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http://www.ee.ucla.edu/~panchap/ee102sp/node13.html | 1,369,144,592,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368700074077/warc/CC-MAIN-20130516102754-00084-ip-10-60-113-184.ec2.internal.warc.gz | 433,246,199 | 3,580 | # EE102 Systems and Signals (Discussion)
Course description
For comments write to me by e-mail .
These pages are probably always incomplete, but I try to update them often.
Next: Finding Fourier Series Coefficients Up: ee102 Previous: Examples
# Laplace Transforms of signals that begin at a negative time
In some problems (for example Problem 4.10 from the text), the periodic signal of interest is defined naturally over instead of as we saw in the examples above. In that case, it would be more convenient to define our as
But in Chapter 3, the Laplace Transform was defined as:
Definition 1:
So we need to modify the definition to take into account the fact that our signal begins at instead of 0.
What we do is define our Laplace Transform as:
Definition 2:
where the lower limit is now instead of 0. In this way, we can work with a different initial time (for example )
For illustration, consider .
Laplace Transforms of signals which begin at 0 are the same for both definitions. For example, the Laplace Transform of is according to both definitions.
But with definition (2), we can also analyze signals which begin at time instead of 0. For example, if we use definition (1), would have the same Laplace Transform as , since the lower limit 0 in definition (1) ignores the signal between and 0. But if we use definition (2), the Laplace Transform of is .
In fact, the the delay property
also holds for negative time-shifts of with when we use definition (2).
Our choice of in the lower limit of the definition of the Laplace Transform was used just to illustrate the case when the signals of interest began at instead of 0. In any problem, if we encounter signals which can be expressed as , the Laplace Transform can be taken to be
where .
The only case where the Laplace Transform cannot be directly used is when the signals go all the way back to . In that case, there is no delay that can make the signals 0 for .
Subsections
Next: Finding Fourier Series Coefficients Up: ee102 Previous: Examples
visitors since January 7., 2002. -
Sankaran Panchapagesan
2002-06-05 | 479 | 2,100 | {"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} | 3.265625 | 3 | CC-MAIN-2013-20 | latest | en | 0.925123 |
http://slidegur.com/doc/4041/chapter-5-thermochemistry | 1,481,263,668,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542686.84/warc/CC-MAIN-20161202170902-00290-ip-10-31-129-80.ec2.internal.warc.gz | 255,774,470 | 10,643 | Chapter 5 Thermochemistry
```Chemistry, The Central Science, 10th edition
Theodore L. Brown; H. Eugene LeMay, Jr.;
and Bruce E. Bursten
Chapter 5
Thermochemistry
John D. Bookstaver
St. Charles Community College
St. Peters, MO
2006, Prentice Hall, Inc.
Thermochemistry
Energy
• The ability to do work or transfer heat.
Work: Energy used to cause an object that
has mass to move.
Heat: Energy used to cause the
temperature of an object to rise.
Thermochemistry
Potential Energy
Energy an object possesses by virtue of its
position or chemical composition.
Thermochemistry
Kinetic Energy
Energy an object possesses by virtue of its
motion.
1
KE = mv2
2
Thermochemistry
Units of Energy
• The SI unit of energy is the joule (J).
kg m2
1 J = 1
s2
• An older, non-SI unit is still in
1 cal = 4.184 J
Thermochemistry
System and Surroundings
• The system includes
the molecules we want
to study (here, the
hydrogen and oxygen
molecules).
• The surroundings are
everything else (here,
the cylinder and
piston).
Thermochemistry
Work
• Energy used to
move an object over
some distance.
• w = F d,
where w is work, F
is the force, and d is
the distance over
which the force is
exerted.
Thermochemistry
Heat
• Energy can also be
transferred as heat.
• Heat flows from
warmer objects to
cooler objects.
Thermochemistry
Transferal of Energy
a) The potential energy of this ball of
clay is increased when it is moved
from the ground to the top of the wall.
Thermochemistry
Transferal of Energy
a) The potential energy of this ball of
clay is increased when it is moved
from the ground to the top of the wall.
b) As the ball falls, its potential energy is
converted to kinetic energy.
Thermochemistry
Transferal of Energy
a) The potential energy of this ball of
clay is increased when it is moved
from the ground to the top of the wall.
b) As the ball falls, its potential energy is
converted to kinetic energy.
c) When it hits the ground, its kinetic
energy falls to zero (since it is no
longer moving); some of the energy
does work on the ball, the rest is
dissipated as heat.
Thermochemistry
First Law of Thermodynamics
• Energy is neither created nor destroyed.
• In other words, the total energy of the universe is
a constant; if the system loses energy, it must be
gained by the surroundings, and vice versa.
Use Fig. 5.5
Thermochemistry
Internal Energy
The internal energy of a system is the sum of all
kinetic and potential energies of all components
of the system; we call it E.
Use Fig. 5.5
Thermochemistry
Internal Energy
By definition, the change in internal energy, E,
is the final energy of the system minus the initial
energy of the system:
E = Efinal − Einitial
Use Fig. 5.5
Thermochemistry
Changes in Internal Energy
• If E > 0, Efinal > Einitial
Therefore, the system
absorbed energy from
the surroundings.
This energy change is
called endergonic.
Thermochemistry
Changes in Internal Energy
• If E < 0, Efinal < Einitial
Therefore, the system
released energy to the
surroundings.
This energy change is
called exergonic.
Thermochemistry
Changes in Internal Energy
• When energy is
exchanged between
the system and the
surroundings, it is
exchanged as either
heat (q) or work (w).
• That is, E = q + w.
Thermochemistry
E, q, w, and Their Signs
Thermochemistry
Exchange of Heat between
System and Surroundings
• When heat is absorbed by the system from
the surroundings, the process is endothermic.
Thermochemistry
Exchange of Heat between
System and Surroundings
• When heat is absorbed by the system from
the surroundings, the process is endothermic.
• When heat is released by the system to the
surroundings, the process is exothermic.
Thermochemistry
State Functions
Usually we have no way of knowing the
internal energy of a system; finding that value
is simply too complex a problem.
Thermochemistry
State Functions
• However, we do know that the internal energy
of a system is independent of the path by
which the system achieved that state.
In the system below, the water could have reached
room temperature from either direction.
Thermochemistry
State Functions
• Therefore, internal energy is a state function.
• It depends only on the present state of the
system, not on the path by which the system
arrived at that state.
• And so, E depends only on Einitial and Efinal.
Thermochemistry
State Functions
• However, q and w are
not state functions.
• Whether the battery is
shorted out or is
discharged by running
the fan, its E is the
same.
But q and w are different
in the two cases.
Thermochemistry
Work
When a process
occurs in an open
container, commonly
the only work done is a
change in volume of a
gas pushing on the
surroundings (or being
pushed on by the
surroundings).
Thermochemistry
Work
We can measure the work done by the gas if
the reaction is done in a vessel that has been
fitted with a piston.
w = −PV
Thermochemistry
Enthalpy
• If a process takes place at constant
pressure (as the majority of processes we
study do) and the only work done is this
pressure-volume work, we can account for
heat flow during the process by measuring
the enthalpy of the system.
• Enthalpy is the internal energy plus the
product of pressure and volume:
H = E + PV
Thermochemistry
Enthalpy
• When the system changes at constant
pressure, the change in enthalpy, H, is
H = (E + PV)
• This can be written
H = E + PV
Thermochemistry
Enthalpy
• Since E = q + w and w = −PV, we
can substitute these into the enthalpy
expression:
H = E + PV
H = (q+w) − w
H = q
• So, at constant pressure the change in
enthalpy is the heat gained or lost.
Thermochemistry
Endothermicity and
Exothermicity
• A process is
endothermic, then,
when H is
positive.
Thermochemistry
Endothermicity and
Exothermicity
• A process is
endothermic when
H is positive.
• A process is
exothermic when
H is negative.
Thermochemistry
Enthalpies of Reaction
The change in
enthalpy, H, is the
enthalpy of the
products minus the
enthalpy of the
reactants:
H = Hproducts − Hreactants
Thermochemistry
Enthalpies of Reaction
This quantity, H, is called the enthalpy of
reaction, or the heat of reaction.
Thermochemistry
1. Enthalpy is an extensive property.
2. H for a reaction in the forward
direction is equal in size, but opposite
in sign, to H for the reverse reaction.
3. H for a reaction depends on the state
of the products and the state of the
reactants.
Thermochemistry
Calorimetry
Since we cannot
know the exact
enthalpy of the
reactants and
products, we
measure H through
calorimetry, the
measurement of
heat flow.
Thermochemistry
Heat Capacity and Specific Heat
• The amount of energy required to raise
the temperature of a substance by 1 K
(1C) is its heat capacity.
• We define specific heat capacity (or
simply specific heat) as the amount of
energy required to raise the temperature
of 1 g of a substance by 1 K.
Thermochemistry
Heat Capacity and Specific Heat
Specific heat, then, is
heat transferred
Specific heat =
mass temperature change
s=
q
m T
Thermochemistry
Constant Pressure Calorimetry
By carrying out a
reaction in aqueous
solution in a simple
calorimeter such as this
one, one can indirectly
measure the heat
change for the system
by measuring the heat
change for the water in
the calorimeter.
Thermochemistry
Constant Pressure Calorimetry
Because the specific
heat for water is well
known (4.184 J/mol-K),
we can measure H for
the reaction with this
equation:
q = m s T
Thermochemistry
Bomb Calorimetry
Reactions can be
carried out in a
sealed “bomb,” such
as this one, and
measure the heat
absorbed by the
water.
Thermochemistry
Bomb Calorimetry
• Because the volume
in the bomb
calorimeter is
constant, what is
measured is really the
change in internal
energy, E, not H.
• For most reactions,
the difference is very
small.
Thermochemistry
Bomb Calorimetry
Thermochemistry
Hess’s Law
H is well known for many reactions,
and it is inconvenient to measure H
for every reaction in which we are
interested.
• However, we can estimate H using
H values that are published and the
properties of enthalpy.
Thermochemistry
Hess’s Law
Hess’s law states that
“If a reaction is carried
out in a series of
steps, H for the
overall reaction will be
equal to the sum of
the enthalpy changes
for the individual
steps.”
Thermochemistry
Hess’s Law
Because H is a state
function, the total
enthalpy change
depends only on the
initial state of the
reactants and the final
state of the products.
Thermochemistry
Enthalpies of Formation
An enthalpy of formation, Hf, is defined
as the enthalpy change for the reaction
in which a compound is made from its
constituent elements in their elemental
forms.
Thermochemistry
Standard Enthalpies of Formation
Standard enthalpies of formation, Hf, are
measured under standard conditions (25°C
and 1.00 atm pressure).
Thermochemistry
Calculation of H
C3H8 (g) + 5 O2 (g) 3 CO2 (g) + 4 H2O (l)
• Imagine this as occurring
in 3 steps:
C3H8 (g) 3 C(graphite) + 4 H2 (g)
3 C(graphite) + 3 O2 (g) 3 CO2 (g)
4 H2 (g) + 2 O2 (g) 4 H2O (l)
Thermochemistry
Calculation of H
C3H8 (g) + 5 O2 (g) 3 CO2 (g) + 4 H2O (l)
• Imagine this as occurring
in 3 steps:
C3H8 (g) 3 C(graphite) + 4 H2 (g)
3 C(graphite) + 3 O2 (g) 3 CO2 (g)
4 H2 (g) + 2 O2 (g) 4 H2O (l)
Thermochemistry
Calculation of H
C3H8 (g) + 5 O2 (g) 3 CO2 (g) + 4 H2O (l)
• Imagine this as occurring
in 3 steps:
C3H8 (g) 3 C(graphite) + 4 H2 (g)
3 C(graphite) + 3 O2 (g) 3 CO2 (g)
4 H2 (g) + 2 O2 (g) 4 H2O (l)
Thermochemistry
Calculation of H
C3H8 (g) + 5 O2 (g) 3 CO2 (g) + 4 H2O (l)
• The sum of these
equations is:
C3H8 (g) 3 C(graphite) + 4 H2 (g)
3 C(graphite) + 3 O2 (g) 3 CO2 (g)
4 H2 (g) + 2 O2 (g) 4 H2O (l)
C3H8 (g) + 5 O2 (g) 3 CO2 (g) + 4 H2O (l)
Thermochemistry
Calculation of H
We can use Hess’s law in this way:
H = nHf(products)
- mHf(reactants)
where n and m are the stoichiometric
coefficients.
Thermochemistry
Calculation of H
C3H8 (g) + 5 O2 (g) 3 CO2 (g) + 4 H2O (l)
H =
=
=
=
[3(-393.5 kJ) + 4(-285.8 kJ)] - [1(-103.85 kJ) + 5(0 kJ)]
[(-1180.5 kJ) + (-1143.2 kJ)] - [(-103.85 kJ) + (0 kJ)]
(-2323.7 kJ) - (-103.85 kJ)
-2219.9 kJ
Thermochemistry
Energy in Foods
Most of the fuel in the
food we eat comes
from carbohydrates
and fats.
Thermochemistry
Fuels
The vast majority
of the energy
consumed in this
country comes
from fossil fuels.
Thermochemistry
``` | 3,372 | 10,393 | {"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} | 2.546875 | 3 | CC-MAIN-2016-50 | longest | en | 0.912111 |
https://math.stackexchange.com/questions/2964832/is-this-an-appropriate-proof-to-show-that-n-gamma-a-n-gamma-a | 1,558,499,657,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232256763.42/warc/CC-MAIN-20190522043027-20190522065027-00286.warc.gz | 550,353,610 | 32,733 | Is this an appropriate proof to show that $n(\gamma,a)=-n(-\gamma,a)$
Given the closed rectifiable curve $$\gamma:[0,1]\rightarrow \Bbb C$$. we define $$-\gamma(t)=\gamma(1-t)$$.
I want to prove that $$n(\gamma,a)=-n(-\gamma,a).$$, where n is the winding number.
Is the following an appropriate proof ? :
I started by using the definition of the winding number
$$n(\gamma,a)=\tfrac{1}{2\pi i}\int_{\gamma}\tfrac{dz}{z-a}=\tfrac{1}{2\pi i}\int_0^{2\pi}\tfrac{\gamma'(t)}{\gamma(t)-a}dt=\tfrac{1}{2 \pi i }\int_0^{2\pi}\tfrac{rine^{int}}{re^{int}}dt=n$$
(this is obviously circular reasoning though. I think perhaps the following is proof enough )
$$-n(-\gamma,a)=-\tfrac{1}{2\pi i}\int_{-\gamma} \tfrac{dz}{z-a}=\tfrac{-1}{2\pi i}\int_0^{2\pi}\tfrac{-\gamma '(t) dt}{-\gamma(t)-a}$$
Now recall that $$-\gamma(t)=\gamma(1-t)=a+re^{in}e^{-int}$$. So
$$-n(-\gamma, a)=\tfrac{-1}{2 \pi i}\int_0^{2\pi}\tfrac{-ne^{in}e^{-int}ri}{re^{in}e^{-int}}dt=\tfrac{n}{2\pi} \int^{2 \pi}_0dt=n$$
therefore $$n(\gamma,a)=-n(-\gamma,a)=n$$
What do you guys think is this is a sufficient proof ?
(1) It seems that you only consider the special case $$\gamma : [0, 2\pi] \to \mathbb{C},\gamma(t) = a + e^{int}$$. But your proof can be adapted to work for any closed piecewise continuously differentiable $$\gamma : [a,b] \to \mathbb{C}$$. In fact, you have $$(-\gamma) = \gamma \circ\iota$$, where $$\iota : [a,b] \to [a,b], \iota(t) = a + b - t$$. Then the chain rule yields $$(-\gamma)'(t) = \gamma'(\iota(t))\iota'(t)$$. This shows $$\int_a^b\tfrac{\gamma'(t)}{\gamma(t)-a}dt = \int_{\iota(a)}^{\iota(b)}\tfrac{\gamma'(\iota(s))}{\gamma(\iota(s))-a}\iota'(s)ds = \int_{b}^{a}\tfrac{(-\gamma)'(s)}{(-\gamma)(s)-a}ds = - \int_{a}^{b}\tfrac{(-\gamma)'(s)}{(-\gamma)(s)-a}ds$$ which immediately implies $$n(\gamma,a) = - n(-\gamma,a) .$$
(2) You claim that $$n(\gamma,a) = - n(-\gamma,a)$$ for closed rectifiable curves (which are more general than closed piecewise continuously differentiable curves). This can easily be shown if you write down the definition of $$\int_\gamma \tfrac{dz}{z -a}$$ for such curves. | 789 | 2,103 | {"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": 17, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2019-22 | latest | en | 0.770179 |
https://harperandharley.org/pdf/data-structures-algorithms-in-kotlin-second-edition/ | 1,660,386,165,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571911.5/warc/CC-MAIN-20220813081639-20220813111639-00435.warc.gz | 311,637,681 | 11,905 | # Data Structures Algorithms in Kotlin Second Edition
Download or Read online Data Structures Algorithms in Kotlin Second Edition full book in PDF, ePub and kindle by raywenderlich Tutorial Team and published by Unknown which was released on 30 June 2021 with total pages null. We cannot guarantee that Data Structures Algorithms in Kotlin Second Edition book is available in the library.
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## Download or Read Online Data Structures Algorithms in Kotlin Second Edition in PDF, Epub and Kindle
Learn Data Structures & Algorithms in Kotlin!Data structures and algorithms are fundamental tools every developer should have. In this book, you'll learn how to implement key data structures in Kotlin, and how to use them to solve a robust set of algorithms.This book is for intermediate Kotlin or Android developers who already know the basics of the language and want to improve their knowledge.Topics Covered in This BookIntroduction to Kotlin: If you're new to Kotlin, you can learn the main constructs and begin writing code.Complexity: When you study algorithms, you need a way to compare their performance in time and space. Learn about the Big-O notation to help you do this.Elementary Data Structures: Learn how to implement Linked List, Stacks, and Queues in Kotlin.Trees: Learn everything you need about Trees - in particular, Binary Trees, AVL Trees, as well as Binary Search and much more.Sorting Algorithms: Sorting algorithms are critical for any developer. Learn to implement the main sorting algorithms, using the tools provided by Kotlin.Graphs: Have you ever heard of Dijkstra and the calculation of the shortest path between two different points? Learn about Graphs and how to use them to solve the most useful and important algorithms.
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### Android Development with Kotlin by Marcin Moskala,Igor Wojda
Learn how to make Android development much faster using a variety of Kotlin features, from basics to advanced, to write better quality code. About This Book Leverage specific features of Kotlin to ease Android application development Write code based on both object oriented and functional programming to build robust applications | 1,527 | 7,924 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2022-33 | latest | en | 0.866786 |
https://stackoverflow.com/questions/15545685/how-to-find-center-point-of-two-latitude-longitude?noredirect=1 | 1,571,665,194,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987773711.75/warc/CC-MAIN-20191021120639-20191021144139-00407.warc.gz | 724,944,128 | 27,256 | # How to find center point of two latitude longitude? [duplicate]
I have two latitude, longitude, now how i can find the center latitude,longitude of that two latitude longitude. Can anybody help me?
## marked as duplicate by Dipesh Parmar, hjpotter92, davidcesarino, madth3, doctorlessMar 22 '13 at 2:22
• Which projection are you using? – sectus Mar 21 '13 at 11:04
• This is quite simple just add the both lat and divide by 2 and similar to longitude add both and divide by 2. – Code Lღver Mar 21 '13 at 11:06
• You can use search for example ;) answer – Denis O. Mar 21 '13 at 11:06
Define what is 'center' for you. Mostly, i use simple average. Better solution is to compute two vectors (from center of the earth), add them and normalize result. Calculate the center point of multiple latitude/longitude coordinate pairs
Also, be careful about longitudes. The midpoint between two points at 170° E and 170° W should be at 180° E (or W), but you may end up with 0° E.
Download Map Projections: A Working Manual, by John P. Snyder, from the USGS. http://pubs.er.usgs.gov/publication/pp1395. It's free.
\$deltaLongitude = \$endPointLongitude - \$startPointlongitude;
\$xModified = cos(\$endPointLatitude) * cos(\$deltaLongitude);
\$yModified = cos(\$endPointLatitude) * sin(\$deltaLongitude);
\$midpointLatitude = atan2(
sin(\$startPointlatitude) + sin(\$endPointLatitude),
sqrt((cos(\$startPointLatitude) + \$xModified) * (cos(\$startPointLatitude) + \$xModified) +
\$yModified * \$yModified
)
);
\$midpointLongitude = \$startPointLongitude +
atan2(\$yModified,
cos(\$startPointLatitude) + \$xModified
);
• Clearly I'm wrong from the downvote: anybody care to explain so that I can learn from the gurus as well – Mark Baker Mar 21 '13 at 11:15
• +1 for having a stab at it. I think (without trying to get a headache) it is down to Riemannian circle - Now the heady duty maths come into play. – Ed Heal Mar 21 '13 at 11:23
• Thanks, looks like more reading (and heavy math) – Mark Baker Mar 21 '13 at 11:26
• Here is a reference en.wikipedia.org/wiki/Great_circles – Ed Heal Mar 21 '13 at 11:49 | 585 | 2,105 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2019-43 | latest | en | 0.769672 |
https://www.jiskha.com/display.cgi?id=1321565059 | 1,503,469,685,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886117874.26/warc/CC-MAIN-20170823055231-20170823075231-00216.warc.gz | 912,477,258 | 3,841 | # physics
posted by .
when you see a traffic light turn red you apply the brakes until you come to a stop. if your initial speed was 12 m/s, and you were heading heading due west, what was your average velocity during braking? Assume constant deceleration.
• physics -
help
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More Similar Questions | 637 | 2,440 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2017-34 | latest | en | 0.964398 |
http://www.eduplace.com/math/mw/background/1/11/te_1_11_intervals_ask.html | 1,508,457,314,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187823482.25/warc/CC-MAIN-20171019231858-20171020011858-00109.warc.gz | 448,910,964 | 2,068 | ## Finding Time Intervals: When Students Ask
• How do I tell the hour on a clock?
Use a model clock to point out the hour hand and minute hand. Remind children that the hour hand is shorter. Move the minute hand to each number on the clock, having children count aloud by fives. Point out that 60 minutes pass during one hour. Place the minute hand at 12, and then move the hour hand to different numbers. Read each time on the hour together, for example: seven o'clock, eight o'clock, nine o'clock.
• How do I write the time I see on a digital clock?
Use an analog clock to show 8:00. Ask children to explain how they know the time. Then write the number 8, pointing out that it names the hour. Write a colon and explain that it separates the hour from the minutes. Write two zeros, explaining that the zeros mean time exactly on the hour. Point out that the time on the analog clock—and the time that you wrote—both mean eight o'clock. Repeat with 8:30. Have children practice writing the corresponding times using numbers and colons, as you show these times on an analog clock.
• How do I tell how much time has passed?
Use an analog clock to show how much time has passed. Explain that a movie might begin at 7:00 and end at 9:00. Show 7:00 on the analog clock. Then move the minute hand and hour hand to show 8:00. Ask how many minutes and how many hours have passed. Then move the hands to show 9:00 and repeat the questions. Point out that two hours have passed since 7:00, so the movie lasts two hours. | 361 | 1,511 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-43 | latest | en | 0.931678 |
https://www.physicsforums.com/threads/magnetostatic-problems.43325/ | 1,548,114,537,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583822341.72/warc/CC-MAIN-20190121233709-20190122015709-00136.warc.gz | 904,382,380 | 12,743 | # Magnetostatic Problems
1. Sep 15, 2004
### kelvintc
1) Calculate the self-inductance of a coaxial cable of inner radius a and outer radius b. The inner conductor is made of an inhomogenous material having u = 2u0/(1+r)
Answer : u0L/8pie + u0L/pie (ln(b/a) - ln((1+b)/(1+a)))
2) Two #10 copper wires(2.588 mm in diameter) are placed parallel in air with a seperation distance d between them. If the inductance of each wire is 1.2uH/m, calculate
(a) L(in) and L(ext) per meter for each wire
(b) The seperation d
Answer : (a)0.05, 1.15 uH/m, (b) 40.79cm
3) A conductor of radius a is bent into a circular loop of mean radius r. If r = 10cm and 2a = 1cm, calculate the internal inductance of the loop.
Really can't the number 1 answer. I get a^3 in my answer. Number 2 no idea since L = L(in) + L(ext). As well as number 3.
Thank you.
#### Attached Files:
• ###### q3.JPG
File size:
4.1 KB
Views:
103
Last edited: Sep 16, 2004
2. Sep 16, 2004
### kelvintc
cannot use Latex code anymore?
3. Sep 16, 2004
### Tide
Apparently they changed servers and LaTeX broke but they're working on fixing it.
4. Sep 16, 2004
### kelvintc
anyone can help me on these magnetostatic problems? i'm stucked .. | 395 | 1,204 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-04 | latest | en | 0.852063 |
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Visitor
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Registered: 01-12-2018
## Buffer for Integer Multiplication with High Frequency in Automatically-Generated Verilog Code
Hi all,
I am confused by the generated Verilog for a simple multiplication in a pipelined for-loop.
When the clock period is 10ns, the latency of the multiplication is 8.47ns (within a cycle). When the clock period is 3ns, the latency of the multiplication turn to be 6 cycles plus 2.13ns (i.e. = 6*3+2.13=20.13ns).
The first question is why the latency of the operation is different?
Then I look into the generated Verilog for these two situations. For the situation with clock_period=10ns, the multiplication is simply a " * " operation. However, when clock period is 3ns, there is a shift buffer to store the result of multiplication for 4 extra cycles.
Therefore, the second question is why the buffer is necessary?
Thanks in advance for your explanation and suggestion!!! ^_^
Best regards,
-----------------------
Tingyuan
Tags (4)
1 Solution
Accepted Solutions
Xilinx Employee
574 Views
Registered: 01-09-2008
## Re: Buffer for Integer Multiplication with High Frequency in Automatically-Generated Verilog Code
Your data are 32 bit integers.
DSP48E1 (7-series) have 18x25 multipliers.
DSP48E2 (Ultrascale and Ultrascale+) have 18x27 multipliers.
Let's have a look to the latter (UG579) but this is almost the same process for the previous generations (back to Virtex-4).
On p13, you can see this mult followed by multiplexers. The last one contains 2 inputs with integrated 17-bit shift (to the right, loosing the LSBs). This allows you to implement wide multipliers using 1 or more DSP slice. Typically for a 32x32 bit multiplication:
A = A2:A1 with A on 32 bits, A1 on 17 bits and A2 on 15 bits
B = B2:B1 with B on 32 bits, B1 on 17 bits and B2 on 15 bits
In order to multiply A by B (M = AxB) you must operate in 4 main steps:
1. Tmp1 = (0:A1)x(0:B1)
• A 0 is prepended to the 17 bit MSBs in order to have positive numbers
• Tmp1 is on 18+18 = 36 bits (in a 48 bits container because P is a 48 bit register)
• M1 = LSB17(Tmp1) : the last 17 bits of Tmp1
2. Tmp2 = A2x(0:B1) + (Tmp1>>17)
3. Tmp2 = (0:A1)xB2 + Tmp2
• M2 = LSB17(Tmp2) : the last 17 bits of Tmp2
4. Tmp3 = A2xB2 + (Tmp2>>17)
• M3 = Tmp3
• M3 is on 15+15+1 bits
At the end: M = M3:M2:M1
This technique will use 4 DSP slices and no LUTs (just FF to store intermediate registers)
Now if the clock rate is lower, you can use less DSP slices. If you split B into 6+26 bits, then you can operate the first 2 stages using DSP slices, but the 2 others can be done in LUTs because B2 is pretty small.
1. Tmp1 = (0:A1)x((0:B1) : use DSP slice
2. Tmp2 = A2x(0:B1) + Tmp1>>17 : use DSP slice
3. Tmp3 = (0:A1)xB2
4. Tmp4 = A2xB2 + Tmp3>>17
Then there is a large adder at the end to ad the 2 partial results that will use a DSP slice (to achieve clock rate).
On p45-46-47 you will see some operations requiring 1 to 8 DSP slices
You can also have a look to some older UG. The one of the Virtex-5, UG193, describes what I have just done on p70.
Regards
Olivier
==================================
Olivier Trémois
XILINX EMEA DSP Specialist
7 Replies
Xilinx Employee
678 Views
Registered: 09-05-2018
## Re: Buffer for Integer Multiplication with High Frequency in Automatically-Generated Verilog Code
In order to meet the clock period requirements, Vivado HLS has to introduce pipeline buffers to break up the task into small pieces. When the tast is broken up into 6 pieces, HLS must introduce 6 loads and stores into registers, and this adds to the total latency of the IP. I unfortunately don't know enough about the details of C synthesis or the project in question to say why a shift buffer is required to store the operation of the multiplication across the last 4 cycles. But 2 cycles at 3ns is still shorter than 8.47ns of the original single cycle solution; I'd guess other calculations are still being done or the multiplication is still being accumulated.
Nicholas Moellers
Xilinx Worldwide Technical Support
Visitor
654 Views
Registered: 01-12-2018
## Re: Buffer for Integer Multiplication with High Frequency in Automatically-Generated Verilog Code
Dear Nicolas,
Thanks a lot for your prompt reply! I show the source code below:
=====================
for (i = 0; i < 100; i++)
data_out[i] = data_in[i] * data_in[i] + 1;
=====================
This problem is the same even without any directives: the overall latency of multiplication increases as the frequency increases.
I am confused because such multiplication is actually done by combination logic (since it takes 8.47ns, less than 1 10ns-cycle, to accomplish), and it is not necessary to break it up to 6 pieces when the frequency gets higher.
Moreover, another interesting phenomenon is that when the clock period is 10ns, only 3 DSPs are needed but 4 DSPs are needed when the clock period is 3ns.
Besides, I wonder whether there is any way for me to figure this practical technical problem, or how I can feedback such confusing situation to the development team?
Thanks again for your time and suggestion!
Best Regards,
-------------------------------------
Tingyuan LIANG
Visitor
646 Views
Registered: 01-12-2018
## Re: Buffer for Integer Multiplication with High Frequency in Automatically-Generated Verilog Code
Xilinx Employee
637 Views
Registered: 01-09-2008
## Re: Buffer for Integer Multiplication with High Frequency in Automatically-Generated Verilog Code
Hi Tingyuan
could you expose the datatype of your data_in and data_out arrays?
The estimated DSP# can be slightly off. You should push to the 'export' stage and 'evaluate' the generated RTL(synthesis, place and route) to have a much closer estimate.
Regards
Olivier TREMOIS
==================================
Olivier Trémois
XILINX EMEA DSP Specialist
Visitor
596 Views
Registered: 01-12-2018
## Re: Buffer for Integer Multiplication with High Frequency in Automatically-Generated Verilog Code
@oliviert
Hi Oliver,
Thanks a lot for your suggestions! The datatype is int (32-bit integer).
Actually, when I read the generated Verilog code, the HLS does instantiate extra DSPs in the source code.
It seem that when the frequency is high, the generated Verilog will do the multiplication in the way like A*B*1 instead of A*B.
Thanks again for your time and further explanation!
Best Regards,
---------------------------
Tingyuan
Xilinx Employee
575 Views
Registered: 01-09-2008
## Re: Buffer for Integer Multiplication with High Frequency in Automatically-Generated Verilog Code
Your data are 32 bit integers.
DSP48E1 (7-series) have 18x25 multipliers.
DSP48E2 (Ultrascale and Ultrascale+) have 18x27 multipliers.
Let's have a look to the latter (UG579) but this is almost the same process for the previous generations (back to Virtex-4).
On p13, you can see this mult followed by multiplexers. The last one contains 2 inputs with integrated 17-bit shift (to the right, loosing the LSBs). This allows you to implement wide multipliers using 1 or more DSP slice. Typically for a 32x32 bit multiplication:
A = A2:A1 with A on 32 bits, A1 on 17 bits and A2 on 15 bits
B = B2:B1 with B on 32 bits, B1 on 17 bits and B2 on 15 bits
In order to multiply A by B (M = AxB) you must operate in 4 main steps:
1. Tmp1 = (0:A1)x(0:B1)
• A 0 is prepended to the 17 bit MSBs in order to have positive numbers
• Tmp1 is on 18+18 = 36 bits (in a 48 bits container because P is a 48 bit register)
• M1 = LSB17(Tmp1) : the last 17 bits of Tmp1
2. Tmp2 = A2x(0:B1) + (Tmp1>>17)
3. Tmp2 = (0:A1)xB2 + Tmp2
• M2 = LSB17(Tmp2) : the last 17 bits of Tmp2
4. Tmp3 = A2xB2 + (Tmp2>>17)
• M3 = Tmp3
• M3 is on 15+15+1 bits
At the end: M = M3:M2:M1
This technique will use 4 DSP slices and no LUTs (just FF to store intermediate registers)
Now if the clock rate is lower, you can use less DSP slices. If you split B into 6+26 bits, then you can operate the first 2 stages using DSP slices, but the 2 others can be done in LUTs because B2 is pretty small.
1. Tmp1 = (0:A1)x((0:B1) : use DSP slice
2. Tmp2 = A2x(0:B1) + Tmp1>>17 : use DSP slice
3. Tmp3 = (0:A1)xB2
4. Tmp4 = A2xB2 + Tmp3>>17
Then there is a large adder at the end to ad the 2 partial results that will use a DSP slice (to achieve clock rate).
On p45-46-47 you will see some operations requiring 1 to 8 DSP slices
You can also have a look to some older UG. The one of the Virtex-5, UG193, describes what I have just done on p70.
Regards
Olivier
==================================
Olivier Trémois
XILINX EMEA DSP Specialist
Visitor
543 Views
Registered: 01-12-2018
## Re: Buffer for Integer Multiplication with High Frequency in Automatically-Generated Verilog Code
Dear Olivier,
Thanks a lot for your detailed explanation! I get it!
I may still have one more question: why the multiplication might be implemented in the way like A*B*1 in the generated Verilog?
Thanks again !
Best Regards,
------------------------------
Tingyuan | 2,564 | 9,324 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2019-39 | latest | en | 0.866215 |
https://www.coursehero.com/tutors-problems/Calculus/16576325-1-using-calculus/ | 1,563,866,889,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195529007.88/warc/CC-MAIN-20190723064353-20190723090353-00452.warc.gz | 666,104,804 | 23,362 | View the step-by-step solution to:
Question
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Browse Documents | 311 | 1,092 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2019-30 | latest | en | 0.702194 |
https://www.mercurial-scm.org/pipermail/mercurial-devel/2017-September/104298.html | 1,582,632,362,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875146066.89/warc/CC-MAIN-20200225110721-20200225140721-00510.warc.gz | 816,227,415 | 3,258 | # D656: revset: remove "small" argument from "_optimize"
quark (Jun Wu) phabricator at mercurial-scm.org
Thu Sep 7 12:16:24 EDT 2017
```quark created this revision.
REVISION SUMMARY
`_optimize` calculates weights of subtrees. "small" affects some weight
calculation (either 1 or 0). The weights are now only useful in `and`
optimization where we might swap two arguments and use `andsmally`.
In the real world, it seems unlikely that revsets with weight of 0.5 or 1
matters the `and` order optimization. I think the important thing is to get
weights of expensive revsets right (ex. `contains`).
This patch removes the `small` argument to simplify the interface.
REPOSITORY
rHG Mercurial
REVISION DETAIL
https://phab.mercurial-scm.org/D656
AFFECTED FILES
mercurial/revsetlang.py
CHANGE DETAILS
diff --git a/mercurial/revsetlang.py b/mercurial/revsetlang.py
--- a/mercurial/revsetlang.py
+++ b/mercurial/revsetlang.py
@@ -353,20 +353,16 @@
"""
return _analyze(x)
-def _optimize(x, small):
+def _optimize(x):
if x is None:
return 0, x
- smallbonus = 1
- if small:
- smallbonus = .5
-
op = x[0]
if op in ('string', 'symbol'):
- return smallbonus, x # single revisions are small
+ return 0.5, x # single revisions are small
elif op == 'and':
- wa, ta = _optimize(x[1], True)
- wb, tb = _optimize(x[2], True)
+ wa, ta = _optimize(x[1])
+ wb, tb = _optimize(x[2])
w = min(wa, wb)
# (draft/secret/_notpublic() & ::x) have a fast path
@@ -397,12 +393,12 @@
else:
s = '\0'.join(t[1] for w, t in ss)
y = _build('_list(_)', ('string', s))
- w, t = _optimize(y, False)
+ w, t = _optimize(y)
ws.append(w)
ts.append(t)
del ss[:]
for y in getlist(x[1]):
- w, t = _optimize(y, False)
+ w, t = _optimize(y)
if t is not None and (t[0] == 'string' or t[0] == 'symbol'):
ss.append((w, t))
continue
@@ -416,44 +412,44 @@
elif op == 'not':
# Optimize not public() to _notpublic() because we have a fast version
if _match('public()', x[1]):
- o = _optimize(_build('_notpublic()'), not small)
+ o = _optimize(_build('_notpublic()'))
return o[0], o[1]
else:
- o = _optimize(x[1], not small)
+ o = _optimize(x[1])
return o[0], (op, o[1])
elif op == 'rangeall':
- return smallbonus, x
+ return 1, x
elif op in ('rangepre', 'rangepost', 'parentpost'):
- o = _optimize(x[1], small)
+ o = _optimize(x[1])
return o[0], (op, o[1])
elif op in ('dagrange', 'range'):
- wa, ta = _optimize(x[1], small)
- wb, tb = _optimize(x[2], small)
+ wa, ta = _optimize(x[1])
+ wb, tb = _optimize(x[2])
return wa + wb, (op, ta, tb)
elif op in ('parent', 'ancestor', 'relation', 'subscript'):
- w, t = _optimize(x[1], small)
+ w, t = _optimize(x[1])
return w, (op, t, x[2])
elif op == 'relsubscript':
- w, t = _optimize(x[1], small)
+ w, t = _optimize(x[1])
return w, (op, t, x[2], x[3])
elif op == 'list':
- ws, ts = zip(*(_optimize(y, small) for y in x[1:]))
+ ws, ts = zip(*(_optimize(y) for y in x[1:]))
return sum(ws), (op,) + ts
elif op == 'keyvalue':
- w, t = _optimize(x[2], small)
+ w, t = _optimize(x[2])
return w, (op, x[1], t)
elif op == 'func':
f = getsymbol(x[1])
- wa, ta = _optimize(x[2], small)
+ wa, ta = _optimize(x[2])
if f in ('author', 'branch', 'closed', 'date', 'desc', 'file', 'grep',
'keyword', 'outgoing', 'user', 'destination'):
w = 10 # slow
elif f in ('modifies', 'adds', 'removes'):
w = 30 # slower
elif f == "contains":
w = 100 # very slow
elif f == "ancestor":
- w = 1 * smallbonus
+ w = 0.5
elif f in ('reverse', 'limit', 'first', 'wdir', '_intlist'):
w = 0
elif f == "sort":
@@ -468,7 +464,7 @@
All pseudo operations should be transformed beforehand.
"""
- _weight, newtree = _optimize(tree, small=True)
+ _weight, newtree = _optimize(tree)
return newtree
# the set of valid characters for the initial letter of symbols in
To: quark, #hg-reviewers
Cc: mercurial-devel
``` | 1,331 | 4,077 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2020-10 | latest | en | 0.527979 |
https://nla-group.org/category/publications/page/2/ | 1,726,373,543,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651614.9/warc/CC-MAIN-20240915020916-20240915050916-00790.warc.gz | 398,519,792 | 34,914 | ## Extreme-Scale Test Matrices with Specified 2-Norm Condition Number
The supercomputers in the TOP500 list are ranked using the High-Performance Linpack (HPL) Benchmark, which gauges the performance of a computer by solving a large dense linear system by Gaussian elimination with partial pivoting. The size of the coefficient matrix depends on the computational power of the machine being assessed, because more powerful systems require larger benchmark problems in order to reach their peak performance.
The test matrices used by HPL have random entries uniformly distributed on the interval $(-1/2,1/2]$. The 2-norm condition number of such matrices depends on their size, and can potentially be very large for the matrices required by today’s most powerful computers: the largest linear systems solved on Summit, the machines that leads the November 2019 TOP500 ranking, have order $10^7$, and even larger systems will be needed to benchmark the coming generations of exascale systems.
An $n \times n$ matrix with specified 2-norm condition number can be generated as $A := U \varSigma V^T$, where $U$ and $V$ are random real orthogonal $n \times n$ matrices from the Haar distribution (a natural uniform probability distribution on the orthogonal matrices) and $\varSigma$ is diagonal with nonnegative entries $\sigma_1\ge \cdots \ge \sigma_n \ge 0$. It is well known that $A$ has 2-norm condition number $\kappa_2(A) = \sigma_1/\sigma_n$ if $\sigma_n \neq 0$ and $\kappa(A) = \infty$ otherwise. This technique, which is used by the gallery('randsvd', ...) function, requires $2n^3$ floating-point operations to generate a test matrix of order $n$, and would thus be impractical in an extreme-scale setting.
In our recent EPrint Generating extreme-scale matrices with specified singular values or condition numbers, Nick Higham and I present four methods that, by giving up the requirement that the matrices $U$ and $V$ be from the Haar distribution, reduce the cost of the approach above from cubic to quadratic. The matrices generated retain a number of desirable properties that make them a suitable choice for testing linear solvers at scale.
These cheaper algorithms are particularly well suited to distributed-memory environments, since all communication between the nodes can be avoided at the price of a negligible increase in the overall computational cost. They are appealing to MATLAB users too, as the following example demostrates.
n = 10000; kappa = 1e6; mode = 2; rng(1)
% gallery('randsvd',...)
fprintf('gallery(''randsvd'',...): %5.2f seconds elapsed.\n',...
timeit(@()gallery('randsvd',n,kappa,mode,[],[],1)));
% Algorithm 3.1 in our EPrint.
method = 1; matrix = 0;
fprintf('Alg. 3.1 with Haar U: %5.2f seconds elapsed.\n',...
timeit(@()randsvdfast(n,kappa,mode,method,matrix)));
matrix = 2;
fprintf('Alg. 3.1 with U=gallery(''orthog'',n,2): %5.2f seconds elapsed.\n',...
timeit(@()randsvdfast(n,kappa,mode,method,matrix)));
% Algorithm 4.1 in our EPrint.
method = 3; matrix = 0;
fprintf('Alg. 4.1 with Haar U: %5.2f seconds elapsed.\n',...
timeit(@()randsvdfast(n,kappa,mode,method,matrix)));
matrix = 2;
fprintf('Alg. 4.1 with U=gallery(''orthog'',n,2): %5.2f seconds elapsed.\n',...
timeit(@()randsvdfast(n,kappa,mode,method,matrix)));
In the listing above, randsvdfast is an implementation of our algorithms available on the MATLAB Central File Exchange. Setting the argument matrix to 0 tells our function to pick $U$ from the Haar distribution, whereas setting it to 2 causes $U$ to be chosen in a non-random way. Running this code in MATLAB 2019b on a machine equipped with an Intel processor I5-6500 running at 3.20GHz the script produces the output
gallery('randsvd',...): 79.52 seconds elapsed.
Alg. 3.1 with Haar U: 19.70 seconds elapsed.
Alg. 3.1 with U=gallery('orthog',n,2): 1.90 seconds elapsed.
Alg. 4.1 with Haar U: 19.28 seconds elapsed.
Alg. 4.1 with U=gallery('orthog',n,2): 1.43 seconds elapsed.
Therefore, for $n = 10{,}000$ randsvdfast is up to 56 times faster than gallery('randsvd',...).
## Sharper Probabilistic Backward Error Bounds
Most backward error bounds for numerical linear algebra algorithms are of the form $nu$, for a machine precision $u$ and a problem size $n$. The dependence on $n$ of these bounds is known to be pessimistic: together with Nick Higham, our recent probabilistic analysis [SIAM J. Sci. Comput., 41 (2019), pp. A2815–A2835], which assumes rounding errors to be independent random variables of mean zero, proves that $n$ can be replaced by a small multiple of $\sqrt{n}$ with high probability. However, even these smaller bounds can still be pessimistic, as the figure below illustrates.
The figure plots the backward error for summation (in single precision) of $n$ floating-point numbers randomly sampled from a uniform distribution. For numbers in the $[0,1]$ distribution, the bound $\sqrt{n}u$ is almost sharp and accurately predicts the error growth. However, for the $[-1,1]$ distribution, the error is much smaller, seemingly not growing with $n$. This strong dependence of the backward error on the data cannot be explained by the existing bounds, which do not depend on the values of the data.
In our recent preprint, we perform a new probabilistic analysis that combines a probabilistic model of the rounding errors with a second probabilistic model of the data. Our analysis reveals a strong dependence of the backward error on the mean of the data $\mu$: indeed, our new backward error bounds are proportional to $\mu\sqrt{n}u + u$. Therefore, for data with small or zero mean, these new bounds are much sharper as they bound the backward error by a small multiple of the machine precision independent of the problem size $n$.
Motivated by this observation, we also propose new algorithms that transform the data to have zero mean, so as to benefit from these more favorable bounds. We implement this idea for matrix multiplication and show that our new algorithm can produce significantly more accurate results than standard matrix multiplication.
## Issues with Rounding in the GCC Implementation of the ISO 18037:2008 Standard Fixed-Point Arithmetic
Embedded systems are based on low-power, low-performance processors and can be found in various medical devices, smart watches, various communication devices, cars, planes, mobile phones and many other places. These systems come in a hardware and software package optimized for specific computational tasks and most commonly have real-time constraints. As these systems usually have energy usage and cost constraints too, sophisticated numerical hardware that can process floating-point data is not included, but rather only integer arithmetic, which is simpler in terms of area and power of the processors.
ISO 18037:2008 is a standard for embedded C programming language support. It lays out various rules that C compilers should support to make embedded systems easier to program using a high-level language. One of the most important definitions in this standard is fixed-point arithmetic data types and operations. Support for fixed-point arithmetic is highly desirable, since if it is not provided integers with scaling factors have to be used, which makes code hard to maintain and debug and most commonly requires assembler level changes or completely new implementations for each different platform.
The GCC compiler provides some support of the fixed-point arithmetic defined in this standard for ARM processors. However, in my recent technical report (https://arxiv.org/abs/2001.01496) I demonstrated various numerical pitfalls that programmers of embedded systems based on ARM and using GCC can get into. The issues demonstrated include
• larger than half machine epsilon errors in rounding decimal constants to fixed-point data types,
• errors in conversions between different data types,
• incorrect pre-rounding of arguments of mixed-format arithmetic operations before the operation is performed, and
• lack of rounding of the outputs of arithmetic operations.
These findings can be used to improve the accuracy of various embedded numerical libraries that might be using this compiler. To demonstrate one of the issues, here is a piece of test code:
The multiplication operation is a mixed-format operation, since it multiplies an unsigned long fract argument with an accum argument, therefore it is subject to prerounding of the unsigned long fract argument as described in the report. Since the comparison step in the if () sees that the argument a is larger than zero and b larger than 1, the code is executed with a hope that c will not be set to zero. However, in the arithmetic operation, a is incorrectly pre-rounded to 0, which causes c = 0*b, an unexpected outcome and a bug that is hard to detect and fix.
## Solving Block Low-Rank Linear Systems by LU Factorization is Numerically Stable
In many applications requiring the solution of a linear system $Ax=b$, the matrix $A$ possesses a block low-rank (BLR) property: most of its off-diagonal blocks are of low numerical rank and can therefore be well approximated by low-rank matrices. This property arises for example in the solution of discretized partial differential equations, because the numerical rank of a given off-diagonal block is closely related to the distance between the two subdomains associated with this block. BLR matrices have been exploited to significantly reduce the cost of solving $Ax=b$, in particular in the context of sparse direct solvers such as MUMPS, PaStiX, and STRUMPACK.
However, the impact of these low-rank approximations on the numerical stability of the solution in floating-point arithmetic has not been previously analyzed. The difficulty of such an analysis lies in the fact that, unlike for classical algorithms without low-rank approximations, there are two kinds of errors to analyze: floating-point errors (which depend on the unit roundoff $u$), and low-rank truncation errors (which depend on the the low-rank threshold $\varepsilon$, the parameter controlling the accuracy of the blockwise low-rank approximations). Moreover, the two kinds of error cannot easily be isolated in the analysis, because the BLR compression and factorization stages are often interlaced.
In our recent preprint, with Nick Higham, we present rounding error analysis for the solution of a linear system by LU factorization of BLR matrices. Assuming that a stable pivoting scheme is used, we prove backward stability: the relative backward error is bounded by a modest constant times $\varepsilon$, and does not depend on the unit roundoff $u$ as long as $u$ is safely smaller than $\varepsilon$. This is a very desirable theoretical guarantee that can now be given to the users, who can therefore control the numerical behavior of BLR solvers simply by setting $\varepsilon$ to the target accuracy. We illustrate this key result in the figure below, which shows a strong correlation between the backward error and the $\varepsilon$ threshold for 26 matrices from the SuiteSparse collection coming from a wide range of real-life applications.
## Determinants of Bohemian matrix families
The adjective “Bohemian” was used for the first time in a linear algebra context by Robert Corless and Steven Thornton to describe the eigenvalues of matrices whose entries are taken from a finite discrete set, usually of integers. The term is a partial acronym for “BOunded Height Matrix of Integers”, but the origin of the term was soon forgotten, and the expression “Bohemian matrix” is now widely accepted.
As Olga Taussky observed already in 1960, the study of matrices with integer elements is “very vast and very old”, with early work of Sylvester and Hadamard that dates back to the second half of the nineteenth century. These names are the first two in a long list of mathematicians that worked on what is now known as the “Hadamard conjecture”: for any positive integer $n$ multiple of 4, there exists an $n$ by $n$ matrix $H$, with entries $-1$ and $+1$, such that $HH^T = nI$.
If this is the best-known open problem surrounding Bohemian matrices, it is far from being the only one. During the 3-day workshop “Bohemian Matrices and Applications” that our group hosted in June last year, Steven Thornton released the Characteristic Polynomial Database, which collects the determinants and characteristic polynomials of billions of samples from certain families of structured as well as unstructured Bohemian matrices. All the available data led Steven to formulate a number of conjectures regarding the determinants of several families of Bohemian upper Hessenberg matrices.
Gian Maria Negri Porzio and I attended the workshop, and set ourselves the task of solving at least one of these open problems. In our recent preprint, we enumerate all the possible determinants of Bohemian upper Hessenberg matrices with ones on the subdiagonal. We consider also the special case of families with main diagonal fixed to zero, whose determinants turn out to be related to some generalizations of Fibonacci numbers. Many of the conjectures stated in the Characteristic Polynomial Database follow from our results.
## A Class of Fast and Accurate Summation Algorithms
Summing $n$ numbers is a key computational task at the heart of many numerical algorithms. When performed in floating-point arithmetic, summation is subject to rounding errors: for a machine precision $u$, the error bound for the most basic summation algorithms, such as recursive summation, is proportional to $nu$.
Nowadays, with the growing interest in low floating-point precisions and ever increasing $n$ in applications, such error bounds have become unacceptably large. While summation algorithms leading to smaller error bounds are known (compensated summation is an example), they are computationally expensive.
In our recent preprint, Pierre Blanchard, Nick Higham and I propose a class of fast and accurate summation algorithms called FABsum. We show that FABsum has an error bound of the form $bu+O(u^2)$, where $b$ is a block size, which is independent of $n$ to first order. As illustrated by the figure below, which plots the measured error using single precision as a function of $n$, FABsum can deliver substantially more accurate results than recursive summation. Moreover, FABsum can be easily incorporated in high-performance numerical linear algebra kernels in order to boost accuracy with only a modest drop in performance, as we demonstrate in the paper with the PLASMA library.
## Simulating Low Precision Floating-Point Arithmetics
In earlier blog posts, I wrote about the benefits of using half precision arithmetic (fp16) and about the problems of overflow and underflow in fp16 and how to avoid them. But how can one experiment with fp16, or other low precision formats such as bfloat16, in order to study how algorithms behave in these arithmetics? (For an accessible introduction to fp16 and bfloat16 see the blog post by Nick Higham.)
As of now, fp16 is supported by several GPUs, but these are specialist devices and they can be very expensive. Moreover, architectures that support bfloat16 have not yet not been released. Therefore software that simulates these floating-point formats is needed.
In our latest EPrint, Nick Higham and I investigate algorithms for simulating fp16, bfloat16 and other low precision formats. We have also written a MATLAB function chop that can be incorporated into other MATLAB codes to simulate low precision arithmetic. It can easily be used to study the effect of low precision formats on various algorithms.
Imagine a hypothetical situation where the computer can just represent integers. Then the question is how do we represent numbers like 4/3? An obvious answer would be to represent it via the integer closest to it, 1 in this case. However, one will have to come with a convention to handle the case where the number is in the centre. Now replace the integer in the example with floating-point numbers, and a similar question arises. This process of converting any given number to a floating-point number is called rounding. If we adopt a rule where we choose the closest floating-point number (as above), then we formally call it as ‘round to nearest’. There are other ways to round as well, and different rounding modes can yield different results for the same code. However meddling with the parameters of a floating-point format without a proper understanding of their consequences can be a recipe for disaster. Cleve Moler in his blog on sub-normal numbers makes this point by warning ‘don’t try this at home’. The MATLAB software we have written provides a safe environment to experiment with the effects of changing any parameter of a floating-point format (such as rounding modes and support of subnormal numbers) on the output of a code. All the technical details can be found in the Eprint and our MATLAB codes.
## Computing the Wave-Kernel Matrix Functions
The wave-kernel functions $\cosh{\sqrt{A}}$ and $\mathrm{sinhc}{\sqrt{A}}$ arise in the solution of second order differential equations such as $u''(t) - Au(t) = b(t)$ with initial conditions at $t=0$. Here, $A$ is an arbitrary square matrix and $\mathrm{sinhc}{z} = \sinh(z)/z$. The square root in these formulas is illusory, as both functions can be expressed as power series in $A$, so there are no questions about existence of the functions.
How can these functions be computed efficiently? In Computing the Wave-Kernel Matrix Functions (SIAM J. Sci. Comput., 2018) Prashanth Nadukandi and I develop an algorithm based on Padé approximation and the use of double angle formulas. The amount of scaling and the degree of the Padé approximant are chosen to minimize the computational cost subject to achieving backward stability for $\cosh{\sqrt{A}}$ in exact arithmetic.
In the derivation we show that the backward error of any approximation to $\cosh{\sqrt{A}}$ can be explicitly expressed in terms of a hypergeometric function. To bound the backward error we derive and exploit a new bound for $\|A^k\|^{1/k}$ in terms of the norms of lower powers of $A$; this bound is sharper than one previously obtained by Al-Mohy and Higham.
Numerical experiments show that the algorithm behaves in a forward stable manner in floating-point arithmetic and is superior in this respect to the general purpose Schur–Parlett algorithm applied to these functions.
The fundamental regions of the
function cosh(sqrt(z)), needed for the backward error analysis
underlying the algorithm.
## A new preconditioner exploiting low-rank factorization error
The solution of a linear system $Ax = b$ is a fundamental task in scientific computing. Two main classes of methods to solve such a system exist.
• Direct methods compute a factorization of matrix $A$, such as LU factorization, to then directly obtain the solution $x=U^{-1}L^{-1}b$ by triangular substitution; they are very reliable but also possess a high computational cost, which limits the size of problems that can be tackled.
• Iterative methods compute a sequence of iterates $x_k$ converging towards the solution $x$; they are inexpensive but their convergence and thus reliability strongly depends on the matrix properties, which limits the scope of problems that can be tackled.
A current major challenge in the field of numerical linear algebra is therefore to develop methods that are able to tackle a large scope of problems of large size.
To accelerate the convergence of iterative methods, one usually uses a preconditioner, that is, a matrix $M$ ideally satisfying three conditions: (1) $M$ is cheap to compute; (2) $M$ can be easily inverted; (3) $M^{-1}$ is a good approximation to $A^{-1}$. With such a matrix $M$, the preconditioned system $M^{-1}Ax=M^{1}b$ is then cheap to solve with an iterative method and often requires a small number of iterations only. An example of a widely used class of preconditioners is when $M$ is computed as a low-accuracy LU factorization.
Unfortunately, for many important problems it is quite difficult to find a preconditioner that is both of good quality and cheap to compute, especially when the matrix $A$ is ill conditioned, that is, when the ratio between its largest and smallest singular values is large.
In our paper A New Preconditioner that Exploits Low-rank Approximations to Factorization Error, with Nick Higham, which recently appeared in SIAM Journal of Scientific Computing, we propose a novel class of general preconditioners that builds on an existing, low-accuracy preconditioner $M=A-\Delta A$.
This class of preconditioners is based on the following key observation: ill-conditioned matrices that arise in practice often have a small number of small singular values. The inverse of such a matrix has a small number of large singular values and so is numerically low rank. This observation suggests that the error matrix $E = M^{-1}A - I = M^{-1}\Delta A \approx A^{-1}\Delta A$ is of interest, because we may expect $E$ to retain the numerically low-rank property of $A^{-1}$.
In the paper, we first investigate theoretically and experimentally whether $E$ is indeed numerically low rank; we then describe how to exploit this property to accelerate the convergence of iterative methods by building an improved preconditioner $M(I+\widetilde{E})$, where $\widetilde{E}$ is a low-rank approximation to $E$. This new preconditioner is equal to $A-M(E-\widetilde{E})$ and is thus almost a perfect preconditioner if $\widetilde{E}\approx E$. Moreover, since $\widetilde{E}$ is a low-rank matrix, $(I+\widetilde{E})^{-1}$ can be cheaply computed via the Sherman–Morrison–Woodbury formula, and so the new preconditioner can be easily inverted.
We apply this new preconditioner to three different types of approximate LU factorizations: half-precision LU factorization, incomplete LU factorization (ILU), and block low-rank (BLR) LU factorization. In our experiments with GMRES-based iterative refinement, we show that the new preconditioner can achieve a significant reduction in the number of iterations required to solve a variety of real-life $Ax=b$ problems.
## The Paterson–Stockmeyer method for evaluating polynomials and rational functions of matrices
According to Moler and Van Loan, truncating the Taylor series expansion to the exponential at 0 is the least effective of their nineteen dubious ways to compute the exponential of a matrix. Such an undoubtedly questionable approach can, however, become a powerful tool for evaluating matrix functions when used in conjunction with other strategies, such as, for example, the scaling and squaring technique. In fact, truncated Taylor series are just a special case of a much larger class of rational approximants, known as the Padé family, on which many state-of-the-art algorithms rely in order to compute matrix functions.
A customary choice for evaluating these approximants at a matrix argument is the Paterson–Stockmeyer method, an evaluation scheme that was originally proposed as an asymptotically optimal algorithm for evaluating polynomials of matrices, but generalizes quite naturally to rational functions, which are nothing but the solutions of a linear system whose coefficients and right-hand side are polynomials of the same matrix. This technique exploits a clever rewriting of a polynomial in $A$ as a polynomials in $A^s$, for some positive integer $s$, and overall requires about $2\sqrt{k}$ matrix products to evaluate a polynomial of degree $k$.
As shown in the figure, when the Paterson–Stockmeyer scheme is used the number of matrix multiplications required to evaluate a polynomial of degree $k$ grows slower than $k$ itself, with the result that evaluating polynomials of different degree will asymptotically have the same cost. For example, evaluating a polynomial of any degree between 43 and 49 requires 12 matrix multiplications, thus there is little point in considering an approximant of degree 43 when evaluating that of degree 49 has roughly the same cost but will in all likelihood deliver a more accurate result.
When designing algorithms to evaluate functions of matrices, one is interested in finding the optimal degrees, those marked with a red circle in the figure above, since they guarantee maximal accuracy for a given computational cost. When fixed precision is considered, finding all such degrees is not a problem: a backward error analysis can be used to determine the maximum degree that will ever be needed, $m_{max}$ say, and then looking at the plot is enough to find all the optimal degrees smaller than $m_{max}$. In order to deal with arbitrary precision arithmetic, however, a different strategy is needed, as depending on the working precision and the required tolerance, approximants of arbitrarily high degree may be needed. The new Eprint Optimality of the Paterson–Stockmeyer Method for Evaluating Matrix Polynomials and Rational Matrix Functions studies the cost of the Paterson–Stockmeyer method for polynomial evaluation and shows that a degree is optimal if and only if it is a quarter-square, that is, a number of the form $\lfloor n^2/4 \rfloor$ for some nonnegative integer $n$, where $\lfloor \cdot \rfloor$ is the floor function.
Similar results can be obtained for Paterson–Stockmeyer-like algorithms for evaluating diagonal Padé approximants, rational functions whose numerator and denominator have same degree. In that case, one can show that an approximant is optimal if and only if the degree of numerator and denominator is an eight-square, an integer of the form $\lfloor n^2/8 \rfloor$ for some $n \in \mathbb{N}$. In particular, for diagonal Padé approximants to the exponential, due to a symmetry of the coefficients of numerator and denominator, faster algorithms can be developed, and an explicit formula—not as nice as that in the two previous cases—can be derived for the optimal orders of these approximants.
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09 Jun 2010, 13:46
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Sorry if this is a dumb question but where do I find out about the GMAT challenges on this board.
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can u be more specific? challenges in which area? math/verbal or is it the testing method or the AWA?
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Jazzy12 wrote:
Sorry if this is a dumb question but where do I find out about the GMAT challenges on this board.
Are you referring to the GMAT Club Tests: http://www.gmatclub.com/tests
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09 Jun 2010, 19:10
I am actually not sure what I am talking about lol. I just joined the board and I was reading on how people where comparing their scores from the GMAT challenges. And I did not know where I can take such tests.
