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## Factorials and their applications
We will see some of the important properties of factorials.
A thorough understanding of Factorials is important because they play a pivotal role not only in the Concepts surrounding Numbers but also other
important topics like Permutation and Combination
## What is highest power?
Suppose you have a number N= x2y. here the highest power of x in N will be 2 and the highest power of y in N will be 1.
Questions based on this can be categorized based on the nature of the number whose highest power we are nding in the factorial, i.e.
Solution:
Solution:
## So highest power of 5 in 100! = 24
ALTERNATIVE METHOD
100/5+100/52 =20+4=24 (We take upto 52 as it is the highest power of 5 which is less than 100)
Solution:
## 1) Factorize the number into primes.
2) Find the highest power of all the prime numbers in that factorial using the previous method.
## 3) Take the least power.
ILLUSTRATION:
33) To find the highest power of 10 in 100!
Soltuion:
Factorize 10=5*2.
## 1.Highest power of 5 in 100! =24
2. Highest power of 2 in 100! =97 Therefore, the answer will be 24, because to get a 10, you need a pair of 2 and 5, and only 24 such pairs are
available.
Solution:
12=22 *3
## 100/2= 50;50/2= 25;25/2= 12;12/2= 6;6/2= 3;3/2= 1
Highest power of 2 = 50 + 25 + 12 + 6 + 3 + 1 = 97
## So highest power of 22=97/2= 48 (out of 97 2s only 48 can make 22)
Now for the highest power of 3 100/3= 33;33/3= 11;11/3= 3;3/3= 1; Highest power of 3 = 48
Highest power of 12 = 48
## II) Number of zeros in the end of a factorial or a product
In base 10, number of zeros in the end depends on the number of 10s; i.e. effectively, on the number of 5s (10=52.
if you see the previous example, you will get a lesser number of 5s than 2s.
that is why we need to calculate only the number of 5s) Similarly, for base 12, the number of zeros will depend on the number 12 itself.
In base N, number of zeroes in the end is nothing but the highest power of N in that product.
## We need to nd the highest power of 10 in 13! = highest power of 5 in 13!
This is because there will be lesser number of 5s present as compared to 2s (illustrated in eg _____ above)
## Number of 2s= 13/2=6;6/2=3 ;3/2=1.
total =6+3+1=10 Number of 5s =13/5=2. number of zeroes will thus depend only on the number of 5s = 2
Soluiton
## We need to nd the highest power of 10 in 25!
Highest power of 10 will be the highest power of 5 in 25! As explained in the previous example 25/5=5;5/5=1.
## number of zeroes = 5+1=6
37) Find the number of zeroes in the end of 15! in base 12.
Solution:
Solution
## 39) Find the number of factors of 6!
Solution:
STEP 1: Prime factorize 6! i.e. nd out the highest power of all prime factors till 6 ( i.e. 2,3 and 5). 6! = 24*32*51
STEP 2: Then use the formula N=amxbn (a, b are the prime factors).
Then number of factors= (m+1)(n+1) The number of factors= (4+1)(2+1)(1+1) =30 Answer=30
## 40) Find the number of factors of 12!
Solution:
STEP 1: Prime factorize 12! i.e. nd out the highest power of all prime factors till 12 ( i.e. 2,3,5,7,11). 12! = 210*35*52*7*11
STEP2: Then use the formula N=am*bn(a, b are the prime factors) Then number of factors= (m+1)(n+1)
Solution:
option (a)
## APPLICATION QUESTION BASED ON FACTORIAL
42) How many natural numbers are there such that their factorials are ending with 5 zeroes?
Soluiton
10! is 1x2x3x4x(5)x6x7x8x9x(25).
From this we can see that highest power of 5 till 10! is 2( as there are 2 ves) Continuing like this, 10!-14!, highest power of 5 will be 2.
The next 5 will be obtained at 15 = (53).
Therefore, from 15! To 19! The highest power of 5 will be 3. 20!-24! Highest Power = 4 In 25,
we are getting one extra ve, as 25=55. Therefore, 25! to 29!, we will get highest power of 5 as 6.
The answer to the question is therefore, 0. There are no natural numbers whose factorials end with 5 zeroes.
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Free CAT Prep (http://byjus.com/free-cat-prep/) | 1,838 | 5,673 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-10 | latest | en | 0.835565 |
http://mathhelpforum.com/advanced-statistics/105230-question-about-regression-equations.html | 1,481,063,932,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542002.53/warc/CC-MAIN-20161202170902-00181-ip-10-31-129-80.ec2.internal.warc.gz | 155,155,592 | 9,646 | 1. ## Question about Regression Equations
I'm working on an assignment using Minitab and the last question is giving me some trouble.
We have a list of data regarding states and their percentages of exercise and diabetes.
The question is as follows:
For the regression using exercise as the predictor variable (X) and diabetes as the predicted variable (Y), if 25% of respondents for a certain state reported that they exercised 3 or more times a week, what would the regression equation predict would be the % of respondents that report having been diagnosed with diabetes?
I can assume the teacher wants an actual % and not just an estimation (like the people diagnosed with diabetes would decrease). But I'm not sure how to solve this (using an equation or just using Minitab).
Thanks!
EDIT: Okay so I figured out that I need to put numbers back into the regression equation. Which is Y= -.2522x + 15.22
I assume I put 3 in for x but I have no idea where the 25% comes in.
HALLLP.
2. updated | 223 | 1,004 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-50 | longest | en | 0.946769 |
http://math.stackexchange.com/questions/64679/for-how-many-functions-f-is-fx2-x2/64687 | 1,464,516,898,000,000,000 | text/html | crawl-data/CC-MAIN-2016-22/segments/1464049278887.13/warc/CC-MAIN-20160524002118-00237-ip-10-185-217-139.ec2.internal.warc.gz | 183,869,457 | 18,021 | # For how many functions $f$ is $f(x)^{2}=x^{2}$?
How many functions $f$ are there that satisfy $f(x)^{2}=x^{2}$ for all $x$?
My text (Spivak's Calculus; chapter 7 problem 7) asks this question for continuous $f$, for which the answer is, of course 4:$$f(x)=x$$ $$f(x)=-x$$ $$f(x)=\lvert x \rvert$$ $$f(x)=-\lvert x \rvert,$$ and I want to make sure I'm correct that if $f$ does not have to be continuous, there are infinitely many: any piecwise combination of them (infinitely many of which are one of the above, and infinitely many of which are not).
-
You are correct. In fact there are $2^{2^\omega}=2^\mathfrak{c}$ such functions if they need not be continuous. For each $A\subseteq \mathbb{R}$ define $f_A$ as follows: $$f_A(x)=\begin{cases}x,&x\in A\\-x,&x\notin A\end{cases}$$ Then $f_A(x)^2=x^2$ for all $x\in\mathbb{R}$. It’s easy to see that if $A\setminus\{0\}\ne B\setminus\{0\}$, then $f_A\ne f_B$, so we have one such function for every subset of $\mathbb{R}\setminus\{0\}$. Finally, every function with the desired property is such a function: if $f(x)^2=x^2$ for all $x\in \mathbb{R}$, then $f=f_A$, where $A =$ $\{x\in\mathbb{R}:f(x)=x\}$.
If you limit yourself to piecewise continuous functions, there are only $2^\omega=\mathfrak{c}$ of them: there are $(2^\omega)^\omega=2^\omega$ ways to choose the partition points between the pieces, $2^\omega$ ways to choose $x$ or $-x$ on each interval, and $2^\omega$ ways to choose the values at the endpoints, for a total of $(2^\omega)^3=2^\omega$ functions.
-
@Jonas: Yep, that was sloppy of me; fixed. Thanks. – Brian M. Scott Sep 15 '11 at 4:37
@Brian: Excellent, thanks! There is some notation I'm not familiar with ($\omega$?) and other's I'm guessing at ($\setminus$) but I get the gist and this fully answers my question. – raxacoricofallapatorius Sep 15 '11 at 13:21
@raxacoricofallapatorius: $\omega$ is the smallest infinite cardinal number; you may know it under the alias $\aleph_0$. $A\setminus B$ is the set of things that are in $A$ but not $B$; in older and elementary texts it’s often written $A-B$. $\mathbb{R}\setminus\{0\}$ is the set of all non-zero real numbers. – Brian M. Scott Sep 15 '11 at 17:39
It's really only $2^{2^\omega-1}$, since the apparent choice is illusory in the case of $x=0$. – MJD Oct 8 '12 at 17:50
If $f$ doesn't have to be continuous, you can just choose $f(x)$ to be $|x|$ or $-|x|$ arbitrarily at each $x$, so there are lots of possibilities for $f$ (most of which I would not describe as "piecewise" combinations of the 4 continuous solutions).
-
This is pretty much the answer I was looking for (it's exactly "my level") and I'd love to vote it the answer, but Brian's ended up being more complete and the one that's probably most widely useful. Thanks! – raxacoricofallapatorius Sep 15 '11 at 13:23
Let $s$ be an arbitrary function $\mathbb{R}\to\{-1,+1\}$. Then $f=s\cdot\mathrm{id}$ satisfies the property: $(s(x)x)^2=x^2$. Indeed, this exhausts all possible functions, as $s$ can be constructed by looking at $x/f(x)$ (and making an arbitrary choice when $x=0$).
There are uncountably many of these functions, in fact $2^\mathfrak{c}$ where $\mathfrak{c}=2^{\aleph_0}$ is the cardinality of the continuum.
- | 1,036 | 3,235 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2016-22 | latest | en | 0.835824 |
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## Available Class Times
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HSC Mathematics Extension 2 extends the boundaries of mathematics notably beyond the expected standard at the high school level, and begins to scrape the surface of first year university. The course is intended to challenge students who have excelled in Year 11 Mathematics Extension 1 to think more deeply and creatively in proving powerful mathematical results and working through more intense computation. Although the course is only treated as assumed knowledge for degrees involving a mathematics major, it is still recommended for many students interested in STEM-based degrees as the coverage of university level content provides a small, but effective head start to their coursework. It is commonly regarded as one of the hardest high-school mathematics courses in the country.
## Studying Mathematics Extension 2 Year 12, at TuteSmart
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## HSC Mathematics Extension 2 Year 12 Curriculum
Introduction, N1.1 Use the complex number system N1.1: Represent and use complex numbers in Cartesian form, N1.2: Represent and use complex numbers in the complex plane and in polar (mod-arg) N1.2: Represent and use complex numbers in the complex plane and in polar (mod-arg) form, N1.3: Other representations of complex numbers N1.2: Prove and use the basic identities involving modulus and argument N2.1: Solving equations with complex numbers (De Moivre’s theorem and real quadratic equations) N2.1: Solving equations with complex numbers (Complex quadratic equations and real polynomials) N2.2: Addition, subtraction of complex numbers as vectors in the complex plane; the geometric relationship of the conjugate, multiplication by i and by a real constant N2.2: Examine and use the geometric interpretation of multiplying complex numbers, including rotation and dilation P1: The nature of proof (Proof on inequalities) N2.2: Identify subsets of the complex plane determined by relations P1: The nature of proof (Language of proof) Test 1: Complex Numbers, Proofs P1: The nature of proof (Examples and counter-examples, inequalities) P1: The nature of proof (Inequalities) Introduction?/P2: Further proof by mathematical induction (different initial values, different increments, sigma notation ) P2: Further proof by mathematical induction (divisibility, inequalities, calculus) P2: Further proof by mathematical induction (probability, first-order recurrences, geometry) Spare lesson: Advice + Worksheet V1.1: Introduction to three-dimensional vectors, V1.2: Define and use the magnitude and scalar product in three dimensions Half Yearly Exam Review of Half Yearly Exam V1.2: Prove geometric results in the plane and in three dimensions V1.3: Use Cartesian coordinates in 2D and 3D space; spheres; uses vector equations V1.3 Understand and use the vector equation of a line and line segments; connection to y=mx+c (2D case) V1.3: Determine when lines in vector form are parallel or perpendicular; determine when a given point lies on a given line C1: Further integration (integration by substitution, quadratic denominators) C1: Further integration (partial fractions, integration by parts) C1: Further integration (recurrences) Harder integration, Revision - Student poll? M1.1: Simple harmonic motion M1.1: Simple harmonic motion/ More mechanics ATAR Notes Trials ATAR Notes Trials Review M1.2: Use formulae for acceleration and Newton's laws Continued Trials Review + Small worksheet M1.2: Examine constant and non constant acceleration; outside of projectiles and simple harmonic M1.3: Solve problems involving resisted motion of a particle moving along a horizontal line M1.3: Solve problems involving the motion of a particle moving vertically (upwards or downwards) in a resisting medium and under the influence of gravity M1.4: Solve problems involving projectiles in a variety of contexts M1.4: Solve problems involving projectile motion in a resisting medium and under the influence of gravity Exam Tips + Worksheet Revision Revision Revision Revision Final Lesson
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Year 12 | 1,063 | 5,111 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2021-21 | longest | en | 0.94863 |
https://www.netwasgroup.us/oil-well/ellipse-of-uncertainty.html | 1,586,243,147,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371675859.64/warc/CC-MAIN-20200407054138-20200407084638-00539.warc.gz | 1,004,409,145 | 7,251 | ## Ellipse Of Uncertainty
The concept of an ellipse at any point in a drill hole to quantify the possible locations of the point is illustrated with the following example.
Suppose the measured depth is 2500 feet as shown in the sketch and that systematic errors (errors which regularly re-occur and are not compensating) at any survey station are 1° in azimuth, 0.5° in inclination, and 1 part in 500 for measured depth. Further, suppose the inclination and azimuth are thought to be 30° and 90°, respectively.
EAST
NORTH A MD sin I
EAST
EAST
VERTICAL
The expected location of the bottom of the drill hole is
The actual location of the bottom of the drill hole could be anywhere within a volume shown as an ellipsoid in the sketch. The dimensions of the ellipsoid are given by the variations of variables of inclination, azimuth, and measured depth. The maximum dimensions of the ellipsoid and consequently the position of uncertainty which is called the "ellipse of uncertainty" are
+8 -1
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### Responses
• isengar greenhand
What is the cone of uncertainity in directional drilling?
7 years ago
• Susanne
What is ellipse of uncertainty?
6 years ago
• katja pusila
What is oil and gas well elipsoid?
5 years ago
• ireneo
What is the ellipsoid of uncertainty?
5 years ago
• Vanessa
What is ellipsoid of uncertainty in directional survey?
5 years ago
• pietro romano
Why the ellipsoid of uncertainty is not round?
2 years ago
• aija
Why is directional uncertainty an elipse?
2 years ago
• kate
Why error surface of mwd survey is elliptical?
1 year ago
• HARRIET
What is ellipses of uncertainty?
1 year ago
• anna
Why does ellipse of uncertainty larger when drilling east west?
11 months ago | 519 | 2,261 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-16 | latest | en | 0.938215 |
https://www.snapxam.com/problems/39433445/9-1-3-2-0-5-1-1-3 | 1,590,813,431,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347407289.35/warc/CC-MAIN-20200530040743-20200530070743-00555.warc.gz | 913,892,696 | 7,816 | # Step-by-step Solution
Go!
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## Step-by-step explanation
Problem to solve:
$9^{\frac{1}{3}}\left(\sqrt{2}+1\right)^{\frac{1}{3}}\left(27-\sqrt{162}\right)^{\frac{1}{6}}$
Learn how to solve multiplication of numbers problems step by step online.
$\sqrt[3]{9}\cdot \left(\sqrt{2}+1\right)^{\frac{1}{3}}\cdot \left(27-1\cdot \sqrt{162}\right)^{\frac{1}{6}}$
Learn how to solve multiplication of numbers problems step by step online. Multiply 9^(1/3)(2^0.5+1)^(1/3)*(27-162^0.5)^(1/6). Divide 1 by 3. The square root of 2 is \sqrt{2}. Calculate the power \sqrt[3]{9}. Add the values \sqrt{2} and 1.
$4.346$
### Problem Analysis
$9^{\frac{1}{3}}\left(\sqrt{2}+1\right)^{\frac{1}{3}}\left(27-\sqrt{162}\right)^{\frac{1}{6}}$
### Main topic:
Multiplication of numbers
~ 0.02 seconds | 454 | 1,024 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2020-24 | latest | en | 0.492694 |
https://oleksii.shmalko.com/20200817173305/ | 1,695,286,259,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233505362.29/warc/CC-MAIN-20230921073711-20230921103711-00547.warc.gz | 501,173,150 | 6,184 | # ๐Monoid
tags
ยง Category Theory
In Category Theory, Monoid is a Category with one object.
\begin{tikzcd} m \arrow[loop, "f", distance=2em, leftarrow, swap] \arrow[loop, "g", distance=4em, leftarrow, swap] \arrow[loop, "g \circ f", distance=6em, leftarrow, swap] \end{tikzcd}
In Set Theory, Monoid is a set of values with
• an associative binary operator $*$
• a unit value $\exists e \ni \forall a.\; e * a = a = a *e$
These definitions are equivalent.
• set of values โ set of morphisms
• associative binary operator โ composition of morphisms
• unit value โ identity morphism | 184 | 599 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 2, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2023-40 | longest | en | 0.609632 |
https://niniloos.com/latest-greatest-common-factor-worksheet/ | 1,620,931,580,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991943.36/warc/CC-MAIN-20210513173321-20210513203321-00250.warc.gz | 439,493,078 | 8,127 | # Latest greatest common factor worksheet info
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Latest Greatest Common Factor Worksheet. In other words the HCF is the largest of all the common factors. Prime factorization Other contents. When we multiply 2 cdot 2 cdot 3 we get 12 which is the greatest common factor. The GCF of two numbers is the largest number that is a factor of both numbers.
Greatest Common Factor Worksheets Greatest Common Factors Common Factors Graphing Worksheets From pinterest.com
A large collection of GCF worksheets is meticulously drafted for students in grade 5 through grade 8. This twelve-question worksheet addresses prime and composite numbers factors common factors and greatest common factor. HCF - Highest Common Factor Math WorksheetsPrintables PDF for kids. Greatest Common Factor Finding the factors. Make sure the final actors are prime numbers. We can see that the common prime factors are 22 and 3.
### This worksheet helps fourth graders build a strong foundation in GCF concepts as they work with fractions and equivalence.
The number set ranges number of problems and more are customizable on the PDF. Download and print these GCF worksheets to find the GCF of two numbers three. Find the common factors of 18 and 24. Worksheets for the greatest common factor GFC and least common multiple LCM of numbers. GCF is also known as greatest common divisor GCD highest common factor HCF greatest common measure GCM or highest common divisor HCD. A step-by-step tutorial plus examples and practice problems help clarify the process.
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Greatest Common Factor GCF Other contents. The problems in the left-hand column mirror those in the right-hand column so you could use one side. A step-by-step tutorial plus examples and practice problems help clarify the process. You can choose the number. Greatest common factor GCF Grade 5 Factoring Worksheet Find the greatest common factor of the two numbers shown.
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Since 10 2 5 2 2 1. Jul 2 2016 - Grab these GCF worksheets that contain exercises like greatest common factor for the set of numbers prime factorization and Venn diagram to find GCF. These factoring worksheets are pdf files. When we multiply 2 cdot 2 cdot 3 we get 12 which is the greatest common factor. Download and print these GCF worksheets to find the GCF of two numbers three.
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These exercises have students find all numbers which are factors in two composite numbers. Greatest Common Factor GCF Add to my workbooks 44 Download file pdf Embed in my website or blog Add to Google Classroom. Of numerical coefficients. In other words the HCF is the largest of all the common factors. The number set ranges number of problems and more are customizable on the PDF.
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Greatest Common Factor Worksheet. Greatest Common Factor Other contents. The number set ranges number of problems and more are customizable on the PDF. 18 24 1 x 18 1 x 24 2 x 9 2 x 12 3 x 6 3 x 8 4 x 4 x 6 5 x 5 x 6 rpts 6 rpts So the common factors of 18 and 24 are 1 2 3 and 6. GCF is also known as greatest common divisor GCD highest common factor HCF greatest common measure GCM or highest common divisor HCD.
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Printable worksheets for practice determining greatest common factors. 18 24 1 x 18 1 x 24 2 x 9 2 x 12 3 x 6 3 x 8 4 x 4 x 6 5 x 5 x 6 rpts 6 rpts So the common factors of 18 and 24 are 1 2 3 and 6. This worksheet helps fourth graders build a strong foundation in GCF concepts as they work with fractions and equivalence. Greatest common factor GCF Grade 5 Factoring Worksheet Find the greatest common factor of the two numbers shown. A large collection of GCF worksheets is meticulously drafted for students in grade 5 through grade 8.
Source: pinterest.com
In other words the HCF is the largest of all the common factors. This twelve-question worksheet addresses prime and composite numbers factors common factors and greatest common factor. Find numbers which are factors of two composites. Greatest Common Factor Finding the factors. Greatest common factor worksheets GCF of two numbers up to 30 923 KiB 1212 hits GCF of two numbers up to 50 978 KiB 1225 hits.
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These factoring worksheets are pdf files. You can create free printable worksheets for finding the greatest common factor GFC and least common multiple LCM of up to 6 different numbers. Jul 2 2016 - Grab these GCF worksheets that contain exercises like greatest common factor for the set of numbers prime factorization and Venn diagram to find GCF. The worksheets can be made in PDF or html formats and are customizable with lots of options. Add to my workbooks 1 Embed in my website or blog Add to Google Classroom.
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16122020 Thus the highest common factor of 25a 15ab The HCF. These exercises have students find all numbers which are factors in two composite numbers. Greatest Common Factor GCF Add to my workbooks 44 Download file pdf Embed in my website or blog Add to Google Classroom. We can see that the common prime factors are 22 and 3. You can create free printable worksheets for finding the greatest common factor GFC and least common multiple LCM of up to 6 different numbers.
Source: pinterest.com
Useful as guided practice independent practice homework or assessment. Worksheets for the greatest common factor GFC and least common multiple LCM of numbers. Since 10 2 5 2 2 1. Greatest Common Factor Finding the factors. Jul 2 2016 - Grab these GCF worksheets that contain exercises like greatest common factor for the set of numbers prime factorization and Venn diagram to find GCF.
Source: pinterest.com
Find numbers which are factors of two composites. The GCF of two numbers is the largest number that is a factor of both numbers. A large collection of GCF worksheets is meticulously drafted for students in grade 5 through grade 8. Worksheets for the greatest common factor GFC and least common multiple LCM of numbers. Greatest common factor GCF Grade 5 Factoring Worksheet Find the greatest common factor of the two numbers shown.
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In other words the HCF is the largest of all the common factors. Greatest Common Factor Other contents. The common factors or of 12 and 18 are 1 2 3 and 6. Greatest common factor GCF Grade 5 Factoring Worksheet Find the greatest common factor of the two numbers shown. 18 24 1 x 18 1 x 24 2 x 9 2 x 12 3 x 6 3 x 8 4 x 4 x 6 5 x 5 x 6 rpts 6 rpts So the common factors of 18 and 24 are 1 2 3 and 6.
Source: pinterest.com
The number set ranges number of problems and more are customizable on the PDF. Greatest common factor worksheets GCF of two numbers up to 30 923 KiB 1212 hits GCF of two numbers up to 50 978 KiB 1225 hits. The worksheets can be made in PDF or html formats and are customizable with lots of options. Children will learn how to identify the greatest common factor of two numbers in this engaging math worksheet. Printable worksheets for practice determining greatest common factors.
Source: pinterest.com
Printable worksheets for practice determining greatest common factors. Now we need to pick the common prime factors. Download and print these GCF worksheets to find the GCF of two numbers three. A large collection of GCF worksheets is meticulously drafted for students in grade 5 through grade 8. You can create free printable worksheets for finding the greatest common factor GFC and least common multiple LCM of up to 6 different numbers.
Source: pinterest.com
We can see that the common prime factors are 22 and 3. Of literal coefficients The highest common factor of s s is s. Greatest common factor GCF Grade 5 Factoring Worksheet Find the greatest common factor of the two numbers shown. This twelve-question worksheet addresses prime and composite numbers factors common factors and greatest common factor. A large collection of GCF worksheets is meticulously drafted for students in grade 5 through grade 8.
Source: pinterest.com
16122020 Thus the highest common factor of 25a 15ab The HCF. Greatest Common Factor GCF Other contents. Greatest common factor worksheets GCF of two numbers up to 30 923 KiB 1212 hits GCF of two numbers up to 50 978 KiB 1225 hits. The GCF of 10 2 2. Greatest common factor of two numbers less than 50 Below are six versions of our grade 5 math worksheet on finding the greatest common factor GCF of two numbers between 1-50.
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https://byjus.com/question-answer/2-under-root-3-plus-under-root-3-is-equals-to/ | 1,675,211,406,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499891.42/warc/CC-MAIN-20230131222253-20230201012253-00140.warc.gz | 169,865,831 | 22,342 | Question
# $2\sqrt{3}+\sqrt{3}$ is equal to
Open in App
Solution
## Find the value of$2\sqrt{3}+\sqrt{3}$.It is an example of adding two irrational numbers.On taking out$\sqrt{3}$ as the common factor$\begin{array}{rcl}2\sqrt{3}+\sqrt{3}& =& \left(2+1\right)\sqrt{3}\\ & =& 3\sqrt{3}\end{array}$$\therefore$$2\sqrt{3}+\sqrt{3}$ is equal to $3\sqrt{3}$.
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45 | 143 | 379 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 7, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2023-06 | latest | en | 0.675622 |
http://archive.org/stream/HistoryOfTheTheoryOfNumbersI/TXT/00000233.txt | 1,472,195,107,000,000,000 | text/html | crawl-data/CC-MAIN-2016-36/segments/1471982295358.50/warc/CC-MAIN-20160823195815-00258-ip-10-153-172-175.ec2.internal.warc.gz | 13,521,308 | 7,477 | # Full text of "History Of The Theory Of Numbers - I"
## See other formats
```224 HlSTOKY OF THE THEORY OF NUMBERS. [CHAP. VIII
L. Poinsot10 gave the proof due to Crelle.9
J. A. Grunert11 proceeded by induction from n — l to n, making use of the first part of Lagrange's proof. D. A. da Silva12 gave a proof.
NUMBER OF ROOTS OP HIGHER CONGRUENCES. G. Libri16 found that/(z, y, . . .)=0 (mod m) has
1 * 4 hr1 2kirf . . . 2kirf\ — S £ . . A S cos — -+i sin — -f mx~av~c U-o ™ m J
sets of solutions such that a^x^b, c^y^d, ____ The total number of sets of solutions is
155... l+e«2£+co.^+ . . . +cos 2(»-
V. A. Lebesgue17 proved that if p is a prime we obtain as follows the residue modulo p of the number Sk of sets of solutions of F(XI, . . ., sn)=0 (mod p), in which each xt- is chosen from 0, 1, . . ., p — 1, and F is a polynomial with integral coefficients. Let SA be the sum of the coefficients of the terms Axf. . .xk° of the expansion of Fp~l in which each of the exponents a, . . ., g is a multiple >0 of p-1. Then Sk= (-1) *+1 SA (mod p).
Henceforth, let p = /im-j-l. First, let F=zm-a. In F*"1 the coefficient of x"«»-i-»> is (P;1) ( - a)n= an (mod p) . The exponent of x will be a multiple >0 of p — 1 only when n = k(p — l)/d, for A; = 0, 1,. . ., d— 1, where dis the g. c. d. of m and p - 1. Thus ^sSa*^"1^ (mod p), while evidently /Si<p. According as a{p~1)/d= 1 or not, we get Si = d or 0.
Next, let F=-xm-aym-b. Set c=ai/m+Z>. In (xn-c)p-1 we omit the terms in which the exponent of x is not a multiple >0 of p — 1 and also the xm(P-i) not containing y. Since the arithmetical coefficient is =1 as in the first case, we get
chyjn(p-l-h) i C2h^n(p—l-2h) \ i^m-Dh^mk
In this, we replace ckh by those terms of (aym+b)hh in which the exponents are multiples >0 of p — 1, viz.,
Set 7/ = l, and sum for k = l, . . ., m — 1; we get — S2 (mod p). It is shown otherwise that S2 is a multiple < mp of m.
To these two cases is reduced the solution of
(1) F-aiXJm+. . .+ajfc£jtms=ci (mod p
"Jour, de MathSmatiques, 10, 1845, 12-15.
"Klttgel's Math. Worterbuch, 5, 1831, 1069-71.
"Proprietaries . . . Congruencias binomias, Lisbon, 1854. Cf. C. Alasia, Rivista di fisica, mat.
e ec. nat., 4, 1903, p. 9. "M6m. divers Savants Ac. Sc. de 1'Institut de France (Math.), 5, 1838, 32 (read 1825). Jour.``` | 900 | 2,398 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-36 | latest | en | 0.812298 |
https://www.khanacademy.org/math/algebra-home/alg-basic-eq-ineq/alg-solving-equations/v/representing-a-relationship-with-a-simple-equation | 1,674,903,421,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499541.63/warc/CC-MAIN-20230128090359-20230128120359-00092.warc.gz | 852,472,896 | 98,699 | If you're seeing this message, it means we're having trouble loading external resources on our website.
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## Algebra (all content)
### Unit 2: Lesson 2
Why we do the same thing to both sides of an equation
# Representing a relationship with an equation
Equations are about relationships (no, not girlfriends and boyfriends!) between the two sides of the equation. Let's again use a scale example to help us understand. Created by Sal Khan.
## Want to join the conversation?
• Why is learning geometry so important if all we need is algebra to prosper?
• Algebra is another branch of mathematics
• How do you find the balance if the right side doesn't have an 1 kg and a unknown kg like in the video. Or in simpler terms what can you do when you can't simply cross out the same units?
• Then you have to add some sort of mass (either 1kg's or unknowns) to both sides to isolate the unknown on the left side. If you have x+2=0 you will subtract 2 from both sides and you get x=-2. In algebra, the unknown can be negative, and either side of the equation can be negative as well. So x=-2 is completely ok. However it's not ok to say in word problems that you have a mass of -2kg.See?
• How do you find the equation for this relation:
{(1,2), (2,9), (3,28), (4,65), (5,126)}
For lower grade students, it really can be a problem.
Since 1^3=1, 2^3=8,3^3=27,4^3=64,5^3=125，（x^3 means x*x*x）
We will get y=x^3+1 in this question. (x,x^3+1)
• What is a mathematical relationship? (I will look, but first jotting it here)
• The word relationship means "how items are connected". The same definition applies in math. A relationship shows how numbers or variables are connected. For example, an equation where the value of one variable determines (calculates) the value of another variable is a relationship.
• If he used a symbol to represent the unknown number like x.Will the other mass must be known as another symbol to make make sense or not. I would like a fully detailed answer to this.
• You can obviously represent the unknown with a variable of your choice.
But in the above case , sal represented the unknown with a question mark just to make it clear that its an unknown .
Here , we knew the mass of the right side so that is why we didn't represented it by any variable . We were asked to find the mass of that unknown ( question mark ) on the left side .
And if we would have also represented the other know mass with a variable , then we would need atleast two equations to solve for the 2 unknown variables .
Hope this clears your doubt !!
• Can you use an X instead of a question mark?
• Yes! You can use anything, even a silly doodle, to represent an unknown.
• So, whatever you do to the right hand side, you have to do the same to the left?
• Yes. Otherwise the "equation" will no longer remain "equal".
Just think: if we have 5=2+3, and someone says, "I don't like the number 2 because it is an even number, so I'm going to remove it from the equation," erases it, and says, "Now there are only odd numbers in the equation, and I like it very much." Unfortunately, the "equation" will become 5=3. 5 of course does not equal 3. Since 2 was subtracted from the right hand side, 2 must also be subtracted from the left.
• is algebra the same thing as math or a type of math what is the catagory of algebra | 856 | 3,433 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2023-06 | latest | en | 0.917807 |
rvtournament.com | 1,718,784,459,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861806.64/warc/CC-MAIN-20240619060341-20240619090341-00378.warc.gz | 448,562,403 | 25,120 | Select Page
You can review your progress in RV Tournament on the My History page.
Rounds
Under the Rounds tab, you’ll find a list of all of the rounds that you’ve participated in. For each round you’ll see the date, whether or not you chose the correct target image, and how many points you gained or lost (based on the confidence level you submitted for that round). If you wrote or sketched on the Notes page for that round, you can tap the magnifying glass icon to view your notes. You can also tap the icon showing an arrow in a window, to view that round’s two image choices, and to see which one you picked and which one was the correct target image.
Statistics
Under the Stats tab, you’ll find some statistics describing your performance in RV Tournament.
Listed first are the number of rounds you’ve completed, and the percentage of those where you’ve chosen the correct target image.
Below these are several more stats, that only become available once you’ve completed at least 10 rounds. These stats are more complex, and are aimed at those players who are more scientifically or mathematically minded. They are briefly described on the page itself, but below you can find more detailed explanations.
Z Score
This is a measure of how well you’ve performed, compared to what would be expected by random chance alone. Because there are two images to choose from, players on average would be expected to choose the correct target image in 50% of rounds, just by chance, without any remote viewing. If you choose correctly in 50% of rounds, your z-score will be 0. If you perform worse than 50%, your z-score will be negative, and if you perform better than 50%, your z-score will be positive.
Even though players, on average, would get 50% of rounds correct by chance, in reality an individual player would be unlikely to get exactly 50% correct. Imagine flipping a coin 100 times — it’s unlikely that you’d get exactly 50 heads and 50 tails. In fact, on average, you’d get 5 more or less than 50. In statistics, this amount by which an individual result, on average, differs from the mean result, is called the standard deviation. Your z-score is just a count of how many standard deviations distant your performance is from the mean.
Here’s an example. Say you’ve played 100 rounds. Just as with 100 coin flips, the standard deviation is 5. So if you got 55 of those 100 rounds correct, you are one standard deviation away from the mean, and so your z-score is 1.0. If you got 60 correct, you’d be two standard deviations away from the mean, so your z-score would be 2.0. Say you only got 35 correct, then you’d be three standard deviations away from the mean in the other direction, so your z-score would be -3.
The higher your z-score, the less likely it could have happened by chance alone. Say you’ve played 100 rounds and your z-score is 3. That means your performance is 3 standard deviations higher than the mean, so you’ve gotten 65 out of those 100 rounds correct. That’s very unlikely to happen by chance alone!
Probability
This is the probability of performing at least as well as you have, by random chance alone. It’s based on your z-score.
A z-score of 0 means you’re doing exactly as well as would be expected, on average, by chance — that is, you’ve gotten 50% of rounds correct. So, if your z-score is 0, then the probability of performing that well (or better) by chance will of course be 50% — 50% are expected to perform better than average, while the other 50% are expected to perform worse than average.
The higher your z-score, the less likely it is to be a result of chance alone. For example a z-score of 1 has a 15.9% probability of occurring by chance, a z-score of 2 has a 2.3% probability of occurring by chance, and a z-score of 3 has only a 0.1% probability of occurring by chance!
Note that your z-score, and so also the probability it represents, depends a lot on how many rounds you’ve played. If you’ve played 10 rounds, it’s not too unlikely that you’ll get 8 of them correct by chance. But if you’ve played 100 rounds, it’s very unlikely that you’d get 80 of them correct by chance. So, to build up a high z-score (and a low probability of getting the same results by chance), you’ll need to maintain good performance over many rounds.
Z Score with Confidence
Your regular z-score treats each round like a coin flip. If you choose the correct target image, it counts as a 1; if you choose the incorrect image it counts as a 0. Your z-score with confidence, however, takes into consideration the degree of confidence that you submit with your choice. If you choose the correct image with 100% confidence, it counts as 1; and if you choose the incorrect image with 100% confidence, it counts as a 0. But if you were to choose the correct image with confidence of 50%, then it would count as 0.75; if you were to choose the incorrect image with 50% confidence it would count as 0.25. If you submit your choice as 0% confidence in either image, it counts as 0.5.
Some remote viewing experiments have shown better results when participants are able to submit a measure of confidence along with their choice. It will be interesting to see if RV Tournament players will tend to have higher values for their Z Score with Confidence, than they do for their normal Z Score.
Probability with Confidence
This is the probability of performing at least as well as you have, taking your confidence submissions into consideration, by random chance alone. It’s based on your Z Score with Confidence, in the same way that the normal Probability is based on the normal Z Score. | 1,277 | 5,628 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2024-26 | longest | en | 0.954444 |
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## 1.6.1 Summary of Degree of Freedom Table
Before we look at more complicated problems, we need to have a systematic procedure for analyzing interconnected units. We need particularly to have some way to determine which units in a plant can be (and should be) analyzed first. Then we want to establish an order for further analysis. The modules and units considered in the last sections could all be analyzed completely. In many systems, we will find that only a few units can be treated in that way. A system (module, unit or plant) that can be analyzed completely to determine all pertinent flow rates and reaction rates is said to be completely determined. Such a system is also said to have a Degree of Freedom of 0.
The degree of freedom for a unit is a measure of how many specifications we can (or need to) make so that all unknown flows and reaction rates can be found. The larger the degree of freedom for a unit, the less we know about it. Since flow and reaction rates are the unknowns while balances, given compositions, flows, flow ratios and conversions and splitter restrictions all reduce the uncertainty, the differences are what we call the degree of freedom.
Just as each unit in a plant has unknowns (flow and reaction rates) that we may try to solve for, the overall plant has a similar set of unknowns. In fact all we need to do is look strictly at those flow streams that enter and leave the plant and forget the streams that lead from one unit to another. We may have to lump reaction rates if the same reaction occurs in more than one unit because the overall system does not care where a reaction takes place. We also can find much information that is useless in our analysis of the overall system. Only if the information specifies something about the plant feed and product streams can it be used to reduce the degree of freedom for the overall system.
An entirely different problem is connected with analyzing the "process". In this case we look at all the unknowns in the plant. This includes the flows of all compounds in all streams and the rates of all reactions in all units. This time the same reaction is counted as many times as it occurs. In addition all information is usable.
The degree of freedom for the overall system and for the process are difficult to find unless you are careful about the meaning of each term in them. The following table shows a precise way to find the degree of freedom for each unit, the overall system and the process.
``` Unit 1 Unit 2 ..... Overall Process
Flow Variables FV1 FV2 FVO FVP
Reaction Variables R1 R2 RO RP
Balances B1 B2 BO BP
Given Compositions CP1 CP2 CPO CPP
Given Flows F1 F2 FO FP
Given Flow Ratios FR1 FR2 FRO FRP
Given Conversions CV1 CV2 CVO CVP
Splitter Restrictions S1 S2 SO SP
----- ----- ----- -----
Net Degree of Freedom °F1 °F2 °FO °FP
```
Specific definitions of the entries in the table are given next. Some of these are obvious, others can be quite confusing unless you are careful to think precisely about what they represent.
#### Flow Variables
FVk= the sum of the number of compounds that are involved in the balance equations in all the streams entering or leaving unit k. FVO= the sum of the number of compounds that are involved in the balance equations in all the streams entering or leaving the overall system. FVP= the sum of the number of compounds that are involved in the balance equations in all the streams anywhere in the system including those entering or leaving the overall system as well as those connecting the units.
#### Reaction Variables
Rk= the number of independent reactions occurring inside unit k. RO= the number of independent reactions occurring inside the overall system. Do not count each reaction more than once even if it occurs in more than one unit. RP= sum (Rk)
#### Balances
Bk= the number of compounds that are found in the flow streams in or out of unit k or take part in the reactions in the unit. Bk= the number of compounds that are found in the flow streams in or out of the overall system or take part in the reactions inside the system. BP= sum(Bk)
#### Specified Compositions
CPk= the number of compositions given for any flow stream into or out of unit k. The maximum number that is counted in any one stream is one less than the number of compounds in that stream. CPO= the number of compositions given for any flow stream into or out of the overall system. CPP= the number of compositions given for any flow stream anywhere in the system.
#### Specified Flow Rates
Fk= the number of flow rates given for the streams in or out of unit k. FO= the number of flow rates given for the streams in or out of the overall system. FP= the number of flow rates given for all the streams anywhere in the system.
#### Specified Flow Ratios
FRk= the number of flow ratios of streams into or out of unit k. Note that both streams in the ratio must be involved with unit k to be counted. FRO= the number of flow ratios of streams into or out of the overall system. Note that both streams in the ratio must be ones that enter or leave the overall system to be counted. FRP= the number of flow ratios given for streams anywhere in the system.
#### Specified Conversions
CVk= the number of conversions specified for reactants fed to unit k. Do not count a conversion of 100% if the reactant that is the basis of the conversion does not appear in any exit stream from the unit. CVO= the number of conversions specified for reactants fed to the overall system. Do not count a conversion of 100% if the reactant that is the basis of the conversion does not appear in any exit stream from the system. CVP= sum(CVk)
#### Splitter Restrictions
Sk= 0 unless the unit is a stream splitter. If it is a splitter, then it is (number streams leaving - 1)(number of compounds in the stream -1). SP= sum(Sk)
#### Degree of Freedom
°Fk = FVk + Rk - (Bk + CPk + Fk + FRk + CVk + Sk)
°FO = FVO + RO - (BO + CPO + FO + FRO + CVO + SO)
°FP = FP + RP - (BP + CPP + FP + FRP + CVP + SP)
### Degree of Freedom Table
Unit Name:
Unit 1
Unit 2
Unit 3
Unit 4
Unit 5
Unit 6
Over-all
Process
Flow
Variables
Reaction
Variables
Balances
Given
Compositions
Given Flows
Given Flow
Ratios
Given
Conversions
Splitter
Restrictions
Net °F
This is a typical degree of freedom table for a system with no more than 6 units.
## 1.6.2 Using the Modules to Simulate a Plant
Three procedures will be demonstrated in this section of the notes for simulating the steady state behavior of an entire plant. Each procedure uses the module functions shown in the previous section of the notes. The three methods are:
1. Interactive Use of the Modules
2. Using a General Purpose Driver
3. Writing a Program for a Specific Plant
Each method has its advantages and disadvantages. The modules may be readily combined to simulate almost any steady state chemical system. As soon as we encounter systems with a large number of units, we find that it is tedious to keep track of the order of the units and the information that needs to be given as we proceed through the system. It also becomes tedious if we find that a trial and error solution must be used to find some of the flow or reaction variables. Thus we need to adopt one of the last two procedures to be efficient. If we have only a few systems to analyze, a specific program written to simulate each system may be the best approach. This becomes less practical as the number of systems to be analyzed grows. Then we will find general purpose drivers will be very useful.
Two general purpose programs will be demonstrated in this chapter:
Function
Used For
1
drive
Driver function for combining modules. (in ~ceng301/matlab/massmods)
2
vweg2
Vector root finding module. (in ~ceng301/matlab/misc)
### Interactive Use of the Modules
Multi-unit systems can be simulated by combining several of the modular programs. To illustrate this, let's do Example Problem 3.13 in the text. Let the streams be numbered as in Figure 3.7 in the text. The chlorination reactor has two inlet streams and two outlet streams, but react can handle only one inlet and one outlet stream. Therefore, streams 1 and 2 must be mixed; then this mixture can be reacted. Afterwards the outlet from react can be separated using sep.
The new system will have 6 streams. The seven components are:
``` Component Name Formula
1 Benzene C6H6
2 Chlorobenzene C6H5Cl
3 Dichlorobenzene C6H4Cl2
4 Trichlorobenzene C6H3Cl3
5 Tetrachlorobenzene C6H2Cl4
6 Chlorine Cl2
7 Hydrogen Chloride HCl```
Using start301 to set the compounds and reactions, select New Session, followed by Mass Balances Only. We then wish to use the CENG 301 database since all the compounds we are going to use can be found in the database.
```>> start301
Welcome to CENG301's start301!!
Then click on the appropriate choice in the menu bar
1: Click (1) to start a new session
Input the name of your new file: chloros
The output file name is: chloros
Input the number of compounds: 7
The number of compounds is: 7
Enter the name of compound # 1: C6H6
Enter the name of compound # 2: C6H5Cl
Enter the name of compound # 3: C6H4Cl2
Enter the name of compound # 4: C6H3Cl3
Enter the name of compound # 5: C6H2Cl4
Enter the name of compound # 6: Cl2
Enter the name of compound # 7: HCl
Enter the number of reactions: 4
Enter the coefficients for each compound in the same order that
the compounds are listed. Coefficients for reactants should be
Negative, and coefficients for products should be positive
Enter the coefficients for each compound in reaction # 1
C6H6 C6H5Cl C6H4Cl2 C6H3Cl3 C6H2Cl4 Cl2 HCl
-1 1 0 0 0 -1 1
Enter the coefficients for each compound in the same order that
the compounds are listed. Coefficients for reactants should be
Negative, and coefficients for products should be positive
Enter the coefficients for each compound in reaction # 2
C6H6 C6H5Cl C6H4Cl2 C6H3Cl3 C6H2Cl4 Cl2 HCl
0 -1 1 0 0 -1 1
Enter the coefficients for each compound in the same order that
the compounds are listed. Coefficients for reactants should be
Negative, and coefficients for products should be positive
Enter the coefficients for each compound in reaction # 3
C6H6 C6H5Cl C6H4Cl2 C6H3Cl3 C6H2Cl4 Cl2 HCl
0 0 -1 1 0 -1 1
Enter the coefficients for each compound in the same order that
the compounds are listed. Coefficients for reactants should be
Negative, and coefficients for products should be positive
Enter the coefficients for each compound in reaction # 4
C6H6 C6H5Cl C6H4Cl2 C6H3Cl3 C6H2Cl4 Cl2 HCl
0 0 0 -1 1 -1 1
Do you want to check to see if the coefficients are correct?
type "y" for yes or simply press enter to move on:
Your reactions are currently as follows:
1) C6H6 + Cl2 --> C6H5Cl + HCl
2) C6H5Cl + Cl2 --> C6H4Cl2 + HCl
3) C6H4Cl2 + Cl2 --> C6H3Cl3 + HCl
4) C6H3Cl3 + Cl2 --> C6H2Cl4 + HCl
Type "y" to change an equation, or press "enter" to continue:
Here are your compounds' formulae and names:
No. Formula Name
----------------------------------------
1 C6H6 benzene
2 C6H5Cl chlorobenzene
3 C6H4Cl2 m-dichlorobenzene
4 C6H3Cl3 trichlorobenzene
5 C6H2Cl4 tetrachlorobenzene
6 Cl2 chlorine
7 HCl hydrogen chloride
----------------------------------------
1) C6H6 + Cl2 --> C6H5Cl + HCl
2) C6H5Cl + Cl2 --> C6H4Cl2 + HCl
3) C6H4Cl2 + Cl2 --> C6H3Cl3 + HCl
4) C6H3Cl3 + Cl2 --> C6H2Cl4 + HCl
Enter the number of streams: 6
The variables for your compounds have now been created,
you may continue, or come back later and reload the same data.
```
The flow of Benzene in stream 1 is 1000 and of Chlorine in stream 2 is 3600. These may be set by:
```>> ns(1,1)=1000;
>> ns(2,6)=3600;```
Mix streams 1 and 2 to get stream 5.
`>> mix([1 2],5)`
and we see:
```>> showm([1 2],5,10,1)
Compound Inlet | Outlet
Stream 1 2 Total | 5
benzene 1000.0 0.0 1000.0 | 1000.0
chlorine 0.0 3600.0 3600.0 | 3600.0
Total 1000.0 3600.0 4600.0 | 4600.0 ```
Then, stream 5 will be the inlet to the reactor. From material balances the rates of the reactions can be found (as in the text) to be:
` 990 920 800 50`
React stream 5 to get stream 6 and display the result:
```>> react([990 920 800 50],5,6)
>> showm(5,6,10,1)
Compound Inlet | Outlet
Stream 5 | 6
benzene 1000.0 | 10.0
chlorobenzene 0.0 | 70.0
m-dichlorobenzen 0.0 | 120.0
C6H3Cl3 0.0 | 750.0
C6H2Cl4 0.0 | 50.0
chlorine 3600.0 | 840.0
hydrogen chlorid 0.0 | 2760.0
Total 4600.0 | 4600.0 ```
Now, stream 6 must be separated into streams 3 and 4. All of the chlorine and HCl goes into stream 3 and the rest goes to stream 4:
```>> sep([0 0 0 0 0 1 1],6,[3 4])
>> showm(6,3:4,10,1)
Compound Inlet | Outlet
Stream 6 | 3 4 Total
benzene 10.0 | 0.0 10.0 10.0
chlorobenzene 70.0 | 0.0 70.0 70.0
m-dichlorobenzen 120.0 | 0.0 120.0 120.0
C6H3Cl3 750.0 | 0.0 750.0 750.0
C6H2Cl4 50.0 | 0.0 50.0 50.0
chlorine 840.0 | 840.0 0.0 840.0
hydrogen chloride 2760.0 | 2760.0 0.0 2760.0
Total 4600.0 | 3600.0 1000.0 4600.0```
Let's check to see that stream 4 has the compositions given in the problem.
```>> 100*ns(4,:)/sum(ns(4,:))
ans =
1 7 12 75 5 0 0```
It does. The flowrate of primary product, C6H3Cl3 , is seen in stream 4 to be 750.
## Go on to Page 2
[Go to previous section: 1.5| Go to next Chapter: 2.0] | 3,952 | 14,587 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2014-35 | longest | en | 0.963058 |
https://www.ademcetinkaya.com/2022/09/buy-sell-or-hold-bld-stock-forecast.html | 1,686,257,275,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224655143.72/warc/CC-MAIN-20230608204017-20230608234017-00554.warc.gz | 698,851,571 | 59,605 | Accurate stock market prediction is of great interest to investors; however, stock markets are driven by volatile factors such as microblogs and news that make it hard to predict stock market index based on merely the historical data. The enormous stock market volatility emphasizes the need to effectively assess the role of external factors in stock prediction. Stock markets can be predicted using machine learning algorithms on information contained in social media and financial news, as this data can change investors' behavior. We evaluate TopBuild prediction models with Modular Neural Network (Market News Sentiment Analysis) and Ridge Regression1,2,3,4 and conclude that the BLD stock is predictable in the short/long term. According to price forecasts for (n+3 month) period: The dominant strategy among neural network is to Buy BLD stock.
Keywords: BLD, TopBuild, stock forecast, machine learning based prediction, risk rating, buy-sell behaviour, stock analysis, target price analysis, options and futures.
## Key Points
1. How do you decide buy or sell a stock?
2. How do you pick a stock?
3. What are the most successful trading algorithms?
## BLD Target Price Prediction Modeling Methodology
Stock market predictions are one of the challenging tasks for financial investors across the globe. This challenge is due to the uncertainty and volatility of the stock prices in the market. Due to technology and globalization of business and financial markets it is important to predict the stock prices more quickly and accurately. Last few years there has been much improvement in the field of Neural Network (NN) applications in business and financial markets. Artificial Neural Network (ANN) methods are mostly implemented and play a vital role in decision making for stock market predictions. We consider TopBuild Stock Decision Process with Ridge Regression where A is the set of discrete actions of BLD stock holders, F is the set of discrete states, P : S × F × S → R is the transition probability distribution, R : S × F → R is the reaction function, and γ ∈ [0, 1] is a move factor for expectation.1,2,3,4
F(Ridge Regression)5,6,7= $\begin{array}{cccc}{p}_{a1}& {p}_{a2}& \dots & {p}_{1n}\\ & ⋮\\ {p}_{j1}& {p}_{j2}& \dots & {p}_{jn}\\ & ⋮\\ {p}_{k1}& {p}_{k2}& \dots & {p}_{kn}\\ & ⋮\\ {p}_{n1}& {p}_{n2}& \dots & {p}_{nn}\end{array}$ X R(Modular Neural Network (Market News Sentiment Analysis)) X S(n):→ (n+3 month) $\stackrel{\to }{R}=\left({r}_{1},{r}_{2},{r}_{3}\right)$
n:Time series to forecast
p:Price signals of BLD stock
j:Nash equilibria
k:Dominated move
a:Best response for target price
For further technical information as per how our model work we invite you to visit the article below:
How do AC Investment Research machine learning (predictive) algorithms actually work?
## BLD Stock Forecast (Buy or Sell) for (n+3 month)
Sample Set: Neural Network
Stock/Index: BLD TopBuild
Time series to forecast n: 15 Sep 2022 for (n+3 month)
According to price forecasts for (n+3 month) period: The dominant strategy among neural network is to Buy BLD stock.
X axis: *Likelihood% (The higher the percentage value, the more likely the event will occur.)
Y axis: *Potential Impact% (The higher the percentage value, the more likely the price will deviate.)
Z axis (Yellow to Green): *Technical Analysis%
## Conclusions
TopBuild assigned short-term B3 & long-term B2 forecasted stock rating. We evaluate the prediction models Modular Neural Network (Market News Sentiment Analysis) with Ridge Regression1,2,3,4 and conclude that the BLD stock is predictable in the short/long term. According to price forecasts for (n+3 month) period: The dominant strategy among neural network is to Buy BLD stock.
### Financial State Forecast for BLD Stock Options & Futures
Rating Short-Term Long-Term Senior
Outlook*B3B2
Operational Risk 4062
Market Risk5238
Technical Analysis5884
Fundamental Analysis5540
Risk Unsystematic5230
### Prediction Confidence Score
Trust metric by Neural Network: 73 out of 100 with 595 signals.
## References
1. K. Boda and J. Filar. Time consistent dynamic risk measures. Mathematical Methods of Operations Research, 63(1):169–186, 2006
2. D. White. Mean, variance, and probabilistic criteria in finite Markov decision processes: A review. Journal of Optimization Theory and Applications, 56(1):1–29, 1988.
3. Dimakopoulou M, Athey S, Imbens G. 2017. Estimation considerations in contextual bandits. arXiv:1711.07077 [stat.ML]
4. Hastie T, Tibshirani R, Tibshirani RJ. 2017. Extended comparisons of best subset selection, forward stepwise selection, and the lasso. arXiv:1707.08692 [stat.ME]
5. Rosenbaum PR, Rubin DB. 1983. The central role of the propensity score in observational studies for causal effects. Biometrika 70:41–55
6. A. Y. Ng, D. Harada, and S. J. Russell. Policy invariance under reward transformations: Theory and application to reward shaping. In Proceedings of the Sixteenth International Conference on Machine Learning (ICML 1999), Bled, Slovenia, June 27 - 30, 1999, pages 278–287, 1999.
7. Candès EJ, Recht B. 2009. Exact matrix completion via convex optimization. Found. Comput. Math. 9:717
Frequently Asked QuestionsQ: What is the prediction methodology for BLD stock?
A: BLD stock prediction methodology: We evaluate the prediction models Modular Neural Network (Market News Sentiment Analysis) and Ridge Regression
Q: Is BLD stock a buy or sell?
A: The dominant strategy among neural network is to Buy BLD Stock.
Q: Is TopBuild stock a good investment?
A: The consensus rating for TopBuild is Buy and assigned short-term B3 & long-term B2 forecasted stock rating.
Q: What is the consensus rating of BLD stock?
A: The consensus rating for BLD is Buy.
Q: What is the prediction period for BLD stock?
A: The prediction period for BLD is (n+3 month) | 1,449 | 5,839 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 2, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2023-23 | latest | en | 0.933577 |
http://www.numbersaplenty.com/70000000 | 1,579,372,247,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250593295.11/warc/CC-MAIN-20200118164132-20200118192132-00076.warc.gz | 263,541,015 | 4,234 | Search a number
70000000 = 27577
BaseRepresentation
bin1000010110000…
…01110110000000
311212201100221121
410023001312000
5120410000000
610540202024
71506663430
oct413016600
9155640847
1070000000
1136571044
121b539314
131166b825
1494222c0
15622ab1a
hex42c1d80
70000000 has 128 divisors (see below), whose sum is σ = 199218240. Its totient is φ = 24000000.
The previous prime is 69999989. The next prime is 70000027. The reversal of 70000000 is 7.
It is a happy number.
It is a sliding number, since 70000000 = 20000000 + 50000000 and 1/20000000 + 1/50000000 = 0.000000070000000.
It is a tau number, because it is divible by the number of its divisors (128).
It is an ABA number since it can be written as A⋅BA, here for A=7, B=10.
It is a Harshad number since it is a multiple of its sum of digits (7).
It is a super Niven number, because it is divisible the sum of any subset of its (nonzero) digits.
It is a nialpdrome in base 10.
It is a congruent number.
It is an unprimeable number.
It is a polite number, since it can be written in 15 ways as a sum of consecutive naturals, for example, 9999997 + ... + 10000003.
Almost surely, 270000000 is an apocalyptic number.
70000000 is a gapful number since it is divisible by the number (70) formed by its first and last digit.
It is an amenable number.
It is a practical number, because each smaller number is the sum of distinct divisors of 70000000, and also a Zumkeller number, because its divisors can be partitioned in two sets with the same sum (99609120).
70000000 is an abundant number, since it is smaller than the sum of its proper divisors (129218240).
It is a pseudoperfect number, because it is the sum of a subset of its proper divisors.
70000000 is an frugal number, since it uses more digits than its factorization.
70000000 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 56 (or 14 counting only the distinct ones).
The product of its (nonzero) digits is 7, while the sum is 7.
The square root of 70000000 is about 8366.6002653408. The cubic root of 70000000 is about 412.1285299809.
The spelling of 70000000 in words is "seventy million", and thus it is an aban number and an uban number. | 668 | 2,227 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2020-05 | latest | en | 0.905121 |
https://myassignmentguru.com/assignments/acst201-today-is-1-january-2019-mqu-bank-is-offering-a-y-year-1500000-loan-product/ | 1,718,760,686,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861796.49/warc/CC-MAIN-20240618234039-20240619024039-00672.warc.gz | 360,064,665 | 17,747 | Throughout this assessment you will need to consider y years. To calculate y, you will use your student ID number. Let r be the integer remainder after dividing your student ID number by 25. Let y = r + 20. For example, if the ID number is 12345678, then 12345678 = 493827×25+3, r = 3 and y = 23. You should use this value for y throughout the assessment.
