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https://timedate.org/if-today-is-friday-before-99-days
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# If Today Is Friday, what was the day before 99 days
If today is friday,what was the date before 99 days ago ?
Thursday
## How do we solve it ?
A week has 7 days " Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday ".
99 / 7 = 14 ( Remainder = 1)
So you can say that before multiples of 7 days, the day will repeat itself as friday.
Hence before 98 days Friday would occur again.
So we calculate before the remainder (1) , the answer is Thursday.
Simple Logic = And whatever the remainder obtained is, deduct that to the day.
### Weekdays
If 99 days for other weekdays.
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## Einstein
Why Does E=mc2? Brian Cox & Jeff Forshaw
1. Space & Time
2. The Speed of Light
3. Special Relativity
4. Spacetime
5. Why Does E=mc2?
6. And Why Should We Care?
7. The Origin of Mass
8. Warping Spacetime
The objective is to describe Einstein’s theory of space & time
Energy & mass are interchangeable & the speed of light is the exchange rate
1. Space & Time
a. Hermann Minkowski (Einstein’s teacher) thought of space & time being blended
b. E=mc2 Belongs to special relativity
c. Speed of light (c) = 299,792,458 metres per second
i. Sun 8 mins
ii. Width of Milky Way = 100,000 light years
iii. Andromeda is 2 million light years away
iv. Edge of the observable universe is > 10 billion light years away
v. Earth moves at about 67,000 mph around the sun
vi. Sun travels at 486,000 mph around the Milky Way (226m years to orbit once)
d. C is the cosmic speed limit
e. Galileo’s principle of relativity – there is no such thing as absolute motion
2. The Speed of Light
a. Faraday showed that magnetism & electricity are connected
b. Maxwell’s waves showed the relationship between electric & magnetic fields travel at c – there is a link btw electromagnetic fields & c; c is a constant of nature
c. 1881 Michelson & Morley experiment showed there to be no difference in c in any direction at any time of the year
d. That we can never catch up with a beam of light no matter how fast we travel infers that there is no absolute time
3. Special Relativity
a. Light is a symbiosis of electric & magnetic fields
b. 2 axioms:
i. Light always travels through empty space at the same speed regardless of the motion of the source or the observer
ii. No experiment can ever be performed that is capable of identifying absolute motion
c. Thought experiment of the light clock (two mirrors bouncing a beam of light; the mirrors are 1m apart so the light takes 6.67 nanoseconds for a round trip)
i. If light clock on a train & observed from a station the light has to travel further for the station observer. The Newtonian world says the light speeds up due to the motion of the train, so the same time for an observer on the station compared to one on a train. But if light cannot speed up it will take longer from the perspective of the station observer. This implies time passes at different rates depending upon how we are moving.
ii. The clock slows by an amount known as γ (gamma) = 1/(1-(v2/c2))
γ is always >1 because v/c will be <1 (i.e. the speed of the clock will be <c)
When v is very small relative to c, γ is close to 1
At speeds of 90% of c, the time stretching factor is >2, i.e. time would be halved
iii. Experimental proof comes from muons. At rest they live for 2.2 micro-seconds before turning into an electron & a pair of neutrinos
At a speed of 99.94% of c their life extended as Einstein’s theory predicted to 60 micro-seconds
iv. Time is malleable & so is space as objects shrink when they move
4. Spacetime
a. Einstein’s theory can be constructed almost entirely using the language of geometry
b. 3 Concepts:
i. Invariance
1. Rotational invariance: The laws of nature do not change if we spin around. If they are unchanged irrespective of the direction in which we face then there exists a quantity that is conserved called angular momentum [The moon moves 4cm farther away from the earth every year due to friction from the sea on the surface slowing the earth’s spin; & so the earth day lengthens by two-thousandths of a second every century. The conservation of angular momentum is transferred to the moon which speeds up in its orbit around the earth]
2. Translational invariance: The laws of nature are universal
Spacetime is the merging of space & time into a single entity & is invariant. It is an absolute
ii. Causality: the order of cause & effect cannot be reversed
1. Time can be measured in metres if we take any time interval & multiply it by a calibrating speed, such as the speed of light.
2. If S is the distance between points A & B where space is on the x-axis & time on the y-axis, we get S2 = (ct)2 + x2 or S2 = (ct)2 – x2
3. The latter is a hyperbola & always in an area of a spacetime diagram that is above lines of 45° (the lightcone), & so respects the causality concept
4. C is the cosmic speed limit. Nothing can travel faster than c because if it did it could be used to transmit information that could violate the principle of cause & effect
iii. Distance
1. Distances in spacetime are invariant
2. Everything moves over spacetime at the same speed, c
3. Something moving in space uses up some of its fixed quota of spacetime & so leaves less for its motion through time
5. Why Does E=mc2?
a. It seems that light itself is important in the structure of the universe
b. Momentum = Mass x Speed (p=mv)
c. More mass takes a greater force to move it (F=ma)
d. Mass is assumed to be an invariant quantity
e. Law of conservation of energy (unlike momentum, energy has no direction)
f. Kinectic energy = ½mv2
g. Mass & energy are potentially interchangeable
h. The spacetime momentum vector shows us that in the space direction we have the old law of momentum conservation with a tweak for things moving close to the c
i. Along the time direction of the vector we get a new version of the law of conservation of energy, where ½mv2 gets replaced with mc2 so that even an object standing still has energy associated with it, E= mc2
j. Energy & mass are different manifestations of the same thing
k. The spacetime momentum vector combines energy, mass & momentum
l. Matter can pop in & out of existence
m. If mass is zero then its speed is c
n. Light has no mass as far as we know
o. The energy locked away inside even quite small masses is mind-bogglingly large. A city of 100k people requires just one-trillionth of a gram of matter to be converted into energy every second, i.e. 3kg will last 100 years.
6. And Why Should We Care?
a. Mass is a measure of the latent energy stored up within matter
b. The more latent energy something has, the more massive it is
c. For anything to happen at all in the universe, energy & mass must be continually sloshing back & forth
d. Mass can be measured in electron volts. One eV is the amount of energy an electron gets when it is accelerated through a potential difference of 1 volt. It is divided by c2 to turn it into a measure of mass.
i. Proton = 938,272,013 eV/c2
ii. Electron 510,998 eV/c2
iii. Yet a Hydrogen atom has a mass that is 13.6 eV/c2 less than the sum of a proton & electron
iv. Note that the electron has mass
v. This ‘negative energy’ is the amount of effort to dismantle the atom; its binding energy
vi. There are smaller mass differences between the hydrogen & the sum of the proton & electron, as the electron can be in a different orbit
1. The next difference is 10.2 eV/c2
2. The differences are discrete
3. The electron with the smallest mass has the electron closer to the proton (‘ground state’)
e. Left alone, a heavier thing will turn into a lighter thing if at all possible
f. Heavier versions of the hydrogen atom shed mass by emitting single particles of light (but isn’t light supposed to massless?) The excess energy is carried away by a photon
i. Our eyes are photon detectors
g. All of the atoms in nature come in a tower of energies (or masses), depending on where the electrons are, & since there is more than a single electron in every atom except hydrogen, the light emitted from them spans all the colours of the rainbow & beyond, which is why the world is colourful.
h. Molecules are less massive than their comprising atoms as it takes energy to break them apart
i. Uranium has 92 protons & 146 neutrons (in its most stable, naturally occurring form). Its half-life is 4.5 billion years, i.e. in this time half of the atoms in a lump of uranium will have spontaneously split up into lighter things, the heaviest of which is lead, & energy liberated as a result (fission)
i. There exists a mineral known as zircon. It naturally incorporates uranium into its crystalline structure, but not lead. So any lead comes from the radioactive decay of uranium & permits high precision dating.
j. When we bring two protons together one of the protons will spontaneously turn itself into a neutron & the excess positive electric charge is shed as a particle called a positron. They are +ve charged electrons. There is also a neutrino emitted. The proton & neutron come together under the influence of the strong nuclear force as a deuteron & the emission of a positron is called radioactive beta decay. (Fussion).
k. The energy released in a nuclear reaction is typically a million times the energy released in a chemical reaction
l. Temperature is essentially nothing more than a measure of the average speed of things
i. At 10,000 °C electrons are ripped from their orbits around nuclei, leaving behind a gas of protons & electrons known as plasma
ii. At 10 million °C the plasma gets transformed into a star as nuclear fusion takes place. The deuterons fuse with protons to produce helium & the energy released in the star shinning.
iii. The sun loses 4m tons of mass every second as it converts 600m tons of hydrogen into helium every second
iv. When the hydrogen starts to run out the star begins to collapse. At 100 million °C the helium begins to fuse. They form beryllium (4 protons & 4 neutrons)
v. Neutrinos flow out from the sun. About 100 billion pass through every square cm of the earth every second.
vi. Once the helium is exhausted the collapse begins again. At 500 million the °C carbon fuses. It produces heavier elements all the way up to iron.
vii. Neutron stars are more massive than the sun, but the size of a city. They emit beams of radiation known as pulsars. They can become black holes.
7. The Origin of Mass
a. The sun is several thousand times less efficient than the human body at converting mass to energy. 1kg of the sun generates 1/5,000 watts, whereas the human body generates more than 1 watt per kg.
b. The master equation of the Standard Model explains how every particle interacts with every other particle – except gravity
i. The first part explains the kinetic energy carried by W & Z particles, the photon & the gluon
ii. The second part how every particle in the universe interacts with each other & the kinetic energies of all matter particles
iii. The third part is a fix to the problem of gauge symmetry demanding no mass
c. The standard model assumes the existence of:
i. 6 types of quark
ii. 3 types of charged lepton (of which the electron is one)
iii. 3 types of neutrino
iv. For every particle there is a corresponding anti-particle
d. Protons are made mainly of 2 up quarks & one down quark
e. Neutrons of two up & one down quark
f. Up & down quarks & electrons are the predominant particles of everyday matter
g. The W & Z particles, the photon & the gluon are responsible for interactions between particles. They carry the force of interaction.
h. The photon carries the force between electrically charged particles, such as electrons & quarks. It mediates the electromagnetic field.
i. The gluon mediates the force that ‘glues’ protons together inside the atomic nucleus
j. The W particle is responsible for the interaction that turns a proton into a neutron during the formation of a deutron in stars. The force carried by the W & Z particles is extremely weak.
k. As it seems necessary to have only 4 particles to build a universe, the existence of the other 8 is a mystery
l. Particles have wavelike characteristics while remaining particles
m. Electrodynamics: the interaction of light & matter
n. The standard model unifies electromagnetism & the weak force. They are different manifestations of the same thing.
o. The masses of the W & Z particles were predicted before they were discovered
p. Nature is symmetrical – Gauge symmetry
q. An electron & anti-electron (positron) bang into each other & annihilate to produce a single photon (so we start with some mass & end up with no mass?)
r. For every observed matter particle there are around 100 billion photons
s. The stuff of the universe that makes up stars, planets & people is only a tiny residue left over after the grand annihilation of mass that took place early on in the universe’s history
t. We don’t know why the universe is not just filled with light
u. The reason that protons, electrons & photons dominate the stuff of the veryday world is they have nothing to decay into. Up & down quarks are the lightest quarks. The electron is the lightest lepton, & the photon has no mass.
v. The Higgs mechanism solves the problem of Gauge symmetry demanding no mass & the fact that there is mass
i. It is what slows things from moving at c & is responsible for the origin of mass
ii. The Higgs field holds back quarks & leptons, but allows photons t pass unimpeded
8. Warping Spacetime
a. Is spacetime warped & curved differently from place to place in the universe?
b. Einstein’s journey to special relativity was triggered by the simple question: what would it mean if c were the same for all observers?
His journey to general relativity begins with another simple question: why do all things fall to the ground with the same acceleration?
c. In a falling lift there is no experiment to distinguish between the possibility you are plummeting to the earth or floating in space. This universality of freefall is called the principle of equivalence
d. Gravity is stronger the closer to the centre of the earth you are; the gravity at your feet is stronger than at your head
e. The force of gravity is nothing more than a signal that spacetime itself is curved
f. Since gravity is found in the vicinity of matter, we might conclude that spacetime is warped in the vicinity of matter, & since E=mc2, energy
g. The sun is a massive object that distorts spacetime. The earth is moving freely through spacetime but the warping of spacetime makes the earth go in circles.
h. Straight-lines in spacetime are called Geodesics
i. Gravity is geometry & all things move along straight-lines unless they are knocked off course
j. Pulsars (spinning neutron stars) were discovered in 1967. There is only one known instance of 2 pulsars circling each other (discovered in 2004). They cause gravitational waves.
k. Clocks run faster in weak gravitational fields
l. GPS: 24 satellites 20,000km up. Their spacetime is warped so that their clocks speed up at a rate of 45 microseconds per day. But because they travel at 14,000 kmph, time dilation means their clocks slow by 7 microseconds per day. The net effect of would mean the system would be out by more than 10km per day. Hence the satellite clocks are made to run 38 microseconds slower.
m. General relativity is not compatible with quantum theory
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### Merge branch 'master' of Lephenixnoir/sprite-optimizer into master
`@ -32,8 +32,9 @@ def print_stats(img):` ``` ``` `def get_lines(img):` ` """Generate all potential lines the image contains"""` ` lines = []` ` pixels = [(x, y) for x in range(img.width) for y in range(img.height) if img.getpixel((x, y)) == 0]` ` lines = set()` ` lines_uniq = set()` ` pixels = {(x, y) for x in range(img.width) for y in range(img.height) if img.getpixel((x, y)) == 0}` ``` ``` ` i, n = 0, len(pixels) * len(pixels)` ``` ``` `@ -42,53 +43,21 @@ def get_lines(img):` ` # for each pair of pixels, get the line` ` for a in pixels:` ` for b in pixels:` ` line = list(bresenham(a[0], a[1], b[0], b[1]))` ` is_in = True` ` for p in line:` ` if p not in pixels:` ` is_in = False` ` break` ` line = tuple(bresenham(*a, *b))` ` line_uniq = tuple(sorted(line))` ``` ``` ` # if image contains the line, put it on the array` ` if is_in:` ` lines.append(line)` ` if set(line).issubset(pixels) and line_uniq not in lines_uniq:` ` lines.add(line)` ` lines_uniq.add(line_uniq)` ` i += 1` ` if progress:` ` print("\rGet lines: {:.1%}".format(i / n), end = "")` ` if progress:` ` print("\rGet lines: complete")` ``` ``` ` if progress:` ` print("\rGet lines: complete")` ` print("{} lines found".format(len(lines)))` ` return lines` ``` ``` ``` ``` `def removing_doubles(lines):` ` """Remove lines that are symetric"""` ` results = []` ` n = len(lines)` ``` ``` ` if progress:` ` print("Remove duplicated lines:", end = "")` ` # for each line, see if it's already in the output array` ` # beware not to change the orientation of the line (bresenham is not symetric)` ` # TODO: optimize a bit this operation` ` for i, l in enumerate(lines):` ` s = sorted(l)` ` same = False` ` for o in results:` ` if sorted(o) == s:` ` same = True` ` break` ` if same == False:` ` results.append(l)` ` if progress:` ` print("\rRemove double lines: {:.1%}".format(i / n), end = "")` ` if progress:` ` print("\rRemove double lines: complete")` ``` ``` ` if progress:` ` print("{} lines kept".format(len(results)))` ` return results` ` return list(lines)` ``` ``` ``` ``` `def removing_useless(lines):` `@ -99,18 +68,20 @@ def removing_useless(lines):` ` if progress:` ` print("Remove useless lines:", end = "")` ` # for each line, see if there is a line that contains every pixel of it` ` for i, l in enumerate(lines):` ` lines_set = [ set(l) for l in lines ]` ``` ``` ` for i, l in enumerate(lines_set):` ` inclusions = 0` ` # others are all lines that are not l` ` others = (x for x in lines if x != l)` ` for k in others:` ` if len(list(set(l).intersection(set(k)))) == len(l):` ` for j, k in enumerate(lines_set):` ` if i == j: continue` ` if l.issubset(k):` ` inclusions += 1` ` break` ` # or len(l) == 1 : we keep single pixels to complete the image if necessary` ` # TODO: do some tests to see if it's worth or not` ` if inclusions == 0 or len(l) == 1:` ` results.append((len(l), l))` ` results.append((len(l), lines[i], l))` ` if progress:` ` print("\rRemove useless lines: {:.1%}".format(i / n), end = "")` ` if progress:` `@ -123,22 +94,24 @@ def removing_useless(lines):` ``` ``` `def get_best_solution(img, lines):` ` """Compute an optimized solution. The magic part of the algorithm"""` ` px_left = [(x, y) for x in range(img.width) for y in range(img.height) if img.getpixel((x, y)) == 0]` ` px_left = {(x, y) for x in range(img.width) for y in range(img.height)` ` if img.getpixel((x, y)) == 0}` ` results = []` ` n = len(px_left)` ``` ``` ` if progress:` ` print("Draw:", end = "")` ` # while the entier image has not been drown` ` while len(px_left):` ` while px_left:` ` # define the length of lines` ` lines = [(sum([1 for p in l if p in px_left]) - len(l)/(2*max(img.size)), l) for n, l in lines]` ` lines = [(len(l_set.intersection(px_left)) - len(l)/(2*max(img.size)),` ` l, l_set) for n, l, l_set in lines]` ` # sort them by length` ` lines = sorted(lines)` ` # pop the longest` ` (p, line) = lines.pop()` ` (p, line, line_set) = lines.pop()` ` # define the pixels that are not covered by any lines` ` px_left = [p for p in px_left if p not in line]` ` px_left = px_left.difference(line_set)` ` results.append((line[0], line[-1]))` ` if progress:` ` print("\rDraw: {:.0%}".format(1 - len(px_left)/n), end="")` `@ -222,7 +195,6 @@ if __name__ == "__main__":` ` if args.info:` ` print_stats(image)` ` lines = get_lines(image)` ` lines = removing_doubles(lines)` ` lines = removing_useless(lines)` ` lines = get_best_solution(image, lines)` ` code = generate_code(lines, args)`
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A307188 Positions of 0's in the square spiral in A275609. 6
%I
%S 0,9,11,14,16,18,20,22,30,47,49,51,53,59,61,63,66,68,70,72,74,76,88,
%T 90,92,100,117,119,121,123,125,127,137,139,141,147,149,151,153,155,
%U 158,160,162,178,180,182,184,186,194,196,198,210,219,221,223,225,227
%N Positions of 0's in the square spiral in A275609.
%C This sequence and A307189-A307192 together partition the nonnegative numbers.
%C Comment from _N. J. A. Sloane_, Mar 31 2019 (Start)
%C Both Rémy Sigrist's illustration and mine show the same block structure. A block B(r,s) will mean an array of r X s 0's separated by rows and columns of blanks. Here is B(3,5), with blanks represented by dots:
%C ......
%C 0.0.0.
%C ......
%C 0.0.0.
%C ......
%C 0.0.0.
%C The dots are always to the right of or above the 0's.
%C The block B(r,s) has size 2r X 2s.
%C The block containing the center point is a B(3,5).
%C Looking at the blocks along the x-axis, moving from West to East, the successive r-values are:
%C ... 23, 19, 15, 11, 7, (3), 5, 9, 13, 17, 21, ...
%C Looking at the blocks along the y-axis, moving from South to North, the successive s-values are:
%C ... 24, 20, 16, 12, 8, (5), 6, 10, 14, 18, 22, ...
%C Except at the central block B(3,5), the differences between adjacent r- and s-values always have magnitude 4.
%C Furthermore, as we move North, each block is displaced by one step to the East from the block below it. As we move East, each block is displaced by one step to the South from the block to the West of it.
%C These displacements can be seen in both illustrations. In Sigrist's illustration, notice how the blocks gradually shift to the East along the y-xis, and to the South along along the x-axis.
%C (End)
%H Alois P. Heinz, <a href="/A307188/b307188.txt">Table of n, a(n) for n = 1..20000</a> (first 1240 terms from N. J. A. Sloane)
%H Rémy Sigrist, <a href="/A307188/a307188.png">Colored scatterplot of (x,y) such that A275609(x,y) = 0 and -500 <= x <= 500 and -500 <= y <= 500</a> (where the color is function of (x mod 2) + 2 * (y mod 2))
%H N. J. A. Sloane, <a href="/A307188/a307188.txt">Central portion of spiral in A275609 showing just the 0's</a> [Entries 1,2,3,4 in the spiral have been replaced by dots. The central 0 has been changed to an X. The illustration shows cells (x,y) with -35 <= x <= 35, -33 <= y <= 36.]
%Y Cf. A275609, A307189-A307192.
%K nonn
%O 1,2
%A _N. J. A. Sloane_, Mar 28 2019
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## 579 - Clock Hands
Moderator: Board moderators
saiful_islam
New poster
Posts: 6
Joined: Tue Feb 08, 2011 7:41 pm
### Re: 579 - ClockHands
ok
Last edited by saiful_islam on Fri Mar 30, 2012 10:07 pm, edited 2 times in total.
Dark Lord
New poster
Posts: 2
Joined: Mon Sep 27, 2010 11:27 am
### Re: 579 - ClockHands
Your code is giving these output.
input:
1:59
1:01
-0.000
-36893488.....something.
But the output should be:
65.500
24.500
Try to use fabs function from math.h
remember angle is always <=180.000
saiful_islam
New poster
Posts: 6
Joined: Tue Feb 08, 2011 7:41 pm
### Re: 579 - ClockHands
tnx Bro ACC
20112685
New poster
Posts: 1
Joined: Thu Jan 19, 2012 1:07 pm
### Re: 579 - ClockHands(WHY WA??SOS)
#include<stdio.h>
#include<math.h>
int main(void)
{
int h,m;
double a;
while(scanf("%i:%i",&h,&m)!=EOF && (h||m))
{
a=fabs(5.5*m-30*h);
if(a-180>1e-9)
a=360-a;
printf("%.3f\n",a);
}
return 0;
}
AT-2012
New poster
Posts: 1
Joined: Sun May 27, 2012 9:58 pm
### Re: 579 - ClockHands
Plz!! Someone help me out wid this code. I got WA.
579: Code
#include<stdio.h>
#include<math.h>
int main()
{
float h,m,ang,n;
while(scanf("%f:%f",&h,&m)==2)
{
if(h==0 && m==0) break;
ang=((h*60+m)/2-(m*6));
if(ang>180) ang=360-ang;
if(ang<0) ang=fabs(ang);
printf("%.3f\n",ang);
}
return 0;
}
brianfry713
Guru
Posts: 5947
Joined: Thu Sep 01, 2011 9:09 am
Location: San Jose, CA, USA
### Re: 579 - ClockHands
Check input and AC output for thousands of problems on uDebug!
shellexecutor
New poster
Posts: 2
Joined: Sat Sep 08, 2012 3:36 pm
### Re: 579 - ClockHands
Float is OK. I use float and got AC.
moxlotus
New poster
Posts: 31
Joined: Sat Sep 17, 2011 6:47 am
### 579 TLE
The java code is getting TLE, can someone tell me how can this piece of code be optimized ?
The equivalent C code has passed the judge without any problem.
Code: Select all
``````import java.util.*;
import java.text.DecimalFormat;
import java.io.IOException;
public class ClockHands
{
public static void main(String[] args) throws IOException
{
String[] arr = line.split("[:\\s]+");
int hr = Integer.parseInt(arr[0]);
int min= Integer.parseInt(arr[1]);
double hrDegree, minDegree, diff;
DecimalFormat df = new DecimalFormat("0.000");
while(hr > 0 || min > 0)
{
if(hr == 12) hr = 0;
hrDegree = ((hr+(min/60.0))/12.0) * 360;
minDegree = (min/60.0) * 360;
diff = hrDegree > minDegree ? hrDegree-minDegree : minDegree-hrDegree;
if(diff > 180) diff = 360 - diff;
System.out.printf("%s\n", df.format(diff));
arr = line.split("[:\\s]+");
hr = Integer.parseInt(arr[0]);
min = Integer.parseInt(arr[1]);
}
}
}
``````
brianfry713
Guru
Posts: 5947
Joined: Thu Sep 01, 2011 9:09 am
Location: San Jose, CA, USA
### Re: 579 TLE
Use Class Main. Java is too slow for some problems.
Check input and AC output for thousands of problems on uDebug!
moxlotus
New poster
Posts: 31
Joined: Sat Sep 17, 2011 6:47 am
### Re: 579 TLE
Actually got a few friends helped me optimized the code.
Apparently the use of StringBuilder and printing it out all at once is much faster than calling System.out.printf.
The performance gain is more than 100%
brianfry713
Guru
Posts: 5947
Joined: Thu Sep 01, 2011 9:09 am
Location: San Jose, CA, USA
### Re: 579 TLE
You could also try BufferedWriter.
Check input and AC output for thousands of problems on uDebug!
cyberdragon
New poster
Posts: 20
Joined: Fri Aug 30, 2013 5:42 am
### 579 - ClockHands
3rd case in the sample input.
How 8:10 outputs 175, It should be 180 ?
brianfry713
Guru
Posts: 5947
Joined: Thu Sep 01, 2011 9:09 am
Location: San Jose, CA, USA
### Re: 579 - ClockHands
The big hand at 10 is 60 degrees to the right of 12.
The small hand at 8:10 is 115 degrees to the left of 12.
Check input and AC output for thousands of problems on uDebug!
This Is ERFAN
New poster
Posts: 6
Joined: Thu Dec 04, 2014 11:40 pm
### Re: 579 - Clock Hands
Getting WA...plz help me out.
Code: Select all
``````#include<stdio.h>
int main()
{
float a,b;
char c;
while(scanf("%f %c %f",&a,&c,&b)==3)
{
if(a==0&&b==0) return 0;
float d,d1,d2,d3;
if(b==0)
{
if(a<=6) d=a*5*6;
else
d=(12-a)*5*6;
}
else
{
d1=b/2;
d2=b/5;
if(a==d2) d3=0;
else if(a>d2) d3=(a-d2)*5*6;
else if(a<d2) d3=(d2-a)*5*6;
if(a>d2) d=d3+d1;
else if(a==d2) d=d1;
else d=d3-d1;
if(d>180) d=360-d;
}
printf("%0.3f\n",d);
}
return 0;
}
``````
brianfry713
Guru
Posts: 5947
Joined: Thu Sep 01, 2011 9:09 am
Location: San Jose, CA, USA
### Re: 579 - Clock Hands
Try using double instead of float.
Check input and AC output for thousands of problems on uDebug!
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| 2.84375
| 3
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CC-MAIN-2019-35
|
latest
|
en
| 0.551481
|
https://rdrr.io/github/bdwilliamson/nova/man/measure_accuracy.html
| 1,718,871,821,000,000,000
|
text/html
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crawl-data/CC-MAIN-2024-26/segments/1718198861916.26/warc/CC-MAIN-20240620074431-20240620104431-00330.warc.gz
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|
# measure_accuracy: Estimate the classification accuracy In bdwilliamson/nova: Perform Inference on Algorithm-Agnostic Variable Importance
measure_accuracy R Documentation
## Estimate the classification accuracy
### Description
Compute nonparametric estimate of classification accuracy.
### Usage
``````measure_accuracy(
fitted_values,
y,
full_y = NULL,
C = rep(1, length(y)),
Z = NULL,
ipc_weights = rep(1, length(y)),
ipc_fit_type = "external",
ipc_eif_preds = rep(1, length(y)),
ipc_est_type = "aipw",
scale = "logit",
na.rm = FALSE,
nuisance_estimators = NULL,
a = NULL,
...
)
``````
### Arguments
`fitted_values` fitted values from a regression function using the observed data (may be within a specified fold, for cross-fitted estimates). `y` the observed outcome (may be within a specified fold, for cross-fitted estimates). `full_y` the observed outcome (not used, defaults to `NULL`). `C` the indicator of coarsening (1 denotes observed, 0 denotes unobserved). `Z` either `NULL` (if no coarsening) or a matrix-like object containing the fully observed data. `ipc_weights` weights for inverse probability of coarsening (IPC) (e.g., inverse weights from a two-phase sample) weighted estimation. Assumed to be already inverted. (i.e., ipc_weights = 1 / [estimated probability weights]). `ipc_fit_type` if "external", then use `ipc_eif_preds`; if "SL", fit a SuperLearner to determine the IPC correction to the efficient influence function. `ipc_eif_preds` if `ipc_fit_type = "external"`, the fitted values from a regression of the full-data EIF on the fully observed covariates/outcome; otherwise, not used. `ipc_est_type` IPC correction, either `"ipw"` (for classical inverse probability weighting) or `"aipw"` (for augmented inverse probability weighting; the default). `scale` if doing an IPC correction, then the scale that the correction should be computed on (e.g., "identity"; or "logit" to logit-transform, apply the correction, and back-transform). `na.rm` logical; should `NA`s be removed in computation? (defaults to `FALSE`) `nuisance_estimators` not used; for compatibility with `measure_average_value`. `a` not used; for compatibility with `measure_average_value`. `...` other arguments to SuperLearner, if `ipc_fit_type = "SL"`.
### Value
A named list of: (1) the estimated classification accuracy of the fitted regression function; (2) the estimated influence function; and (3) the IPC EIF predictions.
bdwilliamson/nova documentation built on Feb. 1, 2024, 10:04 p.m.
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CC-MAIN-2024-26
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latest
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en
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https://answercult.com/questions/question/how-do-sail-boats-actually-work/
| 1,660,686,007,000,000,000
|
text/html
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crawl-data/CC-MAIN-2022-33/segments/1659882572581.94/warc/CC-MAIN-20220816211628-20220817001628-00240.warc.gz
| 117,755,208
| 19,821
|
# How do sail boats actually work?
59 views
0
The sails always seems to be somewhat parallel to the direction of the boat, but if the wind is blowing sideways, should that push the boat off course?
In: 9
The next time you’re at a pool, grab one of those floaty board things, shove it deep underwater, and try to wiggle it around.
You’ll notice that it’s really easy to move the board along its thin directions, but really hard to move it along the flat, wide direction.
Sailboats (and most other boats actually) are like that board. They have long thin underwater bits that make it very hard for them to be pushed sideways.
___
Another fun thing about sailboats is that the most intuitive way for them to work, with the wind pushing directly into back of the sail to move it forwards, is actually the slowest way to sail.
The fastest sailing is when you’re travelling directly perpendicular to the wind.
The reason for this is that (modern) sailboats use their sails as *wings* instead of like parachutes.
The wind pushes the sail into a nice curved shape, just like the wing of a plane. Then the pressure differential between the outwardly curved and inwardly curved sides is what pushes the sailboat forwards.
This is what allows sailboats to sail upwind. The wind is *pushing* them back, but the pressure difference across the sail makes the boat move forwards anyway.
Imagine a sailboat like an airplane tipped sideways with one wing in the air and one in the water – the one in the air is the sail and the one in the water is the keel.
The boat moves because the wind is pushing against the water resistance of the keel; the result is low air pressure in front of the sail and high air pressure behind it, combined with water pressure on the lee side of the keel, but no resistance forward or aft.
So all the pressures even out by the boat moving forward.
Sails act like a wing most of the time. The air moving over them generates lift*. Boats have a keel or a dagger/centreboard that stops them drifting sideways to an extent, but they do still drift slightly.
*If a boat is heading away from the wind, the primary mechanism of propulsion may just be the wind pushing the sail.
Okay, so you’re probably looking at sloop rigged sailboats, they look like this: ⛵️
Now, when they’re heading downwind, they can use a sail called a spinnaker or push the mainsail and jib out to roughly parallel to the boat to go faster. They actually do go perpendicular when the wind is pushing them from behind.
So when going upwind (not straight upwind, typically about 45 degrees off of straight into the wind and the sails are off to an angle that creates better lift, but let’s ignore that for now) the sails SHOULD only push the boat downwind, but there’s a clever way to deal with that: a keel. The keel is a straight board in the water that’s parallel with the boat.
So let’s say wind is coming from the top of your phone, straight down. The sail is parallel with the boat/keel, moving 50 degrees to the left of going straight up, with 90 degrees being straight left. (Draw it out if it helps) There’s a component of force down yes, and that should push the boat back, but there’s a leftward component that is bigger than the down component. So the sailboat will go left more than down/back.
Now remember the keel, it’s in water which has much more friction than air. So the keel helps make the boat move in only two directions, forward or back. So whatever force is pushing on the sail pushed down to the keel, and is equalized out. The only part of the force that is allowed to push the boat forward is the remaining leftward force, which pushes the boat left, but that gets equalized out too by the keel, so the boat slips forward in the keel direction too.
In the end the boat is pushed left, but because of the keel, goes forward and left, along that 50 degree bearing.
It’s like a toy car on a string.
In the water, there’s a few fins that work like the tires of the toy car. Above the water, there’s wind that works like somebody pulling the string.
It’s not an exact comparison, but this is ELI5.
When the string is pulled, the car wants to roll in the direction of the tires, not always directly towards the person pulling the string.
The tires can turn much like the rudder of the boat to steer.
Beyond this, things can become slightly more complex when sailing perpendicular to the wind or towards where the wind is blowing from. (But not directly into the wind, that doesn’t work.)
This is where the angle of the sail becomes quite important. The force of the wind acting on the sails and the force of water acting on the fins can squeeze the boat (in a sense).
Like pinching a bead in your fingertips, the bead can be launched away while neither of your fingers are applying force in the launch direction. It’s two separate forces working together to create a new force in a different direction.
0 views
0
The sails always seems to be somewhat parallel to the direction of the boat, but if the wind is blowing sideways, should that push the boat off course?
In: 9
The next time you’re at a pool, grab one of those floaty board things, shove it deep underwater, and try to wiggle it around.
You’ll notice that it’s really easy to move the board along its thin directions, but really hard to move it along the flat, wide direction.
Sailboats (and most other boats actually) are like that board. They have long thin underwater bits that make it very hard for them to be pushed sideways.
___
Another fun thing about sailboats is that the most intuitive way for them to work, with the wind pushing directly into back of the sail to move it forwards, is actually the slowest way to sail.
The fastest sailing is when you’re travelling directly perpendicular to the wind.
The reason for this is that (modern) sailboats use their sails as *wings* instead of like parachutes.
The wind pushes the sail into a nice curved shape, just like the wing of a plane. Then the pressure differential between the outwardly curved and inwardly curved sides is what pushes the sailboat forwards.
This is what allows sailboats to sail upwind. The wind is *pushing* them back, but the pressure difference across the sail makes the boat move forwards anyway.
Imagine a sailboat like an airplane tipped sideways with one wing in the air and one in the water – the one in the air is the sail and the one in the water is the keel.
The boat moves because the wind is pushing against the water resistance of the keel; the result is low air pressure in front of the sail and high air pressure behind it, combined with water pressure on the lee side of the keel, but no resistance forward or aft.
So all the pressures even out by the boat moving forward.
Sails act like a wing most of the time. The air moving over them generates lift*. Boats have a keel or a dagger/centreboard that stops them drifting sideways to an extent, but they do still drift slightly.
*If a boat is heading away from the wind, the primary mechanism of propulsion may just be the wind pushing the sail.
Okay, so you’re probably looking at sloop rigged sailboats, they look like this: ⛵️
Now, when they’re heading downwind, they can use a sail called a spinnaker or push the mainsail and jib out to roughly parallel to the boat to go faster. They actually do go perpendicular when the wind is pushing them from behind.
So when going upwind (not straight upwind, typically about 45 degrees off of straight into the wind and the sails are off to an angle that creates better lift, but let’s ignore that for now) the sails SHOULD only push the boat downwind, but there’s a clever way to deal with that: a keel. The keel is a straight board in the water that’s parallel with the boat.
So let’s say wind is coming from the top of your phone, straight down. The sail is parallel with the boat/keel, moving 50 degrees to the left of going straight up, with 90 degrees being straight left. (Draw it out if it helps) There’s a component of force down yes, and that should push the boat back, but there’s a leftward component that is bigger than the down component. So the sailboat will go left more than down/back.
Now remember the keel, it’s in water which has much more friction than air. So the keel helps make the boat move in only two directions, forward or back. So whatever force is pushing on the sail pushed down to the keel, and is equalized out. The only part of the force that is allowed to push the boat forward is the remaining leftward force, which pushes the boat left, but that gets equalized out too by the keel, so the boat slips forward in the keel direction too.
In the end the boat is pushed left, but because of the keel, goes forward and left, along that 50 degree bearing.
It’s like a toy car on a string.
In the water, there’s a few fins that work like the tires of the toy car. Above the water, there’s wind that works like somebody pulling the string.
It’s not an exact comparison, but this is ELI5.
When the string is pulled, the car wants to roll in the direction of the tires, not always directly towards the person pulling the string.
The tires can turn much like the rudder of the boat to steer.
Beyond this, things can become slightly more complex when sailing perpendicular to the wind or towards where the wind is blowing from. (But not directly into the wind, that doesn’t work.)
This is where the angle of the sail becomes quite important. The force of the wind acting on the sails and the force of water acting on the fins can squeeze the boat (in a sense).
Like pinching a bead in your fingertips, the bead can be launched away while neither of your fingers are applying force in the launch direction. It’s two separate forces working together to create a new force in a different direction.
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| 4,013
|
This site is updated infrequently. For up-to-date information, please visit the new OCaml website at ocaml.org.
Re: [Caml-list] Generalized Algebraic Datatypes
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Date: 2010-10-29 (15:54) From: Jacques Le Normand Subject: Re: [Caml-list] Generalized Algebraic Datatypes
```Assuming I understand this syntax, the following currently valid type
definition would have two interpretations:
type 'a t = IntLit of 'a constraint 'a = int
One interpretation as a standard constrained ADT and one interpretation as a
GADT. We could use another token, other than constraint, for example:
type 'a t = IntLit of 'a : 'a = int
to which I have no objections. As you pointed out, though, the current
syntax is more concise.
cheers,
--Jacques
On Fri, Oct 29, 2010 at 10:32 AM, Dario Teixeira <darioteixeira@yahoo.com>wrote:
> Hi,
>
> > I am pleased to announce an experimental branch of the O'Caml compiler:
> > O'Caml extended with Generalized Algebraic Datatypes. You can find more
> > information on this webpage:
>
> I have a couple of questions regarding the syntax you've chosen for GADT
> declaration. For reference, let's consider the first example you've
> provided:
>
> type _ t =
> | IntLit : int -> int t
> | BoolLit : bool -> bool t
> | Pair : 'a t * 'b t -> ('a * 'b) t
> | App : ('a -> 'b) t * 'a t -> 'b t
> | Abs : ('a -> 'b) -> ('a -> 'b) t
>
>
> constructors are portrayed as being like functions. While this does make
> sense in Haskell, in Ocaml it feels a bit out of place, because you cannot,
> for example, partially apply a type constructor.
>
> Also, note that in all the variant declarations the final token is 't'.
> Are there any circumstances at all where a GADT constructor will not end
> by referencing the type being defined? If there are not, then this syntax
> imposes some syntactic salt into the GADT declaration.
>
> I know this is not the sole syntax that was considered for GADTs in Ocaml.
> Xavier Leroy's presentation in CUG 2008 shows a different one, which even
> though slightly more verbose, does have the advantage of being more
> "Camlish".
> Is there any shortcoming to the 2008 syntax that resulted in it being
> dropped
> in favour of this new one?
>
> Best regards,
> Dario Teixeira
>
>
>
>
>
```
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https://financememos.com/2017/10/07/a-note-on-use-of-geometric-and-arithmetic-averages-of-daily-stock-returns/
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# A Note on Use of Geometric and Arithmetic Averages of Daily Stock Returns
Question: The table below has price data and daily return data for Vanguard fund VB. Calculate the arithmetic and geometric averages of the daily return data. Show that the geometric average accurately reflects the relationship between the initial and final stock price and the arithmetic average does not accurately explain this relationship.
Daily Price and Returns For Vanguard Fund VB Date Adjusted Close Daily Return 7/1/16 115.480674 7/5/16 113.99773 0.987158509 7/6/16 114.744179 1.006547929 7/7/16 114.913373 1.001474532 7/8/16 117.202487 1.019920345 7/11/16 118.128084 1.007897418 7/12/16 119.451781 1.011205608 7/13/16 119.10344 0.997083836 7/14/16 119.262686 1.001337039 7/15/16 119.402023 1.00116832 7/18/16 119.63093 1.001917112 7/19/16 119.202965 0.996422622 7/20/16 119.959369 1.006345513 7/21/16 119.481646 0.996017627 7/22/16 120.297763 1.00683048 7/25/16 120.019083 0.997683415 7/26/16 120.616248 1.004975584 7/27/16 120.347522 0.997772058 7/28/16 120.536625 1.001571308 7/29/16 120.894921 1.002972507 8/1/16 120.735675 0.998682773 8/2/16 119.12335 0.986645828
Analysis: The table below presents calculation of the two averages and the count of return days. The product of the initial value of the ETF, the pertinent average and the count of return days is the estimate of the final value. Estimates of final ETF value are calculated for both the arithmetic average and the geometric average and these estimates are compared to the actual value of the stock on the final day in the period.
Understanding The Difference Between Arithmetic Mean and Geometric Mean Returns Statistic Value Note Arithmetic Average of Daily Stock Change Ratio 1.001506208 Average function Geometric Average of Daily Stock Change Ratio 1.001479966 Geomean function Count of Return Days 21 Count Function Estimate of final value based on arithmetic average 119.1889153 Initial Value x Arithmetic Return Average x Count Days Estimate of final value based on geometric average 119.12335 Initial Value X Geometric Return Average x Count Days Ending Value 119.12335 Copy from data table
There is another way to show that the daily return should be modeled with the geometric mean rather than arithmetic mean. The average daily return of the stock is (FV/IV)(1/n) – 1 where FV is final value and IV is initial value and n is the number of market days in the period, which for this problem is 21.
Using this formula we find the daily average holding period return is 0.001479966. Note that 1 minus the geometric mean of the daily stock price ratio is also 0.001479966.
The geometric mean gives us the correct holding period return.
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# Design and Analysis of Inventory Model for Quadratic Trapezoidal Type Demand under Partial Backlogging
DOI : 10.17577/IJERTV8IS030195
Text Only Version
#### Design and Analysis of Inventory Model for Quadratic Trapezoidal Type Demand under Partial Backlogging
Pravat Kumar Sukla
1. S. College , Koksara, Kalahandi, odisha
India
Rajani Ballav Dasp
Anjan Kumar Rana2
Pragati College, Bhabanipatana, Kalahandi, Odisha,
India
Ex Reader in mathematics Revenshaw University, at present Visiting Faculty Revenshaw
University, Cuttack. India
Abstract:- In this paper, we consider the inventory model for perishable items with quadratic trapezoidal type demand rate, that is, the demand rate is a piecewise quadratic function. The model consider allows for shortages and the demand is partially backlogged. The model is solved analytically by minimizing the total inventory cost. The result is illustrated with numerical example for the model.
Keywords: Quadratic ttrapezoidal demand. Deterioration. Shortages. Partial backlogging
1. INTRODUCTION
Deteriorating items are very common thing in our daily life situation. In recent years, many researchers have studied inventory models for deteriorating items, however, academia has not reached a consensus on the definition of the deteriorating items. According to the study of Wee (1993), deteriorating items refers to the items that become decayed, damaged, evaporative, expired, invalid, devaluation and so on through time. According to the definition, deteriorating items can be classified in to two categories. The first category refers to the items that become decayed, damaged, evaporative, or expired through time, like meat, vegetables, fruit, medicine, flowers and so on; the other category refers to the items that lose part or total value through time because of new technology or the introduction of alternatives, like computer chips, mobile phones, fashion and seasonal goods and so on. The inventory problem of deteriorating items was first studied by Whitin (1957), he studied fashion items deteriorating at the end of the storage period. Then Ghare and Schrader (1963) concluded in their study that the consumption of the deteriorating items was closely relative to a negative exponential function of time. Various authors (Deng et al. (2007), Cheng and Wang (2009), Cheng et al. (2011), Hung (2011)) studied inventory models for deteriorating items in various aspects.
In world business market, demand has been always one of the most key factors in the decisions relating to the inventory and production activities. There are mainly two
categories demands in the present studies, one is deterministic demand and the other is stochastic demand. Various formations of consumption tendency have been studied, such as constant demand (Padmanabhan and Vrat (1990), Sukla (2012), , Sukla and sahu (2008) Chung and Lin (2001), Benkherouf et al. (2003), Chu et al (2004)), level-dependent demand (Giri and Choudhuri (1998), Chung et al. (2000), Bhattacharya (2005), Wu et al. (2006)), price dependent demand (Wee and Law (1999), Abad (1996, 2001)), time dependent demand (Resh et al. (1976), Henery (1979), Sachan (1984), Dave (1989), Teng
(1996), Teng et al. (2002), Skouri and Papachristos (2002), Panda, Sahoo, and Sukla (2012) ), Panda, Sahoo, and Sukla (2013) , Sett et al. (2013), Shah, , Chaudhari and Jani (2015), Shah, , Chaudhari and Jani (2016), Mishra et al. (2013)) and time and price dependent demand (Wee (1995)). Among them, ramp type demand is a special type of time dependent demand. Hill (1995), one of the pioneers, developed an inventory model with ramp type demand that begins with a linear increasing demand until to
the turning point, denoted as , proposed by previous
researchers, then it becomes a constant demand. There has been a movement towards developing this type of inventory system for minimum cost and maximum profit problems. Several authors: Mandal and Pal (1998) focused on deteriorating items. Wu et al. (1999) were concerned with backlog rates relative to the waiting time. Wu and Ouyang (2000) tried to build an inventory system under two replenishment policies: starting with shortage or without shortage. Panda, Sahoo and sukla (2013), Wu (2001) considered the deteriorated items satisfying Weibull distribution. Giri et al (2003) dealt with more generalized three parameter Weibull deterioration distribution. Deng (2005) extended the inventory model of Wu et al. (1999) for the situation where the in-stock period is shorter
than . Manna and Chaudhuri (2006) set up a model
where the deterioration is dependent on time. Panda et al. (2007) constructed an inventory model with a comprehensive ramp type demand. Deng et al. (2007)
contributed to the revision of Mandal and Pal (1998), and Wu and Ouyang (2000). Panda et al. (2008) examined the cyclic deterioration items. Wu et al. (2008) studied the maximum profit problem with the stock-dependent selling rate. They developed two inventory models all related to the conversion of the ramp type demand, and then examined the optimal solution for each case. However, in a realistic product life cycle, demand is increasing with time during the growth phase. Then, after reaching its peak, the demand becomes stable for a finite time period called the maturity phase. Thereafter, the demand starts decreasing with time and eventually reaching zero or constant.
In this work, we extend Hills ramp type demand rate to quadratic trapezoidal type demand rate. Such type of demand pattern is generally seen in the case of any fad or seasonal goods coming to market. The demand rate for such items increases quadratic-ally with the time up to certain time and then ultimately stabilizes and becomes constant, and finally the demand rate approximately decreases to a constant, and then begins the next replenishment cycle. We think that such type of demand rate is quite natural and useful in real world market situation. One can think that our work may provide a solid foundation for the future study of this kind of important inventory models with quadratic trapezoidal type demand rate and preservation technology
2. ASSUMPTION AND NOTATIONS
• k2 is the inventory holding cost per unit per unit of time.
• k3 is the shortage cost per unit per unit of time.
• S is the maximum inventory level for the ordering cycle, such that S=I(0).
• Q is the ordering quantity per cycle.
• A1(t1) is the average total cost per unit time under
the condition t1 1 .
• A2(t1) is the average total cost per unit time, for
1 t1 2 .
• A3(t1) is the average total cost per unit time, for
2 t1 T
3. MATHEMATICAL AND THEORETICAL RESULTS
Here, we consider the deteriorating inventory model with demand rate is trapezoidal type quadratic function. Replenishment occurs at time t =0 when the inventory level attains its maximum. For t [0, t1 ] , the inventory level reduces due to both demand and deterioration. At time t1, the inventory level reaches zero, then shortage is allowed to occur during the interval (t1, T), and all of the demand during the shortage period (t1, T) is completely backlogged. The total amount of backlogged items is replaced by the next replenishment. The rate of change of the inventory during the stock period [0, t1] and shortage period (t1, T) is governed by the following differential equations:
The fundamental assumption and notations used in this paper are given as follows: The demand rate, R(t), which is positive and consecutive, is assumed to be a quadratic
dI (t) I (t) R(t) 0 , 0 t t
dt 1
dI (t)
, (2)
trapeoidal type function of time, that is
R(t) 0 , t1 t T , (3)
b t c t 2 , t , dt
1 1 1
with boundary condition I(0)=S and I(t1)=0. One can think
R(t) R0 ,
1 t 2 ,
about t1, t1 may occur within
[0, ] or [ , ]
b t c t 2 , t T
1 1 2
2 2 2
(1)
or[ 2 , T ] . Hence in this paper we are going to discuss all three possible cases.
Chose b , c1, b and c2 such a way that b t c t 2 should
Case 1:
0 t
1 2 2 2 1 1
not be negative for 2 t T . Where 1 is the time point changing from the increasing quadratic demand to constant demand, and 2 is the time point changing from
The quadratic trapezoidal type market demand and constant rate of deterioration, the inventory level gradually diminishes during the period [0, t1] and ultimately reaches to zero at time t=t1. Then, from equations (2) and (3), we
the constant demand to the decreasing demand.
• Replenishment rate is infinite, thus replenishment
have
dI (t) I (t) b t c t 2 0 , 0 t t
(4)
is instantaneous. 1 1 1
• I(t) is the inventory level at any time t, dt
0 t T .
dI (t) b t c t 2 0 , t t
(5)
• T is the fixed length of each ordering cycle.
dt 1 1 1 1
• is the constant rate of deterioration,
dI (t) R
0 ,
t
(6)
0 1.
• t1 is the time when the inventory level reaches
zero.
1
1
• t * is an optimal point.
• k0 is the fixed ordering cost per order.
• k1 is the cost of each deteriorated item.
dt 0 1 2
dI (t) b t c t 2 0 , t T (7)
dt 2 2 2
Now solving the differential equations (4) (7) with the condition I(t1)=0 and continuous property of I(t), we get
b t c t 2 b 2c t 2c
c t 2 b 2c t 2c
I (t) 1 1 1 1 1 1 1 1 e
(t1 t )
1 1 1 1 1 1 e (t1 t )
2
3
t1
2 3
b t c t 2
b 2c t 2c
0
b t c t 2
b 2c t 2c
dt
1 1 1 1 1
1 1 1 1 1
, 0 t t1
2 3
(8)
b t
• c t 2
3
3
b 2c t
2 3
4
4
2c
I (t) (t 2 t 2 ) b1 (t 3 t 3 ) c1 , t
t
1 1 1 1
2
1 1 1
1 (e t1 1)
1 2 1 3 1
b
1 (9)
c
b1
2c1
c1
b1 2 c1 t 3
I (t) R t (t 2 2 ) 1 (t 3 2 3 ) 1 ,
2
3 t1
2
2 t1
3 1
(14)
0 1 1 2 1 1 3
1 t 2
I (t) (t 2 2 ) b1 (t 3 2 3 ) c1
(10)
The total shortage quantity during the interval [t1, T], say BT, is
1 1 2 1 1 3
,
b c
T
BT I (t)dt
(t 2 2 ) 2 (t 3 2 3 ) 2 t1
2 2 2 3
1 2 T
2 t T
(11)
I (t)dt I (t)dt I (t)dt
The beginning inventory level can be computed as
t1 1 2
b 2c
1 b c
2
3
1
2 1 3
S I (0) 1 1 (et1 1)
(12)
(t 2 t 2 )
t1
1 (t 3 t 3 ) 1 dt
b t c t 2 2c t
t
1 1 1 1 1 1 e 1 2
2
R t (t 2 2 ) b1 (t 3 2 3 ) c1
0
1 1 2 1
1 3 dt
The total number of items which is perish in the interval [0, t1], say DT, is
1
2 2 b1 3
3 c1
t1 t1
2
T (t1 1 ) 2 (t1 21 ) 3
DT S R(t)dt S (b1t c1t
)dt
b
dt
c
0 0 2 (t 2 2 ) 2 (t 3 2 3 ) 2
b 2c t
2 2
2 3
1 1 (e 1 1)
2 3
b1 t 2 ( t ) b1 ( 3 t 3 )
b t c t 2 2c t
b t 2
(13)
c t 3
2 1 1 1 6 1 1
1 1 1 1 1 1 et1 1 1 1 1 c c
2 2 3
1 t 3 (
t ) 1 ( 4 t 4 )
3 1 1 1 12 1 1
The total amounts of inventory carried during the R b
interval [0, t1], say CT, is
0 ( 2 2 ) 1 t 2 (
)
t1 2 1 2 2 1 2 1
CT I (t)dt
b1 2 (
) c1 t 3 (
)
0 2 1 2
1 3 1 2 1
2c1 3 (
) b1 t 2 (T
) b1 2 (T )
3 1 2
1 2 1
2 2 1 2
c1 (t 3 2 3 )(T
) b2 (T 3 3 ) b2 2 (T
) c2 (T 4 4 ) 2c2 3 (T )
(15)
3 1 1 2 6
2 2 2
2 12
2 3 2 2
The average total cost per unit time for 0 t1 1 is given by
The total back order amount at the end of the cycle is
1 b1 (t*2 2 ) c1 (t*3 2 3 )
A1 (t1 ) T [k0 k1 DT k2CT k3 BT ] (16)
1 2 1 1 3 1
1
(21)
The first order derivative of A1 (t1 ) with respect
b2 (T 2 2 ) c2 (T 3 2 3 )
to t is as follows:
2 2 3 2
1 Therefore, the optimal order quantity, denoted by
1
1
Q* , is Q* S * , where
S * denote the
T
T
dA1 (t1 ) 1 k
k2 (et1 1) k (t
T )
optimal value of S.