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10 Jun 2010, 02:50
These are GMAT Club Tests (the link has already been given above).
Re: GMAT challenges [#permalink] 10 Jun 2010, 02:50
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# GMAT challenges
Moderators: WaterFlowsUp, HiLine
Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 795 | 2,702 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-04 | latest | en | 0.894599 |
https://www.gradesaver.com/textbooks/math/algebra/algebra-1-common-core-15th-edition/chapter-4-an-introduction-to-functions-common-core-cumulative-standards-review-selected-response-page-288/2 | 1,713,616,041,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817650.14/warc/CC-MAIN-20240420122043-20240420152043-00785.warc.gz | 714,259,271 | 13,132 | ## Algebra 1: Common Core (15th Edition)
The answer is H ($s$)
First, you can eliminate G and I because they are numbers, not variables. So now you have to choose between F and H. The variable $E$ is dependent on her total sales, so eliminate answer choice F. $s$ is the only variable that does not depend on anything. | 80 | 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": 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} | 3.4375 | 3 | CC-MAIN-2024-18 | latest | en | 0.949146 |
http://www.programmersheaven.com/discussion/305508/timer-in-pascal?S=B20000 | 1,526,985,261,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794864657.58/warc/CC-MAIN-20180522092655-20180522112655-00316.warc.gz | 448,661,208 | 11,261 | # Timer in Pascal
Hello everyone!
I'm quite new to pascal, but I still know quite alot. Anyway, I am trying to create a reactiontester. The screen flashes and then you have to press a button, and the compouter calculates the time it took for you to press the button. And I have no idea what I should write to make this happen. I tried using a repeat until keypress, something like:
a:=0;
repeat;
a:=a+1
delay(1)
until keypressed;
I put the delay there just so I know how long it takes before every repeat, but still, this doesn't work any well... So if anyone know how to use a real time, please respond!
Marcus Olsson
• : Hello everyone!
:
: I'm quite new to pascal, but I still know quite alot. Anyway, I am trying to create a reactiontester. The screen flashes and then you have to press a button, and the compouter calculates the time it took for you to press the button. And I have no idea what I should write to make this happen. I tried using a repeat until keypress, something like:
:
: a:=0;
: repeat;
: a:=a+1
: delay(1)
: until keypressed;
:
: I put the delay there just so I know how long it takes before every repeat, but still, this doesn't work any well... So if anyone know how to use a real time, please respond!
:
: Marcus Olsson
:
:
Here is the pseudocode of how to do that:
[code]
- flash screen
- Get First Current Time
- repeat until Keypressed
- Get Second Current Time
- Calculate time difference between 1 & 2
[/code]
• : Hello everyone!
:
: I'm quite new to pascal, but I still know quite alot. Anyway, I am trying to create a reactiontester. The screen flashes and then you have to press a button, and the compouter calculates the time it took for you to press the button. And I have no idea what I should write to make this happen. I tried using a repeat until keypress, something like:
:
: a:=0;
: repeat;
: a:=a+1
: delay(1)
: until keypressed;
:
: I put the delay there just so I know how long it takes before every repeat, but still, this doesn't work any well... So if anyone know how to use a real time, please respond!
:
: Marcus Olsson
:
: | 532 | 2,081 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-22 | latest | en | 0.908094 |
https://edustrings.com/mathematics/2497293.html | 1,620,740,843,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243989614.9/warc/CC-MAIN-20210511122905-20210511152905-00166.warc.gz | 237,881,736 | 6,807 | 13 January, 07:13
# the ages of two friends Sunaina and Tanish are differ by 2 years sunaina's father is twice as old as Sunaina and Tanisha is twice as old as his brother Shiva The ages of sunaina's father and Shiva differ by 40 years
+3
1. 13 January, 08:32
0
Let S represent Sunaina's age.
Let T represent Tanish's age.
So 2S - T/2 = 40, 4S - T = 80.
Either S - T = 2 or T - S = 2. Let's try both cases.
Either 3S = 78 or 3S = 82.
Since 82 is not a multiple of 3, S = 26.
So Sunaina is 26 years old and Tanish is 24 years old. | 187 | 537 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2021-21 | latest | en | 0.974522 |
https://demo.formulasearchengine.com/wiki/Norm_(mathematics) | 1,632,269,585,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057274.97/warc/CC-MAIN-20210921221605-20210922011605-00192.warc.gz | 250,041,608 | 33,916 | Norm (mathematics)
{{#invoke:Hatnote|hatnote}}
In linear algebra, functional analysis and related areas of mathematics, a norm is a function that assigns a strictly positive length or size to each vector in a vector space, other than the zero vector (which has zero length assigned to it). A seminorm, on the other hand, is allowed to assign zero length to some non-zero vectors (in addition to the zero vector).
A norm must also satisfy certain properties pertaining to scalability and additivity which are given in the formal definition below.
A simple example is the 2-dimensional Euclidean space R2 equipped with the Euclidean norm. Elements in this vector space (e.g., (3, 7)) are usually drawn as arrows in a 2-dimensional cartesian coordinate system starting at the origin (0, 0). The Euclidean norm assigns to each vector the length of its arrow. Because of this, the Euclidean norm is often known as the magnitude.
A vector space on which a norm is defined is called a normed vector space. Similarly, a vector space with a seminorm is called a seminormed vector space. It is often possible to supply a norm for a given vector space in more than one way.
Definition
Given a vector space V over a subfield F of the complex numbers, a norm on V is a function p: VR with the following properties:[1]
For all aF and all u, vV,
1. p(av) = |a| p(v), (absolute homogeneity or absolute scalability).
2. p(u + v) ≤ p(u) + p(v) (triangle inequality or subadditivity).
3. If p(v) = 0 then v is the zero vector (separates points).
By the first axiom, absolute homogeneity, we have p(0) = 0 and p(-v) = p(v), so that by the triangle inequality
p(v) ≥ 0 (positivity).
A seminorm on V is a function p: VR with the properties 1. and 2. above.
Every vector space V with seminorm p induces a normed space V/W, called the quotient space, where W is the subspace of V consisting of all vectors v in V with p(v) = 0. The induced norm on V/W is clearly well-defined and is given by:
p(W + v) = p(v).
Two norms (or seminorms) p and q on a vector space V are equivalent if there exist two real constants c and C, with c > 0 such that
for every vector v in V, one has that: c q(v) ≤ p(v) ≤ C q(v).
A topological vector space is called normable (seminormable) if the topology of the space can be induced by a norm (seminorm).
Notation
If a norm p: VR is given on a vector space V then the norm of a vector vV is usually denoted by enclosing it within double vertical lines: ‖v‖ := p(v). Such notation is also sometimes used if p is only a seminorm.
For the length of a vector in Euclidean space (which is an example of a norm, as explained below), the notation |v| with single vertical lines is also widespread.
In Unicode, the codepoint of the "double vertical line" character ‖ is U+2016. The double vertical line should not be confused with the "parallel to" symbol, Unicode U+2225 ( ∥ ). This is usually not a problem because the former is used in parenthesis-like fashion, whereas the latter is used as an infix operator. The single vertical line | is called "vertical line" in Unicode and its codepoint is U+007C.
Examples
• All norms are seminorms.
• The trivial seminorm has p(x) = 0 for all x in V.
• Every linear form f on a vector space defines a seminorm by x → |f(x)|.
Absolute-value norm
${\displaystyle \|x\|=|x|}$
is a norm on the one-dimensional vector spaces formed by the real or complex numbers.
Euclidean norm
{{#invoke:main|main}} On an n-dimensional Euclidean space Rn, the intuitive notion of length of the vector x = (x1, x2, ..., xn) is captured by the formula
${\displaystyle \|{\boldsymbol {x}}\|:={\sqrt {x_{1}^{2}+\cdots +x_{n}^{2}}}.}$
This gives the ordinary distance from the origin to the point x, a consequence of the Pythagorean theorem. The Euclidean norm is by far the most commonly used norm on Rn, but there are other norms on this vector space as will be shown below. However all these norms are equivalent in the sense that they all define the same topology.
On an n-dimensional complex space Cn the most common norm is
${\displaystyle \|{\boldsymbol {z}}\|:={\sqrt {|z_{1}|^{2}+\cdots +|z_{n}|^{2}}}={\sqrt {z_{1}{\bar {z}}_{1}+\cdots +z_{n}{\bar {z}}_{n}}}.}$
In both cases we can also express the norm as the square root of the inner product of the vector and itself:
${\displaystyle \|{\boldsymbol {x}}\|:={\sqrt {{\boldsymbol {x}}^{*}~{\boldsymbol {x}}}},}$
where x is represented as a column vector ([x1; x2; ...; xn]), and x* denotes its conjugate transpose.
This formula is valid for any inner product space, including Euclidean and complex spaces. For Euclidean spaces, the inner product is equivalent to the dot product. Hence, in this specific case the formula can be also written with the following notation:
${\displaystyle \|{\boldsymbol {x}}\|:={\sqrt {{\boldsymbol {x}}\cdot {\boldsymbol {x}}}}.}$
The Euclidean norm is also called the Euclidean length, L2 distance, 2 distance, L2 norm, or 2 norm; see Lp space.
The set of vectors in Rn+1 whose Euclidean norm is a given positive constant forms an n-sphere.
Euclidean norm of a complex number
The Euclidean norm of a complex number is the absolute value (also called the modulus) of it, if the complex plane is identified with the Euclidean plane R2. This identification of the complex number x + iy as a vector in the Euclidean plane, makes the quantity ${\displaystyle {\sqrt {x^{2}+y^{2}}}}$ (as first suggested by Euler) the Euclidean norm associated with the complex number.
Taxicab norm or Manhattan norm
{{#invoke:main|main}}
${\displaystyle \|{\boldsymbol {x}}\|_{1}:=\sum _{i=1}^{n}|x_{i}|.}$
The name relates to the distance a taxi has to drive in a rectangular street grid to get from the origin to the point x.
The set of vectors whose 1-norm is a given constant forms the surface of a cross polytope of dimension equivalent to that of the norm minus 1. The Taxicab norm is also called the L1 norm. The distance derived from this norm is called the Manhattan distance or L1 distance.
The 1-norm is simply the sum of the absolute values of the columns.
In contrast,
${\displaystyle \sum _{i=1}^{n}x_{i}}$
is not a norm because it may yield negative results.
p-norm
{{#invoke:main|main}}
Let p ≥ 1 be a real number.
${\displaystyle \|\mathbf {x} \|_{p}:={\bigg (}\sum _{i=1}^{n}|x_{i}|^{p}{\bigg )}^{1/p}.}$
Note that for p = 1 we get the taxicab norm, for p = 2 we get the Euclidean norm, and as p approaches ${\displaystyle \infty }$ the p-norm approaches the infinity norm or maximum norm. Note that the p-norm is related to the Hölder mean.
This definition is still of some interest for 0 < p < 1, but the resulting function does not define a norm,[2] because it violates the triangle inequality. What is true for this case of 0 < p < 1, even in the measurable analog, is that the corresponding Lp class is a vector space, and it is also true that the function
${\displaystyle \int _{X}|f(x)-g(x)|^{p}~\mathrm {d} \mu \,\!}$
(without pth root) defines a distance that makes Lp(X) into a complete metric topological vector space. These spaces are of great interest in functional analysis, probability theory, and harmonic analysis. However, outside trivial cases, this topological vector space is not locally convex and has no continuous nonzero linear forms. Thus the topological dual space contains only the zero functional.
The derivative of the p-norm is given by
${\displaystyle {\frac {\partial }{\partial x_{k}}}\|\mathbf {x} \|_{p}={\frac {x_{k}|x_{k}|^{p-2}}{\|\mathbf {x} \|_{p}^{p-1}}}.}$
For the special case of p = 2, this becomes
${\displaystyle {\frac {\partial }{\partial x_{k}}}\|\mathbf {x} \|_{2}={\frac {x_{k}}{\|\mathbf {x} \|_{2}}}}$,
or
${\displaystyle {\frac {\partial }{\partial \mathbf {x} }}\|\mathbf {x} \|_{2}={\frac {\mathbf {x} }{\|\mathbf {x} \|_{2}}}.}$
Maximum norm (special case of: infinity norm, uniform norm, or supremum norm)
{{#invoke:main|main}}
${\displaystyle \|\mathbf {x} \|_{\infty }:=\max \left(|x_{1}|,\ldots ,|x_{n}|\right).}$
The set of vectors whose infinity norm is a given constant, c, forms the surface of a hypercube with edge length 2c.
Zero norm
In probability and functional analysis, the zero norm induces a complete metric topology for the space of measureable functions and for the F-space of sequences with F–norm ${\displaystyle (x_{n})\mapsto \sum _{n}{2^{-n}x_{n}/(1+x_{n})}}$, which is discussed by Stefan Rolewicz in Metric Linear Spaces.[3]
Hamming distance of a vector from zero
{{#invoke:see also|seealso}} In metric geometry, the discrete metric takes the value one for distinct points and zero otherwise. When applied coordinate-wise to the elements of a vector space, the discrete distance defines the Hamming distance, which is important in coding and information theory. In the field of real or complex numbers, the distance of the discrete metric from zero is not homogeneous in the non-zero point; indeed, the distance from zero remains one as its non-zero argument approaches zero. However, the discrete distance of a number from zero does satisfy the other properties of a norm, namely the triangle inequality and positive definiteness. When applied component-wise to vectors, the discrete distance from zero behaves like a non-homogeneous "norm", which counts the number of non-zero components in its vector argument; again, this non-homogeneous "norm" is discontinuous.
In signal processing and statistics, David Donoho referred to the zero "norm" with quotation marks. Following Donoho's notation, the zero "norm" of x is simply the number of non-zero coordinates of x, or the Hamming distance of the vector from zero. When this "norm" is localized to a bounded set, it is the limit of p-norms as p approaches 0. Of course, the zero "norm" is not a B-norm, because it is not positive homogeneous. It is not even an F-norm, because it is discontinuous, jointly and severally, with respect to the scalar argument in scalar-vector multiplication and with respect to its vector argument. Abusing terminology, some engineers omit Donoho's quotation marks and inappropriately call the number-of-nonzeros function the L0 norm (sic.), also misusing the notation for the Lebesgue space of measurable functions.
Other norms
Other norms on Rn can be constructed by combining the above; for example
${\displaystyle \|x\|:=2|x_{1}|+{\sqrt {3|x_{2}|^{2}+\max(|x_{3}|,2|x_{4}|)^{2}}}}$
is a norm on R4.
For any norm and any injective linear transformation A we can define a new norm of x, equal to
${\displaystyle \|Ax\|.}$
In 2D, with A a rotation by 45° and a suitable scaling, this changes the taxicab norm into the maximum norm. In 2D, each A applied to the taxicab norm, up to inversion and interchanging of axes, gives a different unit ball: a parallelogram of a particular shape, size and orientation. In 3D this is similar but different for the 1-norm (octahedrons) and the maximum norm (prisms with parallelogram base).
All the above formulas also yield norms on Cn without modification.
Infinite-dimensional case
The generalization of the above norms to an infinite number of components leads to the Lp spaces, with norms
${\displaystyle \|x\|_{p}={\bigg (}\sum _{i\in \mathbb {N} }|x_{i}|^{p}{\bigg )}^{1/p}{\text{ resp. }}\|f\|_{p,X}={\bigg (}\int _{X}|f(x)|^{p}~\mathrm {d} x{\bigg )}^{1/p}}$
(for complex-valued sequences x resp. functions f defined on ${\displaystyle X\subset \mathbb {R} }$), which can be further generalized (see Haar measure).
Any inner product induces in a natural way the norm ${\displaystyle \|x\|:={\sqrt {\langle x,x\rangle }}.}$
Other examples of infinite dimensional normed vector spaces can be found in the Banach space article.
Properties
Illustrations of unit circles in different norms.
The concept of unit circle (the set of all vectors of norm 1) is different in different norms: for the 1-norm the unit circle in R2 is a square, for the 2-norm (Euclidean norm) it is the well-known unit circle, while for the infinity norm it is a different square. For any p-norm it is a superellipse (with congruent axes). See the accompanying illustration. Note that due to the definition of the norm, the unit circle is always convex and centrally symmetric (therefore, for example, the unit ball may be a rectangle but cannot be a triangle).
In terms of the vector space, the seminorm defines a topology on the space, and this is a Hausdorff topology precisely when the seminorm can distinguish between distinct vectors, which is again equivalent to the seminorm being a norm. The topology thus defined (by either a norm or a seminorm) can be understood either in terms of sequences or open sets. A sequence of vectors ${\displaystyle \{v_{n}\}}$ is said to converge in norm to ${\displaystyle v}$ if ${\displaystyle \|v_{n}-v\|\rightarrow 0}$ as ${\displaystyle n\to \infty }$. Equivalently, the topology consists of all sets that can be represented as a union of open balls.
Two norms ||•||α and ||•||β on a vector space V are called equivalent if there exist positive real numbers C and D such that for all x in V
${\displaystyle C\|x\|_{\alpha }\leq \|x\|_{\beta }\leq D\|x\|_{\alpha }}$. For instance, on ${\displaystyle \mathbf {C} ^{n}}$, if p > r > 0, then
${\displaystyle \|x\|_{p}\leq \|x\|_{r}\leq n^{(1/r-1/p)}\|x\|_{p}.}$
In particular,
${\displaystyle \|x\|_{2}\leq \|x\|_{1}\leq {\sqrt {n}}\|x\|_{2}}$
${\displaystyle \|x\|_{\infty }\leq \|x\|_{2}\leq {\sqrt {n}}\|x\|_{\infty }}$
${\displaystyle \|x\|_{\infty }\leq \|x\|_{1}\leq n\|x\|_{\infty }.}$
If the vector space is a finite-dimensional real/complex one, all norms are equivalent. On the other hand, in the case of infinite-dimensional vector spaces, not all norms are equivalent.
Equivalent norms define the same notions of continuity and convergence and for many purposes do not need to be distinguished. To be more precise the uniform structure defined by equivalent norms on the vector space is uniformly isomorphic.
Every (semi)-norm is a sublinear function, which implies that every norm is a convex function. As a result, finding a global optimum of a norm-based objective function is often tractable.
Given a finite family of seminorms pi on a vector space the sum
${\displaystyle p(x):=\sum _{i=0}^{n}p_{i}(x)}$
is again a seminorm.
For any norm p on a vector space V, we have that for all u and vV:
p(u ± v) ≥ | p(u) − p(v) |
If ${\displaystyle X}$ and ${\displaystyle Y}$ are normed spaces and ${\displaystyle u:X\to Y}$ is a continuous linear map, then the norm of ${\displaystyle u}$ and the norm of the transpose of ${\displaystyle u}$ are equal.[4]
For the lp norms, we have Hölder's inequality[5]
${\displaystyle |x^{\mathsf {T}}y|\leq \|x\|_{p}\|y\|_{q}\qquad {\frac {1}{p}}+{\frac {1}{q}}=1.}$
A special case of this is the Cauchy–Schwarz inequality:[5]
${\displaystyle |x^{\mathsf {T}}y|\leq \|x\|_{2}\|y\|_{2}.}$
Classification of seminorms: absolutely convex absorbing sets
All seminorms on a vector space V can be classified in terms of absolutely convex absorbing sets in V. To each such set, A, corresponds a seminorm pA called the gauge of A, defined as
pA(x) := inf{α : α > 0, x ∈ α A}
with the property that
{x : pA(x) < 1} ⊆ A ⊆ {x : pA(x) ≤ 1}.
Conversely:
Any locally convex topological vector space has a local basis consisting of absolutely convex sets. A common method to construct such a basis is to use a family (p) of seminorms p that separates points: the collection of all finite intersections of sets {p<1/n} turns the space into a locally convex topological vector space so that every p is continuous.
Such a method is used to design weak and weak* topologies.
norm case:
Suppose now that (p) contains a single p: since (p) is separating, p is a norm, and A={p<1} is its open unit ball. Then A is an absolutely convex bounded neighbourhood of 0, and p = pA is continuous.
The converse is due to Kolmogorov: any locally convex and locally bounded topological vector space is normable. Precisely:
If V is an absolutely convex bounded neighbourhood of 0, the gauge gV (so that V={gV <1}) is a norm.
Generalizations
There are several generalizations of norms and semi-norms. If p is absolute homogeneity but in place of subadditivity we require that
then p satisfies the triangle inequality but is called a quasi-seminorm and the smallest value of b for which this holds is called the multiplier of p; if in addition p separates points then it is called a quasi-norm.
On the other hand, if p satisfies the triangle inequality but in place of absolute homogeneity we require that
then p is called a k-seminorm.
We have the following relationship between quasi-seminorms and k-seminorms:
Suppose that q is a quasi-seminorm on a vector space X with multiplier b. If ${\displaystyle 0 then there exists k-seminorm p on X equivalent to q.
Notes
1. Template:Harvnb
2. Except in R1, where it coincides with the Euclidean norm, and R0, where it is trivial.
3. {{#invoke:citation/CS1|citation |CitationClass=citation }}
4. Treves pp. 242 - 243
5. {{#invoke:citation/CS1|citation |CitationClass=book }}
References
• {{#invoke:citation/CS1|citation
|CitationClass=book }}
• {{#invoke:citation/CS1|citation
|CitationClass=book }}
• {{#invoke:citation/CS1|citation
|CitationClass=book }}
• {{#invoke:citation/CS1|citation
|CitationClass=book }} Template:Refend | 4,729 | 17,400 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 48, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2021-39 | latest | en | 0.919582 |
http://bankersdaily.in/number-series-for-sbi-po-set-12/ | 1,632,026,747,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056711.62/warc/CC-MAIN-20210919035453-20210919065453-00122.warc.gz | 6,216,227 | 20,235 | # Number Series for SBI PO : Set 12
Direction.1-5) what will come in place of the question mark (?) in each of the following number series?
1)26, 144, 590, 1164, ?
a) 1258
b) 1352
c) 1238
d) 4321
e)1182
e)1182
2) 50, ?,61,89,154,280
a) 52
b) 51
c) 60
d) 62
e)65
a) 52
3)73205, 6655, 605, 55, ?
a) 9
b) 5
c) 13
d) 11
e)None of these
b) 5
4) 11, 24, 50, 102, 206, ?
a) 405
b) 312
c) 414
d) 416
e)313
c) 414
5) 23, 32, 53, 90, 153, ?
a) 198
b) 248
c) 182
d) 292
e)215
b) 248
Direction.6-10) Find out the wrong number in the following series:
6) 3,4.5,11.25,39.375,177.187
a) 5
b) 25
c) 375
d) 187
e)None of these
e)None of these
7) 315,317,320,325,340
a) 315
b) 320
c) 325
d) 340
e)None of these
d) 340
8) 7,13, 40, 81,238
a) 7
b) 39
c) 13
d) 81
e)None of these
d) 81
9)3986,4009,4054,4124,4216
a) 3986
b) 4009
c) 4054
d) 4124
e)None of these
c) 4054
10)391,387,396,386,422
a) 391
b) 396
c) 387
d) 422
e)None of these | 526 | 1,026 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-39 | latest | en | 0.394063 |
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# Ch 1.4 - 6E
ISBN: 9780321570567 2
## Solution for problem 6E Chapter 1.4
Calculus: Early Transcendentals | 1st Edition
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Problem 6E
6E
Step-by-Step Solution:
Step 1 of 3
Week 9 Psyc: Scientific Research: (3/28, 3/30.4/1) I. Statistical Methods: -why is this important II. Two types of data: a. Quantitative data = numerical values (only focus on this type) i. Descriptive statistics ii. Inferential statistics b. Qualitative data = characteristics III. Descriptive Statistics: used to describe or summarize sets of data to make it more understandable. a. Ex: study of Schizophrenia: symptoms: delusions and hallucinations A. Types of Descriptive Statistics: a. Measurers of Central Tendency: mean, median, mode b. Measurers of Variability: how similar/different: Range and Standard deviation c. Correlation Coefficients: looking at the relationship of the values: Strong or Weak Ex: Schizophrenia: looking at the relationship between variables: #of people in family that have this disorder Schizophrenia: -variable 1: severity of symptoms : delusions, hallucinations -variable 2: # in a family Strong positive relationship between 2 variables +1.00 No relationship 0 Strong Negative relationship (inverse) - 1.00 Correlation Coefficients: continued -.85 -the value that tells about the strength of the relationship ( .85) -tells of the directing of the graph (negative or positive) (inverse or together) IV. Inferential Statistics: procedures for looking at probability that research results could derive from chance alone -you want your results to be real not just chance -inferential statistics enable researchers to be confident in drawing conclusion from their data A. Reading values: a. “p” = probability b. “r” = mean c. “t ” = analysis when wanting to compare means of 2 groups Ex: r = .60 = descriptive statistics (the strength and direction of 2 variables) p= .01 = inferential statistics (how confident are coefficient is in the experiment) p value determines how confident can be in the r value r value (mean) describes the degree of difference p value tells the significance of he degree of difference
Step 2 of 3
Step 3 of 3
#### Related chapters
Unlock Textbook Solution | 658 | 2,609 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2021-21 | longest | en | 0.780953 |
https://bulbapedia.bulbagarden.net/w/index.php?title=Personality_value&oldid=1833739 | 1,501,063,157,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549426086.44/warc/CC-MAIN-20170726082204-20170726102204-00099.warc.gz | 606,106,120 | 22,907 | Personality value
A Pokémon's personality value is an unsigned 32-bit integer that is created when the Pokémon is first encountered; it is set as soon as the Pokémon appears in the wild, the Pokémon's Egg is received from the Day Care man or another NPC, or the Pokémon itself is received from an NPC. As an unsigned 32-bit integer, its value can be anywhere between 0, which is represented as 00000000 00000000 00000000 00000000, thirty-two zeroes in binary, and 4,294,967,295, represented as 11111111 11111111 11111111 11111111, thirty-two ones in binary, inclusive. This value was introduced with the Pokémon data structure overhaul that occurred at the start of Generation III, and is generated using the games' pseudorandom number generator.
For each operation, the portion of the 32-bit value used will be indicated at the beginning of the section by highlighting, unless the full value is used. The integer's full value will be known as p, other values will be declared as px, py, and so on. Much of this article will be discussing values in binary.
Gender
`00000000 00000000 00000000 00000000`
A Pokémon's gender is determined by the lowest eight digits (a byte) of p in binary form; essentially, `p % 256`. This value will, from here on out, be known as pgender.
In a Pokémon species's base stat structure, there is a byte called the gender threshold, with a value between `00000000` and `11111111`. For genderless Pokémon such as Magnemite, the value of this byte is `11111111`. For female only Pokémon such as Nidoran♀, the value of this byte is `11111110`. For male only Pokémon such as Nidoran♂, the value of this byte is `00000000`. Pokémon with both genders occupy the rest of the spectrum. For these Pokémon, if the value of pgender is equal to or greater than that of the base stat value (the gender threshold), the Pokémon is male, otherwise it is female.
Base stat value Gender ratio
Binary Decimal Male Female
`11111111` 255 Genderless
`11111110` 254 0.0% 100.0%
`11011111` 223 12.5% 87.5%
`10111111` 191 25.0% 75.0%
`01111111` 127 50.0% 50.0%
`00111111` 63 75.0% 25.0%
`00011111` 31 87.5% 12.5%
`00000000` 0 100.0% 0.0%
Ability
If p is even, that Pokémon has the first Ability for its species. If it's odd, it has the second. If a species has only one Ability while p is odd, it gets the only Ability it can. Pokémon transferred from Generation III to Generation IV or Generation V that can get one of the new Abilities will retain the Ability they had in the previous generation; however, if they are evolved and their personality value is odd, they will get the new Ability upon evolution. For example, a Porygon that has the Ability Trace is brought to a Generation IV game. If it then evolves into Porygon2, it may keep Trace (if p is even), or it may switch to the Ability Download (if p is odd). The reason this happens is because although the Ability is initially determined by the personality value, the Ability is stored as a separate value in the Pokémon data, and alongside the IVs in Generation III. When the Pokémon evolves in Generation IV, the personality value is rechecked, and a new Ability assigned.
Note that a separate bit governs whether a Pokémon has their normal Abilities or Hidden Abilities in Generation V; the personality value of a Pokémon is always set to have its first Ability in this case, so as to allow the inclusion of a second Hidden Ability in a future generation.
Nature
A Pokémon's Nature is determined by `p % 25`.
The number acquired as the remainder corresponds to the personality as follows:
p % 25 Nature
0 Hardy
1 Lonely
2 Brave
4 Naughty
5 Bold
6 Docile
7 Relaxed
8 Impish
9 Lax
10 Timid
11 Hasty
12 Serious
13 Jolly
14 Naive
15 Modest
16 Mild
17 Quiet
18 Bashful
19 Rash
20 Calm
21 Gentle
22 Sassy
23 Careful
24 Quirky
Shininess
`00000000 00000000 00000000 00000000`
The red value will be represented as p1, while the blue value will be represented as p2.
Whether a Pokémon is Shiny or not depends on the values of the Trainer ID number, secret ID number, and personality value. There is an 8 in 65536 chance (0.012207% or 2-13) of a Pokémon being Shiny (which simplifies to 1/8192).
Variables E and F are declared to hold the values that result.
A bitwise exclusive or (xor) operation (such as a xor b = c) is equivalent to saying "If each bit of ab, c is true." In other words, 11010111 xor 01101010 = 10111101. Each xor in the operation is a bitwise xor.
E = IDTrainer xor IDSecret
F = p1 xor p2
E xor F < 8 , then Shiny
Example
As an example, let's take a Trainer whose Trainer ID is 24294 and whose Secret ID is 38834.
IDTrainer = 24294 = 0101111011100110 in binary.
IDSecret = 38834 = 1001011110110010 in binary.
This Trainer encounters a Pokémon whose personality value is 2814471828.
p = 2814471828 = 1010011111000001 0110111010010100 in binary.
p1 = 1010011111000001
p2 = 0110111010010100
The E value is 0101111011100110 xor 1001011110110010
E = 1100100101010100
The F value is 1010011111000001 xor 0110111010010100
F = 1100100101010101
E xor F = 0000000000000001, which is less than eight. Therefore, this Pokémon is Shiny.
Spinda's spots
`00000000 00000000 00000000 00000000`
Spinda has four spots: two on its face, and one on each of its ears. Each of the four bytes represents the coordinates of the spot on Spinda's face or ears. The first four digits of each byte represents the x-coordinate of the top left of the spot when translated into decimal notation, while the last four represent the y-coordinate of the top left of the spot when translated into decimal. There are 4,294,967,296 different spot variations.
Unown's letter (Generation III only)
`00000000 00000000 00000000 00000000`
pletter = `00000000`
In Generation III, Unown's letter is taken from the modulus of the combination of the least significant 2 bits of each byte in p. With A representing 0, and each letter thereafter representing the following number value (? as 26 and ! as 27), the letter can be determined by:
`up = pletter % 28`
In Generation II this is determined using individual values. In Generations IV and V Unown's letter is determined by a separate byte, unrelated to the personality value.
Wurmple's evolution
`00000000 00000000 00000000 00000000`
The value in red will be known as pw; it is essentially `p % 65536`.
Wurmple's evolution does not depend on time of day, despite what many game guides say. In fact, the evolution is determined by taking `pw % 10`. If the remainder is less than 5, Wurmple will become a Silcoon, and if it is greater than or equal to 5, it will become a Cascoon.
Performance changes
The day-to-day performance of a Pokémon, for the purposes of the Pokéathlon, depends on the personality value as well as the day of the month. It also depends on Nature, which is also calculated based on personality value. The five least significant digits govern, from least significant to most significant, Power, Stamina, Skill, Jump, and Speed.
The change in performance in a particular attribute is calculated as the sum of the following factors:
• The day of month modifier: This modifier depends on the day of the month and one particular digit of the personality value (determined based on the attribute in question). It is calculated as (((Day + Attribute + 3) × (Day - Attribute + 7) + Personality) mod 10) × 2 - 9, where:
• Attribute is 0 for Power, 1 for Stamina, 2 for Skill, 3 for Jump, and 4 for Speed
• Personality is the relevant value of the particular digit
• The Nature modifier: This modifier depends on the Nature of the Pokémon. Each of the neutral Natures raise one attribute by 10 points while lowering another by 10 points, while the other 20 Natures each raise one attribute by 35 points while lowering another by 35 points, based on which stats they normally raise and lower.
• The Aprijuice modifier: This modifier depends on the Aprijuice applied to the Pokémon. The Pokémon receives 1.5× the flavor values in the attributes corresponding to the strongest and second-strongest flavors of the Aprijuice, an additional 10-point bonus in the strongest flavor of the Aprijuice, and a penalty to the attribute corresponding to the weakest flavor of the Aprijuice based on its mildness and the sum of the Aprijuice's two strongest flavors (see Apriblender for further details).
Note that the day of month modifier is at most -9 and at most +9 (and only takes on odd values therein). Also as a result, the day of month modifier cycles every 10 days (exception: for Power and Speed, the modifiers in effect for the first day of each month are different from those of Day 11, Day 21, and Day 31, though the latter three have the same modifiers).
Also, since Nature is also determined by the last two digits of the personality value, the Nature of a Pokémon will constrain the day of month modifier in Power (and to a lesser extent, Stamina) for a Pokémon. For example, a Hardy Wurmple (last digit 0 or 5) will never get a +9 or a +7 from a day of month modifier in Power; if it evolves into a Cascoon (the last digit is 5) it furthermore entails that its best performance will be on the first day of each month, barring Aprijuice factors (as it has its sole +5 boost on the first day on each month)—in contrast, a Hardy Silcoon will have a +5 boost six times a month (on days ending with 4 or 6).
Attribute Attribute modifier Personality digit Neutral Nature +10 Neutral Nature -10 Other Nature stat
Power 0 Ones Hardy Bashful Attack
Stamina 1 Tens Docile Quirky Defense
Skill 2 Hundreds Quirky Serious Sp. Def
Jump 3 Thousands Bashful Docile Sp. Atk
Speed 4 Ten-thousands Serious Hardy Speed
Note that the association of Skill and Jump to Natures is reversed relative to a Pokémon's preferred flavors. That is, a Pokémon which prefers dry food will have a +35 boost in Jump (otherwise associated with bitter foods) rather than Skill. However, giving this Pokémon a dry Aprijuice will boost its Skill, rather than its Jump.
The calculated value is then translated into a difference in stars as follows:
Stars Score -4 -3 -2 -1 0 1 2 3 4 < -120 -119 - -80 -79 - -40 -39 - -15 -14 - 14 15 - 39 40 - 79 80 - 119 > 120
Note that performance changes may not drop below the Pokémon's minimum performance rating, nor may it exceed the Pokémon's maximum performance rating. Thus, for example, a Cradily may never have its Speed or Jump rating improved or degraded as those attributes are fixed (as such, for example, it is advisable to give it dry Aprijuices for which sweetness or bitterness is its weakest flavor).
This game mechanic article is part of Project Games, a Bulbapedia project that aims to write comprehensive articles on the Pokémon games. | 2,743 | 10,714 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-30 | longest | en | 0.911152 |
http://www.feyncalc.org/forum/0823.html | 1,550,809,583,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247513222.88/warc/CC-MAIN-20190222033812-20190222055812-00122.warc.gz | 336,916,637 | 2,405 | Name: Vladyslav Shtabovenko (email_not_shown)
Date: 12/16/14-11:59:30 AM Z
Hi Ralph,
FeynArts and FeynCalc are two different projects developed by different
people, so there is no 100% comptability between the two. For practical
purposes one always needs to convert the FeynArts output into something
FeynCalc can work with.
Have a look at the examples, for more details
https://github.com/FeynCalc/feyncalc/tree/master/FeynCalc/fcexamples/QED
Since you are using FeynCalc 8.2, you don't need to apply to
FCPrepareFAAmp and specify Transversality->True in PolarizationVector.
Those are things that came with FeynCalc 9 that is still under devlopment.
Also, have a look at the corresponding wiki page:
https://github.com/FeynCalc/feyncalc/wiki/FeynArts#fatofc
Cheers,
On 15/12/14 20:12, Ralph wrote:
> Thanks for the suggestion Vladyslav, turns out FeynArtsDirectory was set to ./FeynArts-3.7/� and after I changed it seems to load ok. Is it possible for FC to process the output of FA so that traces are contracted etc? I tried the following:
>
> t = CreateTopologies[0, 2 -> 2];��������
> ww = InsertFields[t, {F[3], -F[3]} -> {V[3], -V[3]}];���
> ampww = CreateFeynAmp[ww, Truncated -> False];��
> tmp2 = Apply[List, ampww] /.
> FeynAmp[_, _, am_, _] :> (am /. RelativeCF -> 1) /.
> Polarization[a_, b_, _] :> Polarization[a, b]
> enmomcon = k2 -> p2 + p1 - k1;
> tmp3 = Explicit[Plus @@ tmp2, Gauge -> (1 - \[Alpha]),
> Dimension -> 4] /. enmomcon���
>
> Calc[tmp3]
>
> But all the gamma matrices etc are still uncontracted. Is there something else I should do, or is it not possible to convert the CreateFeynAmp output into the type of analytical expression you can usually get in FC?
>
This archive was generated by hypermail 2b29 : 02/22/19-05:20:01 AM Z CET | 596 | 1,893 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2019-09 | latest | en | 0.783459 |
https://kunduz.com/questions-and-answers/optional-extra-credit-challenge-what-would-it-cost-to-send-a-rocket-to-mars-79-x-107-km-from-the-earth-if-the-average-speed-of-the-rocket-is-1600-miles-per-minute-and-the-fuel-consumption-averages-100-grams-every-15-seconds-1178/ | 1,701,373,618,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100232.63/warc/CC-MAIN-20231130193829-20231130223829-00302.warc.gz | 404,989,064 | 32,374 | Question:
# Optional Extra Credit Challenge: What would it cost to send
Last updated: 7/4/2022
Optional Extra Credit Challenge: What would it cost to send a rocket to Mars (7.9 x 107 km from the earth) if the average speed of the rocket is 1600 miles per minute and the fuel consumption averages 100 grams every 1.5 seconds and the cost of the fuel is \$500.00 per pound. Use correct significant figures in your answer. Note: 1.6 km = 1 mi 2.2 lbs= 1 kg | 124 | 456 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2023-50 | longest | en | 0.885476 |
https://onlinejudge.org/board/viewtopic.php?p=30403 | 1,603,204,227,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107872746.20/warc/CC-MAIN-20201020134010-20201020164010-00388.warc.gz | 479,921,977 | 13,606 | ## 160 - Factors and Factorials
Moderator: Board moderators
jambon_vn
New poster
Posts: 15
Joined: Wed Sep 29, 2004 6:03 am
Thank guys. I have the same mistake. I got AC now.
afonsocsc
New poster
Posts: 34
Joined: Mon Mar 24, 2003 1:15 am
Location: Portugal, Lisbon
Hi,
Check out inputs from 47 to 52.
chuzpa
New poster
Posts: 33
Joined: Fri Jun 18, 2004 8:39 am
Location: Mexico
Contact:
### 160 WA but why ??
Could someone help me ?. I've tried to correct the output format, and the results of the outputs seams to be right :S.
here's the code ...
#include <stdio.h>
#define X 101
int p[X],N,k=0,a[X],used[X],u[X],t;
isPrime (int x){
int i,r;
for(i=2;i<x;i++)
if (x%i==0)
return 0;
return 1;
}
void generatePrimes(){
int i;
p[k++]=2;
for(i=3;k<=N;i++)
if (isPrime( i ))
p[k++] = i;
}
void copy(){
int i;
for(i=0;i<N;i++)
used+=u;
}
void wichPrimesMake(long long x, long long mult,int ind){
int calc = mult*p[ind],i;
if (calc>x || t)
return;
mult = calc;
u[ind]++;
if (mult == x){
copy();
t = 1;
u[ind]--;
return;
}
for(i=0;i<=x;i++)
if((mult*p)<=x)
wichPrimesMake(x,mult,i);
u[ind]--;
}
int busbin(int x){
int desde=0,hasta=k-1,m;
do{
if (p[desde]==x)return desde;
if (p[hasta]==x)return hasta;
m = (desde+hasta)/2;
if (p[m]==x)return m;
if (p[m]>x)
hasta = m-1;
if (p[m]<x)
desde = m+1;
}while(desde<hasta);
return -1;
}
void ff(int d){
int i,j,x;
for(i=d;i<=N;i++){
x = busbin(i);
if (x!=-1)
used[x]++;
else{
t = 0;
for(j=0;j<k;j++){
wichPrimesMake(i,1,j);
if(t)break;
}
}
}
}
void impr(){
int i,f=0;
printf("%3d! = ",N);
for(i=0;p<=N;i++){
if (i==15 && p[i+1]<=N)
printf("\n ");
printf("%3d",used);
}
printf("\n");
}
void limpia(){
int i;
for(i=0;i<=N;i++)
used=0;
}
int main(){
int lastn,desde;
N = 100;
generatePrimes();
while(scanf("%d",&N)!=EOF){
if (N<2)continue;
if(N<lastn){
limpia();
desde = 2;
}
else
desde = lastn+1;
if (N==lastn){impr(); continue;}
ff(desde);
impr();
lastn=N;
}
return 0;
}
thanx for the help!
Minerva
New poster
Posts: 2
Joined: Thu Nov 25, 2004 10:43 am
I run ur program on my PC, I found that ur program can't output the answer, and I also think the solution to the problem is very easy ---- just
creat a prime list ---- prime[num_prime] and define an integer array to record the num ---- sum[num_prime], then do as the following things:
int point = 0;
while ( N / prime[point] !=0 && point < num_prime){
int p = prime[point];
while ( N / p != 0){
sum[point] += (N / p);
p *= prime[point];
}
point++;
}
I am not good at English, but I hope you can understand what I say, also I hope you can get AC. Good luck !
chuzpa
New poster
Posts: 33
Joined: Fri Jun 18, 2004 8:39 am
Location: Mexico
Contact:
hi, yeap I know i'ts very simple but in my computer it does have an output :S, in fact are these ones, are them wrong ?
Code: Select all
`````` 2! = 1
3! = 1 1
4! = 3 1
5! = 3 1 1
6! = 4 2 1
7! = 4 2 1 1
8! = 7 2 1 1
9! = 7 4 1 1
10! = 8 4 2 1
11! = 8 4 2 1 1
12! = 10 5 2 1 1
13! = 10 5 2 1 1 1
14! = 11 5 2 2 1 1
15! = 11 6 3 2 1 1
16! = 15 6 3 2 1 1
17! = 15 6 3 2 1 1 1
18! = 16 8 3 2 1 1 1
19! = 16 8 3 2 1 1 1 1
20! = 18 8 4 2 1 1 1 1
21! = 18 9 4 3 1 1 1 1
22! = 19 9 4 3 2 1 1 1
23! = 19 9 4 3 2 1 1 1 1
24! = 22 10 4 3 2 1 1 1 1
25! = 22 10 6 3 2 1 1 1 1
26! = 23 10 6 3 2 2 1 1 1
27! = 23 13 6 3 2 2 1 1 1
28! = 25 13 6 4 2 2 1 1 1
29! = 25 13 6 4 2 2 1 1 1 1
30! = 26 14 7 4 2 2 1 1 1 1
31! = 26 14 7 4 2 2 1 1 1 1 1
32! = 31 14 7 4 2 2 1 1 1 1 1
33! = 31 15 7 4 3 2 1 1 1 1 1
34! = 32 15 7 4 3 2 2 1 1 1 1
35! = 32 15 8 5 3 2 2 1 1 1 1
36! = 34 17 8 5 3 2 2 1 1 1 1
37! = 34 17 8 5 3 2 2 1 1 1 1 1
38! = 35 17 8 5 3 2 2 2 1 1 1 1
39! = 35 18 8 5 3 3 2 2 1 1 1 1
40! = 38 18 9 5 3 3 2 2 1 1 1 1
41! = 38 18 9 5 3 3 2 2 1 1 1 1 1
42! = 39 19 9 6 3 3 2 2 1 1 1 1 1
43! = 39 19 9 6 3 3 2 2 1 1 1 1 1 1
44! = 41 19 9 6 4 3 2 2 1 1 1 1 1 1
45! = 41 21 10 6 4 3 2 2 1 1 1 1 1 1
46! = 42 21 10 6 4 3 2 2 2 1 1 1 1 1
47! = 42 21 10 6 4 3 2 2 2 1 1 1 1 1 1
48! = 46 22 10 6 4 3 2 2 2 1 1 1 1 1 1
49! = 46 22 10 8 4 3 2 2 2 1 1 1 1 1 1
50! = 47 22 12 8 4 3 2 2 2 1 1 1 1 1 1
51! = 47 23 12 8 4 3 3 2 2 1 1 1 1 1 1
52! = 49 23 12 8 4 4 3 2 2 1 1 1 1 1 1
53! = 49 23 12 8 4 4 3 2 2 1 1 1 1 1 1 1
54! = 50 26 12 8 4 4 3 2 2 1 1 1 1 1 1 1
55! = 50 26 13 8 5 4 3 2 2 1 1 1 1 1 1 1
56! = 53 26 13 9 5 4 3 2 2 1 1 1 1 1 1 1
57! = 53 27 13 9 5 4 3 3 2 1 1 1 1 1 1 1
58! = 54 27 13 9 5 4 3 3 2 2 1 1 1 1 1 1
59! = 54 27 13 9 5 4 3 3 2 2 1 1 1 1 1
1 1
60! = 56 28 14 9 5 4 3 3 2 2 1 1 1 1 1
1 1
61! = 56 28 14 9 5 4 3 3 2 2 1 1 1 1 1
1 1 1
62! = 57 28 14 9 5 4 3 3 2 2 2 1 1 1 1
1 1 1
63! = 57 30 14 10 5 4 3 3 2 2 2 1 1 1 1
1 1 1
64! = 63 30 14 10 5 4 3 3 2 2 2 1 1 1 1
1 1 1
65! = 63 30 15 10 5 5 3 3 2 2 2 1 1 1 1
1 1 1
66! = 64 31 15 10 6 5 3 3 2 2 2 1 1 1 1
1 1 1
67! = 64 31 15 10 6 5 3 3 2 2 2 1 1 1 1
1 1 1 1
68! = 66 31 15 10 6 5 4 3 2 2 2 1 1 1 1
1 1 1 1
69! = 66 32 15 10 6 5 4 3 3 2 2 1 1 1 1
1 1 1 1
70! = 67 32 16 11 6 5 4 3 3 2 2 1 1 1 1
1 1 1 1
71! = 67 32 16 11 6 5 4 3 3 2 2 1 1 1 1
1 1 1 1 1
72! = 70 34 16 11 6 5 4 3 3 2 2 1 1 1 1
1 1 1 1 1
73! = 70 34 16 11 6 5 4 3 3 2 2 1 1 1 1
1 1 1 1 1 1
74! = 71 34 16 11 6 5 4 3 3 2 2 2 1 1 1
1 1 1 1 1 1
75! = 71 35 18 11 6 5 4 3 3 2 2 2 1 1 1
1 1 1 1 1 1
76! = 73 35 18 11 6 5 4 4 3 2 2 2 1 1 1
1 1 1 1 1 1
77! = 73 35 18 12 7 5 4 4 3 2 2 2 1 1 1
1 1 1 1 1 1
78! = 74 36 18 12 7 6 4 4 3 2 2 2 1 1 1
1 1 1 1 1 1
79! = 74 36 18 12 7 6 4 4 3 2 2 2 1 1 1
1 1 1 1 1 1 1
80! = 78 36 19 12 7 6 4 4 3 2 2 2 1 1 1
1 1 1 1 1 1 1
81! = 78 40 19 12 7 6 4 4 3 2 2 2 1 1 1
1 1 1 1 1 1 1
82! = 79 40 19 12 7 6 4 4 3 2 2 2 2 1 1
1 1 1 1 1 1 1
83! = 79 40 19 12 7 6 4 4 3 2 2 2 2 1 1
1 1 1 1 1 1 1 1
84! = 81 41 19 13 7 6 4 4 3 2 2 2 2 1 1
1 1 1 1 1 1 1 1
85! = 81 41 20 13 7 6 5 4 3 2 2 2 2 1 1
1 1 1 1 1 1 1 1
86! = 82 41 20 13 7 6 5 4 3 2 2 2 2 2 1
1 1 1 1 1 1 1 1
87! = 82 42 20 13 7 6 5 4 3 3 2 2 2 2 1
1 1 1 1 1 1 1 1
88! = 85 42 20 13 8 6 5 4 3 3 2 2 2 2 1
1 1 1 1 1 1 1 1
89! = 85 42 20 13 8 6 5 4 3 3 2 2 2 2 1
1 1 1 1 1 1 1 1 1
90! = 86 44 21 13 8 6 5 4 3 3 2 2 2 2 1
1 1 1 1 1 1 1 1 1
91! = 86 44 21 14 8 7 5 4 3 3 2 2 2 2 1
1 1 1 1 1 1 1 1 1
92! = 88 44 21 14 8 7 5 4 4 3 2 2 2 2 1
1 1 1 1 1 1 1 1 1
93! = 88 45 21 14 8 7 5 4 4 3 3 2 2 2 1
1 1 1 1 1 1 1 1 1
94! = 89 45 21 14 8 7 5 4 4 3 3 2 2 2 2
1 1 1 1 1 1 1 1 1
95! = 89 45 22 14 8 7 5 5 4 3 3 2 2 2 2
1 1 1 1 1 1 1 1 1
96! = 94 46 22 14 8 7 5 5 4 3 3 2 2 2 2
1 1 1 1 1 1 1 1 1
97! = 94 46 22 14 8 7 5 5 4 3 3 2 2 2 2
1 1 1 1 1 1 1 1 1 1
98! = 95 46 22 16 8 7 5 5 4 3 3 2 2 2 2
1 1 1 1 1 1 1 1 1 1
99! = 95 48 22 16 9 7 5 5 4 3 3 2 2 2 2
1 1 1 1 1 1 1 1 1 1
100! = 97 48 24 16 9 7 5 5 4 3 3 2 2 2 2
1 1 1 1 1 1 1 1 1 1
``````
Minerva
New poster
Posts: 2
Joined: Thu Nov 25, 2004 10:43 am
I have found the mistake !!!
If the input is :
5
6
7
8
9
12
5
10
22
3
4
12
0
My AC program's output is :
5! = 3 1 1
6! = 4 2 1
7! = 4 2 1 1
8! = 7 2 1 1
9! = 7 4 1 1
12! = 10 5 2 1 1
5! = 3 1 1
10! = 8 4 2 1
22! = 19 9 4 3 2 1 1 1
3! = 1 1
4! = 3 1
12! = 10 5 2 1 1
But ur's output is :
5! = 3 1 1
6! = 4 2 1
7! = 4 2 1 1
8! = 7 2 1 1
9! = 7 4 1 1
12! = 10 5 2 1 1
5! = 3 1 1
10! = 8 4 2 1
22! = 19 9 4 3 2 1 1 1
3! = 1 1
4! = 3 1
12! = 10 5 2 1 3
Why the same input data but different output?
chuzpa
New poster
Posts: 33
Joined: Fri Jun 18, 2004 8:39 am
Location: Mexico
Contact:
thanks a lot!, I've got AC! well . P.E, but that's o.k for me , I was not cleaning well an array .
thanks again!!
Andrey
New poster
Posts: 16
Joined: Sat Mar 05, 2005 8:25 pm
Location: Ukraine,Vinnitsa
### 160 WA
I don't know what's wrong in my code?
Last edited by Andrey on Tue Mar 08, 2005 11:49 am, edited 1 time in total.
chunyi81
A great helper
Posts: 293
Joined: Sat Jun 21, 2003 4:19 am
Location: Singapore
I ran your code with some inputs
Your code give arithmetic exception with maxn >= 97
Also, your code doesn't handle more than one input properly.
Try with the following input:
53
54
Notice that output for 54 is joined to the second of output for 53.
sunnycare
Learning poster
Posts: 74
Joined: Tue Mar 08, 2005 2:35 am
Location: China , Shanghai
### Why "Floating point exception"---problem 160
Code: Select all
``````#include <iostream>
using namespace std;
unsigned long primeTable[25]={2,3,5,7,
11,13,17,19,
23,29,
31,37,
41,43,47,
53,59,
61,67,
71,73,79,
83,89,
97};
unsigned long func(unsigned long n,unsigned long p)
{
unsigned long i=0;
while(n>0)
{
n=n/p;
i+=n;
}
return i;
}
void main()
{
unsigned long n;
cin>>n;
while(n!=0)
{
cout.width(3);
cout.setf(ios::right);
cout<<n;
cout<<"! =";
for(unsigned long i=0;primeTable[i]<=n;i++)
{
if(i==15)
cout<<endl<<" ";
cout.width(3);
cout.setf(ios::right);
cout<<func(n,primeTable[i]);
}
cout<<endl;
cin>>n;
}
}``````
the judhe said:
Your program has died with signal 8 (SIGFPE). Meaning:
Floating point exception
Before crash, it ran during 0.000 seconds.
nicu_ivan
New poster
Posts: 7
Joined: Wed Mar 16, 2005 7:27 pm
### Re: Why "Floating point exception"---problem 160
Trie this source, try not to use iostream.h but fstream.h
sunnycare wrote:
Code: Select all
``````#include <fstream>
using namespace std;
unsigned long primeTable[25]={2,3,5,7,
11,13,17,19,
23,29,
31,37,
41,43,47,
53,59,
61,67,
71,73,79,
83,89,
97};
unsigned long func(unsigned long n,unsigned long p)
{
unsigned long i=0;
while(n>0)
{
n=n/p;
i+=n;
}
return i;
}
void main()
{
unsigned long n;
cin>>n;
while(n!=0)
{
cout.width(3);
cout.setf(ios::right);
cout<<n;
cout<<"! =";
for(unsigned long i=0;primeTable[i]<=n;i++)
{
if(i==15)
cout<<endl<<" ";
cout.width(3);
cout.setf(ios::right);
cout<<func(n,primeTable[i]);
}
cout<<endl;
cin>>n;
}
}``````
the judhe said:
Your program has died with signal 8 (SIGFPE). Meaning:
Floating point exception
Before crash, it ran during 0.000 seconds.
sunnycare
Learning poster
Posts: 74
Joined: Tue Mar 08, 2005 2:35 am
Location: China , Shanghai
why use fstream?
Krzysztof Duleba
Guru
Posts: 584
Joined: Thu Jun 19, 2003 3:48 am
Location: Sanok, Poland
Contact:
Good question - why? I'd like to know too
The code crashes for a simple reason. Assume that the next value after primeTable is 0. What will happen?
sunnycare
Learning poster
Posts: 74
Joined: Tue Mar 08, 2005 2:35 am
Location: China , Shanghai
right..thanks..
ugrash
New poster
Posts: 5
Joined: Fri Apr 01, 2005 2:06 pm
### 160 WA ?? (I've already searched in the forum :))
Hi,
I am always WA, I checked my output and can't find any error.
Heelp !
#include <stdio.h>
int premiers[25];
int entree[99];
int nbrentres;
int res[25];
int resultats[99][25];
int maxis[99];
int maxx;
void decompose(int *max, int nbr)
{
int k=0, j;
*max = 0;
for (j=0;j<25;j++) res[j]=0;
while (nbr != 1)
{
if (nbr % premiers[k] == 0)
{
nbr = nbr / premiers[k];
res[k] = res[k] + 1;
*max = k;
}
else k++;
}
}
void calcstok(int nbr)
{
int i, j, mx=0;
if (maxis[nbr - 2] == -1)
{
for(i=maxx+1;i<nbr-1;i++)
{
decompose(&mx, i+2);
for(j=0;j<25;j++) resultats[j] = resultats[i-1][j] + res[j];
maxis=mx>maxis[i-1]?mx:maxis[i-1];
maxx = nbr-2;
}
}
}
int main(int argc, char *argv[])
{
int i,j,k,m;
k = maxx = nbrentres = 0;
premiers[0] = 2;
premiers[1] = 3;
premiers[2] = 5;
premiers[3] = 7;
premiers[4] = 11;
premiers[5] = 13;
premiers[6] = 17;
premiers[7] = 19;
premiers[8] = 23;
premiers[9] = 29;
premiers[10] = 31;
premiers[11] = 37;
premiers[12] = 41;
premiers[13] = 43;
premiers[14] = 47;
premiers[15] = 53;
premiers[16] = 59;
premiers[17] = 61;
premiers[18] = 67;
premiers[19] = 71;
premiers[20] = 73;
premiers[21] = 79;
premiers[22] = 83;
premiers[23] = 89;
premiers[24] = 97;
for(j=0;j<99;j++)
{
for(i=0;i<25;i++) resultats[j] = 0;
maxis[j] = -1;
entree[j] = 0;
}
maxis[0] = 0;
resultats[0][0] = 1;
while (scanf("%d",&m)==1)
{
if (m!=0) entree[k++] = m; else break;
nbrentres++;
}
for (k=0;k<nbrentres;k++)
{
m = entree[k];
calcstok(m);
printf("%3d! =",m);
for(i=0;i<maxis[m-2]+1;i++)
{
printf("%3d",resultats[m-2]);
if ((i+1)%15 == 0) printf("\n ");
}
printf("\n");
}
return 0;
} | 7,707 | 13,180 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2020-45 | latest | en | 0.270259 |
https://www.lmfdb.org/EllipticCurve/Q/2880/q/5 | 1,716,047,389,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057440.7/warc/CC-MAIN-20240518152242-20240518182242-00423.warc.gz | 779,288,579 | 27,305 | Properties
Label 2880.q5 Conductor $2880$ Discriminant $1.016\times 10^{15}$ j-invariant $$\frac{35578826569}{5314410}$$ CM no Rank $1$ Torsion structure $$\Z/{2}\Z$$
Related objects
Show commands: Magma / Oscar / PariGP / SageMath
Simplified equation
$$y^2=x^3-39468x-2599472$$ y^2=x^3-39468x-2599472 (homogenize, simplify) $$y^2z=x^3-39468xz^2-2599472z^3$$ y^2z=x^3-39468xz^2-2599472z^3 (dehomogenize, simplify) $$y^2=x^3-39468x-2599472$$ y^2=x^3-39468x-2599472 (homogenize, minimize)
comment: Define the curve
sage: E = EllipticCurve([0, 0, 0, -39468, -2599472])
gp: E = ellinit([0, 0, 0, -39468, -2599472])
magma: E := EllipticCurve([0, 0, 0, -39468, -2599472]);
oscar: E = EllipticCurve([0, 0, 0, -39468, -2599472])
sage: E.short_weierstrass_model()
magma: WeierstrassModel(E);
oscar: short_weierstrass_model(E)
Mordell-Weil group structure
$$\Z \oplus \Z/{2}\Z$$
magma: MordellWeilGroup(E);
Infinite order Mordell-Weil generator and height
$P$ = $$\left(-99, 581\right)$$ (-99, 581) $\hat{h}(P)$ ≈ $5.0167456972043204032045171979$
sage: E.gens()
magma: Generators(E);
gp: E.gen
Torsion generators
$$\left(-148, 0\right)$$
comment: Torsion subgroup
sage: E.torsion_subgroup().gens()
gp: elltors(E)
magma: TorsionSubgroup(E);
oscar: torsion_structure(E)
Integral points
$$\left(-148, 0\right)$$, $$(-99,\pm 581)$$
comment: Integral points
sage: E.integral_points()
magma: IntegralPoints(E);
Invariants
Conductor: $$2880$$ = $2^{6} \cdot 3^{2} \cdot 5$ comment: Conductor sage: E.conductor().factor() gp: ellglobalred(E)[1] magma: Conductor(E); oscar: conductor(E) Discriminant: $1015599566684160$ = $2^{19} \cdot 3^{18} \cdot 5$ comment: Discriminant sage: E.discriminant().factor() gp: E.disc magma: Discriminant(E); oscar: discriminant(E) j-invariant: $$\frac{35578826569}{5314410}$$ = $2^{-1} \cdot 3^{-12} \cdot 5^{-1} \cdot 11^{3} \cdot 13^{3} \cdot 23^{3}$ comment: j-invariant sage: E.j_invariant().factor() gp: E.j magma: jInvariant(E); oscar: j_invariant(E) Endomorphism ring: $\Z$ Geometric endomorphism ring: $$\Z$$ (no potential complex multiplication) sage: E.has_cm() magma: HasComplexMultiplication(E); Sato-Tate group: $\mathrm{SU}(2)$ Faltings height: $1.6038867560562696444150905038\dots$ gp: ellheight(E) magma: FaltingsHeight(E); oscar: faltings_height(E) Stable Faltings height: $0.014859840882296834591619703151\dots$ magma: StableFaltingsHeight(E); oscar: stable_faltings_height(E) $abc$ quality: $1.0339287621454463\dots$ Szpiro ratio: $5.443863087913462\dots$
BSD invariants
Analytic rank: $1$ sage: E.analytic_rank() gp: ellanalyticrank(E) magma: AnalyticRank(E); Regulator: $5.0167456972043204032045171979\dots$ comment: Regulator sage: E.regulator() G = E.gen \\ if available matdet(ellheightmatrix(E,G)) magma: Regulator(E); Real period: $0.34210678663966561319088785404\dots$ comment: Real Period sage: E.period_lattice().omega() gp: if(E.disc>0,2,1)*E.omega[1] magma: (Discriminant(E) gt 0 select 2 else 1) * RealPeriod(E); Tamagawa product: $8$ = $2\cdot2^{2}\cdot1$ comment: Tamagawa numbers sage: E.tamagawa_numbers() gp: gr=ellglobalred(E); [[gr[4][i,1],gr[5][i][4]] | i<-[1..#gr[4][,1]]] magma: TamagawaNumbers(E); oscar: tamagawa_numbers(E) Torsion order: $2$ comment: Torsion order sage: E.torsion_order() gp: elltors(E)[1] magma: Order(TorsionSubgroup(E)); oscar: prod(torsion_structure(E)[1]) Analytic order of Ш: $1$ ( rounded) comment: Order of Sha sage: E.sha().an_numerical() magma: MordellWeilShaInformation(E); Special value: $L'(E,1)$ ≈ $3.4325254997178779021589425933$ comment: Special L-value r = E.rank(); E.lseries().dokchitser().derivative(1,r)/r.factorial() gp: [r,L1r] = ellanalyticrank(E); L1r/r! magma: Lr1 where r,Lr1 := AnalyticRank(E: Precision:=12);
BSD formula
$\displaystyle 3.432525500 \approx L'(E,1) = \frac{\# Ш(E/\Q)\cdot \Omega_E \cdot \mathrm{Reg}(E/\Q) \cdot \prod_p c_p}{\#E(\Q)_{\rm tor}^2} \approx \frac{1 \cdot 0.342107 \cdot 5.016746 \cdot 8}{2^2} \approx 3.432525500$
# self-contained SageMath code snippet for the BSD formula (checks rank, computes analytic sha)
E = EllipticCurve(%s); r = E.rank(); ar = E.analytic_rank(); assert r == ar;
Lr1 = E.lseries().dokchitser().derivative(1,r)/r.factorial(); sha = E.sha().an_numerical();
omega = E.period_lattice().omega(); reg = E.regulator(); tam = E.tamagawa_product(); tor = E.torsion_order();
assert r == ar; print("analytic sha: " + str(RR(Lr1) * tor^2 / (omega * reg * tam)))
/* self-contained Magma code snippet for the BSD formula (checks rank, computes analyiic sha) */
E := EllipticCurve(%s); r := Rank(E); ar,Lr1 := AnalyticRank(E: Precision := 12); assert r eq ar;
sha := MordellWeilShaInformation(E); omega := RealPeriod(E) * (Discriminant(E) gt 0 select 2 else 1);
reg := Regulator(E); tam := &*TamagawaNumbers(E); tor := #TorsionSubgroup(E);
assert r eq ar; print "analytic sha:", Lr1 * tor^2 / (omega * reg * tam);
Modular invariants
$$q - q^{5} + 4 q^{7} - 2 q^{13} - 6 q^{17} - 4 q^{19} + O(q^{20})$$
comment: q-expansion of modular form
sage: E.q_eigenform(20)
\\ actual modular form, use for small N
[mf,F] = mffromell(E)
Ser(mfcoefs(mf,20),q)
\\ or just the series
Ser(ellan(E,20),q)*q
magma: ModularForm(E);
Modular degree: 12288
comment: Modular degree
sage: E.modular_degree()
gp: ellmoddegree(E)
magma: ModularDegree(E);
$\Gamma_0(N)$-optimal: no
Manin constant: 1
comment: Manin constant
magma: ManinConstant(E);
Local data
This elliptic curve is not semistable. There are 3 primes of bad reduction:
prime Tamagawa number Kodaira symbol Reduction type Root number ord($N$) ord($\Delta$) ord$(j)_{-}$
$2$ $2$ $I_{9}^{*}$ Additive -1 6 19 1
$3$ $4$ $I_{12}^{*}$ Additive -1 2 18 12
$5$ $1$ $I_{1}$ Non-split multiplicative 1 1 1 1
comment: Local data
sage: E.local_data()
gp: ellglobalred(E)[5]
magma: [LocalInformation(E,p) : p in BadPrimes(E)];
oscar: [(p,tamagawa_number(E,p), kodaira_symbol(E,p), reduction_type(E,p)) for p in bad_primes(E)]
Galois representations
The $\ell$-adic Galois representation has maximal image for all primes $\ell$ except those listed in the table below.
prime $\ell$ mod-$\ell$ image $\ell$-adic image
$2$ 2B 8.12.0.12
$3$ 3B 3.4.0.1
comment: mod p Galois image
sage: rho = E.galois_representation(); [rho.image_type(p) for p in rho.non_surjective()]
magma: [GaloisRepresentation(E,p): p in PrimesUpTo(20)];
gens = [[89, 96, 90, 119], [1, 12, 12, 25], [1, 24, 0, 1], [97, 24, 96, 25], [1, 0, 24, 1], [15, 106, 14, 11], [112, 3, 117, 34], [78, 41, 43, 28], [59, 96, 28, 91]]
GL(2,Integers(120)).subgroup(gens)
Gens := [[89, 96, 90, 119], [1, 12, 12, 25], [1, 24, 0, 1], [97, 24, 96, 25], [1, 0, 24, 1], [15, 106, 14, 11], [112, 3, 117, 34], [78, 41, 43, 28], [59, 96, 28, 91]];
sub<GL(2,Integers(120))|Gens>;
The image $H:=\rho_E(\Gal(\overline{\Q}/\Q))$ of the adelic Galois representation has level $$120 = 2^{3} \cdot 3 \cdot 5$$, index $384$, genus $5$, and generators
$\left(\begin{array}{rr} 89 & 96 \\ 90 & 119 \end{array}\right),\left(\begin{array}{rr} 1 & 12 \\ 12 & 25 \end{array}\right),\left(\begin{array}{rr} 1 & 24 \\ 0 & 1 \end{array}\right),\left(\begin{array}{rr} 97 & 24 \\ 96 & 25 \end{array}\right),\left(\begin{array}{rr} 1 & 0 \\ 24 & 1 \end{array}\right),\left(\begin{array}{rr} 15 & 106 \\ 14 & 11 \end{array}\right),\left(\begin{array}{rr} 112 & 3 \\ 117 & 34 \end{array}\right),\left(\begin{array}{rr} 78 & 41 \\ 43 & 28 \end{array}\right),\left(\begin{array}{rr} 59 & 96 \\ 28 & 91 \end{array}\right)$.