Today is 1 January 2019. MQU bank is offering a y year \$1,500,000 loan product from the beginning of 2019 to its customers.
a. This loan product requires customers to make monthly repayments. Payments will be made at the beginning of each month with an amount of \$9,000. Use the Goal Seek to find the implied annual nominal rate of interest payable monthly (i.e., j12) charged by MQU bank. Assume that there is an annual fee of \$500 paid on 1 January of each year for the first 10 years (the first one is paid on 1 January 2019) and there is an annual fee of \$1000 paid on 1 July of each year for the remaining term of the loan. The first loan payment is made today.
b. For this loan product, each customer has an option to choose a 3-year interest-only period. Here is the option detail
• Customers can borrow \$1,500,000 on 1 January 2019 and make the repayments that only cover the interest on the borrowed amount (i.e., \$1,500,000) during the interest-only period (i.e., first three years). The interest rate is j12 = 6% for the interest-only period.
• After the interest only period, customers’s repayments will be paid at the beginning of each month with an amount of \$9,500 for the remaining y− 3 years.
• Assume that there is an annual fee of \$700 paid on 1 January of each year during the loan term (i.e., y years). The first one is paid on 1 January 2019
Calculate the monthly repayment amount for the interest-only period and use the Goal Seek to find the implied annual nominal rate of borrowing cost payable monthly (i.e., j12) charged by MQU bank.
c. Based on the result of part b, find the following equivalent nominal interest rates: j2, j4 and j365. Use a bar or column chart to plot these rates.
Click on Buy Solution and make payment. All prices shown above are in USD. Payment supported in all currencies. Price shown above includes the solution of all questions mentioned on this page. Please note that our prices are fixed (do not bargain).
After making payment, solution is available instantly.Solution is available either in Word or Excel format unless otherwise specified. | 591 | 2,468 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2024-26 | latest | en | 0.955799 |
https://fresherbell.com/quizdiscuss/python/what-will-be-the-output-of-the-following-python-co3213 | 1,723,354,315,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640975657.64/warc/CC-MAIN-20240811044203-20240811074203-00353.warc.gz | 211,463,894 | 8,591 | # Quiz Discussion
What will be the output of the following Python code?
['f', 't'][bool('spam')]
Course Name: Python
• 1]
Error
• 2]
No output
• 3]
f
• 4]
t
##### Solution
No Solution Present Yet
#### Top 5 Similar Quiz - Based On AI&ML
Quiz Recommendation System API Link - https://fresherbell-quiz-api.herokuapp.com/fresherbell_quiz_api
# Quiz
1
Discuss
What will be the output of the following Python code?
if (9 < 0) and (0 < -9):
print("hello")
elif (9 > 0) or False:
print("good")
else:
print("bad")
• 1]
• 2]
good
• 3]
hello
• 4]
error
##### Solution
2
Discuss
To find the decimal value of 1111, that is 15, we can use the function:
• 1]
int(‘1111’,2)
• 2]
int(1111,2)
• 3]
int(‘1111’,10)
• 4]
int(1111,10)
##### Solution
3
Discuss
What will be the output of the following Python expression?
int(1011)?
• 1]
1101
• 2]
13
• 3]
11
• 4]
1011
##### Solution
4
Discuss
What will be the output of the following Python expression?
~100?
• 1]
-100
• 2]
100
• 3]
-101
• 4]
101
##### Solution
5
Discuss
What will be the value of the following Python expression?
4+2**5//10
• 1]
0
• 2]
77
• 3]
7
• 4]
3
##### Solution
6
Discuss
What will be the output of the following Python code?
class Truth:
pass
x=Truth()
bool(x)
• 1]
error
• 2]
false
• 3]
true
• 4]
pass
##### Solution
7
Discuss
What will be the output of the following Python code?
l=[1, 0, 2, 0, 'hello', '', []]
list(filter(bool, l))
• 1]
[1, 2, ‘hello’]
• 2]
[1, 0, 2, ‘hello’, ”, []]
• 3]
[1, 0, 2, 0, ‘hello’, ”, []]
• 4]
Error
##### Solution
8
Discuss
Which among the following list of operators has the highest precedence?
+, -, **, %, /, <<, >>, |
• 1]
%
• 2]
|
• 3]
**
• 4]
<<, >>
##### Solution
9
Discuss
Which of the following expressions is an example of type conversion?
• 1]
3 + 7
• 2]
5.0 + 3
• 3]
5.3 + 6.3
• 4]
4.0 + float(3)
##### Solution
10
Discuss
What is the value of the following Python expression?
bin(0x8)
• 1]
‘0b1000’
• 2]
1000
• 3]
8
• 4]
‘0bx1000’
# Quiz | 734 | 2,060 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2024-33 | latest | en | 0.5551 |
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Vocabulary Cards and Word Walls Revised: June 29, 2011 Important Notes for Teachers: The vocabulary cards in this file match the Common Core, the math curriculum adopted by the Utah State Board of Education, | 5,910 | 26,136 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2018-30 | longest | en | 0.661393 |
http://slideplayer.com/slide/2846346/ | 1,503,540,172,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886126017.3/warc/CC-MAIN-20170824004740-20170824024740-00651.warc.gz | 364,330,546 | 19,054 | # Mathematics in the Prep School. Objectives To give you an overview of how Maths is taught in the Prep School To inform you of important changes.
## Presentation on theme: " Mathematics in the Prep School. Objectives To give you an overview of how Maths is taught in the Prep School To inform you of important changes."— Presentation transcript:
Mathematics in the Prep School
Objectives To give you an overview of how Maths is taught in the Prep School To inform you of important changes to the National Curriculum in Maths To give you some practical ideas of how to help your child succeed in Maths To show you the ActiveLearn interactive platform and how it supports learning
Overview Overall aim: for children to enjoy learning in Maths Every lesson/ activity differentiated to suit the learner Concrete materials used to develop understanding- not just tricks Teaching a range of strategies- mental and written- what works for the child? Towards problem solving- group work
National Curriculum Move to exposing children to fractions and decimals earlier More ‘Secondary Maths’ in Year 6 eg. long division Practice, practice, practice- more repetition of key number facts eg. times-tables No Calculators in KS2 SATS- mental methods and efficient written methods as a focus At Finborough, we take the best bits!
How to help….. Take the two handouts (targets and websites) and use these at home Make it enjoyable- our main aims should be the same Good old fashioned times-table and number bonds practice is always useful eg. timing, races, rhymes Involve children in the time, DIY measurements, logic games eg. cards, dominoes etc. The answer is ___. What’s the question? Give children a pair of compasses, a ruler, a protractor- explore symmetry and making shapes or patterns
ActiveLearn http://www.activelearnprimary.co.uk http://www.activelearnprimary.co.uk
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Similar presentations | 508 | 2,131 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2017-34 | longest | en | 0.874479 |
https://mathbench.umd.edu/modules/misc_scaling/page23.htm | 1,718,623,603,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861701.67/warc/CC-MAIN-20240617091230-20240617121230-00049.warc.gz | 341,272,916 | 2,744 | # The derivation of the 3/4 scaling law
So, our new formula for "b" is:
Based on some of the principles above (and some we haven't presented yet) - the authors suggest that size (and by size, they really mean the volume of the transport network) is related to three main factors: 1) the number of times the network branches from the top level to the final level (N), 2) how the branches decrease in radius from level to level ( a value they call γ), and 3) how the branches decrease in length from level to level (a value they call β). They then show that size (volume) is proportional to the quantity (γβ2)-N. So using that information, we can do some further manipulations and come up with a new equation:
Finally, the authors show us that γ is equal to n-1/3 and β is equal to n-1/2, so we can do one final manipulation:
Whew!!!! We finally arrived at our final destination - that "b" should be equal to 3/4!!! | 229 | 917 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2024-26 | latest | en | 0.926912 |
http://www.ionizationx.com/index.php?PHPSESSID=718kq9o8kc19c519ca4vev1j96&topic=1251.5;wap2 | 1,670,190,901,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710980.82/warc/CC-MAIN-20221204204504-20221204234504-00225.warc.gz | 79,770,062 | 1,994 | Stanley Meyer > Stan Meyers system 3
Start here your topics on the water injector & ambient airgasprocessor system
<< < (2/27) > >>
sebosfato:
exactly
check what i said
sebosfato:
0,000007.4 liters * 6000 injections in 1 minute * 60minutes =2,6 liters per hour maintaining 3000 rpm he clearly stated this
sebosfato:
about the energy content he say that in one gallon of water you have 1,66 lb of hydrogen and in one gallon of gasoline you have 0,66 if you divide 1,66/0,66= 2,666 considering the contaminants he said 10 % nitrogen present on water should give you the famous 2,5 times more powerful
this is a comparison by weight
outlawstc:
your right i made a mistake... i will go over it tomorrow and fix it my brains tired.. have a nasty lil sinus cold and been reading to much today.. thanks for posting the by weight comparison..
sebosfato:
ok
we could also state 2,6 liters of water *1847 gas expansion = 4800 liters of gas in one hour
at 3000rpm about 80 liters of gas per minute or 1,3 liters of gas per second
injections per second 3000/2*4/60 = 100 counting all injectors
so for every injection of gas in the case you would need 1,3 liter / 100 = 0,013 liters of oxygen hydrogen per injection
i estimated that each injector would works thus 25 times for second so if we have 4 cycles the time you would have to inject this gas inside the engine would be like 1/100 of a second or 0,01 second or 10 milliseconds remember at 3000 rpm if you double rpm the time to inject will be halved. | 409 | 1,514 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2022-49 | latest | en | 0.856733 |
https://www.enotes.com/homework-help/simplify-give-answer-with-non-negative-exponents-448787 | 1,516,547,184,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084890771.63/warc/CC-MAIN-20180121135825-20180121155825-00773.warc.gz | 865,407,091 | 9,339 | # Simplify, give answer with non-negative exponents. (2x-y)^2
justaguide | Certified Educator
The expression (2x - y)^2 has to be simplified.
Use the expansion (a + b)^2 = a^2 + 2*a*b + b^2
(2x - y)^2
= (2x)^2 - 2*(2x)*y + y^2
= 4x^2 - 4xy + y^2
The simplified form of (2x - y)^2 = 4x^2 - 4xy + y^2 | 132 | 305 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-05 | latest | en | 0.713416 |
https://www.teachoo.com/3382/744/Ex-10.1--4---In-Fig.--identify-the-vectors.-(i)-coinitial/category/Ex-10.1/ | 1,726,447,300,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651668.26/warc/CC-MAIN-20240915220324-20240916010324-00622.warc.gz | 937,463,749 | 21,551 | Ex 10.1
Chapter 10 Class 12 Vector Algebra
Serial order wise
### Transcript
Ex 10.1, 4 In Fig 10.6 (a square), identify the following vectors. (i) coinitialTwo or more vectors are said to be cointial if they start from the same point. Here, 𝑎 ⃗ and 𝑑 ⃗ start from the same point P. So, 𝒂 ⃗ and 𝒅 ⃗ are cointial. Ex 10.1, 4 In Fig 10.6 (a square), identify the following vectors. (ii) EqualTwo or more vectors are said to be equal if they have the same magnitude same direction. Since the given figure is a square, magnitudes of 𝑎 ⃗ , 𝑏 ⃗ , 𝑐 ⃗ and 𝑑 ⃗ are equal. And direction of 𝑏 ⃗ is same as direction of 𝑑 ⃗ Thus, 𝒃 ⃗ and 𝒅 ⃗ are equal. Ex 10.1, 4 In Fig (a square), identify the following vectors. (iii) Collinear but not equalTwo of more vectors are said to be collinear if they are parallel to the same line. 𝑏 ⃗ and 𝑑 ⃗ are parallel to line 𝑚 ⃗ ; hence collinear. But 𝑏 ⃗ = 𝑑 ⃗ 𝑎 ⃗ and 𝑐 ⃗ are parallel to line 𝑛 ⃗ ; hence collinear. But 𝑎 ⃗ ≠ 𝑐 ⃗ , since they are opposite in direction. ∴ 𝒂 ⃗ and 𝒄 ⃗ are collinear but not equal. | 393 | 1,041 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2024-38 | latest | en | 0.933041 |
https://www.physicsforums.com/threads/is-it-possible-to-fly-horizontally-using-augmented-fly-pack.921044/ | 1,558,716,538,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232257699.34/warc/CC-MAIN-20190524164533-20190524190533-00143.warc.gz | 884,552,177 | 18,685 | # Is it possible to fly horizontally using augmented fly pack?
#### Tabaristiio
Assume there is a wingless augmented flight pack such as a jetpack attached to a person's body. Assuming that there is sufficient amount of power.
will it be possible according to the laws of physics to fly in a horizontal position instead of a vertical position?
What are the requirements for someone to be able to fly horizontally using a wingless augmented flight pack?
This question ignores irrelevant factors such as health risks and etc. It's just asking if it's possible within the laws of physics to fly horizontally using a wingless flight pack.
Related General Engineering News on Phys.org
#### FactChecker
Gold Member
2018 Award
The thrust would have two components -- one part vertical and one part horizontal. As long as the vertical part is equal to the force of gravity, the flight would be horizontal. That is a lot easier if the jetpack exhaust is pointing down. If it was pointing more horizontally, it would have to be a huge jet to have a downward component great enough to oppose gravity. Then the horizontal component and resulting horizontal speed of flight would be like a rocket.
#### cosmik debris
In skydiving (ski-jumping too) it is possible to shape one's body in the general shape of an aerofoil and gain considerable lift, one can even notice stall. Whether there is sufficient lift at non-injurious speeds I'm not sure.
Cheers
#### Tabaristiio
The thrust would have two components -- one part vertical and one part horizontal. As long as the vertical part is equal to the force of gravity, the flight would be horizontal. That is a lot easier if the jetpack exhaust is pointing down. If it was pointing more horizontally, it would have to be a huge jet to have a downward component great enough to oppose gravity. Then the horizontal component and resulting horizontal speed of flight would be like a rocket.
Yes, I was referring to flying horizontally with the body in a horizontal position during flight. So does everything else you wrote still apply?
#### jbriggs444
Homework Helper
Yes, I was referring to flying horizontally with the body in a horizontal position during flight. So does everything else you wrote still apply?
So the jet pack is horizontal -- purely horizontal thrust. And the person's body is horizontal. No lift from the person's body. No wings allowed. So there is no vertical thrust of any sort. Then the jet pack and human will fall like a rock and hit the ground. Possibly miles ahead if the launch was from the top of a cliff.
The only way out would be to reach orbital velocity. Move horizontally fast enough and the ground curves downward beneath you as fast as you fall. That would burn you to a cinder if it didn't tear you apart first. But we are told to ignore such issues.
#### FactChecker
Gold Member
2018 Award
Yes, I was referring to flying horizontally with the body in a horizontal position during flight. So does everything else you wrote still apply?
The jet exhaust would have to tilt down. It needs a vertical component large enough to counteract gravity. It seems impractical to me. Wings are very important for horizontal flight.
#### Tabaristiio
The jet exhaust would have to tilt down. It needs a vertical component large enough to counteract gravity. It seems impractical to me. Wings are very important for horizontal flight.
What do you mean by a vertical component? Could you provide an example of something like that please?
#### Tabaristiio
So the jet pack is horizontal -- purely horizontal thrust. And the person's body is horizontal. No lift from the person's body. No wings allowed. So there is no vertical thrust of any sort. Then the jet pack and human will fall like a rock and hit the ground. Possibly miles ahead if the launch was from the top of a cliff.
The only way out would be to reach orbital velocity. Move horizontally fast enough and the ground curves downward beneath you as fast as you fall. That would burn you to a cinder if it didn't tear you apart first. But we are told to ignore such issues.
The only rule is that the person's body is horizontal during flight and no wings are allowed. The jetpack itself can be arranged in anyway required in order to accomplish this and accommodate the body. If possible, it's allowed for the jetpack to be arranged in a way where it can do enable the horizontal fight. This includes enabling vertical thrust.
How do other wingless aircrafts fly horizontally? Assuming enough power exists, can't human augmented jet packs / flight packs also fly like that? Or is it mandatory for the aircraft to have to travel at orbital speeds in order to accomplish this?
#### CWatters
Homework Helper
Gold Member
The only rule is that the person's body is horizontal during flight and no wings are allowed. The jetpack itself can be arranged in anyway required in order to accomplish this and accommodate the body. If possible, it's allowed for the jetpack to be arranged in a way where it can do enable the horizontal fight. This includes enabling vertical thrust.,
Sure that would work. Point the jets down at some angle and some of the thrust stops him falling and some pushes him along horizontally. Take a look at the Harrier jump jet and it's vectored thrust. Yes it has wings but at slow speeds these have no lifting effect.
#### FactChecker
Gold Member
2018 Award
What do you mean by a vertical component? Could you provide an example of something like that please?
Any force can be divided into parts of force in different directions. The force of a jetpack can be divided into the force upward and the forces sideways (horizontally). You can call those the components of the total force.
"Is it possible to fly horizontally using augmented fly pack?"
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# m238sp06t1z - Dierential Equations test one Tuesday x 1...
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Differential Equations test one Tuesday, April 25, 2006 1 . Given the function y = 4+ x 2 sin s 2 ds ( a ) Calculate y (2). ( b ) Calculate dy dx . 2 . Is the function y = x + 1 x a solution of dy dx + y 2 = 3+ x 2 ? 3 . Determine whether each of the following equations is linear or non-linear. ( a ) dy dt = t - y 2 ( b ) t dy dt = e t y - cos t 2 ( c ) y dy dt = 4 t - y 4 . Which of the following equations is exact? ( a ) 3 x 2 - 4 y dx +(2 y - 4 x ) dy = 0 ( b ) dy dx + x 2 y = x 5 . Given the initial value problem t dy dt + 3 t - 2 y = 4 t + 1 and y (3) = 5 determine the largest t -interval over which a unique solution is guaranteed to exist. 6 . Given the initial value problem dy dx = y 3 + x 3 y and y (3) = 0 can one be certain that a unique solution exists?
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page two 7 . Use Euler’s method (i.e. the tangent-line method) with step-size h = 1 / 2 to fill in the following table for the initial value problem dy dx = 2 x - y 2 and y (1) = 2 x y dy dx approximate Δ y 1 2 1 . 5 8 . Solve each of the following differential equations or initial value problems:
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# ECO 403 MACROECONOMICS Assignment NO. 2 Discussion and Solution Due Date: 02 February, 2016
MACROECONOMICS (ECO 403)
ASSIGNMENT # 02
Marks 10 DUE DATE: 02 FEBRUARY, 2016
GRACE PERIOD: AVAILABLE
The Case:
Brazil, an economy in South American region, is growing rapidly. One of the reasons of this growth is exploration of natural resources like gold, iron ore, manganese, nickel, phosphates, platinum, tin, uranium, petroleum, hydropower and many others. Major industries based on these natural resources are promoting business environment which help to create macroeconomic stability. Government of Brazil wants to analyze the macroeconomic stability of the economy. Aggregate expenditures are important part of macroeconomic analysis. Aggregate expenditures of Brazil economy are consisting of expenditures by household, business and government. These expenditures are also known as ‘Planned Expenditures’. It can be denoted in an equation as:
Planned Aggregate Expenditures = AE = C + I + G
Keeping in view the above scenario of Brazil economy where Keynesian cross analysis will apply. Consumption function of Brazil economy is given by an equation:
C = 450 + 0.5(Y-T)
Where
. 0.5 is MPC
. Y represents income = Actual Expenditures
. T shows taxes
We Assume
. Income = Actual Expenditures = Y = 1500
. Planned investment = I = 200
. Government Expenditure = G = 200
. Taxes = T = 200
Requirements:
Keeping in view the above Scenario and given information of Brazil Economy,
Calculate the following:
1. Consumption Level (C)
2. Planned Aggregate Expenditures (AE)
3. Government Purchases Multiplier, if MPC is 0.5
4. Tax Multiplier, if MPC is 0.5
Marks: (2.5+2.5+2.5+2.5)
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### Replies to This Discussion
Our main purpose here discussion not just Solution
We are here with you hands in hands to facilitate your learning and do not appreciate the idea of copying or replicating solutions.
The Case:
Brazil, an economy in South American region, is growing rapidly. One of the
reasons of this growth is exploration of natural resources like gold, iron ore,
manganese, nickel, phosphates, platinum, tin, uranium, petroleum, hydropower
and many others. Major industries based on these natural resources are
promoting business environment which help to create macroeconomic stability.
Government of Brazil wants to analyze the macroeconomic stability of the
economy. Aggregate expenditures are important part of macroeconomic analysis.
Aggregate expenditures of Brazil economy are consisting of expenditures by
household, business and government. These expenditures are also known as
‘Planned Expenditures’. It can be denoted in an equation as:
Planned Aggregate Expenditures = AE = C + I + G
Keeping in view the above scenario of Brazil economy where Keynesian cross
analysis will apply. Consumption function of Brazil economy is given by an
equation:
C = 450 + 0.5(Y-T)
Where
0.5 is MPC
Y represents income = Actual Expenditures
T shows taxes
We Assume
Income = Actual Expenditures = Y = 1500
Planned investment = I = 200
Government Expenditure = G = 200
Taxes = T = 200
Requirements:
Keeping in view the above Scenario and given information of Brazil Economy,
Calculate the following:
1. Consumption Level (C)
2. Planned Aggregate Expenditures (AE)
3. Government Purchases Multiplier, if MPC is 0.5
4. Tax Multiplier, if MPC is 0.5
Marks: (2.5+2.5+2.5+2.5)
Is assignment ka solution lecture no 26 me hey right? tax multiplier key formuly sy tax multiplier ajaigaa baki ka koi betady plzzz
kindly send solution through tax multiplier method...at least discussion tou start ho jayay ge...plz
MACROECONOMICS (ECO 403)
ASSIGNMENT # 02
Hakim Zeeshan (MC-150200141)
Requirements:
Keeping in view the above Scenario and given information of Brazil Economy,
Calculate the following:
1. Consumption Level (C)
2. Planned Aggregate Expenditures (AE)
3. Government Purchases Multiplier, if MPC is 0.5
4. Tax Multiplier, if MPC is 0.5
1. Consumption Level = C
= C (Y-T)
450+0.5 (1500-200)
450 + 650
1100
C = 1100
2. PAE = C + I + G
1100 + 200 + 200
1500
PAE = 1500
3. GPM, if MPC = 0.5
= 1/1.MPC
2
GPM = 2
4. Tax Multiplier
= -MPC/1-MPC
-0.5/1-0.5
1
Tax multiplier = 1
Remember me in your prayers :)
thnx
no thanx bro
zeeshan bro love u but i think there is a little amendment in part three whose formula is written wrong
Y/T = 1/1.MPC =2
Y= 2G
SO I THINK ANSWER IS 2G NOT 2
WHAT DO U SAY
ANS 4 IS ALSO WRONG, IT SHOULD BE -1
yes it should be -1.. written by mistake :)
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3 | 1,828 | 6,901 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-34 | latest | en | 0.884528 |
https://www.mbatious.com/topic/1006/quant-boosters-vikas-saini-set-4/44 | 1,558,842,485,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232258621.77/warc/CC-MAIN-20190526025014-20190526051014-00127.warc.gz | 859,938,880 | 19,371 | # Quant Boosters - Vikas Saini - Set 4
• 14 mod 4 = 2.
We need to find unit digit number just from 1 to 9,then multiply by 10.
Unit digit of 1,5,6 is 1,5,6 respectively.
Unit digit of 2,3,7,8 is 4,9,9,4 respectively.
Unit digit of 4 and 9 is 6 and 1.
Add all = 1+5+6+4+9+9+4+6+1 = 45.
We will multiply it by 10 because unit digit will be as same as 1 to 9 of 11 to 19,21 to 29 and so on.
Now 45 x 10 = 450.
Hence unit digit 0.
• Q18) Find unit digit of ((1997)^41)^401
• Unit digit is 7.We have to use cycle of 4.
41^401 mod 4 = 1.
7^1 = 7.
Unit digit is 7.
• Q19) Find the last two digits of 147^912
• Concept - To get unit digit number we see only unit digit whereas to get last two digits, we see only last two digits.
If any number’s unit digit is 1, then to get last two digits we apply this formula
(xyz1)^abc = (c x z)/1 (last two digit)
Then unit digit is nothing but 1.
Ten’s digit = z x c.
Examples -
Find the last two digits of 21^3.
Unit digit is 1.
Ten’s digit = 2 x 3 = 6.
Hence last two digits of 21^3 is 61.
(21^3 = 9261).
Find the last two digit of 91^2.
Unit digit is 1.
Ten’s digit = 9 x 2 = 18 = 8.
Hence last two digits of 91^2 = 81.
(91^2 = 8281)
Shortcut approaches we will apply here :-
1. Try to get unit digit 1, so that we could apply above written formula.
Though it’s possible in case of odd integers only.
2. We need to find remainder of power after dividing by 40 only.
3. In case of even number, we shall try to make power of 2.
2^10 = 24 (last two digits)
2^20k = 76(last two digits)
In our given question, We need to focus on 47 from 147.
We know for 7, cycle of 4 works.
7^4 = 1(unit digit)
Hence 47^4 = 81(last two digits).
912 mod 40 = 32.
32 = 4 x 8.
Therefore 47^912 = 81^8.
We have unit digit 1,we can apply formula here.
Unit digit = 1.
Ten’s digit = 8 x 8 = 64 = 4.
Hence last two digits = 41.
• Q20) Find the last two digits of 296^144
• We take only 96.
As we can’t get 1 in any power of 6.
So 144 mod 40 = 24.
Therefore 296^144
= 96^24
= 16^12
= ((2)^4)^12
= 2^48
= 2^(20k+8 )
= 2^20k x 2^8
= 76 x 56
= 56.
Hence last two digits are 56.
• Q21) Find last two digits of 488^488
• We take 88 from 488.
488 mod 40 = 8.
= 88^8
= (2^3 x 11 )^8
= 2^24 x 11^8
= 2^(20K+4) x 11^8
= 2^20k x 2^4 x 11^8
= 76 x 16 x 81
= 16 x 81
= 96
Hence last two digits 96.
• Q22) Find the last two digits of 297^455
• We take 97 from 297.
455 mod 40 = 15.
Therefore 297^455
= 97^15
= (97)^4k + 3
= (97^4)^3 x 97^3
=81^3 x 97^2 x 97
= 41 x 09 x 97
= 69 x 97
= 93.
Hence 93.
• Q23) In first 252 natural numbers, how many times we need to write number 4?
• Unit place = 10+10+5=25.
Ten’s place = 10+10+10 = 30.
Total times = 25 + 30 = 55.
• Q24) In first 252 natural numbers, how many numbers contain digit 4?
• Unit digit = 10 + 10 + 5 = 25.
Ten’s digit = 10 + 10 + 10 = 30.
Unit & Ten’s digit both = 3 (44,144,244)
Total numbers = 25 + 30 - 3 = 52.
• Q25) In book has 100 pages, how many digits have been used to number the pages ?
• 1- 9 = 9 digits
10 – 99 = 2 x 90 = 180 digits.
100 = 3 digits.
Total = 9 + 180 + 3 = 192.
• Q26) There are 3 consecutive natural numbers such that square of the second is 19 more than 9 times of sum of first & third number. Find second number.
• Let 3 consecutive numbers are n-1, n and n+1.
=> n^2 – 19 = 9(n -1+n+1)
=> n^2 – 19 = 9 x 2n
=> n^2 – 18n + 19 = 0.
=> (n-19) (n+1) = 0.
=> n = 19, -1.
Second number is 19.
• Q27) Let N = 999999 ... 36 times. How many 9’s are there in N^2 ?
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robinborchelt.blogspot.com | 1,600,933,403,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400214347.17/warc/CC-MAIN-20200924065412-20200924095412-00610.warc.gz | 101,335,650 | 15,117 | ## Friday, March 30, 2012
### Counting by 5s
The topic of counting by 5s has come up recently for Robin in Saxon Math K. This has been quite challenging for him. He needs a lot of repetition to learn things like this. At the same time he has excellent pattern recognition skills. We're supposed to use nickels, but it was just confusing him terribly. So I got out the hundred number chart and showed him how to put a nickel on each multiple of 5. He understood how to do that immediately, and once he got that he had no problem figuring out how many nickels it takes, for example, to make 35 cents.
Including a bi-lingual retelling of Jack and the Beanstalk.
And a book about a boy who assisted the Wright Brothers at Kitty Hawk.
Robin jumped on a trampoline, walked, biked, played kick-ball, used a batting cage, and played miniature golf (Happy Birthday, Patrick!)
## Friday, March 23, 2012
### Metal Detector
Robin found a magnifying glass/metal detector kit at the thrift store for 90 cents. We weren't sure it would work but he put it together with some help from his dad and lo and behold it works quite well! We read books about meadowlands, lions, tigers and sharks. We looked up ligers and tigons. We also had a discussion about sextuplets (Robin's sister is very interested in multiple babies).
Robin continues to work with Saxon Math K. This week's topics included tangrams, and we worked on counting a lot.
Robin walked, roller bladed, and biked.
## Friday, March 16, 2012
### Magnetism
Robin got really into an experiment we did with magnetism:
The paperclip is defying gravity (well, sort of)! We also visited the Battle Creek Cypress Swamp in Calvert County with our homeschooling group. And read a book about owls.
We celebrated Pi Day on 3/14. Robin worked on Saxon Math K.
I read to Robin and he read to me! This is still so exciting. We 'catch' him reading things occasionally, too, although he claims he is just 'guessing'.
Robin and his brother had quite a jam session with a drum and the piano.
Robin walked and rode his bike.
## Friday, March 9, 2012
### Marble Run
This morning, after arriving at our house at 6:30 in the morning from our 2 week trip to Texas (we drove at night) Robin and his brother were inspired by an episode of Curious George to take our recycling and some tape and design a marble run. I am especially impressed because they did it ALL by themselves, while I was unpacking and decompressing from the 36 hours in the car. (Excuse the lousy photo - I forgot the camera was on manual!) Robin did several puzzles, including one up to 550 pieces.
We spent the last two weeks on vacation, driving to Texas (through Virginia, Tennessee, & Arkansas) and visiting family in Austin and Roby. Robin was quite interested in mapping our trip.
He learned about the history of Lady Bird Lake in Austin, TX.
Robin walked a lot, including in neighborhoods and along Lady Bird Lake. He also did a lot of bouncing on giant trampolines (trust me, this is a major workout.) He used his grandparent's apartment building's exercise room, too.
We visited the Texas Memorial Museum, a Natural Science museum including fossils and minerals from Texas (and dinosaur footprints!) Robin also went to the Zilker Botanical Gardens in Austin. | 812 | 3,305 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-40 | latest | en | 0.97327 |
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(ECON214)[2009](s)midterm~752^_10239 - Econ 214 Spring 2009...
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1 Econ 214 Spring, 2009 Midterm Exam. Instructor: Yan YU Name:__________________ I.D. number:_____________________ Total Score: ________________________ General Instructions: This is a close-book examination. You have 120 minutes. Please check if you have a total of 6 pages including this cover page. You may use calculators.
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2 I. Multiple choices (5 points each, 35 points total) 1. (2.3) Suppose the demand for X is given by Q X d = 100 - 2P X + 4P Y + 10M + 2A, where P X represents the price of good X, P Y is the price of good Y, M is income and A is the amount of advertising on good X. Good X is a. An inferior good b. A normal good c. A Giffin good d. A complement 2. (3.7) Which of the following provides a measure of the overall fit of a regression? a. T-statistic b. F-statistic c. R-square d. The F-statistic and R-square 3. (3.2) When the own price elasticity of good X is -3.5 then total revenue can be increased by a. Increasing the price b. Decreasing the quantity supplied c. Decreasing the price d. Neither increase price, decrease price nor decrease quantity supplied 4. (2.3, 3.4, 3.5, 3.8) The management of Local Cinema has estimated the monthly
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(ECON214)[2009](s)midterm~752^_10239 - Econ 214 Spring 2009...
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Ask a homework question - tutors are online | 519 | 1,860 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2017-22 | latest | en | 0.839034 |
https://www.ncatlab.org/nlab/show/prequantized+Lagrangian+correspondence | 1,623,759,484,000,000,000 | application/xhtml+xml | crawl-data/CC-MAIN-2021-25/segments/1623487621273.31/warc/CC-MAIN-20210615114909-20210615144909-00107.warc.gz | 838,216,353 | 38,325 | # nLab prequantized Lagrangian correspondence
Classical Mechanics via Prequantum Correspondences
This entry is a sub-chapter of geometry of physics. See there for background
The following is effectively a derivation of, and an introduction to, classical mechanics by studying correspondences in what is called (as we will explain) the slice topos over the moduli stack of prequantum line bundles. One such correspondence in this slice topos is precisely a prequantized Lagrangian correspondence and the reader looking for just these should skip ahead to the section The classical action functional prequantizes Lagrangian correspondences. But for completeness and to introduce the technology used here, we start with introducing also more basic concepts, such as phase space etc.
## Surveys, textbooks and lecture notes
#### Symplectic geometry
symplectic geometry
higher symplectic geometry
# Classical Mechanics via Prequantum Correspondences
## Model Layer
Here we discuss the notion of prequantized Lagrangian correspondences and how it serves to embed traditional classical mechanics in its formulation as Hamiltonian mechanics and Lagrangian mechanics into the general context of local prequantum field theory and its motivic quantization.
A traditional notion is that of a plain Lagrangian correspondence, which is a Lagrangian submanifold of the Cartesian product of a symplectic manifold and another one, with opposite symplectic structure. As discussed there, plain Lagrangian correspondences naturally generalize symplectomorphisms – hence transformations between phase spaces in physics – via correspondences of symplectic manifolds.
But in prequantum field theory proper, and in particular with an eye towards geometric quantization, one considers the prequantization of these symplectic manifolds by lifting them to prequantum circle bundles with principal connection. The notion of prequantized Lagrangian correspondence is the refinement of that of plain Lagrangian correspondence which does properly respect and reflect this prequantization information: a prequantized Lagrangian correspondence is a Lagrangian subspace as before, but now equipped with an explicit gauge transformation between the pullbacks of the two prequantum circle bundles to the correspondence space.
We discuss below how the concept of prequantized Lagrangian correspondences neatly induces and unifies ingredients and aspects of classical mechanics, notably the Hamiltonian mechanics of symplectic manifolds and the Lagrangian mechanics of action functionals associated to it via the Legendre transform. Specifically, below in The classical action prequantizes Hamiltonian correspondences we see that prequantized Lagrangian correspondences are diagrams which schematically express this data as follows:
$\array{ && {{space\,of} \atop {trajectories}} \\ & {}^{\mathllap{{initial}\atop {values}}}\swarrow && \searrow^{\mathrlap{{Hamiltonian} \atop {evolution}}} \\ phase\,space_{in} && \swArrow_{{action} \atop {functional}} && phase \,space_{out} \\ & {}_{\mathllap{{prequantum}\atop {bundle}_{in}}}\searrow && \swarrow_{\mathrlap{{prequantum} \atop {bundle}_{out}}} \\ && {{2\text{-}group} \atop {of\,phases}} } \,.$
This describes a diagram in what is called the slice topos of smooth sets over the moduli stack of prequantum circle bundles. Once formulated this way, there is an evident refinement to higher moduli stacks of prequantum n-bundles.
For $n = 2$ we show how the notion of prequantized Lagrangian correspondence is – still naturally in the context of local prequantum field theory – further refined from the context of symplectic manifolds to that of Poisson manifolds. Specifically, this is obtained by realizing that prequantized Lagrangian correspondences are really naturally to be regarded as correspondences-of-correspondences in a 2-category of correspondences, where now the new lower-order correspondences are instead boundary field theories for a 2d Chern-Simons theory (a non-perturbative Poisson sigma-model).
### Phase spaces and symplectic manifolds
Given a physical system, one says that its phase space is the space of its possible (“classical”) histories or trajectories. The first two of Newton's laws of motion say that trajectories of physical systems are (typically) determined by differential equations of second order, and therefore these spaces of trajectories are (typically) equivalent to initial value data of 0th and of 1st derivatives. In physics this data (or rather its linear dual) is referred to as the canonical coordinates and the canonical momenta, respectively, traditionally denoted by the symbols “$q$” and “$p$”. But being coordinates, these are actually far from being canonical in the mathematical sense; all that has invariant meaning is, locally, the surface element $\mathbf{d}p \wedge \mathbf{d}q$ spanned by a change of coordinates and momenta.
So far this says that a physical phase space is mathematically formalized by a sufficiently smooth manifold $X$ which is equipped with a closed and non-degenerate differential 2-form $\omega \in \Omega^2_{\mathrm{cl}}(X)$, hence by a symplectic manifold $(X,\omega)$.
The non-degeneracy of a symplectic form encodes the special property (as we will make explicit below) that (time) evolution of coordinates and momenta is uniquely induced by an action functional/Hamiltonian generating the evolution. This is however famously not the case for systems with gauge equivalences, hence such systems which have configurations that are nominally different but nevertheless physically equivalent. Presence of such gauge equivalences is not the exception, but the rule for physical systems, and therefore we want to include this case.
In the presence of gauge equivalences, the phase space form $\omega$ is still a closed differential 2-form, it just need not be non-degenerate anymore. While in such a case the pair $(X,\omega)$ could just be called a smooth manifold equipped with a closed differential 2-form}, it is traditional to call this a pre-symplectic manifold in order to amplify the indented use as a model for phase spaces. (Some authors demand that a pre-symplectic form be a closed form with constant rank, but here this technical condition will not be relevant and will not be considered.)
###### Example
The sigma-model describing the propagation of a particle on the real line $\mathbb{R}$ has as phase space the plane $\mathbb{R}^2 = T^\ast \mathbb{R}$ and as symplectic form its canonical volume form. Traditionally the two canonical coordinate functions on this phase space are denoted $q,p \;\colon\; \mathbb{R}^2 \longrightarrow \mathbb{R}$ (called the “canonical coordinate” and the “canonical momentum”, respectively), and in terms of these the symplectic form in this example is $\omega = \mathbf{d} q \wedge \mathbf{d} p$.
When dealing with spaces $X$ that are equipped with extra structure, such as $\omega \in \Omega^2_{\mathrm{cl}}(X)$, then it is useful to have a universal moduli space for these structures, and this will be central for our developments here. So we need a “smooth set$\mathbf{\Omega}^2_{\mathrm{cl}}$ of sorts, characterized by the property that there is a natural bijection between smooth closed differential 2-forms $\omega \in \Omega^2_{\mathrm{cl}}(X)$ and smooth maps $X \longrightarrow \mathbf{\Omega}^2_{\mathrm{cl}}$. Of course such a universal moduli spaces of closed 2-forms does not exist in the category of smooth manifolds. But it does exist canonically if we slightly generalize the notion of “smooth set” suitably.
###### Definition
A smooth set or smooth 0-type $X$ is
1. an assignment to each $n \in \mathbb{N}$ of a set, to be written $X(\mathbb{R}^n)$ and to be called the set of smooth maps from $\mathbb{R}^n$ into $X$,
2. an assignment to each ordinary smooth function $f : \mathbb{R}^{n_1} \to \mathbb{R}^{n_2}$ between Cartesian spaces of a function of sets $X(f) : X(\mathbb{R}^{n_2}) \to X(\mathbb{R}^{n_1})$, to be called the pullback of smooth functions into $X$ along $f$;
such that
1. this assignment respects composition of smooth functions;
2. this assignment respects the covering of Cartesian spaces by open disks: for every good open cover $\{\mathbb{R}^n \simeq U_i \hookrightarrow \mathbb{R}^n\}_i$, the set $X(\mathbb{R}^n)$ of smooth functions out of $\mathbb{R}^n$ into $X$ is in natural bijection with the set $\left\{ (\phi_i)_i \in \prod_i X(U_i) \;|\; \forall_{i,j}\; \phi_i|_{U_{i} \cap U_j} = \phi_j|_{U_{i} \cap U_j} \right\}$ of tuples of smooth functions out of the patches of the cover which agree on all intersections of two patches.
For more on this see at geometry of physics in the section Smooth sets.
While the formulation of this definition is designed to make transparent its geometric meaning, of course equivalently but more abstractly this says the following:
###### Definition
Write CartSp for the category of Cartesian spaces with smooth functions between them, and consider it as a site by equipping it with the coverage of good open covers. A smooth set or smooth 0-type is a sheaf on this site. The topos of smooth 0-types is the category of sheaves
$\mathrm{Smooth}0\mathrm{Type} \coloneqq \mathrm{Sh}(\mathrm{CartSp}) \,.$
In the following we will abbreviate the notation to
$\mathbf{H} \coloneqq \mathrm{Smooth}0\mathrm{Type} \,.$
The topos of prop. also has another site of definition.
###### Proposition
Write $SmoothMfd$ for the category of smooth manifolds regarded as a site with the standard Grothendieck topology of open covers. There is an equivalence of categories
$Sh(SmoothMfd) \simeq \mathrm{Smooth}0\mathrm{Type} \,.$
###### Proof
The canonical inclusion $CartSp\hookrightarrow SmoothMfd$ is readily seen to be a dense subsite. (This is just the statement that – by definition – every smooth manifold may be covered by Cartesian spaces.) The statement hence follows by the comparison lemma.
For the discussion of presymplectic manifolds, we need the following two examples.
###### Example
Every smooth manifold $X \in \mathrm{SmoothManifold}$ becomes a smooth 0-type by the assignment
$X \;\colon\; n \mapsto C^\infty(\mathbb{R}^n, X) \,.$
This construction extends to a full embedding of smooth manifolds into smooth sets
$SmoothManifold \hookrightarrow \mathbf{H}$
###### Proof
This follows via prop. by the Yoneda lemma.
###### Example
For $p \in \mathbb{N}$, write $\mathbf{\Omega}^p_{\mathrm{cl}}$ for the smooth set whose $n$-dimensional plots are smooth differential p-forms on $\mathbb{R}^n$:
$\mathbf{\Omega}^p_{\mathrm{cl}} \colon \mathbb{R}^n \mapsto \Omega^p_{\mathrm{cl}}(\mathbb{R}^n)$
and which sends a smooth function $f \colon \mathbb{R}^{n_1} \to \mathbb{R}^{n_2}$ to the pullback of differential forms along this map
$\array{ \mathbb{R}^{n_1} &\mapsto& \Omega^p_cl(\mathbb{R}^{n_1}) \\ \downarrow^{\mathrlap{f}} && \uparrow^{\mathrlap{f^\ast}} \\ \mathbb{R}^{n_2} &\mapsto& \Omega^p_cl(\mathbb{R}^{n_2}) } \,.$
For more on this example see at geometry of physics in the section Differential forms.
This solves the moduli problem for closed smooth differential forms:
###### Proposition
For $p \in \mathbb{N}$ and $X \in SmoothManifold \hookrightarrow Smooth0Type$, there is a natural bijection
$\mathbf{H}(X,\mathbf{\Omega}^p_{\mathrm{cl}}) \simeq \Omega^p_{\mathrm{cl}}(X)$
between morphisms (of smooth sets)
$X \longrightarrow \Omega^2_{cl}$
and smooth closed 2-forms
$\omega \in \Omega^2_{cl}(X)$
on $X$.
###### Proof
This follows via prop. by the Yoneda lemma.
So a presymplectic manifold $(X,\omega)$ is equivalently a map of smooth sets of the form
$\omega \;\colon\; X \longrightarrow \mathbf{\Omega}^2_{\mathrm{cl}} \,.$
### Canonical transformations and symplectomorphisms
An equivalence between two phase spaces, hence a re-expression of the “canonical coordinates” and “canonical momenta”, is called a canonical transformation in physics. Mathematically this is a symplectomorphism.
###### Definition
Given two symplectic manifolds $(X_1, \omega_1)$ and $(X_2, \omega_2)$ (which might be two copies of one single symplectic manifold), a symplectomorphism between them
$f \;\colon\; (X_1, \omega_1) \longrightarrow (X_2, \omega_2)$
is a diffeomorphism
$f \;\colon\; X_1 \longrightarrow X_2$
of the underlying smooth manifolds, such that the pullback of the second symplectic form along $f$ equals the first,
$f^\ast \omega_2 = \omega_1 \,.$
The above formulation of pre-symplectic manifolds as maps into a moduli space of closed differential 2-forms yields the following formulation of symplectomorphisms, which is very simple in itself, but contains in it the seed of an important phenomenon:
###### Proposition
A symplectomorphism $f \colon (X_1, \omega_2) \longrightarrow (X_2, \omega_2)$ as above is, under the identification of prop. , equivalently a commuting diagram in $\mathbf{H}$ of the form
$\array{ X_1 && \stackrel{f}{\longrightarrow}&& X_2 \\ & {}_{\mathllap{\omega_1}}\searrow && \swarrow_{\mathrlap{\omega_2}} \\ && \Omega^2_{cl} } \,.$
###### Proof
This is the naturality of the Yoneda lemma:
By prop. the identification of $\omega_i \in \Omega^2_{cl}(X_i)$ with $\omega_i \colon X \to \Omega^2_{cl}$ is via the natural equivalence
$Hom_{Smooth0Type}(X,\mathbf{\Omega}^2_{cl}) \stackrel{\simeq}{\longrightarrow} \Omega^2_{cl}(X) \,.$
This being natural means that for every morphism $X_1 \stackrel{f}{\longrightarrow} X_2$ there is a commuting diagram of the form
$\array{ Hom(X_2,\mathbf{\Omega}^2_{cl}) &\stackrel{\simeq}{\longrightarrow}& \Omega^2_{cl}(X_2) \\ \downarrow^{\mathrlap{(-) \circ f}} && \downarrow^{\mathrlap{f^\ast}} \\ Hom(X_1,\mathbf{\Omega}^2_{cl}) &\stackrel{\simeq}{\longrightarrow}& \Omega^2_{cl}(X_1) }$
where on the left we have the pre-composition operation on morphisms and on the right we have, by example , the pullback of differential forms.
Consider then the element
$(X_2 \stackrel{\omega_2}{\to} \mathbf{\Omega}^2_{cl}) \in Hom(X_2,\mathbf{\Omega}^2_{cl})$
in the top left set in this diagram. Sending it along the top and right maps yields the pullback of differential forms $f^\ast \omega_2 \in \Omega^2_{cl}(X_1)$. On the other hand, sending it along the left and bottom maps yields the differential form represented by the composite morphism $(X_1 \stackrel{f}{\to} X_2 \stackrel{\omega_1}{\to}\mathbf{\Omega}^2_{cl})$. Commutativity of the above naturality diagram means that these two elements of $\Omega^2_{cl}(X_1)$ coincide. This is the claim to be proven.
Situations like this are naturally interpreted in a slice topos:
###### Definition
For $A \in \mathbf{H}$ any smooth set, the slice topos $\mathbf{H}_{/A}$ is the category whose objects are objects $X \in \mathbf{H}$ equipped with maps $X \to A$, and whose morphisms are commuting diagrams in $\mathbf{H}$ of the form
$\array{ X &&\longrightarrow&& Y \\ & \searrow && \swarrow \\ && A }$
###### Definition
Write $\mathrm{SymplManifold}$ for the category with presymplectic smooth manifolds as objects and symplectomorphisms, def. , betwen them as morphisms
###### Proposition
The construction of prop. which sends a smooth symplectic manifold $(X,\omega)$ to the classifying morphism of smooth sets $(X \stackrel{\omega}{\longrightarrow}\mathbf{\Omega}^2_{cl})$ regarded as an object in the slice topos, def. constitutes a full and faithful functor
$SymplManifold \hookrightarrow Smooth0Type_{/\mathbf{\Omega}^2_{\mathrm{cl}}}$
of pre-symplectic manifolds with symplectomorphisms between them into the slice topos of smooth sets over the smooth moduli space of closed differential 2-forms.
By prop. .
### Trajectories and Lagrangian correspondences
A symplectomorphism clearly puts two symplectic manifolds “in relation” to each other. But it does so also in the formal sense of relations in mathematics. Recall:
###### Definition
For $X,Y \in$ Set two sets, a relation $R$ between elements of $X$ and elements of $Y$ is a subset of the Cartesian product set
$R \hookrightarrow X \times Y \,.$
More generally, for $X, Y \in \mathbf{H}$ two objects of a topos (such as the topos of smooth sets), then a relation $R$ between them is a subobject of their Cartesian product
$R \hookrightarrow X \times Y \,.$
In particular any function induces the relation$y$ is the image of $x$”:
###### Example
For $f \;\colon\; X \longrightarrow Y$ a function, its induced relation is the relation which is exhibited by the graph of $f$
$graph(f) \coloneqq \left\{ (x,y) \in X \times Y \;|\; f(x) = y \right\}$
canonically regarded as a subobject
$graph(f) \hookrightarrow X \times Y \,.$
Hence in the context of classical mechanics, in particular any symplectomorphism $f \;\colon\; (X_1, \omega_1) \longrightarrow (X_2, \omega_2)$ induces the relation
$graph(f) \hookrightarrow X_1 \times X_2 \,.$
Since we are going to think of $f$ as a kind of “physical process”, it is useful to think of the smooth set $graph(f)$ here as the space of trajectories of that process. To make this clearer, notice that we may equivalently rewrite every relation $R \hookrightarrow X \times Y$ as a diagram of the following form:
$\array{ && R \\ & {}^{\mathllap{i_X}}\swarrow && \searrow^{\mathrlap{i_Y}} \\ X && && Y } \;\; = \;\; \array{ && R \\ && \downarrow \\ && X \times Y \\ & {}^{\mathllap{p_X}}\swarrow && \searrow^{\mathrlap{p_Y}} \\ X && && Y }$
reflecting the fact that every element $(x \sim y) \in R$ defines an element $x = i_X(x \sim y) \in X$ and an element $y = i_Y(x \sim y) \in Y$.
Then if we think of $R = graph(f)$ we may read the relation as “there is a trajectory from an incoming configuration $x_1$ to an outgoing configuration $x_2$
$\array{ && graph(f) \\ & {}^{\mathllap{incoming}}\swarrow && \searrow^{\mathrlap{outgoing}} \\ X_1 && && X_2 } \,.$
Notice here that the defining property of a relation as a subset/subobject translates into the property of classical physics that there is at most one trajectory from some incoming configuration $x_1$ to some outgoing trajectory $x_2$ (for a fixed parameter time interval at least, we will formulate this precisely in the next section when we genuinely consider Hamiltonian correspondences).
In a more general context one could consider there to be several such trajectories, and even a whole smooth set of such trajectories between given incoming and outgoing configurations. Each such trajectory would “relate” $x_1$ to $x_2$, but each in a possible different way. We can also say that each trajectory makes $x_1$ correspond to $x_2$ in a different way, and that is the mathematical term usually used:
###### Defininition
For $X, Y \in \mathbf{H}$ two spaces, a correspondence between them is a diagram in $\mathbf{H}$ of the form
$\array{ && Z \\ & \swarrow && \searrow \\ X && && Y }$
with no further restrictions. Here $Z$ is also called the correspondence space.
An equivalence between two such correspondences is an equivalence $Z_1 \stackrel{\simeq}{\to}Z_2$ that gives a commuting diagram of the form
$\array{ && Z_1 \\ & {}^{}\swarrow && \searrow^{} \\ X &&\downarrow^{\mathrlap{\simeq}} && Y \\ & {}_{}\nwarrow && \nearrow_{} \\ && Z_2 }$
Correspondences between $X$ any $Y$ with such equivalences between them form a groupoid. (See at geometry of physics the section Essence of gauge theory: Groupoids and basic homotopy 1-type theory for more on this.) Hence we write
$Corr\left(\mathbf{H}\right)(X,Y) \in Grpd \,.$
###### Example
The correspondence induced by the graph of a function $f \colon X \to Y$ as in example is equivalent, in the sense of def. , to the correspondence
$\array{ && X \\ & {}^{\mathllap{id}}\swarrow && \searrow^{\mathrlap{f}} \\ X && && Y } \,.$
The equivalence
$\array{ && X \\ & {}^{\mathllap{id}}\swarrow && \searrow^{\mathrlap{f}} \\ X &&\downarrow^{\mathrlap{\simeq}} && Y \\ & {}_{\mathllap{i_X}}\nwarrow && \nearrow_{\mathrlap{i_Y}} \\ && graph(f) }$
is induced by
$X \stackrel{\simeq}{\longrightarrow} graph(f)$
$x \mapsto (x,f(x))$
Moreover, if we think of correspondences as modelling spaces of trajectories, then it is clear that their should be a notion of composition:
###### Definition
Given two consecutive correspondences, then their composite is the correspondence obtained by forming the fiber product of the two coincident morphisms:
$\left( \array{ && Y_1 &&&& Y_2 \\ & \swarrow && \searrow && \swarrow && \searrow \\ X_1 && && X_2 && && X_3 } \right) \;\;\;\; \mapsto \;\;\;\; \left( \array{ && Y_1 \circ_{X_2} Y_2 \\ & \swarrow && \searrow \\ X_1 && && X_3 } \right) \,.$
###### Remark
Heuristically, the composite space of trajectories $Y_1 \circ_{X_2} Y_2$ should consist precisely of those pairs of trajectories $( f, g ) \in Y_1 \times Y_2$ such that the endpoint of $f$ is the starting point of $g$. The space with this property is precisely the fiber product of $Y_1$ with $Y_2$ over $X_2$, denoted $Y_1 \underset{X_2}{\times} Y_2$ (also called the pullback of $Y_2 \longrightarrow X_2$ along $Y_1 \longrightarrow X_2$ and then abbreviated $(pb)$):
$\left( \array{ && Y_1 \circ_{X_2} Y_2 \\ & \swarrow && \searrow \\ X_1 && && X_3 } \right) \;\;\; = \;\;\; \left( \array{ && && Z_1 \underset{Y}{\times} Z_2 \\ && & \swarrow && \searrow \\ && Z_1 && (pb) && Z_2 \\ & \swarrow && \searrow && \swarrow && \searrow \\ X && && Y && && Z } \right) \,.$
Hence given a topos $\mathbf{H}$, correspondences between its objects form a category which composition the fiber product operation, where however the collection of morphisms between any two objects is not just a set, but is a groupoid (the groupoid of correspondences between two given objects and equivalences between them).
One says that correspondences form a (2,1)-category
$Corr(\mathbf{H}) \in (2,1)Cat \,.$
But for most purposes here, the reader unwilling to enter higher category theory can, to good approximation, pretend that correspondences form an ordinary category.
One reason for formalizing this notion of correspondences so much in the present context that it is useful now to apply it not just to the ambient topos $\mathbf{H}$ of smooth sets, but also to its slice topos $\mathbf{H}_{/\mathbf{\Omega}_{cl}^2}$ over the universal moduli space of closed differential 2-forms.
To see how this is useful in the present context, notice the following basic observation:
###### Definition
Given a symplectic manifold $(X,\omega)$, then a submanifold
$L \hookrightarrow X$
is called
###### Proposition
Let $f \colon X_1 \to X_2$ be a smooth function between smooth manifolds and let
$\array{ && graph(f) \\ && \downarrow \\ && X_1 \times X_2 \\ & {}^{\mathllap{p_1}}\swarrow && \searrow^{\mathrlap{p_2}} \\ X_1 && && X_2 }$
be the induced correspondence. If $\omega_1$ and $\omega_2$ are symplectic forms on $X_1$ and $X_2$, respectively, then $p_1^\ast \omega_1 - p_2^\ast \omega_2$ is a pre-symplectic form on $X_1 \times X_2$, and $f$ is a symplectomorphism precisely if $graph(f) \hookrightarrow X_1 \times X_2$ is a Lagrangian submanifold.
To capture this phenomenon, one traditionally sets:
###### Definition
For $(X_1,\omega_1)$ and $(X_2,\omega_2)$ two symplectic manifolds (not necessarily of the same dimension), an isotropic correspondence or Lagrangian correspondence between them is a correspondence of the underlying manifolds
$\array{ && R \\ && \downarrow \\ && X_1 \times X_2 \\ & {}^{\mathllap{p_1}}\swarrow && \searrow^{\mathrlap{p_2}} \\ X_1 && && X_2 }$
such that the correspondence space $R \hookrightarrow X_1 \times X_2$ is an isotropic submanifold or Lagrangian submanifold, respectively of the product symplectic manifold given by
$(X_1 \times X_2 , p_1^\ast \omega_1 - p_2^\ast \omega_2) \,.$
###### Proposition
Under the identification of prop. , isotropic correspondences as in def. are equivalent to diagrams of smooth sets of the form
$\array{ && R \\ & {}^{\mathllap{p_1}}\swarrow && \searrow^{\mathrlap{p_2}} \\ X_1 && \swArrow_= && X_2 \\ & {}_{\mathllap{\omega_1}}\searrow && \swarrow_{\mathrlap{\omega_2}} \\ && \Omega^2_{cl} } \,.$
This in turn is equivalent to being a correspondence in the slice topos $\mathbf{H}_{/\Omega^2_{cl}}$, def. , under the identification of prop. .
###### Proof
By prop. the commutativity of this diagram says precisely that on $R$ we have
$p_1^\ast \omega_1 = p_2^\ast \omega_2$
hence
$p_1^\ast \omega_1 - p_2^\ast \omega_2 = 0 \,.$
Therefore we have:
###### Proposition
For $(X_1, \omega_2)$ and $(X_2, \omega_2)$ two symplectic manifolds, there is a full embedding
$LagrangianCorrespondences\left(\left(X_1,\omega_1\right), \left(X_2, \omega_2\right)\right) \hookrightarrow Corr\left(\mathbf{H}_{/\mathbf{\Omega}^2_{cl}}\right)\left(\left(X_1,\omega_1\right), \left(X_2, \omega_2\right)\right)$
of the Lagrangian correspondences into the space of correspondences between the two manifolds as objects in the slice topos over the universal moduli space of closed differential 2-forms.
###### Example
The graph of a function $f \colon X_1\to X_2$ between symplectic manifold $(X_i, \omega_i)$ is a Lagrangian correspondence precisely if $f$ is a symplectomorphism.
