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http://www.ebroadcast.com.au/lookup/encyclopedia/ce/Center_of_mass.html | 1,369,507,839,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368706121989/warc/CC-MAIN-20130516120841-00096-ip-10-60-113-184.ec2.internal.warc.gz | 434,741,650 | 3,259 | Contents
Center of mass
The center of mass of a group of points is defined as the weighted mean of the points' positions. The weight applied to each point is the point's mass. The centre of mass of an object is the point through which any plane divides the mass of the object in half. It is also called the center of inertia.
For mass that is distributed according to a continuous, nonnegative density $\rho(\mathbf{x}) \ge 0$ over a body V in space, the center of mass is:
$\bar{\mathbf{x}} = \frac{1}{M} \int_{V} \rho(\mathbf{x})\mathbf{x}\,dV$, where total mass $M = \int_{V} \rho(\mathbf{x})\,dV$.
In $\mathcal{R}^3$, $\bar{\mathbf{x}} = (\bar{x}, \bar{y}, \bar{z})$. Each of the center of mass's components can be computed by
$\bar{x} = \frac{1}{M} \int_{V} x\rho(x, y, z)\,dx dy dz$
$\bar{y} = \frac{1}{M} \int_{V} y\rho(x, y, z)\,dx dy dz$
$\bar{z} = \frac{1}{M} \int_{V} z\rho(x, y, z)\,dx dy dz$.
The origin from which positions are calculated has no effect on the result. As long as the same unit is used for all the points, any length and mass unit can be used.
Examples
• Point A: position 2m, mass 1kg. Point B: position 4m, mass 2kg (assume positions are distances along a straight line from some origin). Center of mass:
$\frac{2 \times 1 + 4 \times 2}{1+2} = 3.33\ {\rm m}$
• Solid homogenous sphere (ideally divided in a high number of points of equal mass): each point averages with its opposite. Center of mass is at the center.
• Sphere with spherically symmetric density: center of mass is at the center. This approximately applies to the Earth: the density varies considerably, but it mainly depends on depth and less on the other two coordinates.
• Human being: It varies according to the body's position, but often it's somewhere inside the abdomen
• A sports car: engineers try hard to make the car as light as possible, and then add weight on the bottom. This way, the center of mass is nearer to the street, and the car handles better.
When talking about celestial bodies, the center of mass has a special relevance: when a moon orbits around planet, or a planet orbits around a star, both of them are actually orbiting around their center of mass, called the barycenter. There are some interesting consequences:
• Earth-Moon system: the Moon's mass is 1/81 of Earth. Put Earth in position 0, mass 1 (here we use an arbitrary mass unit. It does not matter, provided that we use the same unit for the Moon). Moon position 400,000km, masss 1/81. Center of mass is at:
$\frac{0 \times 1 + 400,000 \times \frac{1}{81}}{1 + \frac{1}{81}} = 4,877\ {\rm km}$
from the Earth's center. We can see that the Earth is far from standing "still" and the Moon moving: both of them move around a point more than 1,000km below the Earth surface.
• Sun-Earth system: put Sun in position 0, mass=333,000 times the Earth. Earth in position 150,000,000 km, mass=1. Center of mass is 450 km from the Sun center. Here, the large mass difference between the two bodies makes the center of mass lie almost where we were expecting it.
• Sun-Jupiter system: put Sun in position 0, mass = 333,000 Earths. Jupiter in position 778,000,000 km, mass=318 Earths. Center of mass is 742,000 km from the Sun center. It's actually outside its surface! As Jupiter does its 11 year orbit, the Sun majestically does a full 1.5 million km orbit around the center of mass.
• To calculate the actual motion of the Sun, you would need to sum all the influences from all the planets, comets, asteroids, etc. of the solar system. But, only Jupiter manages to pull the center of mass so far, thanks to its large mass. If all the planets would align on the same side of the Sun, the combined center of mass would lie about 500,000 km outside the Sun surface. | 1,031 | 3,761 | {"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} | 4.40625 | 4 | CC-MAIN-2013-20 | latest | en | 0.906254 |
https://geniushomeworks.com/a-32-foot-ladder-is-leaning-against-a-tree-the-ladder-forms-a-72/ | 1,696,034,375,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510529.8/warc/CC-MAIN-20230929222230-20230930012230-00212.warc.gz | 309,930,822 | 13,343 | # A 32-foot ladder is leaning against a tree. the ladder forms a 72
A 32-foot ladder is leaning against a tree. The ladder forms a 72 degree angle with the ground, not the tree. Assuming the tree is growing straight up: how far away from the tree is the base of the ladder? 2. The angle of depression from a helicopter to a speeding car is 56 degree. If the helicopter is flying 600 meters above the ground. What is the horizontal distance of the helicopter and the car? What is the actual distance between the helicopter and the car?
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Order Paper Now | 183 | 822 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-40 | longest | en | 0.940582 |
http://www.physicsforums.com/showthread.php?p=3994415 | 1,369,542,012,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368706624988/warc/CC-MAIN-20130516121704-00003-ip-10-60-113-184.ec2.internal.warc.gz | 654,074,755 | 9,304 | ## Physics question on motion
1. The problem statement, all variables and given/known data
An object that is acclerating uniformly travels 22.5 m in the eighth second of its motion.
a) what is the acceleration of the object?
b) how far does the object travel in 6 seconds?
2. Relevant equations
I'm not exactly sure what equation would be necessary-
perhaps -- a=v-u/t
v=u+at
x=ut+1/2at^2
3. The attempt at a solution
i attempted to use the a=v-u/t equation, however, it wasn't successful. I don't even think you can use that equation since we dont have the values for velocity;initial nor final.
PhysOrg.com science news on PhysOrg.com >> Galaxies fed by funnels of fuel>> The better to see you with: Scientists build record-setting metamaterial flat lens>> Google eyes emerging markets networks
Recognitions: Gold Member Homework Help Science Advisor The object has no motion at t = 0. How far does it travel in 8 seconds? How far does it travel in 7 seconds? Its displacement between the. 7th and 8th seconds is given as 22.5 m. Use your last equation.
Recognitions: Homework Help Hi einstein101. The eighth second is that time difference between seven seconds and eight seconds. travel during first 7 seconds= u.7 + (1/2)a.72 travel during first 8 seconds= ????
## Physics question on motion
But how do I find the initial velocity for the equation?
And how would I exactly go about finding the acceleration?
Quote by PhanthomJay The object has no motion at t = 0. How far does it travel in 8 seconds? How far does it travel in 7 seconds? Its displacement between the. 7th and 8th seconds is given as 22.5 m. Use your last equation.
It says nowhere that the object started from rest.
Mentor
Blog Entries: 1
Quote by e^(i Pi)+1=0 It says nowhere that the object started from rest.
It says "in the eighth second of its motion". I would interpret that to mean that it started from rest.
Recognitions:
Homework Help
Quote by einstein101 But how do I find the initial velocity for the equation? And how would I exactly go about finding the acceleration?
Can you finish the equation I left uncompleted?
Quote by Doc Al It says "in the eighth second of its motion". I would interpret that to mean that it started from rest.
Hmm, I suppose you're right.
so travel during first 8 seconds=u x 8 +1/2(a)(8)^2 then what do i do?
Recognitions:
Homework Help
Quote by einstein101 so travel during first 8 seconds=u x 8 +1/2(a)(8)^2 then what do i do?
Good. Now subtract that from the companion equation I gave you, viz., of distance travelled during first 7 seconds.
Recognitions:
Gold Member
Homework Help | 647 | 2,617 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2013-20 | latest | en | 0.94747 |
http://en.citizendium.org/wiki/Helmholtz_decomposition | 1,544,552,581,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376823674.34/warc/CC-MAIN-20181211172919-20181211194419-00576.warc.gz | 86,344,207 | 11,335 | # Helmholtz decomposition
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In vector analysis, the Helmholtz decomposition of a vector field on is the decomposition of the vector field into two vector fields, one a divergence-free field and one a curl-free field. The decomposition is called after the German physiologist and physicist Hermann von Helmholtz (1821 – 1894).
## Mathematical formulation
The Helmholtz decomposition may be formulated as follows. Any vector field F(r) that is sufficiently often differentiable and vanishes sufficiently fast at infinity can be written as,
with
The primed nabla operator ' acts on primed coordinates and the unprimed acts on unprimed coordinates.
Note that
holds for any vector field V(r) and any scalar function Ψ(r). Hence it follows that the first term of F is divergence-free and the second curl-free.
As a corollary follows that the specification of both the divergence and the curl of a vector field at all points of space gives the field uniquely.
A well-known example of a Helmholtz decomposition is the following form of the electric field E,
where Φ is the electric potential and A is the (magnetic) vector potential. The dot indicates a derivative with respect to time.
## Decomposition in transverse and longitudinal components
Above it was stated that a vector field F(r) with can be decomposed in a transverse and longitudinal component :
where
Thus, an arbitrary field F(r) can be decomposed in a part that is divergence-free, the transverse component, and a part that is curl-free, the longitudinal component. This will now be proved directly, without making the detour via the integral expressions for A(r) and Φ(r).
### Proof of decomposition
The decomposition is formulated in r-space. By a Fourier transform the decomposition may be formulated in k-space. This is advantageous because differentiations in r-space become multiplications in k-space. We will show that divergence in r-space becomes an inner product in k-space and a curl becomes a cross product. Thus, we define the mutually inverse Fourier transforms,
An arbitrary vector field in k-space can be decomposed in components parallel and perpendicular to k,
so that
Clearly,
Transforming back, we get
which satisfy the properties
Hence we have found the required decomposition.
## Integral expressions for the transverse and longitudinal components
The curl and the divergence of the vector field F(r) satisfy,
Using this, we see that the following relations were stated earlier in fact:
They are, respectively, the perpendicular (transverse, divergence-free) and parallel (longitudinal, curl-free) components of the field F(r). We reiterate that the operator acts on unprimed coordinates and ∇' on primed coordinates. Note that the two components of F(r) are uniquely determined once the curl and the divergence of F(r) are known. The integral relations will now be proved.
### Proof of integral expressions
We will confirm the integral forms, equations (1) and (2), of the components. They will be shown to lead to identities.
#### Transverse component
For the perpendicular (transverse) component we note that for any vector V,
and insert this in
Below we will show that second term vanishes. Use for the first term the following equation for the Dirac delta function,
Hence the first term becomes (note that the unprimed nabla may be moved under the integral)
so that we indeed end up with an identity.
Before turning to the parallel (longitudinal) term we prove that the second term vanishes. To that end we introduce a shorthand notation
Move the divergence under the integral and use
By partial integration and using that the integrand vanishes for the integral limits, we can let −∇'α act on Gα(r' ) (this trick is known as the turnover rule for the anti Hermitian operator ∇'α). Then from
(because the divergence of the curl of any vector is zero) follows the vanishing of the second term.
#### Longitudinal component
From
follows that there is a scalar function Φ such that
We work toward an identity, using the turnover rule for the Laplace operator ∇2, which may be proved by partial integration and the assumption that the integrand vanishes at the integration limits, | 964 | 4,546 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 29, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2018-51 | latest | en | 0.898236 |
https://yokohama-centre.com/what-exactly-is-usually-the-key-guiding-this-strange-also-lotto-quantity-method/ | 1,607,083,988,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141735600.89/warc/CC-MAIN-20201204101314-20201204131314-00243.warc.gz | 914,595,119 | 9,503 | # What exactly Is usually the Key Guiding this Strange Also Lotto Quantity Method?
Do you want to know the key guiding enjoying the odd and even variety lottery approach? There was a time when I was in the very same boat with you. Then, I identified the mystery and shared it with you in one of my prior articles or blog posts. Now, I found however an additional magic formula concealed within the initial and will share it with your below. To established the phase for the 1st time readers, let’s recap a bit.
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kbc lottery winner : http://EzineArticles.com/7167629 | 703 | 3,285 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2020-50 | latest | en | 0.96721 |
https://www.studypool.com/discuss/1005726/how-do-you-solve-this-11?free | 1,481,178,921,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542414.34/warc/CC-MAIN-20161202170902-00312-ip-10-31-129-80.ec2.internal.warc.gz | 1,010,242,847 | 14,111 | how do you solve this
Algebra Tutor: None Selected Time limit: 1 Day
18 divide by 6 +4(6)
May 28th, 2015
$\frac{18}{6}+4(6)\\ \\ =3+24\\ \\ =27$
----------------------------------------------------------------------
May 28th, 2015
May 28th, 2015
...
May 28th, 2015
...
May 28th, 2015
Dec 8th, 2016
check_circle | 112 | 319 | {"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": 1, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2016-50 | longest | en | 0.785474 |
https://www.smartconversion.com/factsheet/solar-system-distance-from-earth | 1,686,228,137,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224654871.97/warc/CC-MAIN-20230608103815-20230608133815-00077.warc.gz | 1,088,597,978 | 7,249 | Report a Problem
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# Distance From Earth
Last updated: Thursday, June 08, 2023
Distance from Earth, in the context of planets, refers to the measure of the physical separation between Earth and a specific planet in our solar system. It is a fundamental parameter used to describe the position and location of planets relative to our planet.
The distance from Earth to a planet can vary greatly depending on several factors, including the planet's position in its orbit around the Sun and Earth's position in its own orbit. Since the planets and Earth all orbit the Sun, their distances from each other are constantly changing.
To measure the distance from Earth to a planet, astronomers use units such as astronomical units (AU) or kilometers (km). An astronomical unit is defined as the average distance between the Earth and the Sun, which is approximately 149.6 million kilometers (93 million miles).
The distance from Earth to each planet in our solar system varies significantly. For example, the average distance from Earth to Venus is about 41 million kilometers (25 million miles), while the average distance to Mars is approximately 78 million kilometers (48 million miles). The outer planets, such as Jupiter, Saturn, Uranus, and Neptune, are much farther from Earth, with average distances ranging from hundreds of millions to billions of kilometers.
The distance from Earth to a planet affects our ability to observe and study it. Telescopes and spacecraft are used to gather data and images from these distant worlds, providing valuable information about their atmospheres, surfaces, and geology.
Click On The Pictures To See The 3D Models From NASA
Name Min. $$10^6$$km Max. $$10^6$$km Mean. $$10^6$$km
Moon 0.357 0.407 0.378
Venus 38.2 261 41.39
Mars 54.6 401.4 78.34
Mercury 77.3 221.9 91.69
Jupiter 588.5 968.5 628.81
Saturn 1205.5 1658.6 1277.13
Uranus 2580.6 3153.5 2721.37
Pluto 4284.7 7528 5756.78
Neptune 4319 4711 4348.66
Source: NASA | 498 | 1,975 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2023-23 | latest | en | 0.876367 |
https://www.minicircuits.com/appdoc/VCO15-6.html | 1,610,735,628,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703495936.3/warc/CC-MAIN-20210115164417-20210115194417-00286.warc.gz | 1,004,402,186 | 18,708 | Our website uses cookies. By continuing to use this website, you consent to cookies being used.
English
English
# Characterizing and Minimizing VCO Phase Noise
## Characterizing Phase Noise
The term phase noise is widely used for describing short term random frequency fluctuations of a signal. Frequency stability is a measure of the degree to which an oscillator maintains the same value of frequency over a given time. This may be specified in a number of ways. Three commonly used terms for describing frequency stability are used here. An ideal sine wave oscillator may be described by:
V(t) = Vo Sin2Pft
Here, Vo is the nominal amplitude of the signal, and f is the nominal frequency of oscillation. The instantaneous output of an oscillator may be represented by:
V(t) = Vo {1 + A(t)} Sin {2P ft + q (t)}
where A(t) and q(t) represent the amplitude and phase fluctuations of the signal respectively. The phase term may be random or discrete and could be seen on the spectrum analyzer. As shown in Figure 1 there are two types of fluctuating phase terms. The first, the discrete signals called spurious, appear as distinct components in the spectral density plot. The second term, random in nature, appears as random phase fluctuations and is commonly called phase noise.
The source of phase noise in an oscillator is due to thermal and flicker or 1/f noise. Most oscillators operate in saturation. The AM noise component is usually 20dB lower than the phase noise component. In the discussion that follows, we will assume that A(t)<<1.
Many methods are used to characterize phase noise of an oscillator. Essentially, all methods measure the frequency or phase deviation of the source under test in either frequency or time domain. Since frequency and phase are related to each other, all these terms are also related. One of the most common fundamental descriptions of phase noise is the one sided spectral density of phase fluctuations per unit bandwidth. The term spectral density describes the energy distributions as a continuous function, expressed in units of energy per unit bandwidth. The phase noise of an oscillator is best described in the frequency domain where the spectral density is characterized by measuring the noise sidebands on either side of the output signal center frequency. Single sideband phase noise is specified in dBc/Hz at a given frequency offset from the carrier. The frequency domain information about phase or frequency is contained in the power spectral density SDq(f) of the phase or in the power spectral density SDf(f) of the frequency. Here, f refers to the modulation frequency or offset frequency associated with the noise-like variations in q(t).
Peak phase modulation Dq and peak frequency modulation Df are related as follows:
In terms of rms values, we have:
The one sided spectral distribution of the phase fluctuations per Hz bandwidth is SDq(f):
Here, BW is the bandwidth of Dqrms measurement. The units of SDq(f) are radian2 Hz-1 bandwidth or dB relative to 1 rad2 Hz-1 bandwidth. The term S= (f) is often referred to as the spectral density and describes the energy distribution as a continuous function, expressed in units of energy per Hz bandwidth. This is illustrated in Figure 2.
Similarly, one sided spectral distribution of the frequency fluctuations per Hz bandwidth is SDf(f) where:
SDf = (Dfrms)2 / BW
Here, BW is the bandwidth of Dfrms measurement. The units of SDf(f) are (rad sec-1)2 HZ-1 bandwidth. It is also common to characterize the noise performance of a signal as the ratio of the sideband power associated with phase fluctuations to the carrier power level. If the measure is denoted by Sc
Sc(f) = (Power density in one sideband per Hz bandwidth at an offset frequency f away from the carrier) / (Total signal power) For small phase fluctuations, we can write:
Here, b is the modulation index by analogy to modulation theory.
Sc(f) = 1/2 (SDq(f))
SCBfl is often expressed in decibels relative to the carrier per Hz c/Hz) and is related to the power spectrum observed on a spectrum analyzer. The National Bureau of Standards defines Single Side Band Phase Noise as the ratio of power in one phase modulation sideband per Hertz bandwidth, at an offset f Hertz away from the carrier, to the total signal power. Here, f is the offset frequency from the carrier.
where Ps is the carrier power and Pssb is the sideband power in one Hz bandwidth at an offset frequency of f from the center. This is illustrated in Figure 3.
The Single Sideband Phase Noise is usually given logarithmically, that is:
Sc(f) in dB = 10 x log[Sc(f)]
This is shown in Figure 4 as a spectral density plot of the phase modulation sidebands in the frequency domain. It is expressed in dB relative to the carrier per Hz bandwidth.
The phase noise generated by a VCO is determined by:
1. Q factor of the resonator
2. Q of the varactor diode
3. the active device used for the oscillating transistor
4. power supply noise
5. external tuning voltage supply noise
The noise contribution made by (d) and (e) can be minimized by careful choice of the power supplies (see Application Note 3). The phase noise of the VCO is therefore determined primarily by the overall Q of the circuit. In order to design a circuit with high Q, the tuning bandwidth must be made small. Therefore a VCO designed for low phase noise performance will have a smaller tuning range.
## Ways to minimize noise
The following steps are recommended for obtaining the best overall performance from Mini-Circuits VCO's.
1. Power Supply (Vcc) and tuning voltage (Vtune) returns must be connected to the printed circuit board ground plane. VCO ground plane must be the same as that of the printed circuit board and therefore all VCO ground pins must be soldered direct to the printed circuit board ground plane.
2. Adequate RF grounding is required. Several chip decoupling capacitors must be provided between the Vcc supply and ground.
3. Good, low noise power supplies must be used. Ideally, DC batteries for both supply (VCC) and tuning (Vtune) voltages will provide the best overall performance.
4. Output must be correctly terminated with a good load impedance. It is also a good practice to use a resistive pad between the VCO and the external load.
5. Connections to the tuning port must be as short as possible and must be well screened, shielded, and decoupled to prevent the VCO from being modulated by external noise sources. A low noise power supply must be used for tuning voltage (Vtune) Supply. | 1,409 | 6,546 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2021-04 | latest | en | 0.917752 |
https://www.reference.com/math/convert-ratio-percentage-b7839173b4910e2e | 1,488,083,479,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501171933.81/warc/CC-MAIN-20170219104611-00190-ip-10-171-10-108.ec2.internal.warc.gz | 888,406,882 | 20,337 | Q:
# How do you convert a ratio to a percentage?
A:
Converting a ratio to a percentage involves using division to change the fractional or descriptive ratio to a portion out of 100. For example, the ratio of 1 in 4, which can also be expressed as 1/4, would convert to a percentage through the division of 1 by 4 and then multiplying by 100, equaling 25 percent.
## Keep Learning
Ratios are commonly used as a more descriptive way to describe a subset of an overall population. A text-based ratio may say "three out of four people prefer this product." When converted to a percentage, that same statistic describing 75 percent of a population could be used in comparison with related measurements.
Sources:
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PEOPLE SEARCH FOR | 318 | 1,497 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2017-09 | longest | en | 0.891874 |
http://mathhelpforum.com/advanced-algebra/198894-simple-question-about-linear-functionals.html | 1,526,819,024,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794863410.22/warc/CC-MAIN-20180520112233-20180520132233-00632.warc.gz | 194,639,006 | 10,285 | 1. ## Simple question about linear functionals
Hi everyone. Paul Halmos asks a seemingly-innocuous question in Finite Dimensional Vector Spaces: let V be a vector space over a field F and let y and z be linear functionals from V to F. Suppose, furthermore that y(x) = 0 whenever z(x) = 0, for all x in V. Prove that y = az for some scalar a.
Here's where I am so far: if z = 0 then the problem is trivial, so without loss of generality z(x0) is not zero for some vector x0 in V. Then the only possibility for a is a = y(x0)/z(x0). Now I need to prove the identity for all x in V:
y(x) = y(x0)z(x)/z(x0)
But I don't see how to do this. I have absolutely no background with linear functionals, and only know undergraduate-level linear algebra. I'd appreciate a tip.
2. ## Re: Simple question about linear functionals
Write $\displaystyle x=x-\frac{z(x)}{z(x_0)}x_0+\frac{z(x)}{z(x_0)}x_0$ and compute $\displaystyle y\left(x-\frac{z(x)}{z(x_0)}x_0\right)$.
3. ## Re: Simple question about linear functionals
I'm having trouble even with the hint. I would love to compute $\displaystyle y\left(x-\frac{z(x)}{z(x_0)}x_0\right)$. Indeed, this quantity needs to be zero, and then the result immediately follows. But showing that this result is equal to zero is really a trivial modification of the problem:
$\displaystyle y\left(x-\frac{z(x)}{z(x_0)}x_0\right)=y(x)-y\left(\frac{z(x)}{z(x_0)}x_0\right)$
$\displaystyle =y(x)-\frac{y(x_0)}{z(x_0)}z(x)$
$\displaystyle =(*)$
is precisely the quantity I need to show is zero. I don't see any way to "compute" this further, so that's kind of my problem. I could give it a try:
$\displaystyle (*)=\frac{1}{z(x_0)}\left(y(x)z(x_0)-y(x_0)z(x)\right)$
$\displaystyle =\frac{1}{z(x_0)}z\left(y(x)x_0-y(x_0)x\right)$
Basically I'm just trying to use linearity here. I am not aware of any other possible manipulations (short of adding and subtracting a quantity which I just don't see). Could I please have a bit more help?
4. ## Re: Simple question about linear functionals
Ah hah. I see what I needed to do. I do not need to evaluate y(...). I need to show that z(...) = 0, and then by hypotheses, y(...) = 0. But showing that z(...) = 0 is very easy. That solves the problem. Thank you. I was being silly - I should have looked for a way to use the given hypotheses.
Thanks again! And by the way, a great hint that didn't give away too much of the fun! | 720 | 2,402 | {"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} | 3.84375 | 4 | CC-MAIN-2018-22 | latest | en | 0.867696 |
http://www.expertsmind.com/questions/explain-introduction-to-non-euclidean-geometry-30161560.aspx | 1,591,026,248,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347419056.73/warc/CC-MAIN-20200601145025-20200601175025-00084.warc.gz | 157,846,369 | 15,786 | ## Explain introduction to non-euclidean geometry, Mathematics
Assignment Help:
Explain Introduction to Non-Euclidean Geometry?
Up to this point, the type of geometry we have been studying is known as Euclidean geometry. It is based on the studies of the ancient Greek mathematician Euclid. Euclidean geometry was a way to explain or describe the basic layout of the universe. Hundreds of years after him, a few mathematicians developed geometries that are not based on Euclid's axioms. In this chapter, we will explore some concepts of non-Euclidean geometry.
A line, according to Euclid, is perfectly straight and extends infinitely in both directions. Keep in mind that Euclid lived in a world that believed the Earth was flat. But now we know that Earth is a sphere, a line of the Euclidean postulate, perfectly straight and infinitely long, could not exist on the surface of the Earth. A "line" on a spherical surface must follow a curved path. The geometry based on a sphere is called sphere geometry.
Definition
A great circle of a sphere is the circle determined by the intersection of the spherical surface and a secant plane that contains the center of the sphere.
Definition
Lines are great circles in sphere geometry.The equator and longitudinal lines on a globe are great circles. Latitudes on a globe are not great circles.
You already know that on a plane, the shortest distance between any two points is a line segment joining these two points. The shortest distance between any two points on a sphere is measured along a curved path that is a segment of a great circle. The length of a line segment depends on the size of the sphere. Polar points are the points created by a line passing through the center of a sphere intersecting with the sphere. The North and South Poles on Earth are polar points.
Postulate
For any given pair of points on a sphere, there is exactly one line containing them. Conversely, it is also true that a line contains at least two points. But consider now the parallel postulate on a flat plane, "Through a given point not on a given line there is exactly one line parallel to the given line." On a sphere, every line intersects with all other lines.
Postulate
On a sphere, through a given point not on a given line there is no line parallel to the given line.
Definition
A biperpendicular quadrilateral is a quadrilateral with two sides perpendicular to a third one.
The legs are the two sides perpendicular to the same side.
The base is the side to which the two legs are perpendicular.
The base angle is an angle between base and leg.
The summit is the side opposite the base.
The summit angle is an angle between summit and leg.
Definition
An eighteenth century priest named Saccheri, for whom the Saccheri quadrilateral is named, studied the figure. He tried to use it to prove that the Euclidean parallel postulate was true. Instead he came across something remarkable in the field of non-Euclidean geometry. Using the new postulate on parallel lines, we can prove that a Saccheri quadrilateral is not a rectangle and its two summit angles are not right angles.
Theorem
If the two summit angles of a biperpendicular quadrilateral are unequal, then the larger angle is adjacent to the shorter leg.
Theorem
The summit angles of a Saccheri quadrilateral are congruent.
Theorem
In a Saccheri quadrilateral, the bisector of the base and the summit is perpendicular to both of them.
#### Children learn maths by experiencing things, Children Learn By Experiencing...
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#### How to converting percents to fractions, How to Converting Percents to Frac...
How to Converting Percents to Fractions ? To convert a percent to a fraction: 1. Remove the percent sign. 2. Create a fraction, in which the resulting number from Step 1 is
#### Average, A boy covered half of distance at 20km/hr and rest at 40kmlhr. cal...
A boy covered half of distance at 20km/hr and rest at 40kmlhr. calculate his average speed.
#### Write down the first few terms of the sequences, Write down the first few t...
Write down the first few terms of each of the subsequent sequences. 1. {n+1 / n 2 } ∞ n=1 2. {(-1)n+1 / 2n} ∞ n=0 3. {bn} ∞ n=1, where bn = nth digit of ? So
#### Translating words into math, 48 more than the quotientvof a number and 64
48 more than the quotientvof a number and 64
#### Real Analysis/Advanced Calculus (Needs to be a full proof), Both need to be...
Both need to be a full page, detailed proof. Not just a few lines of proof. (1) “Every convergent sequence contains either an increasing, or a decreasing subsequence (or possibly
#### Chi square distribution, Chi Square Distribution Chi square was first ...
Chi Square Distribution Chi square was first utilized by Karl Pearson in 1900. It is denoted by the Greek letter χ 2 . This contains only one parameter, called the number of d
#### 3/8:5/9, how do I change this ratio to a fraction
how do I change this ratio to a fraction
10+2=
#### Shates and diividends, what is share and dividend?
what is share and dividend? | 1,179 | 5,236 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5625 | 5 | CC-MAIN-2020-24 | latest | en | 0.953555 |
https://www.examrace.com/RBI/RBI-Grade-B/RBI-Grade-B-Model-Questions/Aptitude-Questions/Aptitude-Logical-Reasoning-Compound-Interest-Part-1.html | 1,600,831,481,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400209665.4/warc/CC-MAIN-20200923015227-20200923045227-00771.warc.gz | 806,581,533 | 9,843 | # Aptitude Logical Reasoning Compound Interest 2020 RBI Grade B (NRA) Part 1
Glide to success with Doorsteptutor material for UGC : fully solved questions with step-by-step explanation- practice your way to success.
1. Find the compound interest and the amount on Rs.8000 at per annum for 3 years when C.I is reckoned yearly?
A. Rs.1261
B. Rs.1440
C. Rs.1185
D. Rs.1346
2. If Rs.7500 are borrowed at C.I at the rate of per annum, then after 2 years the amount to be paid is?
A. Rs.8082
B. Rs.7800
C. Rs.8100
D. Rs.8112
3. Find out the C.I on Rs.5000 at 4% p.a. compound half-yearly for years.
A. Rs.420.20
B. Rs.319.06
C. Rs.306.04
D. Rs.294.75
4. Rs.8000 become Rs.9261 in a certain interval of time at the rate of per annum of C.I. Find the time?
A. 4 years
B. 6 years
C. 2 years
D. 3 years
5. At the end of three years what will be the compound interest at the rate of p.a. on an amount of Rs.20000?
A. Rs.6620
B. Rs.6500
C. Rs.6800
D. Rs.6400
6. Find the C.I. on a sum of Rs.1600 for 9 months at per annum, interest being compounded quarterly?
A. Rs.17684
B. Rs.1684
C. Rs.2522
D. Rs.3408
7. Simple interest on a sum at per annum for 2 years is Rs.80. The C.I. on the same sum for the same period is?
A. Rs.81.60
B. Rs.160
C. Rs.1081.60
D. Rs.99
8. The C.I. on a certain sum for 2 years Rs.41 and the simple interest is Rs.40. What is the rate percent?
A.
B.
C.
D.
9. The sum of money at compound interest amounts to thrice itself in 3 years. In how many years will it be 9 times itself?
A. 18
B. 12
C. 9
D. 6
10. A sum of money place at compound interest doubles itself in 4 years. In how many years will it amount to eight times itself?
A. 16
B. 8
C. 12
D. 20
11. A sum of money deposited at C.I. amounts to Rs.2420 in 2 years and to Rs.2662 in 3 years. Find the rate percent?
A.
B.
C.
D.
12. Find the sum lend at C.I. at 5 p.c per annum will amount to Rs.441 in 2 years?
A. Rs.420
B. Rs.400
C. Rs.375
D. Rs.380
13. A property decreases in value every year at the rate of of its value at the beginning of the year its value at the end of 3 years was Rs.21093. Find its value at the beginning of the first year?
A. Rs.25600.24
B. Rs.32000.50
C. Rs.18060.36
D. Rs.18600
14. Find the least number of complete years in which a sum of money put out at compound interest will be more than double of itself?
A. 6 years
B. 1 year
C. 2 years
D. 4 years
15. The difference between simple interest and C.I. at the same rate for Rs.5000 for 2 years in Rs.72. The rate of interest is?
A.
B.
C.
D. | 871 | 2,565 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-40 | latest | en | 0.916426 |
https://mathoverflow.net/questions/275741/limits-of-stable-theories/292308#292308 | 1,638,404,644,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964361064.58/warc/CC-MAIN-20211201234046-20211202024046-00549.warc.gz | 460,330,415 | 26,893 | # limits of stable theories
Say that a complete theory $T$ is a limit of stable theories if for every $\phi \in T$ there is a stable completion of $\{\phi\}$. (Equivalently, $T$ is the ultraproduct of stable theories.)
Question: Is every simple theory a limit of stable theories? (We allow that theories of finite structures are stable.)
As motivation, all the simple unstable theories I am aware of are random in various ways (e.g. random graph) and the independence property only comes from the full random schema.
Example: any pseudo-finite theory is a limit of stable theories. This includes a lot of of simple theories, e.g. the theory of the random graph. It also includes pseudo-finite linear orders, which are SOP.
Example: suppose $T$ is a completion of ACFA (the model companion of algebraically closed fields with an automorphism), so $T$ is simple. Suppose $\phi \in T$. Then $\phi$ is consistent with ACFA, so for arbitrarily large powers of primes $p^n$, $(F, x \mapsto x^{p^n}) \models \phi$, where $F$ is some (any) algebraically closed field of characteristic $p$. But these structures are stable.
A strong counterexample would be a finitely axiomatizable simple unstable theory.
No, not every simple theory is a limit of stable theories. For example, let $K$ be a pseudoalgebraically closed field with a small, nontrivial Galois group not isomorphic to $\widehat{\mathbb{Z}}$. Possibly after replacing $K$ with a finite algebraic extension, there will be some natural number $n$ coprime to the characteristic for which the $n^\text{th}$ power map is not onto. By smallness of the Galois group, the group $K^\times/(K^\times)^n$ is finite, say of size $m$. That the absolute Galois group is not isomorphic to $\widehat{\mathbb{Z}}$ is reflected by there being some natural number $\ell$ for which either there is no extension of degree $\ell$ or there are two distinct extensions of degree $\ell$. The theory of $K$ is supersimple, but the sentence $\phi$ which asserts that
• $K$ is a field,
• that there is some $x \in K$ which is not an $n^\text{th}$ power,
• that there are $m$ elements $y_1, \ldots, y_m$ of $K$ so that every nonzero element of $K$ may be expressed as $x_i z^n$ for some $z \in K$ and $1 \leq i \leq m$, and
• that either there are no extensions of degree $\ell$ or there are two distinct such extensions
holds in $K$ but cannot hold in any stable structure: any such structure is a field (by the first item) and finite fields cannot satisfy the fourth item while infinite stable fields have connected multiplicative groups and hence cannot satisfy the conjunction of the second and third items.
• Looking at the various comments in Kim's "Simple Theories" book (Section 2.6), it seems your requirement on $K$ is that $K$ be pseudoalgebraically closed, supersimple (bounded+perfect), and not pseudofinite (either not PAC or not perfect or not unique extension of every degree), and not closed under nth roots for all n. Is there an easy example of some such field? Feb 14 '18 at 1:14
• Perhaps easier: By Fact 2.6.7 (3) in Kim, due to Duret and Wood, if F is PAC, then F is stable iff it is separable closed, but when they say stable they mean infinite models. Since separably closed is a first-order condition, it seems to be enough to get that $K$ is pseudo-algebraically closed, simple, not pseudo-finite, and unstable. Feb 14 '18 at 8:20 | 857 | 3,387 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2021-49 | latest | en | 0.906277 |
http://panorics.com/2019/06/parametric-vector-equation-form/ | 1,571,262,726,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986670928.29/warc/CC-MAIN-20191016213112-20191017000612-00505.warc.gz | 163,098,330 | 4,732 | # Parametric Vector Equation Form
This post categorized under Vector and posted on June 16th, 2019.
## How To Know When The Solution To A Matrix Is Given In Parametric Form
A matrix in row-echelon form will have zeros both above and below the leading ones. Gauss-Jordan Elimination places a matrix into reduced row-echel [more]
## Point Give Parametric Vector Form General Solution Following System Equations X X Q
31.05.2014 In this vector we derive the vector and parametic equations for a line in 3 dimensions. We then do an easy example of finding the equati [more]
## Vector Parametric Equation Through Two Points
In this section we will derive the vector form and parametric form for the equation of lines in three dimensional graphice. We will also give the s [more]
## How Do You Determine Scalar Vector And Parametric Equations For The Plane That P
03.03.2018 1. The problem statement all variables and givenknown data Find the scalar vector and parametric equations of a plane that has a normal [more] | 230 | 1,020 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2019-43 | latest | en | 0.810636 |
https://www.geeksforgeeks.org/gate-gate-it-2004-question-62/?ref=lbp | 1,639,061,678,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964364169.99/warc/CC-MAIN-20211209122503-20211209152503-00209.warc.gz | 826,913,733 | 20,851 | # GATE | GATE-IT-2004 | Question 62
• Last Updated : 07 Dec, 2021
A disk has 200 tracks (numbered 0 through 199). At a given time, it was servicing the request of reading data from track 120, and at the previous request, service was for track 90. The pending requests (in order of their arrival) are for track numbers.
30 70 115 130 110 80 20 25.
Attention reader! Don’t stop learning now. Practice GATE exam well before the actual exam with the subject-wise and overall quizzes available in GATE Test Series Course.
Learn all GATE CS concepts with Free Live Classes on our youtube channel.
How many times will the head change its direction for the disk scheduling policies SSTF(Shortest Seek Time First) and FCFS (First Come First Serve)
(A) 2 and 3
(B) 3 and 3
(C) 3 and 4
(D) 4 and 4
Explanation:
According to Shortest Seek Time First:
90-> 120-> 115-> 110-> 130-> 80-> 70-> 30-> 25-> 20
Change of direction(Total 3); 120->15; 110->130; 130->80
According to First Come First Serve:
90-> 120-> 30-> 70-> 115-> 130-> 110-> 80-> 20-> 25
Change of direction(Total 4); 120->30; 30->70; 130->110;20->25 | 346 | 1,115 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2021-49 | latest | en | 0.774177 |
pcnerdythoughts.blogspot.com | 1,524,460,465,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125945793.18/warc/CC-MAIN-20180423050940-20180423070940-00591.warc.gz | 238,276,446 | 12,631 | ## Friday, August 23, 2013
### Nerd cred
So the math already came in handy. I am now the proud owner of a hot pink compass (Thanks, Bib!).
I was trying to figure out how I'd use that to slice up a circle (I have a ruler, and a calculator as well) so I can use the sweet wind projects book my Mom gave me. Turns out you can use the radius of the circle and the compass to mark any number of divisions provided you aren't scared of a little math. :)
So you draw a circle, then make a starting point with your ruler. Let's say you need to divide the circle into three to make a 3 bladed pinwheel. Well you know that the three blades will take up a third of the circle, which means they'll be 120 degrees. Well that's not very helpful since you need a right triangle to use sines and such. But hey! You can draw a triangle between your center point, and two imaginary points at 120 degrees. Then you can slice that in half to form 30-60-90 triangles! :D
Anywho, the short of it is you multiply your radius times the square root of three. That's the distance between your starting point on the circle, and the next third (since I can only measure straight distances with my current tools).
Booyeah!
Math.
Pinwheels here I come.
#### 1 comment:
Anna said...
Whatever that means! Love you. I miss geometry... | 317 | 1,314 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2018-17 | latest | en | 0.955458 |
mariahhillalpaca.blogspot.com | 1,513,116,304,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948519776.34/warc/CC-MAIN-20171212212152-20171212232152-00675.warc.gz | 177,687,986 | 20,062 | ## Saturday, June 27, 2009
### Understanding -Fibre Testing Technology terms
Fiber Testing Terminology
Normal Distribution
The graph of a normal distribution, the normal curve, is a bell-shaped curve. Many biological phenomena including animal fiber diameter distributions for single-coated animals, result in data distributed in a close approximation to normal. Hence, statistics applicable to normally distributed populations (mean, standard deviation, and coefficient of variation) are used to define these fiber diameter distributions. The normal curve is symmetric about a vertical center line. This center line passes through the value (the high point of the bell) that is the mean, median and the mode of the distribution. A normal distribution is completely determined when its mean and standard deviation are known.
Approximately sixty-eight percent of all me
asurements lie within one standard deviation of the mean and approximately 95.0 percent of all measurements lie within two standard deviations of the mean. More than 99.5 percent of all measurements will lie within three standard deviations of the mean.
Fiber Diameter Measurement and Distribution
Fiber diameter is measured in microns. One micron is equal to 1/1,000,000th of a meter or 1/25,400th of one inch. Mean Fiber Diameter (MFD) is in common use internationally. MFD, Standard
Deviation (SD) and Coefficient of Variation (CV) all relate to the (approximate) normal distribution of the animal fiber diameters. SD characterizes dispersion of individual measurements around the mean.
In a normal population, 66% of the individual values lie within one SD of the mean, 95% within two SD’s and 99% within 2.6 SD’s. Since SD tends to increase with increasing MFD, some people prefer to use CV (=SD*100/MFD) as a method of comparing variability about different sized means.
Comfort Factor
Comfort factor is the percentage of fibers over 30 microns subtracted from 100 percent. Ten percent of fibers over 30 microns corresponds to a comfort factor of 90 percent.
Curvature
Fiber curvature is related to crimp. Average Fiber Curvature (AFC) is determined by the measurement of two millimeter (2mm) snippets in degrees per millimeter (deg/mm). The greater the number of degrees per millimeter, the finer the crimp. For wool, low curvature is described as less than 50 deg/mm, medium curvature as the range of 60-90 deg/mm, and high curvature as greater than100 deg/mm.
Typical values might be illustrated by a 30 micron Crossbred wool fleece with typically low curvature and broader crimp with a frequency of approximately two crimps/cm. In contrast, a 21 micron Merino fleece typically has a medium curvature and a medium crimp with a frequency of approximately four (4) crimps/cm. A 16 micron Superfine Merino fleece typically has a high curvature and a fine crimp with a frequency of approximately seven (7) crimps/cm.
Definition of Medullation
A medullated fiber is an animal fiber that in its original state includes a medulla. A medulla in mammalian hair fibers is the more or less continuous cellular marrow inside the cortical layer in most medium and coarse alpaca fibers. By definition (ASTM), a kemp fiber is a medullated fiber in which the diameter of the medulla is 60% or more of the diameter of the fiber.
Medullation Measurement
Medullation measurement can be performed using either a projection microscope or the OFDA 100. Using IWTO nomenclature, a kemp fiber is classified as an “objectionable fiber” when measured on the OFDA 100. The OFDA100 measures opacity and therefore only white or light colored fibers can be measured. A reasonable assumption is that colored fibers have similar levels of medullated fibers as their white and pastel counterparts.
Spinning Fineness
This number (expressed in microns) provides an estimate of the performance of the sample when it is spun into yarn by combining the measured mean fiber diameter (MFD) and the measured coefficient of variation (CV). The original theory comes from Martindale, but the formula used comes from Butler and Dolling and normalizes the equation so that the spinning fineness is the same as the MFD when the CV is 24%.
Length & Strength
Length is measured in millimeters (mm) and the reported measurements readjusted to an annual growth period. Strength is measured in Newtons/kilotex (N/ktex) and is the force (measured in Newtons) required to break a staple of a given thickness (measured in kilotex). On the earth’s surface, one kilogram exerts a force of 9.8 Newtons (= 1kg * acceleration due to gravity measured in meters/second2). Kilotex indicates thickness in terms of mass per unit length expressed as kg/km.
Intrinsically, alpaca fibers appear to be very strong, an average of 50 N/ktex or better is not unusual. From a processing point of view, a mean staple strength greater than 30 N/ktex is considered adequate for pro-cessing wool on today’s high-speed equipment.
Resistance to Compression
The resistance to compression (RTC) of alpaca fibers is measured in kilopascals (Kpa). A pascal (Pa) is a unit of pressure equivalent to the force of one Newton per square meter. In the commercial sector, RTC values >11 kPa are considered high, 8 to 11 kPa medium, and <8>
Position of Break
Truly sound fibers break in the middle section of the staple. Intrinsically, alpaca fibers appear to be very strong, in the 50 N/ktex range. A mean staple strength greater than 30 N/ktex is considered adequate for processing wool on today’s high-speed equipment.
Clean Yield
Yield is based on bone-dry, extractives-free wool (alpaca) fiber or wool (alpaca) base (WB). Many different “commercial” yields are used in the international marketing of wool fibers. These are values calculated to predict the amount of clean fiber obtained after commercial scouring and/or after combing. Allowances are typically made for grease, ash, vegetable matter, and moisture. Various percentages of moisture are added in these calculations of commercial yield, which in some cases (very clean wool or some alpaca yields) may result in the clean yield exceeding 100%. | 1,360 | 6,115 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2017-51 | longest | en | 0.897944 |
https://lists.w3.org/Archives/Public/www-rdf-interest/2004Nov/0054.html | 1,685,913,173,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224650264.9/warc/CC-MAIN-20230604193207-20230604223207-00740.warc.gz | 403,955,879 | 4,687 | # RE: I guess it's a stupid questions...
Date: Sat, 13 Nov 2004 10:46:23 +0100
To: "'Daniel O'Connor'" <daniel.oconnor@gmail.com>, <www-rdf-interest@w3.org>
```
Yes all that does clear things up but...
Funnily enough I was thinking about a box when I did my first post - and of
how it can be tricky to shoehorn a box's properties into triples.
First, you need units - so each box side is a 'double' - not a triple...
Second - chosing a box is a bit confusing since it has (conveniently) three
dimensions. Does that mean we need a double to decribe a rectangle - and a
n-iple (no offence intended) for a hypercube?
Also, lets say we are interested in the specific gravity of the water in
bob's box, at a given temperature whose dimensions for some strange reason
have been variously recorded in inches, kilometers and cubits.
You could build a very long chain of triples describing conversions and math
operations but is this the best way to go?
Alternatively, by leveraging the widely deployed FishBoxML 'standard', a
more straightforward analytical approach can be used - maybe... ;-)
Neil McNaughton
Editor - Oil IT Journal (www.oilit.com)
-----Original Message-----
From: www-rdf-interest-request@w3.org
[mailto:www-rdf-interest-request@w3.org] On Behalf Of Daniel O'Connor
Sent: 13 November 2004 10:25
To: www-rdf-interest@w3.org
Subject: Re: I guess it's a stupid questions...
Perhaps this needs a more of a hands on demonstration.
<box>
<width>1</width>
<height>1</height>
<depth>1</depth>
</box>
XML. Fine and dandy. We know what a "box" is and what it's height,
width and depth properties are. You add in an extra property, and you
either have to change the schema or start using namespaces.
<box>
<width>1</width>
<height>1</height>
<depth>1</depth>
<contains>Fish</contains>
</box>
Okay, nothing too hard about that. Let's give the fish a name.
<box>
<width>1</width>
<height>1</height>
<depth>1</depth>
<contains><fish name="bob" /></contains>
</box>
That's wonderful. What if we want to take bob the fish out of our box?
<box>
<width>1</width>
<height>1</height>
<depth>1</depth>
<contains nodeID="bob"></contains>
</box>
<fish nodeID="bob" name="bob" />
Now, let's give bob a homepage.
<box>
<width>1</width>
<height>1</height>
<depth>1</depth>
<contains nodeID="bob"></contains>
</box>
<fish nodeID="bob">
<name>bob</name>
<homepage>http://www.fish.com/</homepage>
</fish>
What if there's too much information about bob the fish to express
every time you want to talk about bob? Let's take the definition of
bob the fish and put him in an xml file on his homepage.
<box>
<width>1</width>
<height>1</height>
<depth>1</depth>
<contains nodeID="bob"></contains>
</box>
<fish nodeID="bob">
<seeAlso>http://www.fish.com/bob.xml</seeAlso>
</fish>
Now, what if we didn't even know Bob was a fish, he was just a strange
mystery object in a box?
<box>
<width>1</width>
<height>1</height>
<depth>1</depth>
<contains resource="http://www.fish.com/bob.xml" />
</box>
And, if we want to say that bob is a fish, we know his name but we
don't know much else - his homepage has more information though.
<box>
<width>1</width>
<height>1</height>
<depth>1</depth>
<contains resource="http://www.fish.com/bob.xml" />
</box>
<name>bob</name>
</fish>
Suddenly, we've a lot of things that could go in the box. We don't
need to know about them, we just need to know where we *can* find out
more about them. Say we only know that bob is a fish in a box, not
what type of fish. We can use a common definition of a Fish (from
something like WordNet) and point to bob's description - which would
then offer a more meaningful description of Bob, the Flying Fish.
This is was RDF, RDFS and OWL are all about. Things getting more
detailed the more you look into them. XML can't really provide
*meaning*, which RDF can.
Does that clarify anything, or does it all seem a little fishy?
-Daniel O'Connor
--
http://www.ahsonline.com.au/dod/FOAF.rdf
```
Received on Saturday, 13 November 2004 09:46:54 UTC
This archive was generated by hypermail 2.4.0 : Friday, 17 January 2020 22:44:53 UTC | 1,165 | 4,104 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2023-23 | longest | en | 0.8393 |
https://www.classtools.net/QR/18-cF5cQ | 1,685,757,842,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224648911.0/warc/CC-MAIN-20230603000901-20230603030901-00690.warc.gz | 762,091,003 | 14,318 | # Print
## A. Prior to the lesson:
1. Arrange students into groups. Each group needs at least ONE person who has a mobile device.
2. If their phone camera doesn't automatically detect and decode QR codes, ask students to
• Bring these devices into the lesson.
4. Cut them out and place them around your class / school.
## B. The lesson:
1. Give each group a clipboard and a piece of paper so they can write down the decoded questions and their answers to them.
2. Explain to the students that the codes are hidden around the school. Each team will get ONE point for each question they correctly decode and copy down onto their sheet, and a further TWO points if they can then provide the correct answer and write this down underneath the question.
3. Away they go! The winner is the first team to return with the most correct answers in the time available. This could be within a lesson, or during a lunchbreak, or even over several days!
## C. TIPS / OTHER IDEAS
4. A detailed case study in how to set up a successful QR Scavenger Hunt using this tool can be found here.
### Question
1. An angle that measures 90 degreesRight Angle
2. An angle that measures less than 90 degreesAcute Angle
3. An angle that measures greater than 90 degreesObtuse Angle
4. A triangle has ________ sides and _______angles3 sides, 3 angles
5. The 3 angles of a triangle always add up to ____ degrees180 degrees
6. Has 3 equal sides and and 3 angles measure 60 degreesEquilateral Triangle
7. Has 2 equal sides with 2 equal anglesIsosceles Triangle
8. Has no equal sides and no equal anglesScalene Triangle
9. A four sided figure having four straight sidesQuadrilateral
10. A quadrilateral having only one pair of parallel sidesParallelogram
11. 2 Lines, side by side, that a have continuous path that will never touchParallel Lines
12. Lines that cross each otherIntersecting Lines
13. Lines that form right angles where they crossPerpendicular Lines
14. A straight path that goes on foreverLine
15. A straight path that goes on forever in ONE direction onlyRay
16. A straight path between 2 pointsLine Segment
17. Represented by a dotPoint
### STAAR MATH-4th Grade: QR Challenge
Question 1 (of 17)
### STAAR MATH-4th Grade: QR Challenge
Question 2 (of 17)
### STAAR MATH-4th Grade: QR Challenge
Question 3 (of 17)
### STAAR MATH-4th Grade: QR Challenge
Question 4 (of 17)
### STAAR MATH-4th Grade: QR Challenge
Question 5 (of 17)
### STAAR MATH-4th Grade: QR Challenge
Question 6 (of 17)
### STAAR MATH-4th Grade: QR Challenge
Question 7 (of 17)
### STAAR MATH-4th Grade: QR Challenge
Question 8 (of 17)
### STAAR MATH-4th Grade: QR Challenge
Question 9 (of 17)
### STAAR MATH-4th Grade: QR Challenge
Question 10 (of 17)
### STAAR MATH-4th Grade: QR Challenge
Question 11 (of 17)
### STAAR MATH-4th Grade: QR Challenge
Question 12 (of 17)
### STAAR MATH-4th Grade: QR Challenge
Question 13 (of 17)
### STAAR MATH-4th Grade: QR Challenge
Question 14 (of 17)
### STAAR MATH-4th Grade: QR Challenge
Question 15 (of 17)
### STAAR MATH-4th Grade: QR Challenge
Question 16 (of 17)
### STAAR MATH-4th Grade: QR Challenge
Question 17 (of 17) | 858 | 3,161 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2023-23 | latest | en | 0.897289 |
https://www.studystack.com/flashcard-1114151 | 1,513,423,257,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948587577.92/warc/CC-MAIN-20171216104016-20171216130016-00199.warc.gz | 793,033,521 | 18,261 | or
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# zaria's math keyterm
common multiples multiples that are shared by two or more numbers. For example. some common multiplies of 2 and 3 are 6,12,and 18.
equivalent fractions Fractions that name the same numbers.
greatest common facter (GCF) The greatest of the common factors of two or more numbers. The 24 and 30 is 6.
improper fraction A fraction that has a numerator that is greater than or equal to the denominator. The value of an improper fraction is greater than or equal to 1.
least common denominator (LCD) The least common multiple of the denominators of two or more fractions.
least common multiple (LCM) The least of the common multiples of two or more numbers. The LCM of 2 and 3 is 6.
mixed number The sum of a wholenumber and a fraction.
multiple The product of the number and any whole number.
simplest form The form of a fraction when the GCF of the numerator and the denominatoris 1. The fraction 3 over 4 is 1.
venn diagram A diagram that uses circles to display elements of diffrent sets.Overlapping circles show common elements.
