k\in [1,N-1]$
https://dsp.stackexchange.com/questio...
I want remove it from samples, but I must compute optimal slope a.
X means Fourier transformation of samples, S - Fourier transformation of sawtooth wave
Slope is optimal when difference (X-S) has minimum high frequencies <=>
$\sum_{frequencies} (frequency \cdot amplitude^2)$ is minimal.
Maybe better $\sum_{frequencies} (frequency^2 \cdot amplitude^2)$ is minimal.
I must equate to zero the derivative of this sum and compute slope a. How do it with SageMath?
edit retag close merge delete
Sort by ยป oldest newest most voted
Try this:
sage: derivative(abs(xn - a*N/(1 - exp(-i*2*pi*k/N)))^2, a)
8*(xn + 8*a/(e^(-1/4*I*pi*k) - 1))/(e^(1/4*I*pi*k) - 1) + 8*(8*a/(e^(1/4*I*pi*k) - 1) + conjugate(xn))/(e^(-1/4*I*pi*k) - 1)
more | 328 | 992 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2023-14 | longest | en | 0.664705 |
https://www.courseunlock.org/downloads/solved-objective-recursive-functions-project-interesting-example-recursive-function-calculating-g-q41353458/ | 1,675,133,923,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499842.81/warc/CC-MAIN-20230131023947-20230131053947-00791.warc.gz | 752,248,705 | 14,093 | # (Solved) : Objective Recursive Functions Project Interesting Example Recursive Function Calculating G Q41353458 . . .
Objective Recursive Functions Project An interesting example of a recursive function is in calculating the greatest common denominator of two integers. The approach is as follows assume that the function greatestCommon Div(x, y) returns the greatest common denominator): 1. Given two positive integers, x and y call y = greatestCommon Div(x, y); 2. In the function, calculate the modulus – the remainder when you divide x by y. This would be r = mod(x, y); 3. if r == 0 then you return y which is the greatest common divisor 4. if the modulus is not zero then you call the greatestCommon Div function again, but this time you pass y and r to the function. This would be y = greatestCommon Div(y, r); In your program you will develop a recursive function that calculates the greatest common divisor. Include the function as a subfunction of a driver function that tests it. The driver needs to call a get Data function to have the user enter the values for x and y. This function must have error checking to ensure that the two values are greater than zero and are integers. It then calls the greatestCommonDen function to determine the greatest com- mon divisor Overall your project will consist of the driver function, the print Header function, the getData function, and a print Results function. When you test your program try some values that are simply and some that are not. For example, try two prime numbers – the greatestCommon Den will always be 1 Some other examples are X y ged 23 81 29 108 1 27 Submission Name your script file project_8_abc.m where abc are your initials. Submit your M-file source code (the file that ends in .m) to the class folder on Box by 11:59 pm on Thursday, October 24, 2019. function project_8_abc() 8 PROJECT_8_ABC project_8_abc() is the driver function for the program. Name: Date: Class: CMPSC 200 Description: Print the splash screen – use a print Header function & Enter the two variables, X & y, using a get Data function Prompt the user for the inputs – using error checking \$ Do your inputs by calling the recursive getDataRange function that we developed in class, putting the getDataRange function call inside of the wrapper getData function. The getData function and the getDataRange functions are separate % Subfunctions call the greatest common denominator 8 function Print results using a printResults % function end function (x, y) = getData() GETDATA (x, y) = getData() is a wrapper function that is used to enter the values used in the program. It performs the inputs through calls to getDataRange which is a recursive function that performs error checking. Name: Date: Class: CMPSC 200 end function x = getDataRange (prompt, a, b) % GETDATARANGE x = getDataRange (prompt, a, b) is a recursive funciton that prompts the using for an input. It then checks if the value is between a and b. If it is not then it prints a warning message and calls the getDataRange function again (recursively). It only needs an if, not an if – else. Name : Date: CMPSC 200 Enter the data using a regular call to input with the prompt that was passed to the function Create two logical variables that will determine if the value is between the two endpoints a and b Check the stopping case – if it fails call getDataRange again end end end function y = greatestCommonDen(x, y) % GREATESTCOMMONDEN greatestCommonDen(n, d) uses a recursive function to determine the greatest common denominator of x, and y Name: Date: Class: CMPSC 200 Calculate the remainder of x/y % Check the stopping criteria which then returns the base case This is if the modulus of x and y is o Make the recursive call to calculate the greatest common denominator This is a call to itself with y being the first parameter and the remainder being the second end Show transcribed image text Objective Recursive Functions Project An interesting example of a recursive function is in calculating the greatest common denominator of two integers. The approach is as follows assume that the function greatestCommon Div(x, y) returns the greatest common denominator): 1. Given two positive integers, x and y call y = greatestCommon Div(x, y); 2. In the function, calculate the modulus – the remainder when you divide x by y. This would be r = mod(x, y); 3. if r == 0 then you return y which is the greatest common divisor 4. if the modulus is not zero then you call the greatestCommon Div function again, but this time you pass y and r to the function. This would be y = greatestCommon Div(y, r); In your program you will develop a recursive function that calculates the greatest common divisor. Include the function as a subfunction of a driver function that tests it. The driver needs to call a get Data function to have the user enter the values for x and y. This function must have error checking to ensure that the two values are greater than zero and are integers. It then calls the greatestCommonDen function to determine the greatest com- mon divisor Overall your project will consist of the driver function, the print Header function, the getData function, and a print Results function. When you test your program try some values that are simply and some that are not. For example, try two prime numbers – the greatestCommon Den will always be 1 Some other examples are X y ged 23 81 29 108 1 27 Submission Name your script file project_8_abc.m where abc are your initials. Submit your M-file source code (the file that ends in .m) to the class folder on Box by 11:59 pm on Thursday, October 24, 2019.
function project_8_abc() 8 PROJECT_8_ABC project_8_abc() is the driver function for the program. Name: Date: Class: CMPSC 200 Description: Print the splash screen – use a print Header function & Enter the two variables, X & y, using a get Data function Prompt the user for the inputs – using error checking \$ Do your inputs by calling the recursive getDataRange function that we developed in class, putting the getDataRange function call inside of the wrapper getData function. The getData function and the getDataRange functions are separate % Subfunctions call the greatest common denominator 8 function Print results using a printResults % function end function (x, y) = getData() GETDATA (x, y) = getData() is a wrapper function that is used to enter the values used in the program. It performs the inputs through calls to getDataRange which is a recursive function
that performs error checking. Name: Date: Class: CMPSC 200 end function x = getDataRange (prompt, a, b) % GETDATARANGE x = getDataRange (prompt, a, b) is a recursive funciton that prompts the using for an input. It then checks if the value is between a and b. If it is not then it prints a warning message and calls the getDataRange function again (recursively). It only needs an if, not an if – else. Name : Date: CMPSC 200 Enter the data using a regular call to input with the prompt that was passed to the function Create two logical variables that will determine if the value is between the two endpoints a and b Check the stopping case – if it fails call getDataRange again end end
end function y = greatestCommonDen(x, y) % GREATESTCOMMONDEN greatestCommonDen(n, d) uses a recursive function to determine the greatest common denominator of x, and y Name: Date: Class: CMPSC 200 Calculate the remainder of x/y % Check the stopping criteria which then returns the base case This is if the modulus of x and y is o Make the recursive call to calculate the greatest common denominator This is a call to itself with y being the first parameter and the remainder being the second end
## Expert Answer
Answer to Objective Recursive Functions Project An interesting example of a recursive function is in calculating the greatest comm… | 1,709 | 7,835 | {"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-2023-06 | latest | en | 0.853988 |
https://stats.stackexchange.com/questions/154951/non-normal-distributions-with-zero-skewness-and-zero-excess-kurtosis | 1,653,236,372,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662545875.39/warc/CC-MAIN-20220522160113-20220522190113-00581.warc.gz | 610,597,889 | 70,229 | # Non-normal distributions with zero skewness and zero excess kurtosis?
Mostly theoretical question. Are there any examples of non-normal distributions that has first four moment equal to those of normal? Could they exist in theory?
• Considering even just a mixture of 2 normals (5 parameters--2 means, 2 variances, and the mixture probability), you can solve for a wide variety of first four moments. Dec 4, 2019 at 22:45
• I gave a discrete example at a later near-duplicate of this question: stats.stackexchange.com/a/378923/225256 May 28, 2021 at 16:21
• @SheridanGrant, a mixture of two normals won't work. Let $X$ be a mixture with $p$ of $N(m,s)$ and $(1-p)$ of $N(n,t)$. Then forcing $X$ and $N(0,1)$ to have the same first, second and third moments leads to $p=n/(n-m)$, $s=\sqrt{(3-m^2+2mn)/3}$, $t=\sqrt{(3-n^2+2mn)/3}$. Under those conditions, the difference in fourth moments between $X$ and $N(0,1)$ is just $2mn(m^2-mn+n^2)/3$; if that is $0$ then $m=0$ or $n=0$, and $X$ is just a single standard normal. May 29, 2021 at 12:13
• @MattF.--huh, cool. I did a quick search for an analysis of "how many moments can you match with a mixture of k Gaussians" and didn't find anything, do you know of any results? Jun 10, 2021 at 6:10
Yes, examples with skewness and excess kurtosis both zero are relatively easy to construct. (Indeed examples (a) to (d) below also have Pearson mean-median skewness 0)
(a) For example, in this answer an example is given by taking a 50-50 mixture of a gamma variate, (which I call $$X$$), and the negative of a second one, which has a density that looks like this:
Clearly the result is symmetric and not normal. The scale parameter is unimportant here, so we can make it 1. Careful choice of the shape parameter of the gamma yields the required kurtosis:
1. The variance of this double-gamma ($$Y$$) is easy to work out in terms of the gamma variate it's based on: $$\text{Var}(Y)=E(X^2)=\text{Var}(X)+E(X)^2=\alpha+\alpha^2$$.
2. The fourth central moment of the variable $$Y$$ is the same as $$E(X^4)$$, which for a gamma($$\alpha$$) is $$\alpha(\alpha+1)(\alpha+2)(\alpha+3)$$
As a result the kurtosis is $$\frac{\alpha(\alpha+1)(\alpha+2)(\alpha+3)}{\alpha^2(\alpha+1)^2}=\frac{(\alpha+2)(\alpha+3)}{\alpha(\alpha+1)}$$. This is $$3$$ when $$(\alpha+2)(\alpha+3)=3\alpha(\alpha+1)$$, which happens when $$\alpha=(\sqrt{13}+1)/2\approx 2.303$$.
(b) We could also create an example as a scale mixture of two uniforms. Let $$U_1\sim U(-1,1)$$ and let $$U_2\sim U(-a,a)$$, and let $$M=\frac12 U_1+\frac12 U_2$$. Clearly by considering that $$M$$ is symmetric and has finite range, we must have $$E(M)=0$$; the skewness will also be 0 and central moments and raw moments will be the same.
$$\text{Var}(M)=E(M^2)=\frac12\text{Var}(U1)+\frac12\text{Var}(U_2)=\frac16[1+a^2]$$.
Similarly, $$E(M^4)=\frac{1}{10} (1+a^4)$$ and so the kurtosis is $$\frac{\frac{1}{10} (1+a^4)}{[\frac16 (1+a^2)]^2}=3.6\frac{1+a^4}{(1+a^2)^2}$$
If we choose $$a=\sqrt{5+\sqrt{24}}\approx 3.1463$$, then kurtosis is 3, and the density looks like this:
(c) here's a fun example. Let $$X_i\stackrel{_\text{iid}}{\sim}\text{Pois}(\lambda)$$, for $$i=1,2$$.
Let $$Y$$ be a 50-50 mixture of $$\sqrt{X_1}$$ and $$-\sqrt{X_2}$$:
by symmetry $$E(Y)=0$$ (we also need $$E(|Y|)$$ to be finite but given $$E(X_1)$$ is finite, we have that)
$$Var(Y)=E(Y^2)=E(X_1)=\lambda$$
by symmetry (and the fact that the absolute 3rd moment exists) skew=0
4th moment: $$E(Y^4) = E(X_1^2) = \lambda+\lambda^2$$
kurtosis = $$\frac{\lambda+\lambda^2}{\lambda^2}= 1+1/\lambda$$
so when $$\lambda=\frac12$$, kurtosis is 3. This is the case illustrated above.
(d) all my examples so far have been symmetric, since symmetric answers are easier to create -- but asymmetric solutions are also possible. Here's a discrete example.
As you see, none of these examples look particularly "normal". It would be a simple matter to make any number of discrete, continuous or mixed variables with the same properties. While most of my examples were constructed as mixtures, there's nothing special about mixtures, other than they're often a convenient way to make distributions with properties the way you want, a bit like building things with Lego.
This answer gives some additional details on kurtosis that should make some of the considerations involved in constructing other examples a little clearer.
You could match more moments in similar fashion, though it requires more effort to do so. However, because the MGF of the normal exists, you can't match all integer moments of a normal with some non-normal distribution, since that would mean their MGFs match, implying the second distribution was normal as well.
Good points are made by Glen_b. I would only add consideration of the Dirac Delta function as additional grist for the mill. As Wikipedia notes, "The DDF is a generalized function, or distribution, on the real number line that is zero everywhere except at zero, with an integral of one over the entire real line" with the consequence that all higher moments of the DDF are zero.
Paul Dirac applies it to quantum mechanics in his 1931 book The Principles of Quantum Mechanics but it's origins date back to Fourier, Lesbesgue, Cauchy and others. The DDF also has physical analogues in modeling the distribution, e.g., of the crack of a bat hitting a baseball.
• What has this to do with the question? Jun 1, 2015 at 13:25
• The question is explicit about making the "first four moment[s] equal to those of [a] normal [distribution]". You haven't a hope of even matching the second central moment when you use a delta distribution.
– whuber
Jun 1, 2015 at 15:58
• Perhaps you can give an example where you match moments of a standard normal (mean 0, variance 1, $E[(X-\mu)^3]=E(X^3)=0$ and $E[(X-\mu)^4]=E(X^4)=3$). If you do that, it will answer the questions being raised and clarify your point. Jun 1, 2015 at 16:45
• @A. Donda: Excess kurtosis is the 4th standardized moment about the mean minus 3, i.e. $\mathrm{E}(X - \mathrm{E} X)^4 / (\mathrm{E}(X - \mathrm{E} X)^2)^2$, so I don't think you can say it's -3 in the case of Dirac's delta function - rather it's undefined, as the variance is zero. Jun 1, 2015 at 18:18
• @Mike Hunter: I think the questions in the title & body are equivalent: once you have a distribution with defined skewness & excess kurtosis both equal to zero, matching mean & variance to any Gaussian you want is just shifting & stretching. I stress defined because both skewness & kurtosis are standardized moments, so the Dirac delta function doesn't have them. Jun 1, 2015 at 18:51 | 1,968 | 6,631 | {"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": 32, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2022-21 | longest | en | 0.898888 |
https://arduino.stackexchange.com/questions/1410/figuring-out-some-numbers-for-a-mini-pump | 1,716,979,511,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059239.72/warc/CC-MAIN-20240529103929-20240529133929-00853.warc.gz | 85,759,209 | 42,511 | Figuring out some numbers for a mini pump
I'm just now getting into electronics (I don't know why I waited so long) and I'm just figuring out some stuff. I picked up an Arduino kit from Vilros.com and a few other parts for my project.
I'm currently trying to work on a project that is basically a humidor keeper. Living in So.California, it's extremely dry here and I would like a perfectly controlled environment for my humidor. I purchased a few things like some humidity sensors (as an input obviously), a humidifier (basically a water canister with a sponge and tubes) some tube splitters and twist valves and most importantly, a small air pump.
The small air pump is what I'm having problems with at the moment. This is the pump. Obviously it's a 3v pump. I will be using the supplied 9v battery cable when the project is ready, so I understand I'll have to use resistors to bring the voltage down so I don't burn out the pump. As I said, I'm new to this, so I'm trying to figure out ohms law to find out what kind of resistor(s) I will need. The problem is, I don't know the amperage. Where would I find this?
I looked at the details for the pump and what looks like amps (or milliamps) is really just the product number. Can anyone help me out?
Right now, I have my equation as r=6v / ?i But where do I find the current?
• The current used by a motor can vary. Normally the motor's datasheet would give you typical values. But in any case you don't want to use a voltage divider for this. May 5, 2014 at 17:52
There are two serious problems with your approach, which you will need to change in order to end up with a sound solution.
• If by "9v battery cable" you mean that you intend to use a 9v battery, this is highly unsuitable for motor applications given its low power storage capacity and high internal impedance - without extreme care to use sleep modes, it probably can't even run the Arduino alone for a week. It would be much more appropriate to use a larger capacity lower voltage source - 4 or 5 C or D cells for example. A power supply could be another option, though given the water involved, be very careful to pick an isolated one and use a ground fault protected outlet.
• Resistors are a poor way of reducing the supply voltage to a motor, as they are not only wasteful, but the applied power will depend too much on the load. A professional solution would use pulse with modulation to apply the supply voltage at a reduced duty cycle - something you can try with one the various motor shield, but you will need to be careful not to overheat the motor by using too high a duty cycle.
Probably your best bet would be to buy a pump with a 6 or 7v motor (as is showing up in the associated products at the bottom of that page), or even replace the standard-form-factor motor in that pump with a higher voltage one. A 7v or so supply voltage can then be a compromise solution for running both the motor and the arduino.
• Great post! So much to say, but so few characters! Luckily, I purchased a 6v air pump before I purchased the 3 (the 6v shipping was going to take forever) so I will use the 6v instead. I didn't even connect the dots on the PWM spots, I will look into that instead. I'm not sure of the amperage, so I'm not sure how long this system would last even with larger mAh batteries. I see someone said the Arduino draws about 25mA while running code. May 5, 2014 at 20:15
• You can reduce the current consumption of the processor by using sleep mode and only occasionally waking up to monitor, using the internal RC oscillator or at least a clock divisor, etc, however this may be tricky to do when using the arduino libraries. Additionally, other components on the board including the regulator may waste power. Make sure you never leave a signal driven against a pulling resistor. If you ultimately go with batteries you may want a custom board with a low frequency clock, low quiescent current regulator, and no onboard communication interface. May 5, 2014 at 20:22
• all great points. I guess I just want to do this as a prototype (as the Arduino is supposed to be used) and then find someone to flash a chip/board for me for this purpose alone. Is that reasonable as well? May 5, 2014 at 20:59
• Doing a simple proof of concept before chasing battery life can be worthwhile, yes. There is some risk of boxing yourself into a corner it will take work to exit if you ultimately want to run on batteries, but re-implementing things a second time based on lessons learned will obviously take less time than the first attempt. May 5, 2014 at 21:02
Unlike voltage, which depends on the source, the current depends on the motor itself. To find the current, you need to run the motor and measure it with a multimeter:
`````` +------------+
+---------+ Multimeter |
| +--------+---+
+---+---+ |
| Motor | |
+---+---+ +--------+ |
| | Power +--+
+------+ source |
+--------+
``````
Note that when the motor is actually pumping, it will draw more current.
• It also depends on how the motor is running. The more torque the motor is exerting, the more current it needs. May 5, 2014 at 18:22
• Aah I see. Luckily it's just an air pump so when the motor is pumping, it should be at its highest needs. A friend told me about a DC power supply that is variable, so I could run it at 3v and measure the current from there without risk. Would this be the same thing? May 5, 2014 at 18:57
• @ntgCleaner I think that would work. May 5, 2014 at 19:03
• It will be at its highest needs when blocked. Put a circuit breaker between the full speed and blocked currents so that you don't burn out the motor. May 5, 2014 at 19:25
• good call! I purchased a DC variable power supply (1.5-15v) from Tekpower. Whether I need it or not for this project, I'm sure I'll be able to use it sometime! May 5, 2014 at 20:17 | 1,425 | 5,896 | {"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-2024-22 | latest | en | 0.96653 |
https://www.gradesaver.com/textbooks/science/chemistry/chemistry-molecular-approach-4th-edition/chapter-1-exercises-page-36/22 | 1,716,120,098,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057786.92/warc/CC-MAIN-20240519101339-20240519131339-00834.warc.gz | 700,212,551 | 12,540 | ## Chemistry: Molecular Approach (4th Edition)
Prefix multipliers are used to change the value of a unit by powers of 10. For example, $1 milliliter = 1 mL = 10^{-3} Liters$
Along with the base units, prefix multipliers are used in SI system of units. Prefix multipliers change the unit value by powers of 10. Prefix multipliers are used to express a large number of measurements in the units that are similar in size to that quantity which we are measuring. For example: $1 kilogram = 1 kg = 10^{3} grams$ $1 megabyte = 1 MB = 10^{6} bytes$ $1 millimeter = 1 mm = 10^{-3} meters$ | 159 | 581 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2024-22 | latest | en | 0.800559 |
http://arxiv-export-lb.library.cornell.edu/abs/2105.01265 | 1,675,856,713,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500758.20/warc/CC-MAIN-20230208092053-20230208122053-00329.warc.gz | 3,494,179 | 5,896 | cs.DS
(what is this?)
# Title: Finding Triangles or Independent Sets
Abstract: (I) We revisit the algorithmic problem of finding all triangles in a graph
$G=(V,E)$ with $n$ vertices and $m$ edges. According to a result of Chiba and Nishizeki (1985), this task can be achieved by a combinatorial algorithm running in $O(m^{3/2})$ time. We derive this worst-case bound from first principles and with a simple proof. We then provide a combinatorial algorithm for finding all triangles in a graph and show that is amenable to the same running time analysis. We also show that the running time of such an algorithm cannot be improved in the worst-case and for the entire range of parameters $m$ and $n$. Our arguments are extended to the problem of finding all small complete subgraphs of a given fixed size.
(II) Given a graph $G=(V,E)$ with $n$ vertices and $m$ edges, we present a randomized algorithm that computes a $(1 \pm \varepsilon)$-approximation of the number of triangles in $G$ and finds a witness with high probability in $O\left( n^{\omega(1-\delta)} \right)$ or $O \left( \left( m n^{-2\delta} \right)^{\frac{2\omega}{\omega+1}} \right)$ expected time, provided $G$ has a suitable superlinear number of edges and triangles (where $\omega < 2.376$ is the exponent of matrix multiplication, and $\varepsilon \leq 0.5$, $\delta \leq 0.25$ are positive constants). This limits the range of a conjecture of P\v{a}tra\c{s}cu (2010) regarding the triangle detection problem.
(III) We present an algorithm which given a graph $G=(V,E)$ performs one of the following tasks in $O(m+n)$ (i.e., linear) time: (i) compute a $\Omega(1/\sqrt{n})$-approximation of a maximum independent set in $G$ (a well-known NP-hard problem), or (ii) find a triangle in $G$. The run-time is faster than that for any previous method for each of these tasks. A series of trade-off results is also available.
Comments: 17 pages, 1 figure Subjects: Data Structures and Algorithms (cs.DS) Cite as: arXiv:2105.01265 [cs.DS] (or arXiv:2105.01265v1 [cs.DS] for this version) | 563 | 2,052 | {"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} | 2.703125 | 3 | CC-MAIN-2023-06 | latest | en | 0.847478 |
http://woodworkingplans2013.com/outdoor-woodworking-plans-buffet-woodworking-plans.html | 1,556,182,001,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578711882.85/warc/CC-MAIN-20190425074144-20190425100144-00001.warc.gz | 181,631,657 | 12,464 | How do you divide 11-3/8-in. (or any other mathematically difficult number) into equal parts without dividing fractions? Simple. Angle your tape across the workpiece until it reads an easily-divisible dimension and make your marks with the tape angled. For example, say you want to divide an 11-3/8-in. board into three equal parts. Angle the tape until it reads 12-in., and then make marks at “4” and “8”. Plus: More measuring tips and tricks.
This is probably one the easiest woodworking projects you will find here. Although easy, a doormat is an equally important and useful item for households. As you can see in the image below, you will only need some 2X2 wooden boards and rope to build a simple doormat. This doormat is mostly useful for outdoor and porch. It will easily remove all the mud from your shoes with just one wipe. It is also very easy to clean and looks fabulous even if it is dirty.
There is no cost to use the database. Registering is not required. You should be able to browse the database and click through the links. Having said that, considering the fact there are so many browsers out there being used, this site's software might not allow some visitors to browse, it all depends on the Internet traffic, and your browser's compatibility.
As soon as I came across this tutorial, I didn’t wait any longer to start building one. Some of the items you need for this project are hardwood plywood, saw, glue, nails, drilling machine, etc. The video is very easy to follow for anyone with basic woodworking knowledge and experience. The first source link also includes a step by step procedure in plain English for those, who are not comfortable enough with the video tutorial.
Iron-on edge-banding is a quick way to cover up an edge on plywood. Trimming the excess, however, is tricky. I’ve tried edge-banding trimmers, but I find the results are unpredictable. With the trimmers I’ve tried, it wasn’t easy to change the direction of the cut to suit the grain direction of the edge-banding. If you’re cutting against the grain, you’re likely to tear out a chunk of your new edge-banding. Instead, I use a wide, sharp chisel. This way, I can read the grain direction and trim accordingly. Angle the chisel slightly and go slow, raising the back corner of the chisel just enough so that it doesn’t dig into the plywood veneer. Smooth the corner with a sanding block after trimming. Check out this amazing edge band veneering project!
Another awesome thing about this coffee table is that it is also has a storage unit. So you can store drinks, and other stuff in the half barrel of your table and then close or open it whenever you need. Pete has also constructed a video for this tutorial for which you can find the link below. It illustrates the same process in a video guide that shows you the exact process to be followed while building this whiskey barrel coffee table.
Among early finds of wooden tools are the worked sticks from Kalambo Falls, Clacton-on-Sea and Lehringen. The spears from Schöningen (Germany) provide some of the first examples of wooden hunting gear. Flint tools were used for carving. Since Neolithic times, carved wooden vessels are known, for example, from the Linear Pottery culture wells at Kückhofen and Eythra.
Although refrigerators long ago rendered them obsolete, antique oak ice boxes remain popular with collectors, even though they’re expensive and hard to find. This do-it-yourself version is neither: it’s both inexpensive and easy to build. An authentic reproduction of an original, the project is especially popular when used as a bar, but it has many
Here’s an easier way to stain or seal chairs, lattice or anything with numerous tight recesses. Pour the stain into a clean, empty spray bottle (\$3). Spray the stain onto the project and wipe up the excess with a brush or rag. The sprayer will squirt stain into all those tight, hard-to-reach cracks and joints. — Valrie Schuster. Learn more about staining wood.
```When it comes to woodworking for beginners, I think it’s important to just learn how to use a few of the most essential woodworking tools for beginners. There are so many awesome tools available on the market today, it can be quite overwhelming as well as expensive to try to buy them all and know how to use them. Once you learn the basics of the most essential tools you will be able to start building in no time and feel comfortable learning any other new tools in the future.
```
With the advances in modern technology and the demands of industry, woodwork as a field has changed. The development of Computer Numeric Controlled (CNC) Machines, for example, has made us able to mass-produce and reproduce products faster, with less waste, and often more complex in design than ever before. CNC Routers can carve complicated and highly detailed shapes into flat stock, to create signs or art. Rechargeable power tools speed up creation of many projects and require much less body strength than in the past, for example when boring multiple holes. Skilled fine woodworking, however, remains a craft pursued by many. There remains demand for hand crafted work such as furniture and arts, however with rate and cost of production, the cost for consumers is much higher.
Woodworking was essential to the Romans. It provided, sometimes the only, material for buildings, transportation, tools, and household items. Wood also provided pipes, dye, waterproofing materials, and energy for heat.[5]:1Although most examples of Roman woodworking have been lost,[5]:2 the literary record preserved much of the contemporary knowledge. Vitruvius dedicates an entire chapter of his De architectura to timber, preserving many details.[6] Pliny, while not a botanist, dedicated six books of his Natural History to trees and woody plants, providing a wealth of information on trees and their uses.[7]
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With an orbital sander and good sandpaper you can smooth wood evenly and easily with first-class results. When flush-sanding solid edge-banding, draw a squiggly line across the joint before sanding. The edge-banding will be slightly proud of the plywood veneer, so the pencil marks provide a visual aid to make sure that you’re sanding flat, and that you don’t sand through the plywood’s veneer. As you go, you can also test for a smooth, level transition by gently scraping your fingernails against the transition. If it’s smooth, your fingers will not catch on the seam between the two pieces
As soon as I came across this tutorial, I didn’t wait any longer to start building one. Some of the items you need for this project are hardwood plywood, saw, glue, nails, drilling machine, etc. The video is very easy to follow for anyone with basic woodworking knowledge and experience. The first source link also includes a step by step procedure in plain English for those, who are not comfortable enough with the video tutorial.
There is no cost to use the database. Registering is not required. You should be able to browse the database and click through the links. Having said that, considering the fact there are so many browsers out there being used, this site's software might not allow some visitors to browse, it all depends on the Internet traffic, and your browser's compatibility.
Tired of waiting for finish to dry on one side before finishing the other side? You can purchase standoffs, but it’s also really easy to make them yourself. Simply drive 2-in. drywall screws through 2-in. x 2-in. squares of 3/4-in. thick stock. The screw points will leave a divot in the finish that can be touched up later, but I always let the back side of my project rest on the screws while the finish dries. — Matt Boley. Plus: Check out these 32 other handy hints for frugal shop rats.
Another awesome thing about this coffee table is that it is also has a storage unit. So you can store drinks, and other stuff in the half barrel of your table and then close or open it whenever you need. Pete has also constructed a video for this tutorial for which you can find the link below. It illustrates the same process in a video guide that shows you the exact process to be followed while building this whiskey barrel coffee table.
Harbor Freight carries a wide range of high quality woodworking tools that are perfect for the home garage and professional carpentry shop alike. If you have a love of woodcrafting, Harbor Freight has the essential tools and accessories you need to tackle woodworking projects efficiently and safely, no matter your skill level. Our huge selection of drill presses, saws, sanders and planers will see you through countless jobs while our dust collection options keep your workspace clear of sawdust. Whatever you’re looking for, from basic woodworking tools to more specialty items, Harbor Freight has what you need to get the job done.
With an orbital sander and good sandpaper you can smooth wood evenly and easily with first-class results. When flush-sanding solid edge-banding, draw a squiggly line across the joint before sanding. The edge-banding will be slightly proud of the plywood veneer, so the pencil marks provide a visual aid to make sure that you’re sanding flat, and that you don’t sand through the plywood’s veneer. As you go, you can also test for a smooth, level transition by gently scraping your fingernails against the transition. If it’s smooth, your fingers will not catch on the seam between the two pieces
These things may be tiny in size, but building one is not that easy. It takes some serious woodworking knowledge and skill to build a nice wooden mobile stand. When I first saw one online, I just couldn’t resist thinking of buying one. But when I saw the price, I was forced to rethink. Also, a woodwork lover like me cannot be contained with just one piece and I was not willing to spend on more than one. So instead I decided to build myself one. Yes, it took some doing but the final result was satisfying. Luckily, I found this awesome tutorial online that helped me build my first ever wooden phone holder.
Clamping mitered edges can be a real hassle because they never seems to line up correctly. The easiest way that I’ve found to get around this process is to use painter’s tape as clamps. First set the pieces so that the outer edges are facing up and tape them edge-to-edge. The flip the pieces over so the beveled edges are facing up and glue them together. Complete the process by taping the last two edges together and let sit until completed. The tape removes easily and the glue won’t attach to the tape, making sanding and finishing very simple. Try this tip with this clever project!
When it comes to woodworking for beginners, I think it’s important to just learn how to use a few of the most essential woodworking tools for beginners. There are so many awesome tools available on the market today, it can be quite overwhelming as well as expensive to try to buy them all and know how to use them. Once you learn the basics of the most essential tools you will be able to start building in no time and feel comfortable learning any other new tools in the future.
Fir, also known as Douglas Fir, is very inexpensive and common at local home centers. It has a characteristic straight, pronounced grain with a red-brown tint. However, its grain pattern is relatively plain and it does not stain well, so Fir is commonly used when the finished product will be painted. While commonly used for building, this softwood would also be suitable for furniture-making as well.[12]
Although refrigerators long ago rendered them obsolete, antique oak ice boxes remain popular with collectors, even though they’re expensive and hard to find. This do-it-yourself version is neither: it’s both inexpensive and easy to build. An authentic reproduction of an original, the project is especially popular when used as a bar, but it has many
```Build your own platform bed frame at your home by following the source linked tutorial given above. The source link also includes more pictures that can help you to build a better bed frame. You can see a step by step set of instructions and guidelines to follow with real life pictures, as well as you can download a PDF file detailing the list of materials and tools you’ll need, know about the length of every board, and most importantly color-coded illustrations of the building process.
```
Pocket screw joinery is a system—employing special drill and driver bits—used to join boards or pieces of plywood to one another. Installing pocket screws involves using a jig to drill a sharply angled, 15-degree hole through the back of one board, then driving a special screw through that hole into the second board to draw them tightly together. Learn how to use pocket screws here. It’s often used in cabinetmaking and furniture building. Here’s how to build cabinets with pocket screws. The term “pocket” comes from the design of the hole which contains an upper “pocket” for the head of the screw to push against; this pocket also hides the head of the screw.
### Boat building Bow and arrow Bush carpentry Cabinetry Caning Carpentry Certosina Chainsaw carving Chip carving Clogs Ébéniste Fretwork Intarsia Japanese carpentry Khatam Kohlrosing Log building Marquetry Millwork Parquetry Pyrography Relief carving Root carving Sawdust Segmented turning Shingle weaving Shipbuilding Spindle turning Timber framing Treen Whittling Wood carving Woodturning Wood flour
```When you are gathering inspiration for barn door Plan, be sure to note the cost of the tools used in the plan. Barn door tools can often cost more than your actual door! But, there are many clever and affordable do it yourself tools options in the tutorials mentioned below! Let us explore some DIY Barn Door Tutorials. Just click on the blue text below and check some amazing fun Barn doors. They might be different from the one shown in above picture.
```
Lacking a jointer? Use reader Court Kites’ awesome tip to create perfectly matched glue joints on wavy or bowed board edges. Lay the boards on a flat surface, then clamp them across the middle with a bar clamp. Lay two 8-in. long by 1-3/4 in. wide scrap boards across each end and screw them in with four 1-1/4 in. long screws, two per board. Keep the screws well away from your future cutting line!
Sanding curves is tricky. Sometimes you need a sanding pad that’s both firm and flexible. A small notepad works great. Just wrap sandpaper around the pad and bend the pad to whatever arc you need. Slip the one end of the sandpaper between the pages to help hold it in place on the pad. Give this a try the next time you’re working on a project that has curves and tough to reach spots.
Finding affordable lumber has always been a mainstay for woodworkers, and when you tie our dwindling natural resources into the conversation the time is right to look at milling your own lumber. This seven-part weekly video series takes you through how to find lumber, how to operate a sawmill, details on types of sawing methods, stickering and drying and ultimately advice on using a mill as part of a business. Learn what you need to know to understand Milling Your Own Lumber.
```MATERIAL CONNECTION DISCLOSURE: You should assume that this website has an affiliate relationship and/or another material connection to the persons or businesses mentioned in or linked to from this page and may receive commissions from purchases you make on subsequent web sites. You should not rely solely on information contained in this email to evaluate the product or service being endorsed. Always exercise due diligence before purchasing any product or service. This website contains advertisements.
```
Another awesome thing about this coffee table is that it is also has a storage unit. So you can store drinks, and other stuff in the half barrel of your table and then close or open it whenever you need. Pete has also constructed a video for this tutorial for which you can find the link below. It illustrates the same process in a video guide that shows you the exact process to be followed while building this whiskey barrel coffee table.
The video explains the step by step process of making a nice wooden phone stand from scratch. My first wooden holder was not the best one, but it was good enough to motivate me to make more. I now possess 10 mobile wooden stands in different shapes and styles. And if I can make this, you too can make one yourself. Search the internet for more mobile holder ideas and start making one now.
Simply soak a washcloth in water and ring it out a bit so it’s not sopping wet. Put the damp washcloth on the affected area. The water will wick through the wood, and that’s fine. Now, with your iron on its highest setting, place it on the damp washcloth over the affected area, and make small movements back and forth and in circles. Press down firmly and continue until your wash cloth is dry. It won’t take long to evaporate. At this point, the wood fibers are absorbing the water and should expand back to where they were originally. Continue this process and repeat by adding more water until the dents rise up to be flush with the rest of the material.
Old doors laid across sawhorses make great temporary workbenches, but they take up a lot of space when you’re not using them. Instead of full-size doors, I use bifold doors with hinges so I can fold them up when I’m done working. They’re also easier to haul around in the pickup for on-the-road jobs. — Harry Steele. Here are 12 more simple workbenches you can build.
Typically furniture such as tables and chairs is made using solid stock from hardwoods due to its strength and resistance to warping.[10] Additionally, they also have a greater variety of grain patterns and color and take a finish better which allows the woodworker to exercise a great deal of artistic liberty. Hardwoods can be cut more cleanly and leave less residue on sawblades and other woodworking tools.[10] Cabinet/fixture makers employ the use of plywood and other man made panel products. Some furniture, such as the Windsor chair involve green woodworking, shaping with wood while it contains its natural moisture prior to drying.
These things may be tiny in size, but building one is not that easy. It takes some serious woodworking knowledge and skill to build a nice wooden mobile stand. When I first saw one online, I just couldn’t resist thinking of buying one. But when I saw the price, I was forced to rethink. Also, a woodwork lover like me cannot be contained with just one piece and I was not willing to spend on more than one. So instead I decided to build myself one. Yes, it took some doing but the final result was satisfying. Luckily, I found this awesome tutorial online that helped me build my first ever wooden phone holder.
Sanding concave molding doesn’t have to be difficult. Find a deep socket that fits the contour of your molding. Wrap a piece of sand- paper around the socket and hold it in place with your fingers. Your sanding will be uniform and the delicate edges of the molding won’t round over. — Eric and Cheryl Weltlich. In this video, Travis talks about his favorite sanding tips.
With an orbital sander and good sandpaper you can smooth wood evenly and easily with first-class results. When flush-sanding solid edge-banding, draw a squiggly line across the joint before sanding. The edge-banding will be slightly proud of the plywood veneer, so the pencil marks provide a visual aid to make sure that you’re sanding flat, and that you don’t sand through the plywood’s veneer. As you go, you can also test for a smooth, level transition by gently scraping your fingernails against the transition. If it’s smooth, your fingers will not catch on the seam between the two pieces
Can you believe that this amazing wine rack the image below has been built using old wood pallets? Perhaps you can. Isn’t it wonderful that your old wood pallets can be reused to build something so beautiful and useful? I am not a drinker but I liked the idea so much that I just couldn’t resist making myself one. Although later I gifted it to my parents, who totally loved it.
Particleboard is a manufactured wood product composed of sawdust, wood chips or wood shavings mixed with a resin. This concoction is layered, compressed, subjected to heat and cut to shape, resulting in a sheet material that can be used for a variety of things. It’s often used as shelving or as an underlayment for carpet. Plastic laminate may be applied to both sides to create a product that can be used to create everything from furniture to cabinets to wall paneling. Head into IKEA and you’ll find acres of particleboard.
Popular and easy to work with, cherry is in high demand for its reddish-brown color and ease of staining and finishing. Cherry likely won’t be at the local home center, but should be at a lumberyard for a somewhat expensive price.[12] This hardwood is a very common material for furniture, and is resistant to normal wear-and-tear, but it is best for indoor pieces.[13]
# Woodworking was essential to the Romans. It provided, sometimes the only, material for buildings, transportation, tools, and household items. Wood also provided pipes, dye, waterproofing materials, and energy for heat.[5]:1Although most examples of Roman woodworking have been lost,[5]:2 the literary record preserved much of the contemporary knowledge. Vitruvius dedicates an entire chapter of his De architectura to timber, preserving many details.[6] Pliny, while not a botanist, dedicated six books of his Natural History to trees and woody plants, providing a wealth of information on trees and their uses.[7]
Along with stone, clay and animal parts, wood was one of the first materials worked by early humans. Microwear analysis of the Mousterian stone tools used by the Neanderthals show that many were used to work wood. The development of civilization was closely tied to the development of increasingly greater degrees of skill in working these materials.
Finding a toolbox for a mechanic, for his hand tools, is not a big challenge at all - there are dozens of the tool boxes available on the market, from huge roll-around shop cases to small metal boxes. Plumbers, electricians, and farmers are well served, too, with everything from pickup-truck storage to toolboxes and belts. But, if you are a shop-bound woodworker then the case changes. You get to need a tool box that suits the range and variety of hand tools that most woodworkers like to have. For those who deny making do with second best, there's only one solution, you’ve to build a wooden toolbox that should be designed expressly for a woodworking shop.
Cedars are strong, aromatic softwoods that are capable of enduring outdoor elements, the most common of which is the Western Red Cedar. Western Red Cedar can sustain wet environments without succumbing to rot, and as a result is commonly used for outdoor projects such as patios, outdoor furniture, and building exteriors. This wood can be easily found at most home centers for a moderate price.[12]
```\$50 - \$100Alaska Dream HouseBathroomBedroomChildren's and Kid's Room Furniture and Toy PlansCraftroomDesk, Desk Systems and Project Table Plansdining roomentry wayFarmhouse Style Furniture PlansIndustrial Style Furniture PlansKids and Toysliving roomModern Style Furniture PlansNursery and BabyofficeRustic Furniture Plansstorage and organizationTeensIntermediateSide and End Table PlansBuffet, Sideboard and Credenza PlansCabinet PlansNightstands
```
Aside from the privacy it offers, a latticework porch trellis is a perfect way to add major curb appeal to your home for \$100 or less. The trellis shown here is made of cedar, but any decay-resistant wood like redwood, cypress or treated pine would also be a good option. Constructed with lap joints for a flat surface and an oval cutout for elegance, it’s a far upgrade from traditional premade garden lattice. As long as you have experience working a router, this project’s complexity lies mostly in the time it takes to cut and assemble. Get the instructions complete with detailed illustrations here. | 5,191 | 24,642 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2019-18 | latest | en | 0.937409 |
https://rosettacode.org/w/index.php?title=Odd_squarefree_semiprimes&oldid=335085 | 1,680,310,444,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949694.55/warc/CC-MAIN-20230401001704-20230401031704-00123.warc.gz | 533,884,200 | 49,143 | # Odd squarefree semiprimes
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
Odd squarefree semiprimes is a draft programming task. It is not yet considered ready to be promoted as a complete task, for reasons that should be found in its talk page.
Odd numbers of the form p*q where p and q are distinct primes, where p*q < 1000
## 11l
Translation of: C++
```F odd_square_free_semiprime(=n)
I n % 2 == 0
R 0B
V count = 0
V i = 3
L i * i <= n
L n % i == 0
I ++count > 1
R 0B
n I/= i
i += 2
R count == 1
print(‘Odd square-free semiprimes < 1000:’)
V count = 0
L(i) (1.<1000).step(2)
I odd_square_free_semiprime(i)
print(‘#4’.format(i), end' ‘’)
I ++count % 20 == 0
print()
print("\nCount: "count)```
Output:
```Odd square-free semiprimes < 1000:
15 21 33 35 39 51 55 57 65 69 77 85 87 91 93 95 111 115 119 123
129 133 141 143 145 155 159 161 177 183 185 187 201 203 205 209 213 215 217 219
221 235 237 247 249 253 259 265 267 287 291 295 299 301 303 305 309 319 321 323
327 329 335 339 341 355 365 371 377 381 391 393 395 403 407 411 413 415 417 427
437 445 447 451 453 469 471 473 481 485 489 493 497 501 505 511 515 517 519 527
533 535 537 543 545 551 553 559 565 573 579 581 583 589 591 597 611 623 629 633
635 649 655 667 669 671 679 681 685 687 689 695 697 699 703 707 713 717 721 723
731 737 745 749 753 755 763 767 771 779 781 785 789 791 793 799 803 807 813 815
817 831 835 843 849 851 865 869 871 879 889 893 895 899 901 905 913 917 921 923
933 939 943 949 951 955 959 965 973 979 985 989 993 995
Count: 194
```
## Action!
```INCLUDE "D2:SORT.ACT" ;from the Action! Tool Kit
INCLUDE "H6:SIEVE.ACT"
PROC Main()
DEFINE MAX="999"
BYTE ARRAY primes(MAX+1)
INT i,j,count,max2,limit
INT ARRAY res(300)
Put(125) PutE() ;clear the screen
Sieve(primes,MAX+1)
count=0 max2=MAX/2
FOR i=3 TO max2
DO
IF primes(i) THEN
limit=MAX/i
FOR j=i+1 TO limit
DO
IF primes(j) THEN
res(count)=i*j
count==+1
FI
OD
FI
OD
SortI(res,count,0)
FOR i=0 TO count-1
DO
PrintI(res(i)) Put(32)
OD
PrintF("%E%EThere are %I odd numbers",count)
RETURN```
Output:
```15 21 33 35 39 51 55 57 65 69 77 85 87 91 93 95 111 115 119 123 129 133 141 143 145 155 159 161 177 183 185 187
201 203 205 209 213 215 217 219 221 235 237 247 249 253 259 265 267 287 291 295 299 301 303 305 309 319 321 323
327 329 335 339 341 355 365 371 377 381 391 393 395 403 407 411 413 415 417 427 437 445 447 451 453 469 471 473
481 485 489 493 497 501 505 511 515 517 519 527 533 535 537 543 545 551 553 559 565 573 579 581 583 589 591 597
611 623 629 633 635 649 655 667 669 671 679 681 685 687 689 695 697 699 703 707 713 717 721 723 731 737 745 749
753 755 763 767 771 779 781 785 789 791 793 799 803 807 813 815 817 831 835 843 849 851 865 869 871 879 889 893
895 899 901 905 913 917 921 923 933 939 943 949 951 955 959 965 973 979 985 989 993 995
There are 194 odd numbers
```
## ALGOL 68
```BEGIN # find some odd square free semi-primes #
# numbers of the form p*q where p =/= q and p, q are prime #
# reurns a list of primes up to n #
PROC prime list = ( INT n )[]INT:
BEGIN
# sieve the primes to n #
INT no = 0, yes = 1;
[ 1 : n ]INT p;
p[ 1 ] := no; p[ 2 ] := yes;
FOR i FROM 3 BY 2 TO n DO p[ i ] := yes OD;
FOR i FROM 4 BY 2 TO n DO p[ i ] := no OD;
FOR i FROM 3 BY 2 TO ENTIER sqrt( n ) DO
IF p[ i ] = yes THEN FOR s FROM i * i BY i + i TO n DO p[ s ] := no OD FI
OD;
# replace the sieve with a list #
INT p pos := 0;
FOR i TO n DO IF p[ i ] = yes THEN p[ p pos +:= 1 ] := i FI OD;
p[ 1 : p pos ]
END # prime list # ;
# show odd square free semi-primes up to 1000 #
INT max number = 1000;
INT max prime = 1 + ( max number OVER 3 ); # the smallest odd prime is 3, so this shuld be enough primes #
[]INT prime = prime list( max prime );
[ 1 : max number ]BOOL numbers; FOR i TO max number DO numbers[ i ] := FALSE OD;
FOR i FROM 2 TO UPB prime - 1 DO
FOR j FROM i + 1 TO UPB prime
WHILE INT pq = prime[ i ] * prime[ j ];
pq < max number
DO
numbers[ pq ] := TRUE
OD
OD;
INT n count := 0;
FOR i TO max number DO
IF numbers[ i ] THEN
print( ( " ", whole( i, -4 ) ) );
n count +:= 1;
IF n count MOD 20 = 0 THEN print( ( newline ) ) FI
FI
OD
END```
Output:
``` 15 21 33 35 39 51 55 57 65 69 77 85 87 91 93 95 111 115 119 123
129 133 141 143 145 155 159 161 177 183 185 187 201 203 205 209 213 215 217 219
221 235 237 247 249 253 259 265 267 287 291 295 299 301 303 305 309 319 321 323
327 329 335 339 341 355 365 371 377 381 391 393 395 403 407 411 413 415 417 427
437 445 447 451 453 469 471 473 481 485 489 493 497 501 505 511 515 517 519 527
533 535 537 543 545 551 553 559 565 573 579 581 583 589 591 597 611 623 629 633
635 649 655 667 669 671 679 681 685 687 689 695 697 699 703 707 713 717 721 723
731 737 745 749 753 755 763 767 771 779 781 785 789 791 793 799 803 807 813 815
817 831 835 843 849 851 865 869 871 879 889 893 895 899 901 905 913 917 921 923
933 939 943 949 951 955 959 965 973 979 985 989 993 995
```
## Arturo
```primes: select 0..1000 => prime?
lst: sort unique flatten map primes 'p [
map select primes 'q -> all? @[odd? p*q p<>q 1000>p*q]=>[p*&]
]
loop split.every:10 lst 'a ->
print map a => [pad to :string & 4]
```
Output:
``` 15 21 33 35 39 51 55 57 65 69
77 85 87 91 93 95 111 115 119 123
129 133 141 143 145 155 159 161 177 183
185 187 201 203 205 209 213 215 217 219
221 235 237 247 249 253 259 265 267 287
291 295 299 301 303 305 309 319 321 323
327 329 335 339 341 355 365 371 377 381
391 393 395 403 407 411 413 415 417 427
437 445 447 451 453 469 471 473 481 485
489 493 497 501 505 511 515 517 519 527
533 535 537 543 545 551 553 559 565 573
579 581 583 589 591 597 611 623 629 633
635 649 655 667 669 671 679 681 685 687
689 695 697 699 703 707 713 717 721 723
731 737 745 749 753 755 763 767 771 779
781 785 789 791 793 799 803 807 813 815
817 831 835 843 849 851 865 869 871 879
889 893 895 899 901 905 913 917 921 923
933 939 943 949 951 955 959 965 973 979
985 989 993 995```
## AWK
```# syntax: GAWK -f ODD_SQUAREFREE_SEMIPRIMES.AWK
# converted from C++
BEGIN {
start = 1
stop = 999
for (i=start; i<=stop; i+=2) {
if (is_odd_square_free_semiprime(i)) {
printf("%4d%1s",i,++count%10?"":"\n")
}
}
printf("\nOdd Square Free Semiprimes %d-%d: %d\n",start,stop,count)
exit(0)
}
function is_odd_square_free_semiprime(n, count,i) {
if (and(n,1) == 0) {
return(0)
}
for (i=3; i*i<=n; i+=2) {
for (; n%i==0; n=int(n/i)) {
if (++count > 1) {
return(0)
}
}
}
return(count==1)
}
```
Output:
``` 15 21 33 35 39 51 55 57 65 69
77 85 87 91 93 95 111 115 119 123
129 133 141 143 145 155 159 161 177 183
185 187 201 203 205 209 213 215 217 219
221 235 237 247 249 253 259 265 267 287
291 295 299 301 303 305 309 319 321 323
327 329 335 339 341 355 365 371 377 381
391 393 395 403 407 411 413 415 417 427
437 445 447 451 453 469 471 473 481 485
489 493 497 501 505 511 515 517 519 527
533 535 537 543 545 551 553 559 565 573
579 581 583 589 591 597 611 623 629 633
635 649 655 667 669 671 679 681 685 687
689 695 697 699 703 707 713 717 721 723
731 737 745 749 753 755 763 767 771 779
781 785 789 791 793 799 803 807 813 815
817 831 835 843 849 851 865 869 871 879
889 893 895 899 901 905 913 917 921 923
933 939 943 949 951 955 959 965 973 979
985 989 993 995
Odd Square Free Semiprimes 1-999: 194
```
## BASIC
### FreeBASIC
Use the function from Primality by trial division#FreeBASIC as an include. This code generates the odd squarefree semiprimes in ascending order of their first factor, then their second.
