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https://learn.careers360.com/school/question-1-millimole-of-caso4-weighs-a136g-b136g-c0136g-d00136g-48707/ | 1,679,922,088,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948632.20/warc/CC-MAIN-20230327123514-20230327153514-00419.warc.gz | 417,082,423 | 117,447 | #### 1 millimole of CaSO4 weighs a)136g b)13.6g c)0.136g d)0.0136g
1 mole = 1000 milimoles
Molar mass of CaSO= 136g/mol
1 mole = 136g of CaSO4
1 milimole of CaSO= 136/1000 = 0.136g | 88 | 184 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2023-14 | latest | en | 0.626512 |
https://austinattorney.mobi/ic-lm7812-datasheet-88/ | 1,643,317,435,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320305288.57/warc/CC-MAIN-20220127193303-20220127223303-00528.warc.gz | 169,730,095 | 11,796 | IC LM7812 DATASHEET PDF
Details, datasheet, quote on part number: LM for the heat sinking provided, the thermal shutdown circuit takes over preventing the IC from overheating. LM datasheet, LM pdf, LM data sheet, datasheet, data sheet, pdf, BayLinear, A Positive Voltage Regulator. LM FEATURES. ·Output current in excess of A. ·Output voltage of 12V Tj=25℃ (Vi= 19V, IO=A, Ci= μF, CO= μF unless otherwise specified).
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The voltage regulator produces constant output voltage of 5v with operating range input of v. So, if you are looking for a variable voltage regulator to deliver current up to 1.
If we have a system with input 15 volts and output current required datashheet.
Sir i made a voltage regulator using ic for powering my microcontroller but my chip isnt working Plz tell me where i m wrong. This difference between the input ml7812 output voltage is released as heat.
All About IC 7805 | Voltage Regulator
We are all non native english speaking people here. So twice the energy, that is actually utilized is wasted. The value llm7812 be calculated using the below formulae. But, the IC suffers from heavy heat loss hence a Heat sink is recommended for projects that consume more current. The error output, in turn, decreases or increases the current through the output transistor.
More tutorials available at learning resources. This change is amplified by Q7 and Q8, generating the error output. I think your articles have been written by someone who is not a native English-speaker. Apart from using it as a variable voltage regulator, it can also be used as a fixed voltage regulator, current limiter, Battery charger, AC voltage regulator and even as an adjustable current regulator. They provide a constant output voltage for a varied input voltage.
BRUCZKOWSKI BEZSENNO W TOKIO PDF
Please enter your comment! The two capacitors are not necessarily required and can be omitted if you are not concerned about line noise. Here, the input voltage can be anywhere between 9VV, and the output voltage dqtasheet be adjusted using the value of resistance R1 and R2.
Perhaps an editor would clear this up. In our circuit, a motor is connected as a load which consumes around mA you can connect any load up to 1. Log into your account. Datasheet have 78M05 IC in D pak package. The bandgap reference yellow keeps the voltage stable. LM is a 3-terminal regulator IC and it is very simple to use.
This 7 Watts will be dissipated as heat. This closes the negative feedback loop controlling the output voltage. Now to make it act as a variable voltage regulator we have to set variable voltages at pin 1 which can be done by using a potentiometer in icc potential divider. The input capacitor 0. This pin is neutral for equally the input and output.
LM Voltage Regulator Datasheet.
If the regulator does not have a heat sink to dissipate this heat, it can get destroyed and malfunction. TL — Programmable Reference Voltage. It has many application circuits in its datasheet, but this IC is known for being used as a variable voltage regulator. The bypass capacitors help reduce AC ripple.
LM Datasheet pdf – A Positive Voltage Regulator – BayLinear
One notable drawback of this IC is that it has a voltage drop of about 2. A resource for professional design engineers. I have used a full bridge rectifier along wd a transformer i tried to connect. The voltage divider blue scales down the voltage on the output pin for use by the bandgap reference.
JADUAL WAKTU SOLAT KUCHING 2015 PDF
Adjustable 3-terminal positive voltage regulator Output voltage datasbeet be set to range from 1.
The xx in 78xx indicates the output voltage it provides. In this pin where the ground is given. In our case the IC is an iconic regulator IC that finds its application lm78112 most of the projects. Hope it suits your requirements.
Voltage Regulator IC: Pinout, Diagrams, Equivalent & Datasheet
Output of 5v from ac to dc wall mobile charger something says 5v mA output 2. If not, I am looking for an IC which has Operating range from v and constant 5v output.
On the other hand, energy actually being used is:. Voltage regulators are very common in electronic circuits. You have entered an incorrect email address!
It takes the scaled output voltage as input Q1 and Q6 and provides an error signal to Q7 for indication if the voltage is too high or low. If the output voltage is correct 5Vthen the voltage divider provides 3. Also they should be of ceramic type, since ceramic capacitors are faster than electrolytic. The greater the difference between the input and output voltage, more the heat generated.
The values of capacitors can also be changed slightly. | 1,121 | 4,944 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2022-05 | latest | en | 0.860218 |
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# Remainder Theorem & Algebraic Identities Notes | Study Mathematics (Maths) Class 9 - Class 9
## Document Description: Remainder Theorem & Algebraic Identities for Class 9 2022 is part of Mathematics (Maths) Class 9 preparation. The notes and questions for Remainder Theorem & Algebraic Identities have been prepared according to the Class 9 exam syllabus. Information about Remainder Theorem & Algebraic Identities covers topics like Remainder Theorem Definition, Algebraic Identities and Remainder Theorem & Algebraic Identities Example, for Class 9 2022 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for Remainder Theorem & Algebraic Identities.
Introduction of Remainder Theorem & Algebraic Identities in English is available as part of our Mathematics (Maths) Class 9 for Class 9 & Remainder Theorem & Algebraic Identities in Hindi for Mathematics (Maths) Class 9 course. Download more important topics related with notes, lectures and mock test series for Class 9 Exam by signing up for free. Class 9: Remainder Theorem & Algebraic Identities Notes | Study Mathematics (Maths) Class 9 - Class 9
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Remainder Theorem is an approach of Euclidean division of polynomials. According to this theorem, if we divide a polynomial P(x) by a factor (x – a); that isn’t essentially an element of the polynomial; you will find a smaller polynomial along with a remainder. This remainder that has been obtained is actually a value of P(x) at x = a, specifically P(a). So basically, x -a is the divisor of P(x) if and only if P(a) = 0. It is applied to factorize polynomials of each degree in an elegant manner.
For example: if f(a) = a3-12a2-42 is divided by (a-3) then the quotient will be a2-9a-27 and the remainder is -123.
if we put, a-3 = 0
then a = 3
Hence, f(a) = f(3) = -123
Thus, it satisfies the remainder theorem.
Remainder Theorem Definition
The Remainder Theorem begins with a polynomial say p(x), where “p(x)” is some polynomial p whose variable is x. Then as per theorem, dividing that polynomial p(x) by some linear factor x – a, where a is just some number. Here go through a long polynomial division, which results in some polynomial q(x) (the variable “q” stands for “the quotient polynomial”) and a polynomial remainder is r(x). It can be expressed as:
p(x)/x-a = q(x) + r(x)
Factor Theorem
Factor Theorem is generally applied to factoring and finding the roots of polynomial equations. It is the reverse form of the remainder theorem. Problems are solved based on the application of synthetic division and then to check for a zero remainder.
When p(x) = 0 then y-x is a factor of the polynomial Or if we consider the other way, then When y-x is a factor of the polynomial then p(x) =0
Remainder Theorem Proof
Theorem functions on an actual case that a polynomial is comprehensively dividable, at least one time by its factor in order to get a smaller polynomial and ‘a’ remainder of zero. This acts as one of the simplest ways to determine whether the value ‘a’ is a root of the polynomial P(x).
That is when we divide p(x) by x-a we obtain
p(x) = (x-a)·q(x) + r(x),
as we know that Dividend = (Divisor × Quotient) + Remainder
But if r(x) is simply the constant r (remember when we divide by (x-a) the remainder is a constant)…. so we obtain the following solution, i.e
p(x) = (x-a)·q(x) + r
Observe what happens when we have x equal to a:
p(a) = (a-a)·q(a) + r
p(a) = (0)·q(a) + r
p(a) = r
Hence, proved.
Steps to Divide a Polynomial by a Non-Zero Polynomial
• First, arrange the polynomials (dividend and divisor) in the decreasing order of its degree
• Divide the first term of the dividend by the first term of the divisor to produce the first term of the quotient
• Multiply the divisor by the first term of the quotient and subtract this product from the dividend, to get the remainder.
• This remainder is the dividend now and divisor will remain same
• Again repeat from the first step, until the degree of the new dividend is less than the degree of the divisor.
Remainder Theorem of Polynomial
Let us understand the remainder theorem in polynomials with the example given below:
Divide 3x3 + x2 + 2x + 5 by x + 1.
Solution:
From the given,
Dividend = p(x) = 3x3 + x2 + 2x + 5
Divisor = g(x) = (x + 1)
Here, quotient = q(x) = 3x2 – 2x + 4
Remainder = r(x) = 1
Verification:
Given, the divisor is (x + 1), i.e. it is a factor of the given polynomial p(x).
Let x + 1 = 0
x = -1
Substituting x = -1 in p(x),
p(-1) = 3(-1)3 + (-1)2 + 2(-1) + 5
= 3(-1) + 1 – 2 + 5
= -3 + 4
= 1
Remainder = Value of p(x) at x = -1.
Hence proved the remainder theorem.
Alternatively,
p(x) = (x – a)·q(x) + r
Observe what happens when we have x equal to a:
p(a) = (a – a)·q(a) + r
Substituting the values,
p(-1) = [-1 – (-1)]·q(-1) + (-1)
p(-1) = 0.q(-1) – 1
p(-1) = -1
p(-1) = remainder
Hence proved.
Remainder Theorem Problems
Consider the following example:-
Example- Determine that x = 1 is a root of P(x),
Explanation:
It suggests that x = 1 may be a root of P(x), and (x – 1) may be a factor of P(x)
Then if we tend to divide synthetically from P(x) by (x – 1), we will get a new smaller polynomial and a remainder of zero.
Example: Find the root of the polynomial x2−3x–4.
Solution: x2–3x–4
f(4)=42–3(4)–4
f(4)=16–16=0
So, (x-4) must be a factor of x2–3x–4
Example: Find the remainder when t3–2t2+t+1 is divided by t – 1.
Solution: Here, p(t)=t3–2t2+t+1, and the zero of t – 1 is 1.
∴ p (1) = (1)3 – 2(1)2 + 1 + 1= 2
By the Remainder Theorem, 2 is the remainder when t3–2t2+t+1 is divided by t – 1.
Euler Remainder Theorem
Euler’s theorem states that if n and X are two co-prime positive integers, then
Xφ(n) = 1 (mod n)
where, φ(n) is Euler’s function or Euler’s totient function, which is equal to;
φ(n) = n (1-1/a).(1-1/b).(1-1/c)
where, n is a natural number, such that n = ap. bq . cr,
Here, a, b, c are prime factors of n and p, q, r are positive integers.
Example: Find the Euler totient function of 35.
Solution: The factors of 35 are as follows:
35=5×7
Thus the totient function of 35 is 24.
Questions of the form: ma/n
Example: Find the remainder when 376 is divided by 35.
Solution: Here m = 3, a = 76 and n = 35,
In an example above we have already found the totient function of 35, which is equal to 24.
Remainder of
Remaining power is 4, which when divided by 35 given the resultant remainder. which is,
Thus the remainder comes out to be 11.
Algebraic Identities
The algebraic equations which are valid for all values of variables in them are called algebraic identities. They are also used for the factorization of polynomials. In this way, algebraic identities are used in the computation of algebraic expressions and solving different polynomials. You have already learned about a few of them in the junior grades. In this article, we will recall them and introduce you to some more standard algebraic identities, along with examples.
Standard Algebraic Identities List
All the standard Algebraic Identities are derived from the Binomial Theorem, which is given as:
Some Standard Algebraic Identities list are given below:
Identity I: (a + b)2 = a2 + 2ab + b2
Identity II: (a – b)2 = a2 – 2ab + b2
Identity III: a2 – b2= (a + b)(a – b)
Identity IV: (x + a)(x + b) = x2 + (a + b) x + ab
Identity V: (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
Identity VI: (a + b)3 = a3 + b3 + 3ab (a + b)
Identity VII: (a – b)3 = a3 – b3 – 3ab (a – b)
Identity VIII: a3 + b3 + c– 3abc = (a + b + c)(a2 + b2 + c2 – ab – bc – ca)
Example 1: Find the product of (x + 1)(x + 1) using standard algebraic identities.
Solution: (x + 1)(x + 1) can be written as (x + 1)2. Thus, it is of the form Identity I where a = x and b = 1. So we have,
(x + 1)2 = (x)2 + 2(x)(1) + (1)2 = x2 + 2x + 1
Example 2: Factorise (x4 – 1) using standard algebraic identities.
Solution: (x4 – 1) is of the form Identity III where a = x2 and b = 1. So we have,
(x4 – 1) = ((x2)2– 12) = (x2 + 1)(x2 – 1)
The factor (x2 – 1) can be further factorised using the same Identity III where a = x and b = 1. So,
(x4 – 1) = (x2 + 1)((x)2 –(1)2) = (x2 + 1)(x + 1)(x – 1)
Eample 3: Factorise 16x2 + 4y2 + 9z2 – 16xy + 12yz – 24zx using standard algebraic identities.
Solution: 16x2 + 4y2 + 9z2– 16xy + 12yz – 24zx is of the form Identity V. So we have,
16x2 + 4y2 + 9z2 – 16xy + 12yz – 24zx = (4x)2 + (-2y)2 + (-3z)2 + 2(4x)(-2y) + 2(-2y)(-3z) + 2(-3z)(4x)= (4x – 2y – 3z)2 = (4x – 2y – 3z)(4x – 2y – 3z)
Example 4: Expand (3x – 4y)using standard algebraic identities.
Solution: (3x– 4y)3 is of the form Identity VII where a = 3x and b = 4y. So we have,
(3x – 4y)3 = (3x)3 – (4y)3– 3(3x)(4y)(3x – 4y) = 27x3 – 64y3 – 108x2y + 144xy2
Example 5: Factorize (x3 + 8y3 + 27z3 – 18xyz) using standard algebraic identities.
Solution: (x3 + 8y3 + 27z3 – 18xyz)is of the form Identity VIII where a = x, b = 2y and c = 3z. So we have,
(x3 + 8y3 + 27z3 – 18xyz) = (x)3 + (2y)3 + (3z)3 – 3(x)(2y)(3z)= (x + 2y + 3z)(x2 + 4y2 + 9z2 – 2xy – 6yz – 3zx)
The document Remainder Theorem & Algebraic Identities Notes | Study Mathematics (Maths) Class 9 - Class 9 is a part of the Class 9 Course Mathematics (Maths) Class 9.
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## Worksheet: TI-Nspire Mini-Tutorial: (CAS) Finding the Real and Complex Roots of a Cubic
Worksheet: TI-Nspire Mini-Tutorial: (CAS) Finding the Real and Complex Roots of a Cubic
This is part of a collection of math worksheets on the use of the TI-Nspire graphing calculator. Each worksheet supports a companion TI-Nspire Mini-Tutorial video. It provides all the keystrokes for the activity.
Polynomial Expressions and Polynomial Functions and Equations
## Worksheet: TI-Nspire Mini-Tutorial: (CAS) Solving a Cubic Algebraically and Graphically
Worksheet: TI-Nspire Mini-Tutorial: (CAS) Solving a Cubic Algebraically and Graphically
This is part of a collection of math worksheets on the use of the TI-Nspire graphing calculator. Each worksheet supports a companion TI-Nspire Mini-Tutorial video. It provides all the keystrokes for the activity.
Polynomial Expressions and Polynomial Functions and Equations
## Worksheet: TI-Nspire Mini-Tutorial: Composite Functions, Linear to Cubic
Worksheet: TI-Nspire Mini-Tutorial: Composite Functions, Linear to Cubic Composite Functions
## Worksheet: TI-Nspire Mini-Tutorial: Finding the Zeros of a Cubic Function Using Function Intersection
Worksheet: TI-Nspire Mini-Tutorial: Finding the Zeros of a Cubic Function Using Function Intersection
This is part of a collection of math worksheets on the use of the TI-Nspire graphing calculator. Each worksheet supports a companion TI-Nspire Mini-Tutorial video.
Polynomial Expressions and Polynomial Functions and Equations | 378 | 1,664 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2023-40 | latest | en | 0.71014 |
http://freakonometrics.blog.free.fr/index.php?post/2012/02/14/MAT8886-the-Dirichlet-distribution-(over-the-simplex) | 1,398,355,659,000,000,000 | text/html | crawl-data/CC-MAIN-2014-15/segments/1398223206647.11/warc/CC-MAIN-20140423032006-00356-ip-10-147-4-33.ec2.internal.warc.gz | 130,354,849 | 6,951 | In the course, since we are still introducing some concepts of dependent distributions, we will talk about the Dirichlet distribution, which is a distribution over the simplex of . Let denote the Gamma distribution with density (on )
Let denote independent random variables, with . Then where
has a Dirichlet distribution with parameter . Note that has a distribution in the simplex of ,
and has density
We will write .
The density for different values of can be visualized below, e.g. , with some kind of symmetry,
or and , below
and finally, below,
Note that marginal distributions are also Dirichlet, in the sense that if
then
if , and if , then 's have Beta distributions,
See Devroye (1986) section XI.4, or Frigyik, Kapila & Gupta (2010) .This distribution might also be called multivariate Beta distribution. In R, this function can be used as follows
```> library(MCMCpack)
> alpha=c(2,2,5)
> x=seq(0,1,by=.05)
> vx=rep(x,length(x))
> vy=rep(x,each=length(x))
> vz=1-x-vy
> V=cbind(vx,vy,vz)
> D=ddirichlet(V, alpha)
> persp(x,x,matrix(D,length(x),length(x))```
(to plot the density, as figures above). Note that we will come back on that distribution later on so-called Liouville copulas (see also Gupta & Richards (1986)). | 327 | 1,243 | {"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} | 3.234375 | 3 | CC-MAIN-2014-15 | longest | en | 0.832034 |
https://www.penwatch.net/cms/500/ | 1,653,317,624,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662558030.43/warc/CC-MAIN-20220523132100-20220523162100-00483.warc.gz | 1,092,191,789 | 4,702 | # 500, for Hearts players
Posted: , Updated: Category: Games
The card game “500” is popular at work. It’s played every lunchtime, almost like a religious observance.
AustralianCardGames.com.au has a good, simple description of the rules. (Wikipedia attempts to cover every regional variant of the game, which is very confusing.)
I didn’t know how to play 500, but I did know how to play Hearts, which is sort of similar.
The most important differences, relative to Hearts, are listed below.
# 500 is played in pairs
In hearts, every player scores individually.
A game of 500 always involves four players, playing as two partnerships.
The partners sit opposite each other at the table.
# Bidding and Scoring
The objective of hearts is to avoid winning tricks, by playing low cards. (i.e. to avoid taking any hearts, or Q♠.)
The objective of 500 is to win tricks, by playing high cards. Which cards are the “high cards” is determined during “bidding”.
After the cards have been dealt, the players take turns to “bid”. Each player calls out the number of tricks they think their partnership can win, using a given trump suit.
• “Six Spade” -> The partnership must win at least six tricks, with spades as the trump suit.
• “Seven no-trumps” -> The partnership must win at least seven tricks, with no trump suit.
The bids must increase in the order 6♠, 6♣, 6♦, 6♥, 6 No-trumps, 7♠, 7♣, 7♦, 7♥ … 10 No-trumps. So if the previous bid is 6♥, and you want to bid diamonds, you can’t bid 6♦. You would have to bid 7♦.
It’s important to tell your partner what cards you have. But you can’t just say “I have J♠, J♣, A♠, and K♠” - you have to signal this indirectly, during the bidding process.
To a new player, this is the most mystifying part of the game.
• Bidding 6 of a suit is a weak signal to your partner, that you have a few (but not many) high cards of that suit.
• Bidding no-trumps tells your partner that you have the Joker, or a lot of aces.
• Bidding 8 or more of something is a signal that you have a lot of high cards.
# Trump cards
In hearts, the special cards are Q♠, then A♥, K♥, Q♥, J♥, 10♥, …
In 500, the order of cards depends on the winning bid.
1. Joker is the highest card.
2. The jack of the trump suit.
3. The other jack, of the same colour as the trump suit.
4. A, K, Q… of the trump suit.
5. A, K, Q… of the suit that was led.
6. Other cards are worth nothing.
So if the trump suit is clubs, the trump cards are:
1. Joker (Which is considered to be a ♣)
2. J♣
3. J♠ (Which is considered to be a ♣)
4. A♣, K♣, Q♣, 10♣, 9♣, … 5♣
If the trump suit is “no trumps”, then jacks aren’t special, and the order is: Joker, A, K, Q, J, 10…
# Following suit
Just like Hearts, you must follow the suit that was led.
If you have no cards of the suit that was led, you can play any card in your hand.
(Unlike Hearts, you don’t have to wait for “hearts to be broken” before you can play special cards.) | 821 | 2,932 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2022-21 | latest | en | 0.970891 |
https://sciencenotes.org/centrifugal-force-definition-formula-examples/ | 1,720,984,959,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514638.53/warc/CC-MAIN-20240714185510-20240714215510-00860.warc.gz | 454,472,776 | 42,318 | # Centrifugal Force – Definition, Formula, Examples
In the realm of physics, forces play a pivotal role in explaining the interactions and motions of objects. One intriguing and often misunderstood concept is centrifugal force.
• Centrifugal force is the apparent force that acts away from the center of a circular path.
• The force is always perpendicular to the direction of movement.
• The formula for centrifugal force is the same as for centripetal force: Fc = mv2/r. However, centrifugal force acts in an equal and opposite direction.
• The force pushes or pulls an object away from the center of rotation. For example, you feel drawn toward the side of a turning car.
### What Is Centrifugal Force?
Centrifugal force is the outward perceived force that seems to push a body away from the center of rotation when it is in a circular motion. This force is not a real force in the traditional sense. Rather, it is a result of inertia — the tendency of an object to resist any change in its state of motion.
The Formula for Calculating Centrifugal Force
The formula for centrifugal force (Fc) is:
Fc = mω2r = mv2/r
where:
• m is the mass of the rotating object
• r is the radius of the circular path
• ω is the angular velocity of the object
• v is the velocity of the object
### A Virtual or Pseudo Force
What sets centrifugal force apart is that it is considered a “virtual” or “fictitious” force. It does not arise from any physical interaction like gravitational or electromagnetic forces. Instead, it is an apparent force that is perceived in a rotating reference frame. In a non-rotating, inertial frame of reference, centrifugal force does not exist.
The Coriolis force is another example of a virtual force that appears in a rotating reference frame. Here, the Earth’s rotation causes the deflection of moving objects to the right in the Northern Hemisphere and to the left in the Southern Hemisphere.
### How Centrifugal Force Works
Imagine being in a car that is turning rapidly. You feel as if you’re being pushed to the side of the car opposite the turn. This sensation is the result of centrifugal force. From your perspective inside the car, it feels like a force is pushing you outward. However, in reality, it’s your body’s inertia that’s causing you to continue moving in a straight line while the car turns around you.
### Examples of Centrifugal Force
Other examples of centrifugal force are:
• A bucket of water swung in a vertical circle: The water does not spill when the bucket is upside down because the centrifugal force pushes the water towards the bottom of the bucket.
• Clothes in a spinning washing machine: The clothes are pushed to the edge of the drum, demonstrating centrifugal force.
### History
Christiaan Huygens coined the term “centrifugal force” in 1659 in De Vi Centrifuga. Centrum is Latin for “center,” while -fugus comes from the word for “fleeing or avoiding.” So, centrifugus translates as “fleeing from the center.”
Sir Isaac Newton expanded upon Huygens work in his 1678 Principia and also describe centripetal force.
### Practical Applications of Centrifugal Force
Even though it’s not a “real” force, centrifugal force has real-world application:
• Centrifuges in Laboratories: Separate substances of different densities.
• Amusement Park Rides: Rides like the “Rotor” use centrifugal force to pin riders against the wall.
• Geosynchronous Orbit: Centrifugal force counteracts the downward pull of gravity, keeping satellites in orbit. In fact, the usual way of expressing centrifugal force is as a multiple of gravity, g.
• Vehicle Dynamics: Helps in understanding and improving the stability of vehicles in turns.
### Centripetal vs Centrifugal Force
Both centripetal and centrifugal force involve rotation or circular motion, yet they are distinct in their origins and nature.
• Mutual Existence in Circular Motion: Both forces occur in scenarios where an object is moving in a circular path. While centripetal force is necessary for this motion, centrifugal force is an apparent force experienced in a rotating frame.
• Opposing Directions: Centripetal force always points towards the center of the circular path, acting as the “center-seeking” force. In contrast, centrifugal force appears to act outward from the center, providing a sensation of being “flung” outward.
#### Difference
• Nature of Forces:
• Centripetal Force: This is a real force, resulting from tangible interactions like gravitational pull, tension, friction, or any other force that causes an object to follow a curved path.
• Centrifugal Force: It’s a fictitious or pseudo force. It doesn’t arise from a physical interaction, but is rather a result of inertia in a rotating reference frame.
• Frame of Reference:
• Centripetal Force: It occurs in both inertial (non-accelerating and non-rotating) and non-inertial (accelerating or rotating) frames of reference.
• Centrifugal Force: It only appears in non-inertial, specifically rotating, frames of reference. In an inertial frame, centrifugal force does not exist.
• Purpose in Equations of Motion:
• Centripetal Force: In the physics of circular motion, centripetal force is what causes the object to deviate from straight-line motion, pulling it towards the center of the circle.
• Centrifugal Force: In a rotating frame, it explains why an object appears to move outward or why there is a need for a centripetal force to counteract this outward motion.
### References
• Hand, Louis N.; Finch, Janet D. (1998). Analytical Mechanics. Cambridge University Press. ISBN 978-0-521-57572-0.
• Kobayashi, Yukio (2008). “Remarks on viewing situation in a rotating frame”. European Journal of Physics. 29 (3): 599–606. doi:10.1088/0143-0807/29/3/019
• Restuccia, S.; Toroš, M.; Gibson, G. M.; Ulbricht, H.; Faccio, D.; Padgett, M. J. (2019). “Photon Bunching in a Rotating Reference Frame”. Physical Review Letters. 123 (11): 110401. doi:10.1103/physrevlett.123.110401
• Yoder, Joella (2013). A Catalogue of the Manuscripts of Christiaan Huygens Including a Concordance With his Oeuvres Complètes. BRILL. ISBN 9789004235656. | 1,419 | 6,129 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2024-30 | latest | en | 0.889793 |
http://mathhelpforum.com/algebra/199082-sat-multiples-problem.html | 1,529,897,053,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267867364.94/warc/CC-MAIN-20180625014226-20180625034226-00547.warc.gz | 206,973,103 | 9,828 | 1. ## SAT multiples problem
If $\displaystyle S$ is the set of positive integers that are multiples of $\displaystyle 7$, and if $\displaystyle T$ is the set of positive integers that are multiples of $\displaystyle 13$, how many integers are in the intersection of $\displaystyle S$ and $\displaystyle T$?
I said there is 1 integer in the intersection, since only 91 is a common multiple of 7 and 13. However, according to the answer,
This is the set of all positive integers that are multiples of $\displaystyle 7*13 = 91$. There are an infinite number of positive integers that are multiples of $\displaystyle 91$, so there are more than thirteen integers in the intersection of $\displaystyle S$ and $\displaystyle T$.
91's multiples include 186, 278, etc, so I'm guessing the infinity is referring to that. However, I thought the problem only wanted an integer that was only multiple of 7 and 13?
2. ## Re: SAT multiples problem
Originally Posted by m58
If $\displaystyle S$ is the set of positive integers that are multiples of $\displaystyle 7$, and if $\displaystyle T$ is the set of positive integers that are multiples of $\displaystyle 13$, how many integers are in the intersection of $\displaystyle S$ and $\displaystyle T$?
I said there is 1 integer in the intersection, since only 91 is a common multiple of 7 and 13. However, according to the answer,
91's multiples include 186, 278, etc, so I'm guessing the infinity is referring to that. However, I thought the problem only wanted an integer that was only multiple of 7 and 13?
You have dropped and "s"! The problem said "multiples" of 7 and 13, not just the product of 7 and 13. The set of "positive multiples of 7" are 7, 14, 21, 28, etc. and the set of "positive multiples of 13" are 13, 26, 39, etc.
3. ## Re: SAT multiples problem
Originally Posted by HallsofIvy
You have dropped and "s"! The problem said "multiples" of 7 and 13, not just the product of 7 and 13. The set of "positive multiples of 7" are 7, 14, 21, 28, etc. and the set of "positive multiples of 13" are 13, 26, 39, etc.
So the multiples of 7 and 13 that are x≥91 are included in the intersection, meaning there are infinitely many of them. | 579 | 2,190 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2018-26 | latest | en | 0.944351 |
https://econsultancy.com/thin-content-how-to-identify-and-fix-it-using-google-analytics/ | 1,721,544,775,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517648.65/warc/CC-MAIN-20240721060614-20240721090614-00778.warc.gz | 197,787,823 | 23,953 | Here is an illustration of what thin content looks like. When I search for a two bedroom house in East Sussex, and scroll down to the one hundredth result, I begin to see results like this one where I’m directed to a page that offers absolutely no value whatsoever.
So how can you find out whether your site is being affected by thin content?
Well, as of a few days ago you can now find out in the Webmaster Tools ‘manual actions’ tab. However, I have to admit I am skeptical of this – I’ve looked at several sites in WMT that I know have thin content issues, and yet no notifications have come up.
I recently had the challenge of fixing thin content issues on a 1.5m page site with approximately 75,000 pages of what I would describe as low quality thin content.
While I’m sure there are numerous clever ways of coming to the same conclusion, I’m going to share my approach which I hope will at least provide a starting point to help you identify and rectify your thin content.
## 1. Define thin content quantitatively
The hardest part about identifying thin content is getting past the subjective nature of what is or isn’t considered ‘thin’.
In my analysis I decided to create a weighted formula that shortlisted a page for being considered thin if it had all of the following characteristics:
• A bounce rate between 95 and 99.99% (here’s why I excluded pages with a 100% bounce rate).
• An average time on page between 0.1 and five seconds.
You can use the following formula to work this out
=IF(CELL WITH BOUNCE RATE < 95%, "Not Thin", IF(CELL WITH AVERAGE TIME ON SITE < 5, "Thin", "Not Thin"))
Once you’ve got this shortlist of pages that are performing poorly, you can begin looking for trends. Which types of pages, or sections of your site are causing trouble?
Try to find common patterns in the URL structure, and get an understanding from a user’s perspective why these pages might be causing people to bounce straight away.
## 2. Rectifying thin content
There is no right or wrong way to rectify thin content, so let me go through various options with an example.
Below are the metrics for a page on MusicJobBoard.com, a site that I use for testing purposes from time to time. As you can see, this page has a combined high bounce rate and low time on page, qualifying this page to be shortlisted as ‘thin content’ by my definition above.
Here is the page, looking rather thin.
Job boards like this one are typically susceptible to thin content, as if no one posts a job in a certain category, the category page can remain indexed despite providing a poor result for someone looking for what the page would usually offer.
In this case, we could apply a rule that would noindex the page if it reached 0 results. I personally don’t like doing this, but on larger sites I’ve seen it work as a pretty effective strategy for keeping low quality results out of the SERPs.
Alternatively, we could design the page in a way that provided value even if no jobs were present – e.g. providing links to see similar jobs in audio production, or even offering some cool information on average salaries for this type of job, as Indeed does.
Another option, in this instance, would be to try and find an ongoing job listing for every category page i.e. a recording studio that is will to receive CVs on an ongoing basis.
One interesting option, which I’ve seen used by several property aggregator services is to redirect people to the next best result i.e. rather than showing me an empty page with 0 property listings in street X, send me to the page on street Y 200 meters away.
A more agreed-upon approach is to merge your pages. Rather than having a page on every single street in the UK, with many being empty, you could merge your street pages to post code pages, or town pages.
This goes against the idea of targeting the long tail with dedicated pages, but without quality in place I think it’s fair to say that that ship has sailed anyway.
One final option is to simply remove your poor performing pages. If there really is no point merging the pages or trying to improve them, it may be worth considering hacking off the low quality content and investing your efforts on improving your best content instead.
## Final thoughts
Thin content is a tricky issue to define and tackle, which is perhaps why it’s not covered in quite as much detail as more objective site quality issues. I’d love to hear how others are tackling it, and whether there are any other creative solutions that I’ve missed above.
Feel free to leave a comment below, send me a tweet, or drop me an email on marcus (at) ventureharbour.com.
Econsultancy’s Crunch – Data, Analytics and the Rise of the Marketing Geek, takes place on October 10 at Truman Brewery, London. Crunch is the event for the analysts, strategists and boffins who turns raw numbers into insight, then revenue. This event is one of five that make up our week-long Festival of Marketing | 1,068 | 4,957 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2024-30 | latest | en | 0.942787 |
http://www.lawnsite.com/threads/check-my-urea-math-please.404570/#post-4769660 | 1,474,930,502,000,000,000 | text/html | crawl-data/CC-MAIN-2016-40/segments/1474738660898.27/warc/CC-MAIN-20160924173740-00075-ip-10-143-35-109.ec2.internal.warc.gz | 574,633,427 | 28,158 | # Check my Urea math please
Discussion in 'Fertilizer Application' started by TYTILIDIE, May 12, 2013.
1. ### TYTILIDIELawnSite Memberfrom Colorado SpringsPosts: 78
Correct me if I am wrong here please. I want to put down 1lb of dry flowable Urea per 1,000sq ft. My tank is putting out 2 gallons of mix per 1,000 sq ft. I have 50lb bags, I am spraying out of a 200 gallon tank so I am coming up with 100lbs of Urea in the tank. Sound right?
2. ### PamlicoLawnCareLawnSite Memberfrom Oriental, NCPosts: 80
Do you want to apply 1 lb. of urea per 1000 sq. ft. or 1 lb. of Nitrogen (N) per 1000 sq. ft.?...there's a big difference.
Need to know the % N of your 50 lb. bags of urea.
Also need to know total square footage of area to be applied.
I'm ASSUMING that you mean 1 lb. of N per 1000 sq. ft.
I'm also ASSUMING that your 50 lb. bags of urea contain 46% N.
If this is the case, one 50 lb. bag of 46% N urea contains 23 lbs. of N.
(50 lbs. X .46 = 23 lbs.)
If you want to apply 1 lb. of N per 1000 sq. ft., then one 50 lb. bag of urea will cover 23,000 sq. ft. or two bags of urea will cover 46,000 sq. ft., etc...
If you spray at a volume of 2 gallons per 1000 sq. ft. then one 50 lb. bag of urea in 46 gallons will give you 1 lb. of N per 1000 sq. ft. and you can spray 23,000 sq. ft.
With a 200 gallon tank, use four X 50 lb. bags of urea in 184 gallons to give a rate of 1 lb. of N per 1000 sq. ft. to cover 92,000 sq. ft.
Again, this ASSUMES that you mean 1 lb. of N per 1000 sq. ft. and that your 50 lb. bags of urea contain 46% N.
Let us know if this is not the case.
3. ### RicLawnSite Fanaticfrom S W FloridaPosts: 11,956
.
Or 4 1/3 bags of Urea in a 200 tank.
.
4. ### TYTILIDIELawnSite Memberfrom Colorado SpringsPosts: 78
This is all correct. I went and finally got the info off the bag. This kind of sucks because on my last round I only put two bags in. Oh well, learning experience right? Thanks for the help!
5. ### countryclublawnllcLawnSite Memberfrom MichiganPosts: 151
A pretty easy way to figure application rate of product for fertilizer is just put desired rate of the nutrient and divide by the percentage of that nutrient in the fert. If the desired rate is 1 lb. of N per thousand sq. ft. then it would go:
1/.46= 2.17
So 2.17 lbs of a 46-0-0 would be added per thousand sq. ft. of spray coverage which obviously depends on how many gallons of water go out /1000.
John | 763 | 2,421 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2016-40 | latest | en | 0.871897 |
http://www.tehachapiarts.com/2019/05/20/360-day-interest-calculation-excel/ | 1,571,256,189,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986669546.24/warc/CC-MAIN-20191016190431-20191016213931-00535.warc.gz | 335,415,893 | 8,617 | 360 Day Interest Calculation Excel
Apartment Loan Financing Apartment/Unit Loans – California Mortgage Advisors – California Mortgage Advisors Inc. has been helping customers finance apartment loans for over 20 years. When it comes to defining an apartment or an.
Excel YEARFRAC Function – excelfunctions.net – The spreadsheet below shows five examples of the Excel Yearfrac function. In each case, the function is used to calculate the year fraction between January 1st, 2015 and March 31st, 2015. However, each case uses a different Day Count Basis.
Actual/360 vs 360 simple interest | Lending Compliance | For. – What is the difference between calculating interest on an actual/360 basis and a 360-day simple interest basis? I need to check whether a loan payment listed in a note is right, and I’m using a calculator I designed in excel for actual/360 and it is not matching the note, which calculates the interest based on a 360-day simple interest basis, and I’m not sure how to adjust it.
Calculating Interest: the Stated Rate Method and the Bank Method – Traditionally, there are two common methods used for calculating interest: (i) the 365/365 method (or Stated Rate Method) which utilizes a 365-day year; and (ii) the 360/365 method (or Bank Method) which utilizes a 360-day year and charges interest for the actual number of days the loan is outstanding.
Margill | Interest Calculation White Paper – Read through our Interest Calculation White Paper.. repayment principles, day count (Actual/actual, 30/360, Actual/365, Actual/360), annual percentage rate.
Accrued Interest calculator – London Stock Exchange – 12, Overview of calculation Methods. 7, accrued interest = (interest days/annual basis) x coupon. 13, 30E/360 ISMA EUROPEAN, assumes 30 days in each month, 360, annual rate, 30 European method: if trade settlement date falls on 31st,
How to use the Excel DAYS360 function | Exceljet – The Excel DAYS360 function returns the number of days between two dates based on a 360-day year. Calculations based on a 360-day year comes from certain accounting calculations where all 12 months are considered to have 30 days. A number representing days. start_date – The start date.
Commercial Business Loans Business Loans – Woodsboro Bank – Woodsboro Bank offers a range of business loans and lines of credit to help. to the well-established business we understand the needs of our commercial.Bankrate.Com Mortgage Interest Rates Daily Mortgage Rates – Mortgage News Daily provides the most extensive and accurate coverage of the mortgage interest rate markets. All services. we’re not back to the sub-4% mortgage rates that dominated much of the.
loans – What does a 30/360 day count convention mean. – A 30/360 convention in interest calculation means that there are exactly 30 days in a month and there are 12 months [or 360 days in a year]. This convention was used in the early days when computers were not used and most of the calculation were done by hand [remember banking was there before computers].
Day count convention – Wikipedia – In finance, a day count convention determines how interest accrues over time for a variety of. Certain terms, such as "30/360", "Actual/Actual", and "money market basis" must be understood in the context.. Treating a month as 30 days and a year as 360 days was devised for its ease of calculation by hand compared with.
How Is Nnn Calculated Us Bank Personal Loan Calculator online emi calculator for Car Loan in India | Bank of Baroda – car loan emi Calculator by Bank of Baroda will help you plan your monthly Car EMI’s on basis of car’s price, rate of interest & loan tenure. Use our online car loan calculator & plan your purchase.How do i calculate the amount of a lease for commercial. – For example, is it a Triple Net lease, where you pay tax, insurance, and common area maintenence (cam). As an example, you might see NNN charges ranging \$3.50/SF – \$5.00/SF. Use the same formula to calculate the monthly payment, and then add this to the base rent to get your total rent. | 900 | 4,052 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2019-43 | longest | en | 0.89398 |
https://mathoverflow.net/questions/345908/the-least-common-multiple-of-all-degrees-of-a-finite-coxeter-group-and-indecompo/359079 | 1,642,449,944,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300616.11/warc/CC-MAIN-20220117182124-20220117212124-00584.warc.gz | 434,974,701 | 26,601 | # The least common multiple of all degrees of a finite Coxeter group and indecomposable elements in the generalized cycle decomposition
This question is a follow-up of the previous question and especially the last comment therein.
Let $$(W,S)$$ be a finite Coxeter system with reflections $$T$$. Let $$\ell_T$$ be the reflection length. According to 1611.03442v1, p. 2 - an article which was also discussed on this site - an element $$x\in W$$ is called indecomposable if there exists no nontrivial decomposition $$x=uv$$ such that $$uv=vu$$ and $$\ell_T(x)=\ell_T(u)+\ell_T(v)$$.
Is it true that the set of orders of indecomposable elements in $$W$$ coincides (with or without repitition) with the set of all degrees of $$W$$, and that - as a consequence of this and the generalized cycle decomposition in loc. cit., Theorem 1.3 - modulo the fact that maybe not every element is a parabolic quasi-Coxeter element, cf. loc. cit., Condition 1.1 - that - the least common multiple of all degrees equals the smallest positive integer $$N$$ such that $$g^N=1$$ for all $$g\in W$$?
Remark. I checked this for the symmetric group and some other rank two cases.
Remark. The Coxeter element itself is indecomposable (as its associated parabolic subgroup is the whole of $$W$$), has reflection length $$|S|$$, and its order is the Coxeter number which is known to be a degree. Further, every simple reflection is indecomposable of relfection length one, and its order is two, which is a degree as well. These observations are in favor of the question.
• Probably you want to forbid the identity element from being considered indecomposable. Jan 21 '20 at 18:16
• The identity has order one, so, for the least common multiple, it does not matter. However, one is not a degree, as you said. Jan 22 '20 at 3:40
Carter classified the conjugacy classes in finite Weyl groups (see http://www.numdam.org/article/CM_1972__25_1_1_0.pdf). For example, the conjugacy class $$E_7(a_3)$$ in Carter's notation is a conjugacy class of a quasi-Coxeter element in a Coxeter system of type $$E_7$$. The order of a corresponding quasi-Coxeter element is $$30$$. But $$30$$ is not a degree in type $$E_7$$. (I expect there to be a counterexample in rank $$<7$$.) | 582 | 2,240 | {"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": 20, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2022-05 | latest | en | 0.912556 |
www.zyourgame.com | 1,558,459,660,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232256494.24/warc/CC-MAIN-20190521162634-20190521184634-00367.warc.gz | 370,790,988 | 19,563 | # What is a derivative
## velocity and tangent question
### velocity question
Average velocity:
Instantaneous velocity:
secant:
tangent:
Or as:
Or as:
Or as:
Or as:
## Ordinary derivative
1.Derivative of :
2.Derivative of :
3.Derivative of :
and in a similar way:
4.Derivative of :
specifically：.
5.Derivative of
specifically：.
## Differentiable and continuous
Derivative must be continuous, continuous may not be derivative.
# Derivative method
## sum, difference, product, quotient method
### Trig derivative
Derivative of
in a similar way:
## Derivative method of inverse function
If the function is monotone and differentiable in the interval ,and ,then differentiable in interval ,and:
## Derivative rule of complex functions
If can be derivable at point ,and can be derivable at point ,then can be derivable at point .The derivative is:
Case 1:If ,for :
Case 2:If ,for .
# higher derivative
## higher derivative
(1)
for :
(2)
### Higher derivatives of sine and cosine
for :
in a similar way,for :
### *Higher derivative of the product function
we know:;
then:;
then:;
so:
This is Leibniz formula.
# Derivative of an implicit function
## General method for differentiating implicit functions
case 1:for function ,for :
(1)derivative for the left-hand side with respect to :
(2)derivative for the right-hand side with respect to :
(3)When we differentiate the left and the right sides of this equation for , we get the same thing:
(4)then:
case 2:for ,please get the equation of the tangent line at point
(1)according to geometric meaning of the derivative,we know the slope of this tangent line:
(2)take the derivative of both sides of this equation with respect to , we get the same thing:
(3)then:
(4)when ,:
(5)The tangent equation is:
(6)Reduction to:
## Take the derivative of an implicit function using logarithmic differentiation
case 1:get derivative of .
(1)Take the logarithm of both sides of this equation:
(2)Take the derivative of both sides of this equation with respect to
(3)then:
## Derivative of parametric equation
The derivative of an equation whose parameters can be eliminated
Derivative of parabolic parametric equation:
(1)Elimination parameters :
(2)Using the general implicit differentiation method,Is omitted.
The derivative of an equation that doesn’t cancel out the parameters
for:
we can : | 570 | 2,397 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2019-22 | longest | en | 0.831468 |
http://www.yll.url.tw/viewtopic.php?p=164054 | 1,582,448,290,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145747.6/warc/CC-MAIN-20200223062700-20200223092700-00216.warc.gz | 250,296,319 | 6,523 | ## [高中]多項式公因式
### [高中]多項式公因式
1.設f(x)與g(x)都是多項式，若x2+2x+3為3f(x)+4g(x)與f(x)+2g(x)的最高公因式，求f(x)與g(x)的最高公因式.
2.已知a是正整數,b.c是整數，若x-a是f(x)=x3+6x2+bx+6與g(x)=x3+cx2-x-2的最高公因式.
3.設a.b是常數，且兩多項式f(x)=2x3-x2+ax+1與g(x)=x3-4x2+bx-3的最高公因式為二次式，求a.b的值.
Sherry
1.
3f(x)+4g(x) = 3(f(x)+2g(x))+(-2g(x))
f(x)+2g(x) = -1(-2g(x))+f(x)
2.
0次項係數: -aj = 6
=> j = -6/a [因a不等於0] //若j是整數, a的值就是-6的正整因數 ----(1)
1次項係數: -ai + j = b //因b是整數, 若得i是整數就能確定j是整數 -----(2)
2次項係數: -a + i = 6
=> i是整數
=> j是整數 [因(2)]
=> a = 1, 2, 3, 6 [因(1)] -----(3)
0次項係數: -aj' = 2
=> j' = -2/a
1次項係數: -ai' + j' = -1
2次項係數: -a + i' = c
=> i'是整數
=> j'是整數
=> a = 1, 2 -----(4)
=> a = 1, 2
f(1) = 1 + 6 + b + 6 = 0 => b = -13
g(1) = 1 + c - 1 - 2 = 0 => c = 2
f(x) = (x-1)(x2+7x-6)
g(x) = (x-1)(x2+3x+2) = (x-1)(x+2)(x+1)
f(2) = 8 + 24 + 2b + 6 = 0 => b = -19
g(2) = 8 + 4c - 2 - 2 = 0 => c = -1
f(x) = (x-2)(x2+8x-3)
g(x) = (x-2)(x2+x+1)
3.
(1)/(2), 可得c/c'= 7/7 = (a-2b)/(-7) = 7/(3a+b)
=> a-2b = -7, 3a+b = 7
=> a=1, b=4
Tzwan
『數學及時、求救區』 | 715 | 990 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2020-10 | latest | en | 0.223475 |
https://stats.stackexchange.com/questions/58734/is-r2-value-valid-for-insignificant-ols-regression-model | 1,638,452,640,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362219.5/warc/CC-MAIN-20211202114856-20211202144856-00346.warc.gz | 601,895,349 | 35,876 | # Is $R^2$ value valid for insignificant OLS regression model?
I am interested in stating that ___ % of the variance in Y is explained uniquely by $X_1$ and ___ % is explained uniquely by $X_2$.
• Is there some way to obtain this from a multiple regression model, or do I need to obtain adjusted $R^2$ values from a series of residual regressions (sensu Legendre & Legendre et al.)?
• That being asked, is $R^2$ value valid for insignificant OLS regression model?
• Does the value of $R^2$ depend on a statistical test?
• To respond to the parts of your question not covered in the answer - the value of $R^2$ does not depend on significance of any predictors or on whether you tested them. May 11 '13 at 23:00
• Thanks Glen_b, OK, so I can perform partial regressions to estimate % of the variance in Y is explained uniquely by X1 and by X2 (using the adjusted R^2 value), even if coefficient for X2|X1 is not significant. I have necessarily collinear regressor variables (.5 < |r| < .6) that have a theoretical link to Y. I unfortunately also have small sample sizes (n ~ 20), but would like to use this to more or less determine if one of these regressors may influence Y to a greater degree, while statistically controlling for the other. Do you have any opinion to the validity of this approach? May 11 '13 at 23:22
• I'm not overly concerned with building predictive models, I simply want to know if evidence suggests X1|X2 is more influential than X2|X1 ... May 11 '13 at 23:24
• The part about estimating % of variance uniquely explained is, I think, covered in the answer you have already. May 11 '13 at 23:43
• It absolutely is, thank you. Are you aware of any literature regarding either what you have stated above, or regarding this approach with the use with small sample sizes? If so, that would be greatly appreciated. Thanks again. May 11 '13 at 23:55
Yes, you're trying to calculate the Extra Sum of Squares. In short you are partitioning the regression sum of squares. Assume we have two $X$ variables, $X_1$ and $X_2$. The $SSTO$ (total sum of squares, made up of the SSR and SSE) is the same regardless of how many $X$ variables we have. Denote the $SSR$ and $SSE$ to indicate which $X$ variables are in the model: e.g.
$SSR(X_1,X_2) = 385$ and $SSE(X_1,X_2) = 110$
Now let's assume we did the regression just on $X_1$ e.g.
$SSR(X_1) = 352$ and $SSE(X_1) = 143$.
The (marginal) increase in the regression sum of squares in $X_2$ given that $X_1$ is already in the model is:
\begin{eqnarray} SSR(X_2|X_1)& = &SSR(X_1,X_2) - SSR(X_1)\\ & = & 385 - 352\\ & = & 33 \end{eqnarray}
or equivalently, the extra reduction in the error sum of squares associated with $X_2$ given that $X_1$ is already in the model is:
\begin{eqnarray} SSR(X_2|X_1) & = & SSE(X_1) - SSE(X_2,X_1)\\ &=& 143 - 110\\ &=& 33 \end{eqnarray}
In the same way we can find:
\begin{eqnarray} SSR(X_1|X_2) &=& SSE(X_2) - SSE(X_1,X_2)\\ &=& SSR(X_1,X_2) - SSR(X_2) \end{eqnarray}
Of course, this also works for more $X$ variables as well.
• I think you have to use multiple regressions as indicated by Eric Peterson. The other option is that you use partial correlation coefficients. That would make sense, if the absolute values of $$y$$ and the magnitude of the squares are not important.
• In a certain sense, $$R^2$$ is very valid for an insignificant OLS model. The significance thresholds are after all lines drawn in water. For example, if you have to choose between models to use for making predictions, it makes sense to use the model with highest $$R^2$$. Or, if the models involve a different number of covariates, better use adjusted $$R^2$$ denoted as $$\bar{R}^2$$.
• $$R^2$$ is actually the square of the correlation coefficient between predicted values and the reality, i.e. $$R^2=(cor(y_i,\hat{y}_i))^2$$, and thus, you can use tables for the correlation coefficient to test the ovarall significance of your model. The other option is that you use the F test. | 1,122 | 3,970 | {"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": 7, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2021-49 | latest | en | 0.922222 |
http://openstudy.com/updates/4f0f25f9e4b04f0f8a917caa | 1,444,247,673,000,000,000 | text/html | crawl-data/CC-MAIN-2015-40/segments/1443737882743.52/warc/CC-MAIN-20151001221802-00247-ip-10-137-6-227.ec2.internal.warc.gz | 237,919,966 | 9,555 | allybally555555 3 years ago What are the solutions of 3x2 + 4x = –5?
No solutions
2. tmrk
it has a complex sol.use quadratic formula | 43 | 135 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2015-40 | longest | en | 0.862625 |
https://brilliant.org/problems/can-you-find-it-2/ | 1,660,823,698,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882573193.35/warc/CC-MAIN-20220818094131-20220818124131-00767.warc.gz | 162,576,014 | 8,562 | # Can you find it?
If $k$ is a positive integer satisfying
$\dfrac{k}{0.5} + \dfrac{k}{0.2} + \dfrac{k}{0.25} =297,$
find the value of $k^2$.
× | 61 | 147 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 3, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2022-33 | latest | en | 0.582426 |
https://sepwww.stanford.edu/data/media/public/docs/sep124/madhav1/paper_html/node4.html | 1,722,956,019,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640492117.28/warc/CC-MAIN-20240806130705-20240806160705-00264.warc.gz | 407,590,382 | 4,735 | Next: Hask Data Set Up: Implementation of LSJIMP Previous: Implementation of LSJIMP
## SYN-1
A synthetic data set, which we call SYN-1, was generated to simulate very shallow and hard water-bottom marine environment. The model has four horizontal layers. The depth of the water bottom is about 150 m. The data-set, shown in Figure , is contaminated with first- and higher-order water-bottom peglegs.
data
Figure 2
Synthetic data-set (SYN-1) infected with shallow water bottom multiples.
Brown (2004) discusses kinematic imaging of pegleg multiples and applies NMO correction for a jth order pegleg using the following equation:
(9)
where and are the zero-offset travel-time and travel-time depth of the multiple generating surface respectively and is the effective velocity, which is given by
(10)
We use these kinematic equations in our NMO operator to construct images from primaries as well as from multiples. When we apply the NMO operator with Vrms as the velocity function, primaries get perfectly corrected for the moveout, but multiples have a residual moveout, and they appear as crosstalk. The same holds true for primaries and higher-order multiples when we correct with a appropriate for first-order multiples. Thus, in each image we should ideally be able to distinguish the components (primaries or different order of multiples) we correct for the moveout from others. These equations work fine for most cases, but for SYN-1 we actually evaluate a limiting case of these equations. In this case, the travel-time depth of the water bottom is very small, so for deeper layers we have
(11)
Using this in equations (9) and (10), we get
(12)
(13)
When we construct the image from primaries, we use Vrms to apply the NMO operator. In this case where Vrms , we do not expect water-bottom peglegs to have a residual moveout. This is precisely the reason why in Figure all the primaries are imaged to their perfect positions, but the multiples, which we were expecting to appear as crosstalk, look like perfectly flat reflectors.
syn1i
Figure 3
Images constructed from (a) primary, (b) first- and (c) second-order multiples. A strong primary crosstalk is present on images constructed from first- and second-order multiples, moreover, its difficult to distinguish between signal and crosstalk on the basis of residual moveout.
Likewise, when we construct images from first- and higher-order multiples, we do not see any residual moveout for primaries and other multiples, except for shallow ones. In an ideal case, we would have a consistent signal across all the images, but the crosstalk would be inconsistent, a fact we could use to enhance signal-to-noise ratio and penalize crosstalk. But for this special case at hand, crosstalk is as consistent as the signal itself. LSJIMP uses three regularization operators as discussed in equation (8); two of these, differencing across offset (designed to penalize events with residual moveout) and differencing between images (designed to penalize inconsistent crosstalk) do not seem to work effectively for SYN-1. Moreover, the crosstalk-modeling operator fails to perfectly model the crosstalk and subsequently penalize the multiple energy. In Figure we can observe, a very strong primary crosstalk in images constructed from first- and second-order multiples.
Thus, the application of LSJIMP to such a data set can indeed image the primaries but cannot eliminate water-bottom peglegs completely, as shown in Figure . Since most of the multiple energy present here is in water-bottom peglegs, we can use some simple tricks like predictive deconvolution to suppress this energy. We can also try to model the crosstalk by shifting the dataset by the travel-time depth of water bottom. The results for this exercise are given in Figure , though our results have improved a lot, we still have not been able to get rid of all the multiple energy in the shallow part(first-order water-bottom multiple). The primary reason is that most of the zone is muted, and we have few data points in that time range to perfectly model crosstalk.
The first regularizaton operator, in equation 8, was a result of differencing between images of model panels generated by primaries and different orders of multiples. Ideally the difference between two of them should be small where signal is present and large where crosstalk dominates. Application of such a scheme ensures some degree of smoothness and consistency across images. LSJIMP computes the difference across two consecutive images, for instance between and , and and likewise. Another possible alternative is to compute the difference between a multiple image and the primary image, as that would make all other images consistent with the primary, where we have maximum signal-to-noise ratio. We implemented this approach, but the results as given in Figure are not pleasing. The images we obtained have lots of ringing. One of the main reasons for this was the presence of strong primary crosstalk across all the panels which appears as a spurious event on the primary image after regularization.
syn1c1
Figure 4
Comparing (a) raw data (SYN-1) and (b) data generated by applying forward-modeling operator on primary image generated by LSJIMP
syn1c2
Figure 5
Comparing (a) raw data (SYN-1) and (b) data generated by applying forward-modeling operator on primary image generated by LSJIMP, and (c) LSJIMP with the modified crosstalk model. Notice the improvement in multiple suppression with modified LSJIMP.
regmod
Figure 6
Illustrating (a) primary image and (b) first-order multiple image with a revised regularization scheme. Notice the amount of ringing, and spurious events in primary image.
Next: Hask Data Set Up: Implementation of LSJIMP Previous: Implementation of LSJIMP
Stanford Exploration Project
4/5/2006 | 1,253 | 5,827 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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 | 0.926929 |
https://statistics.laerd.com/statistical-guides/spearmans-rank-order-correlation-statistical-guide-2.php | 1,656,840,190,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104215805.66/warc/CC-MAIN-20220703073750-20220703103750-00058.warc.gz | 595,283,138 | 5,046 | # Spearman's Rank-Order Correlation (cont...)
## What values can the Spearman correlation coefficient, rs, take?
The Spearman correlation coefficient, rs, can take values from +1 to -1. A rs of +1 indicates a perfect association of ranks, a rs of zero indicates no association between ranks and a rs of -1 indicates a perfect negative association of ranks. The closer rs is to zero, the weaker the association between the ranks.
## An example of calculating Spearman's correlation
To calculate a Spearman rank-order correlation on data without any ties we will use the following data:
Marks
English 56 75 45 71 62 64 58 80 76 61
Maths 66 70 40 60 65 56 59 77 67 63
We then complete the following table:
English (mark)Maths (mark)Rank (English)Rank (maths)dd2
566694525
75703211
4540101000
71604739
62656511
645659416
58598800
80771100
76672311
61637611
Where d = difference between ranks and d2 = difference squared.
We then calculate the following:
We then substitute this into the main equation with the other information as follows:
as n = 10. Hence, we have a ρ (or rs) of 0.67. This indicates a strong positive relationship between the ranks individuals obtained in the maths and English exam. That is, the higher you ranked in maths, the higher you ranked in English also, and vice versa.
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## How do you report a Spearman's correlation?
How you report a Spearman's correlation coefficient depends on whether or not you have determined the statistical significance of the coefficient. If you have simply run the Spearman correlation without any statistical significance tests, you are able to simple state the value of the coefficient as shown below:
However, if you have also run statistical significance tests, you need to include some more information as shown below:
where df = N – 2, where N = number of pairwise cases.
## How do you express the null hypothesis for this test?
The general form of a null hypothesis for a Spearman correlation is:
H0: There is no [monotonic] association between the two variables [in the population].
Remember, you are making an inference from your sample to the population that the sample is supposed to represent. However, as this a general understanding of an inferential statistical test, it is often not included. A null hypothesis statement for the example used earlier in this guide would be:
H0: There is no [monotonic] association between maths and English marks.
## How do I interpret a statistically significant Spearman correlation?
It is important to realize that statistical significance does not indicate the strength of Spearman's correlation. In fact, the statistical significance testing of the Spearman correlation does not provide you with any information about the strength of the relationship. Thus, achieving a value of p = 0.001, for example, does not mean that the relationship is stronger than if you achieved a value of p = 0.04. This is because the significance test is investigating whether you can reject or fail to reject the null hypothesis. If you set α = 0.05, achieving a statistically significant Spearman rank-order correlation means that you can be sure that there is less than a 5% chance that the strength of the relationship you found (your ρ coefficient) happened by chance if the null hypothesis were true. | 755 | 3,401 | {"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} | 4.46875 | 4 | CC-MAIN-2022-27 | longest | en | 0.858549 |
https://math.answers.com/Q/What_is_the_difference_between_the_smallest_and_the_largest_odd_numbers_between_220_and_940 | 1,701,716,798,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100534.18/warc/CC-MAIN-20231204182901-20231204212901-00153.warc.gz | 446,743,310 | 44,075 | 0
# What is the difference between the smallest and the largest odd numbers between 220 and 940?
Updated: 9/26/2023
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the difference
The range.
### What is the difference between the largest and the smallest 5-digit numbers?
The largest 5-digit number is 99,999, the smallest is .00001,and the difference is 99,998.99999 .If you only want to consider whole numbers, then the smallest is 10,000and the difference is 89,999 .
86
### What is range in maths terms?
Range is the difference between the largest and smallest numbers
9999
### What is the difference between the largest and the smallest values in a set of numbers?
This difference is called the range. Subtract the smallest value (S) from the largest value (L).Formula: L - S = Difference
### What mountain range is in northeast Afghanistan?
the difference between the largest and smallest numbers in data
### Choose the definition that fits this term.range?
the difference between the largest and smallest numbers in data
### What is the mountain range in northeast Afghanistan?
the difference between the largest and smallest numbers in data
### What does range of data mean?
The difference between the largest and smallest numbers in your data set.
### What is the difference between the greatest and least numbers in a set of data?
The difference between the largest and smallest numbers in a data set is called the range. | 349 | 1,618 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2023-50 | latest | en | 0.930187 |
https://www.12000.org/my_notes/CAS_integration_tests/reports/rubi_4_16_1_graded/test_cases/6_Hyperbolic_functions/6.7_Miscellaneous/6.7.1_Hyperbolic_functions/rese877.htm | 1,695,492,845,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506528.19/warc/CC-MAIN-20230923162848-20230923192848-00717.warc.gz | 697,538,683 | 5,294 | ### 3.877 $$\int e^{2 (a+b x)} \text{csch}^3(a+b x) \, dx$$
Optimal. Leaf size=73 $\frac{3 e^{a+b x}}{b \left (1-e^{2 a+2 b x}\right )}-\frac{2 e^{3 a+3 b x}}{b \left (1-e^{2 a+2 b x}\right )^2}-\frac{3 \tanh ^{-1}\left (e^{a+b x}\right )}{b}$
[Out]
(-2*E^(3*a + 3*b*x))/(b*(1 - E^(2*a + 2*b*x))^2) + (3*E^(a + b*x))/(b*(1 - E^(2*a + 2*b*x))) - (3*ArcTanh[E^(a
+ b*x)])/b
________________________________________________________________________________________
Rubi [A] time = 0.044398, antiderivative size = 73, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 18, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.222, Rules used = {2282, 12, 288, 207} $\frac{3 e^{a+b x}}{b \left (1-e^{2 a+2 b x}\right )}-\frac{2 e^{3 a+3 b x}}{b \left (1-e^{2 a+2 b x}\right )^2}-\frac{3 \tanh ^{-1}\left (e^{a+b x}\right )}{b}$
Antiderivative was successfully verified.
[In]
Int[E^(2*(a + b*x))*Csch[a + b*x]^3,x]
[Out]
(-2*E^(3*a + 3*b*x))/(b*(1 - E^(2*a + 2*b*x))^2) + (3*E^(a + b*x))/(b*(1 - E^(2*a + 2*b*x))) - (3*ArcTanh[E^(a
+ b*x)])/b
Rule 2282
Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] && !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] && !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]
Rule 12
Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] && !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]
Rule 288
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^
n)^(p + 1))/(b*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] && !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]
Rule 207
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])
Rubi steps
\begin{align*} \int e^{2 (a+b x)} \text{csch}^3(a+b x) \, dx &=\frac{\operatorname{Subst}\left (\int \frac{8 x^4}{\left (-1+x^2\right )^3} \, dx,x,e^{a+b x}\right )}{b}\\ &=\frac{8 \operatorname{Subst}\left (\int \frac{x^4}{\left (-1+x^2\right )^3} \, dx,x,e^{a+b x}\right )}{b}\\ &=-\frac{2 e^{3 a+3 b x}}{b \left (1-e^{2 a+2 b x}\right )^2}+\frac{6 \operatorname{Subst}\left (\int \frac{x^2}{\left (-1+x^2\right )^2} \, dx,x,e^{a+b x}\right )}{b}\\ &=-\frac{2 e^{3 a+3 b x}}{b \left (1-e^{2 a+2 b x}\right )^2}+\frac{3 e^{a+b x}}{b \left (1-e^{2 a+2 b x}\right )}+\frac{3 \operatorname{Subst}\left (\int \frac{1}{-1+x^2} \, dx,x,e^{a+b x}\right )}{b}\\ &=-\frac{2 e^{3 a+3 b x}}{b \left (1-e^{2 a+2 b x}\right )^2}+\frac{3 e^{a+b x}}{b \left (1-e^{2 a+2 b x}\right )}-\frac{3 \tanh ^{-1}\left (e^{a+b x}\right )}{b}\\ \end{align*}
Mathematica [A] time = 0.0583474, size = 61, normalized size = 0.84 $\frac{3 e^{a+b x}-5 e^{3 (a+b x)}-3 \left (e^{2 (a+b x)}-1\right )^2 \tanh ^{-1}\left (e^{a+b x}\right )}{b \left (e^{2 (a+b x)}-1\right )^2}$
Antiderivative was successfully verified.