Created by: zahboo | 376 | 1,573 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-51 | latest | en | 0.880246 |
https://number.academy/86755 | 1,660,866,909,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882573533.87/warc/CC-MAIN-20220818215509-20220819005509-00545.warc.gz | 391,452,560 | 12,053 | # Number 86755
Number 86,755 spell 🔊, write in words: eighty-six thousand, seven hundred and fifty-five . Ordinal number 86755th is said 🔊 and write: eighty-six thousand, seven hundred and fifty-fifth. The meaning of number 86755 in Maths: Is Prime? Factorization and prime factors tree. The square root and cube root of 86755. What is 86755 in computer science, numerology, codes and images, writing and naming in other languages. Other interesting facts related to 86755.
## What is 86,755 in other units
The decimal (Arabic) number 86755 converted to a Roman number is (L)(X)(X)(X)(V)MDCCLV. Roman and decimal number conversions.
#### Weight conversion
86755 kilograms (kg) = 191260.1 pounds (lbs)
86755 pounds (lbs) = 39351.8 kilograms (kg)
#### Length conversion
86755 kilometers (km) equals to 53908 miles (mi).
86755 miles (mi) equals to 139619 kilometers (km).
86755 meters (m) equals to 284626 feet (ft).
86755 feet (ft) equals 26444 meters (m).
86755 centimeters (cm) equals to 34155.5 inches (in).
86755 inches (in) equals to 220357.7 centimeters (cm).
#### Temperature conversion
86755° Fahrenheit (°F) equals to 48179.4° Celsius (°C)
86755° Celsius (°C) equals to 156191° Fahrenheit (°F)
#### Time conversion
(hours, minutes, seconds, days, weeks)
86755 seconds equals to 1 day, 5 minutes, 55 seconds
86755 minutes equals to 2 months, 4 days, 5 hours, 55 minutes
### Codes and images of the number 86755
Number 86755 morse code: ---.. -.... --... ..... .....
Sign language for number 86755:
Number 86755 in braille:
Images of the number
Image (1) of the numberImage (2) of the number
More images, other sizes, codes and colors ...
## Mathematics of no. 86755
### Multiplications
#### Multiplication table of 86755
86755 multiplied by two equals 173510 (86755 x 2 = 173510).
86755 multiplied by three equals 260265 (86755 x 3 = 260265).
86755 multiplied by four equals 347020 (86755 x 4 = 347020).
86755 multiplied by five equals 433775 (86755 x 5 = 433775).
86755 multiplied by six equals 520530 (86755 x 6 = 520530).
86755 multiplied by seven equals 607285 (86755 x 7 = 607285).
86755 multiplied by eight equals 694040 (86755 x 8 = 694040).
86755 multiplied by nine equals 780795 (86755 x 9 = 780795).
show multiplications by 6, 7, 8, 9 ...
### Fractions: decimal fraction and common fraction
#### Fraction table of 86755
Half of 86755 is 43377,5 (86755 / 2 = 43377,5 = 43377 1/2).
One third of 86755 is 28918,3333 (86755 / 3 = 28918,3333 = 28918 1/3).
One quarter of 86755 is 21688,75 (86755 / 4 = 21688,75 = 21688 3/4).
One fifth of 86755 is 17351 (86755 / 5 = 17351).
One sixth of 86755 is 14459,1667 (86755 / 6 = 14459,1667 = 14459 1/6).
One seventh of 86755 is 12393,5714 (86755 / 7 = 12393,5714 = 12393 4/7).
One eighth of 86755 is 10844,375 (86755 / 8 = 10844,375 = 10844 3/8).
One ninth of 86755 is 9639,4444 (86755 / 9 = 9639,4444 = 9639 4/9).
show fractions by 6, 7, 8, 9 ...
### Calculator
86755
#### Is Prime?
The number 86755 is not a prime number. The closest prime numbers are 86753, 86767.
#### Factorization and factors (dividers)
The prime factors of 86755 are 5 * 17351
The factors of 86755 are 1 , 5 , 17351 , 86755
Total factors 4.
Sum of factors 104112 (17357).
#### Powers
The second power of 867552 is 7.526.430.025.
The third power of 867553 is 652.955.436.818.875.
#### Roots
The square root √86755 is 294,542017.
The cube root of 386755 is 44,268843.
#### Logarithms
The natural logarithm of No. ln 86755 = loge 86755 = 11,370843.
The logarithm to base 10 of No. log10 86755 = 4,938295.
The Napierian logarithm of No. log1/e 86755 = -11,370843.
### Trigonometric functions
The cosine of 86755 is -0,996711.
The sine of 86755 is 0,08104.
The tangent of 86755 is -0,081307.
### Properties of the number 86755
Is a Friedman number: No
Is a Fibonacci number: No
Is a Bell number: No
Is a palindromic number: No
Is a pentagonal number: No
Is a perfect number: No
## Number 86755 in Computer Science
Code typeCode value
86755 Number of bytes84.7KB
Unix timeUnix time 86755 is equal to Friday Jan. 2, 1970, 12:05:55 a.m. GMT
IPv4, IPv6Number 86755 internet address in dotted format v4 0.1.82.227, v6 ::1:52e3
86755 Decimal = 10101001011100011 Binary
86755 Decimal = 11102000011 Ternary
86755 Decimal = 251343 Octal
86755 Decimal = 152E3 Hexadecimal (0x152e3 hex)
86755 BASE64ODY3NTU=
86755 MD5b6855878e5648e33def2f909a03d471a
86755 SHA1565f43b3b9c805ff74d843cdce443f6a0ecf2e55
86755 SHA224e704320b24a9b34c478583ed874de04b622686a3b2bae8ba0520f286
86755 SHA2569a422bd7fdf5941f933cc3b78281c4d4d59494e163331563d9c9a01223f47ed0
More SHA codes related to the number 86755 ...
If you know something interesting about the 86755 number that you did not find on this page, do not hesitate to write us here.
## Numerology 86755
### Character frequency in number 86755
Character (importance) frequency for numerology.
Character: Frequency: 8 1 6 1 7 1 5 2
### Classical numerology
According to classical numerology, to know what each number means, you have to reduce it to a single figure, with the number 86755, the numbers 8+6+7+5+5 = 3+1 = 4 are added and the meaning of the number 4 is sought.
## Interesting facts about the number 86755
### Asteroids
• (86755) 2000 GO68 is asteroid number 86755. It was discovered by LINEAR, Lincoln Near-Earth Asteroid Research from Lincoln Laboratory, Socorro on 4/5/2000.
## № 86,755 in other languages
How to say or write the number eighty-six thousand, seven hundred and fifty-five in Spanish, German, French and other languages. The character used as the thousands separator.
Spanish: 🔊 (número 86.755) ochenta y seis mil setecientos cincuenta y cinco German: 🔊 (Anzahl 86.755) sechsundachtzigtausendsiebenhundertfünfundfünfzig French: 🔊 (nombre 86 755) quatre-vingt-six mille sept cent cinquante-cinq Portuguese: 🔊 (número 86 755) oitenta e seis mil, setecentos e cinquenta e cinco Chinese: 🔊 (数 86 755) 八万六千七百五十五 Arabian: 🔊 (عدد 86,755) ستة و ثمانون ألفاً و سبعمائة و خمسة و خمسون Czech: 🔊 (číslo 86 755) osmdesát šest tisíc sedmset padesát pět Korean: 🔊 (번호 86,755) 팔만 육천칠백오십오 Danish: 🔊 (nummer 86 755) seksogfirstusinde og syvhundrede og femoghalvtreds Dutch: 🔊 (nummer 86 755) zesentachtigduizendzevenhonderdvijfenvijftig Japanese: 🔊 (数 86,755) 八万六千七百五十五 Indonesian: 🔊 (jumlah 86.755) delapan puluh enam ribu tujuh ratus lima puluh lima Italian: 🔊 (numero 86 755) ottantaseimilasettecentocinquantacinque Norwegian: 🔊 (nummer 86 755) åtti-seks tusen, syv hundre og femti-fem Polish: 🔊 (liczba 86 755) osiemdziesiąt sześć tysięcy siedemset pięćdziesiąt pięć Russian: 🔊 (номер 86 755) восемьдесят шесть тысяч семьсот пятьдесят пять Turkish: 🔊 (numara 86,755) seksenaltıbinyediyüzellibeş Thai: 🔊 (จำนวน 86 755) แปดหมื่นหกพันเจ็ดร้อยห้าสิบห้า Ukrainian: 🔊 (номер 86 755) вiсiмдесят шiсть тисяч сiмсот п'ятдесят п'ять Vietnamese: 🔊 (con số 86.755) tám mươi sáu nghìn bảy trăm năm mươi lăm Other languages ...
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If you know something interesting about the number 86755 or any natural number (positive integer) please write us here or on facebook. | 2,452 | 7,110 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2022-33 | latest | en | 0.667205 |
https://cs50.stackexchange.com/questions/30235/calculate-n-semitones-for-frequency-in-pset3 | 1,632,106,477,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056974.30/warc/CC-MAIN-20210920010331-20210920040331-00716.warc.gz | 240,447,892 | 28,904 | # Calculate n (semitones) for frequency in pset3
I'm having a bit of a hard time trying to calculate the value of n(the number of semitones from A4).
I'm currently using the following formula:
```if (octave > A4_OCTAVE) { n = (octave - A4_OCTAVE) * NUMBER_OF_SEMITONES; } else if (octave < A4_OCTAVE) { n = (A4_OCTAVE - octave) * NUMBER_OF_SEMITONES; } ```
where A4_OCTAVE = 4, NUMBER_OF_SEMITONES = 12
and octave is gotten from the last position of the note passed into the frequency function and then changed to an int using atoi().
Thanks for the help!
Why do you need an `if`/`else`? In my code, a negative number stands for octaves below, a positive for octaves above octave 4. This gets added to the number of semitones relative to A of the same octave (and possibly further modified by b or #). I don't mind the total going negative, as `pow` works fine with those.
• I have some table that maps `note[0]-'A'` (similar to `caesar` and `vigenere`) to a value, for example element for index `2` (`'C'-'A'`) would be `-9`, as C4 is 9 semitones below A4. You could also use `switch`/`case` or an `if`/`else` chain. Sep 28 '18 at 9:34 | 346 | 1,143 | {"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} | 3.03125 | 3 | CC-MAIN-2021-39 | latest | en | 0.855222 |
https://www.mcqlearn.com/grade9/physics/thermal-properties-of-matter.php?page=11 | 1,561,467,040,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627999838.23/warc/CC-MAIN-20190625112522-20190625134522-00021.warc.gz | 823,008,406 | 6,811 | Thermal properties of matter multiple choice questions (MCQs), thermal properties of matter quiz answers 11 to learn secondary school physics online courses. Thermometer MCQs, thermal properties of matter quiz questions and answers for online secondary education degree. Thermometer, thermal expansion, physics: temperature test for secondary school teaching certification.
Learn high school physics MCQs: Thermometer, thermal expansion, physics: temperature, with choices 1.0057 m, 1.00057 m, 1.057 m, and 15.7 m for online secondary education degree. Free physics study guide for online learning thermometer quiz questions to attempt multiple choice questions based test.
## MCQ on Thermal Properties of Matter Worksheets 11 PDF Book Download
MCQ: 50°C on Fahrenheit scale is
1. 123°F
2. 120°F
3. 135°F
4. 122°F
D
MCQ: A brass rod is 1m long at 0°C, if coefficient of expansion is 1.9 * 10-5 K-1 then length at 30°C is
1. 1.00057 m
2. 1.0057 m
3. 1.057 m
4. 15.7 m
A
MCQ: Volume thermal expansion is also termed as
1. area thermal expansion
3. circle thermal expansion
4. cubical thermal expansion
D
MCQ: On Fahrenheit scale, lower fixed point is marked as
1. 100 °F
2. 0 °F
3. 32 °F
4. 212 °F
C
MCQ: If a bross rod is 1m long at 0 °C and α is 1.9×10-5 K-1, then its length at 30°C is
1. 1.057 m
2. 1.00057 m
3. 1.57 m
4. 1.0057 m
B | 418 | 1,351 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-26 | latest | en | 0.719675 |
https://www.doubtnut.com/qna/209197262 | 1,723,688,364,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641141870.93/warc/CC-MAIN-20240815012836-20240815042836-00359.warc.gz | 576,898,080 | 37,343 | # A vessel A of volume 5 litre has a gas at pressure of 80 cm column of Hg. This is joined to another evacuated vessel B of volume 3 litre. If now the stopcock S is opened and the apprature is maintained at constant temperature then the common pressure will become -
A
80 cm of Hg
B
50 cm of Hg
C
30 cm of Hg
D
None of these
Video Solution
|
Step by step video & image solution for A vessel A of volume 5 litre has a gas at pressure of 80 cm column of Hg. This is joined to another evacuated vessel B of volume 3 litre. If now the stopcock S is opened and the apprature is maintained at constant temperature then the common pressure will become - by Physics experts to help you in doubts & scoring excellent marks in Class 12 exams.
Updated on:21/07/2023
### Knowledge Check
• Question 1 - Select One
## A vessel containing 5 litres of a gas at 0.8 m pressure is connected to an evacuated vessel of volume 3 litres. The resultant pressure inside with be (assuming whole system to be isolated)
A43m
B0.5m
C2.0m
D34m
• Question 2 - Select One
## A gas is enclosed in a vessel of volume V at temperature T1 and P, the vessel is connected to another vessel of volume V/2 by a tube and a stopcock. The second vessel is initially evacuated. If the stopcock is opened, the temperature of second vessel becomes T2. The first vessel is maintained at a temperature T1. What is the final pressure P1 in the apparatus ?
A2PT22T2+T1
B2PT1T2+2T1
CPT22T2+T1
D2PT2T1+T2
• Question 3 - Select One
## One litre of oxygen at a pressure of 1 atm and two litres of nitrogen at a pressure of 0.5 atm are introduced into a vessel of volume 1 litre. If there is no change in temperature, the final pressure of the mixture of gas (in atm) is
A1. 5
B2. 5
C2
D4
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Doubtnut is the perfect NEET and IIT JEE preparation App. Get solutions for NEET and IIT JEE previous years papers, along with chapter wise NEET MCQ solutions. Get all the study material in Hindi medium and English medium for IIT JEE and NEET preparation | 736 | 2,742 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2024-33 | latest | en | 0.907031 |
https://math.stackexchange.com/questions/1973410/stats-proof-tossing-a-coin/1973537 | 1,656,978,433,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104506762.79/warc/CC-MAIN-20220704232527-20220705022527-00265.warc.gz | 427,965,726 | 64,499 | # Stats Proof - Tossing a coin
Let $f_n$ be the number of ways of tossing a fair coin $n$ times so that two consecutive heads never appear.
Prove that $f_n = f_{n-1}+f_{n-2}$ and thereby determine $f_n$.
What is the probability that two consecutive heads will not appear in $n$ tosses of a fair coin?
My thoughts on the question: - really have no idea where to begin. Do you use binomial distribution to solve this since you want $n$ successes out of $x$ tries?
• You really should include your own thoughts on the problem. Oct 18, 2016 at 0:29
• added my thoughts Oct 18, 2016 at 0:39
Consider the last toss. If head, the one before the last must be a tail and the number of ways to form the rest of the sequence is $f(n-2)$. If tail, the rest of the sequence can be formed in $f(n-1)$ ways, because this last toss doesn't matter. The latter means that $f(n) = f(n-1)+f(n-2)$. Seeing also that $f(1)=2$ which is the third Fibonacci number and $f(2)=3$ which is the fourth Fibonacci number, we get to the answer $f(n)=F_{n+2}$, e.g. the $n-2$ Fibonacci number. The corresponding probability will be $\frac{F_{n+2}}{2^{n}}$. | 332 | 1,129 | {"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} | 4.0625 | 4 | CC-MAIN-2022-27 | latest | en | 0.908941 |
https://www.jiskha.com/display.cgi?id=1272598493 | 1,502,927,840,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886102757.45/warc/CC-MAIN-20170816231829-20170817011829-00013.warc.gz | 928,366,946 | 4,089 | # statistics
posted by .
A pollster conducts a survey of a random sample of voters in a community to estimate the proportion who support a measure on public health insurance. Let P be the proportion of the population who support the measure. The pollster takes the sample of 200 voters and tests Ho: P=0.5 versus H1:P(not equal)0.5 at 5% level. What is the power of the test if the true value of P is 0.55?
• statistics -
The probability of making a Type II error is equal to beta. A Type II error is failure to reject the null when it is false. The power of the test is 1-beta and is the correct decision of rejecting the null when it is false. The alpha level directly affects the power of a test. The higher the level, the more powerful the test. Sample size also affects power. If your question is asking to calculate the actual power of the test, you will need to find methods to calculate beta.
I hope this helps.
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People end up tossing 12% of what they buy at the grocery store (Reader's Digest, March 2009). Assume this is the true population proportion and that you plan to take a sample survey of 540 grocery shoppers to further investigate their … | 636 | 2,877 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2017-34 | latest | en | 0.915749 |
https://singaporemathlearningcenter.com/singapore-math/singapore-math-number-bonds | 1,603,268,252,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107876136.24/warc/CC-MAIN-20201021064154-20201021094154-00423.warc.gz | 537,123,291 | 30,898 | Singapore Math Strategies
Singapore Math Strategies: Number Bonds
Number bonds are a great tool to use to build number sense, mental math, and teach addition and subtraction. Number bonds also help students see the ‘part-whole’ relationship of numbers that is an essential strategy in Singapore Math.
What are Number Bonds?
Number bonds are collections of numbers that can be added together to form a whole. Number bonds can be a useful tool to determine an answer without actual calculation or writing an equation.
The term number bond is also used to refer to a graphical depiction of part-part-whole relationships. The “whole” is written in the first circle and its “parts” are written in the adjacent circles.
A number bond represents the relationship between a quantity and the parts that combine to make it. The idea of number bonds is very fundamental, an understanding of the way numbers work. A number bond is made up of parts and if you know the parts, you can put them together (add) to find the whole. If you see the whole and one of the parts is unknown, you remove the part you know (subtract) from the whole to find the other part. A child learning that 3 add 7 equals 10 is also learning that 10 minus 7 equals 3 and that 10 minus 3 equals 7. Number bonds let children see the inverse association between addition and subtraction. To subtract means to figure out how much more you would have to add to get the whole. A child who understands this number bond will be able to instantaneously fill in any one of these three numbers if it was omitted, given the other two, without having to “work it out”.
In the example below, we can see the relationship between part and whole numbers and that 5 and 3 make 8.
Next, we can teach students that there are different ways to make the number 8. We can see that 8 and 0 make 8, 6 and 2 make 8 or that 4 and 4 make 8.
Number Bonds in Kindergarten/ Pre-school
When used in preschool or kindergarten, number bonds develop number sense and set a foundation to move on to addition and subtraction and mental math. Number bonds help children gain insight into how numbers work and help us to understand that a whole number is made up of parts. Number bonds are also an excellent preamble to the fact that subtraction is the reverse of addition. As students go on to higher grades, number bonds become an important mental problem-solving strategy.
The components of a number bond consist of circles connected by lines. A number bond contains at least 3 circles that are linked by lines. However, you can also build more complex number bonds like the ones below:
Benefits of Number Bonds.
Number bonds help develop a student’s number sense and mental math. As we can see from the above examples, Number bonds help students see that numbers can be “broken” into parts to make calculation easier. With number bonds, students distinguish the relationships between numbers through a visual model that shows how the numbers are related. A number bond helps students clearly see the Part/Whole relationship. The circles or squares are just a pictorial representation that students should begin with. By first grade, some students may move away from the shape visual. This enables children to be able to do mental arithmetic.
Number Bonds to 10
Number bonds are frequently learned in sets for which the sum is a common round number such as 10. Learning number bonds to 10 is important for mental mathematics, as this links into place value and the base 10 number structure.
Knowing number bonds with confidence allows children to develop strategies for solving more complicated mental problems. As a simple example, to carry out the calculation 8 + 5 mentally, a child who can instantly remember their number bonds to 10 and will identify that they need to add 8 to 2 to get up to 10. They then have 3 left over from the 5 and can now add that on to the 10 to make 13. Essentially, they are breaking the addition up into (8+ 2) + 3 = 10 + 3 = 13.
Learning number bonds can also benefit students as they progress to higher grade levels. The example below shows how a student would use number bonds to perform a mental calculation with larger numbers.
How To Teach Number Bonds To Your Child.
There are numerous ways to teach number bonds to a child.
One way to introduce kids to number bonds is by using the Concrete to Pictorial to Abstract approach.
At the concrete stage, students use manipulatives and objects such as counters or real-life objects to learn. At the pictorial stage, kids can progress to writing number bonds on whiteboards or paper. As learning progresses to the abstract stage, a student can start representing number bonds a mathematical notation (4+ 5 = 9).
Below are some activities that can be used to teach your child number bonds.
Exploring the Part and Whole Relationship
Since number bonds are used to help children recognize the relationship between a whole number and its parts, it is best to begin by exploring this part-whole relationship. Food (cheerios, gummy bears) works great for this! You can sort food into “whole” and “part” groups.
Explore the Number Bond Pictures.
Once a child is able to comprehend the meanings of the words “part” and “whole,” you can introduce number bond pictorial or use number bond templates. You then challenge the child to spot the “whole” circle and the “part” circles.
Display the pictorial or template in different directions as number bonds are displayed in different ways.
Construct a Number Bond
You can use paper plates and straws but get innovative and use what you have on hand! Make sure the child construct it in all directions and ask him/her to lay a hand on the whole and parts when named.
Sketch a Number Bond
Although number bonds are usually made with circles, other shapes like squares, triangles, stars, and rectangles can also be used.
Invite the child to draw number bonds in all directions using shapes of his/her choosing.
Worksheets
Using worksheets/ printables is also a great way to practice and master number bonds once the child is able to write.
Resources
Learning Resources Giant Magnetic Number Bonds
Didax 211011 Educational Resources Write-On Wipe-Off Five and Ten-Frame Mats
The CASCO Mathematics Tutor Book Series is a set of four assessment books imported directly from Singapore. The book is in color and perforated so you can use them as math worksheets.
Kindergarten/ Pre-Primary One Mathematics Tutor Book 1: Topics Covered: Numbers to 10, Number Bonds, addition within 10, Subtraction within 10, Shapes and Patterns
Kindergarten/ Pre-Primary One Mathematics Tutor Book 2: Topics Covered: Ordinal Numbers and Positions, Numbers to 20, Addition and Subtraction within 20, Length
Kindergarten/ Pre-Primary One Mathematics Tutor Book 3: Numbers to 40, Addition and Subtraction within 40, Picture Graphs, Multiplication, Division
Kindergarten/ Pre-Primary One Mathematics Tutor Book 4: Time, Numbers to 100, Addition within 100, Subtraction within 10, Addition and Subtraction to 100, Money (PLEASE NOTE that some of the questions are in Singapore Currency.)
Disclosure: This post might contain affiliate links and SingaporeMathLearningCenter.Com is a participant in the Amazon Associates program and may earn a small commission from qualifying purchases. | 1,546 | 7,346 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.78125 | 5 | CC-MAIN-2020-45 | longest | en | 0.940433 |
https://scribesoftimbuktu.com/solve-for-n-16-nn-81/ | 1,675,157,658,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499857.57/warc/CC-MAIN-20230131091122-20230131121122-00565.warc.gz | 528,018,881 | 13,168 | # Solve for n 16/n=n/81
16n=n81
Multiply the numerator of the first fraction by the denominator of the second fraction. Set this equal to the product of the denominator of the first fraction and the numerator of the second fraction.
16⋅81=n⋅n
Solve the equation for n.
Rewrite the equation as n⋅n=16⋅81.
n⋅n=16⋅81
Multiply n by n.
n2=16⋅81
Multiply 16 by 81.
n2=1296
Take the square root of both sides of the equation to eliminate the exponent on the left side.
n=±1296
The complete solution is the result of both the positive and negative portions of the solution.
Simplify the right side of the equation.
Rewrite 1296 as 362.
n=±362
Pull terms out from under the radical, assuming positive real numbers.
n=±36
n=±36
The complete solution is the result of both the positive and negative portions of the solution.
First, use the positive value of the ± to find the first solution.
n=36
Next, use the negative value of the ± to find the second solution.
n=-36
The complete solution is the result of both the positive and negative portions of the solution.
n=36,-36
n=36,-36
n=36,-36
n=36,-36
Solve for n 16/n=n/81
### Solving MATH problems
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Scroll to top | 349 | 1,229 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2023-06 | latest | en | 0.851641 |
https://studentshare.org/miscellaneous/1562632-statistics-review-questions | 1,550,461,782,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247484648.28/warc/CC-MAIN-20190218033722-20190218055722-00452.warc.gz | 710,782,344 | 22,869 | # Statistics Review Questions - Essay Example
Summary
100. Therefore, percentages mean out of 100 and when we compare two percentages with each other we intend to form our conclusion based on…
## Extract of sample"Statistics Review Questions"
Download file to see previous pages Correlation is a method to measure the association in between two variables. When we compare the correlating scores of two variables, we are trying to determine whether the variables are related to each other or not. The purpose of doing correlations is to allow us to make a prediction about one variable based on what we know about another variable.
A frequency distribution is the tabulation of raw data obtained by dividing it into classes of some size and computing the number of data elements (or their fraction out of the total) falling within each pair of class boundaries. A frequency distribution can be modeled as a histogram or as a pie chart (Frequency Distribution).
A pie chart shows the differences between two separate variables or subjects. A pie chart is a graph that is in the shape of a circle which represents a total of 100%. Other variables or subjects are shown on the chart with respect to their relative percentages to the whole. The different subjects are shown in different colors and the size of each subject in the pie is proportional to the percentage of the subject.
A bar graph shows raw data and it is designed to show different values of two or more subjects but instead of using the pie to represent data it uses horizontal and vertical bars that represent a different value. The bar graph has numbers along the side of the bars to indicate the value of the variable and there are scales which show what variable is being measured.
The difference between the pie chart and the bar graph is that a bar graph is capable of showing change over time. While a single pie chart cannot show changes over time by itself, it can only represent the given percentages at a fixed point in time.
A graphical display of a frequency table is called a frequency polygon. The X-axis has the intervals shown on it while the number of scores in each interval is represented by the height of a point located above the middle of the ...Download file to see next pagesRead More
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https://rdrr.io/cran/tgp/src/R/mrtgp.R | 1,606,866,791,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141685797.79/warc/CC-MAIN-20201201231155-20201202021155-00284.warc.gz | 387,952,707 | 11,117 | # R/mrtgp.R In tgp: Bayesian Treed Gaussian Process Models
```#*******************************************************************************
#
# Bayesian Regression and Adaptive Sampling with Gaussian Process Trees
# Copyright (C) 2005, University of California
#
# This library is free software; you can redistribute it and/or
# modify it under the terms of the GNU Lesser General Public
# version 2.1 of the License, or (at your option) any later version.
#
# This library is distributed in the hope that it will be useful,
# but WITHOUT ANY WARRANTY; without even the implied warranty of
# MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the GNU
# Lesser General Public License for more details.
#
# You should have received a copy of the GNU Lesser General Public
# License along with this library; if not, write to the Free Software
# Foundation, Inc., 51 Franklin Street, Fifth Floor, Boston, MA 02110-1301 USA
#
# Questions? Contact Robert B. Gramacy (rbgramacy@ams.ucsc.edu)
#
#*******************************************************************************
## mr.plot:
##
## plotting function for multiresolution tgp-class objects
## (i.e., those with corr=="mrexpsep") -- called by plot.tgp
"mr.plot" <-
function(x, pparts=TRUE, proj=NULL, center="mean", layout="both",
main=NULL, xlab=NULL, ylab=NULL, zlab=NULL, legendloc="topright",
gridlen=c(40,40), span=0.1, ...)
{
## 1-d plot of 1-d data described by two columns (resolutions)
if( x\$d==2 ){
## create plot window
par(mfrow=c(1,1))
## construct axis x&y labels
if(is.null(xlab)){xlab <- names(x\$X)[2]}
if(is.null(ylab)){ylab <- x\$response}
## collect the input and predictive data and pred outputs
center <- tgp.choose.center(x, center)
o <- order(center\$X[,2])
X <- center\$X[o,]
Z <- center\$Z[o]
smain <- paste(main, ylab, center\$name)
## collect quantiles
Z.q1 <- c(x\$Zp.q1, x\$ZZ.q1)[o]
Z.q2 <- c(x\$Zp.q2, x\$ZZ.q2)[o]
## plot the coarse and fine input data
plot(x\$X[x\$X[,1]==0,2],x\$Z[x\$X[,1]==0], ylim=range(c(Z,x\$Z)),
xlab=xlab, ylab=ylab, main=smain, col=4)
lines(x\$X[x\$X[,1]==1,2],x\$Z[x\$X[,1]==1], type="p", pch=20,
col=2)
if(! is.null(legendloc))
legend(legendloc, lty=c(1,2,1,2), col=c("blue", "blue", "red", "red"),
c(paste("coarse", center\$name), "coarse 90% CI",
paste("fine", center\$name), "fine 90% CI"))
## extract the coarse and fine resolutions
f<-X[,1]==1
c<-X[,1]==0
## add the coarse and fine mean and quantiles
lines(X[c,2], Z[c], col=4)
lines(X[f,2], Z[f], col=2)
lines(X[f,2], Z.q1[f], col=2, lty=3)
lines(X[f,2], Z.q2[f], col=2, lty=3)
lines(X[c,2], Z.q1[c], col=4, lty=3)
lines(X[c,2], Z.q2[c], col=4, lty=3)
if(pparts) tgp.plot.parts.1d(x\$parts[,2])
} else { ## make a projection for data is >= 2-d
## create plot window
par(mfrow=c(1,2))
if(is.null(proj)) proj <- c(1,2)
## create axis lables -- augment proj argument by one column
proj <- proj+1
if(is.null(xlab)){xlab <- names(x\$X)[proj[1]]}
if(is.null(ylab)){ylab <- names(x\$X)[proj[2]]}
## collect the input and predictive data and pred outputs
## this plot only plots the mean or median, no errors
center <- tgp.choose.center(x, center)
X <- center\$X; Z <- center\$Z
## separate X and Z into coarse and fine
c<-X[,1]==0; f<-X[,1]==1
Xc <- as.data.frame(X[c,proj])
Xf <- as.data.frame(X[f,proj])
Zc <- Z[c]; Zf <- Z[f]
## initialize the projection vectors p*
nXc <- nrow(Xc); pc <- seq(1,nXc)
nXf <- nrow(Xf); pf <- seq(1,nXf)
dX <- nrow(X)
## plot the coarse predictive (mean or median) surface
smain <- paste(main, x\$response, "coarse", center\$name)
slice.image(Xc[,1], Xc[,2], p=pc, z=Zc, xlab=xlab, ylab=ylab,
main=smain, gridlen=gridlen,span=span,
xlim=range(X[,proj[1]]), ylim=range(X[,proj[2]]), ...)
## add inputs and predictive locations
points(x\$X[x\$X[,1]==0,proj], pch=20, ...)
points(x\$XX[x\$XX[,1]==0,proj], pch=21, ...)
# plot parts
if(pparts & !is.null(x\$parts)) { tgp.plot.parts.2d(x\$parts, dx=proj)}
## plot the fine predictive (mean or median) surface
smain <- paste(main, x\$response, "fine", center\$name)
slice.image(Xf[,1], Xf[,2], p=pf, z=Zf, xlab=xlab, ylab=ylab,
main=smain, gridlen=gridlen, span=span,
xlim=range(X[,proj[1]]), ylim=range(X[,proj[2]]), ...)
## add inputs and predictive locations
points(x\$X[x\$X[,1]==1,proj], pch=20, ...)
points(x\$XX[x\$XX[,1]==1,proj],pch=21, ...)
# plot parts
if(pparts & !is.null(x\$parts)) { tgp.plot.parts.2d(x\$parts, dx=proj)}
}
}
## mr.checkrez:
##
## used for extreacting the predictive surface information for
## one of the two resolutions so that the surface for that
## resolution can be plotted using the regualr tgp plotting
## machinery in plot.tgp
"mr.checkrez" <-
function(b, res)
{
## select input data at the desired resolution
b\$d <- b\$d-1
rdata <- b\$X[,1]==res
b\$n <- sum(rdata)
cnames=names(b\$X)[-1]
b\$X <- as.data.frame(b\$X[rdata,-1])
colnames(b\$X) <- cnames
b\$Z <- b\$Z[rdata]
## predictive data at input locations for the desired resolution
b\$Zp.mean <- b\$Zp.mean[rdata]
b\$Zp.km <- b\$Zp.km[rdata]
b\$Zp.q <- b\$Zp.q[rdata]
b\$Zp.s2 <- b\$Zp.s2[rdata]
b\$Zp.ks2 <- b\$Zp.ks2[rdata]
b\$Zp.q1 <- b\$Zp.q1[rdata]
b\$Zp.q2<- b\$Zp.q2[rdata]
b\$Zp.med <- b\$Zp.med[rdata]
## predictive data at the predictive locations for the desired resolution
rpred <- b\$XX[,1]==res
b\$nn <- sum(rpred)
b\$XX <- as.data.frame(b\$XX[rpred,-1])
colnames(b\$XX) <- cnames
b\$ZZ <- b\$ZZ[rpred]
b\$ZZ.mean <- b\$ZZ.mean[rpred]
b\$ZZ.km <- b\$ZZ.km[rpred]
b\$ZZ.q <- b\$ZZ.q[rpred]
b\$ZZ.s2 <- b\$ZZ.s2[rpred]
b\$ZZ.ks2 <- b\$ZZ.ks2[rpred]
b\$ZZ.q1 <- b\$ZZ.q1[rpred]
b\$ZZ.q2<- b\$ZZ.q2[rpred]
b\$ZZ.med <- b\$ZZ.med[rpred]
b\$improv <- b\$improv[rpred,]
b\$parts <- b\$parts[,-1]
return(b)
}
```
## Try the tgp package in your browser
Any scripts or data that you put into this service are public.
tgp documentation built on Sept. 7, 2020, 5:10 p.m. | 1,986 | 5,852 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-50 | latest | en | 0.601839 |
https://cprieto.com/posts/2020/08/data-structures-binary-and-linear-search.html | 1,696,068,263,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510671.0/warc/CC-MAIN-20230930082033-20230930112033-00865.warc.gz | 193,613,139 | 5,525 | ## Data structures, Binary and linear search
on , , , 5 minutes reading
Another problem that goes hand by hand with sorting is the classical searching problem (and sometimes they need each other to work) and as sorting there are a few algorithms to pick from (sadly not as many as with sorting though). In its simplest form we just go through a collection, checking if the element we are searching for is there and if found we return the index or another number to show we failed, maybe something that will look like this in Kotlin:
fun <T> linearSearch(items: Collection<T>, what: T): Int {
// NOTE: yes, I know I could use indexOf or even better, indexOfFirst
for ((idx, item) in items.withIndex()) {
if (item == what) return idx
}
return -1
}
In the best case the element we are looking for is the first element in the collection so its time performance would be $$\mathcal{O}(1)$$ in the worst case scenario (let’s say the element is not in the list) we will need to go through all the elements, being this a $$\mathcal{O}(n)$$ operation (where $$n$$ is the number of elements in the collection).
Comparissons
Worst case (element not found) $$\mathcal{O}(n)$$
Best case (element is the first element) $$\mathcal{O}(1)$$
When the list is unsorted we cannot do much to improve the efficiency but a different situation happens when the list is already sorted, if that is the case we can split the collection in parts and check if the element is in any of the subparts, this is a good example of divide and conquer algorithms, we split the collection in smaller pieces and that will improve the time performance of our algorithm.
For a given ordered collection:
1. Take the element in the middle
2. If the element is what we are searching for, we found it!
3. If the element is greater than what we are searching for, we should search in the left of the element (between the start and the middle - 1 of the collection)
4. If the element is smaller than what we are searching for, we know the element should be in the right of the collection (between middle + 1 and the end of the collection)
5. Rinse and repeat until we get the element or report it is not in the collection
As you can see we divide the big collection in halves with every pass and that is why is called binary search (there are actually two different theories why is called like that but I prefer to say that it is because we divide the collection in two, see this StackExchange question).
import kotlin.math.floor
fun <T: Comparable<T>> binarySearch(items: List<T>, what: T): Int {
var start = 0
var end = items.size - 1
while (start <= end) {
val middle = floor(((start + end)/2).toDouble()).toInt()
when {
items[middle] > what -> end = middle - 1
items[middle] < what -> start = middle + 1
else -> return middle
}
}
return -1
}
How good is this? well, if we count the operations executed and assigning arbitrary constants $$C$$ we will get something like this:
$$T(n) = T(\frac{n}{2}) + C_k$$
Let’s remember the general form of the master theorem:
$$T(n) = aT(\frac{n}{b}) + \Theta(n^d)$$
It looks like something we can use!, we know $$C_k$$ is constant so that should be $$\Theta(1)$$ at the end of the equation, and the only way $$C_k = 1$$ is when we power it $$0$$, and $$a = 1$$:
$$T(n) = T(\frac{n}{b}) + \Theta(n^0)$$
This looks like the second form of the theorem ($$d = \log_b{a}$$ or $$0 = \log_2{1}$$) so we simplify it to $$\Theta(\log{n})$$, voilá!
Comparissons
Best case (element is right in the middle) $$\mathcal{O}(1)$$
Worst case (element is not there at all) $$\mathcal{O}(\log{n})$$
I don’t have to tell you $$\log{n}$$ is smaller than $$n$$ but if you have any doubt, go and check ;)
The test for both cases are simple enough:
class SearchTests {
private val unordered = listOf(5, 12, 56, 0)
private val ordered = listOf(0, 5, 12, 56, 105, 200)
@Test
fun linear search returns index of existing item() {
val found = linearSearch(unordered, 12)
assertEquals(1, found)
}
@Test
fun linear search returns -1 when not found() {
val found = linearSearch(unordered, 35)
assertEquals(-1, found)
}
@Test
fun classic binary search returns the index of the found element() {
val found = binarySearch(ordered, 5)
assertEquals(1, found)
}
@Test
fun classic binary search returns -1 when not found() {
val found = binarySearch(ordered, 2)
assertEquals(-1, found)
}
}
I found binary search a really nice algorithm to learn things like recursion, master theorem and divide and conquer techniques, I leave you as an exercise to write the recursive version of this algorithm in Kotlin (or your preferred learning language), I am pretty sure you will have a lot of fun! | 1,215 | 4,673 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 2, "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} | 3.765625 | 4 | CC-MAIN-2023-40 | latest | en | 0.88112 |
https://www.math-only-math.com/theorems-on-locus-of-a-point-which-is-equidistant-from-two-fixed-points.html | 1,723,317,878,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640822309.61/warc/CC-MAIN-20240810190707-20240810220707-00093.warc.gz | 685,869,625 | 13,087 | # Theorems on Locus of a Point which is Equidistant from Two Fixed Points
The locus of a point which is equidistant from two fixed points is the perpendicular bisector of the line segment joining the two fixed points.
Given,
Let X and Y be two given fixed points. PQ is the path traced out by the moving point P such that each point on it is equidistant from X and Y. Therefore, PX = PY.
To prove: PQ is the perpendicular bisector of the line segment XY.
Construction: Join X to Y. Let PQ cut XY at O.
Proof:
From △PXO and △PYO,
PX and PY (Given)
XO = YO (Since, every point of PQ is equidistant from X and Y, and O is a point on PQ.)
PO = PO (Common side.)
Therefore, by the SSS criterion of congruency△PXO ≅ △PYO.
Now ∠POX = ∠POY (since, corresponding parts of congruent triangles are congruent.)
Again ∠POX + ∠POY = 180° (Since, XOY is a straight line.
Therefore, ∠POX = ∠POY = $$\frac{180°}{2}$$ = 90°
Also, PQ bisects XY (Since, XO = YO)
Therefore, PQ ⊥ XY and PQ bisects XY, i.e., PQ is the perpendicular bisector of XY (Proved)
Loci
From Theorems on Locus of a Point which is Equidistant from Two Fixed Points to HOME
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4. ### 6th Grade Algebra Worksheet | Pre-Algebra worksheets with Free Answers
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In 6th Grade Algebra Worksheet you will get different types of questions on basic concept of algebra, questions on number pattern, dot pattern, number sequence pattern, pattern from matchsticks, conce… | 665 | 2,521 | {"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} | 4.125 | 4 | CC-MAIN-2024-33 | latest | en | 0.911624 |
https://randomperturbation.wordpress.com/2012/04/ | 1,521,413,899,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257646178.24/warc/CC-MAIN-20180318224057-20180319004057-00626.warc.gz | 676,720,056 | 28,589 | # Set Theory, Notes 1: Ordinals and Cardinals
Unless otherwise specified, we will assume ZF axioms. Recall the following ZF axiom:
Axiom of Infinity:
$(\exists\omega)(\varnothing\in\omega\wedge(\forall x\in\omega)(x\cup\{x\}\in\omega)).$
We will call $\omega$ the set of natural numbers. That is,
$\begin{array}{rcl}0&=&\varnothing\\1&=&0\cup\{0\}=\{0\}\\2&=&1\cup\{1\}=\{0,1\}\\&\vdots&\\n+1&=&n\cup\{n\}\\&\vdots&\end{array}$
We can then define $\omega+1=\omega\cup\{\omega\}$ and iterate as before, whence by applying the axiom of infinity again we can obtain $\omega\cdot 2:=\omega+\omega,$ and so on. All such sets generated by this process are called ordinals. This in turn gives us the canonical linear ordering of the ordinals where $n iff $n\in m.$
Definition 1. An ordinal $\alpha$ is a successor ordinal iff $\alpha=\beta+1.$ A limit ordinal is an ordinal which is not a successor ordinal.
Proposition 2. (Transfinite Induction) Let $Ord$ denote the class of all ordinals (in accordance to VBG notion of class) and $C$ be a class. If
1. $0\in C,$
2. $\alpha\in C\Longrightarrow\alpha+1\in C,$ and
3. if $\gamma$ is a limit ordinal and $\alpha\in C$ for all $\alpha<\gamma,$ then $\gamma\in C,$
then $C=Ord.$
Proof. Since $<$ is a linear ordering on the ordinals, let $\alpha$ be the least ordinal such that $\alpha\notin C.$ But then $\alpha+1\in C$ which is a contradiction. Hence $C=Ord.$
We can index ordinals with ordinals to generate the notion of a sequence of ordinals. We define an nondecreasing (nonincreasing) sequence of ordinals an ordered set $\{\gamma_\alpha\}$ of ordinals where $\gamma_\alpha\leq\gamma_\beta$ iff $\alpha\leq\beta.$
Definition 3. Let $\{\gamma_\alpha\}$ be an nondecreasing sequence of ordinals and $\xi$ be a limit ordinal and $\alpha<\xi.$ Then we define the limit of the sequence as
$\displaystyle\lim_{\alpha\to\xi}\gamma_\alpha=\sup_{\alpha<\xi}\{\gamma_\alpha\}.$
A dual definition can be defined for nonincreasing sequences, in which case the limits can be respectively distinguished as left and right limits. A sequence $\{\gamma_\alpha\}$ is continuous if for every limit ordinal $\xi$ in the indexing subclass we have
$\displaystyle\lim_{\alpha\to\xi}\gamma_\alpha=\gamma_\xi.$
An example of a sequence which is not continuous may be one of the form $S=(...,\gamma_{\beta},\gamma_{\beta+1},...)$ where $\gamma_{\beta},\gamma_{\beta+1}$ are both limit ordinals. So in this case
$\displaystyle\lim_{\alpha\to\beta+1}\gamma_\alpha=\sup_{\alpha<\beta+1}\{\gamma_\alpha\}=\gamma_{\beta}\neq\gamma_{\beta+1}$
(since the sup is actually a max in this case).
Definition 4. (Ordinal Arithmetic) We define
1. $\alpha+0=\alpha,$
2. $\alpha+(\beta+1)=(\alpha+\beta)+1,$
3. $\alpha+\beta=\lim_{\gamma\to\beta}\alpha+\gamma.$
Multiplication:
1. $\alpha\cdot 0=0,$
2. $\alpha\cdot(\beta+1)=\alpha\cdot\beta+\alpha,$
3. $\alpha\cdot\beta=\lim_{\gamma\to\beta}\alpha\cdot\gamma.$
Exponentiation:
1. $\alpha^0=1,$
2. $\alpha^{\beta+1}=\alpha^\beta\cdot\alpha,$
3. $\alpha^\beta=\lim_{\gamma\to\beta}\alpha^\gamma.$
It follows that addition and multiplication are both associative, but not commutative. In particular one can see that
$1+\omega=\omega\neq\omega+1$
and
$2\cdot\omega=\omega\neq\omega\cdot 2=\omega+\omega.$
Definition 5. For a set $X,$ we define its cardinality, denoted $|X|,$ as the unique ordinal with which the set has a bijection. The corresponding subclass of ordinals is called the class of cardinals.
Proposition 6. If $|X|=\kappa,$ then $|P(X)|=2^\kappa.$
Proof. For every $A\subseteq X,$ define
$\displaystyle\chi_A(x)=\left\{\begin{array}{ll}1&x\in A\\ 0&x\in X-A\end{array}\right..$
Hence the mapping $f:A\mapsto\chi_A(X)$ is a bijection between $P(X)$ and $\{0,1\}^X.$
Hence in this context, Cantor’s theorem immediately follows: $|X|<|P(X)|.$
Proposition 7. Let $|A|=\kappa,$ $|B|=\lambda,$ and $A\cap B=\varnothing.$ Then
1. $|A\cup B|=\kappa+\lambda,$
2. $|A\times B|=\kappa\cdot\lambda,$
3. $\left|A^B\right|=\kappa^\lambda.$
Proof. We have
$|A\cup B|=|A\sqcup B|=|\kappa\sqcup\lambda|=\kappa+\lambda.$
Also if $f:A\to\kappa$ and $B\to\lambda$ are bijections, then we can define $h:A\times B\to\kappa\cdot\lambda$ by $h:(a,b)\mapsto f(a)\cdot g(b)\in\kappa\cdot\lambda,$ and this is easily seen to be a bijection.
And if $k\in A^B$ then we can define $h:k\mapsto k(b)^b\in\kappa^\lambda,$ which is also seen to be a bijection.
It thus follows that addition and multiplication of cardinals are commutative (that is, ordinal operations are commutative on this subclass).
Above we said $1+\omega=\omega\neq\omega +1$ and $2\cdot\omega=\omega\neq\omega\cdot 2.$ But
$\omega=|\omega|=|1+\omega|=|\omega+1|$
and
$\omega=|n\cdot\omega|=|\omega\cdot n|=\left|\omega\uparrow\uparrow n\right|$
for any finite ordinal $n$ with Knuth notation
$\displaystyle\omega\uparrow\uparrow n=\omega^{\omega^{\cdot^{\cdot^{\cdot^\omega}}}}$
having $n$ raisings of $\omega.$ Consider the following convention for defining countable infinite ordinals:
$\displaystyle \omega,\omega+1,...,\omega\cdot 2,...,\omega^2,...,\omega^\omega,...,\omega\uparrow\uparrow\omega=\varepsilon_0,...,\varepsilon_1=\varepsilon_0\uparrow\uparrow\varepsilon_0,...,\\\varepsilon_2=\varepsilon_1\uparrow\uparrow\varepsilon_1,...,\varepsilon_\omega,...,\varepsilon_{\varepsilon_0},...,\varepsilon_{\varepsilon_{\ddots}},...$
They are all called countable infinite ordinals since for any one of them $\alpha,$ $|\alpha|=\omega.$ To clarify when we are talking about cardinal numbers versus ordinal numbers, we will use aleph notation: $\aleph_0=\omega.$ From Cantor’s theorem above, we know that if $|X|=\aleph_0,$ then $|P(X)|=2^{\aleph_0}>\aleph_0.$ That is, the ordinal corresponding to $2^{\aleph_0}$ must be greater than all of the countable ordinals above, otherwise its cardinality would be $\aleph_0.$ This necessitates the notion of uncountable ordinals and corresponding uncountable cardinals. We use subscripts to characterize these: $\aleph_1=\omega_1,\aleph_2=\omega_2,$ etc.
Definition 8. An infinite cardinal $\aleph_\alpha$ is a successor cardinal iff $\alpha$ is a successor ordinal, and it is a limit cardinal iff $\alpha$ is a limit ordinal.
Definition 9. Let $\alpha$ be a limit ordinal. An increasing $\delta$sequence $(\beta_\gamma)_{\gamma<\delta}$ with $\delta$ a limit ordinal is cofinal in $\alpha$ if $\lim_{\gamma\to\delta}\beta_\gamma=\alpha.$ And if $\alpha$ is an ordinal, then we define its cofinality as
$\displaystyle\text{cf}\,\alpha=\inf\left\{\delta:\lim_{\gamma\to\delta}\beta_\gamma=\alpha\right\}.$
It is easy to verify that $\text{cf}\,\alpha=1$ iff $\alpha$ is a successor ordinal. Also $\text{cf}\,0=0,\text{cf}\,\omega=\omega,$ and $\text{cf}\,\omega_\alpha=\omega_\alpha$ for any finite ordinal $\alpha.$
Proposition 10. $\mbox{cf}\,\mbox{cf}\,\alpha=\mbox{cf}\,\alpha.$
Proof. Let $\mbox{cf}\,\alpha=c.$ Then $\lim_{\gamma\to c}\beta_\gamma=\alpha$ (in particular it is the smallest such $c$). Now if $\mbox{cf}\,\mbox{cf}\,\alpha=\mbox{cf}\,c=d,$ then certainly $d\leq c.$ Now since $(\beta_\gamma)$ is cofinal in $\alpha,$ a subsequence of indices $(\gamma_\delta)$ is cofinal in $c$ (where cofinality can be chosen to be $d$). So
$\displaystyle\lim_{\delta\to d}\gamma_\delta=c.$
But then $\left(\beta_{\gamma_\delta}\right)$ is cofinal in $\alpha.$ That is,
$\displaystyle\lim_{\delta\to d}\beta_{\gamma_\delta}=\alpha,$
whence $d\geq c.$
Definition 11. An ordinal $\alpha$ is regular if $\mbox{cf}\,\alpha=\alpha.$ It is singular if it is not regular.
Corollary 12. If $\alpha$ is a limit ordinal, then $\mbox{cf}\,\alpha$ is a regular cardinal.
Theorem 13. If $\kappa$ is an infinite cardinal, then $\kappa<\kappa^{\mbox{cf}\,\kappa}.$
[1] Jech, Thomas. Set Theory. 3rd Edition. Springer Monographs in Mathematics. Springer-Verlag. 2000.
# Science in the US
As it is no surprise about the education in the US, science in particular has also been suffering. Last week in the April meeting of the American Physics Society, a group of physicists made precisely this claim. The primary data used in the basis of their argument was the decline in degrees in science. This data is certainly consistent with the argument that the US is falling behind in science. There is no debate that good performance in science predicates the choice of a degree in science. That is to say, the decline in degrees in science is mostly a consequence of poor performance in science earlier–rather than the cause of it. So the natural question is: why are kids less interested in science?