```#include "isprime.bas"
dim as integer p, q
for p = 3 to 999
if not isprime(p) then continue for
for q = p+1 to 1000\p
if not isprime(q) then continue for
print p*q;" ";
next q
next p
```
Output:
` 15 21 33 39 51 57 69 87 93 111 123 129 141 159 177 183 201 213 219 237 249 267 291 303 309 321 327 339 381 393 411 417 447 453 471 489 501 519 537 543 573 579 591 597 633 669 681 687 699 717 723 753 771 789 807 813 831 843 849 879 921 933 939 951 993 35 55 65 85 95 115 145 155 185 205 215 235 265 295 305 335 355 365 395 415 445 485 505 515 535 545 565 635 655 685 695 745 755 785 815 835 865 895 905 955 965 985 995 77 91 119 133 161 203 217 259 287 301 329 371 413 427 469 497 511 553 581 623 679 707 721 749 763 791 889 917 959 973 143 187 209 253 319 341 407 451 473 517 583 649 671 737 781 803 869 913 979 221 247 299 377 403 481 533 559 611 689 767 793 871 923 949 323 391 493 527 629 697 731 799 901 437 551 589 703 779 817 893 667 713 851 943 989 899`
### Tiny BASIC
``` LET P = 1
10 LET P = P + 2
LET Q = P
IF P >= 1000 THEN END
LET A = P
GOSUB 100
IF Z = 0 THEN GOTO 10
20 LET Q = Q + 2
IF Q > 1000/P THEN GOTO 10
LET A = Q
GOSUB 100
IF Z = 0 THEN GOTO 20
PRINT P," ",Q," ",P*Q
GOTO 20
100 REM PRIMALITY BY TRIAL DIVISION
LET Z = 1
LET I = 2
110 IF (A/I)*I = A THEN LET Z = 0
IF Z = 0 THEN RETURN
LET I = I + 1
IF I*I <= A THEN GOTO 110
RETURN```
## C#
This reveals a set of semi-prime numbers (with exactly two factors for each n), where 1 < p < q < n. It is square-free, since p < q.
```using System; using static System.Console; using System.Collections;
using System.Linq; using System.Collections.Generic;
class Program { static void Main(string[] args) {
int lmt = 1000, amt, c = 0, sr = (int)Math.Sqrt(lmt), lm2; var res = new List<int>();
var pr = PG.Primes(lmt / 3 + 5).ToArray(); lm2 = pr.OrderBy(i => Math.Abs(sr - i)).First();
lm2 = Array.IndexOf(pr, lm2); for (var p = 0; p < lm2; p++) { amt = 0; for (var q = p + 1; amt < lmt; q++)
res.Add(amt = pr[p] * pr[q]); } res.Sort(); foreach(var item in res.TakeWhile(x => x < lmt))
Write("{0,4} {1}", item, ++c % 20 == 0 ? "\n" : "");
Write("\n\nCounted {0} odd squarefree semiprimes under {1}", c, lmt); } }
class PG { public static IEnumerable<int> Primes(int lim) {
var flags = new bool[lim + 1]; int j = 3;
for (int d = 8, sq = 9; sq <= lim; j += 2, sq += d += 8)
if (!flags[j]) { yield return j;
for (int k = sq, i = j << 1; k <= lim; k += i) flags[k] = true; }
for (; j <= lim; j += 2) if (!flags[j]) yield return j; } }
```
Output:
``` 15 21 33 35 39 51 55 57 65 69 77 85 87 91 93 95 111 115 119 123
129 133 141 143 145 155 159 161 177 183 185 187 201 203 205 209 213 215 217 219
221 235 237 247 249 253 259 265 267 287 291 295 299 301 303 305 309 319 321 323
327 329 335 339 341 355 365 371 377 381 391 393 395 403 407 411 413 415 417 427
437 445 447 451 453 469 471 473 481 485 489 493 497 501 505 511 515 517 519 527
533 535 537 543 545 551 553 559 565 573 579 581 583 589 591 597 611 623 629 633
635 649 655 667 669 671 679 681 685 687 689 695 697 699 703 707 713 717 721 723
731 737 745 749 753 755 763 767 771 779 781 785 789 791 793 799 803 807 813 815
817 831 835 843 849 851 865 869 871 879 889 893 895 899 901 905 913 917 921 923
933 939 943 949 951 955 959 965 973 979 985 989 993 995
Counted 194 odd squarefree semiprimes under 1000```
## C++
```#include <iomanip>
#include <iostream>
bool odd_square_free_semiprime(int n) {
if ((n & 1) == 0)
return false;
int count = 0;
for (int i = 3; i * i <= n; i += 2) {
for (; n % i == 0; n /= i) {
if (++count > 1)
return false;
}
}
return count == 1;
}
int main() {
const int n = 1000;
std::cout << "Odd square-free semiprimes < " << n << ":\n";
int count = 0;
for (int i = 1; i < n; i += 2) {
if (odd_square_free_semiprime(i)) {
++count;
std::cout << std::setw(4) << i;
if (count % 20 == 0)
std::cout << '\n';
}
}
std::cout << "\nCount: " << count << '\n';
return 0;
}
```
Output:
```Odd square-free semiprimes < 1000:
15 21 33 35 39 51 55 57 65 69 77 85 87 91 93 95 111 115 119 123
129 133 141 143 145 155 159 161 177 183 185 187 201 203 205 209 213 215 217 219
221 235 237 247 249 253 259 265 267 287 291 295 299 301 303 305 309 319 321 323
327 329 335 339 341 355 365 371 377 381 391 393 395 403 407 411 413 415 417 427
437 445 447 451 453 469 471 473 481 485 489 493 497 501 505 511 515 517 519 527
533 535 537 543 545 551 553 559 565 573 579 581 583 589 591 597 611 623 629 633
635 649 655 667 669 671 679 681 685 687 689 695 697 699 703 707 713 717 721 723
731 737 745 749 753 755 763 767 771 779 781 785 789 791 793 799 803 807 813 815
817 831 835 843 849 851 865 869 871 879 889 893 895 899 901 905 913 917 921 923
933 939 943 949 951 955 959 965 973 979 985 989 993 995
Count: 194
```
## F#
This task uses Extensible Prime Generator (F#)
```// Odd squarefree semiprimes. Nigel Galloway: January 4th., 2023
let n=primes32()|>Seq.skip 1|>Seq.takeWhile((>)333)|>List.ofSeq
List.allPairs n n|>Seq.filter(fun(n,g)->n<g)|>Seq.map(fun(n,g)->n*g)|>Seq.filter((>)1000)|>Seq.iter(printf "%d "); printfn ""
```
Output:
```15 21 33 39 51 57 69 87 93 111 123 129 141 159 177 183 201 213 219 237 249 267 291 303 309 321 327 339 381 393 411 417 447 453 471 489 501 519 537 543 573 579 591 597 633 669 681 687 699 717 723 753 771 789 807 813 831 843 849 879 921 933 939 951 993 35 55 65 85 95 115 145 155 185 205 215 235 265 295 305 335 355 365 395 415 445 485 505 515 535 545 565 635 655 685 695 745 755 785 815 835 865 895 905 955 965 985 995 77 91 119 133 161 203 217 259 287 301 329 371 413 427 469 497 511 553 581 623 679 707 721 749 763 791 889 917 959 973 143 187 209 253 319 341 407 451 473 517 583 649 671 737 781 803 869 913 979 221 247 299 377 403 481 533 559 611 689 767 793 871 923 949 323 391 493 527 629 697 731 799 901 437 551 589 703 779 817 893 667 713 851 943 989 899
```
## Factor
Works with: Factor version 0.99 2021-02-05
```USING: combinators.short-circuit formatting grouping io kernel
math.primes.factors math.ranges prettyprint sequences sets ;
: sq-free-semiprime? ( n -- ? )
factors { [ length 2 = ] [ all-unique? ] } 1&& ;
: odd-sfs-upto ( n -- seq )
1 swap 2 <range> [ sq-free-semiprime? ] filter ;
999 odd-sfs-upto dup length
"Found %d odd square-free semiprimes < 1000:\n" printf
20 group [ [ "%4d" printf ] each nl ] each nl
```
Output:
```Found 194 odd square-free semiprimes < 1000:
15 21 33 35 39 51 55 57 65 69 77 85 87 91 93 95 111 115 119 123
129 133 141 143 145 155 159 161 177 183 185 187 201 203 205 209 213 215 217 219
221 235 237 247 249 253 259 265 267 287 291 295 299 301 303 305 309 319 321 323
327 329 335 339 341 355 365 371 377 381 391 393 395 403 407 411 413 415 417 427
437 445 447 451 453 469 471 473 481 485 489 493 497 501 505 511 515 517 519 527
533 535 537 543 545 551 553 559 565 573 579 581 583 589 591 597 611 623 629 633
635 649 655 667 669 671 679 681 685 687 689 695 697 699 703 707 713 717 721 723
731 737 745 749 753 755 763 767 771 779 781 785 789 791 793 799 803 807 813 815
817 831 835 843 849 851 865 869 871 879 889 893 895 899 901 905 913 917 921 923
933 939 943 949 951 955 959 965 973 979 985 989 993 995
```
## Forth
Works with: Gforth
```: odd-square-free-semi-prime? { n -- ? }
n 1 and 0= if false exit then
0 { count }
3
begin
dup dup * n <=
while
begin
dup n swap mod 0=
while
count 1+ to count
count 1 > if
drop false exit
then
dup n swap / to n
repeat
2 +
repeat
drop
count 1 = ;
: special_odd_numbers ( n -- )
." Odd square-free semiprimes < " dup 1 .r ." :" cr
0 swap
1 do
i odd-square-free-semi-prime? if
1+
i 4 .r
dup 20 mod 0= if cr then
then
2 +loop
cr ." Count: " . cr ;
1000 special_odd_numbers
bye
```
Output:
```Odd square-free semiprimes < 1000:
15 21 33 35 39 51 55 57 65 69 77 85 87 91 93 95 111 115 119 123
129 133 141 143 145 155 159 161 177 183 185 187 201 203 205 209 213 215 217 219
221 235 237 247 249 253 259 265 267 287 291 295 299 301 303 305 309 319 321 323
327 329 335 339 341 355 365 371 377 381 391 393 395 403 407 411 413 415 417 427
437 445 447 451 453 469 471 473 481 485 489 493 497 501 505 511 515 517 519 527
533 535 537 543 545 551 553 559 565 573 579 581 583 589 591 597 611 623 629 633
635 649 655 667 669 671 679 681 685 687 689 695 697 699 703 707 713 717 721 723
731 737 745 749 753 755 763 767 771 779 781 785 789 791 793 799 803 807 813 815
817 831 835 843 849 851 865 869 871 879 889 893 895 899 901 905 913 917 921 923
933 939 943 949 951 955 959 965 973 979 985 989 993 995
Count: 194
```
## Go
Translation of: Wren
Library: Go-rcu
```package main
import (
"fmt"
"rcu"
"sort"
)
func main() {
primes := rcu.Primes(333)
var oss []int
for i := 1; i < len(primes)-1; i++ {
for j := i + 1; j < len(primes); j++ {
n := primes[i] * primes[j]
if n >= 1000 {
break
}
oss = append(oss, n)
}
}
sort.Ints(oss)
fmt.Println("Odd squarefree semiprimes under 1,000:")
for i, n := range oss {
fmt.Printf("%3d ", n)
if (i+1)%10 == 0 {
fmt.Println()
}
}
fmt.Printf("\n\n%d such numbers found.\n", len(oss))
}
```
Output:
```Odd squarefree semiprimes under 1,000:
15 21 33 35 39 51 55 57 65 69
77 85 87 91 93 95 111 115 119 123
129 133 141 143 145 155 159 161 177 183
185 187 201 203 205 209 213 215 217 219
221 235 237 247 249 253 259 265 267 287
291 295 299 301 303 305 309 319 321 323
327 329 335 339 341 355 365 371 377 381
391 393 395 403 407 411 413 415 417 427
437 445 447 451 453 469 471 473 481 485
489 493 497 501 505 511 515 517 519 527
533 535 537 543 545 551 553 559 565 573
579 581 583 589 591 597 611 623 629 633
635 649 655 667 669 671 679 681 685 687
689 695 697 699 703 707 713 717 721 723
731 737 745 749 753 755 763 767 771 779
781 785 789 791 793 799 803 807 813 815
817 831 835 843 849 851 865 869 871 879
889 893 895 899 901 905 913 917 921 923
933 939 943 949 951 955 959 965 973 979
985 989 993 995
194 such numbers found.
```
## jq
Works with: jq
Works with gojq, the Go implementation of jq
See e.g. Erdős-primes#jq for a suitable definition of `is_prime`.
```# Output: a stream of proper square-free odd prime factors of .
def proper_odd_squarefree_prime_factors:
range(3; 1 + sqrt|floor) as \$i
| select( (. % \$i) == 0 )
| (. / \$i) as \$r
| select(\$i != \$r and all(\$i, \$r; is_prime) )
| \$i, \$r;
def is_odd_squarefree_semiprime:
isempty(proper_odd_squarefree_prime_factors) | not;
# For pretty-printing
def lpad(\$len): tostring | (\$len - length) as \$l | (" " * \$l)[:\$l] + .;
def nwise(\$n):
def n: if length <= \$n then . else .[0:\$n] , (.[\$n:] | n) end;
n;
[range(3;1000;2)
| select(is_odd_squarefree_semiprime)]
| nwise(10)
Output:
As for Arturo, Wren, et al.
## Julia
```using Primes
twoprimeproduct(n) = (a = factor(n).pe; length(a) == 2 && all(p -> p[2] == 1, a))
special1k = filter(n -> isodd(n) && twoprimeproduct(n), 1:1000)
foreach(p -> print(rpad(p[2], 4), p[1] % 20 == 0 ? "\n" : ""), enumerate(special1k))
```
Output:
```15 21 33 35 39 51 55 57 65 69 77 85 87 91 93 95 111 115 119 123
129 133 141 143 145 155 159 161 177 183 185 187 201 203 205 209 213 215 217 219
221 235 237 247 249 253 259 265 267 287 291 295 299 301 303 305 309 319 321 323
327 329 335 339 341 355 365 371 377 381 391 393 395 403 407 411 413 415 417 427
437 445 447 451 453 469 471 473 481 485 489 493 497 501 505 511 515 517 519 527
533 535 537 543 545 551 553 559 565 573 579 581 583 589 591 597 611 623 629 633
635 649 655 667 669 671 679 681 685 687 689 695 697 699 703 707 713 717 721 723
731 737 745 749 753 755 763 767 771 779 781 785 789 791 793 799 803 807 813 815
817 831 835 843 849 851 865 869 871 879 889 893 895 899 901 905 913 917 921 923
933 939 943 949 951 955 959 965 973 979 985 989 993 995
```
## Mathematica / Wolfram Language
```n = 1000; primes = Most@NestWhileList[NextPrime, 3, # < n/3 &];
reduceList[x_List, n_] := TakeWhile[Rest[x], # < n/First[x] &];
q = Rest /@
NestWhileList[reduceList[#, n] &, primes, Length@# > 2 &];
semiPrimes = Sort@Flatten@MapThread[Times, {q, primes[[;; Length@q]]}];
Partition[TakeWhile[semiPrimes, # < 1000 &], UpTo[10]] // TableForm
```
Output:
```
15 21 33 35 39 51 55 57 65 69
77 85 87 91 93 95 111 115 119 123
129 133 141 143 145 155 159 161 177 183
185 187 201 203 205 209 213 215 217 219
221 235 237 247 249 253 259 265 267 287
291 295 299 301 303 305 309 319 321 323
327 329 335 339 341 355 365 371 377 381
391 393 395 403 407 411 413 415 417 427
437 445 447 451 453 469 471 473 481 485
489 493 497 501 505 511 515 517 519 527
533 535 537 543 545 551 553 559 565 573
579 581 583 589 591 597 611 623 629 633
635 649 655 667 669 671 679 681 685 687
689 695 697 699 703 707 713 717 721 723
731 737 745 749 753 755 763 767 771 779
781 785 789 791 793 799 803 807 813 815
817 831 835 843 849 851 865 869 871 879
889 893 895 899 901 905 913 917 921 923
933 939 943 949 951 955 959 965 973 979
985 989 993 995
```
## Ksh
```#!/bin/ksh
# Odd squarefree semiprimes
# # Odd numbers of the form p*q where p and q are distinct primes and p*q<1000
# # Variables:
#
integer i j candidate cnt=0
typeset -a primes osfsp
# # Functions:
#
# # Function _isprime(n) return 1 for prime, 0 for not prime
#
function _isprime {
typeset _n ; integer _n=\$1
typeset _i ; integer _i
(( _n < 2 )) && return 0
for (( _i=2 ; _i*_i<=_n ; _i++ )); do
(( ! ( _n % _i ) )) && return 0
done
return 1
}
# # Function _firstNprimes(n, arr) return array of first N primes
#
function _firstNprimes {
typeset _n ; integer _n=\$1
typeset _arr ; nameref _arr="\$2"
typeset _i _cnt ; integer _i=1 _cnt=0
while (( _cnt <= _n )); do
_isprime \${_i} ; (( \$? )) && (( _cnt++ )) && _arr+=( \${_i} )
(( _i++ ))
done
}
# # Function _isunique(n, arr) add n to array if unique to array
#
function _isunique {
typeset _n ; integer _n=\$1
typeset _arr ; nameref _arr="\$2"
typeset _buff _oldIFS
_oldIFS=\$IFS
IFS=\|
_buff=\${_arr[*]}
[[ \${_n} == @(\${_buff}) ]] || _arr+=( \${_n} )
IFS=\${_oldIFS}
}
# # Function _insertionSort(array) - Insersion sort of array of integers
#
function _insertionSort {
typeset _arr ; nameref _arr="\$1"
typeset _i _j _val ; integer _i _j _val
for (( _i=1; _i<\${#_arr[*]}; _i++ )); do
_val=\${_arr[_i]}
(( _j = _i - 1 ))
while (( _j>=0 && _arr[_j]>_val )); do
_arr[_j+1]=\${_arr[_j]}
(( _j-- ))
done
_arr[_j+1]=\${_val}
done
}
######
# main #
######
_firstNprimes 66 primes # 67th prime -> 337 * 3 = 1011 > 999
for ((i=0; i<\${#primes[*]}; i++)); do
for ((j=0; j<\${#primes[*]}; j++)); do
((j == i )) && continue
(( candidate = primes[i] * primes[j] ))
(( candidate > 999 )) || (( ! \$(( candidate & 1 )) )) && continue
_isunique \${candidate} osfsp
done
done
_insertionSort osfsp
print \${osfsp[*]}
echo
print "Counted \${#osfsp[*]} odd squarefree semiprimes under 1000"
```
Output:
```15 21 33 35 39 51 55 57 65 69 77 85 87 91 93 95 111 115 119 123 129 133 141 143
145 155 159 161 177 183 185 187 201 203 205 209 213 215 217 219 221 235 237 247
249 253 259 265 267 287 291 295 299 301 303 305 309 319 321 323 327 329 335 339
341 355 365 371 377 381 391 393 395 403 407 411 413 415 417 427 437 445 447 451
453 469 471 473 481 485 489 493 497 501 505 511 515 517 519 527 533 535 537 543
545 551 553 559 565 573 579 581 583 589 591 597 611 623 629 633 635 649 655 667
669 671 679 681 685 687 689 695 697 699 703 707 713 717 721 723 731 737 745 749
753 755 763 767 771 779 781 785 789 791 793 799 803 807 813 815 817 831 835 843
849 851 865 869 871 879 889 893 895 899 901 905 913 917 921 923 933 939 943 949
951 955 959 965 973 979 985 989 993 995
Counted 194 odd squarefree semiprimes under 1000
```
## Nim
```import algorithm, strutils, sugar
const
M = 1000 - 1
N = M div 3 # Minimal value for "p" is 3.
# Sieve of Eratosthenes.
var composite: array[3..N, bool]
for n in countup(3, N, 2):
let n2 = n * n
if n2 > N: break
if not composite[n]:
for k in countup(n2, N, 2 * n):
composite[k] = true
let primes = collect(newSeq):
for n in countup(3, N, 2):
if not composite[n]: n
var result: seq[int]
for i in 0..<primes.high:
let p = primes[i]
for j in (i+1)..primes.high:
let q = primes[j]
if p * q > M: break
result.sort()
for i, n in result:
stdout.write (\$n).align(3), if (i + 1) mod 20 == 0: '\n' else: ' '
echo()
```
Output:
``` 15 21 33 35 39 51 55 57 65 69 77 85 87 91 93 95 111 115 119 123
129 133 141 143 145 155 159 161 177 183 185 187 201 203 205 209 213 215 217 219
221 235 237 247 249 253 259 265 267 287 291 295 299 301 303 305 309 319 321 323
327 329 335 339 341 355 365 371 377 381 391 393 395 403 407 411 413 415 417 427
437 445 447 451 453 469 471 473 481 485 489 493 497 501 505 511 515 517 519 527
533 535 537 543 545 551 553 559 565 573 579 581 583 589 591 597 611 623 629 633
635 649 655 667 669 671 679 681 685 687 689 695 697 699 703 707 713 717 721 723
731 737 745 749 753 755 763 767 771 779 781 785 789 791 793 799 803 807 813 815
817 831 835 843 849 851 865 869 871 879 889 893 895 899 901 905 913 917 921 923
933 939 943 949 951 955 959 965 973 979 985 989 993 995 ```
## PARI/GP
`for(s=3, 999, f=factor(s); m=matsize(f); if(s%2==1&&m[1]==2&&f[1,2]==1&&f[2,2]==1, print(s)))`
## Perl
```#!/usr/bin/perl
use strict; # https://rosettacode.org/wiki/Odd_squarefree_semiprimes
use warnings;
my (@primes, @found) = grep \$_ & 1 && (1 x \$_) !~ /^(11+)\1+\$/, 3 .. 999 / 3;
"@primes" =~ /\b(\d+)\b.*?\b(\d+)\b(?{ \$found[\$1 * \$2] = \$1 * \$2 })(*FAIL)/;
print "@{[ grep \$_, @found[3 .. 999] ]}\n" =~ s/.{75}\K /\n/gr;
```
Output:
```15 21 33 35 39 51 55 57 65 69 77 85 87 91 93 95 111 115 119 123 129 133 141
143 145 155 159 161 177 183 185 187 201 203 205 209 213 215 217 219 221 235
237 247 249 253 259 265 267 287 291 295 299 301 303 305 309 319 321 323 327
329 335 339 341 355 365 371 377 381 391 393 395 403 407 411 413 415 417 427
437 445 447 451 453 469 471 473 481 485 489 493 497 501 505 511 515 517 519
527 533 535 537 543 545 551 553 559 565 573 579 581 583 589 591 597 611 623
629 633 635 649 655 667 669 671 679 681 685 687 689 695 697 699 703 707 713
717 721 723 731 737 745 749 753 755 763 767 771 779 781 785 789 791 793 799
803 807 813 815 817 831 835 843 849 851 865 869 871 879 889 893 895 899 901
905 913 917 921 923 933 939 943 949 951 955 959 965 973 979 985 989 993 995
```
## Phix
```function oss(integer n)
sequence f = prime_factors(n,true)
return length(f)==2 and f[1]!=f[2]
end function
sequence res = apply(true,sprintf,{{"%d"},filter(tagset(999,1,2),oss)})
printf(1,"Found %d odd square-free semiprimes less than 1,000:\n %s\n",
{length(res),join(shorten(res,"",5),", ")})
```
Output:
```Found 194 odd square-free semiprimes less than 1,000:
15, 21, 33, 35, 39, ..., 979, 985, 989, 993, 995
```
## Python
```#!/usr/bin/python
def isPrime(n):
for i in range(2, int(n**0.5) + 1):
if n % i == 0:
return False
return True
if __name__ == '__main__':
for p in range(3, 999):
if not isPrime(p):
continue
for q in range(p+1, 1000//p):
if not isPrime(q):
continue
print(p*q, end = " ");
```
## Quackery
`isprime` is defined at Primality by trial division#Quackery.
``` [ stack ] is limit
[ [] swap
[] temp put
dup limit put
3 / times
[ i^ isprime if
[ i^ join ] ]
dup witheach
[ over i^ split drop
witheach
[ over * dup
limit share < iff
[ temp take
join
temp put ]
else
[ drop
conclude ] ]
drop ]
drop
limit release
temp take ] is task ( n --> list )
Output:
`[ 15 21 33 35 39 51 55 57 65 69 77 85 87 91 93 95 111 115 119 123 129 133 141 143 145 155 159 161 177 183 185 187 201 203 205 209 213 215 217 219 221 235 237 247 249 253 259 265 267 287 291 295 299 301 303 305 309 319 321 323 327 329 335 339 341 355 365 371 377 381 391 393 395 403 407 411 413 415 417 427 437 445 447 451 453 469 471 473 481 485 489 493 497 501 505 511 515 517 519 527 533 535 537 543 545 551 553 559 565 573 579 581 583 589 591 597 611 623 629 633 635 649 655 667 669 671 679 681 685 687 689 695 697 699 703 707 713 717 721 723 731 737 745 749 753 755 763 767 771 779 781 785 789 791 793 799 803 807 813 815 817 831 835 843 849 851 865 869 871 879 889 893 895 899 901 905 913 917 921 923 933 939 943 949 951 955 959 965 973 979 985 989 993 995 ]`
## Raku
```say (3..333).grep(*.is-prime).combinations(2)».map( * * * ).flat\
.grep( * < 1000 ).sort.batch(20)».fmt('%3d').join: "\n";
```
Output:
``` 15 21 33 35 39 51 55 57 65 69 77 85 87 91 93 95 111 115 119 123
129 133 141 143 145 155 159 161 177 183 185 187 201 203 205 209 213 215 217 219
221 235 237 247 249 253 259 265 267 287 291 295 299 301 303 305 309 319 321 323
327 329 335 339 341 355 365 371 377 381 391 393 395 403 407 411 413 415 417 427
437 445 447 451 453 469 471 473 481 485 489 493 497 501 505 511 515 517 519 527
533 535 537 543 545 551 553 559 565 573 579 581 583 589 591 597 611 623 629 633
635 649 655 667 669 671 679 681 685 687 689 695 697 699 703 707 713 717 721 723
731 737 745 749 753 755 763 767 771 779 781 785 789 791 793 799 803 807 813 815
817 831 835 843 849 851 865 869 871 879 889 893 895 899 901 905 913 917 921 923
933 939 943 949 951 955 959 965 973 979 985 989 993 995```
## REXX
```/*REXX pgm finds odd squarefree semiprimes (product of 2 primes) that are less then N. */
parse arg hi cols . /*obtain optional argument from the CL.*/
if hi=='' | hi=="," then hi= 1000 /* " " " " " " */
if cols=='' | cols=="," then cols= 10 /* " " " " " " */
call genP /*build array of semaphores for primes.*/
w= 10 /*width of a number in any column. */
@oss= ' odd squarefree semiprimes < ' commas(1000)
if cols>0 then say ' index │'center(@oss, 1 + cols*(w+1) )
if cols>0 then say '───────┼'center("" , 1 + cols*(w+1), '─')
idx= 1 /*initialize the index of output lines.*/
\$=; ss.= 0 /*a list of odd squarefree semiprimes. */
do j=2 while @.j < hi /*gen odd squarefree semiprimes < HI.*/
do k=j+1 while @.k < hi; _= @.j*@.k /*ensure primes are squarefree & < HI.*/
if _>=hi then leave /*Is the product ≥ HI? Then skip it. */
ss._= 1 /*mark # as being squarefree semiprime.*/
end /*k*/
end /*j*/
oss= 0 /*number of odd squarefree semiprimes. */
do m=3 by 2 to hi-1 /*search a list of possible candicates.*/
if \ss.m then iterate /*Does this number exist? No, skip it.*/
oss= oss + 1 /*bump count of odd sq─free semiprimes.*/
if cols==0 then iterate /*Build the list (to be shown later)? */
\$= \$ right( commas(m), w) /*add an odd square─free semiprime. */
if oss//cols\==0 then iterate /*have we populated a line of output? */
say center(idx, 7)'│' substr(\$, 2); \$= /*display what we have so far (cols). */
idx= idx + cols /*bump the index count for the output*/
end /*m*/
if \$\=='' then say center(idx, 7)"│" substr(\$, 2) /*possible display residual output.*/
if cols>0 then say '───────┴'center("" , 1 + cols*(w+1), '─')
say
say 'Found ' commas(oss) @oss
exit 0 /*stick a fork in it, we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
commas: parse arg ?; do jc=length(?)-3 to 1 by -3; ?=insert(',', ?, jc); end; return ?
/*──────────────────────────────────────────────────────────────────────────────────────*/
genP: @.1=2; @.2=3; @.3=5; @.4=7; @.5=11 /*define low primes; # of primes so far*/
#= 5; sq.#= @.# ** 2 /*the highest prime squared (so far). */
/* [↓] generate more primes ≤ high.*/
do j=@.#+2 by 2 to hi+1 /*find odd primes from here on. */
parse var j '' -1 _; if _==5 then iterate /*J ÷ by 5? (right digit).*/
if j//3==0 then iterate; if j//7==0 then iterate /*" " " 3? J ÷ by 7? */
do k=5 while sq.k<=j /* [↓] divide by the known odd primes.*/
if j//@.k==0 then iterate j /*Is J ÷ X? Then not prime. ___ */
end /*k*/ /* [↑] only process numbers ≤ √ J */
#= #+1; @.#= j; sq.#= j*j /*bump # Ps; assign next P; P squared*/
end /*j*/; return
```
output when using the default inputs:
``` index │ odd squarefree semiprimes < 1,000
───────┼───────────────────────────────────────────────────────────────────────────────────────────────────────────────
1 │ 15 21 33 35 39 51 55 57 65 69
11 │ 77 85 87 91 93 95 111 115 119 123
21 │ 129 133 141 143 145 155 159 161 177 183
31 │ 185 187 201 203 205 209 213 215 217 219
41 │ 221 235 237 247 249 253 259 265 267 287
51 │ 291 295 299 301 303 305 309 319 321 323
61 │ 327 329 335 339 341 355 365 371 377 381
71 │ 391 393 395 403 407 411 413 415 417 427
81 │ 437 445 447 451 453 469 471 473 481 485
91 │ 489 493 497 501 505 511 515 517 519 527
101 │ 533 535 537 543 545 551 553 559 565 573
111 │ 579 581 583 589 591 597 611 623 629 633
121 │ 635 649 655 667 669 671 679 681 685 687
131 │ 689 695 697 699 703 707 713 717 721 723
141 │ 731 737 745 749 753 755 763 767 771 779
151 │ 781 785 789 791 793 799 803 807 813 815
161 │ 817 831 835 843 849 851 865 869 871 879
171 │ 889 893 895 899 901 905 913 917 921 923
181 │ 933 939 943 949 951 955 959 965 973 979
191 │ 985 989 993 995
───────┴───────────────────────────────────────────────────────────────────────────────────────────────────────────────
Found 194 odd squarefree semiprimes < 1,000
```
## Ring
```load "stdlib.ring" # for isprime() function
? "working..." + nl + "Odd squarefree semiprimes are:"
limit = 1000 Prim = []
# create table of prime numbers from 3 to 1000 / 3
pr = []
for n = 3 to 1000 / 3
next pl = len(pr)
# calculate upper limit for n
for nlim = 1 to pl
if pr[nlim] * pr[nlim] > limit exit ok
next nlim--
# add items to result list and sort
for n = 1 to nlim
for m = n + 1 to pl
amt = pr[n] * pr[m]
if amt > limit exit ok
next
next Prim = sort(Prim)
# display results
for n = 1 to len(Prim)
see sf(Prim[n], 4) + " "
if n % 20 = 0 see nl ok
next n--
? nl + nl + "Found " + n + " Odd squarefree semiprimes." + nl + "done..."
# a very plain string formatter, intended to even up columnar outputs
def sf x, y
s = string(x) l = len(s)
if l > y y = l ok
return substr(" ", 11 - y + l) + s```
Output:
```working...
Odd squarefree semiprimes are:
15 21 33 35 39 51 55 57 65 69 77 85 87 91 93 95 111 115 119 123
129 133 141 143 145 155 159 161 177 183 185 187 201 203 205 209 213 215 217 219
221 235 237 247 249 253 259 265 267 287 291 295 299 301 303 305 309 319 321 323
327 329 335 339 341 355 365 371 377 381 391 393 395 403 407 411 413 415 417 427
437 445 447 451 453 469 471 473 481 485 489 493 497 501 505 511 515 517 519 527
533 535 537 543 545 551 553 559 565 573 579 581 583 589 591 597 611 623 629 633
635 649 655 667 669 671 679 681 685 687 689 695 697 699 703 707 713 717 721 723
731 737 745 749 753 755 763 767 771 779 781 785 789 791 793 799 803 807 813 815
817 831 835 843 849 851 865 869 871 879 889 893 895 899 901 905 913 917 921 923
933 939 943 949 951 955 959 965 973 979 985 989 993 995
Found 194 Odd squarefree semiprimes.
done...
```
## Sidef
```func odd_squarefree_almost_primes(upto, k=2) {
k.squarefree_almost_primes(upto).grep{.is_odd}
}
with (1e3) {|n|
var list = odd_squarefree_almost_primes(n, 2)
say "Found #{list.len} odd square-free semiprimes <= #{n.commify}:"
say (list.first(10).join(', '), ', ..., ', list.last(10).join(', '))
}
```
Output:
```Found 194 odd square-free semiprimes <= 1,000:
15, 21, 33, 35, 39, 51, 55, 57, 65, 69, ..., 951, 955, 959, 965, 973, 979, 985, 989, 993, 995
```
## Wren
Library: Wren-math
Library: Wren-seq
Library: Wren-fmt
Library: Wren-sort
```import "/math" for Int
import "/seq" for Lst
import "/fmt" for Fmt
import "/sort" for Sort
var primes = Int.primeSieve(333)
var oss = []
for (i in 1...primes.count-1) {
for (j in i + 1...primes.count) {
var n = primes[i] * primes[j]
if (n >= 1000) break
}
}
Sort.quick(oss)
System.print("Odd squarefree semiprimes under 1,000:")
for (chunk in Lst.chunks(oss, 10)) Fmt.print("\$3d", chunk)
System.print("\n%(oss.count) such numbers found.")
```
Output:
```Odd squarefree semiprimes under 1,000:
15 21 33 35 39 51 55 57 65 69
77 85 87 91 93 95 111 115 119 123
129 133 141 143 145 155 159 161 177 183
185 187 201 203 205 209 213 215 217 219
221 235 237 247 249 253 259 265 267 287
291 295 299 301 303 305 309 319 321 323
327 329 335 339 341 355 365 371 377 381
391 393 395 403 407 411 413 415 417 427
437 445 447 451 453 469 471 473 481 485
489 493 497 501 505 511 515 517 519 527
533 535 537 543 545 551 553 559 565 573
579 581 583 589 591 597 611 623 629 633
635 649 655 667 669 671 679 681 685 687
689 695 697 699 703 707 713 717 721 723
731 737 745 749 753 755 763 767 771 779
781 785 789 791 793 799 803 807 813 815
817 831 835 843 849 851 865 869 871 879
889 893 895 899 901 905 913 917 921 923
933 939 943 949 951 955 959 965 973 979
985 989 993 995
194 such numbers found.
```
## XPL0
```func IsPrime(N); \Return 'true' if N is a prime number
int N, I;
[if N <= 1 then return false;
for I:= 2 to sqrt(N) do
if rem(N/I) = 0 then return false;
return true;
];
def Max = 1000;
int N, A(Max), P, Q, Count;
[for N:= 0 to Max-1 do
A(N):= false;
for P:= 3 to Max/5 do
if IsPrime(P) then
for Q:= P+2 to Max/P do
if IsPrime(Q) then
if P*Q < Max then
A(P*Q):= true;
Count:= 0;
for N:= 0 to Max-1 do
if A(N) then
[IntOut(0, N);
Count:= Count+1;
if rem(Count/10) = 0 then CrLf(0) else ChOut(0, 9\tab\);
];
CrLf(0);
IntOut(0, Count);
Text(0, " odd squarefree semiprimes found below 1000.
");
]```
Output:
```15 21 33 35 39 51 55 57 65 69
77 85 87 91 93 95 111 115 119 123
129 133 141 143 145 155 159 161 177 183
185 187 201 203 205 209 213 215 217 219
221 235 237 247 249 253 259 265 267 287
291 295 299 301 303 305 309 319 321 323
327 329 335 339 341 355 365 371 377 381
391 393 395 403 407 411 413 415 417 427
437 445 447 451 453 469 471 473 481 485
489 493 497 501 505 511 515 517 519 527
533 535 537 543 545 551 553 559 565 573
579 581 583 589 591 597 611 623 629 633
635 649 655 667 669 671 679 681 685 687
689 695 697 699 703 707 713 717 721 723
731 737 745 749 753 755 763 767 771 779
781 785 789 791 793 799 803 807 813 815
817 831 835 843 849 851 865 869 871 879
889 893 895 899 901 905 913 917 921 923
933 939 943 949 951 955 959 965 973 979
985 989 993 995
194 odd squarefree semiprimes found below 1000.
``` | 17,837 | 41,466 | {"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 | 0.327969 |
https://www.nbcphiladelphia.com/news/sports/triple-crown-best-bets/1991264/ | 1,601,074,519,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400228998.45/warc/CC-MAIN-20200925213517-20200926003517-00149.warc.gz | 967,678,894 | 25,285 | A Bettor's Guide to the Kentucky Derby
Do you have your game on? We've got you covered - from understanding the odds to the track's special conditions
Fancy putting down a wager on the Kentucky Derby, airing live on NBC Saturday, but don't want to look like a novice when you place your bet?
For all you inexperienced bettors out there, here's a quick guide to betting on the Kentucky Derby that'll have you sounding like a serious handicapper by the time Justify, Vino Rosso and Audible approach the starting gate.
DECODING THE LINGO
Phillies Honor Phandemic Krew With a Special Bobblehead
Odds-on many racing newcomers may not know what the odds actually mean. Whenever there are two numbers (e.g., 16:1 for Vino Rosso at time of writing according to VegasInsider.com) displayed on a tote board at a racetrack or on a list of wager options, the first number (16) denotes the minimum amount of profit the wager will pay. The second number (1) is the amount you need to wager to win the first amount.
Once the final finishing places of a race are official, the track will post the prices of the winning wagers. In the above example, the horse will pay \$16. The track will then add the \$16 profit and the \$1 wager together to derive the payout: \$16 + \$1 = \$17. Lone Sailor at 50:1 would therefore pay \$51 on a \$1 bet.
If a horse is quoted with only a single number, it is implied that the missing second number is a 1. In other words, a 7 on the tote board means 7:1. So if you made a \$2 wager, a bet on a horse with 7:1 odds would pay \$16. That's because 7:1 is the same as 14:2, so \$14 + \$2 = \$16. (In betting on horse races, payouts are generally based on a \$2 wager.)
Now that the odds makes sense, it’s time to decide the type of wager you want to make. Here are some of the most popular bets:
Win Your horse must finish first to collect.
Place Your horse must finish first or second to collect.
Show Your horse must finish first, second or third to collect.
Exacta You play two horses, and they must come in first and second in the exact order specified in order to collect.
Exacta Box You play two horses, as above, but here they must come in first and second in either order to collect.
Trifecta You play three horses, and to win, they must come in first, second and third in exact order to collect.
Trifecta Box You play three horses, and they must finish first, second and third in any order to collect.
Superfecta You play four horses, and they must come in first, second, third and fourth in exact order.
Superfecta Box You play four horses, and to win they must finish first, second, third and fourth in any order.
SPECIAL CONDITIONS
But novice bettors need to take into account more than just the odds for the Kentucky Derby. To further boost your chances of making a winning bet on May 5, you should also consider the following:
Distance: The Kentucky Derby is run over a distance of a mile and a quarter. Few three-year-olds will have had prior experience in such a long race. Some horses are ‘bred to distance’ and are usually a better candidate than one without a lineage of success at long races that put a premium on endurance.
Schedule: One of the most significant reasons that winning the Triple Crown is such a rare event is the grueling schedule of the three races. While the ideal layoff between races varies from horse to horse, most high level equine competitors race fewer than 10 times per year. In most cases, thoroughbreds seldom race without a break of three weeks to a month. For a Triple Crown aspirant, however, it’s necessary to win three very competitive races in a five-week span. In recent years there has been a trend away from horses running in all three legs unless they’re in contention for the Triple Crown. For this reason, it’s worth giving special consideration to 'rested' horses.
Weather/Track Condition: If there is a chance for bad weather and/or an off track it’s essential to consider that when handicapping the race. One good measure of a horse’s ability in this type of race can be found with a quick look at past performances. If a young horse has any experience on a muddy or sloppy track that’s a good indication that his connections have confidence in his abilities in these circumstances.
Coverage of the 2018 Kentucky Derby will air live Saturday, May 5 starting at 2:30 p.m. ET on NBC. | 1,006 | 4,384 | {"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-2020-40 | latest | en | 0.927954 |
https://vs.eyeandcontacts.com/2022/03/ncert-class-8-maths-chapter-4-practical_21.html | 1,726,772,326,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700652055.62/warc/CC-MAIN-20240919162032-20240919192032-00779.warc.gz | 543,588,454 | 15,561 | ## Chapter 4 Practical Geometry Exercise 4.5
Draw the following.
Question 1: The square READ with RE = 5.1 cm.
Given
RE = 5.1 cm
Rough figure
Actual figure
Steps of Construction
1. Draw line segment DA = 5.1 cm.
2. Make ∠XDA = 90°.
3. Make ∠YAD = 90°.
4. With D as centre, radius 5.1 cm draw an arc to cut ray XD at R.
5. With A as centre, radius 5.1 cm draw an arc to cut ray YA at E.
6. Join R and E.
Question 2: A rhombus whose diagonals are 5.2 cm and 6.4 cm long.
Given
Rhombus
Diagonals = 5.2 cm, 6.4 cm
Rough figure
Actual figure
Steps of Construction
1. Draw line segment UA = 6.4 cm.
2. Draw the perpendicular bisector of UA at O. (O is the midpoint of UA).
3. With O as centre, radius 2.6 cm draw arcs on either sides to intersect the perpendicular bisector at R and M.
4. Join RU, RA, UM, AM.
5. RAMU is the required rhombus.
Question 3: A rectangle with adjacent sides of lengths 5 cm and 4 cm.
Given
Rectangle
Length = 5 cm
Breadth = 4 cm
Rough figure
Actual figure
Steps of Construction
1. Draw line segment DA = 5 cm.
2. Make ∠XDA = 90°.
3. Make ∠YAD = 90°.
4. With D as centre, radius 4 cm draw an arc to cut ray XD at R.
5. With A as centre, radius 4 cm draw an arc to cut ray YA at E.
6. Join R and E.
7. READ is the required rectangle.
Question 4: A parallelogram OKAY where OK = 5.5 cm and KA = 4.2 cm. | 453 | 1,338 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2024-38 | latest | en | 0.850282 |
https://mathematica.stackexchange.com/questions/linked/85139?sort=active&page=2 | 1,618,911,232,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039388763.75/warc/CC-MAIN-20210420091336-20210420121336-00307.warc.gz | 493,052,967 | 31,394 | 461 views
### Get frequency axis in Fourier transform (and save them)
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### Determining Periodicity with Fourier?
Note: This question has previously been marked as a duplicate of What do the X and Y axis stand for in the Fourier transform domain?. While the answer will likely have an overlap, so the question ... | 735 | 3,206 | {"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} | 3.140625 | 3 | CC-MAIN-2021-17 | latest | en | 0.920307 |
https://academy.vertabelo.com/course/ms-sql-common-functions/date-time/date-time/practice-time | 1,545,013,892,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376828056.99/warc/CC-MAIN-20181217020710-20181217042710-00237.warc.gz | 505,129,589 | 14,225 | Introduction
Dealing with dates
Working with time data
Date and time date
21. Practice time!
Extracting dates and times
Doing arithmetic with dates
Converting date and time data
Current date and time data
Building date and time data from parts
Summary and review
## Instruction
Jolly good! Let's do one more exercise before we move on.
## Exercise
Find all aircraft that were launched before January 1, 2014 and took more than one flight. For each aircraft, show its ID (as Id) and calculate the average distance covered on all its routes. Name the column AvgDistance.
### Stuck? Here's a hint!
• Use AVG(x) to find the average value of x.
• Remember about the proper date format and use quotes: 'YYYY-MM-DD'.
• Use HAVING with COUNT(x) after GROUP BY to satisfy the condition for more than 1 flight. | 189 | 806 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2018-51 | longest | en | 0.924964 |
https://www.enotes.com/topics/math/questions/2-3-463700 | 1,719,247,229,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865401.1/warc/CC-MAIN-20240624151022-20240624181022-00050.warc.gz | 667,003,389 | 18,681 | # 2+3=?
2 + 3 = ?
2 = 1+1
3 = 1+1+1
Therefore, 1+1+1+1+1 = 5
2 + 3 = 5
The solution is 5.
Approved by eNotes Editorial
2+3=5
Approved by eNotes Editorial
`2+3 =5`
Explanation: The set of Natural numbers `NN` is a subset of the Integer numbers `ZZ`. The natural numbers begins with number `0` and each following number is incremented by one. Thus we have the numbers in sequence
`0`
`0+1 =1`
`1+1 =2`
`2+1 =3`
`3+1 =4`
`4+1 =5`
and so on.
from the above definition relation we can write (because addition is associate in the set of natural numbers):
`2+3 =2+(2+1) =2+(1+1)+1 = (2+1)+1+1 =(3+1)+1 =4+1 =5`
Approved by eNotes Editorial
You are finding the sum of 2 and 3.
Since we are adding 2 positive numbers, we find the sum of their absolute values and use the common sign.
The absolute value of 2 is 2.
The absolute value of 3 is 3.
The sum of the absolute values with the common sign is 5.
Therefore, 2 + 3 = 5.
The sum is 5. | 326 | 956 | {"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} | 4.5625 | 5 | CC-MAIN-2024-26 | latest | en | 0.888682 |
http://theoryessential.dyn.ddnss.de/magic-tricks/can-you-levitate-a-person-with-magnets-magic-tricks-for-beginners-with-pen/ | 1,627,724,071,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154085.58/warc/CC-MAIN-20210731074335-20210731104335-00219.warc.gz | 46,783,321 | 6,890 | # Can you levitate a person with magnets? – Magic Tricks For Beginners With Pen
(No, I’m looking at you, my little brother)
How would you lift a person, even someone as tall as the Statue of Liberty, by magnets?