###### Proof
Under the identification of
$\array{ && graph(f) \\ & {}^{\mathllap{p_1}}\swarrow && \searrow^{\mathrlap{p_2}} \\ X_1 && && X_2 }$
### Hamiltonian (time evolution) trajectories and Hamiltonian correspondences
An important class of symplectomorphisms are the following
###### Definition
Let $(X,\omega)$ be a symplectic manifold. The induced Poisson bracket $\{-,-\}$ takes a smooth function $H \in C^\infty(X)$ (the “Hamiltonian”) to the derivation $\{H,-\}$ on $C^\infty(X)$. This is equivalently a vector field $v_H \in \Gamma T X$, the corresponding Hamiltonian vector field.
A Hamiltonian symplectomorphisms from a symplectic manifold $(X,\omega)$ to itself, is a symplectomorphism $X \to X$ which is the flow of a Hamiltonian vector field for some parameter “time” $t \in \mathbb{R}$
$\exp( t \{H,-\}) \;\colon\; (X,\omega) \longrightarrow (X,\omega) \,.$
We call a Lagrangian correspondence, def. , induced from Hamiltonian symplectomorphisms a Hamiltonian correspondences.
###### Remark
Under the interpretation of correspondences as spaces of trajectories as in example , the smooth correspondence space of a Hamiltonian correspondence is naturally identified with the space of classical trajectories of the Hamiltonian dynamics of $H$
$Fields_{traj}^{class}(t) = graph\left( \exp(t\{H,-\}) \right)$
in that
1. every point in the space corresponds uniquely to a trajectory of parameter time length $t$ characterized as satisfying the equations of motion as given by Hamilton's equations for $H$;
2. the two projection maps to $X$ send a trajectory to its initial and to its final configuration, respectively.
###### Remark
Forming Hamiltonian correspondences consitutes a functor from 1-dimensional cobordisms with Riemannian structure to the category of correspondences in the slice topos:
$\exp((-)\{H,-\}) \;\colon\; Bord^{Riem}_1 \longrightarrow Corr_1(\mathbf{H}_{/\Omega^2})$
since for all (“time”) parameter valued $t_1, t_2 \in \mathbb{R}$ we have a composition (by fiber product) of correspondences exhibited by the following pasting diagram:
$\array{ &&&& graph\left(\exp\left(\left(t_1+t_2\right)\right) \left\{H,-\right\} \right) \\ && & \swarrow && \searrow \\ && graph\left(\exp\left(t_1\right)\left\{H,-\right\}\right) && (pb) && graph\left(\exp\left(t_2 \left\{H,-\right\}\right)\right) \\ & \swarrow && \searrow & & \swarrow && \searrow \\ X && && X && && X \\ & \searrow &&& \downarrow &&& \swarrow \\ &&&& \Omega^2_{cl} } \,.$
### The kinetic action and Planck’s constant
To naturally see why there would be any Hamiltonian associated to a (to some) symplectomorphism in the first place, we step back and consider local trivializations or local potentials for symplectic forms. Doing so turns out to give rise to what in physics is called the kinetic action, what in the context of geometric quantization is called prequantization and what in mathematics is called lifting to differential cohomology. All these concepts arise directly from the following simple consideration.
Given a pre-symplectic form $\omega \in \Omega^2_{\mathrm{cl}}(X)$, by the Poincaré lemma there is a good open cover $\{U_i \hookrightarrow X\}_i$ such that one can find smooth differential 1-forms $\theta_i \in \Omega^1(U_i)$ such that these are local trivializations/potentials for the symplectic form on each patch $U_i$ of the cover:
$\mathbf{d}\theta_i = \omega_{|U_i} \,.$
Physically such a 1-form is (up to a factor of 2) a choice of kinetic energy density called a kinetic Lagrangian $L_{\mathrm{kin}}$ (below in example we connect this statement to a maybe more familiar formla):
$\theta_i = 2 L_{\mathrm{kin}, i} \,.$
###### Example
Consider the phase space $(\mathbb{R}^2, \; \omega = \mathbf{d} q \wedge \mathbf{d} p)$ of example . Since $\mathbb{R}^2$ is a contractible topological space we consider the trivial covering ($\mathbb{R}^2$ covering itself) since this is already a good covering in this case. Then all the $\{g_{i j}\}$ are trivial and the data of a prequantization consists simply of a choise of 1-form $\theta \in \Omega^1(\mathbb{R}^2)$ such that
$\mathbf{d}\theta = \mathbf{d}q \wedge \mathbf{d}p \,.$
A standard such choice is
$\theta = - p \wedge \mathbf{d}q \,.$
Then given a trajectory $\gamma \colon [0,1] \longrightarrow X$ which satisfies Hamilton's equation for a standard kinetic energy term, then $(p \mathbf{d}q)(\dot\gamma)$ is this kinetic energy of the particle which traces out this trajectory.
Given a path $\gamma : [0,1] \to X$ in phase space, its kinetic action $S_{\mathrm{kin}}$ is supposed to be the integral of $L_{\mathrm{kin}}$ along this trajectory. In order to make sense of this in the generality where there is no globally defined $\theta$, there need to be functions $g_{i j} \in C^\infty(U_i \cap U_j, \mathbb{R})$ for each double intersection of patches of the cover, such that these the local $\theta$‘s differ on these double intersection only by the total derivative (de Rham differential $\mathbf{d}$ ) of these functions:
$\theta_j|_{U_j} - \theta_i|_{U_i} = \mathbf{d}g_{i j} \,.$
One then finds (from the theory of Cech cohomology) that if on triple intersections these functions satisfy
$g_{ij} + g_{j k} = g_{i k}$
then there is a well defined action functional
$S_{\mathrm{kin}}(\gamma) \in \mathbb{R}$
obtained by dividing $\gamma$ into small pieces that each map to a single patch $U_i$, integrating $\theta_i$ along this piece, and adding the contribution of $g_{i j}$ at the point where one switches from using $\theta_i$ to using $\theta_j$. Technically this is called the holonomy or parallel transport of the $(\mathbb{R},+)$-principal connection which is defined by the data $(\{\theta_i\}, \{g_{i j}\} )$.
However, requiring this condition on triple overlaps as an equation between $\mathbb{R}$-valued functions makes the local patch structure trivial: if this is possible then one can in fact already find a single $\theta \in \Omega^1(X)$ and functions $h_i \in C^\infty(U_i, \mathbb{R})$ such that $\theta_i = \theta|_{U_i} + \mathbf{d}h_i$. This has the superficially pleasant effect that the action is simply the integral against this globally defined 1-form, $S_{\mathrm{kin}} = \int_{[0,1]} \gamma^\ast L_{\mathrm{kin}}$, but it also means that the pre-symplectic form $\omega$ is exact, which is not the case in many important examples. (In more abstract terms what this is saying is that every $(\mathbb{R},+)$-principal bundle over a manifolds is trivializable.)
On the other hand, what really matters in prequantum physics is not the action functional $S_{\mathrm{kin}} \in \mathbb{R}$ itself, but the exponentiated action
$\exp\left( \tfrac{i}{\hbar} S \right) \in \mathbb{R}/(2\pi \hbar)\cdot\mathbb{Z} \,,$
which takes values in the quotient of the additive group of real numbers by integral multiples of Planck's constant $2\pi \hbar$.
In more detail, consider the canonical inclusion
$\mathbb{Z} \hookrightarrow \mathbb{R}$
of the integers as an addiditve subgroup of the real numbers. Strictly speaking what appears in physics is the real line on which a unit is chosen as part of the identification of mathematical formalism with physical reality, one should really consider all possible additive group homomorphisms $\mathbb{Z}\to \mathbb{R}$. These are parameterized by
$h \in (\mathbb{R}- \{0\}) \hookrightarrow \mathbb{R}$
$(-)\cdot h \;\colon\; \mathbb{Z} \longrightarrow \mathbb{R}$
and this “physical unit$h$ is what is called Planck’s constant.
In particular the induced circle group is identified as the quotient of $\mathbb{R}$ by $h \mathbb{Z}$, in this sense
$U(1) \simeq \mathbb{R}/h \mathbb{Z}$
and under this identification its quotient map is expressed in terms of the exponential function $\exp \colon z \mapsto \sum_{k = 0}^\infty \frac{z^k}{k!} \in \mathbb{C}$ as
$\exp(2 \pi \tfrac{i}{h}(-)) = \exp(\tfrac{i}{\hbar} (-)) \;\colon\; \mathbb{R} \longrightarrow U(1) \,,$
where
$\hbar \coloneqq h/2\pi \,.$
The resulting short exact sequence is the real exponential exact sequence
$0 \to \mathbb{Z} \longrightarrow \mathbb{R} \stackrel{\exp(\tfrac{i}{\hbar}(-))}{\longrightarrow} U(1) \to 0 \,.$
This is the source of the ubiquity of the expression $\exp(\tfrac{i}{\hbar} (-))$ in quantum physics, say in the path integral, where the exponentiated action functional appears as $\exp(\tfrac{i}{\hbar} S)$.
### Pre-Quantization and Differential cohomology
By the above discussion, for the exponentiated kinetic action functional to be well defined, one only needs that the equation $g_{i j} + g_{j k} = g_{i k}$ on triple intersection holds modulo addition of an integral multiple of Planck's constant $h = 2\pi \hbar$.
If this is the case, then one says that the data $(\{\theta_i\}, \{g_{i j}\})$ defines equivalently
on $X$, with curvature the given symplectic 2-form $\omega$.
Such data is called a pre-quantization of the symplectic manifold $(X,\omega)$. Since it is the exponentiated action functional $\exp(\frac{i}{\hbar} S)$ that enters the quantization of the given mechanical system (for instance as the integrand of a path integral), the prequantization of a symplectic manifold is indeed precisely the data necessary before quantization.
Therefore, in the spirit of the above discussion of pre-symplectic structures, we would like to refine the smooth moduli space of closed differential 2-forms to a moduli space of prequantized differential 2-forms.
Again this does naturally exist if only we allow for a good notion of “space”. An additional phenomenon to be taken care of now is that while pre-symplectic forms are either equal or not, their pre-quantizations can be different and yet be equivalent:
because there is still a remaining freedom to change this data without changing the exponentiated action along a closed path: we say that a choice of functions $h_i \in C^\infty(U_i, \mathbb{R}/(2\pi\hbar)\mathbb{Z})$ defines an equivalence between $(\{\theta_i\}, \{g_{i j}\})$ and $(\{\tilde \theta_i\}, \{\tilde g_{i j}\})$ if $\tilde \theta_i - \theta_i = \mathbf{d}h_i$ and $\tilde g_{i j} - g_{i j} = h_j - h_i$.
This means that the space of prequantizations of $(X,\omega)$ is similar to an orbifold: it has points which are connected by gauge equivalences: there is a groupoid of pre-quantum structures on a manifold $X$.
In just the same way then that above we found a smooth moduli space $\mathbf{\Omega}^2_{cl}$ of closed differential 2-forms, one can find a smooth groupoid (for more on this see at geometry of physics the section Smooth homotopy types ), which we denote
$\mathbf{B}U(1)_{\mathrm{conn}} \in \mathbf{H}$
###### Proposition
The smooth groupoid $\mathbf{B}U(1)_{\mathrm{conn}}$ is characterized as follows,
1. For $X$ a smooth manifold, maps
$\nabla \colon X \longrightarrow \mathbf{B}U(1)_{conn}$
are equivalent to the above prequantum data $(\{\theta_i\}, \{g_{i j}\})$ on $X$;
2. for $\nabla_1, \nabla_2 \colon X \longrightarrow \mathbf{B}U(1)_{conn}$ two such maps, homotopies
$\array{ & \nearrow \searrow^{\mathrlap{\nabla_1}} \\ X & \Downarrow & \mathbf{B}U(1)_{conn} \\ & \searrow \nearrow_{\mathrlap{\nabla_2}} }$
between these are equivalent to the above gauge transformations $(\{h_i\})$ between this data
$(\theta_2)_i - (\theta_1)_i = \mathbf{d} \tfrac{\hbar}{i} log (h_i) \,.$
###### Proposition
There is a universal curvature map, a morphism of smooth groupoids
$F \;\colon\; \mathbf{B}U(1)_{\mathrm{conn}} \longrightarrow \mathbf{\Omega}^2_{\mathrm{cl}}$
which is such that for $\nabla \colon X \longrightarrow \mathbf{B}U(1)_{conn}$ a $U(1)$-principal connection, the composite
$F_\nabla \;\colon\; X \stackrel{\nabla}{\longrightarrow} \mathbf{B}U(1)_{conn} \stackrel{F_{(-)}}{\longrightarrow} \mathbf{\Omega}^2_{cl}$
is its curvature 2-form.
Hence this is the map that sends $(\{\theta_i\}, \{g_{i j}\})$ to $\omega$ with $\omega|_{U_i} = \mathbf{d}\theta_i$.
Therefore:
###### Definition
A prequantization of a symplectic manifold $(X,\omega)$ is – if it exists – a choice of circle group-principal connection $\nabla$ on $X$ whose curvature 2-form is the given symplectic form
$F_\nabla = \omega \,.$
In terms of the classifying morphism of differential forms as in prop. this reads as follows.
###### Proposition
Given a presymplectic manifold $(X,\omega)$, regarded equivalently as an object $(X \stackrel{\omega}{\longrightarrow} \mathbf{\Omega}^2_{cl}) \in \mathbf{H}_{/\mathbf{\Omega}^2_{cl}}$ by prop. , then a prequantization of $(X,\omega)$, def. , is equivalently a choice of lift $\nabla$ in
$\array{ X &\stackrel{\nabla}{\longrightarrow}& \mathbf{B}U(1)_{conn} \\ & {}_{\mathllap{\omega}}\searrow & \downarrow^{\mathrlap{F_{(-)}}} \\ && \mathbf{\Omega}^2_{cl} } \,.$
Phrased this way, there is an evident concept of prequantization of Lagrangian correspondences:
###### Definition
Given prequantized symplectic manifolds $(X_i,\nabla_i)$ as in prop. , and given a Lagrangian correspondence as in prop. , then a prequantization of this correspondence is a lift of the whole diagram through the universal curvature map of prop. :
$\array{ && Z \\ & \swarrow && \searrow \\ X_1 && && X_2 \\ & {}_{\mathllap{\omega_1}}\searrow && \swarrow_{\mathrlap{\omega_2}} \\ && \mathbf{\Omega}^2_{cl} } \;\;\;\; \mapsto \;\;\;\; \array{ && Z \\ & \swarrow && \searrow \\ X_1 && \swArrow_{\simeq} && X_2 \\ & {}_{\mathllap{\nabla_1}}\searrow && \swarrow_{\mathrlap{\nabla_2}} \\ && \mathbf{B}U(1)_{conn} \\ && \downarrow^{\mathrlap{F}} \\ && \mathbf{\Omega}^2_{cl} } \,.$
###### Remark
This means in words that a prequantized Lagrangian correspondence is a prequantization of the in- and out-going symplectic manifolds together with a choice of equivalence/gauge transformation between the two prequantum circle bundles pulled back to the correspondences space.
###### Remark
By duality in the smmetric monoidal (2,1)-category of correspondences, a prequantized Lagragian correspondence is equivalently a diagram of the form
$\array{ && Y \\ & \swarrow && \searrow \\ \ast && \swArrow && X_1 \times X_2 \\ & \searrow && \swarrow_{\mathrlap{\nabla_2- \nabla}} \\ \\ && \mathbf{B}U(1)_{conn} }$
hence a trivialization of the product of one prequantum bundle with the negative (the inverse under tensor product) of the other, on the correspondence space.
### Hamiltonian flows, the Legendre transform and the Hamilton-Jacobi action
###### Proposition
Consider the phase space $(\mathbb{R}^2, \; \omega = \mathbf{d} q \wedge \mathbf{d} p)$ of example equipped with its canonical prequantization by $\theta = p \, \mathbf{d}q$ from example ,
Then smooth 1-parameter flows of this data via prequantized correspondences, def. ,
$t \;\;\;\; \mapsto \;\;\;\; \array{ X && \stackrel{f_t}{\longrightarrow} && X \\ & {}_{\mathllap{\theta}} \searrow & \swArrow_{ F_t } & \swarrow_{\mathrlap{\theta}} \\ && \mathbf{B}U(1)_{conn} }$
are in bijection with smooth functions $H \colon \mathbb{R}^2 \longrightarrow \mathbb{R}$.
This bijection works by regarding $H$ as a Hamiltonian, def. , and assigning the flow $f_t = \exp(t \{H,-\})$ of its Hamiltonian vector field
$t \;\; \mapsto \;\; \array{ X && \stackrel{\exp(t \{H,-\})}{\longrightarrow} && X \\ & {}_{\mathllap{\theta}} \searrow & \swArrow_{\exp( \tfrac{i}{\hbar} S_t )} & \swarrow_{\mathrlap{\theta}} \\ && \mathbf{B}U(1)_{conn} } \,,$
where the prequantization is given by
• $S_t \;\colon\; \mathbb{R}^2 \longrightarrow \mathbb{R}$ is the Hamilton-Jacobi action of the classical trajectories induced by $H$,
• which is the integral $S_t = \int_{0}^t L \, d t$ of the Lagrangian $L \,d t$ induced by $H$,
• which is the Legendre transform of the Hamiltonian
$L \coloneqq p \frac{\partial H}{\partial p} - H \;\colon\; \mathbb{R}^2 \longrightarrow \mathbb{R} \,.$
###### Proof
By prop. the prequantum filler of the diagram is given by a function $F_t =\exp(\tfrac{i}{\hbar} S_t)$ satisfying
$f_t^\ast \theta - \theta = -\mathbf{d}S_t \,.$
By standard Lie theory a smooth such 1-parameter flow is fixed by its derivative by $t$. For the above equation this yields
$\mathcal{L}_v \theta = -\mathbf{d}L$
where
1. $v \in \Gamma(T X)$ is the vector field of the flow $t\mapsto f_t$;
2. $\mathcal{L}_v$ is the Lie derivative along $v$;
3. $L \coloneqq \frac{\partial S}{\partial t}$.
By Cartan's magic formula this equation is equivalent to
$\iota_v \omega = -\mathbf{d}L - \mathbf{d} \iota_v \theta \,.$
This is the symplectic form of Hamilton's equations for $v$ and says that
$H \coloneqq - L - \iota_v \theta$
is a Hamiltonian that makes $v$ a Hamiltonian vector field. The correction term is
\begin{aligned} \iota_v \theta &= \iota_v ( p \, \mathbf{d}q ) \\ & = p \partial_v q \\ \end{aligned} \,.
But since $v$ is Hamiltonian, this is given by one component of Hamilton's equations $\iota_v (\mathbf{d}p \wedge \mathbf{d}q) = \mathbf{d}H$ saying that $\partial_v q = \frac{\partial H}{\partial p}$.
Hence in summary the flow is Hamiltonian and the pre-quntum filler is the choice of Hamiltonian $H$ specified by
$\frac{\partial S}{\partial t} = L = p \frac{\partial H}{\partial p} - H \,.$
###### Remark
In particular, this induces a functor
$\exp(\tfrac{i}{\hbar} S) \;\colon\; Bord_1^{Riem} \longrightarrow Corr_1(\mathbf{H}_{/\mathbf{B}U(1)_{conn}}) \,.$
In summary, prop. and remark say that a prequantized Lagrangian correspondence is conceptually of the following form
$\array{ && {{space\,of} \atop {trajectories}} \\ & {}^{\mathllap{{initial}\atop {values}}}\swarrow && \searrow^{\mathrlap{{Hamiltonian} \atop {evolution}}} \\ phase\,space_{in} && \swArrow_{{action} \atop {functional}} && phase \,space_{out} \\ & {}_{\mathllap{{prequantum}\atop {bundle}_{in}}}\searrow && \swarrow_{\mathrlap{{prequantum} \atop {bundle}_{out}}} \\ && {{2-group} \atop {of\,phases}} } \,.$
###### Remark
The proof of prop. recovers, from general abstract input, precisely all the ingredients known in physics as canonical transformations.
The proposition says that the slice topos $\mathbf{H}_{/\mathbf{B}U(1)_{conn}}$ unifies classical mechanics in its two incarnations as Hamiltonian mechanics and as Lagrangian mechanics, where the relation between the two via the Legendre transform is exhibited by the homotopies that fill diagrams in the slice topos over $\mathbf{B}U(1)_{conn}$.
Hamiltonian$\leftarrow$ Legendre transform $\rightarrow$Lagrangian
Lagrangian correspondenceprequantizationprequantized Lagrangian correspondence
### Heisenberg group and Poisson bracket
Above we have interpreted maps $f \colon X \to Y$ as correspondences between $X$ and $Y$ by taking the correspondence space to be the graph of $f$. There is also another natural way to regard maps as correspondences: we may simply take $X$ as the correspondence space, take the left map out of it to be the identity and the right map to be $f$ itself:
$\left( X \stackrel{f}{\longrightarrow} Y \right) \;\; \mapsto \;\; \left( \array{ && X \\ & {}^{\mathllap{id}}\swarrow && \searrow^{\mathrlap{f}} \\ X && && Y } \right) \,.$
Consider now those correspondences which are equivalences (isomorphisms) in the category of correspondences $Corr_1(\mathbf{H})$. If we forget the smooth structure on everything and consider just correspondences of the underlying sets, hence $Corr_1(Set)$, then it is easy to see that under the cardinality map correspondences are given by matrices with cardinality entries and composition of correspondence by fiber product induces matrix multiplication.
Therefore for a correspondence to be an equivalence-transformation it has to be of the form above, induced by a direct map, which in addition is an equivalence $f \colon X \stackrel{\simeq}{\longrightarrow} Y$.
###### Proposition
Let $(X,\omega)$ be a symplectic manifold and choose any prequantization $(L,\nabla)$, thought of, via remark , as an object in the slice (2,1)-topos, $\nabla \in \mathbf{H}_{/\mathbf{B}U(1)_{conn}}$. Then
See (hgp 13)
For some reason, the quantomorphism group which is the Lie integration of the Poisson bracket is less famous than the Heisenberg group that sits inside it:
###### Remark
Suppose that $(X,\omega)$ itself has the structure of a group (for instance if $(X,\omega)$ is a symplectic vector space such as $(\mathbb{R}^{2n}, \sum_i p_i \mathbf{d}q^i)$ ), then the subgroup of the quantomorphism group whose underlying diffeomorphisms are given by the action of $X$ is the Heisenberg group of $X$.
### Hamiltonian actions and moment maps
For $G$ a Lie group, a Hamiltonian action of $G$ on $(X,\omega)$ is equivalently an action by prequantized Lagrangian correspondences, hence a group homomorphism
$G \longrightarrow \mathbf{Aut}_\nabla(Corr_1(\mathbf{H}_{/\mathbf{B}U(1)_{conn}})) \,.$
The Lie differentiation of this is the corresponding moment map.
See (hgp 13)
## Semantic Layer
We now discuss the above constructions more abstractly in cohesive topos theory.
under construction
given $V$ then the delooping $\mathbf{B} \mathbf{Aut}(V)$ of its automorphism group $\mathbf{Aut}(V)$ is the 1-image factorization of the name $\ast \stackrel{\vdash V}{\longrightarrow} Type$ of $V$
$\ast \stackrel{}{\longrightarrow} \mathbf{B}\mathbf{Aut}(V) \hookrightarrow Type \,.$
for the slice over $\mathbf{B}U(1)_{conn}$ this needs to be subjected to differential concretification
(…)
## Syntactic Layer
We now discuss the above constructions yet more abstractly in homotopy type theory.
(…)
prequantization is lift through dependent sum along the universal curvature map of prop.
$\underset{F_{(-)}}{\sum} \;\colon\; \mathbf{H}_{/\mathbf{B}U(1)_{conn}} \longrightarrow \mathbf{H}_{/\mathbf{\Omega}^2_{cl}}$
(…)
## References
As far as it is not covered by traditional material, the above discussion is taken from
For more and related references see there and see at motivic quantization.
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# M01 Question 18
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Re: 3 Digit Numbers Divisible by 7 (and others) [#permalink]
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29 Dec 2010, 01:40
How many of the three-digit numbers are divisible by 7?
(A) 105
(B) 106
(C) 127
(D) 128
(E) 142
GENERALLY:
$$# \ of \ multiples \ of \ x \ in \ the \ range = \frac{Last \ multiple \ of \ x \ in \ the \ range \ - \ First \ multiple \ of \ x \ in \ the \ range}{x}+1$$.
For example: how many multiples of 5 are there between -7 and 35, not inclusive?
Last multiple of 5 IN the range is 30;
First multiple of 5 IN the range is -5;
$$\frac{30-(-5)}{5}+1=8$$.
OR:
How many multiples of 7 are there between -28 and -1, not inclusive?
Last multiple of 7 IN the range is -7;
First multiple of 7 IN the range is -21;
$$\frac{-7-(-21)}{7}+1=3$$.
Back to the original question:
Last 3-digit multiple of 7 is 994;
First 3-digit multiple of 7 is 105;
So # of 3-digt multiples of 7 is $$\frac{994-105}{7}+1=128$$.
Answer: D.
gettinit wrote:
this strategy should work for most of these types of problems. I have never seen an exclusive scenario.
The above approach will work for both inclusive and exclusive cases, which is also.
hirendhanak wrote:
gettinit wrote:
this strategy should work for most of these types of problems. I have never seen an exclusive scenario.
Also to figure out those 3 digit numbers divisible by 35 I would just use the same formula.
Thanks for your reply, i understand that i can find the nos with the same formula
But i wish to understand whethr the same is possible with set theory.. because the q could be asked as a DS question in some different format also
What do you mean by the red part?
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Re: 3 Digit Numbers Divisible by 7 (and others) [#permalink]
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02 Jan 2011, 01:13
Bunuel wrote:
How many of the three-digit numbers are divisible by 7?
(A) 105
(B) 106
(C) 127
(D) 128
(E) 142
GENERALLY:
$$# \ of \ multiples \ of \ x \ in \ the \ range = \frac{Last \ multiple \ of \ x \ in \ the \ range \ - \ First \ multiple \ of \ x \ in \ the \ range}{x}+1$$.
For example: how many multiples of 5 are there between -7 and 35, not inclusive?
Last multiple of 5 IN the range is 30;
First multiple of 5 IN the range is -5;
$$\frac{30-(-5)}{5}+1=8$$.
OR:
How many multiples of 7 are there between -28 and -1, not inclusive?
Last multiple of 7 IN the range is -7;
First multiple of 7 IN the range is -21;
$$\frac{-7-(-21)}{7}+1=3$$.
Back to the original question:
Last 3-digit multiple of 7 is 994;
First 3-digit multiple of 7 is 105;
So # of 3-digt multiples of 7 is $$\frac{994-105}{7}+1=128$$.
Answer: D.
Hi Bunuel,
I'm a little confused. The general formula above, does it give us the number of integers inclusive of the upper and lower limits or non inclusive of the limits?
For example: what is the number of integers between 1 and 5, inclusive? The answer is 5, Which is also substantiated by the formula above.
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Re: 3 Digit Numbers Divisible by 7 (and others) [#permalink]
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02 Jan 2011, 02:38
gmatpapa wrote:
Hi Bunuel,
I'm a little confused. The general formula above, does it give us the number of integers inclusive of the upper and lower limits or non inclusive of the limits?
For example: what is the number of integers between 1 and 5, inclusive? The answer is 5, Which is also substantiated by the formula above.
You should read the solution carefully: it doesn't matter whether the the range is inclusive or exclusive, the formula gives you the correct answer in any case.
How many multiples of 5 are there between 5 and 35, not inclusive?
Last multiple of 5 IN the range is 30 (not 35 but 30 as 35 is not IN the range);
First multiple of 5 IN the range is 10 (not 5 but 10 as 5 is not IN the range);
$$\frac{30-10}{5}+1=5$$: 10, 15, 20, 25 and 30.
How many multiples of 5 are there between 5 and 35, inclusive?
Last multiple of 5 IN the range is 35;
First multiple of 5 IN the range is 5;
$$\frac{35-5}{5}+1=7$$: 5, 10, 15, 20, 25, 30 and 35.
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Re: 3 Digit Numbers Divisible by 7 (and others) [#permalink]
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02 Jan 2011, 11:17
A quick demonstration of the formula and why it is intuitive:
How many numbers are there from 3 to 7 (inclusive)?
3, 4, 5, 6, 7 i.e. 5 numbers
When we do (7 - 3 = 4), we remove 3 from the list.
I I I I I I I : 7 sticks
[strike]I I I[/strike] I I I I : (7 - 3 = 4) sticks
The third stick has been removed but we need it to be included too. So we add 1 to the result 4 + 1 = 5 sticks
These 5 sticks are: I I I I I
Now, how many numbers are there from 3 to 7 (exclusive)?
4, 5, 6 i.e. 3 numbers
When we do (7 - 3 = 4), we still have 7 in the list.
I I I I I I I : 7 sticks
[strike]I I I[/strike] I I I I : (7 - 3 = 4) sticks
The seventh stick is still there which has to be removed. So we subtract 1 from the result 4 - 1 = 3 sticks
These 3 sticks are: I I I
or you can think of the exclusive case as (inclusive case - 2)
You say in inclusive case: (Last term - First term + 1): Both end terms are included here.
Then you subtract 2 to remove both the end terms and you get: (Last term - First term + 1) - 2 = Last term - First term - 1
I have also explained this in the post: http://gmatquant.blogspot.com/2010/11/four-prongs.html
Check out the subtraction part.
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Re: M01 Question 18 [#permalink]
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04 Feb 2011, 11:27
Bunuel wrote:
jayasimhaperformer wrote:
a 7*15=105
b 106/7=15.1
c 127/7=18.1
d 128/7=18.12 aprox
e 142/7=20.--- aprox
so a is right & a straight ans as its div by 7*15
Hi, and welcome to Gmat Club.
It seems that you misinterpreted the question.
The question is "How many of the three-digit numbers are divisible by 7?" It's not necessary the answer itself to be divisible by 7
GENERALLY:
$$# \ of \ multiples \ of \ x \ in \ the \ range = \frac{Last \ multiple \ of \ x \ in \ the \ range \ - \ First \ multiple \ of \ x \ in \ the \ range}{x}+1$$.
For example: how many multiples of 5 are there between -7 and 35, not inclusive?
Last multiple of 5 IN the range is 30;
First multiple of 5 IN the range is -5;
$$\frac{30-(-5)}{5}+1=8$$.
OR:
How many multiples of 7 are there between -28 and -1, not inclusive?
Last multiple of 7 IN the range is -7;
First multiple of 7 IN the range is -21;
$$\frac{-7-(-21)}{7}+1=3$$.
Back to the original question:
Last 3-digit multiple of 7 is 994;
First 3-digit multiple of 7 is 105;
So # of 3-digt multiples of 7 is $$\frac{994-105}{7}+1=128$$.
Answer: D.
Hope it helps.
Hello Bunuel, nice explanation as usual
is there any shortcut to find last and first multiples? for example how to find 994 ad 105 of 7?
or we have to use the long method of division?
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Re: M01 Question 18 [#permalink]
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04 Feb 2011, 12:03
Bunuel wrote:
tinki wrote:
Hello Bunuel, nice explanation as usual
is there any shortcut to find last and first multiples? for example how to find 994 ad 105 of 7?
or we have to use the long method of division?
thx 4 response
Well, it depends multiple of which number you want to find and the range you are looking in. Common sense, and divisibility rules should help you in this but sometimes 'trial and error' is also a good method. Check for example divisibility rules here: math-number-theory-88376.html
i did not know about 7 and 11 . nowhere seen the rule before.
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Re: M01 Question 18 [#permalink]
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22 Sep 2011, 06:27
hockeyplaya182 wrote:
1) Take all numbers between 1 and 999 divisible by 7: 999/7=142
2) Subtract the number that are not three digits: 99/7=14
142-14=128
Great explanation. Thanks.
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Re: M01 Question 18 [#permalink]
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22 Sep 2011, 10:07
Here is a simple Solution :
The count of numbers<=999 divisible by seven = rounded[999/7] = 142
The number of 2 digit numbers divesible by 7 = rounded[99/7] = 14
The number of 3 digit numbers divisible by 7 = 142 - 14 = 128
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Re: 3 Digit Numbers Divisible by 7 (and others) [#permalink]
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22 Sep 2011, 10:30
james12345 wrote:
I'm kept my lame questions to a minimum. I see why plus 1 work for inclusive. I don't see for exclusive why minus 1 would work. If both of the end number "excluded" are actually multiples of the number in question it changes it. but that should be the case for inclusive as well. in other words, how many multiples of 7 between 8 and 22 exclusive means 9 to 21 which means the answer is 2. 21-9 is 12 which divided by 7 is 1 so plus 1 = 2. however 7 to 28 exclusive means 8-27 = 19 which is 2 7s which is actually correct.
It makes me very nervous to apply formulas without being crystal clear as to why they're right. Maybe I've misunderstood part of the premise of the question.
The key point is that in a simple subtraction, we are acting inclusive in one side and exclusive in the other side. For example, when we say 10-6=4, we have not considered 6 itself (exclusive in one side) but we have considered 10 (inclusive in the other side). This is the rule. So, if you figure that out, you see that when GMAT says exclusive, we have to just exclude the side which was not originally excluded. Since one side is not excluded in a normal subtraction, we just subtract 1, not 2. On the other hand, when GMAT says inclusive, we have already included one side, so we just add 1, not 2, to include the other side.
Hope it helps
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Re: M01 Question 18 [#permalink]
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24 Sep 2012, 05:18
Ans is 128.
14*7 = 98 next will be a 3 digit factor..
and 994 /7 = 142 next will be a 4 digit..
so 142 -14 = 128
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Re: M01 Question 18 [#permalink]
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24 Sep 2012, 05:22
Total 3 digit nos 100 to 999
among them total digits div.by 7 ={(last digit div.by 7 Minus 1st digit div by 7)/7 }+1
{(994-105)/7}+1 = 128
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Re: M01 Question 18 [#permalink]
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24 Sep 2012, 08:47
T
How do we find out the largest 3 digit number that is divisible by 7 ?
Well one way is trial and error, but that is not too efficient.
Rather , let's divide 999 by 7 to find the remainder. Subtract the remainder from 999 and voila !
For eg : When 999 is divided by , the remainder is 5. Subtract from 999, the answer is 994.
If m/n leaves "r" as the remainder, then m-r is divisible by n.
What is the logic behind this ? Well when you divided a number by another divisor, the remainder represents the piece that is not divisible by the divisor. Remove
the offending remainder from the quotient and the new quotient will be divisible by the divisor
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Re: M01 Question 18 [#permalink]
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27 Feb 2013, 03:54
bb wrote:
Is something not clear in the Official Explanation?
Approach One: Divide all of the three-digit numbers $$999 - 100 + 1 = 900$$ (Don't forget to add 1 to get the number of all the 3-digit numbers) by 7, which is 128.57, and then round it off to 128.
Approach Two:
Find the first and the last multiples of 7 WITHIN the range (which is between 105 and 994). Find the positive difference, divide by 7, and add 1:
$$\frac{994-105}{7}+1=128$$
Hi,
I don't understand how and why approach 1 works. Please help explain why it works and provide some examples.
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Re: M01 Question 18 [#permalink]
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23 Sep 2013, 08:34
bb wrote:
Is something not clear in the Official Explanation?
Approach One: Divide all of the three-digit numbers $$999 - 100 + 1 = 900$$ (Don't forget to add 1 to get the number of all the 3-digit numbers) by 7, which is 128.57, and then round it off to 128.
Approach Two:
Find the first and the last multiples of 7 WITHIN the range (which is between 105 and 994). Find the positive difference, divide by 7, and add 1:
$$\frac{994-105}{7}+1=128$$
Be careful using approach one because it can lead to wrong answers. It is best used as an approximation because it will be within 1 of the correct answer.
How many two-digit numbers that start with 3 are divisible by 7: 39 - 30 + 1 = 10. Divide by 7 to get 1.43. Round down to 1 (Correct because we have 35 only.)
How many two-digit numbers that start with 4 are divisible by 7: 49 - 40 + 1 = 10. Divide by 7 to get 1.43. Round down to 1 (Incorrect because we have 42 and 49.)
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Re: M01 Question 18 [#permalink]
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23 Sep 2013, 20:05
How many of the three-digit numbers are divisible by 7?
(A) 105
(B) 106
(C) 127
(D) 128
(E) 142
Firstly,find the first 3 digit number that is divisible by 7 ie 105 and the last 3 digit number that is divisible by 7 is 994.
so, series become like this:
105,112,.....................994.
l = a + (n-1) d
994 = 105+(n-1)7
994-105+7=7n
n=896/7=128
so, right answer is
128.
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Re: M01 Question 18 [#permalink] 23 Sep 2013, 20:05
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# M01 Question 18
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Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 5,289 | 17,277 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2017-43 | latest | en | 0.887599 |
http://www.noahbprince.com/calculus/351/ | 1,519,564,322,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891816462.95/warc/CC-MAIN-20180225130337-20180225150337-00612.warc.gz | 496,532,774 | 9,711 | # Happy Approximation to Pi Day!
March 14 is one of my least favorite days of the year. While most math teachers revel in Pi Day, I spend my 3/14s frustrated that we waste an opportunity to pay homage to one of the most important numbers in all of mathematics making endless nods to the fact that the name of a certain letter of the Greek alphabet, when mispronounced, is homophonous with a baked dessert. The entire premise of Pi Day is flawed anyway – it is based on $\pi$ being $3.14$, but that is only an approximation. So instead, I propose that we celebrate Approximation to Pi Day.
There are some really great approximations to $\pi$ out there. And there need to be: since the decimal expansion of $\pi$ goes on forever with no repeating pattern, if we didn’t have a decent estimate, we’d never be able to deal with $\pi$ at all. Schoolchildren all learn $22/7$ as their first estimate of $\pi$, and it’s not bad – since $\pi=3.1415926535\ldots$ and $\frac{22}{7}=3.1428571428\ldots$, this estimate at least gets the first two decimal places right. Other estimates are far more accurate. I wrote about The Wallis Product a while back, which can be used to estimate $\pi$ to any desired accuracy level.
One of the forgotten estimates of $\pi$ is that it is approximately $\sqrt{10}$. The estimate $\pi\approx\sqrt{10}$ is so accurate that in ancient times the two numbers were thought to be equal. Using modern calculators, we know that $\sqrt{10}=3.1622776601\ldots$, so it is a slight overestimate of $\pi$, but without those tools, it can be quite difficult to see that $\pi$ and $\sqrt{10}$ are distinct at all, let alone discern which is the larger. One reason it is hard to compare $\pi$ and $\sqrt{10}$ is that they tend to measure different things. We use $\pi$ to measure areas of lengths of curves, but $\sqrt{10}$, if anything, is usually the length of a side of a square. It would be helpful if we ever saw a $\pi^2$ in a formula somewhere so that we could just compare it to $10$. Luckily, $\pi^2$ does show up in a couple of places. For example, the volume of a $4$-dimensional hypersphere with radius $r$ is $\dfrac{\pi^2}{2}r^4$. But since $4$-dimensional space is hard to wrap our heads around, it might be easier to use a fact about infinite series: $$1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+\cdots =\dfrac{\pi^2}{6}.$$ On the left side, we add up the reciprocals of the squares of all the positive integers, and after adding all of the infinitely many terms, we simply end up with $\pi^2/6$! If we want to know how $\pi$ and $\sqrt{10}$ compare, we can instead try to figure out how $1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+\cdots$ compares to $\dfrac{10}{6}$, or $\dfrac{5}{3}=1.66\overline{6}$.
Infinite sums are notoriously hard to evaluate exactly, so oftentimes, we rely only on estimates of them. With this particular sum, we can observe that
$$\dfrac{1}{6^2}+\dfrac{1}{7^2}+\dfrac{1}{8^2}+\dfrac{1}{9^2}+\cdots < \dfrac{1}{6\cdot 5}+\dfrac{1}{7\cdot 6}+\dfrac{1}{8\cdot 7}+\dfrac{1}{9\cdot 8}+\cdots.$$
Each term of the sum on the right can be split into a difference of two terms:
$$\dfrac{1}{6^2}+\dfrac{1}{7^2}+\dfrac{1}{8^2}+\dfrac{1}{8^2}+\cdots < \left(\dfrac{1}{5}-\dfrac{1}{6}\right)+\left(\dfrac{1}{7}-\dfrac{1}{6}\right)+\left(\dfrac{1}{8}-\dfrac{1}{7}\right)+\left(\dfrac{1}{8}-\dfrac{1}{9}\right)+\cdots.$$
When we drop the parentheses on the right side, everything cancels except the $\dfrac{1}{5}$, so we can say with confidence that $\dfrac{1}{6^2}+\dfrac{1}{7^2}+\dfrac{1}{8^2}+\dfrac{1}{9^2}+\cdots<\dfrac{1}{5}$. Therefore,
$$1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+\dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+\cdots< 1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+\dfrac{1}{5^2}+\dfrac{1}{5} =\dfrac{5989}{3600}=1.6636\overline{1}.$$
So, the infinite sum, the one equal to $\dfrac{\pi^2}{6}$, ends up being less than $\dfrac{5}{3}$, so $\pi^2<10$, and thus $\pi<\sqrt{10}$. | 1,325 | 3,956 | {"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} | 4.09375 | 4 | CC-MAIN-2018-09 | latest | en | 0.932914 |
https://math.stackexchange.com/questions/1139730/number-of-cyclic-subgroups-order-p2-in-mathbbz-p-times-mathbbz-p-tim | 1,561,559,872,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560628000353.82/warc/CC-MAIN-20190626134339-20190626160339-00468.warc.gz | 516,262,753 | 34,579 | # Number of cyclic subgroups order $p^2$ in $\mathbb{Z}_p \times \mathbb{Z}_p \times \mathbb{Z}_{p^2}$
Let $$G={ {<a>}_{p} \times {<b>}_{p} \times {<c>}_{p^2}} \cong \mathbb{Z}_p \times \mathbb{Z}_p \times \mathbb{Z}_{p^2} \text{, p is prime}$$ There are $p^3-1$ elements with order $p$, $p^2(p^2-p)$ elements with order $p^2$.
So my questions are:
1) How many cyclic subgroups order $p^2$ (like $\mathbb{Z}_{p^2}, \text{ not } \mathbb{Z}_p \times \mathbb{Z}_p$) are in G? (As i know, number of all subgroups order $p^2$ in G are $2p^2+p+1$)
2) How many subgroups order $p$ are in $\mathbb{Z}_p \times \mathbb{Z}_{p^2}$
Lemma: Let $n$ be number of the elements of order $m$ then there are $$\dfrac{n}{\phi(m)}$$ cylic subgroup of order $m$.
By the lemma, There are $\dfrac{p^2(p^2-p)}{\phi(p^2)}=p^2$ cyclic subgroup of order $p^2$. | 342 | 837 | {"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} | 3.484375 | 3 | CC-MAIN-2019-26 | latest | en | 0.652347 |
http://www.oreillynet.com/cs/user/view/cs_msg/45806?page=last&x-order=date | 1,406,313,690,000,000,000 | text/html | crawl-data/CC-MAIN-2014-23/segments/1405997894689.94/warc/CC-MAIN-20140722025814-00171-ip-10-33-131-23.ec2.internal.warc.gz | 728,017,215 | 11,404 | # Women in Technology
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Article: SQL Data Types Subject: how do we find nth highest in SQL Date: 2004-10-13 04:11:13 From: sunrek Response to: how do we find nth highest in SQL can u please help me
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• how do we find nth highest in SQL
To find the nth highest salary as below
select max(a.sal) from emp a where &n=(select count(b.sal) from emp b where (a.sal output-
put the value of n and get the highest salary that u have give the number i;e 4,5,6,10 etc.
thanx,
Haribrat
• how do we find nth highest in SQL
Query for Nth Max (Highest)
select sal from emp t
where &n = (select count(sal)
from (select distinct sal from emp)
where t.sal<=sal);
<b>Query for Nth Min (Lowest)
select sal from emp t
where &n = (select count(sal)
from (select distinct sal from emp)
where t.sal>=sal);
• query for nth max
2007-08-25 02:05:46 kashif094 [View]
select * from emp e where 1=(select count(distinct sal) from emp where e.sal<=sal)
hi i am kashif can anyone tell me what is the concept of 1 in this query?
• query for nth max
2007-10-25 05:02:21 jolly_cet [View]
1 specifies the position..
• how do we find nth highest in SQL
Query for Nth Max (Highest)
select sal from emp t
where &n = (select count(sal)
from (select distinct sal from emp)
where t.sal<=sal);
<b>Query for Nth Min (Lowest)
select sal from emp t
where &n = (select count(sal)
from (select distinct sal from emp)
where t.sal>=sal);
• how do we find nth highest in SQL
2005-01-18 21:03:32 hiexplain [View]
thanks a lot
"select sal from emp t
where &n = (select count(sal)
from (select distinct sal from emp)
where t.sal<=sal);"
it is working fine
could u please let me know how it works...
thanks and reguards
surendra
• how do we find nth highest in SQL
2005-10-17 23:31:59 Apurva_Sharma [View]
Hi, | 555 | 1,821 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2014-23 | longest | en | 0.854624 |
https://www.telerik.com/forums/averaging-monthly-sum-in-crosstab | 1,638,870,641,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363337.27/warc/CC-MAIN-20211207075308-20211207105308-00132.warc.gz | 1,104,204,244 | 27,680 | This is a migrated thread and some comments may be shown as answers.
Averaging Monthly Sum in Crosstab
5 Answers 152 Views
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Rod asked on 12 Nov 2012, 07:13 PM
I have a crosstab tab that sums up the amount sold during a month. This works very well. I want to average the monthly sum but only using month that have a non-zero amount sold.
For example:
Product May Jun Jul Avg
123 10 9 8 9
124 15 - 10 12.5
125 6 10 12 9.333
126 - - 7 7
So product 124 is divided by 2 instead of 3 and product 123 divided by 1.
I have looked at the forums and used google but I cannot figure out how to do this. Any help would be greatly appreciated.
ROD>
5 Answers, 1 is accepted
0
Peter
Telerik team
answered on 13 Nov 2012, 09:13 AM
Hi ROD,
Our suggestion is to use a conditional expression as shown in the following example:
= Avg(IIF(Fields.Value<>0, Fields.Value, null))
All the best,
Peter
the Telerik team
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0
Rod
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answered on 13 Nov 2012, 08:25 PM
Peter,
I understand what you are proposing, what I cannot seem to figure out is how to count the non-zero columns. I am able to use EXEC to get the total sum for the row just fine.
Please see the tonnagereport.jpg. I need to count the non-zero columns in the red circled sum for my average. I cannot seem to access that value easily.
Thank you,
ROD>
0
Peter
Telerik team
answered on 14 Nov 2012, 11:32 AM
Hi Rod,
You can accomplish your requirement with a custom aggregate function that counts only the months that have value bigger than zero as shown the following example:
= Sum(Fields.MTWeight)/CountX(Fields.Month, Fields.MTWeight)
`namespace ``Telerik.Reporting.Examples.CSharp`
`{`
` ``using ``System;`
` ``using ``System.Collections.Generic;`
` ``using ``System.Text;`
` ``using ``Telerik.Reporting.Expressions;`
` ``class ``CountX : IAggregateFunction`
` ``{`
` ``List<``int``> validMonths;`
` ``void ``IAggregateFunction.Init()`
` ``{`
` ``this``.validMonths = ``new ``List<``int``>();`
` ``}`
` ``void ``IAggregateFunction.Accumulate(``object``[] values)`
` ``{`
` ``var month = (``int``)values[0];`
` ``var value = (``int``)values[1];`
` ``if ``(value > 0)`
` ``{`
` ``if ``(!``this``.validMonths.Contains(month))`
` ``{`
` ``this``.validMonths.Add(month);`
` ``}`
` ``}`
` ``}`
` ``object ``IAggregateFunction.GetValue()`
` ``{`
` ``return ``this``.validMonths.Count;`
` ``}`
` ``void ``IAggregateFunction.Merge(IAggregateFunction aggregateFunction)`
` ``{`
` ``var countX = (CountX)aggregateFunction;`
` ``foreach``(var month ``in ``countX.validMonths)`
` ``{`
` ``if ``(!``this``.validMonths.Contains(month))`
` ``{`
` ``this``.validMonths.Add(month);`
` ``}`
` ``}`
` ``}`
` ``}`
`}`
Kind regards,
Peter
the Telerik team
HAPPY WITH TELERIK REPORTING? Do you feel that it is fantastic? Or easy to use? Or better than Crystal Reports? Tell the world, and help fellow developers! Write a short review about Telerik Reporting and Telerik Report Designer in Visual Studio Gallery today!
0
Rod
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Rank 1
answered on 14 Nov 2012, 05:16 PM
Peter,
Thank you. I was making this much more complicated that it had to be. I was able to get the result I needed with a much more simplified expression.
I added a column to the right outside of the group and used the following expression:
`=Sum(Fields.MTWeight)/CountDistinct(Fields.RequiredDate.Month)`
The CountDistinct works since no data will be returned for that month if there are no sales. I have not tested this but there could be an issue if more than 12 months of data is pulled which I think would be unusual for this client.
It was your formula for accessing the month field that gave me the idea. Thank you so much for the super quick responses. I will have one happy client now.
ROD>
0
May Zin
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answered on 23 Dec 2016, 12:00 PM
Hi!
This example is very good answer for us but I have a few difficulty for use this example.
When I use this example on crystal report, I face the error on avg formula.
IIF(Fields.Value<>0, Fields.Value, null)
Above the formula, How can I write to attach "the average function" on crystal report?
When I wirte, I face the error "Does not match with avg formula".
So Give me guideline how to write the formula on the crystal.
Thanks. I hope the reply letter from you.
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or | 1,414 | 5,264 | {"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} | 2.828125 | 3 | CC-MAIN-2021-49 | latest | en | 0.883657 |
http://mathhelpforum.com/calculus/209020-relate-rates-problem.html | 1,527,233,634,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867046.34/warc/CC-MAIN-20180525063346-20180525083346-00414.warc.gz | 182,273,703 | 10,942 | 1. ## Relate Rates Problem
Here is the question I am having trouble with:
The sun is passing over a 100 m tall building. The angle θ made by the sun with the ground is increasing at a rate of pi/20 rads/min. At what rate is the length of the shadow of the building changing when the shadow is 60 m long? Give your answer in exact values.
So far I have got:
100cosfata=x
dx/dt=-100(cosfata)(dfata/dx)
Dont no where to get cosfata from.
2. ## Re: Relate Rates Problem
"Increasing at a rate of rads/min?"
3. ## Re: Relate Rates Problem
Oh my gosh I have read this problem so many times I can just picture the numbers when they are not even there. Ugh.
4. ## Re: Relate Rates Problem
The sun is passing over a 100 m tall building. The angle θ made by the sun with the ground is increasing at a rate of pi/20 rads/min. At what rate is the length of the shadow of the building changing when the shadow is 60 m long?
$\displaystyle \cot{\theta} = \frac{x}{100}$
$\displaystyle -\csc^2{\theta} \cdot \frac{d\theta}{dt} = \frac{1}{100} \cdot \frac{dx}{dt}$
5. ## Re: Relate Rates Problem
But how do I find a value for -csc^2fata
6. ## Re: Relate Rates Problem
Originally Posted by mjo
But how do I find a value for -csc^2fata
review your basic right triangle trig ...
$\displaystyle \csc{\theta} = \frac{hypotenuse}{opposite} = \frac{\sqrt{60^2+100^2}}{100}$
square the result and change its sign to get the value of
$\displaystyle -\csc^2{\theta}$
btw ... $\displaystyle \theta$ is prounounced "theta" , not "fata"
7. ## Re: Relate Rates Problem
I found my original problem. Thanks guys | 471 | 1,596 | {"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} | 4.125 | 4 | CC-MAIN-2018-22 | latest | en | 0.803218 |
http://www.cs.rice.edu/~javaplt/311/Notes/04/07.5.html | 1,369,355,739,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368704117624/warc/CC-MAIN-20130516113517-00085-ip-10-60-113-184.ec2.internal.warc.gz | 406,440,397 | 2,796 | Lecture 7.5: Data in the lambda calculus
So far we have worked with a lambda calculus that only has integers as constants, but no operations on these constants. In fact, we only need one constant in order to use the lambda calculus for all kinds of computation. The constant is only needed to provide an easy way to observe the result of computation.
Natural numbers (or "non-negative integers") can be represented using what is called the church encoding. The idea is to represent any number N as functions that take two arguments, and then return the result of applying the first argument N times to the second argument. The following functions provide a convenient way to encode numbers using this strategy:
```let rec build n =
if n = 0 then Vr(0) else
Ap(Vr(1), build(n-1))
let writer n =
Lm(Lm(build n))
```
To decode numbers, it is tempting to write a function that takes apart a lambda term. This strategy, however, would be two fragile, because it would require the result to have a very specific syntactic form. A better strategy is implemented by the following function:
```let rec reader_step env e =
let v = ev6 ((V_Cn 0)::env) e in
match v with
V_Lm(env', e') -> 1 + (reader_step env' e')
| _ -> 0
let reader e =
let v = ev6 [] (Ap(Ap(e,Lm(Lm(Vr 1))),(Cn 0))) in
match v with
V_Lm(env', e') -> 1 + reader_step env' e'
| _ -> 0
```
Homework Assignment:
How can we define increment, addition, multiplication, exponentiation and subtraction with this representation? We covered all but the last in class today. Don't spend too much time on the last. Instead, bring to class next time a piece of paper showing what 2+3, 2*3, and 2^3 evaluate to.
Think about how we can represent pairs and sums (unions) in a similar way, but you don't have to bring in anything on this. | 454 | 1,789 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2013-20 | latest | en | 0.911213 |
http://betterlesson.com/lesson/resource/2867373/multiple-addends | 1,487,802,824,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501171053.19/warc/CC-MAIN-20170219104611-00072-ip-10-171-10-108.ec2.internal.warc.gz | 27,504,312 | 18,772 | ## Multiple Addends - Section 2: Activities
Unit 3: Targeted Skills Interventions
Lesson 13 of 20
## Big Idea: And, and, and, and...
Print Lesson
Standards:
Subject(s):
Math, Number Sense and Operations, fluency, review, Operations, Intervention
180 minutes
### Erin Doughty
##### Similar Lessons
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Environment: Suburban | 272 | 1,098 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2017-09 | latest | en | 0.831504 |
https://www.archivemore.com/what-is-the-ratio-for-the-volumes-of-two-similar-cylinders/ | 1,716,411,177,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058568.59/warc/CC-MAIN-20240522194007-20240522224007-00699.warc.gz | 593,907,675 | 7,562 | ## What is the ratio for the volumes of two similar cylinders?
Answer and Explanation: The ratio of their heights and radii of two similar cylinders is 2:3 .
## What is ratio of volumes?
Ratio of Volumes (cube units) If two solids are similar, the ratio of their volumes is equal to the cube of the ratio of their corresponding sides. (Note that volume is not a “length” measurement – it is a 3-D measurement.)
## How do you find similarity ratios?
If two triangles are similar, their similarity ratio is the ratio between a side length in the first triangle and the corresponding side length in the second triangle.
## What is ratio of similarity?
The RATIO OF SIMILARITY between any two similar figures is the ratio of any pair of corresponding sides. Simply stated, once it is determined that two figures are similar, all of their pairs of corresponding sides have the same ratio.
## What is the similarity ratio of similar triangles?
When two triangles are similar, the reduced ratio of any two corresponding sides is called the scale factor of the similar triangles. In Figure 1 , Δ ABC∼ Δ DEF. Figure 1 Similar triangles whose scale factor is 2 : 1. The ratios of corresponding sides are 6/3, 8/4, 10/5.
## How do you find the similarity ratio of two rectangles?
For two rectangles to be similar, their sides have to be proportional (form equal ratios). The ratio of the two longer sides should equal the ratio of the two shorter sides. However, the left ratio in our proportion reduces. We can then solve by cross multiplying.
## What is the ratio of the areas of the two rectangles?
If the ratio of the perimeters of two rectangles is 4:7, then the ratio of their areas must be 16:49. If two quadrilaterals are similar, then their areas must be in the same ratio as the square of the ratio of their perimeters. smaller rectangle has area of 28 un2, what is the area of the larger rectangle.
## Are these two rectangles similar?
Rectangles are similar if the length of the corresponding sides form a proportion This proportion is called scale factor. To find if two rectangles are similar, we need to find the ratio between the corresponding sides. If the ratios are equal, then the rectangles are similar. | 483 | 2,228 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.625 | 5 | CC-MAIN-2024-22 | latest | en | 0.935998 |
http://chaotic-flow.com/saas-metrics-viral-growth-trumps-saas-churn/?replytocom=18931 | 1,545,171,252,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376829812.88/warc/CC-MAIN-20181218204638-20181218230638-00283.warc.gz | 52,394,128 | 17,923 | SaaS Metrics
# SaaS Metrics | Viral Growth Trumps SaaS Churn
Everybody wants their Internet startup to go viral. But, just what does going viral mean? In his book, The Tipping Point, Malcolm Gladwell spells out the mechanics of how ideas spread virally by modeling the roles of key individuals that he calls connectors, mavens and salesmen (highly recommended on the Chaotic Flow blogroll Worthy Reads). When it comes to the Internet, Josh Kopelman eruditely points out that you can’t go viral by bolting it on as a last minute marketing program. You must apply SaaS Top Ten Do #5 and build viral growth into the product.
The aspiration of this post is not to add to the complexity of these theories of viral growth, but to uncover the simplicity of viral growth through a little mathematics. This is the second post in a series on SaaS metrics that explores the impact of viral growth in SaaS using a simple heuristic model with the goal of extending the list of SaaS Metrics Rules of Thumb started in the first post in the series regarding SaaS churn.
The mechanics of viral growth vary greatly by product, customer, market, and even culture. But, the mathematics pretty much boil down to the singular idea expressed so well in the kitschy Faberge shampoo commercial from the 70’s and 80’s.
Viral growth is customer growth that is proportionate to the number of customers.
Cn + 1 = ( 1 + g ) x Cn
All the connectors, mavens, salesmen and friends are rolled into the little “g” (for growth rate) in the formula above, which states that the number of customers in the time period n + 1 is equal to the number of customers in the time period n, plus a multiple of those same customers. You can think of g as the percentage of friends who actually told two friends who actually then went out and bought some shampoo divided by the amount of time it took them all to complete this circuit.