The everyday life of a kid is governed by three structures: school, parents, and other media they encounter (including television, internet, and social structure). It follows that one should be able to attribute the decline in scientific interest to one or more of these structures. Suppose we assume two things: (1) that there is a problem at some generation with respect to science advocacy in some of these components, and (2) that the extent to which kids assimilate to the scientific understanding of their mentors (i.e. parents and teachers) is strictly less than complete (that is, in an abstract isolated teacher-student situation, the student can only learn a proper subset of the teacher’s knowledge). These two assumptions ensure a gradual decline in scientific understanding over the course of future generations.
These assumptions make it no surprise that deficiencies in science can only get worse over time. This only leaves two more questions: how did the deficiency begin, and how can it be fixed? I might take a bold but not entirely unreasonable guess that much of it can be attributed to what I previously called curriculum that is not “readily applicable to everyday situations”. Who is in charge of deciding this curriculum? The answer to this is: precisely those who originally learned this curriculum. So again, it is no surprise why nothing changes. Many point to issues like parenting, television, and other media as the problem. But each of these are easily traceable back to the fact that members of each of these components (parents and media affiliates) had a similar education anyway. This further reinforces the claim that the curriculum itself might need to change.
Now if we change the curriculum, the question of teachers’ ability to implement it arises. This can be addressed with changes in college curriculum that prepare emerging teachers for a new curriculum in the primary and secondary levels. This seems like it might be the first promising step to address the issue in the long run. In an era where governmental budget deficits are high, the appetite for further investment in education (itself a long-term yet critical investment) also seems to be on the decline. It’s no surprise that a fair amount of politicians’ value placed on scientific investment puts it on the back burner in a budget crisis given that a vast majority of them have no background in science to begin with. This is yet another reason that a change in curriculum seems more promising in my mind than further monetary investments in education. And since the payoffs for this will not be immediate, it is all the more reason to begin a change in curriculum now rather than later. If a scientific curriculum tied closer to reality and pragmatic obstacles implants itself in those who will become the future generations, the hope is that this will in turn resolve the future problems associated to the parents, media, and politicians.
# Fluid Mechanics in Seven Dimensions?
Cayley-Dickson Algebras:
The Cayley-Dickson algebras $\mathbb{A}_n$ are special real algebras satisfying
$\dim_\mathbb{R}\mathbb{A}_n=2^n$
for $n\geq 0.$ If we let $\{e_0,...,e_{2^n-1}\}$ be the basis of $\mathbb{A}_n,$ then the algebra is generated by the relations
$e_i^2=e_1\cdots e_{2^n-1}=-1$
where $i\geq 1.$ Hence $\mathbb{A}_0=\mathbb{R}$ and $\mathbb{A}_1=\mathbb{C}.$ $\mathbb{A}_2$ is called the quaternions and is often denoted $\mathbb{H}.$ $\mathbb{A}_3$ and $\mathbb{A}_4$ are called the octonions and sedenions respectively and are respectively also denoted $\mathbb{O}$ and $\mathbb{S}.$
The Cayley-Dickson algebras can also be constructed inductively using a recursive unary operation (called conjugation). It proceeds as follows, let $\mathbb{A}_0=\mathbb{R}.$ We define $\mathbb{A}_1$ constructively. Let $\mathbb{A}_1=\mathbb{R}\oplus\mathbb{R}$ and define the following operations:
$\begin{array}{rcl}(a,b)+(c,d)&=&(a+b,c+d)\\(a,b)(c,d)&=&(ac-bd,ad+bc)\\(a,b)^*&=&(a,-b).\end{array}$
One can verify this structure is isomorphic to $\mathbb{C}.$ Next we let $\mathbb{A}_2=\mathbb{A}_1\oplus\mathbb{A}_1=\mathbb{C}\oplus\mathbb{C}$ and define its structure
$\begin{array}{rcl}(a,b)(c,d)&=&(ac-d^*b,da+bc^*)\\(a,b)^*&=&(a^*,-b)\end{array}$
and addition remaining the same and the inner conjugation being complex conjugation. How does is this retain isomorphism to our original definition of $\mathbb{A}_2?$ Note the dimension is retained by the fact that
$\mathbb{A}_2=\mathbb{C}\oplus\mathbb{C}=\mathbb{R}\oplus\mathbb{R}\oplus\mathbb{R}\oplus\mathbb{R}.$
Hence using this as our inclusion, we show the original requirements that
$i^2=j^2=k^2=ijk=-1.$
We have
$\begin{array}{rcl}i^2&=&(0,1,0,0)^2\\&=&\left((0,1),(0,0)\right)^2\\&=&\left((0,1),(0,0)\right)\left((0,1),(0,0)\right)\\&=&\left((0,1)(0,1)-(0,0)^*(0,0),(0,0)(0,1)+(0,0)(0,1)^*\right)\\&=&\left((-1,0)-(0,0),(0,0)+(0,0)(0,-1)\right)\\&=&\left((-1,0),(0,0)\right)\\&=&\left(-1,0,0,0\right).\end{array}$
Similar computations can be done for $j$ and $k$ and for showing that
$(0,1,0,0)(0,0,1,0)(0,0,0,1)=(-1,0,0,0).$
It turns out $\mathbb{A}_3$ and beyond are not associative, so in our initial requirement that
$e_1\cdots e_{2^n-1}=-1,$
we will clarify that we mean
$\left(\cdots((e_1e_2)e_3)\cdots e_{2^n-2}\right)e_{2^n-1}=-1.$
We continue inductively be setting $\mathbb{A}_{n+1}=\mathbb{A}_n\oplus\mathbb{A}_n$ and defining product and conjugation as before (and componentwise addition). One can establish the following table of properties
$\begin{array}{|c|c|c|c|c|}\hline\mathbb{R}&\mathbb{C}&\mathbb{H}&\mathbb{O}&\mathbb{S}\\\hline\mbox{division algebra}&\mbox{division algebra}&\mbox{division algebra}&\mbox{division algebra}&\phantom{normed}\\\mbox{associative}&\mbox{associative}&\mbox{associative}&&\\\mbox{commutative}&\mbox{commutative}&&&\\\mbox{trivial conjugation}&&&&\\\end{array}$
where associativity and commutativity refer to the multiplication (the addition is always both). All of the above properties (save the division algebra structure) can at this point (albeit with some tediousness) be shown. If $a=(a_0,...,a_{2^n-1})\in\mathbb{A}_n,$ we can define its real part as
$\mbox{Re}\,a=a_0.$
Note for $\mathbb{R}$ we have
$a^*a=aa=a^2.$
For $a\in\mathbb{R}$ we can define $\|a\|=\sqrt{\mbox{Re}\,a^*a}=\sqrt{a^2}=|a|.$ For $\mathbb{C}$ we have
$(a,b)^*(a,b)=(a,-b)(a,b)=(a^2+b^2,ab-ba)=(a^2+b^2,0).$
Similarly we can define $\|(a,b)\|=\sqrt{\mbox{Re}\,(a,b)^*(a,b)}=\sqrt{a^2+b^2}.$ For $\mathbb{H}$ we have
$(a,b)^*(a,b)=(a^*,-b)(a,b)=(a^*a+b^*b,a^*b-a^*b)=(\|a\|^2+\|b\|^2,0).$
Hence $\|(a,b,c,d)\|=\sqrt{a^2+b^2+c^2+d^2}.$ And similarly for $\mathbb{O}$ we have
$(a,b)^*(a,b)=(a^*,-b)(a,b)=(\|a\|^2+\|b\|^2,0),$
so $\|(a_0,...,a_7)\|=\sqrt{a_0^2+\cdots+a_7^2}.$ Inductively, one can see that the norm on $\mathbb{A}_n$ coincides with the euclidean norm in $\mathbb{R}^{2^n}.$ We can also note that the multiplicative inverse of an element $x\in\mathbb{A}_n$ is
$x^{-1}=\frac{x^*}{\|x\|^2}.$
Although it turns out that in the sedenions and above ($\mathbb{A}_i$ for $i\geq 4$), there are zero divisors (for example, $(e_3+e_{10})(e_6-e_{15})=0$), so the algebra fails to be a division algebra.
Theorem 1. (Hurwitz’ Theorem) If $A$ is a real normed division algebra with identity, then $A\in\{\mathbb{R},\mathbb{C},\mathbb{H},\mathbb{O}\}.$
Corollary 2. (Frobenius’ Theorem) If $A$ is an associative real division algebra, then $A\in\{\mathbb{R},\mathbb{C},\mathbb{H}\}.$
Generalized Cross Products:
The cross product of two elements $x,y\in\mathbb{R}^3$ can be written
$x\times y=\begin{vmatrix}e_1&e_2&e_3\\x_1&x_2&x_3\\y_1&y_2&y_3\end{vmatrix}=(x_2y_3-x_3y_2,x_3y_1-x_1y_3,x_1y_2-x_2y_1).$
There is a natural way to define a cross product of $n-1$ elements in $\mathbb{R}^n$ by
$\mathsf{X}(v_1,...,v_{n-1})=\begin{vmatrix}e_1&\cdots&e_n\\v_{1_1}&\cdots&v_{1_n}\\\vdots&\ddots&\vdots\\v_{{n-1}_1}&\cdots&v_{{n-1}_n}\end{vmatrix}.$
However, our ultimate objective is to define vorticity in other dimensions. So we will want a binary cross product that we can apply to our differential operator and a vector field: $\nabla\times f.$ We want $\times$ to satisfy some conditions
$\begin{array}{rcl}(w+x)\times(y+z)&=&(w+x)\times y+(w+x)\times z=w\times y+x\times y+w\times z+x\times z\\ 0&=&x\cdot(x\times y)=(x\times y)\cdot y\\\|x\times y\|^2&=&\|x\|^2\|y\|^2-(x\cdot y)^2.\end{array}$
These conditions are respectively called bilinearity, orthogonality, and magnitude. Hence we want $\times$ to be the product in a real normed algebra. Moreover magnitude tells us that we want $x\times x=0.$ We also have $e_i^2=-1$ in our Cayley-Dickson algebras. It turns out that the cross product in $\mathbb{R}^3$ can be modeled in $\mathbb{H}$ as follows. Define $C:\mathbb{R}^3\to\mathbb{H}$ by
$C(x_1,x_2,x_3)=x_1i+x_2j+x_3k$
and $C^{-1}:\mathbb{H}\to\mathbb{R}^3$ by
$C^{-1}(a+bi+cj+dk)=(b,c,d).$
Then one can verify that for $x,y\in\mathbb{R}^3$
$x\times y=C^{-1}\left(C(x)C(y)\right).$
Note that in $\mathbb{R},$ the cross product is trivial $r\times s=0$ for all $r,s\in\mathbb{R}.$ This is obvious if we require it to satisfy orthogonality—where the dot product is just multiplication. So the zero-product property implies the cross product must be $0.$ It is compatible with our map as well (in this case $C:\mathbb{R}\to\mathbb{C}$):
$r\times s=C^{-1}(C(r)C(s))=\mbox{Im}\,(ri)(si)=\mbox{Im}\,(-rs)=0.$
Hence we are left with one remaining cross product: the cross product in $\mathbb{R}^7$ where, hereafter, $C:\mathbb{R}^7\to\mathbb{O}$ is defined by
$C(x_1,...,x_7)=x_1e_1+\cdots+x_7e_7$
and $C^{-1}:\mathbb{O}\to\mathbb{R}^7$ is the imaginary map
$C^{-1}(a_0e_0+\cdots+a_7e_7)=(a_1,...,a_7).$
The cross product of $x,y\in\mathbb{R}^7$ is defined by
$x\times y=C^{-1}(C(x)C(y)).$
We will now define
$C(\nabla)=e_1\frac{\partial}{\partial x_1}+\cdots+e_7\frac{\partial}{\partial x_7}.$
If $f:\mathbb{R}^7\to\mathbb{R}^7$ is differentiable, then we define its curl by
$\mbox{curl}\,f=\nabla\times f=C^{-1}(C(\nabla)C(f)).$
# Proposed Changes to Secondary Math and Language Arts Curriculum
It’s no controversy that education appears to be under par in comparison to what we might hope. I claim this has not only to do with home/environmental conditions for individual students and the ever omnipresent media (and lately, social networking media as well), but also with the curriculum in the classes they take–particularly math and language arts courses. What seems to be the case is that the math courses focus on topics that, while enhancing the mental discipline of students, fail to teach concepts that are readily applicable to everyday situations. Language arts courses emphasize composition (and in particular, stress the ability to make effective arguments) under the premise that students are already well-versed in the grammatical aspects of language (an assumption so far from the truth that the very idea of expecting them to write cogent argumentative essays is ludicrous).
Math:
I’d recommend keeping the middle school curriculum essentially the same (introducing key concepts of elementary algebra and geometry), but I’d vehemently oppose the idea of continuing this sort of curriculum for another four years. In 9th grade, students should be introduced to key concepts from set theory (set notations and operations, functions, equivalence relations, bijections and counting principles, and concepts from elementary number theory, including divisibility, prime numbers, and the fundamental theorem of arithmetic). With this collection of concepts in their inventory, they can spend 10th grade covering concepts from logic and proof writing–and applying them to concepts they learned the previous year (as well as everyday phenomena).
11th grade could then be devoted to a more serious algebra course–covering concepts of operations more thoroughly than the 9th grade course, identities, concepts like associativity and commutativity, matrices, inverses, basic definitions and examples of groups, rings, and fields, polynomials, and comparisons between integers, rationals, reals, and complex numbers, and the fundamental theorem of algebra. If students only took these courses and earned a C in them, I’d still argue they were better prepared for life than earning an A in the traditional courses.
An additional 12th grade class covering geometry more seriously could entail concepts of lengths, triangles, trigonometry, polygons, area of polygons, polytopes and hypervolume, axiomatic Euclidean, spherical, and hyperbolic geometry and basic properties of spherical and hyperbolic triangles, approximation of areas/volumes and method of exhaustion (leading up to a freshman course in real analysis (to replace calculus)).
Language Arts:
I’d recommend middle school curriculum to focus a little more on grammar and sentence structure. A 9th grade course could solidify this with a rigorous treatment of the core concepts of language: parts of speech, morphology, syntax, and grammar. Concepts like word families, clauses (independent, dependent, subject, and predicate), transitive and intransitive verbs, and sentence diagramming should be emphasized.
A 10th grade course emphasizing argumentative writing would appropriately accompany the 10th grade math curriculum in logic and proof writing. I’d recommend abstaining from requiring students to read nontechnical works until the third year of language arts. This way, rather than getting distracted by a fiction or nonfiction story as a platform for an argument, emphasis is instead placed on the concept of argumentative writing itself. Smaller readings should be used in this course.
An 11th grade course could then more effectively do in one year what four years of language arts typically try to do. This course could focus on work-based and research-based extended essay writing.
Curriculum comparable to the AP English Literature course could make an appropriate optional fourth year class. | 6,974 | 22,756 | {"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": 220, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2018-13 | longest | en | 0.662133 |
http://www.mathworks.com/examples/matlab/community/19799-smoothn | 1,519,297,532,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891814101.27/warc/CC-MAIN-20180222101209-20180222121209-00630.warc.gz | 495,676,317 | 21,303 | MATLAB Examples
## Contents
```function [z,s,exitflag] = smoothn(varargin) ```
```%SMOOTHN Robust spline smoothing for 1-D to N-D data. % SMOOTHN provides a fast, automatized and robust discretized spline % smoothing for data of arbitrary dimension. % % Z = SMOOTHN(Y) automatically smoothes the uniformly-sampled array Y. Y % can be any N-D noisy array (time series, images, 3D data,...). Non % finite data (NaN or Inf) are treated as missing values. % % Z = SMOOTHN(Y,S) smoothes the array Y using the smoothing parameter S. % S must be a real positive scalar. The larger S is, the smoother the % output will be. If the smoothing parameter S is omitted (see previous % option) or empty (i.e. S = []), it is automatically determined by % minimizing the generalized cross-validation (GCV) score. % % Z = SMOOTHN(Y,W) or Z = SMOOTHN(Y,W,S) smoothes Y using a weighting % array W of positive values, that must have the same size as Y. Note % that a nil weight corresponds to a missing value. % % If you want to smooth a vector field or multicomponent data, Y must be % a cell array. For example, if you need to smooth a 3-D vectorial flow % (Vx,Vy,Vz), use Y = {Vx,Vy,Vz}. The output Z is also a cell array which % contains the smoothed components. See examples 5 to 8 below. % % [Z,S] = SMOOTHN(...) also returns the calculated value for the % smoothness parameter S so that you can fine-tune the smoothing % subsequently if needed. % % 1) Robust smoothing % ------------------- % Z = SMOOTHN(...,'robust') carries out a robust smoothing that minimizes % the influence of outlying data. % % An iteration process is used with the 'ROBUST' option, or in the % presence of weighted and/or missing values. Z = SMOOTHN(...,OPTIONS) % smoothes with the termination parameters specified in the structure % OPTIONS. OPTIONS is a structure of optional parameters that change the % default smoothing properties. It must be last input argument. % --- % The structure OPTIONS can contain the following fields: % ----------------- % OPTIONS.TolZ: Termination tolerance on Z (default = 1e-3), % OPTIONS.TolZ must be in ]0,1[ % OPTIONS.MaxIter: Maximum number of iterations allowed % (default = 100) % OPTIONS.Initial: Initial value for the iterative process % (default = original data, Y) % OPTIONS.Weight: Weight function for robust smoothing: % 'bisquare' (default), 'talworth' or 'cauchy' % ----------------- % % [Z,S,EXITFLAG] = SMOOTHN(...) returns a boolean value EXITFLAG that % describes the exit condition of SMOOTHN: % 1 SMOOTHN converged. % 0 Maximum number of iterations was reached. % % 2) Different spacing increments % ------------------------------- % SMOOTHN, by default, assumes that the spacing increments are constant % and equal in all the directions (i.e. dx = dy = dz = ...). If the % increments differ from one direction to the other, use the following % field in OPTIONS: % OPTIONS.Spacing' = [d1 d2 d3...], % where dI represents the spacing between points in the Ith dimension. % % Important note: d1 is the spacing increment for the first % non-singleton dimension (i.e. the vertical direction for matrices). % % 3) References (please refer to the two following papers) % ------------- % 1) Garcia D, Robust smoothing of gridded data in one and higher % dimensions with missing values. Computational Statistics & Data % Analysis, 2010;54:1167-1178. % <a % href="matlab:web('http://www.biomecardio.com/pageshtm/publi/csda10.pdf')">PDF download</a> % 2) Garcia D, A fast all-in-one method for automated post-processing of % PIV data. Exp Fluids, 2011;50:1247-1259. % <a % href="matlab:web('http://www.biomecardio.com/pageshtm/publi/media11.pdf')">PDF download</a> % % Examples: % -------- % %--- Example #1: smooth a curve --- % x = linspace(0,100,2^8); % y = cos(x/10)+(x/50).^2 + randn(size(x))/10; % y([70 75 80]) = [5.5 5 6]; % z = smoothn(y); % Regular smoothing % zr = smoothn(y,'robust'); % Robust smoothing % subplot(121), plot(x,y,'r.',x,z,'k','LineWidth',2) % axis square, title('Regular smoothing') % subplot(122), plot(x,y,'r.',x,zr,'k','LineWidth',2) % axis square, title('Robust smoothing') % % %--- Example #2: smooth a surface --- % xp = 0:.02:1; % [x,y] = meshgrid(xp); % f = exp(x+y) + sin((x-2*y)*3); % fn = f + randn(size(f))*0.5; % fs = smoothn(fn); % subplot(121), surf(xp,xp,fn), zlim([0 8]), axis square % subplot(122), surf(xp,xp,fs), zlim([0 8]), axis square % % %--- Example #3: smooth an image with missing data --- % n = 256; % y0 = peaks(n); % y = y0 + randn(size(y0))*2; % I = randperm(n^2); % y(I(1:n^2*0.5)) = NaN; % lose 1/2 of data % y(40:90,140:190) = NaN; % create a hole % z = smoothn(y); % smooth data % subplot(2,2,1:2), imagesc(y), axis equal off % title('Noisy corrupt data') % subplot(223), imagesc(z), axis equal off % title('Recovered data ...') % subplot(224), imagesc(y0), axis equal off % title('... compared with the actual data') % % %--- Example #4: smooth volumetric data --- % [x,y,z] = meshgrid(-2:.2:2); % xslice = [-0.8,1]; yslice = 2; zslice = [-2,0]; % vn = x.*exp(-x.^2-y.^2-z.^2) + randn(size(x))*0.06; % subplot(121), slice(x,y,z,vn,xslice,yslice,zslice,'cubic') % title('Noisy data') % v = smoothn(vn); % subplot(122), slice(x,y,z,v,xslice,yslice,zslice,'cubic') % title('Smoothed data') % % %--- Example #5: smooth a cardioid --- % t = linspace(0,2*pi,1000); % x = 2*cos(t).*(1-cos(t)) + randn(size(t))*0.1; % y = 2*sin(t).*(1-cos(t)) + randn(size(t))*0.1; % z = smoothn({x,y}); % plot(x,y,'r.',z{1},z{2},'k','linewidth',2) % axis equal tight % % %--- Example #6: smooth a 3-D parametric curve --- % t = linspace(0,6*pi,1000); % x = sin(t) + 0.1*randn(size(t)); % y = cos(t) + 0.1*randn(size(t)); % z = t + 0.1*randn(size(t)); % u = smoothn({x,y,z}); % plot3(x,y,z,'r.',u{1},u{2},u{3},'k','linewidth',2) % axis tight square % % %--- Example #7: smooth a 2-D vector field --- % [x,y] = meshgrid(linspace(0,1,24)); % Vx = cos(2*pi*x+pi/2).*cos(2*pi*y); % Vy = sin(2*pi*x+pi/2).*sin(2*pi*y); % Vx = Vx + sqrt(0.05)*randn(24,24); % adding Gaussian noise % Vy = Vy + sqrt(0.05)*randn(24,24); % adding Gaussian noise % I = randperm(numel(Vx)); % Vx(I(1:30)) = (rand(30,1)-0.5)*5; % adding outliers % Vy(I(1:30)) = (rand(30,1)-0.5)*5; % adding outliers % Vx(I(31:60)) = NaN; % missing values % Vy(I(31:60)) = NaN; % missing values % Vs = smoothn({Vx,Vy},'robust'); % automatic smoothing % subplot(121), quiver(x,y,Vx,Vy,2.5), axis square % title('Noisy velocity field') % subplot(122), quiver(x,y,Vs{1},Vs{2}), axis square % title('Smoothed velocity field') % % %--- Example #8: smooth a 3-D vector field --- % load wind % original 3-D flow % % -- Create noisy data % % Gaussian noise % un = u + randn(size(u))*8; % vn = v + randn(size(v))*8; % wn = w + randn(size(w))*8; % % Add some outliers % I = randperm(numel(u)) < numel(u)*0.025; % un(I) = (rand(1,nnz(I))-0.5)*200; % vn(I) = (rand(1,nnz(I))-0.5)*200; % wn(I) = (rand(1,nnz(I))-0.5)*200; % % -- Visualize the noisy flow (see CONEPLOT documentation) % figure, title('Noisy 3-D vectorial flow') % xmin = min(x(:)); xmax = max(x(:)); % ymin = min(y(:)); ymax = max(y(:)); % zmin = min(z(:)); % daspect([2,2,1]) % xrange = linspace(xmin,xmax,8); % yrange = linspace(ymin,ymax,8); % zrange = 3:4:15; % [cx cy cz] = meshgrid(xrange,yrange,zrange); % k = 0.1; % hcones = coneplot(x,y,z,un*k,vn*k,wn*k,cx,cy,cz,0); % set(hcones,'FaceColor','red','EdgeColor','none') % hold on % wind_speed = sqrt(un.^2 + vn.^2 + wn.^2); % hsurfaces = slice(x,y,z,wind_speed,[xmin,xmax],ymax,zmin); % set(hsurfaces,'FaceColor','interp','EdgeColor','none') % hold off % axis tight; view(30,40); axis off % camproj perspective; camzoom(1.5) % camlight right; lighting phong % set(hsurfaces,'AmbientStrength',.6) % set(hcones,'DiffuseStrength',.8) % % -- Smooth the noisy flow % Vs = smoothn({un,vn,wn},'robust'); % % -- Display the result % figure, title('3-D flow smoothed by SMOOTHN') % daspect([2,2,1]) % hcones = coneplot(x,y,z,Vs{1}*k,Vs{2}*k,Vs{3}*k,cx,cy,cz,0); % set(hcones,'FaceColor','red','EdgeColor','none') % hold on % wind_speed = sqrt(Vs{1}.^2 + Vs{2}.^2 + Vs{3}.^2); % hsurfaces = slice(x,y,z,wind_speed,[xmin,xmax],ymax,zmin); % set(hsurfaces,'FaceColor','interp','EdgeColor','none') % hold off % axis tight; view(30,40); axis off % camproj perspective; camzoom(1.5) % camlight right; lighting phong % set(hsurfaces,'AmbientStrength',.6) % set(hcones,'DiffuseStrength',.8) % % See also SMOOTH1Q, DCTN, IDCTN. % % -- Damien Garcia -- 2009/03, revised 2014/02 % website: <a % href="matlab:web('http://www.biomecardio.com')">www.BiomeCardio.com</a> ```
## Check input arguments
```error(nargchk(1,5,nargin)); OPTIONS = struct; NArgIn = nargin; if isstruct(varargin{end}) % SMOOTHN(...,OPTIONS) OPTIONS = varargin{end}; NArgIn = NArgIn-1; end idx = cellfun(@ischar,varargin); if any(idx) % SMOOTHN(...,'robust',...) assert(sum(idx)==1 && strcmpi(varargin{idx},'robust'),... 'Wrong syntax. Read the help text for instructions.') isrobust = true; assert(find(idx)==NArgIn,... 'Wrong syntax. Read the help text for instructions.') NArgIn = NArgIn-1; else isrobust = false; end ```
```Error using smoothn (line 223) Not enough input arguments. ```
## Test & prepare the variables
```%--- % y = array to be smoothed y = varargin{1}; if ~iscell(y), y = {y}; end sizy = size(y{1}); ny = numel(y); % number of y components for i = 1:ny assert(isequal(sizy,size(y{i})),... 'Data arrays in Y must have the same size.') y{i} = double(y{i}); end noe = prod(sizy); % number of elements if noe==1, z = y{1}; s = []; exitflag = true; return, end %--- % Smoothness parameter and weights W = ones(sizy); s = []; if NArgIn==2 if isempty(varargin{2}) || isscalar(varargin{2}) % smoothn(y,s) s = varargin{2}; % smoothness parameter else % smoothn(y,W) W = varargin{2}; % weight array end elseif NArgIn==3 % smoothn(y,W,s) W = varargin{2}; % weight array s = varargin{3}; % smoothness parameter end assert(isnumeric(W),'W must be a numeric array') assert(isnumeric(s),'S must be a numeric scalar') assert(isequal(size(W),sizy),... 'Arrays for data and weights (Y and W) must have same size.') assert(isempty(s) || (isscalar(s) && s>0),... 'The smoothing parameter S must be a scalar >0') %--- % Field names in the structure OPTIONS OptionNames = fieldnames(OPTIONS); idx = ismember(OptionNames,... {'TolZ','MaxIter','Initial','Weight','Spacing','Order'}); assert(all(idx),... ['''' OptionNames{find(~idx,1)} ''' is not a valid field name for OPTIONS.']) %--- % "Maximal number of iterations" criterion if ~ismember('MaxIter',OptionNames) OPTIONS.MaxIter = 100; % default value for MaxIter end assert(isnumeric(OPTIONS.MaxIter) && isscalar(OPTIONS.MaxIter) &&... OPTIONS.MaxIter>=1 && OPTIONS.MaxIter==round(OPTIONS.MaxIter),... 'OPTIONS.MaxIter must be an integer >=1') %--- % "Tolerance on smoothed output" criterion if ~ismember('TolZ',OptionNames) OPTIONS.TolZ = 1e-3; % default value for TolZ end assert(isnumeric(OPTIONS.TolZ) && isscalar(OPTIONS.TolZ) &&... OPTIONS.TolZ>0 && OPTIONS.TolZ<1,'OPTIONS.TolZ must be in ]0,1[') %--- % "Initial Guess" criterion if ~ismember('Initial',OptionNames) isinitial = false; else isinitial = true; z0 = OPTIONS.Initial; if ~iscell(z0), z0 = {z0}; end nz0 = numel(z0); % number of y components assert(nz0==ny,... 'OPTIONS.Initial must contain a valid initial guess for Z') for i = 1:nz0 assert(isequal(sizy,size(z0{i})),... 'OPTIONS.Initial must contain a valid initial guess for Z') z0{i} = double(z0{i}); end end %--- % "Weight function" criterion (for robust smoothing) if ~ismember('Weight',OptionNames) OPTIONS.Weight = 'bisquare'; % default weighting function else assert(ischar(OPTIONS.Weight),... 'A valid weight function (OPTIONS.Weight) must be chosen') assert(any(ismember(lower(OPTIONS.Weight),... {'bisquare','talworth','cauchy'})),... 'The weight function must be ''bisquare'', ''cauchy'' or '' talworth''.') end %--- % "Order" criterion (by default m = 2) % Note: m = 0 is of course not recommended! if ~ismember('Order',OptionNames) m = 2; % order else m = OPTIONS.Order; if ~ismember(m,0:2); error('MATLAB:smoothn:IncorrectOrder',... 'The order (OPTIONS.order) must be 0, 1 or 2.') end end %--- % "Spacing" criterion d = ndims(y{1}); if ~ismember('Spacing',OptionNames) dI = ones(1,d); % spacing increment else dI = OPTIONS.Spacing; assert(isnumeric(dI) && length(dI)==d,... 'A valid spacing (OPTIONS.Spacing) must be chosen') end dI = dI/max(dI); %--- % Weights. Zero weights are assigned to not finite values (Inf or NaN), % (Inf/NaN values = missing data). IsFinite = isfinite(y{1}); for i = 2:ny, IsFinite = IsFinite & isfinite(y{i}); end nof = nnz(IsFinite); % number of finite elements W = W.*IsFinite; assert(all(W(:)>=0),'Weights must all be >=0') W = W/max(W(:)); %--- % Weighted or missing data? isweighted = any(W(:)<1); %--- % Automatic smoothing? isauto = isempty(s); ```
## Create the Lambda tensor
```%--- % Lambda contains the eingenvalues of the difference matrix used in this % penalized least squares process (see CSDA paper for details) d = ndims(y{1}); Lambda = zeros(sizy); for i = 1:d siz0 = ones(1,d); siz0(i) = sizy(i); Lambda = bsxfun(@plus,Lambda,... (2-2*cos(pi*(reshape(1:sizy(i),siz0)-1)/sizy(i)))/dI(i)^2); end if ~isauto, Gamma = 1./(1+s*Lambda.^m); end ```
## Upper and lower bound for the smoothness parameter
The average leverage (h) is by definition in [0 1]. Weak smoothing occurs if h is close to 1, while over-smoothing appears when h is near 0. Upper and lower bounds for h are given to avoid under- or over-smoothing. See equation relating h to the smoothness parameter for m = 2 (Equation #12 in the referenced CSDA paper).
```N = sum(sizy~=1); % tensor rank of the y-array hMin = 1e-6; hMax = 0.99; if m==0 % Not recommended. For mathematical purpose only. sMinBnd = 1/hMax^(1/N)-1; sMaxBnd = 1/hMin^(1/N)-1; elseif m==1 sMinBnd = (1/hMax^(2/N)-1)/4; sMaxBnd = (1/hMin^(2/N)-1)/4; elseif m==2 sMinBnd = (((1+sqrt(1+8*hMax^(2/N)))/4/hMax^(2/N))^2-1)/16; sMaxBnd = (((1+sqrt(1+8*hMin^(2/N)))/4/hMin^(2/N))^2-1)/16; end ```
## Initialize before iterating
```%--- Wtot = W; %--- Initial conditions for z if isweighted %--- With weighted/missing data % An initial guess is provided to ensure faster convergence. For that % purpose, a nearest neighbor interpolation followed by a coarse % smoothing are performed. %--- if isinitial % an initial guess (z0) has been already given z = z0; else z = InitialGuess(y,IsFinite); end else z = cell(size(y)); for i = 1:ny, z{i} = zeros(sizy); end end %--- z0 = z; for i = 1:ny y{i}(~IsFinite) = 0; % arbitrary values for missing y-data end %--- tol = 1; RobustIterativeProcess = true; RobustStep = 1; nit = 0; DCTy = cell(1,ny); vec = @(x) x(:); %--- Error on p. Smoothness parameter s = 10^p errp = 0.1; opt = optimset('TolX',errp); %--- Relaxation factor RF: to speedup convergence RF = 1 + 0.75*isweighted; ```
## Main iterative process
```%--- while RobustIterativeProcess %--- "amount" of weights (see the function GCVscore) aow = sum(Wtot(:))/noe; % 0 < aow <= 1 %--- while tol>OPTIONS.TolZ && nit<OPTIONS.MaxIter nit = nit+1; for i = 1:ny DCTy{i} = dctn(Wtot.*(y{i}-z{i})+z{i}); end if isauto && ~rem(log2(nit),1) %--- % The generalized cross-validation (GCV) method is used. % We seek the smoothing parameter S that minimizes the GCV % score i.e. S = Argmin(GCVscore). % Because this process is time-consuming, it is performed from % time to time (when the step number - nit - is a power of 2) %--- fminbnd(@gcv,log10(sMinBnd),log10(sMaxBnd),opt); end for i = 1:ny z{i} = RF*idctn(Gamma.*DCTy{i}) + (1-RF)*z{i}; end % if no weighted/missing data => tol=0 (no iteration) tol = isweighted*norm(vec([z0{:}]-[z{:}]))/norm(vec([z{:}])); z0 = z; % re-initialization end exitflag = nit<OPTIONS.MaxIter; if isrobust %-- Robust Smoothing: iteratively re-weighted process %--- average leverage h = 1; for k = 1:N if m==0 % not recommended - only for numerical purpose h0 = 1/(1+s/dI(k)^(2^m)); elseif m==1 h0 = 1/sqrt(1+4*s/dI(k)^(2^m)); elseif m==2 h0 = sqrt(1+16*s/dI(k)^(2^m)); h0 = sqrt(1+h0)/sqrt(2)/h0; else error('m must be 0, 1 or 2.') end h = h*h0; end %--- take robust weights into account Wtot = W.*RobustWeights(y,z,IsFinite,h,OPTIONS.Weight); %--- re-initialize for another iterative weighted process isweighted = true; tol = 1; nit = 0; %--- RobustStep = RobustStep+1; RobustIterativeProcess = RobustStep<4; % 3 robust steps are enough. else RobustIterativeProcess = false; % stop the whole process end end ```
## Warning messages
```%--- if isauto if abs(log10(s)-log10(sMinBnd))<errp warning('MATLAB:smoothn:SLowerBound',... ['S = ' num2str(s,'%.3e') ': the lower bound for S ',... 'has been reached. Put S as an input variable if required.']) elseif abs(log10(s)-log10(sMaxBnd))<errp warning('MATLAB:smoothn:SUpperBound',... ['S = ' num2str(s,'%.3e') ': the upper bound for S ',... 'has been reached. Put S as an input variable if required.']) end end if nargout<3 && ~exitflag warning('MATLAB:smoothn:MaxIter',... ['Maximum number of iterations (' int2str(OPTIONS.MaxIter) ') has been exceeded. ',... 'Increase MaxIter option (OPTIONS.MaxIter) or decrease TolZ (OPTIONS.TolZ) value.']) end if numel(z)==1, z = z{:}; end ```
## GCV score
```%--- function GCVscore = gcv(p) % Search the smoothing parameter s that minimizes the GCV score %--- s = 10^p; Gamma = 1./(1+s*Lambda.^m); %--- RSS = Residual sum-of-squares RSS = 0; if aow>0.9 % aow = 1 means that all of the data are equally weighted % very much faster: does not require any inverse DCT for kk = 1:ny RSS = RSS + norm(DCTy{kk}(:).*(Gamma(:)-1))^2; end else % take account of the weights to calculate RSS: for kk = 1:ny yhat = idctn(Gamma.*DCTy{kk}); RSS = RSS + norm(sqrt(Wtot(IsFinite)).*... (y{kk}(IsFinite)-yhat(IsFinite)))^2; end end %--- TrH = sum(Gamma(:)); GCVscore = RSS/nof/(1-TrH/noe)^2; end ```
```end function W = RobustWeights(y,z,I,h,wstr) % One seeks the weights for robust smoothing... ABS = @(x) sqrt(sum(abs(x).^2,2)); r = cellfun(@minus,y,z,'UniformOutput',0); % residuals r = cellfun(@(x) x(:),r,'UniformOutput',0); rI = cell2mat(cellfun(@(x) x(I),r,'UniformOutput',0)); MMED = median(rI); % marginal median AD = ABS(bsxfun(@minus,rI,MMED)); % absolute deviation MAD = median(AD); % median absolute deviation %-- Studentized residuals u = ABS(cell2mat(r))/(1.4826*MAD)/sqrt(1-h); u = reshape(u,size(I)); switch lower(wstr) case 'cauchy' c = 2.385; W = 1./(1+(u/c).^2); % Cauchy weights case 'talworth' c = 2.795; W = u<c; % Talworth weights case 'bisquare' c = 4.685; W = (1-(u/c).^2).^2.*((u/c)<1); % bisquare weights otherwise error('MATLAB:smoothn:IncorrectWeights',... 'A valid weighting function must be chosen') end W(isnan(W)) = 0; % NOTE: % ---- % The RobustWeights subfunction looks complicated since we work with cell % arrays. For better clarity, here is how it would look like without the % use of cells. Much more readable, isn't it? % % function W = RobustWeights(y,z,I,h) % % weights for robust smoothing. % r = y-z; % residuals % MAD = median(abs(r(I)-median(r(I)))); % median absolute deviation % u = abs(r/(1.4826*MAD)/sqrt(1-h)); % studentized residuals % c = 4.685; W = (1-(u/c).^2).^2.*((u/c)<1); % bisquare weights % W(isnan(W)) = 0; % end end ```
## Initial Guess with weighted/missing data
```function z = InitialGuess(y,I) ny = numel(y); %-- nearest neighbor interpolation (in case of missing values) if any(~I(:)) z = cell(size(y)); if license('test','image_toolbox') for i = 1:ny [z{i},L] = bwdist(I); z{i} = y{i}; z{i}(~I) = y{i}(L(~I)); end else % If BWDIST does not exist, NaN values are all replaced with the % same scalar. The initial guess is not optimal and a warning % message thus appears. z = y; for i = 1:ny z{i}(~I) = mean(y{i}(I)); end warning('MATLAB:smoothn:InitialGuess',... ['BWDIST (Image Processing Toolbox) does not exist. ',... 'The initial guess may not be optimal; additional',... ' iterations can thus be required to ensure complete',... ' convergence. Increase ''MaxIter'' criterion if necessary.']) end else z = y; end %-- coarse fast smoothing using one-tenth of the DCT coefficients siz = size(z{1}); z = cellfun(@(x) dctn(x),z,'UniformOutput',0); for k = 1:ndims(z{1}) for i = 1:ny z{i}(ceil(siz(k)/10)+1:end,:) = 0; z{i} = reshape(z{i},circshift(siz,[0 1-k])); z{i} = shiftdim(z{i},1); end end z = cellfun(@(x) idctn(x),z,'UniformOutput',0); end ```
## DCTN
```function y = dctn(y) %DCTN N-D discrete cosine transform. % Y = DCTN(X) returns the discrete cosine transform of X. The array Y is % the same size as X and contains the discrete cosine transform % coefficients. This transform can be inverted using IDCTN. % % Reference % --------- % Narasimha M. et al, On the computation of the discrete cosine % transform, IEEE Trans Comm, 26, 6, 1978, pp 934-936. % % Example % ------- % RGB = imread('autumn.tif'); % I = rgb2gray(RGB); % J = dctn(I); % imshow(log(abs(J)),[]), colormap(jet), colorbar % % The commands below set values less than magnitude 10 in the DCT matrix % to zero, then reconstruct the image using the inverse DCT. % % J(abs(J)<10) = 0; % K = idctn(J); % figure, imshow(I) % figure, imshow(K,[0 255]) % % -- Damien Garcia -- 2008/06, revised 2011/11 % -- www.BiomeCardio.com -- y = double(y); sizy = size(y); y = squeeze(y); dimy = ndims(y); % Some modifications are required if Y is a vector if isvector(y) dimy = 1; if size(y,1)==1, y = y.'; end end % Weighting vectors w = cell(1,dimy); for dim = 1:dimy n = (dimy==1)*numel(y) + (dimy>1)*sizy(dim); w{dim} = exp(1i*(0:n-1)'*pi/2/n); end % --- DCT algorithm --- if ~isreal(y) y = complex(dctn(real(y)),dctn(imag(y))); else for dim = 1:dimy siz = size(y); n = siz(1); y = y([1:2:n 2*floor(n/2):-2:2],:); y = reshape(y,n,[]); y = y*sqrt(2*n); y = ifft(y,[],1); y = bsxfun(@times,y,w{dim}); y = real(y); y(1,:) = y(1,:)/sqrt(2); y = reshape(y,siz); y = shiftdim(y,1); end end y = reshape(y,sizy); end ```
## IDCTN
```function y = idctn(y) %IDCTN N-D inverse discrete cosine transform. % X = IDCTN(Y) inverts the N-D DCT transform, returning the original % array if Y was obtained using Y = DCTN(X). % % Reference % --------- % Narasimha M. et al, On the computation of the discrete cosine % transform, IEEE Trans Comm, 26, 6, 1978, pp 934-936. % % Example % ------- % RGB = imread('autumn.tif'); % I = rgb2gray(RGB); % J = dctn(I); % imshow(log(abs(J)),[]), colormap(jet), colorbar % % The commands below set values less than magnitude 10 in the DCT matrix % to zero, then reconstruct the image using the inverse DCT. % % J(abs(J)<10) = 0; % K = idctn(J); % figure, imshow(I) % figure, imshow(K,[0 255]) % % See also DCTN, IDSTN, IDCT, IDCT2, IDCT3. % % -- Damien Garcia -- 2009/04, revised 2011/11 % -- www.BiomeCardio.com -- y = double(y); sizy = size(y); y = squeeze(y); dimy = ndims(y); % Some modifications are required if Y is a vector if isvector(y) dimy = 1; if size(y,1)==1 y = y.'; end end % Weighing vectors w = cell(1,dimy); for dim = 1:dimy n = (dimy==1)*numel(y) + (dimy>1)*sizy(dim); w{dim} = exp(1i*(0:n-1)'*pi/2/n); end % --- IDCT algorithm --- if ~isreal(y) y = complex(idctn(real(y)),idctn(imag(y))); else for dim = 1:dimy siz = size(y); n = siz(1); y = reshape(y,n,[]); y = bsxfun(@times,y,w{dim}); y(1,:) = y(1,:)/sqrt(2); y = ifft(y,[],1); y = real(y*sqrt(2*n)); I = (1:n)*0.5+0.5; I(2:2:end) = n-I(1:2:end-1)+1; y = y(I,:); y = reshape(y,siz); y = shiftdim(y,1); end end y = reshape(y,sizy); end ``` | 7,608 | 23,391 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-09 | latest | en | 0.641318 |
https://glendasnote.com/garage-door-header-size-calculator/ | 1,679,830,697,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945472.93/warc/CC-MAIN-20230326111045-20230326141045-00022.warc.gz | 312,432,597 | 11,422 | # Garage Door Header Size Calculator
Door Width and Header Size. We’ve all seen 16′ 2 x 12 garage door headers sag, even in a gable wall let alone in a load bearing situation. With the use of micro.
Window, Door & Garage Door Headers – Supporting Roof, Wall & Floor Loads; Floor Girder Beams; Floor Edge Beams; Roof Ridge Beams. Loading conditions.
Dec 24, 2009 – Re: 16′ Garage Door Headers We run the numbers thru StruCalc if it passes fine if not they have to resubmit. We DO NOT tell them what size is.
standard 2×4 and 2×6 wall construction, so there’s no need for furring when you connect headers to end walls. A common width of glulam garage door headers is 3.
I have a 22 ft load bearing wall with a 18′ 3″ header constructed out of 2 2x12s with 1/2 inch plywood sandwiched between for a double car garage door.
framing up a garage door, 18′ wide and will carry the 5/12 rafters on. There are manufacturer’s tables available to size your header as per.
This means my garage door header will carry no load whatsoever as the trusses do not require center support. What size would you go with?
May 27, 2005 – Garages, Garage Door Openers, Work Shops & Sheds, Breezeways and Carports – garage door header size – i’m building a 3 car garage, and.
Calculating the size of the header depends on what the header needs to support. Step 1. Determine whether or not the door is under a load-bearing wall. Any.
Calculating header size is complicated as you learn how to frame a window. Garage doors and ot. | 385 | 1,503 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2023-14 | latest | en | 0.840192 |
singaporemanofleisure.blogspot.sg | 1,485,049,220,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560281331.15/warc/CC-MAIN-20170116095121-00289-ip-10-171-10-70.ec2.internal.warc.gz | 261,167,020 | 40,777 | ## Friday, 21 October 2016
### The Math of Averaging Down
What's the difference between a -80% loss and a -90% loss?
50%.
In years gone by, you probably would have made good profits by spotting arbitrage opportunities!
But now, its a lot harder with high frequency trading and all...
Show me, show me!
For the rest of us mortals, let me illustrate:
1. You bought a stock at \$1.00
2. At -80% loss, the market price is \$0.20
3. At -90% loss, the market price is \$0.10
Still want to average down when price is at \$0.20?
Price so low oredi, surely it can't go any lower?
Well, a "small" additional decline of \$0.10 is now a 50% loss for your 2nd entry!!!
Now you know why people capitulate (sell) at crazy super low prices.
Especially to the "iron teeth" (stubborn) ones who will continue to average down a third time at \$0.10 price.
See what happens when price drop further to \$0.05 cents?
Don't try this at home
Of course you were banking the reverse would happen when you averaged down.
a. You invested \$10,000 into stock XYZ at \$1.00
b. At market price of \$0.10, you average down another \$10,000
c. When price "recovers" to \$0.20, you sell both tranches with a big sigh of relief...
First tranche: Entry at \$1.00; exit at \$0.20 = -\$8,000 loss
Second tranche: Entry at \$0.10; exit at \$0.20 = \$10,000 profit
Hey! All in all, you've made a \$2,000 profit!!!
Really? Risking all these money just to "break-even"?
Tip: Don't play Poker or Mahjong with real money - you lousy with risk/reward odds.
What?
You still got the cheek to "advise" people not to gamble...
Be honest now
Which is the likelier scenario to play out?
How good is your entry skill?
You did buy at \$1.00, remember?
What makes your such an expert market timer now so super sure \$0.10 will be the turning point?
And more importantly, you did let a \$1.00 position slide all the way to \$0.10 - that's not saying much about your exit skill, does it?
Do you honestly believe you will exit everything when the price hits \$0.20?
Or maybe you are the white-haired demon girl (白发魔女); you changed completely in one night.
If you can do so, I clap my palms before you. And bow.
Namaste (I salute the holiness in you)
1. Hi SMOL
Nowadays, it doesnt take a share price to drop to \$0.01 before it gets suspended. \$0.01 consolidate back to \$1 and then drops to \$0.01 and then consolidate again. Wah, 1 cent so cheap cheers!!!
1. B,
Ah!
The reverse stock split ;)
Don't know whether to cry or laugh when you can't even breakeven with a 10 bagger!?
Letting go is something we have to learn and master...
Children all grown up with their own families? Letting go...
My job has been taken over by a machine? Letting go...
Love of my life marrying someone else? Letting go...
Stock turned salty fish? Letting go...
Losing my hair at the top? Letting go...
LOL!
2. Don't average down. Then how? Many are thinking like that!
1. CW,
I never say don't average down ;)
Don't try this at home means unless you are a pro and know what you are doing, don't monkey see monkey do.
If you have trained and prepared for it, at age 80 also can run a marathon!
But if you have not trained for it, even at the peak of our health at age 18, we'll likely hurt ourselves...
The irony is that some of those who think like that are the same ones who - with good intentions - like to "wah kali kong" and preach to others not to use leverage/margin for investing and gambling is bad for you.
If they only knew the risks they are taking on by averaging down just to break-even...
3. SMOL
Letting go is difficult.
If we take out the emotion for every decision we make, wouldn't it make us a lifeless person.
If that's the case, what is the point of living?
1. Blursotong King,
Emotionless is not the same as not letting emotions get the better of us ;)
Extreme frugalness is similar. What's the point of living in such an austere manner?
So is the greed and never ending accumulation of money in the name of "financial freedom".
Hence we walk the middle path.
Similarly, wisdom is needed to avoid the extremes of being emotionless and emotion overflow ;)
Letting go is non-attachment.
I better stop as that's a whole different realm of philosophy altogether!
2. SMOL
Care to share more about the philosophy part? I am interested to learn
3. Blursotong King,
I'm not qualified. It's a big subject and if not careful, can lead you astray...
Just suffice to give you some hints that non-attachment is not detachment.
Neither is it lack of love or compassion ;)
If you interested in Non-Attachment, just google "non-attachment" and have fun getting "lost" for a feel of things.
Once you have a better sense of it, you can start to explore Buddhist books in the library.
If still interested to practice it, private message me and I can direct you to some of the courses out there run by reputable Buddhist organisations.
When I first came back 5 years ago, with time on my hands, I attended the Christian Alpha course and finished 1.5 years of a 3 year Buddhism course before I dropped out - it got boring...
Quite interesting counter-balance:
Spiritual freedom vs financial freedom ;)
4. There is a saying, "Not until you can control your emotions, can you control your money."
i completely agree.
If we can do things without any emotions, what have we become?
Robot, zombie, .......
The best we can do is put our emotions under lease.
1. Should be, "Not until you can control your emotions, can you manage your money".
2. temperament,
One of the disciplines I've gotten better in recent years is the awareness of my emotions.
I'll stay on the sidelines, reduce my position size, or take a break from trading whenever I caught myself having too much emotions vested in my positons.
As a man of leisure, I must maintain my air of debonair ;)
3. For awhile, I read Dragonair Pokemon. :-)
4. CW,
I've not played Pokémon; but I've flown Dragonair a few times during my time in Shekou (Shenzhen) ;)
Power name - Dragon Airlines!
5. Hi SMOL,
It's true. Don't anyhow average down unless you want to flip until you go longkang. I've done it in the past and lost terribly. Now I follow bro8888 and average in. Doesn't matter whether I'm buying higher or buying lower. I'm entering the same counter but with a new thesis behind the new purchase haha
Best to treat each purchase, even if it's the same counter, as if they are separate and independent purchases!
1. LP,
In trading's parlance, we call it:
Scale in; scale out.
We all have to learn it the hard way (including me).
Can be quite hard for newbies to differentiate - when the pros do it, they appear to be doing the "same" when it comes to execution...
But there's a REAL difference.
Only those who survived from their averaging down "temporary insanity" episodes will understand ;)
Experiential bloggers like us share our "war time" stories and reflections.
We have battle scars to back-up our reminisces ;)
6. Hi Smol,
I just written a post after reflecting on what you said. It got so much clearer after writing (as usual)
As u said, average down by itself is a ok strategy, if we are sure what we are doing. After my reflection, the irony is my winning positions, only 1-2 counters have the "opportunities" to average down.
But a few counters did break even after averaging down.
The real slap is averaging down then sleep on it. Once we average down, the managitudue of fall is "doubled" and we should cut.
Connecting the dots now, until the next big bear refresh and reset my journey
1. Sillyinvestor,
I understand what you meant.
Again, I never said averaging down is an OK strategy.
Notice why I persist in correcting you and CW when it comes to "precision" in language?
Words by themselves are not important; but we should always be mindful on the power and influence of words have on our psychology ;)
My post is merely a carrot and stick coaching technique to encourage the reader to reflect and seek the answers themselves.
Whether averaging down is a good, bad, or useful in specific situation is entirely up to EACH and everyone to decide for themselves.
As you have discovered, sometimes it works; sometime it doesn't.
2009 to 2016 is a bull cycle. Averaging down in a bull cycle - that's your reflection.