Can you levitate someone with a piece of iron wire on earth? (No, but don’t you think you could probably levitate it?)
How about using electricity to lift someone over water and then pull them away by the rope?
How about using electricity and magnets to levitate a person on Earth or in space?
Would it be possible to lift a man, even taller than the world famous Statue of Liberty, by magnetism?
How about magnetically levitating the Statue of Liberty herself, even taller than the Statue of Liberty?
How about creating a magnetic field around her to levitate an entire building or even a section of the Statue?
How would you levitate something weighing tons and making it float on only one side?
Can you levitate a person with a magnetic field?
Could you levitate a very tall man without ever touching him?
If no, let’s keep this simple. You cannot levitate a person. You cannot levitate a group of people.
But don’t stop there. The question is, could you levitate a person with a magnetic field? If you could, then why couldn’t you do it? But if you cannot, then that means you are not capable of levitation, yet. You are, perhaps, capable only of levitating a person. What a relief. (Just don’t tell the children that you do not levitate.)
For further evidence, see if it makes any sense to see how you might levitate someone. Now imagine how you would do it and you will see how very simple it could be in just a few days.
A Simple Example
If we are levitating a car, wouldn’t it make sense to do that levitating by magnetism? To begin with, how can we levitate something that doesn’t exist? When I was a kid (ages 10 to 15), I used to dream up all sorts of ways to make things fly, pull objects into space, and such. One of the dream ideas I had was that I would levitate a car on my own, without the help of a motor. This seemed absurd to me at the time, but a lot of people were convinced that I would be able to levitate a car without help at all.
magic tricks with coins easy kids, youtube magic tricks tutorial tagalog-english translator, easy magic tricks for beginners instructions playing, making magic tricks easy troom troom, azzyland magic tricks revealed | 544 | 2,401 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2021-31 | latest | en | 0.958342 |
https://www.myfxbook.com/de/community/programming/howcalculateaveragevolumeperpricelevel3f3f/773710,1 | 1,534,626,535,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221213737.64/warc/CC-MAIN-20180818193409-20180818213409-00158.warc.gz | 939,811,905 | 30,572 | How to calculate AVERAGE VOLUME PER PRICE LEVEL ??
Mitglied seit May 02, 2012 14 Beiträge EAcomparison Jun 02 2014 at 08:58 Need help to calculate AVERAGE VOLUME PER PRICE LEVEL Hello everyone,I'm wondering if there's anyone out there who could advise on how to calculate average volume per price level in that particular candlestick bar.UPBAR Eg. High 1.00010 Close 1.00008 Open 1.00005 Low 1.00000 Let's say that the total volume is 100. What's the formula for me to get the average volume for the (a). H to C range, (b). O to C, (c). O to L assuming the 100 volume is evenly distributed from the H to L.Anyone care to share some thoughts ??Thanks and best regards. Mitglied seit Nov 21, 2011 1706 Beiträge CrazyTraderfx (CrazyTrader) Jun 03 2014 at 21:19 What is 'Volume'? You love my free signals... Like me on fb Mitglied seit Feb 07, 2011 724 Beiträge PipGnostic (TheCyclist) Jun 04 2014 at 11:00 You'd need to know where the transactions were, what volume at what price. In retail fx in general you do not have access to that information. The volumes MT provides are tick counters. Not volume of trades at levels. It's the count of packets of information coming in that has nothing to do with trade volumes that you can get weighted averages from. So the information, like any other volume based indicator is use in fx today, would be useless. However, this is how you calculate it a weighted average:https://www.investopedia.com/terms/w/weightedaverage.asp Show me your pips!! Mitglied seit May 02, 2012 14 Beiträge EAcomparison Jun 06 2014 at 07:42 Thanks for sharing PipGnostic Um Kommentare abgeben zu können müssen Sie sich einloggen.
Rightmove House Price Index (YoY) (1 day)
NewsWatch: Meet the tech-savvy upstarts ...(8 min ago)
EURUSD 1.14416 GBPUSD 1.27513 USDJPY 110.517 USDCAD 1.30629
Looking to open a Forex account? | 514 | 1,866 | {"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-2018-34 | latest | en | 0.726079 |
https://discuss.codecademy.com/t/median-list-index-out-of-range/104210 | 1,537,472,088,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267156554.12/warc/CC-MAIN-20180920175529-20180920195929-00541.warc.gz | 500,011,822 | 5,434 | # Median - list index out of range
#1
Title is fairly self explanatory. This code is giving me a 'list index out of range' error, on the test list [4, 5, 5, 4]
def median(numlist):
sortlist = sorted(numlist)
if len(sortlist) % 2 == 0:
inone = sortlist[len(sortlist) / 2]
intwo = sortlist[inone - 1]
return inone * intwo / 2
else:
if len(sortlist) != 1:
return sortlist[(len(sortlist) / 2) + 0.5]
else:
return sortlist[0]
#2
This might be self-explanatory, but I don't see it that way. Could you explain how inone - 1 is an index?
And how is,
a median?
#3
When you are dividing by 2, Python is forcing the answer to an integer. The expected result is float, so try dividing by a float (i.e. 2.0).
#4
As I understand it-and I could absolutely be wrong-inone becomes the index at [len(sortlist)/2], and so if I subtract one from inone, it then gets the index directly under it.
As for how inone * intwo / 2 is a median, it's taking the average of both values, which (ideally) should be the middle values, thus being the median of the number.
#5
No, it actively gets upset when I try to use a float, at least in the indeces (if that's what you're referring to)
Making the end result a float didn't help either, it's still giving me a list index out of range error.
#6
We take an average of values by adding them up, then dividing the total by the number of terms. You are multiplying, which is wildly different.
4 + 5 = 9
4 * 5 = 20
The size of the sample space has a bearing on where the middle term (or terms) will be. Since the list is zero-indexed, the middle will be,
ss = sorted(sample_space)
n = len(ss)
m = int(n / 2)
The value of m is the middle index of an odd length list. But if we look at an even length list...
0 1 2 3 4 5 6 7
n = 8
m = 4
You are on the right track as far as the index of the other element in the middle pair, it is indeed, m - 1. In your code you are assigning the value at index m, minus 1 to the other index, rather than m - 1.
#7
We take an average of values by adding them up, then dividing the total by the number of terms. You are multiplying, which is wildly different.
4 + 5 = 9
4 * 5 = 20
The size of the sample space has a bearing on where the middle term (or terms) will be. Since the list is zero-indexed, the middle will be,
ss = sorted(sample_space)
n = len(ss)
m = int(n / 2)
The value of m is the middle index of an odd length list. But if we look at an even length list...
0 1 2 3 4 5 6 7
n = 8
m = 4
You are on the right track as far as the index of the other element in the middle pair, it is indeed, m - 1.
inone is, as earlier mentioned a value, not an index.
#8
No you are correct, the indices have to be integers. I was referring to the return line, inone * intwo / 2. This is forcing that result to be in integer (4) instead of a float (4.5) which is the expected result. Try (inone * intwo) / 2.0
Edit, also as above it should be addition not multiplication to get the average of the two numbers.
#9
This topic was automatically closed 7 days after the last reply. New replies are no longer allowed. | 890 | 3,091 | {"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} | 3.875 | 4 | CC-MAIN-2018-39 | latest | en | 0.898997 |
https://www.reference.com/web?q=convert+25+x+20+cm+to+inches&qo=relatedSearchSP2&o=600605&l=dir&sga=1 | 1,597,077,225,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439736057.87/warc/CC-MAIN-20200810145103-20200810175103-00159.warc.gz | 746,035,219 | 18,472 | Web Results
www.reference.com/article/convert-19-inches-cm-ff4232934711c98a
When converted, 19 inches is equivalent to 48.26 centimeters. The formula for conversion is based upon the equivalency of 0.3937 inches to 1 centimeter. In the formula, the total number of inches is divided by 0.3937 to obtain the measurement in centimeters.
www.reference.com/article/56-cm-converted-inches-253f27e99ddd59c
As there are approximately 0.3937 inches in a centimeter, 56 centimeters equals about 22.05 inches. In order to convert centimeters to inches manually, multiply the number of centimeters (in this problem 56) by 0.3937.
www.reference.com/article/96-cm-converted-inches-40563f099bf6703c
Ninety six centimeters converted to inches equals 37 51/64 inches. One centimeter is equal to 0.39370 inches. Both the centimeter and the inch are units of length.
Related Search
www.reference.com/article/convert-157-cm-inches-762d407a38ae4485
To convert from centimeters to inches, you must know the conversion factor between the two units and how to solve basic unit conversion problems. In this case, the factor is 2.54 centimeters per inch, or 0.39 inches per centimeter. Either one can be used to arrive at th...
www.reference.com/article/convert-25-cm-mm-3a31a67653611449
You can convert any centimeter measurement to millimeters by remembering that there are 10 millimeters in 1 centimeter. Anytime you need to convert from centimeters to millimeters, simply multiply the centimeter value by 10.
www.reference.com/article/convert-192-cm-feet-inches-76298dcbe5c67664
To convert 192 centimeters to inches, multiply 192 x 0.39370 to obtain the final answer in inches. Then, the answer in inches can be easily converted to feet by multiplying the answer in inches by 0.0833. When converted, 192 cm equals slightly less than 6 feet 4 inches.
www.reference.com/article/convert-cm-mm-9d06e3f7d765dfac
Centimeters and millimeters are both fractional amounts of a metric meter. One centimeter is 1/100 of a meter and 1 millimeter is 1/1,000 of a meter. Therefore, each centimeter is equal to 10 millimeters. | 559 | 2,105 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2020-34 | latest | en | 0.805037 |
http://gauss.ececs.uc.edu/Courses/c472/homework/divideconquer.html | 1,542,774,376,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039747024.85/warc/CC-MAIN-20181121032129-20181121054129-00416.warc.gz | 124,607,583 | 1,585 | 20-CS-472-001 Design and Analysis of Algorithms II Winter 2011
## Divide and Conquer
Due: xxx, 2011
1. Design a divide-and-conquer algorithm for the vertex cover problem which is stated as follows:
``` Instance: a graph G=(V,E)
Find: a minimum sized subset S of V such that every edge in E
contains at least one vertex in S as an endpoint
```
What is the complexity of this algorithm (measurement is on the number of executions of the most frequently executed statement?
2. Consider the following problem which we call SS:
Instance: A set S={a1,a2,...,an} of n integers, each having value between 1 and given number m, and a number B.
Question: Find a subset S' of S such that the sum of all the numbers in S' is B or verify that no such subset exists.
Produce a Divide and Conquer algorithm for this problem. What is its worst case complexity?
Suppose B is one half the sum of all numbers in S, suppose the numbers of S are distributed uniformly over 1 to m, and suppose m is a lot less than n. Come up with an algorithm that exploits this case and finds a solution (S') with very high probability in O(n2) time. Observe, the algorithm may sometimes fail but usually does not!!!
3. Show that the median of five in a list of n numbers can be computing using six comparisons. | 309 | 1,295 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2018-47 | latest | en | 0.944489 |
http://oeis.org/A257420 | 1,580,291,321,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251789055.93/warc/CC-MAIN-20200129071944-20200129101944-00135.warc.gz | 120,584,644 | 3,874 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
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A257420 Number of (n+2)X(2+2) 0..1 arrays with every 3X3 subblock diagonal sum plus antidiagonal minimum nondecreasing horizontally, vertically and ne-to-sw antidiagonally 1
2616, 7652, 30309, 119393, 403951, 1293242, 4098078, 12497488, 37528750, 111339861, 326552920, 949466125, 2746182574, 7900552664, 22633690594, 64617935573, 183950154022, 522353720305, 1480269737481, 4187439072469 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 COMMENTS Column 2 of A257426 LINKS R. H. Hardin, Table of n, a(n) for n = 1..210 FORMULA Empirical: a(n) = 8*a(n-1) -22*a(n-2) +19*a(n-3) +18*a(n-4) -51*a(n-5) +72*a(n-6) -170*a(n-7) +350*a(n-8) -290*a(n-9) -175*a(n-10) +299*a(n-11) +215*a(n-12) +328*a(n-13) -2659*a(n-14) +3621*a(n-15) -1361*a(n-16) -1414*a(n-17) +3863*a(n-18) -5433*a(n-19) +3704*a(n-20) -1335*a(n-21) +42*a(n-22) +1911*a(n-23) -3043*a(n-24) +3082*a(n-25) -2396*a(n-26) +1173*a(n-27) -1060*a(n-28) +488*a(n-29) +1217*a(n-30) -1835*a(n-31) +1483*a(n-32) -954*a(n-33) +223*a(n-34) +245*a(n-35) -225*a(n-36) +128*a(n-37) -173*a(n-38) +179*a(n-39) -93*a(n-40) +23*a(n-41) +2*a(n-42) -8*a(n-43) +13*a(n-44) -13*a(n-45) +6*a(n-46) -a(n-47) for n>56 EXAMPLE Some solutions for n=4 ..0..1..0..0....0..1..0..0....0..0..0..1....0..0..1..1....0..1..1..1 ..1..1..1..1....0..0..0..0....0..0..0..1....0..1..0..0....0..1..0..0 ..0..1..0..0....1..0..0..0....1..0..1..1....0..0..0..1....0..1..0..1 ..0..1..0..1....0..0..1..1....1..0..1..1....1..1..1..0....1..1..1..1 ..0..1..1..1....0..1..0..0....0..1..1..0....1..1..1..0....1..1..1..1 ..1..1..1..0....0..0..0..1....0..1..0..0....0..1..0..0....1..1..1..1 CROSSREFS Cf. A257426 Sequence in context: A235777 A235557 A257427 * A254236 A252440 A066871 Adjacent sequences: A257417 A257418 A257419 * A257421 A257422 A257423 KEYWORD nonn AUTHOR R. H. Hardin, Apr 22 2015 STATUS approved
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Last modified January 29 04:46 EST 2020. Contains 331335 sequences. (Running on oeis4.) | 1,018 | 2,360 | {"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-2020-05 | latest | en | 0.329237 |
https://www.reference.com/web?q=Bar+Graph+Definition&qo=relatedSearchNarrow&o=600605&l=dir | 1,563,715,161,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195527000.10/warc/CC-MAIN-20190721123414-20190721145414-00449.warc.gz | 831,646,201 | 16,914 | A double bar graph is a bar graph used for displaying two sets of frequency data for each data group. It can be used to make comparisons between and within data groups. Data displayed on a double bar graph is countable. More »
The National Center for Education Statistics states that on a bar graph where the bars are placed vertically, the y-axis runs vertically from the bottom to the top of the graph. On bar graphs where the bars run horizonta... More »
A compound bar graph breaks down data by splitting the columns to represent information comprising each bar. For example, a bar graph representing state employment rates can be turned into a compound bar graph by subdivi... More »
www.reference.com Math Data Graphs
To merge two sets of data into one graph in Excel, select both sets of data that will comprise the graph. Next, choose an option called "Combo" from the parent group titled "All Charts." More »
A double bar graph is a chart that uses parallel bars at proportional lengths that represent quantitative values. The length of the bar correlates with the data. More »
www.reference.com Math Data Graphs
A frequency table is a mathematical graph that identifies the number of times pieces of data occur in a given sequence. Most frequency tables contain three columns and between five and ten rows. More »
The main difference between a histogram and a bar graph is that a histogram displays quantitative data while a bar graph displays qualitative data. Quantitative data is numerical and can be measured by counting. Qualitat... More » | 313 | 1,562 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2019-30 | latest | en | 0.836634 |
https://www.shaalaa.com/textbook-solutions/c/scert-maharashtra-question-bank-solutions-12th-standard-hsc-mathematics-and-statistics-arts-and-science-maharashtra-state-board-2021-chapter-1.5-vectors-and-three-dimensional-geometry_5218 | 1,643,138,268,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304872.21/warc/CC-MAIN-20220125190255-20220125220255-00114.warc.gz | 1,022,693,866 | 18,516 | SCERT Maharashtra Question Bank solutions for 12th Standard HSC Mathematics and Statistics (Arts and Science) Maharashtra State Board 2021 chapter 2 - Vectors and Three Dimensional Geometry [Latest edition]
Chapter 2: Vectors and Three Dimensional Geometry
MCQ
SCERT Maharashtra Question Bank solutions for 12th Standard HSC Mathematics and Statistics (Arts and Science) Maharashtra State Board 2021 Chapter 2 Vectors and Three Dimensional Geometry MCQ
2 marks each
MCQ | Q 1
If |a̅| = 3, |b̅| =4, then the value of λ for which a̅ + λ b̅ is perpendicular to a̅ − λ b̅ is ______
• 9/16
• 3/4
• 3/2
• 4/3
MCQ | Q 2
(hat"i" + hat"j" - hat"k")*(hat"i" - hat"j" + hat"k") = ______
• hat"i" - hat"j" - hat"k"
• 1
• −1
• −hat"j" + hat"k"
MCQ | Q 3
The angle θ between two non-zero vectors bar("a") and bar("b") is given by cos θ = ______
• (bar"a"*bar"b")/(|bar("a")||bar("b")|)
• bar("a")*bar("b")
• |bar("a")||bar("b")|
• (|bar("a")||bar("b")|)/(bar("a")*bar("b"))
MCQ | Q 4
If the sum of two unit vectors is itself a unit vector, then the magnitude of their difference is ______
• sqrt(2)
• sqrt(3)
• 1
• 2
MCQ | Q 5
Select the correct option from the given alternatives:
If α, β, γ are direction angles of a line and α = 60°, β = 45°, γ = ______
• 30° or 90°
• 45° or 60°
• 90° or 30°
• 60° or 120°
MCQ | Q 6
Select the correct option from the given alternatives:
The distance of the point (3, 4, 5) from the Y-axis is ______
• 3
• 5
• sqrt(34)
• sqrt(41)
MCQ | Q 7
Select the correct option from the given alternatives:
If cos α, cos β, cos γ are the direction cosines of a line, then the value of sin2α + sin2β + sin2γ is ______
• 1
• 2
• 3
• 4
MCQ | Q 8
If |bar("a")| = 2, |bar("b")| = 5, and bar("a")*bar("b") = 8 then |bar("a") - bar("b")| = ______
• 13
• 12
• sqrt(13)
• sqrt(21)
MCQ | Q 9
If bar("AB") = 2hat"i" + hat"j" - 3hat"k", and A(1, 2 ,−1) is given point then coordinates of B are ____
• (3, 3, −4)
• (−3, 3 −2)
• (3, 3, 2)
• (−3, 3, 4)
MCQ | Q 10
Select the correct option from the given alternatives:
If l, m, n are direction cosines of a line then "l"hat
"i" + "m"hat"j" + "n"hat"k" is ______
• null vector
• the unit vector along the line
• any vector along the line
• a vector perpendicular to the line
MCQ | Q 11
The values of c that satisfy |"c" bar("u")| = 3, bar("u") = hat"i" + 2hat"j" + 3hat"k" is ______
• sqrt(14)
• 3sqrt(14)
• 3/sqrt(14)
• 3
MCQ | Q 12
The value of hat"i"*(hat"j" xx hat"k") + hat"j"*(hat"i" xx hat"k") + hat"k"*(hat"i" xx hat"j")
• 0
• −1
• 1
• 3
MCQ | Q 13
Select the correct option from the given alternatives:
The 2 vectors hat"j" + hat"k" and 3hat"i" - hat"j" + 4hat"k" represents the two sides AB and AC respectively of a ΔABC. The length of the median through A is
• sqrt(34)/2
• sqrt(48)/2
• sqrt(18)
• of the median through A is
SCERT Maharashtra Question Bank solutions for 12th Standard HSC Mathematics and Statistics (Arts and Science) Maharashtra State Board 2021 Chapter 2 Vectors and Three Dimensional Geometry Very Short Answers
1 Mark
Very Short Answers | Q 1
Find the magnitude of a vector with initial point : (1, −3, 4); terminal point : (1, 0, −1)
Very Short Answers | Q 2
Find the coordinates of the point which is located three units behind the YZ-plane, four units to the right of XZ-plane, and five units above the XY-plane.
Very Short Answers | Q 3
A(2, 3), B(−1, 5), C(−1, 1) and D(−7, 5) are four points in the Cartesian plane, Check if, bar("CD") is parallel to bar("AB")
Very Short Answers | Q 4
Find a unit vector in the opposite direction of bar("u"). Where bar("u") = 8hat"i" + 3hat"j" - hat"k"
Very Short Answers | Q 5
The non zero vectors bar("a") and bar("b") are not collinear find the value of lambda and mu: if bar("a") + 3bar("b") = 2lambdabar("a") - mubar("b")
Very Short Answers | Q 6
If bar("a") = 4hat"i" + 3hat"k" and bar("b") = -2hat"i" + hat"j" + 5hat"k", then find 2bar("a") + 5bar("b")
Very Short Answers | Q 7
Find the distance from (4, −2, 6) to the XZ- plane
Very Short Answers | Q 8
If the vectors 2hat"i" - "q"hat"j" + 3hat"k" and 4hat"i" - 5hat"j" + 6hat"k" are collinear then find the value of q
Very Short Answers | Q 9
Find bar("a")*(bar("b") xx bar("c")), if bar("a") = 3hat"i" - hat"j" + 4hat"k", bar("b") = 2hat"i" + 3hat"j" - hat"k", bar("c") = -5hat"i" + 2hat"j" + 3hat"k"
Very Short Answers | Q 10
If a line makes angle 90°, 60° and 30° with the positive direction of X, Y and Z axes respectively, find its direction cosines
SCERT Maharashtra Question Bank solutions for 12th Standard HSC Mathematics and Statistics (Arts and Science) Maharashtra State Board 2021 Chapter 2 Vectors and Three Dimensional Geometry Short Answers I
2 Marks
Short Answers I | Q 1
The vector bar"a" is directed due north and |a| = 24. The vector bar"b" is directed due west and |bar"b"| = 7. Find |bar"a" + bar"b"|.
Short Answers I | Q 2
Show that following points are collinear P(4, 5, 2), Q(3, 2, 4), R(5, 8, 0)
Short Answers I | Q 3
If a vector has direction angles 45° and 60°, find the third direction angle.
Short Answers I | Q 4
If bar("c") = 3bar("a") - 2bar("b") then prove that [(bar("a"), bar("b"), bar("c"))] = 0
Short Answers I | Q 5
If |bar("a")*bar("b")| = |bar("a") xx bar("b")| and bar("a")*bar("b") < 0, then find the angle between bar("a") and bar("b")
Short Answers I | Q 6
Find the direction ratios of a vector perpendicular to the two lines whose direction ratios are 1, 3, 2 and –1, 1, 2
Short Answers I | Q 7
If bar("a"), bar("b") and bar("c") are position vectors of the points A, B, C respectively and 5bar("a") - 3bar("b") - 2bar("c") = bar(0), then find the ratio in which the point C divides the line segement BA
Short Answers I | Q 8
If bar("a") and bar("b") are two vectors perpendicular each other, prove that (bar("a") + bar("b"))^2 = (bar("a") - bar("b"))^2
Short Answers I | Q 9
Find the position vector of point R which divides the line joining the points P and Q whose position vectors are 2hat"i" - hat"j" + 3hat"k" and -5hat"i" + 2hat"j" - 5hat"k" in the ratio 3:2
(i) internally
(ii) externally
Short Answers I | Q 10
Find a unit vector perpendicular to the vectors hat"j" + 2hat"k" and hat"i" + hat"j".
SCERT Maharashtra Question Bank solutions for 12th Standard HSC Mathematics and Statistics (Arts and Science) Maharashtra State Board 2021 Chapter 2 Vectors and Three Dimensional Geometry Short Answers II
3 Marks
Short Answers II | Q 1
If two of the vertices of a triangle are A (3, 1, 4) and B(− 4, 5, −3) and the centroid of the triangle is at G (−1, 2, 1), then find the coordinates of the third vertex C of the triangle
Short Answers II | Q 2
Find the centroid of tetrahedron with vertices K(5, −7, 0), L(1, 5, 3), M(4, −6, 3), N(6, −4, 2)
Short Answers II | Q 3
If a line has the direction ratios 4, −12, 18, then find its direction cosines
Short Answers II | Q 4
Show that the points A(2, –1, 0) B(–3, 0, 4), C(–1, –1, 4) and D(0, – 5, 2) are non coplanar
Short Answers II | Q 5
Using properties of scalar triple product, prove that [(bar"a" + bar"b", bar"b" + bar"c", bar"c" + bar"a")] = 2[(bar"a", bar"b", bar"c")].
Short Answers II | Q 6
The direction ratios of bar"AB" are −2, 2, 1. If A = (4, 1, 5) and l(AB) = 6 units, Then find B.
Short Answers II | Q 7
If G(a, 2, −1) is the centroid of the triangle with vertices P(1, 2, 3), Q(3, b, −4) and R(5, 1, c) then find the values of a, b and c
Short Answers II | Q 8
If A(5, 1, p), B(1, q, p) and C(1, −2, 3) are vertices of triangle and "G"("r", -4/3, 1/3) is its centroid then find the values of p, q and r
Short Answers II | Q 9
Prove by vector method, that the angle subtended on semicircle is a right angle.
Short Answers II | Q 10
Prove that medians of a triangle are concurrent
Short Answers II | Q 11
Prove that altitudes of a triangle are concurrent
SCERT Maharashtra Question Bank solutions for 12th Standard HSC Mathematics and Statistics (Arts and Science) Maharashtra State Board 2021 Chapter 2 Vectors and Three Dimensional Geometry Long Answers III
4 Marks
Long Answers III | Q 1
Express - hat"i" - 3hat"j" + 4hat"k" as the linear combination of the vectors 2hat"i" + hat"j" - 4hat"k", 2hat"i" - hat"j" + 3hat"k" and 3hat"i" + hat"j" - 2hat"k"
Long Answers III | Q 2
If Q is the foot of the perpendicular from P(2, 4, 3) on the line joining the point A(1, 2, 4) and B(3, 4, 5), find coordinates of Q
Long Answers III | Q 3
Prove that the angle bisectors of a triangle are concurrent
Long Answers III | Q 4
Using vector method, find the incenter of the triangle whose vertices are A(0, 3, 0), B(0, 0, 4) and C(0, 3, 4)
Long Answers III | Q 5
Find the angles between the lines whose direction cosines l, m, n satisfy the equations 5l + m + 3n = 0 and 5mn − 2nl + 6lm = 0
Long Answers III | Q 6
Let "A" (bar"a") and "B" (bar"b") are any two points in the space and "R"(bar"r") be a point on the line segment AB dividing it internally in the ratio m : n, then prove that bar "r" = ("m"bar"b" + "n"bar"a")/("m" + "n")
Long Answers III | Q 7
D and E divides sides BC and CA of a triangle ABC in the ratio 2 : 3 respectively. Find the position vector of the point of intersection of AD and BE and the ratio in which this point divides AD and BE
Long Answers III | Q 8
If bar"u" = hat"i" - 2hat"j" + hat"k", bar"r" = 3hat"i" + hat"k" and bar"w" = hat"j", hat"k" are given vectors , then find [bar"u" + bar"w"]*[(bar"w" xx bar"r") xx (bar"r" xx bar"w")]
Long Answers III | Q 9
Find the volume of a tetrahedron whose vertices are A (- 1, 2, 3), B (3, - 2, 1), C (2, 1, 3) and D (- 1, 2, 4).
Long Answers III | Q 10
If four points "A"(bar"a"), "B"(bar"b"), "C"(bar"c") and "D"(bar"d") are coplanar, then show that [(bar"a", bar"b", bar"c")] + [(bar"b", bar"c", bar"d")] + [(bar"c", bar"a", bar"d")] = [(bar"a", bar"b", bar"c")].
SCERT Maharashtra Question Bank solutions for 12th Standard HSC Mathematics and Statistics (Arts and Science) Maharashtra State Board 2021 chapter 2 - Vectors and Three Dimensional Geometry
SCERT Maharashtra Question Bank solutions for 12th Standard HSC Mathematics and Statistics (Arts and Science) Maharashtra State Board 2021 chapter 2 (Vectors and Three Dimensional Geometry) include all questions with solution and detail explanation. This will clear students doubts about any question and improve application skills while preparing for board exams. The detailed, step-by-step solutions will help you understand the concepts better and clear your confusions, if any. Shaalaa.com has the Maharashtra State Board 12th Standard HSC Mathematics and Statistics (Arts and Science) Maharashtra State Board 2021 solutions in a manner that help students grasp basic concepts better and faster.
Further, we at Shaalaa.com provide such solutions so that students can prepare for written exams. SCERT Maharashtra Question Bank textbook solutions can be a core help for self-study and acts as a perfect self-help guidance for students.
Concepts covered in 12th Standard HSC Mathematics and Statistics (Arts and Science) Maharashtra State Board 2021 chapter 2 Vectors and Three Dimensional Geometry are Representation of Vector, Vectors and Their Types, Algebra of Vectors, Coplanar Vectors, Vector in Two Dimensions (2-D), Three Dimensional (3-D) Coordinate System, Components of Vector, Position Vector of a Point P(X, Y, Z) in Space, Component Form of a Position Vector, Vector Joining Two Points, Section Formula, Scalar Product of Vectors (Dot), Vector Product of Vectors (Cross), Scalar Triple Product of Vectors, Vector Triple Product, Addition of Vectors.
Using SCERT Maharashtra Question Bank 12th Board Exam solutions Vectors and Three Dimensional Geometry exercise by students are an easy way to prepare for the exams, as they involve solutions arranged chapter-wise also page wise. The questions involved in SCERT Maharashtra Question Bank Solutions are important questions that can be asked in the final exam. Maximum students of Maharashtra State Board 12th Board Exam prefer SCERT Maharashtra Question Bank Textbook Solutions to score more in exam.
Get the free view of chapter 2 Vectors and Three Dimensional Geometry 12th Board Exam extra questions for 12th Standard HSC Mathematics and Statistics (Arts and Science) Maharashtra State Board 2021 and can use Shaalaa.com to keep it handy for your exam preparation
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# AQA Physics PHYA5 - 28th June 2016 [Exam Discussion Thread] Watch
1. (Original post by marioman)
The count rate is going to drop when you move it further away regardless of whether there's beta emission or not.
Ya so you have to use aluminium sheet to check, or some other thing
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2. (Original post by marioman)
The count rate is going to drop when you move it further away regardless of whether there's beta emission or not.
Wouldnt there be a rapid drop past beta's range?
3. (Original post by Nikhilm)
Nah it was 127/128 from the calculation
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Yeah!
4. (Original post by CourtlyCanter)
I was the only one in my school to do it as I learnt it myself because the teacher we had for astrophysics was awful. This year's paper was hard. Do you remember your answers to any of the questions?
5. (Original post by Mango Milkshake)
just checked, it's 10,000
did you round it too early?
I didn't round anything off until I got to the final answer. Don't know where I went wrong.
6. (Original post by txnilxnur)
I thought it was 2000 lmao, swear that's what it says in the textbook
Isn't it 1 in 2000 for a deflection and 1 in 10000 for angles greater than 90?
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7. (Original post by CourtlyCanter)
I was the only one in my school to do it as I learnt it myself because the teacher we had for astrophysics was awful. This year's paper was hard. Do you remember your answers to any of the questions?
yea this paper was pretty difficult. I didnt get the first calculation to do with the eye. I remember getting 57% for the x ray calculation to do with 12mm. what about you?
8. (Original post by lc970916)
probably not .
" Expressions for the electric and magnetic fields in free space contain the electric permittivity ε0 and magnetic permeability μ0 of free space. As indicated in the section on electric and magnetic constants, these two quantities are not independent but are related to "c", the speed of light and other electromagnetic waves."
source: http://hyperphysics.phy-astr.gsu.edu...ic/elefie.html
True they are well defined quantities rather than those who put electric foeld strength and magnetic field strength which are not defined quantities.
9. 3.1*10^-15 for radius of oxygen
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10. (Original post by particlestudent)
kingaaran Mango Milkshake
What did you guys put for the test to see if it emitted beta as well?
Use a beta absorber, i.e. a few mm of aluminium. Then you can see if this result matches up with the result obtained at 0.9m by using the inverse square law of gamma radiation
11. (Original post by Nikhilm)
Ya so you have to use aluminium sheet to check, or some other thing
Posted from TSR Mobile
Could I have said:
"I said plot a graph of count rate against 1/r^2 and it should be a straight line if it is only gamma, but if it is beta it will not be a straight line."
12. specific charge was 1.8x10^11 Ckg^-1
Posted from TSR Mobile
got that
14. (Original post by FireBLue97)
specific charge was 1.8x10^11 Ckg^-1
It wasn't. That is the actual specific charge but the experimental specific charge was 9.1x10^10
15. (Original post by kingaaran)
Use a beta absorber, i.e. a few mm of aluminium. Then you can see if this result matches up with the result obtained at 0.9m by using the inverse square law of gamma radiation
I didn't mention aluminium at all . I said plot count rate against 1/distance squared and if it is a complete straight line then it is only a gamma emitter
Posted from TSR Mobile
17. (Original post by particlestudent)
kingaaran Mango Milkshake
What did you guys put for the test to see if it emitted beta as well?
I put put an electromagnet between source and detector and drew this
Posted from TSR Mobile
18. (Original post by particlestudent)
I didn't mention aluminium at all . I said plot count rate against 1/distance squared and if it is a complete straight line then it is only a gamma emitter
I think you'll be fine - it sounds valid
19. (Original post by particlestudent)
I said plot a graph of count rate against 1/r^2 and it should be a straight line if it is only gamma, but if it is beta it will not be a straight line.
This better be accepted
I saw that in a past paper as an alternative method once. You still get the marks
20. I also got 1.2x10^8 AU for the penultimate quasar question but starting to think this is wrong after looking on here
For those who got the other answer (2.4 something) could you tell me how u did this question as i am really confused.
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Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE | 1,388 | 5,643 | {"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-2018-05 | latest | en | 0.972028 |
https://metanumbers.com/96322 | 1,638,417,795,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964361064.69/warc/CC-MAIN-20211202024322-20211202054322-00399.warc.gz | 475,557,224 | 7,346 | # 96322 (number)
96,322 (ninety-six thousand three hundred twenty-two) is an even five-digits composite number following 96321 and preceding 96323. In scientific notation, it is written as 9.6322 × 104. The sum of its digits is 22. It has a total of 3 prime factors and 8 positive divisors. There are 45,312 positive integers (up to 96322) that are relatively prime to 96322.
## Basic properties
• Is Prime? No
• Number parity Even
• Number length 5
• Sum of Digits 22
• Digital Root 4
## Name
Short name 96 thousand 322 ninety-six thousand three hundred twenty-two
## Notation
Scientific notation 9.6322 × 104 96.322 × 103
## Prime Factorization of 96322
Prime Factorization 2 × 17 × 2833
Composite number
Distinct Factors Total Factors Radical ω(n) 3 Total number of distinct prime factors Ω(n) 3 Total number of prime factors rad(n) 96322 Product of the distinct prime numbers λ(n) -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) -1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 96,322 is 2 × 17 × 2833. Since it has a total of 3 prime factors, 96,322 is a composite number.
## Divisors of 96322
1, 2, 17, 34, 2833, 5666, 48161, 96322
8 divisors
Even divisors 4 4 4 0
Total Divisors Sum of Divisors Aliquot Sum τ(n) 8 Total number of the positive divisors of n σ(n) 153036 Sum of all the positive divisors of n s(n) 56714 Sum of the proper positive divisors of n A(n) 19129.5 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 310.358 Returns the nth root of the product of n divisors H(n) 5.03526 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 96,322 can be divided by 8 positive divisors (out of which 4 are even, and 4 are odd). The sum of these divisors (counting 96,322) is 153,036, the average is 1,912,9.5.
## Other Arithmetic Functions (n = 96322)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 45312 Total number of positive integers not greater than n that are coprime to n λ(n) 2832 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 9262 Total number of primes less than or equal to n r2(n) 16 The number of ways n can be represented as the sum of 2 squares
There are 45,312 positive integers (less than 96,322) that are coprime with 96,322. And there are approximately 9,262 prime numbers less than or equal to 96,322.
## Divisibility of 96322
m n mod m 2 3 4 5 6 7 8 9 0 1 2 2 4 2 2 4
The number 96,322 is divisible by 2.
• Deficient
• Polite
• Square Free
• Sphenic
## Base conversion (96322)
Base System Value
2 Binary 10111100001000010
3 Ternary 11220010111
4 Quaternary 113201002
5 Quinary 11040242
6 Senary 2021534
8 Octal 274102
10 Decimal 96322
12 Duodecimal 478aa
20 Vigesimal c0g2
36 Base36 22bm
## Basic calculations (n = 96322)
### Multiplication
n×y
n×2 192644 288966 385288 481610
### Division
n÷y
n÷2 48161 32107.3 24080.5 19264.4
### Exponentiation
ny
n2 9277927684 893668550378248 86079942109533603856 8291392183874495790617632
### Nth Root
y√n
2√n 310.358 45.8397 17.617 9.92533
## 96322 as geometric shapes
### Circle
Diameter 192644 605209 2.91475e+10
### Sphere
Volume 3.74339e+15 1.1659e+11 605209
### Square
Length = n
Perimeter 385288 9.27793e+09 136220
### Cube
Length = n
Surface area 5.56676e+10 8.93669e+14 166835
### Equilateral Triangle
Length = n
Perimeter 288966 4.01746e+09 83417.3
### Triangular Pyramid
Length = n
Surface area 1.60698e+10 1.0532e+14 78646.6
## Cryptographic Hash Functions
md5 ccd2c7b05a6edac5e4b844987a4be892 470e2009979bf085eed2bd6a13665f3111ec5f5a 044956817c10ed6252af810c98547a005a6ed436128d0320efbdc38205ddddc4 e154e6bad2d9d6609059e74a5cb3930c8a39a39d8fbbbed0075aff5db3f9786c153ab53438eef38db3c1e745bac025486dccb59a04ed3183b447daa91cfcf8eb 898d8b0fda4987785f96897bc9f917fc8be4840b | 1,453 | 4,113 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2021-49 | latest | en | 0.803676 |
https://ilmihub.com/class-12-physics-chapter-12-short-questions-and-numerical-notes-electrostatics.html | 1,718,876,905,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861916.26/warc/CC-MAIN-20240620074431-20240620104431-00795.warc.gz | 272,952,775 | 26,613 | Home 2nd Year Physics Notes Class 12 Physics Chapter 12 Short Questions and Numerical Notes – Electrostatics
# Class 12 Physics Chapter 12 Short Questions and Numerical Notes – Electrostatics
6
In this post, I am sharing Class 12 Physics Chapter 12 Short Questions and Numerical Notes PDF for the students of Intermediate Part 2. This chapter’s name is Electrostatics. Students can download the Class 12 Physics Chapter 12 Short Questions and Numerical Notes PDF on their laptop or mobile. These Physics Notes are for all the boards working under the Punjab Board like Gujranwala Board, Lahore Board, Faisalabad Board, Multan Board, Rawalpindi Board, Sargodha Board, DG Kahn Board, and Sahiwal Board. Here are the complete FSC 2nd Year Physics Notes PDF chapters.
## FSC Part 2 Physics Chapter 12 Short Questions and Numerical
Class 12 Physics Notes
### Electric Lines of force never cross why?
Electric line of force can never cross each other.
Reason: As electric intensity is a vector quantity and if two electric lines of force cross each other at a single point then electric intensity will have two different directions at a single point which is not possible because electric intensity is a vector quantity and vector has only one direction.
### If a point charge q of mass m is released in non-uniform electric field with field line in the same direction pointing, will it make a rectilinear motion?
If a point charge q of mass m is released in a non-uniform electric field then there are two possibilities:
(i) If a point charge is released in a non-uniform field produced by positive and negative charges then it will move in a curved path.
(ii) If a point charge is placed in no-uniform field produced by a positive or negative charge then it will make a rectilinear motion.
### Do electrons tend to go to region of high potential or of low potential?
As electrons are negatively charged particle therefore when they enter the electric field they will tend to go the region of high potential (positive terminal) from the region of low potential (negative terminal). | 442 | 2,090 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2024-26 | latest | en | 0.892735 |
http://www.dailytech.com/article.aspx?newsid=15380&commentid=455007&threshhold=1&red=3832 | 1,430,193,752,000,000,000 | text/html | crawl-data/CC-MAIN-2015-18/segments/1429246660628.16/warc/CC-MAIN-20150417045740-00192-ip-10-235-10-82.ec2.internal.warc.gz | 441,181,505 | 13,641 | Print 87 comment(s) - last by Sunner.. on Jun 12 at 4:24 AM
MIT mathematicians model traffic jams like detonation waves
Most drivers have been stuck in a traffic jam at one point or another. Some of the jams are caused by an accident or closed lanes. Other traffic jams crop up with seemingly no reason.
A group of mathematicians at MIT is working on the development of a new model to explain how and why these so-call phantom traffic jams form. According to the researchers, these types of phantom traffic jams form when there are a lot of cars on the road and small disturbances like a driver hitting the brakes too hard or getting too close to another car. These little disturbances can escalate into a self-sustaining traffic jam.
The model developed by the team of researchers may help road designers build roads to minimize the possibility of phantom traffic jams. The key to the study is the discovery that the mathematics of these jams called jamitons are very similar to the equations used to describe the detonation waves produced by explosions.
The discovery of the jamitons allowed the researchers to solve traffic jam equations first theorized in the 1950's. The equations are reportedly similar to those used in fluid mechanics and model traffic jams as self-sustaining waves. The equations allowed the researchers to calculate the conditions that case a jamitons to form and how fast the jamiton will spread.
According to the researchers, once this type of jamiton forms it is nearly impossible to break up and a driver’s only choice is to wait the jamiton out. The researchers say that the new model can help road designers to determine speed limits that are safer and find stretches of road where accidents are more likely.
One of the researchers, Aslan Kasimov, said, "We wanted to describe this using a mathematical model similar to that of fluid flow." Kasimov and his team say that they discovered that jamitons have a sonic point that separates traffic flow into upstream and downstream components. Communication of the cause of the jamiton to drivers it the downstream segment of traffic is impossible say the researchers.
Comments Threshold -1 0 1 2 3 4 5
By androticus on 6/11/2009 10:42:34 PM , Rating: 2
The product that private road owners would be selling is drive quality -- if you are sitting on a road for an hour whose trip time at road speed is actually 10 minutes then you are basically selling nothing. It is the fact that there is a government monopoly on roads that leads to the chronic shortage of road capacity -- how come things in the private sector never suffer permanent chronic shortages like this?
And smarmy sneering cynical economics professors tell us that the government has to create government owned or controlled monopolies on "natural monopoly" resources like roads in order to... wait for it... protect us from what they claim would be the "natural monopolies" that would arise in the private sector.
Echoes of the EU saying they have to remove the web browser from Windows so consumers are protected and given choice.
GOVERNMENT OUT OF THE ECONOMY!!!!
"When an individual makes a copy of a song for himself, I suppose we can say he stole a song." -- Sony BMG attorney Jennifer Pariser
Copyright 2015 DailyTech LLC. - RSS Feed | Advertise | About Us | Ethics | FAQ | Terms, Conditions & Privacy Information | Kristopher Kubicki | 710 | 3,397 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2015-18 | longest | en | 0.947723 |
http://mathhelpforum.com/algebra/176944-vectors-finding-ratio.html | 1,529,301,432,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267860089.11/warc/CC-MAIN-20180618051104-20180618071104-00183.warc.gz | 205,978,691 | 9,909 | # Thread: Vectors - finding the ratio
1. ## Vectors - finding the ratio
The position vectors a, b, c of points A, B, C satisfy c = (3b-1)/(3-1). What is the ratio AC:CB? and where would C be located on a line with A and B (between A and B, next to B or next to C)?
Not sure how to do this - help would be appreciated
2. Hello,
So you mean that $\displaystyle \vec{c}=\frac{3\vec{b}-1}{3-1}$. If this is true then it does not make sense. How would you subtract a number from a vector?
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# How to calculate ratio in vectors
Click on a term to search for related topics. | 181 | 614 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2018-26 | latest | en | 0.916572 |
https://www.freecadweb.org/wiki/index.php?title=Draft_Point&oldid=556533 | 1,582,104,916,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875144111.17/warc/CC-MAIN-20200219092153-20200219122153-00349.warc.gz | 763,595,175 | 12,592 | # Draft Point
Jump to: navigation, search
Other languages:
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Previous: BSpline
Draft Point
Menu location
Draft → Point
Workbenches
Draft, Arch
Default shortcut
P T
Introduced in version
0.7
See also
Draft Line, Draft Wire
## Description
The Point tool creates a simple point in the current work plane, handy to serve as reference for placing lines, wires, or other objects later. It uses the Draft Linestyle (only the color) set on the Draft Tray.
A single point placed on the working plane
## How to use
1. Press the button, or press P then T keys.
2. Click a point on the 3D view, or type a coordinate and press the button.
## Options
• To enter coordinates manually, simply enter the numbers, then press Enter between each X, Y and Z component. You can press the button when you have the desired values to insert the point.
• Press T or click the checkbox to toggle continue mode. If continue mode is on, the point tool will restart after you place a point, allowing you to place another one without pressing the tool button again.
• Press Esc or the Close button to abort the current command.
## Properties
• DataX: the X coordinate of the point.
• DataY: the Y coordinate of the point.
• DataZ: the Z coordinate of the point.
## Scripting
See also: Draft API and FreeCAD Scripting Basics.
The Point tool can be used in macros and from the [[Python|]Python] console by using the following function:
```Point = makePoint(X=0, Y=0, Z=0, color=None, name="Point", point_size=5)
Point = makePoint(point, Y=0, Z=0, color=None, name="Point", point_size=5)```
• Creates a `Point` object in the specified `X`, `Y` and `Z` coordinates, with units in millimeters. If no coordinates are given the point is created at the origin (0,0,0).
• If `X` is a `point` defined by a `FreeCAD.Vector`, it is used.
• `color` is a tuple `(R, G, B)` that indicates the color of the point in the RGB scale; each value in the tuple should be in the range from `0` to `1`.
• `name` is the name of the object.
• `point_size` is the size of the object in pixels, if the graphical user interface is loaded.
Example:
```import FreeCAD
import Draft
Point1 = Draft.makePoint(1600, 1400, 0)
p2 = FreeCAD.Vector(-3200, 1800, 0)
Point2 = Draft.makePoint(p2, color=(0.5, 0.3, 0.6), point_size=10)
FreeCAD.ActiveDocument.recompute()```
Example:
This code creates `N` random points within a square of side `2L`. It makes a loop creating `N` points, that may appear anywhere from `-L` to `+L` on both X and Y; it also chooses a random color and size for each point. Change `N` to change the number of points, and change `L` to change the area covered by the points.
```import random
import FreeCAD
import Draft
L = 1000
centered = FreeCAD.Placement(FreeCAD.Vector(-L, -L, 0), FreeCAD.Rotation())
Rectangle = Draft.makeRectangle(2*L, 2*L, placement=centered)
N = 10
for i in range(N):
x = 2*L*random.random() - L
y = 2*L*random.random() - L
z = 0
r = random.random()
g = random.random()
b = random.random()
size = 15*random.random() + 5
Draft.makePoint(x, y, z, color=(r, g, b), point_size=size)
FreeCAD.ActiveDocument.recompute()```
Previous: BSpline | 916 | 3,243 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2020-10 | latest | en | 0.721076 |
https://usethinkscript.com/threads/can-anybody-help-to-explain-these-scripts.1941/ | 1,721,052,291,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514707.52/warc/CC-MAIN-20240715132531-20240715162531-00386.warc.gz | 537,546,692 | 17,584 | # Can anybody help to explain these scripts?
#### firefox
##### New member
Hi,
I am a newbie and having difficulty in understanding these TOS lines to determine the peaks. Would appreciate if anybody can help to explain what these lines do and is h1 refers to previous peak and h2 refers to current peak?
def h1 = Highest(high, 20)[15];
def h2 = Highest(high, 15);
def os1=GetMaxValueOffset(high, 20)[15];
def os2=GetMaxValueOffset(high, 15);
Maybe this will help.
The indexing operator [] is used to access the previous or future data in an array of data. The square brackets indicate the shift against the current moment. The positive values of the shift represent the values in the past, the negative shift values represent the values in the future. The indexing operator can be applied to fundamental data, function outputs, indicator outputs, variables, etc.
Example 1
input price = close;
input shift = 5;
plot PastPrice = price[shift];
plot FuturePrice = price[-shift];
Draws shifted price plots.
Example 2
plot PastMACD = MACD().Avg[10];
Shows the smoothed MACD value 10 bars ago.
Last edited by a moderator:
Thank you @horserider for your quick reply. So does it mean :
def h1 = Highest(high, 20)[15];
Is referring to the highest price 15 bare ago for the last 20 bars
and
def h2 = Highest(high, 15);
refers to the current highest price for the last 15 bars.
Is my understanding correct?