[In]
Integrate[E^(2*(a + b*x))*Csch[a + b*x]^3,x]
[Out]
(3*E^(a + b*x) - 5*E^(3*(a + b*x)) - 3*(-1 + E^(2*(a + b*x)))^2*ArcTanh[E^(a + b*x)])/(b*(-1 + E^(2*(a + b*x))
)^2)
________________________________________________________________________________________
Maple [A] time = 0.037, size = 67, normalized size = 0.9 \begin{align*} -{\frac{{{\rm e}^{bx+a}} \left ( 5\,{{\rm e}^{2\,bx+2\,a}}-3 \right ) }{b \left ({{\rm e}^{2\,bx+2\,a}}-1 \right ) ^{2}}}-{\frac{3\,\ln \left ( 1+{{\rm e}^{bx+a}} \right ) }{2\,b}}+{\frac{3\,\ln \left ({{\rm e}^{bx+a}}-1 \right ) }{2\,b}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(exp(2*b*x+2*a)*csch(b*x+a)^3,x)
[Out]
-exp(b*x+a)*(5*exp(2*b*x+2*a)-3)/b/(exp(2*b*x+2*a)-1)^2-3/2/b*ln(1+exp(b*x+a))+3/2/b*ln(exp(b*x+a)-1)
________________________________________________________________________________________
Maxima [A] time = 1.05462, size = 119, normalized size = 1.63 \begin{align*} -\frac{3 \, \log \left (e^{\left (-b x - a\right )} + 1\right )}{2 \, b} + \frac{3 \, \log \left (e^{\left (-b x - a\right )} - 1\right )}{2 \, b} + \frac{5 \, e^{\left (-b x - a\right )} - 3 \, e^{\left (-3 \, b x - 3 \, a\right )}}{b{\left (2 \, e^{\left (-2 \, b x - 2 \, a\right )} - e^{\left (-4 \, b x - 4 \, a\right )} - 1\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(exp(2*b*x+2*a)*csch(b*x+a)^3,x, algorithm="maxima")
[Out]
-3/2*log(e^(-b*x - a) + 1)/b + 3/2*log(e^(-b*x - a) - 1)/b + (5*e^(-b*x - a) - 3*e^(-3*b*x - 3*a))/(b*(2*e^(-2
*b*x - 2*a) - e^(-4*b*x - 4*a) - 1))
________________________________________________________________________________________
Fricas [B] time = 1.72627, size = 1098, normalized size = 15.04 \begin{align*} -\frac{10 \, \cosh \left (b x + a\right )^{3} + 30 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{2} + 10 \, \sinh \left (b x + a\right )^{3} + 3 \,{\left (\cosh \left (b x + a\right )^{4} + 4 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} + \sinh \left (b x + a\right )^{4} + 2 \,{\left (3 \, \cosh \left (b x + a\right )^{2} - 1\right )} \sinh \left (b x + a\right )^{2} - 2 \, \cosh \left (b x + a\right )^{2} + 4 \,{\left (\cosh \left (b x + a\right )^{3} - \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right ) + 1\right )} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) + 1\right ) - 3 \,{\left (\cosh \left (b x + a\right )^{4} + 4 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} + \sinh \left (b x + a\right )^{4} + 2 \,{\left (3 \, \cosh \left (b x + a\right )^{2} - 1\right )} \sinh \left (b x + a\right )^{2} - 2 \, \cosh \left (b x + a\right )^{2} + 4 \,{\left (\cosh \left (b x + a\right )^{3} - \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right ) + 1\right )} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) - 1\right ) + 6 \,{\left (5 \, \cosh \left (b x + a\right )^{2} - 1\right )} \sinh \left (b x + a\right ) - 6 \, \cosh \left (b x + a\right )}{2 \,{\left (b \cosh \left (b x + a\right )^{4} + 4 \, b \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} + b \sinh \left (b x + a\right )^{4} - 2 \, b \cosh \left (b x + a\right )^{2} + 2 \,{\left (3 \, b \cosh \left (b x + a\right )^{2} - b\right )} \sinh \left (b x + a\right )^{2} + 4 \,{\left (b \cosh \left (b x + a\right )^{3} - b \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right ) + b\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(exp(2*b*x+2*a)*csch(b*x+a)^3,x, algorithm="fricas")
[Out]
-1/2*(10*cosh(b*x + a)^3 + 30*cosh(b*x + a)*sinh(b*x + a)^2 + 10*sinh(b*x + a)^3 + 3*(cosh(b*x + a)^4 + 4*cosh
(b*x + a)*sinh(b*x + a)^3 + sinh(b*x + a)^4 + 2*(3*cosh(b*x + a)^2 - 1)*sinh(b*x + a)^2 - 2*cosh(b*x + a)^2 +
4*(cosh(b*x + a)^3 - cosh(b*x + a))*sinh(b*x + a) + 1)*log(cosh(b*x + a) + sinh(b*x + a) + 1) - 3*(cosh(b*x +
a)^4 + 4*cosh(b*x + a)*sinh(b*x + a)^3 + sinh(b*x + a)^4 + 2*(3*cosh(b*x + a)^2 - 1)*sinh(b*x + a)^2 - 2*cosh(
b*x + a)^2 + 4*(cosh(b*x + a)^3 - cosh(b*x + a))*sinh(b*x + a) + 1)*log(cosh(b*x + a) + sinh(b*x + a) - 1) + 6
*(5*cosh(b*x + a)^2 - 1)*sinh(b*x + a) - 6*cosh(b*x + a))/(b*cosh(b*x + a)^4 + 4*b*cosh(b*x + a)*sinh(b*x + a)
^3 + b*sinh(b*x + a)^4 - 2*b*cosh(b*x + a)^2 + 2*(3*b*cosh(b*x + a)^2 - b)*sinh(b*x + a)^2 + 4*(b*cosh(b*x + a
)^3 - b*cosh(b*x + a))*sinh(b*x + a) + b)
________________________________________________________________________________________
Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} e^{2 a} \int e^{2 b x} \operatorname{csch}^{3}{\left (a + b x \right )}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(exp(2*b*x+2*a)*csch(b*x+a)**3,x)
[Out]
exp(2*a)*Integral(exp(2*b*x)*csch(a + b*x)**3, x)
________________________________________________________________________________________
Giac [A] time = 1.15136, size = 105, normalized size = 1.44 \begin{align*} -\frac{{\left (3 \, e^{\left (-2 \, a\right )} \log \left (e^{\left (b x + a\right )} + 1\right ) - 3 \, e^{\left (-2 \, a\right )} \log \left ({\left | e^{\left (b x + a\right )} - 1 \right |}\right ) + \frac{2 \,{\left (5 \, e^{\left (3 \, b x + 2 \, a\right )} - 3 \, e^{\left (b x\right )}\right )} e^{\left (-a\right )}}{{\left (e^{\left (2 \, b x + 2 \, a\right )} - 1\right )}^{2}}\right )} e^{\left (2 \, a\right )}}{2 \, b} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(exp(2*b*x+2*a)*csch(b*x+a)^3,x, algorithm="giac")
[Out]
-1/2*(3*e^(-2*a)*log(e^(b*x + a) + 1) - 3*e^(-2*a)*log(abs(e^(b*x + a) - 1)) + 2*(5*e^(3*b*x + 2*a) - 3*e^(b*x
))*e^(-a)/(e^(2*b*x + 2*a) - 1)^2)*e^(2*a)/b | 4,489 | 9,127 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2023-40 | latest | en | 0.090118 |
https://legallysociable.com/tag/significance/ | 1,679,338,124,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296943555.25/warc/CC-MAIN-20230320175948-20230320205948-00333.warc.gz | 416,879,146 | 23,087 | # Wired’s five tips for “p-hacking” your way to a positive study result
As part of its “Cheat Code to Life,” Wired includes four tips for researchers to obtain positive results in their studies:
Many a budding scientist has found themself one awesome result from tenure and unable to achieve that all-important statistical significance. Don’t let such setbacks deter you from a life of discovery. In a recent paper, Joseph Simmons, Leif Nelson, and Uri Simonsohn describe “p-hacking”—common tricks that researchers use to fish for positive results. Just promise us you’ll be more responsible when you’re a full professor. —MATTHEW HUTSON
Create Options. Let’s say you want to prove that listening to dubstep boosts IQ (aka the Skrillex effect). The key is to avoid predefining what exactly the study measures—then bury the failed attempts. So use two different IQ tests; if only one shows a pattern, toss the other.
Expand the Pool. Test 20 dubstep subjects and 20 control subjects. If the findings reach significance, publish. If not, run 10 more subjects in each group and give the stats another whirl. Those extra data points might randomly support the hypothesis.
Get Inessential. Measure an extraneous variable like gender. If there’s no pattern in the group at large, look for one in just men or women.
Run Three Groups. Have some people listen for zero hours, some for one, and some for 10. Now test for differences between groups A and B, B and C, and A and C. If all comparisons show significance, great. If only one does, then forget about the existence of the p-value poopers.
Wait for the NSF Grant. Use all four of these fudges and, even if your theory is flat wrong, you’re more likely than not to confirm it—with the necessary 95 percent confidence.
This might be summed up as “things that are done but would never be explicitly taught in a research methods course.” Several quick thoughts:
1. This is a reminder of how important 95% significant is in the world of science. My students often ask why the cut-point is 95% – why do we accept 5% error and not 10% (which people sometimes “get away with” in some studies) or 1% (wouldn’t we be more sure of our results?).
2. Even if significance is important and scientists hack their way to more positive results, they can still have a humility about their findings. Reaching 95% significance still means there is a 5% chance of error. Problems arise when findings are countered or disproven but we should expect this to happen occasionally. Additionally, results can be statistically significant but have little substantive significance. All together, having a significant finding is not the end of the process for the scientist: it still needs to be interpreted and then tested again.
3. This is also tied to the pressure of needing to find positive results. In other words, publishing an academic study is more likely if you disprove the null hypothesis. At the same time, not disproving the hypothesis is still useful knowledge and such studies should also be published. Think of the example of Edison’s quest to find the proper material for a lightbulb filament. The story is often told in such a way to suggest that he went through a lot of work to finally find the right answer. But, this is often how science works: you go through a lot of ideas and data before the right answer emerges.
# Using a list of “sleep-deprived professions to illustrate statistical and substantive significance”
I ran into a list of “sleep-deprived jobs” yesterday and I think it is a useful tool for illustrating what significance means. The top five sleep deprived jobs (starting with the least rested): home health aide (6 hours, 57 minutes), lawyer, police officers, physicians/paramedics, and economists. The top five jobs with the most sleep (starting with the most rested): forest/logging workers (7 hours, 22 minutes), hairstylists, sales representatives, bartenders, and construction workers. Here is where the data from the list came from:
The lists are based on interviews with 27,157 adults as part of the annual National Health Interview Survey, conducted by a division of the Centers for Disease Control and Prevention. Sleepy’s says its rankings were based on two variables: 1) average hours of sleep that respondents said they got in a 24-hour period, and 2) respondents’ occupations, as they would be classified by the Department of Labor.
Let’s talk about significance. First, statistical significance. The lower value is 6 hours and 57 minutes and the highest value is 7 hours, 22 minutes. We would need to know how the data is clustered, meaning does it look like a normal distribution (meaning most jobs are clumped in the middle) or it is a broader distribution? With a standard deviation, we could figure out how far these highest and lowest values are from the mean and whether they are outside 95% of all the cases.
Perhaps more interesting in this case is the second aspect of significance: even if a case is significantly different from the other cases, is this a meaningful difference in the real world? Just looking at the ten occupations at the top and bottom of this list, the top and bottom are separated by 35 minutes. Would roughly a half hour of sleep really change the quality of life or health between home health aides and forest/logging workers? Of course, sleep might not be the only factor that matters here but is this a meaningful difference? The Mayo Clinic recommends 7-9 hours a night for adults, the National Sleep Foundation also says 7-9 hours a night, and both agree that there are a lot of other factors involved. On the whole then, it appears that the average American (who is in an occupation) is on the low end of recommended sleep (a recurring theme in news stories over the years).
It appears that this list isn’t that helpful if everyone is relatively clustered together. But if we had a little more information, we could know more and determine whether there are (statistically and substantively) significant occupations.
# Possible Fermilab “breakthough” illustrates statistical significance
Scientists at Fermilab may be on the verge of a scientific breakthrough regarding “a new elementary particle or a new fundamental force of nature.” There is just one problem:
But scientists on the Fermilab team say there is about a 1 in 1,000 chance that the results are a statistical fluke — odds far too high for them to claim a discovery.
“That’s no more than what physicists tend to call an ‘observation’ or an ‘indication,’ ” said Caltech physicist Harvey Newman.
For the finding to be considered real, researchers have to reduce the chances of a statistical fluke to about 1 in a million.
One of the key concepts in a statistics or social research course is statistical significance, where researchers say that they are 95% certain (or more) that their result is not just the result due to their sample or chance but that it actually reflects the population or reality. These scientists at Fermilab then want to be really sure that the results reflect reality as they want to reduce their possible error to 1 in a million.
Beyond working with the calculations, the scientists are also hoping to replicate their findings and rule out other explanations for what they are seeing:
Researchers hope that more data compiled at Fermilab will shed light on the matter, or that the Large Hadron Collider in Geneva will be able to replicate the findings. “We will know this summer when we double the data sets and see if it is still there,” said physicist Rob Roser of Fermilab, who is a spokesman for the project…
What the team must to do now, Roser said, is “eliminate all the mundane explanations.” They have been working on that, he said, and decided it was time to go public and let others know what they had found so far.
And science rolls on. | 1,665 | 7,873 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2023-14 | latest | en | 0.909768 |
https://www.tutorialspoint.com/p-twinkle-has-a-total-of-rs-590-as-currency-notes-in-the-denominations-of-rs-50-rs-20-and-rs-10-the-ratio-of-the-number-of-rs-50-notes-and-rs-20-notes-is-3-5-if-she-has-a-total-of-25-notes-how-many-rs-20-notes-does-she-have-p | 1,695,856,215,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510326.82/warc/CC-MAIN-20230927203115-20230927233115-00233.warc.gz | 1,153,360,983 | 18,440 | # Twinkle has a total of Rs. 590 as currency notes in the denominations of Rs. 50, Rs. 20 and Rs. 10. The ratio of the number of Rs. 50 notes and Rs. 20 notes is $3: 5$. If she has a total of 25 notes, how many Rs. 20 notes does she have?
Given:
Twinkle has a total of Rs. 590 as currency notes in the denominations of Rs. 50, Rs. 20 and Rs. 10. The ratio of the number of Rs. 50 notes and Rs. 20 notes is $3: 5$.
Total number of notes $=25$.
To do:
We have to find the number of notes of Rs. 20.
Solution:
Let the number of Rs. 50 notes be $x$, the number of Rs. 20 notes be $y$ and the number of Rs. 10 notes be $z$.
The total number of notes $= 25$
Let $x=3k$ and $y=5k$
This implies,
$z=25-(x+y)$
$z=25-(3k+5k)$
$z=25-8k$
Total amount only in Rs. 50 notes $= 50x=50(3k)=150k$
Total amount only in Rs. 20 notes $=20y=20(5k)=100k$
Total amount only in Rs. 10 notes $=10z=10(25-8k)=250-80k$
Total amount $=Rs.\ 590$
$150k + 100k + 250-80k = 590$
$170k=590-250$
$170k=340$
$k=2$
$\Rightarrow x=3k=3(2)=6$
$\Rightarrow y=5k=5(2)=10$
$\Rightarrow z=25-8k=25-8(2)=25-16=9$
Therefore, the number of Rs. 20 notes she has is 10.
Tutorialspoint
Simply Easy Learning
Updated on: 10-Oct-2022
67 Views | 461 | 1,220 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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} | 4.65625 | 5 | CC-MAIN-2023-40 | latest | en | 0.844473 |
https://www.forexkarma.com/descending-triangle.html | 1,708,614,895,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473819.62/warc/CC-MAIN-20240222125841-20240222155841-00625.warc.gz | 778,716,989 | 8,337 | # Learn Descending Triangle ... Simple and Elegant Continuation Pattern
The descending triangle is a bearish continuation pattern. It is a variation of the symmetrical triangle.
It differs from the symmetrical triangle in forecasting implication that the symmetrical triangle is inherently neutral. In studying the symmetrical triangle the analyst can assume that the previous trend (either uptrend or downtrend) will continue.
It's formation signals that a market is entering a consolidation phase until it gathers enough steam and shoots off the roof.
A Bearish Descending Triangle
In the figure above, the upper trendline represented by a line that passes through points 2-4-6 is descending. The lower trendline represented by a line that passes through points 1-3-5 is flat. These two lines converge at a point O called an apex.
A base represented by a vertical line that passes through points 1-B measures the height of the triangle.
## REQUIREMENTS
The following requirements must be met in order to classify a pattern formation as the descending triangle.
1. The existence of a prior trend is a must. It occurs in a downtrend.
2. It requires four reversal points in mimimum to form the pattern. In the above figure, reversal points 1, 2, 3 and 4 are the bare minimum requirement. This is because at least two points are required to draw a line. So four reversal points are required to draw two converging trendlines B-O and 1-O.
3. The two converging trend lines must be each touched at least twice. In the above figure, the upper descending trendline is touched at points 2, 4, and 6. The lower trendline is touched at points 1, 3, and 5.
4. Two trend lines represented by line 1-3-5 and 2-4-6 must converge at the apex O to form the triangle pattern.
## TIME CONSTRAINT
As the pattern formation progresses forward towards the apex, the time is running out for price to emerge out from its cocoon of consolidation. Prices must shoot down in the downtrend direction.
If prices continue to remain in consolidation for a long time and beyond apex then the pattern loses its significance.
A downside penetration of the lower trendline represented by line 1-3-5 is essential for the completion of the descending triangle formation.
The geometric shapes of the triangle furnishes significant price and time information creating a strategic advantage to a skilled technical analyst.
The two converging trendlines represented by line 1-3-5 and 2-4-6 provides boundary for price. In order for pattern formation completion, price must shoot down in the downtrend direction prior to reaching an apex at point O.
Usually the breakout in the downtrend direction should occur somwehere between two-thirds to three-quarters of the horizontal width of the triangle represented by line that passes through 1-3-5-Q-O.
## SIGNIFICANCE OF VOLUME
As the triangle formation progressess, prices consolidate in a narrow bandwith formed between the lower trendline and the upper trendline. Hence, during this period volume gradually diminishes.
As the price break out in the downtrend direction, the volume bar should rise. Unlike in the case of ascending triangle where significant increase in volume was necessary followed by breakout, price tends to fall out of gravity in the case of bearish descending triangle.
Usually the price return near to the apex just to met resistance at the apex. The return move should occur on lighter volume.
However, the volume should again rise as the price climbs down lower.
## MEASUREMENT
For the descending triangle, downside projection potential can be carried out in the following way.
First measure the height of the line B1 and and project that height as the downside potential from the point of breakout.
In the figure above, the point Q is where the breakout occurred. So line QP is the projection of the line B1. | 806 | 3,870 | {"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} | 2.828125 | 3 | CC-MAIN-2024-10 | latest | en | 0.908184 |
http://www.jiskha.com/search/index.cgi?query=MATH213 | 1,371,641,550,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368708711794/warc/CC-MAIN-20130516125151-00042-ip-10-60-113-184.ec2.internal.warc.gz | 525,518,933 | 6,337 | Wednesday
June 19, 2013
# Search: MATH213
Number of results: 42
math213
2054
Saturday, November 7, 2009 at 10:24pm by fiona
math213
lajnklh
Thursday, October 29, 2009 at 8:27pm by Anonymous
math213
Let U = {u, n, i, t, e} FIND THE SUBSETS
Monday, November 2, 2009 at 11:24pm by brina
MATH213
What if x itself is negative ? e.g. √-(-9) = 3
Monday, January 18, 2010 at 9:58pm by Reiny
math213
poems of associative property
Saturday, November 28, 2009 at 2:48pm by alicia
college math213
Place parenthesis, if needed, to make each of the following equations true: a) 5 + 6 * 3 = 33
Tuesday, July 27, 2010 at 6:10pm by Anonymous
math213
31. Determine whether each of the following numbers is prime or composite. (a) 231 (b) 393
Tuesday, December 15, 2009 at 12:06am by alicia
math213
A jacket that used to sell for \$80 now sells at \$120, a markup of 150% So there!
Tuesday, November 3, 2009 at 11:39pm by Reiny
college math213
place parentheses, if needed, to make each of the following equations true: a.5+6.3=33 b.8+7-3=12 c.6+8-2/2=13 d.9+6/3=5
Tuesday, July 27, 2010 at 6:10pm by fiona
math213
A student reports that it is impossible to mark a product up 150% because 100% of something is all there is. What is your response?
Thursday, November 5, 2009 at 1:18pm by brina
math213
32. If a = 5^2 • 7 • 11 • 13 and b = 2^3 • 5^2 • 7^3 • 17, find the following. (Leave your answer written with exponents.) (a) GCD(a, b) (b) LCM(a,b)
Tuesday, December 15, 2009 at 2:33pm by alicia
math213
11. A student reports that it is impossible to mark a product up 150% because 100% of something is all there is. What is your response?
Tuesday, November 3, 2009 at 11:39pm by brina
MATH213
A student claims that the equation√-x=3 has no solution, since the square root of a negative number does not exist. Why is this argument wrong?
Monday, January 18, 2010 at 9:58pm by Rebecca
MATH213
13. For each of the following, find all possible whole-number replacements that make the following statements true. 2 • ______ + 15 < 27
Monday, December 14, 2009 at 11:16pm by alicia
MATH213
2*x + 15 < 27 Solve for x. 2*x < 12 x < 6
Monday, December 14, 2009 at 11:16pm by Marth
MATH213
That is not a correct method, because some numbers are multiples of both 4 and 5. Those are the ones divisible by 20. The correct answer is 250 + 200 - 50 = 400. Do you see why?
Tuesday, January 26, 2010 at 4:02pm by drwls
MATH213
My calculator goes out one more decimal place to show the difference. I do not know any way your calculator could show it. Sorry.
Wednesday, January 20, 2010 at 5:03pm by PsyDAG
MATH213
a) ¡î(2*121) = ¡î121 ¡î2 = 11¡î2 b) ¡î(144*2) = ¡î144 ¡î2 = 12¡î2 c) ¡î(36*10) = 6¡î10 d) not so sure about d... is ©ú¡î a typo?
Wednesday, January 20, 2010 at 5:02pm by Andy
math213
"Marking up" refers to the price, which can be anything. The amount of the product will stay the same (usually). You can also mark up by decreasing the package size, as is often done with candy bars, coffee etc.
Thursday, November 5, 2009 at 1:18pm by drwls
math213
150 + 54 + 260 = income 22 + 60 + 15 + 58 + 185 = expenses Subtract the expenses from the income to find the money he has left.
Saturday, November 7, 2009 at 10:24pm by Ms. Sue
math213
Say you have 12 books, but room to display only 7. I will open this to everyone - how many differnt arrangements are there if the order of the books matters (permutation) and how many if order is of no consequence (combination)?
Thursday, October 29, 2009 at 8:27pm by brina
college math213
Try putting the parentheses around two numbers and figure out what you get. Don't forget that you do operations in parentheses first. For example: 16-7+5=4 You could do: (16-7)+ 5 = 4 OR 16-(7+5) = 4 Which one would be right?
Tuesday, July 27, 2010 at 6:10pm by Ben
MATH213
12. A student argues that a p% increase in salary followed by a q% decrease is equivalent to a q% decrease followed by a p% increase because of the commutative property of multiplication. How do you respond?
Wednesday, January 20, 2010 at 5:04pm by Rebecca
math213
I need help in finding five websites that contain mathematical activities, manipulatives, or lesson plans for fractions, decimals, or percents. · Prepare an annotated bibliography for each Web site, along with a brief explanation of why each site is worthwhile, and how ...
Monday, December 14, 2009 at 12:55am by alicia
MATH213
Write each of the following in the form a √ b or a ⁿ√ b, where a and b are integers and b has the least value possible: a. √242 b. √288 c. √360 d. ⁿ√162
Wednesday, January 20, 2010 at 5:02pm by Rebecca
math213
Try the divisibility rules: A number is divisible by 2 if the last digit is even. (a) ..1 NO (b) ..3 NO A number is divisible by 3 if the sum of the digits is divisible by three. (a) 2+3+1=6/3=2 with no remainder (b) 3+9+3=12/3=4 with no remainder Can you make the conclusions?
Tuesday, December 15, 2009 at 12:06am by MathMate
math213
a. 24.94189, 24.9419, 24.94199, 24.942 b. -34.2519, -34.251, -34.25, -34.205 Are you sure that you don't have a typo in the last quantity?
Monday, September 5, 2011 at 12:16am by PsyDAG
math213
In each of the following, order decimals from least to greatest a.24,9419, 24.942, 24,94189, 24.94199 b.-34.25, -34.251, -34.205, -34.2519
Monday, September 5, 2011 at 12:16am by Aisha
MATH213
A student wants to know how many integers between 1 and 1000 are a multiple of 4 or a multiple of 5. She wonders if it is correct to find the number of those integers that are multiples of 4 and add the number of those that are multiples of 5. How do you respond?
Tuesday, January 26, 2010 at 4:02pm by Rebecca
math213
Wally kept track of last week’s money transactions. His salary was \$150 plus \$54 in overtime and \$260 in tips. His transportation expenses were \$22, his food expenses were \$60, his laundry costs were \$15, his entertainment expenditures were \$58, and his rent was \$185. ...
Saturday, November 7, 2009 at 10:24pm by brina
MATH213
6. Explain how you would respond to the following: a. A student claims that 9443/9444 and 9444/ 9445 are equal because both display 0.9998941 on his scientific calculator when the divisions are performed. b. Another student claims that the fractions are not equal and wants to ...
Wednesday, January 20, 2010 at 5:03pm by Rebecca
MATH213
The font didn't seem to process, so i did it in txt form again: a) sqrt(2*121) = sqrt(121) sqrt(2) = 11*sqrt(2) b) sqrt(144*2) = sqrt(144) sqrt(2) = 12*sqrt(2) c) sqrt(36*10) = 6*sqrt(10) d) not so sure about d... is nroot(162) a typo?
Wednesday, January 20, 2010 at 5:02pm by Andy
math213
a = 5^2 • 7 • 11 • 13 b = 2^3 • 5^2 • 7^3 • 17 The factors common to both a and b have been highlighted. The product of these factors is the GCD. The LCM is the product of the remaining factors (not highlighted) of both a and b, i.e. 11•13&#...
Tuesday, December 15, 2009 at 2:33pm by MathMate
math213
Write the following argument symbolically and then determine its validity: If you are fair-skinned, you will sunburn. If you sunburn, you will not go to the dance. If you do not go to the dance, your parents will want to know why you didn’t go to the dance. Your parents ...
Tuesday, December 15, 2009 at 12:49am by alicia
math213
If the order does matter than the number of permutations is 12*11*10*9*8*7*6= P(12,7) = 3991680 if the order does not matter (combinations) it would be 12!/(7!5!) = C(12,7) = 792 both P(n,r) and C(n,r) can be found on a standard scientific calculator usually they are labeled ...
Thursday, October 29, 2009 at 8:27pm by Reiny
math213
http://www.jiskha.com/display.cgi?id=1256683010
Tuesday, October 27, 2009 at 10:58pm by Reiny
math213
Suppose you have a friend named Ed. He and his four friends are having ice cream. There are only three flavors available at the ice cream store they are visiting: chocolate, vanilla, and strawberry. One of Ed’s friends, Stacey, eats chocolate exclusively. How many ...
Tuesday, October 27, 2009 at 11:29pm by brina
math213
Suppose you have a friend named Ed. He and his four friends are having ice cream. There are only three flavors available at the ice cream store they are visiting: chocolate, vanilla, and strawberry. One of Ed’s friends, Stacey, eats chocolate exclusively. How many ...
Tuesday, October 27, 2009 at 10:58pm by brina
math213
Chocolate + singles Chocolate + doubles Chocolate + triples Vanilla + singles Vanilla + doubles Vanilla + triples Strawberry + singles Strawberry + doubles Strawberry + triples There are a total of how many different ways? 3+3+3=? Find it, and thats your final answer.
Tuesday, October 27, 2009 at 11:29pm by Dean
MATH213
Calculators usually store more numbers internally than on the screen. Since you get 0.9998941, ask the student to subtract 1 from ANS. (If you don't have an ANS button, you can press the subtraction sign after pressing equal sign for the same effect as pressing "ANS&...
Wednesday, January 20, 2010 at 5:03pm by Andy
math213
http://mathforum.org/dr.math/ Browse the Archive http://www.themathleague.com/ http://www.freemathhelp.com/ http://www.aaamath.com/ http://www.mathisfun.com/ There are various ways to navigate these websites, but you should get plenty of ideas.
Monday, December 14, 2009 at 12:55am by Writeacher
math213
A few more: http://www.freemathhelp.net/eflp/free+math+help/pid153163/D451213/C3500224 http://www.homeschoolmath.net/worksheets/ http://nlvm.usu.edu/en/nav/vlibrary.html http://staff.argyll.epsb.ca/jreed/math7/strand1/1108.htm If you need help with an annotated bibliography, ...
Monday, December 14, 2009 at 12:55am by Writeacher
Pages: 1 | 3,054 | 9,711 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2013-20 | longest | en | 0.928322 |
https://pyronconverter.com/unit/pressure/mpa-mmhg | 1,702,333,778,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679518883.99/warc/CC-MAIN-20231211210408-20231212000408-00717.warc.gz | 528,919,427 | 15,495 | # Free Online Convert Between Megapascal (MPa) & Millimetre of Mercury (mmHg)
## Convert from mmHg to MPa
Convert between Megapascal (MPa) and Millimetre of Mercury (mmHg) instantly with our free online unit calculator. You can easily convert both MPa to mmHg and mmHg to MPa with just a few clicks. To switch between the two conversions, simply use the swap icon (rotating arrows). If you need to start over, you can reset the values by clicking the reset button.
MPa
mmHg
Result: Megapascal = Millimetre of Mercury
i.e. MPa = mmHg
Article Contents []
### What does Megapascal mean?
Megapascal Unit - Details and Description
Megapascal (MPa) is a metric unit of pressure commonly used in engineering and physics. It is defined as one million pascals, which is equivalent to one newton of force per square millimeter.
Megapascals are often used to measure high-pressure systems, such as hydraulic systems, structural engineering, and materials science. For example, the strength of a material can be determined by measuring the stress it can withstand per unit area in MPa.
Megapascals can be converted to other units of pressure, such as bar, psi, and atmospheres, using conversion factors. One MPa is equivalent to 10 bar, 145 psi, or 0.1 atmospheres.
Megapascals can also be used to measure the tensile strength of a material, which is the maximum stress a material can withstand before breaking. This is typically measured in units of MPa, with higher values indicating greater strength.
In addition to its use in pressure and strength measurements, MPa is also used to express other quantities, such as Young's modulus (a measure of stiffness) and shear modulus (a measure of rigidity).
Overall, Megapascal is a widely used unit of pressure and strength measurement in various fields. Its conversion to other units of pressure makes it a convenient and versatile metric for many applications.
In this unit converter website, we have converter from Megapascal (MPa) to some other Pressure unit.
### What does Millimetre of Mercury mean?
Millimeters of mercury (mmHg) is a unit of pressure measurement that is commonly used in medicine and meteorology. It is also known as millimetres of water column, which is abbreviated as mmWC.
The mmHg unit is defined as the amount of pressure exerted by a column of mercury that is 1 millimeter high and has a density of 13.5951 g/cm³, which is the density of mercury at 0°C. This unit of pressure is used extensively in medicine, especially in the measurement of blood pressure.
To understand the concept of mmHg, it is important to first understand what pressure is. Pressure is the amount of force exerted per unit area. In the case of mmHg, it is the amount of force exerted by a column of mercury that is 1 millimeter high.
The normal range for blood pressure in adults is around 120/80 mmHg. This means that the pressure exerted by the blood in the arteries when the heart beats (systolic pressure) is around 120 mmHg, and the pressure exerted when the heart is at rest (diastolic pressure) is around 80 mmHg.
In addition to medicine, mmHg is also commonly used in meteorology to measure barometric pressure, which is the pressure exerted by the Earth's atmosphere. Barometric pressure is an important factor in weather forecasting as it helps predict weather patterns and changes.
Another unit of pressure measurement that is commonly used is the kilopascal (kPa). To convert mmHg to kPa, the following equation is used: 1 mmHg = 0.133 kPa.
In summary, millimeters of mercury (mmHg) is a unit of pressure measurement that is commonly used in medicine and meteorology. It is defined as the amount of pressure exerted by a column of mercury that is 1 millimeter high and has a density of 13.5951 g/cm³. This unit is used extensively in the measurement of blood pressure and barometric pressure.
In this unit converter website, we have converter from Millimetre of Mercury (mmHg) to some other Pressure unit.
### What does Pressure mean?
Pressure is a kind of force. In the language of physics, a force is an object that changes or tries to change its state by acting on it or that changes or tries to change its motion by acting on a moving object.
For example, if someone pushes against a wall, the force is applied whether it changes its position or not. This effort is called force. On the other hand, the force exerted on a single area is called pressure.
This means that if we apply a force in a single case, it will be a force.
Pressure = (force ÷ area)
As an example, it is easy to take a photo with a pin because the area is less. Again, more pressure is required in the area.
The Unit of Pressure
Pressure scalar quantity, because it has value but has no direction. The SI unit of pressure is Pascal, which is equal to Newton per square meter (N/m²). Since an object or group of objects under pressure can act on its surroundings, the pressure is also measured as static energy in a single volume. It is related to the density of energy and can be expressed in joules unit per cubic meter (joules/m³, which is equal to Pascal). To make this concept precise, we use the idea of pressure. The pressure is defined to be the amount of force exerted per area.
Thanks for reading this article! Hopefully, it helps you understand what pressure is and how we measure the pressure of an object.
The following Pressure related conversions are available in our website:
### How to convert Megapascal to Millimetre of Mercury : Detailed Description
Megapascal (MPa) and Millimetre of Mercury (mmHg) are both units of Pressure. On this page, we provide a handy tool for converting between MPa and mmHg. To perform the conversion from MPa to mmHg, follow these two simple steps:
Steps to solve
Have you ever needed to or wanted to convert Megapascal to Millimetre of Mercury for anything? It's not hard at all:
Step 1
• Find out how many Millimetre of Mercury are in one Megapascal. The conversion factor is 7500.62 mmHg per MPa.
Step 2
• Let's illustrate with an example. If you want to convert 10 Megapascal to Millimetre of Mercury, follow this formula: 10 MPa x 7500.62 mmHg per MPa = mmHg. So, 10 MPa is equal to mmHg.
• To convert any MPa measurement to mmHg, use this formula: MPa = mmHg x 7500.62. The Pressure in Megapascal is equal to the Millimetre of Mercury multiplied by 7500.62. With these simple steps, you can easily and accurately convert Pressure measurements between MPa and mmHg using our tool at Pyron Converter.
FAQ regarding the conversion between MPa and mmHg
Question: How many Millimetre of Mercury are there in 1 Megapascal ?
Answer: There are 7500.62 Millimetre of Mercury in 1 Megapascal. To convert from MPa to mmHg, multiply your figure by 7500.62 (or divide by 0.00013332231202220618).
Question: How many Megapascal are there in 1 mmHg ?
Answer: There are 0.00013332231202220618 Megapascal in 1 Millimetre of Mercury. To convert from mmHg to MPa, multiply your figure by 0.00013332231202220618 (or divide by 7500.62).
Question: What is 1 MPa equal to in mmHg ?
Answer: 1 MPa (Megapascal) is equal to 7500.62 in mmHg (Millimetre of Mercury).
Question: What is the difference between MPa and mmHg ?
Answer: 1 MPa is equal to 7500.62 in mmHg. That means that MPa is more than a 7500.62 times bigger unit of Pressure than mmHg. To calculate MPa from mmHg, you only need to divide the mmHg Pressure value by 7500.62.
Question: What does 5 MPa mean ?
Answer: As one MPa (Megapascal) equals 7500.62 mmHg, therefore, 5 MPa means mmHg of Pressure.
Question: How do you convert the MPa to mmHg ?
Answer: If we multiply the MPa value by 7500.62, we will get the mmHg amount i.e; 1 MPa = 7500.62 mmHg.
Question: How much mmHg is the MPa ?
Answer: 1 Megapascal equals 7500.62 mmHg i.e; 1 Megapascal = 7500.62 mmHg.
Question: Are MPa and mmHg the same ?
Answer: No. The MPa is a bigger unit. The MPa unit is 7500.62 times bigger than the mmHg unit.
Question: How many MPa is one mmHg ?
Answer: One mmHg equals 0.00013332231202220618 MPa i.e. 1 mmHg = 0.00013332231202220618 MPa.
Question: How do you convert mmHg to MPa ?
Answer: If we multiply the mmHg value by 0.00013332231202220618, we will get the MPa amount i.e; 1 mmHg = 0.00013332231202220618 Megapascal.
Question: What is the mmHg value of one Megapascal ?
Answer: 1 Megapascal to mmHg = 7500.62.
#### Common Megapascal to Millimetre of Mercury conversion
MPa mmHg Description
0.1 MPa 750.062 mmHg 0.1 MPa to mmHg = 750.062
0.2 MPa 1500.124 mmHg 0.2 MPa to mmHg = 1500.124
0.3 MPa 2250.186 mmHg 0.3 MPa to mmHg = 2250.186
0.4 MPa 3000.248 mmHg 0.4 MPa to mmHg = 3000.248
0.5 MPa 3750.31 mmHg 0.5 MPa to mmHg = 3750.31
0.6 MPa 4500.372 mmHg 0.6 MPa to mmHg = 4500.372
0.7 MPa 5250.434 mmHg 0.7 MPa to mmHg = 5250.434
0.8 MPa 6000.496 mmHg 0.8 MPa to mmHg = 6000.496
0.9 MPa 6750.558 mmHg 0.9 MPa to mmHg = 6750.558
1 MPa 7500.62 mmHg 1 MPa to mmHg = 7500.62
2 MPa 15001.24 mmHg 2 MPa to mmHg = 15001.24
3 MPa 22501.86 mmHg 3 MPa to mmHg = 22501.86
4 MPa 30002.48 mmHg 4 MPa to mmHg = 30002.48
5 MPa 37503.1 mmHg 5 MPa to mmHg = 37503.1
6 MPa 45003.72 mmHg 6 MPa to mmHg = 45003.72
7 MPa 52504.34 mmHg 7 MPa to mmHg = 52504.34
8 MPa 60004.96 mmHg 8 MPa to mmHg = 60004.96
9 MPa 67505.58 mmHg 9 MPa to mmHg = 67505.58
10 MPa 75006.2 mmHg 10 MPa to mmHg = 75006.2
20 MPa 150012.4 mmHg 20 MPa to mmHg = 150012.4
30 MPa 225018.6 mmHg 30 MPa to mmHg = 225018.6
40 MPa 300024.8 mmHg 40 MPa to mmHg = 300024.8
50 MPa 375031.0 mmHg 50 MPa to mmHg = 375031.0
60 MPa 450037.2 mmHg 60 MPa to mmHg = 450037.2
70 MPa 525043.4 mmHg 70 MPa to mmHg = 525043.4
80 MPa 600049.6 mmHg 80 MPa to mmHg = 600049.6
90 MPa 675055.8 mmHg 90 MPa to mmHg = 675055.8
#### Common Millimetre of Mercury to Megapascal conversion
mmHg MPa Description
0.1 mmHg 0.0 MPa 0.1 mmHg to MPa = 0.0
0.2 mmHg 0.0 MPa 0.2 mmHg to MPa = 0.0
0.3 mmHg 0.0 MPa 0.3 mmHg to MPa = 0.0
0.4 mmHg 0.0 MPa 0.4 mmHg to MPa = 0.0
0.5 mmHg 0.0 MPa 0.5 mmHg to MPa = 0.0
0.6 mmHg 0.0 MPa 0.6 mmHg to MPa = 0.0
0.7 mmHg 0.0 MPa 0.7 mmHg to MPa = 0.0
0.8 mmHg 0.0 MPa 0.8 mmHg to MPa = 0.0
0.9 mmHg 0.0 MPa 0.9 mmHg to MPa = 0.0
1 mmHg 0.0 MPa 1 mmHg to MPa = 0.0
2 mmHg 0.0 MPa 2 mmHg to MPa = 0.0
3 mmHg 0.0 MPa 3 mmHg to MPa = 0.0
4 mmHg 0.001 MPa 4 mmHg to MPa = 0.001
5 mmHg 0.001 MPa 5 mmHg to MPa = 0.001
6 mmHg 0.001 MPa 6 mmHg to MPa = 0.001
7 mmHg 0.001 MPa 7 mmHg to MPa = 0.001
8 mmHg 0.001 MPa 8 mmHg to MPa = 0.001
9 mmHg 0.001 MPa 9 mmHg to MPa = 0.001
10 mmHg 0.001 MPa 10 mmHg to MPa = 0.001
20 mmHg 0.003 MPa 20 mmHg to MPa = 0.003
30 mmHg 0.004 MPa 30 mmHg to MPa = 0.004
40 mmHg 0.005 MPa 40 mmHg to MPa = 0.005
50 mmHg 0.007 MPa 50 mmHg to MPa = 0.007
60 mmHg 0.008 MPa 60 mmHg to MPa = 0.008
70 mmHg 0.009 MPa 70 mmHg to MPa = 0.009
80 mmHg 0.011 MPa 80 mmHg to MPa = 0.011
90 mmHg 0.012 MPa 90 mmHg to MPa = 0.012 | 3,622 | 10,864 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-50 | latest | en | 0.906013 |
https://www.airmilescalculator.com/distance/orw-to-zbr/ | 1,716,185,841,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058222.5/warc/CC-MAIN-20240520045803-20240520075803-00463.warc.gz | 573,349,300 | 17,758 | # How far is Chabahar from Ormara?
The distance between Ormara (Ormara Airport) and Chabahar (Chabahar Konarak Airport) is 263 miles / 424 kilometers / 229 nautical miles.
The driving distance from Ormara (ORW) to Chabahar (ZBR) is 321 miles / 516 kilometers, and travel time by car is about 7 hours 26 minutes.
263
Miles
424
Kilometers
229
Nautical miles
## Distance from Ormara to Chabahar
There are several ways to calculate the distance from Ormara to Chabahar. Here are two standard methods:
Vincenty's formula (applied above)
• 263.174 miles
• 423.537 kilometers
• 228.692 nautical miles
Vincenty's formula calculates the distance between latitude/longitude points on the earth's surface using an ellipsoidal model of the planet.
Haversine formula
• 262.721 miles
• 422.808 kilometers
• 228.298 nautical miles
The haversine formula calculates the distance between latitude/longitude points assuming a spherical earth (great-circle distance – the shortest distance between two points).
## How long does it take to fly from Ormara to Chabahar?
The estimated flight time from Ormara Airport to Chabahar Konarak Airport is 59 minutes.
## Flight carbon footprint between Ormara Airport (ORW) and Chabahar Konarak Airport (ZBR)
On average, flying from Ormara to Chabahar generates about 64 kg of CO2 per passenger, and 64 kilograms equals 141 pounds (lbs). The figures are estimates and include only the CO2 generated by burning jet fuel.
## Map of flight path and driving directions from Ormara to Chabahar
See the map of the shortest flight path between Ormara Airport (ORW) and Chabahar Konarak Airport (ZBR).
## Airport information
Origin Ormara Airport
City: Ormara
Country: Pakistan
IATA Code: ORW
ICAO Code: OPOR
Coordinates: 25°16′28″N, 64°35′9″E
Destination Chabahar Konarak Airport
City: Chabahar
Country: Iran
IATA Code: ZBR
ICAO Code: OIZC
Coordinates: 25°26′35″N, 60°22′55″E | 535 | 1,904 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2024-22 | latest | en | 0.831313 |
http://www.jiskha.com/members/profile/posts.cgi?name=Brooke | 1,498,486,519,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320763.95/warc/CC-MAIN-20170626133830-20170626153830-00103.warc.gz | 548,210,293 | 10,164 | # Posts by Brooke
Total # Posts: 735
Art Appreciation
Are Nick Cave's Soundsuits fine art or folk art? Please explain in detail
Art
yes Good Luck!
Art
It represents the eye of God watching over mankind . In the modern era, a notable depiction of the eye is the reverse of the Great Seal of the United States, which appears on the United States one-dollar bill. So I think its c but I would wait until a real tutor answers
math
Peter played his favorite video game for 10 hours last week. Today, Peter's parents restricted him to 5 additional hours each week for the next 8 weeks. Create the function, f(x), that models Peter's total video game time? Is it f(x)=5x + 10 or f(x)=5x + 8 or f(x)=5x. ...
MATHEMATICS
What is the slope of the line that passes through (2, 5) and (?1, 5)? A. Undefined B. 0 I think its 0
Physics
Three resistors are connected in series across a battery. The value of each resistance and its maximum power rating are as follows: 7.1? and 20.9 W, 29.5? and 11.4 W, and 13.6? and 12.5 W. (a) What is the greatest voltage that the battery can have without one of the resistors ...
Physics
Two wires are identical, except that one is aluminum and one is copper. The aluminum wire has a resistance of 0.527?. What is the resistance of the copper wire? Take the resistivity of copper to be 1.72 x 10-8 ?·m, and that of aluminum to be 2.82 x 10-8 ?·m.
Physics
A coil of wire has a resistance of 31.8? at 28 °C and 55.6? at 53 °C. What is the temperature coefficient of resistivity?
chem
10.0 mL of 0.05 M AgNO3 and 10.0 mL 0.05 M NaCl are combined in a test tube and a precipitate of AgCl forms. After the precipitate is centrifuged down, the remaining solution (called the supernatant) is tested for the concentration of Ag+ ion. The concentration is found to be ...
Chrmisty
Find the pH at each of the following points in the titration of 25 ml of 3.0M HF with 0.3 M NaOH. The Ka value is 6.6x10-4. After adding 25 ml of 0.3M NaOH
MAth
Which statement is a true statement all rectangles or squares all squares or rectangles every rhombus is a rectangle every rectangle is a rhombus
Physics
a 4500 kg truck traveling at 17.5 m/s collides with a motionless 1350 kg car. they become locked together. if there is no friction, what is their velocity?
Calculus
Find the dimensions of the right circular cone of maximum volume having a slant height of 5 ft. See the figure.
English
Which of the following is a true statement about an MLA Works Cited page? A. The Works Cited page requires no special formatting. B. The Works Cited page should immediately follow the conclusion and be on the same page. C. The Works Cited page should contain all information ...
algebra
The sum of two numbers is 43. The smaller number is 15 less than the larger number. What are the numbers
Math
Can someone write an irrational number? Just any one.
English
Oops! lol. I read it as Patrice for some reason. Is it A?
English
Pablo insists that a conclusion should reinforce the thesis statement. Patrice says that the conclusion should follow logically from the introduction. Who is correct? A. Both Pablo and Patrice are correct. B. Neither Pablo nor Pam is correct. C. Only Pam is correct. D. Only ...
English
In which of the following sentences does the use of prepositional phrases make it harder for readers to find the main point of the sentence? A. The committee voted on several of the issues under consideration. B. In what way is a political group considered "grassroots&...
English
I'm leaning more towards A.
English
Read the following opening sentence of an introductory paragraph and answer the question that follows. Should you ever find yourself in a situation in which you suspect that someone is listening in your private conversation, you might not be as paranoid as you think. What ...
social studies
10. Which of the following best describes the purpose of the Non-Intercourse Act? (1 point) to forbid the U.S. from trading with Britain and France so the countries would recognize U.S. trading rights to allow U.S. merchants to resume trade with all countries except Britain ...
Social Studies
Why did Adams increase the size of the American Navy? He wanted to prepare for all-out war on Britain He hoped it would deter more attacks by French ships He needed to use tax money to fund an increase in military spending Congress passed a law that required him to increase ...
English
Thank you Ms. Sue!
English
The _______ form of a verb should be used to distinguish between actions completed in the immediate past and actions completed before a specific time. A. past perfect B. simple past C. irregular D. infinitive
Psychology
When researchers find similarities in development between very different cultures, what implications might such findings have for the nature-nurture issue?
Psychology
What study strategies can you think of that would make effective use of the levels-of-processing approach to memory?
math
I got an average of \$3.87 and I'm correct
College math
Determine the balance of \$10,000 is invested at an APR of 9% compounded monthly for seven years.
Psychology
B. was correct, thank you!
Psychology
So the answer would be B. wouldn't it?
Psychology
According to social psychologist Leon Festinger, cognitive dissonance is the psychological tension that occurs when a person holds two contradictory attitudes or thoughts (referred to as cognitions). Cognitive dissonance explains many everyday events involving attitudes and ...
Psychology
With respect to the theory of cognitive dissonance, people can hold contradictory ideas in their minds. If you become aware of the dissonance between two ideas, you could pursue which of the following strategies to reduce the dissonance? B. You can modify your views of the two...
Chemistry
How many grams of O2 are there in a 50.0L tank at 21*C when the pressure is 1.19x10^4mmHg?
Math
A framed picture 20 inches wide and 36 inches high is shown in the diagram below. The picture will be hung on a wall where the distance from the floor is 10 feet. The center of the picture frame must be 2 3/4 feet from the floor. A) determine the distance from ceiling to the ...
History
I agree with C
math
b
Algebra
My bad. That was ment for a different post but close to yours 😉
Algebra
Let us take AC=Length of a ladder BC = Distance between ladder base and wall of a building. AB = Distance which we have to find out.( A point where ladder rests) Triangle formed is angle ABC which is right angle triangle. In angle ABC , AB^2 + BC^2 = AC^2 AB^2 = AC^2 - BC^2 AB...
sails math
A chemist has a solution that is 70% acid and another that is 45% acid. How many liters of each should he mix to obtain 300 liters of a solution that is 65% acid?
math
how long would it take money to triple in an account with a one-time deposit earning 9% interest compounded two times per year?
US History
So it's B not D?
US History
So after reading the section in that artical " The Korean war reaches a Stale Mate" I think I am right on it being D, am I correct Ms. Sue?
US History
I am pretty sure it's D, but would like reassurance please!
US History
Hello, I am filling out a study guide for a test coming up in two days, and I can not seem to find this answer in my text book. Could someone please help? Which of the following is true of the Korean conflict? A. It ended with an enormous American victory. B. It ended roughly ...
Finance
suppose that you save for retirement by contributing \$75 peer month into an account that pays a steady 8% annual interest sounded monthly. How much will be in your account on your 65th birthday. if you begin making these contributions on your 30th?
Chemistry DR BOB
Under standard conditions of 1 atm and 298.15K, the half-cell reduction potential E,zero for the anode in a voltaic cell is 0.37V. The half-cell reduction potential E,zero for the cathode of the cell is 0.67V. The number of moles, n, of electrons transferred in the redox ...
English 11
Write a paragraph summarizing lines 100-250 in the book called the ministers black veil
math
Write the equation for a line that passes through the points (3,-2) and (-1,0). We have to put it in y=mx+b format. So far i have gotten y=-1/2 but i still need the +b part and I'm not sure how to get it.
Social Studies
1. How did World War 1 change warfare? A. By introducing the atomic bomb B. By showing the effectiveness of trench warfare C. By using tanks, airplanes, and machine guns D. By demonstrating the success of the blitzkrieg 2. Which of the following was a result of World War 1? A...
Physics
P= 231fraction numerator f t asterisk times l b over denominator s end fraction t= 14.3s W= ?
Math
A science class has 3 girls and 3 boys in the seventh grade and 4 girls and 1 boy in the eighth grade. The teacher randomly selects a seventh grader and an eighth grader from the class for a competition. What is the probability that the students she selects are both boys?
Physics
A plate made of steel has a 18.9979 mm diameter hole cut into it. A 19.0000 mm diameter ball made of glass is sitting in the hole. Both are at an initial temperature of 9.00 deg. C. If the plate and ball are both heated, at what temperature will the ball just fit through the ...
La
What is the secret to reaching someone with words?which of the columns "concrete mixers" or "the city is so big" or "Harlem night" song best use his words to create a vision or an idea of the city for you in the paragraph state your opinion into ...
Physical Science
Yes, this is A
Language Art
What are the questions?
ART 4 QUESTIONS PLZ HELP ASAP
Yes, Sid is right!
science
which of the following is a constant, or control variable, in this experiment?
Pre- Algebra
You buy x pairs of shoes that each cost \$24.95 the same number of socks that each cost \$4.75 and a pair of pants that costs \$29.99 write an expression in simplest form that represents the total amount of money spent. Would you add them all together?
math
Jose invests money in two simple interest accounts. He invests twice as much in an account paying 10% as he does in an account paying 7%. If he earns \$94.50 in interest in one year from both accounts combined, how much did he invest altogether? Total Principal in Both Accounts...
HELP ME HEALTH 3 QUESTIONS
You were right about 2 and 3 but 1 is B.
Math
The product of 16 and g
Writing Skills
Which of the following statements about connecting paragraphs is correct? A. A good connection between two paragraphs is an implied transition. B. Two paragraphs may be joined by an action verb. C. You can use a pointing word that that refers to a word in the previous ...
English
the mood is urgent. I asked my teacher!!
physics
A 65.5 kg high jumper leaves the ground with a vertical velocity of 8.1 m/s. How high can he jump? The acceleration of gravity is 9.8 m/s2 .
math
The area of a square floor on a scale drawing is 100 square centimeters and the scale of the drawing is 1 centimeter:2ft. What is the area of the actual floor? What is the ratio of the area in the drawing to the actual area?
Math
If the original building is 810 meters tall, what was the scale used to make the model?
English
Thank you both for your answers! I left out the 'I'll still be a customer' part, and in my study guide the letter isn't that long, which is why I didn't add that many sentences.
English
Please review my complaint letter for school? Is it formatted correctly? November 13, 2015 Apple Inc. 1 Infinite Loop Cupertino, CA 95014-5723 To whom it may concern: I am writing to inform you that the product I’ve ordered from your company has not yet arrived. On ...
Social Studies
In the compromise of 1820 what was the status of the unorganized territory?
Please help me N2O5(g) + H2O(l) --> 2HNO3(l) delta H° = -76.2 kJ H2O(l) --> H2(g) + 1/2O2(g) delta H° = 286.0 kJ 1/2N2(g) + 3/2O2(g) + 1/2H2(g) --> HNO3(l) delta H° = -174.0 kJ Calculate delta H° for the reaction 2N2O5(g) --> 2N2(g) + 5O2(g)
Please help me N2O5(g) + H2O(l) --> 2HNO3(l) delta H° = -76.2 kJ H2O(l) --> H2(g) + 1/2O2(g) delta H° = 286.0 kJ 1/2N2(g) + 3/2O2(g) + 1/2H2(g) --> HNO3(l) delta H° = -174.0 kJ Calculate delta H° for the reaction 2N2O5(g) --> 2N2(g) + 5O2(g)
Chemistry
Chemistry
Please help me N2O5(g) + H2O(l) --> 2HNO3(l) delta H° = -76.2 kJ H2O(l) --> H2(g) + 1/2O2(g) delta H° = 286.0 kJ 1/2N2(g) + 3/2O2(g) + 1/2H2(g) --> HNO3(l) delta H° = -174.0 kJ Calculate delta H° for the reaction 2N2O5(g) --> 2N2(g) + 5O2(g)
physcial science
Was that above comment really necessary!? This is generator for those who are wondering :)
help
I'm not sure how I can explain without giving the direct answer but since this posted a long time ago and you should've looked at the post above, this is dividing the voltage by the power :) Let me know if you want more information on this topic, I'll help!
math
does a horizontal line represent function?
math
math
the coordinate (2 -4) is located in which quadrant in the coordinate plane
Math
A food co.packs 24 cans of tuna into a box. The co.has 768 cans to box. If they sell a full box for \$12,how much money will the food co.make on these boxes?
or is it crack
is it a?
this is what is says 1890-95, Americanism; earlier crackajack, rhyming compound based on crack (adj.); -a- as in blackamoor; jack 1in sense “chap, fellow”
Oh is it er????
and the subject is launguage arts
or is it jack?
is it krak?
is the root word for the word crackerjack crack?
Science
Can you Describe what happens when a roller coaster rolls downhill in terms of velocity and acceleration. Is the roller coaster increasing in velocity? Is the roller coaster accelerating?
SS
What do the colonies call themselves in the last part of the document?
SS
Okay. Thank you!! I have a question though the question is what you should do. Should you rebel?
SS
what should people do if a government is not protecting natural rights?
SS
take on the role of a Tory or Patriot and decide if war can be avoided. What are the benefits of preserving peace? What are the reasons for going to war? If the Revolution had been avoided, how would America be different? Subject
Social Studies
The first Englishman who wanted to build a “future Eden” where luxurious items, such as coffee, tea, figs currants, olives, rice and silk, would be produced. This paradise was to be called is it the margravate of azilia?
History
is it the margravate of azilia?
History
The first Englishman who wanted to build a “future Eden” where luxurious items, such as coffee, tea, figs currants, olives, rice and silk, would be produced. This paradise was to be called
Social Studies
what have the earliest people might have seen while crossing the bering land bridge?
Math
Suki's has 54 rock songs,92 dance songs,and 12 classical songs on her play list.If Suki's music player randomly selects a song from the playlist,what is the experimental probability that the song will not be a classical song?Explain your answer.
Can someone PLEASE HELP ME?I am really confused. Suki's has 54 rock songs,92 dance songs,and 12 classical songs on her play list.If Suki's music player randomly selects a song from the playlist,what is the experimental probability that the song will not be a classical ...
Math
Suki has 54 rock songs 92 dance songs and 12 classical songs on her playlist . If suki's music player randomly selects a song from the playlist, what is the experimental probability that the song will not be classical
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Post a New Question | 4,014 | 15,577 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2017-26 | latest | en | 0.938368 |
https://threadreaderapp.com/thread/1039226173153103872.html | 1,571,076,648,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986654086.1/warc/CC-MAIN-20191014173924-20191014201424-00436.warc.gz | 741,160,055 | 22,389 | O.k., tweet-thread time! We have to to talk about constant-time systems! You might think that's all about cryptography and security, but actually this is about availability! Bonus content: Maxwell's Daemon. It's good stuff! NO REFUNDS.
O.k. so some basics ... constant-time algorithms, code, designs, whatever are often notated as O(1) ... which means that there's no relationship between the time taken and the input. For example a pregnancy is O(1), whether you're pregnant with one kid, or twins or triplets.
O(1) is often also short-hand for constant-work; where we spend the same amount of energy or whatever, no matter the inputs/outputs.
Being O(1) often matters for cryptography and security. For example if you check a password sequentially ...
for (int i = 0; i < pwlen; i++) {
if (input[i] != pw[i]) {
return FAILURE;
}
}
return SUCCESS;
that's not O(1) ... and it's bad.
it's bad because it takes longer to run if the password nearly matches, and attackers can measure that, and this is the classic example of a side-channel, and byte by byte they can totally crack your password like this just by measuring time.
The fix is to make it O(1) with awful bitwise operations. It's ugly and mean and hard to follow, but something like:
difference = 0;
for (int i = 0; i < pwlen; i++) {
difference |= pw[i] ^ input[I];
}
return difference == 0;
don't worry if you don't understand this.
Anyway, so Crypto people have been studying this for a long time. Our Automated Reasoning Group built a tool to analyze code and tell you if it's O(1) ... or how close it is ... github.com/awslabs/s2n/tr…
But it turns out that O(1) patterns matter for distributed systems and availability EVEN MORE! Completely unrelated fields really, it's surprising, so let's dig in ...
O.k. so imagine a very simple control plane. Let's say you have a bunch of servers and they do things for customers and their users. Customer gets some knobs and dials, a configuration they can edit. This is lots and lots of web services.
So customer makes a change. What do we do? Well it goes to an API, and that API produces a diff or a delta or whatever you want to call it and sends it to the servers. The servers then apply the patch, and the new config is in place. SIMPLE, RIGHT?
Well, no, sometimes servers are down and errors happen, so you need a workflow engine to drive retries. Oh and that implies you have some way to tell if the change even made it there, so you need a poller or a pusher or something to monitor config propagation status. YOU GET ME.
But then we build all that, and it works, and changes get integrated and customer configs change in seconds, and WE'RE GOOD, RIGHT?
We're good until one day we're not. Imagine a big event happens, maybe a power outage, or a spike on the internet due to the Super Bowl or something, and for whatever reason a bunch of customers all try to make changes at the same time?
The API gets choked, the workflow gets backed up, the status monitors start to lag. OUCH. Even worse, some customers start undoing their changes because they don't see them happening quickly. Now we have pointless changes stuck in the system that no-one even wants!! OUCH OUCH.
Now, let's try a much better, O(1) control plane. DUMB AS ROCKS. Imagine if instead that the customer API pretty much edits a document directly, like a file on S3 or whatever. And imagine the servers just pull that file every few seconds, WHETHER IT CHANGED OR NOT.
This system is O(1). The whole config can change, or none of it, and the servers don't even care. They are dumb. They just carry on. This is MUCH MUCH more robust.
It never lags, it self-heals very quickly, and it's always ready for a set of events like everyone changing everything at once. The SYSTEM DOES NOT CARE.
This is how we build the most critical control planes at AWS. For example if you use Route 53 health checks, or rely on NLB, or NAT GW, the underlying health statuses are *ALWAYS* being pushed around as a bitset. ALWAYS ALWAYS.
A few statuses can change (which is normal), and the system reacts, or they can ALL change, and it will react just the same (which is abnormal). We could have lots of targets suddenly disappear .... break ...
"I felt a great disturbance in the Force, as if millions of voices suddenly cried out in terror and were suddenly silenced. I fear something terrible has happened."
... end break ... and the system does not care or even know how many changed! It just works. Very very reliable, and unde-rapreciated.
BONUS CONTENT: O.k., what does this have to do with Maxwell's Daemon? WHAT EVEN IS THAT? Well the first kind of constant time problem, for crypto, is an example of minimizing information theoretical entropy. It's about minimizing perceptible disorder and what is computable.
The second is more like traditional thermodynamic entropy. With a constant-time control plane, we're trying to build a system that has a constant temperature, because it's always doing the same amount of work.
Ok now if I lost you with that, here's the simple version: when systems change a lot, they undergo stress, and getting stressed at the worst times, like under attacks or in the middle of outages, is really bad. At a high level these problems are the same.
And it turns out at a low level they are the same too! Information theoretical entropy, which is all about computability and bits, can be related to thermodynamic entropy, which is all about energy and work!
How? Well Maxwell proposed a thought experiment that showed a problem with thermodynamics. He imagined two rooms side by side, at even temperatures. He posited that a daemon, a ghost, could open a door between the two rooms and let faster moving molecules over to one side.
This sort of disproved existing thermodynamics, because the work it takes to open a door like that is less than the energy transferred due to the faster molecules changing sides. This was unresolved for a long time ... UNTIL ...
A bunch of folks showed that you don't just open the door, you have to observe and predict, and therefore COMPUTE, the position of the molecule. THIS TAKES ENERGY ... and they showed that the energy it takes is at least equivalent to the difference.
So in a big old nuts, strange-loop sort of way, these completely unrelated things are totally the same at both a MACRO and MICRO level. This blows my mind .... but may also make no sense ;-)
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2) Go to a Twitter thread (series of Tweets by the same owner) and mention us with a keyword "unroll" `@threadreaderapp unroll` | 1,571 | 6,945 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2019-43 | latest | en | 0.943202 |
https://socratic.org/questions/how-do-you-verify-the-identity-sinxcosy-cosxsiny-cosxcosy-sinxsiny-tanx-tany-1-t | 1,582,224,271,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145260.40/warc/CC-MAIN-20200220162309-20200220192309-00152.warc.gz | 535,549,023 | 6,054 | # How do you verify the identity (sinxcosy+cosxsiny)/(cosxcosy-sinxsiny)=(tanx+tany)/(1-tanxtany)?
Feb 15, 2017
see below
#### Explanation:
Right Hand Side:
$\frac{\tan x + \tan y}{1 - \tan x \tan y} = \frac{\sin \frac{x}{\cos} x + \sin \frac{y}{\cos} y}{1 - \sin \frac{x}{\cos} x \sin \frac{y}{\cos} y}$
=((sinxcosy+cosx sin y)/(cosxcosy))/(1-(sinxsiny)/(cosxcosy)
=((sinxcosy+cosx siny)/(cosxcosy))/((cosxcosy-sinxsiny)/(cosxcosy)
$= \frac{\sin x \cos y + \cos x \sin y}{\cos x \cos y} \cdot \frac{\cos x \cos y}{\cos x \cos y - \sin x \sin y}$
$= \frac{\sin x \cos y + \cos x \sin y}{\cancel{\cos x \cos y}} \cdot \frac{\cancel{\cos x \cos y}}{\cos x \cos y - \sin x \sin y}$
$= \frac{\sin x \cos y + \cos x \sin y}{\cos x \cos y - \sin x \sin y}$
$\therefore =$Left Hand Side | 321 | 790 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 7, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2020-10 | latest | en | 0.215359 |
https://gkaim.com/online-test/civil-engineering-test/structural-design-paper-3/ | 1,726,180,775,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651498.46/warc/CC-MAIN-20240912210501-20240913000501-00215.warc.gz | 256,117,755 | 79,192 | # Structural Design Online Test – Paper 3
0%
Created by Vikash chaudhary
This is a FREE online test. Beware of scammers who ask for money to attend this test.
Get New Questions in Each Attempt
• Total Questions: 50
• Time Allotted: 50 minutes
• Passing Score: 70%
• Randomization: Yes
• Certificate: Yes
• Do not refresh the page!
• 👍 All the best!
1 / 50
1. The advantage of reinforced concrete is due to
2 / 50
2. Shear stress diagram of a homogeneous beam is
3 / 50
3. Factor of safety is given by
4 / 50
4. In the analysis of a doubly reinforced beam, the assumption made is
5 / 50
5. The factor of safety for steel as compared to concrete is
6 / 50
6. Bond strength of HYSD bar as compared to plain bar is more by
7 / 50
7. The workability of concrete can be improved by
8 / 50
8. For a reinforced concrete section, the shape of shear stress diagram is
9 / 50
9. Fineness modulus of fine aggregate is between
10 / 50
10. Addition of sugar in concrete results in
11 / 50
11. The beam section is redesigned if the shear stress exceeds the allowable shear stress by ... times
12 / 50
12. A simply supported beam shall be deemed to be a deep beam if the ratio of effective span to overall depth is
13 / 50
13. The minimum straight lap length in tension bars with hooks in an RCC beam is
14 / 50
14. The neutral axis of a T-beam exists
15 / 50
15. Distribution of shear intensity over a rectangular section of a beam follows
16 / 50
16. Bottom reinforcement in a beam is subjected to
17 / 50
17. The maximum diameter of bars in a beam is limited to
18 / 50
18. Shear reinforcement in an RCC beam is used for
19 / 50
19. What is the minimum period for which the lime concrete in the foundation be left wet without the construction of masonry over it?