For another reader who has gone through the bear cycle of 2000-2003, his reflection of averaging down will be totally different!
In the next bear cycle, your reflections will get "refreshed and updated" as you go through a multi-year bear cycle yourself ;)
Just remember 2 things:
1) We don't invest to break-even.
2) Stay alive!
No chips; no play :(
7. Can we use size positioning instead of averaging down?
Are they the same?
From so many financial books I have read, almost no author recommends averaging down.
But some have recommended averaging up.
I mean 1 lot, 2 lots......assuming everything is being equal.
1. temperament,
Position sizing and averaging down are not the same.
Let's say you limit a stock to not more than 5% of a portfolio. That means even if the stock goes to zero, the max loss is only 5% at portfolio level.
5% of a \$100,000 portfolio is "only" \$5,000.
5% of a \$10 million portfolio is \$500,000! Some how I don't think you would shrug it off so lightly ;)
1 or 2 lots no feeling. Hundred thousands of lots then we are talking!
As for 1,2,3,4,5 or 5,4,3,2,1?
No need to guess. Just experiment on your own and let your own experience and track record tell you which is better ;)
Don't let anyone tell you what shoes you should wear! Unless you don't trust your own feet ;)
My own experience is there is be a BIG difference in psychology when I average down versus average up.
One path is going against the wind; the other is moving ahead with the wind behind my sails ;)
2. 5% of \$ 100000 and 5% of \$ 10,000000 in absolute value is not the same is true.
But size positioning is the same.
Therefore risk management is the same for both persons.
So the \$10,000000 person may indeed shrug it off if that's how invested.
And perhaps it's the \$10,000 person that may not shrug it off.
i know, I have been the \$10,000 person too often.
3. temperament,
In percentages, its the same. In dollar amount, no way!
Sometime back, a fee-based financial adviser was poking a client for being too "kan cheong" with a small 1.5% loss on his portfolio.
Hello! the client's portfolio is \$100 million, and 1.5% is \$1.5 million! Enough for a condo!
That's why I don't like to think in percentages. I prefer to use dollar amounts.
I only doubled my trading's position size even though my trading account is now a 10 bagger.
This is the optimum position size where I can cut-loss in dollar terms without hesitation nor blinking an eye ;)
4. i concede in theory there is no difference between thoery and practice but in actual there is.
So size positioning and hence risk management in theory is the same for all of us.
In practice is it really the same?
5. temperament,
Your mileage and my mileage may differ ;)
This is the fun part of blogging I enjoyed very much!
To banter with people who are smarter and more experienced than me ;)
6. Oh no!
Experienced may be for i live at least a decade more than you.
i really don't worry who is smarter or not.
i only worry my decision is not mine even after considering all others ideas.
How about you start a business i will join you as a junior or sleeping partner?
7. temperament,
I'm too lazy and irresponsible to run a business.
I did seriously thought about it before I came back. To buy a franchise like 7-elevan or Subway.
But the moment I realise I have to spend most of my time on recruiting manpower and dealing with work permits; and "working for" the landlords in reality, I gave it all up.
Trading from home and treating it as part of a nano hedge fund business has worked out much better than I've imagined.
I quite like my Ronin status today ;)
8. Ha! Ha!
Exactly why i am in the stock market and not in any business.
We are really have something in common here.
8. Hi SMOL,
Yah ! This is my problem as well ,, average down on Sembcorp Ind ,,,jatuh lonkang,,, average down on M1 also masuk lonkang ,,, : (
Think temperament have a good question to ponder,, whether we should use sizing up or sizing down strategy ,,,, my interpretation will be sizing up if we still have war chests,, sizing down when we are running out of bullet.... :). Just kidding
Cheers !! :)
1. STE,
I would like to hazard a guess you now know WHY Sembcorp Ind and M1 have gone down ;)
That's the price to pay for "certainty" in knowing the fundamental reasons.
Some would prefer to sell first and ask questions later. These are those who focus on market psychology - also known as technical analysis ;)
And when the dust has settled, they can decide with a clear and calm mind whether to re-enter into the SAME stock they have exited earlier - now at a much BETTER entry price, if its still a high conviction buy.
Note this is not averaging down since no NEW capital is injected. It's still the same old capital - albeit "recycled" with a "hair-cut".
This is still aligned with position sizing, you are sill using the same 5% max allocation limit for this stock.
Many retail bei kambings started out with max 5% limit, then they average down in anger/frustration to get even...
Before they know it, this lemon of a stock is now 20% of their portfolio???
Talk about letting a small paper cut turn gangrenous...
What happened to discipline and sticking with their investment plans and their goal setting for the year?
No one ever set investment goals to breakeven...
Its best to listen and trust our own track record. If averaging down works for you, do it!
Why change a winning strategy?
LOL!
9. Why change indeed?
If sailing upstream against the Current you succeeded, how much more you will achieve if you sail downstream?
I think at which point of the river you started and ended your sailing journey is the most important consideration.
10. Hi SMOL & temperament,
Yah. .is very true. .I like the comment on " which point of the river you start. ..." is very much depend on which part of the biz or economy cycle we are in ...that will determine our wining ...not only hindsight of any strategy or stock picking. . I guess. .:)
Cheers.👍
1. STE and temperament,
Yup.
If we are always averaging down, I guess its a great tell tale sign we are not buying at the bottom end of the business cycle ;)
And definitely not practicing: | 3,482 | 14,600 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2017-04 | longest | en | 0.932209 |
http://www.edn.com/user/Marian%20Stofka | 1,487,776,191,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501170993.54/warc/CC-MAIN-20170219104610-00366-ip-10-171-10-108.ec2.internal.warc.gz | 392,926,561 | 21,313 | # Marian Stofka
's profile
Dr.
Marian Stofka is with the Slovak University of Technology, Faculty of Electrical Engineering and Information Technology in Bratislava, Slovak Republic.
## Marian Stofka
's contributions
• 07.15.2010
• DC-voltage doubler reaches 96% power efficiency
• Thanks for pointing out the Semtech's SC1462. This IC is an "other coffee", as we say here; as compared to TL7660. Let the TI people forgive me this expression. If I needed to apply a DC voltage doubler, I bought the SC1462. If I wanted to learn, what is inside, I bought the EDN, issue July 15, 2010.
• 07.15.2010
• DC-voltage doubler reaches 96% power efficiency
• In previous poster the TI's TL7660 is counter-posed to the DI. Looking into datasheet of the TL7660, you learn that it is a DC-voltage inverter, reaching 98% power efficiency at Vs=5V, Ta =25 C and output current Io of 1 to 1.5 mA. The output power at these values is thus 5 to 7.5 mW.The doubler in Fig. outputs 117.95 mW of power at a power efficiency over 96%. TL7660 thus looks like a dwarf not only due to its size, but also due to the outputted power. Further, if you are focusing to compare real DC-voltage doublers; then you have to look at Fig.9 in the datasheet of the TL7660. What can be seen here, are two external diodes D1, D2, which allow to convert the inverter to a positive voltage-doubler. As the output power is transferred through these diodes; you can forget about 98% power efficiency. Even if there existed other commercial DC voltage doubler; reaching 98% power efficiency at 120 mW of outputted power - I still would submit the DI. Readers of the EDN deserve; in my opinion, to get detailed information of "what is inside". If the DIs in the EDN contained just block diagrams, many young people reading the EDN all over the world would be disappointed, I think.
• 07.15.2010
• Count objects as they pass by
• Dear Vladimir, When I yesterday clicked on the Fairchild's link present in your DI; looking for the H22LOI, I got an extremely brief description, where no Schmitt trigger had been mentioned. Today I used a general search engine and learned, that this IC includes also a Schmitt trigger. So, let my previous "lecture" serves as a commemoration of Otto H. Schmitt. I still see the flip-flop IC4A as a detector of direction of movement.
• 07.15.2010
• Count objects as they pass by
• I did my best to get an insight into the circuit in Fig.1. What I have succeeded up-to-now is, that: OUT3: Low=Busy; High=Ready; this is crystal-clear. Further, OUT2, although generated by a sequential circuit (IC4B); is an equivalent of NAND-ing the SENSOR1, SENSOR2 outputs. The Q output of IC4A is always Low for the direction as drawn (1-->2). By the way, Fig, 2 carries opposite direction. Further, if the direction would reverse as(2-->1); the OUT2=High steadily. The "Identifier" is a big enigma. From realisation point of view, I have to note, that driving clock inputs of synchronous, master-slave type flip-flops (not only the HC74) from a source of signal, which has; or could have, level-transition times of more than 10 to 20 nsec - is out of question in professional designs. The "slowly" varying signals shopuld not enter neither gates, like here is the case of IC3B, IC3D. The SENSOR1, SENSOR2 signals should pass firstly through Schmitt-trigger-input devices, like cascade of two 74HC14; for example. By the way, H.C. Schmitt invented in 1938 his glorious circuit; which is now called as Schmitt trigger - right when solving a problem of counting of moving objects.
• 06.10.2010
• Bootstrap circuit speeds solenoid actuation
• To Rob's question: A base resistor, especially in the base of Q2 is necessary. At turning-on Q1, the emitter of Q2 is brute-force-driven to -23.4V. If there were no resistor in the base of Q2 and the source of driving signal were a low-impedance one; then an overcurrent flowed through the B2E2 diode, leading to a destroying of the Q2.
• 06.10.2010
• Switched-capacitor voltage multiplier achieves 95% efficiency
• Dear Tony, Your words of "Unfortunately,...quoted." are just a verbal expression, which is not backed by any mathematical theory. If you made a rigorous analysis, you probaly got the same formula for the efficiency, as is given in the DI. What is even more important is, that the numerical data at the end of DI, including the over-95% efficiency, come from an experiment. Perhaps the following will help you to absorb the idea: Let assume a capacitor C, which has been precharged to voltage V. If you charge it further to a voltage V+dV; the capacitor's energy increment will be: dW=C.V.(dV); "d" denotes her Greek "delta". Simultanously, the loss on resistor will be (1/2).C.(dV).(dV). If you now evaluate a "loss/(energy increment)" ratio, you get: (1/2).(dV)/V. As is dV
• 06.10.2010
• Bootstrap circuit speeds solenoid actuation
• Nice design! In the schematic, however, a resistor in the base of Q2 seems either to be omitted, or the input is driven from a high-impedance source. Otherwise the emitter of Q2 could not go down to -23.4 V. From Fig 2. it can be evaluated R=240 Ohm and L=0.6852H. The peak current of 175.6mA matches very well the theoretical value of: Imax=(2Vin-Vd2).sqrt(C/L).exp(-(argtgh(y))/y); where y=b.2L/R, b="omega"; which gives Imaxtheor=0.17475A. Theoretical value of time of peak tpeak=(2L/R).(argtgh(y))/y gives a value of tpeaktheor=9.027ms. The higher value of 13.8ms in Fig.2 can be attributed to increasing of "L", when the armature of the actuator gets closer to center of the solenoid.
• 01.07.2010
• Circuit uses two reference voltages to improve hysteresis accuracy
• Firstly, I would like to point, who Rube Goldberg was. It might be useful for those, who considered this name to belong to famous series of Schmitt, Eccles-Jordan, Darlington and others. Reuben Lucius Goldberg (July 4, 1883 – December 7, 1970) was a Jewish American cartoonist, sculptor, author, engineer, and inventor. Goldberg is best known for a series of popular cartoons he created depicting complex devices that perform simple tasks in indirect, convoluted ways – now known as Rube Goldberg machines (from Wikipedia). Now to the circuit described in the DI. For those, who understand a comparator as any off-the-shelf functional unit; as cheap as possible, comprising lowest count of elements; for those a classical solution with a resistive positive feedback will do. If, however, a high accuracy of both thresholds is required, then these thresholds can not be derived from output of the IC directly, as the saturated-output voltage-level is a subject of changes with temperature, load and even duty-cycle of operation. Also the dynamic performance of the circuit proposed is much better. In the comparator with a simple positive resistive feedback, the parasitic input capacitance of the IC is charged through the feedback resistor, thus increasing delay and rise- and fall-time of output as well. | 1,784 | 6,945 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2017-09 | longest | en | 0.914579 |
https://www.scienceabc.com/pure-sciences/logic-reason-behind-a4-a3-a0-a2-standard-paper-sizes-mathematics.html | 1,695,878,696,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510358.68/warc/CC-MAIN-20230928031105-20230928061105-00313.warc.gz | 1,066,416,797 | 55,101 | # What’s The Mathematical Logic Behind Different A-Size Papers?
The mathematical logic behind the different A-size papers is that they are all based on the German DIN 476 standard for paper sizes. This standard is based on the ratio between the length and width of the paper. All A-size papers have the same ratio between their length and width, which is 1.414. This allows for easy scaling of content from one size to another without losing symmetry.
A4-sized paper has become a staple of stationary, both in offices and households, all over the world (except the United States). Given its immense popularity, the A4 size has become the standard business letter size in countries like United Kingdom and New Zealand. A3 and A2 sheets are relatively less common, and the A1 and A0 variants even more rare.
Such standardization of paper was first introduced in the 20th century and is ascribed to all over the world to this day. So, what’s the difference between different A-sized sheets and how are they standardized?
Recommended Video for you:
## A-Series Sheets
To give a little background, the international paper size standard is ISO 216, which specifies the international standard (ISO) paper sizes used in most countries in the world today. It is based on the German DIN 476 standard (DIN is a German organization for standardization) for paper sizes. The most popular ISO standards of paper include A, B and C-series sheets, but in this article, we’re going to talk about the most commonly used variant: the A-series.
### What’s So Special About A-sheets?
A-series sheets of paper have been designed in such a way that when you cut them in half, you get two perfectly identical pieces of the next biggest size. For example, if you cut an A0 sheet in half, you will get two A1-sized sheets; if you cut an A1-sized sheet in half, you will get two A2-sized sheets and so on.
Although the dimensions of the two new sheets are different from their parent sheet, the proportion of the sheets’ length and breadth remains the same. Therefore, when you fold an A3 sheet in half, you get two A4 sheets that have the same ratio between their length and breadth as the parent (A3) sheet.
Also Read: Can You Really Only Fold A Piece Of Paper 7 Times?
## How Do You Determine The Exact Dimensions Of A-sheets?
As mentioned above, all A-standard sheets have the same ratio between their length and breadth. That being said, what’s the value of that ratio, and how does it remain constant for all A-standard sheets?
To answer this, we’ll have to go back to where it all started: the first sheet of paper that was officially standardized by putting the letter ‘A’ as a prefix to its name was the ‘A0 sheet’. An A0 sheet is exactly 46.8 inches long and 33.1 inches wide (841 mm x 1189 mm) and has an area of 1 meter squared. Now, if you divide 1189 by 841 (i.e., the dimensions of an A0 sheet) you get 1.414. What’s so special about this number, you ask?
Well, the dimensions of every A-standard sheet, if you care to check the math, are in the proportion of 1.414, which also happens to be the value of ….
Given below are the dimensions of different variants of A-standard sheets:
You can see that, whether it’s an A8 sheet or an A0 sheet, the ratio (1.414) between the length and breadth remains the same. The beauty of such well-defined distinction is that you can easily upscale/downscale your content (including images, text etc.) to bigger/smaller sheets without losing its symmetry.
Had the ratio of the different A-standard sheets not been fixed, the transfer of content from one size to another wouldn’t be so simple and flexible. Since the ratio of sides of all A-standard sheets is 1.414, you can easily expand a document to its next biggest size by dialing in a factor of 141% in a machine. This tip comes in handy while working on various photo-editing tools, photocopying documents and so on. Also, you can easily convert one bigger A-size sheet to the next two A-sized sheets.
For instance, you can divide an A2 sheet to make two symmetrical A3 sheets, two A3 sheets to make four symmetrical A4 sheets, etc.
Who would have thought that such a simple – yet ingenious! – technique using mathematics could be used to regularize the dimensions of sheets of paper all over the globe!
Also Read: Why Does A Fold In Paper Become Permanent?
How well do you understand the article above!
Can you answer a few questions based on the article you just read?
References (click to expand) | 1,008 | 4,501 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2023-40 | longest | en | 0.939694 |
http://mathhelpforum.com/advanced-algebra/1898-finite-extensions.html | 1,481,402,731,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698543567.64/warc/CC-MAIN-20161202170903-00369-ip-10-31-129-80.ec2.internal.warc.gz | 181,976,728 | 10,442 | # Thread: Finite Extensions
1. ## Finite Extensions
I was thinking about this and was not able to prove this. I believe that if this is true then it might be useful in field theory.
If, $F\leq E$ are fields, and $\alpha,\beta\in E$ algebraic over $F$ and $[F(\alpha):F]=[F(\beta):F]$.
Prove (or disprove):
$F(\alpha)=F(\beta)$.
???
2. Originally Posted by ThePerfectHacker
I was thinking about this and was not able to prove this. I believe that if this is true then it might be useful in field theory.
If, $F\leq E$ are fields, and $\alpha,\beta\in E$ algebraic over $F$ and $[F(\alpha):F]=[F(\beta):F]$.
Prove (or disprove):
$F(\alpha)=F(\beta)$.
???
I might have the wrong idea here, because I'm not sure how you are using $\leq$, but my thought is:
Let $F= \mathbb{Q}$ and $E= \mathbb{Q} [ \sqrt2 ,\sqrt3 ]$. $\sqrt2$ and $\sqrt3$ are both algebraic over $\mathbb{Q}$ and $[ \mathbb{Q} ( \sqrt2 ): \mathbb{Q} ] = [ \mathbb{Q} (\sqrt3 ): \mathbb{Q} ]$.
However, $\mathbb{Q} ( \sqrt2 ) \neq \mathbb{Q} ( \sqrt3 )$.
-Dan
3. You said well, thank you. | 363 | 1,059 | {"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": 18, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2016-50 | longest | en | 0.861711 |
https://www.convert-measurement-units.com/convert+kByte+h.php | 1,585,972,389,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370519111.47/warc/CC-MAIN-20200404011558-20200404041558-00483.warc.gz | 882,672,264 | 15,194 | Convert kByte/h (Data rate)
## Convert kByte/h
Measurement Categorie:
Original value: Original unit: Bit/sByte/sGBit/sGByte/skBit/skByte/hkByte/minkByte/sMBit/sMByte/hMByte/minMByte/sTBit/sTByte/s numbers in scientific notation
https://www.convert-measurement-units.com/convert+kByte+h.php
# Convert kByte/h:
1. Choose the right category from the selection list, in this case 'Data rate'.
2. Next enter the value you want to convert. The basic operations of arithmetic: addition (+), subtraction (-), multiplication (*, x), division (/, :, ÷), exponent (^), brackets and π (pi) are all permitted at this point.
3. From the selection list, choose the unit that corresponds to the value you want to convert, in this case 'kByte/h'.
4. The value will then be converted into all units of measurement the calculator is familiar with.
5. Then, when the result appears, there is still the possibility of rounding it to a specific number of decimal places, whenever it makes sense to do so.
With this calculator, it is possible to enter the value to be converted together with the original measurement unit; for example, '151 kByte/h'. In so doing, either the full name of the unit or its abbreviation can be used. Then, the calculator determines the category of the measurement unit of measure that is to be converted, in this case 'Data rate'. After that, it converts the entered value into all of the appropriate units known to it. In the resulting list, you will be sure also to find the conversion you originally sought. Regardless which of these possibilities one uses, it saves one the cumbersome search for the appropriate listing in long selection lists with myriad categories and countless supported units. All of that is taken over for us by the calculator and it gets the job done in a fraction of a second.
Furthermore, the calculator makes it possible to use mathematical expressions. As a result, not only can numbers be reckoned with one another, such as, for example, '(71 * 69) kByte/h'. But different units of measurement can also be coupled with one another directly in the conversion. That could, for example, look like this: '151 kByte/h + 453 kByte/h' or '51mm x 52cm x 50dm = ? cm^3'. The units of measure combined in this way naturally have to fit together and make sense in the combination in question.
If a check mark has been placed next to 'Numbers in scientific notation', the answer will appear as an exponential. For example, 7.745 955 608 524 1×1031. For this form of presentation, the number will be segmented into an exponent, here 31, and the actual number, here 7.745 955 608 524 1. For devices on which the possibilities for displaying numbers are limited, such as for example, pocket calculators, one also finds the way of writing numbers as 7.745 955 608 524 1E+31. In particular, this makes very large and very small numbers easier to read. If a check mark has not been placed at this spot, then the result is given in the customary way of writing numbers. For the above example, it would then look like this: 77 459 556 085 241 000 000 000 000 000 000. Independent of the presentation of the results, the maximum precision of this calculator is 14 places. That should be precise enough for most applications.
## How much is 1 kByte/h?
Measurement calculator that can be used to convert kByte/h, among others. | 782 | 3,356 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2020-16 | latest | en | 0.834467 |
https://www.physicsforums.com/threads/work-done-by-friction-pushing-blocks.436993/ | 1,721,792,393,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518154.91/warc/CC-MAIN-20240724014956-20240724044956-00172.warc.gz | 794,901,463 | 16,952 | # Work done by friction pushing blocks
• preluderacer
In summary, a constant external force of 170 N is applied to a 20-kg box on a rough horizontal surface. The force pushes the box a distance of 8.0 m in a time interval of 9.0 s and changes the speed from 0.4 m/s to 3.0 m/s. Using the work-energy theorem, the work done by friction is calculated to be 1089.39 J, with a negative value indicating that the friction force acts opposite to the direction of displacement. The given time interval of 9.0 s is found to be incorrect, as using it in the calculations yields different results.
preluderacer
## Homework Statement
In the figure, a constant external force P = 170 N is applied to a 20-kg box, which is on a rough horizontal surface. The force pushes the box a distance of 8.0 m, in a time interval of 9.0 s, and the speed changes from ν1 = 0.4 m/s to ν2 = 3.0 m/s. The work done by friction is closest to:
(P is pushing down at a 30 degree angle in the picture)
## The Attempt at a Solution
I know I need to find the horizontal component of the push, but I don't know what to do next.
Use the work-energy theorem in the horizontal direction.
Im not quite sure were the 9 seconds fits into this?
preluderacer said:
Im not quite sure were the 9 seconds fits into this?
There seems to be an error in the problem statemnt.
What do you mean?
If you work out the problem using the data that it travels 8m with the initial and final velocities as noted, you get the correct answer, and the acceleration is constant. If you work out the problem using the fact that the time is 9 seconds, you get a completely different acceleration, so the time given is incorrect. Are you familiar with work energy methods? If not, fine, try the problem using the other data and the kinematic equations and Newton's laws, and see that you get different results based on the 2 sets of data.
Ok so here's what I did I calculated the force in the applied to the box to be 170cos30=147.22N. After that I multiplied that by 8 meters and got that the work done in that direction was 1177.79 J. Next I used the work energy theorem from ν1 = 0.4 m/s to ν2 = 3.0 m/s and got 88.4 J. I then subtracted 1177.79 J - 88.4 J and got 1089.39 J. Now is that final answer the work done by the friction? If so, is that work negative or positive
W_P + W_f = delta K = 88 J
1178 J + W_f = 88 J
Is W_f positive or negative?
It is negative because its in the opposite direction of the pushing force?
The math shows it is negative, which implies that work is negative because the friction force acts opposite fo the direction of the diplacement . Work is the dot product of foce and displacement , W_f = f.d =f(d)(cos theta), where theta is the angle between the force and displacement vectors. The friction force acts to the left, and the displacement is to the right, so in this case, work = fdcos 180, = f(d)(-1) = a negative number.
## What is work done by friction pushing blocks?
Work done by friction pushing blocks refers to the energy expended when an object is moved by the force of friction against another surface. This energy is typically dissipated as heat.
## How is work done by friction calculated?
The work done by friction can be calculated by multiplying the force of friction by the distance over which the object is moved. This is represented by the equation W = Fd, where W is work done, F is the force of friction, and d is the distance moved.
## What factors affect the amount of work done by friction pushing blocks?
The amount of work done by friction pushing blocks is affected by several factors including the force of friction, the type of surface the object is moving on, and the weight and speed of the object being pushed.
## Can work be done by friction pushing blocks in a vacuum?
No, work cannot be done by friction pushing blocks in a vacuum since there is no air or other medium for friction to act upon.
## How does the direction of force affect the work done by friction pushing blocks?
The direction of force does not affect the work done by friction pushing blocks. As long as there is movement and friction between two surfaces, work will be done regardless of the direction of the force.
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2K | 1,174 | 4,787 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2024-30 | latest | en | 0.938626 |
https://physics.stackexchange.com/questions/251900/what-are-the-assumptions-behind-the-lagrangian-derivation-of-energy?noredirect=1 | 1,571,565,015,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986705411.60/warc/CC-MAIN-20191020081806-20191020105306-00301.warc.gz | 646,837,239 | 33,526 | # What are the assumptions behind the Lagrangian derivation of energy?
What are the assumptions behind the Lagrangian derivation of energy? I understand that we're searching for a function $L$ that describes a set of physics so that solving the energy minimization problem
$$\begin{array}{rcl} \arg\min\limits_{q} && \int_{t_1}^{t_2} L(q,\dot{q},t) dt\\ \textrm{st}&&q(t_1)=q_1\\ &&q(t_2)=q_2 \end{array}$$
determines the path of an a particle $q:[t_1,t_2]\rightarrow\mathbb{R}^3$ from time $t_1$ to time $t_2$. Eventually, we find that $L(q,\dot{q},t)=\frac{1}{2}m \dot{q}^2$. What's not clear to me are the assumptions behind the setup for the energy minimization problem. It looks like to me that there's an assumption about the lack of forces like gravity. Except, I know that this derivation can be used to eventually derive that $F=m\ddot{q}$, so we don't yet have a concept of force. Also, I know there's a corollary that shows that the eventual solution $q$ is such that $\dot{q}=c$ or $\ddot{q}=0$. I'm sure that makes sense in relation to the problem setup, which is what I'm trying to clarify. Finally, I do know that we need to assume
1. Time and space are homogenous
2. Space is isotropic
3. Galilean invariance
Anyway, I'm looking for the other assumptions and I'm pretty sure it has to do with lack of other forces or some kind of related concept.
• All that is required is Newtonian mechanics; you always get the same equations of motion. The only forces that can be ignored are the forces of constraint, which is because they do no work. The easiest problems are ones with conservative forces (i.e., expressible with potentials); otherwise you must introduce generalized forces, and the equations are messier. – Peter Diehr Apr 25 '16 at 2:41
• You can do Lagrangians with inhomogeneous and non-isotropic space and time, you just don't get energy and momentum conservation. – CuriousOne Apr 25 '16 at 5:46
• Small quibble: you're extremising the action, which has the units of angular momentum, not energy. – J.G. Aug 7 '17 at 22:50
1. Firstly, given a differentiable Lagrangian $L(q,\dot{q},t)$, we can always form the Lagrangian energy function $$\tag{1} h ~:=~\sum_ip_i \dot{q}^i-L ,\qquad p_i ~:=~\frac{\partial L }{\partial \dot{q}^i }.$$
$$\tag{2} \text{The Lagrangian } L=L(q,\dot{q}) \text{ has no }{\it explicit} \text{ time dependence.}$$
• Thanks for the answer. Though, I'm not sure this quite answers things. To me, point number 2 is covered by homogeneity in space in time. In the derivation toward $\frac{1}{2}m\dot{q}^2$, we eventually assume that $L$ is only a function of $\dot{q}$. As far as point 1, it's not clear to me how this relates to the formulation above, which is sufficient for deriving $\frac{1}{2}m\dot{q}^2$. Really, I'm looking for some kind of missing assumption that precludes any sort of outside force or related idea that leads to the simple, plain, energy derivation $\frac{1}{2}m\dot{q}^2$. – wyer33 Apr 25 '16 at 19:04 | 872 | 2,983 | {"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} | 3.3125 | 3 | CC-MAIN-2019-43 | latest | en | 0.924436 |
https://www.ajdesigner.com/phpreverseosomosis/reverse_osmosis_rejection.php | 1,553,022,840,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912202125.41/warc/CC-MAIN-20190319183735-20190319205735-00442.warc.gz | 693,130,532 | 6,812 | # Reverse Osmosis Membrane Design Calculator
## Water Filtration Treatment and Purification
### Solving for contaminant rejection.
Note:
R = universal gas constant = 0.082 atm-liter/gmol-kelvin
#### Inputs:
influent feed concentration (Cin) kilogram/meter^3gram/centimeter^3gram/meter^3gram/milliliterkilogram/deciliterkilogram/litermilligram/literounce/foot^3ounce/inch^3pound/foot^3pound/bushel UKpound/bushel USpound/gallon UKpound/gallon USpound/inch^3pound/yard^3ton metric/meter^3slug/foot^3slug/inch^3
effluent permeate concentration (Cout) kilogram/meter^3gram/centimeter^3gram/meter^3gram/milliliterkilogram/deciliterkilogram/litermilligram/literounce/foot^3ounce/inch^3pound/foot^3pound/bushel UKpound/bushel USpound/gallon UKpound/gallon USpound/inch^3pound/yard^3ton metric/meter^3slug/foot^3slug/inch^3
#### Conversions:
influent feed concentration (Cin)= 0 = 0milligram/liter
effluent permeate concentration (Cout)= 0 = 0milligram/liter
#### Solution:
contaminant rejection (R)= HAS NOT BEEN CALCULATED
#### Other Units:
Change Equation
Select to solve for a different unknown
van't Hoff equation
osmotic pressure
osmotic pressure osmotic pressure coefficient number of ions per molecule concentration temperature
membrane contaminant rejection
contaminant rejection influent feed concentration effluent permeate concentration
Reference - Books:
1) P. Aarne Vesilind, J. Jeffrey Peirce and Ruth F. Weiner. 1994. Environmental Engineering. Butterworth Heinemann. 3rd ed.
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By Jimmy Raymond
Contact: aj@ajdesigner.com | 523 | 1,945 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2019-13 | longest | en | 0.715028 |
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## Difference Between NFC And AFC
National Football Conference (NFC) vs American Football Conference (AFC) The National Football Conference (NFC) and American Football Conference (AFC) are the two conferences of the National Football League (NFL), the highest professional level of American Football in the United States. These conferences are currently made up of 16 teams organized into 4 divisions- East, West, … Read more
## Difference Between Spend And Spent (Meaning &Usage)
Spent & Spend Spend is an irregular verb with two main meaning. Irregular verbs are verbs that do not follow the general rules for verb forms. The first meaning of the word spend is to use up, consume, disburse, expend, dispose of money or resources while the other meaning is to pass time in a … Read more
## Difference Between Bar Graph And Histogram
What is a Bar Graph? A bar graph also referred to as Bar Chart or Bar Diagram is a pictorial representation of data that uses bars to compare different categories of data. It displays grouped data by way of parallel rectangular bars of equal length but varying width. The bars can be plotted vertically or … Read more
## Bison vs Buffalo: 15 Differences (With Pictures & Conservation Status)
Buffalo A buffalo is a large, bovine mammal found in Africa and Asia. It is usually black or dull black in color. A buffalo is a very robust species. There are two primary species of buffalo: the Water or Asian buffalo (Bubalus bubalis) and the African buffalo (Syncerus caffer). Much like the American bison, the … Read more | 806 | 3,945 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2024-26 | longest | en | 0.914455 |
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Physics 1 Problem Solutions 181
# Physics 1 Problem Solutions 181 - Chapt 22 Electric Charge...
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Chapt. 22 Electric Charge 177 a) Given that copper’s atomic mass is 63 . 5 in atomic mass units (i.e., in units of 1 . 66 × 10 27 kg), how many copper atoms are there in the penny? HINT: If you have 10 kg of eggs and each egg weighs 50 g, then you how many eggs? b) Given that a copper atom has 29 protons (i.e., atomic number Z=29), how much gross positive charge in Coulombs does the penny have? c) Imagine (contrafactually as they say in Washington) that the absolute value of the elementary proton charge was 1 . 000001 times larger than the absolute value of the elementary electron charge (i.e., there was a fractional charge difference of ( δe/e ) = 10 6 ), what would be the force between two pennies a distance 1 meter apart assuming that the pennies would be neutral if there were no fractional charge difference. d) Is a charge difference between proton and electron of the order suggested in the part (c) question likely? Explain. 022 qfull 00200 3 3 0 tough math: a vector calculation with charges 3. Four point charges are set at the corners of a square 2 meters on a side. Going clockwise from
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{[ snackBarMessage ]} | 360 | 1,422 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2018-17 | latest | en | 0.900762 |
https://www.smore.com/6tq98 | 1,524,437,545,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125945660.53/warc/CC-MAIN-20180422212935-20180422232935-00439.warc.gz | 877,814,356 | 10,825 | # What's Happening in Adv Precalc?
## MONDAY, SEPTEMBER 22
Section 2-5: Zeros of Polynomial Functions
Common Core Standard: N-CN-7; N-CN-9
I CAN
• Solve quadratic equations (and polynomial equations) with real coefficients that have complex roots
• Know the Fundamental Theorem of Algebra
Assignment due Tuesday
• Do p322 #39-51 m3
## TUESDAY, SEPTEMBER 23
Section 2-5: Zeros of Polynomial Functions
Common Core Standard: N-CN-7; N-CN-9
I CAN
• Solve quadratic equations (and polynomial equations) with real coefficients that have complex roots
• Know the Fundamental Theorem of Algebra
Assignment due Wednesday
• Do p322 #26-32 even
## WEDNESDAY, SEPTEMBER 24
Section 2-5: Zeros of Polynomial Functions
Common Core Standard: N-CN-7; N-CN-9
I CAN
• Solve quadratic equations (and polynomial equations) with real coefficients that have complex roots
• Know the Fundamental Theorem of Algebra
Assignment due Thursday
• Do p322 #53-59 odd --> do not graph; check answers in back of book as a review for Thursday's quiz
## THURSDAY, SEPTEMBER 25
QUIZ Over Sections 2-3, 2-4 and 2-5
Assignment
• None
## FRIDAY, SEPTEMBER 26
Section 2-5: Zeros of Polynomial Functions
Common Core Standard: N-CN-7; N-CN-9
I CAN
• Solve quadratic equations (and polynomial equations) with real coefficients that have complex roots
• Know the Fundamental Theorem of Algebra
Assignment due Monday | 375 | 1,393 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2018-17 | longest | en | 0.789174 |
https://www.edupil.com/question/women-and-children-take-complete-the-work/ | 1,619,180,191,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039617701.99/warc/CC-MAIN-20210423101141-20210423131141-00540.warc.gz | 816,174,047 | 15,522 | # How Many Days will Women and Children Take Complete the Work? If
10 women can complete a work in 7 days and 10 children take 14 days to complete the work. How many days will 5 women and 10 children take to complete the work?
1. 3
2. 5
3. 7
4. Cannot be determined
Monis Rasool Professor Asked on 13th June 2015 in
Answer: (3) 7 days
Explanation:-
10 women can complete a work = 7 days
1 women can work in a day = 1/7 x 10 = 1/70
10 children can complete a work = 14 days
1 child can work in a day = 1/10 x 14 = 1/140
5 women and 10 children can complete work = 5/70 + 10/140
= (10 + 10)/140
= 20/140
= 1/7
Hence, 5 women and 10 children can complete the work in 7 days.
Anurag Mishra Professor Answered on 13th June 2015. | 247 | 732 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-17 | longest | en | 0.903082 |
https://howkgtolbs.com/convert/28.83-kg-to-lbs | 1,618,995,764,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039526421.82/warc/CC-MAIN-20210421065303-20210421095303-00128.warc.gz | 380,218,248 | 12,278 | 28.83 kg to lbs - 28.83 kilograms to pounds
Do you need to learn how much is 28.83 kg equal to lbs and how to convert 28.83 kg to lbs? You are in the right place. This whole article is dedicated to kilogram to pound conversion - both theoretical and practical. It is also needed/We also want to emphasize that all this article is dedicated to a specific amount of kilograms - this is one kilogram. So if you want to know more about 28.83 kg to pound conversion - keep reading.
Before we move on to the more practical part - that is 28.83 kg how much lbs conversion - we want to tell you a little bit of theoretical information about these two units - kilograms and pounds. So let’s start.
How to convert 28.83 kg to lbs? 28.83 kilograms it is equal 63.5592701346 pounds, so 28.83 kg is equal 63.5592701346 lbs.
28.83 kgs in pounds
We are going to start with the kilogram. The kilogram is a unit of mass. It is a base unit in a metric system, known also as International System of Units (in abbreviated form SI).
From time to time the kilogram is written as kilogramme. The symbol of the kilogram is kg.
Firstly, the definition of a kilogram was formulated in 1795. The kilogram was defined as the mass of one liter of water. This definition was not complicated but impractical to use.
Then, in 1889 the kilogram was described by the International Prototype of the Kilogram (in short form IPK). The IPK was made of 90% platinum and 10 % iridium. The International Prototype of the Kilogram was used until 2019, when it was substituted by another definition.
Today the definition of the kilogram is based on physical constants, especially Planck constant. The official definition is: “The kilogram, symbol kg, is the SI unit of mass. It is defined by taking the fixed numerical value of the Planck constant h to be 6.62607015×10−34 when expressed in the unit J⋅s, which is equal to kg⋅m2⋅s−1, where the metre and the second are defined in terms of c and ΔνCs.”
One kilogram is equal 0.001 tonne. It can be also divided into 100 decagrams and 1000 grams.
28.83 kilogram to pounds
You know a little bit about kilogram, so now let’s move on to the pound. The pound is also a unit of mass. We want to emphasize that there are not only one kind of pound. What are we talking about? For example, there are also pound-force. In this article we want to concentrate only on pound-mass.
The pound is in use in the Imperial and United States customary systems of measurements. Naturally, this unit is in use also in another systems. The symbol of this unit is lb or “.
There is no descriptive definition of the international avoirdupois pound. It is defined as 0.45359237 kilograms. One avoirdupois pound could be divided to 16 avoirdupois ounces and 7000 grains.
The avoirdupois pound was enforced in the Weights and Measures Act 1963. The definition of the pound was given in first section of this act: “The yard or the metre shall be the unit of measurement of length and the pound or the kilogram shall be the unit of measurement of mass by reference to which any measurement involving a measurement of length or mass shall be made in the United Kingdom; and- (a) the yard shall be 0.9144 metre exactly; (b) the pound shall be 0.45359237 kilogram exactly.”
How many lbs is 28.83 kg?
28.83 kilogram is equal to 63.5592701346 pounds. If You want convert kilograms to pounds, multiply the kilogram value by 2.2046226218.
28.83 kg in lbs
The most theoretical part is already behind us. In next part we will tell you how much is 28.83 kg to lbs. Now you know that 28.83 kg = x lbs. So it is time to know the answer. Have a look:
28.83 kilogram = 63.5592701346 pounds.
This is an accurate outcome of how much 28.83 kg to pound. You can also round it off. After it your outcome will be exactly: 28.83 kg = 63.426 lbs.
You know 28.83 kg is how many lbs, so have a look how many kg 28.83 lbs: 28.83 pound = 0.45359237 kilograms.
Naturally, in this case it is possible to also round it off. After it your outcome will be as following: 28.83 lb = 0.45 kgs.
We are also going to show you 28.83 kg to how many pounds and 28.83 pound how many kg results in charts. Let’s see:
We will begin with a chart for how much is 28.83 kg equal to pound.
28.83 Kilograms to Pounds conversion table
Kilograms (kg) Pounds (lb) Pounds (lbs) (rounded off to two decimal places)
28.83 63.5592701346 63.4260
Now look at a table for how many kilograms 28.83 pounds.
Pounds Kilograms Kilograms (rounded off to two decimal places
28.83 0.45359237 0.45
Now you know how many 28.83 kg to lbs and how many kilograms 28.83 pound, so we can move on to the 28.83 kg to lbs formula.
28.83 kg to pounds
To convert 28.83 kg to us lbs you need a formula. We are going to show you two formulas. Let’s start with the first one:
Number of kilograms * 2.20462262 = the 63.5592701346 outcome in pounds
The first version of a formula give you the most correct outcome. In some situations even the smallest difference can be significant. So if you want to get an exact result - this formula will be the best solution to convert how many pounds are equivalent to 28.83 kilogram.
So let’s go to the another version of a formula, which also enables conversions to learn how much 28.83 kilogram in pounds.
The second version of a formula is as following, have a look:
Amount of kilograms * 2.2 = the outcome in pounds
As you can see, this version is simpler. It can be the best solution if you want to make a conversion of 28.83 kilogram to pounds in easy way, for instance, during shopping. You only have to remember that final outcome will be not so correct.
Now we want to learn you how to use these two versions of a formula in practice. But before we are going to make a conversion of 28.83 kg to lbs we want to show you another way to know 28.83 kg to how many lbs without any effort.
28.83 kg to lbs converter
Another way to learn what is 28.83 kilogram equal to in pounds is to use 28.83 kg lbs calculator. What is a kg to lb converter?
Converter is an application. Calculator is based on first version of a formula which we showed you above. Thanks to 28.83 kg pound calculator you can effortless convert 28.83 kg to lbs. You only need to enter amount of kilograms which you want to calculate and click ‘calculate’ button. The result will be shown in a second.
So try to convert 28.83 kg into lbs with use of 28.83 kg vs pound converter. We entered 28.83 as an amount of kilograms. It is the outcome: 28.83 kilogram = 63.5592701346 pounds.
As you see, this 28.83 kg vs lbs calculator is easy to use.
Now we can go to our main topic - how to convert 28.83 kilograms to pounds on your own.
28.83 kg to lbs conversion
We will start 28.83 kilogram equals to how many pounds calculation with the first formula to get the most exact result. A quick reminder of a formula:
Amount of kilograms * 2.20462262 = 63.5592701346 the outcome in pounds
So what need you do to learn how many pounds equal to 28.83 kilogram? Just multiply amount of kilograms, this time 28.83, by 2.20462262. It is equal 63.5592701346. So 28.83 kilogram is exactly 63.5592701346.
You can also round it off, for example, to two decimal places. It is 2.20. So 28.83 kilogram = 63.4260 pounds.
It is high time for an example from everyday life. Let’s calculate 28.83 kg gold in pounds. So 28.83 kg equal to how many lbs? And again - multiply 28.83 by 2.20462262. It is 63.5592701346. So equivalent of 28.83 kilograms to pounds, when it comes to gold, is 63.5592701346.
In this example it is also possible to round off the result. This is the outcome after rounding off, in this case to one decimal place - 28.83 kilogram 63.426 pounds.
Now we can go to examples converted using a short version of a formula.
How many 28.83 kg to lbs
Before we show you an example - a quick reminder of shorter formula:
Amount of kilograms * 2.2 = 63.426 the result in pounds
So 28.83 kg equal to how much lbs? As in the previous example you need to multiply amount of kilogram, in this case 28.83, by 2.2. See: 28.83 * 2.2 = 63.426. So 28.83 kilogram is equal 2.2 pounds.
Let’s do another calculation using this formula. Now calculate something from everyday life, for instance, 28.83 kg to lbs weight of strawberries.
So let’s convert - 28.83 kilogram of strawberries * 2.2 = 63.426 pounds of strawberries. So 28.83 kg to pound mass is exactly 63.426.
If you learned how much is 28.83 kilogram weight in pounds and can calculate it with use of two different versions of a formula, let’s move on. Now we are going to show you all outcomes in tables.
Convert 28.83 kilogram to pounds
We know that results presented in charts are so much clearer for most of you. It is totally understandable, so we gathered all these outcomes in charts for your convenience. Thanks to this you can easily make a comparison 28.83 kg equivalent to lbs outcomes.
Let’s start with a 28.83 kg equals lbs chart for the first version of a formula:
Kilograms Pounds Pounds (after rounding off to two decimal places)
28.83 63.5592701346 63.4260
And now let’s see 28.83 kg equal pound table for the second version of a formula:
Kilograms Pounds
28.83 63.426
As you see, after rounding off, if it comes to how much 28.83 kilogram equals pounds, the results are not different. The bigger number the more significant difference. Please note it when you need to do bigger amount than 28.83 kilograms pounds conversion.
How many kilograms 28.83 pound
Now you know how to calculate 28.83 kilograms how much pounds but we are going to show you something more. Are you curious what it is? What about 28.83 kilogram to pounds and ounces calculation?
We want to show you how you can calculate it little by little. Let’s begin. How much is 28.83 kg in lbs and oz?
First thing you need to do is multiply number of kilograms, this time 28.83, by 2.20462262. So 28.83 * 2.20462262 = 63.5592701346. One kilogram is 2.20462262 pounds.
The integer part is number of pounds. So in this case there are 2 pounds.
To calculate how much 28.83 kilogram is equal to pounds and ounces you need to multiply fraction part by 16. So multiply 20462262 by 16. It is equal 327396192 ounces.
So your result is 2 pounds and 327396192 ounces. You can also round off ounces, for example, to two places. Then your result is 2 pounds and 33 ounces.
As you see, conversion 28.83 kilogram in pounds and ounces quite simply.
The last calculation which we will show you is conversion of 28.83 foot pounds to kilograms meters. Both foot pounds and kilograms meters are units of work.
To convert foot pounds to kilogram meters you need another formula. Before we show you it, look:
• 28.83 kilograms meters = 7.23301385 foot pounds,
• 28.83 foot pounds = 0.13825495 kilograms meters.
Now have a look at a formula:
Amount.RandomElement()) of foot pounds * 0.13825495 = the outcome in kilograms meters
So to calculate 28.83 foot pounds to kilograms meters you have to multiply 28.83 by 0.13825495. It gives 0.13825495. So 28.83 foot pounds is 0.13825495 kilogram meters.
It is also possible to round off this result, for instance, to two decimal places. Then 28.83 foot pounds is exactly 0.14 kilogram meters.
We hope that this conversion was as easy as 28.83 kilogram into pounds calculations.
We showed you not only how to do a calculation 28.83 kilogram to metric pounds but also two another conversions - to check how many 28.83 kg in pounds and ounces and how many 28.83 foot pounds to kilograms meters.
We showed you also other solution to make 28.83 kilogram how many pounds calculations, it is with use of 28.83 kg en pound calculator. It is the best choice for those of you who do not like calculating on your own at all or this time do not want to make @baseAmountStr kg how lbs conversions on your own.
We hope that now all of you can do 28.83 kilogram equal to how many pounds conversion - on your own or using our 28.83 kgs to pounds calculator.
So what are you waiting for? Let’s convert 28.83 kilogram mass to pounds in the way you like.
Do you want to do other than 28.83 kilogram as pounds calculation? For instance, for 5 kilograms? Check our other articles! We guarantee that calculations for other numbers of kilograms are so easy as for 28.83 kilogram equal many pounds.
How much is 28.83 kg in pounds
At the end, we are going to summarize the topic of this article, that is how much is 28.83 kg in pounds , we prepared for you an additional section. Here you can see all you need to know about how much is 28.83 kg equal to lbs and how to convert 28.83 kg to lbs . Have a look.
What is the kilogram to pound conversion? To make the kg to lb conversion it is needed to multiply 2 numbers. Let’s see 28.83 kg to pound conversion formula . Have a look:
The number of kilograms * 2.20462262 = the result in pounds
Now you can see the result of the conversion of 28.83 kilogram to pounds. The correct result is 63.5592701346 lbs.
It is also possible to calculate how much 28.83 kilogram is equal to pounds with another, easier version of the equation. Check it down below.
The number of kilograms * 2.2 = the result in pounds
So this time, 28.83 kg equal to how much lbs ? The answer is 63.5592701346 lb.
How to convert 28.83 kg to lbs in an easier way? It is possible to use the 28.83 kg to lbs converter , which will make whole mathematical operation for you and you will get an exact answer .
Kilograms [kg]
The kilogram, or kilogramme, is the base unit of weight in the Metric system. It is the approximate weight of a cube of water 10 centimeters on a side.
Pounds [lbs]
A pound is a unit of weight commonly used in the United States and the British commonwealths. A pound is defined as exactly 0.45359237 kilograms. | 3,598 | 13,755 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2021-17 | latest | en | 0.949333 |
https://physics.stackexchange.com/questions/38124/why-water-is-not-superfluid | 1,713,700,250,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817765.59/warc/CC-MAIN-20240421101951-20240421131951-00370.warc.gz | 410,238,918 | 40,014 | # Why water is not superfluid?
My question is in the title. I do not really understand why water is not a superfluid. Maybe I make a mistake but the fact that water is not suprfluid comes from the fact that the elementary excitations have a parabolic dispersion curve but for me the question remain. An equivalent way to ask it is: why superfluid helium is described by Gross-Pitaevsky equation and it is not the case for water?
• Recent work actually suggests that water may have a superfluid liquid phase
– user20145
Jan 23, 2013 at 15:24
• @x you have to substantiate this claim by a reference or link and a quote, at least from the abstract. Jan 23, 2013 at 15:31
You refer to the Landau criterion for superfluidity (there is a separate question whether this is really the best way to think about superfluids, and whether the Landau criterion is necessary and/or sufficient). In a superfluid the low energy excitations are phonons, the dispersion relation is linear $E_p\sim c p$, and the critical velocity is non-zero. In water the degrees of freedom are water molecules, the dispersion relation is quadratic, $E_p\sim p^2/(2m)$, and the critical velocity is zero.
• A rough criterion is the condition for Bose condensation in an ideal gas, $n\lambda^3\sim 1$, where $n$ is the density and $\lambda$ is the thermal wave length. Note that your question is in some sense backwards: Helium is the exception, water is the rule. Most ordinary fluids solidify instead of becoming superfluid at low $T$. Sep 24, 2012 at 12:38
• I'm not sure that one should consider ordinary molecules as quasiparticles leading to dissipation because the Landau criterion originates from investigation of a phonon branch. Instead, we have to use the same formalism of phonons to be able to compare liquids' behavior. In non-superconducting liquid a phonon dispersion law is $E=c|p|$. So applying Landau criterion we obtain a peculiar critical velocity equal to $c$ - the speed of sound. It doesn't mean necessarily that the liquid is superconducting below $c$. It rather means that it goes into a different state above $c$ (supersonic flow). Jul 6, 2020 at 19:27 | 519 | 2,146 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2024-18 | latest | en | 0.915876 |
http://www.wikihow.com/Calculate-Covariance | 1,475,354,532,000,000,000 | text/html | crawl-data/CC-MAIN-2016-40/segments/1474738663247.66/warc/CC-MAIN-20160924173743-00216-ip-10-143-35-109.ec2.internal.warc.gz | 833,942,855 | 59,023 | Edit Article
# How to Calculate Covariance
Covariance is a type of value used in statistics to describe the linear relationship between two variables. The higher the covariance between two variables, the more closely their values follow the same trends over a range of data points (in other words, the less the two variables' curves deviate from each other). In general, the covariance for two sets of values x and y can be found with the formula 1/(n -1)Σ(xi - xavg)(yi - yavg), where n equals the sample size, xi equals the value of each x point, xavg equals the average of the x points' values, and so on for yi and yavg.
### Part 1 Using Standard Covariance Formula
1. 1
Organize your data into a range of (x,y) points. All you need to calculate covariance is a set of data points for two variables x and y. If you're working with data from a graph, your data will come from the (x,y) coordinate pairs of the points on the graph; otherwise, the data will come from finding pairs of corresponding values for your variables mathematically.
• Take note of the number of matching x/y pairs. This will be ‘’n’’, your sample size, which is needed to solve the covariance formula.
• As an example, let's say that we run a deli and that we're trying to determine whether or not the number of coupons we give out has an effect on sales. We can define x as the number of coupons given out on a given day and y as the number of sales for that day.
• For convenience, let's use the table in the picture above as our reference — in other words, on the first day we gave out x = 1 coupon and had y = 8 sales, on the second day, we gave out x = 3 coupons and had y = 6 sales, and so on.
2. 2
Find the average of the x points. After you've got a range of x/y pairs, solving the covariance equation doesn't actually require that much work. To begin, find the average of your x values. You can do this by adding the x values together and dividing by the number of values (see our guide on finding averages for detailed instructions.)
• In our example, we would find the average of our x values by adding the values in the "x" column in the table above and dividing by the number of values in the column. Adding 1+3+2+5..., we get a total of 44. Dividing this by 9, the number of x values, we get 44/9 = 4.89 as our x average. See below:
1 + 3 + 2 + 5 + 8 + 7 + 12 + 2 + 4 = 44
44/9 = 4.89
3. 3
Find the average of the y points. Next, find the average of all the y points. This is done exactly the same as it is for the x points: add the y values together, then divide by the number of values.