1
1
dt1
3 1
Case-II, 1 t1 2
1 1
1 1
2
2
(b1t1 c t )
The necessary condition for
(17)
A1 (t1 ) to be
For the time period t1 [1 , 2 ] , then, the differential equations governing the inventory model can be expressed as follows:
minimized, is
dA1 (t1 ) 0 , that is
dI (t) I (t) b t c t 2 0 , 0 t
dt1
1 k
dt 1 1
1
(22)
k 2 (e t1 1) k (t
T )
dI (t)
T 1
3 1
(18)
I (t) R0 0 , 1 t t1
dt
(23)
(b t
c t 2 ) 0
dI (t)
1 1 1 1 R0 0 , t1 t 2
(24)
This implies that dt
1
1
k2 t
dI (t) 2
k
(e
1 1) k3 (t1 T ) 0
(19)
dt b2t c2t
0 , 2 t T
(25)
Let p(t ) k k2 (et1 1) k (t
T ) (20)
Solving differential equations (22) to (25), using
1 1 3 1
I(t1)=0, we get
R
b b t c t 2
b 2c t 2c
Since
I (t) 0 et1 1 e1 et 1 1 1 1 1
k
2
2 3
p(0) k T 0, p(T ) k 2 (eT 1) 0
and
3 1
2c1 1 e ( 1 t ) 2c1 e ( 1 t ) , 0 t
k
2 3 1
p(t ) k 1 et1 k
0 , it implies
(26)
1 1 3
that p(t1) is a strictly monotonically increasing function and equation (19) has a unique solution at
t * , for t* (0, T ) . Therefore, we have
I (t) R0 (e (t1 t ) 1) ,
1
t t1
(27)
1 1 I (t) R (t
t), t1 t 2
(28)
Property-1
The constant deteriorating rate of an inventory model with quadratic trapezoidal type demand rate under the time interval 0 t1 1 ,
A (t ) attains its minimum at t t* , where
0 1
I (t) R t b2 (t 2 2 ) c2 (t 3 2 3 )
0 1 2 2 3 2
, 2 t T
1 1 1 1
(29)
p(t* ) 0 if t*
. On the other hand,
The beginning inventory can be computed as
1 1 1
A (t ) attains is minimum at t* if
1 1 1 1
t* .
1 1
S I (0)
R0 et1
• b1
2
e1
• b1
2
2c1
3
2c1 1 e1
2
2c1 e1
3
(30)
The total amount of items which is perish within the time interval [0, t1] is
t
t
R t b2 (t 2 2 )
2 T 0 1 2
2 T 0 1 2
2
1 R0 (t1 t)dt
dt
DT S R(t)dt
t1 2 c2 (t 3 2 3 )
0
S (b1t c1t )dt R dt
S (b1t c1t )dt R dt
1 t1
3
R
2
0
0
2 R t (
t ) 0 ( 2 t 2 )
0 1
0 1 2 1
2 2 1
R0 et1 b1
2c11 2c1 e1 R (t
)
• R t (T
) b2 (T 3 3 )
2
2 3
0 1 1
0 1 2 6 2
1 2 2 3
(31)
b2 2 (T ) c2 (T 4 4 )
b1 c
2 1
2 1
3
3
1 1
2 2 2 12 2
2 3
(33)
2c
The total amount of inventory carried during the time interval [0, t1] is
2 3 (T ) 3 2 2
t1
CT I (t)dt
0
Now, the average total cost per unit time under the
condition 1 t1 2 , can be obtained as
1
1 t1
A2 (t1 ) T [k0 k1 DT k2CT k3 BT ] (34)
I (t)dt I (t)dt
0 1
The first order derivative of A2 (t1 ) with respect to
t1 isgiv
R t b
t
b t c t 2
0 e
1 1 e
1 e
1 1
dA2 (t1 ) R0 k
k2 (et1 1) k (t
T )
2
dt T 1
3 1
b
b
1
1
• 2c1t
2
2c1
3
dt
1
(35)
0
The required necessary condition for
A2 (t1 ) to be
2c11 e ( 1 t ) 2c1 e ( 1 t )
dA (t )
2 3
minimized is 2 1 0 , that is
dt
R (t t )
R (t t )
t
1
k2 t
1
0 (e 1 1) dt
k1 (e 1 1) k3 (t1 T ) 0
(36)
1
k
b R t b
Let p(t ) k 2 (et1 1) k (t
T ) ,
3
3
2
2
3
3
1 0 e 1 1 e 1
1 1 3 1
b 2c 2c 2c
(37)
1 1 1 1 e 1 1 1 e 1
k
2 3
4 4
since p(t ) k 1 et1 k
0 , which
R0 R0 (t )
(32)
1 1 3
2 1 1
implies that p(t1 ) is strictly monotonically
The total amount of shortage during the interval [t1, T]
T
BT I (t)dt
t1
2 T
I (t)dt I (t)dt
increasing function during the interval
1 t1 2 .
Property-2
The constant deteriorating rate of an inventory model with quadratic trapezoidal type demand function during the time interval 1 t1 2 ,
t1 2
A (t ) attains its minimum at t* if
2 1 1 1
t* and A (t )
attains its minimum
1 1 2 1
1
1
2
2
at t*
if 2
t* .
1
1
Now, we can calculate the total amount of back- order quantity at the end of the cycle is
R t * b2 (T 2 2
• c2
t1
3 3
2 0 1 2
) (T
2 3
2 2 )
(38)
DT S R(t)dt
Therefore, the optimal order quantity denoted by 0
Q* is Q* S* , where
2
S * denotes the
1
S (b t c t 2 )dt
2 t1
R dt b t c t 2 )dt
optimal vale of S.
Case-III 2 t1 T
1 1
0
b 2c t 2c
b t c t2
0 2 2
1 2
For the time interval t
[
,T ) , then, the
I (t) 1 1 1 1 1
1 2 2 3
3
3
differential equations governing the inventory model can be expressed as follows:
2
b t c t
b t c t
b
b
2 1 2 1 2
2
2
2c2t1
2
2c2 e (t1 t )
dI (t) I (t) b t c t 2 0 , 0 t
(39)
dt 1 1 1
b 2c b t 2
dI (t)
1 1 2 1
c2t
• b2
2c2t1 2c2 et1
I (t) R0 0 , 1 t 2
dt
(40)
2 3 2 3
dI (t) I (t) b t c t 2 0 , t t
(41)
b1 2c11 2c1 1
dt 2 2 2 1
2 3
e
e
dI (t) 2
b2 2c2 2
2c2
b1 2
c1 3
dt b2t c2t
0 , t1 t T
(42)
2
3 e
2
2
1 3 1
Solving the differential equations (39)- (42) with I(t1)=0, b c
we can get
R ( ) 2 (t 2 2 ) 2 (t 3 3 )
b 2c
2c
2c
b 2c
0 2 1 2 1
2 3 1 2
2 2 2 2 e ( 2 t ) 1 1 1 1 e ( 1 t ) ,
(48)
2
3
3
2
The total amount of inventory carried during the
0 t 1
(43)
time interval [0, t1] is
t1
b t
• c t 2 b
C I (t)dt
2 1 2 1 2 T
2
2
I (t) R0
e (t1 t ) 0
2c t 2c
1 2 t1
2 1 2
2 3
I (t)dt
0
I (t)dt
1
I (t)dt
2
b2 2c2 2 2c2 e ( 2 t ) ,
t
(44)
2 3 1 2
b 2c t 2c b t c t 2
b t c t 2 b 2c t
2c
b 2c t
2c b t c t 2 ,
1 1 1 1
3
3
e
e
1
1
2
2
2
I (t) 2 1
2 1 2
2 1
2 e (t1 t ) 2
2 2 2 2
b t
• c t
b 2c t
2c
2 2
3
2 3
2 1 2 1 2
2
2 1 2
2 3
(t1 t )
2 t t1
b c
(45)
1
b 2c
2c
dt
I (t)
2 (t 2 t 2 ) 2 (t 3 t 3 ) , t
t T
0 2
2 2
2 e ( 2 t )
2 1 3 1 1
2
3
(46)The total amount of inventory level at the beginning
2c
b 2c
can be computed as
1 1 1 1 e ( 1 t )
b 2c
3 2
S I (0) 1 1
2 3
R b t
• c t 2 b
2c t
2c
2 0 2 1 2 1 2 2 1 2 e (t1 t )
b2t1 c2t1 b2
2c2t1
2c2
t1
2
2 2
3
2 2
3
e
dt
1 b2 2c2 2 2c2 e ( 2 t )
b2 2c2 2
2c
2
2
e
2c
2
b1 2c11 1
1
1
/p>
2
e
e
3
2
3
3
2
t1 2 2
2
2
(47)
b2t1 c2t1 b2 2c2t1 2c2 e (t1 t ) b2 2c2t 2c2 b2t c2t dt
The total amount of items which is perish within the time interval [0, t1] is
2
2 3
2 3
b t c t2
2c t b
2c et1 1
k1
t
2 1 2 1 2 1 2 2
since p (t1 )
k1 e 1
k3 0 , which
3
3
2
b 2c 2c e2 1
implies that p(t1 ) is strictly monotonically
2 2 2 2
increasing function within the interval
2 3
t [ , T ] .
2c b 2c e1 1 b 2 c 3 b
2c
1 2
c 2
1 1 1 1 1 1 1 1 1 1 1 1 1
3
2 2
3 2
3 2
Property-3
In this case, the inventory model under the
R0 ( ) b2 (t 2 2 ) c2 (t 3 3 )
condition 2 t1 T , A3 (t1 ) attains its
2 1
2 1
2 3 1 2
minimum at t t* , where
1 1
1 1
b2
2c2 (t
) c2 (t 2 2 )
(49)
p(t* ) 0 if t* . On the other hand,
2
3 1
2 2 1 2
1 2 1
A (t )
attains its minimum
Total quantity of shortage during the time interval [t1, T] is 3 1
1
1
1
1
2
2
2
2
T at t*
if t*
.Now, we can calculate the
BT I (t)dt
total back-order quantity at the end of the cycle is
1
1
2 (t t )
2 (t t )
t b2 (T 2 t * 2 ) c2 (t *3 T 3 ) .
T b c
3 2 1 3 1
2 3 3 dt
2 3 3 dt
2 2 2
1
t1
3 (t t1 )
Therefore, the optimal order quantity, denoted by Q* , is Q* S * , where S * denotes the
b2 t 2 (T t ) b2 (T 3 t
2 1 1 6
c t3
2 1 (T t1)
3
3 ) c2 (T 4 t 4 )
1 12 1
(50)
3
optimal value of S.From the above three cases, we can derive the following results
Result-1
An inventory model having constant deteriorating
Then, the total average cost per unit time under the time interval 2 t1 T , can be written as
rate with quadratic trapezoidal type demand, the optimal replenishment time is * and
A (t ) 1 [k k D k C k B ] (51)
t
t
1
1
A (t ) attains its minimum at t t* if and only
3 1 T 0 1 T 2 T 3 T
1 1 1 1
if t* . On the other hand, A (t ) attains its
The first order derivative of
A (t ) with respect to t is as 1 1 2 1
3 1 1
minimum at t t* if and only if t*
follows:
1 1 1 1 2
and A (t ) attains its minimum at t t* if and
dA3 (t1 ) 1 k k2 et1 1) k (t T )(b t c t 2 ) 3 1 1 1
dt T 1
(
3 1 2 1 2 1
only if
t* , where t * is the unique solution of
1
2 1 1
The required necessary condition for minimized is
dA (t )
(52)
A3 (t1 ) to be
equation p(t1 ) 0 .
Example 1
We can consider suitable values of the following
3 1 0 , that is
dt
parameters as follows: T= 20 weeks, 1 = 6
1 weeks,
2 =15 weeks, b1=10 unit, c1= 5 unit,
1 k
k2 (et1 1) k (t
T )
b2=20 unit, c2= 2 unit, 0.1, k0=\$220, k1= \$3
T
T
1
1
3 1
(53)
per unit, k2=\$12 per unit, k3=\$4 per unit. By Using MATHEMATICA 8.0 the above data, we can find
1 1 1 1
1 1 1 1
(b t c t 2 ) 0
p(1 ) =168.1206>0, the optimal replenishment
time t * =3.41 weeks, the optimal order quantity
This implies that 1
t
t
k
Q*, for each ordering cycle, is 3576.478 unit and
k1 2 (e 1 1) k3 (t1 T ) 0
(54)
the minimum cost A (t * )=\$4688.2
1 1
Let p(t ) k k2 (et1 1) k (t
T ) ,(55)
1 1
1 1
3 1
parameter
b1
Table
1 1
1 1
0 0 t1 Q ( +50 3.53 3674.30 4894 + 25 3.50 3664.02 4883 + 20 3.45 3584.112 4730 +10 3.41 3576.478 4688 -10 3.30 3576.478 4626
0 0 t1 Q ( +50 3.53 3674.30 4894 + 25 3.50 3664.02 4883 + 20 3.45 3584.112 4730 +10 3.41 3576.478 4688 -10 3.30 3576.478 4626
A t)
.2
.5
.1
.2
.4
+ 50 3.86 3665.4 4867.3
+ 25 3.77 3584.7 4856.9
+ 20 3.66 3563.2 4852.2
+10 3.512 3533.4 4847.3
-10 3.487 3489.7 4842.2
-20 3.415 3477.8 4834.6
-25 3.330 3465,4 4830.7
-50 3.311 3443.6 4822.4
-20 3.27 3560.9 4610.7 -25 3.2 3284.37 4577.6 + 50 4.265 2824.5 3346.8 -50 3.17 3225.33 4566.9 + 25 3.880 2753.8 3384.2
#### .
+50 3.07 3888.237 4331.4
+25 3.31 3424.66 4874.7
+20 3.35 3234.050 3814.8
+10 3.44 3021.12 3 3665.3
c1 -10 3.73 2734.464 3396.0
+ 20 3.825 2757.2 3391.1
+10 3.654 2631.8 3513.6
k1 -10 3.057 2566.3 3544.2
-20 2.879 2108.1 3573.2
-25 2.533 2015.5 3604.9
-50 1.865 1994.2 3604.9
-20 3.82 2494.19 3123.4 -25 4.26 2124.38 3068.8 + 50 2.936 2724.6 3674.2 -50 4.41 1912.01 2894.4 +25 3.833 2675.7 3624.1
+ 50 2.99 2887.633 4843.2
+10 3.63 3476.768 4655.6
b2 -10 3.69 3534.739 4596.7
+20 3.714 2634.3 3600.5
+ 10 3.685 2536.3 3557.4
k2 -10 3.612 2222.1 3487.2
– 20 3.467 2185.2 3426.2
-25 3.345 2105.4 3385.1
-50 3.292 1538.7 3114.8
+25 3.48 3331.54 4771.3 +20 3.55 3379.18 4733.8
-20 3.76 3581.35 4534.2 . -25 3.88 3636.42 4412.5 +50 4.265 2824.5 3346.8 -50 4.75 3675.55 4385.4 + 25 3.884 2753 3384.2
50 3.645 3658.4 4855.0
25 3.583 3497.2 4839.7
20 3.572 3484.5 4834.6
10 3.557 3448.1 4826.4
c2 -10 3.534 3436.3 4823.1
+ 20 3.825 2657.2 3391.1
+ 10 3.654 2631.8 3513.6
3 -20 2.879 2108.1 3573.2 -25 2.533 2015.5 3604.9 -50 1.865 19994.2 3688.3
3 -20 2.879 2108.1 3573.2 -25 2.533 2015.5 3604.9 -50 1.865 19994.2 3688.3
k -10 3.057 2566.3 3544.2
performed by changing the parameter -50%, -25%, -20, –
-20 3.475 3401.2 4818.5 -25 3.461 3382.3 4806.9 -50 3.391 3232.4 4806.9 In the above table some sensitivity analysis of the model is
-20 3.475 3401.2 4818.5 -25 3.461 3382.3 4806.9 -50 3.391 3232.4 4806.9 In the above table some sensitivity analysis of the model is
10, 10%,20%, 25%, and 50%, taking one at time and keeping the remaining parameters unaltered.
CONCLUSION 4
In a realistic product life cycle, demand is increasing with time during the growth phase. Then, after reaching its peak, the demand becomes stable for a finite time period called the maturity phase. Thereafter, the demand starts decreasing with time. Therefore, in this paper, we study the inventory model for constant deteriorating items with quadratic trapezoidal demand rate. We proposed an inventory replenishment policy for this type of inventory model. From the market information, we find that the quadratic trapezoidal type demand rate is more realistic than ramp type demand rate, constant demand rate and other time dependent demand rate Our paper provides an interesting topic for the future study of such kind of important inventory models, and at the same time, the following problems can be considered for future research work (1) How about the inventory model starting with shortages? (2) How about the inventory model with time dependent deteriorating rate instead of constant deteriorating rate?
5. REFERENCES:
1. Abad, P. L. (1996). Optimal pricing and lot sizing under conditions of perish ability and partial back ordering. Management Science, 42, 1093-1104.
2. Abad, P. L. (2001). Optimal price and order size for a reseller under partial back ordering. Computers and Operations Research, 28, 53-65.
3. Benkherouf, L., Bluemner, A.and Aggoun, L. (2003). A diffusion inventory model for deteriorating items. Applied Mathematics and Computation, 138, 21-39.
4. Bhattacharya, D. K. (2005). On multii item inventory. European Journal of Operational Research, 162, 786-791.
5. Cheng, M., Zhang, B. and Wang, G. (2011). Optimal policy for deteriorating items with trapezoidal type demand and partial backlogging. Applied Mathematical Modeling, 35, 3552-3560.
6. Chu, P., Yang, K. L., Liang, S. K. and Niu, T. (2004). Note on inventory model with a mixture of back orders and lost sales. European Journal of Operational Research, 159, 470-475.
7. Chung, K. J., Chu, P. and Lan, S. P. (2000). A note on EOQ models for deteriorating items under stock dependent selling rate. European Journal of Operational Research, 124, 550-559.
8. Chung, K. J. and Lin, C. N. (2001). Optimal inventory replenishment models for deteriorating items taking account of time discounting. Computers and Operations Research, 28, 67- 83.
9. Chung, M. and Wang, G. (2009). A note on the inventory model for deteriorating items with trapezoidal type demand rate. Computers and Industrial Engineering, 56, 1296-1300.
10. Dave, U. (1989). A deterministic lot size inventory model with shortages and a linear trend in demand. Naval Research Logistics, 36, 507-514.
11. Deng, S. (2005). Improved inventory models with ramp type demand and Weibull deterioration. International Journal of Information and Management Sciences, 16, 79-86.
12. Deng, P. S., Lin, R. and Peter, Chu. P. (2007). A note on inventory models for deteriorating items with ramp type demand rate. European Journal of Operational Research, 178, 112-120.
13. Gobinda Chandra Panda, Satyajit Sahoo1, Pravat Kumar Sukla,(2012) Analysis of Constant Deteriorating Inventory Management with Quadratic Demand Rate , American Journal of Operational Research 2012, 2(6): 98-10
14. G. Ch panda S sahoo., P.K sukla., (2013), A note on inventory model for ameliorating items with time dependent second order demand rate. LogForum 9 (1), 43-49
15. Ghare, P. M. and Schrader, G. P. (1963). A model for an exponentially decaying inventory, Journal of Industrial Engineering, 14, 5.
16. Giri, B. C. and Chaudhuri, K. S. (1998). Deterministic models of perishable inventory with stock dependent demand rate and non- linear holding cost. European Journal of Operational Research, 105, 467-474.
17. Giri, B. C., Jalan, A. K. and Chaudhuri, K. S. (2003). Economic order quantity model with Weibull deterioration distribution, shortage and ramp type demand, International Journal of system and science, 34, 237-243.
18. Hennery, R. J. (1979). Inventory replenishment policy for increasing demand, Journal of The Operational Research Society, 30, 611-617.
19. Hill, R. M. (1995). Inventory models for increasing demand followed by level demand. Journal of The Operational Research Society, 46, 1250-1259.
20. Hung, K. C. (2011). An inventory model with generalized type demand, deterioration and back order rates. European Journal of Operational Research, 208, 239-242.
21. Mandal, B. and Pal, A. K. (1998). Order level inventory system with ramp type demand rate for deteriorating items. Journal of Interdisciplinary Mathematics, 1, 49-66.
22. Manna, S. K. and Chaudhuri, K. S. (2006). An EOQ model with ramp type demand rate, time dependent deterioration rate, unit production cost and shortages. European Journal of Operational Research, 171, 557-566.
23. Mishra, V. K., Singh, L. S., Kumar, R. (2013). An inventory model for deteriorating items with time dependent demad and time varying holding cost under partial backlogging. Journal of Industrial Engineering International, 9, 4-8.
24. Padmanabhan, G. (1990). Inventory model with a mixture of back-orders and lost sales. International Journal of System Science, 21, 1721-1726.
25. P.K.Sukla(2012) An Inventory Ordering Policy Using Constant Deteriorating Items With Constant Demand ,IJERT Vol. 1 Issue 7, September
26. P.K.Sukla S.K.Sahu(2008) ,A note for weibul deteriorating model with time varying demand and partial back-ordering.Acta Cienia Indica,vol,xxxiv M,No,4, 1673-1670.
27. Panad, S., Senapati, S. and Basu, M. (2008). Optimal replenishment policy for perishable seasonal products in a season with ramp type time dependent demand.Computers and Industrial Engineering, 54, 301-314.
28. Panda, S., Saha, S. and Basu, M. (2007). An EOQ model with generalized ramp type demand and Weibull distribution deterioration. Asia-Pacific Journal of Operations Research, 24, 93-109.
29. Resh, M., Friedman, M. and Barbosa, L. C. (1979). On a general solution of the deterministic lot size problem with time proportional demand. Operations Research, 24, 718-725.
30. Sachan, R. S. (1984). On policy inventory model for deteriorating items with time proportional demand. Journal of The Operational Research Society, 35, 1013-1019.
31. Set, B. K., Sarkar, B. and Goswami, A. (2013). A two-warehouse inventory model with increasing demand and time varying deterioration. Scientia Iranica E, 19, 1969-1977.
32. Teng, J. T. (1996). A deterministic replenishment model with linear trend in demand. Operations Research Letters, 19, 33-41.
33. Teng, J. T., Chang, H. J., Dye, C. Y. and Hung, C. H. (2002). An optimal replenishment policy for deteriorating items with time varying demand and partial backlogging. Operations Research Letters, 30, 387-393.
34. Wee, H. M. (1993). Economic production lot size model for deteriorating items with partial back ordering. Computer and Industrial Engineering, 24, 449-458.
35. Wee, H. M. (1995). Joint price and replenishment policy for deteriorating inventory with declining market. International Journal of Production Economics, 40, 163-171.
36. Wee, H. M. and Law, S. T. (1999). Economic production lot size for deteriorating items taking account of the time value of money. Computers and Operations Research, 26, 545-558.
37. Whitin, T. M. (1957). Theory of inventory management. Princeton University Press, Princeton NJ, 62-72.
38. Wu, J. W., Lin, C., Tan, B. and Lee, W. C. (1999). An EOQ with ramp type demand rate for deteriorating items with Weibull deterioration. International Journal of Information and Management Sciences, 10, 41-51.
39. Wu, K. S. (2001). An EOQ inventory model for items with Weibull distribution deterioration, ramp type demand rate and partial back logging. Production Planning and Control, 12, 787- 793.
40. Wu, K. S. and Ouyang, L. Y. (2000). A replenishment policy for deteriorating items with ramp type demand rate. Proceeding of National Science Council ROC (A), 24, 279-286.
41. Wu, K. S., Ouyang, L. Y. and Yang, C. T. (2006). An optimal replenishment policy for non-instantaneous deteriorating items with stock dependent demand and partial back logging. International Journal of Production Economics, 101, 369-384.
42. Wu, K. S., Ouyang, L. Y. and Yang, C. T. (2008). Retailers optimal ordering policy for deteriorating items with ramp type demand under stock dependent consumption rate. International Journal of Information and Management Sciences, 19, 245-262.
43. Shah, Nita H, Chaudhari Urmila and Jani Mrudul Y (2015), Optimal Policies for Time-Varying Deteriorating item with Preservation Technology under Selling Price and Trade Credit Dependent Quadratic demand in a Supply Chain, ,International Journal of Applied and Computational Mathematics 3(2) 363- 379.
44. Shah, Nita H, Chaudhari Urmila and Jani Mrudul Y (2016), Impact of future price increase on ordering policies for deteriorating polices for deteriorating items under under quadratic demand. 7(3), 423-436.
BIOGRAPHY
1. Pravat kumar sukla is working as a lecturer in mathematics Panchayat samiti college koksara, kalahandi, odisha, INDIA. He has obtained his M.sc and Ph.D in mathematics from Sambalpur University, odisha, India .he has published 11 research papers in different national and international journals. His research interests include inventory control.
3. Rajani Ballav Dash is Ex Reader in mathematics Revenshaw University, at present Visiting Faculty Revenshaw University, Cuttack. He has a retired principal,
S.C.S (Autonomous) college, Puri, odisha Ex- Visiting Faculty Institute of mathematics and Applications, bbsr
,odisha.Ten students have completed Ph.D under guidance. He has published 50 research papers different national and international journals and NO of text books -14. His research interests include inventory control and Numerical Analysics, Operations Research (Theorical and applied), Graph Theory and fracture mechanics.
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## 1523316000000
1,523,316,000,000 (one trillion five hundred twenty-three billion three hundred sixteen million) is an even thirteen-digits composite number following 1523315999999 and preceding 1523316000001. In scientific notation, it is written as 1.523316 × 1012. The sum of its digits is 21. It has a total of 16 prime factors and 252 positive divisors. There are 406,214,400,000 positive integers (up to 1523316000000) that are relatively prime to 1523316000000.
## Basic properties
• Is Prime? No
• Number parity Even
• Number length 13
• Sum of Digits 21
• Digital Root 3
## Name
Short name 1 trillion 523 billion 316 million one trillion five hundred twenty-three billion three hundred sixteen million
## Notation
Scientific notation 1.523316 × 1012 1.523316 × 1012
## Prime Factorization of 1523316000000
Prime Factorization 28 × 3 × 56 × 126943
Composite number
Distinct Factors Total Factors Radical ω(n) 4 Total number of distinct prime factors Ω(n) 16 Total number of prime factors rad(n) 3808290 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 0 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 1,523,316,000,000 is 28 × 3 × 56 × 126943. Since it has a total of 16 prime factors, 1,523,316,000,000 is a composite number.
## Divisors of 1523316000000
252 divisors
Even divisors 224 28 14 14
Total Divisors Sum of Divisors Aliquot Sum τ(n) 252 Total number of the positive divisors of n σ(n) 5.06778e+12 Sum of all the positive divisors of n s(n) 3.54446e+12 Sum of the proper positive divisors of n A(n) 2.01102e+10 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 1.23423e+06 Returns the nth root of the product of n divisors H(n) 75.7483 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 1,523,316,000,000 can be divided by 252 positive divisors (out of which 224 are even, and 28 are odd). The sum of these divisors (counting 1,523,316,000,000) is 5,067,777,631,616, the average is 201,102,286,96.,888.
## Other Arithmetic Functions (n = 1523316000000)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 406214400000 Total number of positive integers not greater than n that are coprime to n λ(n) 25388400000 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 56370827024 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 406,214,400,000 positive integers (less than 1,523,316,000,000) that are coprime with 1,523,316,000,000. And there are approximately 56,370,827,024 prime numbers less than or equal to 1,523,316,000,000.
## Divisibility of 1523316000000
m n mod m 2 3 4 5 6 7 8 9 0 0 0 0 0 4 0 3
The number 1,523,316,000,000 is divisible by 2, 3, 4, 5, 6 and 8.
• Abundant
• Polite
• Practical
• Frugal
## Base conversion (1523316000000)
Base System Value
2 Binary 10110001010101100101101011000010100000000
3 Ternary 12101121221111210122211010
4 Quaternary 112022230231120110000
5 Quinary 144424222344000000
6 Senary 3123445100320520
8 Octal 26125455302400
10 Decimal 1523316000000
12 Duodecimal 20728b543140
20 Vigesimal 2ja1g50000
36 Base36 jfsv0k5c
## Basic calculations (n = 1523316000000)
### Multiplication
n×i
n×2 3046632000000 4569948000000 6093264000000 7616580000000
### Division
ni
n⁄2 7.61658e+11 5.07772e+11 3.80829e+11 3.04663e+11
### Exponentiation
ni
n2 2320491635856000000000000 3534842036765618496000000000000000000 5384681432077654904852736000000000000000000000000 8202571380386804959040650392576000000000000000000000000000000
### Nth Root
i√n
2√n 1.23423e+06 11506.1 1110.96 273.249
## 1523316000000 as geometric shapes
### Circle
Diameter 3.04663e+12 9.57128e+12 7.29004e+24
### Sphere
Volume 1.48067e+37 2.91602e+25 9.57128e+12
### Square
Length = n
Perimeter 6.09326e+12 2.32049e+24 2.15429e+12
### Cube
Length = n
Surface area 1.39229e+25 3.53484e+36 2.63846e+12
### Equilateral Triangle
Length = n
Perimeter 4.56995e+12 1.0048e+24 1.31923e+12
### Triangular Pyramid
Length = n
Surface area 4.01921e+24 4.16585e+35 1.24378e+12
## Cryptographic Hash Functions
md5 f98afa512564fd2d5c21c489ef14af47 f8d32c908858f7fe4e56990420cc33526f53e2ef abbc29f601d4e9b5cae02618a0f0f72f9614226233ceeab7a269b2418ac46778 6f9e803558dab6804869fa1a99aac0269a14dc9cec07aca9ed9b78b59006e9f5750a5330459f33fc215c52930a23250cf569a3f644480c9917a4d09eabb2afbb d97d808ad78f245ff2c9049af930e4f50a4309f8
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0
424
6
I need help... I did most of my hw, but these just can't? IDK. but...
Among all fractions $x$ that have a positive integer numerator and denominator and satisfy $$\frac{9}{11} \le x \le \frac{11}{13},$$ which fraction has the smallest denominator?
AND
Evaluate: $\dfrac{(x+y)^2 - (x-y)^2}{y}$ for $x=6$, $y \not= 0$.
Dec 2, 2018
#1
+4569
+1
1. First, try to change the x in the middle, to a fraction. Let's say x/y. We then multiply both sides by b, to get 9b/11 and 11b/13. Now, the trick here is to find a positive integer that satisfies and is less than 1, since 11/13 is less than 1.
Dec 2, 2018
#2
+6185
+2
deleted
.
Dec 2, 2018
edited by Rom Dec 3, 2018
edited by Rom Dec 3, 2018
#3
+4569
+2
Not quite, Rom. Continuing with my method and plugging the values of b, we get b=6, so the answer is $$\boxed{\frac{5}{6}}.$$ Can someone verify this?
tertre Dec 3, 2018
#4
+111321
+1
9/11 = .8181.....
5/6 = .8333....
11/13 = .846....
Mmmmm.....it looks like you could be correct, tertre.....
CPhill Dec 3, 2018
edited by CPhill Dec 3, 2018
edited by CPhill Dec 3, 2018
edited by CPhill Dec 3, 2018
#5
+111321
+3
Here might be another way to look at this
Notice that the fractions seem to have the form n / [ n + 2] where n is an integer
Suppose that there exists a fraction such that
9/11 < n / [n + 2] < 11/13
We can split this into two inequalities
Looking at the inequality on the left....we have
9[n + 2] < 11n
9n + 18 < 11n
18 < 2n
9 < n or
n > 9
Looking at the inequality on the right.... we have
13(n) < 11[n + 2]
13n < 11n + 22
2n < 22
n < 11
This implies that
9 < n < 11
So...n = 10 is the only integer that satisfies this
And n + 2 = 12
So....the fraction is 10/12 = 5/6 ......just as tertre found!!!
Good job, tertre !!!!
Dec 3, 2018
edited by CPhill Dec 3, 2018
#6
+4569
+1
Thank you, CPhill! Great solution!
tertre Dec 3, 2018
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### AP Physics 1 Week 7 (July 23 - 29, 2017)
Part 1: Algebra (Rearrange)
1. solve for l
2. solve for x
3. solve for v
4. solve for VR
5. solve for q
Part 2: Algebra (Inverse and Direct Relationships)
6. The rate of flow of water emerging from the end of a circular pipe under a fixed pressure is directly proportional to the square of the radius of the pipe. If a pipe of two inch radius emits 864 cubic inches of water per second, at what rate will a pipe of 2.25 inches radius emit water under the same pressure? (DBW, p224, #6)
7. The power required to drive a nautical vessel varies directly as the cube of its speed. If 450 horsepower are needed to run the Bombay Queen at 6 knots, what horsepower is required to drive her at 8 knots? (DBW, p224, #7)
8. The maximum deflection of a beam is directly proportional to the cube of its length. If a maximum deflection of 0.002 centimeters occurs in a beam 10 meters long, find the deflection in a 6 meter beam. (DBW, p224, #8)
9. A square plate 3 inches wide has a moment of intertia of 6.75. What is the moment of inertia of a square plate 2 inches wide, if the moment of inertia varies directly as the fourth power of the width of the plate? (DBW, p224, #9)
10. The crushing load of a circular pillar with a 10 inch diameter is 200 tons. Find the crushing load of a pillar of the same height and material but with a diameter of 14 inches, if the rushing load is directly proportional to the fourth power of the diameter.
(DBW, p224, #10)
Part 3: Geometry/Trigonometry
(For Physics, make sure your calculator is set in degrees)
Pythagorean Theorem (math version) c2 = a2 + b2, (physics version) r2 = x2 + y2
r = hypotenuse of the right triangle
x = horizontal or (east-west) component of the triangle
y = vertical or (north-south) component of the triangle
q = the reference angle (theta) always between the x axis and the hypotenuse
horizontal component x = r*cos(q) vertical component y = r*sin(q)
11. Jerry walks 1 block (217 feet) east along a vacant lot and then 2 blocks (400 feet) north to a friends house. Pam starts at the same point and walks diagonally through the vacant lot coming out at the same point as Jerry. How far did Pam walk?
(DBW, p 393, #1)
12. In a naval maneuver two ships rendezvous at position A. One then proceeds east 10 miles and north 14 miles to position B. At what bearing or direction (angle) should the second ship head to meet the first ship at position B? (DBW, p 394, #3)
13. A jet heads due west at 627 miles per hour. If a 25 mile per hour north wind is blowing, what is the planes ground speed and course (direction)? (DBW, p 394, #4)
14. At what bearing or direction (angle) and speed would a pilot head if he wants to fly due north at 345 miles per hour when a 40 mile per hour west wind is blowing?
(DBW, p 394, #5)
15. At what bearing and speed should the navigator direct the captain of a ship to head, if the captain wants to steam straight ahead at 18.2 knots when seas of 7.4 knots are hitting the ship broadside? (DBW, p 394, #6)
Part 4: Word Problems
16. Tonya's outboard can drive her boat at 7 mph in still water. It takes her 10 minutes
more to reach her friends camp 4 miles up the river that to return to her camp
down river. What is the speed of the current? (DBW, p201, 39)
17. Jack was standing on a direct line between Tom and the point on the surface of the earth where a bolt of lightning struck. If Tom heard the sound of the thunder
associated with the lightning 5.5 seconds after it struck, how far was he standing
from Jack if Jack was standing 4480 feet from the point where the bolt struck.
Jack heard the thunder 1.5 seconds before Tom. (DBW, p217, #13)
18. At noon a ship sailed east from Key West. Two hours later another ship traveling
4 mph faster sailed due south from that port. At 10 pm the same (Key West
time), the ships were 100 miles apart. What was the average speed of each?
(DBW, p296, #9)
19. Park City is 23 miles directly south of Adobe Flats. A train running due west at 40
mph left Park City at the same time another train traveling due south at the same speed left Adobe Flats. How soon were the trains 17 miles apart?
(DBW, p297, #10)
20. When a favorable wind caused an increase of 20 mph over the usual speed of the
plane, the pilot made the 980 mile trip between 2 cities in 7 minutes less time.
Find the usual speed of the plane. (DBW, p326, #8)
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# American Institute of Mathematical Sciences
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Stabilization of discrete-time Markovian jump systems with partially unknown transition probabilities
November 2011, 16(4): 1185-1195. doi: 10.3934/dcdsb.2011.16.1185
## Necessary and sufficient conditions for stability of impulsive switched linear systems
1 Department of Mathematics and Statistics, Curtin University of Technology, Perth, WA 6845 2 Department of Mathematics and Statistics, Curtin University of Technology, GPO Box U 1987, Perth, W.A. 6845 3 School of Information Science and Engineering, Central South University, Changsha, 410083, China
Received October 2010 Revised June 2011 Published August 2011
This paper addresses fundamental stability problems of impulsive switched linear systems, featuring given impulsive switching time intervals and switching rules. First, based on the state dynamical behaviors, we construct a new state transition-like matrix, called an impulsive-type state transition (IST) matrix. Then, based on the IST matrix and Lyapunov stability theory, necessary and sufficient conditions for the uniform stability, uniform asymptotic stability, and exponential stability of impulsive switched linear systems are established. These stability conditions require the testing on the IST matrix of the impulsive switched linear systems. The results can be reduced to those for switched linear systems without impulsive effects, and also to those for impulsive linear systems without switchings.
Citation: Honglei Xu, Kok Lay Teo, Weihua Gui. Necessary and sufficient conditions for stability of impulsive switched linear systems. Discrete & Continuous Dynamical Systems - B, 2011, 16 (4) : 1185-1195. doi: 10.3934/dcdsb.2011.16.1185
##### References:
show all references
##### References:
[1] Elena K. Kostousova. State estimation for linear impulsive differential systems through polyhedral techniques. Conference Publications, 2009, 2009 (Special) : 466-475. doi: 10.3934/proc.2009.2009.466 [2] Ugo Boscain, Grégoire Charlot, Mario Sigalotti. Stability of planar nonlinear switched systems. Discrete & Continuous Dynamical Systems - A, 2006, 15 (2) : 415-432. doi: 10.3934/dcds.2006.15.415 [3] Philippe Jouan, Said Naciri. Asymptotic stability of uniformly bounded nonlinear switched systems. Mathematical Control & Related Fields, 2013, 3 (3) : 323-345. doi: 10.3934/mcrf.2013.3.323 [4] Moussa Balde, Ugo Boscain. Stability of planar switched systems: The nondiagonalizable case. Communications on Pure & Applied Analysis, 2008, 7 (1) : 1-21. doi: 10.3934/cpaa.2008.7.1 [5] Alexander Pimenov, Dmitrii I. Rachinskii. Linear stability analysis of systems with Preisach memory. Discrete & Continuous Dynamical Systems - B, 2009, 11 (4) : 997-1018. doi: 10.3934/dcdsb.2009.11.997 [6] Xiang Xie, Honglei Xu, Xinming Cheng, Yilun Yu. Improved results on exponential stability of discrete-time switched delay systems. Discrete & Continuous Dynamical Systems - B, 2017, 22 (1) : 199-208. doi: 10.3934/dcdsb.2017010 [7] Michael Basin, Pablo Rodriguez-Ramirez. An optimal impulsive control regulator for linear systems. Numerical Algebra, Control & Optimization, 2011, 1 (2) : 275-282. doi: 10.3934/naco.2011.1.275 [8] Xueyan Yang, Xiaodi Li, Qiang Xi, Peiyong Duan. Review of stability and stabilization for impulsive delayed systems. Mathematical Biosciences & Engineering, 2018, 15 (6) : 1495-1515. doi: 10.3934/mbe.2018069 [9] Lin Du, Yun Zhang. $\mathcal{H}_∞$ filtering for switched nonlinear systems: A state projection method. Journal of Industrial & Management Optimization, 2018, 14 (1) : 19-33. doi: 10.3934/jimo.2017035 [10] Hongyan Yan, Yun Sun, Yuanguo Zhu. A linear-quadratic control problem of uncertain discrete-time switched systems. Journal of Industrial & Management Optimization, 2017, 13 (1) : 267-282. doi: 10.3934/jimo.2016016 [11] Canghua Jiang, Kok Lay Teo, Ryan Loxton, Guang-Ren Duan. A neighboring extremal solution for an optimal switched impulsive control problem. Journal of Industrial & Management Optimization, 2012, 8 (3) : 591-609. doi: 10.3934/jimo.2012.8.591 [12] Stepan Sorokin, Maxim Staritsyn. Feedback necessary optimality conditions for a class of terminally constrained state-linear variational problems inspired by impulsive control. Numerical Algebra, Control & Optimization, 2017, 7 (2) : 201-210. doi: 10.3934/naco.2017014 [13] Xiaojun Zhou, Chunhua Yang, Weihua Gui. State transition algorithm. Journal of Industrial & Management Optimization, 2012, 8 (4) : 1039-1056. doi: 10.3934/jimo.2012.8.1039 [14] Daniel Alpay, Eduard Tsekanovskiĭ. Subclasses of Herglotz-Nevanlinna matrix-valued functtons and linear systems. Conference Publications, 2001, 2001 (Special) : 1-13. doi: 10.3934/proc.2001.2001.1 [15] Peter Howard, Bongsuk Kwon. Spectral analysis for transition front solutions in Cahn-Hilliard systems. Discrete & Continuous Dynamical Systems - A, 2012, 32 (1) : 125-166. doi: 10.3934/dcds.2012.32.125 [16] Evelyn Buckwar, Girolama Notarangelo. A note on the analysis of asymptotic mean-square stability properties for systems of linear stochastic delay differential equations. Discrete & Continuous Dynamical Systems - B, 2013, 18 (6) : 1521-1531. doi: 10.3934/dcdsb.2013.18.1521 [17] Mikhail Gusev. On reachability analysis for nonlinear control systems with state constraints. Conference Publications, 2015, 2015 (special) : 579-587. doi: 10.3934/proc.2015.0579 [18] Shengji Li, Chunmei Liao, Minghua Li. Stability analysis of parametric variational systems. Numerical Algebra, Control & Optimization, 2011, 1 (2) : 317-331. doi: 10.3934/naco.2011.1.317 [19] Behrouz Kheirfam. Multi-parametric sensitivity analysis of the constraint matrix in piecewise linear fractional programming. Journal of Industrial & Management Optimization, 2010, 6 (2) : 347-361. doi: 10.3934/jimo.2010.6.347 [20] Eugen Stumpf. Local stability analysis of differential equations with state-dependent delay. Discrete & Continuous Dynamical Systems - A, 2016, 36 (6) : 3445-3461. doi: 10.3934/dcds.2016.36.3445
2018 Impact Factor: 1.008
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## Contents
Gunning Fog, Flesch Reading Ease, and Flesch-Kincaid are reading level algorithms that can be helpful in determining how readable your content is. Reading level algorithms only provide a rough guide, as they tend to reward short sentences made up of short words. Whilst they're rough guides, they can give a useful indication as to whether you've pitched your content at the right level for your intended audience.
## Interpreting the Results
This service analyses the readability of all rendered content. Unfortunately, this will include navigation items, and other short items of content that do not make up the part of the page that is intended to be the subject of the readability test. These items are likely to skew the results. The difference will be minimal in situations where the copy content is much larger than the navigation items, but documents with little content but lots of navigation items will return results that aren't correct.
Philip Chalmers of Benefit from IT provided the following typical Fog Index scores, to help ascertain the readability of documents.
Typical Fog Index Scores
Fog Index Resources
6 TV guides, The Bible, Mark Twain
8 - 10 Most popular novels
10 Time, Newsweek
11 Wall Street Journal
14 The Times, The Guardian
Over 20 Only government sites can get away with this, because you can't ignore them.
Over 30 The government is covering something up
The following table contains the readability results for https://promus32.blogspot.com/.
Summary Value
Total sentences 232
Total words 1155
Average words per Sentence 4.98
Words with 1 Syllable 484
Words with 2 Syllables 317
Words with 3 Syllables 145
Words with 4 or more Syllables 209
Percentage of word with three or more syllables 30.65%
Average Syllables per Word 2.07
Gunning Fog Index 14.25
## Gunning-Fog Index
The following is the algorithm to determine the Gunning-Fog index.
• Calculate the average number of words you use per sentence.
• Calculate the percentage of difficult words in the sample (words with three or more syllables).
• Add the totals together, and multiply the sum by 0.4.
• Algorithm: (average_words_sentence + number_words_three_syllables_plus) * 0.4
The result is your Gunning-Fog index, which is a rough measure of how many years of schooling it would take someone to understand the content. The lower the number, the more understandable the content will be to your visitors. Results over seventeen are reported as seventeen, where seventeen is considered post-graduate level.
The following is the algorithm to determine the Flesch Reading Ease.
• Calculate the average number of words you use per sentence.
• Calculate the average number of syllables per word.
• Multiply the average number of syllables per word multiplied by 84.6 and subtract it from the average number of words multiplied by 1.015.
• Subtract the result from 206.835.
• Algorithm: 206.835 - (1.015 * average_words_sentence) - (84.6 * average_syllables_word)
The result is an index number that rates the text on a 100-point scale. The higher the score, the easier it is to understand the document. Authors are encouraged to aim for a score of approximately 60 to 70.
The following is the algorithm to determine the Flesch-Kincaid grade level.
• Calculate the average number of words you use per sentence.
• Calculate the average number of syllables per word.
• Multiply the average number of words by 0.39 and add it to the average number of syllables per word multiplied by 11.8.
• Subtract 15.50 from the result.
• Algorithm: (0.39 * average_words_sentence) + (11.8 * average_syllables_word) - 15.9
The result is the Flesch-Kincaid grade level. Like the Gunning-Fog index, it is a rough measure of how many years of schooling it would take someone to understand the content. Negative results are reported as zero, and numbers over twelve are reported as twelve.
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Desmond Robertson
2023-02-26
Derive relationship between Moment of inertia , angular velocity and torque
Kamari Potts
Angular momentum is given by,
$L=mv×r=\left(m{r}^{2}\right)\left(\frac{v}{r}\right)=I\omega$
I = moment of inertia = mr^2, ω = v/r = angular velocity.
Do you have a similar question?
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# By referring to the Universal Declaration of Human Rights as "purely programmatic" (line 49) in nature, the author mo...
question # 14
can someone please explain to me why answer choice b is somehow incorrect? the last sentence in the passage states just that "moons in solar system s4 all orbit the planet alpha" I got b
Replies
AndreaK on January 28, 2020
I am seeing video lessons for logical reasoning appear on my end when I attempt to view the question you posted about.
Is this what you're referring to? If so, if you are able to specify which video (and where within) or example from the lesson you're having trouble with, I'm happy to help you out!
Skylar on January 28, 2020
As you note, the passage concludes that "the moons in solar system S4 all orbit the planet Alpha." Our task is to find the answer choice that guarantees that this conclusion is true based off the premise that every moon orbits some planet in its solar system. Currently, this is a flawed argument because there is no support connecting the premise to the conclusion. Namely, how do we know that the moons in S4 all orbit Alpha? What's to say that they don't all orbit the planet Beta? Or what's to say that some of them orbit Beta and others orbit Alpha? We need to find an answer choice that will explain why Alpha is the specific planet that all of these moons orbit.
(B) is incorrect because it fails to specify why Alpha is the specific planet all the moons orbit. (B) states that "every moon in S4 orbits the same planet," but the identify of this "same planet" is left unspecified. Let's say that every moon in S4 orbits is Beta - this would make (B) true, but it would not make the conclusion of the passage (that all the moons in S4 order Alpha) true. This means that (B) does not guarantee the conclusion.
Does that make sense? Please let us know if you have any additional questions!
AndreaK on January 28, 2020
Oh, my bad. I see now you specified it was question 14!
So, this is a sufficient assumption question. That means, it's our goal to guarantee the truth of the conclusion. Our correct answer choice is going to give us the piece of information we need to do that.
Our conclusion is that the moons in solar system S4 *ALL* orbit planet alpha. Our evidence for that is: Any moon, by definition, orbits some planet in a solar system.
So, with just that piece of evidence (that any moon orbits some planet in a solar system), how are we then supposed to guarantee that ALL the moons in S4 orbit SPECIFICALLY the planet ALPHA?
Think about the information we don’t have about the solar system in question. For one, we don’t know if alpha is the only planet. If it were, then by definition, all the moons would have to orbit alpha. However, what if it isn't the only planet? If it’s not the only planet in S4, then maybe some or all of those moons are orbiting around another planet.
If our goal is guarantee that the author’s conclusion (the moons in solar system S4 *ALL* orbit planet ALPHA) is true, specifying that alpha is the ONLY planet in S4 would do the trick. Because then, we know those moons have no where else to orbit so they must all be orbiting alpha.
That gets us to answer choice C.
In answer choice B, “Every moon in S4 orbits the same planet†the problem is we don’t actually know what planet that is. They language says they all orbit “the same planet.†Well, what if that planet isn’t alpha? We would need to know that planet was alpha for this to work, which is why C is correct.
I see how this question could be tricky. My guess is you understood the concept, but didn’t realize you were relying on an unstated assumption in your reasoning. The assumption is what you need to isolate in your thinking, as opposed to using it (when it may be unwarranted to do so) to make other deductions with. The LSAT is all about bringing those unstated, sometimes obvious seeming assumptions to light. If you can spot the assumption you’ve targeted the place the right answer lies. Different questions stems have you manipulate assumptions in different ways.
Hope this helps!
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Malcolm Webb wrote on 2013-01-29 UTCBelowAverage ★★
```I give the below average rating because your description does not give me enough information to play the game. I make the following points:
A) The opening scenario is unnecessary and inappropriate. A force of commandos & terrorists would not bring their King into a battle, nor would the forces be identical. Chess dates from the time when Kings actually rode out to battle. A besieged castle would be a better description.
B) The 3-D board has all the white squares above or below the other white squares in a vertical column; the black squares are similarly arranged. This means that a Rook moving up or down (like a lift in a multi-storey building) would stay in cells of the same colour. Thus if you were looking at the building from the front, you would not see a checkered board; instead you would see four vertical columns of white squares alternating with four vertical columns of black squares. Was this your intention?
C) This being so, I do not understand how the Bishop moves when changing levels. In order to not change colour it would have to move "triagonally" (that is, through the corner of the cells), not diagonally. It would have to follow the same path you illustrate for pawn captures when they are going up or down. An illustration of the Bishop's move would help.
D) From your description, the Queen's move is identical to the Rook's move, namely an orthogonal move in all six directions. The only difference is that the Queen has the additional power to place an enemy King in check if they are both on the same level with no other enemy piece on the same level. Was this your intention?
E) Going from your description the checkmate position you illustrate is not checkmate at all. I presume that the Black Queen is the one under the White King. The Black Rook does not protect the Black Queen as it is in a diagonal line from the Queen. The White King could move to any of the five squares surrounding it on the same level without being threatened by the Black Queen or by any other piece. It could also move upwards diagonally or "triagonally" and not be threatened. The only square that it could not move to on the level above it would be the square directly above itself. In total I see eleven cells to which the King could move to escape check.
Perhaps illustration of the moves of the Rook, Bishop & Queen would assist, as I can only go by your verbal description.
The game will take quite a long time to play. This is not necessarily a fault. However it was for this reason that Ferdinand Maack created the original 3D chess as a 5 x 5 x 5 game.```
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## Discrete Electronic Warfare Signal Processing
Nov 10, 2015 - radar parameter measurement and processing in EW receivers need to be ... using random modulator and pre-integrator (RMPI) states that.
Defence Science Journal, Vol. 65, No. 6, November 2015, pp. 472-476, DOI : 10.14429/dsj.65.8414 2015, DESIDOC
Discrete Electronic Warfare Signal Processing using Compressed Sensing Based on Random Modulator Pre-integrator M. Sreenivasa Rao*, Chandan C. Mishra#, K. Krishna Naik#,!, and K. Maheshwara Reddy* Defence Avionics Research Establishment, Bangalore - 560 093, India # Defence Institute of Advanced Technology, Pune - 411 025, India ! E-mail: [email protected]
*
The RMPI provides with increased bandwidth and resolution without utilising superior ADCs below Nyquist rate. The RMPI is linear and time invariant within a sample period5,6. The paper is organised as RMPI theory, application to EW receiver, Simulation and recovery of radar signals. 2. THEORY OF RANDOM MODULATOR PREINTEGRATOR The input signal x(t) is randomised by the random sequence which spreads the spectrum over the entire bandwidth. The RMPI encodes compressed samples by modulating the input signal x(t) with a PN sequence p(t) within ±1s. Its chipping rate must be faster than the Nyquist rate of the input signal. The purpose of modulation is to provide randomness necessary for successful CS recovery. The modulation is followed by an Integrator with impulse response h(t). Finally, the signal is sampled at rate M using a traditional low rate ADC. The model used for RMPI theory is as shown in Fig. 1. The signal x(t) is defined on the interval [0,T] decomposes according to some set of N orthonormal basis functions {pn(t)} as given in the following inner product. x(t ) = ∑ n =1 xn , pn (t ) N
Figure1. The model used for RMPI analysis1.