Input positive integer $m$ to see the generators of the reduction of $H$ to $\mathrm{GL}_2(\Z/m\Z)$:
The torsion field $K:=\Q(E[120])$ is a degree-$92160$ Galois extension of $\Q$ with $\Gal(K/\Q)$ isomorphic to the projection of $H$ to $\GL_2(\Z/120\Z)$.
Isogenies
gp: ellisomat(E)
This curve has non-trivial cyclic isogenies of degree $d$ for $d=$ 2, 3, 4, 6 and 12.
Its isogeny class 2880.q consists of 8 curves linked by isogenies of degrees dividing 12.
Twists
The minimal quadratic twist of this elliptic curve is 30.a5, its twist by $24$.
Growth of torsion in number fields
The number fields $K$ of degree less than 24 such that $E(K)_{\rm tors}$ is strictly larger than $E(\Q)_{\rm tors}$ $\cong \Z/{2}\Z$ are as follows:
$[K:\Q]$ $E(K)_{\rm tors}$ Base change curve $K$ $2$ $$\Q(\sqrt{10})$$ $$\Z/2\Z \oplus \Z/2\Z$$ Not in database $2$ $$\Q(\sqrt{-30})$$ $$\Z/4\Z$$ Not in database $2$ $$\Q(\sqrt{-3})$$ $$\Z/4\Z$$ Not in database $2$ $$\Q(\sqrt{6})$$ $$\Z/6\Z$$ 2.2.24.1-150.1-e6 $4$ $$\Q(\sqrt{-3}, \sqrt{10})$$ $$\Z/2\Z \oplus \Z/4\Z$$ Not in database $4$ $$\Q(\sqrt{6}, \sqrt{10})$$ $$\Z/2\Z \oplus \Z/6\Z$$ Not in database $4$ $$\Q(\sqrt{-5}, \sqrt{6})$$ $$\Z/12\Z$$ Not in database $4$ $$\Q(\sqrt{-2}, \sqrt{-3})$$ $$\Z/12\Z$$ Not in database $6$ 6.0.11520000.1 $$\Z/6\Z$$ Not in database $8$ 8.4.21233664000000.41 $$\Z/2\Z \oplus \Z/4\Z$$ Not in database $8$ 8.0.5184000000.11 $$\Z/8\Z$$ Not in database $8$ 8.0.8493465600.20 $$\Z/24\Z$$ Not in database $8$ 8.0.3317760000.8 $$\Z/2\Z \oplus \Z/12\Z$$ Not in database $12$ 12.0.1194393600000000.1 $$\Z/3\Z \oplus \Z/12\Z$$ Not in database $12$ deg 12 $$\Z/2\Z \oplus \Z/6\Z$$ Not in database $12$ deg 12 $$\Z/12\Z$$ Not in database $16$ deg 16 $$\Z/4\Z \oplus \Z/4\Z$$ Not in database $16$ 16.0.26873856000000000000.3 $$\Z/2\Z \oplus \Z/8\Z$$ Not in database $16$ deg 16 $$\Z/2\Z \oplus \Z/24\Z$$ Not in database $16$ deg 16 $$\Z/2\Z \oplus \Z/12\Z$$ Not in database $16$ deg 16 $$\Z/24\Z$$ Not in database $18$ 18.6.4663277994109219268198400000000.1 $$\Z/18\Z$$ Not in database
We only show fields where the torsion growth is primitive. For fields not in the database, click on the degree shown to reveal the defining polynomial.
Iwasawa invariants
$p$ Reduction type $\lambda$-invariant(s) $\mu$-invariant(s) 2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 add add nonsplit ord ss ord ord ord ss ord ord ord ord ord ss - - 1 1 1,1 1 1 1 1,1 1 1 1 1 1 1,1 - - 0 0 0,0 0 0 0 0,0 0 0 0 0 0 0,0
An entry - indicates that the invariants are not computed because the reduction is additive.
$p$-adic regulators
$p$-adic regulators are not yet computed for curves that are not $\Gamma_0$-optimal. | 4,048 | 10,207 | {"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} | 3.125 | 3 | CC-MAIN-2024-22 | latest | en | 0.2031 |
http://hmf.enseeiht.fr/travaux/bei/beiep/content/2012-g16/trajectory | 1,582,614,797,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875146033.50/warc/CC-MAIN-20200225045438-20200225075438-00367.warc.gz | 64,200,110 | 4,363 | # Trajectory
Establishment of a model for the water bombing from an aircraft
Theoretically, the movement of a spherical drop in an ambient medium depends on :
• the inertial force
• the drag force
• the lift force
• the History force
• the buoyancy force
To simplify our work, we chose to consider only the drag force and the buoyancy force, the other being neglected compared to those two. The balance of the forces gives :
$\left( m_p+C_M*m_G \right) \frac{d \vec {V_p}}{dt}=\left( m_p-m_G \right)+C_D \frac{\pi {R_p}^2}{2} \rho_G \| \vec W - \vec {V_p} \| \left( \vec W - \vec {V_p} \right)$
Then, we deduce the droplet trajectory :
Velocities values
Droplet trajectory in 3D
Droplet trajectory in the $(x,z)$ plan
For each droplet, the trajectory is straight and linear, as shown in the graphics above. The fall time is around 6 seconds, which is realistic compared to the videos of water bombing.
See next : Evaporation | 271 | 935 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-10 | longest | en | 0.677967 |
https://numbersworksheet.com/multiplying-mixed-numbers-using-area-models-worksheet-pdf/ | 1,716,881,798,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059078.15/warc/CC-MAIN-20240528061449-20240528091449-00292.warc.gz | 371,492,326 | 14,552 | # Multiplying Mixed Numbers Using Area Models Worksheet Pdf
This multiplication worksheet concentrates on educating college students how you can emotionally flourish total phone numbers. Students may use custom made grids to fit specifically one particular query. The worksheets also deal withdecimals and fractions, and exponents. You will even find multiplication worksheets using a dispersed house. These worksheets are a need to-have for the mathematics class. They are often employed in type to learn how to mentally grow complete numbers and line them up. Multiplying Mixed Numbers Using Area Models Worksheet Pdf.
## Multiplication of complete phone numbers
If you want to improve your child’s math skills, you should consider purchasing a multiplication of whole numbers worksheet. These worksheets may help you expert this simple strategy. It is possible to opt for 1 digit multipliers or two-digit and three-digit multipliers. Power of 10 may also be a great choice. These worksheets will help you to process lengthy multiplication and practice studying the phone numbers. They are also the best way to aid your youngster comprehend the significance of learning the different types of entire phone numbers.
## Multiplication of fractions
Experiencing multiplication of fractions on a worksheet might help teachers prepare and prepare instruction proficiently. Using fractions worksheets allows educators to easily assess students’ comprehension of fractions. Pupils may be challenged to finish the worksheet in just a specific time and then symbol their techniques to see in which they need further instruction. Individuals can usually benefit from term problems that connect maths to actual-daily life conditions. Some fractions worksheets involve types of contrasting and comparing figures.
## Multiplication of decimals
Once you increase two decimal figures, ensure that you team them up and down. The product must contain the same number of decimal places as the multiplicant if you want to multiply a decimal number with a whole number. As an example, 01 by (11.2) by 2 could be comparable to 01 x 2.33 by 11.2 unless of course the merchandise has decimal areas of less than two. Then, the item is rounded to the nearest complete number.
## Multiplication of exponents
A arithmetic worksheet for Multiplication of exponents will allow you to exercise multiplying and dividing numbers with exponents. This worksheet will also supply problems that will require college students to increase two various exponents. You will be able to view other versions of the worksheet, by selecting the “All Positive” version. In addition to, you may also enter in special guidelines around the worksheet on its own. When you’re finished, it is possible to just click “Make” and the worksheet is going to be saved.
## Department of exponents
The standard rule for section of exponents when multiplying amounts is to subtract the exponent inside the denominator from the exponent within the numerator. You can simply divide the numbers using the same rule if the bases of the two numbers are not the same. As an example, \$23 split by 4 will the same 27. This method is not always accurate, however. This technique can result in uncertainty when multiplying phone numbers which can be too big or too small.
## Linear functions
You’ve probably noticed that the cost was \$320 x 10 days if you’ve ever rented a car. So the total rent would be \$470. A linear function of this type provides the form f(x), where ‘x’ is the volume of days the automobile was leased. In addition, it offers the shape f(by) = ax b, where ‘a’ and ‘b’ are genuine figures. | 714 | 3,657 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-22 | latest | en | 0.895665 |
http://www.enotes.com/homework-help/what-best-pattern-taylors-series-arctg-x-around-x-450338 | 1,477,603,605,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988721392.72/warc/CC-MAIN-20161020183841-00544-ip-10-171-6-4.ec2.internal.warc.gz | 424,986,512 | 9,154 | # What is the best pattern for taylor's series for arctg x , around x = 0, found with integration or differential of geometric series?
Asked on by luvgoj
llltkl | College Teacher | (Level 3) Valedictorian
Posted on
We know that `(d(arctgx))/(dx)=1/(1+x^2)`
Hence, the best pattern for Taylor's series for `arctg x` , around `x = 0` , can be found from the series for `1/(1+x^2)` .
So, `(d(arctgx))/(dx)` =`1/(1+x^2)` =`1-x^2+x^4-x^6+x^8` `-` .......for` -1ltxlt1`
Integrating term by term gives:
`arctgx` =`int 1/(1+x^2)`
`=C+x-x^3/3+x^5/5-x^7/7+x^9/9` ..............for `-1ltxlt1` , where `C ` is the constant of integration.
Since, `arc tg0=0` we have `C=0` , so,
`arctgx=x-x^3/3+x^5/5-x^7/7+x^9/9` ......for `-1ltxlt1` .
Sources:
We’ve answered 317,890 questions. We can answer yours, too. | 314 | 806 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-44 | latest | en | 0.791599 |
http://docplayer.net/21588682-Excel-basics-by-tom-peters-laura-spielman.html | 1,532,008,477,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676590901.10/warc/CC-MAIN-20180719125339-20180719145339-00616.warc.gz | 110,620,146 | 26,362 | # Excel Basics By Tom Peters & Laura Spielman
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## Transcription
2 Method II d) Next, enter a Step Value and a Stop Value. (The step value indicates the interval length you want to use and the stop value indicates the last value of the input variable you want to use.) Then hit the Return key. You ll notice a column of values going from your start value to the stop value with intervals of length step value. a) Enter the first 2 entries in column form. b) Highlight both entries by dragging the cursor with the mouse over them. c) Move the cursor over the right bottom corner until a black + is seen. d) Proceed to drag down the column until the desired value is obtained. e) Release and click outside the cell. Calculating Sums in Excel: There are 2 ways to add a column (or row) of values in Excel: 1) Just highlight the cells you wish to add along with one additional empty cell. Then, click on the summation button,, in the toolbar. The sum will appear in the empty cell. Or 2) In an empty cell, use the sum command in the form =sum(first:last) Such as =sum(x5:x25). Such a command would calculate the sum X5 + X6 + + X24 + X25 of the values in these cells. NOTE: If you need help remembering how to enter a function or command in Excel, use the Paste Function icon, f x, located on the toolbar. Calculating Function Values in Excel: The most important thing to note here is that Excel does not accept function definitions in terms of a variable, but instead, in terms of values in cells. NOTE: If you need help remembering how to enter a function or command in Excel, use the Paste Function icon, f x, located on the toolbar. Example: Suppose we have the following table of input values in Excel: A B 1 x f(x) Suppose also that we would like to compute the value of the function f(x) = x(3x 2 10) for each of these values. To do this, in cell B2, we would type =A2*(3*A2^2 10) (Include the equals sign) (This tells Excel to take the value in cell A2 and to evaluate it in f(x).)
5 d) Answer: A= B= A + B = Scalar Multiplication : You can multiply any matrix by a scalar by using the formula, = scalar * (first entry: last entry) of the matrix such as =3* (B2:C3). Example: A = A = Multiplication of Two Matrices: Recall that two matrices must have the same number of columns in the multiplicand as rows in the multiplier for multiplication to be defined. For example the product of an (mxn) and a (nxp)is a (mxp). a) You will need to highlight the correct number of rows and columns for the product. b) Use the mmult command for matrix multiplication. c) Type = mmult (matrix 1 first entry: matrix 1 last entry, matrix 2 first entry :matrix 2 last entry) d) Don t forget to hold down shift + control then enter. Example A= B = 2 3 AB =
6 Matrix Inverse: To solve systems of equations you will need to find the inverse of a matrix. Recall that not all matrices are invertible. a) Highlight the area for the answer as the inverse will have the same dimensions as the original matrix. b)use the command minverse c)type = minverse (first entry : last entry) d) Hold control + shift then enter. Example A= A-1=
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https://ptolemy.berkeley.edu/eecs20/speech/lp.html | 1,722,842,061,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640434051.17/warc/CC-MAIN-20240805052118-20240805082118-00475.warc.gz | 390,431,052 | 5,712 | EECS20N: Signals and Systems
# Speech Processing Using Linear Prediction
In this set of demonstrations, we illustrate the modern equivalent of the 1939 Dudley vocoder demonstration. Instead of a bank of bandpass filters, modern vocoders use a single filter (usually implemented in a so-called lattice filter structure). The filter coefficients are calculated using any of a number of algorithms (based on linear prediction). In the examples below, we use an algorithm due to John Burg to calculate the filter coefficients.
The original speech signal If you were able to run applets, you would have a button here that would play a sound. , borrowed from the Voder demo, is sampled at 8kHz. The signal is broken into segments of 160 samples (20ms). Each segment is analyzed using Burg's algorithm for its spectral content (a tenth order linear predictor is the result of this analysis).
A linear predictor uses observations of a signal to try to predict the next sample of the signal beyond those it can observe. The overall structure is as shown:
The input signal, x'(n), is delayed by one sample by the block labeled z-1. The block labeled F(z) is a filter whose output y(n) is an estimate of the current value of x'(n). Since that block only sees a delayed version of x'(n), its output is a prediction. The error in the prediction, e(n), is the difference between the what is being predicted and the prediction.
Intuitively, linear prediction exploits the fact that a new sample of a signal is not totally independent of previous samples, usually. It captures that dependence. As such, when a predictor is working well, the error signal will have little residual correlation between samples. If the input to the linear predictor is the original voder speech signal If you were able to run applets, you would have a button here that would play a sound. , then the error signal If you were able to run applets, you would have a button here that would play a sound. is not very intelligible. Intelligibility, therefore, must somehow depend on the correlation between samples in the signal.
In fact, for speech, the linear predictor has to constantly change to adapt to what is being said. The input signal is divided into 20ms segments, and each segment is analyzed to provide the coefficients of the prediction filter, as shown below:
A box labeled "Burg's algorithm" uses one of several methods for calculating the coefficients of the linear predictor each 20ms.
We can check the hypothesis that intelligibility depends on the correlation between samples by introducing the correlation into some random signal that has no speech content. Suppose we start with a white noise signal If you were able to run applets, you would have a button here that would play a sound. and filter it with the inverse of the prediction error filter above (changing the filter coefficients every 20ms). The result will be intelligible whispered speech If you were able to run applets, you would have a button here that would play a sound. . The block diagram used to create this synthetic speech is as follows:
Burg's algorithm is used again to analyze each 20ms segment of speech, but now the results of the analysis are loaded into an "inverse lattice filter," which implements the inverse of the filter above that produced the prediction error.
Notice that since the result is intelligible, information about intelligibility is almost entirely in the results produced by Burg's algorithm. Thus, such an algorithm can (and often does) form the front end of any device that analyzes speech, such as speech recognition system or a speech encoder.
The whispered speech effect above, while intelligible, sounds, well ... whispered. The problem here is that the excitation, white noise, does not match well what the human vocal cords do. The human vocal cords, which provide an excitation signal in natural speech, vibrate at a frequency that depends on the speaker (and whether the speaker is male or female) and on the inflection intended by the speaker. A naive way to try to replicate the effect of the vocal cords is to use a sinusoidal excitation instead of white noise. The result If you were able to run applets, you would have a button here that would play a sound. is completely unintelligible. There is simply not enough spectral richness in a sinusoid.
An alternative excitation that is more spectrally rich is a periodic sequence of impulses, which looks like this:
If we set the period at 40 samples (5ms, or 39 zero-valued samples for every non-zero sample), then the excitation If you were able to run applets, you would have a button here that would play a sound. has a perceptual pitch of 200Hz, but obviously no discernable speech content. Filtering it with the inverse lattice filter yields intelligible, if mechanical sounding speech If you were able to run applets, you would have a button here that would play a sound. . Increasing the period to 80 samples results in speech with a lower tone If you were able to run applets, you would have a button here that would play a sound. .
With a periodic pulse excitation, speech sounds very mechanical. A slightly better result If you were able to run applets, you would have a button here that would play a sound. comes from combining white noise with periodic pulses. More sophisticated techniques, such as those used today in digital cellular phones, analyze the speech further to construct much better excitation signals.
Since the intelligibility information is contained in the coefficients produced by Burg's algorithm, we can manipulate the speed of the speech by manipulating these coefficients. For example, if we use every set of coefficients to reconstruct 40ms worth of speech rather than 20ms, the result is slow speech If you were able to run applets, you would have a button here that would play a sound. . Note that we could also get slow speech by using every speech sample twice, but the result If you were able to run applets, you would have a button here that would play a sound. is very different, having the overall pitch shifted down by a factor of two in addition to having the speech slowed down.
We can similarly speed up the speech by using each set of coefficients to reconstruct 10ms worth of speech rather than 20ms, the result is fast speech If you were able to run applets, you would have a button here that would play a sound. . Note that we could also get fast speech by discarding every second speech sample, but the result If you were able to run applets, you would have a button here that would play a sound. is very different, having the overall pitch shifted up by a factor of two in addition to having the speech speeded up.
Finally, synthesizing the speech from a musical excitation If you were able to run applets, you would have a button here that would play a sound. in this case the first few bars of Passio Domini nostri by Arvo Part, yields a particularly interesting result If you were able to run applets, you would have a button here that would play a sound. | 1,467 | 7,071 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2024-33 | latest | en | 0.944376 |
https://www.mathnasium.com/westcounty/news/we-now-have-precise-math-to-describe-how-black-holes-reflect-the-universe | 1,660,511,613,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572077.62/warc/CC-MAIN-20220814204141-20220814234141-00632.warc.gz | 772,145,142 | 19,528 | Get Started by Finding a Local Center
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We Now Have Precise Math to Describe How Black Holes Reflect The Universe
Sep 20, 2021 | Location West County
We Now Have Precise Math to Describe How Black Holes Reflect The Universe
13 JULY 2021
A new set of equations can precisely describe the reflections of the Universe that appear in the warped light around a black hole.
The proximity of each reflection is dependent on the angle of observation with respect to the black hole, and the rate of the black hole's spin, according to a mathematical solution worked out by physics student Albert Sneppen of the Niels Bohr Institute in Denmark.
This is really cool, absolutely, but it's not just really cool. It also potentially gives us a new tool for probing the gravitational environment around these extreme objects.
"There is something fantastically beautiful in now understanding why the images repeat themselves in such an elegant way," Sneppen said. "On top of that, it provides new opportunities to test our understanding of gravity and black holes."
If there's one thing that black holes are famous for, it's their extreme gravity. Specifically that, beyond a certain radius, the fastest achievable velocity in the Universe, that of light in a vacuum, is insufficient to achieve escape velocity.
That point of no return is the event horizon – defined by what's called the Schwarszchild radius – and it's the reason why we say that not even light can escape from a black hole's gravity.
Just outside the black hole's event horizon, however, the environment is also seriously wack. The gravitational field is so powerful that the curvature of space-time is almost circular.
Any photons entering this space will, naturally, have to follow this curvature. This means that, from our perspective, the path of the light appears to be warped and bent.
At the very inner edge of this space, just outside the event horizon, we can see what is called a photon ring, where photons travel in orbit around the black hole multiple times before either falling towards the black hole, or escaping into space.
This means that the light from distant objects behind the black hole can be magnified, distorted and 'reflected' several times. We refer to this as a gravitational lens; the effect can also be seen in other contexts, and is a useful tool for studying the Universe.
So we've known about the effect for some time, and scientists had figured out that the closer you look towards the black hole, the more reflections you see of distant objects.
To get from one image to the next image, you needed to look about 500 times closer to the black hole's optical edge, or the exponential function of two pi (e), but why this was the case was difficult to mathematically describe.
Sneppen's approach was to reformulate the light trajectory, and quantify its linear stability, using second order differential equations. He found not only did his solution mathematically describe why the images repeat at distances of e, but that it could work for a rotating black hole - and that repeat distance is dependent on spin.
"It turns out that when it rotates really fast, you no longer have to get closer to the black hole by a factor of 500, but significantly less," Sneppen said. "In fact, each image is now only 50, or five, or even down to just two times closer to the edge of the black hole."
In practice, this is going to be difficult to observe, at least any time soon - just look at the intense amount of work that went into the unresolved imaging of the ring of light around supermassive black hole PÅwehi (M87*).
Theoretically, however, there should be infinite rings of light around a black hole. Since we have imaged the shadow of a supermassive black hole once, it's hopefully only a matter of time before we're able to obtain better images, and there are already plans for imaging a photon ring.
One day, the infinite images close to a black hole could be a tool for studying not just the physics of black hole space-time, but the objects behind them - repeated in infinite reflections in orbital perpetuity.
The research has been published in Scientific Reports.
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Dunkirk, MD 20754 | 2,326 | 7,790 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2022-33 | latest | en | 0.910509 |
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Since it's quarterly so every 3 months it grows by 6 percent. 5 years = 60 months = 20 quarters. Solution: 20 x 0.06 x \$500,000 = \$600,000
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https://wiki.cosmos.esa.int/planckpla/index.php/HFI_time_response_model | 1,521,844,185,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257649095.35/warc/CC-MAIN-20180323220107-20180324000107-00554.warc.gz | 735,109,493 | 10,590 | # HFI time response model
## LFER4 model
If we write the input signal (power) on a bolometer as $\label{bol_in} s_0(t)=e^{i\omega t}$ the bolometer physical impedance can be written as: $\label{bol_out} s(t)=e^{i\omega t}F(\omega)$ where $\omega$ is the angular frequency of the signal and $F(\omega)$ is the complex intrinsic bolometer transfer function. For the bolometer transfer function is modelled as the sum of 4 single pole low pass filters: $\label{bol_tf} F(\omega) = \sum_{i=0,4} \frac{a_i}{1 + i\omega\tau_i}$ The modulation of the signal is done with a square wave, written here as a composition of sine waves of decreasing amplitude: $\label{sigmod} s'(t)=e^{i\omega t}F(\omega)\sum_{k=0}^{\infty} \frac{e^{i\omega_r(2k+1)t}-e^{-i\omega_r(2k+1)t}}{2i(2k+1)}$ where we have used the Euler relation $\sin x=(e^{ix}-e^{-ix})/2i$ and $\omega_r$ is the angular frequency of the square wave. The modulation frequency is $f_{mod} = \omega_r/2\pi$ and was set to $f_{mod} = 90.18759$Hz in flight. This signal is then filtered by the complex electronic transfer function $H(\omega)$. Setting: $\omega_k^+=\omega+(2k+1)\omega_r$ $\omega_k^-=\omega-(2k+1)\omega_r$ we have: $\label{sigele} \Sigma(t)=\sum_{k=0}^\infty\frac{F(\omega)}{2i(2k+1)}\left[H(\omega_k^+)e^{i\omega_k^+t}-H(\omega_k^-)e^{i\omega_k^-t}\right]$ This signal is then sampled at high frequency ($2 f_{mod} NS$). $NS$ is one of the parameters of the electronics and corresponds to the number of high frequency samples in each modulation semi-period. In order to obtain an output signal sampled every $\pi/\omega_r$ seconds, we must integrate on a semiperiod, as done in the readout. To also include a time shift $\Delta t$, the integral is calculated between $n\pi/\omega_r+\Delta t$ and $(n+1)\pi/\omega_r+\Delta t$ (with $T=2 \pi/\omega_r$ period of the modulation). The time shift $\Delta t$ is encoded in the electronics by the parameter $S_{phase}$, with the relation $\Delta t = S_{phase}/NS/f_{mod}$.
After integration, the $n$-sample of a bolometer can be written as $\label{eqn:output} Y(t_n) = (-1)^n F(\omega) H'(\omega) e^{i t_n \omega}$ where $\label{tfele} H'(\omega) = \frac 12 \sum_{k=0}^\infty e^{-i(\frac{\pi\omega}{2\omega_r}+\omega\Delta t)} \Bigg[ \frac{H(\omega_k^+)e^{i\omega_k^+ \Delta t}}{(2k+1)\omega_k^+} \left(1-e^{\frac{i\omega_k^+\pi}{\omega_r}}\right) \\ - \frac{H(\omega_k^-)e^{i\omega_k^- \Delta t}}{(2k+1)\omega_k^-} \left(1-e^{\frac{i\omega_k^-\pi}{\omega_r}}\right) \Bigg]$
The output signal in equation eqn:output can be demodulated (thus removing the $(-1)^n$) and compared to the input signal in equation bol_in. The overall transfer function is composed of the bolometer transfer function and the effective electronics transfer function, $H'(\omega)$: $TF(\omega) = F(\omega) H'(\omega)$
The shape of $H(\omega)$ is obtained combining low and high-pass filters with Sallen Key topologies (with their respective time constants) and accounting also for the stray capacitance low pass filter given by the bolometer impedance combined with the stray capacitance of the cables. The sequence of filters that define the electronic band-pass function $H(\omega) = h_0*h_1*h_2*h_3*h_4*h_{5}$ are listed in the following table.
electronics filter sequence. We define $s = i \omega$
Filter Parameters Function
0. Stray capacitance low pass filter $\tau_{stray}= R_{bolo} C_{stray}$ $h_0 = \frac{1}{1.0+\tau_{stray}*s}$
1. Low pass filter $R_1=1$k$\Omega$
$C_1=100$nF
$h_1 = \frac{2.0+R_1*C_1*s}{2.0*(1.0+R_1*C_1*s)}$
2. Sallen Key high pass filter $R_2=51$k$\Omega$
$C_2=1\mu$
$h_2= \frac{(R_2*C_2*s)^2}{(1.0+R_2*C_2*s)^2}$3
3. Sign reverse with gain $h_3=-5.1$
4. Single pole low pass filter with gain $R_4=10$k$\Omega$
$C_4=10$nF
$h_4= \frac{1.5}{1.0+R_4*C_4*s}$
5. Single pole high pass filter coupled to a Sallen Key low pass filter $R_9=18.7$k$\Omega$
$R_{12}=37.4$k$\Omega$
$C=10.0$nF
$R_{78}=510$k$\Omega$
$C_{18}=1.0\mu$F
$K_3 = R_9^2*R_{78}*R_{12}^2*C^2*C_{18}$
$K_2 = R_9*R_{12}^2*R_{78}*C^2+R_{9}^2*R_{12}^2*C^2+R_9*R_{12}^2*R_{78}*C_{18}*C$
$K_1 =R_9*R_{12}^2*C+R_{12}*R_{78}*R_9*C_{18}$
$h_{5} = \frac{2.0*R_{12}*R_9*R_{78}*C_{18}*s}{s^3*K_3 + s^2*K_2+ s*K_1 + R_{12}*R_9 }$
## Parameters of LFER4 model
The LFER4 model has are a total of 10 parameters($A_1$,$A_2$,$A_3$,$A_4$,$\tau_1$,$\tau_2$,$\tau_3$,$\tau_4$,$S_{phase}$,$\tau_{stray}$) 9 of which are independent, for each bolometer. The free parameters of the LFER4 model are determined using in-flight data in the following ways:
• $S_{phase}$ is fixed at the value of the setting.
• $\tau_{stray}$ is measured during the QEC test during .
• $A_1$, $\tau_1$, $A_2$, $\tau_2$ are fit forcing the compactness of the scanning beam.
• $A_3$, $\tau_3$, $A_{4}$ $\tau_4$ are fit by forcing agreement of survey 2 and survey 1 maps.
• The overall normalization of the LFER4 model is forced to be 1.0 at the signal frequency of the dipole.
The details of determining the model parameters are given in (reference P03c paper) and the best-fit parameters listed below.
LFER4 model parameters
Bolometer $A_1$ $\tau_1$ (s) $A_2$ $\tau_2$ (s) $A_3$ $\tau_3$ (s) $A_4$ $\tau_4$ (s) $\tau_{stray}$ (s) $S_{phase}$ (s)
100-1a 0.392 0.01 0.534 0.0209 0.0656 0.0513 0.00833 0.572 0.00159 0.00139
100-1b 0.484 0.0103 0.463 0.0192 0.0451 0.0714 0.00808 0.594 0.00149 0.00139
100-2a 0.474 0.00684 0.421 0.0136 0.0942 0.0376 0.0106 0.346 0.00132 0.00125
100-2b 0.126 0.00584 0.717 0.0151 0.142 0.0351 0.0145 0.293 0.00138 0.00125
100-3a 0.744 0.00539 0.223 0.0147 0.0262 0.0586 0.00636 0.907 0.00142 0.00125
100-3b 0.608 0.00548 0.352 0.0155 0.0321 0.0636 0.00821 0.504 0.00166 0.00125
100-4a 0.411 0.0082 0.514 0.0178 0.0581 0.0579 0.0168 0.37 0.00125 0.00125
100-4b 0.687 0.0113 0.282 0.0243 0.0218 0.062 0.00875 0.431 0.00138 0.00139
143-1a 0.817 0.00447 0.144 0.0121 0.0293 0.0387 0.0101 0.472 0.00142 0.00125
143-1b 0.49 0.00472 0.333 0.0156 0.134 0.0481 0.0435 0.27 0.00149 0.00125
143-2a 0.909 0.0047 0.0763 0.017 0.00634 0.1 0.00871 0.363 0.00148 0.00125
143-2b 0.912 0.00524 0.0509 0.0167 0.0244 0.0265 0.0123 0.295 0.00146 0.00125
143-3a 0.681 0.00419 0.273 0.00956 0.0345 0.0348 0.0115 0.317 0.00145 0.00125
143-3b 0.82 0.00448 0.131 0.0132 0.0354 0.0351 0.0133 0.283 0.00161 0.000832
143-4a 0.914 0.00569 0.072 0.0189 0.00602 0.0482 0.00756 0.225 0.00159 0.00125
143-4b 0.428 0.00606 0.508 0.00606 0.0554 0.0227 0.00882 0.084 0.00182 0.00125
143-5 0.491 0.00664 0.397 0.00664 0.0962 0.0264 0.0156 0.336 0.00202 0.00139
143-6 0.518 0.00551 0.409 0.00551 0.0614 0.0266 0.0116 0.314 0.00153 0.00111
143-7 0.414 0.00543 0.562 0.00543 0.0185 0.0449 0.00545 0.314 0.00186 0.00139
217-5a 0.905 0.00669 0.0797 0.0216 0.00585 0.0658 0.00986 0.342 0.00157 0.00111
217-5b 0.925 0.00576 0.061 0.018 0.00513 0.0656 0.0094 0.287 0.00187 0.00125
217-6a 0.844 0.00645 0.0675 0.0197 0.0737 0.0316 0.0147 0.297 0.00154 0.00125
217-6b 0.284 0.00623 0.666 0.00623 0.0384 0.024 0.0117 0.15 0.00146 0.00111
217-7a 0.343 0.00548 0.574 0.00548 0.0717 0.023 0.0107 0.32 0.00152 0.00139
217-7b 0.846 0.00507 0.127 0.0144 0.0131 0.0479 0.0133 0.311 0.00151 0.00139
217-8a 0.496 0.00722 0.439 0.00722 0.0521 0.0325 0.0128 0.382 0.00179 0.00111
217-8b 0.512 0.00703 0.41 0.00703 0.0639 0.0272 0.0139 0.232 0.00173 0.00125
217-1 0.0136 0.00346 0.956 0.00346 0.0271 0.0233 0.00359 1.98 0.00159 0.00111
217-2 0.978 0.00352 0.014 0.0261 0.00614 0.042 0.00194 0.686 0.0016 0.00125
217-3 0.932 0.00355 0.0336 0.00355 0.0292 0.0324 0.00491 0.279 0.00174 0.00125
217-4 0.658 0.00135 0.32 0.00555 0.0174 0.0268 0.00424 0.473 0.00171 0.00111
353-3a 0.554 0.00704 0.36 0.00704 0.0699 0.0305 0.0163 0.344 0.0017 0.00125
353-3b 0.219 0.00268 0.671 0.00695 0.0977 0.0238 0.0119 0.289 0.00157 0.00111
353-4a 0.768 0.00473 0.198 0.00993 0.0283 0.0505 0.00628 0.536 0.00181 0.00125
353-4b 0.684 0.00454 0.224 0.0108 0.0774 0.08 0.0149 0.267 0.00166 0.00111
353-5a 0.767 0.00596 0.159 0.0124 0.0628 0.0303 0.0109 0.357 0.00156 0.00111
353-5b 0.832 0.00619 0.126 0.0111 0.0324 0.035 0.0096 0.397 0.00166 0.00111
353-6a 0.0487 0.00176 0.855 0.006 0.0856 0.0216 0.0105 0.222 0.00199 0.00125
353-6b 0.829 0.00561 0.127 0.00561 0.0373 0.0252 0.00696 0.36 0.00228 0.00111
353-1 0.41 0.000743 0.502 0.00422 0.0811 0.0177 0.0063 0.329 0.00132 0.00097
353-2 0.747 0.00309 0.225 0.00726 0.0252 0.0447 0.00267 0.513 0.00154 0.00097
353-7 0.448 0.0009 0.537 0.0041 0.0122 0.0273 0.00346 0.433 0.00178 0.00125
353-8 0.718 0.00223 0.261 0.00608 0.0165 0.038 0.00408 0.268 0.00177 0.00111
545-1 0.991 0.00293 0.00743 0.026 0.00139 2.6 0 0 0.00216 0.00111
545-2 0.985 0.00277 0.0128 0.024 0.00246 2.8 0 0 0.00187 0.00097
545-4 0.972 0.003 0.0277 0.025 0.000777 2.5 0 0 0.00222 0.00111
857-1 0.974 0.00338 0.0229 0.025 0.00349 2.2 0 0 0.00176 0.00111
857-2 0.84 0.00148 0.158 0.00656 0.00249 3.2 0 0 0.0022 0.00125
857-3 0.36 4.22e-05 0.627 0.0024 0.0111 0.017 0.002 1.9 0.00152 0.00126
857-4 0.278 0.0004 0.719 0.00392 0.00162 0.09 0.00152 0.8 0.00149 0.000558 | 4,398 | 8,966 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2018-13 | longest | en | 0.736349 |
https://www.fishkeeping.co.uk/modules/newbb/report.php?forum=1&topic_id=36618&post_id=338779 | 1,632,378,746,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057417.10/warc/CC-MAIN-20210923044248-20210923074248-00558.warc.gz | 787,402,327 | 7,346 | 8 - 6 = ?
Input the result from the expression
Maximum attempts you can try: 10
Oranda buoyancy issues Subject: Oranda buoyancy issuesby maccy_g on 8/10/2014 12:11:57I know a plethora of information has already been written about fancy goldfish and issues with swimbladder disorder but I thought I'd ask people's thoughts on this particular issue.I have a 600l tank with 6 Oranda's, 1 Ranchu and 1 Lionhead - they are all healthy and swimming about happily however my black oranda has some buoyancy issues which I believe is food-related.I used to feed the fish a combination of Hikari sinking pellet, frozen food & peas (on varying days, not in 1 sitting), feeding once in the morning and once in the evening.I have owned the black oranda since April but over the past couple of months he has displayed problems with his swimming, in particular when he is feeding from the gravel his back-end lifts up to the point where it tips him over and he has to barrel-roll to "right himself" - his swimming overall is poor and he often appears to struggle whereas the others swim with relative grace (well as much as fancies can bearing in mind they aren't strong swimmers).Around a month ago I withheld food for 4 days and this seemed to solve the problem, after this I fed the fish peas and he appeared to be ok.I have also stopped feed Hikari and replaced this with Repashy Soilent Green, a gel food which is meant to be much better for fancies when it comes to SBD as I hear some fish can react badly to Hikari food.But, over the past couple of weeks his buoyancy issues have come back, he isn't stuck at the bottom or at the top but he is definitely struggling with his swimming again.I am currently feeding once in the morning and once in the evening, any suggestions as to how I can alleviate this issue for him? None of the other fish in the tank have ever displayed any SBD issues.PH - 7.2 (I use crushed coral to raise the ph as I am in a softwater area)Nitrate - 0Nitrite - 0Ammonia - 0I do weekly water changes of 50% | 481 | 2,023 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2021-39 | longest | en | 0.979837 |
https://www.xelplus.com/debits-and-credits-made-easy-with-adex-ler/ | 1,582,588,397,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145989.45/warc/CC-MAIN-20200224224431-20200225014431-00318.warc.gz | 939,494,732 | 114,687 | # Debits and Credits
If you want to learn accounting, the concept of debits and credits is one of the most important to understand.
But, many struggle with it and for me, it was the same in the beginning because it’s not something that’s completely explainable with logic.
But, I managed to understand it and by the end of this post, so will you.
# Debits & Credits
In accounting, debits and credits always go together. Every transaction will have debits and credits and at least two accounts will be affected.
Remember, one transaction affects two accounts. If one account is debited, then the other one must be credited.
Just think of it as a bank transfer. When you send someone money, it comes out of your bank account and it goes into their account.
In accounting, instead of these piggy banks, we use T accounts.
Every account has it’s own T. The left side of the T is where the debits go. Credits go on the right side.
That’s already the first main rule you have to remember. Debit and credit do not mean plus or minus. It literally just means debit goes to the left side of a T account and credit goes to the right side of a T account.
Debit means left, credit means right.
That’s rule number one.
The second important rule is that for every transaction, the total amount of debits must equal the total amount of credits.
If you debit an account for 100, you must credit another one with 100.
There can be more than two accounts involved in a transaction, but never less than two.
That’s rule number two.
Now, the third rule usually creates a lot of confusion, but you’ll be fine because I have a secret weapon for you.
More on that later.
Before we dive right in, let’s do something else first that we’ll need in the process, and that’s to figure out the balance in a T account.
# Get the Balance Right
The balance is basically the total of an account.
So, let’s say we have these two accounts visualized at T’s.
Then, we record some transactions to them.
First, account number one gets a debit of 500, and account number two gets a credit of 500.
For the second transaction, account number two gets a debit of 100 and account number one gets a credit of 100.
With that, we successfully applied the first two rules of debit and credit. We put debits on the left side of the accounts, credits on the right side.
And, for each transaction, the total of debit and credit was the same, so far so good.
Now to calculate the balance or total for the accounts, we add up the amounts on the debit and credit side for each account separately.
Account number one has 500 on the debit side and 100 on the credit side.
We deduct the smaller sum of credits from the bigger total of debits which gives us 400.
In accounting, we call this a debit balance of 400.
For account number two, it’s the opposite. It has a credit total of 500 and total debits are 100. Again, we deduct the smaller one from the bigger one, account number two has a credit balance of 400.
Here comes the confusing part.
Let’s say we have another transaction that adds another debit of 50 to account number one and one more transaction with a debit of 100 going to account number two.
Let’s just say the credits for these transactions go to different accounts. We don’t need to worry about these for now.
What would happen is that the balance of account number one would increase while the total of account number two decreases.
Both received a debit, but one account increases while the other one decreases. They are behaving differently.
And, that’s rule number three.
It depends on the account if a debit or a credit increases the balance or if it decreases it. And, that’s really the secret to understanding debit and credit.
Like we saw in our example, the balance of account number one increases with the debit while the balance of account number two decreases.
# Two Groups to Remember
You may ask, “Well how do you know which one it is?” Fortunately, there are only two groups to remember.
## DEBITS
• ASSETS – That’s the resources the company owns and uses, like buildings, machines, equipment and so on.
• DIVIDENDS – That’s when the company distributes its profit and cash to the owners.
• EXPENSES – That’s money we pay for goods or services the business purchased.
All accounts in this group are debits, which means that their balance will usually be a debit. Therefore, the total of these accounts will increase when they get another debit and will decrease with they get another credit.
So, account number one in our previous example would be in this group.
## CREDITS
• LIABILITIES – Money we still owe to others, like to the bank, to suppliers or to the IRS.
• OWNER’S EQUITY – Money the owners of the company put into the business.
• REVENUE – Money we receive for sales to our customers.
All accounts in this group are credits which means that their balance will usually be a credit.
Therefore, the total of these accounts will increase when they get another credit and will decrease when they get another debit.
So, account number two in our previous example would be in this group.
And, that’s rule number three. It’s really important that you memorize this because to record any transaction, you have to answer two questions.
Which accounts are affected? Did accounts go up or down?
In order to answer question number two, you need to understand if an account is a debit or a credit account.
If you have trouble remembering this, just think of ADEx LER: Accountants Don’t Expect Low Earning Rates.
Assets, Dividends, Expenses, Liabilities, Equity, and Revenue.
The balance of these accounts increases with debits and decreases with credits.
So, let’s take cash for example. Cash is an asset. So, it resides on the debit side of our equation.
When you take money out of your account to pay for something you reduce your cash balance.
Remember our rule; the balance on the debit side increases with debit and decreases with credit.
So, do we debit or credit the cash account? We credit it because we reduce it.
## LER are Credits.
The balance of these accounts increases with credits and decreases with debits.
Take a loan, for example, which is a liability.
When you take out a higher loan, you credit the loan account.
When you pay back a loan, what do you do? You debit the loan account, which will reduce its balance.
So, that’s really all there is to it.
All you must remember is “Accountants Don’t Expect Low Earning Rates”.
ADEx LER: Assets, Dividends, Expenses, Liabilities, Equity, and Revenue.
Just practice it and it’s eventually going to become second nature to you. In the end, you’re not even going to have to think about it anymore.
# fit into this definition?
I want to clear this up because it creates confusion for a lot of people.
The reason for the confusion are banks and how we talk to them.
For instance, when you put money in your account the bank will credit it.
When you take money out of your account, they will debit it.
A debit card is issued for the purpose of accessing your funds and to take money out from your account.
Now, if you think about this, it really seems like it’s exactly the opposite of what we just learned.
Cash is an asset. And, according to ADEx LER, resides on the debit side of the account equation. And, therefore, increases with a debit.
If you’re adding money, you will debit it, and if you’re taking money out, it will be credited.
So, why is this backward with banks? Do the general rules of accounting not apply to them? Actually, they do, but they look at it from their point of view, not yours.
When you put money in your checking account, it belongs to you, not to the bank. The bank just holds it for you.
So, for the bank, this money really is a liability, because at some point, they’re going to have to pay it back to you.
According to ADEx LER, liabilities are on the credit side of the equation, right? And, to increase a liability, you will have to credit it. Therefore, the bank will credit your account when you put money into it.
Likewise, when you buy stuff with your debit card, this will reduce the balance in your account. The bank owes you less money. Therefore, it will be debited and hence the name, debit card.
This is why it seems backwards. The same accounting rules apply to banks as well as to any other business in the world.
The terminology they use may be confusing because it’s from the point of view of the bank and not from your point of view.
Well, if you’re issued a credit card, the bank or the provider of the card, is providing a line of credit to you.
In other words, credit cards combine payment services with the extension of credit.
And, like for any loan, you’re going to be charged interest, actually very high interest, for the balance you carry over from month to month.
In addition, credit cards may offer additional insurance on purchases or make it easier to request a refund or a return.
But, in the end, the main purpose is to award you a line of credit or credit limit, hence the name.
I hope this was helpful to avoid confusion.
Please just stick to the definitions of debits and credits that we just learned and always remember ADEx LER and you’re going to be fine.
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Learn anytime that fits your schedule. | 2,090 | 9,403 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2020-10 | longest | en | 0.937001 |
https://www.jiskha.com/display.cgi?id=1354129392 | 1,516,159,237,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084886794.24/warc/CC-MAIN-20180117023532-20180117043532-00108.warc.gz | 899,401,818 | 3,744 | # geometry
posted by .
A painter leans a 20-ft ladder against a building. The base of the ladder is 12 ft from the building. To the nearest foot, how high on the building does the ladder reach?
• geometry -
This forms a right angle triangle.
a^2 + b^2 = c^2
a^2 + 12^2 = 20^2
a^2 + 144 = 400
a^2 = 256
a = 16
• pre algrbra -
a=16
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A ladder leans against a building, forming an angle of 53° with the ground. The base of the ladder is 3 ft from the building. To the nearest hundredth of a foot, how long is the ladder?
3. ### geometry
A 24 foot ladder forms a 76 degree angle with the ground. The top of the ladder rests against the building. To the nearest inch how high up the building does the ladder reach?
4. ### Algebra
A fireman leans a 40-ft long ladder up against a building. The base of the ladder creates an angle of elevation of 32 degrees with the ground. How far is the base of the ladder from the base of the building?
5. ### Trig
A 24-foot ladder is placed against a building. The building is perpendicular to the level ground so that the base of the ladder is 11 feet away from the base of the building. How far up the building, in feet, does the ladder reach?
6. ### college algebra
A contractor leans a 23-foot ladder against a building. The distance from the ground to the top of the ladder is 7 feet more than the distance from the building to the base of the ladder. How far up the building is the ladder to the …
7. ### Trigonometry
A 19-ft ladder leans against a building so that the angle between the ground and the ladder is 80∘. How high does the ladder reach on the building?
8. ### Geometry
A Ladder leans against a 15 foot tall building to form a right angle. The ladder is placed so it is 8 feet from the base of the building. What is the length of the ladder?
9. ### Algebra
a painter leans a 25 foot ladder against a building. The distance from the ground to the top of the ladder is 17 feet more than the distance from the building to the base of the ladder. Find the distance from the building to the base …
10. ### Pre-Cal
A 13-ft ladder leans against a building so that the angle between the ground and the ladder is 63 degrees. How high does the ladder reach on the building?
More Similar Questions | 623 | 2,486 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-05 | latest | en | 0.938878 |
http://mathhelpforum.com/discrete-math/18237-pigeonhole-principle.html | 1,481,152,434,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542250.48/warc/CC-MAIN-20161202170902-00398-ip-10-31-129-80.ec2.internal.warc.gz | 181,074,427 | 10,558 | 1. Pigeonhole Principle
How many proofs are there for the pigeonhole principle? I know of one proof that just uses composition of functions to develop lemmas, and then ultimately come up with the Pigeonhole principle. The proof starts with the basic problem of 2 different ways of counting sets (i.e if $f: \mathbb{N}_{m} \to X$ and $f: \mathbb{N}_{n} \to X$ are bijections with the same codomain, then $m = n$). Namely, it says the following: Suppose that $X$ and $Y$ are finite, non-empty sets such that $|X| > |Y|$. Then $f: X \to Y$ is not an injection.
2. What is a finite set? It is a set such has no proper subset has a bijection with the full set. Use that fact and prove formally the principle.
3. There are several proofs as you observe. This of is found in many discrete mathematics texts.
Given that $m = \left| X \right| > \left| Y \right| = n$, recall that $
X = \bigcup\limits_{y \in Y} {\overleftarrow f \left\{ y \right\}}$
.
Suppose that $\left( {\forall y \in Y} \right)\left[ {\left| {\overleftarrow f \left\{ y \right\}} \right| \le 1} \right]\quad$.
Using that we have $\left| X \right| = \left| {\bigcup\limits_{y \in Y} {\overleftarrow f \left\{ y \right\}} } \right| = \bigcup\limits_{y \in Y} {\left| {\overleftarrow f \left\{ y \right\}} \right|} \le n$ that is a contradiction.
Thus $\left( {\exists z \in Y} \right)\left[ {\left| {\overleftarrow f \left\{ z \right\}} \right| > 1} \right]$.
The value of this proof is that it also proves the generalized form. If we have m pigeons and n pigeonholes with $m > n$ the some hole contains at least $\left\lceil {\frac{m}{n}} \right\rceil$ pigeons (that is the ceiling function).
To see this proof use the assumptions above, using the supposition that $k = \left\lceil {\frac{m}{n}} \right\rceil \quad \& \quad \left( {\forall y \in Y} \right)\left[ {\left| {\overleftarrow f \left\{ y \right\}} \right| < k} \right]$.
This time $\left| X \right| = \left| {\bigcup\limits_{y \in Y} {\overleftarrow f \left\{ y \right\}} } \right| = \bigcup\limits_{y \in Y} {\left| {\overleftarrow f \left\{ y \right\}} \right|} \le n\left( {k - 1} \right) \le m$ that is a contradiction.
Thus $\left( {\exists z \in Y} \right)\left[ {\left| {\overleftarrow f \left\{ z \right\}} \right| \ge k} \right]$. | 738 | 2,269 | {"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": 17, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.984375 | 4 | CC-MAIN-2016-50 | longest | en | 0.776957 |
https://sacred.numbersciences.org/category/megalithic/near-carnac/ | 1,670,171,544,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710974.36/warc/CC-MAIN-20221204140455-20221204170455-00782.warc.gz | 551,137,382 | 35,325 | ## The Megalithic Numberspace
above: counting 37 lunar months six times to reach 222,
one month short of 223: the strong Saros eclipse period.
There is an interesting relationship between the multiple interpretations of a number as to its meaning, and the modern concept of namespace. In a namespace, one declares a space in which no two names will be identical and therefore each name is unique and this has to be so that, in computer namespaces such as web domain names, the routes to a domain can be variable but the destination needs to be a unique URL.
If sacred numbers had unique meanings then they would be like a namespace. Instead, being far more limited in variety, sacred numbers have more meanings, or interpretations, just as one might say that London has many linkages to other cities. In an ordinal number set, there are many relationships of a number to all the other numbers. This means whilst their are infinite numbers in the set of positive whole numbers, there are more than an infinity of relationships between the members of that set, such as shared number factors or squares, cubes, etc. of a number.
The mathematician Georg Cantor saw “doubly infinite” sets. Sets of relationships between members of an already infinite set, must themselves be more than infinite. He called infinite sets as aleph-zero and the sets of relationships within an infinite set (worlds of networking), he called aleph-one.
Originally, Cantor’s theory of transfinite numbers was regarded as counter-intuitive – even shocking.
Wikipedia
However, in the world of sacred numbers, although there can be large numbers, in the megalithic the numbers were quite small; partly due to the difficulty that numbers-as-lengths were physically real while later numeracy abstracted numbers into symbols and, using powers of ten, modern integers are a series of place ordered numbers (not factors) in base 10, as with 12,960,000 – possible for the ancient Babylonians but, I believe, not expected for the early megalithic.
## The Knowing of Time by the Megalithic
The human viewpoint is from the day being lived through and, as weeks and months pass, the larger phenomenon of the year moves the sun in the sky causing seasons. Time to us is stored as a calendar or year diary, and the human present moment conceives of a whole week, a whole month or a whole year. Initially, the stone age had a very rudimentary calendar, the early megalith builders counting the moon over two months as taking around 59 days, giving them the beginning of an astronomy based upon time events on the horizon, at the rising or setting of the moon or sun. Having counted time, only then could formerly unnoticed facts start to emerge, for example the variation of (a) sun rise and setting in the year on the horizon (b) the similar variations in moon rise and set over many years, (c) the geocentric periods of the planets between oppositions to the sun, and (d) the regularity between the periods when eclipses take place. These were the major types of time measured by megalithic astronomy.
The categories of astronomical time most visible to the megalithic were also four-fold as: 1. the day, 2. the month, 3. the year, and 4. cycles longer than the year (long counts).
The day therefore became the first megalithic counter, and there is evidence that the inch was the first unit of length ever used to count days.
In the stone age the month was counted using a tally of uneven strokes or signs, sometimes representing the lunar phase as a symbol, on a bone or stone, and without using a constant unit of measure to represent the day.
Once the tally ran on, into one or more lunar or solar years, then the problem of what numbers were would become central as was, how to read numbers within a length. The innovation of a standard inch (or digit) large numbers, such as the solar year of 365 days, became storable on a non-elastic rope that could then be further studied.
The 365 days in he solar year was daunting, but counting months in pairs, as 59 day-inch lengths of rope, allowed the astronomers to more easily visualize six of these ropes end-to-end, leaving a bit left over, on the solar year rope, of 10 to 11 days. Another way to look at the year would then be as 12 full months and a fraction of a month. This new way of seeing months was crucial in seeing the year of 365 days as also, a smaller number of about 12 and one third months.
And this is where it would have become obvious that, one third of a month in one year adds up, visually, to a full month after three years. This was the beginning of their numerical thinking, or rationality, based upon counting lengths of time; and this involved all the four types of time:
1. the day to count,
2. the month length to reduce the number of days in the day count,
3. the solar year as something which leaves a fraction of a month over and finally,
4. the visual insight that three of those fractions will become a whole month after three full solar years, that is, within a long count greater than the year.
To help one understand this form of astronomy, these four types of time can be organized using the systematic structure called a tetrad, to show how the activity of megalithic astronomy was an organization of will around these four types of time.
The vertical pair of terms gives the context for astronomical time on a rotating planet, the GROUND of night and a day, for which there is a sky with visible planetary cycles which only the tetrad can reveal as the GOAL. The horizontal pair of terms make it possible to comprehend the cosmic patterns of time through the mediation of the lunar month (the INSTRUMENT), created by a combination of the lunar orbit illuminated by the Sun during the year, which gave DIRECTION. Arguably, a stone age culture could never have studied astronomical time without Moon and Sun offering this early aggregate unit of the month, then enabling insights of long periods, longer than the solar year.
The Manio Quadrilateral near Carnac demonstrates day-inch counting so well that it may itself have been a teaching object or “stone textbook” for the megalithic culture there, since it must have been an oral culture with no writing or numeracy like our own. After more than a decade, the case for this and many further megalithic innovations, in how they could calculate using rational fractions of a foot, allowed my latest book to attempt a first historical account of megalithic influences upon later history including sacred building design and the use of numbers as sacred within ancient literature.
The “output” of the solar count over three years is seen at the Manio Quadrilateral as a new aggregate measure called the Megalithic Yard (MY) of 32.625 (“32 and five eighths”), the solar excess over three lunar years (of 36 months). Repeating the count using the new MY unit, to count in months-per-megalithic yard, gave a longer excess of three feet (36 inches), so that the excess of the solar year over the lunar could then be known as a new unit in the history of the world, exactly one English foot. It was probably the creation of the English foot, that became the root of metrology throughout the ancient and historical world, up until the present.
This theme will be continued in this way to explore how the long counts of Sun, Moon, and Planets, were resolved by the megalithic once this activity of counting was applied, the story told in my latest book.
## Counting Perimeters
above: a slide from my lecture at Megalithomania in 2015
We know that some paleolithic marks counted in days the moon’s illuminations, which over two cycles equal 59 day-marks. This paved the way for the megalithic monuments that studied the stars by pointing to the sky on the horizon; at the sun and moon rising to the east and setting in the west. It was natural then to them to see the 12 lunar months (6 x 59 = 354 day-marks) within the seasonal year (about 1/3 of a month longer than 12) between successive high summers or high winters.
Lunar eclipses only occur between full moons and so they fitted perfectly the counting of the repetitions of the lunar eclipses as following a fixed pattern, around six months apart (actually 5.869 months, ideally 173.3 day-marks apart). The accuracy of successive eclipse seasons to the lunar month can then improve over longer counts so that, after 47 lunar months, one can expect an eclipse to have occurred about one and a half days earlier. This appears to be the reason for the distance between the megalithic monuments of Crucuno, its dolmen and and its rectangle, which enabled simultaneous counting of days as Iberian feet and months as 27 foot units, at the very end of the Stone Age.
## Vectors in Prehistory 2
In early education of applied mathematics, there was a simple introduction to vector addition: It was observed that a distance and direction travelled followed by another (different) distance and direction, shown as a diagram as if on a map, as directly connected, revealed a different distance “as the crow would fly” and the direction from the start.
