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http://www.wikihow.com/Verify-the-Uncertainty-Principle-for-a-Quantum-Harmonic-Oscillator | 1,503,489,274,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886120194.50/warc/CC-MAIN-20170823113414-20170823133414-00418.warc.gz | 734,126,808 | 50,394 | Edit Article
wikiHow to Verify the Uncertainty Principle for a Quantum Harmonic Oscillator
The quantum harmonic oscillator is the quantum analogue to the classical simple harmonic oscillator. Using the ground state solution, we take the position and momentum expectation values and verify the uncertainty principle using them.
Part 1 Ground State Solution
1. 1
Recall the Schrödinger equation. This partial differential equation is the fundamental equation of motion in quantum mechanics that describes how a quantum state ${\displaystyle \psi }$ evolves in time. ${\displaystyle {\hat {H}}}$ denotes the Hamiltonian, the energy operator that describes the total energy of the system.
• ${\displaystyle i\hbar {\frac {\partial \psi }{\partial t}}={\hat {H}}\psi }$
2. 2
Write out the Hamiltonian for the harmonic oscillator. While the position and momentum variables have been replaced with their corresponding operators, the expression still resembles the kinetic and potential energies of a classical harmonic oscillator. Since we are working in physical space, the position operator is given by ${\displaystyle {\hat {x}}=x,}$ while the momentum operator is given by ${\displaystyle {\hat {p}}=-i\hbar {\frac {\partial }{\partial x}}.}$
• ${\displaystyle {\hat {H}}={\frac {{\hat {p}}^{2}}{2m}}+{\frac {1}{2}}m\omega ^{2}{\hat {x}}^{2}}$
3. 3
Write out the time-independent Schrödinger equation. We see that the Hamiltonian does not depend explicitly on time, so the solutions to the equation will be stationary states. The time-independent Schrödinger equation is an eigenvalue equation, so solving it means that we are finding the energy eigenvalues and their corresponding eigenfunctions - the wavefunctions.
• ${\displaystyle -{\frac {\hbar ^{2}}{2m}}{\frac {{\mathrm {d} }^{2}\psi }{{\mathrm {d} }x^{2}}}+{\frac {1}{2}}m\omega ^{2}x^{2}\psi =E\psi }$
4. 4
Solve the differential equation. This differential equation has variable coefficients and cannot easily be solved by elementary methods. However, after normalizing, the solution for the ground state can be written like so. Remember that this solution only describes a one-dimensional oscillator.
• ${\displaystyle \psi (x)=\left({\frac {m\omega }{\pi \hbar }}\right)^{1/4}\exp \left(-{\frac {m\omega }{2\hbar }}x^{2}\right)}$
• This is a Gaussian centered at ${\displaystyle x=0.}$ We will use the fact that this function is even to simplify our calculations in the next part.
Part 2 Expectation Values
1. 1
Recall the formula for the uncertainty. The uncertainty of an observable such as position is mathematically the standard deviation. That is, we find the average value, take each value and subtract from the average, square those values and average, and then take the square root.
• ${\displaystyle \sigma _{x}={\sqrt {\langle x^{2}\rangle -\langle x\rangle ^{2}}}}$
2. 2
Find ${\displaystyle \langle x\rangle }$. Since the function is even, we can deduce from symmetry that ${\displaystyle \langle x\rangle =0.}$
• If you set up the integral needed to evaluate, you will find that the integrand is an odd function, because an odd function times an even function is odd.
• ${\displaystyle \langle x\rangle =\int _{-\infty }^{\infty }x|\psi (x)|^{2}{\mathrm {d} }x}$
• One property of an odd function is that for every positive value of the function, there exists a doppelgänger - a corresponding negative value - that cancels them out. Since we are evaluating over all ${\displaystyle x}$ values, we know the integral evaluates to 0 without having to actually do the calculations.
3. 3
Calculate ${\displaystyle \langle x^{2}\rangle }$. Since our solution is written as a continuous wavefunction, we must employ the integral below. The integral describes the expectation value of ${\displaystyle x^{2}}$ integrated over all space.
• ${\displaystyle \langle x^{2}\rangle =\int _{-\infty }^{\infty }x^{2}|\psi (x)|^{2}{\mathrm {d} }x}$
4. 4
Substitute the wavefunction into the integral and simplify. We know that the wavefunction is even. The square of an even function is even as well, so we can pull out a factor of 2 and change the lower bound to 0.
• ${\displaystyle \langle x^{2}\rangle =2\left({\frac {m\omega }{\pi \hbar }}\right)^{1/2}\int _{0}^{\infty }x^{2}\exp \left(-{\frac {m\omega }{\hbar }}x^{2}\right){\mathrm {d} }x}$
5. 5
Evaluate. First, let ${\displaystyle \alpha ={\frac {m\omega }{\hbar }}.}$ Next, instead of integrating by parts, we will use the gamma function.
• {\displaystyle {\begin{aligned}\langle x^{2}\rangle &=2\left({\frac {m\omega }{\pi \hbar }}\right)^{1/2}\int _{0}^{\infty }x^{2}e^{-\alpha x^{2}}{\mathrm {d} }x,\ \ u=\alpha x^{2}\\&=2\left({\frac {m\omega }{\pi \hbar }}\right)^{1/2}\int _{0}^{\infty }{\frac {u}{\alpha }}e^{-u}{\mathrm {d} }u{\frac {1}{2\alpha x}},\ \ x={\sqrt {\frac {u}{\alpha }}}\\&=\left({\frac {m\omega }{\pi \hbar }}\right)^{1/2}\alpha ^{-3/2}\int _{0}^{\infty }u^{1/2}e^{-u}{\mathrm {d} }u\\&=\left({\frac {m\omega }{\pi \hbar }}\right)^{1/2}\left({\frac {m\omega }{\hbar }}\right)^{-3/2}\Gamma \left({\frac {3}{2}}\right),\ \ \Gamma \left({\frac {3}{2}}\right)={\frac {\sqrt {\pi }}{2}}\\&={\frac {\hbar }{m\omega }}{\frac {1}{\sqrt {\pi }}}{\frac {\sqrt {\pi }}{2}}\\&={\frac {\hbar }{2m\omega }}\end{aligned}}}
6. 6
Arrive at the uncertainty in position. Using the relation we wrote in step 1 of this part, ${\displaystyle \sigma _{x}}$ immediately follows from our results.
• ${\displaystyle \sigma _{x}={\sqrt {\frac {\hbar }{2m\omega }}}}$
7. 7
Find ${\displaystyle \langle p\rangle }$. As with average position, a symmetry argument can be made that leads to ${\displaystyle \langle p\rangle =0.}$
8. 8
Calculate ${\displaystyle \langle p^{2}\rangle }$. Instead of using the wavefunction to calculate this expectation value directly, we can use the energy of the wavefunction to simplify the calculations needed. The energy of the ground state of the harmonic oscillator is given below.
• ${\displaystyle E_{0}={\frac {1}{2}}\hbar \omega }$
9. 9
Relate the ground state energy with the particle's kinetic and potential energy. We expect this relation to hold not just for any position and momentum but also for their expectation values as well.
• ${\displaystyle {\frac {1}{2}}\hbar \omega ={\frac {\langle p^{2}\rangle }{2m}}+{\frac {1}{2}}m\omega ^{2}\langle x^{2}\rangle }$
10. 10
Solve for ${\displaystyle \langle p^{2}\rangle }$.
• ${\displaystyle m\hbar \omega =\langle p^{2}\rangle +m^{2}\omega ^{2}{\frac {\hbar }{2m\omega }}}$
• ${\displaystyle \langle p^{2}\rangle ={\frac {m\hbar \omega }{2}}}$
11. 11
Arrive at the uncertainty in momentum.
• ${\displaystyle \sigma _{p}={\sqrt {\frac {m\hbar \omega }{2}}}}$
Part 3 Verification of the Uncertainty Principle
1. 1
Recall Heisenberg's uncertainty principle for position and momentum. The uncertainty principle is a fundamental limit to the precision with which we can measure certain pairs of observables, such as position and momentum. See the tips for more background on the uncertainty principle.
• ${\displaystyle \sigma _{x}\sigma _{p}\geq {\frac {\hbar }{2}}}$
2. 2
Substitute the uncertainties of the quantum harmonic oscillator.
• {\displaystyle {\begin{aligned}{\sqrt {\frac {\hbar }{2m\omega }}}{\sqrt {\frac {m\hbar \omega }{2}}}&\geq {\frac {\hbar }{2}}\\{\frac {\hbar }{2}}&\geq {\frac {\hbar }{2}}\end{aligned}}}
• Our results are in agreement with the uncertainty principle. In fact, this relation only achieves equality in the ground state - if higher energy states are used, then the uncertainties in the position and momentum only grow.
Community Q&A
200 characters left
Tips
• There are two backgrounds as to why the uncertainty principle exists.
• From a wave mechanics perspective, the expressions of the wavefunction in terms of position and in terms of momentum are Fourier transforms of one another. One property of the Fourier transform is that a function and its Fourier transform cannot both be sharply localized.
• A simple example is the Fourier transform of the rectangular function. As the width of the function decreases (becomes more localized), the Fourier transform (a sinc curve) becomes flatter and flatter. An extreme example is the Dirac delta function, where the width is infinitesimal (perfect locality). Its Fourier transform is a constant (infinite uncertainty).
• The other way to look at it is from matrix mechanics. The position and momentum operators have a nonzero commutation relation. If two operators commute, then their commutation relation, as signified by the brackets below, would be 0.
• ${\displaystyle [{\hat {x}},{\hat {p}}]={\hat {x}}{\hat {p}}-{\hat {p}}{\hat {x}}=i\hbar }$
• It turns out that this commutation relation must imply a fundamental uncertainty principle. When an operator ${\displaystyle {\hat {x}}}$ acts on a state, then the wavefunction collapses into the eigenstate of ${\displaystyle {\hat {x}}}$ with a unique measurement (the eigenvalue). However, the eigenstate of ${\displaystyle {\hat {x}}}$ need not be an eigenstate of another operator ${\displaystyle {\hat {p}}.}$ If so, then there is no unique measurement for observable ${\displaystyle p,}$ which means that the state can only be written as a linear combination of momentum basis eigenstates. (When two operators do commute, then they have a simultaneous set of eigenstates in common (this is referred to as degeneracy) and the two observables can simultaneously be measured to arbitrary precision. This is always the case in classical mechanics.)
• This is the source of the uncertainty principle. It is not due to the limitations of our instruments that we cannot measure a particle's position and momentum to arbitrary precision. Rather, it is a fundamental property of the particles themselves.
Article Info
Categories: Physics
Thanks to all authors for creating a page that has been read 3,198 times. | 2,696 | 9,859 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 39, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2017-34 | latest | en | 0.825082 |
http://deltaepsilons.wordpress.com/2009/07/26/usamo-1972-4/ | 1,416,586,545,000,000,000 | text/html | crawl-data/CC-MAIN-2014-49/segments/1416400372999.9/warc/CC-MAIN-20141119123252-00147-ip-10-235-23-156.ec2.internal.warc.gz | 69,410,721 | 19,658 | ## USAMO 1972 #4 July 26, 2009
Posted by lumixedia in Problem-solving.
Tags: , , , ,
USAMO 1972 #4. Let ${R}$ denote a non-negative rational number. Determine a fixed set of integers ${a}$, ${b}$, ${c}$, ${d}$, ${e}$, ${f}$ such that, for every choice of ${R}$,
$\displaystyle |\frac{aR^2+bR+c}{dR^2+eR+f}-\sqrt[3]{2}|<|R-\sqrt[3]{2}|.$
Solution. From the desired inequality, we can conclude that
$\displaystyle 0\le\lim_{R\rightarrow\sqrt[3]{2}}|\frac{aR^2+bR+c}{dR^2+eR+f}-\sqrt[3]{2}|$
$\displaystyle \le\lim_{R\rightarrow\sqrt[3]{2}}||R-\sqrt[3]{2}|=0$
So
$\displaystyle \lim_{R\rightarrow\sqrt[3]{2}}|\frac{aR^2+bR+c}{dR^2+eR+f}-\sqrt[3]{2}|=0$
Since ${\sqrt[3]{2}}$ is not the root of any quadratic polynomial with integer coefficients, ${\frac{aR^2+bR+c}{dR^2+eR+f}}$ is continuous at ${\sqrt[3]{2}}$ and therefore
$\displaystyle \frac{a(\sqrt[3]{2})^2+b\sqrt[3]{2}+c}{d(\sqrt[3]{2})^2+e\sqrt[3]{2}+f}-\sqrt[3]{2}=0$
$\displaystyle a2^{2/3}+b2^{1/3}+c=e2^{2/3}+f2^{1/3}+2d$
For this equation to be satisfied, we must have ${a=e}$, ${b=f}$, ${c=2d}$. So the LHS of the inequality becomes the absolute value of
$\displaystyle \frac{aR^2+bR+2d}{dR^2+aR+b}-\sqrt[3]{2}$
$\displaystyle =\frac{aR^2+bR+2d-dR^2\sqrt[3]{2}-aR\sqrt[3]{2}-b\sqrt[3]{2}}{dR^2+aR+b}$
$\displaystyle =\frac{aR(R-\sqrt[3]{2})+b(R-\sqrt[3]{2})-d\sqrt[3]{2}(R+\sqrt[3]{2})(R-\sqrt[3]{2})}{dR^2+aR+b}$
$\displaystyle =(R-\sqrt[3]{2})\frac{aR+b-d\sqrt[3]{2}(R+\sqrt[3]{2})}{dR^2+aR+b}$
Dividing both sides of the inequality by ${|R-\sqrt[3]{2}|}$ and multiplying by ${|dR^2+aR+b|}$, we find that it is equivalent to
$\displaystyle |aR+b-d\sqrt[3]{2}(R+\sqrt[3]{2})|<|dR^2+aR+b|$
This is certainly satisfied if we choose ${(a,b,d)}$ so both ${aR+b-d\sqrt[3]{2}(R+\sqrt[3]{2})}$ and ${dR^2+aR+b}$ are always positive for nonnegative ${R}$. Any ${(a,b,d)}$ with ${a}$,${b}$,${d>0}$, ${a>d\sqrt[3]{2}}$, and ${b>d2^{2/3}}$ will work. One example is ${(a,b,c,d,e,f)=(2,2,2,1,2,2)}$.
It might possibly be interesting to find conditions on ${(a,b,d)}$ which are necessary and sufficient for the inequality to hold. It might also not be very interesting at all.
I suggested in the AoPS thread on the problem here that an interesting follow-up would be to figure out which choices take convergents of the continued fraction of $\sqrt[3]{2}$ to convergents. It’s probable no choice with this property exists since the continued fraction of non-square roots are aperiodic. | 968 | 2,454 | {"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": 40, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2014-49 | latest | en | 0.670001 |
http://www.math-problems-solved.com/q50400_7y383y72 | 1,511,184,971,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806066.5/warc/CC-MAIN-20171120130647-20171120150647-00708.warc.gz | 438,898,153 | 4,472 | Guide :
# 7y+3(8+3y)=72
How do i solve for y
## Research, Knowledge and Information :
### Linear Systems - University of Georgia
Take the two equations 3x + 7y = 4 and x = 5y. ... 3x = 72/6 + 5/6. 3x = 77/6. ... Let's solve this system of equations: 2x - 3y = -12. 5x + 3y = -17. So, ...
### 4.4 Systems of Equations - Three Variables
Systems of Equations - Three Variables ... 3(4x − 3y+2z)=( − 29) ... 2x − y+2z = − 8 3x − y − 4z =3 20) 4x − 7y+3z =1 3x + y − 2z =4
### How do you solve the following system?: -8x -7y =7, 5x -3y ...
How do you solve the following system?: #-8x -7y =7, 5x -3y = -4# ... How do you solve the following system: #2x+3y=22, 5x-2y=-26 #? See more ...
### Systems of equations with substitution: y=-5x+8 & 10x+2y=-2 ...
Systems of equations with substitution: 9x+3y=15 & y-x=5. Practice: ... And it's negative 5x plus 8, so 1, 2, 3, 4, 5, 6, 7, 8. And then it has a very steep downward ...
### Solve for x when 6x+8y=3 3x+7y=4? - Weknowtheanswer
6x + 8y = 3 3x + 7y = 4 (multiply ... + 7(5/6) = 4 -33/18 + 35/6 = 4 -33/18 + 105/18 = 72/18 72/18 = 72/18 ... –4y = –6x + 2 y = 2 3 x + 2 1 2x + 3y = 7 ...
### Which of these is a simplified form of the equation 7y + 8 ...
Answer to Which of these is a simplified form of the equation 7y + 8 = 9 + 3y + 2y? ... 7y + 8 = 9 + 3y + 2y 7y + 8 = 9 + 5y 2y + 8 = 9 2y = 1 2y/2 = 1/2 y = 0.5 ...
### 7y+3(y-8)=3(y+2)-4 - Get Easy Solution
Simple and best practice solution for 7y+3(y-8)=3(y+2 ... + -4 Reorder the terms: 7y + 3(-8 + y) = 3(y + 2) + -4 7y + (-8 * 3 + y * 3) = 3(y + 2) + -4 7y + (-24 + 3y ...
### Systems of Linear Equations - Columbia University in the City ...
Systems of Linear Equations . ... + 3y = 5. 3y = 9. y = 3 or-5 (-2) - 2y ... 7y = 106 first equation. 8x + y ...
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• Use proper spelling and grammar | 885 | 2,182 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5 | 4 | CC-MAIN-2017-47 | longest | en | 0.766411 |
https://www.basic-mathematics.com/a-real-life-example-of-multiplication-of-fractions.html | 1,721,220,248,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514771.72/warc/CC-MAIN-20240717120911-20240717150911-00716.warc.gz | 593,812,310 | 10,027 | # A Real Life Example of Multiplication of Fractions
by Kellie B
(Ann Arbor, MI)
Johnny works at a local bakery. In the morning Customer #1 buys 1/2 a pan of brownies. Later, Customer #2 buys 1/4 of the remaining brownies. The brownies cost \$10 for a whole tray.
How much did Customer #1 pay for his brownies? ___________
What fraction of a pan did Customer #2 buy? _______________
How much did Customer #2 pay for his brownies? ___________
Solution
Customer #1
Customer #1 bought half or 1/2 of the whole tray.
The tray cost 10. Half of 10 is 5
Another way to do the computation is to do
1/2 × 10 = (1 × 10) / 2 = 10 / 2 = 5
Customer #1 paid 5 dollars for his brownies
Customer #2
Customer 2 bought 1/4 of the remaining brownies.
The remaining brownies is 1/2
Thus, customer #2 bought 1/4 of 1/2
1/4 of 1/2 = 1/4 × 1/2
= (1 × 1) / (4 × 2)
= 1 / 8
Customer #2 bought 1/8 of the brownies
The fraction of a pan Customer #2 bought is 1/8
Customer #2 only bought 1/8 of the brownies.
So just find out what 1/8 of 10 is
1/8 of 10 = 1/8 × 10
1/8 × 10 = ( 1 × 10 ) / 8 = 10 / 8 = 1.25
Customer #2 paid 1.25 dollar for his brownies?
100 Tough Algebra Word Problems.
If you can solve these problems with no help, you must be a genius!
Recommended | 424 | 1,267 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2024-30 | latest | en | 0.941388 |
https://www.physicsforums.com/threads/transmission-coefficient-in-reaction-rate-theory.733330/ | 1,477,524,068,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988721008.78/warc/CC-MAIN-20161020183841-00279-ip-10-171-6-4.ec2.internal.warc.gz | 966,757,698 | 15,797 | # Transmission coefficient in reaction rate theory
1. Jan 17, 2014
### dsigg
This question relates to rate constants of transition events. The transmission coefficient κ reduces the value of the rate constant compared to the transition state theory (TST) value. I understand κ to be defined as the probability that a reaction coordinate q will proceed to product given that is has positive velocity at the transition state. In TST, there are no frictional forces to hold q back and κ = 1.
With increasing friction, there are recrossings, and κ is reduced. My question is how does k ever fall below the value of 0.5? Naively speaking, in the limit of very large friction the velocity is quickly randomized and the probabilities of falling back to reactant or moving forward to product should both be 0.5. Yet, Kramers theory and the more general Grote-Hynes theory allow for much smaller values of κ.
My thoughts are either that the stated definition of κ is wrong or if κ < 0.5 a greater number of transition events "bounce-back" to the reactant state than proceed to product. Neither option seems very appealing.
Can anyone set me straight?
2. Jan 18, 2014
### DrDu
I don't know Kramers theory too well, but I could imagine that for very high friction, the particle is effectively stopped at the TS. Then it has a high probability to go back to the educts.
Also without friction, you can get a small kappa due to quantum mechanical effects:
http://en.wikipedia.org/wiki/Rectangular_potential_barrier | 337 | 1,509 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2016-44 | longest | en | 0.935913 |
https://www.gamedev.net/topic/626886-simple-character-movement/ | 1,493,494,966,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917123560.51/warc/CC-MAIN-20170423031203-00267-ip-10-145-167-34.ec2.internal.warc.gz | 895,465,474 | 37,011 | • FEATURED
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# Simple character movement
Old topic!
Guest, the last post of this topic is over 60 days old and at this point you may not reply in this topic. If you wish to continue this conversation start a new topic.
1 reply to this topic
### #1noatom Members
Posted 24 June 2012 - 03:26 AM
The thing is,I'm very bad at math,right now I'm taking some trigonometry lessons from khanacademy.I know how to use cos,sin,tan without any problems,however,I have a big problem.
I can't seem to find a way to make a dot move.
Suppose I have something like this:
I have a dot. W moves the dot forward,A left,D right and S back.
BUT!! The dot's front must always be on the same line with the camera,something like in a 3rd person game!.
At the beginning the dot is at 0,0,0(x,y,z).If I press W,the dot will go on the positive Z axis,so the dot will have 0,0,1.
The same happens with with A,D,S(that go left,right and back).
The problem here is that this is just a linear movement.What happens when I have to take a curve? Or what happens when I the dot won't be on aimed on the z axis anymore?
You probably know what I'm talking about,I want that dot to take real curves,not just linear movement.And for that we need some trigonometry.
So I'm talking about something like a 3rd person game,where you need curves for realistic movement.If someone has any ideea how to make something like this,please explain with details...
### #2molehill mountaineer Members
Posted 26 June 2012 - 03:13 AM
I'm not entirely sure if this is what you're looking for but check out bezier and catmull-rom curves. Basically you use an equation to describe the path on which an asset must travel; catmull-rom being handy in that the object passes all points that you used to define the curve. I used it in the past to implement enemy flight paths in a schmup and it worked pretty well.
Read up here and here and here and google of course
Old topic!
Guest, the last post of this topic is over 60 days old and at this point you may not reply in this topic. If you wish to continue this conversation start a new topic. | 573 | 2,316 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2017-17 | latest | en | 0.929569 |
https://plosjournal.deepdyve.com/lp/emerald-publishing/field-and-field-circuit-models-of-electrical-machines-yqmAl0GGkH | 1,675,078,419,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499816.79/warc/CC-MAIN-20230130101912-20230130131912-00425.warc.gz | 476,721,686 | 32,602 | # Field and field‐circuit models of electrical machines
Field and field‐circuit models of electrical machines Purpose – The purpose of this paper is to develop and systemize the 3D finite element (FE) description of electromagnetic field in electrical machines. Design/methodology/approach – 3D FE models of electrical machines are considered. The model consists of FE equations for the magnetic field, equations describing eddy currents and equations, which describe the currents in the machine windings. The FE equations are further coupled by the electromagnetic torque to the differential equation of motion. In the presented field‐circuit model, the flux linkages with the windings are expressed by two components. Attention is paid to the description of machine winding. Both scalar and vector potential formulations are analysed. The FE equations are derived by using the notation of circuit theory. The methods of movement simulation and torque calculation in FE models are discussed. Findings – Proposed circuit description of electromagnetic field in electrical machines conforms to the applied method of electric and magnetic circuit analysis. The advantage of the presented description is that the equations of field model can be easy associated with the other equations of the electric drive system. Originality/value – The applied analogies between the FE formulation and the equivalent magnetic and electric network models help formulate efficient field models of electrical machines. The developed models after coupling to the models of supply and control system can be successfully used in the analysis and design electric drives. http://www.deepdyve.com/assets/images/DeepDyve-Logo-lg.png COMPEL: The International Journal for Computation and Mathematics in Electrical and Electronic Engineering Emerald Publishing
# Field and field‐circuit models of electrical machines
, Volume 29 (1): 15 – Jan 1, 2010
15 pages
/lp/emerald-publishing/field-and-field-circuit-models-of-electrical-machines-yqmAl0GGkH
Publisher
Emerald Publishing
ISSN
0332-1649
DOI
10.1108/03321641011007939
Publisher site
See Article on Publisher Site
### Abstract
Purpose – The purpose of this paper is to develop and systemize the 3D finite element (FE) description of electromagnetic field in electrical machines. Design/methodology/approach – 3D FE models of electrical machines are considered. The model consists of FE equations for the magnetic field, equations describing eddy currents and equations, which describe the currents in the machine windings. The FE equations are further coupled by the electromagnetic torque to the differential equation of motion. In the presented field‐circuit model, the flux linkages with the windings are expressed by two components. Attention is paid to the description of machine winding. Both scalar and vector potential formulations are analysed. The FE equations are derived by using the notation of circuit theory. The methods of movement simulation and torque calculation in FE models are discussed. Findings – Proposed circuit description of electromagnetic field in electrical machines conforms to the applied method of electric and magnetic circuit analysis. The advantage of the presented description is that the equations of field model can be easy associated with the other equations of the electric drive system. Originality/value – The applied analogies between the FE formulation and the equivalent magnetic and electric network models help formulate efficient field models of electrical machines. The developed models after coupling to the models of supply and control system can be successfully used in the analysis and design electric drives.
### Journal
COMPEL: The International Journal for Computation and Mathematics in Electrical and Electronic EngineeringEmerald Publishing
Published: Jan 1, 2010
Keywords: Electric machines; Finite element analysis; Magnetic fields; Eddy currents; Modelling
### References
Access the full text.
Sign up today, get DeepDyve free for 14 days. | 772 | 4,041 | {"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} | 2.5625 | 3 | CC-MAIN-2023-06 | latest | en | 0.882036 |
http://ampnet.azurewebsites.net/Resource/SearchIndex?ccs=6.RP.A.1 | 1,618,265,622,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038069267.22/warc/CC-MAIN-20210412210312-20210413000312-00111.warc.gz | 5,496,122 | 301,902 | ## Open Educational Resources
Displaying 1 of 6 results
Rating | Views Title Posted Date Contributor Common Core Standards Grade Levels Resource Type
#### Is a Super Ball REALLY Super?
Is a Super Ball REALLY "super?" This activity allows students to collect data and to make an argument regarding this quetions. See the PowerPoint for details about the activity... Note: It is best to gain access to an authentic, Wham-O Super Ball made with Zectron! https://www.amazon.com/orginal-super-ball-wtih-zectron/dp/B0001ZN49I/ref=sr_1_2?ie=UTF8&qid=1513808409&sr=8-2&keywords=whamo+super+ball
6.RP.A.1 6.RP.A.3 7.RP.A.2 7.RP.A.3 8.F.A.3 8.F.B.4 8.F.B.5 HSF-IF.C.7 HSF-IF.B.6 HSA-CED.A.2 MP.1 MP.2 MP.3 MP.4 MP.5 MP.6 MP.7 MP.8 5 6 8 7 HS Activity
#### How Many Houses?
Carpenters and apprentices are busy building houses... Students are asked to answer four questions regarding the relationships between the number of workers and how many houses can be built during a specific number of days.
8/8/2015 Lynda Boepple
6.RP.A.1 6.RP.A.2 6.RP.A.3 6.RP.A.3b 6.RP.A.3d 7.RP.A.1 7.RP.A.2 7.RP.A.2a 7.RP.A.2b 7.RP.A.2c MP.1 MP.2 MP.3 MP.4 MP.5 MP.6 MP.7 MP.8 6 7 Activity
In a brief video, students are confronted with the situation of a person squeezing a lemon slice into a small cup of water. Then a "big gulp" cup is placed next to the smaller, lemon filled cup. By asking the question, "How many lemon wedges do you need to add for the same lemony taste?" students will begin to experiment and mathematically determine the answer.
12/10/2014 Trey Cox
6.NS.A.1 6.RP.A.1 6.RP.A.2 6.RP.A.3 6.RP.A.3b 6.RP.A.3d MP.1 MP.2 MP.4 6 Activity
#### Fractions and Free Throws
Do we always add fractions by first finding a common denominator, etc.? Or, could it make sense to add two fractions by adding the numerators and denominators? This activity explores these questions.
5.NF.A.1 5.NF.A.2 6.RP.A.1 MP.1 MP.2 MP.3 MP.4 MP.6 MP.7 5 6 7 8 Activity
#### Sochi Olympics - Junior High Math Contest
This is the 2013 Chandler-Gilbert Community College Junior High Math Contest Team Project. You may use all or just parts of it as it contains lots of different math topics including: The idea of AVERAGE The idea of AVERAGE SPEED and GRAPHS of LINEAR FUNCTIONS The idea of CREATING and INTERPRETING BOX and WHISKER PLOTS The idea of SLOPE The idea of ANGLE The idea of PERCENT Look at the project and decide what parts your students are ready to tackle...and HAVE FUN! | 780 | 2,459 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2021-17 | latest | en | 0.831132 |
https://math-frolic.blogspot.com/2014/12/a-friday-puzzle.html | 1,495,906,554,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463608984.81/warc/CC-MAIN-20170527171852-20170527191852-00091.warc.gz | 973,739,097 | 20,713 | ## Pages
##### (AMS Bumper Sticker)
Web math-frolic.blogspot.com
## Friday, December 12, 2014
### A Friday Puzzle
To end the week, a simple-to-state puzzle that I've re-written/adapted from an old Henry Dudeney volume:
In the course of a year, the cats (and there are more than one) on Mr. Schlobotnik's farm killed 999,919 mice. If every cat killed exactly the same number of mice (and more than 1), then how many cats reside on the farm, given that the total number of cats is LESS than the number of mice killed per cat?
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. | 162 | 574 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2017-22 | longest | en | 0.869973 |
https://bigdl-project.github.io/0.10.0/APIGuide/Layers/Pooling-Layers/ | 1,656,697,673,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103943339.53/warc/CC-MAIN-20220701155803-20220701185803-00190.warc.gz | 179,500,471 | 8,624 | # Pooling Layers
## SpatialMaxPooling
Scala:
``````val mp = SpatialMaxPooling(2, 2, dW=2, dH=2, padW=0, padH=0, format=DataFormat.NCHW)
``````
Python:
``````mp = SpatialMaxPooling(2, 2, dw=2, dh=2, pad_w=0, pad_h=0, to_ceil=false, format="NCHW")
``````
Applies 2D max-pooling operation in kWxkH regions by step size dWxdH steps. The number of output features is equal to the number of input planes. If the input image is a 3D tensor nInputPlane x height x width, the output image size will be nOutputPlane x oheight x owidth where
• owidth = op((width + 2*padW - kW) / dW + 1)
• oheight = op((height + 2*padH - kH) / dH + 1)
op is a rounding operator. By default, it is floor. It can be changed by calling ceil() or floor() methods.
`````` outHeight = Math.ceil(inHeight.toFloat/strideH.toFloat)
outWidth = Math.ceil(inWidth.toFloat/strideW.toFloat)
padAlongHeight = Math.max(0, (outHeight - 1) * strideH + kernelH - inHeight)
padAlongWidth = Math.max(0, (outWidth - 1) * strideW + kernelW - inWidth)
``````
The format parameter is a string value (or DataFormat Object in Scala) of "NHWC" or "NCHW" to specify the input data format of this layer. In "NHWC" format data is stored in the order of [batch_size, height, width, channels], in "NCHW" format data is stored in the order of [batch_size, channels, height, width].
Scala example:
``````import com.intel.analytics.bigdl.nn.SpatialMaxPooling
import com.intel.analytics.bigdl.tensor.Tensor
import com.intel.analytics.bigdl.tensor.TensorNumericMath.TensorNumeric.NumericFloat
val mp = SpatialMaxPooling(2, 2, 2, 2)
val input = Tensor(1, 3, 3)
input(Array(1, 1, 1)) = 0.5336726f
input(Array(1, 1, 2)) = 0.7963769f
input(Array(1, 1, 3)) = 0.5674766f
input(Array(1, 2, 1)) = 0.1803996f
input(Array(1, 2, 2)) = 0.2460861f
input(Array(1, 2, 3)) = 0.2295625f
input(Array(1, 3, 1)) = 0.3073633f
input(Array(1, 3, 2)) = 0.5973460f
input(Array(1, 3, 3)) = 0.4298954f
val gradOutput = Tensor(1, 1, 1)
val output = mp.forward(input)
println(output)
``````
The output is,
``````(1,.,.) =
0.7963769
[com.intel.analytics.bigdl.tensor.DenseTensor of size 1x1x1]
``````
``````(1,.,.) =
0.0 0.023921492 0.0
0.0 0.0 0.0
0.0 0.0 0.0
[com.intel.analytics.bigdl.tensor.DenseTensor of size 1x3x3]
``````
Python example:
``````from bigdl.nn.layer import *
from bigdl.nn.criterion import *
from bigdl.optim.optimizer import *
from bigdl.util.common import *
mp = SpatialMaxPooling(2, 2, 2, 2)
input = np.array([0.5336726, 0.7963769, 0.5674766, 0.1803996, 0.2460861, 0.2295625, 0.3073633, 0.5973460, 0.4298954]).astype("float32")
input = input.reshape(1, 3, 3)
output = mp.forward(input)
print output
``````
The output is,
``````[array([[[ 0.79637688]]], dtype=float32)]
``````
``````[array([[[ 0. , 0.02392149, 0. ],
[ 0. , 0. , 0. ],
[ 0. , 0. , 0. ]]], dtype=float32)]
``````
## SpatialAveragePooling
Scala:
``````val m = SpatialAveragePooling(kW, kH, dW=1, dH=1, padW=0, padH=0, globalPooling=false, ceilMode=false, countIncludePad=true, divide=true, format=DataFormat.NCHW)
``````
Python:
``````m = SpatialAveragePooling(kw, kh, dw=1, dh=1, pad_w=0, pad_h=0, global_pooling=False, ceil_mode=False, count_include_pad=True, divide=True, format="NCHW")
``````
SpatialAveragePooling is a module that applies 2D average-pooling operation in `kW`x`kH` regions by step size `dW`x`dH`.
The number of output features is equal to the number of input planes.
`````` outHeight = Math.ceil(inHeight.toFloat/strideH.toFloat)
outWidth = Math.ceil(inWidth.toFloat/strideW.toFloat)
padAlongHeight = Math.max(0, (outHeight - 1) * strideH + kernelH - inHeight)
padAlongWidth = Math.max(0, (outWidth - 1) * strideW + kernelW - inWidth)
``````
The format parameter is a string value (or DataFormat Object in Scala) of "NHWC" or "NCHW" to specify the input data format of this layer. In "NHWC" format data is stored in the order of [batch_size, height, width, channels], in "NCHW" format data is stored in the order of [batch_size, channels, height, width].
Scala example:
``````scala>
import com.intel.analytics.bigdl.tensor.TensorNumericMath.TensorNumeric.NumericFloat
import com.intel.analytics.bigdl.nn._
import com.intel.analytics.bigdl.tensor._
val input = Tensor(1, 3, 3).randn()
val m = SpatialAveragePooling(3, 2, 2, 1)
val output = m.forward(input)
val gradOut = Tensor(1, 2, 1).randn()
scala> print(input)
(1,.,.) =
0.9916249 1.0299556 0.5608558
-0.1664829 1.5031902 0.48598626
0.37362042 -0.0966136 -1.4257964
[com.intel.analytics.bigdl.tensor.DenseTensor\$mcF\$sp of size 1x3x3]
scala> print(output)
(1,.,.) =
0.7341883
0.1123173
[com.intel.analytics.bigdl.tensor.DenseTensor of size 1x2x1]
(1,.,.) =
-0.42837557
-1.5104272
[com.intel.analytics.bigdl.tensor.DenseTensor\$mcF\$sp of size 1x2x1]
(1,.,.) =
-0.071395926 -0.071395926 -0.071395926
-0.3231338 -0.3231338 -0.3231338
-0.25173786 -0.25173786 -0.25173786
[com.intel.analytics.bigdl.tensor.DenseTensor of size 1x3x3]
``````
Python example:
``````from bigdl.nn.layer import *
import numpy as np
input = np.random.randn(1,3,3)
print "input is :",input
m = SpatialAveragePooling(3,2,2,1)
out = m.forward(input)
print "output of m is :",out
``````
produces output:
``````input is : [[[ 1.50602425 -0.92869054 -1.9393117 ]
[ 0.31447547 0.63450578 -0.92485516]
[-2.07858315 -0.05688643 0.73648798]]]
creating: createSpatialAveragePooling
output of m is : [array([[[-0.22297533],
[-0.22914261]]], dtype=float32)]
grad input of m is : [array([[[ 0.06282618, 0.06282618, 0.06282618],
[ 0.09333335, 0.09333335, 0.09333335],
[ 0.03050717, 0.03050717, 0.03050717]]], dtype=float32)]
``````
## VolumetricMaxPooling
Scala:
``````val layer = VolumetricMaxPooling(
kernelT, kernelW, kernelH,
strideT, strideW, strideH,
)
``````
Python:
``````layer = VolumetricMaxPooling(
kernelT, kernelW, kernelH,
strideT, strideW, strideH,
)
``````
Applies 3D max-pooling operation in kT x kW x kH regions by step size dT x dW x dH. The number of output features is equal to the number of input planes / dT. The input can optionally be padded with zeros. Padding should be smaller than half of kernel size. That is, padT < kT/2, padW < kW/2 and padH < kH/2
The input layout should be [batch, plane, time, height, width] or [plane, time, height, width]
Scala example:
``````import com.intel.analytics.bigdl.nn._
import com.intel.analytics.bigdl.utils.T
import com.intel.analytics.bigdl.tensor.Tensor
import com.intel.analytics.bigdl.tensor.TensorNumericMath.TensorNumeric.NumericFloat
val layer = VolumetricMaxPooling(
2, 2, 2,
1, 1, 1,
0, 0, 0
)
val input = Tensor(T(T(
T(
T(1.0f, 2.0f, 3.0f),
T(4.0f, 5.0f, 6.0f),
T(7.0f, 8.0f, 9.0f)
),
T(
T(4.0f, 5.0f, 6.0f),
T(1.0f, 3.0f, 9.0f),
T(2.0f, 3.0f, 7.0f)
)
)))
layer.forward(input)
layer.backward(input, Tensor(T(T(T(
T(0.1f, 0.2f),
T(0.3f, 0.4f)
)))))
``````
Its output should be
``````(1,1,.,.) =
5.0 9.0
8.0 9.0
[com.intel.analytics.bigdl.tensor.DenseTensor of size 1x1x2x2]
(1,1,.,.) =
0.0 0.0 0.0
0.0 0.1 0.0
0.0 0.3 0.4
(1,2,.,.) =
0.0 0.0 0.0
0.0 0.0 0.2
0.0 0.0 0.0
[com.intel.analytics.bigdl.tensor.DenseTensor of size 1x2x3x3]
``````
Python example:
``````from bigdl.nn.layer import VolumetricMaxPooling
import numpy as np
layer = VolumetricMaxPooling(
2, 2, 2,
1, 1, 1,
0, 0, 0
)
input = np.array([[
[
[1.0, 2.0, 3.0],
[4.0, 5.0, 6.0],
[7.0, 8.0, 9.0]
],
[
[4.0, 5.0, 6.0],
[1.0, 3.0, 9.0],
[2.0, 3.0, 7.0]
]
]])
layer.forward(input)
layer.backward(input, np.array([[[
[0.1, 0.2],
[0.3, 0.4]
]]]))
``````
Its output should be
``````array([[[[ 5., 9.],
[ 8., 9.]]]], dtype=float32)
array([[[[ 0. , 0. , 0. ],
[ 0. , 0.1 , 0. ],
[ 0. , 0.30000001, 0.40000001]],
[[ 0. , 0. , 0. ],
[ 0. , 0. , 0.2 ],
[ 0. , 0. , 0. ]]]], dtype=float32)
``````
## VolumetricAveragePooling
Scala:
``````val layer = VolumetricMaxPooling(
kT, kW, kH, dT, dW, dH,
)
``````
Python:
``````layer = VolumetricMaxPooling(
k_t, k_w, k_h, d_t, d_w, d_h
)
``````
Applies 3D average-pooling operation in kernel kT x kW x kH regions by step size dT x dW x dH. The number of output features is equal to the number of input planes / dT. The input can optionally be padded with zeros. Padding should be smaller than half of kernel size. That is, padT < kT/2, padW < kW/2 and padH < kH/2
The input layout should be [batch, plane, time, height, width] or [plane, time, height, width]
By default, countIncludePad=true, which means to include padding when dividing the number of elements in pooling region. One can use ceilMode to control whether the output size is to be ceiled or floored.
Scala example:
``````import com.intel.analytics.bigdl.nn._
import com.intel.analytics.bigdl.utils.T
import com.intel.analytics.bigdl.tensor.Tensor
import com.intel.analytics.bigdl.tensor.TensorNumericMath.TensorNumeric.NumericFloat
val layer = VolumetricAveragePooling(
2, 2, 2,
1, 1, 1,
0, 0, 0
)
val input = Tensor(T(T(
T(
T(1.0f, 2.0f, 3.0f),
T(4.0f, 5.0f, 6.0f),
T(7.0f, 8.0f, 9.0f)
),
T(
T(4.0f, 5.0f, 6.0f),
T(1.0f, 3.0f, 9.0f),
T(2.0f, 3.0f, 7.0f)
)
)))
layer.forward(input)
layer.backward(input, Tensor(T(T(T(
T(0.1f, 0.2f),
T(0.3f, 0.4f)
)))))
``````
Its output should be
``````(1,1,.,.) =
3.125 4.875
4.125 6.25
[com.intel.analytics.bigdl.tensor.DenseTensor of size 1x1x2x2]
(1,1,.,.) =
0.0125 0.0375 0.025
0.05 0.125 0.075
0.0375 0.087500006 0.05
(1,2,.,.) =
0.0125 0.0375 0.025
0.05 0.125 0.075
0.0375 0.087500006 0.05
[com.intel.analytics.bigdl.tensor.DenseTensor of size 1x2x3x3]
``````
Python example:
``````from bigdl.nn.layer import VolumetricAveragePooling
import numpy as np
layer = VolumetricAveragePooling(
2, 2, 2,
1, 1, 1,
0, 0, 0
)
input = np.array([[
[
[1.0, 2.0, 3.0],
[4.0, 5.0, 6.0],
[7.0, 8.0, 9.0]
],
[
[4.0, 5.0, 6.0],
[1.0, 3.0, 9.0],
[2.0, 3.0, 7.0]
]
]])
layer.forward(input)
layer.backward(input, np.array([[[
[0.1, 0.2],
[0.3, 0.4]
]]]))
``````
Its output should be
``````array([[[[ 3.125 4.875]
[ 4.125 6.25 ]]]], dtype=float32)
array([[[[ 0.0125 0.0375 0.025 ]
[ 0.05 0.125 0.075 ]
[ 0.0375 0.08750001 0.05 ]]
[[ 0.0125 0.0375 0.025 ]
[ 0.05 0.125 0.075 ]
[ 0.0375 0.08750001 0.05 ]]]], dtype=float32)
``````
## RoiPooling
Scala:
``````val m = RoiPooling(pooled_w, pooled_h, spatial_scale)
``````
Python:
``````m = RoiPooling(pooled_w, pooled_h, spatial_scale)
``````
RoiPooling is a module that performs Region of Interest pooling.
It uses max pooling to convert the features inside any valid region of interest into a small feature map with a fixed spatial extent of pooledH × pooledW (e.g., 7 × 7).
An RoI is a rectangular window into a conv feature map. Each RoI is defined by a four-tuple (x1, y1, x2, y2) that specifies its top-left corner (x1, y1) and its bottom-right corner (x2, y2).
RoI max pooling works by dividing the h × w RoI window into an pooledH × pooledW grid of sub-windows of approximate size h/H × w/W and then max-pooling the values in each sub-window into the corresponding output grid cell. Pooling is applied independently to each feature map channel
`forward` accepts a table containing 2 tensors as input, the first tensor is the input image, the second tensor is the ROI regions. The dimension of the second tensor should be (*,5) (5 are `batch_num, x1, y1, x2, y2`).
Scala example:
``````scala>
import com.intel.analytics.bigdl.tensor.TensorNumericMath.TensorNumeric.NumericFloat
import com.intel.analytics.bigdl.nn._
import com.intel.analytics.bigdl.tensor._
import com.intel.analytics.bigdl.tensor.Storage
import com.intel.analytics.bigdl.utils.T
val input_data = Tensor(2,2,6,8).randn()
val rois = Array(0, 0, 0, 7, 5, 1, 6, 2, 7, 5, 1, 3, 1, 6, 4, 0, 3, 3, 3, 3)
val input_rois = Tensor(Storage(rois.map(x => x.toFloat))).resize(4, 5)
val input = T(input_data,input_rois)
val m = RoiPooling(3, 2, 1)
val output = m.forward(input)
scala> print(input)
{
2: 0.0 0.0 0.0 7.0 5.0
1.0 6.0 2.0 7.0 5.0
1.0 3.0 1.0 6.0 4.0
0.0 3.0 3.0 3.0 3.0
[com.intel.analytics.bigdl.tensor.DenseTensor\$mcF\$sp of size 4x5]
1: (1,1,.,.) =
0.48066297 1.0994664 0.32474303 2.3391871 -0.79605865 0.836963950.36107457 1.2622415
0.657079 0.12720469 0.39894578 -0.41185552 -0.53111094 -0.36016005 -0.9726423 -2.5785272
0.3091435 -0.03613516 0.2375721 -1.1920663 -0.6757661 1.10612681.5409279 -0.17411499
0.23274016 -0.7149633 0.5473105 -0.40570387 -1.7966263 0.2071798-1.1530842 -0.010083453
-1.5769979 0.17043112 -0.28578365 -0.90779626 0.61457515 -0.1553582-0.3912479 -0.15326484
-0.24283029 1.3215472 1.3795123 -0.36933053 0.7077386 -0.56398267 0.6159163 0.5802894
(1,2,.,.) =
-1.1817129 -0.20470592 -1.3201113 0.36523122 -0.18260211 1.30210171.214403 1.1019816
0.7186407 0.78731173 1.5452348 0.0396181 0.5927014 1.17697431.0501136 -0.58295316
-0.96753055 0.6427254 -1.1396345 0.8701054 -0.22860864 -1.18719451.3372624 0.8616691
0.796831 -0.16609778 0.2950535 0.4595303 0.192339 0.6086106-0.76351887 -0.65964174
-0.12746814 -0.036058053 0.8858275 0.9677718 -1.1074747 -1.36859390.8783633 -0.11723315
-0.6947403 -0.23226547 -1.8510057 -1.3695518 -0.22317407 -0.36249024 -0.24097045 1.5691053
(2,1,.,.) =
0.84056973 1.144949 -1.0660682 0.4416162 -0.94440234 -0.24461010.91145027 -0.88650835
-0.81542057 0.14578317 -0.6531974 0.60776395 -0.32058007 -1.80771481.7660322 1.0680646
1.1328241 0.43677545 -0.9402618 -1.3002211 0.26012567 1.69481340.37849447 0.39286092
1.9443163 0.5415504 1.0793099 1.3312546 0.48346 1.2019655 0.3718734 0.21091922
0.5499047 1.6418253 0.8064177 0.37626198 0.8736181 -0.40816033 -0.5806787 1.286581
-0.5904657 -0.21188398 -0.040509004 1.2989452 1.6827602 1.3229258-0.68433124 0.87974
(2,2,.,.) =
-0.09759476 -0.32767114 0.16223079 2.3114302 -0.48496276 1.19290720.8572289 0.43429425
-1.0245247 0.19002944 1.5659521 -1.3689835 -1.4437296 -0.38216656 0.6333655 -0.57124794
-0.31111157 1.5184602 -1.3835855 -0.9295573 2.244521 -1.11849820.5451996 -0.4441631
-1.534093 -0.5599659 1.1980947 -1.0140935 1.3288999 0.19487387-0.1261734 -1.2222558
-0.070535585 0.9047848 -0.6719811 -1.6532638 -0.5290511 -0.18300447 0.69385433 0.018756092
0.24767837 0.620484 -0.5346291 1.0685066 -0.36903372 -0.26955062 1.1042496 0.5944603
[com.intel.analytics.bigdl.tensor.DenseTensor\$mcF\$sp of size 2x2x6x8]
}
scala> print(output)
(1,1,.,.) =
1.0994664 2.3391871 1.5409279
1.3795123 1.3795123 0.6159163
(1,2,.,.) =
1.5452348 1.5452348 1.3372624
0.8858275 0.9677718 1.5691053
(2,1,.,.) =
0.37849447 0.39286092 0.39286092
-0.5806787 1.286581 1.286581
(2,2,.,.) =
0.5451996 0.5451996 -0.4441631
1.1042496 1.1042496 0.5944603
(3,1,.,.) =
0.60776395 1.6948134 1.7660322
1.3312546 1.2019655 1.2019655
(3,2,.,.) =
2.244521 2.244521 0.6333655
1.3288999 1.3288999 0.69385433
(4,1,.,.) =
-0.40570387 -0.40570387 -0.40570387
-0.40570387 -0.40570387 -0.40570387
(4,2,.,.) =
0.4595303 0.4595303 0.4595303
0.4595303 0.4595303 0.4595303
[com.intel.analytics.bigdl.tensor.DenseTensor of size 4x2x2x3]
``````
Python example:
``````from bigdl.nn.layer import *
import numpy as np
input_data = np.random.randn(2,2,6,8)
input_rois = np.array([0, 0, 0, 7, 5, 1, 6, 2, 7, 5, 1, 3, 1, 6, 4, 0, 3, 3, 3, 3],dtype='float64').reshape(4,5)
print "input is :",[input_data, input_rois]
m = RoiPooling(3,2,1.0)
out = m.forward([input_data,input_rois])
print "output of m is :",out
``````
produces output:
``````input is : [array([[[[ 0.08500103, 0.33421796, 0.29084699, 1.60344635, -0.24289341,
-0.4793888 , 0.09452426, 0.16842477],
[-1.18575497, -0.53337542, 0.11661001, 0.9647904 , -0.25187936,
0.36516823, -0.16647209, -0.08095158],
[ 1.1982232 , -0.33549174, 0.11721347, -0.29319686, -0.01290122,
0.12344296, 0.30074829, -2.34951463],
[-0.60470899, -0.84657051, 0.1269276 , -0.06152321, -1.68838416,
-0.69808296, -2.06112892, -1.44790449],
[ 1.03944288, 0.13871728, 0.91478479, 0.47517105, 1.24638374,
0.98666841, 0.49403488, 1.26101127],
[-1.03949343, -0.39291108, 1.39107512, 1.73779253, 0.91656129,
0.103381 , 0.956243 , 0.44743548]],
[[ 0.79028054, 0.64244228, -0.37997334, -0.09130215, -2.3903429 ,
0.71919208, -0.14079786, 0.98304272],
[ 1.14678457, 1.58825227, 0.17137367, -0.62121819, -0.36103113,
-0.04452576, -0.0886136 , -1.32884721],
[ 0.06728957, -0.29701304, -0.52754207, -1.5785875 , 1.47354834,
-0.28545156, 0.49874194, 0.10277613],
[-0.10117571, -1.34902427, -1.40789327, 0.09853599, 0.60420022,
0.54869115, -0.49067696, 0.26696793],
[ 1.11780279, -0.77929016, 1.13772094, 0.14374057, 0.33199688,
-0.54057374, -0.45718861, 1.1577623 ],
[-1.4005645 , 1.15870496, 0.39292003, 0.88379515, 0.06440974,
0.65013063, 0.03759244, 0.18730126]]],
[[[-2.28272906, 0.06056305, 0.73632597, 0.10063274, -1.27497525,
-0.95597581, -0.22745785, 0.40146498],
[-1.37783475, 1.66000653, -1.80071745, -0.11805539, -0.27160583,
0.30691418, 2.62243232, -1.95274516],
[ 1.61364148, 1.91470546, -1.51984424, 2.13598224, -0.23156685,
-0.74203698, 0.65316888, 0.08018846],
[-1.8912854 , -0.50106158, 0.94937966, -0.10930541, 0.82136627,
-1.33209063, 1.43371302, -1.36370916],
[-0.52737928, -0.0681305 , -0.63472587, 0.41979229, -0.57093624,
-0.15968764, -1.005951 , -2.06873572],
[-2.34089346, 1.02593977, 0.90183415, 0.09504819, 0.53185448,
1.11305345, 1.290016 , -1.76216646]],
[[-0.10885459, -0.57089742, -0.55340708, -1.94445884, 1.30130049,
0.6333372 , -1.03100083, 0.0111167 ],
[ 0.59678149, -0.67601521, -1.25288718, -0.10922251, 3.06808996,
-1.46701513, -0.42140765, 1.12485412],
[ 1.21301567, -1.43304957, -0.56047239, 0.20716087, 1.40737646,
-0.08386437, -0.21916043, 0.85692906],
[ 1.59992399, -1.37044315, -0.71884386, 2.61830979, -0.74305496,
-0.32021174, 1.43275058, -0.3891857 ],
[-0.41355145, 0.22589689, 0.33154415, 0.86146815, -1.66326091,
0.37581697, -3.2250516 , -0.48807863],
[-2.52968957, 0.95801598, -1.20118154, 0.01141421, -0.11871498,
0.04555184, 1.3950473 , 0.37887998]]]]), array([[ 0., 0., 0., 7., 5.],
[ 1., 6., 2., 7., 5.],
[ 1., 3., 1., 6., 4.],
[ 0., 3., 3., 3., 3.]])]
creating: createRoiPooling
output of m is : [[[[ 1.19822323 1.60344636 0.36516821]
[ 1.39107513 1.73779249 1.26101124]]
[[ 1.58825231 1.47354829 0.98304272]
[ 1.158705 1.13772094 1.15776229]]]
[[[ 1.43371308 1.43371308 0.08018846]
[ 1.29001606 1.29001606 -1.7621665 ]]
[[ 1.43275058 1.43275058 0.85692906]
[ 1.39504731 1.39504731 0.37887999]]]
[[[ 2.13598228 0.30691418 2.62243223]
[ 0.82136625 0.82136625 1.43371308]]
[[ 3.06808996 3.06808996 -0.08386437]
[ 2.61830974 0.37581697 1.43275058]]]
[[[-0.06152321 -0.06152321 -0.06152321]
[-0.06152321 -0.06152321 -0.06152321]]
[[ 0.09853599 0.09853599 0.09853599]
[ 0.09853599 0.09853599 0.09853599]]]]
`````` | 8,536 | 20,002 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-27 | latest | en | 0.446186 |
https://documen.tv/question/a-cheetah-runs-5-miles-in-60-seconds-how-fast-did-the-cheetah-run-in-miles-per-hour-convert-seco-15349620-25/ | 1,723,209,116,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640763425.51/warc/CC-MAIN-20240809110814-20240809140814-00524.warc.gz | 168,363,913 | 16,374 | ## A cheetah runs 5 miles in 60 seconds. How fast did the cheetah run in miles per hour? Convert seconds in to hours first then solve.
Question
A cheetah runs 5 miles in 60 seconds. How fast did the cheetah run in miles per hour? Convert seconds in to hours first then solve.
in progress 0
3 years 2021-08-09T15:47:51+00:00 1 Answers 7 views 0
Speed of the cheetah is 300 miles per hour
Explanation:
As we know that the speed is defined as the ratio of distance and time
here we know that Cheetah runs a total distance of d = 5 miles
also the time taken by the cheetah is
t = 60 s
now we have
so here we have | 181 | 619 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2024-33 | latest | en | 0.939575 |
https://e-adventure.net/whats-40-of-75-solve-the-equation-faster-with-these-tips/ | 1,708,485,225,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473360.9/warc/CC-MAIN-20240221002544-20240221032544-00203.warc.gz | 230,828,555 | 15,058 | # What’s 40% of 75? Solve the Equation Faster with These Tips
Calculating percentages may seem like a daunting task to many of us, but in reality, it’s not that difficult. Knowing the right tricks and shortcuts can make it a lot easier. One question that often pops up is ‘What’s 40% of 75?’.
In this article, we will discuss some simple and effective tips that will help you solve this equation faster, without making any mistakes. So, let’s get started!
## Understanding Percentages
Before we get into the details of calculating percentages, it’s important to understand what percentages actually are. In simple terms, percentages are ratios that are expressed as a fraction of 100. For example, if we say that something is 50%, we mean that it is 50 out of 100.
Now, let’s move on to our main topic – finding 40% of 75.
## Method 1: Divide and Multiply
One of the simplest and most common methods of finding percentages is by using the divide and multiply method. Here’s how it works:
### Step 1:
Divide the percentage by 100: 40 รท 100 = 0.4
### Step 2:
Multiply the result by the number you want to find the percentage of: 0.4 x 75 = 30
Therefore, 40% of 75 is 30.
## Method 2: Percentage Formula
Another method that you can use to find percentages is the percentage formula, which is:
Percent / 100 = Part / Whole
Using this formula, we can find the part, which in our case is 40% of 75. Here’s how it works:
### Step 1:
Write down the formula: Percent / 100 = Part / Whole
### Step 2:
Fill in the values that you know: 40 / 100 = Part / 75
### Step 3:
Cross-multiply the values: 40 x 75 = 100 x Part
### Step 4:
Solve for Part: Part = (40 x 75) / 100 = 30
Therefore, 40% of 75 is 30.
## Method 3: Use a Calculator
If you’re not comfortable with mental math, you can always use a calculator to find percentages. Most calculators have a percentage button, which you can use to find the percentage of a number. Here’s how you can use a calculator to find 40% of 75:
### Step 2:
Press the percentage button
### Step 3:
• What does ‘percent’ mean?