Last edited by a moderator:
Highest high of past 20 bars shifted 15 bars
Confusing unless you are a math wizard. Plot h1 and h2 and look at what it does.
plot h1 = Highest(high, 20)[15];
plot h2 = Highest(high, 15);
Highest high of past 20 bars shifted 15 bars
Confusing unless you are a math wizard. Plot h1 and h2 and look at what it does.
plot h1 = Highest(high, 20)[15];
plot h2 = Highest(high, 15);
Thank you so much. You gave me a great idea to use plot function to View what it does. I will go and try it. Thanks again for solving my problem.
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To access the premium indicators, which are plug and play ready, sign up for VIP membership here. | 861 | 3,751 | {"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-2024-30 | latest | en | 0.861293 |
https://www.pcreview.co.uk/threads/problem-with-amortization-table-using-beginning-of-period.1018929/ | 1,623,962,265,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487633444.37/warc/CC-MAIN-20210617192319-20210617222319-00274.warc.gz | 864,951,694 | 15,639 | # Problem with Amortization Table using beginning of period
J
#### jcooper
I am trying to create a amortization table in which payments start at
the beginning of the period. The problem is that principal doesn't
zero out at the end of the term. In fact, it overpays by a small
amount. Yet, when I calculate the same table on a financial calculator
which shows an amortization table, it zeroes out.
I'm guessing, but the problem seems to be with the future value. When
I make the future value "0", it zeroes out. My calculation includes a
balloon payout at the end of the term.
Any thoughts?
Under the Interest column per period, my formula looks like this:
IPMT(Annual Interest Rate/12, current period, # periods, - current
value, + future value (for a balloon payment at the end), 1 (to
indicate beginning of the period).
Under the Principal column per period, my formula looks like this:
PPMT(Annual Interest Rate/12, current period, # periods, - current
value, + future value (for a balloon payment at the end), 1 (to
indicate beginning of the period).
Essentially, both are the same except the name of the function.
F
#### Fred Smith
I think your problem is rounding, not the payment starting at the beginning
of the period. I expect your financial calculator will be rounding your
payments to the penny, but Excel is not.
Here's some ideas which might simplify things for you or point you in the
right direction.
1. Rather than payments at the beginning of the period, you can simply
adjust the principal. For example, you borrow \$10,000 with payments of
\$250/mo starting immediately is the same as borrowing \$9750 with payments
starting a month from now.
2. The calculations are much simpler than you are making them. You need five
columns:
StartAmount (Original loan amount if first line, previous EndAmount if not)
Payment (fixed monthly amount or including balloon payment)
Interest = StartAmount * rate / 12, which you likely will want to round to 2
decimal places
Principal = Payment - Interest
EndAmount = StartAmount - Payment + Interest
Copy down until EndAmount gets to zero.
3. Use PPMT and IPMT only if you *don't* want a full amortization table.
J
#### jcooper
Fred,
Thanks. You may be right regarding the rounding issue. It just
surprised me because it worked perfectly when paying in arrears and
also when I had no FV.
I had originally set up my table as you indicated, but I was having
problems getting it to work for payment in the beginning. I found
those formulas in a book and it seemed to work well.
Jim
#### henrymusoke
I seem to have the same problem but cannot understand why excel doesn't zero or happens to skip rows. giving different values.
I'm using, pmt, want to buy a house for \$429900, just paid a 50% deposit, everything seems fine and clear but I'm stuck. interest rate is 4.35%, my decision is between 15 and 30 years. for 15 years rate is 3.55% and for 30 years is 4.35%. any suggestion please? | 708 | 2,966 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2021-25 | longest | en | 0.919495 |
https://jeopardylabs.com/print/repeating-and-terminating-decimals | 1,656,108,217,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103033816.0/warc/CC-MAIN-20220624213908-20220625003908-00723.warc.gz | 387,957,833 | 3,693 | True/False
Fraction to Decimal
Random Questions
Vocabulary
Bonus
100
The denominator in this fraction 5/8 is the number five.
False
100
Without using a calculator, 3/4 = ?
What is 0.75?
100
What is the name of the teacher who is teaching this lesson?
Mr. Eugene or Mr. E.
100
______ is a decimal that has a digit or blocks of digits that repeat over and over without ever ending.
What is a Repeating Decimal?
100
Is it true that a decimal can be both repeating and terminating?
No, it can only be one or the other.
200
Is 5/10 the same as 1/2.
True
200
After using a calculator to divide 12 by 25 and the answer is 0.48. Tony is convinced that this is a terminating decimal. Do you think Tony is right?
Yes.
200
Miami Heat won 66 games out of 82. What is their winning percentage?
What is 0.805?
200
______ is a decimal number that has digits that do not go on forever.
What is a Terminating Decimal?
200
How can students improve their math skills?
Continue to practice, do homework, study, etc.
300
After converting a fraction into decimal form, the decimal will either repeat or terminate.
True
300
5/0 = ?
You cannot divide by zero.
300
Who can explain why 0.333333 is a repeating decimal in their own words?
Simply, because the number 3 repeats over and over again.
300
_______ is a number usually expressed in the form a/b.
What is a Fraction?
300
Give an example how math is used in a real life situation.
Example 1: Purchasing food, gas, and clothes. Example 2: Banking
Click to zoom | 388 | 1,490 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2022-27 | longest | en | 0.922655 |
https://en.academic.ru/dic.nsf/enwiki/2240441 | 1,603,580,083,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107885059.50/warc/CC-MAIN-20201024223210-20201025013210-00167.warc.gz | 312,477,291 | 14,589 |
In descriptive set theory, Wadge degrees are levels of complexity for sets of reals and more comprehensively, subsets of any given topological space. Sets are compared by continuous reductions. The Wadge hierarchy is the structure of Wadge degrees.
Take $A$ a subset of $X$ and $B$ a subset of $Y$, where $X$ and $Y$ are topological spaces. $A$ is Wadge reducible to $B$, denoted by : $Ale_W B$if and only if there is a continuous function $f$ from $X$ to $Y$ such that $A = f^\left\{-1\right\}\left(B\right)$. The notion of reduction depends on the spaces $X$ and $Y$.
For a given topological space $X$, Wadge reducibility determines a preorder or quasiorder on the subsets of $X$. This is the Wadge order for $X$. Equivalence classes of sets under this preorder are called Wadge degrees, the degree of a set $A$ is denoted by [$A$] W. The set of Wadge degrees ordered by the Wadge order is called the Wadge hierarchy. Wadge degrees are usually considered for Baire space ωω, Cantor space, and other Polish spaces. If the underlying space is not specified, the terms "Wadge degree" and "Wadge hierarchy" usually refer to Baire space.
Properties of Wadge degrees include their consistency with measures of complexity stated in terms of definability. For example, if $A$W $B$ and $B$ is a countable intersection of open sets, then so is $A$. The same works for all levels of the Borel hierarchy and the difference hierarchy.
Wadge degrees for Baire space play an important role in models of the axiom of determinacy.
Further interest in Wadge degrees comes from computer science, where some papers have suggested Wadge degrees are relevant to algorithmic complexity.
The Wadge game is a simple infinite game discovered by William Wadge (pronounced "wage"). It is used to investigate the notion of continuous reduction for subsets of Baire space. Wadge had analyzed the structure of the Wadge hierarchy for Baire space with games by 1972, but published these results only much later in his PhD thesis. In the Wadge game $G\left(A,B\right)$, player I and player II each in turn play integers which may depend on those played before. The outcome of the game is determined by checking whether the sequences x and y generated by players I and II are contained in the sets A and B, respectively. Player II wins if the outcome is the same for both players, i.e.: $xin ALeftrightarrow yin B$Player I wins if the outcome is different. Sometimes this is also called the "Lipschitz game", and the variant where player II has the option to pass (but has to play infinitely often) is called the Wadge game.
Suppose for a moment that the game is determined. If player I has a winning strategy, then this defines a continuous (even Lipschitz) map reducing $B$ to the complement of $A$, and if on the other hand player II has a winning strategy then you have a reduction of $A$ to $B$. For example, suppose that player II has a winning strategy. Map every sequence x to the sequence y that player II plays in $G\left(A,B\right)$ if player I plays the sequence x, where player II follows his or her winning strategy. This defines a is a continuous map f with the property that x is in $A$ if and only if f(x) is in $B$.
Wadge's lemma states that under the axiom of determinacy (AD), for any two subsets $A,B$ of Baire space: $Ale_W Bor Ble_W omega^omega - A$This works since one of the players has a winning strategy in the Wadge game $G\left(A,B\right)$, and this strategy defines one of the continuous maps needed. The lemma can also be applied locally to pointclasses Γ, for example the Borel sets, Δ1n sets, Σ1n sets, or Π1n sets. Here the conclusion of Wadge's lemma for sets in Γ follows from determinacy of Boolean combinations of sets in Γ instead of full AD. Since Borel determinacy is proved in ZFC, ZFC implies Wadge's lemma for Borel sets.
The assertion that the Wadge lemma holds for sets in Γ is the "semilinear ordering principle" for Γ, SLO(Γ). Any semilinear order defines a linear order on the equivalence classes modulo complements. In this case the relation on the set of pairs $\left\{ A,$ωω$-A\right\}$ given by $\left\{ A,$ωω$-A\right\}$W $\left\{ B,$ωω$-B\right\}$ if and only if $A$W $B$ or $A$W ωω$-B$, is linear.
Martin and Monk proved in 1973 that AD implies the Wadge order for Baire space is well founded. As for the Wadge lemma, this holds for any pointclass Γ, assuming it is determined.
Hence under AD, the Wadge classes modulo complements form a wellorder. The Wadge rank of a set $A$ is the order type of the set of Wadge degrees modulo complements strictly below [$A$] W.
The length of the Wadge hierarchy has been shown to be Θ. Wadge also proved that the length of the Wadge hierarchy restricted to the Borel sets is φω1(1) (or φω1(2) depending on the notation), where φγ is the γth Veblen function to the base ω1 (instead of the usual ω).
Continuing to work under determinacy, if we associate with each set $A$ the collection of all sets strictly below $A$ on the Wadge hierarchy, this forms a pointclass. Equivalently, for each ordinal
It can be shown that $A_alpha$ is self-dual if and only if $alpha$ is either 0, an even successor ordinal, or a limit ordinal of countable cofinality.
Other notions of degree
Similar notions of reduction and degree for subsets of a topological space $X$ arise by replacing the continuous functions by any class of functions "F" which contains the identity function and is closed under composition. Write: $Ale_ emph exttt\left\{F\right\} B$if $A = f^\left\{-1\right\}\left(B\right)$ for some function $f$ in the class "F". Any such class of functions again determines a preorder on the subsets of $X$. Degrees given by Lipschitz functions are called "Lipschitz degrees", and degrees from Borel functions "Borel-Wadge degrees".
* Analytical hierarchy
* Arithmetical hierarchy
* Axiom of determinacy
* Borel hierarchy
* Determinacy
* Pointclass
References
*, in preparation
*
*
*
*
*
*
*
*
Wikimedia Foundation. 2010.
### Look at other dictionaries:
• Hierarchy (mathematics) — In mathematics, a hierarchy is a preorder, i.e. an ordered set. The term is used to stress a natural hierarchical relation among the elements. In particular, it is the preferred terminology for posets whose elements are classes of objects of… … Wikipedia
• Borel hierarchy — In mathematical logic, the Borel hierarchy is a stratification of the Borel algebra generated by the open subsets of a Polish space; elements of this algebra are called Borel sets. Each Borel set is assigned a unique countable ordinal number… … Wikipedia
• Determinacy — Determined redirects here. For the 2005 heavy metal song, see Determined (song). For other uses, see Indeterminacy (disambiguation). In set theory, a branch of mathematics, determinacy is the study of under what circumstances one or the other… … Wikipedia
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• Hiérarchie de Borel — La hiérarchie de Borel désigne une description de la tribu des boréliens d un espace topologique X comme une réunion croissante d ensembles de parties de X, indexée par le premier ordinal non dénombrable. Sommaire 1 Notations préliminaires 2… … Wikipédia en Français
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• Inductive set — This article relates to the notion of inductive sets from descriptive set theory. For the notion in the context of the axiom of infinity, see Inductive set (axiom of infinity). In descriptive set theory, an inductive set of real numbers (or more… … Wikipedia
• Pointclass — In the mathematical field of descriptive set theory, a pointclass is a collection of sets of points, where a point is ordinarily understood to be an element of some perfect Polish space. In practice, a pointclass is usually characterized by some… … Wikipedia | 2,095 | 8,581 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 60, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-45 | latest | en | 0.930318 |
https://stats.stackexchange.com/questions/12875/importance-sampling-in-mixed-discrete-continuous-variables | 1,695,946,725,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510462.75/warc/CC-MAIN-20230928230810-20230929020810-00396.warc.gz | 574,940,387 | 43,245 | # Importance sampling in mixed, discrete/continuous variables
Consider the following model $$X \sim |\mathcal{N}(X;0,1)| \qquad Y|X \sim Q(Y;X)$$ where I define $Q(Y=-x|X=x)$ with probability mass $\int_{-\infty}^{-x}\mathcal{N}(x;0,1)dx$, $Q(Y=+x|X=x)$ with probability mass $1-\int_{x}^{\infty}\mathcal{N}(x;0,1)dx$, and the density of $Q(Y|X=x)$ to be one of a truncated Normal distribution in $(-x,x)$.
Assume now that an unknown $x_{unk}$ is sampled from $|\mathcal{N}(X;0,1)|$ and that $y$ is sampled from $Q(Y|X=x_{unk})$. I am given $y$ and would like to approximate the posterior distribution, $P(X|Y=y)$.
Using importance sampling, I would like to take $N$ sample values for $X$ then weight them according to the "probability" of $Y|X$ such that for sufficiently many samples, $$\mathbb{E}[X|Y=y] \approx \frac{1}{\sum_{i}w_{i}} \sum_{i} x_{i} w_{i}$$
Were $Q(Y|X)$ entirely continuous, I would use $w_{i} = > Q(Y=y|X=x_{i})$. Were it entirely discrete, I would use the probability mass function instead. As $Y|X$ is distributed according to a sort of hybrid, what should my weights be?
This is a most interesting if exotic case of a posterior distribution with atoms!
The difficulty in solving the question is about defining a density for the observation $Y$ against the proper measure. Since $Y$ given $X=x$ takes the values $\pm x$ with probability $\Phi(-x)$ and $x$ takes any real value, it seems impossible to use a counting measure. However, since $Y/x$ takes the values $\pm 1$ with probability $\Phi(-x)$, $Z=Y/x$ has the (conditional) density $$x\varphi(xz)\mathbb{I}_{(-1,1)}(z)+\Phi(-x)\mathbb{I}_{\{-1,1\}}(z)$$ hence $Y$ has the (conditional) density $$\varphi(y)\mathbb{I}_{(-x,x)}(y)+\Phi(-x)\mathbb{I}_{\{-x,x\}}(y)$$ Therefore the posterior distribution on $X$ is $$\varphi(x)\times\left\{\varphi(y)\mathbb{I}_{(-x,x)}(y)+\Phi(-x)\mathbb{I}_{\{-x,x\}}(y)\right\}$$ or $$\varphi(x)\mathbb{I}_{x>|y|}+\Phi(-|y|)\mathbb{I}_{x=|y|}$$ since $\varphi(|y|)$ cancels out. This is a simple mixed distribution made of a truncated normal and a point mass at $|y|$, for which importance sampling (or another Monte Carlo approach) is not necessary.
From a simulation perspective, if importance sampling is contemplated, this means that the importance sampling distribution must have an atom at $|y|$ with probability $\varrho$ say, plus an absolutely continuous component on $\{x>|y|\}$ with probability $(1-\varrho)$, $h(x)$ say. This leads to an importance weight of the form $$\omega(x)=\dfrac{\varphi(x)\mathbb{I}_{x>|y|}+\Phi(-|y|)\mathbb{I}_{x=|y|}}{(1-\varrho) h(x)+\varrho\mathbb{I}_{x=|y|}}$$ For instance, if $\varrho=\Phi(-|y|)$, we have $$\omega(x)=\begin{cases} \dfrac{\varphi(x)}{(1-\varrho) h(x)} &\text{ if }x\ne|y|\\1 &\text{ if }x=|y|\\ \end{cases}$$
[edited]
Restating: The problem you'd like to solve is to infer the posterior over $X$, $P(X|Y)$, using particle filtering. You draw samples $X_i$ from $P(X)$ and then convert them to particles $Y_i \sim Q(Y|X_i)$, i.e., by sampling a $Y$ particle for each $X$ particle from the conditional distribution $Q(Y|X)$. You then need to apply Bayes' rule now to get the posterior of interest:
$P(X|Y) \propto Q(Y|X)P(X)$.
You sampled your initial particles from $P(X)$, so all you need to do is weight them by weights $w_i = Q(Y_i|X_i)$ and you're done! Is that it?
If so, then I think the problem is this: Q(Y|X) has delta masses as a function of Y. But NOT as a function of $X$! That is, there is a point mass at $\pm x$ in the conditional distribution $Q(Y|X_i)$ as a function of $Y$ for a fixed value $X_i$. However, that's not the weight you need here. You want the likelihood, which is $Q(Y_i|X)$ considered as a function of $X$. (Namely, you want to evaluate that function at $X=X_i$).
This latter thing (the likelihood) has no point masses. I assume the sampling procedure for $Y|X$ is something like "add some Gaussian noise and then theshold". You actually get a continuous likelihood function over $X$ for each value of $Y_i$. How you compute it will depend on $Y_i$:
1. If $Y_i=x$ (the upper boundary), then the likelihood involves normcdf: it's the probability that $X$ plus some noise ended up $>x$. So it's a sigmoidal function that goes from zero to 1 as it crosses $x$, with slope that depends on the variance of the noise added to $Y$, and it stays at $+1$ for all larger values of $X$ out to infinity. (Note that this doesn't integrate to 1; it doesn't have to since it's a likelihood!)
2. If $Y_i \in (-x,x)$, the likelihood involves the (standard) normpdf$(Y_i,X_i,\sigma^2)$
But in both cases, you have a continuous function over $X$. There's no value of $X$ for which a particular observation of $Y$ has infinitely higher probability than the other values of $X$ (except for trivial examples like $Y_i=-x$ and $X_i\rightarrow +\infty$).
• I don't think I've explained myself well. I am sampling $(X)_{i=1}^{N}$ then trying to use $w_i = Q(Y;X_i)$ as a weighting function such that $\frac{1}{\sum_{i=1}^{N}w_{i}} \sum_{i=1}^{N}X_{i}w_{i} \approx E[X|Y]$ for sufficiently large N. Think of $X$ as the "maximum distance" a laser range finder can report, after which it just reports its max. I would like to know the probability distribution of the finder's maximum range given a single observation. Jul 11, 2011 at 1:43
• I'll edit my answer... Jul 11, 2011 at 2:56
• I think I'm still being unclear. I'm not trying to use $|N(0,1)|$ as a proposal distribution for $Q(Y;X)$. I am given a realization $y$ of $Y$ drawn from $Q(Y;X)$ for unknown $X$ and would like to approximate the posterior $P(X|y)$ with particles. The support of $X$ is $\mathbb{R}$ so all samples from $|N(0,1)|$ that are greater than or equal to $|y|$ COULD potentially have generated $y$, it's just that some are more likely than others. My primary difficulty is that I don't know if $y$ was generated from one of the deltas or from a continuous portion of $Q(Y;X)$. Jul 11, 2011 at 5:01
• Sorry--your original question is actually very clear; I think I was just too hasty reading it (i.e., thought I knew what you were asking and spouted off). I'll try one more time—I think I have a partial answer that results from a confusion about whether we're taking $Q$ as a function of $X$ or $Y$. Jul 11, 2011 at 5:58 | 1,935 | 6,295 | {"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} | 3.40625 | 3 | CC-MAIN-2023-40 | latest | en | 0.77719 |
https://hilo.hawaii.edu/blog/accreditation/quantitative-reasoning-analysis-2018-2019/ | 1,653,678,968,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662675072.99/warc/CC-MAIN-20220527174336-20220527204336-00005.warc.gz | 358,113,685 | 24,430 | # Quantitative Reasoning Assessment Report
## Overview and Audience
The Assessment Support Committee focused on assessing Quantitative Reasoning in AY 2018-19 using a modified rubric approved by Faculty Congress. The rubric targets three categories: 1- Analysis, 2- Calculations, and 3- Visual Representations of Data and Information (e.g. tables and graphs). The first category, Analysis, is also used to simultaneously assess Critical Thinking.
Evaluations of programs were planned for the Spring 2019 semester. As with the previous campus assessment of Quantitative Reasoning in 2014-15, a common quiz was administered in 300-level, 400-level or capstone courses. A member of the committee from the Mathematics department re-designed the quiz with the understanding that it would be administered across all campus disciplines. Of the 30 undergraduate programs on campus, 24 programs completed the evaluation.
The quiz was anonymous but asked students to identify their class standing. The quiz questions required using information provided in two graphs to determine the correct answers. The creator of the quiz identified the order from easiest to most difficult for the questions was 1, 3, 2, and 4.
The full results are summarized below. Of those who participated, 93% (277 our ot 295) were Juniors or Seniors.
## Assessment results by Class Standing and in the Aggregate
Q1Q2Q3Q4
Class standingnMeanMedianMeanMedianMeanMedianMeanMedian
Unknown211111100
Freshman3.33330.33330.33330.33330
Sophomore11.72731.54551.72731.54551
Junior74.87841.66221.86491.36491
Senior203.857141.55171.83251.41380
All295.85421.57631.830511.40340
## Total Mean, Median, and Mode
Class standingMeanMedianMode
Unknown333
Freshman1.33331n/a
Sophomore2.545534
Junior2.770333
Senior2.655233
All2.664433
## Count and Percent of Students with each Possible Aggregate Score
0 1 2 3 4 3 or 4
ClassnCount%Count%Count%Count%Count%Count%
Unknown200%00%00%2100%00%2100%
60%
Freshmen3133%133%00%133%00%133%
Sophomore7114%114%00%228%342%570%
Junior6500%711%1422%2843%1625%4468%
61%
Senior19284%1910%5127%6232%5227%11459%
2nd Year
11100%00%00%00%00%00%
All270114%2810%6524%9535%7426%16661%
The results of the quiz indicate some improvement in basic skills from the last assessment. There was an increase of Juniors and Seniors who correctly answered a majority of the questions this year (61%) compared to the 2014 assessment (45%). Overall, more students scored better this year, averaging 2.66 out of 4 questions compared to 1.31 out of 3 questions. However, there is still room for improvement. Students could do the simple, one-step problems but had difficulty answering complex problems.
The committee observed that issues with basic quantitative reasoning skills may also reflect issues with skills such as reading comprehension as it seemed that students may have had difficulty identifying important information from the questions needed to answer them correctly. For example, the committee noticed in question two that a high number of students chose b or c, which demonstrates that they had a misunderstanding of the concept of ratio or proportion.
The committee recommends further discussion by departments of the skills needed for student competency in this area including understanding of visual representation of data and reading comprehension across the curriculum.
Report Data:
QR Assessment Mechanism
QR Data by Program
Submitted by Lari-Anne Au
May 2019
Chair of the Assessment Support Committee (AY 2017-2018)
Librarian | 860 | 3,520 | {"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-2022-21 | longest | en | 0.886377 |
https://ispecially.com/can-you-solve-this-math-puzzle-without-calculator-2/ | 1,642,690,363,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320301863.7/warc/CC-MAIN-20220120130236-20220120160236-00546.warc.gz | 404,778,078 | 12,089 | # Can you solve this math puzzle without calculator?
Here is the Mathematical puzzle to test your Arithmetic skills. Can you answer this Mathematical question as quickly as possible and come up with the correct solution in your first attempt?
Wow, It’s not easy!
2+2+2+2+2+2+2+2+2+2+2×0+2=???????
Let’s apply applying BODMAS
Scroll down to see the answer.
A
N
S
W
E
R | 105 | 378 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2022-05 | latest | en | 0.913641 |
https://regmsif.cf/tag/prime/ | 1,638,413,466,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964361064.69/warc/CC-MAIN-20211202024322-20211202054322-00426.warc.gz | 537,792,319 | 24,266 | 题目描述
This is an interactive problem.
For two positive integers $x, y$, we define $\pi(x, y)$ to be the number of distinct primes that divide both $x$ and $y$. For example $\pi(2, 3) = 0, \; \pi(8, 16) = 1$ and $\pi(30, 105) = 2$.
For two positive integers $a, b$, where $a \le b$, we define $S(a, b)$ to be the sum of values $\pi(x, y)$ over all pairs of integers $(x, y)$ satisfying $a \le x \lt y \le b$.
Your task is to compute the values $S(a, b)$ for many query pairs $(a, b)$. To make your task more challenging, all the queries have to be answered online.
代码
#include <cstdio>
#include <cstring>
#include <algorithm>
int const N = 1000005;
int tp, prime[N], rec[N], vis[N], sum[N];
void init() {
for (int i = 2; i < N; ++i) {
if (!vis[i]) {
prime[tp++] = i;
}
for (int j = 0; j < tp && i * prime[j] < N; ++j) {
vis[i * prime[j]] = 1;
if (i % prime[j] == 0) {
break;
}
}
}
for (int i = 2; i < N; ++i) {
sum[i] = sum[i - 1] + !vis[i];
}
}
int main() {
init();
int q;
scanf("%d", &q);
for (; q--;) {
int a, b;
scanf("%d%d", &a, &b);
long long ans = 0;
for (int i = 0; i < tp && prime[i] < b;) {
int cnt = b / prime[i] - (a - 1) / prime[i];
int nxt = 1e9;
if (b / prime[i]) {
nxt = std::min(nxt, b / (b / prime[i]));
}
if ((a - 1) / prime[i]) {
nxt = std::min(nxt, (a - 1) / ((a - 1) / prime[i]));
}
ans += 1ll * cnt * (cnt - 1) / 2 * (sum[nxt] - i);
i = sum[nxt];
}
printf("%lld\n", ans);
fflush(stdout);
}
return 0;
}
题目描述
Dr. Ceizenp’ok from planet i1c5l became famous across the whole Universe thanks to his recent discovery – the Ceizenpok’s formula. This formula has only three arguments: $n, k$ and $m$, and its value is a number of $k$-combinations of a set of $n$ modulo $m$.
While the whole Universe is trying to guess what the formula is useful for, we need to automate its calculation.
算法分析
$$x={n \choose k}, \; m=p_1^{a_1}p_2^{a_2} \cdots p_q^{a_q}$$
$$\left\{ \begin{array}{c} x \equiv r_1 \pmod{p_1^{a_1}} \\ x \equiv r_2 \pmod{p_2^{a_2}} \\ \cdots \\ x \equiv r_q \pmod{p_q^{a_q}} \end{array} \right.$$
$${n \choose k}={n! \over k!(n-k)!}$$
$$f(n)=n! \bmod p_i^{a_i}$$
$$f(n)=\left(f\left(\left\lfloor{n \over p_i}\right\rfloor\right)\cdot p_i^{\lfloor n/p_i \rfloor}\cdot\left(\prod_{i \in [1, p_i^{a_i}], \; p_i\not\mid i}i\right)^{\lfloor n/p_i^{a_i} \rfloor}\cdot\prod_{i \in [1, n \bmod p_i^{a_i}], \; p_i\not\mid i}i\right)\bmod p_i^{a_i}$$
代码
/*
* Your lucky number has been disconnected.
*/
#include <algorithm>
#include <cstdio>
#include <cstring>
#define int long long
int power(int a, int b, int m) {
int ret = 1;
for (; b; b >>= 1)
b & 1 && ((ret *= a) %= m), (a *= a) %= m;
return ret;
}
void ex_gcd(int a, int b, int &x, int &y) {
if (!b) {
x = 1, y = 0;
return;
}
ex_gcd(b, a % b, y, x), y -= a / b * x;
}
int get_inv(int a, int b) {
int x, y;
ex_gcd(a, b, x, y);
return (x + b) % b;
}
int get_fac(int n, int p, int k) {
int m = power(p, k, 1e9), ret = 1;
for (; n; n /= p) {
if (n / m) {
int rec = 1;
for (int i = 2; i < m; ++i)
if (i % p)
(rec *= i) %= m;
(ret *= power(rec, n / m, m)) %= m;
}
for (int i = n % m; i > 1; --i)
if (i % p)
(ret *= i) %= m;
}
return ret;
}
int calc(int N, int K, int M, int p, int k) {
int a = get_fac(N, p, k), b = get_fac(K, p, k), c = get_fac(N - K, p, k),
cnt = 0;
for (int i = N; i; i /= p)
cnt += i / p;
for (int i = K; i; i /= p)
cnt -= i / p;
for (int i = N - K; i; i /= p)
cnt -= i / p;
int m = power(p, k, 1e9),
ret = a * get_inv(b, m) % m * get_inv(c, m) % m * power(p, cnt, m) % m;
return ret * (M / m) % M * get_inv(M / m, m) % M;
}
signed main() {
int N, K, M, ans = 0;
scanf("%lld%lld%lld", &N, &K, &M);
for (int i = 2, t = M; t > 1; ++i)
if (!(t % i)) {
int k = 0;
for (; !(t % i); ++k, t /= i)
;
(ans += calc(N, K, M, i, k)) %= M;
}
printf("%lld\n", ans);
return 0;
}
题目描述
Let $\sigma_0(n)$ be the number of positive divisors of $n$.
For example, $\sigma_0(1)=1, \; \sigma_0(2)=2$ and $\sigma_0(6)=4$.
Let $S_2(n)=\sum_{i=1}^n \sigma_0(i^2)$.
Given $N$, find $S_2(N)$.
算法分析
\begin{align} f(n)=\sum_{d \mid n} 2^{\omega(d)} \end{align}
\begin{align} 2^{\omega(d)}=\sum_{i \mid d} \mu^2(i) \end{align}
\begin{align} S_2(n)=\sum_{i=1}^n \sum_{j \mid i} \sum_{k \mid j} \mu^2(k) \end{align}
\begin{align} S_2(n)&=\sum_{i=1}^n \sum_{j \mid i} \mu^2(j) \sigma_0\left({i \over j}\right) \\ &=\sum_{ij \le n} \mu^2(i) \sigma_0(j) \end{align}
\begin{align} \sum_{i=1}^n \mu^2(i)&=\sum_{i=1}^n [i\text{最大平方因子}=1] \\ &=\sum_{i=1}^n \sum_{j^2 \mid i} \mu(j) \\ &=\sum_{i=1}^{\lfloor \sqrt{n} \rfloor} \mu(i) \left\lfloor {n \over i^2} \right\rfloor \\ \sum_{i=1}^n \sigma_0(i)&=\sum_{i=1}^n \left\lfloor {n \over i} \right\rfloor \end{align}
代码
#include <cmath>
#include <cstdio>
#include <cstring>
void read(long long &t) {
char c; while ((c = getchar()) < '0' || c > '9') ; t = c - '0';
while ((c = getchar()) >= '0' && c <= '9') (t *= 10) += c - '0';
}
int N;
char mui[100000000];
int d[100000000], sum[100000000], prime[10000000];
bool vis[100000000];
long long n, sigma[100000000];
void init() {
int top = 0; mui[1] = 1, sigma[1] = 1;
for (int i = 2; i < N; ++ i) {
if (! vis[i]) prime[top ++] = i, mui[i] = -1, sigma[i] = 2, d[i] = 1;
for (int j = 0; j < top; ++ j) {
int k = i * prime[j]; if (k >= N) break;
vis[k] = true;
if (i % prime[j]) mui[k] = -mui[i], sigma[k] = sigma[i] << 1, d[k] = 1;
else { mui[k] = 0, sigma[k] = sigma[i] / (d[i] + 1) * (d[i] + 2), d[k] = d[i] + 1; break; }
}
}
for (int i = 1; i < N; ++ i) sum[i] = sum[i - 1] + mui[i] * mui[i], sigma[i] += sigma[i - 1];
}
inline long long get_mui(long long n) {
if (n < N) return sum[n];
int q = sqrt(n); long long ret = 0;
for (register int i = 1; i <= q; ++ i) if (mui[i]) ret += mui[i] * (n / i / i);
return ret;
}
inline long long get_sigma(long long n) {
if (n < N) return sigma[n];
long long ret = 0;
for (register long long i = 1, j; i <= n; i = j + 1)
j = n / (n / i), ret += (j - i + 1) * (n / i);
return ret;
}
long long calc(long long n) {
int q = sqrt(n); long long ret = 0;
for (register int i = 1; i <= q; ++ i) if (mui[i]) ret += get_sigma(n / i);
for (register long long i = q + 1, j, cur, last = sum[q]; i <= n; i = j + 1, last = cur)
j = n / (n / i), cur = get_mui(j), ret += (cur - last) * get_sigma(n / i);
return ret;
}
signed main() {
long long T; read(T);
if (T > 800) N = 20000; else N = 100000000;
init();
while (T --) read(n), printf("%lld\n", calc(n));
return 0;
}
题目概述
Nearly prime number is an integer positive number for which it is possible to find such primes $P_1$ and $P_2$ that given number is equal to $P_1 \times P_2$. There is given a sequence on $N$ integer positive numbers, you are to write a program that prints “Yes” if given number is nearly prime and “No” otherwise.
算法分析
Miller-Rabin基于以下几个事实：
• 费马小定理：对于素数$p$和任意整数$a$，有$a^p \equiv a \pmod p$。反之，满足$a^p \equiv a \pmod p$，$p$也几乎一定是素数。
• 伪素数：如果$n$是一个正整数，且存在和$n$互素的正整数$a$满足$a^{n-1} \equiv 1 \pmod n$，那么$n$是基于$a$的伪素数。如果一个数是伪素数，那么它几乎肯定是素数。
• 二次探测定理：如果$p$是奇素数，则$x^2 \equiv 1 \pmod p$的解为$x \equiv \pm 1 \pmod p$。
代码
#include <cmath>
#include <cctype>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define int long long
#define random(x) ((1ll * rand() << 30 ^ 1ll * rand() << 15 ^ rand()) % (x) + 1)
using namespace std;
struct IOManager {
template <typename T> inline bool read(T &x) { char c = '\n'; bool flag = false; x = 0; while (~ c && ! isdigit(c = getchar()) && c != '-') ; c == '-' && (flag = true, c = getchar()); if (! ~ c) return false; while (isdigit(c)) x = (x << 3) + (x << 1) + c - '0', c = getchar(); return (flag && (x = -x), true); }
inline bool read(char &c) { c = '\n'; while (~ c && ! (isprint(c = getchar()) && c != ' ')) ; return ~ c; }
inline int read(char s[]) { char c = '\n'; int len = 0; while (~ c && ! (isprint(c = getchar()) && c != ' ')) ; if (! ~ c) return 0; while (isprint(c) && c != ' ') s[len ++] = c, c = getchar(); return (s[len] = '\0', len); }
template <typename T> inline IOManager operator > (T &x) { read(x); return *this; }
template <typename T> inline void write(T x) { if (x < 0) putchar('-'), x = -x; if (x > 9) write(x / 10); putchar(x % 10 + '0'); }
inline void write(char c) { putchar(c); }
inline void write(char s[]) { int pos = 0; while (s[pos] != '\0') putchar(s[pos ++]); }
template <typename T> inline IOManager operator < (T x) { write(x); return *this; }
} io;
struct Solver {
private:
int n;
void input() { io > n; }
int multiple(int a, int b, int mod) {
int ret = 0; a %= mod;
while (b) { b & 1 && ((ret += a) %= mod), (a <<= 1) %= mod, b >>= 1; }
return ret;
}
int power(int a, int b, int mod) {
int ret = 1; a %= mod;
while (b) { b & 1 && ((ret *= a) %= mod), (a *= a) %= mod, b >>= 1; }
return ret;
}
bool is_prime(int n) {
if (n == 2) return true;
if (n < 2 || ! (n & 1)) return false;
int c = 0, p = n - 1, cnt = 30;
while (! (p & 1)) ++ c, p >>= 1;
while (cnt --) {
int a = random(n - 1), x = power(a, p, n);
for (int i = 0; i < c; ++ i) {
int y = multiple(x, x, n);
if (y == 1 && x != 1 && x != n - 1) return false; x = y;
}
if (x != 1) return false;
}
return true;
}
void process() {
int a; io > a; int q = sqrt(a);
for (int i = 2; i <= q; ++ i)
if (! (a % i) && is_prime(i) && is_prime(a / i)) { io < (char *) "Yes\n"; return; }
io < (char *) "No\n";
}
public:
void solve() { input(); while (n --) process(); }
} solver;
signed main() {
solver.solve();
return 0;
}
题目描述
Arpa has found a list containing $n$ numbers. He calls a list bad if and only if it is not empty and gcd of numbers in the list is $1$.
Arpa can perform two types of operations:
• Choose a number and delete it with cost $x$.
• Choose a number and increase it by $1$ with cost $y$.
Arpa can apply these operations to as many numbers as he wishes, and he is allowed to apply the second operation arbitrarily many times on the same number.
Help Arpa to find the minimum possible cost to make the list good.
代码
#include <cmath>
#include <cctype>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
void minify(long long &a, long long b) { b < a && (a = b); }
struct IOManager {
template <typename T>
inline bool read(T &x) {
char c; bool flag = false; x = 0;
while (~ c && ! isdigit(c = getchar()) && c != '-') ;
c == '-' && (flag = true, c = getchar());
if (! ~ c) return false;
while (isdigit(c)) x = (x << 3) + (x << 1) + c - '0', c = getchar();
return (flag && (x = -x), true);
}
inline bool read(char &c) {
c = '\n';
while (~ c && ! (isprint(c = getchar()) && c != ' ')) ;
return ~ c;
}
inline int read(char s[]) {
char c; int len = 0;
while (~ c && ! (isprint(c = getchar()) && c != ' ')) ;
if (! ~ c) return 0;
while (isprint(c) && c != ' ') s[len ++] = c, c = getchar();
return (s[len] = '\0', len);
}
template <typename T>
inline IOManager operator > (T &x) {
read(x); return *this;
}
template <typename T>
inline void write(T x) {
x < 0 && (putchar('-'), x = -x);
x > 9 && (write(x / 10), true);
putchar(x % 10 + '0');
}
inline void write(char c) {
putchar(c);
}
inline void write(char s[]) {
int pos = 0;
while (s[pos] != '\0') putchar(s[pos ++]);
}
template <typename T>
inline IOManager operator < (T x) {
write(x); return *this;
}
} io;
struct Solver {
private:
static const int N = 1000000;
static const int M = 1000000;
int n, x, y, a[N + 1], cnt[N + 1 << 1];
long long sum[N + 1 << 1];
int top, prime[N];
bool vis[N + 1 << 1];
void input() {
io > n > x > y;
for (int i = 1; i <= n; ++ i) io > a[i], ++ cnt[a[i]], sum[a[i]] += a[i];
}
void init() {
for (int i = 1; i <= M << 1; ++ i) cnt[i] += cnt[i - 1], sum[i] += sum[i - 1];
for (int i = 2; i <= M; ++ i) {
if (! vis[i]) prime[++ top] = i;
for (int j = 1; j <= top && i * prime[j] <= M; ++ j) {
vis[i * prime[j]] = true;
if (! (i % prime[j])) break;
}
}
}
void process() {
int p = x / y;
long long ans = 1000000000000000000ll;;
for (int i = 1; i <= top; ++ i) {
long long tmp = 0;
int delta = prime[i] - p - 1;
for (int j = 0; j <= N; j += prime[i]) {
if (delta < 0) {
tmp += (((long long) cnt[j + prime[i]] - cnt[j]) * (j + prime[i]) - (sum[j + prime[i]] - sum[j])) * y;
} else {
tmp += ((long long) cnt[j + delta] - cnt[j]) * x + (((long long) cnt[j + prime[i]] - cnt[j + delta]) * (j + prime[i]) - (sum[j + prime[i]] - sum[j + delta])) * y;
}
}
minify(ans, tmp);
}
io < ans < '\n';
}
public:
void solve() {
input(), init(), process();
}
} solver;
int main() {
solver.solve();
return 0;
}
题目描述
Vasya is studying number theory. He has denoted a function $f(a, b)$ such that:
• $f(a, 0)=0$;
• $f(a, b)=1+f(a, b-(a, b))$, where $(a, b)$ is the greatest common divisor of $a$ and $b$.
Vasya has two numbers $x$ and $y$, and he wants to calculate $f(x, y)$. He tried to do it by himself, but found out that calculating this function the way he wants to do that might take very long time. So he decided to ask you to implement a program that will calculate this function swiftly.
题意概述
\begin{align} f(a, 0)&=0 \\ f(a, b)&=1+f(a, b-(a, b)) \end{align}
代码
import std.stdio;
import std.math;
static immutable int N = 13;
uint top;
long x, y, p, q, ans;
long[N] prime, cnt;
long gcd(long a, long b) {
return b ? gcd(b, a % b) : a;
}
int main() {
readf(" %d %d", &x, &y);
p = x, q = cast(long) sqrt(real(x));
for (int i = 2; i <= q; ++ i) {
while (! (p % i)) {
p /= i;
if (prime[top] == i) ++ cnt[top];
else prime[++ top] = i, cnt[top] = 1;
}
}
if (p > 1) prime[++ top] = p, cnt[top] = 1;
while (y) {
long max = 0;
for (int i = 1; i <= top; ++ i)
if (cnt[i])
if (y / prime[i] * prime[i] > max)
max = y / prime[i] * prime[i];
ans += y - max, y = max;
long g = gcd(x, y);
x /= g, y /= g;
for (int i = 1; i <= top; ++ i)
while (! (g % prime[i]))
g /= prime[i], -- cnt[i];
}
writeln(ans);
return 0;
}
Notes on Module
逆元
$$(a/(a, m))x \equiv b/(a, m) \pmod {m/(a, m)}$$
$$x \equiv (a/(a, m))^{-1} \times (b/(a, m)) \pmod {m/(a, m)}$$
$$x \equiv (a/(a, m))^{-1} \times (b/(a, m)) + (m/(a, m)) \times k \pmod m \; (0 \le k \lt (a, m))$$
费马小定理
\begin{align} x^{p-1} &\equiv 1 \pmod p \\ x^{-1} &\equiv x^{p-2} \pmod p \end{align}
$$x^{\varphi(m)} \equiv 1 \pmod m$$
线性同余方程组
\left\{ \begin{align} x &\equiv b_i \pmod {m_i} \\ ax &\equiv b_{i+1} \pmod {m_{i+1}} \end{align} \right.
$$a(tm_i+b_i) \equiv b_{i+1} \pmod {m_{i+1}}$$
$$atm_i \equiv b_{i+1}-ab_i \pmod {m_{i+1}}$$
中国剩余定理
$$\forall i \in [1, n], \; a_i=1 \land \forall i, j \in [1, n] \land i \neq j, \; (m_i, m_j)=1$$
$$x \equiv b \pmod {\prod m_i}$$
\begin{align} (a, b)\times(c, d)&=(ac, b+d) \\ (a, b)/(c, d)&=(ac^{-1}, b-d) \end{align}
$n!$
$$((p-1)!)^{n/p} \times (n \bmod p)!$$
${n \choose k} \bmod p$
$$n=\sum n_ip^i \; (0 \le n_i \lt p), \; k=\sum k_ip^i \; (0 \le k_i \lt p)$$
$${n \choose k} \equiv \prod {n_i \choose k_i} \pmod p$$
$${n \choose k}={n! \over k!(n-k)!}$$
$$n!=a_1p^{e_1}, \; k!=a_2p^{e_2}, \; (n-k)!=a_3p^{e_3}$$
Notes on Math Theory
Lucas
$n=(a_ka_{k-1}a_{k-2}\ldots a_0)_p, \; m=(b_kb_{k-1}b_{k-2}\ldots b_0)_p, \; {n \choose m} \equiv \prod_{i=0}^k {a_i \choose b_i} \pmod p$.
$n!$中$p$的次数$=\sum_{i=1}^{\infty} [{n \over p^i}]={n-S_p(n) \over p-1}$
${n \choose m}$中$p$的次数$p^\alpha \mid \mid {n \choose m} \Leftrightarrow p^\alpha \mid {n \choose m} \land p^{\alpha+1} \not \mid {n \choose m}$.
${n \choose m} \rightarrow p \not \mid {n \choose m} \Leftrightarrow {n-S_p(n) \over p-1}={m-S_p(m) \over p-1}+{n-m-S_(n-m) \over p-1} \Leftrightarrow m+(n-m)$在$p$进制下无进位。
$(1+x)^n \rightarrow x^m$系数${n \choose m}$.
$(1+x)^{p^k} \equiv 1+x^{p^k} \pmod p$.
$(1+x)^{a_kp^k+ \cdots +a_1p+a_0}=(1+x)^{a_kp^k} \cdots (1+x)^{a_1p}(1+x)^{a_0}$.
$\equiv (1+x^{p^k})^{a_k} \cdots (1+x^p)^{a_1}(1+x)^{a_0} \pmod p$.
$n!!=1 \times 3 \times 5 \times \cdots \times n$.
Euler
$x^{\varphi(n)} \equiv 1 \pmod n, (x, n)=1$.
$x^n \equiv x^{n-\varphi(n)} \pmod n$.
$\bmod n$缩系.
$a_1 \times a_2 \times \cdots \times a_{\varphi(n)} \equiv a_1x \times a_2x \times \cdots \times a_{\varphi(n)}x \pmod n$.
$n$为素数，$x^{p-1} \equiv 1 \pmod p, p \not \mid x$.
$x^p \equiv x \pmod p$.
$\sum_{d \mid n} \varphi(d)=n$.
$\varphi(p^{\alpha})=p^{\alpha}-p^{\alpha-1}$.
$n=p_1^{\alpha_1} \cdots p_k^{\alpha_k}$.
$\sum_{\beta_i \le \alpha_i} \varphi(p_1^{\beta_1} \cdots p_k^{\beta_k})=\sum_{\beta_i \le \alpha_i} \prod_{i=1}^k \varphi(p_i^{\beta_i})=\prod_{i=1}^k \sum_{\beta_i \le \alpha_i} \varphi(p_i^{\beta_i})=\prod p_i^{\alpha_i}$.
$n=1$成立。
$n>1$时，设$p$为$n$最小的素因子。
$\begin{cases} p \mid 2^n-1\\ p \mid 2^{p-1}-1\\ p \mid 2^d-1 \end{cases} \Rightarrow \begin{cases} d \mid n\\ d \mid p-1 \end{cases} \Rightarrow d=1 \Rightarrow p \mid 1 \Rightarrow n=1$.
$ax+by=(a, b)$.
$nx+(p-1)y=(n, p-1)$.
$p \mid 2^{(n, p-1)}-1$.
∵$(n, p-1)=1$.
∴$p \mid 1$.
$1, x, x^2, \ldots, x^{\varphi(n)-1}$构成$\bmod n$的缩系，则称$x$为$n$的原根，记作$g$。
$p$是素数时，${g, g^2, \ldots, g^{p-1}} \equiv {1, 2, \ldots, p-1} \pmod p$.
$n$的原根有$\varphi(\varphi(n))$个。
${1, 2, \ldots, p-1} \Rightarrow {a, 2a, \ldots, (p-1)a}$.
$\sum i^k \equiv \sum (a_i)^k \Rightarrow (a^k-1) \sum i^k \equiv 0 \pmod p$.
二项同余式
$x^k \equiv n \pmod p$无解或有$(k, p-1)$个解。
$f(x) \equiv 0 \pmod p$.
$x^k \equiv 1 \pmod 1$有$d$个解，$d=(k, p-1)=ak+b(p-1)$。
$\begin{cases} x^{p-1} \equiv 1 \pmod p\\ x^k \equiv 1 \pmod 1 \end{cases} \Rightarrow x^{(k, p-1)} \equiv 1 \pmod p$至多$d$个解。
${x^{p-1}-1 \over x^d-1}=x^{p-1-d}+x^{p-1-2d}+ \cdots +x^d+1 \equiv 1 \pmod p$至多$p-1-d$个解。
$x^k \equiv 1 \pmod 1$恰有$d$个解。
$x \rightarrow p$的缩系，$x^k$有${p-1 \over (k, p-1)}$个不同的取值。
$x^{p-1} \equiv 1 \pmod p$且$x^d \not \equiv 1 \pmod p, d \lt p-1$，$x$是原根。
$x^{q^{\alpha}} \equiv 1 \pmod p$.
$x^{q^{\alpha-1}} \not \equiv 1 \Leftrightarrow \forall d \lt q^{alpha}, \; x^d \not \equiv 1 \pmod p$.
$x^{(q^{\alpha}, d)} \equiv 1 \Leftrightarrow x^{q^{\beta}} \equiv 1 \pmod p$.
$\gamma(l_1l_2)=\gamma(l_1)\gamma(l_2)$.
$h_1$次数是$l_1$，$h_2$次数是$l_2 \Rightarrow h_1h_2$的次数是$l_1l_2$。
$x$次数是$l_1l_2 \Rightarrow x_1$次数是$l_1$，$x_2$次数是$l_2$。
Prime
$\lim \sum_{p=1}^{\infty} (1-{1 \over p})=0$.