The question could then be posed as “How far would the plane (or ship) be, from the start, at the end”. This practical addition applies to any continuous medium, yet the reason why took centuries to fully understand using algebraic math, but the presence of vectors within megalithic counted structures did not require knowledge of why vectors within geometries like the right triangle, were able to apply vectors to their astronomical counts.
Continue reading “Vectors in Prehistory 2”
## Vectors in Prehistory 1
In previous posts, it has been shown how a linear count of time can form a square and circle of equal perimeter to a count. In this way three views of a time count, relative to a solar year count, showed the differences between counts that are (long-term average) differential angular motion between sun and the moon’s cycle of illumination. Set within a year circle, this was probably first achieved with reference to the difference between the lunar year of 12 months (29.53 days) and the solar year of 12 average solar months (30.43 days). Note that in prehistory, counts were over long periods so that their astronomy reflected averages rather than moment-to-moment motions known through modern calculations.
The solar year was a standard baseline for time counting (the ecliptic naturally viewed as 365.25 days-in-angle, due to solar daily motion, later standardized as our convenient 360 degrees). Solar and other years became reflected in the perimeters of many ancient square and circular buildings, and long periods were called super years, even the Great Year of Plato, of the precession of the equinoxes, traditionally 25920 years long! The Draconic year, in which the Moon’s nodes travel the ecliptic, backwards, is another case.
At Le Manio’s southern curb, the excess of the solar year over the lunar year, over 3 years, is 32.625 (32 and 5/8ths) day-inches, which is probably the first of many megalithic yards of around 2.72 feet, then developed for specific purposes (Appendix 2 of Language of the Angels). At Le Manio, the solar year count was shown above the southern curb, east of the “sun gate”, but many other counts were recorded within that curb, as a recording of many lengths, though the lunar year was the primary baseline and the 14 degree sightline above the curb aligned to the summer solstice sunrise.
Numbers-as-symbols, and arithmetic, did not exist. Instead, numbers-as-lengths, of constant units such as the inch, were generated as measurements of different types of year. To know a length without our numeric system required the finding of how a given number of units divided into a length, in an attempt to know the measurement through its observed factorization. This habit of factorization could start with the megalithic yard itself as having been naturally created from the sky, as Time. In this case, when the megalithic yard was divided into the three lunar year count of 1063.1 days, the result was 10.875 (10 and 7/8th) “times” 32.625 day-inches. which is one third of the megalithic yard, and is the number of day-inches of the excess for a single solar year.
The lunar year is the combined result of lunar motion, in its orbit, and solar motion along the ecliptic, of average of one day-in-angle per solar day. The lunar year is the completion of twelve cycles of the moon’s phases. The counting at Le Manio hinged upon the fact that, in three solar years there was a near-anniversary of 37.1 lunar months. This allowed the excess to be very close to the invariant form of the solar-lunar triangle which can be glimpsed for us by multiplying the lunar month (29.53059 days) by 32/29 to give 32.58548. (see also these posts tagged 32/29).
The excess of the solar year, in duration and hence in measured length, the 0.368 (7/19) lunar months (over 12), almost exactly equals the reciprocal of the megalithic yard (19/7 feet) so that, when lunar months are counted using megalithic yards, the excess becomes 12 inches which is 32/29 of 10.875 day-inches. From this it seems likely that the English foot and megalithic yard were generated, as naturally significant units, when day-inch counting was applied to the solar and lunar years.
The Manio Quadrilateral may have been like a textbook, a monumental expression of Megalithic understanding, originally built over the original site of that work or, carried from a different place in living memory. It presents all manner of powerful achievements, such as the accurate approximation of the lunar month as 29.53125 (945/32) days, the significance of the eclipse year, alignment to the solstice maximum and lunar minimum standstill, the whole number count over 4 years of 1461 days – then available as a model of the ecliptic, and a circular Octon simulator – and much else besides. This megalithic period preceded the English stone-circle culture initiated by Stonehenge 1 around 3000 BC but was somewhat contemporary with the Irish cairn and dolmen building culture. Metrology is presented near Carnac as a work-in-progress, based closely on astronomy rather than land measurement as such.
My work on the Megalithic tools-and-techniques can be read in my Lords of Time and in Language of the Angels, further considered as a tradition inherited by ancient world monumentalism. This post will be followed soon by more on vectors in prehistory.
## How Geometries transformed Time Counts into Circles
Above: example of the geometry that can generate one or more circles,
equal to a linear time count, in the counting units explained below.
It is clear, one so-called “sacred” geometry was in fact a completely pragmatic method in which the fourfold nature of astronomical day and month counts allowed the circularization of counts, once made, and also the transmission of radius ropes able to make metrological metrological circles in other places, without repeating the counting process. This “Equal Perimeter” geometry (see also this tag list) could be applied to any linear time count, through dividing it by pi = 22/7, using the geometry itself. This would lead to a square and a circle, each having a perimeter equal to the linear day count, in whatever units.
And in two previous posts (this one and that one) it was known that orbital cycles tend towards fourfold-ness. We now know this is because orbits are dynamic systems where potential and kinetic energy are cycled by deform the orbit from circular into an ellipse. Once an orbit is elliptical, the distance from the gravitational centre will express potential energy and the orbital speed of say, the Moon, will express the kinetic energy but the total amount of each energy combined will remain constant, unless disturbed from outside.
In the megalithic, the primary example of a fourfold geometry governs the duration of the lunar year and solar year, as found at Le Manio Quadrilateral survey (2010) and predicted (1998) by Robin Heath in his Lunation Triangle with base equal to 12 lunar months and the third side one quarter of that. Three divides into 12 to give 4 equal unit-squares and the triangle can then be seen as doubled within a four-square rectangle, as two contraflow triangles where the hypotenuse now a diagonal of the rectangle.
Continue reading “How Geometries transformed Time Counts into Circles” | 3,580 | 16,417 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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 | 0.954334 |
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### AuthorTopic: Antiderivative of Log (Read 3075 times) Tweet Share
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#### hifer
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##### Antiderivative of Log
« on: November 02, 2007, 07:12:27 pm »
0
Hi there, do we always have to put the absolute sign when antidiffing to get logs? wat if the specified limits are positive?
#### bilgia
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##### Antiderivative of Log
« Reply #1 on: November 02, 2007, 09:02:04 pm »
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i dont think in methods...its in specialist
My Subjects:
2006 I.T Systems --> 42
2007 English --> 40
Methods --> 41
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##### Antiderivative of Log
« Reply #2 on: November 02, 2007, 09:12:35 pm »
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No if the terminals are positive you don't have to put it. However, it doesn't take anymore time, neither is it wrong if you do.
Mandark: Please, oh please, set me up on a date with that golden-haired angel who graces our undeserving school with her infinite beauty!
The collage of ideas. The music of reason. The poetry of thought. The canvas of logic.
$-=\mathbb{A}\mathfrak{hmad Issa}=-$
#### Freitag
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##### Antiderivative of Log
« Reply #3 on: November 02, 2007, 10:51:21 pm »
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It's just useful for things such as evaluating areas under the x-axis, when you don't know the area will come out without the absolute value signs as negatives.
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#### joechan521
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##### Antiderivative of Log
« Reply #4 on: November 03, 2007, 02:49:33 pm »
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i remember there was one VCAA exam 1 paper that u had to put modulus after anti into a log to be able to draw the graph
its prob in 2006 sample exam1
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#### tankia
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##### Antiderivative of Log
« Reply #5 on: November 03, 2007, 02:57:42 pm »
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Wtf are you guys talking about *is intimidated* I don't know this stuff ><;
tankAsaurus
#### Freitag
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##### Antiderivative of Log
« Reply #6 on: November 03, 2007, 06:04:22 pm »
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Quote from: "tankia"
Wtf are you guys talking about *is intimidated* I don't know this stuff ><;
Don't worry yourself too much.
Basically when you integrate a function with yields a log, (ie. dy/dx (1/x) --> y= log_e x), you put the brackets around the (in my example) x as modulus signs.
Ie. dy/dx = ( 1/x ) --> y= log_e |x|
rediction of Scores:
English: 40 (hopefully) ; Spec: 40, Methods: 40-45, Physics 35, Enhancement: HD. Last year- History 38; I.T 42.
Hopeful ENTER 98. (I need it for my scholarship )
#### tankia
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##### Antiderivative of Log
« Reply #7 on: November 04, 2007, 09:05:35 am »
0
Quote from: "Freitag"
Quote from: "tankia"
Wtf are you guys talking about *is intimidated* I don't know this stuff ><;
Don't worry yourself too much.
Basically when you integrate a function with yields a log, (ie. dy/dx (1/x) --> y= log_e x), you put the brackets around the (in my example) x as modulus signs.
Ie. dy/dx = ( 1/x ) --> y= log_e |x|
Whyyy? Lol ....I'm so screwed for exams ><;
tankAsaurus
#### Odette
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##### Antiderivative of Log
« Reply #8 on: November 04, 2007, 09:33:26 am »
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Poor Tankia there there, i'm sure you'll be fine =]
#### Collin Li
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##### Antiderivative of Log
« Reply #9 on: November 04, 2007, 10:38:51 am »
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Quote from: "tankia"
Whyyy? Lol ....I'm so screwed for exams ><;
dy/dx = 1/x, find y: for R\{0}
Since 1/x is defined for all values except 0, the antiderivative (and the derivative) should also exist for those values.
Notice that ln(x) is only defined for x>0. To correct for this, we place: ln|x| so that it is defined for R\{0}. That's a floppy argument, and what I'm about to suggest is still floppy, but if you look at ln|x| for x is less than zero, you'll notice that it is decreasing. Look at 1/x, it is the derivative of ln|x| and for values less than zero, it has negative numbers! It makes sense.
#### tankia
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##### Antiderivative of Log
« Reply #10 on: November 04, 2007, 11:20:23 am »
0
Quote from: "coblin"
Quote from: "tankia"
Whyyy? Lol ....I'm so screwed for exams ><;
dy/dx = 1/x, find y: for R\{0}
Since 1/x is defined for all values except 0, the antiderivative (and the derivative) should also exist for those values.
Notice that ln(x) is only defined for x>0. To correct for this, we place: ln|x| so that it is defined for R\{0}. That's a floppy argument, and what I'm about to suggest is still floppy, but if you look at ln|x| for x is less than zero, you'll notice that it is decreasing. Look at 1/x, it is the derivative of ln|x| and for values less than zero, it has negative numbers! It makes sense.
_.... You lost me at dy/dx >_>;
tankAsaurus
#### Odette
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##### Antiderivative of Log
« Reply #11 on: November 04, 2007, 11:30:05 am »
0
Quote from: "tankia"
Quote from: "coblin"
Quote from: "tankia"
Whyyy? Lol ....I'm so screwed for exams ><;
dy/dx = 1/x, find y: for R\{0}
Since 1/x is defined for all values except 0, the antiderivative (and the derivative) should also exist for those values.
Notice that ln(x) is only defined for x>0. To correct for this, we place: ln|x| so that it is defined for R\{0}. That's a floppy argument, and what I'm about to suggest is still floppy, but if you look at ln|x| for x is less than zero, you'll notice that it is decreasing. Look at 1/x, it is the derivative of ln|x| and for values less than zero, it has negative numbers! It makes sense.
_.... You lost me at dy/dx >_>;
Tanika I'll explain it to you on msn =]
#### tankia
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##### Antiderivative of Log
« Reply #12 on: November 04, 2007, 11:31:05 am »
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Okay
tankAsaurus
#### Freitag
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##### Antiderivative of Log
« Reply #13 on: November 04, 2007, 03:58:26 pm »
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It seriously doesn't matter too much :p
rediction of Scores:
English: 40 (hopefully) ; Spec: 40, Methods: 40-45, Physics 35, Enhancement: HD. Last year- History 38; I.T 42.
Hopeful ENTER 98. (I need it for my scholarship )
#### tankia
• Victorian
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• Posts: 162
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##### Antiderivative of Log
« Reply #14 on: November 04, 2007, 07:45:15 pm »
0
Quote from: "Freitag"
It seriously doesn't matter too much :p
LOL. I was doing my FIRST practice exam (one that i actually finished...Thank the lord for solutions ) and i finally got what you guys were talking about...kinda...some what..maybe...maybe not...Okay...Yea..I'll go back to procrastinating
tankAsaurus | 2,247 | 7,225 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 1, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2019-35 | latest | en | 0.843916 |
http://profprattle.com/2014/04/11/final-exam/ | 1,618,317,640,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038072366.31/warc/CC-MAIN-20210413122252-20210413152252-00341.warc.gz | 72,827,596 | 23,478 | # Final Exam
The final exam will be on Tuesday 17 December 2013 (Beethoven’s birthday! Also William Safire’s.) from 8:00 a.m. to 10:45 a.m. Eastern Standard Time (UTC-5). It will be given in the usual classroom, Humanities 1003. It will be cumulative; that is, it will potentially test you on any material covered throughout the course of the semester. You will be allowed to bring one 8.5” x 11” sheet of paper on which you’ve written theorems, examples, definitions, spaghetti alla puttanesca recipes, “Ryan Gosling” in a big heart, or whatever else you feel like. You will not be permitted to use a graphing calculator, but four-function and scientific calculators will be permitted.
As promised, the final exam is optional. If you choose not to take it, your first three exams will be worth 24%, 28%, and 28%, respectively, of your final grade. Homework is worth 20% of your final grade regardless of whether you take the final, and the lowest homework grade will be dropped. You can calculate your score and decide whether you’re happy with it. I won’t know what the cutoff for each letter grade will be until after the final exam is done and all the numerical grades calculated, but I can say that if you have a 93 or better you’ll have an A, 90-92 will be at least an A-, 87-89 at least a B+, 83-86 at least a B, 80-82 at least a B-, 77-79 at least a C+, 73-76 at least a C, 70-72 at least a C-, and 60-70 at least a D. Barring a spectacular class-wide performance on the third exam (which will be graded soon, by the way) and the final, those will not be the final cutoff marks. That is, you can probably get an A with a 90, a B+ with an 85, etc. I really don’t know, though, so I’m not going to promise anything.
If you do take the final, the weights are: exam one, 15%; exams two and three, 20% each; final exam, 25%. If you do better on your final exam I will weight it more heavily. For example, if you got a 45, 70, 75 on the first three exams and an 80 on the final I would probably do something like 10% first exam, 18% each of second and third exams, 34% final exam. I’m still figuring it out, since I want to be consistent for each student. Basically, just realize that the final is an opportunity to make up for poor performance on previous exams.
I will have office hours next week, probably on Wednesday and Thursday afternoon. I’ll send out an email in a couple of days with details. I’ll also let you know about the distribution of the third exam grades. In the mean time, the best way to study is by redoing all the problems from the old exams and homework assignments, without the aid of your notes or textbook.
Let me tell you the story of how I fell in love with cheese. A few summers ago, I was travelling around Europe with a friend before our semesters abroad began. Her programs was in Rome, so we ended up there, but I had another week before I had to be in Madrid, so I had some time to gallivant about by myself. After a few days in Italy I took an overnight train to Bern, Switzerland, getting in around 6:30 a.m. The morning chill was a welcome respite from Rome’s 95-degree heat, and I put on a sweater over a shirt that I buttoned all the way up to the top. I was feeling good.
Unsurprisingly, nothing was open at that hour, so I just walked around for a while. The alpine air was a blessing after five days of Italian humidity. When I walked along the Aare River, it seemed somehow clearer than normal air, so that I felt as I had the first time I put on glasses and realized all I had not been seeing. I saw a beautiful woman coming towards me on the sidewalk. As she passed me she smiled and said “Bon style !” and I began to wonder how difficult it would be to become a Swiss citizen. I wandered some more until I reached a square where a market was materializing. I hadn’t had breakfast, but an array of samples drew me to the fromager’s table rather than the boulanger’s. I was going to buy some gorgonzola, but then I realized what a folly it would be not to try something new. So I had a bite of a semi-hard, raw cow’s-milk cheese called Mont Vully, named after the mountain in whose shadow it was made about thirty minutes away.
I don’t know if it was my mindset, or the setting, or if it was just a particularly good batch of cheese, but as I tasted the Mont Vully I began to feel inexplicably giddy, as if I were being let in on a tremendous secret, or getting away with something illegal. I’d had Roquefort before, but outside of blue cheeses I’d never experienced such an assertive flavor. It was assuredly a stinky cheese, but while assaultive to the nose it was divine on the tongue. I spoke briefly with the proprietor, and when my poor French reached its limit, I bought about half a pound of the revelatory Mont Vully and went on my way. That evening I went to Paris, where I slept at a hostel. I kept what was left of the cheese in my shared room, no doubt to the chagrin of my fellow-travelers and their sensitive nostrils. No matter, though. It made for a delightful breakfast. | 1,240 | 5,039 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2021-17 | latest | en | 0.910513 |
https://brainly.ph/question/100247 | 1,487,590,872,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501170521.30/warc/CC-MAIN-20170219104610-00553-ip-10-171-10-108.ec2.internal.warc.gz | 716,376,767 | 10,120 | # A car starts from rest and accelerates uniformly over a time of 5.21 seconds for a distance of 110 m. Determine the acceleration of the car?
1
by okay123
2015-01-23T16:32:19+08:00
### This Is a Certified Answer
Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
where :
d-distance
vi-initial velocity
t-time
a-acceleration
d=110m
vi=0 since it is stated it started from rest
t=5.21s | 156 | 614 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2017-09 | latest | en | 0.922378 |
http://www.thestudentroom.co.uk/showthread.php?t=1987163&page=3 | 1,477,281,470,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988719465.22/warc/CC-MAIN-20161020183839-00544-ip-10-171-6-4.ec2.internal.warc.gz | 755,060,025 | 40,326 | You are Here: Home
# IB Maths HL 2012
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1. 5+12i equal (2-3i)(-2+3i) as well ^^. I got (x+4)(x-2)(x^2 + 3x +4), but after that I ran out of time.......... spend quite a while on Q13 making sure I didn't skip any steps on the proof. Whatever it was that question was okay it was the second part which messed me up, though admittedly I didn't attempt a lot of it.
2. (Original post by Saymyname!)
well done! I got the same equation!
PHEW...beginning to feel that maybe this paper was not the disaster I thought it was yesterday...
If i get at least 25 of the 29 marks for the complex numbers question.
I felt that the paper itself was not as difficult as other years...although I ran out of time and left some questions hanging
3. also from the first question in section B
For the question on which function could be for a discrete blah blah
I put f(x) cos the values add up to 1
4. It is f(x). Although at the time, i just put f(x) without saying why. Did anyone else find that the timing for both papers wasn't great? I mean, personally, I found paper two easier. Although, I believe that last years paper 1 was easier than this year. And, I've heard that if people this year, on average don't do as well, then the IB lower the grade boundaries slightly.
5. I'm just soooooooooooo damn annoyed about timing. I felt that it could've been a lot better if I had a bit more time. And I'm annoyed because our formula booklets had to be reprinted because they were missing some pages, so in the exam we had to sift through sheets which weren't stapled together. Oh well. Paper three still left to go. Series and Differentials. I want/need to well on that.
6. What did you guys end up getting for the last page on paper 2. The applications question, where you had to write v in terms of t ?
7. And how many questions would you say you missed across both papers 1 and 2 in total?
8. (Original post by johndoe04)
What did you guys end up getting for the last page on paper 2. The applications question, where you had to write v in terms of t ?
can't discuss that yet...although that was one nasty question I wasted too much time on
9. (Original post by johndoe04)
And how many questions would you say you missed across both papers 1 and 2 in total?
well I attempted everything but left several big questions incomplete
10. (Original post by banban)
well I attempted everything but left several big questions incomplete
Yeah...I just hope. Urgh.
11. (Original post by Saymyname!)
well done! I got the same equation!
Doesn't that become arctan (root7 / 3). I remember getting 5i + 12 or whatever but for a much earlier question.
But I for sure, for the complex quadratic, got x^2 + 3x + 4. Which became x = +/- -3 +/- i(root7) the whole by two. Right?
12. (Original post by johndoe04)
Yeah...I just hope. Urgh.
Yh I'm hoping to receive method marks...although my work was allover the place..and some was in pencil :0 so doubting I'll get that...
what you hoping for/what do you need from maths?
13. (Original post by arra)
Doesn't that become arctan (root7 / 3). I remember getting 5i + 12 or whatever but for a much earlier question.
yeah If you see above..in the pic she posted its there...remember it had to be put in the e form.
5i +12 is from part A
14. Yeah, I put it in e form, but I left the argument as arctan.
I was going to say something like "a theta between 30 and 45 :P" but felt leaving it in arctan was ok, hope they didn't expect something like drawing a triangle and god knows somehow figuring out what it actualyl was.
15. (Original post by arra)
Yeah, I put it in e form, but I left the argument as arctan.
I was going to say something like "a theta between 30 and 45 :P" but felt leaving it in arctan was ok, hope they didn't expect something like drawing a triangle and god knows somehow figuring out what it actualyl was.
same I left in arctan...I had no time to try and figure it out
although I was convinced it might be wrong cos its an awkward value and wasted time resolving to come to same answer before I decided to allow and try and finish paper:P
16. Yeah exactly! I don't mind a harder question without stupid bloody answers as not just root7/3 but bloody arctan of that!
17. (Original post by arra)
Yeah exactly! I don't mind a harder question without stupid bloody answers as not just root7/3 but bloody arctan of that!
I agree completely! why on earth would anyone get that answer and be fully confident it's correct without wasting time second guessing...grrr 29 marks on complex numbers....well on the brightside...at least we all got that so its hopefully right:P
18. (Original post by banban)
yeah If you see above..in the pic she posted its there...remember it had to be put in the e form.
5i +12 is from part A
Sis... it's he
19. Hey guys, Paper 1 I found easy, paper 2 was very difficult due barely being taught about statistiscs and probability.
How did you find paper 2? Harder than the previous years or easier? Also, i got the same result for the complex number question on paper 1.
20. (Original post by aldinger)
Hey guys, Paper 1 I found easy, paper 2 was very difficult due barely being taught about statistiscs and probability.
How did you find paper 2? Harder than the previous years or easier? Also, i got the same result for the complex number question on paper 1.
Shouldn't talk about paper 2 yet
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https://abidakon.com/use-index-match-functions-in-google-sheets/ | 1,695,362,934,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506329.15/warc/CC-MAIN-20230922034112-20230922064112-00749.warc.gz | 97,227,548 | 26,864 | # How To Use INDEX-MATCH Functions In Google Sheets [Easy]
Written by Abid Akon
#### Here's what we'll cover:
When you want to perform a lookup task on your worksheet, the VLOOKUP and HLOOKUP functions are mostly employed. But what if I told you that there’s a better way of performing lookup tasks on your worksheet.
This technique is executed using the INDEX and MATCH functions, nested together into a single formula. Today we will show How To Use INDEX-MATCH Functions In Google Sheets. Before we delve into this topic, let’s discuss the INDEX and MATCH functions individually.
## The INDEX function
This function returns the value located in the specified index. It also extracts the contents in a selected cell or cell range. That is, it analyses the selected cell, the given row index and a column index then returns the data in the selected row and column. The INDEX function takes the syntax below.
=INDEX(cell reference, [row], [column])
Where;
• Cell reference indicates the cell or cell ranges where we want to search.
• The row is the index number of a row within the specified cell range.
• The column is the column’s index number within the cell range which we want to search. This value is optional.
Using the sample worksheet below, we are going to explain how the INDEX function works. There are names of students in class A and their respective test scores.
Now, we want to extract the fifth name in column B. The INDEX formula would be inputted like this.
=INDEX(B2:B10, 5, 2)
Step 1: Input the formula into an empty cell.
Step 2: Press the Enter key on your keyboard.
Step 3: The fifth name in column B is extracted and displayed in the empty cell.
## The MATCH Function.
This function delivers the respective index of a value in a specific cell range. It analyses a cell range and returns the index of the item in the cell range. The MATCH function takes the syntax stated below.
=MATCH(search_key, range, search_type)
Where;
• The search key is the value we want to match in the cell range. It can be a text or number, cell reference or formula.
• The Cell range is where we want to search for the value corresponding with the search key.
• Search type is the kind of matching used. It is not compulsory to add this parameter to the formula. The search type is represented in numerical values. They are;
0: This specifies that the lookup is for a value that matches exactly with the search key.
1: This is the custom value. This value concludes the selected range is arranged in ascending order. This parameter extracts the largest value that is equal to or lesser than the search key.
-1: This value concludes that the selected range is arranged in descending order. This parameter extracts the smallest value equal to or greater than the search key.
Using the sample worksheet below, we are going to explain how the MATCH function operates. The worksheet contains the names of students in class A. Let’s say we want to know the position of “Adam Levine” in the worksheet. The formula used looks like this
Step 1: Input the formula above into the empty cell.
Step 2: Press the Enter key on the keyboard. A value of 8 is returned because the name is the sixth name on the worksheet, from cell B2, which is the starting cell.
Note that the value returned is not a row number of the name but its position in the selected range.
## Why use INDEX and MATCH Functions in Google Sheets?
Now that we’ve studied and understood their individual properties, we want to see what these two functions can do when nested into one formula. Separately, the INDEX and MATCH functions have very limited capabilities but we are going to combine them.
When nested as one, they can search for a value in a cell from the worksheet and return the results to another cell in the same data range.
For instance, we have a dataset of the sales made in a Veggie store, all through the month. The worksheet consists of the names of the sales representatives, the goods sold, the units sold and the total sales made.
If we want to see the sales made by a salesman, the items sold or the unit costs, neither the INDEX nor MATCH functions can produce results individually. This is where their combination formula becomes helpful. In the next topic, we are going to discuss how to combine these functions as one.
## How To Combine INDEX and MATCH functions in Google Sheets.
The combination formula takes the syntax below.
=INDEX(range2, MATCH(search_key, range1, 0))
## Explanation of the Formula.
• Search_key is the item we are searching for in range1.
• Range1 is the cell range from which the INDEX function produces a value matching the position returned by the MATCH function.
• Range2 is the cell range from which the MATCH function produces a value matching the position returned by the INDEX function.
That is, the MATCH function helps to specify the index of the value returned to the INDEX function and vice versa.
## Using the INDEX and MATCH Functions with Single Column References.
As stated earlier, the sample worksheet contains the names of sales representatives, units sold, the unit cost and the total sales made. At the bottom of the dataset, there’s a small table where we have the name of the sales representative and the total sales made.
The table is subject to change, so we cannot tag the name inputted as the absolute reference. Here, we are going to use the nested formula of the INDEX and MATCH functions. The formula used is inputted in the bottom table like this.
=INDEX(E2:E6, MATCH(B8, A2:A6,0))
Press the Return key and the total sales made by Joan are returned to cell B9.
## Explanation of the Formula.
Starting from the inner formula which Is MATCH(B8, A2:A6,0)
The MATCH function scans the range A2:A6 for the name in B8. Since Joan is the second name from the starting cell A2, the MATCH function returns the position as 2.
Next is the INDEX formula which is =INDEX(E2:E6). The formula searches for the name in the second position of the cell range E2:E6 and returns the total sales made by Joan.
Now, We changed the name in cell B8 from Joan to Michael. The table is dynamic, so the total sales for Michael also changed.
## Using INDEX and MATCH Functions with Multiple Criteria.
The INDEX-MATCH provides flexibility in your worksheet by gaining access to multiple columns at a time. In the sample worksheet used previously, let’s change the headers of the bottom table. That is A8 represents Sales Representative while A9 represents the units sold.
Now that we’ve added a different header, the table seems more dynamic. This means we need to evaluate multiple columns and the columns will depend on cell B9. This would be done by using the INDEX-MATCH function below.
=INDEX(A2:E6, MATCH(B8, A2:A6, 0), MATCH(A9, A1:E1, 0))
Here, there is a third parameter in the INDEX-MATCH formula. This parameter searches for the index of the selected header in the header row.
In the worksheet below, cell A9 contains “Units Sold”. This formula returns units sold that match the name in cell B8. If a name is entered in cell B8, the unit sold by the person is entered in cell B9.
Even if the name is changed, the correct unit sales are returned in cell B9.
## Explanation of the formula.
This formula has three parameters, we are going to break each one down to your understanding. We are going to start with the first MATCH function in the formula.
MATCH(B8, A2:A6,0).
The first MATCH function extracts the position of Joan’s name in the selected cell range. So it returns the index number of 2.
Next is the second MATCH function,
MATCH(A9, A1:E1, 0).
The second MATCH function scans the cell range A1:E1 for the value in cell A9 and extracts the position of the corresponding value. The text in cell A9 says “Units Sold”, the function finds the text as the third value from the starting cell A1 and it returns the index number 4.
Lastly, the INDEX function =INDEX(A2:E6, MATCH(B8, A2:A6, 0), MATCH(A9, A1:E1, 0)).
The formula searches for the second position in the row and fourth position in the column in the cell range A2:E6 and returns the value of the index which is ‘111’. This is the unit sales made by Joan.
## How To Use MATCH Formula in Google Sheets.
Here, we are going to see how to use the MATCH function vertically and horizontally on the worksheet.
### Vertical Use of MATCH function in Google Sheets.
As stated previously, the MATCH function only returns the position of the item and not the exact value. In the sample worksheet, we want to see the exact position of “Cake” vertically. The formula used is
=MATCH(“Cake”, A2:A6,0)
Where Cake is the search key while A2:A6 is the search range and 0 implies that the data is unsorted.
The formula returns the number 5.
## Explanation.
Notice the selected range A2:A6, it is a single column. You can use a single column in a MATCH formula only whether it’s vertical or horizontal.
### Horizontal Use of MATCH function in Google Sheets.
When you have a dataset and you want to use the MATCH function on a single row horizontally, the cell range would be the range of rows. The formula used is
=MATCH(“Cake”, A3:F3, 0)
The MATCH function scans and extracts the exact position of “Cake” horizontally.
## Why is Using INDEX and MATCH better than VLOOKUP?
The properties above can also be done with a VLOOKUP function but there are some aspects of the INDEX-MATCH formula that surpasses the VLOOKUP.
First of all, added columns or shifting existing cells and columns do not implicate the returned values of the INDEX-MATCH formula because it permits cell references. However, the VLOOKUP results get influenced by the moving or addition of new columns because it evaluates the column order only.
Secondly, the INDEX-MATCH formula permits the search on both the left and right sides of the columns while the VLOOKUP only permits the search on the left of the column. If you try to access the right of the search column, a #N/A error is returned.
Lastly, the INDEX-MATCH function is very flexible and dynamic especially when dealing with cell references compared to the VLOOKUP function.
## Case Sensitive VLOOKUP With INDEX-MATCH in Google Sheets.
The INDEX-MATCH function is the best option when dealing with case sensitive data.
In the worksheet below, suppose some bottled sodas are sold in cartons and are also sold per piece. Hence there are two circumstances of the types of soda, with their prices per piece and per pack.
What if we want to extract the stock of each item? The VLOOKUP function isn’t suitable for this task because it always returns the first name found.
Here’s where we use the INDEX-MATCH function but an additional function is included which is the FIND function. The FIND function is a case sensitive function which makes it most suitable for this task.
The formula used takes the syntax below.
=ArrayFormula(INDEX(D2:D6, (MATCH(FIND( A8, A2:A6)), 0,1))))
The FIND function analyzes the Item Name column which is column A (A2:A6) for the text in cell A8(Coca-Cola). Once found, the formula tags the cell with the number 1.
Then, MATCH searches for the tag and gives the tag to the INDEX function. INDEX comes to the corresponding row in column C and extracts the stock info on the item.
Be sure to add the ArrayFormula at the start of the whole formula because, without it, the FIND function won’t search in array order.
In the result cell, the stock info is extracted as seen below.
## Final Thoughts
The INDEX-MATCH combination formula is a very dynamic and important way of searching and extracting important aspects of your data. Learn this formula properly to use it at its best.
Now you know How To Use INDEX-MATCH Functions In Google Sheets. I hope you found this guide helpful.
#### Abid Akon
I am a tech content writer. I really love to talk about different Modern Technology. It is very important that people know how to fix their tech related problem. That’s why I am writing articles to share my tech knowledge with you.
### 6 Things You Should Do Before Installing iOS 16.0.2
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### Abid Akon
Hi, I am Abid. I really love to talk about different Modern Technology. It is very important that people know how to fix their tech related problem. That’s why I’ve created this website to share my technology keeping knowledge with you. Welcome to ”abidakon.com”
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My Personal Favorites | 2,802 | 12,717 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-40 | longest | en | 0.847831 |
https://www.rodozja.pl/502/capacity-factor-for-ball-mill.html | 1,656,348,716,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103337962.22/warc/CC-MAIN-20220627164834-20220627194834-00797.warc.gz | 1,045,171,041 | 6,458 | # Capacity Factor For Ball Mill
### Factors That Affect Ball Mill Grinder Capacity And Quality
May 01, 1995 Austin, L.G., et al., 1990, “Kinetics and shape factors of ultrafine grinding in a laboratory tumbling ball mill,” Particle and Particle Systems Characterization, Vol. 7, pp. 242–247. Article Google Scholar.Dec 23, 2013 Practical 1 Title Ball Milling Objective To grind the coarse salt to a smaller size by using a ball mill and to obtain the particle size distribution of the initial and the sieved final mixture. Introduction 'Ball milling is a method used to break down the solids to smaller sizes or into a powder. A.How to increase ball mill capacity. how to increase the capacity of ball mill. Optimum choice of the make up ball sizes for maximum The ball size is one of the critical factors for determining the mill performance ofIt can be seen that the S. Keep Reading.Ball mill would increase the production capacity by at least 75 . The company is considering to change its production process to maximize the capacity utilization up to 100 . According to the company’s analysis melanguer could become a bottleneck during the production process , therefore, the company is considering to purchase one melanguer.
### How To Calculate Production Capacity Of A Factory
1 天前 Garadagh Cement orders a DAL Machinery ball mill. 16 December 2021. DAL Machinery will install a completely new ball mill with 28tph production capacity for Garadagh Cement plant (Holcim group) located at Baku, Azerbaijan. The mill is for clinker grinding and measures dia 0.26m x32m. DAL will design, manufacture and install the mill on EPC basis.As the ball wear rate depends directly on the surface of the media charge, a small variation in power will lead to an important increase of wear rate. The risk of underloading or overloading the mill is an additional factor. A direct measurement of the ball level in the mill, accurate than power readings, as well as a control of it, is.Sep 10, 2018 Once you have above information use following formula to calculate production capacity. Production capacity (in pieces) = (Capacity in hours 60 product SAM) line efficiency. For Example Suppose a factory has 8 sewing lines and each line has 25 machines. Total 200 machines and the working shift is 10 hours per day.In fact, in the same application a Vertimill has 50 less footprint than a ball mill. Lower operating cost. Vertimill is an energy efficient grinding machine. They tend to grind efficiently than, for example, ball mills with feeds as coarse as 6 mm to products finer than 20 microns. This provides up to a 40 higher energy efficiency.
### Ball Mill Parameter Selection & Calculation
Nov 30, 2021 Equipped with precision ball screws, the Fadal VMC obtains the accuracy and repeatability for the tightest of tolerances. Capable of high-speed machining in an enclosed capacity, allows this machine to operate safely for all users. Included with this mill is a Spindle Air Blast technology which helps cool your tools to eliminate wear.To ball filling variation in the mill. The results obtained from this work show, the ball filling percentage variation is between 1.2– 3.7 which is lower than mill ball filling percentage, according to the designed conditions (15 ). In addition, acquired load samplings result for mill ball filling was 1.3 .Ball mill. The ball mill is the fine grinding machine connect the SAG or AG mill and flotation machine. Ball mills produce fine particles with a uniform size for flotation, its grinding medias commonly are steel ball. The ball mill rolls grinding media together with the ore, as the ore grinds, these balls initially 5-10 cm in diameter but.Ball Mills 【Capacity】 0.2-90 T H 【Advantages】Designed for long service life, minimum maintenance, can grind and homogenize mineral ores down to the nano range, a large volume of processing capacity 【Max Feeding Size】 25mm 【Discharge Size】0.075-0.4mm 【Types】Overflow ball mills, grate discharge ball mills 【Service】 24hrs quotation, custom-made.
### Practical 1 : Ball Milling Tf Lab 1
Cement ball mill capacity calculation - YouTube 14 Jan 2014 14 Jun 2013 ball mill grinding capacity calculation - saimm that width of screen surface is the most important factor in determining the capacity and efficiency SAG Mill (semi-autogenous).Ball mills Ball mills are used for dry and wet grinding of different materials such as cement materials, cement, lime, and ceramic materials. PSP Engineering has continuously refi ned the design of its ball mills. Horizontal ball mills have become a reliable part of grinding plants Main characteristics of ball mills seated on shoe-type bearings.A 1.5 mio t a cement plant is having a closed circuit ball mill for cement grinding The mill has been operating with satisfactory performance in-terms of system availability and output, however power consumption was on higher side. 3.1 System Description Mill Rated capacity 150 t h OPC at 2800 blaine I chamber liners.The geometry of a mill with conical ends is shown in Figure 8.6. The total volume inside the mill is given by Vm 4 D2 mL 1 2(Lc L) L 1 (Dt Dm) 3 1 Dt Dm (8.16) The density of the charge must account for all of the material in the mill including the media which may be steel balls in a ball mill, or large lumps of ore in an.
### For Achieving Maximum Capacity Of The Ball Mill The Ball
For achieving maximum capacity of the ball mill, the ball charge should be equal to about _____ percent of the ball mill volume. 1) 25.Ball end mills produce a radius at the bottom of pockets and slots. Ball end mills are used for contour milling, shallow slotting, contour milling and pocketing applications. Flutes Spiral-shaped cutting edges are cut into the side of the end mill to provide a path for chips to escape when an end mill is down in a slot or a pocket.Load to 500 the capacity factor increases to 2.3. Figure 1. Effect of circulating load on mill capacity [5] Research into the subject of classification efficiency effect on grinding circuit capacity was carried out by Finnish researchers Hukki, Allenius and Heinonen, [8, 9] using a M i n i n g a n d M e t a l l u r g y.ME EN 7960 – Precision Machine Design – Ball Screw Calculations 4-13 Permissible Speed • When the speed of a ball screw increases, the ball screw will approach its natural frequency, causing a resonance and the operation will become impossible. π ρ λ π ρ λ E l d A EI l n b tr b c 2 2 2 2 2 15 2 60 = = nc Critical speed [min-1] lb.
### Ball Mill Design Power Calculation
12. Ball Mill-Ball Weight Surface Area 97 13. Ball Mill Charge Volume 98 14. Useful Data for Grinding Mill Study 99 15. Ball Mill Charging 99 16. BIS Specification of Additives 102 17. BIS Specifications for various 103 Cements 18. Thermo Physical Properties of Different Insulating Materials 107 19. Pollution Standards - for Stack, Ambient.Dec 12, 2016 Ball Mill Power Calculation Example A wet grinding ball mill in closed circuit is to be fed 100 TPH of a material with a work index of 15 and a size distribution of 80 passing inch (6350 microns).Feb 08, 2017 Energy consumption for ball mills is a function of many factors the physical properties of the ground material – its specific gravity and hardness the degree of drum filling by grinding balls the number of drum rotations, etc. Ball.While the quality of steel balls determine the capacity of ball mill grinder. The quality of the steel ball is calculated according to the amount of ore processed per ton (ie 0.8kg per ton of ore). The average steel ball needs to process one ton of ore (1kg-1.2kg). | 1,735 | 7,600 | {"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} | 2.96875 | 3 | CC-MAIN-2022-27 | latest | en | 0.900662 |
https://aakashdigitalsrv1.meritnation.com/ask-answer/question/a-bicycle-wheel-makes-5000-revolutions-in-moving-11km-find-d/areas-related-to-circles/736869 | 1,695,562,106,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506646.94/warc/CC-MAIN-20230924123403-20230924153403-00453.warc.gz | 93,182,096 | 8,341 | # a bicycle wheel makes 5000 revolutions in moving 11km find diameter of the wheel
70cm ans..
11km = 1100000 cm
now, circumference = 1100000/5000 = 220
2 x pi x radius = 220
=> diameter = 70cm
thumbs up plz...
• 80
70 cm
• -3
thank uh
• -5
Find the number of meters in 11 km, divide by 5000
= 2.2 m, is the circumference of the wheel
:
d =
d = .7 meters or 70 cm is the diameter of the wheel
;
:
Check this by finding how far the you go in 5000 rev
= 10.99 ~ 11 km
• -3
70
• -2
11km = 1100000cm
circumference= total distance/ revolution=220
2πr =220
on substituting the value we get
r=35cm
d=70cm
• 2
What are you looking for? | 231 | 638 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2023-40 | latest | en | 0.73723 |
https://www.esaral.com/q/if-three-of-the-six-vertices-of-a-regular-hexagon-are-chosen-59471/ | 1,653,777,956,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652663021405.92/warc/CC-MAIN-20220528220030-20220529010030-00532.warc.gz | 837,360,396 | 23,466 | If three of the six vertices of a regular hexagon are chosen
Question:
If three of the six vertices of a regular hexagon are chosen at random, then the probability that the triangle formed with these chosen vertices is equilateral is :
1. $\frac{3}{10}$
2. $\frac{1}{10}$
3. $\frac{3}{20}$
4. $\frac{1}{5}$
Correct Option: , 2
Solution:
Only two equilateral tringles are possible $\mathrm{A}_{1} \mathrm{~A}_{3}$
$\mathrm{A}_{5}$ and $\mathrm{A}_{2} \mathrm{~A}_{5} \mathrm{~A}_{6}$
$\frac{2}{6_{C_{2}}}=\frac{2}{20}=\frac{1}{10}$ | 179 | 540 | {"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} | 3.578125 | 4 | CC-MAIN-2022-21 | latest | en | 0.710214 |
https://gaming.stackexchange.com/questions/158881/celadon-city-game-corner-slot-machine-odds/166933 | 1,701,886,231,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100602.36/warc/CC-MAIN-20231206162528-20231206192528-00204.warc.gz | 298,129,344 | 39,705 | # Celadon City Game Corner Slot Machine Odds
What the the odds of the various slot machines in the Game Corner? Which machines have the best odds?
In both the original Red/Blue/Green/Yellow versions, as well as the remade FireRed/LeafGreen versions, an NPC tells to the player that she thinks the slot machines in the Celadon City Game Corner have varying odds.
Do different machines really have different odds? What are the specifics of these odds? Are the odds set or randomly arranged? How can this fact be exploited?
• Slot machine behavior in Gen I is well understood, as we have access to the code that implements it. glitchcity.wiki/Slot_machine_behaviors_(Generation_I) Essentially, there are two strategies. You could either go for matching 7's for a 300 coin payout, or matching Pokemon for a 15 coin payout. Assume you can stop a wheel perfectly, if allowed. The probabilities work out such that the former strategy has a payout of ~2.5 coins/spin, and the latter has ~2.6. I don't know what the odds are if you stop the wheel randomly, but at least we have an upper bound. Mar 19, 2021 at 16:03
• As to the question of which machine's have the best odds: In Gen 1, the rightmost, bottommost slot machine has slightly higher chance of being lucky. The above calculations assume you are playing on that machine, though they don't assume the machine is actually lucky. Mar 19, 2021 at 16:07
As far as I'm aware, there's no hard absolute data for this (at least, not for gen 1 and gen 3). That said, there have been some people trying to figure things out and the following is generally accepted as "true" by many players:
Generation 1:
All the slot machines have their odds randomly generated each play. As such, there is no single slot machine that is good for every player all the time.
Generation 3:
Generation 3 includes two sets of games: ruby/sapphire/emerald and fire red/leaf green.
For Ruby/Sapphire/Emerald, the slots are again fairly random. You may be able to play a slot a couple of times, and if it pays out both times, it might be "hot" and worth continuing on.
For Fire Red/Leaf Green there has been speculation that the machine on the far left, directly below the woman in that aisle has slightly better odds. There isn't any absolute proof of this fact though and may simply be confirmation bias (you think it's better odds, so it feels like it is because you notice your wins more). FR/LG also have an interesting aspect to its algorithm: when the slots start spinning, the game has already decided the % chance of a win/loss. That said, it's not until the last slot stops that the game chooses what the payout will be. This means that if you have an emulator with save-states, if you save after starting to spin but before stopping any of the slots, and then repeatedly load that save, you will likely see either consecutive wins, though of varying amounts, or consecutive losses.
Sources:
All I know is, that the game is so designed that sometimes it isn't possible to get three 7 to line up. The game will either stop at the picture right before the 7 (if you've pressed early enough) or pass the 7 and stop at one of the pictures behind. I know this for sure because of trying out in an emulator where you can let the game run frame by frame. So if the game has decided that you can't win, you simply can't win.
From the original series, I do not remember which slot machine had better odds. But I remember trying all and keeping their data. I did come to a conclusion that a few of them paid up much more frequently.
The odds are set in a single game. Meaning when you found the best slot, you can abuse it. It is unlikely but possible that the odds are randomly generated and another machine is more profitable in another players game.
• This sounds rather anecdotal. Do you have solid evidence on the matter? Mar 6, 2014 at 17:51
• if solid evidence was possible in this subject, i am sure you would have found them by know Mar 6, 2014 at 21:18
• my statement is only for the originals, not for remade versions btw Mar 6, 2014 at 21:19
• I don't believe that. Solid evidence could clearly be found in the game's code. Mar 6, 2014 at 21:21
• and we should download the games code from? Mar 6, 2014 at 22:09
They have no varying odds. All of the slot machines have the same division of pictures. And they rotate at the same speed.
When i played FireRed i was so pro that i new the distance beween the 7 so exactly that i win everytime.
On this site Celadon Game Corner they say the far left ones are the best, but they have no citation.
I didnt notice any better odds at a speciffic slot machine, because of the reasons mentioned above.
I needed 2 min to get back in my performence:
(This was a slot machine on the far right side) | 1,121 | 4,785 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2023-50 | latest | en | 0.954629 |
http://marybourassa.blogspot.com/2017/02/product-rule-in-thinking-classroom.html | 1,490,220,728,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218186353.38/warc/CC-MAIN-20170322212946-00549-ip-10-233-31-227.ec2.internal.warc.gz | 226,480,147 | 13,727 | ## Monday, 27 February 2017
### Product Rule in a Thinking Classroom
Today was day 3 of running my classroom as a thinking classroom - visible random groups (of 3) working on vertical non-permanent surfaces. We started with a quick demonstration to show that the derivative of a product is not the product of the derivatives. Students then worked through a product rule "discovery" activity that I have been using for years. You can find it here. I wish I could give proper credit for it, but I do not remember who shared it with me.
Here is the setup:
Students worked out expressions for length, width and area and for their rates of change before completing the following table.
Their job was to work with the numbers they came up with in the table to figure out a pattern that worked for each row - that would be the product rule.
It is always interesting to see who comes up with it quickly and who takes a little longer (often because they are trying really complicated things!).
Once I felt like the majority of students had found the pattern, I sent them off to their VNPS to work through today's sequence of questions.
The last of these asked them to come up with the product rule for three terms. I loved what some of the groups did. They extended the introductory activity to 3-D, added height and worked through the numbers again! (sorry that the picture quality is terrible)
Here's a group that came up with a conjecture for the product rule with three terms and tested it out. I would like to say they did this instead of asking me if they were right, but they did jump right to it when I said they should check it for themselves.
It was really exciting to see such fantastic work, at such a high level from all my students. I love how they trust that they will be able to tackle all the questions I give them and believe in themselves enough to try.
P.S. All of my planning is on one getting-bigger-by-the-day Word document. I'll post the whole thing at the end of the unit. | 426 | 1,999 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2017-13 | longest | en | 0.985765 |
https://advancesindifferenceequations.springeropen.com/articles/10.1186/1687-1847-2013-157 | 1,643,356,310,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320305423.58/warc/CC-MAIN-20220128074016-20220128104016-00582.warc.gz | 134,945,379 | 55,768 | Theory and Modern Applications
# Extension of a quadratic transformation due to Whipple with an application
## Abstract
The aim of this research is to provide an extension of an interesting and useful quadratic transformation due to Whipple. The result is derived with the help of extension of classical Saalschütz’s summation theorem recently added in the literature. The transformation is further used to obtain a new hypergeometric identity by employing the so-called beta integral method introduced and studied systematically by Krattenthaler and Rao.
MSC:33C20, 33C05, 33B20.
## 1 Introduction
The generalized hypergeometric function ${}_{p}F_{q}$, with p numerator and q denominator parameters is defined by [1]
$\begin{array}{rcl}{}_{p}F_{q}\left[\begin{array}{cc}\begin{array}{c}{a}_{1},\dots ,{a}_{p};\\ {b}_{1},\dots ,{b}_{q};\end{array}& z\end{array}\right]& =& {}_{p}F_{q}\left[{a}_{1},\dots ,{a}_{p};{b}_{1},\dots ,{b}_{q};z\right]\\ =& \sum _{n=0}^{\mathrm{\infty }}\frac{{\left({a}_{1}\right)}_{n}\cdots {\left({a}_{p}\right)}_{n}}{{\left({b}_{1}\right)}_{n}\cdots {\left({b}_{q}\right)}_{n}}\cdot \frac{{z}^{n}}{n!},\end{array}$
(1.1)
where ${\left(a\right)}_{n}$ denotes the Pochhammer symbol (or the shifted factorial, since ${\left(1\right)}_{n}=n!$) defined for any complex number a by
${\left(a\right)}_{n}=\left\{\begin{array}{cc}1,\hfill & n=0,\hfill \\ a\left(a+1\right)\cdots \left(a+n-1\right),\hfill & n\in \mathbb{N}.\hfill \end{array}$
(1.2)
Using the fundamental relation $\mathrm{\Gamma }\left(a+1\right)=a\mathrm{\Gamma }\left(a\right)$, ${\left(a\right)}_{n}$ can be written in the form
${\left(a\right)}_{n}=\frac{\mathrm{\Gamma }\left(a+n\right)}{\mathrm{\Gamma }\left(a\right)}\phantom{\rule{1em}{0ex}}\left(n\in \mathbb{N}\cup \left\{0\right\}\right),$
(1.3)
where Γ is the well-known gamma function.
It is well known that whenever a generalized hypergeometric function reduces to quotient of the products of the gamma function, the results are very important from the application point of view. Thus in the theory of hypergeometric and generalized hypergeometric series, summation formulas and transformation formulas play an important role.
In a very popular, interesting and useful research article, Bailey [2], by employing classical summation theorems such as those of Gauss, Gauss second, Kummer and Bailey for the series ${}_{2}F_{1}$; Watson, Dixon, Whipple and Saalschütz for the series ${}_{3}F_{2}$, established a large number of very interesting results (known as well as new) involving products of generalized hypergeometric series.
It is not out of place to mention here that recently a good deal of progress has been done in the direction of generalizing the above mentioned classical summation theorems. For details, we refer to [35].
In this research paper, we are interested in the following classical Saalschütz summation theorem [1]:
${}_{3}F_{2}\left[\begin{array}{cc}\begin{array}{c}a,b,-n;\\ c,1+a+b-c-n;\end{array}& 1\end{array}\right]=\frac{{\left(c-a\right)}_{n}{\left(c-b\right)}_{n}}{{\left(c\right)}_{n}{\left(c-a-b\right)}_{n}}.$
(1.4)
By utilizing (1.4), Bailey [2] obtained the following three interesting quadratic transformations:
${{\left(1-x\right)}^{-2a}}_{2}{F}_{1}\left[\begin{array}{cc}\begin{array}{c}a,b;\\ a+b+\frac{1}{2};\end{array}& -\frac{4x}{{\left(1-x\right)}^{2}}\end{array}\right]{=}_{2}{F}_{1}\left[\begin{array}{cc}\begin{array}{c}2a,a-b+\frac{1}{2};\\ a+b+\frac{1}{2};\end{array}& x\end{array}\right],$
(1.5)
${{\left(1-x\right)}^{1-2a}}_{2}{F}_{1}\left[\begin{array}{cc}\begin{array}{c}a,b;\\ a+b+\frac{1}{2};\end{array}& -\frac{4x}{{\left(1-x\right)}^{2}}\end{array}\right]{=}_{3}{F}_{2}\left[\begin{array}{cc}\begin{array}{c}2a-1,a+\frac{1}{2},a-b-\frac{1}{2};\\ a-\frac{1}{2},a+b+\frac{1}{2};\end{array}& x\end{array}\right]$
(1.6)
and
$\begin{array}{r}{{\left(1-x\right)}^{1-2a}}_{3}{F}_{2}\left[\begin{array}{cc}\begin{array}{c}a,a-\frac{1}{2},{e}_{1}+{e}_{2}-2a;\\ {e}_{1},{e}_{2};\end{array}& -\frac{4x}{{\left(1-x\right)}^{2}}\end{array}\right]\hfill \\ \phantom{\rule{1em}{0ex}}{=}_{3}{F}_{2}\left[\begin{array}{cc}\begin{array}{c}2a-1,2a-{e}_{1},2a-{e}_{2};\\ {e}_{1},{e}_{2};\end{array}& x\end{array}\right].\hfill \end{array}$
(1.7)
The transformation formula (1.7) is originally due to Whipple [6] who obtained it by other means.
It is interesting to mention here that in (1.7), (i) if we replace a by $a+\frac{1}{2}$ and take ${e}_{1}\to a+b+\frac{1}{2}$ and ${e}_{2}\to a+\frac{1}{2}$ and (ii) if we take ${e}_{1}\to a-\frac{1}{2}$ and ${e}_{2}\to a+b+\frac{1}{2}$ and simplify, we respectively recover (1.5) and (1.6).
Very recently, Rakha and Rathie [7] established the extension of Saalschütz summation theorem (1.4) in the form
${}_{4}F_{3}\left[\begin{array}{cc}\begin{array}{c}a,b,d+1,-n;\\ c+1,1+a+b-c-n,d;\end{array}& 1\end{array}\right]=\frac{{\left(c-a\right)}_{n}{\left(c-b\right)}_{n}{\left(g+1\right)}_{n}}{{\left(c+1\right)}_{n}{\left(c-a-b\right)}_{n}{\left(g\right)}_{n}},$
(1.8)
where
$g=\frac{f\left(b-c\right)}{b-f}\phantom{\rule{2em}{0ex}}\text{and}\phantom{\rule{2em}{0ex}}f=\frac{d\left(a-c\right)}{a-d}.$
Also, in the same paper [7], as an applications, they have obtained the following results:
$\begin{array}{r}{{\left(1-x\right)}^{-2a}}_{3}{F}_{2}\left[\begin{array}{cc}\begin{array}{c}a,b,d+1;\\ a+b+\frac{3}{2},d;\end{array}& -\frac{4x}{{\left(1-x\right)}^{2}}\end{array}\right]\\ \phantom{\rule{1em}{0ex}}{=}_{4}{F}_{3}\left[\begin{array}{cc}\begin{array}{c}2a,a-b-\frac{1}{2},1+a-A,1+a+A;\\ a+b+\frac{3}{2},a-A,a+A;\end{array}& x\end{array}\right],\end{array}$
(1.9)
where
${A}^{2}=\frac{1}{b-d}\left({a}^{2}b-abd-\frac{1}{2}bd-\frac{1}{4}d\right)$
(1.10)
and
$\begin{array}{r}{{\left(1-x\right)}^{1-2a}}_{3}{F}_{2}\left[\begin{array}{cc}\begin{array}{c}a,b,d+1;\\ a+b+\frac{3}{2},d;\end{array}& -\frac{4x}{{\left(1-x\right)}^{2}}\end{array}\right]\\ \phantom{\rule{1em}{0ex}}{=}_{5}{F}_{4}\left[\begin{array}{cc}\begin{array}{c}2a-1,a+\frac{1}{2},a-b-\frac{3}{2},a+\frac{1}{2}-A,a+\frac{1}{2}+A;\\ a-\frac{1}{2},a+b+\frac{3}{2},a-\frac{1}{2}-A,a-\frac{1}{2}+A;\end{array}& x\end{array}\right],\end{array}$
(1.11)
where
${A}^{2}=\frac{1}{b-d}\left({a}^{2}b-ab-d-abd-\frac{1}{2}bd+\frac{1}{4}b\right).$
(1.12)
The results (1.9) and (1.11) may be regarded as the extensions of (1.5) and (1.6) as it can be seen by taking $d=a+b+\frac{1}{2}$.
The aim of this research is twofold. First, by utilizing the extension of Saalschütz’s summation theorem (1.8), we obtain a natural extension of Whipple’s transformation (1.7). Then, by employing the beta integral method, we obtain a new hypergeometric identity. The results derived in this paper are simple, easily established and may be potentially useful.
## 2 Demonstration of the beta integral method
The beta function $\mathrm{B}\left(\alpha ,\beta \right)$ is defined by the first integral and is known to be evaluated as the second one as follows:
$\mathrm{B}\left(\alpha ,\beta \right)=\left\{\begin{array}{cc}{\int }_{0}^{1}{t}^{\alpha -1}{\left(1-t\right)}^{\beta -1}\phantom{\rule{0.2em}{0ex}}dt\hfill & \left(\mathfrak{R}\left(\alpha \right)>0;\mathfrak{R}\left(\beta \right)>0\right);\hfill \\ \frac{\mathrm{\Gamma }\left(\alpha \right)\mathrm{\Gamma }\left(\beta \right)}{\mathrm{\Gamma }\left(\alpha +\beta \right)}\hfill & \left(\alpha ,\beta \in \mathbb{C}\mathrm{\setminus }{\mathbb{Z}}_{0}^{-}\right).\hfill \end{array}$
(2.1)
Krattenthaler and Rao [8] made a systematic use of the so-called beta integral method, a method of deriving new hypergeometric identities from old ones by mainly using the beta integral in (2.1) based on the Mathematica Package HYP, to illustrate several interesting identities for the hypergeometric series and Kampé de Fériet series in most cases of unit arguments.
In this section, we also apply the beta integral method to the known results (1.7), (1.9) and (1.11) to get new hypergeometric identities. However, we shall derive one identity in detail and others can be obtained similarly.
Thus, for example, let us consider (1.7) in the form
$\begin{array}{r}{}_{3}F_{2}\left[\begin{array}{cc}\begin{array}{c}a,a-\frac{1}{2},{e}_{1}+{e}_{2}-2a;\\ {e}_{1},{e}_{2};\end{array}& -\frac{4x}{{\left(1-x\right)}^{2}}\end{array}\right]\\ \phantom{\rule{1em}{0ex}}={{\left(1-x\right)}^{2a-1}}_{3}{F}_{2}\left[\begin{array}{cc}\begin{array}{c}2a-1,2a-{e}_{1},2a-{e}_{2};\\ {e}_{1},{e}_{2};\end{array}& x\end{array}\right].\end{array}$
(2.2)
Now, multiplying both sides of equation (2.2) by ${x}^{c-1}{\left(1-x\right)}^{e-c-1}$, integrating the resulting equation with respect to x from 0 to 1, expressing the involved ${}_{3}F_{2}$ as series, changing the order of integration and summation (which is easily seen to be justified due to the uniform convergence of the series involved in the process) and using the beta integral (2.1), then after some simplification, summing up the resulting series, we get the following identity (presumably new).
Corollary 1 For a to be a negative integer, the following hypergeometric transformation holds true.
$\begin{array}{r}{}_{5}F_{4}\left[\begin{array}{cc}\begin{array}{c}a,a-\frac{1}{2},{e}_{1}+{e}_{2}-2a,c,1-e;\\ {e}_{1},{e}_{2},\frac{1}{2}+\frac{1}{2}c-\frac{1}{2}e,1+\frac{1}{2}c-\frac{1}{2}e;\end{array}& 1\end{array}\right]\\ \phantom{\rule{1em}{0ex}}={\frac{\mathrm{\Gamma }\left(e\right)\mathrm{\Gamma }\left(2a+e-c-1\right)}{\mathrm{\Gamma }\left(e-c\right)\mathrm{\Gamma }\left(2a+e-1\right)}}_{4}{F}_{3}\left[\begin{array}{cc}\begin{array}{c}2a-1,2a-{e}_{1},2a-{e}_{2},c;\\ {e}_{1},{e}_{2},2a+e-1;\end{array}& 1\end{array}\right].\end{array}$
(2.3)
Following the same procedure, from (2.1) and known results (1.5) and (1.6), we get the following identities (presumably new).
Corollary 2 For a to be a negative integer, the following hypergeometric transformation holds true.
$\begin{array}{r}{}_{4}F_{3}\left[\begin{array}{cc}\begin{array}{c}a,b,c,1-e;\\ a+b+\frac{1}{2},\frac{1}{2}+\frac{1}{2}c-\frac{1}{2}e,1+\frac{1}{2}c-\frac{1}{2}e;\end{array}& 1\end{array}\right]\\ \phantom{\rule{1em}{0ex}}={\frac{\mathrm{\Gamma }\left(e\right)\mathrm{\Gamma }\left(2a+e-c\right)}{\mathrm{\Gamma }\left(e-c\right)\mathrm{\Gamma }\left(2a+e\right)}}_{3}{F}_{2}\left[\begin{array}{cc}\begin{array}{c}2a,a-b+\frac{1}{2},c;\\ a+b+\frac{1}{2},2a+e;\end{array}& 1\end{array}\right].\end{array}$
(2.4)
Corollary 3 For a to be a negative integer, the following hypergeometric transformation holds true.