Received 04 February 2015, revised 27 October 2015, online published 10 November 2015
472
(1)
Rao, et al.: Discrete EW Signal Processing using Compressed Sensing Based on RMPI
where pn(t) satisfy = 1 for n = m and 0 otherwise. The number of basis functions that represent all the waveforms {x(t)} may be infinite, xn = n ∆T, and n = 0,1,2…..N-1; i.e. N may equal ∞. Using the set of basic functions, the function x(t) maps to a set of N real numbers in a vector {xi}; these real-valued scalar coefficients assemble into an N-dimensional real-valued vector. X = [ X 1 , X 2 , X 3 ,... ... ... .. X N ] (2) The discrete signal x(n) and p(n) are discrete samples of x(t) and p(t). Random chip sequence pc(t) is periodic chipping sequence13. Discrete sequence p(n)={p0, p1, p2 , p3 - - - - -;pN}, p(n) ∈ {+1;-1} to be the pseudo-random bit sequence (PRBS). Discrete samples of x(t) are given by x(n) = w(t − nT ) ∫
n
n −1
where
n
n −1
x(t )dt
x(t )dt
(3)
is the average value of x(t) on the nth
interval giving a discrete value of x(n). w (t-nT) is a rectangular window function having magnitude one in interval [-∆T/2, ∆T/2] and 0 elsewhere. The frequency domain representation using Discrete Fourier Transform of input signal x(n) is given by N −1 c(n) = ∑ n = 0 q (n) h(mH − 1) (4) The frequency domain representation using Discrete Fourier Transform of signal p(n) is given by P ( K ) = ∑ n = 0 p ( n) e N −1
−j
2 πkn N
, k = 0,1,..... N − 1
(5)
Discrete signal model after the sampler is given by q(n)= p(n). x(n) (Time domain product) (6) The time domain product will have samples at the time instants of input signal at the rate of p(n) and will have the bandwidth of the sum of the input signal and p(n) pulse4. Hence, after the sampler the pulse output is spread and spreading is proportional to bandwidth p(n) sequence. Q(K)= P(K) ©X(K) (Circular convolution)
(7)
Q( K ) = ∑ 1= 0 X (1) | P ( K − 1) |N (Periodic convolution) (8) The discrete frequency domain samples are compressed. The input signal is spreading in the time domain and compressing in the frequency domain. The input signal is sampled with base vectors; sparse coefficients are extracted and input signal recovery is called compressed sensing. Since p (n) is truncated the sequence q(n) also must be converged. The reason of convergence for discrete sequence q(n) is important for stability and analysis using transform theory. In CS theory, inner product can be implemented by using discrete components and transform coding. The current transform coding theory suggests that time, frequency localisation and high resolution scaling of frequency domain to study the new signal properties. DFT coefficients are periodic and non-coherent. Optimum base coefficients with high resolution frequency domain scaling can be obtained using curve lets transform. The literature survey suggests that the use of directional transforms like discrete curve lets transform and shear-lets transform to study the physical signal new properties in CS receiver applications. Integrate the modulator output over discrete periodic time sequence with N −1
period N. The number of output samples is ‘m’ and sampling rate is M and the integrator impulse response is h (mM). The output of the integrator is convolution of input signal amplitude samples q(n) and impulse response of integrator h(mM). c(n) = q (n) ∗ h(mM ) (9) c(n) = ∑ n = 0 q (n) h(mM − 1) N −1
(10)
C ( K ) = Q( K ) . H ( K )
(11)
The time domain sequence c(n) can be obtained through IDFT of C(K). c ( n) =
1 N
N −1
C(K ) e k =0
j 2 πkn N
, k = 0,1,..... N − 1
(12)
Digitise the integrator output by sampling with a low-rate ADC, at the end of each time interval. y (m) = ∑ n = 0 c(n) δ(n − mT ) = c(mT ) m
(13)
The y(m) samples are m sparse and m
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login home contents what's new discussion bug reports help links subscribe changes refresh edit
http://en.wikipedia.org/wiki/Tensor_product
A tensor product is "the most general bilinear operation" available in a specified domain of computation, satisfying:
(1)
(2)
(3)
We can use the domain constructor Sum SandBoxSum
fricas
)lib SUM
Sum is now explicitly exposed in frame initial
Sum will be automatically loaded when needed from
/var/aw/var/LatexWiki/SUM.NRLIB/SUM
First we can define some recursive operations on the polynomials
fricas
scanPoly(p,n) == _
(p=0 => 0; mapMonomial(leadingMonomial(p),n)+scanPoly(reductum p,n))
Type: Void
fricas
mapMonomial(p,n) == _
monomial(coefficient(p,degree p),scanIndex(degree(p),n))$SMP(Integer,Sum(Symbol,Symbol)) Type: Void fricas scanIndex(p,n) == _ (zero? p => 0$IndexedExponents(Sum(Symbol,Symbol)); _
if n=1 then in1(leadingSupport(p))$Sum(Symbol,Symbol) _ else in2(leadingSupport(p))$Sum(Symbol,Symbol) _
)$IndexedExponents(Sum(Symbol,Symbol))+ _ scanIndex(reductum(p),n)) Type: Void For example: fricas -- functions are first compiled here -- scanPoly(x,1) fricas Compiling function scanIndex with type (IndexedExponents(Symbol), Integer) -> IndexedExponents(Sum(Symbol,Symbol)) fricas Compiling function mapMonomial with type (Polynomial(Integer), Integer) -> SparseMultivariatePolynomial(Integer,Sum(Symbol, Symbol)) fricas Compiling function scanPoly with type (Polynomial(Integer),Integer) -> SparseMultivariatePolynomial(Integer,Sum(Symbol,Symbol)) fricas Compiling function scanPoly with type (Polynomial(Integer),Integer) -> SparseMultivariatePolynomial(Integer,Sum(Symbol,Symbol)) fricas Compiling function scanPoly with type (Variable(x),Integer) -> SparseMultivariatePolynomial(Integer,Sum(Symbol,Symbol)) (4) Type: SparseMultivariatePolynomial?(Integer,Sum(Symbol,Symbol)) injects the polynomial x in to the tensor product. So now the full tensor product is just: fricas tensorPoly(p,q) == _ scanPoly(p,1)*scanPoly(q,2) Type: Void For example: fricas p:=2*x^2+3 (5) Type: Polynomial(Integer) fricas q:=5*x*y+7*y+11 (6) Type: Polynomial(Integer) fricas r:=tensorPoly(p,q) fricas Compiling function tensorPoly with type (Polynomial(Integer), Polynomial(Integer)) -> SparseMultivariatePolynomial(Integer,Sum( Symbol,Symbol)) (7) Type: SparseMultivariatePolynomial?(Integer,Sum(Symbol,Symbol)) fricas monomials(r) (8) Type: List(SparseMultivariatePolynomial?(Integer,Sum(Symbol,Symbol))) Demonstrating the axioms (1) (2) and (3) of the tensor product: fricas w:= 13*y^2+17*y+19 (9) Type: Polynomial(Integer) fricas test( tensorPoly(p+q,w) = (tensorPoly(p,w) + tensorPoly(q,w)) ) (10) Type: Boolean fricas test( tensorPoly(p,q+w) = (tensorPoly(p,q) + tensorPoly(p,w)) ) (11) Type: Boolean fricas test( tensorPoly(p,23*w) = 23*tensorPoly(p,w) ) (12) Type: Boolean fricas test( tensorPoly(23*p,w) = 23*tensorPoly(p,w) ) (13) Type: Boolean I suppose that we could give an inductive proof that this implementation of the tensor product of polynomials is correct ... but for now lets take this demonstration as reassurance. Re-coding the interpreter functions as library package. spad )abbrev package TPROD TensorProduct IE ==> IndexedExponents(VAR) IEP ==> IndexedExponents(Sum(VAR,VAR)) SMP ==> SparseMultivariatePolynomial(R,Sum(VAR,VAR)) TensorProduct(R:Ring, VAR: OrderedSet, P:PolynomialCategory(R,IE,VAR)): with _\_/: (P,P) -> SMP == add scanIndex(x:IE,n:Integer):IEP == zero? x => 0 monomial(leadingCoefficient(x), _ if n=1 then in1(leadingSupport(x))$Sum(VAR,VAR) _
else in2(leadingSupport(x))$Sum(VAR,VAR) _ ) + scanIndex(reductum(x),n) mapMonomial(p:P,n:Integer):SMP == monomial(coefficient(p,degree p),scanIndex(degree(p),n))$SMP
scanPoly(p:P,n:Integer):SMP ==
p=0 => 0
_\_/(p:P, q:P) : SMP == scanPoly(p,1)*scanPoly(q,2)
Compiling FriCAS source code from file
using old system compiler.
TPROD abbreviates package TensorProduct
------------------------------------------------------------------------
initializing NRLIB TPROD for TensorProduct
compiling into NRLIB TPROD
compiling local scanIndex : (IndexedExponents VAR,Integer) -> IndexedExponents Sum(VAR,VAR)
Time: 0.03 SEC.
compiling local mapMonomial : (P,Integer) -> SparseMultivariatePolynomial(R,Sum(VAR,VAR))
Time: 0.01 SEC.
compiling local scanPoly : (P,Integer) -> SparseMultivariatePolynomial(R,Sum(VAR,VAR))
Time: 0 SEC.
compiling exported \/ : (P,P) -> SparseMultivariatePolynomial(R,Sum(VAR,VAR))
Time: 0 SEC.
(time taken in buildFunctor: 0)
;;; *** |TensorProduct| REDEFINED
;;; *** |TensorProduct| REDEFINED
Time: 0 SEC.
Warnings:
[1] scanIndex: not known that (OrderedSet) is of mode (CATEGORY domain (IF (has VAR (Finite)) (IF (has VAR (Finite)) (ATTRIBUTE (Finite)) noBranch) noBranch) (IF (has VAR (Monoid)) (IF (has VAR (Monoid)) (ATTRIBUTE (Monoid)) noBranch) noBranch) (IF (has VAR (AbelianMonoid)) (IF (has VAR (AbelianMonoid)) (ATTRIBUTE (AbelianMonoid)) noBranch) noBranch) (IF (has VAR (CancellationAbelianMonoid)) (IF (has VAR (CancellationAbelianMonoid)) (ATTRIBUTE (CancellationAbelianMonoid)) noBranch) noBranch) (IF (has VAR (Group)) (IF (has VAR (Group)) (ATTRIBUTE (Group)) noBranch) noBranch) (IF (has VAR (AbelianGroup)) (IF (has VAR (AbelianGroup)) (ATTRIBUTE (AbelianGroup)) noBranch) noBranch) (IF (has VAR (OrderedAbelianMonoidSup)) (IF (has VAR (OrderedAbelianMonoidSup)) (ATTRIBUTE (OrderedAbelianMonoidSup)) noBranch) noBranch) (IF (has VAR (OrderedSet)) (IF (has VAR (OrderedSet)) (ATTRIBUTE (OrderedSet)) noBranch) noBranch) (SIGNATURE selectsum ((Union (: acomp VAR) (: bcomp VAR)) $)) (SIGNATURE in1 ($ VAR)) (SIGNATURE in2 ($VAR))) [2] mapMonomial: not known that (OrderedSet) is of mode (CATEGORY domain (IF (has VAR (Finite)) (IF (has VAR (Finite)) (ATTRIBUTE (Finite)) noBranch) noBranch) (IF (has VAR (Monoid)) (IF (has VAR (Monoid)) (ATTRIBUTE (Monoid)) noBranch) noBranch) (IF (has VAR (AbelianMonoid)) (IF (has VAR (AbelianMonoid)) (ATTRIBUTE (AbelianMonoid)) noBranch) noBranch) (IF (has VAR (CancellationAbelianMonoid)) (IF (has VAR (CancellationAbelianMonoid)) (ATTRIBUTE (CancellationAbelianMonoid)) noBranch) noBranch) (IF (has VAR (Group)) (IF (has VAR (Group)) (ATTRIBUTE (Group)) noBranch) noBranch) (IF (has VAR (AbelianGroup)) (IF (has VAR (AbelianGroup)) (ATTRIBUTE (AbelianGroup)) noBranch) noBranch) (IF (has VAR (OrderedAbelianMonoidSup)) (IF (has VAR (OrderedAbelianMonoidSup)) (ATTRIBUTE (OrderedAbelianMonoidSup)) noBranch) noBranch) (IF (has VAR (OrderedSet)) (IF (has VAR (OrderedSet)) (ATTRIBUTE (OrderedSet)) noBranch) noBranch) (SIGNATURE selectsum ((Union (: acomp VAR) (: bcomp VAR))$)) (SIGNATURE in1 ($VAR)) (SIGNATURE in2 ($ VAR)))
Cumulative Statistics for Constructor TensorProduct
Time: 0.04 seconds
finalizing NRLIB TPROD
Processing TensorProduct for Browser database:
--->/usr/local/lib/fricas/target/x86_64-unknown-linux/../../src/algebra/TENSOR.spad-->TensorProduct((\/ ((SparseMultivariatePolynomial R (Sum VAR VAR)) P P))): Not documented!!!!
; compiling file "/var/aw/var/LatexWiki/TPROD.NRLIB/TPROD.lsp" (written 31 JUL 2013 03:47:24 PM):
; /var/aw/var/LatexWiki/TPROD.NRLIB/TPROD.fasl written
; compilation finished in 0:00:00.025
------------------------------------------------------------------------
TensorProduct is now explicitly exposed in frame initial
TensorProduct will be automatically loaded when needed from
/var/aw/var/LatexWiki/TPROD.NRLIB/TPROD
fricas
test( p\/q = r )
(14)
Type: Boolean
fricas
test( (p+q) \/ w = (p\/w) + (q\/w) )
(15)
Type: Boolean
fricas
test( p \/ (q+w) = (p\/q) + (p\/w) )
(16)
Type: Boolean
fricas
test( p \/ (23*w) = 23*(p\/w) )
(17)
Type: Boolean
fricas
test( (23*p) \/ w = 23*(p\/w) )
(18)
Type: Boolean
Here's another way to write this - maybe better this way as first step to express associativity of the tensor product.
)abbrev package TPROD2 TensorProduct2
IE1 ==> IndexedExponents(VAR1)
IE2 ==> IndexedExponents(VAR2)
S ==> Sum(VAR1,VAR2)
IEP ==> IndexedExponents(S)
SMP ==> SparseMultivariatePolynomial(R,S)
TensorProduct2(R:Ring, VAR1: OrderedSet, VAR2: OrderedSet, P:PolynomialCategory(R,IE1,VAR1), Q:PolynomialCategory(R,IE2,VAR2)): with
_\_/: (P,Q) -> SMP
scanIndex1(x:IE1):IEP ==
zero? x => 0
monomial(leadingCoefficient(x), in1(leadingSupport(x))$S) + scanIndex1(reductum(x)) scanIndex2(x:IE2):IEP == zero? x => 0 monomial(leadingCoefficient(x), in2(leadingSupport(x))$S) + scanIndex2(reductum(x))
mapMonomial1(p:P):SMP ==
monomial(coefficient(p,degree p),scanIndex1(degree(p)))$SMP mapMonomial2(q:Q):SMP == monomial(coefficient(q,degree q),scanIndex2(degree(q)))$SMP
scanPoly1(p:P):SMP ==
p=0 => 0
scanPoly2(q:Q):SMP ==
q=0 => 0
_\_/(p:P, q:Q) : SMP == scanPoly1(p)*scanPoly2(q)
Compiling FriCAS source code from file
using old system compiler.
TPROD2 abbreviates package TensorProduct2
------------------------------------------------------------------------
initializing NRLIB TPROD2 for TensorProduct2
compiling into NRLIB TPROD2
compiling local scanIndex1 : IndexedExponents VAR1 -> IndexedExponents Sum(VAR1,VAR2)
Time: 0 SEC.
compiling local scanIndex2 : IndexedExponents VAR2 -> IndexedExponents Sum(VAR1,VAR2)
Time: 0 SEC.
compiling local mapMonomial1 : P -> SparseMultivariatePolynomial(R,Sum(VAR1,VAR2))
Time: 0 SEC.
compiling local mapMonomial2 : Q -> SparseMultivariatePolynomial(R,Sum(VAR1,VAR2))
Time: 0 SEC.
compiling local scanPoly1 : P -> SparseMultivariatePolynomial(R,Sum(VAR1,VAR2))
Time: 0 SEC.
compiling local scanPoly2 : Q -> SparseMultivariatePolynomial(R,Sum(VAR1,VAR2))
Time: 0 SEC.
compiling exported \/ : (P,Q) -> SparseMultivariatePolynomial(R,Sum(VAR1,VAR2))
Time: 0 SEC.
(time taken in buildFunctor: 0)
;;; *** |TensorProduct2| REDEFINED
;;; *** |TensorProduct2| REDEFINED
Time: 0 SEC.
Warnings:
[1] scanIndex1: not known that (OrderedSet) is of mode (CATEGORY domain (IF (has VAR1 (Finite)) (IF (has VAR2 (Finite)) (ATTRIBUTE (Finite)) noBranch) noBranch) (IF (has VAR1 (Monoid)) (IF (has VAR2 (Monoid)) (ATTRIBUTE (Monoid)) noBranch) noBranch) (IF (has VAR1 (AbelianMonoid)) (IF (has VAR2 (AbelianMonoid)) (ATTRIBUTE (AbelianMonoid)) noBranch) noBranch) (IF (has VAR1 (CancellationAbelianMonoid)) (IF (has VAR2 (CancellationAbelianMonoid)) (ATTRIBUTE (CancellationAbelianMonoid)) noBranch) noBranch) (IF (has VAR1 (Group)) (IF (has VAR2 (Group)) (ATTRIBUTE (Group)) noBranch) noBranch) (IF (has VAR1 (AbelianGroup)) (IF (has VAR2 (AbelianGroup)) (ATTRIBUTE (AbelianGroup)) noBranch) noBranch) (IF (has VAR1 (OrderedAbelianMonoidSup)) (IF (has VAR2 (OrderedAbelianMonoidSup)) (ATTRIBUTE (OrderedAbelianMonoidSup)) noBranch) noBranch) (IF (has VAR1 (OrderedSet)) (IF (has VAR2 (OrderedSet)) (ATTRIBUTE (OrderedSet)) noBranch) noBranch) (SIGNATURE selectsum ((Union (: acomp VAR1) (: bcomp VAR2)) $)) (SIGNATURE in1 ($ VAR1)) (SIGNATURE in2 ($VAR2))) [2] mapMonomial1: not known that (OrderedSet) is of mode (CATEGORY domain (IF (has VAR1 (Finite)) (IF (has VAR2 (Finite)) (ATTRIBUTE (Finite)) noBranch) noBranch) (IF (has VAR1 (Monoid)) (IF (has VAR2 (Monoid)) (ATTRIBUTE (Monoid)) noBranch) noBranch) (IF (has VAR1 (AbelianMonoid)) (IF (has VAR2 (AbelianMonoid)) (ATTRIBUTE (AbelianMonoid)) noBranch) noBranch) (IF (has VAR1 (CancellationAbelianMonoid)) (IF (has VAR2 (CancellationAbelianMonoid)) (ATTRIBUTE (CancellationAbelianMonoid)) noBranch) noBranch) (IF (has VAR1 (Group)) (IF (has VAR2 (Group)) (ATTRIBUTE (Group)) noBranch) noBranch) (IF (has VAR1 (AbelianGroup)) (IF (has VAR2 (AbelianGroup)) (ATTRIBUTE (AbelianGroup)) noBranch) noBranch) (IF (has VAR1 (OrderedAbelianMonoidSup)) (IF (has VAR2 (OrderedAbelianMonoidSup)) (ATTRIBUTE (OrderedAbelianMonoidSup)) noBranch) noBranch) (IF (has VAR1 (OrderedSet)) (IF (has VAR2 (OrderedSet)) (ATTRIBUTE (OrderedSet)) noBranch) noBranch) (SIGNATURE selectsum ((Union (: acomp VAR1) (: bcomp VAR2))$)) (SIGNATURE in1 ($VAR1)) (SIGNATURE in2 ($ VAR2)))
Cumulative Statistics for Constructor TensorProduct2
Time: 0 seconds
finalizing NRLIB TPROD2
Processing TensorProduct2 for Browser database:
--->-->TensorProduct2(constructor): Not documented!!!!
--->-->TensorProduct2((\/ ((SparseMultivariatePolynomial R (Sum VAR1 VAR2)) P Q))): Not documented!!!!
--->-->TensorProduct2(): Missing Description
; compiling file "/var/aw/var/LatexWiki/TPROD2.NRLIB/TPROD2.lsp" (written 31 JUL 2013 03:47:24 PM):
; /var/aw/var/LatexWiki/TPROD2.NRLIB/TPROD2.fasl written
; compilation finished in 0:00:00.035
------------------------------------------------------------------------
TensorProduct2 is now explicitly exposed in frame initial
TensorProduct2 will be automatically loaded when needed from
/var/aw/var/LatexWiki/TPROD2.NRLIB/TPROD2
fricas
test( p\/q = r )
(19)
Type: Boolean
fricas
test( (p+q) \/ w = (p\/w) + (q\/w) )
(20)
Type: Boolean
fricas
test( p \/ (q+w) = (p\/q) + (p\/w) )
(21)
Type: Boolean
fricas
test( p \/ (23*w) = 23*(p\/w) )
(22)
Type: Boolean
fricas
test( (23*p) \/ w = 23*(p\/w) )
(23)
Type: Boolean
Associativity of the tensor product means these two expressions should be identical:
fricas
(p\/q)\/w
(24)
Type: SparseMultivariatePolynomial?(Integer,Sum(Sum(Symbol,Symbol),Symbol))
fricas
p\/(q\/w)
(25)
Type: SparseMultivariatePolynomial?(Integer,Sum(Symbol,Sum(Symbol,Symbol)))
Subject: Be Bold !! ( 14 subscribers )
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Light Globe as Inrush Current Limiter for Vectrix VX1?
Could a light globe be used as the resistor in the Inrush Current Limiter (ICL)?
The circle with "40W" written in it represents a 240V, 40W incandescent light bulb.
The ammeter is optional, I would use it just to be able to see what residual current is still flowing when the globe stops to glow.
I believe the globe would actually light up whilst the capacitors on the motor controller are being pre-charged.
Here are some calculations regarding this:
40W = 240VAC x 0.16666A (Current flow through filament at 240V grid voltage)
240V / 0.166666A = 1.44kΩ (Resistance of HOT globe filament)
140VDC / 1.44kOhm = 97.2mA
The calculations are only approximate, for starters, maybe a RMS correction (factor 1.414) needs to be thrown in somewhere??? Comments, please!
.
Another factor is the variable resistance of the filament, depending on it's temperature:
From Wikipedia: http://en.wikipedia.org/wiki/Incandescent_light_bulb
Electrical characteristics
Incandescent lamps are nearly pure resistive loads with a power factor of 1. This means the actual power consumed (in watts) and the apparent power (in volt-amperes) are equal. The actual resistance of the filament is temperature-dependent. The cold resistance of tungsten-filament lamps is about 1/15 the hot-filament resistance when the lamp is operating. For example, a 100-watt, 120-volt lamp has a resistance of 144 ohms when lit, but the cold resistance is much lower (about 9.5 ohms) [40] [41]. ....
For a 100-watt, 120 volt general-service lamp, the current stabilizes in about 0.10 seconds, and the lamp reaches 90% of its full brightness after about 0.13 seconds. [42]
.
The resistance of the cold filament might then be: 1.44kΩ / 15 = 96Ω
The current through that resistance, at 140VDC battery voltage, would be: 140V / 96Ω = 1.458A.
Please check these maths for me and comment if you can!
Re: Light Globe as Inrush Current Limiter for Vectrix VX1?
Choices, choices.....
This information may be used entirely at your own risk.
There is always a way if there is no other way!
Re: Light Globe as Inrush Current Limiter for Vectrix VX1?
I have tested the light bulb as ICL option and found a 15W @ 240V globe the best option.
But the exact measurement of inrush currents is tricky due to their very short duration; maybe a 40W (@240V) globe is better, but I think they both work.
This information may be used entirely at your own risk.
There is always a way if there is no other way!
Re: Light Globe as Inrush Current Limiter for Vectrix VX1?
Permanently installed connectors allow easy connection of a 320Ω (15W@240V) incandescent bulb to prevent damage due to inrush current.
More details there: https://www.endless-sphere.com/forums/viewtopic.php?p=163141#p163141
This information may be used entirely at your own risk.
There is always a way if there is no other way!
Re: Light Globe as Inrush Current Limiter for Vectrix VX1?
A low resistance to start and higher later is not prefered, the other way around would be better, but even a simple fixed resistor (10 to 100 ohms) would likely work as well. One just wants the "R" to be high enough to limit the V/R = I current to perhaps 10 amps or less, yet pass enough current (R low enough) to "equalize" in a few seconds (perhaps under 10 seconds). Since the current flows for only a short time, a high wattage (I*I*R = Watts) resistor is not really necessary. A 100 ohm resistor of 20 (or perhaps even 10) watts is likely to be sufficient.
So, for a fixed installation, one could just "cut" the wire going through the battery at any convenient spot, insert a switch (with heavy duty contacts, maybe 200 amps), then parallel the switch with the resistor. Then, open the switch before disconnecting the Andersons, and close the switch (about 20 seconds) after reconnecting the Andersons.
But, the proposed wiring, across the Andersons as shown above, should also "work", it just leaves one side connected. Better to use a 2-pole switch and disconnect both sides. Perhaps enen better is another (smaller) connector pair in place of the switch, if physical separation is desired. Then, plug in the smaller connectors first, wait some seconds, and then plug in the larger Anderson connectors.
Edit: I see that the good discussion in the "endless sphere" forum also suggests this simple small-connector (and one resistor) solution.
Cheers, Gary
XM-5000Li, wired for cell voltage measuring and logging.
Re: Light Globe as Inrush Current Limiter for Vectrix VX1?
Thank you for your thoughts on this!
A low resistance to start and higher later is not prefered, the other way around would be better, but even a simple fixed resistor (10 to 100 ohms) would likely work as well.
The resistance of a light bulb does actually drop down again when the inrush current decreases and the filament cools down. The filament reduces it's temperature and resistance and ends up at with a resistance similar to the starting resistance. It does not heat the other parts of the globe up very much (due to the short duration of the inrush current), so that the filament has an effective heat sink.
One just wants the "R" to be high enough to limit the V/R = I current to perhaps 10 amps or less, yet pass enough current (R low enough) to "equalize" in a few seconds (perhaps under 10 seconds). Since the current flows for only a short time, a high wattage (I*I*R = Watts) resistor is not really necessary. A 100 ohm resistor of 20 (or perhaps even 10) watts is likely to be sufficient.
This is most likely correct, but has a disadvantage: The inrush current will be so short that it would be easy to miss it, even if an indicator LED or something else was being used to confirm correct connection of the ICL.
An ICL which does not give feedback of proper operation will soon cause problems.
An ICL will be needed for many repairs to Vectrix scooters, but the Vectux is to my knowledge the only VX-1 with an option to install permanent tabs for ICL connection easily, because of the Manual-BMS (M-BMS) cables already in place.
The much more common approach to use an ICL will be to use temporary, insulation piercing connectors to connect to the cables coming from the rear battery to the blue Andersons connector, and normal clamps to connect to the tabs where the front battery connects to the front end of the Andersons connector. This unfortunately creates much opportunity for poor connection quality, or no connection at all, or connection to the wrong cell of the front battery. The cables could also be crossed over inadvertently, causing a 64V dead short the moment that the Andersons connector is closed. Not pretty!
A simple resistor will work if it is connected properly, but there is no easy and reliable way of knowing if it is connected properly. Once you hear a spark when you put the Andersons connector together it is too late, the fuse will have been exposed to an inrush current of several hundred amps, or worse!
In my opinion any ICL needs to have an indicator showing that it is working and that the inrush current has passed. A light bulb does both nicely and cheaply! Just see it light up and slowly fade out and you know that it is connected correctly.
Each of the cables used should also be fused to prevent disasters due to inadvertent connection to the wrong places.
The tabs of the M-BMS in my Vectux are fused at the junction to the cell, so I did not include another fuse into the light globe ICL circuit (I did include fuses in the resistor ladder ICL device discussed on another page because of it's high potential for operator error, and one of the 500mA fuses + the 800mA fuse in my DMM did actually blow "somehow").
So, for a fixed installation, one could just "cut" the wire going through the battery at any convenient spot, insert a switch (with heavy duty contacts, maybe 200 amps), then parallel the switch with the resistor. Then, open the switch before disconnecting the Andersons, and close the switch (about 20 seconds) after reconnecting the Andersons.
But, the proposed wiring, across the Andersons as shown above, should also "work", it just leaves one side connected. Better to use a 2-pole switch and disconnect both sides. Perhaps enen better is another (smaller) connector pair in place of the switch, if physical separation is desired. Then, plug in the smaller connectors first, wait some seconds, and then plug in the larger Anderson connectors.
Edit: I see that the good discussion in the "endless sphere" forum also suggests this simple small-connector (and one resistor) solution.
The two batteries need to be separated completely to enable lifting of the rear battery out of the frame.
There is very little space for a switch, I think the installation of a permanent ICL connection point is a better solution. I did not install the light globe ICL permanently because the vibrations during scooter use will probably damage it.
Permanent installation of an ICL would also introduce an additional danger: The battery and motor controller might erroneously deemed to be "safe" after disconnection of the Andersons connector, but if an ICL was still connected, then the capacitors on the MC will remain fully charged and dangerous for months, until the battery is empty.
This information may be used entirely at your own risk.
There is always a way if there is no other way!
Re: Light Globe as Inrush Current Limiter for Vectrix VX1?
Another advantage of this type of ICL is that it will show that there is a fault if the fuse had failed due to some actual fault which made it necessary for the fuse to open.
In such circumstances any ICL which gives feedback that the capacitor charging has finished would NOT give such feedback, indicating that something is wrong and that the Andersons connector must not be closed.
The light globe would stay on, indicating the short / fault condition.
To reliably detect such states, a lower wattage light globe would be preferable over a higher wattage globe.
See this thread for an example: http://visforvoltage.org/forum/7846-vectrix-died-22000kshelp#comment-45204
This information may be used entirely at your own risk.
There is always a way if there is no other way!
Re: Light Globe as Inrush Current Limiter for Vectrix VX1?
Here is a link to a youtube video showing very nicely how to use a light=globe as ICL:
Thank's to MitchJi for digging it up and to whoever made it!
http://visforvoltage.org/forum/9866-videos-access-encoder-planetary-gear#comment-54766
This information may be used entirely at your own risk.
There is always a way if there is no other way!
Re: Light Globe as Inrush Current Limiter for Vectrix VX1?
What do the better qualified members of the forum think about "elevating" the light globe ICL method to some sort of "well tested and reliable" status?
This information may be used entirely at your own risk.
There is always a way if there is no other way!
Re: Light Globe as Inrush Current Limiter for Vectrix VX1?
Using a light globe as an ICL seems like a really good method to me. My Vectrix's main fuse blew last week. I was maintaining 50 km/h up a slight incline, and then suddenly lost all throttle/regen, and the dashboard lights went out a few seconds later, just like every other person's fuse that's blown. I've now removed the blown fuse (and did it without removing the rear battery!) and ordered an FWX-200A fuse to replace it.
I'm planning on using a 15-25W (depending on what's available) globe bulb as an ICL. If anyone has additional advice please let me know.
Thanks to everyone who has tested this, especially Mik!
Re: Light Globe as Inrush Current Limiter for Vectrix VX1?
Hello, I just want to whatch the video but this web page does not let me do it because I have to log in but I have already done that. Thats why I am wriying right now.
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# Analysis of a propeller in compressible, steady flow
Research output: Contribution to journalArticlepeer-review
## Abstract
THIS Synoptic describes an analytical method for solving the equation governing the inviscid, irrotational, compressible, potential flow about a propeller. The equation and boundary conditions are transferred to a noninertial system of coordinates rotating with the propeller, in which the basic problem becomes a steady one. The solution method takes advantage of the linearity of the model by superposing a “compressible”THIS Synoptic describes an analytical method for solving the equation governing the inviscid, irrotational, compressible, potential flow about a propeller. The equation and boundary conditions are transferred to a noninertial system of coordinates rotating with the propeller, in which the basic problem becomes a steady one. The solution method takes advantage of the linearity of the model by superposing a “compressible” solution to the potential equation on an “incompressible” wake solution. In addition, the boundary conditions are satisfied by dividing the flowfield at the propeller plane, solving the equations separately ahead of and behind this plane, and enforcing continuity matching conditions. Applying the final boundary condition yields an infinite-series integral equation for the unknown circulation distribution. A lifting-line method is used to produce numerical results. Presented results establish the effect of compressibility on the induced field solution to the potential equation on an “incompressible” wake solution. In addition, the boundary conditions are satisfied by dividing the flowfield at the propeller plane, solving the equations separately ahead of and behind this plane, and enforcing continuity matching conditions. Applying the final boundary condition yields an infinite-series integral equation for the unknown circulation distribution. A lifting-line method is used to produce numerical results. Presented results establish the effect of compressibility on the induced field.
Original language English (US) 1555-1556 2 AIAA journal 28 9 https://doi.org/10.2514/3.25252 Published - Sep 1990
## ASJC Scopus subject areas
• Aerospace Engineering
## Fingerprint
Dive into the research topics of 'Analysis of a propeller in compressible, steady flow'. Together they form a unique fingerprint.
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Address 11533 Slater Ave, Fountain Valley, CA 92708 (714) 638-8655 http://www.tristarcomputer.us
# example of systematic error in physics Cerritos, California
You may need to take account for or protect your experiment from vibrations, drafts, changes in temperature, electronic noise or other effects from nearby apparatus. Random errors often have a Gaussian normal distribution (see Fig. 2). Random vs. Random Errors > 5.2.
Further Reading Introductory: J.R. Lag time and hysteresis (systematic) - Some measuring devices require time to reach equilibrium, and taking a measurement before the instrument is stable will result in a measurement that is generally The Gaussian normal distribution. Sometimes a correction can be applied to a result after taking data to account for an error that was not detected.
The precision is limited by the random errors. Blunders should not be included in the analysis of data. Percent difference: Percent difference is used when you are comparing your result to another experimental result. Advanced: R.
You can read off whether the length of the object lines up with a tickmark or falls in between two tickmarks, but you could not determine the value to a precision Failure to calibrate or check zero of instrument(systematic) - Whenever possible, the calibration of an instrument should be checked before taking data. Observational. For example, unpredictable fluctuations in line voltage, temperature, or mechanical vibrations of equipment.
For the error estimates we keep only the first terms: DR = R(x+Dx) - R(x) = (dR/dx)x Dx for Dx ``small'', where (dR/dx)x is the derivative of function R with Instrument resolution (random) - All instruments have finite precision that limits the ability to resolve small measurement differences. Systematic errors, by contrast, are reproducible inaccuracies that are consistently in the same direction. Examples of systematic errors caused by the wrong use of instruments are: errors in measurements of temperature due to poor thermal contact between the thermometer and the substance whose temperature is
Plot the measured points (x,y) and mark for each point the errors Dx and Dy as bars that extend from the plotted point in the x and y directions. Re-zero the instrument if possible, or measure the displacement of the zero reading from the true zero and correct any measurements accordingly. Systematic Errors Not all errors are created equal. Note: a and b can be positive or negative, i.e.
Fitting a Straight Line through a Series of Points Frequently in the laboratory you will have the situation that you perform a series of measurements of a quantity y at different Example: Say quantity x is measured to be 1.00, with an uncertainty Dx = 0.10, and quantity y is measured to be 1.50 with uncertainty Dy = 0.30, and the constant It is a good rule to give one more significant figure after the first figure affected by the error. If the graph does not cut the expected intercept, the shift is probably due to systematic error.Next: Zero Error, Accuracy and Precision Previous: Random Errors Back To Measurement (A Level) shares
That means some measurements cannot be improved by repeating them many times. Repeated measurements produce a series of times that are all slightly different. Here, we list several common situations in which error propagion is simple, and at the end we indicate the general procedure. Sometimes you will encounter significant systematic errors in your experiments.
Hence: s » ¼ (tmax - tmin)
is an reasonable estimate of the uncertainty in a single measurement. Next, draw the steepest and flattest straight lines, see the Figure, still consistent with the measured error bars. If you have a calculator with statistical functions it may do the job for you. Note that systematic and random errors refer to problems associated with making measurements.
The experimenter may measure incorrectly, or may use poor technique in taking a measurement, or may introduce a bias into measurements by expecting (and inadvertently forcing) the results to agree with These calculations are also very integral to your analysis analysis and discussion. The two quantities are then balanced and the magnitude of the unknown quantity can be found by comparison with the reference sample. Random errors: These are errors for which the causes are unknown or indeterminate, but are usually small and follow the laws of chance.
Blunders A final source of error, called a blunder, is an outright mistake. Username E-mail pbIxrgZSVD Recent Forum Topics storage effective resistance rotational mechanicsTop Posts & Pages How To Read A Vernier Caliper How To Read A Micrometer Screw Gauge O Level Physics UY1: Doing so often reveals variations that might otherwise go undetected. A similar effect is hysteresis where the instrument readings lag behind and appear to have a "memory" effect as data are taken sequentially moving up or down through a range of
They vary in random vary about an average value. It measures the random error or the statistical uncertainty of the individual measurement ti: s = Ö[SNi=1(ti - átñ)2 / (N-1) ].
About two-thirds of all the measurements have a deviation Null or balance methods involve using instrumentation to measure the difference between two similar quantities, one of which is known very accurately and is adjustable. the equation works for both addition and subtraction.
Multiplicative Formulae When the result R is calculated by multiplying a constant a times a measurement of x times a measurement of
One of the best ways to obtain more precise measurements is to use a null difference method instead of measuring a quantity directly. Register Forgotten Password Cancel Register For This SiteA password will be e-mailed to you. Random errors can be evaluated through statistical analysis and can be reduced by averaging over a large number of observations. If you want to judge how careful you have been, it would be useful to ask your lab partner to make the same measurements, using the same meter stick, and then
Independent errors cancel each other with some probability (say you have measured x somewhat too big and y somewhat too small; the error in R might be small in this case). Systematic errors cannot be detected or reduced by increasing the number of observations, and can be reduced by applying a correction or correction factor to compensate for the effect. This calculation will help you to evaluate the relevance of your results. The accuracy will be given by the spacing of the tickmarks on the measurement apparatus (the meter stick).
Gross personal errors, sometimes called mistakes or blunders, should be avoided and corrected if discovered. Taylor, An Introduction to Error Analysis, Oxford UP, 1982. For instance, you may inadvertently ignore air resistance when measuring free-fall acceleration, or you may fail to account for the effect of the Earth's magnetic field when measuring the field of The following are some examples of systematic and random errors to consider when writing your error analysis.
Random vs. These variations may call for closer examination, or they may be combined to find an average value.
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# BME100 f2016:Group11 W1030AM L3
BME 100 Fall 2016 Home
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# LAB 3 WRITE-UP
## Inferential Stats
Based on the data obtained through the two tailed tests, it can be confirmed that the spree did not relay accurate information when it tested the temperature of the subjects. Although the average values were similar, the variance between the data were off by nearly 96%. This can further be proven when comparing the p-value obtained from the temperature data chart, which was p=0.01, to the known p-value of p=0.05. Because the measured p-value was 16% lower than the known value, the spree's data is known to be unreliable. The data obtained from measuring the heart rate, however, was extremely accurate. The compared p-value is p=0.72, and since it is more than 5%, it shows that the spree is capable of relaying accurate information in terms of heart rate. When analyzing the graphs, the similarity between the spree and the gold-standard become more apparent, considering the averages and the standard deviations are nearly identical.
## Design Flaws and Recommendations
We are evaluating the Spree headband’ claims of accurate heart rate and body temperature measurements. We are using a pulse oximeter and an internal thermometer as the ‘gold standard’ in this comparisons. A gold standard allows us to compare how by what magnitude does the Spree headband’s measurements differ.
For the heartrate, the averages were about the same and the scatterplots trend line was similar as well, albeit there was some discrepancy within the plots. The heartrate is fairly accurate on the Spree but not up to the Gold Standard.
On the other hand, the temperature measurement seems to be off by a larger margin with constant spree temperature, not matching up with the actual temperature. This means that even though the thermometer is reading different temperature, the spree is only reading one, despite the changes. This indicates a shortcoming in terms of the Spree’s temperature measuring capabilities. Sources of error in the Spree headband could be due to improper placement of the headband, so human error or the location or capabilities of the sensor being diminished due to the increased physical and its production of movement and sweat.
## Experimental Design of Own Device
Our device will be tested on how well it filters waste from the blood. The pH of blood is a good way to measure how much waste is in the blood. Normally the pH of blood is around 7.4 when the kidneys are functioning at their full potential. Dialysis takes over the function of the kidney and must keep the blood at that pH. The experiment would contain a group of 50 participants, with FDA approval, suffering from stage 4 kidney disease or kidney failure. The experiment will be a paired t-test, where the participant's blood is taken before and after using the device. Participants will need to get vascular access surgery to connect to the device 3 weeks prior to the experiment. Before the experiment, the participant will be taken off any kind of dialysis for 24 hours leading up to the test. This will make sure that there is as much waste that has accumulated in the participant as there would be in between dialysis treatment. The participant's blood will then be taken and the pH measured. Then the participant will put on the device, wear it for 24 hours and then have their blood taken and measured. Twenty-four hours gives The device adequate time to filter the blood and make sure it is functioning at all times. The two measures of pH of the blood will then be compared to the average standard pH of blood (7.4) to see if the device is properly functioning. We can analyze the results to see not only if the device filtered the blood but how well the blood was filtered.
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# Chapter 3 Electromagnetism
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### Chapter 3 Electromagnetism
1. 1. SUCCESS IS DIRECTLY PROPORTIONAL TO NOBLE VALUES
2. 2. By the end of lesson, you will able to: State what an electromagnet is. Draw the magnetic field pattern due to a current in a: a) straight wire b) coil c) solenoid Factors affecting the strength of magnetic field of an electromagnet. Describe applications of electromagnets.
3. 3. An electromagnet is a temporary magnet when current is passed through the wire winding a soft iron core.
4. 4. The pattern of magnetic field depends on shape of the conductor. The direction of magnetic field depends on direction of the current.
5. 5. A magnetic field pattern can be represented by field lines that show the shape of the field. Current flow
6. 6. The right-hand grip rule
7. 7. At the centre of the coil: The magnetic field is the strongest because magnetic field lines are close together The field patterns is straight at right angle
8. 8. The direction of magnetic fieldinside the solenoid The pattern of the is opposite to the magnetic field of a direction outside solenoid is similar solenoid. to that of a bar magnet.
9. 9. The thumb points towards north pole of the magnetic field. Other fingers indicate the direction of the Right hand grip current in therule for solenoid solenoid.
10. 10. Draw a magnetic field pattern and direction N S
11. 11. Electric bell Magnetic relay Telephone Circuit earpiece breaker
12. 12. Soft iron armatureCurrent flow attracted towards the through magnetic iron core and Spring return the solenoid 1 disconnects from the iron armature to itswhen switch contact. original positionis pressed. 3 and circuit closed 6 again.Soft iron core in solenoid become 2 Contact 5) Circuit iselectromagnet 5 open and no . current flows and electromagnet lose it magnetism. 4 Hammer hits the gong very quickly to produce sound.
13. 13. Use as a switch to turn on high voltage appliances such as air conditioner to prevents direct contact with human.
14. 14. Current flow through high The bottom of spring voltage at 2nd circuit. contact is bent upwards 5 4 Spring contact 3The iron armature attracts towards electromagnet . 2The iron core magnetised to become electromagnet. 1 When 1st switch is closed, current flow in solenoid.
15. 15. SUCCESS IS DIRECTLY PROPORTIONAL TO NOBLE VALUES
16. 16. By the end of lesson, you will able to: Describe what happens to a current-carrying conductor in a magnetic field. Draw the pattern of the combined magnetic field due to a current-carrying conductor in a magnetic field. Describe how a current-carrying conductor in a magnetic field experiences a force. Explain the factors affecting magnitude of force. Describe how a current-carrying coil in a magnetic field experiences a turning force. Describe how a direct current motor works. State factors affect the speed of rotation of an electric motor.
17. 17. A magnetic force is produced when a current- carrying conductor is in a magnetic field.
18. 18. The direction of magnetic force, F acting on the wire can be determine by using Fleming’s Left-hand Rule.
19. 19. - N + SBy using Fleming’s Left-hand Rule, determine: Flow of current Direction of magnetic field Direction of magnetic force
20. 20. N Direction of currentDirection of Force S
21. 21. N Direction of Force Direction of currentS
22. 22. S Direction of currentDirection of Force N
23. 23. S N The direction of magnetic field is parallel to the direction of current. The short wire stays at rest.
24. 24. When current-carrying conductor is in a magnetic field of permanent magnet, the interaction between two magnetic field produce a force on the conductor. The direction of magnetic field, the direction of current and direction of force are perpendicular to each other. Direction of force Direction of magnetic field Direction of current
25. 25. 1. Permanent 2. Current-carrying 3. The two field magnet produced conductor interact to a uniform, parallel produced a produced a magnetic field. circular magnetic resultant field. magnetic field. N S N S
26. 26. The two field interact to produced a resultant magnetic field known as Catapult Field. Region of weaker magnetic field- 2 fields act in oppositedirection Region of strong The interaction between magnetic field two magnetic field - 2 fields act in the produce a force on the same direction. conductor
27. 27. + Magnetic Magnetic field N S of Current- field ofpermanent carrying magnet conductor Region of weaker magnetic field Region of strongmagnetic field
28. 28. Magnetic Magnetic field field of permanent N S + of Current- carrying magnet conductor Region of Region of stronger weaker magnetic N S magnetic field field - 2 fields- 2 fields act in theact in sameopposite direction.direction Direction of force Catapult field
29. 29. Magnitude of current• Current can be increased by increasing the e.m.f of power supply // using thicker wire of same length// shorter wire.• The larger the current in conductor, the larger the force acting on it. Strength of magnetic field• Stronger magnetic field can be produced by using more powerful magnets or by placing the magnet closer to each other to narrow the gap between the poles of the magnets.• The stronger the strength of magnetic field, the larger the force acting on it.
30. 30. If a current carrying coil is placed in a magnetic field, a pair of forces will be produced on the coil. This is due to the interaction of the magnetic field of the permanent magnet and the magnetic field of the current carrying coil.
31. 31. carrying coil.
32. 32. Carbon brush: To contact with the commutator so the current from the battery enters the coil. Spring: Push the brush so it will always contact with the commutator. Split ring commutator: To ensure that the forces on the coil turn the coil in one direction only.
33. 33. The direct current motor uses the turning effect on a current- carrying coil in a magnetic field. Electric motor converts electrical energy to kinetic energy.
34. 34. Magnitude of current The higher the magnitude of current, the higher the speed of rotation of electric motor. Number of turns of the coil The higher the number of turns of coil, the higher the speed of rotation of electric motor. Strength of the magnetic field The higher the strength of magnetic field, the higher the speed of rotation of electric motor.
35. 35. SUCCESS IS DIRECTLY PROPORTIONAL TO NOBLE VALUES
36. 36. Electromagnetic induction is the production of an electric current by a changing magnetic field. Induced current only produced when there is relative motion between the conductor and the magnetic field lines.
37. 37. o Moving a straight wire • Moving a permanent quickly across a magnet towards one end magnetic field of a solenoid. between two permanent magnets.
38. 38. a)Direction of induced current in a straight wire can be determine by using Fleming’s right hand rule.
39. 39. Determine the direction of current.
40. 40. a)Direction of induced current in a solenoid can be determine by using Lenz’s law. Lenz’s Law: States that the direction of the induced current in a solenoid is such that its magnetic effect always oppose the change producing it.
41. 41. Lenz’s law is an example of the Principle of Conservation of Energy. When the magnet or solenoid is moved against the opposing force, work is done. Therefore mechanical energy is converted to electrical energy.
42. 42. The magnitude of the induced e.m.f is directly proportional to the rate at which the conductor cuts through the magnetic field lines.The size of the induced e.m.f. and thus the inducedcurrent can be increased by:•moving the magnet or the solenoid at a higher speed.• increasing the number of turns on the solenoid.• increasing the strength of the magnetic field throughthe use of a stronger magnet.
43. 43. A generator is essentially the opposite of a motor which converts mechanical energy to electrical energy.
44. 44. The two ends of the coil are connected to two slip rings which rotate with the coil.
45. 45. Coil AB move downwards and coil CD move upwards. Induced current flows from D to C and from B to A. In external circuit, current flows from P to Q. The galvanometer deflected to the right. Side AB and CD are moving parallel to the magnetic field and thus no induced current is produced. The galvanometer returns to zero.
46. 46. Coil CD move downwards and coil AB move upwards. Induced current flows from A to B and from C to D. In external circuit, current flows from Q to P. The galvanometer deflected to the left. Side AB and CD are moving parallel to the magnetic field and thus no induced current is produced. The galvanometer returns to zero.
47. 47. Angle of rotation/ 0900 1800 2700 3600
48. 48. Direct current (d.c) Alternate current (a.c)A direct current is a current that flows An alternating current is a currentin one direction only in a circuit. which flows to and fro in two opposite directions in a circuit.The magnitude of a direct current It changes its direction periodically.may be:(a) constant(b) changes with time
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Warning! This set of interlinked pages is very much under construction. If you are already familiar with the subject matter, I would appreciate any help you might provide. If you are seeking information on the subject matter, I would be interested to hear what you would ultimately like this page to contain. Thank you.
Gutzwiller Trace Formula ___ [link] ___ Selberg Trace Formula | | [link] | | | | [link] | | Riemann's zeta function ___ [link] ___ Riemann-Weil Explicit Formula | | [link] | | | | [link] | | Connes'(hypothetical) trace formula ___ [link] ___ (Deninger's) Lefschetz Trace Formulae
On p.1 of C. Deninger's "Lefschetz trace formulas and explicit formulae in analytic number theory" we read
"...In particular, we were lead to a Lefschetz trace formula...If we specialized to the motive M(chi) of an algebraic Hecke character chi which for simplicity we suppose to be unitary, [the LTF] implies the following relation
delta(Phi(0)+Phi)(1)) - Sum rho Phi(rho) = Sumscript p(Phi;Chi)
This looks very similar to the version of the Riemann-Weil Explicit Formula which Deninger gives in the first paragraph ("It takes the following form if chi is unitary and normalized.")
delta(Phi(0)+Phi(1)) - SumrhoPhi(rho) = alpha F(0) + Sumscript p Wscript p (F;chi)
Here F is a rapidly decreasing function on the reals, Phi is its Fourier transform, rho are nontrivial zeros of L(chi, s), delta is 0 or 1 (depending on the triviality of chi as a character), script p are places of the number field K.
Wscript p(F;chi) are local terms. The field K completes at place script p locally as Kscript p. alpha is a constant depending on K and chi, but not on F.
So to summarise: Deninger's investigation of global arithmetic cohomology theory lead to a Lefschetz trace formula. By then specialising to the motive of an algebraic Hecke character chi, he was lead to a definition of local traces which produces a relation analogous to the Riemann-Weil explicit formula. This suggests that the local terms Wscript p in the explicit formula should be interpreted as traces. In particular, Deninger argues that they are traces of an operator on certain infinite dimensional spaces script Fscript p(M(chi)).
To summarise the summary: A particular Lefschetz trace formula, appropriately specialized, produces a relation analogous to the Riemann-Weil Explicit Formula.
I have yet to grasp how Deninger then precedes from this observation.
Andreas Juhl wrote to me "The ideal case would be to understand the explicit formulas (as of the Riemann zeta function) as Lefschetz formulas. And as you know, this was also the idea of Connes' recent efforts."
back to trace formulae, explicit formulae and number theory page
archive tutorial mystery new search home contact
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# Can the quadratic formula be used when factorizing a denominator?
I’m doing partial fractions and need to factorize the denominator. They are quadratic. However there are some that aren’t so easy to factorize and my first choice was to use the quadratic equation to find the roots however comparing my answer with the correct one the signs are different. Is the quadratic formula only to be used when the equation is equal to zero? The answer used another method of factorizing that didn’t involve equating anything to zero and I can’t find anything about it online. Where did I go wrong? My denominator is:
$$-3z^2 -4z-1$$
the correct answer is: $$-(3z+1)(z+1)$$
while if I do this using the quadratic formula I get: $$(3z+1)(z+1)$$
however if I factorize the negative sign then use the quadratic formula I get the correct answer which is confusing to me.
• It's not very clear what you mean, maybe you could give a practical example and write some of the terms that you want to factor? – Matti P. Oct 29 '18 at 12:20
• I have edited the question. – Ian Oct 29 '18 at 12:27
Suppose I have a mystery quadratic $$p(x) = -2x^2 + x + 1$$, and want to factorise it using the quadratic formula. I can find the roots easily enough:
$$x = \frac{-1 \pm \sqrt{1 + 8}}{-4}$$ and so the roots are $$-\frac{1}{2}$$ and $$1$$, and so I think that the polynomial should factorise as $$p(x) = (x - 1)(x+\frac{1}{2})$$. But this is wrong, since I forgot that $$p(x)$$ and $$2p(x)$$ and $$\frac{-p(x)}{9}$$ and so on all have the same roots! And so knowing the two roots only determines $$p(x)$$ up to scaling. But it's easy to find the right scaling factor, by just looking at the $$-2x^2$$ term. So I arrive at the actual answer, of $$p(x) = -2x^2 + x + 1 = -2(x-1)(x + \frac{1}{2})$$
So after you find the roots, remember to multiply the whole thing by the right number to fix up the $$x^2$$ term.
• what is this phenomenon called? where can I read up more on it? – Ian Oct 29 '18 at 12:36
• I don't know if it's called anything particular. You just have to notice that scaling a function changes the function but not the roots, since if $p(\alpha) = 0$, then $2 p(\alpha) = 0$ also. But $p$ and $2p$ are not the same function. So knowing the roots is not enough to determine the quadratic, you have to know something about how steep it is. – Joppy Oct 29 '18 at 12:38
• does this only apply if the coefficient of the $x^2$ is non zero???? – Ian Oct 29 '18 at 12:39
• If the coefficient of the $x^2$ is zero, then you have a linear function $ax + b$, not a quadratic. – Joppy Oct 29 '18 at 12:40
• so as a rule of thumb, once the coefficient isn't 1 I should multiply by the coefficient after I use the quadratic formula? – Ian Oct 29 '18 at 12:41
I'm guessing, what you have is the following fraction: $$f(z) = \frac{N(z)}{D(z)},$$ where $$D(z) = -3z^2-4z-1$$.
Basically, you may write $$D(z) = -(3z^2+4z+1) = -D_2(z)$$.
Finally, you can decompose the fraction $$f$$ as $$f(z) = \frac{N(z)}{D(z)}$$ or as $$f(z) = \frac{-N(z)}{D_2(z)}$$.
In other words, you'll get the same partial fraction in any case (you'll keep only the two factors in the denominators, and push the minus sign to the numerator)
Factoring in $$\mathbb Q[x]$$ is unique up to associates. The units in $$\mathbb Q[x]$$ are all nonzero rational numbers. So $$(x + \frac 12)$$, $$(2x+1), -(2x+1), (4x+2)$$ and so on are all considered equivalent factors.
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• Go To
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• Practice and Assignment problems are not yet written. As time permits I am working on them, however I don't have the amount of free time that I used to so it will take a while before anything shows up here.
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Home / Differential Equations / Laplace Transforms / Solving IVP's with Laplace Transforms
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### Section 4.5 : Solving IVP's with Laplace Transforms
It’s now time to get back to differential equations. We’ve spent the last three sections learning how to take Laplace transforms and how to take inverse Laplace transforms. These are going to be invaluable skills for the next couple of sections so don’t forget what we learned there.
Before proceeding into differential equations we will need one more formula. We will need to know how to take the Laplace transform of a derivative. First recall that $$f^{(n)}$$ denotes the $$n^{\mbox{th}}$$ derivative of the function $$f$$. We now have the following fact.
#### Fact
Suppose that $$f$$, $$f'$$, $$f''$$,… $$f^{(n-1)}$$ are all continuous functions and $$f^{(n)}$$ is a piecewise continuous function. Then,
$\mathcal{L}\left\{ {{f^{\left( n \right)}}} \right\} = {s^n}F\left( s \right) - {s^{n - 1}}f\left( 0 \right) - {s^{n - 2}}f'\left( 0 \right) - \cdots - s{f^{\left( {n - 2} \right)}}\left( 0 \right) - {f^{\left( {n - 1} \right)}}\left( 0 \right)$
Since we are going to be dealing with second order differential equations it will be convenient to have the Laplace transform of the first two derivatives.
\begin{align*}\mathcal{L}\left\{ {y'} \right\} & = sY\left( s \right) - y\left( 0 \right)\\ \mathcal{L}\left\{ {y''} \right\} & = {s^2}Y\left( s \right) - sy\left( 0 \right) - y'\left( 0 \right)\end{align*}
Notice that the two function evaluations that appear in these formulas, $$y\left( 0 \right)$$ and $$y'\left( 0 \right)$$, are often what we’ve been using for initial condition in our IVP’s. So, this means that if we are to use these formulas to solve an IVP we will need initial conditions at $$t = 0$$.