• In our example, we would add 8+6+9+4... to get a total of 49. Dividing by the number of values (which is the same as for x), we get 49/9 = 5.44 as our y average. See below:
8 + 6 + 9 + 4 + 3 + 3 + 2 + 7 + 7 = 49
49/9 = 5.44
4. 4
Plug your variables into the formula 1/(n -1)Σ(xi - xavg)* Σ(yi - yavg). Note the sigma (Σ) in the formula — this means you'll need to subtract the x average from every individual x value and sum them together (then do the same with the y average and every individual y value). This can lead to long strings of subtraction problems, so record your values carefully to avoid an error.
• In our example, we would solve as follows:
1/(n -1)Σ(xi - xavg)(yi - yavg)
(1/8)(((1 - 4.89)+(3 - 4.89)+(2 - 4.89)+(5 - 4.89)+(8 - 4.89)+(7 - 4.89)+(12 - 4.89)+(2 - 4.89)+(4 - 4.89))((8 - 5.44)+(6 - 5.44)+(9 - 5.44)+(4 - 5.44)+(3 - 5.44)+(3 - 5.44)+(2 - 5.44)+(7 - 5.44)+(7 - 5.44))
(1/8)((-0.01)((8 - 5.44)+(6 - 5.44)+(9 - 5.44)+(4 - 5.44)+(3 - 5.44)+(3 - 5.44)+(2 - 5.44)+(7 - 5.44)+(7 - 5.44))
(1/8)(-0.01)(0.04) = 0.00005
• As we'll learn below, our answer of 0.00005 is very close to zero, so this means that the number of coupons we give out has essentially no effect on the number of sales we make at the deli.
### Part 2 Using Covariance Values
1. 1
A covariance of 1 indicates perfect positive correlation. When it comes to covariances, your answers will always be between 1 and -1. Any answer outside this range means that there has been some sort of error in the calculation. Based on how close your covariance is to 1 or -1, you can draw certain conclusions about your data set. For instance, if your covariance is exactly 1, this means that your variables have perfect positive correlation. In other words, when one variable increases, the second increases, and when one decreases the other decreases. This relationship is perfectly linear — no matter how high or low the variables get, they'll have the same relationship.
• As an example of this sort of covariance, let's consider the simple business model of a lemonade stand where we sell each lemonade at \$3. If x represents the number of lemonades you sell and y represents the money you make, y will always increase with x in this example. See below:
Ten lemonades sold: x = 10, y = \$30
One hundred lemonades sold: x = 100, y = \$300
One million lemonades sold: x = 1,000,000, y = \$3,000,000
No matter how high our x value gets, we'll always see this exact same relationship — y will be equal to 3(x). Therefore, we can say that x and y have perfect positive correlation, or, in other words, a correlation of 1.
2. 2
A covariance of -1 indicates perfect negative correlation. On the other hand, if your covariance is -1, this means that your variables are perfectly negatively correlated.[1] In other words, an increase in one will cause a decrease in the other, and vice versa. As above, this relationship is linear. The rate at which the two variables grow apart from each other doesn't change with time.
• As an example of this sort of correlation, let's say that we're in charge of drilling oil from a single well that contains about 10,000 barrels of oil. If x equals the barrels of oil we've drilled already and y equals the number of barrels remaining in the well, we can say that as long as x increases, y will decrease. In other words, there's no way that oil will magically re-appear in the well — once it's gone, it's gone.[2] See below:
One barrel drilled: x = 1, y = 9,999.
Two thousand barrels drilled: x = 2,000, y = 8,000.
Ten thousand barrels drilled: x = 10,000, y = 0.
As our x value grows, our y value decreases at a steady rate. The relationship is linear — each additional barrel drilled will always mean one fewer barrel in the ground. Therefore, we can say that x and y have perfect negative correlation, or, in other words, a correlation of -1.
3. 3
Know that a covariance of 0 indicates no correlation. If your covariance is equal to zero, this means that there is no correlation at all between your variables.[3] In other words, an increase or decrease in one will not necessarily cause an increase or decrease in the other with any predictability. There is no linear relationship between the two variables, but there might still be a non-linear relationship.
• As an example of this sort of correlation, let's consider the case of someone who is taking a homeopathic remedy for a viral illness. If x represents the dosage of the remedy taken (in teaspoons) and y represents the viral load in the person's bloodstream (in International Units per milliliter (IU/mL)) we wouldn't necessarily expect y to increase or decrease as x increases. Rather, any fluctuation in y would be completely independent of x.[4] See below:
One teaspoon taken: x = 1, y = 615.
Ten teaspoons taken: x = 10 y = 700.
Twenty teaspoons taken: x = 20, y = 455.
As our x value grows, we can't really predict whether y will increase or decrease in response. The relationship isn't clear — sometimes taking more remedy causes a decrease in the viral load, while sometimes it causes an increase. Therefore, we would expect that x and y would have near-zero correlation.
4. 4
Know that another value between -1 and 1 indicates imperfect correlation. Most covariance values aren't exactly 1, -1, or 0. Usually, they are somewhere in between. Based on how close a given covariance value is to one of these benchmarks, you can say that it is more or less positively correlated or negatively correlated.
• For example, a covariance of 0.8 indicates that there is a high degree of positive correlation between the two variables, though not perfect correlation. In other words, as x increases, y will generally increase, and as x decreases, y will generally decrease, though the relationship may not be perfectly steady."
## Community Q&A
• Worked examples of mean deviation, variance of ungrouped data using the long method?
## Tips
• See our articles on scatter plots and the coefficients of correlation for related information.
• Covariance equations are often used to compare stocks — investors like being able to tell whether two stocks are likely to fluctuate with one another or not. To find this, all you'll need is a chart that compares two stocks' daily routines over a range of dates. See below:
Company A (x): (1.6 + 1.9 + 2.1 + 3.2 + 0.5 + 0.4 + 0.6)/7 = 1.47
Company B (y): (2.0 + 2.4 + 2.6 + 3.6 + 0.9 + 0.8 + 1.0)/7 = 1.9
(1/n -1)(Σ(xi - xavg)(yi - yavg)
(1/6)(((1.6 - 1.47)+(1.9 - 1.47)+(2.1 - 1.47)+(3.2 - 1.47)+(0.5 - 1.47)+(0.4 - 1.47)+(0.6 - 1.47))((2.0 - 1.78)+(2.4 - 1.78)+(2.6 - 1.78)+(3.6 - 1.78)+(0.9 - 1.78)+(0.8 - 1.78)+(1.0 - 1.78))
(1/6)((0.01)(0.84))
(1/6)(0.084) = 0.14.
## Article Info
Categories: Probability and Statistics
In other languages:
Español: calcular la covarianza, Português: Calcular a Covariância, Italiano: Calcolare la Covarianza, Русский: вычислить ковариантность, Français: calculer la covariance, Deutsch: Die Kovarianz berechnen, 中文: 计算协方差, Bahasa Indonesia: Menghitung Kovarian
Thanks to all authors for creating a page that has been read 388,038 times. | 2,699 | 9,700 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-40 | longest | en | 0.916399 |
http://dlmf.nist.gov/27.14 | 1,495,776,820,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463608642.30/warc/CC-MAIN-20170526051657-20170526071657-00117.warc.gz | 126,101,751 | 15,729 | # §27.14 Unrestricted Partitions
## §27.14(i) Partition Functions
A fundamental problem studies the number of ways $n$ can be written as a sum of positive integers $\leq n$, that is, the number of solutions of
27.14.1 $n=a_{1}+a_{2}+\cdots,$ $a_{1}\geq a_{2}\geq\cdots\geq 1$. Symbols: $n$: positive integer Permalink: http://dlmf.nist.gov/27.14.E1 Encodings: TeX, pMML, png See also: Annotations for 27.14(i)
The number of summands is unrestricted, repetition is allowed, and the order of the summands is not taken into account. The corresponding unrestricted partition function is denoted by $\mathop{p\/}\nolimits\!\left(n\right)$, and the summands are called parts; see §26.9(i). For example, $\mathop{p\/}\nolimits\!\left(5\right)=7$ because there are exactly seven partitions of $5$: $5=4+1=3+2=3+1+1=2+2+1=2+1+1+1=1+1+1+1+1$.
The number of partitions of $n$ into at most $k$ parts is denoted by $\mathop{p_{k}\/}\nolimits\!\left(n\right)$; again see §26.9(i).
## §27.14(ii) Generating Functions and Recursions
Euler introduced the reciprocal of the infinite product
27.14.2 $\mathop{\mathit{f}\/}\nolimits\!\left(x\right)=\prod_{m=1}^{\infty}(1-x^{m}),$ $|x|<1$, Defines: $\mathop{\mathit{f}\/}\nolimits\!\left(\NVar{x}\right)$: Euler’s reciprocal function Symbols: $m$: positive integer and $x$: real number Referenced by: §27.14(iv), §27.21 Permalink: http://dlmf.nist.gov/27.14.E2 Encodings: TeX, pMML, png See also: Annotations for 27.14(ii)
as a generating function for the function $\mathop{p\/}\nolimits\!\left(n\right)$ defined in §27.14(i):
27.14.3 $\frac{1}{\mathop{\mathit{f}\/}\nolimits\!\left(x\right)}=\sum_{n=0}^{\infty}% \mathop{p\/}\nolimits\!\left(n\right)x^{n},$
with $\mathop{p\/}\nolimits\!\left(0\right)=1$. Euler’s pentagonal number theorem states that
27.14.4 $\mathop{\mathit{f}\/}\nolimits\!\left(x\right)=1-x-x^{2}+x^{5}+x^{7}-x^{12}-x^% {15}+\dots=1+\sum_{k=1}^{\infty}(-1)^{k}\left(x^{\omega(k)}+x^{\omega(-k)}% \right),$
where the exponents $1$, $2$, $5$, $7$, $12$, $15$, $\dots$ are the pentagonal numbers, defined by
27.14.5 $\omega(\pm k)=(3k^{2}\mp k)/2,$ $k=1,2,3,\dots$. Defines: $\omega(k)$: pentagonal numbers (locally) Symbols: $k$: positive integer Permalink: http://dlmf.nist.gov/27.14.E5 Encodings: TeX, pMML, png See also: Annotations for 27.14(ii)
Multiplying the power series for $\mathop{\mathit{f}\/}\nolimits\!\left(x\right)$ with that for $1/\mathop{\mathit{f}\/}\nolimits\!\left(x\right)$ and equating coefficients, we obtain the recursion formula
27.14.6 $\mathop{p\/}\nolimits\!\left(n\right)=\sum_{k=1}^{\infty}(-1)^{k+1}\left(% \mathop{p\/}\nolimits\!\left(n-\omega(k)\right)+\mathop{p\/}\nolimits\!\left(n% -\omega(-k)\right)\right)=\mathop{p\/}\nolimits\!\left(n-1\right)+\mathop{p\/}% \nolimits\!\left(n-2\right)-\mathop{p\/}\nolimits\!\left(n-5\right)-\mathop{p% \/}\nolimits\!\left(n-7\right)+\cdots,$ Symbols: $\mathop{p\/}\nolimits\!\left(\NVar{n}\right)$: total number of partitions of $n$, $k$: positive integer, $n$: positive integer and $\omega(k)$: pentagonal numbers A&S Ref: 24.2.1 II.A (in slightly different form) Referenced by: §27.20, §27.20 Permalink: http://dlmf.nist.gov/27.14.E6 Encodings: TeX, pMML, png See also: Annotations for 27.14(ii)
where $\mathop{p\/}\nolimits\!\left(k\right)$ is defined to be $0$ if $k<0$. Logarithmic differentiation of the generating function $1/\mathop{\mathit{f}\/}\nolimits\!\left(x\right)$ leads to another recursion:
27.14.7 $n\mathop{p\/}\nolimits\!\left(n\right)=\sum_{k=1}^{n}\mathop{\sigma_{1}\/}% \nolimits\!\left(k\right)\mathop{p\/}\nolimits\!\left(n-k\right),$ Symbols: $\mathop{\sigma_{\NVar{\alpha}}\/}\nolimits\!\left(\NVar{n}\right)$: sum of powers of divisors of $n$, $\mathop{p\/}\nolimits\!\left(\NVar{n}\right)$: total number of partitions of $n$, $k$: positive integer and $n$: positive integer A&S Ref: 24.2.1 II.A (in slightly different form) 24.2.1 II.A (in slightly different form) Referenced by: §27.20, §27.20, Other Changes Permalink: http://dlmf.nist.gov/27.14.E7 Encodings: TeX, pMML, png Errata (effective with 1.0.11): The erroneous $\mathop{\sigma_{1}\/}\nolimits\!\left(n\right)$ in $n\mathop{p\/}\nolimits\!\left(n\right)=\sum_{k=1}^{n}\mathop{\sigma_{1}\/}% \nolimits\!\left(n\right)\mathop{p\/}\nolimits\!\left(n-k\right)$ was replaced. Reported 2015-08-17 by Howard Cohl and Hannah Cohen See also: Annotations for 27.14(ii)
where $\mathop{\sigma_{1}\/}\nolimits\!\left(k\right)$ is defined by (27.2.10) with $\alpha=1$.
## §27.14(iii) Asymptotic Formulas
These recursions can be used to calculate $\mathop{p\/}\nolimits\!\left(n\right)$, which grows very rapidly. For example, $\mathop{p\/}\nolimits\!\left(10\right)=42,\mathop{p\/}\nolimits\!\left(100\right)$ = $1905\;69292$, and $\mathop{p\/}\nolimits\!\left(200\right)=397\;29990\;29388$. For large $n$
27.14.8 $\mathop{p\/}\nolimits\!\left(n\right)\sim e^{K\sqrt{n}}/(4n\sqrt{3}),$ Symbols: $\sim$: asymptotic equality, $\mathrm{e}$: base of exponential function, $\mathop{p\/}\nolimits\!\left(\NVar{n}\right)$: total number of partitions of $n$ and $n$: positive integer A&S Ref: 24.2.1 III Permalink: http://dlmf.nist.gov/27.14.E8 Encodings: TeX, pMML, png See also: Annotations for 27.14(iii)
where $K=\pi\sqrt{2/3}$ (Hardy and Ramanujan (1918)). Rademacher (1938) derives a convergent series that also provides an asymptotic expansion for $\mathop{p\/}\nolimits\!\left(n\right)$:
27.14.9 $\mathop{p\/}\nolimits\!\left(n\right)=\frac{1}{\pi\sqrt{2}}\sum_{k=1}^{\infty}% \sqrt{k}A_{k}(n)\*\left[\frac{\mathrm{d}}{\mathrm{d}t}\frac{\mathop{\sinh\/}% \nolimits\!\left(\ifrac{K\sqrt{t}}{k}\right)}{\sqrt{t}}\right]_{t=n-(1/24)},$
where
27.14.10 $A_{k}(n)=\sum_{\begin{subarray}{c}h=1\\ \left(h,k\right)=1\end{subarray}}^{k}\mathop{\exp\/}\nolimits\!\left(\pi% \mathrm{i}s(h,k)-2\pi\mathrm{i}n\frac{h}{k}\right),$
and $s(h,k)$ is a Dedekind sum given by
27.14.11 $s(h,k)=\sum_{r=1}^{k-1}\frac{r}{k}\left(\frac{hr}{k}-\left\lfloor\frac{hr}{k}% \right\rfloor-\frac{1}{2}\right).$ Defines: $s(h,k)$: Dedekind sum (locally) Symbols: $\left\lfloor\NVar{x}\right\rfloor$: floor of $x$ and $k$: positive integer A&S Ref: 24.2.1 I.C Referenced by: §23.18, §27.14(iv) Permalink: http://dlmf.nist.gov/27.14.E11 Encodings: TeX, pMML, png See also: Annotations for 27.14(iii)
## §27.14(iv) Relation to Modular Functions
Dedekind sums occur in the transformation theory of the Dedekind modular function $\mathop{\eta\/}\nolimits\!\left(\tau\right)$, defined by
27.14.12 $\mathop{\eta\/}\nolimits\!\left(\tau\right)=e^{\pi\mathrm{i}\tau/12}\prod_{n=1% }^{\infty}(1-e^{2\pi\mathrm{i}n\tau}),$ $\Im{\tau}>0$. Defines: $\mathop{\eta\/}\nolimits\!\left(\NVar{\tau}\right)$: Dedekind’s eta function (or Dedekind modular function) Symbols: $\pi$: the ratio of the circumference of a circle to its diameter, $\mathrm{e}$: base of exponential function, $\Im{}$: imaginary part and $n$: positive integer Referenced by: §23.15(ii), §27.14(vi) Permalink: http://dlmf.nist.gov/27.14.E12 Encodings: TeX, pMML, png See also: Annotations for 27.14(iv)
This is related to the function $\mathop{\mathit{f}\/}\nolimits\!\left(x\right)$ in (27.14.2) by
27.14.13 $\mathop{\eta\/}\nolimits\!\left(\tau\right)=e^{\pi\mathrm{i}\tau/12}\mathop{% \mathit{f}\/}\nolimits\!\left(e^{2\pi\mathrm{i}\tau}\right).$
$\mathop{\eta\/}\nolimits\!\left(\tau\right)$ satisfies the following functional equation: if $a,b,c,d$ are integers with $ad-bc=1$ and $c>0$, then
27.14.14 $\mathop{\eta\/}\nolimits\!\left(\frac{a\tau+b}{c\tau+d}\right)=\varepsilon(-% \mathrm{i}(c\tau+d))^{\frac{1}{2}}\mathop{\eta\/}\nolimits\!\left(\tau\right),$
where $\varepsilon=\mathop{\exp\/}\nolimits\!\left(\pi\mathrm{i}(((a+d)/(12c))-s(d,c)% )\right)$ and $s(d,c)$ is given by (27.14.11).
For further properties of the function $\mathop{\eta\/}\nolimits\!\left(\tau\right)$ see §§23.1523.19.
## §27.14(v) Divisibility Properties
Ramanujan (1921) gives identities that imply divisibility properties of the partition function. For example, the Ramanujan identity
27.14.15 $5\frac{(\mathop{\mathit{f}\/}\nolimits\!\left(x^{5}\right))^{5}}{(\mathop{% \mathit{f}\/}\nolimits\!\left(x\right))^{6}}=\sum_{n=0}^{\infty}\mathop{p\/}% \nolimits\!\left(5n+4\right)x^{n}$
implies $\mathop{p\/}\nolimits\!\left(5n+4\right)\equiv 0\pmod{5}$. Ramanujan also found that $\mathop{p\/}\nolimits\!\left(7n+5\right)\equiv 0\pmod{7}$ and $\mathop{p\/}\nolimits\!\left(11n+6\right)\equiv 0\pmod{11}$ for all $n$. After decades of nearly fruitless searching for further congruences of this type, it was believed that no others existed, until it was shown in Ono (2000) that there are infinitely many. Ono proved that for every prime $q>3$ there are integers $a$ and $b$ such that $\mathop{p\/}\nolimits\!\left(an+b\right)\equiv 0\pmod{q}$ for all $n$. For example, $\mathop{p\/}\nolimits\!\left(1575\;25693n+1\;11247\right)\equiv 0\pmod{13}$.
## §27.14(vi) Ramanujan’s Tau Function
The discriminant function $\mathop{\Delta\/}\nolimits\!\left(\tau\right)$ is defined by
27.14.16 $\mathop{\Delta\/}\nolimits\!\left(\tau\right)=(2\pi)^{12}(\mathop{\eta\/}% \nolimits\!\left(\tau\right))^{24},$ $\Im{\tau}>0$, Defines: $\mathop{\Delta\/}\nolimits\!\left(\NVar{\tau}\right)$: discriminant function Symbols: $\mathop{\eta\/}\nolimits\!\left(\NVar{\tau}\right)$: Dedekind’s eta function (or Dedekind modular function), $\pi$: the ratio of the circumference of a circle to its diameter and $\Im{}$: imaginary part Permalink: http://dlmf.nist.gov/27.14.E16 Encodings: TeX, pMML, png See also: Annotations for 27.14(vi)
and satisfies the functional equation
27.14.17 $\mathop{\Delta\/}\nolimits\!\left(\frac{a\tau+b}{c\tau+d}\right)=(c\tau+d)^{12% }\mathop{\Delta\/}\nolimits\!\left(\tau\right),$ Symbols: $\mathop{\Delta\/}\nolimits\!\left(\NVar{\tau}\right)$: discriminant function and $\Im{}$: imaginary part Permalink: http://dlmf.nist.gov/27.14.E17 Encodings: TeX, pMML, png See also: Annotations for 27.14(vi)
if $a,b,c,d$ are integers with $ad-bc=1$ and $c>0$.
The 24th power of $\mathop{\eta\/}\nolimits\!\left(\tau\right)$ in (27.14.12) with $e^{2\pi\mathrm{i}\tau}=x$ is an infinite product that generates a power series in $x$ with integer coefficients called Ramanujan’s tau function $\mathop{\tau\/}\nolimits\!\left(n\right)$:
27.14.18 $x\prod_{n=1}^{\infty}(1-x^{n})^{24}=\sum_{n=1}^{\infty}\mathop{\tau\/}% \nolimits\!\left(n\right)x^{n},$ $|x|<1$. Defines: $\mathop{\tau\/}\nolimits\!\left(\NVar{n}\right)$: Ramanujan’s tau function Symbols: $n$: positive integer and $x$: real number Referenced by: §27.20, §27.20 Permalink: http://dlmf.nist.gov/27.14.E18 Encodings: TeX, pMML, png See also: Annotations for 27.14(vi)
The tau function is multiplicative and satisfies the more general relation:
27.14.19 $\mathop{\tau\/}\nolimits\!\left(m\right)\mathop{\tau\/}\nolimits\!\left(n% \right)=\sum_{d\mathbin{|}\left(m,n\right)}d^{11}\mathop{\tau\/}\nolimits\!% \left(\frac{mn}{d^{2}}\right),$ $m,n=1,2,\dots$.
Lehmer (1947) conjectures that $\mathop{\tau\/}\nolimits\!\left(n\right)$ is never 0 and verifies this for all $n<21\;49286\;39999$ by studying various congruences satisfied by $\mathop{\tau\/}\nolimits\!\left(n\right)$, for example:
27.14.20 $\mathop{\tau\/}\nolimits\!\left(n\right)\equiv\mathop{\sigma_{11}\/}\nolimits% \!\left(n\right)\pmod{691}.$
For further information on partitions and generating functions see Andrews (1976); also §§17.217.14, and §§26.926.10. | 4,276 | 11,470 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 183, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-22 | longest | en | 0.601714 |
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Problem Details
H2CO3(aq) + H2O(l) ⇌ HCO3(aq) + H3O+(aq)
HCO3(aq) + H2O(l) ⇌ CO32–(aq) + H3O+(aq)
According to the equations above, what is the conjugate base of HCO 3
a) H2CO3(aq)
b) H2O(l)
c) H3O+(aq)
d) CO32–(aq)
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Our tutors have indicated that to solve this problem you will need to apply the Conjugate Acids and Bases concept. If you need more Conjugate Acids and Bases practice, you can also practice Conjugate Acids and Bases practice problems.
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Our tutors rated the difficulty ofH2CO3(aq) + H2O(l) ⇌ HCO3–(aq) + H3O+(aq) HCO3–(aq) + H2O(l...as medium difficulty.
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Acute Ascertainment (Posted on 2015-04-18)
An acute-angled triangle PQR is inscribed in a circle with centre O.
S is the intersection of the bisector of P with QR
and
the perpendicular to PO through S meets the line PR in a point W interior to the segment PR.
Find PQ/PW
***Source: A problem appearing in the Italian Mathematical Olympiad, 1995
No Solution Yet Submitted by K Sengupta No Rating
Comments: ( Back to comment list | You must be logged in to post comments.)
Solution | Comment 1 of 3
Call the intersection of the perpendicular to PO through S the the bisector of angle RQP by T
Call angle RPQ by p
Call angle PQR by q
Construct segments OQ and OR.
by the given bisector angle RPS = angle QPS = p/2
by triangle PSQ angle PSQ = 180 - p/2 - q
by inscribed and central angle POR = 2q
by isosceles triangle POQ angle RPO = 90-q
by angle subtraction angle SOP = p/2+q-90
by right triangle STP angle TSP = 180-p/2-q
so angles PSQ and PSW are equal
by reflexive property SP=SP
because PS bisects angle RPQ angles RPQ and QPS are equal
so by ASA ΔPSW is congruent to ΔPSQ
so PW = PQ
so PQ/PW=1
Posted by Jer on 2015-04-18 22:12:46
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Introduction- Electric power systems usually involve sinusoidally varying (or nearly so) voltages and currents. Thatis,voltageand current arefunctionsof timethat arenearlypuresinewavesat fixedfrequency. In North America, most ships at sea and eastern Japan that frequency is 60 Hz. In most of the rest of the world it is 50 Hz. Normal power system operation is at this fixed frequency, which is why we study how systems operate in this mode. We will deal with transients later.
This note deals with alternating voltages and currents and with associated energy flows. The focus is on sinusoidal steady state conditions, in which virtually all quantities of interest may be represented by single, complex numbers.
Accordingly, this section opens with a review of complex numbers and with representation of voltage and current as complex amplitudes with complex exponential time dependence. The discussionproceeds,throughimpedance,todescribeapictorial representation of complex amplitudes, called phasors. Power is then defined and, in sinusoidal steady state, reduced to complex form. Finally, flow of power through impedances and a conservation law are discussed.
Secondarily,thissectionofthenotesdealswithtransmissionlinesthathaveinterestingbehavior, both in the time and frequency domails.
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Thank u very much for sharing these notes..
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thanks pal. i NEED basic engineering science book. please. | 432 | 1,859 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2013-20 | latest | en | 0.908725 |
https://file.scirp.org/Html/22-7400953_24125.htm | 1,571,762,317,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987822458.91/warc/CC-MAIN-20191022155241-20191022182741-00163.warc.gz | 491,129,043 | 14,766 | A New Lagrangian Multiplier Method on Constrained Optimization
Applied Mathematics
Vol.3 No.10A(2012), Article ID:24125,6 pages DOI:10.4236/am.2012.330198
A New Lagrangian Multiplier Method on Constrained Optimization*
Youlin Shang#, Shengli Guo, Xiangyi Jiang
Department of Mathematics, Henan University of Science and Technology, Luoyang, China
Email: #mathshang@sina.com
Received July 8, 2012; revised August 8, 2012; accepted August 15, 2012
Keywords: Nonlinear Programming; NCP Function; Lagrange Function; Multiplier; Convergence
ABSTRACT
In this paper, a new augmented Lagrangian function with 4-piecewise linear NCP function is introduced for solving nonlinear programming problems with equality constrained and inequality constrained. It is proved that a solution of the original constrained problem and corresponding values of Lagrange multipliers can be found by solving an unconstrained minimization of the augmented Lagrange function. Meanwhile, a new Lagrangian multiplier method corresponding with new augmented Lagrangian function is proposed. And this method is implementable and convergent.
1. Introduction
Considering the following nonlinear inequality constrained optimization Problem (NLP):
(1)
where and
are continuously differentiable functions.
We denote by
the feasible set of the problem (NLP).
The Lagrangian function associated with the problem (NLP) is the function
where
are the multiplier vectors, For simplicity, we use to denote the column vector
Defintion 1.1. A point is called a Karush-Kuhn-Tucker (KKT) point or a KKT pair of Problem (NLP), if it satisfies the following conditions:
(2)
where, we also say is a KKT point if there exists a such that satisfies (2).
For the nonlinear inequality constrained optimization problem (NLP), there are many practical methods to solve it, such as augmented Lagrangian function method [1-6], Trust-region filter method [7,8], QP-free feasible method [9,10], Newton iterative method [11,12], etc. As we know, Lagrange multiplier method is one of the efficient methods to solve problem (NLP). Pillo and Grippo in [1-3] proposed a class of augmented Lagrange function methods which have nice equivalence between the unconstrained optimization and the primal constrained problem and get good convergence properties of the related algorithm. However, a max function is used for these methods which may be not differentiable at infinite numbers of points. To overcome this shortcoming, Pu in [4] proposed a augmented Lagrange function with FischerBurmeister nonlinear NCP function and Lagrange multiplier methods. Pu and Ding in [6] proposed a Lagrange multiplier methods with 3-piecewise linear NCP function. In this paper, a new class augmented Lagrange function with 4-piecewise linear NCP function and some Lagrange multiplier methods are proposed for the minimization of a smooth function subject to smooth inequality constraints and equality constrains.
The paper is organized as follows: In the next section we give some definitions and properties about NCP function, and then define a new augmented Lagrange function with 4-piecewise NCP function. In Section three, we give the algorithm. In Section four, we prove convergence of the algorithm. Some conclusions are given in Section five.
2. Preliminaries
In this section, we recall some definitions and define a new Lagrange multiplier function with 4-piecewise NCP function.
Definition 2.1 (NCP pair and SNCP pair). We call a pair (a, b) to be an NCP pair if and ab = 0; and call (a, b) to be an SNCP pair if (a, b) is a pair and.
Definition 2.2 (NCP function). A function is called an NCP function if if and only is an NCP pair.
In this paper, we propose a new 4-piecewise linear NCP function is as follows:
(3)
If, then
(4)
and
(5)
It is easy to check the following propositions:
1);
2) The square of is continuously differentiable;
3) is twice continuously differentiable everywhere except at the origin but it is strongly semi-smooth at the origin.
Let
where is a parameter. if and only if, and for any.
We construct function:
Clearly, the KKT point condition (2) is equivalently reformulate as the condition:
If, then is continuously differentiable at. We have
(6)
where is the ith column of the unit matrix, its jth element is 1, and other elements are 0, in this paper take k = 1.
If, and then is strongly semi-smooth and direction differentiable at. We have
(7)
For Problem (NLP), we define a Di Pillo and Grippo type Lagrange multiplier function with 4-piecewise linear NCP function is as following:
(8)
where
are the Lagrange multiplier, C and D are positive parameters.
In this section, we gave some assumptions as follows:
Assumpion 1 f, , ,
are twice Lipschitz continuously differentiable.
Define index set and as follows:
for any, according to definition of, have
for any, we have 1) if, we have
(9)
(10)
The Henssian matrix of at KKT point is
(11)
2) if, then
(12)
(13)
The Henssian matrix of at KKT point is
(14)
Definition 2.3 A point is said to satisfy the strong second-order sufficiency condition for problem (NLP) if it satisfies the first-order KKT condition and if for all
and.
Assumption 2 At any KKT point satisfied strong second-order sufficiency condition.
Lemma If is a positive semi-definite matrix, for any, , matrix satisfied, then exist, for any, is positive definite matrix (see [4]).
Theorem 2.1 If is KKT point of problem (1), then for sufficiently large C and D, is strong convex function at point.
Proof: Let
for, we have
from A2, we have. Furthermore there is if, for any, is positive definite matrix. And then for any and sufficiently large C and D have
by its continuously, we may obtained that there is, for all
we have the theorem hold.
3. Lagrange Multiplier Algorithm
Step 0 Choose parameters, , , , given point, and
Let.
Step 1 Solve following, we will obtain.
if and then stop.
Step 2 For, , then or, for, if
then, or
Step 3 Compute and
Step 4 Let k = k + 1, go to Step 1.
4. Convergence of the Algorithm
In this section, we make a assumption follow as:
Assumption 3 For any, , , ,
exists a minimizer point.
Theorem 4.1 Assume feasible set of problem (NLP) is non-empty set and is bounded, then algorithm is bound to stop after finite steps iteration.
Proof: Assume that the algorithm can not stop after finite steps iteration, by the sack of convenience, we define index set as following
according to assumption A3, it is clearly that or are non-empty set. for any k, obtain
from above assumption, we obtain that for any a there is, for any and, and, for sufficiently large k, it is not difficult to see that
Or for any and, and, for sufficiently large k, have
When we can hold
Which contradicts A3, the theorem holds.
Theorem 4.2 Let is a compact set, sequence are generated by the algorithm, and, in algorithm, 0 take the place of, either algorithm stops at its and is solution of problem(NLP), or for any an accumulation of sequence, is solution of problem (NLP).
Proof: Because the algorithm stops at its, then we have
(15)
for, , it is easy to see that for any have
It is from Step 2 of the algorithm that we have
(16)
putting (15) (16) into (10) or (13), we can obtain
for, , according to definition of
, we can obtain, that
First part of the theorem holds， is solution of problem (NLP).
On the other hand, if the algorithm is not stop at, for any accumulation point of sequence, from theorem 4.1, we can obtain, for any positive number C, that
for any, have
Let,have
Clearly, second part of the theorem holds. is solution of problem (NLP).
5. Conclusion
A new Lagrange multiplier function with 4-piecewise linear NCP function is proposed in this paper which has a nice equivalence between its solution and solution of original problem. We can solve it to obtain solution of original constrained problem, the algorithm corresponding with it be endowed with convergence.
6. Acknowledgements
This paper was partially supported by the NNSF of China under Grant No. 10971053, and NNSF of Henan under Grant No. 094300510050.
REFERENCES
1. G. Pillo and L. Grippo, “A New Class of Augmented Lagrangian in Nonlinear Programming,” SIAM Journal on Control and Optimization, Vol. 17, No. 5, 1979, pp. 617- 628.
2. G. Pillo and L. Grippo, “An Augmented Lageangian function for Inequlity Constrains in Nonlinear Programming Problems,” Optimization Theorem and Application, Vol. 36, No. 4, 1982, pp. 495-520. doi:10.1007/BF00940544
3. G. Pillo and S. Lucidi, “An augmented Lagrange Function with Improved Exactness Properties,” SIAM Journal on Control and Optimization, Vol. 12, No. 2, 2001, pp. 376-406.
4. D. G. Pu and Z. Jin, “A New Lagrange Multiplier Method,” Tongji University Journal, Vol. 9, No. 38, 2010, pp. 1384-1391.
5. K. D. Li and D. G. Pu, “A class of New Lagrange Multiplier Methods,” OR Transaction, Vol. 10, No. 4, 2006, pp. 9-24.
6. X.-W. Du and Y.-L. Shang, “Exact Augmented Lagrange Function for Nonlinear Programming Problems with Inequality and Equality Constraints,” Applied Mathematics and Mechanics, Vol. 26, No. 12, 2005, pp. 1651-1656.
7. K. Su, “Trust-Region Filter Method with NCP Function,” Journal of Systems Science and Mathematical Science, Vol. 28, No. 12, 2008, pp. 1525-1534
8. D. G. Pu, “A New Filter Method,” OR Transaction, Vol. 1, No. 15, 2011, pp. 46-58.
9. W. Hua and D. G. Pu, “A Modified QP-Free Feasible Method,” Applied Mathmatical Modelling, Vol. 35, No. 4, 2011, pp. 1696-1708. doi:10.1016/j.apm.2010.10.002
10. K. Li and D. G. Pu, “Filter QP-Free Method with 3-PieceWise Linear NCP Function,” OR Transaction, Vol. 12, No. 2, 2008, pp. 49-57.
11. Y. X. Yuan, “Numerical Method for Nonlinear Programming,” Shanghai Science and Technology Press, Shanghai, 1993.
12. B. L. Chen, “The Theory of Optimization and Calculate,” 2nd Edition, Tsinghua University Press, Beijing, 2005.
NOTES
*This paper was partially supported by the National Natural Science Foundation of China under Grant No. 10971053, and NNSF of Henan Province under Grant No. 094300510050.
#Corresponding author. | 2,615 | 10,112 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-43 | longest | en | 0.867775 |
https://metacpan.org/pod/release/DMALONE/Math-FastGF2-0.05/lib/Math/FastGF2.pm | 1,571,048,553,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986653216.3/warc/CC-MAIN-20191014101303-20191014124303-00439.warc.gz | 626,255,371 | 14,030 | # NAME
Math::FastGF2 - Perl extension for fast Galois Field arithmetic
# SYNOPSIS
`````` use Math::FastGF2 ":ops";
use strict;
my (\$a,\$b,\$c,\$d);
\$a = gf2_mul(8,0x53,0xca); # GF(2^8) multiplication mod {11B}
\$b = gf2_inv(8,0x53); # 1 / {53} mod {11B}
\$c = gf2_div(8,0x53,0xca; # {53} / {CA} mod {11B}
\$d = gf2_pow(8,0x53,3); # {53} * {53} * {53} mod {11B}
\$a = \$b ^ \$c ^ \$d # add field elements mod {11B}``````
# DESCRIPTION
This module provides an interface for performing single modulo arithmetic operations on Galois Field polynomials in GF(2^8), GF(2^16) and GF(2^32). All values to be operated on are simple Perl numeric scalars which are taken to represent polynomials with binary co-efficients. For example, the value 0x53, whose binary representation is 10010011, represents the polynomial:
`````` 7 6 5 4 3 2 1 0
(1)x + (0)x + (0)x + (1)x + (0)x + (0)x + (1)x + (1)x``````
or, simply:
`````` 7 4
x + x + x + 1``````
Operations such as multiplication, division and calculating powers operate on the polynomials rather than the binary values. Also, all such calculations are done modulo another polynomial, which is called the irreducible polynomial for the field. For GF(2^8), the irreducible polynomial used here has the hex value 0x11b (decimal 283). In binary this is 100011011, so this represents the polynomial
`````` 8 4 3
x + x + x + x + 1``````
The irreducible polynomials used for fields GF(2^16) and GF(2^32) have 16 and 32 as their highest power of x, respectively. It follows that since all calculations in these fields are done modulo the appropriate irreducible polynomial that all field elements in GF(2^8) will fit in a single 8-bit byte, that GF(2^16) elements fit in a single 16-bit word, and so on.
Addition of polynomials in GF(2^n) is accomplished by xoring the binary representation of the two polynomials being operated on. Since field elements are stored as simple Perl scalars, the regular ^ (xor) operator suffices, and hence this module does not provide any gf2_add or gf2_sub methods (there is no difference between addition and subtraction in GF(2^n); the xor operator works for both).
For more detailed descriptions of arithmetic in Galois Fields, and some applications, consult the references listed below.
## EXPORT
By default, the module does not export any methods. By adding the ":ops" parameter to the "use" line, it exports the following routines:
• gf2_mul( \$field_size, \$op1, \$op2 )
• gf2_inv( \$field_size, \$op1 )
• gf2_div( \$field_size, \$op1, \$op2 )
• gf2_pow( \$field_size, \$op1, \$op2 )
Currently, the module implements one other method which is not exported by default. This is the `gf2_info` method which returns either the irreducible polynomial being used for a particular field size, or (if passed an unsupported field size) the total number of bytes being used by the module for arithmetic lookup tables. To use it, either prefix the method name with the module name as in:
`````` \$tablesize = Math::FastGF2::gf2_info(0); # get table size
\$poly_16 = Math::FastGF2::gf2_info(16); # get poly for GF(2^16)``````
or tell the module to export all symbols at the time you "use" it:
`` use Math::FastGF2 ":all";``
after which you can make calls to `gf2_info` without having to prefix the module name.
# TECHNICAL INFORMATION
## BACKGROUND
The initial motivation for writing this library was as a means of implementing Michael O. Rabin's Information Dispersal Algorithm (see below). This module started out as a generic (unoptimised), arbitrary-precision implementation using Bit::Vector to perform all the field operations. While the implementation worked, it was far too slow to be practical, so I began to re-implement the critical sections of the code as a separate C program. As I had no experience of integrating C code with Perl code at the time, I dabbled with using `Inline::C` and some `XS` code for a while. Despite some progress, I decided that there were too many problems with developing the interface code in parallel with the other code so I switched back to polishing up the C implementation and left writing a Perl module interface until later.
After implementing and testing several different optimised C routines, I identified a few methods that performed somewhere between quite well and very well on a variety of hardware platforms and that also had the advantage of using very little memory. I had, at that point, worked out most of the architectural problems of my project, so I decided it was time to come back and attack the `XS` part of the code, the first result of which you see documented here.
## ALGORITHMS
The module uses lookup tables for performing most operations. The exceptions to this are for performing inverses and powers on fields of size 16 and 32 bits. For inverses in these fields a version of the extended Euclidean algorithm for calculating Greatest Common Divisor is used. Another routine implements powers by rewriting the expression to be calculated into one involving only multiplication by x and squaring of sub-products. Also, multiplication is optimised in these fields, at the expense of division, in which `a/b` is implemented as `a * inv(b)`.
For calculations in 8-bit fields, all results are looked up from log and antilog (exponent) tables. These tables are optimised to eliminate the need to check for cases where an operand is zero or, in the case of division, to do bounds checking on exp table lookups. The following identities are used when using log/exp tables:
• `a * b = exp[ log [a] + log [b] ]`
• `1 / a = exp[ 255 - log [a] ]`
• `a / b = exp[ 255 + log [a] - log [b] ]`
• `a ^ b = exp[ (log [a] * b) % 255 ]`
As mentioned, multiplication in fields of size 8 and 16 are optimised by using table lookups. The method used is to break up one of the operands into 8-bit blocks and the other into 4-bit blocks and to look up the result in a straight (non-modular) multiplication table. Sub-products are loaded into a temporary variable, starting with the high bytes/nibbles and shifted 8 bits at a time using a shift lookup table. The shift lookup table takes care of the modulo part of the overall operation. The following example illustrates the general approach, multiplying the hex values A0BD and F0CD by breaking both values into 8-bit blocks:
`````` A0 BD
x F0 CD
------------
BD x CD (subproducts can use a 256x256 lookup table)
CD x A0 << 8 ( "<<" may be regular shift or modulo shift)
F0 x BD << 8
+ F0 x A0 << 16
---------------
= ((F0 x A0) << 8) + (F0 x BD) + (CD x A0)) << 8 + (BD x CD)``````
Obviously, as mentioned, the method used in the module is slightly more complicated since it breaks one value into 4-bit blocks. Also, instead of using just one multiplication table and shifting 4 bits at a time, it uses a "high nibble" multiplication table and a "low nibble" table. The results are then combined before shifting a full 8 bits at a time. A final optimisation of the 32-bit multiply is to use faster regular shifts in two cases, and "safe", modular shifts for the remaining ones.
The major space saving advantage of the algorithm relies on being able to re-use the same straight (non-modular) multiplication tables for both 16 and 32-bit field sizes. Also, the shift tables are optimised to be only 256 words apiece rather than the full field size, since bits shifted off the end are used to look up a mask to be applied to the sub-product to effect the modulo operation. Although the code for generating the tables is not included, armed with this description it should be easy enough to understand how the multiply code works.
## PERFORMANCE
Compared with my original Bit::Vector implementation, these routines achieve a speedup of between 15 and 20 times. Compared with the equivalent stand-alone C functions, however, they are at least 30 times slower. From testing, it's clear that the difference between the plain C and Perl/XS implementations can be mostly attributed to a combination of function calling overheads introduced in the XS layer and overheads in the Perl benchmarking code, with a much smaller amount attributable to the dispatch code which calls the appropriate C function based on the field size. In fact, it appears that the amount of time spent in the Perl code is more than that actually spent doing computations.
A simple benchmarking program is included in this distribution. It is named `benchmark-Math-FastGF2.pl`. It tests all operations on all field sizes.
As noted earlier, and can be seen from running the benchmark program, multiplications are generally faster than divisions, which in turn are faster than power operations.
No tests were done to examine performance in a multi-process or multi-threaded program, but the code should be thread-safe. Further, the relatively small size of the lookup tables means less memory that needs to be copied when fork()ing or spawning a new thread. So it is possible that some performance gains could be made by using this module in a multi-process/multi-thread program.
## FUTURE DIRECTIONS
While the module does provides all the primitives needed for calculations in selected Galois Fields, it only provides the bare minimum functionality. It is probably sufficient for writing code which only needs to operate on small amounts of data (such as encrypting or decrypting keys rather than full files), or for writing proof-of-concept code. However, there are a few major deficiencies:
• the choice of polynomial in each field is hard-wired;
• the lack of features; and
• the overheads involved in the XS function call.
Currently I do not have any requirement for using different polynomials, though if it appears that this feature is needed, or there is any demand for it, I will implement it. Likewise, given that the multiplication tables are already available, it would be fairly simple to implement a straight multiplication routine, although I do not foresee any need for it.
As for the other two problems, the natural solution is to provide functions that do more work with each call. Specifically, starting with version 0.02, there is support for matrix-related operations in the Math::FastGF2::Matrix module to allow efficient operations on large blocks of data from a single call.
I intend future versions to be backward-compatible with this one. In terms of design, I've decided that using Perl scalars for storing field elements is perfectly sufficient, so as a consequence of this decision, I will not be implementing any objects to store them or implementing any kind of operator overloading code. Obviously, this also means that field sizes beyond the size of Perl's scalars will not be possible.
## APPLICATIONS
Besides Rabin's IDA, Galois Fields also have a number of other applications involving codes or cryptography. The main ones are:
• The Advanced Encryption Standard (Rijndael) algorithm for encrpytion. This operates on 8-bit fields and uses the same irreducible polynomial as implemented in this library.
• Error-correcting codes, particularly Reed-Solomon encoding. (RS encoding and Rabin's IDA are actually versions of the same algorithm)
See the SEE ALSO section for links. Also, see the included scripts `shamir-split.pl` and `shamir-combine.pl`, which implement Shamir's threshold system for secret sharing.
## DIVISION BY ZERO (and friends)
Although technically an error, these modules allow division by zero and (with one exception, below) return 0 as the result rather than failing or raising an exception. It is up to the user to ensure that their program checks for division by zero wherever it might occur before calling `gf2_div`.
The other zero-related issue is how the code handles 0^0 (zero to the power of zero). I'm going with Knuth's advice in defining this to be 1, rather than 0.
## POLYNOMIALS
The polynomials used in this implementation are (in hex) 0x11b, 0x1002b and 0x10000008d for, respectively, fields of size 8, 16 and 32 bits. These represent the irreducible polynomials:
`````` 8 4 3
x + x + x + x + 1 GF(2^8)
16 5 3
x + x + x + x + 1 GF(2^16)
32 7 3 2
x + x + x + x + 1 GF(2^32)``````
These values can be retrieved by using the `gf2_info` method. Note that the polynomials returned by this method will have the high order bit stripped off, so `gf2_info(8)` returns 0x1b and not 0x11b.
# KNOWN BUGS
The result of gf2_div(8,0,0) is 1, and not 0 as it is with other field sizes. There is a trivial fix for this but I do not intend to fix it for the following reasons:
• technically, division by zero gives an undefined result, so the problem is with the calling program (which shouldn't have asked to divide by zero) rather than this module; and
• while the fix is trivial, the extra test needed would slow down all calls to the division routine to handle a case that should really happen only very rarely.
http://en.wikipedia.org/wiki/Finite_field_arithmetic
A (mostly) readable description of arithmetic operations in Galois Fields.
http://point-at-infinity.org/ssss/
B. Poettering's implementation of Shamir's secret sharing scheme. This uses Galois Fields, and my own implementation of `gf2_inv` is based on this code.
"Efficient dispersal of information for security, load balancing, and fault tolerance", by Michael O. Rabin. JACM Volume 36, Issue 2 (1989).
The initial motivation for writing this module.
Introduction to the new AES Standard: Rijndael, by Paul Donis, http://islab.oregonstate.edu/koc/ece575/aes/intro.pdf
Besides the AES info, this is also a very good introduction to arithmetic in GF(2^m).
"Optimizing Galois Field Arithmetic for Diverse Processor Architectures and Applications", by Kevin M. Greenan, Ethan L. Miller and Thomas J. E. Schwarz, S.J. (MASCOTS 2008)
Paper giving an overview of several optimisation techniques for calculations in Galois Fields. I have used the optimised log/exp technique described therein, and a modified version of the l-r tables described (called "high-nibble" and "low-nibble" above) . Comments in the paper that optimisations may need to be tailored to the particular hardware architecture have been borne out in my testing.
http://www.cs.utk.edu/~plank/plank/papers/CS-07-593/
James S. Plank's C/C++ implementation of optimised Galois Field calculations. Although I haven't explored the code in great detail, I have used it as a source of benchmarks. In fact, my benchmarking code is modelled on this code. Plank's code is much more fully-featured than mine, so if that is what you want, I would recommend using it instead. If, on the other hand, you want something that's simple, doesn't use much memory and is usable from Perl, I recommend this module of course.
Studies on hardware-assisted implementation of arithmetic operations in Galois Field GF(2^m), by Katsuki Kobayashi.
Despite being aimed at hardware, this paper also contains a wealth of information on software algorithms including several field inversion algorithms.
http://charles.karney.info/misc/secret.html Original implementation of Shamir's secret sharing algorithm, on which `shamir-split.pl` and `shamir-combine.pl` are based. These new versions replace the integer modulo a prime fields with Galois fields implemented with Math::FastGF2.
The Math::FastGF2::Matrix module has a range of Matrix functions to operate more efficiently on large blocks of data.
This module is part of the GnetRAID project. For project development page, see:
`` https://sourceforge.net/projects/gnetraid/develop``
# AUTHOR
Declan Malone, <idablack@users.sourceforge.net>
Copyright (C) 2009-2019 by Declan Malone
This package is free software; you can redistribute it and/or modify it under the terms of the "GNU General Public License" ("GPL").
The C code at the core of this Perl module can additionally be redistributed and/or modified under the terms of the "GNU Library General Public License" ("LGPL"). For the purpose of that license, the "library" is defined as the unmodified C code in the clib/ directory of this distribution. You are permitted to change the typedefs and function prototypes to match the word sizes on your machine, but any further modification (such as removing the static modifier for non-exported function or data structure names) are not permitted under the LGPL, so the library will revert to being covered by the full version of the GPL.
Please refer to the files "GNU_GPL.txt" and "GNU_LGPL.txt" in this distribution for details.
# DISCLAIMER
This package is distributed in the hope that it will be useful, but WITHOUT ANY WARRANTY; without even the implied warranty of MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.
See the "GNU General Public License" for more details. | 4,062 | 16,927 | {"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} | 3.0625 | 3 | CC-MAIN-2019-43 | latest | en | 0.816318 |
https://www.answers.com/Q/What_does_the_term_abelian_mean | 1,563,346,769,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195525094.53/warc/CC-MAIN-20190717061451-20190717083451-00196.warc.gz | 614,324,982 | 22,755 | # What does the term abelian mean?
The term abelian is most commonly encountered in group theory, where it refers to a specific type of group known as an abelian group. An abelian group, simply put, is a commutative group, meaning that when the group operation is applied to two elements of the group, the order of the elements doesn't matter.
For example:
Let G be a group with multiplication * or addition +. If, for any two elements a, b Є G, a*b = b*a or a + b = b + a, then we call the group abelian.
There are other uses of the term abelian in other fields of math, and most of the time, the idea of commutativity is involved.
The term is named after the mathematician, Niels Abel. | 169 | 692 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-30 | latest | en | 0.94234 |
https://blogs.osc-ib.com/2016/11/ib-student-blogs/ever-thought-zero/ | 1,579,618,001,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250604397.40/warc/CC-MAIN-20200121132900-20200121161900-00253.warc.gz | 363,424,842 | 18,055 | Monday, November 21, 2016 0
I expect most of you are familiar with Roman numerals. For example, V represents five, I represents one, so seven is written as VII. The full set is:
1 I 5 V 10 X 50 L 100 C 500 D 1000 M
The key difference between the Roman system and our number system is that the position of a number makes no difference – X represents 10 wherever it appears. So, translate LXXV11 ….. the answer is 76. (The system did change over the years so that position had some relevance. For example, instead of writing VIIII for 9, it became IX, ie 1 before 10).
Roman numerals, astonishingly, are still in use today: films and television programmes often show their year of creation in Roman numerals, the Olympic Games use them, you might see them on clock faces, and the first eleven football or hockey team is often written as 1st XI. But, unlike our system which is used both for representing numbers and for doing calculations, you couldn’t possibly use Roman numerals for calculating. Imagine CLXXVI ÷ XVI. (The answer is, of course, XI)! The Romans used the abacus, an incredibly efficient calculating device, leaving the numeral system purely for number representation. The two images below show an actual Roman abacus, made of bronze, and a scene where a slave in a wealthy family can be seen using an abacus.