Like churn, viral growth scales with the number of customers.
When the viral growth rate exceeds the churn rate,
growth explodes through the churn limit.
If you read the previous SaaS metrics post on SaaS churn, you might recognize this formula, because it is identical to the churn formula only the negative churn rate -a has been replaced with the positive viral growth rate g. Thankfully, we can skip over the algebra this time and jump to the solution, simply by replacing -a with (g-a). This quick slight-of-hand gives us the formula for the number of customers in the time period n, Cn, that incorporates viral growth as well as churn.
Cn = b⁄(g-a) x ( ( 1 + g -a )n -1)
In this formula, ”b” is the baseline constant customer acquisition rate prior to either viral growth or SaaS churn kicking in. The mirror-like relationship between viral growth and SaaS churn in the formula above leads us to our next SaaS metrics rule of thumb.
### SaaS Metrics Rule-of-Thumb #3 – Viral Growth Trumps SaaS Churn
The previous SaaS Metrics Rule-of-Thumb #2 claimed that in order to break through the churn limit, new customer acquisition growth must outpace churn. Because churn increases in direct proportion to the number of customers, the surest approach is to drive growth at a higher rate that also increases in proportion to the number of customers, i.e., viraly. Moreover, investors generally expect companies to increase revenue on a percentage basis year over year. Holding products and prices constant, this again requires viral growth of your customer base. Viral growth can come from many sources, but I like to classify it into the following three distinct stages.
### Stage 1 Viral Growth – Brute Force Sales and Marketing (small g)
In any given industry, most companies will spend a rather fixed percentage of revenue on sales and marketing, regardless of the size of the company. When the effectiveness of these efforts scales in proportion to the level of spending (which is clearly not always the case), they will drive growth at a rate proportionate to revenue. In a SaaS business, this will also be proportionate to the number of customers. Hence, it is possible to drive growth in proportion to the number of customers simply through nuts and bolts sales and marketing. This stage is arguably not viral growth in the usual marketing sense of the words, but it does meet the strict mathematical definition where customer growth is proportionate to the number of customers. This in fact is the beautiful irony of the Faberge shampoo ad, which was a very effective nuts and bolts marketing campaign, but was probably less successful at getting even one friend to tell one friend about the product, let alone two.
### Stage 2 Viral Growth – Customer Advocacy (modest g)
When you are successful at driving word-of-mouth and getting your customers to recommend purchasing your product to new prospects you reach stage 2, the most commonly understood variation of viral growth made famous by the Faberge ad. SaaS customers hold great potential to be advocates, because they confirm their commitment day after day as they continue to use your product and month after month after month as they send in their renewal payments. Driving viral growth through active customer engagement that turns customers into advocates and advocates into evangelists should be high on the agenda of every SaaS marketing plan.
### Stage 3 Viral Growth – Ecosystem Buzz (BIG g)
The ecosystem for your product extends beyond your customer base to include all potentially interested parties such as press, analysts, bloggers, social media hounds, partners, vendors, investors, employees, and perhaps even the general public. Most SaaS compaines engage in public relations, tracking down and pitching Mr. Gladwell’s mavens and connectors in the hopes that they will pass the word on to their respective audiences and networks. But, true and lasting stage 3 ecosystem buzz is invariably built upon strong stage 2 customer advocacy. Twitter is a great curent example. If it weren’t for the dedicated and comparatively small group of Silicon Valley Web 2.0 junkies that were the early adopters of Twitter, you would not be able to follow CNN tweets today.
As previously mentioned, this is the second post in this series on SaaS metrics. In the next post, I’ll start adding revenue and costs to the model to explore the impact of viral growth and SaaS churn on SaaS profitability over time.
SaaS Metric Math Notes
The discrete model presented above can also be treated as a continuous model represented by the linear first order differential equation: C'(t) = b – a C(t) + gC(t) with the solution: C(t) = b⁄(g-a) ( e(g-a)t – 1 ). The graph above is plotted using this continuous solution. In the discrete model, the factors for viral growth and SaaS churn represent change over a specified constant period of time, e.g., a month or year, whereas in the continuous approach they represent instantaneous change.
For a rapidly growing SaaS business, where year over year growth hides the change from quarter to quarter or even month to month, the continuous model is better suited for estimating actual SaaS metrics. From a practical point of view, you just have to be careful not to put overly averaged growth or churn percentages into the formulas, e.g., for the SaaS churn limit b/a. The distinction really only matters for rates of 30% or more, in which case the continuous rate is given by the formula of gcont = -log(1+gavg).
Going forward, I’ll be dropping the discrete model entirely and will present only graphical solutions to the continuous model, because the continuous metric model is easier to manage and produces more elegant solutions. But, I will not be presenting the calculus required, so as not to put my loyal readers to sleep. Happy to share it with anyone who asks for it by email.
### Check out the rest of the SaaS Metrics Rules-of-Thumb
• […] York created a mathematical model showing how growth rates and churn rates interact in viral marketing. Here’s an […]
• George smith says:
Hi Joel,
This is great by the way. I really like your model, but I was wondering if you’d thought perhaps of incorporating the capacity of the viral universe? i.e the assumption been that. total number of customers can only grow up to a certain point, which probably means that the growth rate (g) or the baseline growth (b) will have to be adjusted for time.
• Hi George,
I’ve looked at it, but did not include it in the post simply because most companies worry mostly about how to get growth going, as opposed to what happens when it slows.
The solution is a logistic function as the kind found in population growth (as I’m guessing you know). Basically, starts exponential and then flattens out as the target market is converted to customers.
The most relevant part for SaaS would be the value that it flattens out to,
which is Target Market Size x ( 1 – a / g ). Without churn, viral growth would
eventually consume the entire market. But, just as in the case where growth is flat
churn places a limit on growth.
This is the most intuitive point to remember: churn chases the acquisition rate until it kills it (Rule #1). Acquisition growth must outpace churn for the company to keep growing (Rule #2), and since churn is exponential decay, the only way to really outpace it permanently is exponential growth (Rule #3)…otherwise churn will catch it. And, when the total churn churn rate x # customers = acquisition rate….growth stops. For the limited population scenario, growth slows below the exponential threshold as there are fewer and fewer customers to convert. And, stops when the churn rate equals the acquisition rate at the “churn limit” above.
• […] SaaS Metrics Rule-of-Thumb #3 – Viral Growth Trumps SaaS Churn […]
• […] I also like this one from Joel York with some insights on relationship of churn:http://chaotic-flow.com/saas-met…Via Ryan Withop.Comment Loading… • Post • 7:10am Ryan Withop, Data Mining […] | 2,123 | 9,950 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-51 | latest | en | 0.933352 |
https://www.mcadcentral.com/threads/twisted-wire-model.3230/ | 1,723,357,762,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640975657.64/warc/CC-MAIN-20240811044203-20240811074203-00606.warc.gz | 670,975,994 | 17,127 | Continue to Site
# twisted wire model
#### boydt
##### New member
I am trying to model 2 wires that are realistically twisted together in wildfire.
Not sure if wildfire is the same but this is the method i used in creating one in version 2001
Create a surface var sec sweep... use norm to origin traj
I sketched both my trajectories...
Trajectory 1: Main trajectory
So draw it away from the actual central point of rotation you want. This will all need to be worked out!
And through relations will be rotated.
View attachment 274
And your second trajectory. See image below (trajectory near botton of piccy)
View attachment 275
In sketcher draw a line from the origin at 45 degrees and dimension the length of it
View attachment 276
got to in the main tool bar at the top....sketcher- relations
sd#=10*360*trajpar
sd#=angle dimension
10= no of rotations
RESULTS:
View attachment 277
Repeat for the other side:
View attachment 278
The first traj is on the datum planes
the second traj is just above, see image below
View attachment 279
Then just add a swept blend on both surface edges...
View attachment 280
Hope this helps???
Just think about the position of your trajectories, you can always draw them and modify them later?
Also if your wires collide this will be because of where you located your 45 degrees in the section... so change it about, above/ below the datums, just remember to dimension it! eg
View attachment 281
Skiddy,
Thanks for the detailed explaination. There is a model on Pro/E Central you can download of a twisted pair for an example. Perhaps yours?
Couldn't the rotated sweep that defines the edges used to sweep the actual wire diameter be done with a single sketch line that extends an equal amount from each side of the origin tragectory?
-Bernie-
Skiddy - Thanks for the post!!!!!!!!! I've been trying to figure that out for months!!!
Quote
Couldn't the rotated sweep that defines the edges used to sweep the actual wire diameter be done with a single sketch line that extends an equal amount from each side of the origin tragectory?
Ive never experimented with this as a var sec sweep to me is generally 2 trajectories! But it would be a brilliant idea... and one i may look into! | 503 | 2,250 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2024-33 | latest | en | 0.897895 |
https://www.examples.com/maths/roman-numerals-1-to-10.html | 1,722,677,034,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640365107.3/warc/CC-MAIN-20240803091113-20240803121113-00896.warc.gz | 620,015,451 | 20,384 | # Roman Numerals 1 to 10
Created by: Team Maths - Examples.com, Last Updated: July 18, 2024
## Roman Numerals 1 to 10
Roman numerals, originating from ancient Rome, are a numeral system used throughout history and still prevalent in various applications today. The Roman numerals from 1 to 10 are represented as follows: I (1), II (2), III (3), IV (4), V (5), VI (6), VII (7), VIII (8), IX (9), and X (10). This system uses combinations of letters from the Latin alphabet to signify values and is used in contexts such as clock faces, book chapters, and movie sequels. Understanding Roman numerals provides insight into historical numbering systems and enhances one’s ability to interpret a wide range of modern and historical references.
## Rules to Write Roman Numerals 1 to 10
1. Basic Symbols:
• I = 1
• V = 5
• X = 10
2. Repetition:
• A symbol can be repeated up to three times to add values. For example, II = 2 and III = 3.
• When a smaller numeral appears before a larger numeral, it is added to the larger numeral. For example, VI = 6 (5 + 1) and VII = 7 (5 + 2).
4. Subtraction:
• When a smaller numeral appears before a larger numeral, it is subtracted from the larger numeral. For example, IV = 4 (5 – 1) and IX = 9 (10 – 1).
## Solved Problems
1. Convert Roman numeral III to an integer:
• Solution: III = 1 + 1 + 1 = 3
2. Convert Roman numeral VII to an integer:
• Solution: VII = 5 + 1 + 1 = 7
3. Convert Roman numeral IV to an integer:
• Solution: IV = 5 – 1 = 4
4. Convert Roman numeral IX to an integer:
• Solution: IX = 10 – 1 = 9
5. Convert Roman numeral X to an integer:
• Solution: X = 10
6. Convert 2 to a Roman numeral:
• Solution: 2 = I + I = II
7. Convert 6 to a Roman numeral:
• Solution: 6 = V + I = VI
8. Convert 8 to a Roman numeral:
• Solution: 8 = V + I + I + I = VIII
9. Convert 5 to a Roman numeral:
• Solution: 5 = V
10. Convert 1 to a Roman numeral:
• Solution: 1 = I
Understanding Roman numerals from 1 to 10 is fundamental for grasping the basics of this ancient numbering system. The numerals I, II, III, IV, V, VI, VII, VIII, IX, and X represent the numbers 1 through 10, respectively. The system uses a combination of letters from the Latin alphabet with specific rules for repetition, addition, and subtraction to form these numbers. Mastery of these numerals provides a solid foundation for interpreting Roman numerals in various modern contexts, such as in clocks, book chapters, and historical documents, enhancing both historical knowledge and numerical literacy.
Text prompt | 696 | 2,530 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.6875 | 5 | CC-MAIN-2024-33 | latest | en | 0.787564 |
http://lavalle.pl/planning/node692.html | 1,660,253,714,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571502.25/warc/CC-MAIN-20220811194507-20220811224507-00261.warc.gz | 29,950,518 | 3,685 | #### Simplifying the inertia matrix
Now the inertia matrix will be considered more carefully. It is a symmetric matrix, which can be expressed as
(13.93)
For each , the entry is called a moment of inertia. The three cases are
(13.94)
(13.95)
and
(13.96)
The remaining entries are defined as follows. For each such that , the product of inertia is
(13.97)
and .
One problem with the formulation so far is that the inertia matrix changes as the body rotates because all entries depend on the orientation . Recall that it was derived by considering as a collection of infinitesimal particles in the translating frame. It is possible, however, to express the inertia matrix in the body frame of . In this case, the inertia matrix can be denoted as because it does not depend on the orientation of with respect to the translational frame. The original inertia matrix is then recovered by applying a rotation that relates the body frame to the translational frame: , in which is a rotation matrix. It can be shown (see Equation (2.91) and Section 3.2 of [994]) that after performing this substitution, (13.92) simplifies to
(13.98)
The body frame of must have its origin at the center of mass ; however, its orientation has not been constrained. For different orientations, different inertia matrices will be obtained. Since captures the physical characteristics of , any two inertia matrices differ only by a rotation. This means for a given , all inertia matrices that can be defined by different body frame orientations have the same eigenvalues and eigenvectors. Consider the positive definite quadratic form , which represents the equation of an ellipsoid. A standard technique in linear algebra is to compute the principle axes of an ellipsoid, which turn out to be the eigenvectors of . The lengths of the ellipsoid axes are given by the eigenvalues. An axis-aligned expression of the ellipsoid can be obtained by defining , in which is the matrix formed by columns of eigenvectors. Therefore, there exists an orientation of the body frame in which the inertia matrix simplifies to
(13.99)
and the diagonal elements are the eigenvalues. If the body happens to be an ellipsoid, the principle axes correspond to the ellipsoid axes. Moment of inertia tables are given in many texts [690]; in these cases, the principle axes are usually chosen as the axis of the body frame because they result in the simplest expression of .
Steven M LaValle 2020-08-14 | 524 | 2,473 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2022-33 | latest | en | 0.933762 |
https://planetmath.org/inducedpartialorderonanalexandroffspace | 1,716,744,531,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058972.57/warc/CC-MAIN-20240526170211-20240526200211-00396.warc.gz | 394,945,743 | 5,091 | # induced partial order on an Alexandroff space
Let $X$ be a $\mathrm{T}_{0}$, Alexandroff space. For $A\subseteq X$ denote by $A^{o}$ the intersection of all open neighbourhoods of $A$. Define a relation $\leq$ on $X$ as follows: for any $x,y\in X$ we have $x\leq y$ if and only if $x\in\{y\}^{o}$. This relation will be called the induced partial order on $X$.
$(X,\leq)$ is a poset.
Proof. Of course $x\in\{x\}^{o}$ for any $x\in X$. Thus $\leq$ is reflexive.
Assume now that $x\leq y$ and $y\leq x$ for some $x,y\in X$. Assume that $x\neq y$. Then, since $X$ is a $\mathrm{T}_{0}$ space, there is an open set $U$ such that $x\in U$ and $y\not\in U$ or there is an open set $V$ such that $y\in V$ and $x\not\in V$. Both cases lead to contradiction, because we assumed that $x\in\{y\}^{o}$ and $y\in\{x\}^{o}$. Thus every open neighbourhood of one element must also contain the other. Thus $\leq$ is antisymmetric.
Finally assume that $x\leq y$ and $y\leq z$ for some $x,y,z\in X$. Since $y\in\{z\}^{o}$, then $\{z\}^{o}$ is an open neighbourhood of $y$ and thus $\{y\}^{o}\subseteq\{z\}^{o}$. Therefore $x\in\{z\}^{o}$, so $\leq$ is transitive, which completes the proof. $\square$
Proposition 2. Let $X,Y$ be two, $\mathrm{T}_{0}$, Alexandroff spaces and $f:X\to Y$ be a function. Then $f$ is continuous if and only if $f$ preserves the induced partial order.
Proof. ,,$\Rightarrow$” Assume that $f$ is continuous and suppose that $x,y\in X$ are such that $x\leq y$. We wish to show that $f(x)\leq f(y)$, so assume this is not the case. Let $A=\{f(y)\}^{o}$. Then $f(x)\not\in A$. But $A$ is open, so $f^{-1}(A)$ is also open (because we assumed that $f$ is continuous). Furthermore $y\in f^{-1}(A)$ and because $x\leq y$, then $x\in f^{-1}(A)$, but this implies that $f(x)\in A$. Contradiction.
,,$\Leftarrow$” Assume that $f$ preserves the induced partial order and let $U\subseteq Y$ be an open subset. Let $x\in U$. Then for any $y\leq x$ we have $f(y)\leq f(x)$ (because $f$ preserves the induced partial order) and since $\{f(x)\}^{o}\subseteq U$ (because $U$ is open and $\{f(x)\}^{o}$ is the smallest open neighbourhood of $f(x)$) we have that $f(y)\in U$. Thus
$\{x\}^{o}=\{y\in X\ |\ y\leq x\}\subseteq f^{-1}(U)$
which implies that $f^{-1}(U)$ is open because $f^{-1}(A)$ contains a small neighbourhood of each point. This completes the proof. $\square$
Title induced partial order on an Alexandroff space InducedPartialOrderOnAnAlexandroffSpace 2013-03-22 18:45:55 2013-03-22 18:45:55 joking (16130) joking (16130) 4 joking (16130) Derivation msc 54A05 | 907 | 2,580 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 75, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2024-22 | latest | en | 0.807832 |
http://stackoverflow.com/questions/2616154/calculate-intersection-between-two-segments-in-a-symmetric-way | 1,406,602,680,000,000,000 | text/html | crawl-data/CC-MAIN-2014-23/segments/1406510264575.30/warc/CC-MAIN-20140728011744-00036-ip-10-146-231-18.ec2.internal.warc.gz | 269,151,340 | 16,461 | # calculate intersection between two segments in a symmetric way
When using the usual formulas to calculate intersection between two 2D segments, ie here, if you round the result to an integer, you get non-symmetric results.
That is, sometimes, due to rounding errors, I get that `intersection(A,B)!=intersection(B,A)`.
The best solution is to keep working with floats, and compare the results up to a certain precision. However, I must round the results to integers after calculating the intersection, I cannot keep working with floats.
My best solution so far was to use some full order on the segments in the plane, and have `intersection` to always compare the smaller segment to the larger segment.
Is there a better method? Am I missing something?
-
Any reason that you can't use, say, symmetricIntersection(A,B)=(intersection(A,B)+intersection(B,A))/2 ? Floating point addition is commutative. – brainjam Apr 12 '10 at 15:21
@brainjam, it's really nice and mathematically sound, but it requires me to calculate intersection twice, and to rely on floating point rounding, so I'm not sure it's much better than ordering on segments in practice. Anyhow thumbs up! Liked it. – Elazar Leibovich Apr 12 '10 at 15:42
@brainjam, once error creeps in, any operation can amplify the error. So if already comparing `X`==`Y` came out false, `X`==`(X+Y)/2` will likely also come out false. Moreover, in cases when `X`==`Y` could have come out true, `X`==`(X+Y)/2` may come out false (e.g when `X` and `Y` can be represented without error, but `X+Y` can no longer be represented without error.) – vladr Apr 12 '10 at 15:46
You do not want to compare segment lengths.
Also, I assume that when comparing `intersection(A',B')` to `intersection(B",A")` it is understood that `A'`'s coordinates are representationally identical to `A"`'s (idem for `B'` and `B"`), or else there is no solution.
This being said, consider segments `[PQ]` and `[RS]`, where `P`, `Q`, `R` and `S` are points in the plane. You want the intersections of the segments:
• `[PQ]` `[RS]`
• `[QP]` `[RS]`
• `[PQ]` `[SR]`
• `[QP]` `[SR]`
• `[RS]` `[PQ]`
• `[SR]` `[PQ]`
• `[RS]` `[QP]`
• `[SR]` `[QP]`
... to always return the same coordinate pair.
Ordering, first of the endpoints on each segment, then of the segments themselves (based on each segments' least endpoint), is the only solution that guarantees reproducible results. The ordering itself can be computationally trivial, e.g. `P<Q` iff `P.x < Q.x || P.x == Q.x && P.y < Q.y`, although branching could get expensive if dealing with millions of segments (see how you can make use of SIMD to swap segment coordinates in-place if possible to generate the ordering.)
-
Thanks, that's what I thought of. However, if there was a method that always choose the computational path that gives the least precision error, that would be symmetric, and would reduce the precision error. Maybe it is not possible to do that though. – Elazar Leibovich Apr 11 '10 at 16:51
@Vlad, You start off with "You do not want to compare segment lengths". Why do you say that? Is it computational cost, or something else? – brainjam Apr 12 '10 at 14:51
@brainjam, while I can't speak for Vlad, I think he don't want the precision lost and the computational cost of the square root function. – Elazar Leibovich Apr 12 '10 at 15:43
@barinjam, multiple reasons, both computational cost and correctness. In the end what you want is for the oredering of non-commutative terms in the intersection formula local.wasp.uwa.edu.au/~pbourke/geometry/lineline2d to be the same irrespective of the order in which the 4 endpoint `(x,y)` pairs are presented, which can be resolved by ordering them (the points.) Segment length is a naive criterion which does not guarantee non-commutative term ordering in the calculation which ultimately generates the errors. – vladr Apr 12 '10 at 15:43 | 1,005 | 3,886 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2014-23 | latest | en | 0.909873 |
alexfleischer-84755.medium.com | 1,637,980,535,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358078.2/warc/CC-MAIN-20211127013935-20211127043935-00562.warc.gz | 165,537,668 | 32,209 | # ROI : Return On Investment
## CUBEWISE : IBM TM1 / Planning Analytics Business Partner
IBM Planning Analytics powered by TM1 is a business performance management software . In 2020, Cubewise asked me to imagine a puzzle and there would be a prize for the winner. Let me share this story.
Aim : Show that within the Artificial Intelligence buzz Machine Learning is not everything. Optimization matters.
Even before getting interested in Artificial Intelligence (AI), I have always enjoyed puzzles. French mathematical games (see trophy below left) for 30+ years. Moreover I managed to solve 50+ IBM Ponder This challenges.
In 2019, I met some IBM business partners in Vienna to explain the value of decision optimization.
IBM Planning Analytics / TM1 automates planning, budgeting, and forecasting.
Rapid development and deployment of decision optimization models using mathematical and constraint programming.
Cubewise, enjoyed my presentation and even took some pictures during my presentation.
Later on, at their own customer events, it was my turn to take some pictures of them pitching CPLEX on top of PA.
I was very happy with that but that was only the beginning.
In 2020, they made me very happy. They asked me a puzzle (Not solve a puzzle this time, but write the challenge that will make other people crazy, not me!)
And not only for fun : the winner would get a real prize : a trip!
CUBE+WISE=MORE
Use each number between 0 and 9 only once, to replace the letters in the sentence.
The sum must be correct, you must use each number EXACTLY once. With the correct sum, there are many solutions. But which solution has the highest value for the letters R, O and I?
Target: Maximize the value of ROI. There is only 1 correct largest value for ROI. What number (like 123) corresponds to the letters ROI?
One can solve this challenge with a pencil but it takes some time.
Let me take the opportunity to explain how easy it is to solve this puzzle with IBM CPLEX.
Within IBM CPLEX you may rely on OPL and write
`using CP;...dvar int CUBE;dvar int WISE;dvar int MORE;dvar int ROI;maximize ROI;subject to {CUBE+WISE==MORE;ROI==100*R+10*O+I;CUBE == 1000*C+100*U+10*B+1*E;WISE == 1000*W+100*I+10*S+1*E;MORE == 1000*M+100*O+10*R+1*E;// all letters are differentallDifferent(append(C,U,B,E,W,I,S,M,O,R));}`
Which is very easy to read and gives the solution in less than 1 second.
You may do the same with Python.
`CUBE=1000*C+100*U+10*B+EWISE=1000*W+100*I+10*S+EMORE=1000*M+100*O+10*R+EROI=100*R+10*O+I#constraintsmdl.add(CUBE+WISE==MORE)mdl.add(mdl.all_diff(C,U,B,E,W,I,S,M,O,R))#objectivemdl.maximize(ROI)#solvemsol=mdl.solve()`
You may also use a more generic model:
`string maxobjective="ROI";string equation="CUBE+WISE==MORE";`
that will build the OPL model and solve this.
I used Constraint Programming solver within CPLEX. But Mathematical Programming works too both in OPL and Python.
Enumeration in OPL, Python or Javascript work fine too for this tiny challenge. For real business problems that could take years or even much more time. This approach cannot scale.
Thanks again Cubewise and congrats Cansu Agrali for winning this contest.
Let me quote Cubewise as why complete enumeration is not enough in real life:
## Decision Optimization
“Considering we have a solid plan in place and we have an accurate vision of the future, there is only one thing left to do: execute it ! The reality is that we need to execute this plan in an environment that is full of constraints, conflicting targets, and thousands of options to consider. Decision Optimization focuses on what route is best to follow within the constraints we have, given the targets we set in a specific priority.
Today, CPLEX is the best performing solution to tackle these mathematical algorithm challenges. As a bonus, it integrates perfectly with TM1.”
No Optimization within AI is as bad as using only your reptile brain instead of your primate brain. Not wise when you want to get our of a maze.
Optimization Expert at IBM Europe. His expertise is in computer science, mathematics, artificial intelligence and Optimization. Sup Aéro graduate 1996.
Optimization Expert at IBM Europe. His expertise is in computer science, mathematics, artificial intelligence and Optimization. Sup Aéro graduate 1996. | 1,026 | 4,313 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2021-49 | longest | en | 0.952185 |
http://matheducators.stackexchange.com/questions/1550/optimization-problems-that-todays-students-might-actually-encounter | 1,469,740,145,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257828322.57/warc/CC-MAIN-20160723071028-00061-ip-10-185-27-174.ec2.internal.warc.gz | 162,292,750 | 40,772 | # Optimization problems that today's students might actually encounter?
Our students are not fencing in farm fields, cutting wires and folding them, or designing windows, so they are often uninspired by the optimization problems we give them. They seem like something that "someone, somewhere" might use, but the examples feel distant.
What are good examples of constrained optimization problems (perhaps not simple!) that today's students might actually encounter in their lives?
If your goal is to find problems that are more easily accessible, see also the sister question What are easy examples from daily life of contrainted optimization?
-
There aren't any. There may be situations where it's possible to apply optimization to solve a problem you've encountered, but in none of these cases is it honestly worth the effort of solving the problem analytically. I optimize path lengths every day when I walk across the grass on my way to classes, but I'm not going to get out a notebook and calculate an optimal route just to save myself twelve seconds of walking every morning. Mathematics beyond basic arithmetic is simply not useful in ordinary life. But I'm not sure if that's exactly what you mean. – Jack M Apr 11 '14 at 15:45
@JackM That's an extremely depressing way to think about mathematics, and I'm sorry you think that way. – Chris Cunningham Apr 11 '14 at 16:07
@ChrisCunningham What's depressing is the notion that mathematics has to be relevant to practical life in order to be interesting. – Jack M Apr 11 '14 at 16:52
I think you've misjudged my agenda here. Mathematics is beautiful on its own, but optimization problems could additionally be relevant. It sounds like someone once told you that the purpose of mathematics is to apply to the world, and you were very displeased. I don't think anyone here is advocating such a position -- I'm certainly not! – Chris Cunningham Apr 11 '14 at 19:00
I wonder how Dr. Pangloss would answer. We live in the best of all possible worlds, but there is not a really good answer to this question. But Calculus of Variations is a really cool subject! – user52817 May 11 '15 at 18:31
There aren't any. There may be situations where it's possible to apply optimization to solve a problem you've encountered, but in none of these cases is it honestly worth the effort of solving the problem analytically. I optimize path lengths every day when I walk across the grass on my way to classes, but I'm not going to get out a notebook and calculate an optimal route just to save myself twelve seconds of walking every morning. Mathematics beyond basic arithmetic is simply not useful in ordinary life. But I'm not sure if that's exactly what you mean. – JackM
To some extent, I agree with this comment. With few exceptions, mathematics beyond basic arithmetic is simply not useful in everyday life. Students know this, and you'll have trouble convincing them otherwise.
Because of this, I've always found "everyday"-style calculus problems a little artificial. Consider the following problem from Stewart's Calculus: Concepts and Contexts.
A fence 8 ft tall runs parallel to a tall building at a distance of 4 ft from the building. What is the length of the shortest ladder that will reach from the ground over the fence to the wall of the building?
The proper response to this question is: who cares? Is there any reason to calculate this length precisely? Why would anyone ever use calculus to compute this? If you have an actual building and an actual ladder, you could just try it and see if the ladder fits. If you don't have a specific ladder in mind (e.g. you are buying a ladder), the thing to do would be to draw the situation on paper and then use a ruler to estimate the minimum length. Of course, it's neat that you can use calculus to solve this problem precisely, but this is more of a curiosity than a legitimate application.
Chris specifically mentions the farmer fence problem, the wire-cutting problem, and the Norman window problem as not relevant to the students' lives. I agree—none of these problems are relevant. But it's not because the students aren't farmers, or wire-cutters, or architects. Even in a class full of future farmers, the fence problem would still be bad, because farmers don't use calculus to plan their fences.
## Good Optimization Problems
What calculus is useful for is science, economics, engineering, industrial operations, finance, and so forth. That is, it's useful for all the things that make our society run. Most students who take calculus at a university are planning to go into one of these fields, so calculus will be relevant in their lives—specifically in their future studies and in their professions.
Here's something that's closer to a real-life optimization problem:
When a critically damped RLC circuit is connected to a voltage source, the current $I$ in the circuit varies with time according to the equation $$I \;=\; \biggl(\frac{V}{L}\biggr)te^{-Rt/(2L)}$$ where $V$ is the applied voltage, $L$ is the inductance, and $R$ is the resistance (all of which are constant).
Suppose an RLC circuit with a resistance of 30.0 volt/amp and an inductance of 0.400 volt·sec/amp is attached to a 12.0-volt voltage source. Find the maximum current that will occur in the circuit.
This is at least close to something that a physics or engineering student might actually come across in their future studies. It's real in a way that the farmer fence problem isn't, and even students who don't plan to study physics can sense that this is a legitimate application. (By the way, if you have good students, you might even ask them to come up with a formula for the maximum current, without giving them specific numbers for $V$, $L$ and $R$. This has the advantage that it can't simply be solved using a graphing calculator.)
Of course, this isn't actually a constrained optimization problem—it's just an optimization problem. I'm not actually aware of any place in science that simple constrained optimization problems arise, although there are examples from economics (maximizing utility), finance (optimal portfolios), and industrial design (e.g. shape of a can type problems). When I cover constrained optimization in calculus, I usually stick to industrial-type problems (best cans, best shipping crates/boxes, best pipeline across a river, etc.), but that's probably just because I don't know enough about economics or finance to make up problems that involve them.
Finally, I should mention that I've never found the optimization portion of Calculus I particularly compelling. It's good to introduce the idea of optimization, but setting the derivative equal to zero isn't actually a very useful optimization technique by itself. It only really works for simple formulas—for anything complicated it just replaces one essentially numerical problem (finding the maximum of a function) with another (finding roots of a function). I agree that it should be covered, but it's far from the most important application of calculus.
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Let's turn the table a bit. Suppose that it takes you 2 minutes to calculate that grass-field path that will save you 12 seconds every time you walk through the field. Then after 10 days of classes it'll pay off. You can reap the benefits for years to come. I jest but this is the kind of thing optimization problems are good for. You take the time to calculate the optimization once and you continuously benefit from it from then on. – Aleks Vlasev Apr 30 '14 at 4:00
Here's the example I had which inspired me to post the question in the first place:
The game League of Legends was the most-played PC game, in number of hours played, in North America and Europe in 2012. There is a good chance that League of Legends is a part of many of your students' daily life, especially if you are teaching engineering calculus. It doesn't have any sexual or deviant themes -- it is a game about teams of "champions" fighting in an arena -- so I've found it to be reasonable to bring it up in class. The reason to go this route is that many of your students might actually be encountering this optimization problem every day. Here's the problem:
In League of Legends, a player's Effective Health when defending against physical damage is given by $E = \frac{H(100+A)}{100}$, where $H$ is health and $A$ is armor.
(1) Health costs 2.5 gold per unit, and Armor costs 18 gold per unit. You have 3600 gold, and you need to optimize the effectiveness E of your health and armor to survive as long as possible against the enemy team's attacks. How much of each should you buy?
You can actually go further if you want to make it more complicated (and more accurate to what actually happens in the game!)
(2) Ten minutes into the game, you have 1080 health and 10 armor. You have only 720 gold to spend, and again Health costs 2.5 gold per unit while Armor costs 18 gold per unit. Again the goal is to maximize the effectiveness E. Notice that you don't want to maximize the effectiveness of what you purchase -- you want to maximize the effectiveness E of your resulting health and armor. How much of each should you buy?
And the last one is much more challenging, but it is something that even professional players of League of Legends (these exist!) can get incorrect:
(3) Thirty minutes into the game, you have 2000 health and 50 armor. You have 1800 gold to spend, and again Health costs approximately 2.5 gold per unit while Armor costs approximately 18 gold per unit. Again the goal is to maximize the effectiveness E of your resulting health and armor. How much of each should you buy?
My favorite part of this problem is that the game is actually more complicated than this -- in reality you need to defend against both physical and magic damage by buying both armor and magic resistance, which adds another variable. So the "punchline" of the exercise is that you can't actually play the most popular PC game properly until you understand Lagrange multipliers. I'll see you in Multivariable Calculus.
(1) You do not spend equal money on A and H: $E = 3H - \frac{1}{720}H^2$ so the maximum is at $H = 1080$, plug back in for $A = 50$.
(2) One way to do this is to realize from number 1 that we know an optimal configuration is $H = 1080$ and $A = 50$, so right now we have way too much health and not enough armor. The answer to this is that we should spend all the money on 40 armor, to get exactly back to the optimized answer to #1.
(3) If H and A are the amount they plan to buy, the effectiveness is $E = \frac{(H + 2000)(100 + (50 + A))}{100}$ since they started with 2000 and 50 respectively. The critical point appears at H = -100, so the maximum actually occurs at one of the endpoints, not at the critical point at all. Again the player should spend all the money on armor.
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I'd like to point out a disadvantage of this example: PC games are, in general, played much more by the male students, so this could have the effect of exacerbating any preexisting notions that math and science classes are "boys' clubs." I'm not sure how important this is, but it is worth considering. – Chris Cunningham Apr 11 '14 at 19:07
Both PC games and math have these preexisting notions and both are to a great extend true, so by disregarding one you should disregard both notions or disregard neither and accept both. Oh well, either way, a lot of examples in mathematics are about building stuff, but I haven't seen anybody complain yet that construction is a male thing to do (and the gender ratio is even far more strongly male in construction work than it is in gaming or maths) – David Mulder Apr 11 '14 at 22:49
Additionally let me add that I think gaming is one of the greatest possible cases mathematics can use. Why? Because it has simpler models than (useful) physics, yet complex and relevant enough models to be useful for students. True, a part of the students will be quite alienated by the cases, but on average I think students tend to apply their mathematics skill by far the most in games (you would be surprised what crazy spreadsheets highschoolers are able to construct if it is to win some game) – David Mulder Apr 11 '14 at 22:51
(1/4) It's true, and important to keep in mind, that League of Legends players skew heavily male [1], but I'm not so confident in the generalization that "PC games are, in general, played much more by the male students." Almost exactly half of PC gamers in the US are women, although that proportion may change when you specialize to university students [2]. – Vectornaut May 10 '15 at 1:07
(2/4) Gender demographics can vary wildly from game to game. For example, in 2009, women of age 25 and older accounted for more than half of play minutes in World of Warcraft, although they were outnumbered by male players in the same age group [3]. Around the same time, women made up only a fifth of EverQuest 2 players [4]. – Vectornaut May 10 '15 at 1:08
This is from a post of mine on Math Stackexchange.
When I was in college, I owned three lamps and had a dark apartment. I kept trying to position them in different areas of the room, but it was still dark. Then I decided to model the problem with math: given three light sources of equal strength in a rectangular solid of a room, where can you place them to maximize the average value of the amount of light reaching a point in the solid?
Then I realized that the asymptotes at the light bulbs themselves would make the average diverge.
Recently, I thought of a reformulation:
Given three light sources in a rectangular solid of a room (assume height of ten feet, width of 15 and length if 25),all at height 6 ft., where should they be placed to maximize the average value (Edit: Minimum value) of the intensity of the light on the plane at height 3.5 ft.? (Since the couches are that height).
Alternatively, if someone can give a more accurate reformulation of this problem (perhaps involving reflections of the wall) and solve it, that would be much cooler.
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To illustrate what I said in the comments, although this is certainly an interesting problem, and easy for a student to relate to their own life, what I think would be actually harmful is telling a student that being able to solve a problem like this would be actually useful to them (although I don't think that's what the answerer is saying). Say it's interesting, but don't say with a straight face that it's useful - the student isn't that stupid, and you'll only lower their opinion of you, and cement their point of view that math is pointless. – Jack M Apr 11 '14 at 18:36
@JackM Unfortunately, you are wrong. I had struggled with this problem with weeks, and after getting an answer on MathSE, I implemented their solution and significantly improved my lighting. My wife and I use it as a guide every time we rearrange furniture. – Brian Rushton Apr 11 '14 at 18:55
No student is going to consider "you'll slightly (or even significantly) improve your lighting one day", or anything similar, as enough of a justification for soldiering through a class that they hate for another four years. It's better to simply try to communicate to them that the most interesting aspects in mathematics aren't in how it can solve real-world problems. Problems inspired by reality can be useful for that, but it seems counterproductive to take the attitude that the problems are interesting because they're useful. – Jack M Apr 11 '14 at 19:03
@jack-m, the problems are interesting because they are useful somewhere. Some are interesting in "real life", even simple, or simplified, math can shed light on daily real problems. Those quite interest me, they seem to interest a few other people here, Chris that started this question, Brian in this reply, and so on. And I bet it interested them, as it did me, when I was a student. So simply because they do not interest you, don't extrapolate that 'no students' will enjoy them. Some will, many, or a few, I don't know, but some will for sure. – Rolazaro Azeveires Apr 14 '14 at 2:07
An answer depends on what you mean with their lives. I think, it is more likely that students will face such problems in their working (not academic) life.
The Netflix Prize example
My example is the Netflix Prize, which was awarded by Netflix in the mid 2000s. At that time, Netflix was a DVD rental service and users did rate the movies they have seen. Netflix had some algorithm which gave recommendations which movies a particular user could like and awarded 1 mio $to the team which improved the algorithm by 10%. The rating was as follows: For all users$u\in U$and all movies$m\in M$, there is a (possible) rating$r(u,m)\in\lbrace 0,1,2,3,4,5\rbrace$. There is a subset$K\subset U\times M$, for which the rating function is known. Netflix did a partition of$K$into$K_{\text{known-to-everyone}}$and$K_{\text{known-to-netflix}}$. The problem was then: Find an approximating rating function$r_{\text{appr}}$such that$\Vert r_{\text{appr}}(u,m)-r(u,m)\Vert_{K_{\text{known-to-netflix}}}$subject to$r_{\text{appr}}(u,m)=r(u,m)$on$K_{\text{known-to-everyone}}$. That means, everyone got the known data and should make some model to make good predictions (depending on a specific given norm) on unknown data only known to Netflix (who by that data found out which algorithm was good). The documentation of the best algorithm is published; some other teams published their approach as well and used methods from constrained optimization. I was attending a talk where someone presented his solution which uses a low rank approximations and optimization methods on manifolds. Why could that be important to students in their future lives? The prize was awarded and the million is gone. But I can image that such "data fitting" task are common in insurance economy where they have a lot of data and want to find models in order to derive unknown data, which will be important for the insurance company to model their prices. - The original classic real-life example is to lay out a shop floor with multiple workstations and workers, so as to minimize travel distance of the work piece given a required sequence of operations. This can be generalized to consider a standard work load distributed amongst more than one type of workpiece, with different sequences of operations, and different times spent at each workstation. Some links: Any number of additional links could probably be found by Googling "Lean Kai-Zen" and "optimizing shop floor layout" Update: A second classic example is the optimization of paper-flow through an accounting department. Two other examples were these techniques were used to great effect in WWII are noted here (my emphasis): While performing an analysis of the methods used by RAF Coastal Command to hunt and destroy submarines, one of the analysts asked what colour the aircraft were. As most of them were from Bomber Command they were painted black for night-time operations. At the suggestion of CC-ORS a test was run to see if that was the best colour to camouflage the aircraft for daytime operations in the grey North Atlantic skies. Tests showed that aircraft painted white were on average not spotted until they were 20% closer than those painted black. This change indicated that 30% more submarines would be attacked and sunk for the same number of sightings.[19] As a result of these findings Coastal Command changed their aircraft to using white undersurfaces. A Warwick in the revised RAF Coastal Command green/dark grey/white finish Other work by the CC-ORS indicated that on average if the trigger depth of aerial-delivered depth charges (DCs) were changed from 100 feet to 25 feet, the kill ratios would go up. The reason was that if a U-boat saw an aircraft only shortly before it arrived over the target then at 100 feet the charges would do no damage (because the U-boat wouldn't have had time to descend as far as 100 feet), and if it saw the aircraft a long way from the target it had time to alter course under water so the chances of it being within the 20-foot kill zone of the charges was small. It was more efficient to attack those submarines close to the surface when the targets' locations were better known than to attempt their destruction at greater depths when their positions could only be guessed. Before the change of settings from 100 feet to 25 feet, 1% of submerged U-boats were sunk and 14% damaged. After the change, 7% were sunk and 11% damaged. (If submarines were caught on the surface, even if attacked shortly after submerging, the numbers rose to 11% sunk and 15% damaged). Blackett observed "there can be few cases where such a great operational gain had been obtained by such a small and simple change of tactics".[20] - As I mentioned in my other answer, it highly depends what is meant by their lives. Here is a every day problem which I, at least, often encounter: If you do the dishes and finish, there is a lot of foam in your sink. Arrange the position and strength of the water-tap in order to let the foam disappear as fast as possible. The problem sounds easy, but is indeed an optimal control problem with partial differential equations (Navier-Stokes) and boundary control, which is not very easy to solve :) - Ha! This is amazing. Maybe I should remove the calculus tag from my question to encourage more such answers. – Chris Cunningham Apr 11 '14 at 19:03 Not at all relevant to the question, but there is a non-mathematical answer to your foam problem. Dissolve one or two Gas-X tablets in the dish water. The simethicone in the Gas-X is an anti-foaming agent. It will prevent formation of foam, and it will make any existing foam disappear quickly. – Tristan Apr 11 '14 at 19:51 Or sprinkle some salt on top. – JPBurke Apr 12 '14 at 20:56 Here's an optimisation problem that really should have solutions readily available, but hasn't: Consider a SatNav that tells you the best way to get from A to B. Now assume the distance is quite long, and you won't be able to get from A to B without going to a petrol station. So what I want: Find the shortest way from A to B visiting any petrol station within X miles of my current location. They all are capable of showing you nearby petrol stations, but that's not what I want if I'm 200 miles away from B and have fuel for 150 miles in the tank. By the way: Finding the shortest path from A to B is actually an interesting problem in itself. SatNavs (at least the cheap ones that you will tend to use) are not capable of actually finding what I would consider good solutions, which take into account time, fuel consumption, wear and tear on the car, driver stress. (For example a short journey with many turns where I have to give way to other drivers means I stop and accelerate a lot and have more stress). - Even more interesting, find me a petrol station between A and B taking into account the cost of petrol at each station, how much petrol/time is used to get to the station, how much petrol I have at present and how a full tank reduces MPG. – Ian Apr 19 '14 at 13:11 A lot of examples exist in industry. • Manufacturing plants use optimization to figure out how to best run their machinery, buy raw materials, ship finished goods, etc. • Airlines and other passenger transportation services use optimization to determine their schedules. • The cargo transportation industry (trucking, trains, etc.) uses optimization to determine how to transport goods as quickly and inexpensively as possible. I had an entire series of classes in my undergraduate program centered around an example of optimization in transportation. (I was part of a program that combined computer science and business, so the administrators had a lot of leeway designing classes that went together.) We had to write a program for a fictitious train company that determined how to transport orders in the cheapest way possible using the simplex algorithm. We had to write our own implementation of the simplex solver in a programming class and learned about the algorithm in detail in a math-heavy business course. - +1 for mentioning that you found an example rich and compelling enough to build a series of classes around. Creating good problems is as much a writing task as it is a mathematical one. I imagine it's pretty easy to come up with dry, boring examples of linear programming problems, but those are worse than useless from an educational point of view. – Vectornaut May 10 '15 at 3:19 Edit (4/29/14): A popular media piece in the New York Times on rent division gives a great example of how Sperner's Lemma can be used for fair division problems (as related to cake cutting, the history of which can be found in a pasted excerpt at the end of my response here). The AMM article drawn upon is: Su, F. E. (1999). Rental Harmony: Sperner's Lemma in Fair Division. American mathematical monthly, 106(10), 930-942. Link (no paywall). A March 2014 PLOS paper describes a mathematical model using optimal control theory and a system of ODEs to build an app that combats jet-lag. A popular media piece can be found here, with the headline: A new mathematical model can cut jet-lagged time in half. Even if the details elude students, this is certainly a problem that today's students might run into; moreover, they can now take part in testing the app to determine if it works. Excerpt from the latter link: [Biological mathematician] Forger admits that the schedules his model spits out may sometimes seem counterintuitive, but intuition, he says, is not the point. "We're trying to move the science beyond your grandmother's advice of 'wake up late' or 'avoid carbohydrates' to something that can be rigorously tested," he said. "All I know is these schedules are optimal according to the mathematics." Forger spent 10 years building his model based on data collected from sleep studies done at Harvard and the University of Michigan. He had no idea what type of schedules the model would come up with when he first started, so he was pleasantly surprised that according to the math, the best way to beat jet lag is to adjust the time of your dawn and dusk each day. See more in the LA Times write-up or in the methods section (p. 10) of the actual paper. In case the cake-division discussion below does not suffice, there is the question of how one physically carries out the division for a different food: matzah. See the short Wired article here, and perhaps the question of how to break matzah optimally (for two people: in half) can provide a real-life optimization problem. I would probably start by looking to literature on mathematical modeling. An example source is the COMAP book "For All Practical Purposes" which includes sections on optimal entry (in elections), optimal production policies, and optimal schedules. There are plenty of other problems that don't have the word "optimal" in their name (e.g., bin packing). The book provides a wealth of examples; another nice one is fair division, which can be applied to scenarios ranging from company mergers to inheritance to organ transplant policies. Some of these are probably a bit removed from students' daily lives, but others are run into frequently. For instance, you could read over the literature on cake-division. For three cake eaters, there is the Steinhaus Proportional Procedure; for four or more, there is the Banach-Knaster proportional procedure. The book contains a sixty-year history of cake cutting. I have pasted an image of this below (written up as p. 421 in the eighth edition); the word "California" is, for some reason, omitted between the two columns: I have included only one actual excerpt here, but I recommend this book - and other materials on mathematical modeling, e.g., COMAP's Mathematical Modeling Handbook - as sources for optimization problems encountered (either literally, such as when you want to divide up a cake, or mathematically, when you see another problem to which you can apply these models/methods). - On a similar note, it may be a good idea to provide students with an open-ended mathematical modeling problem which seeks the maximal or minimal value of some quantity. Students will ultimately be forced to use calculus to determine the maximum or minimum of this function. – Shivam Sarodia Apr 12 '14 at 3:49 This answer is turning into an incredible resource. Thanks so much for this. – Chris Cunningham Apr 29 '14 at 14:34 @ChrisCunningham Surprising to see fair-division problems repeatedly cropping up: cake-division, then matzah-division, now rent-division. (The Baader-Meinhof Phenomenon in action...) If I start to accrue other (non-division) examples, then I may have to post another response in order to keep this one from becoming unwieldy! – Benjamin Dickman Apr 29 '14 at 14:43 Here's a problem that I found interesting: I listen to audiobooks. When you import them from CD, you get lots of tiny chapters (often 2 or 3 minutes) which is not very practical. Also, audiobooks can be very long - some are 30 hours - which is also not very practical. I prefer audiobooks to be split into individual volumes of not more than 8 hours, and chapters around 10 minutes. Assume that you can combine consecutive chapters into one chapter (chapters are indivisible), and that you can combine consecutive chapters into volumes. Problem 1: Define an optimisation problem. Find a function that, if minimised, produces audiobooks in the way that I (or you) prefer them. I'd want volumes as long as possible, but never more than 8 hours (unless a chapter is longer than 8 hours and it becomes unavoidable), and volumes of approximately equal length. I'd want chapter length around ten minutes, and preferably equal lengths. Defining a function to minimise is not trivial. Problem 2: Find an optimal solution according to the function to be minimised. Decide how important execution speed is (hint: Actually creating an audiobook according to the solution takes a long time). Obviously the way you combine chapters affects the possible lengths of volumes. Problem 3: Does the optimal solution actually match my/your preferences? If not, is it because finding a good solution was hard, or because the optimisation problem wasn't chosen well? If needed, go back to problem 1. - +1 For adding Problem 3, that's a very important part of "real world" optimization that often gets left out of exercises. – Ian Apr 12 '14 at 19:13 How critical is it that the chapters stay in the correct order? I wouldn't want an audiobook that jumps from chapter 3 to chapter 10. – Nzall Apr 14 '14 at 12:08 Finance has lots of good examples. Here is one version of the portfolio optimization problem. Suppose two assets have mean excess returns$\mu, \mu'$, standard deviations$\sigma, \sigma'$and a correlation$\rho$. You have cash to distribute among these two assets in proportions$a, a'$where$a+a'=1$. What values of$a,a'$maximize the Sharpe ratio of your investment: $$\frac{a\mu + a'\mu'}{\sqrt{a^2 \sigma^2 + a'^2\sigma'^2 + 2\rho\, aa'\sigma\sigma'}}?$$ - Can't believe no one's mentioned the coke-can problem! You are the designer of drinks cans for Coke, how can you make them the most money? Pretty quickly it comes out that you want a cylinder of volume 330ml with minimal surface area (use the least metal possible to enclose your beverage). It's a classic problem and one kids can very easily relate to on all number of levels, especially if you hype it a little bit. The interesting punchline is that Coke cans don't look at all like your design (which is predictably as close to a cube as a cylinder can get), for reasons that transpire to be aesthetic. But paint cans totally do though... This is because the metal must be significantly thicker in this case. - When someone swallows a dose of a drug, it doesn't go into their bloodstream all at once. What will the drug's peak blood concentration be, and when will it be reached? If the drug is caffeine, which is absorbed and eliminated by first-order kinetics, its blood concentration$c$rises and falls over time along the curve $$c(t) = \tfrac{D}{1-\beta/\alpha}\left(e^{-\beta t} - e^{-\alpha t}\right),$$ where$\alpha$is the absorption rate,$\beta$is the elimination rate, and$D$is the size of the dose. The expression for the time of the peak is very pretty. Once you work it out, you can check your answer on p. 108 of this study, where the authors use it to help interpret their data. For alcohol, which is eliminated by zeroth-order kinetics, the differential equation for the blood concentration curve$c(t)$is simple enough for a first-year calculus student to solve: $$c'(t) = D\alpha e^{-\alpha t} - \beta.$$ After checking your work against Equation 1 from this study, you can find the peak time as before. In case the study links break, here they are for the record: • Seng et al. Population pharmacokinetics of caffeine in healthy male adults using mixed-effects models. Journal of Clinical Pharmacy and Therapeutics (2009). • Uemura et al. Individual differences in the kinetics of alcohol absorption and elimination: A human study. Forensic Science, Medicine, and Pathology (2005). Although finding the blood concentration curve for caffeine is probably beyond first-year calculus students, it might be a fun exercise for differential equations students. It's given by the linear system $$\left[ \begin{array}{r} s' \\ c'\end{array} \right] = \left[ \begin{array}{r} -\alpha & 0 \\ \alpha & -\beta \end{array} \right] \left[ \begin{array}{r} s \\ c\end{array} \right],$$ where$s$is the concentration of caffeine in the drinker's digestive tract. - Creating a timetable, all students should understand the issues. However mathematical solutions don’t work as well as computer sci solutions. Choosing a set of options, so as to get a degree while having the lease dead time, the dead time being when you are waiting for the next activity to start, but don’t have long enough to do someone useful. In the US when degrees can take a undefined number of years, deciding how much part time work to do, so as to minimise the debt level at age 30. More part time work, increase the time to get the degree, but decreases the debt earn per year. Completing the degree sooner, allows the repayment of the debts to start sooner. - A real-life optimization problem: It takes about 20 clicks and 2 minutes to shoot a wolf. 200 dead wolves will get you enough experience points to get to the next level. It takes about 50 clicks and 10 minutes to steal gold from a palace. 50 burglaries will get you enough experience points to get to the next level. However, you would get carpal tunnel syndrome after 3000 clicks which would force you to stop playing for a day. What is the shortest possible time you can get to the next level? - I imagine that people who farm gold for a living face some version of this problem every day, and that the solutions they find can have a serious effect on their quality of life. – Vectornaut May 10 '15 at 3:11 The concept of impedance matching is one that is occurring in your daily life right now if you're using a wired internet connection. There are more sophisticated and less sophisticated examples of impedance matching. A very conceptually simple one is the following. Suppose that a fixed voltage is applied to resistors$x$and$y$, which are in series. The resistance$y$is fixed. You are free to choose any value of$x$in order to maximize the power dissipated in$x$. The power is$V^2x/(x+y)^2$. Another example from everyday life is that you want to throw a baseball as far as possible. The range is$R=(2v^2/g)\sin\theta\cos\theta$, where$v$is the fixed speed at which you can throw, and$\theta$is the angle at which you throw the ball, relative to the horizontal. At what angle should you throw? - The baseball problem feels pretty contrived to me, but when you replace the baseball with a cannonball, it becomes a problem that was studied intensely by the military engineers of 18th-century Europe. I'd guess that the maximum range of a cannon falls far short of the one you'd compute this way, because air resistance apparently has a big effect on cannon shots, but this computation at least gives you an easy upper bound. Chapter 13 of Steele and Dorland's The Heirs of Archimedes looks like a great resource for this. – Vectornaut May 14 '15 at 22:12 Profit maximation! It's easy to understand, not so hard to calculate and even testable. Let them sell lemonade or whatever at the next school festival and see, if they can find the optimal price for the lemonade. - This isn't really a math problem, as it involves a lot of knowledge about customer habits and preferences. – Jack M Apr 11 '14 at 18:38 @JackM That's no reason not to apply a mathematical model. And there are easy enough models usable for constrained optimization problems. – Toscho Apr 11 '14 at 19:09 Sure. I'd be interested in a longer answer if you have any specific ideas. – Jack M Apr 11 '14 at 19:12 The most simple, usable model is:$\text{Numbers sold}=f(\text{price})$with a linear function$f$with negative linear coefficient. The result will be quadratic function$\text{Profit}=g(\text{price})$with negative main coefficient. – Toscho Apr 11 '14 at 19:15 How about figuring out the dimensions of a cylindrical can of a specific volume that will minimize the surface area (and therefore minimize the amount of metal material needed to make up the can)? If you've taught calculus, you probably recognize this as one of the classic textbook optimization examples along with fences against barns and picture frames and the like, but the cylindrical can example seems to me the most practically relevant out of those classics. Note that a rectangular prism shape would make the optimization solution too boring, and any variation in which a container has an open top just seems too contrived, since stuff is never shipped in a container with an open top. - Many daily life and OR problems take the following form. One has a container of fixed size and a collection of objects each with a value, and size less than that of the container. The goal of the optimization is to decide what items to take which maximizes the value of what gets taken subject to the size constraint. Problems of this kind are known as knapsack problems, and a problem of this kind is solved every time one takes an airplane trip and also when the International Space Station has supplies sent. - No one has mentioned shopping directly, although various economic optimization problems have appeared in this thread. You have$20 to spend for your breakfast budget. See how oatmeal (or your favorite breakfast option) is packaged at different prices, whether by single serving envelope, small container at one price, large container at another. What is your quantity to optimize, and how will it affect your purchase? Flavor, convenience, wastage are some factors to consider, but even if the oatmeal is the same in each packaging, one of them will have the lowest ratio of price to pound. One can take this example in many directions and still keep to the shopping domain.
Gerhard "Come See My Shoe Collection" Paseman, 2015.05.11
- | 8,744 | 39,157 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2016-30 | latest | en | 0.965024 |
https://www.esaral.com/q/find-the-derivative-of-the-following-functions-74457 | 1,726,226,639,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651513.89/warc/CC-MAIN-20240913101949-20240913131949-00768.warc.gz | 697,255,021 | 11,711 | # Find the derivative of the following functions
Question:
Find the derivative of the following functions (it is to be understood that abcdp, q, r and s are fixed non-zero constants and m and n are integers): $\frac{4 x+5 \sin x}{3 x+7 \cos x}$
Solution:
Let $f(x)=\frac{4 x+5 \sin x}{3 x+7 \cos x}$
By quotient rule,
$f^{\prime}(x)=\frac{(3 x+7 \cos x) \frac{d}{d x}(4 x+5 \sin x)-(4 x+5 \sin x) \frac{d}{d x}(3 x+7 \cos x)}{(3 x+7 \cos x)^{2}}$
$=\frac{(3 x+7 \cos x)\left[4 \frac{d}{d x}(x)+5 \frac{d}{d x}(\sin x)\right]-(4 x+5 \sin x)\left[3 \frac{d}{d x} x+7 \frac{d}{d x} \cos x\right]}{(3 x+7 \cos x)^{2}}$
$=\frac{(3 x+7 \cos x)(4+5 \cos x)-(4 x+5 \sin x)(3-7 \sin x)}{(3 x+7 \cos x)^{2}}$
$=\frac{12 x+15 x \cos x+28 \cos x+35 \cos ^{2} x-12 x+28 x \sin x-15 \sin x+35 \sin ^{2} x}{(3 x+7 \cos x)^{2}}$
$=\frac{15 x \cos x+28 \cos x+28 x \sin x-15 \sin x+35\left(\cos ^{2} x+\sin ^{2} x\right)}{(3 x+7 \cos x)^{2}}$
$=\frac{35+15 x \cos x+28 \cos x+28 x \sin x-15 \sin x}{(3 x+7 \cos x)^{2}}$ | 471 | 1,013 | {"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} | 4.46875 | 4 | CC-MAIN-2024-38 | latest | en | 0.270963 |
https://www.convertunits.com/from/hundred+cubic+foot+of+natural+gas/to/zeptojoule | 1,660,495,599,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572043.2/warc/CC-MAIN-20220814143522-20220814173522-00164.warc.gz | 621,203,482 | 17,155 | ## ››Convert hundred cubic foot of natural gas to zeptojoule
hundred cubic foot of natural gas zeptojoule
How many hundred cubic foot of natural gas in 1 zeptojoule? The answer is 9.1979396615158E-30.