While Laplace transforms are particularly useful for nonhomogeneous differential equations which have Heaviside functions in the forcing function we’ll start off with a couple of fairly simple problems to illustrate how the process works.
Example 1 Solve the following IVP. $y'' - 10y' + 9y = 5t,\hspace{0.25in}y\left( 0 \right) = - 1\,\,\,\,\,\,\,y'\left( 0 \right) = 2$
Show Solution
The first step in using Laplace transforms to solve an IVP is to take the transform of every term in the differential equation.
$\mathcal{L}\left\{ {y''} \right\} - 10\mathcal{L}\left\{ {y'} \right\} + 9\mathcal{L}\left\{ y \right\} = \mathcal{L}\left\{ {5t} \right\}$
Using the appropriate formulas from our table of Laplace transforms gives us the following.
${s^2}Y\left( s \right) - sy\left( 0 \right) - y'\left( 0 \right) - 10\left( {sY\left( s \right) - y\left( 0 \right)} \right) + 9Y\left( s \right) = \frac{5}{{{s^2}}}$
Plug in the initial conditions and collect all the terms that have a $$Y(s)$$ in them.
$\left( {{s^2} - 10s + 9} \right)Y\left( s \right) + s - 12 = \frac{5}{{{s^2}}}$
Solve for $$Y(s)$$.
$Y\left( s \right) = \frac{5}{{{s^2}\left( {s - 9} \right)\left( {s - 1} \right)}} + \frac{{12 - s}}{{\left( {s - 9} \right)\left( {s - 1} \right)}}$
At this point it’s convenient to recall just what we’re trying to do. We are trying to find the solution, $$y(t)$$, to an IVP. What we’ve managed to find at this point is not the solution, but its Laplace transform. So, in order to find the solution all that we need to do is to take the inverse transform.
Before doing that let’s notice that in its present form we will have to do partial fractions twice. However, if we combine the two terms up we will only be doing partial fractions once. Not only that, but the denominator for the combined term will be identical to the denominator of the first term. This means that we are going to partial fraction up a term with that denominator no matter what so we might as well make the numerator slightly messier and then just partial fraction once.
This is one of those things where we are apparently making the problem messier, but in the process we are going to save ourselves a fair amount of work!
Combining the two terms gives,
$Y\left( s \right) = \frac{{5 + 12{s^2} - {s^3}}}{{{s^2}\left( {s - 9} \right)\left( {s - 1} \right)}}$
The partial fraction decomposition for this transform is,
$Y\left( s \right) = \frac{A}{s} + \frac{B}{{{s^2}}} + \frac{C}{{s - 9}} + \frac{D}{{s - 1}}$
Setting numerators equal gives,
$5 + 12{s^2} - {s^3} = As\left( {s - 9} \right)\left( {s - 1} \right) + B\left( {s - 9} \right)\left( {s - 1} \right) + C{s^2}\left( {s - 1} \right) + D{s^2}\left( {s - 9} \right)$
Picking appropriate values of $$s$$ and solving for the constants gives,
\begin{align*} & s = 0 & 5 & = 9B & \Rightarrow \hspace{0.25in}B & = \frac{5}{9}\\ & s = 1 & 16 & = - 8D & \Rightarrow \hspace{0.25in}D & = - 2\\ & s = 9 & 248 & = 648C & \Rightarrow \hspace{0.25in}C & = \frac{{31}}{{81}}\\ & s = 2 & 45 & = - 14A + \frac{{4345}}{{81}} & \Rightarrow \hspace{0.25in}A & = \frac{{50}}{{81}}\end{align*}
Plugging in the constants gives,
$Y\left( s \right) = \frac{{\frac{{50}}{{81}}}}{s} + \frac{{\frac{5}{9}}}{{{s^2}}} + \frac{{\frac{{31}}{{81}}}}{{s - 9}} - \frac{2}{{s - 1}}$
Finally taking the inverse transform gives us the solution to the IVP.
$y\left( t \right) = \frac{{50}}{{81}} + \frac{5}{9}t + \frac{{31}}{{81}}{{\bf{e}}^{9t}} - 2{{\bf{e}}^t}$
That was a fair amount of work for a problem that probably could have been solved much quicker using the techniques from the previous chapter. The point of this problem however, was to show how we would use Laplace transforms to solve an IVP.
There are a couple of things to note here about using Laplace transforms to solve an IVP. First, using Laplace transforms reduces a differential equation down to an algebra problem. In the case of the last example the algebra was probably more complicated than the straight forward approach from the last chapter. However, in later problems this will be reversed. The algebra, while still very messy, will often be easier than a straight forward approach.
Second, unlike the approach in the last chapter, we did not need to first find a general solution, differentiate this, plug in the initial conditions and then solve for the constants to get the solution. With Laplace transforms, the initial conditions are applied during the first step and at the end we get the actual solution instead of a general solution.
In many of the later problems Laplace transforms will make the problems significantly easier to work than if we had done the straight forward approach of the last chapter. Also, as we will see, there are some differential equations that simply can’t be done using the techniques from the last chapter and so, in those cases, Laplace transforms will be our only solution.
Let’s take a look at another fairly simple problem.
Example 2 Solve the following IVP. $2y'' + 3y' - 2y = t{{\bf{e}}^{ - 2t}},\hspace{0.25in}y\left( 0 \right) = 0\,\,\,\,\,\,\,y'\left( 0 \right) = - 2$
Show Solution
As with the first example, let’s first take the Laplace transform of all the terms in the differential equation. We’ll the plug in the initial conditions to get,
\begin{align*}2\left( {{s^2}Y\left( s \right) - sy\left( 0 \right) - y'\left( 0 \right)} \right) + 3\left( {sY\left( s \right) - y\left( 0 \right)} \right) - 2Y\left( s \right) & = \frac{1}{{{{\left( {s + 2} \right)}^2}}}\\ \left( {2{s^2} + 3s - 2} \right)Y\left( s \right) + 4 & = \frac{1}{{{{\left( {s + 2} \right)}^2}}}\end{align*}
Now solve for $$Y(s)$$.
$Y\left( s \right) = \frac{1}{{\left( {2s - 1} \right){{\left( {s + 2} \right)}^3}}} - \frac{4}{{\left( {2s - 1} \right)\left( {s + 2} \right)}}$
Now, as we did in the last example we’ll go ahead and combine the two terms together as we will have to partial fraction up the first denominator anyway, so we may as well make the numerator a little more complex and just do a single partial fraction. This will give,
\begin{align*}Y\left( s \right) & = \frac{{1 - 4{{\left( {s + 2} \right)}^2}}}{{\left( {2s - 1} \right){{\left( {s + 2} \right)}^3}}}\\ & = \frac{{ - 4{s^2} - 16s - 15}}{{\left( {2s - 1} \right){{\left( {s + 2} \right)}^3}}}\end{align*}
The partial fraction decomposition is then,
$Y\left( s \right) = \frac{A}{{2s - 1}} + \frac{B}{{s + 2}} + \frac{C}{{{{\left( {s + 2} \right)}^2}}} + \frac{D}{{{{\left( {s + 2} \right)}^3}}}$
Setting numerator equal gives,
\begin{align*} - 4{s^2} - 16s - 15 & = A{\left( {s + 2} \right)^3} + B\left( {2s - 1} \right){\left( {s + 2} \right)^2} + C\left( {2s - 1} \right)\left( {s + 2} \right) + D\left( {2s - 1} \right)\\ & = \left( {A + 2B} \right){s^3} + \left( {6A + 7B + 2C} \right){s^2} + \left( {12A + 4B + 3C + 2D} \right)s\\ & \hspace{2.25in} + 8A - 4B - 2C - D\end{align*}
In this case it’s probably easier to just set coefficients equal and solve the resulting system of equation rather than pick values of $$s$$. So, here is the system and its solution.
\left. {\begin{aligned} & {s^3}: & A + 2B & = 0\\& {s^2}: & 6A + 7B + 2C & = - 4\\ & {s^1}: & 12A + 4B + 3C + 2D & = - 16\\& {s^0}: & 8A - 4B - 2C - D & = - 15\end{aligned}} \right\}\hspace{0.25in} \Rightarrow \hspace{0.25in}\begin{aligned}A & = - \frac{{192}}{{125}} & B & = \frac{{96}}{{125}}\\ C & = - \frac{2}{{25}} & D & = - \frac{1}{5}\end{aligned}
We will get a common denominator of 125 on all these coefficients and factor that out when we go to plug them back into the transform. Doing this gives,
$Y\left( s \right) = \frac{1}{{125}}\left( {\frac{{ - 192}}{{2\left( {s - \frac{1}{2}} \right)}} + \frac{{96}}{{s + 2}} - \frac{{10}}{{{{\left( {s + 2} \right)}^2}}} - \frac{{25\frac{{2!}}{{2!}}}}{{{{\left( {s + 2} \right)}^3}}}} \right)$
Notice that we also had to factor a 2 out of the denominator of the first term and fix up the numerator of the last term in order to get them to match up to the correct entries in our table of transforms.
Taking the inverse transform then gives,
$y\left( t \right) = \frac{1}{{125}}\left( { - 96{{\bf{e}}^{\frac{t}{2}}} + 96{{\bf{e}}^{ - 2t}} - 10t{{\bf{e}}^{ - 2t}} - \frac{{25}}{2}{t^2}{{\bf{e}}^{ - 2t}}} \right)$
Example 3 Solve the following IVP. $y'' - 6y' + 15y = 2\sin \left( {3t} \right),\hspace{0.25in}y\left( 0 \right) = - 1\,\,\,\,\,\,\,y'\left( 0 \right) = - 4$
Show Solution
Take the Laplace transform of everything and plug in the initial conditions.
\begin{align*}{s^2}Y\left( s \right) - sy\left( 0 \right) - y'\left( 0 \right) - 6\left( {sY\left( s \right) - y\left( 0 \right)} \right) + 15Y\left( s \right) & = 2\frac{3}{{{s^2} + 9}}\\ \left( {{s^2} - 6s + 15} \right)Y\left( s \right) + s - 2 & = \frac{6}{{{s^2} + 9}}\end{align*}
Now solve for $$Y(s)$$ and combine into a single term as we did in the previous two examples.
$Y\left( s \right) = \frac{{ - {s^3} + 2{s^2} - 9s + 24}}{{\left( {{s^2} + 9} \right)\left( {{s^2} - 6s + 15} \right)}}$
Now, do the partial fractions on this. First let’s get the partial fraction decomposition.
$Y\left( s \right) = \frac{{As + B}}{{{s^2} + 9}} + \frac{{Cs + D}}{{{s^2} - 6s + 15}}$
Now, setting numerators equal gives,
\begin{align*} - {s^3} + 2{s^2} - 9s + 24 & = \left( {As + B} \right)\left( {{s^2} - 6s + 15} \right) + \left( {Cs + D} \right)\left( {{s^2} + 9} \right)\\ & = \left( {A + C} \right){s^3} + \left( { - 6A + B + D} \right){s^2} + \left( {15A - 6B + 9C} \right)s + 15B + 9D\end{align*}
Setting coefficients equal and solving for the constants gives,
\left. \begin{aligned}& {s^3}: & A + C & = - 1\\ & {s^2}: & - 6A + B + D & = 2\\ & {s^1}: & 15A - 6B + 9C & = - 9\\ & {s^0}: & 15B + 9D & = 24\end{aligned}\right\}\hspace{0.25in} \Rightarrow \hspace{0.25in}\begin{aligned} A & = \frac{1}{{10}} & B & = \frac{1}{{10}}\\ C & = - \frac{{11}}{{10}} & D & = \frac{5}{2}\end{aligned}
Now, plug these into the decomposition, complete the square on the denominator of the second term and then fix up the numerators for the inverse transform process.
\begin{align*}Y\left( s \right) & = \frac{1}{{10}}\left( {\frac{{s + 1}}{{{s^2} + 9}} + \frac{{ - 11s + 25}}{{{s^2} - 6s + 15}}} \right)\\ & = \frac{1}{{10}}\left( {\frac{{s + 1}}{{{s^2} + 9}} + \frac{{ - 11\left( {s - 3 + 3} \right) + 25}}{{{{\left( {s - 3} \right)}^2} + 6}}} \right)\\ & = \frac{1}{{10}}\left( {\frac{s}{{{s^2} + 9}} + \frac{{1\frac{3}{3}}}{{{s^2} + 9}} - \frac{{11\left( {s - 3} \right)}}{{{{\left( {s - 3} \right)}^2} + 6}} - \frac{{8\frac{{\sqrt 6 }}{{\sqrt 6 }}}}{{{{\left( {s - 3} \right)}^2} + 6}}} \right)\end{align*}
Finally, take the inverse transform.
$y\left( t \right) = \frac{1}{{10}}\left( {\cos \left( {3t} \right) + \frac{1}{3}\sin \left( {3t} \right) - 11{{\bf{e}}^{3t}}\cos \left( {\sqrt 6 t} \right) - \frac{8}{{\sqrt 6 }}{{\bf{e}}^{3t}}\sin \left( {\sqrt 6 t} \right)} \right)$
To this point we’ve only looked at IVP’s in which the initial values were at $$t = 0$$. This is because we need the initial values to be at this point in order to take the Laplace transform of the derivatives. The problem with all of this is that there are IVP’s out there in the world that have initial values at places other than $$t = 0$$. Laplace transforms would not be as useful as it is if we couldn’t use it on these types of IVP’s. So, we need to take a look at an example in which the initial conditions are not at $$t = 0$$ in order to see how to handle these kinds of problems.
Example 4 Solve the following IVP. $y'' + 4y' = \cos \left( {t - 3} \right) + 4t,\hspace{0.25in}y\left( 3 \right) = 0\,\,\,\,\,\,\,y'\left( 3 \right) = 7$
Show Solution
The first thing that we will need to do here is to take care of the fact that initial conditions are not at $$t = 0$$. The only way that we can take the Laplace transform of the derivatives is to have the initial conditions at $$t = 0$$.
This means that we will need to formulate the IVP in such a way that the initial conditions are at $$t = 0$$. This is actually fairly simple to do, however we will need to do a change of variable to make it work. We are going to define
$\eta = t - 3\hspace{0.25in} \Rightarrow \hspace{0.25in}\,\,\,t = \eta + 3$
$y''\left( t \right) + 4y'\left( t \right) = \cos \left( {t - 3} \right) + 4t$
Notice that we put in the $$\left( t \right)$$ part on the derivatives to make sure that we get things correct here. We will next substitute in for $$t$$.
$y''\left( {\eta + 3} \right) + 4y'\left( {\eta + 3} \right) = \cos \left( \eta \right) + 4\left( {\eta + 3} \right)$
Now, to simplify life a little let’s define,
$u\left( \eta \right) = y\left( {\eta + 3} \right)$
Then, by the chain rule, we get the following for the first derivative.
$u'\left( \eta \right) = \frac{{du}}{{d\eta }} = \frac{{dy}}{{dt}}\frac{{dt}}{{d\eta }} = y'\left( {\eta + 3} \right)$
By a similar argument we get the following for the second derivative.
$u''\left( \eta \right) = y''\left( {\eta + 3} \right)$
The initial conditions for $$u\left( \eta \right)$$ are,
\begin{align*}u\left( 0 \right) & = y\left( {0 + 3} \right) = y\left( 3 \right) = 0\\ u'\left( 0 \right) & = y'\left( {0 + 3} \right) = y'\left( 3 \right) = 7\end{align*}
The IVP under these new variables is then,
$u'' + 4u' = \cos \left( \eta \right) + 4\eta + 12,\hspace{0.25in}u\left( 0 \right) = 0\,\,\,\,\,\,\,\,u'\left( 0 \right) = 7$
This is an IVP that we can use Laplace transforms on provided we replace all the $$t$$’s in our table with $$\eta$$’s. So, taking the Laplace transform of this new differential equation and plugging in the new initial conditions gives,
\begin{align*}{s^2}U\left( s \right) - su\left( 0 \right) - u'\left( 0 \right) + 4\left( {sU\left( s \right) - u\left( 0 \right)} \right) & = \frac{s}{{{s^2} + 1}} + \frac{4}{{{s^2}}} + \frac{{12}}{s}\\ \left( {{s^2} + 4s} \right)U\left( s \right) - 7 & = \frac{s}{{{s^2} + 1}} + \frac{{4 + 12s}}{{{s^2}}}\end{align*}
Solving for $$U(s)$$ gives,
\begin{align*}\left( {{s^2} + 4s} \right)U\left( s \right) & = \frac{s}{{{s^2} + 1}} + \frac{{4 + 12s + 7{s^2}}}{{{s^2}}}\\ U\left( s \right) & = \frac{1}{{\left( {s + 4} \right)\left( {{s^2} + 1} \right)}} + \frac{{4 + 12s + 7{s^2}}}{{{s^3}\left( {s + 4} \right)}}\end{align*}
Note that unlike the previous examples we did not completely combine all the terms this time. In all the previous examples we did this because the denominator of one of the terms was the common denominator for all the terms. Therefore, upon combining, all we did was make the numerator a little messier and reduced the number of partial fractions required down from two to one. Note that all the terms in this transform that had only powers of $$s$$ in the denominator were combined for exactly this reason.
In this transform however, if we combined both of the remaining terms into a single term we would be left with a fairly involved partial fraction problem. Therefore, in this case, it would probably be easier to just do partial fractions twice. We’ve done several partial fractions problems in this section and many partial fraction problems in the previous couple of sections so we’re going to leave the details of the partial fractioning to you to check. Partial fractioning each of the terms in our transform gives us the following.
\begin{align*}\frac{1}{{\left( {s + 4} \right)\left( {{s^2} + 1} \right)}} & = \frac{{\frac{1}{{17}}}}{{s + 4}} + \frac{1}{{17}}\left( {\frac{{ - s + 4}}{{{s^2} + 1}}} \right)\\ \frac{{4 + 12s + 7{s^2}}}{{{s^3}\left( {s + 4} \right)}} & = \frac{1}{{{s^3}}} + \frac{{\frac{{11}}{4}}}{{{s^2}}} + \frac{{\frac{{17}}{{16}}}}{s} - \,\frac{{\frac{{17}}{{16}}}}{{s + 4}}\end{align*}
Plugging these into our transform and combining like terms gives us
\begin{align*}U\left( s \right) & = \frac{1}{{{s^3}}} + \frac{{\frac{{11}}{4}}}{{{s^2}}} + \frac{{\frac{{17}}{{16}}}}{s} - \frac{{\frac{{273}}{{272}}}}{{s + 4}} + \frac{1}{{17}}\left( {\frac{{ - s + 4}}{{{s^2} + 1}}} \right)\\ & = \frac{{1\frac{{2!}}{{2!}}}}{{{s^3}}} + \frac{{\frac{{11}}{4}}}{{{s^2}}} + \,\frac{{\frac{{17}}{{16}}}}{s} - \frac{{\frac{{273}}{{272}}}}{{s + 4}} + \frac{1}{{17}}\left( {\frac{{ - s}}{{{s^2} + 1}} + \frac{4}{{{s^2} + 1}}} \right)\end{align*}
Now, taking the inverse transform will give the solution to our new IVP. Don’t forget to use $$\eta$$’s instead of $$t$$’s!
$u\left( \eta \right) = \frac{1}{2}{\eta ^2} + \frac{{11}}{4}\eta + \frac{{17}}{{16}} - \frac{{273}}{{272}}{{\bf{e}}^{ - 4\eta }} + \frac{1}{{17}}\left( {4\sin \left( \eta \right) - \cos \left( \eta \right)} \right)$
This is not the solution that we are after of course. We are after $$y(t)$$. However, we can get this by noticing that
$y\left( t \right) = y\left( {\eta + 3} \right) = u\left( \eta \right) = u\left( {t - 3} \right)$
So, the solution to the original IVP is,
\begin{align*}y\left( t \right) & = \frac{1}{2}{\left( {t - 3} \right)^2} + \frac{{11}}{4}\left( {t - 3} \right) + \frac{{17}}{{16}} - \frac{{273}}{{272}}{{\bf{e}}^{ - 4\left( {t - 3} \right)}} + \frac{1}{{17}}\left( {4\sin \left( {t - 3} \right) - \cos \left( {t - 3} \right)} \right)\\ y\left( t \right) & = \frac{1}{2}{t^2} - \frac{1}{4}t - \frac{{43}}{{16}} - \frac{{273}}{{272}}{{\bf{e}}^{ - 4\left( {t - 3} \right)}} + \frac{1}{{17}}\left( {4\sin \left( {t - 3} \right) - \cos \left( {t - 3} \right)} \right)\end{align*}
So, we can now do IVP’s that don’t have initial conditions that are at $$t = 0$$. We also saw in the last example that it isn’t always the best to combine all the terms into a single partial fraction problem as we have been doing prior to this example.
The examples worked in this section would have been just as easy, if not easier, if we had used techniques from the previous chapter. They were worked here using Laplace transforms to illustrate the technique and method.
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# DES and AES cryptography
I have a question that I can't answer:
A plain text has been encrypted 3 times in a DES method with 3 different keys 56 bit each, after it the text was encrypted for the forth time using a AES method using a 128 bit key, an attacker is planning to decrypt the text using meet in the middle method, how many encryptions and decryption approximately he need to do.
1. $2^{56}$
2. $2^{112}$
3. $2^{128}$
4. $2^{168}$
5. $2^{184}$
6. $2^{240}$
7. $2^{296}$
• I don't like this question. There is no way you can perform the encryption in the first place because the block sizes don't match. I presume that the answer is $2^{168}$ because that's the number of block encrypts required for DES EEE, and the number of AES decrypts is negligible compared to that (and you probably cannot split it any further because AES + single DES decrypt is a higher number than that). Jun 27, 2019 at 11:46
Assuming this is ECB mode encryption, the workload of a known-plaintext meet-in-the-middle attack will depend strongly on the plaintext here.
If the plaintext contains some non-identical 128 bit blocks that partly coincide on a DES-block, the adversary can reverse the final AES encryption with $$2^{128}$$ AES decryptions work factor. They can then execute a standard meet-in-the-middle attack on 3DES using parallel collision search (easy compared to the preceding step). See van Oorschot/Wiener for details on the techniques used.
If the plaintext consists only of non-identical 64-bit blocks, the adversary can use the same idea (look for collisions of 64-bit blocks in the ciphertext after AES decryption) to at least rule out some AES keys. They should be able to rule out the vast majority of AES keys when the size of all encrypted plaintexts significantly exceeds the birthday bound for a 64-bit block cipher.
If the plaintext is short and does not contain repetitions of DES blocks that are not also repetitions of AES blocks, there may be no better idea than to perform a meet in the middle attack between 3DES and AES using parallel collision search. This should be doable in time about $$\approx 2^{212}/(\sqrt{w} m)$$ on a machine with $$2w$$ AES blocks of memory and $$m$$ processors (see again the parallel collision search paper).
I believe the the MITM attack for 3DES is $$2^{112}$$, so if MITM attack is between the 3DES and the 4th AES, the most expensive brute force attack between the two is the AES one, which is $$2^{128}$$.
So all it takes is $$2^{128}$$ encryptions / decryptions.
Note that I am trying to learn this myself too!
• Not quite: the MITM attack for 3DES assumes the attacker has corresponding plaintext and ciphertext blocks; with an AES at the end, he doesn't have the ciphertext block after then 3DES Nov 21, 2017 at 14:52
• I am thinking this is similar to MITM for 2DES, where the adversary has the cipher text at the end as well as in between the 2DES, and the attack involves a lookup table of all possible keys.
– pip
Nov 21, 2017 at 14:59
• @poncho I think I get your point now. I need 2^128 to get the cipter text at the end of the 3DES (inv AES), and another 2^112 to do the MITM on the 3DES itself.
– pip
Nov 21, 2017 at 15:42
• It's a bit tougher than that; even with $2^{128}$ work, you won't have any way to determine which AES-decryption is the correct one, and so throwing $2^{128}$ work will yield $2^{128}$ possible 3DES ciphertexts, not the 1 that MITM assumes... Nov 21, 2017 at 15:52
• I believe the point to establish the theoretical number of encryption/descriptions need for this attack, instead of how practical the attack will be. I think $2^{110}$ is already consider secure.
– pip
Nov 21, 2017 at 18:26
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# In turbulent flow in a pipe
This question was previously asked in
MPSC AMVI Official Paper 2: Set A/2013
View all MPSC Motor Vehicle Inspector Papers >
1. Shear stress varies linearly with radius
2. head loss varies linearly with a flow rate
3. Fluid particles move in a straight line
4. Reynolds number is less than 1000
Option 1 : Shear stress varies linearly with radius
Free
ST 1: General Knowledge
3106
20 Questions 20 Marks 20 Mins
## Detailed Solution
Explanation:
Shear stress in turbulent flow:
Perhaps the first thought that comes to mind is to determine the shear stress in an analogous manner to laminar flow from
$$τ =\mu\frac{du}{dr}$$
where u (r) is average velocity profile for turbulent flow.
But the experimental studies show that this is not the case, and the shear stress is much larger due to the turbulent fluctuations.
Therefore, it is convenient to think of the turbulent shear stress as consisting of two parts:
The laminar component, which accounts for the friction between layers in the flow direction
$$τ_{laminar} =\mu\frac{du}{dr}$$
And the turbulent component, which accounts for the friction between the fluctuating fluid particles and the fluid body.
Total shear stress in a flowing fluid in a pipe is given by:
$$τ=τ_v\;+\;τ_t$$
where τv = shear stress due to viscosity and τt = shear stress due to turbulence.
The shear stress can be treated as linear in both axes i.e zero at the centre and varying linearly up to the wall.
Reynolds Expression for Turbulent Shear stress:
Reynold's through his experiment developed an expression between two layers of a fluid at a small distance apart, which is given as:
$$τ_t=\rho{u'}{v'}$$
where u', v' = fluctuating component of velocity in the direction of x and y respectively.
u' and v' are both varying and hence τ will also vary. To calculate τ, time average is done on both sides of the equation.
$$τ_t=\bar{\rho}\bar{u}'\bar{v}'$$
Due to difficulty in calculating time-varying velocities, Prandtl presented a mixing length hypothesis that can be used to express turbulent shear stress in terms of measurable quantities.
According to Prandtl, the mixing length l, is that distance between two layers in the transverse direction such that lumps of fluid particles from one layer could reach the other layer and the particles are mixed in the other layer in such a way that the momentum of the particles in the direction of x is same.
$$u'=l\frac{du}{dy}\;and\;v'=l\frac{du}{dy}$$
The above equation gives velocity fluctuation in x and y direction in terms of mixing length l.
$$\bar{u}\;\times\;\bar{v}=l\frac{du}{dy}\;\times\;l\frac{du}{dy}\Rightarrow\;l^2(\frac{du}{dy})^2$$
∴ turbulence shear stress is given by:
$$τ_t=\bar{\rho}\bar{u}'\bar{v}'\Rightarrow\rho{l^2}(\frac{du}{dy})^2$$
Total shear stress acting on the fluid is:
$$τ=τ_v\;+\;τ_t$$
$$τ=\mu\frac{du}{dy}\;+\;\rho{l^2}(\frac{du}{dy})^2$$
In the case of turbulent flow, viscous shear stress is negligible except near the boundary. Hence, it can be assumed that shear stress is purely dominated by $$τ=\rho{l^2}(\frac{du}{dy})^2.$$
∵ mixing length l is the distance measured in the transverse direction (perpendicular to flow) where lumps reach the other layer, and particles are mixed in the longitudinal direction (parallel to flow), the shear stress can be treated as linear in both axes i.e zero at centre and varying linearly up to the wall.
Important Points
Laminar flow Turbulent flow Laminar flow is the flow in which fluid particles flow in layers. Each layer moves smoothly past the adjacent layer with little or no intermixing. Turbulent flow is the flow in which fluid particles flow in a random fashion where intermixing/intermingling is present. The entire fluid layers move parallel to each other. They can’t cross each other. Fluid layers are not parallel to each other. They can cross each other. Reynolds’s number for a pipe is less than 2000 for laminar flow. Reynolds’s number for a pipe is more than 4000 for turbulent flow.
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# Sumif for merged Collums
#### Henry_paul_616
##### New Member
Hey guys. I hope you're doing well. I'm Brazilian, intern and I'm having a problem using the SUMIF function to add the cost value of a metal alloy. I would like the function to add the values of the costs that are comprised by the date that is compressed by merged columns. The SUMIF function only returns the first cost and does not add the second cost which would total 144000 dollars. My formula has to add the cost if the league and date are what I want. How can I resolve this issue?
Attached is the excel file.
Can you help me?
#### Attachments
• 12.1 KB Views: 4
#### John Jairo V
##### Well-Known Member
Hi, Henry!
Two options for you.
1► With helper row.
PHP:
``[B2] : =IF(B3,B3,A2)``
Drag it to the right until H2
PHP:
``[F11] : =SUMIFS(B6:H6,B4:H4,C8,B2:H2,E11)``
2►Without helper row.
PHP:
``[F11] : =SUMPRODUCT(--(B4:H4=C8),--(LOOKUP(COLUMN(B3:H3),COLUMN(B3:H3)/(B3:H3<>""),B3:H3)=E11),B6:H6)``
Check file. Blessings!
#### Attachments
• 13 KB Views: 7
#### Peter Bartholomew
##### Well-Known Member
I attempted this problem with 365, not because it is easy but rather because it is surprisingly challenging.
The rational solution is to use a helper row, as shown by @John Jairo V .
I chose to write a Lambda function to return the result for a particular date and alloy.
Code:
``````Costλ
LET(
runDate, SCAN(0, date, LAMBDA(p, d, IF(d > 0, d, p))),
selectedCost, IF((runDate = aDate) * (alloy = anAlloy), cost),
SUM(selectedCost)
)
));``````
The variable 'runDate' holds the dates with the merged cells filled across.
The fact that the result is not a range prevents the use of SUMIFS, so 'selectedCost holds costs that match the criteria.
The result is summed.
The worksheet formula uses the Lambda function but with the added complexity of performing the calculation for an array of dates.
Code:
``````WorksheetFormula
= MAP(selectedDate, Costλ(selectedAlloy))``````
#### Attachments
• 18.1 KB Views: 3
#### m9vukyem
##### Member
=IF(B3,B3,A2)
how to use in formula
#### Peter Bartholomew
##### Well-Known Member
That is all very well but, apparently, more than half of the spreadsheets ever written contain merged cells. Mind you, the majority of workbooks do not contain formulas, so perhaps it doesn't matter!
In 365, the SCAN helper function allows fill down, with or without merged cells.
Code:
``````WorksheetFormula
= SCAN(0, date, Fillλ)
Fillλ
= LAMBDA(p, d, IF(d > 0, d, p));``````
so all is not lost.
#### Henry_paul_616
##### New Member
Hi, Henry!
Two options for you.
1► With helper row.
PHP:
``[B2] : =IF(B3,B3,A2)``
Drag it to the right until H2
PHP:
``[F11] : =SUMIFS(B6:H6,B4:H4,C8,B2:H2,E11)``
2►Without helper row.
PHP:
``[F11] : =SUMPRODUCT(--(B4:H4=C8),--(LOOKUP(COLUMN(B3:H3),COLUMN(B3:H3)/(B3:H3<>""),B3:H3)=E11),B6:H6)``
Check file. Blessings!
Thank You!!!!
#### Henry_paul_616
##### New Member
I attempted this problem with 365, not because it is easy but rather because it is surprisingly challenging.
The rational solution is to use a helper row, as shown by @John Jairo V .
I chose to write a Lambda function to return the result for a particular date and alloy.
Code:
``````Costλ
LET(
runDate, SCAN(0, date, LAMBDA(p, d, IF(d > 0, d, p))),
selectedCost, IF((runDate = aDate) * (alloy = anAlloy), cost),
SUM(selectedCost)
)
));``````
The variable 'runDate' holds the dates with the merged cells filled across.
The fact that the result is not a range prevents the use of SUMIFS, so 'selectedCost holds costs that match the criteria.
The result is summed.
The worksheet formula uses the Lambda function but with the added complexity of performing the calculation for an array of dates.
Code:
``````WorksheetFormula
= MAP(selectedDate, Costλ(selectedAlloy))``````
View attachment 80176
Thank You!!!
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# 4.51 hours in hours and minutes
## Result
4.51 hours equals 4 hours and 30.6 minutes
You can also convert 4.51 hours to minutes.
## Converter
Four point five one hours is equal to four hours and thirty point six minutes.
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# How can I edge slide a vertex to a specific coordinate, using Python?
With Python, how would I implement an edge slide of a vertex to a specific coordinate? For example, sliding a vertex along an edge loop to a X coordinate of, say, 10.
The Vertex Slide doesn't have a feature (currently Dec 2015) to limit the extent of a slide for any given axis. This means that there's no script-able way to do it from the operator by simply passing a few parameters. This doesn't mean an alternative doesn't exist, or several. Some code will have to be written though, and to make it usable as a tool also some UI code.
Projecting a vertex of an edge towards a Face on the YZ plane would have the same effect as limiting a vertex's sliding till it reached a given axis limit. (ie x = 10)
We can find the intersection of a line on the infinite plane using a built-in geometry function mathutils.geometry.intersect_line_plane . We then use the plane_nol=(1,0,0) to show the plane extends only in the YZ axes, and plane_co=(your_x, 0, 0) to give the plane an origin (this is the axis constraint)
# alias it first, for convenience here
intersect_l_p = mathutils.geometry.intersect_line_plane
intersect_l_p(line_a, line_b, plane_co, plane_no, no_flip=False)
and then overwrite the vertex location with the intersection location.
Let's take the example above where I have a plane on the X axis at about 4.2 units. and the Vertex i want to move along the arrow till it hits the plane.
### Non-UI version.
This example assumes we have a mesh object in edit-mode
import bpy
import bmesh
import mathutils
def extend_vertex(limit_axis='x', coordinate=4.2):
obj = bpy.context.edit_object
me = obj.data
bm = bmesh.from_edit_mesh(me)
verts = bm.verts
try:
v1, v2 = [v for v in bm.verts if v.select]
except:
print('need two vertices selected, or one edge')
bm.free()
return
plane_idx = {'x': 0, 'y': 1, 'z': 2}.get(limit_axis)
plane_co, plane_no = [0,0,0], [0,0,0]
plane_no[plane_idx] = 1
plane_co[plane_idx] = coordinate
intersect_l_p = mathutils.geometry.intersect_line_plane
new_co = intersect_l_p(v1.co, v2.co, plane_co, plane_no, False)
# move closest of the two vertices
A_len = (v1.co-new_co).length
B_len = (v2.co-new_co).length
if A_len < B_len:
v1.co = new_co
else:
v2.co = new_co
bmesh.update_edit_mesh(me, True)
extend_vertex(limit_axis='x', coordinate=4.2)
warning:
This doesn't take into account the object's unapplied Transforms (origin may not coincide with the world's (0,0,0), you may have scaling, rotation, translation), easy enough to fix - but it complicates the example code a little.
To take matrix_world into account, do something like this: vertex_towards_plane.py
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https://www.gradesaver.com/textbooks/math/algebra/algebra-1-common-core-15th-edition/chapter-8-polynomials-and-factoring-8-4-multiplying-soecial-cases-got-it-page-505/2
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## Algebra 1: Common Core (15th Edition)
The area of the walkway is $(16x+64) ft^{2}$.
Find the total area of the patio and walkway. $(x+8)^{2}$ ...square the binomial. $=x^{2}+2(x)(8)+8^{2}$ ...simplify. $=x^{2}+16x+64$ Find the area of the patio The area of the patio is $x\cdot x$, or $x^{2}$. Find the area of the walkway. Area of walkway=Total area - Area of patio $A=(x^{2}+16x+64)-x^{2}$ $=(x^{2}-x^{2})+16x+64$ ...add like terms. $=16x+64$ The area of the walkway is $(16x+64) ft^{2}$.
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http://www.mathworks.com/help/dsp/ref/nco.html?nocookie=true
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# NCO
Generate real or complex sinusoidal signals
## Library
Signal Operations
`dspsigops`
Sources
`dspsrcs4`
## Description
The block appears in `dspsigops` with Phase increment Source parameter set to `Input port`.
In addition, the block also appears in `dspsrsc4` library with Phase increment Source parameter set to ```Specify via dialog```.
The NCO block generates a multichannel real or complex sinusoidal signal, with independent frequency and phase in each output channel. The amplitude of the created signal is always 1. The block implements the algorithm as shown in the following diagram:
The implementation of a numerically controlled oscillator (NCO) has two distinct parts. First, a phase accumulator accumulates the phase increment and adds in the phase offset. In this stage, an optional internal dither signal can also be added. The NCO output is then calculated by quantizing the results of the phase accumulator section and using them to select values from a lookup table. Since the lookup table contains a finite set of entries, in its normal mode of operation, the NCO block allows the adder's numeric values to overflow and wrap around. The Fixed-Point infrastructure then causes overflow warnings to appear on the command line. This overflow is of no consequence.
Given a desired output frequency F0, calculate the value of the Phase increment block parameter with
$phaseincrement=\left(\frac{{F}_{0}\cdot {2}^{N}}{{F}_{s}}\right)$
where N is the accumulator word length and
${F}_{s}=\frac{1}{{T}_{s}}=\frac{1}{sampletime}$
The frequency resolution of an NCO is defined by
$\Delta f=\frac{1}{{T}_{s}\cdot {2}^{N}}\text{Hz}$
Given a desired phase offset (in radians), calculate the Phase offset block parameter with
$phaseoffset=\frac{{2}^{N}\cdot desiredphaseoffset}{2\pi }$
The spurious free dynamic range (SFDR) is estimated as follows for a lookup table with ${2}^{P}$ entries, where P is the number of quantized accumulator bits:
This block uses a quarter-wave lookup table technique that stores table values from 0 to π/2. The block calculates other values on demand using the accumulator data type, then casts them into the output data type. This can lead to quantization effects at the range limits of a given data type. For example, consider a case where you would expect the value of the sine wave to be –1 at π. Because the lookup table value at that point must be calculated, the block might not yield exactly –1, depending on the precision of the accumulator and output data types.
The NCO block supports real inputs only. All outputs are real except for the output signal in `Complex exponential` mode.
To produce a multichannel output, specify a vector quantity for the Phase increment and Phase offset parameters. Both parameters must have the same length, which defines the number of output channels. Each element of each vector is applied to a different output channel.
## Fixed-Point Data Types
The following diagram shows the data types used within the NCO block.
• You can set the accumulator and output data types in the block dialog as discussed in Dialog Box below.
Note: The lookup table for this block is constructed from double-precision floating-point values. Thus, the maximum amount of precision you can achieve in your output is 53 bits. Setting the word length of the Output data type to values greater than 53 bits does not improve the precision of your output.
• The phase increment and phase offset inputs must be integers or fixed-point data types with zero fraction length.
• You specify the number of quantized accumulator bits in the Number of quantized accumulator bits parameter.
• The phase quantization error word length is equal to the accumulator word length minus the number of quantized accumulator bits, and the fraction length is zero.
## Examples
The NCO block is used in the GSM Digital Down ConverterGSM Digital Down Converter product example. Open this example by typing `dspddc` at the MATLAB® command line.
You can also try the following example. Design an NCO source with the following specifications:
• Desired output frequency
• Frequency resolution
• Spurious free dynamic range
• Sample period
• Desired phase offset $\pi /2$
1. Calculate the number of required accumulator bits from the equation for frequency resolution:
$\begin{array}{l}\Delta f=\frac{1}{{T}_{s}\cdot {2}^{N}}\text{Hz}\\ 0.05=\frac{1}{\frac{1}{8000}\cdot {2}^{N}}\text{Hz}\\ N=18\end{array}$
Note that N must be an integer value. The value of N is rounded up to the nearest integer; 18 accumulator bits are needed to accommodate the value of the frequency resolution.
2. Using this best value of N, calculate the frequency resolution that will be achieved by the NCO block:
3. Calculate the number of quantized accumulator bits from the equation for spurious free dynamic range and the fact that for a lookup table with ${2}^{P}$ entries, P is the number of quantized accumulator bits:
4. Select the number of dither bits. In general, a good choice for the number of dither bits is the accumulator word length minus the number of quantized accumulator bits; in this case 4.
5. Calculate the phase increment:
$\begin{array}{l}phaseincrement=\mathrm{round}\left(\frac{{F}_{0}\cdot {2}^{N}}{{F}_{s}}\right)\\ phaseincrement=\mathrm{round}\left(\frac{510\cdot {2}^{18}}{8000}\right)\\ phaseincrement=16712\end{array}$
6. Calculate the phase offset:
$\begin{array}{l}phaseoffset=\frac{{2}^{accumulatorwordlength}\cdot desiredphaseoffset}{2\pi }\\ phaseoffset=\frac{{2}^{18}\cdot \frac{\pi }{2}}{2\pi }\\ phaseoffset=65536\end{array}$
7. Type `ex_nco_exampleex_nco_example` at the MATLAB command line to open the following model:
The NCO block in the model is populated with the specifications and quantities you just calculated. The output word length and fraction length depend on the constraints of your hardware; this example uses a word length of 16 and a fraction length of 14. You can verify that the specifications of this problem have been met by looking at the NCO Characterization pane of the NCO block.
8. Experiment with the model to observe the effects on the output shown on the Spectrum Analyzer. For example, try turning dithering on and off, and try changing the number of dither bits.
## Dialog Box
The Main pane of the NCO dialog appears as follows.
Phase increment source
Choose how you specify the phase increment. The phase increment can come from an input port or from the dialog.
• If you select `Input port`, the inc port appears on the block icon.
• If you select `Specify via dialog`, the Phase increment parameter appears.
Phase increment
Specify the phase increment. Only integer data types, including fixed-point data types with zero fraction length, are allowed. The dimensions of the phase increment are dictated by those of the phase offset:
• When you specify the phase offset on the block dialog box, the phase increment must be a scalar or a vector with the same length as the phase offset. The block applies each element of the vector to a different channel, and therefore the vector length defines the number of output channels.
• When you specify the phase offset via an input port, the offset port treats each column of the input as an independent channel. The phase increment length must equal the number of columns in the input to the offset port.
This parameter is visible only if you set the Phase increment source parameter to ```Specify via dialog```.
Phase offset source
Choose how you specify the phase offset. The phase offset can come from an input port or from the dialog.
• If you select `Input port`, the offset port appears on the block icon.
• If you select `Specify via dialog`, the Phase offset parameter appears.
When you specify the phase offset via an input port, it can be a scalar, vector, or a full matrix. The block treats each column of the input to the offset port as an independent channel. The number of channels in the phase offset must match the number of channels in the data input. For each frame of the input, the block can apply different phase offsets to each sample and channel. Only integer data types, including fixed-point data types with zero fraction length, are allowed.
Phase offset
Specify the phase offset. When you specify the phase offset using this parameter rather than via an input port, it must be a scalar or vector with the same length as the phase increment. Scalars are expanded to a vector with the same length as the phase increment. Each element of the phase offset vector is applied to a different channel of the input, and therefore the vector length defines the number of output channels. Only integer data types, including fixed-point data types with zero fraction length, are allowed.
This parameter is visible only if `Specify via dialog` is selected for the Phase offset source parameter.
Add internal dither
Select to add internal dithering to the NCO algorithm. Dithering is added using the PN Sequence Generator from the Communications System Toolbox™ product.
Number of dither bits
Specify the number of dither bits.
This parameter is visible only if Add internal dither is selected.
Quantize phase
Select to enable quantization of the accumulated phase.
Number of quantized accumulator bits
Specify the number of quantized accumulator bits. This determines the number of entries in the lookup table. The number of quantized accumulator bits must be less than the accumulator word length.
This parameter is visible only if Quantize phase is selected.
Show phase quantization error port
Select to output the phase quantization error. When you select this, the Qerr port appears on the block icon.
This parameter is visible only if Quantize phase is selected.
Output signal
Choose whether the block should output a `Sine`, `Cosine`, ```Complex exponential```, or `Sine and cosine` signals. If you select `Sine and cosine`, the two signals output on different ports.
Sample time
Specify the sample time in seconds when the block is acting as a source. When either the phase increment or phase offset come in via block input ports, the sample time is inherited and this parameter is not visible.
Samples per frame
Specify the number of samples per frame. When the value of this parameter is `1`, the block outputs a sample-based signal. When the value is greater than 1, the block outputs a frame-based signal of the specified size. In frame-based mode, the phase increment and phase offset can vary from channel to channel and from frame to frame, but they are constant along each channel in a given frame.
When the phase offset input port exists, it has the same frame status as any output port present. When the phase increment input port exists, it does not support frames.
This parameter is only visible if either Phase increment source or Phase offset source is set to `Specify via dialog`.
The Data Types pane of the NCO dialog appears as follows.
Rounding mode
The rounding mode used for this block when inputs are fixed point is always `Floor`.
Overflow mode
The overflow mode used for this block when inputs are fixed point is always `Wrap`.
Accumulator
Specify the word length of the accumulator data type. The fraction length is always zero; this is an integer data type.
Output
Specify the output data type.
• Choose `double` or `single` for a floating-point implementation.
• When you select `Binary point scaling`, you can enter the word length and the fraction length of the output, in bits.
Note: The lookup table for this block is constructed from double-precision floating-point values. Thus, the maximum amount of precision you can achieve in your output is 53 bits. Setting the word length of the Output data type to values greater than 53 bits does not improve the precision of your output.
The NCO Characterization pane of the NCO dialog appears as follows.
The NCO Characterization pane does not have any parameters. Instead, it provides you with details on the NCO signal currently being implemented by the block:
• `Number of data points for lookup table` — The lookup table is implemented as a quarter-wave sine table. The number of lookup table data points is defined by
• `Quarter wave sine lookup table size` — The quarter wave sine lookup table size is defined by
• `Theoretical spurious free dynamic range` — The spurious free dynamic range (SFDR) is calculated as follows for a lookup table with ${2}^{P}$ entries:
## Supported Data Types
PortSupported Data Types
inc
• Fixed point with zero fraction length
• 8-, 16-, and 32-bit signed integers
• 8-, 16-, and 32-bit unsigned integers
offset
• Fixed point with zero fraction length
• 8-, 16-, and 32-bit signed integers
• 8-, 16-, and 32-bit unsigned integers
sin
• Double-precision floating point
• Single-precision floating point
• Fixed point
Qerr
• 8-, 16-, and 32-bit signed integers
• 8-, 16-, and 32-bit unsigned integers
## See Also
PN Sequence Generator Communications System Toolbox Sine Wave DSP System Toolbox Digital Down-Converter DSP System Toolbox Digital Up-Converter DSP System Toolbox
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https://oeis.org/A077066
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The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A077066 Squarefree kernel of prime(n) + 1. 3
3, 2, 6, 2, 6, 14, 6, 10, 6, 30, 2, 38, 42, 22, 6, 6, 30, 62, 34, 6, 74, 10, 42, 30, 14, 102, 26, 6, 110, 114, 2, 66, 138, 70, 30, 38, 158, 82, 42, 174, 30, 182, 6, 194, 66, 10, 106, 14, 114, 230, 78, 30, 22, 42, 258, 66, 30, 34, 278, 282, 142, 42, 154, 78, 314, 318, 166 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 COMMENTS a(A077067(n)) = A077067(n). LINKS Reinhard Zumkeller, Table of n, a(n) for n = 1..10000 FORMULA a(n) = A007947(A008864(n)). EXAMPLE a(25) = rad(prime(25)+1) = rad(97+1) = rad(2*7^2) = 14. PROG (PARI) a(n)=my(f=factor(prime(n)+1)[, 1]); prod(i=1, #f, f[i]) \\ Charles R Greathouse IV, Aug 21 2013 (Haskell) a077066 = a007947 . a008864 -- Reinhard Zumkeller, Sep 04 2013 CROSSREFS Cf. A077063. Sequence in context: A129628 A245691 A071044 * A324890 A180240 A065228 Adjacent sequences: A077063 A077064 A077065 * A077067 A077068 A077069 KEYWORD nonn AUTHOR Reinhard Zumkeller, Oct 23 2002 STATUS approved
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Last modified January 24 01:05 EST 2020. Contains 331178 sequences. (Running on oeis4.)
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https://jp.mathworks.com/matlabcentral/profile/authors/3246794
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#### Overview of this book
Data Forecasting and Segmentation Using Microsoft Excel guides you through basic statistics to test whether your data can be used to perform regression predictions and time series forecasts. The exercises covered in this book use real-life data from Kaggle, such as demand for seasonal air tickets and credit card fraud detection. You’ll learn how to apply the grouping K-means algorithm, which helps you find segments of your data that are impossible to see with other analyses, such as business intelligence (BI) and pivot analysis. By analyzing groups returned by K-means, you’ll be able to detect outliers that could indicate possible fraud or a bad function in network packets. By the end of this Microsoft Excel book, you’ll be able to use the classification algorithm to group data with different variables. You’ll also be able to train linear and time series models to perform predictions and forecasts based on past data.
Preface
Part 1 – An Introduction to Machine Learning Functions
Free Chapter
Chapter 1: Understanding Data Segmentation
Chapter 2: Applying Linear Regression
Chapter 3: What is Time Series?
Part 2 – Grouping Data to Find Segments and Outliers
Chapter 4: Introduction to Data Grouping
Chapter 5: Finding the Optimal Number of Single Variable Groups
Chapter 6: Finding the Optimal Number of Multi-Variable Groups
Chapter 7: Analyzing Outliers for Data Anomalies
Part 3 – Simple and Multiple Linear Regression Analysis
Chapter 8: Finding the Relationship between Variables
Chapter 9: Building, Training, and Validating a Linear Model
Chapter 10: Building, Training, and Validating a Multiple Regression Model
Part 4 – Predicting Values with Time Series
Chapter 11: Testing Data for Time Series Compliance
Chapter 12: Working with Time Series Using the Centered Moving Average and a Trending Component
Chapter 13: Training, Validating, and Running the Model
Other Books You May Enjoy
# Summary
The forecast calculation depends entirely on the quality of autocorrelation in the data. If present values are dependent on past data, then the data will give good future predictions for the time series. The Durbin-Watson probe to check the level of autocorrelation of the time series tells us how good the prediction will be by measuring the influence of past data on the current values.
The season component depends on the CMA distance to the data. The season component is determined by the forecast as a factor to move the trend (linear regression) up or down, depending on the cycles of the time series. Comparing the forecast time-series line chart with the original data gives us an idea of how accurate the model's prediction is.
In this chapter, we learned to use the CMA to smooth the peaks and troughs of the seasonal values over the years. The CMA helps to calculate the seasonal trend weight that leads the regression line up and down for the monthly dependable forecasting...
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Contributor III
Using Aggr function to display Customers with C-Materials only
Dear all,
first of all: this is my first post here and I think this is a great community!
As I have been struggling with this issue for quite a while now, I decided to register myself, hoping you can help me solve my issue.
I assume it's not a big thing, though, I am stuck here.
Plot:
I created an ABC-analysis in order to classify materials based on their importance with regards to total sales volume.
I have Customer data available to match these materials to the respective customers.
What I need to achieve:
As a matter of fact, there are some customers, which receive A, B & C-classified materials. However, I want to display those customers only which receive "C"-classified materials only.
I assume, aggr function will help here. Unfortunately, I do not know how to use it.
felipeflamenco
Ship-To ABC_Material count(distinct(Material)) Customer 1 A 1 Customer 2 A 2 Customer 3 A 1 Customer 4 A 5 Customer 4 B 2 Customer 4 C 1 Customer 5 A 4 Customer 5 A 3 Customer 6 A 4 Customer 6 A 4 Customer 7 A 9 Customer 7 B 1 Customer 8 A 5 Customer 9 A 7 Customer 9 B 1 Customer 10 A 10 Customer 10 B 4 Customer 10 C 1 Customer 11 A 3 Customer 12 A 1 Customer 12 C 1 Customer 13 A 1 Customer 14 A 1 Customer 15 B 1 Customer 15 C 1 Customer 16 A 1 Customer 17 C 1 Customer 18 A 2 Customer 19 A 2 Customer 19 B 2 Customer 20 A 1 Customer 20 C 3
1 Solution
Accepted Solutions
Contributor III
Hi all,
I found a workaround.
I classified all customers into A, AB, AC, ABC, B, BC or C-customers.
This is how I did it:
1. Create a Pivot Table
2. Dimension: [Ship-to]
3. Expression:
if(count({<ABC_Material={'A'}>}distinct(Material))>=1,'A')
&
if(count({<ABC_Material={'B'}>}distinct(Material))>=1,'B')
&
if(count({<ABC_Material={'C'}>}distinct(Material))>=1,'C')
Thanks a lot for your support!
Regards
11 Replies
Do you want to show the customer which is 'C'..
Try this
Dimension : Ship-to
Expression : =Only({<ABC_Material={'C'}>}[Ship-To])
if you want to show the count..
Dimension : Ship-to
Exp : =Count({<ABC_Material={'C'}>}[Ship-To])
Contributor III
Hi Settu,
thank you. This is an interesting approach.
However, these expressions just help me identify customers which receive "C" materials.
What I want to achieve, though, is to identify those customers which receive "C" materials only.
Any further suggestions?
Thank you and best regards
Hi,
Try this
=if(ABC_Material='C', Aggr(Only({<ABC_Material={'C'}>}[Ship-To]),[Ship-To]))
Contributor III
Hi Settu,
thanks a lot! At first I thought that's what I was actually looking for.. unfortunately, it's not. 🙂
Still an intersting string.
Contributor III
Hi,
Try this in 3 expressions:
sum (CountMaterial) ,
sum (TOTAL <Material> CountMaterial) Total Material (header)
sum (TOTAL {\$<Material={'C'}>} CountMaterial) Total C (header)
Kind Regards
see atachment
Contributor III
Hi Marcellino,
thank you.
So, I'll go with
Dimension: [Ship-To]
How would I need to set the commas between the different parameters for the Expressions correctly? The way you have presented it, will not work. Also, putting a sum() before a count() won't work either.
Regards
Master
hi
Use Count({<ABC_Material={'C'}>}[Ship-To])
Contributor III
Hi Neetha,
thank you. Unfortunately, this will help me identify customers which receive "C" materials. Yet, I want to identify only those customers which receive "C" materials only.
--> So, the result of the expression for this small example should be "Customer 17" only.
Contributor III
Hi,
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# Vegetable Light Curves
In this lesson, students observe the surface of rotating potatoes to help them understand how astronomers can sometimes determine the shape of asteroids from variations in reflective brightness.