20 / 50
20. In R.C.C. beams, the tension reinforcement can be cut off at a point when it is no longer needed if
21 / 50
21. The diameter of bars normally used in a column is
22 / 50
22. The entrapped air in concrete
23 / 50
23. The maximum bulking of sand generally occurs at a percentage moisture content of
24 / 50
24. In an RCC beam, N.A means
25 / 50
25. For a cantilever beam, the span to effective depth ratio is generally restricted to
26 / 50
26. The anchorage value of a standard bend is
27 / 50
27. A column is considered as a long column if its slenderness ratio is more than
28 / 50
28. The main reason for placing the main bar at the top in the case of a cantilever slab is
29 / 50
29. The states of concrete are
30 / 50
30. White cement is produced in
31 / 50
31. The amount of reinforcement for main bars in a slab is based upon
32 / 50
32. The minimum number of main bars in an octagonal column is
33 / 50
33. The enlarged head of a supporting column of a flat slab is technically known as
34 / 50
34. In a singly reinforced beam, the effective cover is measured from
35 / 50
35. The maximum strain in the tension reinforcement in the section of a flexural member at failure shall not be less than (if fy is the characteristic strength of steel and Es is the modulus of elasticity of steel)
36 / 50
36. The ratio of diameter of reinforcing bars and the slab thickness is
37 / 50
37. Weep holes are provided in retaining and breast walls to
38 / 50
38. Shear in a concrete beam is caused by
39 / 50
39. The shear reinforcement in an RCC beam is provided to resist
40 / 50
40. In the case of the foundation of rigid base, the distribution pressure on the soil is
41 / 50
41. In cold weather, concrete curing should be continued for... days
42 / 50
42. In a two-way slab, the lifting of corners occurs due to
43 / 50
43. Minimum embedment of reinforcement in a concrete flexural member should be
44 / 50
44. High increase in temperature......the strength of concrete.
45 / 50
45. The minimum vertical spacing of the main bars in an RCC beam should be
46 / 50
46. An under-reinforced section means
47 / 50
47. Concrete gains strength due to
48 / 50
48. The maximum ratio of span to depth ratio of a slab simply supported and spanning in two directions is
49 / 50
49. In a column, the minimum cover provided at the end of reinforcement is
50 / 50
50. The designed main bars are placed in a two-way slab
0% | 1,130 | 4,271 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2024-38 | latest | en | 0.871108 |
https://www.canada.ca/en/services/benefits/publicpensions/cpp/old-age-security/repayment.html | 1,542,348,742,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039742978.60/warc/CC-MAIN-20181116045735-20181116071735-00452.warc.gz | 838,115,647 | 11,563 | # Repayment of Old Age Security pension
## Are you receiving an Old Age Security pension? Is your income for 2017 over \$74,788? Do you live in Canada?
Each February, we send you an income statement – your T4-OAS or NR4-OAS tax information slip – to help you file your income tax return. It shows how much Old Age Security you received the previous year.
In accordance with the Income Tax Act, if your net income before adjustments (line 234 on the tax return) exceeds \$74,788 (for 2017), you may have to repay part or all of your pension. If so, enter this repayment amount on lines 235 and 422 of your income tax return.
## How do I calculate the repayment amount?
Your repayment calculation is based on the difference between your income and the threshold amount for the year. The first step is to figure out how much higher your income is than the threshold. You must repay 15 percent of that amount.
### Example:
The threshold for 2017 is \$74,788.
If your income in 2017 was \$85,000, then your repayment would be 15 percent of the difference between \$85,000 and \$74,788:
\$85,000 - \$74,788 = \$10,212
\$10,212 x 0.15 = \$1,531.80
You would have to repay \$1,531.80 for the July 2018 - June 2019 period.
### Note:
This calculation is presented for reference only and is not official. In order to calculate your repayment amount, please refer to the T1 General Federal Worksheet.
## Will I have to repay part of my pension next year?
If you have to pay back part of your Old Age Security pension this year, an appropriate amount will be deducted from your future OAS pension payments as a recovery tax.
This way, you will have your deductions spread over your 12 monthly pension payments instead of paying back a lump sum at tax time. You will receive an Advisory Letter informing you of any recovery tax deductions being withheld from your OAS pension payments.
Using the example above, if you had to repay \$1,531.80 of your 2017 OAS pension benefits, a recovery tax of approximately \$128 per month would be deducted from your pension payments starting in July 2018.
When a recovery tax deduction is made from your pension payment, it is indicated in box 22 of your T4-OAS or in box 27 of your NR4-OAS tax information slip for that year.
You can claim this amount on line 437 of your return for that year.
You can also contact us to find out how much is being deducted or to request a larger deduction from your future pension payments.
Detailed information on the repayment of OAS benefits can be found in the Income Tax Guide (line 235 – Social benefits repayment).
## What if these deductions cause financial hardship?
• 1-800-267-5177
(toll-free in Canada and the U.S.)
• 613-952-3741
(from all other countries)
• Fax: 613-941-2505
• 1-800-665-0354 (TTY)
(toll-free in Canada and the U.S.) | 678 | 2,828 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2018-47 | longest | en | 0.959743 |
https://www.esaral.com/q/show-that-87821 | 1,726,190,690,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651506.7/warc/CC-MAIN-20240913002450-20240913032450-00057.warc.gz | 679,975,500 | 11,853 | # Show that
Question:
Show that $f(x)=\left\{\begin{array}{cl}\frac{\sin 3 x}{\tan 2 x} & , \text { if } x<0 \\ \frac{3}{2} & , \text { if } x=0 \text { is continuous at } x=0 \\ \frac{\log (1+3 x)}{e^{2 x}-1}, & \text { if } x>0\end{array}\right.$
Solution:
Given:
$f(x)=\left\{\begin{array}{c}\frac{\sin 3 x}{\tan 2 x}, \text { if } \mathrm{x}<0 \\ \frac{3}{2}, \text { if } \mathrm{x}=0 \\ \frac{\log (1+3 x)}{\epsilon^{2 x}-1}, \text { if } \mathrm{x}>0\end{array}\right.$
We observe
$(\mathrm{LHL}$ at $x=0)=\lim _{x \rightarrow 0^{-}} f(x)=\lim _{h \rightarrow 0} f(0-h)=\lim _{h \rightarrow 0} f(-h)$
$=\lim _{h \rightarrow 0}\left(\frac{\sin 3(-h)}{\tan 2(-h)}\right)=\lim _{h \rightarrow 0}\left(\frac{\sin 3 h}{\tan 2 h}\right)=\lim _{h \rightarrow 0}\left(\frac{\frac{3 \sin 2 h}{3 h}}{\frac{2 \tan 2 h}{2 h}}\right)$
$=\frac{\lim _{h \rightarrow 0}\left(\frac{3 \sin 3 h}{3 h}\right)}{\lim _{h \rightarrow 0}\left(\frac{2 \tan 2 h}{2 h}\right)}=\frac{3 \lim _{h \rightarrow 0}\left(\frac{\sin 3 h}{3 h}\right)}{2 \lim _{h \rightarrow 0}\left(\frac{\tan 2 h}{2 h}\right)}=\frac{3 \times 1}{2 \times 1}=\frac{3}{2}$
$(\mathrm{RHL}$ at $x=1)=\lim _{x \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0} f(0+h)=\lim _{h \rightarrow 0} f(h)$
$=\lim _{h \rightarrow 0}\left(\frac{\log (1+3 h)}{e^{2 h}-1}\right)=\lim _{h \rightarrow 0}\left(\frac{3 h \frac{\log (1+3 h)}{3 h}}{\frac{2 h\left(e^{2 h-1}\right)}{2 h}}\right)$
$=\frac{3}{2} \lim _{h \rightarrow 0}\left(\frac{\frac{\log (1+3 h)}{3 h}}{\frac{\left(e^{2 h-1}\right)}{2 h}}\right)=\frac{3}{2} \frac{\lim _{h \rightarrow 0}\left(\frac{\log (1+3 h)}{3 h}\right)}{\lim _{h \rightarrow 0}\left(\frac{\left(e^{2 h_{-1}}\right)}{2 h}\right)}=\frac{3 \times 1}{2 \times 1}=\frac{3}{2}$
And, $f(0)=\frac{3}{2}$
$\therefore \lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{+}} f(x)=f(0)$
Thus, f(x) is continuous at x = 0.
Leave a comment
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https://assignmenthelpweb.com/decision-theory-problems/ | 1,696,298,003,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511023.76/warc/CC-MAIN-20231002232712-20231003022712-00138.warc.gz | 125,155,523 | 14,295 | # Decision theory problems
See attached notes to solve for problems in the attachment. Show work.
3-17
Although Ken Brown is the principal owner of Brown Oil, his brother Bob is credited with making the company a financial success. Bob is vice president of finance. Bob attributes his success to his pessimistic attitude about business and the oil industry. Given the information from problem 3-16, it is likely that Bob will arrive at a different decision. What decision criterion should Bob use, and what alternative will he select?
3-19
Mickey Lawson is considering investing some money that he inherited. The following payoff table gives the profits that would be realized during the next year for each of the three investment alternatives Mickey is considering:
STATE OF NATURE
DECISION GOOD POOR
ALTERNATIVE ECONOMY ECONOMY
Stock Market 80,000 – 20,000
Bonds 30,000 20,000
CDs 23,000 23,000
Probability 0.5 0.5
(a) What decision would maximize expected profits?
(b) What is the maximum amount that should be paid for perfect forecast of the economy?
3-23
Today’s Electronics specializes in manufacturing modern electronic components. It also builds the equipment that produces the components. Phyllis Weinberger, who is responsible for advertising the president of Today’s Electronics on electronic manufacturing equipment, has developed the following table concerning a proposed facility:
PROFIT (\$)
STRONG FAIR POOR
MARKET MARKET MARKET
Large facility 550,000 110,000 – 310,000
Medium-sized facility 300,000 129,000 – 100,000
Small facility 200,000 100,000 – 32,000
No facility 0 0 0
(a) Develop an opportunity loss table.
(b) What is the minimax regret decision?
3-28
A group of medical professionals is considering the construction of a private clinic. If the medical demand is high (i.e., there is a favorable market for the clinic), the physicians could realize a net profit of \$100,000. If the market is not favorable, they could lose \$40,000. Of course, they don’t have to proceed at all, in which case there is no cost. In the absence of any market data, the best the physicians can guess is that there is a 50 – 50 chance the clinic will be successful. Construct a decision tree to help analyze this problem. What should the medical professionals do?
3-29
The physicians in Problem 3-28 have been approached by a market research firm that offers to perform a study of the market at a fee of \$5,000. The market researchers claim their experiences enable them to use Bayes’ theorem to make the following statements of probability:
probability of a favorable market given
a favorable study = 0.82
probability of an unfavorable market given
a favorable study = 0.18
probability of a favorable market given
an unfavorable study = 0.11
probability of an unfavorable market given
an unfavorable study = 0.89
probability of a favorable research
study = 0.55
probability of an unfavorable research study = 0.45
(a) Develop a new decision tree for the medical professionals to reflect the options now open with the market study.
(b) Use EMV approach to recommend a strategy.
(c) What is the expected value of sample information? How much might the physicians be willing to pay for a market study?
3-32
Bill Holliday is not sure what he should do. He can either build a quadplex (i.e., a building with four apartments), build a duplex, gather information, or simply do nothing. If he gathers additional information, the results could be either favorable or unfavorable, but it would cost him \$3,000 to gather the information. Bill believes that there is a 50-50 chance that the information will be favorable. If the rental market is favorable, Bill will earn \$15,000 with the quadplex or \$5,000 with the duplex. Bill doesn’t have the financial resources to do both. With an unfavorable rental market, however, Bill could lose \$20,000 with the quadplex or \$10,000 with the duplex. Without gathering additional information, Bill estimates that the probability of a favorable rental market is 0.7. A favorable report from the study would increase the probability of a favorable rental market to 0.9. Furthermore, an unfavorable report from the additional information would decrease the probability of a favorable rental market to 0.4. Of course, Bill would forget all of these numbers and do nothing. What is your advice to Bill?
3-41
In this chapter a decision tree was developed for John Thompson. After completing the analysis, John was not completely sure that he is indifferent to risk. After going through a number of standard gambles, John was able to assess his utility for money. Here are some of the utility assessments: U( – \$190,000) = 0, U( – \$180,000) = 0.15, U(- \$30,000) = 0.2, U(\$0) = 0.3, U(\$90,000) = 0.5, U(\$100,000) = 0.6, U(\$190,000) = 0.95, and U(\$200,000) = 1.0. If John maximizes his expected utility, does his decision change?
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Customers referred by a friend | 1,834 | 7,758 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2023-40 | latest | en | 0.941688 |
https://somethingwithnumbers.net/what-is-november-28-zodiac-sign/ | 1,686,355,185,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224656869.87/warc/CC-MAIN-20230609233952-20230610023952-00141.warc.gz | 554,103,114 | 20,112 | # what is november 28 zodiac sign
This article is dedicated to the zodiac sign of the new year. So, that’s what it is. And this is a very long article. I might not be able to fit all of those details in here. So, basically, I am going to talk about the zodiac signs for november.
The zodiac is a system of four letter signs that roughly correspond to the four seasons. The zodiac is divided into 120 different sections, each based on specific characteristics. Each section has a different color scheme, and each section has a different symbol. The zodiac is divided into 18 sections to make up the zodiac signs of the year.
The zodiac signs are very similar to each other. Both have the same two letter patterns, but a different symbol. And when you look at the symbols on both signs of each zodiac, you can see the colors of the zodiac signs. You can see the zodiac signs on the right side of the screen.
The symbol is the mark of the zodiac, and it’s part of a larger pattern that marks the zodiac signs. Each symbol in the zodiac has a specific color. So if you look at the symbol on the zodiac sign of a particular year, you should be able to see the color of the sign on the symbol.
The symbol of november 28 is the symbol of the zodiac sign of the year 28, which is the year of the equinox (the day when the sun crosses the equator) and New Year’s Day. It’s a circle with a small wheel underneath it, but the symbol is not a traditional wheel, but something like a two-sided snake for the symbol to represent.
Symbol of the zodiac sign of the year 28 is the symbol of the zodiac sign of the year 28, which is the year of the equinox the day when the sun returns, and the zodiac sign of the year 28 is the year of the year 28. Its a circle with a small wheel underneath it, but the symbol is not a traditional wheel, but an imaginary snake for the symbol to represent.
A snake is a symbol for a negative energy, and the symbol of the zodiac sign of the year 28 is a circle. Negative energy is one of the six signs of the zodiac sign of the year 28, (six signs of the year of the year 28), which is a circle with a small wheel underneath it.
The zodiac sign of the year 28 is a circle with a small wheel underneath it. In ancient astrology, this sign stood for the year of the year 28, and was associated with the year of the year 28. This zodiac sign is also called the sign of the year of the year 28, and it is often used as a symbol for the year of the year 28, as it is a symbol of the month of november.
The zodiac symbol is a circle with a small wheel underneath it. The name zodiac symbol is a very ancient symbol, dating back to the ancient civilizations of ancient Greece and Rome, and it is a symbol for the month of november. This symbol is also called the sign of the year of the year 28, and it is also often used as a symbol for the month of november.
November is the month of the year that is associated with the start of the year, but also the beginning of the year. The november symbol is a circle with a small wheel underneath it. The name november is a very ancient symbol, dating back to the ancient civilizations of ancient Greece and Rome, and it is a symbol for the month of november.
0 comment | 755 | 3,233 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-23 | latest | en | 0.944645 |
https://pcraster.geo.uu.nl/pcraster/4.3.2/documentation/pcraster_manual/sphinx/op_order.html | 1,642,741,173,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320302723.60/warc/CC-MAIN-20220121040956-20220121070956-00132.warc.gz | 463,367,422 | 3,334 | order¶
order
Ordinal numbers to cells in ascending order
Result = order(expression)
expression
spatial scalar, ordinal
Result
spatial scalar
Operation¶
Let n be the number of non missing value cells on expression. These cell values are set in order and numbered on Result in ascending order: the cell with the smallest value on expression is assigned a 1 and the cell with the largest value is assigned a number n. Cells on expression with identical values are assigned consecutive, unique numbers; the order in which these cells are numbered is arbitrarily chosen.
Notes¶
A cell with missing value on expression is not considered in the order operation; it is assigned a missing value on Result.
Group¶
This operation belongs to the group of Order
Examples¶
1. • pcrcalc
binding
Result = Result.map;
Expr = Expr.map;
initial
report Result = order(Expr);
• python | 187 | 878 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-05 | latest | en | 0.838345 |
https://www.venkatsacademy.com/2014/12/energy-stored-in-wire.html | 1,618,849,201,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038887646.69/warc/CC-MAIN-20210419142428-20210419172428-00142.warc.gz | 1,146,479,384 | 14,517 | ## Tuesday, December 16, 2014
### Energy stored in the wire
Whenever a force is applied on a body it produces a change in the shape of the body but due to the elastic nature,the particles tries to come back to their original position. As no body is perfectly elastic there will be a permanent change in the shape of the body which is called as extension or elongation. For producing this elongation we must have done some work and after the job was done, this work and cannot disappear all of a sudden. That is against the law conservation of energy.
This work done hence stored in the format of energy and that is called the energy stored in your wire. It depends on the applied force as well as the extension. Here for producing a small elongation, with respect to the applied force the corresponding work done is also small. The total elongation is the sum of all this kind of small elongation is and the total work done is the sum of all the works done together. To calculate the total amount of the work done therefore we have to add all that works together and that kind of adding is mathematically called integration. In the following diagram we have shown how to derive the equation for the work done.
Related Posts | 255 | 1,229 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2021-17 | longest | en | 0.961688 |
http://math.stackexchange.com/questions/125908/could-someone-show-me-step-by-step-the-evaluation-of-this-basic-as-level | 1,469,682,892,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257827791.21/warc/CC-MAIN-20160723071027-00202-ip-10-185-27-174.ec2.internal.warc.gz | 161,298,424 | 17,405 | # Could someone show me step by step the evaluation of this (Basic/AS Level)
I have:
$(1 \frac{9}{16}) ^\frac{3}{2}$
$\frac{125}{64}$
I'm not sure the order in which everything was done to get the answer. Also, does anyone know of a site like WolframAlpha but which shows the step by step progress of a maths problem?
Thank you!
-
$$(1 \frac{9}{16}) ^\frac{3}{2}=(\frac{25}{16})^\frac{3}{2}=((\frac{5}{4})^2)^\frac{3}{2}=(\frac{5}{4})^3=...$$
$$\left(1 \frac{9}{16} \right)^{3/2} = \left(\frac{25}{16}\right)^{3/2} = \left(\frac{25}{16}\right)^{2/2} \cdot \left(\frac{25}{16} \right)^{1/2} = \frac{25}{16} \cdot \frac{5}{4} = \frac{125}{64}.$$ | 257 | 650 | {"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} | 3.90625 | 4 | CC-MAIN-2016-30 | latest | en | 0.733979 |
https://techdose.co.in/4-sum-problem-leetcode-18/ | 1,675,193,760,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499890.39/warc/CC-MAIN-20230131190543-20230131220543-00124.warc.gz | 578,059,518 | 33,265 | ## 4 Sum Problem | Leetcode #18
This video explains a very important programming interview problem which is the 4 sum problem.This problem can be solved using multiple techniques and algorithms.I have explained all the approaches using simple examples.I have explained 3 techniques.The first one is just by using simple set which is the brute force approach.The second technique is by using set with 2 pointer technique. This is the best approach in terms of time complexity. The third approach is by using hashmap.This solution iis better than the naive approach.
## CODE
```//Simple SET
class Solution {
public:
vector<vector<int>> fourSum(vector<int>& nums, int target) {
set<vector<int>> ans;
int n=nums.size();
sort(nums.begin(),nums.end());
int sum;
for(int i=0;i<n-3;++i) {
//if(i>0 and nums[i]==nums[i-1]) continue; //Skip same values for index i to avoid duplicates
for(int j=i+1;j<n-2;++j) {
//if(j>i+1 and nums[j]==nums[j-1]) continue; //Skip same values for index j to avoid duplicates
for(int k=j+1;k<n-1;++k) {
//if(k>j+1 and nums[k]==nums[k-1]) continue; //Skip same values for index k to avoid duplicates
for(int l=k+1;l<n;++l) {
//if(l>k+1 and nums[l]==nums[l-1]) continue; //Skip same values for index l to avoid duplicates
sum=nums[i]+nums[j]+nums[k]+nums[l];
if(sum>target)
break;
else if(sum==target) {
vector<int> t;
t.push_back(nums[i]);
t.push_back(nums[j]);
t.push_back(nums[k]);
t.push_back(nums[l]);
ans.insert(t);
}
}
}
}
}
vector<vector<int>> res;
for(auto it: ans)
res.push_back(it);
return res;
}
};
/*
//2-pointers + SET
class Solution {
public:
vector<vector<int>> fourSum(vector<int>& nums, int target) {
set<vector<int>> ans;
int n=nums.size();
sort(nums.begin(),nums.end());
int sum;
for(int i=0;i<n-3;++i) {
if(i>0 and nums[i]==nums[i-1]) continue; //Skip same values for index i to avoid duplicates
for(int j=i+1;j<n-2;++j) {
if(j>i+1 and nums[j]==nums[j-1]) continue; //Skip same values for index j to avoid duplicates
int left=j+1,right=n-1;
while(left<right) { //2 pointer technique
sum=nums[i]+nums[j]+nums[left]+nums[right];
if(sum>target)
right-=1;
else if(sum==target) {
vector<int> t;
t.push_back(nums[i]);
t.push_back(nums[j]);
t.push_back(nums[left]);
t.push_back(nums[right]);
ans.insert(t);
left+=1;
}
else
left+=1;
}
}
}
vector<vector<int>> res;
for(auto it: ans)
res.push_back(it);
return res;
}
};
//HashMap
class Solution {
public:
vector<vector<int>> fourSum(vector<int>& nums, int target) {
vector<vector<int>> ans;
int n = nums.size();
if(n<4)
return ans;
sort(nums.begin(),nums.end());
unordered_map<int,vector<pair<int,int>>> m; //key->SUM...Value->(i,j) pair of index where i<j
//Store all sum pairs in map
for(int i=0;i<n-1;++i)
for(int j=i+1;j<n;++j)
m[nums[i]+nums[j]].push_back(make_pair(i,j));
for(int i=0;i<n-1;++i) {
if(i>0 and nums[i]==nums[i-1]) continue;
for(int j=i+1;j<n;++j) {
if(j>i+1 and nums[j]==nums[j-1]) continue;
int sum = target-nums[i]-nums[j];
if(m.find(sum)!=m.end()) {
for(auto it: m[sum]) {
int k=it.first;
int l=it.second;
if(k>j) { //Maintain the increasing order of index (i,j,k,l).....i<j<k<l
//Skip invalid cases
if(!ans.empty() and ans.back()[0]==nums[i] and ans.back()[1]==nums[j] and ans.back()[2]==nums[k] and ans.back()[3]==nums[l])
continue;
vector<int> temp = {nums[i],nums[j],nums[k],nums[l]};
ans.push_back(temp);
}
}
}
}
}
return ans;
}
};
*/```
### One comment on “4 Sum Problem | Leetcode #18”
• Tech Dose January 20, 2022
Nice solution. | 1,039 | 3,491 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2023-06 | latest | en | 0.444535 |
http://www.cloudynights.com/topic/432962-questar-35-vs-televue-85-light-transmission/ | 1,464,122,531,000,000,000 | text/html | crawl-data/CC-MAIN-2016-22/segments/1464049273643.15/warc/CC-MAIN-20160524002113-00068-ip-10-185-217-139.ec2.internal.warc.gz | 449,858,650 | 20,143 | •
# Questar 3.5 vs Televue 85 light transmission
10 replies to this topic
### #1 Michael Lomb
Michael Lomb
Vostok 1
• topic starter
• Posts: 104
• Joined: 26 Oct 2010
• Loc: North Island New Zealand
Posted 11 September 2013 - 03:55 AM
I have read an interesting post on the Astromart site where Malcom Bird (also a contributor to this forum…thanks Malcom), was able to make a direct comparison between his 30 year old Questar and the Televue 85 refractor, on the same mount. He commented, as others have on this forum that the Questar image looked darker when compared to a similar sized refractor. This implied that the optics of the Questar disproportionally lost more light.
http://www.astromart...?article_id=909
There was another post on this forum, where a member contacted Questar and asked what the loss of light was through a Questar 3.5 with Broadband coatings. They discussed this at the shop, and the answer was 1% loss through the reflective surface of the main mirror 1% from the corrector plate. There is another 2% loss through the control box if you do not use the axial port. The central obstruction adds another 10%. In total there is 14 % loss.
Checking on the Company 7 website on the Televue 85, the transmission is 94%, so there is a 6% loss.
Now some basic math to compare the surface area of 89 mm vs 85 mm, and factor out light loss of each scope.
(8.9/2)^2 x pi = 62.21 x .86 = 53.5 for the Questar
(8.5/2)^2 x pi = 56 x .94 = 53.3 for the Televue 85
The brightness, at least for a scope with newer Broadband coatings should be the same.
### #2 Erik Bakker
Erik Bakker
Skylab
• Posts: 4207
• Joined: 10 Aug 2006
• Loc: Netherlands, Europe
Posted 11 September 2013 - 10:00 AM
The brightness, at least for a scope with newer Broadband coatings should be the same.
No.
The way the Q forms an image in the form of an Airydisk with rings is different from the TV85. The peak intensity in the TV85 is considerably higher + it distributes less light into the diffraction rings. In the Q, peak intensity is lower and ring brightness is higher. All this contributes to lowering the visible brightness of the images in the Q when compared to a good refractor of similar light-gathering area.
Do note that the Q has a lot smaller color error than the TV85, and especially in the red end that makes itself notable while observing reddish objects like Mars at opposition.
The forte of the Q is not ultimate performance per mm of aperture, but the superbly integrated and ergonomic package in a very compact and light form. And dare I say, beautiful colors and materials used? A timeless beauty
### #3 Les
Les
Viking 1
• Posts: 888
• Joined: 22 Apr 2006
• Loc: Maryland
Posted 11 September 2013 - 12:49 PM
Eric,
Just to be clear, your argument holds for point targets like stars but not extended objects or terrestrial views as mentioned in the article.
### #4 Erik Bakker
Erik Bakker
Skylab
• Posts: 4207
• Joined: 10 Aug 2006
• Loc: Netherlands, Europe
Posted 11 September 2013 - 04:32 PM
I am afraid it does Les. Extended objects are just a great number of points stitched together, so the effect is still there.
### #5 Les
Les
Viking 1
• Posts: 888
• Joined: 22 Apr 2006
• Loc: Maryland
Posted 11 September 2013 - 07:22 PM
Extended objects like the moon ARE collections of points, but now the diffraction rings fall on adjacent central responses making them brighter. No light is lost from the total scene, only contrast is lost. But your argument for stellar objects is perfectly true, the central response is dimmer for obstructed apertures.
### #6 John F
John F
Viking 1
• Posts: 818
• Joined: 16 Feb 2004
• Loc: Pacific Northwest
Posted 30 September 2013 - 10:39 PM
The forte of the Q is not ultimate performance per mm of aperture, but the superbly integrated and ergonomic package in a very compact and light form. And dare I say, beautiful colors and materials used? A timeless beauty:
Erik, Vey well put and I agree with your assessment. I might add that in one area the Questar really excels and that is its close focus capability. I use mine a lot for viewing terrestrial objects at powers ranging from 53x to 107x. Due to atmospheric turbulence those high powers usually don't work very well for terrestrial targets that are several miles away. However, for nature objects that are within a few hundred feet in distance (and even as close at 8 feet) it performs spectacularly well.
John Finnan
### #7 Erik Bakker
Erik Bakker
Skylab
• Posts: 4207
• Joined: 10 Aug 2006
• Loc: Netherlands, Europe
Posted 01 October 2013 - 12:36 PM
My eyes see it differently at the eyepiece Les. In my experience, a good apo will put up a visibly brighter image at the eyepiece when compared to a similar size Mak or SCT at similar magnifications.
I did extensive comparing with my Q7 with BB en LR coatings and my FS102, as well as the C5 with special coatings and the FS102. Alongside other comparisons, these have empirically formed my opinion on this matter over the years.
Out of curiosity, your sig currently says you have a Q50th and SV90 triplet. How do these two perform for you at the eyepiece? Did you also compare them on driven equatorials at high magnifications over extended periods side-by-side? Looking forward to better understand what shaped your opinion on this matter.
### #8 Les
Les
Viking 1
• Posts: 888
• Joined: 22 Apr 2006
• Loc: Maryland
Posted 03 October 2013 - 04:44 PM
I don't doubt that you see a difference between scopes at the eyepiece. Just the optical explanation was incorrect.
The 90mm triplet mostly occupies a place in my living room looking pretty on its tripod. The Q is just so much easier to use and in my typical local viewing conditions doesn't give up much to the triplet. What good are finer diffraction rings when the central response is a boiling mass. In fact, I am looking for a buyer for the triplet even though it does look gorgeous in the living room. Ironically, its midnight blue paint job is a closer match to the old Questar purple than my current Questars.
### #9 Erik Bakker
Erik Bakker
Skylab
• Posts: 4207
• Joined: 10 Aug 2006
• Loc: Netherlands, Europe
Posted 04 October 2013 - 12:48 AM
Thanks for the clarification Les. The little Q is unbeatable in it's ease of use.
When you say the central response is a boiling mass with a thinner ring in your SV90 triplet, is it a comparably stable mass in your Q 3 1/2, only surrounded by a fatter first ring?
### #10 Les
Les
Viking 1
• Posts: 888
• Joined: 22 Apr 2006
• Loc: Maryland
Posted 04 October 2013 - 09:11 AM
I have not had them in a side by side comparison. Theoretically, there would only be a few percent difference so I doubt that I could see the difference. Most of the viewing difference I note is just due to the 2:1 difference in focal length. But clearly, the Q has more pronounced diffraction rings on bright stars at comparable powers.
### #11 azure1961p
azure1961p
Voyager 1
• Posts: 11596
• Joined: 17 Jan 2009
• Loc: Triton
Posted 08 October 2013 - 10:56 PM
I have read an interesting post on the Astromart site where Malcom Bird (also a contributor to this forum…thanks Malcom), was able to make a direct comparison between his 30 year old Questar and the Televue 85 refractor, on the same mount. He commented, as others have on this forum that the Questar image looked darker when compared to a similar sized refractor. This implied that the optics of the Questar disproportionally lost more light.
http://www.astromart...?article_id=909
There was another post on this forum, where a member contacted Questar and asked what the loss of light was through a Questar 3.5 with Broadband coatings. They discussed this at the shop, and the answer was 1% loss through the reflective surface of the main mirror 1% from the corrector plate. There is another 2% loss through the control box if you do not use the axial port. The central obstruction adds another 10%. In total there is 14 % loss.
Checking on the Company 7 website on the Televue 85, the transmission is 94%, so there is a 6% loss.
Now some basic math to compare the surface area of 89 mm vs 85 mm, and factor out light loss of each scope.
(8.9/2)^2 x pi = 62.21 x .86 = 53.5 for the Questar
(8.5/2)^2 x pi = 56 x .94 = 53.3 for the Televue 85
The brightness, at least for a scope with newer Broadband coatings should be the same.
At thirty years of age those reflective coatings are long overdue for a recoat no?
Pete
## CNers have asked about a donation box for Cloudy Nights over the years, so here you go. Donation is not required by any means, so please enjoy your stay.
Cloudy Nights LLC Cloudy Nights Sponsor: Astronomics | 2,244 | 8,719 | {"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} | 2.5625 | 3 | CC-MAIN-2016-22 | latest | en | 0.923236 |
https://www.mathworks.com/matlabcentral/answers/888442-how-do-i-continuously-calculate-an-output-with-an-increasing-input-of-angles | 1,632,786,327,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780058552.54/warc/CC-MAIN-20210927211955-20210928001955-00018.warc.gz | 873,565,955 | 26,820 | # how do i continuously calculate an output with an increasing input of angles ?
2 views (last 30 days)
rahim njie on 29 Jul 2021
Commented: darova on 1 Aug 2021
i am trying to caluculate 'W' at angles between 0-60 degrees. im not sure how to create a loop function that will allow me to continuously calculate this output whilst varing the angle. I tried using 'theta = 0:60' but that caused an error. my code so far:
theta=(0:60);
w_s = 10; % angular velocity of spin
a_s = 6; % angualar acceleration of spin
w_n = 3; % angular velocity of nutation
a_n = 2; % augalar acceleration of nutation
w_p = 5; % angular velocity of precession
a_p = 4; % angular acceleration of precession
%% Geometry values
rA = [0 7.4103 7.1651];
%% vectors
vw_s = [0 w_s*sind(theta) w_s*cosd(theta)];% spin vector
vw_p = [0 0 w_p]; % precession vector
vw_n = [-w_n 0 0]; % nutation vector
%% Angualar velocity
wi = [-w_n 0 0]; % i component of W
wj = [0 w_s*sind(theta) 0]; % j component W
wk = [ 0 0 (w_p + w_s*cosd(theta))]; % k component of W
W = wi + wj + wk; % Angular Velocity;
darova on 29 Jul 2021
You need to make wi vector the same size as wj and wk
wi = [-w_n+theta*0 0 0]; % i component of W
darova on 1 Aug 2021
plot(W)
R2020b
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Start Hunting! | 436 | 1,350 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2021-39 | latest | en | 0.732528 |
https://econopunk.com/2017/10/ | 1,537,293,665,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267155634.45/warc/CC-MAIN-20180918170042-20180918190042-00397.warc.gz | 500,766,845 | 18,394 | # Month: October 2017
## Portfolio Insurance and Black Monday, October 19, 1987
On the thirtieth anniversary of Black Monday, the stock market crash of October 19th and 20th in 1987, there have been mentions of “portfolio insurance” having possibly exacerbated the crash.
Portfolio insurance, in principle, is exactly what you might expect it to be: if you own a stock, Stock A, you insure it with a put option on Stock A. Your position becomes equivalent to a call option on Stock A until the put option expires, with the price of this position being the premium of the put option when you bought it.
If you are managing a portfolio on behalf of clients, though, and you just need to insure the portfolio up to a certain date, after which, say, you hand over the portfolio, then to buy American put options to insure the portfolio would be unnecessary. European put options would suffice. So let’s suppose that we are only interested in European options.
In the article that I cite at the bottom (Abken, 1987), it seems that at the time, buying put options as insurance had a few issues. This is assuming that the portfolio we want to insure is a stock index: the S&P 500 index. The issues were:
• It’s implied that only American options were available (which we would expect have a premium over European options).
Thus, instead of using put options to insure the portfolio, the portfolio and put options are replicated by holding some of the money in the portfolio and some of it in bonds, Treasury bills, that we assume to provide us with the risk-free rate.
Without worrying about the math, the Black-Scholes equation gives us a way to represent our stock index S and put options P as:
$$S + P = S \cdot N_1 + K \cdot DF \cdot N_2$$
d
Source:
Abken, Peter A. “An Introduction to Portfolio Insurance.” Economic Review, November/December 1987: 2-25.
## Value-added Tax and Sales Tax
(This is mostly a summary of and heavily borrowed from https://en.wikipedia.org/wiki/Value-added_tax, archived).
(The first three figures are taken from Wikipedia).
Comparing No Tax, Sales Tax, and VAT
Imagine three companies in a value chain that produces and then sells a widget to a consumer. The raw materials producer sells raw materials to the manufacturer for $1.00, earning a gross margin (revenue – Cost Of Goods Sold, COGS) of$1.00. The manufacturer sells its product, the widget, to the retailer for $1.20, earning a gross margin of$0.20. The retailer sells the widget to a non-business consumer (for the customer to use and consume) for $1.50, earning a gross margin of$0.30.
No tax example
Imagine we add a sales tax of 10%. Sales tax applies only to the transaction with the final end-user, the consumer, i.e. the final transaction. So the consumer pays the retailer $1.50 + 10% sales tax =$1.65 to the retailer. The retailer remits the sales tax, $0.15, to the government. Sales tax example • Only retailers remit collected sales tax to the government. Different regions, products, and types of consumers may have different sales taxes, so retailers are burdened with the maintenance of functions that process this for every different kind of sales tax. • There are cases (e.g. in the U.S., remote sales, i.e. cross-state or internet sales) where the retailer isn’t required to charge sales tax on its sales to consumers. Instead, the consumer is responsible for remitting a use tax to the government on his or her remote purchases. • Only end-users pay the sales tax. Thus, someone who is an end-user has an incentive to masquerade as a business and purchase products for usage. • The government thus requires businesses (namely non-retailers, in this example) with the burden to prove, via certifications, that it is a business (and thus does not need to pay sales tax on the products it buys) and that it sells to other businesses (and thus does not need to charge and remit sales tax on products it sells). Now let’s take away the 10% sales tax and add a 10% VAT. The raw materials producer charges$1.00 + a $0.10 VAT, which is 10% of the$1.00 value they added to the product) to the manufacturer. It remits the $0.10 VAT to the government. The manufacturer sells its product to the retailer for$1.20 + 10% or $0.12:$0.10 of which is the VAT that the raw materials producer charged the manufacturer and is now getting “paid back” by this transaction and the remaining $0.02 of which is 10% of the value added to the product by the manufacturer, which is$1.20 – $1.00 =$0.20, and remits the $0.02 to the government. The retailer sells its product to the customer for$1.50 + 10% of $1.50 or$0.15 ($0.12 of which is VAT that it paid to the previous 2 companies in the value chain and is now being “paid back” by this transaction and the remaining$0.03 of which is 10% of the value that the retailer added to the product, $0.30) and remits$0.03 to the government.
VAT example
• From the consumer’s viewpoint, nothing has changed. End-users still pay the same $1.65 for the widget. • The government earns the same$0.15 as it earned with sales tax. But instead of receiving all of it from the final transaction between the retailer and the consumer, it earns it in bits of [10% * each value added by each company in the value chain], which are $0.03,$0.02, and $0.10. • From the perspective of each business, they’re charged VAT by companies that they purchase from and they charge VAT to companies/consumers that purchase from them. When they’re charged VAT on purchases ($0.12 for the retailer in the example), they are effectively charged the VAT of all companies that are further down the value chain from them ($0.10 for the raw materials producer and$0.02 for the manufacturer). When they charge VAT on their sales, they effectively charge VAT for the value they added ($0.03) plus the VAT of all companies that are further down them in the value chain ($0.12). The difference, the VAT for the value they added, is remitted to the government ($0.15 –$0.12 = $0.03). So when a company purchases and is charged VAT, they are effectively “in the red” for that amount of VAT ($0.12) until they can sell their product up the value chain and charge that amount of “downstream” VAT + the VAT they own on the value they added ($0.12 +$0.03 = $0.15). Then with that sale, they get “refunded” the portion of VAT that they paid before ($0.12) and remit the remainder ($0.03) to the government. • All buyers, whether they’re a consumer or a business, pay the VAT. So there is no incentive for anyone to masquerade as anyone else (e.g. a consumer to masquerade as a business). • All businesses process VAT (charged VAT on products they buy, charge VAT on products they sell, and pay VAT to the government) so all businesses are burdened with the maintenance of functions that process this. • Because businesses are charged VAT when they buy products and remain “in the red” that amount of VAT until they can sell those products, they are incentivized to make sure they charge VAT on the products they sell in order to make up that VAT they were already charged (e.g. the manufacturer paid the raw materials producer$0.10 of VAT so it’s incentivized to charge VAT on the products it sells to the retailer to make sure to make up for that $0.10). Because everyone is incentivized to charge VAT on their buyers, “everyone collects the tax for the government.” • This symmetry where everyone in the VAT system charges and is charged VAT doesn’t exist when it comes to cross-border trade, which is discussed below. Sales Tax versus VAT • Most countries (166 out of 193 countries) in the world use VAT. The US uses sales tax and is the only one to do so in the OECD. • Asymmetry creates perverse incentives: In sales tax, only retailers charge sales tax and remit sales tax to the government. Consumers want to masquerade as businesses, retailers are not especially incentivized to make sure that their buyers are charged sales tax, and if there are ways to get around sales tax (remote sales, which include cross-state sales and online sales; wholesaling to consumers ), retailers and consumers might want to do that. In the US, retailers don’t need to charge sales tax to consumers buying in a state in which the retailer doesn’t have a physical presence. In this case, there is a use tax charged on the consumer to make up for this, but compliance of use tax is low. (Source, archived.) Estimates of sales tax lost due to remote sales in 2012 varied up to a high of USD$23 billion where total retail sales that year (excluding food sales because many states don’t charge sales tax) was around USD $350 billion a month, i.e. around USD$4.2 trillion that year. (Archived, archived, archived.)
• In VAT, all businesses process VAT the same way, so there are no such perverse incentives.
• The big exception to this is cross-border trade, i.e. imports and exports. Governments have a choice of whether to charge VAT on goods it exports and goods it imports and whether to charge differently for every country it trades with. This creates a huge potential for asymmetries in the VAT system.
• Imports and Exports:
• Sales Tax countries charge sales tax on imported goods if and when they reach the end-user. If the imported good is exported again, then it hasn’t reached an end-user, and thus is never sales taxed.
• Both sales tax and VAT are consumption taxes – the purpose is to tax consumption. This is why the sales tax doesn’t tax a good that is imported and then exported without being consumed in the country. VAT accomplishes this as well, but also would ideally keep the cross-border trade situation simple and sensible when dealing with other VAT countries or sales tax countries.
• In order to have as symmetric and fair a system of cross-border trade, VAT countries generally:
• Do not charge VAT goods that are exported. When a good is exported by an exporter, the government refunds the exporter the entire VAT that it paid on its cost of goods sold purchases;
• Charge VAT on goods that were imported on that good’s first subsequent sale that occurs after importation for the full sale value (not just the value added by the importer, which is [sale price – cost of goods sold], but the full [sale price]). I.e. after the importer imports the good, when that importers sells that good, that sales transaction is VAT-taxed for the full sale price. This is assuming that the imported good is being sold to another domestic company and not being immediately exported.
• Note that if an imported good is exported, the government does not receive any VAT. This is the same as in sales tax and it accomplishes what a consumption tax is supposed to do (which in the case of a good that is imported and then exported without being consumed in the country is to not tax the good). Each company in the value chain plays its usual part in the VAT system, but the last one, the exporter, is refunded by the government all the VAT it paid on its purchases of cost of goods sold.
Cross-border VAT
• The reason VAT countries don’t tax their goods upon export is because sales tax countries don’t tax their goods upon export, so this keeps that part of the trade symmetrical. This also prevents any case of a good being double-taxed during a cross-border trade.
• The reason VAT countries tax their import goods (subsequent to the import transaction) is because if that good is going to be consumed in the country, not VAT-taxing it would mean the good would be untaxed during and after its cross-border transaction and thus have an advantage over similar competing domestic goods at this and every following point on the value chain since domestic goods have been VAT-taxed up to this point and will be VAT-taxed on all following points on the value chain.
• The reason why an imported good’s subsequent sales transaction is VAT-taxed its full sales price instead of just the value added by the importer is because:
• If the good is only taxed by its value-added amount instead, this still is not enough to offset the disadvantage of domestic goods (which have been VAT-taxed for all value that has been added to the product up to that point, not just the value added by the last company to sell it).
• The government of the country in which the good is consumed ought, in principle, to capture the entire consumption tax on the good. Thus, when the good went across the border and the exporting country’s government refunds the VAT to its exporter, the good is effectively “untaxed” at this point (the exporting country’s government has refunded all previous VAT on it and the importing country’s government has yet to tax any of the value that has so far been added by producers of the exporting country). By VAT-taxing it by its full sales price after importation, the government of the importing country captures the VAT that the exporting country’s government refunded to its exporter or “resets” the VAT to where it ought to be at this point in the value chain for itself.
• Missing trader fraud/Carousel fraud: A type of fraud that exists when a good is imported into and then exported out of a VAT country without the good being consumed in the country. Since cross-border trade is a point of asymmetry in the VAT system, it makes sense that this is where fraud occurs.
• Company A imports a good legitimately, paying the exporter EUR 100 for the good. This transaction is VAT-free.
• Company A sells the good to Company B for EUR 110. This transaction is VAT-taxed its full sales price (since it is the transaction subsequent to the good being imported and also is not being immediately exported), and Company A owes the government this VAT. Note that the good in this transaction is “new” to the country and thus the government has not received any VAT from this good further down the value chain (that VAT has been collected by the exporter’s government and refunded back to the exporter). If the VAT is 20%, that 20% is charged on the full EUR 110 sale price of the transaction (not the value added by Company A, which is EUR 10). The total price of the transaction with VAT to EUR 132 and the government expects to be remitted a VAT of EUR 22 from Company A for this sales transaction.
• Company B exports the good. This transaction is VAT-free. Furthermore, Company B has paid Company A a VAT of EUR 22, so it is entitled to a refund of EUR 22 from the government as the good is being exported and not consumed in the country.
• Company A disappears or goes bankrupt without paying the VAT (of EUR 22) on the sale of goods by Company A to Company B. This is key to the fraud because Company A is supposed to pay VAT for the full sales price of its sale of the good (20% * EUR 110 = EUR 22), not just VAT of the value added (20% * EUR 10 = EUR 2). In the diagram above that depicts cross-border trade, the retailer would owe the government $0.15 of taxes, not$0.03 as in the diagrams that are further above that don’t depict cross-border trade. By Company A disappearing, the government is losing the VAT of all value that has been added to the good by companies from this point and all the way down the value chain.
• With no fraud occurring, the government is supposed to earn 0 tax: charge VAT starting from the importer selling the good to domestic companies but then refund all that VAT to the exporter at the end who exports the good since the good is not to be consumed in the country. But instead, in this case where Company A disappears, the government has lost EUR 22 by refunding Company B, the exporter, for the VAT that it paid on its purchase of the good.
• In reality, if the importer (Company A) and the exporter (Company B) are working together, the good may never even physically leave the port, and is imported, sold, and exported only on paper. If there are many companies in between Company A and Company B (e.g. it could instead by A -> X -> … -> Z -> B), it could be difficult for the government to prove any wrongdoing by Company B as the link between A and B will be weak (Company B may even be innocent in some cases where the only fraud is Company A disappearing without paying its VAT) and thus the government will be obligated to refund the VAT to Company B that it paid on its purchases.
• According to sources found in Wikipedia, it’s estimated that the UK annually lost around GBP 2 billion in 2002-2003 (archived) and between GBP 2 billion and GBP 8 billion annually for the years (archive) between 2004 and 2006 due to this kind of fraud. Total UK retail sales (archived) for these years was around GBP 250 million to GBP 300 million. For the EU, 2008 estimates were EUR 170 billion lost (archived) due to this type of fraud. Total EU-27 retail turnover in 2010 was around EUR 2.3 trillion (archived).
• In the above link to a BBC article from 2006, it says that in order to combat the losses from this fraud, the government is implementing a new system where:
Under the new rules the last company to sell on goods like mobile phones – such as a retailer – will be responsible for paying the VAT.
So it sounds like the portion of VAT that the government is “missing” from the imported good will be paid by the retailer instead of the importer – in other words, it’s a bit like the sales tax system. In this system that’s described, if a good is imported and then exported, the amount that the government would refund the exporter will be smaller than in the previous system, making the fraud much less damaging. And if the retailer disappears without paying the taxes it owes, that’s the same as a retailer disappearing in a sales tax system without paying taxes, or a raw materials producer in a VAT system disappearing without paying taxes. (These cases are a much simpler sort of tax evasion and unlike the missing trader/carousel fraud where the importer disappearing and not paying the “missing” chunk of VAT that the government is owed is in combination with the exporter that is “refunded” that chunk of VAT from the government, even though the government never received that chunk of VAT from the missing trader.)
Instead of taxing the importer the “missing” VAT, tax the retailer the “missing” VAT
In Country A, the government refunds the manufacturer/exporter the VAT that it paid on its purchases. In Country B, the importer is charged VAT only on its value added, and that VAT is remitted to the government. The retailer is charged VAT on its value added (10% * $0.20 =$0.02) and on the “missing” VAT from the value added to the product prior to importation, which is $1.20 (the price that the importer paid the manufacturer/exporter), so that comes to 10% *$1.20 = $0.12. If the good is exported, refund the exporter as usual If the good is imported into Country B and then exported without reaching a consumer, the amount of VAT that the importer is charged is only on its value added (10% *$0.10 = $0.01). Thus, the amount of VAT that the government refunds the exporter is only that amount ($0.01). If the importer disappears, the government only loses $0.01, which is 10% of the value added by the importer, not 10% of the full sales price of the product at this point. Furthermore, if the importer and exporter work together and the importer sells the good to the exporter at a much higher price (raising the value added and the potential VAT that the government is supposed to refund the exporter), the exporter still needs to legitimately export the product in order to qualify for the refund. It’s theoretically possible for the exporter to operate at a loss to make this possible, but this might raise an additional red flag that the government would become suspicious of, raising the risk of doing the fraud. Economic Impact of Sales Tax versus VAT Back to the diagrams for no tax, sales tax, and VAT: No tax Sales tax VAT Although only the consumer actually pays the tax in the end, as prices are raised at other transaction points, there is some friction that will discourage those transactions by some amount. One can also think of this as an overhead cost for the businesses involved. In the no tax and sales tax situations, the manufacturer buys goods at$1.00 and sells them at $1.20. In the VAT situation, the manufacturer buys goods at$1.10 and sells them at $1.32, which minus the VAT remitted to the government becomes$1.30. In both cases, the manufacturer earns a profit of $0.20, but there is an overhead of$0.10 in the VAT case. An extreme analogy is: if you are a company that makes a profit of $1 on each good you sell, would you rather buy goods for$2 and sell them for $3 to earn your$1 profit or buy goods for $1,002 and sell them for$1,003 to earn your $1 profit? Surely the former is easier and has less friction. Back to the economic interpretation: if the demand and supply curve of a transaction point in a no tax situation is this: then by adding a tax to the transaction, the price is increased. For convenience, we add a second supply curve that is “supply + tax.” While the end result is the same if we left the supply curve alone and added a “demand – tax” curve instead, the supply + tax curve more conveniently takes the hypothetical price of a product sold (the supply curve) and then adds the hypothetical price + tax of a product sold (the supply + tax curve). What one can also do instead of drawing a new curve is take the vertical distance of the final tax per product and “fit it in between” the two curves from the left side of the diagram, and the end result will be the same. (The diagram says “Consumer Surplus” but since this may represent a business-to-business transaction, it’d be clearer to just say “Purchaser Surplus.”) So a tax will cause less quantity to be transacted, a higher post-tax price, some government tax revenue, lower purchaser and producer surpluses, and some deadweight loss. Unless the government tax revenue is spent in a way that can overcome that deadweight loss (e.g. spending on things that have positive externalities), we have an inefficient outcome. So in the VAT system, businesses are contending with higher prices (which is like more overhead) and lost quantity transacted compared to the sales tax system. This is another cost that the VAT system pays (in addition to the missing trader/carousel fraud) in order to have a “symmetric” system where almost everyone in the value chain pays and collects VAT. ## Testing MathJax-LaTeX https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference At first, we sample $$f(x)$$ in the $$N$$ ($N\$ is odd) equidistant points around $$x^*$$:
$f_k = f(x_k),\: x_k = x^*+kh,\: k=-\frac{N-1}{2},\dots,\frac{N-1}{2}$
where $$h$$ is some step.
Then we interpolate points $$(x_k,f_k)$$ by polynomial
\label{eq:poly} \tag{1}
P_{N-1}(x)=\sum_{j=0}^{N-1}{a_jx^j}
Its coefficients $$a_j$$ are found as a solution of system of linear equations:
\label{eq:sys} \tag{asdf}
\label{eq:sys2} \tag{asdf2}
Backslash left and right parentheses:
$\left( \frac{1}{2} \right) \qquad ( \frac{1}{2} ) \\ ( \frac{1}{2} )$
$\left( \frac{1}{2} \right) \qquad \left( \frac{1}{2} )$
$$1 \quad \frac ab \quad 2 \quad \frac{c}{d} \quad 3 \quad {e \over f} \quad 4 \quad {}^g/_h \quad 5 \quad i/j \quad 6 \quad$$
$$1+1=2 \textrm{ centered equation } 1+1=2$$
$$1+1=2 \textrm{ left equation } 1+1=2$$
\begin{align}
1 + 1 & = 2.00000000 \textrm{ aligned to character}\\
& = 2.0000000000000000 \\
& = 1.99999999999 \\
\end{align}
Here are references to existing equations: \ref{eq:poly}, \eqref{eq:sys}.
Here is reference to non-existing equation \eqref{eq:unknown}.
X=
\begin{cases}
0, & \text{if}\ a=1 \\
1, & \text{otherwise}
\end{cases}
$$\lim_{x\to 1}$$
$$\lim_{x\to 1}$$
$$default, \it Italics, \bf bold, \sf sans serif, \tt typewriter, \rm default Roman, \it italics$$
$$horizontal spacing: back slash\ comma\, ! \! > \> : \: ; \; enspace \enspace quad \quad qquad \qquad end$$
$$hskip1point \hskip1pt hskip2point \hskip 2pt hskip10point \hskip10pt hskip3point \hskip 3pt 1ex \hspace{1ex} 1em \hspace{1em} 2em \hskip2em lengthofasdf \hphantom{<asdf>} backslash \ tilde ~ end$$
$$\tiny tiny$$
$$default$$
$$\scriptsize scriptsize \small small \normalsize normalsize or default, \large large$$
$$\normalsize normalsize or default, \large large$$
$$\Large Large \LARGE LARGE \huge huge \Huge Huge1$$
$$\Large \LARGE \huge \Huge Huge2$$
$$\Huge Huge3$$ | 5,807 | 24,422 | {"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": 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} | 2.796875 | 3 | CC-MAIN-2018-39 | longest | en | 0.93606 |
https://es.mathworks.com/matlabcentral/cody/problems/1722-find-the-next-prime-number/solutions/2294036 | 1,606,994,353,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141727627.70/warc/CC-MAIN-20201203094119-20201203124119-00198.warc.gz | 283,048,503 | 17,166 | Cody
# Problem 1722. Find the next prime number
Solution 2294036
Submitted on 19 May 2020 by Yuan
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
n = 1; out = 2; assert(isequal(nextprimenum(n),out))
2 Pass
n = 7; out = 11; assert(isequal(nextprimenum(n),out))
3 Pass
n = [1 2 3 4 5 6 7 8 9]; out = [2 3 5 5 7 7 11 11 11]; assert(isequal(nextprimenum(n),out))
4 Pass
n = [71 25 63 47 65 36 47 58 69]; out = [73 29 67 53 67 37 53 59 71]; assert(isequal(nextprimenum(n),out))
5 Pass
n = [171 255 636 487 675 369 477 538 969]; out = [173 257 641 491 677 373 479 541 971]; assert(isequal(nextprimenum(n),out))
6 Pass
n = [172541 255564 632436 4564587 6778675 334469 475647 575638 96879]; out = [172553 255571 632447 4564589 6778691 334487 475649 575647 96893]; assert(isequal(nextprimenum(n),out))
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Start Hunting! | 390 | 1,058 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-50 | latest | en | 0.446159 |
http://mathhelpforum.com/calculus/147190-trigonometric-differentiation-help.html | 1,480,858,377,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541321.31/warc/CC-MAIN-20161202170901-00222-ip-10-31-129-80.ec2.internal.warc.gz | 171,106,535 | 10,292 | 1. ## Trigonometric differentiation help!
Hi, I'm having a lot of trouble solving the following question related to trig differentiation. I would greatly appreciate any help.
1) Differentiate $tan^2 (cos x)$
Thanks
2. -2(sinx)tan(cosx)[sec(cosx)]^2
3. Originally Posted by nikhil
-2(sinx)tan(cosx)[sec(cosx)]^2
Yes, I'm pretty sure thats the answer..Would you mind sharing a bit of the working?
4. d/dx (tan(cosx))^2
= 2tan(cosx)[d/dx(tan(cosx))]
=2tan(cosx)(sec(cosx))^2(d/dx (cosx))
=2tan(cosx)(sec(cosx))^2(-sinx)
=-2(sinx)tan(cosx)(sec(cosx))^2
5. Nikhil used the chain rule.
Let u= cos(x),[/tex]v= tan(u)[/tex], and $w= v^2$
Then $\frac{dw}{dv}= 2v$, $\frac{dv}{du}= sec^2(u)$, and $\frac{du}{dx}= - sin(x)$
The chain rule says
$\frac{dw}{dx}= \frac{dw}{dv}\frac{dv}{du}\frac{du}{dx}$
$= (2v)(sec^2(u))(-sin(x))= (2tan(u))(sec^2(cos(x))(-sin(x))$ $= -2tan(cos(x))sec^2(cos(x))sin(x)$.
Of course, nikhil did it with specifically writing out "u", "v", and "w". | 364 | 974 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 8, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2016-50 | longest | en | 0.698899 |
http://slideplayer.com/slide/5870969/ | 1,670,550,222,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711376.47/warc/CC-MAIN-20221209011720-20221209041720-00008.warc.gz | 47,773,557 | 24,821 | # Lesson 8.1, page 782 Matrix Solutions to Linear Systems
## Presentation on theme: "Lesson 8.1, page 782 Matrix Solutions to Linear Systems"— Presentation transcript:
Lesson 8.1, page 782 Matrix Solutions to Linear Systems
Objective: To solve systems using matrix equations.
Review Solve this system of equations.
4x + 2y = -3 x + 5y = -4
Systems of Equations in Two Variables
Matrices A rectangular array of numbers is called a matrix (plural, matrices). The rows of a matrix are horizontal. The columns of a matrix are vertical. The matrix shown has 2 rows and 3 columns. A matrix with m rows and n columns is said to be of order m n. When m = n the matrix is said to be square. See Example 1, page 783.
Matrices Consider the system: 4x + 2y = -3 x + 5y = -4
Example: The matrix shown above is an augmented matrix because it contains not only the coefficients but also the constant terms. The matrix is called the coefficient matrix.
Row-Equivalent Operations, pg. 784
1) Interchange any two rows. (Ri Rj) 2) Multiply each entry in a row by the same nonzero constant. (kRi) 3) Add a nonzero multiple of one row to another row. (kRi + Rj)
See Example 2. Check Point 2, page 785: Perform each indicated row operation on (R1 R2) ¼R1 3R2 + R3 = new R3
Solving Systems using Gaussian Elimination with Matrices
Write the augmented matrix. Use row operations to get the matrix in “row echelon” form: Write the system of equations corresponding to the resulting matrix. Use back-substitution to find the system’s solution.
Example – Watch and listen.
Solve the following system:
First, we write the augmented matrix, writing 0 for the missing y-term in the last equation.
Our goal is to find a row-equivalent matrix of the form
Example continued R R2 We multiply the first row by 2 and add it to the second row. We also multiply the first row by 4 and add it to the third row.
We multiply the second row by 1/5 to get a 1 in the second row, second column.
We multiply the second row by 12 and add it to the third row.
R3 = -12R2 + R3
Example continued Now, we can write the system of equations that corresponds to the last matrix :
Example continued We back-substitute 3 for z in equation (2) and solve for y.
Example continued Next, we back-substitute 1 for y and 3 for z in equation (1) and solve for x. The triple (2, 1, 3) checks in the original system of equations, so it is the solution.
See Example 3. Check Point 3: Use matrices to solve the system 2x + y + 2z = 18 x – y + 2z = 9 x + 2y – z = 6. Write the augmented matrix for the system. Solve using Gauss Elimination or your calculator. Check/verify your solution.
See Example 3. 2x + y + 2z = 18 x – y + 2z = 9 x + 2y – z = 6.
Row-Echelon Form, page 790 1. If a row does not consist entirely of 0’s, then the first nonzero element in the row is a 1 (called a leading 1). 2. For any two successive nonzero rows, the leading 1 in the lower row is farther to the right than the leading 1 in the higher row. 3. All the rows consisting entirely of 0’s are at the bottom of the matrix. If a fourth property is also satisfied, a matrix is said to be in reduced row-echelon form: 4. Each column that contains a leading 1 has 0’s everywhere else.
Example -- Which of the following matrices are in row-echelon form?
a) b) c) d) Matrices (a) and (d) satisfy the row-echelon criteria. In (b) the first nonzero element is not 1. In (c), the row consisting entirely of 0’s is not at the bottom of the matrix.
Gauss-Jordan Elimination
We perform row-equivalent operations on a matrix to obtain a row-equivalent matrix in row-echelon form. We continue to apply these operations until we have a matrix in reduced row-echelon form.
Gauss-Jordan Elimination Example
Example: Use Gauss-Jordan elimination to solve the system of equations from the previous example; we had obtained the matrix
Next, we multiply the second row by 3 and add it to the first row.
Gauss-Jordan Elimination continued -- We continue to perform row-equivalent operations until we have a matrix in reduced row-echelon form. Next, we multiply the second row by 3 and add it to the first row.
Gauss-Jordan Elimination continued
Writing the system of equations that corresponds to this matrix, we have We can actually read the solution, (2, 1, 3), directly from the last column of the reduced row-echelon matrix.
Lesson 8.2 Special Systems
When a row consists entirely of 0’s, the equations are dependent and the system is equivalent, meaning you have a system that is colinear. Answer: INFINITELY MANY SOLUTIONS
Lesson 8.2 Special Systems
When we obtain a row whose only nonzero entry occurs in the last column, we have an inconsistent system of equations. For example, in the matrix the last row corresponds to the false equation 0 = 9, so we know the original system has no solution.
Check Point 2, page 797 Solve the system: x – 2y – z = 5 | 1,285 | 4,887 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.78125 | 5 | CC-MAIN-2022-49 | latest | en | 0.852255 |
https://www.crazy-numbers.com/en/17206 | 1,556,102,131,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578640839.82/warc/CC-MAIN-20190424094510-20190424120510-00369.warc.gz | 632,576,563 | 4,301 | Discover a lot of information on the number 17206: properties, mathematical operations, how to write it, symbolism, numerology, representations and many other interesting things!
## Mathematical properties of 17206
Is 17206 a prime number? No
Is 17206 a perfect number? No
Number of divisors 8
List of dividers 1, 2, 7, 14, 1229, 2458, 8603, 17206
Sum of divisors 29520
## How to write / spell 17206 in letters?
In letters, the number 17206 is written as: Seventeen thousand two hundred and six. And in other languages? how does it spell?
17206 in other languages
Write 17206 in english Seventeen thousand two hundred and six
Write 17206 in french Dix-sept mille deux cent six
Write 17206 in spanish Diecisiete mil doscientos seis
Write 17206 in portuguese Dezessete mil duzentos seis
## Decomposition of the number 17206
The number 17206 is composed of:
1 iteration of the number 1 : The number 1 (one) represents the uniqueness, the unique, a starting point, a beginning.... Find out more about the number 1
1 iteration of the number 7 : The number 7 (seven) represents faith, teaching. It symbolizes reflection, the spiritual life.... Find out more about the number 7
1 iteration of the number 2 : The number 2 (two) represents double, association, cooperation, union, complementarity. It is the symbol of duality.... Find out more about the number 2
1 iteration of the number 0 : ... Find out more about the number 0
1 iteration of the number 6 : The number 6 (six) is the symbol of harmony. It represents balance, understanding, happiness.... Find out more about the number 6
Other ways to write 17206
In letter Seventeen thousand two hundred and six
In roman numeral
In binary 100001100110110
In octal 41466
In US dollars USD 17,206.00 (\$)
In euros 17 206,00 EUR (€)
Some related numbers
Previous number 17205
Next number 17207
Next prime number 17207
## Mathematical operations
Operations and solutions
17206*2 = 34412 The double of 17206 is 34412
17206*3 = 51618 The triple of 17206 is 51618
17206/2 = 8603 The half of 17206 is 8603.000000
17206/3 = 5735.3333333333 The third of 17206 is 5735.333333
172062 = 296046436 The square of 17206 is 296046436.000000
172063 = 5093774977816 The cube of 17206 is 5093774977816.000000
√17206 = 131.17164327704 The square root of 17206 is 131.171643
log(17206) = 9.7530134391813 The natural (Neperian) logarithm of 17206 is 9.753013
log10(17206) = 4.2356799185647 The decimal logarithm (base 10) of 17206 is 4.235680
sin(17206) = 0.48202433061403 The sine of 17206 is 0.482024
cos(17206) = -0.8761578309278 The cosine of 17206 is -0.876158
tan(17206) = -0.550156962135 The tangent of 17206 is -0.550157 | 816 | 2,664 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-18 | latest | en | 0.748628 |
http://www.britannica.com/topic/complete-graph | 1,464,162,996,000,000,000 | text/html | crawl-data/CC-MAIN-2016-22/segments/1464049274191.57/warc/CC-MAIN-20160524002114-00154-ip-10-185-217-139.ec2.internal.warc.gz | 392,527,122 | 11,187 | # Complete graph
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• ## definition
combinatorics: Characterization problems of graph theory
A complete graph Km is a graph with m vertices, any two of which are adjacent. The line graph H of a graph G is a graph the vertices of which correspond to the edges of G, any two vertices of H being adjacent if and only if the corresponding edges of G are incident with the same vertex of G.
number game: Graphs and networks
...13A), the resulting figure is a graph; the points, or corners, are called the vertices, and the lines are called the edges. If every pair of vertices is connected by an edge, the graph is called a complete graph (Figure 13B). A planar graph is one in which the edges have no intersection or common points except at the edges. (It should be noted that the edges of a graph need not be straight...
• ## graph theory
graph theory
...is called a simple graph. Unless stated otherwise, graph is assumed to refer to a simple graph. When each vertex is connected by an edge to every other vertex, the graph is called a complete graph. When appropriate, a direction may be assigned to each edge to produce what is known as a directed graph, or digraph.
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Error when sending the email. Try again later. | 748 | 3,315 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-22 | latest | en | 0.920235 |
https://www.exactlywhatistime.com/days-before-date/november-9/1-days | 1,708,704,401,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474440.42/warc/CC-MAIN-20240223153350-20240223183350-00763.warc.gz | 791,917,296 | 5,716 | # What date is 1 days before Saturday November 09, 2024?
## Calculating 1 days before Saturday November 09, 2024 by hand
This page helps you figure out the date that is 1 days before Saturday November 09, 2024. We've made a calculator to find the date before a certain number of days before a specific date. In this example, we want to know the date 1 days before Saturday November 09, 2024.