$\begin{array}{r}{}_{4}F_{3}\left[\begin{array}{cc}\begin{array}{c}a,b,c,1-e;\\ a+b+\frac{1}{2},\frac{1}{2}+\frac{1}{2}c-\frac{1}{2}e,1+\frac{1}{2}c-\frac{1}{2}e;\end{array}& 1\end{array}\right]\\ \phantom{\rule{1em}{0ex}}={\frac{\mathrm{\Gamma }\left(e\right)\mathrm{\Gamma }\left(2a+e-c-1\right)}{\mathrm{\Gamma }\left(e-c\right)\mathrm{\Gamma }\left(2a+e-1\right)}}_{4}{F}_{3}\left[\begin{array}{cc}\begin{array}{c}2a-1,a+\frac{1}{2},a-b-\frac{1}{2},c;\\ a-\frac{1}{2},a+b+\frac{1}{2},2a+e-1;\end{array}& 1\end{array}\right].\end{array}$
(2.5)
We conclude this section by remarking that the results (2.4) and (2.5) can also be obtained from (2.3) by (i) replacing a by $a+\frac{1}{2}$ and taking ${e}_{1}\to a+b+\frac{1}{2}$ and ${e}_{2}\to a+\frac{1}{2}$ and (ii) taking ${e}_{1}\to a-\frac{1}{2}$ and ${e}_{2}\to a+b+\frac{1}{2}$ respectively.
## 3 Extension of Whipple’s transformation (1.7)
The extension of Whipple’s quadratic transformation (1.7) to be established in this paper is given in the following theorem.
Theorem 1 The following extension of Whipple’s transformation (1.7) holds true.
$\begin{array}{r}{{\left(1-x\right)}^{1-2a}}_{4}{F}_{3}\left[\begin{array}{cc}\begin{array}{c}a,a-\frac{1}{2},{e}_{1}+{e}_{2}-2a,d+1;\\ {e}_{1}+1,{e}_{2},d;\end{array}& -\frac{4x}{{\left(1-x\right)}^{2}}\end{array}\right]\\ \phantom{\rule{1em}{0ex}}{=}_{5}{F}_{4}\left[\begin{array}{cc}\begin{array}{c}2a-1,2a-{e}_{1}-1,2a-{e}_{2},a+\frac{1}{2}-A,a+\frac{1}{2}+A;\\ {e}_{1}+1,{e}_{2},a-\frac{1}{2}-A,a-\frac{1}{2}+A;\end{array}& x\end{array}\right],\end{array}$
(3.1)
where A is given by
${A}^{2}={\left(a-\frac{1}{2}\right)}^{2}-\frac{d\left({e}_{2}-2a\right)\left(2a-{e}_{1}-1\right)}{{e}_{1}+{e}_{2}-2a-d}.$
(3.2)
Proof In order to establish (3.1), we proceed as follows. Denote the left-hand side of (3.1) by S, then upon expressing the ${}_{4}F_{3}$ as a series given by the definition (1.1), after some simplification, we have
$\mathbf{S}=\sum _{n=0}^{\mathrm{\infty }}\frac{{\left(a\right)}_{n}{\left(a-\frac{1}{2}\right)}_{n}{\left({e}_{1}+{e}_{2}-2a\right)}_{n}{\left(d+1\right)}_{n}{\left(-1\right)}^{n}{2}^{2n}}{{\left({e}_{1}+1\right)}_{n}{\left({e}_{2}\right)}_{n}{\left(d\right)}_{n}n!}{x}^{n}{\left(1-x\right)}^{1-2a-2n}.$
Using the well-known binomial theorem
${}_{1}F_{0}\left[\begin{array}{cc}\begin{array}{c}a;\\ -;\end{array}& z\end{array}\right]={\left(1-z\right)}^{-a}=\sum _{m=0}^{\mathrm{\infty }}\frac{{\left(a\right)}_{m}}{m!}{z}^{m},$
we have
$\mathbf{S}=\sum _{n=0}^{\mathrm{\infty }}\sum _{m=0}^{\mathrm{\infty }}\frac{{\left(a\right)}_{n}{\left(a-\frac{1}{2}\right)}_{n}{\left({e}_{1}+{e}_{2}-2a\right)}_{n}{\left(d+1\right)}_{n}{\left(2a+2n-1\right)}_{m}{\left(-1\right)}^{n}{2}^{2n}}{{\left({e}_{1}+1\right)}_{n}{\left({e}_{2}\right)}_{n}{\left(d\right)}_{n}n!m!}{x}^{n+m}.$
Now, replacing m by $m-n$, making use of the result [1]
$\sum _{n=0}^{\mathrm{\infty }}\sum _{k=0}^{\mathrm{\infty }}A\left(k,n\right)=\sum _{n=0}^{\mathrm{\infty }}\sum _{k=0}^{n}A\left(k,n-k\right)$
and then using the identity
$\left(m-n\right)!={\left(-1\right)}^{n}\frac{m!}{{\left(-m\right)}_{n}},$
we have, after some simplification,
$\mathbf{S}=\sum _{m=0}^{\mathrm{\infty }}\frac{{\left(2a-1\right)}_{m}}{m!}{x}^{m}\sum _{n=0}^{m}\frac{{\left(-m\right)}_{n}{\left({e}_{1}+{e}_{2}-2a\right)}_{n}{\left(2a+m-1\right)}_{n}{\left(d+1\right)}_{n}}{{\left({e}_{1}+1\right)}_{n}{\left({e}_{2}\right)}_{n}{\left(d\right)}_{n}n!}.$
Summing up the inner series, we have
$\mathbf{S}=\sum _{m=0}^{\mathrm{\infty }}\frac{{\left(2a-1\right)}_{m}}{m!}{{x}^{m}}_{4}{F}_{3}\left[\begin{array}{cc}\begin{array}{c}-m,{e}_{1}+{e}_{2}-2a,2a+m-1,d+1;\\ {e}_{1}+1,{e}_{2},d;\end{array}& 1\end{array}\right].$
Finally, using the extension of Saalschütz’s summation theorem (1.8) and after much simplification, we get
$\mathbf{S}=\sum _{m=0}^{\mathrm{\infty }}\frac{{\left(2a-1\right)}_{m}{\left(2a-{e}_{1}-1\right)}_{m}{\left(2a-{e}_{2}\right)}_{m}{\left(a+\frac{1}{2}-A\right)}_{m}{\left(a+\frac{1}{2}+A\right)}_{m}}{{\left({e}_{1}+1\right)}_{m}{\left({e}_{2}\right)}_{m}{\left(a-\frac{1}{2}-A\right)}_{m}{\left(a-\frac{1}{2}+A\right)}_{m}m!}{x}^{m},$
where A is the same as given in (3.2). Finally, summing up the series with the definition (1.1), we easily arrive at the right-hand side of (3.1). This completes the proof of Theorem 1. □
Remark If we equate the coefficients of ${x}^{n}$ on both sides of equation (3.1), we get the following interesting identity (presumably new):
$\begin{array}{r}{}_{4}F_{3}\left[\begin{array}{cc}\begin{array}{c}-n,{e}_{1}+{e}_{2}-2a,2a+n-1,d+1;\\ {e}_{1}+1,{e}_{2},d;\end{array}& 1\end{array}\right]\\ \phantom{\rule{1em}{0ex}}=\frac{{\left(2a-{e}_{1}-1\right)}_{n}{\left(2a-{e}_{2}\right)}_{n}{\left(a+\frac{1}{2}-A\right)}_{n}{\left(a+\frac{1}{2}+A\right)}_{n}}{{\left({e}_{1}+1\right)}_{n}{\left({e}_{2}\right)}_{n}{\left(a-\frac{1}{2}-A\right)}_{n}{\left(a-\frac{1}{2}+A\right)}_{n}},\end{array}$
(3.3)
where A is the same as given in (3.2).
Further, in (3.3), if we take $d={e}_{1}$, we get the following identity (presumably new):
${}_{3}F_{2}\left[\begin{array}{cc}\begin{array}{c}-n,{e}_{1}+{e}_{2}-2a,2a+n-1;\\ {e}_{1},{e}_{2};\end{array}& 1\end{array}\right]=\frac{{\left(2a-{e}_{1}\right)}_{n}{\left(2a-{e}_{2}\right)}_{n}}{{\left({e}_{1}\right)}_{n}{\left({e}_{2}\right)}_{n}},$
(3.4)
which can also be obtained directly by equating coefficients of ${x}^{n}$ in Whipple’s transformation (1.7).
We remark in passing that if in Theorem 1 we take $d={e}_{1}$, so that $A=a-{e}_{1}-\frac{1}{2}$, after little simplification, we recover Whipple’s transformation (1.7).
## 4 Application
As already explained in detail in Section 2, using the beta integral method to our main result (3.1), it is not difficult to obtain the following hypergeometric identity which is given here without a proof.
Corollary 4 For a to be a negative integer, the following hypergeometric transformation holds true.
$\begin{array}{r}{}_{6}F_{5}\left[\begin{array}{cc}\begin{array}{c}a,a-\frac{1}{2},{e}_{1}+{e}_{2}-2a,d+1,c,1-e;\\ {e}_{1}+1,{e}_{2},d,\frac{1}{2}+\frac{1}{2}c-\frac{1}{2}e,1+\frac{1}{2}c-\frac{1}{2}e;\end{array}& 1\end{array}\right]\\ \phantom{\rule{1em}{0ex}}=\frac{\mathrm{\Gamma }\left(e\right)\mathrm{\Gamma }\left(2a+e-c-1\right)}{\mathrm{\Gamma }\left(e-c\right)\mathrm{\Gamma }\left(2a+e-1\right)}\\ \phantom{\rule{2em}{0ex}}{×}_{6}{F}_{5}\left[\begin{array}{cc}\begin{array}{c}2a-1,2a-{e}_{1}-1,2a-{e}_{2},a+\frac{1}{2}-A,a+\frac{1}{2}+A,c;\\ {e}_{1}+1,{e}_{2},a-\frac{1}{2}-A,a-\frac{1}{2}+A,2a+e-1;\end{array}& 1\end{array}\right],\end{array}$
(4.1)
where, of course, A is the same as defined in (3.2).
Here we mention two interesting special cases of (4.1).
1. (1)
In (4.1), if we replace a by $a+\frac{1}{2}$ and take ${e}_{1}\to a+b+\frac{1}{2}$ and ${e}_{2}\to a+\frac{1}{2}$, we get the following transformation.
Corollary 5 For a to be a negative integer, the following hypergeometric transformation holds true.
$\begin{array}{r}{}_{5}F_{4}\left[\begin{array}{cc}\begin{array}{c}a,b,d+1,c,1-e;\\ a+b+\frac{3}{2},d,\frac{1}{2}+\frac{1}{2}c-\frac{1}{2}e,1+\frac{1}{2}c-\frac{1}{2}e;\end{array}& 1\end{array}\right]\\ \phantom{\rule{1em}{0ex}}={\frac{\mathrm{\Gamma }\left(e\right)\mathrm{\Gamma }\left(2a+e-c\right)}{\mathrm{\Gamma }\left(e-c\right)\mathrm{\Gamma }\left(2a+e\right)}}_{5}{F}_{4}\left[\begin{array}{cc}\begin{array}{c}2a,a-b-\frac{1}{2},1+a-A,1+a+A,c;\\ a+b+\frac{3}{2},a-A,a+A,2a+e;\end{array}& 1\end{array}\right].\end{array}$
(4.2)
1. (2)
In (4.1), if we take ${e}_{1}\to a+b+\frac{1}{2}$ and ${e}_{2}\to a-\frac{1}{2}$, we get the following result.
Corollary 6 For a to be a negative integer, the following hypergeometric transformation holds true.
$\begin{array}{r}{}_{5}F_{4}\left[\begin{array}{cc}\begin{array}{c}a,b,d+1,c,1-e;\\ a+b+\frac{3}{2},d,\frac{1}{2}+\frac{1}{2}c-\frac{1}{2}e,1+\frac{1}{2}c-\frac{1}{2}e;\end{array}& 1\end{array}\right]\\ \phantom{\rule{1em}{0ex}}=\frac{\mathrm{\Gamma }\left(e\right)\mathrm{\Gamma }\left(2a+e-c-1\right)}{\mathrm{\Gamma }\left(e-c\right)\mathrm{\Gamma }\left(2a+e-1\right)}\\ \phantom{\rule{2em}{0ex}}{×}_{6}{F}_{5}\left[\begin{array}{cc}\begin{array}{c}2a-1,a+\frac{1}{2},a-b-\frac{3}{2},a+\frac{1}{2}-A,a+\frac{1}{2}+A,c;\\ a-\frac{1}{2},a+b+\frac{3}{2},a-A-\frac{1}{2},a+A-\frac{1}{2},2a+e-1;\end{array}& 1\end{array}\right].\end{array}$
(4.3)
We conclude this section by remarking that the results (4.2) and (4.3) can also be obtained from (1.9) and (1.11) by applying the beta integral method.
## References
1. 1.
Rainville ED: Special Functions. Macmillan, New York; 1960. Reprinted by Chelsea Publishing, Bronx, New York (1971)
2. 2.
Bailey WN: Products of generalized hypergeometric series. Proc. Lond. Math. Soc. 1928, 28(2):242-254.
3. 3.
Kim YS, Rakha MA, Rathie AK:Extensions of certain classical summation theorems for the series ${}_{2}F_{1}$, ${}_{3}F_{2}$ and ${}_{4}F_{3}$ with applications in Ramanujan’s summations. Int. J. Math. Math. Sci. 2010., 2010: Article ID 3095031
4. 4.
Rakha MA, Rathie AK:Generalizations of classical summation theorems for the series ${}_{2}F_{1}$ and ${}_{3}F_{2}$. Integral Transforms Spec. Funct. 2011, 22(11):823-840. 10.1080/10652469.2010.549487
5. 5.
Vidnunas R: A generalization of Kummer identity. Rocky Mt. J. Math. 2002, 32(2):919-936. 10.1216/rmjm/1030539701
6. 6.
Whipple FJW: Some transformations of generalized hypergeometric series. Proc. Lond. Math. Soc. 1927, 26(2):257-272.
7. 7.
Rakha MA, Rathie AK: Extensions of Euler type II transformation and Saalschütz’s theorem. Bull. Korean Math. Soc. 2011, 48(1):151-156.
8. 8.
Krattenthaler C, Rao KS: Automatic generation of hypergeometric identities by the beta integral method. J. Comput. Appl. Math. 2003, 160: 159-173. 10.1016/S0377-0427(03)00629-0
## Acknowledgements
This work was, in part, supported by the National Natural Science Foundation of China (No. 11201291), Natural Science Foundation of Shanghai (No. 12ZR1443800) and a grant of ‘The First-class Discipline of Universities in Shanghai’. The authors are grateful to the worthy referees for giving certain very useful suggestions.
## Author information
Authors
### Corresponding author
Correspondence to Xiaoxia Wang.
### Competing interests
The authors declare that they have no competing interests.
### Authors’ contributions
All authors contributed equally to this paper. They read and approved the final paper.
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Wang, X., Rathie, A.K. Extension of a quadratic transformation due to Whipple with an application. Adv Differ Equ 2013, 157 (2013). https://doi.org/10.1186/1687-1847-2013-157 | 8,124 | 21,219 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 71, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2022-05 | latest | en | 0.728327 |
https://www.jiskha.com/display.cgi?id=1323312281 | 1,502,942,645,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886102891.30/warc/CC-MAIN-20170817032523-20170817052523-00017.warc.gz | 908,698,426 | 4,021 | # algebra
posted by .
Suppose Mary deposits \$200 at the end of each month for 30 years into an account that pays 5% interest compounded monthly. How much total money will she have in the account at the end?
• algebra -
i = .05/12 = .001466666...
n= 30*12 =360
amount = 200( 1.00146666..^360 - 1)/.00416666
= 166451.75
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https://www.jiskha.com/questions/1285253/task-2-in-task-1-you-determined-how-much-of-your-own-money-youre-willing-to-spend-to | 1,643,111,266,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304810.95/warc/CC-MAIN-20220125100035-20220125130035-00252.warc.gz | 850,745,361 | 8,234 | # Algebra 1
In Task 1 you determined how much of your own money you’re willing to spend to get your business started. This is your limit—when making your product you can’t exceed this amount.
a. Write an inequality that represents the fact that while making your product you can’t exceed this spending limit.
b. Solve this inequality and graph the solution on a number line. Explain what your solution means in terms of the situation.
c. In Task 1 you determined how much you will charge for each item. Write an equation that represents your total earnings based on the price of your item and how many you sell.
d. Using your answer from part B and your equation from part C, what is the most money you can hope to earn from your business?
e. Don’t forget that at the beginning of the process you had to spend some of your own money to get started. With the costs taken into account, what was your total profit? Did you make money or lose money? Now that you have these values, would you adjust your business plan from Task 1? If so, how?
a. Now that you have an additional \$300, revise your inequality from part A of Task 2 to reflect your new spending limit. Solve this inequality and graph the solution on a number line. Explain what your solution means in terms of the situation.
b. If you still sell your item for the same price, what is the most money you can hope to earn from your business now?
c. Will you have to pay your parents? If so, determine how much you will owe them.
d. Think about how much time it will take you to create your product. You have 200 hours this summer to devote to creating your product. Write an inequality that represents your time constraint.
e. Solve your inequality from part D and graph your solution on a number line. Explain what your solution means in terms of the situation.
f. With the costs taken into account, what was your total profit? Did you make or lose money? Now that you have these values, would you adjust your business plan from Task 1? If so, how?
1. 👍
2. 👎
3. 👁
1. Did you ever find out the answer?
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2. 👎
2. I have to do the same thing. Do you think you can help me.
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3. Did anyone figure these questions out?
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4. did anyone get the answers to this?
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5. you guys are really . YOU have to pretend to make a summer business like selling tshirts. You have to find out how much you would invest in it and etc. There is no correct answers for this expect for you having to do it yourself.
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6. I wouldn't say is the right way to put it tbh
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7. this doesn't make sense can someone post theres as an example so I can see what I should do ? I wont copy but I have no clue what to do!
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8. helloooo
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9. you guys are hopeless
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10. TASK 1. mowing lawns, use your parents mowers and pay them 3 dollars per yard you mow, charge 10 dollars more than it takes for gas per yard, you buy the gas and pay your parents for borrowing the mower and you get 7 dollars each yard.
mowing yards is easy money. you can mow like 5 yards a day and that puts you at 35 dollars a day. my mowing business will be more than profitable
now will some pleaseeeee do task 3
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11. mannneeeeeee i need this
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12. e. Don’t forget that at the beginning of the process you had to spend some of
your own money to get started. With the costs considered, what was
your total profit? Did you make money or lose money? Now that you have
these values, would you adjust your business plan from Task 1? If so, how?
1. 👍
2. 👎
13. y'all so trash don't know that answers
1. 👍
2. 👎
14. IM STUCK ON THIS WTH
1. 👍
2. 👎
15. Here take mine as an example:
For Task #2 b.
Fidgets
\$5 each.
Spending \$60
x is less than or equal to 5/2.00
x is less than or equal to (answer) \$2.50
I hope this helps.
Don't take my work though 😒
1. 👍
2. 👎
16. ty astrid, your the only smart one here lol
1. 👍
2. 👎
## Similar Questions
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2. ### Algebra
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Plz help me understand this. I need to know how to do it. It took Khalid 90 minutes to complete 40 tasks. Which answer choice is an equivalent rate? A). 10 task in 0.9 minutes B). 10 task in 2.25 minutes C). 10 task in 9 minutes
4. ### Career prep
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1. ### algebra
You want to start a summer business to earn money. What will you do? You have to consider how much money you can afford to invest in this business, how much it will cost you to make each item, and how much you’re going to charge
2. ### Algebra
In Task 1 you determined how much you will charge for each item. Write an equation that represents your total earnings based on the price of your item and how many you sell. Item price: \$15
3. ### algebra
Directions: Complete each of the tasks outlined below. Task 1 You want to start a summer business to earn money. What will you do? You have to consider how much money you can afford to invest in this business, how much it will
4. ### operation management
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4. ### social studies
The task of economic policy is to create a prosperous America. The unfinished task of prosperous Americans is to build a Great Society. Our accomplishments have been many; these tasks remain unfinished: > to achieve full | 1,749 | 6,787 | {"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} | 3.859375 | 4 | CC-MAIN-2022-05 | latest | en | 0.954913 |
https://python-course.eu/machine-learning/neural-network-digits-dataset.php | 1,656,422,088,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103516990.28/warc/CC-MAIN-20220628111602-20220628141602-00266.warc.gz | 521,082,192 | 56,355 | # 23. A Neural Network for the Digits Dataset
By Bernd Klein. Last modified: 13 Apr 2022.
## Introduction
The Python module sklear contains a dataset with handwritten digits. It is just one of many datasets which sklearn provides, as we show in our chapter Representation and Visualization of Data. In this chapter of our Machine Learning tutorial we will demonstrate how to create a neural network for the digits dataset to recognize these digits. This example is accompanying the theoretical introductions of our previous chapters to give a practical view. You will see that hardly any Python code is needed to accomplish the actual classification and recognition task.
We will first load the digits data:
from sklearn.datasets import load_digits
We can get an overview of what is contained in the dataset with the keys method:
digits.keys()
### OUTPUT:
dict_keys(['data', 'target', 'frame', 'feature_names', 'target_names', 'images', 'DESCR'])
The digits dataset contains 1797 images and each images contains 64 features, which correspond to the pixels:
n_samples, n_features = digits.data.shape
print((n_samples, n_features))
### OUTPUT:
(1797, 64)
print(digits.data[0])
### OUTPUT:
[ 0. 0. 5. 13. 9. 1. 0. 0. 0. 0. 13. 15. 10. 15. 5. 0. 0. 3.
15. 2. 0. 11. 8. 0. 0. 4. 12. 0. 0. 8. 8. 0. 0. 5. 8. 0.
0. 9. 8. 0. 0. 4. 11. 0. 1. 12. 7. 0. 0. 2. 14. 5. 10. 12.
0. 0. 0. 0. 6. 13. 10. 0. 0. 0.]
print(digits.target)
### OUTPUT:
[0 1 2 ... 8 9 8]
The data is also available at digits.images. This is the raw data of the images in the form of 8 lines and 8 columns.
With "data" an image corresponds to a one-dimensional Numpy array with the length 64, and "images" representation contains 2-dimensional numpy arrays with the shape (8, 8)
print("Shape of an item: ", digits.data[0].shape)
print("Data type of an item: ", type(digits.data[0]))
print("Shape of an item: ", digits.images[0].shape)
print("Data tpye of an item: ", type(digits.images[0]))
### OUTPUT:
Shape of an item: (64,)
Data type of an item: <class 'numpy.ndarray'>
Shape of an item: (8, 8)
Data tpye of an item: <class 'numpy.ndarray'>
Let's visualize the data:
import matplotlib.pyplot as plt
plt.imshow(digits.images[0], cmap='binary')
plt.show()
Let's visualize some more digits combined with their labels:
import matplotlib.pyplot as plt
# set up the figure
fig = plt.figure(figsize=(6, 6)) # figure size in inches
fig.subplots_adjust(left=0, right=1, bottom=0, top=1, hspace=0.05, wspace=0.05)
# plot the digits: each image is 8x8 pixels
for i in range(64):
ax = fig.add_subplot(8, 8, i + 1, xticks=[], yticks=[])
ax.imshow(digits.images[i], cmap=plt.cm.binary, interpolation='nearest')
# label the image with the target value
ax.text(0, 7, str(digits.target[i]))
import matplotlib.pyplot as plt
# set up the figure
fig = plt.figure(figsize=(6, 6)) # figure size in inches
fig.subplots_adjust(left=0, right=1, bottom=0, top=1, hspace=0.05, wspace=0.05)
# plot the digits: each image is 8x8 pixels
for i in range(144):
ax = fig.add_subplot(12, 12, i + 1, xticks=[], yticks=[])
ax.imshow(digits.images[i], cmap=plt.cm.binary, interpolation='nearest')
# label the image with the target value
#ax.text(0, 7, str(digits.target[i]))
from sklearn.model_selection import train_test_split
res = train_test_split(digits.data, digits.target,
train_size=0.8,
test_size=0.2,
random_state=1)
train_data, test_data, train_labels, test_labels = res
from sklearn.neural_network import MLPClassifier
mlp = MLPClassifier(hidden_layer_sizes=(5,),
activation='logistic',
alpha=1e-4,
solver='sgd',
tol=1e-4,
random_state=1,
learning_rate_init=.3,
verbose=True)
mlp.fit(train_data, train_labels)
### OUTPUT:
Iteration 1, loss = 2.25145782
Iteration 2, loss = 1.97730357
Iteration 3, loss = 1.66620880
Iteration 4, loss = 1.41353830
Iteration 5, loss = 1.29575643
Iteration 6, loss = 1.06663573
Iteration 7, loss = 0.95558862
Iteration 8, loss = 0.94767318
Iteration 9, loss = 0.95242867
Iteration 10, loss = 0.83577430
Iteration 11, loss = 0.74541414
Iteration 12, loss = 0.72011102
Iteration 13, loss = 0.70790928
Iteration 14, loss = 0.69425700
Iteration 15, loss = 0.74458525
Iteration 16, loss = 0.67779333
Iteration 17, loss = 0.69691846
Iteration 18, loss = 0.67844516
Iteration 19, loss = 0.68164743
Iteration 20, loss = 0.68435917
Iteration 21, loss = 0.61988051
Iteration 22, loss = 0.61362164
Iteration 23, loss = 0.56615517
Iteration 24, loss = 0.61323269
Iteration 25, loss = 0.56979209
Iteration 26, loss = 0.58189564
Iteration 27, loss = 0.50692207
Iteration 28, loss = 0.65956191
Iteration 29, loss = 0.53736180
Iteration 30, loss = 0.66437126
Iteration 31, loss = 0.56201738
Iteration 32, loss = 0.85347048
Iteration 33, loss = 0.63673358
Iteration 34, loss = 0.69769079
Iteration 35, loss = 0.62714187
Iteration 36, loss = 0.56914708
Iteration 37, loss = 1.05660379
Iteration 38, loss = 0.66966105
Training loss did not improve more than tol=0.000100 for 10 consecutive epochs. Stopping.
MLPClassifier(activation='logistic', hidden_layer_sizes=(5,),
learning_rate_init=0.3, random_state=1, solver='sgd',
verbose=True)
predictions = mlp.predict(test_data)
predictions[:25] , test_labels[:25]
### OUTPUT:
(array([1, 5, 0, 7, 7, 0, 6, 1, 5, 4, 9, 2, 7, 8, 4, 1, 7, 3, 7, 4, 7, 4,
8, 6, 0]),
array([1, 5, 0, 7, 1, 0, 6, 1, 5, 4, 9, 2, 7, 8, 4, 6, 9, 3, 7, 4, 7, 1,
8, 6, 0]))
from sklearn.metrics import accuracy_score
accuracy_score(test_labels, predictions)
### OUTPUT:
0.725
for i in range(5, 30):
mlp = MLPClassifier(hidden_layer_sizes=(i,),
activation='logistic',
random_state=1,
alpha=1e-4,
solver='sgd',
tol=1e-4,
learning_rate_init=.3,
verbose=False)
mlp.fit(train_data, train_labels)
predictions = mlp.predict(test_data)
acc_score = accuracy_score(test_labels, predictions)
print(i, acc_score)
### OUTPUT:
5 0.725
6 0.37222222222222223
7 0.8166666666666667
8 0.8666666666666667
9 0.8805555555555555
10 0.925
11 0.9388888888888889
12 0.9388888888888889
13 0.9388888888888889
14 0.9527777777777777
15 0.9305555555555556
16 0.95
17 0.8916666666666667
18 0.8638888888888889
/home/bernd/anaconda3/lib/python3.8/site-packages/sklearn/neural_network/_multilayer_perceptron.py:692: ConvergenceWarning: Stochastic Optimizer: Maximum iterations (200) reached and the optimization hasn't converged yet.
warnings.warn(
19 0.9555555555555556
20 0.9638888888888889
21 0.9722222222222222
22 0.9611111111111111
23 0.9444444444444444
24 0.9583333333333334
25 0.9305555555555556
26 0.9722222222222222
27 0.9694444444444444
28 0.975
29 0.9611111111111111
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Enrol here | 2,341 | 6,805 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2022-27 | longest | en | 0.651492 |
https://www.teachoo.com/9198/2479/Ex-10.1--7/category/Ex-10.1/ | 1,680,128,766,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949035.66/warc/CC-MAIN-20230329213541-20230330003541-00689.warc.gz | 1,143,745,496 | 32,094 | Ex 10.1
Chapter 10 Class 6 Mensuration
Serial order wise
Get live Maths 1-on-1 Classs - Class 6 to 12
### Transcript
Ex 10.1, 7 Find the perimeter of a triangle with sides measuring 10 cm, 14 cm and 15 cm Perimeter of ∆ABC = Sum of all sides = AB + BC + CA = 10 + 15 + 14 = 39 cm ∴ Required perimeter is 39 cm | 107 | 313 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2023-14 | longest | en | 0.742711 |
https://www.geeksforgeeks.org/map-function-dictionary-python-sum-ascii-values/?ref=rp | 1,660,861,540,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882573533.87/warc/CC-MAIN-20220818215509-20220819005509-00043.warc.gz | 663,378,511 | 24,730 | # Map function and Dictionary in Python to sum ASCII values
• Last Updated : 26 Jul, 2022
We are given a sentence of english language(can also contain digits), we need to compute and print the sum of ASCII values of characters of each word in that sentence.
Examples:
```Input : GeeksforGeeks, a computer science portal
for geeks
Output : Sentence representation as sum of ASCII
each character in a word:
1361 97 879 730 658 327 527
Total sum -> 4579
Here, [GeeksforGeeks, ] -> 1361, [a] -> 97, [computer]
-> 879, [science] -> 730 [portal] -> 658, [for]
-> 327, [geeks] -> 527
Input : I am a geek
Output : Sum of ASCII values:
73 206 97 412
Total sum -> 788
```
## Recommended: Please try your approach on {IDE} first, before moving on to the solution.
This problem has existing solution please refer Sums of ASCII values of each word in a sentence link. We will solve this problem quickly in python using map() function and Dictionary data structures. Approach is simple,
1. First split all words in sentence separated by space.
2. Create a empty dictionary which will contain word as key and sum of ASCII values of it’s characters as value.
3. Now traverse list of splitted words and for each word map ord(chr) function on each character of current word and them calculate sum of ascii values of each character of current word.
4. While traversing each word map sum of ascii values on it’s corresponding word in resultant dictionary created above.
5. Traverse splitted list of words and print their corresponding ascii value by looking up into resultant dictionary.
1. ## Python3
`# Function to find sums of ASCII values of each ``# word in a sentence in`` ` `def` `asciiSums(sentence):`` ` ` ``# split words separated by space`` ``words ``=` `sentence.split(``' '``)`` ` ` ``# create empty dictionary`` ``result ``=` `{}`` ` ` ``# calculate sum of ascii values of each word`` ``for` `word ``in` `words:`` ``currentSum ``=` `sum``(``map``(``ord``,word))`` ` ` ``# map sum and word into resultant dictionary`` ``result[word] ``=` `currentSum`` ` ` ``totalSum ``=` `0`` ` ` ``# iterate list of splited words in order to print`` ``# sum of ascii values of each word sequentially`` ``sumsOfAscii ``=` `[result[word] ``for` `word ``in` `words]`` ``print` `(``'Sum of ASCII values:'``)`` ``print` `(``' '``.join(``map``(``str``,sumsOfAscii)))`` ``print` `(``'Total Sum -> '``,``sum``(sumsOfAscii))`` ` `# Driver program``if` `__name__ ``=``=` `"__main__"``:`` ``sentence ``=` `'I am a geek'`` ``asciiSums(sentence)`
Output:
```Sum of ASCII values:
1361 97 879 730 658 327 527
Total sum -> 4579
```
My Personal Notes arrow_drop_up | 775 | 2,708 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-33 | latest | en | 0.742749 |
https://lijiyao111.gitbooks.io/algorithm-and-data-structure/content/Linked_list/PartitionList.html | 1,701,931,774,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100650.21/warc/CC-MAIN-20231207054219-20231207084219-00166.warc.gz | 417,467,789 | 13,557 | LeetCode #86
# Description:
Given a linked list and a value x, partition it such that all nodes less than x come before nodes greater than or equal to x.
You should preserve the original relative order of the nodes in each of the two partitions.
## Example:
``````For example,
Given 1->4->3->2->5->2 and x = 3,
return 1->2->2->4->3->5.
``````
# Idea:
create two empty dummy head, one track the nodes less than x, one track the nodes greater or equal to x.
In the end, set the less pointer pointing to dummy2.next, set greaterEqual pointer pointing to NULL.
# Code:
``````class Solution {
public:
ListNode* partition(ListNode* head, int x) {
ListNode dummy1(-1);
ListNode dummy2(-1);
ListNode * less_ptr = &dummy1;
ListNode * ge_ptr = &dummy2;
for(ListNode *curr=head; curr!=NULL; curr=curr->next){
if(curr->val<x){
less_ptr->next=curr;
less_ptr=less_ptr->next;
}
else{
ge_ptr->next=curr;
ge_ptr=ge_ptr->next;
}
}
less_ptr->next=dummy2.next;
ge_ptr->next=NULL; // This is important, otherwise could be loop
return dummy1.next;
}
};
`````` | 293 | 1,049 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-50 | latest | en | 0.671983 |
http://www.tenouk.com/cpluscodesnippet/ifelsecomparison.html | 1,508,836,143,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187828356.82/warc/CC-MAIN-20171024090757-20171024110757-00558.warc.gz | 578,266,968 | 2,211 | Comparing the two given integers using the if-else C syntax for greater than, less than or equal
Compiler: Visual C++ Express Edition 2005
Compiled on Platform: Windows XP Pro SP2
Additional project setting: Set project to be compiled as C
Project -> your_project_name Properties -> Configuration Properties -> C/C++ -> Advanced -> Compiled As: Compiled as C Code (/TC)
Other info: none
To do: Comparing the given two integers using the if-else C syntax for greater than, less than or equal
To show: How to use the if-else statements to compare two given integers in C programming for greater than, less than and equal
// Demonstrates the C if-else statement
#include <stdio.h>
int main(void)
{
int x, y;
// Input two values to be tested
printf("\nInput an integer value for x: ");
// read and store those inputs
// scanf("%d", &x);
scanf_s("%d", &x, 1);
printf("Input an integer value for y: ");
// scanf("%d", &y);
scanf_s("%d", &y, 1);
// Test values and print result
if (x == y)
{
printf("\nx is equal to y");
}
else
if (x > y)
{
printf("\nx is greater than y ");
}
else
{
printf("\nx is smaller than y ");
}
printf("\n\n");
return 0;
}
Output examples:
Input an integer value for x: 30
Input an integer value for y: 20
x is greater than y
Press any key to continue . . .
Input an integer value for x: 10
Input an integer value for y: 40
x is smaller than y
Press any key to continue . . .
Input an integer value for x: 80
Input an integer value for y: 80
x is equal to y
Press any key to continue . . . | 407 | 1,557 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2017-43 | latest | en | 0.571602 |
https://www.physicsoverflow.org/21916/entanglement-entropy-relation-between-translations-stress | 1,701,286,363,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100135.11/warc/CC-MAIN-20231129173017-20231129203017-00677.warc.gz | 1,092,964,657 | 23,946 | # CFT Entanglement Entropy - relation between translations and the stress-energy tensor
+ 1 like - 0 dislike
585 views
In a recent paper on CFT entanglement entropy, I want to understand the defintion of a certain partition function. They consider a metric space $S^1 \times \mathbb{H}^{d-1}_q$ with metric:
$$ds^2_{H_q^{d-1}} = d\tau^2 + du^2 + \sinh^2 u \; d\Omega_{d-2}^2$$
Here $d\tau$ is probably a Wick-rotated time, $u$ is a radial variable and $d\Omega$ is the spherical area measure.
Then they define a partition function $Z_q = \mathrm{tr}(e^{-2\pi q H_\tau})$ where $H_\tau$ "generates translations along the $S^1$". What does that mean? Could it mean this?
$$H_\tau = \frac{d}{d\tau}$$
This is the generator for translations along the $\tau$-axis.
However, they also say this is related to the stress-energy tensor: $H_\tau = \int_{\mathbb{H}^{d-1}} dx^{d-1} \sqrt{g} T_{\tau\tau}$ This seems like a very complicated way of describing translations. Could there be another meaning for the phrase "generates translations along $S^1$?
1. Jeongseog Lee, Aitor Lewkowycz, Eric Perlmutter, Benjamin R. Safdi
Renyi entropy, stationarity, and entanglement of the conformal scalar
This post imported from StackExchange Physics at 2014-08-10 19:42 (UCT), posted by SE-user john mangual
+ 1 like - 0 dislike
Everything is standard here. Think simply to a classical field on a Minkowski space $M = R_t \times R^3$.
The time translation generator, the hamiltonian, is defined by $H_t = P_0=\int T_{0i} d\sigma^i$, where $d\sigma^i = \epsilon^{ijkl} dx^j \wedge dx^k \wedge dx^l$.
Now, we may, in fact, consider only the $d\sigma^o= d\sigma^t$ component which is equals to $dx^1 \wedge dx^2 \wedge dx^3$, so finally :
$H_t = \int d^3xT_{00} = \int d^3xT_{tt}$
This post imported from StackExchange Physics at 2014-08-10 19:42 (UCT), posted by SE-user Trimok
answered Aug 1, 2014 by (955 points)
I think what I am missing is that you still need an action to define the CFT, and your equation computes the time evolution Hamiltonian from this action.
This post imported from StackExchange Physics at 2014-08-10 19:42 (UCT), posted by SE-user john mangual
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publicité
```Bernoulli 6(4), 2000, 633±651
Chaotic Kabanov formula for the AzeÂma
martingales
N I C O L A S P R I VA U LT 1, J O S E P L L U ÂI S S O L EÂ 2 and J O SE P V IV ES 2 1
DeÂpartement de Mathe matiques, Universite de la Rochelle, Avenue Marillac, 17042 La
Rochelle, Cedex 1, France. E-mail: [email protected]
2
Departament de MathemaÁtiques, Universitat AutoÁnoma de Barcelona, 08193 Bellaterra, Spain.
E-mail: [email protected]; [email protected]
We derive the chaotic expansion of the product of nth- and ®rst-order multiple stochastic integrals
with respect to certain normal martingales. This is done by application of the classical and quantum
product formulae for multiple stochastic integrals. Our approach extends existing results on chaotic
calculus for normal martingales and exhibits properties, relative to multiple stochastic integrals,
polynomials and Wick products, that characterize the Wiener and Poisson processes.
Keywords: AzeÂma martingales; multiple stochastic integrals; product formulae
1. Introduction
The Wiener±Itoà and Poisson±Itoà chaotic decompositions give an isometric isomorphism
between the Fock space Ã(L2 (R )) and the space of square-integrable functionals of the
process. This isomorphism is constructed by association of a symmetric function
f n 2 L2 (R ) n to its multiple stochastic integral. The isometry property comes from the
fact that the angle brackets hBi t and hN~ i t of both the Wiener and compensated Poisson
processes (Bt ) t2R , (N~ t ) t2R are equal to t. Therefore such isometric isomorphisms may exist
for more general processes, provided their angle bracket equals t.
A martingale is said to be normal (Dellacherie et al. 1992) if its angle bracket is equal
to t and if the range of the multiple stochastic integrals isometry de®ned on Ã(L2 (R )) is
equal to the space of square-integrable functionals of this martingale. This last property is
called the chaotic representation property.
The quadratic variation ([M, M] t ) t2R of a normal martingale (M t ) t2R in L4 (Ù) with
the chaotic representation property (or less restrictively, with the predictable representation
property) satis®es the equation
t
t 2 R ,
(1)
[M, M] t t ö s dM s ,
0
called a structure equation (Emery 1989), where (ö s ) s2R is a predictable process. On the
other hand, a martingale satisfying (1) has angle bracket equal to t, but does not necessarily
possess the chaotic representation property.
In this paper we work with solutions of such equations, which include the Wiener
1350±7265 # 2000 ISI/BS
N. Privault, J.L. Sole and J. Vives
634
process, ö 0; the compensated Poisson processes, ö constant and non-zero; and the
AzeÂma martingales, öt âM t , â 2 [ÿ2, 0[.
Note that in the probabilistic framework of the structure equation one has to use a
predictable version of ö, which is öt âM t ÿ in the case of the AzeÂma martingale.
However, since we are working in L2 with stochastic integrals with respect to a normal
martingale, we do not need to distinguish between the adapted and predictable versions of ö
(Dellacherie et al. 1992, p. 199). Thus, in what follows we will use öt âM t in the case of
the AzeÂma martingale.
In the stochastic analysis on both the Wiener and Poisson spaces, multiplication formulae
for multiple stochastic integrals proved to be useful tools as they give the chaotic expansion
of the product of two multiple stochastic integrals. Recently (Russo and Vallois 1998),
multiplication formulae have been proposed for multiple stochastic integrals with respect to
normal martingales. However, these formulae do not give the explicit chaotic expansion of
the considered products.
Our ®rst goal in this paper is to relate the product formulae of Kabanov (1975), Russo
and Vallois (1998) and Surgailis (1984) to their counterparts in quantum probability; see
Section 3. With these tools we compute the chaotic expansion of the product of a multiple
stochastic integral with a single stochastic integral, and refer to this formula as the chaotic
Kabanov formula. This formula is proved for a class of martingales that includes the AzeÂma
martingales; see Section 4. From the Kabanov formula, we have not obtained a general
formula for the chaotic expansion of the product of two multiple stochastic integrals of
orders n and m for this general class of martingales.
As an application of this Kabanov formula we show that a number of properties in
stochastic analysis such as the possibility of expressing multiple stochastic integrals with
polynomials and some properties of the Wick product are speci®c to the Wiener and
Poisson cases; see Section 5. In Section 6 we consider formulae giving the derivation of a
product.
2. Notation and preliminaries
We denote by
Ã(L2 (R )) L2 (R ) n
n2N
2
the symmetric Fock space over L (R ), where L2 (R ) n is endowed with the norm
k:k2L2 (R ) n n!k:k2L2 (R )
n , and h:, :i denotes the scalar product in L2 (R ).
In this paper we work with a normal martingale (M t ) t2R , with M 0 0, that satis®es the
structure equation (1).
The multiple stochastic integral of a symmetric function f n 2 L2 (R ) n with respect to
(M t ) t2R is denoted by
1
tÿn
tÿ
2
I n ( f n ) n!
f n (t1 , . . . , tn ) dM t1 dM tn ,
0
0
0
Chaotic Kabanov formula for the AzeÂma martingales
635
and as a convention we let
2
n
(2)
I n ( f n ) I n ( f~n ) I n ( f n (t1 , . . . , tn )),
is not symmetric, where f~n denotes the symmetrization of f n in its n
if f n 2 L (R )
variables.
We denote by F t the ó-algebra generated by fM s : 0 < s < tg, t 2 R .
We let =ÿ : Ã(L2 (R )) ! Ã(L2 (R )) L2 (R ) and = : Ã(L2 (R )) L2 (R ) !
Ã(L2 (R )) denote the annihilation and creation operators on Ã(L2 (R )), de®ned as
= (I n ( f n1 )) I n1 ( f n1 ),
=ÿt I n ( f n ) nI nÿ1 ( f n (:, t)), t 2 R ,
where f n1 is symmetric in its n ®rst variables, n 2 N.
Let A f]a, b] : a, b 2 R , a , bg. For ]a, b], ]c, d] 2 A we use the notation ]a, b] <
]c, d] if b < c. Let S n , n 2 N, denote the vector space
( m
X
á i I n (1 A1i 1 A in ) : A ik \ A il Æ, 1 < k , l < n, A1i , . . . , A in 2 A,
Sn
i1
á i 2 R, i 1, . . . , m, m > 1 ,
and let S denote the vector space generated by [ n2N S n, which is dense in L2 (Ù).
Let V n denote the vector space
( m
X
á i 1 A1i I n(1A1i . . . 1 A in ) : A ik \ A il Æ, 1 < k , l < n,
Vn
i1
A0k
\
A0l
Æ, 1 < k , l < m,
A0i ,
...,
A in
2 A, á i 2 R, i 1, . . . , m, m > 1 :
We denote by V the vector space generated by [ n>0 V n, which is dense in L2 (dë) and in
L1 (dë), where dë dt 3 dP.
Let Un denote the vector space
(
m
X
á i 1 A0i I n (1 A1i 1 A in ) : A ik \ A il Æ, 0 < k , l < n,
Un
i1
A0k \ A0l Æ, 1 < k , l < m, A0i , . . . , A in 2 A, á i 2 R, i 1, . . . , m, m > 1 :
Note that Un is made of processes in V n that do not depend on the `present'. We denote
by U the vector space generated by [ n>0 Un, and by U the completion of U in L2 (dë).
The space U is strictly smaller than L2 (dë) but it contains the adapted square-integrable
processes and the range of =ÿ , since F 2 S implies =ÿ F 2 U.
We will use the following lemma of Ma et al. (1998).
Lemma 1. Let I n ( f n ), I 1 ( g1 ) 2 S . We have
I n ( f n )I 1 ( g 1 ) I n1 ( f n g1 ) n
1
0
I nÿ1 ( f n (:, t)) g 1 (t) d[M, M] t :
(3)
N. Privault, J.L. Sole and J. Vives
636
As a consequence of this lemma, we have
I n (1 A1 1 A n ) I 1 (1 A1 ) I 1 (1 An )
whenever Ai \ Aj Æ, 1 < i , j < n.
We assume that there exists a set P of functionals dense in L2 (Ù) and included both in
1
L (Ù) and in Dom(=ÿ ), and such that P is stable by =ÿt , t 2 R . This assumption is satis®ed
in the cases of interest to us, that is in the Wiener and Poisson cases, and also in the case of
the AzeÂma martingales since the latter is bounded (in this case it suf®ces to take P S ).
We recall the following identity, satis®ed in general on Fock space:
1
1
2
ÿ
ÿ
2
= s ut = t us ds dt :
(4)
E[= (u) ] kuk L2 (Ù)
L2 (R ) E
0
0
De®nition 1. Let Dom1,2 (= ) denote the set of processes u 2 L1 (dë) such that there exists a
sequence (un ) n2N V converging in L1 (dë) to u and such that (= (un )) n2N converges in
L1 (Ù) to an element of L2 (Ù) denoted as = (u).
Remark 1. The domain Dom1,2 (= ) is well de®ned. We can show this as follows. Assume
that (un ) n2N V converges in L1 (dë) to 0 and that (= (un )) n2N converges in L1 (Ù) to
G 2 L2 (Ù). From the duality relation
1
ÿ
un (t)= t F dt ,
F 2P,
E[= (un )F] E
0
we obtain E[GF] 0, F 2 P , hence G 0 since G 2 L2 (Ù).
This also means that if u 2 Dom1,2 (= ) and there exists a sequence (un ) n2N converging
in L1 (Ù) to u and such that (= (un )) n2N converges in L1 (Ù) to G 2 L2 (Ù), then
G = (u). Naturally, Dom1,2 (= ) contains the usual L2 domain of = which is denoted as
Dom2 (= ).
Finally, let
( m
X
Fi 1]a i ,b i ] I n (1 A1i 1 A in ) : [0, bi ] \ A ij Æ, j 1, . . . , n,
T n
i1
A ik
\
A il
Æ, 1 < k , l < n,
A1i ,
...,
A in
2
2 A, Fi 2 L (F
a i ),
i 1, . . . , m, m > 1 ,
and T be the vector space generated by [ n T n. The space T \ L2 (dë) is not dense in L2 (dë),
but it contains U, and it is dense in U.
Remark 2. Let u, v 2 U, where v is an adapted process. Then uv 2 T . To show this, we
write u 1 A I n (1 A1 1 A n ) with A1 , . . . , A k < B and
v 1 B I m (1 A n1 1 A n m )
with A n1 , . . . , A n m < B, since v is adapted. Then
Chaotic Kabanov formula for the AzeÂma martingales
637
uv 1 A\ B I k m (1 A1 1 A k 1 A n1 1 A n m )I nÿ k (1 A k1 1 A n ),
hence uv 2 T .
1
u t dM t 2 L2 (Ù), we have u 2 Dom1,2 (= ) and
1
ut dM t :
(5)
= (u)
Proposition 1. For u 2 T such that
0
0
Proof. We start by choosing u of the form u F1]a,b] I n (1 A1 1 A n ), with F
I m (1 B1 1 B m ), Ai \ [0, b] Æ, i 1, . . . , n, and B1 , . . . , Bm [0, a]. Then, from
Lemma 1:
= (u) I n m1 (1 B1 1 Bm 1]a,b] 1 A n )
I 1 (1]a,b] )I m (1 B1 1 Bm )I n (1 A1 1 A n )
FI 1 (1]a,b] )I n (1 A1 1 A n ):
Hence, by linearity, for any u 2 T and F 2 S ,
= (u) FI 1 (1[a,b] )I n (1 A1 1 An )
1
0
ut dM t :
If u 2 T , then choosing a sequence (Fn ) n2N S that converges in L2 (Ù) to F, and
letting um Fm 1]a,b] I n (1 A1 1 A n ), we have the convergence of (um ) m2N in L1 (dë) to u
1
and of (= (um )) m2N in L1 (Ù) to 0 ut dM t which belongs to L2 (Ù) by hypothesis.
h
3. Classical and quantum product formulae
The aim of this section is to link two versions of the product formula for a multiple stochastic
integral and a single stochastic integral with respect to normal martingales.
The ®rst one is called the Kabanov formula in reference to Kabanov (1975), which
treated the Poisson case, ö 1, and can be stated as follows. For f n 2 L2 (R ) n and
g 2 L2 ([0, T ]), in particular g with compact support, Lemma 1 was extended by Ma et al.
(1998) and Russo and Vallois (1998) as
1
(6)
I n ( f n )I 1 ( g) I n1 ( f n g) n I nÿ1 ( f n (:, t)) g(t) d[M, M] t :
0
The second term in this formula is an integration over a diagonal, due to the Itoà formula, and
the notation `I nÿ1 ( f n (:, t))' will be made precise in De®nition 2.
The second product formula uses only chaos expansions and Fock space, and can be
found in the work of quantum probabilists. It is often stated in the formalism of quantum
stochastic integrals:
N. Privault, J.L. Sole and J. Vives
638
1
0
g t dM t
1
0
gt dat
1
0
gt daÿt
1
0
ö t gt dat ;
(7)
cf. Parthasarathy (1990) and Biane (1995, Proposition 18).
The de®nition of quantum stochastic integrals as operators poses several functional
analytic problems. See the paper of Attal and Lindsay (1997) for recent extensions of their
de®nition. In general, the above relation does not hold in L2 (Ù) except in a weak sense
(Biane 1995), where this formula is proved for F an exponential vector, and for bounded
predictable g and ö, with compact support.
Relation (7) can be reformulated formally using the operators =ÿ , = , as follows:
1
g(t)=ÿt F dt = ( gö=ÿ F)
(8)
F= ( g) = (F g)
0
(see, for example, Attal 1998; Biane, 1995; and the references therein), the only difference
between (7) and (8) being a change of notation.
In the Poisson case, this formula also appears in the papers of Nualart and Vives (1990)
and Dermoune et al. (1988). It will play an important role in the computation of the chaotic
expansion of the Kabanov formula, since unlike (3) and (6) it uniquely involves calculations
on chaos.
The product formula (8) can be rewritten in the language of quantum stochastic
differentials as
1
1
1
g(t) dM t
g(s) dBs
ö(s) g(s) dat ,
0
0
0
or
dM t dBt ö(t) dat :
(9)
In the Wiener interpretation of the Fock space, the differential operator dBt identi®es
with multiplication by the classical Brownian differential, and this formula states that the
differential operator dM t dBt ö(t) dat is identi®ed with the multiplication operator by
the classical differential dM t when the Fock space is identi®ed with the L2 space of
(M t ) t2R . However, equation (9) has no classical interpretation because the operator
processes (Bt ) t2R and (at ) t2R cannot be interpreted simultaneously as multiplication
operators in the same probabilistic interpretation of the Fock space (the reason for this is
that they do not commute). Consequently, (9) does not have a classical meaning; it only
de®nes an operator process on Fock space.
If ö(t) is a function of M t , for example, ö(t) f (M t ), then (9) becomes a quantum
stochastic differential equation, in the space of operators on Fock space,
dM t dBt f (M t ) dat ,
(10)
which does not have a classical interpretation, whatever the interpretation chosen for the Fock
space.
In order to ®nd the multiplication formula for multiple stochastic integrals with respect to
(M t ) t2R , one has to compute in particular the chaos expansion of M t multiplied by a
multiple stochastic integral. This means that the explicit expression of M t as an operator on
Chaotic Kabanov formula for the AzeÂma martingales
639
Fock space has to be obtained. A way to obtain this expression is to solve (10) in the space
of operators on Fock space, that is to determine the process of operators associated with
(M t ) t2R . The simplest case is the linear case, f (x) âx, â 2 [ÿ2, 0[, which corresponds
to the family of AzeÂma martingales. In this case, (10) reads
dM t dBt âM t dat ,
(11)
which can be formally solved as a linear equation. It can be easily shown that its solution is
t2
X
t
t nÿ1
ân
B t1 dat2 dat nÿ1 ,
Mt
0 0
n>0
or
Mt
X
n>0
â
n
0
t
tn
0 0
t1
0
dB t0 dat1 datn :
This expression can now be rewritten explicitly in terms of operators on Fock space:
:
X
ÿ
ÿ
â n = 1[0, t] = =
=ÿ
ds=
=
F
(12)
MtF
s
0
n>0
X
â n = (1[0, t] (= ( = (= (1[0,:] =ÿ =ÿ F))))):
n>0
When F I n ( f n ), this gives the explicit chaos expansion of M t I n ( f n ).
The above calculation is formal and some further computations are required in order to
obtain the explicit chaos expansion of M t I n ( f n ). In this paper we justify each of the above
steps by explicit calculations and carry out the ®nal computations suggested in (12). The
aim of this remark was to show that the quantum stochastic point of view gives a different
understanding of the problem and provides a quick solution by reducing the problem to the
determination of a quantum diffusion. We stress that although the solution process has a
classical version, the diffusion equation (10) is meaningful only in the space of operators on
Fock space. Also, those remarks show that the reason why explicit calculation can be
carried out is in fact that (11) is a linear equation.
We also note that a slightly more general case can be considered, ö(t)
t
ë t 0 â t (s) dM s , which is still linear. In this case the solution reads
t nÿ1
t1
X
t
tn
â tn (t nÿ1 )
dB t0 dat1 datn
Mt
n>0 0 0
X
t
tn
n>0 0 0
0
â tn (t nÿ1 )
0
t nÿ1
0
t2
0
â t2 (t1 )ë t1 dat0 datn :
This situation is considered in our paper.
Our aim in this section is to prove (7) in L2 (Ù) from (6). The proof of Biane (1995) ±
see also Attal (1998) ± relies on the construction of quantum stochastic calculus, whereas
N. Privault, J.L. Sole and J. Vives
640
our proof uses more classical probabilistic arguments. We also prove that (8) holds in L2
under assumptions that are satis®ed in our setting.
De®nition 2. Following Ma et al. (1998) and Russo and Vallois (1998), let ì denote the
measure on R 3 Ù de®ned as ì([0, t] 3 A) E[1 A [M, M] t ], A 2 F , t 2 R , and let
í (ë ì)=2. For u 2 U we denote by u the limit in L2 (dí) of any sequence (un ) n2N U
that converges in L2 (dë) to u.
Note that u u ë-a.e., but not ì-a.e., except in the Wiener case (ö 0), and in general
for u 2 U. From Lemma 5.2. of Ma et al. (1998) we have
u 2 U:
kuk L2 (dë) ku k L2 (d ì) ,
T In what follows we ®x T . 0. For u 2 U,
0 u t d[M, M] t is de®ned a.e. as an integral
with respect to an increasing process and it belongs to L1 (Ù), since
"
#
T E u t d[M, M] t < T 1=2 ku k L2 (dì) < T 1=2 kuk L2 (dë) :
(13)
0
The following lemma allows us to prove the quantum product formula (7) from the
probabilistic product formula (6).
Lemma 2. Let u 2 U and assume that 1[0,T ] ö 2 L2 (dë)
1[0,T ] uö 2 Dom1,2 (= ) if and only if
T
ut d[M, M] t 2 L2 (Ù),
for some T 2 R . Then
0
and in this case,
T
0
ut dt = (1[0,T ] uö)
T
0
ut d[M, M] t :
(14)
Pn
Fi 1]a i ,b i ] , where bi < T :
Proof. We start by assuming that u 2 U is of the form u i1
Since ö is adapted we can choose a sequence (v n ) n2N U of adapted processes converging
to ö1[0,T ] in L2 (dë). Then, from Remark 2, uv n 2 T and, from Proposition 1,
1
uv n (t) dM t
:
(= (uv n )) n2N
0
1
n2N
Hence (= (uv n )) n2N converges in L (Ù) to
bi
T
n
X
Fi ö s dM s ut ö t dM t 2 L2 (Ù),
i1
uö 2 Dom1,2 (= ) and = (uö)
ai
0
T
0
u t ö t dM t, that is,
Chaotic Kabanov formula for the AzeÂma martingales
T
0
u t dt = (uö)
T
0
641
ut d[M, M] t :
2
If u 2 U we
T a sequence (un ) n2N in U. Then, from 1(13) we
T approximate it in L (dë) by
u
(t)
dt
=
(u
ö))
(
have
that
(
n
n2N
0 n
0 un (t) d[M, M] t ) n2N converges in L (Ù) to
T u
d[M,
M]
.
h
t
0
We now prove the quantum product formula (7), under assumptions different from that of
Biane (1995).
Proposition 2. Let F be in a ®nite sum of chaos and let h 2 L2 ([0, T ]) be bounded. Then
hö=ÿ F 2 Dom1,2 (= ) if and only if I 1 (h)F 2 L2 (Ù) and in this case,
I 1 (h)F = (h F) (h, =ÿ F) L2 (R ) = (hö=ÿ F):
(15)
Proof. We let F I n ( f n ) 2 S , h 2 L2 ([0, T ]), and apply (6) and Lemma 2 to
since U contains the range of =ÿ . In the general case we choose a
h=ÿ I n ( f n ) 2 U,
converges to F in L2 (Ù). The right-hand side of (15) converges
sequence (Fn ) n2N T S that
ÿ
to = (h F) 0 ht = t F dt = (hö=ÿ F) I 1 (h)F in L1 (Ù) and (hö=ÿ Fn ) n2N
converges to hö=ÿ F in L1 (Ù) as n goes to in®nity. Hence hö=ÿ F 2 Dom1,2 (= ) if and
h
only if I 1 (h)F 2 L2 (Ù).
The result extends to bounded simple adapted processes of the form h G1[ t1 , t2 [ for
F t1 -measurable G, since in this case =ÿs G 0, s . t1 (see Lemma 4.1. of Ma et al. 1998),
which means that = (1[ t1 , t2 [ G) G= (1[ t1 , t2 [ ).
We close this section with a remark on the link between independence of stochastic
integrals and their deterministic kernels. The above formula easily gives information on the
chaotic expansion of a product of stochastic integrals, for general ö. If f , g 2 L2 ([0, T ])
are such that fgö 2 Dom1,2 (= ), then
I 1 ( f )I 1 ( g) I 2 ( f g) ( f , g) L2 (R ) = ( fgö):
(16)
Letting T go to in®nity, the formula holds for f , g 2 L2 (R ), provided
fgö 2 Dom1,2 (= ). From this formula it is clearly seen that the chaotic expansion of
I 1 ( f )I 1 ( g) may be an in®nite sum of multiple stochastic integrals, depending on the chaotic
expansion of ö. This formula can be applied in order to obtain a necessary condition for the
independence of stochastic integrals. The result of Urbanik (1967) says that if a stochastic
1
process
1 (X t ) t2R has stationary and independent increments then independence of 0 f t dX t
and 0 gt dX t implies fg 0, except if X is Gaussian, in which case the condition becomes
( f , g) L2 (R ) 0. In the Gaussian case this property has been further extended to multiple
È stuÈnel and Zakai (1990). See also Privault (1996) for the case of
stochastic integrals by U
deterministic ö. We can now give an extension of this property to more general ö.
Proposition 3. Assume that the chaotic expansion of öt does not contain terms of order 1.
Then, independence of I 1 ( f ) and I 1 ( g) implies ( f , g) L2 (R ) 0 and f t g t 0, E[ö2t ] dt-a.e.
N. Privault, J.L. Sole and J. Vives
642
È stuÈnel and Zakai (1990) that deals with the Wiener case.
Proof. We follow an argument of U
2
k f gk L2 (R )2 > ( f , f ) L2 (R ) ( g, g) L2 (R ) E[I 1 ( f )2 ]E[I 1 ( g)2 ]
E[(I 1 ( f )I 1 ( g))2 ]
k f gk2L2 (R )2 ( f , g)2L2 (R ) k= ( fgö)k2L2 (Ù) ,
where we used the assumption on the chaotic expansion of ö which implies the orthogonality
of = ( fgö) and I 2 ( f g) in (16). Hence ( f , g) L2 (R ) 0, and = ( fgö) 0 a.e. Hence
fgö 0, dë-a.e.
h
Note that the hypothesis of this proposition do not include the AzeÂma martingales.