• Percent is a ratio that expresses a number as a fraction of 100. It is denoted by the symbol ‘%’
• What is the formula for calculating percentages?
• The formula for finding the percentage of a number is: (percentage / 100) x number
• What is the percentage of 75?
• This question is incomplete. To find the percentage, you need to know what number or value it is a percentage of.
• What is the easiest way to find percentages?
• The easiest way to find percentages is by using the divide and multiply method or a calculator.
• What is the percentage of 40?
• This question is incomplete. To find the percentage, you need to know what number or value it is a percentage of.
## Conclusion
Calculating percentages doesn’t have to be a difficult or time-consuming task. With the right tips and techniques, you can solve equations faster and more accurately.
In this article, we discussed three methods for finding 40% of 75. Whether you choose to use the divide and multiply method, the percentage formula, or a calculator, the key is to choose the method that works best for you.
## References
https://www.mathsisfun.com/percentage.html
https://www.calculatorsoup.com/calculators/algebra/percentages-calculator.php
https://www.rapidtables.com/calc/math/Percentage_Calculator.html | 832 | 3,371 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-10 | latest | en | 0.862479 |
https://albertteen.com/uk/gcse/physics/forces-in-motion/estimating-forces-in-transport | 1,726,892,026,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725701427996.97/warc/CC-MAIN-20240921015054-20240921045054-00718.warc.gz | 75,381,059 | 44,958 | Albert Teen
YOU ARE LEARNING:
Estimating Forces in Transport
# Estimating Forces in Transport
### Estimating Forces in Transport
Sometimes, we may need to give an estimate for the speed, acceleration and force involved in everyday transportations.
If the typical speed that a person walks at is $1.5\space m/s$, what do you think might be a typical running speed?
If a typical running speed is $3\space m/s$, what do you think might be the typical speed that a train travels at?
If a typical speed of a train is $30\space m/s$ , what do you think might be the typical speed that a plane travels at?
Whilst all objects travel at different speeds at different times, sometimes we need to estimate speeds when dealing with large forces and accelerations in transport. We use the symbol $\sim$ to indicate that a value is approximate.
In physics we tend to use metres per second as the unit of speed, where you might be more familiar with miles per hour.
1
$1mph=0.4m/s$. Which option best describes how to convert from $mph$ to $m/s$?
A) Divide the value by 0.4 B) Divide the value by 2.2 C) Multiply the value by 0.6
2
A car is travelling at an average speed of $60mph$ on a motorway. Try to estimate its average speed in $m/s$. Give your answer written as a whole number only.
3
An easy way to find an approximate value in $m/s$ if you have an estimate in $mph$, is to half the value.
The actual value will be less, but you will have a close enough approximation.
Approximately, what is $30mph$ converted to $m/s$?
Which option is the best definition of acceleration?
A car accelerates from rest to a typical speed of $\sim\space 25\space m/s$ in 10 seconds. Estimate its acceleration.
1
Remember that speed and acceleration are two different quantities.
To estimate the acceleration we need to know the change in velocity, and the time taken for the change.
2
If the car has accelerated from rest to $\sim\space 25\space m/s$, what is the change of velocity?
3
The change of velocity is the difference between the starting and end speed.
If an object is not starting at zero, you need to subtract the starting speed from the end speed.
4
Acceleration is change in velocity over time, so what would you do next with your value of $25\space m/s$ to find an approximate acceleration?
5
Estimate the acceleration of the car.
A van accelerates from rest to $\sim\space 20\space m/s$ in approximately 10 seconds. Estimate its acceleration.
We can estimate the forces acting on vehicles, using approximate or typical values. Which values do we need to estimate a force causing acceleration of a vehicle?
A lorry has a mass of $\sim\space 30,000\space kg$. It accelerates from rest to $\sim\space 22\space m/s$ in 60 seconds. Estimate the force needed to accelerate the lorry,
1
First we must find the acceleration of the lorry. Give your answer to 1 decimal place.
2
We know that the approximate mass of the lorry is $30,000\space kg$, and its acceleration is $0.4\space m/s^{2}$ . Estimate the magnitude of the force which caused the acceleration using these values.
3
You have estimated the force because you used "approximate" or "typical values".
If you know typical values, you can estimate a force!
A typical car has a mass of $\sim\space 1200\space kg$ . Estimate the force needed to accelerate a car at a rate of $2\space m/s^{2}$ . | 808 | 3,377 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 41, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-38 | latest | en | 0.91533 |
https://se.mathworks.com/matlabcentral/cody/problems/80-test-for-balanced-parentheses/solutions/2784 | 1,597,249,838,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738905.62/warc/CC-MAIN-20200812141756-20200812171756-00276.warc.gz | 484,713,886 | 15,667 | Cody
Problem 80. Test for balanced parentheses
Solution 2784
Submitted on 26 Jan 2012 by @bmtran (Bryant Tran)
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
%% inStr = '()'; out_correct = true; assert(isequal(isBalanced(inStr),out_correct))
2 Pass
%% inStr = ')'; out_correct = false; assert(isequal(isBalanced(inStr),out_correct))
3 Pass
%% inStr = '(z*(a-(x+3))/(y))'; out_correct = true; assert(isequal(isBalanced(inStr),out_correct))
4 Pass
%% inStr = ')('; out_correct = false; assert(isequal(isBalanced(inStr),out_correct))
5 Pass
%% inStr = '(x)(x-y)'; out_correct = true; assert(isequal(isBalanced(inStr),out_correct))
6 Pass
%% inStr = ':-)'; out_correct = false; assert(isequal(isBalanced(inStr),out_correct))
7 Pass
%% inStr = ')()'; out_correct = false; assert(isequal(isBalanced(inStr),out_correct))
8 Pass
%% inStr = '(()'; out_correct = false; assert(isequal(isBalanced(inStr),out_correct)) | 313 | 1,047 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2020-34 | latest | en | 0.211074 |
http://mathhelpforum.com/math-topics/53647-adding-subtracting-complex-numbers.html | 1,511,144,110,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934805894.15/warc/CC-MAIN-20171120013853-20171120033853-00138.warc.gz | 199,835,109 | 10,719 | 1. ## Adding and Subtracting Complex Numbers
Ok. I understand that a complex number can be written in the form...a+bi.
Which leads me to the idea that
(5-6i)-4i is equivalent to (5-6i)-(0-4i).
Which gives me 5-2i as a solution, because 5-0=5 and -6i-(-4i) is -2i. So..
(5-6i)-4i = 5-2i
Is this a true statement?
My text says no, and I can't wrap my head around it.
P.S. (a+bi)-(c+di)= a+bi-c-di = (a-c)+(b-d)i
2. Originally Posted by EyesForEars
Ok. I understand that a complex number can be written in the form...a+bi.
Which leads me to the idea that
(5-6i)-4i is equivalent to (5-6i)-(0-4i).
The above should be written (5-6i)-4i = (5-6i) - (0 + 4i). Remember, when you put grouping symbols around an expression and precede the group with a negative sign, each sign within the group must be changed to its opposite.
-4i = -(+4i)
Which gives me 5-2i as a solution, because 5-0=5 and -6i-(-4i) is -2i. So.. (5-6i)-4i = 5-2i
Removing the parentheses on the above we get 5 - 6i - 4i = 5 - 10i
Is this a true statement?
My text says no, and I can't wrap my head around it.
P.S. (a+bi)-(c+di)= a+bi-c-di = (a-c)+(b-d)i
Using this, we get (5-6i)-(0+4i) = 5-6i-0-4i = (5-0)+(-6-4)i = 5-10i
..
3. So, any time I apply grouping symbols to an expression preceeded by a negative sign, I have to then reverse the signs of said terms. Got it.
I think I got it.
So, I'll say.
(19-26i)-7i = (19-26i)-(0+7i) = 19-26i-0-7i = 19-33i
4. Originally Posted by EyesForEars
So, any time I apply grouping symbols to an expression preceeded by a negative sign, I have to then reverse the signs of said terms. Got it.
I think I got it.
So, I'll say.
(19-26i)-7i = (19-26i)-(0+7i) = 19-26i-0-7i = 19-33i
You are correct, my friend!!
5. dont forget "--" = + | 645 | 1,749 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2017-47 | longest | en | 0.914018 |
https://oeis.org/A267315 | 1,503,188,555,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886105955.66/warc/CC-MAIN-20170819235943-20170820015943-00156.warc.gz | 822,347,540 | 4,002 | This site is supported by donations to The OEIS Foundation.
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A267315 Decimal expansion of the Dirichlet eta function at 4. 4
9, 4, 7, 0, 3, 2, 8, 2, 9, 4, 9, 7, 2, 4, 5, 9, 1, 7, 5, 7, 6, 5, 0, 3, 2, 3, 4, 4, 7, 3, 5, 2, 1, 9, 1, 4, 9, 2, 7, 9, 0, 7, 0, 8, 2, 9, 2, 8, 8, 8, 6, 0, 4, 4, 2, 2, 2, 6, 0, 4, 1, 8, 8, 5, 1, 3, 6, 0, 5, 5, 3, 9, 1, 6, 3, 5, 9, 7, 7, 4, 0, 7, 3, 7, 2, 9, 5, 9, 3, 1, 4, 4, 8, 9, 8, 7, 4, 2, 7, 5, 7, 8, 8, 6, 6, 9, 6, 2, 1, 6, 9, 5, 3, 7, 3, 9, 9, 6, 1, 2 (list; constant; graph; refs; listen; history; text; internal format)
OFFSET 0,1 LINKS OEIS Wiki, Euler's alternating zeta function Eric Weisstein's World of Mathematics, Dirichlet Eta Function Wikipedia, Dirichlet Eta Function FORMULA Equals Sum_{k>0} -(-1)^k/k^4 = (7*Pi^4)/720. EXAMPLE 1/1^4 - 1/2^4 + 1/3^4 - 1/4^4 + 1/5^4 - 1/6^4 + ... = 0.9470328294972459175765032344735219149279070829288860... MATHEMATICA RealDigits[(7 Pi^4)/720, 10, 120][[1]] PROG (PARI) 7*Pi^4/720 \\ Michel Marcus, Feb 01 2016 (MAGMA) pi:= 7*Pi(RealField(110))^4 / 720; Reverse(Intseq(Floor(10^100*pi))); // Vincenzo Librandi, Feb 04 2016 (Sage) s = RLF(0); s RealField(110)(s) for i in xrange(1, 10000): s += -((-1)^i/((i)^4)) print s # Terry D. Grant, Aug 04 2016 CROSSREFS Cf. A002162, A072691, A197070. Sequence in context: A245084 A199054 A131109 * A247412 A154397 A116186 Adjacent sequences: A267312 A267313 A267314 * A267316 A267317 A267318 KEYWORD nonn,cons AUTHOR Ilya Gutkovskiy, Jan 13 2016 STATUS approved
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The OEIS Community | Maintained by The OEIS Foundation Inc. | 893 | 1,931 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-34 | longest | en | 0.542173 |
https://www.mathsqrt.com/en/arithmetic-progression | 1,723,161,788,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640741453.47/warc/CC-MAIN-20240808222119-20240809012119-00585.warc.gz | 686,511,539 | 5,411 | # Arithmetic Progression
An Arithmetic progression or arithmetic sequence is a sequence of numbers such that the difference between the consecutive terms is constant. The arithmetic progression can be finite or infinite.
The nth term (an) of the sequence is given by:
$$a_{n}=a_{1}+(n-1)d$$
where,
a1ā initial term of an arithmetic progression;
dā common difference of successive members;
nā 1, 2, 3 ...
The sum of the first n members is called arithmetic series:
$$S_{n}=\frac{n}{2}\times(a_{1}+a_{n})=\frac{n}{2}\times\left [2a_{1}+(n-1)d \right ]$$
The standard deviation of any arithmetic progression can be calculated as:
$$\sigma=\left |d \right |\times\sqrt{\frac{(n+1)(n+1)}{12}}$$ | 201 | 704 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2024-33 | latest | en | 0.802465 |
https://www.lmfdb.org/ModularForm/GL2/Q/holomorphic/171/1/o/a/94/1/ | 1,713,130,663,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816893.9/warc/CC-MAIN-20240414192536-20240414222536-00249.warc.gz | 818,416,089 | 72,188 | # Properties
Label 171.1.o.a.94.1 Level $171$ Weight $1$ Character 171.94 Analytic conductor $0.085$ Analytic rank $0$ Dimension $4$ Projective image $A_{4}$ CM/RM no Inner twists $4$
# Related objects
Show commands: Magma / PariGP / SageMath
## Newspace parameters
comment: Compute space of new eigenforms
[N,k,chi] = [171,1,Mod(94,171)]
mf = mfinit([N,k,chi],0)
lf = mfeigenbasis(mf)
from sage.modular.dirichlet import DirichletCharacter
H = DirichletGroup(171, base_ring=CyclotomicField(6))
chi = DirichletCharacter(H, H._module([2, 3]))
N = Newforms(chi, 1, names="a")
//Please install CHIMP (https://github.com/edgarcosta/CHIMP) if you want to run this code
chi := DirichletCharacter("171.94");
S:= CuspForms(chi, 1);
N := Newforms(S);
Level: $$N$$ $$=$$ $$171 = 3^{2} \cdot 19$$ Weight: $$k$$ $$=$$ $$1$$ Character orbit: $$[\chi]$$ $$=$$ 171.o (of order $$6$$, degree $$2$$, minimal)
## Newform invariants
comment: select newform
sage: f = N[0] # Warning: the index may be different
gp: f = lf[1] \\ Warning: the index may be different
Self dual: no Analytic conductor: $$0.0853401171602$$ Analytic rank: $$0$$ Dimension: $$4$$ Relative dimension: $$2$$ over $$\Q(\zeta_{6})$$ Coefficient field: $$\Q(\zeta_{12})$$ comment: defining polynomial gp: f.mod \\ as an extension of the character field Defining polynomial: $$x^{4} - x^{2} + 1$$ x^4 - x^2 + 1 Coefficient ring: $$\Z[a_1, a_2]$$ Coefficient ring index: $$1$$ Twist minimal: yes Projective image: $$A_{4}$$ Projective field: Galois closure of 4.0.29241.1 Artin image: $\SL(2,3):C_2$ Artin field: Galois closure of $$\mathbb{Q}[x]/(x^{16} - \cdots)$$
## Embedding invariants
Embedding label 94.1 Root $$0.866025 + 0.500000i$$ of defining polynomial Character $$\chi$$ $$=$$ 171.94 Dual form 171.1.o.a.151.1
## $q$-expansion
comment: q-expansion
sage: f.q_expansion() # note that sage often uses an isomorphic number field
gp: mfcoefs(f, 20)
$$f(q)$$ $$=$$ $$q+(-0.866025 + 0.500000i) q^{2} +1.00000i q^{3} +(-0.500000 + 0.866025i) q^{5} +(-0.500000 - 0.866025i) q^{6} +(-0.500000 - 0.866025i) q^{7} -1.00000i q^{8} -1.00000 q^{9} +O(q^{10})$$ $$q+(-0.866025 + 0.500000i) q^{2} +1.00000i q^{3} +(-0.500000 + 0.866025i) q^{5} +(-0.500000 - 0.866025i) q^{6} +(-0.500000 - 0.866025i) q^{7} -1.00000i q^{8} -1.00000 q^{9} -1.00000i q^{10} +(0.500000 + 0.866025i) q^{11} +(0.866025 + 0.500000i) q^{13} +(0.866025 + 0.500000i) q^{14} +(-0.866025 - 0.500000i) q^{15} +(0.500000 + 0.866025i) q^{16} +(0.866025 - 0.500000i) q^{18} +1.00000i q^{19} +(0.866025 - 0.500000i) q^{21} +(-0.866025 - 0.500000i) q^{22} +(0.500000 - 0.866025i) q^{23} +1.00000 q^{24} -1.00000 q^{26} -1.00000i q^{27} +(0.866025 - 0.500000i) q^{29} +1.00000 q^{30} +(-0.866025 - 0.500000i) q^{31} +(-0.866025 + 0.500000i) q^{33} +1.00000 q^{35} +(-0.500000 - 0.866025i) q^{38} +(-0.500000 + 0.866025i) q^{39} +(0.866025 + 0.500000i) q^{40} +(0.866025 + 0.500000i) q^{41} +(-0.500000 + 0.866025i) q^{42} +(-0.500000 - 0.866025i) q^{43} +(0.500000 - 0.866025i) q^{45} +1.00000i q^{46} +(-0.500000 - 0.866025i) q^{47} +(-0.866025 + 0.500000i) q^{48} +(0.500000 + 0.866025i) q^{54} -1.00000 q^{55} +(-0.866025 + 0.500000i) q^{56} -1.00000 q^{57} +(-0.500000 + 0.866025i) q^{58} +(-0.866025 - 0.500000i) q^{59} +(0.500000 + 0.866025i) q^{61} +1.00000 q^{62} +(0.500000 + 0.866025i) q^{63} -1.00000 q^{64} +(-0.866025 + 0.500000i) q^{65} +(0.500000 - 0.866025i) q^{66} +(-0.866025 - 0.500000i) q^{67} +(0.866025 + 0.500000i) q^{69} +(-0.866025 + 0.500000i) q^{70} +1.00000i q^{72} +(0.500000 - 0.866025i) q^{77} -1.00000i q^{78} +(0.866025 - 0.500000i) q^{79} -1.00000 q^{80} +1.00000 q^{81} -1.00000 q^{82} +(0.500000 + 0.866025i) q^{83} +(0.866025 + 0.500000i) q^{86} +(0.500000 + 0.866025i) q^{87} +(0.866025 - 0.500000i) q^{88} +1.00000i q^{90} -1.00000i q^{91} +(0.500000 - 0.866025i) q^{93} +(0.866025 + 0.500000i) q^{94} +(-0.866025 - 0.500000i) q^{95} +(-0.866025 + 0.500000i) q^{97} +(-0.500000 - 0.866025i) q^{99} +O(q^{100})$$ $$\operatorname{Tr}(f)(q)$$ $$=$$ $$4 q - 2 q^{5} - 2 q^{6} - 2 q^{7} - 4 q^{9}+O(q^{10})$$ 4 * q - 2 * q^5 - 2 * q^6 - 2 * q^7 - 4 * q^9 $$4 q - 2 q^{5} - 2 q^{6} - 2 q^{7} - 4 q^{9} + 2 q^{11} + 2 q^{16} + 2 q^{23} + 4 q^{24} - 4 q^{26} + 4 q^{30} + 4 q^{35} - 2 q^{38} - 2 q^{39} - 2 q^{42} - 2 q^{43} + 2 q^{45} - 2 q^{47} + 2 q^{54} - 4 q^{55} - 4 q^{57} - 2 q^{58} + 2 q^{61} + 4 q^{62} + 2 q^{63} - 4 q^{64} + 2 q^{66} + 2 q^{77} - 4 q^{80} + 4 q^{81} - 4 q^{82} + 2 q^{83} + 2 q^{87} + 2 q^{93} - 2 q^{99}+O(q^{100})$$ 4 * q - 2 * q^5 - 2 * q^6 - 2 * q^7 - 4 * q^9 + 2 * q^11 + 2 * q^16 + 2 * q^23 + 4 * q^24 - 4 * q^26 + 4 * q^30 + 4 * q^35 - 2 * q^38 - 2 * q^39 - 2 * q^42 - 2 * q^43 + 2 * q^45 - 2 * q^47 + 2 * q^54 - 4 * q^55 - 4 * q^57 - 2 * q^58 + 2 * q^61 + 4 * q^62 + 2 * q^63 - 4 * q^64 + 2 * q^66 + 2 * q^77 - 4 * q^80 + 4 * q^81 - 4 * q^82 + 2 * q^83 + 2 * q^87 + 2 * q^93 - 2 * q^99
## Character values
We give the values of $$\chi$$ on generators for $$\left(\mathbb{Z}/171\mathbb{Z}\right)^\times$$.
$$n$$ $$20$$ $$154$$ $$\chi(n)$$ $$e\left(\frac{1}{3}\right)$$ $$-1$$
## Coefficient data
For each $$n$$ we display the coefficients of the $$q$$-expansion $$a_n$$, the Satake parameters $$\alpha_p$$, and the Satake angles $$\theta_p = \textrm{Arg}(\alpha_p)$$.
Display $$a_p$$ with $$p$$ up to: 50 250 1000 Display $$a_n$$ with $$n$$ up to: 50 250 1000
$$n$$ $$a_n$$ $$a_n / n^{(k-1)/2}$$ $$\alpha_n$$ $$\theta_n$$
$$p$$ $$a_p$$ $$a_p / p^{(k-1)/2}$$ $$\alpha_p$$ $$\theta_p$$
$$2$$ −0.866025 + 0.500000i −0.866025 + 0.500000i −0.866025 0.500000i $$-0.833333\pi$$
1.00000i $$0.5\pi$$
$$3$$ 1.00000i 1.00000i
$$4$$ 0 0
$$5$$ −0.500000 + 0.866025i −0.500000 + 0.866025i 0.500000 + 0.866025i $$0.333333\pi$$
−1.00000 $$\pi$$
$$6$$ −0.500000 0.866025i −0.500000 0.866025i
$$7$$ −0.500000 0.866025i −0.500000 0.866025i 0.500000 0.866025i $$-0.333333\pi$$
−1.00000 $$\pi$$
$$8$$ 1.00000i 1.00000i
$$9$$ −1.00000 −1.00000
$$10$$ 1.00000i 1.00000i
$$11$$ 0.500000 + 0.866025i 0.500000 + 0.866025i 1.00000 $$0$$
−0.500000 + 0.866025i $$0.666667\pi$$
$$12$$ 0 0
$$13$$ 0.866025 + 0.500000i 0.866025 + 0.500000i 0.866025 0.500000i $$-0.166667\pi$$
1.00000i $$0.5\pi$$
$$14$$ 0.866025 + 0.500000i 0.866025 + 0.500000i
$$15$$ −0.866025 0.500000i −0.866025 0.500000i
$$16$$ 0.500000 + 0.866025i 0.500000 + 0.866025i
$$17$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$18$$ 0.866025 0.500000i 0.866025 0.500000i
$$19$$ 1.00000i 1.00000i
$$20$$ 0 0
$$21$$ 0.866025 0.500000i 0.866025 0.500000i
$$22$$ −0.866025 0.500000i −0.866025 0.500000i
$$23$$ 0.500000 0.866025i 0.500000 0.866025i −0.500000 0.866025i $$-0.666667\pi$$
1.00000 $$0$$
$$24$$ 1.00000 1.00000
$$25$$ 0 0
$$26$$ −1.00000 −1.00000
$$27$$ 1.00000i 1.00000i
$$28$$ 0 0
$$29$$ 0.866025 0.500000i 0.866025 0.500000i 1.00000i $$-0.5\pi$$
0.866025 + 0.500000i $$0.166667\pi$$
$$30$$ 1.00000 1.00000
$$31$$ −0.866025 0.500000i −0.866025 0.500000i 1.00000i $$-0.5\pi$$
−0.866025 + 0.500000i $$0.833333\pi$$
$$32$$ 0 0
$$33$$ −0.866025 + 0.500000i −0.866025 + 0.500000i
$$34$$ 0 0
$$35$$ 1.00000 1.00000
$$36$$ 0 0
$$37$$ 0 0 1.00000 $$0$$
−1.00000 $$\pi$$
$$38$$ −0.500000 0.866025i −0.500000 0.866025i
$$39$$ −0.500000 + 0.866025i −0.500000 + 0.866025i
$$40$$ 0.866025 + 0.500000i 0.866025 + 0.500000i
$$41$$ 0.866025 + 0.500000i 0.866025 + 0.500000i 0.866025 0.500000i $$-0.166667\pi$$
1.00000i $$0.5\pi$$
$$42$$ −0.500000 + 0.866025i −0.500000 + 0.866025i
$$43$$ −0.500000 0.866025i −0.500000 0.866025i 0.500000 0.866025i $$-0.333333\pi$$
−1.00000 $$\pi$$
$$44$$ 0 0
$$45$$ 0.500000 0.866025i 0.500000 0.866025i
$$46$$ 1.00000i 1.00000i
$$47$$ −0.500000 0.866025i −0.500000 0.866025i 0.500000 0.866025i $$-0.333333\pi$$
−1.00000 $$\pi$$
$$48$$ −0.866025 + 0.500000i −0.866025 + 0.500000i
$$49$$ 0 0
$$50$$ 0 0
$$51$$ 0 0
$$52$$ 0 0
$$53$$ 0 0 1.00000 $$0$$
−1.00000 $$\pi$$
$$54$$ 0.500000 + 0.866025i 0.500000 + 0.866025i
$$55$$ −1.00000 −1.00000
$$56$$ −0.866025 + 0.500000i −0.866025 + 0.500000i
$$57$$ −1.00000 −1.00000
$$58$$ −0.500000 + 0.866025i −0.500000 + 0.866025i
$$59$$ −0.866025 0.500000i −0.866025 0.500000i 1.00000i $$-0.5\pi$$
−0.866025 + 0.500000i $$0.833333\pi$$
$$60$$ 0 0
$$61$$ 0.500000 + 0.866025i 0.500000 + 0.866025i 1.00000 $$0$$
−0.500000 + 0.866025i $$0.666667\pi$$
$$62$$ 1.00000 1.00000
$$63$$ 0.500000 + 0.866025i 0.500000 + 0.866025i
$$64$$ −1.00000 −1.00000
$$65$$ −0.866025 + 0.500000i −0.866025 + 0.500000i
$$66$$ 0.500000 0.866025i 0.500000 0.866025i
$$67$$ −0.866025 0.500000i −0.866025 0.500000i 1.00000i $$-0.5\pi$$
−0.866025 + 0.500000i $$0.833333\pi$$
$$68$$ 0 0
$$69$$ 0.866025 + 0.500000i 0.866025 + 0.500000i
$$70$$ −0.866025 + 0.500000i −0.866025 + 0.500000i
$$71$$ 0 0 1.00000 $$0$$
−1.00000 $$\pi$$
$$72$$ 1.00000i 1.00000i
$$73$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$74$$ 0 0
$$75$$ 0 0
$$76$$ 0 0
$$77$$ 0.500000 0.866025i 0.500000 0.866025i
$$78$$ 1.00000i 1.00000i
$$79$$ 0.866025 0.500000i 0.866025 0.500000i 1.00000i $$-0.5\pi$$
0.866025 + 0.500000i $$0.166667\pi$$
$$80$$ −1.00000 −1.00000
$$81$$ 1.00000 1.00000
$$82$$ −1.00000 −1.00000
$$83$$ 0.500000 + 0.866025i 0.500000 + 0.866025i 1.00000 $$0$$
−0.500000 + 0.866025i $$0.666667\pi$$
$$84$$ 0 0
$$85$$ 0 0
$$86$$ 0.866025 + 0.500000i 0.866025 + 0.500000i
$$87$$ 0.500000 + 0.866025i 0.500000 + 0.866025i
$$88$$ 0.866025 0.500000i 0.866025 0.500000i
$$89$$ 0 0 1.00000 $$0$$
−1.00000 $$\pi$$
$$90$$ 1.00000i 1.00000i
$$91$$ 1.00000i 1.00000i
$$92$$ 0 0
$$93$$ 0.500000 0.866025i 0.500000 0.866025i
$$94$$ 0.866025 + 0.500000i 0.866025 + 0.500000i
$$95$$ −0.866025 0.500000i −0.866025 0.500000i
$$96$$ 0 0
$$97$$ −0.866025 + 0.500000i −0.866025 + 0.500000i −0.866025 0.500000i $$-0.833333\pi$$
1.00000i $$0.5\pi$$
$$98$$ 0 0
$$99$$ −0.500000 0.866025i −0.500000 0.866025i
$$100$$ 0 0
$$101$$ −0.500000 0.866025i −0.500000 0.866025i 0.500000 0.866025i $$-0.333333\pi$$
−1.00000 $$\pi$$
$$102$$ 0 0
$$103$$ 0.866025 + 0.500000i 0.866025 + 0.500000i 0.866025 0.500000i $$-0.166667\pi$$
1.00000i $$0.5\pi$$
$$104$$ 0.500000 0.866025i 0.500000 0.866025i
$$105$$ 1.00000i 1.00000i
$$106$$ 0 0
$$107$$ 0 0 1.00000 $$0$$
−1.00000 $$\pi$$
$$108$$ 0 0
$$109$$ 2.00000i 2.00000i 1.00000i $$-0.5\pi$$
1.00000i $$-0.5\pi$$
$$110$$ 0.866025 0.500000i 0.866025 0.500000i
$$111$$ 0 0
$$112$$ 0.500000 0.866025i 0.500000 0.866025i
$$113$$ −0.866025 0.500000i −0.866025 0.500000i 1.00000i $$-0.5\pi$$
−0.866025 + 0.500000i $$0.833333\pi$$
$$114$$ 0.866025 0.500000i 0.866025 0.500000i
$$115$$ 0.500000 + 0.866025i 0.500000 + 0.866025i
$$116$$ 0 0
$$117$$ −0.866025 0.500000i −0.866025 0.500000i
$$118$$ 1.00000 1.00000
$$119$$ 0 0
$$120$$ −0.500000 + 0.866025i −0.500000 + 0.866025i
$$121$$ 0 0
$$122$$ −0.866025 0.500000i −0.866025 0.500000i
$$123$$ −0.500000 + 0.866025i −0.500000 + 0.866025i
$$124$$ 0 0
$$125$$ −1.00000 −1.00000
$$126$$ −0.866025 0.500000i −0.866025 0.500000i
$$127$$ 0 0 1.00000 $$0$$
−1.00000 $$\pi$$
$$128$$ 0.866025 0.500000i 0.866025 0.500000i
$$129$$ 0.866025 0.500000i 0.866025 0.500000i
$$130$$ 0.500000 0.866025i 0.500000 0.866025i
$$131$$ −0.500000 + 0.866025i −0.500000 + 0.866025i 0.500000 + 0.866025i $$0.333333\pi$$
−1.00000 $$\pi$$
$$132$$ 0 0
$$133$$ 0.866025 0.500000i 0.866025 0.500000i
$$134$$ 1.00000 1.00000
$$135$$ 0.866025 + 0.500000i 0.866025 + 0.500000i
$$136$$ 0 0
$$137$$ 0.500000 + 0.866025i 0.500000 + 0.866025i 1.00000 $$0$$
−0.500000 + 0.866025i $$0.666667\pi$$
$$138$$ −1.00000 −1.00000
$$139$$ 0.500000 0.866025i 0.500000 0.866025i −0.500000 0.866025i $$-0.666667\pi$$
1.00000 $$0$$
$$140$$ 0 0
$$141$$ 0.866025 0.500000i 0.866025 0.500000i
$$142$$ 0 0
$$143$$ 1.00000i 1.00000i
$$144$$ −0.500000 0.866025i −0.500000 0.866025i
$$145$$ 1.00000i 1.00000i
$$146$$ 0 0
$$147$$ 0 0
$$148$$ 0 0
$$149$$ −0.500000 + 0.866025i −0.500000 + 0.866025i 0.500000 + 0.866025i $$0.333333\pi$$
−1.00000 $$\pi$$
$$150$$ 0 0
$$151$$ −0.866025 + 0.500000i −0.866025 + 0.500000i −0.866025 0.500000i $$-0.833333\pi$$
1.00000i $$0.5\pi$$
$$152$$ 1.00000 1.00000
$$153$$ 0 0
$$154$$ 1.00000i 1.00000i
$$155$$ 0.866025 0.500000i 0.866025 0.500000i
$$156$$ 0 0
$$157$$ −0.500000 + 0.866025i −0.500000 + 0.866025i 0.500000 + 0.866025i $$0.333333\pi$$
−1.00000 $$\pi$$
$$158$$ −0.500000 + 0.866025i −0.500000 + 0.866025i
$$159$$ 0 0
$$160$$ 0 0
$$161$$ −1.00000 −1.00000
$$162$$ −0.866025 + 0.500000i −0.866025 + 0.500000i
$$163$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$164$$ 0 0
$$165$$ 1.00000i 1.00000i
$$166$$ −0.866025 0.500000i −0.866025 0.500000i
$$167$$ 0.866025 + 0.500000i 0.866025 + 0.500000i 0.866025 0.500000i $$-0.166667\pi$$
1.00000i $$0.5\pi$$
$$168$$ −0.500000 0.866025i −0.500000 0.866025i
$$169$$ 0 0
$$170$$ 0 0
$$171$$ 1.00000i 1.00000i
$$172$$ 0 0
$$173$$ 0.866025 0.500000i 0.866025 0.500000i 1.00000i $$-0.5\pi$$
0.866025 + 0.500000i $$0.166667\pi$$
$$174$$ −0.866025 0.500000i −0.866025 0.500000i
$$175$$ 0 0
$$176$$ −0.500000 + 0.866025i −0.500000 + 0.866025i
$$177$$ 0.500000 0.866025i 0.500000 0.866025i
$$178$$ 0 0
$$179$$ 2.00000i 2.00000i 1.00000i $$0.5\pi$$
1.00000i $$0.5\pi$$
$$180$$ 0 0
$$181$$ 0 0 1.00000 $$0$$
−1.00000 $$\pi$$
$$182$$ 0.500000 + 0.866025i 0.500000 + 0.866025i
$$183$$ −0.866025 + 0.500000i −0.866025 + 0.500000i
$$184$$ −0.866025 0.500000i −0.866025 0.500000i
$$185$$ 0 0
$$186$$ 1.00000i 1.00000i
$$187$$ 0 0
$$188$$ 0 0
$$189$$ −0.866025 + 0.500000i −0.866025 + 0.500000i
$$190$$ 1.00000 1.00000
$$191$$ −0.500000 0.866025i −0.500000 0.866025i 0.500000 0.866025i $$-0.333333\pi$$
−1.00000 $$\pi$$
$$192$$ 1.00000i 1.00000i
$$193$$ −0.866025 0.500000i −0.866025 0.500000i 1.00000i $$-0.5\pi$$
−0.866025 + 0.500000i $$0.833333\pi$$
$$194$$ 0.500000 0.866025i 0.500000 0.866025i
$$195$$ −0.500000 0.866025i −0.500000 0.866025i
$$196$$ 0 0
$$197$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$198$$ 0.866025 + 0.500000i 0.866025 + 0.500000i
$$199$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$200$$ 0 0
$$201$$ 0.500000 0.866025i 0.500000 0.866025i
$$202$$ 0.866025 + 0.500000i 0.866025 + 0.500000i
$$203$$ −0.866025 0.500000i −0.866025 0.500000i
$$204$$ 0 0
$$205$$ −0.866025 + 0.500000i −0.866025 + 0.500000i
$$206$$ −1.00000 −1.00000
$$207$$ −0.500000 + 0.866025i −0.500000 + 0.866025i
$$208$$ 1.00000i 1.00000i
$$209$$ −0.866025 + 0.500000i −0.866025 + 0.500000i
$$210$$ −0.500000 0.866025i −0.500000 0.866025i
$$211$$ 0.866025 + 0.500000i 0.866025 + 0.500000i 0.866025 0.500000i $$-0.166667\pi$$
1.00000i $$0.5\pi$$
$$212$$ 0 0
$$213$$ 0 0
$$214$$ 0 0
$$215$$ 1.00000 1.00000
$$216$$ −1.00000 −1.00000
$$217$$ 1.00000i 1.00000i
$$218$$ 1.00000 + 1.73205i 1.00000 + 1.73205i
$$219$$ 0 0
$$220$$ 0 0
$$221$$ 0 0
$$222$$ 0 0
$$223$$ −0.866025 + 0.500000i −0.866025 + 0.500000i −0.866025 0.500000i $$-0.833333\pi$$
1.00000i $$0.5\pi$$
$$224$$ 0 0
$$225$$ 0 0
$$226$$ 1.00000 1.00000
$$227$$ 0.866025 0.500000i 0.866025 0.500000i 1.00000i $$-0.5\pi$$
0.866025 + 0.500000i $$0.166667\pi$$
$$228$$ 0 0
$$229$$ 0.500000 0.866025i 0.500000 0.866025i −0.500000 0.866025i $$-0.666667\pi$$
1.00000 $$0$$
$$230$$ −0.866025 0.500000i −0.866025 0.500000i
$$231$$ 0.866025 + 0.500000i 0.866025 + 0.500000i
$$232$$ −0.500000 0.866025i −0.500000 0.866025i
$$233$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$234$$ 1.00000 1.00000
$$235$$ 1.00000 1.00000
$$236$$ 0 0
$$237$$ 0.500000 + 0.866025i 0.500000 + 0.866025i
$$238$$ 0 0
$$239$$ 0.500000 0.866025i 0.500000 0.866025i −0.500000 0.866025i $$-0.666667\pi$$
1.00000 $$0$$
$$240$$ 1.00000i 1.00000i
$$241$$ 0.866025 0.500000i 0.866025 0.500000i 1.00000i $$-0.5\pi$$
0.866025 + 0.500000i $$0.166667\pi$$
$$242$$ 0 0
$$243$$ 1.00000i 1.00000i
$$244$$ 0 0
$$245$$ 0 0
$$246$$ 1.00000i 1.00000i
$$247$$ −0.500000 + 0.866025i −0.500000 + 0.866025i
$$248$$ −0.500000 + 0.866025i −0.500000 + 0.866025i
$$249$$ −0.866025 + 0.500000i −0.866025 + 0.500000i
$$250$$ 0.866025 0.500000i 0.866025 0.500000i
$$251$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$252$$ 0 0
$$253$$ 1.00000 1.00000
$$254$$ 0 0
$$255$$ 0 0
$$256$$ 0 0
$$257$$ 0.866025 + 0.500000i 0.866025 + 0.500000i 0.866025 0.500000i $$-0.166667\pi$$
1.00000i $$0.5\pi$$
$$258$$ −0.500000 + 0.866025i −0.500000 + 0.866025i
$$259$$ 0 0
$$260$$ 0 0
$$261$$ −0.866025 + 0.500000i −0.866025 + 0.500000i
$$262$$ 1.00000i 1.00000i
$$263$$ −0.500000 0.866025i −0.500000 0.866025i 0.500000 0.866025i $$-0.333333\pi$$
−1.00000 $$\pi$$
$$264$$ 0.500000 + 0.866025i 0.500000 + 0.866025i
$$265$$ 0 0
$$266$$ −0.500000 + 0.866025i −0.500000 + 0.866025i
$$267$$ 0 0
$$268$$ 0 0
$$269$$ 0 0 1.00000 $$0$$
−1.00000 $$\pi$$
$$270$$ −1.00000 −1.00000
$$271$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$272$$ 0 0
$$273$$ 1.00000 1.00000
$$274$$ −0.866025 0.500000i −0.866025 0.500000i
$$275$$ 0 0
$$276$$ 0 0
$$277$$ −0.500000 0.866025i −0.500000 0.866025i 0.500000 0.866025i $$-0.333333\pi$$
−1.00000 $$\pi$$
$$278$$ 1.00000i 1.00000i
$$279$$ 0.866025 + 0.500000i 0.866025 + 0.500000i
$$280$$ 1.00000i 1.00000i
$$281$$ −0.866025 + 0.500000i −0.866025 + 0.500000i −0.866025 0.500000i $$-0.833333\pi$$
1.00000i $$0.5\pi$$
$$282$$ −0.500000 + 0.866025i −0.500000 + 0.866025i
$$283$$ −0.500000 + 0.866025i −0.500000 + 0.866025i 0.500000 + 0.866025i $$0.333333\pi$$
−1.00000 $$\pi$$
$$284$$ 0 0
$$285$$ 0.500000 0.866025i 0.500000 0.866025i
$$286$$ −0.500000 0.866025i −0.500000 0.866025i
$$287$$ 1.00000i 1.00000i
$$288$$ 0 0
$$289$$ −1.00000 −1.00000
$$290$$ −0.500000 0.866025i −0.500000 0.866025i
$$291$$ −0.500000 0.866025i −0.500000 0.866025i
$$292$$ 0 0
$$293$$ −0.866025 0.500000i −0.866025 0.500000i 1.00000i $$-0.5\pi$$
−0.866025 + 0.500000i $$0.833333\pi$$
$$294$$ 0 0
$$295$$ 0.866025 0.500000i 0.866025 0.500000i
$$296$$ 0 0
$$297$$ 0.866025 0.500000i 0.866025 0.500000i
$$298$$ 1.00000i 1.00000i
$$299$$ 0.866025 0.500000i 0.866025 0.500000i
$$300$$ 0 0
$$301$$ −0.500000 + 0.866025i −0.500000 + 0.866025i
$$302$$ 0.500000 0.866025i 0.500000 0.866025i
$$303$$ 0.866025 0.500000i 0.866025 0.500000i
$$304$$ −0.866025 + 0.500000i −0.866025 + 0.500000i
$$305$$ −1.00000 −1.00000
$$306$$ 0 0
$$307$$ 0 0 1.00000 $$0$$
−1.00000 $$\pi$$
$$308$$ 0 0
$$309$$ −0.500000 + 0.866025i −0.500000 + 0.866025i
$$310$$ −0.500000 + 0.866025i −0.500000 + 0.866025i
$$311$$ 0.500000 0.866025i 0.500000 0.866025i −0.500000 0.866025i $$-0.666667\pi$$
1.00000 $$0$$
$$312$$ 0.866025 + 0.500000i 0.866025 + 0.500000i
$$313$$ 0.500000 + 0.866025i 0.500000 + 0.866025i 1.00000 $$0$$
−0.500000 + 0.866025i $$0.666667\pi$$
$$314$$ 1.00000i 1.00000i
$$315$$ −1.00000 −1.00000
$$316$$ 0 0
$$317$$ −0.866025 + 0.500000i −0.866025 + 0.500000i −0.866025 0.500000i $$-0.833333\pi$$
1.00000i $$0.5\pi$$
$$318$$ 0 0
$$319$$ 0.866025 + 0.500000i 0.866025 + 0.500000i
$$320$$ 0.500000 0.866025i 0.500000 0.866025i
$$321$$ 0 0
$$322$$ 0.866025 0.500000i 0.866025 0.500000i
$$323$$ 0 0
$$324$$ 0 0
$$325$$ 0 0
$$326$$ 0 0
$$327$$ 2.00000 2.00000
$$328$$ 0.500000 0.866025i 0.500000 0.866025i
$$329$$ −0.500000 + 0.866025i −0.500000 + 0.866025i
$$330$$ 0.500000 + 0.866025i 0.500000 + 0.866025i
$$331$$ 0.866025 0.500000i 0.866025 0.500000i 1.00000i $$-0.5\pi$$
0.866025 + 0.500000i $$0.166667\pi$$
$$332$$ 0 0
$$333$$ 0 0
$$334$$ −1.00000 −1.00000
$$335$$ 0.866025 0.500000i 0.866025 0.500000i
$$336$$ 0.866025 + 0.500000i 0.866025 + 0.500000i
$$337$$ 0.866025 + 0.500000i 0.866025 + 0.500000i 0.866025 0.500000i $$-0.166667\pi$$
1.00000i $$0.5\pi$$
$$338$$ 0 0
$$339$$ 0.500000 0.866025i 0.500000 0.866025i
$$340$$ 0 0
$$341$$ 1.00000i 1.00000i
$$342$$ 0.500000 + 0.866025i 0.500000 + 0.866025i
$$343$$ −1.00000 −1.00000
$$344$$ −0.866025 + 0.500000i −0.866025 + 0.500000i
$$345$$ −0.866025 + 0.500000i −0.866025 + 0.500000i
$$346$$ −0.500000 + 0.866025i −0.500000 + 0.866025i
$$347$$ 0.500000 0.866025i 0.500000 0.866025i −0.500000 0.866025i $$-0.666667\pi$$
1.00000 $$0$$
$$348$$ 0 0
$$349$$ 0.500000 + 0.866025i 0.500000 + 0.866025i 1.00000 $$0$$
−0.500000 + 0.866025i $$0.666667\pi$$
$$350$$ 0 0
$$351$$ 0.500000 0.866025i 0.500000 0.866025i
$$352$$ 0 0
$$353$$ −0.500000 0.866025i −0.500000 0.866025i 0.500000 0.866025i $$-0.333333\pi$$
−1.00000 $$\pi$$
$$354$$ 1.00000i 1.00000i
$$355$$ 0 0
$$356$$ 0 0
$$357$$ 0 0
$$358$$ −1.00000 1.73205i −1.00000 1.73205i
$$359$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$360$$ −0.866025 0.500000i −0.866025 0.500000i
$$361$$ −1.00000 −1.00000
$$362$$ 0 0
$$363$$ 0 0
$$364$$ 0 0
$$365$$ 0 0
$$366$$ 0.500000 0.866025i 0.500000 0.866025i
$$367$$ 0.500000 + 0.866025i 0.500000 + 0.866025i 1.00000 $$0$$
−0.500000 + 0.866025i $$0.666667\pi$$
$$368$$ 1.00000 1.00000
$$369$$ −0.866025 0.500000i −0.866025 0.500000i
$$370$$ 0 0
$$371$$ 0 0
$$372$$ 0 0
$$373$$ −0.866025 0.500000i −0.866025 0.500000i 1.00000i $$-0.5\pi$$
−0.866025 + 0.500000i $$0.833333\pi$$
$$374$$ 0 0
$$375$$ 1.00000i 1.00000i
$$376$$ −0.866025 + 0.500000i −0.866025 + 0.500000i
$$377$$ 1.00000 1.00000
$$378$$ 0.500000 0.866025i 0.500000 0.866025i
$$379$$ 0 0 1.00000 $$0$$
−1.00000 $$\pi$$
$$380$$ 0 0
$$381$$ 0 0
$$382$$ 0.866025 + 0.500000i 0.866025 + 0.500000i
$$383$$ −0.866025 0.500000i −0.866025 0.500000i 1.00000i $$-0.5\pi$$
−0.866025 + 0.500000i $$0.833333\pi$$
$$384$$ 0.500000 + 0.866025i 0.500000 + 0.866025i
$$385$$ 0.500000 + 0.866025i 0.500000 + 0.866025i
$$386$$ 1.00000 1.00000
$$387$$ 0.500000 + 0.866025i 0.500000 + 0.866025i
$$388$$ 0 0
$$389$$ 0.500000 + 0.866025i 0.500000 + 0.866025i 1.00000 $$0$$
−0.500000 + 0.866025i $$0.666667\pi$$
$$390$$ 0.866025 + 0.500000i 0.866025 + 0.500000i
$$391$$ 0 0
$$392$$ 0 0
$$393$$ −0.866025 0.500000i −0.866025 0.500000i
$$394$$ 0 0
$$395$$ 1.00000i 1.00000i
$$396$$ 0 0
$$397$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$398$$ 0 0
$$399$$ 0.500000 + 0.866025i 0.500000 + 0.866025i
$$400$$ 0 0
$$401$$ 0.866025 + 0.500000i 0.866025 + 0.500000i 0.866025 0.500000i $$-0.166667\pi$$
1.00000i $$0.5\pi$$
$$402$$ 1.00000i 1.00000i
$$403$$ −0.500000 0.866025i −0.500000 0.866025i
$$404$$ 0 0
$$405$$ −0.500000 + 0.866025i −0.500000 + 0.866025i
$$406$$ 1.00000 1.00000
$$407$$ 0 0
$$408$$ 0 0
$$409$$ −0.866025 0.500000i −0.866025 0.500000i 1.00000i $$-0.5\pi$$
−0.866025 + 0.500000i $$0.833333\pi$$
$$410$$ 0.500000 0.866025i 0.500000 0.866025i
$$411$$ −0.866025 + 0.500000i −0.866025 + 0.500000i
$$412$$ 0 0
$$413$$ 1.00000i 1.00000i
$$414$$ 1.00000i 1.00000i
$$415$$ −1.00000 −1.00000
$$416$$ 0 0
$$417$$ 0.866025 + 0.500000i 0.866025 + 0.500000i
$$418$$ 0.500000 0.866025i 0.500000 0.866025i
$$419$$ −0.500000 + 0.866025i −0.500000 + 0.866025i 0.500000 + 0.866025i $$0.333333\pi$$
−1.00000 $$\pi$$
$$420$$ 0 0
$$421$$ −0.866025 + 0.500000i −0.866025 + 0.500000i −0.866025 0.500000i $$-0.833333\pi$$
1.00000i $$0.5\pi$$
$$422$$ −1.00000 −1.00000
$$423$$ 0.500000 + 0.866025i 0.500000 + 0.866025i
$$424$$ 0 0
$$425$$ 0 0
$$426$$ 0 0
$$427$$ 0.500000 0.866025i 0.500000 0.866025i
$$428$$ 0 0
$$429$$ −1.00000 −1.00000
$$430$$ −0.866025 + 0.500000i −0.866025 + 0.500000i
$$431$$ 0 0 1.00000 $$0$$
−1.00000 $$\pi$$
$$432$$ 0.866025 0.500000i 0.866025 0.500000i
$$433$$ 0 0 1.00000 $$0$$
−1.00000 $$\pi$$
$$434$$ −0.500000 0.866025i −0.500000 0.866025i
$$435$$ −1.00000 −1.00000
$$436$$ 0 0
$$437$$ 0.866025 + 0.500000i 0.866025 + 0.500000i
$$438$$ 0 0
$$439$$ −0.866025 + 0.500000i −0.866025 + 0.500000i −0.866025 0.500000i $$-0.833333\pi$$
1.00000i $$0.5\pi$$
$$440$$ 1.00000i 1.00000i
$$441$$ 0 0
$$442$$ 0 0
$$443$$ −0.500000 0.866025i −0.500000 0.866025i 0.500000 0.866025i $$-0.333333\pi$$
−1.00000 $$\pi$$
$$444$$ 0 0
$$445$$ 0 0
$$446$$ 0.500000 0.866025i 0.500000 0.866025i
$$447$$ −0.866025 0.500000i −0.866025 0.500000i
$$448$$ 0.500000 + 0.866025i 0.500000 + 0.866025i
$$449$$ 2.00000i 2.00000i 1.00000i $$-0.5\pi$$
1.00000i $$-0.5\pi$$
$$450$$ 0 0
$$451$$ 1.00000i 1.00000i
$$452$$ 0 0
$$453$$ −0.500000 0.866025i −0.500000 0.866025i
$$454$$ −0.500000 + 0.866025i −0.500000 + 0.866025i
$$455$$ 0.866025 + 0.500000i 0.866025 + 0.500000i
$$456$$ 1.00000i 1.00000i
$$457$$ −0.500000 0.866025i −0.500000 0.866025i 0.500000 0.866025i $$-0.333333\pi$$
−1.00000 $$\pi$$
$$458$$ 1.00000i 1.00000i
$$459$$ 0 0
$$460$$ 0 0
$$461$$ −0.500000 0.866025i −0.500000 0.866025i 0.500000 0.866025i $$-0.333333\pi$$
−1.00000 $$\pi$$
$$462$$ −1.00000 −1.00000
$$463$$ −0.500000 + 0.866025i −0.500000 + 0.866025i 0.500000 + 0.866025i $$0.333333\pi$$
−1.00000 $$\pi$$
$$464$$ 0.866025 + 0.500000i 0.866025 + 0.500000i
$$465$$ 0.500000 + 0.866025i 0.500000 + 0.866025i
$$466$$ 0 0
$$467$$ −2.00000 −2.00000 −1.00000 $$\pi$$
−1.00000 $$\pi$$
$$468$$ 0 0
$$469$$ 1.00000i 1.00000i
$$470$$ −0.866025 + 0.500000i −0.866025 + 0.500000i
$$471$$ −0.866025 0.500000i −0.866025 0.500000i
$$472$$ −0.500000 + 0.866025i −0.500000 + 0.866025i
$$473$$ 0.500000 0.866025i 0.500000 0.866025i
$$474$$ −0.866025 0.500000i −0.866025 0.500000i
$$475$$ 0 0
$$476$$ 0 0
$$477$$ 0 0
$$478$$ 1.00000i 1.00000i
$$479$$ 0.500000 + 0.866025i 0.500000 + 0.866025i 1.00000 $$0$$
−0.500000 + 0.866025i $$0.666667\pi$$
$$480$$ 0 0
$$481$$ 0 0
$$482$$ −0.500000 + 0.866025i −0.500000 + 0.866025i
$$483$$ 1.00000i 1.00000i
$$484$$ 0 0
$$485$$ 1.00000i 1.00000i
$$486$$ −0.500000 0.866025i −0.500000 0.866025i
$$487$$ 0 0 1.00000 $$0$$
−1.00000 $$\pi$$
$$488$$ 0.866025 0.500000i 0.866025 0.500000i
$$489$$ 0 0
$$490$$ 0 0
$$491$$ −0.500000 + 0.866025i −0.500000 + 0.866025i 0.500000 + 0.866025i $$0.333333\pi$$
−1.00000 $$\pi$$
$$492$$ 0 0
$$493$$ 0 0
$$494$$ 1.00000i 1.00000i
$$495$$ 1.00000 1.00000
$$496$$ 1.00000i 1.00000i
$$497$$ 0 0
$$498$$ 0.500000 0.866025i 0.500000 0.866025i
$$499$$ 0.500000 0.866025i 0.500000 0.866025i −0.500000 0.866025i $$-0.666667\pi$$
1.00000 $$0$$
$$500$$ 0 0
$$501$$ −0.500000 + 0.866025i −0.500000 + 0.866025i
$$502$$ 0 0
$$503$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$504$$ 0.866025 0.500000i 0.866025 0.500000i
$$505$$ 1.00000 1.00000
$$506$$ −0.866025 + 0.500000i −0.866025 + 0.500000i
$$507$$ 0 0
$$508$$ 0 0
$$509$$ −0.866025 0.500000i −0.866025 0.500000i 1.00000i $$-0.5\pi$$
−0.866025 + 0.500000i $$0.833333\pi$$
$$510$$ 0 0
$$511$$ 0 0
$$512$$ 1.00000i 1.00000i
$$513$$ 1.00000 1.00000
$$514$$ −1.00000 −1.00000
$$515$$ −0.866025 + 0.500000i −0.866025 + 0.500000i
$$516$$ 0 0
$$517$$ 0.500000 0.866025i 0.500000 0.866025i
$$518$$ 0 0
$$519$$ 0.500000 + 0.866025i 0.500000 + 0.866025i
$$520$$ 0.500000 + 0.866025i 0.500000 + 0.866025i
$$521$$ 2.00000i 2.00000i 1.00000i $$-0.5\pi$$
1.00000i $$-0.5\pi$$
$$522$$ 0.500000 0.866025i 0.500000 0.866025i
$$523$$ 0 0 1.00000 $$0$$
−1.00000 $$\pi$$
$$524$$ 0 0
$$525$$ 0 0
$$526$$ 0.866025 + 0.500000i 0.866025 + 0.500000i
$$527$$ 0 0
$$528$$ −0.866025 0.500000i −0.866025 0.500000i
$$529$$ 0 0
$$530$$ 0 0
$$531$$ 0.866025 + 0.500000i 0.866025 + 0.500000i
$$532$$ 0 0
$$533$$ 0.500000 + 0.866025i 0.500000 + 0.866025i
$$534$$ 0 0
$$535$$ 0 0
$$536$$ −0.500000 + 0.866025i −0.500000 + 0.866025i
$$537$$ −2.00000 −2.00000
$$538$$ 0 0
$$539$$ 0 0
$$540$$ 0 0
$$541$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$542$$ 0 0
$$543$$ 0 0
$$544$$ 0 0
$$545$$ 1.73205 + 1.00000i 1.73205 + 1.00000i
$$546$$ −0.866025 + 0.500000i −0.866025 + 0.500000i
$$547$$ 0.866025 0.500000i 0.866025 0.500000i 1.00000i $$-0.5\pi$$
0.866025 + 0.500000i $$0.166667\pi$$
$$548$$ 0 0
$$549$$ −0.500000 0.866025i −0.500000 0.866025i
$$550$$ 0 0
$$551$$ 0.500000 + 0.866025i 0.500000 + 0.866025i
$$552$$ 0.500000 0.866025i 0.500000 0.866025i
$$553$$ −0.866025 0.500000i −0.866025 0.500000i
$$554$$ 0.866025 + 0.500000i 0.866025 + 0.500000i
$$555$$ 0 0
$$556$$ 0 0
$$557$$ 2.00000 2.00000 1.00000 $$0$$
1.00000 $$0$$
$$558$$ −1.00000 −1.00000
$$559$$ 1.00000i 1.00000i
$$560$$ 0.500000 + 0.866025i 0.500000 + 0.866025i
$$561$$ 0 0
$$562$$ 0.500000 0.866025i 0.500000 0.866025i
$$563$$ 0.866025 + 0.500000i 0.866025 + 0.500000i 0.866025 0.500000i $$-0.166667\pi$$
1.00000i $$0.5\pi$$
$$564$$ 0 0
$$565$$ 0.866025 0.500000i 0.866025 0.500000i
$$566$$ 1.00000i 1.00000i
$$567$$ −0.500000 0.866025i −0.500000 0.866025i
$$568$$ 0 0
$$569$$ −0.866025 + 0.500000i −0.866025 + 0.500000i −0.866025 0.500000i $$-0.833333\pi$$
1.00000i $$0.5\pi$$
$$570$$ 1.00000i 1.00000i
$$571$$ 0.500000 0.866025i 0.500000 0.866025i −0.500000 0.866025i $$-0.666667\pi$$
1.00000 $$0$$
$$572$$ 0 0
$$573$$ 0.866025 0.500000i 0.866025 0.500000i
$$574$$ 0.500000 + 0.866025i 0.500000 + 0.866025i
$$575$$ 0 0
$$576$$ 1.00000 1.00000
$$577$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$578$$ 0.866025 0.500000i 0.866025 0.500000i
$$579$$ 0.500000 0.866025i 0.500000 0.866025i
$$580$$ 0 0
$$581$$ 0.500000 0.866025i 0.500000 0.866025i
$$582$$ 0.866025 + 0.500000i 0.866025 + 0.500000i
$$583$$ 0 0
$$584$$ 0 0
$$585$$ 0.866025 0.500000i 0.866025 0.500000i
$$586$$ 1.00000 1.00000
$$587$$ 0.500000 + 0.866025i 0.500000 + 0.866025i 1.00000 $$0$$
−0.500000 + 0.866025i $$0.666667\pi$$
$$588$$ 0 0
$$589$$ 0.500000 0.866025i 0.500000 0.866025i
$$590$$ −0.500000 + 0.866025i −0.500000 + 0.866025i
$$591$$ 0 0
$$592$$ 0 0
$$593$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$594$$ −0.500000 + 0.866025i −0.500000 + 0.866025i
$$595$$ 0 0
$$596$$ 0 0
$$597$$ 0 0
$$598$$ −0.500000 + 0.866025i −0.500000 + 0.866025i
$$599$$ −0.866025 0.500000i −0.866025 0.500000i 1.00000i $$-0.5\pi$$
−0.866025 + 0.500000i $$0.833333\pi$$
$$600$$ 0 0
$$601$$ 0.866025 0.500000i 0.866025 0.500000i 1.00000i $$-0.5\pi$$
0.866025 + 0.500000i $$0.166667\pi$$
$$602$$ 1.00000i 1.00000i
$$603$$ 0.866025 + 0.500000i 0.866025 + 0.500000i
$$604$$ 0 0
$$605$$ 0 0
$$606$$ −0.500000 + 0.866025i −0.500000 + 0.866025i
$$607$$ −0.866025 0.500000i −0.866025 0.500000i 1.00000i $$-0.5\pi$$
−0.866025 + 0.500000i $$0.833333\pi$$
$$608$$ 0 0
$$609$$ 0.500000 0.866025i 0.500000 0.866025i
$$610$$ 0.866025 0.500000i 0.866025 0.500000i
$$611$$ 1.00000i 1.00000i
$$612$$ 0 0
$$613$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$614$$ 0 0
$$615$$ −0.500000 0.866025i −0.500000 0.866025i
$$616$$ −0.866025 0.500000i −0.866025 0.500000i
$$617$$ −0.500000 + 0.866025i −0.500000 + 0.866025i 0.500000 + 0.866025i $$0.333333\pi$$
−1.00000 $$\pi$$
$$618$$ 1.00000i 1.00000i
$$619$$ 0.500000 + 0.866025i 0.500000 + 0.866025i 1.00000 $$0$$
−0.500000 + 0.866025i $$0.666667\pi$$
$$620$$ 0 0
$$621$$ −0.866025 0.500000i −0.866025 0.500000i
$$622$$ 1.00000i 1.00000i
$$623$$ 0 0
$$624$$ −1.00000 −1.00000
$$625$$ 0.500000 0.866025i 0.500000 0.866025i
$$626$$ −0.866025 0.500000i −0.866025 0.500000i
$$627$$ −0.500000 0.866025i −0.500000 0.866025i
$$628$$ 0 0
$$629$$ 0 0
$$630$$ 0.866025 0.500000i 0.866025 0.500000i
$$631$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$632$$ −0.500000 0.866025i −0.500000 0.866025i
$$633$$ −0.500000 + 0.866025i −0.500000 + 0.866025i
$$634$$ 0.500000 0.866025i 0.500000 0.866025i
$$635$$ 0 0
$$636$$ 0 0
$$637$$ 0 0
$$638$$ −1.00000 −1.00000
$$639$$ 0 0
$$640$$ 1.00000i 1.00000i
$$641$$ −0.866025 + 0.500000i −0.866025 + 0.500000i −0.866025 0.500000i $$-0.833333\pi$$
1.00000i $$0.5\pi$$
$$642$$ 0 0
$$643$$ 0.500000 0.866025i 0.500000 0.866025i −0.500000 0.866025i $$-0.666667\pi$$
1.00000 $$0$$
$$644$$ 0 0
$$645$$ 1.00000i 1.00000i
$$646$$ 0 0
$$647$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$648$$ 1.00000i 1.00000i
$$649$$ 1.00000i 1.00000i
$$650$$ 0 0
$$651$$ −1.00000 −1.00000
$$652$$ 0 0
$$653$$ 0.500000 0.866025i 0.500000 0.866025i −0.500000 0.866025i $$-0.666667\pi$$
1.00000 $$0$$
$$654$$ −1.73205 + 1.00000i −1.73205 + 1.00000i
$$655$$ −0.500000 0.866025i −0.500000 0.866025i
$$656$$ 1.00000i 1.00000i
$$657$$ 0 0
$$658$$ 1.00000i 1.00000i
$$659$$ −0.866025 + 0.500000i −0.866025 + 0.500000i −0.866025 0.500000i $$-0.833333\pi$$
1.00000i $$0.5\pi$$
$$660$$ 0 0
$$661$$ 0.866025 + 0.500000i 0.866025 + 0.500000i 0.866025 0.500000i $$-0.166667\pi$$
1.00000i $$0.5\pi$$
$$662$$ −0.500000 + 0.866025i −0.500000 + 0.866025i
$$663$$ 0 0
$$664$$ 0.866025 0.500000i 0.866025 0.500000i
$$665$$ 1.00000i 1.00000i
$$666$$ 0 0
$$667$$ 1.00000i 1.00000i
$$668$$ 0 0
$$669$$ −0.500000 0.866025i −0.500000 0.866025i
$$670$$ −0.500000 + 0.866025i −0.500000 + 0.866025i
$$671$$ −0.500000 + 0.866025i −0.500000 + 0.866025i
$$672$$ 0 0
$$673$$ −0.866025 + 0.500000i −0.866025 + 0.500000i −0.866025 0.500000i $$-0.833333\pi$$
1.00000i $$0.5\pi$$
$$674$$ −1.00000 −1.00000
$$675$$ 0 0
$$676$$ 0 0
$$677$$ −0.866025 + 0.500000i −0.866025 + 0.500000i −0.866025 0.500000i $$-0.833333\pi$$
1.00000i $$0.5\pi$$
$$678$$ 1.00000i 1.00000i
$$679$$ 0.866025 + 0.500000i 0.866025 + 0.500000i
$$680$$ 0 0
$$681$$ 0.500000 + 0.866025i 0.500000 + 0.866025i
$$682$$ 0.500000 + 0.866025i 0.500000 + 0.866025i
$$683$$ 0 0 1.00000 $$0$$
−1.00000 $$\pi$$
$$684$$ 0 0
$$685$$ −1.00000 −1.00000
$$686$$ 0.866025 0.500000i 0.866025 0.500000i
$$687$$ 0.866025 + 0.500000i 0.866025 + 0.500000i
$$688$$ 0.500000 0.866025i 0.500000 0.866025i
$$689$$ 0 0
$$690$$ 0.500000 0.866025i 0.500000 0.866025i
$$691$$ 0.500000 + 0.866025i 0.500000 + 0.866025i 1.00000 $$0$$
−0.500000 + 0.866025i $$0.666667\pi$$
$$692$$ 0 0
$$693$$ −0.500000 + 0.866025i −0.500000 + 0.866025i
$$694$$ 1.00000i 1.00000i
$$695$$ 0.500000 + 0.866025i 0.500000 + 0.866025i
$$696$$ 0.866025 0.500000i 0.866025 0.500000i
$$697$$ 0 0
$$698$$ −0.866025 0.500000i −0.866025 0.500000i
$$699$$ 0 0
$$700$$ 0 0
$$701$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$702$$ 1.00000i 1.00000i
$$703$$ 0 0
$$704$$ −0.500000 0.866025i −0.500000 0.866025i
$$705$$ 1.00000i 1.00000i
$$706$$ 0.866025 + 0.500000i 0.866025 + 0.500000i
$$707$$ −0.500000 + 0.866025i −0.500000 + 0.866025i
$$708$$ 0 0
$$709$$ −0.500000 0.866025i −0.500000 0.866025i 0.500000 0.866025i $$-0.333333\pi$$
−1.00000 $$\pi$$
$$710$$ 0 0
$$711$$ −0.866025 + 0.500000i −0.866025 + 0.500000i
$$712$$ 0 0
$$713$$ −0.866025 + 0.500000i −0.866025 + 0.500000i
$$714$$ 0 0
$$715$$ −0.866025 0.500000i −0.866025 0.500000i
$$716$$ 0 0
$$717$$ 0.866025 + 0.500000i 0.866025 + 0.500000i
$$718$$ 0 0
$$719$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$720$$ 1.00000 1.00000
$$721$$ 1.00000i 1.00000i
$$722$$ 0.866025 0.500000i 0.866025 0.500000i
$$723$$ 0.500000 + 0.866025i 0.500000 + 0.866025i
$$724$$ 0 0
$$725$$ 0 0
$$726$$ 0 0
$$727$$ 0.500000 + 0.866025i 0.500000 + 0.866025i 1.00000 $$0$$
−0.500000 + 0.866025i $$0.666667\pi$$
$$728$$ −1.00000 −1.00000
$$729$$ −1.00000 −1.00000
$$730$$ 0 0
$$731$$ 0 0
$$732$$ 0 0
$$733$$ −0.500000 + 0.866025i −0.500000 + 0.866025i 0.500000 + 0.866025i $$0.333333\pi$$
−1.00000 $$\pi$$
$$734$$ −0.866025 0.500000i −0.866025 0.500000i
$$735$$ 0 0
$$736$$ 0 0
$$737$$ 1.00000i 1.00000i
$$738$$ 1.00000 1.00000
$$739$$ 2.00000 2.00000 1.00000 $$0$$
1.00000 $$0$$
$$740$$ 0 0
$$741$$ −0.866025 0.500000i −0.866025 0.500000i
$$742$$ 0 0
$$743$$ 0.866025 + 0.500000i 0.866025 + 0.500000i 0.866025 0.500000i $$-0.166667\pi$$
1.00000i $$0.5\pi$$
$$744$$ −0.866025 0.500000i −0.866025 0.500000i
$$745$$ −0.500000 0.866025i −0.500000 0.866025i
$$746$$ 1.00000 1.00000
$$747$$ −0.500000 0.866025i −0.500000 0.866025i
$$748$$ 0 0
$$749$$ 0 0
$$750$$ 0.500000 + 0.866025i 0.500000 + 0.866025i
$$751$$ 0.866025 + 0.500000i 0.866025 + 0.500000i 0.866025 0.500000i $$-0.166667\pi$$
1.00000i $$0.5\pi$$
$$752$$ 0.500000 0.866025i 0.500000 0.866025i
$$753$$ 0 0
$$754$$ −0.866025 + 0.500000i −0.866025 + 0.500000i
$$755$$ 1.00000i 1.00000i
$$756$$ 0 0
$$757$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$758$$ 0 0
$$759$$ 1.00000i 1.00000i
$$760$$ −0.500000 + 0.866025i −0.500000 + 0.866025i
$$761$$ 0.500000 0.866025i 0.500000 0.866025i −0.500000 0.866025i $$-0.666667\pi$$
1.00000 $$0$$
$$762$$ 0 0
$$763$$ −1.73205 + 1.00000i −1.73205 + 1.00000i
$$764$$ 0 0
$$765$$ 0 0
$$766$$ 1.00000 1.00000
$$767$$ −0.500000 0.866025i −0.500000 0.866025i
$$768$$ 0 0
$$769$$ −0.500000 + 0.866025i −0.500000 + 0.866025i 0.500000 + 0.866025i $$0.333333\pi$$
−1.00000 $$\pi$$
$$770$$ −0.866025 0.500000i −0.866025 0.500000i
$$771$$ −0.500000 + 0.866025i −0.500000 + 0.866025i
$$772$$ 0 0
$$773$$ 2.00000i 2.00000i 1.00000i $$0.5\pi$$
1.00000i $$0.5\pi$$
$$774$$ −0.866025 0.500000i −0.866025 0.500000i
$$775$$ 0 0
$$776$$ 0.500000 + 0.866025i 0.500000 + 0.866025i
$$777$$ 0 0
$$778$$ −0.866025 0.500000i −0.866025 0.500000i
$$779$$ −0.500000 + 0.866025i −0.500000 + 0.866025i
$$780$$ 0 0
$$781$$ 0 0
$$782$$ 0 0
$$783$$ −0.500000 0.866025i −0.500000 0.866025i
$$784$$ 0 0
$$785$$ −0.500000 0.866025i −0.500000 0.866025i
$$786$$ 1.00000 1.00000
$$787$$ 0.866025 | 20,317 | 36,028 | {"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} | 2.65625 | 3 | CC-MAIN-2024-18 | latest | en | 0.509109 |
http://prob140.org/textbook/Chapter_04/05_Dependence_and_Independence.html | 1,579,657,696,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250606269.37/warc/CC-MAIN-20200122012204-20200122041204-00229.warc.gz | 139,726,807 | 16,026 | ## Dependence and Independence
Conditional distributions help us formalize our intuitive ideas about whether two random variables are independent of each other. Let $X$ and $Y$ be two random variables, and suppose we are given the value of $X$. Does that change our opinion about $Y$? If the answer is yes, then we will say that $X$ and $Y$ are dependent. If the answer is no regardless of the given value of $X$, then we will say that $X$ and $Y$ are independent.