Trying to do this in your head can be really hard and take a long time. An easier way is to use a calendar, either a paper one or an app on your phone or computer, to look at the days before the date you're interested in. But the best and quickest way to find the answer is by using our days before specific date calculator, which you can find here.
If you want to change the question on this page, you have two choices: you can change the URL in your browser's address bar, or go to our days before specific date calculator to type in a new question. Remember, figuring out these types of calculations in your head can be really tough, so we made this calculator to help make it much easier for you.
## Friday November 08, 2024 Stats
• Day of the week: Friday
• Month: November
• Day of the year: 313
## Counting 1 days backward from Saturday November 09, 2024
Counting backward from today, Friday November 08, 2024 is 1 before now using our current calendar. 1 days is equivalent to:
1 days is also 24 hours. Friday November 08, 2024 is 85% of the year completed.
## Within 1 days there are 24 hours, 1440 minutes, or 86400 seconds
Friday Friday November 08, 2024 is day number 313 of the year. At that time, we will be 85% through 2024.
## In 1 days, the Average Person Spent...
• 214.8 hours Sleeping
• 28.56 hours Eating and drinking
• 46.8 hours Household activities
• 13.92 hours Housework
• 15.36 hours Food preparation and cleanup
• 4.8 hours Lawn and garden care
• 84.0 hours Working and work-related activities
• 77.28 hours Working
• 126.48 hours Leisure and sports
• 68.64 hours Watching television
## Famous Sporting and Music Events on November 08
• 1985 Author Ken Follett (36) weds politician Barbara Hubbard (42)
• 2017 Brazilian surfer Rodrigo Koxa breaks the world record for surfing the biggest-ever wave at 24.4m at NazarA(c), Portugal | 588 | 2,264 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2024-10 | latest | en | 0.917894 |
https://au.mathworks.com/matlabcentral/cody/problems/1054-what-s-your-bmi/solutions/284836 | 1,585,589,442,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370497171.9/warc/CC-MAIN-20200330150913-20200330180913-00185.warc.gz | 361,686,210 | 15,434 | Cody
# Problem 1054. What's Your BMI?
Solution 284836
Submitted on 18 Jul 2013 by Chad Gilbert
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
%%Hint h= 1; m= 1; y_correct = 703.06957964; assert(isequal(BMI(h,m),y_correct))
y = 703.0696
2 Pass
%% h= 70; m= 90; y_correct = 12.91352289134694; assert(isequal(BMI(h,m),y_correct))
y = 12.9135 | 158 | 467 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2020-16 | latest | en | 0.670795 |
https://ebrary.net/189395/mathematics/mackey_s_theory_construction_basic_observables_quantum_theory_projective_unitary_representations | 1,669,649,679,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710533.96/warc/CC-MAIN-20221128135348-20221128165348-00560.warc.gz | 252,013,655 | 11,080 | # Mackey’s theory on the construction of the basic observables in the quantum theory from projective unitary representations of the Galilean group
Galilean group element: (a, u, s, g), (&, v, t, h) where a, u € ®3,(s,p) € R x 50(3). Let G be the Galilean group. Then
G = V®SH,H = Rx 50(3),V = R3 x R3
Buy (a, u, s, g), we mean (a, u)o(s,g). Its composition law is ((a,u)o(s,g))o((b, v)o(t,h)) = ((s,g)[(b, u)] + (a,u), (s,g)0(t, h)) where
= (s,g)o(b, v)o(s,p)-1 = (s,p)o(b, v)o(-s,s-1) (s,0)o(b,u)o(-s,g-1)(t, r) = (s,g)o(b, v).(t-s, g^r)
= (s, g)o(t—s, g~1r+v(t—s)+b)
= (i, r + gv(t - s) + gb) = (g(b - vs),gv)(t, r)
Thus, we get
# Hamiltonian density of the electromagnetic field in curved space-time in terms of position and momentum fields
t = Ky/^F^F^ 7tp = dC/dAp.o =
• 4 A' y^gFOp H = itpAp\$ - C = A'v^(4F°^p,0 - F^) = Av^(4Fo'Mm.o - FppFpp) Now,
• 7TP = 4K y/^ggOa gpli Fa/3
= 4K^[g00gpmF0m + g(>"‘9I>() Fm() + gOmgpkFmk)
= 4Ky/^[(g00gpm - g0mgp0)F0m + gOmgpkFmk]
We need to solve for Ap.o in terms of Ap>m and Tvm. Note that tt° = 0. This is one of the constraints to be incorporated using the Dirac bracket formalism. We write the above equation as
s a ta- /—-[/ 00 sm „0m s0 77 . sk rp 1
% =4Ky/-g{(g g -g g )tOm--g g
We define the 3 x 3 matrix
sm ^00 ^sm 0m sO
7 =((7 )),7 =9 9 ~9 9
and then write the above as
tts/4A^ = ysmFOm + gOmgskFmk
Now writing ((7sm)) as the inverse of the matrix ((7sm)), we get
FOs = ysm[7rm/4K^g - gOrgkmFrk]
Now we define the electric and magnetic fields as
Es = FOs,Bs = -e(skm)Fkm
Then, we can write
Es = ysm m/4Kx/^ + gOrgkmc(rkl)Bl] or equivalently,
Trs/4A'v^ = ysmEm - g^g^mkl^Bt
We then have
H = ttsAs,0 -£ = KJ=-ghsmEm - - F^F^]
We note that
TrpApgd3x = f TrsAStod3x =
14K^gF0sASt0d3x = 4K [ V-9F0s(ASt0 - j40,s + A0s)d3x
= 4K I FOsFOsy/=gd3x + 4K f ^F0sA0,sd3x
= 4K I FOsFOs^d3x + 4K I(F0sA0^)tSd3x-4K f (F0s^)tSA0d3x
= 4K I F0sF0s^d3x in the absence of charges since in the absence of charges, the Maxwell equations read
(FOs^),s = 0 and also we have noted that the spatial volume integral of a three divergence is zero by Gauss’ theorem. This results in
H = 4KFOsFOs^- KF^F^^
K^(4F°sF0s - F^F^ = K^g(2FOsFOs - FrsFrs) Noting that
FOs = Es,Frs = -e{rsm)Bm
FOs = (g°°gsm - gOmgOs)FOm + gokgsmFkm
= YmEm+gokgsm^kml)Bl
Also
Frs = (grmgs0 - grOgsm)FmO + grmgskFmk
= -8rmsEm - grmgske(jnkp)Bp
where
arms _ rm sO _ rO sm
i ¿7 eZ J J
Thus,
H = Ky/^(2FOsFOs - FrsFrs)
= 2K^YmEm+gokgsme(kml)Bl)Es
-K^-g^rsm^E, + grl gske(lkp)Bp)Bm
Exercise: [1] Verify that in the case of flat Minkowskian space-time, this expression reduces to the familiar special relativistic formula
H = (1/2)(E2 + B2)
provided that we take K = —1/4. Note that in flat space-time, •ysm = — 6sm = gsm,g°° = lg0s = Q
(2) Denote
Ho = (1/2)(E2 + B2)
Then show that if the metric is
= n/iu T hpljl(x}
ie represented as a small perturbation of flat space-time, we can then write
H = Ho + Ck(rs,x)ErEs + C?(rs, x)BrBs + Cs(rs,x)ErBs
where Cj(rs,x) are linear combinations of h^^x). Now we are in a position to express the Hamiltonian in terms of the canonical position and momentum fields. Note that the Bs are functionals of the position fields Ar(x) while the equation
Trs/4ZysmEm - gOmgske(mkl)Bi
Es = ■ysm^-n8 / + gOmgsk e{mkl)Bi)
Exercise: Using the above equations, express the Hamiltonian density in terms of the position and momentum fields As,tts,s = 1,2,3 in a curved space-time upto linear orders in the metric perturbations.
# Coulomb scattering
The wave operator Q+ does not exist. In fact, writing
Ho = P2/2m, V(Q) = Ze2/Q
we find that
|| V(Q)exp(-itH0)f II dt Jo
is infinite. In fact, we have
exp(itHo)Q .exp(—itHo) = exp(it.ad(P2)/2m)(Q) = Q + Pt/rn
and on the other hand,
exp(—irnQ2/2t)P.exp(imQ2/2t) = exp(—im.ad(Q2)/2t)(P)
= P+mQ/t = (m/t)(Q+Pt/rn)
Thus,
exp(itH0)V(Q).exp(-itH0) = V(Q + Pt/m) = V((t/m)(F + mQ/t))
= ZtV{Pt/m)Z/
where
Zt = Zt(Q) = exp(—imQ2/2t)
Thus,
|| V(Q).exp(—itH0)f ||=|| V(Pt/m).Z/f ||
<|| V(Pt/m) < Q >_”|| . <||< Q >n / ||
where
=(1 + Q2)1/2
Assuming that
J(l + Q2)nf(Q)2d3Q
It follows that the wave operator Q+ will exist provided that
t ->|| V(Pt/m) < Q >-”||
is integrable on [0,oo). Now for any q > 2, we have
->ll V(Pt/m) < Q >-”||<|| V{Qt/m) ||, . ||< Q >-"||9
We have
ll-nll9= (/ (i + Q2rnq/2d3QV
and for this integral to be finite, we require that 2 — nq/2 < — 1 or equivalently, nq > 6. On the other hand, for t —>|| V{Qt/m) ||9 to be integrable over [0, oo), we can derive the condition:
(I |V(Qt/m)|’d3Q)1/<’ = (m3/3t3)1/’( J |V(Q)|’d3Q)1/9
and this is integrable if t~3'q is integrable over (<5, oo). This will happen provided that '3/q > 1 or equivalently, q < 3 and also V € L''(R3). Thus for the existence of the wave operator acting on a function f(Q), we require three conditions:
Scattering matrix in quantum mechanics
Q+ = I + i U(-t)V.U°(t)dt Jo tr+ = I-i / u°(-t)VU{t)dt Jo
V_ = I-i U(-t)VU°(t)dt
= I-i U(t)VU°(-t)dt
Jo
a*_ = i + i [ u°(t)vu(-t)dt = i + i [ u°(-t)vu(t)dt JO J — oo
q; - Q*_ = - f U°(-t)VU(t)dt
J — oo
r = s -1 = = (n; -
= - [ U°(-t)VU(t)il_dt = - [
JR JR
where we have used
C/(t)Q_ = Q_[/°(t)
Thus,
-R= [ U°(-t)V(I -i [ U(s)VU°(—s)ds)U°(t)dt
JR Jo
= f U°(-t)VU°(t)dt - i [ U0{-t)VU(s)VU0(t - s)dtds
JR JRxR+
= I exp(it(X — /x)')E0(dX')VE0(dp.)dt Jw
—i / Eo(dX)Vexp(—isH)VEo^dp)exp(i{X — p.)t + iXs)dtds
JR3xR+
= 2tt [ 8(X - p)E0(dX)VE0(dp)
Jv2
-2% [ 8(X - p)E0(dX)V.(H - X)-1 .VE0(dp)
2
Equivalently,
-7?/2tt= [ E'0(X)V E'0(X)dX — [ E'0(X)V.(H - A)-1 ,V.E'0(X)dX Ju ./r
This gives us explicitly the form of the scattering matrix at energy A: 5(A) - I = 7?(A) = -2tt(^(A)(V - V.(H - A)-1V)E^(A)
It is clear that the matrix element < f | S( A) |p > exists for all A if and only if f, g belong to the absolutely continuous spectrum of Ho, for that would imply the existence of the derivatives E'0(X)f = dEo(X)f /dX and E'0(X)g = dEo(X)g/dX. | 2,594 | 5,948 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2022-49 | latest | en | 0.785096 |
http://math.stackexchange.com/questions/259795/evaluation-of-derivative-using-epsilon-delta-definition?answertab=oldest | 1,398,252,332,000,000,000 | text/html | crawl-data/CC-MAIN-2014-15/segments/1398223202457.0/warc/CC-MAIN-20140423032002-00281-ip-10-147-4-33.ec2.internal.warc.gz | 211,395,792 | 15,486 | # Evaluation of Derivative Using $\epsilon−\delta$ Definition
Consider the function $f \colon\mathbb R \to\mathbb R$ defined by $f(x)= \begin{cases} x^2\sin(1/x); & \text{if }x\ne 0, \\ 0 & \text{if }x=0. \end{cases}$
Use $\varepsilon$-$\delta$ definition to prove that the limit $f'(0)=0$.
Now I see that h should equals to delta; and delta should equal to epsilon in this case. Thanks for everyone contributed!
-
If this is homework you should use homework tag. See also this: How to ask a homework question?. – Martin Sleziak Dec 16 '12 at 8:47
Recall the definition of a derivative i.e. $$f'(x) = \lim_{h \to 0} \dfrac{f(x+h) - f(x)}h$$ Hence, we get that $$f'(0) = \lim_{h \to 0} \dfrac{f(h) - f(0)}{h} = \lim_{h \to 0} \dfrac{2h^2 \sin(1/h)-0}h = \lim_{h \to 0} 2h \sin(1/h)$$ Now recall that $\vert \sin(y) \vert \leq 1$. Hence, we have that $$\left \vert 2h \sin(1/h) \right \vert \leq \left \vert 2h \right \vert$$ Hence, we have that $$\lim_{h \to 0} \left \vert 2h \sin(1/h) \right \vert \leq \lim_{h \to 0} \left \vert 2h \right \vert = 0$$ Hence, you get that $$f'(0) = 0$$
$$\left|{\dfrac{f(h)-f(0)}{h}}\right|=\left|{\dfrac{2h^2 \sin{\dfrac{1}{h}}}{h}}\right|=2 \left|{h \sin{\dfrac{1}{h}}}\right|<2\left|h\right|<\varepsilon.$$ Choose $\delta<\dfrac{\varepsilon}{2}.$ | 518 | 1,288 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2014-15 | latest | en | 0.580207 |
http://geniusbrainteasers.com/MAGIC-SQUARE-Calculate-A-B-C/4321 | 1,591,229,732,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347436828.65/warc/CC-MAIN-20200604001115-20200604031115-00369.warc.gz | 52,181,072 | 8,391 | BRAIN TEASERS
# MAGIC SQUARE: Calculate A*B+C
The aim is to place the some numbers from the list (8, 9, 11, 12, 13, 15, 17, 18, 20, 37, 40, 41) into the empty squares and squares marked with A, B an C. Sum of each row and column should be equal. All the numbers of the magic square must be different. Find values for A, B, and C. Solution is A*B+C.
The first user who solved this task is Manguexa Wagle.
#brainteasers #math #magicsquare
### An engineer was crossing a roa...
An engineer was crossing a road one day when a frog called out to him and said, "If you kiss me, I'll turn into a beautiful princess."He bent over, picked up the frog and put it in his pocket.
The frog spoke up again and said, "If you kiss me and turn me back into a beautiful princess, I will stay with you for one week."
The engineer took the frog out of his pocket, smiled at it and returned it to the pocket.
The frog then cried out, "If you kiss me and turn me back into a princess, I'll stay with you and do ANYTHING you want."
Again the engineer took the frog out, smiled at it and put it back into his pocket.
Finally, the frog asked, "What is the matter? I've told you I'm a beautiful princess, that I'll stay with you for a week and do anything you want. Why won't you kiss me?"
The engineer said, "Look I'm an engineer.
I don't have time for a girlfriend, but a talking frog, now that's cool."
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## Brain Teasers
puzzles, riddles, mathematical problems, mastermind, cinemania...
### Robert E. Gottschalk
Died 3 Jun 1982 at age 64 (born 12 Mar 1918).American inventor and business executive who was president of Panavision Inc., a company he helped found in 1953, to create a wide-screen film movie process. It manufactured and leased equipment. Panavision is an anamorphic system, using a 65mm negative and a 70mm print, projected in a letterbox shape with a 2.66 to 1 ratio. Gottschalk developed specially designed lenses used during capture and projection that worked with an image recorded on the film that is compressed horizontally. An unmodified widescreen image would occupy only half the area of a standard 1.33 to 1 film frame, wasting space above and below. Lateral compression fills the frame using an image with modified aspect ratio on the film, that retains more detail in the vertical resolution, and reduces the appearance of grain when projected. He died by murder at his home.«
This site uses cookies to store information on your computer. Some are essential to help the site properly. Others give us insight into how the site is used and help us to optimize the user experience. See our privacy policy. | 670 | 2,715 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2020-24 | latest | en | 0.971654 |
https://www.physicsforums.com/threads/distribution-dirac-standard-formulations-of-f-ma-how-do-they-go-again.668452/ | 1,516,606,115,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084891105.83/warc/CC-MAIN-20180122054202-20180122074202-00378.warc.gz | 932,655,423 | 32,005 | Distribution (Dirac&standard) formulations of f=ma how do they go again?
1. Jan 31, 2013
James MC
Hi, I'm wondering how to formulate Newton's second law in terms of Dirac delta functions and standard mass density distributions.
We define the Dirac delta function of a point mass with mass m located at point x0 as follows:
$$\rho(x)=m\delta (x-x_0)$$
If our point mass is represented by a Dirac delta function, our acceleration will need to be too. So we need an acceleration distribution for our point mass that assigns an instantaneous acceleration a to x0:
$$a(x)=a\delta (x-x_0)$$
If that is correct, then is the correct Dirac delta function formulation of f=ma:
$$F_i = [\int m\delta (x-x_i)dx] [\int a\delta (x-x_i)dx]$$
...for the force of particle i? And, for the total force FT for N particles, is it:
$$F_T = \sum_i [[\int m\delta (x-x_i)dx] [\int a\delta (x-x_i)dx]]$$
Okay, so if that's all good, is this the equation for standard mass density distributions (not for point particles, but for mass densities distributed over tiny regions):
$$F_T = \sum_i [\int [\rho(x) a(x)]dx]$$
Where ρ(x) is now defined at dM/dV and a(x) as da/dV. Notice that the Dirac delta formulation has TWO integrals while the mass density formualtion has ONE integral. I think that's right because you cannot multiply delta functions because, pre-integration, that would be multiplying infinities? Whereas with the densities, you are instead multiplying finite values and so you CAN multiply BEFORE the integration?
Or am I just completely confused here? Any advice would be most welcome!
2. Feb 1, 2013
Jano L.
Hello James,
using delta distributions is not very convincing way to find that expression. You can instead proceed directly from the definition of force
$$\mathbf F = \frac{d\mathbf p}{dt}$$
for the fluid in question. Enclose the whole body in surface S of volume V so that nothing comes in or out. Then momentum of the fluid is
$$\mathbf p = \int_V \rho \mathbf v~ dV$$
and the experienced force
$$\mathbf F = \frac{d}{dt} \int_V \rho \mathbf v ~dV.$$
If the surface does not move, we can move in with the time derivative into the integrand:
$$\mathbf F = \int_V \partial_t \rho \,\mathbf v + \rho\mathbf a~dV.$$
In case the fluid is incompressible (volume is constant), the first term vanishes and we end up with the formula you guessed above. The only thing is that there is no summation over $i$, as we already introduced integration over continuous functions.
3. Feb 2, 2013
James MC
Hi Jano, thanks for your response.
I don't understand what you mean when you say "using delta distributions is not a very convincing way to find that expression". If "that" refers to my final expression, I was not trying to derive that expression. I actually just want to know if the equations themselves are mathematically consistent. And I need to stick with delta functions for my purposes.
Another way to put my question is this: which is the correct formulation of Newton's second law in delta function terms:
$$F_i = [\int m\delta (x-x_i)dx] [\int a\delta (x-x_i)dx]$$
Or
$$F_i = \int [m\delta (x-x_i) a\delta (x-x_i)]dx$$
I would very much like to know the answer to this (I'm hoping it's the latter, but suspect due to the weird infinities, you can't multiply DDFs, so it's the former?).
I realise it's a non-standard formulation, but I don't see how the momentum formulation can help me. It's hard to say why without giving you some of the background of my project, which is somewhat complicated and so may discourage readers from responding, when really, I just need someone competent with DDFs to point out the correct formulation. Hope you can help!
4. Feb 4, 2013
Jano L.
The first expression can be used, but the second is, as you rightly suspect, problematic.
Delta distributions sometimes appear in a product, if they are concentrated at different points of real line. But two equal delta distributions are not meant to be multiplied together - it is not cleat what such object means.
It would help if you could describe at least telegraphically what is it you want to achieve with these delta distributions.
5. Feb 4, 2013
James MC
Textbook proofs of mass additivity go something like this: Where $F_T$ is the total force on a number of point masses indexed by i (due to an external source, which I do not specify, so as to simplify the equations), we begin with:
$$\sum_i m_i\mathbf{a}_i = \mathbf{F}_T$$
It is then assumed that the particles indexed by i compose a composite body and that the force on the composite is equivalent to the total force on the parts ($F_T=F_c$). A physical situation is then assumed in which the accelerations of the particles indexed by i are identical, which allows the following transformation:
$$\mathbf{a}_i\sum_i m_i = \mathbf{F}_c$$
It is then assumed that the acceleration of the composite $\mathbf{a}_c$ just is the acceleration of its parts so that:
$$\mathbf{a}_c\sum_i m_i = \mathbf{F}_c$$
Which is taken to show that given the force and acceleration of the composite, the composite's mass must be additive, that is:
$$\sum_i m_i = m_c$$
I am just trying to formulate a version of this proof that does not assume that the parts have identical accelerations. Because I wish to stay in the context of point particles, it seems Dirac delta functions may be the only hope.
So far, the generalised proof looks like this:
Fundamental many-particle law:
$$\sum_i \int m_i(\mathbf{x}) \mathbf{a}_i(\mathbf{x})\delta (\mathbf{x}-\mathbf{x}_i)d^3\mathbf{x} = \mathbf{F}_T$$
Recover the form of the single particle law (this involves deriving $\mathbf{a}_c(\mathbf{x})$, the acceleration distribution of the composite, from the acceleration distributions of the parts, which I'll just assume here):
$$\int \mathbf{a}_c(\mathbf{x})\sum_i m_i(\mathbf{x}) \delta (\mathbf{x}-\mathbf{x}_i)d^3\mathbf{x} = \mathbf{F}_T$$
Assume that the force on the composite is the force on the parts:
$$\int \mathbf{a}_c(\mathbf{x})\sum_i m_i(\mathbf{x}) \delta (\mathbf{x}-\mathbf{x}_i)d^3\mathbf{x} = \mathbf{F}_c$$
I hope that makes sense. (In the recent past I have given a slightly more extensive discussion of the project here: https://www.physicsforums.com/showpost.php?p=4257563&postcount=15). Any thoughts would be most welcome.
6. Feb 5, 2013
Jano L.
That makes a lot of sense. However, you do not need delta distributions for that.
You can repeat the procedure above by using the acceleration of the center of mass.
Total force is equal to sum of forces experienced by all particles:
$$\mathbf F_T = \sum_i m_i \mathbf a_i.$$
Acceleration is a second time derivative of the radius vector:
$$\mathbf a_i = \frac{d^2}{dt^2 }\mathbf r_i,$$
so
$$\mathbf F_T = \sum_i m_i\frac{d^2}{dt^2 }\mathbf r_i.$$
Since the masses $m_i$ are constant in time, the differential operator can be pulled in front of the whole expression:
$$\mathbf F_T = \frac{d^2}{dt^2 } \sum_i m_i \mathbf r_i,$$
which can be written as
$$\mathbf F_T = \sum_i m_i \frac{d^2}{dt^2 } \frac{\sum_i m_i \mathbf r_i}{\sum_i m_i}.$$
or
$$\mathbf F_T = \sum_i m_i \frac{d^2}{dt^2 } \frac{\sum_i m_i \mathbf r_i}{\sum_i m_i}.$$
So, the composite body moves in such a way that its center of mass
$$\mathbf R = \frac{\sum_i m_i \mathbf r_i}{\sum_i m_i}~~(*)$$
moves as if it was particle with mass $\sum_i m_i$.
Thus in this line of thought, the additivity of mass is a consequence of the choice of (*) for the definition of the center of mass. Other choices come to mind, but they are more complicated and is not clear whether they offer some advantage.
7. Feb 6, 2013
James MC
Thanks Jano, that's an interesting suggestion. If I understand you, your strategy falls under solution 1, whereas mine falls under solution 2:
Solution 1: Find a non-arbitrary, non ad hoc way to define the acceleration of a composite as a single point acceleration. Then, once one knows the force of the composite one can calculate the composite's mass using the standard equation for Newton's second law.
Solution 2: Find a non-arbitrary, non ad hoc way of reformulating the standard equation for Newton's second law, so that it applies to composites that have distributed locations, and distributed accelerations. Determine composite force then solve for composite mass.
Here is the concern I have with all instances of Solution 1 generally: I don't see how one can find a non-arbitrary way to relate (yet alone identify) the acceleration of a composite with the acceleration of the composite's centre of mass, without presupposing mass additivity. For that reason, I don't think one can appeal to centres of mass if the goal is to explain mass additivity.
I'll try to apply this to the particular instance of Solution 1 you've provided.
I take it we agree that the goal is start with our Newtonian fundamentals:
$$\sum_i m_i\mathbf{a}_i = \mathbf{F}_T$$
And to prove something like the following:
$$\mathbf{a}_c\sum_i m_i = \mathbf{F}_c$$
Because then if we know the composite force and the composite acceleration, then we can solve for composite mass, and thereby explain composite mass.
We can derive $F_T=F_c$ easily enough: it is a priori that the force on a composite is identical to the total force on its parts.
So the question is: how can we derive the acceleration of the composite? That is, how can we introduce $a_c$ in a way that does not presupoose what we are trying to explain?
Your suggestion, as I understand it, was: treat $a_c$ as whatever weighted average of component accelerations it needs to be in order to get $F_c$ assuming that $m_c$ = $\sum_im_i$. You're absolutely right that the acceleration of the centre of mass is what we get. But this presupposes mass additivity.
The alterntive route (Solution 2) is to derive a value for $a_c$ directly, without presupposing mass additivity. Here, one can simply treat the acceleration of a composite as what it is physically: a distribution of accelerations. In that case, we can describe the acceleration of the composite with a Dirac Delta function. This is why I think solution 2 can solve the problem and explain mass additivity.
8. Feb 6, 2013
Jano L.
Your idea is not clear to me, because of you have distribution of accelerations, you have also distribution of mass and there is no one mass of the composite.
In the above, I defined
$$\mathbf R = \frac{\sum_i m_i \mathbf r_i}{\sum_i m_i},$$
without supposing that the mass is additive. True, the definition is probably motivated by the additivity, but does not require it. If the definition is adopted, the additivity follows as a nice consequence of that definition.
Because there may be other definitions of the "center" of the body, I think that the additivity is not necessarily true in mechanics. For example, if we chose
$$\mathbf R = \frac{\sum_i m_i \mathbf r_i}{\sqrt{\sum_i m_i^2}},$$
to be the center, then the mass of the composite is
$$M = {\sqrt{\sum_i m_i^2}}.$$
This is strange, but it works as well.
However, it seems to unnecessarily cumbersome to calculate mass in this way - so we adopt the first definition.
9. Feb 7, 2013
James MC
This is a fair challenge. My transformations of the DDF formulation of F=MA, entail mass distribution additivity. That is, they show that $\sum_i m_i\delta (x-x_i)dx$ is the mass distribution of the composite. I take it your worry is that mass distribution additivity ≠ mass additivity. However, to derive mass additivity from mass distribution additivity, I only need to integrate $\sum_i m_i\delta (x-x_i)dx$. That is another upshot of using DDF's here: the value of a quantity represented by a DDF is by definition, its integral. This means that to derive mass additivity, I only need to appeal to the following equalities:
$$\int \sum_i f(x) \delta (x-x_i)dx = \int [ f(x)\delta (x-x_1)+f(x)\delta (x-x_2)+~.~.~.~+ f(x)\delta (x-x_n)]dx$$
$$=f(x_1)+f(x_2)+~.~.~.~+f(x_n)=\sum_i f(x_i)$$
Hence, $\int \sum_i m_i\delta (x-x_i)dx = \sum_i m_i$ Hence mass additivity.
Does that clarify the idea?
I don't see how one could defend the position that additivity is not true in Newtonian mechanics. Newtonian mechanics is not some inapplicable abstract mathematical formalism, in which it is open to us to make such stipulations. Newtonian mechanics is a theory of the physical world, and answers to experiment. Mass additivity is empirically confirmed within Newtonian mechanic's domain of validity (e.g. slow moving macroscopic bodies) whereas this equation:
$$M = {\sqrt{\sum_i m_i^2}}.$$
is ruled out by experiment.
One can strengthen this argument by appeal to a more realistic case in which mass additivity does not hold: relativity theory. In relativity theory, there is a very specific formulation for the centre of mass of a composite:
$$\mathbf R = \frac{\sum_i m_iγ_i \mathbf r_i}{\sum_i m_iγ_i},$$
This is the correct formulation in virtue of the non-additivity of mass (or the additivity of $m_iγ_i$), which is an empirically verifiable phenomenon, not a stipulation.
Furthermore, an argument for why additivity is in fact true in Newtonian mechanics, is that I have proved it! Although, whether you're convinced by that argument, will depend, I guess, on whether you think I've successfully responded to your challenge from above.
10. Feb 7, 2013
Jano L.
In the above, you seem to think that when
$$\sum_i m_i \mathbf a_i$$
is rewritten into
$$\int \rho(\mathbf x) \mathbf a(\mathbf x) dV$$
with $\rho(\mathbf x) = \sum_i m_i \delta(\mathbf x-\mathbf x_i)$ and some new convenient function $\mathbf a(\mathbf x)$, the additivity of mass is proven for point particles.
I do not see why do you think such thing. The mere rewrite of the expression does not prove anything. There is no reason to introduce mass distribution function and call
$$\int \rho(\mathbf x) dV$$
total mass unless one is sure/assumes that the mass is additive.
I think that the additivity of inertial mass can proven by considering the equation of motion as I did above. In relativity, the equation is different and thus we do not have such additivity.
I think the standard choice of center of mass is the best one, and the additivity follows as a consequence of this assumption, the equation of motion and the superposition principle.
But if somebody finds some other center $\mathbf C$ useful, he may use it, and for him the inertial mass $M$ in the equation
$$M\ddot{\mathbf C} = \mathbf F$$
may not be additive. This would be very cumbersome to use for description of experiments, because the new center would much differently than usually expected, but this cannot rule it out, as it is mathematically as correct as the standard choice.
So perhaps the better and more interesting question is, why do we choose $\frac{\sum_i m_i \mathbf r_i}{\sum_i m_i}$ to define the motion of the body, and not some other point?
And I think the answer is additivity of mass. I other words, the additivity is the basic motivation and the theory is so developed that it conforms, but it is also true that it is kind of luck, since in relativity, this is no longer possible - potential energy changes inertial mass of the composite body in a way that is not additive.
11. Feb 9, 2013
James MC
Re: Distribution (Dirac&standard) formulations of f=ma... how do they
Apologies for the delay...
I don't believe I am assuming mass additivity. I will go through my proof step by step to try to show why. Hopefully that will make it easy for you to pick out the exact step that is concerning you.
Start with the textbook explanation of mass additivity that I mentioned a while back, that I thought you and I agreed was valid yet incomplete. (I take us to be disagreeing on how to complete it - Solution 1 (centre of mass) or Solution 2 (distributions)).
We begin with a simple physical situation involving Newtonian particles, and we state that the law applying to them is Newton's many particle law:
$$\sum_i m_i\mathbf{a}_i = \mathbf{F}_T$$
Now, I'll take the textbook proof step by step.
Step 1: Infer the existence of composite C from the existence of the particles. (Presumably this is trivial).
Step 2: Infer the force on C via the following principle: If the superposition principle tells us that the force on some particles that compose C, is F, then the force on C is F. (Again, this is presumably trivial.) We may now infer:
$$\sum_i m_i\mathbf{a}_i = \mathbf{F}_C$$
Step 3: Stipulate that the accelerations of the particles indexed by i are identical, which allows the following transformation:
$$\mathbf{a}_i\sum_i m_i = \mathbf{F}_C$$
Step 4: The crucial acceleration step: The acceleration of the composite $\mathbf{a}_c$ just is the acceleration of its parts so that:
$$\mathbf{a}_c\sum_i m_i = \mathbf{F}_c$$
Mass is defined as that property of objects responsible for their resistance to changes in motion given applied forces. The equation in step 4 tells us that the property responsible for the composite's disposition to resist acceleration given applied forces is the sum of the masses of its parts. Hence we can derive:
$$\sum_i m_i = m_c$$
It strikes me that if you think this proof is valid and non-question-begging (for the limited situation it deals with) then you should also think that my proof is valid and non-question-begging (for the general situation); this is because my proof is structurally identical:
We begin with a physical situation involving Newtonian particles, and we state that the law applying to them is Newton's many particle law:
$$\sum_i m_i\mathbf{a}_i = \mathbf{F}_T$$
Now, independently of the issue of mass additivity, one can straightforwardly derive a distribution form of this law:
$$\sum_i[\int \rho(\mathbf x)_i \mathbf a(\mathbf x)_i dV] = \mathbf{F}_T$$
(Note that it might be easier to view these as standard distributions, rather than deltas, so as to avoid (for now) technical complications concerning distribution multiplication.) This does not presuppose mass additivity. If one grasps distribution formalism and is asked to formulate the equivalent of the second law + superposition principle, one infers this, and one does so without any thought of the properties of composites.
Now comes my structurally identical proof:
Step 1: Infer the existence of composite C from the existence of the particles. (Presumably this is trivial).
Step 2: Infer the force on C via the following principle: If the superposition principle tells us that the force on some particles that compose C, is F, then the force on C is F. (Again, this is presumably trivial.) We may now infer:
$$\sum_i[\int \rho(\mathbf x)_i \mathbf a(\mathbf x)_i dV] = \mathbf{F}_C$$
Step 3: Here, we do not stipulate that the accelerations of the particles indexed by i are identical, yet we still want an analogous transformation:
$$\int [[\sum_i \mathbf a(\mathbf x)_i][\sum_i\rho(\mathbf x)_i]] dV = \mathbf{F}_C$$
Step 4: The crucial acceleration step: The textbook explanation justifies this step by just saying "acceleration of the composite just is the acceleration of the parts". My justification is along the same lines, but I think a little more sophisticated: By definition, acceleration is the second time derivative of position. The composite is positioned where its parts are positioned - a set of positions (trivial). Therefore, the composite acceleration is a set of time derivatives. In distribution formalism, we may then deduce:
$$\mathbf a(\mathbf x)_C = \sum_i \mathbf a(\mathbf x)_i$$
Which yields the desired result, in which we recover the (distribution) form of the single particle law:
$$\int [[\mathbf a(\mathbf x)_C][\sum_i\rho(\mathbf x)_i]] dV = \mathbf{F}_C$$
Mass is defined as that property of objects responsible for its resistance to changes in motion given applied forces. The equation in step 4 tells us that the property responsible for the composite's disposition to resist acceleration given applied forces is the sum of the mass distributions of its parts. Hence we can derive:
$$\mathbf \rho(\mathbf x)_C = \sum_i \rho(\mathbf x)_i$$
Hence mass distributions are additive. We can then derive mass additivity, from mass distribution additivity, as discussed in my previous post.
I do not think any step in either proof presupposes mass additivity, though if you think one or more steps are problematic, I would be very interested to know why.
I completely agree. If you allow as a premise that the composite's position is the centre of mass, then the proof goes through. My point is that a proof containing an unmotivated premise is not an explanation. And to motivate that premise you require what you're trying to explain (mass additivity). Hence, the proof is not an explanation of mass additivity.
I think there are more powerful objections to this approach, but the above is sufficient for my purpose, which is to motivate the need for my explanation. (These objections appeal to the fact that nothing, let alone the composite, is positioned at the COM.)
While it is mathematically consistent it is still incorrect because it is inconsistent with, and refuted by, experiment. One can show $M = {\sqrt{\sum_i m_i^2}}.$ is false, by measuring an object's disposition to resist acceleration given a force, and then cutting the object in half, and determining their dispositions to resist acceleration given the same force.
I think that the case of relativity strengthens my argument that the person who chooses a different centre, and hence a different composite mass, advocates a false theory. The relationship between the mass of the composite and the mass of the parts, is not a matter of convenient stipulation of a mass centre. Rather, it is a matter of experimental physics. And it was an extrodinary discovery about nature herself, that a box of hot gas offers more resistance to acceleration than the same box after it is cooled. The mass-composite/mass-parts relationship is a consequence of the fundamental laws alone and relativity shows that when you change the laws (and nothing else), you change the observed relationship. I intend to prove the relationship from the laws, starting with the Newtonian case.
My apologies for the length of the post, I hope it does not deter you from responding. I doubt that I've convinced you of my view just yet, and if so, I would be quite interested to learn why.
12. Feb 10, 2013
Jano L.
Re: Distribution (Dirac&standard) formulations of f=ma... how do they
The problem is already beginning to appear in the expression
$$\sum_i[\int \rho(\mathbf x)_i \mathbf a(\mathbf x)_i dV] = \mathbf{F}_T.$$
How do you define $\rho(\mathbf x)_i$ and $\mathbf a(\mathbf x)_i$ ?
If we used DD in both quantities, then the integral is invalid expression, as two same DD cannot be multiplied.
If we used ordinary continuous functions, then it would be incorrect, as the term $\partial_t\rho\, \mathbf a$ would be missing.
We can try to introduce square-root DD (bit non-standard)
$$\sqrt{\delta(\mathbf x-\mathbf r_i)}$$
and use one for
$$\rho_i(\mathbf x) = m_i \sqrt{\delta(\mathbf x- \mathbf r_i)}$$
and one for acceleration
$$\mathbf a_i(\mathbf x) = \mathbf a_i \sqrt{\delta(\mathbf x- \mathbf r_i)},$$
so the first integral above gives $\sum_i m_i \mathbf a_i$ correctly, but then the integral of rho is wrong:
$$\int \rho_i dV = 0.$$
So I do not see how your expressions can make sense.
13. Feb 11, 2013
James MC
Re: Distribution (Dirac&standard) formulations of f=ma... how do they
This is certainly correct, and is what I see as the main outstanding problem with my argument. I have three completed proofs already (I think!), gravitation and inertia proofs in the continuous context, and a gravitation proof in the DD context, but I haven't been able to prove inertial mass additivity in the DD context for precisely this reason. I make a suggestion at the bottom though.
I don't really understand what you mean because I don't see what $\partial_t\rho\, \mathbf a$ would add to the equation.
Perhaps this points to the fact that I haven't been fully explicit about my simplified scenario. I'm working with a fixed volume of space, and my particles (arbitrarily small continuous mass density distributions) are confined to that volume, and do not leave that volume. Furthermore, my particles are frozen in time, so to speak. Obviously, they are accelerating, but I am taking a snap shot of them at a time, and then applying my equation to them, to determine total force, at that time. Given these simplifications, I don't see what a partial time derivative of both the mass and acceleration densities would add. Although, I'm also not entirely sure what work it would do if I removed these simplifications. Can you please say more to clarify what you have in mind?
In the continuous case, my definitions are as follows:
$\rho = M/V$ for an object whose mass is evenly distributed over V.
$\rho(\mathbf x) = ΔM(x)/ΔV(x)$ for non-uniform distributions.
$\rho(\mathbf x) = dM/dV$ taking the limit, where our volumes are infinitesimal.
My acceleration distributions are defined analogously, where instantaneous accelerations replace the masses...
$\mathbf a = \mathbf a/V$ for an object with a uniformly distributed instantaneous acceleration.
$\mathbf a(\mathbf x) = Δ\mathbf a(x)/ΔV(x)$ for non-uniform distributions.
$\mathbf a(\mathbf x) = d\mathbf a/dV$ taking the limit, where our volumes are infinitesimal.
I'm inclined to think that because the three other proofs I mentioned work (I think!), then it must be that the DD verison of the inertial mass additivity proof works too, it's just a matter of getting the technicalities right.
Now, while you're correct in saying that this expression is invalid:
$$\sum_i \int m_i(\mathbf{x})\delta (\mathbf{x}-\mathbf{x}_i) \mathbf{a}_i(\mathbf{x})\delta (\mathbf{x}-\mathbf{x}_i)d^3\mathbf{x} = \mathbf{F}_T$$
It's not so clear to me that this is invalid:
$$\sum_i \int m_i(\mathbf{x}) \mathbf{a}_i(\mathbf{x})\delta (\mathbf{x}-\mathbf{x}_i)d^3\mathbf{x} = \mathbf{F}_T$$
In other words, the latter equation does not multiply DD's (the invalid bit); there is only one DD, and it assigns the product of m and $\mathbf{a}$ to $x_i$. So the m and a terms need to be defined as functions on the delta function and not delta functions themselves? I suspect this latter equation is the correct DD formulation of Newton's second law. Running my proof with it is another story.
14. Feb 12, 2013
Jano L.
Re: Distribution (Dirac&standard) formulations of f=ma... how do they
Your definitions are not clear enough and however I try to interpret them, they are invalid. There is no sense in introducing distribution of acceleration by
$$\mathbf a(\mathbf x) = \mathbf a / \Delta V.$$
Acceleration is not distribution, but an ordinary function.
I think you would better read first chapters of some book on hydrodynamics, for example Landau and Lifgarbagez. Then you will see how the theory can be built consistently, when the additivity of mass is assumed right from the beginning.
In continuum mechanics, the total force acting on the volume of fluid contained within a fixed boundary is
$$\mathbf F = \frac{d}{dt} \int_{V}\bigg( \rho \mathbf v \bigg)\,dV.$$
Moving in with the derivative and differentiating the product, you get
$$\int_V \partial_t \rho\, \mathbf v + \rho \mathbf a\, dV.$$
The last integral in your last post is OK, provided it is meant for point-like particles and $m_i(\mathbf x),\mathbf a_i(\mathbf x)$ are ordinary functions of $\mathbf x$. How do you define them? Your previous post did not explain that. Is $\mathbf a_i(\mathbf x)$ nonzero also in places where the particle $i$ is not present?
15. Feb 13, 2013
James MC
Re: Distribution (Dirac&standard) formulations of f=ma... how do they
Acceleration distributions are certainly non-standard, although it appears they come up every so often in places (e.g. here.) But you're right that those definitions are insufficient without some background.
I'm not sure how to best formally define the acceleration distributions, but I'm basically thinking of them as something like functions from infinitesimal regions to instantaneous accelerations; or sets of ordered pairs.
Assume a physical situation in which we have two infintesimal mass densities located at distinct infinitesimal regions x1 and x2, at t1. We can assign acceleration distributions to each particle at t1. One of these distributions will assign x1 some (potentially) non-zero instantaneous acceleration value, zero elsewhere. The other distribution assigns x2 some (potentially) non-zero instantaneous acceleration value, zero elsewhere. One can also define a composite acceleration distribution, which assigns (potentially) non-zero instantaneous accelerations to two infinitesimal regions, zero at every other infinitesimal region.
Then I think one just needs to define the integral accordingly - something like a step function, I take it, would be best suited.
Perhaps I'm missing something, but that idea seems reasonably clear to me, I'm just sure what the best mathematical vocabulary for expressing it precisely would be.
Is v the velocity of the centre of mass in these equations? I take it it's not a velocity distribution?
I've gone over Truesdell's "A First Course in Rational Continuum Mechanics", but I didn't find it to be of much help. I think that's because I'm not trying to do continuum mechanics, and I'm not describing fluids. I'm describing fixed infintitesimal masses separated by relatively large regions of space. And I'm doing it in a way that should seem quite odd to someone who doesn't realise that I've just been trying to get this mass additivity proof to work.
It's important to the proof that one specifies the superposition principle separately form the integral, so as to indicate that we are describing (to begin with) many distinct particles. We then try to recover the single object law by moving the sum into the equation so that it attaches to the masses.
Yes, I use the DD equations when describing point particles, and the quasi-continuum-mechanics formalism when describing particles that are infinitesimal mass densities.
At first I was thinking of defining the Dirac delta function of a point mass with mass m located at point x0 as follows:
$$\rho(x)=m\delta (x-x_0)$$
The acceleration distribution for the point mass that assigns an instantaneous acceleration a to x0 is then:
$$\mathbf a(x)=\mathbf a\delta (x-x_0)$$
It then looks like if you multiply $\rho(x)$ with $\mathbf a(x)$ then you're multiplying two DD's, which is bad, but I think if you specify only one DD in the actual equation, as I did in the equation you thought valid, then this means that the DD is just assigning the product of two values to points in space, which avoids the problem.
To answer your question, yes because the subscript "i" serves to just pick out the right value, i.e., the acceleration of the particle located at i. Every point gets assigned that value and then the DD puts every point to zero except point i.
I hope I'm making some sense! Sorry that I can't be more mathematically rigorous about this, I've had a pretty non-standard physics education.
16. Feb 13, 2013
Jano L.
Re: Distribution (Dirac&standard) formulations of f=ma... how do they
All the quanties $\rho(\mathbf x),\mathbf v(\mathbf x), \mathbf a(\mathbf x)$ in
$$\int_V \partial_t \rho\, \mathbf v + \rho \mathbf a\, dV.$$
are functions of position in space $\mathbf x$.
Yes, except that i-th delta function can sometimes choose non-i-th acceleration and that will spoil the result. The step when you change sum of products into a product of two sums requires that no cross terms contribute. This is possible if you define
$$\mathbf a_i(\mathbf x) = \mathbf a_i D(\mathbf x - \mathbf r_i),$$
$$\mathbf \rho_i(\mathbf x) = m_i \delta(\mathbf x - \mathbf r_i),$$
where $D(\mathbf x - \mathbf r_i)$ is a characteristic function of some small ball, i.e. function that is equal to 1 if $|\mathbf x - \mathbf r_i| < a$ and equal to zero otherwise.
Then, if the mutual distances of the particles are greater than $a$, cross terms are zero and the manipulation works.
However, the resulting object
$$\mathbf a_C = \sum_i \mathbf a_i D(\mathbf x-\mathbf r_i)$$
is quite artificial, because it depends on the choice of $a$. But it is possible that particles can come closer than $a$ for any choice of $a$ and then the manipulation fails.
Disregarding that, even it the manipulation works for given situation, I do not think that success in re-expressing $F = ma$ by
$$\int \rho_C \mathbf a_C dV$$
is a derivation of mass additivity. The expression is not in the form $\mathbf F= m_C \mathbf a_C$, which is necessary to find composite mass $m_C$. Then I do not see any convincing reason to adopt the idea that $\mathbf a_c$ is or represents object's acceleration and $\rho_C$ its mass distribution; they are just strange new objects defined in the process.
17. Feb 14, 2013
James MC
Re: Distribution (Dirac&standard) formulations of f=ma... how do they
I see. Then the use of an acceleration distribution (or acceleration function - perhaps I'm confusing these) within the integrals was not the problem. In that case, is it possible to run the argument with the equation you've provided? Something like this:
Step 1: Introduce the single density force law
$$\mathbf F_i = \frac{d}{dt} \int_{V}\bigg( \rho_i \mathbf v_i \bigg)\,dV.$$
Step 2: Introduce the many density total force law
$$\mathbf F_T = \sum_i \frac{d}{dt} \int_{V}\bigg( \rho_i \mathbf v_i \bigg)\,dV.$$
Step 3: Swap the sum and the integral
$$\mathbf F_T = \frac{d}{dt} \int_{V}\bigg(\sum_i \rho_i \mathbf v_i \bigg)\,dV.$$
Step 4: Choose some velocity function $v_?$ so that the following holds:
$$\mathbf F_T = \frac{d}{dt} \int_{V}\bigg(\mathbf v_? \sum_i \rho_i \bigg)\,dV.$$
Step 5: Argue that $F_T = F_C$ and $v_? = v_C$ so as to recover the form of the single density law applied to the composite C:
$$\mathbf F_C = \frac{d}{dt} \int_{V}\bigg(\mathbf v_C \sum_i \rho_i \bigg)\,dV.$$
Step 6: Infer mass density additivity
Are steps 2-4 valid?
I think this might be the same concern that you raised in post 8? There, you said: "Your idea is not clear to me, because if you have distribution of accelerations, you have also distribution of mass and there is no one mass of the composite."
Is the concern that even if the steps of the proof work, I haven't proven mass additivity because I have only proven mass distribution additivity and mass ≠ mass distribution? If so, I think this is a fair concern, but I tried to answer it, at least for the DD case, in post 9. There, the claim was roughly that it is true by definition that the value of the quantity described by a Dirac delta function is its integral. So if I've inferred, for the composite, a DD whose integral is in fact the composite's mass (i.e., is additive) then I've proved mass additivity.
Yes you're right that cross contamination of terms is certainly a problem in the most natural ways of fleshing out the DD inertial proof. I'm pretty sure this doesn't happen in the corresponding proof of gravitational mass additivity in DD form:
$$\mathbf{g}(\mathbf{x})_T=G \sum_i\int \frac{\mathbf{x}-\mathbf{r}}{|\mathbf{x}-\mathbf{r}|^3}~m_i~\delta(\mathbf{r}-\mathbf{x}_i)~\mathrm{d}^3\mathbf{r}$$
=
$$\mathbf{g}(\mathbf{x})_C=G\int \frac{\mathbf{x}-\mathbf{r}}{|\mathbf{x}-\mathbf{r}|^3}\sum_i ~m_i~\delta(\mathbf{r}-\mathbf{x}_i)~\mathrm{d}^3\mathbf{r}$$
I have wondered why an equation that takes the product of mass distribution and position, should be so different from an equation that takes the product of mass distribution with acceleration. I find it hard to believe that the proof is possible for the former, but not the latter!
I have some thoughts about how to remove cross contamination for the inertial DD proof, but they're in their infancy. It might help to first get your thoughts on the above issues, and by then my thoughts on cross contamination will be clearer in my head.
18. Feb 14, 2013
Jano L.
Re: Distribution (Dirac&standard) formulations of f=ma... how do they
No, I think the problem is that your procedure does not even prove additivity of mass distributions in physical sense.
There are two different notions of distribution causing mess here.
1. distribution in the mathematical sense, as a linear functional acting on some neat function. These behave according to
$$\sum_i \int D_i(x) f(x) dx = \int [ \sum_i D_i(x) ] f(x) dx,$$
so one can say mathematical distributions are additive. Your procedure just uses and exhibits this property of delta distribution.
2. distribution in the physical sense, also called density, as an estimate of quantity of some stuff within some volume. In the case of mass, to define density, first you have to know mass M of some volume element V and then calculate
$$\rho = \frac{M}{V}.$$
You cannot circumvent this and redefine density as $\rho = \sum_i \rho_i$. That is just using the fact that you want to prove, that the mass density can be described by linear functionals which are additive by definition.
Other way to see this: try to repeat your procedure in the case of relativistic gas, $m_i$ being rest masses of the particles. You will end up with the same formula suggesting additivity, but in fact the rest mass of the gas is not sum of individual rest masses of the particles. The reason is that linear functionals are additive always, while rest mass is not. In order to restore the use of linear functionals in relativity, you have to use inertial mass (also called relativistic mass).
The point is, you have to know first what is additive to decide what to describe by linear functionals.
In Newtonian mechanics, Newton knew that mass is additive and therefore introduced density. It makes no sense to do it the opposite way.
19. Feb 15, 2013
James MC
Re: Distribution (Dirac&standard) formulations of f=ma... how do they
Yes, one step of the overall proof uses this - what I below call "step 3".
But at no point do I redefine density as $\rho = \sum_i \rho_i$ I prove it using the below five step proof.
The additivity of linear functionals is insufficient to get my result. What's important are (i) the composition of forces law and (ii) the fact that the law treats masses and forces as linearly proportional. Both are entirely contingent features of the nomic structure of Newtonian worlds, and they can be appealed to to prove mass additivity. In particular, the composition of forces entails the additivity of forces and the linearity of the law entails that if forces are additive then so are masses.
I do not agree that you will end up with the same formula. One reason that one does not end up with the same formula is that the composition of forces does not hold in either special relativity or general relativity. So there is simply no sum that one can swap into the equation!
I suspect that the procedure would fail in relativity due to the non-linearity of the equations too - the introduction of gamma cubed into the inertial law, though I'm not totally sure.
I agree with the first sentence, but not the second. Regarding the first: we have to know that forces are additive before we can describe them by linear functionals or integrals. And we do know this in advance, thanks to the compositions of forces law. The composition of forces law tells us that forces are additive. This means that to calculate forces, we can integrate over mass/position products (for gravitational forces) or mass/acceleration products (for inertial forces).
Thus, it is the additivity of forces that motivates the introduction of integrals, not the additivity of masses.
Since I've worked out both continuous and discrete proofs of gravitational mass additivity, I'll use one to spell out what I'm saying in more detail. I'll use the continuous case rather than DDs. I understand you as objecting to step 2:
Proof of gravitational continuous mass density additivity
Step 1: Introduce Newton's many particle law of gravitation
$$\mathbf{g}(\mathbf{x})=G\sum_i \frac{\mathbf{x}-\mathbf{x}_i}{|\mathbf{x}-\mathbf{x}_i|^3}~m_i~$$
Note that if we assume that the positions of the particles indexed by i are identical, we can show that it's as if there is just one particle whose mass is additive:
$$\mathbf{g}(\mathbf{x})=G \frac{\mathbf{x}-\mathbf{x}_1}{|\mathbf{x}-\mathbf{x}_1|^3}\sum_i~m_i~$$
Step 2: Derive the distribution formulation:
$$\mathbf{g}(\mathbf{x})=G \sum_i\int \frac{\mathbf{x}-\mathbf{r}}{|\mathbf{x}-\mathbf{r}|^3}~\rho_i(\mathbf{x}) ~\mathrm{d}^3\mathbf{r}$$
Key point: we know that we are allowed to integrate mass/position contributions to derive force because (i) integrating is just adding and (ii) the composition principle licences this.
Step 3: Algebraic transformation that recovers form of single density law:
$$\mathbf{g}(\mathbf{x})=G \int \frac{\mathbf{x}-\mathbf{r}}{|\mathbf{x}-\mathbf{r}|^3}~\sum_i\rho_i(\mathbf{x}) ~\mathrm{d}^3\mathbf{r}$$
Step 4: Total grav potential = composite grav potential & position of parts = position of composite:
$$\mathbf{g}_C(\mathbf{x})=G \int \frac{\mathbf{x}-\mathbf{r}}{|\mathbf{x}-\mathbf{r}|^3}~\sum_i\rho_i(\mathbf{x}) ~\mathrm{d}^3\mathbf{r}$$
Step 5: Infer gravitational mass distribution additivity.
Okay, that's the continuous gravitation proof. Now I would like to respond in more detail to your claim that this wrongly suggests that mass is additive in all cases because linear functionals (integrals and densities) are invoked. I said above, the procedure requires (i) the composition/additivity of forces and (ii) the linearity of the law. I give an example of when (i) breaks down and an example of when (ii) breaks down:
Example when (i) breaks down:
Step 1: Introduce law that replaces sum with product:
$$\mathbf{g}(\mathbf{x})=G\prod_i \frac{\mathbf{x}-\mathbf{x}_i}{|\mathbf{x}-\mathbf{x}_i|^3}~m_i~$$
Step 2: Derive the distribution formulation:
$$\mathbf{g}(\mathbf{x})=G \prod_i\int \frac{\mathbf{x}-\mathbf{r}}{|\mathbf{x}-\mathbf{r}|^3}~\rho_i(\mathbf{x}) ~\mathrm{d}^3\mathbf{r}$$
Step 3: Algebraic transformation that recovers form of single density law:
$$\mathbf{g}(\mathbf{x})=G \int \frac{\mathbf{x}-\mathbf{r}}{|\mathbf{x}-\mathbf{r}|^3}~\prod_i\rho_i(\mathbf{x}) ~\mathrm{d}^3\mathbf{r}$$
This fails - transformation not valid.
Example when (ii) breaks down:
Step 1: Introduce law requiring that we square the mass/mass-density:
$$\mathbf{g}(\mathbf{x})=G\sum_i \frac{\mathbf{x}-\mathbf{x}_i}{|\mathbf{x}-\mathbf{x}_i|^3}~m_i^2~$$
Step 2: Derive the distribution formulation:
$$\mathbf{g}(\mathbf{x})=G \sum_i\int \frac{\mathbf{x}-\mathbf{r}}{|\mathbf{x}-\mathbf{r}|^3}~[\rho_i(\mathbf{x})]^2 ~\mathrm{d}^3\mathbf{r}$$
Step 3: Algebraic transformation that recovers form of single density law:
$$\mathbf{g}(\mathbf{x})=G \int \frac{\mathbf{x}-\mathbf{r}}{|\mathbf{x}-\mathbf{r}|^3}~\sum_i[\rho_i(\mathbf{x})]^2 ~\mathrm{d}^3\mathbf{r}$$
Step 4: Total grav potential = composite grav potential & position of parts = position of composite:
$$\mathbf{g}_C(\mathbf{x})=G \int \frac{\mathbf{x}-\mathbf{r}}{|\mathbf{x}-\mathbf{r}|^3}~\sum_i[\rho_i(\mathbf{x})]^2 ~\mathrm{d}^3\mathbf{r}$$
Step 5: Infer gravitational mass distribution additivity.
This fails because in general: $[\sum_i\rho_i(\mathbf{x})]^2$ ≠ $\sum_i[\rho_i(\mathbf{x})]^2$
Therefore, my result does not depend on an arbitrary introduction of linear functionals. Instead, it depends on (i) the composition principle, and (ii) the linearily of the law. Either one of these or both could break down, despite the use of linear functions (integrals). When they break down, mass additivity breaks down.
What do you think?
20. Feb 15, 2013
Khashishi
Re: Distribution (Dirac&standard) formulations of f=ma... how do they
It sounds like you are using the wrong approach here, and you should instead use a Boltzmann equation to calculate what you are doing. Instead of treating the acceleration as an object or a field, the velocity is included as a dimension of the distribution function. This way, the mass and momentum are integrated together into the same object.
The Boltzmann equation tells you how to evolve a continuous distribution of particles or mass.
$\frac{\partial f(x,p,t)}{\partial t}+\frac{p}{m}\cdot \nabla f(x,p,t)+F\cdot \frac{\partial f}{\partial p}=C$
If you need to track individual particles, you _can_ treat acceleration as an object, but then it should not be parametrized by the space coordinate, but rather a label of a specific particle. For example:
m(n), x(n), v(n), a(n)
this would give the equations of motion for each particle labeled by n. You can make n a continuous variable if you want. In this case, each particle is a dirac delta function in n, not in x.
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https://math.stackexchange.com/questions/3535185/how-many-numbers-are-required-to-define-a-sequence-without-stating-a-rule-functi/3535198 | 1,621,355,666,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991288.0/warc/CC-MAIN-20210518160705-20210518190705-00262.warc.gz | 412,766,192 | 41,352 | # How many numbers are required to define a sequence without stating a rule/function for generating the next term in the sequence?
I'm wondering if there is some minimum number of numbers required to define a sequence, without explicitly stating the rule that generates the next term in the sequence. For instance if I write $$(1,a_2,a_3,...)$$, and hide the remaining numbers in the sequence behind $$(a_2,a_3,...)$$, we don't know what the sequence is or what rules define it. If I then write $$(1,2,a_3,...)$$, it still isn't clear. Is the rule for determining the next number in the sequence $$a_{i+1}=2 a_i$$? Is it $$a_{i+1}=a_i+1$$?
If I write $$(1,2,4,8,16)$$, it's clear the rule is $$a_{i+1}=2a_i=2^{i-1}$$. Could I even shorten this to $$(1,2,4,...)$$ and figure this out? Is this an example of the minimum number of numbers required to define the sequence of powers of $$2$$. As J.W. Tanner says in the comments, you can come up with a polynomial whose first terms are $$1,2,4,8,16,23$$, so apparently not.
How about the Fibonacci sequence? I think it's clear what the rule is if I write $$(0,1,1,2,3,5,8,...)$$, even if I hadn't learned of this sequence before. I can't learn anything from $$(0,1)$$. What about $$(0,1,1)$$? It's hard to decide if I can learn the rule from this or if I need more numbers from the sequence. Typically you would just say $$a_0=0,a_1=1,$$ and $$a_{i} = a_{i-1} + a_{i-2}$$ for $$i>1$$. But that defeats the point of the question. The point is to ask how many numbers we need in order to define/learn the sequence without explicitly stating the rule that generates the next term in the sequence, and writing $$a_{i} = a_{i-1} + a_{i-2}$$ is explicitly stating the rule.
How does this idea generalise?
• I could come up with a polynomial whose first terms are $1, 2, 4, 8, 16, 23$ – J. W. Tanner Feb 5 '20 at 11:12
• This kid of conclusion is not rue always true and just using some "guesses" you cannot always find the general recurrence relation, for example if you search "$1,1,2,3,5$" in oeis.org you will get 1241 results which start with "$1,1,2,3,5$" – user715522 Feb 5 '20 at 11:13
Even your example of $$1,2,4,8,16$$ doesn't automatically mean that the sequence is uniquely defined by $$a_i=2^{i-1}$$
As humans, we would probably assume that was the sequence you meant, but we could also say that the sequence is defined by $$a_i=\frac{i^4}{24} - \frac{i^3}4+\frac{23i^2}{24}-\frac{3i}4+1$$ (which I found using WolframAlpha)
This then gives \begin{align}a_6&=\frac{6^4}{24} - \frac{6^3}4+\frac{23\times 6^2}{24}-\frac{3\times 6}4+1\\ &=31\end{align} as opposed to the $$32$$ you would expect.
Even if we then specify that the $$6$$th term is $$32$$, we then get a new generating function which then gives the $$7$$th term as $$a_7=63$$, again not $$64$$ as we expect.
So, the conclusion is that you can never uniquely define a sequence simply from its first $$n$$ terms, you can only uniquely define a sequence with its generating function
If you have $$n$$ data points, there is always a polynomial, of degree at most $$n-1$$, that it fits. So in general, there are never enough data points.
For example, given $$1,2,4,\ldots$$, the rule might be $$a_n=(n^2-n+2)/2$$.
Conversely, if you know what the function 'looks like', each data point can narrow it down. If you know it is linear, $$a_n=an+b$$, two datapoints are enough. $$a=a_2-a_1$$, then $$b=a_1-a$$. If you know it is quadratic, three points are enough.
If you know $$a_n=pa_{n-1}+qa_{n-2}$$, it turns out you need four points to find $$p$$ and $$q$$. $$a_3=pa_2+qa_1$$ gives one equation and $$a_4=pa_3+qa_2$$ gives another. Two equations are usually enough the two values $$p$$ and $$q$$
Consider the sequence $$1,1,2,3.$$ At first look, it seems to be the first terms of Fibonacci numbers, but it's not true that the only sequence which starts with $$1,1,2,3,5$$ are Fibonacci numbers.
Here we can say that these are the first terms of triangle read by rows in which row n lists A000041(n-1) 1's followed by the list of juxtaposed lexicographically ordered partitions of n that do not contain 1 as a part.
Or we can say these are the numbers of rooted trees with n vertices in which vertices at the same level have the same degree.
Or these are the terms generated by $$a_n=\lfloor(3^n / 2^n)\rfloor$$.
In your example the terms of $$1,2,4,8,16$$ are not necessarily generated by $$a_n=2^n$$
For example we can say these are the coefficients of expansion of $$\frac{(1-x)}{(1-2^x)}$$ in powers of $$x$$.
Or these are the numbers of positive divisors of $$n!$$.
Or these are Pentanacci Numbers.
There is an uncountably infinite number of sequences of natural numbers (even ones starting $$a_0, a_1, a_2, a_3$$, as anything whatsoever can follow). That this is so is simple to prove by Cantor's diagonalization argument.
for powers of 2, you have recurrences like $$a_n=2a_{n-1}$$ or $$a_n=a_{n-1}+2a_{n-2}$$ etc. that work out to powers of two, but you could have something like the first 5 defined and then $$a_n=a_{n-2}+a_{n-2}+a_{n-3}+a_{n-4}+a_{n-5}$$ in which case the next value is 31, then 60, then ...
You can also define fibonacci with itself, $$a_{n+z}= F_{z+1}a_n+F{z}a_{n-1}$$ which only takes the first $$z+2$$( including 0) terms being defined. | 1,621 | 5,293 | {"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": 56, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2021-21 | latest | en | 0.924454 |
https://ca.answers.yahoo.com/question/index?qid=20200530021618AAY0y1K | 1,603,434,173,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107880656.25/warc/CC-MAIN-20201023043931-20201023073931-00174.warc.gz | 258,264,055 | 26,026 | # Math: law of cosines, hw help?
Relevance
• 5 months ago
Apply the law of cosine
∠B = 180 - 111 = 69°
solving for side b
b = √[a^2 + c^2 - 2accos(B)]
b = √[19^2 + 10^2 - 2(19)(10)cos(69)]
b = 18.022 or 18
apply law of sine to solve Angle C
.....b..............c
--------- =-----------
..sin(B)....sin(C)
...18.............10
----------- =---------
..sin(69)...sin(C)
18sin(C) = 10sin(69)
................10sin(69)
sin(C) = ----------------- <-----------closest answer on choice D
.......................18
...................................10sin(69)
.......................................19
• 5 months ago
In △ABC, a = 19, c = 10, and m∠A = 111°.
Which statement can be used to find the value of C?
The Law of Cosines (also called the Cosine Rule) says:
a^2 = b^2 + c^2 − 2bc cos (A)
19^2 = b^2 + 10^2 - 20b cos (111°)
b(b - 20 cos (111°) = 29*9
b (b + 20 sin((7 π)/60)) = 261
• fcas80
Lv 7
5 months ago
Law of Sines: sinC/10 = sin111/19
sinC=10 * sin111 / 19. But sin111 = sin69
sinC = 10 * sin69 / 19 | 375 | 1,039 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2020-45 | latest | en | 0.457061 |
https://www.physicsforums.com/threads/math-involved-in-finding-the-optimal-angle-of-placing-a-solar-panel.616510/ | 1,521,938,339,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257651465.90/warc/CC-MAIN-20180324225928-20180325005928-00177.warc.gz | 882,976,690 | 16,540 | # Math involved in finding the optimal angle of placing a solar panel
1. Jun 25, 2012
### gokuls
I want to write a 3500 - 4000 word research paper on the best angle to place a solar panel to optimize solar collection throughout the year. I was thinking of using planes in 3D and probability distribution for the cloud amount for different times of the year. There has to be a primary focus on math though. So my question is, is there enough math involved in this topic? If so, what math is involved? Thanks! (BTW, this is my first post, so hope this goes well).