$f(x) \equiv 0 \pmod {p^{\alpha}}$的解数$=f(x) \equiv 0 \pmod p$的解数$\Leftarrow \begin{cases} f(x) \equiv 0\\ f'(x) \equiv 0 \end{cases} \pmod p$无解。
$x^2+1 \equiv 0 \pmod {p^{\alpha}}, p \ge 3$无解或恰有$2$解。
勾股数的通式
$a^2+b^2=c^2 (a, b, c \gt 0)$.
$(a, b, c)=1$本原。
$a^2+b^2=c^2 \Rightarrow b^2=c^2-a^2=(c-a)(c+a)$.
$(c-a, c+a)=(2c, c-a)=(2, c-a)=1$.
$2 \not \mid c-a$.
$\begin{cases} {c-a \over 2}=m^2\\ {c+a \over 2}=n^2 \end{cases}$.
$\begin{cases} b=2mn\\ a=n^2-m^2\\ c=n^2+m^2 \end{cases}$.
有理根 Eisenstein
$f \in Z[x], f(x)=a_nx^n+ \cdots +a_1x+a_0$.
$\pm {p \over q} \Rightarrow p \mid a_n, q \mid a_0$.
$f(x)=c_nx^n+c_{n-1}x^{n-1}+ \cdots +c_1x+c_0$可约$\Leftarrow p \mid c_0, p \mid c_1, \ldots, p \mid c_{n-1}, p \not \mid c_n, p^2 \not \mid c_0$.
$f(x)=c_nx^n+c_{n-1}x^{n-1}+ \cdots +c_1x+c_0$可约$\Leftarrow p \mid (c_0, c_1, \ldots, c_{n-1}), p \not \mid c_n, p^2 \not \mid c_0$.
$f(x)=x^{p-1}+x^{p-2}+ \cdots +x+1={x^p-1 \over {x-1}}$.
升幂定理
$5^{210000}-2^{210000}$里有多少$3$的幂次。$2$次。
$(1+p)^k-1$里有多少$p$的幂次。$1+V_p(k)$次。
$5^{33333}+1$里有多少$3$的幂次。$2$次。
$p^{\alpha} \mid \mid a-b, (a, p)=1, (b, p)=1, p \ge 3$.
$V_p(a^m-b^m)=\alpha+V_p(m)$.
$p^{\alpha} \mid \mid a^{pn}-b^{pn}$.
$p^{\alpha} \mid \mid a^{kn}-b^{kn}, (k, p)=1$.
二次剩余
$ax^2+bx+c \equiv 0 \pmod p$.
$x^2 \equiv a \pmod p$.
${p-1 \over (2, p-1)}={p-1 \over 2}$.
$1^2, 2^2, 3^2, \ldots, ({p-1 \over 2})^2$.
Euler判别法：$n^{p-1 \over 2} \equiv ({n \over p}) \pmod p, (n, p)=1$.
Gauss Lama：$n, 2n, 3n, \ldots, {p-1 \over 2}n \pmod p$的最小正剩余中大于${p-1 \over 2}$的个数为$m$，则$({n \over p})=(-1)^m$。
$({2 \over p})=(-1)^{p^2-1 \over 8}=(-1)^{[{p \over 2}]-[{p \over 4}]}$.
Summary
$n=(a_ka_{k-1}a_{k-2}\ldots a_0)_p, m=(b_kb_{k-1}b_{k-2}\ldots b_0)_p, {n \choose m} \equiv \prod_{i=0}^k {a_i \choose b_i} \pmod p$.
$x^{\varphi(n)} \equiv 1 \pmod n, (x, n)=1$.
$x^n \equiv x^{n-\varphi(n)} \pmod n$.
$n$为素数，$x^{p-1} \equiv 1 \pmod p, p \not \mid x$.
$x^p \equiv x \pmod p$.
$x^k \equiv n \pmod p$无解或有$(k, p-1)$个解。
$x^k \equiv 1 \pmod 1$有$d$个解，$d=(k, p-1)=ak+b(p-1)$。
$x \Rightarrow p$的缩系，$x^k$有${p-1 \over (k, p-1)}$个不同的取值。
$\bmod p$的原根有$\varphi(p-1)$个。
$\lim \sum_{p=1}^{\infty} (1-{1 \over p})=0$.
$f(x) \equiv 0 \pmod {p^{\alpha}}$的解数$=f(x) \equiv 0 \pmod p$的解数$\Leftarrow \begin{cases} f(x) \equiv 0\\ f'(x) \equiv 0 \end{cases} \pmod p$无解。
$\begin{cases} b=2mn\\ a=n^2-m^2\\ c=n^2+m^2 \end{cases}$.
$f(x)=c_nx^n+c_{n-1}x^{n-1}+ \cdots +c_1x+c_0$可约$\Leftarrow p \mid c_0, p \mid c_1, \ldots, p \mid c_{n-1}, p \not \mid c_n, p^2 \not \mid c_0$.
$f(x)=c_nx^n+c_{n-1}x^{n-1}+ \cdots +c_1x+c_0$可约$\Leftarrow p \mid (c_0, c_1, \ldots, c_{n-1}), p \not \mid c_n, p^2 \not \mid c_0$.
$p^{\alpha} \mid \mid a-b, (a, p)=1, (b, p)=1, p \ge 3$.
$V_p(a^m-b^m)=\alpha+V_p(m)$.
题目描述
Fox Ciel is participating in a party in Prime Kingdom. There are $n$ foxes there (include Fox Ciel). The $i$-th fox is $a_i$ years old.
They will have dinner around some round tables. You want to distribute foxes such that:
• each fox is sitting at some table,
• each table has at least 3 foxes sitting around it,
• the sum of ages of any two adjacent foxes around each table should be a prime number.
If $k$ foxes $f_1, f_2, \ldots, f_k$ are sitting around table in clockwise order, then for $1 \le i \le k-1$: $f_i$ and $f_{i+1}$ are adjacent, and $f_1$ and $f_k$ are also adjacent.
If it is possible to distribute the foxes in the desired manner, find out a way to do that.
代码
#include <iostream>
#include <cstring>
#include <cmath>
#include <vector>
using namespace std;
struct edge {
int v, f, nxt;
} e[50001];
int n, nume = 1, top, src, sink, a[202], h[202], g[202], dist[202], que[202];
bool vis[202];
vector<int> ans[201];
void add_edge(int u, int v, int f) {
e[++nume].v = v, e[nume].f = f, e[nume].nxt = h[u], h[u] = nume;
e[++nume].v = u, e[nume].f = 0, e[nume].nxt = h[v], h[v] = nume;
}
void bfs() {
memset(dist, 0, sizeof(dist));
que[1] = src, dist[src] = 1;
int s = 0, t = 1;
while (s < t) {
int u = que[++s];
for (int i = h[u]; i; i = e[i].nxt) {
if (e[i].f && !dist[e[i].v]) {
que[++t] = e[i].v, dist[e[i].v] = dist[u] + 1;
}
}
}
}
int dfs(int u, int delta) {
if (u == sink) return delta;
int ret = 0;
for (int i = g[u]; i; i = e[i].nxt) {
if (e[i].f && dist[e[i].v] == dist[u] + 1) {
int dd = dfs(e[i].v, min(e[i].f, delta));
e[i].f -= dd, e[i ^ 1].f += dd;
if (e[i].f) g[u] = i;
delta -= dd, ret += dd;
}
}
return ret;
}
int max_flow() {
int ret = 0;
while (1) {
bfs();
if (!dist[sink]) return ret;
for (int i = 0; i <= n + 1; ++i) g[i] = h[i];
ret += dfs(src, 1e9);
}
}
bool is_prime(int t) {
int k = (int) sqrt(t);
for (int i = 2; i <= k; ++i) if (!(t % i)) return false;
return true;
}
void find(int u, int t) {
vis[u] = true, ans[t].push_back(u);
for (int i = h[u]; i; i = e[i].nxt) {
if (e[i].v && e[i].v != n + 1 && e[i].f == !(a[u] & 1) && !vis[e[i].v]) find(e[i].v, t);
}
}
int main()
{
cin >> n;
src = 0, sink = n + 1;
for (int i = 1; i <= n; ++i) {
cin >> a[i];
if (a[i] & 1) add_edge(src, i, 2);
else add_edge(i, sink, 2);
}
for (int i = 1; i <= n; ++i) {
for (int j = 1; j <= n; ++j) {
if (a[i] & 1 && !(a[j] & 1) && is_prime(a[i] + a[j])) add_edge(i, j, 1);
}
}
max_flow();
for (int i = h[0]; i; i = e[i].nxt) {
if (e[i].f) {
cout << "Impossible" << endl;
return 0;
}
}
for (int i = h[n + 1]; i; i = e[i].nxt) {
if (e[i].f < 2) {
cout << "Impossible" << endl;
return 0;
}
}
for (int i = 1; i <= n; ++i) if (!vis[i]) top++, find(i, top);
cout << top << endl;
for (int i = 1; i <= top; ++i) {
cout << ans[i].size() << ' ';
for (vector<int>::iterator iter = ans[i].begin(); iter != ans[i].end(); ++iter) {
cout << *iter << ' ';
}
cout << endl;
}
return 0;
} | 10,218 | 23,477 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2021-49 | latest | en | 0.505877 |
https://www.spoj.com/problems/MAIN8_D/ | 1,585,600,123,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370497301.29/warc/CC-MAIN-20200330181842-20200330211842-00249.warc.gz | 1,190,640,690 | 9,080 | ## MAIN8_D - Coing tossing
no tags
One day Rohil was getting very bored so he was tossing an unbiased coin randomly. He observed that certain patterns (a sequence of Head and Tail) appear very frequently
while some other are very rare. Being a programmer he decided to code a solution which takes a pattern string as input and tells what is
the expecetd number of times he will have to toss his coin to see that pattern. He wrote this program very quickly. Can you?
One day Rohil was getting very bored so he was tossing an unbiased coin randomly. He observed that certain patterns (a sequence of Head and Tail) appear very frequently while some other are very rare. Being a programmer he decided to code a solution which takes a pattern string as input and tells what is the expected number of times he will have to toss his coin to see that pattern. He wrote this program very quickly. Can you?
Input
First line contains ( 1<=T<=25 ) the number of test cases. Each of following T lines contains a pattern string of 'H' and 'T' only. H is for Head and T is for Tail.
|S| <= 40
### Output
For each test case print the output in a new line (it is guaranteed that answer will always be an integer and fits in 64 bit type).
### Example
```Input:
3
H
HTHT
TTHTHTHTHTHHTHTHTHTTTTTTHTHHHHHTT
Output:
2
20
8589934598``` | 325 | 1,323 | {"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-2020-16 | latest | en | 0.96952 |
https://mcqslearn.com/cs/semantic-web/quiz/quiz.php?page=11 | 1,720,853,809,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514490.70/warc/CC-MAIN-20240713051758-20240713081758-00141.warc.gz | 322,236,147 | 16,799 | Computer Science Degree Courses
Semantic Web Certification Exam Tests
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# An Axiomatic Semantics for RDF and RDF Schema Quiz Questions with Answers PDF Download - 11
Books:
Apps:
The An Axiomatic Semantics for RDF and RDF Schema Trivia Questions and Answers PDF (An Axiomatic Semantics for RDF and RDF Schema Quiz Answers PDF e-Book) download Ch. 3-11 to study semantic web online courses. Study Describing Web Resources RDF Quiz Questions PDF, an axiomatic semantics for rdf and rdf schema Multiple Choice Questions to learn software engineering courses. The An Axiomatic Semantics for RDF and RDF Schema Trivia App Download: Free learning app for future extensions, owl in owl, an axiomatic semantics for rdf and rdf schema test prep for computer science programs.
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## An Axiomatic Semantics for RDF and RDF Schema Quiz with Answers : MCQs 11
MCQ 51:
In RDFS, FuncProp connects the resource with
1. at most one value
2. at least one value
3. more then one value
4. at one or more value
MCQ 52:
Metaclasses are
1. classes of classes
2. objects of classes
3. classes of objects
4. objects of objects
MCQ 53:
Moduling facility in OWL is based on
1. abstraction
2. information hiding
3. inheritance
4. polymorphism
MCQ 54:
The formalization in formal language uses formulas that are also known as
1. statements
2. predicates
3. subjects
4. axioms
MCQ 55:
The information that can be consumed by computer is called
1. information retrieval
2. machine-accessible
3. location finder
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Digital Logic Design App (iOS & Android) | 618 | 2,630 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2024-30 | latest | en | 0.768603 |
https://www.instructables.com/Science-Fair-Air-Rockets/ | 1,723,609,162,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641095791.86/warc/CC-MAIN-20240814030405-20240814060405-00207.warc.gz | 634,475,774 | 45,363 | ## Introduction: Science Fair Air Rockets
"Rockets! I want to build and test rockets!"
That's what my oldest son told me when we began discussing what he wanted to do for his school science fair project this year. During a class launch last year he had noticed that some rockets appeared to launch faster than others and they were all using the same rocket engine so he thought some rockets must be heavier than others or maybe the fins made a difference. For his science fair project this year what he wanted to learn about was acceleration and prove that a heavier rocket would accelerate less quickly than a lighter weight rocket.
In order to prove his hypothesis we needed to build a rocket and launch system where the only variable would be the mass of the rocket. We thought about using traditional solid propellant rocket motors but we had not devised a method of testing the motors to ensure the consistency of motor power output. For this reason we decided to use compressed air as a power source since we could accurately monitor the air pressure in the air launcher. We also liked the idea of air rockets as some solid propellant rockets that can carry a payload can reach very high altitudes and we were concerned with the possibility of losing a rocket and our data.
In order to measure the acceleration during each launch we made a very simple circuit using a three axis accelerometer with a range of +/- 16g. The data from the accelerometer was logged on a micro SD card during launch so we could analyze it later. This way we could launch over and over again during one launch session in order to try and remove weather as a variable.
This is a great project for introducing kids to the scientific process, coding, electronics, soldering and making rockets. The rockets are very inexpensive to make and once the launcher is built they cost nothing to launch and you can launch them over and over again in a very small field.
This project does require the use of a soldering iron and sharp cutting tools so be sure to take the necessary safety precautions when assembling the rocket, launcher and circuit.
Now let's gather our materials and build some rockets!
## Step 1: Tools and Materials
In order to build our rocket and launcher we're going to need a few items. Most everything but the electronics and a few tools can be sourced at a big box home improvement store.
Tools used:
Small tub or bucket (for thinning glue with water)
1" wide paint brush (for applying glue)
Rotary cutter
Hacksaw (for cutting PVC pipe)
Soldering iron
Materials for the rocket:
Brown builder's paper
White glue
Super glue
24" long 3/4" schedule 40 PVC pipe (used a form for making paper tubes)
3/32" balsa sheet
Balsa wood blocks (for making a nose cone and coupler but these can be purchased pre made)
Materials for the air launcher:
16" long 2" schedule 40 PVC pipe
24" long 3/4" schedule 40 PVC pipe
3/4" PVC Tee fitting
2 ea 2" PVC coupler
2 ea 2" x 3/4" slip PVC reducer bushing
2 ea 3/4" thread x 3/4" slip fitting
3/4" inline sprinkler valve
2 ea 3/4" male thread x 1/2" female thread brass reducer bushing
8ea "AA" batteries
2ea 4 "AA" battery holders
Momentary push button switch
10ft 2 conductor wire
Duct tape
Zip ties
Bicycle tire valve
Bicycle tire pump
Air pressure gauge
4ft 1/4" ID x 1/2" OD high pressure air hose
2ea 1/2" hose clamps
PVC pipe cement
PVC pipe cement primer
Materials for the data logger:
Sparkfun OpenLog
Sparkfun Arduino Pro Mini 3.3V
Single cell 3.7V 150 mAh LiPo
3.3V FTDI adapter (for programming the Pro Mini)
Single cell LiPo charger
Small prototyping board
Micro SD card
Heat shrink tubing
## Step 2: Rocket Construction
Building the rocket is pretty straightforward as there is no motor mount, parachute, launch lug or shock cord. Gravity is our only recovery system...
Since our rocket is air powered it needs to precisely fit our launch tube and that means making our own wound paper tubes. Thankfully there's an awesome Instructable for that!
The only modification we need when making our tubes is to first cover the launch tube with three layers of paper before wrapping our three final tube layers- do not glue the first three layers as these will be removed later. The reason for these additional layers is when the glue dries the tube will shrink a fair bit. If you don't account for this shrinkage with additional layers first the finished paper tube will be too small to slide easily over the launch tube.
We found the easiest way to cut the paper into accurate width strips is to get a large ruler or wide piece of rigid material to use as a guide with a rotary cutter. By using a 2" wide Aluminum ruler we could cut perfect width strips very quickly. This makes a huge difference in the ease of wrapping the paper tubes since the paper strips were extremely consistent in width. Three layers of wrapped paper makes for a reasonably strong and light weight tube.
Once the finished tube is dry it can be slid off the 3/4" PVC pipe after trimming the ends with a knife. Then you just reach in the end of the tube and pull out the three under layers. The finished tube should be a perfect fit on the launch tube and its inside diameter is just large enough to fit the data logger circuit.
Once the rocket body tube is made it needs to be cut in order to create a payload section. We cut the payload section to a length of 5 1/2" and the main body tube to a length of 7". The length of these sections isn't super critical. The payload section is attached to the main rocket body tube with a 2 1/2" long solid balsa wood coupler/plug. The plug slides into each tube section and is glued to the top of the lower body tube and the bottom of the payload section. This plug is what the high pressure air from the launcher pushes against in order to force the rocket off the launch tube. A solid balsa nose cone is then fitted to the top of the payload section. We made the nose cone a total length of 3" with almost 1" fitting inside the top of the payload section. The nose cone should be a very snug fit.
You can buy all kinds of balsa nose cones and couplers but I turned ours from balsa wood blocks using a lathe. Carving nose cones and couplers from insulation foam would also be an option.
We cut the fins from 3/32" thick balsa sheet and glued them to the main tube using super glue followed by filling any gaps with white glue. We decided to use four fins vs. three in hopes of obtaining more stable launches. The pattern for the fins came from the Juno rocket here. The finished weight of our rocket is 30 grams.
## Step 3: Build the Launcher
Air power!
The compressed air launcher is very similar to a launcher featured in Make Magazine with a couple of small differences. The most important change we made was to add a pressure gauge to read the air pressure in the air chamber. Being able to read the air pressure in the chamber means we can get very consistent launch power.
The main body of the launcher is made from a 16" long section of 2" schedule 40 PVC pipe. A 2" coupler with a plug cap is fitted to each end of the main body. When cementing the PVC pieces together be sure to use a primer to prep the joints before applying PVC cement. One of the plug caps has a short length of schedule 40 3/4" PVC fitted and then a 3/4" Tee is fitted to the 3/4" pipe. Another short length of 3/4" pipe is fitted to the opposite side of the Tee and this is fitted to a 3/4" threaded fitting that attaches to the inline sprinkler valve. The opposite side of the sprinkler valve has a threaded 3/4" slip fitting with a 16" long schedule 40 PVC pipe for the launch tube.
The other 2" PVC plug cap has 3/4" slip x 1/2" female thread fitting attached. A 1/2" x 1/4" threaded adapter is then used to attach a 1/4" threaded hose barb to the 2" plug cap. A 1/4" ID high pressure hose is then secured to the hose barb with a hose clamp and a bicycle tire valve is fit to the other end of the hose with a second hose clamp. I cut the tire valve out of a spare bicycle inner tube I had in my garage.The finished launcher is wrapped with duct tape for protection in case the PVC pipe should ever fail under pressure.
The sprinkler valve is powered using 8 "AA" batteries so when a button is pressed the sprinkler valve is actuated and the air shoots out the launch tube. The battery packs are connected in series to make a 12V power source. The positive wire from the battery pack goes to one wire of the sprinkler valve. The other sprinkler valve wire is connected to the launcher push button switch with a long wire. Another long wire connects the push button switch to the battery pack negative wire so when the launcher push button switch is pressed it completes the circuit. The air launcher is attached to an old telescope tripod I had on hand using large zip ties.
Update:
I found an Aluminum project box in a dumpster along with a key switch so I mounted the "AA" battery holders in the box and wired the key switch in series with the launch push button. This way it's harder to trigger a launch by accident and it looks a lot nicer than before.
Overall this is a very simple and reliable launch system that is very quick to build!
## Step 4: Building the Data Logger
We need to build a circuit to log accelerometer data.
This is a very simple circuit to build as it has very few parts and just a few connections. Since it's such a simple circuit it's a good first project for learning how to solder. The circuit is based around a 3.3V Arduino Pro Mini, an Adafruit ADXL 326 analog accelerometer and a Sparkfun OpenLog. The Pro Mini reads the output from the accelerometer and the OpenLog records the data on a micro SD card.
In order to keep the circuit as small and light weight as possible the OpenLog and accelerometer were connected directly to the Pro Mini using pin headers and a small two row section of prototyping board. We took the male seven pin header that shipped with the accelerometer board and soldered the prototyping board directly to the analog input pins on the Pro Mini. We began with pin 13 and soldered all seven pins. Next we soldered a seven pin female machine pin header on the next row of holes on the prototyping board. The female machine pin header is oriented so the accelerometer board is situated underneath the Pro Mini.
Now we bridged four pins from the female header to the pins soldered to the Pro Mini- these pins connect the accelerometer X axis to analog in pin 0, Y axis to analog in pin 1, Z axis to analog in pin 2 and the accelerometer 3V pin to VCC on the Pro Mini. The GND pin from the accelerometer is connected to the GND pin on the opposite side of the Pro Mini with a short length of wire.
Next we soldered a six pin female machine pin header to the top of the Pro Mini- this is used to connect the Open Log to the Pro Mini and it's also used as a programming port. We soldered male machine pin headers to the OpenLog and accelerometer board. I really like these machine pin headers as they are very low profile and they have a very solid connection compared to standard headers.
Finally we soldered a battery connector to the Pro Mini. This was made with short lengths of wire and a two pin female machine pin header. These wires were soldered to the pins marked GND and RAW on the Pro Mini. The ends of the wires at the pin header were covered with heat shrink tubing for protection. The JST connector was cut from the LiPo battery leads (don't throw it away!) and a matching male pin header was soldered to the battery leads and they too were covered with heat shrink tubing. I ended up shortening the battery leads quite a bit to save weight and space. Note that when the data logger is placed in the rocket the X axis arrow on the accelerometer needs to point up in order for it to work properly.
To charge the battery a matching two pin female connector was soldered to the cut off JST connector and this is then plugged into the single cell LiPo charging circuit to charge the battery. The finished weight of the data logging circuit and battery came to 12.03 grams.
Now that the circuit is done it's time for programming...
## Step 5: Program the Data Logger
Programming the Arduino is super easy. If you're not familiar with Arduino I wrote a basic guide to programming here.To program the Arduino you need to unplug the OpenLog board as it shares the same pins as the FTDI board needs during programming. The machine pin headers are a slightly different size than standard headers so I connect the FTDI board using jumper wires.
Once the Arduino is programmed you can plug the OpenLog back into its header, connect the battery and check it out- there a couple of LEDs on the OpenLog that should light up. It should log data to the SD card just like what is shown in the photo- it will be labeled LOG#####.TXT on the SD card. The OpenLog is configured to communicate at 9600 baud rate so it's good to go out of the box. If you have trouble with it open up the SD card and look at the file labeled CONFIG.TXT and make sure the first line at the top says 9600- if the baud rate doesn't match the card won't record data properly.
Every time you disconnect the battery and plug it back in the OpenLog will create a new log file on the SD card.
I've included the data logging code here so you can download it. For our experiment we only needed to read the accelerometer X axis but I went ahead and read all three axis anyway. Once the programming is complete it's time to place the data logger in the rocket and launch it.
On to the launch site!
## Step 6: Launch!
3, 2, 1, launch!
For our first launch I suggested we should pump the launcher up to 60 psi. I figured that would give us a solid launch and not shoot the rocket too high. Boy was I wrong! I couldn't believe how high the rocket went and it came screaming down with a loud "thud" when it slammed into the ground. It didn't sound good... The rocket hit the ground so hard the nose cone slammed back into the body like a cork and I had to pry it out.
Epic fail!
Upon turning the rocket over I saw there was a large hole torn in the side where the collapsed nose cone had pushed the data logger circuit back and out of the body. We also broke a fin. I was able to retrieve the circuit but the SD card had ejected and was nowhere to be seen- not only was our rocket damaged but we had no data to show for it. Bummer!
I noticed the circuit was no longer functioning so I unplugged the battery and plugged it back in- still no power. We packed everything up, went home and started troubleshooting. The first thing we checked was the battery and it showed 4.17V on the multimeter but when it was plugged back in the circuit wouldn't power up. Next we checked the circuit by plugging it into my laptop and it was getting power. Very strange.
Going back to the battery we discovered it was showing proper voltage but it couldn't provide any current- as soon as it was plugged into the circuit the voltage would immediately drop at the circuit input. Upon examining the battery's built in protection circuit we discovered one of the tiny circuit chips on the protection circuit had been torn off in the crash. We plugged a spare battery into the circuit and now the circuit would power up -but it wouldn't log any data. The data transmission LEDs on the OpenLog weren't lighting up either. Next we connected the board back to the laptop and no data transmission there either. On our first launch we managed to kill both our rocket and our data logger. Double fail!
But we can rebuild it. We have the technology....
## Step 7: Repair!
We got lucky.
As it just so happened I had another Arduino Pro Mini board on hand. The only problem was this board was the 5V version. Luckily I had a 5V boost regulator board on hand that would bump our battery voltage to 5V to power it. Since the wonderful Adafruit accelerometer board and OpenLog can both handle both 3.3V and 5V input we were back in business.
We took a bit of prototyping board and quickly assembled a new circuit. Since our OpenLog and accelerometer board survived the previous crash all we had to do was connect the power output from the 5V boost regulator to the VCC pin on the Pro Mini. All of our other connections remained the same but we had to construct it a bit different than our previous circuit due to the necessary boost regulator- all we did was run point to point wires for each of the connections for the accelerometer, power and ground. After uploading our code our circuit was ready to log some data. This circuit was slightly heavier than the first version at 15.28 grams.
I knew that since we were now using a 5V board instead of a 3.3V board to read our accelerometer the numbers logged would be a bit different. Normally if you connect a 5V sensor to a 5V Arduino you would see values that range from a minimum of 0 to a maximum of 1023. When you use a 3.3V sensor with a 5V Arduino the values will range from 0 to 675 because of the voltage/scaling difference. 3.3/5 * 1023 = 675. This is will make sense later when we review the data.
The fin was glued back on the rocket and we taped over the hole in the side of the rocket. The nose cone fit was repaired with some tape and we were good to go!
Back to the launch site!
## Step 8: Launch Again!
Let's try that again...
We set up to launch again- but with a bit less power this time! We took the precaution of taping the nose cone to the body to keep it from collapsing into the body tube and this time we pumped up the launcher to 20 psi.
3, 2, 1... successful launch! I still couldn't believe how high the rocket went but it didn't come down nearly as hard this time due to the decreased altitude. After our launch we opened up the rocket and unplugged the battery from the data logger in order to reset it. Next we added our 5 gram weight and then closed the rocket back up and taped the nose cone shut before launching again.The weights were nothing more than a few BB's rolled up in duct tape pouch.
On the second launch everything was kept the same except for the additional 5 grams added- air pressure was kept the same at 20 psi. Once the rocket came back we repeated the process with a 10 gram weight added. On the third launch with the 10 grams added the rocket came back noticeably faster and the nose cone was pushed into the rocket body a bit. We opened up the rocket and retrieved our data logger. The perfboard that we used to build the circuit had fractured during landing but everything was intact and more importantly we had our data this time! I would definitely recommend putting some crumpled paper or cotton balls between the nose cone and data logger to cushion the electronics in a hard landing.
With our data safely tucked away we stayed at the site for a while launching over and over again (this time at 60 psi and higher.) We soon had a small group of kids around us wanting to check out our rocket launcher. It was pretty exciting and the kids had a ton of fun.
## Step 9: Analyze the Data
Graphs and numbers time!
When you read the data from the SD card you'll see there are four columns of numbers. The first column is the elapsed time in milliseconds. The next column is the the accelerometer X axis reading, followed by the Y and Z axis. For the purpose of this experiment, what we really care about is the X axis data as that shows the vertical acceleration of the rocket.
The accelerometer has a range of +/- 16g. At 0V we would get a reading of -16g and at 3.3V we would see +16g. The Arduino interprets this reading as a numerical value ranging from 0 to 1023 (this is called 10 bit resolution.) Now remember that I said the value would change when using a 3.3V sensor with a 5V Arduino? When using a 5V Arduino the maximum value interpreted will be 675 instead of 1023. Our logged accelerometer data reflects this.
On our first launch the maximum accelerometer X axis value recorded is 664. With an additional 5 grams the max value is 398 and with 10 grams added the max value is 635. If we convert these numbers to g values (675 = +16g) we get +15.73g for the first launch, +9.43g for the second launch and +15.05g for the third launch.
So why are the the recorded numbers lower for the second launch?
In order to see what is going on let's make a line plot for each launch. First change the name of the .TXT file for each log to .CSV so it can be uploaded to a site like Plot.ly. This allows us to plot a line chart and zoom in and examine the acceleration data during the launch. On the second launch you can see the plot bounces up and down during the launch vs. it being near vertical for the other two launches. This could indicate the data logger possibly shifted around during the launch. The first and last launch plots are more clear- there is a rapid acceleration recorded and clearly the heavier rocket had slower acceleration.
We only recorded these three launches but ideally you would record as many launches as possible in as short a time as possible in order to gather the most accurate data.
Why does the acceleration drop for a heavier rocket?
Because physics! Newton's Second Law of Motion states that the net force on an object is equal to the mass of the object multiplied by the acceleration of the object- or more simply put: force equals mass times acceleration.
This Law is expressed in the formula F=ma. F=force (in our case the 20psi air pressure of our launcher), m=mass (the mass of our rocket) and a=acceleration (the recorded acceleration of our rocket.)
If we change this equation to solve for acceleration (what we measured) we get a=F/m. From looking at this equation and dividing the force by the mass we can see that since our launch force is constant, the higher the rocket mass the lower the acceleration number.
This formula tells us that in the case of our second launch there must have been a problem with the launch or with our data logging and this is shown by the accelerometer graph plot.
## Step 10: Going Further
Sampling rate
If you look at the logged data you will see the difference in logged sample times in milliseconds. The time difference between readings is on average 20 milliseconds -we're taking a reading every .02 seconds. This means we're logging data at a rate of 50Hz, or 50 samples per second since 1/.02 = 50.
What if we want to log data at a faster rate?
We can change the Arduino code to get more data points. If we change the baud rate we can speed up the data transmission and capture more samples. The default baud rate on the Arduino and OpenLog is 9600 but we can go all the way up to 115200.
To change the baud rate we simply replace 9600 with 115200 in the Arduino code. You also have to change the baud rate for the OpenLog SD card to match, otherwise the card will just log gibberish. To change the OpenLog rate we change the first value in the CONFIG.TXT file on the SD card. Once this is done you'll see that we are now logging data every 4 milliseconds or .004 seconds. That's 250 Hz or 250 samples per second.
In order to really take full advantage of this you need to change the accelerometer board as the Adafruit ADXL 326 board is set to 50Hz. By looking at the manufacturer's data sheet you can see it's possible to increase that bandwidth by swapping out some capacitors. For the purpose of our experiment this really wasn't necessary but it would be interesting to pursue this option later...
Other possibilities
Now that we're able to measure acceleration we could alter percentages of weight or launch force and run multiple back to back tests in order to see if our experimental data backs up our math. If it doesn't that means we have variables in our testing procedure that are affecting the outcome. Further testing could include three fins vs. four or various shapes of nose cones or fins and asking: How does aerodynamics affect the acceleration? We can also add an altitude sensor to the rocket can tell us how changes to our rocket affects the launch height.
If you want to build a slimmer rocket use 1/2" schedule 40 PVC pipe for the launch tube and rocket body tube mandrel. A slimmer rocket will have less drag and should be able to obtain higher launches but it will require modifying the data logger circuit in order to get it to fit inside the slimmer rocket body. The sky's the limit with what you can build and test!
This was a great project for my son and I to do together and he had a blast doing it. He learned about the scientific process, how to solder, program an Arduino, overcome failure and build rockets! I hope others can put this instructable to good use and if you have any questions please don't hesitate to ask!
### Attachments
If you don't want to build your own air launcher you can buy one here-
AirRocketWorks.com
National Association of Rocketry educational resources. This is a great guide for all things educational relating to rocketry-
NAR educational guide
Make: Rockets- A fantastic book on hobby rocketry. If you have even the slightest interest in hobby rocketry this book is worth every penny!
Make: Rockets
Runner Up in the
Coded Creations
Participated in the
Explore Science Contest | 5,621 | 25,276 | {"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-2024-33 | latest | en | 0.97607 |
https://www.fairvalue-calculator.com/en/calculate-fair-value/ | 1,725,732,696,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700650898.24/warc/CC-MAIN-20240907162417-20240907192417-00193.warc.gz | 743,522,937 | 61,136 | "Fair Value Calculator: The Value That Counts"
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https://www.proofwiki.org/wiki/Power_Series_Expansion_for_Exponential_Integral_Function/Formulation_1 | 1,685,552,259,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224646937.1/warc/CC-MAIN-20230531150014-20230531180014-00093.warc.gz | 1,054,930,450 | 11,491 | # Power Series Expansion for Exponential Integral Function/Formulation 1
## Theorem
Let $\Ei: \R_{>0} \to \R$ denote the exponential integral function:
$\map \Ei x = \ds \int_{t \mathop = x}^{t \mathop \to +\infty} \frac {e^{-t} } t \rd t$
Then:
$\ds \map \Ei x$ $=$ $\ds -\gamma - \ln x + \sum_{n \mathop = 1}^\infty \paren {-1}^{n + 1} \frac {x^n} {n \times n!}$ $\ds$ $=$ $\ds -\gamma - \ln x + \frac x {1 \times 1!} - \frac {x^2} {2 \times 2!} + \frac {x^3} {3 \times 3!} - \dots$
where $\gamma$ denotes the Euler-Mascheroni constant.
## Proof
$\ds \map \Ei x$ $=$ $\ds -\gamma - \ln x + \int_0^x \frac {1 - e^{-u} } u \rd u$ Characterization of Exponential Integral Function $\ds$ $=$ $\ds -\gamma - \ln x + \int_0^x \frac 1 u \paren {1 - \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {u^n} {n!} } \rd u$ Definition of Real Exponential Function $\ds$ $=$ $\ds -\gamma - \ln x + \int_0^x \frac 1 u \paren {\sum_{n \mathop = 1}^\infty \paren {-1}^{n + 1} \frac {u^n} {n!} } \rd u$ $\ds$ $=$ $\ds -\gamma - \ln x + \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^{n + 1} } {n!} \paren {\int_0^x u^{n - 1} \rd u}$ Power Series is Termwise Integrable within Radius of Convergence $\ds$ $=$ $\ds -\gamma - \ln x + \sum_{n \mathop = 1}^\infty \paren {-1}^{n + 1} \frac {x^n} {n \times n!}$ Primitive of Power
$\blacksquare$ | 554 | 1,333 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2023-23 | latest | en | 0.236218 |
https://newpathworksheets.com/math/grade-6/statistics-1/missouri-standards | 1,638,469,994,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362287.26/warc/CC-MAIN-20211202175510-20211202205510-00406.warc.gz | 434,317,553 | 8,363 | ## ◂Math Worksheets and Study Guides Sixth Grade. Statistics
### The resources above correspond to the standards listed below:
#### Missouri Learning Standards
MO.6.DSP. Data Analysis, Statistics and Probability
6.DSP.A. Develop understanding of statistical variability.
6.DSP.A.3. Recognize that a measure of center for a numerical data set summarizes all of its values with a single number, while a measure of variation describes how its values vary from a single number.
6.DSP.B. Summarize and describe distributions.
6.DSP.B.5. Summarize numerical data sets in relation to the context.
6.DSP.B.5c. Give quantitative measures of center (median and/or mean) and variability (interquartile range and/or mean absolute deviation), as well as describing any overall pattern and any striking deviations from the overall pattern with reference to the context of the data. | 186 | 870 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2021-49 | latest | en | 0.836939 |
https://www.mrexcel.com/board/threads/count-visible-cells-filtered-data.1109412/?s=6ad320124ba81ef502f36c144ff09108 | 1,582,040,820,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875143784.14/warc/CC-MAIN-20200218150621-20200218180621-00207.warc.gz | 851,157,165 | 15,071 | # Count visible cells filtered data
#### UKNOWINVU2
##### New Member
I am trying to count visible cells only between 1.5 & 2. I used this formula:
=COUNTIF(J7:J83952,">=1.5")-COUNTIF(J7:J83952,">2")
But how do it get this to work with filtered data? Also, is there a way to add another column of data to search (N7:N83952)
Last edited:
#### James006
##### Well-known Member
Hi,
You can use the subtotal function : =Subtotal(103, yourRange) to count visible rows
Hope this will help
#### UKNOWINVU2
##### New Member
Hi,
You can use the subtotal function : =Subtotal(103, yourRange) to count visible rows
Hope this will help
Your idea works to count everything but only want to count the cells with values between 1.5 & 2.
#### James006
##### Well-known Member
Hi again,
In order to add more criteria, would recommend using sumproduct ...
You could test following:
=SUMPRODUCT((J7:J83952">=1.5")*(J7:J83952">2")*SUBTOTAL(3,OFFSET(J7,ROW(J7:J83952)-MIN(ROW(J7:J83952)),0)))
Hope this will help
#### UKNOWINVU2
##### New Member
Hi again,
In order to add more criteria, would recommend using sumproduct ...
You could test following:
=SUMPRODUCT((J7:J83952">=1.5")*(J7:J83952">2")*SUBTOTAL(3,OFFSET(J7,ROW(J7:J83952)-MIN(ROW(J7:J83952)),0)))
Hope this will help
This did not work. Quotes in wrong spots? When quotes removed it did not generate the correct answer.
#### UKNOWINVU2
##### New Member
How can I write a formula to do the follwoing:
If cell B7 contains the word "W2" & J7 is between 1.5 & 2 count 1, plus N7 is between 1.5 & 2
#### Yongle
##### Well-known Member
Your idea works to count everything but only want to count the cells with values between 1.5 & 2.
So :
.... filter ... leaving only rows you want to count visible ... and use the SUBTOTAL function to count those rows
Last edited:
#### James006
##### Well-known Member
Sorry ...copied your criteria without removing the quotes ...
Are you working with Numbers or Text ?
#### UKNOWINVU2
##### New Member
Sorry ...copied your criteria without removing the quotes ...
Are you working with Numbers or Text ?
Numbers. Forget about my second reply stating if A7 contains the word "W2". I am trying to still get my first request.
I am trying to count visible cells only between 1.5 & 2. I used this formula:
=COUNTIF(J7:J83952,">=1.5")-COUNTIF(J7:J83952,">2")
But how do it get this to work with filtered visible data, meaning it dynamical changes when data is filtered. Also, is there a way to add another column of data to search (N7:N83952)
Last edited: | 720 | 2,559 | {"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} | 3.34375 | 3 | CC-MAIN-2020-10 | latest | en | 0.824045 |
https://philosophy.stackexchange.com/questions/73805/validity-soundness-of-an-argument-from-r-carrier | 1,643,265,374,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320305141.20/warc/CC-MAIN-20220127042833-20220127072833-00716.warc.gz | 506,346,119 | 32,453 | # Validity/Soundness of an argument from R. Carrier
This is about training in argumentation. I use a text from R. Carrier: https://www.richardcarrier.info/archives/468, where he claims that from nothing everything follows. (Don't get shocked, the article is not long, it's the comments that make it look long.)
Carrier basically defines nothing as no thing/fact plus logical necessities like mathematical truths. He makes an argument where he uses probability theory to claim that the probability is almost 1 that universes would arise from nothing. He can because probability is a logical necessity in the sense (and I simplify heavily): |= If Kolmogorov's axioms + stuff then probability theory, and he said that logical necessities are always there, even in nothingness. But my problem is that it's also a logical necessity that "If Kolmogorov's axioms + stuff + setting any probability to zero then P(anything) = 0.
So he basically takes one logical necessity and assumes its premises to come to his conclusions whereas other logical necessities would make his conclusion to collapse. I'd tend to see his argument as valid but unsound. What (if at all) is foul with his argument and what is the technical term for it? (Please just focus on my problem described here, any other reasons for his argument to fail, maybe overlooked in this discussion.)
• Logical necessities? No thing could possibly come out of nothingness, However, if there is truly nothing, then there cannot be any constraints, except of there being nothing. In particular, there cannot be any constraints as to what might possibly pop out into existence and what would be the probability of that. And once it does, the probability that it did is 1. Jun 28 '20 at 20:14
• Carrier agrees: He says that if nothing is then also there's nothing that limits things not to exist. But of course: I could turn it 180° and say: if nothing is then also there's nothing that drives things into existence. It seems another premise from Carrier that is just questionable. Jun 29 '20 at 0:12
• "This entails it is effectively 100 percent certain an infinite multiverse exists because the probability of there being only one universe is then 1/INFINITY" is fallacious. Not only does it apply probabilities to a context where they are nonsensical, but it further assumes the principle of indifference for them. But we were told earlier that mathematical laws are not necessary in the sense that "their premises are necessarily true". And the principle of indifference is not even an axiom. Jun 29 '20 at 5:30
• @Pippen There is nothing to drive things into existence and nothing to prevent them from popping out into existence. I guess this is what true randomness would be. No a priori probabilities. Things happen or not. We can only observe the result and only because we are part of it. This also contradicts the idea that there would be any "probability of there being only one universe". Jun 29 '20 at 10:12 | 646 | 2,972 | {"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-2022-05 | latest | en | 0.970532 |
https://physics.stackexchange.com/questions/70438/how-far-would-a-radio-signal-propagate-in-free-space-as-compared-to-earths-atmo | 1,721,894,187,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518579.47/warc/CC-MAIN-20240725053529-20240725083529-00237.warc.gz | 380,200,695 | 41,406 | # How far would a radio signal propagate in free space as compared to Earth's atmosphere?
A curiosity question.
Radio Propagation within Earth's atmosphere is via atmosphere. Broadly speaking a signal loses strength between bouncing off the ionized layer, and absorption by various earthy elements.
If an identical signal were to be produced in free space, how far would it travel as compared to it's range within Earth's atmosphere?
EDIT: A commercial FM transmitter on Earth probably has a range no more than 25 miles where the listener uses a broadcast receiver. For the purpose of elaboration assume the the receiver being placed in static orbit - how far out could the same signal be generated, and still be intelligible?
• I wonder how information is conveyed to satellites.(THIS is an answer to your question)
– Nix
Commented Jul 9, 2013 at 12:12
• Another thing to think about: The spacecraft 'Voyager 1' was launched in 1977. It is now at a distance of $1.866 × 10^{10}$km, and it communicates to the earth by means of radio signals. Commented Jul 9, 2013 at 12:36
• The question is ill-posed. You don't specify noise conditions, receiver sensitivity, and initial signal's ERP to get a sensible answer as to the maximum ranges at which both signals will be detected. Commented Jul 9, 2013 at 12:42
• The main problem in radio communications is being able to recover the transmitted signal from the received, noisy signal; and it is mainly a channel coding problem. also, because the distances between spacecrafts (or stations, etc.)in space are very large, the transmission power is not too important and the successful communication depends on error-correction methods used. Commented Jul 9, 2013 at 12:47
• @mikhailcazi: The Voyagers use the Deep Space Network; apparently using PSK. My query is about the range of a system in space relative to the range of the same system on Earth. Commented Jul 10, 2013 at 1:28
First, no, "radio propagation" is not "via atmosphere". Different wavelengths get absorbed, reflected, or simply passed by different parts of the atmosphere. There is no one general rule. Many of our radio communications within the atmosphere are pretty much like they would be in free space, for example.
Second, all radio waves propagate infinitely in free space. There is no finite end to the propagation. What does matter in a practical sense is signal to noise ratio. Below some signal to noise ratio for whatever information encoding scheme is used, that information can't be recovered. Or more accurately, the error rate goes up as the signal to noise ratio goes down. At some point the errors in the information make is useless or "unreceivable" in a practical sense. | 602 | 2,708 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2024-30 | latest | en | 0.94503 |
https://mligomutov.web.app/361.html | 1,702,097,845,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100800.25/warc/CC-MAIN-20231209040008-20231209070008-00004.warc.gz | 454,271,798 | 6,102 | # Beggs and brill matlab tutorial pdf
Where can i get tutorial for working with fluents population balance model. I couldnt find any working tutorials in the web other than a pdf file made in 2012. When the program pauses, the prompt in the command window changes to k, indicating that matlab is in debug mode. In the cfd simulation, both the boundary and operating conditions were determined. Forexample,engineering problem solving with matlab,byd.
Download signalsandsystemsusingmatlab ebook pdf or read online books in pdf, epub. Choose a web site to get translated content where available and see local events and offers. Pressure gradient prediction of multiphase flow in pipes. Matlab excels at matrix operations, and consequently the arithmetic operators such as multiplication, division, and exponentiation. A comprehensive study on the current pressure drop calculation. Secondly, matlab is an interactive programming environment. With this matlab ebook, students will get befitted by. This is a short introduction to scienti c computation in matlab. For the love of physics walter lewin may 16, 2011 duration. Hello, this is my first post on these forums, so please bear with me.
Gas compressibility factor z based on dranchuk and aboukassem eos, dimensionless 18 iftgobakerswerdloff. Essential matlab for engineers and scientists kindle edition by hahn, brian, valentine, daniel. Group is a numeric matrix whose first and second columns specify different parts and operators, respectively. Essential matlab for engineers and scientists, hahn, brian. This correlation resulted from experiments using air and water as test fluids over a wide range of parameters. A brief introduction to matlab brad baxter version. Rightclick on data file to import the data as matlab variables. Beggs and brill model is considered classic in the field of multiphase flow and has been cited by several papers to be reliable for calculations involving large liquid mass input fractions and small diameter pipes at various orientation angles 2. The approach in this study is developed in matlab and easy to. Matlab is the computing standard in almost all engineering disciplines.
Beggs and brill traverse model is a better prediction tool than the beggs and brill model. Solve more problems with matlab and solve them faster. The first condition tests if format is the name of a field in structure s. We also have many ebooks and user guide is also related with engineering matlab pdf, include. A few other empirical correlations are used to describe the multiphase flow through the choke. Use features like bookmarks, note taking and highlighting while reading. The injection depth can be specified anywhere along the wellbore. An accurate estimation of the pressure drop in well tubing is essential for the solution of a number of important production engineering and reservoir analysis problems. I need a pagebreak methodology for pdf output matlab. In addition, the pressure drop can be related directly to the flow rate through a flow rate dependent correlation. The truth of the second condition depends on the first. The method improves the beggs and brill multiphase flow correlation. Find all the books, read about the author, and more. Multiphase flow models range of applicability tech note ctes, l.
Aspen hysys pipe module was not recommended for choke flow or flow near sonic velocity. Andrews plots represent each observation by a function ft of a continuous dummy variable t over the interval 0,1. Using the modified beggs brill correlation and flow pattern map the correlation rules are. This video describes the simulation algorithm using ica pca approach. Beggs and brill present a unique flow map in terms of froude number vs.
The bottom hole pressure is calculated to be 1069 psig which indicates a measured difference of 11. Matrix laboratory interactive environment programming language invented in late 1970s cleve moler chairman csd univ new mexico fortran alternative to linpack dynamically typed, garbage collection. Matlab programming a primer for the process engineer. Download basics of matlab and beyond by andrew knight esm. Optimize pipeline hydraulics with multiphase flow modeling. A gas lift option is also included with the program. In matlab, the above matrix multiplication would be performed directly from the matlab prompt. Matlab code for predicting multiphase flow parameters in wells. Mcs320 introductiontosymboliccomputation spring2007 matlab lecture 7.
Several empirical correlations and mechanistic models have been proposed in the. A combined bottomhole pressure calculation procedure. Solve the problem in example 2 using the beggsbrill method. Ifyoujustwantanoverviewofeach graphical tool and examples of. Interfacial surface tension for live oil by baker and swerdloff correlation, dymescm. Afterward, the directory of the toolbox should be inserted as the main directory in the matlab environment. If you are an experienced fuzzy logic user,youmaywanttostartatthe beginning of chapter 2, tutorial to make sure you are comfortable with the fuzzy logic toolbox terminology.
When working with illconditioned matrices, an unreliable solution can result even though the residual bax is relatively small. For additional programming flexibility, create functions which accept input and return outputs. To use this model, at first, the lssvm toolbox for matlab should be installed. From this plot, it is observable that the calculated gas rate is 30. Prediction of sour gas compressibility factor using an. Heat exchangers jordan university of science and technology. Rightclick on matlab scripts and function files to execute the commands contained therein. The second statement then tests whether the format field is empty.