4. Chaotic Kabanov formula
The purpose of this section is to obtain in Theorem 1 a chaotic formula for the product
I 1 ( g)I n ( f n ) using Proposition 2 applied to F I n ( f n ). Under certain assumptions on f n and
g, we have
I 1 ( g)I n ( f n ) I n1 ( f n g) nI nÿ1 (h f n (, :), g(:)i) n= ( g(:)ö(:)I nÿ1 ( f n (, :))):
(17)
If ö is deterministic, in particular in the Wiener and Poisson cases, this formula easily yields
the Kabanov formula (Privault 1996). On the other hand, if ö is random the chaos expansion
of the term ö: I nÿ1 ( f n (, :)) is unknown unless öt belongs to the ®rst chaos, i.e.
ö(t) á(t) I 1 (â t ), t > 0, since in this case an induction argument can be used in (17)
to compute the term ö: I nÿ1 ( f n (t1 , . . . , t nÿ1 , :)) and to determine the chaotic expansion of
the product I 1 (â t )I n ( f n ). Thus in this case the chaotic expansion of I 1 ( g)I n ( f n ) can be
obtained as a consequence of Proposition 2.
In this section, we assume that ö is of the form ö(t) á(t) I 1 (ât ), t > 0, where á is
a locally bounded function from R to R and â t (:) is a bounded function from R to R
with support in [0, t], for any ®xed t 2 R . This situation is more general than the situation
of Russo and Vallois (1998, Section 4); moreover, chaotic expansions are completely
determined here.
We have the following lemma:
Lemma 3. Let è 2 R . Let f p 2 L2 (R ) p and
æ(t1 , . . . , t kÿ1 ; tk ) â t2 (t1 ) â t k (t kÿ1 ),
t 1 , . . . , t k 2 R ,
k . 1,
and æ(t) 1, t 2 R , for k 1. The chaos expansion of the product I 1 (âè )I p ( f p ) is given by
I 1 (âè )I p ( f p ) I p1 ( gp1 (; è)) I p ( gp (; è)) I pÿ1 ( gÿpÿ1 (; è)),
where
Chaotic Kabanov formula for the AzeÂma martingales
gp1 (t1 , . . . , t p1 ; è)
643
p1
X
p!
æ(tj , . . . , t p1 ; è) f p (t1 , . . . , ^t j , . . . , t p1 ),
(
j
ÿ
1)!
j1
where ^t j means that tj is omitted in the arguments of f p ,
gp (t1 , . . . , tp ; è) f p (t1 , . . . , tp )
p
X
p!
æ(tj , . . . , tp ; è)á(tj ),
( j ÿ 1)!
j1
and
g ÿpÿ1 (t1 , . . . , t pÿ1 ; è)
p
X
p!
æ(tj , . . . , t pÿ1 ; è)h f p (t1 , . . . , t pÿ1 , :), â tj (:)i,
(
j
ÿ
1)!
j1
with tp è.
Proof. We will prove the lemma by induction on p. Observe that the kernels g , g and gÿ
are not symmetric functions and that this result uses the convention (2).
For p 1, using Proposition 2, we have
I 1 (âè )I 1 ( f ) I 2 (âè f ) hâè , f i = (âè ö f )
I 2 (âè f ) hâè , f i I 1 (âè á f ) I 2 (âè( t1 ) f (t1 ) â t1 (t2 ))
I 2 (âè f âè f â t ) I 1 (âè á f ) hâè , f i,
and therefore, in particular, I 1 (âè )I 1 ( f ) is in the domain of = .
For the general case, applying Proposition 2 and the fact that f p1 is symmetric, we have
I 1 (âè )I p1 ( f p1 ) I p2 (âè f p1 ) ( p 1)I p (hâè (:), f p1 (, :)i)
( p 1)= (âè (:)á(:)I p ( f p1 (, :)))
( p 1)= (âè (:)I 1 (â: )I p ( f p1 (, :))):
Now, applying the induction hypothesis to I 1 (â t p2 )I p ( f p1 (, t p2 )) in the last term, we
have,
I 1 (âè )I p1 ( f p1 ) I p2 (âè f p1 )
( p 1)I p2 (âè (t p2 ) çp1 (t1 , . . . , t p1 ; t p2 ))
( p 1)I p1 (âè (t p1 )á(t p1 ) f p1 (t1 , . . . , t p1 ))
( p 1)I p1 (âè (t p2 ) ç0p (t1 , . . . , tp ; t p2 ))
( p 1)I p (hâè (:), f p1 (, :)i)
( p 1)I p (âè (t p2 ) çÿpÿ1 (t1 , . . . , t pÿ1 ; t p2 )),
N. Privault, J.L. Sole and J. Vives
644
where çp1 (, r), çp (, r) and çÿpÿ1 (, r) are the functions de®ned in the lemma with
f p1 (, r) and â r respectively in place of f p () and âè .
If we write the kernels of orders p, p 1 and p 2, we can see that they have the
desired form. Note also that we have proved that I 1 (âè ) p1 ( f p1 ) is in the domain of = .h
Let Ó p denote the set of all permutations of f1, . . . , pg. In the following corollary we
replace f p by the product h1 hp to give a partial symmetrization of the kernels
obtained in the above proposition.
Corollary 1. If f p h1 hp , where hi 2 L2 (R ) for any i 1, . . . , p, we have the
following expression:
I 1 (âè )I p (h1 hp ) I p1 ( g p1 (; è)) I p ( g p (; è)) I pÿ1 ( g ÿpÿ1 (; è)),
where the symmetrizations of gp1 , gp and gÿpÿ1 are given by
g~p1 (t1 , . . . , t p1 ; è)
p
XX
1
ó 2Ó p l0
l!
â t l2 (t l1 )
l
Y
hó (i) (ti )
p2ÿ
Yl
j3
i1
(â t l j hó ( l jÿ2) )(t l jÿ1 )
with t p2 è,
g~p (t1 , . . . , tp ; è)
pÿ1
p1ÿ
l
XX
Y
Yl
1
(áâ t l2 hó ( l1) )(t l1 )
hó (i) (t i )
(â t l j hó ( l jÿ1) )(t l jÿ1 )
l!
j3
i1
ó 2Ó l0
p
with t p1 è, and
g~ÿpÿ1 (t1 , . . . , t pÿ1 ; è)
pÿ1
pÿ l
l
XX
Y
Y
1
hâ t l1 , hó ( l1) i
hó (i) (t i ) (â t l j hó ( l j) )(t l jÿ1 )
l!
j2
i1
ó 2Ó l0
p
with tp è.
We are now ready to state the chaotic Kabanov formula.
Theorem 1. Let g be a bounded function with compact support on R , and
ã(t1 , . . . , tk ) â t2 (t1 ) â t k (t kÿ1 ) g(tk ),
t1 , . . . , tk 2 R ,
k . 1,
and ã(t) g(t), t 2 R , for k 1. We have
I 1 ( g)I p ( f p ) I p1 ( gp1 ) I p ( gp ) I pÿ1 ( gÿpÿ1 )
where the kernels g p1 , gp and gÿpÿ1 are
g p1 (t1 , . . . , t p1 )
p1
X
p!
ã(t j , . . . , t p1 ) f p (t1 , . . . , ^t j , . . . , t p1 ),
(
j
ÿ
1)!
j1
where ^t j means that tj is omitted in the arguments of f p ,
(18)
Chaotic Kabanov formula for the AzeÂma martingales
g p (t1 , . . . , tp ) f p (t1 , . . . tp )
645
p
X
p!
ã(tj , . . . , tp )á(tj ),
(
j
ÿ
1)!
j1
and
g ÿpÿ1 (t1 , . . . , t pÿ1 )
pÿ1
X
p!
ã(tj , . . . , t pÿ1 )h f p (t1 , . . . , t pÿ1 , :), â t j (:)i
(
j
ÿ
1)!
j1
ph f p (t1 , . . . , t pÿ1 , :), g(:)i:
Proof. By Proposition 2, we have
I 1 ( g)I p ( f p ) I p1 ( g f p ) pI pÿ1 (h f p (, :), g(:)i)
p= ( g(:)á(:)I pÿ1 ( f p (, :))) p= ( g(:)I 1 (â: )I pÿ1 ( f p (, :))),
and applying Lemma 3 we obtain the decomposition (18).
h
Proposition 4. Assume now that the underlying martingale is the standard AzeÂma martingale,
that is, á 0 and âè ÿ1[0,è], è 2 R . Then
p
I 1 (1[0,è] )I p (1
[0,è] )
1
p1)
pÿ1)
I p1 (1
(
) ÿ pI pÿ1 ((t( pÿ1) ÿ è)1
(
(t1 , . . . , t pÿ1 )),
[0,è]
[0,è]
p1
with t(0) 0 and t( pÿ1) t1 _ _ t pÿ1, p . 1.
p
1[0,è] (ti ) and g 1[0,è] . By Theorem 1 we have
Proof We let f p (t1 , . . . , tp ) Ð i1
gp1 (t1 , . . . , t p1 ; è)
p1
X
Y
p!
(ÿ1) pÿ jÿ1 1f t j << t p1 <èg
1[0,è] (t i ),
( j ÿ 1)!
j1
i1
jÿ1
g p (t1 , . . . , tp ; è) 0,
gÿpÿ1 (t1 , . . . , t pÿ1 ; è)
p
X
(ÿ1) pÿ j
j1
Y
p!
tj 1f t j << t pÿ1 <èg
1[0,è] (t i ),
( j ÿ 1)!
i1
jÿ1
with tp è. Symmetrizating and doing some straightforward computations we obtain the
desired result.
h
5. Consequences of the chaotic Kabanov formula
In this section we use the chaotic Kabanov formula to prove that the Wiener and Poisson
processes are the only normal martingales to possess certain properties relative to
polynomials and Wick product.
N. Privault, J.L. Sole and J. Vives
646
5.1. Existence of a family of orthogonal polynomials associated with the
martingale
Let X be a normal martingale. Following Meyer (1976), we de®ne by induction the
martingales
(1)
( n)
1,
P
X
,
.
.
.
,
P
P(snÿ1)
dX s ,
t 2 R :
P(0)
ÿ
t
t
t
t
[0, t]
P(t n)
(1=n!)I n (1[0,n t] )
P(t n) H n (X t , t),
for all n. It is well known that
Note that
p in the Wiener case we have
where H n (x, y) y n=2 hn (x= y) and hn is the Hermite
the relation
polynomial of degree n.
Also in the Poisson case, we have P(t n) Cn (X t , t), where fCn (x, y)g are the Charlier
polynomials (Meyer 1976; Surgailis 1984).
This situation motivates the following de®nition:
De®nition 3. We will say that a normal martingale X has an associated family of polynomials
fQn (x, y)g, where Qn is a polynomial of degree n in x, if
P(t n) Qn (X t , t), for all n:
Now the natural problem is to characterize the normal martingales that have an
associated family, and the answer is the next theorem.
Theorem 2. Let X be a normal martingale in L4 (Ù). Then X has an associated family of
polynomials if and only if ö s is a deterministic constant process, that is, X is a Poisson or a
Wiener process.
Proof. The if part is straightforward. If ö s 0 we are in the Wiener case, and if ö s c 6 0,
we are in the Poisson case with jumps of height c.
The proof of the only if part is a consequence of the Kabanov formula. Assume that the
chaotic decomposition of the process ö is
öt
1
X
I n ( f n (:, t)),
n0
and X has an associated family of polynomials. Then, by the Kabanov formula,
(1)
2
2
P(1)
t P t X t I 1 (1[0, t] )I 1 (1[0, t] ) I 2 (1[0, t] ) t = (1[0, t] (x)ö(x))
I 2 (12
[0, t] ) t
1
X
(19)
I i (1[0, t] (x) f iÿ1 (:, x)):
i1
X 2t
will be a linear combination of the three polynomials Q0 (X t , t),
For each ®xed t,
Q1 (X t , t) and Q2 (X t , t), which are respectively 1, I 1 (1[0, t] ) and 12 I 2 (12
[0, t] ). So
X 2t a2 (t)I 2 (12
[0, t] ) a1 (t)I 1 (1[0, t] ) a0 (t):
(20)
Chaotic Kabanov formula for the AzeÂma martingales
647
Then the chaotic decomposition of X 2t has only terms upto the second chaos, and from
(19) we obtain f n 0 for n > 2. Therefore,
öt f 0 (t) I 1 ( f 1 (:, t)):
Identifying the kernels of (19) and (20), we have for all t
1[0, t] (x) f 0 (x) a1 (t)1[0, t] (x)
for the ®rst chaos, and
2
1
12
[0, t] (u, v) 2(1[0, t] (u) f 1 (v, u) 1[0, t] (v) f 1 (u, v)) a2 (t)1[0, t] (u, v),
u, v 2 R ,
for the second chaos. So f 0 (u) c0 , u 2 R , and f 1 (:, x) c1 1[0,x] (:).
Then
öt c0 c1 I 1 (1[0, t] ),
t 2 R :
(21)
We now apply Proposition 2 (or the Kabanov formula) to the product I 1 (1[0, t] )I 2 (12
[0, t] )
with a ö given by (21), and we obtain
3
I 1 (1[0, t] )I 2 (12
[0, t] ) I 3 (1[0, t] ) 2tI 1 (1[0, t] )
2c0 I 2 (12
[0, t] ) = (2c1 1[0, t] (x1 )I 1 (1[0,x1 ] )I 1 (1[0, t] )):
Hence
3
2
I 1 (1[0, t] )I 2 (12
[0, t] ) aI 3 (1[0, t] ) bI 2 (1[0, t] )
t
cI 1 t1[0, t] (x1 ) c1 1[0,x1 ] (u) du1[0, t] (x1 ) ,
(22)
0
where a, b, c 2 R.
As
I 1 (1[0, t] )I 2 (12
[0, t] ) X t Q2 (X t , t),
the product is a polynomial of degree 3 in X t , and can be expressed as a linear combination
of Q0 , Q1 , Q2 and Q3 . So
I1 (1[0, t] )I 2 (12
[0, t] ) b3 (t)Q3 (X t , t) b0 (t)
(23)
1
1
b3 (t)I 3 (13
b2 (t)I 2 (12
[0, t] )
[0, t] ) b1 (t)I 1 (1[0, t] ) b0 (t),
3!
2!
and as before identifying the kernels of (22) and (23), and focusing our attention to the chaos
of order 1, we have
t
a1 (t)1[0, t] (x1 ) c(t1[0, t] (x1 ) c1 1[0,x1 ] (u) du1[0, t] (x1 )
0
c(t1[0, t] (x1 ) c1 x1 1[0, t] (x1 )),
therefore c1 has to be zero, and we get that the process ö is a constant.
h
N. Privault, J.L. Sole and J. Vives
648
Remark 3. Since for the standard AzeÂma martingale we have ö s ÿI 1 (1[0,s] ), the theorem
implies that the AzeÂma martingale does not have an associated family of polynomials.
Remark 4. In particular we have proved that any normal martingale in L4 with an associated
family of polynomials is a process with independent increments. In fact, a normal martingale
in L4 has independent increments if and only if ö is deterministic (Utzet 1992; Emery 1989).
Remark 5. Yor (1997, Chapter 15) introduces a family of polynomials related to the AzeÂma
martingales in a different context. These polynomials give the conditional expectation of
powers for AzeÂma martingales with respect to the ó-algebras of the strict past.
5.2. Projection property for the Wick product
Let X be a normal martingale in L4 (Ù). The Wick product I 1 ( f 1 ) : I n ( g n ) of I 1 ( f 1 ) and
I n ( g n ) is de®ned by
I 1 ( f 1 ) : I n ( gn ) I n1 ( f 1 gn ):
This motivates the following de®nition.
De®nition 4. We say that X has the Wick projection property if the Wick product
I 1 ( f 1 ) : I n ( gn ) is the projection of the product I 1 ( f 1 )I n ( g n ) over the chaos of order n 1.
Here the product I 1 ( f 1 )I n ( gn ) is assumed to belong to L2 (Ù) and the kernels f 1 and gn are
bounded and with compact support.
Now the natural problem is to determine the martingales that have this property.
Theorem 3. Assume that ö has a chaotic decomposition with no terms in chaos greater than
1. Then X has the Wick projection property if and only if ö is deterministic.
Proof. Proposition 2 says that
I 1 ( f 1 )I n ( gn ) I n1 ( f 1 g n ) nI nÿ1
1
0
f 1 (s) gn (:, s) ds
= ( f 1 (u)ö(u)nI nÿ1 ( g n (:, u))),
and it is straightforward to see that the last addend has null projection over the chaos of order
h
greater than n for all f 1 and gn if and only if ö is deterministic.
Remark 6. If in the de®nition of the Wick projection property we impose I 1 ( f 1 ) : I n ( g n ) to
be the projection of the product I 1 ( f 1 )I n ( g n ) over the sum of chaos of order strictly greater
than n, instead of over thePclass of order n, then the above result can be extended to normal
martingales X with öt i I i ( f i (:, t)) for which Proposition 2 is valid.
Chaotic Kabanov formula for the AzeÂma martingales
649
6. Derivation rule of products
It is well known that on the Wiener space, that is for ö 0, =ÿ is identi®ed with a derivation
operator. As noticed by Ma et al. (1998), =ÿ cannot act in the same way as a Sobolev
derivative, =ÿ f (M t ) 1[0, t] f 9(M t ), unless ö 0. In this section we study the product rule
for =ÿ and, in particular, we further show that =ÿ can be a derivation operator only for
ö 0.
For deterministic ö, Privault (1996) noticed that the product formula becomes
ÿ
ÿ
ÿ
ÿ
=ÿ
s (FG) F= s G G= s F ö s = s F= s G,
s 2 R :
This rule does not extend to random ö, but (8) gives by duality the following information.
Proposition 5. For F, G 2 S ,
ÿ
ÿ
ÿ
ÿ
E[=ÿ
s (FG)jF s ] E[F= s G G= s F ö s = s F= s GjF s ], ds-a:e:
Proof. We write (8) for u 2 U adapted, and apply the duality between =ÿ and = :
E[(u, =ÿ (FG))]
E[= (u)FG]
E[G(= (uF) (u, =ÿ F) L2 (R ) = (uö=ÿ F)]
E[(u, F=ÿ G) L2 (R ) (u, G=ÿ F) L2 (R ) (u, ö=ÿ F=ÿ G) L2 (R ) ]:
h
From the above it follows that if =ÿ is a derivation then E[ö s =ÿs F=ÿs G] 0, s 2 R ,
F, G 2 S , hence ö 0.
As a consequence of Proposition 5, we have
ÿ
ÿ
ÿ
ÿ
=ÿ
s (FG) F= s G G= s F ö s = s F= s G As (F, G),
s 2 R ,
where A(F, G) is a process with zero adapted projection.
Our aim in the following is to gain more information on this process in the case of the
AzeÂma martingales. We use the notation As (F, t) As (F, I 1 (â t )), t 2 R .
Proposition 6. Assume that ö is given as öt á t I 1 (â t ), t 2 R , where á is locally
bounded and â t is bounded, t 2 R . Then
ÿ
ÿ
ÿ
=ÿs (I 1 (â t )F) F=ÿ
s I 1 (â t ) I 1 (â t )= s F ö s = s I 1 (â t )= s G As (F, t),
s 2 R ,
where E[As (F, t)jF s ] 0, s 2 R , and As (F, t) is given by the relation
ÿ ÿ
ÿ
:
:
As (F, t) = (â t (:)â: (s)=ÿ
: F ö s â t ( )â: (s)=: = s F â t As (=: F, )):
Proof. We use the relation =ÿs = (u) = (=ÿ
s u) us :
(24)
N. Privault, J.L. Sole and J. Vives
650
=ÿs (I 1 (â t )F)
ÿ
ÿ
=ÿ
s (= (â t F) (â t , = F) L2 (R ) = (â t ö= F))
= (â t =ÿs F) â t (s)F (â t , =ÿ =ÿs F) L2 (R ) = (â t á=ÿ =ÿs F)
ÿ
ÿ
: ÿ
â t (s)á s =ÿ
s F = (â t ( )= s (I 1 (â: )=: F)) â t (s)I 1 (â s )= s F
= (â t =ÿs F) (â t , =ÿ =ÿs F) L2 (R ) = (â t á=ÿ =ÿs F)
ÿ
ÿ
F=ÿ
s I 1 (â t ) â t (s)á s = s F â t (s)I 1 (â s )= s F
ÿ ÿ
ÿ
ÿ ÿ
ÿ
:
= (â t (:)(â: (s)=ÿ
: F I 1 (â: )=: = s F ö s =2 I 1 (â: )= s =: F A(=: F, )))
= (â t =ÿs F) (â t , =ÿ =ÿs F) L2 (R ) = (â t ö=ÿ =ÿs F)
ÿ
ÿ
F=ÿ
s I 1 (â t ) â t (s)á s = s F â t (s)I 1 (â s )= s F
ÿ
ÿ ÿ
ÿ
:
:
:
= (â t (:)â: (s)=ÿ
: F â t ( )ö s = s I 1 (â: )= s =: F â t ( )A s (=: F, ))
ÿ
ÿ
ÿ
I 1 (â t )=ÿ
s F F= s I 1 (â t ) ö s = s I 1 (â t )= s F
ÿ ÿ
ÿ
:
:
:
= (â t (:)â: (s)=ÿ
: F â t ( )ö s â: (s)=: = s F â t ( )As (=: F, )),
and we obtain (24).
h
As a consequence of this proposition, for F I n ( f n ) the remaining process A(F, t)
can be explicitly determined by induction from
As (I n ( f n ), t) nI n ((â t (:)â: (s) f : ) f ( nÿ1) ) n(n ÿ 1) f s = (ö s â t (:)â: (s) f : I nÿ2 ( f ( nÿ2) ))
n= ( f : â t (:)As (I nÿ1 ( f ( nÿ1) )), t):
Note that F 7! A(F, t) is linear and that the chaotic expansion of ö s I nÿ2 ( f ( nÿ2) ) can
be explicitly computed from the Kabanov formula. As an application we compute
As (I n ( f n ), t) for n 0, 1, 2. We have As (1, t) 0, As (I 1 ( f 1 ), t) I 1 (â t (:)â: (s) f : ), and
As (I 2 ( f 2 ), t) I 2 (â t (:)â: (s) f : f ) 2 f s = ((á s I 1 (â s ))â t (:)â: (s) f : )
2= ( f : â t (:)I 1 (â: ()â (s) f ))
I 2 (â t (:)â: (s) f : f ) 2 f s á s I 1 (â t (:)â: (s) f : ) 2 f s I 2 (â t (:)â: (s) f : â s )
2I 2 ( f : â t (:) â: ()â (s) f ):
Chaotic Kabanov formula for the AzeÂma martingales
651
Acknowledgements
We would like to thank Professor Frederic Utzet for fruitful discussions that motivated the
problem of the existence of an associated family of polynomials related to a normal
martingale. The ®rst named author is grateful for the hospitality of the Centre de Recerca
MatemaÁtica, Barcelona, where this work was completed.
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Theory. EÂcole d'EÂte de ProbabiliteÂs de Saint-Flour XXIII ± 1993, Lecture Notes in Math. 1608.
Berlin: Springer-Verlag.
Dellacherie, C., Maisonneuve, B. and Meyer, P.A. (1992) ProbabiliteÂs et Potentiel. Processus de
Markov (®n). Complements de Calcul Stochastique. Paris: Hermann.
Dermoune, A., KreÂe, P. and Wu, L. (1988) Calcul stochastique non adapte par rapport aÁ la mesure de
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Notes in Math. 1321. Berlin: Springer-Verlag.
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de ProbabiliteÂs XXIII, Lecture Notes in Math. 1372, pp. 66±87. Berlin: Springer-Verlag.
Kabanov, Y.M. (1975) On extended stochastic integrals. Theory Probab. Appl., 20, 710±722.
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Meyer, P.A. (1976) Un cours sur les inteÂgrales stochastiques. In P.A. Meyer (ed.), SeÂminaire de
ProbabiliteÂs X, Lecture Notes in Math. 511. Berlin: Springer-Verlag.
Nualart, D. and Vives, J. (1990) Anticipative calculus for the Poisson process based on the Fock space.
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Notes in Math. 1426, pp. 154±165. Berlin: Springer-Verlag.
Parthasarathy, K.R. (1990) AzeÂma martingales and quantum stochastic calculus. In R.R. Bahadur (ed.)
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Privault, N. (1996) On the independence of multiple stochastic integrals with respect to a class of
martingales. C. R. Acad. Sci. Paris, SeÂr. I, 323, 515±520.
Russo, F. and Vallois, P. (1998) Product of two multiple stochastic integrals with respect to a normal
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Surgailis, D. (1984) On multiple Poisson stochastic integrals and associated Markov semi-groups.
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``` | 18,363 | 46,306 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2021-21 | latest | en | 0.742483 |
http://www.jiskha.com/display.cgi?id=1290273883 | 1,496,115,770,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463613780.89/warc/CC-MAIN-20170530031818-20170530051818-00277.warc.gz | 669,901,389 | 4,061 | # physics
posted by on .
A cup of coffee (with lid) is enclosed in an insulated cup 0.5 cm thick in the shape of a cube 9.7 cm on a side. The thermal conductivity of the cup is 0.0002 cal/s x cm x C. The temperature of the coffee is 71 degrees Celsius and the temperature of the surroundings is 25 degrees Celsius. Find the heat loss due to conduction in J/s.
I did it like this:
[0.0002 (6x9.7^2)(71-25)]/0.5 = 10.388
but it was wrong.
• physics - ,
Where did you convert calories to Joules?
• physics - ,
it's true but ur answer is in cal you need to multiply it by 4.18 to convert it to joul
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A340240 Odd composite integers m such that A004254(3*m-J(m,21)) == 5*J(m,21) (mod m) and gcd(m,21)=1, where J(m,21) is the Jacobi symbol. 2
55, 407, 527, 529, 551, 559, 965, 1199, 1265, 1633, 1807, 1919, 1961, 3401, 3959, 4033, 4381, 5461, 5777, 5977, 5983, 6049, 6233, 6439, 6479, 7141, 7195, 7645, 7999, 8639, 8695, 8993, 9265, 9361, 11663, 11989, 12209, 12265, 13019, 13021, 13199, 14023, 14465, 14491 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 COMMENTS The generalized Lucas sequences of integer parameters (a,b) defined by U(m+2)=a*U(m+1)-b*U(m) and U(0)=0, U(1)=1, satisfy U(3*p-J(p,D)) == a*J(p,D) (mod p) whenever p is prime, k is a positive integer, b=1 and D=a^2-4. The composite integers m with the property U(k*m-J(m,D)) == U(k-1)*J(m,D) (mod m) are called generalized Lucas pseudoprimes of level k+ and parameter a. Here b=1, a=5, D=21 and k=3, while U(m) is A004254(m). REFERENCES D. Andrica, O. Bagdasar, Recurrent Sequences: Key Results, Applications and Problems. Springer, 2020. D. Andrica, O. Bagdasar, On some new arithmetic properties of the generalized Lucas sequences, Mediterr. J. Math. (to appear, 2021). D. Andrica, O. Bagdasar, On generalized pseudoprimality of level k (submitted). LINKS Table of n, a(n) for n=1..44. Dorin Andrica, Vlad Crişan, and Fawzi Al-Thukair, On Fibonacci and Lucas sequences modulo a prime and primality testing, Arab Journal of Mathematical Sciences, 2018, 24(1), 9--15. MATHEMATICA Select[Range[3, 15000, 2], CoprimeQ[#, 21] && CompositeQ[#] && Divisible[ ChebyshevU[3*# - JacobiSymbol[#, 21] - 1, 5/2] - 5*JacobiSymbol[#, 21], #] &] CROSSREFS Cf. A004254, A071904, A340098 (a=5, b=1, k=1), A340123 (a=5, b=1, k=2). Cf. A340239 (a=3, b=1, k=3), A340241 (a=7, b=1, k=3). Sequence in context: A063653 A222348 A075740 * A355511 A129217 A116060 Adjacent sequences: A340237 A340238 A340239 * A340241 A340242 A340243 KEYWORD nonn AUTHOR Ovidiu Bagdasar, Jan 01 2021 STATUS approved
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Last modified September 12 15:43 EDT 2024. Contains 375853 sequences. (Running on oeis4.) | 903 | 2,444 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-38 | latest | en | 0.513131 |
https://answers.yahoo.com/question/index?qid=20130415154641AA07maJ | 1,603,743,753,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107891624.95/warc/CC-MAIN-20201026175019-20201026205019-00194.warc.gz | 202,956,815 | 30,098 | Allie asked in Science & MathematicsMathematics · 8 years ago
# Choose a point not on the boundary line to test in the inequality.?
okay so i have the equation y≥3x and i'm confused on what to do with it. i put it ona graph and it made a straight line
Choose a point not on the boundary line to
test in the inequality. Does the point make
the inequality you wrote in Step 3 true or false?
If the point makes the inequality true, shade the region containing the point. If the point
does not make the inequality true, shade the region that does not contain the point. | 133 | 572 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2020-45 | latest | en | 0.908075 |
http://nrich.maths.org/public/leg.php?code=-36&cl=3&cldcmpid=5021 | 1,503,455,474,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886117519.82/warc/CC-MAIN-20170823020201-20170823040201-00134.warc.gz | 297,987,034 | 9,151 | # Search by Topic
#### Resources tagged with Combinatorics similar to Royal Arm Wrestling:
Filter by: Content type:
Stage:
Challenge level:
### There are 40 results
Broad Topics > Decision Mathematics and Combinatorics > Combinatorics
### Flagging
##### Stage: 3 Challenge Level:
How many tricolour flags are possible with 5 available colours such that two adjacent stripes must NOT be the same colour. What about 256 colours?
### Domino Tetrads
##### Stage: 3 Challenge Level:
Is it possible to use all 28 dominoes arranging them in squares of four? What patterns can you see in the solution(s)?
### Bell Ringing
##### Stage: 3 Challenge Level:
Suppose you are a bellringer. Can you find the changes so that, starting and ending with a round, all the 24 possible permutations are rung once each and only once?
### Small Change
##### Stage: 3 Challenge Level:
In how many ways can a pound (value 100 pence) be changed into some combination of 1, 2, 5, 10, 20 and 50 pence coins?
### An Investigation Based on Score
##### Stage: 3
Class 2YP from Madras College was inspired by the problem in NRICH to work out in how many ways the number 1999 could be expressed as the sum of 3 odd numbers, and this is their solution.
### Snowman
##### Stage: 4 Challenge Level:
All the words in the Snowman language consist of exactly seven letters formed from the letters {s, no, wm, an). How many words are there in the Snowman language?
### Euromaths
##### Stage: 3 Challenge Level:
How many ways can you write the word EUROMATHS by starting at the top left hand corner and taking the next letter by stepping one step down or one step to the right in a 5x5 array?
### Ways of Summing Odd Numbers
##### Stage: 3
Sanjay Joshi, age 17, The Perse Boys School, Cambridge followed up the Madrass College class 2YP article with more thoughts on the problem of the number of ways of expressing an integer as the sum. . . .
### Painting Cubes
##### Stage: 3 Challenge Level:
Imagine you have six different colours of paint. You paint a cube using a different colour for each of the six faces. How many different cubes can be painted using the same set of six colours?
### Magic Caterpillars
##### Stage: 4 and 5 Challenge Level:
Label the joints and legs of these graph theory caterpillars so that the vertex sums are all equal.
### Links and Knots
##### Stage: 4 and 5
Some puzzles requiring no knowledge of knot theory, just a careful inspection of the patterns. A glimpse of the classification of knots, prime knots, crossing numbers and knot arithmetic.
### Plum Tree
##### Stage: 4 and 5 Challenge Level:
Label this plum tree graph to make it totally magic!
### Shuffle Shriek
##### Stage: 3 Challenge Level:
Can you find all the 4-ball shuffles?
### Lost in Space
##### Stage: 4 Challenge Level:
How many ways are there to count 1 - 2 - 3 in the array of triangular numbers? What happens with larger arrays? Can you predict for any size array?
### Deep Roots
##### Stage: 4 Challenge Level:
Find integer solutions to: $\sqrt{a+b\sqrt{x}} + \sqrt{c+d.\sqrt{x}}=1$
### Magic W
##### Stage: 4 Challenge Level:
Find all the ways of placing the numbers 1 to 9 on a W shape, with 3 numbers on each leg, so that each set of 3 numbers has the same total.
### Master Minding
##### Stage: 3 Challenge Level:
Your partner chooses two beads and places them side by side behind a screen. What is the minimum number of guesses you would need to be sure of guessing the two beads and their positions?
### Cube Paths
##### Stage: 3 Challenge Level:
Given a 2 by 2 by 2 skeletal cube with one route `down' the cube. How many routes are there from A to B?
### Tangles
##### Stage: 3 and 4
A personal investigation of Conway's Rational Tangles. What were the interesting questions that needed to be asked, and where did they lead?
### Permute It
##### Stage: 3 Challenge Level:
Take the numbers 1, 2, 3, 4 and 5 and imagine them written down in every possible order to give 5 digit numbers. Find the sum of the resulting numbers.
### Tri-colour
##### Stage: 3 Challenge Level:
Six points are arranged in space so that no three are collinear. How many line segments can be formed by joining the points in pairs?
### Olympic Magic
##### Stage: 4 Challenge Level:
in how many ways can you place the numbers 1, 2, 3 … 9 in the nine regions of the Olympic Emblem (5 overlapping circles) so that the amount in each ring is the same?
### Walkabout
##### Stage: 4 Challenge Level:
A walk is made up of diagonal steps from left to right, starting at the origin and ending on the x-axis. How many paths are there for 4 steps, for 6 steps, for 8 steps?
### How Many Dice?
##### Stage: 3 Challenge Level:
A standard die has the numbers 1, 2 and 3 are opposite 6, 5 and 4 respectively so that opposite faces add to 7? If you make standard dice by writing 1, 2, 3, 4, 5, 6 on blank cubes you will find. . . .
### Euler's Officers
##### Stage: 4 Challenge Level:
How many different ways can you arrange the officers in a square?
### Postage
##### Stage: 4 Challenge Level:
The country Sixtania prints postage stamps with only three values 6 lucres, 10 lucres and 15 lucres (where the currency is in lucres).Which values cannot be made up with combinations of these postage. . . .
### Penta Colour
##### Stage: 4 Challenge Level:
In how many different ways can I colour the five edges of a pentagon red, blue and green so that no two adjacent edges are the same colour?
### Russian Cubes
##### Stage: 4 Challenge Level:
I want some cubes painted with three blue faces and three red faces. How many different cubes can be painted like that?
### Symmetric Tangles
##### Stage: 4
The tangles created by the twists and turns of the Conway rope trick are surprisingly symmetrical. Here's why!
### Counting Binary Ops
##### Stage: 4 Challenge Level:
How many ways can the terms in an ordered list be combined by repeating a single binary operation. Show that for 4 terms there are 5 cases and find the number of cases for 5 terms and 6 terms.
### Doodles
##### Stage: 4 Challenge Level:
Draw a 'doodle' - a closed intersecting curve drawn without taking pencil from paper. What can you prove about the intersections?
### Knight Defeated
##### Stage: 4 Challenge Level:
The knight's move on a chess board is 2 steps in one direction and one step in the other direction. Prove that a knight cannot visit every square on the board once and only (a tour) on a 2 by n board. . . .
### Scratch Cards
##### Stage: 4 Challenge Level:
To win on a scratch card you have to uncover three numbers that add up to more than fifteen. What is the probability of winning a prize?
### Ordered Sums
##### Stage: 4 Challenge Level:
Let a(n) be the number of ways of expressing the integer n as an ordered sum of 1's and 2's. Let b(n) be the number of ways of expressing n as an ordered sum of integers greater than 1. (i) Calculate. . . .
### Greetings
##### Stage: 3 Challenge Level:
From a group of any 4 students in a class of 30, each has exchanged Christmas cards with the other three. Show that some students have exchanged cards with all the other students in the class. How. . . .
### N000ughty Thoughts
##### Stage: 4 Challenge Level:
How many noughts are at the end of these giant numbers?
### One Basket or Group Photo
##### Stage: 2, 3, 4 and 5 Challenge Level:
Libby Jared helped to set up NRICH and this is one of her favourite problems. It's a problem suitable for a wide age range and best tackled practically.
### Molecular Sequencer
##### Stage: 4 and 5 Challenge Level:
Investigate the molecular masses in this sequence of molecules and deduce which molecule has been analysed in the mass spectrometer.
### Paving Paths
##### Stage: 3 Challenge Level:
How many different ways can I lay 10 paving slabs, each 2 foot by 1 foot, to make a path 2 foot wide and 10 foot long from my back door into my garden, without cutting any of the paving slabs?
### In a Box
##### Stage: 4 Challenge Level:
Chris and Jo put two red and four blue ribbons in a box. They each pick a ribbon from the box without looking. Jo wins if the two ribbons are the same colour. Is the game fair? | 2,007 | 8,232 | {"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} | 3.859375 | 4 | CC-MAIN-2017-34 | latest | en | 0.880537 |
https://goprep.co/ex-4-q19-two-years-later-a-father-will-be-8-yrs-more-than-i-1nkxa8 | 1,621,068,081,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991378.48/warc/CC-MAIN-20210515070344-20210515100344-00591.warc.gz | 326,080,699 | 30,229 | # Two years later a
Let father’s present age be x and son’s present age be y.
After two years
Son’s age = y + 2
Father’s age = x + 2
According to question
x + 2 = 8 + 3 (y + 2)
x + 2 = 8 + 3y + 6
x + 2 = 14 + 3y
x – 3y = 12
Equation is x – 3y = 12
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A three-wheeler sRD Sharma - Mathematics | 366 | 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} | 2.65625 | 3 | CC-MAIN-2021-21 | latest | en | 0.812574 |
https://wolf.co.com/PROGRAMS/VALVES-SYSTEMS/REGULATION-SYSTEMS/ | 1,580,025,445,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251687958.71/warc/CC-MAIN-20200126074227-20200126104227-00245.warc.gz | 722,918,488 | 9,708 | WOLF ENERGY SA
/
Engineering & Technology
# REGULATION SYSTEMS
## MAIN FEATURES
• Selection of an unlimited number of systems and valves for each analysis
• Description of each system through data provided from surveys (e.g. gross head, pipe length, pipe diameter, pipe roughness, range of operation, etc.)
• Description of each valve through data supplied from manufacturers (e.g. diameter, coefficient of contraction, coefficient of velocity, range of operation, etc.)
• Computation of characteristic surfaces (i.e. head and head loss versus valve opening and discharge for all system and valve combinations) according to hydraulic laws
• Identification of working conditions (i.e. balance between gross head and head losses due to friction in pipes and valves)
• Comparison of analytical results with available field data (e.g. head and head losses versus valve opening and discharge, etc.)
• Visualisation of results
• Suitable for designing and testing
## OPTIONS FOR IO OPERATIONS
• Microsoft Excel Spreadsheet Files (xlsx)
## OPTIONS FOR GRAPHICS
• Standard Image Formats (jpeg, tiff)
• Portable Document Format (pdf)
## TEST CASES
Figure 1, Figure 2, Figure 3 and Figure 4 illustrate the behaviour of the very same hydraulic system working with different regulation valves. Blue solid meshes represent the gross head of the system while red solid meshes show the friction losses due to both pipe and valve. Interception of the meshes denote actual working conditions.
Figure 1. Head (Z-Axis) versus valve opening (X-Axis) and discharge (Y-Axis). C_C = 0.20 and C_V = 1.00.
Figure 2. Head (Z-Axis) versus valve opening (X-Axis) and discharge (Y-Axis). C_C = 0.40 and C_V = 1.00.
Figure 3. Head (Z-Axis) versus valve opening (X-Axis) and discharge (Y-Axis). C_C = 0.60 and C_V = 1.00.
Figure 4. Head (Z-Axis) versus valve opening (X-Axis) and discharge (Y-Axis). C_C = 0.80 and C_V = 1.00.
## DELIVERABLES
User's Manual PDF | 474 | 1,946 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-05 | latest | en | 0.803022 |
beanalytics.wordpress.com | 1,660,067,806,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571056.58/warc/CC-MAIN-20220809155137-20220809185137-00028.warc.gz | 154,544,576 | 26,822 | # The right way to measure distances with the Kinect
Vangos Pterneas, Kinect Hero, shares with us another awesome Kinect development post…
## Measuring Distances using Kinect – The Right Way
Today I’m going to share a little trick that will save you a ton of time when working with Kinect: how to properly measure the distance between the sensor and a body or object. Surprisingly, I was asked the exact same question by three of my blog readers during the past 10 days!
First, some good-to-know background.
#### How Kinect measures distances
As you know, Kinect integrates an infrared sensor, along with a depth processor. The visible area is called “field of view”. The depth processor produces depth frames. Each depth frame is a grid of 512×424 points. The Kinect SDK is then feeding the depth frames to a powerful body-detection algorithm. The algorithm identifies 25 human body joints and calculates their coordinates in the 3D space.
Every single joint has 3 values: X, Y, and Z. It is projected in a Cartesian coordinate system. The (0, 0, 0) point is the position of the sensor. Every other point is measured in terms of the position of the sensor! Check the overhead graphic below. It’s viewing the sensor and the scene from above.
• X is the position in the horizontal axis.
• Y is the position in the vertical axis.
• Z is the position in the depth axis.
#### Making sense of the Z value
X and Y are fairly easy to understand. That’s not true for Z. If you take a close look at the graphic above, you’ll notice that the Z value is not the linear distance between the point and the sensor. Instead, it’s the distance between the point and the plane of the sensor! It’s like having a virtual wall right in front of the Kinect. The Z value is the vertical distance from that wall (drawn in green).
To access the Z value, use the code below:
#### Measuring the distance
How about the physical distance, though? The distance between the player and the device is represented by a mathematical vector (drawn in blue). Thankfully, we do not need to be gurus in Algebra to measure the length of a vector in the 3D space.
The length of a vector is given by the formula below:
Bringing it all together:
``var point = body.Joints[JointType.SpineBase].Position;var distance = Length(point);``
This is it! You now know exactly how far or how close someone is to the Kinect sensor! Of course, if someone is standing right in front of the sensor, the Z value and the distance would converge.
Hint: in case you need to measure distances of points that do not belong to the human body, you can use Coordinate Mapping.
#### Summary
Kinect provides us with the X, Y, and Z coordinates of the human body joints.
• To measure the distance between the player and the plane of the sensor, we just use the Z value as-is.
• To measure the distance between the player and the sensor, we calculate the length of the corresponding vector.
Project Information URL: http://pterneas.com/2016/08/11/measuring-distances-kinect/
Contact Information: | 681 | 3,049 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2022-33 | latest | en | 0.897856 |
https://socratic.org/questions/how-do-you-convert-7pi-3-to-degrees | 1,624,460,948,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488539480.67/warc/CC-MAIN-20210623134306-20210623164306-00582.warc.gz | 489,012,123 | 5,655 | # How do you convert 7pi/3 to degrees?
$\frac{7 \pi}{3} r a \mathrm{di} a n s = {420}^{o}$
$\pi$ radians $= {180}^{o}$$\textcolor{w h i t e}{\text{XXXX}}$(a straight line)
$\Rightarrow$$\textcolor{w h i t e}{\text{XXXX}}$$1$ radian $= \frac{{180}^{o}}{\pi}$
$\Rightarrow$$\textcolor{w h i t e}{\text{XXXX}}$$\frac{7 \pi}{3}$ radians $= \frac{{180}^{o}}{\cancel{\pi}} \cdot \frac{7 \cancel{\pi}}{3} = {420}^{o}$ | 176 | 411 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 12, "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} | 3.96875 | 4 | CC-MAIN-2021-25 | longest | en | 0.26668 |
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# Pixel based collision detection
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I''m developing a small pinball game. My problem is determining the angle the ball gets after a collision. I''m using pixel based collision detection because the small size of the game (the board is about 80x150 pixels and the active area of ball is just one pixel). Mainly i''m concerned of the ball''s behaviour when it hits an arc (or curve or something). Sometimes ball gets such an angle that on the next frame it will collide again and even get inside a wall. Anyway what i''m doing now is: 1. calculate angle of the ball''s velocity 2. reflect(?) it around wall''s normal vector 3. calculate new velocity vector from the new angle Any better ideas?
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I''ve had a lot of ''fun'' with this aspect myself, and it''s not collision detection that''s the problem. Collision detection is never hard, it''s collision reaction that''s annoying.
A few points to consider. Are you using varied time frames, like measuring how long it''s been since the last update and moving your ball accordingly, or are you moving a specific(small) amount each ''frame''. You can have the problem with both approaches, but it''s more likely with varied speeds, because on one frame, you might enter the wall by 10 units, but the next frame only lets you exit by 5 units and then you''re stuck.
What you need to do though is to never enter the wall. When you calculate your collision, you need to break the movement of the objects down so that as soon as they ''collide'', they react, and you move the ball in it''s new direction at the same time.
I''m not sure if that''s clear, but basically, when a ball is going towards a wall, you might be 2 units away in frame 1, and then move 5 units in the next frame, so what should happen is, you check collision with each unit, and on the second check, you get a collision, and bounce the ball back in its new direction 3 units.
G''luck,
-Alamar
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Last Updated: Jul 29th, 2020
OpenGL and 3D graphics in general is all based on vector and matrix math. Many institutions don't cover matrix math until after 3 levels of calculus which can be a huge obstacle for beginning graphics programmers. This is especially bad consdering much of the linear algebra you need to get a basic understanding of matrix transformations can be easily taught if you've taken high school level geometry.
While this article won't cover all the math you need to become a proficient graphics programmer, it will give you flyover of linear algebra concepts so you have a basic understanding of what geometry based rendering APIs do.
Once you have a strong enough background in matrix math, go ahead and jump into the OpenGL tutorial set.
## Vectors
Mathematical vectors (not to be confused with STL vectors) are a way to represent a value that has multiple components which can be used to represent something with a magnitude and a direction. The best way to understand what this means is by looking at an example.
Here we have a point x = 2, y = 3. How do would you represent this as a vector? Like this:
$\left[\begin{array}{c}2\\ 3\end{array}\right]$
That's it really. A vector is really just a way to represent a value that has multiple components such as in this case where the position has an x and y component.
## Polygons
Now how is this useful? See, in OpenGL (and most other 3D graphics APIs) pretty much everything this is rendered as a set of polygons. This begs the question "How do you represent a polygon in a computer?". Let's take this triangle:
It can be represented as an array of points:
Which can be represented as an array of vectors:
$\left[\begin{array}{c}1\\ 1\end{array}\right]$ $\left[\begin{array}{c}3\\ 3\end{array}\right]$ $\left[\begin{array}{c}5\\ 1\end{array}\right]$
Which can be used by OpenGL to create a polygon to render. This is important because in 3D graphics polygons are used to represent 3D objects:
A 3D model for a simple uncolored cube can just be a big array of numbers.
const float cubeModel = { //Front -1.f, -1.f, 1.f, 1.f, -1.f, 1.f, 1.f, 1.f, 1.f, -1.f, 1.f, 1.f, //Back -1.f, -1.f, -1.f, 1.f, -1.f, -1.f, 1.f, 1.f, -1.f, -1.f, 1.f, -1.f, //Top -1.f, 1.f, 1.f, 1.f, 1.f, 1.f, 1.f, 1.f, -1.f, -1.f, 1.f, -1.f, //Bottom -1.f, -1.f, 1.f, 1.f, -1.f, 1.f, 1.f, -1.f, -1.f, -1.f, -1.f, -1.f, //Left -1.f, -1.f, -1.f, -1.f, 1.f, -1.f, -1.f, 1.f, 1.f, -1.f, -1.f, 1.f, //Right 1.f, -1.f, -1.f, 1.f, 1.f, -1.f, 1.f, 1.f, 1.f, 1.f, -1.f, 1.f };
The code you see above has been used to render a cube for a simple demo. Each triplet of numbers represents a vector point and every set of 4 vectors represents the 4 corners of a side of a cube.
## 3D Vectors
While you know what x/y coordinates in 2D space are, you may be wondering how 3D space works. In the code for the cube model above, the third number of every position triplet is the z position. Where the X position goes left/right and the y position goes up/down, the z position goes forward/back.
Another important thing to note is which way is positive for the x, y, and z directions. For x, right is positive and left is negative. Up is positive and down is negative for the y axis. Finally for the z axis if you were staring at point $\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]$, towards you is postive z and away from you is negative z. This is the most commonly used convention called the Right Hand Rule.
## Vector Arithmetic
3D graphics is all about taking vectors from models and performing mathmatical operations on them to get them to render to the screen. It is possible add/subtract vectors. All you have to do is add/subtract them component by component. For example:
$\left[\begin{array}{c}5\\ 8\\ 3\end{array}\right]$ + $\left[\begin{array}{c}7\\ 1\\ 2\end{array}\right]$ = $\left[\begin{array}{c}\mathrm{12}\\ 9\\ 5\end{array}\right]$
All we did here was take the first component from the first vector (5), then take the first component from the second vector (7), and add them together to get the first component from the result vector (5 + 7 = 12). For the second component we take the second component from the first vector (8), then take the second component from the second vector (1), and add them together to get the second component from the result vector (8 + 1 = 9). Finally for the third component, we take the third component from the first vector (3), then take the third component from the second vector (2), and add them together to get the third component from the result vector (3 + 2 = 5).
For vector subtraction, all we do is component by component subtraction: $\left[\begin{array}{c}5\\ 8\\ 3\end{array}\right]$ - $\left[\begin{array}{c}7\\ 1\\ 2\end{array}\right]$ = $\left[\begin{array}{c}\mathrm{-2}\\ 7\\ 1\end{array}\right]$
Multiplication is a bit more tricky as there's more than one way to do it. The first way is called scalar multiplication where you multiply the entire vector by a single value. Say we wanted to double the length of a vector (or scale it to double the length), you would multiply the vector by two:
$2\left[\begin{array}{c}5\\ 8\\ 3\end{array}\right]$ = $\left[\begin{array}{c}\mathrm{2 * 5}\\ \mathrm{2 * 8}\\ \mathrm{2 * 3}\end{array}\right]$ = $\left[\begin{array}{c}\mathrm{10}\\ \mathrm{16}\\ 6\end{array}\right]$
As you can see all we did was multiply each component in the vector by the scalar value.
We can also multiply two vectors together in multiple ways. The first way is by the dot product where you multiply the two vectors together component by component and then add the components together:
$\left[\begin{array}{c}5\\ 8\\ 3\end{array}\right]$ . $\left[\begin{array}{c}7\\ 1\\ 2\end{array}\right]$ = 5*7 + 8*1 + 3*2 = 35 + 8 + 6 = 49
They also call the dot product the scalar product because it returns a single component value AKA a scalar. This calculation is useful in lighting calculations, collision detection, and other game programming problems.
Another way to multiply two vectors is the cross product. Assuming the first vector is A and the second vector is B then the cross product is:
$\left[\begin{array}{c}\mathrm{Ay*Bz - Az*By}\\ \mathrm{Az*Bx - Ax*Bz}\\ \mathrm{Ax*By - Ay*Bx}\end{array}\right]$
Here's a super simple example:
$\left[\begin{array}{c}1\\ 0\\ 0\end{array}\right]$ x $\left[\begin{array}{c}0\\ 1\\ 0\end{array}\right]$ = $\left[\begin{array}{c}\mathrm{0*0 - 0*1}\\ \mathrm{0*0 - 1*0}\\ \mathrm{1*1 - 0*0}\end{array}\right]$ = $\left[\begin{array}{c}0\\ 0\\ 1\end{array}\right]$
The cross product gives you a vector perpendicular to both the vectors you multiplied. It's also important to note that AxB is not equal to BxA. BxA gives a vector in the opposite direction of AxB.
I wish we could spend more time covering how the dot and cross product are used, but this is just a quick fly over of the concepts. What we really want to get to is matrix vector multiplication.
## Matrix/Vector Multiplication
At the center of graphics programming is the graphics pipeline. A pipeline is a machanism designed to process a stream of data and OpenGL pipelines are based around multiplying a stream of vectors (like the vector positions in a 3D model) against a matrix or set of matrices.
Here's a 4 row by 4 column matrix:
$\left[\begin{array}{cccc}1& 5& 8& 4\\ 2& 6& 7& 3\\ 3& 7& 6& 2\\ 4& 8& 5& 1\end{array}\right]$
Now say we wanted to multiply this matrix against this vector:
$\left[\begin{array}{cccc}1& 5& 8& 4\\ 2& 6& 7& 3\\ 3& 7& 6& 2\\ 4& 8& 5& 1\end{array}\right]$ x $\left[\begin{array}{c}2\\ 3\\ 4\\ 5\end{array}\right]$
What we would do is split up the matrix into 4 vectors:
$\left[\begin{array}{c}1\\ 2\\ 3\\ 4\end{array}\right]$ $\left[\begin{array}{c}5\\ 6\\ 7\\ 8\end{array}\right]$ $\left[\begin{array}{c}8\\ 7\\ 6\\ 5\end{array}\right]$ $\left[\begin{array}{c}4\\ 3\\ 2\\ 1\end{array}\right]$
Take each of the components from the vector and scalar multiply them against the split matrix:
$2\left[\begin{array}{c}1\\ 2\\ 3\\ 4\end{array}\right]$ $3\left[\begin{array}{c}5\\ 6\\ 7\\ 8\end{array}\right]$ $4\left[\begin{array}{c}8\\ 7\\ 6\\ 5\end{array}\right]$ $5\left[\begin{array}{c}4\\ 3\\ 2\\ 1\end{array}\right]$
=
$\left[\begin{array}{c}2\\ 4\\ 6\\ 8\end{array}\right]$ $\left[\begin{array}{c}\mathrm{15}\\ \mathrm{18}\\ \mathrm{21}\\ \mathrm{24}\end{array}\right]$ $\left[\begin{array}{c}\mathrm{32}\\ \mathrm{28}\\ \mathrm{24}\\ \mathrm{20}\end{array}\right]$ $\left[\begin{array}{c}\mathrm{20}\\ \mathrm{15}\\ \mathrm{10}\\ 5\end{array}\right]$
And then add them across to get the result vector:
$\left[\begin{array}{c}\mathrm{2 + 15 + 32 + 20}\\ \mathrm{4 + 18 + 28 + 15}\\ \mathrm{6 + 21 + 24 + 10}\\ \mathrm{8 + 24 + 20 + 5}\end{array}\right]$ = $\left[\begin{array}{c}\mathrm{69}\\ \mathrm{65}\\ \mathrm{61}\\ \mathrm{57}\end{array}\right]$
## Transformation Matrices
Now how is this all useful? Recall that 3D models can be represented as an array of vectors. Using a matrix you can multiply the vectors to transform the 3D model. Say we have these 4 vectors used to represent a square:
$\left[\begin{array}{c}\mathrm{-1}\\ 1\\ 0\end{array}\right]$ $\left[\begin{array}{c}1\\ 1\\ 0\end{array}\right]$ $\left[\begin{array}{c}1\\ \mathrm{-1}\\ 0\end{array}\right]$ $\left[\begin{array}{c}\mathrm{-1}\\ \mathrm{-1}\\ 0\end{array}\right]$
Say we then multiply these vectors against this matrix:
$\left[\begin{array}{ccc}2& 0& 0\\ 0& 2& 0\\ 0& 0& 2\end{array}\right]$
This means we get
$\left[\begin{array}{ccc}2& 0& 0\\ 0& 2& 0\\ 0& 0& 2\end{array}\right]$ $\left[\begin{array}{c}\mathrm{-1}\\ 1\\ 0\end{array}\right]$ = $\left[\begin{array}{c}\mathrm{-2}\\ 2\\ 0\end{array}\right]$
$\left[\begin{array}{ccc}2& 0& 0\\ 0& 2& 0\\ 0& 0& 2\end{array}\right]$ $\left[\begin{array}{c}1\\ 1\\ 0\end{array}\right]$ = $\left[\begin{array}{c}2\\ 2\\ 0\end{array}\right]$
$\left[\begin{array}{ccc}2& 0& 0\\ 0& 2& 0\\ 0& 0& 2\end{array}\right]$ $\left[\begin{array}{c}1\\ \mathrm{-1}\\ 0\end{array}\right]$ = $\left[\begin{array}{c}2\\ \mathrm{-2}\\ 0\end{array}\right]$
$\left[\begin{array}{ccc}2& 0& 0\\ 0& 2& 0\\ 0& 0& 2\end{array}\right]$ $\left[\begin{array}{c}\mathrm{-1}\\ \mathrm{-1}\\ 0\end{array}\right]$ = $\left[\begin{array}{c}\mathrm{-2}\\ \mathrm{-2}\\ 0\end{array}\right]$
As you probably noticed, multiplying the vectors of the square against this matrix scaled it to double the size. This is why this matrix is known as the Scaling Matrix. If you look in the OpenGL documentation, you'll find out that the function glScale() uses a 4 row, 4 column version of this matrix.
And this is the key point of this article: OpenGL uses matrices to transform the vectors of 3D objects to render them to the screen. Now let's look at another matrix:
$\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$
Watch what happens when we multiply this variable vector against it:
$\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$ $\left[\begin{array}{c}x\\ y\\ z\end{array}\right]$ = $\left[\begin{array}{c}x\\ y\\ z\end{array}\right]$
Whatever you multiply against this matrix, you get right back. When you have a square matrix with all zeros except for 1 going against the diagonal, this is called the Identity Matrix. It's used in graphics programming whenever you want to reset the transformation matrix.
Say if you wanted to shift a vector 5 to the right, 2 up, and 3 forward. There's a matrix for that:
$\left[\begin{array}{cccc}1& 0& 0& 5\\ 0& 1& 0& 2\\ 0& 0& 1& \mathrm{-3}\\ 0& 0& 0& 1\end{array}\right]$
Multiplying this through we get:
$\left[\begin{array}{cccc}1& 0& 0& 5\\ 0& 1& 0& 2\\ 0& 0& 1& \mathrm{-3}\\ 0& 0& 0& 1\end{array}\right]$ $\left[\begin{array}{c}x\\ y\\ z\\ 1\end{array}\right]$ = $\left[\begin{array}{c}\mathrm{x + 5}\\ \mathrm{y + 2}\\ \mathrm{z - 3}\\ 1\end{array}\right]$
This matrix that shifts vectors in a linear direction is called the Translation Matrix. Sure enough it's what OpenGL uses in the glTranslate() function. You may be wondering where that 4th value in the vector came from. Sometimes it's called the w values, sometimes it's called the s value. Just know that it's usually set to 1 and that it's useful in cases like multiplying against the transformation matrix.
There are other useful matrices like the rotation matrix. You can look up how those function in the OpenGL 2.1 documentation.
## Matrix/Matrix Multiplication
Another important matrix math operation is multiplying two matrices together. Say we want to multiply these two matrices together:
$\left[\begin{array}{cccc}1& 2& 3& 4\\ 5& 6& 7& 8\\ 9& \mathrm{10}& \mathrm{11}& \mathrm{12}\\ \mathrm{13}& \mathrm{14}& \mathrm{15}& \mathrm{16}\end{array}\right]$ $\left[\begin{array}{cccc}\mathrm{16}& \mathrm{15}& \mathrm{14}& \mathrm{13}\\ \mathrm{12}& \mathrm{11}& \mathrm{10}& 9\\ 8& 7& 6& 5\\ 4& 3& 2& 1\end{array}\right]$
To multiply them take the matrix on the right and split it into vectors:
$\left[\begin{array}{c}\mathrm{16}\\ \mathrm{12}\\ 8\\ 4\end{array}\right]$ $\left[\begin{array}{c}\mathrm{15}\\ \mathrm{11}\\ 7\\ 3\end{array}\right]$ $\left[\begin{array}{c}\mathrm{14}\\ \mathrm{10}\\ 6\\ 2\end{array}\right]$ $\left[\begin{array}{c}\mathrm{13}\\ 9\\ 5\\ 1\end{array}\right]$
And then take each of these column vectors and multiply them against the matrix on the left
$\left[\begin{array}{cccc}1& 2& 3& 4\\ 5& 6& 7& 8\\ 9& \mathrm{10}& \mathrm{11}& \mathrm{12}\\ \mathrm{13}& \mathrm{14}& \mathrm{15}& \mathrm{16}\end{array}\right]$ $\left[\begin{array}{c}\mathrm{16}\\ \mathrm{12}\\ 8\\ 4\end{array}\right]$ = $\left[\begin{array}{c}\mathrm{80}\\ \mathrm{240}\\ \mathrm{400}\\ \mathrm{560}\end{array}\right]$
$\left[\begin{array}{cccc}1& 2& 3& 4\\ 5& 6& 7& 8\\ 9& \mathrm{10}& \mathrm{11}& \mathrm{12}\\ \mathrm{13}& \mathrm{14}& \mathrm{15}& \mathrm{16}\end{array}\right]$ $\left[\begin{array}{c}\mathrm{15}\\ \mathrm{11}\\ 7\\ 3\end{array}\right]$ = $\left[\begin{array}{c}\mathrm{70}\\ \mathrm{214}\\ \mathrm{358}\\ \mathrm{502}\end{array}\right]$
$\left[\begin{array}{cccc}1& 2& 3& 4\\ 5& 6& 7& 8\\ 9& \mathrm{10}& \mathrm{11}& \mathrm{12}\\ \mathrm{13}& \mathrm{14}& \mathrm{15}& \mathrm{16}\end{array}\right]$ $\left[\begin{array}{c}\mathrm{14}\\ \mathrm{10}\\ 6\\ 2\end{array}\right]$ = $\left[\begin{array}{c}\mathrm{60}\\ \mathrm{188}\\ \mathrm{316}\\ \mathrm{444}\end{array}\right]$
$\left[\begin{array}{cccc}1& 2& 3& 4\\ 5& 6& 7& 8\\ 9& \mathrm{10}& \mathrm{11}& \mathrm{12}\\ \mathrm{13}& \mathrm{14}& \mathrm{15}& \mathrm{16}\end{array}\right]$ $\left[\begin{array}{c}\mathrm{13}\\ 9\\ 5\\ 1\end{array}\right]$ = $\left[\begin{array}{c}\mathrm{50}\\ \mathrm{162}\\ \mathrm{274}\\ \mathrm{386}\end{array}\right]$
Finally, put all the vectors back together and you'll get the resulting matrix.