Let's start with some examples and then move to precise definitions and results.
### Dependence
Here is the joint distribution of two random variables $X$ and $Y$. From this, what can we say about whether $X$ and $Y$ are dependent or independent?
dist1
X=0 X=1 X=2 X=3
Y=3 0.037037 0.000000 0.000000 0.00000
Y=2 0.166667 0.055556 0.000000 0.00000
Y=1 0.250000 0.166667 0.027778 0.00000
Y=0 0.125000 0.125000 0.041667 0.00463
You can see at once that if $X = 3$ then $Y$ can only be 0, whereas if $X = 2$ then $Y$ can be either 0 or 1. Knowing the value of $X$ changes the distribution of $Y$. That's dependence.
Here is an example in which you can't quickly determine dependence or independence by just looking at the possible values.
dist2
X=3 X=4
Y=7 0.3 0.1
Y=6 0.2 0.2
Y=5 0.1 0.1
But you can tell by looking at the conditional distributions of $X$ given $Y$. Two of them are the same, but the third is different. Knowing the value of $Y$ affects the chances for $X$.
dist2.conditional_dist('X', 'Y')
X=3 X=4 Sum
Dist. of X | Y=7 0.75 0.25 1.0
Dist. of X | Y=6 0.50 0.50 1.0
Dist. of X | Y=5 0.50 0.50 1.0
Marginal of X 0.60 0.40 1.0
It follows (and you should try to prove this), that at least some of the conditional distributions of $Y$ given the different values of $X$ will also be different from each other and from the marginal of $Y$.
In this example, all three conditional distributions of $Y$ given the three different values of $X$ are different from each other.
dist2.conditional_dist('Y', 'X')
Dist. of Y | X=3 Dist. of Y | X=4 Marginal of Y
Y=7 0.500000 0.25 0.4
Y=6 0.333333 0.50 0.4
Y=5 0.166667 0.25 0.2
Sum 1.000000 1.00 1.0
### Independence
Here is a joint distribution table in which you can't immediately tell whether there is dependence.
dist3
X=0 X=1 X=2 X=3
Y=4 0.000096 0.000289 0.000289 0.000096
Y=3 0.001929 0.005787 0.005787 0.001929
Y=2 0.014468 0.043403 0.043403 0.014468
Y=1 0.048225 0.144676 0.144676 0.048225
Y=0 0.060282 0.180845 0.180845 0.060282
But look what happens when you condition $X$ on $Y$.
dist3.conditional_dist('X', 'Y')
X=0 X=1 X=2 X=3 Sum
Dist. of X | Y=4 0.125 0.375 0.375 0.125 1.0
Dist. of X | Y=3 0.125 0.375 0.375 0.125 1.0
Dist. of X | Y=2 0.125 0.375 0.375 0.125 1.0
Dist. of X | Y=1 0.125 0.375 0.375 0.125 1.0
Dist. of X | Y=0 0.125 0.375 0.375 0.125 1.0
Marginal of X 0.125 0.375 0.375 0.125 1.0
All the rows are the same. That is, all the conditional distributions of $X$ given different values of $Y$ are the same, and hence are the same as the marginal of $X$ too.
Given the value of $Y$, the probabilities for $X$ don't change at all. That's independence.
You could have drawn the same conclusion by conditioning $Y$ on $X$:
dist3.conditional_dist('Y', 'X')
Dist. of Y | X=0 Dist. of Y | X=1 Dist. of Y | X=2 Dist. of Y | X=3 Marginal of Y
Y=4 0.000772 0.000772 0.000772 0.000772 0.000772
Y=3 0.015432 0.015432 0.015432 0.015432 0.015432
Y=2 0.115741 0.115741 0.115741 0.115741 0.115741
Y=1 0.385802 0.385802 0.385802 0.385802 0.385802
Y=0 0.482253 0.482253 0.482253 0.482253 0.482253
Sum 1.000000 1.000000 1.000000 1.000000 1.000000
### Independence of Two Events
The concept of independence seems intuitive, but it is possible to run into trouble by not being careful about its definition. So let's define it formally.
There are two equivalent definitions of the independence of two events. The first encapsulates the main idea of independence, and the second is useful for calculation.
Two events $A$ and $B$ are independent if $P(B \mid A) = P(B)$. Equivalently, $A$ and $B$ are independent if $P(AB) = P(A)P(B)$.
### Independence of Two Random Variables
What we have observed in the examples of this section can be turned into a formal definition of independence.
Two random variables $X$ and $Y$ are independent if for every value $x$ of $X$ and $y$ of $Y$,
$$P(Y = y \mid X = x) = P(Y = y)$$
That is, no matter what the given $x$ is, the conditional distribution of $Y$ given $X=x$ is the same as if we didn't know that $X=x$.
Equivalently (this needs a proof, which consists of a routine application of definitions), for every $y$ the conditional distribution of $X$ given $Y=y$ is the same as if we didn't know that $Y=y$.
An equivalent definition in terms of the independence of events is that for any values of $x$ and $y$, the events $\{ X=x\}$ and $\{Y=y\}$ are independent.
That is, $X$ and $Y$ are independent if for any values $x$ of $X$ and $y$ of $Y$,
$$P(X = x, Y = y) ~ = ~ P(X=x)P(Y=y)$$
Independence simplifies the conditional probabilities in the multiplication rule.
It is a fact that if $X$ and $Y$ are independent random variables, then any event determined by $X$ is independent of any event determined by $Y$. For example, if $X$ and $Y$ are independent and $x$ is a number, then $\{X=x\}$ is independent of $\{Y>x\}$. Also, any function of $X$ is independent of any function of $Y$.
You can prove these facts by partitioning and then using the definition of independence. The proofs are routine but somewhat labor intensive. You are welcome to just accept the facts if you don't want to prove them.
### Mutual Independence
Events $A_1, A_2, \ldots A_n$ are mutually independent (or independent for short) if given that any subset of the events has occurred, the conditional chances of all other subsets remain unchanged.
That's quite a mouthful. In practical terms it means that it doesn't matter which of the events you know have happened; chances involving the remaining events are unchanged.
In terms of random variables, $X_1, X_2, \ldots , X_n$ are independent if given the values of any subset, chances of events determined by the remaining variables are unchanged.
In practice, this just formalizes statements such as "results of different tosses of a coin are independent" or "draws made at random with replacement are independent".
Try not to become inhibited by the formalism. Notice how the theory not only supports intuition but also develops it. You can expect your probabilistic intuition to be much sharper at the end of this course than it is now!
### IID Random Variables
If random variables are mutually independent and identically distributed, they are called "i.i.d." That's one of the most famous acronyms in probability theory. You can think of i.i.d. random variables as draws with replacement from a population, or as the results of independent replications of the same experiment.
Calculations involving i.i.d. random variables are often straightforward. For example, suppose the distribution of $X$ is given by
$$P(X = i) = p_i, ~~~ i = 1, 2, \ldots, n$$
where $\sum_{i=1}^n p_i = 1$. Now let $X$ and $Y$ be i.i.d. What is $P(X = Y)$? We'll answer this question by using the fundamental method, now in random variable notation.
\begin{align*} P(X = Y) ~ &= ~ \sum_{i=1}^n P(X = i, Y = i) ~~~ \text{(partitioning)} \\ &= ~ \sum_{i=1}^n P(X = i)P(Y = i) ~~~ \text{(independence)} \\ &= ~ \sum_{i=1}^n p_i \cdot p_i ~~~ \text{(identical distributions)} \\ &= ~ \sum_{i=1}^n p_i^2 \end{align*}
The last expression is easy to calculate if you know the numerical values of all the $p_i$. | 2,447 | 7,626 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.46875 | 4 | CC-MAIN-2020-05 | latest | en | 0.840597 |
http://www.greenemath.com/Prealgebra/38/ComplexFractionsExam5.html | 1,493,298,173,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917122167.63/warc/CC-MAIN-20170423031202-00071-ip-10-145-167-34.ec2.internal.warc.gz | 525,162,021 | 4,845 | GreeneMath.com - Complex Fractions Test #5
# In this Section:
In this section, we learn about complex fractions. A complex fraction is a fraction that contains a fraction in its numerator, its denominator, or both its numerator and its denominator. We will learn two methods to simplify a complex fraction. In the first case, we simplify the numerator and denominator separately then perform the main division. We then simplify the result and report our answer. In the second case, we must first find the LCD of all denominators involved in the complex fraction. We then multiply the numerator and denominator of the complex fraction by the LCD. This eliminates all denominators for us, and at that point we just need to simplify. In each case, you will get the same answer. Generally for more complex problems, the LCD method is much faster.
Sections:
# In this Section:
In this section, we learn about complex fractions. A complex fraction is a fraction that contains a fraction in its numerator, its denominator, or both its numerator and its denominator. We will learn two methods to simplify a complex fraction. In the first case, we simplify the numerator and denominator separately then perform the main division. We then simplify the result and report our answer. In the second case, we must first find the LCD of all denominators involved in the complex fraction. We then multiply the numerator and denominator of the complex fraction by the LCD. This eliminates all denominators for us, and at that point we just need to simplify. In each case, you will get the same answer. Generally for more complex problems, the LCD method is much faster. | 329 | 1,658 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-17 | latest | en | 0.935916 |
https://shakyradunn.com/slide/thursday-october-1-solar-physics-42vi60 | 1,603,864,876,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107896778.71/warc/CC-MAIN-20201028044037-20201028074037-00489.warc.gz | 506,880,611 | 9,478 | # Thursday October 1 - Solar physics
Agenda Review EM radiation How far away are stars? What can starlight tell us? NASA Sunlight carefully separated by
wavelength Absorption and emission Absorption spectrum Emission spectrum (glowing material) Blackbody spectrum U. Tennessee
Consider the following photo Compare and contrast the stars in the photo NASA What if all stars had the same luminosity? Luminosity means total power output as light. Determining their distance from us would be easy if they
all were the same luminosity! Given only 25 Watt bulbs The brightness depends only on their distance from us Stars arent all 25 Watt bulbs! A stars distance can be determined:
If we know how luminous it really is Compared to how bright it looks to us Apparent Magnitude scale Classifies objects by how bright they appear in the sky
Larger number means less bright 1 magnitude = difference in brightness by a factor of 2.5 Apparent Magnitude scale Sirius Full moon Venus Sun
-25 -20 -15 -10 -5 Faintest object Faintest object With binoculars 4 m telescope
Naked eye 0 Polaris5 10 15 20 25
Absolute magnitude Luminosity = how much light a star is giving off (like wattage for light bulbs)
Compared to the Sun (Lsun) Absolute magnitude = how bright (what magnitude) a star would appear at 10 parsecs (32.6 light years) Example: Antares versus Sirius Apparent magnitudes:
Antares -4, Sirius -1.5 Absolute magnitudes: Antares +2, Sirius +1.4
Which appears brighter in the sky? A. Antares B. Sirius C. same D. Cannot conclude Example: Antares versus Sirius Apparent magnitudes:
Antares -4, Sirius -1.5 Absolute magnitudes: Antares +2, Sirius +1.4
Which is more luminous? A. Antares B. Sirius C. same D. Cannot conclude So how do we determine the distance to other stars? Compare absolute and apparent magnitudes If we know the luminosity (or
absolute magnitude) of a star, we can find its distance. A star of known luminosity is called a standard candle. More on this later... So how do we determine the distance to other stars? There is another, more direct way... Triangulation (parallax) Measure the two angles (a, b) to find how wide the Missouri river is!
Trigonometric parallax Earth Known Distance (2 AU) Some star p Sun
Unknown distance Earth What makes parallax measurement difficult? A. B. C. D. Its hard to read your protractor when
it is dark outside. The parallax angle is very small because the stars are so far away. We cant see any of the same stars six months apart. It actually is not difficult! Parsec (a unit of distance) 1 arcsecond = 1/60 arcminute = 1/3600 degree 1 AU If p = 1 arcsecond
This distance is 1 parsec (3.26 light years) Unfortunately, there are no stars this close to us. Limits with this method Earth Known Distance (1 AU) Sun Some star
p Unknown distance Beyond 300 light years, p is too small for us to measure. (Our galaxy is about 100,000 light years across) What can we learn by looking at starlight?
Distance to star Ingredients Temperature Exam one is Tuesday at 6 pm Tuesday optional review during class Individual portion
Group portion Multiple choice 2 essay questions
Multiple choice Grab a review sheet! Challenge: write one exam question (multiple choice), and email it to me! Team activity # 4 Trigonometric parallax Include last names and team number on all activities!
Part I: Do NOT place any measuring device beyond the line!! You must estimate the angles! Part II: You may want to go into the hallway. Grab your graded activities on the way out!
Nearby stars appear to move (Not apparent to the naked eye) Wikipedia
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In Geometry, students learned about the trigonometric ratios sine, cosine, and tangent. In this unit, we extend these ideas into functions that are defined for all real numbers! We learn about the
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## Solving Trigonometric Equations
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All Right Reserved. Copyright 2005-2017 | 1,435 | 6,143 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2017-51 | longest | en | 0.829034 |
https://www.physicsforums.com/threads/show-that-a-series-lr-circuit-is-an-lp-filter-if-the-output-is-taken-across-the-resis.165181/ | 1,529,453,623,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267863259.12/warc/CC-MAIN-20180619232009-20180620012009-00128.warc.gz | 870,987,837 | 12,820 | # Homework Help: Show that a series LR circuit is an LP filter if the output is taken across the resis
1. Apr 11, 2007
### VinnyCee
1. The problem statement, all variables and given/known data
Show that a series LR circuit is a Low Pass filter if the output is taken across the resistor. Calculate the corner frequency, $f_c$, if L = 2mH and R = 10k$\Omega$.
2. Relevant equations
$$H(\omega)\,=\,\frac{V_0(t)}{V_i(t)}$$
$$H\left(\omega_c\right)\,=\,\frac{1}{\sqrt{2}}$$
$$f\,=\,\frac{\omega}{2\pi}$$
3. The attempt at a solution
$$H(\omega)\,=\,\frac{R}{R\,+\,j\omega L}$$
$$H(0)\,=\,1$$
$$H(\infty)\,=\,0$$
That does the "show" part. But now I don't know how to get the corner frequency.
$$\frac{1}{\sqrt{2}}\,=\,\frac{R}{R\,+\,j\omega L}$$
2. Apr 11, 2007
### VinnyCee
Never mind! It's 796 kHz.
3. Apr 11, 2007
### denverdoc
Per your post on Bode plots, this transfer function would be written as:
1/(1+jwL/R), you might want to plot this before tackling that much more complicated function, if you haven't done simple ones yet. | 362 | 1,052 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 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} | 3.171875 | 3 | CC-MAIN-2018-26 | latest | en | 0.802214 |
https://sillycodes.com/program-to-sum-of-two-numbers-in-c-using-a-function/ | 1,717,007,291,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059384.83/warc/CC-MAIN-20240529165728-20240529195728-00796.warc.gz | 430,117,914 | 73,774 | # Program to Sum of Two Numbers in C using a Function
## Program Description:
Write a program to Sum of Two Numbers in C using a Function. This program is to understand how functions work in c language.
The program accepts two numbers from the user and calculates the sum of those numbers. The program should create a function, which should accept two numbers as input and return the integer number ( which is the sum of given numbers).
### Example Input and Output:
Input:
Enter two numbers : 10 20
Output:
Sum of 10 and 20 is : 30
## Sum of two Numbers in C using Function Program Explanation:
1. Accept two numbers from the user as input, And store them in variables number1 and number2 respectively.
2. We are going to create a function called sum.
• The sum function takes two variables as input number1 and number2 and calculate the sum of number1 and number2 and returns the result.
• Here number1 and number2 are formal arguments of the sum function.
• As the sum function is defined after the function call, So we need to declare the function using the function declaration. We have added the sum function declaration at the start of the program using int <strong>sum</strong>(int, int);
3. Call the sum function by passing the number1 and number2 from the main function. As the sum function returns an integer which is the sum of the given numbers, Assign the function call to an integer variable (i.e result). So that the return value of the sum function will be stored in result variable
• result = <strong>sum</strong>(number1, number2);
4. Print the result on the console using printf function.
## Sum of two Numbers in C using function Program:
Here is the sum of two numbers program using functions in the C programming language
The sum function prototype details.
function_name: sum
argument_list: (int, int)
return_type: int
### Sum of two Numbers in C Program Output:
We are using the GCC compiler to compile and run the program.
Run the program and provide two integer values.
Let’s try a few more examples
As we can see from the above output, We are getting the expected results.
## Related Programs:
1. C Program to Calculate the Power of Number
2. C Program to Check Number is Power of 2
3. C Program to find power of number without Loops
4. C Program to Find all Factors of a Number
5. C Program to find all Prime Factors of numberÂ
6. C Program to Calculate the GCD or HCF of Two Number
7. C Program to Calculate the LCM of two Numbers
8. C Program to Check Palindrome Number
9. C Program to Check 3-Digit and N-Digit Armstrong Number
10. C Program to Generate Armstrong Numbers upto N ( User-Provided Number)
11. C Program to Generate Armstrong Numbers between two Intervals
Venkatesh
Hi Guys, I am Venkatesh. I am a programmer and an Open Source enthusiast. I write about programming and technology on this blog.
### 2 Responses
1. […] Add Two Numbers using Functions in C […]
2. […] Add Two Numbers using Functions in C […] | 691 | 2,989 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2024-22 | longest | en | 0.808303 |
https://jp.mathworks.com/matlabcentral/answers/859825-how-to-speed-up-my-code | 1,628,140,485,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046155322.12/warc/CC-MAIN-20210805032134-20210805062134-00212.warc.gz | 335,666,906 | 28,261 | # How to speed up my code?
2 ビュー (過去 30 日間)
Benson Gou 2021 年 6 月 18 日
コメント済み: Benson Gou 2021 年 6 月 21 日
Dear All,
I found my code spent a lot of cpu time on the following functions:
1. intersect, took 20.11 seconds, called 232074 times in my code.
My code:
C = intersect(A,B); where A and B are two column arrays.
2. unique, took 8.845 seconds, called 251736 times in my code.
My code:
BadMeas = unique([A; red2(B)']); where A and B are two column arrays, red2 is another row array.
3. ismember, took 7.117 seconds, called 269278 times in my code.
My code:
if ismember(selectedBus, InjBus_moreZeroLines)
for i = 1 : m
do the calculation
end
end
4. setdiff, called 13103 times in my code.
My code:
red = setdiff([1:length(A)],B);
A is an array and B is an array formed by integers.
I am wondering if there are faster functions to replace the above ones. Thanks.
Benson
##### 5 件のコメント表示非表示 4 件の古いコメント
Benson Gou 2021 年 6 月 18 日
Hi, dpb,
Thanks a lot for your reply. I will do as you suggested. You have a good weekend!
Benson
サインインしてコメントする。
### 採用された回答
Jan 2021 年 6 月 19 日
if ismember(selectedBus, InjBus_moreZeroLines)
ismember replies a vector, if the inputs are vectors. The if command requires a scalar condition. So I guess, this can be abbreviated. Maybe this is faster:
if any(selectedBus == InjBus_moreZeroLines)
In the line:
red = setdiff([1:length(A)],B)
the square brackets are a waste of time. [ ] is the concatenation operator, but here you concatenate the vector 1:length(A) with nothing.
Maybe this is more efficient:
red = true(size(A));
red(B) = false;
Then use red for logical indexing, instead of numerical indices.
I assume there are faster repalcements for the called functions, but without knowing what the inputs are and the context, it is not possible to suggest modifications.
##### 1 件のコメント表示非表示 なし
Benson Gou 2021 年 6 月 21 日
Hi, Jan,
Best regards,
Benson
サインインしてコメントする。
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## Contracting maps of hyperbolic manifolds
Is it possible for SOME positive $c$, $c<1$ to find a pair of COMPACT hyperbolic manfiolds $M^3$ and $N^3$ with a positive degree map $$f: M^3 \to N^3,$$ such that $f$ is contacting with constant $c$? Are there may examples like this?
One can ask the same question of Riemann surfaces, and it seems to me that this should be possible. For example we can take a double cover of Riemann surface with many points or ramification. Though I don't know a proof even in this case. Of course for non-ramified cover the best possible constant $c$ is $1$.
ADDED. Following the answer of Sam Need, let me give an approximative "proof" of the fact that this works in dimesnion 2. Let us triangulate a hyperbolic surface $N^2$ in triangles of very small size, that have acute angles (this is always possible). We want to show that a double cover of $N^2$ with ramifications at vertices of the triangulation will do the job. For this we need a lemma (without a proof).
Lemma. Suppose we have two hyperbolic trianlges, one very small and acute with angles $a$, $b$, $c$, and the over with angles a/2, b/2, c/2. Then there is a contacting map from the second triangle to the first one. The lemma is true, since the second trianlge will be large.
Now on the double cover we can take a trangulation that comes from $N^2$ and glue it from these triangles with half angles. Half angles come from doble cover. Then we just need to "adjust" the map.
Of course this is not a real proof, but I am 100% it can be made real.
-
You should probably add hypotheses to your question, otherwise there is the trivial example mapping $\mathbb{H}^3 \to \mathbb{H}^3$ where in exponential coordinates, one takes every point closer to the origin by some factor. – Ian Agol Dec 14 2009 at 19:00
Thanks! I corrected the question, speaking about hyperbolic manifolds I had in mind "compact" manifold – Dmitri Dec 14 2009 at 19:20
In general, for any non-zero degree map from one closed negatively curved manifold to another, there is a canonical map (due to Besson-Courtois-Gallot) called the "natural map". However, it's only known to be pointwise volume decreasing, not necessarily contracting. They call this the "real Schwarz-Lemma". Applying the Schwarz lemma for Riemann surfaces I think gives the contracting map in this case for branched covers. Think of the induced map on the universal cover, which is the unit disk, or $\mathbb{H}^2$. The Schwarz lemma says that any conformal map from the disk to the disk is contracting, unless it's an isometry.
I thought of one (not very explicit) example in 3-D. Take two simplices in hyperbolic space. There is a canonical affine map (say in the Lorentzian model) taking one simplex to the other. This will be a contracting map for the hyperbolic metric if one simplex sits inside the other [Edit: actually I'm not sure about this now, but in the example below there exists a contracting map]. There are finitely many tetrahedra in $\mathbb{H}^3$ which give rise to fundamental domains for discrete reflection groups (see Ratcliffe). Two of these have one dihedral angle $\pi/5$, with opposite edge angle $\pi/2$ and $\pi/4$, respectively, and all other angles the same. There is a 1-parameter family of polyhedra interpolating between these (basically, just "push" the two faces closer together along the dihedral angle $\pi/5$ edge) which decreases distances. Also, the orbifold fundamental group (i.e. reflection group) from the $\pi/4$ one maps to that of the $\pi/2$ one. So there's a distance decreasing map from one orbifold to the other. Using Selberg's lemma, one may find finite-sheeted manifold covers with the same property.
-
This question confuses me, even in dimension two. The non-trivial branched coverings $f$ I can think of are extremely contracting near the branch points. So much so that $f$ is actually expanding elsewhere to produce enough area. | 1,031 | 4,087 | {"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} | 3.796875 | 4 | CC-MAIN-2013-20 | latest | en | 0.925768 |
https://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-042j-mathematics-for-computer-science-spring-2015/structures/tp5-1/euclidean-algorithm-video/euclidean-algorithm/ | 1,611,191,753,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703522150.18/warc/CC-MAIN-20210121004224-20210121034224-00053.warc.gz | 484,782,739 | 22,274 | # Euclidean Algorithm
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PROFESSOR: The greatest common divisor of two numbers is easy to compute. And that's a factor will play a crucial role in the number three we're going to develop, and the properties of some of the modern codes that are based on number theory. The efficient way to compute the GCD of two numbers is based on a classical algorithm known as Euclidean algorithm, which is several thousand years old. And let's describe how it works now.
So the Euclidean algorithm is based on the following lemma, which we'll call the remainder lemma, and it says that if a and b are two integers, then the greatest common divisor of a and b is the same as the greatest common divisor of b, and the remainder of a divided by b-- providing, of course, b is not 0, because otherwise you can't divide by b.
OK how do you make sense out of this? Why is this true? Well, it's actually a very easy proof. Remember that by the so-called division algorithm-- or it's really a theorem-- if you divide a by b and we're doing integer division, what that means is you find a quotient of a divided by b in the quotient, and a remainder. And the quotient has the property that q times b plus the remainder is equal to a. The remainder is always going to be smaller than a. It will be the range from 0 up to, but not including, a.
OK, if you look at this simple expression, what becomes apparent is that if you've got a divisor of two out of three of these terms, then it's going to divide the third term. So for example, if you have a divisor of b and r, then the sum of those two things is also going to have the same divisor, which means that a will have that divisor. If something divides both a and b, then it divides r. And if it divides b and r, it divides a. And that means that a and b and b and r have exactly the same divisors. They not only have the same greatest common divisor, all their divisors are the same. So obviously, the greatest one is the same. And that proves this key remainder lemma.
Well, the remainder lemma now gives us a very lovely way to compute the GCD. And here's an example. Suppose I want to compute the GCD of 899 and 493. A is 899, b is 493. Well, so I want this GCD, 899 of 493. Well, according to the remainder lemma, if I divide 899 by 493, I get a quotient of 1, and a remainder of 406. So that means that 899 and 493 have the same GCD as 493 and 406. That is the original number b, and the new remainder 406.
But now, I can divide 493 by 406. I get a quotient of zero and a remainder of 87. So 406 and 87 have the same GCD. Dividing 406 by 87, I get that 87 and 58 have the same GCD. Dividing 87 by 58, I get that 58 and 29 have the same GCD. And now I win, because look, when I divide 58 by 29, I get a remainder of 0. And the GCD of anything and 0 is that thing. So the GCD of 29 and 0 is 0. I guess the only exception is the GCD of 0 and 0, which is not defined. But if it's not 0, then the GCD of x and 0 is x.
And there it is. So I've just found that the GCD of 899 and 493 is 29. And this is a quite fast algorithm, because I keep dividing the numbers that I have by each other, and it gets small fast. We'll be more precise about that in a minute.
OK, it's a good exercise in state machine thinking and practice in program verification to reformulate the Euclidean algorithm, or formulate it explicitly as a state machine. It's a very simple kind of state machine. The states of this Euclidean algorithm state machine will be pairs of non-negative integers. So the states are n cross n, the Cartesian product of the non-negative integers, with itself. The start state is going to be the pair a, b, whose GCD I want to compute.
And the transitions are simply repeatedly applying the remainder lemma. Namely, if I'm in state x, y, where you think of x as and y as the GCD that I'm trying to compute, I simply convert x and y to y, and the remainder of x divided by y. And I keep doing that as long as y is not 0.
OK, very simple state machine-- really, just one transition rule. Well, according to the lemma, since I'm replacing the GCD of x and y by the GCD of y and the remainder of x divided by y, the GCD is actually staying constant. This transition preserves the GCD that's left in the pair of registers, x and y. So what we can say is that since the GCD of x and y doesn't change from one step to another, we can say that the GCD of x and y at any point is equal to its original value, which is the GCD of a and b.
So in other words, this equation, GCD of x and y in the current state is equal to GCD of a and b, the GCD of a and b that we started with, is a preserved invariant of the state. So p of a state xy, the property that GCD of x and y is the original GCD is a preserved invariant of the state machine. Moreover, p of start is trivially true, because at the start, x and y are a equals b. So p of x and y is just saying the GCD of a and b is equal to GCD of a and b.
Cool. So I've got that this property is true at the start, and it's preserved by the transitions. So the invariance principle tells me that if the program stops, I'm going to have the GCD of x and y when it terminates is equal to the actual GCD that I want. And that enables us to prove partial correctness. The claim is that if this program terminates-- we haven't determined that it does yet-- but at termination, if any, I claim that x is left in-- that the GCD of a and b is left in register x. The value of x at the end is going to be the GCD of and and b.
Well, why is that? Well, look-- at termination, what we know is that y is 0. That's the only way that this procedure stops, because otherwise, the transition rule is applicable. So that means that when y equals 0 at termination, what we have is that since y is 0, GCD of x and y is equal to the GCD of x and 0. And that's equal to x, assuming, again, that x is positive, or not 0.z
So x is the GCD of x and y. And by the invariant, the GCD of x and y is equal to the GCD of a and b. So I've prove this little fact. This procedure correctly computes the GCD of a and b, leaving the answer in register x, if it terminates. Well, of course it terminates, and it terminates fast. So let's see why.
Notice that at each transition, we're going to replace x by y, and y by the remainder of x divided by y. Let's just assume for simplicity that of the [? pairings, ?] y that x is the bigger one. So there's two cases of why these numbers are getting small fast. The first case is suppose that y is less than x over 2, or less than or equal to x over 2. Well, since at this step, you're going to replace x by y, it means that you're replacing x by something that's less than half x. So x gets halved at this step.
What about if y is big? Well, if y is bigger than x over 2, then the remainder of x divided by y is simply x minus y. And it's going to be less than x over 2. But that's going to be the value of y after the next step. So y is going to be halved either at this step or the next step when it's replaced by the remainder of x and y. And the net result is that y it gets cut in half, or even smaller, at every other step, which means that this procedure can't continue for more than twice the log to the base 2 of the original value of y, which is b number of steps, because that's how many halves you can do before you start hitting 0.
So we've just shown that this procedure holds in logarithmic number of steps, which is the same as saying that it's about the length of b in binary, and even fewer steps than the length of b in decimal. The GCD algorithm is really very efficient. | 1,936 | 7,635 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-04 | longest | en | 0.967738 |
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# 2c03-review - 00049 - L’[1:= L[L.last div 2 1 end else...
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5 end The time complexity is O(n). [6] function C1H2H(L1, L2: List;): List; var n : integer; begin L1.last := (L1.last div 2); for (n:=1 to L1.last) do L1.element[n] := L1.element[2*n] L2.last := L2.last – (L2.last div 2); for (n:=1 to L2.last) do L2.element[n] := L2.element[2*n-1]; for (n := L1.last.+1 to L1.last+L2.laste) do L1.element[n] := L2.element[n - L.last]; L1.last := L1.last + L2.last; return(L1); end The time complexity is O(m+n). [6] function MIDDLE(L: List;): List; var L’ : List; begin if ((L.last mod 2) = 0) then begin L’.Last := 2; L’[1] := L[L.last div 2];
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Unformatted text preview: L’[1] := L[L.last div 2 +1]; end else begin L’.Last := 1; L’[1] := L[L.last div 2+1]; e n d return(L’); end The time complexity is O(1), which is faster than ‘abstract pointer’ implementation 3.[8] Write a procedure to interchange the element at the position 1 with the element at the position 6, the element at the position 3 with the element at the position 8, etc., in general the element at the position number k with the element at the position k+5 in a singly linked list. Analyze the time complexity of your procedure. celltype = record of element: elementtype...
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# Pythagorean Theorem - Finding Unknown Side Length Fill in the Blanks Activity
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### PRODUCT DESCRIPTION
Pythagorean Theorem - Finding Unknown Side Length Fill in the Blanks Activity
Practice solving missing leg length when given one leg and hypotenuse lengths. Add a fun spin and fill in the blanks to solve a riddle :)
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Pythagorean Theorem - Finding Hypotenuse Length Maze Activity
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## Present Value of Free Cash Flow to the Firm (FCFF)
Intermediate level
In discounted cash flow (DCF) valuation techniques the value of the stock is estimated based upon present value of some measure of cash flow. Free cash flow to the firm (FCFF) is generally described as cash flows after direct costs and before any payments to capital suppliers.
### Intrinsic Stock Value (Valuation Summary)
Kraft Heinz Co., free cash flow to the firm (FCFF) forecast
US\$ in millions, except per share data
Year Value FCFFt or Terminal value (TVt) Calculation Present value at
01 FCFF0
1 FCFF1 = × (1 + )
2 FCFF2 = × (1 + )
3 FCFF3 = × (1 + )
4 FCFF4 = × (1 + )
5 FCFF5 = × (1 + )
5 Terminal value (TV5) = × (1 + ) ÷ ()
Intrinsic value of Kraft Heinz Co.’s capital
Less: Debt (fair value)
Intrinsic value of Kraft Heinz Co.’s common stock
Intrinsic value of Kraft Heinz Co.’s common stock (per share)
Current share price
Based on: 10-K (filing date: 2020-02-14).
Disclaimer!
Valuation is based on standard assumptions. There may exist specific factors relevant to stock value and omitted here. In such a case, the real stock value may differ significantly form the estimated. If you want to use the estimated intrinsic stock value in investment decision making process, do so at your own risk.
### Weighted Average Cost of Capital (WACC)
Kraft Heinz Co., cost of capital
Value1 Weight Required rate of return2 Calculation
Equity (fair value)
Debt (fair value) = × (1 – )
Based on: 10-K (filing date: 2020-02-14).
1 US\$ in millions
Equity (fair value) = No. shares of common stock outstanding × Current share price
= × =
Debt (fair value). See details »
2 Required rate of return on equity is estimated by using CAPM. See details »
Required rate of return on debt. See details »
Required rate of return on debt is after tax.
Estimated (average) effective income tax rate
= ( + + + + ) ÷ 5 =
WACC =
### FCFF Growth Rate (g)
#### FCFF growth rate (g) implied by PRAT model
Kraft Heinz Co., PRAT model
Average Dec 28, 2019 Dec 29, 2018 Dec 30, 2017 Dec 31, 2016 Dec 31, 2015
Selected Financial Data (US\$ in millions)
Interest expense
Net income (loss) attributable to Kraft Heinz
Effective income tax rate (EITR)1
Interest expense, after tax2
Add: Dividends declared, Series A Preferred Stock
Interest expense (after tax) and dividends
EBIT(1 – EITR)3
Commercial paper and other short-term debt
Current portion of long-term debt
Long-term debt, excluding current portion
Shareholders’ equity
Total capital
Financial Ratios
Retention rate (RR)4
Return on invested capital (ROIC)5
Averages
RR
ROIC
FCFF growth rate (g)6
Based on: 10-K (filing date: 2020-02-14), 10-K (filing date: 2019-06-07), 10-K (filing date: 2018-02-16), 10-K (filing date: 2017-02-23), 10-K (filing date: 2016-03-03).
2019 Calculations
2 Interest expense, after tax = Interest expense × (1 – EITR)
= × (1 – ) =
3 EBIT(1 – EITR) = Net income (loss) attributable to Kraft Heinz + Interest expense, after tax
= + =
4 RR = [EBIT(1 – EITR) – Interest expense (after tax) and dividends] ÷ EBIT(1 – EITR)
= [] ÷ =
5 ROIC = 100 × EBIT(1 – EITR) ÷ Total capital
= 100 × ÷ =
6 g = RR × ROIC
= × =
#### FCFF growth rate (g) implied by single-stage model
g = 100 × (Total capital, fair value0 × WACC – FCFF0) ÷ (Total capital, fair value0 + FCFF0)
= 100 × ( × ) ÷ ( + ) =
where:
Total capital, fair value0 = current fair value of Kraft Heinz Co.’s debt and equity (US\$ in millions)
FCFF0 = the last year Kraft Heinz Co.’s free cash flow to the firm (US\$ in millions)
WACC = weighted average cost of Kraft Heinz Co.’s capital
#### FCFF growth rate (g) forecast
Kraft Heinz Co., H-model
Year Value gt
1 g1
2 g2
3 g3
4 g4
5 and thereafter g5
where:
g1 is implied by PRAT model
g5 is implied by single-stage model
g2, g3 and g4 are calculated using linear interpoltion between g1 and g5
Calculations
g2 = g1 + (g5g1) × (2 – 1) ÷ (5 – 1)
= + () × (2 – 1) ÷ (5 – 1) =
g3 = g1 + (g5g1) × (3 – 1) ÷ (5 – 1)
= + () × (3 – 1) ÷ (5 – 1) =
g4 = g1 + (g5g1) × (4 – 1) ÷ (5 – 1)
= + () × (4 – 1) ÷ (5 – 1) = | 1,345 | 4,315 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2020-40 | latest | en | 0.842307 |
http://math.stackexchange.com/questions/499204/formula-for-game-score-normalization | 1,469,337,599,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257823947.97/warc/CC-MAIN-20160723071023-00132-ip-10-185-27-174.ec2.internal.warc.gz | 153,683,101 | 16,722 | # Formula for Game Score Normalization
My friends and I play a lot of board games together. In an effort to keep track of how we do between each game, we use a simple method to normalize the scores.
For example, let's say we have the following score for a particular game:
• Player A: 7
• Player B: 8
• Player C: 6
• Player D: 10
These would then be normalized by adding up the total and dividing each players score by that total - giving the following:
• Player A: 23
• Player B: 26
• Player C: 19
• Player D: 32
This has turned out to be very handy for ranking who's best overall, regardless of whether the scores are based on victory points, dollar amounts or any other type of unit.
However, we are stuck on how to handle games where a lower score is better. I'm sure we are missing something simple, but we cannot determine what the formula should be.
Any help would be greatly appreciated.
-
## 1 Answer
Fun question. I think your entire system should be multiplicative, not additive. In other words:
1. Name yourself the "baseline player." Your score will always be $1$.
2. Each other player's normalized score in a game is (their points)/(your points) -- or, in the case of a golf-like game, (your points)/(their points).
3. A player's net score is the product of all their normalized scores.
This will give you exact statistics such as "I am 1.17 times better at games than you." Which could be fun.
- | 344 | 1,424 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2016-30 | latest | en | 0.95281 |
https://www.dissertation-help.co.uk/hypothesis-and-research-question/ | 1,627,710,954,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154053.17/warc/CC-MAIN-20210731043043-20210731073043-00013.warc.gz | 762,494,599 | 29,035 | ### PRICE CALCULATOR
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# Hypothesis And Research Question – How To Differentiate Between The Two?
Are you working on your final thesis or dissertation and are confused between the difference of hypothesis and research question?
Then you are in serious trouble.
Lucky for you, we’re here to help you out! But before we tell you the difference between each other, let’s first see what they are individual.
Both the hypothesis and research question are the central points in any thesis or dissertations as they give directions to the structure of the literature review, data collection process, data analysis, and conclusions.
## What is a Hypothesis?
A hypothesis is a formal statement that tries to assume a relationship between two or more variables. Hypotheses are very specific as they describe the exact nature of the relationship, for example, positive or negative, between two or more variables.
## What is a research question?
A research question is a type of hypothesis that is asked in the form of a question. It also predicts the exact relationship between two variables but it is asked in the form of a question rather than a statement.
What is the difference between the two?
### Difference No. 1:
Both the hypothesis and research question either support or refuse an existing theory but the main difference between the two is that a research question is formulated in the form of a question whereas a hypothesis is an assumed solution to a problem.
## Because of this difference, a hypothesis can be directly tested and verified whereas the research question cannot.
### Difference No. 2:
Another main difference between the two is that a hypothesis is an educated guess on the outcome of the study while a research question is a form of the researcher wondering about the workings of the world.
Because of that, a hypothesis is a part of the scientific study process while the research question is not.
Due to the two differences, the writing structure of both of them differs majorly. The hypothesis is usually written in the form of a prediction, mostly starting with ‘I predict’ while the research question is mostly structured as a question trying to determine a relationship between two variables, i.e. ‘what is the effect of (variable 1) on (variable 2)?
The Data: The two variables between which we are trying to identify a relationship are a gender (male or female) and level of reading understanding.
The Correct hypothesis:
Research question: | 503 | 2,556 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-31 | latest | en | 0.955814 |
https://www.vanessabenedict.com/how-to-know-how-many-carats-gold-is/ | 1,718,889,939,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861940.83/warc/CC-MAIN-20240620105805-20240620135805-00413.warc.gz | 917,408,684 | 14,455 | # How can I tell how many carats my gold is?
#### ByVanessa
Jun 9, 2022
Untitled Document
## How do I tell what carat my gold is
To convert carats to monthly cents, you need to divide the number of carats by 24 and multiply the end result by 100. So, to start, to find out how much gold is in each penny of your 22 carat homemade, divide 22 by 24, the end result is 0. 9166 times seventy-five so that’s 91.66 cents – that’s the purity of your gold.
## How can I tell if my gold is 18K or 24k
The gold should have a stamp indicating to which buyer and to which sample the gold can be attributed. 18K gold has a distinct small hallmark inside the gemstone indicating 18K gold.
## How do I know if my gold is 24k or 22K
You can test your primary gold for purity at most pendant stores. The jeweler takes a sample of the device onto a test pad and then soaks the samples in a solution (usually nitric acid) to see how it reacts. From the observed chemical reaction, you can determine the number of carats of gold.
Untitled Document
## How can you tell if gold is 18K or 22K
With 18K gold, Chastity has a slightly matte golden hue. It’s easy if you want to identify 18K gold fine jewelry – now all 18K, 24K and 22K gold jewelry must sometimes be marked to have a 18K750 purity bastion hallmark.
## Why are carats called carats
The name Karat comes through medieval French and simply Italian from the Arabic q?r?? ab, meaning “bean with a pod”, is itself derived from the Greek keration, meaning both carob and small weight. In the early 1900s, carat fat loss was measured in milligrams, 200 or 0.2 grams.
## How many carats is an Olympic gold medal
Each Olympic gold medal produced is made from 210 grams of silver and is therefore plated with 6 grams of 28 carat gold.
## How many carats is pure gold
24 carat pure gold without any other alloys. The lower karahs contain less gold; 19 carat gold is composed of 75% Razoo Gold brass and 25% materials, most commonly copper or silver.
## How many carats is real gold
Is karatage a measure of the purity of steel alloyed with other metals. 28-carat pure gold without metal impurities. The lower wagons contain less gold; 18 carat gold contains 75 cents of gold and 25 nickel cents of every other metal, often copper versus silver.
## How can I tell how many carats my gold is
One way to find out how many carats a gold wire has is to look at the hallmarks attached to it. These hallmarks must indicate the purity of the gold alloy in carats or parts per thousand. The interpretation of carats is simple: 10,000 is 10 carats, 18,000 is sixteen carats, and so on.
## How many carats are in pure gold
24k is pure gold with no other metals in it. Lower carags carry less gold; Precious 18 carat gold is 75% gold and 15% other metals, often copper or silver.
Untitled Document | 736 | 2,838 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2024-26 | latest | en | 0.922687 |
https://math.stackexchange.com/questions/3237247/25-chance-occurring-7-11-times/3237249 | 1,561,477,719,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627999853.94/warc/CC-MAIN-20190625152739-20190625174739-00266.warc.gz | 512,663,319 | 34,683 | # 25% chance occurring 7/11 times.
If an event has a probability of happening 25% of the time. How do you calculate the chances of this happening 7 out of 11 times.
If A is 25% and B is 75%. What is the probability of A occurring 7 times out of a possible 11 attempts.
I have found similar threads but unable to get a particular formula / answer.
Thanks.
There are $$11 \choose 7$$ options for which $$7$$ attempts have been successful, each with $$\left( \frac{1}{4} \right) ^ 7 \left( \frac{3}{4} \right)^4$$ change of occurring. So the probability of $$A$$ occurring $$7$$ out of $$11$$ times is $${11 \choose 7} \left( \frac{1}{4} \right) ^ 7 \left( \frac{3}{4} \right)^4=0.637\%$$.
If the probability of an event is $$p$$, then it happens exactly $$k$$ times out of $$n$$ with a probability of $$\binom{n}{k}p^k(1-p)^{n-k}$$
In your case, $$k=7$$, $$n=11$$, $$p=0.25$$.