# Measuring Angular Size and Distance
This is an activity about measuring angular size and understanding the solar and lunar proportions that result in solar eclipses. Learners will use triangles and proportions to create a shoebox eclipse simulator. They will then apply what they learn... (View More)
# Log Algebra
In this activity, students solve exponential equations where the unknown is contained in the exponent. Students learn that taking base-10 or base-2 logs pulls down the exponent, allowing the unknown to be isolated and solved. This activity is... (View More)
# Scientific Notation
In this activity students convert antilogs to logs, and logs to antilogs using scientific notation as an intermediate step. They will thereby develop a look-up table for solving math problems by using logarithms. This is activity D2 in the "Far Out... (View More)
# Base-Two Slide Rule
In this activity, students construct base-two slide rules that add and subtract base-2 exponents (log distances), in order to multiply and divide corresponding powers of two. Students use these slide rules to generate both log and antilog equations,... (View More)
# Inverse Functions
In this activity students use log tapes and base-two slide rules as references to graph exponential functions and log functions in base-10 and base-2. Students discover that exponential and log functions are inverse, reflecting across the y = x axis... (View More)
# Multiplying Slide Rule
In this activity students construct multiplying slide rules scaled in Base-10 exponents and use them to calculate products and quotients. They will come to appreciate that super numbers (exponents, orders of magnitude and logarithms) play by... (View More)
# Adding Slide Rule
In this activity, students construct adding slide rules, scaled with linear calibrations like ordinary rulers. Students learn to move these scales relative to each other in ways that add and subtract distances, thus calculating sums and differences.... (View More)
# String Calculator
In this activity students add and subtract log distances on their Log Tapes to discover that the corresponding numbers multiply and divide. This will lead them to an experiential understanding of the laws of logarithms. This is activity B2 in the... (View More)
# Slide-Rule Graph
In this activity students use their Log Tapes as a reference for ordered pairs, and graph positive numbers as a function of their base-10 logarithms. They extend each plotted point to the vertical axis, thereby generating a logarithmic scale that... (View More)
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Systems of Equations
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by Lonnie Myers
| 4 Questions
Note from the author:
Solving systems of equations algebraically
1
1 pt
Two car-rental agencies are competing for business. Mertz charges \$35 a day and \$0.15 per mile, and Mavis charges \$20 a day and \$0.45 per mile. Write an equation that represents the charges to rent a car at each agency.
2
2 pts
Solve the systems of equations. Show all work and write the point of intersection as an ordered pair
3
1 pt
What is the point of intersection? Explain what it means.
4
1 pt
Which agency should you choose? Why?
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# Describing subsets of sample spaces exercise | Probability and Statistics | Khan Academy | Summary and Q&A
180.0K views
March 8, 2015
by
Describing subsets of sample spaces exercise | Probability and Statistics | Khan Academy
## TL;DR
Analyzing sample spaces and probabilities using examples from Harry Potter and a game called "Fire-Water-Sponge."
## Install to Summarize YouTube Videos and Get Transcripts
### Q: How is the probability of selecting a wand made of holly or unicorn hair calculated?
To calculate the probability, we count the number of outcomes that involve holly or unicorn hair and divide it by the total number of equally likely outcomes, which is 20. In this case, there are 8 outcomes with holly or unicorn hair, resulting in a probability of 8/20 or 40%.
### Q: What determines the subset of outcomes in "Fire-Water-Sponge" where the friend either wins or there is a tie?
The subset in this case consists of outcomes where the friend either wins or there is a tie. We exclude outcomes where the player wins. In the highlighted outcomes (1, 3, 4, 5, 7, and 8), the friend either wins or there is a tie, leading to the subset that satisfies this condition.
### Q: How many possible outcomes are there in "Fire-Water-Sponge"?
In "Fire-Water-Sponge," there are 9 possible outcomes since each player has 3 choices (fire, water, or sponge), resulting in a total of 3 * 3 = 9 possibilities.
### Q: What determines the subset of outcomes where there is not a tie in "Fire-Water-Sponge"?
The subset consists of outcomes where there is not a tie. Since ties occur when both players choose the same object (fire-fire, water-water, or sponge-sponge), we exclude these outcomes from the subset. This leaves us with outcomes where someone wins (outcome 1, 3, 4, 5, 7, or 8).
## Summary & Key Takeaways
• The video discusses how to describe sets and subsets of sample spaces using examples from Harry Potter and a game called "Fire-Water-Sponge."
• In the Harry Potter example, the video calculates the probability of selecting a wand made of holly or unicorn hair.
• In the "Fire-Water-Sponge" example, the video analyzes the subset of outcomes where the friend either wins or there is a tie.
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# If the numerator of a fraction is decreased 25 percent and the denomi
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If the numerator of a fraction is decreased 25 percent and the denomi [#permalink]
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18 Feb 2018, 23:10
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If the numerator of a fraction is decreased 25 percent and the denominator of that fraction is increased 25 percent, then the difference between the resulting and the original fractions represents what percentage decrease?
(A) 40%
(B) 45%
(C) 50%
(D) 60%
(E) 75%
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If the numerator of a fraction is decreased 25 percent and the denomi [#permalink]
### Show Tags
18 Feb 2018, 23:14
Bunuel wrote:
If the numerator of a fraction is decreased 25 percent and the denominator of that fraction is increased 25 percent, then the difference between the resulting and the original fractions represents what percentage decrease?
(A) 40%
(B) 45%
(C) 50%
(D) 60%
(E) 75%
Let the original fraction will be $$\frac{x}{y}$$
If the numerator is decreased 25% and the denominator is increased 25%,
the resultant fraction becomes $$\frac{0.75x}{1.25y} = \frac{3}{5}*\frac{x}{y}$$(60% of its original value)
Therefore, the percentage decrease should be 40%(Option A)
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If the numerator of a fraction is decreased 25 percent and the denomi [#permalink]
### Show Tags
20 Feb 2018, 01:34
Bunuel wrote:
If the numerator of a fraction is decreased 25 percent and the denominator of that fraction is increased 25 percent, then the difference between the resulting and the original fractions represents what percentage decrease?
(A) 40%
(B) 45%
(C) 50%
(D) 60%
(E) 75%
Assigning numbers here works well.
Original fraction
Use numbers that yield easy results when multiplied by $$\frac{3}{4}$$ (.75) and $$\frac{5}{4}$$ (1.25)
Let the original fraction = $$\frac{100}{100}$$ (which = 1)
New fraction
The numerator is decreased 25 percent. The denominator increases by 25 percent. New fraction:
$$\frac{75}{125}$$ (which =$$\frac{3}{5}$$)
Percent decrease?
Percent decrease generally:
$$\frac{|New-Old|}{Old}*100$$
Percent decrease here $$\frac{|\frac{3}{5}-1|}{1}*100)=(|-\frac{2}{5}|*100)=(0.4*100)=$$
40 percent decrease
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Re: If the numerator of a fraction is decreased 25 percent and the denomi [#permalink]
### Show Tags
22 Feb 2018, 09:17
Bunuel wrote:
If the numerator of a fraction is decreased 25 percent and the denominator of that fraction is increased 25 percent, then the difference between the resulting and the original fractions represents what percentage decrease?
(A) 40%
(B) 45%
(C) 50%
(D) 60%
(E) 75%
We can let the original fraction = n/d and the new fraction = 0.75n/1.25d = (3n/4)/(5d/4) = 12n/20d = 3n/5d
Now we can determine the percent decrease:
[(3n/5d) - (5n/5d)]/(n/d) x 100
-2n/5d x d/n x 100 = -2/5 x 100 = -40, a 40% decrease
Alternate Solution:
Let’s let the original fraction be 1/1. When the numerator is decreased by 25% and the denominator is increased by 25%, the new fraction is 0.75/1.25 = 75/125 = 3/5.
We use the percent change formula (new - old)/old x 100%:
(3/5 -1)/1 x 100% = -2/5 x 100% = -40 percent, which is a 40% decrease.
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Re: If the numerator of a fraction is decreased 25 percent and the denomi [#permalink] 22 Feb 2018, 09:17
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# SHIMMERING HUBCAP – Tints, Symmetry, Contrast
Students use a compass to design a hubcap divided into 7 equal parts, and colour it with neon paint and glitter glue.
120 Minutes
Language Arts
Mathematics
Visual Arts
#### Vocabulary
contrast hubcap symmetry tint
#### Materials
Crayola Neon Paint Crayola Paint Brushes Crayola Construction Paper - Black Crayola Sketchbooks Crayola Scissors Crayola Glitter Glue Pencils Rulers Compasses Erasers Water Containers Plastic Container Lids for Palettes Paper Towels
## Steps
### Step One
1. Make a plan drawing in your sketchbook.
2. Draw a large circle.
3. Follow the instructions to divide it into 7 equal parts. (Downloads - DividingCircleSevenParts.pdf)
4. Design a hubcap that represents you in some way.
5. Shade in the negative spaces of your design.
6. These will remain black when the design is painted.
### Step Two
1. Draw a large circle on the black construction paper.
2. Divide it into 7 equal parts.
3. Cut out a template for the main shape of your design.
4. Trace around the template in each section of the circle.
### Step Three
1. Use several layers of neon paint to complete the design.
2. Paint white in the middle of the shapes.
3. Paint a colour around the edges of the shapes.
4. Blend the colours where they meet to go from light to dark.
5. Repeat this process several times as each layer dries.
### Step Four
1. Trim the paper to make equal borders around the circle.
2. Add glitter glue to parts of the design to add highlights and sparkle.
## Learning Goals
Students will be able to:
• follow instructions to divide a circle into 7 equal parts;
• create a personal hubcap design;
• use tints to create the illusion of depth;
• demonstrate technical accomplishment and creativity;
• support their ideas with evidence found in the artworks.
## Extensions
Have students:
• explore ways to use a compass to create circular geometric designs based on different divisions, for example, threefold - tenfold symmetries;
• use coloured watercolour paints to colour their designs;
• teach another student how to divide a circle into a specific number of equal parts.
## Prepare
1. Prior to this lesson provide time for students to practice using a compass.
2. Review or teach the concept of symmetry and circular symmetry.
3. Download and display the Colour, Contrast and Shape posters available on this website.
- review or teach the element of shape – positive, negative
- review or teach the principle of contrast – strong differences
- review or teach the element of tints – colour with white added to it
4. Download images of hubcaps from the Internet, for example,
Hubcap1
Hubcap2
Hubcap3
5. Download and copy the How to Divide a Circle into 7 Equal Parts worksheet, enough for each student to have one. (Downloads – DividingCircleSevenParts.pdf)
## Introduction
1. View and discuss images of hubcaps. Invite students to share what they know about hubcaps, for example,
- hubcaps cover the rim of a wheel
- they protect the central part of the car’s wheel
- they can be made of plastic or metal
- some race cars have non-rotating hubcaps that stay in the same position even when the car is moving
2. View examples of hubcaps and ask students to decide which one they like the best.
3. Discuss why they like certain designs more than others.
- angles of shapes
- negative shapes
- contrasting shapes
- colour
- feeling design evokes
4. Brainstorm ways to make a hubcap design that reflects something about the designer, e.g.,
- colour
- shapes and angles
- positive and negative shapes
- feeling of speed, or safety
5. Introduce the challenge.
## Activities
### The Challenge
1. Follow instructions to divide a circle into 7 equal parts.
2. Create a personal hubcap design.
3. Use tints to create the illusion of depth.
4. Demonstrate technical accomplishment and creativity.
5. Support your ideas with evidence found in the artworks.
### The Process
1. Make sure everyone understands the challenge.
2. Establish success criteria with your students, for example,
I know I am successful when I have:
- used a compass to draw a circle
- divided the circle into 7 equal parts
- created a personal hubcap design
- used tints to create the illusion of depth
- kept the paper in good condition
3. Guide students through the steps outlined in this lesson plan.
4. Observe students as they work.
5. Provide individual assistance and encouragement.
## Sharing
1. Place students into groups of about 6.
2. Ask them to view the hubcap designs and to share thoughts about the works.
3. During the discussion include references to:
Design - How does the design make you feel? Why?
- Colour - What effect do the colours have on the overall design?
- Negative Shapes - What effect do the negative shapes have on the overall design?
- Technical Accomplishment - Where can you see that the artist has paid attention to detail?
- Message - What does the design tell you about the person who created it?
4. Ask volunteers to share some ideas with the whole class.
## Assessment
1. Observe students as they work – thoughtful focus, discriminating, seeking more information, elaborating, experimenting.
2. Observe students as they discuss their artworks – speaks with a clear voice, looks at audience while speaking, holds painting to the side, provides accurate information, answers questions from the audience effectively.
3. Observe students as they listen – looks at presenter, asks effective questions, supports ideas with evidence found in the artwork.
4. Use a checklist to track progress. (Downloads - Hubcap_tracking.pdf)
5. Have students use the self-assessment form to evaluate their work. (Downloads - ShimmeringHubcap_self-assessment.pdf)
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## Trains
A 108m long train starts from rest (at t=0) and accelerates uniformly. At the same time (at t=0), a car moving with a constant speed in the same direction reaches the back end of the train. At t=12s the car reaches the front of the train. However, the train continues to speed up and pulls ahead of the car. At t=32s, the car is left behind the train. Determine, a. the car's speed b. the train's acceleration Please provide detailed answers instead of 4 equations and 4 unknowns
• At t=12: 12v - 108 = 0.5*a*12^2
12v - 108 = 72a
At t=32: 32v = 0.5*a*32^2
32v = 512a
v = 16a
Substitute this into the first equation
12v - 108 = 4.5v
7.5v = 108
v = 14.4 m/s
14.4 = 16a
a = 0.9 m/s^2
• Anonymous commented
can you provide more detail
• Anonymous commented
can someone explain this in more detail?
• Anonymous commented
how does 12v - 108 = 72a become 12v - 108 = 4.5v when v = 16a?
Get homework help
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You are on page 1of 1
Monte Vista Elementary
1 st grade – Ms. Pryor
http://apryorknowledge.weebly.com/
4-08-2019
LITERACY focus standard MATH focus standard
CCSS: RI 1.2 Students will identify the CCSS: Number and Operations in Base Ten
main topic and retell key details of a text. 1.NBT.4 Add within 100, including adding a two-digit
Students identify the author’s most important point number and a one-digit number, and adding a two-digit
in an informational text. They will also choose key number and a multiple of 10, using concrete models or
details that support their choice. drawings and strategies based on place value, properties of
operations, and/or the relationship between addition and
This week’s focus text is: subtraction; relate the strategy to a written method and
What if You Had Animal Teeth? explain the reasoning used. Understand that in adding two-
by Sandra Markle digit numbers, one adds tens and tens, ones and ones; and
sometimes it is necessary to compose a ten.
Phonics Skills (week 5.3)
vowel sound “oo” as in book
-ed, -ing, -s
Practice Makes Progress!
Read for 15 minutes each night and fill-in
reading chart in yellow folder (with date).
Practice reading and saying high frequency
words, and weekly phonics focus words.
Important Information:
Complete homework pages. Kona Ice fundraiser this Friday. Please send in
Istation Home (ISIP Reading, on-line books, money only Monday-Thursday.
and ISIP Math tabs) Las Cruces Space Festival 4/7 – 4/13.
Spring Holiday days off from school are Friday,
4/19 and Monday, 4/22.
Habit of the Week: Our class raised \$744 for Jog-a-thon! Wow!!
MVES Raffle begins today. See information
Habit 2:
sheet. Each student gets 5 tickets to sell.
Begin With the The school is purchasing butterfly eggs for us.
End in Mind I am asking families for \$1 - \$2 each so we can
also have ladybug and praying mantis eggs. We
Have a Plan will be starting the insect and life cycle unit in a
few weeks. Thank you for your help.
Words to Practice:
among another
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# Leap year calculator
I'm creating a program to determine whether or not a given year is a leap year. I just wanted to get anyone's feedback on it. I am aware of the calendar module in Python. I just wanted to try it from scratch.
year=int(raw_input("Input year:"))
if (year%4==0 and year%100==0 and year%400==0)or (year%4==0 and year%100 !=0 and year%400==0)or(year%4==0 and year%100 !=0 and year%400!=0):
print str(year)+" is a leap year"
else:
print str(year)+" is NOT a leap year"
• Try pylint on your code Nov 26, 2014 at 10:40
## Logic
Consider the if condition, which I have reproduced below with different alignment for ease of viewing…
if (year%4==0 and year%100==0 and year%400==0)or \
(year%4==0 and year%100!=0 and year%400==0)or \
(year%4==0 and year%100!=0 and year%400!=0):
…
It's clear that the first two lines can be collapsed due to redundancy. As for the third line, any number that is not divisible by 100 is automatically not divisible by 400 as well.
if (year%4==0 and year%400==0)or \
(year%4==0 and year%100!=0):
…
That can be further simplified to:
if (year%4==0 and (year%100!=0 or year%400==0)):
…
## Formatting
Please take care to follow Python formatting conventions, especially the consistent indentation by four spaces, especially since whitespace is significant in Python.
year = int(raw_input("Input year: "))
if year % 4 == 0 and (year % 100 != 0 or year % 400 == 0):
print str(year) + " is a leap year"
else:
print str(year) + " is NOT a leap year"
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Home > Standard Error > What Does The Standard Error Of The Mean Represent
# What Does The Standard Error Of The Mean Represent
## Contents
So 1 over the square root of 5. Standard error: meaning and interpretation. We aim to elucidate the distinctions between SD and SEM and to provide proper usage guidelines for both, which summarize data and describe statistical results.Keywords: Standard deviation, Standard error of the Related articles Related pages: Calculate Standard Deviation Standard Deviation . http://3cq.org/standard-error/what-does-the-standard-error-of-measurement-represent.php
However, if the sample size is very large, for example, sample sizes greater than 1,000, then virtually any statistical result calculated on that sample will be statistically significant. Footer bottom Explorable.com - Copyright © 2008-2016. If the interval calculated above includes the value, “0”, then it is likely that the population mean is zero or near zero. By using this site, you agree to the Terms of Use and Privacy Policy.
## Standard Error Formula
Therefore, it is essential for them to be able to determine the probability that their sample measures are a reliable representation of the full population, so that they can make predictions An experiment must be conducted on the entire population to acquire a more accurate confirmation of a hypothesis, but it is essentially impossible to survey an entire population. The standard error can be computed from a knowledge of sample attributes - sample size and sample statistics.
Let's say the mean here is 5. And it turns out, there is. Let's see if it conforms to our formula. Difference Between Standard Error And Standard Deviation And this time, let's say that n is equal to 20.
Nagele P. Standard Error Vs Standard Deviation The determination of the representativeness of a particular sample is based on the theoretical sampling distribution the behavior of which is described by the central limit theorem. The effect size provides the answer to that question. recommended you read As an example of the use of the relative standard error, consider two surveys of household income that both result in a sample mean of \$50,000.
I don't necessarily believe you. Standard Error Symbol Here are the key differences: • The SD quantifies scatter — how much the values vary from one another.• The SEM quantifies how precisely you know the true mean of the The two concepts would appear to be very similar. We keep doing that.
## Standard Error Vs Standard Deviation
When the samples of the same sample size are repeatedly and randomly taken from the same population, they are different each other because of sampling variation as well as sample means This Site The computations derived from the r and the standard error of the estimate can be used to determine how precise an estimate of the population correlation is the sample correlation statistic. Standard Error Formula more... Standard Error Regression Ecology 76(2): 628 – 639. ^ Klein, RJ. "Healthy People 2010 criteria for data suppression" (PDF).
Because the 5,534 women are the entire population, 23.44 years is the population mean, μ {\displaystyle \mu } , and 3.56 years is the population standard deviation, σ {\displaystyle \sigma } http://3cq.org/standard-error/what-is-the-difference-between-standard-deviation-and-standard-error.php The standard error is an important indicator of how precise an estimate of the population parameter the sample statistic is. In regression analysis, the term "standard error" is also used in the phrase standard error of the regression to mean the ordinary least squares estimate of the standard deviation of the The variability of a statistic is measured by its standard deviation. Standard Error Of The Mean Definition
However, many statistical results obtained from a computer statistical package (such as SAS, STATA, or SPSS) do not automatically provide an effect size statistic. Korean J Anesthesiol. 2014;67:20–25. [PMC free article] [PubMed]13. Note: the standard error and the standard deviation of small samples tend to systematically underestimate the population standard error and deviations: the standard error of the mean is a biased estimator navigate here What's your standard deviation going to be?
But it's going to be more normal. Standard Error Of Proportion If σ is not known, the standard error is estimated using the formula s x ¯ = s n {\displaystyle {\text{s}}_{\bar {x}}\ ={\frac {s}{\sqrt {n}}}} where s is the sample That in turn should lead the researcher to question whether the bedsores were developed as a function of some other condition rather than as a function of having heart surgery that
## Journal of the Royal Statistical Society.
So I have this on my other screen so I can remember those numbers. It is particularly important to use the standard error to estimate an interval about the population parameter when an effect size statistic is not available. Low S.E. Standard Error Excel Sokal and Rohlf (1981)[7] give an equation of the correction factor for small samples ofn<20.
As shown above, we found that some of the studies have inappropriately used the SD, SEM and confidence intervals in reporting their statistical results. It is the variance -- the SD squared -- that doesn't change predictably, but the change in SD is trivial and much much smaller than the change in the SEM.)Note that Get All Content From Explorable All Courses From Explorable Get All Courses Ready To Be Printed Get Printable Format Use It Anywhere While Travelling Get Offline Access For Laptops and his comment is here There was also a study that used the mean in the text but showed an interquartile range in the graphs.
Because you use the word "mean" and "sample" over and over again. It is useful to compare the standard error of the mean for the age of the runners versus the age at first marriage, as in the graph. Here, we would take 9.3. Use of the standard error statistic presupposes the user is familiar with the central limit theorem and the assumptions of the data set with which the researcher is working.
However, while the standard deviation provides information on the dispersion of sample values, the standard error provides information on the dispersion of values in the sampling distribution associated with the population This interval is a crude estimate of the confidence interval within which the population mean is likely to fall. In this way, the standard error of a statistic is related to the significance level of the finding. So this is the variance of our original distribution.
However, different samples drawn from that same population would in general have different values of the sample mean, so there is a distribution of sampled means (with its own mean and And if we did it with an even larger sample size-- let me do that in a different color. Lee H, Shon YJ, Kim H, Paik H, Park HP. The standard error estimated using the sample standard deviation is 2.56.
And it actually turns out it's about as simple as possible. View Mobile Version Warning: The NCBI web site requires JavaScript to function. The standard error of the mean (SEM) (i.e., of using the sample mean as a method of estimating the population mean) is the standard deviation of those sample means over all The variance or SD includes the differences of the observed values from the mean (Fig. 1); thus, these values represent the variation of the data [1,2,3].
Mahler DL. BMJ. 2005;331:903. [PMC free article] [PubMed]4. The standard error of a statistic is therefore the standard deviation of the sampling distribution for that statistic (3) How, one might ask, does the standard error differ from the standard
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# RS Aggarwal Class 8 Solutions Chapter 7 Factorisation Ex 7C
## RS Aggarwal Class 8 Solutions Chapter 7 Factorisation Ex 7C
These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 7 Factorisation Ex 7C.
Other Exercises
Question 1.
Solution:
x2 + 8x + 16
= (x)2 + 2 × x × 4 + (4)2
= (x + 4)2
Question 2.
Solution:
x2 + 14x + 49
= (x)2 + 2 × x × 7 + (7)2
= (x + 7)2 Ans.
Question 3.
Solution:
1 + 2x + x2
= (1)2 + 2 × 1 × x + (x)2
= (1 + x)2 Ans.
Question 4.
Solution:
9 + 6z + z2
= (3)2 + 2 x 3 x z + (z)2
= (3 + z)2 Ans.
Question 5.
Solution:
x2 + 6ax + 9a2
= (x)2 + 2 × x × 3a + (3a)2
= (x + 3a)2 Ans.
Question 6.
Solution:
4y2 + 20y + 25
= (2y)2 + 2 x 2y x 5 + (5)2
= (2y2 + 5)2
{ ∵ a2 + 2ab + b2 = (a + b)2}
Question 7.
Solution:
36a2 + 36a + 9
= 9 [4a2 + 4a + 1]
= 9 [(2a)2 + 2 x 2a x 1 + (1)2]
= 9 [2a + 1]2
Question 8.
Solution:
9m2 + 24m + 16
= (3m)2 + 2 x 3m x 4 + (4)2
= (3m + 4)2
{ ∵ a2 + 2ab + b2 = (a + b)2}
Question 9.
Solution:
z2 + z + $$\\ \frac { 1 }{ 4 }$$
= (z)2 + 2 x z x $$\\ \frac { 1 }{ 2 }$$ + $${ \left( \frac { 1 }{ 2 } \right) }^{ 2 }$$
= $${ \left( z+\frac { 1 }{ 2 } \right) }^{ 2 }$$
Question 10.
Solution:
49a2 + 84ab + 36b2
= (7a)2 + 2 x 7a x 6b + (6b)2
{ ∵ a2 + 2ab + b2 = (a + b)2}
= (7a + 6b)2
Question 11.
Solution:
p2 – 10p + 25
= (p)2 – 2 x p x 5 + (5)2
= (p – 5)2
{ ∵ a2 – 2ab + b2 = (a – b)2}
Question 12.
Solution:
121a2 – 88ab + 16b2
= (11a)2 – 2 x 11a x 4b + 4(b)2
= (11a – 4b)2
Question 13.
Solution:
1 – 6x + 9x2
= (1)2 – 2 x 1 x 3x + (3x)2
= (1 – 3x)2
{ ∵ a2 – 2ab + b2 = (a – b)2}
Question 14.
Solution:
9y2 – 12y + 4
= (3y)2 – 2 x 3y x 2 + (2)2
{ ∵ a2 – 2ab + b2 = (a – b)2}
= (3y – 2)2
Question 15.
Solution:
16x2 – 24x + 9
= (4x)2 – 2 x 4x x 3 + (3)2
= (4x – 3)2 Ans.
Question 16.
Solution:
m2 – 4mn + 4n2
= (m)2 -2 x m x 2n + (2n)2
= (m – 2n)2 Ans.
Question 17.
Solution:
a2b2 – 6abc + 9c2
= (ab)2 – 2 x ab x 3c + (3c)2
= (ab – 3c)2 Ans.
Question 18.
Solution:
m4 + 2m2n2 + n4
= (m2)2 + 2m2n2 + (n2)2
= (m2 + n2)2
{ ∵ a2 + 2ab + b2 = (a + b)2}
Question 19.
Solution:
(l + m)2 – 4lm
= l2 + m2 + 2lm – 4lm
= l2 + m2 – 2lm
= l2 – 2lm + m2
= (l – m)2
{ ∵ a2 – 2ab + b2 = (a – b)}
Hope given RS Aggarwal Solutions Class 8 Chapter 7 Factorisation Ex 7C are helpful to complete your math homework.
If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.
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Cody
# Problem 18. Bullseye Matrix
Solution 1797506
Submitted on 25 Apr 2019 by Parker Sheaffer
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
n = 5; a = [3 3 3 3 3; 3 2 2 2 3; 3 2 1 2 3; 3 2 2 2 3; 3 3 3 3 3]; assert(isequal(bullseye(n),a));
a = 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 a = 3 3 3 3 3 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 a = 3 3 3 3 3 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 3 3 3 3 3 a = 3 3 3 3 3 3 0 0 0 0 3 0 0 0 0 3 0 0 0 0 3 3 3 3 3 a = 3 3 3 3 3 3 0 0 0 3 3 0 0 0 3 3 0 0 0 3 3 3 3 3 3 a = 3 3 3 3 3 3 0 0 0 3 3 0 0 0 3 3 0 0 0 3 3 3 3 3 3 a = 3 3 3 3 3 3 0 0 0 3 3 0 1 0 3 3 0 0 0 3 3 3 3 3 3 a = 3 3 3 3 3 3 0 0 0 3 3 0 1 0 3 3 2 2 2 3 3 3 3 3 3 a = 3 3 3 3 3 3 2 2 2 3 3 0 1 0 3 3 2 2 2 3 3 3 3 3 3 a = 3 3 3 3 3 3 2 2 2 3 3 2 1 0 3 3 2 2 2 3 3 3 3 3 3 a = 3 3 3 3 3 3 2 2 2 3 3 2 1 2 3 3 2 2 2 3 3 3 3 3 3 a = 3 3 3 3 3 3 2 2 2 3 3 2 1 2 3 3 2 2 2 3 3 3 3 3 3
2 Pass
n = 7; a = [4 4 4 4 4 4 4; 4 3 3 3 3 3 4; 4 3 2 2 2 3 4; 4 3 2 1 2 3 4; 4 3 2 2 2 3 4; 4 3 3 3 3 3 4; 4 4 4 4 4 4 4]; assert(isequal(bullseye(n),a))
a = 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 a = 4 4 4 4 4 4 4 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 a = 4 4 4 4 4 4 4 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 4 4 4 4 4 4 4 a = 4 4 4 4 4 4 4 4 0 0 0 0 0 0 4 0 0 0 0 0 0 4 0 0 0 0 0 0 4 0 0 0 0 0 0 4 0 0 0 0 0 0 4 4 4 4 4 4 4 a = 4 4 4 4 4 4 4 4 0 0 0 0 0 4 4 0 0 0 0 0 4 4 0 0 0 0 0 4 4 0 0 0 0 0 4 4 0 0 0 0 0 4 4 4 4 4 4 4 4 a = 4 4 4 4 4 4 4 4 0 0 0 0 0 4 4 0 0 0 0 0 4 4 0 0 0 0 0 4 4 0 0 0 0 0 4 4 0 0 0 0 0 4 4 4 4 4 4 4 4 a = 4 4 4 4 4 4 4 4 0 0 0 0 0 4 4 0 0 0 0 0 4 4 0 0 1 0 0 4 4 0 0 0 0 0 4 4 0 0 0 0 0 4 4 4 4 4 4 4 4 a = 4 4 4 4 4 4 4 4 0 0 0 0 0 4 4 0 0 0 0 0 4 4 0 0 1 0 0 4 4 0 2 2 2 0 4 4 0 0 0 0 0 4 4 4 4 4 4 4 4 a = 4 4 4 4 4 4 4 4 0 0 0 0 0 4 4 0 2 2 2 0 4 4 0 0 1 0 0 4 4 0 2 2 2 0 4 4 0 0 0 0 0 4 4 4 4 4 4 4 4 a = 4 4 4 4 4 4 4 4 0 0 0 0 0 4 4 0 2 2 2 0 4 4 0 2 1 0 0 4 4 0 2 2 2 0 4 4 0 0 0 0 0 4 4 4 4 4 4 4 4 a = 4 4 4 4 4 4 4 4 0 0 0 0 0 4 4 0 2 2 2 0 4 4 0 2 1 2 0 4 4 0 2 2 2 0 4 4 0 0 0 0 0 4 4 4 4 4 4 4 4 a = 4 4 4 4 4 4 4 4 3 3 3 3 3 4 4 3 2 2 2 3 4 4 3 2 1 2 3 4 4 3 2 2 2 3 4 4 3 3 3 3 3 4 4 4 4 4 4 4 4
| 2,098
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| 3.1875
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CC-MAIN-2020-16
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latest
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en
| 0.518001
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https://www.theguardian.com/world/2005/nov/03/gender.golf
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| 932,643,981
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# Gavyn Davies does the maths
Are men really that much better at golf?
Male golfers, to their immense discredit, have worked themselves into a lather about the recent decision to allow women to enter the British Open Championship from next year onwards. Women are already permitted to enter professional male tournaments in the US, but so far this has only been attempted on a handful of occasions. So we do not yet know whether the top women, such as Annika Sorenstam and the 16-year-old "phenom" Michelle Wie, will actually be able to hold their own against the men.
One way of finding out is to allow women to enter hundreds of male tournaments, and see how they get on. A much quicker way is to deduce the answer by examining some statistics. These show that the average male tournament pro is much better at every aspect of the game than his female counterpart. For example, the median male drives the ball 288yds, compared with 246yds for the median woman. This makes an enormous difference, as does the fact that the male takes 1.75 putts per green, while the woman takes 1.84 putts - enough to make a difference of more than six shots in a 72-hole tournament. If an average woman pro entered male events, she would never make the cut and would win no money.
However, Sorenstam and Wie are not just average women. They are as dominant in their own game as Tiger Woods has been in his. If Annika were to play on the men's tour, she would do rather well. Admittedly, her distance off the tee is only 264yds, which would place her 199th out of the 200 men on the US PGA tour. But distance is not everything. Annika hits the fairway on 73% of the holes she plays, which would place her 5th on the men's tour list for accuracy. And, crucially, she takes only 1.75 putts per green, which would place her 20th among the men.
These statistics would earn Annika a lot of money if she played against the men - which, presumably, is why some of them are running scared. Stephen Shmanske at California State University has published an econometric analysis, explaining male and female earnings on the tour by skill differences such as driving and putting. If you substitute Annika's stats into these equations, you find that she would earn around \$34,000 (£19,200) per tournament, which would put her a little above halfway on the men's overall money list in America.
Does this mean that she might ever win a Major in the men's game? It is improbable, but not impossible. If she were ranked, say, 80th on the US money list, her world ranking would be around 106th. Few men ranked as low as that have ever won a major, but it does happen occasionally. You may have forgotten Ben Curtis, Todd Hamilton and Shaun Micheel, but all of them have won Majors from virtually unranked positions in the past three years.
The correct title of the British Open is simply the Open Championship, precisely because it is open to all comers, and is therefore the world championship in all but name. What on earth is the use of a world championship that excludes just over half of the human race from entering?
Topics
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CC-MAIN-2020-10
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https://en.wikipedia.org/wiki/Quadratic_classifier
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In statistics, a quadratic classifier is a statistical classifier that uses a quadratic decision surface to separate measurements of two or more classes of objects or events. It is a more general version of the linear classifier.
## The classification problem
Statistical classification considers a set of vectors of observations x of an object or event, each of which has a known type y. This set is referred to as the training set. The problem is then to determine, for a given new observation vector, what the best class should be. For a quadratic classifier, the correct solution is assumed to be quadratic in the measurements, so y will be decided based on
${\displaystyle \mathbf {x^{T}Ax} +\mathbf {b^{T}x} +c}$
In the special case where each observation consists of two measurements, this means that the surfaces separating the classes will be conic sections (i.e., either a line, a circle or ellipse, a parabola or a hyperbola). In this sense, we can state that a quadratic model is a generalization of the linear model, and its use is justified by the desire to extend the classifier's ability to represent more complex separating surfaces.
Quadratic discriminant analysis (QDA) is closely related to linear discriminant analysis (LDA), where it is assumed that the measurements from each class are normally distributed.[1] Unlike LDA however, in QDA there is no assumption that the covariance of each of the classes is identical.[2] When the normality assumption is true, the best possible test for the hypothesis that a given measurement is from a given class is the likelihood ratio test. Suppose there are only two groups, with means ${\displaystyle \mu _{0},\mu _{1}}$ and covariance matrices ${\displaystyle \Sigma _{0},\Sigma _{1}}$ corresponding to ${\displaystyle y=0}$ and ${\displaystyle y=1}$ respectively. Then the likelihood ratio is given by
${\displaystyle {\text{Likelihood ratio}}={\frac {{\sqrt {2\pi |\Sigma _{1}|}}^{-1}\exp \left(-{\frac {1}{2}}(\mathbf {x} -{\boldsymbol {\mu }}_{1})^{T}\Sigma _{1}^{-1}(\mathbf {x} -{\boldsymbol {\mu }}_{1})\right)}{{\sqrt {2\pi |\Sigma _{0}|}}^{-1}\exp \left(-{\frac {1}{2}}(\mathbf {x} -{\boldsymbol {\mu }}_{0})^{T}\Sigma _{0}^{-1}(\mathbf {x} -{\boldsymbol {\mu }}_{0})\right)}}
for some threshold ${\displaystyle t}$. After some rearrangement, it can be shown that the resulting separating surface between the classes is a quadratic. The sample estimates of the mean vector and variance-covariance matrices will substitute the population quantities in this formula.
## Other
While QDA is the most commonly-used method for obtaining a classifier, other methods are also possible. One such method is to create a longer measurement vector from the old one by adding all pairwise products of individual measurements. For instance, the vector
${\displaystyle [x_{1},\;x_{2},\;x_{3}]}$
would become
${\displaystyle [x_{1},\;x_{2},\;x_{3},\;x_{1}^{2},\;x_{1}x_{2},\;x_{1}x_{3},\;x_{2}^{2},\;x_{2}x_{3},\;x_{3}^{2}].}$
Finding a quadratic classifier for the original measurements would then become the same as finding a linear classifier based on the expanded measurement vector. This observation has been used in extending neural network models;[3] the "circular" case, which corresponds to introducing only the sum of pure quadratic terms ${\displaystyle \;x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+\cdots \;}$ with no mixed products (${\displaystyle \;x_{1}x_{2},\;x_{1}x_{3},\;\ldots \;}$), has been proven to be the optimal compromise between extending the classifier's representation power and controlling the risk of overfitting (Vapnik-Chervonenkis dimension).[4]
For linear classifiers based only on dot products, these expanded measurements do not have to be actually computed, since the dot product in the higher-dimensional space is simply related to that in the original space. This is an example of the so-called kernel trick, which can be applied to linear discriminant analysis as well as the support vector machine.
## References
### Citations
1. ^ Tharwat, Alaa (2016). "Linear vs. quadratic discriminant analysis classifier: a tutorial". International Journal of Applied Pattern Recognition. 3 (2): 145. doi:10.1504/IJAPR.2016.079050. ISSN 2049-887X.
2. ^ "Linear & Quadratic Discriminant Analysis · UC Business Analytics R Programming Guide". uc-r.github.io. Retrieved 2020-03-29.
3. ^ Cover TM (1965). "Geometrical and Statistical Properties of Systems of Linear Inequalities with Applications in Pattern Recognition". IEEE Transactions on Electronic Computers. EC-14 (3): 326–334. doi:10.1109/pgec.1965.264137.
4. ^ Ridella S, Rovetta S, Zunino R (1997). "Circular backpropagation networks for classification". IEEE Transactions on Neural Networks. 8 (1): 84–97. doi:10.1109/72.554194. PMID 18255613. href IEEE: [1].
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https://profiles.doe.mass.edu/mcas/achievement_level.aspx?linkid=32&orgcode=07170010&orgtypecode=6&
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# Colrain Central
Districts Schools
Did You Know:
## Next Generation MCAS Tests 2022Percent of Students at Each Achievement Level for Colrain Central
Grade and Subject Meeting or Exceeding Expectations % Exceeding Expectations % Meeting Expectations % Partially Meeting Expectations % Not Meeting Expectations % No. of Students Included Part. Rate % Avg. Scaled Score Avg.SGP Included in Avg. SGP Ach.Pctl School State School State School State School State School State GRADE 03 - ENGLISH LANGUAGE ARTS 33 44 0 6 33 38 58 41 8 15 12 100 492 N/A N/A 34 GRADE 03 - MATHEMATICS 50 41 0 6 50 35 25 39 25 20 12 100 490 N/A N/A 40 GRADE 04 - ENGLISH LANGUAGE ARTS 38 4 34 46 16 7 N/A 5 GRADE 04 - MATHEMATICS 42 6 37 40 17 7 N/A 5 GRADE 05 - ENGLISH LANGUAGE ARTS 9 41 0 5 9 36 64 46 27 13 11 100 480 N/A 9 7 GRADE 05 - MATHEMATICS 18 36 0 4 18 32 64 48 18 16 11 93 482 N/A 9 20 GRADE 05 - SCIENCE AND TECH/ENG 9 43 0 7 9 36 64 40 27 18 11 93 479 N/A N/A 12 GRADE 06 - ENGLISH LANGUAGE ARTS 50 41 21 8 29 33 43 36 7 22 14 100 507 N/A 11 89 GRADE 06 - MATHEMATICS 57 42 0 5 57 37 36 43 7 15 14 100 505 N/A 11 81 GRADES 03 - 08 - ENGLISH LANGUAGE ARTS 30 41 7 6 23 35 52 42 18 17 44 100 491 55 25 37 GRADES 03 - 08 - MATHEMATICS 41 39 0 6 41 33 39 43 20 17 44 98 492 69 25 44 GRADES 05 & 08 - SCIENCE AND TECH/ENG 9 42 0 6 9 36 64 40 27 18 11 93 479 N/A N/A 12
Click on any Grade and Subject rows in the table to view Achievement Level in graph.
Note: School achievement percentiles (1-99) compare each group’s average scaled score to the average scaled scores of the same group from all public schools across the state. Only students enrolled in the same school since October 1 are included in calculations, and groups with fewer than 10 students do not receive percentiles.
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http://en.wikipedia.org/wiki/Euler's_laws
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# Euler's laws of motion
(Redirected from Euler's laws)
"Euler's first law" and "Euler's second law" redirect to here. For other uses, see Euler (disambiguation).
In classical mechanics, Euler's laws of motion are equations of motion which extend Newton's laws of motion for point particle to rigid body motion.[1] They were formulated by Leonhard Euler about 50 years after Isaac Newton formulated his laws.
## Overview
### Euler's first law
Euler's first law states that the linear momentum of a body, p (also denoted G) is equal to the product of the mass of the body m and the velocity of its center of mass vcm: [1][2][3]
$\mathbf p = m \mathbf v_{\rm cm}$.
Internal forces between the particles that make up a body do not contribute to changing the total momentum of the body.[4] The law is also stated as:[4]
$\mathbf F = m \mathbf a_{\rm cm}$.
where acm = dvcm/dt is the acceleration of the centre of mass and F = dp/dt is the total applied force on the body. This is just the time derivative of the previous equation (m is a constant).
### Euler's second law
Euler's second law states that the rate of change of angular momentum L (also denoted H) about a point that is fixed in an inertial reference frame or the mass center of the body, is equal to the sum of the external moments of force (torques) M (also denoted τ or Γ) about that point:[1][2][3]
$\mathbf M = {d\mathbf L \over dt}$.
Note that the above formula holds only if both M,L are computed with respect to a fixed inertial frame or a frame parallel to the inertial frame but fixed on the center of mass. For rigid bodies translating and rotating in only 2d, this can be expressed as:[5]
$\mathbf M = \mathbf r_{\rm cm} \times \mathbf a_{\rm cm} m + I \boldsymbol{\alpha}$,
where rcm is the position vector of the center of mass with respect to the point about which moments are summed, α is the angular acceleration of the body, and I is the moment of inertia. See also Euler's equations (rigid body dynamics).
## Explanation and derivation
The density of internal forces at every point in a deformable body are not necessarily equal, i.e. there is a distribution of stresses throughout the body. This variation of internal forces throughout the body is governed by Newton's second law of motion of conservation of linear momentum and angular momentum, which normally are applied to a mass particle but are extended in continuum mechanics to a body of continuously distributed mass. For continuous bodies these laws are called Euler’s laws of motion. If a body is represented as an assemblage of discrete particles, each governed by Newton’s laws of motion, then Euler’s equations can be derived from Newton’s laws. Euler’s equations can, however, be taken as axioms describing the laws of motion for extended bodies, independently of any particle structure.[6]
The total body force applied to a continuous body with mass m, mass density ρ, and volume V, is the volume integral integrated over the volume of the body:
$\mathbf F_B=\int_V\mathbf b\,dm = \int_V\mathbf b\rho\,dV$
where b is the force acting on the body per unit mass (dimensions of acceleration, misleadingly called the "body force"), and dm = ρdV is an infinitesimal mass element of the body.
Body forces and contact forces acting on the body lead to corresponding moments of force (torques) relative to a given point. Thus, the total applied torque M about the origin is given by
$\mathbf M= \mathbf M_B + \mathbf M_C$
where MB and MC respectively indicate the moments caused by the body and contact forces.
Thus, the sum of all applied forces and torques (with respect to the origin of the coordinate system) in the body can be given as the sum of a volume and surface integral:
$\mathbf F = \int_V \mathbf a\,dm = \int_V \mathbf a\rho\,dV = \int_S \mathbf{t} dS + \int_V \mathbf b\rho\,dV$
$\mathbf M = \int_S \mathbf r \times \mathbf t dS + \int_V \mathbf r \times \mathbf b\rho\,dV.$
where t = t(n) is called the surface traction, integrated over the surface of the body, in turn n denotes a unit vector normal and directed outwards to the surface S.
Let the coordinate system (x1, x2, x3) be an inertial frame of reference, r be the position vector of a point particle in the continuous body with respect to the origin of the coordinate system, and v = dr/dt be the velocity vector of that point.
Euler’s first axiom or law (law of balance of linear momentum or balance of forces) states that in an inertial frame the time rate of change of linear momentum p of an arbitrary portion of a continuous body is equal to the total applied force F acting on the considered portion, and it is expressed as
\begin{align} \frac{d\mathbf p}{dt} &= \mathbf F \\ \frac{d}{dt}\int_V \rho\mathbf v\,dV&=\int_S \mathbf t dS + \int_V \mathbf b\rho \,dV. \\ \end{align}
Euler’s second axiom or law (law of balance of angular momentum or balance of torques) states that in an inertial frame the time rate of change of angular momentum L of an arbitrary portion of a continuous body is equal to the total applied torque M acting on the considered portion, and it is expressed as
\begin{align} \frac{d\mathbf L}{dt} &= \mathbf M \\ \frac{d}{dt}\int_V \mathbf r\times\rho\mathbf v\,dV&=\int_S \mathbf r \times \mathbf t dS + \int_V \mathbf r \times \mathbf b\rho\,dV. \\\end{align}
The derivatives of p and L are material derivatives.
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https://www.daniweb.com/programming/software-development/threads/493598/can-anyone-explain-this-code-to-me
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I came across this code of firefly algorithm and need to use it in my major. If any of you could explain the working of the functions, it would be of great help to me. Thanks in advance.
``````#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<math.h>
#include<time.h>
#include<string.h>
#include<memory.h>
#defineDUMP 1
#defineMAX_FFA 1000
#defineMAX_D 1000
usingnamespacestd;
int D = 1000; // dimension of the problem
int n = 20; // number of fireflies
intMaxGeneration; // number of iterations
intNumEval; // number of evaluations
int Index[MAX_FFA]; // sort of fireflies according to fitness values
doubleffa[MAX_FFA][MAX_D]; // firefly agents
doubleffa_tmp[MAX_FFA][MAX_D]; // intermediate population
double f[MAX_FFA]; // fitness values
double I[MAX_FFA]; // light intensity
doublenbest[MAX_FFA]; // the best solution found so far
doublelb[MAX_D]; // upper bound
doubleub[MAX_D]; // lower bound
double alpha = 0.5; // alpha parameter
doublebetamin = 0.2; // beta parameter
doublegama = 1.0; // gamma parameter
doublefbest; // the best objective function
typedefdouble (*FunctionCallback)(double sol[MAX_D]);
/*benchmark functions */
doublecost(double sol[MAX_D]);
doublesphere(double sol[MAX_D]);
/*Write your own objective function */
FunctionCallback function = &cost;
// optionally recalculate the new alpha value
doublealpha_new(double alpha, intNGen)
{
double delta; // delta parameter
delta = 1.0-pow((pow(10.0, -4.0)/0.9), 1.0/(double) NGen);
return (1-delta)*alpha;
}
// initialize the firefly population
voidinit_ffa()
{
inti, j;
double r;
// initialize upper and lower bounds
for (i=0;i<D;i++)
{
lb[i] = 0.0;
ub[i] = 2.0;
}
for (i=0;i<n;i++)
{
for (j=0;j<D;j++)
{
r = ( (double)rand() / ((double)(RAND_MAX)+(double)(1)) );
ffa[i][j]=r*(ub[i]-lb[i])+lb[i];
}
f[i] = 1.0; // initialize attractiveness
I[i] = f[i];
}
}
// implementation of bubble sort
voidsort_ffa()
{
inti, j;
// initialization of indexes
for(i=0;i<n;i++)
Index[i] = i;
// Bubble sort
for(i=0;i<n-1;i++)
{
for(j=i+1;j<n;j++)
{
if(I[i] > I[j])
{
double z = I[i]; // exchange attractiveness
I[i] = I[j];
I[j] = z;
z = f[i]; // exchange fitness
f[i] = f[j];
f[j] = z;
int k = Index[i]; // exchange indexes
Index[i] = Index[j];
Index[j] = k;
}
}
}
}
// replace the old population according the new Index values
voidreplace_ffa()
{
inti, j;
// copy original population to temporary area
for(i=0;i<n;i++)
{
for(j=0;j<D;j++)
{
ffa_tmp[i][j] = ffa[i][j];
}
}
// generational selection in sense of EA
for(i=0;i<n;i++)
{
for(j=0;j<D;j++)
{
ffa[i][j] = ffa_tmp[Index[i]][j];
}
}
}
voidfindlimits(int k)
{
inti;
for(i=0;i<D;i++)
{
if(ffa[k][i] <lb[i])
ffa[k][i] = lb[i];
if(ffa[k][i] >ub[i])
ffa[k][i] = ub[i];
}
}
voidmove_ffa()
{
inti, j, k;
double scale;
double r, beta;
for(i=0;i<n;i++)
{
scale = abs(ub[i]-lb[i]);
for(j=0;j<n;j++)
{
r = 0.0;
for(k=0;k<D;k++)
{
r += (ffa[i][k]-ffa[j][k])*(ffa[i][k]-ffa[j][k]);
}
r = sqrt(r);
if(I[i] > I[j]) // brighter and more attractive
{
double beta0 = 1.0;
beta = (beta0-betamin)*exp(-gama*pow(r, 2.0))+betamin;
for(k=0;k<D;k++)
{
r = ( (double)rand() / ((double)(RAND_MAX)+(double)(1)) );
doubletmpf = alpha*(r-0.5)*scale;
ffa[i][k] = ffa[i][k]*(1.0-beta)+ffa_tmp[j][k]*beta+tmpf;
}
}
}
findlimits(i);
}
}
voiddump_ffa(int gen)
{
cout<<"Dump at gen= "<< gen <<" best= "<<fbest<<endl;
}
/* display syntax messages */
voidhelp()
{
cout<<"Syntax:"<<endl;
cout<<" Firefly [-h|-?] [-l] [-p] [-c] [-k] [-s] [-t]"<<endl;
cout<<" Parameters: -h|-? = command syntax"<<endl;
cout<<" -n = number of fireflies"<<endl;
cout<<" -d = problem dimension"<<endl;
cout<<" -g = number of generations"<<endl;
cout<<" -a = alpha parameter"<<endl;
cout<<" -b = beta0 parameter"<<endl;
cout<<" -c = gamma parameter"<<endl;
}
intmain(intargc, char* argv[])
{
inti;
int t = 1; // generation counter
// interactive parameters handling
for(inti=1;i<argc;i++)
{
if((strncmp(argv[i], "-h", 2) == 0) || (strncmp(argv[i], "-?", 2) == 0))
{
help();
return0;
}
elseif(strncmp(argv[i], "-n", 2) == 0) // number of fireflies
{
n = atoi(&argv[i][2]);
}
elseif(strncmp(argv[i], "-d", 2) == 0) // problem dimension
{
D = atoi(&argv[i][2]);
}
elseif(strncmp(argv[i], "-g", 2) == 0) // number of generations
{
MaxGeneration = atoi(&argv[i][2]);
}
elseif(strncmp(argv[i], "-a", 2) == 0) // alpha parameter
{
alpha = atof(&argv[i][2]);
}
elseif(strncmp(argv[i], "-b", 2) == 0) // beta parameter
{
betamin = atof(&argv[i][2]);
}
elseif(strncmp(argv[i], "-c", 2) == 0) // gamma parameter
{
gama = atof(&argv[i][2]);
}
else
{
cerr<<"Fatal error: invalid parameter: "<<argv[i] <<endl;
return -1;
}
}
// firefly algorithm optimization loop
// determine the starting point of random generator
srand(1);
// generating the initial locations of n fireflies
init_ffa();
#ifdef DUMP
dump_ffa(t);
#endif
while(t <= MaxGeneration)
{
// this line of reducing alpha is optional
alpha = alpha_new(alpha, MaxGeneration);
// evaluate new solutions
for(i=0;i<n;i++)
{
f[i] = function(ffa[i]); // obtain fitness of solution
I[i] = f[i]; // initialize attractiveness
}
// ranking fireflies by their light intensity
sort_ffa();
// replace old population
replace_ffa();
// find the current best
for(i=0;i<D;i++)
nbest[i] = ffa[0][i];
fbest = I[0];
// move all fireflies to the better locations
move_ffa();
#ifdef DUMP
dump_ffa(t);
#endif
t++;
}
cout<<"End of optimization: fbest = "<<fbest<<endl;
return0;
}
// FF test function
doublecost(double* sol)
{
double sum = 0.0;
for(inti=0;i<D;i++)
sum += (sol[i]-1)*(sol[i]-1);
return sum;
}
doublesphere(double* sol) {
int j;
double top = 0;
for (j = 0; j < D; j++) {
top = top + sol[j] * sol[j];
}
}
``````
Well, to begin with, this is a C++ program, not a C program, and a poorly written one at that. The header formats indicate that it probably dates back to the late 1990s, when the then-new C++ standard was still being formalized - either that, or the programmer didn't understand the standard correctly. Several critical values are hard-coded in rather than parameterized, making the implementation brittle. The choice of bubblesort for the sorting algorithm is questionable at best, though with the small range it probably isn't very performance-critical. Finally, the code is pretty badly mangled, with the return types running into the function names and so forth, which means that without many fixes it wouldn't compile cleanly.
While it could fairly easily be re-written in C, you would be a lot better off writing a new implementation of the algorithm yourself. You would waste more time trying to fix this than you would starting over from scratch.
Thanks for the early reply Schol-R-LEA. Also I wanted to know, seeing a source code how can you tell if it's C or C++ when they are so similar ? I might get down votes on this weird question but I really want to know. It would be of great help if you could find me as easily understandable code for the same.
Well, the first thing to look at is the headers, both the names and the format of them. A C program will have headers in the form of `foo.h`, whereas in (modern) C++, the standard library headers will be `foo` without the extension. Now, this isn't a foolproof method, as there's a lot of pre-standard C++ code floating around, but if you see a header without an extension, it is presumably C++.
Also, in C++, the C library headers should be prepended with a `c`, making (for example) `<stdio.h>` into `<cstdio>`. Again, a lot of older code doesn't match this standard, and even some current code isn't correct in this regard, but if it is like that, it is definitely C++ rather than C.
Next, the headers used may be different; C++ adds a large number of new libraries which use the C++ facilities that don't exist in C. The main ones to look for are `<iostream>`, `<fstream>`, `<string>`, `<vector>` and `<algorithm>`.