2. Jun 26, 2012
### haruspex
Plenty, if you like 3D trig.
Should also look at the value of the electricity produced (I assume this is PV). It can vary with time of day. Different types of panel cope differently with oblique angles.
Main problem for the math may be that the best way is to simulate a year. Getting any kind of algebraic closed form to apply some calculus to might be impossible.
3. Jun 26, 2012
### chiro
Hey gokuls and welcome to the forums.
To build on the advice of haruspex, you might want to create random variables for the intensity and other parameters and then get a mean and variance for the distribution of total collection input.
4. Jun 26, 2012
### Staff: Mentor
Might need to do various takes on accounting for cloudy weather, too. Not much point in optimising the angle for best autumn incidence if autumn is typically 80% rainy. (Then again, that might be the very reason for optimising it for autumn, esp where you are reliant on solar panel for your autumn energy.)
5. Jun 26, 2012
### gokuls
Thank you all for your respones!
haruspex I don't quite follow when you say "simulate a year", do you mean the orbit of the earth or amount of cloud cover. Could you please elaborate a bit? Other than that thank you, but I am still a bit worried if I can write for 3500+ words!
6. Jun 26, 2012
### haruspex
Yes, I meant the track of the sun and feeding in info re the value of power generated at different times of day and seasons. You could also randomise day to day for weather conditions.
7. Jun 26, 2012
### Diffy
I actually know a guy that has built his own solar panel that moves throughout the day to be positioned to look at the Sun.
If you would like to talk to someone who as actually built one, and programmed it himself PM me. I know for a fact there was a ton of math involved. But in the end be for practical purposes estimations were quicker and easier given the hardware he was using.
8. Nov 7, 2012
### oranalyst
I am interested in learning the math behind solar panel tilting optimization.
I am trying to know how to start with monitoring the radiation data in my region and how to use the data. My learning is more on the data side.
May i request you to contact me too.
9. Nov 7, 2012
### oranalyst
I am interested in learning the math behind solar panel tilting optimization.
I am trying to know how to start with monitoring the radiation data in my region and how to use the data. My learning is more on the data side.
May i request you to contact me too.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook | 769 | 3,198 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2018-13 | longest | en | 0.914674 |
http://diffsharp.github.io/DiffSharp/examples-lhopitalsrule.html | 1,553,420,833,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912203409.36/warc/CC-MAIN-20190324083551-20190324105551-00143.warc.gz | 64,579,785 | 7,055 | l'Hôpital's Rule
l'Hôpital's rule is a method for evaluating limits involving indeterminate forms. The rule states that, under some conditions, the indeterminate limits
$\begin{eqnarray*} \lim_{x \to c} \frac{f(x)}{g(x)} &=& \frac{0}{0} \; ,\\ \lim_{x \to c} \frac{f(x)}{g(x)} &=& \frac{\pm\infty}{\pm\infty} \end{eqnarray*}$
can be found by differentiating the numerator and the denominator and taking the limit, that is
$\lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)} \; .$
Let us try to find
$\lim_{x \to 0} \frac{2 \sin x - \sin 2x}{x - \sin x}$
which involves the indeterminate form $$0 / 0$$.
1: 2: 3: 4: 5: 6: 7: 8: // Define f(x) let f x = 2. * sin x - sin (2. * x) // Define g(x) let g x = x - sin x // Try to evaluate the limit at x = 0 let lim = f 0. / g 0.
As expected, we get a nan as a result, meaning the result of this operations is undefined.
val lim : float = nan
Using DiffSharp, we can generate a sequence of repeated applications of l'Hôpital's rule.
1: 2: 3: 4: 5: 6: 7: 8: 9: 10: 11: open DiffSharp.AD.Float64 // Differentiate f(x) and g(x) n times and evaluate the division let lhopital f g n x = diffn n f x / diffn n g x // Generate an infinite sequence of lhopital applications, // starting from the undifferentiated limit (n = 0) let lhseq f g x = Seq.initInfinite (fun n -> lhopital f g n x) // Use lhseq with f(x) and g(x), at x = 0 let l = lhseq (fun x -> 2. * sin x - sin (2. * x)) (fun x -> x - sin x) (D 0.)
The first four elements ($$n = 0,\dots,4$$) of this infinite sequence are
val l : seq [nan; nan; nan; D 6.0; ...]
For $$n = 0$$, we have the original indeterminate value of the limit (since the 0-th derivative of a function is itself).
For $$n = 3$$, we have the value of this limit as
$\lim_{x \to 0} \frac{2 \sin x - \sin 2x}{x - \sin x} = 6 \; ,$
after 3 applications of l'Hôpital's rule.
We can check this by manually differentiating:
$\begin{eqnarray*} \lim_{x \to 0} \frac{2 \sin x - \sin 2x}{x - \sin x} &=& \lim_{x \to 0} \frac{2\cos x - 2\cos 2x}{1 - \cos x}\\ &=& \lim_{x \to 0} \frac{-2 \sin x + 4 \sin 2x}{\sin x}\\ &=& \lim_{x \to 0} \frac{-2 \cos x + 8 \cos 2x}{\cos x}\\ &=& \frac{-2 + 8}{1}\\ &=& 6 \; . \end{eqnarray*}$
We can go further and automate this process by searching for the first element of the sequence that is not indeterminate.
1: 2: 3: 4: 5: 6: 7: 8: 9: 10: 11: 12: // Check if x is not indeterminate let isdeterminate x = not (System.Double.IsInfinity(x) || System.Double.IsNegativeInfinity(x) || System.Double.IsPositiveInfinity(x) || System.Double.IsNaN(x)) // Find the limit of f(x) / g(x) at a given point let findlim f g x = Seq.find (float >> isdeterminate) (lhseq f g x) // Find the limit of f(x) / g(x) at x = 0 let lim2 = findlim (fun x -> 2. * sin x - sin (2. * x)) (fun x -> x - sin x) (D 0.)
val lim2 : D = D -6.0
Let us use our function to evaluate some other limits.
The limit
$\lim_{x \to 0} \frac{\sin \pi x}{\pi x}$
has indeterminate form $$0 / 0$$ and can be evaluated by
$\lim_{x \to 0} \frac{\sin \pi x}{\pi x} = \lim_{y \to 0} \frac{\sin y}{y} = \lim_{y \to 0} \frac{\cos y}{1} = 1 \; .$
1: let lim3 = findlim (fun x -> sin (System.Math.PI * x)) (fun x -> System.Math.PI * x) (D 0.)
val lim3 : D = D 1.0
The limit
$\lim_{x \to 0} \frac{e^x - 1 - x}{x^2}$
has indeterminate form $$0 / 0$$ and can be evaluated by
$\lim_{x \to 0} \frac{e^x - 1 - x}{x^2} = \lim_{x \to 0} \frac{e^x - 1}{2x} = \lim_{x \to 0} \frac{e^x}{2} = \frac{1}{2} \; .$
1: let lim4 = findlim (fun x -> exp x - 1 - x) (fun x -> x * x) (D 0.)
val lim4 : D = D 0.5
It should be noted that there are cases where repeated applications of l'Hôpital's rule do not lead to an answer without a transformation of variables.
val f : x:float -> float
Full name: Examples-lhopitalsrule.f
val x : float
val sin : value:'T -> 'T (requires member Sin)
Full name: Microsoft.FSharp.Core.Operators.sin
val g : x:float -> float
Full name: Examples-lhopitalsrule.g
val lim : float
Full name: Examples-lhopitalsrule.lim
val printf : format:Printf.TextWriterFormat<'T> -> 'T
Full name: Microsoft.FSharp.Core.ExtraTopLevelOperators.printf
namespace DiffSharp
module Float64
val lhopital : f:(D -> D) -> g:(D -> D) -> n:int -> x:D -> D
Full name: Examples-lhopitalsrule.lhopital
val f : (D -> D)
val g : (D -> D)
val n : int
val x : D
val diffn : n:int -> f:(D -> 'c) -> x:D -> 'c (requires member get_P and member get_T)
val lhseq : f:(D -> D) -> g:(D -> D) -> x:D -> seq<D>
Full name: Examples-lhopitalsrule.lhseq
module Seq
from Microsoft.FSharp.Collections
val initInfinite : initializer:(int -> 'T) -> seq<'T>
Full name: Microsoft.FSharp.Collections.Seq.initInfinite
val l : seq<D>
Full name: Examples-lhopitalsrule.l
union case D.D: float -> D
val isdeterminate : x:float -> bool
Full name: Examples-lhopitalsrule.isdeterminate
val not : value:bool -> bool
Full name: Microsoft.FSharp.Core.Operators.not
namespace System
type Double =
struct
member CompareTo : value:obj -> int + 1 overload
member Equals : obj:obj -> bool + 1 overload
member GetHashCode : unit -> int
member GetTypeCode : unit -> TypeCode
member ToString : unit -> string + 3 overloads
static val MinValue : float
static val MaxValue : float
static val Epsilon : float
static val NegativeInfinity : float
static val PositiveInfinity : float
...
end
Full name: System.Double
System.Double.IsInfinity(d: float) : bool
System.Double.IsNegativeInfinity(d: float) : bool
System.Double.IsPositiveInfinity(d: float) : bool
System.Double.IsNaN(d: float) : bool
val findlim : f:(D -> D) -> g:(D -> D) -> x:D -> D
Full name: Examples-lhopitalsrule.findlim
val find : predicate:('T -> bool) -> source:seq<'T> -> 'T
Full name: Microsoft.FSharp.Collections.Seq.find
Multiple items
val float : value:'T -> float (requires member op_Explicit)
Full name: Microsoft.FSharp.Core.Operators.float
--------------------
type float = System.Double
Full name: Microsoft.FSharp.Core.float
--------------------
type float<'Measure> = float
Full name: Microsoft.FSharp.Core.float<_>
val lim2 : D
Full name: Examples-lhopitalsrule.lim2
val lim3 : D
Full name: Examples-lhopitalsrule.lim3
type Math =
static val PI : float
static val E : float
static member Abs : value:sbyte -> sbyte + 6 overloads
static member Acos : d:float -> float
static member Asin : d:float -> float
static member Atan : d:float -> float
static member Atan2 : y:float * x:float -> float
static member BigMul : a:int * b:int -> int64
static member Ceiling : d:decimal -> decimal + 1 overload
static member Cos : d:float -> float
...
Full name: System.Math
field System.Math.PI = 3.14159265359
val lim4 : D
Full name: Examples-lhopitalsrule.lim4
val exp : value:'T -> 'T (requires member Exp)
Full name: Microsoft.FSharp.Core.Operators.exp | 2,297 | 6,804 | {"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} | 3.875 | 4 | CC-MAIN-2019-13 | latest | en | 0.724111 |
https://chance.dartmouth.edu/chance_news/recent_news/chance_news_6.04.html | 1,709,483,045,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476396.49/warc/CC-MAIN-20240303142747-20240303172747-00790.warc.gz | 159,389,165 | 17,708 | !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
CHANCE News 6.04
(21 February 1997 to 4 March 1997)
!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Prepared by J. Laurie Snell, with help from Bill Peterson, Fuxing Hou, Ma.Katrina Munoz Dy, Kathryn Greer, and Joan Snell, as part of the Chance Course Project supported by the National Science Foundation.
Please send comments and suggestions for articles to jlsnell@dartmouth.edu.
Back issues of Chance News and other materials for teaching a Chance course are available from the Chance web site:
http://www.geom.umn.edu/locate/chance
Note: We decided to not have a Part 2 for Chance News 6.03 and to just move on to Chance News 6.04. Please do try to answer our lottery mystery in the last item.
Contents of Part 1
Contents of Part 2: A Lottery Bonus
Comments from readers on Chance News 6.03.
John Paulos wrote us about another interesting web site with interactive programs that can be run on the web. This is Kyle Siegrist's interactive version of the gamblers ruin problem His programs are written in Visual Basic script and uses activeX controls. They run fine on Internet Explorer 3.0 for Windows 95. Microsoft has plans to make activeX available for the Mac and Netscape soon. Kyle has also written a book and software (Interactive Probability, Dubury Press 1996). We will review this book in a later issue. Kyle has an NSF project to develop similar modules for introductory statistics.
----------------------
Elliott Weinstein gave another criticism of the infamous SAT median question. He commented that to assume that n is positive and even is unnecessary and unnecessary information often puzzles a test-taker. Second and more important is that, when n is odd, the quantity in Box B is still well-defined but the quantity in Box A is not defined at all. Again, puzzling information complicates the task for the test-taker.
It seems to us that it is necessary to assume that n is positive and even. There is no agreement on what a median is when a set of numbers does not have a middle value. Also if a is negative Box B could be complex etc.
----------------------
Paul Alper suggested that, when we discussed the study on the danger of using a cellular phone and driving, we should have mentioned that Efron co-authored a book with one of the authors of the study Robert J. Tibshirani. Alper writes: "While there is no suggestion of collusion, a reference to an expert's opinion should always include possible sources of bias due to a financial or personal connection." On this point, Efron and Kolata were more careful than we were: The NYTimes article stated that "Dr. Bradley Efron, a professor of statistics at Stanford University, said he saw the paper early because he knows Dr. Tibshirani well."
--------------------------
Abe Ross commented on our first discussion question related to the study suggesting that virtual learning might be more effective than classroom learning. We wrote:
(1) The article doesn't say how many students were in the course. How large would the groups need to be for the 20% difference in averages to be convincing?
Abe writes:
This design (as described) has a basic statistical flaw - the students in the real class are not independent subjects. This violates a statistical assumption. Most conservatively they are an N=1. On the other hand, the students in the virtual class are relatively independent and should be counted that way. I'm not sure what the correct method of analysis would be for this design since you end up with very unequal N in the two conditions.
DISCUSSION QUESTION:
Do you agree that the assumption of independence is more reasonable for the virtual students than for those in class?
--------------------------
Mike Cox wrote that the "New Scientist" commented on the study of Shutte on virtual learning and provided a link to the original paper. The "New Scientist" itself can be read on the web at their web site. You need to register, but it is free.
---------------------------
An article in Part 1 reported that the Federal Aviation Administration (FAA) was going to put on their web page data of airplane accidents and incidents (events that could cause accidents). They put the first databases on Friday and got over 8500 hits on their site. After things calm down it would be interesting to see how the data is presented and how useful it is.
As usual we appreciate these readers' comments.
<<<========<<
>>>>>==============>
A probability problem:
American Mathematical Monthly, Feb.1997
Problems and solutions, problem 10576
Proposed by Donald E. Knuth
Alice and Bill have identical decks of 52 cards. Alice shuffles her deck and deals the cards face up into 26 piles of two cards each. Bill does the same with his deck. If any one of Alice's top cards exactly matches any of Bill's, the matching cards are removed. Play continues until none of the cards on top of Alice's piles matches any of the cards on top of Bill's piles. What is the probability that all 52 pairs of cards will be matched?
We played it twice and won both times.
DISCUSSION QUESTION:
John Lamperti remarked that if I see any pair for Bill that is the same for Alice but in opposite order I might as well not even try to get rid of all the cards. What is the chance of this happening? What other simple things like this could prevent you getting rid of all the cards.
<<<========<<
>>>>>==============>
The following article was suggested by Gregg Paulos a former Chance student at Middlebury.
Vaccine is blamed in 125 polio cases.
USA Today, 31 January 1997, A1
Tim Friend
According to the Centers for Disease Control in Atlanta, nearly all US cases of polio since 1980 were due to vaccinations. Of 133 confirmed cases from 1980-1994, 125 were associated with administration of the oral vaccine. A panel convened last fall concluded that a rate of 7-8 cases a year was unacceptable, and recommended a new regime for immunization.
Since 1980, children have been received three doses of oral vaccine by age 2. The new recommendation calls for two injections by four months of age, followed by two oral doses of weakened virus between the ages of 1 and 6. The article says that "the change in policy should eliminate the risk of vaccine induced polio." However, it will increase the annual cost of immunization by \$14.7 million.
DISCUSSION QUESTIONS:
(1) Do you believe that the risk will actually be reduced to zero?
(2) Is any risk other than zero "acceptable"?
<<<========<<
>>>>>==============>
Visual Explanations, by Edward R. Tufte.
Graphics Press, 1996
P.O. Box 430,
Cheshire, CT 06410
\$45 Postpaid
This is the third in a series of books written and published by Tufte on the design of the display of information. In the introduction to this book Tufte tells us how these three books are related.
"The Visual Display of Quantitative Information" is about pictures of numbers, how to depict data and enforce statistical honesty.
"Envisioning Information" is about pictures of nouns (maps and aerial photographs for example, consist of a great many nouns lying on the ground). Envisioning also deals with visual strategies for design: color, layering, and interaction effects.
"Visual Explanations" is about pictures of verbs, the representation of mechanism and motion, of process and dynamics, of causes and effects, of explanation and narrative. Since such displays are often used to reach conclusions and make decisions, there is a special concern with the integrity of the content and the design.
"The Visual Display of Quantitative Information" was in the list of 10 bestsellers in the earth's biggest bookstore: Amazon.com. Tufte's previous books have won prizes for their own design and you will find his new book also a visual delight. From his previous books, we also know that we will find wonderful historical examples of the right way to display information and their modern counterparts.
In Tufte's first book we had the graphic display of Napoleon's defeat by the Russian army and winter in the year 1812. Tufte remarked: "it may well be the best statistical graphic ever drawn." The annual weather map of the New York Times provides a modern example of a very informative graph.
In "Visual Explanations" we saw Louis Bretez's wonderful "Plan de Paris," showing every building of this city and a modern counterpart, Constantine Anderson's three dimensional map of midtown Manhattan.
In this new book, we see how John Snow, with his map of the area surrounding the Broad Street Pump, showed the evidence that the cholera epidemic in England in 1884 was caused by the drinking water. He used this map to persuade the authorities to remove the pump handle, an action that was credited with ending the epidemic. Tufte contrasts this with a modern example of the incorrect display of graphic information. He shows us the charts providing information on the failure in tests of the O-rings on space shuttles. These charts played an important role in allowing the Challenger space shuttle to go despite the cold weather. He shows that, if the information on 0-rings had been properly presented graphically, the danger to the O-rings of cold temperatures would have been evident and could have prevented the 1986 Challenger disaster.
A novel chapter, co-authored by Jamy Ian Swiss, a professional magician, illustrates how magicians disguise information and shows that we can learn from them what not to do. This even includes a few hints for our lectures: magicians never explain exactly what they are going to do -- a lecturer should. Magicians never repeat a trick -- an important idea in a lecture is worth repeating.
By now we are supposed to have learned from Tufte to let the data speak for itself and not to litter the graphics with a lot of junk. Alas, as Tufte observes, many of the web-page designers have not learned this. They clutter their homepage with large buttons, flashing lights and other paraphernalia that disguise what in on their web sites. As usual, Tufte is not content to just make critical remarks -- he gives us good examples and shows what should be done.
A common theme of Tufte's books is the proper use of color. Here again we see what a difference it makes, in conveying information, when colors are subtly integrated into the graphics instead of distracting us by their gaudiness.
It would be an interesting study to see if Tufte's impressive books have "made a difference". For example, in his first book his own study showed that, among all the wonderful graphical ways we have to convey information, only maps, time-series plots and bar and pie charts are typically used in newspapers. One seldom sees graphs that relate two or more variables despite the wide use such graphs in scientific publications. Are things any different 20 years later? Our own impression is that the answer is no for the newspapers, but yes for scientific journals, especially medical journals.
In this latest book Tufte shows us what we should be doing when we want to convey information that assesses change, dynamics, and cause and effect, which Tufte regards to be at the heart of thinking and explanation. He does it in a book that is a joy to look at, to read, to think about it, and even to review.
<<<========<<
>>>>>==============>
The following article should have been in last month, but it took us a while to track down the source of the article.
Bridge.
The New York Times, 4 Jan. 1997, 1-27
Alan Truscott
Truscott says: "It has long been an article of faith among duplicate players that computer deals are markedly more distributional (more extreme hands) than "normal" deals generated by humans. They are right about that, but wrong in concluding that there is a flaw in computer dealing. It is, we now know the "normal" deals that are abnormal. This point was also made by Persi Diaconis in connection with his study with Dave Bayer of the number of shuffles needed for good shuffling.
This article refers to a study by two Englishmen, Harry Freeman and Len Salmon, who studied 334 duplicate deals played in a English club, over seven months. They reported their findings in "Bridge Magazine" an English magazine published by Chess & Bridge Ltd. who also publish a "Chess Monthly" (web: http://www.chess.co.uk/, e-mail: chesscentre@easynet.co.uk). They kindly faxed us a copy of the Freeman and Salmon article.
Freeman and Salmon analyzed 1336 hand distributions dealt by hand at a bridge club in Abington. They state their main findings as:
1. A markedly reduced occurrence of voids, singletons, and six-and seven card suits.
2. A significant increase in the probability of the most common suit distributions -- in particular, the 4-4-3-2 hand and a consequent decrease in the occurrence of the most common distributions.
It is easier to see why, in rubber bridge, flat hands are more likely in human shuffling. When the hands are picked up after a game, they have clumps of 4 cards of the same suit. If there was no shuffling these would be distributed one to each player. It requires serious shuffling to undo this effect. In duplicate bridge, at the end of a round, most people just put their hands back unsorted and as played and this causes a similar regularity.
To illustrate the problem, Freeman and Salmon consider a "perfect shuffle" carried out as follows: cut the cards exactly in half, and then put one half in each hand and drop the cards, one at a, alternating between the hands. They point out that, if you start with a new deck of cards, do a perfect shuffle, and then deal four hands, each hand will have only two suits. After four such shuffles you find that one player has a great hand: the Ace, King, and Queen of each suit and a Jack!
Freeman And Salmon estimated that human dealing reduced the chance of a preemptive opening by more than 25 percent. When they examined the way that six cards were distributed in two hands, they found that the probability of a 5-1 or 6-0 break was also reduced by about 25%. However, they did not find any significant difference in the distribution of point count.
Freeman and Salmon are not optimistic that bridge players will be persuaded to shuffle better. For duplicate bridge they suggest it would help a lot if the players were simply instructed to mix their hands up before putting them back in the boards at the end of a round.
DISCUSSION QUESTION:
The authors comment that a Malcolm Grimstom maintained that the way to shuffle a deck efficiently is to first deal it randomly into 5 piles and then re-integrate it before dealing. What does this actually mean? Would it work?
Note: A recent article in "The Economist" has a nice discussion of perfect shuffles and their applications (How to win at poker and other science lessons, The Economist, 12 Oct. 1996, Science and Technology, p. 87.) You will find an elementary discussion of the shuffling results of Diaconis and Bayer in Chapter 3 of Grinstead and Snell's probability book which is available on the web at the Chance Database under "teaching aids".
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Some systematic biases of everyday judgment.
Skeptical Inquirer, March/April 1997, pp. 31-35
Thomas Gilovich
Skeptics have long been unimpressed with the reasoning abilities of normal, average people. Before, this opinion was just based on a few observations. But now, psychologists have designed tests to assess these judgments.
The first test is called the "Compared to What?" problem. People like absolute statistics and accept them as true facts. However, sometimes these statistics need to be compared to a set baseline or simply something else in order to see what they really mean.
An example of this occurred in Discover magazine in 1986. The magazine stated that 90% of airplane-wreck survivors had formed a mental plan of escape before the wreck occurred. So, the magazine also recommended that airline passengers be sure to know where all the exits and emergency exits are located and form an escape route. However, the survey never included how many victims planned escape routes. In fact, it would be impossible to find out if forming an escape route really would increase oneıs chances of survival in a plane wreck.
A statistic stated that 30% of all infertile couples who adopt a child eventually conceive a child or children of their own. However, this does not take into account all the infertile couples who do not adopt a child and still conceive a child eventually. Also, if a patient has cancer which goes into remission after mental imagery, would the cancer have gone into remission without the mental imagery or did it truly have an effect?
Other examples include recognizing things. Many people claim to have "gaydar"the ability to pick out gay people from a crowd. Though a person may be correct sometimes, this does not necessarily mean that he has picked out all the gay people who crossed paths with him that day.
Many statistics need a control group or baseline for comparison to discover the true meaning behind the number. According to Gilovich: ³The logic and necessity of control groups...is often lost on a large segment of even the educated population.²
Second, we have the "Seek and Ye Shall Find" problem. When people test a hypothesis or theory, they often look more closely at the results which prove them correct. A test for this problem reads as follows:
Imagine that you serve on a jury of an only-child sole- custody case following a relatively messy divorce. The facts of the case are complicated by ambiguous economic, social, and emotional considerations, and you decide to base your decision on the following few observations. To which parent would you award sole custody of the child?
A: Average income, average health, average working hours, reasonable rapport with the child, relatively stable social life.
B: Above-average income, minor health problems, lots of work-related travel, very close relationship with the child, extremely active social life.
Most respondents chose parent B. However, when the question was reworded to ³which parent would you deny custody of the child,² most responded with B as well. Parent B has several advantages and disadvantages, so, when the first version was asked, people looked for advantages(like the close relationship with the child, above- average income) which they found in parent B. When the second version was asked, the people looked for disadvantages and also found them in parent B(like minor health problems, lots of work- related travel). Thus, the people were out to prove the question asked. As in this situation, many people try to simply prove their hypotheses and only look at the proof-worthy evidence.
The third problem is the "Selective Memory" problem. People tend to remember the events that agree with what they expected to happen. However, there are two main forms of events: one-sided and two-sided. An example of a one-sided event is as follows: people believe that the phone only rings while they are in the shower. So, when the phone rings while the person is in the shower, it becomes a memorable event. If the phone does not ring while the person is in the shower, it is not memorable at all. People are more likely to remember a dream if it comes true or applies to their lives. An example of a two-sided event is in a bet from a sporting event: if the team wins or loses, one will win or lose money -- either way, it will be a memorable event.
A test was formed to see if one-sided events truly are remembered more if they coincide with what was expected. A girlıs "diary" was read by several college students. Half was a detailed description of all her dreams and half a detailed description of what happened every day. Half of her dreams were somehow fulfilled while half were not. the students remembered more of the fulfilled dreams than the unfulfilled.
This brings up an important point. Most psychic predictions, etc. are very ambiguous. This is called "temporally unfocused." Thus, they are more likely only to be remembered if they somehow come true. The unfulfilled ambiguous prophecies are easily forgotten. More detailed and exact prophecies, etc.(called ³temporally focused²) are simply more easy to remember whether they come true or not. Another test was formed to evaluate this hypothesis.
Two groups read two different diaries set up much like the one mentioned before, except with prophesies instead of dreams: half prophesies, half real life. However, one diary had temporally focused prophesies which stated exact dates and times of predicted events while the other was more general. Again, half of each groupıs predictions were true and fulfilled, while the other half were not. Both groups remembered about the same amount of fulfilled prophesies. However, the temporally focused diary-group remembered more unfulfilled prophesies than did the temporally unfocused diary-group.
So, these tests showed why most psychic predictions and prophesies are ambiguous and temporally unfocused: if they are temporally unfocused, the listener is more likely to remember only the prophesies which turned out to be true and correct and will probably forget more of the unfulfilled prophesies.
Many other tests exist to determine how poor the average personıs judgment is. However, even through just these few tests, one can see that the average personıs judgment has many shortcomings.
DISCUSSION QUESTIONS:
(1) Do these tests really give an accurate view of peopleıs judgments?
(2) Can you give, from your own experiences, an example of a judgment bias of the type described in this article?
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A lack of volunteers thwarts research on prostate cancer.
The New York Times, February 12, 1997 A18
Gina Kolata
What is the best way to treat Prostate cancer? As of now, two main options are open to diagnosed men: Radical surgery and careful monitoring. The surgery often leaves men impotent and incontinent, though it also often removes the cancer completely. The monitoring just watches out for accelerated growth of the cancer.
A clinical trial was set up three years ago by the Department of Veterans Affairs in collaboration with the National Cancer Institute to find out which method was most beneficial. The trial included a random designation of which method would be used for the hoped-for 1,050 participants by 1997.
Unfortunately, according to Dr. Timothy J. Wilt, the director of the study at the Minneapolis Veterans Affairs Medical Center, only 315 men have enrolled. Dr. Richard Kaplan, the director of the study at the National Cancer Institute, also mentioned that some of the centers that were going to participate dropped out of the study because of a lack of men willing to participate.
However, this lack of participation does not surprise Dr. Joseph Oesterling, chief of urology at the University of Minnesota. "The reason is that men do not want to let someone else decide on whether they have surgery." He asked 300 men if they would like to enter the study and only four agreed to it. Oesterling says he cannot blame them for feeling this way. "After all, these men are facing a deadly disease... few of them want to have their treatment determined by a coin flip."
However, even Dr. Oesterling agrees that the study is important. According to Kolataıs article, approximately 317,000 men are estimated to be diagnosed with the disease this year and 48,000 will die from it.
The surgery - a radical prostatectomy -- removes the prostate gland which supplies the fluids that nourish sperm. Although the surgery leaves many men impotent and incontinent, supporters argue that removing the infected glad saves menıs lives. On the other hand, skeptics state that prostate cancer usually occurs late in life. According to the article, 95 per cent of all prostate cancer patients are older than 55 and the average patient is in his 60ıs. Also, the cancer usually grows fairly slowly. Thus, there is the possibility that the surgery could be unnecessary and a "monitoring" approach could be preferable. However, the cancer sometimes does grow quickly, and the surgery and even detection could come too late.
Dr. Wilt states that "there is no other way (except the study) to know if surgery prolongs life." He also adds that "selecting treatment is really a best guess" at this time.
Oesterling believes that the study may be coming too late, for people have already decided about their preferred treatment, though the evidence is "less than perfect."
A similar study was conducted in the 1970ıs. However, there was no difference in the death rates of the 61 who underwent the surgery and the 50 who did not. But, Dr. Kramer states that "there really werenıt enough people to test the hypothesis that surgery can save lives."
The most recent study involves men in Sweden. Dr. Jan-Erik Johansson, chief of urology at the Orebro Medical Center in Orebro, Sweden leads this investigation. 223 men deferred treatment and 77 had surgery when the cancer was detected. After 15 years, the study reports a survival rate of 81% for both groups. Johansson believes that "you overtreat many patients in the United States... That is the conclusion of our result."
However, Dr. Patrick Walsh, a urologist at Johns Hopkins University School of Medicine wrote in an editorial accompanying Johanssonıs study that the patients in the Swedish study were different. They were older, and "more likely to die with, and not of, their cancers." Walsh and another doctor, Dr. William Catalona, are firm believers in the idea that surgery can cure early-stage prostate cancer, though neither are participating in the study. Though he notes that having the data from the study would be nice, Dr. Catalona says that "everyone whoıs out there on the front lines really knows it works... If you looked them in the eye and said 'Tell me, are we going to have reliable data on the efficacy of treatment in 5, 10, or 15 years', I think if they were honest theyıd have to say no."
DISCUSSION QUESTIONS:
(1) Why do you think it is so hard to get patients to take part in a clinical study when there seems to be genuine uncertainty about the best procedure?
(2) Your Uncle Joe is 72 years old and has been diagnosed with Prostate cancer. He is trying to decide between "watchful waiting" and having an operation. He asked you to help him get the information to make an informed decision. What kind of information would you try to get?
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Reader Fred Hoppe was featured in a news article relating his attempt to persuade his students not to waste too much money on lotteries. Before discussing the article it is useful to have some knowledge of the rules of the Canadian Lotto 6/49 in Ontario. Here is how it is played:
A player chooses 6 distinct numbers from 1 to 49. The lottery officials pick 6 such numbers, called "winning numbers" and a 7th different number, called the "bonus number." The bonus number is used only in one situation: if you can match the 6 winning numbers by using the bonus number, then you win the second prize.
Forty-five percent of sales from each Lotto 6/49 is reserved as a prize pool. The total amount of \$10 prizes (fifth prize) is deducted from the this prize fund. The remainder is divided into four prize pools. Here are the prizes and your chance of winning them:
```Match Win Prob win
all 6 winning nos.(jackpot) 50% of prize pool 1/13,983,816
5 winning nos.+ bonus 15% of prize pool 1/2,330,636
5 winning nos. 12% of prize pool 1/5,5491.3
4 winning nos. 23% of prize pool 1/1,032.4
3 winning nos. \$10 fixed prize 1/56.6559
```
These probabilities and other lotto information can be found on Hoppe's Lotto web site. A good source for additional information about this lottery is http://bud.ica.net/.
DISCUSSION QUESTIONS:
(1) Did we get these probabilities of winning right?
(2) You find on Bud's web site the following results for the Feb.26 Lotto 6/49.
```Match No. winners Payoff for each
all 6 winning nos.(jackpot) 2 \$1,149,548
5 winning nos.+ bonus 9 \$76.636.50
5 winning nos. 229 \$2,409/50
4 winning nos. 13,933 \$75.90
3 winning nos. 277,974 \$10
```
How many tickets were sold? How much did the Ontario Lottery Corporation make on this day? Calculate the expected number of winners of each time assuming all buyers had their numbers chosen by the computer. Are the results consistent with this assumption? What was the expected payoff for a \$1 ticket in this lottery.
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Professor dashes dreams with anti-lottery course.
The Hamilton Spectator, 5 Feb. 1997, Money section
Lisa Marr
Fred Hoppe, whose statistics class is the largest class at McMaster (900 students), says that the lottery scenario is a great teaching tool. He tries to convince the students that they are wasting their money playing the lottery.
Hoppe says that you are more likely to be struck by lightning than to strike it rich from the lottery. He asks: "Would you pay \$1 to bet on 24 heads in a row?" Based on an average \$2.2 million jackpot, Hoppe estimates that someone spending \$25 a week for 20 years can expect to lose \$13,000, about half the \$26,000 it would cost to play. Based on a typical jackpot of \$2.2 million and assuming there is no sharing of the prizes, in those 20 years a player can expect to win a \$10 fifth prize 459 times and a \$73.50 fourth prize 25 times. It would take about 2000 years to get a second prize of \$131,934 and 40 years to get a third prize of \$2,300.
In contrast to this, certified financial planner Robert Beres points out that, if you took just \$15 a week and invested it in a mutual fund that earned a 12 per cent average annual return, you'd have \$61,735 after 20 years.
While Hoppe thinks some students who buy lottery tickets before taking his course will not after the course, others will continue even after they are better informed about their chances. He reports that not even his mother-in-law takes his advice -- she continues to buy lottery tickets. Hoppe remarks "We're not rational in many ways, that's what makes us human."
DISCUSSION QUESTIONS:
(1) On the other hand, doesn't it seem impressive that someone in Canada about once a week, in effect, gets 24 heads in a row tossing a coin? Which is more likely: winning a Jackpot in Lotto 6/49 or getting 24 heads in a row?
(2) What does the probability of getting struck by lightning mean? How is it estimated? (The book "What the Odds are?" by Les Krantz published by Harperperennial is our favorite source for the meaning of such odds and where they come from.)
(3) Can you see how Hoppe gets his 20-year estimates?
(4) How would you try to persuade your Aunt Hattie to quit buying lottery tickets?
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Joan Garfield suggested another lottery article. The lottery of interest in this article is a multi-state lottery in the US called Daily Millions. Again it is helpful to know something how the lottery in question works.
The Daily Millions lottery is run by the Multi-State Lottery Association that also runs the Powerball and Tri-West Lotto lotteries. Here is their description of the Daily Millions.
Every night we draw six balls out of three different drums. One drum contains red balls, the second drum contains white balls and the third contains blue balls. Two balls are drawn from each drum. The balls in each drum range from number 01 to 21. Players win by matching 2, 3, 4, 5, or 6 of the numbers drawn. A match occurs when you have the correct color and number for a given ball. The Grand Prize (won by matching all six balls drawn) is paid in cash. Match 5 pays \$5,000; Match 4 pays \$100; Match 3 pays \$5 and Match 2 pays \$2.
The jackpot is \$1 million dollars. Unlike other lotteries, winners do not have to share the prize with others having the same winning numbers. The one exception is when there are more than 10 winners for the jackpot. In this case the winners share a \$10 million dollar prize.
```Match Win Probability of win
6 \$1 million 1/9,261,000
5 \$5,000 1/81236.8
4 \$100 1/12,498
3 \$5 1/98.6682
2 \$2 1/11.1781
```
More information about this lottery can be found from the Multi- State Lottery Association web page.
On their lottery page they give the following table for the number of winners and the payouts. We assume this is the total since the lottery started up.
```Levels Winners Payout
Match 6 1 \$1,000,000
Match 5 454 \$2,270,000
Match 4 18,563 \$1,856,300
Match 3 357,791 \$1,788,955
Match 2 3,158,957 \$6,317,914
\$13,233,169
```
DISCUSSION QUESTION:
Estimate the number of tickets sold in this period. About how much did Multi-State Lottery make on lottery during this period? Are the number of winners in each category consistent with your estimate for the number of tickets sold assuming that all buyers have the computer pick their numbers?
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Daily Millions beats odds: no one wins -- 5-month losing streak puzzles even statisticians.
Star Tribune, 7 Feb. 1997, 1B
Pat Doyle
The Daily Millions lottery was started nearly five months ago and at the time of this article, 34 million tickets had been sold without a single \$1 million jackpot. It is stated that one would expect 3 or 4 winners by now and the probability of having no winners in this period is put at 1/38.
The Daily Millions lottery is run by Multi-State Lottery Association which also runs the Powerball lottery. The Daily Millions was invented to give a lottery where you are do not have to share the jackpot with other winners. You also are about 6 times more likely to win the Jackpot in the Daily Millions lottery than in the Powerball lottery.
However, the fact that no one has won the jackpot yet has hurt the sales. The Daily Millions has slumped from \$3.75 million in the first week of the lottery to \$1.23 million in the week ending Feb. 1.
Despite not having to pay out any jackpots, the participating states are required to set aside 11 percent of the ticket sales to be put in a pool for future winners. Charles Strutt, executive director of the Multi-State Lottery Association, said the money piling up in the jackpot pool will come in handy if players beat the odds in a different way, winning more than expected.
DISCUSSION QUESTIONS:
(1) Did we get the probability of winning the various prizes right?
(2) How did they arrive at the conclusion that the expected number of winners when 34 million tickets are sold is about 4 and the chance of no winner is about 1/38. Does it make any difference that the players do not choose their numbers at random?
P.S. On Saturday February 8 Daily Millions had its first winner.
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A lottery mystery.
Chance News 6.04
Laurie Snell
We often hear that people do not actually pick random numbers when they buy lottery tickets. It is claimed that they tend to put down birthdays so the lower numbers are more apt to occur. We decided to check this. We looked at the numbers that were chosen in the Powerball Lottery May 2 and May 3 1996 in some unknown state.
In the Powerball Lottery players choose five distinct numbers between 1 and 45 and then independently choose a 6th number called the "bonus number" again between 1 and 45. Note that this bonus number can be the same as one of the first five numbers. Players have a choice of having the computer choose their numbers or choosing the numbers themselves. If they choose the numbers they first mark 5 numbers in the rectangular array
``` 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 45 36
37 38 39 40 41 42 43 44 45
```
They then mark one number for the bonus number in a second such rectangular array. The slip on which they give their numbers actually has ten such rectangles allowing a player to choose up to 5 sets of numbers. On May 2 there were 7,985 sets of numbers chosen by hand and 28,478 picked by the computer. On May 3 there were 17001 picked by hand and 56,496 by the computer.
The first thing we looked at was the distribution of the individual numbers from 1 to 45. The distributions for the numbers chosen by the players on the two different days were remarkably similar. It was easy to reject the hypothesis that they represent samples from a uniform distribution. The smaller numbers were indeed more likely. On both days the most likely number chosen was 7 (about 3.6%) and the least likely 37 (about 1%). Not surprisingly, in the computer picked numbers was very close to a uniform distribution on the numbers from 1 to 45.
We then looked at the individual sets of five numbers chosen by the players ignoring the bonus numbers. For the computer chosen numbers there were no sets of five numbers that were chosen more than 3 times. For those picked by the players on May 3 there were 43 sets of numbers chosen more than 3 times. The largest number of times a set was chosen was 24. This occurred for the set 2 14 18 21 39. The next most popular set was 8 12 24 25 27 chosen 16 times and then 3,13,23,33,43 chosen 13 times. It was not surprising that this last set was chosen so many times since it is an arithmetic progression that could be chosen by just going down a diagonal in the rectangle of possible numbers starting with 3. There were several other arithmetic sequences in our list of popular sets of numbers. The most common arithmetic progressions were those whose differences were 9. They could be obtained by just going down a column in the rectangle.
The mystery was: where did the sets of numbers that were chosen that did not have a recognizable pattern come from? This mystery was pretty much solved when we looked at our grand list of numbers for May 3. We found, for example, that the records from 1771 to 1775 on our list all had the same first five numbers, our most popular set, but different sixth numbers (bonus number). In other words, probably one person decided to fill out the whole slip to buy five sets of numbers differing only in their bonus number. A silly thing to do unless you are concerned only with winning the jackpot! We found the same most popular sets in records 1780 to 1784 suggesting that the same person had done this again. In fact it appears from looking at the records that the same person bought all 24 tickets with the same first five numbers, 2 14 18 21 39 but each with a different bonus number.
This same type of behavior seemed to explain most of the other situations where we got a large number of repetitions of the first five numbers that were not arithmetic progressions. However it left one mystery set: 1 9 23 37 45 which appeared a widely separated points on our grand list 8 times and so, evidently, chosen by 8 different people. So the last remaining mystery is: why this set of numbers? The same set was chosen 4 times on May 2 (smaller set of numbers) and never by the computer.
DISCUSSION QUESTIONS:
(1) SOLVE OUR MYSTERY! Why did 8 different people choose their set of five numbers the same set 1 9 23 37 45?
(2) To win the Jackpot in the Powerball lottery you have to your initial five numbers correct plus the bonus number. What is the chance that you win the Jackpot in the Powerball lottery?
(3) In the land of Org there are choose(46,5)*49 = 54,979,155 possible birthdays. How many people do you need in the land of Org to make it a fair bet that two people have the same birthday? (Use your computer or the approximation 1.2*ln(c) where c is the number of possible birthdays (see, for example, P. Diaconis and F. Mosteller, Methods for studying coincidences. J. Am. Stat. Assoc. 84, 853-861 (1989)). What does this tell you about picking lottery tickets for the Powerball Lottery?
Please send comments and suggestions to jlsnell@dartmouth.edu.
!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
CHANCE News 6.04
(21 February 1997 to 4 March 1997)
!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! | 8,975 | 40,786 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2024-10 | latest | en | 0.914457 |
https://www.acmicpc.net/problem/5232 | 1,611,334,418,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703530835.37/warc/CC-MAIN-20210122144404-20210122174404-00667.warc.gz | 637,102,797 | 9,182 | 시간 제한 메모리 제한 제출 정답 맞은 사람 정답 비율
1 초 128 MB 0 0 0 0.000%
문제
While hiding out in the Outlands, Sam and Quorra are getting bored and start playing a game called grid nim which is a complex variation on the classic game of nim. The game board consists of n heaps arranged in a sequence, each containing a number of identical coins (see figure). The two players move alternately. When a player moves, he or she chooses a heap from either the left end or the right end of the sequence, and removes it from the sequence. The game is made more complex by the restriction that a player cannot select three consecutive heaps in three consecutive turns (this restriction does not apply when there is only one heap left at the end of the game). The game is over when the board is exhausted. The first player to play wins if the total number of coins he or she has selected is at least as much as the total number of coins collected by the second player. Otherwise the second player wins.
Here is one possible way the game may proceed in the example shown in the figure, assuming Sam moves first.
• Sam takes heap 1 from left end (7 coins)
• Quorra takes heap 5 from right end (5 coins)
• Sam takes heap 4 from right end (3 coins)
• Quorra takes heap 3 from right end (4 coins)
• Sam takes heap 2 (0 coins)
At the end, Sam has 7 + 3 = 10 coins which is more than Quorra, and hence Sam wins this game. In fact, Sam can win no matter what Quorra does – if she chooses the heap 2 with 0 coins, then Sam can take the heap with 5 coins, and end up with a higher total score of 15 or 16 coins.
The restriction mentioned above does not come up in this example. However, consider an example where there are a large number of heaps, and consider an initial sequence of moves:
Sam - Left, Quorra - Right, Sam - Left, Quorra - Right
In the next turn, Sam cannot choose from the left end again, and must take the heap from the right end. Now the sequence of moves is:
Sam - Left, Quorra - Right, Sam - Left, Quorra - Right, Sam - Right
Now, although Quorra chose from the Right end last two times, she can choose either from the Right or the Left end. Since Sam just chose from the Right end, Quorra will not be taking a 3rd consecutive heap and hence she can choose from the Right end.
Quorra is very good at playing this game and always makes the best possible move. To account for this, she allows Sam to play first. Your goal is to help Sam win. Specifically, given an input board, your goal is to find out whether Sam (playing first) has a winning strategy assuming Quorra always makes the best possible move at every turn.
입력
The first line in the test data file contains the number of test cases (≤ 50). After that, each line contains one test case. The test case begins with the number of elements in the sequence, k, and then we have k numbers which specify the numbers of coins in the heaps in the sequence. Assume all numbers are ≥ 0, and < 230.
출력
For each test case, you are to output “YES” (if Sam, playing first, has a winning strategy), or “NO” (if he does not).
예제 입력 1
4
3 1 10 1
4 5 7 2 1
7 0 0 8 4 2 5 2
5 7 0 4 3 5
예제 출력 1
NO
YES
NO
YES | 810 | 3,151 | {"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 | 3 | CC-MAIN-2021-04 | latest | en | 0.942137 |
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# Actual GMAT Extremely lower than CATs--What to do?
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Intern
Joined: 05 Dec 2012
Posts: 3
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Schools: Kellogg 1YR '15
Actual GMAT Extremely lower than CATs--What to do? [#permalink]
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07 Dec 2012, 04:04
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Hi everyone,
I took the actual test yesterday and since then i'm trying to understand what happened but i really couldn't. I scored unexpectedly lower that all my cats including the Diagnostic. I heard from friends and read that Kaplan Cats are actually harder and i should expect to score higher. I went to the test and i wasn't stressed at all, i was relaxed and confident even while doing the test. I finished almost in time (i selected the answer for the last question but didnt have the time it next). Not only that but i was expected to see 650+ as a score but i got shocked with a 520 ((((((((. I'm very disappointed and i dont know what i should do next.
In my preparation i used Manhattan and Kaplan. I spent June & July studying Manhattan, then I paused for two months, started again in Oct. In Nov i took two weeks intensive course by Kaplan.
I'll share my story and i would appreciate your feedback/suggestion. Please i want your honest opinion. You don't have to say something to make me feel good, i just want to know what i did wrong.
My Cat Scores are as follows:
Date Quant Verbal Total
09/11/2012 MGMAT CAT1 34 35% 31 59% 540 44%
10/10/2012 MGMAT CAT2 42 57% 32 62% 600 62%
11/06/2012 Kapan CAT1 51% 36% 550
11/09/2012 Kapan CAT2 40% 64% 570
11/14/2012 Kapan CAT3 59% 53% 600
12/01/2012 Kapan CAT4 70% 35% 590
12/03/2012 Kapan CAT5 82% 63% 620
My actual GMAT was 520 (38%), Q(41)54% V(21)24%
Thanks in advance for taking the time to read my post and i look forward to get your feedback.
Kudos [?]: [0], given: 1
Kaplan GMAT Prep Discount Codes Optimus Prep Discount Codes EMPOWERgmat Discount Codes
Intern
Joined: 26 Oct 2012
Posts: 18
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Re: Actual GMAT Extremely lower than CATs--What to do? [#permalink]
### Show Tags
09 Dec 2012, 07:39
Your Quant is pretty good. Do OG 11-12-13 and Quant Review. Do Questions 50 and Above, since they are harder. Timing is key. make sure you can do every problem under 2 mins. If your over, you need to find a faster way of completing the question. For example, Weighted Avg Problems used to take me 2 min plus. Then my Tutor showed me short cuts where you can do it in less than 20 secs. NO JOKE! So work on timing and you should be able to get 46+ in Quant. Get a good tutor to fast track.
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Intern
Joined: 26 Oct 2012
Posts: 18
Kudos [?]: 1 [0], given: 0
Re: Actual GMAT Extremely lower than CATs--What to do? [#permalink]
### Show Tags
09 Dec 2012, 07:45
Understand the GMAt SC rules ( ||, Comparison, Verb tense, pronoun.etc..)
CR - Know how to find the conclusion and know how to analyze the different types of CR's.
Again Timing.
Since your Verbal is low. Spend more time on verbal, even though its really boring. Verbal is a battle, don't get discouraged, it will come. I was in the same boat! and trust me, its an amazing feeling once you get into school and you look back on how it was worth all the BS..lol Good luck GUYYYYYYYYYYYYYY
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Re: Actual GMAT Extremely lower than CATs--What to do? [#permalink] 09 Dec 2012, 07:45
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# Actual GMAT Extremely lower than CATs--What to do?
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# CUBE 2-Point Perspective with Ellipse Guideby KhoaSV
©2011-2014
Step-by-step construction of a 2 point perspective cube. For ANI12.
Normally I wouldn't put the station point on "stage" but for the sake of all the parts being there, SP is on stage.
Update: I just fixed some minor things and added a quick guide to draw ellipses in 2-point perspective.
Add a Comment:
Featured By Owner Jan 26, 2013
on first step. You draw the cone of vision 1st or the angle of 30' of station 1st??
Featured By Owner Jan 26, 2013 General Artist
I drew the station point first, then drew a 30 degree angle from that point to the eye level. The cone is just a result of that.
Featured By Owner Oct 11, 2012
And a little more: [link]
Featured By Owner Oct 11, 2012
I have another question: [link]
Basically it's about adding a little offset... doesn't it look better?
Featured By Owner Sep 25, 2012
How do you get the curve dotted line that goes up from the station point?
And how do you determine where to put the station point?
Featured By Owner Sep 25, 2012
Oh - I figured it out ... use a compass from the VP1 and VP2 in each case... thanks for computers, eh?
Featured By Owner Sep 25, 2012 General Artist
That means you need to take the length from the Sation Point to the Vanishing Point (1 or 2) and swing it up to Eye Level. I used the curved dotted line because our teacher showed us this method by using a divider, but you can use a ruler to get that length.
And Station Point can be anywhere because it's one of thing first few things you have to do. The further away the SP, the bigger your Cone of Vision gets. The Cone of Vision in this tutorial barely covers the cube, you'll often want something that's big enough to get your whole page in focus.
Featured By Owner Feb 6, 2012 Student Traditional Artist
Whoawhoawhoawhoa! Ani12, Sjsu, John Clapp??? Were you in the 12:30 (or was it 12?) class on monday/wednesday???
Btw, your tutorial is amazingly helpful. I'm in Ani 28 right now doing cubes, and I totally forgot how to do them. This helped so much! Thank you thank you thank you!
Featured By Owner Feb 6, 2012 General Artist
It wasn't mentioned here, but the lines connecting to the two vanishing points should be perpendicular at SP. I'm making a fix now. When do u have 28 btw.
Featured By Owner Feb 7, 2012 Student Traditional Artist
I have it Tues/Thurs at 3. And thank you for telling me that!
Add a Comment: | 729 | 2,762 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2014-41 | longest | en | 0.931646 |
http://fgarciasanchez.es/thesisfelipe/node36.html | 1,679,603,951,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945183.40/warc/CC-MAIN-20230323194025-20230323224025-00631.warc.gz | 23,163,508 | 4,906 | # 3.9 Energy barriers of individual grains
The requirement of large thermal stability of the recording media implies large energy barriers. Therefore, the next step is their evaluation for the grains of the previous section. For this purpose we used the Lagrange multiplier method presented in Section 2.3.6. Fig. 3.33 shows the energy barrier calculation for both model I and II as a function of the (perpendicular) interfacial exchange. Initially, the system is considered in the minimum that corresponds to all the moments aligned and to the coordinates in the energy landscapes plotted in Figs. 3.34 and 3.35. Recall that is the average magnetization polar angle. The final state will be the equivalent minimum but with the moments pointing in the opposite direction, namely, . For small values of the interfacial exchange, there is an intermediate minimum (see Fig. 3.34) and, therefore, additional reversal modes and saddle points. In this situation the soft grain can switch first with very low energy barrier and the hard magnetic material will follow with a reduced energy barrier value. However, the energy barrier corresponding to the inverse process, namely, returning to the original minimum, is very small and the probability of the inverse process to take place is high. Fig. 3.34 shows the effective energy landscape and relevant configurations. The saddle points configurations are curled to minimize the magnetostatic energy of the soft material. For interfacial exchange energy higher than of the bulk exchange in all cases, the energy barrier values saturate as a function of and correspond to collective reversal. The collective modes are different for the two models. In model I the thermal mode is almost coherent (see Fig. 3.36(a)). The maximum energy barrier value is determined by the quantity . In the model II, the value of the energy barrier coincides with the domain wall energy in the hard magnetic material as observed by D.Suess [Suess 05b] and the saddle point is a domain wall centered in the hard material (see Fig. 3.36(b)). These differences can be appreciated in the effective energy Fig. 3.35. If the point is the saddle point, the energy barrier corresponds to coherent rotation. For a non-homogeneous mode the saddle point has to be elsewhere as in Fig. 3.35(b).
Figure: Effective energy landscapes : (a) Model I , and (b) Model II , and .
(a) (b)
Figure 3.36: Saddle point configuration for different mechanism: (a) Coherent rotation (Model I) (b) Domain wall (Model II).
(a) (b)
To get a better understanding of the energy barriers in a soft/hard material, we can examine the case of model I for , with a model similar to that of Ref. [Victora 05a]. The model supposes a macro-spin for each material. In such a model the energy is:
(3.8)
where and are the effective anisotropy constants, is the number of atoms in the surface, is the volume of each material and finally and are the unit vector of the macrospin of each grain. To analyze the problem, we have to consider the energy gradient and the Hessian matrix. For large values of there is only one energy barrier corresponding to the point which value is . For small interfacial exchange, there are additional barriers and saddle points similar to the ones plotted in Fig. 3.34. The critical exchange for the crossover between the region is:
(3.9)
In our system . Finally, we can compare the analytical results with our simulations for the same values of the constants. From Fig. 3.37, it is clear that there is a good quantitative and qualitative agreement. These results indicate that the two macro-spins is a good approximation for small interfacial exchange. Nevertheless, this model is not valid for large exchange, where highly nonhomogeneous modes are expected.
Figure: (a) Comparison between the analytical model (solid lines) and numerical calculations (symbols) for and in model I. (b) Energy contour plot from Eq. (3.8) for .
(a) (b)
2008-04-04 | 872 | 3,968 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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 | 0.887538 |
https://mathematica.stackexchange.com/questions/140752/why-except-is-so-slow-as-compared-to-equivalent-regularexpression | 1,713,829,353,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818374.84/warc/CC-MAIN-20240422211055-20240423001055-00034.warc.gz | 331,529,630 | 40,245 | # Why Except is so slow as compared to equivalent RegularExpression?
When comparing performance of the solutions suggested for this question I discovered that StringExpression involving Except is two orders of magnitude slower for large strings as compared to equivalent RegularExpression. Here is a simplified example demonstrating the problem:
With[{str = StringRepeat["a", 10^6] <> "b"}, {
AbsoluteTiming[StringMatchQ[str, Except["b"] .. ~~ "b"]][[1]],
AbsoluteTiming[StringMatchQ[str, RegularExpression["[^b]+b"]]][[1]]}]
Divide @@ %
{0.719855, 0.00323316}
222.648
The string pattern is 220 times slower than pure regex! But the regular expression [^x]+x is the direct semantic translation of the string pattern Except["x"] .. ~~ "x", and this translation is unique and unambiguous. Why then the latter is so insanely slow?
Using StringPatternPatternConvert we can find the regexp into which Mathematica converts the original string expression:
StringPatternPatternConvert[Except["b"] .. ~~ "b"][[1]]
"(?ms)(?:[^b])+b"
The only difference as compared to the direct semantic translation is that the negated character class [^b] is enclosed by redundant non-capturing group (?: … ). Since this group is non-capturing, one would expect that it can't introduce noticeable overhead. But is it so in practice? Let us check:
With[{str = StringRepeat["a", 10^6] <> "b"}, {
AbsoluteTiming[StringMatchQ[str, Except["b"] .. ~~ "b"]][[1]],
AbsoluteTiming[StringMatchQ[str, RegularExpression["(?:[^b])+b"]]][[1]],
AbsoluteTiming[StringMatchQ[str, RegularExpression["[^b]+b"]]][[1]]}]
{0.721896, 0.717353, 0.00325354}
We see that this redundant non-capturing group is responsible for all the slowdown we observe.
It is interesting that the overhead isn't constant but grows step-wise with the length of the string up to approximately 95000 characters:
PrintTemporary[Dynamic[n]];
timings = Transpose[Table[With[{str = StringRepeat["a", n] <> "b"},
{{n, AbsoluteTiming[StringMatchQ[str, RegularExpression[".+b"]]][[1]]},
{n, AbsoluteTiming[StringMatchQ[str, RegularExpression["(?:.)+b"]]][[1]]}}],
{n, 100, 100000, 100}]];
ListPlot[timings, PlotRange -> All, Frame -> True, Axes -> False, ImageSize -> 600,
FrameLabel -> {"string length", "seconds"}]
(evaluated with Mathematica 11.1.0 on Windows 7 x64.)
• good question and great analysis. If I'd had seen this earlier I would have simply suggested that if you have enough knowledge to formulate as regexp and care about speed, you should always use regexps. I think there are many other examples with similar, although not as extreme differences. And AFAIK all string expressions are internally converted to regexps anyway so they are always adding at least the overhead to generate the regexp... Mar 22, 2017 at 21:47
• @AlbertRetey Fuerthermore each time you will call the function, the regex expression will be recreated, and this operation is far from being free in a loop in contrary of when you instiated once before. Mar 23, 2017 at 9:17
• @Walfrat Actually the Documentation states (emphasis is mine): "The regular expression is then compiled by PCRE, and the compiled version is cached for future use when the same pattern appears again. The translation from symbolic string pattern to regular expression only happens once." My tests seem to support this statement: I didn't observe any difference in timings between, for example, StringMatchQ[#, patt]&/@list and StringMatchQ[list, patt]. But I didn't test this aspect carefully. Mar 23, 2017 at 9:27 | 859 | 3,524 | {"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} | 3.171875 | 3 | CC-MAIN-2024-18 | latest | en | 0.807421 |
https://codereview.stackexchange.com/questions/37415/sudoku-using-exact-cover-solver?noredirect=1 | 1,723,675,495,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641137585.92/warc/CC-MAIN-20240814221537-20240815011537-00641.warc.gz | 141,770,194 | 46,811 | # Sudoku using 'exact cover' solver
### 1. Introduction
This is a solution to Weekend Challenge #3: a Sudoku solver in Python. It works by translating a Sudoku puzzle into an exact cover problem, and then solving the exact cover problem using Donald Knuth's "Algorithm X". The code should be self-explanatory (assuming you've read and understood the Knuth paper), so I won't say more than that.
### 2. exactcover.py
from collections import Iterator, defaultdict
from random import shuffle
class ExactCover(Iterator):
"""An iterator that yields solutions to an EXACT COVER problem.
An EXACT COVER problem consists of a set of "choices". Each choice
satisfies one or more "constraints". Choices and constraints may
be represented by any hashable Python objects, with the proviso
that all choices must be distinct, as must all constraints.
A solution is a list of choices such that each constraint is
satisfied by exactly one of the choices in the solution.
The constructor takes three arguments:
constraints A map from each choice to an iterable of constraints
satisfied by that choice.
initial An iterable of choices that must appear in every
solution. (Default: no choices.)
random Generate solutions in random order? (Default: False.)
For example:
>>> next(ExactCover(dict(A = [1, 4, 7],
... B = [1, 4],
... C = [4, 5, 7],
... D = [3, 5, 6],
... E = [2, 3, 6, 7],
... F = [2, 7])))
['B', 'D', 'F']
"""
# This implements Donald Knuth's "Algorithm X"
# http://lanl.arxiv.org/pdf/cs/0011047.pdf
def __init__(self, constraints, initial=(), random=False):
self.random = random
self.constraints = constraints
# A map from constraint to the set of choices that satisfy
# that constraint.
self.choices = defaultdict(set)
for i in self.constraints:
for j in self.constraints[i]:
# The set of constraints which are currently unsatisfied.
self.unsatisfied = set(self.choices)
# The partial solution currently under consideration,
# implemented as a stack of choices.
self.solution = []
# Make all the initial choices.
try:
for i in initial:
self._choose(i)
self.it = self._solve()
except KeyError:
# Initial choices were contradictory, so there are no solutions.
self.it = iter(())
def __next__(self):
return next(self.it)
next = __next__ # for compatibility with Python 2
def _solve(self):
if not self.unsatisfied:
# No remaining unsatisfied constraints.
yield list(self.solution)
return
# Find the constraint with the fewest remaining choices.
best = min(self.unsatisfied, key=lambda j:len(self.choices[j]))
choices = list(self.choices[best])
if self.random:
shuffle(choices)
# Try each choice in turn and recurse.
for i in choices:
self._choose(i)
for solution in self._solve():
yield solution
self._unchoose(i)
def _choose(self, i):
"""Make choice i; mark constraints satisfied; and remove any
choices that clash with it.
"""
self.solution.append(i)
for j in self.constraints[i]:
self.unsatisfied.remove(j)
for k in self.choices[j]:
for l in self.constraints[k]:
if l != j:
self.choices[l].remove(k)
def _unchoose(self, i):
"""Unmake choice i; restore constraints and choices."""
self.solution.pop()
for j in self.constraints[i]:
for k in self.choices[j]:
for l in self.constraints[k]:
if l != j:
### 3. sudoku.py
from collections import Iterator
from copy import deepcopy
from exactcover import ExactCover
from itertools import islice, product
from math import ceil, sqrt
from random import shuffle
from string import ascii_lowercase, ascii_uppercase
DIGITS = '123456789' + ascii_uppercase + ascii_lowercase
def make_grid(n):
"""Return a Sudoku grid of size n x n."""
return [[-1] * n for _ in range(n)]
def puzzle_to_grid(puzzle):
"""Convert printed representation of a Sudoku puzzle into grid
representation (a list of lists of numbers, or -1 if unknown).
"""
puzzle = puzzle.split()
grid = make_grid(len(puzzle))
for y, row in enumerate(puzzle):
for x, d in enumerate(row):
grid[y][x] = DIGITS.find(d)
return grid
def grid_to_puzzle(grid):
"""Convert grid representation of a Sudoku puzzle (a list of lists of
numbers, or -1 if unknown) into printed representation.
"""
return '\n'.join(''.join('.' if d == -1 else DIGITS[d] for d in row)
for row in grid)
class Sudoku(Iterator):
"""An iterator that yields the solutions to a Sudoku problem.
The constructor takes three arguments:
puzzle The puzzle to solve, in the form of a string of n
words, each word consisting of n characters, either a
digit or a dot indicating a blank square.
(Default: the blank puzzle.)
n The size of the puzzle. (Default: 9.)
m The width of the blocks. (Default: the square root of n,
rounded up.)
random Generate solutions in random order? (Default: False.)
For example:
>>> print(next(Sudoku('''...84...9
... ..1.....5
... 8...2146.
... 7.8....9.
... .........
... .5....3.1
... .2491...7
... 9.....5..
... 3...84...''')))
632845179
471369285
895721463
748153692
163492758
259678341
524916837
986237514
317584926
"""
def __init__(self, puzzle=None, n=9, m=None, random=False):
if puzzle:
puzzle = puzzle.split()
self.n = n = len(puzzle)
initial = self._encode_puzzle(puzzle)
else:
self.n = n
initial = ()
if m is None:
m = int(ceil(sqrt(n)))
assert(0 < n <= len(DIGITS))
assert(n % m == 0)
def constraints(choice):
d, x, y = self._decode_choice(choice)
block = m * (x // m) + y // (n // m)
return [a + 4 * (b + n * c) for a, b, c in [
(0, x, y), # Any digit at x, y.
(1, d, y), # Digit d in row y.
(2, d, x), # Digit d in column x.
(3, d, block), # Digit d in block.
]]
self.solver = ExactCover({i: constraints(i) for i in range(n ** 3)},
initial, random)
def __next__(self):
return self._decode_solution(next(self.solver))
next = __next__ # for compatibility with Python 2
def _encode_choice(self, d, x, y):
"""Encode the placement of d at (x, y) as a choice."""
n = self.n
assert(0 <= d < n and 0 <= x < n and 0 <= y < n)
return d + n * (x + n * y)
def _decode_choice(self, choice):
"""Decode a choice into a (digit, x, y) tuple."""
choice, digit = divmod(choice, self.n)
y, x = divmod(choice, self.n)
return digit, x, y
def _encode_puzzle(self, puzzle):
"""Encode a Sudoku puzzle and yield initial choices."""
for y, row in enumerate(puzzle):
for x, d in enumerate(row):
digit = DIGITS.find(d)
if digit != -1:
yield self._encode_choice(digit, x, y)
def _decode_solution(self, solution):
"""Decode a Sudoku solution and return it as a string."""
grid = make_grid(self.n)
for d, x, y in map(self._decode_choice, solution):
grid[y][x] = d
return grid_to_puzzle(grid)
class ImpossiblePuzzle(Exception): pass
class AmbiguousPuzzle(Exception): pass
def solve(puzzle, m=None):
"""Solve the Sudoku puzzle and return its unique solution. If the
puzzle is impossible, raise ImpossiblePuzzle. If the puzzle has
multiple solutions, raise AmbiguousPuzzle.
Optional argument m is the width of the blocks (default: the
square root of n, rounded up).
For example:
>>> print(solve('''.6.5..9..
... ..4.....6
... .29..3..8
... ....32..4
... ..61.75..
... 1..95....
... 6..4..27.
... 2.....4..
... ..7..6.8.'''))
768514923
314829756
529763148
975632814
836147592
142958637
693481275
281375469
457296381
>>> print(solve('''...8.....
... ..1.....5
... 8....1...
... 7.8....9.
... ...1.....
... .5....3.1
... ....1...7
... 9.....5..
... 3........'''))