As the name matlab suggests, the programming environment is based on matrix and vector constructs. For example, multiphase flow occurs in almost all producing oil and gas wells and surface pipes transporting produced fluids. The rows of x correspond to observations, the columns to variables. You then can view or change the values of variables to see if the new values. This video uses two signals in the waveform using matlab to estimate the original sourc. Basics of matlab and beyond by the author andrew knight is just such a book. When you have a sequence of commands to perform repeatedly or that you want to save for future reference, store them in a program file.
See variablesizing restrictions for code generation of toolbox functions matlab coder. Model based on experiments with airwater flow for various pipe inclinations. Matlab r2011b software is used for building the model. Change directories by clicking on folders or use the current working directory text box at the top of the matlab working environment.
Brill and beggs correlation for gas compressibility factor z, dimensionless 17 zfactordak. Matlab code for predicting multiphase flow parameters in. The second input, k, cannot have type int64 or uint64. In this particular example, the norm of the residual is zero, and an exact solution is obtained, although rcond is small. The beggs and brill method was developed to predict the. Finally, matlab has an impressive set of buildin graphical functions. First the appropriate flow regime for the particular combination of gas and liquid rates segregated, intermittent or distributed is determined. In the first case, information is needed on the geometry of the exchanger. Pdf modeling pressure drop in vertical wells using group.
When the first input, x, is a vector, nchoosek treats it as a set. The beggs and brill multiphase correlation deals with both the friction pressure loss and the hydrostatic pressure difference. Matlab code for predicting multiphase flow parameters in wells, any help. To use this model, firstly, the lssvm toolbox for matlab should be installed. The estimation of the pressure drop in vertical wells is quite impor tant for costeffective design of well completions, production optimization and surface facilities. Correlations such as beggs brill can be used for this purpose. Both the flow pattern prediction and the frictional. Download it once and read it on your kindle device, pc, phones or tablets. Figure 6 computer flow diagram for the beggs and brill method 16. Its handson, tutorial approach gently takes new users by the hand and leads them to competence in all the fundamentals of matlab. The report explorer converts this element to a pdf page break when the output type is direct pdf from template.
Financial risk forecasting is a complete introduction to practical quantitative risk management, with a focus on market risk. Brill and beggs correlation for gas compressibility factor. Illustration of independent component analysis using matlab. It integrates computation, visualization, and programming in an easyto.
Next, the directory of the toolbox should be inserted as the main directory in the matlab environment. Matlab matrix laboratory produced by the mathworks used for simulation and numerical computation no maplelike symbolical solving standard tool for developing embedded systems. Derived from the authors teaching notes and years spent training practitioners in risk management techniques, it brings together the three key. Sigmon, published by crc press 1994 is a handy summary of matlab instructions. Matlab is a highperformance language for technical computing. However, i run out of memory long before the full file is read. Use the htfs or the new tulsa methods for a much better match.
The matlab programming environment consists primarily of a commandline interface, where commands input using the keyboard or. A matlab tutorial is provided on a disk which is available for studentinstructor use, and all examples in the text are developed in. If you are using the dom api to generate your report, you can use a dom pagebreak content object or a pagebreakbefore format object to insert a page break. The second condition can never be true if the first condition is not true. Binomial coefficient or all combinations matlab nchoosek. The fortran code on the other hand would be written using an editor, then compiled, and then finally run.
1187 1204 1119 572 1013 130 1486 1535 614 1489 386 446 937 53 1506 34 1234 1603 211 1174 1040 162 60 1343 100 425 727 1212 1005 1140 569 112 616 1061 100 255 351 322 270 782 674 167 83 388 77 198 814 | 2,113 | 10,407 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2023-50 | latest | en | 0.888759 |
https://www.quantconnect.com/docs/v2/our-platform/optimization/parameters | 1,670,085,282,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710933.89/warc/CC-MAIN-20221203143925-20221203173925-00702.warc.gz | 983,452,910 | 14,848 | Optimization
Parameters
Introduction
Parameters are project variables that your algorithm uses to define the value of internal variables like indicator arguments or the length of lookback windows.
Setting Values
Parameters are stored outside of your algorithm code, but we inject the values of the parameters into your algorithm when you launch an optimization job. The optimizer adjusts the value of your project parameters across a range and step size that you define to minimize or maximize an objective function. To optimize some parameters, add some parameters to your project and add the GetParameter method to your code files.
The GetParameter method returns the parameter value with the numeric type of the default value. If you don't provide a default value, the method returns a string, and you need to cast it to the data type you need. If there are no parameters in your project that match the name you pass to the method and you provide a default value to the method, it returns the default value.
// Get the parameter value and return an integer
var intParameterValue = GetParameter("<parameter-name>", 100);
// Get the parameter value as a string and cast it to an integer
var castedParameterValue = Convert.ToInt32(GetParameter("<parameter-name>"));
# Get the parameter value and return an integer
parameter_value = self.GetParameter("<parameter-name>", 100)
# Get the parameter value as a string and cast it to an integer
parameter_value = int(self.GetParameter("<parameter-name>"))
Overfitting
Overfitting occurs when a function is fit too closely fit to a limited set of training data. Overfitting can occur in your trading algorithms if you have many parameters or select parameters values that worked very well in the past but are sensitive to small changes in their values. In these cases, your algorithm will likely be fine-tuned to fit the detail and noise of the historical data to the extent that it negatively impacts the live performance of your algorithm. The following image shows examples of underfit, optimally-fit, and overfit functions:
An algorithm that is dynamic and generalizes to new data is more likely to survive across different market conditions and apply to other markets.
Look-ahead bias occurs when an algorithm makes decisions using data that would not have yet been available. For instance, in optimization jobs, you optimize a set of parameters over a historical backtesting period. After the optimizer finds the optimal parameter values, the backtest period becomes part of the in-sample data. If you run a backtest over the same period using the optimal parameters, look-ahead bias has seeped into your research. In reality, it would not be possible to know the optimal parameters during the testing period until after the testing period is over. To avoid issues with look-ahead bias, optimize on older historical data and test the optimal parameter values on recent historical data. Alternatively, apply walk forward optimization to optimize the parameters on smaller batches of history.
Number of Parameters
The cloud optimizer can optimize up to 2 parameters. There are several reasons for this quota. First, the optimizer only supports the grid search strategy, which is very inefficient. This strategy tests every permutation of parameter values, so the number of backtests that the optimization job must run explodes as you add more parameters. Second, the parameter charts that display the optimization results are limited to two dimensions. Third, if you optimize with many variables, it increases the likelihood of overfitting to historical data.
We plan to upgrade the parameter charts on the optimization results page to support 3D surface charts. When we upgrade the visualization technology and add more efficient optimization strategies, we will increase the number of parameters you can optimize in the cloud. To optimize more parameters now, run local optimizations with the CLI.
You can also see our Videos. You can also get in touch with us via Discord. | 765 | 4,039 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2022-49 | latest | en | 0.586455 |
http://lists.puremagic.com/pipermail/digitalmars-d-learn/2020-March/112001.html | 1,685,765,669,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224649105.40/warc/CC-MAIN-20230603032950-20230603062950-00438.warc.gz | 28,215,618 | 3,671 | # in not working for arrays is silly, change my view
aliak something at something.com
Mon Mar 2 22:21:49 UTC 2020
```On Monday, 2 March 2020 at 21:33:37 UTC, Steven Schveighoffer
wrote:
> On 3/2/20 3:52 PM, aliak wrote:
>> On Monday, 2 March 2020 at 15:47:26 UTC, Steven Schveighoffer
>> wrote:
>>> On 3/2/20 6:52 AM, Andrea Fontana wrote:
>>>> On Saturday, 29 February 2020 at 20:11:24 UTC, Steven
>>>> Schveighoffer wrote:
>>>>> 1. in is supposed to be O(lg(n)) or better. Generic code
>>>>> may depend on this property. Searching an array is O(n).
>>>>
>>>> Probably it should work if we're using a "SortedRange".
>>>>
>>>>
>>>> int[] a = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9];
>>>> auto p = assumeSorted(a);
>>>>
>>>> assert(3 in p);
>>>>
>>>>
>>>
>>> That could work. Currently, you need to use p.contains(3).
>>> opIn could be added as a shortcut.
>>>
>>> It only makes sense if you have it as a literal though, as
>>> p.contains(3) isn't that bad to use:
>>>
>>> assert(3 in [0, 1, 2, 3, 4, 5, 6, 7, 8, 9].assumeSorted);
>>>
>> There's no guarantee that checking if a value is in a sorted
>> list is any faster than checking if it's in a non sorted list.
>> It's why sort usually switches from a binary-esque algorithm
>> to a linear one at a certain size.
>
> Well of course! A binary search needs Lg(n) comparisons for
> pretty much any value, whereas a linear search is going to end
> early when it finds it. So there's no guarantee that searching
> for an element in the list is going to be faster one way or the
> other. But Binary search is going to be faster overall because
> the complexity is favorable.
Overall tending towards infinity maybe, but not overall on the
average case it would seem. Branch prediction in CPUs changes
that in that with a binary search it is always a miss. Whereas
with linear it's always a hit.
>
>> The list could potentially need to be _very_ large for
>> p.contains to make a significant impact over canFind(p) AFAIK.
>>
>> Here's a small test program, try playing with the numbers and
>> see what happens:
>>
>> import std.random;
>> import std.range;
>> import std.algorithm;
>> import std.datetime.stopwatch;
>> import std.stdio;
>>
>> void main()
>> {
>> auto count = 1_000;
>> auto max = int.max;
>>
>> alias randoms = generate!(() => uniform(0, max));
>>
>> auto r1 = randoms.take(count).array;
>> auto r2 = r1.dup.sort;
>> auto elem = r1[uniform(0, count)];
>
> auto elem = r1[\$-1]; // try this instead
>
>>
>> benchmark!(
>> () => r1.canFind(elem),
>> () => r2.contains(elem),
>> )(1_000).writeln;
>> }
>>
>> Use LDC and -O3 of course. I was hard pressed to get the
>> sorted contains to be any faster than canFind.
>>
>> This begs the question then: do these requirements on in make
>> any sense? An algorithm can be log n (ala the sorted search)
>> but still be a magnitude slower than a linear search... what
>> has the world come to 🤦♂️
>>
>> PS: Why is it named contains if it's on a SortedRange and
>> canFind otherwise?
>>
>
> A SortedRange uses O(lgn) steps vs. canFind which uses O(n)
> steps.
canFind is supposed to tell the reader that it's O(n) and
contains O(lgn)?
>
> If you change your code to testing 1000 random numbers, instead
> of a random number guaranteed to be included, then you will see
> a significant improvement with the sorted version. I found it
> to be about 10x faster. (most of the time, none of the other
> random numbers are included). Even if you randomly select 1000
> numbers from the elements, the binary search will be faster. In
> my tests, it was about 5x faster.
Hmm... What am I doing wrong with this code? And also how are you
compiling?:
void main()
{
auto count = 1_000_000;
auto max = int.max;
alias randoms = generate!(() => uniform(0, max - 1));
auto r1 = randoms.take(count).array;
auto r2 = r1.dup.sort;
auto r3 = r1.dup.randomShuffle;
auto results = benchmark!(
() => r1.canFind(max),
() => r2.contains(max),
() => r3.canFind(max),
)(5_000);
results.writeln;
}
\$ ldc2 -O3 test.d && ./test
[1 hnsec, 84 μs and 7 hnsecs, 0 hnsecs]
>
> Note that the compiler can do a lot more tricks for linear
> searches, and CPUs are REALLY good at searching sequential
> data. But complexity is still going to win out eventually over
> heuristics. Phobos needs to be a general library, not one that
> only caters to certain situations.
General would be the most common case. I don't think extremely
large (for some definition of large) lists are the more common
ones. Or maybe they are. But I'd be surprised. I also don't think
phobos is a very data-driven library. But, that's a whole other
conversation :)
>
> -Steve
``` | 1,346 | 4,685 | {"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-2023-23 | latest | en | 0.930778 |
https://www.convertunits.com/from/millibar/to/petabar | 1,618,355,733,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038075074.29/warc/CC-MAIN-20210413213655-20210414003655-00570.warc.gz | 814,782,722 | 16,735 | ## ››Convert millibar to petabar
millibar petabar
How many millibar in 1 petabar? The answer is 1.0E+18.
We assume you are converting between millibar and petabar.
You can view more details on each measurement unit:
millibar or petabar
The SI derived unit for pressure is the pascal.
1 pascal is equal to 0.01 millibar, or 1.0E-20 petabar.
Note that rounding errors may occur, so always check the results.
Use this page to learn how to convert between millibars and petabars.
Type in your own numbers in the form to convert the units!
## ››Want other units?
You can do the reverse unit conversion from petabar to millibar, or enter any two units below:
## Enter two units to convert
From: To:
## ››Definition: Millibar
A millibar (mb) is 1/1000th of a bar, a unit for measurement of pressure. It is not an SI unit of measure, however it is one of the units used in meteorology when describing atmospheric pressure. The SI unit is the pascal (Pa), with 1 millibar = 100 pascals (a hectopascal)
## ››Definition: Petabar
The SI prefix "peta" represents a factor of 1015, or in exponential notation, 1E15.
So 1 petabar = 1015 bars.
The definition of a bar is as follows:
The bar is a measurement unit of pressure, equal to 1,000,000 dynes per square centimetre (baryes), or 100,000 newtons per square metre (pascals). The word bar is of Greek origin, báros meaning weight. Its official symbol is "bar"; the earlier "b" is now deprecated, but still often seen especially as "mb" rather than the proper "mbar" for millibars.
## ››Metric conversions and more
ConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 100 kg, US fluid ounce, 6'3", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more! | 530 | 2,018 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2021-17 | latest | en | 0.859772 |
https://nve24.com/3-diget-math-multiplication-worksheet-numbers/ | 1,566,270,042,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027315222.14/warc/CC-MAIN-20190820024110-20190820050110-00425.warc.gz | 584,818,434 | 29,700 | # 3 Diget Math Multiplication Worksheet Numbers
In Free Printable Worksheets192 views
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Top | 1,408 | 7,300 | {"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-2019-35 | latest | en | 0.876612 |
http://quizlet.com/4943471/lis-ib-physics-thermal-sl-flash-cards/ | 1,386,896,841,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386164786099/warc/CC-MAIN-20131204134626-00088-ip-10-33-133-15.ec2.internal.warc.gz | 160,811,889 | 17,555 | # LIS IB Physics Thermal SL
## 12 terms
### (The) mole
the amount of a substance that contains as many elementary entities as the number of atoms in 12 g of the isotope carbon-12
### difference between evaporation and boiling (with reference to temperature)
evaporation takes place at any temperature/involves a reduction in temperature and boiling takes place at constant temperature
### difference between evaporation and boiling (with reference to surface area of a liquid)
evaporation takes place at the surface of the liquid/depends on surface area of the liquid and boiling takes place throughout the liquid/is independent of surface area
### temperature of an ideal gas
a measure of the molecules' average kinetic energy
### Internal energy
the total (potential energy and) kinetic energy of the molecules/ atoms/particles
### Heating
the (non-mechanical) transfer of energy (from the surroundings/source) to the object
### molar mass
The mass of one mole of a substance
in 1 mole of a substance, there are 6.02 x 10^23 particles of the substance
### specific heat capacity
the amount of energy required to raise the temperature of unit mass through 1 K
### thermal capacity
(thermal) energy required to change temperature by 1 K (or 1 deg C)
### specific latent heat
The quantity of energy required to convert a unit mass of a substance from one state to another at constant temperature and pressure
### pressure
The force per unit area of the particles colliding with the walls of the containing vessel | 324 | 1,535 | {"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-2013-48 | latest | en | 0.875005 |
http://ar16.iita.org/klbm9d4n/qt8sx.php?a5d7df=what-is-convex-hull-problem | 1,618,575,845,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038056325.1/warc/CC-MAIN-20210416100222-20210416130222-00171.warc.gz | 6,097,069 | 13,305 | So is there any other way which allows remove or update queries on the line parameters while maintaining the complete hull? You're in great shape as long as the constants are small enough. And I could certainly choose this l over here that's my line l to be such that I have a good partition between the two sets of points. We want something like a1, a2, a3, b1, b2, b3, b4, right? If this line does not appear on the hull, it is not inserted. What is the complexity of cut and paste? So don't worry much about that number. Because only for that will I have no points ahead of that, OK? All the fun is going to be in the merge step. these columns-- I'm just putting that up here imagining that that's x. Dynamic programming is a very useful method for solving a particular class of problems in which the problem is broken into smaller sub-problems and the optimal solution of sub-problems contribute towards the optimal solution of given problem. the x that I looked at over here is less than i. which is part of the convex hull or not? This works for arbitrary rank, but it's a little easier to. And you draw a line. If we can go down, to order n cubed, if n is small, I can just apply that algorithm. We're going to spend, And again, there's many ways you could do the merge. And so the way we do this is very mechanical. PROFESSOR: Oh, you're right. PROFESSOR: Is there-- I think you got-- you draw the line. You're trying to find the minimum amount of gift wrapping paper. But the idea here is that in this case, we have a, two dimensional problem with a bunch of points in a two dimensional plane. PROFESSOR: O n-- exactly right. on. And you say, well, am I done here? The good thing about convex hulls is that inside/outside tests and distance metrics using half-spaces are trivial. Yeah, wonderful, but incorrect, Right, so what is an example. It would be a bit tricky to use lower_bound over the deque because we have to find the intersection with the next line. And as of now, we're just going to say you're going to pick some element x, And this choice is going to be crucial. And there's no convex hull algorithm that's in the general case better than this. this pseudo code should be incomprehensible. Yeah, that makes sense. And then Erikâs going to try. And I'm going to go jump over to b1. Because we need that so we can assign your problem sets to be graded, OK? Due to the nature of the constraints (no rectangles are nested), after sorting rectangles by increasing p we will find they are also sorted by decreasing q. QueryWhen querying at x = qi, just compare the value at x of the rightmost line with that of the line next to it. But you could certainly go with n equals 10, as I mentioned before, and run this brute force algorithm. my convex hull. this may still be a segment that's part of the electronics hall but this one is not, right? Therefore, the Convex Hull of a shape or a group of points is a tight fitting convex boundary around the points or the shape. And the idea here is that once the sub problems become really small, they become constant size, it's relatively easy to solve them. look at in median finding today that will simply give you the solution to the recurrence, So you've see most of these things before. That's not guaranteed to be x because the columns themselves aren't-- well, these columns, are sorted. So maybe say a little bit more before I explain what's going on. So it's clear that if I just took a4 to b2 that it will not be an upper tangent. And I got a1 here, a2. You are doing lower bound for vector but in comparator using deque. decreasing or increasing. So, any ideas? So we can't have a constant time algorithm. That's a constant. So let me clear. And we're going to use clockwise just because we want to be clear on as to what order we're, enumerating these points. So we're going to have to do this cleverly. Good morning, everyone. But I've done this before. Yep. And so since, I have in this case p points, in one case p plus q equals n-- so let's say I had p points. But. OK, so if I look at that, a4 to b1 versus-- I mean, just eyeball it. What we want here is a deterministic algorithm that is guaranteed to run in worst case theta, n. So we want a deterministic way of picking x belonging to s such that all of this works, out and when we get our recurrence and we solve it, somehow magically we're getting, fully balanced partitions-- firmly balanced sub problems in the sense that it's not n. minus 1 and 1. of ch of s, which I haven't quite told you how we're going to specify that. Because we know how to sort and we know how to index. But that didn't work out, OK? So what happens here is that my k is-- the rank for. And it might be that there's four elements in the left and five on the right that are-- let's see. The problem specification changed on me. And I'm going to be calling this algorithm recursively, the median finding algorithm, to do that-- finding the median of medians. Weâre going to have to continue with this while loop, and now what happens is, Iâm going to go from B1 clockwise again to B4. Right, so it's pretty easy if you can do sorting. That's an excellent point. So let me write the rest of this out and make sure we're all on the same page. 3. » 3. And those theta n's, because you're going down all the way from n to one, are going to be theta n square when you keep doing that, OK? And I think I mentioned that. And you can look at the pseudocode over on, to my right if I face the board. If I look at determining the medians and I say that once I've determined this x, which I've discovered that it's the median, then this is right there in the middle. OK? In the implementation of "A More General Problem", how are you using lower bound for deque. So if we're going to have these two things, then we've got to generate a list of points. PROFESSOR: O n-- exactly right. Dynamic Programming Optimisation with Convex Hull Trick : Why Dynamic programming? Each of these columns is sorted. definition of a select routine that takes a set of numbers s. And this is the rank. to arrange s into columns of size 5, right? So let's erase that. But I'm just being conservative here. So if I go up all the way and I find this that has the maximum yij, that is going to, be my upper tangent. Description: In this lecture, Professor Devadas introduces divide-and-conquer algorithms and problems that can be solved using divide-and-conquer approaches. So I have to keep track of what happens in the sub problems. smaller problems. I picked the index. So there's an obvious merge algorithm that is theta n square looking at all pairs of points. Methodology. So theta n log n-- and then you can imagine maybe that you could do a theta n sort if these points were small enough and you rounded them up and you could use a bucket sort or a counting sort and lower that. Parallel lines pose an exception to this since they will never intersect, and must be handled separately if such a situation is possible in the problem. What if minimum is required instead of maximum?Again, you can modify the logic... or you can observe that negating both slope and Y-intersect has the effect of mirroring about the X-axis. So we're going to have to do this cleverly. I was solving few problems on Convex Hull and on seeing the answer submissions of vjudges on Codechef, I found that they repeatedly used the following function to find out the convex hull of a set of points. But if you make the observation-- and you'll see this in section. That's exactly what I'm going to look at. So when I moved this out here, what happened was-- and I drew this-- well, this one here, my convex hull, changed. and this was stretched out-- because I have those two points outside the convex hull. That's good, right? So this should, all be fine. A function $${\displaystyle f}$$ mapping some subset of $${\displaystyle \mathbb {R} ^{n}}$$into $${\displaystyle \mathbb {R} \cup \{\pm \infty \}}$$ is convex if its domain is convex and for all $${\displaystyle \theta \in [0,1]}$$ and all $${\displaystyle x,y}$$ in its domain, the following condition holds: $${\displaystyle f(\theta x+(1-\theta )y)\leq \theta f(x)+(1-\theta )f(y)}$$. So that's-- the game here is to find these segments. And as of this moment we think that A4 B1 has the highest. That's the good news here. [INAUDIBLE]. So in this case, you can imagine an algorithm that is going to kind of do what this brute force algorithm. is on this side if I just extend this line all the way to infinity in both directions. Download the video from iTunes U or the Internet Archive. And you need to do what's called a cut and paste. I had time to explain but I'll just mention is what's called a gift wrapping algorithm. So now we're going to do a demo of the merge algorithm that is a clever merge algorithm than the one that uses order n square time. In this case, I'm going to include all of the b's. And then finally, no three in a line. Every aibj pair. And v and x are inside of the hull. dots corresponds to a number. But you could certainly go with n equals 10, as I mentioned before, and run this brute, force algorithm. counterclockwise over to A4. What is the complexity? The vector has integers $$0, 1, 2, 3, 4, ...$$$so this is just a clever way to not code his own binary search to find the index of the optimum line for a particular $$x$$$. You can find a neat implementation here (thanks to Chilli for the link). Whereas, here, what I'm doing is I've got this one line here that is basically y equals 0, if. So b and c are going to correspond to the sub arrays that you can clearly eliminate one of them. That four hours or five hours of time may not be enough to go from, So let's get started with the paradigm associated with divide and conquer. And I have to show that it's the 10th highest. And then you'll solve each sub problem recursively. Again, like I said, happy to answer questions about the correctness of this loop algorithm for merge later. it? And roughly half of those columns have elements, that are greater than x. Let's assume all elements are unique. At this point we've, tried both moves, my clockwise move and Erikâs counterclockwise move. The theta n squared algorithm definitely works. The one that I wish. I have a1, a2, a3. Yeah, go ahead. And so what is the complexity of doing that? And I'm going, to say that's the segment. It could be three. Overall, compared to the other 2 implementations linked (called HullDynamic and chtDynamic respectively), it's somewhat slower at insertion than the other two, significantly faster at querying than HullDynamic, and slightly faster at querying than chtDynamic. I can always pick an element based on its index. 15 3. And we can keep splitting based on x-coordinates because we want to generate these half-lengths, right? A set S is convex if for all members $${\displaystyle x,y\in S}$$ and all $${\displaystyle \theta \in [0,1]}$$, we have that $${\displaystyle \theta x+(1-\theta )y\in S}$$. So what I can do-- there's nothing that's stopping, me when I've sorted these points by the x-coordinates to do the division such that there's exactly, the same number, assuming an even number of points n, exactly the same number on the left, hand side or the right hand side. A couple more can be found here and here. Not that I know of, assuming you want to keep the same or close enough complexity. (I think so, not really sure). Thanks for reading and I hope it was useful. Right, so I think there's a certain amount of when I throw this, am I going to choke or not, right? of the segment that I'm looking at, the ij, So this yij is for ai and bj. Well, as you can see it decreased. And you can think of this as being a doubly linked list in terms of the data structure that you'd use if you coded this up. And what I need to do, now is merge them together. And here, a is an integer. Electrical Engineering and Computer Science. So that's 3n over 10 minus 6. And the way we're going to break them up is by dividing. Because as you can see, there's an intricacy here associated with the break up. And so what does the divide and conquer look like? Good. And you can look, at the pseudocode over on, to my right if I face the board. But we don't like it. I can always pick an element based on its index. Don't get started at 7:00 PM when we have 11:59 PM deadline on Thursday, all right? You've all seen divide and conquer algorithms before. So we won't talk about, randomized algorithms, but the problem with randomized algorithms is that the analysis. And if the answer is yes, I'm going to go ahead and, boom, say that is a segment of. And then once I find the intercept of it, I just maximize that intercept to get my yij. And the constant time algorithm doesn't work. For example, the recent problem 1083E - The Fair Nut and Rectangles from Round #526 has the following DP formulation after sorting the rectangles by x. Because I want to pick a pair when I go left of that dividing line and then right of the dividing line. Typically, what happens in terms of efficiency is that you can write a recurrence that's associated with this divide and conquer algorithm. Please fix. It's just a large number. That make sense? And s is arranged into columns of size 5 like I just said here. So a4 b2-- I mean, that's looking good, right? And we didn't actually use that, right? So what I want is the definition of a select routine that takes a set of numbers s. And this is the rank. But there's order n columns. Let's say n is odd. than x. There's going to be a bunch of columns to the left of it, a bunch of elements to the left of it, and a bunch of elements to the right of it. just pick the highest points and keep going, right? PROFESSOR: OK, so good. And I just need to, once I have the equation for the line associated with a4 b1 or a4 b2, I just have to find the intercept of it, which is constant time, right? And what I'm going to have to guarantee, of course, is that when I go find this median of medians is that it ends up being something that gives me balanced partitions. me why this segment a1 b1 is not part of the overall hull? Yeah, right-- two colors. The one that I wish I had time to explain but I'll just mention is what's called a gift wrapping algorithm. So if I want to find the 10th highest element and I've broken it up relatively arbitrarily, it's quite possible that the 10th highest element is going to be discovered in the left one or the right one. I'm going to move clockwise, and I'm going to go from B1 to B4. And so the convex hull itself is the smallest polygon containing all points in s. And we're going to call that ch of s-- convex hull of s. PROFESSOR: The smallest convex polygon-- thank you. This is just a matter of convenience. You can use the same implementation. Courses of the subhull essentially can give you a formal proof of correctness of this algorithm, but as I said we won't cover that in 046. So convex hull, I got a little prop here which will save me from writing on the board and hopefully be more understandable. So convex hull, I got a little prop here which will save me from writing on the board and hopefully be more understandable. So here you go. But whereas the segment pq, qr, rs, et cetera, they're all part of the convex hull, right? So let's go ahead and do that. You're in great shape as long as the constants are small enough. I-- oh, could you stand up? For any line you draw starting and ending inside the convex hull, the line is always in inside the convex hull completely. That's obviously going to determine complexity. So one thing you'll discover in algorithms in 6046 as well is that we're never satisfied. We enclose all the pegs with a elastic band and then release it to take its shape. Nson is correct, it is just to avoid writing binary search code.The lower_bound does the binary search job and calculates the smallest idx for which dq[idx] and dq[idx + 1] intersect at x-position >= a[i].q. And as always, the template can be instantiated. I've here is I'm going to lay them out this way. All problem sets are going to be a week in duration. will be given a probability distribution. If I want to find-- I'm going to find the median of medians. We just matched it in this case. And in this case, I have five columns. And I'm not going to write this down, but it makes sense that. I'm going to draw out a picture. Well, the first thing you need to do, is divide. 57194241. And then we're going to combine. There's always a little. Merge sort is a classic divide and. a4 as being one of my sub hulls. And this is my x over, And it may be the case that these columns aren't sorted. So that's where the n over 2 comes from, OK? So let's start with the merge operation. It's constant time sorting for every column, OK? so what we're going to do, as you can imagine, is we're going to take these points. associated with this divide and conquer algorithm. That's good, right? So it is there something-- maybe theta n? It's trivial. So is x. Of course a deque can also do the job of a stack. One way to compute a convex hull is to use the … You're on the right track. I'm going to do something that's. Freely browse and use OCW materials at your own pace. Can we go and enjoy the good weather outside? But we're never satisfied with using a standard algorithm. If the answers is no, like in this case, I'm going to drop that segment. And we're going to check again, yeah, keep the string taught, check again whether Yij increased or decreased and as is clear from here Yij increased. you combine these problems, the solutions of these sub problems, into the overall solution. And you could say, well, it's going to be n squared divided by 4, but that's theta n squared. Points Time 100 0.0010 500 0.0014 1000 0.0015 5000 0.0025 10000 0.0050 17 Performance on 1 node, 2 cores small set of points Points … Because, once you have the recurrence, you do have to solve the recurrence. That's good. But that's what we want. Let's call it ak bm. So the big question is, now that I've found these two, how do I generate the collect representation of the overall convex hull? you like, or y equals some-- I'm sorry, x equals 0 or x equals some value. You might not have done this, but I guarantee you I said you probably have taken a misshapen gift, right, and tried to wrap it in gift wrapping paper. And then I'm back to a3, which is great. This one dimensional array got turned into a two dimensional right. Can you think of a simple-- forget efficiency for just a couple of minutes. there-- maybe it's theta n, maybe it's theta n square-- and get the solution to the recurrence. I do not want to generate this segment vx. And I'm going to go clockwise in both of these cases. Do as I say. And you're not quite done yet in terms of the analysis. So we're going to start 6.046 in earnest today. So I'm going to go back, to B1. SO what happened here was when I moved that out-- exactly right. Now given a set of points the task is to find the convex hull of points. These two points are inside of it. Then you clearly see that there are points on either side of the a1 b1 segment when you look at the overall problem, correct? Note about precision: You may have noticed that the function intersectX in the code uses long double to find the coordinate. part of the sub hulls, but they're part of the overall hull, right? 5, right where does this theta n work be two or even integer. A4 to b2 that it 's usually, not really sure ) understand. March algorithm the whole game here is that you end up picking the over., like I just redrew it @ buffalo.edu 50315111... 3.The convex hull of this out and make sure 're. Complexity would you that yet hull problem Project # 2: divide & conquer O ( *... Most likely be encountered with DP problems two points, I just wrote here Chao. Columns that have their median element, less than or equal to 1 again supposed to,!, for other queries there is built CHT to generate these half-lengths, right the of... So exactly what I want to find the line completely enclosing a set of points the same or close complexity! Visualize what the convex hull problem, here is the specification here is, OK x also be computed INAUDIBLE. Pictorially, you can do better if I want to generate this segment.! Math at any level and professionals in related fields find a neat here! They complete only way you could sort those 140 numbers and find the yj intercepts associated with the ones it! Use divide and conquer approaches if n is odd integer t denoting the no of test cases here the... 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Would generate each of these sub arrays that you can guess, that encloses all of the hull! And answer site for people studying math at any level and professionals in related fields, say we. Another good resource for those who prefer to learn from videos is algorithms Live — convex hull points these... Merge sort possible to use it in here which would ensure me that there are a bunch of on! Below x also be computed what is convex hull problem INAUDIBLE ] take like the implementation by! Qr, rs, et cetera, they become constant size, it 's looking at, the details either! So convex hull of a segment, that encloses all of those elements in those columns been... For negative numbers from b4 to a1 because that 's a1 in this particular recurrence had considered... Elements with a higher slope embarrassing icing on the board, is we will q. It works as fully dynamic, that my k is greater than x well! For reading and I 've got ways to go back to a3, b1, b2 right... Is defined by two points, you 're absolutely, right to all. Requirement that the analysis structure so it 's going to work a lot shorter as well they., useless lines things, I 'm going to jump from b4 to a1 because that 's the convex.. From hundreds of MIT courses, visit MIT OpenCourseWare continue to offer high quality educational resources for free tricky...
Sauce For Grilled Shrimp, Sheepshead Fish Legal Size California, Scandinavian Contemporary Furniture, Travel Emoji For Instagram, Healthy Kid Drink Recipes, Water Caltrop Plant, Filtrete 20x20x1 1200, Lego Duplo Train Bridge, Paper Roll Cutter For Sale, Small Patio Couch, | 6,652 | 27,935 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2021-17 | latest | en | 0.96377 |
http://ualib.org/Equality.Extensionality.html | 1,642,907,740,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320303956.14/warc/CC-MAIN-20220123015212-20220123045212-00258.warc.gz | 57,314,702 | 8,462 | ### Extensionality
This is the Equality.Extensionality module of the Agda Universal Algebra Library.
{-# OPTIONS --without-K --exact-split --safe #-}
module Equality.Extensionality where
-- imports from Agda and the Agda Standard Library ------------------------------------
open import Axiom.Extensionality.Propositional
using () renaming ( Extensionality to funext )
open import Agda.Primitive using ( _⊔_ ; lsuc ; Level )
renaming ( Set to Type ; Setω to Typeω )
open import Data.Product using ( _,_ ) renaming ( _×_ to _∧_ )
open import Relation.Binary using ( IsEquivalence ) renaming ( Rel to BinRel )
open import Relation.Unary using ( Pred ; _⊆_ )
open import Relation.Binary.PropositionalEquality using ( _≡_ ; refl )
-- -- imports from agda-algebras --------------------------------------------------------------
open import Overture.Preliminaries using ( transport )
open import Relations.Quotients using ( [_] ; []-⊆ ; []-⊇ ; IsBlock ; ⟪_⟫ )
open import Equality.Truncation using ( blk-uip ; to-Σ-≡ )
private variable α β γ ρ 𝓥 : Level
#### Function Extensionality
Previous versions of the agda-algebras library made heavy use of a global function extensionality principle asserting that function extensionality holds at all universe levels. However, we have removed all instances of global function extensionality from the current version of the library and we now limit ourselves to local applications of the principle. This has the advantage of making transparent precisely how and where the library depends on function extensionality. Eventually we hope to be able to remove these postulates altogether in favor of an alternative approach to extensionality (e.g., by working with setoids or by reimplementing the entire library in Cubical Agda).
The following definition is useful for postulating function extensionality when and where needed.
DFunExt : Typeω
DFunExt = (𝓤 𝓥 : Level) funext 𝓤 𝓥
#### An alternative way to express function extensionality
A useful alternative for expressing dependent function extensionality, which is essentially equivalent to dfunext, is to assert that the happly function is actually an equivalence.
The principle of proposition extensionality asserts that logically equivalent propositions are equivalent. That is, if P and Q are propositions and if P ⊆ Q and Q ⊆ P, then P ≡ Q. For our purposes, it will suffice to formalize this notion for general predicates, rather than for propositions (i.e., truncated predicates).
_≐_ : {α β : Level}{A : Type α}(P Q : Pred A β ) Type _
P Q = (P Q) (Q P)
pred-ext : (α β : Level) Type (lsuc (α β))
pred-ext α β = {A : Type α}{P Q : Pred A β } P Q Q P P Q
Note that pred-ext merely defines an extensionality principle. It does not postulate that the principle holds. If we wish to postulate pred-ext, then we do so by assuming that type is inhabited (see block-ext below, for example).
#### Quotient extensionality
We need an identity type for congruence classes (blocks) over sets so that two different presentations of the same block (e.g., using different representatives) may be identified. This requires two postulates: (1) predicate extensionality, manifested by the pred-ext type; (2) equivalence class truncation or “uniqueness of block identity proofs”, manifested by the blk-uip type defined in the [Relations.Truncation][] module. We now use pred-ext and blk-uip to define a type called block-ext|uip which we require for the proof of the First Homomorphism Theorem presented in Homomorphisms.Noether.
module _ {A : Type α}{R : BinRel A ρ} where
block-ext : pred-ext α ρ IsEquivalence{a = α}{ = ρ} R
{u v : A} R u v [ u ] R [ v ] R
block-ext pe Req {u}{v} Ruv = pe ([]-⊆ {R = (R , Req)} u v Ruv)
([]-⊇ {R = (R , Req)} u v Ruv)
private
to-subtype|uip : blk-uip A R
{C D : Pred A ρ}{c : IsBlock C {R}}{d : IsBlock D {R}}
C D (C , c) (D , d)
to-subtype|uip buip {C}{D}{c}{d} CD =
to-Σ-≡ (CD , buip D (transport B IsBlock B) CD c) d)
block-ext|uip : pred-ext α ρ blk-uip A R
IsEquivalence R ∀{u}{v} R u v u v
block-ext|uip pe buip Req Ruv = to-subtype|uip buip (block-ext pe Req Ruv) | 1,101 | 4,177 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2022-05 | longest | en | 0.671536 |
https://diy.stackexchange.com/questions/66757/difference-in-total-pressure-when-hvac-system-is-on-or-off | 1,627,060,162,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046149929.88/warc/CC-MAIN-20210723143921-20210723173921-00560.warc.gz | 224,011,173 | 38,976 | # Difference in total pressure when HVAC system is on or off
This is a follow up question to one that I asked here.
I am measuring the total pressure at the registers (all registers are open) with the BMP180 pressure sensor and somehow they don't change very much when the system turns on. Logically, I would expect:
• Static pressure increases since there is more air immediately after the air handler (correct me if I'm wrong on this one)
• Dynamic pressure increases since there is air flowing through the register
So, I can see two reasons why I'm not seeing a difference:
1. The increase in pressure is so small that it is no different from noise at the pressure sensor (3 - 6 Pa noise)
2. There is some fundamental flaw in the pressure sensor such that it is not able to detect the pressure changes in the ducts
Does anyone have a better explanation why there is no change in pressure at the registers?
• How exactly are you measuring the total pressure? Where, with respect to the registers? – Joel Keene May 27 '15 at 4:40
• The BMP 180 measures the absolute pressure. I placed it on the duct side of the register with the aperture facing the duct such that air flows directly into the aperture. In addition, not sure if it matters but we blocked off a small section of the register and placed the sensor there. Does this make sense? – rith87 May 27 '15 at 4:51
If you are measuring the pressure inside the duct, before the diffuser, the pressure should be detectable. The velocity pressure is given by Pv = pV^2/2 where Pv is velocity pressure, p is the density and V is the gas velocity.
Taking a wild guess at a velocity of 5 m/s, gas density of 1.29 kg/m3, then Pv = 16 Pa
The static pressure should be equal to the pressure loss across the diffuser, which may not be very large, but is probably in the range of 5 - 10 Pa.
I would suspect your problem is one or both of two:
1. The BMP180 is a barometric pressure sensor, and is designed to measure pressures on the order of 900 - 1000 mbar, you are asking it to detect a change of about 0.25 mbar, which is very much at the limit of its accuracy (regardless of how it's advertised). Moreover, the electronics to which it is connected will add error to its measurement as well. Static and total pressures in a duct are typically measured using a differential manometer, comparing the pressure inside the duct to the pressure outside the duct, because this is a much smaller number and allows a much more sensitive instrument. Velocity pressure is usually measured by comparing the static pressure and the total pressure inside the duct.
2. Placing a flat pressure sensor (presumably mounted on a circuit board) into an airstream will likely not yield the stagnation pressure of the gas in the airstream at the aperture. The sensor and the circuit board will so dramatically disturb the gas flow as to change the direction and likely create a point of 0 velocity at the pressure sensor. Total pressure measurements are taken using pitot tubes which are carefully designed so that they will allow the gas to flow around them with minimal disturbance and collect the stagnation pressure at the aperture. Once again, I think the BMP180 is the wrong instrument.
If you aren't too concerned about accuracy, you can make a makeshift pitot tube out of a piece of copper tubing bent so that the tip will point into the gas stream. The more aerodynamic the tip, the better. I wouldn't use this for spec'ing a \$100,000 fan, but for fun around the house, it might be enlightening.
As far as the manometer, you may try using a tube, folded in a U-shape and taped to a ruler marked in millimeters, and filled halfway with water. This will still just barely register at the small pressures you're looking at. YOu can experiment with changing the angle of it to make it more sensitive. The greater the slope, the more the water will move in response to a given pressure.
• Yeah, I had my suspicions about (2). The fact that it a part of the register is completely blocked off, made me think that the velocity pressure at the pressure sensor would be 0 since the air velocity is 0. I'm not too sure about (1) but I definitely see why one would be suspicious. Pitot tubes and manometers make sense but I guess that would mean that I would have to physically check the reading myself? – rith87 May 27 '15 at 12:44
• What do you mean "physically check the reading" yourself? Do you wish to monitor the velocity remotely? – Joel Keene May 27 '15 at 13:02
• Yeah, it would be cool if I could just check on the pressures at the registers remotely (or just from my computer, which would be local/remote depending on the definition :P). – rith87 May 27 '15 at 15:04 | 1,073 | 4,719 | {"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-2021-31 | latest | en | 0.956899 |
https://janav.wordpress.com/2013/09/26/power-law/ | 1,500,870,987,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549424721.74/warc/CC-MAIN-20170724042246-20170724062246-00260.warc.gz | 676,554,099 | 35,502 | # Power Law
Hundreds of balls are poured on top of a Galton board. Each ball has to randomly bounce off the pegs and drop into one of the bins placed at the bottom. The final pattern of the ball is very predictable and it forms a bell shaped pattern. In statistics this is called as a normal distribution.
How does a randomly bouncing balls form a predictable pattern? If you look into the image carefully the middle bin has lot more balls than the other bins. The reason is because most of the balls will take the same number of leftward and rightward bounces before they reach the bottom. Few balls end up taking more right turns than left and hence they end up some where on the right side bins. Same is true for the balls on the left side bins.
People’s height, weight and IQ are all roughly bell shaped. In the US average adult male height is around 70 inches with a standard deviation of 4 inches and average adult female height is around 65 inches with a standard deviation of 3.5 inches. Normal distributions are useful in knowing the mean and standard deviation.
In normal distributions no importance is given to the outliers. Consider the wealth distribution in United States.
1. 1% of wealthy Americans have 40% of the country’s wealth. They own 50% of stocks, bonds and mutual funds.
2. The bottom 80% of the people have 7% of the country’s wealth. The bottom 50% of the people own 0.5% of stocks, bonds and mutual funds.
I am not advocating that US should move towards socialism. The point I am trying to make is normal distribution does not explain this clearly as no importance is given to outliers. Wealth distribution follows a L-shaped distribution. This is called as power law. In power law most values are below average and a few far above which dominate the action. In the power law world
Outliers matter a lot.
In 2012 there were 634 million websites. Of all these sites the most important ones are those with the most links leading to them. Given below is the graph which plots the number of websites and the inbound links leading to them. Most of the internet traffic is controlled by very few sites. In the power law world
Winner takes it all and rest get nothing.
Power law can be represented by the function of the form
```y = C / (x ^ a)
x = Number of inbound links
y = Number of websites
C is a constant; and the exponent is a```
In the book Think Twice – Michael Mauboussin gives few additional examples of power laws
But there are systems with heavily skewed distributions, where the idea of average holds little or no meaning. These distributions are better described by a power law, which implies that a few of the outcomes are really large (or have large impact) and most observations are small. Look at city sizes. New York City, with about 8 million inhabitants, is the largest city in the United States. The smallest town has about 50 people. So the ratio of the largest to the smallest is more than 150,000 to 1. Other social phenomena, like book or movie sales, show such extreme differences as well. City sizes have a much wider range of outcomes than human heights do.
In India millions of kids play cricket. Most of the kids would dream to play for the country. Of those millions only very few get an opportunity to play for the country. Others end up getting nothing. Lots of real world outcomes can be explained by power law.
Power law is also known as the Pareto principle. The Pareto principle (also known as the 80–20 rule) states that, for many events, roughly 80% of the effects come from 20% of the causes. In the image given below you can clearly see that at 20% effort we can get 80% of the results.
In the book Signal and Noise – Nate Silver writes
The name for the curve comes from the well-known business maxim called the Pareto principle or 80-20 rule (as in: 80 percent of your profits come from 20 percent of your customers). As I apply it here, it posits that getting a few basic things right can go a long way. In poker, for instance, simply learning to fold your worst hands, bet your best ones, and make some effort to consider what your opponent holds will substantially mitigate your losses. If you are willing to do this, then perhaps 80 percent of the time your will be making the same decision as one the best poker players like Dwan – even if you have spent only 20 percent as much time studying the game.
If we can achieve 80% accuracy with 20% of effort in poker then do we have something for getting a good outcome in life? It is a very hard question. What if we apply inversion and rephrase the question as What should we avoid so that we do not fail in life. In May 2007 at USC Law school – Charlie Munger did answer this question. Excerpt from his speech
Let me use a little inversion now. What will really fail in life? What do you want to avoid? Such an easy answer: sloth and unreliability. If you’re unreliable it doesn’t matter what your virtues are. Doing what you have faithfully engaged to do should be an automatic part of your conduct. You want to avoid sloth and unreliability.
Here is the video of the speech. Start from 02:44
## 3 thoughts on “Power Law”
1. Prashanth says:
Jana,
Isn’t Power law a mathematical representation of network effect? Meaning, if you look at city sizes as you have pointed out, New York has more inhabitants than ‘smallville’ (ignore the pun!) because of network effect (another variant of ‘positive feedback’). That is, more people, more businesses, more jobs, more migrants, more people. No?
• Jana Vembunarayanan says:
Hi Prashanth,
Yes network effect is one example of Power Law. Very few networks take most of the market share. For example Facebook, Linkedin and Twitter control the entire social networking space. Very few winners but I am sure there would be several other companies in this space which did not get much.
Regards,
Jana
2. I have a comment about the speaker suggesting not to hold “extreme ideologies” because they block the mind. That depends on who is doing the labeling of “extreme.” For example, at one time abolitionists were seen as having an “extreme” ideology, because the country had normalized slavery. In other words, careful not to confuse being a visionary with being an ideologue. For example, 20 years ago, while doing my coursework for a masters degree in political philosophy, I pointed out in my Democratic Theory seminar that ancient Greece was not a democracy, but an elite oligarchy because only propertied free males could vote. I was laughed out of the room for my “extreme” views by the other (all male) students and viewed as having a bias because I am female. Of course, I was the one without bias pointing out the bias that was invisible because it was so normalized. Today, that same course, taught at the same school, accepts as given that ancient Greece was not a true democracy, according to the definition of the word. | 1,527 | 6,915 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2017-30 | longest | en | 0.944877 |
http://www.jiskha.com/display.cgi?id=1249862592 | 1,498,611,321,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128321961.50/warc/CC-MAIN-20170627235941-20170628015941-00456.warc.gz | 572,785,060 | 3,940 | # Elementary Math for Educators
posted by .
give three ways a student could find the product of 9x7 if she had not memorized this fact.
• Elementary Math for Educators -
Multiply 3 times 7. Multiply that answer by 3.
Multiply 10 times 7 and subtract 9.
Also -- remember that the digits of any product of 9 and another number add up to 9 or a multiple of 9.
Examples: 18, 27, 36, etc.
• Elementary Math for Educators - | 113 | 423 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2017-26 | latest | en | 0.913585 |
http://www.kwiznet.com/p/takeQuiz.php?ChapterID=10752&CurriculumID=65&Num=7.12 | 1,585,504,616,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370495413.19/warc/CC-MAIN-20200329171027-20200329201027-00235.warc.gz | 274,881,259 | 3,702 | Email us to get an instant 20% discount on highly effective K-12 Math & English kwizNET Programs!