Fourth, C++ has namespaces, but C does not; so, if you see a reference to `using namespace foo;`, it will be C++. The same goes for the scoping operator, `::`, which is used to indicate membership in either a class or a namespace.
Finally, if it uses `cin` and `cout` for input and output, it is going to be C++.
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# Matrices problem
Printable View
• Aug 17th 2012, 09:59 AM
ineedhelplz
Matrices problem
Hi, I'm having trouble figuring this out:
A is an nxn matrix. It's given that A2=A, A=\=0, A=\=I (I is nxn)
prove or disprove: Ax=0 has only the trivial solution
I tried to disprove that with the determinant:
det(A2)=det(A)
2det(A)=det(A)
det(A)=0
Hence A is singular
BUT im not sure that this det(A2)=2det(A) is right, and I can't fathom why A=\=0, A=\=I (I is nxn) is given, so I suspect I'm mistaken.
help is very welcomed
• Aug 17th 2012, 12:47 PM
ModusPonens
Re: Matrices problem
Your solution is incorrect, because det(A^2)=det(A)^2. But this is the right track. Can you follow from here?
• Aug 17th 2012, 02:46 PM
Deveno
Re: Matrices problem
consider A =
[1 0]
[0 0].
• Aug 17th 2012, 03:01 PM
ineedhelplz
Re: Matrices problem
Thanks to the both of you!
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Mathbox for Thierry Arnoux < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > Mathboxes > acunirnmpt2f Structured version Visualization version GIF version
Theorem acunirnmpt2f 30414
Description: Axiom of choice for the union of the range of a mapping to function. (Contributed by Thierry Arnoux, 7-Nov-2019.)
Hypotheses
Ref Expression
acunirnmpt.0 (𝜑𝐴𝑉)
acunirnmpt.1 ((𝜑𝑗𝐴) → 𝐵 ≠ ∅)
aciunf1lem.a 𝑗𝐴
acunirnmpt2f.c 𝑗𝐶
acunirnmpt2f.d 𝑗𝐷
acunirnmpt2f.2 𝐶 = 𝑗𝐴 𝐵
acunirnmpt2f.3 (𝑗 = (𝑓𝑥) → 𝐵 = 𝐷)
acunirnmpt2f.4 ((𝜑𝑗𝐴) → 𝐵𝑊)
Assertion
Ref Expression
acunirnmpt2f (𝜑 → ∃𝑓(𝑓:𝐶𝐴 ∧ ∀𝑥𝐶 𝑥𝐷))
Distinct variable groups: 𝑥,𝑓,𝐴 𝐵,𝑓 𝐶,𝑓,𝑥 𝑓,𝑗,𝜑,𝑥
Allowed substitution hints: 𝐴(𝑗) 𝐵(𝑥,𝑗) 𝐶(𝑗) 𝐷(𝑥,𝑓,𝑗) 𝑉(𝑥,𝑓,𝑗) 𝑊(𝑥,𝑓,𝑗)
Proof of Theorem acunirnmpt2f
Dummy variables 𝑐 𝑦 𝑘 are mutually distinct and distinct from all other variables.
StepHypRef Expression
1 simplr 768 . . . . . 6 ((((𝜑𝑥𝐶) ∧ 𝑦 ∈ ran (𝑗𝐴𝐵)) ∧ 𝑥𝑦) → 𝑦 ∈ ran (𝑗𝐴𝐵))
2 vex 3472 . . . . . . 7 𝑦 ∈ V
3 eqid 2822 . . . . . . . 8 (𝑗𝐴𝐵) = (𝑗𝐴𝐵)
43elrnmpt 5805 . . . . . . 7 (𝑦 ∈ V → (𝑦 ∈ ran (𝑗𝐴𝐵) ↔ ∃𝑗𝐴 𝑦 = 𝐵))
52, 4ax-mp 5 . . . . . 6 (𝑦 ∈ ran (𝑗𝐴𝐵) ↔ ∃𝑗𝐴 𝑦 = 𝐵)
61, 5sylib 221 . . . . 5 ((((𝜑𝑥𝐶) ∧ 𝑦 ∈ ran (𝑗𝐴𝐵)) ∧ 𝑥𝑦) → ∃𝑗𝐴 𝑦 = 𝐵)
7 nfv 1915 . . . . . . . . 9 𝑗𝜑
8 acunirnmpt2f.c . . . . . . . . . 10 𝑗𝐶
98nfcri 2967 . . . . . . . . 9 𝑗 𝑥𝐶
107, 9nfan 1900 . . . . . . . 8 𝑗(𝜑𝑥𝐶)
11 nfcv 2979 . . . . . . . . 9 𝑗𝑦
12 nfmpt1 5140 . . . . . . . . . 10 𝑗(𝑗𝐴𝐵)
1312nfrn 5801 . . . . . . . . 9 𝑗ran (𝑗𝐴𝐵)
1411, 13nfel 2993 . . . . . . . 8 𝑗 𝑦 ∈ ran (𝑗𝐴𝐵)
1510, 14nfan 1900 . . . . . . 7 𝑗((𝜑𝑥𝐶) ∧ 𝑦 ∈ ran (𝑗𝐴𝐵))
16 nfv 1915 . . . . . . 7 𝑗 𝑥𝑦
1715, 16nfan 1900 . . . . . 6 𝑗(((𝜑𝑥𝐶) ∧ 𝑦 ∈ ran (𝑗𝐴𝐵)) ∧ 𝑥𝑦)
18 simpllr 775 . . . . . . . . 9 ((((((𝜑𝑥𝐶) ∧ 𝑦 ∈ ran (𝑗𝐴𝐵)) ∧ 𝑥𝑦) ∧ 𝑗𝐴) ∧ 𝑦 = 𝐵) → 𝑥𝑦)
19 simpr 488 . . . . . . . . 9 ((((((𝜑𝑥𝐶) ∧ 𝑦 ∈ ran (𝑗𝐴𝐵)) ∧ 𝑥𝑦) ∧ 𝑗𝐴) ∧ 𝑦 = 𝐵) → 𝑦 = 𝐵)
2018, 19eleqtrd 2916 . . . . . . . 8 ((((((𝜑𝑥𝐶) ∧ 𝑦 ∈ ran (𝑗𝐴𝐵)) ∧ 𝑥𝑦) ∧ 𝑗𝐴) ∧ 𝑦 = 𝐵) → 𝑥𝐵)
2120ex 416 . . . . . . 7 (((((𝜑𝑥𝐶) ∧ 𝑦 ∈ ran (𝑗𝐴𝐵)) ∧ 𝑥𝑦) ∧ 𝑗𝐴) → (𝑦 = 𝐵𝑥𝐵))
2221ex 416 . . . . . 6 ((((𝜑𝑥𝐶) ∧ 𝑦 ∈ ran (𝑗𝐴𝐵)) ∧ 𝑥𝑦) → (𝑗𝐴 → (𝑦 = 𝐵𝑥𝐵)))
2317, 22reximdai 3297 . . . . 5 ((((𝜑𝑥𝐶) ∧ 𝑦 ∈ ran (𝑗𝐴𝐵)) ∧ 𝑥𝑦) → (∃𝑗𝐴 𝑦 = 𝐵 → ∃𝑗𝐴 𝑥𝐵))
246, 23mpd 15 . . . 4 ((((𝜑𝑥𝐶) ∧ 𝑦 ∈ ran (𝑗𝐴𝐵)) ∧ 𝑥𝑦) → ∃𝑗𝐴 𝑥𝐵)
25 acunirnmpt2f.2 . . . . . . . 8 𝐶 = 𝑗𝐴 𝐵
26 acunirnmpt2f.4 . . . . . . . . . 10 ((𝜑𝑗𝐴) → 𝐵𝑊)
2726ralrimiva 3174 . . . . . . . . 9 (𝜑 → ∀𝑗𝐴 𝐵𝑊)
28 dfiun3g 5813 . . . . . . . . 9 (∀𝑗𝐴 𝐵𝑊 𝑗𝐴 𝐵 = ran (𝑗𝐴𝐵))
2927, 28syl 17 . . . . . . . 8 (𝜑 𝑗𝐴 𝐵 = ran (𝑗𝐴𝐵))
3025, 29syl5eq 2869 . . . . . . 7 (𝜑𝐶 = ran (𝑗𝐴𝐵))
3130eleq2d 2899 . . . . . 6 (𝜑 → (𝑥𝐶𝑥 ran (𝑗𝐴𝐵)))
3231biimpa 480 . . . . 5 ((𝜑𝑥𝐶) → 𝑥 ran (𝑗𝐴𝐵))
33 eluni2 4817 . . . . 5 (𝑥 ran (𝑗𝐴𝐵) ↔ ∃𝑦 ∈ ran (𝑗𝐴𝐵)𝑥𝑦)
3432, 33sylib 221 . . . 4 ((𝜑𝑥𝐶) → ∃𝑦 ∈ ran (𝑗𝐴𝐵)𝑥𝑦)
3524, 34r19.29a 3275 . . 3 ((𝜑𝑥𝐶) → ∃𝑗𝐴 𝑥𝐵)
3635ralrimiva 3174 . 2 (𝜑 → ∀𝑥𝐶𝑗𝐴 𝑥𝐵)
37 acunirnmpt.0 . . . . 5 (𝜑𝐴𝑉)
38 aciunf1lem.a . . . . . . 7 𝑗𝐴
39 nfcv 2979 . . . . . . 7 𝑘𝐴
40 nfcv 2979 . . . . . . 7 𝑘𝐵
41 nfcsb1v 3879 . . . . . . 7 𝑗𝑘 / 𝑗𝐵
42 csbeq1a 3869 . . . . . . 7 (𝑗 = 𝑘𝐵 = 𝑘 / 𝑗𝐵)
4338, 39, 40, 41, 42cbvmptf 5141 . . . . . 6 (𝑗𝐴𝐵) = (𝑘𝐴𝑘 / 𝑗𝐵)
44 mptexg 6966 . . . . . 6 (𝐴𝑉 → (𝑘𝐴𝑘 / 𝑗𝐵) ∈ V)
4543, 44eqeltrid 2918 . . . . 5 (𝐴𝑉 → (𝑗𝐴𝐵) ∈ V)
46 rnexg 7600 . . . . 5 ((𝑗𝐴𝐵) ∈ V → ran (𝑗𝐴𝐵) ∈ V)
47 uniexg 7451 . . . . 5 (ran (𝑗𝐴𝐵) ∈ V → ran (𝑗𝐴𝐵) ∈ V)
4837, 45, 46, 474syl 19 . . . 4 (𝜑 ran (𝑗𝐴𝐵) ∈ V)
4930, 48eqeltrd 2914 . . 3 (𝜑𝐶 ∈ V)
50 id 22 . . . . . 6 (𝑐 = 𝐶𝑐 = 𝐶)
5150raleqdv 3392 . . . . 5 (𝑐 = 𝐶 → (∀𝑥𝑐𝑗𝐴 𝑥𝐵 ↔ ∀𝑥𝐶𝑗𝐴 𝑥𝐵))
5250feq2d 6480 . . . . . . 7 (𝑐 = 𝐶 → (𝑓:𝑐𝐴𝑓:𝐶𝐴))
5350raleqdv 3392 . . . . . . 7 (𝑐 = 𝐶 → (∀𝑥𝑐 𝑥𝐷 ↔ ∀𝑥𝐶 𝑥𝐷))
5452, 53anbi12d 633 . . . . . 6 (𝑐 = 𝐶 → ((𝑓:𝑐𝐴 ∧ ∀𝑥𝑐 𝑥𝐷) ↔ (𝑓:𝐶𝐴 ∧ ∀𝑥𝐶 𝑥𝐷)))
5554exbidv 1922 . . . . 5 (𝑐 = 𝐶 → (∃𝑓(𝑓:𝑐𝐴 ∧ ∀𝑥𝑐 𝑥𝐷) ↔ ∃𝑓(𝑓:𝐶𝐴 ∧ ∀𝑥𝐶 𝑥𝐷)))
5651, 55imbi12d 348 . . . 4 (𝑐 = 𝐶 → ((∀𝑥𝑐𝑗𝐴 𝑥𝐵 → ∃𝑓(𝑓:𝑐𝐴 ∧ ∀𝑥𝑐 𝑥𝐷)) ↔ (∀𝑥𝐶𝑗𝐴 𝑥𝐵 → ∃𝑓(𝑓:𝐶𝐴 ∧ ∀𝑥𝐶 𝑥𝐷))))
57 acunirnmpt2f.d . . . . . 6 𝑗𝐷
5857nfcri 2967 . . . . 5 𝑗 𝑥𝐷
59 vex 3472 . . . . 5 𝑐 ∈ V
60 acunirnmpt2f.3 . . . . . 6 (𝑗 = (𝑓𝑥) → 𝐵 = 𝐷)
6160eleq2d 2899 . . . . 5 (𝑗 = (𝑓𝑥) → (𝑥𝐵𝑥𝐷))
6238, 58, 59, 61ac6sf2 30378 . . . 4 (∀𝑥𝑐𝑗𝐴 𝑥𝐵 → ∃𝑓(𝑓:𝑐𝐴 ∧ ∀𝑥𝑐 𝑥𝐷))
6356, 62vtoclg 3542 . . 3 (𝐶 ∈ V → (∀𝑥𝐶𝑗𝐴 𝑥𝐵 → ∃𝑓(𝑓:𝐶𝐴 ∧ ∀𝑥𝐶 𝑥𝐷)))
6449, 63syl 17 . 2 (𝜑 → (∀𝑥𝐶𝑗𝐴 𝑥𝐵 → ∃𝑓(𝑓:𝐶𝐴 ∧ ∀𝑥𝐶 𝑥𝐷)))
6536, 64mpd 15 1 (𝜑 → ∃𝑓(𝑓:𝐶𝐴 ∧ ∀𝑥𝐶 𝑥𝐷))
Colors of variables: wff setvar class Syntax hints: → wi 4 ↔ wb 209 ∧ wa 399 = wceq 1538 ∃wex 1781 ∈ wcel 2114 Ⅎwnfc 2960 ≠ wne 3011 ∀wral 3130 ∃wrex 3131 Vcvv 3469 ⦋csb 3855 ∅c0 4265 ∪ cuni 4813 ∪ ciun 4894 ↦ cmpt 5122 ran crn 5533 ⟶wf 6330 ‘cfv 6334 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1797 ax-4 1811 ax-5 1911 ax-6 1970 ax-7 2015 ax-8 2116 ax-9 2124 ax-10 2145 ax-11 2161 ax-12 2178 ax-ext 2794 ax-rep 5166 ax-sep 5179 ax-nul 5186 ax-pow 5243 ax-pr 5307 ax-un 7446 ax-reg 9044 ax-inf2 9092 ax-ac2 9874 This theorem depends on definitions: df-bi 210 df-an 400 df-or 845 df-3or 1085 df-3an 1086 df-tru 1541 df-ex 1782 df-nf 1786 df-sb 2070 df-mo 2622 df-eu 2653 df-clab 2801 df-cleq 2815 df-clel 2894 df-nfc 2962 df-ne 3012 df-ral 3135 df-rex 3136 df-reu 3137 df-rmo 3138 df-rab 3139 df-v 3471 df-sbc 3748 df-csb 3856 df-dif 3911 df-un 3913 df-in 3915 df-ss 3925 df-pss 3927 df-nul 4266 df-if 4440 df-pw 4513 df-sn 4540 df-pr 4542 df-tp 4544 df-op 4546 df-uni 4814 df-int 4852 df-iun 4896 df-iin 4897 df-br 5043 df-opab 5105 df-mpt 5123 df-tr 5149 df-id 5437 df-eprel 5442 df-po 5451 df-so 5452 df-fr 5491 df-se 5492 df-we 5493 df-xp 5538 df-rel 5539 df-cnv 5540 df-co 5541 df-dm 5542 df-rn 5543 df-res 5544 df-ima 5545 df-pred 6126 df-ord 6172 df-on 6173 df-lim 6174 df-suc 6175 df-iota 6293 df-fun 6336 df-fn 6337 df-f 6338 df-f1 6339 df-fo 6340 df-f1o 6341 df-fv 6342 df-isom 6343 df-riota 7098 df-om 7566 df-wrecs 7934 df-recs 7995 df-rdg 8033 df-en 8497 df-r1 9181 df-rank 9182 df-card 9356 df-ac 9531 This theorem is referenced by: aciunf1lem 30415
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# Differential equation help
1. Feb 12, 2016
### theone
I understand what is in the picture http://postimg.org/image/u5ib33kzb/
but the book goes on to say that the solution is thus of the form
$X_n = a_n sin \frac{n \pi x}{l}$
How does putting $β=σ^2=\frac{n^2π^2}{l^2}$ into (6.37) result in that?
2. Feb 12, 2016
### RUber
I will assume that you have a differential equation that looks like:
$x'' +\beta x = 0$
with boundary conditions:
$x(0)=x(l) = 0$
The general solution for the differential equation is
$x = A \sin( \sqrt{\beta} t ) + B \cos(\sqrt{\beta} t)$
And the boundary condition at $t=0$ forces B to go to zero and the boundary condition at $t = l$ forces $\beta$ to be the form you have above.
3. Feb 12, 2016
### theone
thats right, the differential equation is (X is X(x), a function of x) :
$X'' + \beta X = 0$
Assuming a general solution of $X(x) = A e^{ -\sqrt{-\beta}x} + B e^{+\sqrt{-\beta} x}$, that $\sqrt{-\beta}$ is complex (ie. $\beta =σ^2$) , and that the boundary conditions are $X(0)=0$ and $X(l)=0$, they found that $σ=\frac{n\pi}{l}$
What I want to know is how putting $σ=\frac{n\pi}{l}$ into the general solution results in $X_n=a_n\sin\frac{n\pi x}{l}$
Or how their general solution is equivalent to yours?
Last edited: Feb 12, 2016
4. Feb 12, 2016
### pasmith
$$\cos x = \frac{e^{ix} + e^{-ix}}2 \\ \sin x = \frac{e^{ix} - e^{-ix}}{2i}$$
5. Feb 12, 2016
### RUber
Applying your first boundary condition tells you that A = -B, giving $X(x) = A\left(e^{-i\sigma x}- e^{i\sigma x}\right)$
Noting what pasmith wrote above, this is equivalent to $C \sin (\sigma x )$.
Then, since any sigma of the form given can be a solution, your full solution might be an infinite sum:
$X(x) =\sum_{n=1}^\infty X_n(x) = \sum_{n=1}^\infty a_n \sin(\sigma_n x )$
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Cody
# Problem 7. Column Removal
Solution 1994456
Submitted on 28 Oct 2019 by bruce collett
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
A = [1 2 3; 4 5 6]; n = 2; B_correct = [1 3; 4 6]; assert(isequal(column_removal(A,n),B_correct))
A = 1 3 4 6 B = 1 3 4 6
2 Pass
A = magic(4); n = 3; B = [16 2 13; 5 11 8; 9 7 12; 4 14 1]; B_correct = B; assert(isequal(column_removal(A,n),B_correct))
A = 16 2 13 5 11 8 9 7 12 4 14 1 B = 16 2 13 5 11 8 9 7 12 4 14 1
3 Pass
A = 1:10; n = 7; B_correct = [1 2 3 4 5 6 8 9 10]; assert(isequal(column_removal(A,n),B_correct))
A = 1 2 3 4 5 6 8 9 10 B = 1 2 3 4 5 6 8 9 10
### Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!
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# Bharathidasan University BCA Syllabus I Year
PRACTICAL – I : C PROGRAMMING LAB
1. Solution of a Quadratic Equation (all cases).
2. Sum of Series (Sine, Cosine, e
x
).
3. Ascending and Descending order of numbers using arrays (use it to find largest
and smallest numbers).
4. Sorting of names in alphabetical order.
5. Matrix Operations (Addition, Subtraction, Multiplication – use functions).
6. Finding factorials, generating Fibonacci Numbers using recursive functions.
7. String Manipulation without using String functions (String length, String
Comparison, String copy, Polidrome checking, counting words and lines in
strings – use function pointers).
8. Bisection and Newton-Raphson method
9. Lagranges Interpolation formula.
10. Gauss Elimination Method.
11. Euler and Runge-Kutta (II order only) methods.
12. Trapezoidal and Simpson’s 1/3rd
Rule.
13. Mean, Standard Deviation, Variance.
14. Correlation – regression coefficients.
15. Creation and Processing of Sequential files for payroll and Mark list preparation
(use structures for Record Description).
Reference: www.bdu.ac.in
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# X+y=14 y- x= 4 slovecgraphically
1
by anni3
2016-01-27T18:22:46+05:30
from First equation:
X+Y=14
SO, X=14-Y
we know that:
Y-X=4 (Substituting X value from first equation)
Y-(14-Y)
Y-14+Y=4
Y+Y=4+14 (transposing 14 to RHS and Y to LHS)
2Y=18
therefore Y=18/2
=9
Therefore:X=14-Y
=14-9
=5
Therefore X=5 and Y= 9
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# Thread: TI-89 Titanium: SOLVE AN EQUATION
1. ## TI-89 Titanium: SOLVE AN EQUATION
The equation is this: ln(x-1)=2x-3
When i solve for x, the calculator says: More solutions may exist.
and it does not give me any solutions.
What should i do to get the results?
2. Originally Posted by tupac
The equation is this: ln(x-1)=2x-3
When i solve for x, the calculator says: More solutions may exist.
and it does not give me any solutions.
What should i do to get the results?
Graph it.
From the graph, there is no intersection point.
Thus, the solution to the equation you asked about will be complex.
On the TI89, use
Code:
csolve(ln(x-1)=2x-3,x)
You will get 1 solution.
I hope this clarifies things!
--Chris
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# Looking for a particular algorithm for numerical integration
Consider the following differential equation $$p(t) = \frac{\partial q(t)}{\partial t}$$ where $t \in (0,\infty)$. I have a build a code that spits out values of the function $p$ for the variable $t$.
Now, I'm looking for a scheme that will analyse these values of $p$ in order to determine $q$. Does anybody have any suggestions? The main problem is that if I would calculate $q(t) = \int p(t) \; \mathrm{d}t$, then I'll end up with just a number (i.e. the area under the graph), but I need to know the actual function of $q(t)$.
I've cross-posted this question here: https://math.stackexchange.com/questions/1326854/looking-for-a-particular-algorithm-for-numerical-integration#comment2695909_1326854
• This is an introductory problem in every course on numerical methods, and is covered in every book on numerical methods. What have you tried so far where you got stuck? Commented Jun 16, 2015 at 12:09
• I looks like you've posted this same question in at least three places. I think it would be best if these questions could be merged by an admin. Please do not cross post. scicomp.stackexchange.com/questions/19944/… math.stackexchange.com/questions/1326854/… stackoverflow.com/questions/30856354/… Commented Jun 16, 2015 at 12:54
• If you integrate p(t) you will not get 'just a number' (the area) but will get instead a function of time t (which you should expect as you set this equal to q(t)...). You only get an area (or 'just a number') if you are integrating between two specific values Commented Jun 16, 2015 at 19:32
I will answer for the simplest case:
$\frac{dy}{dt} = -\lambda{y}, \hspace{4mm} y(0)=1$
Note: You need an initial condition which you did not specify in your original question.
In this simple problem $q(t) = y(t)$ and $p(t) = -\lambda{q(t)}$. If you use a basic forward Euler finite difference we get:
$\frac{y^{n+1}-y^{n}}{dt} = -\lambda{y^{n}}, \hspace{4mm} y^{0} = 1$
Which gives the recurrence relation:
$y^{n+1} = (1-dt\lambda{})y^{n}, \hspace{4mm} y^{0}=1$
You now just march forward using a specific dt to get y at time t. Obviously there is much more to the story. Here we have used the simplest time-stepping scheme (forward Euler). We have also used a very simple and specific example, but hopefully this will get you started. The basic procedure is to first approximate the derivative using a discrete formula (i.e. forward Euler) and then march forward in time starting from the initial condition.
Your confusion mainly seems to be with calculus, not with numerical methods. Specifically,
• make sure you understand the difference between an indefinite integral and a definite integral.
• consider what the initial conditions or boundary conditions are for your problem. Perhaps you know $q(t_0) = q_0$ for a given time $t_0$.
• Write the integral formulation again, explicitly involving the boundary conditions.
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# Using VPython program to calculate the E-field
• Comp Sci
## Homework Statement
Charge a ring of radius R=5.0cm laying in the x-y plane to 50nC.
Create a VPython program that will allow you to calculate the E-field due to the ring anywhere in space.
## Homework Equations
E_ring=kQz/(R^2+z^2)^(3/2), point P above the xy-plane
## The Attempt at a Solution
from visual import *
C = pi*.10 # meters
Q = 50.0*10**-9 #Coulombs
k = 8.99*10**9 # Nm^2/C^2
E = vector(0.0,0.0,0.0) # N/C
Lambda = Q/C # linear charge density
s=vector(0.025,0.0,0.025)
# we are looking for the E-field due to the ring at anywhere, I think we would have dD,
# but I don't know how to define it.
ds = vector(D/10000,D/10000,D/10000)
dq = mag(Lambda*ds)
while s.x < L+D:
rate=(10000)
dE = k*dq/(mag(s)**2)*norm(s)
E = E + dE
s = s + ds
print 'sx=',s.x,'dE=',dE,'E=',E
print 'E=',E,'N/C'[/B]
robphy
Homework Helper
Gold Member
Does your formula for the ring work everywhere? Or just on the axis?
It might better to model ring of charge as a set of equally spaced point charges on the ring.
Then at the point of interest, find the electric-field vector there as a vector sum of electric-field contributions from each point charge.
You can use your formula to check at a point where it is valid.
Does your formula for the ring work everywhere? Or just on the axis?
It might better to model ring of charge as a set of equally spaced point charges on the ring.
Then at the point of interest, find the electric-field vector there as a vector sum of electric-field contributions from each point charge.
You can use your formula to check at a point where it is valid.
The formula only model on the z-axis.
I don't know how to model ring of charge as a set of equally spaced point charges on the ring?
Can you give me a hit?
robphy
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## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)
(a) $\omega = 88.3~rad/s$ (b) The fraction of the kinetic energy that is rotational kinetic energy is $\frac{2}{7}$ (which is equal to 0.286)
(a) We can find the height $h$ of the incline: $\frac{h}{L} = sin(\theta)$ $h = L~sin(\theta)$ $h = (2.1~m)~sin(25^{\circ})$ $h = 0.89~m$ We can use conservation of energy to solve this question. The potential energy at the top of the incline will be converted into translational kinetic energy and rotational kinetic energy at the bottom of the incline. We can find the speed at the bottom. $KE_{trans}+KE_{rot} = PE$ $\frac{1}{2}Mv^2+\frac{1}{2}I~\omega^2 = PE$ $\frac{1}{2}Mv^2+\frac{1}{2}(\frac{2}{5}MR^2)(\frac{v}{R})^2 = PE$ $\frac{1}{2}Mv^2+\frac{1}{5}Mv^2 = Mgh$ $\frac{7}{10}v^2 = gh$ $v = \sqrt{\frac{10~gh}{7}}$ $v = \sqrt{\frac{(10)(9.80~m/s^2)(0.89~m)}{7}}$ $v = 3.53~m/s$ We can find the angular velocity; $\omega = \frac{v}{R}$ $\omega = \frac{3.53~m/s}{0.040~m}$ $\omega = 88.3~rad/s$ (b) $KE_{rot} = \frac{1}{5}Mv^2$ $KE = \frac{7}{10}Mv^2$ We can find the fraction of the kinetic energy that is rotational kinetic energy. $\frac{KE_{rot}}{KE} = \frac{\frac{1}{5}Mv^2}{\frac{7}{10}Mv^2}$ $\frac{KE_{rot}}{KE} = \frac{2}{7}$ The fraction of the kinetic energy that is rotational kinetic energy is $\frac{2}{7}$ (which is equal to 0.286).
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# Solving MATLAB ODE with Matrix Coefficients: Help Requested
• MATLAB
• christy
In summary, the author is asking for help with a 2nd order linear system. They reduced the system to 1st order and are having difficulty solving it with matrix coefficients. However, there are other methods available.
christy
Hi,
I have tried everything to figure this out in MATLAB and I can't so I was wondering if anyone could help me please.
I have a 2nd order linear system that I reduced to 1st order such that
q = dz/dt
and A.(dq/dt) = Gq(t) + f(t)
where A and G are matrices...I can't figure out how to solve this in MATLAB with having matrix coefficients. Can someone please give me some hints?
A and G are generated matrices and are constant for this problem.
Thanks
You can solve systems of equations and systems of systems of equations the same way.
If q is size N x 1, and z is size N x 1, then define
y = [q; z] which is size 2N x 1
Using a simple forward Euler scheme,
y[n+1] = y[n] + dt*[f1(q,z,t(n)) ; f2(q,z,t(n))]
where f2 = q[n] and f1 = inv(A)*(G*q[n] + f(tn))
G is size N x N, q is N x 1, and f is N x 1, so [f1; f2] is size 2N x 1, just like y. So all the sizes work fine. You're just solving one really big system. Then you can make some operator C which is size 1 x N to pick out the ouput you want, like z_1(t), or whatever.
Alternatively, you could keep the two equations separate and solve
q[n+1] = q[n] + dt*inv(A)*(G*q[n] + f(tn))
and
z[n+1] = z[n] + dt*(q[n])
Now you're solving two systems, each size N x 1.
If you still don't like that, you can break up q and z into N equations each.
Since q = [q_1; q_2; q_3;...;q_N], you can write it as
q_1[n+1] = q_1[n]+dt*((inv(A)*G)(1,:)*q[n]+inv(A)f_1(tn))
q_2[n+1] = q_2[n]+dt*((inv(A)*G)(2,:)*q[n]+inv(A)f_2(tn))
.
.
.
and do the same for z to get 2N individual equations to solve.
So the bottom line is, you can either stick all of your equations, each of which may be working on a vector, into one giant vector and solve that, or you can break it up into any number of equations and solve them all individually stepping through time.
Last edited:
Dear all,
I have a equation that goes like this...
d^2/dt^2[x] + f(t) x - gamma * d/dt [x] where f(t) is a function of t. I have a numerical form of f(t), but no analytical form, so i want to pass this whole vector into my function...but I am able to pass only a single value...My program is as follows,
Main.m
clear all
close all;
clc;
s1 = input('Input File name without any *.txt extention :','s');
input_file_name = [s1 '.txt'];
time = Data(:,1)'; % Convert to row vecor
init = [0,0.5*2*pi]; % Initial condition
f =1.42E27*Data(:,2); % f(t) -- only numerical form is available
[t,y]=ode45(@eqn,time,init,[],f); % If u remove f from here and write this as %exp(-t/2E-12) inside the fun eqn.m, then, it works
plot(t,y(:,1))
eqn.m
function rk = eqn(t,y,f)
gamma= 2*pi*0.1E12;
rk = [y(2); - (f y(1)) -gamma*y(2)];
Could anyone help me out to solve this problem...
with best regards
kamaraju
## What is MATLAB ODE with Matrix Coefficients?
MATLAB ODE with Matrix Coefficients is a method of solving ordinary differential equations (ODEs) using a matrix representation of the coefficients. This approach is particularly useful for systems of ODEs with multiple variables.
## How does MATLAB ODE with Matrix Coefficients work?
In this method, the coefficients of the ODE are represented as a matrix, and the ODE is transformed into a matrix equation. This matrix equation can then be solved using MATLAB's built-in functions, such as the "ode45" function.
## What are the advantages of using MATLAB ODE with Matrix Coefficients?
Using this method can often lead to more efficient and accurate solutions compared to other numerical methods. It also allows for easy integration with other MATLAB functions and programs.
## What types of ODEs can be solved using MATLAB ODE with Matrix Coefficients?
This method can be used to solve a wide range of ODEs, including linear and nonlinear systems, first-order and higher-order equations, and equations with constant or variable coefficients.
## Are there any limitations to using MATLAB ODE with Matrix Coefficients?
While this method is powerful and versatile, it may not be suitable for all types of ODEs. Additionally, it may require some familiarity with MATLAB and matrix operations to effectively use this approach.
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# Regarding Apopolis meteor 99942.
#### Orion Humans
There is a claim that the calculation of the Apopolis meteor could have an error.
The link provide the calculation and the possibility of a real impact.
#### billslugg
He proposes making a one meter diameter plastic disc and measuring the circumference. I have already done this, but on a 5.5 meter drum. We used 5.5 meter steam drums in the paper machine. They were cut to a diameter +/- 0.01 mm. We measured the diameter with a standard pi tape and it verified exactly. He is not correct in his assertion.
COLGeek
#### promytius
If I may, those videos were unenlightening; anyone who has ever cut wood or foam knows you get an imperfect surface, and the notion of wrapping a piece of metal around a cut edge to "precisely" measure anything cannot be a perfect measurement, and pulling and stretching to get the 'correct' results is unimpressive, and as I watched, there was no point to it, I saw no result that had anything to do with an asteroid. How is a physical expression of Pi superior to a computer generated image that can be rendered to a precision impossible with real materials; why he is using physical things to prove a mathematical point is beyond me. Look at the small difference he's trying to show; the lines have width! the metal stretches, the surfaces and composition of the materials is different - wood is hard to compress, foam is easy. While he was very detailed and specific about what he used and how he used them, it did not seem to lead to anything, I may have missed it. Watching him manipulate the band to show his expected results was humorous to watch.
So, as an outsider, what does cutting circles have to do with an asteroid?
Kepler's Golden triangle? Kepler? What would Archimedes have to say? How did Kepler claim it?
So I cannot see how a physical demonstration of pi is in any way equal to or superior to a mathematical one, where precision is at a place where pi can be precisely measured to beyond 1,000,000 digits? Finally, I cannot relate that to an asteroid. Thanks.
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is this a valid strategy for options?
Discussion in 'Options' started by beachallstar, Jul 22, 2010.
1. beachallstar
i have been developing a simple strategy for trend following with options and i was wondering if anyone could post some advice:
The valid point of using a trend indicator is: the stock is already gaining, so its doing what you want,appreciating in value.
The downside is if you trend follow with stock,and the market/stock turns against you you can get out with a very small(5%) loss
The downside with options is you can be wiped out very very easily.....so my strategy is like this
say i bought 25 calls PEP sept at 65 for 1.15 and 25 puts at 62.50 for 1.14 and PEP was trading at 64.20. would that be enough time for the stock to diverge in either direction to off set the call/put hedge and still profit?
I was also wondering if anyone knows and sites that screen delta/gamma so i can optimize this approach with a more volatile stock?
2. ForexForex
It's easy to calculate:
Total cost for calls and puts = \$2.29
Call strike \$65.00 + \$2.29 = \$67.90
Put strike \$62.50 - \$2.29 = 60.21
Commission = ?????????
The stock must trade higher than \$67.90 or lower than \$60.21 to profit, plus enough to offset the commission. Looks like you need about a 5% move in either direction to break even.
3. MTE
ForexForex, your calculation applies ONLY at expiration.
beachallstar, this strategy is called a strangle and you can find a lot of info on it either here in the Options forum or using Google.
4. spindr0
In a strangle, initially, what one side makes, the other side loses. Add time decay on two sides and you have a position that makes money only if the underlying moves. The longer it takes to move, the larger the move that you need to break even.
AFAIK, optimizing delta/gamma isn't going to help you much. Figuring out which stocks will move enough to offset 4 commissions (open and close 2 sides), lost B/A slippage and up to 2 months of time decay is your problem. Do that and you make money.
5. JJacksET4
Nope. Strangles are when you use different strike prices - for example a 45 put and a 50 call. Straddles are when the strike price is the same (50 put/50 call).
Buying or selling has no bearing on the name (other then short/long).
JJacksET4
6. JPope
makes sense, thanks
7. beachallstar
thanks for the help!
So does anyone know which works better,a straddle or a strangle? Is that dependant more on market enviornment? Because with a reasonable strangle you can be slightly more out of the money and get more leverage for your dollar.
But a straddle might have a higher probability because the strike price is the same for the call as it is for the put,so your only gap would be the bid/ask spread.
Does anyone trade options with 8 weeks of time value? That seems to me to give you more opportunity to offset your insurance side of your play and see profit.
8. KINGOFSHORTS
Timeline depends on the circumstance, Options are one of the most complex instruments to trade but in my opinion the most rewarding financially and intellectually stimulating if you get deep and get a good understanding, and a few years of experience.
Anyhow I have traded options with only a week and some days left to expiration, and sometimes LEAPS etc.. it all depends on the situation.
But you need to get to step one, and understand the underlying well before you even go to step two which is options.
Learn to crawl then walk before you go and run. Then it takes years.
I only focus on a very small handful of underlying, same ones for years, you do not want to be a jack of all trades master of none.
10. trefoil
You're betting on a volatility spike. Best done, in my experience, a week or two before earnings, if you notice that volatility on the options is not out of line with normal despite earnings coming up soon.
Not an easy trade. Can be a huge moneymaker on a big miss or upside surprise, of course. Also, if volatility moves up before earnings, you could wind up making money even before earnings come out. Then you have to sit and wonder whether to cash it in or see if actual earnings wind up being a big enough surprise to make you even more money.
#10 Jul 25, 2010
WHILE YOU'RE HERE, TAKE A MINUTE TO VISIT SOME OF OUR SPONSORS:
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# Visualizing the Orion Constellation
I have something different from others!
It showed an error when I code"%matplotlib notebook".
And I search google, I found I haven’t installed the ipympl yet.
So I installed it and change the code to “%matplotlib widget”, it’s done!
here the line: https://github.com/matplotlib/ipympl
First
``````%matplotlib widget
from matplotlib import pyplot as plt
from mpl_toolkits.mplot3d import Axes3D
``````
Second
``````# Orion
x = [-0.41, 0.57, 0.07, 0.00, -0.29, -0.32,-0.50,-0.23, -0.23]
y = [4.12, 7.71, 2.36, 9.10, 13.35, 8.13, 7.19, 13.25,13.43]
z = [2.06, 0.84, 1.56, 2.07, 2.36, 1.72, 0.66, 1.25,1.38]
``````
Third
I have a question.
Why? Is it because “add_subplot” is kind of attribute?
``````fig = plt.figure()
plt.scatter(x,y)
plt.show()
``````
Fourth
``````fig_3d = plt.figure()
constellation3d = plt.scatter(x,y,z,color="black")
plt.show
``````
1 Like
Good stuff.
In regards to your question, add_subplot is a method of your figure instance. If you’ve not spent much time working with classes then this could be confusing and you may need to do some background reading or lessons on how they work. If you have then you’ll need to spend a little time nosing around how matplotlib woks under the hood since it’s not immediately obvious (or at least it wasn’t when I first started using it).
There are ways to create subplots without having a figure instance in the first place since one is created with the function call anyway, e.g. `fig1, ax1 = plt.subplots(1, 1)`
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# Advanced Work Rate GMAT Problem
#### Work Rate Problem Introduction &Challenges
Hi guys! Today we’re going to look at a super challenging work rate problem. This is one of those keystone problems, where if you have this problem really, really down then it indicates that you won’t have a problem with any work rate problem, speed-distance problem whatsoever.
The challenges in this problem are several. First off, you don’t have any base number to work with. That is, it’s all done in percentages and while a lot of times this can be an asset for running a scenario, here the scenario can really trip us up, as we’ll see. We also are being given different movements that are happening in different directions and it’s not entirely clear from the language in this problem that they are happening in different directions. Finally, there’s a notational issue that both the problem and the answer choices are in percentage and yet with work-rate, speed-distance, many times it’s best to work with fractions. So we have the option to notationally shift over to fractions and back. So let’s dive in.
#### Machine Efficiency (& Inverse)
On its surface, this is a textbook GMAT problem and what we’re being asked for is relatively straightforward. That is, there’s not a large interpretive part to understanding what we’re being asked for. The challenge here is how we’re going to get it. What would be the best way to approach this, in my opinion, for most people is to break apart the two different changes that are happening in the problem: the machine efficiency and the length of the production process and understand what each of these two tweaks is doing to your overall problem. Let’s begin with the production time.
The length of the production is decreasing by 25. So the pivot question is how much time is this saving us? And you’re right, it’s on the surface, it’s 25%. But we want to take the inverse of this because we don’t care about the time that it’s saving us, even though the question’s asking that. Rather, what we need to do is figure out how much total time we’re taking at the start versus with these tweaks and see how that’s affected so we’re going to have to invert everything and instead of saying: okay I’m saving 25% of my time, it’s taking only 75% of the time. If we do this in fractions we’re saving 1/4 so it takes us 3/4’s as long to do the same thing. Let’s take that three quarters put it to the side for a second.
#### Machine Speed
The second part is a little more tricky! Our speed is increasing by 1/3 but that’s the inverse that is when we go faster the time we’re going to spend decreases. I’ll say that once more: the faster we’re going the less time we’re spending. So what does increasing our speed by a third do for us? Well instead of making 3 things, in the same amount of time we can make 4 things. We start out with a unit and we’re adding another third. Three parts, three, four parts.
So once again we are taking only 3/4s of the time to do the same thing due to the increased speed so we only have to spend 3/4s of the amount of time for the increased efficiency and only three quarters of the amount of time for the speed. We put those together multiplicatively and we find that it’s going to take us 9/16 of the time that we used to spend doing in order. The flip of that where we invert, is that we’re saving 7/16 of our time or a little under half. You’ll see from the answer choices that if you’re attuned, you’re limited to the 56.25% and/or 62.5%.
And if you’re familiar with your eighths being 50 and 62.5 as two of the of the eighths, then it’s got to be that 56 number. So there’s a bit of known numbers feeding into this but ultimately the challenge is doing these inversions and recognizing that these problems are built for fractions. Not just this one but just about every work-rate problem.
#### Deal With Work Rate Problems Fractionally
You want to deal with it fractionally because it’s these ratios of time, the numerator to the denominator that allow us to do things flexibly and to flip stuff around. 4/3s to 3/4s as we saw in this problem. Notice here that running a scenario, especially if you choose a generic number like: Okay well let’s say I make 100 widgets is going to get you into a ton of problems because it’s going to be extraordinarily complex from a calculation standpoint.
You can think about it in advance and choose really good numbers but in doing so you’ll have circumvented the solution path and logically push yourself into saying well wait this is going to be 16ths of something. So for just about everyone out there I would say put it into fractions think carefully and deeply about what each of these two switches does and then put it all together the way you would any other problem. Post your questions below. Subscribe to our channel! And come visit us anytime at www.apexgmat.com. I’ll look forward to seeing you guys again soon.
For another gmat work rate problem, try this Car Problem.
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CFD Online Discussion Forums (https://www.cfd-online.com/Forums/)
- Main CFD Forum (https://www.cfd-online.com/Forums/main/)
- - Orr Sommerfeld equation (https://www.cfd-online.com/Forums/main/98748-orr-sommerfeld-equation.html)
thegodfather March 18, 2012 02:12
Orr Sommerfeld equation
My question is very fundamental.
why, at all do we need to have a separate equation that predicts the growth or decay of perturbations in a steady base flow..? If at all, there are any perturbations in the flow, why can't the resultant motion of fluid be predicted using the usual Navier Stokes equations??
truffaldino March 18, 2012 05:27
Because Navier-Stokes equation is too complicated to be solved analytically. This is the main problem of all fluid dynamics. That is why we still know almost nothing exact about turbulence.
thegodfather March 18, 2012 07:43
thanks a lot for replying... but as far as i know, even Orr - Sommerfeld equation is a fourth order linear differential equation which cannot be solved analytically for most cases, and various numerical methods have to be resorted to solve it... In that case would it not be better to bear with a little more complexity in terms of better numerical methods and solve the full N-S equations?
truffaldino March 18, 2012 07:57
As you said Orr-Sommerfeld is a linear system, while Navier-Stokes is nonlinear and shows chaotic behavior (turbulence) which is difficult analyse even numerically (DNS is done on supercomputers).
In the case of Orr-Somerfeld you have a possibility of qualitative analysys of transition to turbulence. You can see how perturbations amplify etc.
thegodfather March 18, 2012 08:04
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# How would you break geometry proofs down into the necessary steps?
Nov 21, 2015
There are a few methodologies (see below), but the three most important ones are practice, practice and practice.
#### Explanation:
Proof is a set of logical steps that lead you from the given preposition statements to a statement to be proven.
In a way, it is like finding a way in a labyrinth from the starting point (given) to the ending point (to be proven).
One of the most universal methodologies of finding these steps is analysis-based approach.
This method starts from the ending point and determines which step might be leading to it. Finding one, we find another step that leads to the step we found. Unraveling this backwards step by step we have to get to the beginning statement given as the premises.
After this process (analysis) is completed, we should be able to prove our statement by going from the beginning reversing each step as we go until we reach a conclusion.
For example, let's prove that in an isosceles triangle $\Delta A B C$ ($A B \cong A C$) two medians, $B M$ and $C N$ to congruent sides are congruent.
Analysis
(Step 1) In order to prove congruence of two segments, we can include them into congruent triangles.
(Step 2) Let's choose triangles $\Delta A B M$ and $\Delta A C N$. Proving their congruence will suffice to state that $B M \cong C N$.
(Step 3) In order to prove congruence of triangles $\Delta A B M$ and $\Delta A C N$ we can find three pairs of congruent elements of these triangles.
3.1. They have a common angle $\angle B A C$
3.2. Congruent sides $A B$ in $\Delta A B M$ and $A C$ in $\Delta A C N$ (since both are legs of isosceles triangle $\Delta A B C$)
3.3. Congruent sides $A M$ in $\Delta A B M$ and $A N$ in $\Delta A C N$ (since they all are equal to half of a side $A B$).
This completes our analysis. But this is not the proof yet.
The real proof is the way backwards.
Proof
(reverse the logic from the given preposition to a statement to be proven)
1. Since $A B = A C$,
=> $A M = \frac{A C}{2} = \frac{A B}{2} = A N$
2. Since $A B = A C$ and $A M = A N$ and $\angle B A C$ is common for triangles $\Delta A B M$ and $\Delta A C N$,
=> $\Delta A B M$ ~= $\Delta A C N$
3. Since $\Delta A B M$ ~= $\Delta A C N$,
=> $B M = C N$
End of proof.
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# Incidental Parameters Problem with Binary Response Data and Unobserved Individual Effects
December 5, 2013
By
(This article was first published on Econometrics by Simulation, and kindly contributed to R-bloggers)
It is a well known problem that in some models as the number of observations becomes large, econometric estimators fail to converge on consistent estimators. The leading case of this is when estimating a binary response model with panel data with potential “fixed effects” correlated with the explanatory variables appearing in the population.
One method that is typically implemented by researchers is to observe inviduals or organizations over multiple periods of time. This is called panel data. Within the context of panel data it is often assumed unobserved effects: genetics, motivation, business structure, and whatever other unobservables that might be correlated with the explanatory variables are unchanging over time.
If we do then it is relatively easy to remove the effect of unobservables from our analysis. In my previous post I demonstrate 3 distinct but equivalent methods for accomplishing this task when our structural model is linear.
However, when our model is a binary response variables (graduate from college or not, get married or not, take the job or not, ect.) it is usually no longer logically consistent to stick with a linear model.
In addition, not all of our remidies which worked for the linear model provide consistent estimators for non-linear models. Let us see this in action.
First we will start with generating the data as we did in the December 4th post.
`# Let's say: x = x.base + fe set.seed(2) nperson <- 500 # Number of personsnobs <- 5 # Number of observations per person fe.sd <- 1 # Spefify the standard deviation of the fixed effedx.sd <- 1 # Specify the base standard deviation of x # First generate our data using the time constant effectsconstantdata <- data.frame(id=1:nperson, fe=rnorm(nperson)) # We expand our data by nobsfulldata <- constantdata[rep(1:nperson, each=nobs),] # Add a time index, first define a group apply function# that applies by group index.gapply <- function(x, group, fun) { returner <- numeric(length(group)) for (i in unique(group)) returner[i==group] <- get("fun")(x[i==group]) returner} # Using the generalized apply function coded abovefulldata\$t <- gapply(rep(1,length(fulldata\$id)), group=fulldata\$id, fun=cumsum) # Now we are ready to caculate the time variant xsfulldata\$x <- fulldata\$fe + rnorm(nobs*nperson) # This is were our simulation diverges from a linear model. # Let us define the linear component as:lc <- .5*fulldata\$x + .5*fulldata\$fe # Now we will define the logit probability asfulldata\$pl <- exp(lc)/(1+exp(lc)) # Now we define the probit probability asfulldata\$pp <- pnorm(lc) cor(fulldata\$pl,fulldata\$pp) plot(fulldata\$pl,fulldata\$pp, main="Probit and Logit Probabilities", xlab="Logit", ylab="Probit") plot(sort(lc), sort(fulldata\$pl), main="Probit and Logit Probabilities", ylab="P(y=1|x,a)", xlab="xb+a", type="l", lwd=2, lty=3) lines(sort(lc), sort(fulldata\$pp), lwd=2, lty=2)legend(-3.3,.8,c("logit","probit"), lty=c(3,2), lwd=2) `
`# We can see the probit tends to have a shorter range in which the `
`# action is happening. `
` # We should really think of the probit and the logit as now# two different sets of data. Now let's generate out outcomes.fulldata\$yl <- fulldata\$pl>runif(nobs*nperson)fulldata\$yp <- fulldata\$pp>runif(nobs*nperson) # Let's try to estimate our parameters with the logit.glm(yl ~ x, data = fulldata, family = "binomial") # We can see our estimator is upwards biased as we expect. # Now we will try to inlcude fixed effects as if we were not# aware of the incidental parameters problem.glm(yl ~ x+factor(id), data = fulldata, family = "binomial") # We can see, that including a matrix of dummy variables# seems to actually make our estimator worse. # Instead let's try the remaining fix that is available to us# from the previous post listing 3 fixes. # We will include an average level of the explanatory variable# for each individual. This is referred to as the # Chamberlain-Munlak device.fulldata\$xmean <- ave(fulldata\$x, group=fulldata\$id) glm(yl ~ x+xmean, data = fulldata, family = "binomial")# We can see that including an average effect significantly# reduces the inconsistency in the estimator. # Now, let's see what happens if we do the same things in the# probit model.glm(yp ~ x, data = fulldata, family = binomial(link = "probit"))# Probit experiences similar upward bias to that of the logit. glm(yp ~ x+factor(id), data = fulldata, family = binomial(link = "probit")) glm(yp ~ x+xmean, data = fulldata, family = binomial(link = "probit"))# Interestingly, including the Chamberlain-Munlak device in the probit# though theoretically inconsistent does seem produce estimates# comparably good as including the device with the logit at least# in the sample sizes simulated here.`
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시간 제한메모리 제한제출정답맞힌 사람정답 비율
1 초 256 MB73372758.696%
## 문제
A team of up-to four robots is going to deliver parts in a factory floor. The floor is organized as a rectangular grid where each robot ocupies a single square cell. Each robot is represented by an integer from 1 to 4 and can move in the four orthogonal directions (left, right, up, down). However, once set in motion, a robot will stop only when it detects a neighbouring obstacle (i.e. walls, the edges of the factory or other stationary robots). Robots do not move simultaneously, i.e. only a single robot moves at each time step.
The goal is to compute an efficient move sequence such that robot 1 reaches a designed target spot; this may require moving other robots out of the way or to use them as obstacles for “ricocheting” moves.
Consider the example given above, on the right, where the gray cells represent walls, X is the target location and 1 , 2 mark the initial positions of two robots. One optimal solution consists of the six moves described below.
Note that the move sequence must leave robot 1 at the target location and not simply pass through it (the target does not cause robots to stop — only walls, edges and other robots).
Given the description of the factory floor plan, the initial robot and target positions, compute the minimal total number of moves such that robot 1 reaches the target position.
## 입력
The first line contains the number of robots n, the width w and height h of the factory floor in cells, and an upper-bound limit ℓ on the number of moves for searching solutions.
The remaining h lines of text represent rows of the factory floor with exactly w characteres each representing a cell position:
• W a cell occupied by a wall;
• X the (single) target cell;
• 1,2,3,4 initial position of a robot;
• ’.’ an empty cell.
## 출력
The output should be the minimal number of moves for robot 1 to reach the target location or NO SOLUTION if no solution with less than or equal the given upper-bound number of moves exists.
## 제한
• 1 ≤ n ≤ 4
• max(w, h) ≤ 10
• w, h ≥ 1
• 1 ≤ ℓ ≤ 10
## 예제 입력 1
2 5 4 10
.2...
...W.
WWW..
.X.1.
## 예제 출력 1
6
## 예제 입력 2
1 5 4 10
.....
...W.
WWW..
.X.1.
## 예제 출력 2
NO SOLUTION
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What if the Twin Paradox use the "day time" and "night time" on earth as their age reference?
I've searched this site, I found a similar question here but not exactly like mine.
So I can't understand the Twin Paradox when I use the "day" (bright time) and "night" (dark time) on earth as the age reference.
In the Twin Paradox it is said :
the traveling twin is younger upon return than the Earth-bound twin.
Suppose the twin are 40 years old and can communicate instantly during the journey and the travelling twin calculate his age from his brother on earth standard. Say, when the travelling twin reach a star, the remain-on-earth twin told him that 7300 days and 7300 nights have passed on earth. (So the age of both of the twin are the same, 60 years old if in our common standard).
When the travelling twin return and almost landed on earth, the remain-on-earth twin told him that (for example), 14600 days and 14600 nights have passed on earth. So they still have the same age, which is 80 years old if in our common standard.
So my question is:
What does it mean of "the travelling twin is younger" at the time they meet each other again - while each of them think the same that his age is 29200 days and 29200 nights ? (80 years old if in our common standard).
Is the answer like below ?
Yes, their age is the same, but their appearance is different---> the travelling twin LOOK much much younger than his remain-on-earth brother.
• the cannot communicate instantly… Commented Nov 29, 2021 at 5:35
• The answers are a bit technical and you should refer to them. But I want to address explicitly your point with the figures you have in mind. The traveller looks younger because is younger. His/her inner time, say heart beat count, is less than the same count if s/he would not have moved. This inner time remains short as compared to the age shown by the passport for the rest of his/her life. Commented Nov 29, 2021 at 11:07
• The tell their ages only by their wristwatches (which also count years). There is no day & night in outer space! Commented Nov 29, 2021 at 11:46
The paradox is that if the travelling twin takes a clock with him and consults that clock, instead of just letting the brother at home tell him what time it is, that clock will record less time having passed than what the brother on Earth tells him has passed.
It doesn't matter what kind of clock is used. It could be mechanical, a quartz crystal oscillator, a water clock, or whatever. He could even just count how many times he has to shave his beard. Because the problem isn't that the clock behaves differently on the moving ship. What's happening is that less time actually passes for the brother (and everything else) on the ship compared to on Earth.