We assume you are converting between hundred cubic foot of natural gas and zeptojoule.
You can view more details on each measurement unit:
hundred cubic foot of natural gas or zeptojoule
The SI derived unit for energy is the joule.
1 joule is equal to 9.1979396615158E-9 hundred cubic foot of natural gas, or 1.0E+21 zeptojoule.
Note that rounding errors may occur, so always check the results.
Use this page to learn how to convert between hundred cubic foot of natural gas and zeptojoules.
Type in your own numbers in the form to convert the units!
## ››Quick conversion chart of hundred cubic foot of natural gas to zeptojoule
1 hundred cubic foot of natural gas to zeptojoule = 1.0872E+29 zeptojoule
2 hundred cubic foot of natural gas to zeptojoule = 2.1744E+29 zeptojoule
3 hundred cubic foot of natural gas to zeptojoule = 3.2616E+29 zeptojoule
4 hundred cubic foot of natural gas to zeptojoule = 4.3488E+29 zeptojoule
5 hundred cubic foot of natural gas to zeptojoule = 5.436E+29 zeptojoule
6 hundred cubic foot of natural gas to zeptojoule = 6.5232E+29 zeptojoule
7 hundred cubic foot of natural gas to zeptojoule = 7.6104E+29 zeptojoule
8 hundred cubic foot of natural gas to zeptojoule = 8.6976E+29 zeptojoule
9 hundred cubic foot of natural gas to zeptojoule = 9.7848E+29 zeptojoule
10 hundred cubic foot of natural gas to zeptojoule = 1.0872E+30 zeptojoule
## ››Want other units?
You can do the reverse unit conversion from zeptojoule to hundred cubic foot of natural gas, or enter any two units below:
## Enter two units to convert
From: To:
## ››Definition: Zeptojoule
The SI prefix "zepto" represents a factor of 10-21, or in exponential notation, 1E-21.
So 1 zeptojoule = 10-21 joules.
The definition of a joule is as follows:
The joule (symbol J, also called newton meter, watt second, or coulomb volt) is the SI unit of energy and work. The unit is pronounced to rhyme with "tool", and is named in honor of the physicist James Prescott Joule (1818-1889).
## ››Metric conversions and more
ConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 100 kg, US fluid ounce, 6'3", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more! | 786 | 2,683 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2022-33 | latest | en | 0.498516 |
https://www.mapleprimes.com/users/Mac%20Dude/answers?page=11 | 1,722,975,316,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640508059.30/warc/CC-MAIN-20240806192936-20240806222936-00460.warc.gz | 676,558,234 | 77,200 | ## 1561 Reputation
12 years, 286 days
## Works in Maple 15...
I copy-pasted your statements above into Maple 15 and I am getting the expected plots. The first one is from -10..10 (Maple-picked default); the second one is from -1..1 as expected. Will try Maple 17 later.
Are you sure there isn't something in your code that is not in your post? Or that somehow plot got redefined? You can test for that by copy-pasting the line into a new sheet.
M.D.
Edited: added question about redefined plot function.
## It's the GUI, stupid...
Here is what I get on Maple 15 in the Standard interface:
restart;A := <a,b,c>;
a := 1: b := 2: c := 3:
A;
B:=convert(A, list);
A;
B;
seq(A[i],i=1..3);
The output from the seq command as well as the result for B indicate that in fact A has been set... weird.
Exporting this as Maple Input and reading it into CLI Maple (same installation) I get this:
> A := <a,b,c>;
[a]
[ ]
A := [b]
[ ]
[c]
> a := 1: b := 2: c := 3:
> A;
[1]
[ ]
[2]
[ ]
[3]
> B:=convert(A, list);
B := [a, b, c]
> A;
[1]
[ ]
[2]
[ ]
[3]
> B;
[1, 2, 3]
> seq(A[i],i=1..3);
1, 2, 3
The output from the convert function is a bit weird, otherwise this is as I would have expected. Note that convert produces unexpected output in both cases (GUI and CLI).
Can't be too cautious, can't we, when using computer-generated results.
Mac Dude.
## ArrayTools:-RandomArray...
M:=ArrayTools:-RandomArray(1000,1000,distribution=normal);
?
On my machine (PowerMac G5 running Maple 15; both ancient by modern standards) this is practically instant.
It seems to be preferable to just about any of the other methods mentioned, except that you get to scale the numbers yourself to the range you want; and you get to figure out what range they cover to begin with as the docs don't actually say that.
Mac Dude
## Simplified code...
(I did this before Carl's and Markiyan's answers, so I might as well post it)
restart;
with(geom3d):
with(plots):
TruncatedIcosahedron(football,point(C,(0,0,0)),1):
PlotFootball:=proc(tr::float)
local i::integer;
display(seq(polygonplot3d(Matrix(faces(football)[i]),axes=none,scaling=constrained,transparency=tr),i=1..32)):
end proc:
PlotFootball(0.00);
The whole loop can be combined in the seq. Note that seq in general will be faster than a do loop.
M. D.
I have no clue what an "abecedarian" is, but assuming it is not a fancy word for "analphabet", I suggest some reading material:
A book I found most helpful was "Maple: An Introduction and Reference" by Michael Kofler. It is dated, but what is in there is mostly still valid and it goes beyond just rehashing the company documentation. And you can get it for pennies from Amazon. It will help you starting from zero knowledge.
The second one I like is the "Maple Programming Guide" available in pdf from MapleSoft. This is useful beyond programming as a number of concepts important even for interactive use are discussed. It starts roughly where Kofler's book ends.
There are also introductory guides from Maplesoft which I have mostly skipped in favor of experimenting. That does not mean I don't consider them good, but rather that my style of learning is different.
Edgardo Cheb-Terrab has written a brief introduction I also like (it came too late for me but I gave it to students). Search MaplePrimes for it.
Other than that, experiment and peruse the help facility. All CAS have a learning curve; Maple is no exception. If the help files let you down, read the above and/or ask here. And accept that initially you can solve problems faster by hand. The investment of time using Maple is worth it, though, as that will change as you get proficient.
M.D.
## Programming Guide is the way to go...
I second Tom and acre's suggestions to peruse the Programming Guide. It was the single most beneficial "book" for me, even for online use of Maple.
As you get into programming Maple, pay close attention to Maple's Modules and friends. While some learning is involved, it will make your life so much easier as your projects and programs grow in scope and size. Wrapping your code in modules (and possibly submodules) has proven for me to be a most powerful way to keep things organized and readable.
Inside your modules, a set of relatively short procedures with defined tasks will provide you with high-level functions that you can then connect together in your main program to do whatever it is you want to achieve. This is a more effective way of using Maple than writing long, monolithic code.
Another one of Maple's important functions is seq(), which in many cases can replace loops while executing faster and leading to more compact code. I remember seq() being slightly confusing initially; again, the effort spent on learning it is well worth it.
Ok, I shut up here. Maple is actually a wonderful programming language even if execution could be faster. It is my tool of choice these days unless I do large-scale simulation work that is cpu-time critical.
M.D.
## There is a number of ways you can do thi...
There is a number of ways you can do this. Myself, I would package the routines in a module and load that using with(modulename), somewhat like this:
MName:=module() option package
export A,B,C;
A:=proc(arglist)
...
end proc;
B:=proc(arglist)
...
end proc;
C:=proc(arglist)
...
end proc;
end module;
LibraryTools:-Save(MName,cat(libname[1],"/MName.mla")); # libname[1] has the directory where you want Maple to look for your packages.
A can use B and/or C and will know about them. There is also the "thismodule:-" prefix, but I doubt you need this here. The Maple Programming Guide explains all this; it is worth reading it.The "LibraryTools:-Save" function is necessary to create and save the .mla file Maple needs to load using with(MName).
Modules are a very powerful Maple constructs but they do take a little effort to use correctly. The reward is your ability to write your code in a very organized way, keeping things well structured and clear even when you build relatively large programs. They also allow you to write Maple code in a somewhat object-oriented fashion, further enhancing readability and maintainability of your code.
Mac Dude
## x11?...
If you run Maple in a (X)terminal, window is not supported as a plot device. You can try the x11 device if you run Maple in an xterm. The Maplet device seems to work, but you may have to plot "inside" the Maplet: RTFM (I have never done that).
I would wonder why, if plotting is involved, you use Maple in a terminal and not the GUI. Yes, the GUI is a bit clunky, but its plotting abilities are vastly superior to those of CLI-Maple.
M.D.
## for i from 1 to 10 do for j f...
for i from 1 to 10 do
for j from 1 to 10 do
# stuff (i,j)
end do; # j
end do; # i
You know, there is rather nice online help in Maple? Try it. In your sheet, just say
?for;
M.D.
## A different approach...
If all you want is some random numbers for a given pdf, you may use the proc Rarbit I have uploaded below. The comments should make it clear how to use it. While Maple's RandomTools routines maybe more efficient for some pdfs, for others---in particular those with gaps like yours and also those with constantly changing parameters---they do not work all that well, and the simple-minded approach in Rarbit actually works better. I wrote it for a problem I worked on a while back. Note that later I found out that the use of MersenneTwister in this form is not efficient, but I haven't had the need to fix that yet. It only becomes an issue if you need gazillions of random numbers.
Hope this helps,
Mac Dude
Rarbit.mw
## Load CUDA first...
You need to load CUDA first, using with(CUDA);
Then CUDA is available as a module and you can do your check above.
On my system I get an Error: CUDA not supported on the current system... :-(.
Mac Dude.
## Try update...
Try
DocumentTools[SetProperty]("PerformanceTable",update);
Per the online help this "forces an update".
"value" is not a property of the datatable object, so cannot be set. The refresh=true option works with other embedded components (I use it to update things like progress gauges in do loops), but maybe the datatable is different. It does appear the datatable elements can only be set by setting the associated Matrix elements.
Maybe you need to both set the refresh option and update.
Mac Dude
## not a Package...
The code your link refers to is not a Packasge in the Maple sense, but rather a collection of procedures. You should be able to read it in like this after you saved the code in a file code.mpl:
You can make a true Maple Package out of this code by wrapping it in a module like this:
Code:=module() option package;
export Help,S,...; # sequence of the names of all the procs in the code
(content of the file)
end module;
To make this accessible to Maple you first execute the above, which defines Code as a package. You then put it somewhere accessible:
LibraryTools:-Save(Code,cat(libname[1],"/Code.mla"));
where libname is a list of directories Maple will search for Code.mla.
You do this once and then you can load your package in any worksheet using with:
with(Code);
and it will all be there. Note that in this worksheet, libname also has to include the directory where Code.mla got saved. I do this in my .mapleinit file like this:
libname:="/Applications/Math_Calc/Maple 15/My_Maple_Libraries",libname:
whice prepends my own directory to whatever Maple needs, thus my stuff goes first.
HTH,
M.D.
## Answer my own question...
Ok, I can actually answer my own question:
As the term "module factory" implies, the generated procs are methods of a module that gets instantiated when the module factory routine is called. The module of course is persistent as it is a structure assigned to a name. The proc then gets called using the name:-proc() call. So the stored value has to be local to the instantiated module, not the proc, in order to be persistent.
Oh well, it's at least a year since I last used the module-factory approach. I guess the gray matter isn't as persistent a store as it used to be... :-).
M.D.
## need 64bit Maple...
Given Carl's comment it appears you need to get a copy of 64bit Maple. AFAIK that means Maple 17 or later.By your heading you run Maple 13. It's process is probably limited to something like 2 GB (depending on your compute platform). Can you use a sparse matrix? Maple knows how to do that; but I do not. RTFM.
M.D.
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| 2,560 | 10,525 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2024-33 | latest | en | 0.919468 |
https://www.ademcetinkaya.com/2023/01/lonmvct-molten-ventures-vct-plc.html | 1,679,559,548,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945030.59/warc/CC-MAIN-20230323065609-20230323095609-00202.warc.gz | 705,991,251 | 60,961 | Outlook: MOLTEN VENTURES VCT PLC is assigned short-term Ba1 & long-term Ba1 estimated rating.
Dominant Strategy : Sell
Time series to forecast n: 01 Feb 2023 for (n+4 weeks)
Methodology : Modular Neural Network (News Feed Sentiment Analysis)
## Abstract
MOLTEN VENTURES VCT PLC prediction model is evaluated with Modular Neural Network (News Feed Sentiment Analysis) and Stepwise Regression1,2,3,4 and it is concluded that the LON:MVCT stock is predictable in the short/long term. According to price forecasts for (n+4 weeks) period, the dominant strategy among neural network is: Sell
## Key Points
1. What statistical methods are used to analyze data?
2. Short/Long Term Stocks
3. Is now good time to invest?
## LON:MVCT Target Price Prediction Modeling Methodology
We consider MOLTEN VENTURES VCT PLC Decision Process with Modular Neural Network (News Feed Sentiment Analysis) where A is the set of discrete actions of LON:MVCT stock holders, F is the set of discrete states, P : S × F × S → R is the transition probability distribution, R : S × F → R is the reaction function, and γ ∈ [0, 1] is a move factor for expectation.1,2,3,4
F(Stepwise Regression)5,6,7= $\begin{array}{cccc}{p}_{a1}& {p}_{a2}& \dots & {p}_{1n}\\ & ⋮\\ {p}_{j1}& {p}_{j2}& \dots & {p}_{jn}\\ & ⋮\\ {p}_{k1}& {p}_{k2}& \dots & {p}_{kn}\\ & ⋮\\ {p}_{n1}& {p}_{n2}& \dots & {p}_{nn}\end{array}$ X R(Modular Neural Network (News Feed Sentiment Analysis)) X S(n):→ (n+4 weeks) $\begin{array}{l}\int {r}^{s}\mathrm{rs}\end{array}$
n:Time series to forecast
p:Price signals of LON:MVCT stock
j:Nash equilibria (Neural Network)
k:Dominated move
a:Best response for target price
For further technical information as per how our model work we invite you to visit the article below:
How do AC Investment Research machine learning (predictive) algorithms actually work?
## LON:MVCT Stock Forecast (Buy or Sell) for (n+4 weeks)
Sample Set: Neural Network
Stock/Index: LON:MVCT MOLTEN VENTURES VCT PLC
Time series to forecast n: 01 Feb 2023 for (n+4 weeks)
According to price forecasts for (n+4 weeks) period, the dominant strategy among neural network is: Sell
X axis: *Likelihood% (The higher the percentage value, the more likely the event will occur.)
Y axis: *Potential Impact% (The higher the percentage value, the more likely the price will deviate.)
Z axis (Grey to Black): *Technical Analysis%
## IFRS Reconciliation Adjustments for MOLTEN VENTURES VCT PLC
1. An entity may use practical expedients when measuring expected credit losses if they are consistent with the principles in paragraph 5.5.17. An example of a practical expedient is the calculation of the expected credit losses on trade receivables using a provision matrix. The entity would use its historical credit loss experience (adjusted as appropriate in accordance with paragraphs B5.5.51–B5.5.52) for trade receivables to estimate the 12-month expected credit losses or the lifetime expected credit losses on the financial assets as relevant. A provision matrix might, for example, specify fixed provision rates depending on the number of days that a trade receivable is past due (for example, 1 per cent if not past due, 2 per cent if less than 30 days past due, 3 per cent if more than 30 days but less than 90 days past due, 20 per cent if 90–180 days past due etc). Depending on the diversity of its customer base, the entity would use appropriate groupings if its historical credit loss experience shows significantly different loss patterns for different customer segments. Examples of criteria that might be used to group assets include geographical region, product type, customer rating, collateral or trade credit insurance and type of customer (such as wholesale or retail)
2. Rebalancing does not apply if the risk management objective for a hedging relationship has changed. Instead, hedge accounting for that hedging relationship shall be discontinued (despite that an entity might designate a new hedging relationship that involves the hedging instrument or hedged item of the previous hedging relationship as described in paragraph B6.5.28).
3. Fluctuation around a constant hedge ratio (and hence the related hedge ineffectiveness) cannot be reduced by adjusting the hedge ratio in response to each particular outcome. Hence, in such circumstances, the change in the extent of offset is a matter of measuring and recognising hedge ineffectiveness but does not require rebalancing.
4. To calculate the change in the value of the hedged item for the purpose of measuring hedge ineffectiveness, an entity may use a derivative that would have terms that match the critical terms of the hedged item (this is commonly referred to as a 'hypothetical derivative'), and, for example for a hedge of a forecast transaction, would be calibrated using the hedged price (or rate) level. For example, if the hedge was for a two-sided risk at the current market level, the hypothetical derivative would represent a hypothetical forward contract that is calibrated to a value of nil at the time of designation of the hedging relationship. If the hedge was for example for a one-sided risk, the hypothetical derivative would represent the intrinsic value of a hypothetical option that at the time of designation of the hedging relationship is at the money if the hedged price level is the current market level, or out of the money if the hedged price level is above (or, for a hedge of a long position, below) the current market level. Using a hypothetical derivative is one possible way of calculating the change in the value of the hedged item. The hypothetical derivative replicates the hedged item and hence results in the same outcome as if that change in value was determined by a different approach. Hence, using a 'hypothetical derivative' is not a method in its own right but a mathematical expedient that can only be used to calculate the value of the hedged item. Consequently, a 'hypothetical derivative' cannot be used to include features in the value of the hedged item that only exist in the hedging instrument (but not in the hedged item). An example is debt denominated in a foreign currency (irrespective of whether it is fixed-rate or variable-rate debt). When using a hypothetical derivative to calculate the change in the value of such debt or the present value of the cumulative change in its cash flows, the hypothetical derivative cannot simply impute a charge for exchanging different currencies even though actual derivatives under which different currencies are exchanged might include such a charge (for example, cross-currency interest rate swaps).
*International Financial Reporting Standards (IFRS) adjustment process involves reviewing the company's financial statements and identifying any differences between the company's current accounting practices and the requirements of the IFRS. If there are any such differences, neural network makes adjustments to financial statements to bring them into compliance with the IFRS.
## Conclusions
MOLTEN VENTURES VCT PLC is assigned short-term Ba1 & long-term Ba1 estimated rating. MOLTEN VENTURES VCT PLC prediction model is evaluated with Modular Neural Network (News Feed Sentiment Analysis) and Stepwise Regression1,2,3,4 and it is concluded that the LON:MVCT stock is predictable in the short/long term. According to price forecasts for (n+4 weeks) period, the dominant strategy among neural network is: Sell
### LON:MVCT MOLTEN VENTURES VCT PLC Financial Analysis*
Rating Short-Term Long-Term Senior
Outlook*Ba1Ba1
Income StatementCaa2B3
Balance SheetCC
Leverage RatiosCaa2B3
Cash FlowBa1C
Rates of Return and ProfitabilityBaa2B3
*Financial analysis is the process of evaluating a company's financial performance and position by neural network. It involves reviewing the company's financial statements, including the balance sheet, income statement, and cash flow statement, as well as other financial reports and documents.
How does neural network examine financial reports and understand financial state of the company?
### Prediction Confidence Score
Trust metric by Neural Network: 89 out of 100 with 591 signals.
## References
1. Tibshirani R. 1996. Regression shrinkage and selection via the lasso. J. R. Stat. Soc. B 58:267–88
2. Matzkin RL. 2007. Nonparametric identification. In Handbook of Econometrics, Vol. 6B, ed. J Heckman, E Learner, pp. 5307–68. Amsterdam: Elsevier
3. R. Sutton and A. Barto. Introduction to reinforcement learning. MIT Press, 1998
4. Angrist JD, Pischke JS. 2008. Mostly Harmless Econometrics: An Empiricist's Companion. Princeton, NJ: Princeton Univ. Press
5. Künzel S, Sekhon J, Bickel P, Yu B. 2017. Meta-learners for estimating heterogeneous treatment effects using machine learning. arXiv:1706.03461 [math.ST]
6. Dimakopoulou M, Athey S, Imbens G. 2017. Estimation considerations in contextual bandits. arXiv:1711.07077 [stat.ML]
7. Angrist JD, Pischke JS. 2008. Mostly Harmless Econometrics: An Empiricist's Companion. Princeton, NJ: Princeton Univ. Press
Frequently Asked QuestionsQ: What is the prediction methodology for LON:MVCT stock?
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A: The prediction period for LON:MVCT is (n+4 weeks) | 2,344 | 9,776 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 2, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2023-14 | longest | en | 0.801214 |
https://datascienceparichay.com/article/python-convert-octal-to-integer/ | 1,721,130,750,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514745.49/warc/CC-MAIN-20240716111515-20240716141515-00428.warc.gz | 164,514,101 | 36,482 | Python – Convert Octal String to Integer
In this tutorial, we will look at how to convert an octal number (as a string) to an integer in Python with the help of examples.
How to convert octal string to integer in Python?
You can use the Python built-in `int()` function to convert an octal string to an integer in Python. Pass the oct string as an argument. The following is the syntax –
```# oct string to int
int(oct_str, 8)```
The `int()` function also takes an optional argument, `base` to identify the base of the value passed. For example, use `16` as the base to convert hex string, use `8` as the base for an octal string, etc.
Although, the `int()` function is capable of inferring the base from the string passed, it’s a good practice to explicitly specify the base.
Examples
Let’s look at some examples of using the above syntax to convert an oct string to an int.
Oct string with prefix to int
Let’s convert an octal number (as a string) with the prefix `'0o'` to an integer using the `int()` function.
```# oct string representing 1024
oct_str = '0o2000'
# oct to int
num = int(oct_str, 8)
print(num)```
Output:
`1024`
Here we have an oct string representing the number `1024` with the prefix `'0o'`. We use the `int()` function to convert the oct string to an integer. You can see that we get `1024` as the output.
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Oct string without prefix to int
Let’s now try to convert an oct string without a prefix to int using the `int()` function. You don’t need to change any arguments as the `int()` function is capable to handle the oct string irrespective of the presence of the prefix.
```# oct string representing 1024 without '0o' prefix
oct_str = '2000'
# oct to int
num = int(oct_str, 8)
print(num)```
Output:
`1024`
Here we pass the oct string without the `'0o'` prefix to the `int()` function with `base` as 8. We get the same output as above.
For more on the Python `int()` function, refer to its documentation.
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Author
• Piyush is a data professional passionate about using data to understand things better and make informed decisions. He has experience working as a Data Scientist in the consulting domain and holds an engineering degree from IIT Roorkee. His hobbies include watching cricket, reading, and working on side projects.
Scroll to Top | 672 | 2,743 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2024-30 | latest | en | 0.758114 |
https://au.mathworks.com/matlabcentral/cody/problems/44332-resizing-matrices/solutions/1277564 | 1,603,197,563,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107872686.18/warc/CC-MAIN-20201020105000-20201020135000-00649.warc.gz | 222,184,928 | 16,959 | Cody
# Problem 44332. Resizing Matrices
Solution 1277564
Submitted on 27 Sep 2017 by Zikobrelli
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
x = [ 1 10 5; 3 2 9; 6 10 2; 10 10 5]; y_correct = [ 1 6 10 10 5 2; 3 10 2 10 9 5] assert(isequal(your_fcn_name(x),y_correct))
y_correct = 1 6 10 10 5 2 3 10 2 10 9 5 y = 1 6 10 10 5 2 3 10 2 10 9 5
### Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting! | 226 | 594 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2020-45 | latest | en | 0.700709 |
https://www.askiitians.com/forums/Mechanics/10/23866/newtons-laws-of-motion.htm | 1,685,843,455,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224649348.41/warc/CC-MAIN-20230603233121-20230604023121-00344.warc.gz | 720,146,789 | 33,953 | # Three blocks of masses 1kg,4kg,2kg are placed on a frictionless horizontal plane surface.two forces 120N and 50N are acting on the surface.120 N acts on the 1kg mass particle and 50 N acts in the opposite direction on 2kg masses.Find accelaration of the system.
509 Points
12 years ago
net force on the system=120-50=70N
(total mass)(accleration)=70
7a=70
a=10m/s2 ans
Chiranjeevi .P
29 Points
12 years ago
a=F/m
a=(120-50)g/(1+4+2)
a=70g/7
a=10g
SAGAR SINGH - IIT DELHI
879 Points
12 years ago
Dear student,
In such type of questions first of all find the net force on the system.
Here in the question,
Net force on the system=120-50=70N
(total mass)(accleration)=70
7a=70
a=10m/s2
Hence acceleration of the system = 10 m/s2
All the best.
Win exciting gifts by answering the questions on Discussion Forum. So help discuss any query on askiitians forum and become an Elite Expert League askiitian. | 284 | 920 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2023-23 | latest | en | 0.74394 |
https://mathhelpforum.com/tags/style/ | 1,581,990,100,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875143455.25/warc/CC-MAIN-20200217235417-20200218025417-00496.warc.gz | 474,433,565 | 15,234 | # style
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6. ### AP Style Question, Help?
A particle moves along a line so that at any time t its position is given by x(t)=2pi t + cos 2pi t. a. Find the velocity at time t. b. Find the acceleration at time t. c. What are the values at t, 0<t<3 for which the particle is at rest? d. What is the maximum velocity? I would really...
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Ok, so you might know the game of Risk. If you don't here is a summary of the relevant rules. The attacking player may, if attacking with more than 3 armies, throw 3 dice to descide the results of a battle. The defender may, if the defending territory holds 2 armies, defend with 2 dice. The...
10. ### Prove By Induction (Sigma Style)
\sum_{i = 1}^n i^3 = \frac{n^2(n+ 1)^2}{4}? | 831 | 3,253 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2020-10 | longest | en | 0.897217 |
https://forum.allaboutcircuits.com/threads/eplain-how-that-works.105442/ | 1,506,456,547,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818696681.94/warc/CC-MAIN-20170926193955-20170926213955-00359.warc.gz | 644,002,474 | 16,947 | # Eplain how that works
Discussion in 'Homework Help' started by TylerDurden13, Jan 6, 2015.
1. ### TylerDurden13 Thread Starter New Member
Dec 15, 2014
3
0
Hi i have a homework to school and i need to help , beacause i dont know how that circuits works
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2. ### ErnieM AAC Fanatic!
Apr 24, 2011
7,906
1,789
Work it from top left to right.
First, the input not made by a switch (can't read the letter very top left) will control IC30A and IC31A to light a red or green LED. That works separate from all else.
The top left has 8 inputs, each input is inverted and all 1 signals flow down.
8 of these 16 signals are routed to 16 NAND gates starting with IC1 and going down. Each gate will capture one input state which causes the gate output to go low.
The 16 NAND outputs are directed to more NAND gates at IC20A and below. The NAND output goes high when any of the inputs go low, or any of the state is met. These drives the segments of the display turning a segment ON when they go low.
The letters a-g on the seven segment are standard designation (Google them).
You now should be able make a table where you write down the states selected by IC1-IC19, then which segments are lit by what states.
Hopefully a pattern will emerge you can see. If not, come on back and post your work.
That is what is does. WHY it does what it does is an open question.
3. ### WBahn Moderator
Mar 31, 2012
20,081
5,670
Could you reduce the size of your image? I haven't been able to open it successfully, but I saw a snippet of it momentarily and it looks like it is huge. Open it in something like Paint and just resize it to be, say, 400 pixels wide. That will probably reduce it by a factor of 50 or more.
4. ### djsfantasi AAC Fanatic!
Apr 11, 2010
3,486
1,245
I second that. Trying to open it hung my system.... I was afraid it was infected.
5. ### WBahn Moderator
Mar 31, 2012
20,081
5,670
Okay, I downloaded it and was able to open it. 12000 pixels wide! But I can see that it has so much in it that 400 pixels won't do it. I scaled it down to 1600 pixels which is probably about as far as is reasonable to go and attached the result.
This almost looks like a Boolean monotone function -- or a set of them, actually -- but I doubt that is the intent.
File size:
223 KB
Views:
39
6. ### WBahn Moderator
Mar 31, 2012
20,081
5,670
A slightly different explanation from ErnieM's (but should be equivalent) is to start from the right hand side. Note that in order for a particular segment to light up, the control signal to it must go LO. So that means that for most of the NAND gates in the right-hand column they are logically AND gates followed by an inverter in order to match the logic to the needs of the display.
Let's focus on the top NAND gate in the right-hand column, IC20A, which is driving segment A. In order for segment 'a' to light, all four of the inputs to IC20A have to be HI.
So now let's consider what each one of those inputs represents. If we look at the top NAND gate in the left-hand column, we see that it has eight inputs and that each input is tied to either the inverted or the non-inverted version of one of the eight input signals. As a NAND gate, it is HI except when all of it's inputs are HI. Thus there is just one choice of input signals that will result in its output going LO. This is called a minterm. Each of the 16 NAND gates in the left-hand column thus picks off one combination of inputs that will prevent a segment from lighting up (provided that signal is applied to the input of the corresponding NAND gate in the right-hand column). Thus there are four combinations of inputs that will prevent segment 'a' from lighting while all other combinations will allow it to light.
Note that segments 'b' and 'c' are slightly different in that the inverters following the NAND gates make it so that instead of the selected minterms causing the segment not to light, any of those minterms will cause that segment TO light up.
This almost looks like some kind of game where people have to try to find a combination of switch settings that either cause all of the segments to light up or all of the segments to be dark, with perhaps the input at the far left being a selection between two different games? Just a wild guess.
7. ### TylerDurden13 Thread Starter New Member
Dec 15, 2014
3
0
Thank you very much ...! | 1,095 | 4,390 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2017-39 | latest | en | 0.944229 |
http://www.annaunivedu.in/2012/10/yma001-applied-mathematics-i-syllabus.html | 1,539,781,018,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583511173.7/warc/CC-MAIN-20181017111301-20181017132801-00148.warc.gz | 427,479,263 | 42,741 | ## Saturday, October 13, 2012
### YMA001 APPLIED MATHEMATICS - I SYLLABUS | ANNA UNIVERSITY BCA 1ST SEM SYLLABUS REGULATION 2010 2011 2012-2013
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YMA001 APPLIED MATHEMATICS - I SYLLABUS | ANNA UNIVERSITY BCA 1ST SEM SYLLABUS REGULATION 2010 2011 2012-2013 BELOW IS THE ANNA UNIVERSITY FIRST SEMESTER B.C.A. (BACHELOR OF COMPUTER APPLICATIONS) DEPARTMENT SYLLABUS, TEXTBOOKS, REFERENCE BOOKS,EXAM PORTIONS,QUESTION BANK,PREVIOUS YEAR QUESTION PAPERS,MODEL QUESTION PAPERS, CLASS NOTES, IMPORTANT 2 MARKS, 8 MARKS, 16 MARKS TOPICS. IT IS APPLICABLE FOR ALL STUDENTS ADMITTED IN THE YEAR 2011 2012-2013 (ANNA UNIVERSITY CHENNAI,TRICHY,MADURAI,TIRUNELVELI,COIMBATORE), 2010 REGULATION OF ANNA UNIVERSITY CHENNAI AND STUDENTS ADMITTED IN ANNA UNIVERSITY CHENNAI DURING 2010
YMA001 APPLIED MATHEMATICS - I L T P C
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Expansion of Sin n Cos n in terms of Sin and Cos - Expansion of Sinn ; Cosn in
terms of sines and cosines of multiples of , hyperbolic functions. Inverse hyperbolic functions.
UNIT II MATRICES 12
Rank of matrix - consistency and inconsistency of a system of linear equations – Eigen values
and Eigen vectors – Properties - Cayley Hamilton theorem – Reduction of Quadratic form to
Canonical form by Orthogonal reduction.
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REFERENCES :
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2 Kandasamy.P., Thilagavathy.K. and Gunavathy.K. – “Engineering Mathematics, Volume –
I”, S.Chand & Co., New Delhi, 2001. | 663 | 2,196 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2018-43 | longest | en | 0.630385 |
https://vdict.com/voltage,6,0,0.html | 1,618,239,880,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038067870.12/warc/CC-MAIN-20210412144351-20210412174351-00393.warc.gz | 559,190,175 | 14,392 | Search in: Word
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Computing (FOLDOC) dictionary (also found in English - Vietnamese, English - English (Wordnet), French - Vietnamese)
voltage
electronics (Or "potential difference", "electro-motive
force" (EMF)) A quantity measured as a signed difference
between two points in an electrical circuit which, when
divided by the resistance in Ohms between those points,
gives the current flowing between those points in Amperes,
according to Ohm's Law. Voltage is expressed as a signed
number of Volts (V). The voltage gradient in Volts per metre
is proportional to the force on a charge.
Voltages are often given relative to "earth" or "ground" which
is taken to be at zero Volts. A circuit's earth may or may
not be electrically connected to the actual earth.
The voltage between two points is also given by the charge
present between those points in Coulombs divided by the
capacitance in Farads. The capacitance in turn depends on
the dielectric constant of the insulators present.
Yet another law gives the voltage across a piece of circuit as
its inductance in Henries multiplied by the rate of change
of current flow through it in Amperes per second. | 280 | 1,201 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2021-17 | latest | en | 0.950729 |
http://www.qacollections.com/How-to-Identify-Shapes-Like-Rhombuses | 1,529,289,046,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267859923.59/warc/CC-MAIN-20180618012148-20180618032148-00180.warc.gz | 481,231,477 | 5,702 | # How to Identify Shapes Like Rhombuses?
A rhombus resembles a slanted square creating the appearance of a diamond. A rhombus is a polygon, meaning it is a closed plane figure with straight sides and the same amount of angles as sides. A ... Read More »
http://www.ehow.com/how_8374575_identify-shapes-like-rhombuses.html
Top Q&A For: How to Identify Shapes Like Rhombuses
## Geometry: How to Identify Shapes?
Geometry is a branch of mathematics that deals, in part, with two-dimensional and three-dimensional figures. These figures can be categorized by their attributes such as sides, angles, faces, edges... Read More »
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## Different Eye Shapes?
Eyes, like bodies, come in a variety of shapes and sizes. To get the most out of your eye shape, learn to use the proper types of eye makeup and the right colors. Figure out contouring and shading ... Read More »
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## What do different diamond shapes mean?
Although round diamonds account for over 75% of diamond sales, according to WeddingChannel.com, there are several other diamond shapes to choose from. Each cut has its own specific meaning, expres... Read More »
Related Questions | 355 | 1,572 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-26 | latest | en | 0.890778 |
http://mathforum.org/t2t/discuss/message.taco?thread=4215&n=2 | 1,524,582,302,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125946721.87/warc/CC-MAIN-20180424135408-20180424155408-00556.warc.gz | 197,157,936 | 3,124 | Teacher2Teacher Q&A #4215
Teachers' Lounge Discussion: How to teach perfect numbers to 6 graders?
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From: r!chard tchen
To: Teacher2Teacher Public Discussion
Date: 2005082423:15:30
Subject: Re: divisibility -6th Gr math
On 2005082419:22:58, Rita Pena wrote: >Let me ask you guys this, why bother providing this web site, and >even have a section to "ask a question" and when you hit enter it >brings you to 20 million links except the one that we need & no where >did we find the answer to a "simple" question, I mean really why >bother???? > >Question we were trying to find was: what is the divisibility Rule?? Hi Rita, If you haven't already, check out Ask Dr. Math, the Forum's ask-an-expert service for math students and their teachers. This sister service of Teacher2Teacher offers a searchable archive of questions and answers, plus the Frequently Asked Question Divisibility rules: How can you tell whether a number is divisible by another number (leaving no remainder) without actually doing the division? Why do 'divisibility rules' work? http://mathforum.org/dr.math/faq/faq.divisibility.html Yours, - Richard Tchen The Math Forum 3210 Cherry Street Philadelphia, PA 19104-2713 tel1: (215)895-1080 tel2: (800)756-7823 facs: (215)895-2964 http://mathforum.org/
Teacher2Teacher - T2T ® | 388 | 1,414 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-17 | longest | en | 0.82243 |
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# In a room filled with 7 people, 4 people have exactly 1
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In a room filled with 7 people, 4 people have exactly 1 friend in the room and 3 people have exactly 2 friends in the room. If two individuals are selected from the room at random, what is the probability that those two individuals are NOT friends?
A. 5/21
B. 3/7
C. 4/7
D. 5/7
E. 16/21
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23 Jun 2006, 20:47
Pretend that we have A B C D E F G and AB,CD,EG,EF,GF are groups of friends.
So the needed prob must be 5/7C2 = 5/21
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24 Jun 2006, 11:56
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Who is John Galt?
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Manager
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24 Jun 2006, 18:50
Yeap, you got it Mr. Please show me the way, bcs I'm lost in this one
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Who is John Galt?
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24 Jun 2006, 19:48
[quote="X & Y"]mmmm noup. Please try again. [/quote]
It is my mistake. The explaination is right,huh?
But i calculated the prob that the chosen 2 people are friends.
It is obvious that the needed prob must be 1-5/21=16/21
E it is
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24 Jun 2006, 21:28
(E)
Total ways 2 individiuals are selected - 7C2 = 21
Total pair of friends -
4 friends have exactly 1 friend - 2 pairs
3 friends have exactly 2 - 3 pairs
= 5
prob. that 2 individuals are friends - 5/21
prob. not friends - 1-5/21 = 16/21 = (E)
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24 Jun 2006, 22:55
Quote:
(E)
Total ways 2 individiuals are selected - 7C2 = 21
Total pair of friends -
4 friends have exactly 1 friend - 2 pairs
3 friends have exactly 2 - 3 pairs
= 5
prob. that 2 individuals are friends - 5/21
prob. not friends - 1-5/21 = 16/21 = (E)
Thanks for explanation.
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25 Jun 2006, 11:21
mine approach is pretty much same as sgrover's.
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25 Jun 2006, 11:21
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# MA8391 Question Paper Probability and Statistics Regulation 2017 Anna University
## MA8391 Question Paper Probability and Statistics Regulation 2017
5. What are the parameters and statistics in sampling BTL -1 Remembering
6. State level of significance. BTL -1 Remembering
MA8391 Question Paper Probability and Statistics
A random sample of 25 cups from a certain coffee dispensing
machine yields a mean x = 6.9 occurs per cup. Use 0.05 level of
significance to test, on the average, the machine dispense μ = 7.0
ounces against the null hypothesis that, on the average, the machine
dispenses
μ < 7.0 ounces. Assume that the distribution of ounces per cup is
normal, and that the variance is the known quantity σ2=0.01 ounces
MA8391 Question Paper Probability and Statistics
Twenty people were attacked by a disease and only 18 were survived.
The hypothesis is set in such a way that the survival rate is 85% if
attacked by this disease. Will you reject the hypothesis that it is more
at 5% level. MA8391 Question Paper Probability and Statistics
In a large city A, 20 percent of a random sample of 900 school boys
had a slight physicaldefect. In another large city B, 18.5 percent of a
random sample of 1600 school boys had somedefect. Is the
difference between the proportions significant?
A standard sample of 200 tins of coconut oil gave an average weight
of 4.95 kg with a standard deviation of 0.21 kg. Do we accept that
the net weight is 5 kg per tin at 5% level of significance?
Write down the formula of test statistic‘t’ to test the significance of
difference between the means.
Subject Name Probability and Statistics Subject Code MA8391 Regulation 2017 File PDF | 409 | 1,693 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2024-38 | latest | en | 0.922681 |
https://boards.straightdope.com/t/e-mc2/146671 | 1,720,799,272,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514404.71/warc/CC-MAIN-20240712125648-20240712155648-00144.warc.gz | 107,755,034 | 8,940 | # E=mc2 ??
Energy can not be created nor destroyed…Einstein said that energy can become matter and matter can become energy…but the sum total of energy + matter is always equal. Right?
Could some one explain what else this equation means but like in simple terms?
It means that a small amount of matter can be converted into a large amount of energy (Nuclear fission/fusion). Or that a large amount of energy can be used to create a small amount of matter.
See this Cite
What unit is “c” in? Miles?
Doesn’t matter (no pun intended:cool: ). If c is in miles per hour and m is in kilograms, then E will be in kilogram-miles per hour. The units used on one side of the equation will determine the units to use on the other side.
Doesn’t matter (no pun intended:cool: ). If c is in miles per hour and m is in kilograms, then E will be in kilogram-miles squared per hour squared. The units used on one side of the equation will determine the units to use on the other side.
Darn, wasn’t quick enuff to stop my first, stupider, post
Er, I’ve had this explained to me before, but what is an “hour squared”?
Also… what is a kilogram-mile?
Thats what I dont understand… If it takes a small amount of matter to make alot of energy then how can they always be equal in the universe? Or did I get that part wrong?
Im so totally confused…
I’ll check that site out…thax.
Umm… well, an hour sqaured is an hour times an hour. And a kilogram mile is a kilogram times a mile. shrugs Nothing too it, really. But also not very intuitively useful units.
I think you’re confused.
E=MC[sup]2[/sup] just means matter and energy can be converted back and forth into each other.
It seems like you are assuming all the matter in the Universe is equal to all the energy in the Universe. That doesn’t make sense because matter and energy are equivalent…matter=energy=matter=energy…just different forms of the same thing (like liquid water versus ice water…both still water).
A sample of many.
Whack is correct. When you are studying Einstein’s equations, it’s helpful to think of matter and energy as two aspects of the same thing, commonly called mass-energy. The amount of mass-energy in the Universe is a constant, no matter what percentage of it exhibits which set of properties at the moment.
E=mc[sup]2[/sup] is like all physical equations: dimensionless. The dimensions used on one side determine the dimensions derived for the other side.
Distance per time squared means distance per time per time. It’s a rate of acceleration. For example, g is 9.8 meters per second squared. That means every second, something falling on Earth (and exhibiting the acceleration defined by g) gets 9.8 meters per second faster.
0 s == 0 m/s
1 s == 9.8 m/s
2 s == 19.6 m/s
3 s == 29.4 m/s
etc.
One more.
Oh, and one more thing. Mass and energy are not the same thing. In special relativity there’s a thing called the energy momentum four vector which in some respects is as basic as you can get. In any case, energy is the time component of the four vector, momentum is the space component and mass is the magnitude of the four vector. This is all summed up in the equation:
E[sup]2[/sup] = m[sup]2[/sup]c[sup]4[/sup] + p[sup]2[/sup]c[sup]2[/sup]
E = energy
m = mass
p = momentum
c does not equal 1
Despite your further explanation you lost me. Just when you think you get a handle on something leave it to a physicist to screw it up for you .
I’m really hoping we’re getting caught on semantics here.
Stop me if this analogy is inappropriate: Liquid water and Frozen water are not the same thing but they are different forms of the same thing…water.
VERY loosely speaking I thought energy and matter worked the same way. From your own quote of yourself if you explode a nuclear device in a vault and convert mass to energy (or vice versa) from outside the vault all measurements will be the same. Less matter, more energy…doesn’t matter. Just as if I melted my ice to a liquid then turned the liquid into a gas…I still have the same amount of water and have merely changed its aspect.
What have I missed that doesn’t require 8 years post grad work to understand?
From Ring:
Whoa! How did you write the exponent? I was answering a math question and had to use carrots.
Now I see from my post that it doesn’t work for me even when I quote!
To get really strange, the new quantum theories hypothesize that at the sub-atomic level energy and matter are made up of the same stuff. For instance, in string theory, one of the more popular quantum explanations, the very smallest unit of anything is the string (hence the name) which can be visualized as a rubber-band with a certain number of whole waves in it. The number of waves, or the frequency of the band determines whether you get photon, graviton, proton boson, muon etc . . .
This is of course a VERY simplified explanation. For more in depth info in terms even I could understand I would recommend:
[ul]
[li]The Dancing Wu Li Masters[/li][li]In Search of Schroedingers Cat[/li][/ul]
They’re called carets. ^ is a caret. Note spelling.
Try the [sup] tag. Type x[sup]2[/sup] to get the desired effect.
The [sub] tag does the right thing, too.
I think a lot of people get hung up on this because they think mass and energy are things. But they’re not things – they’re properties of a system. This is not to say they’re not intimately related, they are. If you study the equation:
m[sup]2[/sup] = E[sup]2[/sup] - p[sup]2[/sup]
You can see that if the momentum is zero the mass of the system equals the energy of the system, or to put it another way – mass is equal to the energy of the system that cannot be transformed away.
If you wanted to, you could abandon the concept of mass altogether and just talk about the magnitude of the energy momentum four-vector, but the concept of mass is so ingrained that I doubt this will ever happen. | 1,413 | 5,893 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2024-30 | latest | en | 0.904173 |
https://www.realestateinvestmentsoftwareblog.com/cap-rate-formula/?goback=.gde_35901_member_197817910 | 1,726,094,472,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651405.61/warc/CC-MAIN-20240911215612-20240912005612-00253.warc.gz | 897,196,948 | 8,788 | # How the Cap Rate Formula Serves Three Real Estate Investing Purposes
The cap rate formula is undoubtedly the first formula most real estate agents and investors learn when they initially become engaged with real estate investing.
Real estate investors, agents, appraisers, property tax assessors and others that evaluate real estate investment property typically all use cap rate in one form or the other. So it makes sense that they would learn how to calculate it.
Still, some may not be aware that the cap rate formula can be used to serve three useful purposes when dealing with real estate investment property. So in this article, I thought it would be helpful to show you.
#### Cap Rate Overview
Cap rate is a rate of return used in real estate investing to determine the present value (price) of a real estate investment based upon its future benefits (net operating income).
But that’s the technical definition. In practice, it expresses the relationship between a property’s value and its net operating income (NOI) for the current or coming year.
Although alone it does not provide a true picture of a property’s profitability, it does provide a good first-glance look at a property’s ability to pay its own way. As such, it is one of the most popular returns used for real estate investing.
#### Cap Rate Formulas
As stated, you can use the cap rate formula to achieve three useful purposes associated with real estate investment property.
##### 1. Calculate a property’s cap rate
When you want to know what capitalization rate a rental property recently sold, for instance, you would use that property’s net operating income and sale price to make the computation.
Capitalization Rate = Net Operating Income / Sale Price
Example: You just saw a newly listed apartment complex for \$900,000 that could be what your real estate investor is looking for if its caps above 7.2%. You learn that its NOI is \$61,400 and thereafter discover it caps at 6.82%.
Source: iCalculator by ProAPOD
##### 2. Calculate a property’s reasonable value
By transposing the formula you can compute a property’s estimated value. In preparation for a listing presentation, for instance, you can use the net operating income estimated for that property and the prevailing capitalization rate for similar sold properties to suggest what the property should be worth.
Value = Net Operating Income / Capitalization Rate
Example: You have been asked to list an office complex at what you feel to be a reasonable market value. You know that its NOI is \$75,240 and that similar-type buildings have recently sold for an average 5.8% cap rate. You suggest a listing price of \$1,297,241.
Source: iCalculator by ProAPOD
##### 3. Calculate a property’s net operating income
By transposing the formula again you can compute a property’s net operating income. In cases where you are given a specified price and capitalization rate you can determine what the net operating income should be.
Net Operating Income = Capitalization Rate x Value
Example: You just called for a marketing package on a strip center listed at \$2,500,000 showing a cap rate of 7.75%. You are expecting to see the property generating an NOI of \$193,750.
Source: iCalculator by ProAPOD | 674 | 3,259 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-38 | latest | en | 0.934027 |
https://www.codingbroz.com/branch-reset-groups-hackerrank-solution/ | 1,721,487,857,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763515300.51/warc/CC-MAIN-20240720144323-20240720174323-00439.warc.gz | 617,177,866 | 57,055 | # Branch Reset Groups – HackerRank Solution
In this post, we will solve Branch Reset Groups HackerRank Solution. This problem (Branch Reset Groups) is a part of HackerRank Regex series.
NOTE – `Branch reset group is supported by Perl, PHP, Delphi and R.`
(?!regex)
A branch reset group consists of alternations and capturing groups. (?|(regex1)|(regex2))
Alternatives in branch reset group share same capturing group.
You have a test string S.
Your task is to write a regex which will match S, with following condition(s):
• SÂ consists of 8 digits.
• S must have “—“, “-“, “.” or “:” separator such that string S gets divided in 4 parts, with each part having exactly two digits.
• SÂ string must have exactly one kind of separator.
• Separators must have integers on both sides.
Valid S
``````12-34-56-78
12:34:56:78
12---34---56---78
12.34.56.78``````
Invalid S
``````1-234-56-78
12-45.78:10``````
Note
This is a regex only challenge. You are not required to write any code.
You only have to fill the regex pattern in the blank (`_________`).
## Branch Reset Groups – HackerRank Solution
### PHP
```<?php
\$Regex_Pattern = '/^\d{2}(-(?:--)?|\.|:)\d{2}\1\d{2}\1\d{2}\$/'; //Do not delete '/'. Replace __________ with your regex.
\$handle = fopen ("php://stdin","r");
\$Test_String = fgets(\$handle);
if(preg_match(\$Regex_Pattern, \$Test_String, \$output_array)){
print ("true");
} else {
print ("false");
}
fclose(\$handle);
?>
```
Note: This problem (Branch Reset Groups) is generated by HackerRank but the solution is provided by CodingBroz. This tutorial is only for Educational and Learning purpose. | 459 | 1,634 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2024-30 | latest | en | 0.760669 |
https://www.swfilters.com/mathematical-operators-in-java/ | 1,571,264,588,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986670928.29/warc/CC-MAIN-20191016213112-20191017000612-00028.warc.gz | 1,111,513,227 | 12,200 | # Mathematical Operators In Java
At the same time, Sun Microsystems was pushing for Java to become integrated into web browsers. For 32-bit integer types, optimal semantics are available using bitwise operators (the only case.
Organism With Two Different Alleles Relating to two alleles of a gene pair in a heterozygote that are both fully expressed. When alleles for both white and red are present in a carnation, for example, the result is a pink carnation since both alleles are codominant. The sons of queen bees, ants, and wasps (insects collectively known as Hymenoptera) come in two kinds. The difference is all to do with their genetics. Humans are “diploid” organisms. If this gene. 6.
Operators and Precedence of operators are as important in computer languages as. Java truncates the result of integer division so that the results are always.
Arithmetic Expressions allow us to perform mathematical operations within Java. +. Such expressions can be used for basic math and even more complex.
2 Biology Model Question Paper Syllabus and Model Question Papers. Applied Sciences. Biology (New Model Paper) · Biology (Model Paper for HSSC-II/A/2018 and onward). Dec 29, 2018. Answer Key (English Medium). Answer Key (Tamil Medium). 2. 12th English Model Question Paper (2018-19) – English Medium | Tamil. Tartu Semiotics Library 3 Tartu Semiootika Raamatukogu 3 Тартуская библиотека семиотики 3 Tartu semiotik biblioteket 3 Hoffmeyerit lugedes, bioloogiat ümber sõnastades Читая Хоффмейера, переосмысливая биологию Læs Hoffmeyer — nytænk biologien University of Tartu
Java Arithmetic Operators. Java arithmatic operators are used to perform addition, subtraction, multiplication, and division. They act as basic mathematical operations.
QCL also supports many built in functions that one can find in other common programming languages, such as Java, C++, and Python. These functions include mathematical functions. QCL also allows for.
A perfect example of piping is killing specific processes using Get-Process java | Stop-Process. Numerical values can be assigned or modified using standard math operators like +, -, *, and / as in.
Answer: Yes. The meaning of operators and parentheses is about the same in electronic calculators and in Java. But Java does integer and floating point math,
Introduction. Each language-specific implementation of ReactiveX implements a set of operators. Although there is much overlap between implementations, there are also some operators that are only implemented in certain implementations.
Nov 23, 2018. A quick guide to mathematical and aggregate operators from the RxJava library.
JavaScript provides all the features of a fully comprehensive object-orientated, event driven language, such as Java or C++. than manually setting a variable ourselves. Most operators are.
May 11, 2018. This post shows examples of the Java mixed data type division rules, All integer values ( byte , short and int ) in an arithmetic operations (+,
Mathematical morphology (MM) is a theory and technique for the analysis and processing of geometrical structures, based on set theory, lattice theory, topology, and random functions.MM is most commonly applied to digital images, but it can be employed as well on graphs, surface meshes, solids, and many other spatial structures. Topological and geometrical continuous-space concepts such as.
Jul 20, 2017. Introduction. Mathematical operations are among the most fundamental and universal features of any programming language. In JavaScript.
Mar 4, 2019. Java Math Class provides useful methods for performing the math's operations like exponential, logarithm, roots and trigonometric equations.
The Microsoft Simple Encrypted Arithmetic Library was made open source in December. This offers a particular advantage where data is stored in cloud environments, and the cloud operators never have.
It is fortunate, then, that.NET 2.0 ships with Reflection.Emit for writing IL (the equivalent of Java’s “bytecode” in.NET. like the following in an expression tree (for use as a compiler): person.
The reason Java got the "wrong" answer was because of Operator Precedence. Java treats some mathematical symbols as more important than others. It sees.
It uses an event-driven, nonblocking I/O model that makes it lightweight and efficient, compared to, say, Java Web Pages or ASP.Net. Sass adds inheritance (@extend), variables, nesting, mix-ins,
Operators. Every variable in JavaScript is casted automatically so any operator between two variables will always give some kind of result. The addition operator
Java Arithmetic Operators. Java arithmatic operators are used to perform addition, subtraction, multiplication, and division. They act as basic mathematical operations.
The Bitwise Operators. Java defines several bitwise operators, which can be applied to the integer types, long, int, short, char, and byte. Bitwise operator works on bits and performs bit-by-bit operation.
Numeric Operators. In the last section, we created and initialized some variables. Now let’s look at Java’s facilities for numeric operations. They will look very familiar to.
The operators +, -, *, /, and % give us a way to add, subtract, multiply, divide, and "mod" values together in java, but having to implement some of the more.
Jun 26, 2015. Continue learning the fundamentals of Java programming, with this Java 101. The additive operators increase or decrease a numeric value.
Operator overloading is necessary to allow user-defined numeric types, such as complex , to be used reasonably. Without.
They also used the Java computer language to program the remote controls. The camp was designed to boost students’ interest and confidence in science, technology, engineering and math and show them.
Darwin Lake Near Matlock Ecology Top Down Control Peer Reviewed Articles On Diabetes The research was published in the journal Environment International, a peer-reviewed science journal on environment. Lu Xiangfeng, one of the researchers, said the study would benefit policy making. Below are the latest The Lancet Diabetes & Endocrinology articles published online ahead of print; All research papers have been peer-reviewed and published via our fast-track process within 4. If ever there were an herb that puts existential
Well organized and easy to understand Web building tutorials with lots of examples of how to use HTML, CSS, JavaScript, SQL, PHP, Python, Bootstrap, Java and XML.
Nov 21, 2017. The Java programming language has around 30 operators as summarized in the. The following table lists all arithmetic operators in Java:.
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What is math module in Python? The math module is a standard module in Python and is always available. To use mathematical functions under this module, you have to import the module using import math. It gives access to the underlying C library functions.
MP = (P*(I/12))/(1-Math.pow((1/(1+(I/12))),(T*12))); // Display the monthly. Just do "double P = 0;" 3) Use specific package names to import and not wild card operators. import java.io.FileReader;.
Join David Gassner for an in-depth discussion in this video, Using mathematical operators and the Math class, part of Java 8 Essential Training.
Well organized and easy to understand Web building tutorials with lots of examples of how to use HTML, CSS, JavaScript, SQL, PHP, Python, Bootstrap, Java and XML.
What Thermodynamic Condition Must Be Met The thermodynamics of computation seeks to identify the principled. This is the condition that a thermodynamically reversible process must meed. Thus, as. To appreciate need in its full manifestation one must examine both need and fulfilment. The need can only appear and evolve when the conditions exist for them to be met, and not vice versa. The. Second, every cycle of expansion and contraction creates entropy through natural thermodynamic processes. the consensus model to the
Numeric Operators. In the last section, we created and initialized some variables. Now let’s look at Java’s facilities for numeric operations. They will look very familiar to.
Sure you can program in Scala just like you would in Java and get all the advantages of the cleaner. However it would be nice to be able to use standard mathematical operators like A*B and AT when.
In Java, you need to be aware of the type of the result of a binary (two-argument) arithmetic operator. If either operand is of type double , the other is converted to.
Shorthand operations do not add any feature to the Java programming language. (So it's not a big deal). Division assignment, x /= 4;, x = x / 4;. %=, Remainder.
Java Exact Arithmetic Operations Support in Math Class. By Lokesh Gupta | Filed Under: Java 8. Java 8 has brought many awesome features for java.
MXNet abstracts much of the complexity involved in implementing neural networks, is highly performant and scalable, and offers APIs across popular programming languages such as Python, C++, Java, R,
In digital computer programming, a bitwise operation operates on one or more bit patterns or binary numerals at the level of their individual bits.It is a fast and simple action, directly supported by the processor, and is used to manipulate values for comparisons and calculations. On simple low-cost processors, typically, bitwise operations are substantially faster than division, several.
Possible remedies to the problem including ensuring SIM cards use state-of-the-art cryptography and also using Java virtual machines that restrict applets’ access to certain information. GSMA said it.
Well organized and easy to understand Web building tutorials with lots of examples of how to use HTML, CSS, JavaScript, SQL, PHP, Python, Bootstrap, Java and XML.
New algorithms are presented in this paper which were used for a generic implementation in software of the IEEE 754R decimal floating-point arithmetic. packages for DFP arithmetic include IBM’s.
Java provides a rich set of operators to manipulate variables. Arithmetic operators are used in mathematical expressions in the same way that they are used in.