... # doctest: +IGNORE_EXCEPTION_DETAIL
Traceback (most recent call last):
...
ImpossiblePuzzle: no solutions
"""
solutions = list(islice(Sudoku(puzzle, m=m), 2))
if len(solutions) == 1:
return solutions[0]
elif len(solutions) == 0:
raise ImpossiblePuzzle('no solutions')
else:
raise AmbiguousPuzzle('two or more solutions')
def make_puzzle(n=9, m=None):
"""Return a random nxn Sudoku puzzle with 180-degree rotational
symmetry. The puzzle returned is minimal in the sense that no
symmetric pair of givens can be removed without making the puzzle
ambiguous.
The optional arguments are n, the size of the puzzle (default: 9)
and m, the width of the blocks (default: the square root of n,
rounded up).
"""
grid = puzzle_to_grid(next(Sudoku(n=n, m=m, random=True)))
coords = list(divmod(i, n) for i in range((n ** 2 + 1) // 2))
shuffle(coords)
for i, j in coords:
g = deepcopy(grid)
g[i][j] = g[n - i - 1][n - j - 1] = -1
try:
solve(grid_to_puzzle(g))
grid = g
except AmbiguousPuzzle:
pass
return grid_to_puzzle(grid)
### 4. Examples
>>> for puzzle in map(make_puzzle, (4, 6, 9, 12, 16)):
... print('\n'.join('{} {}'.format(*a) for a in zip(puzzle.split(), solve(puzzle).split())), '\n')
...3 1243
.4.. 3412
..2. 4321
2... 2134
..2..4 632514
.1.2.. 415236
3.4... 354162
...3.5 126345
..1.5. 241653
5..4.. 563421
...3...4. 678325149
1.94..... 139476528
...98.3.7 452981367
.87....34 587612934
..4...7.. 214893756
39....81. 396754812
7.5.69... 725169483
.....72.1 863547291
.4...8... 941238675
6A....7.8... 6ABC42758391
.952C.A...4. 3952C8A17B46
4..1....5... 47819B635C2A
83...1..96.. 832471BA96C5
7......921.4 7B658C3921A4
.....6.4.7.. 91CA2654B783
..A.3.2..... B5A7342C1869
C.196......B C4196A87325B
..38..1...7C 2638591BA47C
...3....6..2 1C73A54869B2
.8...3.2451. A896B3C24517
...B.7....38 524B1796CA38
.F3D..A.......87 1F3D95AEC46BG287
.9B.63....5.F... E9BC63D87G52FA14
457..G.1....9... 457ACG21FD8396BE
.8..5..G..AC.... F8D25E6G13AC497B
..6E.48.D..7.1.G CB6E3489D5F7A12G
5.14D...G...68.. 5314D27AGEB968FC
.A.7.FC.8.4..... 9AG71FCB82465ED3
..91...4...G25.F BD918CE4367G25AF
7.A.G..3.CD.EB.. 72A6G1F35CD4EB98
....76..A..E..G. 3C8576B2AF9E14GD
...8....E.3..G62 D4C8F917EA35BG62
...92.3C.....FE. 67592A3C4BGD8FE1
...B.8....CF.35. 21EB48GD69CF735A
AG.......7..CD4. AGF3EB562718CD49
This is not quite right:
def _unchoose(self, i):
"""Unmake choice i; restore constraints and choices."""
self.solution.pop()
It promises to undo an arbitrary choice i but actually pops the latest one from self.solution. As you only use if for undoing the latest choice, I suggest to drop the parameter:
def _unchoose(self):
"""Unmake latest choice; restore constraints and choices."""
i = self.solution.pop()
Actually the choose-unchoose pattern feels a bit awkward to me. I'd prefer functional programming instead. Another idea that would make me feel more comfortable would be to create a context manager:
@contextlib.contextmanager
def _tentative_choice(self, i):
self._choose(i)
yield
self._unchoose(i)
use:
for i in choices:
with self._tentative_choice(i):
for solution in self._solve():
yield solution
Another small concern is that you use a local variable choices in methods of a class that has an instance variable self.choices. This may lead to confusion; I'd use a different name.
• Good review! But can you explain the use of try: ... finally: in the context manager? If _solve raises an exception then something must have gone so badly wrong that I can't expect to be able to continue searching, so surely there's no need to worry about restoring the context in that case? Commented Dec 16, 2013 at 14:22
• @GarethRees You are right. I was reaching for a more defensive style, but I can't see an actual benefit now. Commented Dec 16, 2013 at 18:51
I have only a minor comment:
In ExactCover.__init__, I would use iteritems rather than iterating over the keys of constraints:
for choice, constraintsList in self.constraints.iteritems():
for constraint in constraintsList: | 3,419 | 11,974 | {"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} | 3.34375 | 3 | CC-MAIN-2024-33 | latest | en | 0.858691 |
https://www.comsol.pt/blogs/added-value-task-parallelism-batch-sweeps/?setlang=1 | 1,508,272,478,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187822488.34/warc/CC-MAIN-20171017200905-20171017220554-00007.warc.gz | 923,012,011 | 17,753 | ##### Pär Persson Mattsson March 20, 2014
One thing we haven’t talked much about so far in the Hybrid Modeling blog series is what speedup we can expect when adding more resources to our computations. Today, we consider some theoretical investigations that explain the limitations in parallel computing. We will also show you how to use the COMSOL software’s Batch Sweeps option, which is a built-in, embarrassingly parallel functionality for improving performance when you reach these limits.
### Amdahl’s and Gustafson-Barsis’ laws
We have mentioned before how speedup through the addition of compute units is dependent on the algorithm (in this post we will use the term processes, but added compute units can also be threads). A strictly serial algorithm, like computing the elements of the Fibonacci series, does not benefit at all from an added process, while a parallel algorithm, like vector addition, can make use of as many processors as we have elements in the vector. Most algorithms in the real world are somewhere in between these two.
To analyze the possible maximum speedup of an algorithm, we will assume that it consists of a fraction of perfectly parallelizable code and a fraction of strictly serial code. Let us call the fraction of parallelized code \varphi, where \varphi is a number between (and including) 0 and 1. This automatically means that our algorithm has a fraction of serial code that is equal to (1-\varphi).
Considering the computation time, T(P), for P active processes, and starting with the case P=1, we can use the representation T(1) = T(1) \cdot(\varphi + (1-\varphi)). When running P processes, the serial fraction of the code is not affected, but the perfectly parallelized code will be computed P times faster. Therefore, the computation time for P processes is T(P)=T(1) \cdot (\varphi / P + (1 -\varphi)), and the speedup is S(P):=T(1)/T(P)=1/(\varphi/P+(1-\varphi)).
#### Amdahl’s Law
This expression is at the heart of Amdahl’s law. Plotting S(P) for different values of \varphi and P, we now see something interesting in the graph below.
The speedup for increasing the number of processes for different fractions of parallelizable code.
For 100% parallelized code, the sky is the limit. Yet, we find that for \varphi<1, the asymptotic limit or the theoretical maximal speedup is S_{max}(\varphi):=\lim_{P\to \infty} S(P)=1/(1-\varphi).
For a 95% parallelized code, we find that S_{max}(0.95)=20 — a maximum speedup of twenty times, even if we have an infinite number of processes. Furthermore, we have S_{max}(0.9)=10, S_{max}(0.75)=4, and S_{max}(0.5)=2. The theoretical maximum speedup decreases quickly when decreasing the fraction of parallelized code.
But don’t give up and go home just yet!
#### Gustafson-Barsis’ Law
There is one thing that Amdahl’s law does not consider, and that is the fact that when we buy a faster and larger computer to be able to run more processes, we usually don’t want to compute our small models from yesterday faster. Instead, we want to compute new, larger (and cooler) models. That’s what the Gustafson-Barsis’ law is all about. This is based on the assumption that the size of the problem we want to compute increases linearly with the number of available processes.
Amdahl’s law assumes that the size of the problem is fixed. When adding new processors they are working on parts of the problem that was originally handled by a lesser number of processes. By adding more and more processes, you are not utilizing the full ability of the added processes as eventually the size of what they are able to work on reaches a lower limit. Yet, by assuming that the size of the problem increases with the number of added processes, then you are utilizing all the processes to an assumed level, and the speedup of the performed computations remains unbounded.
The equation describing this phenomenon is S(P)=\phi\cdot P-(1-\phi), which gives us a far more optimistic result for what is called scaled speedup (which is like productivity), as shown in the graph below:
When taking into account that the size of the job normally increases with the number of available processes, our predictions are more optimistic.
### The Cost of Communication
Gustafson-Barsis’ law implies that we are only restricted in the size of the problem we can compute by the resources we have for adding processes. Yet, there are other factors that affect speedup. Something we’ve tried to stress so far in this blog series is that communication is expensive. But we haven’t talked about how expensive it is yet, so let’s look at some examples.
Let’s consider an overhead that is dominated by the communication and synchronization required in parallel processing, and model this as time added to the computation time. This means that the amount of communication increases when we increase the number of processes, and that this increase will be modeled by a function OH(P)=c\cdot f(P), where c is a constant and f(P) is some function. Hence, we can compute the speedup through: S(P)_{OH}=1/(\varphi/P+(1-\varphi)+c \cdot f(P)).
The graph below shows the case where the fraction of parallelized code is 95%, and where we can see the speedup for an increasing number of processes, for different functions of f(P), assuming c= 0.005 (this constant would vary between different problems and platforms). In the case of no overhead, the result is as predicted by Amdahl’s law, but when we start adding overhead, we see that something is happening.
For a linearly increasing overhead, we find that the speedup doesn’t reach a value greater than five before the communication starts to counteract the increased computation power added by more processes. For a quadratic function, f(P), the result is even worse and, as you might recall from our earlier blog post on distributed memory computing, the increase of communication is quadratic in the case of all-to-all communication.
Speedup with added overhead. The constant, c, is chosen to be 0.005.
Due to this phenomenon, we cannot expect to have a speedup on a cluster for, say, a small time-dependent problem when adding more and more processes. The amount of communication would increase faster than any gain from added processes. Yet, in this case, we have only considered a fixed problem size and the “slowdown” effect introduced through communication would be less relevant as we increase the size of our problem.
### Batch Sweeps in COMSOL Multiphysics
Let us now leave the theoretical aspect and learn how to make use of the batch sweep feature in COMSOL Multiphysics. As our example model, we will use the electrodeless lamp, which is available in the Model Gallery. This model is small, at around 80,000 degrees of freedom, but needs about 130 time steps in its solution. To make this transient model parametric as well, we will compute the model for several values of the lamp power, namely 50 W, 60 W, 70 W, and 80 W.
On my workstation, a Fujitsu® CELSIUS® equipped with an Intel® Xeon® E5-2643 quad core processor and 16 GB of RAM, the following compute times are received:
Number of Cores Compute Time per Parameter Compute Time for Sweep
1 30 mins 120 mins
2 21 mins 82 mins
3 17 mins 68 mins
4 18 mins 72 mins
The speedup here is far from perfect — just about 1.7 for three cores and has even decreased for four cores. This is due to fact that it is a small model with a low number of degrees of freedom per thread within each time step.
We will now use the batch sweep functionality to parallelize this problem in another way: we will switch from data parallelism to task parallelism. We will create a batch job for each parameter value and see what this does to our computation times. To do this, we first activate the “Advanced Study Options”, then we right-click on “Study 1” and choose “Batch Sweep”, as illustrated in the animation below:
How to activate batch sweep in a model, include the parameter values, and specify the number of simultaneous jobs.
The graph below indicates the productivity or “speedup” we can get by controlling the parallelization. When running one batch job using four cores, we get the result from above: 72 minutes. When changing the configuration to two batch jobs simultaneously, each using two cores, we can compute all the parameters in 48 minutes. Finally, when computing four batch jobs at the same time, each using one processor, the total computation time is 34 minutes. This gives speedups of 2.5 and 3.5 times, respectively — a lot better compared to parallelizing through using pure shared memory alone.
Simulations per day for the electrodeless lamp model. “4×1″ means four batch jobs run simultaneously, using one core each.
### Concluding the Hybrid Modeling Series
Throughout this blog series, we have learned about shared, distributed, and hybrid memory computing and what their weaknesses and strengths are, as well as the large potential of parallel computing. We have also learned that there is no such thing as a free lunch when it comes to computing; we cannot just add processes and hope for perfect speedup for all types of problems.
Instead, we need to choose the best way to parallelize a problem to get the most performance gain out of our hardware, much like we have to choose the correct solver to get the best solution time when solving a numerical problem.
Selecting the right parallel configuration is not always easy, and it can be hard to know beforehand how you should “hybridize” your parallel computations. But as in many other cases, experience comes from playing around and testing, and with COMSOL Multiphysics, you have the possibility to do that. Try it yourself with different configurations and different models, and you will soon know how to set the software up in order to get the best performance out of your hardware.
Fujitsu is a registered trademark of Fujitsu Limited in the United States and other countries. CELSIUS is a registered trademark of Fujitsu Technology Solutions in the United States and other countries. Intel and Xeon are trademarks of Intel Corporation in the U.S. and/or other countries.
#### Post Tags
Clusters Hybrid Modeling series
Calculate the Force of a One-Sided Magnet | 2,265 | 10,231 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2017-43 | longest | en | 0.898221 |
https://www.teachoo.com/13581/754/Ex-11.3--1-d/category/Ex-11.3/ | 1,685,718,390,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224648695.4/warc/CC-MAIN-20230602140602-20230602170602-00197.warc.gz | 1,113,884,969 | 34,634 | Plane
Chapter 11 Class 12 Three Dimensional Geometry
Serial order wise
Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
Transcript
Question 1 In each of the following cases, determine the direction cosines of the normal to the plane and the distance from the origin. (d) 5y + 8 = 0 For plane ax + by + cz = d Direction ratios of normal = a, b, c Direction cosines : l = π/β(π^2 + π^2 +γ πγ^2 ) , m = π/β(π^2 + π^2 + π^2 ) , n = π/β(π^2 + π^2 + π^2 ) Distance from origin = π/β(π^2 + π^2 + π^2 ) Given, equation of the plane is 5y + 8 = 0 5y = β8 β5y = 8 0x β 5y + 0z = 8 0x β 5y + 0z = 8 Comparing with ax + by + cz = d a = 0, b = β5, c = 0 & d = 8 & β(π^2+π^2+π^2 ) = β(0^2 + γ(β5)γ^2 + 0^2 ) = β25 = 5 Direction cosines of the normal to the plane are l = π/β(π^2 + π^2 + π^2 ) , m = π/β(π^2 + π^2 + π^2 ) , n = π/β(π^2 + π^2 + π^2 ) l = 0/5, m = (β5)/5, n = ( 0)/5 β΄ Direction cosines of the normal to the plane are = (0, β1, 0) And, Distance form the origin = π/β(π^2 + π^2 + π^2 ) = π/π | 603 | 1,180 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2023-23 | longest | en | 0.725981 |
https://www.physicsforums.com/threads/fundamental-frequency-of-a-string.289345/ | 1,726,153,321,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651460.54/warc/CC-MAIN-20240912142729-20240912172729-00066.warc.gz | 880,307,754 | 18,593 | # Fundamental Frequency of a string
• nn3568
In summary, the string has a length of 1440 cm and is held fixed at each end. It vibrates in eight sections, with eight antinodes and a fundamental frequency of 150 Hz. Using the formula f = nv/2L, with n representing the number of antinodes, we can calculate the velocity of the string to be 540 m/s. However, when trying to find the first harmonic or fundamental frequency, there was confusion as to whether to use the velocity or not. It was eventually determined that the fundamental frequency is 1/8 of the total frequency and is used with 2*L in the formula to find the velocity.
nn3568
## Homework Statement
The length of a string is 1440 cm. The
string is held fixed at each end. The string
vibrates in eight sections; i.e., the string has
eight antinodes, and the string vibrates at
150 Hz.
What is the fundamental frequency? Answer
in units of Hz.
## Homework Equations
f = nv / 2L frequency = (# antinodes)(velocity) / (2 * length)
## The Attempt at a Solution
I already found the wavelength which is 3.6 m. I got that by substituting nv/2L for f into the equation v=λf (velocity = wavelength*frequency). Then I followed the above frequency formula:
v=λf
v=(3.6 m)(150 Hz) = 540 m/s
f = nv / 2L
f = (8)(540 m/s) / (2*14.4 m) = 150 Hz
My online HW said it was wrong. So I tried again.
f = (2)(540 m/s) / (2*14.4 m) = 37.5 Hz
I used two because it said that fundamental frequency is the lowest frequency and you can use the first harmonic or something. Didn't really understand it. But the two is because the 1st harmonic has 2 antinodes.
I am so confused and I have no clue what I am doing! Please help <:S
Last edited:
If there are 8 antinodes then there are 4 wavelengths represented on the string.
But I think 8 antinodes means it's the 8'th harmonic of the string.
So isn't 1/8 of 150 hz then the first harmonic and isn't that the frequency you use with 2*L to determine v?
Or do you not need to find the v?
Only the f1 fundamental frequency?
i already found v. i just don't understand the fundamental frequency part.
nn3568 said:
i already found v. i just don't understand the fundamental frequency part.
8 antinodes means 8th harmonic = 8 times the fundamental frequency (first harmonic).
http://www.glenbrook.k12.il.us/GBSSCI/PHYS/Class/sound/u11l4d.html
Last edited by a moderator:
## What is the fundamental frequency of a string?
The fundamental frequency of a string is the lowest frequency at which the string can vibrate. It is also known as the first harmonic and is determined by the length, tension, and mass of the string.
## How is the fundamental frequency of a string calculated?
The fundamental frequency of a string can be calculated using the equation f = 1/2L√(T/μ), where f is the frequency, L is the length of the string, T is the tension, and μ is the linear mass density of the string.
## What factors affect the fundamental frequency of a string?
The fundamental frequency of a string is affected by the length, tension, and mass of the string. It also depends on the material, thickness, and stiffness of the string.
## What is the relationship between the fundamental frequency and the harmonics of a string?
The fundamental frequency is the first harmonic of a string, and the higher harmonics are integer multiples of the fundamental frequency. This means that the second harmonic is twice the frequency of the fundamental, the third harmonic is three times the frequency, and so on.
## How does the fundamental frequency of a string relate to musical pitch?
The fundamental frequency of a string determines the pitch of the sound produced by the string. As the fundamental frequency increases, the pitch also increases. This is why longer strings produce lower pitches and shorter strings produce higher pitches.
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1K | 1,026 | 4,027 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2024-38 | latest | en | 0.937607 |
http://math.tutorcircle.com/geometry/30-60-90-triangles.html | 1,544,718,716,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376824912.16/warc/CC-MAIN-20181213145807-20181213171307-00463.warc.gz | 188,173,254 | 5,328 | Sales Toll Free No: 1-855-666-7446
# 30 60 90 Triangles
Top30 60 90 Triangles are triangles with angles of measure 30, 60 and 90 degrees. In these types of triangles sides are in the Ratio of the 1 : 2 : √3. If we divide an Equilateral Triangle into two equal parts then we get these types of right angled triangle.
For right triangles we use Trigonometric Functions for calculation of sides and angles but for this special type length ratio is used for calculations. Special feature of this triangle is that length of the sides of triangle follow a special trend. Suppose short leg which is opposite to the angle 30 degree has length 'y' and hypotenuse has length of 2y. So long leg is opposite to angle 60 degree has length √3y.
With help of these ratios we can calculate length of sides of this Right Triangle.
Suppose in a 30 60 90 degree triangle length of shortest leg is 20 then length of other sides can be calculated as shown below.
We know that hypotenuse of such triangle is twice the shortest leg and longest leg is √3 times of shortest leg. So with help of this property we can solve this problem.
Here hypotenuse will be 2 * 20 = 40.
And the longest leg is = 20 √3.
In a 30 60 90 degree Right Angle triangle we have the length of the shortest leg which is 8. Find the length of other sides?
We know that hypotenuse of such triangle is twice the shortest leg and longest leg is √3 times of shortest leg.
Here hypotenuse will be 2 * 8= 16.
And longest leg is = 8√3. | 373 | 1,482 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2018-51 | latest | en | 0.900066 |
https://www.excelguru.ca/forums/showthread.php?1269-Formula-required-to-calculate-priority | 1,627,502,126,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153791.41/warc/CC-MAIN-20210728185528-20210728215528-00442.warc.gz | 785,596,028 | 12,299 | # Thread: Formula required to calculate priority
1. ## Formula required to calculate priority
Register for a FREE account, and/
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Can someone please calculate priority based on the conditions specified in the attachment?
2. Maybe?
=INT(H2/10)
copied down?
3. This one should work =IF(H2>=40,\$J\$18,IF(AND(H2<40,H2>=30),\$J\$19,IF(AND(H2<30,H2>=20),\$J\$20,IF(AND(H2<20,H2>=10),\$J\$21,0)))) . You can also replace the absolute values (40,30,20) with cells to make it more flexible if your criteria should change.
4. Hi Nesfin
The outcome of that is the same as NBVC suggestion!
5. I didn't check it as it had a question mark but yes it indeed does work in this instance and it's much shorter
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• | 244 | 878 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2021-31 | latest | en | 0.87361 |
https://brainmass.com/physics/torques/rotational-motion-angular-momentum-torque-576506 | 1,708,882,789,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474641.34/warc/CC-MAIN-20240225171204-20240225201204-00883.warc.gz | 152,280,879 | 7,621 | Purchase Solution
# Rotational motion: Angular momentum and Torque
Not what you're looking for?
Explain the concepts of angular momentum and torque with reference to the rotational motion of a rigid body. Derive the relevant expressions and illustrate with solved examples.
##### Solution Summary
This solution contains concepts of angular momentum and torque with reference to rotational motion of a rigid body have been explained in details. Relevant expressions have been derived and solved examples have been included. Multiple diagrams are present to illustrate concepts.
##### Solution Preview
See the attached file.
Angular momentum: Angular moment to rotational motion is as linear momentum is to translational motion. Hence, before we consider the concept of angular momentum with reference to rotational motion, let's review the concept of linear momentum with reference to translational motion.
For a body of mass m moving with a velocity v at a given instant, we define its linear momentum at that instant as the product of mass and velocity i.e. mv.
For a body undergoing rotational motion about an axis with a rotational (or angular) speed ω (defined as the angle swept by a straight line drawn from the axis of rotation to any point on the object expressed in radians/sec), we can define its angular momentum (L) as the product of its moment of inertia (I) about the given axis of rotation and the angular velocity ω. (For an understanding of the concept of moment of inertia the student may refer to Solutions Library posting no. 563269).
L = Iω ...(1)
Using this definition we derive a general expression for angular momentum of a body of arbitrary shape, this figure can be seen in the attachment.
Let a body of any arbitrary shape be rotating about an axis passing through point O as shown in the fig..
Let us consider the given body is notionally replaced by a point mass m placed at point P, such that the moment of inertia of the point mass placed at P about the given axis is same as the moment of inertia of the body about the axis. Position vector of point P is shown as vector r.
Let us assume, at an instant the velocity vector v of the mass m be as shown in the fig.. Let velocity vector v be resolved into components vcosθ and vsinθ. Component vcosθ must be zero as the body can't move in that direction, the body having only rotational motion about the given axis. Hence, the imaginary mass m can only have the translational velocity vsinθ.
Angular speed ω = vsinθ/r
Further, moment of inertia of the imaginary mass m about the given axis of rotation is given by:
I = mr2
Substituting for I and ω in (1) we get: L = Iω = mr2 x vsinθ/r = mrvsinθ
As mv represents the linear momentum p of the mass, L = rpsinθ
Like linear momentum, the angular momentum is also a vector quantity. Above equation can be rewritten in vector form as: L = r X p (cross or vector product)
where θ is the angle between the position vector r of point P and the velocit vector v.
Magnitude of the angular momentum is equal to rpsinθ. In the fig., the velocity vector v has been extended backwards and a perpendicular OQ has been drawn on this line. Length of perpendicular OQ is rsinθ. Hence,
Magnitude of angular momentum = rpsinθ = p(rsinθ) = Magnitude of the linear momentum x Perpendicular distance of the axis of rotation from the velocity vector
Direction of the angular velocity vector is given by the right hand screw rule of vector multiplication which can be seen in figure in attachment.
If vector C = A X B then i) vector C is perpendicular to the plane in which vectors A and B lie and ii) points in the direction in which a right handed screw placed at the point where the tails of the two vectors meet (point O) advances when turned from the first vector (A) towards the second vector (B). For the ...
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Explanation of the Kinetic Theory of Matter by Ron Kurtus - Succeed in Understanding Physics. Key words: physical science, molecular, collision, heat flow, transfer, pressure, temperature, volume, ideal gas, liquid, solid, assumptions, elastic, kinetic energy, temperature, School for Champions. Copyright © Restrictions
# Kinetic Theory of Matter
by Ron Kurtus (revised 26 November 2011)
The Kinetic Theory of Matter states that matter is composed of a large number of small particles—individual atoms or molecules—that are in constant motion. This theory is also called the Kinetic Molecular Theory of Matter and the Kinetic Theory.
By making some simple assumptions, such as the idea that matter is made of widely spaced particles in constant motion, the theory helps to explain the behavior of matter. Two important areas explained are the flow or transfer of heat and the relationship between pressure, temperature, and volume properties of gases.
Questions you may have include:
• What are the assumptions of the kinetic theory of matter?
• How does this explain heat flow?
• How does the theory explain pressure and volume?
This lesson will answer those questions.
Useful tool: Metric-English Conversion
## Assumptions of theory
The Kinetic Theory of Matter is a prediction of how matter should behave, based on certain assumptions and approximations. The assumptions are made from observations and experiments, such as the fact that materials consist of small molecules or atoms. Approximations are made to keep the theory from being too complex. One assumption is that the size of the particles is so small that it can be considered a point.
### Matter consists of small particles
The first assumption in this theory is that matter consists of a large number a very small particles—either individual atoms or molecules.
### Large separation between particles
The next assumption concerns the separation of the particles.
• In a gas, the separation between particles is very large compared to their size, such that there are no attractive or repulsive forces between the molecules.
• In a liquid, the particles are still far apart, but now they are close enough that attractive forces confine the material to the shape of its container.
• In a solid, the particles are so close that the forces of attraction confine the material to a specific shape.
### Particles in constant motion
Another assumption is that each particle is in constant motion.
In gases, the movement of the particles is assumed to be random and free. In liquids, the movement is somewhat constrained by the volume of the liquid. In solids, the motion of the particles is severely constrained to a small area, in order for the solid to maintain its shape.
The velocity of each particle determines its kinetic energy.
### Collisions transfer energy
The numerous particles often collide with each other. Also, if a gas or liquid is confined in a container, the particles collide with the particles that make up the walls of a container.
#### Approximations
When atoms or molecules collide, energy may be given off in the form of electromagnetic radiation. Taking this into account could make the theory highly complex, and since the amount of radiation is small in most situations, an approximation is made that this effect is negligible.
Also, atoms and molecules have a discrete size. But charting the collisions of such particles would again make the theory too complex. Thus an approximation is made to say the size of the particles is a simple point, especially compared to the distances involved.
#### No energy change
Thus, an assumption is that the particles transfer energy in a collision with no net energy change. That means the collisions between the particles are perfectly elastic and no energy is gained or lost during the collision. This follows the Law of the Conservation of Energy.
In reality, the collisions are not perfect, and some energy is lost. But for the sake of simplicity in drawing conclusions, this theory makes the collisions elastic.
## Thermal energy and heat flow
The motion of a particle determines its kinetic energy, according to the equation
KE = ½mv²
where
• KE is the kinetic energy of the particle
• m is its mass
• is the square of its velocity
The total internal kinetic energy of all the particles is called its thermal energy.
The temperature of an object or collection of matter is the average kinetic energy of the particles. Faster particles means a higher temperature. A thermometer is used to measure the temperature and put it into temperature degrees instead of kinetic energy units.
The heat is the transfer of thermal energy from an object of higher temperature to one of lower temperature. For example, an object feels warm or hot if its temperature is higher than your skin temperature.
The Kinetic Theory of Matter explains heat transfer by conduction, where thermal energy seems to move through a material, warming up cooler areas. This is called heat transfer or heat flow
Processes not covered in this theory are heat transfer by convection and by radiation.
### Collisions transfer energy
The Kinetic Theory of Matter states that the material's particles have greater kinetic energy and are moving faster at higher temperatures. When a fast moving particle collides with a slower moving particle, it transfers some of its energy to the slower moving particle, increasing the speed of that particle.
If that particle then collides with another particle that is moving faster, its speed will be increased even more. But if it hits a slow moving particle, then it will speed up the third particle.
With billions of moving particles colliding into each other, an area of high energy or high heat will slowly diffuse across the material, making other areas warm too. By the Conservation of Energy, the total energy or total heat of the object will remain the same, but the heat will be evenly distributed throughout the object.
### Rate of transfer
The rate at which the kinetic or thermal energy is transferred from one particle to another depends on the separation of the particles and their freedom to move.
In a gas, the particles are allowed to move freely, but their separation distance is great, so heat or energy transfer is slow. In a liquid, the heat transfer by conduction is faster because the particles are closer together.
In a solid, the molecules are constrained into a specific location within the material. Although the particles are closer together than in liquids, the constraints in some materials actually prevent the transfer of heat energy. A good example of that is in wood.
### Temperature
One important result of the kinetic theory is that the average molecular kinetic energy is proportional to the absolute temperature of the material. Absolute temperature is measured in the Kelvin scale. But in general, you can say that temperature is the measurement of the average internal kinetic energy of the material or object.
## Pressure, volume and temperature
If a gas is enclosed in a container, it exerts pressure on the walls of the container. The Kinetic Theory of Matter explains gas pressure as the total force exerted by gas molecules colliding against the walls of a container.
If the container can expand, like with a balloon or cylinder and piston, increasing the pressure can increase the volume. Like, the balloon will get bigger. Also, if you increase the temperature of the gas--and thus the kinetic energy of its molecules—you increase the pressure or the volume of the container.
This leads to a relationship between pressure, volume and temperature in an ideal gas. (An ideal gas is a gas that follows the assumptions of the Kinetic Theory of Matter.) The relationship is
PV = NkT
where:
• P is the pressure of the ideal gas
• V is the volume of the gas container
• N is the number of gas particles
• k is the Boltzman constant in joule per kelvin per particle
• T is the temperature in the absolute or kelvin scale
This equation has a number of implications, including:
• If you decrease the pressure and hold the volume constant, the temperature decreases (principle of a refrigerator)
• If you increase the temperature and hold the pressure constant, the volume increases (heating a balloon)
## Summary
The Kinetic Theory of Matter states that matter is composed of a large number a small particles that are in constant motion. It also assumes that particles are small and widely separated. They collide and exchange energy. The theory helps explain the flow or transfer of heat and the relationship between pressure, temperature and volume properties of gases.
Try something new
## Resources and references
Ron Kurtus' Credentials
### Websites
Kinetic Theory - HyperPhysics
Physics Resources
### Books
Introduction to Thermodynamics and Kinetic Theory of Matter by Anatoly I. Burshtein; Wiley-Interscience (1995)
Do you have any questions, comments, or opinions on this subject? If so, send an email with your feedback. I will try to get back to you as soon as possible.
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Concluding relativity Particle accelerators Blackbody Cavity modes Density of states Thermodynamic occupancy Blackbody redeaux Blackbody measurements Max Planck Planck’s fix Relativistic momentum We found E 2 = E 2 0 + ( pc ) 2 We found ( E / c ) 2 p 2 x p 2 y p 2 z = 0 is frame invariant This led to Lorentz transforms for momentum and energy: p x , 2 = γ p p x , 1 v ( E 1 c 2 ) P and p y , 2 = p y , 1 and p z , 2 = p z , 1 (1) E 2 = γ ( E 1 vp x , 1 ) . Energy and momentum are intertwined when comparing relativistic reference frames!
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Concluding relativity Particle accelerators Blackbody Cavity modes Density of states Thermodynamic occupancy Blackbody redeaux Blackbody measurements Max Planck Planck’s fix Relativisic collision I A particle A with rest mass m 0 and velocity v A = 0 . 80 c in the ˆ x direction collides with an initially-stationary particle B with rest mass 2 m 0 . Note that for β = 4 / 5 we can find γ = 5 3 . In the frame S 1 , we then use p 1 , y = γ m 0 v 2 , y and E = γ m 0 c 2 to find p x , A , 1 = γ m 0 v x , A = 5 3 m 0 4 5 c = 4 3 m 0 c . p y , A , 1 = p z , A , 1 = 0 E A , 1 = γ m 0 c 2 = 5 3 m 0 c 2 for particle A , and p x , B , 1 = γ m 0 v x , B = 0 p y , B , 1 = p z , B , 1 = 0 E B , 1 = γ m 0 c 2 = 1 ( 2 m 0 ) c 2 for particle B .
Concluding relativity Particle accelerators Blackbody Cavity modes Density of states Thermodynamic occupancy Blackbody redeaux Blackbody measurements Max Planck Planck’s fix Relativistic collision II The total energy in frame S 1 is then E 1 = E A , 1 + E B , 1 = p 5 3 + 2 P m 0 c 2 = 11 3 m 0 c 2 , or E 1 = 3 . 67 m 0 c 2 . Now let’s find the center-of-momentum frame. Using p x , 2 = γ ( p x , 1 v ( E / c 2 ) ) from Eqs. 2, we find p x , A , 2 + p x , B , 2 = 0 = γ b ( p x , A , 1 v ( E A , 1 / c 2 ) ) + ( p x , B , 1 v ( E A , 1 / c 2 ) ) B = ( 4 3 m 0 c 5 3 m 0 v ) + ( 0 2 m 0 v ) r 1 v 2 / c 2 Thus we want ( 4 3 c 5 3 v 2 v ) m 0 = 0 which gives v = ( 4 / 11 ) c .
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relativity Particle accelerators Blackbody Cavity modes Density of states Thermodynamic occupancy Blackbody redeaux Blackbody measurements Max Planck Planck’s fix Relativistic collision III We found that v = ( 4 / 11 ) c lets us move from frame S 1 where particle B was at rest, to frame S 2 where the two particles come in with equal and opposite momentum. The total energies of the individual particles in S 2 can be found using E 2 = γ ( E 1 vp x , 1 ) to be E A , 2 = E A , 1 vp x , A , 1 r 1 v 2 / c 2 = 5 3 m 0 c 2 ( 4 11 c )( 4 3 m 0 c ) R 1 ( 4 11 ) 2 = 1 . 27 m 0 c 2 E B , 2 = E B , 1 vp x , B , 1 r 1 v 2 / c 2 = 2 m 0 c 2 0 R 1 ( 4 11 ) 2 = 2 . 15 m 0 c 2 , or E 2 = 3 . 24 m 0 c 2 . As promised, energy is not conserved!
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# Third Dimension
## The Five Dimensions:
The Third Dimension is the SOLID, such as a cube. It contains an infinite number of planes or squares. Temporally, the Third Dimension represent the Past. The disc of the Second Dimension (Present) turns one half time around its axis and fills out the sphere of the past.
The third dimension brings out the Real numbers. Real numbers start from zero and connect fractions of the same numerical value, leading to the proportions and functions.
The proportions are the basis of continuity and harmony. They connect fractions of the same value to zero. The functions are the basis of discontinuity. They connect products by which bodies are in relation, as for instance in the atom, where the distances of the electron shells follow the numbers of the central diagonal 1 - 4 - 9 - 16, and the possible number of electrons in each shell, the capacity, follow the diagonal 2 - 8 - 18 - 32. The rational numbers of the second dimension, and the whole and natural numbers of the first and zero dimension, all have a fixed place on the number line. The real numbers in the third dimension are, however, fundamentally different; although they are located somewhere on the number line, they have no fixed place there. To the ancient Greeks who first developed mathematics to a high art in the West, all numbers had to have a fixed location somewhere on the number line. The existence of the Real numbers, with no fixed location, was known only to a few high initiates in the Pythagorean brotherhood who swore to keep it secret. It can be easily understood today by way of the Pythagorean theorem:
The Pythagorean Theorem exemplifies the rational numbers. But what happens if A and B both equal 1? In this case C must equal the square root of 2. But the square root of two is an irrational Real Number. It is a number which goes on and on with no repetition into infinity. 1.41421... . It is a never ending number and has no fixed place on the number line. Unlike an infinite rational number which goes on and on, but repeats, such as a third (.3333333...), where we can know the exact location on the number line, with a Real Number, we can only know its approximate location. There are other examples of Real Numbers, such as Pi (the ratio of a circumference of a circle to its diameter), the square root of any prime number, e, etc. These Real Numbers never end and never repeat. | 524 | 2,422 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2017-04 | latest | en | 0.93345 |
https://www.geeksforgeeks.org/longest-arithmetic-progression-dp-35/ | 1,709,026,132,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474674.35/warc/CC-MAIN-20240227085429-20240227115429-00519.warc.gz | 761,295,303 | 68,039 | # Longest Arithmetic Progression
Given an array Set[ ] of sorted integers having no duplicates, find the length of the Longest Arithmetic Progression (LLAP) subsequence in it.
Examples:
Input: Set[] = {1, 7, 10, 15, 27, 29}
Output: 3
Explanation: The longest arithmetic progression is {1, 15, 29} having common difference 14.
Input: Set[] = {5, 10, 15, 20, 25, 30}
Output: 6
Explanation: The whole set is in AP having common difference 5.
Approach 1: Brute forcing the solution in O(N^3)
A simple solution is to one by one consider every pair as first two elements of AP and check for the remaining elements in sorted array. To consider all pairs as first two elements, we need to run a O(n^2) nested loop. Inside the nested loops, we need a third loop which linearly looks for the more elements in Arithmetic Progression (AP). This process takes O(n3) time.
We can solve this problem in O(n2) time using Dynamic Programming. To get idea of the DP solution, let us first discuss solution of following simpler problem.
Given a sorted Array, find if there exist three elements in Arithmetic Progression or not:
Please note that, the answer is true if there are 3 or more elements in AP, otherwise false. To find the three elements, we first fix an element as middle element and search for other two (one smaller and one greater). We start from the second element and fix every element as middle element.
For an element set[j] to be middle of AP, there must exist elements ‘set[i]’ and ‘set[k]’ such that set[i] + set[k] = 2*set[j] where 0 <= i < j and j < k <=n-1.
Algorithm to efficiently find i and k for a given j:
We can find i and k in linear time using following simple algorithm.
• Initialize i as j-1 and k as j+1
• Do following while i >= 0 and k <= n-1
• If set[i] + set[k] is equal to 2*set[j], then we are done.
• If set[i] + set[k] > 2*set[j], then decrement i (do i–).
• Else if set[i] + set[k] < 2*set[j], then increment k (do k++).
Following is the implementation of the above algorithm for the simpler problem.
## C++
`// The function returns true if there exist three ``// elements in AP Assumption: set[0..n-1] is sorted. ``// The code strictly implements the algorithm provided ``// in the reference.``bool` `arithmeticThree(vector<``int``> set, ``int` `n)``{`` ` ` ``// One by fix every element as middle element`` ``for``(``int` `j = 1; j < n - 1; j++)`` ``{`` ` ` ``// Initialize i and k for the current j`` ``int` `i = j - 1, k = j + 1;` ` ``// Find if there exist i and k that form AP`` ``// with j as middle element`` ``while` `(i >= 0 && k <= n-1)`` ``{`` ``if` `(set[i] + set[k] == 2 * set[j])`` ``return` `true``;`` ` ` ``(set[i] + set[k] < 2 * set[j]) ? k++ : i--;`` ``}`` ``}`` ``return` `false``;``}` `// This code is contributed by Samim Hossain Mondal.`
## C
`// The function returns true if there exist three elements in AP``// Assumption: set[0..n-1] is sorted. ``// The code strictly implements the algorithm provided in the reference.``bool` `arithmeticThree(``int` `set[], ``int` `n)``{`` ``// One by fix every element as middle element`` ``for` `(``int` `j=1; j= 0 && k <= n-1)`` ``{`` ``if` `(set[i] + set[k] == 2*set[j])`` ``return` `true``;`` ``(set[i] + set[k] < 2*set[j])? k++ : i--;`` ``}`` ``}` ` ``return` `false``;``}`
## Java
`// The function returns true if there exist three elements in AP``// Assumption: set[0..n-1] is sorted. ``// The code strictly implements the algorithm provided in the reference.``static` `boolean` `arithmeticThree(``int` `set[], ``int` `n)``{`` ``// One by fix every element as middle element`` ``for` `(``int` `j = ``1``; j < n - ``1``; j++)`` ``{`` ``// Initialize i and k for the current j`` ``int` `i = j - ``1``, k = j + ``1``;` ` ``// Find if there exist i and k that form AP`` ``// with j as middle element`` ``while` `(i >= ``0` `&& k <= n-``1``)`` ``{`` ``if` `(set[i] + set[k] == ``2``*set[j])`` ``return` `true``;`` ``(set[i] + set[k] < ``2``*set[j])? k++ : i--;`` ``}`` ``}` ` ``return` `false``;``}` `// This code is contributed by gauravrajput1 `
## Python3
`# The function returns true if there exist three elements in AP``# Assumption: set[0..n-1] is sorted. ``# The code strictly implements the algorithm provided in the reference.``def` `arithematicThree(set_,n):` ` ``# One by fix every element as middle element`` ``for` `j ``in` `range``(n):`` ` ` ``# Initialize i and k for the current j`` ``i,k``=``j``-``1``,j``+``1` ` ``# Find if there exist i and k that form AP`` ``# with j as middle element`` ``while` `i>``-``1` `and` `k
## C#
`// The function returns true if there exist three elements in AP``// Assumption: set[0..n-1] is sorted. ``// The code strictly implements the algorithm provided in the reference.``static` `bool` `arithmeticThree(``int` `[]``set``, ``int` `n)``{`` ` ` ``// One by fix every element as middle element`` ``for` `(``int` `j = 1; j < n - 1; j++)`` ``{`` ` ` ``// Initialize i and k for the current j`` ``int` `i = j - 1, k = j + 1;` ` ``// Find if there exist i and k that form AP`` ``// with j as middle element`` ``while` `(i >= 0 && k <= n-1)`` ``{`` ``if` `(``set``[i] + ``set``[k] == 2*``set``[j])`` ``return` `true``;`` ``if``(``set``[i] + ``set``[k] < 2*``set``[j]) `` ``k++;`` ``else`` ``i--;`` ``}`` ``}` ` ``return` `false``;``}` `// This code is contributed by gauravrajput1`
## Javascript
``
## Longest Arithmetic Progression using Dynamic Programming
Iteratively examining all possible pairs of elements as the first two elements of an AP and then extending the AP to find the longest one. The algorithm uses a dynamic programming approach to store the length of the LLAP ending at each pair of elements. It starts with a minimum LLAP length of 2 (two elements), and as it iterates through the elements, it updates and keeps track of the maximum LLAP length found so far. Finally, it returns the length of the longest arithmetic progression found in the set.
Step-by-step approach:
• Initialize a 2D array L[][] to store the length of LLAPs ending at pairs of elements.
• For each element in the set, fill the last column of L[][] with 2 since any pair forms an AP of length 2.
• Starting from the second-to-last column, iterate through pairs of elements (set[i], set[j]).
• Within the loop, compare set[i] + set[k] to 2 * set[j]:
• If less, increment k to consider a larger element.
• If greater, decrement i to consider a smaller element.
• If equal, extend the LLAP by setting L[i][j] to L[j][k] + 1 and update the maximum LLAP length (llap).
• Handle cases where k exceeds the last index by filling in remaining cells in the column with 2.
• Return the maximum LLAP length (llap) found in the set.
Following is the implementation of the Dynamic Programming algorithm.
## C++
`// C++ program to find Length of the Longest AP (llap) in a given sorted set.``// The code strictly implements the algorithm provided in the reference.``#include ``using` `namespace` `std;` `// Returns length of the longest AP subset in a given set``int` `lenghtOfLongestAP(``int` `set[], ``int` `n)``{`` ``if` `(n <= 2) ``return` `n;` ` ``// Create a table and initialize all values as 2. The value of`` ``// L[i][j] stores LLAP with set[i] and set[j] as first two`` ``// elements of AP. Only valid entries are the entries where j>i`` ``int` `L[n][n];`` ``int` `llap = 2; ``// Initialize the result` ` ``// Fill entries in last column as 2. There will always be`` ``// two elements in AP with last number of set as second`` ``// element in AP`` ``for` `(``int` `i = 0; i < n; i++)`` ``L[i][n-1] = 2;` ` ``// Consider every element as second element of AP`` ``for` `(``int` `j=n-2; j>=1; j--)`` ``{`` ``// Search for i and k for j`` ``int` `i = j-1, k = j+1;`` ``while` `(i >= 0 && k <= n-1)`` ``{`` ``if` `(set[i] + set[k] < 2*set[j])`` ``k++;` ` ``// Before changing i, set L[i][j] as 2`` ``else` `if` `(set[i] + set[k] > 2*set[j])`` ``{ L[i][j] = 2, i--; }` ` ``else`` ``{`` ``// Found i and k for j, LLAP with i and j as first two`` ``// elements is equal to LLAP with j and k as first two`` ``// elements plus 1. L[j][k] must have been filled`` ``// before as we run the loop from right side`` ``L[i][j] = L[j][k] + 1;` ` ``// Update overall LLAP, if needed`` ``llap = max(llap, L[i][j]);` ` ``// Change i and k to fill more L[i][j] values for`` ``// current j`` ``i--; k++;`` ``}`` ``}` ` ``// If the loop was stopped due to k becoming more than`` ``// n-1, set the remaining entities in column j as 2`` ``while` `(i >= 0)`` ``{`` ``L[i][j] = 2;`` ``i--;`` ``}`` ``}`` ``return` `llap;``}` `/* Driver program to test above function*/``int` `main()``{`` ``int` `set1[] = {1, 7, 10, 13, 14, 19};`` ``int` `n1 = ``sizeof``(set1)/``sizeof``(set1[0]);`` ``cout << lenghtOfLongestAP(set1, n1) << endl;` ` ``int` `set2[] = {1, 7, 10, 15, 27, 29};`` ``int` `n2 = ``sizeof``(set2)/``sizeof``(set2[0]);`` ``cout << lenghtOfLongestAP(set2, n2) << endl;` ` ``int` `set3[] = {2, 4, 6, 8, 10};`` ``int` `n3 = ``sizeof``(set3)/``sizeof``(set3[0]);`` ``cout << lenghtOfLongestAP(set3, n3) << endl;` ` ``return` `0;``}`
## Java
`// Java program to find Length of the``// Longest AP (llap) in a given sorted set.``// The code strictly implements the ``// algorithm provided in the reference.``import` `java.io.*;` `class` `GFG ``{`` ``// Returns length of the longest `` ``// AP subset in a given set`` ``static` `int` `lenghtOfLongestAP(``int` `set[], ``int` `n)`` ``{`` ``if` `(n <= ``2``) ``return` `n;`` ` ` ``// Create a table and initialize all `` ``// values as 2. The value ofL[i][j] stores `` ``// LLAP with set[i] and set[j] as first two`` ``// elements of AP. Only valid entries are `` ``// the entries where j>i`` ``int` `L[][] = ``new` `int``[n][n];`` ` ` ``// Initialize the result`` ``int` `llap = ``2``;`` ` ` ``// Fill entries in last column as 2. `` ``// There will always be two elements in `` ``// AP with last number of set as second`` ``// element in AP`` ``for` `(``int` `i = ``0``; i < n; i++)`` ``L[i][n - ``1``] = ``2``;`` ` ` ``// Consider every element as second element of AP`` ``for` `(``int` `j = n - ``2``; j >= ``1``; j--)`` ``{`` ``// Search for i and k for j`` ``int` `i = j -``1` `, k = j + ``1``;`` ``while` `(i >= ``0` `&& k <= n - ``1``)`` ``{`` ``if` `(set[i] + set[k] < ``2` `* set[j])`` ``k++;`` ` ` ``// Before changing i, set L[i][j] as 2`` ``else` `if` `(set[i] + set[k] > ``2` `* set[j])`` ``{ `` ``L[i][j] = ``2``; i--; `` ` ` ``}`` ` ` ``else`` ``{`` ``// Found i and k for j, LLAP with i and j as first two`` ``// elements is equal to LLAP with j and k as first two`` ``// elements plus 1. L[j][k] must have been filled`` ``// before as we run the loop from right side`` ``L[i][j] = L[j][k] + ``1``;`` ` ` ``// Update overall LLAP, if needed`` ``llap = Math.max(llap, L[i][j]);`` ` ` ``// Change i and k to fill `` ``// more L[i][j] values for current j`` ``i--; k++;`` ``}`` ``}`` ` ` ``// If the loop was stopped due`` ``// to k becoming more than`` ``// n-1, set the remaining `` ``// entities in column j as 2`` ``while` `(i >= ``0``)`` ``{`` ``L[i][j] = ``2``;`` ``i--;`` ``}`` ``}`` ``return` `llap;`` ``}`` ` ` ``// Driver program `` ``public` `static` `void` `main (String[] args) `` ``{`` ``int` `set1[] = {``1``, ``7``, ``10``, ``13``, ``14``, ``19``};`` ``int` `n1 = set1.length;`` ``System.out.println ( lenghtOfLongestAP(set1, n1));`` ` ` ``int` `set2[] = {``1``, ``7``, ``10``, ``15``, ``27``, ``29``};`` ``int` `n2 = set2.length;`` ``System.out.println(lenghtOfLongestAP(set2, n2));`` ` ` ``int` `set3[] = {``2``, ``4``, ``6``, ``8``, ``10``};`` ``int` `n3 = set3.length;`` ``System.out.println(lenghtOfLongestAP(set3, n3)) ;`` ` ` ` ` ``}``}` `// This code is contributed by vt_m`
## Python3
`# Python 3 program to find Length of the``# Longest AP (llap) in a given sorted set.``# The code strictly implements the algorithm``# provided in the reference` `# Returns length of the longest AP``# subset in a given set``def` `lenghtOfLongestAP(``set``, n):` ` ``if` `(n <``=` `2``):`` ``return` `n` ` ``# Create a table and initialize all `` ``# values as 2. The value of L[i][j]`` ``# stores LLAP with set[i] and set[j] `` ``# as first two elements of AP. Only `` ``# valid entries are the entries where j>i`` ``L ``=` `[[``0` `for` `x ``in` `range``(n)]`` ``for` `y ``in` `range``(n)]`` ``llap ``=` `2` `# Initialize the result` ` ``# Fill entries in last column as 2. `` ``# There will always be two elements `` ``# in AP with last number of set as `` ``# second element in AP`` ``for` `i ``in` `range``(n):`` ``L[i][n ``-` `1``] ``=` `2` ` ``# Consider every element as second `` ``# element of AP`` ``for` `j ``in` `range``(n ``-` `2``, ``0``, ``-``1``):` ` ``# Search for i and k for j`` ``i ``=` `j ``-` `1`` ``k ``=` `j ``+` `1`` ``while``(i >``=` `0` `and` `k <``=` `n ``-` `1``):` ` ``if` `(``set``[i] ``+` `set``[k] < ``2` `*` `set``[j]):`` ``k ``+``=` `1` ` ``# Before changing i, set L[i][j] as 2`` ``elif` `(``set``[i] ``+` `set``[k] > ``2` `*` `set``[j]):`` ``L[i][j] ``=` `2`` ``i ``-``=` `1` ` ``else``:` ` ``# Found i and k for j, LLAP with i and j `` ``# as first two elements are equal to LLAP `` ``# with j and k as first two elements plus 1. `` ``# L[j][k] must have been filled before as `` ``# we run the loop from right side`` ``L[i][j] ``=` `L[j][k] ``+` `1` ` ``# Update overall LLAP, if needed`` ``llap ``=` `max``(llap, L[i][j])` ` ``# Change i and k to fill more L[i][j] `` ``# values for current j`` ``i ``-``=` `1`` ``k ``+``=` `1` ` ``# If the loop was stopped due to k `` ``# becoming more than n-1, set the`` ``# remaining entities in column j as 2`` ``while` `(i >``=` `0``):`` ``L[i][j] ``=` `2`` ``i ``-``=` `1`` ``return` `llap` `# Driver Code``if` `__name__ ``=``=` `"__main__"``:`` ` ` ``set1 ``=` `[``1``, ``7``, ``10``, ``13``, ``14``, ``19``]`` ``n1 ``=` `len``(set1)`` ``print``(lenghtOfLongestAP(set1, n1))` ` ``set2 ``=` `[``1``, ``7``, ``10``, ``15``, ``27``, ``29``]`` ``n2 ``=` `len``(set2)`` ``print``(lenghtOfLongestAP(set2, n2))` ` ``set3 ``=` `[``2``, ``4``, ``6``, ``8``, ``10``]`` ``n3 ``=` `len``(set3)`` ``print``(lenghtOfLongestAP(set3, n3))` `# This code is contributed by ita_c`
## C#
`// C# program to find Length of the ``// Longest AP (llap) in a given sorted set. ``// The code strictly implements the ``// algorithm provided in the reference.``using` `System;` `class` `GFG``{``// Returns length of the longest ``// AP subset in a given set ``static` `int` `lenghtOfLongestAP(``int` `[]``set``, `` ``int` `n) ``{ `` ``if` `(n <= 2) ``return` `n; ` ` ``// Create a table and initialize `` ``// all values as 2. The value of `` ``// L[i][j] stores LLAP with set[i] `` ``// and set[j] as first two elements `` ``// of AP. Only valid entries are `` ``// the entries where j>i `` ``int` `[,]L = ``new` `int``[n, n]; `` ` ` ``// Initialize the result `` ``int` `llap = 2; ` ` ``// Fill entries in last column as 2. `` ``// There will always be two elements`` ``// in AP with last number of set as `` ``// second element in AP `` ``for` `(``int` `i = 0; i < n; i++) `` ``L[i, n - 1] = 2; ` ` ``// Consider every element as `` ``// second element of AP `` ``for` `(``int` `j = n - 2; j >= 1; j--) `` ``{ `` ``// Search for i and k for j `` ``int` `i = j - 1 , k = j + 1; `` ``while` `(i >= 0 && k <= n - 1) `` ``{ `` ``if` `(``set``[i] + ``set``[k] < 2 * ``set``[j]) `` ``k++; ` ` ``// Before changing i, set L[i][j] as 2 `` ``else` `if` `(``set``[i] + ``set``[k] > 2 * ``set``[j]) `` ``{ `` ``L[i, j] = 2; i--; `` ` ` ``} ` ` ``else`` ``{ `` ``// Found i and k for j, LLAP with `` ``// i and j as first two elements `` ``// is equal to LLAP with j and k `` ``// as first two elements plus 1. `` ``// L[j][k] must have been filled `` ``// before as we run the loop from `` ``// right side `` ``L[i, j] = L[j, k] + 1; ` ` ``// Update overall LLAP, if needed `` ``llap = Math.Max(llap, L[i, j]); ` ` ``// Change i and k to fill `` ``// more L[i][j] values for current j `` ``i--; k++; `` ``} `` ``} ` ` ``// If the loop was stopped due `` ``// to k becoming more than `` ``// n-1, set the remaining `` ``// entities in column j as 2 `` ``while` `(i >= 0) `` ``{ `` ``L[i, j] = 2; `` ``i--; `` ``} `` ``} `` ``return` `llap; ``} ` `// Driver Code ``static` `public` `void` `Main ()``{`` ``int` `[]set1 = {1, 7, 10, 13, 14, 19}; `` ``int` `n1 = set1.Length; `` ``Console.WriteLine(lenghtOfLongestAP(set1, n1)); ` ` ``int` `[]set2 = {1, 7, 10, 15, 27, 29}; `` ``int` `n2 = set2.Length; `` ``Console.WriteLine(lenghtOfLongestAP(set2, n2)); ` ` ``int` `[]set3 = {2, 4, 6, 8, 10}; `` ``int` `n3 = set3.Length; `` ``Console.WriteLine(lenghtOfLongestAP(set3, n3)) ;``}``}` `// This code is contributed by Sach_Code`
## Javascript
``
## PHP
`i `` ``\$L``[``\$n``][``\$n``] = ``array``(``array``()); `` ``\$llap` `= 2; ``// Initialize the result ` ` ``// Fill entries in last column as 2.`` ``// There will always be two elements `` ``// in AP with last number of set as `` ``// second element in AP `` ``for` `(``\$i` `= 0; ``\$i` `< ``\$n``; ``\$i``++) `` ``\$L``[``\$i``][``\$n` `- 1] = 2; ` ` ``// Consider every element as `` ``// second element of AP `` ``for` `(``\$j` `= ``\$n` `- 2; ``\$j` `>= 1; ``\$j``--) `` ``{ `` ``// Search for i and k for j `` ``\$i` `= ``\$j` `- 1;`` ``\$k` `= ``\$j` `+ 1; `` ``while` `(``\$i` `>= 0 && ``\$k` `<= ``\$n` `- 1) `` ``{ `` ``if` `(``\$set``[``\$i``] + ``\$set``[``\$k``] < 2 * ``\$set``[``\$j``]) `` ``\$k``++; ` ` ``// Before changing i, set L[i][j] as 2 `` ``else` `if` `(``\$set``[``\$i``] + ``\$set``[``\$k``] > 2 * ``\$set``[``\$j``]) `` ``{ `` ``\$L``[``\$i``][``\$j``] = 2;`` ``\$i``--; } ` ` ``else`` ``{ `` ``// Found i and k for j, LLAP with `` ``// i and j as first two elements `` ``// is equal to LLAP with j and k `` ``// as first two elements plus 1.`` ``// L[j][k] must have been filled `` ``// before as we run the loop from`` ``// right side `` ``\$L``[``\$i``][``\$j``] = ``\$L``[``\$j``][``\$k``] + 1; ` ` ``// Update overall LLAP, if needed `` ``\$llap` `= max(``\$llap``, ``\$L``[``\$i``][``\$j``]); ` ` ``// Change i and k to fill more `` ``// L[i][j] values for current j `` ``\$i``--; `` ``\$k``++; `` ``} `` ``} ` ` ``// If the loop was stopped due to k `` ``// becoming more than n-1, set the`` ``// remaining entities in column j as 2 `` ``while` `(``\$i` `>= 0) `` ``{ `` ``\$L``[``\$i``][``\$j``] = 2; `` ``\$i``--; `` ``} `` ``} `` ``return` `\$llap``; ``} ` `// Driver Code``\$set1` `= ``array``(1, 7, 10, 13, 14, 19); ``\$n1` `= sizeof(``\$set1``); ``echo` `lenghtOfLongestAP(``\$set1``, ``\$n1``),``"\n"``; ` `\$set2` `= ``array``(1, 7, 10, 15, 27, 29); ``\$n2` `= sizeof(``\$set2``); ``echo` `lenghtOfLongestAP(``\$set2``, ``\$n2``),``"\n"``; ` `\$set3` `= ``array``(2, 4, 6, 8, 10); ``\$n3` `= sizeof(``\$set3``); ``echo` `lenghtOfLongestAP(``\$set3``, ``\$n3``),``"\n"``; ` `// This code is contributed by Sach_Code``?>`
Output
```4
3
5
```
Time Complexity: O(n2
Auxiliary Space: O(n2)
Previous
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https://www.qsl.net/wb6tpu/si-list4/0098.html | 1,553,250,053,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912202642.32/warc/CC-MAIN-20190322094932-20190322120932-00158.warc.gz | 878,224,850 | 5,029 | # RE: [SI-LIST] : Chassis hole opening and frequencies
From: Ray Waugh ([email protected])
Date: Mon Jan 10 2000 - 11:08:35 PST
Doug...
Have you ever seen a parabolic dish antenna which uses a mesh (instead
of a solid metal surface)? It lets the wind and rain through. As a
rule of thumb, keep the holes in the mesh under a tenth wavelength in
dimension.
I'd suggest doing as the RF instrument people do -- cut a hole in your
chassis large enough for your fan, and mount a mesh over it. To avoid
having the mesh act as a RF radiator, make certain that it is
continuously attached to the rim of the chassis.
You could probably build and test a dummy chassis quicker than you could
do a proper analysis.
Ray
------------------------------------------------------
Raymond W. Waugh - WSD Diode Applications
E-mail: [email protected]
USPS : Agilent Technologies
Wireless Semiconductor Division
39201 Cherry Street, MS NK20
Newark, California 94560
------------------------------------------------------
-----Original Message-----
Sent: Thursday, December 30, 1999 1:39 PM
To: [email protected]
Subject: Re: [SI-LIST] : Chassis hole opening and frequencies
Doug,
In antenna theory, a microstrip (for example) that is fed (by an ideal
infinite rise-time source) in one end and
(a) shorted at the other end will radiate at the frequency that
corresponds
to a quarter wavelength
(b) open at the other end will radiate at the frequency that corresponds
to a half wavelength
So if our microstrip is 12 cm long and it is shorted at one end, we
should
be able to radiate 625 MHz pretty well. If you imagine your ideal
source
is in the hole of the chassis and connected across the longest dimension
of the hole, we have the same thing set up in (a) above. Our source is
referenced to ground and the short is, effectively, ground, too. Does
this
make any sense???
So my long-winded answer is a "quarter wavelength of the longest
dimension
that defines the hole."
This should get you very much in the ballpark.
WARNING::DEVIATION FROM QUESTION::WARNING---------------**************
Now the next question is: What if I have 2 or 3 or a whole matrix of
holes
(air holes for cooling for instance) in my chassis. How will THAT
I can indirectly answer that question as follows:
We can get SE (shield effectiveness) from:
20*log(wavelength/2*max_length)
where max_length is the maximum dimension of the slot or aperture.
So the SE of a 100 mil opening at 5 GHz is 21.45 dB-->watch units!
Mulitple apertures reduce the shielding effectiveness. Lets call it
MA_n.
The amount of reduction depends on (1) the spacing between the
apertures,
(2) the frequency, and (3) the number of apertures. When apertures of
equal
size are placed close together (less than a half wavelength), the
reduction
in shielding effectiveness is approximately proportional to the square
root
of the number of apertures n:
MA_n= -10*log(n)
So now our total shielding effectiveness, SE_tot = SE + MA_n
Example: 4 100 mils holes at 5 GHz provides a reduction in shielding of
6 dB
if the holes are less than a half wavelength or 1.18 inches apart.
So now our SE = 21.45 + (-6) = 15.45 dB at 5 GHz.
I know, I know, it is more than you wanted to know.
----->Chris
At 09:45 AM, you wrote:
>As I recall, there is a relationship between a hole in a chassis and the
>frequencies that can pass through that opening. I recall that the longest
>dimension of the hole defines the wavelength, or quarter wavelength, or
>something, of the lowest frequency than can conveniently enter or escape
>through the opening.
>
>Can anyone give me the correct relationship?
>
>Thanks.
>
>And Happy New Year to All...........
>
>Doug Brooks
>and all of us here at UltraCAD
>
>
>
>
>
>.
>************************************************************
>See our updated message re in-house seminars on our web page
>.
>Doug Brooks, President [email protected]
>
>
>**** To unsubscribe from si-list: send e-mail to
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# (7) 2 sin11°15' is equal to-(A)√2-√2+√2 (B) √2 -√2-√2 (C) √2+√2+√2/2 (D) √2+√2-√2/2(8) if 60°+a & 60°-a are roots of sin^2 x+bsinx + c =0 , then -(A)4b^2 + 3=12c(B)4b + 3 =12c(C)4b^2 - 3=-12c(D)4b^2 - 3=12c
Arun
25763 Points
2 years ago
Dear Lisa
sin (60 + a) + sin (60 – a) = – b
2 sin 60 * cos a = – b
sqrt (3) * cos a = – b
now
sin (60 + a) sin (60 – a) = c
sin^2 60 – sin^2 a = c
sin^2 a = ¾ – c
1 – b^2 / 3 = ¾ – c
1 2 – 4 b^2 = 9 – 12c
4b^2 – 3 = 12C | 381 | 818 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-25 | latest | en | 0.632082 |
https://yellowcomic.com/what-transports-ultraviolet-dye-through-the-a-c-system/ | 1,652,693,500,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662510097.3/warc/CC-MAIN-20220516073101-20220516103101-00392.warc.gz | 1,288,254,707 | 4,874 | The refrigerant transports ultraviolet dye through the air conditioning system. It works on the principle in i beg your pardon a tiny measure that fluorescyellowcomic.comt shade is infused into and also circled via a functioning structure of an wait conditioner. The refrigerant blyellowcomic.comd will escape and also get collection at all the break destinations.
You are watching: What transports ultraviolet dye through the a/c system?
refrigerantThe refrigerant does every the work. It"s a strategy in which a small measure the fluorescyellowcomic.comt shade is infused into and circled v a working structure of an A/C The refrigerant blyellowcomic.comd will escape and also gather at all break destinations.
18.5 grams that Nitrogyellowcomic.com gas and also 26.7 grams the Oxygyellowcomic.com gas type how countless grams that Dinitrogyellowcomic.com Pyellowcomic.comtoxide if the reaction is only 80%
29.38 grams
Explanation:
The reaction is:
2N₂(g) + 5O₂(g) → 2N₂O₅ (1)
We must calculate the number of moles the N₂ and O₂:
Now, we need to find the limiting reactant. Indigenous reaction (1) we have actually that 2 mole of N₂ react v 5 moles of O₂:
If we have actually 0.66 moles of N₂ and also we require 0.34 mole to react through O₂, thyellowcomic.com the limiting is O₂.
We have the right to calculate the number of moles of N₂O₅ produced:
Now, we have the right to calculate the theoretical massive of N₂O₅:
Finally, if the reaction is just 80% efficiyellowcomic.comt thyellowcomic.com the mass of N₂O₅ produced is:
Therefore, are developed 29.38 grams the N₂O₅.
I hope it helps you!
5 0
7 month ago
PLZ need HLP! How plenty of milliliters that alcohol room in 167 mL of an 85.0% (v/v) alcohol solution?
ryzh <129>
141.95 hope it helps
4 0
7 months ago
What is the fixed of 3.01 x 10^23 molecules of oxygyellowcomic.com?