Edit
The thing which I don't understand is : "a clock". Because to me, "a clock" is something like this : ...
This isn't how physicists define a clock. Currently we use the frequency of oscillation of the cesium atom as the reference to define the second. That means you build an oscillator locked to a particular absorption line of a sample of cesium gas, and when it oscillates 9,192,631,770 times, that's one second.
But it could work with any kind of clock. Say the two brothers build or buy identical digital wristwatches, that keep time based on a quartz crystal oscillator. Each time the quartz crystal oscillates 32,768 times, one second is recorded.
After the travelling brother returns, the earth-bound brother's wristwatch has counted off 315,360,000 seconds (10 years). The wristwatch that travelled on the space ship might have counted off 320 million seconds, or 600 million seconds, or 315 billion seconds, depending how fast the space ship travelled.
The reason for this is not because wristwatches malfunction when they travel on space ships. It is because more time passed for the space ship and everything in it.
It would work the same if they used mechanical spring clocks, cesium cell atomic clocks, pendulum-driven grandfather clocks (assuming the spaceship is designed to produce exactly 1 g of artificial gravity), etc. Even if the brothers just count how many times their hearts beat (assuming they do similar amounts of exercise, etc., to keep their hearts beating at a uniform rate).
The Earth-Sun system is just one choice of a system with a periodic behavior that can be used to indicate time. If you (some kind of super-powered alien) built a second Earth-Sun system and put it in a ginormous space ship and sent it on the same path as the brother's space ship, when it got back the Earth on the space ship would have revolved more times around the sun than the Earth that was left behind and didn't travel.
• Thank you for the answer, Photon. But I'm sorry as I'm unable to understand it. You wrote : the travelling twin takes a clock with him. The thing which I don't understand is : "a clock". Because to me, "a clock" is something like this : a clock is a measurement for earth rotation. People divide it to 24, 12 for day time and 12 for night time. Each of those 24, people call it "an hour". Then from "an hour", people divide it to 60, called "a minute". Then from "a minute", people divide it to 60, called "a second". So, how is the clock reference for the travelling-twin ? Commented Dec 13, 2021 at 10:41
• @karma A clock in relativistic experiments needs to be portable from frame to frame. Someone zipping around on a rocket ship can't measure their "local time" from the earth's rotation, since that "clock" exists in a different reference frame. If a clock is moving relative to you, it won't measure time in your own reference frame. A twin on a rocket ship who uses the earth's rotation as a clock is effectively reading the wristwatch of someone still on earth, but that doesn't reflect the passage of time on the rocket. Commented Dec 13, 2021 at 18:21
• @NuclearHoagie, thank you for the explanation. I'm sorry as I'm still unable to understand what does it mean "the passage of time" ? I mean, if we are on earth measure the "passage of time" from the earth's rotation (say, for example : 3 nights and 3 days have passed) - how the one on the rocket measure the "passage of time" ? Commented Dec 25, 2021 at 5:44
• @karma, are you being deliberately obtuse? I suggested several ways in my answer: "a quartz crystal oscillator, a water clock," "mechanical spring clocks, cesium cell atomic clocks, pendulum-driven grandfather clocks, etc.", counting how many times he shaves his beard. Commented Dec 25, 2021 at 6:16
It happens because in spacetime different paths between two events can have different lengths through time.
It is analogous to different lengths of paths between two points in space. If you drive around three edges of a square and I drive along the fourth edge, although we start and end at the same point, our odometers will show that we have covered different distances.
Likewise, the travelling twin takes a route through time that is shorter than the route taken by the stay-at-home twin, so when they meet, their clocks show that they have travelled different distances through time.
• Thank you for the answer, Marco. But I'm sorry I'm unable to understand it. You wrote : the travelling twin takes a route through time that is shorter than the route taken by the stay-at-home twin. What does it mean "through time" ? Suppose there is never a clock invention (so, there is no a thing which we call "a watch" with it's second needle), while the stay-at-home twin count his age by counting how many days and nights have passed on earth, how the travelling-twin count his age ? Commented Dec 13, 2021 at 10:30
Yes, I think you have it right. If you define the age of a person as the amount of time that has passed on the Earth since they were born, then the twins will of course always be the same age, because they were born on the same day. However the returning twin will have experienced less time passing, and will therefore be younger than his brother physically, mentally and in basically every other sense of the word, as Marco Ocram and The Photon have already explained very well.
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https://www.studymode.com/essays/Freezing-Point-Depression-Determination-Science-Lab-Experi-65009914.html
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# Freezing Point Depression Determination SCIENCE LAB EXPERIMENT
Good Essays
This paper is a full determination for certain chemicals and their boiling points. It lists some already but given the atomic numbers of any material this project includes a conversion and calculation chart to find the freezing point of most any material. GOOD LUCK!
Abstract:
In this lab we determined the freezing point, and Kf, of pure 2,4,dichloralbenzne as well as a 2,4,dichloralbenzne/biphenyl solution. We used this information to determine the molar mass of an unknown (#24) by the 3rd step in the experiment which was a 2,4,dichloralbenzne/unknown solution. All of the above we charted the time temperature for the later calculations.
I. Introduction
A. Background
This experiment shows how one determines the temperature-composition diagram for a two-component system. The procedure will consist of obtaining cooling curves for the pure substances and a number of their mixtures. A cooling curve is constructed by melting a sample, then allowing it to cool, measuring the temperature at regular intervals. When only melt is present, there is a constant cooling rate. As the solid begins to form, the system remains at a constant temperature until the melt is completely converted to solid. The eutectic composition is that at which two solids crystallize out in a ratio equal to that of the melt, and the cooling curve obtained would have the same characteristics as that of a pure substance. The eutectic temperature is the melting point of such a mixture.. The addition of impurity to each of the pure components decreases the freezing point so that two curves are obtained which intersect at the eutectic point.
The freezing point of a solvent depends upon the concentration of the dissolved solute and the nature of the solvent. If the dissolved solute is a nonelectrolyte, then the decrease in the freezing point, DELTA T, is proportional to the molality, m,( moles of solute per kg of solvent) of a dilute solution according to the equation:
DELTA T = Kfm
Where Kf is the
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# How does buoyancy cause this wheel to spin?
1. May 12, 2010
### RyanC83
Hi all,
In the attached image I have a cylinder shaped wheel with a air pocket embedded near its edge. The wheel has the same density as water and is fully submerged in a tank of water. Also the wheel is only free to spin about its center axis. If the wheel is rotated down the air pocket will cause the wheel to spin upward but how?
Buoyancy is defined as an upward acting force, caused by fluid pressure.
The wheel is under pressure from all side with a greater pressure from the bottom caused by gravity but I'm not understanding how fluid pressure can cause the wheel to spin upward?
If a ping pong ball is fully submerged in a tank of water the water is able to flow under the ball to give it lift. but in this wheel example the wheel is not lifting but spinning so water is not flowing at all.
Maybe someone could help me understand this a little better.
Thanks!
#### Attached Files:
• ###### Buoyancy.jpg
File size:
32.7 KB
Views:
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2. May 12, 2010
### Staff: Mentor
It doesn't. What would cause the wheel to turn is its unbalanced weight: The solid side weighs more than the side with the air pocket.
3. May 12, 2010
### RyanC83
Thanks Doc AI
That was the answer I was looking for. I assumed the solid side would have a neutral buoyancy (the wheel having the same density as water) but if buoyancy does not effect the side with the air pocket it will not effect the solid side ether. So the wheel behaves the same whether it is fully submerged underwater or not.
Thanks very much!
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# What's the idea behind this, assuming it is ideal?
The circuit is a biased push-pull. The diodes keep the input signal above and below the input signal so it is always conducting. The problem is that I don't see why one of the two transistors would not conduct in a semicycle.
If the input is $5.3V$ then above D1 the voltage is $5.9V$. The output after the drop is then $5.3V$. On the other side, below D2 the voltage is $4.7V$, which is below the output voltage, making the php transistor Q2 conduct. I know I'm missing something, but I don't know where. In the book (The Art of Electronics) it says that in the positive swing Q2 shouldn't conduct, but I can't figure out why.
simulate this circuit – Schematic created using CircuitLab
• What V_ee value? During the positive phase, I am pretty sure that Q2 isn't in the saturated zone therefore it cannot conduct in theory.I would need to re-check my book to be sure. The idea behind this set-up from a Box POV, is that it give you a gain in currrent. Commented Aug 4, 2015 at 14:59
• @MathieuL I believe that the idea is to make a voltage buffer with (very) low power consumption, compared to a class A configuration. Commented Aug 4, 2015 at 19:03
Because it's an output stage, it won't make sense without a load. Try looking at it with a 100 ohm resistor from V_out to ground.
When Vin is 0, Vout will also be zero, but there will be a (hopefully small) amount of current flowing through Q1 and Q2.
Now let Vin jump to 1 volt. The anode of D1 will be at about 1.6 volts, and the output at about 1 volt. The cathode of D2 will be at about 0.4 volts, and the emitter-base of Q2 will be reverse biased, so no current will flow through it.
Likewise, if Vin equals -1.0, the output will be pulled down to -1, and Q1 will be cut off.
The advantage of putting the diodes in, rather than just tying the two bases together, is that the output will respond smoothly to Vin variations of less than +/- .6 volts. This is critically important in avoiding crossover distortion when the output moves through zero.
EDIT - Although I realize that this still doesn't explain why one transistor conducts but the other doesn't. Think of it this way. With an appreciable load, the emitter current of the transistor which supplies the load current at the specified polarity is greater than the draw with zero Vin. This means that the emitter/base voltage is greater, too. In turn, this reduces the base/emitter voltage at the other transistor. Since base currents are exponential wrt base voltage, this reduction in base voltage causes a major reduction in emitter current. Thus, only the transistor supplying load current conducts much, and in the process starves the one not conducting.
• Thanks. That might be my problem. On the second case, Q1 should shut off. But the collector is at V_cc, the base is at -0.4 and the emitter at -1. Doesn't that make the base-emitter voltage positive, and thus Q1 conducts? Commented Aug 4, 2015 at 5:19
• @OFRBG - see edit. Commented Aug 4, 2015 at 5:43
• Great. So what explains that one transistor conducts and the other one almost doesn't is the exponential relation between V_BE and the collector current, as long as the load draws as much current as possible? Commented Aug 4, 2015 at 5:49
• Maybe as a strange extra doubt, would you be able to tell why they determined so nonchalantly that the transistors were cutoff without using Ebers Moll? Just for curiosity. Commented Aug 4, 2015 at 5:52
The bias network (resistors and diodes) makes a current flow through both transistors at 0V out. If the diodes and transistors are thermally closely coupled, the current is reasonably stable with temperature (the diodes change similarly to the Vbe of the transistors).
When the amplifier is producing a small to moderate output, the both transistors still conduct significant current at all times (one more than the other, but some bias current flows through both). If the input approaches the rails, then the respective transistor will stop conducting.
Here is what the collector current in the PNP transistor in the above schematic (diodes replaced by diode-connected transistors) looks like with +/-12V supplies and a 5Vpp 1kHz input and 500 ohm load (10K resistors used so bias current is about 1.5mA).
Here is what it looks like with a 10Vpp input:
And the output waveform with 10pp input:
• So in the end both transistors are always sourcing/sinking current but just switch intensity depending on the polarity established with the load? (Assuming the load is not necessarily grounded.) Commented Aug 4, 2015 at 5:41
• The transistor that's not driving load current will conduct the bias current or less (down to about zero for large output voltages near the supply rails- as above in the middle graph). The other one will conduct that current plus the load current. If the load is not grounded then this bias scheme might not be ideal in general, but it's fine for grounded load or a bridging amplifier (where outputs are driven antiphase to double the output voltage across the load). Commented Aug 4, 2015 at 6:06
Part of your confusion might come from the fact that it is not a very good circuit. The bias current in the transistors is not well-controlled. As drawn, shoot-through current from one transistor to the other is determined by the difference between the diode forward voltages and base-emitter forward voltages divided by the transistor intrinsic emitter resistances. These are not well-controlled.
Do a Google search of 'class a-b rubber diode' the show how it is done properly. Oh - I call them V_be multipliers, not rubber diodes.
It's an output stage. The diodes are for a bias voltage, so the transistors work in "class AB" mode.
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# Learn a Fact: 3 x 3
Follow steps 1-6 to master this fact.
You're learning
3
x3
9
3
x3
9
3
X 2
6
3
X 4
12
## This picture and story will help you remember 3 x 3 = 9
3 (Tree) x 3 (Tree) = 9 (Line)
3x3
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### Author Topic: Understanding the VIC (Read 47592 times)
0 Members and 1 Guest are viewing this topic.
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##### Re: Understanding the VIC
« Reply #40 on: December 12, 2011, 21:11:13 pm »
what voltages are you guys achieving across the water cell ?
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##### Re: Understanding the VIC
« Reply #41 on: December 12, 2011, 21:56:08 pm »
If the cells polarity switched with each pulse, wouldn't that be equivalent to the steam resonator? In the sales manual Stan states that there is both a static and alternating field being applied to the cell. I don't understand how that one would work?
Yes, I was just about to point that out. In the Steam Resonator circuit, the gate pulses make the transistor switch the polarity on the plate, much like a H-bridge circuit. The voltage acts much like AC, but pulsed in a way to limit current. As you can test with AC voltage, it will not fracture the water but it will heat it up pretty fast. With tests that I have done with 120vac, I was able to super heat the water from room temp. to over 220 degrees in a matter of minutes. Here is a scope shot of what the output of the Steam Resonator pulses should look like.
(http://www.globalkast.com/images/tonywoodside/steam_resonator_pulse.jpg)
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##### Re: Understanding the VIC
« Reply #42 on: December 12, 2011, 22:23:45 pm »
Oh wow, that's awesome.
Yeah the polarity switching is something I was really questioning myself over for that very reason.
What I want to understand now though is why the WFC does not generate heat when there is collision occuring,
How exactly does that differ from the WFC?
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##### Re: Understanding the VIC
« Reply #43 on: December 12, 2011, 23:08:28 pm »
it came from Stan's stuff. The core material also came from Stan's collection. I just need to wrap the secondary so I can test it.
So does this mean that you know someone inside Quad City Innovate?
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##### Re: Understanding the VIC
« Reply #44 on: December 13, 2011, 00:24:28 am »
no, lol...but one of my guys has been in contact with Chris @ Quad City Innovations. I got the bobbin, SS wire and core from John Bostick for testing. I'm pretty sure he told me it came from Stan's stuff, but of course you should be able to confirm this.
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##### Re: Understanding the VIC
« Reply #45 on: December 13, 2011, 00:59:46 am »
Well that's cool! I thought when you said it came from Stans stuff you got it from Don. Chris is the chief technology guy correct? Do you guys know if they are having any success with Stan's stuff?
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##### Re: Understanding the VIC
« Reply #46 on: December 13, 2011, 04:46:05 am »
Tony,if you got that core from John,I thought it looked familiar,I made those cores.That metal was some material I bought,and a few people here bought some.I spent alot of time making 4 sets of them.Hope it works out for your testing.
Don
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##### Re: Understanding the VIC
« Reply #47 on: December 13, 2011, 06:12:03 am »
Yeah I bought some from you don, took your advice as well and
used a guillotine paper cutter to cut out the e and I sections, it
worked well. Thanks for all your contributions.
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# 8.4.2.1.2 The Scale Tab
This tab provides controls for axis scale, including the axis scale range, axis type, major and minor tick positions.etc. For 2D graphs, you can separately customize the Horizontal and Vertical; for 3D graphs, there are three axes (X, Y and Z) to be customized.
Since Origin 2023, once you have shown both right Y and left Y axis, and set Rescale Each Y Independently in the Show tab, there will be two seperate vertical icons Left and Right to let you cusotmize the left Y axis scale and right Y axis scale seperately.
## Layer Selection
Before customizing the scaling elements, you can use the Layer list to switch between layers in a multi-layer graph.
## From
Set the initial scale value in this text box.
## To
Set the final scale value in this text box.
## Type
This drop-down list will be grayed out if it is not supported for a specified plot type. In addition, some of the following options may not be available for some plot types.
Linear Standard linear scale: X'=X. Base 10 logarithmic scale: X'=log(X). Represents the inverse of a cumulative Gaussian distribution: X'=norminv(X/100). Plotting a cumulative Gaussian distribution produces a sigmoidally-shaped curve. This curve, when displayed on a probability scale, appears as a straight line. Since probabilities are expressed as percentages, all values must fall between 0 and 100. The probability scale range is 0.0001 to 99.999. Like the probability scale, a sigmoidally-shaped curve plots as a straight line. In this case, however, the scale is linear, and the increment between tick marks is exactly one standard deviation. The value "5" on the scale shows the mean, or 50% probability. "6" is one standard deviation away, etc. Reciprocal scale, where X'=1/X. Offset reciprocal scale, where X'=1/(X+offset). Offset is defined as 273.14, where 273.14 is the absolute temperature for 0° C. Logit=ln(Y/(100-Y)). As with the probability and probit scales, a sigmoidally shaped curve plots as a straight line. Natural log scale (base e logarithmic scale). Base 2 logarithmic scale. Double logarithmic reciprocal scale: X'=ln(-ln(1-X)). User-defined axis space by Direct Formula and Inverse Formula Direct Formula Enter a formula to define a one-dimensional space transformation for the axis. Inverse Formula Enter a formula to define a one-dimensional inverse space transformation for the axis. Note: This formula must be the exact inverse of Direct Formula. The From value of this axis cannot exceed the range of the Inverse Formula. Two formulas are using x to represent the variable. For example: Direct Formula: x*x-4 Inverse Formula: sqrt(x+4) Use the X values of the first plot in the current layer as tick labels. This scale type is linear except it takes X values as tick labels and eliminates the extra (like weekends and holidays in Financial Plots). If the Scale range(set by From and To controls) is larger than the actual X data range, the excess part will show as linear scale according to the Major and Minor Ticks settings.
A Note on Log10 Scales: If the log scale range is within one decade, the ticks and grids will be linear. The LabTalk system variable @TL determines whether to use linear tick marks by the following relation: 10 * log10(max/min) <= value Thus, to support linear tick locations for two decades, set this variable to 14. For example, you could enter the following in the Script window: @TL = 14 ; The default value for @TL is 10.
## Symmetrical Log Scale
This option is shown, when Scale Type is Log10, Ln or Log2.
Generally, Origin only supports positive values for the Log10, Ln or Log2 scale. Check this Symmetrical Log Scale option, it will support positive and negative values on the log scale.
When this option is available, these two options Linear Range Threshold and Linear Range Length are shown in the dialog.
### Linear Range Threshold
There is not zero value in the log plot, so you can use Linear Range Threshold to define a range of values near zero within which the plot is linear, to avoid having the plot go to infinity around zero.
The default value of Linear Range Threshold is 1.
### Linear Range Length
Specify ratio of one major tick interval of current log scale. The default value is 1
## Rescale
Fixed The axis is not rescalable. If you try to change the scale or perform an operation which changes the scale (for example, using the Scale In tool), Origin preserves the From and To values. If both axes in a 2D graph layer are set to Fixed, using the Scale In tool opens a dialog box asking if you want to change to Normal mode and rescale. Click Yes to temporarily override the scaling restriction. The axis is rescalable. Alter the axis scale and use the Scale In tool (for 2D graph layers) without restriction. If you are adding or removing datasets from the graph using the Plot Setup or Layer Management dialogs, you will need to select the Rescale check box in those dialogs, if you want to automatically adjust scales on data change. This option is similar to the Normal option, but also allows Origin to automatically scale the axis to accommodate plotted data, if necessary. Note: If you hide/unhide a data plot the axis will not rescale automatically even though the Rescale is set to Auto. Set system variable @PAR = 1 to trigger auto-rescale for such cases. The From value of the axis is fixed and can only be changed by editing the From text box value in the Axis dialog box, while the To value of the axis is in Normal mode. The From value of axis is in Normal mode, while the To value of the axis is fixed and can only be changed by editing the To text box value in the Axis dialog box. The From value of axis is fixed, while the To value of axis is allows Origin to automatically rescale. The From value of axis is allows Origin to automatically rescale, while the To value of axis is fixed.
## Rescale Margin(%)
When rescaling graph axes, this setting can be used to pad dataset minimum and maximum values by some percentage of the difference between the minimum and maximum data values in the N dimension. Because other factors also enter into the resulting scale -- rounding of minimum and maximum values, the scale increment, minor ticks, etc. -- Rescale Margin is best treated as an approximate value.
• If you set Rescale Margin(%) = 0 and set Minor Ticks (By Counts) = 0, then a Normal rescaling will result in From and To values that correspond to dataset minimum and maximum values, i.e. the resulting axis scale will add no padding to minimum and maximum values.
• The default method of padding dataset minimum and maximum values should generally suffice but for more precise control, users can manipulate the value of system variable @RRT. When Rescale Margin(%) exceeds the value of @RRT, the default method is used. When Rescale Margin(%) is less than the value of @RRT, then dataset minimum and maximum values are padded by quantity n * (data.max - data.min), where n = Rescale Margin(%), to arrive at scale From and To values.
• This control is also available for polar axes (both angular and radial axis) and radar chart axes. It is not supported for graphs created from a matrix.
## Reverse
Check this box to reverse the From and To scale values.
## Major Ticks
### Type
By Increment Set the major ticks of this axis by increment specified in the Value text box. Tick labels are placed at each major tick mark. For example, type 10 to display a major tick mark at every tenth value. Set the total major tick number displayed on this axis by the absolute number specified in the Count text box. Only show the ticks at the minimum and maximum of the X scale. Set the location of the major ticks using an existing dataset or a series of space-separated numbers in the Position combo box. Use values in a column label row to label major ticks. Set label row in the Column Label drop-down list. (See the fourth box chart in this list for an example.)
### Value
This option is only available when By Increment is selected for Type. Set the major tick increment for this axis in the associated text box. If the scale units for this axis are time series values, then the value in the Increment text box must indicate the increment in appropriate time series terms. The permissible time series step increment units and their allowable abbreviations are:
Increment Abbreviation s m h d w mo q y
For time series graphs, set the size of the increment by typing a number followed by an increment unit. For example, 1month sets a major tick increment of one month. 4Q sets a major tick increment of four quarters. (Do not leave a space between the number and the increment unit.)
### Count
This option is only available when By Counts is selected for Type. Specify the desired absolute number of total major ticks in the associated text box.
Notes: The maximum number of major ticks is controlled on the Axis tab of the Options dialog box (Preference: Options). You will not necessarily see this number of ticks used as Origin will still try to display reasonable numbers for Tick Labels. It is also allowed to use minus value for major tick numbers, so as to make the tick label numbers rounded-up. In case of negative major tick numbers, Origin will try to get the best range to make sure the tick labels are integers and the total tick number will try to be closest to the absolute value which was set, but not exactly the same.
### Position
This option is only available when By Custom Positions is selected for Type. Select the dataset as the major tick locations in this drop-down list. Also you can enter the desired dataset name or a series of numbers separated by Space.
In the Position combo box, the most recently active worksheet columns are listed first. So, instead of scrolling down, close this dialog box, click on the desired book and sheet, then reopen this dialog box.
### Column Label
This option is only available when By Plot Column Label is selected for Type. This combo box list Auto and the available column label rows (Long Name, Units, Comments etc.)
• Auto: Determined by plot's own. It mainly means index.
For Box/Waterfall plot, it means following the selection in the X position drop-down list of the Plot Details dialog.
### Anchor Tick
This option is available when By Increment or By Counts is selected for Type. Specify the major tick you want to make sure to show on axis.
For example, if your X axis range runs from -3 to 3 and increment is 2. and you want to make sure 1 shows . Set Anchor Tick to be 1, the major ticks show at -3, -1, 1, 3.
Notes: When using calendar-accurate date values as tick labels, the option can be used to specify the value of the first major tick label to be displayed, and at what value subsequent minor ticks should fall. by entering text with format below Specify both a major and minor tick value, include a comma after the major tick value, followed by the minor tick value. For example, if your X axis range runs from 1/1/99 to 12/31/99, you could specify that the first major tick label fall at 1/4/99, and that all subsequent minor tick marks fall on a Monday, by entering any of the following in the Anchor Tick text box: 1/4/99, Monday 1-4-99, Mon Jan 4 1999, M
## Minor Ticks
### Type
By Counts Set the number of minor ticks displayed between adjacent major ticks specified in the Count text box. For example, select By Increment for major tick Type, type 1 in the Value text box, and type 1 in the minor tick Count text box to set the tick sub-step to 0.5. One minor tick will display between each pair of major ticks. Set the location of the minor ticks using an existing dataset or a series of space-separated numbers in the Position combo box.
### Count
This option is only available when By Counts is selected for Type. Specify the desired absolute number of minor ticks in the associated text box. If the scale units for this axis are time series values, then Origin uses the number to automatically determine the most appropriate minor tick labels.
### Position
This option is only available when By Custom Positions is selected for Type. Select the dataset as the Minor tick locations in this drop-down list. Also you can enter the desired dataset name or a series of numbers separated by Space.
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# Kilogram Force/Millimeter to Dyne/Inch Converter
1 Kilogram Force/Millimeter = 24908891 Dyne/Inch
## One Kilogram Force/Millimeter is Equal to How Many Dyne/Inch?
The answer is one Kilogram Force/Millimeter is equal to 24908891 Dyne/Inch and that means we can also write it as 1 Kilogram Force/Millimeter = 24908891 Dyne/Inch. Feel free to use our online unit conversion calculator to convert the unit from Kilogram Force/Millimeter to Dyne/Inch. Just simply enter value 1 in Kilogram Force/Millimeter and see the result in Dyne/Inch.
Manually converting Kilogram Force/Millimeter to Dyne/Inch can be time-consuming,especially when you don’t have enough knowledge about Surface Tension units conversion. Since there is a lot of complexity and some sort of learning curve is involved, most of the users end up using an online Kilogram Force/Millimeter to Dyne/Inch converter tool to get the job done as soon as possible.
We have so many online tools available to convert Kilogram Force/Millimeter to Dyne/Inch, but not every online tool gives an accurate result and that is why we have created this online Kilogram Force/Millimeter to Dyne/Inch converter tool. It is a very simple and easy-to-use tool. Most important thing is that it is beginner-friendly.
## How to Convert Kilogram Force/Millimeter to Dyne/Inch (kgf/mm to dyn/in)
By using our Kilogram Force/Millimeter to Dyne/Inch conversion tool, you know that one Kilogram Force/Millimeter is equivalent to 24908891 Dyne/Inch. Hence, to convert Kilogram Force/Millimeter to Dyne/Inch, we just need to multiply the number by 24908891. We are going to use very simple Kilogram Force/Millimeter to Dyne/Inch conversion formula for that. Pleas see the calculation example given below.
$$\text{1 Kilogram Force/Millimeter} = 1 \times 24908891 = \text{24908891 Dyne/Inch}$$
## What Unit of Measure is Kilogram Force/Millimeter?
Kilogram force per millimeter is a unit of measurement for surface tension. Surface tension is considered to be equal to one kilogram force per millimeter if the force along a line of one millimeter length, where the force is parallel to the surface but perpendicular to the line is equal to one kilogram force.
## What is the Symbol of Kilogram Force/Millimeter?
The symbol of Kilogram Force/Millimeter is kgf/mm. This means you can also write one Kilogram Force/Millimeter as 1 kgf/mm.
## What Unit of Measure is Dyne/Inch?
Dyne per inch is a unit of measurement for surface tension. Surface tension is considered to be equal to one dyne per inch if the force along a line of one inch length, where the force is parallel to the surface but perpendicular to the line is equal to one dyne.
## What is the Symbol of Dyne/Inch?
The symbol of Dyne/Inch is dyn/in. This means you can also write one Dyne/Inch as 1 dyn/in.
## How to Use Kilogram Force/Millimeter to Dyne/Inch Converter Tool
• As you can see, we have 2 input fields and 2 dropdowns.
• From the first dropdown, select Kilogram Force/Millimeter and in the first input field, enter a value.
• From the second dropdown, select Dyne/Inch.
• Instantly, the tool will convert the value from Kilogram Force/Millimeter to Dyne/Inch and display the result in the second input field.
## Example of Kilogram Force/Millimeter to Dyne/Inch Converter Tool
Kilogram Force/Millimeter
1
Dyne/Inch
24908891
# Kilogram Force/Millimeter to Dyne/Inch Conversion Table
Kilogram Force/Millimeter [kgf/mm]Dyne/Inch [dyn/in]Description
1 Kilogram Force/Millimeter24908891 Dyne/Inch1 Kilogram Force/Millimeter = 24908891 Dyne/Inch
2 Kilogram Force/Millimeter49817782 Dyne/Inch2 Kilogram Force/Millimeter = 49817782 Dyne/Inch
3 Kilogram Force/Millimeter74726673 Dyne/Inch3 Kilogram Force/Millimeter = 74726673 Dyne/Inch
4 Kilogram Force/Millimeter99635564 Dyne/Inch4 Kilogram Force/Millimeter = 99635564 Dyne/Inch
5 Kilogram Force/Millimeter124544455 Dyne/Inch5 Kilogram Force/Millimeter = 124544455 Dyne/Inch
6 Kilogram Force/Millimeter149453346 Dyne/Inch6 Kilogram Force/Millimeter = 149453346 Dyne/Inch
7 Kilogram Force/Millimeter174362237 Dyne/Inch7 Kilogram Force/Millimeter = 174362237 Dyne/Inch
8 Kilogram Force/Millimeter199271128 Dyne/Inch8 Kilogram Force/Millimeter = 199271128 Dyne/Inch
9 Kilogram Force/Millimeter224180019 Dyne/Inch9 Kilogram Force/Millimeter = 224180019 Dyne/Inch
10 Kilogram Force/Millimeter249088910 Dyne/Inch10 Kilogram Force/Millimeter = 249088910 Dyne/Inch
100 Kilogram Force/Millimeter2490889100 Dyne/Inch100 Kilogram Force/Millimeter = 2490889100 Dyne/Inch
1000 Kilogram Force/Millimeter24908891000 Dyne/Inch1000 Kilogram Force/Millimeter = 24908891000 Dyne/Inch
# Kilogram Force/Millimeter to Other Units Conversion Table
ConversionDescription
1 Kilogram Force/Millimeter = 9806.65 Newton/Meter1 Kilogram Force/Millimeter in Newton/Meter is equal to 9806.65
1 Kilogram Force/Millimeter = 98.07 Newton/Centimeter1 Kilogram Force/Millimeter in Newton/Centimeter is equal to 98.07
1 Kilogram Force/Millimeter = 9.81 Newton/Millimeter1 Kilogram Force/Millimeter in Newton/Millimeter is equal to 9.81
1 Kilogram Force/Millimeter = 0.00980665 Newton/Micrometer1 Kilogram Force/Millimeter in Newton/Micrometer is equal to 0.00980665
1 Kilogram Force/Millimeter = 0.00000980665 Newton/Nanometer1 Kilogram Force/Millimeter in Newton/Nanometer is equal to 0.00000980665
1 Kilogram Force/Millimeter = 8967.2 Newton/Yard1 Kilogram Force/Millimeter in Newton/Yard is equal to 8967.2
1 Kilogram Force/Millimeter = 2989.07 Newton/Foot1 Kilogram Force/Millimeter in Newton/Foot is equal to 2989.07
1 Kilogram Force/Millimeter = 249.09 Newton/Inch1 Kilogram Force/Millimeter in Newton/Inch is equal to 249.09
1 Kilogram Force/Millimeter = 9.81 Kilonewton/Meter1 Kilogram Force/Millimeter in Kilonewton/Meter is equal to 9.81
1 Kilogram Force/Millimeter = 0.0980665 Kilonewton/Centimeter1 Kilogram Force/Millimeter in Kilonewton/Centimeter is equal to 0.0980665
1 Kilogram Force/Millimeter = 0.00980665 Kilonewton/Millimeter1 Kilogram Force/Millimeter in Kilonewton/Millimeter is equal to 0.00980665
1 Kilogram Force/Millimeter = 0.00000980665 Kilonewton/Micrometer1 Kilogram Force/Millimeter in Kilonewton/Micrometer is equal to 0.00000980665
1 Kilogram Force/Millimeter = 9.80665e-9 Kilonewton/Nanometer1 Kilogram Force/Millimeter in Kilonewton/Nanometer is equal to 9.80665e-9
1 Kilogram Force/Millimeter = 8.97 Kilonewton/Yard1 Kilogram Force/Millimeter in Kilonewton/Yard is equal to 8.97
1 Kilogram Force/Millimeter = 2.99 Kilonewton/Foot1 Kilogram Force/Millimeter in Kilonewton/Foot is equal to 2.99
1 Kilogram Force/Millimeter = 0.24908891 Kilonewton/Inch1 Kilogram Force/Millimeter in Kilonewton/Inch is equal to 0.24908891
1 Kilogram Force/Millimeter = 9806650 Millinewton/Meter1 Kilogram Force/Millimeter in Millinewton/Meter is equal to 9806650
1 Kilogram Force/Millimeter = 98066.5 Millinewton/Centimeter1 Kilogram Force/Millimeter in Millinewton/Centimeter is equal to 98066.5
1 Kilogram Force/Millimeter = 9806.65 Millinewton/Millimeter1 Kilogram Force/Millimeter in Millinewton/Millimeter is equal to 9806.65
1 Kilogram Force/Millimeter = 9.81 Millinewton/Micrometer1 Kilogram Force/Millimeter in Millinewton/Micrometer is equal to 9.81
1 Kilogram Force/Millimeter = 0.00980665 Millinewton/Nanometer1 Kilogram Force/Millimeter in Millinewton/Nanometer is equal to 0.00980665
1 Kilogram Force/Millimeter = 8967200.76 Millinewton/Yard1 Kilogram Force/Millimeter in Millinewton/Yard is equal to 8967200.76
1 Kilogram Force/Millimeter = 2989066.92 Millinewton/Foot1 Kilogram Force/Millimeter in Millinewton/Foot is equal to 2989066.92
1 Kilogram Force/Millimeter = 249088.91 Millinewton/Inch1 Kilogram Force/Millimeter in Millinewton/Inch is equal to 249088.91
1 Kilogram Force/Millimeter = 9806650000 Micronewton/Meter1 Kilogram Force/Millimeter in Micronewton/Meter is equal to 9806650000
1 Kilogram Force/Millimeter = 98066500 Micronewton/Centimeter1 Kilogram Force/Millimeter in Micronewton/Centimeter is equal to 98066500
1 Kilogram Force/Millimeter = 9806650 Micronewton/Millimeter1 Kilogram Force/Millimeter in Micronewton/Millimeter is equal to 9806650
1 Kilogram Force/Millimeter = 9806.65 Micronewton/Micrometer1 Kilogram Force/Millimeter in Micronewton/Micrometer is equal to 9806.65
1 Kilogram Force/Millimeter = 9.81 Micronewton/Nanometer1 Kilogram Force/Millimeter in Micronewton/Nanometer is equal to 9.81
1 Kilogram Force/Millimeter = 8967200760 Micronewton/Yard1 Kilogram Force/Millimeter in Micronewton/Yard is equal to 8967200760
1 Kilogram Force/Millimeter = 2989066920 Micronewton/Foot1 Kilogram Force/Millimeter in Micronewton/Foot is equal to 2989066920
1 Kilogram Force/Millimeter = 249088910 Micronewton/Inch1 Kilogram Force/Millimeter in Micronewton/Inch is equal to 249088910
1 Kilogram Force/Millimeter = 980665000 Dyne/Meter1 Kilogram Force/Millimeter in Dyne/Meter is equal to 980665000
1 Kilogram Force/Millimeter = 9806650 Dyne/Centimeter1 Kilogram Force/Millimeter in Dyne/Centimeter is equal to 9806650
1 Kilogram Force/Millimeter = 980665 Dyne/Millimeter1 Kilogram Force/Millimeter in Dyne/Millimeter is equal to 980665
1 Kilogram Force/Millimeter = 980.66 Dyne/Micrometer1 Kilogram Force/Millimeter in Dyne/Micrometer is equal to 980.66
1 Kilogram Force/Millimeter = 0.980665 Dyne/Nanometer1 Kilogram Force/Millimeter in Dyne/Nanometer is equal to 0.980665
1 Kilogram Force/Millimeter = 896720076 Dyne/Yard1 Kilogram Force/Millimeter in Dyne/Yard is equal to 896720076
1 Kilogram Force/Millimeter = 298906692 Dyne/Foot1 Kilogram Force/Millimeter in Dyne/Foot is equal to 298906692
1 Kilogram Force/Millimeter = 24908891 Dyne/Inch1 Kilogram Force/Millimeter in Dyne/Inch is equal to 24908891
1 Kilogram Force/Millimeter = 1000000 Gram Force/Meter1 Kilogram Force/Millimeter in Gram Force/Meter is equal to 1000000
1 Kilogram Force/Millimeter = 10000 Gram Force/Centimeter1 Kilogram Force/Millimeter in Gram Force/Centimeter is equal to 10000
1 Kilogram Force/Millimeter = 1000 Gram Force/Millimeter1 Kilogram Force/Millimeter in Gram Force/Millimeter is equal to 1000
1 Kilogram Force/Millimeter = 1 Gram Force/Micrometer1 Kilogram Force/Millimeter in Gram Force/Micrometer is equal to 1
1 Kilogram Force/Millimeter = 0.001 Gram Force/Nanometer1 Kilogram Force/Millimeter in Gram Force/Nanometer is equal to 0.001
1 Kilogram Force/Millimeter = 914400 Gram Force/Yard1 Kilogram Force/Millimeter in Gram Force/Yard is equal to 914400
1 Kilogram Force/Millimeter = 304800 Gram Force/Foot1 Kilogram Force/Millimeter in Gram Force/Foot is equal to 304800
1 Kilogram Force/Millimeter = 25400 Gram Force/Inch1 Kilogram Force/Millimeter in Gram Force/Inch is equal to 25400
1 Kilogram Force/Millimeter = 1000 Kilogram Force/Meter1 Kilogram Force/Millimeter in Kilogram Force/Meter is equal to 1000
1 Kilogram Force/Millimeter = 10 Kilogram Force/Centimeter1 Kilogram Force/Millimeter in Kilogram Force/Centimeter is equal to 10
1 Kilogram Force/Millimeter = 0.001 Kilogram Force/Micrometer1 Kilogram Force/Millimeter in Kilogram Force/Micrometer is equal to 0.001
1 Kilogram Force/Millimeter = 0.000001 Kilogram Force/Nanometer1 Kilogram Force/Millimeter in Kilogram Force/Nanometer is equal to 0.000001
1 Kilogram Force/Millimeter = 914.4 Kilogram Force/Yard1 Kilogram Force/Millimeter in Kilogram Force/Yard is equal to 914.4
1 Kilogram Force/Millimeter = 304.8 Kilogram Force/Foot1 Kilogram Force/Millimeter in Kilogram Force/Foot is equal to 304.8
1 Kilogram Force/Millimeter = 25.4 Kilogram Force/Inch1 Kilogram Force/Millimeter in Kilogram Force/Inch is equal to 25.4
1 Kilogram Force/Millimeter = 1000000 Pond/Meter1 Kilogram Force/Millimeter in Pond/Meter is equal to 1000000
1 Kilogram Force/Millimeter = 10000 Pond/Centimeter1 Kilogram Force/Millimeter in Pond/Centimeter is equal to 10000
1 Kilogram Force/Millimeter = 1000 Pond/Millimeter1 Kilogram Force/Millimeter in Pond/Millimeter is equal to 1000
1 Kilogram Force/Millimeter = 1 Pond/Micrometer1 Kilogram Force/Millimeter in Pond/Micrometer is equal to 1
1 Kilogram Force/Millimeter = 0.001 Pond/Nanometer1 Kilogram Force/Millimeter in Pond/Nanometer is equal to 0.001
1 Kilogram Force/Millimeter = 914400 Pond/Yard1 Kilogram Force/Millimeter in Pond/Yard is equal to 914400
1 Kilogram Force/Millimeter = 304800 Pond/Foot1 Kilogram Force/Millimeter in Pond/Foot is equal to 304800
1 Kilogram Force/Millimeter = 25400 Pond/Inch1 Kilogram Force/Millimeter in Pond/Inch is equal to 25400
1 Kilogram Force/Millimeter = 1000 Kilopond/Meter1 Kilogram Force/Millimeter in Kilopond/Meter is equal to 1000
1 Kilogram Force/Millimeter = 10 Kilopond/Centimeter1 Kilogram Force/Millimeter in Kilopond/Centimeter is equal to 10
1 Kilogram Force/Millimeter = 1 Kilopond/Millimeter1 Kilogram Force/Millimeter in Kilopond/Millimeter is equal to 1
1 Kilogram Force/Millimeter = 0.001 Kilopond/Micrometer1 Kilogram Force/Millimeter in Kilopond/Micrometer is equal to 0.001
1 Kilogram Force/Millimeter = 0.000001 Kilopond/Nanometer1 Kilogram Force/Millimeter in Kilopond/Nanometer is equal to 0.000001
1 Kilogram Force/Millimeter = 914.4 Kilopond/Yard1 Kilogram Force/Millimeter in Kilopond/Yard is equal to 914.4
1 Kilogram Force/Millimeter = 304.8 Kilopond/Foot1 Kilogram Force/Millimeter in Kilopond/Foot is equal to 304.8
1 Kilogram Force/Millimeter = 25.4 Kilopond/Inch1 Kilogram Force/Millimeter in Kilopond/Inch is equal to 25.4
1 Kilogram Force/Millimeter = 2204.62 Pound Force/Meter1 Kilogram Force/Millimeter in Pound Force/Meter is equal to 2204.62
1 Kilogram Force/Millimeter = 22.05 Pound Force/Centimeter1 Kilogram Force/Millimeter in Pound Force/Centimeter is equal to 22.05
1 Kilogram Force/Millimeter = 2.2 Pound Force/Millimeter1 Kilogram Force/Millimeter in Pound Force/Millimeter is equal to 2.2
1 Kilogram Force/Millimeter = 0.0022046226218488 Pound Force/Micrometer1 Kilogram Force/Millimeter in Pound Force/Micrometer is equal to 0.0022046226218488
1 Kilogram Force/Millimeter = 0.0000022046226218488 Pound Force/Nanometer1 Kilogram Force/Millimeter in Pound Force/Nanometer is equal to 0.0000022046226218488
1 Kilogram Force/Millimeter = 2015.91 Pound Force/Yard1 Kilogram Force/Millimeter in Pound Force/Yard is equal to 2015.91
1 Kilogram Force/Millimeter = 671.97 Pound Force/Foot1 Kilogram Force/Millimeter in Pound Force/Foot is equal to 671.97
1 Kilogram Force/Millimeter = 56 Pound Force/Inch1 Kilogram Force/Millimeter in Pound Force/Inch is equal to 56
1 Kilogram Force/Millimeter = 35273.96 Ounce Force/Meter1 Kilogram Force/Millimeter in Ounce Force/Meter is equal to 35273.96
1 Kilogram Force/Millimeter = 352.74 Ounce Force/Centimeter1 Kilogram Force/Millimeter in Ounce Force/Centimeter is equal to 352.74
1 Kilogram Force/Millimeter = 35.27 Ounce Force/Millimeter1 Kilogram Force/Millimeter in Ounce Force/Millimeter is equal to 35.27
1 Kilogram Force/Millimeter = 0.0352739621 Ounce Force/Micrometer1 Kilogram Force/Millimeter in Ounce Force/Micrometer is equal to 0.0352739621
1 Kilogram Force/Millimeter = 0.0000352739621 Ounce Force/Nanometer1 Kilogram Force/Millimeter in Ounce Force/Nanometer is equal to 0.0000352739621
1 Kilogram Force/Millimeter = 32254.51 Ounce Force/Yard1 Kilogram Force/Millimeter in Ounce Force/Yard is equal to 32254.51
1 Kilogram Force/Millimeter = 10751.5 Ounce Force/Foot1 Kilogram Force/Millimeter in Ounce Force/Foot is equal to 10751.5
1 Kilogram Force/Millimeter = 895.96 Ounce Force/Inch1 Kilogram Force/Millimeter in Ounce Force/Inch is equal to 895.96
1 Kilogram Force/Millimeter = 70931.61 Poundal/Meter1 Kilogram Force/Millimeter in Poundal/Meter is equal to 70931.61
1 Kilogram Force/Millimeter = 709.32 Poundal/Centimeter1 Kilogram Force/Millimeter in Poundal/Centimeter is equal to 709.32
1 Kilogram Force/Millimeter = 70.93 Poundal/Millimeter1 Kilogram Force/Millimeter in Poundal/Millimeter is equal to 70.93
1 Kilogram Force/Millimeter = 0.070931611876605 Poundal/Micrometer1 Kilogram Force/Millimeter in Poundal/Micrometer is equal to 0.070931611876605
1 Kilogram Force/Millimeter = 0.000070931611876605 Poundal/Nanometer1 Kilogram Force/Millimeter in Poundal/Nanometer is equal to 0.000070931611876605
1 Kilogram Force/Millimeter = 64859.87 Poundal/Yard1 Kilogram Force/Millimeter in Poundal/Yard is equal to 64859.87
1 Kilogram Force/Millimeter = 21619.96 Poundal/Foot1 Kilogram Force/Millimeter in Poundal/Foot is equal to 21619.96
1 Kilogram Force/Millimeter = 1801.66 Poundal/Inch1 Kilogram Force/Millimeter in Poundal/Inch is equal to 1801.66
1 Kilogram Force/Millimeter = 98066500000 Erg/Square Meter1 Kilogram Force/Millimeter in Erg/Square Meter is equal to 98066500000
1 Kilogram Force/Millimeter = 9806650 Erg/Square Centimeter1 Kilogram Force/Millimeter in Erg/Square Centimeter is equal to 9806650
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# Strategies and Odds at Roulette
## Knowing the different combinations roulette and know how to calculate the odds is the only way to win at roulette. Take a look here.
Each round of the game is unknown and the outcome is never determined by the previous round. After eight consecutive blacks in fact, a black man has the same chance to get out of the red.
It can be argued that even for a long period of time is the red black have the same chance of winning.In fact, they are both winners. First of all we have to reason, because each round of the game has the same chances of winning.
Although the results may vary in the course of the game, we might see out the number “laggard” and quickly doubled our bets, the famous Martingale system, used by most novice players. In order to simplify the calculations, we can arrogate that the roulette wheel has no zeros, but only 18 reds and 18 blacks.
Consider that:
For example, on 1 million revolutions of the roulette in theory you could have 500,000 red and 500,000 blacks.
now, we know that they do not come in turn one at a time, one would expect to see two, three, four or more consecutive of one of the two colors.
According to this analysis we consider 8 consecutive blacks as in the example.
If you were to count all the cases in which should go out eight blacks in a row, followed by a red, we also find, too, the same sequence of eight blacks in a row, followed by one black (9 blacks in a row).
You can also find the same number of consecutive red, making it equal to the total of reds and blacks. If you have a simulator roulette on our main computer, try it.
This means that 8 consecutive blacks, have the same ability to output 8 consectutivi red, with the percentage stops at 50%.
Therefore, if you bet on red, after blacks in a row, or if scommetiamo after 8 consecutive red on black, we would have the same winning percentage (50%). So, no matter how many times in a row bait black or red, because for every probability, for example, 8, 10 or 20 blacks in a row, followed by a red, there will be the same possibility of 8, 10 or 20 blacks of row followed by a black man.
You do not know just which of the two sequences will be released first. These are likely to take in the roulette.
After eight blacks in a row, the chances of winning are still at 50 and 50 if you bet on red or black.
With the number “latecomer” doubles your chances of winning:
It is a known fact that many players are waiting four blacks in a row before playing and then start pointing to double your winnings on red up to four times (or more) if they are at a loss.
In doing so, the players believe they have an advantage over the other on the odds of 4 blacks without betting. In fact, if you use this strategy, there are many possibilities to conclude advantage over the other.
We might be wrong, but let’s look at these numbers:
Starting from a bet, if you double your bet up to four times on red and win, you would win a bark (a token). If you lose pointing up to the fourth episode, you may lose up to 15 chips (1 +2 +4 +8 = 15).All this makes us understand that in 16 attempts lost, you will lose all 15 previous wins.
Now, try the same strategy to double, but now betting on red and black. That is, we do not know what came out earlier, we aim at will on both the red on black and on different tables (all folders).
We know that if we lost we can double our bet, up to 4 times. For a long time we could get the same results as the same color or a table and begin to raise our bet.
The fact is that it is the probability of release of a color, but you can lose up to 4 times in a row on a bet given to 50 and 50%, ie 1/2 x 1/2 x 1/2 x 1 / 2 = 1/16. On average you will win 15 times in a row and lose once every 16 attempts. The ultile net is 0.
The Martingale system is in fact based on the probability of losing infinite times in a row.
Although in theory it is infallible, the Martingale system works positively if the player has a bankroll almost unlimited (for example the Casino in Monte Carlo, Monaco, the red can go up to 39 times in a row), and if the casino does not limit the maximum value of the single bet.
If you run out of money or reached the limit of the house (the casino), you can lose a lot, with no possibility of recovery.
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# W.A.P. C++
## WAP | CPP | Programs in C++ | C++ Solutions
### write a program to display fibonacci series upto n terms
Fibonacci series:The first two Fibonacci numbers are 0 and 1, and each remaining number is the sum of the previous two
e.g.: 0 1 1 2 3 5 8 13 21 ....
A number n is taken from user and the series is displayed upto nth term.
Select To use this code as it is.. select and copy paste this code into code.cpp file :)
1. #include<iostream.h>
2. #include<conio.h>
3. void main()
4. {
5. clrscr();
6. int a=0,b=1,c=0,n;
7. cout<<"Enter the number of terms you wanna see: ";
8. cin>>n;
9. cout<<a<<" "<<b<<" ";
10. for(int i=1;i<=n-2;i++)
11. {
12. c=a+b;
13. a=b;
14. b=c;
15. //Coding by: Snehil Khanor
16. //http://WapCPP.blogspot.com
17. cout<<c<<" ";
18. }
19. getch();
20. }
Saalim's innovations
in FOR statement "i<=n" will come :)
Deepak_Pappu
if i<=n is taken then the no of terms displayed is 17...!
ap
#include
using namespace std;
class fib
{
int i,n,f1,f2,f3;
public:
void get();
void put();
};
void fib::get()
{
cout<<"enter the no.";
cin>>n;
}
void fib::put()
{
f1=0,f2=1;
cout<<f1<<endl<<f2<<endl;
i=0;
while(i<n)
{
f3=f1+f2;
cout<<endl<<f3<<endl;
f1=f2;
f2=f3;
++i;
}
}
int main()
{
fib a1;
a1.get();
a1.put();
}
~
Kaushik Raja
#define fibonacci
#include
void main()
{
int a,b,c;
for(a=0,b=1,c=0;c<10;c=a+b,a=b,b=c)
cout<<c;
}
AM I CORRECT? BUT ITS NOT FOR N TERMS.. O/P S BASED ON VALUE OF C... PLS REPLY FOR ANY CORRECTION
Anonymous
can u specify #fibonacci jus like dat?? sorry if this is silly
Vishwa
use for(int i=0;i<=n-2;i++)
its show perfect series
Anonymous
The C Programs
the programs on this site are so useful for computer science students..........C program to convert given number of days into years,weeks and days
preeya
#include
#include
#include
int main()
{
int a,b,c,n;
a=0; b=1;
clrscr();
cout<<”enter number of terms (n)=”;
cin>>n;
cout<<Fibonacci series is”<<end1;
cout<<a<<setw(4)<<b;
int count=3;
while (count<=n)
{
c=a+b;
cout<<setw(4)<<c;
a=b;
b=c;
count++;
}
getch();
return 0;
}
Anonymous
can u plz post a program of fibonacci series using a another function
void fib(int n)
sameer gupta
#include
#include
void main()
{
clrscr();
{
int n,a=0,b=1,c,i;
cout<<"enter the value of n";
cin>>n;
cout<<"fibonacci series is=";
for(i=0;i<n;i++)
{
if(i<=1)
{
c=i;
}
else
c=a+b;
a=b;
b=c;
}
cout<<c<<",";
}
getch();
}
Anonymous
u r already printing the first two numbers
Jasir
See this one.. little more advanced..
Anonymous
actually whoever the programmer is doesn't know anything related to programming... i am atleast a 100 times better programmer....
Subash.M
#include
#include
void main()
{
int n,i,a=0,b=1,series;
cout<<"enter fibonnaci series";
for(i=0;i<n;i++)
{
if(i<=1)
series=i;
else
{
series=a+b;
a=b;
b=series;
}
cout<<series<<endl;
}
}
Hitesh Kumar
C++ Program to Generate Fibonacci series
Fibonacci Series is in the form of 0, 1, 1, 2, 3, 5, 8, 13, 21,...... To find this series we add two previous terms/digits and get next term/number.
Aswathy K.R
int a=-1,b=1,i,c,n;
cout<"enter limit:";
cin>>n;
for(i=0;i<n;i++)
{
c=a+b;
cout<<c<<"\n";
a=b;
b=c;
}
Venkat Shanthi
Very informative article.Thank you author for posting this kind of article .
http://www.wikitechy.com/view-article/Fibonacci-series-program-in-cpp
Both are really good,
Cheers,
Venkat
## Search
Here you'll find a wide range of programs' solution ranging from beginer level to advanced level.
All programs here are made, compiled and tested by me..and are running absolutely fine.. still if you find any bug in any program do let me know :)
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https://math.stackexchange.com/questions/2346846/in-tg-what-is-the-purpose-of-the-set-axiom
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# In TG, what is the purpose of the "set axiom"
Recently I have been reading about Tarski-Grothendieck set theory, and have been impressed by its short axiomatisation, inclusion of inaccessible cardinals, and capability of supporting category theory without proper classes. http://mizar.org/JFM/pdf/tarski.pdf.
However, I am somewhat confused by its first axiom, which states that everything is a set. It then formulates extensionality by saying that two sets are equal if they have the same members, and Tarski's axiom by saying that every set is a member of a Grothendieck universe. This seems unnecessarily complex to me. Using the definition of a set as a "class that is a member of a class", wouldn't the statements ∀ab (a = b ⇔ ∀c (cacb)) and ∀aU (aU), where U is a Grothendieck universe, prove that if an object has no members, it would be the empty set, and that anything else would of course be a set, as it would both have members and be a member of a Grothendieck universe.
Am I misunderstanding something, or is there some other reason for using the set axiom?