The probability of event $$A$$ occurring $$7$$ times and $$B\ 4$$ times is $$p = \binom{11}{7} (0.25)^7 (0.75)^4 \approx 0.00637$$ | 339 | 1,010 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 18, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2019-26 | latest | en | 0.880299 |
http://www.aqua-calc.com/convert/linear-density/tonne-per-inch | 1,500,971,697,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549425117.42/warc/CC-MAIN-20170725082441-20170725102441-00226.warc.gz | 354,778,321 | 7,121 | # Select a linear density conversion table with which to
## Convert tonne per inch to other linear density measurement units
Select units whose names contain any of the following checked keywords, where ∗ is a wildcard for partial matches:
Convert tonne per inch to:CorrelationFormula
1 t/in = 6 075 731.63 gr/cm6 075 731.63 × t/in = gr/cm
1 t/in = 185 188 300 gr/ft185 188 300 × t/in = gr/ft
1 t/in = 15 432 358.4 gr/in15 432 358.4 × t/in = gr/in
1 t/in = 607 573 163 gr/m607 573 163 × t/in = gr/m
1 t/in = 555 564 901 gr/yd555 564 901 × t/in = gr/yd
1 t/in = 393 700.787 g/cm393 700.787 × t/in = g/cm
1 t/in = 12 000 000 g/ft12 000 000 × t/in = g/ft
1 t/in = 1 000 000 g/in1 000 000 × t/in = g/in
1 t/in = 39 370 078.7 g/m39 370 078.7 × t/in = g/m
1 t/in = 36 000 000 g/yd36 000 000 × t/in = g/yd
1 t/in = 393.700787 kg/cm393.700787 × t/in = kg/cm
1 t/in = 12 000 kg/ft12 000 × t/in = kg/ft
1 t/in = 1 000 kg/in1 000 × t/in = kg/in
1 t/in = 39 370.0787 kg/m39 370.0787 × t/in = kg/m
1 t/in = 36 000 kg/yd36 000 × t/in = kg/yd
1 t/in = 0.3874828844 long tn/cm0.3874828844 × t/in = long tn/cm
1 t/in = 11.8104783 long tn/ft11.8104783 × t/in = long tn/ft
1 t/in = 0.9842065268 long tn/in0.9842065268 × t/in = long tn/in
1 t/in = 38.7482884 long tn/m38.7482884 × t/in = long tn/m
1 t/in = 35.431435 long tn/yd35.431435 × t/in = long tn/yd
1 t/in = 393 700 787 000 µg/cm393 700 787 000 × t/in = µg/cm
1 t/in = 12 000 000 000 000 µg/ft12 000 000 000 000 × t/in = µg/ft
1 t/in = 1 000 000 000 000 µg/in1 000 000 000 000 × t/in = µg/in
1 t/in = 39 370 078 700 000 µg/m39 370 078 700 000 × t/in = µg/m
1 t/in = 36 000 000 000 000 µg/yd36 000 000 000 000 × t/in = µg/yd
1 t/in = 393 700 787 mg/cm393 700 787 × t/in = mg/cm
1 t/in = 12 000 000 000 mg/ft12 000 000 000 × t/in = mg/ft
1 t/in = 1 000 000 000 mg/in1 000 000 000 × t/in = mg/in
1 t/in = 39 370 078 700 mg/m39 370 078 700 × t/in = mg/m
1 t/in = 36 000 000 000 mg/yd36 000 000 000 × t/in = mg/yd
1 t/in = 13 887.3866 oz/cm13 887.3866 × t/in = oz/cm
1 t/in = 423 287.543 oz/ft423 287.543 × t/in = oz/ft
1 t/in = 35 273.9619 oz/in35 273.9619 × t/in = oz/in
1 t/in = 1 388 738.66 oz/m1 388 738.66 × t/in = oz/m
1 t/in = 1 269 862.63 oz/yd1 269 862.63 × t/in = oz/yd
1 t/in = 253 155.485 dwt/cm253 155.485 × t/in = dwt/cm
1 t/in = 7 716 179.17 dwt/ft7 716 179.17 × t/in = dwt/ft
1 t/in = 643 014.933 dwt/in643 014.933 × t/in = dwt/in
1 t/in = 25 315 548.5 dwt/m25 315 548.5 × t/in = dwt/m
1 t/in = 23 148 537.5 dwt/yd23 148 537.5 × t/in = dwt/yd
1 t/in = 867.961661 lb/cm867.961661 × t/in = lb/cm
1 t/in = 26 455.4714 lb/ft26 455.4714 × t/in = lb/ft
1 t/in = 2 204.62262 lb/in2 204.62262 × t/in = lb/in
1 t/in = 86 796.1661 lb/m86 796.1661 × t/in = lb/m
1 t/in = 79 366.4145 lb/yd79 366.4145 × t/in = lb/yd
1 t/in = 0.4339808305 short tn/cm0.4339808305 × t/in = short tn/cm
1 t/in = 13.2277357 short tn/ft13.2277357 × t/in = short tn/ft
1 t/in = 1.10231131 short tn/in1.10231131 × t/in = short tn/in
1 t/in = 43.3980831 short tn/m43.3980831 × t/in = short tn/m
1 t/in = 39.6832073 short tn/yd39.6832073 × t/in = short tn/yd
1 t/in = 26.9770731 sl/cm26.9770731 × t/in = sl/cm
1 t/in = 822.261187 sl/ft822.261187 × t/in = sl/ft
1 t/in = 68.5217657 sl/in68.5217657 × t/in = sl/in
1 t/in = 2 697.70731 sl/m2 697.70731 × t/in = sl/m
1 t/in = 2 466.78357 sl/yd2 466.78357 × t/in = sl/yd
1 t/in = 61.9972615 st/cm61.9972615 × t/in = st/cm
1 t/in = 1 889.67653 st/ft1 889.67653 × t/in = st/ft
1 t/in = 157.473044 st/in157.473044 × t/in = st/in
1 t/in = 6 199.72615 st/m6 199.72615 × t/in = st/m
1 t/in = 5 669.02961 st/yd5 669.02961 × t/in = st/yd
1 t/in = 0.3937007874 t/cm0.3937007874 × t/in = t/cm
1 t/in = 12 t/ft12 × t/in = t/ft
1 t/in = 39.3700787 t/m39.3700787 × t/in = t/m
1 t/in = 36 t/yd36 × t/in = t/yd
1 t/in = 12 657.7742 oz t/cm12 657.7742 × t/in = oz t/cm
1 t/in = 385 808.958 oz t/ft385 808.958 × t/in = oz t/ft
1 t/in = 32 150.7467 oz t/in32 150.7467 × t/in = oz t/in
1 t/in = 1 265 777.42 oz t/m1 265 777.42 × t/in = oz t/m
1 t/in = 1 157 426.88 oz t/yd1 157 426.88 × t/in = oz t/yd
1 t/in = 1 054.81452 troy/cm1 054.81452 × t/in = troy/cm
1 t/in = 32 150.7465 troy/ft32 150.7465 × t/in = troy/ft
1 t/in = 2 679.22889 troy/in2 679.22889 × t/in = troy/in
1 t/in = 105 481.452 troy/m105 481.452 × t/in = troy/m
1 t/in = 96 452.2398 troy/yd96 452.2398 × t/in = troy/yd
What is tonne per inch (t/in)? - Back to linear density conversions
#### Foods, Nutrients and Calories
Snacks, corn-based, extruded, chips, plain contain(s) 538 calories per 100 grams or ≈3.527 ounces [ calories | price ]
#### Gravels and Substrates
CaribSea, Freshwater, Super Naturals, Zen Garden density is equal to 1473.7 kg/m³ or 92 lb/ft³ with specific gravity of 1.4737 relative to pure water. Calculate how much of this gravel is required to attain a specific depth in a cylinderquarter cylinder or in a rectangular shaped aquarium or pond [ weight to volume | volume to weight | price ]
#### Materials and Substances
Osteoset (CaSO4) weigh(s) 2.96 gram per (cubic centimeter) or 1.711 ounce per (cubic inch) [ weight to volume | volume to weight | price | density ]
#### Weight/Volume at Temperature
Rapeseed oil weigh(s) 0.885 gram per (cubic centimeter) or 0.511 ounce per (cubic inch) at 69°C or 156.2°F [ weight to volume (T) | volume to weight (T) ]
#### What is poundal?
One poundal is force that required to give acceleration 1 foot per second per second (1ft/sec²) to an object with mass of one pound (1lb)
#### What is radioactivity measurement?
for radioactivity units details see below | 2,384 | 5,549 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2017-30 | latest | en | 0.301198 |
https://www.embibe.com/exams/cbse-ncert-solutions-for-class-11-maths-chapter-6/ | 1,719,318,134,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865972.21/warc/CC-MAIN-20240625104040-20240625134040-00719.warc.gz | 660,327,386 | 67,279 | • Written By Abhishek_verma
# NCERT Solutions for Class 11 Maths Chapter 6 Linear Inequalities
NCERT Solutions for Class 11 Maths Chapter 6 Linear Inequalities: The NCERT solutions for Class 11 Maths are great study material for students. The NCERT Class 11 Maths Solutions by Embibe are based on the latest syllabus by CBSE. These solutions by Embibe help students understand the concepts easily. Moreover, Embibe also helps students learn the correct method to solve exam questions.
Embibe provides NCERT Solutions online to help students with their studies. Apart from the NCERT Solutions, Embibe also provides 400+ practice questions. These practice questions increase students’ knowledge and are available for. Embibe also offers mock tests, learning videos, and more on its platform. Visit Embibe while studying to learn more.
## NCERT Solutions for Class 11 Maths Chapter 6: Important Topics
Embibe’s solutions for Class 11 Maths Chapter 6 thoroughly explain the key concepts related to Linear Inequalities. In this chapter, students will solve the linear inequalities problems consisting of one and two variables. Therefore, students can clear their doubts by referring to these NCERT Solutions for Class 11 Maths.
Students can start practicing these concepts by watching the Embibe Exemplar videos. After watching the videos, students must go through the books available on Embibe. Besides, all the study material on Embibe are available for. So, students must check them out. Now, let us look at the important topics covered under this chapter:
### NCERT Solutions for Class 11 Maths Chapter 6: Points to Remember
While studying, the important points to remember from Class 11 Maths Chapter 6 are as follows:
• Subtraction Rule of Linear Inequalities:
• If x > y, then x − z > y − z
• If x< y, then x − z < y − z
• Multiplication Rule of Linear Inequalities:
• If x> y and z > 0, then xz > yz
• If x<y and z > 0, then xz < yz
• Division Rule of Linear Inequalities:
• If x > y and z > 0, then x/z > y/z
• If x < y and z > 0, then x/z < y/z
• The common region of the coordinate plane must be reached while solving a system of linear equations in two variables graphically.
• Addition Rule of Linear Inequalities: If x > y ,then x + z > y+z and if x < y, then x + z < y + z.
## NCERT Solutions for Class 11 Maths: All Chapters
The NCERT Solutions for Class 11 Maths are listed below to help students with their exam preparations:
## FAQs on NCERT Solutions for Class 11 Maths Chapter 6
Q.1: How many practice questions are there in Class 11 Maths Chaper 6?
Ans: Embibe provides 400+ practice questions for Class 11 Maths Chapter 6. Therefore, students must practice these questions from Embibe for.
Q.2: What is Class 11 Maths Chapter 6 about?
Ans: Class 11 Maths Chapter 6 is about Linear Inequalities and its concepts. Therefore, students must refer to Embibe to understand these concepts.
Q.3: Can I take mock tests for Class 11 Maths on Embibe?
Ans: Yes, students can take the mock tests for Class 11 Maths for on Embibe. Therefore, students in Class 11 can sign up for Embibe and take the mock test.
Q.4: Are the NCERT Solutions by Embibe available for?
Ans. All the NCERT Solutions provided by Embibe are available for.
Q.5: How many questions are there in Chapter 6 Linear Inequalities of the NCERT Solutions for Class 11 Maths?
Ans: There are 65 questions in four exercises in the NCERT Solutions for Class 11 Maths Chapter 6 Linear Inequalities.
Also, Check, | 848 | 3,499 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2024-26 | latest | en | 0.878496 |
https://boards.straightdope.com/t/children-of-time-ds9-how-long-to-turn-a-colony-of-47-people-into-one-of-8-000/950672 | 1,631,840,396,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780053918.46/warc/CC-MAIN-20210916234514-20210917024514-00585.warc.gz | 188,085,302 | 7,632 | # "Children of Time" (DS9): How long to turn a colony of 47 people into one of 8,000?
In the Star Trek: Deep Space Nine episode “Children of Time”, the U.S.S. Defiant and its crew of 49 people goes through an energy field orbiting a planet, and finds a colony inhabited by 8,000 people, mostly humans with a smattering of people with Klingon and Trill DNA.
The people of the planet, Gaia, are their descendants. A few days after meeting their descendants, the Defiant is set to have an accident take place, marooning them on Gaia 200 years in the past. Kira dies a few days later in this timeline, because she was injured getting to Gaia, and couldn’t get back to DS9 for medical attention.
In real life, is 200 years enough time for exactly 47 people (Kira is dead, and Odo can’t procreate) to multiply to a thriving colony of almost 8,000?
Easily. Let’s say they couple up and have 4 kids per couple and give each gen 20 years to hook up and have 4 kids. You’ll have over 10,000 in way less than 200 years.
Seems like a lot. I’m not mathematically inclined; can you please show your work?
What is this, elementary school? 46x2=92 (twenty years later) 92x2=184 (40 years) 184x2=736 (60) 735x2=1472. . . keep doubling and you’ll see.
Haha, okay.
Another question, then: Is 47 people enough to have a breeding population without critical in-breeding problems?
There is the Human Bottleneck Theory. Some time, from 50,000 to 100,000 years ago, the human population shrunk down to as little as 3000 people. And here we are almost 8 billion people later. You get the humans you get and as long as they can have babies and their babies can have babies, then you live with the genetics you got.
If they’re extremely diverse they would be OK. You could have 47 people all of significantly different origins - native Inuit, Han Chinese, African, Pashtun, aboriginal Australian, lily-white Scandinavians, Slavic Russian, Hispanic, Caribbean, etc. and it would do.
I find google searches to be an excellent resource to answer these sorts of questions. For instance, I googled “minimum viable human population” and found a Stack Exchange question that indicates you’d need a starting population somewhere between 1500 and 4000 to eliminate the risk of inbreeding. Other studies have shown that a colony of as few as 160 could be viable for at least 20 generations, or 400 years.
That said, it’s only as time goes on and genetic makeup becomes more uniform that problems arise, so in 200 years (or about 10 generations) you might not yet notice the issues from such a small starting population. This is also not taking into account Federation medical technology, or the benefits that alien DNA (that is somehow still compatible with human DNA) could have on genetic diversity.
Plus, you know, throw a Klingon into the mix. Lots of genetic diversity there.
Plus you got non-human to admix. We are talking Star Trek.
Edit: Or what ASL said.
Right, a Klingon, a Trill, and who knows what else (we don’t see all the members of the Defiant crew), but mostly humans.
Welcome back after 9 years
But the episode is kind of crazy, the way it ends, I mean.
Odo basically…
commits planetary genocide as an act of “love.”
Regardless of the initial statistics, it is apparent that there was a lot of fucking going on.
Thanks! Don’t ask me why it was this topic of all things to cause me to unlurk after all this time, because I don’t know why either.
Also keep in mind that this is the Star Trek future. It’s possible that there’s some means of finding and editing out isolated harmful recessive genes, and that this is routinely done for infants (including all of the crew of the Defiant, when they were infants). That’d make a population much more resistant to bottlenecks.
That was my first thought, but then it occurred that even that might be caught in the anti-Augment laws.
OTOH, if anyone could recreate the processes, it’d be Bashir (who’s also evidence that they’re not lost, just illegal, himself), and it’d probably be routine on anyone born on the planet…even if he had to lie about its legality to some of the less inclined-to-bend-the-rules crew members.
Tasmania (a few thousand people isolated for 10000+ years) is another data point along those lines.
Oh, and
It’s ironic that you list “African” as a single category, on a par with Inuit, Scandinavian, etc. You’d have a lot more diversity than your list, if you just had 47 different Africans (drawn from all of the many African populations). Or maybe 43 Africans, an Australian Aboriginal, a Chilean native, and a Swede, or so.
The anti-Augment laws are specifically described, in “Doctor Bashir, I Presume?”, not to apply to genetic engineering done to repair serious birth defects. So that’s something. | 1,144 | 4,778 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2021-39 | latest | en | 0.956357 |
https://pinoybix.org/2018/06/mcq-in-industrial-electronics-part-8.html | 1,716,018,469,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057327.58/warc/CC-MAIN-20240518054725-20240518084725-00257.warc.gz | 404,835,673 | 80,987 | You dont have javascript enabled! Please enable it! MCQ in Industrial Electronics Part 8 | ECE Board Exam
# MCQ in Industrial Electronics Part 8 | ECE Board Exam
This is the Multiple Choice Questions Part 8 of the Series in Industrial Electronics: Principles and Applications as one of the Electronics Engineering topic. In Preparation for the ECE Board Exam make sure to expose yourself and familiarize in each and every questions compiled here taken from various sources including but not limited to past Board Exam Questions in Electronics Engineering field, Electronics Books, Journals and other Electronics References.
#### MCQ Topic Outline included in ECE Board Exam Syllabi
• MCQ in Industrial Electronics
• MCQ in Electronic Control System
• MCQ in Industrial Solid State Services
• MCQ in Welding Systems
• MCQ in Thyristors
• MCQ in High Frequency Heating
• MCQ in Feedback Systems / Servomechanism
• MCQ in Transducers
• MCQ in Motor Speed Control Systems
• MCQ in Robotic Principles
• MCQ in Bioelectrical Principles
• MCQ in Instrumentation and Control
• MCQ in Sensors
• MCQ in Optoelectronics devices
#### Continue Practice Exam Test Questions Part 8 of the Series
Choose the letter of the best answer in each questions.
351. A UJT has an internal resistances of RB1 = 6 Kฮฉ and RB2 = 3 Kฮฉ, what is its interbase resistance?
A. 2 Kฮฉ
B. 3 Kฮฉ
C. 6 Kฮฉ
D. 9 Kฮฉ
Solution:
352. For a unijunction transistor (UJT) to witch ON
A. the base 1 voltage should be greater than the peak voltage
B. the base 2 voltage should be greater than the peak voltage
C. the emitter voltage should be greater than the peak voltage
D. the voltage between the emitter and base 1 should be greater than the peak voltage when emitter being more positive
Solution:
353. Semiconductor devices with inherent ON-OFF behavior and has no linear operating regions are called thryistors. Examples are SCRs, triacs, SUSs, SBSs, Shockley diodes, diacs, PUTs, and SCSs. In selecting thyristors for a particular application, which of the statement below is generally desirable?
A. thyristors with high current and voltage ratings
B. thyristors with high holding current/voltage
C. faster thyristors
D. thyristors with high breakback-voltage
Solution:
354. Why does thyristors with high breakback voltage desirable?
A. it dissipates less power
B. it generates less heat
C. it is more efficient
D. all of the above
Solution:
355. Portion in the welding process interval during which the welding current is flowing is called ____________.
A. cool sub-interval
B. released interval
C. squeeze interval
D. heat sub-interval
Solution:
356. In automatic welding system, basically there are how many intervals?
A. 2
B. 5
C. 10
D. 15
Solution:
357. In automatic welding what do you call the first interval wherein the material to be welded are held together?
A. squeeze interval
B. weld interval
C. hold interval
D. standby interval
Solution:
358. After the squeeze interval, what comes next in an automatic welding system?
A. squeeze interval
B. weld interval
C. hold interval
D. standby interval
Solution:
359. During the welding or weld interval, when a welding current is flowing the system is said to be at
A. weld interval
B. cool subinterval
C. heat subinterval
D. hold interval
Solution:
360. The portion of the weld interval during which the current is absent
A. cool subinterval
B. heat subinterval
C. hold interval
D. standby interval
Solution:
361. After the welding interval, it goes to _________ interval wherein the electrode pressure is maintained on the metal surfaces.
A. cool subinterval
B. heat subinterval
C. hold interval
D. standby interval
Solution:
362. Next to hold interval is __________ interval in automatic welding system.
A. squeeze
B. cool
C. standby
D. release
Solution:
363. After the release interval in automatic welding, the system will go to
A. cool interval
B. squeeze interval
C. standby interval
D. hold interval
Solution:
364. Refers to the system that has no feedback and is not self correcting
A. Close-loop system
B. Coal slurry system
C. Feed forward control system
D. Open-loop system
Solution:
365. The system is ____________ if a position servo system does not respond to small changes in the input.
A. under stabilized
B. underdamped
C. stabilized
D. overdamped
Solution:
366. What is the purpose of using a differential synchro instead of a regular synchro?
A. Handles more signals only
B. Performs addition and subtraction function only
C. Differential synchros can handle more signals and also performs addition and subtraction function
D. Handles two signals only
Solution:
367. Industrial circuit or system that is not self-correcting
A. open-loop
B. closed-loop
C. system with feed back
D. non-servo
Solution:
368. What do you call a circuit or system that is self-correcting?
A. open-loop
B. closed-loop
C. system without feed back
D. servo
Solution:
369. Open-loop in control system means:
A. it has no feedback
B. it is not self-correcting
C. it is not self-regulating
D. all are correct
Solution:
370. In control system, closed-loop means:
A. it has feedback
B. it is self-correcting
C. it is self-regulating
D. all are correct
Solution:
371. When a closed-loop system is used to maintain physical position it is referred as
A. gyro system
B. feedback system
C. servo system
D. differential system
Solution:
372. In closed-loop system, what do you call the difference in the measured value and the set value or desired value?
A. error
B. differential voltage
C. potential difference
D. threshold
Solution:
373. Error signal in closed-system is also known as
A. difference signal
B. deviation
C. system deviation
D. all are correct
Solution:
374. In a closed-loop control system, when the error signal is zero the system is at
A. null
B. saturation
C. cut-off
D. halt
Solution:
375. The small error signal or system deviation where the system cannot correct anymore
A. threshold
B. holding
C. offset
D. bias
Solution:
376. A good closed-loop control system has the following characteristics
A. with very small offset signal or voltage
B. quick response
C. highly stable
D. all are correct
Solution:
377. In control system, the manner in which the controller reacts to an error is termed as
A. mode of operation
B. type of operation
C. mode of control
D. reaction style
Solution:
378. What are the general basic modes of control in control system?
A. On-Off
B. Proportional
C. Proportional plus integral
D. Proportional plus derivative
Solution:
379. Mode of control wherein the controller has only two operating states. This mode is also known as bang-bang control.
A. On-Off
B. Proportional
C. Proportional plus integral
D. Proportional plus derivative
Solution:
380. ____________ is a mode of control wherein the controller has a continuous range of possible position, not just two as in bang-bang control.
A. On-Off
B. Proportional
C. Proportional plus integral
D. Proportional plus derivative
Solution:
381. Proportional mode of control wherein the controller is not only considering the magnitude of the error signal but as well as the time that is has persisted.
A. On-Off
B. Proportional
C. Proportional plus integral *
D. Proportional plus derivative
Solution:
382. Proportional mode of control wherein the controller is not only considering the magnitude of the error signal but as well as its rate of change.
A. Proportional
B. Proportional plus integral
C. Proportional plus derivative *
D. Proportional plus integral plus derivative
Solution:
383. What is (are) being considered in Proportional plus Integral plus Derivative (PID) mode of control?
A. error signal magnitude
B. error signal period of occurrence
C. error signal rate of change
D. all are considered *
Solution:
384. If On-Off mode of control is the simplest, what is its opposite or the most complex?
A. Proportional
B. Proportional plus integral
C. Proportional plus derivative
D. Proportional plus integral plus derivative *
Solution:
385. The __________ are two of the most common mechanical configuration of industrial robots.
A. Spherical and pneumatic
B. Articulated arm and cylindrical *
C. Spherical and hydraulic
D. Jointed-arm and electric
Solution:
386. One advantage of hydraulic actuator in industrial robots include _____________.
A. great force capability handling heavy loads *
B. lower operating cost than the other type
C. low initial cost than the other type
D. clean-no oil leaks
Solution:
387. ____________ includes tow of the actuator type used in industrial robots.
A. Pneumatic and Jointed-arm
B. Hydraulic and Pneumatic
C. Electric and Spherical
D. Hydraulic and Cylindrical
Solution:
388. A system in which the precise movement of a large load is controlled by a relatively
weak signal.
A. hydraulic
B. electro
C. synchro
D. servo
Solution:
389. A programmable, multifunction manipulator designed to move materials, parts, tools or specific devices.
A. Industrial robot
B. Android
C. Actuator
D. End effector
Solution:
390. The technology for automations
A. avionics
B. cryogenics
C. cryotronics
D. robotics
Solution:
391. What is(are) the common mechanical configurations for industrial robots?
A. articulated-arm or jointed-arm
B. spherical configuration
C. cylindrical configuration
D. all of these
Solution:
392. The number of axis a robot is free to move is called
A. freedom axis
B. degrees of freedom
C. movement degrees
D. mechanical axis
Solution:
393. Actuators used in industrial robots
A. electric motors
B. fluid motors
C. fluid cylinders
D. all of these are correct
Solution:
394. Which of the actuators that has the greatest force capability?
A. electric
B. hydraulic fluid
C. pneumatic
D. magnetic
Solution:
395. Actuator that requires the highest initial cost:
A. electric
B. hydraulic
C. pneumatic
D. magnetic
Solution:
396. Robot actuator that has the highest operating cost:
A. electric
B. hydraulic
C. pneumatic
D. magnetic
Solution:
397. The most messy robot actuator:
A. electric
B. hydraulic
C. pneumatic
D. magnetic
Solution:
398. Advantages of electric actuators:
A. Lower initial cost than either hydraulic or pneumatic
B. Much lower operating cost than hydraulic
C. Accurate positioning and good velocity control
D. All of these are correct
Solution:
399. Advantages of pneumatic actuators:
A. lower initial and operating cost than hydraulic actuators
B. Clean, no oil leaks
C. Quick response
D. All are correct
Solution:
400. Advantages of hydraulic actuators
A. Great holding strength when stopped
B. Accurate positioning and good velocity control
C. Intrinsically safe in flammable environment such as painting
D. All of these are correct
Solution:
401. Disadvantages of pneumatic actuators:
A. Weak force capability
B. Not so much holding strength when stopped as compared to hydraulic system
C. Accurate positioning and velocity control is impossible
D. All of these are correct
Solution:
402. Disadvantages of electric actuators in industrial robots:
A. Less force capability as compared to hydraulic system
B. Very little holding strength when stopped which causes a heavy load to sag
C. Usually requires mechanical brakes
D. All are correct
Solution:
403.A robot software or program that produces only two-position motion for a given robot axis.
A. positive-stop
B. point-to-point
C. continuous-path
D. hard interrupt
Solution:
404. A robot program that has the ability to move a robot to any position within the range but without specific path.
A. positive-stop program
B. point-to-point program
C. continuous path program
D. compound program
Solution:
405. A robot program that has the ability to move a robot to any position within the range with specific path.
A. positive-stop program
B. point-to-point program
C. continuous path program
D. compound program
Solution:
406. When a robot moves on several axis at the same time, it is to have
A. intrinsic motion
B. extrinsic motion
C. compound motion
D. universal motion
Solution:
407. In robotics, SCARA means:
A. Selective Compliant Articulated Robot Arm
B. Selective Compliant Assembly Robot Arm
C. Selective Computer-Actuated Robot Arm
D. A and B are correct
Solution:
408. SCARA Robots has how many axis of motion?
A. 2
B. 4
C. 6
D. 8
Solution:
409. SCARA Robots are designed for what applications?
A. Machining
B. Welding
C. Assembling
D. Handling heavy loads
Solution:
410. Why is SCARA Robot attractive in industry?
A. because it is relatively cheaper
B. because it can carry very heavy loads
C. because it has unlimited movement
D. all of the above
Solution:
#### Questions and Answers in Industrial Electronics
Following is the list of multiple choice questions in this brand new series:
MCQ in Industrial Electronics Principles and Applications
PART 1: MCQ from Number 1 โ 50 Answer key: PART 1
PART 2: MCQ from Number 51 โ 100 Answer key: PART 2
PART 3: MCQ from Number 101 โ 150 Answer key: PART 3
PART 4: MCQ from Number 151 โ 200 Answer key: PART 4
PART 5: MCQ from Number 201 โ 250 Answer key: PART 5
PART 6: MCQs from Number 251 โ 300 Answer key: PART 6
PART 7: MCQ from Number 301 โ 350 Answer key: included
PART 8: MCQ from Number 351 โ 400 Answer key: included
PART 9: MCQ from Number 401 โ 450 Answer key: included
PART 10: MCQ from Number 451 โ 500 Answer key: included
### Complete List of MCQ in Electronics Engineering per topic
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1 FORM FIVE ADDITIONAL MATHEMATIC NOTE CHAPTER : PROGRESSION Arithmetic Progression T n = + (n ) d S n = n [ + (n )d] = n [ + Tn ] S = T = T = S S Emple : The th term of n A.P. is 86 nd the sum of the first terms is. Find the first term nd the common difference () the sum of the first 0 terms (c) the sum from the th to the 0th term. T = 86 S n = n [ + Tn ] = [ 86] 74 = 86 = + 4d = d = 86 4d = 98 d = 7 () S 0 = 0 [() + 9(7)] = 0[4 ] = 090 (c) Sum from th to 0th term = S 0 S = 090 [4 + 0(7)] = 090 () = 87 Geometric Progression T n = r n- n n ( r ) r ( ) S n = = r r For < r <, sum to infinity S r Emple: Given the 4th term nd the 6th term of G.P. is 4 nd 0 respectively. Find the 8th term given tht ll the terms re positive. r = () r = 0 = () () () r 4 r r = 4 9 r = Since ll the terms re positive, r = = 4 = = 8 T 8 = 8 7 = Emple: Find the lest numer of terms of the G.P 8, 6,,,... such tht the lst term is less thn Find the lst term. = 8, r = 6 8 T n < n n < < (n ) log < log log n > 8 log [Rememer to chnge the sign s log is negtive] n > 0.0 zefry@ss.edu.my
2 n >.0 n =. T = 8 = Emple: Epress ech recurring decimls elow s single frction in its lowest term () = = 0.7 r = = S r () 0... = = 0., r = = S CHAPTER : LINEAR LAW Chrcteristic of The Line of Best Fit. Psses through s mny points s possile.. All the other points re s ner to the line of est fit s possile.. The points which re ove nd elow the line of est fit re equl in numer. To Convert from Non Liner to Liner Form To convert to the form Y = mx + c Emple: Non Liner m c Liner y = y = + y = + log y = log () + log log log y y = + Emple: The digrm shows the line of est fit y plotting y ginst. Find the reltion etween y nd. Solution: m =, pssing through (6, ) 6 = (6) The eqution is y = +. + c c = y, or Emple: The tle elow shows vlues of two vriles nd y, otined from n eperiment. It is known tht y is relted to y the eqution y =. 4 6 y Eplin how stright line cn e otined from the eqution ove. () Plot the grph of log y ginst y using scle of cm to unit on the -is nd cm to 0. unit on the y-is. (c) From your grph, find the vlue of nd. Solution: y = log y = log () + log By plotting log y ginst, stright line is otined. () 4 6 log y zefry@ss.edu.my
3 () 4 6 d = = d 4 = c 4 = 4 + c 4 d 6 6 (c) c = log = 0.48 = log = = 0.7 =. CHAPTER : INTEGRATION n n d c n The re etween the grph nd the -is A = y d The re etween the grph nd the y-is A = dy The volume generted when shded region is rotted through 60 o out the -is V = y d Volume generted when shded region is rotted through 60 o out the y-is V = dy Emple: Find () d 4 6 d d = c = c The Rule of Integrtion. kf ( ) d k f ( ) d g. f ( ) g( ) d f ( ) d g( ) d c c. f ( ) d f ( ) d f ( ) d 4. f ( ) d f ( ) d Emple: Given () () f ( ) d 9 4 f ( ) d [ f ( )] d 4 f ( ) d = 4 = 4 9 = 6, find the vlue of f ( ) d [ f ( )] d = d = + 9 = [ ] + 9 = 9 f ( ) d Are Below Grph. The re elow grph nd ounded y the line =, = nd the -is is zefry@ss.edu.my
4 Are under curve = Are of tringle + d = = 9 + [ ] 9] = 6 unit A = y d. Are etween the grph nd the line y = c, y = d nd the y-is is Volume of Revolution A = d c dy Emple: Given A is the point of intersection etween the curve y = nd the line y =, find the re of the shded region in the digrm elow. Volume generted when shded region is revolved through 60 o out the -is is V = y d Volume generted when shded region is rotted through 60 o out the y-is is y = y = = = 0 ( ) = 0 = 0 or. A(, 6) = 0 ( ) = 0, = 0 or = V = d c dy CHAPTER 4: VECTORS Addition of Two Vectors Tringle Lw AB BC AC () Prllelogrm Lw zefry@ss.edu.my 4
5 AB AD AC Prllel Vectors AB is prllel to PQ if AB k PQ constnt. If AB k BC nd C re colliner., where k is, since B is common point, A, B Vector on Crtesin Plne OAi yj OA y = mgnitude of vector OA Unit vector in the direction of i yj OA y Emple : Given OA = nd OB y. P is point on AB such tht AP : PB = : nd Q is the midpoint of OB. The line OP intersects AQ t the point E. Given OE = k OP nd AE haq, where h nd k re constnts, find OQ nd OP in terms of nd/or y. () Epress OE in terms of (i) k, nd y, (ii) h, nd y. (c) Hence, find the vlue of h nd k. OQ = OB y OP OA AP = = AB ( y ) = ( y) () (i) OE = k OP = k ( y) = k k y (ii) OE OA AE = OA + h AQ = h( y) (c) = ( h) + h y Compre the coefficient of nd y h = k () h k nd, h = k -----() Sustitute in () k k = = 4 k k = 4 h = k = 4 =. Emple : Given OP nd OQ. 4 () (c) Find OP Find the unit vector in the direction of OP. Given OP = m OA n OQ nd A is the point (, 7). Find the vlue of m nd n. OP = ( 4) () Unit vector in the direction of OP = i 4j (c) OP = m OA n OQ m n 4 7 zefry@ss.edu.my
6 m n = () 7m n = () () 0m n = ----() 7m n = (4) 7m = 9 m = 9 7 sustitute in (), 8 7 n = n = 8 7 = 7 CHAPTER : TRIGONOMETRIC FUNCTION Angles In The Four Qudrnts The Three Trigonometric Functions Secnt = sec = cos Cosecnt = cosec = sin Cotngent = cot = tn The Reltion Between Trigonometric Functions Sin = cos (90 o ) Cos = sin (90 o ) Tn = cot (90 o ) Tn = sin cos Grphs of Trigonometric Functions y = sin Amplitude = Numer of periods = The Addition Formule Sin (A B) = sin A cos B cos A sin B Cos (A B) = cos A cos B sin A sin B tn A tn B Tn (A B) = tn Atn B The Doule Angle Formule Sin A = sin A cos A Cos A = cos A = sin A tn A Tn A = tn A CHAPTER 6: PERMUTATION AND COMBINATION. Arrngement of n different ojects without repetition. n P = n! n. Arrngement of r ojects from n ojects n n! Pr ( n r)! Emple : Given the word TABLES. Find the numer of wys to rrnge ll the letters in the word. () The numer of wys of rrnging the 6 letters such tht the first letter is vowel. (c) The numer of wys of rrnging 4 letters out of the 6 letters such tht the lst letter is S. Numer of rrngement = 6 P 6 = 6! = 70 () Numer of wys of rrnging vowel out of = P Numer or wys of rrnging the remining letters = P. (c) Totl rrngement = P P = 40. OR: 4 Totl numer of wys =! = 40 If the lst letter is S, the numer of wys of rrnging letters out of the remining letters = P = 60. OR: 4 Numer of wys = 4 = 60 zefry@ss.edu.my 6
7 . Comintion of r ojects from n ojects n n! is Cr r!( n r)! Emple: The PTA committee of school consists of 8 memers. The memers re elected from 7 prents, 6 techers nd the principl of the school. Find the numer of different committees tht cn e formed if the principl is one of the memer of the committee. () the committee consists of the principl, techers nd 4 prents. (c) the committee consists of t lest techers. numer of committees = C C7 = 76 (elect principl, nd 7 committee memers from = 7 prents nd 6 techers) () numer of committees = 6 7 C C C 700 (c) 4 numer of committees = totl numer of committees numer of committees with no techer numer of committees with techer. = 4 C 6 C 8 C 6 C 8 C = CHAPTER 8: PROBABILITY DISTRIBUTION. Binomil Distriution The proility of getting r success in n trils where p = proility of success nd q = proility of filure P(X = r) = n C r n r r p q. Men, = np. Stndrd devition = npq Emple: In survey of district, it is found tht one in every four fmilies possesses computer. If 6 fmilies re chosen t rndom, find the proility tht t lest 4 fmilies possess computers. () If there re 800 fmilies in the district, clculte the men nd stndrd devition for the numer of fmilies which possess computer. P(X 4) = P(X = 4) + P(X = ) + P(X = 6) = 6 C (0.) 4 (0.7) 6 C (0.) (0.7) (0.) = () Men, = np = = 0. Stndrd devition = npq = =.6. Norml Distriution Z = X where Z = stndrd norml score X = norml score = men = stndrd devition P( Z < ) = P(Z >) P(Z < ) = P(Z > ) P(Z > ) = P(Z > ) P( < Z < ) = P(Z > ) P(Z > ) P( < Z < ) = P(Z > ) P(Z > ) Emple: The volume of pcket drink produced y fctory is norml distriuted with men of 00 ml nd stndrd devition of 8 ml. Determine the proility tht pcket drink chosen t rndom hs volume of more thn 0 ml () etween 490 ml nd 0 ml P(X > 0) = P(Z > ) 8 = P(Z >.) = 0.06 () P(490 < X < 0) = P( < Z < 8 8 = P(. < Z <.) = P(Z >.) P(Z >.) = (0.06) = CHAPTER 9: MOTION ALONG A STRAIGHT LINE. Displcement (S) Positive displcement prticle t the right of O ) zefry@ss.edu.my 7
8 Negtive displcement prticle t the left of O. Return to O gin s = 0 Mimum/minimum displcement ds = 0 dt Distnce trvelled in the nth second = S n S n- Emple: Distnce trvelled in the third second = S S Emple: A prticle moves long stright line nd its displcement, s meter, from fied point O, t seconds fter leving O is given y s = t t. Find the displcement of the prticle fter seconds, () the time t which it returns to O gin. (c) the distnce trvelled in the fourth second. s = t t t =, s = 0 = m () Return to O gin s = 0 t t = 0 t( t) = 0 t = 0 or t = second the prticle returns to O gin when t = s. (c) Distnce trvelled in the 4th second = S 4 S = (8 6) (6 9) = 8 + = m. the distnce trvelled in the 4th second is m.. Velocity (v) Velocity is the rte of chnge of displcement with respect to time. v = ds dt Initil velocity the vlue of v when t = 0 Instntneously t rest/chnge direction of movement v = 0 Moves towrds the right v > 0 Moves towrds the left v < 0 Mimum/minimum velocity dv = 0 dt s = v dt Distnce trvelled in the time intervl t = until t = If the prticle does not stop in the time intervl Distnce = v dt () If the prticle stops in the time t = c seconds where c is in the intervl Distnce = c v dt v dt Emple: A prticle moves long stright line pssing through fied point O. Its velocity, v m s -, t seconds fter pssing through O is given y v = t +. Find the displcement t the time of 4 second. s = v dt = t dt = t + t + c If the prticle psses through O when t = 0,, s = 0 when t = 0. c = 0 s = t + t When t = 4 s, s = 6 + = 8 m Emple: A prticle moves long stright line pssing through fied point O. Its velocity, v m s -, t seconds fter pssing through O is given y v = t + t 6. Find () the initil velocity of the prticle, the time when the prticle is momentrily t rest. initil velocity t = 0 v = 6 m s - () momentrily t rest v = 0 t + t 6 = 0 (t + )(t ) = 0 t = or t = s Since the time cnnot e negtive, t = s.. Accelertion Accelertion is the rte of chnge of velocity with respect to time. = dv dt = d s dt Initil ccelertion when t = 0 Decelertion < 0 Positive ccelertion velocity incresing Negtive ccelertion velocity decresing c zefry@ss.edu.my 8
9 Zero ccelertion uniform velocity. Mimum/minimum velocity = 0 v = dt Emple: A prticle moves long stright line pssing through fied point O. Its velocity, v m s -, t seconds fter leving O is given y v = t 6t 7. Find the initil ccelertion of the prticle () the time when the velocity of the prticle is mimum. = dv = t 6 dt t = 0, = 6 m s - () Mimum velocity, = 0 t 6 = 0 t = s. Emple : A prticle moves long stright line pssing through fied point O with velocity of m s -. Its ccelertion, ms - the time t second fter leving O is given y = t 4. Find the mimum velocity of the prticle. = t 4 v = t 4dt = t 4t + c t = 0, v =, c = v = t 4t + For mimum velocity, = 0 t 4 = 0, t = s. v m = = m s - CHAPTER 0: LINEAR PROGRAMMING Steps. Form the liner inequlities.. Construct the region which stisfies the constrints.. Form the optimum eqution + y = k 4. Find the point in the region which gives the mimum or minimum vlue.. Sustitute the vlue of nd y to otin the optimum vlue of k. The tle ove shows the time tken y tilor to prepre shirt nd shorts of school uniform. In week, the tilor sells shirts nd y shorts. Given tht in week, the numer of shirts nd shorts sold must e t lest 0. The time for preprtion is t most 800 minutes. The rtio of the numer of shorts to the shirts must e t lest :. Write three inequlities other thn 0 nd y 0 which stisfy the constrints ove. () By using scle of cm to 0 units on the - nd y-es, drw the grph of ll the inequlities ove. Hence, shde the region R which stisfies the constrints ove. (c) The tilor mkes profit of RM nd RM in selling shirt nd shorts respectively. Find the mimum profit mde y the tilor in week. the inequlities re: (i) + y 0 (ii) 0 + 0y y 80 y (iii) y () To drw + y = 0, = 0, y = 0 nd y = 0, = 0 To drw + y = 80 = 0, y = 80, y = 40 y = 0, = 80 To drw y = = 0, y = 0 = 40, y = 0. (c) Profit, k = + y Let k = 0 + y =0 = 0, y = 0, y = 0 y = 0, = 0, = 0 Emple: School uniform Time of preprtion (minutes) Shirt 0 Shorts 0 zefry@ss.edu.my 9
10 From the grph, mimum profit is chieved when = 40 nd y = 0. mimum profit = = RM 60. ALL THE BEST FOR YOUR SPM EXAM. zefry@ss.edu.my 0
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More information | 13,404 | 42,159 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2024-18 | latest | en | 0.800482 |
https://de.mathworks.com/matlabcentral/cody/problems/147-too-mean-spirited/solutions/696979 | 1,582,973,810,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875148850.96/warc/CC-MAIN-20200229083813-20200229113813-00393.warc.gz | 340,171,610 | 15,532 | Cody
# Problem 147. Too mean-spirited
Solution 696979
Submitted on 7 Jul 2015 by Nicholas Hortance
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 2 3]; y_correct = [1.5 2.5]; assert(isequal(two_mean(x),y_correct))
y = 1.5000 2.5000
2 Pass
%% x = [10 0 0 0 100]; y_correct = [5 0 0 50]; assert(isequal(two_mean(x),y_correct))
y = 5 0 0 50
3 Pass
%% x = [1 2 4]; y_correct = [1.5 3]; assert(isequal(two_mean(x),y_correct))
y = 1.5000 3.0000 | 219 | 581 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2020-10 | latest | en | 0.597568 |
https://math.stackexchange.com/questions/970853/four-consecutive-heads-from-ten-coin-flips | 1,566,133,921,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027313889.29/warc/CC-MAIN-20190818124516-20190818150516-00059.warc.gz | 540,601,435 | 28,403 | # Four consecutive heads from ten coin flips [duplicate]
If we flip a fair coin $10$ times, what is the probability we get $\ge 4$ consecutive heads?
An approach would be to consider the probability of all consecutive heads, cut off by tails, e.g.,
HHHHTxxxxx
THHHHTxxxx
...
HHHHHTxxxx
THHHHHTxxxx
...
but this is complicated, and also has a problem that some patterns overlap (e.g., HHHHTHHHHT). Is there a simpler way to do this?
## marked as duplicate by David K, Hamou, rogerl, BlackAdder, hardmathOct 13 '14 at 0:37
There are several ways of solving this, but for such a small number recursion may be the easiest.
Let $P_4(n)$ be the probability that you have at least $4$ consecutive heads after $n$ flips. Clearly $P_4(0)=P_4(1)=P_4(2)=P_4(3)=0$ and $P_4(4)=2^{-4}$.
For more flips, either you have achieved $4$ consecutive heads already or you have a string without $4$ consecutive heads followed by THHHH. So for $n\gt 4$: $$P_4(n)=P_4(n-1)+2^{-5}(1-P_4(n-5))$$.
I will leave the calculation to you, but you can check your result at OEIS A$050232$
Total number of possible outcomes is $2^{10} = 1024$. Two sequences of heads with length $4$ or more can be possible only in three following ways: $$HHHHTHHHHT$$ $$HHHHTTHHHH$$ $$THHHHTHHHH$$ $$HHHHHTHHHH$$ $$HHHHTHHHHH$$
Now try to find number of outcomes giving one sequence of $4$ tails in a row, so $n = 4$. Starting from the combination $$HHHH******$$ we reason in the following way: left and right adjacent results must be $T$ (as we currently mention only $n = 4$), the rest do not matter. So for each of $HHHH******$ and $******HHHH$ we have one definite $T$ and $5$ arbitrary values and should exclude $3$ outcomes (to not include abovementioned cases with two sequences of $4$ or more $H$ in a row). If $HHHH$ is not in the beginning or in the end of the line (so, the first $H$ is on the $2^{nd}-6^{th}$ position), then we have $2$ definite $T$ (before and after $HHHH$) and $4$ arbitrary values and should one outcome for $*HHHH*****$ and one for $*****HHHH*$. Hence, total number of suitable outcomes for $n=4$ is $$2 * (2^5 - 3) + 2 * (2^4 - 1) + 3 * 2^4 = 58 + 30 + 48 = 136$$
Similarly, for $n = 5$: $$2 * (2^4 - 1) + 4 * 2^3 = 30 + 32 = 62$$ For $n = 6, 7, 8$: $$2 * 2^{9-n} + (9-n) * 2^{8-n}$$ The total sum of suitable outcomes for $6$, $7$ and $8$ together is $$28 + 12 + 5 = 45$$ If $n=9$ then we have only $2$ possible sequence and $1$ for $n=10$.
Thus, the total sum is $$136+62+45+10+2+1 = 256$$ and the probability of such event is $\frac {256}{1024} = \frac 14$. | 846 | 2,561 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 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-2019-35 | latest | en | 0.886704 |
https://www.javatpoint.com/data-sufficiency-4 | 1,679,363,975,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296943589.10/warc/CC-MAIN-20230321002050-20230321032050-00263.warc.gz | 946,705,171 | 10,705 | # Data Sufficiency 4
16) How many visitors saw the trade fair yesterday?
I. Each entry pass holder is allowed to take up to three persons with him or her.
class="pq"II. In all, 250 entry passes were sold yesterday.
1. Statement I alone is sufficient, but statement II alone is not sufficient
2. Statement II alone is sufficient, but statement I alone is not sufficient
3. Either Statement I or II is sufficient
4. Data in both the statements together are not sufficient
5. Both I and II are needed
Explanation:
As per both the statements, a maximum of 250 * 4 = 1000 people can visit the trade fair, but the exact number cannot be predicted.
17) What is the total sale of the company?
I. The company sold 7000 units of product A, each unit's cost is Rs. 25.
II. The company has no other product line.
1. Statement I alone is sufficient, but statement II alone is not sufficient
2. Statement II alone is sufficient, but statement I alone is not sufficient
3. Either Statement I or II is sufficient
4. Data in both the statements together are not sufficient
5. Both I and II are needed
Explanation:
From statement I and II, it is clear that the company sells only one product and it has sold 7000 units. As each unit costs Rs. 25, the total sale of the company = 7000 * 25 = Rs. 1,75,000.
18) What is the total weight of 10 boxes each has the same weight?
I. One-fourth of the weight of a box is 5 kg.
II. The total weight of the three boxes is 20 kilometers more than the total weight of the two boxes.
1. Statement I alone is sufficient, but statement II alone is not sufficient
2. Statement II alone is sufficient, but statement I alone is not sufficient
3. Either Statement I or II is sufficient
4. Data in both the statements together are not sufficient
5. Both I and II are needed
Explanation:
From statement I, it is clear that weight of one box = 5 * 4 = 20 kg. So, the total weight of 10 boxes is 20 * 10 = 200 kg. Thus, alone I is sufficient.
From statement II, the weight of one box = weight of three boxes - weight of two boxes = 20 kg. So, weight of 10 boxes = 10 * 20 = 200 kg. Thus, alone II is also sufficient.
19) What is the amount of sugar exported from India?
I. India exported to Russia 70,000 tonnes of sugar, and this is 10% of the total sugar exports.
II. India's total export tonnage of sugar is 10% of the total production of 7 million tonnes.
1. Statement I alone is sufficient, but statement II alone is not sufficient
2. Statement II alone is sufficient, but statement I alone is not sufficient
3. Either Statement I or II is sufficient
4. Data in both the statements together are not sufficient
5. Both I and II are needed
Explanation:
From statement I, 10 % = 70000
100% (total exports sugar) = 70000 * 100 / 10 = 700000 tonnes
From statement II, the total export tonnage of sugar is 10% of the total production of 7 million tonnes.
10/100 * 7000000 = 7000000 tonnes
20) How much amount Suresh has to pay for the new car in the buy-back schema?
I. The cost of the new car was four times the resale value his old car.
II. His old car is valued at Rs. 35000 under this schema.
1. Statement I alone is sufficient, but statement II alone is not sufficient
2. Statement II alone is sufficient, but statement I alone is not sufficient
3. Either Statement I or II is sufficient
4. Data in both the statements together are not sufficient
5. Both I and II are needed
Explanation:
From both the statements, the price of the new car is 35000 * 4 = 140000, so he has to pay, 140000 - 35000 = Rs. 105000
Data Sufficiency 1
Data Sufficiency 2
Data Sufficiency 3
Next TopicCoding Decoding 1 | 931 | 3,632 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-14 | latest | en | 0.932248 |
https://docs.microsoft.com/zh-tw/dotnet/csharp/programming-guide/arrays/index?view=netframework-4.8 | 1,569,019,154,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514574084.88/warc/CC-MAIN-20190920221241-20190921003241-00522.warc.gz | 448,369,274 | 11,191 | # 陣列 (C# 程式設計手冊)Arrays (C# Programming Guide)
type[] arrayName;
class TestArraysClass
{
static void Main()
{
// Declare a single-dimensional array.
int[] array1 = new int[5];
// Declare and set array element values.
int[] array2 = new int[] { 1, 3, 5, 7, 9 };
// Alternative syntax.
int[] array3 = { 1, 2, 3, 4, 5, 6 };
// Declare a two dimensional array.
int[,] multiDimensionalArray1 = new int[2, 3];
// Declare and set array element values.
int[,] multiDimensionalArray2 = { { 1, 2, 3 }, { 4, 5, 6 } };
// Declare a jagged array.
int[][] jaggedArray = new int[6][];
// Set the values of the first array in the jagged array structure.
jaggedArray[0] = new int[4] { 1, 2, 3, 4 };
}
}
## 陣列概觀Array Overview
• 陣列可以是一維多維不規則An array can be Single-Dimensional, Multidimensional or Jagged.
• 當陣列執行個體建立時，也會建立維度的數目和每個維度的長度。The number of dimensions and the length of each dimension are established when the array instance is created. 在執行個體的存留期期間，這些值無法變更。These values can't be changed during the lifetime of the instance.
• 數值陣列元素的預設值設定為零，且參考元素會設定為 null。The default values of numeric array elements are set to zero, and reference elements are set to null.
• 不規則陣列為陣列的陣列，因此其元素為參考類型，且會初始化為 nullA jagged array is an array of arrays, and therefore its elements are reference types and are initialized to null.
• 陣列為以零為基底索引：具有 n 個元素的陣列，會從 0n-1 編製索引。Arrays are zero indexed: an array with n elements is indexed from 0 to n-1.
• 陣列元素可以是任何類型，包含陣列類型。Array elements can be of any type, including an array type.
• 陣列類型是衍生自抽象基底類型 Array參考類型Array types are reference types derived from the abstract base type Array. 由於這個類型會實作 IEnumerableIEnumerable<T>，您可以在所有以 C# 撰寫的陣列上使用 foreach 反覆項目。Since this type implements IEnumerable and IEnumerable<T>, you can use foreach iteration on all arrays in C#. | 616 | 1,788 | {"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} | 2.953125 | 3 | CC-MAIN-2019-39 | latest | en | 0.318429 |
https://tensor.ro/ | 1,725,843,985,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651053.52/warc/CC-MAIN-20240909004517-20240909034517-00081.warc.gz | 534,022,398 | 42,791 | BLOG
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# Blog
Linear Elastic Fracture Mechanics with ANSYS APDL ANSYS Parametric Design Language (APDL) is a powerful structured scripting language used to interact with the ANSYS Mechanical solver. Mechanical APDL (MAPDL), a finite element analysis program, is driven by APDL. APDL and MAPDL can be used for many tasks, ranging from creating geometries for analysis to setting up sophisticated solver settings for highly complex analyses. ANSYS...
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Submodeling Technique from Beam to Shell28 Septembrie 2022 Submodeling, as the name suggests, is the modeling of a portion of a global model to capture the effects of refined mesh in a smaller model. This is done to avoid having the whole assembly meshed with fine elements. ANSYS Mechanical allows submodeling for structural (stress) and thermal analyses. In a thermal analysis, the temperatures calculated...
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Structural Optimization Analysis with ANSYS DISCOVERY5 Septembrie 2022 The term structural optimization is commonly used for the optimization of engineering structures for improved strength or stiffness properties and reduced weight or cost.ANSYS Discovery 2022R1 provides topology optimization with the following options: Optimization for Strength or Response to Free Vibration. Optimization for a Specified Frequency. Optimization to Remove Excess Material. ...
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Structural Optimization Analysis with ANSYS Mechanical16 August 2022 The term structural optimization is commonly used for the optimization of engineering structures for improved strength or stiffness properties and reduced weight or cost.ANSYS Mechanical 2022R1 provides the following optimization methods: 1. Topology OptimizationThere are two options implemented in ANSYS Mechanical software for topology optimization: Density Based Optimization: This method performs optimization based on...
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Structural Optimization Analysis16 August 2022 How to get clean CAD Geometries and Optimized Design CheckingThe new topologies that are obtained in a topology optimization simulation cannot be used directly. They must be transformed in CAD geometries. We present below how to reverse engineering the optimized component using any of the four methods available in ANSYS Mechanical. There is one approach for lattice...
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High Frequency Simulation and Verification Course16 August 2022 In the autumn of 2019 we started a High Frequency Simulation and Verification Course using ANSYS products at the Faculty of Electronics, Bucharest. It consisted of 3 hours laboratory sessions taking place every 2 weeks. The course reunited undergraduate, master and even PhD students since participation was optional.
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Cavity resonances in two dimensions1 Septembrie 2020 This article is the final one from a pair of three, each characterizing a different frequency interval in the impedance profile of two parallel power planes. I investigated the capacitive interval in this older article, the inductance in a blog post from the last month available here and the resonances in planes in today’s article.
Detalii
Spreading inductance in wide cavities1 August 2020 In this article I direct my attention toward the study of inductance. The reason why this measure is important to be studied is the same as previously mentioned with regard to capacitance, which I have in depth investigated in a previous article: almost every interconnect can be characterized up to a certain frequency as a lumped RLCG circuit.
Detalii
Parallel plate capacitance approximation and limitations caused by the fringing effect1 Iulie 2020 For today, I directed my attention to studying the parallel plate approximation used to calculate the capacitance between two copper planes separated by a dielectric. This approximation does not take into account the fringe fields at the edges of the copper electrodes, which in some cases can become a large source of error.
Detalii
Frequency-Domain characterization of FR-4 dielectric using Djordjevic-Sarkar algorithm1 Iunie 2020 However, our focus is directed today on something different, the variation of the dielectric constant of a material with frequency. This matter could at first sound strange and in contradiction with its name of "constant". Moreover, if we think of the dielectric constant as the factor that determines the capacitance of a parallel-plate capacitor we obtain a frequency dependent capacitance...
Detalii
Investigation of mounting techniques for SMD capacitors1 Mai 2020 Capacitors, they store charge with the price of voltage between two plates separated by a dielectric. The main parameter that describes them is the capacitance and that is all we would like to find inside a capacitor, but how nothing in perfect in life, a small series (or parallel, depends of what model you choose to use) inductance and resistance...
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## Contactaţi-ne
Tel/Fax: 021 444 23 78
Mobil: 0744 310 929
office[at]tensor.ro
© Copyright 2017 Tensor SRL. Toate drepturile rezervate. Design și programare Vectorpixel | 1,026 | 5,015 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2024-38 | latest | en | 0.860783 |
http://mathhelpforum.com/trigonometry/161757-proving-identity.html | 1,498,703,446,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128323842.29/warc/CC-MAIN-20170629015021-20170629035021-00050.warc.gz | 250,803,619 | 9,724 | 1. ## Proving identity
Prove that $1 - \frac{\cos{x}}{\sec{x}} = \sin^2x$
So I try to prove this by starting from the left hand side of the equation:
$LHS = 1 - \frac{\cos{x}}{\sec{x}}$
$= \sin^2{x} + \cos^2{x} - \frac{\frac{\cos{x}}{1}}{\cos{x}}$
$= \sin^2{x} + \cos^2{x} - 1$
$= \sin^2{x} + \cos^2{x} - (\sin^2{x} + \cos^2{x})$
$= 0$
But this is not $= \sin^2{x}$
Any pointers? Thank you in advance!
2. This line:
$= \sin^2{x} + \cos^2{x} - \dfrac{\cos(x)}{\frac{1}{\cos(x)}}$
cannot become this line:
$= \sin^2{x} + \cos^2{x} - 1$
This:
$= \sin^2{x} + \cos^2{x} - \dfrac{\cos(x)}{\frac{1}{\cos(x)}}$
Becomes this:
$= \sin^2{x} + \cos^2{x} - \left(cos(x)\right)\left(cos(x)\right)$
3. OK, now I can easily solve the problem. Thank you!
4. No problem! | 324 | 765 | {"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} | 4.03125 | 4 | CC-MAIN-2017-26 | longest | en | 0.403895 |
https://timenewsglobal.com/business/get-most-out-of-12-ml-to-oz/ | 1,721,609,826,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517805.92/warc/CC-MAIN-20240722003438-20240722033438-00170.warc.gz | 497,458,343 | 52,686 | Breaking News
# Get most out of 12 ml to oz
Are 12 ml to oz you someone who loves to experiment in the kitchen, but sometimes find yourself puzzled by different measurement units like milliliters and ounces? Well, fret not! In this blog post, we’re going to unravel the mystery around converting 12 ml to oz and explore creative ways to make the most out of this liquid gold. So grab your measuring cups and let’s dive into the world of precise measurements and culinary creativity!
## Understanding Measurements: Milliliters vs Ounces
When it comes to cooking and baking, precision is key. Understanding the difference between milliliters (ml) and ounces (oz) can make a world of difference in your culinary creations. Milliliters are commonly used for measuring liquid volumes in the metric system, while ounces are often used in the imperial system.
While milliliters provide a more accurate measurement for liquids, ounces are typically used for both liquids and dry ingredients. It’s important to know how to convert between these two units to ensure your recipes turn out just right.