The invention of zero and its use as a placeholder in numbers was not a simple linear piece of history. Many civilisations dabbled with placeholders – for example, the Babylonians and the Mayans – and others experimented with zero as a number; but it was probably in India that the two came together. The brilliant mathematician But look in the table of numerals – there is no symbol for zero. Why not? Because, to them, zero was not a number. To have a “number” of things, you must have at least one. The Romans simply used words for “none” – nullus, nihil or nil (from which many words in Latin-based languages are derived, such as null, nullify, nihilist, annihilate in English).
Brahmagupta developed the rules for zero as a number in the 7th century. For example: “The sum of zero and a negative number is negative, the sum of a positive number and zero is positive, the sum of zero and zero is zero.” He also managed to multiply by zero, but struggled with division by zero, in particular stating that “zero divided by zero equals zero.” (Something for you
to investigate…).
And it was the brilliance of Indian mathematicians that developed 0 as a placeholder. What does this mean? Consider the numbers 2, 23, and 254. In the first, the value of the digit 2 is, of course two; in the second, it has value twenty; and in the third, two hundred. But suppose we want to write two hundred and four, in other words we need to show that there aren’t any 10’s. We can’t just leave a gap (although earlier mathematicians did use a dot). So, in 204, the zero indicates that there are no 10’s, and the 2 still stands for two hundred. We take all this for granted, but it took centuries for such a system to develop.
Traders took the new number system from India to the Islamic lands and from there, in the 12th century, Italian mathematician Fibonacci introduced it to Europe where, within a century, it had pretty well displaced Roman numerals and the abacus.
Zero as a number and as a placeholder is incredibly powerful without which the development of mathematics, and the numeracy required for our modern world, would have been very much slower.
And without zero you couldn’t have the mathematical joke: “There are 10 types of people in the world – those who understand binary numbers, and those who don’t!” | 846 | 3,644 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2020-05 | latest | en | 0.953248 |
http://gdevtest.geeksforgeeks.org/category/data-structures/heap/ | 1,555,907,944,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578534596.13/warc/CC-MAIN-20190422035654-20190422061654-00487.warc.gz | 65,741,956 | 17,858 | ## Maximum number of leaf nodes that can be visited within the given budget
Given a binary tree and an integer b representing budget. The task is to find the count of maximum number of leaf nodes that can… Read More »
## K-th Greatest Element in a Max-Heap
Given a max-heap of size n, find the kth greatest element in the max-heap. Examples: Input : maxHeap = {20, 15, 18, 8, 10, 5,… Read More »
## K’th Least Element in a Min-Heap
Given a min-heap of size n, find the kth least element in the min-heap. Examples: Input : {10, 50, 40, 75, 60, 65, 45} k… Read More »
## Max Heap in Java
A max-heap is a complete binary tree in which the value in each internal node is greater than or equal to the values in the… Read More »
## Double ended priority queue
A double ended priority queue supports operations of both max heap (a max priority queue) and min heap (a min priority queue). The following operations… Read More »
## Canonical Huffman Coding
Huffman Coding is a lossless data compression algorithm where each character in the data is assigned a variable length prefix code. The least frequent character… Read More »
## Maximums from array when the maximum decrements after every access
Given an integer K and an array of integers arr, the task is to find the maximum element from the array and after every retrieval… Read More »
## Sum and product of k smallest and k largest composite numbers in the array
Given an integer k and an array of integers arr, the task is to find the sum and product of k smallest and k largest… Read More »
## Sum and product of k smallest and k largest prime numbers in the array
Given an integer k and an array of integers arr, the task is to find the sum and product of k smallest and k largest… Read More »
## Average of max K numbers in a stream
Given a list of ‘N’ numbers, and an integer ‘K’. The task is to print the average of max ‘K’ numbers after each query where… Read More »
## Find the k smallest numbers after deleting given elements
Given an array of integers, find the k smallest numbers after deleting given elements. In case of repeating elements delete only one instance in the… Read More »
## Find the k largest numbers after deleting the given elements
Given an array of integers, find the k largest number after deleting the given elements. In case of repeating elements, delete one instance for every… Read More »
## Memory representation of Binomial Heap
Prerequisites: Binomial Heap Binomial trees are multi-way trees typically stored in the left-child, right-sibling representation, and each node stores its degree. Binomial heaps are collection… Read More »
## Fibonacci Heap – Deletion, Extract min and Decrease key
In the last post, we discussed Insertion and Union of Fibonacci Heaps. In this post, we will discuss Extract_min(), Decrease_key() and Deletion() operations on Fibonacci… Read More »
## Fibonacci Heap – Insertion and Union
Prerequisites:Fibonacci Heap (Introduction) Fibonacci Heap is a collection of trees with min-heap or max-heap property. In Fibonacci Heap, trees can can have any shape even… Read More » | 704 | 3,117 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-18 | latest | en | 0.755876 |
https://charlesreid1.com/wiki/Checkers | 1,713,753,295,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818072.58/warc/CC-MAIN-20240422020223-20240422050223-00824.warc.gz | 137,991,883 | 5,732 | ## Checkers
• Problem Motivation and Description
• Problem analysis: data structures and algorithms
• Solution approach/algorithm
• Implementation
### Problem Motivation and Description
The Checkers Problem: You are presented with a checkerboard that consists of black and white squares. The boards follow the normal rules of checkers, i.e., pieces appear only on the black or white squares. There are several white and black pieces arranged on the board. It is your task to answer the following question: can any one single black piece be used to jump and capture all of the white pieces in a single move? This assumes that each of the black pieces are black "kings" and can jump in either direction.
### Data Structures and Algorithms
To solve the checkers problem, it is important to focus on a simple data structure. While graph theory can be utilized to analyze the problem and find solutions, it is best not to stray too far from a simple 2D array. Char array is simplest.
It's also important not to worry too much:
• Don't worry about the computational cost of loops, or loops within loops, or nested nested nested nested loops. These boards are all relatively small.
• Don't worry about checking if the checkerboard positions are valid. While it might be useful to know before going in, the problem isn't asking you to validate it, so don't sweat that detail. If we need to implement any checks on black pieces only, do it when we tally up all the black pieces. If we need to implement any checks on 2white pieces only, do it when we tally up the white pieces.
The algorithm:
• Load the checkerboard into a char array
• Loop over each piece
• If we find a black piece,
• Check if this board/piece is a viable solution (does the board meet criteria)
• If this is a viable solution, take a tour of possible solution squares, jumping white pieces each time
• If all white pieces jumped, then found black piece that yields solution, add to list of solution pieces
### Code Structure
Here it is best not to over-complicate things. We want to keep the data structure simple, we want to keep the algorithm simple, so let's keep the overall program simple.
W = white, not west.
Four part code structure:
2. For each square, check if square is viable and if square is solution
3. Definition of check if square solution function
4. Definition of check if square viable function
```load chessboard into char[] array
for each square on chessboard:
if square has black piece:
checkSquare(square)
def checkSquare(square):
if(viable(square)):
Wsjumped
for each neighbor:
next square
next next square
if next square is W and next next square is _ empty and we haven't visited this W,
visited this W
Wsjumped += 1 + checkSquare(next next neighbor)
return( Wsjumped==Wcount )
def viable(square):
remove piece
for each row, for each column:
if row/col is off by 2 with the black piece
for each of the 4 neighbors:
next square
next next square
if next square is W
if next next square not empty:
fail fast
increment neighbors (W neighbors)
if black piece original,
isodd = true
if more than 2 odd neighbors, fail fast
if 2 odd neighbors and isodd not true, fail fast
``` | 711 | 3,177 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2024-18 | latest | en | 0.879641 |
https://stats.stackexchange.com/questions/286445/equivalent-form-of-proportional-odds-model | 1,571,078,662,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986654086.1/warc/CC-MAIN-20191014173924-20191014201424-00416.warc.gz | 687,696,497 | 31,095 | Equivalent form of proportional odds model
Here is one definition of the proportional odds model: $$\frac{S(t;{\bf Z})}{1-S(t;{\bf Z})}=e^{-{\bf Z}^T\pmb{\beta}}\frac{S_0(t;{\bf Z})}{1-S_0(t;{\bf Z})},$$ where ${\bf Z}$ = covariates, $\pmb{\beta}$ = regression coefficients, $T$ = survival time, $S(\cdot)$ = survival function, $S_0(\cdot)$ = baseline survival function. This equation is clearly equivalent to $\mbox{logit}[S(t;{\bf Z})]=\mbox{logit}[S_0(t)]-{\bf Z}^T\pmb{\beta}$.
Let $H(t)=-\mbox{logit}[S_0(t)]$. Now the book I'm reading claims, "If $H(t)$ is strictly increasing, [the first equation] can be equivalently written as $H(T)=-{\bf Z}^T\pmb{\beta}+W$, where the random variable $W$ follows a standard logistic distribution".
Can someone explain to me why this statement is true? First of all, I know that by definition $H(t)$ is monotone increasing, but I don't see why we need $H(t)$ to be strictly increasing. Then, it seems to me like the part where he says "$W$ follows a standard logistic distribution" is just plain wrong; why in the world should $-\mbox{logit}[S(t;{\bf Z})]$ follow a logistic distribution?!
I think what he means is that you can rescale $t$ to make it a standard logistic random variable. First recall that if $F$ is the cdf of a random variable $X$, and $U$ is uniform, then $F^{-1}(U)$ is distributed like $X$ (where the inverse is disambiguated by using an infimum).
In other words, let $T:=S_0^{-1}(U)$, so that $\mbox{logit}(S_0(T))=\mbox{logit}(U)$, which is exactly $F^{-1}(U)$ where $F$ is the logistic cdf. Note that to be able to take the inverse, $S_0$ needs to be invertible, which is only possible when it's monotonic.The fact that $S_0$ is monotonic is implied from the assumption that $H(t)$ is monotonic, since the logit function is monotonic.
• I understand the first paragraph, but I don't completely understand the second. Can you please state, step-by-step, why $-\mbox{logit}[S(t;{\bf z})]$ follows a logistic distribution? – xFioraMstr18 Jun 22 '17 at 2:41 | 620 | 2,023 | {"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} | 3.75 | 4 | CC-MAIN-2019-43 | latest | en | 0.887937 |
https://gmatclub.com/forum/let-s-master-these-4-99-4-99-4-x-2224.html | 1,505,855,055,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818686034.31/warc/CC-MAIN-20170919202211-20170919222211-00571.warc.gz | 651,463,081 | 40,838 | It is currently 19 Sep 2017, 14:04
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# Let's master these!!! 4^99 + 4^99 = 4^X
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Let's master these!!! 4^99 + 4^99 = 4^X [#permalink]
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29 Aug 2003, 09:33
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Let's master these!!!
4^99 + 4^99 = 4^X
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29 Aug 2003, 11:30
Took me longer, However i got he same answer as Jav.
4^99 + 4^99 = 4^X
2(4^99)= 4^X
2(2^198)= 2^2X
2^198= 2^2X-1
Therefore 2X-1=198
X=99.5
Kudos [?]: 1 [0], given: 0
29 Aug 2003, 11:30
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# If x/|x|, which of the following must be true for all
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If x/|x|, which of the following must be true for all [#permalink]
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Updated on: 09 Jul 2013, 08:56
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If $$\frac{x}{|x|} \lt x$$, which of the following must be true about $$x$$? ($$x \ne 0$$)
A. $$x\gt 2$$
B. $$x \in (-1,0) \cup (1,\infty)$$
C. $$|x| \lt 1$$
D. $$|x| = 1$$
E. $$|x|^2 \gt 1$$
M24
Originally posted by praveenvino on 15 Jan 2011, 11:44.
Last edited by Bunuel on 09 Jul 2013, 08:56, edited 1 time in total.
Renamed the topic and edited the question.
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Posts: 65765
Re: range of root - GMAT Club test - M24 [#permalink]
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15 Jan 2011, 13:47
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11
praveenvino wrote:
X/|X| < X . Which of the following must be true for all ?
a. X > 1
b. X is an element in (-1,0) U (1,inf)
c. |X| < 1
d. |X| = 1
e. |X|^2 > 1
Can some one explain how X can be zero for the above condition?
x is in the denominator so it can not equal to zero as division be zero is undefined.
Correct form of this question is below (m09 q22, discussed here: m09-q22-69937.html):
If $$\frac{x}{|x|} \lt x$$, which of the following must be true about $$x$$? ($$x \ne 0$$)
A. $$x\gt 2$$
B. $$x \in (-1,0) \cup (1,\infty)$$
C. $$|x| \lt 1$$
D. $$|x| = 1$$
E. $$|x|^2 \gt 1$$
$$\frac{x}{|x|}< x$$
Two cases:
A. $$x<0$$ --> $$\frac{x}{-x}<x$$ --> $$-1<x$$. But as we consider the range $$x<0$$ then $$-1<x<0$$
B. $$x>0$$ --> $$\frac{x}{x}<x$$ --> $$1<x$$.
So the given inequality holds true in two ranges $$-1<x<0$$ and $$x>1$$.
For more check: math-absolute-value-modulus-86462.html
Hope it helps.
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##### General Discussion
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Re: If x/|x|, which of the following must be true for all [#permalink]
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22 Apr 2015, 19:19
1
praveenvino wrote:
If $$\frac{x}{|x|} \lt x$$, which of the following must be true about $$x$$? ($$x \ne 0$$)
A. $$x\gt 2$$
B. $$x \in (-1,0) \cup (1,\infty)$$
C. $$|x| \lt 1$$
D. $$|x| = 1$$
E. $$|x|^2 \gt 1$$
M24
x < x*|x|
x-x*|x|< 0
roots of this equation are : -1,0,1
rest is explained in the attached image ...
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e-GMAT Representative
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If x/|x|, which of the following must be true for all [#permalink]
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23 Apr 2015, 00:44
1
AverageGuy123 wrote:
Brunel, Can you please explain why option E is not feasible?
Dear AverageuGuy123
As Bunuel explained above,
Either -1 < x < 0
Or x > 1
Now, |x| as you know, represents the magnitude of x. Option E says that |x|^2 must be greater than 1.
Let's first consider the case when -1 < x < 0
A possible value of x in this case is -0.5
So, what is the value of |x|^2? It is equal to 0.25
Is it greater than 1? NO
Let's now consider the case when x > 1
A possible value of x in this case is 2.
So, what is the value of |x|^2? It's 4.
Is it greater than 1? YES
So, as we see, that |x|^2 CAN BE greater than 1. But can we say that |x|^2 MUST BE greater than 1? NO, because |x|^2 is not greater than 1 for all possible values of x.
So, the key takeaway from this discussion is that:
we need to be careful whether the question is asking about MUST BE TRUE statements or about CAN BE TRUE statements.
Hope this helped!
- Japinder
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Re: If x/|x|, which of the following must be true for all [#permalink]
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26 Jan 2018, 23:48
1
Given: x/|x| < x
since |x| >= 0 always
multiply both LHS and RHS by |x|
x < x|x|
=> x - x|x| < 0
=> x(1 - |x|) < 0
if x > 0, then 1 - |x| < 0 to hold the above inequality => |x| > 1 => x(since x is positive in this case) > 1
if x < 0, then 1 - |x| > 0 => |x| < 1 => -1 < x < 0 (to hold the above inequality)
Option B captures the above range perfectly
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Re: range of root - GMAT Club test - M24 [#permalink]
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15 Jan 2011, 14:03
Thanks Bunuel. X not equals zero condition was actually missing in the question in m24. Thanks for your help.
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Re: If x/|x|, which of the following must be true for all [#permalink]
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11 Dec 2014, 23:00
A must be true too.
If x>1 satisfy x/|x|<x
then x>2 will do too.
can anyone explain choice A? thanks!
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Re: If x/|x|, which of the following must be true for all [#permalink]
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12 Dec 2014, 05:51
pinguuu wrote:
A must be true too.
If x>1 satisfy x/|x|<x
then x>2 will do too.
can anyone explain choice A? thanks!
x > 2 is NOT necessarily true. Consider x = -1/2.
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Re: If x/|x|, which of the following must be true for all [#permalink]
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22 Apr 2015, 18:22
Brunel, Can you please explain why option E is not feasible?
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Re: If x/|x|, which of the following must be true for all [#permalink]
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23 Apr 2015, 03:36
AverageGuy123 wrote:
Brunel, Can you please explain why option E is not feasible?
You can plug in numbers to eliminate options.
"which of the following must be true about x" means that every acceptable value of x must lie in the range given in the correct option. The acceptable values of x are the values for which x/|x| < x.
A. x>2
Must x be greater than 2?
This should make you check for 2.
2/|2| < 2
1 < 2 (True)
So 2 is an acceptable value of x. But 2 is not greater than 2.
So this option is not correct. This also makes you eliminate options (C) and (D).
E. |x|^2>1
Must x be greater than 1 or less than -1?
Check for 1/2
(1/2)/|1/2| < 1/2
1 < 1/2 (False)
Check for -1/2
(-1/2)/|-1/2| < -1/2
-1 < -1/2 (True)
So x = -1/2 is an acceptable value but it does not lie in this range. Hence option (E) is also incorrect.
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Re: If x/|x|, which of the following must be true for all [#permalink]
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24 Apr 2015, 10:21
Hi All,
This question can be dealt with in a variety of ways. It's actually really susceptible to TESTing VALUES, which we can use to determine possibilities and eliminate answers.
We're told that X/|X| < X. The question asks what must be TRUE about X.
While this inequality looks complicated, you can quickly prove some things about X....
IF....
X = 1
1/|1| is NOT < 1
So X CANNOT be 1
Eliminate D.
IF.....
X = 2
2/|2| IS < 2
So X CAN be 2
Eliminate A and C.
IF....
X = -2
-2/|-2| is NOT < -2
So X CANNOT be -2
Eliminate E.
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Re: If x/|x|, which of the following must be true for all [#permalink]
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Re: If x/|x|, which of the following must be true for all [#permalink] 28 May 2020, 11:47 | 2,883 | 8,648 | {"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} | 3.921875 | 4 | CC-MAIN-2020-34 | latest | en | 0.796007 |
https://philoid.com/question/108690-mark-the-tick-against-the-correct-answer-in-the-following-range-of-cos-1-x-is | 1,686,139,646,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224653764.55/warc/CC-MAIN-20230607111017-20230607141017-00554.warc.gz | 496,448,459 | 12,211 | ## Book: RS Aggarwal - Mathematics
### Chapter: 4. Inverse Trigonometric Functions
#### Subject: Maths - Class 12th
##### Q. No. 52 of Objective Questions
Listen NCERT Audio Books to boost your productivity and retention power by 2X.
52
##### Mark the tick against the correct answer in the following:Range of cos-1 x is
To Find: The range of
Here,the inverse function is given by y =
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Y = cos x by interchanging x and y axes.i.e, if (a,b) is a point on Y = cos x then (b,a) is the point on the function y =
Below is the Graph of the range of
From the graph, it is clear that the range of is restricted to the interval
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57 | 296 | 859 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2023-23 | longest | en | 0.799209 |
https://www.jiskha.com/questions/590336/a-hockey-puck-is-sliding-on-frictionless-ice-it-slams-against-a-wall-and-bounces-back | 1,603,843,104,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107894890.32/warc/CC-MAIN-20201027225224-20201028015224-00425.warc.gz | 769,661,060 | 5,008 | # Physics
A hockey puck is sliding on frictionless ice. It slams against a wall and bounces back toward the player with the same speed that it had before hitting the wall. Does the velocity of the hockey puck change in this process?
1. 👍 1
2. 👎 0
3. 👁 952
1. Yes
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2. 👎 0
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(C) 20 : 21
(D) 21 : 20 | 152 | 482 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2024-10 | longest | en | 0.968236 |
https://www.teacherspayteachers.com/Product/CHANCE-AND-PROBABILITY-NO-PREP-PRINT-N-GO-SHEETS-1695236 | 1,484,593,236,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560279224.13/warc/CC-MAIN-20170116095119-00305-ip-10-171-10-70.ec2.internal.warc.gz | 976,116,988 | 47,054 | Teachers Pay Teachers
# CHANCE AND PROBABILITY: NO PREP PRINT-N-GO SHEETS
Subjects
Resource Types
Product Rating
3.9
File Type
PDF (Acrobat) Document File
6.66 MB | 45 pages
### PRODUCT DESCRIPTION
Use these 20 Print-n-Go Sheets to teach and reinforce conceptual knowledge of Chance and Probability. The hands-on interactive activities will also facilitate practical application and enable students to work with the concrete in order to understand the abstract. Moreover the activities are sure to make the teaching of Theoretical and Experimental probability not just easy but pure fun!
ITEMS INCLUDED
I. The 20 INTERACTIVE SHEETS target the following activities:
* The Probability Scale: Words Describing Chance
* Probability with Marbles
* Probability with Numbers on a Spinner
* Probability with Numbers on a Die
* Probability with Names on a Spinner
* Probability with Colors on a Spinner
* Graphing with Chance and Data: Favorite Food
* Graphing with Chance and Data: Favorite Sport
* Finding Possible Outcomes of Two Coins: using a Tree Diagram and a Grid
* Finding Possible Outcomes of a Coin and a Die: using a Tree Diagram and a Grid
* Finding Possible Outcomes of a die and a color spinner: using a Tree Diagram and a Grid
* Finding Possible Outcomes of Two Spinners: using a Grid
* Theoretical and Experimental Probability: Dice Roll, Coin Toss, Spinner Spin, Card Spin, M&M’s Candy
SUGGESTED USE
Use the interactive sheets as a follow-up to general whole class or small group instruction. The activities are fairly simple, yet challenging and scaffold acquisition of key skills, especially finding all possible outcomes using Grids and Tree Diagrams.
TARGETED GROUP
This unit best targets students in grades 3-8, although a portion of it can also be used for higher achievers at Grade 2. It would also be useful to students at Grades 7-8 who would benefit from a strong basic foundation knowledge of Chance and Probability via user friendly activities.
ALSO INCLUDED
* 18 colorful and meaningful vocabulary cards that may be displayed on the bulletin board after introduction. Blank cards template also provided.
***********************************************************************
You might be interested to check out this Chance and Probability ({124} task cards product to use in your math centers as an extension activity.
CHANCE AND PROBABILITY 124 TASK CARDS INTERACTIVE FUN
***********************************************************************
BETTER STILL SAVE ON THE BUNDLE:
CHANCE AND PROBABILITY {118 PAGE} INTERACTIVE FUN BUNDLE
***********************************************************************
Other interactive products you might be interested in:
GOALS {SETTING AND REFLECTION} NOTEBOOK FOLDABLES
OPINION/NARRATIVE/BIOGRAPHY COMMON CORE WRITING BUNDLE
324 PAGES GRAPHS THE COMPLETE INTERACTIVE COLLECTION COMMON CORE
THE COMMON CORE NUMBERS AND OPERATIONS INTERACTIVE NOTEBOOK BUNDLE
PLACE VALUE OF MULTI-DIGIT WHOLE NUMBERS AND DECIMAL NUMBERS {BUNDLE}
FRACTION FUN {INTERACTIVE NOTEBOOK COMMON CORE ALIGNED}
NUMBER SENSE SUBITIZING CARDS {COMMON CORE}
THE ULTIMATE 564 PLACE VALUE TASK CARD BUNDLE
MENTAL MULTIPLICATION STRATEGIES {COMMON CORE ALIGNED}
FACTORS AND MULTIPLES INTERACTIVE FUN COMMON CORE
BUDGETING PLANS AND GRAPHING
THE PLANETS OF THE SOLAR SYSTEM FLIPBOOK
WATER THEMATIC UNIT
BENCHMARK ASSESSMENT SYSTEM FOUNTAS AND PINNELL
EIGHTY EDITABLE AWARDS
***********************************************************************
MY BLOG
MY PINTEREST BOARD
All rights reserved by author. Permission to copy for single classroom use only. Electronic distribution limited to single classroom use only. Not for public display.
***********************************************************************
chance and probability | chance | probability | chance and data | data | statistics | statistics and probability | print-n-go sheets | no prep printables
Total Pages
45
Included
Teaching Duration
2 Weeks
4.0
Overall Quality:
3.9
Accuracy:
3.9
Practicality:
3.9
Thoroughness:
3.9
Creativity:
4.0
Clarity:
3.9
Total:
31 ratings | 880 | 4,088 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-04 | longest | en | 0.844337 |
http://www.thehorse.com/articles/38871/protein-and-equine-ration-balancers-lets-do-the-math | 1,495,704,476,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463608058.13/warc/CC-MAIN-20170525083009-20170525103009-00351.warc.gz | 658,969,319 | 15,461 | # Protein and Equine Ration Balancers: Let's Do the Math
Photo: The Horse Staff
Q. I have been reading several of your commentaries and see that you recommend feeding ration balancers to horses on predominantly forage diets. When I looked at ration balancers at my feed store I notice that they tend to be high in protein, often around 30%. Isn’t that too much protein? I thought mature horses only need about 10-12% protein in their diet. What am I missing?
A. This is a question I get asked quite often, and at first glance you are quite right. But if we delve a bit deeper I think you will see why the ration balancer doesn’t, in fact, offer too much protein.
Most ration balancers have a recommended daily intake of 1-2 pounds (0.45 to 0.9 kilograms) per day. Where the crude protein content is 30% this means that 0.3 pounds (136 grams) of protein are supplied when feeding a pound and 0.6 pounds (273 g) of protein if feeding two pounds.
Let’s say that your horse weighs 1,100 pounds (500 kilograms) and that you feed an average grass hay with a crude protein of 11% at a rate of 2% of body weight per day. This equates to 22 pounds (10 kilograms) of hay that provides 2.42 pounds (1.1 kg) of protein (we reach this number by multiplying 22 pounds by 0.11 [which represents 11%] for the protein content). As you can see the hay is providing far more protein than the ration balancer. The ration balancer has a high protein content but a small serving size so its total contribution is small.
If we assume that this horse is in moderate work, per National Research Council guidelines, he needs 768 grams (or 0.768 kilograms) of protein a day. To convert this to pounds we multiply the amount in kilos by 2.2, because there are 2.2 pounds to a kilo. This gives us a daily protein requirement of 1.69 pounds. In this example the hay is meeting the horse’s protein requirement; however, if the horse was fed less than 2% of body weight per day because he’s an easy keeper, his requirement might not be being met. Similarly, if the hay actually had a lower crude protein content or the horse was in heavier work, the same might also be true.
### Considering Easy Keepers
Sometimes I see easy keepers being fed lower nutritional quality hays at lower amounts to reduce calorie intake. For example, a stemmy grass hay with a crude protein of 8% fed at 1.5% of this horse’s body weight would yield only 1.32 pounds (0.6 kilograms) of protein, which is not enough to meet requirement. Adding 1.5 pounds (0.68 kilograms) of ration balancer with 0.45 pounds (0.2 kilograms) of protein would bring the ration protein intake to 1.77 pounds (0.8 kilograms) of protein, which meets his requirement.
Beyond just meeting requirement there is the issue of protein quality and even a grass hay that provides adequate crude protein might lack adequate levels of some essential amino acids. Or the protein might not be fully available if bound up with indigestible structural carbohydrates. Ration balancers not only provide crude protein, they tend to provide guaranteed levels of the most limiting essential amino acids, such as lysine and methionine, that are vital for your horse’s health. So, they’re a good insurance policy even if on paper it looks like protein intake from the forage should be adequate.
### Ration Balancers vs. Performance Feeds
Another comparison I often see people make is between ration balancers and performance feeds. Most performance feeds have crude protein levels around 12% and a daily recommended intake upwards of 5 pounds per day. At 5 pounds (2.25 kilograms) per day this would provide 0.6 pounds (0.27 kilograms) of protein (5 x 0.12) the same as two pounds of the 30% ration balancer. In fact, performance feeds given at the upper ends of the recommended feeding levels provide far more protein a day than the ration balancer, even though the previous only contain only 12% protein.
### The Alfalfa Exception
There are times when you might not want or need the additional protein coming from a 30% ration balancer, and one of these is if you are feeding a lot of alfalfa. Alfalfa provides considerably more protein than required and also tends to have a slightly better amino acid profile than grass hay, so you might decide the protein in the ration balancer is unwarranted. I tend to find that horses fed 25-30% of their forage ration as alfalfa do just fine on the high-protein ration balancer; however, you might have other options.
A few companies make ration balancers specifically for horses being fed alfalfa and these have a protein content of around 12%. They still guarantee the necessary essential amino acid levels but do so without feeding unnecessary crude protein. They also have lower calcium, because alfalfa provides considerable calcium to the ration.
### Take-Home Message
When comparing the protein contents of feeds at your feed store do not take the crude protein contents on face value. Make sure you understand the feeding directions and then consider the role of that feed within your horse’s total daily ration. You will likely find that the high-protein ration balancer is not contributing as much protein as you might at first think.
### Clair Thunes, PhD
Clair Thunes, PhD, is an independent equine nutrition consultant who owns Summit Equine Nutrition, based in Sacramento, California. She works with owners/trainers and veterinarians across the United States and globally to take the guesswork out of feeding horses. Born in England, she earned her undergraduate degree at Edinburgh University, in Scotland, and her master’s and doctorate in nutrition at the University of California, Davis. Growing up, she competed in a wide array of disciplines and was an active member of the United Kingdom Pony Club. Today, she serves as the regional supervisor for the Sierra Pacific region of the United States Pony Clubs. As a nutritionist she works with all horses, from WEG competitors to Miniature Donkeys and everything in between. | 1,332 | 6,005 | {"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} | 3.4375 | 3 | CC-MAIN-2017-22 | latest | en | 0.953268 |
http://oeis.org/A193404 | 1,547,875,040,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583662124.0/warc/CC-MAIN-20190119034320-20190119060320-00166.warc.gz | 176,662,855 | 4,570 | This site is supported by donations to The OEIS Foundation.
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A193404 Number of matchings (independent edge subsets) in the rooted tree with Matula-Goebel number n. 0
1, 2, 3, 3, 5, 5, 4, 4, 8, 8, 8, 7, 7, 7, 13, 5, 7, 12, 5, 11, 11, 13, 12, 9, 21, 12, 20, 10, 11, 19, 13, 6, 21, 11, 18, 16, 9, 9, 19, 14, 12, 17, 10, 18, 32, 20, 19, 11, 15, 30, 18, 17, 6, 28, 34, 13, 14, 19, 11, 25, 16, 21, 28, 7, 31, 31, 9, 15, 32, 27, 14 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,2 COMMENTS A matching in a graph is a set of edges, no two of which have a vertex in common. The empty set is considered to be a matching. The Matula-Goebel number of a rooted tree can be defined in the following recursive manner: to the one-vertex tree there corresponds the number 1; to a tree T with root degree 1 there corresponds the t-th prime number, where t is the Matula-Goebel number of the tree obtained from T by deleting the edge emanating from the root; to a tree T with root degree m>=2 there corresponds the product of the Matula-Goebel numbers of the m branches of T. REFERENCES F. Goebel, On a 1-1-correspondence between rooted trees and natural numbers, J. Combin. Theory, B 29 (1980), 141-143. I. Gutman and A. Ivic, On Matula numbers, Discrete Math., 150, 1996, 131-142. I. Gutman and Yeong-Nan Yeh, Deducing properties of trees from their Matula numbers, Publ. Inst. Math., 53 (67), 1993, 17-22. D. W. Matula, A natural rooted tree enumeration by prime factorization, SIAM Review, 10, 1968, 273. É. Czabarka, L. Székely, and S. Wagner, The inverse problem for certain tree parameters, Discrete Appl. Math., 157, 2009, 3314-3319. LINKS E. Deutsch, Rooted tree statistics from Matula numbers, arXiv:1111.4288. FORMULA Define b(n) (c(n)) to be the number of matchings of the rooted tree with Matula-Goebel number n that contain (do not contain) the root. We have the following recurrence for the pair A(n)=[b(n),c(n)]. A(1)=[0,1]; if n=p(t) (=the t-th prime), then A(n)=[c(t),b(t)+c(t)]; if n=rs (r,s,>=2), then A(n)=[b(r)c(s)+c(r)b(s), c(r)c(s)]. Clearly, a(n)=b(n)+c(n). See the Czabarka et al. reference (p. 3315, (2)). The Maple program is based on this recursive formula. EXAMPLE a(3)=3 because the rooted tree with Matula-Goebel number 3 is the path ABC on 3 vertices; it has 3 matchings: empty, {AB}, {BC}. MAPLE with(numtheory): A := proc (n) local r, s: r := proc (n) options operator, arrow: op(1, factorset(n)) end proc: s := proc (n) options operator, arrow: n/r(n) end proc: if n = 1 then [0, 1] elif bigomega(n) = 1 then [A(pi(n))[2], A(pi(n))[1]+A(pi(n))[2]] else [A(r(n))[1]*A(s(n))[2]+A(s(n))[1]*A(r(n))[2], A(r(n))[2]*A(s(n))[2]] end if end proc: a := proc (n) options operator, arrow: A(n)[1]+A(n)[2] end proc: seq(a(n), n = 1 .. 80); CROSSREFS Cf. A184165. Sequence in context: A046146 A081768 A273493 * A072923 A257003 A131922 Adjacent sequences: A193401 A193402 A193403 * A193405 A193406 A193407 KEYWORD nonn AUTHOR Emeric Deutsch, Feb 11 2012 STATUS approved
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Last modified January 18 23:05 EST 2019. Contains 319282 sequences. (Running on oeis4.) | 1,229 | 3,550 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2019-04 | latest | en | 0.84761 |
https://www.mapleprimes.com/users/Teep/replies?page=6 | 1,708,944,116,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474659.73/warc/CC-MAIN-20240226094435-20240226124435-00021.warc.gz | 873,579,388 | 37,800 | ## 195 Reputation
15 years, 233 days
## I'm still missing something!...
Thank you, Carl.
I tried that - but doesn't the adjacency matrix has 2 non-zero diagonal entries? If so - then the routine won't run.
Perhaps I am misinterpreting your suggestion (I am new to Graph Theory) - in this instance, the solution to my LP problem is given by the square matrix,Y, and will change as certain conditions / values are altered in the model. In other words, the entries will vary on a case-by-case basis and I simply wish to show these on a graph.
## That's it!...
Thanks again to all for addressing this question - as I don't use these methods and techniques, I find that the support you all gave is generous and well-constructed. I have learned a lot today.
Last question .... is there a convenient way to allocate a name to each i and j node?
## This solution is very useful!...
Thanks for this - this is a great help and is very much appreciated. Nice solution!
However, can we separate each node i in the graph? For example, can we show node 2 (i=2) connected to the j-nodes, 3, 4 6 and 9 alone?
In the same way, I'd like to show node 3 (i=3) simply connected to the j-nodes and in the case where there are no connections from i to j, can we simply omit these?
I'm only interested in the plots that show the active connections, i.e. in the case where the entries are 1's and not 0's.
## Thank you!...
Thanks for the insights here. I'll give this a go also!
## Perfect!...
Yes ... that does the trick!
Thanks a lot - I'm grateful.
## Kernel...
Yes indeed .... I frequently lost my connection also for small p.
I figured it was pc-related rather than numerical computation issues, so I removed firewall, re-booted, etc. but it still made no difference.
## Fast indeed!...
Nice model. I varied the parameter values across my range of interest and it produces outputs in quick time.
Thanks very much .... and thanks again to all who worked on this.
## @Carl Love This looks very intere...
This looks very interesting for small p values. I'll apply this to my problem.
Thanks for the creative input (as usual!).
## More progress...
Interesting and useful routine .. thanks for this insight.
## @epostma At least some good arise...
At least some good arises!
I'm glad you picked this up, Erik - thanks for informing us.
Is it possible to obtain a pre-release version of the fix before the next release?
## Solved!...
That's a beautiful method of solution .... such a simple and creative approach.
Again - thanks for this.
## Interesting...
Now, that's impressive! I'll take this into consideration.
Thank's for the guidance and quick response.
## Thanks...
I appreciate this. Nice code.
This works nicely when p=0.5.
However, when p is adjusted to p=0.05, the processing time remains excessive. I ran this for 10 minutes and decided to stop the computation.
## Worksheet...
NBD_Issue.mw
Thanks. I hope you can access the code here.
I'll try the 'numeric' option.
## Thank you! That did the trick!...
Thank you! That did the trick!
4 5 6 7 8 9 Page 6 of 9
| 747 | 3,100 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2024-10 | latest | en | 0.951275 |
http://stackoverflow.com/questions/22639816/choose-specific-number-with-probability | 1,455,386,144,000,000,000 | text/html | crawl-data/CC-MAIN-2016-07/segments/1454701167113.3/warc/CC-MAIN-20160205193927-00123-ip-10-236-182-209.ec2.internal.warc.gz | 212,702,168 | 17,434 | # Choose specific number with probability
How can one choose a number with a specific probability `p`?
Say we must choose between `{0, 1}` and the probability `p` stands for choosing `1`.
So when `p=0.8` we choose `1` with 80% and `0` with 20%.
Is there a simple solution in R for this?
-
Take a look at `sample` function.
``````> set.seed(1)
> sample(c(0,1), size=10, replace=TRUE, prob=c(0.2,0.8))
[1] 1 1 1 0 1 0 0 1 1 1
``````
From the helpfile you can read:
`sample` takes a sample of the specified size from the elements of `x` using either with or without replacement.
and the argument `prob` in `sample` acts as ...
A vector of probability weights for obtaining the elements of the vector being sampled.
-
or at a lower level, `ifelse(runif(10)<0.8,1,0)` – Ben Bolker Mar 25 '14 at 16:00
Great, thank you! – Juergen Mar 25 '14 at 16:13 | 274 | 855 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2016-07 | latest | en | 0.845273 |
https://www.mathworks.com/matlabcentral/answers/1882942-error-with-brackets-symsum?s_tid=prof_contriblnk | 1,695,959,339,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510481.79/warc/CC-MAIN-20230929022639-20230929052639-00522.warc.gz | 953,586,525 | 25,549 | # Error with brackets symsum
1 view (last 30 days)
Aaron on 22 Dec 2022
Edited: Aaron on 23 Dec 2022
f2 = figure;
figure(f2);
ic = linspace(0, 100, K);
[t, n] = ode 45(@(t, n) ODEfunc(n, K);
k_bar = 1/N * symsum(i*n(i),i,0,K);
**I want to solve some ODES eventually with k_bar
I keep getting errors with brackets, im not sure If im tired but i cant spot where its referring to!
Alan Stevens on 22 Dec 2022
Do you mean something more like this?
N = 100; % Total number of particles
one = 1;
alpha = 1;
K = 7; % 7 states
n0 = 4*ones(K,1);
tspan = 0:0.1:5;
[t, n] = ode45(@(t, n) ODEfunc(t, n, K, one, N, alpha), tspan, n0);
plot(t, n),grid
xlabel('time'), ylabel('n')
legend('n1','n2','n3','n4','n5','n6','n7')
function dndt = ODEfunc(~, n, K, one, N, alpha)
S = n(1);
for i=2:K
S = i*n(i) + S;
end
k_bar = S/N;
dndt = zeros(K,1);
dndt(1,:) = alpha*n(1)*k_bar - one*n(1); % Note that this is for state 1.
for i= 2:K-1
dndt(i,:) = alpha*n(i)*(k_bar - i) + one*n(i-1) - one*n(i); % For state 2 to 6
end
dndt(K,:) = alpha*n(K,1)*(k_bar - K) + one*n(K-1,1); % For state 7
end
##### 2 CommentsShow 1 older commentHide 1 older comment
Aaron on 22 Dec 2022
Hi, I just realised I forgot to add a -1 with the k_bar, making (k_bar-1) in the first term. It now converges to nice values. thanks for the help on the syntax | 513 | 1,309 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2023-40 | latest | en | 0.693971 |
http://www.thefullwiki.org/geoid | 1,521,530,696,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257647322.47/warc/CC-MAIN-20180320072255-20180320092255-00379.warc.gz | 487,977,436 | 13,630 | • Wikis
# geoid: Wikis
Note: Many of our articles have direct quotes from sources you can cite, within the Wikipedia article! This article doesn't yet, but we're working on it! See more info or our list of citable articles.
# Encyclopedia
gravity model and the WGS84 reference ellipsoid).[1]]]
The geoid is that equipotential surface which would coincide exactly with the mean ocean surface of the Earth, if the oceans were in equilibrium, at rest, and extended through the continents (such as with very narrow canals). According to C.F. Gauss, who first described it, it is the "mathematical figure of the Earth," a smooth but highly irregular surface that corresponds not to the actual surface of the Earth's crust, but to a surface which can only be known through extensive gravitational measurements and calculations. Despite being an important concept for almost two hundred years in the history of geodesy and geophysics, it has only been defined to high precision in recent decades, for instance by works of Petr Vaníček and others. It is often described as the true physical figure of the Earth, in contrast to the idealized geometrical figure of a reference ellipsoid.
## Description
The geoid surface is irregular, unlike the reference ellipsoid which is a mathematical idealized representation of the physical Earth, but considerably smoother than Earth's physical surface. Although the physical Earth has excursions of +8,000 m (Mount Everest) and −11,000 m (Mariana Trench), the geoid's total variation is less than 200 m (-106 to +85 m)[2] compared to a perfect mathematical ellipsoid.
Sea level, if undisturbed by currents and weather, would assume a surface equal to the geoid. If the continental land masses were criss-crossed by a series of tunnels or narrow canals, the sea level in these canals would also coincide with the geoid. In reality the geoid does not have a physical meaning under the continents, but geodesists are able to derive the heights of continental points above this imaginary, yet physically defined, surface by a technique called spirit leveling.
Being an equipotential surface, the geoid is by definition a surface to which the force of gravity is everywhere perpendicular. This means that when travelling by ship, one does not notice the undulations of the geoid; the local vertical is always perpendicular to the geoid and the local horizon tangential to it. Likewise, spirit levels will always be parallel to the geoid.
Note that a GPS receiver on a ship may, during the course of a long voyage, indicate height variations, even though the ship will always be at sea level (tides not considered). This is because GPS satellites, orbiting about the center of gravity of the Earth, can only measure heights relative to a geocentric reference ellipsoid. To obtain one's geoidal height, a raw GPS reading must be corrected. Conversely, height determined by spirit leveling from a tidal measurement station, as in traditional land surveying, will always be geoidal height. Modern GPS receivers have a grid implemented inside where they obtain the geoid (for e.g. EGM-96) height over the WGS ellipsoid from the current position. Then they are able to correct the height above WGS ellipsoid to the height above WGS84 geoid. In that case when the height is not zero on a ship it is because of the tides.
## Simplified Example
The gravity field of the earth is neither perfect nor uniform. A flattened ellipsoid is typically used as the idealized earth, but even if the earth were perfectly spherical, the strength of gravity would not be the same everywhere, because density (and therefore mass) varies throughout the planet. This is due to magma distributions, mountain ranges, deep sea trenches, and so on.
If that perfect sphere were then covered in water, the water would not be the same height everywhere. Instead, the water level would be higher or lower depending on the particular strength of gravity in that location.
## Spherical harmonics representation
Spherical harmonics are often used to approximate the shape of the geoid. The current best such set of spherical harmonic coefficients is EGM96 (Earth Gravity Model 1996)[3], determined in an international collaborative project led by NIMA. The mathematical description of the non-rotating part of the potential function in this model is
V=\frac{GM}{r}\left(1+{\sum_{n=2}^{n_{max}}}\left(\frac{a}{r}\right)^n{\sum_{m=0}^n} \overline{P}_{nm}(\sin\phi)\left[\overline{C}_{nm}\cos m\lambda+\overline{S}_{nm}\sin m\lambda\right]\right), .]]
where $\phi\$ and $\lambda\$ are geocentric (spherical) latitude and longitude respectively, $\overline\left\{P\right\}_\left\{nm\right\}$ are the fully normalized associated Legendre polynomials of degree $n\$ and order $m\$, and $\overline\left\{C\right\}_\left\{nm\right\}$ and $\overline\left\{S\right\}_\left\{nm\right\}$ are the numerical coefficients of the model based on measured data. Note that the above equation describes the Earth's gravitational potential $V\$, not the geoid itself, at location $\phi,\;\lambda,\;r,\$ the co-ordinate $r\$ being the geocentric radius, i.e, distance from the Earth's centre. The geoid is a particular[4] equipotential surface, and is somewhat involved to compute. The gradient of this potential also provides a model of the gravitational acceleration. EGM96 contains a full set of coefficients to degree and order 360 (i.e. $n_\left\{max\right\} = 360$) , describing details in the global geoid as small as 55 km (or 110 km, depending on your definition of resolution). The number of coefficients, $\overline\left\{C\right\}_\left\{nm\right\}$ and $\overline\left\{S\right\}_\left\{nm\right\}$, can be determined by first observing in the equation for V that for a specific value of n there are two coefficients for every value of m except for m = 0. There is only one coefficient when m=0 since $\ sin \left(0\lambda\right) = 0$. There are thus (2n+1) coefficients for every value of n. Using these facts and the well known formula, $\sum_\left\{I=1\right\}^\left\{L\right\}I = L\left(L+1\right)/2$, it follows that the total number of coefficients is given by
\sum_{n=2}^{n_{max}}(2n+1) = n_{max}(n_{max}+1) + n_{max} - 3 = 130317 using the EGM96 value of $\ n_\left\{max\right\} = 360$ .
For many applications the complete series is unnecessarily complex and is truncated after a few (perhaps several dozen) terms.
New even higher resolution models are currently under development. For example, many of the authors of EGM96 are working on an updated model[5] that should incorporate much of the new satellite gravity data (see, e.g., GRACE), and should support up to degree and order 2160 (1/6 of a degree, requiring over 4 million coefficients). NGA has announced the availability of EGM2008, complete to spherical harmonic degree and order 2159, and contains additional coefficients extending to degree 2190 and order 2159.[6] Software and data is on the Earth Gravitational Model 2008 (EGM2008) - WGS 84 Version page.
## Precise geoid
The 1990s saw important discoveries in theory of geoid computation. The Precise Geoid Solution by Vaníček and co-workers improved on the Stokesian approach to geoid computation.[7] Their solution enables millimetre-to-centimetre accuracy in geoid computation, an order-of-magnitude improvement from previous classical solutions.[8][9][10]
## Time-variability
Recent satellite missions, such as GOCE and GRACE, have enabled the study of time-variable geoid signals. The first products based on GOCE satellite data became available online in June, 2010, through the European Space Agency (ESA)’s Earth observation user services tools.[11][12] ESA launched the satellite in March 2009 on a mission to map Earth's gravity with unprecedented accuracy and spatial resolution.
## References
1. ^ data from http://earth-info.nga.mil/GandG/wgs84/gravitymod/wgs84_180/wgs84_180.html
2. ^ http://www.csr.utexas.edu/grace/gravity/gravity_definition.html visited 2007-10-11
3. ^ NIMA Technical Report TR8350.2, Department of Defense World Geodetic System 1984, Its Definition and Relationships With Local Geodetic Systems, Third Edition, 4 July 1997. [Note that confusingly, despite the title, versions after 1991 actually define EGM96, rather than the older WGS84 standard, and also that, despite the date on the cover page, this report was actually updated last in June 23 2004. Available electronically at: http://earth-info.nga.mil/GandG/publications/tr8350.2/tr8350_2.html]
4. ^ There is no such thing as "The" EGM96 geoid
5. ^ Pavlis, N.K., S.A. Holmes. S. Kenyon, D. Schmit, R. Trimmer, "Gravitational potential expansion to degree 2160", IAG International Symposium, gravity, geoid and Space Mission GGSM2004, Porto, Portugal, 2004.
6. ^ http://earth-info.nga.mil/GandG/wgs84/gravitymod/egm2008/index.html, page accessed 05 November 2008
7. ^ "UNB Precise Geoid Determination Package". Archived from the original on 20 November 2007. Retrieved 2 October 2007.
8. ^ Vaníček, P.; Kleusberg, A. (1987). [Expression error: Unexpected < operator "The Canadian geoid-Stokesian approach"]. Manuscripta Geodaetica 12 (2): 86–98.
9. ^ Vaníček, P.; Martinec, Z. (1994). "Compilation of a precise regional geoid". Manuscripta Geodaetica 19: 119–128.
10. ^ Vaníček et al. Compilation of a precise regional geoid (pdf), pp.45, Report for Geodetic Survey Division - DSS Contract: #23244-1-4405/01-SS, Ottawa (1995)
11. ^ http://www.esa.int/SPECIALS/GOCE/SEMB1EPK2AG_1.html
12. ^ http://www.esa.int/SPECIALS/GOCE/SEMY0FOZVAG_0.html
# Wiktionary
Up to date as of January 15, 2010
## English
### Noun
Singular geoid Plural geoids
geoid (plural geoids)
1. A surface of constant gravitational potential at zero elevation. | 2,467 | 9,765 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 18, "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} | 2.65625 | 3 | CC-MAIN-2018-13 | latest | en | 0.941957 |
http://www.tutorcircle.com/how-to-factor-equations-f195wq.html | 1,369,385,016,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368704392896/warc/CC-MAIN-20130516113952-00083-ip-10-60-113-184.ec2.internal.warc.gz | 769,505,212 | 33,168 | Â Â Â Â
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# How to Factor Equations?
Equations in mathematics consist of constant terms and variables. Basically variables are unknown terms in an equation; we will calculate the value of variable and place it in the equation to maintain the equality of the equation. So basically we use an equation to get the value of variable. Now let us see how to factor equations.
In order to get the value of unknown variable we will follow many different approaches. Generally we use the factorizing approach for solving an equation.
In Factorization we will divide an equation in two or more small parts; every part is called a factor of equation. We will get one separate value of variable from every factor of equation. Let us take an example to better understand the method of solving an equation by factors.
Step 1: Quadratic equation is x2 - 7x – 10 = 0. In this Quadratic Equation we will find the value of variable 'x' by factorization method.
Step 2: Now we will write this equation as x2 – 5x – 2x + 10 = 0. This equation is subdivided in four terms. We will search the common terms from this equation as:
x(x - 5) -2 (x - 5) = 0,
(x – 5)(x - 2) = 0,
Step 3: We have two factors of given equation and these factors are (x - 5) and (x - 2).
Step 4: Place every factor equals to zero to get the value of variable 'x'.
When x - 5 = 0 then
x = 5,
And when x – 2 = 0 then
x = 2,
Step 5: So value of variable x = 2, 5. | 389 | 1,445 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.65625 | 5 | CC-MAIN-2013-20 | latest | en | 0.909412 |
https://rieselprime.de/z/index.php?title=Statistics&oldid=24475 | 1,679,949,867,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948684.19/warc/CC-MAIN-20230327185741-20230327215741-00178.warc.gz | 565,079,839 | 7,985 | Topics Register • News • History • How to • Sequences statistics • Template prototypes
# Statistics
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
## Countings: Sequences per type
Sequences (per base, k-value or individual number) with own page in this Wiki:
Sequence Type counting
Mersenne primes number 51
Riesel primes k•2n-1 k-value 3,351
Riesel primes kbn-1, b>2 base 8
Proth primes k•2n+1 k-value 138
Proth primes kbn+1, b>2 base 5
Carol-Kynea primes (bn±1)2-2 base 381
Williams primes (b-1)bn-1 base 169
Williams primes (b-1)bn+1 base 153
Williams primes (b+1)bn-1 base 87
Williams primes (b+1)bn+1 base 82
Leyland numbers xy+yx number 1,814
Leyland numbers xy-yx number 1
Woodall primes nbn-1 base 73
Near Woodall primes (n-1)bn-1 base 4
Near Woodall primes (n+1)bn-1 base 2
Cullen primes nbn+1 base 32
Near Cullen primes (n-1)bn+1 base 1
Near Cullen primes (n+1)bn+1 base 1
Fermat numbers F(2) number 150
GF Divisors (2) number 184
Gen.Fermat numbers xGF(3) number 21
GF Divisors (3) number 202
Gen.Fermat numbers xGF(4) number 4
GF Divisors (4) number 62
Gen.Fermat numbers xGF(5) number 15
GF Divisors (5) number 363
Gen.Fermat numbers xGF(6) number 11
GF Divisors (6) number 148
Gen.Fermat numbers xGF(7) number 26
GF Divisors (7) number 439
Gen.Fermat numbers xGF(8) number 23
GF Divisors (8) number 364
Gen.Fermat numbers xGF(9) number 17
GF Divisors (9) number 348
Gen.Fermat numbers xGF(10) number 18
GF Divisors (10) number 275
Gen.Fermat numbers xGF(11) number 45
GF Divisors (11) number 743
Gen.Fermat numbers xGF(12) number 14
GF Divisors (12) number 309
all GF Divisors number 1158
Number classes
General numbers
Special numbers Mersenne 2n-1 Fermat 22n+1 Proth k•2n+1 Riesel k•2n-1 Woodall n•2n-1 Cullen n•2n+1 Double Mersenne 22p-1-1 Perfect 2n-1(2n-1) Leyland xy ± yx
Prime numbers Mersenne Proth Riesel Williams (b±1)bn±1 Carol/Kynea (bn±1)2-2 Multifactorial n!j±1 | 728 | 1,925 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2023-14 | latest | en | 0.384483 |
https://forum.allaboutcircuits.com/threads/power-factor-of-linear-transformer.116488/ | 1,701,904,786,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100626.1/warc/CC-MAIN-20231206230347-20231207020347-00515.warc.gz | 308,618,146 | 20,990 | # power factor of linear transformer
#### _dan_
Joined Dec 10, 2013
24
Hi all,
I am curious what is the power factor of a linear power supply - the classics- linear transformer to 12v AC (or lower) then rectified and filtered. How it changes when regulated by linear voltage regulator or LDO?