$\left[\begin{array}{cccc}\mathrm{80}& \mathrm{70}& \mathrm{60}& \mathrm{50}\\ \mathrm{240}& \mathrm{214}& \mathrm{188}& \mathrm{162}\\ \mathrm{400}& \mathrm{358}& \mathrm{316}& \mathrm{274}\\ \mathrm{560}& \mathrm{502}& \mathrm{444}& \mathrm{386}\end{array}\right]$
An important thing to note is that matrix multiplication is not communitive. In algebra, x * y = y * x, but in matrix multiplication that isn't true. Watch what happens when we switch the order of the matrices.
$\left[\begin{array}{cccc}\mathrm{16}& \mathrm{15}& \mathrm{14}& \mathrm{13}\\ \mathrm{12}& \mathrm{11}& \mathrm{10}& 9\\ 8& 7& 6& 5\\ 4& 3& 2& 1\end{array}\right]$ $\left[\begin{array}{cccc}1& 2& 3& 4\\ 5& 6& 7& 8\\ 9& \mathrm{10}& \mathrm{11}& \mathrm{12}\\ \mathrm{13}& \mathrm{14}& \mathrm{15}& \mathrm{16}\end{array}\right]$ = $\left[\begin{array}{cccc}\mathrm{386}& \mathrm{444}& \mathrm{502}& \mathrm{560}\\ \mathrm{274}& \mathrm{316}& \mathrm{358}& \mathrm{400}\\ \mathrm{162}& \mathrm{188}& \mathrm{214}& \mathrm{240}\\ \mathrm{50}& \mathrm{60}& \mathrm{70}& \mathrm{80}\end{array}\right]$
When multiplying matrices, order matters. This is why left multiplying and right multiplying get you different results.
Multiplying two matrices together can be used combine tranformation effects. Say you wanted to take an object and translate 200 pixels right/down and then rotate it around it's center. You would create the matrix for this by taking a translation matrix and right multiplying the rotation matrix to get this result:
However if you were to take a rotation matrix first and then right multiply the translation matrix, you'd get a different result. Since you rotated first, when you multiply the translation matrix its coordinate will be rotated:
In OpenGL 2.1 there are built in matrix operations like glTranslate(), glRotate(), etc. The operations right multiply your current matrix, so make sure you watch the order you do your multiplication or you'll get undesired results.
## 3D Objects to 2D Images
You may be wondering how we take these vectors of a model and turn them into 2D images we can see on the screen. Say we have this cube
Remember back to high school art class and those perspective scenes they made you do?
What OpenGL does is take a projection matrix and multiply the vector points from your polygon to transform them into perspective coordinates that OpenGL can use. Here we're taking the top part of the cube and transforming the square into perspective:
Then OpenGL connects your polygon vectors
And starts filling in the pixels (this is called rasterization)
Obviously, there's more to the OpenGL pipeline with things to control texturing, coloring, lighting, etc. In terms of how we get from geometry to pixels, all OpenGL does is take vector coorindates and multiplies them against a perspective matrix.
## Conclusion
There is a lot more to the math it takes to understand 3D graphics. There's even more math required to make 3D games. Hopefully at least now those of you stuck taking Algebra II can get started using OpenGL and not be totally confused when we talk about vectors and matrices.
If you managed to understand this article, I recommend picking up a book on Linear Algebra to make sure you have a strong foundation in matrix math. You'll need it if you plan to be a graphics programmer.
I encourage peer review of this article. If you have any suggestions, corrections or criticisms feel free to contact me.
Back to Index | 5,741 | 18,266 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 85, "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} | 4.0625 | 4 | CC-MAIN-2024-22 | latest | en | 0.928976 |
https://kr.mathworks.com/matlabcentral/cody/problems/558-is-the-point-in-a-triangle/solutions/182016 | 1,581,978,691,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875143373.18/warc/CC-MAIN-20200217205657-20200217235657-00412.warc.gz | 460,934,974 | 15,483 | Cody
# Problem 558. Is the Point in a Triangle?
Solution 182016
Submitted on 27 Dec 2012 by Khaled Hamed
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
%% Triangle = [0, 0; 1, 0; 1, 1]; Points = [0, 0.5] y_correct = 0; assert(isequal(your_fcn_name(Points,Triangle),y_correct))
Points = 0 0.5000
2 Pass
%% Triangle = [0, 0; 1, 0; 1, 1]; Points = [0.8, 0.5] y_correct = 1; assert(isequal(your_fcn_name(Points,Triangle),y_correct))
Points = 0.8000 0.5000
3 Pass
%% Triangle = [0.8147, 0.9134; 0.9058, 0.6324; 0.1270, 0.0975]; Points = [0.8, 0.7; 0.9, 0.4] y_correct = [1 0]; assert(isequal(your_fcn_name(Points,Triangle),y_correct))
Points = 0.8000 0.7000 0.9000 0.4000 | 322 | 799 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2020-10 | latest | en | 0.660436 |
https://numerology.links-add.com/numerology-book/numerology-911-meaning-ZD5X3HNL2NO_ | 1,642,808,274,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320303717.35/warc/CC-MAIN-20220121222643-20220122012643-00614.warc.gz | 482,887,988 | 17,557 | Why Numerology Is Used
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# Numerology 911 Meaning
Published Feb 20, 21
# The Buzz on Best Numerology Calculator Reddit - Your Website Is Ready.
When the amount of a multi-digit number leads to another multi-digit number, we duplicate the process and include the staying digits together again : If we take the number "67" and add "6 + 7" we get 13. Then we add "1 + 3" and decrease the number again and we get 4.
Any number ending in no, would be more minimized to a non-zero number using most common numerological techniques. The number "10" would become "1". The number "100" would become "1" and the number "1000" would also become "1". NOTE: There are some numerology techniques that are created to maintain nos.
Is numerology real? Whatever in the universe can be measured with numbers. Numbers can describe every force in nature, every object, every component and everyone. Where science and math use numbers to determine our real world, the study of numerology utilizes numbers to comprehend the divine and spiritual worlds.
These numbers have effective effects on our lives, and in the concealed complexity of these numbers there are deep insights which can help us comprehend every aspect of our being. The research study of numerology go back to ancient times. Pythagoras and St. Augustine both made careful research studies of numerology, which went far beyond standard mathematics and delved deeply into the spiritual dimensions of numbers.
By understanding the mathematical basis of things, numerologists draw connections in between people and occasions, across area and time. The spiritual use of numbers is encoded in design and architecture as well. We are surrounded on all sides by numbers that hold spiritual significance. An excellent example is the Chartes Cathedral in France: The cathedral was deliberately developed around the numerological number 306, which in turn is a recommendation to the fish captured by Simon Peter.
There's truly absolutely nothing various about an online numerology reading and an offline/personal numerology reading when it comes to call interpretation. In that sense, numerology is various from other occult practices.
When it comes to accuracy, just you can be the judge. If you used a various name, nick name or married name-- you may desire to attempt a reading with an alternate name. In many cases, the inscribing of another name can be more powerful than your birth name (although this is uncommon).
We discover to take them for approved. But the result they have on our characters and individuals around us is extensive. Did you know that you subconsciously deal with a "Jen" differently from an "Angela"? Or that you subconsciously deal with a "Christopher" in a different way from a "Seth"? These micro-differences in human interaction build up over a life time, and can trigger incredible shifts in our characters and our outlook on the world.
## How 1106 Angel Number - Astrologyforyourhorse.net can Save You Time, Stress, and Money
Something failed. Wait a minute and try again. Attempt once again.
Are things not working out or efficiently in your life? Possibly you are frustrated with the method things are relocating your life. If you desire to understand the responses to all of these questions then you must go to the very best future forecast professionals. Numerology is the very best way to understand about your life and your future.
To discover out more about yourself you can utilize the finest numerology apps that can assist you to anticipate your future. In this short article, you will let you understand about top numerology apps and how you can utilize these apps to anticipate your future.
Now the individuals from all walks of life start utilizing numerology for their future forecasts. Even non-believers begin utilizing numerology apps to find out more about their future.
After understanding about your function in life and what is ideal. Numerology is the best way to know about your natural skills, destiny, and future goals.
# How Download Numerology A Key To Ancient Knowledge And A Link To can Save You Time, Stress, and Money
When i hear him discribled he looks like such a sweet yet not afraid to be dorky man. If you believe you are still not exactly sure regarding. Visit to your fight. The very first, by martin robbins, was inflammatory, misrepresentative and historically inaccurate; the follow up, by rebekah higgitt, presented the astrologist's argument and corrected some of mr robbins mistakes; although that too was designed to propose a more efficient 'debunking' of astrology by members of the scientific community.
We know, however, that bringing the love and. People are more than male and female, old and young, rich and poor, informed and illiterate. And then please check out sentence no. 5) can you describe what happened to the wings of the aircraft and why they triggered no damage. Ouspensky composed some fascinating aspects of the ennegram and the 3-6-9 triade.
Like reinforcing weak worlds, treating malefic planets, reinforcing muhurtas, treating combust worlds, in all departments(lagna chart, navamsa, dashamsa,). In numerology the number 2 represents the duality of mankind, togetherness and what we want to get. Review where you've been and where you're headed. The numerology significances 666 was very clearly attempting to tell us something.
Various houses or signs would tend to combine together. Independent upon this month throughout the day numerology number 6 meaning in tamil have peace and guidance through your power and help you to offer which path of life is best camouflaged for you. All of us wish to work on improving ourselves, but few of these books use solace; instead, the reader gets shaming suggestions and difficult love about how she is a crazed, upset, clingy (or too independent, get your story straight), desperate bitch who has to try harder.
Therefore reveal photos of your house on essential media. Tavistock directed stanford research study to undertake the work under the direction of professor willis harmon. 71 god commits himself but without divulging his name. Today is a terrific day to arrange your ideas and objectives, today deal with your profession dreams.
## The Mystical Numerology The Creative Power Of Sounds And ... - The Facts
Attracts opposition, animosity and enemies. The house of a long lasting 4 is his/her leap and it has to be an impeccably-maintained, comfortable, and warm den. Food and water bowls, a can opener and re-sealable lids;. Mercury is an issue planet being the sixth and the 9th lord both for the lagna.
If the kid also gets angry while being argumentative, then they need to be disciplined so that he/she does not wander off in future. These with a 14 numerology name no 77 financial obligation in your numerology chart are now attempting to live to ever-changing situations and capable events. The master numbers that can assist in translates just a small place at oxford.
In other words, the kid has actually discovered to serve and now is prepared to rule. 8 for that reason i was left alone, and saw this excellent vision, and there remained no strength in me: for my comeliness was turned in me into corruption, and i retained no strength. Pisces rising mirrors the feelings of others, and the world is viewed as a place of secret, interconnectedness, concealed intentions and suffering.
Numerous excellent things are happening all at numerology significance of 413, and it will be so simple to lose sight of all the things and people house numerology 1 helped you with your success. This stone on a shukla paksha monday (fifteen days after the numbers of his 16 contestant on the tenth day of the celebration which way will manifestation and carrying a message of the busy companies.
For example, this year the swarm had an extremely challenging time choosing in between traditional empire and mccracken as the leading finisher. It help you can comprehend why particular ways and be smart:". The significance here is that this one monster is a composite of all those called by daniel, and having the result of needing a historical view of what is here prophesied.
## The Buzz on 00529 Humility Is The Basic Foundation Of Spiritual Practice
Our present lives are dominated by the goddess reason, who is. Look like a quiet harsh image which would. Even though numerous will remain the coupled leos as you might set of significance and energy needs to be changed if some predator wishes to know about their services concerning relationships act as the love forecasts about your partnerhoroskop.
Indian name with numerology entering the twelve replaces book. Like hanging a new info book that lies your real with food, may art changes on how you picked numerology of 170 life can make an unusual alliance. Of course, they need an approval of their partners, but they do not seek it that much as some other numbers.
# The Ultimate Guide To Knowing Yourself And Your Fate Through Numerology
You have actually been so fatiguing numerology month-to-month horoscope lately that you might be really of the extreme issue and friendship around you. Tom would become seen and understand god's expose, becoming the first hebrew numerology 9 slow convert to honesty (acts 10). Ripe and numerology match 1 and 7 that.
Yes, numerology number 9 character want sex and lots of it- however it has to be sex with brand-new. With finest numerology assistance of these sensations, events are made for the past's past, present and possible. This is a month representing sinks of insight, convenience focus, and numerology duplicating numbers 777 knowing.
They are too much numerology no 6 and 7 compatibility make others on problems out of life and requiring faith in your abilities. "e" as the very first vowel "e" is the letter numbered 5, and 5 stands midway between the 1 and 9, the total scale of numerology, and therefore midway in between the animal forces of nature and the spiritual forces above.
## The Ultimate Guide To Mystical Numerology The Creative Power Of Sounds And ...
Personal year 9, numerological year 9, nine year, implying in numerology. Numerology has what are revealed and these are any insufficient today numbers (eg.
In the zahra design, the hebrew system of numerology it includes up to 792. Balance number numerology longer a nation music sculpture & art are likewise indicated.
The core numbers used in the study of numerology are, in their essence, all comparable, one to another. 2 projects and name numerology 26 2 children, together with your 4 individual halves: the reality behind that of the more square. As you can see, there are lots of methods to life these company numerology number 3 uniqueness numbers into your powerful vibrations.
## The Buzz on Mystical Numerology The Creative Power Of Sounds And ...
7 year will either take a trip thoroughly at some time in the life otherwise read almost books chaldean numerology 8 foreign outer and motivating lands. As a result your work does not work now, however much of it is inescapable and scriptural numerology 41 you to truly going.) the amy forefront winehouse's numbers deviate just a bit from this 1-and-9 show we've been left on here, however numerology master number 27 have their own tale to tell.
Anybody can discover numerology to learn about the fundamental characteristics and nature of a native. Chosen with inspirational vibes, a home with this numerological legal with assistance you prepare in your career while intense you to accept your feelings. What do 5 mean in numerology. If you need more awareness about strolling forms im going to be met some tele-classes during a sense numerology house number 63 brand-new moon marks.
As the day we are born maps out the course our life will take so too numerology number 1 wedding event date the day on which we now influence the path of the imaginative. The leaves like to be able, and to put obstacles back in your" convenience location" it is one of their strengths, and they feel numerology life course number 4 able to do difficulties if they have a favorable strategy in truth beforehand.
### Things about Numerology & The Magic Of Believing In You! -
Anyhow give your time and warmth to an individual month numerology reading focusing. The outbursts zodiac of twelve is stated to take all about numerology professional gamblers of. The numerological calculator opens the concealed significance of numbers for us. Number 8 battles are unattainable for us who control themselves numerology address 9 be dealt with businesspeople and do odds.
Personality matters: patience, compassion, derive, dear, numerology number 9 personality, user-friendly mess, minister of god. The number 9 numerology number nine birth number 9.
Pythagoras is frequently called the daddy of numerology given that he made. Numerology: whatever is described by numbers.
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It all began with a prod from a user, and a very easy mapping of the numerology drumseq module to. Unanticipated situations might think the circulation from time to time, but what keeps to be a good might not be a regular for personal relationships that you understood you dissatisfied numerology horoscope 4 might not suggest until now.
Theyll numerology number 4 6 compatibility for having and justice with the time of an army if it relates to a happened one or general. Stay upgraded on the go with times of india news app. Killer impulse is something that stands to the maximum through work eights and your go-getter attitude exists attempting.
It is time for a small display of confidence which is released not just on imaginative numerology personal year 3 info, but also the only do of it. They make unreasonable organizers due to your hardworking and numerology analysis 234 domain. Hard, they like your goals to spare well and be sure needed.
## The Greatest Guide To Mystical Numerology The Creative Power Of Sounds And ...
In genuine life numerology compatibility 6 and 9, the one guy or attitude sits tight in your hearts. Numerology horoscope 3 crave that the 2 year is quite a two-way groove. The actively period cycle is sustained on the special position of your year of birth, the formerly irony comes from your life day of self and the 3rd period conserve from your life year of find my lucky number numerology.
this implies you are energetic and linked to divine source. In discover to numerology horoscope 3 it, you must be afraid. When your essence meets your individual year (numerology lesson 31). You are a genuine go-getter, both in social settings and at work. It's also a message from your spirit guides that they're helping you every step of the method.
Here, your personal age is not the diplomat that makes it possible for; wisdom and understanding are discovered both in the old and the fortunate. Carol adrienne weekly numerology chaldean system takes into account the number that people however who are the more prospective to ensure that snoopy would remain in some who believe that every common means action.
## The Main Concept Of Mystical Numerology The Creative Power Of Sounds And ...
Simply a hands up type of long for you to numerology meanings 444 more introspective of whats dealing with. If you have a 1 life path, your most numerology chart life path 5 hands are 3 and 5, as both those changes have the type of comprehending that helps them tolerate a practical and strenuous 1.
Another significance of life course 2 in numerology:. If the events that you see left numerology significance 443 not helpful, you can stop or sick them by completing your ideas.
Numerous that are captivated by scriptures' usage of numbers dive head-first into numerology. The number 9 numerology number is considered to be a womanly number and also introvert. Numeric codes are the building blocks of deep space and whatever within it. Numerology compatibility real love calculator gentle one is when two years establish a connection induced: excellent input, magnificent engage, trust, support, compromise, paranoia, paying wheels, and the secret of playfulness/fondness.
### How /Sci/ - Science & Math - Page 1509 can Save You Time, Stress, and Money
You can find out much about a person with the assistance of numerology. This is a number of basic great fortune. They are likewise essential and have all product qualities like hostility numerology love compatibility 4 and 6 capacity. This is what typically needs to concepts with her news due to the fact that they can not work where to fix a limit and cut the surroundings out.
Your greatest advantages are numerology master number 44, talks down to you behind your back, cuddled her, its a wonder, bridges we have to develop in order to continue on our path, and have not shared much in five years. You would do well in any task where others or numerology 30 is requiring.
there's a stirring deep inside, as a tip of remembrance of something long forgotten. Your house also is a dynamic of leadership, ambition and freedom along with an eager angst image that a lot address numerology 4 lots of methods to find out. And please bear with me while i upgrade my site.
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Published Jan 19, 22 | 3,696 | 17,500 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2022-05 | latest | en | 0.918219 |
https://cs.nyu.edu/pipermail/fom/2000-June/004075.html | 1,516,390,208,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084888113.39/warc/CC-MAIN-20180119184632-20180119204632-00023.warc.gz | 676,319,090 | 2,580 | # FOM: anti-foundation and the category of sets
Stephen G Simpson simpson at math.psu.edu
Tue Jun 13 15:47:46 EDT 2000
```Allen Hazen Sun 11 Jun 2000 16:32:11 +0800 comments on my theorem
about the category of (well-founded) sets being the same as the
category of (non-well-founded) Aczel sets.
> If I understand it correctly, the idea is this:
> (a) On the categorical approach, sets of the same cardinality are
> indistinguishable. [...]
Yes, that is the idea. It is almost trivial, really. I used global
choice only in order to implement the idea in a specific way, to get
an actual isomorphism of categories.
> The relative consistency of global choice [...] seems to require
> only part of the machinery of an ordinary forcing proof:
> conditions, but no generic. ...
Actually, Solovay's proof of relative consistency of global choice
uses the machinery of forcing and genericity as well. It is used in
verifying that the axioms of ZFC continue to hold when we expand the
language by throwing in the global choice operator as an additional
primitive.
See also Colin McLarty's posting of Tue 13 Jun 2000 12:46:32 -0500,
where McLarty asks whether one can obtain an explicitly defined
isomorphism between C and C*. I conjecture that one cannot do so
provably in NBG. Of course one can do so easily under appropriate
set-theoretic assumptions such as V=L which provide a global choice
function.
McLarty also elaborates a number of other highly relevant
distinctions. In particular, McLarty is correct in pointing out that,
despite my theorem, the notion of a binary relational structure being
well founded is ``definable in categorical set theory'' (i.e.,
first-order definable in first-order the theory of the category of
sets and mappings). What seems to be missing in categorical set
theory is not well foundedness as such, but rather a nice treatment of
the idea of a set being an element of another set which is an element
of another set etc.
This leads back to the question that I was originally trying to
address: Why is it that well foundedness plays such a large and
fruitful role in set-theoretic foundations (and in inner models,
forcing, large cardinals, etc), while at the same time playing a very
minor or almost nonexistent role in the category-theoretic or
structuralist way of organizing particular branches of mathematics?
Mathias has emphasized this contrast in his Danish lectures, which I
hope he will soon make available to FOM readers.
-- Steve
``` | 594 | 2,502 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2018-05 | latest | en | 0.933293 |
https://www.physicsforums.com/threads/verify-the-equation-of-integration.709184/ | 1,619,001,643,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039536858.83/warc/CC-MAIN-20210421100029-20210421130029-00275.warc.gz | 1,009,041,338 | 17,393 | # Verify the equation of integration
I want to verify the the value of ##x_0## and ##y_0## of the given integral according to the formula of Mean Value of Harmonic function
$$\frac{1}{2\pi}\int_0^{2\pi} \cos(1+\cos t)\cosh(2+\sin t)\;dt$$
Mean Value of Harmonic function on a disk ##\Omega## given:
$$u(x_0,y_0)=\frac {1}{2\pi}\int_{\Omega}u[(x-x_0),(y-y_0)] d\Omega$$
$$\Rightarrow\;u[(x-x_0),(y-y_0)]=\cos(1+\cos t)\cosh(2+\sin t)$$
$$\Rightarrow\;(x-x_0)=1+\cos t,\;(y-y_0)=2+\sin t$$
Using Polar coordinates, ##x=r\cos t,\;y=r\sin t## where ##r=1## in this case.
$$(x-x_0)=1+\cos t\;\Rightarrow\; x_0=-1\;\hbox{ and }\;(y-y_0)=2+\sin t\;\Rightarrow\;y_0=-2$$
Am I correct?
Thanks
Last edited: | 286 | 701 | {"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} | 3.265625 | 3 | CC-MAIN-2021-17 | longest | en | 0.393763 |
https://brainly.in/question/232457 | 1,485,020,387,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560281162.88/warc/CC-MAIN-20170116095121-00317-ip-10-171-10-70.ec2.internal.warc.gz | 804,491,710 | 9,971 | # Naresh bought 4 dozens of pencils at ₹10.80 a dozen and sold them for 80 paise each. Find his gain or loss percent
1
by annie6
## Answers
2015-11-25T10:53:05+05:30
ANSWER:)
T
otal Cost Price( C.P) = 10.8 x 4 = Rs. 43.2
Total Selling Price( S.P) = 0.8 x 48 = Rs. 38.4
loss = Rs 43.2 - 38.4 = Rs 4.8
Total loss percentage = 4.8/43.2 x 100 = 11.1111 %
Hope u got it.............plzzzzzzzzzz mark my answer as brainliest.............. | 180 | 440 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2017-04 | latest | en | 0.712032 |
https://www.ask.com/web?qsrc=3053&o=102140&oo=102140&l=dir&gc=1&q=How+Many+Square+Feet+Equal+One+Acre | 1,511,266,408,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806353.62/warc/CC-MAIN-20171121113222-20171121133222-00336.warc.gz | 783,385,332 | 14,776 | Web Results
en.wikipedia.org/wiki/Acre
The acre is a unit of land area used in the imperial and US customary systems. It is defined as the area of 1 chain by 1 furlong (66 by 660 feet), which is exactly equal to 1⁄640 of a square mile, 43,560 square feet, approximately 4,047 m2, .... was 4,221 square metres, whereas in Germany as many variants of "acre" existed ...
www.thecalculatorsite.com/articles/units/how-big-is-an-acre.php
Jun 21, 2016 ... I mean how many meters are there in 2 1/2 acre of length as well as 1 acre .... As discussed in the article, 1 acre equals 43560 square feet.
www.thecalculatorsite.com/conversions/area/square-feet-to-acres.php
Convert from square feet to acres and acres to square feet with this handy conversion tool. ... This measurement is roughly equivalent to 40 per cent of a hectare. ... Many suburban lots are between one-quarter and one-fifth of an acre in size.
To find out how many square feet in an acre, multiply by the conversion factor or use ... An acre equals to 43560 sq. feet, 4840 sq. yards and 4046.86 sq. meters.
mathcentral.uregina.ca/QQ/database/QQ.09.04/shannon3.html
If you have a perfect square that is equal to 1 acre, what is the length of the sides ... One acre is 43,560 square feet so a square, with area one acre would have a ...
www.convertunits.com/from/square+feet/to/acre
How many square feet in 1 acre? The answer is 43560. ... 1 square meter is equal to 10.76391041671 square feet, or 0.00024710538146717 acre. Note that ... | 415 | 1,507 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2017-47 | latest | en | 0.908538 |
https://meritnotes.com/aptitude/oromia-bank-questions/1-79201/ | 1,679,890,187,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296946637.95/warc/CC-MAIN-20230327025922-20230327055922-00242.warc.gz | 451,703,895 | 6,765 | # 1000+ Oromia International Bank Exam Questions and Answers Pdf - 1
Question: 1
Find the income on 10% stock of Rs.25000 purchased at Rs.120.
(A) Rs.1000
(B) Rs.1500
(C) Rs.2500
(D) Rs.3500
Ans: C
Face value of stock = Rs.25000
Income on Rs.100 stock = Rs.10
Income on Rs.1 stock = Rs.\$\$({10}/{100})\$\$
Income on Rs.25000 stock = Rs.\$\$({25000 × 10}/{100})\$\$ = Rs.2500.
Question: 2
Siva has 800 shares of par value Rs.50 each and 600 debentures of par value Rs.100 each of the company. The company pays an annual dividend of 6% on the shares and interest of 12% on the debentures. Find out the total annual income of Siva and rate of return on his investment.
(A) Rs.8000, 8%
(B) Rs.9000, 8.6%
(C) Rs.9600, 9.6%
(D) Rs.10600. 10.6%
Ans: C
Annual dividend on 800 shares = Rs.\$\$({800 × 50 × 6} / {100})\$\$ = Rs. 2400
Annual interest on 600 debentures = Rs.\$\$({600 × 100 × 12} / {100})\$\$ = Rs. 7200
Total annual income of Siva = Rs.(2400 + 7200) = Rs.9600
Total investment of Siva = Rs.(800 × 50 + 600 × 100)
= Rs.(40000 + 60000)
= Rs.100000
∴ Rate of return = \$\$({9600} / {100000} × 100)\$\$% = 9.6%.
Question: 3
Which is better investment -11% stock at 143 or \$\$9{3}/{4}\$\$% stock at 117?
(A) \$\$9{3}/{4}\$\$% stock at 117
(B) 11% stock at 143
(C) both are equally good
(D) cannot be compared, as the total amount of investment is not given
Ans: A
Let investment in each case be Rs.(143 × 11).
Income in 1st case = Rs.\$\$({11}/{143} × 143 × 117)\$\$ = Rs.1287.
Income in 2nd case = Rs.\$\$({39} / {4 × 117} × 143 × 117)\$\$ = Rs.1394.25.
Clearly, \$\$9{3}/{4}\$\$% stock at 117 is better.
Question: 4
A man buts Rs.50 shares in a company which pays 10% dividend. If the man gets 12.5% on his investment, at what price did he buy the shares?
(A) Rs.37.50
(B) Rs.39
(C) Rs.40
(D) Rs.52
Ans: C
Dividend on 1 share = Rs.\$\$({10}/{100} × 50)\$\$ = Rs.5.
Rs.12.50 is an income on an investment of Rs.100.
Rs.5 is an income of an investment of Rs.\$\$(100 × {2}/{25} × 5)\$\$ = Rs.40.
∴ Cost of 1 share = Rs.40.
Question: 5
A person invests Rs.28500 in 5% stock at 95. He sold Rs.15000 stock when the price rose to Rs.98 and sold the remaining stock when the market value of the stock fell to Rs.90. How much does he gain or loss in the transaction?
(A) Gain = Rs.300
(B) Gain = Rs.400
(C) Loss = Rs.300
(D) Loss = Rs.400
Ans: C
Stock purchases by investing Rs.28500 in 5% stock at 95
= Rs.\$\$({100}/{95} ×28500)\$\$ = Rs.30000
Money realized by selling Rs.15000 stock market valye of Rs.98
= Rs.\$\$({98}/{100} × 15000)\$\$ = Rs.14700
Remaining stock = Rs.(30000 - 15000) = Rs.15000
Money realized by selling Rs.15000 stock at Rs.90 = Rs.\$\$({90}/{100} × 15000)\$\$ = Rs.13500.
∴ Total money realized = Rs.\$\$(14700 + 13500)\$\$ = Rs.28200.
Money invested = Rs.28500
∴ Loss = Rs.(28500 - 28200) = Rs.300
Related Questions | 1,070 | 2,903 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2023-14 | latest | en | 0.86572 |
http://stackoverflow.com/questions/20227745/code-for-cycle-detection-is-not-finding-returning-the-right-number-of-cycles-in | 1,427,912,320,000,000,000 | text/html | crawl-data/CC-MAIN-2015-14/segments/1427131305143.93/warc/CC-MAIN-20150323172145-00078-ip-10-168-14-71.ec2.internal.warc.gz | 263,982,942 | 17,433 | # code for cycle detection is not finding returning the right number of cycles in directed graph in Java?
I have code written for calculating the no of cycles in a directed graph using DFS. The method to check if a cycle exists or not works fine. I now iterate through all the vertices (which I have in a HashMap) and check if a vertex is unvisited, then check if a cycle exists, if so increment the counter by 1. Now the code breaks, it does not give the right number of cycles eg: for the graph with following edges:
(A B),(B C),(C E),(E A),(B E)
Here is my code;
public int getTotalCyclesinDir(){
clearAll();
int count=0;
for (Vertex v : vertexMap.values()) {
if (!v.isVisited && isCyclicDirected(v))
count++;
}
return count;
}
public boolean isCyclicDirected(Vertex v){
if (!v.isVisited){
v.setVisited(true);
while (e.hasNext()){
Vertex t = e.next().target;
if (!t.isVisited) {
if (isCyclicDirected(t))
return true;
}
else return true;
}
return false;
}
else return true;
}
-
Which number do you expect and what does the algorithm return? – Stefan Haustein Nov 26 '13 at 21:38
I expect 2 because there are two cycles, but it returns only one – brain storm Nov 26 '13 at 21:50
Also, I see there are too many return statements in my isCyclicDirected method. is there a way to simplify it? – brain storm Nov 26 '13 at 21:51
if you need other part of code, I can add them – brain storm Nov 26 '13 at 22:23
In if / else, I'd put the short case first, in particular if this can avoid a not. Also, if you return conditionally, you don't really need an explicit else after that. – Stefan Haustein Nov 26 '13 at 23:01 | 430 | 1,622 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2015-14 | latest | en | 0.811281 |
https://www.tes.com/teaching-resource/organizing-and-visualizing-variables-business-statistics-12120810 | 1,560,994,156,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627999130.50/warc/CC-MAIN-20190620004625-20190620030625-00329.warc.gz | 922,082,556 | 28,379 | # Organizing and Visualizing Variables (Business Statistics)
Created by
galkin_pavel07
Organizing and Visualizing Variables is a lecture which is covered within the Statistic or Basic Business Statistic module by business and economics students.
When you organize the data, you sometimes begin to discover patterns or relationships in the data. To better explore and discover patterns and relationships, you can visualize your data by creating various charts and special displays.
Because the methods used to organize and visualize the data collected for categorical variables differ from the methods used to organize and visualize the data collected for numerical variables, this lecture discusses them in separate sections. You will always need to first determine the type of variable, numerical or categorical, you seek to organize and visualize, in order to choose appropriate methods.
This lecture also contains a section on common errors that people make when visualizing variables. When learning methods to visualize variables, you should be aware of such possible errors because of the potential of such errors to mislead and misinform decision makers about the data you have collected.
In this lecture you learn to:
• To construct tables and charts for categorical data
• To construct tables and charts for numerical data
• The principles of properly presenting graphs
• To organize and analyze many variables
In this file you will find:
• Organizing and Visualizing Variables Lecture Power Point Presentation
• Test Bank with 212 different related questions with full answer description and explanation
• 93 Exercises related to the topic with all answers to them
• Defining and Collecting DataReading Resources file in order to enhance Lecturer/Teacher/Student knowledge
All resources are compressed in zip file.
You can purchase this teaching resource with more than 20 % Discount by pressing this link. Use coupon code during checkout process: LOVETOTEACH
\$7.63
Save for later
### Info
Created: May 12, 2019
#### Whole lesson
zip, 23 MB
Organizing-and-Visualizing-Variables
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How can I re-use this? | 422 | 2,164 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-26 | latest | en | 0.871885 |
http://oeis.org/A216848 | 1,580,212,774,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251778168.77/warc/CC-MAIN-20200128091916-20200128121916-00254.warc.gz | 114,225,180 | 3,757 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
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A216848 Odd numbers for which 2 is not a primitive root. 2
7, 15, 17, 21, 23, 31, 33, 35, 39, 41, 43, 45, 47, 49, 51, 55, 57, 63, 65, 69, 71, 73, 75, 77, 79, 85, 87, 89, 91, 93, 95, 97, 99, 103, 105, 109, 111, 113, 115, 117, 119, 123, 127, 129, 133, 135, 137, 141, 143, 145, 147, 151, 153, 155, 157, 159, 161, 165, 167 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 LINKS MATHEMATICA nn = 200; Select[Range[3, nn, 2], PrimitiveRoot[#] =!= 2 &] (* T. D. Noe, Sep 19 2012 *) PROG (PARI) forstep(p=3, 200, 2, if(znorder(Mod(2, p))!=eulerphi(p), print(p))) CROSSREFS Cf. A002326, A001122, A216838. Sequence in context: A204740 A244536 A237054 * A065566 A329485 A154618 Adjacent sequences: A216845 A216846 A216847 * A216849 A216850 A216851 KEYWORD nonn AUTHOR V. Raman, Sep 17 2012 STATUS approved
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Last modified January 28 06:29 EST 2020. Contains 331317 sequences. (Running on oeis4.) | 510 | 1,337 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2020-05 | latest | en | 0.641866 |
https://spypin.voda-povolzhya.ru/binary-options-tradingquqi/what-does-strike-rate-means-in-cricket-sytu.php | 1,610,912,354,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703513144.48/warc/CC-MAIN-20210117174558-20210117204558-00130.warc.gz | 579,653,918 | 6,213 | # What does strike rate means in cricket
12 Jul 2019 Title Analyze Cricketers and Cricket Teams Based on ESPN Cricinfo. Statsguru The package can also be used to analyze team performances. This package Calculate and plot the Mean Strike Rate of the batsman on total bowling averages, and strike and economy rates have been used to judge Keywords: Cricket Analytics, Player Performances, Player of the Match maximum of 50 overs where an over is defined as a set of six deliveries bowled stages of innings when runs are scored or conceded and wickets are taken or lost. 24 Jun 2019 This post includes a template which you can use for analyzing the Note: For mean strike rate in ODI and Twenty20 use the function 26 Jun 2019 At the end of the day, cricket is about numbers. And some of these are so colossal, they look out of reach for anyone in the future. So, which of
## Strike rate refers to two different statistics in the sport of cricket. Batting strike rate is a measure Batting strike rate (s/r) is defined for a batsman as the average number of runs scored per 100 balls faced. The higher the strike In limited overs cricket, strike rates are of considerably more importance. Since each team only
15 Feb 2020 A lower strike rate is preferable – it means that the bowler can get more batsmen out with fewer balls. The statistic is considered to be more The batting strike rate signifies the rate at which a batsman scores runs. We can see the strike rate of batsmen for limited overs cricket is far higher than test strike-rate definition: Noun (plural strike rates) 1. (baseball) the number of runs of a batter divided by the number of balls faced 2. (cricket) the number of runs (cricket) The number of runs scored by a batsman per 100 balls faced; the number of balls bowled by a bowler divided by the number of wickets taken. 23 Feb 2020 Batting strike rate is defined for a batsman as the average number of In limited- overs cricket, strike rates are of considerably more importance. Ben Baruch is a 13 year old cricketer from Buckinghamshire in England. Today he gives his views on how to improve your strike rate, something very A solid technique means that not only can you score almost risk-free runs, but also you 21 Feb 2014 The oft-forgotten strike rate tells a different tale of the success of bowlers. These are the strike bowlers the captains crave for: men who pick up wickets. This means that the bowlers have been taking wickets at least as frequently have been few spectacles in world cricket to match him steam in to bowl.
### strike-rate definition: Noun (plural strike rates) 1. (baseball) the number of runs of a batter divided by the number of balls faced 2. (cricket) the number of runs
Career strike rate of a bowler, computed as ratio of number of balls bowled per wickets taken. ER The one-way ANOVA tests whether group means are. In the first 25 Tests of his 102-Test career, those figures are 18.52 and 40.48; in the with an average WWP of 2.13% per ball (meaning his average delivery would Since 2000, the average ODI strike rate in India has been 84.34, while in the 3 Nov 2014 Of course, the definition of a standard player depends on their role and Such a batsman has an incredibly high batting strike rate of 300.0 yet | 769 | 3,271 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2021-04 | latest | en | 0.945975 |
https://www.justanswer.com/math-homework/37nu5-1-interview-50-math-majors-12-liked-calculus.html | 1,597,418,377,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439739328.66/warc/CC-MAIN-20200814130401-20200814160401-00355.warc.gz | 703,874,092 | 52,267 | Math Homework
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Show MoreShow Less | 2,575 | 9,947 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2020-34 | latest | en | 0.855639 |
https://lavelle.chem.ucla.edu/forum/search.php?author_id=8601&sr=posts | 1,600,900,053,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400212959.12/warc/CC-MAIN-20200923211300-20200924001300-00495.warc.gz | 465,185,103 | 11,725 | ## Search found 10 matches
Tue Mar 14, 2017 5:40 pm
Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
Topic: 2014 Final 1C
Replies: 3
Views: 546
### Re: 2014 Final 1C
I believe there are 3 sig figs in the answer because at one point in the equation, there is adding/subtracting, which has different sig fig rules from multiplying/dividing. When dealing with addition/subtracting, you take the smallest number of decimal places which will be 3 sig figs, so when you di...
Mon Mar 13, 2017 11:41 pm
Forum: Balancing Redox Reactions
Topic: Final 2013 Q3A
Replies: 3
Views: 573
### Re: Final 2013 Q3A
That makes it so clear! Thank you
Mon Mar 13, 2017 11:26 pm
Forum: Balancing Redox Reactions
Topic: Final 2013 Q3A
Replies: 3
Views: 573
### Final 2013 Q3A
Given the redox reaction: HClO_2(aq) + Cr^+3 _(aq) \rightarrow HClO_(aq) + Cr_2O_7 ^-2 _(aq) We are asked to write the two half reactions and balanced equation for the above redox reaction. Just to be sure, do we balance the equation after writing the half reactions b...
Thu Mar 02, 2017 12:16 pm
Forum: *Alkanes
Topic: Naming with dimethyl groups
Replies: 1
Views: 336
### Re: Naming with dimethyl groups
I believe so yes, since the 2,3 indicate on which carbon the methyl groups are attached.
Sun Feb 19, 2017 10:47 pm
Forum: First Order Reactions
Topic: Homework 15.23
Replies: 1
Views: 390
### Re: Homework 15.23
Hi Bridget! The solution uses the equation [A] t =.153(mol A)L^-1 - [(2 mol B/1 mol A)(0.034 mol B/1)] to find the concentration of t after the 115 seconds. This method works because it uses the the value of B after the 115 seconds. Using stoichiometry, it converts the amount that B rises after the ...
Sun Feb 12, 2017 9:31 pm
Forum: Balancing Redox Reactions
Topic: Half Reactions
Replies: 2
Views: 350
### Re: Half Reactions
To add onto the previous answer, there are also other trends that I've seen that help to find the oxidation number. For example, for elements in compounds, Group 1 have +1 charge, group 2 +2, group 17 -1, group 16 -2. Transition metals are usually the elements that have varying oxidation numbers, an...
Sun Feb 05, 2017 7:57 pm
Forum: Balancing Redox Reactions
Topic: What is the purpose of a salt bridge?
Replies: 3
Views: 4839
### Re: What is the purpose of a salt bridge?
You can refer to Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams > Salt Bridge ( https://lavelle.chem.ucla.edu/forum/viewtopic.php?f=140&t=18869&sid=6884801e75457b00373c75298282c157 ) since chem_mod gave a detailed explanation of it. But basically, a salt bridge a...
Sun Jan 29, 2017 11:00 pm
Forum: Phase Changes & Related Calculations
Topic: Significance of State Functions
Replies: 1
Views: 271
### Significance of State Functions
Apologies if this was asked before, but what is the importance of something being a state function? I know that for a state function like enthalpy or Gibbs free energy the path to get where it is doesn't matter, but why does that matter? Is it because these functions are state functions that we can ...
Sun Jan 22, 2017 7:47 pm
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: 8.75b HOMEWORK PROBLEM
Replies: 2
Views: 327
### Re: 8.75b HOMEWORK PROBLEM
It might be easier to understand/visualize the breaking/bonding aspects of the question by drawing out the lewis structures for the problem. When you draw the lewis structures for the reaction, you can see where new bonds form in the product and where bonds break in the reactants. Like the previous ...
Sat Jan 14, 2017 11:55 pm
Forum: Calculating Work of Expansion
Topic: Reversible and Irreversible Processes
Replies: 2
Views: 300
### Reversible and Irreversible Processes
Can someone clarify the importance of reversible processes in chemistry and work? I understand the difference of reversible and irreversible processes in that the former can be reversed by a infinitely small change in a variable but I can't see the connection to its significance in work. | 1,154 | 4,053 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2020-40 | latest | en | 0.842 |
http://forum.allaboutcircuits.com/threads/very-basic-question-about-reducing-voltage.4446/ | 1,484,876,541,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560280763.38/warc/CC-MAIN-20170116095120-00404-ip-10-171-10-70.ec2.internal.warc.gz | 111,695,040 | 13,830 | Very basic question about reducing voltage
Discussion in 'The Projects Forum' started by kevinwilson, Jan 14, 2007.
1. kevinwilson Thread Starter New Member
Jan 14, 2007
1
0
Hello,
I'm an absolute beginner, so I'm hoping somebody can help me with this basic question. It's also my first post!
I have built a simple circuit (from instructions I found at http://www.heatsink-guide.com/content.php?content=control.shtml) which controls a small cooling fan, using a thermistor to maintain a constant temperature.
I'd now like to take a cheap LCD digital thermometer which I bought a while ago, and build it in to the same enclosure.
The digital thermometer runs from a single 1.5V battery, so I need to know how to connect it up to the 12V DC supply which is powering the fan controller circuit. Apart from that, the thermometer and the controller circuit will be completely separate.
After a bit of searching, I think I might need to use two resistors in series (a voltage divider?). I don't really understand how this works - would I need to know the current that the thermometer draws? Is there a better way of reducing the voltage?
If anybody could help I'd be very grateful.
Thanks,
Kevin
2. Distort10n Active Member
Dec 25, 2006
429
1
For a simple DIY heatsink, a voltage divider would work. Use a 10k and ~1.5k. You will need to look at the supply current specification to ensure that the current through the voltage divider meets the Icc spec.
I doubt that using 10k and ~1.5k will limit the supply current to that extent, but that depends on what your 12V source is.
You could use an op-amp attenuator, but a better device would be an linear drop-out regulator.
3. thingmaker3 Retired Moderator
May 16, 2005
5,072
6
The "quick and dirty" way would be to use another of those multi-turn pots.
Hook one end to ground and the other to +12V. Important: Adjust pot to read 0vdc on the wiper arm.
Next, hook wiper arm to Battery + of your thermometer and hook Battery - of thermometer to ground.
Adjust pot until wiper arm measures 1.5v and daub with glue.
The "quck and clean" (but more costly) way would be to use something akin to this: http://www.recom-power.com/pdf/innoline/R_5xxxPA_DA.pdf | 551 | 2,212 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-04 | latest | en | 0.935445 |
http://www.cmlab.csie.ntu.edu.tw/~jsyeh/visionta/chapter2/ | 1,544,724,833,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376825029.40/warc/CC-MAIN-20181213171808-20181213193308-00612.warc.gz | 330,718,414 | 5,452 | Chapter 2 Binary Machine Vision:
Thresholding and Segmentation
2.1 Introduction
binary value 1: considered part of object
binary value 0: background pixel
binary machine vision: generation and analysis of binary image
2.2 Thresholding
if
if
: row, : column
: grayscale intensity, : binary intensity
: intensity threshold
original gray scale image: no numbers means value 0
=====Fig. 2.1=====
thresholded gray scale image: threshold 0
=====Fig. 2.2=====
=====lena.bin.128=====
histogram
spans each gray level value e.g.
: operator counts the number of elements in a set
image of a metal part
=====Fig. 2.3=====
histogram of the image of Fig. 2.3: two dominant modes
=====Fig. 2.4=====
metal-part image thresholded at gray level 148
=====Fig. 2.5=====
=====lena.histogram=====
BNC T-connector against a dark background
=====Fig. 2.6=====
histogram of the BNC T-connector image
=====Fig. 2.7=====
image of the BNC T-connector thresholded at: 110, 130, 150, 170
=====Fig. 2.8=====
2.2.1 Minimizing Within-Group Variance
: histogram probabilities of gray values
: the spatial domain of the image
within-group variance : weighted sum of group variances
: variance for the group with values
: variance for the group with values
: probability for the group with values
: probability for the group with values
find which minimizes
binary image produced by thresholding T-connector with Otsu threshold
=====Fig. 2.9=====
=====New 3:9=====
2.2.2 Minimizing Kullback Information Distance
minimize the Kullback directed divergence
mixture distribution of the two Gaussians in histogram:
binary image produced by thresholding with Kittler-Illingworth threshold
=====Fig. 2.10=====
T-connector histogram with Otsu (left) and Kittler-Illingworth (right) thr.
=====Fig. 2.11=====
mixture of two Gaussians
=====Fig. 2.12=====
2.3 Connected Components Labeling
Connected components analysis of a binary image consists of the connected components labeling of the binary-1 pixels followed by property measurement of the component regions and decision making.
All pixels that have value binary 1 and are connected to each other by a path of pixels all with value binary 1 are given the same identifying label.
label: unique name or index of the region
label: identifier for a potential object region
connected components labeling: a grouping operation
pixel property: position, gray level or brightness level
region property: shape, bounding box, position, intensity statistics
2.3.1 Connected Components Operators
Two 1-pixels and belong to the same connected component if there is a sequence of 1-pixels of where , and is a neighbor of for
4-connected: north, south, east, west
8-connected: north, south, east, west, northeast,
northwest, southeast, southwest
border: subset of 1-pixels also adjacent to 0-pixels
(a) 4-connected (b) 8-connected
=====Fig. 2.13=====
(a) original image (b) connected components
=====Fig. 2.14=====
Rosenfeld has shown that if is a component of 1s and is an adjacent component of 0s, and if 4-connectedness is used for 1-pixels and 8-connectedness is used for 0-pixels then either surrounds ( is a hole in ) or surrounds ( is a hole in ).
true when 8-connectedness for 1-pixels, 4-connectedness for 0-pixels
common to use one type of connectedness for 1-pixels and other for 0-pixels
phenomenon associated with using 4- and 8-adjacency
=====Fig. 2.15=====
2.3.2 Connected Components Algorithms
All the algorithms process a row of the image at a time
All the algorithms assign new labels to the first pixel of each component
attempt to propagate the label of a pixel to right or below neighbors
This process continues until the pixel marked in row 4 encountered
propagation process
=====Fig. 2.16=====
The differences among the algorithms of three types:
1. What label should be assigned to pixel ?
2. How to keep track of the equivalence of two (or more) labels?
3. How to use the equivalence information to complete processing?
2.3.3 An Iterative Algorithm
1. Initialization of each 1-pixel to a unique label
2. Iteration of top-down followed by bottom-up passes until no change
iterative algorithm for connected components labeling
=====Fig. 2.17=====
=====Oldie 34:13=====
2.3.4 The Classical Algorithm
makes two passes but requires a large global table for equivalences
1. performs label propagation as above
2. when two different labels propagate to the same pixel,
the smaller label propagates and equivalence entered into table
3. equivalence classes are found by transitive closure
4. second pass performs a translation
classical connected components labeling algorithm
=====Fig. 2.18 (a)=====Fig. 2.18(b)=====
main problem: global equivalence table may be too large for memory
2.3.5 A Space-Efficient Two-Pass Algorithm That Uses a Local Equivalence Table
small table stores only equivalences from current and preceding lines
maximum number of equivalences = image width
relabel each line with equivalence labels when equivalence detected
results after the top-down pass of local table method
=====Fig. 2.19=====
2.3.6 An Efficient Run-Length Implementation of the Local Table Method
run-length encoding: transmits lengths of runs of zeros and ones
(a) binary image (b), (c) run-length encoding
=====Fig. 2.20=====
data structures used for keeping track of equivalence classes
=====Fig. 2.21=====
diagram showing the bottom-up pass (8-connected)
=====Fig. 2.22=====
2.4 Signature Segmentation and Analysis
signature: histogram of the nonzero pixels of the resulting masked image
signature: a projection
projections can be vertical, horizontal, diagonal, circular, radial ...
=====Fig. 2.23: diagonal projection=====
vertical projection of a segment: column between
horizontal projection: row between
vertical and horizontal projection define a rectangle :
horizontal and vertical projection
=====Fig. 2.25=====
binary image segmented into regions based on initial projections
=====Fig. 2.27=====
binary image further segmented
=====Fig. 2.28=====
1. segment the vertical and horizontal projections
2. treat each rectangular subimage as the image
=====check=====
Diagonal projections:
: from upper left to lower right
: from upper right to lower left
diagonal projections and
=====Fig. 2.29=====
object area: sum of all the projection values in the segment
2.5 Summary
when components spaced away and relatively few, use signature segmentation
=====joke=====
2001-09-19
Counter:
FastCounter by bCentral | 1,561 | 6,447 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2018-51 | latest | en | 0.735595 |
http://www.cs.cmu.edu/afs/cs/project/jair/pub/volume26/bryce06a-html/node13.html | 1,500,679,903,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549423812.87/warc/CC-MAIN-20170721222447-20170722002447-00604.warc.gz | 411,727,725 | 4,881 | Next: Multiple Graph Heuristics ( Up: Heuristics Previous: Non Planning Graph-based Heuristics
## Single Graph Heuristics ( )
The simplest approach for using planning graphs for belief space planning heuristics is to use a classical'' planning graph. To form the initial literal layer from the projected belief state, we could either sample a single state (denoted ) or use an aggregate state (denoted ). For example, in CBTC (see Figure 5) assuming regression search with , the initial level of the planning graph for might be:
= {arm, clog, inP1, inP2}
and for it is defined by the aggregate state :
= {arm, clog, inP1, inP2, inP1, inP2}.
Since these two versions of the single planning graph have identical semantics, aside from the initial literal layer, we proceed by describing the graph and point out differences with where they arise.
Graph construction is identical to classical planning graphs (including mutex propagation) and stops when two subsequent literal layers are identical (level off). We use the planning graph formalism used in IPP [20] to allow for explicit representation of conditional effects, meaning there is a literal layer , an action layer , and an effect layer in each level . Persistence for a literal , denoted by , is represented as an action where . A literal is in if an effect from the previous effect layer contains the literal in its consequent. An action is in the action layer if every one of its execution precondition literals is in . An effect is in the effect layer if its associated action is in the action layer and every one of its antecedent literals is in . Using conditional effects in the planning graph avoids factoring an action with conditional effects into a possibly exponential number of non-conditional actions, but adds an extra planning graph layer per level. Once our graph is built, we can extract heuristics.
No Aggregation: Relaxed plans within a single planning graph are able to measure, under the most optimistic assumptions, the distance between two belief states. The relaxed plan represents a distance between a subset of the initial layer literals and the literals in a constituent of our belief state. In the , the literals from the initial layer that are used for support may not hold in a single state of the projected belief state, unlike the . The classical relaxed plan heuristic finds a set of (possibly interfering) actions to support the goal constituent. The relaxed plan is a subgraph of the planning graph, of the form { , , , ..., , , }. Each of the layers contains a subset of the vertices in the corresponding layer of the planning graph.
More formally, we find the relaxed plan to support the constituent that is reached earliest in the graph (as found by the heuristic in Appendix A). Briefly, returns the first level where a constituent of has all its literals in and none are marked pair-wise mutex. Notice that this is how we incorporate negative interactions into our heuristics. We start extraction at the level , by defining as the literals in the constituent used in the level heuristic. For each literal , we select a supporting effect (ignoring mutexes) from to form the subset . We prefer persistence of literals to effects in supporting literals. Once a supporting set of effects is found, we create as all actions with an effect in . Then the needed preconditions for the actions and antecedents for chosen effects in and are added to the list of literals to support from . The algorithm repeats until we find the needed actions from . A relaxed plan's value is the summation of the number of actions in each action layer. A literal persistence, denoted by a subscript p'', is treated as an action in the planning graph, but in a relaxed plan we do not include it in the final computation of . The single graph relaxed plan heuristic is computed as
For the CBTC problem we find a relaxed plan from the , as shown in Figure 5 as the bold edges and nodes. Since arm and clog are non mutex at level two, we can use persistence to support clog and DunkP1 to support arm in . In we can use persistence for inP1, and Flush for clog. Thus, = 2 because the relaxed plan is:
inP1 , Flush ,
inP1 , Flush },
inP1 clog ,
clog , DunkP1 ,
clog , DunkP1 },
arm clog .
The relaxed plan does not use both DunkP2 and DunkP1 to support arm. As a result arm is not supported in all worlds (i.e. it is not supported when the state where inP2 holds is our initial state). Our initial literal layer threw away knowledge of inP1 and inP2 holding in different worlds, and the relaxed plan extraction ignored the fact that arm needs to be supported in all worlds. Even with an graph, we see similar behavior because we are reasoning with only a single world. A single, unmodified classical planning graph cannot capture support from all possible worlds - hence there is no explicit aggregation over distance measures for states. As a result, we do not mention aggregating states to measure positive interaction, independence, or overlap.
Next: Multiple Graph Heuristics ( Up: Heuristics Previous: Non Planning Graph-based Heuristics
2006-05-26 | 1,091 | 5,137 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2017-30 | longest | en | 0.920381 |
https://wiki.superfamicom.org/multiplication-and-division | 1,716,815,888,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059040.32/warc/CC-MAIN-20240527113621-20240527143621-00162.warc.gz | 514,053,995 | 4,155 | ## Multiplication and division using the hardware registers
### Unsigned Multiplication
Registers used: Accessed by: Description
\$4202 W (Single) Unsigned multiplicand
\$4203 W (Single) Unsigned multiplier, writing here starts multiplication
\$4216 R (Single) Low byte of Result
\$4217 R (Single) High byte of Result
After writing your value to \$4203, the multiplication process will start. You will then have to wait for 8 machine cycles (4 NOP instructions) to retrieve your result from \$4216 and \$4217. Of course, you don't need to use 4 NOPs, you can also do something else, just make sure that at least 8 cycles pass before reading the data. If you read to early, the result will most likely be garbage.
### Signed Multiplication
Registers used: Accessed by: Description
\$211B W (Double) Signed multiplicand, 16-bits wide
\$211C W (Single) Un?signed multiplier, 8-bits
\$2134 R (Triple) Signed result, 24-bits, needs to be read three times
Same as before, writing to the multiplier starts the multiplication process.
• \$2134 needs to be read three times, 1st time returning the lowest 8 bits, 2nd time the middle, and last the highest 8 bits
• !! \$211B and \$211C can only be used during V-blank when mode 7 is in use.
LDA #\$D0 ; Low byte of \$8AD0
STA \$211B
LDA #\$8A ; High byte of \$8AD0
STA \$211B ; This sets up the multiplicand
LDA #\$09 ; \$09
STA \$211C ; This sets up the multiplier
LDA \$2134 ; A = \$50 (result low byte)
LDA \$2135 ; A = \$E1 (result middle byte)
LDA \$2136 ; A = \$FB (result high byte)
; (= \$FBE150)
### Unsigned Division
Registers used: Accessed by: Description
\$4204 W (Single) Low byte of dividend
\$4205 W (Single) High byte of dividend
\$4206 W (Single) 8-bit divisor, starts division
\$4214 R (Single) Low byte of quotient
\$4215 R (Single) High byte of quotient
\$4216 R (Single) Low byte of remainder
\$4217 R (Single) High byte of remainder
All the aforementioned values are unsigned. When the divisor is written to, the division starts. The program will have to wait for 16 cycles (8 NOPs) before being able to read the results.
LDA #\$01
STA \$4204
LDA #\$01 ; Write \$0101 to dividend
STA \$4205
LDA #\$02 ; Write \$02 to the divisor
STA \$4206
NOP ; Wait for 16 machine cycles
NOP
NOP
NOP
NOP
NOP
NOP
NOP
LDA \$4214 ; A = \$80 (result low byte)
LDA \$4215 ; A = \$00 (result high byte)
LDA \$4216 ; A = \$01, as there is a remainder (remainder low byte)
LDA \$4217 ; A = \$00 (remainder high byte)
You can also choose to make a subroutine to do multiplication in the case that \$211B and \$211C are in use because of mode 7. Code for this can be found at 16-bit multiplication and division | 767 | 2,808 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2024-22 | latest | en | 0.746935 |
https://www.physicsforums.com/threads/gravity-acceleration-centrifuge-gr.374346/ | 1,611,849,463,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610704847953.98/warc/CC-MAIN-20210128134124-20210128164124-00170.warc.gz | 874,025,620 | 24,683 | # Gravity ~ Acceleration ~ Centrifuge & GR
Gravity & Acceleration are said to be fairly equivalent.
This includes their respective effects on time dilation doesn't it?
Is artificial gravity generated by centrifugal motion considered to be fairly equivalent to these?
## Answers and Replies
Related Special and General Relativity News on Phys.org
Hi.
Artificial or natural doesn't matter with the time dilation effect of gravity.
Regards.
Thanks sweet.
What happens if one of these effectively cancels out the same type or a different type?
Does that have any effect on the time dilation due to GR?
Examples of same type cancellation:
1. If you are at the gravitational balance point between two planets (or planet & moon)?
2. If you were at the centre of the Earth?
Example of different type cancellation:
2. If you were on a travelator (flat escalator) around the entire Earth diameter that was moving fast enough to make you weightless?
A.T.
Science Advisor
Gravity & Acceleration are said to be fairly equivalent.
This includes their respective effects on time dilation doesn't it?
Is artificial gravity generated by centrifugal motion considered to be fairly equivalent to these?
Clocks at rest in a rotating frame run at different rates along the radial direction. This is equivalent to gravitational-time-dilation.
In the inertial non-rotating frame their desynchronization is blamed on their different speeds which cause different movement-time-dilation.
What happens if one of these effectively cancels out the same type or a different type?
Does that have any effect on the time dilation due to GR?
Examples of same type cancellation:
1. If you are at the gravitational balance point between two planets (or planet & moon)?
2. If you were at the centre of the Earth?
Gravitational time dilation does not cancel out like gravitational force. The time dilation is more related to the gravitational potential than to the gravitational force. In the examples you gave:
1. The gravitational time dilation at the balance point between two planets is greater than an infinity.
2. The gravitational time dilation at the centre of the Earth is greater than an infinity and also greater than at the surface of the Earth. You can calculate this using the interior Schwarzschild solution.
In a very rough visualisation, you can use the rubber sheet model. The depth of the depression of the rubber sheet is roughly proportional to the gravitational time dilation. If you have two masses on the rubber sheet, the sheet is depressed more between the two masses than at the far edges of the sheet.
Note that I am considering the simplified case of a non rotating Earth. The rotation of the Earth effectively increases the time dilation factor at the surface. See comment below about combining gravitational time dilation and velocity time dilation.
Example of different type cancellation:
2. If you were on a travelator (flat escalator) around the entire Earth diameter that was moving fast enough to make you weightless?
This is essentially the same situation as an orbiting satellite. The time dilation on the satellite is the product of time dilation due to the orbital velocity of the satellite and the gravitational time dilation due to the height of the satellite. In the case of low velocities and non extreme gravity as in the case of an Earth satellite this approximates to the sum of the gravitational and velocity time dilation factors.
Thanks Kev.
Gravitational time dilation does not cancel out like gravitational force. The time dilation is more related to the gravitational potential than to the gravitational force. In the examples you gave:
I thought the potential was represented by the slope and not the depth? Is that wrong?
Just wondering because the slope of the balance point between the two planets has a gradient of 0?
1. The gravitational time dilation at the balance point between two planets is greater than an infinity.
2. The gravitational time dilation at the centre of the Earth is greater than an infinity and also greater than at the surface of the Earth. You can calculate this using the interior Schwarzschild solution.
Infinity?
In a very rough visualisation, you can use the rubber sheet model. The depth of the depression of the rubber sheet is roughly proportional to the gravitational time dilation. If you have two masses on the rubber sheet, the sheet is depressed more between the two masses than at the far edges of the sheet.
Fair enough but I just need that clarification about if gravitational potential is represented by the slope or the depth?
Note that I am considering the simplified case of a non rotating Earth. The rotation of the Earth effectively increases the time dilation factor at the surface. See comment below about combining gravitational time dilation and velocity time dilation.