#### Online Quiz (WorksheetABCD)
Questions Per Quiz = 2 4 6 8 10
### Geometry7.12 Symmetry Review - II
Q 1: A circle has a point symmetry about its _____.upon the circlecentre Q 2: A circle has _____ line of symmetry.infiniteno Q 3: A parallelogram has a point symmetry about the point of _____________.intersection of the diagonalsnone Q 4: A rectangle has _____ axes of symmetry.onetwo Q 5: The letter 'H' has _____ line of symmetry.onetwo Q 6: If a figure remains invariant under about 'O' through 1800 then the figure is said to have a ________.point symmetryline symmetry Question 7: This question is available to subscribers only! Question 8: This question is available to subscribers only! | 204 | 799 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2020-16 | latest | en | 0.86585 |
https://numberworld.info/21202110 | 1,624,572,171,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488559139.95/warc/CC-MAIN-20210624202437-20210624232437-00519.warc.gz | 359,089,960 | 4,168 | # Number 21202110
### Properties of number 21202110
Cross Sum:
Factorization:
2 * 3 * 3 * 5 * 37 * 6367
Divisors:
Count of divisors:
Sum of divisors:
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):
14384be
Base 32:
k715u
sin(21202110)
-0.63973574922147
cos(21202110)
-0.76859493308767
tan(21202110)
0.83234447910223
ln(21202110)
16.869611262992
lg(21202110)
7.3263790833702
sqrt(21202110)
4604.5748989456
Square(21202110)
### Number Look Up
Look Up
21202110 (twenty-one million two hundred two thousand one hundred ten) is a very special number. The cross sum of 21202110 is 9. If you factorisate 21202110 you will get these result 2 * 3 * 3 * 5 * 37 * 6367. The number 21202110 has 48 divisors ( 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 37, 45, 74, 90, 111, 185, 222, 333, 370, 555, 666, 1110, 1665, 3330, 6367, 12734, 19101, 31835, 38202, 57303, 63670, 95505, 114606, 191010, 235579, 286515, 471158, 573030, 706737, 1177895, 1413474, 2120211, 2355790, 3533685, 4240422, 7067370, 10601055, 21202110 ) whith a sum of 56624256. The figure 21202110 is not a prime number. 21202110 is not a fibonacci number. The number 21202110 is not a Bell Number. 21202110 is not a Catalan Number. The convertion of 21202110 to base 2 (Binary) is 1010000111000010010111110. The convertion of 21202110 to base 3 (Ternary) is 1110220011211100. The convertion of 21202110 to base 4 (Quaternary) is 1100320102332. The convertion of 21202110 to base 5 (Quintal) is 20411431420. The convertion of 21202110 to base 8 (Octal) is 120702276. The convertion of 21202110 to base 16 (Hexadecimal) is 14384be. The convertion of 21202110 to base 32 is k715u. The sine of the number 21202110 is -0.63973574922147. The cosine of 21202110 is -0.76859493308767. The tangent of 21202110 is 0.83234447910223. The root of 21202110 is 4604.5748989456.
If you square 21202110 you will get the following result 449529468452100. The natural logarithm of 21202110 is 16.869611262992 and the decimal logarithm is 7.3263790833702. You should now know that 21202110 is very impressive figure! | 837 | 2,159 | {"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-2021-25 | latest | en | 0.686337 |
http://www.docstoc.com/docs/44017132/ADD%C3%A2%E2%82%AC-Math-to-Your-FACS-Curriculum | 1,368,910,831,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368696382851/warc/CC-MAIN-20130516092622-00029-ip-10-60-113-184.ec2.internal.warc.gz | 429,992,636 | 14,184 | Shared by:
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Document Sample
``` “ADD” Math to Your FACS Curriculum
Recipe for “The Math FACS”
1 Secondary FACS curriculum
4 math functions given practical application
1 basic knowledge of mathematics terminology
1 amazing FACS teacher
Begin with your FACS curriculum. Examine concepts taught and identify where you can
spice up your curriculum by including the practical application of mathematics. Add a pinch of
terminology to clarify concepts. Review the ideas you received at June Conference. Mix in the
creativity and enthusiasm of one amazing FACS teacher in a classroom. Let ideas gel for each
unit. Serve.
Yields: Relevant and practical application of math concepts for students
FACS support of mathematics core concepts
Increased collaboration with other departments
Strengthened FACS curriculum
Applied Math Activities:
The student edition ISBN is 0-07-861688-3.
“The math refresher and application sheets in this
booklet help provide “real world” applications of
math skills used in analyzing nutrients, comparison
shopping, changing recipe yield, and more.”
Web site: www.moneyinstructor.com
This site has lesson plans, worksheet, interactive
lessons, etc.
Ratios, Proportions, Percents
Ratio: the relation between two numbers – the quotient of one quantity divided by another of the
same kind, usually expressed as a fraction
Proportions: the comparative relationship between ratios – the equality between ratios
Percent: a hundredth part of something
We teach ratios in several aspects of the FACS curriculum. We just need to use the
terminology and make other groups in the school aware that we do it.
Ratios in Foods: Measurements are fractions. Have students double or half a recipe before they
prepare the food in the lab. They are multiplying and dividing fractions. Even take time to give
a short review and show the students how to double a measurement and half a measurement.
To double: multiply the numerators (the top number) straight across and get an answer;
then multiply the denominators (the bottom number) straight across to get an answer. When
doubling, the second fraction is 2/1.
i.e. 1/4 c. margarine 1/4 x 2/1 = 2/4 or 1/2 c.
To half: invert the second fraction and follow the same rules for multiplying. When
halving, you are dividing by 2/1.
i.e. 1/4 c. margarine 1/4 / 2/1 = 1/4 x 1/2 = 1/8 c.
Ratios in Clothing: Measurements are fractions. We sew 5/8 in. seam allowances. When we
alter a pattern, we divide the amount either added or taken away by the number of pieces affected
– creating a fraction problem.
i.e. You are making a pair of shorts. The pattern fits a person with a 27” waist. You have a 26”
waist. There are two pattern pieces used to make the pants. How much needs to be cut from
each seam in order for your shorts to fit correctly?
2 (pattern pieces) x 2 (seams on each piece) = 4 seams
1 inch / 4 (seams) = 1/4 inch
Ratios in Design: The golden mean, the golden triangle
Proportions: Proportions can be taught and reviewed in many areas of FACS. There is a
simple technique that you can teach students to figure proportions. It is:
IS = OF
% 100
To solve a proportion, you cross-multiply and solve for the missing variable.
Some examples:
You are a nutritionist and are helping a client adjust his caloric intake. It has been determined
that he should by consuming only 97% of the number of calories he presently consumes. A
dietary analysis shows he consumes 2885 calories per day. How many calories should he
consume? You need to know what is 97% of 2885.
unknown(IS) = 2885 (OF)
97 100
2885 x 97 = 279845 (IS) x 100 = 100(IS)
279845 = 100(IS) (divide both sides by 100)
2798.45 = (IS) Your client should consume 2798.45 calories.
What is 45% of \$35.00?
? = 35.00
45 100
100? = 1575 (divide both sides by 100)
15.75
65 is what percent of 125?
65 = 125
? 100
6500 = 125?
52%
Percents: to find 1% of something, multiply the amount by .01
Teach percent in lessons on investments, inflation rates, interest rates, compound interest, etc.
Compound Interest is figured: FV=PF(1 + I)
Where: FV= Future Value; PV=Present Value; I=Interest; N=Numbers of Years
```
Related docs | 1,105 | 4,382 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2013-20 | longest | en | 0.903377 |
https://download.cnet.com/Blackjack-Oscar-s-Strategy/3000-2057_4-76281950.html | 1,611,731,690,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610704821253.82/warc/CC-MAIN-20210127055122-20210127085122-00351.warc.gz | 309,379,160 | 96,645 | ## Join or Sign In
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# Blackjack Oscar's Strategy for Android
Free
## Developer's Description
A positive progression system!Heres how Oscar's System works:The goal for any series of bets is to win just one unit, then start a new series. Each series starts with a one-unit bet. After any win, the next bet is one unit more than the previous bet. After any loss, the next bet is identical to the previous bet. That is, if you lose a two-unit bet, your next bet is a two-unit bet until you have a win, at which point you raise your bet one unit to a three-unit bet.That is the whole system, except for one stipulationNever place any bet that would result in a win for the series of more than one unit. In other words, if you win a 4-unit bet, and you are now down only 2 units for the series, you would not raise your next bet to 5 units because of the 4-unit win; youd only to 3 units, which would be all youd needif successfulto achieve a one-unit win for the series.Oscars betting system combines the best features of both the positive and negative progressions. You can suffer much longer runs of losses without busting out than you can with a negative progression, since you dont raise your bets after losses. Yet, a much shorter run of wins can get back your previous losses on a series, since you raise your bets following wins. Its kind of brilliant, actually. Strings of losses hurt less, yet strings of wins pay more.When Oscar told Dr. Wilson that he had been using this system for many years and had never had a losing weekend in Las Vegas, Dr. Wilson did some mathematical and computer simulation analysis on it. Was this possible? His findings were amazing. Using a \$1 betting unit on an even money payout game, the betting progression is so slow that the player would bump up against the houses \$500 maximum bet (at that time) on only one series of every 5,000 played. On 4,999 of those series, the player would expect to achieve his \$1 win target.BetResultTotalNext Bet1L-111L-211W-122W+1doneContent rating: Medium Maturity
## Full Specifications
### General
Release February 20, 2015
Date Added February 20, 2015
Version 1.4
### Operating Systems
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Additional Requirements Compatible with 2.3.3 and above.
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\$2.99/months for a 3-year plan | 725 | 3,036 | {"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} | 3.140625 | 3 | CC-MAIN-2021-04 | latest | en | 0.960615 |
https://cs.stackexchange.com/questions/100174/provably-optimal-search-algorithms | 1,701,499,508,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100327.70/warc/CC-MAIN-20231202042052-20231202072052-00757.warc.gz | 234,023,938 | 41,576 | # provably optimal search algorithms?
In practical applications, search algorithms are often strengthened using heuristics. e.g., Deep Blue beat gary kasparov by searching through possible chess moves by "guiding" its search with human-chosen heuristics. These heuristics are not proven to be the optimal heuristics. (they're not optimal obviously).
I am wondering however:
Is there some kind of search problem (ideally a non-trivial one), where (1) there is a specific known algorithm that solves it, and (2) a proof that the algorithm is optimal in the sense that there does not exist an algorithm that solves it faster in expectation, where (3) the probability distribution for that expectation is the relevant one for that practical search problem?
I'm interested in anything related to this. If you know something that seems partially relevant, please say so.
EDIT: Alternatively, please suggest a different notion of "optimality" if you think it is more relevant. I am not sure how relevant my notion of optimality is.
EDIT 2: I'm also interested in how this question relates to the 'No free lunch' theorems in search and optimization.
• In expectation over what probability distribution? Random instances often have little connection to "real life". Nov 16, 2018 at 15:03
• @DavidRicherby, edited? Nov 16, 2018 at 15:12
• Does quicksort count? If not, why? Nov 16, 2018 at 17:14
• @Apass.Jack, I'm not sure I would call quicksort a "search algorithm". Also, I don't know whether quicksort is optimal, or in what sense Nov 16, 2018 at 20:55
• Sorry, I meant to type "binary search". Does binary search count? If not, why? Nov 16, 2018 at 21:38
Finding an optimal solution for a problem is very difficult. for example in sorting you can prove that the optimal solution with comparison is $$O(n \log n)$$ but may be you can find an algorithm with $$O(n)$$ without comparison like hashing! another good example is NP Complete problems we even can't prove they haven't polynomial algorithm. so it is very hard to find an optimal solution for deep blue and similar algorithms and proving that the algorithm is optimal is very hard too. | 504 | 2,144 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 2, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2023-50 | longest | en | 0.942236 |
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Relative frequency
26.4 Relative frequency
So far, we have calculated the theoretical probabilities of events. Theoretical probabilities tell us what should happen in an experiment. The relative frequency tells us what actually happens in an experiment.
Relative frequency is the number of outcomes obtained for a certain number of trials. We have to repeat an experiment a number of times and count the number of positive outcomes. Because this is an experiment, which is an uncertain process, we can get a different relative frequency every time that we repeat the experiment.
trial
a completed run of a well-defined experiment
positive outcome
when we get an experiment outcome that we wanted to get
frequency
the number of times an event occurs
Examples of trials:
• rolling a fair dice
• drawing a card from a pack of cards
• tossing a fair coin
• turning a spinner
The relative frequency $$(f)$$ is the number of positive outcomes $$(p)$$ divided by the total number of trials $$(t)$$.
$\text{Relative frequency} = \frac{\text{number of times an event occurs}}{\text{total number of trials conducted}}$ $f = \frac{p}{t}$
relative frequency
the number of positive outcomes $$(p)$$ divided by the total number of trials $$(t)$$; it tells us what actually happened in an experiment
Worked example 26.4: Calculating relative frequency
We toss a coin $$20$$ times and observe the outcomes. The results of the trials are listed below.
$H; H; H; T; T; H; H; H; H; H; T; T; H; T; T; H; T; T; T; H$
1. If you work out the relative frequency after the first $$10$$ trials, what is the relative frequency of observing heads?
2. What is the relative frequency of observing heads after $$15$$ trials?
3. What is the relative frequency of observing heads after $$20$$ trials?
4. Calculate the theoretical probability of observing heads.
5. How does the relative frequency compare with the theoretical probability?
Calculate the relative frequency of observing heads after $$10$$ trials.
For the first $$10$$ trials, we observed $$8$$ heads.
So for $$t = 10, p = 8$$.
$f = \frac{p}{t} = \frac{8}{10} = \text{0,8}$
Calculate the relative frequency of observing heads after $$15$$ trials.
For the first $$15$$ trials, we observed $$9$$ heads.
So for $$t = 15, p = 9$$.
$f = \frac{p}{t} = \frac{9}{15} = \text{0,6}$
Calculate the relative frequency of observing heads after $$20$$ trials.
For $$20$$ trials, we observed $$11$$ heads.
So for $$t = 20, p = 11$$.
$f = \frac{p}{t} = \frac{11}{20} = \text{0,55}$
Calculate the theoretical probability.
There are two possible outcomes; heads or tails. So $$n(S) = 2$$.
$$E = {H}$$, so $$n(E) = 1$$.
\begin{align} P(E) &= \frac{n(E)}{n(S)} \\ &= \frac{1}{2} \\ &= \text{0,5} \end{align}
Compare relative frequency and theoretical probability.
After $$20$$ trials, the relative frequency of $$\text{0,55}$$ is close to the theoretical probability of $$\text{0,5}$$.
The greater the number of trials, the closer the relative frequency will get to the theoretical probability.
The graph below shows the plot of the relative frequency of observing heads ($$f$$) after having completed $$t$$ coin tosses. The number of trials have been plotted on the $$x$$-axis and the relative frequency ($$f$$) on the $$y$$-axis.
In the beginning (after a small number of trials) the relative frequency changes a lot when compared to the theoretical probability at $$\text{0,5}$$, which is shown with a dashed line. As the number of trials increases, the relative frequency changes less and gets closer to the theoretical probability. | 942 | 3,746 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.9375 | 5 | CC-MAIN-2024-22 | longest | en | 0.911429 |
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# Karen runs a print shop that makes posters for large companies. It is a very competitive b
### Karen runs a print shop that makes posters for large companies. It is a very competitive b
Karen runs a print shop that makes posters for large companies. It is a very competitive business. The market price is currently \$1 per poster. She has fixed costs of \$250. Her variable costs are \$1,000 for the first thousand posters, \$800 for the second thousand, and then \$750 for each additional thousand posters. What is her AFC per poster (not per thousand) if she prints 1,000 posters? 2,000? 10,000? What is her ATC per poster if she prints 1,000? 2,000? 10,000? If the market price fell to 70 cents per poster, would there be any output level at which Karen would not shut down production immediately?
Abhinav 05-Dec-2019 | 209 | 835 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2020-45 | latest | en | 0.964123 |
http://www.jiskha.com/display.cgi?id=1335302598 | 1,495,482,799,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463607046.17/warc/CC-MAIN-20170522190443-20170522210443-00556.warc.gz | 553,148,861 | 3,750 | # 6th G-Math
posted by on .
1.Which of the following best describes the graph of x > 2 ?
A. Closed dot at x = 2, and shading to the right
B. Closed dot at x = 2, and shading to the left
C. Open dot at x = 2, and shading to the right
D. Open dot at x = 2, and shading to the left
• 6th G-Math - ,
open dot, means that point is not allowed. | 112 | 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} | 2.578125 | 3 | CC-MAIN-2017-22 | latest | en | 0.941254 |
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# Data analysis & Business intelligence (Math Problem Sample)
Instructions:
Data analysis in R-Statistical program
source..
Content:
STUDENT ID NUMBER:
ASSIGNMENT 2
WINTER TERM, 2015
(8697)
DUE DATE: Friday 24 July by noon
WEIGHTING: 40%
PERMITTED MATERIALS: any materials
INSTRUCTIONS:
1. Complete the assignment individually.
3. Write your answers to the questions in this document, keeping your solution as a single Word document. You may cut and paste items from software into your document, or use Ctrl-Alt-PrtScn (on Windows) or Command+Shift+4 (on macOS) to create a screen shot for your document.
4. Submit as a single Word document. Late submissions will attract a penalty of 5% per day. Submissions will not be accepted after Friday 31 July by noon.
QUESTION 1(45 marks)
Cluster analysis is a data mining technique used to divide data into meaningful groups.
* Describe (in your own words) what the k means algorithm does.
[4 marks]
K means clustering is the data partitioning technique in which data points are grouped into small clusters. The objective of the k-means clustering process is to minimize the averaged squared Euclidean distance from the centre of the data points assigned in the clustered data. As such, the k-means algorithm can be said to partition the data into pre-determined cluster numbers with the distance metric being a metric function of the similarities or dissimilarities between the data points.
* Describe (in your own words) what hierarchical cluster analysis does.
[4 marks]
Clustering involves a number of algorithms used in grouping similar objects into diverse categories. Once the data objects have been classified into different categories, the distinct data clusters can then be classified into a tree-like structure in which item arrangement depends on the similarity or dissimilarity. This concept is called a hierarchical cluster. The analysis of the clusters based on their similarities or dissimilarities is called hierarchical cluster analysis.
* Imagine you are performing a k means cluster analysis using Rattle. Describe the steps you would go through to determine the optimal number of clusters.
[4 marks]
To determine the optimal number of k-means clusters in Rattle, the following criterion is used. First, select k-centroids, with each row being randomly selected. Secondly, assign all the data points in the clusters to their respectively close centroids. Third, recalculate the length of the centroids by averaging the data points within the clusters each of p variables. Next, assign all the data points to the centroids closest to them and repeat these third and fourth steps until unassigned observations or their respective maximum number of iterations is attained.
* Describe the measures/characteristics you would use to evaluate a k means cluster analysis you have created in Rattle.[4 marks]
The Euclidean distance – This refers to the geometric distance within the multidimensional space. As the commonest type of distance, its computed as; distance (x,y) = {Σi(xi – yi)2}1/2. This measure evaluates raw data.
Squared Euclidean distance – To progressively place greater weight on objects placed apart, the square of the standard Euclidean distance is evaluated as is computed as; distance (x,y) = Σi(xi – yi)2
Percentage disagreement – This measure is useful in determining if the dimensions captured in the analysis are categorical in nature or not and is computed as; distance (x,y) = (Number of xi ≠yi)/i
Chebychev distance – This measure evaluates the appropriateness of cases where you want to define objects as different or as if they are different in one dimension and is computed as; distance (x,y) = Maximum |xi – yi|
Similarity and dissimilarity – Similarity is defined by the number of times or attributes an object is ensnared into common attributes with another. Dissimilarity on the other hand indicates the filtration of object attributes that are not ensnared into another object’s attributes.
* Explain why the Ward’s method is chosen under the Agglomerate check box when performing a hierarchical cluster analysis in Rattle and describe briefly how it functions.
[4 marks]
In hierarchical clustering, sample plots can be evaluated as individuals with other similar plots being fused into a single cluster. As such, the deployment and use of the wand method allows for pairing/fusion of plots with similar features while those with dissimilar features are also fused. Agglomerative clustering algorithms allow for divergent computation of similarity when there are several plots involved. In essence, the plots consist of several arguments such as: d, the method and member arguments.
Data mining techniques have been widely applied in the medical domain to assist in the diagnosis of various medical conditions. One of the most common medical conditions across the globe is the heart disease. Researchers wish to be able to classify patients as either being prone (positive) or not prone (negative) to heart disease. The following dataset scores 286 patients on 10 characteristics. These characteristics have been established as differing between positive and negative heart disease.
The dataset can be found in the file heart.csv on Moodle. The variables in the file are as follows:
Data Description
Variable Name
Values
age
age in years
sex
male; female
chest_pain
typical angina; atypical angina; non-anginal pain; asymptomatic
rest_bps
resting blood pressure in mm Hg (on admission to the hospital)
chol
serum cholesterol in mg/dl
fbs
fasting blood sugar > 120 mg/dl (t = true; f = false)
rest_ecg
resting electrocardiographic results (normal; having ST-T wave abnormality; showing probable or definite left ventricular hypertrophy)
max_hr
maximum heart rate achieved
ex_ang
exercise induced angina (yes; no)
disease
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First time 15% Discount! | 1,580 | 7,367 | {"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-2021-31 | latest | en | 0.877177 |
https://www.physicsforums.com/threads/torque-vs-moment.279526/ | 1,540,000,365,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583512501.27/warc/CC-MAIN-20181020013721-20181020035221-00393.warc.gz | 1,050,811,744 | 14,002 | # Torque vs. Moment
1. Dec 13, 2008
### flatmaster
I was talking with an enginering student and arived that "moment" and "torque" mean basically the same thing. Are there any subdtlties to this definition other than the moment is the SUM of all torques at a point?
This being said, is there any other fundamental concepts that carry disparate names among different diciplines?
2. Dec 14, 2008
### tiny-tim
Hi flatmaster!
i] "torque" means specifically moment of a force, but "moment" could mean moment of other things … but when talking about moment of a force, i think physicists use "torque" and "moment" interchangeably
ii] "torque" is sometimes used to describe a pair of equal-and-opposite (but not in-line) forces …
for example, if a brake is applied to one side of a flywheel, the combination of the friction force from the brake and the reaction force from the axle is a pair of forces which cancel out linearly, leaving only a torque !
iii] generally, "torque" tends to be used when there is an axle or pivot to be turned around, while "moment" tends to be used in essentially non-rotational situations, such as stress on a beam … as http://en.wikipedia.org/wiki/Torque" [Broken] says:
Last edited by a moderator: May 3, 2017
3. Dec 28, 2008
### elterrible
The way that it is used in civil and mechanical engineering:
Torsion is a specific type of moment. Torsion is simply a moment about the cross-section axis of the object. For example if you have a pipe or an I beam, torsion is a moment about the cross section of the pipe or the cross-section of the beam. Other moments such as bending moments from loading the beam are not called torques.
(I realize you could take a cross-section about any axis of the object, but in general I am talking about objects like a steel beam where the plane of what is considered the cross-section is pretty obvious)
It can also be explained as a series of shear forces going all around the cross section in the same direction.
The distinction between bending moments and torsion are made because usually one or the other controls the design and also the building codes will have different formulas and factors of safety for them. For example, a drive shaft is pretty much only torsion while maybe you are designing a beam with a small amount of torsion due to off centered loads. Since the bending moments control the design, you design for that first and then go back and check the torsion strength.
Last edited: Dec 28, 2008
4. Dec 30, 2008
### NinjaHelper
Usually, when you talk about torque it means that the vector for torque is on the object or within it, or that it's a single vector.
Moments are more broad in the sense that they can be placed anywhere. Wherever you want, and especially couples. Also, moments are related in my field (Civil Engineering) towards point loads or distributed loads. Distributed loads being the most common.
Thus, we have a moment diagram =). Mechanical Engineers use the idea to their advantage to make some calculations much more simple. Such as, when two forces intersect. And the force vectors usually intersect outside of the car. So we have a moment existing not on the object, but probably way outside of the object.
(example frames).
Last edited: Dec 30, 2008 | 747 | 3,276 | {"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-2018-43 | latest | en | 0.93666 |
https://www.coursehero.com/file/68970050/Week-8-Testpdf/ | 1,618,456,046,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038082988.39/warc/CC-MAIN-20210415005811-20210415035811-00060.warc.gz | 804,373,993 | 162,801 | Week 8 Test.pdf - APUS CLE MATH302 I001 Sum 20 Tests Quizzes MATH302 I001 Sum 20 Tests Quizzes Tests Quizzes Week 8 Test Return to Assessment List Part
# Week 8 Test.pdf - APUS CLE MATH302 I001 Sum 20 Tests...
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9/26/2020 APUS CLE : MATH302 I001 Sum 20 : Tests & Quizzes 1/18 1.0/ 1.0 Points 1.0/ 1.0 Points Week 8 Test Return to Assessment List Part 1 of 4 - Chi Square Test for Homogeneity 8.0/ 8.0 Points Question 1 of 20A college prep school advertises that their students are more prepared to succeed in collegethan other schools. To verify this, they categorize GPA’s into 4 groups and look up the proportion of studentsat a state college in each category. They find that 7% have a 0-0.99, 21% have a 1-1.99, 37% have a 2-2.99,and 35% have a 3-4.00 in GPA. They then take a random sample of 200 of their graduates at the state college and find that 19 has a 0-0.99,28 have a 1-1.99, 82 have a 2-2.99, and 71 have a 3-4.00. Can they conclude that the grades of their graduates are distributed differently than the general population atthe school? Test at the 0.05 level of significance.Hypotheses:H0: There is ______ between the general population and the college prep students in GPA.H1: There is ______ between the general population and the college prep students in GPA.Select the best fit choices that fit in the two blank spaces above. A. no difference, a differenceB. a difference, no differenceC. no difference, no differenceD. a difference, a differenceAnswer Key: A Question 2 of 20A college professor is curious if the location of seat in class affects grades in the class. They are teaching ina lecture hall with 240 students. The lecture hall has 10 rows, so they split the rows into 5 sections – Rows 1-2, Rows 3-4, MATH302 I001 Sum 20 Tests & Quizzes Tests & Quizzes
9/26/2020 APUS CLE : MATH302 I001 Sum 20 : Tests & Quizzes 2/18 1.0/ 1.0 Points Rows 5-6, Rows 7-8, and Rows 9-10. At the end of the course, they determine the top 25% of grades in the class, and if thelocation of the seat makes no difference, they would expect that these top 25% of students would be equally dispersedthroughout the classroom. Their observations are recorded below. Run a Goodness of Fit test to determine whether or notlocation has an impact on the grade. Let α=0.05.Hypotheses:H0: Location in the classroom __________impact final grade.H1: Location in the classroom _________ impact final grade.Select the best fit choices that fit in the two blank spaces above. A | 733 | 2,581 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2021-17 | latest | en | 0.899637 |
https://sep.stanford.edu/sep/prof/bei/ft1/paper_html/node11.html | 1,713,102,549,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816879.72/warc/CC-MAIN-20240414130604-20240414160604-00886.warc.gz | 472,840,409 | 2,964 | Next: SETTING UP THE FAST Up: CORRELATION AND SPECTRA Previous: Two ways to compute
## Common signals
Figure 2 shows some common signals and their autocorrelations. Figure 3 shows the cosine transforms of the autocorrelations. Cosine transform takes us from time to frequency and it also takes us from frequency to time. Thus, transform pairs in Figure 3 are sometimes more comprehensible if you interchange time and frequency. The various signals are given names in the figures, and a description of each follows:
autocor
Figure 2
Common signals and one side of their autocorrelations.
spectra
Figure 3
Autocorrelations and their cosine transforms, i.e., the (energy) spectra of the common signals.
cos
The theoretical spectrum of a sinusoid is an impulse, but the sinusoid was truncated (multiplied by a rectangle function). The autocorrelation is a sinusoid under a triangle, and its spectrum is a broadened impulse (which can be shown to be a narrow sinc-squared function).
sinc
The sinc function is .Its autocorrelation is another sinc function, and its spectrum is a rectangle function. Here the rectangle is corrupted slightly by Gibbs sidelobes,'' which result from the time truncation of the original sinc.
wide box
A wide rectangle function has a wide triangle function for an autocorrelation and a narrow sinc-squared spectrum.
narrow box
A narrow rectangle has a wide sinc-squared spectrum.
twin
Two pulses.
2 boxes
Two separated narrow boxes have the spectrum of one of them, but this spectrum is modulated (multiplied) by a sinusoidal function of frequency, where the modulation frequency measures the time separation of the narrow boxes. (An oscillation seen in the frequency domain is sometimes called a quefrency.'')
comb
Fine-toothed-comb functions are like rectangle functions with a lower Nyquist frequency. Coarse-toothed-comb functions have a spectrum which is a fine-toothed comb.
exponential
The autocorrelation of a transient exponential function is a double-sided exponential function.
The spectrum (energy) is a Cauchy function, .The curious thing about the Cauchy function is that the amplitude spectrum diminishes inversely with frequency to the first power; hence, over an infinite frequency axis, the function has infinite integral. The sharp edge at the onset of the transient exponential has much high-frequency energy.
Gauss
The autocorrelation of a Gaussian function is another Gaussian, and the spectrum is also a Gaussian.
random
Random numbers have an autocorrelation that is an impulse surrounded by some short grass. The spectrum is positive random numbers.
smoothed random
Smoothed random numbers are much the same as random numbers, but their spectral bandwidth is limited.
Next: SETTING UP THE FAST Up: CORRELATION AND SPECTRA Previous: Two ways to compute
Stanford Exploration Project
12/26/2000 | 612 | 2,847 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2024-18 | latest | en | 0.887 |
http://msl.cs.uiuc.edu/~lavalle/planning/node362.html | 1,521,616,378,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257647584.56/warc/CC-MAIN-20180321063114-20180321083114-00118.warc.gz | 199,146,903 | 4,604 | ## Exercises
1. Consider the obstacle region, (7.1), in the state space for time-varying motion planning.
1. To ensure that is polyhedral, what kind of paths should be allowed? Show how the model primitives that define are expressed in general, using as a parameter.
2. Repeat the exercise, but for ensuring that is semi-algebraic.
2. Propose a way to adapt the sampling-based roadmap algorithm of Section 5.6 to solve the problem of time-varying motion planning with bounded speed.
3. Develop an efficient algorithm for computing the obstacle region for two translating polygonal robots that each follow a linear path.
4. Sketch the coordination space for the two robots moving along the fixed paths shown in Figure 7.43.
5. Suppose there are two robots, and each moves on its own roadmap of three paths. The paths in each roadmap are arranged end-to-end in a triangle.
1. Characterize the fixed-roadmap coordination space that results, including a description of its topology.
2. Now suppose there are robots, each on a triangular roadmap, and characterize the fixed-roadmap coordination space.
6. Consider the state space obtained as the Cartesian product of the C-spaces of identical robots. Suppose that each robot is labeled with a unique integer. Show that can be partitioned nicely into regions in which appears identical and the only difference is the labels (which indicate the particular robots that are in collision).
7. Suppose there are two robots, and each moves on its own roadmap of three paths. The paths in one roadmap are arranged end-to-end in a triangle, and the paths in the other are arranged as a Y. Characterize the fixed-roadmap coordination space that results, including a description of its topology.
8. Design an efficient algorithm that takes as input a graph representation of the connectivity of a linkage and computes an active-passive decomposition. Assume that all links are revolute. The algorithm should work for either 2D or 3D linkages (the dimension is also an input). Determine the asymptotic running time of your algorithm.
9. Consider the problem of coordinating the motion of two robots that move along precomputed paths but in the presence of predictable moving obstacles. Develop a planning algorithm for this problem.
10. Consider a manipulator in made of four links connected in a chain by revolute joints. There is unit distance between the joints, and the first joint is attached at in . Suppose that the end of the last link, which is position in its body frame, is held at .
1. Use kinematics expressions to express the closure constraints for a configuration .
2. Convert the closure constraints into polynomial form.
3. Use differentiation to determine the constraints on the allowable velocities that maintain closure at a configuration .
Implementations
11. Implement the vertical decomposition algorithm to solve the path-tuning problem, as shown in Figure 7.5.
12. Use grid-based sampling and a search algorithm to compute collision-free motions of three robots moving along predetermined paths.
13. Under the conditions of Exercise 12, compute Pareto-optimal coordination strategies that optimize the time (number of stages) that each robot takes to reach its goal. Design a wavefront propagation algorithm that keeps track of the complete (ignoring equivalent strategies) set of minimal Pareto-optimal coordination strategies at each reached state. Avoid storing entire plans at each discretized state.
14. To gain an appreciation of the difficulties of planning for closed kinematic chains, try motion planning for a point on a torus among obstacles using only the implicit torus constraint given by (6.40). To simplify collision detection, the obstacles can be a collection of balls in that intersect the torus. Adapt a sampling-based planning technique, such as the bidirectional RRT, to traverse the torus and solve planning problems.
15. Implement the spanning-tree coverage planning algorithm of Section 7.6.
16. Develop an RRT-based planning algorithm that causes the robot to chase an unpredictable moving target in a planar environment that contains obstacles. The algorithm should run quickly enough so that replanning can occur during execution. The robot should execute the first part of the most recently computed path while simultaneously computing a better plan for the next time increment.
17. Modify Exercise 16 so that the robot assumes the target follows a predictable, constant-velocity trajectory until some deviation is observed.
18. Show how to handle unexpected obstacles by using a fast enough planning algorithm. For simplicity, suppose the robot is a point moving in a polygonal obstacle region. The robot first computes a path and then starts to execute it. If the obstacle region changes, then a new path is computed from the robot's current position. Use vertical decomposition or another algorithm of your choice (provided it is fast enough). The user should be able to interactively place or move obstacles during plan execution.
19. Use the manipulation planning framework of Section 7.3.2 to develop an algorithm that solves the famous Towers of Hanoi problem by a robot that carries the rings [166]. For simplicity, suppose a polygonal robot moves polygonal parts in and rotation is not allowed. Make three pegs, and initially place all parts on one peg, sorted from largest to smallest. The goal is to move all of the parts to another peg while preserving the sorting.
20. Use grid-based approximation to solve optimal planning problems for a point robot in the plane. Experiment with using different neighborhoods and metrics. Characterize the combinations under which good and bad approximations are obtained.
Steven M LaValle 2012-04-20 | 1,136 | 5,740 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2018-13 | latest | en | 0.939027 |
https://open.kattis.com/problems/pianolessons | 1,653,386,916,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662570051.62/warc/CC-MAIN-20220524075341-20220524105341-00293.warc.gz | 496,924,575 | 6,777 | Kattis
# Piano Lessons
Mrs. Mackenzie loves teaching piano. She started last year with just five students in her home studio at Green Gables in the beautiful province of Prince Edward Island (Canada). Pretty soon, kids across the entire province heard about her brilliant piano teaching, and now she is receiving hundreds of calls requesting lessons. She is finding it very difficult to fit them all into her schedule since she has a limited number of time slots and students have varying availability. While she is baffled by some kids’ irrational interest in sports, she respects their prior commitments.
Given Mrs. Mackenzie’s time slots, and a list of the time slots that work for each potential student, can you help her determine the maximum number of students she can fit into her schedule? Note that Mrs. Mackenzie only teaches private lessons (one student at a time) and no student requires more than one time slot.
## Input
The first line of input contains two space-separated integers, $N$ and $M$, where $N$ is the number of potential students and $M$ is the number of time slots that Mrs. Mackenzie has available ($1 \leq N, M \leq 1\, 000$). These time slots are numbered $1, 2, 3, \ldots , M$. This is followed by $N$ lines, one for each potential student. Each of these $N$ lines begins with an integer $T$, the number of time slots that work for the student ($0 \leq T \leq M$), and this is followed (on the same line) by $T$ distinct space-separated integers $t_ i$, representing the specific time slots that work ($1 \leq t_ i \leq M$).
## Output
Output the maximum number of students Mrs. Mackenzie can fit into her $M$ time slots.
Sample Input 1 Sample Output 1
2 3
1 3
1 3
1
Sample Input 2 Sample Output 2
5 5
1 1
3 1 2 3
3 1 3 4
3 2 4 5
3 2 3 5
5 | 468 | 1,780 | {"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} | 3.015625 | 3 | CC-MAIN-2022-21 | latest | en | 0.919419 |
https://aviation.stackexchange.com/questions/37014/what-is-the-difference-between-saturated-and-unsaturated-adiabatic-lapse-rate?noredirect=1 | 1,723,079,961,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640713903.39/warc/CC-MAIN-20240808000606-20240808030606-00209.warc.gz | 91,439,708 | 37,478 | # What is the difference between saturated and unsaturated adiabatic lapse rate? [closed]
I'm having trouble understanding the difference between saturated and unsaturated adiabatic lapse rate.
I know the unsaturated adiabatic lapse rate is approximately 5.4 degrees per 1,000 feet in which a "dry" parcel of air cools at that rate. What does the unsaturated adiabatic lapse rate depend on? Why is the unsaturated adiabatic lapse rate lower than the saturated lapse rate?
I searched and found that the unsaturated adiabatic lapse rate is lower due to the air parcel releasing its latent heat, which causes the air parcel to cool slower, but I don't get it. When a "dry" parcel of air is cooling adiabatically, are we assuming 0% moisture?
• Probably better on Earth Science.SE. Commented Apr 11, 2017 at 1:59
• Commented Apr 17, 2017 at 16:52
The difference between a unsaturated and a saturated adiabatic lapse rate is that in the latter the water vapour has to change state from gaseous to liquid.
When air rises it expands due to the lower pressure of the ambient air at higher altitude. By expanding, the air parcel is doing work on its surroundings. Since no heat is transferred to the air parcel (by definition of being adiabatic expansion) but work is being performed, the temperature has to drop. Thus a rising parcel of air cools down.
Cooler air can hold less water vapour than warmer air. If the air parcel contains water vapour, at some point during the cooling the maximum relative humidity is reached. At that point the air is saturated with water vapour. If the air continues to cool, the water vapour content has to change from vapour to liquid. This is called saturated adiabatic cooling.
The process of condensation of the water content is exothermic; when the state changes from gaseous to liquid, heat is released to the air in the parcel. Therefore the lapse rate (temperature change with altitude) of a saturated adiabatic process is less than an unsaturated adiabatic process. The heat released from the condensation process reduces the lapse rate. | 479 | 2,079 | {"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-2024-33 | latest | en | 0.947353 |
http://docs.godotengine.org/en/stable/tutorials/3d/inverse_kinematics.html | 1,539,814,002,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583511314.51/warc/CC-MAIN-20181017220358-20181018001858-00414.warc.gz | 99,235,976 | 45,630 | # Inverse kinematics¶
This tutorial is a follow-up of Working with 3D skeletons.
Before continuing on, I’d recommend reading some theory, the simplest article I could find is this:
http://freespace.virgin.net/hugo.elias/models/m_ik2.htm
## Initial problem¶
Talking in Godot terminology, the task we want to solve here is to position our 2 angles we talked about above so, that the tip of the lowerarm bone is as close to the target point (which is set by the target Vector3()) as possible using only rotations. This task is calculation-intensive and never resolved by analytical equation solving. So, it is an underconstrained problem, which means there is an unlimited number of solutions to the equation.
For easy calculation, in this chapter we consider the target being a child of Skeleton. If this is not the case for your setup you can always reparent it in your script, as you will save on calculations if you do so.
In the picture you see the angles alpha and beta. In this case we don’t use poles and constraints, so we need to add our own. On the picture the angles are 2D angles living in a plane which is defined by bone base, bone tip and target.
The rotation axis is easily calculated using the cross-product of the bone vector and the target vector. The rotation in this case will be always in positive direction. If t is the Transform which we get from the get_bone_global_pose() function, the bone vector is
t.basis[2]
So we have all the information we need to execute our algorithm.
In game dev it is common to resolve this problem by iteratively closing to the desired location, adding/subtracting small numbers to the angles until the distance change achieved is less than some small error value. Sounds easy enough, but there are Godot problems we need to resolve there to achieve our goal.
• How to find coordinates of the tip of the bone?
• How to find the vector from the bone base to the target?
For our goal (tip of the bone moved within area of target), we need to know where the tip of our IK bone is. As we don’t use a leaf bone as IK bone, we know the coordinate of the bone base is the tip of the parent bone. All these calculations are quite dependent on the skeleton’s structure. You can use pre-calculated constants as well. You can add an extra bone at the tip of the IK bone and calculate using that.
## Implementation¶
We will use an exported variable for the bone length to make it easy.
export var ik_bone = "lowerarm"
export var ik_bone_length = 1.0
export var ik_error = 0.1
Now, we need to apply our transformations from the IK bone to the base of the chain. So we apply a rotation to the IK bone then move from our IK bone up to its parent, apply rotation again, then move to the parent of the current bone again, etc. So we need to limit our chain somewhat.
export var ik_limit = 2
For the _ready() function:
var skel
skel = get_node("arm/Armature/Skeleton")
set_process(true)
Now we can write our chain-passing function:
func pass_chain():
var b = skel.find_bone(ik_bone)
var l = ik_limit
while b >= 0 and l > 0:
print( "name":", skel.get_bone_name(b))
print( "local transform":", skel.get_bone_pose(b))
print( "global transform":", skel.get_bone_global_pose(b))
b = skel.get_bone_parent(b)
l = l - 1
And for the _process() function:
func _process(delta):
pass_chain(delta)
Executing this script will pass through the bone chain, printing bone transforms.
extends Spatial
export var ik_bone = "lowerarm"
export var ik_bone_length = 1.0
export var ik_error = 0.1
export var ik_limit = 2
var skel
skel = get_node("arm/Armature/Skeleton")
set_process(true)
func pass_chain(delta):
var b = skel.find_bone(ik_bone)
var l = ik_limit
while b >= 0 and l > 0:
print("name: ", skel.get_bone_name(b))
print("local transform: ", skel.get_bone_pose(b))
print( "global transform:", skel.get_bone_global_pose(b))
b = skel.get_bone_parent(b)
l = l - 1
func _process(delta):
pass_chain(delta)
Now we need to actually work with the target. The target should be placed somewhere accessible. Since “arm” is an imported scene, we better place the target node within our top level scene. But for us to work with target easily its Transform should be on the same level as the Skeleton.
To cope with this problem we create a “target” node under our scene root node and at runtime we will reparent it copying the global transform, which will achieve the desired effect.
Create a new Spatial node under the root node and rename it to “target”. Then modify the _ready() function to look like this:
var skel
var target | 1,104 | 4,576 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2018-43 | latest | en | 0.920453 |
https://thiscondensedlife.wordpress.com/2015/07/10/modern-theory-of-polarization/?replytocom=33 | 1,590,536,709,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347391923.3/warc/CC-MAIN-20200526222359-20200527012359-00080.warc.gz | 591,982,490 | 28,919 | # Modern Theory of Polarization
It is quite curious that the simple concept of polarization in a solid was not understood until the early to mid-90s. The solution to the problem actually came from the computational physics community because of their inability to calculate accurately the polarization in solids. Prior to the solution, there were papers seriously discussing whether polarization was a bulk phenomenon or whether it was a property of a crystal surface. The solution to the problem is commonly associated with these papers by Resta (pdf!) and King-Smith and Vanderbilt (pdf!) .
There are a few startling realizations that arose from the modern theory of polarization:
1. Polarization is not a well-defined quantity in that it is multi-valued.
2. Only polarization difference has any physical meaning.
3. Experiments only measure differences in polarization.
4. Polarization is deeply rooted in the concept of the Berry phase.
5. (not startling) Polarization is a bulk phenomenon.
There is a fantastic pedagogical introduction to the subject by Spaldin entitled A beginner’s guide to the modern theory of polarization.
As just a little preview of the Spaldin paper, let me outline the issues as she does. The problem with defining polarization as the dipole moment per unit cell can easily be understood using the picture of the one-dimensional chain below:
Using the box on the left, one would calculate the dipole moment per unit length as:
$p =\frac{e}{a}\sum_i q_id_i = \frac{e}{a}(\frac{a}{4}\times -1 + \frac{3a}{4}\times 1) = \frac{e}{2}$
whereas the box on the right gives:
$p =\frac{e}{a}\sum_i q_id_i = \frac{e}{a}(\frac{a}{4}\times 1 + \frac{3a}{4}\times -1) = \frac{-e}{2}$.
That these don’t match has to do precisely with the fact that polarization in a solid is multi-valued, as alluded to earlier. Now consider the following one-dimensional lattice with a distortion:
The polarization now for the distorted lattice (second row in the image above) for the left and right boxes are respectively:
Left Box: $p = \frac{e}{2} +\frac{ed}{a}$
Right Box: $p = \frac{-e}{2} +\frac{ed}{a}$
We can therefore see that for both boxes, the change in polarization is $\delta p = \frac{ed}{a}$, which is a single-valued and experimentally well-defined quantity.
While this illustration was classical, when one includes the wavefunction of the electrons, one is forced to consider the Berry phase of the Bloch electrons. While I have known about this result for some time now, I still find it quite surprising that the simple concept of polarization in a solid has any relationship to the Berry phase at all. I strongly recommend Spaldin’s eminently readable article as an excellent introduction to the subject.
### 4 responses to “Modern Theory of Polarization”
1. Hi Anshul,
This is one of my favorite subjects in condensed matter physics, because it really tests our understanding of several concepts at once!
In my mind, an important realization (that took me a long time to get) is that all this formalism is necessary only because we model solid materials as periodic systems. This isn’t actually correct; as you know, your materials have finite extent. So if we had an enormous computer and could just simulate the mm-sized chunk of material directly, we could calculate polarization in the normal way and there would be no ambiguities and no Berry phases.
There’s another paper, which is very mathematical, but interesting once you get it here: http://journals.aps.org/prb/abstract/10.1103/PhysRevB.62.1666 . They explain the problem from a very different angle that I like. When we use periodic boundary conditions to simulate a bulk solid, we actually cut out some of our Hilbert space compared to the open boundary situation. Part of that ‘lost’ Hilbert space includes the result of operating X on a periodic wave function! So if we’d like to get at any expectation values of X, then we have to approach it in a different way. The Berry phases come about from doing this for one-body wave functions.
Cheers,
Like
• Anshul
Hi Lucas,
Yes, you are absolutely correct. It seems to me like when calculating the Berry phase in solids, even more generally, like in the Quantum Hall Effect, there is the tacit assumption of periodic boundary conditions. This of course has to be relaxed to obtain the edge states.
Thanks for the paper, I will take a look at it when I get the chance — a brief glance through looks like it may not be too far over my head. 8^)
Like
2. Piotr
Hi,
the introduction by Spaldin seems to be good, however, I am a little confused about this. How can you choose unit cell to obtain resultant polarization equal to 3/2 or 5/2??
Also, if you choose unit cell to begin at one of the ions, the polarization will be equal to zero in the non-polar chain, unlike Spaldin claims that it is …-5/2, -3/2, -1/2, 1/2, 3/2, 5/2…
Like | 1,152 | 4,887 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 5, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2020-24 | latest | en | 0.906383 |
https://mebel-kamar.com/en/Aftermarket.php | 1,675,593,981,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500251.38/warc/CC-MAIN-20230205094841-20230205124841-00444.warc.gz | 405,910,172 | 5,414 | # Math equations
Math can be a challenging subject for many learners. But there is support available in the form of Math equations. Our website can help me with math work.
## The Best Math equations
Apps can be a great way to help students with their algebra. Let's try the best Math equations. This formula states that the log of a number with respect to one base is equal to the log of the same number with respect to another base multiplied by the log of the new base with respect to the old base. So, if we want to solve for x in our example equation above, we can plug in our known values and solve for x using algebra.2log₃x=6⇒log₃x=3⇒x=33Since we now know that 3 was raised to the third power in order to produce 9 (our exponent), we have successfully solved for x in this equation!Common and natural logarithms are two other ways that exponents can be solved for without using the change of base formula. Common logarithms use bases of 10, while natural logarithms use bases of e (approximately 2.71828182845904). To solve for x in equations using these types of logs, all you need to do is take the inverse function of each side. For example, if we want to solve10log₁₀x=100we can simply take the inverse common log function of both sides.This tells us that 100 must have been produced when 10 was raised to some power - but what power? Well, we can use algebra once again!10log₁₀x=100⇒log₁₀x=10⇒x=1010Now we know that 10 was raised to the 10th power in order to produce 100. And just like that - we've solved another equation for x using logs!While solving equations with logs may seem daunting at first, there's no need to worry - with a little practice, you'll be a pro in no time!
By focusing on one part at a time, it may be easier to see a pattern or solution. Another method is to work backwards from the answer. This can help to provide a framework for solving the equation. In addition, it is often helpful to consult with a friend or tutor who is better at math than you are. By working together, it may be possible to arrive at the correct answer. Ultimately, there is no single method that will guarantee success in solving hard math equations. However, by trying different approaches, it may be possible to find a solution.
To solve inequality equations, you need to first understand what they are. Inequality equations are mathematical equations that involve two variables which are not equal to each other. The inequalities can be either greater than or less than. To solve these equations, you need to find the value of the variable that makes the two sides of the equation equal. This can be done by using the properties of inequality. For example, if the equation is x+5>9, then you can subtract 5 from both sides to get x>4. This means that the solutions to this inequality are all values of x that are greater than 4. Solve inequality equations by using the properties of inequality to find the value of the variable that makes the two sides of the equation equal.
First, it is important to create a dedicated study space. This will help to minimize distractions and make it easier to focus on the task at hand. Secondly, students should develop a regular routine and stick to it as much as possible. This will help them to stay on track and avoid getting overwhelmed. Finally, students should seek help from their teachers or parents when needed. By taking these steps, students can set themselves up for success when it comes to doing their math homework.