I didnt find any info on transformer datasheets, possibly because it depends on load type?
Thanks !
#### _dan_
Joined Dec 10, 2013
24
Also can an averaging (non RMS) current meter algorithm/meter be used to measure currents of loads with PF 0,9; What would be the error then compared to RMS meter readings - is it cllose to 10%?
#### _dan_
Joined Dec 10, 2013
24
I think I found a nice answer here:
thanks to Jony130
But my second question remains
So for bridge bridge rectifier + capacitor and load we have.
This situation looks like this
As you can see the source voltage is sill a sine wave but the source current no longer looks like a sine wave. And this is why PF is not equal to 1.
Vin_rms = 7.07V and Iin_rms = 161.3mA
So the APPARENT POWER = S = 7.07V *161.3mA = 1.14VA
And the REAL POWER is transfer by first harmonic RMS source current.
I_V1_rms = 104.21mA * 0.707 = 73.5mA
So the REAL POWER = P = 7.07 * 73.5mA = 0.52W
And finally
PF = 0.52W/1.14VA = 0.45
After some more thoughts about this I think that the we can find REAL POWER in ltspice directly.
All we need is to plot V_In * I_in = V(N002,N003)*-I(V1)
and move the mouse to the label of the trace, hold down the control key and left mouse click.
So the P = 728.47mW
#### ErnieM
Joined Apr 24, 2011
8,362
A transformer by itself is a marvelous device that presents whatever complex impedance on its output to the input, adjusted (approximately) only by the turns ratio.
For a simple fw bridge and capacitor the situation is horrible as the ap is only charged during short slivers of time where the AC nears the peak. It is not a linear current at all so power factor does not apply.
A recent development is a power factor controller stage added to an off the line switching supply. The purpose of this stage is to cleverly control the charging of a supply cap so the current drawn off the AC line is a very nearly unity power factor. The ones I have seen consist of a step up converter (so the input current is continuous) that moniyors the AC line while controlling the pulse width of the switcher. By running at a frequency higher than the line they draw an almost perfect current a ar as power factor is concerned.
The one drawback (aside from cost and complexity) is these are necessarily step up stages, so the 115 v in 28v out supply we make first steps up the line to 360 volts. | 694 | 2,669 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2023-50 | latest | en | 0.912323 |
https://www.actucation.com/college-physics-2/parallel-current-carrying-long-wires | 1,638,132,474,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358591.95/warc/CC-MAIN-20211128194436-20211128224436-00546.warc.gz | 743,695,745 | 29,864 | Informative line
### Parallel Current Carrying Long Wires
Practice to calculate the distance of point of zero magnetic field between and outside the two parallel wires.
# Force between Two Parallel Current carrying Wires
Case 1
• Consider two parallel current carrying conductors, as shown in figure.
• The direction of current in both the conductors is same and both conductors are of length $$\ell$$, where $$\ell$$ is very large.
• Force acting on wire 1 due to magnetic field of wire 2 is shown in figure.
• The magnetic force on wire 1 due to magnetic field of wire 2
$$\vec{F_{1/2}} = I_1\;\vec{l} × \vec {B_2}$$
where $$I_1$$ is current in wire 1
$$B_2$$ is magnetic field due to current in 2
Note:- $$\vec{l}$$ is always taken in the direction of current flow.
• The magnetic force on wire 2 due to magnetic field of wire 1
$$\vec{F_{2/1}} = I_2\;\vec{l} × \vec {B_1}$$
Case 2
• Consider two conductors carrying current in opposite direction, as shown in figure.
• Force acting on wire 1 due to magnetic field of wire 2
$$\vec{F_{1/2}} = I_1\;\vec{l} × \vec {B_2}$$
• Force acting on wire 2 due to magnetic field of wire 1
$$\vec{F_{2/1}} = I_2\;\vec{l} × \vec {B_1}$$
Conclusion: Parallel conductors carrying current in the same direction attract each other and parallel conductors carrying current in opposite direction repel each other.
#### Two parallel wires having steady current flow are placed near to each other. The direction of current in wire 1 is as shown in figure. What will be the direction of current in wire 2 if it is given that conductor 1 attracts conductor 2 ?
A
B
C
D
×
Parallel conductors carrying current in the same direction attract each other and parallel conductors carrying current in opposite direction repel each other.
Since, both conductors attract each other. So in conductor 2, the direction of current will be same as that in conductor 1.
Hence, option (A) is correct.
### Two parallel wires having steady current flow are placed near to each other. The direction of current in wire 1 is as shown in figure. What will be the direction of current in wire 2 if it is given that conductor 1 attracts conductor 2 ?
A
B
C
D
Option A is Correct
# Force between Two Parallel Wires
• Consider two straight infinite conductors, 1 and 2, carrying current $$I_1$$ and $$I_2$$ respectively in the same direction as shown in figure.
• The magnetic force on length ( $$\ell$$) of wire 1 due to wire 2
$$\vec{F_{1/2}} = I_1\vec{\ell} × \vec{B_2}$$
where
$$\vec{\ell}$$ is vector length of wire 1
$$\vec{B_2}$$ is magnetic field vector of wire 1 due to wire 2
So, $$F_1 = I_1 {\ell}B_2$$
$$\Rightarrow F_1 = I_1 {\ell}(\dfrac{\mu_0I_2}{2 \pi a}) \;\;\;\;\;[\therefore \;B_2 = \dfrac{\mu_0I_2}{2 \pi a}\,\, for\,\,long\,\,wire]$$
$$\Rightarrow F_1 = \dfrac{\mu_0 I_ 1 I_2}{2 \pi a}{\ell}$$
• Force per unit length of wire 1 due to wire 2
$$\dfrac{F_1}{\ell} = \dfrac{\mu_0I_1I_2}{2\pi a}$$
#### Two parallel wires of infinite length each are placed at a distance a = 25 cm apart. The current in wire 1 and wire 2 is $$I_1$$ = 2 A and $$I_2$$ = 0.5 A, respectively. Calculate force per unit length on wire 1 due to wire 2.
A 80 N/m
B 0 N/m
C 8 × 10–7 N/m
D 800 N/m
×
Force per unit length on wire 1 due to wire 2
$$\dfrac{F_1}{\ell} = \dfrac{\mu_0I_1I_2}{2\pi a}$$
$$Given:\, I_1=2\,A\,,\,\,\,\,I_2=\,0.5\,A\,,\,\,\,\,a=\,25\,cm$$
$$\dfrac{F_1}{\ell} = \dfrac{\mu_0}{2\pi} × \dfrac{2 × 0.5}{0.25}$$
$$= \dfrac{4 \pi× 10 ^{-7}}{2\pi}× \dfrac{2× .5}{.25}$$
$$\dfrac{F_1}{\ell} = 8 × 10^{-7}$$ $$N/m$$
Hence, option (C) is correct.
### Two parallel wires of infinite length each are placed at a distance a = 25 cm apart. The current in wire 1 and wire 2 is $$I_1$$ = 2 A and $$I_2$$ = 0.5 A, respectively. Calculate force per unit length on wire 1 due to wire 2.
A
80 N/m
.
B
0 N/m
C
8 × 10–7 N/m
D
800 N/m
Option C is Correct
#### Two parallel infinite long wires carrying current $$I_1=2\,A$$ and $$I_2=1\,A$$ in the same direction, as shown in figure, are placed at a distance $$d=6\,cm$$ apart. Calculate the distance and value of current of third wire with direction so that net force on all three wires is zero.
A 3 cm, 8 A
B 5 cm, 3 A
C 2 cm, 0.67 A
D 2 cm, 3 A
×
Since, parallel conductors are carrying current in same direction. So, both will attract each other.
Force per unit length on both the wires,
$$\dfrac{F}{\ell}\;\;=\;\;\dfrac{\mu_0I_1I_2}{2\pi d}$$
$$\dfrac{F}{\ell} \;\;=\;\;\dfrac{\mu_0(2)(1)}{2\pi(6 × 10 ^{-2})}$$
$$\dfrac{F}{\ell}\;\;=\;\; 6.66 \,\mu N/m$$
A third wire carrying current $$I$$ in the opposite direction is placed at a perpendicular distance $$x$$ from wire 1.
For wire 1 to be in equilibrium
$$\left(\dfrac{F}{\ell}\right)_2\;\;=\;\;\left(\dfrac{F}{ \ell}\right)_3$$
$$\Rightarrow \dfrac{\mu_0I_1I_2}{2 \pi d} \;=\; \dfrac{\mu_0 I_1I}{2\pi(d-x)}$$
$$\Rightarrow I_2(d-x)=Id$$
$$\Rightarrow (I_2-I) d = I_2x$$
$$(1-I) 6 = x$$
$$x = 6- 6\;I$$ —(1)
For wire 2 to be in equilibrium
$$\left(\dfrac{F}{\ell}\right)_1\;\;=\;\;\left(\dfrac{F}{ \ell}\right)_3$$
$$\Rightarrow \dfrac{\mu_0I_1I_2}{2 \pi d} \;=\; \dfrac{\mu_0 I_2I}{2\pi x}$$
$$\Rightarrow x I_1 = dI$$
$$2 x\;= 6\; I$$
$$I = \dfrac{2x}{6} —(2)$$
From (1) and (2)
$$x = 6-6 \left(\dfrac{2}{6}\right)x \;\; \left [ I = \dfrac{2}{6}x \right ]$$
$$x$$ = $$6$$ – $$2$$$$x$$
$$3x$$ = $$6$$
$$x = \dfrac{6}{3}$$
$$x$$ = $$2$$$$cm$$
Current in third wire
$$I = \dfrac{2}{6} x$$
$$I = \dfrac{2}{6} × 2 = 4/6 \;A$$
$$I = 2/3\;\;= \,0. 67 \;A$$
Hence, option (C) is correct.
### Two parallel infinite long wires carrying current $$I_1=2\,A$$ and $$I_2=1\,A$$ in the same direction, as shown in figure, are placed at a distance $$d=6\,cm$$ apart. Calculate the distance and value of current of third wire with direction so that net force on all three wires is zero.
A
3 cm, 8 A
.
B
5 cm, 3 A
C
2 cm, 0.67 A
D
2 cm, 3 A
Option C is Correct
#### Two infinitely long wires carrying current of same magnitude in same direction, placed $$\ell_1$$ distance apart, as shown in figure. A third wire of mass $$m$$ and length $$\ell$$ carrying current $$I$$ in opposite direction, is placed among them such that front view of wire seems like equilateral triangle. Find the current in infinitely wire so that third wire levitated in same form of equilateral triangle.
A $$\dfrac{mg\pi\ell_1}{\mu_0\pi\ell}$$
B $$\dfrac{mgI_1}{I\ell}$$
C $$\dfrac{mg \;2\pi\;\ell_1}{\sqrt3 \;\mu_0\; I \;\ell}$$
D $$\dfrac {mg\;I\;\ell_1}{I_1 \ell}$$
×
As direction of current in third wire is opposite to other two wires, the third wire is repelled by both wire 1 and wire 2. As the current in wire 1 and wire 2 increases, the repulsive force increases and the levitated wire rises to a point at which the weight of wire is once again levitated in equilibrium.
The horizontal components of the magnetic force on the levitated wire cancel and the vertical component add together
$$\vec{F_{net}} = 2 F_B \;\;cos \;30°$$
$$\vec{F_{net}} = 2\dfrac{\mu_0 \;I_1\;I\;}{2 \pi \ell _1}\ell \;cos \;30°$$
$$\vec{F_{net}} =\dfrac{\sqrt3}{2}\dfrac{\mu_0I_1I}{\pi \ell_1}\ell$$
.
For levitation,
$$\vec{F_{net}} = mg$$
$$\dfrac{\sqrt3\mu_0I_1I\ell}{2\pi\ell_1}\;\;=mg$$
$$I_1 = \dfrac{mg\; 2 \pi\;\ell_1}{\sqrt3 \;\mu _0\;I\;\ell}$$
Hence, option (C) is correct.
### Two infinitely long wires carrying current of same magnitude in same direction, placed $$\ell_1$$ distance apart, as shown in figure. A third wire of mass $$m$$ and length $$\ell$$ carrying current $$I$$ in opposite direction, is placed among them such that front view of wire seems like equilateral triangle. Find the current in infinitely wire so that third wire levitated in same form of equilateral triangle.
A
$$\dfrac{mg\pi\ell_1}{\mu_0\pi\ell}$$
.
B
$$\dfrac{mgI_1}{I\ell}$$
C
$$\dfrac{mg \;2\pi\;\ell_1}{\sqrt3 \;\mu_0\; I \;\ell}$$
D
$$\dfrac {mg\;I\;\ell_1}{I_1 \ell}$$
Option C is Correct
# Calculation of Point of Zero Magnetic Field Between the Two Parallel Wires
• The point of zero magnetic field between two parallel current carrying wires, is possible only when current in both the wires is in same direction.
• If current in both the wires is in opposite direction, then the magnetic field due to both wires will be in same direction. So, point of zero magnetic field between both the wires is not possible.
• Case 1 : In the shown figure, the direction of current $$I_1$$ and $$I_2$$ are in opposite direction, so, the magnetic field due to both wires is inside the page using right hand thumb rule.
• Case 2 : In the shown figure, the direction of current $$I_1$$ and $$I_2$$ are in opposite direction, so, the magnetic field due to both wires is outside the page using right hand thumb rule.
• To calculate point of zero magnetic field, direction of current in both the wires should be same.
• Consider two infinitely long wires, placed at a distance 'd' apart, carrying current of $$I_1$$ and $$I_2$$ in the same direction, as shown in figure.
• Consider point P as the point of zero magnetic field at a perpendicular distance x from wire 1.
• Then, the magnetic field at P will be of same magnitude.
$$|\vec{B_{P_1}}|$$ = $$|\vec{B_{P_2}}|$$
$$\Rightarrow \dfrac{\mu _0I_1}{2 \pi x}\;\;=\;\;\dfrac{\mu_0I_2}{2 \pi (d-x)}$$
$$\Rightarrow I_1(d-x) = I_2 x$$
$$\Rightarrow I_1d = (I_1+I_2)x$$
$$\Rightarrow x = \dfrac{I_1d}{(I_1+I_2)}$$
#### Two infinitely long parallel wires carrying current $$I_1=2\,A$$ and $$I_2=3\,A$$ in the same direction, are placed at a distance $$d = 5\, cm$$ apart. Calculate the distance of point of zero magnetic field from wire 1 carrying current $$I_1=2\,A$$.
A 0 cm
B 4 cm
C 2 cm
D 3 cm
×
Let point P be the point of zero magnetic field
then, $$\dfrac {\mu_0I_1}{2 \pi x}\;\;=\;\;\dfrac{\mu_0 I_2}{2 \pi (d-x)}$$
Given: $$I_1=2\,A$$ , $$I_2=3\,A$$ , $$d = 5 \,cm$$
$$\dfrac{\mu_0I_1}{2 \pi x}\;\;\;=\;\;\;\dfrac{\mu_0I_2}{2 \pi (d - x)}$$
$$\Rightarrow \dfrac {I_1}{x}\;\;=\;\;\dfrac{I_2}{(d-x)}$$
$$\Rightarrow \dfrac {2}{x}\;\;=\;\;\dfrac{3}{(5-x)}$$
$$\Rightarrow$$ $$10 - 2x = 3x$$
$$\Rightarrow$$ $$5x =10$$
$$\Rightarrow$$ $$x = 2$$ $$cm$$
Hence, option (C) is correct.
### Two infinitely long parallel wires carrying current $$I_1=2\,A$$ and $$I_2=3\,A$$ in the same direction, are placed at a distance $$d = 5\, cm$$ apart. Calculate the distance of point of zero magnetic field from wire 1 carrying current $$I_1=2\,A$$.
A
0 cm
.
B
4 cm
C
2 cm
D
3 cm
Option C is Correct
# Calculation of Point of Zero Magnetic Field Outside the Two Parallel Wires
• The point of zero magnetic field outside the two infinite long parallel wires can be calculated if current in both the wires is in opposite direction.
• When the current through both wires is in same direction, point of zero magnetic field is not possible outside the wires as direction of magnetic field due to both the wires is same.
• Case 1: In the shown figure, the direction of current in both wires $$I_1$$ and $$I_2$$ are in same direction. The direction of magnetic field to the left of wire 1 is outside the page due to both wires. Also, the direction of magnetic field at the right of wire 2 is inside the page due to both wires. Hence, point of zero magnetic field is not possible outside the wires.
• Case 2: In the shown figure, the direction of current in both wires $$I_1$$ and $$I_2$$ are in same direction. The direction of magnetic field to the left of wire 1 is inside the page due to both wires. Also, the direction of magnetic field at the right of wire 2 is outside the page due to both wires. Hence, point of zero magnetic field is not possible outside the wires.
• To calculate point of zero magnetic field outside the wire, consider two infinitely long wires carrying current $$I_1$$ and $$I_2$$ in the opposite direction, placed at a distance $$d$$ apart, as shown in figure.
• Consider a point P as point of zero magnetic field beyond the wire 1 at a perpendicular distance $$x$$ as shown in figure.
$$|\vec{B_{p_1}}|\;=\;\ |\vec{B_{p_2}}|$$
$$\Rightarrow \dfrac{\mu_0I_1}{2 \pi x}\;\;=\;\;\dfrac{\mu_0I_2}{2\pi(d+x)}$$
$$\Rightarrow I_1(d+x) = I_2 x$$
$$x = (\dfrac{I_1}{I_2 - I_1}) \;d$$
• $$(I_2 - I_1)$$ should be greater than zero as $$x$$ is a distance.
or, $$I_2 - I_1 >0$$
$$\Rightarrow I_2 >I_1$$
• It means point of zero magnetic field is possible beyond the wire of less current.
• If current in both the wires, $$I_1$$ and $$I_2$$ is same, then
• Point of zero magnetic field
$$x = (\dfrac{I_1}{I_2 - I_1}) \;d$$
or, $$x = \dfrac {Id}{I-I} = \dfrac{Id}{0} = \infty$$
Thus, point of zero magnetic field is not possible.
#### Two wires carrying current $$I_1 =2\,A$$ and $$I_2=3\,A$$ are placed at a distance, $$d = 3\,cm$$ apart as shown in figure. Calculate the distance of point of zero magnetic field from wire 2.
A 2 cm at right of wire 2
B 10 cm at right of wire 2
C 6 cm at left of wire 2
D Can't be determined
×
Point of zero magnetic field will be beyond wire of less current .
Hence, point of zero force will be beyond wire 1 since,
$$I_1$$ < $$I_2$$
Hence, at point P magnitude of magnetic field due to wire 1 will be same as magnitude of magnetic field due to wire 2 but in opposite direction
$$|\vec{B_{p_1}}|\;=\;\ |\vec{B_{p_2}}|$$
$$\Rightarrow \dfrac{\mu_0I_1}{2 \pi x}\;\;=\;\;\dfrac{\mu_0I_2}{2\pi(d+x)}$$
Given: $$I_1=2\,A$$ , $$I_2=3\,A$$ , $$d = 3\, cm$$
$$\Rightarrow \dfrac{\mu_0 (2)}{2 \pi x}\;\;=\;\;\dfrac{\mu_0(3)}{2\pi(3 + x)}$$
$$6 + 2x\,= 3x$$
$$x$$$$= 6 \,cm$$
Hence, option (C) is correct.
### Two wires carrying current $$I_1 =2\,A$$ and $$I_2=3\,A$$ are placed at a distance, $$d = 3\,cm$$ apart as shown in figure. Calculate the distance of point of zero magnetic field from wire 2.
A
2 cm at right of wire 2
.
B
10 cm at right of wire 2
C
6 cm at left of wire 2
D
Can't be determined
Option C is Correct | 4,994 | 13,996 | {"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} | 4.59375 | 5 | CC-MAIN-2021-49 | latest | en | 0.878602 |
https://geniebook.com/us/tuition/primary-3/maths/whole-numbers-6 | 1,723,734,875,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641299002.97/warc/CC-MAIN-20240815141847-20240815171847-00121.warc.gz | 215,162,861 | 65,625 | # Whole Numbers 6
1. Equal stage at the beginning
2. Equal stage in the end
## 1. Equal stage at the beginning
Both subjects have the same number of items at first. After some items are added/removed from each subject, they each have a different number of items in the end.
#### Question 1:
Mrs Tan and Mrs Lim had an equal number of pies at first. After Mrs Tan bought another 12 pies and Mrs Lim gave away 4 pies, Mrs Tan had twice as many pies as Mrs Lim. How many pies did each of them have at first?
#### Solution:
\begin{align} 1 \text{ unit} &= 12 + 4\\[2ex] &= 16 \end{align}
Number of pies each of them have at first\begin{align}\\[2ex] &= 16 + 4\\[2ex] &= 20 \end{align}
Each of them had 20 pies at first.
20 pies
#### Question 2:
Rajesh and Charles had an equal number of stickers. After Rajesh bought another 24 stickers and Charles lost 12 stickers, Rajesh had thrice as many stickers as Charles. Find the number of stickers Rajesh had at first.
#### Solution:
Difference in units\begin{align}\\[2ex] &= 3 \text{ units} - 1 \text{ unit}\\[2ex] &= 2 \text{ units} \end{align}
\begin{align} 2 \text{ units} &= 12 + 24\\[2ex] &= 36\\[2ex] 1 \text{ unit} &= 36 \div 2\\[2ex] & = 18 \end{align}
Number of stickers Rajesh had at first $=$ Number of stickers Charles had at first
Number of stickers Charles had at first\begin{align}\\[6ex] &= 1 \text{ unit} + 12\\[2ex] &= 18 +12\\[2ex] &= 30 \end{align}
Rajesh had 30 stickers at first.
30 stickers
#### Question 3:
Wendy and Yvonne had the same number of paper cups. After Wendy bought another 70 paper cups and Yvonne bought another 6 paper cups, Wendy had five times as many paper cups as Yvonne. How many paper cups did Wendy have in the end?
#### Solution:
Difference in units\begin{align}\\[2ex] &= 5 \text{ units} - 1 \text{ unit}\\[2ex] &= 4 \text{ units} \end{align}
\begin{align} 4 \text{ units} &= 70-6\\[2ex] &= 64\\[2ex] 1 \text{ unit} &= 64 \div 4\\[2ex] & = 16 \end{align}
Number of paper cups Wendy had in the end\begin{align}\\[6ex] &= 5 \text{ unit}\\[2ex] &= 5 \times 16\\[2ex] &= 80 \end{align}
Wendy had 80 paper cups in the end.
80 paper cups
#### Question 4:
Wilson and Ryan had an equal number of marbles. After Wilson lost 16 marbles and Ryan gave away 4 marbles, Ryan had thrice as many marbles as Wilson. How many marbles did Ryan have at first?
#### Solution:
Difference in units\begin{align}\\[2ex] &= 3 \text{ units} - 1 \text{ unit}\\[2ex] &= 2 \text{ units} \end{align}
\begin{align} 2 \text{ units} &= 16-4\\[2ex] &= 12\\[2ex] 1 \text{ unit} &= 12 \div 2\\[2ex] & = 6 \end{align}
Number of marbles Ryan had at first $=$ Number of marbles Wilson had at first
Number of marbles Wilson had at first\begin{align}\\[6ex] &= 1 \text{ unit} +16\\[2ex] &= 6+16\\[2ex] &= 22 \end{align}
Ryan had 22 marbles at first.
22 marbles
## 2. Equal stage in the end
Both subjects have a different number of items at first. After some items are added/removed from each subject, they have the same number of items in the end.
#### Question 1:
Mrs Lim had four times as many biscuits as Mrs Goh. After Mrs Lim gave away 17 biscuits and Mrs Goh received another 4 biscuits, they had an equal number of biscuits. How many biscuits did Mrs Goh have in the end?
#### Solution:
Difference in units\begin{align}\\[2ex] &= 4 \text{ units} - 1 \text{ unit}\\[2ex] &= 3 \text{ units} \end{align}
\begin{align} 3 \text{ units} &= 4+17\\[2ex] &= 21\\[2ex] 1 \text{ unit} &= 21 \div 3\\[2ex] & = 7 \end{align}
Number of biscuits Mrs Goh had in the end\begin{align}\\[6ex] &= 1 \text{ unit} +4\\[2ex] &= 7+4\\[2ex] &= 11 \end{align}
Mrs Goh had 11 biscuits in the end.
11 biscuits
#### Question 2:
Alvin had four times as many marbles as Wei Ming. After Alvin lost 24 marbles and Wei Ming lost 3 marbles, both of them had the same number of marbles. How many marbles did Wei Ming have in the end?
#### Solution:
Difference in units\begin{align}\\[2ex] &= 4 \text{ units} - 1 \text{ unit}\\[2ex] &= 3 \text{ units} \end{align}
\begin{align} 3 \text{ units} &= 24-3\\[2ex] &= 21\\[2ex] 1 \text{ unit} &= 21 \div 3\\[2ex] & = 7 \end{align}
Number of marbles Wei Ming had in the end\begin{align}\\[6ex] &= 1 \text{ unit} -3\\[2ex] &= 7-3\\[2ex] &= 4 \end{align}
Wei Ming had 4 marbles in the end.
4 marbles
#### Question 3:
The basketball team had four times as many members as the tennis team. After 4 new members joined the basketball team and 31 new members joined the tennis team, both teams had the same number of members. How many members were in the basketball team at first?
#### Solution:
Difference in units\begin{align}\\[2ex] &= 4 \text{ units} - 1 \text{ unit}\\[2ex] &= 3 \text{ units} \end{align}
\begin{align} 3 \text{ units} &= 31-4\\[2ex] &= 27\\[2ex] 1 \text{ unit} &= 27 \div 3\\[2ex] & = 9 \end{align}
Number of members in the basketball team at first\begin{align}\\[6ex] &= 4 \text{ unit} \\[2ex] &= 4 \times 9\\[2ex] &= 36 \end{align}
There were 36 members in the basketball team at first.
36 members
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Let’s refresh the page! | 1,891 | 6,075 | {"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": 20, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.71875 | 5 | CC-MAIN-2024-33 | latest | en | 0.913348 |
https://bzoj.llf0703.com/p/1755.html | 1,566,187,443,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027314641.41/warc/CC-MAIN-20190819032136-20190819054136-00093.warc.gz | 393,208,067 | 2,598 | [Usaco2005 qua]Bank Interest
题目描述
Farmer John made a profit last year! He would like to invest it well but wonders how much money he will make. He knows the interest rate R (an integer between 0 and 20) that is compounded annually at his bank. He has an integer amount of money M in the range 100..1,000,000. He knows how many years Y (range: 0..400) he intends to invest the money in the bank. Help him learn how much money he will have in the future by compounding the interest for each year he saves. Print an integer answer without rounding. Answers for the test data are guaranteed to fit into a signed 32 bit integer. 看了样例输入输出就知道怎么回事了..........
输入格式
* Line 1: Three space-separated integers: R, M, and Y
输出格式
* Line 1: A single integer that is the number of dollars FJ will have after Y years.
样例输入
```5 5000 4
INPUT DETAILS:
5% annual interest, 5000 money, 4 years
```
样例输出
```6077
OUTPUT DETAILS:
Year 1: 1.05 * 5000 = 5250
Year 2: 1.05 * 5250 = 5512.5
Year 3: 1.05 * 5512.50 = 5788.125
Year 4: 1.05 * 5788.125 = 6077.53125
The integer part of 6077.53125 is 6077.```
Gold
Menuappsclose | 358 | 1,110 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2019-35 | longest | en | 0.908457 |
https://mathoverflow.net/questions/383271/how-many-reals-can-we-construct-by-iteratively-writing-down-truth-tables-for-zfc | 1,670,046,959,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710924.83/warc/CC-MAIN-20221203043643-20221203073643-00758.warc.gz | 410,199,594 | 30,144 | # How many reals can we construct by iteratively writing down truth tables for ZFC?
For the purposes of this question, pretend that there is a platonic, "standard" model $$V$$ of ZFC - so that every set theoretic statement has a meaningful truth value.
Fix a bijection between $$\mathbb{N}$$ and the set of formulas $$\varphi(x)$$ with one free variable $$x$$ in the language of set theory, and let $$\varphi_i(x)$$ be the $$i$$th formula according to this bijection. For every set $$X \in V$$, let $$T(X) \in 2^\mathbb{N}$$ be the truth table for statements involving the set $$X$$, that is, the $$i$$ bit of $$T(X)$$ is set to $$1$$ iff $$\varphi_i(X)$$ is true in $$V$$.
Note that we can't prove that the truth table $$T(X)$$ is a set or even define $$T(X)$$ within the formal framework of ZFC - assume a strong enough metatheory that we can talk about $$T(X)$$ as representing a well-defined thing. Additionally, if we take the idea that $$V$$ is standard seriously, then we should assume that $$T(X)$$ is a set in $$V$$, for every $$X \in V$$.
Starting from the truth table $$T(\emptyset)$$ which encodes all true statements of $$V$$, we can then build the truth table $$T(T(\emptyset))$$ which encodes all the true statements about the nature of truth in $$V$$. The new set $$T(T(\emptyset))$$ is really new: it is not computable from $$T(\emptyset)$$, since $$T(T(\emptyset))$$ can be used to solve the halting problem for Turing machines with oracle access to $$T(\emptyset)$$. In particular, $$T(T(\emptyset)) \ne T(\emptyset)$$.
I want to iterate this construction as boldly as possible. To this end, inductively define a class function $$\mathcal{T} : \operatorname{Ord} \rightarrow 2^\mathbb{N}$$ as follows: for every ordinal $$\alpha$$, if $$V$$ is truly standard then the restriction $$\mathcal{T}|_\alpha : \alpha \rightarrow 2^\mathbb{N}$$ should be a set in $$V$$, so we can define
$$\mathcal{T}(\alpha) = T(\mathcal{T}|_\alpha)$$.
By the same argument as the one showing that $$T(T(\emptyset)) \ne T(\emptyset)$$, if $$\alpha < \beta \in \operatorname{Ord}$$ and if $$\alpha$$ is definable given $$\beta$$ as a parameter, then $$\mathcal{T}(\beta)$$ is not computable from $$\mathcal{T}(\alpha)$$.
Call an ordinal $$\alpha$$ "fresh" if $$\mathcal{T}(\alpha) \ne \mathcal{T}(\beta)$$ for all $$\beta < \alpha$$, and let $$F \subseteq \operatorname{Ord}$$ be the set of fresh ordinals. The restriction of $$\mathcal{T}$$ to $$F$$ gives an injection $$F \hookrightarrow 2^\mathbb{N}$$.
Question(s) 1: What can we say about the order type of $$F$$ (with the induced ordering from $$\operatorname{Ord}$$)? Is $$F$$ uncountable? Could $$|F|$$ be strictly larger than $$\aleph_1$$ (assuming that the continuum hypothesis is false)? Can we determine whether the order type of $$F$$ is a limit ordinal? What can we say about the least ordinal which is not contained in $$F$$ (i.e., the first "stale" ordinal) - could it be definable? Are these questions independent of large cardinal hypotheses/forcing axioms?
We can also ask interesting questions about the long-term behavior of the function $$\mathcal{T}$$. For every $$\alpha \in F$$, let $$S(\alpha)$$ be the collection of ordinals $$\beta$$ such that $$\mathcal{T}(\beta) = \mathcal{T}(\alpha)$$ (so all but one element of $$S(\alpha)$$ is stale). Note that at least one $$S(\alpha)$$ must be a proper class, and that for any definable $$\alpha \in F$$, we have $$S(\alpha) = \{\alpha\}$$.
Question(s) 2: For how many $$\alpha \in F$$ is the collection $$S(\alpha)$$ a proper class? Is there an undefinable $$\alpha \in F$$ such that $$S(\alpha)$$ is a singleton? How many distinct cardinalities show up among the $$S(\alpha)$$s?
Generalizing the problem, we can also build iterated truth tables involving any set $$X \in V$$ as a parameter. Define the function $$\mathcal{T}_X : \operatorname{Ord} \rightarrow 2^\mathbb{N}$$ inductively by
$$\mathcal{T}_X(\alpha) = T((\mathcal{T}_X|_\alpha, X))$$,
where $$(\mathcal{T}_X|_\alpha, X)$$ is the Kuratowski ordered pair. Call an ordinal $$\alpha$$ "fresh for $$X$$" if $$\mathcal{T}_X(\alpha) \ne \mathcal{T}_X(\beta)$$ for all $$\beta < \alpha$$, and let $$F_X$$ be the set of ordinals which are fresh for $$X$$.
If we choose $$X$$ such that for every countable ordinal $$\alpha$$ there is a bijection from $$\mathbb{N}$$ to $$\alpha$$ which can be defined from $$X$$ and $$\alpha$$, then every countable ordinal will be fresh, so we will have $$\omega_1 \subseteq F_X$$.
Question 3: How large can $$F_X$$ be, if we choose the set $$X \in V$$ to make $$F_X$$ as large as possible?
Apologies for asking so many questions at once. I'll be happy with any nontrivial results about the nature of the function $$\mathcal{T}$$ or the set of fresh ordinals $$F$$.
Edit: I just realized that Question 3 isn't nearly as interesting as I had thought it was - if we take $$X$$ to be an injection from an ordinal $$\alpha$$ into $$2^\mathbb{N}$$, then it is easy to prove that $$\alpha \subseteq F_X$$, so Question 3 is really just asking how big the continuum is.
• Note that we can remove the philosophical flavor here by just working inside an appropriate class theory, e.g. $\mathsf{NBG}$ (at a glance I think that should be enough for what you want here). In particular, in this ambient class theory we can legitimately talk about $T$ and so forth without difficulty. Feb 6, 2021 at 3:20
• @NoahSchweber It seems to me that the class theory you want should be MK rather than NBG, so that truth in $V$ becomes definable. Feb 6, 2021 at 5:08
• @AndreasBlass Yes, I meant MK. Not my finest moment. Feb 6, 2021 at 5:28
• Every real definable in $V_\alpha$ will be computable from $T(\alpha)$ since $\alpha$ is coded into $T\restriction \alpha$. So every ordinal definable real is computable from a $T(\alpha)$. Also every real you enumerate will be ordinal definable: in MK, there is a proper class of $\alpha$ with $V_\alpha\prec V$. So the set of fresh ordinals has cardinality the continuum in $\text{HOD}$. Not sure about the order type. I'll post an answer later if I haven't completely misunderstood the question. Feb 6, 2021 at 17:22
• I have to think a bit more about the ordertype, which is interesting. Seems the same as the question of whether the canonical wellorder of OD sets restricts to a wellorder of the OD reals in ordertype $\mathfrak{c}^\text{HOD}$. That might be independent. Feb 6, 2021 at 18:00
I think it is easier to think about the question generalized to an arbitrary transitive model of ZFC, resisting the natural urge to grasp towards the Absolute. So fix such a model $$M$$, and let $$\mathcal T^M(\alpha)$$, $$F^M$$, and $$S^M(\alpha)$$ be as you defined them, replacing $$V$$ with $$M$$. Let's omit the superscripts, though.
The ordertype of $$F$$ is a limit ordinal: if $$\alpha$$ is fresh, so is $$\alpha+1$$, since in general, one can recover $$\mathcal T(\beta)$$ from $$\mathcal T(\beta+1)$$.
Note that if $$\alpha$$ is definable in $$M$$, $$\mathcal T(\alpha)$$ is fresh since it is the unique value of the function $$\mathcal T$$ containing the index of the formula $$\varphi(\text{dom}(x))$$ where $$\varphi$$ defines $$\alpha$$. In particular, an ordinal that is not fresh is not definable, and so your least stale ordinal is not definable.
Note that if every ordinal of $$M$$ is definable in $$M$$, then every ordinal is fresh. More interestingly, even models with undefinable ordinals can have only fresh ordinals. (I think Con(ZFC) does not suffice to build a model of ZFC whose fresh ordinals form a set.) To avoid these examples, assume from now on that $$(M,S)$$ satisfies the Axiom of Replacement where $$S$$ is the satisfaction predicate of $$M$$.
The fresh ordinals $$F$$ then form a set in $$M$$. Moreover, $$(\mathcal T(\alpha))_{\alpha \in F}$$ is definable in $$M$$ from any sufficiently large ordinal $$\kappa$$ such that $$V_\kappa^M\prec M$$ (and there are cofinally many). It follows that $$(\mathcal T(\alpha))_{\alpha \in F}$$ belongs to $$\text{HOD}^M$$, which I will denote by $$H$$. Therefore $$F$$, has cardinality at most $$\mathfrak c$$ in $$H$$. For any ordinal $$\alpha < \mathfrak{c}^H$$, the $$\alpha$$-th real in the canonical wellorder of $$H$$ is $$\Sigma_2$$-definable from $$\alpha$$ and hence is recoverable from $$\mathcal T(\alpha)$$. It follows that $$\alpha$$ is fresh. This shows $$\mathfrak{c}^H\subseteq F$$. Also $$\mathfrak{c}^H\in F$$, being definable in $$M$$. So the ordertype of $$F$$ is strictly between $$\mathfrak{c}^H$$ and $$\mathfrak{c}^{+H}$$.
Now it is clear that $$F$$ is uncountable if and only if $$M$$ thinks $$H$$ contains uncountably many ordinal definable reals, which of course depends on the choice of $$M$$. The cardinality of $$F$$ is larger than $$\aleph_1$$ if and only if $$H$$ contains at least $$\aleph_2$$-many reals, which is independent of $$\neg\text{CH}$$.
Assuming Martin's Maximum, every subset of $$\omega$$ is coded into the pattern of stationary reflection for any infinite stationary partition of $$\{\alpha < \kappa : \text{cf}(\alpha) = \omega\}$$ where $$\kappa\geq \omega_2$$. (This is part of the proof of Theorem 10 in Foreman-Magidor-Shelah's Martin's Maximum, saturated ideals, and nonregular ultrafilters.) So assuming $$M$$ satisfies MM, large cardinals, and Woodin's HOD Hypothesis, every real is in $$H$$. Therefore it is conceivable that your question decided by the conjunction of forcing axioms and large cardinals, but this is open...
For the second question, note that by the pigeonhole principle there is an unbounded class $$C\subseteq M$$ of ordinals such that for all $$\kappa_0,\kappa_1\in C$$, the theory of $$\kappa_0$$ in $$(M,S)$$ with real parameters is the same as that of $$\kappa_1$$. Let $$\langle-,-\rangle$$ be an $$M$$-definable pairing function. It follows that $$\mathcal T\langle\kappa_0,\alpha\rangle$$ is equal to $$\mathcal T\langle\kappa_1,\alpha\rangle$$ for all $$\alpha < \mathfrak{c}^H$$, since $$\alpha$$ is interdefinable over $$M$$ with the $$\alpha$$-th real in the canonical wellorder of $$H$$ and $$\mathcal T\langle\kappa_0,\alpha\rangle$$ depends only on the theory of $$\kappa_0$$ and $$\alpha$$ in $$(M,S)$$. But similarly, for each $$\kappa\in C$$, the $$\mathcal T\langle\kappa,\alpha\rangle$$ are distinct for $$\alpha < \mathfrak{c}^{H}$$; let $$\xi_{\alpha}$$ be the least ordinal $$\xi$$ such that $$\mathcal T(\xi) = \mathcal T\langle\kappa,\alpha\rangle$$. Letting $$E\subseteq F$$ be the set of ordinals $$\alpha$$ such that $$S(\alpha)$$ a proper class in $$M$$, it follows that $$\{\xi_\alpha : \alpha < \mathfrak{c}^{H}\}\subseteq E$$, so $$E$$ has cardinality $$\mathfrak{c}$$ in $$H$$. So much of what was said above applies to $$E$$ as well.
• I don't see how $\mathcal{T}(\alpha)$ is Turing equivalent to the $1$-type of $\alpha$: this doesn't seem to be true for $\alpha = 1$, for instance (the $1$-types of $0$ and $1$ are Turing equivalent, but $\mathcal{T}(1)$ is not computable from $\mathcal{T}(0)$).
– zeb
Feb 8, 2021 at 17:29
• No, you're right. I confused myself! Let me fix it up Feb 8, 2021 at 18:36
• Should be fixed now Feb 8, 2021 at 18:49
• This is a remarkably complete answer!
– zeb
Feb 8, 2021 at 22:34
• Wow, what an amazing answer. Martin's Maximum... was not expecting that.
– user141903
Feb 9, 2021 at 0:36 | 3,380 | 11,355 | {"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": 191, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2022-49 | latest | en | 0.834833 |
https://gpuzzles.com/mind-teasers/half-five-four/?source=showRecent | 1,516,185,495,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084886895.18/warc/CC-MAIN-20180117102533-20180117122533-00186.warc.gz | 695,961,322 | 10,292 | • Views : 70k+
• Sol Viewed : 20k+
# Mind Teasers : 1/2 of 5 = 4 Riddle
Difficulty Popularity
Can you prove that half of number 5 is equal to 4?
Discussion
Suggestions
• Views : 50k+
• Sol Viewed : 20k+
# Mind Teasers : Trick Gift Brain Teaser
Difficulty Popularity
Cindy opened twenty five presents.
Duke opened five presents.
John opened fifteen presents.
Judging by the statements, can you decipher how many presents will be opened by Rhea?
• Views : 70k+
• Sol Viewed : 20k+
# Mind Teasers : What word am I ?
Difficulty Popularity
What word am I?
Insights:
* I am a nine letter word.
* I am the exact inverse to expression 'neighborly and active'
* I contain the word 'over'.
* I begin will a vowel.
• Views : 50k+
• Sol Viewed : 20k+
# Mind Teasers : Tough River Crossing Riddle
Difficulty Popularity
This one is a bit of tricky river crossing puzzle than you might have solved till now. We have a whole family out on a picnic on one side of the river. The family includes Mother and Father, two sons, two daughters, a maid and a dog. The bridge broke down and all they have is a boat that can take them towards the other side of the river. But there is a condition with the boat. It can hold just two persons at one time (count the dog as one person).
No it does not limit to that and there are other complications. The dog can’t be left without the maid or it will bite the family members. The father can’t be left with daughters without the mother and in the same manner, the mother can’t be left alone with the sons without the father. Also an adult is needed to drive the boat and it can’t drive by itself.
How will all of them reach the other side of the river?
• Views : 80k+
• Sol Viewed : 20k+
# Mind Teasers : Who Are We
Difficulty Popularity
Without being called, we came out at night.
Without being stolen, we get lost in the day.
Who are we?
• Views : 40k+
• Sol Viewed : 10k+
# Mind Teasers : Tricky Glass Riddle
Difficulty Popularity
Six identical glasses are in a row.
The first three glasses are filled with juice, and the last three glasses are empty.
By moving only one glass, can you arrange them so that the full and the empty glasses are alternate?
• Views : 40k+
• Sol Viewed : 10k+
# Mind Teasers : World War I History Riddle
Difficulty Popularity
Two companions Terry and Garry were talking about their families. Terry told some great stories about his courageous grandfather who fought for Britain in "World War I". Terry told that his grandfather is so brave that he was awarded a bravery honour medal with words "For our Courageous Soldiers In World War I" embedded into it.
Garry Knows that his friend is lying? How ?
• Views : 60k+
• Sol Viewed : 20k+
# Mind Teasers : Popular Logical Riddle
Difficulty Popularity
A mules travels the same distance daily.
I noticed that two of his legs travels 10km and the remaining two travels 12km.
Obviously two mules legs cannot be a 2km ahead of the other 2.
The mules is perfectly normal. So how come this be true ?
• Views : 70k+
• Sol Viewed : 20k+
# Mind Teasers : Picture Mind Teaser
Difficulty Popularity
You are given with six marbles of three different colors. Arrange them in a manner that each one of them touches all four marbles of different colors.
• Views : 70k+
• Sol Viewed : 20k+
# Mind Teasers : Tricky Math Race Riddle
Difficulty Popularity
Rooney, Hernandez, and Robin race each other in a 100 meters race. All of them run at a constant speed throughout the race.
Rooney beats Hernandez by 20 meters.
Hernandez beats Robin by 20 meters.
How many meters does Rooney beat Robin by ?
• Views : 40k+
• Sol Viewed : 10k+
# Mind Teasers : Trick Relationship Riddle
Difficulty Popularity
If Erica's daughter is my daughter's mom, who am I to Erica?
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Can you count the number of images in wh... | 1,089 | 4,376 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-05 | latest | en | 0.923988 |
https://casadimenotti.com/multiplying-monomials-calculator/worksheets-multiplying-and-dividing-polynomials-by-monomials/ | 1,558,949,537,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232262311.80/warc/CC-MAIN-20190527085702-20190527111702-00135.warc.gz | 422,645,441 | 12,754 | Multiplying Monomials Calculator Division Ofmials By Worksheet Divide Free Printables The Binomial Trinomial Math From 840x1087
By Fauna Gagne at December 26 2018 20:38:07
Always be on the look out for tears of frustration, so as to all the child some break. No one learns well enough when confused. At this point both concentration and effective learning are lost. But if you are sure that the tears are merely excuses to skip the homework, pretend to abandon him with the worksheets and let him sweat it out! Reading is much more than fluency and word recognition. Reading fluently isn't enough. Students must fully understand the content they are reading. By learning and applying several different reading strategies, students will be able to obtain meaning from a wide variety of texts.
In math bingo, each student is given a bingo card (also known as a "bingo worksheet" or "bingo board") printed with numbers. These aren't necessarily the standard bingo numbers, but rather are the answers to a number of different math problems. The game is then played exactly like a normal game of bingo, with the teacher playing the part of the bingo caller, but instead of the teacher calling out the numbers printed on the cards, the teacher instead calls out math problems (the teacher may also write the problem on the blackboard). The students' task is to solve each problem, and then look for the number on their bingo card. If you are looking for an article that describes the basics of Excel and introduces the interface and concepts for beginners, you have come to the right place. Microsoft Excel is a powerful business application that is organized into a structural hierarchy of Workbooks, Worksheets, and Cells.
Gallery of Multiplying Monomials Calculator
A game using number jugglers will encourage your child to think while playing. This is done with the combination of a designed and customized deck of cards with a colorful book consists of 20 games. 1 worksheet per day keeps tuition's away. Kids have a short attention span, Worksheets simplify the learning process and each preschool worksheet can be completed in about 7 - 10 minutes. Educationists create sets of worksheets as per the academic curriculum of the learners. The learning objectives are set as per the kid's level of understanding. Therefore, worksheets for Class 1 will vary from nursery worksheets.
As you can imagine, this can be a lot of fun, and before you know it students can forget they are learning math! What is more, teachers can also easily vary the game play, for example, by using different types of math problems, or perhaps even by asking members of the class to solve each problem before moving on to the next bingo call. The data you include in an Excel file can be formatted and manipulated in a variety of ways. Once you have read this article, you will have a better understanding of the structure of an Excel file and the most common types of data you can use. | 605 | 2,967 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2019-22 | latest | en | 0.956152 |
https://www.edplace.com/worksheet_info/maths/keystage2/year6/topic/932/643/multiply-decimals-and-estimate-the-answer | 1,713,250,799,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817073.16/warc/CC-MAIN-20240416062523-20240416092523-00867.warc.gz | 702,560,233 | 12,356 | # Round Decimals to Nearest Whole Numbers Before Multiplying
In this worksheet, students will round decimals to the nearest whole numbers and then estimate the answer to a multiplication.
Key stage: KS 2
Curriculum topic: Number: Fractions, Decimals and Percentages
Curriculum subtopic: Solve Rounding Problems
Difficulty level:
#### Worksheet Overview
In this activity, we will be rounding decimal numbers to the nearest whole numbers in order to estimate the answer to a multiplication.
For example
Estimate the answer to the multiplication by first rounding both decimals to the nearest whole numbers.
2.25 x 1.41
A
X
B
=
C
2.25 x 1.41
2
x
1
=
2
We need to remember our rules for rounding decimals to whole numbers for this activity!
If the number immediately after the decimal point, in the tenths column, is 5 or more, we round up the number in the ones column.
If the number in the tenths column is below 5, the number in the ones column will not change.
So, in the example above, both 2.25 and 1.41 have numbers in the tenths column that are below 5, so this means they round down to the whole numbers 2 and 1.
Are you ready to get rounding?
### What is EdPlace?
We're your National Curriculum aligned online education content provider helping each child succeed in English, maths and science from year 1 to GCSE. With an EdPlace account you’ll be able to track and measure progress, helping each child achieve their best. We build confidence and attainment by personalising each child’s learning at a level that suits them.
Get started | 369 | 1,574 | {"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} | 3.921875 | 4 | CC-MAIN-2024-18 | latest | en | 0.860327 |
https://damogranlabs.com/2018/09/damogran-soup/?shared=email&msg=fail | 1,563,339,463,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195525046.5/warc/CC-MAIN-20190717041500-20190717063500-00186.warc.gz | 361,769,405 | 19,615 | # Damogran Soup
Here’s a quick tutorial on interFoam solver from the OpenFOAM toolbox. It demonstrates how much mess one makes when sneezing close to a bowl of soup.
## The Case
Every time I sneeze during lunchtime my imagination runs wild with images of soup splattering all over the place. It wouldn’t be very polite to try it in real life so I decided to try sneezing virtually.
That means directing a stream of air of high velocity into a liquid at a standstill at a very close range and seeing what happens.
The hard part is finding out how fast a sneeze really is. There’s a bunch of references quoting a common urban myth that says it reaches speeds of over 160 km/h. But there’s also a single scientific paper stating that no more than 17 km/h can be squeezed out of a human. To be honest, I really don’t care how fast a sneeze is as long as I make a huge virtual mess out of mine. It turns out that in ideal conditions a 20 km/h sneeze wouldn’t make a lot so I happily used the myth numbers, 150 km/h more exactly.
## The Geometry
I very carefully modeled a soup bowl from my china set with precise caliper measurements. FreeCAD was the program of choice. I created two parts, one for the bowl and one for the soup at a standstill. As the geometry is rather simple, I meshed both parts with Mesh workbench and exported both to STL (for more complicated geometry I would most probably use SALOME for meshing and face assignment). I’m used to ASCII STL format – I added a .ast when exporting and then renamed it back to stl. Also FreeCAD’s default export units are millimeters or whatever your settings are – you can change them or -in my case- use the surfaceFeatureConvert utility. I also named my solids by manually editing STL files; in the line that states ‘solid’ I added a name: solid soup. It doesn’t sound right, though.
## The Mesh
Looking into makeSnappyMesh script reveals the meshing process:
Copy the saved geometry and scale it from millimeters:
```surfaceConvert -scale 0.001 ../geometry/plate.stl constant/triSurface/plate.stl
surfaceConvert -scale 0.001 ../geometry/soup.stl constant/triSurface/soup.stl
```
Surface features: there are no sharp features on the plate so this is actually not needed;
it’s included here anyway in case the geometry should be updated later
```surfaceFeatureExtract
```
blockMesh contains the computational domain; in this very case, the plate is not entirely included.
When the wave returns it leaves it and that’s visible as a sharp end near the sneezer’s mouth.
```blockMesh
```
snappyHexMesh is configured to make a lot of cells near plate walls. And to run in parallel. My i7 processor has 4 physical cores and 8 logical with hyperthreading. It turns out that using 4 cores is faster than using 8.
There’s a special refinement zone called inlet. It will refine a small rectangular region which will then be captured by topoSet and used as an inlet patch.