Romashka-Z-Leto <24>
Oxygyellowcomic.com is gas. It has 2 O atoms. Atom mass the 1 O atom is 16 amu. So molecular mass that O2 = 2 x 16 = 32 amu.Now, fixed of 6.023 x 10 ^23 = 32 amu.Therefore fixed of 3.01 x 10^23 = 32 x 3.01 x 10^23 / 6.023 x 10 ^23 = 15.99= ~16 gm
7 0
1 year ago
Using the regular table yellowcomic.comtry of steel below, match the numbers v what they represyellowcomic.comt.
balu736 <363>
Atomic number: 26atomic mass: 56this is every i recognize sorry :(
8 0
11 month ago
Other questions:
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Signup | 712 | 2,440 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-21 | latest | en | 0.84823 |
https://www.chegg.com/homework-help/use-vectors-show-distance-p1-x1-y1-z1-plane-ax-cz-d-b-find-chapter-11-problem-10aae-solution-9780321388506-exc | 1,553,531,023,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912204077.10/warc/CC-MAIN-20190325153323-20190325175323-00558.warc.gz | 723,311,896 | 30,434 | # Student Solutions Manual Part 2 for University Calculus (1st Edition) Edit edition Problem 10AAE from Chapter 11: a. Use vectors to show that the distance from P1(x1, y1, z1)...
We have solutions for your book!
Chapter: Problem:
a. Use vectors to show that the distance from P1(x1, y1, z1) to the plane Ax + By + Cz = D is
b. Find an equation for the sphere that is tangent to the planes x + y + z = 3 and x + y + z = 9 if the planes 2x – y = 0 and 3x – z = 0 pass through the center of the sphere.
Step-by-step solution:
Chapter: Problem:
• Step 1 of 5
(a) Given that and plane is
Let be any point on the given plane
Then and is a vector normal to the given plane.
• Chapter , Problem is solved.
Corresponding Textbook
Student Solutions Manual Part 2 for University Calculus | 1st Edition
9780321388506ISBN-13: 032138850XISBN: Authors:
This is an alternate ISBN. View the primary ISBN for: University Calculus: Alternate 1st Edition Textbook Solutions | 272 | 967 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2019-13 | latest | en | 0.80897 |
https://www.convert-measurement-units.com/convert+m3+Bioethanol+to+Barrel+oil+equivalent.php | 1,685,318,294,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224644571.22/warc/CC-MAIN-20230528214404-20230529004404-00580.warc.gz | 805,735,627 | 13,189 | Convert m³ Bioethanol to boe (m³ Bioethanol to Barrel oil equivalent)
convert-measurement-units.com
m³ Bioethanol into boe
## m³ Bioethanol into Barrel oil equivalent
numbers in scientific notation
https://www.convert-measurement-units.com/convert+m3+Bioethanol+to+Barrel+oil+equivalent.php
# Convert m³ Bioethanol to Barrel oil equivalent (m³ Bioethanol to boe):
1. Choose the right category from the selection list, in this case 'Oil equivalent'.
2. Next enter the value you want to convert. The basic operations of arithmetic: addition (+), subtraction (-), multiplication (*, x), division (/, :, ÷), exponent (^), square root (√), brackets and π (pi) are all permitted at this point.
3. From the selection list, choose the unit that corresponds to the value you want to convert, in this case 'm³ Bioethanol'.
4. Finally choose the unit you want the value to be converted to, in this case 'Barrel oil equivalent [boe]'.
5. Then, when the result appears, there is still the possibility of rounding it to a specific number of decimal places, whenever it makes sense to do so.
With this calculator, it is possible to enter the value to be converted together with the original measurement unit; for example, '34 m3 Bioethanol'. In so doing, either the full name of the unit or its abbreviation can be used. Then, the calculator determines the category of the measurement unit of measure that is to be converted, in this case 'Oil equivalent'. After that, it converts the entered value into all of the appropriate units known to it. In the resulting list, you will be sure also to find the conversion you originally sought. Alternatively, the value to be converted can be entered as follows: '72 m3 Bioethanol to boe' or '46 m3 Bioethanol into boe' or '40 m3 Bioethanol -> Barrel oil equivalent' or '25 m3 Bioethanol = boe' or '3 m3 Bioethanol to Barrel oil equivalent' or '72 m3 Bioethanol into Barrel oil equivalent'. For this alternative, the calculator also figures out immediately into which unit the original value is specifically to be converted. Regardless which of these possibilities one uses, it saves one the cumbersome search for the appropriate listing in long selection lists with myriad categories and countless supported units. All of that is taken over for us by the calculator and it gets the job done in a fraction of a second.
Furthermore, the calculator makes it possible to use mathematical expressions. As a result, not only can numbers be reckoned with one another, such as, for example, '(74 * 18) m3 Bioethanol'. But different units of measurement can also be coupled with one another directly in the conversion. That could, for example, look like this: '34 m3 Bioethanol + 102 Barrel oil equivalent' or '4mm x 87cm x 31dm = ? cm^3'. The units of measure combined in this way naturally have to fit together and make sense in the combination in question.
The mathematical functions sin, cos, tan and sqrt can also be used. Example: sin(π/2), cos(pi/2), tan(90°), sin(90) or sqrt(4).
If a check mark has been placed next to 'Numbers in scientific notation', the answer will appear as an exponential. For example, 1.005 563 262 454 3×1028. For this form of presentation, the number will be segmented into an exponent, here 28, and the actual number, here 1.005 563 262 454 3. For devices on which the possibilities for displaying numbers are limited, such as for example, pocket calculators, one also finds the way of writing numbers as 1.005 563 262 454 3E+28. In particular, this makes very large and very small numbers easier to read. If a check mark has not been placed at this spot, then the result is given in the customary way of writing numbers. For the above example, it would then look like this: 10 055 632 624 543 000 000 000 000 000. Independent of the presentation of the results, the maximum precision of this calculator is 14 places. That should be precise enough for most applications. | 939 | 3,935 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-23 | latest | en | 0.823269 |
www.saamstourism.com | 1,571,517,394,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986697760.44/warc/CC-MAIN-20191019191828-20191019215328-00369.warc.gz | 327,467,776 | 9,700 | # The Run Down on Distance Formula Physics Revealed
You give your fuel economy, distance and gas price and you will quickly get the price of the trip. Inside this problem, you’re requested to come across the stopping distance of the vehicle. Hence, it is considered to be undergoing an acceleration.
## Distance Formula Physics – Is it a Scam?
Likewise, there’s a corresponding selection of object positions that produce images in this range. Velocity lets you know how fast an object is moving in addition to in what direction it’s moving. The rest of the kinds of waves require some kind of medium whereby to propagate.
## Distance Formula Physics – What Is It?
A lever is a simple machine in which one force referred to as the effort is utilized to overcome another force called the load. In order to make certain that the stopping sight distance provided is adequate, we are in need of a more in-depth comprehension of the frictional force. http://www.gse.upenn.edu/faculty_research/centers It’s the large force acting for a small interval of time.
Broadly speaking, elastic collisions are characterized by a substantial velocity change, a gigantic momentum change, a massive impulse, and a large force. Now we’ve got a far better result. If an individual in motion wants to maximize their velocity, then that person must make every attempt to maximize the quantity that they’re displaced from their original position.
You do not have to locate every question correct to find the utmost score (800) for the test. The electric field strength isn’t dependent upon the amount of charge on the test charge. Your iClicker needs to be working to acquire participation credit.
Once more, the secret is to construct a decent free-body diagram. Charge Q functions as a point charge to make an electric field. If you wished to find out the distance between two points you might use the distance formula in that case.
## Using Distance Formula Physics
Several different corrections and tricks need to get precise distances. termpaperwriter For many teams, both definitions will yield the exact value. So it can carry out various calculations to provide you with the results that you demand.
Gravity is giving you an additional push, so you don’t have to do all of the work with the pedals. Imagine you’re pushing a heavy box throughout the room. A team’s recent velocity can be helpful in helping predict how much work can be finished by the team in an upcoming sprint.
## Distance Formula Physics
In practice, air resistance isn’t entirely negligible, and so the initial velocity would want to be somewhat larger than that given to reach the exact same height. The proportion of the velocity to the applied field is known as the mobility. The size of the electric field strength is defined with regard to the way that it is measured.
Velocity pressure is helpful in aerodynamics to learn the way the aerodynamic stress varies, and specifically, when it reaches its highest possible value. The genuine mobility also depends upon the kind of dopant. Physical quantities that are completely specified by just giving out there magnitude are called scalars.
Using data from some other teams can provide help. If additional information is necessary, it is going to become apparent as you proceed even when that extra information is the quantity you’re requested to find. Click the link to find out more.
## Distance Formula Physics Secrets
Torque Calculator Torque is only a rotational force. Distance is a simple notion and x is a simple variable. Now find the entire distance traveled.
The purpose of maximum aerodynamic load is a crucial parameter in many applications, like during spacecraft launch. For the mathematically inclined, it’s caused by carrying out of what’s usually regarded as an indefinite integration. The solution is quite easy.
## A Startling Fact about Distance Formula Physics Uncovered
When you’re solving a problem in which you will need to find one of these variables but you’re lacking another one, you should combine both formulae to get rid of the unknown variable. You just need to know the expression displacement’ for Edexcel. Re-read the above mentioned paragraphs.
## The Distance Formula Physics Chronicles
Notice the words distance and displacement are the sole difference between the 2 definitions. The definition of instantaneous velocity does not mean that time consists of instants. The primary difference is that it’s more difficult to add and subtract the vectors, because you’ve got to use components.
The crucial suggestions to understand are simple. The GPS receiver isn’t just a fine electronic mechanism, it can do a whole bunch of mathematics also. Plotted below is the actual error feature, in contrast to the initial three non-zero provisions of its Taylor series.
## Using Distance Formula Physics
This is called the Magnus effect, more commonly known as a banana kick. You probably know more about the Large Hadron Collider (CERN), the most effective particle accelerator on earth. Most individuals consider speed and velocity to be the very same and might even apply these terms interchangeably.
However, it’s still wonderful to grasp the physics behind it. The typical velocity you’re computing is an ordinary rate. It is one of the major parameters of motion.
Because of this it is necessary for practically any physics lover to comprehend the way that it works and ought to be applied. After all the best thing about physics is the simple fact that it may be utilised to deal with real world difficulties. The type of problem we’ve been studying was considered by many thinkers from the first days of science. | 1,102 | 5,681 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2019-43 | longest | en | 0.910386 |
https://www.mometrix.com/academy/friction/ | 1,701,698,009,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100529.8/warc/CC-MAIN-20231204115419-20231204145419-00600.warc.gz | 980,272,725 | 22,614 | # What is Friction?
Hi, and welcome to this video on friction! Friction is a type of resistance that one object encounters when moving over another object. In this video, we’ll look at the friction model and some examples of different kinds of friction.
First, let’s familiarize ourselves with “normal force”. The normal force is the component exactly perpendicular to where the two surfaces touch. When you stand on the ground, the normal force points up to the sky because the two touching surfaces are the bottoms of your feet and the ground. When you push a box up a ramp, the normal force is perpendicular to the ramp:
In the standard model of friction, we consider friction to be proportional to the normal force. Frictional force F equals mu times N:
F=μN
The proportionality constant, μ, changes depending on if the object is in motion or is standing still. We use the coefficient of static friction when the object is not in motion.
Suppose we have a very heavy box we need to push across the ground. We push lightly, and the box doesn’t move. So we start pushing harder and harder until suddenly the box starts moving! In this example, the box doesn’t move at first because static friction resists the change in motion. As we push harder, static friction increases. However, we reach a point where the box suddenly starts moving. At this point, our model switches to kinetic friction, which is a constant, to describe the frictional forces.
This is our standard model of friction. Static friction is directly proportional to the normal force, and it increases until the object begins to move. Kinetic friction is a constant, and it is less than or sometimes nearly equal to the maximum static friction.
Friction eats up energy you would normally convert into useful energy. We see this in static friction: instead of an object moving as soon as you apply force, friction eats up that energy and resists the change. Then, kinetic friction takes away energy as an object moves. If you slide across the floor, eventually you come to a stop because kinetic friction resists your motion and slows you down.
There may be times when you want to minimize kinetic friction. If you want to slide further across the floor, you can buff the floor with wax to make the floor more slippery. Another option is to spray your feet with cooking oil to create a buffer between the floor and your feet. In this case, the cooking oil acts as a lubricant and decreases the coefficient of kinetic friction.
Another way we can lower friction is by using ball bearings and grease in wheels. The ball bearings help the wheel rotate more smoothly and the grease creates a more slippery surface for the balls to roll over.
Sometimes, we actually want to increase kinetic friction! Let’s say you are developing a new kind of abrasive substance for sandpaper. The abrasive substance creates a high coefficient of friction between a test surface and the paper. We can even do an experiment to pick the most abrasive substance from a range of options. Using the same amount of paper and block of a given mass for each, attach a spring scale to the block and pull until the block starts to move. The maximum read from the scale is correlated with static friction. Then you can pull the block at a constant velocity to measure the kinetic friction.
Mass (g) 400
Reading (g) Estimate Abrasive Static Kinetic μS μK A 255 127 0.64 0.32 B 263 153 0.66 0.38 <------- Highest coefficient of kinetic friction C 258 144 0.65 0.36 D 288 136 0.72 0.34 <------- Highest coefficient of static friction
Now, our model isn’t perfect. We are assuming kinetic friction is always constant. For low speeds, this is mostly true, but once you start going very fast, kinetic friction becomes dependent on velocity. One such example is skydiving. You are flying extremely fast through the air, gaining speed from the moment you jump, but, eventually, your speed is constant. This is called terminal velocity, and it happens because kinetic friction is proportional to the square of the velocity.
Now that we’ve covered the basics of what friction is, here’s an interesting example to test your knowledge.
Engineers who designed the Mars rover Curiosity were very worried about Mars’s atmosphere as the rover was landing on Mars. Why?
1. The rover was going so fast that the kinetic friction from the molecules in the atmosphere could heat up the rover and blow it up.
2. The normal force doesn’t exist on Mars, so friction is infinite.
3. They weren’t worried because the atmosphere doesn’t affect the rover’s landing.
4. Gravity is less on Mars, so the atmosphere doesn’t matter, and they were worried over nothing.
The answer is A. While gravity on Mars is less compared to gravity on Earth, the atmosphere on Mars is still thick enough to cause a fast-moving object to convert energy to heat because of kinetic friction. Engineers designed a special nose cone to protect the rover from the extreme heat. They also used a very specially designed parachute to slow down the rover to decrease the energy loss due to kinetic friction. This was particularly tricky due to the thin atmosphere! The parachute needs enough atmosphere to help slow down, which is really hard to do when there isn’t much of an atmosphere to start with!
Thanks for watching, and happy studying! | 1,112 | 5,343 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2023-50 | latest | en | 0.91014 |
http://mathematicalmysterytour.blogspot.com/2015/11/mr-bs-brain-o-quiz-34.html | 1,532,283,057,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676593438.33/warc/CC-MAIN-20180722174538-20180722194538-00243.warc.gz | 221,308,093 | 14,315 | ## Friday, November 20, 2015
### Mr. B's Brain-O Quiz # 34
11/20/2015
Mr. B’s “BRAIN-O” Quiz
Guaranteed to unclog clogged up brains!
NAME:__________________ DATE:____________ PERIOD:_______
Directions: Carefully read and answer the following questions. Print your answer clearly in the box next to each question. Questions are worth one million nano bonus points each. However, if you get all five correct, we will double the points for a total of ten million nano bonus points.
1. (True or False) Five times four twenty, plus two is twenty-three. 2. Write a mathematical expression that equals 1000 using eight 8s. 3. Kangaroos are very interesting and special animals that live in Australia. What can kangaroos do that no other animal (including humans) can do? 4. There is an five letter word in the English language that over 99% of all high school graduates spell and pronounce wrong every time they write it or use it. What is that word? 5. If one thousand pieces of candy cost ten dollars, how much would ten pieces of candy cost?
“Oh those darned Torpedos. Just go around them.”
- Nobody famous or inspiring ever said this.
WARNING: For Educational Purposes Only. | 293 | 1,184 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2018-30 | latest | en | 0.899111 |
https://www.sparknotes.com/physics/specialrelativity/applications/section2/ | 1,716,594,539,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058751.45/warc/CC-MAIN-20240524214158-20240525004158-00276.warc.gz | 867,035,146 | 58,015 | ### Statement
The so-called 'Twin Paradox' is one of the most famous problems in all of science. Fortunately for relativity it is not a paradox at all. As has been mentioned, Special and General Relativity are both self-consistent within themselves and within physics. We will state the twin paradox here and then describe some of the ways in which the paradox can be resolved.
The usual statement of the paradox is that one twin (call her A) remains at rest on the earth relative to another twin who flies from the earth to a distant star at a high velocity (compared to c). Call the flying twin B. B reaches the star and turns around and returns to earth. The twin on earth (A) will see B's clock running slowly due to time dilation. So if the twins compare ages back on earth, twin B should be younger. However, from B's point of view (in her reference frame) A is moving away at high speed as B moves towards the distant star and later A is moving towards B at high speed as B moves back towards the earth. According to B, then, time should run slowly for A on both legs of the trip; thus A should be younger than B! It is not possible that both twins can be right-the twins can compare clocks back on earth and either A's must show more time than B's or vice-versa (or perhaps they are the same). Who is right? Which twin is younger?
### Resolution
The reasoning from A's frame is correct: twin B is younger. The simplest way to explain this is to say that in order for twin B to leave the earth and travel to a distance star she must accelerate to speed v. Then when she reaches the star she must slow down and eventually turn around and accelerate in the other direction. Finally, when B reaches the earth again she must decelerate from v to land once more on the earth. Since B's route involves acceleration, her frame cannot be considered an inertial reference frame and thus none of the reasoning applied above (such as time dilation) can be applied. To deal with the situation in B's frame we must enter into a much more complicated analysis involving accelerating frames of reference; this is the subject of General Relativity. It turns out that while the B is moving with speed v A's clock does run comparatively slow, but when B is accelerating the A's clocks run faster to such an extent that the overall elapsed time is measured as being shorter in B's frame. Thus the reasoning in A's frame is correct and B is younger.
However, we can also resolve the paradox without resorting to General Relativity. Consider B's path to the distant star lined with many lamps. The lamps flash on and off simultaneously in twin A's frame. Let the time measured between successive flashes of the lamps in A's frame be tA. What is the time between flashes in B's frame? As we learned in Heading the flashes cannot occur simultaneously in B's frame; in fact B measures the flashes ahead of him to occur earlier than the flashes behind him (B is moving towards those lamps ahead of him). Since B is always moving towards the flashes which happen earlier the time between flashes is less in B's frame. In B's frame the distance between flash-events is zero (B is at rest) so ΔxB = 0, thus ΔtA = γ(ΔtB - vΔxB/c2) gives:
ΔtB =
Thus the time between flashes is less in B's frame than in A's frame. N is the total number of flashes that B sees during her entire journey. Both twins must agree on the number of flashes seen during the journey. Thus the total time of the journey in A's frame is TA = NΔtA, and the total time in B's frame is TB = NΔtB = N(ΔtA/γ). Thus:
TB =
Thus the total journey time is less in B's frame and hence she is the younger twin.
All this is fine. But what about in B's frame? Why can't we employ the same analysis of A moving past flashing lamps to show that in fact A is younger? The simple answer is that the concept of 'B's frame' is ambiguous; B in fact is in two different frame depending on her direction of travel. This can be seen on the Minkowski diagram in :
Here is lines of simultaneity in B's frame are sloped one way for the outward journey and the other way for the trip back; this leaves a gap in the middle where A observes no flashes, leading to an overall measurement of more time in A's frame. If the distant star is distance d from the earth in A's frame and the flashes occur at intervals ΔtB in B's frame, then they occur at intervals ΔtB/γ = ΔtA in A's frame, as per the usual time dilation effect (this is the same for inward and outward journeys). Again let the twins agree that there are N total flashes during the journey. The total time is B's frame is then TB = NΔtB and for A, TA = N(ΔtB/γ) + τ where τ is the time where A observes no flashes (see the Minkowski diagram). In B's frame the distance between the earth and the star is (half the total journey time times the speed) which is also equal to d /γ due to the usual length contraction. Thus:
TB = TA = + τ = + t
What is τ? We can see from that the slopes of the lines are ±v/c so the time in which A observes no flashes is ct = 2d tanθ = 2dv/c. Thus:
TA = + = frac2dv
Comparing TA and TB we see TB = TA/γ which is the same result we arrived at above. A measures more time and B is younger. | 1,215 | 5,217 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2024-22 | latest | en | 0.9629 |
https://www.nextgurukul.in/questions-answers-forum/question/academic/A-solid-iron-rectangular-block-of-dimensions-44-m-26-m-and-1-m/18980 | 1,561,641,239,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560628001138.93/warc/CC-MAIN-20190627115818-20190627141818-00354.warc.gz | 826,233,522 | 52,044 | Q&A Forum
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Oct 19, 2013
# A solid iron rectangular block of dimensions 4.4 m, 2.6 m and 1 m
A solid iron rectangular block of dimensions 4.4 m, 2.6 m and 1 m is cast into a hollow cylindrical pipe of internal radius 30 cm and thickness 5 cm. Find the length of the pipe.
Agam Gupta
Member since Sep 21, 2013
Internal radius, r = 30 cm Thickness, w = 5 cm External radius, R = 30 + 5 = 35 cm Here volume of cuboid = volume of hollow cylinder 440 × 260 ×100 = π(R2 – r2)h ⇒ 440 × 260 × 100 = π(352 – 302)h ⇒ 440 × 260 × 100 = π(325)h ∴ h = 11200 cm = 112 m
Pranay Guda
Member since
Internal radius, r = 30 cm Thickness, w = 5 cm External radius, R = 30 + 5 = 35 cm Here volume of cuboid = volume of hollow cylinder 440 × 260 ×100 = π(R2 r2)h ⇒ 440 × 260 × 100 = π(352 302)h ⇒ 440 × 260 × 100 = π(325)h ∴ h = 11200 cm = 112 m
Kishore Kumar
Member since
Internal radius, r = 30 cm Thickness, w = 5 cm External radius, R = 30 + 5 = 35 cm Here volume of cuboid = volume of hollow cylinder 440 × 260 ×100 = π(R2 – r2)h ⇒ 440 × 260 × 100 = π(352 – 302)h ⇒ 440 × 260 × 100 = π(325)h ∴ h = 11200 cm = 112 m
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# Analysis of the speed of convergence - PowerPoint PPT Presentation
Analysis of the speed of convergence. Lionel Artige HEC – Université de Liège 30 january 2010. Neoclassical Production Function. We will assume a production function of the Cobb-Douglas form: F[K(t), L(t), A(t)] = A K(t) α L(t) 1- α
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### Analysis of the speed of convergence
Lionel Artige
HEC – Université de Liège
30 january 2010
Neoclassical Production Function
We will assume a production function of the Cobb-Douglas form:
F[K(t), L(t), A(t)] = A K(t)α L(t)1- α
where K(t) is the physical capital stock at time t, L(t) is labor and A is the constant level of total factor productivity.
This Cobb-Douglas function is homogeneous of degree 1. Therefore, it is possible to write it in intensive form:
f[k(t)] = A k(t)α
where k K/L is the capital-labor ratio.
Exogenous Growth Model: The Solow-Swan Model
The fundamental equation of the Solow-Swan model is the equation of the capital accumulation:
dk
= sAkα – (n + δ)
dt
The growth rate of the capital-labor ratio is:
dk/dt
= sAkα - 1 – (n + δ)
dk
The derivative of this growth rate with respect to k is negative.
Production function: graphical representation
f(k)
This curve is the graphical representation of the production function in intensive form, where k = K/L.
The function f(k) is assumed to concave.
k
f(k)
E
f(k*)
The point A, whose coordinates are (k(0);f(k(0)), is the initial level of the economy.
The point E, whose coordinates are (k*;f(k*)), is the steady state of the function f(k). This is the long-term level of the economy.
A
f[k(0)]
k
k(0)
k*
Growth rate at the steady state
f(k)
E
C
B
f(k*)
The straight line (BC) is the tangent to the point E.
On the graph, the slope of the tangent appears to be 0. This slope is the instantaneous rate of variation of the function f(k) at the value k=k*.
Slope of the tangent = f’(k*)
The growth rate of this economy at the steady state (E) is equal to 0.
A
f[k(0)]
k
k(0)
k*
How to calculate the instantaneous rate of variation ?
The instantaneous rate of variation (or derivative at a point) is
f(k*+ h) – f(k*)
f(k) – f(k*)
lim
lim
f ’(k*) =
=
h 0
k k*
k – k*
h
For our production function, the instantaneous rate of variation at the steady state point is
f ’(k*) = 0
This means that the production per worker does not grow at the steady state.
The level of product per worker is f [k(0)] when k = k(0) and is f (k*) when k = k*.
The difference in level (f (k*) - f [k(0)]) is the increase in product per worker when k increases from k = k(0) to k = k*.
We can also calculate the growth rate (g) of the product per worker when k increases from k = k(0) to k = k*. It is the geometric mean of all the instantaneous rates of variation between k = k(0) to k = k*:
dk/dt
1/n
[f ’(k(0)) × … × f ’(k*)]
for all k [k(0), k*] and n
g =
=
k
n derivatives
where n is the number of compounding. When n compounding is continuous.
Growth rate between A and E
f(k)
E
C
B
f(k*)
The instantaneous rates of variation between k(0) and k* are all different since the function is non-linear.
In fact, the instantaneous rates of variation decrease as k increases from k(0) to k*. The slopes are increasingly weaker.
The growth rate between A and E is the geometric mean of all the instantaneous rates of variation
A
f[k(0)]
k
k(0)
k*
Calculation of average growth rate between A and E
If we know the values for f[k(0)] and f(k*) and the number of periods (e.g. number of years) that elapsed for the economy to go from k(0) to k*, then we can calculate the average growth rate between f[k(0)] and f(k*) :
1/t
R = {ln f(k*) – ln f[k(0)]}
where t is the number of periods = (number of dates – 1). (e.g. 1991, 1992 and 1993 are 3 dates but 1991-1992 and 1992-1993 are two periods).
This average growth rate R is calculated by using the geometric mean where the growth rate compound continuously. If the number of periods is 1, then t = 1 and the growth rate is just the continuous growth rate between f[k(0)] and f(k*).
This continuous growth rate is also called the speed of convergence. The name comes from the result that the steady state E is stable, hence the economy converges to E regardless of its initial start k(0) (except k (0) = 0).
Calculation of average growth rate between A and E (cont.)
We can obviously use the growth rate to calculate the level of f(k*) if we know f[k(0)]:
f(k*) = exp(Rt) f[k(0)]
To sum up, the growth rate of f[k(t)] is a non-linear function of k(t). It decreases as k(t) increases due to the concavity of the production function.
Therefore, for linear estimation purposes, it is necessary to compute a growth rate that is linear in k(t) and could be a reasonable approximation of the true growth rate.
Graphical linear approximation of the growth rate
f(k)
F
E
C
B
f(k*)
Graphically, to approximate the growth rate between the points A and E, one has to draw a line (DF) passing through A and E.
The slope of this straight line gives the approximation of the growth rate of the concave function.
The farther A is located from E, the worse is the approximation. In our graph, the approximation is bad because A is too far from E.
A
f[k(0)]
D
k
k(0)
k*
Analytical linear approximation of the growth rate
To compute an analytical linear approximation of the growth rate, one has to linearize the growth rate function (dk/dt)/k around its steady state. To do so, we apply to this function a Taylor expansion of order 1 around the steady state k* to obtain a linear function:
dk/dt
dk/dt
dk/dt
k
+
(k – k*)
k
k*
k
k = k*
In the Solow-Swan model the growth rate is
dk/dt
= sAkα - 1 – (n + δ)
dk
At the steady state, the growth rate is 0, then :
sAkα - 1 = (n + δ)
Analytical linear approximation of the growth rate
Let us approximate the nonlinear Solow growth rate function by a Taylor polynomial of the first order:
(sAk(t)α - 1 – (n + δ))
dk/dt
(k(t) – k*)
sA(k*)α - 1 – (n + δ)
+
k
k
k(t) = k*
0 + (α – 1)sA(k*)α - 2 (k(t) – k*)
Since sAkα - 1 = (n + δ) at the steady state, we can further simplify to:
k(t) – k*
dk/dt
– (1 – α) (n + δ)
dk
k*
where [(k(t) – k*)/k*] is the rate of variation of k(t) around the steady state. This new growth function is linear in k(t). An increase in k(t) yields a decrease in the growth rate of – (1 – α) (n + δ)/k*.
log – linear approximation of the growth rate
A more convenient way for econometric analysis is to log – linearize the original growth rate function. It allows to interpret the result as a percentage deviation from the steady state. The log – linearization consists in applying a first-order Taylor expansion of log(k) around log(k*).
dk/dt
Let us write
= sAkα - 1 – (n + δ)
in log:
dk
d log k(t)
= sA e(α – 1) log k(t) – (n + δ)
dt
Let us define g[log k(t)] sA e(α – 1) log k(t) – (n + δ). Let us approximate this function:
g[log k(t)]
(log k(t) – log k*)
g[log k(t)] g[log k*] +
log k(t)
log k(t) = log k*
log – linear approximation of the growth rate
g[log k(t)] sA e(α – 1) log k* – (n + δ) + (α – 1) sA e(α – 1) log k* (log k(t) – log k*)
0 + (α – 1) (n + δ) (log k(t) – log k*)
Therefore, the log – linear form of the growth rate function is:
d log k(t)
– (1 – α) (n + δ) (log k(t) – log k*)
dt
d{d log k(t)/dt}
And
– (1 – α) (n + δ)
d log k(t)
d{d log k(t)/dt}
is called the speed of convergence in the
where β –
d log k(t)
economic growth literature. 1% deviation from k* yields a percentage change in the growth rate of k equal to – (1 – α) (n + δ) when the production function is Cobb-Douglas.
log – linear approximation of the growth rate
In fact, we are interested in the growth rate of income per capita rather than in the growth rate of the capital –labor ratio. But, they are the same:
dy(t)/dt
d lny(t)
d ln k(t)α
d [α ln k(t)]
d [α ln k(t)]
d k(t)
.
=
=
=
=
y(t)
dt
dt
dt
dk(t)
dt
dk/dt
= α
dk
y(t)
k(t)α
k(t)α
And y(t) = k(t)α =>
=
=
Taking the log :
y*
y*
(k*)α
y(t)
k(t)
log
a log
=
=> log y(t) – log y* = α [log k(t) – log k*]. Then
y*
k*
d log k(t)
d log y(t)
1
1
– β (log k(t) – log k*) =>
– β
(log y(t) – log y*)
dt
α
α
dt
log – linear approximation of the growth rate
As a result:
d log y(t)
– β (log y(t) – log y*) (1)
dt
The speed of convergence is the same for the income per capita as for the capital-labor ratio.
Equation (1) is a first-order differential equation of the type:
log y’(t) + β log y(t) = β log y*
where log y’(t) is the time derivative of log y(t). It can be solved in four steps:
Let us first define: z(t) log y(t)
First step: Solution of the corresponding homogenous equation z’(t) + β z(t) = 0
z’(t)
z’(t)
= – β
dt = –
β dt
z(t)
z(t)
log z(t) + b1 = – βt + b2
log z(t) = – βt + b where b = b1 +b2
e log z(t) = e – βt + b
z1(t) = e– βt eb
z1(t) = e– βtθ where θ = eb
Second step: Particular solution of the equation z’(t) + β z(t) = β z*
An obvious particular solution is at the steady state where z’(t) = 0, then z2(t) =z*.
Third step: General solution of the equation z’(t) + β z(t) = β z*
This is the sum of the solution of the homogenous equation and the particular solution of our equation:
z(t) = z1(t) + z2(t) = e – βtθ + z* (2)
Fourth step: Final solution of the equation z’(t) + β z(t) = β z*
What is left to do is to give a value for θ. This value can be determined by a value for z(t) at a particular date t. For example, the initial condition is a good candidate: z(0) for t = 0. Then, at t = 0,
z(0) = e0θ + z* θ = z(0) – z*
Substituting in (2) for θ :
z(t) = e – βt [z(0) –z*] + z* z(t) = (1 – e – βt) z* + e – βt z(0)
Eventually, as z(t) log y(t), the solution of our differential equation is
log y(t) = (1 – e – βt) log y* + e – βt log y(0) (3)
where β (1 – α)(n + δ) and y* = (k*)α
If we have data on GDP per capita in an initial date and a terminal date, then we can estimate the speed of convergence β. If we substract log y(0) from both sides of (3) and substitute for y* then
1
log y(t) – log y(0) = (1 – e – βt) log
[log sA – log (n + δ]
+ (1 – e – βt) log y(0)
1 - α
In the Solow-Swan model, the growth of income (left-hand side) is a function of the determinants of the steady state and the initial level of income. | 3,440 | 11,089 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2018-30 | latest | en | 0.840255 |
https://mathematica.stackexchange.com/questions/121421/averaging-over-large-number-of-random-realizations | 1,632,660,766,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057861.0/warc/CC-MAIN-20210926114012-20210926144012-00156.warc.gz | 409,306,930 | 38,036 | # averaging over large number of random realizations
I want to run following for few hundreds of a random variable set. This calculates:
1. generate normal random variable
2. find gain with squaring
3. calculate performance metric
4. find max of min of possible M+1 performance metric when a changes (0,1). I have other case with another parameter b which changes (0,1).
5. find the average for 100 or 1000 random realizations
6. find same for different p values
Finally I want to save it as a table with different p and M.
variable[M_] := RandomVariate[NormalDistribution[0, 1], M + 1];
gain[M_, d_, t_] := variable[M]^2/(d/(M + 1))^t;
perform[p_, a_, m_, M_, d_, t_] :=
1/(M + 1) p ((1 - a)/(2 - a)) a^(
m - 1) (Product[gain[M, d, t][[i]], {i, 1, m}]);
maxminperform[p_, M_, d_, t_] :=
Max[Table[
Min[Table[perform[p, a, m, M, d, t], {m, 1, M + 1, 1}]], {a,
0.00001, 0.999, 0.001}]];
avgperform[p_, M_, d_, t_, itr_] :=
Mean[Table[maxminperform[p, M, d, t], {i, 1, itr}]];
Module[{n = 0.001, η = 0.7, σ = 1, k = 7, d = 6, t = 3,
itr = 5},
Export["table.xls", {Table[{avgperform[p, 2, d, t, itr],
avgperform[p, 4, d, t, itr], avgperform[p, 6, d, t, itr]}, {p,
1, 10, 1}] // N}]]
I understand that this takes really long time when I run.
Can someone help me to faster this code?
• General advice to be found here. This looks like it should benefit from Memoization a great deal, e.g. do you need SetDelayed for variable[M_]?
– gwr
Jul 25 '16 at 11:22
• when you compute Product[gain[][[i]]] , each factor in the product is computed using a newly generated random distribution. Is that your intent? You might want in there Product@@gain[][[;;m]] to compute it only once. It will be faster but obviously change the result. Jul 25 '16 at 15:24 | 579 | 1,752 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2021-39 | latest | en | 0.759195 |
https://study.com/academy/answer/solve-lim-limits-to-0-frac-x-sin-x-1-cos-x.html | 1,575,994,900,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540528457.66/warc/CC-MAIN-20191210152154-20191210180154-00465.warc.gz | 559,081,247 | 21,812 | # Solve \lim \limits_{\to 0} \frac {(x)sin(x)}{1-cos(x)}
## Question:
Solve {eq}\lim \limits_{x\to 0} \frac {x \ sin(x)}{1-cos(x)} {/eq}
## Limits of Functions:
The L-Hopitals rule when is applied to the limit expression of the equation containing the limits will get the simplified expression whose limits are found after direct replacement or we will have to apply the L-Hopitals rule again.
So to find the limit of
{eq}\lim_{x\to 0} \frac{x \ sin(x)}{1-cos(x)}\\ {/eq}
we have to do the following. Applying the L-Hopitals rule will give:
{eq}\lim _{x\to \:0}\left(\frac{\left(x\sin \left(x\right)\right)^{'\:}}{\left(1-\cos \left(x\right)\right)^{'\:}}\right)\\ =\lim _{x\to \:0}\left(\frac{\sin \left(x\right)+x\cos \left(x\right)}{\sin \left(x\right)}\right)\\ {/eq}
Again we have to apply the L-Hopitals rule here as follows:
{eq}\lim _{x\to \:0}\left(\frac{\left(\sin \left(x\right)+x\cos \left(x\right)\right)^{'\:}}{\left(\sin \left(x\right)\right)^{'\:}}\right)\\ =\lim _{x\to \:0}\left(\frac{2\cos \left(x\right)-x\sin \left(x\right)}{\cos \left(x\right)}\right)\\ =\frac{2\cos \left(0\right)-0\cdot \sin \left(0\right)}{\cos \left(0\right)}\\ =2 {/eq}
Now this is the required limit. | 432 | 1,206 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2019-51 | latest | en | 0.587206 |
https://pvilax.net/hockey-teams/frequent-question-how-much-does-an-average-hockey-player-skate-during-a-game.html | 1,656,405,579,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103360935.27/warc/CC-MAIN-20220628081102-20220628111102-00360.warc.gz | 528,600,465 | 16,853 | # Frequent question: How much does an average hockey player skate during a game?
Contents
The typical NHL player skates up to 5 miles per game—or 410 miles in just one season. NHL players have to stop on a dime or accelerate to speeds up to 30 mph.
## How fast do hockey players skate during a game?
The Mechanics of Skating
NHL players can reach speeds in excess of 20 miles (32 km) per hour on the ice. Some speed skaters have been clocked at over 30 miles (48 km) per hour! What makes one player faster than another? A combination of strength and mechanics help a skater move efficiently and quickly on the ice.
## How much does a hockey player run in a game?
On average, a hockey player will run eight to nine kilometres during a match. When compared to footballers, who regularly reach more than 10 kilometres in a game, it is worth remembering that hockey is a 60 minute game where football is 90 minutes.
## How many km does an NHL player skate in a game?
5 miles/ 8 km would be the average for an NHL player.
## How many pairs of skates do NHL players use?
One pair of skates is used in a game situation, the other pair is for practice. You play more games that you practice typically therefore your practice skates slowly get broken in. By the time your game skates are worn out, you already have a second pair that is now broken in and ready to use.
## Who is the fastest NHL player ever?
Here are some of the fastest players to ever play in the NHL.
• Pavel Bure. It’s one thing to have incredible speed. …
• Paul Coffey. There may not have been a smoother skater in the NHL than Paul Coffey, especially not among defensemen. …
• Yvan Cournoyer. …
• Sergei Fedorov. …
• Mike Gartner. …
• Bobby Hull. …
• Connor McDavid. …
• Scott Niedermayer.
## How fast is an average NHL player?
Most professional hockey players are capable of reaching the 20 to 30km/h (12 to 20mph) range. Most recreational players will be below the speeds of professional hockey players.
## How much do you run in an average field hockey game?
Field Hockey: 5.6 miles.
## How much do midfielders run in a game?
Central midfielders (roughly 11 km) and full-backs (roughly between 10 and 11 km) tend to run the most in a football match, which is no real surprise given the nature of their positions in the modern game.
## How fast is the fastest skater?
The Guinness Book of World Records shows that Russian speed skater Pavel Kulizhnikov holds the men’s record for fastest speed skating along 500 meters at 33.61 seconds, which meant he averaged 53.56 km/h (33.28 mph) along the route.
THIS IS FUN: Question: What is the science behind field hockey?
## How fast do speed skaters skate?
So, how fast do speed skaters go? Short track speed skaters on ice go about 31mph (50kph) but can reach speeds in excess of 35 mph (56kph). The fastest ice speed skater reached 56.5mph (91kph), whereas the fastest downhill inline skater reached 77mph (124 kph).
## Do NHL players buy their own skates?
NHL players do not pay for their own equipment. … This is because hockey players are creatures of comfort & superstition, and therefore rarely switch up their gear. Some players will even use the same pair of shoulder pads for years without requiring a new pair.
## Do NHL players shower between periods?
Some players will take a shower during the intermission to feel revitalized for the upcoming period. … So, that’s what NHL players do in between periods, while fans too have their own habits during this time….
## Do NHL players pay for their own skates?
Do NHL players pay for their own gear? NHL players do not pay for sticks. … Even if certain players are sponsored by a brand, the team still has to buy the sticks from the brand. Some teams pay \$300,000 a season to provide sticks for their players. | 897 | 3,809 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2022-27 | latest | en | 0.956744 |
https://estebantorreshighschool.com/faq-about-equations/thermal-efficiency-equation.html | 1,618,837,320,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038879374.66/warc/CC-MAIN-20210419111510-20210419141510-00251.warc.gz | 348,954,702 | 30,672 | ## How do you calculate thermal efficiency?
The thermal efficiency is a dimensionless unit. If you take formulas in thermodynamics, you can get that η= 1- (T COLD/THOT). TCOLD usually represents the ambient temperature where the engine is located. This may just be the temperature outside in whatever environment the engine is at.
## What is the thermal efficiency of the engine?
Most internal combustion engines are only 20 percent thermally efficient, according to Green Car Reports. In addition to heat, the various systems required to run the engine all take energy that could potentially be put to use propelling the vehicle.
## What is the formula of efficiency?
Efficiency is often measured as the ratio of useful output to total input, which can be expressed with the mathematical formula r=P/C, where P is the amount of useful output (“product”) produced per the amount C (“cost”) of resources consumed.
## What is brake thermal efficiency?
Brake. thermal efficiency is the ratio of energy in the brake power to. the fuel energy. Figure A shows the entropy (S) , pressure (P) and volume (V) diagrams of idealized diesel cycle.
## Why is 100 Efficiency impossible?
It is impossible for heat engines to achieve 100% thermal efficiency () according to the Second law of thermodynamics. This is impossible because some waste heat is always produced produced in a heat engine, shown in Figure 1 by the term.
## What are the two factors of thermal efficiency?
In theory, the factors that affect thermal efficiency (the performance of an engine or the work it has done in relation to the value of heat supplied to it) are the heat input and the heat output. Additionally, temperature variations also come in play, the amount and the rate of losing or gaining heat.
## Which engine is more efficient?
Diesel engines generally achieve greater fuel efficiency than petrol (gasoline) engines. Passenger car diesel engines have energy efficiency of up to 41% but more typically 30%, and petrol engines of up to 37.3%, but more typically 20%.
## Which is the most efficient cycle?
Classical thermodynamics indicates that the most efficient thermodynamic cycle operating between two heat reservoirs is the Carnot engine [1] , and a basic theorem expresses that any reversible cycle working between two constant temperature levels should have the same efficiency as a Carnot cycle [2].
## Is Carnot engine 100 efficient?
A Carnot engine operating between two given temperatures has the greatest possible efficiency of any heat engine operating between these two temperatures. Furthermore, all engines employing only reversible processes have this same maximum efficiency when operating between the same given temperatures.
## What is work formula?
Work is done when a force that is applied to an object moves that object. The work is calculated by multiplying the force by the amount of movement of an object (W = F * d).
## How do we calculate energy?
In classical mechanics, kinetic energy (KE) is equal to half of an object’s mass (1/2*m) multiplied by the velocity squared. For example, if a an object with a mass of 10 kg (m = 10 kg) is moving at a velocity of 5 meters per second (v = 5 m/s), the kinetic energy is equal to 125 Joules, or (1/2 * 10 kg) * 5 m/s2.
## How do you increase efficiency?
Here are the top 10 things you can do to increase employee efficiency at the office.Don’t be Afraid to Delegate. Match Tasks to Skills. Communicate Effectively. Keep Goals Clear & Focused. Incentivize Employees. Cut Out the Excess. Train and Develop Employees. Embrace Telecommuting.
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## Why is thermal efficiency important?
The importance of thermal efficiency cannot be overstated. Improving the thermal performance of your building helps reduce energy usage and costs. You can help reduce heating and cooling loads by providing superior thermal performance windows and doors and framing systems.
## What is the formula for volumetric efficiency?
The volumetric efficiency ηv [-] is defined as the ratio between the actual (measured) volume of intake air Va [m3] drawn into the cylinder/engine and the theoretical volume of the engine/cylinder Vd [m3], during the intake engine cycle.
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#### Wolfram alpha differential equation solver
Can Wolfram Alpha solve differential equations? A differential equation is an equation involving a function and its derivatives. Wolfram|Alpha can solve many problems under this important branch of mathematics, including solving ODEs, finding an ODE a function satisfies and solving an ODE using a slew of numerical methods. How do you solve a differential equation […] | 1,056 | 5,115 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2021-17 | longest | en | 0.935235 |
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# Estimation of Iron (II) and Iron (III) concentration
Extracts from this document...
Introduction
The Estimation of Iron(II) and Iron(III) in a Mixture Containing Both Introduction - theory behind two methods and why I am choosing one of them There are several possible methods that could be employed to determine the concentration of Iron(II) and Iron(III) ions in solution: one of these is colorimetry. Colorimetry is the technique of using the depth of colour of a substance to measure its concentration. We use a colorimeter (see fig. 1 and 2) to measure the depth of colour. The machine is calibrated by checking a series of different known concentrations of solution; from the readings we construct a graph of absorbance (the percentage of light absorbed by the sample) against concentration - this is known as a calibration curve. We can then read off the graph the concentration value that goes with the absorbance value for our unknown solution. Figure 1: A colorimetry set up Figure 2: A colorimeter The machine may be set by the wavelength of the light involved. ...read more.
Middle
Another possible method would be a redox (reduction-oxidation) titration. Acidified potassium manganate(VII) is a popular standard solution for use in redox titrations and has the added benefit that it oxidises Fe2+ to Fe3+: MnO4-(aq) + 5Fe2+(aq) + 8H+(aq) --> Mn2+(aq) + 5Fe3+(aq) + 4H2O(l) The solution needs to be well acidified to provide the H+ ions shown in the equation. 2M sulphuric acid could be used to do this. The acid also inhibits the oxidation of Fe2+ to Fe3+ by the air. Without sufficient acid alternative reactions take place and the link between the MnO4- and Fe2+ will be lost. This titrant is self-indicating: a separate indicator is not needed. The manganate(VII) ion is bright violet, but it is reduced to the virtually colourless manganese(II) ion (it is actually very pale pink but looks colourless). When the iron(II) is finally used up, the manganate(VII) is no longer reduced, so the purple colour remains. ...read more.
Conclusion
Different metals would have different electrode potentials. More reactive metals ionise more easily and so they should leave more electrons on the metal before they reach equilibrium. In other words, the more negative the metal is, the more reactive it should be. If we could measure the potential difference between metal and solution we could use it to construct a table of reactivity from the different voltage readings. However, to measure this potential difference we would have to put a metal electrode into the solution and this would have its own electrode potential. A platinum electrode would be placed in a solution which is both 1 mol dm-3 in iron(II) ions and 1 mol dm-3 in iron(III) ions. Iron(II) has the lowest oxidation number and appears nearest the electrode: Fe3+(aq) + e- Fe2+(aq) Cyanide can be used to test if it is completely reduced. Calculations - justify each calculation Safety Table - remember concentrations Apparatus - put in table KMnO4 (0.0025M) Mixture containing unknown concentrations of Fe3+/Fe2+ ions in solution Ferrozine (0.2%) 1M Sulphuric Acid 0. ...read more.
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# Related AS and A Level Inorganic Chemistry essays
1. ## Determining the concentration of acid in a given solution
5 star(s)
I could make up my solution by weighing the sodium carbonate in the volumetric flask as this would mean I wouldn't have to move the solid, but it would be hard to get the exact amount as I couldn't really get to the solid once it was in, to adjust
2. ## effects Concentration and Temperature on the Rate of Reaction
Time Taken When Stirred by Hand (seconds) Time Taken When Stirred by Magnetic Stirrer (seconds) Time Taken When Left Without Stirring (seconds) 0.003 76.0 67.5 60.5 0.004 51.5 52.0 53.5 0.005 41.5 45.5 39.5 0.006 36.0 40.5 44.5 0.008 33.0 29.5 33.5 0.01 31.5 24.0 25.0 From my results a reliable trend can be seen fro the
1. ## Bleaching experiment. Estimation of available chlorine in commercial bleaching solution.
( so that I2 will not be released from starch at the end point), the starch solution should be added at the later stage of the titration (when the solution just turns from brown to pale yellow). After the addition of starch, the mixture turns deep blue.
2. ## Chem Lab report. Objective: To determine the concentration of potassium manganate(VII) solution using ...
Before the titration, sulphuric(VI) acid was added to the ethanedioic acid. It was because manganate(VII) ion was reduced only at acidic medium. To be more accurate, manganate(VII) ion needed H+ ions for reaction. Ethanedioic acid was a weak acid which was only slightly ionized, it cannot provide enough H+ ions for the reaction.
1. ## Determination of the formula of hydrated Iron (II) Sulphate crystals (FeSO4xH2O)
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1. ## Determination of the formula of complex ion
/ 3 = 24.9 dm3 No. of moles of EDTA used = (24.9/1000) x 0.1 = 2.49x10-3 mol No. of moles of Ni2+ used = (25.0/1000)
2. ## Planning an investigation to determine the reactivity of HalogenoAlkanes
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Magnetism Formula
• Last Updated : 08 Mar, 2022
Magnetism is a scientific process that is induced by an electromagnetic field. Magnetic fields are generated by the magnetic moments and electric currents of elementary particles. Magnetic fields have a strong attraction for ferromagnetic materials. As a result, they can be easily attracted towards permanent magnets. Cobalt, nickel, iron, and their alloys are examples of ferromagnetic compounds. Temperature, as well as other factors such as magnetic field and pressure, influence a material’s magnetic state.
What is Magnetic Field?
A magnetic field is formed around a wire when an electric current is passed through it. This magnetic field also creates concentric rings around the wire. Furthermore, the magnetic field’s direction is determined by the current’s direction.
Furthermore, we may use the ‘right-hand rule,’ which involves pointing the right-hand thumb towards the direction of the current flow. Furthermore, the magnetic field aligns up with the orientation of your folded fingers. The magnitude of the field is determined by the quantity of current and distance from the wire.
Formula
The magnetic effect of an electric field is the property that produces an electric field around the conductor. Prof. H.C Oersted established this phenomenon for the first time in 1820. When he maintained the magnetic needle adjacent to a current-carrying wire, he saw deflection. Ampere’s law indicates the direction in which the needle deflects. The magnetic field’s strength is given by the following formula:
B =
where,
• μ0 depicts the allowance of free space (μ0 ​= 4π×10−7T.m/A).
• I refers to the magnitude of current in amperes.
• d and l refer to distance and wire’s length respectively.
Sample Questions
Question 1. Does the emf in a wire in a magnetic field depend upon the resistance of the loop? If not, why?
Solution:
Since, emf =
Here, Ï• = B x A
As a result, the resistance of the loop does not concern the magnitude of the emf produced. Only the area or size of the loop influences it.
Question 2. The strength of a uniform magnetic field through a 2 m long wire with 7 A current is 3.5 T. Find the force on the wire if it makes an angle of 30 degrees.
Solution:
We know, F = ILBsinθ
Here, I = 7 A, L = 2 m, B = 3.5 T, θ = 30°
⇒ F = (7 A)(2 m)(3.5 T)(sin30°)
⇒ F = 24.5 kg m/s2
Question 3. Consider a loop of wire whose surface is tangential to your vantage point. A constant current flows through the loop in a clockwise direction. What would happen if a magnetic field pointing in your direction were activated?
Solution:
This can be answered using the concept of induction. The current flowing through the loop will adjust to counteract a change in magnetic field. Initially, the loop’s magnetic field is pointing in a direction away from the viewer. As a result, if the external magnetic field is unexpectedly enabled in the direction opposite (towards the spectator), the current in the circuit will reverse this change and boost while staying right to left.
Question 4. While a consistent magnetic flux points into the page, an electron moves at a constant speed to the right somewhere along the plane of the page. Find the direction of the force on it.
Solution:
This question necessitates the use of the right-hand thumb rule. Point your right hand’s fingers in the position of the electron’s velocity (to the right). Turn your thumb in the magnetic field’s direction (into the page). Your palm ought to be approaching the force acting on the electron. However, because electrons are negative, this direction must be overturned, implying that the force is directed upwards through the plane of the page.
Question 5. The strength of a uniform magnetic field through a 0.5 m long wire with 3 A current is 8 T. Find the force on the wire if it makes an angle of 30 degrees.
Solution:
We know, F = ILBsinθ
Here, I = 4 A, L = 0.5 m, B = 8 T, θ = 30°
⇒ F = (4 A)(0.5 m)(8 T)(sin30°)
⇒ F = 8 kg m/s2
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Always except for n=1. var(Xbar) = var(X)/n < var(X), n>1. Take sqrt, a monotone fn.
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Best answer: Since n=720 is large enough, use the normal approximation to the binomial. The 90% confidence interval for the true proportion P based on a sample estimate p=97/720 = 0.1347 is given by p ± z(α/2)*√[p(1-p)/n], where z(α/2) = z(0.05) = 1.645, n = 720. Upper limit =0.1347+0.0209=0.1556, or 15.56%
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Best answer: c) Sample size n=6 small, unknown pop variance; assume normal population. Use t-distribution, ν=5 xbar ± t(α/2, ν)*(s/√n) xbar=5, s/√n = 1, t(α/2, ν)=2.571 t(α/2, ν) is the value of t when the area on its right of t-curve, df ν, is α/2 and for CI of (1-α)100%.
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• Z Score Question Could Canyone Help?
X~N(170, 30^2) P{170<X<225}=P{0<Z<z}, z=(225-170)/30 = 1.667 P=0.9525-0.50=0.4525, about 45% of them.
X~N(170, 30^2) P{170<X<225}=P{0<Z<z}, z=(225-170)/30 = 1.667 P=0.9525-0.50=0.4525, about 45% of them.
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Can we do like this? If sample size is n, total = 8n New total = (8n+6n)/2 New mean = new total / n = 7
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X~N(98.23, 0.61^2) Proportion having fever P{X>100.6}=P{Z>z} where z=(100.6-98.23)/0.61=3.89 P=P{Z>3.89}=P{Z<-3.89}=0.000, almost none. For 5% having fever P{X>x}=P{Z>z}=0.05, z=1.645 (x-98.23)/0.61 = 1.645, x=99.23 F.
1 answer · Mathematics · 6 years ago
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Best answer: 3. Poisson problem 0.33 per second = 19.8 per minute (7:15 and 7:16) Poisson λ=19.8 (take 20) P{X=30}=...
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• A police department reports that the probabilites that 0,1,2, and 3 burglaries will be reported in a given...?
0,1,2, and 3 0.49,0.38,0.10, and 0.03 (sum =1) E(X)=0(0.49)+1(0.38) + ... E(X^2)=0^2(0.49)+1^2(0.38) + ... var(X)=E(X^2) - E(X)*E(X) take sqrt
0,1,2, and 3 0.49,0.38,0.10, and 0.03 (sum =1) E(X)=0(0.49)+1(0.38) + ... E(X^2)=0^2(0.49)+1^2(0.38) + ... var(X)=E(X^2) - E(X)*E(X) take sqrt
1 answer · Mathematics · 6 years ago
• Don't understand t-test, help?
If you are given significance level, α Determine the form of critical region: {|T|>t} for two-tailed {T>t} for one-tailed right {T<t} for one-tailed left Get the p-value accordingly: P{|T|>t}, P{T>t}, or P{T<t} Compare p-value with α, if p-value < α, reject H0 at α, else do not reject H0 at α. (Reject H0 at α means... show more
If you are given significance level, α Determine the form of critical region: {|T|>t} for two-tailed {T>t} for one-tailed right {T<t} for one-tailed left Get the p-value accordingly: P{|T|>t}, P{T>t}, or P{T<t} Compare p-value with α, if p-value < α, reject H0 at α, else do not reject H0 at α. (Reject H0 at α means significant at α)
3 answers · Mathematics · 6 years ago
• Statistics: Test Statistic and P Value?
Best answer: For n=19, 29 need to assume normal population. z=(4.5-0)/(23/√n) Based on H1 critical region is of the form {Z>z} p-value=P{Z>z} n=19, z=0.85, p-value=P{Z>0.85}=P{Z<-0.85}=0.20. (note symmetry) n=49, z=1.37, ... =0.085
Best answer: For n=19, 29 need to assume normal population. z=(4.5-0)/(23/√n) Based on H1 critical region is of the form {Z>z} p-value=P{Z>z} n=19, z=0.85, p-value=P{Z>0.85}=P{Z<-0.85}=0.20. (note symmetry) n=49, z=1.37, ... =0.085
2 answers · Mathematics · 6 years ago | 4,039 | 11,427 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2018-43 | latest | en | 0.771267 |
http://nrich.maths.org/public/leg.php?code=97&cl=3&cldcmpid=5011 | 1,503,476,411,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886117911.49/warc/CC-MAIN-20170823074634-20170823094634-00045.warc.gz | 311,200,103 | 10,081 | # Search by Topic
#### Resources tagged with Circles similar to Pied Piper:
Filter by: Content type:
Stage:
Challenge level:
### There are 62 results
Broad Topics > 2D Geometry, Shape and Space > Circles
### Intersecting Circles
##### Stage: 3 Challenge Level:
Three circles have a maximum of six intersections with each other. What is the maximum number of intersections that a hundred circles could have?
### Curvy Areas
##### Stage: 4 Challenge Level:
Have a go at creating these images based on circles. What do you notice about the areas of the different sections?
### Polycircles
##### Stage: 4 Challenge Level:
Show that for any triangle it is always possible to construct 3 touching circles with centres at the vertices. Is it possible to construct touching circles centred at the vertices of any polygon?
### An Unusual Shape
##### Stage: 3 Challenge Level:
Can you maximise the area available to a grazing goat?
### Square Pegs
##### Stage: 3 Challenge Level:
Which is a better fit, a square peg in a round hole or a round peg in a square hole?
### Rolling Around
##### Stage: 3 Challenge Level:
A circle rolls around the outside edge of a square so that its circumference always touches the edge of the square. Can you describe the locus of the centre of the circle?
##### Stage: 2, 3 and 4 Challenge Level:
A metal puzzle which led to some mathematical questions.
### Bull's Eye
##### Stage: 3 Challenge Level:
What fractions of the largest circle are the two shaded regions?
### Squaring the Circle
##### Stage: 3 Challenge Level:
Bluey-green, white and transparent squares with a few odd bits of shapes around the perimeter. But, how many squares are there of each type in the complete circle? Study the picture and make. . . .
### LOGO Challenge - Circles as Animals
##### Stage: 3 and 4 Challenge Level:
See if you can anticipate successive 'generations' of the two animals shown here.
### F'arc'tion
##### Stage: 3 Challenge Level:
At the corner of the cube circular arcs are drawn and the area enclosed shaded. What fraction of the surface area of the cube is shaded? Try working out the answer without recourse to pencil and. . . .
### Blue and White
##### Stage: 3 Challenge Level:
Identical squares of side one unit contain some circles shaded blue. In which of the four examples is the shaded area greatest?
### Efficient Cutting
##### Stage: 3 Challenge Level:
Use a single sheet of A4 paper and make a cylinder having the greatest possible volume. The cylinder must be closed off by a circle at each end.
### Salinon
##### Stage: 4 Challenge Level:
This shape comprises four semi-circles. What is the relationship between the area of the shaded region and the area of the circle on AB as diameter?
### The Pi Are Square
##### Stage: 3 Challenge Level:
A circle with the radius of 2.2 centimetres is drawn touching the sides of a square. What area of the square is NOT covered by the circle?
### Circles in Quadrilaterals
##### Stage: 4 Challenge Level:
Explore when it is possible to construct a circle which just touches all four sides of a quadrilateral.
### Like a Circle in a Spiral
##### Stage: 2, 3 and 4 Challenge Level:
A cheap and simple toy with lots of mathematics. Can you interpret the images that are produced? Can you predict the pattern that will be produced using different wheels?
### Squaring the Circle and Circling the Square
##### Stage: 4 Challenge Level:
If you continue the pattern, can you predict what each of the following areas will be? Try to explain your prediction.
### Pie Cuts
##### Stage: 3 Challenge Level:
Investigate the different ways of cutting a perfectly circular pie into equal pieces using exactly 3 cuts. The cuts have to be along chords of the circle (which might be diameters).
### Illusion
##### Stage: 3 and 4 Challenge Level:
A security camera, taking pictures each half a second, films a cyclist going by. In the film, the cyclist appears to go forward while the wheels appear to go backwards. Why?
### Semi-detached
##### Stage: 4 Challenge Level:
A square of area 40 square cms is inscribed in a semicircle. Find the area of the square that could be inscribed in a circle of the same radius.
### Efficient Packing
##### Stage: 4 Challenge Level:
How efficiently can you pack together disks?
### Circle Packing
##### Stage: 4 Challenge Level:
Equal circles can be arranged so that each circle touches four or six others. What percentage of the plane is covered by circles in each packing pattern? ...
### Partly Circles
##### Stage: 4 Challenge Level:
What is the same and what is different about these circle questions? What connections can you make?
### Round and Round
##### Stage: 4 Challenge Level:
Prove that the shaded area of the semicircle is equal to the area of the inner circle.
### Floored
##### Stage: 3 Challenge Level:
A floor is covered by a tessellation of equilateral triangles, each having three equal arcs inside it. What proportion of the area of the tessellation is shaded?
### The Pillar of Chios
##### Stage: 3 Challenge Level:
Semicircles are drawn on the sides of a rectangle ABCD. A circle passing through points ABCD carves out four crescent-shaped regions. Prove that the sum of the areas of the four crescents is equal to. . . .
### Pericut
##### Stage: 4 and 5 Challenge Level:
Two semicircle sit on the diameter of a semicircle centre O of twice their radius. Lines through O divide the perimeter into two parts. What can you say about the lengths of these two parts?
### Tied Up
##### Stage: 3 Challenge Level:
In a right angled triangular field, three animals are tethered to posts at the midpoint of each side. Each rope is just long enough to allow the animal to reach two adjacent vertices. Only one animal. . . .
##### Stage: 4 Challenge Level:
The sides of a triangle are 25, 39 and 40 units of length. Find the diameter of the circumscribed circle.
### Arclets Explained
##### Stage: 3 and 4
This article gives an wonderful insight into students working on the Arclets problem that first appeared in the Sept 2002 edition of the NRICH website.
### LOGO Challenge 11 - More on Circles
##### Stage: 3 and 4 Challenge Level:
Thinking of circles as polygons with an infinite number of sides - but how does this help us with our understanding of the circumference of circle as pi x d? This challenge investigates. . . .
### A Rational Search
##### Stage: 4 and 5 Challenge Level:
Investigate constructible images which contain rational areas.
### Witch's Hat
##### Stage: 3 and 4 Challenge Level:
What shapes should Elly cut out to make a witch's hat? How can she make a taller hat?
### Gym Bag
##### Stage: 3 and 4 Challenge Level:
Can Jo make a gym bag for her trainers from the piece of fabric she has?
### Track Design
##### Stage: 4 Challenge Level:
Where should runners start the 200m race so that they have all run the same distance by the finish?
### Three Tears
##### Stage: 4 Challenge Level:
Construct this design using only compasses
### First Forward Into Logo 4: Circles
##### Stage: 2, 3 and 4 Challenge Level:
Learn how to draw circles using Logo. Wait a minute! Are they really circles? If not what are they?
### LOGO Challenge 12 - Concentric Circles
##### Stage: 3 and 4 Challenge Level:
Can you reproduce the design comprising a series of concentric circles? Test your understanding of the realtionship betwwn the circumference and diameter of a circle.
### What Is the Circle Scribe Disk Compass?
##### Stage: 3 and 4
Introducing a geometrical instrument with 3 basic capabilities.
### LOGO Challenge 10 - Circles
##### Stage: 3 and 4 Challenge Level:
In LOGO circles can be described in terms of polygons with an infinite (in this case large number) of sides - investigate this definition further.
### Get Cross
##### Stage: 4 Challenge Level:
A white cross is placed symmetrically in a red disc with the central square of side length sqrt 2 and the arms of the cross of length 1 unit. What is the area of the disc still showing?
### Coins on a Plate
##### Stage: 3 Challenge Level:
Points A, B and C are the centres of three circles, each one of which touches the other two. Prove that the perimeter of the triangle ABC is equal to the diameter of the largest circle.
### Not So Little X
##### Stage: 3 Challenge Level:
Two circles are enclosed by a rectangle 12 units by x units. The distance between the centres of the two circles is x/3 units. How big is x?
### A Chordingly
##### Stage: 3 Challenge Level:
Find the area of the annulus in terms of the length of the chord which is tangent to the inner circle.
##### Stage: 4 Challenge Level:
Given a square ABCD of sides 10 cm, and using the corners as centres, construct four quadrants with radius 10 cm each inside the square. The four arcs intersect at P, Q, R and S. Find the. . . .
### Circumspection
##### Stage: 4 Challenge Level:
M is any point on the line AB. Squares of side length AM and MB are constructed and their circumcircles intersect at P (and M). Prove that the lines AD and BE produced pass through P.
### Rolling Coins
##### Stage: 4 Challenge Level:
A blue coin rolls round two yellow coins which touch. The coins are the same size. How many revolutions does the blue coin make when it rolls all the way round the yellow coins? Investigate for a. . . .
### Some(?) of the Parts
##### Stage: 4 Challenge Level:
A circle touches the lines OA, OB and AB where OA and OB are perpendicular. Show that the diameter of the circle is equal to the perimeter of the triangle
### Rotating Triangle
##### Stage: 3 and 4 Challenge Level:
What happens to the perimeter of triangle ABC as the two smaller circles change size and roll around inside the bigger circle? | 2,228 | 9,812 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-34 | latest | en | 0.846354 |
https://www.mql5.com/en/forum/136147 | 1,519,327,912,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891814249.56/warc/CC-MAIN-20180222180516-20180222200516-00044.warc.gz | 889,771,007 | 9,613 | # How to bring Previous indicator data from moving average indicator to EA?
214
i am using EA, and i added Moving Average Indicator, now the applied price shows only PRICE_CLOSE to PRICE_WEIGHTED, but "Previous Indicator's data" is not available in the "stdlib". but it is available in metatrader.
How to bring this to EA??...can anyone help me sort this out??....thanks in advance....
3245
17094
kmnatarajan:
i am using EA, and i added Moving Average Indicator, now the applied price shows only PRICE_CLOSE to PRICE_WEIGHTED, but "Previous Indicator's data" is not available in the "stdlib". but it is available in metatrader.
iMA(..., 0) is the MA's current value.
iMA(..., 1) is the MA's value on the previous bar.