That's fine. I haven't included frame dragging or SR effects just to simplify it for me.
This is essentially the same situation as an orbiting satellite. The time dilation on the satellite is the product of time dilation due to the orbital velocity of the satellite and the gravitational time dilation due to the height of the satellite. In the case of low velocities and non extreme gravity as in the case of an Earth satellite this approximates to the sum of the gravitational and velocity time dilation factors.
I think I read that the GR time dilation is the major component of the time dilation with the SR time dilation only cancelling it a very tiny amount. That's cool. But I just want to understand the non-frame drag GR component a bit better first.
Just as another example.
If you had a centrifugal wheel spinning at 1G and you were in a gravity well near a planet at 1G also then at the 'bottom' of the rotation your net gravity would be 2G and at the top it would be 0G. At the bottom of rotation is your 'potential' going to be 2G or 1G still?
Or, if it is depth related instead of slope related then does this mean that your potential between the 2 planets is the sum of the two opposite potentials at that point? Does this mean that instead of 0 time dilation that it is instead proportional to the sum of the opposite potentials. Or is only the greater potential used as the multiplier?
I thought the potential was represented by the slope and not the depth? Is that wrong?
Just wondering because the slope of the balance point between the two planets has a gradient of 0?
In my crude understanding, "potential" and "potential gradient" are two different things. The potential gradient between the two planets is indeed zero, but it is the potential or depth that is important rather than the slope as far as gravitational time dilation is concerned.
Infinity?
By infinity I was referring to a reference clock that is very far from any gravitational body. In the Schwarzschild metric the coordinate time of a clock "at infinity" is compared to the proper time of a local clock. The "infinity clock" runs at the maximum possible rate and is unaffected by any time dilation. Of course actually getting a clock to infinity is impractical and we can consider the limiting case of a clock very far from any gravitational sources, where the time dilation tends towards zero.
Just as another example.
If you had a centrifugal wheel spinning at 1G and you were in a gravity well near a planet at 1G also then at the 'bottom' of the rotation your net gravity would be 2G and at the top it would be 0G. At the bottom of rotation is your 'potential' going to be 2G or 1G still?
OK, lets clarify the situation a little. Imagine you are far above the North pole and the Earth appears to be rotating anticlockwise. The vertical centrifugal wheel is at the equator. Now if the wheel is rotating clockwise (from your viewpoint above the North pole), the velocity of the wheel at the bottom of the rotation is adding to the velocity of the Earth's rotation so the increased effective velocity is increasing the velocity time dilation factor. The gravitational time dilation is simply what it would normally be at the surface of the Earth and is unaffected by how many g's are experienced by the clock on the wheel. If the wheel is rotating anticlockwise, its velocity at the bottom of the rotation is subtracting from the velocity of the Earth, so there is less velocity time dilation in this case. Note that the total time dilation is dependent on the direction the wheel is rotating in and independent of the number of g's experienced.
To clarify further, consider a collection of different size wheels that all have the same rim velocity, but experience different g forces due to the different radii. All the clocks on the rims of all the wheels will run at the same rate as each other, independent of the g forces measured at the rims of the different wheels.
Or, if it is depth related instead of slope related then does this mean that your potential between the 2 planets is the sum of the two opposite potentials at that point? Does this mean that instead of 0 time dilation that it is instead proportional to the sum of the opposite potentials. Or is only the greater potential used as the multiplier?
If I recall correctly, the total gravitational potential between the two planets is simply the sum of the individual contributions from the two planets and these are not "opposite potentials" and do not cancel out.
Janus
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Science Advisor
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Thanks Kev.
I thought the potential was represented by the slope and not the depth? Is that wrong?
Just wondering because the slope of the balance point between the two planets has a gradient of 0?
potential would be represented by the depth, local strength would be the slope.
Infinity?
I believe he means at at infinity. Gravitational time dilation is measured with respect to an observer an infinite distance from the gravity source.
I think I read that the GR time dilation is the major component of the time dilation with the SR time dilation only cancelling it a very tiny amount.
it depends on the orbit.
Just as another example.
If you had a centrifugal wheel spinning at 1G and you were in a gravity well near a planet at 1G also then at the 'bottom' of the rotation your net gravity would be 2G and at the top it would be 0G. At the bottom of rotation is your 'potential' going to be 2G or 1G still?
Neither, potential isn't measured in gs. Potential is related to how work it would take to lift the object to a different height in the field.
Let's two centrifuges, They have different length arms, and spun such that a clock at the end experiences each feels 1g. The clock on the centrifuge on the longer arm will run slower, even though it feels the same g-force as the other clock. Or you could arrange it so that the speed of the ends of the arms are the same for each centrifuge. In this case both clocks will run at the same rate, even though one will feel a greater g0force than the other.
Or, if it is depth related instead of slope related then does this mean that your potential between the 2 planets is the sum of the two opposite potentials at that point?
One way to look at it would be to use the rubber sheet analogy. The planets make "dips" int he rubber sheet. Between the two dips is a shallow "trough". The depth of the trough at any point represents the potential.
Neither, potential isn't measured in gs. Potential is related to how work it would take to lift the object to a different height in the field.
Let's two centrifuges, They have different length arms, and spun such that a clock at the end experiences each feels 1g. The clock on the centrifuge on the longer arm will run slower, even though it feels the same g-force as the other clock. Or you could arrange it so that the speed of the ends of the arms are the same for each centrifuge. In this case both clocks will run at the same rate, even though one will feel a greater g0force than the other.
Thanks Kev & Janus.
Just a question about yours Janus first because I just need clarification on your answer sorry just because it surprised me.
If you were in 0g space (impossible I imagine but can approach hopefully) inside one of two spinning wheels and one wheel was bigger than the other but they spun with the same G force at the edge then would two equal masses - one on each wheel - weigh the same as each other on those different scales?
Janus
Staff Emeritus
Science Advisor
Gold Member
Thanks Kev & Janus.
Just a question about yours Janus first because I just need clarification on your answer sorry just because it surprised me.
If you were in 0g space (impossible I imagine but can approach hopefully) inside one of two spinning wheels and one wheel was bigger than the other but they spun with the same G force at the edge then would two equal masses - one on each wheel - weigh the same as each other on those different scales?
Yes.
Let's two centrifuges, They have different length arms, and spun such that a clock at the end experiences each feels 1g. The clock on the centrifuge on the longer arm will run slower, even though it feels the same g-force as the other clock. Or you could arrange it so that the speed of the ends of the arms are the same for each centrifuge. In this case both clocks will run at the same rate, even though one will feel a greater g0force than the other.
Relating the above back to gravitational potential, are you saying with the above that:
In case 1 (same g different speed) that it will take different forces to move the same weight an initial tiny identical distance in the same time to observers inside the separate wheels?
In case 2 (different g same speed) that it will take the same force to move the same weight an initial tiny identical distance in the same time to observers inside the separate wheels?
Am I interpreting it right or wrongly?
Just to clarify, I'm talking about the internal wheel observers who don't see themselves as moving just that they are under gravity (albeit it is artificial).
That clarification may be the difference but I'll just check with you.
Potential is related to how work it would take to lift the object to a different height in the field.
One way to look at it would be to use the rubber sheet analogy. The planets make "dips" int he rubber sheet. Between the two dips is a shallow "trough". The depth of the trough at any point represents the potential.
I'm also a little confused about this. For the example where you have two planets you are at the balance point between (or being in a bubble in the centre of the Earth) it takes no more effort to 'move' an object than it would if it were in 0g space.
This is in contrast to the rubber sheet example which would suggest it would be harder to move the object at these points?
Can someone help me with understanding the physical distinction?
So is the amount of space time distortion at a point on the curve, that is due to GR, determined by the slope of the curve that its at or the depth of the point?
Janus
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Science Advisor
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Relating the above back to gravitational potential, are you saying with the above that:
In case 1 (same g different speed) that it will take different forces to move the same weight an initial tiny identical distance in the same time to observers inside the separate wheels?
No
In case 2 (different g same speed) that it will take the same force to move the same weight an initial tiny identical distance in the same time to observers inside the separate wheels?
No
Am I interpreting it right or wrongly?
Just to clarify, I'm talking about the internal wheel observers who don't see themselves as moving just that they are under gravity (albeit it is artificial).
That clarification may be the difference but I'll just check with you.
With the the centrifuges, the potential is equal to the amount of total work needed to move the clock from the edge to the center of the centrifuge. This is what determines what the time dilation will be.
Ich
Science Advisor
So is the amount of space time distortion at a point on the curve, that is due to GR, determined by the slope of the curve that its at or the depth of the point?
In the rubber sheet example:
The depth is the potential. Difference in potential = work per kg = difference in "clock rates".
The slope is gravitational acceleration = gravitational force per kg = work per kg and m lift height. Has no effect on time dilatation (see "clock hypothesis").
Spacetime distortion = curvature = tidal acceleration is the second derivation of depth. Has nothing to do with clock rates either.
Janus
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I'm also a little confused about this. For the example where you have two planets you are at the balance point between (or being in a bubble in the centre of the Earth) it takes no more effort to 'move' an object than it would if it were in 0g space.
This is in contrast to the rubber sheet example which would suggest it would be harder to move the object at these points?
Can someone help me with understanding the physical distinction?
We are talking about the total effort needed to move the object from its point of equilibrium to a point far removed from either planet. Imagine you try and move an object from this position along a line perpendicular to the line joining the planets. The instant you start to move the object, the gravity of the planets will try to pull it back. The total amount of energy needed to lift the object from this "sweet spot" is the potential.
For the hollow of the center of the Earth, it takes work to lift the object from the center to the surface, and then additional energy to lift it to an far distance above the Earth's surface. The far removed clock runs faster than the surface clock, and the surface clock runs faster than the clock at the center.
Now I'm starting to understand it a bit better I think.
For gravity wells it depends upon a point at infinity away because it is only at that point that the potential is 0 because it can be 'lifted' no further.
Then you start running backwards from there and determine the work required to get something from the new measure from point to this original infinity point which has the potential of 0?
Is that correct?
Ich
Science Advisor
You don't need that "potential of 0"; there's no need to define an "absolute clock rate", and it's not always possible. We're always talking about relative clock rates.
Thanks Ich. Sure, I get that. No highest point measurable because it is at infinity. So things are instead measured relative to each other in terms of how deep they are in a space-time well where they are from that points respective infinity. You only need to get the difference between the highest object and the lowest object in space-time for GR purposes.
I can't fathom the underlying physical mechanism that generates this 'instantaneous' relativity at a distance but I accept that it is the process by which time-space relativity is determined.
As everyone is saying it is not weight based in any way, only relative depth based but it is a curiosity to me, I admit, as to how they 'simultaneously' conform to relative depths.
The only physical medium we have is this 'space-time' which I have yet to come to grips with how it has this immaterial pervasive presence.
Space-time is not made of any strange particles is it?
It just is; isn't it?
One way to look at it would be to use the rubber sheet analogy. The planets make "dips" int he rubber sheet. Between the two dips is a shallow "trough". The depth of the trough at any point represents the potential.
Someone on here a little while ago helped me to understand that instead of the 'trough' that you describe that you actually end up with something called a saddle ridge.
That is that the two planets in the example that we are between have their low points shifted towards each other (hence why they move towards each other) but the low points don't combine to create a common trough. Instead a ridge remains between the two low points.
If you add us we create our own little space time dimple. If we are at the balance point then we will sit at the bottom of our dimple and will cause the low gravity points of the two planets to shift a tiny little more towards the common centre; but still without combining.
If we are more to one side of the balance than the other then our little dimple will be a little bit more towards the stronger side than us so we will move towards the stronger side because of this.
But I have another question in relation to that.
I assume you would measure our relative potential as being our height relative to the height of the planets in the overall space-time topology. So that actually makes our potential less than that of the planets. So for us at our point in-between the planets does that mean our clocks due to GR should actually be travelling faster?
Ich
Science Advisor
I can't fathom the underlying physical mechanism that generates this 'instantaneous' relativity at a distance but I accept that it is the process by which time-space relativity is determined.
There is no such mechanism. We were talking about potential; this is not a basic ingredient of GR, it's rather a familiar Newtonian term that can be defined only in a static spacetime in static coordinates. So there is no instantaneous relativity, the terms "potential" and "clock rate" only work with a static background.
Generally, it's all about path length, just as in SR. Two observers take different paths in spacetime, the paths have different length = different elapsed time when they meet again. As long as both observers are at different locations, there is generally no well-defined meaning of simultaneousity or "relative clock rate". It doesn't matter anyway.
turin
Homework Helper
With the the centrifuges, the potential is equal to the amount of total work needed to move the clock from the edge to the center of the centrifuge. This is what determines what the time dilation will be.
Can you describe how you would calculate this? I figure that the amount of work depends on how you move the clock. The clock has some initial kinetic energy due to its circular motion, I'll call it KE. Then, I suppose that we place the clock at the center with zero KE. So, the work done on the clock is W = -KE? I must be missing something. Maybe I'm not allowed to stop the rotation first? Does the Coriolis effect come into play?
Someone on here a little while ago helped me to understand that instead of the 'trough' that you describe that you actually end up with something called a saddle ridge.
That is that the two planets in the example that we are between have their low points shifted towards each other (hence why they move towards each other) but the low points don't combine to create a common trough. Instead a ridge remains between the two low points.
The balance point is the highest point on a transect connecting the centres of the two planets (in the rubber sheet model) but this point is still not as high as points inifinitely far away. In fact if you start at the saddle point at move on a line that is at right angles to the connecting line, you continue to go uphill to a higher (less negative) potential.
But I have another question in relation to that.
I assume you would measure our relative potential as being our height relative to the height of the planets in the overall space-time topology. So that actually makes our potential less than that of the planets.
Your potential is actually greater than that of the planets (but less than the potential at infinity). Potential refers to the potential (possibility) of recovering energy by falling. If you fall your potential energy is converted to kinetic energy which is your falling velocity and when you hit the ground the kinetic energy is converted to sound energy and heat energy in a very rough Newtonian sense.
So for us at our point in-between the planets does that mean our clocks due to GR should actually be travelling faster?
Yes, but it would probably be better to say the clocks are ticking faster than the clocks on the planets (but not as fast as a clock at infinity).
We have been banging on about potential in this thread, as that is probably the easiest to way to visualise things (but bear in mind that the rubber sheet model is a crude analogy) and because the Newtonian term for gravitational potential GM/r appears in the gravitation time dilation factor:
$$\sqrt{1-\frac{2GM}{rc^2}}$$
However as Ich as hinted at, that is not the whole story.
It is interesting to note that the Newtonian expression for escape velocity is:
$$v_e = \sqrt{\frac{2GM}{r}}$$
and when this expression is inserted into the factor above it, the gravitational time dilation factor becomes:
$$\sqrt{1-\frac{v_{e}^2}{c^2}}$$
which has a very obvious analogy with time dilation in Special Relativity.
Could it be that escape velocity is the essential ingredient for determining the time dilation at a given point?
All we have to do now, is figure out what the escape velocity on the rim of a rotating wheel is.
Can you describe how you would calculate this? I figure that the amount of work depends on how you move the clock. The clock has some initial kinetic energy due to its circular motion, I'll call it KE. Then, I suppose that we place the clock at the center with zero KE. So, the work done on the clock is W = -KE? I must be missing something. Maybe I'm not allowed to stop the rotation first? Does the Coriolis effect come into play?
This relates closely to my comments above about the escape velocity. One definition of escape velocity is the velocity a particle requires for its KE to be equal to the gravitational potential energy at its location. Escape velocity is also equal to the terminal velocity of a particle that falls from infinity to a given height. I imagine this "escape velocity" as it applies to a wheel is the velocity a test weight achieves when it slides frictionlessly along a spoke from the centre of the wheel to the rim of the wheel.
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...
Generally, it's all about path length, just as in SR. Two observers take different paths in spacetime, the paths have different length = different elapsed time when they meet again. As long as both observers are at different locations, there is generally no well-defined meaning of simultaneousity or "relative clock rate". It doesn't matter anyway.
How would you define the path lengths in this scenario? On twin descends at a controlled rate deep into a gravitational well. The second twin waits one hundred years and then descends down at the same controlled rate to meet his sibling that is biologically twenty years old. The relative time difference is very physically manifested here.
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Jonathan Scott
Gold Member
Can you describe how you would calculate this? I figure that the amount of work depends on how you move the clock. The clock has some initial kinetic energy due to its circular motion, I'll call it KE. Then, I suppose that we place the clock at the center with zero KE. So, the work done on the clock is W = -KE? I must be missing something. Maybe I'm not allowed to stop the rotation first? Does the Coriolis effect come into play?
In Newtonian gravity for height h in a constant field g, the potential difference, giving the potential energy per mass, is gh. If you divide that by c2, giving gh/c2, you have potential energy per energy, which is the dimensionless fraction by which the time rate changes (assuming that the field is weak enough that the fraction is small compared with 1). That is, a clock which is higher by distance h runs (1+gh/c2) times faster than the lower one.
If you want to treat as rotating system as being like a gravitational field, then for non-relativistic speeds the acceleration g is replaced with $v^2/r$, which is equal to $r \omega^2$ where $\omega$ is the angular velocity). This varies for different r values, so you have to integrate the equivalent of $gh/c^2$ from the centre to the desired radius to get the time dilation fraction.
$$\int_0^r \, \frac{R \omega^2}{c^2} \, dR = \frac{1}{2} \, \frac{r^2 \omega^2}{c^2} = \frac{1}{2} \, \frac{v^2}{c^2}$$
That says your time dilation is approximately $\tfrac{1}{2} (v^2/c^2)$ which is of course the same as the ratio of the classical kinetic energy to the total energy, which is the same as the fraction by which time is dilated according to special relativity for speed v, which is the easier way of doing it. In the relativistic case, the usual full special relativistic time dilation factor applies. | 6,028 | 28,613 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2021-04 | latest | en | 0.939737 |
https://math.stackexchange.com/questions/4126549/recasting-a-dispersion-relation-determinant-0-as-a-matrix-eigenvalue-problem | 1,620,576,813,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243989006.71/warc/CC-MAIN-20210509153220-20210509183220-00609.warc.gz | 405,870,482 | 36,764 | # Recasting a dispersion relation (determinant =0) as a matrix eigenvalue problem for numerical solution
I have a physical problem that involves a dispertion relation between two parameters, the frequency $$\omega$$ and the propagation coefficient $$\beta_{n}$$, where $$n$$ is an integer index. The dispersion equation, whose solution will give us the relation $$\omega=\omega(\beta_{n})$$, is deduced from the system's matrix by requiring that its determinant vanish, $$\det[ M ]=0,$$ where $$M$$ is a Hermitian matrix of order $$2(2N+1)\times 2(2N+1)$$, with $$N$$ being the highest allowed values of the index $$n$$ (basically the truncation of a sum over $$-N\leq n\leq N$$). The elements of the matrix $$M$$ include terms, such as Bessel and Hankel functions, that are functions of $$\omega$$ and $$\beta_{n}$$ (too long and convoluted to practically write here explicitly). Typically such a system is solved numerically for a given $$\omega$$ by forcing the determinant above to vanish and taking the corresponding roots as solutions of $$\beta_{n}$$.
For a given $$\omega$$, when I try to directly solve this determinant =0 equation, to find $$\beta_{n}$$, I run into numerical problems (for example in Mathematica) and so far finding the roots of this determinant directly seem to be hopeless. However, I am aware that a method like Arnoldi's method can find the eigenvalues of a given matrix quickly, and it works for my matrix $$M$$, if I test it with given $$\omega$$ and $$\beta_{n}$$ values.
However, I am not sure how to express my problem as an eigenvalue problem here, so as to take advantage of the Arnoldi method, which gives approximate eigenvalues of the matrix? How can I recast this problem so that the eigenvalues returned would give $$\beta_{n}$$? What is the relation between roots of $$\det[M]$$ and of the eigenvalues of $$M$$? | 466 | 1,858 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 23, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2021-21 | latest | en | 0.904484 |
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# The only gift certificates that a certain store sold
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The only gift certificates that a certain store sold yesterday were worth either $100 each or$10 each. If the store sold a total of 20 gift certificates yesterday, how many gift certificates worth $10 each did the store sell yesterday? (1) The gift certificates sold by the store yesterday were worth a total of between$1,650 and $1,800. (2) Yesterday the store sold more than 15 gift certificates worth$100 each.
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The only gift certificates that a certain store sold yesterday were worth either $100 each or$10 each. If the store sold a total of 20 gift certificates yesterday, how many gift certificates worth $10 each did the store sell yesterday? Say the number of$100 certificates sold was $$x$$, then the number of $10 certificates sold was $$20-x$$. (1) The gift certificates sold by the store yesterday were worth a total of between$1,650 and $1,800 --> $$1,650<100x+10(20-x)<1,800$$ --> $$1,650<90x+200<1,800$$ --> $$1,450<90x<1,600$$ --> $$145<9x<160$$ --> $$16.1<x<17.8$$. Since $$x$$ is an integer then $$x=17$$. Sufficient. (2) Yesterday the store sold more than 15 gift certificates worth$100 each --> x>15. Clearly insufficient.
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Let's say,
Number of gift certificates worth $100 = x Number of gift certificates worth$10 = y
and x+y=20
Using (1),
$$1650<100x+10y<1800$$
or $$165<10x+y<180$$
or $$165<20+9x<180$$ (given, x+y=20)
or $$145<9x<160$$
or $$145/9<x<160/9$$
or $$16.1< x < 17.7$$
Thus, x = 17. Sufficient.
Using (2),
x > 15. Insufficient.
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### Show Tags
25 Jun 2012, 17:46
1
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(1) The gift certificates sold by the store yesterday were worth a total of between $1 ,650 and$1,800.
Suff as only one combination of $100 and$10 GC is possible for total ranging from 1650-1800 i.e 17 ($100) and 3 ($10) total 1730, all other combinations within this range do not meet total number of GC condition
Number of gift certificates worth $10 = y and x+y=20 Statement (1) 1650 < x+y < 1800 This gives us only one possible solution... 1730 Since there has to be 20 gift certificates and x=100$ and y=10$, 1730 is the only number that is between 1650 and 1800 that corresponds with 20 gift certificates Every other solution is out of range e.g. 16 x 100$ + 4 x 10$= 1640$ or 18 x 100$+ 2 x 10$ = 1820$, both out of the range 1650 < x+y < 1800 Statement (2) clearly insufficient since you can have more than one possible solution. _________________ Kudos if you like the post! Failing to plan is planning to fail. Manager Joined: 02 Jun 2011 Posts: 159 Followers: 1 Kudos [?]: 75 [0], given: 11 Re: The only gift certificates that a certain store sold [#permalink] ### Show Tags 26 Jun 2012, 13:52 x+y = 20 st1.) 1650<100x+10y<1800 with ths eqn and x+y = 20 sufficient st.2) x>15 not sufficient Answer A. Math Expert Joined: 02 Sep 2009 Posts: 36552 Followers: 7078 Kudos [?]: 93156 [0], given: 10553 Re: The only gift certificates that a certain store sold [#permalink] ### Show Tags 29 Jun 2012, 03:10 SOLUTION The only gift certificates that a certain store sold yesterday were worth either$100 each or $10 each. If the store sold a total of 20 gift certificates yesterday, how many gift certificates worth$10 each did the store sell yesterday?
Say the number of $100 certificates sold was $$x$$, then the number of$10 certificates sold was $$20-x$$.
(1) The gift certificates sold by the store yesterday were worth a total of between $1,650 and$1,800 --> $$1,650<100x+10(20-x)<1,800$$ --> $$1,650<90x+200<1,800$$ --> $$1,450<90x<1,600$$ --> $$145<9x<160$$ --> $$16.1<x<17.8$$. Since $$x$$ is an integer then $$x=17$$. Sufficient.
(2) Yesterday the store sold more than 15 gift certificates worth \$100 each --> x>15. Clearly insufficient.
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25 Dec 2012, 00:35
Is this a trap question?
That is the intentions of the writer is to make you think that you need 1 and 2 to be able to get the correct answer.
because 2 is obviously not correct and if you weren't able to do 1 properly, you would have to automatically assume that 2 is required.
It seems like it is always advantageous to simply pick the A (or whatever is the harder stem) in these types of questions.
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28 Dec 2012, 17:50
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Yes, this question pretty much wants you to understand that A alone is sufficient, otherwise the fallback answer would be C. St II is obviously not sufficient by itself.
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06 Jul 2014, 02:46
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27 Mar 2016, 18:58
Here is a visual that should help.
Nearly every GMAT question can be solved by algebra, plugging in numbers, both, or a combination of both. If you have time, then it's best to try both.
This particular method is a hybrid of the two: use algebra to get closer to conceptual understanding, then plug in numbers to test.
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The only gift certificates that a certain store sold [#permalink] 27 Mar 2016, 18:58
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# What is the logic to swap two numbers?
## What is the logic to swap two numbers?
Example 1: Program to swap numbers using a temp variable Assigning the value of num1 (this contains the first number) to the temp variable to create the backup of first number. 2. Assigning the value of second variable(this contains the second number) to the variable num1.
How do you swap two elements?
C Program to Swap two Numbers
1. Assign x to a temp variable : temp = x.
2. Assign y to x : x = y.
3. Assign temp to y : y = temp.
How can I swap two numbers without temperature?
Let’s see a simple c example to swap two numbers without using third variable.
1. #include
2. int main()
3. {
4. int a=10, b=20;
5. printf(“Before swap a=%d b=%d”,a,b);
6. a=a+b;//a=30 (10+20)
7. b=a-b;//b=10 (30-20)
8. a=a-b;//a=20 (30-10)
### How do you swap two numbers in a single statement?
First Method:
1. int x = 10;
2. int y = 20;
3. y = (x + y) – (x = y);
4. Console. WriteLine(“X = ” + x + “; Y = ” + y);
How do you swap two numbers in an array?
C Program
1. #include
2. int main()
3. {
4. int x, y, t;// x and y are the swapping variables and t is another variable.
5. printf(“Enter the value of X and Y\n”);
6. scanf(“%d%d”, &x, &y);
7. printf(“before swapping numbers: %d %d\n”,x,y);
8. /*swapping*/
How do you swap two variables?
The bitwise XOR operator can be used to swap two variables. The XOR of two numbers x and y returns a number that has all the bits as 1 wherever bits of x and y differ. For example, XOR of 10 (In Binary 1010) and 5 (In Binary 0101) is 1111 and XOR of 7 (0111) and 5 (0101) is (0010).
#### How do you swap elements in an ArrayList?
In order to swap elements of ArrayList with Java collections, we need to use the Collections. swap() method. It swaps the elements at the specified positions in the list.
How can I check my swap status?
The procedure to check swap space usage and size in Linux is as follows:
1. Open a terminal application.
2. To see swap size in Linux, type the command: swapon -s .
3. You can also refer to the /proc/swaps file to see swap areas in use on Linux.
4. Type free -m to see both your ram and your swap space usage in Linux.
How do you swap in C programming?
Swap Numbers Using Temporary Variable In the above program, the temp variable is assigned the value of the first variable. Then, the value of the first variable is assigned to the second variable. Finally, the temp (which holds the initial value of first ) is assigned to second . This completes the swapping process.
## How do you switch variables in one line?
C program to swap two variables in single line
1. #include
2. void main()
3. {
4. int a = 5;
5. int b = 10;
6. (a ^= b), (b ^= a), (a ^= b);
7. printf(“Swapped values of a and b are %d %d”,a, b);
8. }
How do you swap values between two variables in a single line Python?
In short, the expression: “ a, b = b, a”, first right gets assigned to first left and second right get assigned to second left at the same time therefore swap values of a and b.
How do you swap numbers in an array in C++?
The C++ function std::array::swaps() swap contents of the array. This method takes other array as parameter and exchage contents of the both arrays in linear fashion by performing swap operation on induvisual element of array.
### When do you need to do a ” swap “?
A “SWAP” is when you want to transplant an engine into a vehicle that did not originally come with this type of engine. Generally if your vehicle is pre 1998 than you will be in the “Swap” category and I sell engines for this. Although if your location requires SMOG than I don’t, sorry.
Can you replace a 4.8l engine with a 6.0L?
If you want to replace your 4.8 with a 6.0L and a cam you are still in the “Replacement” category as you do NOT need a complete engine. I do not sell incomplete engines as all you need is a long-block engine as seen here.
Do you need a VIN number for a 4.3L V8?
The 4.3L to V8 “replacement” is bit unique but offers its own set of challenges to be SMOG legal and legit. Your VIN number is embedded in the ECM and needs to be retained. Your VIN number also indicates what engine the vehicle is equipped with. So you need to essential run the new V8 on your existing computer and be retuned.
Ruth Doyle | 1,166 | 4,302 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2023-14 | latest | en | 0.731732 |
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Separation Process Principles- 2n - Seader & Henley - Solutions Manual
# Daltons law find a number of equilibrium stages
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Unformatted text preview: a straight operating line that passes through the operating point (Y0 = 0 , X1 = 0.01) at the bottom of the column, as shown in Fig. 6.11(b) and is drawn tangent to the equilibrium curve, as shown. From Eq. (6-5), the slope of the operating line = L'/V' = 3.0 mol B/mol of steam on a B-free basis. Therefore, (V’')min/L' = 0.33. This is compared to the given operating value of L'/V' = 2.0 or V'/L' = 0.5. Analysis: (continued) (a) Plot of results: (b) Plot of results: Exercise 6.9 (continued) Exercise 6.10 Subject: Stripping of benzene (B) from straw oil by steam at 1 atm (101.3 kPa). Given: Oil enters stripper with 8 mol% benzene. 75% of B is stripped. Pure steam enters. Steam exits with 3 mol% B. Sieve-plate column. Henry's law holds with a B partial pressure of 5.07 kPa when benzene mole fraction in the oil is 10 mol%. Assumptions: Straw oil is not volatile and steam does not condense. Dalton's law. Find: (a) Number of equilibrium stages required. (b) Moles of steam required per 100 moles of benzene-oil feed to stripper. (c) Number of equilibrium stages needed if 85% of B is stripped with same amount of steam as in Part (b). Analysis: With reference to the stripper in Fig. 6.11(b), XN+1 = 8/92 = 0.087 mol B/mol B-free oil Y0 = 0.0 mol B/mol B-free steam YN = 3/97 = 0.0309 mol B/mol B-free steam (b) For 100 moles of benzene-oil feed mixture, have 8 moles of benzene. Amount of benzene stripped = 0.75(8) = 6 moles. Because the exit gas contains 0.0309 mol B/mol B-free steam, the steam rate = 6/0.0309 = 194 moles. Therefore, moles of steam/100 moles of benzene-oil feed = 194 Also, X1 = (8 - 6)/92 = 0.0217 mol B/mol B-free oil (a) Assume Henry's law is given by Eq. (4-32), pB = HxB.. Because pB = 5.07 kPa when xB = 0.10, H = 5.07/0.1 = 50.7 kPa. Convert this to an a Y-X equilibrium equation. For a pressure of 101.3 kPa, assuming Dalton's law, yB = pB/P = 50.7xB/101.3 = 0.5xB . From Eq. (6-1), dropping the benzene subscript, B, and noting that KB = 0.5, KB = 0.5 = Y / (1 + Y ) X / (1 + X ) Solving, Y = 0.5 X 1 + 0.5 X (1) The Y-X plot includes the equilibrium curve given by Eq. (1), and a straight operating line that connects the column terminal points (Y0 = 0.0, X1 = 0.0217) and (YN = 0.0309, XN+1 = 0.087). The corresponding slope of the operating line is (0.0309 - 0.0)/(0.087 - 0.0217) = L'/G' = 0.474. From the plot below, stepping off stages as in Fig. 6.11(b), required N = approximately 3 stages. (c) For 85% stripping of B, amount of B stripped = 0.85(8) = 6.8 moles B. Not stripped is1.2 moles B. Therefore, for the same B-oil and steam feeds, X1 = 1.2/92 = 0.01304 mol B/mol oil and YN =6.8/194 = 0.03505 mol B/mol steam. The Y-X plot for this case is also given below, where the equilibrium curve is the same as in Part (a), and a straight operating line connects the column terminal points (Y0 = 0.0, X1 = 0.01304) and (YN = 0.03505, XN+1 = 0.087). The corresponding slope of the operating line is (0.03505 - 0.0)/(0.087 - 0.01304) = L'/G' = 0.474 (same as in Part (a)). From the plot below, stepping off stages as in Fig. 6.11(b), required N = between 5 and 6 stages....
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http://www.w3resource.com/python-exercises/numpy/python-numpy-exercise-71.php | 1,503,414,032,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886110792.29/warc/CC-MAIN-20170822143101-20170822163101-00213.warc.gz | 697,375,358 | 7,125 | Python NumPy: Create and display every element of an numpy array in Fortran order - w3resource
# Python NumPy: Create and display every element of an numpy array in Fortran order
## Python NumPy: Array Object Exercise-71 with Solution
Write a Python program to create and display every element of an numpy array in Fortran order.
Sample Solution:-
Python Code:
``````import numpy as np
x = np.arange(12).reshape(3, 4)
print("Elements of the array in Fortan array:")
for x in np.nditer(x, order="F"):
print(x,end=' ')
print("\n")
```
```
Sample Output:
```Elements of the array in Fortan array:
0 4 8 1 5 9 2 6 10 3 7 11
```
Python Code Editor:
```import numpy as np
x = np.arange(12).reshape(3, 4)
print("Elements of the array in Fortan array:")
for x in np.nditer(x, order="F"):
print(x,end=' ')
print("\n")```
Improve this sample solution and post your code through Disqus | 248 | 887 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2017-34 | latest | en | 0.624595 |
https://republicofsouthossetia.org/question/the-original-price-of-a-shirt-was-20-it-was-decreased-to-15-what-is-the-percent-decrease-of-the-16098669-83/ | 1,638,370,551,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964360803.6/warc/CC-MAIN-20211201143545-20211201173545-00257.warc.gz | 543,348,981 | 13,009 | ## The original price of a shirt was \$20. It was decreased to \$15. What is the percent decrease of the price of this shirt?
Question
The original price of a shirt was \$20. It was decreased to \$15. What is the percent decrease of the price of this shirt?
in progress 0
1 week 2021-11-22T10:03:40+00:00 1 Answer 0 views 0
25%
Step-by-step explanation:
20 – 15 = 5
5 / 20 = .25
.25 x 100 / 100 =
25/100
25%
Feel free to let me know if you need more help! 🙂 | 150 | 471 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.96875 | 4 | CC-MAIN-2021-49 | latest | en | 0.931802 |
http://www.algebra.com/cgi-bin/show-question-source.mpl?solution=44216 | 1,371,542,492,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368707186142/warc/CC-MAIN-20130516122626-00089-ip-10-60-113-184.ec2.internal.warc.gz | 251,038,998 | 773 | ```Question 63527
X^2+9X+3=0 THE QUADRATIC EQUATION IS
X=(-B+-SQRT[B^2-4AC])/2A
X=(-9+-SQRT[81-4*1*3])/2*1
X=(-9+-SQRT[81-12])/2
X=(-9+-SQRT69)/2
X=(-9+-8.3)/2
X=(-9+8.3)/2
X=-.7/2
X=-.35 SOLUTION
X=(-9-8.3)/2
X=-17.3/2
X=8.65 SOLUTION``` | 142 | 238 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2013-20 | latest | en | 0.173428 |
https://www.sarthaks.com/331095/elevation-in-the-boiling-point-for-1-molal-solution-of-glucose-is-2k?show=331125 | 1,560,656,146,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627997533.62/warc/CC-MAIN-20190616022644-20190616044644-00075.warc.gz | 899,422,385 | 11,958 | # Elevation in the boiling point for 1 molal solution of glucose is 2K.
43 views
Elevation in the boiling point for 1 molal solution of glucose is 2K. The depression in the freezing point for 2 molal solution of glucose in the same solvent is 2K. The relation between Kb and Kf is
(1) Kb = 0.5 K
(2) Kb = 2 Kf
(3) Kb = Kf
(4) Kb = 1.5 Kf
+1 vote
selected
The correct option is (2).
Explanation: | 131 | 403 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-26 | latest | en | 0.898485 |
https://www.winlife.com.au/t0x23gb/752eb3-sympy-greek-symbols | 1,620,787,991,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991693.14/warc/CC-MAIN-20210512004850-20210512034850-00334.warc.gz | 1,132,832,641 | 11,041 | SymPy expressions are built up from symbols, numbers, and SymPy functions. For example if we use the GA module function make_symbols() as follows: For example, the code $\int_a^b f(x) = F(b) - F(a)$ renders inline as ∫abf(x)dx=F(b)−F(a). \frac{\partial f_2}{\partial x_1} & \frac{\partial f_2}{\partial x_2} & ~ \\ sympy.abc does not contain the name foo. Write an expression representing the wave equation in one dimension: $${\partial^2 u\over \partial t^2 } = c^2 { \partial^2 u\over \partial x^2}.$$ Remember that $u$ is a function in two variables. The simplest kind of expression is the symbol. Typing Greek letters with Keyboard Shortcuts To insert Greek letter type Ctrl+G ( Command G on Mac OS ) and then type Latin letter mentioned in the table below. \vdots & ~ & \ddots \\ A matrix can contain any symbolic expression. function import _coeff_isneg, AppliedUndef, Derivative: ... greek_letters_set = frozenset (greeks) _between_two_numbers_p = (re. You can also use symbols('i') instead of Idx('i'). Here are the examples of the python api sympy.symbols taken from open source projects. Indexed symbols can be created with IndexedBase and Idx. values for s in symbols: if s is None: return # common symbols not present! The next step down would be to define the R variables but not make them match the names of the SymPy symbols (so, maybe they’re var1, var2, etc — easily predictable). SymPy objects; _clash2 defines the multi-letter clashing symbols; SymPy - Symbols Symbol Symbols () C, O, S, I, N, E {'C': C, 'O': O, 'Q': Q, 'N': N, 'I': I, 'E': E, 'S': S} {'beta': beta, 'zeta': zeta, 'gamma': gamma, 'pi': pi} (a0, a1, a2, a3, a4) (mark1, mark2, mark3) and _clash is the union of both. However, for Greek letters there are issues. For instance, >>> x, y, z = symbols(’x y z’) creates three symbols representing variables named x, y, and z. objects for those names. In SymPy, we have objects that represent mathematical symbols and mathematical expressions (among other things). All SymPy expressions are immutable. from sympy.abc import foo will be reported as an error because alphabets import greeks: from sympy. until the next SymPy upgrade, where sympy may contain a different set of In this particular instance, for different ways to create a Matrix. To get a symbol named foo, Since most languages targeted will not support symbolic representation it is useful to let SymPy evaluate a floating point approximation (up to a user specified number of digits). This module exports all latin and greek letters as Symbols, so you can If you import them i, j = symbols('i j') Multiple symbols can be defined with symbols() method. If you are dealing with a differential equation, say: SymPy's dsolve can (sometimes) produce an exact symbolic solution. These restrictions allow sympy variable names to represent complex symbols. def _print_Derivative (self, expr): """ Custom printing of the SymPy Derivative class. Created using. ����� SymPy also has a Symbols()function that can define multiple symbols at once. As of the time of writing this, the names C, O, S, I, N, The return is a list of dictionaries, mapping symbols to solutions. Write an Indexed expression for $$A[i, j, k]$$. The printers then try to give an appropriate representation of these objects. def pretty_try_use_unicode (): """See if unicode output is available and leverage it if possible""" try: symbols = [] # see, if we can represent greek alphabet symbols. SymPy symbols are created with the symbols() function. However, if you need more symbols, then your can use symbols(): >>> from both sympy.abc and sympy, the second import will “win”. during sympification if one desires Symbols rather than the non-Symbol Write a matrix expression representing $$Au + Bv,$$ where $A$ and $B$ are $100\times 100$ and $u$ and $v$ are $100 \times 1$. from sympy import Basic, Function, Symbol from sympy.printing.str import StrPrinter class CustomStrPrinter (StrPrinter): """ Examples of how to customize the StrPrinter for both a SymPy class and a user defined class subclassed from the SymPy Basic class. """ from sympy.abc import x, y Symbols can be imported from the sympy.abc module. Undefined functions are created with Function(). Alt-Codes can be typed on Microsoft Operating Systems. Greek alphabet letters & symbols (α,β,γ,δ,ε,...) Greek alphabet letters & symbols Greek alphabet letters are used as math and science symbols. Sympy has a quick interface to symbols for upper and lowercase roman and greek letters: Hence, instead of instantiating Symbol object, this method is convenient. a = Symbol('a') b = Symbol('b') They can be defined with Symbol. >>> sym.pi**2 pi**2 >>> sym.pi.evalf() 3.14159265358979 >>> (sym.pi + sym.exp(1)).evalf() 5.85987448204884. as you see, evalf evaluates … Now take the Jacobian of that matrix with respect to your column vector, to get the original matrix back. SymPy objects can also be sent as output to code of various languages, such as C, Fortran, Javascript, Theano, and Python. names. In SymPy's abc module, all Latin and Greek alphabets are defined as symbols. A useful tool in your toolbelt when manipulating expressions is the solve function. That way, some special constants, like , , (Infinity), are treated as symbols and can be evaluated with arbitrary precision: >>>. This module does not define symbol names on demand, i.e. Later you can reuse existing symbols for other purposes. Beta. Matrices support all common operations, and have many methods for performing operations. core. In [3]: alpha1, omega_2 = symbols('alpha1 omega_2') alpha1, omega_2. In from sympy.abc import ..., you are following a file path: python fetches the module abc.py inside sympy/. The module also defines some special names to help detect which names clash >>> from sympy import symbols >>> x,y,z=symbols ("x,y,z") In SymPy's abc module, all Latin and Greek alphabets are defined as symbols. There are a couple of special characters that will combine symbols. SymPy is an open source computer algebra system written in pure Python. Last updated on Dec 12, 2020. Undefined are useful to state that one variable depends on another (for the purposes of differentiation). J = \begin{bmatrix} Some matrix expression functions do not evaluate unless you call doit. Like in Numpy, they are typically built rather than passed to an explicit constructor. SymPy version 1.0 officially supports Python 2.6, 2.7 and 3.2 3.5. See Matrix? Square root is sqrt. more readable. These can be passed for locals you still need to use Symbol('foo') or symbols('foo'). conveniently do, instead of the slightly more clunky-looking. _clash1 defines all the single letter variables that clash with >>> from sympy import symbols >>> x,y,z=symbols("x,y,z") In SymPy's abc module, all Latin and Greek alphabets are defined as symbols. with the default SymPy namespace. Enclose LaTeX code in dollar signs $...$ to display math inline. Use ** for powers. code such as interactive sessions and throwaway scripts that do not survive To get a symbol named foo, you still need to use Symbol ('foo') or symbols ('foo'). E, and Q are colliding with names defined in SymPy. sticking with one and only one way to get the symbols does tend to make the code Basic Operations, x, y, z = symbols("x y z") To numerically evaluate an expression with a Symbol at a point, we might use subs followed by evalf , but it is more efficient and SymPy - Symbols Symbol . SymPy automatically pretty prints symbols with greek letters and subscripts. \end{bmatrix} \frac{\partial f_1}{\partial x_1} & \frac{\partial f_1}{\partial x_2} & \cdots \\ To make life easier, SymPy provides several methods for constructing symbols. SymPy expressions are built up from symbols, numbers, and SymPy functions, In [2]: x, y, z = symbols('x y z') SymPy automatically pretty prints symbols with greek letters and subscripts. If you want all single-letter and Greek-letter variables to be symbols then you can use the clashing-symbols dictionaries that have been defined there as private variables: _clash1 (single-letter variables), _clash2 (the multi-letter Greek names) or _clash (both single … Alpha. Write a symbolic expression for $$\frac{1}{\sqrt{2\pi\sigma^2} } \; e^{ -\frac{(x-\mu)^2}{2\sigma^2} }.$$ Remember that the function for $e^x$ is exp(x). Symbols : Lyre, Laurel wreath, Python, Raven, Bow and Arrows. These characteristics have led SymPy to become a popular symbolic library for the scientific Python ecosystem. Functions that operate on an expression return a new expression. Solve the following ODE: $$f''(x) + 2f'(x) + f(x) = \sin(x)$$, $$\left ( \alpha_{1}, \quad \omega_{2}\right )$$, $$\sin{\left (x + 1 \right )} - \cos{\left (y \right )}$$, $$- \sin{\left (y \right )} \cos{\left (x + 1 \right )}$$, $$\left[\begin{matrix}1 & 2\\3 & 4\end{matrix}\right]$$, $$\left[\begin{matrix}1\\2\\3\end{matrix}\right]$$, $$\left[\begin{matrix}x\\y\\z\end{matrix}\right]$$, $$\left[\begin{matrix}x + 2 y\\3 x + 4 y\end{matrix}\right]$$, $$\left[\begin{matrix}\cos{\left (x \right )} & 1 & 0\\1 & - \sin{\left (y \right )} & 0\\0 & 0 & 1\end{matrix}\right]$$, $$\left [ - \frac{3}{2} + \frac{\sqrt{21}}{2}, \quad - \frac{\sqrt{21}}{2} - \frac{3}{2}\right ]$$, $$\left [ \left ( \frac{2}{5} + \frac{\sqrt{19}}{5}, \quad - \frac{2 \sqrt{19}}{5} + \frac{1}{5}\right ), \quad \left ( - \frac{\sqrt{19}}{5} + \frac{2}{5}, \quad \frac{1}{5} + \frac{2 \sqrt{19}}{5}\right )\right ]$$, $$f{\left (x \right )} = C_{1} \sin{\left (x \right )} + C_{2} \cos{\left (x \right )}$$, # An unnested list will create a column vector. SymPy symbols are created with the symbols () function. If you see utf-8, then your system supports unicode characters.To print any character in the Python interpreter, use a \u to denote a unicode character and then follow with the character code. Create the following matrix $$\left[\begin{matrix}1 & 0 & 1\\-1 & 2 & 3\\1 & 2 & 3\end{matrix}\right]$$, Now create a matrix representing $$\left[\begin{matrix}x\\y\\z\end{matrix}\right]$$ and multiply it with the previous matrix to get $$\left[\begin{matrix}x + z\\- x + 2 y + 3 z\\x + 2 y + 3 z\end{matrix}\right].$$. String contains names of variables separated by comma or space. you will come across this mathematical entity in later notebooks in this tutorial. It is built with a focus on extensibility and ease of use, through both interactive and programmatic applications. MatrixSymbol("M", n, m) creates a matrix $M$ of shape $n \times m$. ... Mul, Number, S, Symbol: from sympy. For example: renders as f′(a)=limx→af(x)−f(a)x−a See the LaTeX WikiBook for more information (especially the section on mathematics). It exports all latin and greek letters as Symbols, so we can conveniently use them. Letter symbol α. SymPy can also operate on matrices of symbolic dimension ($n \times m$). Enclose LaTeX code in double dollar signs $$...$$to display expressions in a centered paragraph. On the other hand, sympy.abc is the attribute named 'abc' of the module object sympy. from sympy import init_printing, symbols, ln, diff >>> init_printing >>> x, y = symbols ('x y') >>> f = x ** 2 / y + 2 * x-ln (y) >>> diff (f, x) 2⋅x ─── + 2 y >>> diff (f, y) 2 x 1 - ── - ─ 2 y y >>> diff (diff (f, x), y)-2⋅x ──── 2 y core. Letter symbol β. Gamma. The help on inserting Greek letters and special symbols is also available in Help menu. You can give solve an Eq, or if you give it an expression, it automatically assumes that it is equal to 0. Gallery/Store Hours: Wednesday to Saturday 10 am to 4 pm. © Copyright 2020 SymPy Development Team. Greek Letters. Then you don’t need to worry about making sure the user-supplied names are legal variable names for R. I could name a symbol something like: symbol = Symbol('(a**2+b**2)**(-1/2)') but that is not a common way to represent symbols. $$. String contains names of variables separated by comma or space. By voting up you can indicate which examples are most useful and appropriate. The programs shows three ways to define symbols in SymPy. Derivatives are computed with the diff() function, using the syntax diff(expr, var1, var2, ...). Extended Symbol Coding¶. """ self.in_vars = sympy.symbols(in_vars) self.out_vars = sympy.symbols(out_vars) if not isinstance(self.in_vars, tuple): self.in_vars = (self.in_vars,) if not isinstance(self.out_vars, tuple): self.out_vars = (self.out_vars,) self.n_in = len(self.in_vars) self.n_out = len(self.out_vars) self.all_vars = list(self.in_vars) + list(self.out_vars) self.eqns_raw = {} # raw string equations self.eqns_fn = {} # … Hephaestus Symbol. encoding = getattr (sys. The most low-level method is to use Symbol class, as we have been doing it before. containers import Tuple: from sympy. You can freely mix usage of sympy.abc and Symbol/symbols, though For instance, the code for β is 03B2, so to print β the command is print('\u03B2').. Out … You will need to create symbols for sigma and mu. You can freely mix usage of sympy.abc and Symbol / symbols, though sticking with one and only one way to get the symbols does tend to make the code more readable. >>> from sympy.abc import x,y,z However, the names C, O, S, I, N, E and Q are predefined symbols. This is an issue only for * imports, which should only be used for short-lived ^ is the XOR operator. The function init_printing() will enable LaTeX pretty printing in the notebook for SymPy expressions. Matrices are created with Matrix. We recommend calling it at the top of any notebook that uses SymPy. solve solves equations symbolically (not numerically). SymPy canonical form of … values ()) # and atoms symbols += atoms_table. Like solve, dsolve assumes that expressions are equal to 0. Dividing two integers in Python creates a float, like 1/2 -> 0.5. It can also handle systems of equations. Contribute to sympy/sympy development by creating an account on GitHub. In Greek mythology Hephaestus was the god of fire and forging, the husband of … Here we give a (quick) introduction to SymPy. As far as I understand the documentation, all of these are equivalent: x = symbols("x") # or @vars x, Sym("x"), or Sym(:x) And that indeed works for "x". SymPy uses Unicode characters to render output in form of pretty print. You can represent an equation using Eq, like. One of the main extensions in latex_ex is the ability to encode complex symbols (multiple greek letters with accents and superscripts and subscripts) is ascii strings containing only letters, numbers, and underscores. IndexedBase("A") represents an array A and Idx('i') represents an index i. 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It is built with a differential equation, say: SymPy 's abc module, all latin and greek are... Mul, Number, use rational ( 1 ) /2 of the Python api sympy.symbols taken from source... Through the symbols ( 'foo ' ) not contain the name foo return a new expression syntax diff )... In help menu led SymPy to become a popular symbolic library for the purposes of differentiation ) function... Sympy.Abc import x, y symbols can be created with the symbols ( 'alpha1 omega_2 ). Can also use symbols ( ' a ' ) b = Symbol ( 'foo ' ) detect which clash.... Mul, Number, use rational ( 1 ) /2 typically built rather than the non-Symbol for! By comma or space on matrices of symbolic dimension ($ n \times m $of shape$ n m! Are dealing with a differential equation, say: SymPy 's abc module, all and! The second import will “ win ” notebooks in this tutorial assumes you are already familiar with expressions! Creates a float, like 1/2 - > 0.5 solve an Eq, 1/2... 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( for the purposes of differentiation ) to your column vector, to get the original matrix back Hours Wednesday. ) method printing in the notebook for SymPy expressions double dollar signs $a. ) introduction to SymPy defines some special names to represent complex symbols the SymPy Derivative class open source projects an... Common symbols not present examples are most useful and appropriate, numbers, and have many for. Expression functions do not evaluate unless you call doit to 4 pm by creating an account GitHub. Passed for locals during sympification if one desires symbols rather than passed to explicit., Bow and Arrows two integers in Python creates a matrix$ m $of shape n. Among other things ) values ( ) ) # and atoms symbols += atoms_table to. Alpha1, omega_2 if one desires symbols rather than passed to an explicit.. Command is print ( '\u03B2 ' ) or S ( 1, 2 ) or (... Them from both sympy.abc and SymPy functions want a rational Number, use (. ( self, expr ): '' '' Custom printing of the SymPy Derivative class in notebook! Lyre, Laurel wreath, Python, Raven, Bow and Arrows available in help menu if one symbols... Import x, y symbols can be defined with symbols ( ' '. Get a Symbol named foo, you still need to use Symbol class, as we have been it... An equation using Eq, like 1/2 - > 0.5 sympy.abc does contain... Index i a Symbol named foo, you still need to use class. Python, Raven, Bow and Arrows pretty prints symbols with greek as! ( greeks ) _between_two_numbers_p = ( re ) function in SymPy, we have objects represent. Your toolbelt when manipulating expressions is the attribute named 'abc ' of the object. Eq, or if you import them from both sympy.abc and SymPy, code! To create symbols for sigma and mu 's abc module, all latin and greek alphabets are as... By voting up you can reuse existing symbols for other purposes dictionaries, mapping to! Original matrix back Multiple symbols in a single function call to 4 pm as we have objects represent... The printers then try to give an appropriate representation of these objects integers in Python creates float! And subscripts the second import will “ win ” does not contain name... ) function = Symbol ( ' i ' ) b = Symbol ( i... Second import will “ win ” printing of the SymPy Derivative class: to. Dimension sympy greek symbols$ n \times m $of shape$ n \times m $of shape$ \times... That matrix with respect to your column vector, to get a Symbol named,... Done through the symbols function, which may create Multiple symbols can be defined symbols. To make life easier, SymPy provides several methods for performing operations have objects that mathematical! Instance, the code for β is 03B2, so this notebook should as! You still need to create symbols for sigma and mu in double dollar signs ... [. Built up from symbols, numbers, and have many methods for constructing symbols can reuse symbols. Foo will be reported as an error because sympy.abc does not define Symbol on! Through both interactive and programmatic sympy greek symbols pretty prints symbols with greek letters and special is... It automatically assumes that it is equal to 0 vector, to the... Them from both sympy.abc and SymPy, the code for β is 03B2, so to print β the is! Are created with the symbols ( ) will enable LaTeX pretty printing in the notebook for SymPy,! Double dollar signs voting up you can also operate on matrices of symbolic dimension ( \$ \times!:... greek_letters_set = frozenset ( greeks ) _between_two_numbers_p = ( re one variable depends on another ( for purposes. Symbols is also available in help menu tool in your toolbelt when manipulating expressions the... Function, which may create Multiple symbols in a centered paragraph so we conveniently... '' '' Custom printing of the Python api sympy.symbols taken from open source.! ) instead of instantiating Symbol object, this method is convenient three ways define... Also use symbols ( ' i ' ) Multiple symbols can be defined with symbols ( ' '. Expression return a new expression a '' ) represents an index i things ) render in... Contain the name foo be passed for locals during sympification if one desires rather! Dictionaries, mapping symbols to solutions... Mul, Number, use rational ( 1 2! In [ 3 ]: alpha1, omega_2 = symbols ( ) will enable LaTeX printing... To get a Symbol named foo, you still need to use class... With symbols ( ) function, using the syntax diff ( expr, var1, var2,... ) an. | 5,824 | 23,545 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2021-21 | latest | en | 0.724905 |
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1. ## Make x the subject
Hi
Where am I going wrong in trying to rearrange this:
$\displaystyle \frac{x+a}{x+b} = c$
$\displaystyle x+a = c(x+b)$
$\displaystyle \frac{x+a}{c} = x+b$
$\displaystyle \frac{x+a}{c}-b = x$
$\displaystyle x+a - cb = xc$
$\displaystyle a - cb = xc - x$
$\displaystyle \frac{a-cb}{c} = x - x$
Thanks
2. Originally Posted by preid
Hi
Where am I going wrong in trying to rearrange this:
$\displaystyle \frac{x+a}{x+b} = c$
$\displaystyle x+a = c(x+b)$
$\displaystyle \frac{x+a}{c} = x+b$
$\displaystyle \frac{x+a}{c}-b = x$
$\displaystyle x+a - cb = xc$
$\displaystyle a - cb = xc - x$
Ok thus far, although a bit longwinded. But now factor out x on the right side first:
$\displaystyle a-cb=x(c-1)$
and divide both sides by $\displaystyle c-1$ to get:
$\displaystyle \frac{a-cb}{c-1}=x$ | 285 | 860 | {"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} | 4.09375 | 4 | CC-MAIN-2018-43 | latest | en | 0.5818 |
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# How to make an array of dates in Crystal
I need to check for holidays in my report. I put this formula in the report header.
I have tried to make an array of dates in Crystal as follows:
Local DateVar Array Holidays:= MakeArray(Date(2011,01,01),Date(2011,12,25))
I get an error message that says "Result of formula cannot be an array"
How do I format the formula to create the array?
0
DSIGRIS
• 4
• 3
1 Solution
Commented:
Try it this way
Local DateVar Array Holidays:= [Date(2011,01,01),Date(2011,12,25)];
""
Crystal returns the value evaluated in the last line as the result of the formula.
mlmcc
0
Commented:
Not sure what you are doing but here is a formula that does the Number of WorkdDays between 2 dates and account for holidays
mlmcc
0
Author Commented:
I was trying to implement your solution for calculating the difference between two dates.
//Holiday Listing formula to go into the report header of the report.
DateVar Array Holidays := [Date (2003,12,25), Date (2003,12,31)];
But this gives me an error.
0
Commented:
You need a final line that is a scalar value like ""
//Holiday Listing formula to go into the report header of the report.
DateVar Array Holidays := [Date (2003,12,25), Date (2003,12,31)];
" "
mlmcc
0
Author Commented:
Why is that?
0
Commented:
Crystal returns the value from the last statement it executes or evaluates as the value of the formula. In your case the declaration of the array so it tries to return the array as the value which is not allowed.
//Holiday Listing formula to go into the report header of the report.
DateVar Array Holidays := [Date (2003,12,25), Date (2003,12,31)];
numberVar HolidayCount := 2;
That would be fine because it would return 2 as the value.
Formulas on the report need to display something.
mlmcc
0
Author Commented:
Great Job thanks.
0
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A) How many ways can we arrange 12 tiles so that they form a rectangle?
B) Get your calculator and try the following:
What do you get?
C) The A line bus runs every 9 minutes and the B line every 12 minutes. If quite in line at the stop, how much it will take to re-match?
EASY
D) Ana has a deck of cards and can make groups of 2, 3, and 5 without a remainder letter, what are the possible numbers of cards Her can have ?.
E) We have a type of brick of dimensions:
How many bricks needed to build a cube?
F) A type of cicada has a parasite whose life cycle is between 2, 3 and 5 years, this means that when the season ends and the parasite dies may take 2 3 or 5 years to reappear. What should be the life cycle of the cicada to not match this? (Based on a real case).
MEDIUM
G) Divide the clock into 6 parts so that all its parts add up the same.
E) Participants in a parade can be in ordered in groups of 3, 5 and 25 but can not do in groups of 4 or 9.
What is the number of participants if we know that between 1000 and 1250?
H) Given this imaginary situation of love:
How many laps will have to give each wheel for lovers touch again?.
HARD
I) How to transport the 12 dogs and 18 cats so that all cages have to carry the same number of animals and cages are the largest possible.
Note: No sane enough mix dogs and cats.
J) We have a balance and 8 balls of the same weight except for one that weighs more.
How many weighs have to do to know for sure which is the different ball?
VERY HARD
K) Suppose x is any prime greater to 3. Show that its square gives a remainder of 1 when divided by 12. | 423 | 1,625 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2018-22 | latest | en | 0.934114 |
https://web2.0calc.com/questions/please-help_68490 | 1,611,061,222,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703518240.40/warc/CC-MAIN-20210119103923-20210119133923-00245.warc.gz | 613,951,655 | 5,621 | +0
Please help.
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149
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+112
If t is a real number, what is the maximum possible value of the expression -t^2 + 8t -4?
Jun 27, 2020
3+0 Answers
#1
0
Here's a hint: Complete the square.
Jun 27, 2020
#2
+112
0
Ok, thanks
QuestionsBug Jun 27, 2020
#3
+28025
+1
Due to the leading negative on the t^2 term, this is a dome shaped parabola
Maximum will occu at t = -b/2a b = 8 a = -1
use this value of 't' in the equation to find the maximum value ....
Jun 27, 2020 | 183 | 486 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-04 | latest | en | 0.753741 |
https://blog.jpolak.org/?p=1981 | 1,566,223,724,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027314752.21/warc/CC-MAIN-20190819134354-20190819160354-00093.warc.gz | 398,019,511 | 6,926 | # Sum of squares a square?
The sum of squares of two integers can be an integer. For example, $3^2 + 4^2 = 25^2$. Those are also the legs of a right-angle triangle. This example is special in that the numbers $3$ and $4$ are consecutive integers!
Can you find a sum of squares of three consecutive integers that make a square?
No, it's not possible. If $n$ is any integer, then the sum
$$n^2 + (n+1)^2 + (n+2)^2 = 3n^2 + 6n + 5.$$
Now, modulo three, any number squared is either 0 or 1. But this polynomial always outputs integers that are 2 mod 3. Therefore, the sum of three consecutive integers cannot be a perfect square.
Exercise: prove that the sum of squares of four consecutive integers cannot be a perfect square.
Maybe, you might be inclined to think that 2 is somehow special, and that a sum of squares of $k$ consecutive integers cannot be a perfect square if $k \gt 2$. But then you would be surprised at this sum of squares of fifty consecutive integers:
$$7^2 + 8^2 + 9^2 + \cdots + 55^2 + 56^2 = 60025 = 245^2.$$
Cool, right? This is by no means the smallest example. Here is another example of 407 consecutive integers:
$$183^2 + 184^2 + \cdots + 589^2 = 8140^2.$$
But are there infinitely many such examples? | 350 | 1,233 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2019-35 | latest | en | 0.903813 |
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