In this section, I’ll introduce some of Java’s operators. Using operators you can take simple. You can perform multiple operations in a single statement, if you want to. Note that arithmetic.
Operators are the foundation of any programming language. Thus the functionality of C/C++ programming language is incomplete without the use of operators. We can define operators as symbols that helps us to perform specific mathematical and logical.
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Java, C++, JavaScript, PHP, Ruby — time and again. and R — which were also developed for math and science work — by a wide margin. It even supports a Lisp-like macro function for expanding the.
Introduction. Each language-specific implementation of ReactiveX implements a set of operators. Although there is much overlap between implementations, there are also some operators that are only implemented in certain implementations.
In digital computer programming, a bitwise operation operates on one or more bit patterns or binary numerals at the level of their individual bits.It is a fast and simple action, directly supported by the processor, and is used to manipulate values for comparisons and calculations. On simple low-cost processors, typically, bitwise operations are substantially faster than division, several.
Java program to perform basic arithmetic operations of two numbers. Java program asks the user to provide integer inputs to perform mathematical operations.
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One example, Matt Halligan says, is Java: “A lot of people use the latest versions of. a mobile solutions provider that serves operators worldwide like Vodafone, Sprint, and Orange. Previously part. | 2,494 | 12,273 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2019-43 | latest | en | 0.81504 |
https://acm.njupt.edu.cn/problem/NOJ1139 | 1,653,799,489,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652663039492.94/warc/CC-MAIN-20220529041832-20220529071832-00663.warc.gz | 123,602,512 | 12,436 | Preparing NOJ
# My Summary
1000ms 65536K
## Description:
I have a dream that one day in the foyer of the ACM/ICPC World Finals Contest , I could solve all the problems and be the champion . But things always went beyond my mind , we just solved four problems .
We submitted Problem B at 29 minutes and got “ Accept “ , we passed Problem A at 85 minutes , then we got “Wrong” at 94 minutes on Problem G , but we soon got “Accept” at 98 minutes , then we submitted Problem D but got “Wrong” ….. At last , we passed Problem ABCG , and Problem D was still “Wrong” .
Now I have to write a summary about the contest , I read our submission states from PC^2 , and I need to list the problem states which illustrates the “Accept” problems and “Wrong” problems , Please help me to do so .
## Input:
The first line of input is one integer T ( T<=100 ) , giving the number of test cases in the input . Each test case starts with a line containing a positive integer N ( N<=100 ) , representing the number of the submission states. In the next N lines contains an integer t(0<=t<=300),the submission time.Then a character ( from ‘A’ to ‘J’ ) , the ID of submission problem follows . At last is a string ( “Accept” or “Wrong” ) , the states of this submission .
## Output:
For each test case , output two lines , the first line print “Accept” , follow by a colon , a space , then list the name of those accept problems in alphabetical ascending order , if there is no “Accept” problem ( what a shame ! ) , just leaves it blank . The second line is about “Wrong” states , the style of which is similar to the first line . Print a blank line after each test case .
1
10
29 B Accept
85 A Accept
94 G Wrong
98 G Accept
120 D Wrong
170 C Wrong
183 C Accept
190 D Wrong
300 A Accept
300 B Wrong
Accept: ABCG
Wrong: D
## Note:
Info
Provider NOJ
Code NOJ1139
Tags
Submitted 5
Passed 2
AC Rate 40%
Date 04/20/2019 10:03:10
Related
Nothing Yet | 513 | 1,944 | {"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} | 2.921875 | 3 | CC-MAIN-2022-21 | latest | en | 0.830052 |
https://www.hse.ru/en/edu/courses/384746411 | 1,660,494,712,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572043.2/warc/CC-MAIN-20220814143522-20220814173522-00485.warc.gz | 704,167,194 | 13,407 | • A
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Regular version of the site
Master 2020/2021
# Multidimensional Data Analysis
Category 'Best Course for Broadening Horizons and Diversity of Knowledge and Skills'
Area of studies: Applied Mathematics and Informatics
When: 1 year, 3, 4 module
Mode of studies: offline
Instructors: Anuška Ferligoj, Valentina Kuskova
Master’s programme: Applied Statistics with Network Analysis
Language: English
ECTS credits: 8
### Course Syllabus
#### Abstract
This course will take a modern, data-analytic approach to the multiple regression model. Our coverage of the material will emphasize the ways that graphical tools can augment traditional methods for describing how the conditional distribution of a dependent variable changes along with the values of one or more independent variables. The course will examine the basic nature and assumptions of the linear regression model, diagnostic tools for detecting violations of the regression as-sumptions, and strategies for dealing with situations in which the basic assumptions are violated.
#### Learning Objectives
• The goal of the course is to ensure that students understand topics and principles of applied linear models on an advanced level.
• The course gives students an important foundation to develop and conduct their own research as well as to evaluate research of others.
#### Expected Learning Outcomes
• Know new insights into the regression analysis.
• Know various modern extensions to the traditional linear model.
• Know complex methods of aggregating data and dimensionality reduction.
• Know innovative, effective methods for presenting the results from statistical investigations of empirical data.
• Be able to explore the advantages and disadvantages of various linear modeling instruments, and demonstrate how they relate to other methods of analysis.
• Be able to work with major linear modeling programs, especially R and SAS, so that they can use them and interpret their output.
• Be able to develop and/or foster critical reviewing skills of published empirical research using applied statistical methods.
• Be able to criticize constructively and determine existing issues with applied linear models in published work.
• Have an understanding of advanced methods of linear models and related multivariate extensions.
• Have the skill to meaningfully develop an appropriate model for the research question.
• Have the skill to work with statistical software, required to analyze the data.
#### Course Contents
• Examining data
The first session will introduce the main concepts of preparation for multivariate regression analysis. We will discuss graphical displays, data transformations, and other techniques of multi-dimensional exploratory analysis.
• General Linear Models I: The Basics of Least Squares Regression
The session is a review of basic linear models with an extension to multivariate regression. We will review graphical fitting, least-squares fitting, properties of least-squares estimator, statistical inference, regression models in matrix form, vector geometry and vector presentation of the regression model, the data ellipsoid and model fit.
• General Linear Models II: Effective Presentation I
The session will go in much more detail into topics of categorical predictors, fitted values, inter-actions and effect displays, standardization and relative importance of predictors.
• Regression with Categorical Dependent Variables I
This sessions will provide an understanding of limited dependent variables and problems with the OLS models; binary logit and probit models, fitted probabilities and effect displays.
• Regression with Categorical Dependent Variables II
This session covers ordered probit and logit models, Multinomial logit models, Generalized linear models, Poisson models for count data, Diagnostics for generalized linear models.
• Regression diagnostics I: Unusual observations
This session will show how to examine outliers, leverage, and influence observations; hat values and studentized residuals; and case-deletion statistics.
• Regression Diagnostics II: Nonlinearity, Nonnormality, and Heteroskedasticity
This session will focus on residual plots, visual and maximum likelihood methods for determining transformations, weighted least squares to adjust for nonconstant error variance, robust standard errors. It will also look at variance inflation, principle components analysis, collinearity and model selection, and ridge regression.
• Resampling techniques for regression
This session will introduce bootstrapping and jackknifing, as well as cross-validation.
• Nonlinear regression
This session will examine transformable nonlinearity, polynomial regression, orthogonal poly-nomials, and non-linear least squares.
• Nonparametric regression I: Local polynomial regression
This session will cover local regression (LOESS) and its fitting parameters, robust local regres-sion, and degrees of freedom in local regression, M-estimation and iteratively weighted least squares and bounded influence regression.
• Nonparametric Regression II: Smoothing Splines
This session will introduce the concepts of piecewise regression, cubic smoothing splines, thin plates smoothing splines, and their degrees of freedom.
• Additive regression models (GAM) and graphical regression
This session will look at estimation and backfitting, cross-validation for smoothing parameters, GAM for binary dependent variables, vector GAM for ordered dependent variables. It will also introduce the concepts of model checking plots, visualizing regression with more than two pre-dictors, and sequentially combining predictors.
#### Assessment Elements
• Final In-Class or Take-home exam (at the dis-cretion of the instructor)
• Homework Assignments (5 x Varied points)
• In-Class Labs (9-10 x Varied points)
• Quizzes (Best 9 of 10, Varied points)
#### Interim Assessment
• Interim assessment (4 module)
0.5 * Final In-Class or Take-home exam (at the dis-cretion of the instructor) + 0.2 * Homework Assignments (5 x Varied points) + 0.2 * In-Class Labs (9-10 x Varied points) + 0.1 * Quizzes (Best 9 of 10, Varied points)
#### Recommended Core Bibliography
• Berry, W. D., & Sanders, M. S. (2000). Understanding Multivariate Research : A Primer For Beginning Social Scientists. Boulder, Colo: Routledge. Retrieved from http://search.ebscohost.com/login.aspx?direct=true&site=eds-live&db=edsebk&AN=421170
• Brown, B. (2012). Multivariate Analysis for the Biobehavioral and Social Sciences : A Graphical Approach. Hoboken, N.J.: Wiley. Retrieved from http://search.ebscohost.com/login.aspx?direct=true&site=eds-live&db=edsebk&AN=405437
• Chatterjee, S., Hadi, A. S., & Ebooks Corporation. (2012). Regression Analysis by Example (Vol. Fifth edition). Hoboken, New Jersey: Wiley. Retrieved from http://search.ebscohost.com/login.aspx?direct=true&site=eds-live&db=edsebk&AN=959808
• Rencher, A. C., & Christensen, W. F. (2012). Methods of Multivariate Analysis (Vol. Third Edition). Hoboken, New Jersey: Wiley. Retrieved from http://search.ebscohost.com/login.aspx?direct=true&site=eds-live&db=edsebk&AN=472234 | 1,516 | 7,163 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2022-33 | latest | en | 0.846049 |
https://www.sanfoundry.com/principles-of-statics/ | 1,718,991,413,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862132.50/warc/CC-MAIN-20240621160500-20240621190500-00034.warc.gz | 842,853,683 | 18,533 | # Principles of Statics
In this tutorial, you will study about the principles and theories in Statics. Statics is a subtopic of Engineering Mechanics. This tutorial includes the definition of Statics, terms involved in Statics, different types of parameters, their definitions, types and their characteristics.
Contents:
## Statics – Introduction & History
Statics is one of the oldest branches of science prevailing across the world since ages. It is a sub-topic of Mechanics.
Some historical references to Statics are given below:
• The fundamental principles of Statics were used by the Egyptians and Babylonians to encounter the problems occurred in building the famous Pyramids and old temples.
• Earliest articles on this chapter were found written by Archimedes. He formulated the laws of equilibrium of the forces acting on a lever and also some principles in the subject Hydrostatics which are in use today.
• However, the subject saw a debacle after Archimedes and had waited a long period of time until the works of Newton, Varignon and Stevinus to rise again. The modern subject in its present form is derived from the works of these three scientists.
• Definition:
Statics is defined as the subject that deals with the conditions of equilibrium of bodies at rest under the action of several forces.
## Rigid and Non-Rigid Bodies
Statics is most concerned about the rigid and non-rigid bodies and their equilibrium. Physical bodies that one encounters in the design of engineering machines are never absolutely rigid. Although they deform slightly, it is very less in comparison to the size of the body.
• A rigid body is defined as a definite quantity of matter, the parts of which are fixed in position relative to one another. This means that the distance between any two particles remains same with or without the application of load onto the body.
• The physical bodies are not absolutely rigid but deform slightly under the action of loads. If the deformation is minute when compared to its size, then the body is assumed to be a Rigid Body.
• Statics study the analysis of forces acting on a rigid body whereas for the analysis on liquids and gases, Fluid Mechanics is taken into account.
• One such example is the strength of a lever. Under the action of two loads at the ends, the bar bends slightly over the fulcrum. It is shown in the figure.
The figure drawn below shows how rigid and non-rigid bodies act under the action of loads.
In the above figure, the bar in green is rigid since it does not deform under the action of balls. But the red bar got deformed easily under the action of loads. Hence it is a non-rigid body.
## What is a Force?
Force is defined as an action that tends to change the state of rest or motion of a body to which it is applied. It is an external agent that produces change in the movement of a certain body.
• Force acting on a rigid body produces one or both of the following effects:
• Linear Displacement
• Angular Displacement
• There are many types of forces acting on a body at several times at several levels. In some cases, if the forces act on both ends of a body, they create either tension or compression.
• Any kind of thing that applies either a push or a pull on another object is said to be applying some kind of force.
• The unit of force is newton. It is denoted by N.
## Examples of Forces Available in Nature
There are many kinds of forces available in nature. Some of the examples are shown below.
• Gravitational Attraction between a star and a planet: All the planets revolve around their respective star due to gravitational force itself.
• Weight: Weight or self-weight of an object is the gravitational attraction of the earth on it. Weight is the product of mass of the body and acceleration due to gravity. Mass of a body is always constant.
• Hydrostatic Pressure: When a body such as dam impounds on water, the water exerts a force on the body which is distributed on the area of contact. This is called Hydrostatic Pressure.
• Gas Pressure: Any gas confined in a container exerts pressure on the walls of the container. This is called Gas Pressure.
• Wind Pressure: Bodies exposed to wind are subjected to a force distributed over their exposed area, called the wind pressure.
## Characteristics of a Force
For complete definition of a force we must know its characteristics or specifications. They are magnitude, direction and point of application.
• Magnitude: The magnitude of a force denotes the size of force. It gives the numerical value of the effect of force as an external agent on a given body. e.g. 25 N, 45 N etc.
• Direction: The direction of a force is nothing but the passage of quantity of force in away.
• Point of Application of Force: The point of application of force is very important characteristic which determines about where a force does act. It is very must in determining or solving forces into several components.
• Simply one can say that magnitude defines the size, direction denotes the path and the point of application is at which the whole force is assumed to be concentrated.
The point of application of force defines the type of load. Load is nothing but a heavy object which is to be carried or which is to be withstood. Loads are the pressure makers of the engineering objects. Let us see some types of loads.
• Point Load: If the entire magnitude of force is acting over a small area, it is called concentrated force or a point load.
• Distributed Load: If the entire magnitude of the force is distributed over a length or an area or a volume, it is called Distributed Force.
• The concept of point load is hypothetical. This is because, in practical, forces can be transmitted through a definite area only. Nevertheless, this is an idealization that helps us to solve problems more carefully and efficiently.
• The distributed forces normally follow a definite pattern of loading over the entire area/length. It cannot be expressed in terms of length and intensity as that of point load.
• In distributed forces, the magnitude is calculated by multiplying the product of its length to that of its intensity. However, in many cases it is impossible practically.
• Center of Gravity: The center of gravity of a body is an imaginary point at which all the forces on the body appear to act. Hence, by this definition one must take care that the point load acts at the center of gravity of a body or on the axis passing through center of gravity.
## Scalars and Vectors
Every physical quantity is divided into two types based upon the units. They are scalars and vectors.
• Scalar Quantity: A physical quantity which can be represented by a single element of a number field is called Scalar Quantity.
• Vector Quantity: A physical quantity which cannot be represented in a single element form is called Vector Quantity.
Some of the properties of Scalars and Vectors are shown in the below table.
Parameters Scalar Vector
Definition A scalar has only magnitude, but no direction. Vector has both magnitude and direction.
Existence Every scalar is one dimensional. Vector can be 1-D,2-D and 3-D
Resolution Scalar quantity cannot be resolved into components. Vector can be resolved in any direction using sine or cosine or other angles.
Changes Any change in scalar quantity is reflected is the reflection of change in its magnitude. Change in vector quantity means any change either in magnitude or in the direction.
Operations Mathematical operations between two scalar quantities always result in a scalar quantity. Mathematical Operations on two vector quantities may result either in vector quantity or scalar quantity.
Examples Examples are mass, length, energy etc. Examples are displacement, acceleration etc.
The above table shows the differences between a scalar quantity and a vector quantity.
## Representation of a Force
Graphically, a force may be represented as a segment of a straight line. Here, one needs to understand the term “Line of Action”. And since force is a vector, it can be represented graphically. Some of the ways to be followed to represent a force are:
• The straight line represents the line of action of the force 1kN and its length represents its magnitude.
• Line of Action: The direction in which the force tends to move a body to which it is applied is called Line of Action.
• The direction of a force is indicated by placing an arrow head on the straight line.
• Either head or tail may be used to indicate the point of application of a force. All the forces involved must be represented consistently.
The figure given below is a representation of force acting on a point.
In the above figure, we observe that a force of 1kN is acting on an object in two cases. The representation is given as such in the diagram. Magnitude, Direction and Point of Application must be represented as shown above.
## Key Points to Remember
Here are the key points that we need to remember about “Principles of Statics”.
• Statics is one of the oldest branches of science existing in the world since ages.
• A rigid body can be converted into a non-rigid body by applying huge loads but vice-versa is not possible.
• A pull or push on an object is called force. It is applied externally and hence is called eternal agent.
• Some of the practical life examples of force are tractive effort on a locomotive, force of magnetic attraction, resistance due to friction etc.
• A point load is a type of load generally concentrated around a very small area, whereas the distributed load acts over considerable areas or volumes.
• If you multiply scalar with a scalar, then the product is a scalar. If you multiply vector with a scalar, the result is a vector.
• The SI units used by engineers to measure the magnitude are newton (N). It is a derived unit from mass, length and time.
If you find any mistake above, kindly email to [email protected] | 2,014 | 9,914 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2024-26 | latest | en | 0.960423 |
https://braingenie.ck12.org/skills/102484/learn | 1,580,083,426,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251694071.63/warc/CC-MAIN-20200126230255-20200127020255-00135.warc.gz | 360,463,593 | 1,820 | ### Sample Problem
Study the picture below and fill in the blanks.
(1) The can of beer is oz. lighter than the bottle of milk.
(2) The box of juice is lb. oz. heavier than the bottle of milk.
(3) The total weight of the 3 items is lb. oz.
#### Solution
(1) Difference between the can of beer and the bottle of milk: 1 lb. 15 oz. - 1 lb. 10 oz. = 5 oz.
(2) Difference between the box of juice and the bottle of milk: 2 lb. 12 oz. - 1 lb. 15 oz. = 0 lb. 13 oz.
(3) Total weight: 1 lb. 10 oz. + 1 lb. 15 oz. + 2 lb. 12 oz. = 6 lb. 5 oz. | 193 | 543 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2020-05 | latest | en | 0.791854 |
https://forum.ansys.com/forums/topic/sine-wave-3d-equation-curve/ | 1,701,744,752,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100540.62/warc/CC-MAIN-20231205010358-20231205040358-00773.warc.gz | 309,763,309 | 35,868 | ## 3D Design
Topics related to Ansys Discovery and Ansys SpaceClaim.
#### Sine Wave 3d equation curve
TAGGED: ,
• info
Subscriber
Hi, i'm having trouble creating a sine wave in circle with the equation tool.
What would be the equation to create below please?
Forum Moderator
Hello Macsys Engineering
Let me check and get back to you.
Forum Moderator
Hello Macsys Engineering
Apologies for the delay here. I did check for the equation of the 3d sine wave but did not find it as per your image.
Surface equation tool does not have Sine wave, but it has custom option...so if you know the 3D equation for x, y and z value, but for that we would need exact equation.
We have a script for the curve shown below, let me know if this is something you are looking for?
Attached is the script and .scdoc file for your reference. You can go through the script and try the same for the curve you would like. I know you needed a surface, but your curve is not very clear in 3d from the image. You can pull the curves to get a surface body of this.
Thank you.
🛈 This post originally contained file attachments which have been removed in compliance with the updated Ansys Learning Forum Terms & Conditions
• info
Subscriber
Devendra Badgujar , thank you. No apologies needed. If focus on the curve, below is my example made in SW. Basically a cos wrapped in a circle. (in this case half a circle (pi) to solve start-end issue). The formula is in the picture.
I was not able to reproduce this in SC. As an alternative i tried polar coordinates, but also not working. Can you get this curve / formulae working in SC?
Forum Moderator
Macsys Engineering
Thanks for your update, let me try this and get back to you.
Regards.
• mg
Subscriber
Macsys Engineering
A wave washer can be created in several steps with the standard SC tools. Have used this with both SC2022R1 and SC2023R1.
The steps are:
1. define the dimensions of the finished washer (see photo)
2. create a 2d wave with Design | Sketch | Equation | Sine Wave
3. create a 3d cylinder with a diameter and height based on the washer dimensions
4. align the wave start-point with the curved surface of the cylinder
5. imprint the wave on the cylinder surface with the Prepare | Analysis | Wrap tool to make a closed spline
6. pull the spline to create a surface,
7. pull the surface to complete the 3d washer material width
8. add reference marks or planes to enable measurements
Refer to video showing the steps. Notes below refer to marked frames in the video.
Cannot normally see the exact SC internal values for calculated spline point locations, and this can create situations where a tool will not operate as expected, with no error information provided. Findings re this sensitivity are explained in the Notes with a 'Caution:' tag.
Notes:
1. Equation values:
(t) Endpoint: washer OD * pi.
Caution: use as many digits of precision as possible to insure that the wave endpoint exactly aligns with the start-point after wrapping, otherwise the spline loop with be open and may not Pull correctly. Also, in the Equation pane, the entered value is displayed rounded down after the cursor is moved to a new field.
(A) Amplitude = (washer overall height - material thickness) / 2, (this example, 1.5 = (4-1)/2)
(P) Period, determines the quantity of top and bottom bearing points of the washer. (this example, (t) Endpoint value / 2pi = 15 / 4 nodes = 3.75
2. Align wave with cylinder surface: +x move = washer outside radius, +y move = (A) + material thickness (2.5mm in example).
3. In theory, Pull top surface of cylinder to height of the washer - Caution: if any point of the wave is greater than this height, the Wrap operation will fail. In this example, the height is pulled to 4.1mm.
4. After Wrap, a single complete closed spline should appear. Caution: if multiple splines are created, this indicates the spline is not closed due to misalignment of the endpoints; increase the (t)-End precision.
5. To Pull the spline downward parallel to the y-axis, zoom in enough to set the exact height with the LMB only. Caution: if while pulling, the space bar is clicked and the pull dimension (1mm in this example) is entered via the keyboard, followed by enter key, the pull will fail (with unexpected behavior)
6. The Measure and Detail | Dimension tools won't resolve the overall height of the washer (4mm) since they are not planar surfaces. Add planes for determining dimensions between curved surfaces.
7. Caution: most dimensions of the finished washer are not exact, even when all the input values are exact. This is resolved when creating the drawing by changing the dimension precision or 0 or 1 decimal places.
🛈 This post originally contained file attachments which have been removed in compliance with the updated Ansys Learning Forum Terms & Conditions
Forum Moderator
maike thanks for your workaround to this problem. Looks like a good way to do this.
Macsys Engineering please check the following video shared by maike to see if that helps.
I tried the equations that you provided, somehow, they were not working and gives this ''Failed to find cusps'' error.
I have raised a bug 805394 for this, will let you know as I have more information from the developers on this.
🛈 This post originally contained file attachments which have been removed in compliance with the updated Ansys Learning Forum Terms & Conditions
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Thank you very much maike and Devendra Badgujar | 1,310 | 5,628 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2023-50 | latest | en | 0.921544 |
https://root-forum.cern.ch/t/step-function/21767 | 1,660,020,372,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882570901.18/warc/CC-MAIN-20220809033952-20220809063952-00390.warc.gz | 457,266,696 | 4,554 | # Step function
Hello,
I would like to define a TF1 with a step function inside, like:
TF1 *f = new TF1(“f”, " sin(x)*theta(x-2)", -10, 10);
where theta is the typical step function: it returns 0 if (x-2)<0 and it returns 1 if (x-2)>0.
Unfortunately I have not found any step function in TMath neither in other libraries, but only TMath::Floor() and TMath:.Ceil().
Could you help me?
Many thanks
TF1 *f = new TF1(“f”, “((x > 2) ? sin(x) : 0)”, -10, 10);
TF1 *f = new TF1(“f”, “sin(x) * ((x > 2) ? 1 : 0)”, -10, 10); // safe in ROOT 6 only
Thanks! | 202 | 553 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-33 | latest | en | 0.709612 |
https://www.powershow.com/view4/8133dc-ZWFhY/What_are_Kinetic_and_Potential_Energy_powerpoint_ppt_presentation | 1,585,971,457,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370519111.47/warc/CC-MAIN-20200404011558-20200404041558-00206.warc.gz | 1,093,336,030 | 19,275 | # What are Kinetic and Potential Energy? - PowerPoint PPT Presentation
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## What are Kinetic and Potential Energy?
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Title: What are Kinetic and Potential Energy?
1
What are Kinetic and Potential Energy?
2
What is ENERGY?
• Energy is the ability to do work.
• Everything that happens in the world uses energy!
• Most of the time we cant see energy, but it is
• everywhere around us!
3
Energy
• is NEVER created or destroyed!
• can only be STORED or TRANFERRED.
4
This car uses a lot of energy
Batteries store energy!
We get our energy from FOOD!
Even this sleeping puppy is using stored energy.
5
Remember ALL matter is made up of
particles. The particles NEVER stop
moving.
6
How is all energy divided?
All Energy
Potential Energy
Kinetic Energy
7
Potential Energy is
• The energy stored in an object.
• "Potential" simply means the energy has the
ability to do something useful later on.
8
Examples of Potential Energy
A stretched rubber band..
Water at the top of a waterfall..
YoYo held in your hand..
A drawn Bow and Arrow
9
• The higher an object, the more potential energy.
• The more mass an object has, the more potential
energy it has.
10
• Which object has more potential energy?
B
A
11
ANSWER
A
This brick has more mass than the
feather therefore more potential energy!
12
Changing an objects height can change its
potential energy.
• If I want to drop an apple from the top of one of
these three things, where will be the most
potential energy?
A
C
B
13
ANSWER
A
• The higher the object, the more potential energy!
14
Potential Energy Converted to Kinetic Energy
• When stored energy begins to move, the object now
transfers from potential energy into kinetic
energy.
15
Kinetic Energy Is
• The energy of a moving object.
• "Kinetic" means movement!
• When stored energy is being used up, it is making
things move or happen.
16
Examples of Kinetic Energy
17
• The faster the object moves, the more kinetic
energy is produced.
• The greater the mass and speed of an object, the
more kinetic energy there will be.
18
When these objects move at the same speed, which
will have more kinetic energy?
19
ANSWER
The semi- truck has more mass therefore, more
kinetic energy!
20
• An object has the MOST kinetic energy when its
movement is the GREATEST.
• When an object has the LEAST potential energy, it
has the MOST kinetic energy.
21
A water bottle is knocked off a desk. When does
the bottle have the MOST kinetic energy?
• A. At the top of the fall.
• B. In the middle of the fall.
• C. At the bottom of the fall.
22
• C. At the bottom of the fall.
• It has the most kinetic energy when its movement
and speed are greatest, which is at the bottom of
the fall right before it hits the ground.
• When an object has the LEAST potential energy is
when it has the MOST kinetic energy.
BrainPop
23
Roller Coasters
• When does the train on this roller coaster have
the MOST potential energy?
• AT THE VERY TOP!
• The HIGHER the train is lifted by the motor, the
MORE potential energy is produced.
• At the top of the hill the train has a huge
amount of potential energy, but it has very
little kinetic energy.
24
• As the train accelerates down the hill the
potential energy is converted into kinetic
energy.
• There is very little
• potential energy at the bottom of the hill, but
there is a great amount of kinetic energy.
25
• When does the train on this roller coaster have
the MOST kinetic energy?
• (When is it moving the fastest?)
• (When does it have the LEAST potential energy???)
• At the bottom of the tallest hill!
26
• Animated Roller Coaster
27
Notes
• All energy is divided into two types potential
and kinetic.
• Potential Energy The energy stored in an object.
• Kinetic Energy The energy of a moving object.
• Energy is never created or destroyed. It is
always stored or transferred.
About PowerShow.com | 1,082 | 4,445 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-16 | latest | en | 0.894826 |
https://statkat.com/stattest.php?t=5&t2=44&t3=8&t4=20 | 1,670,603,442,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711417.46/warc/CC-MAIN-20221209144722-20221209174722-00461.warc.gz | 579,086,505 | 8,273 | # One sample z test for the mean - overview
This page offers structured overviews of one or more selected methods. Add additional methods for comparisons by clicking on the dropdown button in the right-hand column. To practice with a specific method click the button at the bottom row of the table
One sample $z$ test for the mean
Binomial test for a single proportion
Two sample $z$ test
Logistic regression
Independent variableIndependent variableIndependent/grouping variableIndependent variables
NoneNoneOne categorical with 2 independent groupsOne or more quantitative of interval or ratio level and/or one or more categorical with independent groups, transformed into code variables
Dependent variableDependent variableDependent variableDependent variable
One quantitative of interval or ratio levelOne categorical with 2 independent groupsOne quantitative of interval or ratio levelOne categorical with 2 independent groups
Null hypothesisNull hypothesisNull hypothesisNull hypothesis
H0: $\mu = \mu_0$
Here $\mu$ is the population mean, and $\mu_0$ is the population mean according to the null hypothesis.
H0: $\pi = \pi_0$
Here $\pi$ is the population proportion of 'successes', and $\pi_0$ is the population proportion of successes according to the null hypothesis.
H0: $\mu_1 = \mu_2$
Here $\mu_1$ is the population mean for group 1, and $\mu_2$ is the population mean for group 2.
Model chi-squared test for the complete regression model:
• H0: $\beta_1 = \beta_2 = \ldots = \beta_K = 0$
Wald test for individual regression coefficient $\beta_k$:
• H0: $\beta_k = 0$
or in terms of odds ratio:
• H0: $e^{\beta_k} = 1$
Likelihood ratio chi-squared test for individual regression coefficient $\beta_k$:
• H0: $\beta_k = 0$
or in terms of odds ratio:
• H0: $e^{\beta_k} = 1$
in the regression equation $\ln \big(\frac{\pi_{y = 1}}{1 - \pi_{y = 1}} \big) = \beta_0 + \beta_1 \times x_1 + \beta_2 \times x_2 + \ldots + \beta_K \times x_K$. Here $x_i$ represents independent variable $i$, $\beta_i$ is the regression weight for independent variable $x_i$, and $\pi_{y = 1}$ represents the true probability that the dependent variable $y = 1$ (or equivalently, the proportion of $y = 1$ in the population) given the scores on the independent variables.
Alternative hypothesisAlternative hypothesisAlternative hypothesisAlternative hypothesis
H1 two sided: $\mu \neq \mu_0$
H1 right sided: $\mu > \mu_0$
H1 left sided: $\mu < \mu_0$
H1 two sided: $\pi \neq \pi_0$
H1 right sided: $\pi > \pi_0$
H1 left sided: $\pi < \pi_0$
H1 two sided: $\mu_1 \neq \mu_2$
H1 right sided: $\mu_1 > \mu_2$
H1 left sided: $\mu_1 < \mu_2$
Model chi-squared test for the complete regression model:
• H1: not all population regression coefficients are 0
Wald test for individual regression coefficient $\beta_k$:
• H1: $\beta_k \neq 0$
or in terms of odds ratio:
• H1: $e^{\beta_k} \neq 1$
If defined as Wald $= \dfrac{b_k}{SE_{b_k}}$ (see 'Test statistic'), also one sided alternatives can be tested:
• H1 right sided: $\beta_k > 0$
• H1 left sided: $\beta_k < 0$
Likelihood ratio chi-squared test for individual regression coefficient $\beta_k$:
• H1: $\beta_k \neq 0$
or in terms of odds ratio:
• H1: $e^{\beta_k} \neq 1$
AssumptionsAssumptionsAssumptionsAssumptions
• Scores are normally distributed in the population
• Population standard deviation $\sigma$ is known
• Sample is a simple random sample from the population. That is, observations are independent of one another
• Sample is a simple random sample from the population. That is, observations are independent of one another
• Within each population, the scores on the dependent variable are normally distributed
• Population standard deviations $\sigma_1$ and $\sigma_2$ are known
• Group 1 sample is a simple random sample (SRS) from population 1, group 2 sample is an independent SRS from population 2. That is, within and between groups, observations are independent of one another
• In the population, the relationship between the independent variables and the log odds $\ln (\frac{\pi_{y=1}}{1 - \pi_{y=1}})$ is linear
• The residuals are independent of one another
• Variables are measured without error
Also pay attention to:
• Multicollinearity
• Outliers
Test statisticTest statisticTest statisticTest statistic
$z = \dfrac{\bar{y} - \mu_0}{\sigma / \sqrt{N}}$
Here $\bar{y}$ is the sample mean, $\mu_0$ is the population mean according to the null hypothesis, $\sigma$ is the population standard deviation, and $N$ is the sample size.
The denominator $\sigma / \sqrt{N}$ is the standard deviation of the sampling distribution of $\bar{y}$. The $z$ value indicates how many of these standard deviations $\bar{y}$ is removed from $\mu_0$.
$X$ = number of successes in the sample$z = \dfrac{(\bar{y}_1 - \bar{y}_2) - 0}{\sqrt{\dfrac{\sigma^2_1}{n_1} + \dfrac{\sigma^2_2}{n_2}}} = \dfrac{\bar{y}_1 - \bar{y}_2}{\sqrt{\dfrac{\sigma^2_1}{n_1} + \dfrac{\sigma^2_2}{n_2}}}$
Here $\bar{y}_1$ is the sample mean in group 1, $\bar{y}_2$ is the sample mean in group 2, $\sigma^2_1$ is the population variance in population 1, $\sigma^2_2$ is the population variance in population 2, $n_1$ is the sample size of group 1, and $n_2$ is the sample size of group 2. The 0 represents the difference in population means according to the null hypothesis.
The denominator $\sqrt{\frac{\sigma^2_1}{n_1} + \frac{\sigma^2_2}{n_2}}$ is the standard deviation of the sampling distribution of $\bar{y}_1 - \bar{y}_2$. The $z$ value indicates how many of these standard deviations $\bar{y}_1 - \bar{y}_2$ is removed from 0.
Note: we could just as well compute $\bar{y}_2 - \bar{y}_1$ in the numerator, but then the left sided alternative becomes $\mu_2 < \mu_1$, and the right sided alternative becomes $\mu_2 > \mu_1$.
Model chi-squared test for the complete regression model:
• $X^2 = D_{null} - D_K = \mbox{null deviance} - \mbox{model deviance}$
$D_{null}$, the null deviance, is conceptually similar to the total variance of the dependent variable in OLS regression analysis. $D_K$, the model deviance, is conceptually similar to the residual variance in OLS regression analysis.
Wald test for individual $\beta_k$:
The wald statistic can be defined in two ways:
• Wald $= \dfrac{b_k^2}{SE^2_{b_k}}$
• Wald $= \dfrac{b_k}{SE_{b_k}}$
SPSS uses the first definition.
Likelihood ratio chi-squared test for individual $\beta_k$:
• $X^2 = D_{K-1} - D_K$
$D_{K-1}$ is the model deviance, where independent variable $k$ is excluded from the model. $D_{K}$ is the model deviance, where independent variable $k$ is included in the model.
Sampling distribution of $z$ if H0 were trueSampling distribution of $X$ if H0 were trueSampling distribution of $z$ if H0 were trueSampling distribution of $X^2$ and of the Wald statistic if H0 were true
Standard normal distributionBinomial($n$, $P$) distribution.
Here $n = N$ (total sample size), and $P = \pi_0$ (population proportion according to the null hypothesis).
Standard normal distributionSampling distribution of $X^2$, as computed in the model chi-squared test for the complete model:
• chi-squared distribution with $K$ (number of independent variables) degrees of freedom
Sampling distribution of the Wald statistic:
• If defined as Wald $= \dfrac{b_k^2}{SE^2_{b_k}}$: approximately the chi-squared distribution with 1 degree of freedom
• If defined as Wald $= \dfrac{b_k}{SE_{b_k}}$: approximately the standard normal distribution
Sampling distribution of $X^2$, as computed in the likelihood ratio chi-squared test for individual $\beta_k$:
• chi-squared distribution with 1 degree of freedom
Significant?Significant?Significant?Significant?
Two sided:
Right sided:
Left sided:
Two sided:
• Check if $X$ observed in sample is in the rejection region or
• Find two sided $p$ value corresponding to observed $X$ and check if it is equal to or smaller than $\alpha$
Right sided:
• Check if $X$ observed in sample is in the rejection region or
• Find right sided $p$ value corresponding to observed $X$ and check if it is equal to or smaller than $\alpha$
Left sided:
• Check if $X$ observed in sample is in the rejection region or
• Find left sided $p$ value corresponding to observed $X$ and check if it is equal to or smaller than $\alpha$
Two sided:
Right sided:
Left sided:
For the model chi-squared test for the complete regression model and likelihood ratio chi-squared test for individual $\beta_k$:
• Check if $X^2$ observed in sample is equal to or larger than critical value $X^{2*}$ or
• Find $p$ value corresponding to observed $X^2$ and check if it is equal to or smaller than $\alpha$
For the Wald test:
• If defined as Wald $= \dfrac{b_k^2}{SE^2_{b_k}}$: same procedure as for the chi-squared tests. Wald can be interpret as $X^2$
• If defined as Wald $= \dfrac{b_k}{SE_{b_k}}$: same procedure as for any $z$ test. Wald can be interpreted as $z$.
$C\%$ confidence interval for $\mu$n.a.$C\%$ confidence interval for $\mu_1 - \mu_2$Wald-type approximate $C\%$ confidence interval for $\beta_k$
$\bar{y} \pm z^* \times \dfrac{\sigma}{\sqrt{N}}$
where the critical value $z^*$ is the value under the normal curve with the area $C / 100$ between $-z^*$ and $z^*$ (e.g. $z^*$ = 1.96 for a 95% confidence interval).
The confidence interval for $\mu$ can also be used as significance test.
-$(\bar{y}_1 - \bar{y}_2) \pm z^* \times \sqrt{\dfrac{\sigma^2_1}{n_1} + \dfrac{\sigma^2_2}{n_2}}$
where the critical value $z^*$ is the value under the normal curve with the area $C / 100$ between $-z^*$ and $z^*$ (e.g. $z^*$ = 1.96 for a 95% confidence interval).
The confidence interval for $\mu_1 - \mu_2$ can also be used as significance test.
$b_k \pm z^* \times SE_{b_k}$
where the critical value $z^*$ is the value under the normal curve with the area $C / 100$ between $-z^*$ and $z^*$ (e.g. $z^*$ = 1.96 for a 95% confidence interval).
Effect sizen.a.n.a.Goodness of fit measure $R^2_L$
Cohen's $d$:
Standardized difference between the sample mean and $\mu_0$: $$d = \frac{\bar{y} - \mu_0}{\sigma}$$ Cohen's $d$ indicates how many standard deviations $\sigma$ the sample mean $\bar{y}$ is removed from $\mu_0.$
--$R^2_L = \dfrac{D_{null} - D_K}{D_{null}}$
There are several other goodness of fit measures in logistic regression. In logistic regression, there is no single agreed upon measure of goodness of fit.
Visual representationn.a.Visual representationn.a.
--
Example contextExample contextExample contextExample context
Is the average mental health score of office workers different from $\mu_0 = 50$? Assume that the standard deviation of the mental health scores in the population is $\sigma = 3.$Is the proportion of smokers amongst office workers different from $\pi_0 = 0.2$?Is the average mental health score different between men and women? Assume that in the population, the standard devation of the mental health scores is $\sigma_1 = 2$ amongst men and $\sigma_2 = 2.5$ amongst women.Can body mass index, stress level, and gender predict whether people get diagnosed with diabetes?
n.a.SPSSn.a.SPSS
-Analyze > Nonparametric Tests > Legacy Dialogs > Binomial...
• Put your dichotomous variable in the box below Test Variable List
• Fill in the value for $\pi_0$ in the box next to Test Proportion
-Analyze > Regression > Binary Logistic...
• Put your dependent variable in the box below Dependent and your independent (predictor) variables in the box below Covariate(s)
n.a.Jamovin.a.Jamovi
-Frequencies > 2 Outcomes - Binomial test
• Put your dichotomous variable in the white box at the right
• Fill in the value for $\pi_0$ in the box next to Test value
• Under Hypothesis, select your alternative hypothesis
-Regression > 2 Outcomes - Binomial
• Put your dependent variable in the box below Dependent Variable and your independent variables of interval/ratio level in the box below Covariates
• If you also have code (dummy) variables as independent variables, you can put these in the box below Covariates as well
• Instead of transforming your categorical independent variable(s) into code variables, you can also put the untransformed categorical independent variables in the box below Factors. Jamovi will then make the code variables for you 'behind the scenes'
Practice questionsPractice questionsPractice questionsPractice questions | 3,373 | 12,259 | {"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} | 3.328125 | 3 | CC-MAIN-2022-49 | latest | en | 0.715565 |
http://www.cfd-online.com/Forums/main/11803-matrix-dissipation.html | 1,469,781,218,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257829972.49/warc/CC-MAIN-20160723071029-00308-ip-10-185-27-174.ec2.internal.warc.gz | 363,436,008 | 15,420 | # matrix dissipation
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July 12, 2006, 09:17 matrix dissipation #1 Simon Guest Posts: n/a My main question is: How to implement a matrix dissipation method on an unstructured FV, cell centred, Runge-Kutta sheme? Can anyone give me some references on matrix dissipation applied with unstructured grids? Thanks
July 14, 2006, 02:55 Re: matrix dissipation #2 O. Guest Posts: n/a I assume you are talking about a central discretization + a matrix dissipation term. I don't know any papers, unfortunately. Isn't any upwind scheme a matrix dissipation scheme as well? For example the Roe scheme across an interface can be written as: F = 1/2*(F(Q_l) + F(Q_r)) - 1/2*|A|*(Q_r - Q_l) with Q_l and Q_r beeing the extrapolated values from the left and right side and |A| the Roe matrix. In this case the matrix is based on physical considerations (direction of influence, eigenvalues). I think that you could in principle use any other matrix as well. How you extrapolate Q_l and Q_r defines the order of your scheme. 'hope this helps, or did I miss anything out??
July 14, 2006, 08:38 Re: matrix dissipation #3 Simon Guest Posts: n/a Thanks. Yes I mean "central discretization + a matrix dissipation". My main concern is the implementation/calculation of |A| in an unstructured framework? How does the lack of direction of the grid influences the calculation of |A|?
July 17, 2006, 03:12 Re: matrix dissipation #4 O. Guest Posts: n/a You do have a direction. If you compute the flux across an interface than the interface has a normal and thus you can differentiate between an upwind and a downwind side. But it is true that things can be awkward. Something like the next neighbour in upwind direction is impossible on triangular grids. Unfortunately I have no experience with matrix dissipation schemes. For upwind schemes you can formulate something by using a local direction (e.g. interface normal) and the gradients in the two cells adjacent to your interface (for more than 1st order). If you find/found a good paper on this, I'd be interested in the reference.
July 17, 2006, 08:57 Re: matrix dissipation #5 Cut cell Guest Posts: n/a This is classic, it should answer all you questions: http://hdl.handle.net/2002/13559 which should be the same as: http://library-dspace.larc.nasa.gov/...dle/2002/13559 Cheers Andy
July 18, 2006, 01:59 Re: matrix dissipation #6 O. Guest Posts: n/a That describes a very classic upwind solver.
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Contact Us - CFD Online - Top | 840 | 3,203 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2016-30 | longest | en | 0.89316 |
http://www.nag.com/numeric/fl/nagdoc_fl24/html/G08/g08cdf.html | 1,387,491,721,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1387345768537/warc/CC-MAIN-20131218054928-00034-ip-10-33-133-15.ec2.internal.warc.gz | 585,213,316 | 5,864 | G08CDF (PDF version)
G08 Chapter Contents
G08 Chapter Introduction
NAG Library Manual
# NAG Library Routine DocumentG08CDF
Note: before using this routine, please read the Users' Note for your implementation to check the interpretation of bold italicised terms and other implementation-dependent details.
## 1 Purpose
G08CDF performs the two sample Kolmogorov–Smirnov distribution test.
## 2 Specification
SUBROUTINE G08CDF ( N1, X, N2, Y, NTYPE, D, Z, P, SX, SY, IFAIL)
INTEGER N1, N2, NTYPE, IFAIL REAL (KIND=nag_wp) X(N1), Y(N2), D, Z, P, SX(N1), SY(N2)
## 3 Description
The data consists of two independent samples, one of size ${n}_{1}$, denoted by ${x}_{1},{x}_{2},\dots ,{x}_{{n}_{1}}$, and the other of size ${n}_{2}$ denoted by ${y}_{1},{y}_{2},\dots ,{y}_{{n}_{2}}$. Let $F\left(x\right)$ and $G\left(x\right)$ represent their respective, unknown, distribution functions. Also let ${S}_{1}\left(x\right)$ and ${S}_{2}\left(x\right)$ denote the values of the sample cumulative distribution functions at the point $x$ for the two samples respectively.
The Kolmogorov–Smirnov test provides a test of the null hypothesis ${H}_{0}$: $F\left(x\right)=G\left(x\right)$ against one of the following alternative hypotheses:
(i) ${H}_{1}$: $F\left(x\right)\ne G\left(x\right)$. (ii) ${H}_{2}$: $F\left(x\right)>G\left(x\right)$. This alternative hypothesis is sometimes stated as, ‘The $x$'s tend to be smaller than the $y$'s’, i.e., it would be demonstrated in practical terms if the values of ${S}_{1}\left(x\right)$ tended to exceed the corresponding values of ${S}_{2}\left(x\right)$. (iii) ${H}_{3}$: $F\left(x\right). This alternative hypothesis is sometimes stated as, ‘The $x$'s tend to be larger than the $y$'s’, i.e., it would be demonstrated in practical terms if the values of ${S}_{2}\left(x\right)$ tended to exceed the corresponding values of ${S}_{1}\left(x\right)$.
One of the following test statistics is computed depending on the particular alternative null hypothesis specified (see the description of the parameter NTYPE in Section 5).
For the alternative hypothesis ${H}_{1}$.
• ${D}_{{n}_{1},{n}_{2}}$ – the largest absolute deviation between the two sample cumulative distribution functions.
For the alternative hypothesis ${H}_{2}$.
• ${D}_{{n}_{1},{n}_{2}}^{+}$ – the largest positive deviation between the sample cumulative distribution function of the first sample, ${S}_{1}\left(x\right)$, and the sample cumulative distribution function of the second sample, ${S}_{2}\left(x\right)$. Formally ${D}_{{n}_{1},{n}_{2}}^{+}=\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left\{{S}_{1}\left(x\right)-{S}_{2}\left(x\right),0\right\}$.
For the alternative hypothesis ${H}_{3}$.
• ${D}_{{n}_{1},{n}_{2}}^{-}$ – the largest positive deviation between the sample cumulative distribution function of the second sample, ${S}_{2}\left(x\right)$, and the sample cumulative distribution function of the first sample, ${S}_{1}\left(x\right)$. Formally ${D}_{{n}_{1},{n}_{2}}^{-}=\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left\{{S}_{2}\left(x\right)-{S}_{1}\left(x\right),0\right\}$.
G08CDF also returns the standardized statistic $Z=\sqrt{\frac{{n}_{1}+{n}_{2}}{{n}_{1}{n}_{2}}}×D$, where $D$ may be ${D}_{{n}_{1},{n}_{2}}$, ${D}_{{n}_{1},{n}_{2}}^{+}$ or ${D}_{{n}_{1},{n}_{2}}^{-}$ depending on the choice of the alternative hypothesis. The distribution of this statistic converges asymptotically to a distribution given by Smirnov as ${n}_{1}$ and ${n}_{2}$ increase; see Feller (1948), Kendall and Stuart (1973), Kim and Jenrich (1973), Smirnov (1933) or Smirnov (1948).
The probability, under the null hypothesis, of obtaining a value of the test statistic as extreme as that observed, is computed. If $\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left({n}_{1},{n}_{2}\right)\le 2500$ and ${n}_{1}{n}_{2}\le 10000$ then an exact method given by Kim and Jenrich (see Kim and Jenrich (1973)) is used. Otherwise $p$ is computed using the approximations suggested by Kim and Jenrich (1973). Note that the method used is only exact for continuous theoretical distributions. This method computes the two-sided probability. The one-sided probabilities are estimated by halving the two-sided probability. This is a good estimate for small $p$, that is $p\le 0.10$, but it becomes very poor for larger $p$.
## 4 References
Conover W J (1980) Practical Nonparametric Statistics Wiley
Feller W (1948) On the Kolmogorov–Smirnov limit theorems for empirical distributions Ann. Math. Statist. 19 179–181
Kendall M G and Stuart A (1973) The Advanced Theory of Statistics (Volume 2) (3rd Edition) Griffin
Kim P J and Jenrich R I (1973) Tables of exact sampling distribution of the two sample Kolmogorov–Smirnov criterion ${D}_{mn}\left(m Selected Tables in Mathematical Statistics 1 80–129 American Mathematical Society
Siegel S (1956) Non-parametric Statistics for the Behavioral Sciences McGraw–Hill
Smirnov N (1933) Estimate of deviation between empirical distribution functions in two independent samples Bull. Moscow Univ. 2(2) 3–16
Smirnov N (1948) Table for estimating the goodness of fit of empirical distributions Ann. Math. Statist. 19 279–281
## 5 Parameters
1: N1 – INTEGERInput
On entry: the number of observations in the first sample, ${n}_{1}$.
Constraint: ${\mathbf{N1}}\ge 1$.
2: X(N1) – REAL (KIND=nag_wp) arrayInput
On entry: the observations from the first sample, ${x}_{1},{x}_{2},\dots ,{x}_{{n}_{1}}$.
3: N2 – INTEGERInput
On entry: the number of observations in the second sample, ${n}_{2}$.
Constraint: ${\mathbf{N2}}\ge 1$.
4: Y(N2) – REAL (KIND=nag_wp) arrayInput
On entry: the observations from the second sample, ${y}_{1},{y}_{2},\dots ,{y}_{{n}_{2}}$.
5: NTYPE – INTEGERInput
On entry: the statistic to be computed, i.e., the choice of alternative hypothesis.
${\mathbf{NTYPE}}=1$
Computes ${D}_{{n}_{1}{n}_{2}}$, to test against ${H}_{1}$.
${\mathbf{NTYPE}}=2$
Computes ${D}_{{n}_{1}{n}_{2}}^{+}$, to test against ${H}_{2}$.
${\mathbf{NTYPE}}=3$
Computes ${D}_{{n}_{1}{n}_{2}}^{-}$, to test against ${H}_{3}$.
Constraint: ${\mathbf{NTYPE}}=1$, $2$ or $3$.
6: D – REAL (KIND=nag_wp)Output
On exit: the Kolmogorov–Smirnov test statistic (${D}_{{n}_{1}{n}_{2}}$, ${D}_{{n}_{1}{n}_{2}}^{+}$ or ${D}_{{n}_{1}{n}_{2}}^{-}$ according to the value of NTYPE).
7: Z – REAL (KIND=nag_wp)Output
On exit: a standardized value, $Z$, of the test statistic, $D$, without any correction for continuity.
8: P – REAL (KIND=nag_wp)Output
On exit: the tail probability associated with the observed value of $D$, where $D$ may be ${D}_{{n}_{1},{n}_{2}},{D}_{{n}_{1},{n}_{2}}^{+}$ or ${D}_{{n}_{1},{n}_{2}}^{-}$ depending on the value of NTYPE (see Section 3).
9: SX(N1) – REAL (KIND=nag_wp) arrayOutput
On exit: the observations from the first sample sorted in ascending order.
10: SY(N2) – REAL (KIND=nag_wp) arrayOutput
On exit: the observations from the second sample sorted in ascending order.
11: IFAIL – INTEGERInput/Output
On entry: IFAIL must be set to $0$, $-1\text{ or }1$. If you are unfamiliar with this parameter you should refer to Section 3.3 in the Essential Introduction for details.
For environments where it might be inappropriate to halt program execution when an error is detected, the value $-1\text{ or }1$ is recommended. If the output of error messages is undesirable, then the value $1$ is recommended. Otherwise, if you are not familiar with this parameter, the recommended value is $0$. When the value $-\mathbf{1}\text{ or }\mathbf{1}$ is used it is essential to test the value of IFAIL on exit.
On exit: ${\mathbf{IFAIL}}={\mathbf{0}}$ unless the routine detects an error or a warning has been flagged (see Section 6).
## 6 Error Indicators and Warnings
If on entry ${\mathbf{IFAIL}}={\mathbf{0}}$ or $-{\mathbf{1}}$, explanatory error messages are output on the current error message unit (as defined by X04AAF).
Errors or warnings detected by the routine:
${\mathbf{IFAIL}}=1$
On entry, ${\mathbf{N1}}<1$, or ${\mathbf{N2}}<1$.
${\mathbf{IFAIL}}=2$
On entry, ${\mathbf{NTYPE}}\ne 1$, $2$ or $3$.
${\mathbf{IFAIL}}=3$
The iterative procedure used in the approximation of the probability for large ${n}_{1}$ and ${n}_{2}$ did not converge. For the two-sided test, $p=1$ is returned. For the one-sided test, $p=0.5$ is returned.
## 7 Accuracy
The large sample distributions used as approximations to the exact distribution should have a relative error of less than 5% for most cases.
## 8 Further Comments
The time taken by G08CDF increases with ${n}_{1}$ and ${n}_{2}$, until ${n}_{1}{n}_{2}>10000$ or $\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left({n}_{1},{n}_{2}\right)\ge 2500$. At this point one of the approximations is used and the time decreases significantly. The time then increases again modestly with ${n}_{1}$ and ${n}_{2}$.
## 9 Example
This example computes the two-sided Kolmogorov–Smirnov test statistic for two independent samples of size $100$ and $50$ respectively. The first sample is from a uniform distribution $U\left(0,2\right)$. The second sample is from a uniform distribution $U\left(0.25,2.25\right)$. The test statistic, ${D}_{{n}_{1},{n}_{2}}$, the standardized test statistic, $Z$, and the tail probability, $p$, are computed and printed.
### 9.1 Program Text
Program Text (g08cdfe.f90)
### 9.2 Program Data
Program Data (g08cdfe.d)
### 9.3 Program Results
Program Results (g08cdfe.r)
G08CDF (PDF version)
G08 Chapter Contents
G08 Chapter Introduction
NAG Library Manual | 2,953 | 9,527 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 112, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2013-48 | longest | en | 0.756834 |
http://www.onlinemathlearning.com/equivalent-measurements.html | 1,475,354,366,000,000,000 | text/html | crawl-data/CC-MAIN-2016-40/segments/1474738663247.66/warc/CC-MAIN-20160924173743-00294-ip-10-143-35-109.ec2.internal.warc.gz | 645,865,323 | 9,835 | # Equivalent Measurements
Videos to help Grade 5 students learn how to use whole number multiplication to express equivalent measurements.
Common Core Standards: 5.NBT.5, 5.NBT.7, 5.MD.1, 5.NBT.1, 5.NBT.2
Related Topics:
Lesson Plans and Worksheets for Grade 5
Lesson Plans and Worksheets for all Grades
New York State Common Core Math Module 2, Grade 5, Lesson 13
Conversions: Using different units to show the same measurement.
Conversion Factor: a numerical ratio to express a measurement from one unit to another unit.
Conversions
1 centimeter = 10 millimeters
1 meter = 100 centimeters = 1,000 millimeters
1 kilometer = 1,000 meters
1 gram = 1,000 milligrams
1 kilogram = 1,000 grams
1 pound = 16 ounces
1 ton = 2,000 pounds
1 cup = 8 fluid ounces
1 pint = 2 cups
1 quart = 2 pints
1 gallon = 4 quarts
1 liter = 1,000 milliliters
1 kiloliter = 1,000 liters
1 mile = 5.280 feet
1 mile = 1,760 yards
Lesson 13 Concept Development
Problem 1
15 feet = _____ inches
Lesson 13 Problem Set
Convert. Use your Reference Sheet to remind you of the conversion factors. Show your work.
_____ oz = 54 lb
4 mi = ____ yd = _____ ft
2 qt = _____ pt = _____ fl oz
Lesson 13 Concept Development
Problem 1
15 feet = _____ inches
Problem 2
3 tons 140 pounds = _____ pounds.
Problem 3
______ ounces = 9 pounds 11 ounces.
Problem 4
155 gallons = _____ quarts = _____ pints
Lesson 13 Problem Set
4. Convert. Use your Reference Sheet to remind you of the conversion factors. Show your work.
a. 27 ft = _____ in
g. 3 km 85 m = _____ m
h. 2 qt = _____ pt = _____ fl oz
Lesson 13 Homework
This video demonstrates how to find equivalent measurements using multiplication, tape diagrams and number lines.
1. Complete the chart below with the measurement equivalents.
2. Convert.
a. 18 yd = _____ ft
g. 5 km 14 m = _____ m
2. Convert
c. _____ cm = 64 m
g. 5 km 14 m = _____ m
5. The food pantry distributes 10-oz bags of rice. If three 5-lb bags are donated to the pantry, how many 10-ounce bags can be made?
Rotate to landscape screen format on a mobile phone or small tablet to use the Mathway widget, a free math problem solver that answers your questions with step-by-step explanations.
You can use the free Mathway widget below to practice Algebra or other math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations. | 659 | 2,388 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2016-40 | longest | en | 0.693896 |
http://mathhelpforum.com/algebra/10310-difference-cubes-print.html | 1,511,497,854,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934807084.8/warc/CC-MAIN-20171124031941-20171124051941-00277.warc.gz | 195,344,344 | 4,317 | # Difference of Cubes
• Jan 19th 2007, 08:05 PM
symmetry
Difference of Cubes
I have a real hard time understading the difference of cubes.
However, there are several questions on this topic in the coming June state exam.
How do I play with this topic?
QUESTIONS:
(1) Factor a^3b^3 - 8x^6y^9
(2) x^3 + 27
• Jan 19th 2007, 08:37 PM
ticbol
Quote:
Originally Posted by symmetry
I have a real hard time understading the difference of cubes.
However, there are several questions on this topic in the coming June state exam.
How do I play with this topic?