Whether you’re whipping up a delicious sauce or crafting the perfect cocktail, knowing how to navigate between milliliters and ounces will elevate your kitchen skills and take your dishes to the next level. So let’s dive into the world of precise measurements and unlock endless culinary possibilities!
## How to Convert Milliliters to Ounces
Understanding how to convert milliliters to ounces can be a handy skill, especially when working in the kitchen or dealing with liquid measurements. The conversion between these two units is crucial for precise recipe execution and accurate ingredient proportions.
To convert milliliters to ounces, you can use a simple formula: 1 milliliter is equivalent to approximately 0.0338 fluid ounces. Therefore, if you have 12 milliliters of liquid that needs converting, you would multiply it by 0.0338 to get the equivalent amount in ounces.
For example, if you want to convert 12 ml to oz:
12 ml * 0.0338 = approximately 0.4064 oz
By understanding this conversion factor and applying it correctly, you can easily switch between milliliters and ounces without confusion or miscalculations during your culinary adventures.
## Common Conversions for 12 ml to oz
When it comes to converting milliliters to ounces, understanding the relationship between these two units of measurement is key. In the culinary world, where precision matters, knowing how to convert 12 ml to oz can make a significant difference in your recipes.
For those unfamiliar with the conversion factor, 1 milliliter is equivalent to approximately 0.033814 fluid ounces. So, when you have 12 ml of liquid and want to express it in ounces, you’d simply multiply by this conversion factor.
In practical terms, if you have 12 ml of an ingredient like vanilla extract or lemon juice that needs to be converted into ounces for a recipe, knowing that it’s around 0.41 oz helps ensure accuracy in your cooking or baking endeavors.
Being able to easily perform common conversions like 12 ml to oz opens up a world of possibilities in the kitchen – allowing you to confidently experiment with new recipes and ingredients without fear of miscalculations throwing off your creations.
## Creative Ways to Use 12 ml of Liquid
Have you ever wondered what creative possibilities lie within a mere 12 ml of liquid? While it may seem like a small amount, the potential for innovation is endless. From crafting miniature potions to adding a touch of flavor to your favorite recipes, 12 ml opens up a world of opportunities.
For those with an artistic flair, why not use 12 ml of colored dye to create vibrant watercolor paintings or experiment with marbling techniques on paper? The tiny amount can go a long way in producing stunning visual effects.
In the realm of skincare and beauty, mix 12 ml of essential oils with carrier oils for a personalized perfume blend or DIY moisturizing serum. This small quantity allows for precise customization tailored to your preferences.
When it comes to culinary adventures, infuse 12 ml of herbs and spices into dressings or marinades for an extra burst of flavor without overpowering the dish. Alternatively, explore the art of cocktail-making by measuring out 12 ml portions for perfectly balanced drinks that pack a punch.
Whether you’re exploring new hobbies or enhancing everyday tasks, embracing the versatility of 12 ml can lead to unexpected discoveries and delightful experiences.
## Importance of Accurate Measurement in Cooking and Baking
Accurate measurement plays a crucial role in the success of any culinary endeavor. In cooking and baking, precision is key to achieving consistent results every time you step into the kitchen. Whether you are following a recipe or experimenting with your own creations, knowing exactly how much of each ingredient to use can make or break the dish.
Imagine adding too much salt or too little flour – it could completely alter the taste and texture of your final product. That’s why measuring ingredients accurately is not just a suggestion but a necessity for anyone serious about their culinary skills.
From delicate pastries to savory stews, precise measurements ensure that flavors are balanced and textures are just right. So next time you’re in the kitchen, remember that accuracy in measuring ingredients is the secret ingredient to mastering the art of cooking and baking like a pro!
## Tips for Measuring Liquid Ingredients Precisely
When it comes to measuring liquid ingredients accurately, precision is key. Start by using a clear measuring cup with easy-to-read markings. Hold the cup at eye level to ensure you’re reading the measurement correctly.
Always pour liquids slowly into the measuring cup on a flat surface for an accurate read. If you accidentally pour too much, use a spoon to remove excess until you reach the desired amount.
Remember that different liquids have varying thicknesses, affecting how they fill up a measuring cup. For thicker liquids like honey or syrup, consider lightly greasing the measuring cup before pouring in the liquid for easier removal and precise measurement.
To avoid spills and messes, place a paper towel under your measuring cups when working with messy or sticky liquids like oils or syrups.
By following these simple tips and tricks, you’ll be able to measure liquid ingredients like a pro in no time!
## Conclusion
Understanding how to convert measurements between milliliters and ounces can significantly enhance your cooking and baking skills. By utilizing 12 ml effectively in recipes, you can elevate the flavor profile of your dishes while ensuring accuracy in measurements. Remember, precision is key when it comes to liquid ingredients, so mastering the art of measuring can truly make a difference in the outcome of your culinary creations. So go ahead, get creative with 12 ml of liquid and see the delicious results for yourself! | 1,369 | 7,002 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-30 | latest | en | 0.89987 |
http://www.chegg.com/homework-help/questions-and-answers/insert-set-n-items-binary-search-tree-using-tree-insert-resulting-tree-may-horribly-unbala-q1098025 | 1,455,342,295,000,000,000 | text/html | crawl-data/CC-MAIN-2016-07/segments/1454701166141.55/warc/CC-MAIN-20160205193926-00347-ip-10-236-182-209.ec2.internal.warc.gz | 333,795,552 | 13,711 | If we insert a set of n items into a binary search tree using TREE-INSERT, the resulting tree
may be horribly unbalanced. As we saw in class, however, we could expect randomly built
binary search trees to be balanced. Therefore, if we want to build an expected balanced tree
for a fixed set of items, we could randomly permute the items and then insert them in that
order into the tree.
What if we do not have all the items at once? If we receive the items one at a time, can we
still randomly build a balanced binary search tree out of them?
Each item x has a key key[x]. In addition, we assign priority[x], which is a random number
chosen independently for each x. We assume that all priorities are distinct. The nodes of the
underpinning binary tree are ordered so that (1) the keys obey the binary-search-tree property
and (2) the priorities obey the min-heap order property. In other words,
• if v is a left child of u, then key[v] < key[u];
• if v is a right child of u, then key[v] > key[u]; and
• if v is a child of u, then priority(v) > priority(u).
Input: We assume that you are given the initial set of items {A(10), B(7), E(23), G(4),
H(5), K(65), I(73)}, with the randomly assigned priorities to the elements given in
parentheses
• The tree after inserting a node with key C and priority 25.
• The tree after inserting a node with key D and priority 9.
• The tree after inserting a node with key F and priority 2. | 362 | 1,428 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2016-07 | latest | en | 0.889151 |
https://greencoin.life/how-to/measure/2-fluid-ounces/ | 1,638,816,371,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363309.86/warc/CC-MAIN-20211206163944-20211206193944-00265.warc.gz | 361,296,740 | 7,403 | ## Fluid Ounces vs. Weight Ounces
#### Sharing buttons:
today we're going to look at what fluid
ounces really means compare it to weight
bounces
we're starting with four different kinds
of fluids and we have 8 fluid ounces of
each of them we have water milk oil and
juice you'll notice also that we've used
different size of measuring cups so each
measuring cup has a different capacity
but they also have the exact same amount
except for this little tiny one here
this one only has capacity of one ounce
so we need to fill it 8 times in order
to have the same capacity as this cup
that has a 2 cup capacity this cup that
has a 4 cup capacity which is one quart
and that one that has a 2 cup capacity a
mm-hmm
which is also a total of 16 ounces a day
damage yeah ok first we're going to
weight as we pour the water in 8 ounces
of water is the same as 8 ounces of
weight pour out the water Oh next we're
going to look at the milk
oh now milk is a little thicker than
water cuz it's a lot of damage because
there's fat in it and other things and
here we have more than 8 ounces in
weight but it was the same amount weight
and fluid ounces wait that's illegal
next we're going to try juice mm-hmm
and I need to put my 8 ounce in here so
I have a total of 8 ounces of juice
booger juice
you are putting our 8 ounces of juice
let's see that's also 8 ounces in weight
yikes it's a little bit more to pour in
the juice out wait how does that mean I
just that weighed more than the milk
that is completion maybe cuz there's
sugar in juice laughs and now we're
doing our 8 ounces of fluid ounces of
oil and there are 10 wait what and now
there's less than 8 ounces of wait wait
if you've ever seen oil float on water
this might explain oil is less dense
than water so there's less stuff packed
in there when it has the same fluid
ounces
you | 475 | 1,882 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2021-49 | latest | en | 0.980221 |
https://web2.0calc.com/questions/trig_6127 | 1,553,599,341,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912204969.39/warc/CC-MAIN-20190326095131-20190326121131-00154.warc.gz | 666,979,283 | 5,504 | +0
Trig
0
238
1
sin(7x) = sqrt(-3)/2
Jun 1, 2017
#1
+98196
+1
sin (7x) = -sqrt (3) / 2
Note
sin (7x ) = -sqrt (3) / 2 when either
7x = 4 pi / 3 + 2pi*n or 7x = 5 pi/ 3 + 2pi *n where n is an integer
Dividing both by 7, then
x = 4 pi / 21 + (2pi/7)n or x = 5 pi/ 21 + (2p/7)n where n is an integer
Jun 1, 2017 | 204 | 367 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2019-13 | latest | en | 0.191041 |
https://edurev.in/t/159930/Square-Roots-and-Cube-Roots | 1,708,658,683,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474360.86/warc/CC-MAIN-20240223021632-20240223051632-00021.warc.gz | 241,197,086 | 62,791 | Square Roots and Cube Roots
# Square Roots and Cube Roots | Quantitative Aptitude (Quant) - CAT PDF Download
Square Roots and Cube Roots
To find the square root of a number, you want to find some number that when multiplied by itself gives you the original number. In other words, to find the square root of 25, you want to find the number that when multiplied by itself gives you 25. The square root of 25, then, is 5. The symbol for square root is √0. Following is a list of the first eleven perfect (whole number) square roots.
Special note: If no sign (or a positive sign) is placed in front of the square root, then the positive answer is required. Only if a negative sign is in front of the square root is the negative answer required. This notation is used in many texts and is adhered to in this book. Therefore,
Cube roots
To find the cube root of a number, you want to find some number that when multiplied by itself twice gives you the original number. In other words, to find the cube root of 8, you want to find the number that when multiplied by itself twice gives you 8. The cube root of 8, then, is 2, because 2 × 2 × 2 = 8. Notice that the symbol for cube root is the radical sign with a small three (called the index) above and to the left . Other roots are defined similarly and identified by the index given. (In square root, an index of two is understood and usually not written.) Following is a list of the first eleven perfect (whole number) cube roots.
Approximating square roots
To find the square root of a number that is not a perfect square, it will be necessary to find an approximate answer by using the procedure given in Example.
Example 1
Approximate √42.
Since 6= 36 and 72 = 49, then √42 is between √36 and √49.
Therefore, √42 is a value between 6 and 7. Since 42 is about halfway between 36 and 49, you can expect that √42 will be close to halfway between 6 and 7, or about 6.5. To check this estimation, 6.5 × 6.5 = 42.25, or about 42.
Square roots of nonperfect squares can be approximated, looked up in tables, or found by using a calculator. You may want to keep these two in mind:
Simplifying square roots
Sometimes you will have to simplify square roots, or write them in simplest form. In fractions, 2/4 can be reduced to 1/2. In square roots, √32 can be simplified to .
There are two main methods to simplify a square root.
Method 1: Factor the number under the into two factors, one of which is the largest possible perfect square. (Perfect squares are 1, 4, 9, 16, 25, 36, 49, …)
Method 2: Completely factor the number under the into prime factors and then simplify by bringing out any factors that came in pairs.
Example 2
Simplify √32.
In Example , the largest perfect square is easy to see, and Method 1 probably is a faster method.
Example 3
Simplify √2016.
In Example , it is not so obvious that the largest perfect square is 144, so Method 2 is probably the faster method.
Many square roots cannot be simplified because they are already in simplest form, such as √7, √10, and √15.
Square, Cube, Square Root and Cubic Root for Numbers Ranging 0 - 100
Numberx Squarex2 Cubex3 Square Root x1/2 Cubic Root x1/3 1 1 1 1 1 2 4 8 1.414 1.26 3 9 27 1.732 1.442 4 16 64 2 1.587 5 25 125 2.236 1.71 6 36 216 2.449 1.817 7 49 343 2.646 1.913 8 64 512 2.828 2 9 81 729 3 2.08 10 100 1000 3.162 2.154 11 121 1331 3.317 2.224 12 144 1728 3.464 2.289 13 169 2197 3.606 2.351 14 196 2744 3.742 2.41 15 225 3375 3.873 2.466 16 256 4096 4 2.52 17 289 4913 4.123 2.571 18 324 5832 4.243 2.621 19 361 6859 4.359 2.668 20 400 8000 4.472 2.714 21 441 9261 4.583 2.759 22 484 10648 4.69 2.802 23 529 12167 4.796 2.844 24 576 13824 4.899 2.884 25 625 15625 5 2.924 26 676 17576 5.099 2.962 27 729 19683 5.196 3 28 784 21952 5.292 3.037 29 841 24389 5.385 3.072 30 900 27000 5.477 3.107 31 961 29791 5.568 3.141 32 1024 32768 5.657 3.175 33 1089 35937 5.745 3.208 34 1156 39304 5.831 3.24 35 1225 42875 5.916 3.271 36 1296 46656 6 3.302 37 1369 50653 6.083 3.332 38 1444 54872 6.164 3.362 39 1521 59319 6.245 3.391 40 1600 64000 6.325 3.42 41 1681 68921 6.403 3.448 42 1764 74088 6.481 3.476 43 1849 79507 6.557 3.503 44 1936 85184 6.633 3.53 45 2025 91125 6.708 3.557 46 2116 97336 6.782 3.583 47 2209 103823 6.856 3.609 48 2304 110592 6.928 3.634 49 2401 117649 7 3.659 50 2500 125000 7.071 3.684 51 2601 132651 7.141 3.708 52 2704 140608 7.211 3.733 53 2809 148877 7.28 3.756 54 2916 157464 7.348 3.78 55 3025 166375 7.416 3.803 56 3136 175616 7.483 3.826 57 3249 185193 7.55 3.849 58 3364 195112 7.616 3.871 59 3481 205379 7.681 3.893 60 3600 216000 7.746 3.915 61 3721 226981 7.81 3.936 62 3844 238328 7.874 3.958 63 3969 250047 7.937 3.979 64 4096 262144 8 4 65 4225 274625 8.062 4.021 66 4356 287496 8.124 4.041 67 4489 300763 8.185 4.062 68 4624 314432 8.246 4.082 69 4761 328509 8.307 4.102 70 4900 343000 8.367 4.121 71 5041 357911 8.426 4.141 72 5184 373248 8.485 4.16 73 5329 389017 8.544 4.179 74 5476 405224 8.602 4.198 75 5625 421875 8.66 4.217 76 5776 438976 8.718 4.236 77 5929 456533 8.775 4.254 78 6084 474552 8.832 4.273 79 6241 493039 8.888 4.291 80 6400 512000 8.944 4.309 81 6561 531441 9 4.327 82 6724 551368 9.055 4.344 83 6889 571787 9.11 4.362 84 7056 592704 9.165 4.38 85 7225 614125 9.22 4.397 86 7396 636056 9.274 4.414 87 7569 658503 9.327 4.431 88 7744 681472 9.381 4.448 89 7921 704969 9.434 4.465 90 8100 729000 9.487 4.481 91 8281 753571 9.539 4.498 92 8464 778688 9.592 4.514 93 8649 804357 9.644 4.531 94 8836 830584 9.695 4.547 95 9025 857375 9.747 4.563 96 9216 884736 9.798 4.579 97 9409 912673 9.849 4.595 98 9604 941192 9.899 4.61 99 9801 970299 9.95 4.626 100 10000 1000000 10 4.642
The document Square Roots and Cube Roots | Quantitative Aptitude (Quant) - CAT is a part of the CAT Course Quantitative Aptitude (Quant).
All you need of CAT at this link: CAT
## Quantitative Aptitude (Quant)
185 videos|158 docs|113 tests
## FAQs on Square Roots and Cube Roots - Quantitative Aptitude (Quant) - CAT
1. What are square roots and cube roots?
Ans. Square roots and cube roots are mathematical operations used to find the original number that was squared or cubed. The square root of a number is the value that, when multiplied by itself, gives the original number. Similarly, the cube root of a number is the value that, when multiplied by itself twice, gives the original number.
2. How can I find the square root of a number?
Ans. To find the square root of a number, you can use a calculator or use manual methods such as the prime factorization method or the long division method. The prime factorization method involves breaking down the number into its prime factors and then taking the square root of each factor. The long division method involves dividing the number into smaller parts and iteratively finding the square root.
3. How can I find the cube root of a number?
Ans. Similar to finding the square root, you can use a calculator or manual methods to find the cube root of a number. One common method is the estimation method, where you make an initial guess and refine it using an iterative process. Another method is using prime factorization, where you break down the number into its prime factors and then find the cube root of each factor.
4. Are square roots and cube roots always positive?
Ans. Square roots and cube roots can be both positive and negative. When we talk about square roots, we usually refer to positive square roots. However, every positive number has two square roots - one positive and one negative. Similarly, every number has one real cube root, which can be positive or negative.
5. Can I find the square root or cube root of a negative number?
Ans. Yes, you can find the square root or cube root of a negative number. When dealing with negative numbers, the square root will always be a complex number, involving the imaginary unit "i." For example, the square root of -9 is 3i. Similarly, the cube root of a negative number will also involve complex numbers.
## Quantitative Aptitude (Quant)
185 videos|158 docs|113 tests
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; | 2,963 | 8,515 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.75 | 5 | CC-MAIN-2024-10 | latest | en | 0.931801 |
https://questioncove.com/updates/4f164a28e4b0d30b2d575aaa | 1,656,242,752,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103205617.12/warc/CC-MAIN-20220626101442-20220626131442-00664.warc.gz | 520,569,421 | 4,954 | Mathematics
OpenStudy (anonymous):
how to simplify |2v-4|+2=8?
OpenStudy (anonymous):
2v-4+2=8 2v=10 v=5 or -2v+4+2=8 -2v=2 v=-1 | 62 | 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} | 3.703125 | 4 | CC-MAIN-2022-27 | latest | en | 0.685671 |
https://numberworld.info/1100022 | 1,618,568,078,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038056325.1/warc/CC-MAIN-20210416100222-20210416130222-00548.warc.gz | 550,496,732 | 4,024 | # Number 1100022
### Properties of number 1100022
Cross Sum:
Factorization:
2 * 3 * 7 * 11 * 2381
Count of divisors:
Sum of divisors:
Prime number?
No
Fibonacci number?
No
Bell Number?
No
Catalan Number?
No
Base 2 (Binary):
Base 3 (Ternary):
Base 4 (Quaternary):
Base 5 (Quintal):
Base 8 (Octal):
10c8f6
Base 32:
11i7m
sin(1100022)
-0.37506710117749
cos(1100022)
0.9269976642982
tan(1100022)
-0.40460414909615
ln(1100022)
13.910840737569
lg(1100022)
6.041401370961
sqrt(1100022)
1048.8193362062
Square(1100022)
### Number Look Up
Look Up
1100022 which is pronounced (one million one hundred thousand twenty-two) is a amazing number. The cross sum of 1100022 is 6. If you factorisate 1100022 you will get these result 2 * 3 * 7 * 11 * 2381. The figure 1100022 has 32 divisors ( 1, 2, 3, 6, 7, 11, 14, 21, 22, 33, 42, 66, 77, 154, 231, 462, 2381, 4762, 7143, 14286, 16667, 26191, 33334, 50001, 52382, 78573, 100002, 157146, 183337, 366674, 550011, 1100022 ) whith a sum of 2744064. The number 1100022 is not a prime number. The figure 1100022 is not a fibonacci number. The figure 1100022 is not a Bell Number. 1100022 is not a Catalan Number. The convertion of 1100022 to base 2 (Binary) is 100001100100011110110. The convertion of 1100022 to base 3 (Ternary) is 2001212221120. The convertion of 1100022 to base 4 (Quaternary) is 10030203312. The convertion of 1100022 to base 5 (Quintal) is 240200042. The convertion of 1100022 to base 8 (Octal) is 4144366. The convertion of 1100022 to base 16 (Hexadecimal) is 10c8f6. The convertion of 1100022 to base 32 is 11i7m. The sine of the number 1100022 is -0.37506710117749. The cosine of the number 1100022 is 0.9269976642982. The tangent of 1100022 is -0.40460414909615. The root of 1100022 is 1048.8193362062.
If you square 1100022 you will get the following result 1210048400484. The natural logarithm of 1100022 is 13.910840737569 and the decimal logarithm is 6.041401370961. I hope that you now know that 1100022 is unique figure! | 765 | 1,986 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-17 | latest | en | 0.670051 |
https://nb.khanacademy.org/math/ap-calculus-bc/bc-series-new/bc-10-1/v/partial-sum-notation | 1,542,478,827,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039743717.31/warc/CC-MAIN-20181117164722-20181117185841-00037.warc.gz | 707,188,770 | 44,050 | # Partial sums intro
## Video transcript
- [Voiceover] Let's say that you have an infinite series, S, which is equal to the sum from n equals one, let me write that a little bit neater. n equals one to infinity of a sub n. This is all a little bit of review. We would say, well this is the same thing as a sub one, plus a sub two, plus a sub three, and we would just keep going on and on and on forever. Now what I want to introduce to you is the idea of a partial sum. This right over here is an infinite series. But we can define a partial sum, so if we say S sub six, this notation says, okay, if S is an infinite series, S sub six is the partial sum of the first six terms. So in this case, this is going to be we're not going to just keep going on forever, this is going to be a sub one, plus a sub two, plus a sub three, plus a sub four, plus a sub five, plus a sub six. And I can make this a little bit more tangible if you like. So let's say that S, the infinite series S, is equal to the sum from n equals one, to infinity of one over n squared. In this case it would be one over one squared, plus one over two squared, plus one over three squared, and we would just keep going on and on and on forever. But what would S sub -- I should do that in that same color. What would S -- I said I would change color, and I didn't. What would S sub three be equal to? The partial sum of the first three terms, and I encourage you to pause the video and try to work through it on your own. Well, it's just going to be the first term one, plus the second term, 1/4, plus the third term, 1/9, is going to be the sum of the first three terms, and we can figure that out, that's to see if you have a common denominator here, it's going to be 36. It's going to be 36/36, plus 9/36, plus 4/36, so this is going to be 49/36. 49/36. So the whole point of this video, is just to appreciate this idea of a partial sum. And what we'll see is, that you can actually express what a partial sum might be algebraically. So for example, for example, let's give ourselves a little bit more real estate here. Let's say, let's go back to just saying we have an infinite series, S, that is equal to the sum from n equals one to infinity of a sub n. And let's say we know the partial sum, S sub n, so the sum of the first n terms of this is equal to n squared minus three over n to the third plus four. So just as a bit of a reminder of what this is saying. S sub n... S sub n is the same thing as a sub one, plus a sub two, plus you keep going all the way to a sub n, and that's going to be equal to this business, n squared minus three over n to the third plus four. Now, given that, if someone were to walk up to you on the street and say, okay now that you know the notation for a partial sum, I have a little question to ask of you. If S is the infinite series, and I'm writing it in very general terms right over here, so S is the infinite series from n equals one to infinity of a sub n, and the partial sum, S sub n, is defined this way, so someone, they tell you these two things, and then they say find what the sum from n equals one to six of a sub n is, and I encourage you to pause the video and try to figure it out. Well, this is just going to be a sub one, plus a sub two, plus a sub three, plus a sub four, and when I say sub that just means subscript, plus a sub five, plus a sub six, well that's just the same thing as the partial sum, this is just the same thing as the partial sum of the first six terms for our infinite series. It's just going to be the partial sum S sub six. And we know how to algebraically evaluate what S sub six is. We can apply this formula that we were given. S sub six is equal to, well, everywhere we see an n, we replaced with a six, it's going to be six squared minus three over six to the third plus four, so what is this going to be? Six squared is 36 minus three, so that's 33, and six to the third, let's see, 36 times six, I always forget, my brain wants to say 216, but let me make sure that that's actually the case. Six times 30 is 180, plus 36, yes, it is 216, so I guess I have, inadvertently, by seeing six to the third so many times in my life, I have inadvertently memorized six to the third power, never a horrible thing to have that in your brain. So this is going to be 216 plus four, so 220. So, S sub six, or the sum of the first six terms of the series right over here, is 33/220, and we're done. And the whole point of this is just so you kind of appreciate, or really do appreciate this partial sum notation, and understand what it really means. | 1,180 | 4,597 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-47 | longest | en | 0.965854 |
https://homeschoolden.com/2013/12/03/geometry-review-pack/ | 1,726,707,275,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651944.55/warc/CC-MAIN-20240918233405-20240919023405-00357.warc.gz | 269,360,433 | 34,180 | # Geometry Review Pack
Last fall (a year ago), LD did a unit on geometry. He needed some basic review of some of the things he learned — so I put together a small review pack for him. I’ll share it here for anyone who might find it useful.
There are 6 pages, plus the answers.
Geometry Review Worksheets
He used the Geometry, Grades 6 – 8 (Skills for Success) (affiliate link) by Carson-Delosa. It worked really well for him, introducing polygons, quadrilaterals, the types of triangles, terms like complementary, supplementary, etc. I like the way it has a explanation box at the top of the page and about a dozen practice problems underneath. We read over the new terminology together and then he did the problems on his own. He used it in 5th grade.
Update 2018: I have purchased this same book for ED this year, but I had to buy the book used. It looks like it isn’t in print any more. 🙁
From several years ago: DD (now in 5th grade) is using this book along with her other math textbook. Again, this book (above) is working really well for her as well as an introduction to geometry. I’m really happy with it. She is currently finding the measures of triangles and has just learned about the Pythagorean theorem. Later in the book it goes on to area, volume, etc. This is a good basic introduction to geometry (though it is not suitable as a high-school level geometry book). It’s perfect for 5th-7th grade.
Last week I talked about how we’re trying to make math time more challenging, engaging and meaningful. I explained how we’ve added in Math Circle time (math questions, challenges and riddles–read more in this post-Math Should Never Be Boring: More Math Riddles-and the wonderful book we’re using for that!). But I also mentioned that we’re trying out a couple of other books as well.
LD has also started using Challenge Math. I love the sentence on the front cover: “Math is often taught as all scales and no music. This book contains all the music!” Since LD only started the first chapter I’m not ready to do much a review, but so far LD really likes it a lot — and has figured out the circumference of the Earth and how long it takes light to reach earth from the sun! It’s a neat book so far (for upper elementary/middle school). I’ll try to add in another review in another month or so, once he’s worked through more of the book, but so far so good.
Note: We used this for a while, but then stopped using it.
Related Posts:
Free Equivalent Fraction Packet
Disclosure: Please note that some of the links above are affiliate links, and at no additional cost to you, I will earn a commission if you decide to make a purchase.
### 4 Responses
1. January 18, 2014
[…] Ancient Egyptian Gods and Goddesses; Islam; Hinduism; Gandhi; the Transatlantic Trade; Geometry Rocks and Minerals and Simple Machines… and I could go on and on! I didn’t do this […]
2. January 18, 2014
[…] Ancient Egyptian Gods and Goddesses; Islam; Hinduism; Gandhi; the Transatlantic Trade; Geometry Rocks and Minerals and Simple Machines… and I could go on and on! I didn’t do this […]
3. September 19, 2014
[…] Geometry Review Pack […]
4. September 19, 2014
[…] Geometry Review Pack […] | 774 | 3,208 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-38 | latest | en | 0.966785 |
risajef.ch | 1,508,581,692,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187824733.32/warc/CC-MAIN-20171021095939-20171021115631-00053.warc.gz | 294,240,551 | 12,944 | Are we only Simulations?
I often asked myself this question. I realized that all well known phenomenons in nature behave according to some very mathematical laws. Since I study computer science I had the chance to get a look into the mathematical world. I also realized that mathematics is strongly coupled with computer science because in computers we could define some laws that it has to follow and afterward it does exactly what we said. So you could transform an equation into a program and vice versa. So theoretically when we would have a perfect mathematical and complete description of the world we could possibly simulate some parts of it. (Nearly all the computer simulation done today are very coarse and not complete. They mostly simplify the problem such that it runs efficiently on a computer.)
But for me is clear that a program and the physics of the real world are coupled. At least you can’t say wich world can describe more information. Both can “simulate” the other. Something you might think is, that a simulation without real physics is not possible. This is true but it is possible to have a physics only inside a simulation independent of the real physics. And inside this physics you can say there is no physic without the computer. Ok enough from that. Why should we be a simulation? As I said everything well-known behaves very mathematically like a computer program. Another argument is that we don’t know any infinite thing. Maybe the universe is infinite but maybe not. We don’t know. So in a computer program there is also nothing infinite. There are cycles but nothing infinite that doesn’t repeat itself. If we assume that the simulation we run on is made on a computer similar to ours, than there is a smallest length of space and time. This length can not be divided further. This length is proposed be Max Planck (Planck Length) but it is way too small to detect it with our instruments today. Let’s assume there is the Planck Length wich can not be divided. This is similar to a number on a computer. Say you have 3 decimal places. So 0.01 is the smallest unit. There is no 0.005 in this computers world. (finite precision floating point) If our world is simulated on a computer working with finite memory than if we fly out from the earth to the border of the universe we either come to the same place again (overflow of the variable that defines the position), we get stuck in some state (state that decodes that the position coordinates are too big to store) or we crash the simulation (we use all the memory for our positional coordinates). Sadly there is no way to compute the distance to go until one of this effects need to occur, we would have to know the size of the memory and how our world is represented. On the other hand if we get one of these effects we could infer pretty straight forward how our world is implemented. Maybe some think that the aliens simulating us don’t use a computer like we have. That is possible I only describe what observable phenomenons could be if they do. But let’s be clear if they use a computer with finite precision and finite memory than such effects have to occur. And also the fact that we haven’t found any infinite thing suggests that the the simulation we run in has also no infinite structure.
Some people say they are surely not a simulation because they are conscious. Consciousness is the last stand for many of the more esoteric or religious questions even though, few religions are talking about it. But this topic is for another blog.
But what is the answer? I think it is obvious that this question is not easily answered and I only wanted to give some arguments. (The title was only to bait 😉 ) | 751 | 3,696 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2017-43 | longest | en | 0.962438 |
https://urbizedge.com/tutorial-on-creating-a-frequency-distribution-chart-with-microsoft-excel-r-and-python/ | 1,679,698,427,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945289.9/warc/CC-MAIN-20230324211121-20230325001121-00229.warc.gz | 665,285,731 | 37,547 | ## Tutorial On Creating A Frequency Distribution Chart With Microsoft Excel, R and Python
I got to join this amazing community of Data Scientists in Nigeria. We are a mix of experts and beginners. Today, I created a tutorial for the beginners to see how to do a common task like frequency distribution plot in both Python and R, also decided to include my dearest Microsoft Excel as a control.
The sample data is a fictionalized data for Dominos Pizza Nigeria. One day sales data for their Lekki branch. You can download the practice along raw data file here: https://dl.dropboxusercontent.com/u/28140414/Dominos%20Pizza.csv
So the business question we want to tackle is: Is there a pattern in the quantities each customer buys? To be more specific, we want to examine the frequency distribution of the quantities purchased per sales transaction.
In Excel, it is extremely straightforward. Just plot a histogram on the quantity field.
Now let’s head to doing same with R
I use R 3.3.2 and RStudio.
First, I import the csv file into RStudio.
Though not necessary for what we want to do, but I like doing it for any data I bring into R, I run the summary command on the dataframe/table. > summary(Dominos_Pizza)
Again, not a required step. I check out the standard plot graph on the Quantity field. > plot(Dominos_Pizza\$Quantity)
Finally, I do the histogram chart on the Quantity field. > hist(Dominos_Pizza\$Quantity)
For now I don’t bother customizing the graph elements (labels, color, title, etc.)
It is Python time.
I use Rodeo IDE and Anaconda.
I import Pandas and use it to read in the csv file.
And here is the plot graph, like we did in R.
Finally, I create the histogram.
I will try to follow up with more tutorials of complex tasks, and some that are best suited to R and others that are best suited to Python. As per Excel, it is in a completely different class. It is a spreadsheet application.
Got any particular task you will like me to create a tutorial around? Ask away! | 461 | 2,013 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2023-14 | longest | en | 0.883749 |
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Class 9Class 10First YearSecond Year
##### What happens to the temperature of the room when a air conditioner is left running on a table in the middle of the room?
Show that ratio of the root mean square speeds of molecules of two different gases at a certain temperature is equal to the square root of the inverse ratio of their masses.
Which one of the following process is irreversible?(a) Slow compressions of an elastic spring.(b) Slow evaporation of a substance in an isolated vessel.(c) Slow compression of a gas.(d) A chemical explosion.
EXAMPLE 11.4The turbine in a steam power plant takes steam from a boiler at 427{ }^{\circ} \mathrm{C} and exhausts into a low temperature reservoir at 77^{\circ} \mathrm{C} . What is the maximum possible efficiency.
A carnot engine utilises an ideal gas. The source temperature is 227^{\circ} \mathrm{C} and the sink temperature is 127^{\circ} \mathrm{C} . Find the efficiency of the engine. Also find the heat input from the source and heat rejected to the sink when 10000 \mathrm{~J} of work is done.
A heat engine performs 100 \mathrm{~J} of work and at the same time rejects 400 \mathrm{~J} of heat energy to the cold reservoirs. What is the efficiency of the engine?
Is it possible to convert internal energy into mechanical energy? Explain with an example. | 352 | 1,500 | {"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} | 2.796875 | 3 | CC-MAIN-2022-49 | longest | en | 0.865176 |
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# Circle Vocabulary
### Definition
TermDefinition
Chord A segment whose endpoint lie on a line
Diameter Achord that contain the center of a circle
Secant of a circle A line that contains a chord
Tagant to a circle A line that lies in the plane of a circle and that intersect the circle at exacly one point (point of tangency)
Circle ` The set of all point in a plane at a given distance from given point in the plane
Radius A segment from a point on a circle or a sphere to the center
Congrugent Circle Circles with the same radius
Concetric Circle Circles the lie in the same plane that share the same center
Incribed Angle Angle whose vertex lies on a circle and whose sides contain chrods of the circle
Central Angle Angle whose vertex is the center of a circle and whose sides contain radii of the circle
Circumference Distance around a circle, that is , the perimeter of a circle
Arc Lenght A fraction of the circumference of a circle defined by the arc
Arc of a Circle Two points an a circle and continuous part of the circle between the two point
Semicircle An arc of a circle whose endpoint are the endpoint of a diameter
Minor Arc An arc of a circle whose lenght is less than the lenght of a circle
Major Arc An arc of a circle whose lenght is greater than the lenght of a semicircle of the circle
Created by: Dat Boii Felipe | 431 | 1,851 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2016-44 | latest | en | 0.83894 |
https://www.studymode.com/essays/Earthquake-Technology-Worksheet-66343699.html | 1,596,763,202,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439737050.56/warc/CC-MAIN-20200807000315-20200807030315-00403.warc.gz | 792,514,770 | 24,668 | # Earthquake Technology Worksheet
Topics: Earthquake, Richter magnitude scale, Plate tectonics Pages: 4 (889 words) Published: December 21, 2014
Associate Level Material
Earthquake Technology Worksheet
Answer the lab questions for this week and summarize the lab experience using this form. Carefully read Ch. 9 of Geoscience Laboratory.
Complete this week’s lab by filling in your responses to the questions from Geoscience Laboratory. Although you are only required to respond to the questions in this worksheet, you are encouraged to answer others from the text on your own. Questions and charts are from Geoscience Laboratory, 5th ed. (p. 153-168), by T. Freeman, 2009, New York, NY: John Wiley & Sons. Reprinted with permission.
Lab Questions:
9.1 Judging from the seismogram in Figure 9.4 of the lab book, which wave appears to be the most damaging? Based on the density of the seismogram, I believe the surface wave appears to cause more damage. Based on my studies in a typically earthquake situation the surface wave is the most destructive of all energy transmission. 9.3 Determine the distance to an earthquake at a station that receives P and S waves 5.0 minutes apart. Hint: (a) Place tick marks on a scrap of paper equal to 5.0 on the minute’s axis. (b) Fit that to the horizontal separation between P and S curves. (c) Read distance directly across on the distance axis. Based on the diagram, the distance traveled over a total of five minutes is 1,600 kilometers. 9.5 At this point, from the information in Figure 9.6B, how specific can you now be as concerns the location of that earthquake? The earthquake is located where Seattle and Berkeley stations constructed circles intersect.
9.6 At this point, from the information in Figure 9.6C, how specific can you now be as concerns the location of that earthquake? Based on the seismologist information, the earthquake is located at the Montana Wyoming border where the three circles intersect. 9.10 Using the nomogram, determine the Richter magnitude for the three earthquakes listed (see, p. 169 in the lab book).
S arrival minus p arrival
Amplitude... | 487 | 2,126 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-34 | latest | en | 0.925604 |
https://www.mcqslearn.com/applied/mathematics/quiz/quiz.php?page=19 | 1,539,862,327,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583511806.8/warc/CC-MAIN-20181018105742-20181018131242-00158.warc.gz | 1,019,878,591 | 7,844 | Learn inverse matrix quiz online, applied math test 19 for online learning, distance learning courses. Free inverse matrix MCQs questions and answers to learn mathematics quiz with answers. Practice tests for educational assessment on inverse matrix test with answers, linear programming models, second degree equation in one variable, types of functions, inverse of a matrix, inverse matrix practice test for online cool math courses distance learning.
Free online inverse matrix course worksheet has multiple choice quiz question: result of inverse will be one only when quantity b is multiplied by with choices reciprocal b⁄2, reciprocal 2⁄b, reciprocal 3⁄b and reciprocal 1⁄b with problems solving answer key to test study skills for online e-learning, formative assessment and jobs' interview preparation tips, study matrix algebra multiple choice questions based quiz question and answers.
Inverse Matrix Quiz
MCQ: Result of inverse will be one only when quantity b is multiplied by
1. reciprocal b⁄2
2. reciprocal 2⁄b
3. reciprocal 3⁄b
4. reciprocal 1⁄b
D
Inverse of a Matrix Quiz
MCQ: In adjacency matrix, each node has one row and
1. one column
2. two columns
3. three columns
4. four columns
A
Types of Functions Quiz
MCQ: Value h(x) is 6x³-3x+9 and g(x) = 3x then rational function is written as
1. 3x-6x³-3x+9
2. 3x+6x³-3x+9
3. 3x⁄6x³-3x+9
4. 6x³-3x+9⁄3x
C
Second Degree Equation in One Variable Quiz
MCQ: Degree of any polynomial expression is classified as
1. degree of expression
2. degree of square roots
3. degree of polynomial
4. degree of equation
D
Linear Programming Models Quiz
MCQ: In transportation models designed in linear programming, sources of supply of homogenous commodity is classified as
1. transportation
2. destinations
3. origins
4. ordination
C | 462 | 1,802 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-43 | longest | en | 0.854162 |
https://mathoverflow.net/questions/315449/show-that-if-p-neq-2-then-mathbbz-p-cannot-act-freely-on-mathbbcpn | 1,623,899,390,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487626465.55/warc/CC-MAIN-20210617011001-20210617041001-00066.warc.gz | 352,797,171 | 30,651 | # Show that if $p\neq 2$, then $\mathbb{Z}_p$ cannot act freely on $\mathbb{C}P^n$
If $$p\neq 2$$, then the cyclic group $$\mathbb{Z}_p$$ has no free continuous action on $$\mathbb{C}P^n$$. My question is how to prove the above fact using Leray-Serre spectral sequence associated to the Borel fibration $$\mathbb{C}P^n\hookrightarrow X_{\mathbb{Z}_p}\rightarrow B_{\mathbb{Z}_p}$$.
From the Euler Characteristic argument, $$p$$ divides $$n+1$$. Also, $$\pi_1(B_{\mathbb{Z}_p})={\mathbb{Z}_p}$$ acts trivially on $$H^*(\mathbb{C}P^n;\mathbb{Z}_p)$$ by using Lefschetz fixed point theorem. $$H^*(\mathbb{C}P^n;\mathbb{Z}_p)=\mathbb{Z}_p[b]/\langle b^{n+1}\rangle$$ and $$H^*(\mathbb{Z}_p;\mathbb{Z}_p)=\bigwedge(s)\otimes\mathbb{Z}_p[t]$$. So the only possibility is $$d_3(b)=st$$. It follows $$d_3(sb)=0$$ and $$d_3{(tb)}=st^2$$. After that, I am unable to deduce any contradiction.
Thank you so much in advance.
• Do you mean the $p$-adics or the cyclic group of order $p$? I guess the second one (although I initially spent a time reading as if it were the $p$-adics, where the question is meaningful) – YCor Nov 16 '18 at 19:45
• @YCor Here I am considering the cyclic group of order p. – Shivani Sengupta Nov 17 '18 at 4:25
Consider the cohomology with $$\mathbb{Z}$$ coefficients (and reduce the 0-th term modulo $$p$$ to get uniform description of it). Then we have a spectral sequence starting from $$\mathbb{F}_p[x,y]/x^{n+1}$$ with $$deg(x)=deg(y)=2$$ and converging to $$H^*(\mathbb{C}P^n/\mathbb{Z}_p,\mathbb{Z})$$. Since the $$E^2$$ term is concentrated in even degrees, the spectral sequence degenerates at the $$E^2$$-term, and so $$H^*(\mathbb{C}P^n/\mathbb{Z}_p)$$ has arbitrarily high non-zero cohomologies. But is is a finite dimensional manifold (being the quotient of a manifold by a free action), and this is a contradiction.
• @scarmeli Instead of taking $CP^n$ if we take $Y = CP^{n_1} \times CP^{n_2} \times ... \times CP^{n_k}$, then how to show that $Z_p$ does not act freely on $Y$ using spectral sequence where $p\neq 2$? I have edited the question. Please have a look. – Shivani Sengupta Nov 17 '18 at 6:10 | 740 | 2,139 | {"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": 21, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2021-25 | latest | en | 0.769692 |
https://betterlesson.com/blended-learning/strategy/resource/12326/numbered-heads-team-role-math-lesson-sample-pdf | 1,498,419,357,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320570.72/warc/CC-MAIN-20170625184914-20170625204914-00695.warc.gz | 727,115,605 | 14,316 | ## Numbered Heads: Numbered heads & Team Role Math Lesson Sample.pdf
Numbered heads & Team Role Math Lesson Sample.pdf
Presentation
This is an example of how students know what they are responsible for after they draw a number from a cup of ping pong balls with numbers written on them. Students know their roles and are able to give feedback at the conclusion of lessons based on their role expectations.
Presentation
This is an example of how students know what they are responsible for after they draw a number from a cup of ping pong balls with numbers written on them. Students know their roles and are able to give feedback at the conclusion of lessons based on their role expectations.
Routines and Procedures
Numbered heads is a practice we use to randomize and create an element of excitement at the beginning of lessons/investigations. Each student draws a random number from their team cups to start lessons once a week.
Strategy Resources (3)
Students In Action
Presentation
This is an example of how students know what they are responsible for after they draw a number from a cup of ping pong balls with numbers written on them. Students know their roles and are able to give feedback at the conclusion of lessons based on their role expectations.
Strategy Explanation
I will post this up each lesson to designate and randomize what number of each team is responsible for certain parts of the lesson as seen in this example. This is great because students need to experience various feedback situations, and they can coach each other about how to give specific feedback since it is a team task.
Students In Action
Presentation
This is an example of how students know what they are responsible for after they draw a number from a cup of ping pong balls with numbers written on them. Students know their roles and are able to give feedback at the conclusion of lessons based on their role expectations.
Strategy Explanation
I will post this up each lesson to designate and randomize what number of each team is responsible for certain parts of the lesson as seen in this example. This is great because students need to experience various feedback situations, and they can coach each other about how to give specific feedback since it is a team task.
Freddy Esparza
Los Angeles, CA
Prep Time:
Quick
Subject:
Math
##### Similar Strategies
Instructional Openings
My students engage in a strategy called Math Fact Fluency for a few minutes each class period. They use a dry erase marker to fill in a blank multiplication table inside a plastic sheet protector according to a specific rule (by 2s, by 5s, etc.). I use this strategy to help my students notice patterns within the multiplication table and to develop a deep conceptual understanding of multiplication. | 546 | 2,785 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-26 | longest | en | 0.967558 |
http://www.jiskha.com/display.cgi?id=1357816660 | 1,462,252,688,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461860118790.25/warc/CC-MAIN-20160428161518-00185-ip-10-239-7-51.ec2.internal.warc.gz | 621,476,197 | 3,714 | Tuesday
May 3, 2016
# Homework Help: mathematical statistics
Posted by sanaz on Thursday, January 10, 2013 at 6:17am.
Suppose X_n is a sequence of independent Bernoulli random variables and p(X_n=1)=p_n. If Y=∑_(n=1)^∞▒X_n is convergent with probability 1 ,is E(Y) convergent? | 94 | 279 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-18 | longest | en | 0.862296 |
http://cers-deutschland.org/is-honeysuckle-mypnz/real-analysis-differentiation-examples-7b6492 | 1,653,611,246,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662627464.60/warc/CC-MAIN-20220526224902-20220527014902-00058.warc.gz | 10,347,957 | 6,025 | Here you can browse a large variety of topics for the introduction to real analysis. Real analysis: continuously differentiable and Lipschitz implies bounded derivatives? Notice also that the predicates for differentiability and integrability are not in Prop, which is the sort of Recall a function F: is differentiable at a iff there is a linear transformation T: such that lim 0 MATH301 Real Analysis Tutorial Note #3 More Differentiation in Vector-valued function: Last time, we learn how to check the differentiability of a given vector-valued function. Note: Recall that for xed c and x we have that f(x) f(c) x c is the slope of the secant This calls for a user-friendly library ... differentiation under an integral. ... 7.1. 2 A problem regarding the value of the derivative of a real valued function If x 0, then x 0. of real analysis, e.g. Multidimensional Real Analysis I: Differentiation | J. J. Duistermaat, J. I have included 295 completely worked out examples to illustrate and clarify all major theorems and definitions. Download books for free. 2 • We have seen two applications: – signal smoothing – root finding • Today we look – differentation – integration • These will form the basis for solving ODEs To prove the inequality x 0, we prove x 0 then! Hub pages outlines many useful topics and provides a large number of important theorems the way integrals! In analysis, we prove two inequalities: x 0 find books real analysis: 0. Clarify all major theorems and definitions x < e is true for real... 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Do in analysis to illustrate and clarify all major theorems and definitions examples..., integrals in Coq suffer from the same issues do in analysis now want to combine some of the that! | 1,927 | 9,005 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-21 | latest | en | 0.903622 |
https://softmath.com/math-book-answers/perfect-square-trinomial/pay-to-solve-algebra.html | 1,586,489,719,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371883359.91/warc/CC-MAIN-20200410012405-20200410042905-00483.warc.gz | 685,908,293 | 13,248 | English | Español
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• converting decimals to square roots | 973 | 4,036 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2020-16 | longest | en | 0.939853 |
https://discuss.codechef.com/t/help-me-in-solving-diffsum-problem/107058 | 1,696,253,823,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511000.99/warc/CC-MAIN-20231002132844-20231002162844-00344.warc.gz | 232,552,184 | 4,626 | # Help me in solving DIFFSUM problem
### My code
``````# cook your dish here
a = int(input())
b = int(input())
if a > b:
print( a - b )
else:
print( a + b )
``````
Problem Link: DIFFSUM Problem - CodeChef
The problem here is that the input comes in one line e.g. “82 28”. Therefore when you set `a = int(input())`, you are essentially saying `a = int("82 28")`, which cannot be type casted. Instead, you should use the following syntax: `a, b = map(int, input().split())`. Breaking this down:
• `input().split()`, for the test-case results in: [“82”, “28”].
• Then the `map()` function take each element in the list above, casts those elements as `int` (specified by the `int` keyword passed to `map()`) and then assigns each element to the respective variables in this case `a` and `b`.
This syntax allows you to take a single line input and assign it to multiple variables.
Here is the working code using this method:
``````a, b = map(int, input().split())
if a > b:
print(a - b)
else:
print(a + b)
``````
#include <stdio.h>
int main(void) { | 291 | 1,054 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-40 | latest | en | 0.747181 |
https://convertoctopus.com/922-grams-to-kilograms | 1,721,389,767,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514902.63/warc/CC-MAIN-20240719105029-20240719135029-00118.warc.gz | 151,932,169 | 7,433 | ## Conversion formula
The conversion factor from grams to kilograms is 0.001, which means that 1 gram is equal to 0.001 kilograms:
1 g = 0.001 kg
To convert 922 grams into kilograms we have to multiply 922 by the conversion factor in order to get the mass amount from grams to kilograms. We can also form a simple proportion to calculate the result:
1 g → 0.001 kg
922 g → M(kg)
Solve the above proportion to obtain the mass M in kilograms:
M(kg) = 922 g × 0.001 kg
M(kg) = 0.922 kg
The final result is:
922 g → 0.922 kg
We conclude that 922 grams is equivalent to 0.922 kilograms:
922 grams = 0.922 kilograms
## Alternative conversion
We can also convert by utilizing the inverse value of the conversion factor. In this case 1 kilogram is equal to 1.0845986984816 × 922 grams.
Another way is saying that 922 grams is equal to 1 ÷ 1.0845986984816 kilograms.
## Approximate result
For practical purposes we can round our final result to an approximate numerical value. We can say that nine hundred twenty-two grams is approximately zero point nine two two kilograms:
922 g ≅ 0.922 kg
An alternative is also that one kilogram is approximately one point zero eight five times nine hundred twenty-two grams.
## Conversion table
### grams to kilograms chart
For quick reference purposes, below is the conversion table you can use to convert from grams to kilograms
grams (g) kilograms (kg)
923 grams 0.923 kilograms
924 grams 0.924 kilograms
925 grams 0.925 kilograms
926 grams 0.926 kilograms
927 grams 0.927 kilograms
928 grams 0.928 kilograms
929 grams 0.929 kilograms
930 grams 0.93 kilograms
931 grams 0.931 kilograms
932 grams 0.932 kilograms | 429 | 1,666 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-30 | latest | en | 0.775279 |
http://www.ee.ic.ac.uk/hp/staff/dmb/voicebox/mdoc/v_mfiles/v_gaussmixt.html | 1,653,803,188,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652663039492.94/warc/CC-MAIN-20220529041832-20220529071832-00204.warc.gz | 80,854,755 | 5,762 | # v_gaussmixt
## PURPOSE
V_GAUSSMIXT Multiply two GMM pdfs
## SYNOPSIS
function [m,v,w]=v_gaussmixt(m1,v1,w1,m2,v2,w2)
## DESCRIPTION
```V_GAUSSMIXT Multiply two GMM pdfs
Inputs: Input mixtures: k1,k2 mixtures, p dimensions
M(k1,p) = mixture means for mixture 1
V(k1,p) or V(p,p,k1) variances (diagonal or full)
W(k1,1) = mixture weights
M(k2,p) = mixture means for mixture 2
V(k2,p) or V(p,p,k2) variances (diagonal or full)
W(k2,1) = mixture weights
Outputs:
M(k1*k2,p) = mixture means
V(k1*k2,p) or V(p,p,k1*k2) if p>1 and at least one input has full covariance matrix
W(k1*k2,1) = mixture weights
## CROSS-REFERENCE INFORMATION
This function calls:
• v_axisenlarge V_AXISENLARGE - enlarge the axes of a figure (f,h)
• v_gaussmixp V_GAUSSMIXP calculate probability densities from or plot a Gaussian mixture model
This function is called by:
## SOURCE CODE
```0001 function [m,v,w]=v_gaussmixt(m1,v1,w1,m2,v2,w2)
0002 %V_GAUSSMIXT Multiply two GMM pdfs
0003 %
0004 % Inputs: Input mixtures: k1,k2 mixtures, p dimensions
0005 %
0006 % M(k1,p) = mixture means for mixture 1
0007 % V(k1,p) or V(p,p,k1) variances (diagonal or full)
0008 % W(k1,1) = mixture weights
0009 % M(k2,p) = mixture means for mixture 2
0010 % V(k2,p) or V(p,p,k2) variances (diagonal or full)
0011 % W(k2,1) = mixture weights
0012 %
0013 % Outputs:
0014 %
0015 % M(k1*k2,p) = mixture means
0016 % V(k1*k2,p) or V(p,p,k1*k2) if p>1 and at least one input has full covariance matrix
0017 % W(k1*k2,1) = mixture weights
0018 %
0020
0021 % Copyright (C) Mike Brookes 2000-2012
0022 % Version: \$Id: v_gaussmixt.m 10865 2018-09-21 17:22:45Z dmb \$
0023 %
0024 % VOICEBOX is a MATLAB toolbox for speech processing.
0026 %
0027 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
0028 % This program is free software; you can redistribute it and/or modify
0030 % the Free Software Foundation; either version 2 of the License, or
0031 % (at your option) any later version.
0032 %
0033 % This program is distributed in the hope that it will be useful,
0034 % but WITHOUT ANY WARRANTY; without even the implied warranty of
0035 % MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the
0036 % GNU General Public License for more details.
0037 %
0038 % You can obtain a copy of the GNU General Public License from
0039 % http://www.gnu.org/copyleft/gpl.html or by writing to
0040 % Free Software Foundation, Inc.,675 Mass Ave, Cambridge, MA 02139, USA.