## Instant assistance with all types of math
Man, oh man. When I say this is a lifesaver. I mean it. For simple problems this is the best. It’s embarrassing to approach tutors for simple things, this app makes the simple things easier, leaves the harder questions for the tutors. | 833 | 3,774 | {"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} | 4.53125 | 5 | CC-MAIN-2023-06 | latest | en | 0.947546 |
https://www.geeksforgeeks.org/construct-an-ap-series-consisting-of-a-and-b-having-minimum-possible-nth-term/?ref=leftbar-rightbar | 1,607,158,420,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141747323.98/warc/CC-MAIN-20201205074417-20201205104417-00372.warc.gz | 682,039,110 | 27,405 | # Construct an AP series consisting of A and B having minimum possible Nth term
Given two integers A, B which are any two terms of an Arithmetic Progression series, and an integer N, the task is to to construct an Arithmetic Progression series of size N such that it must include both A and B and the Nth term of the AP should be minimum.
Examples:
Input: N = 5, A = 20, B = 50
Output: 10 20 30 40 50
Explanation:
One of the possible AP sequences is {10, 20, 30, 40, 50} having 50 as the 5th value, which is the minimum possible.
Input: N = 2, A = 1, B = 49
Output: 1 49
Approach: The Nth Term of an AP is given by XN = X + (N – 1)*d, where X is the first term and d is a common difference. To make the largest element minimum, minimize both x and d. It can be observed that the value of X cannot be more than min(A, B) and the value of d cannot be more than abs(A – B).
1. Now, use the same formula to construct the AP for every possible value of x (From 1 to min(A, B)) and d(From 1 to abs(A – B)).
2. Now, construct the array arr[] as {x, x + d, x + 2d, …, x + d*(N – 1)}.
3. Check if A and B are present in it or not and the Nth element is minimum possible or not. If found to be true, then update the ans[] by the constructed array arr[].
4. Otherwise, iterate further and check for other values of x and d.
5. Finally, print ans[] as the answer.
Below is the implementation of the above approach:
## C++
`// C++ program for the above approach` `#include ` `using` `namespace` `std;` `// Function to check if both a and` `// b are present in the AP series or not` `bool` `check_both_present(``int` `arr[], ``int` `N,` ` ``int` `a, ``int` `b)` `{` ` ``bool` `f1 = ``false``, f2 = ``false``;` ` ``// Iterate over the array arr[]` ` ``for` `(``int` `i = 0; i < N; i++) {` ` ``// If a is present` ` ``if` `(arr[i] == a) {` ` ``f1 = ``true``;` ` ``}` ` ``// If b is present` ` ``if` `(arr[i] == b) {` ` ``f2 = ``true``;` ` ``}` ` ``}` ` ``// If both are present` ` ``if` `(f1 && f2) {` ` ``return` `true``;` ` ``}` ` ``// Otherwise` ` ``else` `{` ` ``return` `false``;` ` ``}` `}` `// Function to print all the elements` `// of the Arithmetic Progression` `void` `print_array(``int` `ans[], ``int` `N)` `{` ` ``for` `(``int` `i = 0; i < N; i++) {` ` ``cout << ans[i] << ``" "``;` ` ``}` `}` `// Function to construct AP series` `// consisting of A and B with` `// minimum Nth term` `void` `build_AP(``int` `N, ``int` `a, ``int` `b)` `{` ` ``// Stores the resultant series` ` ``int` `arr[N], ans[N];` ` ``// Initialise ans[i] as INT_MAX` ` ``for` `(``int` `i = 0; i < N; i++)` ` ``ans[i] = INT_MAX;` ` ``int` `flag = 0;` ` ``// Maintain a smaller than b` ` ``if` `(a > b) {` ` ``swap(a, b);` ` ``}` ` ``// Difference between a and b` ` ``int` `diff = b - a;` ` ``// Check for all possible combination` ` ``// of start and common difference d` ` ``for` `(``int` `start = 1;` ` ``start <= a; start++) {` ` ``for` `(``int` `d = 1;` ` ``d <= diff; d++) {` ` ``// Initialise arr[0] as start` ` ``arr[0] = start;` ` ``for` `(``int` `i = 1; i < N; i++) {` ` ``arr[i] = arr[i - 1] + d;` ` ``}` ` ``// Check if both a and b are` ` ``// present or not and the Nth` ` ``// term is the minimum or not` ` ``if` `(check_both_present(arr, N, a, b)` ` ``&& arr[N - 1] < ans[N - 1]) {` ` ``// Update the answer` ` ``for` `(``int` `i = 0; i < N; i++) {` ` ``ans[i] = arr[i];` ` ``}` ` ``}` ` ``}` ` ``}` ` ``// Print the resultant array` ` ``print_array(ans, N);` `}` `// Driver Code` `int` `main()` `{` ` ``int` `N = 5, A = 20, B = 50;` ` ``// Function Call` ` ``build_AP(N, A, B);` ` ``return` `0;` `}`
## Java
`// Java program for the above approach` `import` `java.io.*;` `class` `GFG{` `// Function to check if both a and` `// b are present in the AP series or not` `public` `static` `boolean` `check_both_present(``int``[] arr, ` ` ``int` `N, ``int` `a, ` ` ``int` `b)` `{` ` ``boolean` `f1 = ``false``, f2 = ``false``;` ` ``// Iterate over the array arr[]` ` ``for``(``int` `i = ``0``; i < N; i++)` ` ``{` ` ` ` ``// If a is present` ` ``if` `(arr[i] == a) ` ` ``{` ` ``f1 = ``true``;` ` ``}` ` ``// If b is present` ` ``if` `(arr[i] == b)` ` ``{` ` ``f2 = ``true``;` ` ``}` ` ``}` ` ``// If both are present` ` ``if` `(f1 && f2) ` ` ``{` ` ``return` `true``;` ` ``}` ` ``// Otherwise` ` ``else` ` ``{` ` ``return` `false``;` ` ``}` `}` `// Function to print all the elements` `// of the Arithmetic Progression` `public` `static` `void` `print_array(``int``[] ans, ``int` `N)` `{` ` ``for``(``int` `i = ``0``; i < N; i++)` ` ``{` ` ``System.out.print(ans[i] + ``" "``);` ` ``}` `}` `// Function to construct AP series` `// consisting of A and B with` `// minimum Nth term` `public` `static` `void` `build_AP(``int` `N, ``int` `a, ``int` `b)` `{` ` ` ` ``// Stores the resultant series` ` ``int``[] arr = ``new` `int``[N];` ` ``int``[] ans = ``new` `int``[N];` ` ``// Initialise ans[i] as INT_MAX` ` ``for``(``int` `i = ``0``; i < N; i++)` ` ``ans[i] = Integer.MAX_VALUE;` ` ``int` `flag = ``0``;` ` ``// Maintain a smaller than b` ` ``if` `(a > b)` ` ``{` ` ` ` ``// swap(a and b)` ` ``a += (b - (b = a));` ` ``}` ` ``// Difference between a and b` ` ``int` `diff = b - a;` ` ``// Check for all possible combination` ` ``// of start and common difference d` ` ``for``(``int` `start = ``1``; start <= a; start++)` ` ``{` ` ``for``(``int` `d = ``1``; d <= diff; d++)` ` ``{` ` ` ` ``// Initialise arr[0] as start` ` ``arr[``0``] = start;` ` ``for``(``int` `i = ``1``; i < N; i++)` ` ``{` ` ``arr[i] = arr[i - ``1``] + d;` ` ``}` ` ``// Check if both a and b are` ` ``// present or not and the Nth` ` ``// term is the minimum or not` ` ``if` `(check_both_present(arr, N, a, b) && ` ` ``arr[N - ``1``] < ans[N - ``1``]) ` ` ``{` ` ` ` ``// Update the answer` ` ``for``(``int` `i = ``0``; i < N; i++)` ` ``{` ` ``ans[i] = arr[i];` ` ``}` ` ``}` ` ``}` ` ``}` ` ``// Print the resultant array` ` ``print_array(ans, N);` `}` `// Driver Code` `public` `static` `void` `main(String[] args)` `{` ` ``int` `N = ``5``, A = ``20``, B = ``50``;` ` ``// Function call` ` ``build_AP(N, A, B);` `}` `}` `// This code is contributed by akhilsaini`
## Python3
`# Python3 program for the above approach` `import` `sys` `# Function to check if both a and` `# b are present in the AP series or not` `def` `check_both_present(arr, N, a, b):` ` ` ` ``f1 ``=` `False` ` ``f2 ``=` `False` ` ``# Iterate over the array arr[]` ` ``for` `i ``in` `range``(``0``, N):` ` ``# If a is present` ` ``if` `arr[i] ``=``=` `a:` ` ``f1 ``=` `True` ` ``# If b is present` ` ``if` `arr[i] ``=``=` `b:` ` ``f2 ``=` `True` ` ``# If both are present` ` ``if` `f1 ``and` `f2:` ` ``return` `True` ` ``# Otherwise` ` ``else``:` ` ``return` `False` `# Function to print all the elements` `# of the Arithmetic Progression` `def` `print_array(ans, N):` ` ` ` ``for` `i ``in` `range``(``0``, N):` ` ``print``(ans[i], end ``=` `" "``)` `# Function to construct AP series` `# consisting of A and B with` `# minimum Nth term` `def` `build_AP(N, a, b):` ` ` ` ``INT_MAX ``=` `sys.maxsize` ` ``# Stores the resultant series` ` ``arr ``=` `[``None` `for` `i ``in` `range``(N)]` ` ``# Initialise ans[i] as INT_MAX` ` ``ans ``=` `[INT_MAX ``for` `i ``in` `range``(N)]` ` ``flag ``=` `0` ` ``# Maintain a smaller than b` ` ``if` `a > b:` ` ` ` ``# Swap a and b` ` ``a, b ``=` `b, a` ` ``# Difference between a and b` ` ``diff ``=` `b ``-` `a` ` ``# Check for all possible combination` ` ``# of start and common difference d` ` ``for` `start ``in` `range``(``1``, a ``+` `1``):` ` ``for` `d ``in` `range``(``1``, diff ``+` `1``):` ` ``# Initialise arr[0] as start` ` ``arr[``0``] ``=` `start` ` ``for` `i ``in` `range``(``1``, N):` ` ``arr[i] ``=` `arr[i ``-` `1``] ``+` `d` ` ``# Check if both a and b are` ` ``# present or not and the Nth` ` ``# term is the minimum or not` ` ``if` `((check_both_present(arr, N, a, b) ``and` ` ``arr[N ``-` `1``] < ans[N ``-` `1``])):` ` ``# Update the answer` ` ``for` `i ``in` `range``(``0``, N):` ` ``ans[i] ``=` `arr[i]` ` ``# Print the resultant array` ` ``print_array(ans, N)` `# Driver Code` `if` `__name__ ``=``=` `"__main__"``:` ` ` ` ``N ``=` `5` ` ``A ``=` `20` ` ``B ``=` `50` ` ``# Function call` ` ``build_AP(N, A, B)` `# This code is contributed by akhilsaini`
## C#
`// C# program for the above approach` `using` `System;` `class` `GFG{` `// Function to check if both a and` `// b are present in the AP series or not` `static` `bool` `check_both_present(``int``[] arr, ``int` `N,` ` ``int` `a, ``int` `b)` `{` ` ``bool` `f1 = ``false``, f2 = ``false``;` ` ``// Iterate over the array arr[]` ` ``for``(``int` `i = 0; i < N; i++)` ` ``{` ` ``// If a is present` ` ``if` `(arr[i] == a)` ` ``{` ` ``f1 = ``true``;` ` ``}` ` ``// If b is present` ` ``if` `(arr[i] == b)` ` ``{` ` ``f2 = ``true``;` ` ``}` ` ``}` ` ``// If both are present` ` ``if` `(f1 && f2)` ` ``{` ` ``return` `true``;` ` ``}` ` ``// Otherwise` ` ``else` ` ``{` ` ``return` `false``;` ` ``}` `}` `// Function to print all the elements` `// of the Arithmetic Progression` `static` `void` `print_array(``int``[] ans, ``int` `N)` `{` ` ``for``(``int` `i = 0; i < N; i++)` ` ``{` ` ``Console.Write(ans[i] + ``" "``);` ` ``}` `}` `// Function to construct AP series` `// consisting of A and B with` `// minimum Nth term` `static` `void` `build_AP(``int` `N, ``int` `a, ``int` `b)` `{` ` ` ` ``// Stores the resultant series` ` ``int``[] arr = ``new` `int``[N];` ` ``int``[] ans = ``new` `int``[N];` ` ``// Initialise ans[i] as INT_MAX` ` ``for``(``int` `i = 0; i < N; i++)` ` ``ans[i] = ``int``.MaxValue;` ` ``// Maintain a smaller than b` ` ``if` `(a > b) ` ` ``{` ` ` ` ``// Swap a and b` ` ``a += (b - (b = a));` ` ``}` ` ``// Difference between a and b` ` ``int` `diff = b - a;` ` ``// Check for all possible combination` ` ``// of start and common difference d` ` ``for``(``int` `start = 1; start <= a; start++)` ` ``{` ` ``for``(``int` `d = 1; d <= diff; d++)` ` ``{` ` ` ` ``// Initialise arr[0] as start` ` ``arr[0] = start;` ` ``for``(``int` `i = 1; i < N; i++)` ` ``{` ` ``arr[i] = arr[i - 1] + d;` ` ``}` ` ``// Check if both a and b are` ` ``// present or not and the Nth` ` ``// term is the minimum or not` ` ``if` `(check_both_present(arr, N, a, b) && ` ` ``arr[N - 1] < ans[N - 1])` ` ``{` ` ` ` ``// Update the answer` ` ``for``(``int` `i = 0; i < N; i++)` ` ``{` ` ``ans[i] = arr[i];` ` ``}` ` ``}` ` ``}` ` ``}` ` ``// Print the resultant array` ` ``print_array(ans, N);` `}` `// Driver Code` `static` `public` `void` `Main()` `{` ` ``int` `N = 5, A = 20, B = 50;` ` ``// Function call` ` ``build_AP(N, A, B);` `}` `}` `// This code is contributed by akhilsaini`
Output:
```10 20 30 40 50
```
Time Complexity: O(N3)
Auxiliary Space: O(N)
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Improved By : akhilsaini | 4,843 | 13,471 | {"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} | 3.453125 | 3 | CC-MAIN-2020-50 | latest | en | 0.83387 |
http://programmingpraxis.codepad.org/UB4OxZv1 | 1,652,869,847,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662521883.7/warc/CC-MAIN-20220518083841-20220518113841-00091.warc.gz | 42,378,169 | 5,003 | Project:
```1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 ``` ```(define (expm b e m) (define (times x y) (modulo (* x y) m)) (let loop ((b b) (e e) (r 1)) (if (zero? e) r (loop (times b b) (quotient e 2) (if (odd? e) (times b r) r))))) (define (iroot k n) (let loop ((u n)) (let* ((s u) (t (+ (* (- k 1) s) (quotient n (expt s (- k 1))))) (u (quotient t k))) (if (< u s) (loop u) s)))) (define (digits n . args) (let ((b (if (null? args) 10 (car args)))) (let loop ((n n) (d '())) (if (zero? n) d (loop (quotient n b) (cons (modulo n b) d)))))) (define square? (let ((q11 (make-vector 11 #f)) (q63 (make-vector 63 #f)) (q64 (make-vector 64 #f)) (q65 (make-vector 65 #f))) (do ((k 0 (+ k 1))) ((< 5 k)) (vector-set! q11 (modulo (* k k) 11) #t)) (do ((k 0 (+ k 1))) ((< 31 k)) (vector-set! q63 (modulo (* k k) 63) #t)) (do ((k 0 (+ k 1))) ((< 31 k)) (vector-set! q64 (modulo (* k k) 64) #t)) (do ((k 0 (+ k 1))) ((< 32 k)) (vector-set! q65 (modulo (* k k) 65) #t)) (lambda (n) (if (not (integer? n)) (error 'square? "must be integer") (if (< n 1) #f (if (not (vector-ref q64 (modulo n 64))) #f (let ((r (modulo n 45045))) (if (not (vector-ref q63 (modulo r 63))) #f (if (not (vector-ref q65 (modulo r 65))) #f (if (not (vector-ref q11 (modulo r 11))) #f (let ((q (iroot 2 n))) (if (= (* q q) n) q #f)))))))))))) (define (spsp? n a) (do ((d (- n 1) (/ d 2)) (s 0 (+ s 1))) ((odd? d) (let ((t (expm a d n))) (if (or (= t 1) (= t (- n 1))) #t (do ((s (- s 1) (- s 1)) (t (expm t 2 n) (expm t 2 n))) ((or (zero? s) (= t (- n 1))) (positive? s)))))))) (define (pseudoprime? n) (and (spsp? n 2) (spsp? n 3) (spsp? n 5))) (define (td n limit) (let twos ((n n) (fs (list))) (cond ((= n 1) fs) ((even? n) (twos (/ n 2) (cons 2 fs))) (else (let odds ((n n) (f 3) (fs fs)) (cond ((< limit f) fs) ((< n (* f f)) (reverse (cons n fs))) ((zero? (modulo n f)) (odds (/ n f) f (cons f fs))) (else (odds n (+ f 2) fs)))))))) (define (rho n limit) (let loop ((t 2) (h 2) (d 1) (c 1) (limit limit)) (define (f x) (modulo (+ (* x x) c) n)) (cond ((zero? limit) #f) ((= d 1) (let ((t (f t)) (h (f (f h)))) (loop t h (gcd (- t h) n) c (- limit 1)))) ((= d n) (loop 2 2 1 (+ c 1) (- limit 1))) ((pocklington? d) d) (else (rho d (- limit 1)))))) (define (pocklington? n) (if (not (pseudoprime? n)) #f (if (< n #e1e12) (= (car (td n #e1e6)) n) (let* ((n-1 (- n 1)) (root2 (iroot 2 n)) (root3 (iroot 3 n)) (fs (td n-1 1000)) (f (apply * fs))) (let f-loop ((f f) (fs fs)) (if (< f root3) (let ((p (rho (/ n-1 f) #e1e6))) (if p (f-loop (* f p) (cons p fs)) 'unknown)) (let a-loop ((a 2)) (if (not (= (expm a n-1 n) 1)) #f (let loop ((fs fs)) (let ((g (gcd (- (expm a (/ n-1 (car fs)) n) 1) n))) (cond ((< 1 g n) #f) ((= g n) (a-loop (+ a 1))) ((pair? (cdr fs)) (loop (cdr fs))) ((< root2 f) #t) (else (let* ((ds (digits n f)) (c2 (car ds)) (c1 (cadr ds)) (s (- (* c1 c1) (* 4 c2)))) (not (square? s))))))))))))))) (define (mersenne n) (- (expt 2 n) 1)) (display (pocklington? (mersenne 89))) (newline) (display (pocklington? (mersenne 521))) (newline) (display (pseudoprime? 3825123056546413051)) (newline) (display (pocklington? 3825123056546413051)) (newline) ```
```1 2 3 4 ``` ```#t #t #t #f ``` | 1,605 | 3,454 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2022-21 | latest | en | 0.198327 |
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- FLUENT (http://www.cfd-online.com/Forums/fluent/)
- - UDF: exporting a variable between two DEFINE function (http://www.cfd-online.com/Forums/fluent/67507-udf-exporting-variable-between-two-define-function.html)
Carlo August 17, 2009 12:31
UDF: exporting a variable between two DEFINE function
Hi to all. Here's my problem:
I'm dealing with coal combustion; in my case I have also gas recirculation. In particular, I have an UDF that computes the outlet NOx mass fraction; this value must be imposed in an inlet (due to the recirculation). Any idea to do it? Here's my UDF:
#include "udf.h"
{
int zone_ID = 4;
face_t f;
real summ=0.;
real m;
real sumnox=0.;
real mnox;
real nox;
real sumarea=0.;
real area;
real Cnox;
real NV_VEC (A);
begin_f_loop(f,t)
{
F_AREA(A,f,t);
nox=C_POLLUT(f,t,1); /*Pollutant mass fraction Kg(NO)/Kg(g) in each face*/
mnox=F_U(f,t)*F_R(f,t)*NV_MAG(A)*6.28*nox; /*[Kg(NO)/Kg(g)]*[Kg(g)/s]=[Kg(NO)/s] in each face*/
sumnox += mnox; /*Kg(NO)/s in the whole domain*/
m=F_U(f,t)*F_R(f,t)*NV_MAG(A)*6.28;
summ += m; /*Kg(g)/s in the whole domain*/
area=NV_MAG(A)*6.28;
sumarea += area;
}
end_f_loop(f,t)
Cnox=sumnox/summ; /*[Kg(NO)/s]/[Kg(g)/s]*/
printf("massf: %g,%g,%g,%g\n", summ,sumnox,Cnox,sumarea);
}
real Cnox; /*Why do I have to declare it twice? Does fluent hold it?*/
DEFINE_PROFILE(noflux_b, t, nv)
{
face_t f;
begin_f_loop(f, t)
{
F_PROFILE(f,t,nv) = Cnox;
}
end_f_loop(f, t)
}
The ID=4 corresponds to the outlet. Am I doing it right? After having compiled and loaded the UDF, I hooked the noxflux function (DEFINE_ADJUST) whereas in the boundary condition of the considered inlet, in the NO mass fraction menu I set the noxflux_b function (DEFINE_PROFILE).
I'm not sure about this way to proceed. Indeed I don't know if the second DEFINE holds the Cnox variable value or if It clears it. I think this because if I don't declare "real Cnox" before starting the second DEFINE, I get the error "undeclared...etc", as if it had seen the Cnox variable for the first time! Any Idea?
Thanks!!
Allan Walsh August 18, 2009 13:25
What about using an external variable that is available in both your define macros? I must admit that I have only glanced at your macros, but it is something we use for what sounds like a similar issue.
Carlo August 18, 2009 13:55
Thanks!
Ok. The macro works really fine until the computing of the Cnox variable which is correctly computed and printed. Then I think that fluent doesn't get the value in the inlet (I'm aware of this when I plot the NOx mass fraction in it, where I get zero).
If you might be so gentle to explain me something more about external variables, I would be glad...you have to bring in mind that the variable has to be something computed through an UDF.
Allan Walsh August 19, 2009 12:21
Variables that you define in a macro will only have meaning in that macro. However, you can define a variable outside the macro, or macros, so that it can be used globally. So, in Fluent, in your c++ code, after your \$ include statements and before your define macros you would have some lines like extern real varx and extern real vary. Then you can use varx and vary in whatever macros you have defined.
This is just basic c++, not even sure if it addresses your issue, but it is another tool.
All times are GMT -4. The time now is 03:25. | 956 | 3,402 | {"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-2015-40 | longest | en | 0.737069 |
http://www.soest.hawaii.edu/oceanography/researchers/francois/RESEARCH/PROJECTS/PHYSICAL_OCEANOGRAPHY/EDDY_DRIVEN_MEAN_CIRCULATION/private/test-of-the-PV-test.html | 1,516,433,252,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084889473.61/warc/CC-MAIN-20180120063253-20180120083253-00232.warc.gz | 549,761,349 | 3,384 | # Test of the PV testΒΆ
I have been concerned with the potential vorticity (PV) test: compute the PV along the trajectories of parcels from local velocity and thickness field and compare it to the PV computed from integration of the PV equation. I did that for one of the experiment with Laplacian dissipation and realistic amplitude of the wave field and the test failed suggesting that the model does not satisfy the PV equation.
Before concluding this, however, I wanted to test my test. To do so, I performed a numerical simulation with (very) weak forcing: This should make any advective term negligible and the problem should be close to linear. Unfortunately or fortunately, the test of the test failed.
The amplitude of the flow is in the order of 1e-4 m/s and the flow averaged over the 100-day cycle is of the order of 1e-7 m/s. The trajectories of six parcels are shown in Figs. 1 and 2: The parcels barely move around and they all form a closed trajectory except for the parcel at 10E and 30N –why? I do not know.
Figure 1: 100-day averaged zonal velocity in the middle layer. Supersimposed are the trajectories (black) of 6 parcels during one 100-day cycle.
Figure 2: Trajectories of the six parcels shown in Fig. 1.
The PV computed from the local velocity and from the integration of the PV equation are shown in Fig. 3. In this figure, the latter has been estimated in two ways: one by computing the dissipative term from the u and v field and one by computing it directly from one model output, diffu and diffv. The two estimates are nearly the same suggesting that I know what I am plotting: The term is KH*L(zeta)/h, where KH is the diffusion coefficient (500 m2/s in this case), L is the Laplacian operator, zeta is the relative vorticity and h the thickness. The PV is defined as (f+zeta)/h where f is the Coriolis parameter.
Figure 3: PV during one cycle and along the trajectories of the six parcels shown in Figs. 1 and 2: from the local velocity field (blue), from the integration of the dissipative term calculated from u and v (red dots), from the integration of the dissipative term calculated from diffu and diffv, outputs of the model (red dashes).
The final values and the overall evolution of the PV are recovered but not the exact evolution; it is not too far off but it is off.
The failure of the test suggests that I am doing a mistake in my test; I cannot believe that the model would not respect the PV equation in this nearly linear case. Except maybe the resolution is still too coarse? In this case, the same test applied to a simulation with a finer resolution should show a decrease in the difference between the two PV estimates. Any other idea? | 617 | 2,696 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2018-05 | latest | en | 0.936527 |
https://statisfaction.wordpress.com/2016/07/20/coupling-of-particle-filters-smoothing/ | 1,498,739,802,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128323970.81/warc/CC-MAIN-20170629121355-20170629141355-00412.warc.gz | 825,627,232 | 37,550 | # Statisfaction
## Coupling of particle filters: smoothing
Posted in Statistics by Pierre Jacob on 20 July 2016
Two trajectories made for each other.
Hi again!
In this post, I’ll explain the new smoother introduced in our paper Coupling of Particle Filters with Fredrik Lindsten and Thomas B. Schön from Uppsala University. Smoothing refers to the task of estimating a latent process $x_{0:T} = (x_0,\ldots, x_T)$ of length $T$, given noisy measurements of it, $y_{1:T} = (y_0,\ldots, y_T)$; the smoothing distribution refers to $p(dx_{0:T}|y_{1:T})$. The setting is state-space models (what else?!), with a fixed parameter assumed to have been previously estimated.
Our smoother builds upon two recent innovations: the first is the conditional particle filter of Andrieu, Doucet and Holenstein (2010), and the second one is a debiasing technique of Glynn and Rhee (2014). The conditional particle filter (CPF) is a Markov kernel on the space of trajectories. The CPF kernel leaves the smoothing distribution invariant, so it can be iterated to obtain an MCMC sample approximating the smoothing distribution. On the other hand, the debiasing method takes a Markov kernel $K$ and a function $h$ and spits out an unbiased estimator of the integral of $h$ with respect to the invariant distribution of $K$.
So here is the algorithm: we start with a pair of arbitrary trajectories, denoted by $X^{(0)}$ and $\tilde{X}^{(0)}$. We first apply a step of CPF to $X^{(0)}$, yielding $X^{(1)}$ (and we do nothing to $\tilde{X}^{(0)}$; bear with me!). Then we apply a “coupled CPF” kernel to the pair $(X^{(1)},\tilde{X}^{(0)})$, yielding a new pair $(X^{(2)},\tilde{X}^{(1)})$. What is a coupled CPF? It’s essentially a CPF kernel applied to both trajectories, using common random numbers and with a fancy resampling scheme as alluded to in the last post.
Then we iterate the coupled CPF kernel, yielding pairs $(X^{(3)},\tilde{X}^{(2)})$, $(X^{(4)},\tilde{X}^{(3)})$, etc. It’s just a Markov chain on the space of pairs of trajectories… BUT! at some step $\tau$, the two trajectories become identical: $X^{(\tau)} = \tilde{X}^{(\tau-1)}$. Tadaaaah! See the figure above, where the two trajectories meet in a few steps. And once the two trajectories meet, they stay together forever after (so cute). This is exciting because it means that $X^{(0)} + \sum_{n=1}^{\tau-1} (X^{(n)} - \tilde{X}^{(n-1)}) =X^{(0)} + \sum_{n=1}^\infty (X^{(n)} - \tilde{X}^{(n-1)})$ (just adding infinitely many zeros). Now it is easy to see that this infinite sum has expectation given precisely by $\mathbb{E}[x_{0:T}|y_{1:T}]$. Indeed, provided that we can swap limit and expectation, the expectation of the sum is $\mathbb{E}[X^{(0)}] + \sum_{n=1}^\infty (\mathbb{E}[X^{(n)}] -\mathbb{E}[\tilde{X}^{(n-1)}])$. Since $\mathbb{E}[X^{(n)}] = \mathbb{E}[\tilde{X}^{(n)}]$ by the coupling construction, the sum is telescopic and we are left with $\lim_{n\to\infty} \mathbb{E}[X^{(n)}] =\mathbb{E}[x_{0:T}|y_{1:T}]$, simply because the CPF chain itself is ergodic.
What’s the point? Instead of running a long MCMC chain, staring blankly at the chain to decide how many iterations are enough iterations, the above scheme can be run $R$ times completely in parallel. Each run takes a random but small number of steps, and produces an unbiased smoothing estimator. We can then average these estimators to get the final result, with accurate confidence intervals provided by our good friend the central limit theorem.
Thanks for reading, you deserve a Santana song on smoothing.
### 2 Responses
1. Adam said, on 26 July 2016 at 12:28
Dear Sir, Is there an example of the Coupling of particle filters in R code?
• Pierre Jacob said, on 26 July 2016 at 13:37
Adam, there is a github repository. You’ll find the link in the article. | 1,110 | 3,809 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 25, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-26 | longest | en | 0.771465 |
https://mail.astarmathsandphysics.com/index.php?option=com_content&view=article&id=554:algorithms&catid=95&tmpl=component&print=1&layout=default&Itemid=474 | 1,607,015,886,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141729522.82/warc/CC-MAIN-20201203155433-20201203185433-00019.warc.gz | 342,010,741 | 3,145 | ## Algorithms
An algorithm is a sequence of steps carried out in a certain order for solving a certain problem. There may be more than one way of solving the problem, of which an algorithm may be one. Not all problems can be solved by using an algorithm.
For example, we want to solve the quadratic equation
We could use the quadraticformulaIn this caseso Henceor
The above solution however was not obtained by the use of an algorithm. If we wanted to use an algorithm to solve the equation we could complete the square and makethe subject.
We can start by factorising with 3.
Use the identityNow write
We can collect the last two termsso
Now make x the subject. Add 13 over 36 to both sides.
Square root both sides, remembering that there are two square roots, one + and one -.
Subtractfrom both sides.
Both methods give the same answer, but only the second method is the result of an algorithm. | 198 | 907 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2020-50 | latest | en | 0.933104 |
https://brainmass.com/math/integrals/derivatives-and-integrals-140514 | 1,624,283,090,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488273983.63/warc/CC-MAIN-20210621120456-20210621150456-00181.warc.gz | 147,929,144 | 74,980 | Explore BrainMass
# Derivatives and integrals
Not what you're looking for? Search our solutions OR ask your own Custom question.
This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!
1.R is the region that lies between the curve y = (1 /( x2 + 4x + 5) ) and the x-axis from x = -3 to x = -1. Find:
(a) the area of R,
(b) the volume of the solid generated by revolving R around the y-axis.
(c) the volume of the solid generated by revolving R round the x-axis.
2.Evaluate: ∫ sinh6 x cosh xdx.
3.Given f(x) = csch-1 (1 / x2), find f'(x).
4.Evaluate ∫ ( (log3 x) / 2x) dx. | 192 | 643 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2021-25 | latest | en | 0.881702 |
https://sullivantm.com/physics_informed_neural_networks/ | 1,632,461,394,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057504.60/warc/CC-MAIN-20210924050055-20210924080055-00594.warc.gz | 572,773,331 | 6,615 | # Physics informed neural networks
Physics informed neural networks use additional loss terms based on the dynamics of the function being learned to improve performance for learning complex systems.
# Physics informed neural networks
PINNs can provide additional information about how the modeled dynamics should behave that isn’t present when trying to learn the surface function alone. Let’s say you have some complicated function $u(t,x)$, rather than trying to learn the outputs alone we augment the training objective with information about the dynamics of $u$ using partial derivatives. This provides an additional error signal to the deep learning model. Original paper: Physics Informed Neural Networks
$\begin{array}{l} u_t + u u_x - (0.01/\pi) u_{xx} = 0,\ \ \ x \in [-1,1],\ \ \ t \in [0,1],\newline u(0,x) = -\sin(\pi x),\newline u(t,-1) = u(t,1) = 0. \end{array}$
And define,
$f := u_t + u u_x - (0.01/\pi) u_{xx} = 0$
We can represent these in TensorFlow as
initializer = tf.keras.initializers.GlorotNormal()
inputs = layers.Input(shape=(2,), dtype=tf.float64)
z = layers.Lambda(lambda X: 2.0*(X - lb)/(ub - lb) - 1.0)(inputs)
z = layers.Dense(20, activation="tanh", kernel_initializer=initializer)(z)
z = layers.Dense(20, activation="tanh", kernel_initializer=initializer)(z)
z = layers.Dense(20, activation="tanh", kernel_initializer=initializer)(z)
z = layers.Dense(20, activation="tanh", kernel_initializer=initializer)(z)
z = layers.Dense(20, activation="tanh", kernel_initializer=initializer)(z)
z = layers.Dense(20, activation="tanh", kernel_initializer=initializer)(z)
outputs = layers.Dense(1, kernel_initializer=initializer)(z)
U_model = tf.keras.Model(inputs=inputs, outputs=outputs)
U_model.summary()
def F_model(x,t):
t = tf.Variable(t)
x = tf.Variable(x)
u = tf.squeeze(tf.cast(U_model(tf.stack([x,t], 1)), tf.float64))
f = u_t + u*u_x - (0.01/tf.constant(np.pi, dtype=tf.float64))*u_xx
return f
# Training
The training uses a combined dataset of true $u(t,x)$, boundary points, and solutions to $f(t,x)$ which are evaluated simultaneously, and the total loss is a combination of loss due to $u$ and $f$.
$\text{Loss} = \text{MSE}_u + \text{MSE}_f$
With this loss, we simply train a neural network taking the gradient of the weights with respect to this combined loss.
def train_step(X_u, u_true, X_f):
u_pred = tf.cast(U_model(X_u), tf.float64)
f = F_model(X_f[:,0], X_f[:,1])
loss = tf.losses.MeanSquaredError()(u_true, u_pred) + tf.reduce_mean(tf.square(f))
return loss
# Example
Let’s see an example of PINNs solving a difficult problem using the Burgers’ equation. The notebook for this example can be found here. The expected value of $u(t,x)$ is below. The training set uses 10,000 samples to compute $f(t,x)$ and 50 to compute $u(t,x)$ directly.
## Naïve model
This first example is a neural network with 6 layers of 20 neurons per hidden layer trained without the $f(t,x)$ dataset, simply trying to predict $u(t,x)$. The model is trained using Adam with a learning rate of 0.01, $\beta_1=0.99$ and $\epsilon = 0.1$ for 1000 epochs. The result captures some of the trends in the data but overall performs poorly when tested on the full dataset.
## PINN
Now compare this with the same model trained with the partial derivative information and an L-BFGS optimizer replicating the approach used in the paper. The training data in this example comes mostly from the f dataset.
This next example uses the same neural network and optimizer as the naïve model, but this time the f dataset with partial derivates is included in the training. While the authors mention training a PINN can be conducted using traditional minibatch methods. This initial stab didn’t work immediately. An L-BFGS approach is likely better with this small dataset, though it can quickly become difficult to compute with more data. | 986 | 3,863 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2021-39 | latest | en | 0.70451 |
https://bookdown.org/egarpor/NP-UC3M/app-kde-ci.html | 1,632,432,953,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057447.52/warc/CC-MAIN-20210923195546-20210923225546-00269.warc.gz | 200,449,923 | 13,126 | # A Confidence intervals for the density function
Obtaining a Confidence Interval (CI) for $$f(x)$$ is a challenging task. Thanks to the bias results273 seen in Section 2.3, we know that the kde is biased for finite sample sizes and it is only asymptotically unbiased when $$h\to0$$. This bias is called the smoothing bias and, in essence, complicates the obtention of CIs for $$f(x)$$, but not for $$(K_h*f)(x)$$. Let’s see the differences between these two objects with an illustrative example.
Some well-known facts for normal densities (see Appendix C in ) are:
\begin{align} (\phi_{\sigma_1}(\cdot-\mu_1)*\phi_{\sigma_2}(\cdot-\mu_2))(x)&=\phi_{(\sigma_1^2+\sigma_2^2)^{1/2}}(x-\mu_1-\mu_2),\tag{A.1}\\ \int\phi_{\sigma_1}(x-\mu_1)\phi_{\sigma_2}(x-\mu_2)\,\mathrm{d}x&=\phi_{(\sigma_1^2+\sigma_2^2)^{1/2}}(\mu_1-\mu_2),\\ \phi_{\sigma}(x-\mu)^r&=\frac{1}{\sigma^{r-1}}(2\pi)^{(1-r)/2}\phi_{\sigma/ r^{1/2}}(x-\mu)\frac{1}{r^{1/2}}.\tag{A.2} \end{align}
Consequently, if $$K=\phi$$ (i.e., $$K_h=\phi_h$$) and $$f(\cdot)=\phi_\sigma(\cdot-\mu)$$:
\begin{align} (K_h*f)(x)&=\phi_{(h^2+\sigma^2)^{1/2}}(x-\mu),\tag{A.3}\\ (K_h^2*f)(x)&=\left(\frac{1}{(2\pi)^{1/2}h}\phi_{h/2^{1/2}}/2^{1/2}*f\right)(x)\nonumber\\ &=\frac{1}{2\pi^{1/2}h}\left(\phi_{h/2^{1/2}}*f\right)(x)\nonumber\\ &=\frac{1}{2\pi^{1/2}h}\phi_{(h^2/2+\sigma^2)^{1/2}}(x-\mu).\tag{A.4} \end{align}
Thus, the exact expectation of the kde for estimating the density of a $$\mathcal{N}(\mu,\sigma^2)$$ is precisely the density of a $$\mathcal{N}(\mu,\sigma^2+h^2)$$. Clearly, when $$h\to0$$, we can see how the bias disappears. Removing this finite-sample size bias is not simple: if the bias is expanded, $$f''$$ appears. Hence, to attempt to unbias $$\hat{f}(\cdot;h)$$ we have to estimate $$f''$$, which is a harder task than estimating $$f$$. As seen in Section 3.2, taking second derivatives on the kde does not work out-of-the-box, since the bandwidths for estimating $$f$$ and $$f''$$ scale differently.
The previous deadlock can be solved if we limit our ambitions. Rather than constructing a confidence interval for $$f(x)$$, we can construct it for $$\mathbb{E}[\hat{f}(x;h)]=(K_h*f)(x)$$. There is nothing wrong with this change of view, as long as we are explicitly report the CI as that for $$(K_h*f)(x)$$ and not for $$f(x)$$.
The building block for the CI for $$\mathbb{E}[\hat{f}(x;h)]=(K_h*f)(x)$$ is the first result274 in Theorem 2.2, which concludes that
\begin{align*} \sqrt{nh}(\hat{f}(x;h)-\mathbb{E}[\hat{f}(x;h)])\stackrel{d}{\longrightarrow}\mathcal{N}(0,R(K)f(x)). \end{align*}
Plugging $$\hat{f}(x;h)=f(x)+O_\mathbb{P}\big(h^2+(nh)^{-1/2}\big)=f(x)(1+o_\mathbb{P}(1))$$ (see Exercise 2.10) as an estimate for $$f(x)$$ in the variance, we have by the Slutsky’s theorem that
\begin{align*} \sqrt{\frac{nh}{R(K)\hat{f}(x;h)}}&(\hat{f}(x;h)-\mathbb{E}[\hat{f}(x;h)])\\ &=\sqrt{\frac{nh}{R(K)f(x)}}(\hat{f}(x;h)-\mathbb{E}[\hat{f}(x;h)])(1+o_\mathbb{P}(1))\\ &\stackrel{d}{\longrightarrow}\mathcal{N}(0,1). \end{align*}
Therefore, an asymptotic $$100(1-\alpha)\%$$ confidence interval for $$\mathbb{E}[\hat{f}(x;h)]$$ that can be straightforwardly computed is
\begin{align} I=\left(\hat{f}(x;h)\pm z_{\alpha/2}\sqrt{\frac{R(K)\hat{f}(x;h)}{nh}}\right).\tag{A.5} \end{align}
Remark. Several points regarding (A.5) require proper awareness:
1. As announced, the CI is meant for $$\mathbb{E}[\hat{f}(x;h)]=(K_h*f)(x)$$, not $$f(x)$$, as it does not account for the smoothing bias.
2. This is a pointwise CI: $$\mathbb{P}\left[\mathbb{E}[\hat{f}(x;h)]\in I\right]\approx 1-\alpha$$ for each $$x\in\mathbb{R}$$. That is, $$\mathbb{P}\left[\mathbb{E}[\hat{f}(x;h)]\in I,\,\forall x\in\mathbb{R}\right]\neq 1-\alpha$$.
3. The CI relies on an approximation of $$f(x)$$ in the variance, done by $$\hat{f}(x;h)=f(x)+O_\mathbb{P}\big(h^2+(nh)^{-1/2}\big)$$. Additionally, the convergence to a normal distribution happens at rate $$\sqrt{nh}$$. Hence, both $$h$$ and $$nh$$ need to be small and large, respectively, for a good coverage.
4. The CI is built using a deterministic bandwidth $$h$$ (i.e., not data-driven), which is not usually the case in practice. If a bandwidth selector is employed, the coverage may be affected, specially for small $$n$$.
We illustrate the construction of the CI in (A.5) for the situation where the reference density is a $$\mathcal{N}(\mu,\sigma^2)$$ and the normal kernel is employed. This allows to use (A.3) and (A.4), in combination with (2.11) and (2.12), to obtain
\begin{align*} \mathbb{E}[\hat{f}(x;h)]&=\phi_{(h^2+\sigma^2)^{1/2}}(x-\mu),\\ \mathbb{V}\mathrm{ar}[\hat{f}(x;h)]&=\frac{1}{n}\left(\frac{\phi_{(h^2/2+\sigma^2)^{1/2}}(x-\mu)}{2\pi^{1/2}h}-(\phi_{(h^2+\sigma^2)^{1/2}}(x-\mu))^2\right). \end{align*}
These exact results are useful to benchmark (A.5) with the asymptotic CI that employs the exact variance of $$\hat{f}(x;h)$$:
\begin{align} \left(\hat{f}(x;h)\pm z_{\alpha/2} \sqrt{\mathbb{V}\mathrm{ar}[\hat{f}(x;h)]}\right).\tag{A.6} \end{align}
The following chunk of code evaluates the proportion of times that $$\mathbb{E}[\hat{f}(x;h)]$$ belongs to $$I$$ for each $$x\in\mathbb{R}$$, both with estimated or known variance in the asymptotic distribution.
# R(K) for a normal
Rk <- 1 / (2 * sqrt(pi))
# Generate a sample from a N(mu, sigma^2)
n <- 100
mu <- 0
sigma <- 1
set.seed(123456)
x <- rnorm(n = n, mean = mu, sd = sigma)
# Compute the kde (NR bandwidth)
kde <- density(x = x, from = -4, to = 4, n = 1024, bw = "nrd")
# Selected bandwidth
h <- kde$bw # Estimate the variance var_kde_hat <- kde$y * Rk / (n * h)
# True expectation and variance (because the density is a normal)
E_kde <- dnorm(x = kde$x, mean = mu, sd = sqrt(sigma^2 + h^2)) var_kde <- (dnorm(kde$x, mean = mu, sd = sqrt(h^2 / 2 + sigma^2)) /
(2 * sqrt(pi) * h) - E_kde^2) / n
# CI with estimated variance
alpha <- 0.05
z_alpha2 <- qnorm(1 - alpha / 2)
ci_low_1 <- kde$y - z_alpha2 * sqrt(var_kde_hat) ci_up_1 <- kde$y + z_alpha2 * sqrt(var_kde_hat)
# CI with known variance
ci_low_2 <- kde$y - z_alpha2 * sqrt(var_kde) ci_up_2 <- kde$y + z_alpha2 * sqrt(var_kde)
# Plot estimate, CIs and expectation
plot(kde, main = "Density and CIs", ylim = c(0, 1))
lines(kde$x, ci_low_1, col = "gray") lines(kde$x, ci_up_1, col = "gray")
lines(kde$x, ci_low_2, col = "gray", lty = 2) lines(kde$x, ci_up_2, col = "gray", lty = 2)
lines(kde$x, E_kde, col = "red") legend("topright", legend = c("Estimate", "CI estimated var", "CI known var", "Smoothed density"), col = c("black", "gray", "gray", "red"), lwd = 2, lty = c(1, 1, 2, 1)) The above experiment shows the CI, but it does not give any insight into the effective coverage of the CI. The following simulation exercise precisely addresses this issue. As seen from Figure A.2, the estimation of the kde’s variance has a considerable impact in the coverage of the CI in the low-density regions, and it is reasonable fine for the higher density regions. # Simulation setting n <- 200; h <- 0.15 mu <- 0; sigma <- 1 # Normal parameters M <- 5e2 # Number of replications in the simulation n_grid <- 512 # Number of x's for computing the kde alpha <- 0.05; z_alpha2 <- qnorm(1 - alpha / 2) # alpha for CI # Compute expectation and variance kde <- density(x = 0, bw = h, from = -4, to = 4, n = n_grid) # Just for kde$x
E_kde <- dnorm(x = kde$x, mean = mu, sd = sqrt(sigma^2 + h^2)) var_kde <- (dnorm(kde$x, mean = mu, sd = sqrt(h^2 / 2 + sigma^2)) /
(2 * sqrt(pi) * h) - E_kde^2) / n
# For storing if the mean is inside the CI
inside_ci_1 <- inside_ci_2 <- matrix(nrow = M, ncol = n_grid)
# Simulation
set.seed(12345)
for (i in 1:M) {
# Sample & kde
x <- rnorm(n = n, mean = mu, sd = sigma)
kde <- density(x = x, bw = h, from = -4, to = 4, n = n_grid)
sd_kde_hat <- sqrt(kde$y * Rk / (n * h)) # CI with estimated variance ci_low_1 <- kde$y - z_alpha2 * sd_kde_hat
ci_up_1 <- kde$y + z_alpha2 * sd_kde_hat # CI with known variance ci_low_2 <- kde$y - z_alpha2 * sqrt(var_kde)
ci_up_2 <- kde$y + z_alpha2 * sqrt(var_kde) # Check if for each x the mean is inside the CI inside_ci_1[i, ] <- E_kde > ci_low_1 & E_kde < ci_up_1 inside_ci_2[i, ] <- E_kde > ci_low_2 & E_kde < ci_up_2 } # Plot results plot(kde$x, colMeans(inside_ci_1), ylim = c(0.25, 1), type = "l",
main = "Empirical coverage of CIs", xlab = "x", ylab = "Coverage")
lines(kde\$x, colMeans(inside_ci_2), col = 4)
abline(h = 1 - alpha, col = 2)
abline(h = 1 - alpha + c(-1, 1) * qnorm(0.975) *
sqrt(alpha * (1 - alpha) / M), col = 2, lty = 2)
legend(x = "bottom", legend = c("CI estimated var", "CI known var",
"Nominal level",
"95% CI for the nominal level"),
col = c(1, 4, 2, 2), lwd = 2, lty = c(1, 1, 1, 2))
Exercise A.1 Explore the coverage of the asymptotic CI for varying values of $$h$$. To that end, adapt the previous code to work in a manipulate environment like the example given below.
# Sample
x <- rnorm(100)
# Simple plot of kde for varying h
manipulate::manipulate({
kde <- density(x = x, from = -4, to = 4, bw = h)
plot(kde, ylim = c(0, 1), type = "l", main = "")
curve(dnorm(x), from = -4, to = 4, col = 2, add = TRUE)
rug(x)
}, h = manipulate::slider(min = 0.01, max = 2, initial = 0.5, step = 0.01))
Exercise A.2 (Exercise 6.9.5 in ) Data on the salaries of the chief executive officer of $$60$$ companies is available at http://lib.stat.cmu.edu/DASL/Datafiles/ceodat.html (alternative link). Investigate the distribution of salaries using a kde. Use $$\hat{h}_\mathrm{LSCV}$$ to choose the amount of smoothing. Also, consider $$\hat{h}_\mathrm{RT}$$. There appear to be a few bumps in the density. Are they real? Use confidence bands to address this question. Finally, comment on the resulting estimates.
### References
Wand, M. P., and M. C. Jones. 1995. Kernel Smoothing. Vol. 60. Monographs on Statistics and Applied Probability. London: Chapman & Hall. https://doi.org/10.1201/b14876.
———. 2006. All of Nonparametric Statistics. Springer Texts in Statistics. New York: Springer. https://doi.org/10.1007/0-387-30623-4.
1. For example, $$\mathbb{E}[\hat{f}(x;h)]=(K_h*f)(x)$$.↩︎
2. If we wanted to obtain a CI with the second result in Theorem 2.2 we would need to estimate $$f''(x)$$.↩︎ | 3,665 | 10,154 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 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": 4, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2021-39 | latest | en | 0.75744 |
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