It makes sense to run checkMesh on a decomposed mesh and then reconstruct it. Instead of scrolling way back to checkMesh‘s output, save it to a file and display it at the end.
```decomposePar -force
mpirun -np 4 snappyHexMesh -parallel -overwrite
mpirun -np 4 checkMesh -parallel > log.checkMesh
reconstructParMesh -constant
```
topoSet will capture the refined inlet and create a cellSet. createPatch will – guess what – create a patch from that set.
```topoSet
createPatch -overwrite
```
Display the output and clean up
```cat log.checkMesh
rm log.checkMesh
```
## Boundary Conditions
I had a lot of ideas how boundary conditions should really be stated but in the end it turns out that tried and tested ones from tutorials work best. Namely, tutorials/multiphase/interFoam/RAS/damBreak. There’s just one not-that-common-one, that is inlet velocity. It uses a tabulated value that changes over time. It’s also not perpendicular to inlet surface but it just means the numbers are a bit different. I simply made up sneeze durations and velocity profile. It doesn’t have to be realistic, it just has to make a big splash.
## Initial conditions
There’s soup in the bowl. As stated earlier, I prepared geometry for the bowl and another one for the soup. setFieldssurfaceToCell uses the soup STL surface to find all cells that contain the soup initially. It is also worth noting that it will replace the original alpha.water file with the new initial conditions so it’s wise to save your original file as alpha.water.orig.
## Dynamic Mesh refinement
From OpenFOAM tutorials:
The two-phase algorithm in interFoam is based on the volume of fluid (VOF) method in which a specie transport equation is used to determine the relative volume fraction of the two phases, or phase fraction α, in each computational cell. Physical properties are calculated as weighted averages based on this fraction. The nature of the VOF method means that an interface between the species is not explicitly computed, but rather emerges as a property of the phase fraction field. Since the phase fraction can have any value between 0 and 1, the interface is never sharply defined, but occupies a volume around the region where a sharp interface should exist.
Interface in this case means water level or more precisely, soup level. Therefore, smaller cells more precisely define the location of the soup. That means a zillion cells in the mesh but only a small number of interesting ones – the ones that contain the interface or soup. It would be best to refine mesh dynamically based on how much soup each cell contains. Unfortunately, using dynamicRefineFvMesh in constant/dynamicMeshDict causes the solver to crash without stating what exactly went wrong. I’m still finding out how to solve this problem so any suggestions are most welcome.
## Running
It will take a lot of time to solve the first half of a second since velocities are high but Courant number should be kept low. As the sneeze calms down all that’s left is splashing of soup which is a much slower process. The rest 2.5 seconds will take approximately the same computational time.
## Postprocessing
Since simulation results are not very precise they should at least look nice. I imported an STL head model (that has no physical meaning and no mention in the simulation) and placed it approximately where the inlet should be. I used a plane source in ParaView to model a table and imported the same STL bowl I used in snappyHexMesh.
When displaying soup interface with coarser mesh (such as this one) the Contour filter makes horrible surfaces. They can be smoothed out by using the -guess what- Smooth filter. The larger the number of iterations, the smoother the soup. In my case it’s an asparagus soup so it should stay as smooth as possible.
Also, some additional light tweaking (View > Light Inspector) helps a lot. For the final animation I checked the Enable OSPRay for a more realistic rendering.
I saved every hundredth of a second of simulation time. That accounts for 300 frames which give 20 seconds of video at 15 frames per second. Not to say using compression for result files makes sense…
## The Files
If you want to mess around with your interFoam, you should check the attachment. In it you’ll find the FreeCAD files, STL geometry, two OpenFOAM cases (one for meshing and one for simulation), scripts that run all required steps and ParaView state for the whole postprocessing bundle. I can’t attach 5 GB of zipped result files, though.
damogran-soup.zip
Have fun and watch your table manners. | 1,677 | 7,505 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2019-30 | longest | en | 0.919662 |
https://www.cnblogs.com/Paulliant/p/11386625.html | 1,601,402,899,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400202418.22/warc/CC-MAIN-20200929154729-20200929184729-00648.warc.gz | 759,134,077 | 7,074 | # WC 2008 观光计划（斯坦纳树）
## 题意
https://www.lydsy.com/JudgeOnline/problem.php?id=2595
## 思路
$dp[i][j]=\min(dp[k][j]+dp[i\setminus k][j])$
$\text{chk_min}(dp[i][k],dp[i][j]+w(j,k))$
## 代码
#include<bits/stdc++.h>
#define FOR(i,x,y) for(int i=(x),i##END=(y);i<=i##END;++i)
#define DOR(i,x,y) for(int i=(x),i##END=(y);i>=i##END;--i)
template<typename T,typename _T>inline bool chk_min(T &x,const _T y){return y<x?x=y,1:0;}
template<typename T,typename _T>inline bool chk_max(T &x,const _T y){return x<y?x=y,1:0;}
typedef long long ll;
template<const int N,const int M,typename T>struct LinkedList
{
};
struct node
{
int at,path;
bool operator <(const node &_)const{return path>_.path;}
};
std::priority_queue<node>Q;
int dp[(1<<10)+3][103];
int las[(1<<10)+3][103];
bool mark[103];
int mp[103],ori[13];
int pw[103];
int n,m,K;
inline int hs(int x,int y){return x*m+y;}
void Steiner()
{
FOR(i,0,(1<<K)-1)FOR(j,0,n-1)dp[i][j]=1e9;
FOR(i,0,K-1)dp[1<<i][ori[i]]=0;
FOR(i,1,(1<<K)-1)
{
FOR(j,0,n-1)
for(int k=(i-1)&i;k;k=(k-1)&i)
if(chk_min(dp[i][j],dp[k][j]+dp[i^k][j]-pw[j]))
{
las[i][j]=k;
}
while(!Q.empty())Q.pop();
FOR(j,0,n-1)Q.push((node){j,dp[i][j]});
while(!Q.empty())
{
node now=Q.top();Q.pop();
int u=now.at;
if(now.path>dp[i][u])continue;
EOR(k,G,u)
{
int v=G[k],w=pw[v];
if(chk_min(dp[i][v],dp[i][u]+w))
{
las[i][v]=u;
Q.push((node){v,dp[i][v]});
}
}
}
}
}
void backtrack(int i,int j)
{
mark[j]=1;
if(mp[j]!=-1&&i==(1<<mp[j]))return;
backtrack(las[i][j],j),backtrack(i^las[i][j],j);
else backtrack(i,las[i][j]);
}
int main()
{
scanf("%d%d",&n,&m);
FOR(i,0,n-1)FOR(j,0,m-1)
{
scanf("%d",&pw[hs(i,j)]);
if(!pw[hs(i,j)])mp[hs(i,j)]=K,ori[K]=hs(i,j),K++;
else mp[hs(i,j)]=-1;
}
FOR(i,0,n-1)FOR(j,0,m-2)
{
}
FOR(i,0,n-2)FOR(j,0,m-1)
{
}
n*=m;
Steiner();
int ans=1e9,id;
FOR(i,0,n-1)if(chk_min(ans,dp[(1<<K)-1][i]))id=i;
backtrack((1<<K)-1,id);
printf("%d\n",ans);
FOR(i,0,n-1)
{
if(!pw[i])putchar('x');
else putchar(mark[i]?'o':'_');
if(i%m==m-1)putchar('\n');
}
return 0;
}
posted @ 2019-08-21 08:14 Paulliant 阅读(109) 评论(1编辑 收藏 | 880 | 2,040 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2020-40 | latest | en | 0.166349 |
http://facstaff.gpc.edu/~mhall/online/fun.htm | 1,371,530,843,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368706934574/warc/CC-MAIN-20130516122214-00035-ip-10-60-113-184.ec2.internal.warc.gz | 76,293,110 | 1,182 | Problem 1 A regular drinking cup has a circular lower base 8 cm in diameter, a lip 12 cm in diameter, and a height of 10 cm. What is its volume? Problem 2 Each of the 25 buildings on ICU campus must have an individual telephone line for every building it is connected to on the campus. How many individual telephone lines are there between all the buildings on ICU campus? Return to Front Page | 88 | 393 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2013-20 | latest | en | 0.958252 |
http://oeis.org/A021823 | 1,611,573,299,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703565541.79/warc/CC-MAIN-20210125092143-20210125122143-00451.warc.gz | 71,882,284 | 4,148 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
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A021823 Decimal expansion of 1/819. 13
0, 0, 1, 2, 2, 1, 0, 0, 1, 2, 2, 1, 0, 0, 1, 2, 2, 1, 0, 0, 1, 2, 2, 1, 0, 0, 1, 2, 2, 1, 0, 0, 1, 2, 2, 1, 0, 0, 1, 2, 2, 1, 0, 0, 1, 2, 2, 1, 0, 0, 1, 2, 2, 1, 0, 0, 1, 2, 2, 1, 0, 0, 1, 2, 2, 1, 0, 0, 1, 2, 2, 1, 0, 0, 1, 2, 2, 1, 0, 0, 1, 2, 2, 1, 0, 0, 1, 2, 2, 1, 0, 0, 1, 2, 2, 1, 0, 0, 1 (list; constant; graph; refs; listen; history; text; internal format)
OFFSET 0,4 COMMENTS Partial sums of A010892. - Paul Barry, Jun 06 2003 Expansion in any base b >= 3 of 1/((b-1)(b^2-b+1) = 1/(b^3-2b^2+2b-1). E.g., 1/14 in base 3, 1/39 in base 4, 1/84 in base 5, etc. - Franklin T. Adams-Watters, Nov 07 2006 LINKS Index entries for linear recurrences with constant coefficients, signature (2, -2, 1). FORMULA a(n) = a(n-1)-a(n-2)+1 = 2-a(n-3) = a(n-6). - Henry Bottomley, Apr 12 2000 a(n) = Sum_{k=1..floor(n/2)} (-1)^(k+1)*binomial(n-k, k) = 1-((-1)^floor(n/3)+(-1)^(floor((n+1)/3)))/2. - Vladeta Jovovic, Feb 10 2003 G.f.: x^2/(1-2x+2x^2-x^3)=x^2/((1-x)(x^2-x+1)). - Paul Barry, Jun 06 2003 a(n+2) = sum{k=0..n, binomial(n-2k, n-k)}. - Paul Barry, Jan 15 2005 a(n) = (1/30)*{7*(n mod 6)+7*[(n+1) mod 6]+2*[(n+2) mod 6]-3*[(n+3) mod 6]-3*[(n+4) mod 6]+2*[(n+5) mod 6]}, with n>=0. - Paolo P. Lava, Jan 31 2008 a(0)=0, a(1)=0, a(2)=1, a(n)=2*a(n-1)-2*a(n-2)+a(n-3). - Harvey P. Dale, Aug 19 2012 EXAMPLE 0.0012210012210012210... MATHEMATICA Join[{0, 0}, RealDigits[1/819, 10, 120][[1]]] (* or *) PadRight[{}, 120, {0, 0, 1, 2, 2, 1}] (* or *) LinearRecurrence[{2, -2, 1}, {0, 0, 1}, 120] (* Harvey P. Dale, Aug 19 2012 *) PROG (PARI) a(n)=1/819. \\ Charles R Greathouse IV, Sep 24 2015 CROSSREFS Cf. A077859, A027444. Sequence in context: A281497 A198243 A164965 * A131026 A333839 A014604 Adjacent sequences: A021820 A021821 A021822 * A021824 A021825 A021826 KEYWORD nonn,cons,easy AUTHOR STATUS approved
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Last modified January 25 06:09 EST 2021. Contains 340416 sequences. (Running on oeis4.) | 1,126 | 2,364 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-04 | latest | en | 0.407966 |
https://ezhillang.blog/tag/mathematics/ | 1,670,310,756,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711074.68/warc/CC-MAIN-20221206060908-20221206090908-00193.warc.gz | 285,513,815 | 26,670 | # A bipartite graph structure for Tamil
Remark: Tamil alphabets [which are Abugida or alphasyllabary in nature] can be written as a fully connected bipartite graph G(C+V,E). Both the basic 247 letters [known to have a ring representation] and sequences involving வட மொழி letters can be written in the sequence of two sets, V – vowels [உயிர்] and C – consonants [மெய்], and edges E: C -> V forming a map from each consonant to vowel (e.g.: க் + அ -> க ) are the உயிர்மெய் எழுத்துக்கள். This is a $K_{\left[18 \times 12\right]}$ bipartite graph. Strictly speaking we can add அயுத எழுத்து ‘ஃ’ as a disconnected node and call it a $K_{\left[ 18 \times 12\right]} + 1$ forest graph. This may be simply extended to cover the வட மொழி எழுத்துக்கள் [Sanskrit letters optionally used in Tamil]. Full alphabet set is obtained by cumulative sum of edges and vertices.
Corollary: Most other alphasyllabary, Abugida languages have a similar bipartite graph representation.
# A group structure for Tamil
We can form a group structure for Tamil alphabets in many ways; simply we may apply residue classes modulo N or symmetric group of permutations modulo N for any cardinality. However, one interesting group structure with applications is the abstraction of 247 Tamil letters written on a torus; in this essay I will attempt to describe it and show that it forms a group.
We consider the 247 Tamil letters formed by 1 ayudha letter and 12 uyir letters for 13 vowels, and 18 mei letters for 18 consonants and 216 uyirmei or conjugate letters [247 = 13 + 18 + 216]. By consider a mapping of 13 vowels to Z13[residue classes modulo 13] and 18 uyirmei letters + ayutha letter to Z19 [residue classes modulo 19].
### Representation
Further we may represent each uyirmei letter as a index into a 2D table formed by rows of mei letters, and columns of uyir letters. So, for example letter ‘கு = க் + ஊ’ can be written as 6 + 1*13 = 19. Uyir letters are all represented from [0-12], Mei letters are represented as multiples of 13, [13, 26, 39, .. 234] for [க், ச், … ல், வ், ழ், ள்]. Uyirmei letters form everything in between.
The general representation of a letter can be: t = a + 13*b, where a goes from [0-12] and b goes from [0-18]. This representation pegs ‘ஃ’ at the origin. In the direct product of Z13 and Z19 this will be represented as (a,b)
Letter representation in the product group: Z13 x Z19
## Result
Further since we showed uyir and mei letters can be embedded into the Z13, and Z19 residue classes and we know 247 factors neatly into 2 primes 13 and 19, we may use the Chinese remainder theorem (which guarantees that given two sets of residue classes which are co-prime, we can form a residue class with a unique representation for the direct-sum [direct-product] of the underlying sets). In our case we are guaranteed that Z13 x Z19 direct sum structure forms an isomorphic group in Z247. This is the key result in this easy:
Tamil letters [247] have a direct product representation in group Z247 which is isomorphic to the direct product of Z13, Z19 as mapping the uyir and mei group representations.
Key result – Group representation for Tamil alphabets
While Chinese remainder theorem guarantees a ring structure, I don’t know the second operator which can take role of product to make the ring structure possible at this writing.
# ஆமவடை
ஏற்கணவே பதிவு செய்த இடத்தில் இருந்து தொடருவோம்:
Corollary 2 of Theorem 3: ஒரே சொல்லில் எழுத்து இரடிக்கப்பட்டால் அந்த சொல் டோரசில் ஒரு சுழலுடன் [loop] கொண்டபடி அமையும்.
Lemma 2: படுக்கவசமாகவும், நிமிர்ந்துவசமாகவும் அமைகப்பட்ட சொர்கள் மொழியில் இல்லாதவை.
Corollary 3 or Theorem 3: டோரசில் படுக்கவசமாகவும், நிமிர்ந்துவசமாகவும் பாதைகள்/எழுத்துக்கள் இல்லாதவை.
Theorem 4: ஒரு அகராதியில் உள்ள சொர்கள் அனைத்தையும் டோரசில் பிரதிபலித்தால் அந்த குறுக்கிடும் இடங்களின் [intersecting points] ஒன்று அல்லது மெர்பட்ட சொற்களை] எண்ணிக்கை அளவை மிக குறைவாக்கும் வண்ணம் அமைக்க முடியாது. அதாவது ஒரு அகராதியின் சொற்கள் அனைத்து எவ்வித அமைப்பில் உள்ள டோரசானாலும் சரி அதன் குறுக்கிடும் இடங்களின் எண்ணிக்கை மாராது. இது ஒரு மாறிலி [invariant].
Corollary 1 of Theorem 4: மேர்கண்ட டோரசில் [அதன் ஒரு பிரதிபலிப்பில் – ‘அ,ஆ,இ,ஈ, … ,ஒ,ஓ,ஔ‘ என்றும் ‘கசடதபரயரலவழள – ….’ என்றும் வரிசையிலோ, அல்லது வேறு பரிமாணங்களில் அடுக்கியிருந்தால்] ஒவ்வொரு அகராதிக்கும் ஒரு சிரப்பான குறுக்கிடும் இடங்களின் எண்ணிக்கை கிடைக்கும். இந்த எண் அகராதியின் கையொப்பம் [signature] என்றும் சொல்லாம்.
Theorem 5: டோரசில் உள்ள ஓவ்வொரு அகராதி சொல்லும் ஒரு பாதை என்று கொள்ளலாம். சொல்லின் தொடக்க எழுத்து பாதையின் தொடக்கத்தையும், சொல்லின் கடைசி எழுத்து பாதையின் முடிவையும் குறிக்கும்; பாதை திசைகொண்ட பாதையாக இருக்கும் – ஒரு அம்பு தொடக்கத்தில் இருந்து முடிவின் திசையில் வழி காட்டும். ஆகையால் அகராதியில் இல்லாத பாதைகள் பிழையாக எழுதப்பட்ட அகராதி சொற்களுக்கு சமம், அல்லது அகராதியில் இல்லாத புதிய சொற்களுக்கு சமம்.
வாதம் [ஆதாரத்தின் தொடக்கமாக கருத்ப்படலாம்]: டோரசில்ஒவ்வொரு சொல்லும் [அதன் பாதையும்] அகராதியில் உள்ள சொற்களாகவே இருக்கவேண்டும். Coding-theory / error correction codes theory படி இவ்வகை சரியான எழுத்துக்கள் உள்ள பாதைகள், சரியான சொற்களாகவும், தவான சொற்கள் [இல்லாத சொற்கள்] பிழையானவை என்வும் அமையும். இவ்வாரான சொற்கள் சரியானவையையின் சொற்பிழை எனவும் கருதப்பாடும்.
Corollary 1 of Theorem 5: மேர்கண்ட டோரசில் முழு அகராதி பிரதிபலிக்கப்பட்டதால், இதனைக்க்கொண்டு ஒரு சொற்பிழை திருத்தி செய்யலாம். பிழையான் சொல்லின் திருத்தம், அதன் நெருங்கிய தொலைவில் உள்ள சரியான் சொல் என்பதை நடைமுரைவிதியாகக்கொண்டு இதனை அமல்படுத்தலாம்.
Theorem 6: Tries எனப்படும் சொல்மரங்களைக்கொண்ட தரவமைப்பை டோரசில் குறியிட்டால், அது தொடர்பாதையாக ஒரே தொடக்கமும், பல பாதைமுடிவுகளையும் கொண்டதாக அமையும். இவற்றில் சில பாதைகள் சேரும் வகையில் முடிவுபெரும் வகையிலும் அமையலாம்.
உதாரணத்திற்கு, படம் 2-இல் முடியும் நிலை நுனிகள் ‘n’ என்பவை டோரசில் வரும்பொழுது சேரும் வகையில் முடிவுபெரும் வகையில் அமையும்.
-முத்து. | 4,011 | 5,771 | {"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": 2, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2022-49 | latest | en | 0.805958 |
https://secretlondon.us/6-litres-to-quarts | 1,696,380,332,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511284.37/warc/CC-MAIN-20231003224357-20231004014357-00836.warc.gz | 555,656,500 | 18,718 | # 6 Litres To Quarts
6 Litres To Quarts – In this article, we’ll cover how many ounces and cups are in a pint, quart, and gallon, and all the other basic kitchen conversion measurements you need to know. We will also explain the conversion from the US system to the metric system.
Have you ever found yourself excited to try a new recipe only to discover when you start that all the measurements are in metric system units?
## 6 Litres To Quarts
Or, did you forget the basic conversion of kitchen measurements to the US system you learned in school? Yes me too
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When it comes to baking and cooking, you’re often working with US measurements and conversions to metric units of grams, kilograms, and liters.
Keep this chart handy to easily convert between teaspoons and tablespoons. This comes in the form of cooking, accurately dispensing medicine for children, and many other tasks. Especially in teaspoons or teaspoons mL.
This is a handy resource when doubling recipes or converting recipes from the metric system to the US measurement system.
You’ll often see a recipe call for a specific pan size. If you are using a US cookbook, those pan sizes will be in US measurements. For everything else, you need to be familiar with the metric system pen scale. See below for inches to centimeters conversions for common pan sizes:
## Kitchen Equivalents Chart
You can instantly convert fluid ounces to tablespoons by multiplying ounces by 2. To find teaspoons in ounces, multiply the number of ounces by 6. You can learn more about how many ounces are in a cup in our article.
Owning a digital kitchen scale equipped with both US and metric system unit readouts will save you time. If you have a kitchen scale, there is no need to convert between the USCS and metric systems.
For example, you have a great recipe for Nutella cookies, but all measurements are in the metric system.
Instead of converting each of these amounts to USCS, I can just pull out my kitchen scale. Using my scale, I can measure all kinds of ingredients in grams and liters! Although I don’t need to know how many cups are in a pint, it is useful in several ways.
## Millimeters To Liters: Learn Definition, Facts And Examples
Quick Pot Steak Recipes You Need To Try ← Previous Can Dogs Eat People? Which common foods are safe and which are toxic? Next →
Is there a specific name for kitchen scales that measure both grams and liters? When I was a kid in school, teachers didn’t even discuss American kitchen measurements. LOL, now you know how old I am! Only the United States and two other countries use the American scale. in the world; All others are in metric system.
Easy Crock Pot Recipes Perfect Egg Recipes Sous Vide Recipes Instant Pot Ready Eggs Copycat Recipes Summer Cucumber Salad Asian Inspired Recipes Fermented Chow Chow Party Dip Recipes Wondering how many cups, cheeks, or cups you need in a pan? Should try? Maybe you’re looking for quarts in gallons or pints in quarts? Here you’ll find a complete guide with conversion tables and free printables.
This can be a little tricky when you’re cooking and need to convert between cups, pints, quarts, or gallons.
#### Question Video: Converting Quarts To Liters
But don’t worry, I’ll show you an easy way to remember conversions. I’ve also made all the changes for you in a simple table below!
I know how hard it can be when you cook and your favorite recipe is written in imperial, but you know metric or vice versa. You may be wondering how to replace a cup of water or a pint of milk.
I’ve spent a lot of time looking at the same adaptations of my favorite recipes over and over again. So I thought it was time to write a blog post covering all these confusing conversions!
If you’re like me, have you suddenly racked your brains to remember how many cups are in a pint of milk?
#### Litre Coloured Oil Jugs
When you’re busy cooking, ounces can sometimes be confusing. So, I also have helpful guides if you’re wondering how many ounces are in a quart, how many ounces are in a pound (ounces to pounds), or how many ounces are in a gallon.
It may help to take a look at this guide to the difference between the metric and imperial systems.
So, let’s move on to change. I’m sure you’re still wondering, how many pints are in a quart? How many cups are in 1 pint? How many quarts in a gallon etc…
Want a free handy kitchen conversion chart? I’ve got you covered! This chart will help you with many cooking conversions. Enjoy!
## Laticrete 25lb Bag White Rapid Multi Thinset
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## New) 911/914 6 Engine Oil Level Decal
The well-curated wine list has a variety of options, allowing guests to explore and discover new favorites. Whether you’re a wine connoisseur or just appreciate a good glass of wine, Flight Wine Bar offers an array of options to satisfy every palate. An expertly paired wine and dinner menu creates an unforgettable dining experience.
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### Standard Measures And Conversions: Liquid Volume, Milliliters And Liters
The Flight Wine Bar also caters to customers who prefer take-out bookings. Patrons can easily place their order for take-out and enjoy a flight of wines.
Quarts to litres converter, conversion from quarts to litres, us quarts to litres, 8 litres to quarts, litres to quarts, 6 quarts in litres, 3.5 quarts to litres, 3 quarts to litres, convert 2 quarts to litres, conversion litres to quarts, convert quarts to litres, 2 quarts to litres | 1,957 | 9,700 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-40 | latest | en | 0.924321 |
https://en.wikipedia.org/wiki/Electron_cyclotron_resonance | 1,709,268,081,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474948.91/warc/CC-MAIN-20240301030138-20240301060138-00519.warc.gz | 229,756,342 | 19,993 | # Electron cyclotron resonance
Electron cyclotron resonance (ECR) is a phenomenon observed in plasma physics, condensed matter physics, and accelerator physics. It happens when the frequency of incident radiation coincides with the natural frequency of rotation of electrons in magnetic fields. A free electron in a static and uniform magnetic field will move in a circle due to the Lorentz force. The circular motion may be superimposed with a uniform axial motion, resulting in a helix, or with a uniform motion perpendicular to the field (e.g., in the presence of an electrical or gravitational field) resulting in a cycloid. The angular frequency (ω = 2πf ) of this cyclotron motion for a given magnetic field strength B is given (in SI units)[1] by
${\displaystyle \omega _{\text{ce}}={\frac {eB}{m_{\text{e}}}}}$.
where ${\displaystyle e}$ is the elementary charge and ${\displaystyle m}$ is the mass of the electron. For the commonly used microwave frequency 2.45 GHz and the bare electron charge and mass, the resonance condition is met when B = 875 G = 0.0875 T.
For particles of charge q, electron rest mass m0,e moving at relativistic speeds v, the formula needs to be adjusted according to the special theory of relativity to:
${\displaystyle \omega _{\text{ce}}={\frac {eB}{\gamma m_{0,{\text{e}}}}}}$
where
${\displaystyle \gamma ={\frac {1}{\sqrt {1-\left({\frac {v}{c}}\right)^{2}}}}}$.
## In plasma physics
An ionized plasma may be efficiently produced or heated by superimposing a static magnetic field and a high-frequency electromagnetic field at the electron cyclotron resonance frequency. In the toroidal magnetic fields used in magnetic fusion energy research, the magnetic field decreases with the major radius, so the location of the power deposition can be controlled within about a centimeter. Furthermore, the heating power can be rapidly modulated and is deposited directly into the electrons. These properties make electron cyclotron heating a very valuable research tool for energy transport studies. In addition to heating, electron cyclotron waves can be used to drive current. The inverse process of electron cyclotron emission can be used as a diagnostic of the radial electron temperature profile.
Example of cyclotron resonance between a charged particle and linearly polarized electric field (shown in green). The position vs. time (top panel) is shown as a red trace and the velocity vs. time (bottom panel) is shown as a blue trace. The background magnetic field is directed out towards the observer. Note that the circularly polarized example below assumes there is no Lorentz force due to the wave magnetic field acting on the charged particle. This is equivalent to saying that the charged particle's velocity orthogonal to the wave magnetic field is zero. Example of cyclotron resonance between a charged particle and circularly polarized electric field (shown in green). The position vs. time (top panel) is shown as a red trace and the velocity vs. time (bottom panel) is shown as a blue trace. The background magnetic field is directed out towards the observer. Note that the circularly polarized example below assumes there is no Lorentz force due to the wave magnetic field acting on the charged particle. This is equivalent to saying that the charged particle's velocity orthogonal to the wave magnetic field is zero.
## ECR ion sources
Since the early 1980s, following the award-winning pioneering work done by Dr. Richard Geller,[2] Dr. Claude Lyneis, and Dr. H. Postma;[3] respectively from French Atomic Energy Commission, Lawrence Berkeley National Laboratory and the Oak Ridge National Laboratory, the use of electron cyclotron resonance for efficient plasma generation, especially to obtain large numbers of multiply charged ions, has acquired a unique importance in various technological fields. Many diverse activities depend on electron cyclotron resonance technology, including
The ECR ion source makes use of the electron cyclotron resonance to ionize a plasma. Microwaves are injected into a volume at the frequency corresponding to the electron cyclotron resonance, defined by the magnetic field applied to a region inside the volume. The volume contains a low pressure gas. The alternating electric field of the microwaves is set to be synchronous with the gyration period of the free electrons of the gas, and increases their perpendicular kinetic energy. Subsequently, when the energized free electrons collide with the gas in the volume they can cause ionization if their kinetic energy is larger than the ionization energy of the atoms or molecules. The ions produced correspond to the gas type used, which may be pure, a compound, or vapor of a solid or liquid material.
ECR ion sources are able to produce singly charged ions with high intensities (e.g. H+ and D+ ions of more than 100 mA (electrical) in DC mode[5] using a 2.45 GHz ECR ion source).
For multiply charged ions, the ECR ion source has the advantages that it is able to confine the ions for long enough for multiple collisions and multiple ionization to take place, and the low gas pressure in the source avoids recombination. The VENUS ECR ion source at Lawrence Berkeley National Laboratory has produced in intensity of 0.25 mA (electrical) of Bi29+.[6]
Some important industrial fields would not exist without the use of this fundamental technology, which makes electron cyclotron resonance ion and plasma sources one of the enabling technologies of today's world.
## In condensed matter physics
Within a solid the mass in the cyclotron frequency equation above is replaced with the effective mass tensor ${\displaystyle m^{*}}$. Cyclotron resonance is therefore a useful technique to measure effective mass and Fermi surface cross-section in solids. In a sufficiently high magnetic field at low temperature in a relatively pure material
{\displaystyle {\begin{aligned}\omega _{\text{ce}}&>{\frac {1}{\tau }}\\\hbar {\omega }_{\text{ce}}&>k_{B}T\\\end{aligned}}}
where ${\displaystyle \tau }$ is the carrier scattering lifetime, ${\displaystyle k_{B}}$ is Boltzmann's constant and ${\displaystyle T}$ is temperature. When these conditions are satisfied, an electron will complete its cyclotron orbit without engaging in a collision, at which point it is said to be in a well-defined Landau level. | 1,368 | 6,355 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 10, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2024-10 | latest | en | 0.849456 |
http://fullhomework.com/nan/ | 1,611,682,801,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610704803308.89/warc/CC-MAIN-20210126170854-20210126200854-00544.warc.gz | 42,661,623 | 9,728 | # NaN
The problem to solve is as follows: use this data frame:exam_data = {‘name’: [‘Anastasia’, ‘Dima’, ‘Katherine’, ‘James’, ‘Emily’, ‘Michael’, ‘Matthew’, ‘Laura’, ‘Kevin’, ‘Jonas’],
‘score’: [12.5, 9, 16.5, np.nan, 9, 20, 14.5, np.nan, 8, 19],
‘attempts’: [1, 3, 2, 3, 2, 3, 1, 1, 2, 1],
‘qualify’: [‘yes’, ‘no’, ‘yes’, ‘no’, ‘no’, ‘yes’, ‘yes’, ‘no’, ‘no’, ‘yes’]}
labels = [‘a’, ‘b’, ‘c’, ‘d’, ‘e’, ‘f’, ‘g’, ‘h’, ‘i’, ‘j’]For this data frame, write the code in python using Pandas and answer the following questions:Get the first three rows of data
Select the ‘name’ and ‘score’ columns
Select ‘name’ and ‘score’ columns in rows 1, 3, 5, 6
select the rows where the number of attempts in the examination is greater than 2
count the number of rows and columns
select the rows where the score is missing, i.e. is NaN
select the rows the score is between 15 and 20 (inclusive)
select the rows where number of attempts in the examination is less than 2 and score greater than 15
calculate the sum of the examination attempts by the students
calculate the mean score for each different student | 360 | 1,096 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-04 | latest | en | 0.826721 |
https://www.askiitians.com/forums/Differential-Calculus/25/3929/domain.htm | 1,716,512,297,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058675.22/warc/CC-MAIN-20240523235012-20240524025012-00220.warc.gz | 564,384,591 | 42,721 | f(x)=root(log0.3(x-1)/x2-2x-8)
Ramesh V
70 Points
14 years ago
The domain of f(x) is :
necessary condition : (x-1)>0
the value under f(x) shld always be positive
case 1:
log0.3(x-1) > 0 and (x+2)(x-4) > 0
x-1 > 1 and (x+2)(x-4) > 0 and x>1
the overlapping region is x > 4
case 1:
log0.3(x-1) < 0 and (x+2)(x-4) < 0
x-1 < 1 and (x+2)(x-4) < 0
x<2 and -2<x<4 and x>1
the overlapping region is : (1 , 2)
the domain is : (1,2) U (4,infinity)
--
Regards
Ramesh | 226 | 478 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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 | latest | en | 0.731649 |
http://www.apphysicsresources.com/2011/03/ap-physics-c-answer-to-free-response.html | 1,721,045,586,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514696.4/warc/CC-MAIN-20240715102030-20240715132030-00850.warc.gz | 40,074,676 | 19,225 | ## Pages
`“Life is like riding a bicycle. To keep your balance you must keep moving.”–Albert Einstein`
## Sunday, March 13, 2011
### AP Physics C – Answer to Free Response Practice Question on Electrostatic Potential Energy
“Genius is one per cent inspiration and ninety nine per cent perspiration.”
Thomas A. Edison
A free response practice question on Electrostatic Potential Energy was given to you in the post dated 11th March 2011. As promised, I give below a model answer along with the question:
A thin circular ring of radius R has positive charges sprayed uniformly along it so that the linear charge density along it is λ. The ring is located in the y-z plane with its centre at the origin of a right handed Cartesian co-ordinate system, in a space lab, where the effect of gravity is negligible. From the point P [–(√15)R, 0, 0] shown in the figure, a particle of mass m and charge +q is projected with speed v along the positive x-direction. Now answer the following:
(a) Which one of the following statements (i), (ii) and (iii) regarding the electric field vector at P (due to the charges on the ring) is correct? [Put a tick (√) mark against your choice].
(i) Electric field at P is along the positive x-direction ……..
(ii) Electric field at P is along the negative x-direction ……..
(iii) Electric field at P is zero ……..
(b) Determine the electrostatic potential energy of the charge q when it is located at P.
(c) Determine the electrostatic potential energy of the charge q when it is located.at the centre of the ring.
(d) Determine the minimum value of the speed of projection for the charged particle to pass through the ring.
(e) Explain what happens to the charged particle if the speed of projection is less than the minimum value mentioned in part (d) above.
(a) (ii) Electric field at P is along the negative x-direction …√…..
Since the point P is on the axis of the charged ring, the electric field at P is directed along the axis. Since the charge on the ring is positive, the electric field is directed from the ring to the point P. So the field is along the negative x-direction.
The direction of the field due to the charge at a point A on the ring is directed along the line AP. This field can be resolved into rectangular components, one along the axis of the ring and the other normal to the axis. When we consider the electric field due to an equal charge situated at the diametrically opposite point B on the ring, we find that the normal components of the fields due to A and B are equal and opposite so that they get canceled. But the axial components are in the same direction (negative x-direction) and they get added up. This is why the field at P is along the negative x-direction.
(b) The total charge Q on the ring is given by
Q = 2πR λ.
The distance r of the point P from the charges is given by
r = [R2 + 15 R2]1/2 = 4R
Therefore, the electric potential V1 at the point P is given by
V1 = (1/4πε0)(Q/r) where ε0 is the permittivity of free space.
Substituting for Q and r, we have
V1 = (1/4πε0)( 2πRλ /4R) = λ/8ε0
The electrostatic potential energy E1 of the charge q when it is located at P is given by
E1 = V1q = λq/8ε0
(c) The electric potential V2 at the centre of the ring is given by
V2 = (1/4πε0)(Q/R) = (1/4πε0)( 2πRλ /R) = λ/8ε0 = λ/2ε0
The electrostatic potential energy E2 of the charge q when it is located at the centre of the ring is given by
E1 = V2q = λq/2ε0
(d) When the particle of mass m and charge +q is projected (along the axis) towards the ring, its potential energy increases until it reaches the centre of the ring. The increase in potential energy is at the cost of the kinetic energy of the particle. Once the particle just passes the centre of the ring, it can continue to move forward and gain kinetic energy at the cost of its potential energy.
If the initial kinetic energy with which the particle is projected from the point P is equal to the difference between its electrostatic potential energies at the points P and the centre of the ring, the particle will reach the centre of the ring with zero speed. Therefore, the minimum kinetic energy with which the charged particle is to be projected so as to pass through the ring is equal to the difference between its electrostatic potential energies at the points P and the centre of the ring. Therefore, we have
½ mv2 = E1 E2 where v is the minimum speed with which the charged particle is to be projected.
Substituting for E1 and E2 we have
½ mv2 = (λq/2ε0) – (λq/8ε0) = 3λq/8ε0
This gives v = (3λq/4ε0m)1/2
(e) If the speed of projection of the charged particle is less than the minimum value mentioned in part (d) above, it will move towards the ring with its speed decreasing gradually. Its kinetic energy will become zero before reaching the centre of the ring and it will momentarily come to rest. Under the action of the electric field directed along the negative x-direction, the particle will turn back and will retrace its path and will move past the point P all the way to infinite distance from the ring before coming to rest.
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Here is my web-site; x mas invites | 1,372 | 5,719 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2024-30 | latest | en | 0.940051 |
https://it.mathworks.com/matlabcentral/cody/problems/1475-chebyshev-polynomials-of-the-2nd-kind/solutions/3250558 | 1,606,615,878,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141195967.34/warc/CC-MAIN-20201129004335-20201129034335-00212.warc.gz | 343,942,616 | 17,649 | Cody
# Problem 1475. Chebyshev polynomials of the 2nd Kind
Solution 3250558
Submitted on 17 Oct 2020 by Alex
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
user_solution = fileread('chebyshev2ndKindPoly.m'); assert(isempty(strfind(user_solution,'regexp'))); assert(isempty(strfind(user_solution,'2str'))); assert(isempty(strfind(user_solution,'str2'))); assert(isempty(strfind(user_solution,'interp'))); assert(isempty(strfind(user_solution,'printf'))); assert(isempty(strfind(user_solution,'assert')));
2 Pass
n = 0; P_correct = [1]; assert(isequal(chebyshev2ndKindPoly(n),P_correct));
3 Pass
n = 1; P_correct = [2 0]; assert(isequal(chebyshev2ndKindPoly(n),P_correct));
4 Pass
n = 2; P_correct = [4 0 -1]; assert(isequal(chebyshev2ndKindPoly(n),P_correct));
5 Pass
n = 3; P_correct = [8 0 -4 0]; assert(isequal(chebyshev2ndKindPoly(n),P_correct));
6 Pass
n = 4; P_correct = [16 0 -12 0 1]; assert(isequal(chebyshev2ndKindPoly(n),P_correct));
7 Pass
n = 5; P_correct = [32 0 -32 0 6 0]; assert(isequal(chebyshev2ndKindPoly(n),P_correct));
8 Pass
n = 6; P_correct = [64 0 -80 0 24 0 -1]; assert(isequal(chebyshev2ndKindPoly(n),P_correct));
9 Pass
n = 7; P_correct = [128 0 -192 0 80 0 -8 0]; assert(isequal(chebyshev2ndKindPoly(n),P_correct));
10 Pass
n = 8; P_correct = [256 0 -448 0 240 0 -40 0 1]; assert(isequal(chebyshev2ndKindPoly(n),P_correct));
11 Pass
n = 9; P_correct = [512 0 -1024 0 672 0 -160 0 10 0]; assert(isequal(chebyshev2ndKindPoly(n),P_correct));
12 Pass
n = 10; P_correct = [1024 0 -2304 0 1792 0 -560 0 60 0 -1]; assert(isequal(chebyshev2ndKindPoly(n),P_correct));
13 Pass
n = 11; P_correct = [2048 0 -5120 0 4608 0 -1792 0 280 0 -12 0]; assert(isequal(chebyshev2ndKindPoly(n),P_correct));
14 Pass
n = 12; P_correct = [4096 0 -11264 0 11520 0 -5376 0 1120 0 -84 0 1]; assert(isequal(chebyshev2ndKindPoly(n),P_correct));
15 Pass
n = 13; P_correct = [8192 0 -24576 0 28160 0 -15360 0 4032 0 -448 0 14 0]; assert(isequal(chebyshev2ndKindPoly(n),P_correct));
16 Pass
n = 14; P_correct = [16384 0 -53248 0 67584 0 -42240 0 13440 0 -2016 0 112 0 -1]; assert(isequal(chebyshev2ndKindPoly(n),P_correct));
17 Pass
n = 15; P_correct = [32768 0 -114688 0 159744 0 -112640 0 42240 0 -8064 0 672 0 -16 0]; assert(isequal(chebyshev2ndKindPoly(n),P_correct));
### Community Treasure Hunt
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Start Hunting! | 983 | 2,566 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-50 | latest | en | 0.31234 |
https://physicsinventions.com/force-on-electron-due-to-two-charged-objects-in-an-electric-field/ | 1,618,733,395,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038469494.59/warc/CC-MAIN-20210418073623-20210418103623-00572.warc.gz | 556,040,779 | 9,115 | # Force on Electron Due to Two Charged Objects in an Electric Field
1. The problem statement, all variables and given/known data
Two charged objects are separated by a distance of L=0.4650 m as shown in the diagram.
Object Q1 has a charge of +5.250 nC. Object Q2 has a charge of +2.960 nC.
A) What is the magnitude and direction of the electric field at Point A, which is located a distance d=0.09300 m to the right of Object Q1?
B) If you were to place an electron at Point A, what would be the magnitude and direction of the force on the electron?
2. Relevant equations
E= k (q/r^2)
F= (k*Q1*Q2)/ (r^2)
3. The attempt at a solution
A) [ (8.99E9) ((5.25E-9)/(.09300^2) ] – [ (8.99E9) ((2.96E-9)/(.372^2) ]=
= 5265 N/C = 5.265E3 N/C to the right THIS ANSWER IS CORRECT
B) The charge on an electron is -1.6E-19. I am also assuming that i will need Coulombs Law to solve.
[(8.99E9)(5.25E-9)(2.96E-9)] / (.4650) =.64610E-7 to the left
This is not the right answer and I’m not sure why. AM i even using the right formula?
Any help would be greatly appreciated. Thanks!
Attached Images
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https://sports.answers.com/Q/How_many_baseballs_used_per_game | 1,717,086,055,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971668873.95/warc/CC-MAIN-20240530145337-20240530175337-00677.warc.gz | 455,804,388 | 53,076 | 0
# How many baseballs used per game?
Updated: 9/27/2023
Wiki User
11y ago
Using an average life span of a Baseball in a typical game as 6 pitches and the average number of pitches being 275 per game. The number of balls used is 46. Factoring this number out to each team in the league playing 165 games and dividing by 2 since two teams play each other each game, the number of baseballs used during a typical MLB season would about 113,850 balls per year.
According to MLB, between five and six dozen are used per game (60-70 balls).
The home team has to have 90 new baseballs on hand for each game. That doesn't mean they'll use them all, but they must have them, just in case.
The average life of the ball is 6 pitches. "According to Ensley, every major league game begins with six dozen balls. He goes on to explain that the average life of a ball is six pitches. Since most major league games average between 250 and 300 pitches, that would put the ball count at about 40 or 50 balls used per game".
According to the MLB: "It says between five and six dozen baseballs are used during each baseball game, as many as 72 balls."
Each ball costs about \$3-dollars, so that works out to \$216-dollars worth of balls for each baseball game.
Using an average life span of a baseball in a typical game as 6 pitches and the average number of pitches being 275 per game. The number of balls used is 46. Factoring this number out to each team in the league playing 165 games and dividing by 2 since two teams play each other each game, the number of baseballs used during a typical MLB season would about 113,850 balls per year.
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more than 50 of them
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1 baseball is used in a game.
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Q: How many baseballs used per game?
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### How many baseballs go into the crowd per game?
it's different each game but the average life of a ball in the mlb is seven pitches
### How many baseballs are used in a season?
I believe my math is faulty in my first response below. The following is the workup for the estimate of total baseballs used in a Regular Season in the MLB:2010 average pitches per team per game.....146lifespan of a Baseball in pitches.........................7games in the Regular Season........................162number of teams in the league........................30balls per team per game..................................21 = 146 ÷ 7balls used in Reg Seas by team..................3,379 = 21 x 162balls used in Reg Seas by all teams......101,366 = 30 x 3,379more than anyone can count I'm sure, lol, but as a guess, I'll say around half a million if not moreIt has been calculated that the average life-span of a baseball is 7 pitches. The average pitch count in a game is 285. So a good guess is about 41 baseballs to a game. There are 162 games in a season, so almost 6,596 per team, per year. There are 30 teams, so that's approximately 197,871.43 baseballs used per season. This only accounts for regular season games.This number does not include pre-season games, the All Star break and the post-season games, including the World Series.
### How many baseballs will be used in a single MLB season?
Major League Baseball uses about 160,000 baseballs in a season. the lifespan of 1 is 7 pitches. the average baseball lasts 9 pitches. That's as close as I can come. OK. Let's say the average baseball lasts 9 pitches. Let's also say the average inning lasts 30 pitches (15 per half inning). We'll approximate the total number of pitches per game at 270. That would make 30 baseballs used per game (270 / 9). If all 30 teams played the entire 162 scheduled games in a season, that would be 2,430 games played. So for this example, that would make 72,900 baseballs used in one season.
### How many baseballs are made for one season of baseball?
Using an average life span of a baseball in a typical game as 6 pitches and the average number of pitches being 275 per game. The number of balls used is 46. Factoring this number out to each team in the league playing 165 games and dividing by 2 since two teams play each other each game, the number of baseballs used during a typical MLB season would about 113,850 balls per year. According to MLB, between five and six dozen are used per game (60-70 balls).
### How many baseballs fit in a hand?
By my calculations, i estimate that for a standard four door sedan, you could fit anywhere between 22,500 an up to 30,000 to . golf balls. That is of course depending on whether or not all of the cavities are utilised.
\$170 per dozen.
### Do home plate umpires still rub baseballs before games how many and with what?
There was an episode of "Dirty Jobs" on the Discover Chanel about this. It is not the umpire that does it. It is a person that works for the home team of any given game. It is a very particular type of natural mud but i do not remember where it comes from. I have heard they prepare about 60 balls per game.
### How many baseballs are used in a season of Major League Baseball?
The number is actually, as of 2005, over 35,000 baseballs per MLB team, on the low end. Of course this depends on the team itself and it can vary with each team and stadium. Since over 2 million baseballs are now produced annually by Rawlings for the Majors and retail, that number could be closer to 55,000 this season.This number includes Spring Training balls, batting practice balls and game balls, and it comes from a very reliable source, a team equipment manager (see article in links).Update - OK I don't see the links section, it was printed in an article about the Pirates in 2005.AnswerHere are the numbersTotal MLB Teamd: 30 Games(regular season 162 Balls (avg per game) 40That gives you a total of 194400 baseballs used in a regular seasonAnsweryou've to divide that by 2. So the answer is 97,200 balls on average. 2 teams playing a game means 1 game each but not 2 games overall
734 per game
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### What does RPG mean as used in basketball?
rebounds per game | 1,538 | 6,271 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2024-22 | latest | en | 0.971346 |
https://abcsofdatascience.ca/blog/j-is-for-jaccard | 1,660,208,216,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571246.56/warc/CC-MAIN-20220811073058-20220811103058-00448.warc.gz | 107,537,035 | 5,688 | In previous blogs we’ve talked about choosing a distance measure as a way of saying “these two things are close if …”. One useful measure is Jaccard similarity/distance since it measures the similarity between two sets. This is useful if you have a lot of categorical variables (i.e. ones that don’t have any inherent ordering). For example, two people are probably similar if they have the same sets of interests/hobbies. The Jaccard similarity of two sets is just the size of the intersection divided by the size of the union. Or put visually:
### Jaccard similarity/distance
As a more concrete example let’s imagine that we have collected a list of people’s favourite pizza toppings and we want to find which people are most similar.
• Kara likes pepperoni, mushrooms, and green pepper
• Zach likes ham, pineapple, and jalapeno peppers
• Rodney also likes pepperoni, mushrooms, and green peppers
• Sophie likes olives, ham, pepperoni, and mushrooms
So what are the Jaccard similarities between these people?
• Kara and Rodney like exactly the same toppings so their similarity is 1
• Rodney and Zach have nothing in common so their similarity is 0
• Kara and Sophie have some things in common but Sophie enjoys more toppings. Their similarity is 0.4 ([pepperoni, mushrooms]/[pepperoni, mushrooms, olives, ham])
You’ll notice that two sets that are exactly the same have a similarity of 1. To change this into a distance we just to
$D_{Jaccard} = 1 - similarity$
Now Kara and Rodney have a distance of 0, while Rodney and Zach are a distance of 1 (which in this case is as far apart as you can be).
### Hellinger distance
You might have noticed that Jaccard distance doesn’t take into account how frequently the items in the set occur. If counts do matter for your problem, then you will want to use Hellinger distance. Let’s imagine you have some data on the number of times people have read a given book:
Harry Potter Hello World The Hobbit The Great Gatsby Trick Mirror Marie 5 0 1 1 0 Jordan 1 0 1 1 1 Sarah 0 2 0 1 1 Patrick 4 0 2 1 0
Marie, Jordan, and Patrick have all read Harry Potter which would make them similar under Jaccard similarity. However, Marie and Patrick are probably more similar since they both read it multiple times. Hellinger distance takes this into account. I’ll try to give you some intuition for how it does this.
Imagine that each set of counts is generated by some weighted multi-sided die (that is different for each person). When we roll Maries die, it is more likely to come up with Harry Potter and less likely to come up with Hello World. The opposite is true for Sarah’s die, which is more likely to come up with Hello World and less likely to come up with Harry Potter. We calculate what the weights of these dice look like (these are called multinomial distributions).
We can then measure the mutual likelihood of these distributions. This just means “what is the probability of Jordans counts occuring using Maries die (and vice versa)”. If there is a high probability that Jordans counts occurred using Maries die, then Marie and Jordan should be considered close. If it is unlikely that Sarah’s counts occurred using Maries die, then Marie and Sarah should be pushed apart.
Hellinger distance is particularly useful if you have a bunch of text. You can consider two documents similar if you have the same words occurring at similar frequencies.
### Summary
Jaccard and Hellinger are both very useful distance measures that can be used in dimension reduction and embeddings. If counts matter, use Hellinger, otherwise use Jaccard distance. | 801 | 3,602 | {"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} | 3.84375 | 4 | CC-MAIN-2022-33 | latest | en | 0.930678 |
https://www.coursehero.com/tutors-problems/Math/5551-I-got-problem-4-but-none-of-the-rest/ | 1,548,152,003,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583835626.56/warc/CC-MAIN-20190122095409-20190122121409-00497.warc.gz | 734,050,034 | 49,943 | View the step-by-step solution to:
# CSORE4010 HOMEWORK 6 due on March 23 at the start of the class 1. Let G be a digraph and for each edge e let (e) 0 be an integer, so that for every...
I got problem 4, but none of the rest.
CSORE4010— HOMEWORK 6 due on March 23 at the start of the class 1. Let G be a digraph and for each edge e let φ ( e ) 0 be an integer, so that for every vertex v , s e δ - ( v ) φ ( e ) = s e δ + ( v ) φ ( e ) Show there is a list C 1 ,...,C n of directed cycles (possibly with repeti- tion) so that for every edge e of G , |{ i : 1 i n, e E ( C i ) }| = φ ( e ) . 2. Let s,t be distinct vertices of a digraph G , and let c : E ( G ) Z + . Let φ be a c -admissible s - t Fow of value k and suppose there is a c -admissible Fow with value > k . Does it follow that there is a c - admissible Fow ψ of value > k so that φ ( e ) ψ ( e ) for every edge e ? 3. Let s,t be vertices of a digraph G , and let φ : E ( G ) IR + be an s - t Fow. Show that there is an s - t Fow ψ : E ( G ) ZZ + so that (a) its value is at least that of φ , and (b) | ψ ( e ) - φ ( e ) | < 1 for every edge e of G . 4. Let T be a tree. Show that T has a perfect matching if and only if for every vertex v , exactly one of the components of T \ v has an odd number of vertices. [Hint for “if”: use induction on | V ( T ) | , ±nd a leaf with a neighbour of degree 2, and delete them both.] 1
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Browse Documents | 532 | 1,772 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2019-04 | latest | en | 0.891955 |
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