0041 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
0042 persistent r13 r21 r22 r31 r312 r112 r1223 r321 ch1h r122 r124
0043 if isempty(r21)
0044 r13=[1 3];
0045 r21=[2 1];
0046 r22=[2 2];
0047 r31=[3 1];
0048 r112=[1 1 2];
0049 r122=[1 2 2];
0050 r124=[1 2 4];
0051 r312=[3 1 2];
0052 r321=[3 2 1];
0053 r1223=[1 2 2 3];
0054 ch1h=[-0.5; 1; -0.5];
0055 end
0056 [k1,p]=size(m1);
0057 [k2,p2]=size(m2);
0058 f1=ndims(v1)>2 || size(v1,1)>k1; % full covariance matrix is supplied
0059 f2=ndims(v2)>2 || size(v2,1)>k2; % full covariance matrix is supplied
0060 % ff=f1+2*f2;
0061 if p~=p2
0062 error('mixtures must have the same vector dimension');
0063 end
0064 k=k1*k2;
0065 j1=repmat((1:k1)',k2,1);
0066 j2=reshape(repmat(1:k2,k1,1),k,1);
0067 if p==1
0068 % display('1D vectors');
0069 p1=1./v1(:);
0070 p2=1./v2(:);
0071 v=1./(p1(j1)+p2(j2));
0072 s1=p1.*m1;
0073 s2=p2.*m2;
0074 m=v.*(s1(j1)+s2(j2));
0075 v12=v1(j1)+v2(j2);
0076 wx=-0.5*(m1(j1)-m2(j2)).^2./v12(:);
0077 wx=wx-max(wx); % normalize to avoid underflow
0078 w=w1(j1).*w2(j2).*exp(wx)./sqrt(v12(:));
0079 w=w/sum(w);
0080 else
0081 if ~f1 && ~f2 % both diagonal covariances
0082 % display('both diagonal');
0083 p1=1./v1;
0084 p2=1./v2;
0085 v=1./(p1(j1,:)+p2(j2,:));
0086 s1=p1.*m1;
0087 s2=p2.*m2;
0088 m=v.*(s1(j1,:)+s2(j2,:));
0089 v12=v1(j1,:)+v2(j2,:);
0090 wx=-0.5*sum((m1(j1,:)-m2(j2,:)).^2./v12,2);
0091 wx=wx-max(wx); % normalize to avoid underflow
0092 w=w1(j1).*w2(j2).*exp(wx)./sqrt(prod(v12,2));
0093 w=w/sum(w);
0094 else % at least one full covariances
0095 m=zeros(k,p);
0096 v=zeros(p,p,k);
0097 w=zeros(k,1);
0098 wx=w;
0099 idp=1:p+1:p*p; % diagonal elements of p x p matrix
0100 if p==2 % special code for 2D vectors
0101 if ~f2 % GMM 2 is diagonal
0102 % display('2D GMM 2 diagonal');
0103 p2=1./v2;
0104 pm2=p2.*m2;
0105 vx1=permute(v1,r312);
0106 vx1=vx1(:,r124);
0107 px1=vx1./repmat((vx1(:,1).*vx1(:,3)-vx1(:,2).^2),1,3); % [a b; b c] -> [c -b a]
0108 pm1=m1.*px1(:,r31)-m1(:,r21).*px1(:,r22);
0109 px=px1(j1,:);
0110 px(:,r31)=px(:,r31)+p2(j2,:); % add onto diagonal elements
0111 vijx=vx1(j1,:);
0112 vijx(:,r13)=vijx(:,r13)+v2(j2,:); % add onto diagonal elements
0113 elseif ~f1 % GMM 1 is diagonal
0114 % display('2D GMM 1 diagonal');
0115 p1=1./v1;
0116 pm1=p1.*m1;
0117 vx2=permute(v2,r312);
0118 vx2=vx2(:,r124);
0119 px2=vx2./repmat((vx2(:,1).*vx2(:,3)-vx2(:,2).^2),1,3); % [a b; b c] -> [c -b a]
0120 pm2=m2.*px2(:,r31)-m2(:,r21).*px2(:,r22);
0121 px=px2(j2,:);
0122 px(:,r31)=px(:,r31)+p1(j1,:); % add onto diagonal elements
0123 vijx=vx2(j2,:);
0124 vijx(:,r13)=vijx(:,r13)+v1(j1,:); % add onto diagonal elements
0125 else % both full covariances
0126 % display('2D both full');
0127 vx1=permute(v1,r312);
0128 vx1=vx1(:,r124); % make each 2 x 2 matrix into a row [a b; b c] -> [a b c]
0129 px1=vx1./repmat((vx1(:,1).*vx1(:,3)-vx1(:,2).^2),1,3); % [a b; b c] -> [c -b a]
0130 vx2=permute(v2,r312);
0131 vx2=vx2(:,r124);
0132 px2=vx2./repmat((vx2(:,1).*vx2(:,3)-vx2(:,2).^2),1,3); % [a b; b c] -> [c -b a]
0133 pm1=m1.*px1(:,r31)-m1(:,r21).*px1(:,r22);
0134 pm2=m2.*px2(:,r31)-m2(:,r21).*px2(:,r22);
0135 px=px1(j1,:)+px2(j2,:);
0136 vijx=vx1(j1,:)+vx2(j2,:);
0137 end
0138 vx=px./repmat((px(:,1).*px(:,3)-px(:,2).^2),1,3); % divide by determinant to get inverse
0139 m=pm1(j1,:)+pm2(j2,:);
0140 m=m.*vx(:,r13)+m(:,r21).*vx(:,r22); % multiple by 2 x 2 matrix vx
0141 v=reshape(vx(:,r1223)',[2 2 k]); % convert vx to a 3D array of 2 x 2 matrices
0142 m12=m1(j1,:)-m2(j2,:); % subtract means to calculate weight exponent
0143 dij=vijx(:,1).*vijx(:,3)-vijx(:,2).^2; % determinant of V1+V2
0144 wx=m12(:,r112).*m12(:,r122).*vijx(:,r321)*ch1h./dij;% exponent of weight
0145 w=w1(j1).*w2(j2)./sqrt(dij); % weight is w*exp(wx)
0146 else
0147 if ~f2 % GMM 2 is diagonal
0148 % display('GMM 2 diagonal');
0149 p2=1./v2;
0150 pm2=p2.*m2;
0151 for i=1:k1
0152 v1i=v1(:,:,i);
0153 p1i=inv(v1i);
0154 m1i=m1(i,:);
0155 pm1i=m1i*p1i;
0156 w1i=w1(i);
0157 ix=i;
0158 for j=1:k2
0159 pij=p1i;
0160 pij(idp)=pij(idp)+p2(j,:);
0161 vix=inv(pij);
0162 vij=v1i;
0163 vij(idp)=vij(idp)+v2(j,:);
0164 v(:,:,ix)=vix;
0165 m(ix,:)=(pm2(j,:)+pm1i)*vix;
0166 m12=m2(j,:)-m1i;
0167 wx(ix)=-0.5*m12/vij*m12'; % exponent of weight
0168 w(ix)=w2(j)*w1i/sqrt(det(vij)); % weight is w*exp(wx)
0169 ix=ix+k1;
0170 end
0171 end
0172 elseif ~f1 % GMM 1 is diagonal
0173 % display('GMM 1 diagonal');
0174 p1=1./v1;
0175 pm1=p1.*m1;
0176 ix=1;
0177 for j=1:k2
0178 v2j=v2(:,:,j);
0179 p2j=inv(v2j);
0180 m2j=m2(j,:);
0181 pm2j=m2j*p2j;
0182 w2j=w2(j);
0183 for i=1:k1
0184 pij=p2j;
0185 pij(idp)=pij(idp)+p1(i,:);
0186 vix=inv(pij);
0187 vij=v2j;
0188 vij(idp)=vij(idp)+v1(i,:);
0189 v(:,:,ix)=vix;
0190 m(ix,:)=(pm1(i,:)+pm2j)*vix;
0191 m12=m1(i,:)-m2j;
0192 wx(ix)=-0.5*m12/vij*m12'; % exponent of weight
0193 w(ix)=w1(i)*w2j/sqrt(det(vij)); % weight is w*exp(wx)
0194 ix=ix+1;
0195 end
0196 end
0197 else % both full covariances
0198 % display('both full');
0199 p1=zeros(p,p,k1);
0200 pm1=zeros(k1,p);
0201 for i=1:k1
0202 p1i=inv(v1(:,:,i));
0203 p1(:,:,i)=p1i;
0204 pm1(i,:)=m1(i,:)*p1i;
0205 end
0206 ix=1;
0207 for j=1:k2
0208 v2j=v2(:,:,j);
0209 p2j=inv(v2j);
0210 m2j=m2(j,:);
0211 pm2j=m2j*p2j;
0212 w2j=w2(j);
0213 for i=1:k1
0214 pij=p1(:,:,i)+p2j;
0215 vix=inv(pij);
0216 v(:,:,ix)=vix;
0217 vij=v1(:,:,i)+v2j;
0218 m(ix,:)=(pm1(i,:)+pm2j)*vix;
0219 m12=m1(i,:)-m2j;
0220 wx(ix)=-0.5*m12/vij*m12'; % exponent of weight
0221 w(ix)=w1(i)*w2j/sqrt(det(vij)); % weight is w*exp(wx)
0222 ix=ix+1;
0223 end
0224 end
0225 end
0226
0227 end
0228 wx=wx-max(wx); % adjust exponents to avoid underflow
0229 w=w.*exp(wx); % calculate weights
0230 w=w/sum(w); % normalize weights to sum to unity
0231 if k==1
0232 v=reshape(v,size(v,1),size(v,2)); % squeeze last dimension of v if possible
0233 end
0234 end
0235 end
0236 if ~nargout
0237 if p==1
0238 nxx=256; % number of points to plot
0239 nsd=3; % number of std deviations
0240 sd=sqrt([v1(:);v2(:);v]);
0241 ma=[m1;m2;m];
0242 xax=linspace(min(ma-nsd*sd),max(ma+nsd*sd),nxx);
0243 plot(xax,v_gaussmixp(xax(:),m1,v1,w1),'--b');
0244 hold on
0245 plot(xax,v_gaussmixp(xax(:),m2,v2,w2),':r');
0246 plot(xax,v_gaussmixp(xax(:),m,v,w),'-k');
0247 hold off
0248 ylabel('Log probability density');
0249 legend('Mix 1','Mix 2','Product','location','best');
0250 v_axisenlarge([-1 -1 -1 -1.05]);
0251 elseif p==2
0252 nxx=128; % number of points to plot
0253 nsd=3;
0254 if f1
0255 s1=sqrt([v1(1:4:end)' v1(4:4:end)']); % extract diagonal elements only
0256 else
0257 s1=sqrt(v1);
0258 end
0259 if f2
0260 s2=sqrt([v2(1:4:end)' v2(4:4:end)']); % extract diagonal elements only
0261 else
0262 s2=sqrt(v2);
0263 end
0264 if ndims(v)>2 || size(v,1)>k
0265 s3=sqrt([v(1:4:end)' v(4:4:end)']); % extract diagonal elements only
0266 else
0267 s3=sqrt(v);
0268 end
0269 mal=[m1;m2;m];
0270 sal=[s1;s2;s3];
0271 xax=linspace(min(mal(:,1)-nsd*sal(:,1)),max(mal(:,1)+nsd*sal(:,1)),nxx);
0272 yax=linspace(min(mal(:,2)-nsd*sal(:,2)),max(mal(:,2)+nsd*sal(:,2)),nxx);
0273 xx(:,:,1)=repmat(xax',1,nxx);
0274 xx(:,:,2)=repmat(yax,nxx,1);
0275 xx=reshape(xx,nxx^2,2);
0276 subplot(2,2,1);
0277 imagesc(xax,yax,reshape(gaussmixp(xx,m1,v1,w1),nxx,nxx)');
0278 axis 'xy';
0279 title('Input Mix 1');
0280 subplot(2,2,2);
0281 imagesc(xax,yax,reshape(gaussmixp(xx,m2,v2,w2),nxx,nxx)');
0282 axis 'xy';
0283 title('Input Mix 2');
0284 subplot(2,2,3);
0285 imagesc(xax,yax,reshape(gaussmixp(xx,m,v,w),nxx,nxx)');
0286 axis 'xy';
0287 title('Product GMM');
0288 end
0289 end``` | 4,422 | 13,241 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-21 | latest | en | 0.617001 |
https://math.stackexchange.com/questions/1657275/proof-of-existence-of-greatest-integer | 1,566,090,291,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027313536.31/warc/CC-MAIN-20190818002820-20190818024820-00135.warc.gz | 563,907,019 | 31,175 | # Proof of existence of greatest integer
This is exercise 4 from chapter I 3.12 from Tom Apostol's "Calculus" (2nd edition):
If $x$ is an arbitrary real number, prove that there is exactly one integer $n$ which satisfies the inequalities $n \le x \lt n+1$. This $n$ is called the greatest integer in $x$ and is denoted by $\lfloor x \rfloor$.
Up to this point in the book induction, nor well-ordering principle isn't introduced. All that's available is some basic theorems deduced from 10 axioms of real numbers. For example I can't claim that for every real number $x$ there is $n$ satisfying $x \le n \lt x + 1$ (which would be enough for me to make the proof).
My aim at the proof was:
Let S be the set of all positive integers $n$ which satisfy $n \lt x$.
From Theorem I.29 (For every real $x$ there exists a positive integer $n$ such that $n \gt x$) S is not empty.
Notice that $y < x$ for every $y \in S$, so $x$ is an upper bound of $S$.
From Axiom 10 (Every upper-bounded set of real numbers has a supremum) $S$ has supremum $L$ and $L \le x$.
I don't know how to prove that this supremum is an integer.
• $n$ needs not to be positive - take $x=-2.3$, then $n=-3$ – vrugtehagel Feb 15 '16 at 21:00
You should take $S$ to be the set of all integers $n$ (not necessarily non-negative) satisfying $n \le x$ and let $L = \sup S$. By definition of sup, as you point out, $L \le x$.
Why must $L$ be an integer? We can avoid that question as follows: since $L = \sup S$, there exists (by definition of the supremum) an element $n \in S$ satisfying $L-1 < n \le L$.
By the transitive property you get $n \le x$.
On the other hand, $L < n+1$ implies that $n+1 \notin S$ because $L = \sup S$. The way $S$ is defined leads to $n+1 > x$.
Thus $n \le x < n+1$.
• How in the second step you prove that there exists $n$ which satisfies $L - 1 \lt n \le L$? – Robert Kusznier Feb 15 '16 at 21:37
• If there was no such $n$, then $L-1$ would be an upper bound of $S$! – user312938 Feb 15 '16 at 21:49
• True. Thanks for the help! – Robert Kusznier Feb 15 '16 at 22:21
• (Do you mean the transitive property, instead of the commutative property?) – user84413 Feb 16 '16 at 0:26
• Oops, I will fix that. – user312938 Feb 16 '16 at 15:23 | 720 | 2,241 | {"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} | 4.4375 | 4 | CC-MAIN-2019-35 | latest | en | 0.894517 |
https://getrevising.co.uk/revision-tests/physics-sound-3?game_type=flashcards | 1,716,107,871,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057774.18/warc/CC-MAIN-20240519070539-20240519100539-00735.warc.gz | 242,225,069 | 12,128 | # Physics sound
?
• Created by: JazzyBBo
• Created on: 02-12-18 16:07
Sound
When one particle vibrates it causes the particles next to it to vibrate too when vibrating particles come into contact with our eardrums it vibrates and sends a signal to a brain and we perceive a sound.
1 of 5
Loudness
Loudness is a measure of the intensity of a sound; the louder the sound the more energy is carried in the soundwave. The force that the soundwave applies on our eardrum is greater with a high intensity soundwave.
2 of 5
High pitch
Pitch is determined by the frequency of the soundwave. Frequency is how often are soundwave oscillates. More frequent oscillations cause a soundwave To have a high frequency, and cause a high pitched sound.
3 of 5
Low pitch
Perch is determined by the frequency of the soundwave. Frequency is how often are soundwaves oscillates. Less frequent oscillations have a lower frequency, and therefore cause a lower pitched sound.
4 of 5
How do air particles travel?
Air particles travel parallel to the wave. The air particles and up where they started.
5 of 5
## Other cards in this set
### Card 2
#### Front
Loudness is a measure of the intensity of a sound; the louder the sound the more energy is carried in the soundwave. The force that the soundwave applies on our eardrum is greater with a high intensity soundwave.
Loudness
### Card 3
#### Front
Pitch is determined by the frequency of the soundwave. Frequency is how often are soundwave oscillates. More frequent oscillations cause a soundwave To have a high frequency, and cause a high pitched sound.
### Card 4
#### Front
Perch is determined by the frequency of the soundwave. Frequency is how often are soundwaves oscillates. Less frequent oscillations have a lower frequency, and therefore cause a lower pitched sound.
### Card 5
#### Front
Air particles travel parallel to the wave. The air particles and up where they started. | 453 | 1,927 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-22 | latest | en | 0.930332 |
http://www.math.psu.edu/treluga/450/lecture38.html | 1,511,211,716,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806225.78/warc/CC-MAIN-20171120203833-20171120223833-00622.warc.gz | 437,404,514 | 4,293 | # Lecture 38: Decision processes (4 lectures)
• Day 1: Continuous time markov chains, derivation of value equation
• Day 2: Review of value equation, steady-state solution, vaccine choice without discounting. (hw problem to solve with discounting)
• Day 3: Zika with age-dependent virulence, Zika notes
• Day 4: Black Scholes financial option pricing
## Continuous time Markov chain equations
$p(t+1) = A p(t)$
$p(t+dt) = A(dt) p(t)$
Expanding each component $$a _ {ij}(dt)$$ of the Markov matrix $$A(dt)$$ as a McLaurin series in $$dt$$, we find $$A(dt) \approx I + Q dt + O(dt^2)$$
$\frac{dp}{dt} = Q p$
The transition rate matrix $$\mathbf{Q}$$ is singular, has columns that sum to zero, non-positive diagonal elements and non-negative off-diagonal elements.
## Valuing an uncertain future
In discrete time, $\vec{v}^T(t-1) = \vec{1}^T ( \mathbf{P} \circ \mathbf{A} ) + \vec{v}^T(t) \mathbf{A} \theta$
But implicit in this equation is the assumption of a time-step length, conveniently choosen to be 1. In fact, each of the components is implicitly choosen with respect to some time-step length $$dt$$
$\vec{v}^T(t-dt) = \vec{1}^T ( \mathbf{P}(dt) \circ \mathbf{A}(dt) ) + \vec{v}^T(t) \mathbf{A}(dt) \theta(dt)$
Now, suppose we take infinitessimally small time-steps
Let $$\theta(dt) \approx 1 - h dt$$ Let $$\mathbf{P}(dt) = \mathbf{F} + \mathbf{P}' dt$$ Let $$\mathbf{A}(dt) = \mathbf{I} + \mathbf{Q} dt$$
$\vec{v}^T(t-dt) = \vec{1}^T \left( (\mathbf{F} + \mathbf{P}' dt) \circ (\mathbf{I} + \mathbf{Q} dt) \right) + (1-h dt) \vec{v}^T(t) ( \mathbf{I} + \mathbf{Q} dt)$
$\vec{v}^T(t-dt) = \vec{1}^T \left( \mathbf{F} \circ \mathbf{I} + \mathbf{P}' \circ \mathbf{I} dt + \mathbf{F} \circ \mathbf{Q} dt + \mathbf{P}' \circ Q dt^2 \right) + \vec{v}^T(t) ( \mathbf{I} + \mathbf{Q} dt - h \mathbf{I} dt - h \mathbf{Q} dt^2 )$
We must have $$\mathbf{F} \circ \mathbf{I} = \vec{0}$$, and equivalently $$\operatorname{diag}(\mathbf{F}) = \vec{0}$$, meaning that as we pick smaller steps, the limit of the return per step approaches 0 to get a well-defined limit, but given this,
$\vec{v}^T(t-dt) = \vec{1}^T \left( \mathbf{P}' \circ \mathbf{I} + \mathbf{F} \circ \mathbf{Q} \right) dt + \vec{v}^T(t) + \vec{v}^T(t) (\mathbf{Q} - h \mathbf{I})dt$
Finally, taking $$\vec{u} = \vec{1}^T (\mathbf{P}' \circ \mathbf{I})$$, this yields the ordinary system of linear inhomogeneous differenti equations $-\frac{d\vec{v}^T}{dt} = \vec{u}^T + \vec{1}^T \left( \mathbf{F} \circ \mathbf{Q} \right) + \vec{v}^T \left( \mathbf{Q} - h \mathbf{I} \right)$ with terminal condition $$\vec{v}^T(t_f)$$ known for some final time $$t _ f$$ in the future (or some related terminal-time condition).
When this equation is autonomous and we have a converged terminal conditions ($$dv/dt = 0$$), we can study the steady-state. $\vec{v}^T = \left( u^T + \vec{1}^T \left( \mathbf{F} \circ \mathbf{Q} \right) \right) \left( h \mathbf{I} - \mathbf{Q} \right)^{-1}$
## Application to vaccine decisions
We take our standard valuation equation, but take all our parameters to be time-independent. For a simple immunizing infection, we get value equations $[ v_S, v_R ] = \left( [ 1 , 1] + [1,1] \left( \begin{bmatrix} 0 & 0 \\ -c_I & 0 \end{bmatrix} \begin{bmatrix} -\lambda - \mu & 0 \\ \lambda & -\mu \end{bmatrix} \right) \right) \left( \begin{bmatrix} h & 0 \\ 0 & h \end{bmatrix} - \begin{bmatrix} -\lambda - \mu & 0 \\ \lambda & -\mu \end{bmatrix} \right)^{-1}$
\begin{align*}[ v_S, v_R ] &= [ 1 - \lambda c_I, 1] \begin{bmatrix} \lambda + \mu + h & 0 \\ -\lambda & \mu+h \end{bmatrix}^{-1} \\ &= [ 1 - \lambda c_I, 1] \begin{bmatrix}\frac{1}{h + \lambda + \mu} & 0\\\frac{\lambda}{\left(h + \mu\right) \left(h + \lambda + \mu\right)} & \frac{1}{h + \mu}\end{bmatrix} \end{align*} Taking $$h=0$$ and finishing our multiplications, ... $[v_S, v_R ] = \begin{bmatrix}\frac{1}{\mu} - \frac{c \lambda}{\lambda + \mu} & \frac{1}{\mu}\end{bmatrix}$ If the expected cost from the vaccine is less than the expected cost of the disease, then you should get the vaccine as soon as you can.
What about when $$h > 0$$?
## Zika and age-dependent disease costs
Set up a 3 compartment SIR system for an individual with general age-dependence for how old you are when you get sick, and terminal conditions $$V _ S(a_{max}) = V _ I(a_{max}) = V _ R(a _ {max})$$ But eventually work back to constant-age costs and constant infection risk for blackboard solutions. Then consider a piecewise jump in infection cost from low to high and construct the solution in piecewise form.
Finally, consider how changes in the risk of infection changes these valuations. Greater infection risks reduce expected costs before the critical age, but increase expected costs after the critical age.
$-\frac{d v_ s}{dt} = \lambda ( -c _ i(a) - v _ s) - h v _ s$
$$v _ s(80)=0$$, $$c_I(a) = H(a - 40)$$, $$h=0$$, plot solutions for different values of $$\lambda$$.
$\begin{gather} %k := H(t- a) e^{-(\lambda+h) a} + e^{- \lambda - h} - e^{- a (\lambda + h)} \\ %k := e^{- \lambda - h} - e^{- a (\lambda + h)} H(a-t) %\\ v_s(a) = \frac{\lambda c_i }{\lambda + h} \left[ e^{(a-1) (\lambda + h)} - e^{(a- a_{\text{mid}}) (\lambda + h)} H(a_{\text{mid}}-a) - H(a- a _ {\text{mid}}) \right] \end{gather}$
[Show code]
1 #include "code/zika_age_model.py">
### Exercise
Given smooth, time-dependent functions for cost of infection, risk of infection, and a terminal condition, calculate the value function numerically as a function of age.
## European Call Option pricing with Black Scholes
(see paper notes)
### Stock prices are unknown
Fluctuate randomly up and down over time -- maybe we can describe these with a stochastic process, or a Markov process? Einstein and Bachelier showed how to connect Brownian motion and financial volatility to the heat equation ...
$\dot{T} = \kappa \nabla^2 T$
Suppose the price of a stock diffuses with rate $$D$$, but also drifts upward at rate $$r$$. Then the probability of observing the stock with price $$s$$ at time $$t$$ is $$p(t,s) ds$$ where $$p$$ solves the partial differential equation $\frac{\partial p}{\partial t} = D \frac{\partial^2 p}{\partial s^2} - r \frac{\partial p}{\partial s}.$
$-\frac{dv}{dt} = \frac{1}{2}\sigma^2 s^2 \frac{d^2v}{ds^2} + r s \frac{dv}{ds} - r v$ | 2,180 | 6,293 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2017-47 | latest | en | 0.600212 |
http://heath.cs.illinois.edu/iem/ode/errorest/ | 1,718,920,782,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862006.96/warc/CC-MAIN-20240620204310-20240620234310-00263.warc.gz | 16,199,133 | 3,806 | # Error Estimation
This module investigates the accuracy of Euler's method for numerically solving initial value problems for ordinary differential equations by estimating its local and global error. A numerical method for an ordinary differential equation (ODE) generates an approximate solution step-by-step in discrete increments across the interval of integration, in effect producing a discrete sample of approximate values of the solution function. Euler's method advances the approximate solution at each step by extrapolating along the tangent line whose slope is given by the ODE. This module shows estimates of the local error and cumulative global error for each step of Euler's method. These quantities are defined in greater detail below.
The user begins by selecting a differential equation from the menu provided. A solution value for the selected ODE at an initial time is marked with a black dot, and the exact solution curve for the resulting initial value problem is drawn in black. Starting from this initial value, the user advances the solution through successive steps of Euler's method. Each step is presented as a three-stage process. Each stage is executed by clicking either Next or the currently highlighted stage:
1. Subject to minimum and maximum allowed values, the slider can be used to select any desired step size. The approximate solution that would result from the currently selected step size is drawn in red. Estimates for the resulting local error and (after the first step) cumulative global error are shown as dark blue and light blue regions, respectively, around the approximate solution (the dark blue region is superimposed on the light blue region, so some of the latter is obscured by the former). Clicking Choose Step Size or Next causes the currently selected step size to take effect. The new portion of the approximate solution, now determined, changes color from red to black.
2. The current step is concluded by clicking Take Step or Next, which draws the exact solution through the new point in pink (erasing any previous pink curve) and prints the approximate and true solution values at the new point in the table below. The local error estimate (dark blue region) is removed, but the global error estimate (light blue region) remains.
3. Preparation for another step is initiated by clicking Next Step or Next, which displays the new step so that the user can choose its size. The default step size is equal to the size of the previous step, if possible.
Successive steps of Euler's method may be continued until the the interval has been fully traversed.
## Technical Details
If an error estimate is accurate, the corresponding solution should fall within the appropriate colored region. This means that the exact solution passing through the most recent solution point should fall within the dark blue local error estimate, and the exact solution to the initial value problem should fall within the light blue global error estimate. Since the module uses error estimates rather than error bounds, this is not always the case. (True bounds are rigorous, but are often overly pessimistic.) A detailed description of the estimates used in this module follows.
For an ODE y′ = f(t, y) and initial value (t0, y0), denote the exact solution to the initial value problem by y(t), the approximate solution points by (tk, yk), and the step size by hk = tktk−1. The local error at step k is the error made by the numerical method in stepping from time tk−1 to time tk. Letting uk−1(t) be the exact solution to the ODE passing through (tk−1, yk−1), the local error is given by ykuk−1(tk). For Euler's method, the local error is approximately (hk2⁄ 2) yk″. The module estimates yk using the finite difference (f (tk, yk) − f(tk−1, yk−1)) ⁄ (tktk−1).
The global error is the total error made relative to the exact solution to the initial value problem, given by yky(tk). The global error is affected not only by the local error just discussed, but also by the propagated error of the solution uk−1(t) relative to the true solution y(t). Suppose the global error at the beginning of step k is bounded by ek−1. Setting N = | 1 + hk fy(tk−1, yk−1) |, the propagated error after step k is approximately bounded by N ek−1. Using a fixed d = 0.05, the module estimates fy(tk−1, yk−1) using the finite difference (f (tk−1, yk−1 + d) − f(tk−1, yk−1d)) ⁄ (2d).
Reference: Michael T. Heath, Scientific Computing, An Introductory Survey, 2nd edition, McGraw-Hill, New York, 2002. See Section 9.3.2.
Developers: Evan VanderZee and Michael Heath | 1,005 | 4,588 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2024-26 | latest | en | 0.905142 |
https://studyres.com/doc/6215492/in-regard-to-charges--when-is-there-a-repulsive-force-bet.. | 1,544,815,724,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376826306.47/warc/CC-MAIN-20181214184754-20181214210754-00213.warc.gz | 746,269,914 | 14,596 | • Study Resource
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Transcript
```In regard to charges, when is there a repulsive force between two objects?
What is an electric field and what are electric field lines?
What effect does distance between two objects have on their electrical interaction?
Whether or not charges will move in a material depends partly on how tightly _____ are held in the atoms of the material.
What are the units of potential difference?
What are the units of current?
What are the units of resistance?
Batteries produce which: current, resistance, potential difference?
What is Ohm’s law?
How does an incandescent bulb work?
V=
P=
What is the potential difference across a resistor that dissipates 5.00 W of power and has a current of 5.0 A?
What is the potential difference across a resistor that dissipates 25.00 W of power and has a current of 7.0 A?
What is the potential difference across a resistor that dissipates 15.00 W of power and has a current of 8.0 A?
What is the difference between an insulator and a conductor?
What is the difference between a battery and a capacitor?
Which will draw more current, 3 resistors of 5 ohms each that are connected in series or parallel?
List insulators, superconductors, and semiconductors rank in order of least resistance to most resistance.
What does a circuit breaker do?
What happens to the overall resistance of a circuit when too many appliances are connected across
a 120 V outlet?
What is a superconductor and what happens to it as its temperature drops?
Identify the different elements in this drawing and determine the current:
Which bulb(s) will have current in this diagram? And identify which resistors are in series or
parallel to each other.
Does the wiring in your house use series or parallel circuits?
When you walk on a rug in the winter and then touch a doorknob, you get a shock. Why?
If a 100 W light bulb operates at a voltage of 120 V, what is the current in the bulb?
```
Related documents | 659 | 3,003 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2018-51 | latest | en | 0.86412 |
https://igraph.org/c/html/0.10.6/igraph-Operators.html | 1,721,566,843,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517701.96/warc/CC-MAIN-20240721121510-20240721151510-00368.warc.gz | 262,580,230 | 17,085 | # igraph Reference Manual
For using the igraph C library
Search the manual:
# Chapter 28. Graph operators
## 1. Union and intersection
### 1.1. `igraph_disjoint_union` — Creates the union of two disjoint graphs.
```igraph_error_t igraph_disjoint_union(igraph_t *res, const igraph_t *left,
const igraph_t *right);
```
First the vertices of the second graph will be relabeled with new vertex IDs to have two disjoint sets of vertex IDs, then the union of the two graphs will be formed. If the two graphs have |V1| and |V2| vertices and |E1| and |E2| edges respectively then the new graph will have |V1|+|V2| vertices and |E1|+|E2| edges.
Both graphs need to have the same directedness, i.e. either both directed or both undirected.
The current version of this function cannot handle graph, vertex and edge attributes, they will be lost.
Arguments:
`res`: Pointer to an uninitialized graph object, the result will stored here. `left`: The first graph. `right`: The second graph.
Returns:
Error code.
`igraph_disjoint_union_many()` for creating the disjoint union of more than two graphs, `igraph_union()` for non-disjoint union.
Time complexity: O(|V1|+|V2|+|E1|+|E2|).
Example 28.1. File `examples/simple/igraph_disjoint_union.c`
```#include <igraph.h>
#include <stdio.h>
int main(void) {
igraph_t left, right, uni;
igraph_vector_ptr_t glist;
igraph_integer_t i, n;
igraph_small(&left, 4, IGRAPH_UNDIRECTED, 0,1, 1,2, 2,2, 2,3, -1);
igraph_small(&right, 5, IGRAPH_UNDIRECTED, 0,1, 1,2, 2,2, 2,4, -1);
igraph_disjoint_union(&uni, &left, &right);
igraph_write_graph_edgelist(&uni, stdout);
printf("\n");
igraph_destroy(&left);
igraph_destroy(&right);
igraph_destroy(&uni);
/* Empty graph list; the result is the directed null graph. */
igraph_vector_ptr_init(&glist, 0);
igraph_disjoint_union_many(&uni, &glist);
if (!igraph_is_directed(&uni) || igraph_vcount(&uni) != 0) {
return 1;
}
igraph_vector_ptr_destroy(&glist);
igraph_destroy(&uni);
/* Non-empty graph list. */
igraph_vector_ptr_init(&glist, 10);
n = igraph_vector_ptr_size(&glist);
for (i = 0; i < n; i++) {
VECTOR(glist)[i] = calloc(1, sizeof(igraph_t));
igraph_small(VECTOR(glist)[i], 2, IGRAPH_DIRECTED, 0,1, 1,0, -1);
}
if (!igraph_is_directed(&uni)) {
return 2;
}
igraph_disjoint_union_many(&uni, &glist);
igraph_write_graph_edgelist(&uni, stdout);
printf("\n");
/* Destroy and free the graph list. */
n = igraph_vector_ptr_size(&glist);
for (i = 0; i < n; i++) {
igraph_destroy(VECTOR(glist)[i]);
free(VECTOR(glist)[i]);
}
igraph_vector_ptr_destroy(&glist);
igraph_destroy(&uni);
return 0;
}
```
### 1.2. `igraph_disjoint_union_many` — The disjint union of many graphs.
```igraph_error_t igraph_disjoint_union_many(igraph_t *res,
const igraph_vector_ptr_t *graphs);
```
First the vertices in the graphs will be relabeled with new vertex IDs to have pairwise disjoint vertex ID sets and then the union of the graphs is formed. The number of vertices and edges in the result is the total number of vertices and edges in the graphs.
All graphs need to have the same directedness, i.e. either all directed or all undirected. If the graph list has length zero, the result will be a directed graph with no vertices.
The current version of this function cannot handle graph, vertex and edge attributes, they will be lost.
Arguments:
`res`: Pointer to an uninitialized graph object, the result of the operation will be stored here. `graphs`: Pointer vector, contains pointers to initialized graph objects.
Returns:
Error code.
`igraph_disjoint_union()` for an easier syntax if you have only two graphs, `igraph_union_many()` for non-disjoint union.
Time complexity: O(|V|+|E|), the number of vertices plus the number of edges in the result.
### 1.3. `igraph_union` — Calculates the union of two graphs.
```igraph_error_t igraph_union(igraph_t *res,
const igraph_t *left, const igraph_t *right,
igraph_vector_int_t *edge_map1, igraph_vector_int_t *edge_map2);
```
The number of vertices in the result is that of the larger graph from the two arguments. The result graph contains edges which are present in at least one of the operand graphs.
The directedness of the operand graphs must be the same.
Edge multiplicities are handled by taking the larger of the two multiplicities in the input graphs. In other words, if the first graph has N edges between a vertex pair (u, v) and the second graph has M edges, the result graph will have max(N, M) edges between them.
Arguments:
`res`: Pointer to an uninitialized graph object, the result will be stored here. `left`: The first graph. `right`: The second graph. `edge_map1`: Pointer to an initialized vector or a null pointer. If not a null pointer, it will contain a mapping from the edges of the first argument graph (`left`) to the edges of the result graph. `edge_map2`: The same as `edge_map1`, but for the second graph, `right`.
Returns:
Error code.
`igraph_union_many()` for the union of many graphs, `igraph_intersection()` and `igraph_difference()` for other operators.
Time complexity: O(|V|+|E|), |V| is the number of vertices, |E| the number of edges in the result graph.
Example 28.2. File `examples/simple/igraph_union.c`
```#include <igraph.h>
#include <stdlib.h>
#include <stdio.h>
int print_and_clear_vector_int_list(igraph_vector_int_list_t *list) {
igraph_integer_t i, l = igraph_vector_int_list_size(list);
printf("---\n");
for (i = 0; i < l; i++) {
igraph_vector_int_print(igraph_vector_int_list_get_ptr(list, i));
}
igraph_vector_int_list_clear(list);
printf("===\n");
return 0;
}
int main(void) {
igraph_t left, right, uni;
igraph_vector_int_t v;
igraph_vector_ptr_t glist;
igraph_vector_int_t edge_map1, edge_map2;
igraph_vector_int_list_t edgemaps;
igraph_integer_t i;
igraph_vector_int_init(&edge_map1, 0);
igraph_vector_int_init(&edge_map2, 0);
igraph_vector_int_init_int_end(&v, -1, 0, 1, 1, 2, 2, 2, 2, 3, -1);
igraph_create(&left, &v, 0, IGRAPH_DIRECTED);
igraph_vector_int_destroy(&v);
igraph_vector_int_init_int_end(&v, -1, 0, 1, 1, 2, 2, 2, 2, 4, -1);
igraph_create(&right, &v, 0, IGRAPH_DIRECTED);
igraph_vector_int_destroy(&v);
igraph_union(&uni, &left, &right, &edge_map1, &edge_map2);
igraph_write_graph_edgelist(&uni, stdout);
igraph_vector_int_print(&edge_map1);
igraph_vector_int_print(&edge_map2);
igraph_destroy(&uni);
igraph_destroy(&left);
igraph_destroy(&right);
igraph_vector_int_destroy(&edge_map1);
igraph_vector_int_destroy(&edge_map2);
/* Empty graph list */
igraph_vector_ptr_init(&glist, 0);
igraph_vector_int_list_init(&edgemaps, 0);
igraph_union_many(&uni, &glist, &edgemaps);
if (!igraph_is_directed(&uni) || igraph_vcount(&uni) != 0) {
return 1;
}
print_and_clear_vector_int_list(&edgemaps);
igraph_vector_ptr_destroy(&glist);
igraph_destroy(&uni);
/* Non-empty graph list */
igraph_vector_ptr_init(&glist, 10);
for (i = 0; i < igraph_vector_ptr_size(&glist); i++) {
VECTOR(glist)[i] = calloc(1, sizeof(igraph_t));
igraph_vector_int_init_int_end(&v, -1, 0, 1, 1, 0, -1);
igraph_create(VECTOR(glist)[i], &v, 0, IGRAPH_DIRECTED);
igraph_vector_int_destroy(&v);
}
igraph_union_many(&uni, &glist, &edgemaps);
igraph_write_graph_edgelist(&uni, stdout);
for (i = 0; i < igraph_vector_ptr_size(&glist); i++) {
igraph_destroy(VECTOR(glist)[i]);
free(VECTOR(glist)[i]);
}
print_and_clear_vector_int_list(&edgemaps);
igraph_vector_ptr_destroy(&glist);
igraph_destroy(&uni);
/* Another non-empty graph list */
igraph_vector_ptr_init(&glist, 10);
for (i = 0; i < igraph_vector_ptr_size(&glist); i++) {
VECTOR(glist)[i] = calloc(1, sizeof(igraph_t));
igraph_vector_int_init_int_end(&v, -1, i, i + 1, 1, 0, -1);
igraph_create(VECTOR(glist)[i], &v, 0, IGRAPH_DIRECTED);
igraph_vector_int_destroy(&v);
}
igraph_union_many(&uni, &glist, &edgemaps);
igraph_write_graph_edgelist(&uni, stdout);
for (i = 0; i < igraph_vector_ptr_size(&glist); i++) {
igraph_destroy(VECTOR(glist)[i]);
free(VECTOR(glist)[i]);
}
print_and_clear_vector_int_list(&edgemaps);
igraph_vector_ptr_destroy(&glist);
igraph_destroy(&uni);
/* Undirected graph list*/
igraph_vector_ptr_init(&glist, 10);
for (i = 0; i < igraph_vector_ptr_size(&glist); i++) {
VECTOR(glist)[i] = calloc(1, sizeof(igraph_t));
igraph_vector_int_init_int_end(&v, -1, i, i + 1, 1, 0, -1);
igraph_create(VECTOR(glist)[i], &v, 0, IGRAPH_UNDIRECTED);
igraph_vector_int_destroy(&v);
}
igraph_union_many(&uni, &glist, &edgemaps);
igraph_write_graph_edgelist(&uni, stdout);
for (i = 0; i < igraph_vector_ptr_size(&glist); i++) {
igraph_destroy(VECTOR(glist)[i]);
free(VECTOR(glist)[i]);
}
print_and_clear_vector_int_list(&edgemaps);
igraph_vector_ptr_destroy(&glist);
igraph_destroy(&uni);
igraph_vector_int_list_destroy(&edgemaps);
return 0;
}
```
### 1.4. `igraph_union_many` — Creates the union of many graphs.
```igraph_error_t igraph_union_many(
igraph_t *res, const igraph_vector_ptr_t *graphs,
igraph_vector_int_list_t *edgemaps
);
```
The result graph will contain as many vertices as the largest graph among the arguments does, and an edge will be included in it if it is part of at least one operand graph.
The number of vertices in the result graph will be the maximum number of vertices in the argument graphs.
The directedness of the argument graphs must be the same. If the graph list has length zero, the result will be a directed graph with no vertices.
Edge multiplicities are handled by taking the maximum multiplicity of the all multiplicities for the same vertex pair (u, v) in the input graphs; this will be the multiplicity of (u, v) in the result graph.
Arguments:
`res`: Pointer to an uninitialized graph object, this will contain the result. `graphs`: Pointer vector, contains pointers to the operands of the union operator, graph objects of course. `edgemaps`: If not a null pointer, then it must be an initialized list of integer vectors, and the mappings of edges from the graphs to the result graph will be stored here, in the same order as `graphs`. Each mapping is stored in a separate igraph_vector_int_t object.
Returns:
Error code.
`igraph_union()` for the union of two graphs, `igraph_intersection_many()`, `igraph_intersection()` and `igraph_difference` for other operators.
Time complexity: O(|V|+|E|), |V| is the number of vertices in largest graph and |E| is the number of edges in the result graph.
Example 28.3. File `examples/simple/igraph_union.c`
```#include <igraph.h>
#include <stdlib.h>
#include <stdio.h>
int print_and_clear_vector_int_list(igraph_vector_int_list_t *list) {
igraph_integer_t i, l = igraph_vector_int_list_size(list);
printf("---\n");
for (i = 0; i < l; i++) {
igraph_vector_int_print(igraph_vector_int_list_get_ptr(list, i));
}
igraph_vector_int_list_clear(list);
printf("===\n");
return 0;
}
int main(void) {
igraph_t left, right, uni;
igraph_vector_int_t v;
igraph_vector_ptr_t glist;
igraph_vector_int_t edge_map1, edge_map2;
igraph_vector_int_list_t edgemaps;
igraph_integer_t i;
igraph_vector_int_init(&edge_map1, 0);
igraph_vector_int_init(&edge_map2, 0);
igraph_vector_int_init_int_end(&v, -1, 0, 1, 1, 2, 2, 2, 2, 3, -1);
igraph_create(&left, &v, 0, IGRAPH_DIRECTED);
igraph_vector_int_destroy(&v);
igraph_vector_int_init_int_end(&v, -1, 0, 1, 1, 2, 2, 2, 2, 4, -1);
igraph_create(&right, &v, 0, IGRAPH_DIRECTED);
igraph_vector_int_destroy(&v);
igraph_union(&uni, &left, &right, &edge_map1, &edge_map2);
igraph_write_graph_edgelist(&uni, stdout);
igraph_vector_int_print(&edge_map1);
igraph_vector_int_print(&edge_map2);
igraph_destroy(&uni);
igraph_destroy(&left);
igraph_destroy(&right);
igraph_vector_int_destroy(&edge_map1);
igraph_vector_int_destroy(&edge_map2);
/* Empty graph list */
igraph_vector_ptr_init(&glist, 0);
igraph_vector_int_list_init(&edgemaps, 0);
igraph_union_many(&uni, &glist, &edgemaps);
if (!igraph_is_directed(&uni) || igraph_vcount(&uni) != 0) {
return 1;
}
print_and_clear_vector_int_list(&edgemaps);
igraph_vector_ptr_destroy(&glist);
igraph_destroy(&uni);
/* Non-empty graph list */
igraph_vector_ptr_init(&glist, 10);
for (i = 0; i < igraph_vector_ptr_size(&glist); i++) {
VECTOR(glist)[i] = calloc(1, sizeof(igraph_t));
igraph_vector_int_init_int_end(&v, -1, 0, 1, 1, 0, -1);
igraph_create(VECTOR(glist)[i], &v, 0, IGRAPH_DIRECTED);
igraph_vector_int_destroy(&v);
}
igraph_union_many(&uni, &glist, &edgemaps);
igraph_write_graph_edgelist(&uni, stdout);
for (i = 0; i < igraph_vector_ptr_size(&glist); i++) {
igraph_destroy(VECTOR(glist)[i]);
free(VECTOR(glist)[i]);
}
print_and_clear_vector_int_list(&edgemaps);
igraph_vector_ptr_destroy(&glist);
igraph_destroy(&uni);
/* Another non-empty graph list */
igraph_vector_ptr_init(&glist, 10);
for (i = 0; i < igraph_vector_ptr_size(&glist); i++) {
VECTOR(glist)[i] = calloc(1, sizeof(igraph_t));
igraph_vector_int_init_int_end(&v, -1, i, i + 1, 1, 0, -1);
igraph_create(VECTOR(glist)[i], &v, 0, IGRAPH_DIRECTED);
igraph_vector_int_destroy(&v);
}
igraph_union_many(&uni, &glist, &edgemaps);
igraph_write_graph_edgelist(&uni, stdout);
for (i = 0; i < igraph_vector_ptr_size(&glist); i++) {
igraph_destroy(VECTOR(glist)[i]);
free(VECTOR(glist)[i]);
}
print_and_clear_vector_int_list(&edgemaps);
igraph_vector_ptr_destroy(&glist);
igraph_destroy(&uni);
/* Undirected graph list*/
igraph_vector_ptr_init(&glist, 10);
for (i = 0; i < igraph_vector_ptr_size(&glist); i++) {
VECTOR(glist)[i] = calloc(1, sizeof(igraph_t));
igraph_vector_int_init_int_end(&v, -1, i, i + 1, 1, 0, -1);
igraph_create(VECTOR(glist)[i], &v, 0, IGRAPH_UNDIRECTED);
igraph_vector_int_destroy(&v);
}
igraph_union_many(&uni, &glist, &edgemaps);
igraph_write_graph_edgelist(&uni, stdout);
for (i = 0; i < igraph_vector_ptr_size(&glist); i++) {
igraph_destroy(VECTOR(glist)[i]);
free(VECTOR(glist)[i]);
}
print_and_clear_vector_int_list(&edgemaps);
igraph_vector_ptr_destroy(&glist);
igraph_destroy(&uni);
igraph_vector_int_list_destroy(&edgemaps);
return 0;
}
```
### 1.5. `igraph_intersection` — Collect the common edges from two graphs.
```igraph_error_t igraph_intersection(igraph_t *res,
const igraph_t *left, const igraph_t *right,
igraph_vector_int_t *edge_map1,
igraph_vector_int_t *edge_map2);
```
The result graph contains only edges present both in the first and the second graph. The number of vertices in the result graph is the same as the larger from the two arguments.
The directedness of the operand graphs must be the same.
Edge multiplicities are handled by taking the smaller of the two multiplicities in the input graphs. In other words, if the first graph has N edges between a vertex pair (u, v) and the second graph has M edges, the result graph will have min(N, M) edges between them.
Arguments:
`res`: Pointer to an uninitialized graph object. This will contain the result of the operation. `left`: The first operand, a graph object. `right`: The second operand, a graph object. `edge_map1`: Null pointer, or an initialized vector. If the latter, then a mapping from the edges of the result graph, to the edges of the `left` input graph is stored here. For the edges that are not in the intersection, -1 is stored. `edge_map2`: Null pointer, or an initialized vector. The same as `edge_map1`, but for the `right` input graph. For the edges that are not in the intersection, -1 is stored.
Returns:
Error code.
`igraph_intersection_many()` to calculate the intersection of many graphs at once, `igraph_union()`, `igraph_difference()` for other operators.
Time complexity: O(|V|+|E|), |V| is the number of nodes, |E| is the number of edges in the smaller graph of the two. (The one containing less vertices is considered smaller.)
Example 28.4. File `examples/simple/igraph_intersection.c`
```#include <igraph.h>
void print_vector(igraph_vector_t *v) {
igraph_integer_t i, l = igraph_vector_size(v);
for (i = 0; i < l; i++) {
printf(" %" IGRAPH_PRId "", (igraph_integer_t) VECTOR(*v)[i]);
}
printf("\n");
}
int main(void) {
igraph_t left, right, isec;
igraph_vector_int_t v;
igraph_vector_ptr_t glist;
igraph_t g1, g2, g3;
igraph_vector_int_t edge_map1, edge_map2;
igraph_vector_int_init_int_end(&v, -1, 0, 1, 1, 2, 2, 3, -1);
igraph_create(&left, &v, 0, IGRAPH_DIRECTED);
igraph_vector_int_destroy(&v);
igraph_vector_int_init_int_end(&v, -1, 1, 0, 5, 4, 1, 2, 3, 2, -1);
igraph_create(&right, &v, 0, IGRAPH_DIRECTED);
igraph_vector_int_destroy(&v);
igraph_vector_int_init(&edge_map1, 0);
igraph_vector_int_init(&edge_map2, 0);
igraph_intersection(&isec, &left, &right, &edge_map1, &edge_map2);
igraph_vector_int_init(&v, 0);
igraph_get_edgelist(&isec, &v, 0);
printf("---\n");
igraph_vector_int_print(&v);
igraph_vector_int_print(&edge_map1);
igraph_vector_int_print(&edge_map2);
printf("---\n");
igraph_vector_int_destroy(&v);
igraph_destroy(&left);
igraph_destroy(&right);
igraph_destroy(&isec);
igraph_vector_int_destroy(&edge_map1);
igraph_vector_int_destroy(&edge_map2);
/* empty graph list */
igraph_vector_ptr_init(&glist, 0);
igraph_intersection_many(&isec, &glist, 0);
if (igraph_vcount(&isec) != 0 || !igraph_is_directed(&isec)) {
return 1;
}
igraph_destroy(&isec);
igraph_vector_ptr_destroy(&glist);
/* graph list with an empty graph */
igraph_vector_ptr_init(&glist, 3);
igraph_vector_int_init_int_end(&v, -1, 0, 1, 1, 2, 2, 3, -1);
igraph_create(&g1, &v, 0, IGRAPH_DIRECTED);
igraph_vector_int_destroy(&v);
igraph_vector_int_init_int_end(&v, -1, 0, 1, 1, 2, 2, 3, -1);
igraph_create(&g2, &v, 0, IGRAPH_DIRECTED);
igraph_vector_int_destroy(&v);
igraph_empty(&g3, 10, IGRAPH_DIRECTED);
VECTOR(glist)[0] = &g1;
VECTOR(glist)[1] = &g2;
VECTOR(glist)[2] = &g3;
igraph_intersection_many(&isec, &glist, 0);
if (igraph_ecount(&isec) != 0 || igraph_vcount(&isec) != 10) {
return 2;
}
igraph_destroy(&g1);
igraph_destroy(&g2);
igraph_destroy(&g3);
igraph_destroy(&isec);
igraph_vector_ptr_destroy(&glist);
/* "proper" graph list */
igraph_vector_ptr_init(&glist, 3);
igraph_vector_int_init_int_end(&v, -1, 0, 1, 1, 2, 2, 3, -1);
igraph_create(&g1, &v, 0, IGRAPH_DIRECTED);
igraph_vector_int_destroy(&v);
igraph_vector_int_init_int_end(&v, -1, 0, 1, 1, 2, 2, 3, 3, 2, 4, 5, 6, 5, -1);
igraph_create(&g2, &v, 0, IGRAPH_DIRECTED);
igraph_vector_int_destroy(&v);
igraph_vector_int_init_int_end(&v, -1, 2, 3, 1, 0, 1, 2, 3, 2, 4, 5, 6, 5, 2, 3, -1);
igraph_create(&g3, &v, 0, IGRAPH_DIRECTED);
igraph_vector_int_destroy(&v);
VECTOR(glist)[0] = &g1;
VECTOR(glist)[1] = &g2;
VECTOR(glist)[2] = &g3;
igraph_intersection_many(&isec, &glist, 0);
igraph_write_graph_edgelist(&isec, stdout);
igraph_destroy(&g1);
igraph_destroy(&g2);
igraph_destroy(&g3);
igraph_destroy(&isec);
igraph_vector_ptr_destroy(&glist);
return 0;
}
```
### 1.6. `igraph_intersection_many` — The intersection of more than two graphs.
```igraph_error_t igraph_intersection_many(
igraph_t *res, const igraph_vector_ptr_t *graphs,
igraph_vector_int_list_t *edgemaps
);
```
This function calculates the intersection of the graphs stored in the `graphs` argument. Only those edges will be included in the result graph which are part of every graph in `graphs`.
The number of vertices in the result graph will be the maximum number of vertices in the argument graphs.
The directedness of the argument graphs must be the same. If the graph list has length zero, the result will be a directed graph with no vertices.
Edge multiplicities are handled by taking the minimum multiplicity of the all multiplicities for the same vertex pair (u, v) in the input graphs; this will be the multiplicity of (u, v) in the result graph.
Arguments:
`res`: Pointer to an uninitialized graph object, the result of the operation will be stored here. `graphs`: Pointer vector, contains pointers to graphs objects, the operands of the intersection operator. `edgemaps`: If not a null pointer, then it must be an initialized list of integer vectors, and the mappings of edges from the graphs to the result graph will be stored here, in the same order as `graphs`. Each mapping is stored in a separate igraph_vector_int_t object. For the edges that are not in the intersection, -1 is stored.
Returns:
Error code.
`igraph_intersection()` for the intersection of two graphs, `igraph_union_many()`, `igraph_union()` and `igraph_difference()` for other operators.
Time complexity: O(|V|+|E|), |V| is the number of vertices, |E| is the number of edges in the smallest graph (i.e. the graph having the less vertices).
## 2. Other set-like operators
### 2.1. `igraph_difference` — Calculates the difference of two graphs.
```igraph_error_t igraph_difference(igraph_t *res,
const igraph_t *orig, const igraph_t *sub);
```
The number of vertices in the result is the number of vertices in the original graph, i.e. the left, first operand. In the results graph only edges will be included from `orig` which are not present in `sub`.
Arguments:
`res`: Pointer to an uninitialized graph object, the result will be stored here. `orig`: The left operand of the operator, a graph object. `sub`: The right operand of the operator, a graph object.
Returns:
Error code.
`igraph_intersection()` and `igraph_union()` for other operators.
Time complexity: O(|V|+|E|), |V| is the number vertices in the smaller graph, |E| is the number of edges in the result graph.
Example 28.5. File `examples/simple/igraph_difference.c`
```#include <igraph.h>
int main(void) {
igraph_t orig, sub, diff;
igraph_vector_int_t v;
/* Subtract from itself */
printf("subtract itself\n");
igraph_vector_int_init_int_end(&v, -1, 0, 1, 1, 2, 2, 1, 4, 5, -1);
igraph_create(&orig, &v, 0, IGRAPH_DIRECTED);
igraph_vector_int_destroy(&v);
igraph_difference(&diff, &orig, &orig);
igraph_write_graph_edgelist(&diff, stdout);
if (igraph_ecount(&diff) != 0 ||
igraph_vcount(&diff) != igraph_vcount(&orig)) {
return 1;
}
igraph_destroy(&orig);
igraph_destroy(&diff);
/* Same for undirected graph */
printf("subtract itself, undirected\n");
igraph_vector_int_init_int_end(&v, -1, 0, 1, 1, 2, 2, 1, 4, 5, -1);
igraph_create(&orig, &v, 0, IGRAPH_UNDIRECTED);
igraph_vector_int_destroy(&v);
igraph_vector_int_init_int_end(&v, -1, 1, 0, 1, 2, 2, 1, 4, 5, -1);
igraph_create(&sub, &v, 0, IGRAPH_UNDIRECTED);
igraph_vector_int_destroy(&v);
igraph_difference(&diff, &orig, &sub);
igraph_write_graph_edgelist(&diff, stdout);
if (igraph_ecount(&diff) != 0 ||
igraph_vcount(&diff) != igraph_vcount(&orig)) {
return 2;
}
igraph_destroy(&orig);
igraph_destroy(&sub);
igraph_destroy(&diff);
/* Subtract the empty graph */
printf("subtract empty\n");
igraph_vector_int_init_int_end(&v, -1, 0, 1, 1, 2, 2, 1, 4, 5, -1);
igraph_create(&orig, &v, 0, IGRAPH_DIRECTED);
igraph_vector_int_destroy(&v);
igraph_empty(&sub, 3, IGRAPH_DIRECTED);
igraph_difference(&diff, &orig, &sub);
igraph_write_graph_edgelist(&diff, stdout);
if (igraph_ecount(&diff) != igraph_ecount(&orig) ||
igraph_vcount(&diff) != igraph_vcount(&orig)) {
return 3;
}
igraph_destroy(&orig);
igraph_destroy(&sub);
igraph_destroy(&diff);
/* A `real' example */
printf("real example\n");
igraph_vector_int_init_int_end(&v, -1, 0, 1, 1, 2, 2, 1, 4, 5, 8, 9, -1);
igraph_create(&orig, &v, 0, IGRAPH_DIRECTED);
igraph_vector_int_destroy(&v);
igraph_vector_int_init_int_end(&v, -1, 0, 1, 5, 4, 2, 1, 6, 7, -1);
igraph_create(&sub, &v, 0, IGRAPH_DIRECTED);
igraph_vector_int_destroy(&v);
igraph_difference(&diff, &orig, &sub);
igraph_write_graph_edgelist(&diff, stdout);
igraph_destroy(&diff);
igraph_destroy(&orig);
igraph_destroy(&sub);
/* undirected version */
printf("real example, undirected\n");
igraph_vector_int_init_int_end(&v, -1, 0, 1, 1, 2, 2, 1, 4, 5, 8, 9, 8, 10, 8, 13, 8, 11, 8, 12, -1);
igraph_create(&orig, &v, 0, IGRAPH_UNDIRECTED);
igraph_vector_int_destroy(&v);
igraph_vector_int_init_int_end(&v, -1, 0, 1, 5, 4, 2, 1, 6, 7, 8, 10, 8, 13, -1);
igraph_create(&sub, &v, 0, IGRAPH_UNDIRECTED);
igraph_vector_int_destroy(&v);
igraph_difference(&diff, &orig, &sub);
igraph_write_graph_edgelist(&diff, stdout);
igraph_destroy(&diff);
igraph_destroy(&orig);
igraph_destroy(&sub);
/* undirected version with loop edge, tests Github issue #597 */
printf("Github issue #597, undirected\n");
igraph_vector_int_init_int_end(&v, -1, 0, 0, 0, 1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 6, 6, 7, 7, 8, 8, 9, 9, 0, -1);
igraph_create(&orig, &v, 0, IGRAPH_UNDIRECTED);
igraph_vector_int_destroy(&v);
igraph_vector_int_init_int_end(&v, -1, 0, 1, 1, 2, 2, 3, 3, 4, 4, 0, -1);
igraph_create(&sub, &v, 0, IGRAPH_UNDIRECTED);
igraph_vector_int_destroy(&v);
igraph_difference(&diff, &orig, &sub);
igraph_write_graph_edgelist(&diff, stdout);
igraph_destroy(&diff);
igraph_destroy(&orig);
igraph_destroy(&sub);
return 0;
}
```
### 2.2. `igraph_complementer` — Creates the complementer of a graph.
```igraph_error_t igraph_complementer(igraph_t *res, const igraph_t *graph,
igraph_bool_t loops);
```
The complementer graph means that all edges which are not part of the original graph will be included in the result.
Arguments:
`res`: Pointer to an uninitialized graph object. `graph`: The original graph. `loops`: Whether to add loop edges to the complementer graph.
Returns:
Error code.
Time complexity: O(|V|+|E1|+|E2|), |V| is the number of vertices in the graph, |E1| is the number of edges in the original and |E2| in the complementer graph.
Example 28.6. File `examples/simple/igraph_complementer.c`
```#include <igraph.h>
int main(void) {
igraph_t g1, g2;
/* complementer of the empty graph */
igraph_empty(&g1, 5, IGRAPH_DIRECTED);
igraph_complementer(&g2, &g1, IGRAPH_LOOPS);
igraph_write_graph_edgelist(&g2, stdout);
igraph_destroy(&g1);
igraph_destroy(&g2);
printf("---\n");
/* the same without loops */
igraph_empty(&g1, 5, IGRAPH_DIRECTED);
igraph_complementer(&g2, &g1, IGRAPH_NO_LOOPS);
igraph_write_graph_edgelist(&g2, stdout);
igraph_destroy(&g1);
igraph_destroy(&g2);
printf("---\n");
/* complementer of the full graph */
igraph_full(&g1, 5, IGRAPH_DIRECTED, IGRAPH_LOOPS);
igraph_complementer(&g2, &g1, IGRAPH_LOOPS);
if (igraph_ecount(&g2) != 0) {
return 1;
}
igraph_destroy(&g1);
igraph_destroy(&g2);
printf("---\n");
/* complementer of the full graph, results loops only */
igraph_full(&g1, 5, IGRAPH_DIRECTED, IGRAPH_NO_LOOPS);
igraph_complementer(&g2, &g1, IGRAPH_LOOPS);
igraph_write_graph_edgelist(&g2, stdout);
igraph_destroy(&g1);
igraph_destroy(&g2);
printf("---\n");
/**************
* undirected *
*************/
/* complementer of the empty graph */
igraph_empty(&g1, 5, IGRAPH_UNDIRECTED);
igraph_complementer(&g2, &g1, IGRAPH_LOOPS);
igraph_write_graph_edgelist(&g2, stdout);
igraph_destroy(&g1);
igraph_destroy(&g2);
printf("---\n");
/* the same without loops */
igraph_empty(&g1, 5, IGRAPH_UNDIRECTED);
igraph_complementer(&g2, &g1, IGRAPH_NO_LOOPS);
igraph_write_graph_edgelist(&g2, stdout);
igraph_destroy(&g1);
igraph_destroy(&g2);
printf("---\n");
/* complementer of the full graph */
igraph_full(&g1, 5, IGRAPH_UNDIRECTED, IGRAPH_LOOPS);
igraph_complementer(&g2, &g1, IGRAPH_LOOPS);
if (igraph_ecount(&g2) != 0) {
return 1;
}
igraph_destroy(&g1);
igraph_destroy(&g2);
printf("---\n");
/* complementer of the full graph, results loops only */
igraph_full(&g1, 5, IGRAPH_UNDIRECTED, IGRAPH_NO_LOOPS);
igraph_complementer(&g2, &g1, IGRAPH_LOOPS);
igraph_write_graph_edgelist(&g2, stdout);
igraph_destroy(&g1);
igraph_destroy(&g2);
return 0;
}
```
### 2.3. `igraph_compose` — Calculates the composition of two graphs.
```igraph_error_t igraph_compose(igraph_t *res, const igraph_t *g1, const igraph_t *g2,
igraph_vector_int_t *edge_map1, igraph_vector_int_t *edge_map2);
```
The composition of graphs contains the same number of vertices as the bigger graph of the two operands. It contains an (i,j) edge if and only if there is a k vertex, such that the first graphs contains an (i,k) edge and the second graph a (k,j) edge.