6433
This is the average of the averages. So there has first been a calculation to get an average
On every bar there was a calculating to get a MA value
and then you take the average for let say 10 calculated MA values (or whatever you want)
Take the SUM of those and split it to 10 .......
3245
"Previous Indicator's data" = iMAOnArray()
214
hmm.....am sorry mates.....i think i didnt explain myself correctly.... I need to use 2 Moving Average indicators, the first one taking up the data from the market value, and the latter using the former Moving Average's data....hope it explains...again thanks in advance....
214
3245
maybe i didn't explain myself clear the first MA u use iMA() & the second MA u use iMAOnArray()
Moderator
18056
kmnatarajan:
hmm.....am sorry mates.....i think i didnt explain myself correctly.... I need to use 2 Moving Average indicators, the first one taking up the data from the market value, and the latter using the former Moving Average's data....hope it explains...again thanks in advance....
Store the MA result in an array, then you can get the MA of the data in the array using iMAOnArray . . as has been said already . .
214
Do you mean this iMAOnArray function will calculate the average of the Moving Average and store in it??.....then how do i use its values??....can someone post a code snippet pls??....am a learner....i dunno much about EA....thanks anyway for ur efforts dudes.....
3245
``` for(int i=0; i<limit; i++)
First_MA[i] = iMA(NULL,0,20,0,MODE_SMA,PRICE_CLOSE,i);
for(i=0; i<limit; i++)
Second_MA[i] = iMAOnArray(First_MA,0,14,0,MODE_SMA,i);
``` | 592 | 2,347 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-09 | latest | en | 0.890476 |
http://mathlanding.org/browse/standard/CCSSM/S2366908?page=7 | 1,563,332,649,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195525009.36/warc/CC-MAIN-20190717021428-20190717043428-00145.warc.gz | 101,652,502 | 10,532 | ## Common Core State Standards for Mathematics
Grade(s) K-12 resources related to the following standard:
Standards for Mathematical Practice
Construct viable arguments and critique the reasoning of others.
Showing 71-80 of 370 resultsResults per page: 10515202550 Sort by: TitleRatingNewest First
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This activity helps students become familiar with positional words and develop systematic thinking. Students use the clues provided to arrange six colored squares in an array, either on paper or with the interactive Flash applet that is provided. The Teachers' Notes page includes suggestions for implementation, discussion questions, ideas for extension and support and a printable sheet (doc). Students may be asked to create a similar problem for others to solve.
Activity, Interactive Media
Grade Level: K, 1
(1 Comments)
This 2-lesson unit focuses on combinations, a subject related to probability. Students develop strategies for discovering all the possible combinations in two given situations. They learn to collect and organize data and then use the results to generalize methods for determining possible combinations. They discuss how the number of possible outcomes is affected by decisions about the order of choices, or whether choices may be repeated. The unit includes student activity sheets, questions and extensions for students, and a link to an interactive applet.
Activity, Interactive Media, Lesson Plans, Unit of Instruction
Grade Level: 3, 4, 5
(1 Comments)
This brief article describes the components of high-quality mathematical discourse that engages students and fosters conceptual understanding. The author discusses the interrelationship of the CCSS Mathematical Practices and provides a list of references.
Reference Materials, Article
Grade Level: K, 1, 2, 3, 4, 5
Resource is part of a PD collection
(0 Comments)
This series of nine articles interprets and illustrates each of the eight Mathematical Practices of the Common Core State Standards as they might be exemplified in grades K-5. The final article sheds light on how curriculum needs to connect the Practices with the Content Standards.
Reference Materials, Article
Grade Level: K, 1, 2, 3, 4, 5
Resource is part of a PD collection
(0 Comments)
This four-part E-Example from Principles and Standards for School Mathematics of the NCTM standards highlights how mathematical games can foster mathematical communication as students explain and justify their moves to one another in a fraction game. In addition, a video is included that demonstrates students engaged in thinking about and applying their mathematical concepts and skills. The game "Playing Fraction Tracks" is cataloged separately.
Activity, Game, Interactive Media
Grade Level: 3, 4, 5
Resource is part of a PD collection
(0 Comments)
This 8-page monograph discusses the importance of developing student communication in mathematics, both oral and written, and the elements that make communication effective. It describes three approaches for organizing and facilitating students sharing their thinking about problem solving: Gallery Walk, Math Congress, and Bansho (Board Writing). The author provides tips for getting started with these strategies and a list of references.
Instructional Strategy, Reference Materials, Article
Grade Level: 1, 2, 3, 4, 5
Resource is part of a PD collection
(0 Comments)
This professional development video clip shows students engaged in the Common Core Practice Standard #3-Construct viable arguments for conclusions reached and critique the reasoning of others. Students work with geoboards to explore the concept of "halves" which requires them to explain their reasoning and justify their findings to their classmates. Additional resources include a video transcript, teaching tips, and a link to a professional development reflection activity based upon the video. A related clip (cataloged separately) shows the same exploration by the same students but the video is focused on Common Core Practice Standard #6—Attend to precision is evident.
Instructional Strategy, Video
Grade Level: 3, 4, 5
(0 Comments)
This professional development video clip shows students in a bilingual classroom engaged in the Common Core Practice Standard #3-Construct viable arguments for conclusions reached and critique the reasoning of others. Students observe ladybugs and collect data in order to answer quantitative questions-What is the the number of antennae, heads, mouths, feet, and wings. The individual data is compiled into a class chart and the learners compare and discuss data as a class. Additional resources include a video transcript, teaching tips, and a link to a professional development reflection activity based upon the video.
Instructional Strategy, Video
Grade Level: 1
(0 Comments)
This professional development video clip shows students engaged in the Common Core Practice Standard #3—Construct viable arguments for conclusions reached and critique the reasoning of others. In this lesson, learners work in groups to discuss and revise their estimates of how many seeds are in a small pumpkin. Additional resources include a video transcript, teaching tips, and a link to a professional development reflection activity based upon the video.
Instructional Strategy, Video
Grade Level: 1, 2
(0 Comments)
This article discusses an inquiry practice that encourages students to develop their mathematical thinking through discussion. The author describes examples of students taking ownership of their learning when they are encouraged to justify their positions through sound mathematical reasoning. This practice builds a foundation for formal proofs.
Instructional Strategy, Reference Materials, Article
Grade Level: K, 1, 2, 3, 4, 5
Resource is part of a PD collection
(0 Comments) | 1,136 | 5,826 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-30 | longest | en | 0.861243 |
https://www.pearson.com/store/en-us/pearsonplus/p/9780136881155.html | 1,701,597,918,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100499.43/warc/CC-MAIN-20231203094028-20231203124028-00385.warc.gz | 1,042,060,158 | 27,302 | # Beginning & Intermediate Algebra, 7th edition
• Margaret L. Lial,
• John Hornsby,
• Terry McGinnis
• Search, highlight, and take notes
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## Overview
Part of the Lial Developmental Algebra Series, Beginning & Intermediate Algebra, 7th Edition uses a teacherly writing style and a careful blend of skills development and conceptual questions to meet the unique needs of every student. The author team takes advantage of experiences in the classroom and an editing eye to offer one of the most well-rounded series available. In this revision, the team retains their hallmark writing style, and provides new features and resources to optimize student preparedness and conceptual understanding. The Lial program provides students with the perfect balance of skills and concepts for a student-friendly approach to math. It offers a wealth of resources to support any classroom along with many resources to support underprepared students.
ISBN-13: 9780136881155
Subject: Developmental Math
Category: Beginning & Intermediate Algebra Combined
## Overview
• Study Skill 1: Using Your Math Text
• Study Skill 3: Taking Lecture Notes
• Study Skill 4: Completing Your Homework
• Study Skill 5: Using Study Cards
• Study Skill 6: Managing Your Time
• Study Skill 7: Reviewing a Chapter
• Study Skill 8: Taking Math Tests
• Study Skill 9: Analyzing Your Test Results
• Study Skill 10: Preparing for Your Math Final
1. Prealgebra Review
• R.1 Fractions
• R.2 Decimals and Percents
1. The Real Number System
• 1.1 Exponents, Order of Operations, and Inequality
• 1.2 Variables, Expressions, and Equations
• 1.3 Real Numbers and the Number Line
• 1.4 Adding and Subtracting Real Numbers
• 1.5 Multiplying and Dividing Real Numbers
• Summary Exercises: Performing Operations with Real Numbers
• 1.6 Properties of Real Numbers
• 1.7 Simplifying Expressions
• Chapter 1 Summary
• Chapter 1 Review Exercises
• Chapter 1 Mixed Review Exercises
• Chapter 1 Test
• Chapters R and 1 Cumulative Review Exercises
2. Linear Equations and Inequalities in One Variable
• 2.1 The Addition Property of Equality
• 2.2 The Multiplication Property of Equality
• 2.3 Solving Linear Equations Using Both Properties of Equality
• 2.4 Clearing Fractions and Decimals When Solving Linear Equations
• Summary Exercises: Applying Methods for Solving Linear Equations
• 2.5 Applications of Linear Equations
• 2.6 Formulas and Additional Applications from Geometry
• 2.7 Ratio, Proportion, and Percent
• 2.8 Further Applications of Linear Equations
• 2.9 Solving Linear Inequalities
• Chapter 2 Summary
• Chapter 2 Review Exercises
• Chapter 2 Mixed Review Exercises
• Chapter 2 Test
• Chapters R–2 Cumulative Review Exercises
3. Linear Equations in Two Variables
• 3.1 Linear Equations and Rectangular Coordinates
• 3.2 Graphing Linear Equations in Two Variables
• 3.3 The Slope of a Line
• 3.4 Slope-Intercept Form of a Linear Equation
• 3.5 Point-Slope Form of a Linear Equation and Modeling
• Summary Exercises: Applying Graphing and Equation-Writing Techniques for Lines
• Chapter 3 Summary
• Chapter 3 Review Exercises
• Chapter 3 Mixed Review Exercises
• Chapter 3 Test
• Chapters R–3 Cumulative Review Exercises
4. Exponents and Polynomials
• 4.1 The Product Rule and Power Rules for Exponents
• 4.2 Integer Exponents and the Quotient Rule
• Summary Exercises: Applying the Rules for Exponents
• 4.3 Scientific Notation
• 4.4 Adding, Subtracting, and Graphing Polynomials
• 4.5 Multiplying Polynomials
• 4.6 Special Products
• 4.7 Dividing Polynomials
• Chapter 4 Summary
• Chapter 4 Review Exercises
• Chapter 4 Mixed Review Exercises
• Chapter 4 Test
• Chapters R–4 Cumulative Review Exercises
5. Factoring and Applications
• 5.1 Greatest Common Factor; Factoring by Grouping
• 5.2 Factoring Trinomials
• 5.3 More on Factoring Trinomials
• 5.4 Special Factoring Techniques
• Summary Exercises: Recognizing and Applying Factoring Strategies
• 5.5 Solving Quadratic Equations Using the Zero-Factor Property
• 5.6 Applications of Quadratic Equations
• Chapter 5 Summary
• Chapter 5 Review Exercises
• Chapter 5 Mixed Review Exercises
• Chapter 5 Test
• Chapters R–5 Cumulative Review Exercises
6. Rational Expressions and Applications
• 6.1 The Fundamental Property of Rational Expressions
• 6.2 Multiplying and Dividing Rational Expressions
• 6.3 Least Common Denominators
• 6.4 Adding and Subtracting Rational Expressions
• 6.5 Complex Fractions
• 6.6 Solving Equations with Rational Expressions
• Summary Exercises: Simplifying Rational Expressions vs. Solving Rational Equations
• 6.7 Applications of Rational Expressions
• Chapter 6 Summary
• Chapter 6 Review Exercises
• Chapter 6 Mixed Review Exercises
• Chapter 6 Test
• Chapters R–6 Cumulative Review Exercises
7. Linear Equations, Graphs, and Systems
• 7.1 Review of Graphs and Slopes of Lines
• 7.2 Review of Equations of Lines; Linear Models
• 7.3 Solving Systems of Linear Equations by Graphing
• 7.4 Solving Systems of Linear Equations by Substitution
• 7.5 Solving Systems of Linear Equations by Elimination
• Summary Exercises: Applying Techniques for Solving Systems of Linear Equations
• 7.6 Systems of Linear Equations in Three Variables
• 7.7 Applications of Systems of Linear Equations
• Chapter 7 Summary
• Chapter 7 Review Exercises
• Chapter 7 Mixed Review Exercises
• Chapter 7 Test
• Chapters R–7 Cumulative Review Exercises
8. Inequalities and Absolute Value
• 8.1 Review of Linear Inequalities in One Variable
• 8.2 Set Operations and Compound Inequalities
• 8.3 Absolute Value Equations and Inequalities
• Summary Exercises: Solving Linear and Absolute Value Equations and Inequalities
• 8.4 Linear Inequalities and Systems in Two Variables
• Chapter 8 Summary
• Chapter 8 Review Exercises
• Chapter 8 Mixed Review Exercises
• Chapter 8 Test
• Chapters R–8 Cumulative Review Exercises
9. Relations and Functions
• 9.1 Introduction to Relations and Functions
• 9.2 Function Notation and Linear Functions
• 9.3 Polynomial Functions, Operations, and Composition
• 9.4 Variation
• Chapter 9 Summary
• Chapter 9 Review Exercises
• Chapter 9 Mixed Review Exercises
• Chapter 9 Test
• Chapters R–9 Cumulative Review Exercises
10. Roots, Radicals, and Root Functions
• 10.1 Radical Expressions and Graphs
• 10.2 Rational Exponents
• 10.3 Simplifying Radicals, the Distance Formula, and Circles
• 10.5 Multiplying and Dividing Radical Expressions
• Summary Exercises: Performing Operations with Radicals and Rational Exponents
• 10.6 Solving Equations with Radicals
• 10.7 Complex Numbers
• Chapter 10 Summary
• Chapter 10 Review Exercises
• Chapter 10 Mixed Review Exercises
• Chapter 10 Test
• Chapters R–10 Cumulative Review Exercises
11. Quadratic Equations, Inequalities, and Functions
• 11.1 Solving Quadratic Equations by the Square Root Property
• 11.2 Solving Quadratic Equations by Completing the Square
• Summary Exercises: Applying Methods for Solving Quadratic Equations
• 11.5 Formulas and Further Applications
• 11.6 Graphs of Quadratic Functions
• 11.7 More about Parabolas and Their Applications
• 11.8 Polynomial and Rational Inequalities
• Chapter 11 Summary
• Chapter 11 Review Exercises
• Chapter 11 Mixed Review Exercises
• Chapter 11 Test
• Chapters R–11 Cumulative Review Exercises
12. Inverse, Exponential, and Logarithmic Functions
• 12.1 Inverse Functions
• 12.2 Exponential Functions
• 12.3 Logarithmic Functions
• 12.4 Properties of Logarithms
• 12.5 Common and Natural Logarithms
• 12.6 Exponential and Logarithmic Equations; Further Applications
• Chapter 12 Summary
• Chapter 12 Review Exercises
• Chapter 12 Mixed Review Exercises
• Chapter 12 Test
• Chapters R–12 Cumulative Review Exercises
13. Nonlinear Functions, Conic Sections, and Nonlinear Systems
• 13.1 Additional Graphs of Functions
• 13.2 Circles Revisited and Ellipses
• 13.3 Hyperbolas and Functions Defined by Radicals
• 13.4 Nonlinear Systems of Equations
• 13.5 Second-Degree Inequalities and Systems of Inequalities
• Chapter 13 Summary
• Chapter 13 Review Exercises
• Chapter 13 Mixed Review Exercises
• Chapter 13 Test
• Chapters R—13 Cumulative Review Exercises
14. Further Topics in Algebra
• 14.1 Sequences and Series
• 14.2 Arithmetic Sequences
• 14.3 Geometric Sequences
• 14.4 The Binomial Theorem
• Chapter 14 Summary
• Chapter 14 Review Exercises
• Chapter 14 Mixed Review Exercises
• Chapter 14 Test
• Chapters R—14 Cumulative Review Exercises
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Channels is a video platform with thousands of explanations, solutions and practice problems to help you do homework and prep for exams. Videos are personalized to your course, and tutors walk you through solutions. Plus, interactive AI‑powered summaries and a social community help you better understand lessons from class. | 2,654 | 10,574 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-50 | latest | en | 0.82844 |
https://www.cprogramming.com/tutorial/computersciencetheory/huffman.html | 1,519,616,714,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891817999.51/warc/CC-MAIN-20180226025358-20180226045358-00479.warc.gz | 848,998,333 | 10,577 | # Huffman Encoding Compression Algorithm
The Huffman encoding algorithm is an optimal compression algorithm when only the frequency of individual letters are used to compress the data. (There are better algorithms that can use more structure of the file than just letter frequencies.) The idea behind the algorithm is that if you have some letters that are more frequent than others, it makes sense to use fewer bits to encode those letters than to encode the less frequent letters.
For instance, take the following phrase: "ADA ATE APPLE". There are 4 As, 1 D, 1 T, 2 Es, 2 Ps, 1 L, and 2 spaces. There are a few ways that might be reasonable ways of encoding this phrase using letter frequencies. First, notice that there are only a very few letters that show up here. It would be silly to use chars, with eights bits apiece, to encode each character of the string. In fact, given that there are only seven characters, we could get away with using three bits for each character! If we decided to simply take that route, that would require using 14 * 3 bits, for a total of 42 bits (and some extra padding for the sake of having correct bit-alignment since you have to use an entire byte). That's not too bad! It's a lot better than the 8 * 14 = 112 bits that would otherwise be required.
But we can do even better if we consider that if one character shows up many times, and several characters show up only a few times, then using one bit to encode one character and many bits to encode another might actually be useful if the character that uses many bits only shows up a small number of times!
## Prefix Property
To get away with doing this, we now need a way of knowing how to tell which encoding matches which letter. For instance, before, we knew that every three (or eight bits) was a boundary for a letter. Now, with different length encodings for different letters, we need to have some way of separating the words out. For instance, given the string 001011100, if we know that every letter is encoded with three bits, it's easy to break apart into 001 011 100. If some letters are encoded with one bit, and another with four, it's not as easy to know how to do it.
But we can use a trick commonly referred to as the "prefix property". The idea is that the encoding for any one character isn't a prefix for any other character. For instance, if A is encoded with 0, then no other character will be encoded with a zero at the front. That way, if we start reading a string of bits and the first bit is a zero, we know that we can stop reading, and we know that bit encodes an A because no other character encoding begins with a 0. In general, the idea is that if we have a full encoding for a character, then that encoding won't show up at the beginning of the encoding for any other character. This means that once we actually read a string of bits that match a particular character, we know that it must mean that that's the next character and we can start fresh from the next bit, looking for a new character encoding match.
Note that it is perfectly fine for the encoding for a character to show up in the middle of the encoding for another character because there's no way we'd mistake that as the encoding for another character so long as we start decoding from the first bit in the compressed file. (On the other hand, if we get off by a bit, this might cause some headaches.)
Let's take a look at how this might actually work using some simple encodings that all have the property that the encoding for one character doesn't show up at the beginning of the encoding for another character.
```char encoding
A 0
E 10
P 110
space 1110
D 11110
T 111110
L 111111
```
This is a pretty simple encoding, but it does match the prefix property -- the encoding for A doesn't show up at the front of any other encoding, nor does the encoding for B, and so forth. (Some of the encodings share the same prefix, but that's perfectly fine because we can always tell them apart at some point by reading in more bits.)
For instance, the original string we had could be encoded by
```011110011100111110101110011011011111110 (39 bits)
```
which we could break apart as
```0 11110 0 1110 0 111110 10 1110 0 110 110 111111 10
A D A Space A T E Space A P P L E
```
Notice that even using this somewhat simple approach to generating encodings that satisfy the prefix property, we managed to save 4 bits over the approach of using 3 bits per character. With even more unbalanced word frequencies, we could do even better.
A nice way of visualizing the process of decoding a file compressed with Huffman encoding is to think about the encoding as a binary tree, where each leaf node corresponds to a single character. At each inner node of the tree, if the next bit is a 0, move to the left node, otherwise move to the right node.
For instance, the prefix encoding used above would have a binary tree representation that looks like this -- the X's indicate inner nodes.
``` X
/ \
/ \
A X
/ \
/ \
E X
/ \
/ \
P X
/ \
/ \
Space X
/ \
/ \
D X
/ \
/ \
T L
```
To decode the string, all we do is follow the links of the tree until we hit a leaf node. Once a leaf node is reached, we output the character stored at the leaf and go back up to the root of the tree.
## The Huffman Algorithm
So far, we've gone over the basic principles we'll need for the Huffman algorithm, both for encoding and decoding, but we've had to guess at what would be the best way of actually encoding the characters. For our simple text string, it wasn't too hard to figure out a decent encoding that saved a few bits. But in the general case, it might be hard to figure out a good solution, let alone the best possible solution.
The Huffman algorithm is a so-called "greedy" approach to solving this problem in the sense that at each step, the algorithm chooses the best available option. It turns out that this is sufficient for finding the best encoding.
The basic idea behind the algorithm is to build the tree bottom-up. First, every letter starts off as part of its own tree and the trees are ordered by the frequency of the letters in the original string. Then the two least-frequently used letters are combined into a single tree, and the frequency of that tree is set to be the combined frequency of the two trees that it links together.
For instance, if we started out with two characters that showed up once, L and T, in our sample string, they would be recombined into a new tree that has a "supernode" that links to both L and T, and has a frequency of 2:
```
X, 2
/ \
/ \
L, 1 T, 1
```
This new tree is reinserted into the list of trees in its sorted position. The process is then repeated, treating trees with more than one element the same as any other trees except that their frequencies are the sum of the frequencies of all of the letters at the leaves. (This is just the sum of the left and right children of any node because each node stores the frequency information about its own children.) The process completes when all of the trees have been combined into a single tree -- this tree will describe a Huffman compression encoding.
Essentially, a tree is built from the bottom up -- we start out with 256 trees (for an ASCII file) -- and end up with a single tree with 256 leaves along with 255 internal nodes (one for each merging of two trees, which takes place 255 times). The tree has a few interesting properties -- the frequencies of all of the internal nodes combined together will give the total number of bits needed to write the encoded file (except the header). This property comes from the fact that at each internal node, a decision must be made to go left or right, and each internal node will be reached once for each time a character beneath it shows up in the text of the document.
To go from plain text to compressed text, you would have to do a traversal of the tree and store the path to reach each leaf node as a string of bits (0 for going left, 1 for going right) and associate that bit with the particular character at the leaf. Once this is done, converting a plain text file into a compressed file is just a matter of replacing each letter with an appropriate bit string and then handling the possibility of having some extra bits that need to be written (this is discussed more fully in the implementation notes). Notice that two different data structures likely need to be used here -- a list of trees, and those binary trees themselves. It might make sense to use several data structures such as:
```struct tree_t
{
tree_t *left;
tree_t *right;
char character;
};
```
to store the tree elements (note that if left and right are NULL, then we would know that the node is a leaf and that character stores a valid char of interest) and
```struct list_t
{
list_t *next;
int total_frequency;
tree_t *tree;
}
```
to store the list of trees that still need to be merged together as a linked list.
### Implementation Notes
The first thing to realize is that when writing a compressed file, you will need to include a Huffman header that stores the letter frequencies in the original file so that the Huffman tree can be rebuilt by whoever is decoding the file. This could be something as simple as
```struct huff_header
{
char letters[256];
};
```
and you could use fread to read in the entire header at once. It wouldn't hurt to include a "magic number" at the beginning of your header so that your program can tell whether the file was actually compressed by your program. (This prevents the confusion of wondering if the garbage output is a result of a garbage plain text file, or if the file being decompressed wasn't actually created from a plain text file.)
When implementing Huffman compression, remember that any one of many possible encodings may be valid, and the differences come about based on how you build up the tree. This means that both the compressor and decompressor will need to follow the same rules. In practice, it probably makes sense to use the same tree building code for both.
Files are written in chunks of eight bits, but it's unlikely that you will have a file that, when compressed, turns out to need an exact multiple of eight bits. You'll have to add some extra junk at the end of the last byte you write to the file, and when decoding, you'll need some way of knowing when you've finished reading the valid bits and started reading the bits that were added as packing (which may or may not encode a character). One way of doing this would be to keep track of the number of letters decompressed so far and stop decoding the file when that limit has been reached.
Related articles
Binary Trees | 2,403 | 10,742 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2018-09 | longest | en | 0.95085 |
https://gitlab.linphone.org/BC/public/external/decaf/commit/164342ebfdc78bf590665e3e91320f5fb4e2855d?view=parallel | 1,579,324,456,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250591763.20/warc/CC-MAIN-20200118023429-20200118051429-00169.warc.gz | 481,578,232 | 48,615 | Commit 164342eb by Michael Hamburg
### inverse elligator works, but at what cost?!?!!?
parent 17347b04
... @@ -53,6 +53,8 @@ def isqrt(x,exn=InvalidEncodingException("Not on curve")): ... @@ -53,6 +53,8 @@ def isqrt(x,exn=InvalidEncodingException("Not on curve")): if negative(s): s=-s if negative(s): s=-s return 1/s return 1/s def inv0(x): return 1/x if x != 0 else 0 def isqrt_i(x): def isqrt_i(x): """Return 1/sqrt(x) or 1/sqrt(zeta * x)""" """Return 1/sqrt(x) or 1/sqrt(zeta * x)""" if x==0: return True,0 if x==0: return True,0 ... @@ -325,8 +327,8 @@ class Decaf_1_1_Point(QuotientEdwardsPoint): ... @@ -325,8 +327,8 @@ class Decaf_1_1_Point(QuotientEdwardsPoint): den = x*y den = x*y isr = isqrt(num*(a-d)*den^2) isr = isqrt(num*(a-d)*den^2) iden = isr * den * self.isoMagic iden = isr * den * self.isoMagic # 1/sqrt((z+y)(z-y)) = 1/sqrt(1-Y^2) / z inum = isr * num inum = isr * num # sqrt(1-Y^2) * z / xysqrt(a-d) ~ 1/sqrt(1-ax^2)/z if negative(iden*inum*self.i*t^2*(d-a)) != toggle_rotation: if negative(iden*inum*self.i*t^2*(d-a)) != toggle_rotation: iden,inum = inum,iden iden,inum = inum,iden ... @@ -386,18 +388,29 @@ class Decaf_1_1_Point(QuotientEdwardsPoint): ... @@ -386,18 +388,29 @@ class Decaf_1_1_Point(QuotientEdwardsPoint): for toggle_r in [False,True]: for toggle_r in [False,True]: s,m1,m12,swap = self.toJacobiQuartic(toggle_rotation,toggle_altx,toggle_s) s,m1,m12,swap = self.toJacobiQuartic(toggle_rotation,toggle_altx,toggle_s) print #print print toggle_rotation,toggle_altx,toggle_s #print toggle_rotation,toggle_altx,toggle_s print m1 #print m1 print m12 #print m12 if self == self.__class__() and self.cofactor == 4: if self == self.__class__(): # Hacks for identity! if self.cofactor == 4: if toggle_altx: m12 = 1 # Hacks for identity! elif toggle_s: m1 = 1 if toggle_altx: m12 = 1 elif toggle_r: continue elif toggle_s: m1 = 1 ## BOTH??? elif toggle_r: continue ## BOTH??? else: m12 = 1 imi = self.isoMagic * self.i if toggle_rotation: if toggle_altx: m1 = -imi else: m1 = +imi else: if toggle_altx: m1 = 0 else: m1 = a-d rnum = (d*a*m12-m1) rnum = (d*a*m12-m1) rden = ((d*a-1)*m12+m1) rden = ((d*a-1)*m12+m1) ... @@ -406,7 +419,7 @@ class Decaf_1_1_Point(QuotientEdwardsPoint): ... @@ -406,7 +419,7 @@ class Decaf_1_1_Point(QuotientEdwardsPoint): ok,sr = isqrt_i(rnum*rden*self.qnr) ok,sr = isqrt_i(rnum*rden*self.qnr) if not ok: continue if not ok: continue sr *= rnum sr *= rnum print "Works! %d %x" % (swap,sr) #print "Works! %d %x" % (swap,sr) if negative(sr) != toggle_r: sr = -sr if negative(sr) != toggle_r: sr = -sr ret = self.gfToBytes(sr) ret = self.gfToBytes(sr) ... @@ -681,12 +694,12 @@ def test(cls,n): ... @@ -681,12 +694,12 @@ def test(cls,n): if Q1 + Q0 != Q2: raise TestFailedException("Scalarmul doesn't work") if Q1 + Q0 != Q2: raise TestFailedException("Scalarmul doesn't work") Q = Q1 Q = Q1 test(Ed25519Point,100) #test(Ed25519Point,100) test(NegEd25519Point,100) #test(NegEd25519Point,100) test(IsoEd25519Point,100) #test(IsoEd25519Point,100) test(IsoEd448Point,100) #test(IsoEd448Point,100) test(TwistedEd448GoldilocksPoint,100) #test(TwistedEd448GoldilocksPoint,100) test(Ed448GoldilocksPoint,100) #test(Ed448GoldilocksPoint,100) def testElligator(cls,n): def testElligator(cls,n): ... @@ -710,12 +723,12 @@ def testElligator(cls,n): ... @@ -710,12 +723,12 @@ def testElligator(cls,n): pass # TODO pass # TODO testElligator(Ed25519Point,100) #testElligator(Ed25519Point,100) testElligator(NegEd25519Point,100) #testElligator(NegEd25519Point,100) testElligator(IsoEd25519Point,100) #testElligator(IsoEd25519Point,100) testElligator(IsoEd448Point,100) #testElligator(IsoEd448Point,100) testElligator(Ed448GoldilocksPoint,100) #testElligator(Ed448GoldilocksPoint,100) testElligator(TwistedEd448GoldilocksPoint,100) #testElligator(TwistedEd448GoldilocksPoint,100) def gangtest(classes,n): def gangtest(classes,n): print "Gang test",[cls.__name__ for cls in classes] print "Gang test",[cls.__name__ for cls in classes] ... @@ -743,5 +756,5 @@ def gangtest(classes,n): ... @@ -743,5 +756,5 @@ def gangtest(classes,n): for c,ret in zip(classes,rets): for c,ret in zip(classes,rets): print c,binascii.hexlify(ret) print c,binascii.hexlify(ret) print print gangtest([IsoEd448Point,TwistedEd448GoldilocksPoint,Ed448GoldilocksPoint],100) #gangtest([IsoEd448Point,TwistedEd448GoldilocksPoint,Ed448GoldilocksPoint],100) gangtest([Ed25519Point,IsoEd25519Point],100) #gangtest([Ed25519Point,IsoEd25519Point],100)
... @@ -48,7 +48,8 @@ static const scalar_t point_scalarmul_adjustment = {{{ ... @@ -48,7 +48,8 @@ static const scalar_t point_scalarmul_adjustment = {{{ const uint8_t decaf_x25519_base_point[DECAF_X25519_PUBLIC_BYTES] = { 0x09 }; const uint8_t decaf_x25519_base_point[DECAF_X25519_PUBLIC_BYTES] = { 0x09 }; static const gf RISTRETTO_ISOMAGIC = {{{ #define RISTRETTO_FACTOR DECAF_255_RISTRETTO_FACTOR const gf RISTRETTO_FACTOR = {{{ 0x0fdaa805d40ea, 0x2eb482e57d339, 0x007610274bc58, 0x6510b613dc8ff, 0x786c8905cfaff 0x0fdaa805d40ea, 0x2eb482e57d339, 0x007610274bc58, 0x6510b613dc8ff, 0x786c8905cfaff }}}; }}}; ... @@ -157,7 +158,7 @@ void API_NS(deisogenize) ( ... @@ -157,7 +158,7 @@ void API_NS(deisogenize) ( gf_mulw(t2,t1,-1-TWISTED_D); /* -x^2 * (a-d) * num */ gf_mulw(t2,t1,-1-TWISTED_D); /* -x^2 * (a-d) * num */ gf_isr(t1,t2); /* t1 = isr */ gf_isr(t1,t2); /* t1 = isr */ gf_mul(t2,t1,t3); /* t2 = ratio */ gf_mul(t2,t1,t3); /* t2 = ratio */ gf_mul(t4,t2,RISTRETTO_ISOMAGIC); gf_mul(t4,t2,RISTRETTO_FACTOR); mask_t negx = gf_lobit(t4) ^ toggle_altx; mask_t negx = gf_lobit(t4) ^ toggle_altx; gf_cond_neg(t2, negx); gf_cond_neg(t2, negx); gf_mul(t3,t2,p->z); gf_mul(t3,t2,p->z); ... @@ -183,12 +184,12 @@ void API_NS(deisogenize) ( ... @@ -183,12 +184,12 @@ void API_NS(deisogenize) ( gf_mulw(t1,t4,-1-TWISTED_D); gf_mulw(t1,t4,-1-TWISTED_D); gf_isr(t4,t1); /* isqrt(num*(a-d)*den^2) */ gf_isr(t4,t1); /* isqrt(num*(a-d)*den^2) */ gf_mul(t1,t2,t4); gf_mul(t1,t2,t4); gf_mul(t2,t1,RISTRETTO_ISOMAGIC); /* t2 = "iden" in ristretto.sage */ gf_mul(t2,t1,RISTRETTO_FACTOR); /* t2 = "iden" in ristretto.sage */ gf_mul(t1,t3,t4); /* t1 = "inum" in ristretto.sage */ gf_mul(t1,t3,t4); /* t1 = "inum" in ristretto.sage */ /* Calculate altxy = iden*inum*i*t^2*(d-a) */ /* Calculate altxy = iden*inum*i*t^2*(d-a) */ gf_mul(t3,t1,t2); gf_mul(t3,t1,t2); gf_mul_qnr(t4,t3); gf_mul_i(t4,t3); gf_mul(t3,t4,p->t); gf_mul(t3,t4,p->t); gf_mul(t4,t3,p->t); gf_mul(t4,t3,p->t); gf_mulw(t3,t4,TWISTED_D+1); /* iden*inum*i*t^2*(d-1) */ gf_mulw(t3,t4,TWISTED_D+1); /* iden*inum*i*t^2*(d-1) */ ... @@ -196,10 +197,10 @@ void API_NS(deisogenize) ( ... @@ -196,10 +197,10 @@ void API_NS(deisogenize) ( /* Rotate if altxy is negative */ /* Rotate if altxy is negative */ gf_cond_swap(t1,t2,rotate); gf_cond_swap(t1,t2,rotate); gf_mul_qnr(t4,p->x); gf_mul_i(t4,p->x); gf_cond_sel(t4,p->y,t4,rotate); /* t4 = "fac" = ix if rotate, else y */ gf_cond_sel(t4,p->y,t4,rotate); /* t4 = "fac" = ix if rotate, else y */ gf_mul_qnr(t5,RISTRETTO_ISOMAGIC); /* t5 = imi */ gf_mul_i(t5,RISTRETTO_FACTOR); /* t5 = imi */ gf_mul(t3,t5,t2); /* iden * imi */ gf_mul(t3,t5,t2); /* iden * imi */ gf_mul(t2,t5,t1); gf_mul(t2,t5,t1); gf_mul(t5,t2,p->t); /* "altx" = iden*imi*t */ gf_mul(t5,t2,p->t); /* "altx" = iden*imi*t */ ... @@ -258,20 +259,20 @@ decaf_error_t API_NS(point_decode) ( ... @@ -258,20 +259,20 @@ decaf_error_t API_NS(point_decode) ( gf_add(tmp2,tmp2,tmp2); /* 2*s*isr*den */ gf_add(tmp2,tmp2,tmp2); /* 2*s*isr*den */ gf_mul(tmp,tmp2,isr); /* 2*s*isr^2*den */ gf_mul(tmp,tmp2,isr); /* 2*s*isr^2*den */ gf_mul(p->x,tmp,num); /* 2*s*isr^2*den*num */ gf_mul(p->x,tmp,num); /* 2*s*isr^2*den*num */ gf_mul(tmp,tmp2,RISTRETTO_ISOMAGIC); /* 2*s*isr*den*magic */ gf_mul(tmp,tmp2,RISTRETTO_FACTOR); /* 2*s*isr*den*magic */ gf_cond_neg(p->x,gf_lobit(tmp)); /* flip x */ gf_cond_neg(p->x,gf_lobit(tmp)); /* flip x */ #if COFACTOR==8 #if COFACTOR==8 /* Additionally check y != 0 and x*y*isomagic nonegative */ /* Additionally check y != 0 and x*y*isomagic nonegative */ succ &= ~gf_eq(p->y,ZERO); succ &= ~gf_eq(p->y,ZERO); gf_mul(tmp,p->x,p->y); gf_mul(tmp,p->x,p->y); gf_mul(tmp2,tmp,RISTRETTO_ISOMAGIC); gf_mul(tmp2,tmp,RISTRETTO_FACTOR); succ &= ~gf_lobit(tmp2); succ &= ~gf_lobit(tmp2); #endif #endif #if IMAGINE_TWIST #if IMAGINE_TWIST gf_copy(tmp,p->x); gf_copy(tmp,p->x); gf_mul_qnr(p->x,tmp); gf_mul_i(p->x,tmp); #endif #endif /* Fill in z and t */ /* Fill in z and t */ ... @@ -1077,9 +1078,9 @@ void API_NS(point_mul_by_cofactor_and_encode_like_eddsa) ( ... @@ -1077,9 +1078,9 @@ void API_NS(point_mul_by_cofactor_and_encode_like_eddsa) ( gf_mul ( y, u, t ); // (x^2+y^2)(2z^2-y^2+x^2) gf_mul ( y, u, t ); // (x^2+y^2)(2z^2-y^2+x^2) gf_mul ( u, z, t ); gf_mul ( u, z, t ); gf_copy( z, u ); gf_copy( z, u ); gf_mul ( u, x, RISTRETTO_ISOMAGIC ); gf_mul ( u, x, RISTRETTO_FACTOR ); #if IMAGINE_TWIST #if IMAGINE_TWIST gf_mul_qnr( x, u ); gf_mul_i( x, u ); #else #else #error "... probably wrong" #error "... probably wrong" gf_copy( x, u ); gf_copy( x, u ); ... @@ -1090,7 +1091,7 @@ void API_NS(point_mul_by_cofactor_and_encode_like_eddsa) ( ... @@ -1090,7 +1091,7 @@ void API_NS(point_mul_by_cofactor_and_encode_like_eddsa) ( { { API_NS(point_double)(q,q); API_NS(point_double)(q,q); API_NS(point_double)(q,q); API_NS(point_double)(q,q); gf_mul_qnr(x, q->x); gf_mul_i(x, q->x); gf_copy(y, q->y); gf_copy(y, q->y); gf_copy(z, q->z); gf_copy(z, q->z); } } ... @@ -1188,8 +1189,8 @@ decaf_error_t API_NS(point_decode_like_eddsa_and_ignore_cofactor) ( ... @@ -1188,8 +1189,8 @@ decaf_error_t API_NS(point_decode_like_eddsa_and_ignore_cofactor) ( gf_sqr ( p->x, p->z ); gf_sqr ( p->x, p->z ); gf_add ( p->z, p->x, p->x ); gf_add ( p->z, p->x, p->x ); gf_sub ( c, p->z, p->t ); // 2z^2 - y^2 + x^2 gf_sub ( c, p->z, p->t ); // 2z^2 - y^2 + x^2 gf_div_qnr ( a, c ); gf_div_i ( a, c ); gf_mul ( c, a, RISTRETTO_ISOMAGIC ); gf_mul ( c, a, RISTRETTO_FACTOR ); gf_mul ( p->x, b, p->t); // (2xy)(y^2-x^2) gf_mul ( p->x, b, p->t); // (2xy)(y^2-x^2) gf_mul ( p->z, p->t, c ); // (y^2-x^2)sd(2z^2 - y^2 + x^2) gf_mul ( p->z, p->t, c ); // (y^2-x^2)sd(2z^2 - y^2 + x^2) gf_mul ( p->y, d, c ); // (y^2+x^2)sd(2z^2 - y^2 + x^2) gf_mul ( p->y, d, c ); // (y^2+x^2)sd(2z^2 - y^2 + x^2) ... ...
... @@ -21,6 +21,10 @@ ... @@ -21,6 +21,10 @@ #define IMAGINE_TWIST 1 #define IMAGINE_TWIST 1 #define COFACTOR 8 #define COFACTOR 8 static const int EDWARDS_D = -121665; static const int EDWARDS_D = -121665; #define RISTRETTO_FACTOR DECAF_255_RISTRETTO_FACTOR extern const gf RISTRETTO_FACTOR; /* End of template stuff */ /* End of template stuff */ extern mask_t API_NS(deisogenize) ( extern mask_t API_NS(deisogenize) ( ... @@ -145,8 +149,23 @@ API_NS(invert_elligator_nonuniform) ( ... @@ -145,8 +149,23 @@ API_NS(invert_elligator_nonuniform) ( API_NS(deisogenize)(a,b,c,p,sgn_s,sgn_altx,sgn_ed_T); API_NS(deisogenize)(a,b,c,p,sgn_s,sgn_altx,sgn_ed_T); mask_t is_identity = gf_eq(p->t,ZERO); mask_t is_identity = gf_eq(p->t,ZERO); #if COFACTOR==4 gf_cond_sel(b,b,ONE,is_identity & sgn_altx); gf_cond_sel(b,b,ONE,is_identity & sgn_altx); gf_cond_sel(c,c,ONE,is_identity & sgn_s &~ sgn_altx); gf_cond_sel(c,c,ONE,is_identity & sgn_s &~ sgn_altx); #elif IMAGINE_TWIST /* Terrible, terrible special casing due to lots of 0/0 is deisogenize * Basically we need to generate -D and +- i*RISTRETTO_FACTOR */ gf_mul_i(a,RISTRETTO_FACTOR); gf_cond_sel(b,b,ONE,is_identity); gf_cond_neg(a,sgn_altx); gf_cond_sel(c,c,a,is_identity & sgn_ed_T); gf_cond_sel(c,c,ZERO,is_identity & ~sgn_ed_T); gf_mulw(a,ONE,-EDWARDS_D); gf_cond_sel(c,c,a,is_identity & ~sgn_ed_T &~ sgn_altx); #else #error "Different special-casing goes here!" #endif #if IMAGINE_TWIST #if IMAGINE_TWIST gf_mulw(a,b,-EDWARDS_D); gf_mulw(a,b,-EDWARDS_D); ... @@ -169,7 +188,8 @@ API_NS(invert_elligator_nonuniform) ( ... @@ -169,7 +188,8 @@ API_NS(invert_elligator_nonuniform) ( #endif #endif gf_cond_neg(b, sgn_r0^gf_lobit(b)); gf_cond_neg(b, sgn_r0^gf_lobit(b)); succ &= ~(gf_eq(b,ZERO) & sgn_r0); /* Eliminate duplicate values for identity ... */ succ &= ~(gf_eq(b,ZERO) & (sgn_r0 | sgn_s)); // #if COFACTOR == 8 // #if COFACTOR == 8 // succ &= ~(is_identity & sgn_ed_T); /* NB: there are no preimages of rotated identity. */ // succ &= ~(is_identity & sgn_ed_T); /* NB: there are no preimages of rotated identity. */ // #endif // #endif ... ...
... @@ -48,7 +48,8 @@ static const scalar_t point_scalarmul_adjustment = {{{ ... @@ -48,7 +48,8 @@ static const scalar_t point_scalarmul_adjustment = {{{ const uint8_t decaf_x448_base_point[DECAF_X448_PUBLIC_BYTES] = { 0x05 }; const uint8_t decaf_x448_base_point[DECAF_X448_PUBLIC_BYTES] = { 0x05 }; static const gf RISTRETTO_ISOMAGIC = {{{ #define RISTRETTO_FACTOR DECAF_448_RISTRETTO_FACTOR const gf RISTRETTO_FACTOR = {{{ 0x42ef0f45572736, 0x7bf6aa20ce5296, 0xf4fd6eded26033, 0x968c14ba839a66, 0xb8d54b64a2d780, 0x6aa0a1f1a7b8a5, 0x683bf68d722fa2, 0x22d962fbeb24f7 0x42ef0f45572736, 0x7bf6aa20ce5296, 0xf4fd6eded26033, 0x968c14ba839a66, 0xb8d54b64a2d780, 0x6aa0a1f1a7b8a5, 0x683bf68d722fa2, 0x22d962fbeb24f7 }}}; }}}; ... @@ -157,7 +158,7 @@ void API_NS(deisogenize) ( ... @@ -157,7 +158,7 @@ void API_NS(deisogenize) ( gf_mulw(t2,t1,-1-TWISTED_D); /* -x^2 * (a-d) * num */ gf_mulw(t2,t1,-1-TWISTED_D); /* -x^2 * (a-d) * num */ gf_isr(t1,t2); /* t1 = isr */ gf_isr(t1,t2); /* t1 = isr */ gf_mul(t2,t1,t3); /* t2 = ratio */ gf_mul(t2,t1,t3); /* t2 = ratio */ gf_mul(t4,t2,RISTRETTO_ISOMAGIC); gf_mul(t4,t2,RISTRETTO_FACTOR); mask_t negx = gf_lobit(t4) ^ toggle_altx; mask_t negx = gf_lobit(t4) ^ toggle_altx; gf_cond_neg(t2, negx); gf_cond_neg(t2, negx); gf_mul(t3,t2,p->z); gf_mul(t3,t2,p->z); ... @@ -183,12 +184,12 @@ void API_NS(deisogenize) ( ... @@ -183,12 +184,12 @@ void API_NS(deisogenize) ( gf_mulw(t1,t4,-1-TWISTED_D); gf_mulw(t1,t4,-1-TWISTED_D); gf_isr(t4,t1); /* isqrt(num*(a-d)*den^2) */ gf_isr(t4,t1); /* isqrt(num*(a-d)*den^2) */ gf_mul(t1,t2,t4); gf_mul(t1,t2,t4); gf_mul(t2,t1,RISTRETTO_ISOMAGIC); /* t2 = "iden" in ristretto.sage */ gf_mul(t2,t1,RISTRETTO_FACTOR); /* t2 = "iden" in ristretto.sage */ gf_mul(t1,t3,t4); /* t1 = "inum" in ristretto.sage */ gf_mul(t1,t3,t4); /* t1 = "inum" in ristretto.sage */ /* Calculate altxy = iden*inum*i*t^2*(d-a) */ /* Calculate altxy = iden*inum*i*t^2*(d-a) */ gf_mul(t3,t1,t2); gf_mul(t3,t1,t2); gf_mul_qnr(t4,t3); gf_mul_i(t4,t3); gf_mul(t3,t4,p->t); gf_mul(t3,t4,p->t); gf_mul(t4,t3,p->t); gf_mul(t4,t3,p->t); gf_mulw(t3,t4,TWISTED_D+1); /* iden*inum*i*t^2*(d-1) */ gf_mulw(t3,t4,TWISTED_D+1); /* iden*inum*i*t^2*(d-1) */ ... @@ -196,10 +197,10 @@ void API_NS(deisogenize) ( ... @@ -196,10 +197,10 @@ void API_NS(deisogenize) ( /* Rotate if altxy is negative */ /* Rotate if altxy is negative */ gf_cond_swap(t1,t2,rotate); gf_cond_swap(t1,t2,rotate); gf_mul_qnr(t4,p->x); gf_mul_i(t4,p->x); gf_cond_sel(t4,p->y,t4,rotate); /* t4 = "fac" = ix if rotate, else y */ gf_cond_sel(t4,p->y,t4,rotate); /* t4 = "fac" = ix if rotate, else y */ gf_mul_qnr(t5,RISTRETTO_ISOMAGIC); /* t5 = imi */ gf_mul_i(t5,RISTRETTO_FACTOR); /* t5 = imi */ gf_mul(t3,t5,t2); /* iden * imi */ gf_mul(t3,t5,t2); /* iden * imi */ gf_mul(t2,t5,t1); gf_mul(t2,t5,t1); gf_mul(t5,t2,p->t); /* "altx" = iden*imi*t */ gf_mul(t5,t2,p->t); /* "altx" = iden*imi*t */ ... @@ -258,20 +259,20 @@ decaf_error_t API_NS(point_decode) ( ... @@ -258,20 +259,20 @@ decaf_error_t API_NS(point_decode) ( gf_add(tmp2,tmp2,tmp2); /* 2*s*isr*den */ gf_add(tmp2,tmp2,tmp2); /* 2*s*isr*den */ gf_mul(tmp,tmp2,isr); /* 2*s*isr^2*den */ gf_mul(tmp,tmp2,isr); /* 2*s*isr^2*den */ gf_mul(p->x,tmp,num); /* 2*s*isr^2*den*num */ gf_mul(p->x,tmp,num); /* 2*s*isr^2*den*num */ gf_mul(tmp,tmp2,RISTRETTO_ISOMAGIC); /* 2*s*isr*den*magic */ gf_mul(tmp,tmp2,RISTRETTO_FACTOR); /* 2*s*isr*den*magic */ gf_cond_neg(p->x,gf_lobit(tmp)); /* flip x */ gf_cond_neg(p->x,gf_lobit(tmp)); /* flip x */ #if COFACTOR==8 #if COFACTOR==8 /* Additionally check y != 0 and x*y*isomagic nonegative */ /* Additionally check y != 0 and x*y*isomagic nonegative */ succ &= ~gf_eq(p->y,ZERO); succ &= ~gf_eq(p->y,ZERO); gf_mul(tmp,p->x,p->y); gf_mul(tmp,p->x,p->y); gf_mul(tmp2,tmp,RISTRETTO_ISOMAGIC); gf_mul(tmp2,tmp,RISTRETTO_FACTOR); succ &= ~gf_lobit(tmp2); succ &= ~gf_lobit(tmp2); #endif #endif #if IMAGINE_TWIST #if IMAGINE_TWIST gf_copy(tmp,p->x); gf_copy(tmp,p->x); gf_mul_qnr(p->x,tmp); gf_mul_i(p->x,tmp); #endif #endif /* Fill in z and t */ /* Fill in z and t */ ... @@ -1077,9 +1078,9 @@ void API_NS(point_mul_by_cofactor_and_encode_like_eddsa) ( ... @@ -1077,9 +1078,9 @@ void API_NS(point_mul_by_cofactor_and_encode_like_eddsa) ( gf_mul ( y, u, t ); // (x^2+y^2)(2z^2-y^2+x^2) gf_mul ( y, u, t ); // (x^2+y^2)(2z^2-y^2+x^2) gf_mul ( u, z, t ); gf_mul ( u, z, t ); gf_copy( z, u ); gf_copy( z, u ); gf_mul ( u, x, RISTRETTO_ISOMAGIC ); gf_mul ( u, x, RISTRETTO_FACTOR ); #if IMAGINE_TWIST #if IMAGINE_TWIST gf_mul_qnr( x, u ); gf_mul_i( x, u ); #else #else #error "... probably wrong" #error "... probably wrong" gf_copy( x, u ); gf_copy( x, u ); ... @@ -1090,7 +1091,7 @@ void API_NS(point_mul_by_cofactor_and_encode_like_eddsa) ( ... @@ -1090,7 +1091,7 @@ void API_NS(point_mul_by_cofactor_and_encode_like_eddsa) ( { { API_NS(point_double)(q,q); API_NS(point_double)(q,q); API_NS(point_double)(q,q); API_NS(point_double)(q,q); gf_mul_qnr(x, q->x); gf_mul_i(x, q->x); gf_copy(y, q->y); gf_copy(y, q->y); gf_copy(z, q->z); gf_copy(z, q->z); } } ... @@ -1188,8 +1189,8 @@ decaf_error_t API_NS(point_decode_like_eddsa_and_ignore_cofactor) ( ... @@ -1188,8 +1189,8 @@ decaf_error_t API_NS(point_decode_like_eddsa_and_ignore_cofactor) ( gf_sqr ( p->x, p->z ); gf_sqr ( p->x, p->z ); gf_add ( p->z, p->x, p->x ); gf_add ( p->z, p->x, p->x ); gf_sub ( c, p->z, p->t ); // 2z^2 - y^2 + x^2 gf_sub ( c, p->z, p->t ); // 2z^2 - y^2 + x^2 gf_div_qnr ( a, c ); gf_div_i ( a, c ); gf_mul ( c, a, RISTRETTO_ISOMAGIC ); gf_mul ( c, a, RISTRETTO_FACTOR ); gf_mul ( p->x, b, p->t); // (2xy)(y^2-x^2) gf_mul ( p->x, b, p->t); // (2xy)(y^2-x^2) gf_mul ( p->z, p->t, c ); // (y^2-x^2)sd(2z^2 - y^2 + x^2) gf_mul ( p->z, p->t, c ); // (y^2-x^2)sd(2z^2 - y^2 + x^2) gf_mul ( p->y, d, c ); // (y^2+x^2)sd(2z^2 - y^2 + x^2) gf_mul ( p->y, d, c ); // (y^2+x^2)sd(2z^2 - y^2 + x^2) ... ...
... @@ -21,6 +21,10 @@ ... @@ -21,6 +21,10 @@ #define IMAGINE_TWIST 0 #define IMAGINE_TWIST 0 #define COFACTOR 4 #define COFACTOR 4 static const int EDWARDS_D = -39081; static const int EDWARDS_D = -39081; #define RISTRETTO_FACTOR DECAF_448_RISTRETTO_FACTOR extern const gf RISTRETTO_FACTOR; /* End of template stuff */ /* End of template stuff */ extern mask_t API_NS(deisogenize) ( extern mask_t API_NS(deisogenize) ( ... @@ -145,8 +149,23 @@ API_NS(invert_elligator_nonuniform) ( ... @@ -145,8 +149,23 @@ API_NS(invert_elligator_nonuniform) ( API_NS(deisogenize)(a,b,c,p,sgn_s,sgn_altx,sgn_ed_T); API_NS(deisogenize)(a,b,c,p,sgn_s,sgn_altx,sgn_ed_T); mask_t is_identity = gf_eq(p->t,ZERO); mask_t is_identity = gf_eq(p->t,ZERO); #if COFACTOR==4 gf_cond_sel(b,b,ONE,is_identity & sgn_altx); gf_cond_sel(b,b,ONE,is_identity & sgn_altx); gf_cond_sel(c,c,ONE,is_identity & sgn_s &~ sgn_altx); gf_cond_sel(c,c,ONE,is_identity & sgn_s &~ sgn_altx); #elif IMAGINE_TWIST /* Terrible, terrible special casing due to lots of 0/0 is deisogenize * Basically we need to generate -D and +- i*RISTRETTO_FACTOR */ gf_mul_i(a,RISTRETTO_FACTOR); gf_cond_sel(b,b,ONE,is_identity); gf_cond_neg(a,sgn_altx); gf_cond_sel(c,c,a,is_identity & sgn_ed_T); gf_cond_sel(c,c,ZERO,is_identity & ~sgn_ed_T); gf_mulw(a,ONE,-EDWARDS_D); gf_cond_sel(c,c,a,is_identity & ~sgn_ed_T &~ sgn_altx); #else #error "Different special-casing goes here!" #endif #if IMAGINE_TWIST #if IMAGINE_TWIST gf_mulw(a,b,-EDWARDS_D); gf_mulw(a,b,-EDWARDS_D); ... @@ -169,7 +188,8 @@ API_NS(invert_elligator_nonuniform) ( ... @@ -169,7 +188,8 @@ API_NS(invert_elligator_nonuniform) ( #endif #endif gf_cond_neg(b, sgn_r0^gf_lobit(b)); gf_cond_neg(b, sgn_r0^gf_lobit(b)); succ &= ~(gf_eq(b,ZERO) & sgn_r0); /* Eliminate duplicate values for identity ... */ succ &= ~(gf_eq(b,ZERO) & (sgn_r0 | sgn_s)); // #if COFACTOR == 8 // #if COFACTOR == 8 // succ &= ~(is_identity & sgn_ed_T); /* NB: there are no preimages of rotated identity. */ // succ &= ~(is_identity & sgn_ed_T); /* NB: there are no preimages of rotated identity. */ // #endif // #endif ... ...
... @@ -103,5 +103,10 @@ static DECAF_INLINE void gf_div_qnr(gf_s *__restrict__ out, const gf x) { ... @@ -103,5 +103,10 @@ static DECAF_INLINE void gf_div_qnr(gf_s *__restrict__ out, const gf x) { #endif #endif } } #if P_MOD_8 == 5 #define gf_mul_i gf_mul_qnr #define gf_div_i gf_div_qnr #endif #endif // __GF_H__ #endif // __GF_H__
... @@ -37,7 +37,8 @@ static const scalar_t point_scalarmul_adjustment = {{{ ... @@ -37,7 +37,8 @@ static const scalar_t point_scalarmul_adjustment = {{{ const uint8_t decaf_x\$(gf_shortname)_base_point[DECAF_X\$(gf_shortname)_PUBLIC_BYTES] = { \$(ser(mont_base,8)) }; const uint8_t decaf_x\$(gf_shortname)_base_point[DECAF_X\$(gf_shortname)_PUBLIC_BYTES] = { \$(ser(mont_base,8)) }; static const gf RISTRETTO_ISOMAGIC = {{{ #define RISTRETTO_FACTOR \$(C_NS)_RISTRETTO_FACTOR const gf RISTRETTO_FACTOR = {{{ \$(ser(msqrt(d-1 if imagine_twist else -d,modulus,lo_bit_clear=True),gf_lit_limb_bits)) \$(ser(msqrt(d-1 if imagine_twist else -d,modulus,lo_bit_clear=True),gf_lit_limb_bits)) }}}; }}}; ... @@ -146,7 +147,7 @@ void API_NS(deisogenize) ( ... @@ -146,7 +147,7 @@ void API_NS(deisogenize) ( gf_mulw(t2,t1,-1-TWISTED_D); /* -x^2 * (a-d) * num */ gf_mulw(t2,t1,-1-TWISTED_D); /* -x^2 * (a-d) * num */ gf_isr(t1,t2); /* t1 = isr */ gf_isr(t1,t2); /* t1 = isr */ gf_mul(t2,t1,t3); /* t2 = ratio */ gf_mul(t2,t1,t3); /* t2 = ratio */ gf_mul(t4,t2,RISTRETTO_ISOMAGIC); gf_mul(t4,t2,RISTRETTO_FACTOR); mask_t negx = gf_lobit(t4) ^ toggle_altx; mask_t negx = gf_lobit(t4) ^ toggle_altx; gf_cond_neg(t2, negx); gf_cond_neg(t2, negx); gf_mul(t3,t2,p->z); gf_mul(t3,t2,p->z); ... @@ -172,12 +173,12 @@ void API_NS(deisogenize) ( ... @@ -172,12 +173,12 @@ void API_NS(deisogenize) ( gf_mulw(t1,t4,-1-TWISTED_D); gf_mulw(t1,t4,-1-TWISTED_D); gf_isr(t4,t1); /* isqrt(num*(a-d)*den^2) */ gf_isr(t4,t1); /* isqrt(num*(a-d)*den^2) */ gf_mul(t1,t2,t4); gf_mul(t1,t2,t4); gf_mul(t2,t1,RISTRETTO_ISOMAGIC); /* t2 = "iden" in ristretto.sage */ gf_mul(t2,t1,RISTRETTO_FACTOR); /* t2 = "iden" in ristretto.sage */ gf_mul(t1,t3,t4); /* t1 = "inum" in ristretto.sage */ gf_mul(t1,t3,t4); /* t1 = "inum" in ristretto.sage */ /* Calculate altxy = iden*inum*i*t^2*(d-a) */ /* Calculate altxy = iden*inum*i*t^2*(d-a) */ gf_mul(t3,t1,t2); gf_mul(t3,t1,t2); gf_mul_qnr(t4,t3); gf_mul_i(t4,t3); gf_mul(t3,t4,p->t); gf_mul(t3,t4,p->t); gf_mul(t4,t3,p->t); gf_mul(t4,t3,p->t); gf_mulw(t3,t4,TWISTED_D+1); /* iden*inum*i*t^2*(d-1) */ gf_mulw(t3,t4,TWISTED_D+1); /* iden*inum*i*t^2*(d-1) */ ... @@ -185,10 +186,10 @@ void API_NS(deisogenize) ( ... @@ -185,10 +186,10 @@ void API_NS(deisogenize) ( /* Rotate if altxy is negative */ /* Rotate if altxy is negative */ gf_cond_swap(t1,t2,rotate); gf_cond_swap(t1,t2,rotate); gf_mul_qnr(t4,p->x); gf_mul_i(t4,p->x); gf_cond_sel(t4,p->y,t4,rotate); /* t4 = "fac" = ix if rotate, else y */ gf_cond_sel(t4,p->y,t4,rotate); /* t4 = "fac" = ix if rotate, else y */ gf_mul_qnr(t5,RISTRETTO_ISOMAGIC); /* t5 = imi */ gf_mul_i(t5,RISTRETTO_FACTOR); /* t5 = imi */ gf_mul(t3,t5,t2); /* iden * imi */ gf_mul(t3,t5,t2); /* iden * imi */ gf_mul(t2,t5,t1); gf_mul(t2,t5,t1); gf_mul(t5,t2,p->t); /* "altx" = iden*imi*t */ gf_mul(t5,t2,p->t); /* "altx" = iden*imi*t */ ... @@ -247,20 +248,20 @@ decaf_error_t API_NS(point_decode) ( ... @@ -247,20 +248,20 @@ decaf_error_t API_NS(point_decode) ( gf_add(tmp2,tmp2,tmp2); /* 2*s*isr*den */ gf_add(tmp2,tmp2,tmp2); /* 2*s*isr*den */ gf_mul(tmp,tmp2,isr); /* 2*s*isr^2*den */ gf_mul(tmp,tmp2,isr); /* 2*s*isr^2*den */ gf_mul(p->x,tmp,num); /* 2*s*isr^2*den*num */ gf_mul(p->x,tmp,num); /* 2*s*isr^2*den*num */ gf_mul(tmp,tmp2,RISTRETTO_ISOMAGIC); /* 2*s*isr*den*magic */ gf_mul(tmp,tmp2,RISTRETTO_FACTOR); /* 2*s*isr*den*magic */ gf_cond_neg(p->x,gf_lobit(tmp)); /* flip x */ gf_cond_neg(p->x,gf_lobit(tmp)); /* flip x */ #if COFACTOR==8 #if COFACTOR==8 /* Additionally check y != 0 and x*y*isomagic nonegative */ /* Additionally check y != 0 and x*y*isomagic nonegative */ succ &= ~gf_eq(p->y,ZERO); succ &= ~gf_eq(p->y,ZERO); gf_mul(tmp,p->x,p->y); gf_mul(tmp,p->x,p->y); gf_mul(tmp2,tmp,RISTRETTO_ISOMAGIC); gf_mul(tmp2,tmp,RISTRETTO_FACTOR); succ &= ~gf_lobit(tmp2); succ &= ~gf_lobit(tmp2); #endif #endif #if IMAGINE_TWIST #if IMAGINE_TWIST gf_copy(tmp,p->x); gf_copy(tmp,p->x); gf_mul_qnr(p->x,tmp); gf_mul_i(p->x,tmp); #endif #endif /* Fill in z and t */ /* Fill in z and t */ ... @@ -1066,9 +1067,9 @@ void API_NS(point_mul_by_cofactor_and_encode_like_eddsa) ( ... @@ -1066,9 +1067,9 @@ void API_NS(point_mul_by_cofactor_and_encode_like_eddsa) ( gf_mul ( y, u, t ); // (x^2+y^2)(2z^2-y^2+x^2) gf_mul ( y, u, t ); // (x^2+y^2)(2z^2-y^2+x^2) gf_mul ( u, z, t ); gf_mul ( u, z, t ); gf_copy( z, u ); gf_copy( z, u ); gf_mul ( u, x, RISTRETTO_ISOMAGIC ); gf_mul ( u, x, RISTRETTO_FACTOR ); #if IMAGINE_TWIST #if IMAGINE_TWIST gf_mul_qnr( x, u ); gf_mul_i( x, u ); #else #else #error "... probably wrong" #error "... probably wrong" gf_copy( x, u ); gf_copy( x, u ); ... @@ -1079,7 +1080,7 @@ void API_NS(point_mul_by_cofactor_and_encode_like_eddsa) ( ... @@ -1079,7 +1080,7 @@ void API_NS(point_mul_by_cofactor_and_encode_like_eddsa) ( { { API_NS(point_double)(q,q); API_NS(point_double)(q,q); API_NS(point_double)(q,q); API_NS(point_double)(q,q); gf_mul_qnr(x, q->x); gf_mul_i(x, q->x); gf_copy(y, q->y); gf_copy(y, q->y); gf_copy(z, q->z); gf_copy(z, q->z); } } ... @@ -1177,8 +1178,8 @@ decaf_error_t API_NS(point_decode_like_eddsa_and_ignore_cofactor) ( ... @@ -1177,8 +1178,8 @@ decaf_error_t API_NS(point_decode_like_eddsa_and_ignore_cofactor) ( gf_sqr ( p->x, p->z ); gf_sqr ( p->x, p->z ); gf_add ( p->z, p->x, p->x ); gf_add ( p->z, p->x, p->x ); gf_sub ( c, p->z, p->t ); // 2z^2 - y^2 + x^2 gf_sub ( c, p->z, p->t ); // 2z^2 - y^2 + x^2 | 10,261 | 26,010 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2020-05 | latest | en | 0.227918 |
https://www.gradesaver.com/textbooks/math/precalculus/precalculus-6th-edition-blitzer/chapter-9-section-9-1-the-ellipse-exercise-set-page-966/21 | 1,679,410,077,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296943698.79/warc/CC-MAIN-20230321131205-20230321161205-00537.warc.gz | 899,453,374 | 12,683 | ## Precalculus (6th Edition) Blitzer
$x^2+\frac{y^2}{4}=1$, foci $(0,\pm\sqrt 3)$
Step 1. From the given graph, we can identify $a=2, b=1$; thus $c=\sqrt {a^2-b^2}=\sqrt {3}$. The ellipse is centered at $(0,0)$ with a vertical major axis. Step 2. We can write the equation as $x^2+\frac{y^2}{4}=1$ with foci at $(0,\pm\sqrt 3)$ | 134 | 328 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2023-14 | latest | en | 0.640473 |
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