QUESTIONS:
(1) Factor a^3b^3 - 8x^6y^9
(2) x^3 + 27
You have to memorize the factoring of the forms (x^3 +y^3) and (x^3 -y^3).
x^3 +y^3 = (x+y)(x^2 -xy +y^2) -----(i)
x^3 -y^3 = (x-y)(x^2 +xy +y^2) ------(ii)
(1) (a^3)(b^3) -8(x^6)(y^9)
= (ab)^3 -[2(x^2)(y^3)]^3
= [(ab) -2(x^2)(y^3)]*{(ab)^2 +(ab)[2(x^2)(y^3)] +[2(x^2)(y^3)]^2}
= [ab -2(x^2)(y^3)]*[(a^2)(b^2) +2ab(x^2)(y^3) +4(x^4)(y^6)] ----answer.
(2) x^3 +27
= x^3 +3^3
= (x+3)(x^2 -x*3x +3^2)
• Jan 20th 2007, 12:49 AM
CaptainBlack
Quote:
Originally Posted by symmetry
I have a real hard time understading the difference of cubes.
However, there are several questions on this topic in the coming June state exam.
How do I play with this topic?
QUESTIONS:
(1) Factor a^3b^3 - 8x^6y^9
(2) x^3 + 27
Ticbol says you have to memorise the factorisation of the sum and
difference of two cubes, but in fact if you know that sums or differences
of two cubes are involved you can deduce what the formulas are on the fly.
Consider f(x) = x^3+y^3, if x=-y, then f(x)=0, so (x+y) is a factor of f(x).
so we may write:
x^3+y^3=(x+y)(x^2 + A xy + y^2) = x^3 + A x^2y + xy^2 +yx^2 + Axy^2+y^3
Collect the terms with like powers of x and y:
x^3+y^3=(x+y)(x^2 + A xy + y^2) = x^3 + (A+1) x^2y + (A+1)yx^2 + y^3
So as the two sides are equal (A+1)=0, or A=-1, and:
x^3+y^3=(x+y)(x^2 - xy + y^2).
This could have been obtained using polynomial long division once we know
that (x+y) is a factor of f(x).
The other factorisation can be obtained in a similar manner by observing
that (x-y) is a factor of x^3-y^3, or by replacing y by -y' throughout
RonL
• Jan 20th 2007, 04:41 AM
Soroban
Hello, symmetry!
I assume you know the sum and difference formulas.
. . $\begin{array}{cc}A^3 + B^3 \:=\:(A + B)(A^2 - AB + B^2) \\ A^3 - B^3 \:=\:(A - B)(A^2 + AB + B^2)\end{array}$
. . Understanding them?
. . Memorizing them?
. . . . . . . . . . . . . . . . . . . . . . . $\underbrace{\text{plus or minus}}_{\downarrow}$
First, we must recognize the form: . $A^3 \pm B^3$
. . . . . . . . . . . . . . . . . . . . . . . . $\overbrace{\text{cube}}^{\uparrow}\;\overbrace{\te xt{cube}}^{\uparrow}$
The cubes break up into a linear factor and a quadratic factor.
. . . . . . $\underbrace{(A \;\pm \;B)}_{\text{linear}} \underbrace{(A^2 \;\mp \;AB \;+ \;B^2)}_{\text{quadratic}}$
To memorize the signs, think of the word SOAP.
. . $A^3 \;\pm \;B^3 \:=\:(A \;\pm\;B)\;(A^2 \;\mp \;AB \;+ \;B^2)$
. . . . . . . . . . . . . . $\uparrow\qquad\qquad\;\uparrow\qquad\quad\uparrow$
. . . . . . . . . . . . Same . . Opposite . Always Positive
• Jan 20th 2007, 04:51 AM
topsquark
Actually you don't need to memorize them, though memorization IS handy. As long as you recall that $a^3 \pm b^3$ is divisible by $a \pm b$ I think this provides a good exercise for that long division you mentioned in another thread. :)
-Dan
• Jan 20th 2007, 05:12 AM
CaptainBlack
Quote:
Originally Posted by topsquark
Actually you don't need to memorize them, though memorization IS handy. As long as you recall that $a^3 \pm b^3$ is divisible by $a \pm b$ I think this provides a good exercise for that long division you mentioned in another thread. :)
-Dan
There seems to be a bit of an echo in here today:eek: .
RonL
• Jan 20th 2007, 05:53 AM
symmetry
ok
I want to thank every tutor for your insight and tips.
Yes, my problem has been memorizing difference of cubes formulas.
• Jan 21st 2007, 05:21 AM
topsquark
Quote:
Originally Posted by CaptainBlack
There seems to be a bit of an echo in here today:eek: .
RonL
Oops! I guess it pays to read the details of each post in the thread, which I obviously didn't do here. :o
-Dan | 1,561 | 4,214 | {"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": 11, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2017-47 | longest | en | 0.878924 |
http://www.glencoe.com/sec/math/studytools/cgi-bin/msgQuiz.php4?isbn=0-02-106024-X&chapter=7&stp=yes&headerFile=0 | 1,371,663,898,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368709000375/warc/CC-MAIN-20130516125640-00039-ip-10-60-113-184.ec2.internal.warc.gz | 466,462,270 | 5,694 | 1. The number of calendars a department store sold in a six-month period is shown in the graph. Which statement gives the best assumption about the calendar sales? A. The number of calendars sold remained the same throughout the year. B. Calendar sales decreased during the end of the year. C. Twice as many calendars were purchased in December than in October. D. Most calendars were sold towards the end of the year and the beginning of the new year. Hint 2. Replace ∅ with <, >, or = to make a true sentence.6,473,281 ∅ 6,463,281 A. + B. < C. > D. = Hint 3. Write as a decimal. A. 9.1 B. 0.9 C. 0.09 D. 0.1 Hint 4. What place value word do you use if you read 0.097 in words? A. thousandths B. tens C. tenths D. hundredths Hint 5. Find 43 + 20. A. 63 B. 4,320 C. 23 D. 4.32 Hint 6. Round 48.47 to the ones place. A. 48 B. 48.5 C. 50 D. 49 Hint 7. Name the Addition Property that is shown.19.5 + 19.1 = 19.1 + 19.5 A. Associative Property B. Distributive Property C. Identity Property D. Commutative Property Hint 8. Tell which equation represents a way to find the product of 5 × 29. A. 5 × (30 + 1) B. 5 × (5 + 29) C. 5 × (29 + 1) D. 5 × (30 − 1) Hint 9. Tell which sentence is the best estimate for 1,450 ÷ 29. A. 2,900 ÷ 29 = 10 B. 1,500 ÷ 50 = 30 C. 1,500 ÷ 30 = 50 D. 1,400 ÷ 200 = 70 Hint 10. Write the word form and find the value of 17 − 8. A. 8 less than 17; 9 B. 17 less than 8; 25 C. 17 plus 8; 25 D. 17 divided by 8; 9 Hint 11. Gabriela can type 38 words per minute. What function rule tells how many words she can type in n minutes? A. 38 ÷ n B. n − 38 C. 38 + n D. 38n Hint 12. Solve for y. y − 10 = 1 A. 10 B. 9 C. 8 D. 11 Hint 13. Solve 32 = 4d. A. 36 B. 28 C. 8 D. 128 Hint 14. Name the point with coordinates (1, 3). A. B B. E C. F D. C Hint 15. What are the coordinates of point D? A. (0, 4) B. (4, 0) C. (1, 4) D. (4, 4) Hint 16. Order 4, 16, 8, 7 from least to greatest. A. 16, 8, 7, 4 B. 4, 7, 8, 16 C. 4, 8, 16, 7 D. 16, 4, 7, 8 Hint 17. Find 44 − 28. A. 16 B. 1,232 C. 26 D. 72 Hint 18. The set of data shows the number of times 20 students took public transportation last month.1, 4, 5, 6, 6, 7, 7, 11, 11, 14, 15, 16, 16, 17, 17, 18, 19, 20, 22, 22Make a frequency table using the intervals 0–6, 7–13, 14–20, 21–27. Which interval has the greatest number of data values? A. 7–13 B. 0–6 C. 21–27 D. 14–20 Hint 19. Suppose you have data on the number of commercials played during an hour on 5 radio stations. Which of the following is the most appropriate type of graph to display the data? A. line plot B. line graph C. double bar graph D. double line graph Hint 20. Which choice uses compatible numbers to get an estimate? 2,499 + 1,559 A. 2,400 + 1,500 = 3,900 B. 2,500 + 1,500 = 4,000 C. 2,500 + 1,600 = 4,100 D. 3,000 + 2,000 = 5,000 Hint | 1,090 | 2,770 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2013-20 | latest | en | 0.837615 |
https://www.justintools.com/unit-conversion/area.php?k1=square-hectometers&k2=square-miles | 1,601,481,037,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600402127075.68/warc/CC-MAIN-20200930141310-20200930171310-00017.warc.gz | 868,083,916 | 27,677 | Please support this site by disabling or whitelisting the Adblock for "justintools.com". I've spent over 10 trillion microseconds (and counting), on this project. This site is my passion, and I regularly adding new tools/apps. Users experience is very important, that's why I use non-intrusive ads. Any feedback is appreciated. Thank you. Justin XoXo :)
# AREA Units Conversionsquare-hectometers to square-miles
1 Square Hectometers
= 0.0038610215859254 Square Miles
Category: area
Conversion: Square Hectometers to Square Miles
The base unit for area is square meters (Non-SI/Derived Unit)
[Square Hectometers] symbol/abbrevation: (hm2, sq hm)
[Square Miles] symbol/abbrevation: (mi2, sq mi)
How to convert Square Hectometers to Square Miles (hm2, sq hm to mi2, sq mi)?
1 hm2, sq hm = 0.0038610215859254 mi2, sq mi.
1 x 0.0038610215859254 mi2, sq mi = 0.0038610215859254 Square Miles.
Always check the results; rounding errors may occur.
Definition:
In relation to the base unit of [area] => (square meters), 1 Square Hectometers (hm2, sq hm) is equal to 10000 square-meters, while 1 Square Miles (mi2, sq mi) = 2589988.11 square-meters.
1 Square Hectometers to common area units
1 hm2, sq hm = 10000 square meters (m2, sq m)
1 hm2, sq hm = 100000000 square centimeters (cm2, sq cm)
1 hm2, sq hm = 0.01 square kilometers (km2, sq km)
1 hm2, sq hm = 107639.15051182 square feet (ft2, sq ft)
1 hm2, sq hm = 15500031.000062 square inches (in2, sq in)
1 hm2, sq hm = 11959.900463011 square yards (yd2, sq yd)
1 hm2, sq hm = 0.0038610215859254 square miles (mi2, sq mi)
1 hm2, sq hm = 15500031000062 square mils (sq mil)
1 hm2, sq hm = 1 hectares (ha)
1 hm2, sq hm = 2.4710516301528 acres (ac)
Square Hectometersto Square Miles (table conversion)
1 hm2, sq hm = 0.0038610215859254 mi2, sq mi
2 hm2, sq hm = 0.0077220431718507 mi2, sq mi
3 hm2, sq hm = 0.011583064757776 mi2, sq mi
4 hm2, sq hm = 0.015444086343701 mi2, sq mi
5 hm2, sq hm = 0.019305107929627 mi2, sq mi
6 hm2, sq hm = 0.023166129515552 mi2, sq mi
7 hm2, sq hm = 0.027027151101477 mi2, sq mi
8 hm2, sq hm = 0.030888172687403 mi2, sq mi
9 hm2, sq hm = 0.034749194273328 mi2, sq mi
10 hm2, sq hm = 0.038610215859254 mi2, sq mi
20 hm2, sq hm = 0.077220431718507 mi2, sq mi
30 hm2, sq hm = 0.11583064757776 mi2, sq mi
40 hm2, sq hm = 0.15444086343701 mi2, sq mi
50 hm2, sq hm = 0.19305107929627 mi2, sq mi
60 hm2, sq hm = 0.23166129515552 mi2, sq mi
70 hm2, sq hm = 0.27027151101477 mi2, sq mi
80 hm2, sq hm = 0.30888172687403 mi2, sq mi
90 hm2, sq hm = 0.34749194273328 mi2, sq mi
100 hm2, sq hm = 0.38610215859254 mi2, sq mi
200 hm2, sq hm = 0.77220431718507 mi2, sq mi
300 hm2, sq hm = 1.1583064757776 mi2, sq mi
400 hm2, sq hm = 1.5444086343701 mi2, sq mi
500 hm2, sq hm = 1.9305107929627 mi2, sq mi
600 hm2, sq hm = 2.3166129515552 mi2, sq mi
700 hm2, sq hm = 2.7027151101477 mi2, sq mi
800 hm2, sq hm = 3.0888172687403 mi2, sq mi
900 hm2, sq hm = 3.4749194273328 mi2, sq mi
1000 hm2, sq hm = 3.8610215859254 mi2, sq mi
2000 hm2, sq hm = 7.7220431718507 mi2, sq mi
4000 hm2, sq hm = 15.444086343701 mi2, sq mi
5000 hm2, sq hm = 19.305107929627 mi2, sq mi
7500 hm2, sq hm = 28.95766189444 mi2, sq mi
10000 hm2, sq hm = 38.610215859254 mi2, sq mi
25000 hm2, sq hm = 96.525539648134 mi2, sq mi
50000 hm2, sq hm = 193.05107929627 mi2, sq mi
100000 hm2, sq hm = 386.10215859254 mi2, sq mi
1000000 hm2, sq hm = 3861.0215859254 mi2, sq mi
1000000000 hm2, sq hm = 3861021.5859254 mi2, sq mi
(Square Hectometers) to (Square Miles) conversions | 1,375 | 3,500 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2020-40 | latest | en | 0.753406 |
https://lessonplanet.com/teachers/area-and-circumference-1 | 1,558,696,158,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232257605.76/warc/CC-MAIN-20190524104501-20190524130501-00543.warc.gz | 539,986,843 | 22,910 | # Area and Circumference 1
##### This Area and Circumference 1 assessment also includes:
Have the class describe the relationship of the area and perimeter of inscribed polygons and the area and circumference of the circle. Pupils investigate the relationships of areas and perimeters of inscribed regular polygons and circles as the number of sides increase. The scholars use the area formula for regular polygons to derive a form of the area formula for a circle.
CCSS: Designed
##### Instructional Ideas
• Have individuals that are struggling with the relationships measure the perimeters and the circumferences
##### Classroom Considerations
• Class members should be familiar with finding the area of a regular polygon
• The resource is the first of three that develop the argument for the formulas of circumference and area of a circle
##### Pros
• Includes samples of individual work
• Provides a description of what to expect at different levels of understanding
• None | 190 | 985 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-22 | latest | en | 0.891393 |
http://mathmisery.com/wp/2019/02/03/simple-but-evil-8-probability-distributions/ | 1,582,648,766,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875146123.78/warc/CC-MAIN-20200225141345-20200225171345-00479.warc.gz | 98,688,084 | 9,652 | # Simple But Evil #8 — Probability Distributions
Whether you’re teaching AP Stats or an introductory or senior level probability course for undergrads, you’ll know that students have a difficult time getting their minds wrapped around arithmetic for probability distributions. Here’s a simple example from the world of continuous probability.
Let $$X, Y$$ be iid normal with mean zero and variance 1. Now we’ll have a question like
What is the distribution of $$Z = X – Y$$?
Now a typical course will have students go through the math symbolism mechanics to eventually arrive at $$Z \sim N(0,\sqrt{2})$$, using the notation $$N(\mu,\sigma)$$ rather than $$\sigma^{2}$$. This is all good and well. But there’s often no visual follow up. Or even a sampling follow up.
Do me one favor, ask your students this simple but evil question.
Prove or disprove. Let $$X, Y$$ be iid normal with mean zero and variance 1. If $$Z_{1} = X – Y$$, $$Z_{2} = Y – X$$, and $$Z_{3} = X + Y$$ then $$Z_{1}$$, $$Z_{2}$$, and $$Z_{3}$$ all follow the same distribution.
I like this question for a lot of reasons. Here are a few main reasons (and there are lots of smaller reasons, too many to enumerate).
• Coursework in a probability course is often focused on transformations of distributions while the comparison of one distribution to another is held off for a statistics course. I see the wisdom in the separation, but that doesn’t mean we can’t start preparing students for what’s to come.
• This question can be a headscratcher because sometimes students want to believe that if “$$X$$ and $$Y$$ are the same thing” (that’s the mistake), then their difference should be zero. Ah, but $$X$$ and $$Y$$ are not the same thing. They just have the same distribution but come from two independent, though identically distributed “streams”. I say “streams” because what’s lost in a probability course is the sampling aspect of all these distributions — the thing left for a stats course. We are drawing from $$X$$ and drawing from $$Y$$. So the first draw from $$X$$ and from $$Y$$ don’t produce identical values. The $$X – Y$$ part of the question helps students to see the difference between that and $$X – X$$ or $$Y – Y$$.
• The second bullet point also helps separate the notion of “equality” when it comes to random numbers / processes. There’s equality in value and then there’s equality in distribution!
• Even if you don’t want to convert your probability course into a stats course, this is a great chance for visualizations!
• I will have a hard time believing that you won’t have a long discussion about this!
• Want a part deux? Compare $$\frac{X}{Y}$$ against $$\frac{Y}{X}$$.
### Some Visualizations
Here are some visualizations. [Also, an end of the post plug: remember to subscribe!]
Ok, first it’s good to just take a look at what $$X$$, $$Y$$, and $$X-Y$$ look like sampled (with a smoothed pdf overlaying them). (The histograms are normalized so they integrate to 1, rather than their height being their frequency.)
Next, here are fits for $$X-Y$$ and $$X+Y$$ based on 10,000 samples.
And finally, tables are visualizations too! ($$X-Y$$) is the “diff” column and ($$X + Y$$) is the “sum” column. | 781 | 3,206 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2020-10 | latest | en | 0.920429 |
http://slideplayer.com/slide/2657808/ | 1,506,433,623,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818695726.80/warc/CC-MAIN-20170926122822-20170926142822-00062.warc.gz | 309,387,295 | 25,066 | # Solving the eValue Rubik’s cube
## Presentation on theme: "Solving the eValue Rubik’s cube"— Presentation transcript:
Solving the eValue Rubik’s cube
The simple 5 step method September 2013
Solving the cube We’re going to help you solve your eValue Rubik’s cube in 5 simple steps: We’ll take you through every move you need to make to get your eValue Rubik’s cube back to its original position. 1. Top edges 2. Top corners 3. Middle 4. Bottom corners 5. Bottom edges Click to move on Back
Preparation – Naming the sides of your cube
We’ll be holding the cube so the white face with the eValue logo is always at the top You’ll need to turn the cube as you go through the solution so the front and other sides will change colours As you hold the cube looking at 3 sides, the names we’ll use for these sides are shown below: “K” for Back “L” for Left side “T” for Top “F” for Front “R” for Right side Click to move on Back “B” for Bottom
Preparation – let’s check the moves
We’re going to use some specific ‘cube moves’ in this solution We’ll use images and letters to represent the moves that look like this: The picture of the cube will show you which side to move: R-C “R” Right “L” Left “F” Front “T” Top “B” Bottom Click to move on Back
Preparation – let’s check the moves
The arrow will show you which way to move it and how much: For each side, the direction for clockwise and anticlockwise are as you look at that particular side. And we’ve included the letters to make it easier to read and remember if you want to Turn twice “2” so the side moves half way round Turn once clockwise “C” Turn once anti-clockwise “A” Continue Back
You’ll see arrows to take you through the instructions. Click on them once you’ve done on that page. During the solution, there will be sometimes be more than one choice so just click on the arrow that’s right for you. Arrows look like this: If you make a mistake there’s a “Whoops” button at the top of each page which will help you go back to any of the previous steps. It looks like this: And if you need to go back to the start of the step you’re on, use the “Back” arrow in the bottom left of each screen. Next Whoops! I’m ready to start Back
Step 1 – check for the top edges
Whoops! Look at the top middle edges of your cube. It doesn’t matter which way round you hold it. Answer the question on the right to continue. Are all the top edge cubes in the right place and the right way round? Yes No Back
Step 1 – find the right cube to put it in position
Whoops! Turn the whole cube so that the front top position does not have a properly positioned cube in it. Find the cube that should be in this Top Front position. It will have the 2 colours for the Top and the Front sides. Click on the cube that shows where your target cube is sitting: Top Front If you can see your target cube immediately click one of these: If you can’t see it immediately, it’ll be at the back of the cube and click one of these instead: Top Top Front Front Front Front Front Front Left Left Top Back Right Front Front Front Front Back
Step 1 – Right Top Make the following moves:
Whoops! Make the following moves: Is your target cube the right way round? R-A F-A Yes No Back
Step 1 – Back Top Make the following moves:
Whoops! Make the following moves: Is your target cube the right way round? T-C R-A T-A F-A Yes No Back
Step 1 – Top Left Make the following moves:
Whoops! Make the following moves: Is your target cube the right way round? L-C F-C Yes No Back
Step 1 – Front Right Make the following move:
Whoops! Make the following move: Is your target cube the right way round? F-A Yes No Back
Step 1 – Front Left Make the following moves:
Whoops! Make the following moves: Is your target cube the right way round? F-C Yes No Back
Step 1 – Back Right Make the following moves:
Whoops! Make the following moves: Is your target cube the right way round? R-2 F-A R-2 Yes No Back
Step 1 – Back Left Make the following moves:
Whoops! Make the following moves: Is your target cube the right way round? L-2 F-C L-2 Yes No Back
Step 1 – Front Bottom Make the following move:
Whoops! Make the following move: Is your target cube the right way round? F-2 Yes No Back
Step 1 – Bottom Right Make the following move:
Whoops! Make the following move: Is your target cube the right way round? B-A F-2 Yes No Back
Step 1 – Bottom Back Make the following move:
Whoops! Make the following move: Is your target cube the right way round? B-2 F-2 Yes No Back
Step 1 – Bottom Left Make the following move:
Whoops! Make the following move: Is your target cube the right way round? B-C F-2 Yes No Back
Step 1 – putting the cube in the right way round
Whoops! To get your cube the right way round, make the following moves: Are all your middle edge cubes now in the right position and the right way round? F-A T-C L-A T-A Yes No Back
Step 2 – Finding the top corners
Whoops! Find a top corner cube, which is not currently in the right place. This is your target cube. It will either be currently found as one of the top 4 corner cubes, or one of the bottom 4 corners Choose the next step by answering the question on the right. Is this target cube on the top or the bottom? Top Bottom Back
Step 2 – the target cube is on the top layer
Whoops! Turn the whole cube so your target cube is at the front right. Make the following moves: The cube should now be on the bottom layer. R-A B-A R-C Continue Back
Step 2 – move the cube to the right place
Whoops! Now rotate the bottom layer until the target cube is under the target position: Make the following moves: The cube should now be in the right position Is this target cube the right way round? Yes, and there are no more to do Yes but there are more to do No R-A B-A R-C Back
Step 2 – rotate the cube until it’s the right way round
Whoops! Make the following moves: If the target cube is still not the right way round repeat the above moves once more. Are there any more top corner cubes to complete? R-A B-2 R-C F-C B-2 F-A Yes No Back
Step 3 – Starting point We’re now going to sort out the middle edges of the cube Continue Back
Step 3 – Middle edge Find a middle edge cube that’s not in the right position and the right way round This cube will either be: on the bottom layer in the middle layer but in the wrong place In the middle layer and in the right place, but the wrong way round Is this target cube in the middle or on the bottom? Bottom Middle but in the wrong place Middle, already in the right place but the wrong way round Back
Step 3 – Put middle edge cube on the bottom layer
Whoops! Hold the whole cube so that the middle edge cube you need to move is towards you on the Front Right: Then make the following moves: This will move the cube you’re looking to place onto the bottom layer R-A B-C R-C B-C F-C B-A F-A Continue Back
Step 3 – Place the middle edge cube
Rotate the bottom layer so that the colour of the middle edge cube that’s not on the bottom is next to the same colour as the centre cube Hold the cube so that the matching edges gives one of the following: Where is your middle edge cube now? It’s on the right hand side Top Top Front Front It’s on the front Back
Step 3 – move the cube up from the right hand side
Whoops! Make the following moves Then the following moves: You should now have the middle edge cube in the right place and the right way round: Are there any more middle edge cubes to put in the right place? B-C F-C B-A F-A Yes B-A R-A B-C R-C No Back
Step 3 – move the cube up from the front
Whoops! Make the following moves Then the following moves: You should now have the middle edge cube in the right place and the right way round: Are there any more middle edge cubes to put in the right place? B-A R-A B-C R-C Yes B-C F-C B-A F-A No Back
Step 4 – Bottom corners We’re now going to sort out the bottom corners of the cube Continue Back
Step 4 - Check of now many cubes are in the right place
Pick a corner and rotate the bottom until that corner is in the right place. It doesn’t matter if it’s not the right way round. You’ll find that once this cube is in the right place, you’ll have one of the following: All the bottom corner cubes are in the right place There are 2 out of place cubes, and they’re next to each other There are 2 out of place cubes, and they’re diagonally opposite Front Right Front Right Bottom Bottom I have this I have this I have this Back
Step 4 – Switching cubes next to each other
Whoops! Hold the cube so the 2 out of place cubes are both on the front of the cube Make the following moves: Then the following moves: R-A B-A R-C F-C B-C F-A R-A B-C R-C B-2 Continue Back
Step 4 – Switching cubes diagonally opposite
Whoops! Hold the cube so that one of the out of place cubes is Front Left, and the other is Back Right Make the following moves: Then the following moves: R-A B-A R-C F-C B-2 F-A R-A B-C R-C B-2 Continue Back
Step 4 – Position bottom edge cubes the right way round
The pattern on the bottom will now look like one of the following. Turn the cube round to make the Front side match the right diagram for you Front Right Bottom Front Right Bottom Front Right Bottom Front Right Bottom Front Right Bottom Front Right Bottom Front Right Bottom Continue Back
Step 4 – putting the corner cubes the right way round
Whoops! Make the following moves: Then make the following moves: If the corner cubes are still not the right way round you may have to repeat this set of moves up to 3 times, repositioning the Front side each time. Are the bottom corners now the right way round? R-C B-A R-C B-A Yes R-A B-2 R-C B-2 No Back
Step 5 – bottom edges We’re now going to sort out the bottom edges of the cube – the last step! How many of the bottom edges are already in the right place? None Just one All 4 Back
Step 5 – No edges are in the right place
Whoops! Make the following moves (it doesn’t matter which side is the Front): Then do the following moves: This will make one edge cube be in the right place L-A R-C F-C L-C R-A B-2 L-A R-C F-C L-C R-A Continue Back
Step 5 – One edge cube is in the right place
Whoops! Turn the whole cube so the one bottom edge cube which is in the right place is at the Front. Then make the following moves: Then do the following moves: All 4 cubes are now in the right place. L-A R-C F-C L-C R-A B-2 L-A R-C F-C L-C R-A Continue Back
Step 5 - Check which way round the bottom edge cubes are
You should now have one of the following patterns on the bottom of the cube. Turn the whole cube so the cube has the same Front as shown. Front Right Bottom Front Right Bottom Front Right Bottom I have this I have this I have this Back
Step 5 – checked pattern Make the following moves:
Whoops! Make the following moves: Then the following moves: And finally, the following moves: L-A R-C F-2 L-C R-A B-2 B-2 L-A R-C F-C L-C R-A L-A R-C F-2 L-C R-A B-A Finished Back
Step 5 – “H” pattern Whoops! Make the following moves, the top of the H as the Front of the cube: Then the following moves: And finally, the following moves: L-A R-C F-C L-C R-A B-C L-A R-C F-C L-C R-A B-C B-C L-A R-C F-2 L-C R-A Nearly there Back
Step 5 – “H” pattern continued
Whoops! Make the following moves: And finally, the following moves: L-A R-C F-C L-C R-A B-C L-A R-C F-C L-C R-A B-2 Finished Back
Step 5 – “Pointing” pattern
Whoops! Hold the cube so that the “point” is Front and to the Right. Then make the following moves: Then make the following moves: And finally, make the following moves: L-A R-C F-C L-C R-A B-A L-A R-C L-C R-A F-A B-A L-A R-C F-2 L-C R-A Nearly there Back
Step 5 – “Pointing” pattern continued
Whoops! Turn the whole cube so the one bottom edge cube which is in the right place is at the Front. Then make the following moves: Then do the following moves: L-A R-C F-C L-C R-A B-2 L-A R-C F-C L-C R-A Finished Back
Congratulations! Well done - you have solved your cube!
Tell everyone within shouting distance!
Whoops! Don’t worry if you’ve made a mistake. You’ll still be able to solve the cube. If you aren’t able to undo the moves you’ve just done, you’ll need to start from an earlier step and work through again. Click on the arrow below depending on which step has become undone during your moves 1. Top edges 2. Top corners 3. Middle 4. Bottom corners 5. Bottom edges This one This one This one This one This one | 3,114 | 12,330 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2017-39 | longest | en | 0.86833 |
https://www.onlinemath4all.com/substitution-method-questions-6.html | 1,623,962,958,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487633444.37/warc/CC-MAIN-20210617192319-20210617222319-00396.warc.gz | 846,936,753 | 13,833 | # SUBSTITUTION METHOD QUESTIONS 6
In this page substitution method questions 6 we are going to see solution of first question in the worksheet of substitution method.
What is substitution method ?
Solving system of equation by substitution method, involves solving any one of the given equation for either 'x' or 'y' and plugging that in the other equation and solve that equation for another variable.Substitution method questions 2
Step 1 :
Solve any one of the equations either x = or y =
Step 2 :
Substitute the value that we got from step 1 in the other equation.
Step 3 :
Now we have got the value of any one of the variables x or y.
Step 4 :
Apply this value in step 1 in order to get the value of other variable.
Let us see the following example problem to understand the substitution method. Substitution method questions 6
## Substitution Method Questions 6
Question 6 :
Solve the following equations by substitution method
−3x − 3y = 3 and y = −5x − 17
Solution :
−3x − 3y = 3 -------(1)
y = −5x − 17 -------(2)
Substitute y = -5x - 17 in the first equation
−3x − 3(-5x - 17) = 3
-3x + 15x + 51 = 3
12x + 51 = 3
Subtract by 51 on both sides
12x + 51 - 51 = 3 - 51
12x = -48
Divide by 12 on both sides
12x/12 = -48/12 ==> x = -4
Applying x = -4 in the second equation
y = -5(-4) - 17
y = 20 - 17
y = 3
Hence x = -4 and y = 3 is the solution. Substitution method questions 6
## More practice questions
Question 1 :
Solve the following equations by substitution method
5 x - 3 y - 8 = 0 and 2x - 3 y - 5 = 0
Solution
Question 2 :
Solve the following equations using substitution method
5x - 3y - 8 = 0 and 2x - 3y - 5 = 0
Question 3 :
Solve the following equations by substitution method
y = 6x - 11 and -2x - 3y = -7
Solution
Question 4 :
Solve the following equations by substitution method
2x − 3y = −1 and y = x − 1
Solution
Question 5 :
Solve the following equations by substitution method
y = −3x + 5 and 5x − 4y = −3
Question 6 :
Solve the following equations by substitution method
−3x − 3y = 3 and y = −5x − 17
Solution
Question 7 :
Solve the following equations by substitution method
y = 5x − 7 and −3x − 2y = −12
Solution
Question 8 :
Solve the following equations by substitution method
−4x + y = 6 and −5x − y = 21
Solution
Question 9 :
Solve the following equations by substitution method
2x + y = 20 and 6x − 5y = 12
Solution
Question 10 :
Solve the following equations by substitution method
y = −2 and 4x − 3y = 18
Solution
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Aptitude Test | 1,438 | 5,450 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.84375 | 5 | CC-MAIN-2021-25 | latest | en | 0.676524 |
https://opensourceprojects.org/lets-get-javascript-maths-functions-and-operators/ | 1,553,517,033,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912203947.59/warc/CC-MAIN-20190325112917-20190325134917-00136.warc.gz | 574,454,573 | 28,438 | ## Getting began!
Welcome to a different put up on Code The Net! Initially, I wish to encourage you to observe alongside on this article. It can make it easier to be taught higher, and in addition make it easier to to recollect what you’ve gotten performed. Let’s begin by making a brand new HTML file with a `<script>` tag in it:
As soon as that’s performed, open it up in your net browser and also you’re able to go! (don’t neglect to save lots of and reload the web page each time you make a change)
## Sorts of numbers
There are two predominant kinds of numbers in JavaScript: floats and integers. Integers are complete numbers with out decimals. Listed below are just a few examples:
• `three`
• `zero`
• `156`
Floats are numbers which comprise a decimal. You will need to observe that floats may be complete numbers. Wait whaaat? I believed you stated that integers had been complete numbers? Nicely, stuff like `5.zero` remains to be thought-about a float, as a result of it has a decimal and might be `5.1`.
• `2.225345`
• `zero.zero`
• `42.zero`
For essentially the most half, you gained’t have to fret about these differing kinds as a result of JavaScript will mechanically convert them for you (????). Nonetheless, in some programming languages, you’ll have to do it your self. There are additionally some circumstances in JavaScript the place you will must work with changing stuff between floats and integers.
## Fundamental operators
Let’s begin proper from the start – with the fundamental operations!
Addition in JavaScript is actually easy – all that you must do is put a plus signal between two numbers, identical to in actual life. Strive it out! Add the next to your script, save, then reload the web page:
It’s also possible to add floats and integers within the one expression:
Clearly, the quantity that’s returned will likely be a float. Shifting on!
### Subtraction
Subtraction additionally works simply because it does in actual life. Simple, huh? Listed below are some examples – you may strive them out if you need:
### Multiplication
Multiplication is barely totally different to how you’ll do it on paper. Usually, we might simply use a cross image – the letter `x` on a keyboard. Nonetheless, we are able to’t use that! In programming, we attempt to give every character just one which means. Since `x` is already a letter, we’ve got to make use of one thing else. In JavaScript, we use the asterisk (`*`) image. Listed below are some examples of multiplication in JavaScript:
### Division
Division additionally works in another way to on paper. Whereas there’s a Unicode symbol for division (÷), it isn’t one thing which you could sort simply in your keyboard and isn’t that generally used. As an alternative, we use the slash (`/`) signal to imply ‘divided by’. Listed below are some examples:
I simply wish to spotlight the final one on that checklist:
In maths, something divided by zero is the same as infinity (explanation here). Nonetheless, in JavaScript it may well’t equal to “`infinity`” – in any other case JavaScript would suppose it was a variable! So, we have to capitalize it to make it `Infinity`. This can be a particular worth in JavaScript (not a variable). It doesn’t actually have many use circumstances however is the result of statements just like the one above.
Enjoyable reality: `Infinity - Infinity` in JavaScript does not equal `0`!
### Modulo
Modulo is one thing that you could be not have heard of earlier than, not like the 4 operations above. Put merely, modulo is the the rest when dividing two numbers. It’s performed by placing a `%` signal between the 2 numbers. For instance:
Let’s break it down a bit additional. We have now the numbers `24` and `5`, separated by the modulo (`%`) signal. Which means to calculate the reply in our heads, we’d first must divide `24` by `5` (into teams of 5). Right here, we are able to make 4 teams of 5. Nonetheless, we’d nonetheless have an additional `Four` left over as a the rest. So, the reply is `Four`. Listed below are another examples – should you nonetheless don’t get it, attempt to do the method above on these:
## `Math` capabilities
In addition to the operators above, there are additionally some capabilities that we are able to use to hold out barely extra superior duties. These capabilities usually observe the type of `Math.whateverTheFunctionIs()`. It is because `Math` is an object containing a lot of totally different math-related capabilities. I’ll speak extra about objects in a later article, however you don’t actually have to fret about the way it works for the second. Simply use the syntax that I put right here and also you’ll be advantageous.
### To the ability of
We do that utilizing the `Math.pow(quantity,energy)` operate. For instance, let’s say we wished to sq. the quantity `5`:
Wait whaaat? A operate inside the `alert` operate? Yup. It is because `Math.pow` is a operate that returns something. You may consider it identical to a variable. So as a substitute of `x` being equal to `25`, `Math.pow(5,2)` is the same as `25`.
It’s also possible to go to larger powers if you need (pun meant ???? *sigh*):
### Sq. root & Dice root
You may calculate sq. an dice roots in JavaScript utilizing the `Math.sqrt()` and `Math.cbrt()` capabilities. Listed below are some examples which you’ll check out:
### Rounding
We are able to spherical a quantity to an entire quantity utilizing the `Math.spherical()` operate. Listed below are some examples:
It can spherical up if the decimal a part of the quantity is bigger than or equal to `zero.5`. In any other case it’s going to spherical down.
### Particularly rounding up or down
However what if we wish to particularly spherical up or down? For instance, what if earlier than we wished to spherical `35.82562` downwards? That is the place flooring and ceiling turn out to be useful. `Math.flooring()` rounds the quantity down it doesn’t matter what, whereas `Math.ceil()` rounds the quantity up it doesn’t matter what. Word that `Math.ceil(6)` and even `Math.ceil(6.zero)` wouldn’t spherical as much as `7`. Listed below are some examples:
## Conclusion
That’s it for right this moment! This simply lined some primary math operations, however there are numerous extra. Click here for an inventory of all of the capabilities within the `Math` object (those that begin with `Math.`). In addition to having capabilities, the `Math` object additionally has some static values resembling `Math.PI` – they’re listed on that web page as nicely.
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Code The Web | 1,832 | 8,274 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2019-13 | longest | en | 0.915235 |
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Wherefore, if a straight line, &c. QED PROPOSITION XXVIII. THEOREM. If a straight line falling upon two other straight lines, make the exterior angle equal to the interior and opposite upon the same side of the line ; or make the interior angles upon...
Elements of geometry, based on Euclid, books i-iii - Side 32
av Edward Atkins - 1876 - 119 sider
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## The Elements of Euclid: Viz. the First Six Books, Together with the Eleventh ...
Euclid, Robert Simson - 1806 - 518 sider
...straight line, &c. QED PROP. XXVIII. THEOR. IF a straight line falling upon two other straight lines makes the exterior angle equal to the interior and opposite upon the same side of the line ; or makes the interior angles upon the same side together equal to two right angles ; the two straight...
Uten tilgangsbegrensning - Om denne boken
## The British encyclopedia, or, Dictionary of arts and sciences, Volum 5
William Nicholson - 1809
...the exterior angle equal to the interior and opposite angle on the same side ; and likewise, the two interior angles upon the same side, together, equal to two right angles. If AB, (fig. 5) be parallel to С D, and EF cut them in the points HG, then the angle AHG equals the...
Uten tilgangsbegrensning - Om denne boken
## The British Encyclopedia, Or Dictionary of Arts and Sciences ..., Volum 5
William Nicholson - 1809
...the exterior angle equal to the interior and opposite augle on the same side ; and likewise, the two interior angles upon the same side, together, equal to two right angles. If AB, (fig. 5) be parallel to CD, and £ F cut them in the points HG, then the angle AHG equals the...
Uten tilgangsbegrensning - Om denne boken
## The Elements of Euclid: Viz. the First Six Books, Together with the Eleventh ...
Euclid - 1810 - 518 sider
...CD. Wherefore, if a straight line, &c. QED PROP. XXVIII. THEOR. IF a straight line falling upon two other straight lines make the exterior angle equal...interior and opposite upon the same side of the line; or makes the interior angles upon the same side together equal to two right angles; the two straight lines...
Uten tilgangsbegrensning - Om denne boken
## Pantologia. A new (cabinet) cyclopædia, by J.M. Good, O. Gregory ..., Volum 5
John Mason Good - 1813
...line fall upon two parallel straight lines, it makes the alternate angles equal to one another; and the exterior angle equal to the interior and opposite upon the same side; and likewise the two interior angles upon the same side together equal to two right ansies. Prop. XXX....
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## The Elements of Euclid: Viz. the First Six Books, Together with the Eleventh ...
Euclides - 1816 - 528 sider
...line, &c. QED PROP. XXVIII. THEOR. IF a straight line falling upon two other straight lines, makes the exterior angle equal to the interior and opposite upon the same side of the line; or makes the interior angles upon the same side together equal to two right angles ; the two straight...
Uten tilgangsbegrensning - Om denne boken
## Pantologia. A new (cabinet) cyclopædia, by J.M. Good, O. Gregory ..., Volum 5
John Mason Good - 1813
...these two straight lines »hall be parallel. Prop. XX V 111. Theor. If a straight line falling upon two other straight lines make« the exterior angle equal...interior and opposite upon the same side of the line; or makes the interior angles upon i he same side together equal to two rirhc angles ; the two straight...
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## Elements of Geometry: Containing the First Six Books of Euclid, with a ...
John Playfair - 1819 - 333 sider
...exterior angle equal to the interior and opposite upon the same side of the line; or makes the imerior angles upon the same side together equal to two right angles ; the two straight lines are parallel to one another. Let the straight line EF, which falls upon the two straight lines AB,...
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## Elements of Geometry: Containing the First Six Books of Euclid: With a ...
John Playfair - 1819 - 317 sider
...line fall upon two parallel straight lines, it makes the alternate angles equal to one another ; and the exterior angle equal to the interior and opposite upon the same side ; and likewise the too interior angles upon the same side together equal to two right angles. Let the...
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## The Elements of Euclid: Viz. the First Six Books, Together with the Eleventh ...
Euclid, Robert Simson - 1821 - 516 sider
...line fall upon two parallel straight lines, it makes the alternate angles equal to one another; and the exterior angle equal to the interior and opposite upon the same side; and likewise the two interior angles upon the same side together equal to two right angles.* Let the...
Uten tilgangsbegrensning - Om denne boken | 1,228 | 4,943 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2020-34 | latest | en | 0.609187 |
https://www.daniweb.com/programming/computer-science/threads/488008/help-with-flowchart-from-my-algorithm | 1,547,629,975,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583657097.39/warc/CC-MAIN-20190116073323-20190116095323-00020.warc.gz | 769,486,261 | 12,292 | i have my algorithm but i can't get the flowchart to come out.
``````Main()
Declare Sticker_Price as Real
Declare Discount_Percent as Real
Declare Tax_Percent as Real
Declare Tag_Color as Character
Declare Final_Price as Real
Set Tax_ Percent = 0.07
Final_Price = 0.00
Repeat
Process Get_Price(Sticker_Price)
Process Discount(Tag_Color, Discount_Percent)
Process Calc_Final(Sticker_Price, Tax_Percent, Discount_Percent, Final_Price)
Repeat
Print "Do you want to enter another item (y,n)"
If Answer != 'y' or 'n'
Print "That letter is invalid"
End If
Until Answer == āyā or ānā
Print "The total is \$" + Final_Price
End
Get_Price(Price as Real&)
Repeat
Print "Enter sticker price of item"
Input Price
If Price <0.01
Print "Please enter a price greater than \$0.00"
End if
Until Price >0.00
Exit
Discount(Tag_Color as Character, Discount_Percent as Real&)
Repeat
Print "Enter the first letter of the sales tag color as UPPERCASE (W - B - O - R)"
Input Tag_Color
Case of Tag_Color:
= 'W':
Discount_Percent = 0.00
= 'B':
Discount_Percent = 0.25
= 'O':
Discount_Percent = 0.50
= 'R':
Discount_Percent = 0.75
Otherwise
Print "That is a invalid tag color"
Discount_Percent = -1
End of Case
Until Discount_Percent != -1
Exit
Calc_Final(Sticker_Price as Real, Tax_Percent as Real, Discount_Percent as Real, Final_Price as Real&)
Declare Item_Price as Real
Declare Discount_Amount as Real
Declare Tax_Amount as Real
Discount_Amount = Sticker_Price * Discount_Percent
Tax_Amount = Sticker_Price * Tax_Percent
Discount_Price = Sticker_Price - Discount_Amount
Item_Price = Discount_Price + Tax_Amount
Print "The Item price is \$" + Item_Price + "that is with \$" + Tax_Amount + "and a \$" +
Discount_Amount + "Discount"
Final_Price = Final_Price + Item_Price
Exit
``````
i have my algorithm but i can't get the flowchart to come out.
What does that mean? Are you using a program to create a flow chart for you?
Anyway, here are my observations...
1) Line 15, shouldn't it be `Set Final_Price = 0.00`?
2) Lines 18, 19, and 20, do you really expect to pass variable by references? It is uncommon practice in this case for what you are dong (common practice: prefer a returned value from the function instead).
i thought he wants to convert the program to a flowchat
Oh? Then it is easy, but I am not going to do it for him. :P
Look at examples using case statements. | 656 | 2,378 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2019-04 | latest | en | 0.888124 |
https://ncatlab.org/nlab/show/directed%20topological%20space | 1,708,896,915,000,000,000 | application/xhtml+xml | crawl-data/CC-MAIN-2024-10/segments/1707947474643.29/warc/CC-MAIN-20240225203035-20240225233035-00269.warc.gz | 416,636,333 | 18,701 | Contents
Contents
Idea
A directed topological space is a topological space $X$ in which there is some ‘sense of direction’. This can happen in various different ways and the level of the ‘directedness’ can be different in different situations, so naturally there are several ‘competing’ ideas, but the beginning of a consensus on what the overarching idea is.
If one bases homotopy theory on the idea of a singular simplex or more generally a singular cell of any shape, then there is no way in which a ‘sense of direction’ can be encode. If we have a path in a space we can go along it (traverse it) in either direction, from 0 to 1 or from 1 to 0. From this perspective a directed space is one in which not every singular cell $\Delta^n \to X$ (for $\Delta^n$ the standard topological simplex) is supposed to be traversable in all directions, in some sense: instead these $k$-dimensional paths may have a direction .
As an example one can base the ‘sense of direction’ on a closed preorder or partial order, (that is a pospace),so that the paths from the directed interval $[0,1]$ with the usual order to the space $X$, can only be ‘traversed’ in one direction. Another example which does not fit into this first type would be the directed circle.
In other words, a circle with direction determined by the anticlockwise sense. Again it is easy to see that there are certain paths that respect the direction, ‘directed paths’ whilst others do not.
So far there exists a well-developed theory for a notion of directed spaces $X$ where 1-dimensional paths given by maps $[0,1] \to X$ from the interval into the space are equipped with a direction. See in particular the book by Marco Grandis on Directed Algebraic Topology listed below. Another suggested notion for modelling directed spaces is that of framed spaces, which is tailored towards certain higher categorical applications.
Note that a directed space is like a generalised space; not every directed space need be a space in the traditional sense, in accordance with the red herring principle. As an instance of this, note that Marco Grandis in his book Directed Algebraic Topology handles the directed homotopy of small categories, and of cubical complexes, since this is useful for comparison an interpretation of directed homotopy ‘invariants’.
Directed spaces are studied in directed homotopy theory, a relatively young topic. In generalization of how a topological space has a fundamental groupoid, a directed space has a fundamental category.
Homotopy-theoretic perspective
From a homotopy theoretic perspective one would wish that notions of directed spaces might serve to generalize the homotopy hypothesis – which identifies ordinary (undirected) topological spaces with ∞-groupoids, i.e., with (∞,0)-categories – to a more general context where (∞,0)-categories are generalized to (∞,r)-categories with $r \gt 0$:
An (∞,r)-category in this context might correspond to a $r$-directed topological space , one that comes equipped with a notion of orientation of its $k$-cells for $0 \leq k \leq r$, but was impartial on direction above that dimension.
If such a definition exists, it may need to use filtered topological spaces instead of bare topological spaces.
Even in the absence of a homotopy-theoretic definition of $r$-directed space in this sense, from the perspective of homotopy theory one might take the standpoint of the homotopy hypothesis and define a (nice) $r$-directed space to be an (∞,r)-category, just as it makes good sense and is nowadays common practice in algebraic topology to define a nice topological space to be an ∞-groupoid.
See (n,r)-category for more on that.
Variants
$d$-Spaces
A directed topological space or d-space is pair $(X, d X)$ consisting of a topological space $X$ and a subset $d X \subset C(I,X)$ of continuous maps from the interval $I = [0,1]$ into $X$ – called directed paths or d-paths – satisfying the following conditions:
1. (constant paths) every constant map $I\to X$ is directed,
2. (reparametrisation) $dX$ is closed under composition with increasing maps $I\to I$,
3. (concatenation) $dX$ is closed under path-concatenation: if the d-paths $a, b$ are consecutive in $X$ $(a(1) = b(0))$, then their ordinary concatenation $a+b$ is also a d-path
$(a+b)(t) = a(2t),\,\text{if}\, 0\le t\le \frac{1}{2},$
$(a+b)(t) = b(2t-1),\,\text{if}\, \frac{1}{2}\le t\le 1.$
A morphism of directed topological spaces $f : (X, d X)\to (Y , d Y)$ is a morphism of topological spaces $f: X \to Y$ which preserves directed paths in that for every $\gamma: I \to X$ in $d X$ the path $f_* \gamma : I \stackrel{\gamma}{\to} X \stackrel{f}{\to} Y$ is in $d Y$.
Example
Basic examples of $d$-spaces include:
• The standard directed interval is $I_d = ([0,1], d I)$ with $d I$ the set of all monotonic continuous maps $[0,1] \to [0,1]$ is a d-space
• Any pospace $X$ gives rise to a d-space by taking the directed paths to be, well, directed paths, i.e. continuous order-preserving maps from $I_d$ to $X$.
Many other example can be found in the references.
Streams
A different definition comes from Sanjeevi Krishnan, A Convenient Category of Locally Preordered Spaces, Applied Categorical Structures, 2009, vol. 17, no 5, p. 445-466 (arxiv):
Definition A stream is a tuple $X, \leq_{-}$, where $\leq_{-}$ assigns to each open subset $U \subset X$ a preorder $\leq_U$, such that:
$\leq_{\bigcup_i U_i} = \bigvee_i \leq_{U_i}$
Here, $U_i, i \in I$ is a collection of open sets, and $\bigvee$ is pointwise or of relations.
Remarks
• Morally, a point in a stream is less or equal to another point if it is less or equal in any open set. The relation $\leq_X$ for the whole space does not hold much information.
• In the typical example of the clockwise oriented circle, each point is less or equal to every other one in the relation $\leq_{S^1}$; but for a contractible subset $U \subset S^1$, a point $x$ is less or equal than $y$ if $y$ can be reached from $x$ in a monotonous clockwise path.
• The defining equation of a stream can be thought of as a (co)sheaf condition, and there is indeed a cosheafification result.
• Every d-space gives rise to a stream.
• The category of streams has good properties. In particular, there is an further notion of compactly flowing streams_ extending the notion of compactly generated Hausdorff spaces, and indeed the forgetful functor creates limits and colimits.
• The category of compactly flowing streams is Cartesian closed.
Framed spaces
Motivation
Another way to endow spaces with directions is via framings, i.e. choices of a “basis of the vector space of tangential directions at each point”. Somewhat abstracting this idea, frames may also be thought about in terms of projections, as the following remark motivates.
Remark
Let $V$ be an $n$-dimensional vector space with an inner product $g$. The following structures on $V$ are equivalent.
• An orthonormal frame of $V$, i.e. an ordered sequence of vectors $v_i$, $1 \leq i \leq n$, such that $g(v_i,v_j)=\delta_{ij}$.
• A sequence of surjections $V_i \to V_{i-1}$, $1 \leq i \leq n$, where $V_i$ is an oriented $i$-dimensional vector space (and $V_n = V$).
The two structures are related by setting $v_i$ to be the unit vector spanning $\ker(V_i \to V_{i-1})$ such that $V_{i-1} \oplus v_i$ recovers the orientation of $V_i$ (note that all $V_i$ canonically embed in $V$ as the orthogonal complement of the kernel of the composite map $V \to V_i$).
In the absence of inner products, one cannot speak of orthonormal frames any longer. However, sequences of projections can still be defined, and may be regarded as playing the role of “metric-free orthonormal” frames. (A vaguely analogous line of thinking is that a Morse function $M \to \mathbb{R}$ provides useful “direction” information on $M$, e.g. for the construction of handlebodies, that is ultimately independent from any chosen metric on $M$.) Moreover, this approach offers room for generalization by varying the length of the projection sequence and the vector space dimensions.
Example
The standard orthonormal frame of $n$-dimensional euclidean space $\mathbb{R}^n$ consists for the ordered sequence of vectors $e_1 = (1,0,...,0)$, $e_2 = (0,1,0,...,0)$, …, $e_n = (0,...,0,1)$. By the previous remark, this orthonormal frame is equivalently described by the sequence of projections $\pi_i : \mathbb{R}^i = \mathbb{R}^{i-1} \times \mathbb{R} \to \mathbb{R}^{i-1}$ (each $\mathbb{R}^i$ being endowed with standard orientation).
When forgetting basepoints, then the previous remark and example equally apply to affine spaces, now endowing each point in the space with a basis of frames. Using affine standard framed $\mathbb{R}^n$ (or rather compact contractible patches of it, that interact nicely with the projections, see below) as our “local models” for framed spaces one may define global framed spaces and their maps.
Definition
Terminology
Inductively in $n \in \mathbb{N}$, an $n$-framed patch $U \subset \mathbb{R}^n$ is a non-empty subspace of $\mathbb{R}^n$ with the property that its projection $\pi_n(U)$ is an $(n-1)$-framed patch, such that $\pi_n : U \to \pi_n(U)$ has fibers $\pi^{-1}_n(x)$ of the form $[\gamma_-(u),\gamma_+(u)]$ for two continuous sections $\gamma_\pm : \pi_n(U) \to \pi_n(U) \times \mathbb{R}$. Given two $n$-framed patches $U$ and $V$, a (partial) $n$-framed patch map $F : U \to V$ is a (partial) continuous map which descends along $\pi_n$ to a (partial) $(n-1)$-framed patch map $F_{n-1} : \pi_n (U) \to \pi_n(V)$ such that mappings of fibers $F : \pi^{-1}_n(x) \to \pi^{-1}_n(F_{n-1}(x))$ are monotone.
Note that $n$-framed patches are compact and contractible spaces.
Example
The standard example of an $n$-framed patch is the closed $n$-cube $\mathbf{I}^n = [-1,1]^n \subset \mathbb{R}^n$. However, in general $n$-framed patches need not be “$n$-dimensional”: for instance, the $0$th slice $\{0\} \times \mathbf{I}^{n-1}$ of the $n$-cube is itself an $n$-framed patch, and so are the “$k$-directed intervals” $\mathbf{I}_k := \{0\}^{k-1} \times \mathbf{I} \times \{0\}^{n-k}$.
Definition
Let $X$ be a topological space. Fix $n \in \mathbb{N}$.
• An $n$-framed chart $(U,\gamma)$ in $X$ is an embedding $\gamma : U \hookrightarrow \mathbb{R}^n$ of a subspace $U \subset X$ whose image $\im(\gamma)$ is an $n$-framed patch.
• Two $n$-framed charts $(U,\gamma)$, $(V,\rho)$ in $X$ are compatible if $\rho \circ \gamma^{-1}$ is a partial map of $n$-framed patches.
An $n$-framed space is a space $X$ endowed with an $n$-framing structure $\mathcal{A}$, which is an “atlas” of compatible $n$-framed charts $\{(U_i,\gamma_i)\}$ such that $U_i$ are a locally finite cover of $X$.
Remark
The condition for covers to be locally finite is convenient as it describes the situation of locally finite cell complexes. However, the condition could be replaced, and the definition generalized, in several ways. The above version of the definition should be considered a first approximation to a potentially more general notion of framed spaces.
Maps of framed spaces could be defined along the following lines.
Definition
Given spaces with $n$-framing structure $(X,\mathcal{A}) \to (Y,\mathcal{B})$ then a framed map $F : X \to Y$ is a map such that for any charts $(U,\gamma) \in \mathcal{A}$ and $(V,\rho) \in \mathcal{B}$, $F$ yields a partial map $\rho \circ F \circ \gamma^{-1} : \mathrm{im}(\gamma) \to \mathrm{im}(\rho)$ of $n$-framed patches.
Framed spaces and framed maps are, in a sense, “very rigid” variants of directed spaces. Nonetheless, they are interesting to study as they turn out to have a rich combinatorial theory associated to them. This combinatorial counterpart is particular useful when translating between the topology of directed spaces and the combinatorics of higher categories: an example of this is the definition of manifold diagrams in the language of framed spaces.
Example
Realizations of geometric computads (resp. their functors) have the structure of framed spaces (resp. of framed maps), see Dorn23.
Comparison to $d$-spaces
Remark
One may want to define directed paths in an $n$-framed space as maps from “the framed interval” into that space: but, in fact, there are now $n$ different framed intervals, corresponding to the $n$ possible directions of $n$-framings. These are precisely modelled by the $n$-patches $\mathbf{I}_k$ mentioned in a previous example. Note, there is a framed bijection $\mathbf{I}_k \to \mathbf{I}_j$ if and only if $k \leq j$ (in particular, directions aren’t interchangeable at all: analogously, note that in an n-category a $(n-k)$-morphisms can be (possibly degenerate) $(n-j)$-morphisms only if $k \leq j$). Given a framed space $(X,\mathcal{A})$, and defining $d_k X = \mathrm{Map}_{\mathrm{fr}}(\mathbf{I}_k, (X,\mathcal{A}))$, we thus obtain a filtration of directed paths:
$d_n X \subset d_{n-1} X \subset ... \subset d_1 X \subset \Map(\mathbf{I},X)$
instead of a single path subset $dX \subset \Map(\mathbf{I},X)$, as required in the definition of $d$-spaces.
Remarks
• If we can equip directed spaces with an internal hom, then a directed space with at least one directed path should be a strictly directed object in the category of directed spaces, with respect to the standard directed interval as the interval object, while an ordinary topological space regarded as a directed space should be an undirected object.
References
The above definition of $d$-spaces is from
This has now developed into a book
A discussion of reparameterization of directed paths in directed topological spaces is in
• Ulrich Fahrenberg and Martin Raussen, Reparametrizations of Continuous Paths (arXiv, (blog))
Further references are given in directed homotopy theory.