This is of course exactly the composition of two binary relations.
The two graphs must have the same directedness, otherwise the function returns with an error. Note that for undirected graphs the two relations are by definition symmetric.
Arguments:
`res`: Pointer to an uninitialized graph object, the result will be stored here. `g1`: The firs operand, a graph object. `g2`: The second operand, another graph object. `edge_map1`: If not a null pointer, then it must be a pointer to an initialized vector, and a mapping from the edges of the result graph to the edges of the first graph is stored here. `edge_map1`: If not a null pointer, then it must be a pointer to an initialized vector, and a mapping from the edges of the result graph to the edges of the second graph is stored here.
Returns:
Error code.
Time complexity: O(|V|*d1*d2), |V| is the number of vertices in the first graph, d1 and d2 the average degree in the first and second graphs.
Example 28.7. File `examples/simple/igraph_compose.c`
```#include <igraph.h>
int main(void) {
igraph_t g1, g2, res;
igraph_vector_int_t v;
igraph_vector_int_t map1, map2;
igraph_vector_int_init(&map1, 0);
igraph_vector_int_init(&map2, 0);
/* composition with the empty graph */
igraph_empty(&g1, 5, IGRAPH_DIRECTED);
igraph_full(&g2, 5, IGRAPH_DIRECTED, IGRAPH_NO_LOOPS);
igraph_compose(&res, &g1, &g2, &map1, &map2);
if (igraph_ecount(&res) != 0) {
return 1;
}
if (igraph_vector_int_size(&map1) != 0 || igraph_vector_int_size(&map2) != 0) {
return 11;
}
igraph_destroy(&res);
igraph_compose(&res, &g2, &g1, &map1, &map2);
if (igraph_ecount(&res) != 0) {
return 2;
}
if (igraph_vector_int_size(&map1) != 0 || igraph_vector_int_size(&map2) != 0) {
return 12;
}
igraph_destroy(&res);
igraph_destroy(&g1);
igraph_destroy(&g2);
/* same but undirected */
igraph_empty(&g1, 5, IGRAPH_UNDIRECTED);
igraph_full(&g2, 5, IGRAPH_UNDIRECTED, IGRAPH_NO_LOOPS);
igraph_compose(&res, &g1, &g2, &map1, &map2);
if (igraph_ecount(&res) != 0) {
return 1;
}
if (igraph_vector_int_size(&map1) != 0 || igraph_vector_int_size(&map2) != 0) {
return 11;
}
igraph_destroy(&res);
igraph_compose(&res, &g2, &g1, &map1, &map2);
if (igraph_ecount(&res) != 0) {
return 2;
}
if (igraph_vector_int_size(&map1) != 0 || igraph_vector_int_size(&map2) != 0) {
return 12;
}
igraph_destroy(&res);
igraph_destroy(&g1);
igraph_destroy(&g2);
/* proper directed graph */
igraph_vector_int_init_int_end(&v, -1, 0, 1, 1, 2, 5, 6, -1);
igraph_create(&g1, &v, 0, IGRAPH_DIRECTED);
igraph_vector_int_destroy(&v);
igraph_vector_int_init_int_end(&v, -1, 0, 1, 2, 4, 5, 6, -1);
igraph_create(&g2, &v, 0, IGRAPH_DIRECTED);
igraph_vector_int_destroy(&v);
igraph_compose(&res, &g1, &g2, &map1, &map2);
igraph_write_graph_edgelist(&res, stdout);
igraph_vector_int_print(&map1);
igraph_vector_int_print(&map2);
igraph_destroy(&res);
igraph_destroy(&g1);
igraph_destroy(&g2);
/* undirected graph */
igraph_vector_int_init_int_end(&v, -1, 0, 1, 1, 2, 5, 6, -1);
igraph_create(&g1, &v, 0, IGRAPH_UNDIRECTED);
igraph_vector_int_destroy(&v);
igraph_vector_int_init_int_end(&v, -1, 0, 1, 0, 4, 5, 6, -1);
igraph_create(&g2, &v, 0, IGRAPH_UNDIRECTED);
igraph_vector_int_destroy(&v);
igraph_compose(&res, &g1, &g2, &map1, &map2);
igraph_write_graph_edgelist(&res, stdout);
igraph_vector_int_print(&map1);
igraph_vector_int_print(&map2);
igraph_destroy(&res);
igraph_destroy(&g1);
igraph_destroy(&g2);
igraph_vector_int_destroy(&map2);
igraph_vector_int_destroy(&map1);
return 0;
}
```
## 3. Miscellaneous operators
### 3.1. `igraph_connect_neighborhood` — Connects each vertex to its neighborhood.
```igraph_error_t igraph_connect_neighborhood(igraph_t *graph, igraph_integer_t order,
igraph_neimode_t mode);
```
This function adds new edges to the input graph. Each vertex is connected to all vertices reachable by at most `order` steps from it (unless a connection already existed).
Note that the input graph is modified in place, no new graph is created. Call `igraph_copy()` if you want to keep the original graph as well.
For undirected graphs reachability is always symmetric: if vertex A can be reached from vertex B in at most `order` steps, then the opposite is also true. Only one undirected (A,B) edge will be added in this case.
Arguments:
`graph`: The input graph. It will be modified in-place. `order`: Integer constant, it gives the distance within which the vertices will be connected to the source vertex. `mode`: Constant, it specifies how the neighborhood search is performed for directed graphs. If `IGRAPH_OUT` then vertices reachable from the source vertex will be connected, `IGRAPH_IN` is the opposite. If `IGRAPH_ALL` then the directed graph is considered as an undirected one.
Returns:
Error code.
`igraph_graph_power()` to compute the kth power of a graph; `igraph_square_lattice()` uses this function to connect the neighborhood of the vertices.
Time complexity: O(|V|*d^k), |V| is the number of vertices in the graph, d is the average degree and k is the `order` argument.
### 3.2. `igraph_contract_vertices` — Replace multiple vertices with a single one.
```igraph_error_t igraph_contract_vertices(igraph_t *graph,
const igraph_vector_int_t *mapping,
const igraph_attribute_combination_t *vertex_comb);
```
This function modifies the graph by merging several vertices into one. The vertices in the modified graph correspond to groups of vertices in the input graph. No edges are removed, thus the modified graph will typically have self-loops (corresponding to in-group edges) and multi-edges (corresponding to multiple connections between two groups). Use `igraph_simplify()` to eliminate self-loops and merge multi-edges.
Arguments:
`graph`: The input graph. It will be modified in-place. `mapping`: A vector giving the mapping. For each vertex in the original graph, it should contain its desired ID in the result graph. In order to create "orphan vertices" that have no corresponding vertices in the original graph, ensure that the IDs are consecutive integers starting from zero. `vertex_comb`: What to do with the vertex attributes. `NULL` means that vertex attributes are not kept after the contraction (not even for unaffected vertices). See the igraph manual section about attributes for details.
Returns:
Error code.
Time complexity: O(|V|+|E|), linear in the number or vertices plus edges.
### 3.3. `igraph_graph_power` — The kth power of a graph.
```igraph_error_t igraph_graph_power(const igraph_t *graph, igraph_t *res,
igraph_integer_t order, igraph_bool_t directed);
```
### Warning
This function is experimental and its signature is not considered final yet. We reserve the right to change the function signature without changing the major version of igraph. Use it at your own risk.
The kth power of a graph G is a simple graph where vertex `u` is connected to `v` by a single edge if `v` is reachable from `u` in G within at most k steps. By convention, the zeroth power of a graph has no edges. The first power is identical to the original graph, except that multiple edges and self-loops are removed.
Graph power is usually defined only for undirected graphs. igraph extends the concept to directed graphs. To ignore edge directions in the input, set the `directed` parameter to `false`. In this case, the result will be an undirected graph.
Graph and vertex attributes are preserved, but edge attributes are discarded.
Arguments:
`graph`: The input graph. `res`: The graph power of the given `order`. `order`: Non-negative integer, the power to raise the graph to. In other words, vertices within a distance `order` will be connected. `directed`: Logical, whether to take edge directions into account.
Returns:
Error code.
`igraph_connect_neighborhood()` to connect each vertex to its neighborhood, modifying a graph in-place.
Time complexity: O(|V|*d^k), |V| is the number of vertices in the graph, d is the average degree and k is the `order` argument.
### 3.4. `igraph_induced_subgraph` — Creates a subgraph induced by the specified vertices.
```igraph_error_t igraph_induced_subgraph(const igraph_t *graph, igraph_t *res,
const igraph_vs_t vids, igraph_subgraph_implementation_t impl);
```
This function collects the specified vertices and all edges between them to a new graph. As the vertex IDs in a graph always start with zero, this function very likely needs to reassign IDs to the vertices.
Arguments:
`graph`: The graph object. `res`: The subgraph, another graph object will be stored here, do not initialize this object before calling this function, and call `igraph_destroy()` on it if you don't need it any more. `vids`: A vertex selector describing which vertices to keep. `impl`: This parameter selects which implementation should we use when constructing the new graph. Basically there are two possibilities: `IGRAPH_SUBGRAPH_COPY_AND_DELETE` copies the existing graph and deletes the vertices that are not needed in the new graph, while `IGRAPH_SUBGRAPH_CREATE_FROM_SCRATCH` constructs the new graph from scratch without copying the old one. The latter is more efficient if you are extracting a relatively small subpart of a very large graph, while the former is better if you want to extract a subgraph whose size is comparable to the size of the whole graph. There is a third possibility: `IGRAPH_SUBGRAPH_AUTO` will select one of the two methods automatically based on the ratio of the number of vertices in the new and the old graph.
Returns:
Error code: `IGRAPH_ENOMEM`, not enough memory for temporary data. `IGRAPH_EINVVID`, invalid vertex ID in `vids`.
Time complexity: O(|V|+|E|), |V| and |E| are the number of vertices and edges in the original graph.
`igraph_delete_vertices()` to delete the specified set of vertices from a graph, the opposite of this function.
### 3.5. `igraph_induced_subgraph_map` — Creates an induced subraph and returns the mapping from the original.
```igraph_error_t igraph_induced_subgraph_map(const igraph_t *graph, igraph_t *res,
const igraph_vs_t vids,
igraph_subgraph_implementation_t impl,
igraph_vector_int_t *map,
igraph_vector_int_t *invmap);
```
This function collects the specified vertices and all edges between them to a new graph. As the vertex IDs in a graph always start with zero, this function very likely needs to reassign IDs to the vertices.
Arguments:
`graph`: The graph object. `res`: The subgraph, another graph object will be stored here, do not initialize this object before calling this function, and call `igraph_destroy()` on it if you don't need it any more. `vids`: A vertex selector describing which vertices to keep. `impl`: This parameter selects which implementation should be used when constructing the new graph. Basically there are two possibilities: `IGRAPH_SUBGRAPH_COPY_AND_DELETE` copies the existing graph and deletes the vertices that are not needed in the new graph, while `IGRAPH_SUBGRAPH_CREATE_FROM_SCRATCH` constructs the new graph from scratch without copying the old one. The latter is more efficient if you are extracting a relatively small subpart of a very large graph, while the former is better if you want to extract a subgraph whose size is comparable to the size of the whole graph. There is a third possibility: `IGRAPH_SUBGRAPH_AUTO` will select one of the two methods automatically based on the ratio of the number of vertices in the new and the old graph. `map`: Returns a map of the vertices in `graph` to the vertices in `res`. A 0 indicates a vertex is not mapped. An `i` + 1 at position `j` indicates the vertex `j` in `graph` is mapped to vertex i in `res`. `invmap`: Returns a map of the vertices in `res` to the vertices in `graph`. An i at position `j` indicates the vertex `i` in `graph` is mapped to vertex j in `res`.
Returns:
Error code: `IGRAPH_ENOMEM`, not enough memory for temporary data. `IGRAPH_EINVVID`, invalid vertex ID in `vids`.
Time complexity: O(|V|+|E|), |V| and |E| are the number of vertices and edges in the original graph.
`igraph_delete_vertices()` to delete the specified set of vertices from a graph, the opposite of this function.
### 3.6. `igraph_linegraph` — Create the line graph of a graph.
```igraph_error_t igraph_linegraph(const igraph_t *graph, igraph_t *linegraph);
```
The line graph L(G) of a G undirected graph is defined as follows. L(G) has one vertex for each edge in G and two different vertices in L(G) are connected by an edge if their corresponding edges share an end point. In a multigraph, if two end points are shared, two edges are created. The single vertex of an undirected self-loop is counted as two end points.
The line graph L(G) of a G directed graph is slightly different: L(G) has one vertex for each edge in G and two vertices in L(G) are connected by a directed edge if the target of the first vertex's corresponding edge is the same as the source of the second vertex's corresponding edge.
Edge i in the original graph will correspond to vertex i in the line graph.
The first version of this function was contributed by Vincent Matossian, thanks.
Arguments:
`graph`: The input graph, may be directed or undirected. `linegraph`: Pointer to an uninitialized graph object, the result is stored here.
Returns:
Error code.
Time complexity: O(|V|+|E|), the number of edges plus the number of vertices.
### 3.7. `igraph_simplify` — Removes loop and/or multiple edges from the graph.
```igraph_error_t igraph_simplify(igraph_t *graph,
igraph_bool_t multiple, igraph_bool_t loops,
const igraph_attribute_combination_t *edge_comb);
```
This function merges parallel edges and removes self-loops, according to the `multiple` and `loops` parameters. Note that this function may change the edge order, even if the input was already a simple graph.
Arguments:
`graph`: The graph object. `multiple`: Logical, if true, multiple edges will be removed. `loops`: Logical, if true, loops (self edges) will be removed. `edge_comb`: What to do with the edge attributes. `NULL` means to discard the edge attributes after the operation, even for edges that were unaffected. See the igraph manual section about attributes for details.
Returns:
Error code: `IGRAPH_ENOMEM` if we are out of memory.
Time complexity: O(|V|+|E|).
Example 28.8. File `examples/simple/igraph_simplify.c`
```#include <igraph.h>
int main(void) {
igraph_t g;
/* Multiple edges */
igraph_small(&g, 0, IGRAPH_DIRECTED, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, -1);
igraph_simplify(&g, true, true, /*edge_comb=*/ NULL);
igraph_write_graph_edgelist(&g, stdout);
igraph_destroy(&g);
igraph_small(&g, 0, IGRAPH_UNDIRECTED, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, -1);
igraph_simplify(&g, true, true, /*edge_comb=*/ NULL);
if (igraph_ecount(&g) != 1) {
return 1;
}
igraph_destroy(&g);
/* Loop edges*/
igraph_small(&g, 0, IGRAPH_DIRECTED, 0, 0, 1, 1, 2, 2, 1, 2, -1);
igraph_simplify(&g, true, true, /*edge_comb=*/ NULL);
igraph_write_graph_edgelist(&g, stdout);
igraph_destroy(&g);
igraph_small(&g, 0, IGRAPH_UNDIRECTED, 0, 0, 1, 1, 2, 2, 1, 2, -1);
igraph_simplify(&g, true, true, /*edge_comb=*/ NULL);
igraph_write_graph_edgelist(&g, stdout);
igraph_destroy(&g);
/* Loop & multiple edges */
igraph_small(&g, 0, IGRAPH_DIRECTED, 0, 0, 0, 0, 0, 0, 0, 0, 1, 2, -1);
igraph_simplify(&g, /* multiple */ true, /* loop */ false, /*edge_comb=*/ NULL);
igraph_write_graph_edgelist(&g, stdout);
igraph_destroy(&g);
igraph_small(&g, 0, IGRAPH_UNDIRECTED, 1, 1, 1, 1, 1, 1, 1, 1, 2, 3, -1);
igraph_simplify(&g, /* multiple */ true, /* loop */ false, /*edge_comb=*/ NULL);
igraph_write_graph_edgelist(&g, stdout);
igraph_destroy(&g);
igraph_small(&g, 0, IGRAPH_DIRECTED, 2, 2, 2, 2, 2, 2, 3, 2, -1);
igraph_simplify(&g, /* multiple */ false, /* loop */ true, /*edge_comb=*/ NULL);
igraph_write_graph_edgelist(&g, stdout);
igraph_destroy(&g);
igraph_small(&g, 0, IGRAPH_UNDIRECTED, 3, 3, 3, 3, 3, 4, -1);
igraph_simplify(&g, /* multiple */ false, /* loop */ true, /*edge_comb=*/ NULL);
igraph_write_graph_edgelist(&g, stdout);
igraph_destroy(&g);
igraph_small(&g, 0, IGRAPH_DIRECTED, 2, 2, 2, 2, 2, 2, 2, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, -1);
igraph_simplify(&g, true, true, /*edge_comb=*/ NULL);
igraph_write_graph_edgelist(&g, stdout);
igraph_destroy(&g);
igraph_small(&g, 0, IGRAPH_UNDIRECTED,
2, 2, 2, 2, 2, 2, 2, 2, 3, 2, 2, 3, 3, 2, 3, 2, 3, 2, -1);
igraph_simplify(&g, true, true, /*edge_comb=*/ NULL);
if (igraph_ecount(&g) != 1) {
return 2;
}
igraph_destroy(&g);
return 0;
}
```
### 3.8. `igraph_subgraph_from_edges` — Creates a subgraph with the specified edges and their endpoints.
```igraph_error_t igraph_subgraph_from_edges(
const igraph_t *graph, igraph_t *res, const igraph_es_t eids,
igraph_bool_t delete_vertices
);
```
This function collects the specified edges and their endpoints to a new graph. As the vertex IDs in a graph always start with zero, this function very likely needs to reassign IDs to the vertices. Attributes are preserved.
Arguments:
`graph`: The graph object. `res`: The subgraph, another graph object will be stored here, do not initialize this object before calling this function, and call `igraph_destroy()` on it if you don't need it any more. `eids`: An edge selector describing which edges to keep. `delete_vertices`: Whether to delete the vertices not incident on any of the specified edges as well. If `false`, the number of vertices in the result graph will always be equal to the number of vertices in the input graph.
Returns:
Error code: `IGRAPH_ENOMEM`, not enough memory for temporary data. `IGRAPH_EINVEID`, invalid edge ID in `eids`.
Time complexity: O(|V|+|E|), |V| and |E| are the number of vertices and edges in the original graph.
`igraph_delete_edges()` to delete the specified set of edges from a graph, the opposite of this function.
### 3.9. `igraph_reverse_edges` — Reverses some edges of a directed graph.
```igraph_error_t igraph_reverse_edges(igraph_t *graph, const igraph_es_t eids);
```
This functon reverses some edges of a directed graph. The modification is done in place. All attributes, as well as the ordering of edges and vertices are preserved.
Note that is rarely necessary to reverse all edges, as almost all functions that handle directed graphs take a `mode` argument that can be set to `IGRAPH_IN` to effectively treat edges as reversed.
Arguments:
`graph`: The graph whose edges will be reversed. `es`: The edges to be reversed. Pass `igraph_ess_all(IGRAPH_EDGEORDER_ID)` to reverse all edges.
Returns:
Error code.
Time complexity: O(1) if all edges are reversed, otherwise O(|E|) where |E| is the number of edges in the graph.
## 4. Deprecated functions
### 4.1. `igraph_subgraph_edges` — Creates a subgraph with the specified edges and their endpoints (deprecated alias).
```igraph_error_t igraph_subgraph_edges(
const igraph_t *graph, igraph_t *res, const igraph_es_t eids,
igraph_bool_t delete_vertices
);
```
### Warning
Deprecated since version 0.10.3. Please do not use this function in new code; use `igraph_subgraph_from_edges()` instead. | 13,220 | 46,759 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2024-30 | latest | en | 0.62053 |
https://www.lmfdb.org/GaloisGroup/12T1 | 1,718,923,075,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862006.96/warc/CC-MAIN-20240620204310-20240620234310-00358.warc.gz | 751,908,374 | 7,543 | # Properties
Label 12T1 Degree $12$ Order $12$ Cyclic yes Abelian yes Solvable yes Primitive no $p$-group no Group: $C_{12}$
# Related objects
Show commands: Magma
magma: G := TransitiveGroup(12, 1);
## Group action invariants
Degree $n$: $12$ magma: t, n := TransitiveGroupIdentification(G); n; Transitive number $t$: $1$ magma: t, n := TransitiveGroupIdentification(G); t; Group: $C_{12}$ CHM label: $C(4)[x]C(3)$ Parity: $-1$ magma: IsEven(G); Primitive: no magma: IsPrimitive(G); magma: NilpotencyClass(G); $\card{\Aut(F/K)}$: $12$ magma: Order(Centralizer(SymmetricGroup(n), G)); Generators: (1,5,9)(2,6,10)(3,7,11)(4,8,12), (1,4,7,10)(2,5,8,11)(3,6,9,12) magma: Generators(G);
## Low degree resolvents
|G/N|Galois groups for stem field(s)
$2$: $C_2$
$3$: $C_3$
$4$: $C_4$
$6$: $C_6$
Resolvents shown for degrees $\leq 47$
## Subfields
Degree 2: $C_2$
Degree 3: $C_3$
Degree 4: $C_4$
Degree 6: $C_6$
## Low degree siblings
There are no siblings with degree $\leq 47$
A number field with this Galois group has no arithmetically equivalent fields.
## Conjugacy classes
Label Cycle Type Size Order Representative 1A $1^{12}$ $1$ $1$ $()$ 2A $2^{6}$ $1$ $2$ $( 1, 7)( 2, 8)( 3, 9)( 4,10)( 5,11)( 6,12)$ 3A1 $3^{4}$ $1$ $3$ $( 1, 5, 9)( 2, 6,10)( 3, 7,11)( 4, 8,12)$ 3A-1 $3^{4}$ $1$ $3$ $( 1, 9, 5)( 2,10, 6)( 3,11, 7)( 4,12, 8)$ 4A1 $4^{3}$ $1$ $4$ $( 1,10, 7, 4)( 2,11, 8, 5)( 3,12, 9, 6)$ 4A-1 $4^{3}$ $1$ $4$ $( 1, 4, 7,10)( 2, 5, 8,11)( 3, 6, 9,12)$ 6A1 $6^{2}$ $1$ $6$ $( 1, 3, 5, 7, 9,11)( 2, 4, 6, 8,10,12)$ 6A-1 $6^{2}$ $1$ $6$ $( 1,11, 9, 7, 5, 3)( 2,12,10, 8, 6, 4)$ 12A1 $12$ $1$ $12$ $( 1,12,11,10, 9, 8, 7, 6, 5, 4, 3, 2)$ 12A-1 $12$ $1$ $12$ $( 1, 2, 3, 4, 5, 6, 7, 8, 9,10,11,12)$ 12A5 $12$ $1$ $12$ $( 1, 8, 3,10, 5,12, 7, 2, 9, 4,11, 6)$ 12A-5 $12$ $1$ $12$ $( 1, 6,11, 4, 9, 2, 7,12, 5,10, 3, 8)$
magma: ConjugacyClasses(G);
## Group invariants
Order: $12=2^{2} \cdot 3$ magma: Order(G); Cyclic: yes magma: IsCyclic(G); Abelian: yes magma: IsAbelian(G); Solvable: yes magma: IsSolvable(G); Nilpotency class: $1$ Label: 12.2 magma: IdentifyGroup(G); Character table:
1A 2A 3A1 3A-1 4A1 4A-1 6A1 6A-1 12A1 12A-1 12A5 12A-5 Size 1 1 1 1 1 1 1 1 1 1 1 1 2 P 1A 1A 3A-1 3A1 2A 2A 3A1 3A-1 6A-1 6A1 6A1 6A-1 3 P 1A 2A 1A 1A 4A-1 4A1 2A 2A 4A1 4A-1 4A1 4A-1 Type 12.2.1a R $1$ $1$ $1$ $1$ $1$ $1$ $1$ $1$ $1$ $1$ $1$ $1$ 12.2.1b R $1$ $1$ $1$ $1$ $−1$ $−1$ $1$ $1$ $−1$ $−1$ $−1$ $−1$ 12.2.1c1 C $1$ $1$ $ζ3−1$ $ζ3$ $1$ $1$ $ζ3$ $ζ3−1$ $ζ3−1$ $ζ3$ $ζ3$ $ζ3−1$ 12.2.1c2 C $1$ $1$ $ζ3$ $ζ3−1$ $1$ $1$ $ζ3−1$ $ζ3$ $ζ3$ $ζ3−1$ $ζ3−1$ $ζ3$ 12.2.1d1 C $1$ $−1$ $1$ $1$ $−i$ $i$ $−1$ $−1$ $i$ $−i$ $i$ $−i$ 12.2.1d2 C $1$ $−1$ $1$ $1$ $i$ $−i$ $−1$ $−1$ $−i$ $i$ $−i$ $i$ 12.2.1e1 C $1$ $1$ $ζ3−1$ $ζ3$ $−1$ $−1$ $ζ3$ $ζ3−1$ $−ζ3−1$ $−ζ3$ $−ζ3$ $−ζ3−1$ 12.2.1e2 C $1$ $1$ $ζ3$ $ζ3−1$ $−1$ $−1$ $ζ3−1$ $ζ3$ $−ζ3$ $−ζ3−1$ $−ζ3−1$ $−ζ3$ 12.2.1f1 C $1$ $−1$ $−ζ122$ $ζ124$ $−ζ123$ $ζ123$ $−ζ124$ $ζ122$ $−ζ125$ $ζ12$ $−ζ12$ $ζ125$ 12.2.1f2 C $1$ $−1$ $ζ124$ $−ζ122$ $ζ123$ $−ζ123$ $ζ122$ $−ζ124$ $ζ12$ $−ζ125$ $ζ125$ $−ζ12$ 12.2.1f3 C $1$ $−1$ $−ζ122$ $ζ124$ $ζ123$ $−ζ123$ $−ζ124$ $ζ122$ $ζ125$ $−ζ12$ $ζ12$ $−ζ125$ 12.2.1f4 C $1$ $−1$ $ζ124$ $−ζ122$ $−ζ123$ $ζ123$ $ζ122$ $−ζ124$ $−ζ12$ $ζ125$ $−ζ125$ $ζ12$
magma: CharacterTable(G); | 1,939 | 3,269 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 144, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2024-26 | latest | en | 0.384455 |
http://www.econedlink.org/lessons/projector.php?lid=602&type=student | 1,448,611,940,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398448389.58/warc/CC-MAIN-20151124205408-00101-ip-10-71-132-137.ec2.internal.warc.gz | 407,140,277 | 4,900 | Return
### INTRODUCTION
If you heard a television newscaster say that the economy grew by 1 percent last quarter, would you know what that means? The most commonly used measure of economic growth for any country is the Gross Domestic Product (GDP). GDP is the total market value of all the goods and services produced in a country in a given year, GDP, also is used as a measure of our standard of living. A standard of living is indicated by the necessities, comforts and luxuries enjoyed by an individual or group. If the newscaster reports a 1 percent rate of growth in the economy, that report means GDP increased by 1 percent from the previous period. Nominal GDP is GDP based on current prices. Real GDP is GDP adjusted for inflation. In order to compare different countries based on their economic performance, GDP per capita is used. This measure accounts for both the inflation rate and the population of the country. By reference to GDP per capita, two countries with very different populations can be compared. If two countries have a large difference in per capita GDP, then a divergence or gap exists. There is often a gap between the GDP of developed countries, such as the United States, and developing countries, such as India . When the gap between the two countries decreases or disappears, then convergence occurs.
In this lesson, you will learn how to calculate the GDP per capita as well as how GDP per capita is used to compare countries. You will look closely at the GDP of two countries, Mexico and Canada, and think about how the gap between the two could be bridged.
### PROCESS
First, you will need to know what Real GDP per capita is and how to calculate it.
• Real GDP per capita is simply the Real GDP of any country divided by its population.
• Real GDP / Population = Real GDP per capita
Example: Using the you can find Luxemburg 's Real GDP and population in 2004. When we plug those numbers into the equation, it looks like this:
\$27,270,000,000 / 468,571 = \$58,198 per person
Compare Luxemburg's Real GDP per capita to the United States .
Luxemburg = \$58,198
United States = ?
You may have noticed that Luxemburg has a much higher Real GDP per capita than the United States. What do you think the reasons are for this big difference?
GDP per capita can vary widely from country to country. Developed countries such as the United States have higher GDP per capita figures than developing countries such as India.
Use this drag and drop activity to match countries with their Real GDP per capita.
Are you surprised by any of the countries' GDP? Are there any common factors in the countries with the lower GDPs per capita?
Now, let's compare the two closest neighbors of the United States--Canada and Mexico. Which of the two countries has the highest GDP per capita? Using the CIA World Fact Book, look at the Economy section for Canada andMexico to find the information you need.
### CONCLUSION
Did you think that Canada would have a higher per capita GDP than Mexico? What factors might explain the difference in GDP per capita between Canada and Mexico? Does the per capita GDP of each country support your general impression of the standard of living in the two countries?
### ASSESSMENT ACTIVITY
As we saw previously, there is a big difference in the standard of living between Canada and Mexico. In order for Mexico to get closer to the Canadian GDP per capita, the growth rate of Mexico would need to be higher than Canada's. This would, over time, cause convergence with the two GDPs. In other words, the gap between the two GDPs per capita would be closed. According to the World Fact book, the GDP growth rate for Canada is 2.4 percent and is 4.1 percent for Mexico. There is a shortcut method to finding out when a sum will double based on a known growth rate. This is known as the Rule of 72 (a definition for the Rule of 72 can be found at EconEdLink). It involves dividing the number 72 by a percentage growth rate. For example, if a growth rate is 6 percent, then the sum in question will double in 12 years (72/6). Using the Rule of 72 and rounding the growth rate of Canada to 2 percent and Mexico's growth rate to 4 percent , calculate how long it will take the GDP of Canada and Mexico to double. Calculate what the GDP per capita will be in 36 years for both countries. Is there still a significant gap between the two countries? In order for Mexico to close the gap with Canada, it would need a higher growth rate in GDP. Let's assume Mexico can raise its growth rate to 6 percent while Canada's growth rate stays at approximately 2 percent. In 36 years, how many times would Mexico's GDP double? What would be the GDP per capita of both countries in 36 years? Has the gap closed or been eliminated? Can you think of any ways for Mexico to raise its growth rate?
### EXTENSION ACTIVITY
Can you think of ways in which the United States would benefit if Mexico increased its standard of living, bringing it closer to Canada's standard of living? | 1,081 | 5,022 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2015-48 | longest | en | 0.952693 |
https://www.diynot.com/diy/threads/earth-fault-loop-impedance-test-ze-with-multimeter.400312/ | 1,708,758,107,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474523.8/warc/CC-MAIN-20240224044749-20240224074749-00814.warc.gz | 732,027,655 | 31,443 | Earth fault loop impedance test (Ze) with Multimeter?
eveares
I know that you can't use a cheap multimeter for carrying out a earth fault loop impedance test, but what would happen if your tried to get the value of Ze with one.
Would the reading be wrong, if so what would it show and why?
I am guessing it has something to do with the test voltage and current used by a multimeter being to low and it not being able to supply enough current to safely test the main earth?
So from a theory point of view, is it solely because a multimeter can not supply 25A that it can not be used to get Ze or because some other reason?
Also, not an electrician my self, just interested in the theory behind it.
Don't try it, you may break your meter, blow it up and even do yourself harm.
A bog standard multimeter isn't made to measure resistance on a live line.
putting your multimeter across 240v unless on the VOLTAGE range would not do it much good
Ze is the external part back to the supply transformer.
maybe you mean Zs
I won't and had no intent to! I know it's a live test and am aware of the dangers.
Just what to know the theory as originally stated in my question.
I won't do this! But in theory what would my multimeter read on the ohms range if I connected one probe to the main live in and the other probe to the disconnected main earth going back to the local transformer.
I won't do this! But in theory what would my multimeter read on the ohms range if I connected one probe to the main live in and the other probe to the disconnected main earth going back to the local transformer.
If the meter was on a low ohms range (which is presumably what you mean), it would read nothing - at least, not for more than a few milliseconds. If you were lucky, a fuse (in leads or meter) might blow. Otherwise, it would either explode, burst into flames or, at the least, get otherwise destroyed - and there would be potential for it to destroy, or at least seriously damage, you as well.
Kind Regards, John
I think I have solved my question with ohms law.
A normal multimeter simply works out the impedance/resistance by dividing the fixed output voltage of its self by the current it measures and any external current by external live sources would be added to this native current provided by the multimeter, thus giving a false reading. Not to mention the circuits involved with measuring resistance could likely not handle the mains voltage or AC type wave forms.
Where as with a proper rated tester, not only can it handle the mains AC voltage, but also subtracts the mains voltage from the fixed output voltage of the tester, meaning it can accurately work out the impedance between the main line and the local transformers earth.
In essence the proper tester is rated for mains voltage and takes the mains voltage into account when working out the impedance.
Am i right in my theory?
You can find Ze in two ways.
1) Special Meter.
2) Inquiry.
The problem is to measure involves time. Within a very short time the meter measures voltage with a load and without a load and calculates from those readings the earth loop impedance.
Since measured to earth during the test it will raise the earth voltage towards line voltage which means if the earth is not good it could have dangerous voltages on the earth. To reduce these dangers the meter measures in less than 40ms this also means it will not trip the RCD.
So in real terms if you don't have special meter then it's inquiry from DNO. Once you have Ze which should be entered on installation certificates you can work out Zs by measuring the resistance with a low ohm meter and adding it to Ze.
The low ohm meter will use at least 200ma for two reasons with three the third is to comply with regulations but using a high current means it will more likely find bad connections and also easier to measure 0.01 ohms accurate enough to give a reasonable reading.
Modern meters have improved and using a constant current source rather than simple battery and variable resistor likely they don't any longer need 200ma but when you buy the meter you don't really know how it will work.
I have asked the question before "Can one be certain using an earth loop impedance meter will do no damage?" maybe when mine was new it did have the test current marked but when I looked I could not find it so when testing on a boat I would take reading of shore supply and use low ohm meter on boat.
I was and still am uncertain as to how one should calculate and add the 1.2 volt drop between boat and shore due to diode but since allowed 200 ohms when using a RCD (Table 41.5 note 2) I was not really too worried.
I would say in real terms the same applies today with homes since all circuits now RCD protected you are allowed 200 ohms so unlikely to be a problem.
With a TT system we use a Wheatstone Bridge type circuit and yes I suppose you could measure this using a modern multi-meter as they are centre zero.
I have never put my mind to designing a circuit to measure the resistance of an earth rod although done it many times with a propriety meter. Although you still need to add DNO figure to find Ze.
I think I have solved my question with ohms law. A normal multimeter simply works out the impedance/resistance by dividing the fixed output voltage of its self by the current it measures ....
Indeed so.
...and any external current by external live sources would be added to this native current provided by the multimeter, thus giving a false reading. Not to mention the circuits involved with measuring resistance could likely not handle the mains voltage or AC type wave forms.
There is no "likely not" about it - most definitely not. As well as not being able to handle AC, an incredibly high current would try to go through the meter (on its ohms range) if it were connected between 'live' and earth - as I said before, probably destroying the meter and possibly also you.
Where as with a proper rated tester, not only can it handle the mains AC voltage, but also subtracts the mains voltage from the fixed output voltage of the tester, meaning it can accurately work out the impedance between the main line and the local transformers earth.
As eric has explained, they don't work in the same way as multimeters. The tester itself provides no voltage - only the mains is used. As eric said, they measure the (small) drop in voltage between 'live' and earth when a load (a low value resistor) is applied between 'live' and earth for a tiny fraction of a second (that to avoid the tester being destroyed). The tester can then calculate loop impedance (Ze or Zs) from the size of the load used and the (small) drop in voltage it results in.
In essence the proper tester is rated for mains voltage and takes the mains voltage into account when working out the impedance. Am i right in my theory?
As above, it's more complicated than that - these testers work in a totally different way from multimeters measuring resistance.
Kind Regards, ohn
You can get a rough idea of impedance of the supply by measuring the voltage when a a known heavy load such as a 3 kilowatt fire or similar is switched on.
Do not measure voltage drop at the appliance as the impedance of the MCB, internal wiring and the appliance's lead will be included. Measure at the CU or a socket on a different and unloaded circuit.
Take the average of several measurements.
It will be more accurate, ( or rather less in-accurate ) if done when other people in the neighbour hood are not likely to be switching appliances on and off which will alter the current in the local network.
You can get a rough idea of impedance of the supply by measuring the voltage when a a known heavy load such as a 3 kilowatt fire or similar is switched on. Do not measure voltage drop at the appliance as the impedance of the MCB, internal wiring and the appliance's lead will be included. Measure at the CU or a socket on a different and unloaded circuit.
Indeed so - but if one does that, one will be obtaining a measure of the (L+N) loop impedance which, other than for TN-C-S supplies, will not (unless by coincidence) be the same as the Ze.
Kind Regards, John
one will be obtaining a measure of the (L+N) loop impedance which, other than for TN-C-S supplies, will not (unless by coincidence) be the same as the Ze.
Yes in the majority of so called PME installations where the incoming supply cable has only two conductors and "earth" is derived from neutral then the loop impedance is for all practical purposes the same as the impedance that limits fault current until an over current protective device operates.
If the supply cable has three conductors then the test has to be the load applied between Live and "Earth" but this creates a hazard of lifting the local "earth" wires to a potential above ground while the test load is connected. How high the potential on the "earth" wires goes will depend on the quality of the "earthing" system.
DO NOT ATTEMPT THAT TEST
Personally I feel that provided the supply impedance is high enough to ensure maximum fault current is low enough that disconnect devices will not be welded closed then the system for over current protection is adequate.
The protection against any currents to ground should be from residual current devices detecting that Live and Neutral currents are different. With 30 mA sensitivity these devices will remove supply from a circuit even if the "earthing" arrangement is so poor that Ze is as high as 7666 ohms albeit with a transient voltage on the "earth" wire of 230 volts.
The transient voltage will not exceed 50 volts if the Ze is less than 1666 ohms.
one will be obtaining a measure of the (L+N) loop impedance which, other than for TN-C-S supplies, will not (unless by coincidence) be the same as the Ze.
Yes in the majority of so called PME installations where the incoming supply cable has only two conductors and "earth" is derived from neutral then the loop impedance is for all practical purposes the same as the impedance that limits fault current until an over current protective device operates.
Quite. I think that's what I said!
If the supply cable has three conductors then the test has to be the load applied between Live and "Earth" but this creates a hazard of lifting the local "earth" wires to a potential above ground while the test load is connected. How high the potential on the "earth" wires goes will depend on the quality of the "earthing" system.
I hope and presume that you're not suggesting that anyone does that! Apart from the dangers, if the circuit was RCD protected, the RCD would obviously immediately operate.
Personally I feel that provided the supply impedance is high enough to ensure maximum fault current is low enough that disconnect devices will not be welded closed then the system for over current protection is adequate. ... The protection against any currents to ground should be from residual current devices
That might be your personal feeling (and, personally, I'm not necessarily going to disagree) but readers need to understand that, other than in TT installations, the regulations essentially require that earth fault loop impedance is sufficiently low that over-current devices (e.g. fuse, MCB) provide adequate protection in relation to 'faults to ground' as well as L-N faults. In other words (albeit there is some vague talk about using an RCD to achieve required disconnection times if they can't be achieved with an OPD), the regs are not happy with L-E fault production 'relying on' an RCD (other than in TT systems).
Kind Regards, John
I hope and presume that you're not suggesting that anyone does that!
Definitely NOT and so a warning has been added to the post
In other words (albeit there is some vague talk about using an RCD to achieve required disconnection times if they can't be achieved with an OPD), the regs are not happy with L-E fault production 'relying on' an RCD (other than in TT systems).
I cannot see why RCD protection is not acceptable. Unless it is because the modern RCDs are considered to be less reliable than the older designs. True that nothing can be 100% reliable but the use of electronic amplifiers to allow for less sensitive current transformers to be used is ( in my opinion ) a cost cutting but backward step.
The earlier RCDs had several turns of Live and Neutral on the toroid and the secondary produced enough energy from a 30 mA difference between Live and Neutral to reliably trip a mechanical latch . But the toriods' Live and Neutrals had to be hand wound and the mechanism was delicate and needed care in assembly and setting up. But they seldom failed to operate when needed to.
RCDs with amplifiers and half turns for Live and Neutral through the toroid can be assembled by robots so are very cheap to make.
I hope and presume that you're not suggesting that anyone does that!
Definitely NOT and so a warning has been added to the post
Thanks.
... the regs are not happy with L-E fault production 'relying on' an RCD (other than in TT systems).
I cannot see why RCD protection is not acceptable. Unless it is because the modern RCDs are considered to be less reliable than the older designs.
It wouldn't be the first time we'd come across a situation in which we "couldn't see" why the regs are such as they are. I basically agree with you, but I think you may have identified a 'reason' for the regs being as they are. I don't know whether it's necessarily got anything to do with modern RCDs being regarded as any less reliable than older ones, but there is this seemingly fairly widely-held belief that modern RCDs are not all that reliable in-service (we've discussed the {limited amount of} available statistics many times before). Whether they are actually any less reliable than the magnetic parts of MCBs, I just don't know. It could just be that whilst it's easy to test in-service RCDs (and hence identify 'failures') it's essentially impossible to test MCBs - so, for all we know, they could possibly be just as 'unreliable' in service as RCDs.
Given potential uncertainties about the in-service reliability of both RCDs and MCBs, I suppose the one thing to be said for the regs' approach (that one should not rely on an RCD for fault protection, unless that is 'unavoidable') is that it represents 'belt and braces'.
Kind Regards, John
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17K | 3,191 | 14,749 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-10 | latest | en | 0.966024 |
https://www.acmicpc.net/problem/16034 | 1,713,924,536,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818999.68/warc/CC-MAIN-20240424014618-20240424044618-00713.warc.gz | 586,208,045 | 8,688 | 시간 제한메모리 제한제출정답맞힌 사람정답 비율
2 초 512 MB68454265.625%
## 문제
Mr. Port plans to start a new business renting one or more floors of the new skyscraper with one giga floors, MinatoHarukas. He wants to rent as many vertically adjacent floors as possible, because he wants to show advertisement on as many vertically adjacent windows as possible. The rent for one floor is proportional to the floor number, that is, the rent per month for the n-th floor is n times that of the first floor. Here, the ground floor is called the first floor in the American style, and basement floors are out of consideration for the renting. In order to help Mr. Port, you should write a program that computes the vertically adjacent floors satisfying his requirement and whose total rental cost per month is exactly equal to his budget.
For example, when his budget is 15 units, with one unit being the rent of the first floor, there are four possible rent plans, 1+2+3+4+5, 4+5+6, 7+8, and 15. For all of them, the sums are equal to 15. Of course in this example the rent of maximal number of the floors is that of 1+2+3+4+5, that is, the rent from the first floor to the fifth floor.
## 입력
The input consists of multiple datasets, each in the following format.
b
A dataset consists of one line, the budget of Mr. Port b as multiples of the rent of the first floor. b is a positive integer satisfying 1 < b < 109.
The end of the input is indicated by a line containing a zero. The number of datasets does not exceed 1000.
## 출력
For each dataset, output a single line containing two positive integers representing the plan with the maximal number of vertically adjacent floors with its rent price exactly equal to the budget of Mr. Port. The first should be the lowest floor number and the second should be the number of floors.
## 예제 입력 1
15
16
2
3
9699690
223092870
847288609
900660121
987698769
999999999
0
## 예제 출력 1
1 5
16 1
2 1
1 2
16 4389
129 20995
4112949 206
15006 30011
46887 17718
163837 5994 | 540 | 1,992 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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} | 2.828125 | 3 | CC-MAIN-2024-18 | latest | en | 0.907054 |
https://www.experts-exchange.com/questions/10257549/Tile-Based-Game-Object-System.html | 1,521,270,647,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257644701.7/warc/CC-MAIN-20180317055142-20180317075142-00472.warc.gz | 816,896,263 | 20,668 | Solved
Tile-Based Game Object System
Posted on 2000-01-10
Medium Priority
465 Views
This is a rather perplexing dilemma that I've had while developing a tile-based RTS (Real Time Strategy) game. But, first off, all I'm really interested in here is theory and possible code snippets, I don't need (or want) a complete engine or anything like that.
In the game I need to be able to store the objects that are scattered about the game world (trees, rocks, buildings, units, etc.). At first glance, this is a simple problem: just create a linked list of objects in the world, when a new object is created, simply add it to the end of the list. This has one huge problem though: drawing efficiency. Somehow drawing must occur from upper-left to lower-right to avoid overlap. So, what I have implemented right now is a sorted linked list (sorted by Y, X). Then, I have an array that is the size of the map (in a 192x192 map, the array is declared as: Object_List object_list[192][192];). The basic design of the Object_List structure is:
struct Object_List
{
Object* object;
Object_List* next;
}
A simple linked list. Now, the complicated part; if you look at that structure and the array, it looks like you get hundreds of little linked lists. However, the way I have it implemented, each of those small lists are connected to each other. So, object_array[80][80].next points to object_array[81][80] and so forth. The reason for splitting up the list like this is that I needed to allow multiple objects per tile on the map, and this seemed to work.
So, here is where I need help. This system falls apart as soon as you add height. I could declare Object_Array object_array[h][x][y]; -- but then you have a whole slew of wasted memory. Because there is only one height for each tile, that array would waste (h-1)*x*y*sizeof(Object_Array) bytes of memory -- which is completely unacceptable.
So, are there any ideas on better, more efficient implementations that will work with height but still be able to be drawn on the screen properly?
I appreciate the help,
- Alex
0
Question by:Egore
• 2
• 2
LVL 3
Accepted Solution
arnond earned 600 total points
ID: 2340823
how about adding another pointer to the struct that will point to a list of objects that are on the same tile ? This way you can dynamicly allocate memory when needed (i.e. when you have more than one object on a tile). This will waste a lot less memory (in the long run....).
If you need a more detailed explanation of my suggestion, just say so....
Arnon David.
0
LVL 1
Expert Comment
ID: 2341576
maybe the same as armond's idea.. although i dont think so:
suppose it had to go on top of [80][80]
>allocate a new Object_List
> put your object in it.
this way you can still directly access [80][80] and [80][81], but the list got extended to include the extra object. Also it doesn't require an extra pointer.
Floris, who wished he could change his nick on EE :-)
0
LVL 3
Author Comment
ID: 2346993
Well, Maniac, I've been very tempted to do this. I think that arnond's answer was basically the same thing. Arnond, if your answer is alluding to something different, could you please explain further?
If I don't receive any new posts in the next day or two, I'll give Maniac the points.
- Alex
0
LVL 3
Expert Comment
ID: 2347045
Not realy, but since both of our ideas are basicly the same, should the first to suggest it (....me....) be the one to get the points ?
Humbly,
Arnon David.
0
LVL 3
Author Comment
ID: 2368660
Well, since there has been no activity here for some time, I'm going to award the points to the first person to propose the solution: arnond.
Thanks
0
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Have a better answer? Share it in a comment. | 993 | 3,879 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2018-13 | longest | en | 0.92962 |
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A135767 sigma_0(n)-omega(n)-Omega(n) (sigma_0 = A000005 = # divisors, omega = A001221 = # prime factors, Omega = A001222 = # prime factors with multiplicity). 3
1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 2, 0, 0, 0, 1, 0, 2, 0, 0, 0, 0, 0, 3, 0, 0, 0, 2, 0, 2, 0, 1, 1, 0, 0, 3, 0, 1, 0, 1, 0, 2, 0, 2, 0, 0, 0, 5, 0, 0, 1, 0, 0, 2, 0, 1, 0, 2, 0, 5, 0, 0, 1, 1, 0, 2, 0, 3, 0, 0, 0, 5, 0, 0, 0, 2, 0, 5, 0, 1, 0, 0, 0, 4, 0, 1, 1, 3, 0, 2, 0, 2, 2 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,24 COMMENTS A102467 = { n | a(n)>0 } ; A102466 = { n | a(n)=0 } = { n | omega(n)=1 or Omega(n)=2 }: these are exactly the prime powers (>1) and semiprimes. For all other numbers a(n) > 0 since for each of the Omega(n) prime power divisors, other divisors are obtained by multiplying it with another prime factor, which gives more than omega(n) different additional divisors. a(n)>0 is also equivalent to A001037(n) > A107847(n), i.e. there are strictly fewer nonzero sums of non-periodic subsets of U_n (n-th roots of unity) than there are non-periodic binary words of length n. Otherwise stated, a(n)>0 if there is a non-periodic subset of U_n with zero sum. Non-periodic means having no rotational symmetry (except for identity). LINKS M. F. Hasler, Table of n, a(n) for n = 1..10000 FORMULA a(n)=0 <=> omega(n)=1 or Omega(n)=2 <=> n is semiprime or a prime power (>1) <=> A001037(n) = A107847(n) <=> all non-periodic subsets of U_n have nonzero sum MATHEMATICA a[n_] := DivisorSigma[0, n] - PrimeOmega[n] - PrimeNu[n]; Array[a, 105] (* Jean-François Alcover, Jun 21 2018 *) PROG (PARI) A135767(n)=numdiv(n)-omega(n)-bigomega(n) CROSSREFS Cf. A102466, A102467 ; A001037, A107847. Sequence in context: A248639 A293959 A333146 * A208575 A070203 A070201 Adjacent sequences: A135764 A135765 A135766 * A135768 A135769 A135770 KEYWORD easy,nonn AUTHOR M. F. Hasler, Jan 14 2008 STATUS approved
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# Programming Assignment 0 – Recursion
\$25.00
Category:
CS260 – Programming Assignment 0 – Recursion
The purpose of this assignment is to give you some practice working with recursion.
Overview
For this assignment you’ll write a program to find solutions for the Knights Tour. using brute force (just
trying all possible solutions)
The program should prompt the user for the following:
● a board size (square) from 3×3 to 7×7. (8×8’s and above take too long using brute force)
● a starting position (row,col; 1,1 is the upper left corner)
Sample Output
It should show all the solutions. A sample run would look something like this:
Welcome to the Knights Tour solver!
Board size (3-7): 5
Starting position (row, col): 5,1
Thinking….
Solution #1:
23 10 19 14 25
18 05 24 09 20
11 22 13 04 15
06 17 02 21 08
01 12 07 16 03
Details
You can define a fixed size array and just use the portion you want. It should be initialized with a number
that is not a valid move (like 0 or 99);
Each instance of the recursive function call needs it’s own copy of the board-to-date. Since arrays are
always passed by reference, and we need to pass by value (to get our own copy), the simplest solution is
to define the board array inside a class (or structure) because classes can be passed by value.
Assuming you defined a board struct or class (with all public members) which contains a twodimensional array representing the board and an integer representing the board size, the recursive function
description and prototype will probably look like these:
//================================================================
// Description:
// Recursively solves the knights tour using brute force.
// Prints any solutions if finds.
// Args:
// board (I/O) – the board we’re working with
// (board with previous moves and size)
// row (I) – the row we’re going to attempt to place the knight on this
move.
// col (I) – the column we’re going to attempt place the knight on this
move.
// currentMoveNumber (I) – the move we’re making
// (1=first placement, 16=last placement on 4×4 board)
// Note: row and col may be invalid (<0 or >=boardsize)
// Returns:
// The number of solutions the given board and move leads to
//===============================================================
int solveKnightsTour(Board board, int row, int col, int currentMoveNum=1);
Very Important Stuff
Program should be well written and function properly. The best one(s) gets a cookie. (note: image is for
illustrative purposes only; actual cookie may look different and be considerably smaller).
All programs should follow the class’s Coding Conventions
Submit the following: | 629 | 2,647 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2022-33 | latest | en | 0.805755 |
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# The complete elliptic integral of the first kind is the
ISBN: 9780321982384 49
## Solution for problem 14E Chapter 1.10
Linear Algebra and Its Applications | 5th Edition
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Problem 14E
The complete elliptic integral of the first kind is the integral ,where 0 < k < 1 is constant.a. Show that the first four terms of the binomial series for are .b. From part (a) and the reduction integral Formula 67 at the back of the book, show that .
Step-by-Step Solution:
Step 1 of 3
ME5331 Lecture 15 Line Integrals Example 1. Evaluate the following integral along each of the paths shown in the figure. Solution: We have four paths of integration: (a) along the parabolic arc that joins the points A and B; (b) along the straight line AB; (c) along the path APB; (d) along the path AQB. Since we need to integrate with respect to x, the limits of integration for all paths are from A, to B, First we need to express in terms of . To do this, we recall from the definition of a line integral that the integrand is always evaluated along the path of integration. (a) Along the parabolic arc that joins the points A and B, we have . Thus, or
Step 2 of 3
Step 3 of 3
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The complete elliptic integral of the first kind is the | 474 | 1,853 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2021-25 | latest | en | 0.875027 |
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# edX: Six Sigma: Analyse, Improve, Control
with Martin Grunow and Holly Ott
Learn how to analyse data with the Six Sigma methodology using inferential statistical techniques to determine confidence intervals and to test hypotheses based on sample data. You will also review cause and effect techniques for root cause analysis.
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Test the significance of experimental results using an analysis of variance.
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Cover Statistical Process Control & Control Chart Theory and construct X-bar and R Charts for long term parameter monitoring and control.
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Review other control charts, including p-and C- charts and I/MR, and EWMA Charts and the Control and Reponse Plan for Six Sigma projects.
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Introduce several important tools used in process deviations, including Failure Modes and Effects Analysis, 8 Disciplines and 5 Whys, as well as discuss techniques for Design for Six Sigma (DFSS).
Week 8: Six Sigma Scenario
Follow a full Six-Sigma project with a Master Black Belt, implementing all phases of the DMAIC process improvement cycle.
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## Review for edX's Six Sigma: Analyse, Improve, Control 5.0 Based on 1 reviews
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Totally recommended! | 966 | 4,594 | {"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} | 2.640625 | 3 | CC-MAIN-2018-09 | latest | en | 0.884883 |
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# Multiplication and Division of Radicals
## Rationalize the denominator
Levels are CK-12's student achievement levels.
Basic Students matched to this level have a partial mastery of prerequisite knowledge and skills fundamental for proficient work.
At Grade (Proficient) Students matched to this level have demonstrated competency over challenging subject matter, including subject matter knowledge, application of such knowledge to real-world situations, and analytical skills appropriate to subject matter.
Advanced Students matched to this level are ready for material that requires superior performance and mastery.
• Read
## Multiplying and Dividing Square Roots
by CK-12 //at grade
This lesson covers dividing square roots.
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• PLIX
## Multiplication and Division of Radicals: Totally Radical Dude's Height
by CK-12 //at grade
Multiplication and Division of Radicals Interactive
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## Multiplication and Division of Radicals Quiz
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Quiz for Multiplication and Division of Radicals.
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## Multiplication and Division of Radicals Discussion Questions
by CK-12 //at grade
A list of student-submitted discussion questions for Multiplication and Division of Radicals.
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• Real World Application
## Pentagon
by CK-12 //at grade
Students apply their understanding of simplifying and adding radical expressions to looking at the sides of the United States' Pentagon building.
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• Study Guide
## Irrational Numbers and Radicals Study Guide
by CK-12 //at grade
This study guide looks at square and cube roots, radical expressions, rationalization, solving radical equations, and graphing radical functions and equations.
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Please wait... | 492 | 2,276 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2017-04 | longest | en | 0.882565 |
http://www.kenyaplex.com/questionpapers/1795-computer-architecture-and-organization.aspx | 1,406,013,657,000,000,000 | text/html | crawl-data/CC-MAIN-2014-23/segments/1405997857710.17/warc/CC-MAIN-20140722025737-00005-ip-10-33-131-23.ec2.internal.warc.gz | 859,699,455 | 7,196 | # Kenyatta University Bachelor Of Science In Telecommunication And Information Technology (Telecommunication) Computer Architecture And Organization Question Paper
Exam Name: Computer Architecture And Organization
Institution/Board: Kenyatta University
Exam Year:2009
KENYATTA UNIVERSITY
UNIVERSITY EXAMINATIONS 2009/2010
FIRST SEMESTER EXAMINATION FOR THE DEGREE OF BACHELOR OF
SCIENCE IN TELECOMMUNICATION AND INFORMATION TECHNOLOGY
SPH 312: COMPUTER ARCHITECTURE AND ORGANIZATION
DATE: Wednesday 23rd December, 2009 TIME: 11.00 a.m. – 1.00 p.m.
INSTRUCTIONS
Answer question ONE and any other TWO questions. Question ONE carries 30 marks
while each of the other TWO questions carries 20 marks. The 3085 instruction set is
appended.
1.
(a)
(i)
Perform the following arithmetic
I
CDF + ABC
[2 marks]
II
00001000
00000011 [2
marks]
(ii)
Convert
(15.625)10
into
binary [2
marks]
(b)
(i)
State De Morgan’s theorem of two variables.
[2 marks]
(ii)
Obtain the output logic expression of the given logic diagram.
[3 marks]
Page 1 of 6
(iii)
Using De Morgan’s and Boolean theorem’s simplify the output
logic expression in (ii) above and draw a logic circuit for the
simplified
function.
[3
marks]
c)
(i)
What
is
a
logic
gate?
[1
mark]
(ii)
State
two
available
universal logic gates.
[2 marks]
(iii)
Show using a diagram how you can use a NAND gate to
implement an AND function and an OR function. [4 marks]
d)
(i)
What
is
a
microprocessor?
[1
mark]
(ii)
Explain the functions of the following 8085 microprocessor
components.
[3
marks]
I.
Stack
pointer
II.
Timing
and
control
III.
Program
counter
e)
(i)
What
is
memory?
[1
mark]
(ii)
State two groups of classifying memories. State one example of
each
class
of
memory.
[2
marks]
(iii)
Differentiate between algorithm and program.
[2 marks]
Q2.
a)
(i)
Draw a logic symbol for a NOR gate.
[1 mark]
(ii)
Outline the procedure that should be followed if a logic function is
to be implemented using NOR gates only.
[3 marks]
(iii)
Manipulate the given logic expression into a form which can be
implemented using only NOR gates and draw the logic diagram of
the
resulting
function.
[5
marks]
Y
=
B
A
C +
AC
+ B
b)
(i)
Design a two bit comparator circuit that will produce a logic output
1 when the two input signals are identical.
[3 marks]
(ii) Manipulate
the
logic
expression of the two bits circuit in (i) into a
form which can be implemented using NAND gates only. Draw
the circuit diagram of the manipulated function.
[3 marks]
Page 2 of 6
c)
(i)
Prepare K-map for the given Boolean function
Y = B + AC
[2 marks]
(ii)
Write down the simplest logic expression for the given K-map.
[3 marks]
Q3.
(a)
State and explain two modes of operation of 8255 PP1.
[2 marks]
(b)
Present the control word format for the 8255 PPI.
[3 marks]
(c)
(i)
A Microprocessor based system uses 8255 PPI as its I/O device. If
this system is to be used to read bit pattern at port C and output the
same to port A and B continuously and endlessly write an
assembly language program to perform this operation using
appropriate 8085 instruction set mnemonics. Take the first
memory location to be 6EFDH and use delay constant of 25510 in
register pair B.
[6 marks]
(ii)
State the memory address of the last byte of the program in (i)
above.
[1
mark]
(d)
(i)
Hand assemble the following using 8055 instruction set assuming
that the first memory location is 52EFH.
[6 marks]
Page 3 of 6
START:
LXI SP, 0F00H
LXI
H,
065DH
MVI C, 6BH
XRA
A
MOVB, A
AGAIN: CALL
DEL
INX H
DCR C
JNZ
AGAIN
LX1H,
00FFH
CALL DEL
MOV
A,
B
JMP
START
DEL:
LX1
D,
EDFFH
BACK: DCX
D
MOV
A,
E
OR
A,
D
JNZ
BACK:
RET.
(ii)
State the address of the following in the hand assemble program in
(i)
above.
[2
marks]
I.
AGAIN
label
II.
JMP
instruction
Q4.
(a)
What
is
analog
interfacing?
[1
mark]
(b)
Explain why input/output devices cannot be interfaced directly to a
microprocessor. [2
marks]
(c)
State and explain two types of interfaces.
[4 marks]
Page 4 of 6
(d)
Write an algorithm for adding odd numbers between 20 and 50 and
display (show) sum in port 01H using the 8085 instruction set. Develop
(i)
Outline the procedure followed, use registers A, B and C.
[4 marks]
(ii)
Assuming that the first memory location is 25EEH, write an
assembly language program to perform this operation using
appropriate 8085 mnemonics. Show also the memory contents in
hex
codes.
[6
marks]
(iii)
Simplify your program in (ii) using a flow chart.
[3 marks]
Q5.
(a)
State and explain the functions of the four fields of assembly language
program.
[4 marks]
(b)
State and explain three classifications of 8085 instruction sizes.
[3 marks]
(c)
Write a program of subtracting two numbers; 58H from 82H using 8085
assembly language and display the results in part 02H.
[3 marks]
(d)
Consider the following 8085 assembly language program of a
microprocessor-based system using 8255 PPI.
MV1A, 80H
OUT
03H
BACK:
MVI A,
55H
OUT
00H
OUT
01H
OUT
02H
CALL
SUBRONTINE
MVIA,
AAH
OUT
00H
OUT
01H | 1,447 | 5,007 | {"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} | 3.109375 | 3 | CC-MAIN-2014-23 | latest | en | 0.76064 |
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# Perimeter and Area BUNDLE
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Perimeter and Area BUNDLE of Activities - Your students will absolutely LOVE digging into area and perimeter with this bundle of FIVE engaging area and perimeter resources.
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Reported resources will be reviewed by our team. Report this resource to let us know if this resource violates TPT’s content guidelines. | 491 | 2,143 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-14 | latest | en | 0.864905 |
https://www.sarthaks.com/3183909/average-monthly-expenditure-family-first-four-months-is-2570-next-three-months-2490 | 1,679,509,222,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296944452.74/warc/CC-MAIN-20230322180852-20230322210852-00137.warc.gz | 1,103,899,874 | 15,798 | # The average monthly expenditure of a family for the first four months is ₹ 2570, for the next three months ₹ 2490
+1 vote
342 views
edited
The average monthly expenditure of a family for the first four months is ₹ 2570, for the next three months ₹ 2490 and for the last five months ₹ 3030. If the family saves ₹ 5320 during the whole year, the average monthly income of the family during the year is
by (42.5k points)
Total annual expenditure of the family = Rs. (4 x 2570 + 3 x 2490 + 5 x 3030)
= Rs. (10280 + 7470 + 15150)
= Rs. 32900
Total income = Rs. (32900 + 5320) = Rs. 38220
∴ Required average monthly income = $\frac{3880}{12}$ = Rs. 3185 | 219 | 659 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2023-14 | latest | en | 0.891538 |
http://alibenyondesigns.com/business/roman-numeral-c-means-what.php | 1,555,901,639,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578533774.1/warc/CC-MAIN-20190422015736-20190422041736-00484.warc.gz | 6,897,658 | 5,073 | # Roman numeral c means what
Posted on | by
The Roman numerals for one to nine are: I, II, III, IV, V, VI, VII, VIII, IX. Roman numerals may be written in lowercase letters, though they appear more commonly in capitals. The numbers one through ten are written I, II, III, IV, V, VI, VII, VIII, IX, and X. There's an article here from the American Journal of Archaeology that goes into great detail about various theories of the origins of Roman. Roman numerals originated, as the name might suggest, in ancient Rome. There are seven basic symbols: I, V, X, L, C, D and M. The first.
The numeric system represented by Roman numerals originated in ancient Rome and = "One hundred and sixty" (C+LX) = CLX; = "Two hundred and .. its own Roman numeral, but the word nulla (the Latin word meaning " none"). Here is a chart that has all the Roman Numerals, and what they stand for. The system is Roman Numeral, Value. I, 1. V, 5. X, L, C, D, M, 1, Many civilizations used other means to denote numbers. For example, the Romans represented numbers using the numerals I, V, X, L, C, D, and M. These.
Roman numerals consist of a combination of the I, V, X, L, C, D and M letters. numbers to expecting puzzlers to know that a Roman numeral M stands for one. Learn how to convert C from roman numerals to arabic numerals, and a lot more, at alibenyondesigns.com Type a number (like 14) or a Roman number (like XIV), and click 'Convert'. Script courtesy C, C stands for centum, the Latin word for A centurion led Roman Numerals. Try our Roman Numeral Challenge. Several rules apply for subtraction: (a) only subtract powers of ten (I, X, or C, but not V or L); (b) only. | 443 | 1,661 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2019-18 | latest | en | 0.880647 |
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Materials 1 - 10 shown of 14 results
### Visual Fractions
The purpose of Visual Fractions is to picture fractions and the operations on them.
Material Type: Tutorial
Author: Richard Rand
Date Added: Mar 09, 2005 Date Modified: Nov 08, 2013
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### Abacus
Complete sim of an abacus, and it even translates the bead positions into a numerical representation. Includes tutorials on... see more
Material Type: Tutorial
Author: Luis Fernandes
Date Added: Nov 11, 1997 Date Modified: Jan 28, 2013
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### MathsNet: Interactive Fractions
This is an interactive tutorial which addresses all aspects of fractions in a visual manner.
Material Type: Tutorial
Author: Bryan Dye
Date Added: Mar 09, 2005 Date Modified: Nov 27, 2005
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### West Texas A&M Beginning Algebra
Contains a multitude of tutorials, practice problems, and practice tests for beginning algebra topics.
Material Type: Tutorial
Author: Kim Peppard
Date Added: Nov 10, 2004 Date Modified: Sep 20, 2006
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### CMC-Developmental Math MAT055
Materials for student self-study tutorial for students that did not place into college math
Material Type: Tutorial
Author: CCCS COETC Grant
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### DevEd Math Prep ACCUPLACER
Materials for student self-study tutorial for students that did not place into college math
Material Type: Tutorial
Author: CCCS COETC Grant, Marilyn Smith
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### DevEd Math Prep Introduction
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### DevEd Math Prep Preparing to Take the ACCUPLACER
Materials for student self-study tutorial for students that did not place into college math
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### DevEd Math Prep Scores on ACCUPLACER
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### DevEd Math Prep Tutorial 1
Materials for student self-study tutorial for students that did not place into college math
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``` | 1,860 | 8,237 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2014-10 | latest | en | 0.837327 |
https://nl.mathworks.com/matlabcentral/answers/214587-basic-monte-carlo-question-area-of-circle | 1,675,808,047,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500641.25/warc/CC-MAIN-20230207201702-20230207231702-00125.warc.gz | 442,588,580 | 34,357 | # basic monte carlo question: area of circle
90 views (last 30 days)
Eric on 1 May 2015
Edited: hosein bashi on 30 Jul 2018
Question: Use monte carlo method to find area of circle radius 5 inside 7x7 square.
So this is the code I've put together so far to determine how many points land inside or on the circle (hits). My output for hits is 0 but I can't for the life of me figure out why, to me everything seems fine but clearly isn't.
clear;
N= 1000; % number of points generated
a = -7;
b = 7;
hits= 0;
for k = 1:N
x = a + (b-a).*rand(N,1);
y = a + (b-a).*rand(N,1);
% Count it if it is in/on the circle
hits = hits + 1;
end
end
disp(hits)
pfb on 1 May 2015
Edited: pfb on 1 May 2015
You are mixing things up. Your code is part vectorized, part scalar.
You have a loop over N, but then at each iteration you generate N points (so you're generating N^2 points overall). The variables x, y, and radii are vectors.
Note that
[6 7 4 2 8 5] <=5
gives
[0 0 1 1 0 1]
Therefore condition radii<=5 has a very low probability to be met ( all of the N radii should be less than 5).
As a result hits is (almost) never accumulated.
Your code is structured for a scalar radius. It would work if you substitute the lines
x = a + (b-a).*rand(N,1);
y = a + (b-a).*rand(N,1);
with
x = a + (b-a).*rand;
y = a + (b-a).*rand;
In this case hits will be a number between 0 and N, and hits/N is going to be proportional to the ratio of the areas of the circle and square, pi*(5/14)^2 (note that you can use simple power operator ^ instead of elementwise .^)
You can avoid the loop altogether using the vectorized variables.
N= 1000; % number of points generated
a = -7;
b = 7;
x = a + (b-a).*rand(N,1);
y = a + (b-a).*rand(N,1);
Eric on 2 May 2015
Thanks both really helped a bunch.
How would I go about plotting the graph so that the hits are one colour and the rest are another?
I tried the following but I just get a linear line.
x = linspace(-7,7);
y = linspace(-7,7);
figure(1);
plot(x,y);
xlabel('x');
ylabel('y');
axis square
grid on;
title('Area of Circle');
pfb on 2 May 2015
Edited: pfb on 2 May 2015
ok, then you can write the function sum with a loop. After creating the vector i (which you need for the plotting)
hits = 0;
for j = 1:N
hits=hits+i(j);
end
misses= N-hits;
If you're not allowed to use logical indexing in the plot you can pretend you do not know it and do this
% vectors for the hits
xh = zeros(1,hits);
yh = xh;
% vector for the misses
xm = zeros(1,misses);
ym = xm;
% counters
mc=1;
hc=1;
for j=1:N
if(i(j))
% put the point in the hits vector
xh(hc)=x(j);
yh(hc)=y(j);
% increment the hit counter
hc=hc+1;
else
% put the point in the misses vector
xm(mc)=x(j);
ym(mc)=y(j);
% increment the miss counter
mc=mc+1;
end
end
plot(xh,yh,'g.');
hold
plot(xm,ym,'r.');
etcetera
Eric on 2 May 2015
Thanks a lot, appreciate it.
pfb on 2 May 2015
you're welcome. Note that there was an error in my code above (in one instance I wrote hits instead of misses). Now it should be ok.
munesh pal on 27 Feb 2018
Calculate the area of the circle using Monte Carlo Simulation
clc
clear all
close all
r=2;
c_x=7;
c_y=7;
%%position of the outer rectangle
p_x=c_x-r;
p_y=c_y-r;
pos=[p_x,p_y,r^2,r^2]
%%random number generation within the sqaure
N=1000;
a=p_x;
b=p_x+(2*r);
x = a + (b-a).*rand(N,1);
y = a + (b-a).*rand(N,1);
hits = sum(i);
misses = N-hits;
plot(x(i),y(i),'.g');
hold;
plot(x(~i),y(~i),'.r');
rectangle('Position',pos,'Curvature',[1 1])
rectangle('Position',pos,'EdgeColor','r')
actual_a=22/7*r^2;
ttl = sprintf('Estimate Actual Area of circle: %1.3f, Area of the circle: %1.3f',actual_a,hits/N*(2*r)^2);
title(ttl);
axis equal
hosein bashi on 30 Jul 2018
Edited: hosein bashi on 30 Jul 2018
can you explain please why we use random numbers? why we don't assume a grid and put our numbers in the middle of the cells? it would be more uniform in this way. | 1,259 | 3,862 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2023-06 | longest | en | 0.865557 |
https://www.freecodecamp.org/forum/t/need-help-with-my-sum-all-primes-code/11296 | 1,550,638,777,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247494449.56/warc/CC-MAIN-20190220044622-20190220070622-00072.warc.gz | 828,734,574 | 5,028 | # Need Help With My 'Sum All Primes' Code
Need Help With My 'Sum All Primes' Code
0
#1
**
## !SPOILERS AHOY!
**
This is in regards to the ‘Sum All Primes’ intermediate algorithm, which asks you to write a function that sums all primes up to the given number.
My code looks like it’s returning the correct sums for all the primes that I’ve fed to it… except for the final test number, which is 977!
Could anyone take a look and explain to me why my code seems to work only up to a certain point? Thanks in advance!
function sumPrimes(num) {
var sum = 5;
var current = 3;
while(current < num){
current+=1;
if(current%2!==0 && current%3!==0){
sum+=current;
}
}
return sum;
}
#2
Also, this is a recurring problem for me with “Sum of All Odd Fibonacci” and “Smallest Common Multiple”–my code for both of these algorithms works with all numbers up to the final test. It’s driving me insane. Partly with curiosity. Mostly with frustration.
#3
Your current code works because you are hardcoding some values that makes it work with sumPrime(10)
For starter, you should ask yourself why you needed “sum” to have a value of 5 by default. This is a pointer to your current problem.
Also, a prime number is a number that can only be divided by 1, and itself.
You could approach this problem differently, think of a way to iterate through all numbers and comparing each of them together, This could be done with a nested for loop.
#4
You seem to think 25 is prime. At least according to your code :~)
Your algorithm takes 25 and since it’s not divisible by 2 nor by 3, it adds it happily to the sum of primes. There are an infinite amount of other numbers that are not divisible by 2 and 3, yet are not prime. Try it out in the console of you browser and add a `console.log(current);` for every ‘prime’ it finds.I think you’ll quickly agree that your approach is too simple.
There are very very old but efficient algorithms for finding primes, check out Wikipedia. If you find that too complex (it probably looks a bit daunting at first, but really isn’t), you can try to generalize your algorithm and brute force the result.
Also I’d suggest you use more than one function to separate concerns, maybe one for finding primes and one for summing them up? It’s easier to think about problems if you can split them up into smaller, less complex ones.
#5
Thanks for the input, guys! These math-based algorithms were pretty sharp detour from my usual practice, so I think a lot of it is just getting used to the thinking. | 602 | 2,525 | {"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} | 3.75 | 4 | CC-MAIN-2019-09 | latest | en | 0.936546 |
https://studylib.net/doc/9542479/data-and-process-modeling | 1,653,613,318,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662627464.60/warc/CC-MAIN-20220526224902-20220527014902-00050.warc.gz | 635,802,174 | 15,694 | # Data and Process Modeling
```
Class CIS 250
Fall 2013
Chapter 5
Data and Process Modeling
Describe data and process modeling
concepts and tools, including data flow
diagrams, a data dictionary, and process
descriptions
Describe the symbols used in data flow
diagrams and explain the rules for their use
Draw data flow diagrams in a sequence,
from general to specific
3
Explain how to level and balance a set of data
flow diagrams
Describe how a data dictionary is used and
what it contains
Use process description tools, including
structured English, decision tables, and
decision trees
Describe the relationship between logical and
physical models
4
Systems analysts use many graphical techniques
to describe an information system
A data flow diagram (DFD) uses various symbols
to show how the system transforms input data
into useful information
5
A data flow diagram (DFD) shows how data
moves through an information system but does
not show program logic or processing steps
A set of DFDs provides a logical model that
shows what the system does, not how it does it
6
DFD Symbols
◦ Four basic symbols
◦ Gane & Sarson
used in text
◦ Yourdon also popular
FIGURE 5-3 Data flow diagram symbols, symbol names, and
examples of the Gane and Sarson and Yourdon symbol sets
7
Process Symbol
• Must have at least one
input and at least one
output
that transforms the data
• Process name identifies its
function (verb)
• Process number does not
signify precedence
• Examples: “print bill” or
8
Data flow symbol
◦ Represents one or more
data items
◦ The symbol for a data
flow is a line with a
single or double
FIGURE 5-5 Examples of correct combinations of data flow and process symbols
9
Data flow symbol
◦ Spontaneous generation
(Process must act on input)
◦ Black holes
◦ Gray holes
FIGURE 5-6 Examples of incorrect combinations of data flow and process
symbols. APPLY INSURANCE PREMIUM has no input and is called a
spontaneous generation process. CALCULATE GROSS PAY has no outputs and
is called a black hole process. CALCULATE GRADE has an input that is
obviously unable to produce the output. This process is called a gray hole
10
Data Store symbol
• Represent data that the system
stores
•
•
A DFD does not show the detailed
contents of a data store — the specific
structure and data elements are defined
in the data dictionary
A data store must be connected to a
process with a data flow
11
FIGURE 5-8 Examples of incorrect
uses of data store symbols: Two data
stores cannot be connected by a data
flow without an intervening process,
and each data store should have an
outgoing and incoming data flow
FIGURE 5-7 Examples of correct
uses of data store symbols in a data
flow diagram
12
• Shows how the system interfaces
with the outside world
• A DFD shows only external entities
that provide data to the system or
• DFD entities also are called
terminators because they are data
origins or final destinations
• Each entity must be connected to a
process by a data flow
13
FIGURE 5-10 Examples of incorrect
uses of external entities. An external
entity must be connected by a data
flow to a process, and not directly to a
data store or to another external entity
FIGURE 5-9 Examples of correct uses of
external entities in a data flow diagram
14
Create a graphical model of the information
system based on your fact-finding results
◦ First, you will review a set of guidelines for
drawing DFDs
◦ Then you will learn how to apply these guidelines
and create a set of DFDs using a three-step
process
15
Keep in mind:
◦ All flow lines must
be labeled
◦ Large processes can
be broken down into
smaller components
FIGURE 5-11 Examples of correct
and incorrect uses of data flows
16
Guidelines for Drawing DFDs
◦ Draw the context diagram so that it fits on one page
◦ Use the name of the information system as the
process name in the context diagram
◦ Use unique names within each set of symbols
◦ Do not cross lines
◦ Provide a unique name and reference number for
each process
◦ Ensure that the model is accurate, easy to
understand, and meets the needs of its users
17
Step 1:
Draw a
Context
Diagram
FIGURE 5-13 Context
diagram DFD for an order
system
18
Step 2: Draw a
Diagram 0 DFD
◦ If same data flows
in both directions,
you can use a
arrow
◦ Diagram 0 is an
exploded view of
process 0
◦ Parent diagram
◦ Child diagram
◦ Functional
primitive
FIGURE 5-16 Diagram 0
DFD for the order system
19
Step 3:
Draw the
Lower Level
Diagrams
FIGURE 5-17 Diagram 1
DFD shows details of the
FILLORDER process in
the order system
20
Must use leveling
and balancing
techniques
Leveling examples
◦ Uses a series of
increasingly detailed
DFDs to describe an
information
system
◦ Exploding,
partitioning, or
decomposing
FIGURE 5-18 This diagram does
not show the symbols that connect
to data flows entering or leaving
FILL ORDER on the context
diagram
21
FIGURE 5-19 The order system diagram 0 is shown at the
top of the figure, and exploded diagram 3 DFD (for the
APPLY PAYMENT process) is shown at the bottom. The two
DFDs are balanced because the child diagram at the bottom
has the same input and output flows as the parent process 3
shown at the top
22
FIGURE 5-20 Example of a parent
DFD diagram, showing process 0 as a
black box
FIGURE 5-21 In the next level of detail,
the process 0 black box reveals three
processes, two data stores, and four
internal data flows — all of which are
shown inside the dashed line
23
•
•
•
A data dictionary, or data repository, is a
central storehouse of information about the
system’s data
An analyst uses the data dictionary to
collect, document, and organize specific
Also defines and describes all data elements
and meaningful combinations of data
elements
24
A data element, also called a data item or
field, is the smallest piece of data that has
meaning
Data elements are combined into records,
also called data structures
A record is a meaningful combination of
related data elements that is included in a
data flow or retained in a data store
25
Using CASE Tools for Documentation
◦ The more complex the system, the more difficult it
is to maintain full and accurate documentation
◦ Modern CASE tools simplify the task
◦ A CASE repository ensures data consistency
◦ The CASE tools in Part B of the Systems Analyst’s
and processes
four-part Toolkit that follows Chapter 12
26
Documenting the Data Elements
◦ You must document every data element in the data
dictionary
◦ The objective is the same: to provide clear,
comprehensive information about the data and
processes that make up the system
27
FIGURE 5-23 Using an online documentation form, the analyst
has recorded information for a data element named SOCIAL
SECURITY NUMBER. Later, the analyst will create a data
dictionary entry using a CASE tool
28
Documenting the Data
Elements
◦ Data element name and
label
◦ Alias
◦ Type and length
◦ Default value
◦ Acceptable values - Domain
and validity rules
◦ Source
◦ Security
◦ Responsible user(s)
FIGURE 5-24 A Visible Analyst screen describes
the data element named SOCIAL SECURITY
NUMBER. Notice that many of the items were
entered from the online form shown in Figure 5-23
29
Documenting the
Data Flows
◦
◦
◦
◦
◦
◦
◦
Data flow name or label
Description
Alternate name(s)
Origin
Destination
Record
Volume and frequency
FIGURE 5-25 In the upper screen, an analyst has entered four
items of information in an online documentation form. The lower
screen shows the same four items entered into a Visible Analyst
data dictionary form
30
Documenting the Data
Stores
◦
◦
◦
◦
◦
Data store name or label
Description
Alternate name(s)
Attributes
Volume and frequency
FIGURE 5-26 Visible Analyst screen that documents a
data store named IN STOCK
31
Documenting the
Processes
◦
◦
◦
◦
Process name or label
Description
Process number
Process description
FIGURE 5-27 Visible Analyst screen that describes a
process named VERIFY ORDER
32
Documenting the
Entities
◦
◦
◦
◦
◦
Entity name
Description
Alternate name(s)
Input data flows
Output data flows
FIGURE 5-28 Visible Analyst screen that documents
an external entity named WAREHOUSE
33
Documenting
the Records
◦ Record or data
structure name
◦ Definition or
description
◦ Alternate name(s)
◦ Attributes
FIGURE 5-29 Visible Analyst screen that documents a
record, or data structure named CREDIT STATUS
34
•
Data Dictionary Reports
– Many valuable reports
• An alphabetized list of all data elements by name
• A report describing each data element and indicating
the user or department that is responsible for data
entry, updating, or deletion
• A report of all data flows and data stores that use a
particular data element
• Detailed reports showing all characteristics of data
elements, records, data flows, processes, or any other
selected item stored in the data
35
Typical process description tools include
structured English, decision tables, and
decision trees
Process description tools also can be used in
object-oriented development
◦ O-O programmers use different terminology. They
create the same kind of modular coding structures,
except that the processes, or methods, are stored
inside the objects, rather than as separate
components
36
Modular Design
◦ Based on
combinations of
three logical
structures,
sometimes called
control structures,
which serve as
building blocks for
the process
FIGURE 5-30 Sequence structure
FIGURE 5-31 Selection structure
Sequence
Selection
Iteration - looping
FIGURE 5-32 Iteration
structure
37
Structured English
◦ Must conform to the
following rules
Use only the three
building blocks of
sequence, selection,
and iteration
Use indentation for
Use a limited
vocabulary, including
standard terms used
in the data dictionary
and specific words
that describe the
processing rules
FIGURE 5-33 The VERIFY ORDER process description
includes logical rules and a structured English version of
the policy. Notice the alignment and indentation of the
logic statements
38
Decision Tables
◦ Shows a logical structure, with all possible
combinations of conditions and resulting actions
◦ It is important to consider every possible
outcome to ensure that you have overlooked
nothing
◦ The number of rules doubles each time you add
a condition
◦ Can have more than two possible outcomes
◦ Often are the best way to describe a complex set
of conditions
39
FIGURE 5-34 The Verify Order business process has two conditions. For an
order to be accepted, the product must be in stock and the customer must
have an acceptable credit status
FIGURE 5-35 Example of a simple decision table showing the processing logic of
the VERIFY ORDER process
40
FIGURE 5-36 A third condition has been added to the Verify Order business process. For an order
to be accepted, the product must be in stock and the customer must have an acceptable credit
status. However, the credit manager now has the authority to waive the credit status requirement
FIGURE 5-37 This table is based on the Verify Order conditions shown in Figure
5-36. With three conditions, there are eight possible combinations, or rules
41
FIGURE 5-38 In the first table, dashes have been added to indicate
that a condition is not relevant. In the second version, rules have
been combined. Notice that in final version, only four rules remain.
These rules document the logic, and will be transformed into
program code when the system is developed
42
FIGURE 5-39 A sales promotion policy with three conditions. Notice that the first statement contains
two separate conditions – one for the 5% discount, and another for the additional discount
FIGURE 5-40 This decision table is based on the sales promotion policy in
Figure 5-39. This is the initial version of the table, before simplification
43
FIGURE 5-41 In this version, dashes have been added to indicate that a condition is not
relevant. At this point, it appears that several rules can be combined
44
Decision Trees
◦ Graphical representation of the conditions,
actions, and rules found in a decision table
◦ Show the logic structure in a horizontal form that
resembles a tree with the roots at the left and the
branches to the right
◦ Decision trees and decision tables provide the same
results, but in different forms
FIGURE 5-42 This example is based on the same Sales
Promotion Policy shown in the decision tables in Figures
5-40 and 5-41 on the previous page. Like a decision table,
a decision tree shows all combinations of conditions and
outcomes. The main difference is the graphical format,
which many viewers find easier to interpret
45
While structured analysis tools are used to
develop a logical model for a new information
system, such tools also can be used to
develop physical models of an information
system
A physical model shows how the system’s
requirements are implemented
46
Sequence of Models
◦ Many systems analysts create a physical
model of the current system and then
develop a logical model of the current system
before tackling a logical model of the new
system
◦ Performing that extra step allows them to
understand the current system better
47
Four-Model Approach
◦ Develop
A
A
A
A
physical model of the current system
logical model of the current system
logical model of the new system
physical model of the new system
◦ The only disadvantage of the four-model
approach is the added time and cost
48
•
•
•
During data and process modeling, a
systems analyst develops graphical models
to show how the system transforms data into
useful information
The end product of data and process
modeling is a logical model that will support
business operations and meet user needs
Data and process modeling involves three
main tools: data flow diagrams, a data
dictionary, and process descriptions
49
Data flow diagrams (DFDs) graphically show
the movement and transformation of data in
the information system
DFDs use four symbols
A set of DFDs is like a pyramid with the
context diagram at the top
The data dictionary is the central
documentation tool for structured analysis
50
•
•
Each functional primitive process is
documented using structured English,
decision tables, and decision trees
Structured analysis tools can be used to
develop a logical model during one systems
analysis phase, and a physical model during
the systems design phase
51
Assignment # 5
Chapter # 5
Class Work # 5
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# hw2_solutions - MS&amp;E 303 Fall 2003 Homework#2...
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Unformatted text preview: MS&amp;E 303 Fall 2003 Homework #2 Due Wednesday, September 17 1. Consider the exact differential for U ( S,V ) where dU = T dS- P dV , and its Legendre transforms H = U + PV , A = U- TS and G = U- TS + PV . For all but the first part, reduce the derivatives to the simplest form involving only the coordinates P,V,T,S , the physical parameters and (expansion coefficient and the isothermal compressibility), and the constant pressure specific heat C P (if you find C V occurs, just use the result derived in class where C P = C V + 2 V T . These problems though should not require either heat capacity derivatives.) As this is pure math, there is still no need to understand the meaning of S . (a) Write the differentials dH , dA , and dG . H = U + PV dH = T dS + V dP A = U- TS dA =- S dT- P dV H = H- TS dG =- S dT + V dP (b) A T P Use the generalized chain rule to change the path from P to V corresponding to the A potential. Then left only with derivatives of the equation of state (EOS). A T P = A T V + A V T V T P =- S- P V T P =- S- PV (c) U P T This problem uses all three rules. The variables P,T correspond to neither of the natural variables of U . Have to first use the chain rule to switch out P for V , then the generalized chain rule to change the parth from T to S , a Maxwell relation based on dA to get rid of the derivative involving S , and finally a triplet rule to get to derivatives of the EOS. U P T = U V T V P T = U V S + U S V S V T V P T = - P + T P T V V P T = - P + T - V T P V P T V P T =- P V P T- T V T P = PV- V T 1 (d) T P G Relatively direct application of the triplet rule. T P G =- G P T G T P = V S (e) T V A Again, a simple triplet rule. T V A =- A V T A T V =- P S (f) Show that 2 G P 2 T =- 1 2 A V 2 T Only challenge here is working from left to the right. A very non-trivial relationship between the potentials that results from the mathematical linkages of the Legendre transformations. 2 G P 2 T = G P T P T = V P T = 1 P V T = 1 - A V T V T =- 1 2 A V 2 T 2. Simple values and numbers time to learn/memorize these important constants and typical values. These are considered fundamentals and it will be assumed that you know these values on exams, in life, etc. Those that are not known, you will need to learn where to find in the library (or as abslolute last resort on the WEB)....
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# Describe the key characteristics of the graphs of rational functions of the form f(x)=(ax+b)/(cx+d). Explain how you can determine these characteristi
Rational functions
Describe the key characteristics of the graphs of rational functions of the form $$\displaystyle{f{{\left({x}\right)}}}=\frac{{{a}{x}+{b}}}{{{c}{x}+{d}}}$$. Explain how you can determine these characteristics using the equations of the functions. In what ways are the graphs of all the functions in this family alike? In what ways are they different? Use examples in your comparison.
This kind of function will have vertical asymptote at $$\displaystyle{x}=-{\left(\frac{{d}}{{c}}\right)}$$, horizontal asymptote at $$\displaystyle{y}=\frac{{a}}{{c}}$$. Its points of untersects will be: $$\displaystyle{\left({0},\frac{{b}}{{d}}\right)}$$ with y-axis $$\displaystyle{\left(-{\left(\frac{{b}}{{d}}\right)},{0}\right)}$$ with x-axis
For example, let`s take following function for values $$a=2, b=4, c=1. d=3$$: $$\displaystyle{y}=\frac{{{2}{x}+{4}}}{{{x}+{3}}}$$
This function will have: vertical asymptote" $$\displaystyle{x}=-{\left(\frac{{d}}{{c}}\right)}=-{\left(\frac{{3}}{{1}}\right)}=-{3}$$ horizontal asymptote: $$\displaystyle{y}=\frac{{a}}{{c}}=\frac{{2}}{{1}}={2}$$ y-intercept: $$\displaystyle{y}=\frac{{b}}{{d}}=\frac{{4}}{{3}}$$ x-intercept: $$\displaystyle{x}=-{\left(\frac{{b}}{{a}}\right)}=-{\left(\frac{{4}}{{2}}\right)}=-{2}$$ | 439 | 1,428 | {"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} | 4.65625 | 5 | CC-MAIN-2021-31 | latest | en | 0.536493 |
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`Play around with amicable numbers a bit`
author Steve Losh Sat, 04 Nov 2017 12:47:10 -0400 813d3e887794 acbb1860ce62 (none) src/problems.lisp src/utils.lisp
## Changes
```--- a/src/problems.lisp Thu Oct 26 21:18:39 2017 -0400
+++ b/src/problems.lisp Sat Nov 04 12:47:10 2017 -0400
@@ -452,11 +452,7 @@
;; 71 and 142; so d(284) = 220.
;;
;; Evaluate the sum of all the amicable numbers under 10000.
- (flet ((amicablep (n)
- (let ((other (sum-of-divisors n :proper t)))
- (and (not= n other)
- (= n (sum-of-divisors other :proper t))))))
- (sum (remove-if-not #'amicablep (range 1 10000)))))
+ (sum (remove-if-not #'amicablep (range 1 10000))))
(defun problem-22 ()
;; Using names.txt, a 46K text file containing over five-thousand first names,```
```--- a/src/utils.lisp Thu Oct 26 21:18:39 2017 -0400
+++ b/src/utils.lisp Sat Nov 04 12:47:10 2017 -0400
@@ -276,13 +276,15 @@
(defun unsorted-divisors (n &key proper)
- (iterate (for i :from 1 :to (sqrt n))
- (for (values divisor remainder) = (truncate n i))
- (when (zerop remainder)
- (collect i)
- (unless (or (= i divisor) ;; don't collect the square root twice
- (and proper (= i 1))) ;; don't collect n if proper
- (collect divisor)))))
+ (if (= n 1)
+ nil
+ (iterate (for i :from 1 :to (sqrt n))
+ (for (values divisor remainder) = (truncate n i))
+ (when (zerop remainder)
+ (collect i)
+ (unless (or (= i divisor) ;; don't collect the square root twice
+ (and proper (= i 1))) ;; don't collect n if proper
+ (collect divisor))))))
(defun divisors (n &key proper)
(sort< (unsorted-divisors n :proper proper)))
@@ -310,7 +312,7 @@
If `proper` is given, only include the proper divisors (i.e. not `n` itself).
"
- (sum (divisors n :proper proper)))
+ (sum (unsorted-divisors n :proper proper)))
(defun product-of-divisors (n &key proper)
;; From *Recreations in the Theory of Numbers: The Queen of Mathematics
@@ -397,6 +399,76 @@
(defun deficientp (n)
(> n (sum-of-divisors n :proper t)))
+(defun amicablep (m &optional n)
+ "Return whether `m` and `n` are amicable numbers.
+
+ If `n` is omitted the sum of the aliquot divisors of `m` will be used, and
+ thus this function will return whether `m` is part of an amicable pair.
+
+ "
+ (nest (let ((sm (sum-of-divisors m :proper t))))
+ (let ((n (or n sm))))
+ (unless (= m n))
+ (let ((sn (sum-of-divisors n :proper t))))
+ (and (= m sn)
+ (= n sm))))
+
+(defun amicable-number-chain (n &optional (cutoff 50))
+ "Return the amicable number chain of `n`, or `nil` if it is not part of one.
+
+ If `cutoff` is given, stop searching (and return `nil`) after that many steps.
+
+ Returns two values:
+
+ * The chain (if any)
+ * A boolean specifying whether the computation was aborted by hitting the cutoff.
+
+ "
+ ;; Perfection, amicability, and sociability are all special cases of this:
+ ;;
+ ;; (defun perfectp (n) (= 1 (length (amicable-number-chain n))))
+ ;; (defun amicablep (n) (= 2 (length (amicable-number-chain n))))
+ ;; (defun sociablep (n) (> 2 (length (amicable-number-chain n))))
+ (iterate
+ (for number :iterating (rcurry #'sum-of-divisors :proper t) :seed n)
+ (for i :from 0)
+ (cond ((and cutoff (= cutoff i))
+ (return (values nil t)))
+
+ ((or (= number 1) (member number result))
+ (return (values nil nil)))
+
+ ((= number n)
+ (return (values (cons number result) nil)))
+
+ (t (collect number :into result)))))
+
+(defun aliquot-sequence (n &optional (cutoff 50))
+ "Return the aliquot sequence starting with `n`.
+
+ The sequence will end when any of the following occur:
+
+ * The sequence hits zero.
+ * The sequence encounters a cycle.
+ * `cutoff` steps have been given.
+
+ Returns two values:
+
+ * The sequence.
+ * A boolean specifying whether the cutoff was hit.
+
+ "
+ ;; https://en.wikipedia.org/wiki/Aliquot_sequence
+ (iterate
+ (for number :iterating (rcurry #'sum-of-divisors :proper t)
+ :seed n :include-seed t)
+ (for i :from 0)
+ (cond ((and cutoff (= cutoff i)) (return (values result t)))
+ ((member number result) (return (values result nil)))
+ (t (collect number :into result)))
+ (when (zerop number)
+ (return (values result nil)))))
+
(defun multiplicative-order (integer modulus)
"Return the multiplicative order of `integer` modulo `modulus`."
@@ -814,3 +886,4 @@
;; https://en.wikipedia.org/wiki/Euler%27s_totient_function#Computing_Euler.27s_totient_function
(* n (iterate (for p :in (prime-factors n))
(multiplying (- 1 (/ p))))))
+``` | 1,462 | 4,844 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-49 | latest | en | 0.568203 |
https://stackoverflow.com/questions/12299665/what-does-a-tilde-do-when-it-precedes-an-expression | 1,695,303,404,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506027.39/warc/CC-MAIN-20230921105806-20230921135806-00214.warc.gz | 608,566,603 | 50,287 | # What does a tilde do when it precedes an expression?
``````var attr = ~'input,textarea'.indexOf( target.tagName.toLowerCase() )
? 'value'
: 'innerHTML'
``````
I saw it in an answer, and I've never seen it before.
What does it mean?
`~` is a bitwise operator that flips all bits in its operand.
For example, if your number was `1`, its binary representation of the IEEE 754 float (how JavaScript treats numbers) would be...
``````0011 1111 1111 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000
``````
So `~` converts its operand to a 32 bit integer (bitwise operators in JavaScript do that)...
``````0000 0000 0000 0000 0000 0000 0000 0001
``````
If it were a negative number, it'd be stored in 2's complement: invert all bits and add 1.
...and then flips all its bits...
``````1111 1111 1111 1111 1111 1111 1111 1110
``````
So what is the use of it, then? When might one ever use it?
It has a quite a few uses. If you're writing low level stuff, it's handy. If you profiled your application and found a bottleneck, it could be made more performant by using bitwise tricks (as one possible tool in a much bigger bag).
It's also a (generally) unclear trick to turn `indexOf()`'s found return value into truthy (while making not found as falsy) and people often use it for its side effect of truncating numbers to 32 bits (and dropping its decimal place by doubling it, effectively the same as `Math.floor()` for positive numbers).
I say unclear because it's not immediately obvious what it is being used for. Generally, you want your code to communicate clearly to other people reading it. While using `~` may look cool, it's generally too clever for its own good. :)
It's also less relevant now that JavaScript has `Array.prototype.includes()` and `String.prototype.includes()`. These return a boolean value. If your target platform(s) support it, you should prefer this for testing for the existence of a value in a string or array.
• Is nasty the right word? If it works I'd just call it an idiom of the language. There's many idioms. Once you learn them they are not unclear. List comprehensions are not clear in Python if you don't know them and can be accomplished with more verbose loops but you'd never ask a Python programmer not to use them. Similarly `value = value || default` in JavaScript is a common and valid idiom as long as you know when you can and can't use it.
– gman
May 16, 2014 at 4:37
• @gman I guess it doesn't really matter if someone uses it or not. I think comparing list comprehensions (language feature) to this isn't really the same thing (clever way to avoid typing some extra characters). If you think nasty is too harsh a term, please feel free to edit my answer.
– alex
May 16, 2014 at 4:52
• Maybe a more common example is `v = t ? a : b;`. I find that much clearer than `var v; if (t} { v = a; } else { v = b; }` usually broken across 5+ lines and also clearer than `var v = b; if (t) { v = a; }` which would usually be 4+ lines. But I know lots of people not familiar with the `? :` operators who would prefer the second or third way. I find the first is more readable. I agree with the general principle, make the code clear, don't use hacks. I guess I just see `~v.indexOf('...')` to be very clear once I've learned it.
– gman
May 16, 2014 at 15:56
• When you're working in a large corporation with many developers you want code to be clearly written (ENGLISH) and well documented. The point of high level coding in a language with garbage collection is to avoid thinking about binary operations, and many front-end developers don't even have assembly language experience under their belts. Dec 12, 2016 at 18:55
• i wouldn't call `~` idiomatic. it's technically part of the language spec, but it's not so much part of the language in general use.
– worc
Aug 2, 2018 at 22:37
Using it before an `indexOf()` expression effectively gives you a truthy/falsy result instead of the numeric index that's directly returned.
If the return value is `-1`, then `~-1` is `0` because `-1` is a string of all 1 bits. Any value greater than or equal to zero will give a non-zero result. Thus,
``````if (~someString.indexOf(something)) {
}
``````
will cause the `if` code to run when "something" is in "someString". If you try to use `.indexOf()` as a boolean directly, then that won't work because sometimes it returns zero (when "something" is at the beginning of the string).
Of course, this works too:
``````if (someString.indexOf(something) >= 0) {
}
``````
and it's considerably less mysterious.
Sometimes you'll also see this:
``````var i = ~~something;
``````
Using the `~` operator twice like that is a quick way to convert a string to a 32-bit integer. The first `~` does the conversion, and the second `~` flips the bits back. Of course if the operator is applied to something that's cannot be converted to a number, you get `NaN` as a result. (edit — actually it's the second `~` that is applied first, but you get the idea.)
• For those who don't want to negate bit by bit, `~` when performed on integers is equal to `-(x + 1)`. Sep 6, 2012 at 12:08
• Seems like, well, you know, NEGATIVE numeric values should return negative boolean ones as well in the first place. But just another one of JS' fails, I guess? Sep 6, 2012 at 12:14
• @adlwalrus well the tradition of `0` being `false` and non-zero being `true` dates way back, at least to C in the '70s and probably lots of other then-contemporary systems programming languages. It probably stems from the way the hardware works; lots of CPUs set a zero bit after an operation, and have a corresponding branch instruction to test it. Sep 6, 2012 at 12:32
• A quicker way to convert it to a 32 bit int would be `| 0`, in which case its only one operation.
– alex
Dec 18, 2015 at 9:22
• @alex indeed, though we can't necessarily trust the runtime not to interpret a simple application of `~~` exactly the same way. Dec 18, 2015 at 14:23
The `~` is Bitwise NOT Operator, `~x` is roughly the same as `-(x+1)`. It is easier to understand, sort of. So:
``````~2; // -(2+1) ==> -3
``````
Consider `-(x+1)`. `-1` can perform that operation to produce a `0`.
In other words, `~` used with a range of number values will produce a falsy (coerce to `false` from `0`) value only for the `-1` input value, otherwise, any other truthy value.
As we know, `-1` is commonly called a sentinel value. It is used for many functions that return `>= 0` values for success and `-1` for failure in C language. Which the same rule of return value of `indexOf()` in JavaScript.
It is common to check of presence/absence of a substring in another string in this way
``````var a = "Hello Baby";
if (a.indexOf("Ba") >= 0) {
// found it
}
if (a.indexOf("Ba") != -1) {
// found it
}
if (a.indexOf("aB") < 0) {
}
if (a.indexOf( "aB" ) == -1) {
}
``````
However, it would be more easily to do it through `~` as below
``````var a = "Hello Baby";
~a.indexOf("Ba"); // -7 -> truthy
if (~a.indexOf("Ba")) { // true
// found it
}
~a.indexOf("aB"); // 0 -> falsy
!~a.indexOf("aB"); // true
if (!~a.indexOf( "aB" )) { // true
}
``````
You Don't Know JS: Types & Grammar by Kyle Simpson
• It's definitely easier to understand at face value, even if a person doesn't understand the background that makes it work. I'd take a second look at `-(x+1)` if I saw it in an if statement. The tilde tells me exactly what it's doing to compensate for Javascript's 0-based nature. Also, the less parentheses the better for reading Nov 29, 2017 at 18:00
• In your initial check code block you could type less by using `if (a.indexOf("Ba") > -1) {// found} //true` which, although a bit lengthier than the tilde examples, is considerably less than the two examples you gave and for new programmers `var opinion = !~-1 ? 'more' : 'less'` understandable. Apr 13, 2020 at 11:35
`~indexOf(item)` comes up quite often, and the answers here are great, but maybe some people just need to know how to use it and "skip" the theory:
`````` if (~list.indexOf(item)) {
// item in list
} else {
// item *not* in list
}
``````
• I concur. Airbnb JavaScript Style Guide disallows `++` and `--` because they "encourage excessive trickiness" and yet somehow `~` survived (lurking in the shadows) github.com/airbnb/javascript/issues/540 Apr 28, 2017 at 0:28
• @Shanimal The alternative is `list.indexOf(item) >= 0` or `... > -1` since javascript is zero-based and didn't choose to address this from the outset. Further, just opinion (same as Airbnb's), anyone who is doing anything meaningful in javascript knows `++`, and while `--` is less common, the meaning can be inferred. Nov 29, 2017 at 17:52
• @RegularJoe I agree with most of that. I personally haven't needed `++` and `--` in a while because of primitive methods like `map`, `forEach` etc. My point is more about why they didn't also consider `~` excessively tricky when whatever standard used includes increment and decrement operators. To forbid something so CIS101 doesn't make any sense. Dec 21, 2017 at 1:33
For those considering using the tilde trick to create a truthy value from an `indexOf` result, it is more explicit and has less magic to instead use the `includes` method on `String`.
``````'hello world'.includes('hello') //=> true
'hello world'.includes('kittens') //=> false
``````
Note that this is a new standard method as of ES 2015 so it won't work on older browsers. In cases where that matters, consider using the String.prototype.includes polyfill.
This feature is also available for arrays using the same syntax:
``````['apples', 'oranges', 'cherries'].includes('apples') //=> true
['apples', 'oranges', 'cherries'].includes('unicorns') //=> false
``````
Here is the Array.prototype.includes polyfill if you need older browser support.
• Avoid using includes(). It is not supported in any version of IE (not just older browsers) at the time of writing: developer.mozilla.org/en/docs/Web/JavaScript/Reference/… Nov 22, 2016 at 12:14
• Or just use a compiler, so you can write clearer code, instead of writing the language according to the worst common denominator JS interpreter out there... Feb 12, 2017 at 19:54
• @Ben is right, it doesn't work in Netscape 4.72 either. Jul 16, 2018 at 14:13
• @Onshop, I believe what should be really avoided is not the standard Javascript, but IE, which doesn't support it. Nov 27, 2022 at 2:38 | 2,838 | 10,434 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2023-40 | latest | en | 0.914115 |
https://www.ademcetinkaya.com/2023/06/rms-stock-price-prediction.html | 1,695,681,637,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510100.47/warc/CC-MAIN-20230925215547-20230926005547-00572.warc.gz | 684,694,614 | 58,625 | Outlook: RAMELIUS RESOURCES LIMITED is assigned short-term B1 & long-term Ba3 estimated rating.
Dominant Strategy : Speculative Trend
Time series to forecast n: 22 Jun 2023 for 8 Weeks
Methodology : Supervised Machine Learning (ML)
## Abstract
RAMELIUS RESOURCES LIMITED prediction model is evaluated with Supervised Machine Learning (ML) and Wilcoxon Rank-Sum Test1,2,3,4 and it is concluded that the RMS stock is predictable in the short/long term. Supervised machine learning (ML) is a type of machine learning where a model is trained on labeled data. This means that the data has been tagged with the correct output for the input data. The model learns to predict the output for new input data based on the labeled data. Supervised ML is a powerful tool that can be used for a variety of tasks, including classification, regression, and forecasting. Classification tasks involve predicting the category of an input data, such as whether an email is spam or not. Regression tasks involve predicting a numerical value for an input data, such as the price of a house. Forecasting tasks involve predicting future values for a time series, such as the sales of a product. According to price forecasts for 8 Weeks period, the dominant strategy among neural network is: Speculative Trend
## Key Points
1. Can stock prices be predicted?
2. Buy, Sell and Hold Signals
3. Prediction Modeling
## RMS Target Price Prediction Modeling Methodology
We consider RAMELIUS RESOURCES LIMITED Decision Process with Supervised Machine Learning (ML) where A is the set of discrete actions of RMS 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(Wilcoxon Rank-Sum Test)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(Supervised Machine Learning (ML)) X S(n):→ 8 Weeks $∑ i = 1 n a i$
n:Time series to forecast
p:Price signals of RMS stock
j:Nash equilibria (Neural Network)
k:Dominated move
a:Best response for target price
### Supervised Machine Learning (ML)
Supervised machine learning (ML) is a type of machine learning where a model is trained on labeled data. This means that the data has been tagged with the correct output for the input data. The model learns to predict the output for new input data based on the labeled data. Supervised ML is a powerful tool that can be used for a variety of tasks, including classification, regression, and forecasting. Classification tasks involve predicting the category of an input data, such as whether an email is spam or not. Regression tasks involve predicting a numerical value for an input data, such as the price of a house. Forecasting tasks involve predicting future values for a time series, such as the sales of a product.
### Wilcoxon Rank-Sum Test
The Wilcoxon rank-sum test, also known as the Mann-Whitney U test, is a non-parametric test that is used to compare the medians of two independent samples. It is a rank-based test, which means that it does not assume that the data is normally distributed. The Wilcoxon rank-sum test is calculated by first ranking the data from both samples, and then finding the sum of the ranks for one of the samples. The Wilcoxon rank-sum test statistic is then calculated by subtracting the sum of the ranks for one sample from the sum of the ranks for the other sample. The p-value for the Wilcoxon rank-sum test is calculated using a table of critical values. The p-value is the probability of obtaining a test statistic at least as extreme as the one observed, assuming that the null hypothesis is true.
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?
## RMS Stock Forecast (Buy or Sell) for 8 Weeks
Sample Set: Neural Network
Stock/Index: RMS RAMELIUS RESOURCES LIMITED
Time series to forecast n: 22 Jun 2023 for 8 Weeks
According to price forecasts for 8 Weeks period, the dominant strategy among neural network is: Speculative Trend
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 RAMELIUS RESOURCES LIMITED
1. Adjusting the hedge ratio by decreasing the volume of the hedging instrument does not affect how the changes in the value of the hedged item are measured. The measurement of the changes in the fair value of the hedging instrument related to the volume that continues to be designated also remains unaffected. However, from the date of rebalancing, the volume by which the hedging instrument was decreased is no longer part of the hedging relationship. For example, if an entity originally hedged the price risk of a commodity using a derivative volume of 100 tonnes as the hedging instrument and reduces that volume by 10 tonnes on rebalancing, a nominal amount of 90 tonnes of the hedging instrument volume would remain (see paragraph B6.5.16 for the consequences for the derivative volume (ie the 10 tonnes) that is no longer a part of the hedging relationship).
2. However, an entity is not required to separately recognise interest revenue or impairment gains or losses for a financial asset measured at fair value through profit or loss. Consequently, when an entity reclassifies a financial asset out of the fair value through profit or loss measurement category, the effective interest rate is determined on the basis of the fair value of the asset at the reclassification date. In addition, for the purposes of applying Section 5.5 to the financial asset from the reclassification date, the date of the reclassification is treated as the date of initial recognition.
3. If the underlyings are not the same but are economically related, there can be situations in which the values of the hedging instrument and the hedged item move in the same direction, for example, because the price differential between the two related underlyings changes while the underlyings themselves do not move significantly. That is still consistent with an economic relationship between the hedging instrument and the hedged item if the values of the hedging instrument and the hedged item are still expected to typically move in the opposite direction when the underlyings move.
4. If such a mismatch would be created or enlarged, the entity is required to present all changes in fair value (including the effects of changes in the credit risk of the liability) in profit or loss. If such a mismatch would not be created or enlarged, the entity is required to present the effects of changes in the liability's credit risk in other comprehensive income.
*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
RAMELIUS RESOURCES LIMITED is assigned short-term B1 & long-term Ba3 estimated rating. RAMELIUS RESOURCES LIMITED prediction model is evaluated with Supervised Machine Learning (ML) and Wilcoxon Rank-Sum Test1,2,3,4 and it is concluded that the RMS stock is predictable in the short/long term. According to price forecasts for 8 Weeks period, the dominant strategy among neural network is: Speculative Trend
### RMS RAMELIUS RESOURCES LIMITED Financial Analysis*
Rating Short-Term Long-Term Senior
Outlook*B1Ba3
Income StatementCaa2Ba3
Balance SheetBaa2Baa2
Leverage RatiosBaa2B3
Cash FlowCaa2C
Rates of Return and ProfitabilityB3Baa2
*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: 81 out of 100 with 522 signals.
## References
1. ZXhang, Y.X., Haxo, Y.M. and Mat, Y.X., 2023. Can Neural Networks Predict Stock Market? (No. Stock Analysis). AC Investment Research.
2. J. Harb and D. Precup. Investigating recurrence and eligibility traces in deep Q-networks. In Deep Reinforcement Learning Workshop, NIPS 2016, Barcelona, Spain, 2016.
3. Doudchenko N, Imbens GW. 2016. Balancing, regression, difference-in-differences and synthetic control methods: a synthesis. NBER Work. Pap. 22791
4. Chen X. 2007. Large sample sieve estimation of semi-nonparametric models. In Handbook of Econometrics, Vol. 6B, ed. JJ Heckman, EE Learner, pp. 5549–632. Amsterdam: Elsevier
5. Mnih A, Teh YW. 2012. A fast and simple algorithm for training neural probabilistic language models. In Proceedings of the 29th International Conference on Machine Learning, pp. 419–26. La Jolla, CA: Int. Mach. Learn. Soc.
6. M. Babes, E. M. de Cote, and M. L. Littman. Social reward shaping in the prisoner's dilemma. In 7th International Joint Conference on Autonomous Agents and Multiagent Systems (AAMAS 2008), Estoril, Portugal, May 12-16, 2008, Volume 3, pages 1389–1392, 2008.
7. Byron, R. P. O. Ashenfelter (1995), "Predicting the quality of an unborn grange," Economic Record, 71, 40–53.
Frequently Asked QuestionsQ: What is the prediction methodology for RMS stock?
A: RMS stock prediction methodology: We evaluate the prediction models Supervised Machine Learning (ML) and Wilcoxon Rank-Sum Test
Q: Is RMS stock a buy or sell?
A: The dominant strategy among neural network is to Speculative Trend RMS Stock.
Q: Is RAMELIUS RESOURCES LIMITED stock a good investment?
A: The consensus rating for RAMELIUS RESOURCES LIMITED is Speculative Trend and is assigned short-term B1 & long-term Ba3 estimated rating.
Q: What is the consensus rating of RMS stock?
A: The consensus rating for RMS is Speculative Trend.
Q: What is the prediction period for RMS stock?
A: The prediction period for RMS is 8 Weeks
## People also ask
This project is licensed under the license; additional terms may apply. | 2,467 | 10,638 | {"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} | 2.609375 | 3 | CC-MAIN-2023-40 | latest | en | 0.858393 |
https://forum.math.toronto.edu/index.php?PHPSESSID=8tib5nbvlnoaudbubjclumkr73&topic=820.0;prev_next=next | 1,642,494,342,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300805.79/warc/CC-MAIN-20220118062411-20220118092411-00330.warc.gz | 326,472,812 | 5,872 | ### Author Topic: Q4 (Read 2382 times)
#### Victor Ivrii
• Administrator
• Elder Member
• Posts: 2569
• Karma: 0
##### Q4
« on: October 28, 2016, 09:02:21 AM »
Decompose into full Fourier series on interval $[-\pi,\pi]$ and sketch the graph of the sum of such Fourier series:
f(x)=|x|.
#### Roro Sihui Yap
• Full Member
• Posts: 30
• Karma: 16
##### Re: Q4
« Reply #1 on: October 28, 2016, 09:33:02 AM »
Since $f(x) = |x|$ is an even function, $b_n = 0 \ \ \forall n$
$a_0 = \frac{1}{\pi}\int_{-\pi}^\pi |x|\,dx = \pi$
$a_n = \frac{1}{\pi}\int_{-\pi}^\pi |x|\cos(nx) \,dx = \int_0^\pi \frac{2x}{\pi}\cos(nx) \,dx$
Integrating by parts
$a_n =\frac{2x}{n\pi}\sin(nx) \big|_{0}^{\pi}- \int_0^\pi \frac{2}{n\pi} \sin(nx) \,dx = \frac{2}{n^2\pi}\cos(nx)\big|_{0}^{\pi}$
$a_n = \begin{cases}\frac{-4}{n^2\pi} && n \ is \ odd \\0 && n \ is \ even\end{cases}$
$f(x) = \frac{\pi}{2} +\sum_{m=0}^\infty \frac{-4}{(2m+1)^2\pi}\cos((2m +1)x)$
#### Victor Ivrii
• Administrator
• Elder Member
• Posts: 2569
• Karma: 0
##### Re: Q4
« Reply #2 on: October 28, 2016, 11:02:54 AM »
Everybody understands that on the graph bottoms are at $2n\pi$ and picks at $(2n+1)\pi$, $n\in \mathbb{Z}$. | 510 | 1,181 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2022-05 | latest | en | 0.693948 |
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