Some ‘local combinatorial’ aspects of framed spaces are discussed in:
A global (higher-categorical) perspective on directed spaces is taken in:
Last revised on May 13, 2023 at 12:30:45. See the history of this page for a list of all contributions to it. | 3,836 | 14,052 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 170, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-10 | latest | en | 0.949035 |
https://www.physicsforums.com/threads/magnetic-field-of-gyromotion.790116/ | 1,701,588,058,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100489.16/warc/CC-MAIN-20231203062445-20231203092445-00016.warc.gz | 1,064,455,281 | 15,937 | # Magnetic field of gyromotion
• TESL@
In summary, the conversation discusses the concept of Lorentz force and its effect on the movement of electrons in a magnetic field. The speaker also raises a question about the behavior of a magnet entering a coil and the application of Lenz's law. They mention the possibility of an equilibrium and the calculation of the induced magnetic field. The responder adds that Lenz's law also applies to a magnet moving into a coil, causing a counter emf.
#### TESL@
Hello,
I am stuck with this problem:
An electron beam is injected perpendicular to a magnetic field. The electrons feel Lorentz force and start to revolve. This movement reduces the magnetic field, therefore the gyroradius gets higher, which in turn increases the magnetic field again. So the electrons again get closer, and they keep oscillating. This seems wrong. I have probably made a wrong assumption.
This also applies to a magnet entering a coil. As the magnetic field strength increases inside the coil, a current is driven "resisting" the increment. So does the field strength remain zero (the magnet is still moving), drop to a constant value, or oscillate like I mentioned above?
Thank you.
up!
TESL@ said:
Hello,
I am stuck with this problem:
An electron beam is injected perpendicular to a magnetic field. The electrons feel Lorentz force and start to revolve. This movement reduces the magnetic field, therefore the gyroradius gets higher, which in turn increases the magnetic field again. So the electrons again get closer, and they keep oscillating. This seems wrong. I have probably made a wrong assumption.
There would be some equilibrium. An electric current produces it's own magnetic field. One could calculate a magnetic field due the current of an electron beam and compare the induced field with the applied/imposed external field.
This also applies to a magnet entering a coil. As the magnetic field strength increases inside the coil, a current is driven "resisting" the increment. So does the field strength remain zero (the magnet is still moving), drop to a constant value, or oscillate like I mentioned above?
Thank you.
Lenz's law applies to a magnet moving into a coil (solenoid). There would be backward emf, or counter emf due to the magnetic field of the solenoid current.
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/farlaw.html#c2
When an emf is generated by a change in magnetic flux according to Faraday's Law, the polarity of the induced emf is such that it produces a current whose magnetic field opposes the change which produces it.
vanhees71
Thanks Astronuc.
## 1. What is the magnetic field of gyromotion?
The magnetic field of gyromotion refers to the magnetic field that is generated by a moving charged particle as it rotates or orbits around a central axis. This field is perpendicular to both the particle's velocity and the axis of rotation, and its strength is dependent on the particle's velocity, charge, and mass.
## 2. How is the magnetic field of gyromotion calculated?
The magnetic field of gyromotion can be calculated using the formula B = qv/r, where B is the magnetic field strength, q is the charge of the particle, v is its velocity, and r is the radius of its orbit. This formula is known as the Lorentz force law and is a fundamental principle in electromagnetism.
## 3. What is the significance of the magnetic field of gyromotion?
The magnetic field of gyromotion plays a crucial role in many physical phenomena, such as the motion of charged particles in a magnetic field, the behavior of plasma in fusion reactors, and the formation of celestial bodies like planets and stars. It also has practical applications in technologies such as particle accelerators and MRI machines.
## 4. How does the magnetic field of gyromotion affect charged particles?
The magnetic field of gyromotion can exert a force on charged particles, causing them to change direction and follow a curved path. This phenomenon is known as the Lorentz force and is responsible for the circular motion of particles in a magnetic field. The strength and direction of the force depend on the charge, velocity, and orientation of the magnetic field.
## 5. Can the magnetic field of gyromotion be manipulated?
Yes, the magnetic field of gyromotion can be manipulated by changing the velocity, charge, or direction of the moving charged particles. It can also be altered by adjusting the strength and orientation of external magnetic fields. This manipulation is essential in controlling and studying the behavior of particles in various scientific and technological applications. | 967 | 4,644 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2023-50 | latest | en | 0.931216 |
http://www.uiagrc.com.sg/might-you-probably-pay-off-a-payday-this-is-30/ | 1,721,293,321,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514826.9/warc/CC-MAIN-20240718064851-20240718094851-00535.warc.gz | 54,431,557 | 8,758 | Might you probably pay off A payday this is certainly 3-month loan 3 Months?
## Might you probably pay off A payday that is 3-month loan 3 Months?
Might you probably pay off A payday this is certainly loan that is 3-month Months?
Certain, an extended loan that is payday time that is additional protect the loan down, but it also means greater costs without any additional benefits.
One of several main problems with payday advances is the exceptionally brief repayment terms. With a term that is typical of two weeks, it may very difficult for many individuals to pay for the home loan off on-time.
But recently some loan that is payday have really attempted to offer payday advances with notably longer terms, like ninety days. So may be these a safer bet?
## Why don’t we do a little mathematics.
Therefore you may require financing calculator that you can figure the cost out of a three-month pay day loan. Since we now have actuallyn’t perfected our loan calculator technology yet, we utilized this one.
You will must also learn how much you’re borrowing, additionally it is APR, or apr. The APR actions simply just how much that loan would run you in costs and interest throughout the amount of a 12 months that is complete. It’s a measure this is certainly lets that are standard are going to be making an oranges to oranges cost contrast between loans.
Many loans which can be payday APRs as much as 400per cent (and many have APRS that are, gulp, means greater). But in addition for now, we will use 300% as our APR, so we is likely to make usage of \$1,000 when it comes to loan amount.
Spend the away a \$1,000 cash loan at a 300% APR, you will should pay back \$1,536.90 During the last end of 90 days.
Consequently, may be the known undeniable fact that practical? Perhaps. a few months to protect right straight back \$1,536.90 works off to mortgage loan of \$128.08 each week. But while those real numbers may appear reasonable, the fact is the one thing entirely different.
## Settling a cash that is 3-month in one swelling amount is hard.
When it comes to loans, much much longer re re payment terms are usually better. Longer terms recommend more repayments which are manageable more opportunities to improve your credit score simply by making stated re re repayments on time.
And https://carolinapaydayloans.org/, hey, if you’re able to pay the mortgage off very early, this is certainly great! You will save money on interest.
Nevertheless having a cash that is three-month, a few of these advantages could possibly be totally lacking. To begin, you will have the greater repayments which can be manageable which a fast pay day loan is improbable to obtain.
Unlike installment loans, which break your payment up into a couple of smaller re repayments, payday advances generally rely on lump-sum payment, and that means you invest the mortgage off all in the past.
Analysis suggests that people have actually an arduous time spending their pay check loans right back on time, and lump sum repayment re repayment payment is truly a element that is huge. Spending that loan off in small chunks is easier for them than saving inside the money to be in the entire security.
Which means that, saving up\$1,536.90 over a couple of months will likely to be a great deal harder than simply having to spend \$128.08 when every week.
## You simply can’t save you cash by settling a 3-month cash loan early.
Upcoming, there’s spending your loan off very early to save interest. This can maybe not help numerous advances that are payday as his / her charges and interest are charged at a level that is flat. What this means is the interest does accrue on your own n’t stability over-time. Instead, it is determined up-front and straight away added to your payment amount.
Whenever interest is likely to be charged as a flat-rate, very early payment does maybe not allow you to get any discounts or additional bonuses. Well, okay, it shall allow you to get straight down debt, that can easily be pretty awesome. Nevertheless, if you’re more likely to just simply just take a loan out, you desire the one that can benefit your cash once you consider the long-lasting.
Also making straight down their sky-high rates of interest, spend loans provide hardly any in method of long-lasting advantages day.
## A payday this is certainly 3-month won’t assist your credit.
Finally, the possibilities can be found by you to improve your credit score. Even if a quick payday loan provider have been to report your instalments to the credit agencies, spending the mortgage down within one re that is single could have an inferior sized impact that is g d your rating than investing it well in numerous installments.
But that’s pretty much a m t point, as cash advance providers actually report any payment seldom information in the end.
## Installment loans provide an alternative that is improved.
Since ch sing \$1,500 in the past is just t big an ask for most people, you shall may be best down getting an installment loan. This is certainly that loan that allows one to pay your loan off a bit at any time in selection of smaller, usually planned repayments all of which goes towards both the eye although the major loan volume. You want a loan provider whose loans are manufactured to be paid down the very first time, possibly perhaps not the fifth.
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Top | 1,209 | 5,550 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2024-30 | latest | en | 0.979288 |
https://www.jiskha.com/display.cgi?id=1201988896 | 1,498,646,839,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128323604.1/warc/CC-MAIN-20170628101910-20170628121910-00165.warc.gz | 881,761,531 | 3,995 | # math
posted by .
Consider the following binomials.
A=(x^2+5x) B=(6x+30)
Part 1: Factor each binomial by finding the GCF. Then, add the two factored binomials to make a single expression.
Part 2: Now add the original forms of binomial A and B together to make a trinomial. Factor the trinomial.
Part 3: Do you see something you could do to your answer in Part A to get your answer to Part B? Explain.
I know how to factor the binomials by finding the GCF. But I'm confused in this case for Parts A and B. For B, I think the gcf can be 3 and for A, the GCF can be 5.
• math -
1. The GCF is x + 5
A = (x+5)x and b = 6(x+5)
2. A + B = x^2 + 11 x + 30 = (x+6)(x+5)
3. Looking at the answer in (1), it can also be written (x+5)(x+6), with the GCF as one of the factors. | 258 | 776 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2017-26 | longest | en | 0.903185 |
https://codereview.stackexchange.com/questions/248330/recursive-sudoku-solver-using-python | 1,653,760,283,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652663016949.77/warc/CC-MAIN-20220528154416-20220528184416-00141.warc.gz | 227,053,723 | 71,218 | # Recursive Sudoku solver using Python
A Sudoku solver that works recursively. I'd appreciate your comments about coding style, structure and how to improve it. Thank you very much for your time.
Code structure
The Solver works by accepting a string of 81 digits for the Sudoku puzzle input. Zeros are taken as empty cells. It parses it into a 9x9 Numpy array.
The get_candidates function creates lists of possible digits to fill each cell following Sudoku's rules (no repeating 1-9 digit along rows, columns and 3x3 sub-grids).
The main solver function is solve. First, it discards wrong candidates with the filter-candidates function. "Wrong candidates" are those that when filled to a empty cell, led to another cell having no more candidates elsewhere on the Sudoku grid.
After filtering candidates, fill_singles is called to fill empty cells that have only one remaining candidate. If this process leads to a completely filled Sudoku grid, it's returned as a solution. There's a clause to return None which is used to backtrack changes by the make_guess function. This function will fill the next empty cell with the least quantity of candidates with one of its candidates, a "guess" value. It then recursively calls solve to either find a solution or reach a no-solution grid (in which case solve returns None and the last guess changes are reverted).
from copy import deepcopy
import numpy as np
def create_grid(puzzle_str: str) -> np.ndarray:
"""Create a 9x9 Sudoku grid from a string of digits"""
# Deleting whitespaces and newlines (\n)
lines = puzzle_str.replace(' ','').replace('\n','')
digits = list(map(int, lines))
# Turning it to a 9x9 numpy array
grid = np.array(digits).reshape(9,9)
return grid
def get_subgrids(grid: np.ndarray) -> np.ndarray:
"""Divide the input grid into 9 3x3 sub-grids"""
subgrids = []
for box_i in range(3):
for box_j in range(3):
subgrid = []
for i in range(3):
for j in range(3):
subgrid.append(grid[3*box_i + i][3*box_j + j])
subgrids.append(subgrid)
return np.array(subgrids)
def get_candidates(grid : np.ndarray) -> list:
"""Get a list of candidates to fill empty cells of the input grid"""
def subgrid_index(i, j):
return (i//3) * 3 + j // 3
subgrids = get_subgrids(grid)
grid_candidates = []
for i in range(9):
row_candidates = []
for j in range(9):
# Row, column and subgrid digits
row = set(grid[i])
col = set(grid[:, j])
sub = set(subgrids[subgrid_index(i, j)])
common = row | col | sub
candidates = set(range(10)) - common
# If the case is filled take its value as the only candidate
if not grid[i][j]:
row_candidates.append(list(candidates))
else:
row_candidates.append([grid[i][j]])
grid_candidates.append(row_candidates)
return grid_candidates
def is_valid_grid(grid : np.ndarray) -> bool:
"""Verify the input grid has a possible solution"""
candidates = get_candidates(grid)
for i in range(9):
for j in range(9):
if len(candidates[i][j]) == 0:
return False
return True
def is_solution(grid : np.ndarray) -> bool:
"""Verify if the input grid is a solution"""
if np.all(np.sum(grid, axis=1) == 45) and \
np.all(np.sum(grid, axis=0) == 45) and \
np.all(np.sum(get_subgrids(grid), axis=1) == 45):
return True
return False
def filter_candidates(grid : np.ndarray) -> list:
"""Filter input grid's list of candidates"""
test_grid = grid.copy()
candidates = get_candidates(grid)
filtered_candidates = deepcopy(candidates)
for i in range(9):
for j in range(9):
# Check for empty cells
if grid[i][j] == 0:
for candidate in candidates[i][j]:
# Use test candidate
test_grid[i][j] = candidate
# Remove candidate if it produces an invalid grid
if not is_valid_grid(fill_singles(test_grid)):
filtered_candidates[i][j].remove(candidate)
# Revert changes
test_grid[i][j] = 0
return filtered_candidates
def merge(candidates_1 : list, candidates_2 : list) -> list:
"""Take shortest candidate list from inputs for each cell"""
candidates_min = []
for i in range(9):
row = []
for j in range(9):
if len(candidates_1[i][j]) < len(candidates_2[i][j]):
row.append(candidates_1[i][j][:])
else:
row.append(candidates_2[i][j][:])
candidates_min.append(row)
return candidates_min
def fill_singles(grid : np.ndarray, candidates=None) -> np.ndarray:
"""Fill input grid's cells with single candidates"""
grid = grid.copy()
if not candidates:
candidates = get_candidates(grid)
any_fill = True
while any_fill:
any_fill = False
for i in range(9):
for j in range(9):
if len(candidates[i][j]) == 1 and grid[i][j] == 0:
grid[i][j] = candidates[i][j][0]
candidates = merge(get_candidates(grid), candidates)
any_fill = True
return grid
def make_guess(grid : np.ndarray, candidates=None) -> np.ndarray:
"""Fill next empty cell with least candidates with first candidate"""
grid = grid.copy()
if not candidates:
candidates = get_candidates(grid)
# Getting the shortest number of candidates > 1:
min_len = sorted(list(set(map(
len, np.array(candidates).reshape(1,81)[0]))))[1]
for i in range(9):
for j in range(9):
if len(candidates[i][j]) == min_len:
for guess in candidates[i][j]:
grid[i][j] = guess
solution = solve(grid)
if solution is not None:
return solution
grid[i][j] = 0
def solve(grid : np.ndarray) -> np.ndarray:
"""Recursively find a solution filtering candidates and guessing values"""
candidates = filter_candidates(grid)
grid = fill_singles(grid, candidates)
if is_solution(grid):
return grid
if not is_valid_grid(grid):
return None
return make_guess(grid, candidates)
# # Example usage
# puzzle = """100920000
# 524010000
# 000000070
# 050008102
# 000000000
# 402700090
# 060000000
# 000030945
# 000071006"""
# grid = create_grid(puzzle)
# solve(grid)
$$$$
I was able to improve the performance of the program by about 900% without understanding or changing much of the algorithm in about an hour. Here's what I did:
First of all, you need a benchmark. It's very simple, just time your program
start = time.time()
solve(grid)
print(time.time()-start)
On my computer, it took about 4.5 seconds. This is our baseline.
The next thing is to profile. The tool I chose is VizTracer, which is developed by myself :) https://github.com/gaogaotiantian/viztracer
VizTracer will generate an HTML report(or json that could be loaded by chrome:://tracing) of timeline of your code execution. It looks like this in your original version:
As you can tell, there are a lot of calls on there. The thing we need to do is to figure out what is the bottleneck here. The structure is not complicated, a lot of fill_singles are called, and we need to zoom in to check what's in there.
It's very clear that get_candidates is the function that caused most of the time in fill_singles, which is occupying most of the timeline. So that's the function we want to take a look at first.
def get_candidates(grid : np.ndarray) -> list:
"""Get a list of candidates to fill empty cells of the input grid"""
def subgrid_index(i, j):
return (i//3) * 3 + j // 3
subgrids = get_subgrids(grid)
grid_candidates = []
for i in range(9):
row_candidates = []
for j in range(9):
# Row, column and subgrid digits
row = set(grid[i])
col = set(grid[:, j])
sub = set(subgrids[subgrid_index(i, j)])
common = row | col | sub
candidates = set(range(10)) - common
# If the case is filled take its value as the only candidate
if not grid[i][j]:
row_candidates.append(list(candidates))
else:
row_candidates.append([grid[i][j]])
grid_candidates.append(row_candidates)
return grid_candidates
The thing that caught my eyes first was the end of your nested for loop. You checked whether grid[i][j] is filled. If it is, then that's the only candidate. However, if it's filled, then it has nothing to do with candidates, which you computed very hard in your nested for loop.
So the first thing I did was moving the check to the beginning of the for loop.
for i in range(9):
row_candidates = []
for j in range(9):
if grid[i][j]:
row_candidates.append([grid[i][j]])
continue
# Row, column and subgrid digits
row = set(grid[i])
col = set(grid[:, j])
sub = set(subgrids[subgrid_index(i, j)])
common = row | col | sub
candidates = set(range(10)) - common
row_candidates.append(list(candidates))
This optimization alone cut the running time in half, we are at about 2.3s now.
Then I noticed in your nested for loop, you are doing a lot of redundant set operations. Even row/col/sub only needs to be computed 9 times, you are computing it 81 times, which is pretty bad. So I moved the computation out of the for loop.
def get_candidates(grid : np.ndarray) -> list:
"""Get a list of candidates to fill empty cells of the input grid"""
def subgrid_index(i, j):
return (i//3) * 3 + j // 3
subgrids = get_subgrids(grid)
grid_candidates = []
row_sets = [set(grid[i]) for i in range(9)]
col_sets = [set(grid[:, j]) for j in range(9)]
subgrid_sets = [set(subgrids[i]) for i in range(9)]
total_sets = set(range(10))
for i in range(9):
row_candidates = []
for j in range(9):
if grid[i][j]:
row_candidates.append([grid[i][j]])
continue
# Row, column and subgrid digits
row = row_sets[i]
col = col_sets[j]
sub = subgrid_sets[subgrid_index(i, j)]
common = row | col | sub
candidates = total_sets - common
# If the case is filled take its value as the only candidate
row_candidates.append(list(candidates))
grid_candidates.append(row_candidates)
return grid_candidates
This cut the running time to about 1.5s. Notice that, I have not try to understand your algorithm yet. Thing only thing I did was to use VizTracer to find the function that needs to be optimized and do same-logic transform. I improved performance by about 300% in like 15 minutes.
To this point, the overhead of VizTracer on WSL is significant, so I turned off the C function trace. Only Python functions were left and the overhead was about 10%.
Now the get_candidates was improved(although it can be done better), we need to take a bigger picture of this. What I can observe from VizTracer's result was that fill_singles called get_candidates very frequently, just too many calls. (This is something that's hard to notice on cProfiler)
So the next step was to figure out if we can make fill_singles call get_candidates less often. Here it requires some level of algorithm understanding.
while any_fill:
any_fill = False
for i in range(9):
for j in range(9):
if len(candidates[i][j]) == 1 and grid[i][j] == 0:
grid[i][j] = candidates[i][j][0]
candidates = merge(get_candidates(grid), candidates)
any_fill = True
It looks like here you tried to fill in one blank with only one candidate, and recalculate the candidates of the whole grid, then find the next blank with one candidate. This is a valid method, but this caused too many calls to get_candidates. If you think about it, when we fill in a blank with a number n, all the other blanks with only one candidate that's not n won't be affected. So during one pass of the grid, we could actually try to fill more blanks in, as long as we do not fill in the same number twice. This way, we can call get_candidates less often, which is a huge time consumer. I used a set to do this.
filled_number = set()
for i in range(9):
for j in range(9):
if len(candidates[i][j]) == 1 and grid[i][j] == 0 and candidates[i][j][0] not in filled_number:
grid[i][j] = candidates[i][j][0]
any_fill = True
candidates = merge(get_candidates(grid), candidates)
This brought the running time to 0.9s.
Then I looked at the VizTracer report, I realized fill_singles is almost always called by filter_candidates and the only thing filter_candidates is interested in, is whether fill_singles returns a valid grid. This is an information we might know early, as long as fill_singles finds a position with no candidates. If we return early, we don't need to calculate get_candidates that many times.
So I changed the code structure a little bit, made fill_singles return None if it can't find a valid grid.
Finally I was able to make the run time to 0.5s, which is 900% faster than the original version.
It was actually a fun adventure because I was testing my project VizTracer and tried to figure out if it was helpful to locate the time consuming part. It worked well :)
• That's amazing! Your modifications are quite simple but they vastly improve performance. I'll perform try performing similar analyses in the future. Congratulations on VizTracer! Great tool. I'll use it in the future. Aug 25, 2020 at 3:43
• Thanks! I was hoping that VizTracer can help people on problems like this. My boss once told me, never optimize without profiling. A good profiler is indeed very helpful to improve the performance :) Aug 25, 2020 at 5:17
# Numpyification
get_subgrids essentially rearranges a numpy array with a minimum of numpy. It could be done with numpy itself, for example:
def get_subgrids(grid: np.ndarray) -> np.ndarray:
"""Divide the input grid into 9 3x3 sub-grids"""
swapped = np.swapaxes(np.reshape(grid, (3, 3, 3, 3)), 1, 2)
return np.reshape(swapped, (9, 9))
The downside I suppose is that swapping the middle two axes of a 4D array is a bit mind-bending.
# Performance
Almost all time is spent in get_candidates. I think the reasons for that are mainly:
• It gets called too often. For example, after filling in a cell (such as in fill_singles), rather than recompute the candidates from scratch, it would be faster to merely remove the new value from the candidates in the same row/col/house.
• If a cell is filled, the list of candidates is just the filled-in value, but the expensive set computation is done anyway. That's easy to avoid just by moving those statement inside the if.
# Algorithmic performance
This solver only makes use of Naked Singles as a "propagation technique", adding Hidden Singles is in my experience a very large step towards an efficient solver.
• Thank you very much for your review. I couldn't find of way to numpify the sub-grids and your recommendation is spot on. One question, how did you find that the get_candidates function is the most used? Was it by simple checking of the code or did you use any tool? Thanks again. Aug 23, 2020 at 22:19
• @fabrizzio_gz I used %%prun` in a jupyter notebook, if you're running the code as standalone script you could use one of these Aug 23, 2020 at 22:26
• Great to know. Thanks a lot. Aug 24, 2020 at 3:22 | 3,606 | 14,333 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 1, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2022-21 | longest | en | 0.829645 |
http://www.conversion-website.com/volume/bushel-US-to-ton-freight.html | 1,652,938,458,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662525507.54/warc/CC-MAIN-20220519042059-20220519072059-00748.warc.gz | 72,063,076 | 4,619 | # Bushels (US) to tons (freight) (bu to FT)
## Convert bushels (US) to tons (freight)
Bushels (US) to tons (freight) conversion calculator shown above calculates how many tons (freight) are in 'X' bushels (US) (where 'X' is the number of bushels (US) to convert to tons (freight)). In order to convert a value from bushels (US) to tons (freight) (from bu to FT) simply type the number of bu to be converted to FT and then click on the 'convert' button.
## Bushels (US) to tons (freight) conversion factor
1 bushel (US) is equal to 0.03111140031563 tons (freight)
## Bushels (US) to tons (freight) conversion formula
Volume(FT) = Volume (bu) × 0.03111140031563
Example: Find out how many tons (freight) equal 515 bushels (US).
Volume(FT) = 515 ( bu ) × 0.03111140031563 ( FT / bu )
Volume(FT) = 16.02237116255 FT or
515 bu = 16.02237116255 FT
515 bushels (US) equals 16.02237116255 tons (freight)
## Bushels (US) to tons (freight) conversion table
bushels (US) (bu)tons (freight) (FT)
300.93334200946891
401.2444560126252
501.5555700157815
601.8666840189378
702.1777980220941
802.4889120252504
902.8000260284067
1003.111140031563
1103.4222540347193
1203.7333680378756
1304.0444820410319
1404.3555960441882
1504.6667100473445
1604.9778240505008
1705.2889380536571
1805.6000520568134
1905.9111660599697
2006.222280063126
2106.5333940662823
2206.8445080694386
bushels (US) (bu)tons (freight) (FT)
3009.333420094689
40012.444560126252
50015.555700157815
60018.666840189378
70021.777980220941
80024.889120252504
90028.000260284067
100031.11140031563
110034.222540347193
120037.333680378756
130040.444820410319
140043.555960441882
150046.667100473445
160049.778240505008
170052.889380536571
180056.000520568134
190059.111660599697
200062.22280063126
210065.333940662823
220068.445080694386
Versions of the bushels (US) to tons (freight) conversion table. To create a bushels (US) to tons (freight) conversion table for different values, click on the "Create a customized volume conversion table" button.
## Related volume conversions
Back to bushels (US) to tons (freight) conversion
TableFormulaFactorConverterTop | 746 | 2,124 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-21 | latest | en | 0.654566 |
http://www.hypergeometricaluniverse.com/?p=2898 | 1,371,706,336,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368710313659/warc/CC-MAIN-20130516131833-00094-ip-10-60-113-184.ec2.internal.warc.gz | 492,345,457 | 13,669 | # HyperGeometrical Universe Theory
## June 27, 2010
### Crossing the Ts, dotting the Is
Filed under: Uncategorized — admin @ 6:01 am
The Pioneer Anomaly
The solution to the Pioneer Anomaly can be derived directly from the Hypergeometrical Universe topology. It is our current understanding that only matter curves spacetime, this means that locally the curvature is quite well defined by the inexistence of localized matter and equal to zero, that is, without the obvious occurrence of matter we believe that we live in a locally flat space. The addition of another spatial dimension in the Hypergeometrical Universe theory changes things, since there are now many curvatures to talk about.
Let's see what happens when one measures the speed of light within our standard paradigm. The measure time delay between shooting the laser pulse and measuring the reflection is equal to twice the distance divided by the speed of light.
For the reflected radiation bouncing back from the mirror at rest,
Now if we place the mirror in motion we arrive at the standard derivation of the Doppler Effect:
The Doppler shift is given by:
The derivation is different if one allows v to vary:
Resulting:
The derivation is different still if one allows both c and v to vary:
Resulting:
This means that both accelerations on the speed of light or the actual speed of the Pioneer spacecraft would contribute to the Doppler Shift. This also means that if you are not considering that the effective speed of light vary with distance, you will be missing the contribution.
Under the current paradigm, the time it takes for light to bounce back from a mirror is given by 2L/c and doesn't change with the distance L. This considers that light and the mirror are traveling or existing in the same dimension (3D) which is not the Hypergeometrical Universe Paradigm.
There, light (a spatial modulation of the source of a dilaton field) travels within the 5D spacetime (4D spatial manifold). Anything else just flows within the 3D Hyperspherical Universe, which itself travels through the 4D spatial manifold.
If you consider our topology, it is clear that there is a curvature and that is equal to 1/R0 and it is independent of the existence of matter locally. Remember that a 5D Spacetime has more than one radius of curvature. The RX radius of curvature is R0, and so it is the ФX radius of curvature. The situation changes one consider the τX (τ is the local or proper time). The τX radius of curvature will depend upon which laws do you use to describe your Universe (Gauss Electrostatic Law, Newton's Gravitational Law). If you were to use the Quantum Lagrangian Principle to describe dilator dynamics, then you could use a Cartesian reference system for the local proper time. The usage of Newton's, Gauss' or similar laws to describe dilator dynamics makes it necessary to make local proper spacetime metric to be Lorentzian.
Now let's get down to the Pioneer Anomaly and how do we measure how long the Universe is alive from the observed Pioneer Spacecraft deceleration.
The first thing to consider is what would be the perceived speed of light deceleration resulting from the geometry. To do this, we need to consider how much light has to travel back as a function of the farthest distance traveled by the spacecraft.
This figure shows what would be the path traversed by each pulse emitted by the Pioneer spacecraft. The angle of the light ray is always 45 degrees. The leftmost radial line represents Earth position. This means that when Pioneer is far from Earth by 45 degrees or PI/4*R0, light will never come back (the light ray will be parallel to the Earth radial line. Two parallel lines never meet.
Let's say that at time zero Pioneer is at R0 from the 4D Center of the Universe and L from Earth and let's derive the deceleration from the simple geometry.
Let R(t) be the Radius of the Universe at the time t:
Where time is being measured after a light pulse is emitted from the Pioneer while it is at distance from the 4D Center of the Universe equal to R0.
The Cosmological Angle associated with Pioneer at time t given by:
for calculating the deceleration of the speed of light
And
for calculating the deceleration of the speed of the Pioneer spacecraft
The angle alpha is called Cosmological Angle and it is measured from the actual Center of the Universe. The time that measures the expansion of the Universe is the Cosmological Time φ, but for quasi-relaxed fabric of space (low velocities), the Cosmological Time is a reasonable approximation to the proper time.
The equation for the photon trajectory leaving the Spacecraft and reaching Earth later is given by:
Where y(t,t') is the horizontal distance represented in the figure above of the reflected light beam as a function of time t', which starts counting at the reflection moment. The time t refers to the t governing the expansion of the Universe from some given initial condition (R0, L).
Equating this equation to zero and solving for t', one obtains the time it takes for light to come from Pioneer to Earth at any given time or Pioneer position.
The perceived distance traversed is given by x=c.t':
Taking the second derivative, expanding in Taylor Series and simplifying to obtain the acceleration as:
The observed acceleration is 8.75e-10 m/s2, the speed of light = 299792458 m / s. This yields a value for R0= 2.05E+26 meters. Since the Pioneer velocity is much smaller than c and the second term can be neglected.
Since the 4D observed velocity is m this means that the Universe is years old or 1.5364E+10 years or
15.364 Billion Years Old…J
PerceiveAge=15.364 Billion Years Old…:) (using the Hubble Constant as a measure of the age of the Universe).
The important result is that the Universe is actually older that the age perceived by inspecting the Hubble Constant:
RealAge=15.36*sqrt(2)= 21.65 Billion Years Old:)
One salient point is that the Hypergeometrical Lightspeed Expanding Shockwave Universe expands at the observed speed of light, this means that the material world travels slower than the dilaton field, by a factor of cos(45) or .
This conclusion is very important because it breaks the symmetry between dilators and dilatons, in the sense that they do not travel at the same speed. On the other hand, this velocity relationship is strictly necessary for the Quantum Lagrangian Principle to make all dilators to surf the surrounding dilaton field.
Dilators interact with other dilators through retarded potentials or retarded dilaton fields.
The Beginning of Times
At time zero, the initial macroscopic metric fluctuation (the one that followed the dimensional transition) decayed into a huge number of dilators. Many recombined and released Gamma Rays. The initial radiation blast accelerated the extremely small Universe both radially and tangentially with the same strength. Dilators flowed with the retarded dilaton field in all directions of the 4D spatial manifold, creating the 3D Lightspeed Expanding Shockwave Universe. The 4D Big Bang also synchronized all dilators' spinning. Only dilators that were flush (in perfect overlap) with the 3D Hypersphere at time zero, were accelerated into c.
Up to now, I considered that dilators and dilatons always traveled at the only allowed speed (c).
Of course, I noticed that dilators have inertia with respect to lateral motion but I thought that the radial motion was the natural motion of a metric deformation, thus requiring no initial push to be achieved.
If I accept that the dilaton speed or light speed is actually larger than the observed speed of light then the picture associated with the initial Big Bang changes from a simple decaying process into a tangential explosion which set the 3D Hypersphere to travel at the observed speed of light.
It is clear that a dilaton field cannot accelerate dilators to its speed (sqr(2)*c) because interaction has to occur through RETARDED POTENTIALS. Retarded Potentials always propagate at c (Lightspeed) due to the initial geometry.
The interaction of dilators with the retarded dilaton field from other dilators produces a distribution of tangentially dilators traveling at c which is also a distribution of radially traveling dilators at that same speed. In other words, a lightspeed traveling Hyperspherical Universe and an sqr(2)*c dilaton-dilator interaction form a self-consistent solution to the problem.
This salient point can be summarized as:
The speed of light is SQR(2) times C or light is faster than the speed of light..J
This is due to motion be part of a coherence that was placed in motion at the beginning of times through the interaction of near field dilatons. The interference pattern of near field radially propagating homogeneous dilatons travels along 45 degrees. This is another very salient point.
Cheers,
MP | 1,900 | 8,869 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2013-20 | latest | en | 0.922729 |
https://list.coin-or.org/pipermail/ipopt/2016-September/004333.html | 1,718,340,676,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861521.12/warc/CC-MAIN-20240614043851-20240614073851-00183.warc.gz | 347,527,482 | 3,197 | # [Ipopt] Maximun number of iterations exceeded problem.
Pablo Perez pabloperez555 at hotmail.com
Thu Sep 29 12:03:24 EDT 2016
```Dear all,
I am having a problem using Ipopt: After having modelling a simple dynamic system based on newton rules and discretize it, I am getting a maximun number of iterations exceeded problem. I am not sure if the problem is due to the discretization method (a leapfrog integration method), because using a simple Euler method (look for "##" in the code below) the system convergency is got in a few iterations.
Does someone understand where the problem is? It is worth the extra precission of having into acount the aceleration for calculating the space or the euler method would be good enought for prototyping?
Please, I copy the AMPL code I am using. Take into account that the low number of steps is due to the restricted version of AMPL. Anyway, using a bigger number in Neos server does not solve the problem.
I am not sure if this mailling list is for this kind of questions. If it is not I would be sorry, please, could someone tell me about any forum of Ipopt in particular and optimal control in general?
Pablo
# We have a train hat is at position x=0 at time t=0, and we can
# control the acceleration, a, (which includes braking as well).
#
# N: Number of iterations.
# L: Path length.
# T: Travel time initial guess.
# aU: Upper bound on acceleration.
# aL: Lower bound on acceleration.
# jU: Upper bound on jerk.
# jL: Lower bound on jerk.
# Ra: Drag coeficient (independent).
# Rb: Drag coeficient (v dependant).
# Rc: Drag coeficient (v^2 dependant).
#
# The overall optimal control problem is therefore:
#
# min tf
# s.t. dx/dt(t) = v(t) for t in [0,tf]
# dv/dt(t) = a(t) - Ra - Rb*v(t) - Rc*v(t)^2 for t in [0,tf]
# aL <= dv/dt(t) <= aU for t in [0,tf]
# jL <= d2v/dt2(t) <= jU for t in [0,tf]
# v(t) <= Vmax(x) for t in [0,tf]
#
# x(0) = 0; v(0) = 0
# x(tf) = L; v(tf) = 0
#
# Leapfrog discretization.
# With this, we can state the overall optimization problem below:
option solver ipopt;
# Number of discretization intervals
param N := 1000;
# Final position
param L := 1000;
# Final time
param T := 50;
# Upper and lower bound on acceleration
param aU := 1.15;
param aL := -1.2;
# Upper and lower bound on jerk
param jU := 0.8;
param jL := -0.8;
# Drag factors
param Ra := 5.71428/322;
param Rb := 0.11688/322;
param Rc := 0.00842/322;
# Size of discretization intervals
var tf >= 0 := T;
var h = tf/N;
# Velocity
var x{i in 0..N} >= 0, := i*(L/N);
# Time
var v{i in 0..N} >= 0, := L/T;
# Acceleration
var a{i in 0..N} := 0;
# Objective function
minimize time:
tf;
subject to Amax {i in 0..N}:
a[i] <= if (v[i] <= 11) then (343/322) else (343/322)*(11/v[i]);
subject to Dmax {i in 0..N}:
a[i] >= if (v[i] <= 19) then -(327/322) else -(327/322)*(19/v[i]);
# Differential equation for velocity
subject to dx {i in 0..N-1}:
(x[i+1]-x[i])/h = v[i] + 0.5*(a[i] - Ra - Rb*v[i] - Rc*v[i]^2)*h;
##(x[i+1]-x[i])/h = v[i];
# Differential equation for time
subject to dv {i in 0..N-1}:
(v[i+1]-v[i])/h = a[i] - Ra - Rb*v[i] - Rc*v[i]^2;
# Confort constraints
subject to acel {i in 0..N-1}:
aL <= (v[i+1]-v[i])/h <= aU;
subject to jerk {i in 1..N-1}:
jL <= (v[i+1]-2*v[i]+v[i-1])/h^2 <= jU;
# Boundary conditions for position
subject to x0:
x[0] = 0;
subject to xf:
x[N] = L;
# Boundary conditions for velocity
subject to v0:
v[0] = 0;
subject to vf:
v[N] = 0;
# Speed limit
subject to limit {i in 1..N-1}:
v[i] <= if (x[i] <= 250) then 45/3.6 else if (x[i] <= 700) then 70/3.6 else 30/3.6;
# Solve the optimization problem
solve;
# write the data into a file for gnuplot
for {i in 0..N}
printf : "%16.4e %16.4e %16.4e %16.4e \n", i*h, x[i], 3.6*v[i], 10*a[i] > ..\gnuplot.dat;
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``` | 1,347 | 4,070 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2024-26 | latest | en | 0.893281 |
https://embuchestissues.wordpress.com/tag/manifold/ | 1,521,318,667,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257645310.34/warc/CC-MAIN-20180317194447-20180317214447-00745.warc.gz | 602,102,010 | 14,751 | ## Posts tagged ‘manifold’
### Ten constructions of the cohomology of varieties
When talking about “the” cohomology of mathematical objects, we do not always explicitly mention which cohomology is used, because it is obvious (in cases there is only one possible definition), or because we really don’t care (since as we will see, it is frequent that different definitions lead to equivalent results). The case of differentiable manifolds or algebraic varieties is particularly impressive, since there were a lot of equivalent cohomology theories defined during last century in order to simplify proofs or allow generalisations. Most cohomology theories, if not all, are defined as the cohomology of a complex : i.e. a sequence of vector spaces or modules $(V_n)$ with a boundary map $d$. The kernel of $d$ is called the set of cycles, while the images by $d$ are called boundaries : the cohomology $H^n$ is then the quotient of cycles in $V_n$ by the subspace of boundaries.
Classical topologists, for example, will use preferably (see MacLane, Homology or the book of Allen Hatcher) :
• simplicial (co)homology : it is defined for a triangulated space, i.e. the manifold is cut by curves, surfaces, etc. which make it isomorphic to a sort of polyhedron (a complex); simplicial homology describe non-triviality (holes) in the combinatorial structure of this polyhedron; here the boundary map is really the boundary map.
• singular (co)homology : a more abstract version of simplicial homology; now we consider the set of all possible simplexes (curves, polygons, polyhedra, and their generalisations…) drawn on the manifold; this definition was given by Eilenberg; I don’t know who first defined simplicial homology, but MacLane mentions that Poincaré and Noether gaves important contributions to this theory.
Then come sheaves, which were explicitly defined by Leray (tales for young mathematicians help remember that Leray made considerable efforts as a prisoner in concentration camps to focus his work on especially “useless” subjects to avoid helping Nazis). (more…) | 473 | 2,076 | {"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": 6, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-13 | longest | en | 0.938901 |
https://www.calculatoratoz.com/en/perimeter-of-a-square-when-area-is-given-calculator/Calc-367 | 1,618,916,272,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039388763.75/warc/CC-MAIN-20210420091336-20210420121336-00402.warc.gz | 778,240,510 | 32,467 | 🔍
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## Credits
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## Perimeter of a square when area is given Solution
STEP 0: Pre-Calculation Summary
Formula Used
perimeter = 4*sqrt(Area)
P = 4*sqrt(A)
This formula uses 1 Functions, 1 Variables
Functions Used
sqrt - Squre root function, sqrt(Number)
Variables Used
Area - The area is the amount of two-dimensional space taken up by an object. (Measured in Square Meter)
STEP 1: Convert Input(s) to Base Unit
Area: 50 Square Meter --> 50 Square Meter No Conversion Required
STEP 2: Evaluate Formula
Substituting Input Values in Formula
P = 4*sqrt(A) --> 4*sqrt(50)
Evaluating ... ...
P = 28.2842712474619
STEP 3: Convert Result to Output's Unit
28.2842712474619 Meter --> No Conversion Required
28.2842712474619 Meter <-- Perimeter
(Calculation completed in 00.000 seconds)
## < 11 Other formulas that you can solve using the same Inputs
Diagonal of a Rectangle when breadth and area are given
Diagonal of a Rectangle when length and area are given
diagonal = sqrt(((Area)^2/(Length)^2)+(Length)^2) Go
Side of a Kite when other side and area are given
side_a = (Area*cosec(Angle Between Sides))/Side B Go
Perimeter of rectangle when area and rectangle length are given
perimeter = (2*Area+2*(Length)^2)/Length Go
Angle between the Rectangle Diagonals in terms of Area and Rectangle Diagonal
sin(sinϑ) = 2*Area/(Diagonal)^2 Go
Buoyant Force
buoyant_force = Pressure*Area Go
Diagonal of a Square when area is given
diagonal = sqrt(2*Area) Go
Length of rectangle when area and breadth are given
Breadth of rectangle when area and length are given
Pressure when force and area are given
pressure = Force/Area Go
Stress
stress = Force/Area Go
## < 11 Other formulas that calculate the same Output
Perimeter of a Right Angled Triangle
perimeter = Side A+Side B+sqrt(Side A^2+Side B^2) Go
Perimeter of a rectangle when diagonal and length are given
perimeter = 2*(Length+sqrt((Diagonal)^2-(Length)^2)) Go
Perimeter Of Parallelepiped
perimeter = 4*Side A+4*Side B+4*Side C Go
Perimeter of a Parallelogram
perimeter = 2*Side A+2*Side B Go
Perimeter of a Kite
perimeter = 2*(Side A+Side B) Go
Perimeter of a rectangle when length and width are given
perimeter = 2*Length+2*Width Go
Perimeter of an Isosceles Triangle
perimeter = Side A+2*Side B Go
Perimeter of a Cube
perimeter = 12*Side Go
Perimeter of a square when side is given
perimeter = 4*Side Go
Perimeter of an Equilateral Triangle
perimeter = 3*Side Go
Perimeter of a Rhombus
perimeter = 4*Side Go
### Perimeter of a square when area is given Formula
perimeter = 4*sqrt(Area)
P = 4*sqrt(A)
## What is Perimeter of a Square?
The perimeter of a square is the total length of all the sides of the square. So, we can find the perimeter of a square by adding all its four sides and since a square is a type of rectangle in which the adjacent sides are equal, hence its perimeter will be 4 times its side, i.e. 4 × Side. But for finding the perimeter using the area of a square, we can multiply the square root of its area with the number 4.
## How to Calculate Perimeter of a square when area is given?
Perimeter of a square when area is given calculator uses perimeter = 4*sqrt(Area) to calculate the Perimeter, The perimeter of a square is the total length of all the sides of the square. Perimeter and is denoted by P symbol.
How to calculate Perimeter of a square when area is given using this online calculator? To use this online calculator for Perimeter of a square when area is given, enter Area (A) and hit the calculate button. Here is how the Perimeter of a square when area is given calculation can be explained with given input values -> 28.28427 = 4*sqrt(50).
### FAQ
What is Perimeter of a square when area is given?
The perimeter of a square is the total length of all the sides of the square and is represented as P = 4*sqrt(A) or perimeter = 4*sqrt(Area). The area is the amount of two-dimensional space taken up by an object.
How to calculate Perimeter of a square when area is given?
The perimeter of a square is the total length of all the sides of the square is calculated using perimeter = 4*sqrt(Area). To calculate Perimeter of a square when area is given, you need Area (A). With our tool, you need to enter the respective value for Area and hit the calculate button. You can also select the units (if any) for Input(s) and the Output as well.
How many ways are there to calculate Perimeter?
In this formula, Perimeter uses Area. We can use 11 other way(s) to calculate the same, which is/are as follows -
• perimeter = 2*Length+2*Width
• perimeter = 4*Side
• perimeter = 3*Side
• perimeter = 2*Side A+2*Side B
• perimeter = 4*Side
• perimeter = Side A+2*Side B
• perimeter = Side A+Side B+sqrt(Side A^2+Side B^2)
• perimeter = 12*Side
• perimeter = 2*(Side A+Side B)
• perimeter = 4*Side A+4*Side B+4*Side C
• perimeter = 2*(Length+sqrt((Diagonal)^2-(Length)^2))
Let Others Know | 1,410 | 5,089 | {"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} | 4.125 | 4 | CC-MAIN-2021-17 | latest | en | 0.731876 |
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Posts
## Noncommutative probability
The traditional mathematical axiomatization of probability, due to Kolmogorov, begins with a probability space $P$ and constructs random variables as certain functions $P \to \mathbb{R}$. But start doing any probability and it becomes clear that the space $P$ is de-emphasized as much as possible; the real focus of probability theory is on the algebra of random variables. It would be nice to have an approach to probability theory that reflects this.
Moreover, in the traditional approach, random variables necessarily commute. However, in quantum mechanics, the random variables are self-adjoint operators on a Hilbert space $H$, and these do not commute in general. For the purposes of doing quantum probability, it is therefore also natural to look for an approach to probability theory that begins with an algebra, not necessarily commutative, which encompasses both the classical and quantum cases.
Happily, noncommutative probability provides such an approach. Terence Tao’s notes on free probability develop a version of noncommutative probability approach geared towards applications to random matrices, but today I would like to take a more leisurely and somewhat scattered route geared towards getting a general feel for what this formalism is capable of talking about.
An interesting result that demonstrates, among other things, the ubiquity of $\pi$ in mathematics is that the probability that two random positive integers are relatively prime is $\frac{6}{\pi^2}$. A more revealing way to write this number is $\frac{1}{\zeta(2)}$, where
$\displaystyle \zeta(s) = \sum_{n \ge 1} \frac{1}{n^s}$
is the Riemann zeta function. A few weeks ago this result came up on math.SE in the following form: if you are standing at the origin in $\mathbb{R}^2$ and there is an infinitely thin tree placed at every integer lattice point, then $\frac{6}{\pi^2}$ is the proportion of the lattice points that you can see. In this post I’d like to explain why this “should” be true. This will give me a chance to blog about some material from another math.SE answer of mine which I’ve been meaning to get to, and along the way we’ll reach several other interesting destinations. | 495 | 2,222 | {"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": 10, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2013-48 | longest | en | 0.925699 |
https://www.justintools.com/unit-conversion/energy.php?k1=ton-of-coal-equivalent&k2=kilograms-of-TNT | 1,679,583,316,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945168.36/warc/CC-MAIN-20230323132026-20230323162026-00445.warc.gz | 957,592,819 | 28,162 | Please support this site by disabling or whitelisting the Adblock for "justintools.com". I've spent over 10 trillion microseconds (and counting), on this project. This site is my passion, and I regularly adding new tools/apps. Users experience is very important, that's why I use non-intrusive ads. Any feedback is appreciated. Thank you. Justin XoXo :)
# ENERGY Units Conversionton-of-coal-equivalent to kilograms-of-TNT
1 Ton Of Coal Equivalent
= 7000 Kilograms Of TNT
Embed this to your website/blog
Category: energy
Conversion: Ton Of Coal Equivalent to Kilograms Of TNT
The base unit for energy is joules (Non-SI/Derived Unit)
[Ton Of Coal Equivalent] symbol/abbrevation: (TCE)
[Kilograms Of TNT] symbol/abbrevation: (kgTNT)
How to convert Ton Of Coal Equivalent to Kilograms Of TNT (TCE to kgTNT)?
1 TCE = 7000 kgTNT.
1 x 7000 kgTNT = 7000 Kilograms Of TNT.
Always check the results; rounding errors may occur.
Definition:
In relation to the base unit of [energy] => (joules), 1 Ton Of Coal Equivalent (TCE) is equal to 29288000000 joules, while 1 Kilograms Of TNT (kgTNT) = 4184000 joules.
1 Ton Of Coal Equivalent to common energy units
1 TCE = 29288000000 joules (J)
1 TCE = 29288000 kilojoules (kJ)
1 TCE = 7000000000 calories (cal)
1 TCE = 7000000 kilocalories (kcal)
1 TCE = 1.828009337278E+29 electron volt (eV)
1 TCE = 8135555.5555556 watt hour (Wh)
1 TCE = 6.7178256337987E+27 atomic unit of energy (au)
1 TCE = 7 tons of TNT (tTNT)
1 TCE = 21601720228.033 foot pound force (ft lbf)
1 TCE = 2.9288E+17 ergs (ergs)
Ton Of Coal Equivalentto Kilograms Of TNT (table conversion)
1 TCE = 7000 kgTNT
2 TCE = 14000 kgTNT
3 TCE = 21000 kgTNT
4 TCE = 28000 kgTNT
5 TCE = 35000 kgTNT
6 TCE = 42000 kgTNT
7 TCE = 49000 kgTNT
8 TCE = 56000 kgTNT
9 TCE = 63000 kgTNT
10 TCE = 70000 kgTNT
20 TCE = 140000 kgTNT
30 TCE = 210000 kgTNT
40 TCE = 280000 kgTNT
50 TCE = 350000 kgTNT
60 TCE = 420000 kgTNT
70 TCE = 490000 kgTNT
80 TCE = 560000 kgTNT
90 TCE = 630000 kgTNT
100 TCE = 700000 kgTNT
200 TCE = 1400000 kgTNT
300 TCE = 2100000 kgTNT
400 TCE = 2800000 kgTNT
500 TCE = 3500000 kgTNT
600 TCE = 4200000 kgTNT
700 TCE = 4900000 kgTNT
800 TCE = 5600000 kgTNT
900 TCE = 6300000 kgTNT
1000 TCE = 7000000 kgTNT
2000 TCE = 14000000 kgTNT
4000 TCE = 28000000 kgTNT
5000 TCE = 35000000 kgTNT
7500 TCE = 52500000 kgTNT
10000 TCE = 70000000 kgTNT
25000 TCE = 175000000 kgTNT
50000 TCE = 350000000 kgTNT
100000 TCE = 700000000 kgTNT
1000000 TCE = 7000000000 kgTNT
1000000000 TCE = 7000000000000 kgTNT
(Ton Of Coal Equivalent) to (Kilograms Of TNT) conversions | 961 | 2,553 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2023-14 | latest | en | 0.636692 |
https://webot.org/basic/?search=Solid_mechanics | 1,628,185,012,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046156141.29/warc/CC-MAIN-20210805161906-20210805191906-00676.warc.gz | 607,748,552 | 15,059 | # Solid mechanics
https://en.wikipedia.org/wiki/Solid_mechanics
Solid mechanics, also known as mechanics of solids, is the branch of continuum mechanics that studies the behavior of solid materials, especially their motion and deformation under the action of forces, temperature changes, phase changes, and other external or internal agents.
Solid mechanics is fundamental for civil, aerospace, nuclear, biomedical and mechanical engineering, for geology, and for many branches of physics such as materials science. [1] It has specific applications in many other areas, such as understanding the anatomy of living beings, and the design of dental prostheses and surgical implants. One of the most common practical applications of solid mechanics is the Euler–Bernoulli beam equation. Solid mechanics extensively uses tensors to describe stresses, strains, and the relationship between them.
Solid mechanics is a vast subject because of the wide range of solid materials available, such as steel, wood, concrete, biological materials, textiles, geological materials, and plastics.
## Fundamental aspects
A solid is a material that can support a substantial amount of shearing force over a given time scale during a natural or industrial process or action. This is what distinguishes solids from fluids, because fluids also support normal forces which are those forces that are directed perpendicular to the material plane across from which they act and normal stress is the normal force per unit area of that material plane. Shearing forces in contrast with normal forces, act parallel rather than perpendicular to the material plane and the shearing force per unit area is called shear stress.
Therefore, solid mechanics examines the shear stress, deformation and the failure of solid materials and structures.
The most common topics covered in solid mechanics include:
1. stability of structures - examining whether structures can return to a given equilibrium after disturbance or partial/complete failure
2. dynamical systems and chaos - dealing with mechanical systems highly sensitive to their given initial position
3. thermomechanics - analyzing materials with models derived from principles of thermodynamics
4. biomechanics - solid mechanics applied to biological materials e.g. bones, heart tissue
5. geomechanics - solid mechanics applied to geological materials e.g. ice, soil, rock
6. vibrations of solids and structures - examining vibration and wave propagation from vibrating particles and structures i.e. vital in mechanical, civil, mining, aeronautical, maritime/marine, aerospace engineering
7. fracture and damage mechanics - dealing with crack-growth mechanics in solid materials
8. composite materials - solid mechanics applied to materials made up of more than one compound e.g. reinforced plastics, reinforced concrete, fiber glass
9. variational formulations and computational mechanics - numerical solutions to mathematical equations arising from various branches of solid mechanics e.g. finite element method (FEM)
10. experimental mechanics - design and analysis of experimental methods to examine the behavior of solid materials and structures
## Relationship to continuum mechanics
As shown in the following table, solid mechanics inhabits a central place within continuum mechanics. The field of rheology presents an overlap between solid and fluid mechanics.
Continuum mechanicsThe study of the physics of continuous materials Solid mechanicsThe study of the physics of continuous materials with a defined rest shape. ElasticityDescribes materials that return to their rest shape after applied stresses are removed. PlasticityDescribes materials that permanently deform after a sufficient applied stress. RheologyThe study of materials with both solid and fluid characteristics. Fluid mechanicsThe study of the physics of continuous materials which deform when subjected to a force. Non-Newtonian fluidDo not undergo strain rates proportional to the applied shear stress. Newtonian fluids undergo strain rates proportional to the applied shear stress.
## Response models
A material has a rest shape and its shape departs away from the rest shape due to stress. The amount of departure from rest shape is called deformation, the proportion of deformation to original size is called strain. If the applied stress is sufficiently low (or the imposed strain is small enough), almost all solid materials behave in such a way that the strain is directly proportional to the stress; the coefficient of the proportion is called the modulus of elasticity. This region of deformation is known as the linearly elastic region.
It is most common for analysts in solid mechanics to use linear material models, due to ease of computation. However, real materials often exhibit non-linear behavior. As new materials are used and old ones are pushed to their limits, non-linear material models are becoming more common.
These are basic models that describe how a solid responds to an applied stress:
1. Elasticity – When an applied stress is removed, the material returns to its undeformed state. Linearly elastic materials, those that deform proportionally to the applied load, can be described by the linear elasticity equations such as Hooke's law.
2. Viscoelasticity – These are materials that behave elastically, but also have damping: when the stress is applied and removed, work has to be done against the damping effects and is converted in heat within the material resulting in a hysteresis loop in the stress–strain curve. This implies that the material response has time-dependence.
3. Plasticity – Materials that behave elastically generally do so when the applied stress is less than a yield value. When the stress is greater than the yield stress, the material behaves plastically and does not return to its previous state. That is, deformation that occurs after yield is permanent.
4. Viscoplasticity - Combines theories of viscoelasticity and plasticity and applies to materials like gels and mud.
5. Thermoelasticity - There is coupling of mechanical with thermal responses. In general, thermoelasticity is concerned with elastic solids under conditions that are neither isothermal nor adiabatic. The simplest theory involves the Fourier's law of heat conduction, as opposed to advanced theories with physically more realistic models.
## Timeline
Galileo Galilei published the book " Two New Sciences" in which he examined the failure of simple structures
Leonhard Euler developed the theory of buckling of columns
• 1826: Claude-Louis Navier published a treatise on the elastic behaviors of structures
• 1873: Carlo Alberto Castigliano presented his dissertation "Intorno ai sistemi elastici", which contains his theorem for computing displacement as partial derivative of the strain energy. This theorem includes the method of least work as a special case
• 1874: Otto Mohr formalized the idea of a statically indeterminate structure.
• 1922: Timoshenko corrects the Euler–Bernoulli beam equation
• 1936: Hardy Cross' publication of the moment distribution method, an important innovation in the design of continuous frames.
• 1941: Alexander Hrennikoff solved the discretization of plane elasticity problems using a lattice framework
• 1942: R. Courant divided a domain into finite subregions
• 1956: J. Turner, R. W. Clough, H. C. Martin, and L. J. Topp's paper on the "Stiffness and Deflection of Complex Structures" introduces the name "finite-element method" and is widely recognized as the first comprehensive treatment of the method as it is known today
## References
### Notes
1. ^ Allan Bower (2009). Applied mechanics of solids. CRC press. Retrieved March 5, 2017.
### Bibliography
• L.D. Landau, E.M. Lifshitz, Course of Theoretical Physics: Theory of Elasticity Butterworth-Heinemann, ISBN 0-7506-2633-X
• J.E. Marsden, T.J. Hughes, Mathematical Foundations of Elasticity, Dover, ISBN 0-486-67865-2
• P.C. Chou, N. J. Pagano, Elasticity: Tensor, Dyadic, and Engineering Approaches, Dover, ISBN 0-486-66958-0
• R.W. Ogden, Non-linear Elastic Deformation, Dover, ISBN 0-486-69648-0
• S. Timoshenko and J.N. Goodier," Theory of elasticity", 3d ed., New York, McGraw-Hill, 1970.
• G.A. Holzapfel, Nonlinear Solid Mechanics: A Continuum Approach for Engineering, Wiley, 2000
• A.I. Lurie, Theory of Elasticity, Springer, 1999.
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• Y. C. Fung, Pin Tong and Xiaohong Chen, Classical and Computational Solid Mechanics, 2nd Edition, World Scientific Publishing, 2017, ISBN 978-981-4713-64-1. | 1,899 | 9,007 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-31 | latest | en | 0.928178 |
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