• Unlike NBG, TG does not have a notion of class, so you can't use the term 'class'. Commented Jul 5, 2017 at 4:15
• Formally, there's no point to it, no. In ZF we understand everything in the domain of quantification to be a set, and don't need any special axiom to tell us so. Commented Jul 5, 2017 at 5:05
• I am not at all familiar with TG. But is there a def'n of a Grothendieck universe? I get the impression that sets are the things that are members of Grothendieck universes but that G.-universes are not members of anything. Is that correct? Commented Jul 5, 2017 at 21:34
• @DanielWainfleet Grothendieck universes are sets that contain all members of their members, all powersets of their members, and all subsets of themselves that are of a cardinality corresponding to an ordinal that they contain. They are designed as sets that share some properties with the proper classes of NBG/MK, but they are still sets and may be members of sets. Commented Jul 6, 2017 at 6:08
You're reading it "wrong". They only clarify the context, that all the variables indicate sets, and not some atomic objects like you might want to think about the real numbers as being "numbers" rather than sets.
Note that this is a short paper related to the Mizar system, which is a proof assistant. Since proof assistants are meant to be used by mathematicians who might not subscribe to the notion "everything is a set", it is a good idea to remind them of this fact when it is relevant.
From a set theoretic foundational perspective, yeah, nothing is new by this update. For other people? Well, that might not be the case.
The article you referred to is the second of the two axiomatic articles of the Mizar Mathematical Library. As described on that site:
All other texts undergo verification by Mizar to be correct consequences of those axioms. The Mizar system assists the author of a new text in preparing available terminology and results, verifies the claims of the text and extracts facts and definitions for inclusion into the library.
So the article you referred to (TARSKI) (note that the version you read was human written abstract + latexed version of the human written article for the Mizar system, the latter being able to be understood by the machine) was explicitly written as foundation for a strict computational verification system. As such, you can't simply be
Using the definition of a set as a "class that is a member of a class"
because that is not what it is built on. The axiomatics, as well as Mizar in general, are covered in "Mizar in a Nutshell" by Adam Grabowski et al.
The other one of the two axiomatic articles is literally HIDDEN. The important logical concepts (e.g. $\land$ &, $\lor$ or, $\neg$ not (for predicates) or non (for attributes), $\Rightarrow$ implies, $\Leftrightarrow$ iff, $\forall$ for, $\exists$ ex) are wired directly into the software and grammar of the Mizar language, but explicit modes (what kind of something else something can be, like an object or a set) or explicit predicates (saying that one or several something(s) are in some relation to another or not, like $x=y$ x = y, $x\neq y$ x <> y and $x\in X$ x in X) which basically the foundation to be even able to talk about axioms haven't been declared implicitly in the software, but explicitly in HIDDEN. They just are. And based on that, the axioms in TARSKI could be defined from the programming point of view. Or as it is put in "Mizar in a Nutshell":
[HIDDEN] documents a part of the Mizar axiomatics – it shows how the primitives of set theory are introduced in the Mizar Mathematical Library.
"Mizar in a Nutshell" goes through each of the short axiomatic articles. It states
[TARSKI] defines axiomatic foundations of the Tarski- Grothendieck set theory: extensionality axiom [TARSKI:2] [...]
and goes on with the rest of the article, thereby implicitly stating that TARSKI:1 ("everything is a set") is not directly part of the TG axioms, but simply a part of the Mizar axiomatics. In fact, one could have left objects out of the whole axiomatization and sometimes I wonder if that wouldn't have been better, or just introduce "object" as a different notation of set, but in the end, it basically amounts to the same thing, but improves readability alot, depending on an author wanting to emphasize if a certain "thing" is an object or rather a set (which e.g. can be empty or stand on the right side of an in).
If you are interested in it, the empty set is derived from these axioms here.
First you should look at https://en.wikipedia.org/wiki/Tarski%E2%80%93Grothendieck_set_theory for the axioms in TG. They are written in standard/logic language You can see that there is no set axiom. Second the reference you quoted starts with axiom 2 and a footnote says that axiom 1 (the set axiom which bothers you) has been deleted. You must have read it in another/older paper. This is a translation in machine readable language of the axioms found in Wikipedia. Obviously, if every thing that you can quantify is called "set" , then there is no point in writing/checking this any longer (the predicate set(x) is always true).
• I believe that the paper in the question is the original definition of the TG set theory. It says that it is from 1989. Commented Jul 7, 2017 at 7:13
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# Math
posted by .
What is another way to write the expression 44n? 44÷n?
• Math -
Another way to write 44n would be 44 x n because you are multiplying 44 by n, not dividing them.
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# On measurable functions of two variables
Suppose $f:[0,1]^2\to\mathbb{R}$, $(t,x)\mapsto f(t,x)$, is such that for each $t\in[0,1]$ $f(t,\cdot)$ is Lebesgue measurable on $[0,1]$, and for each $x\in[0,1]$ $f(\cdot,x)$ is continuous everywhere on $[0,1]\ni t$.
1. Does this imply that $f(t,x)$ is measurable on $[0,1]^2$?
2. Does this imply that the function $g(x)=\min\limits_{t\in[0,1]}f(t,x)$ is measurable on $[0,1]$?
-
1. Yes, by continuity in the $t$ variable $f(t,x)=\lim_n f(\lfloor n t\rfloor/n,x)$, which expresses $f$ as the pointwise limit of a sequence of measurable functions.
2. Yes, by continuity in the $t$ variable we have $\min_{t\in[0,1]} f(t,x)=\min_{t\in[0,1]\cap {\mathbb Q}} f(t,x)$, where $\mathbb Q$ means the rational numbers.
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Home
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Problem: Greatest number which can divide 1354 , 1806 , 2762 leaving the same remainder 10 in each case. [ Calculator, Method and examples ]
Solution:
Your problem -> Greatest number which can divide 1354 , 1806 , 2762 leaving the same remainder 10 in each case.
The number divides 1354 and leaves 10 as remainder
:. The number exactly divides 1354 - 10 = 1344
The number divides 1806 and leaves 10 as remainder
:. The number exactly divides 1806 - 10 = 1796
The number divides 2762 and leaves 10 as remainder
:. The number exactly divides 2762 - 10 = 2752
Now, we have to find HCF of 1344, 1796, 2752
Find HCF of 1344,1796,2752
1. Find HCF of (1344,1796)
1
1344|
1 7 9 6
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1 3 4 4
2
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4 5 2
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1 3 4 4
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9 0 4
1
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4 4 0
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4 5 2
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4 4 0
3 6
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1 2
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4 4 0
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3 6
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8 0
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7 2
1
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8
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1 2
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8
2
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4
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8
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8
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0
HCF of (1344, 1796) = 4
2. Now find HCF of 4 and 3^(rd) number 2752
6 8 8
4|
2 7 5 2
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2 4
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3 5
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3 2
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3 2
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3 2
|
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0
HCF of (4, 2752) = 4
:. HCF of given numbers (1344,1796,2752) = 4
:. Required number = HCF of 1344, 1796, 2752 = 4.
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# Pentagon with equal-area Triangles
Many of you have come across a certain type of pentagon:
Let us label a pentagon $ABCDE$. It satisfies that $[ABC]=[BCD]=[CDE]=[DEA]=[EAB]$ where $[XYZ]$ means the area of $\triangle XYZ$.
The question: What is the least amount of obtuse angles the pentagon must have? Prove your claim.
Note by Daniel Liu
6 years, 8 months ago
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I have found a way to construct such a pentagon which has NO obtuse angles: http://i60.tinypic.com/2yocn4w.png
This pentagon has 3 acute angles and 2 reflex angles, none of which are obtuse (an obtuse angle is an angle $\theta$ such that $90 < \theta < 180$).
In the diagram, let $\angle BAE = \angle ABC = x = 30$, $\angle CDE = 2y = arcsin \left(\frac{\sqrt{3}}{4}\right)$, $z = 90+x+y$, $AB = 1$, $AE = BC = m = \frac{sin(x)sin(2y)}{cos^2 (x+y)}$, and $CD = DE = n = \frac{sin(x)}{cos(x+y)}$.
By symmetry we see that $[ABC] = [EAB]$ and $[DEA] = [BCD]$.
Now, we see that $[EAB] = \frac{m*1*sin(x)}{2}$ and, by definition of $z$ and $n$, we have:
$[DEA] = \frac{m*n*sin(z)}{2} = \frac{m*n*sin(90+x+y)}{2} = \frac{m*(n*cos(x+y))}{2} = \frac{m*sin(x)}{2} = [EAB]$
Thus $[DEA] = [EAB]$.
Now by definition of $m$, $n$ and $z$, notice that $m*sin(z) = m*cos(x+y) = \frac{sin(x)sin(2y)}{cos(x+y)} = n*sin(2y)$. Therefore,
$[DEA] = \frac{n*(m*sin(z))}{2} = \frac{n*(n*sin(2y))}{2} = [CDE]$
Therefore, $[ABC] = [BCD] = [CDE] = [DEA] = [EAB]$. Finally we need to show that this pentagon is "anatomically correct" - i.e. that the defined side lengths and angles create a correct pentagon. To do this, we simply need to show that $m*cos(x) + n*sin(y) = \frac{1}{2}$ (so that the horizontal components match up correctly).
$m*cos(x) + n*sin(y) = \frac{sin(x)cos(x)sin(2y) + sin(x)sin(y)cos(x+y)}{cos^2(x+y)}$
$= \frac{\frac{1}{2} * \frac{\sqrt{3}}{2} * \frac{\sqrt{3}}{4} + \frac{1}{2}sin(y)cos(30+y)}{cos^2(30+y)}$
$= \frac{3 + 8sin(y) \left[\frac{\sqrt 3}{2} cos(y) - \frac{1}{2} sin(y) \right]}{16 \left[\frac{\sqrt{3}}{2} cos(y) - \frac{1}{2} sin(y) \right]^2}$
$= \frac{3 + 2\sqrt 3 sin(2y) - 4sin^2(y)}{12cos^2(y) - 8 \sqrt 3 cos(y)sin(y) + 4sin^2(y)}$
$= \frac{3 + \frac{3}{2} - 4sin^2(y)}{12 - 4 \sqrt 3 sin(2y) - 8sin^2(y)}$
$= \frac{\frac{9}{2} - 4sin^2(y)}{9 - 8sin^2(y)} = \frac{1}{2}$
Therefore, this pentagon satisfies the necessary conditions and has no obtuse angles. Therefore the least possible amount of obtuse angles is zero.
- 6 years, 8 months ago
That would work, given that the technical definition of an "obtuse angle" is between 90 and 180 degrees, and that the angle of a vertex in any closed polygon, convex or not, is measured on the inside.
- 6 years, 8 months ago
Ah, I assumed that the pentagon was convex. Nice job finding a concave pentagon with no obtuse angles though.
- 6 years, 8 months ago
Well, the question didn't specify :P Thanks!
- 6 years, 8 months ago
Pentagon
For all those triangles to have equal areas, then all the chords must be parallel to the opposite side. From this, all the angles ${1, 2, 3, 4, 5}$ must be as shown in the figure above. Now, given a vertex, say, $A$, which is the sum of angles ${4, 1, 3}$, for this to be obtuse, then angle $a$ must be $<90$. If this is so, then angles $b$ and $e$ cannot be $<90$. If one of the remaining angles $b, c$ is $<90$, then the other cannot be. Thus, at most, only $2$ vertices can be acute, which means at least $3$ of the triangles must be obtuse.
I'm not sure if this is what is to be proven, but certainly no triangle can have more than one obtuse angle.
- 6 years, 8 months ago
If I may ask, how do you post pictures here?
- 6 years, 8 months ago
I picked this up while working on Brilliant. To post a picture, enter the following Latex coding
!$[...title... ]$ $(...http...address...)$
- 6 years, 8 months ago
Picture
OK I got it! Thanks!!
- 6 years, 8 months ago
Sorry, typoed! I wanted to least amount of obtuse angles the pentagon could have.
- 6 years, 8 months ago
Well, fix it, Daniel. It was a fun problem, once I decided what the problem was. See my motto. Maybe later I'll put up a graphic of such a pentagon with just 3 obtuse angles.
- 6 years, 8 months ago
Huh, I fixed it when I commented. Must have been a glitch or something. Fixed it again.
- 6 years, 8 months ago
Here's the case where such a pentagon has 2 right angle vertices
Right Angles Pentagon
In working this one out, once one vertex was made a right angle, it works out that there is only one degree of freedom to adjust the angle of the other vertex to make it a right angle, so that making any more vertices a right angle was not possible. As per the proof given above.
The figure given here has a lot of interesting proportions.
- 6 years, 8 months ago
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## How much deli meat do I need per person?
As a general rule of thumb, provide 2 to 3 ounces of lunch meat per person.
## How many pounds of deli meat do I need for 25 people?
Tip: Figuring How Much Food to Buy for Parties Cold cuts: generally 2 to 3 ounces of deli meat per serving (25 servings of deli meat would be 3 to 5 pounds of meat ). This is based on what is called a “sandwich slice” rather than shaved. Sandwich slices will go further than shaved as people often overload on shaved.
## How many pounds of cold cuts do I need for 15 people?
Figure on 3 ounces of cold cuts per person. Multiply the number of people you are anticipating as guests by 3 and then divide by 16 to get the number of pounds of cold cuts you need to have for the event. Divide 3 by 16 to determine how many pounds of cold cuts you need per person.
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## How do you ask for deli meat slices?
You Can Ask to See Sample Slices If you want your deli meat and cheese sliced thick, thin or paper-thin, make sure you specify that when you order. Then, ask to see a piece before they slice the whole pound. The slicer can be adjusted so every piece is perfect.
## How many slices is 2 oz of deli meat?
I usually just eat one slice and estimate that it’s about 2 ounces (the slices are ovals, probably 6 inches long, 5 inches wide in the middle).
## How do I order cold cuts?
To order deli meat, tell the counter person how thick you’d like the slice. Next, specify how much you would like based on weight. Deli meat does not keep very long once cut. Store deli meats in the refrigerator, in airtight plastic bags.
## How many slices of Oscar Mayer deli Fresh turkey is 2 oz?
There are 50 calories in 1 serving, 6 slices (2 oz) of Oscar Mayer Deli Fresh Oven Roasted Turkey Breast, 98% Fat-Free, fully cooked.
## How much cheese goes on a sandwich?
cheese as people often overload the sandwich with shredded cheese. So, How much Cheese for sandwiches per person – you should have 1.25 ounces of cheese slices per sandwich per person. How Many ounces is one slice of cheese? One slice of cheese is 0.75 oz.
## How much meat goes on a deli sandwich?
Wondering how much deli meat you should order on your next visit? Here’s a good rule of thumb: 1 pound of deli meat, sliced sandwich style, makes about 5 – 6 sandwiches. Cheese slices should be ordered thinner than deli meat, so you’ll have more cheese slices to layer on. Order away!
You might be interested: FAQ: How Do You Know If Deli Meat Is Nitrate Free?
## How much meat do you need for a charcuterie?
Plan for about 1-2 ounces of meat per person. At the deli counter, ask for your meat selections to be sliced thin (at a 1-2 thickness) so they’re easy to layer.
## What is chipped lunch meat?
Chipped chopped ham or chipped ham is a processed ham luncheon meat made from chopped ham. By chipping or shaving the meat loaf against a commercial meat slicer blade, the resultant thinly sliced product has a different texture and flavor compared to thickly sliced ham.
## How thick can you cut deli meat?
Employees should be thoroughly trained in the operation of deli slicers and should carefully follow manufacturer’s recommendations. Slices meat from shaved (paper-thin) to more than ¼” thick.
## How do deli scales work?
Label Printing Scales, or Deli/Meat scales, are devices that allow you to weigh an item away from the POS and print a specialized price label. These labels are often called “price embedded”, because the price of the weighed package is embedded in the barcode.
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Abutment Design Example to BD 30
The proposed deck consists of 11No. Y4 prestressed concrete beams and concrete deck slab as
shown.
The ground investigation report shows suitable founding strata about 9.5m below the proposed road
level. Test results show the founding strata to be a cohesionless soil having an angle of shearing
resistance (φ) = 30
o
and a safe bearing capacity of 400kN/m
2
.
Backfill material will be Class 6N with an effective angle of internal friction (ϕ') = 35
o
and density (γ)
= 19kN/m
3
.
Critical Reaction Under One
Beam
Total Reaction on Each
Abutment
Nominal
Reaction
(kN)
Ultimate
Reaction
(kN)
Nominal
Reaction
(kN)
Ultimate
Reaction
(kN)
Concrete Deck 180 230 1900 2400
Surfacing 30 60 320 600
HA udl+kel 160 265 1140 1880
45 units HB 350 500 1940 2770
A grillage analysis gave the following reactions for the various load cases:
191kN/m
HA live Load on Deck = 1140 / 11.6 = 98kN/m
HB live Load on Deck = 1940 / 11.6 =
167kN/m
From BS 5400 Part 2 Figures 7 and 8 the minimum and maximum shade air temperatures are -19 and
+37
o
C respectively.
For a Group 4 type strucutre (see fig. 9) the corresponding minimum and maximum effective bridge
Design the fixed and free end cantilever abutments to the 20m span deck shown to carry HA and 45
units of HB loading. Analyse the abutments using a unit strip method. The bridge site is located south
east of Oxford (to establish the range of shade air temperatures).
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temperatures are -11 and +36
o
C from tables 10 and 11.
Hence the temperature range = 11 + 36 = 47
o
C.
From Clause 5.4.6 the range of movement at the free end of the 20m span deck = 47 x 12 x 10
-6
x
20 x 10
3
= 11.3mm.
The ultimate thermal movement in the deck will be ± [(11.3 / 2) γf
3
γf
L
] = ±[11.3 x 1.1 x 1.3 /2] =
± 8mm.
Option 1 - Elastomeric Bearing:
With a maximum ultimate reaction = 230 + 60 + 500 = 790kN then a suitable elastomeric bearing
would be Ekspan's Elastomeric Pad Bearing EKR35:
Shear Deflection = 13.3mm
Shear Stiffness = 12.14kN/mm
Bearing Thickness = 19mm
Note: the required shear deflection (8mm) should be limited to between 30% to 50% of the thickness
of the bearing. The figure quoted in the catalogue for the maximum shear deflection is 70% of the
thickness.
A tolerance is also required for setting the bearing if the ambient temperature is not at the mid range
temperature. The design shade air temperature range will be -19 to +37
o
C which would require the
bearings to be installed at a shade air temperature of [(37+19)/2 -19] = 9
o
C to achieve the ± 8mm
movement.
If the bearings are set at a maximum shade air temperature of 16
o
C then, by proportion the deck will
expand 8x(37-16)/[(37+19)/2] = 6mm and contract 8x(16+19)/[(37+19)/2] = 10mm.
Let us assume that this maximum shade air temperature of 16
o
C for fixing the bearings is specified in
the Contract and design the abutments accordingly.
Horizontal load at bearing for 10mm contraction = 12.14 x 10 = 121kN.
This is an ultimate load hence the nominal horizontal load = 121 / 1.1 / 1.3 = 85kN at each bearing.
Total horizontal load on each abutment = 11 x 85 = 935 kN 935 / 11.6 =
81kN/m.
Alternatively using BS 5400 Part 9.1 Clause 5.14.2.6:
H = AGδ
r
/t
q
Using the Ekspan bearing EKR35
Area = 610 x 420 = 256200mm
2
Nominl hardness = 60 IRHD
Bearing Thickness = 19mm
Shear modulus G from Table 8 = 0.9N/mm
2
H = 256200 x 0.9 x 10
-3
x 10 / 19 = 121kN
This correllates with the value obtained above using the shear stiffness from the manufacturer's data
sheet.
Option 2 - Sliding Bearing:
With a maximum ultimate reaction of 790kN and longitudinal movement of ± 8mm then a suitable
bearing from the Ekspan EA Series would be /80/210/25/25:
Base Plate A dimension = 210mm
Base Plate B dimension = 365mm
Movement ± X = 12.5mm
BS 5400 Part 2 - Clause 5.4.7.3:
Average nominal dead load reaction = (1900 + 320) / 11 = 2220 / 11 = 200kN
Contact pressure under base plate = 200000 / (210 x 365) = 3N/mm
2
As the mating surface between the stainless steel and PTFE is smaller than the base plate then the
pressure between the sliding faces will be in the order of 5N/mm
2
.
From Table3 of BS 5400 Part 9.1 the Coefficient of friction = 0.08 for a bearing stress of 5N/mm
2
Hence total horizontal load on each abutment when the deck expands or contracts = 2220 x 0.08 =
180kN 180 / 11.6 =
16kN/m.
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Traction and Braking Load - BS 5400 Part 2 Clause 6.10:
Nominal Load for HA = 8kN/m x 20m + 250kN = 410kN
Nominal Load for HB = 25% of 45units x 10kN x 4axles = 450kN
450 > 410kN hence HB braking is critical.
Braking load on 1m width of abutment = 450 / 11.6 =
39kN/m.
When this load is applied on the deck it will act on the fixed abutment only.
Skidding Load - BS 5400 Part 2 Clause 6.11:
300 < 450kN hence braking load is critical in the longitudinal direction.
When this load is applied on the deck it will act on the fixed abutment only.
Backfill
For Stability calculations use active
earth pressures = K
a
γ h
K
a
for Class 6N material = (1-Sin35) /
(1+Sin35) = 0.27
Density of Class 6N material =
19kN/m
3
Active Pressure at depth h = 0.27 x 19
x h = 5.13h kN/m
2
Hence Fb = 5.13h
2
/2 = 2.57h
2
kN/m
Surcharge - BS 5400 Part 2 Clause 5.8.2:
10 kN/m
2
2
Assume a surchage loading for the compaction plant to be equivalent to 30 units of HB
Hence Compaction Plant surcharge =
12 kN/m
2
.
For surcharge of w kN/m
2
:
Fs = K
a
w h = 0.27wh kN/m
Backfill +
Construction
surcharge
Backfill + HA
surcharge +
contraction
Backfill + HA
surcharge +
Braking
behind
abutment +
Backfill + HB
surcharge +
Backfill + HA
surcharge +
deck
Fixed
Abutment
Only
Backfill + HA
surcharge +
deck +
Braking on
deck
1) Stability Check
Initial Sizing for Base Dimensions
There are a number of publications that will give guidance on base sizes for free standing cantilever
walls, Reynolds's Reinforced Concrete Designer's Handbook being one such book.
Alternatively a simple spreadsheet will achieve a result by trial and error.
CASE 1 - Fixed Abutment
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Weight
Lever Arm Moment
Stem 163 1.6 261
Base 160 3.2 512
Backfill 531 4.25 2257
Surcharge 52 4.25 221
=
906 = 3251
Density of reinforced concrete = 25kN/m
3
.
Weight of wall stem = 1.0 x 6.5 x 25 = 163kN/m
Weight of base = 6.4 x 1.0 x 25 = 160kN/m
Weight of backfill = 4.3 x 6.5 x 19 = 531kN/m
Weight of surcharge = 4.3 x 12 = 52kN/m
Backfill Force Fb = 0.27 x 19 x 7.5
2
/ 2 = 144kN/m
Surcharge Force Fs = 0.27 x 12 x 7.5 = 24 kN/m
Restoring Effects:
F
Lever Arm Moment
Backfill 144 2.5 361
Surcharge 24 3.75 91
=
168 = 452
Overturning Effects:
Factor of Safety Against Overturning = 3251 / 452 = 7.2 > 2.0 OK.
For sliding effects:
Active Force = Fb + Fs = 168kN/m
Frictional force on underside of base resisting movement = W tan(φ) = 906 x tan(30
o
) = 523kN/m
Factor of Safety Against Sliding = 523 / 168 = 3.1 > 2.0 OK.
Bearing Pressure:
Check bearing pressure at toe and heel of base slab = (P / A) ± (P x e / Z) where P x e is the moment
about the centre of the base.
P = 906kN/m
A = 6.4m
2
/m
Z = 6.4
2
/ 6 = 6.827m
3
/m
Nett moment = 3251 - 452 = 2799kNm/m
Eccentricity (e) of P about centre-line of base = 3.2 - (2799 / 906) = 0.111m
Pressure under base = (906 / 6.4) ± (906 x 0.111 / 6.827)
Pressure under toe = 142 + 15 = 157kN/m
2
< 400kN/m
2
OK.
Pressure under heel = 142 - 15 = 127kN/m
2
Hence the abutment will be stable for Case 1.
F of S
Overturning
F of S
Sliding
Bearing
Pressure at Toe
Bearing
Pressure at Heel
Case 1 7.16 3.09 156 127
Case 2 2.87 2.13 386 5
Case 2a 4.31 2.64 315 76
Case 3 3.43 2.43 351 39
Case 4 4.48 2.63 322 83
Case 5 5.22 3.17 362 81
Case 6 3.80 2.62 378 43
Analysing the fixed abutment with Load Cases 1 to 6 and the free abutment with Load Cases 1 to 5
using a simple spreadsheet the following results were obtained:
Fixed Abutment:
F of S
Overturning
F of S
Sliding
Bearing
Pressure at Toe
Bearing
Pressure at Heel
Free Abutment:
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Case 1 7.15 3.09 168 120
Case 2 2.91 2.14 388 7
Case 2a 4.33 2.64 318 78
Case 3 3.46 2.44 354 42
Case 4 4.50 2.64 325 84
Case 5 5.22 3.16 365 82
It can be seen that the use of elastomeric bearings (Case 2) will govern the critical design load cases
on the abutments. We shall assume that there are no specific requirements for using elastomeric
bearings and design the abutments for the lesser load effects by using sliding bearings.
2) Wall and Base Design
Loads on the back of the wall are calculated using 'at rest' earth pressures. Serviceability and
Ultimate load effects need to be calculated for the load cases 1 to 6 shown above. Again, these are
best carried out using a simple spreadsheet.
Using the Fixed Abutment Load Case 1 again as an example of the calculations:
Wall Design
K
o
= 1 - Sin(ϕ') = 1 - Sin(35
o
) = 0.426
γ
fL
for horizontal loads due to surcharge and backfill from BS 5400 Part 2 Clause 5.8.1.2:
Serviceability = 1.0
Ultimate = 1.5
γ
f3
= 1.0 for serviceability and 1.1 for ultimate (from BS 5400 Part 4 Clauses 4.2.2 and 4.2.3)
Backfill Force Fb on the rear of the wall = 0.426 x 19 x 6.5
2
/ 2 = 171kN/m
Surcharge Force Fs on the rear of the wall = 0.426 x 12 x 6.5 = 33kN/m
At the base of the Wall:
Serviceability moment = (171 x 6.5 / 3) + (33 x 6.5 / 2) = 371 + 107 =
478kNm/m
Ultimate moment = 1.1 x 1.5 x 478 = 789kNm/m
Ultimate shear = 1.1 x 1.5 x (171 + 33) =
337kN/m
Moment
Moment
SLS Live
Moment
ULS
Shear
ULS
Case 1 371 108 790 337
Case 2a 829 258 1771 566
Case 3 829 486 2097 596
Case 4 829 308 1877 602
Case 5 829 154 1622 543
Case 6 829 408 1985 599
Analysing the fixed abutment with Load Cases 1 to 6 and the free abutment with Load Cases 1 to 5
using a simple spreadsheet the following results were obtained for the design moments and shear at
the base of the wall:
Fixed Abutment:
Moment
Moment
SLS Live
Moment
ULS
Shear
ULS
Case 1 394 112 835 350
Case 2a 868 265 1846 581
Case 3 868 495 2175 612
Case 4 868 318 1956 619
Case 5 868 159 1694 559
Free Abutment:
Design for critical moments and shear in Free
Abutment:
Concrete to BS 8500:2006
Use strength class C32/40 with water-cement ratio 0.5 and minimum cement content of 340kg/m
3
for
exposure condition XD2.
Nominal cover to reinforcement = 60mm (45mm minimum cover plus a tolerance
c
of 15mm).
Reinforcement to BS 4449:2005 Grade B500B: f
y
= 500N/mm
2
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Reinforced concrete walls are designed to BS 5400
Part 4 Clause 5.6.
Check classification to clause 5.6.1.1:
Ultimate axial load in wall from deck reactions = 2400
+ 600 + 2770 = 5770 kN
0.1f
cu
A
c
= 0.1 x 40 x 10
3
x 11.6 x 1 = 46400 kN >
5770 design as a slab in accordance with clause 5.4
Bending
BS 5400 Part 4 Clause 5.4.2 for reisitance moments in slabs design to clause 5.3.2.3:
z = {1 - [ 1.1f
y
A
s
) / (f
cu
bd) ]} d
Use B40 @ 150 c/c:
A
s
= 8378mm
2
/m, d = 1000 - 60 - 20 = 920mm
z = {1 - [ 1.1 x 500 x 8378) / (40 x 1000 x 920) ]} d = 0.875d < 0.95d OK
Mu = (0.87f
y
)A
s
z = 0.87 x 500 x 8378 x 0.875 x 920 x 10
-6
= 2934kNm/m > 2175kNn/m OK
Carrying out the crack control calculation to Clause 5.8.8.2 gives a crack width of 0.2mm < 0.25mm.
Also the steel reinforcement and concrete stresses meet the limitations required in clause 4.1.1.3
serviceability requirements are satisfied.
Shear
Shear requirements are designed to BS 5400 clause 5.4.4:
v = V / (bd) = 619 x 10
3
/ (1000 x 920) = 0.673 N/mm
2
No shear reinforcement is required when v < ξ
s
v
c
ξ
s
= (500/d)
1/4
= (500 / 920)
1/4
= 0.86
v
c
= (0.27/γ
m
)(100A
s
/b
w
d)
1/3
(f
cu
)
1/3
= (0.27 / 1.25) x ({100 x 8378} / {1000 x 920})
1/3
x (40)
1/3
=
0.72
ξ
s
v
c
= 0.86 x 0.72 = 0.62 N/mm
s
< 0.673 hence shear reinforcement should be provided, however
check shear at distance H/8 (8.63 / 8 = 1.079m) up the wall.
ULS shear at Section 7H/8 for load case 4 = 487 kN
v = V / (bd) = 487 x 10
3
/ (1000 x 920) = 0.53 N/mm
2
< 0.62
Hence height requiring strengthening = 1.073 x (0.673 - 0.62) / (0.673 - 0.53) = 0.4m < d.
Provide a 500 x 500 splay at the base of the wall with B32 @ 150c/c bars in sloping face.
Early Thermal Cracking
Considering the effects of casting the wall stem onto the base slab by complying with the early
thermal cracking of concrete to BD 28 then B16 horizontal lacer bars @ 150 c/c will be required in
both faces in the bottom half of the wall.
Minimum area of secondary reinforcement to Clause 5.8.4.2 = 0.12% of b
a
d = 0.0012 x 1000 x 920 =
1104 mm
2
/m (use B16 @ 150c/c - A
s
= 1340mm
2
/m)
Base Design
Maximum bending and shear effects in the base slab will occur at sections near the front and back of
the wall. Different load factors are used for serviceability and ultimate limit states so the calculations
need to be carried out for each limit state using 'at rest pressures'
Using the Fixed Abutment Load Case 1 again as an example of the calculations:
CASE 1 - Fixed Abutment Serviceability Limit State
γ
fL
= 1.0 γ
f3
= 1.0
Weight of wall stem = 1.0 x 6.5 x 25 x 1.0 =
163kN/m
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Weight
Lever Arm Moment
Stem 163 1.6 261
Base 160 3.2 512
Backfill 531 4.25 2257
Surcharge 52 4.25 221
=
906 = 3251
Weight of base = 6.4 x 1.0 x 25 x 1.0 = 160kN/m
Weight of backfill = 4.3 x 6.5 x 19 x 1.0 = 531kN/m
Weight of surcharge = 4.3 x 12 x 1.0 = 52kN/m
B/fill Force Fb = 0.426 x 19 x 7.5
2
x 1.0 / 2 =
228kN/m
Surcharge Force Fs = 0.426 x 12 x 7.5 x 1.0 = 38
kN/m
Restoring Effects:
F
Lever Arm Moment
Backfill 228 2.5 570
Surcharge 38 3.75 143
=
266 = 713
Overturning Effects:
Bearing Pressure at toe and heel of base slab = (P / A) ± (P x e / Z)
P = 906kN/m
A = 6.4m
2
/m
Z = 6.4
2
/ 6 = 6.827m
3
/m
Nett moment = 3251 - 713 = 2538kNm/m
Eccentricity (e) of P about centre-line of base = 3.2 - (2538 / 906) = 0.399m
Pressure under base = (906 / 6.4) ± (906 x 0.399 / 6.827)
Pressure under toe = 142 + 53 = 195kN/m
2
Pressure under heel = 142 - 53 = 89kN/m
2
Pressure at front face of wall = 89 + {(195 - 89) x 5.3 / 6.4} = 177kN/m
2
Pressure at rear face of wall = 89 + {(195 - 89) x 4.3 / 6.4} = 160kN/m
2
SLS Moment at a-a = (177 x 1.1
2
/ 2) + ([195 - 177]
x 1.1
2
/ 3) - (25 x 1.0 x 1.1
2
/ 2) = 99kNm/m
(tension in bottom face).
SLS Moment at b-b = (89 x 4.3
2
/ 2) + ([160 - 89] x
4.3
2
/ 6) - (25 x 1.0 x 4.3
2
/ 2) - (531 x 4.3 / 2) - (52
x 4.3 / 2) = -443kNm/m (tension in top face).
CASE 1 - Fixed Abutment Ultimate Limit State
γ
fL
for concrete = 1.15
γ
fL
for fill and surcharge(vetical) = 1.2
γ
fL
for fill and surcharge(horizontal) = 1.5
Weight of wall stem = 1.0 x 6.5 x 25 x 1.15 =
187kN/m
Weight of base = 6.4 x 1.0 x 25 x 1.15 = 184kN/m
Weight of backfill = 4.3 x 6.5 x 19 x 1.2 = 637kN/m
Weight of surcharge = 4.3 x 12 x 1.2 = 62kN/m
Backfill Force Fb = 0.426 x 19 x 7.5
2
x 1.5 / 2 =
341kN/m
Surcharge Force Fs = 0.426 x 12 x 7.5 x 1.5 = 58
kN/m
Restoring Effects:
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Weight
Lever Arm Moment
Stem 187 1.6 299
Base 184 3.2 589
Backfill 637 4.25 2707
Surcharge 62 4.25 264
=
1070 = 3859
F
Lever Arm Moment
Backfill 341 2.5 853
Surcharge 58 3.75 218
=
399 = 1071
Overturning Effects:
Bearing Pressure at toe and heel of base slab = (P / A) ± (P x e / Z)
P = 1070kN/m
A = 6.4m
2
/m
Z = 6.4
2
/ 6 = 6.827m
3
/m
Nett moment = 3859 - 1071 = 2788kNm/m
Eccentricity (e) of P about centre-line of base = 3.2 - (2788 / 1070) = 0.594m
Pressure under base = (1070 / 6.4) ± (1070 x 0.594 / 6.827)
Pressure under toe = 167 + 93 = 260kN/m
2
Pressure under heel = 167 - 93 = 74kN/m
2
Pressure at front face of wall = 74 + {(260 - 74) x 5.3 / 6.4} = 228kN/m
2
Pressure at rear face of wall = 74 + {(260 - 74) x 4.3 / 6.4} = 199kN/m
2
γ
f3
= 1.1
ULS Shear at a-a = 1.1 x {[(260 + 228) x 1.1 / 2] -
(1.15 x 1.1 x 25)} = 260kN/m
ULS Shear at b-b = 1.1 x {[(199 + 74) x 4.3 / 2] -
(1.15 x 4.3 x 25) - 637 - 62} = 259kN/m
ULS Moment at a-a = 1.1 x {(228 x 1.1
2
/ 2) + ([260
- 228] x 1.1
2
/ 3) - (1.15 x 25 x 1.0 x 1.1
2
/ 2)} =
148kNm/m (tension in bottom face).
SLS Moment at b-b = 1.1 x {(74 x 4.3
2
/ 2) + ([199 -
74] x 4.3
2
/ 6) - (1.15 x 25 x 1.0 x 4.3
2
/ 2) - (637 x
4.3 / 2) - (62 x 4.3 / 2)} = -769kNm/m (tension in
top face).
Section a-a Section b-b
ULS
Shear
SLS
Moment
ULS
Moment
ULS
Shear
SLS
Moment
ULS
Moment
Case 1 261 99
147 259 447 768
Case 2a 528 205
302 458 980 1596
Case 3 593 235
340 553 1178 1834
Case 4 550 208 314 495 1003 1700
Case 5 610 241
348 327 853 1402
Case 6 637 255
365 470 1098 1717
Analysing the fixed abutment with Load Cases 1 to 6 and the free abutment with Load Cases 1 to 5
using a simple spreadsheet the following results were obtained:
Fixed Abutment Base:
Section a-a Section b-b
ULS
Shear
SLS
Moment
ULS
Moment
ULS
Shear
SLS
Moment
ULS
Moment
Free Abutment Base:
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Case 1 267 101 151 266 475 816
Case 2a 534 207 305 466 1029 1678
Case 3 598 236
342 559 1233 1922
Case 4 557 211 317 504 1055 1786
Case 5 616 243
351 335 901 1480
Design for shear and bending effects at sections a-a and b-b for the Free Abutment:
Bending
BS 5400 Part 4 Clause 5.7.3 design as a slab for reisitance moments to clause 5.3.2.3:
z = {1 - [ 1.1f
y
A
s
) / (f
cu
bd) ]} d
Use B32 @ 150 c/c:
A
s
= 5362mm
2
/m, d = 1000 - 60 - 16 = 924mm
z = {1 - [ 1.1 x 500 x 5362) / (40 x 1000 x 924) ]} d = 0.92d < 0.95d OK
Mu = (0.87f
y
)A
s
z = 0.87 x 500 x 5362 x 0.92 x 924 x 10
-6
= 1983kNm/m > 1922kNm/m OK
(1983kNm/m also > 1834kNm/m B32 @ 150 c/c suitable for fixed abutment.
For the Serviceability check for Case 3 an approximation of the dead load moment can be obtained by
for Case 3 as 723kNm, thus the live load moment = 1233 - 723 = 510kNm.
Carrying out the crack control calculation to Clause 5.8.8.2 gives a crack width of 0.27mm > 0.25mm
Fail.
This could be corrected by reducing the bar spacing, but increase the bar size to B40@150 c/c as this
is required to avoid the use of links (see below).
Using B40@150c/c the crack control calculation gives a crack width of 0.17mm < 0.25mm OK.
Also the steel reinforcement and concrete stresses meet the limitations required in clause 4.1.1.3
serviceability requirements are satisfied.
Shear
Shear on Toe - Use Fixed Abutment Load Case 6:
By inspection B32@150c/c will be adequate for the
bending effects in the toe (M
uls
= 365kNm <
1983kNm)
Shear requirements are designed to BS 5400 clause
5.7.3.2(a) checking shear at d away from the front
face of the wall to clause 5.4.4.1:
ULS Shear on toe = 1.1 x {(620 + 599) x 0.5 x 0.176
- 1.15 x 1 x 0.176 x 25} = 112kN
v = V / (bd) = 112 x 10
3
/ (1000 x 924) = 0.121 N/mm
2
No shear reinforcement is required when v < ξ
s
v
c
Reinforcement in tension = B32 @ 150 c/c
ξ
s
= (500/d)
1/4
= (500 / 924)
1/4
= 0.86
v
c
= (0.27/γ
m
)(100A
s
/b
w
d)
1/3
(f
cu
)
1/3
= (0.27 / 1.25) x ({100 x 5362} / {1000 x 924})
1/3
x (40)
1/3
=
0.62
ξ
s
v
c
= 0.86 x 0.62 = 0.53 N/mm
s
> 0.121N/mm
s
OK
Shear on Heel - Use Free Abutment Load Case 3:
Shear requirements are designed at the back face of
the wall to clause 5.4.4.1:
Length of heel = (6.5 - 1.1 - 1.0) = 4.4m
ULS Shear on heel = 1.1 x {348 x 0.5 x (5.185 - 2.1)
- 1.15 x 1 x 4.4 x 25 - 1.2 x 4.4 x (8.63 x 19 + 10)}
= 559kN
Using B32@150 c/c then:
v = V / (bd) = 559 x 10
3
/ (1000 x 924) = 0.605 N/mm
2
No shear reinforcement is required when v < ξ
s
v
c
Bridge ABUTMENT DESIGN EXAMPLE http://www.childs-ceng.demon.co.uk/tutorial/abutex.html
9 of 11 6/15/2013 6:41 P
M
ξ
s
= (500/d)
1/4
= (500 / 924)
1/4
= 0.86
v
c
= (0.27/γ
m
)(100A
s
/b
w
d)
1/3
(f
cu
)
1/3
= (0.27 / 1.25) x ({100 x 5362} / {1000 x 924})
1/3
x (40)
1/3
=
0.62
ξ
s
v
c
= 0.86 x 0.62 = 0.53 N/mm
s
< 0.605N/mm
s
Fail
Rather than provide shear reinforcement try increasing bars to B40 @ 150 c/c (also required for
crack control as shown above).
v
c
= (0.27/γ
m
)(100A
s
/b
w
d)
1/3
(f
cu
)
1/3
= (0.27 / 1.25) x ({100 x 8378} / {1000 x 920})
1/3
x (40)
1/3
=
0.716
ξ
s
v
c
= 0.86 x 0.716 = 0.616 N/mm
s
> 0.605N/mm
s
OK
Early Thermal Cracking
Considering the effects of casting the base slab onto the blinding concrete by complying with the early
thermal cracking of concrete to BD 28 then B16 distribution bars @ 250 c/c will be required.
Minimum area of main reinforcement to Clause 5.8.4.1 = 0.15% of b
a
d = 0.0015 x 1000 x 924 =
1386 mm
2
/m (use B20 @ 200c/c - A
s
= 1570mm
2
/m).
HB braking load to BS 5400 clause
6.10 = 25% x 45units x 10kN on each
axle = 112.5kN per axle.
Assume a 45
o
dispersal to the curtain
wall and a maximum dispersal of the
width of the abutment (11.6m) then:
1st axle load on back of abutment =
112.5 / 3.0 = 37.5kN/m
2nd axle load on back of abutment =
112.5 / 6.6 = 17.0kN/m
3rd & 4th axle loads on back of
abutment = 2 x 112.5 / 11.6 =
19.4kN/m
Local Effects
Curtain Wall
This wall is designed to be cast onto the top of the abutment after the deck has been built. Loading
will be applied from the backfill, surcharge and braking loads on top of the wall.
Maximum load on back of abutment = 37.5 + 17.0 + 19.4 = = 73.9kN/m
Bending and Shear at Base of 3m High Curtain Wall
Horizontal load due to HB surcharge = 0.426 x 20 x
3.0 = 25.6 kN/m
Horizontal load due to backfill = 0.426 x 19 x 3.0
2
/ 2
= 36.4 kN/m
SLS Moment = (73.9 x 3.0) + (25.6 x 1.5) + (36.4 x
1.0) = 297 kNm/m (36 dead + 261 live)
ULS Moment = 1.1 x {(1.1 x 73.9 x 3.0) + (1.5 x
25.6 x 1.5) + (1.5 x 36.4 x 1.0)} = 392 kNm/m
ULS Shear = 1.1 x {(1.1 x 73.9) + (1.5 x 25.6) +
(1.5 x 36.4)} = 192kN/m
400 thick curtain wall with B32 @ 150 c/c :
M
ult
= 584 kNm/m > 392 kNm/m OK
SLS Moment produces crack width of 0.21mm < 0.25 OK
ξ
s
v
c
= 0.97 N/mm
2
> v = 0.59 N/mm
2
Shear OK
Back to Abutment Tutorial | Back to Tutorial Index
Last Updated : 28/02/11
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https://holooly.com/solutions-v2-1/two-pipe-flow-with-pump-as-well-as-frictional-and-form-losses-given-two-water-reservoirs-ha-5m-and-hb-15m-connected-by-two-parallel-wrought-ir5m-and-hon-pipes-d1-0-05m-%CE%B51-0-04mm-l1/
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## Q. 4.4
Two-pipe Flow with Pump as well as Frictional and Form Losses
Given two water reservoirs ($\rm H_A$ = 5m and $\rm H_B$ =15m) connected by two parallel wrought-iron pipes ($\rm D_1$ = 0.05m, $ε_1$ = 0.04mm; $\rm L_1$ ≈ 15m, and $\rm D_2$ = 0.10m, $ε_2$ = 0.06mm; $\rm L_2$ ≈ 15m with pump (10 kW; η = 0.8) to convey the water (20ºC; ρ = 988 kg/m³ and μ = 10$^{−3}$ kg/(m·s)). From reservoir A to reservoir B. Take $\rm K_{bend} ≈ 0.3$ and find $\rm Q_1$ and $\rm Q_2$ !
$\rm h_{P,T}=\frac{P_{P,T}}{\dot mg}$ (4.7b,c)
$\rm f^{-1/2}\approx-1.8\ {log}\bigg[{\frac{6.9}{{Re_{D}}}}+\left\lgroup{\frac{\varepsilon /{D}}{3.7}}\right\rgroup ^{1.11}\bigg]$ (4.5e)
Assumptions Sketch • Steady uniform flow • $\rm H_A$ and $\rm H_B$ are constant • Minor losses only in parallel pipes • Turbulent flow, i.e., check that $\rm Re=\frac{vD}{\nu}>4000$ for all pipes Concepts: • Extended Bernoulli with $\rm\mathrm{h}_{\mathrm{pump}}=\frac{\dot{W}_{\mathrm{electr}}\cdot{\eta }}{\rho{\mathrm{~g~}}Q}~(\mathrm{see}\,\mathrm{Eq}.\,(4.7\mathrm{b}))$ • $\rm h_L = h_{L,1} + h_{L,2}$ because Δp is the same for each pipe • Total losses $\rm h_L = h_{friction} + h_{form}$ • Friction factors from Eq. (4.5e) or Moody chart • Flow rate $\rm Q = Q_1 + Q_2$ • Solve system of n equations with n unknowns simultaneously. Alternatively, guess $\rm f_i$ and use Moody chart or correlation
## Verified Solution
With $\rm v_A$ = $\rm v_B$ = 0, $\rm p_A$ = $\rm p_B$ and Δz = $\rm H_B − H_A$ the extended Bernoulli equation yields:
$\rm\mathrm{h}_{\mathrm{pump}}=\mathrm{H}_{\mathrm{B}}-\mathrm{H}_{\mathrm{A}}+\mathrm{h}_{\mathrm{L}}={\frac{{\dot{\rm W}}_{\mathrm{el}}\cdot{ \eta }}{\mathrm{\rho g Q}}}$ (E.4.4.1a, b)
where
$\rm \mathrm{h}_{\mathrm{L,i}}=\left[\left\lgroup f\frac{\mathrm{L}}{\mathrm{D}}\right\rgroup _{\mathrm{i}}+\Sigma\mathrm{K}_{\mathrm{i}}\right] \frac{\mathrm{v}_{\mathrm{i}}^{2}}{2{\mathrm{g}}{}};\quad i=1,2$ (E.4.4.2a, b)
Employing Eq. (4.5e):
$\mathrm{f}_{\mathrm{i}}\approx\left\{-1.8\log\left[{\frac{6.9}{\mathrm{R}\mathrm{e}_{\mathrm{i}}}}+\left\lgroup{\frac{\varepsilon /{\mathrm{D}_{\mathrm{i}}}}{3.7}}\right\rgroup ^{1.11}\right]\right\}^{-2}$ (E.4.4.3)
with
$\rm\mathrm{Re}_{\mathrm{i}}=\frac{\mathrm{v_{i}}\,\mathrm{D}_{\mathrm{i}}}{\mathrm{\nu}}$ (E.4.4.4)
$\mathrm{v_{i}}=\mathrm{Q_{i}}/\left\lgroup{\frac{\pi\,\mathrm{D_{i}}^{2}}{4}}\right\rgroup$ (E.4.4.5a)
and
$\rm Q=Q_1+Q_2$ (E.4.4.5b)
Guess $\rm f_1$ = 0.02 and $\rm f_2$ = 0.018 , which leads with
$\rm\frac{{\varepsilon }_{1}}{{ D}_{1}}=8\times10^{-4}\qquad\mathrm{and}\qquad\frac{{\varepsilon }_{2}}{{ D}_{2}}=6\times10^{-4}$ using (E.4.4.3) to:
$\mathrm{Re}_{1}=3\times10^{5}\qquad\mathrm{and}\qquad\mathrm{Re}_{2}=8\times10^{5}\,.$
Now $\rm v_1$ and $\rm v_2$ can be computed as well as Q and $\rm h_L$ (see Eqs. (E.4.4.5a, b and (E.4.4.2a, b)). Clearly, Eq. (E.4.4.1b) has to be fulfilled, which requires $\rm h_L$ and Q have to match! If not, a new f-value has to be assumed and the calculations have to be repeated. The final result is $\rm Q_1$ = 0.0124 kg/s and $\rm Q_2$ =0.066 kg/s.
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CC-MAIN-2023-14
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latest
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en
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https://www.codewolfy.com/c-language/example/c-program-to-count-numbers-of-digit
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crawl-data/CC-MAIN-2023-14/segments/1679296948632.20/warc/CC-MAIN-20230327123514-20230327153514-00334.warc.gz
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In this example, you will learn to count the number of digits in an integer entered by the user. For example if we input 1234 then program returns 4.
## C Program to Count the Number of Digits Using Loop
``` ```
#include <stdio.h>
int main() {
long long n;
int count = 0;
printf("Enter an positive number: ");
scanf("%lld", &n);
//this loop runs till value of n not equal to 0
while (n != 0) {
n /= 10; // n = n/10
++count;
}
printf("Number of digits: %d", count);
}
```
```
Output :
``` ```
Enter an positive number: 12563
Number of digits: 5
```
```
## C Program to Count the Number of Digits Using User Define Function
``` ```
#include <stdio.h>
int countDigit(long long n)
{
int count = 0;
while (n != 0)
{
n = n / 10;
++count;
}
return count;
}
int main(void)
{
long long n;
printf("Enter an positive number: ");
scanf("%lld", &n);
printf("Number of digits : %d", countDigit(n));
return 0;
}
```
```
Output :
``` ```
Enter an positive number: 12586947
Number of digits : 8
```
```
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https://ptcouncil.net/calculate-the-mass-percent-of-the-solution-assume-a-density-of-1-08-g-ml-for-the-solution/
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L>ptcouncil.net: Calculations involving molality, molarity, density, mass percent, mole portion (Problems #11 - 25)Calculations entailing molality, molarity, density, mass percent, mole fractionProblems #11 - 25Fifteen ExamplesProblems 1 - 10Return to services MenuProblem #11: calculate the molarity and also mole fraction of acetone in a 2.28-molal solution of acetone (CH3COCH3) in ethanol (C2H5OH). (Density that acetone = 0.788 g/cm3; density of ethanol = 0.789 g/cm3.) Assume the the quantities of acetone and ethanol add.
You are watching: Calculate the mass percent of the solution. (assume a density of 1.08 g/ml for the solution.)
Solution because that molarity:Remember, 2.28-molal means 2.28 moles of acetone in 1.00 kilogram that ethanol.1) identify volumes that acetone and ethanol, then complete volume:acetone2.28 mol x 58.0794 g/mol = 132.421 g132.421 g separated by 0.788 g/cm3 = 168.047 cm3ethanol1000 g divided by 0.789 g/cm3 = 1267.427 cm3total volume168.047 + 1267.427 = 1435.474 cm32) determine molarity:2.28 mol / 1.435 l = 1.59 MSolution because that mole fraction:1) identify moles of ethanol:1000 g / 46.0684 g/mol = 21.71 mol2) determine mole fraction of acetone:2.28 / (2.28 + 21.71) = 0.0950Problem #12: calculation the normality the a 4.0 molal sulfuric acid equipment with a density of 1.2 g/mL.Reminders:N = #equivalents / l solution#equivalents = molecular weight / n (n = number of H+ or OH¯ released every dissociation.)molal = moles solute / kg solventSolution:1) identify grams the H2SO4 present:4.0 molal = 4.0 moles H2SO4 / 1000 g solution4.0 mol times 98.09 g/mol = 392.32 g2) determine equivalent load for H2SO4:98.09 g/mol / 2 dissociable hydrogen/mol = 49.05 g/equivalent3) determine # equivalents in 392.32 g:392.32 g time (1 tantamount / 49.05 g) = 8.0 equivalents4) recognize volume of solution:392.32 g + 1000 g = 1392.32 g (total massive of solution)1392.32 g / 1.2 g/mL = 1160.27 mL5) recognize normality:N = 8.0 equivalents / 1.16027 l = 6.9 NProblem #13: An auto antifreeze mixture is do by mix equal volumes of ethylene glycol (d = 1.114 g/mL, molar mass 62.07 g/mol) and also water (d = 1.000 g/mL) at 20.0 °C. The density of the systems is 1.070 g/mL.Express the concentration that ethylene glycol as:(a) volume percent(b) mass percent(c) molarity(d) molality(e) mole fractionSolution come (a):Since the volumes space equal, the volume percent the ethylene glycol is 50%Solution come (b):1) determine the masses that equal volumes (we"ll use 50.0 mL) the the 2 substances:ethylene glycol: (50.0 mL) (1.114 g/mL) = 55.7 gwater: (50.0 mL) (1.000 g/mL) = 50.0 g2) identify percent due to ethylene glycol:55.7 g / 105.7 g = 52.7%Solution to (c):1) recognize moles of ethylene glycol:55.7 g / 62.07 g/mol = 0.89737 mol2) recognize volume the solution:105.7 g / 1.070 g/mL = 98.785 mL3) identify molarity:0.89737 mol / 0.098785 l = 9.08 MSolution to (d):0.89737 mol / 0.050 kg = 17.9 mNote the large difference between the molarity and also the molality.Solution to (e):1) identify moles of water:50.0 g / 18.0 g/mol = 2.777782) identify mole portion for ethylene glycol:0.89737 mol / 3.67515 mol = 0.244Problem #14: What is the percent of CsCl by mass in a 0.0711 M CsCl systems that has a thickness of 1.09 g/mL?Solution:1) identify mass of dissolved CsCl:Let us assume 100.0 mL the solution.MV = grams / molar mass(0.0711 mol/L) (0.100 L) = x / 168.363 g/molx = 1.197 g2) recognize mass of solution:1.09 g/mL time 100.0 mL = 109 gDetermine massive percent the CsCl in solution:1.197 g / 109 g = 1.098%to three sig figs: 1.10%Problem #15: A 8.77 M equipment of one acid, HX, has a thickness of 0.853 g/mL.The acid, HX, has actually a molar mass of 31.00 g/mol. Recognize the molal concentration the this solution, ΧHX (mole portion of HX), and % w/w (percent by mass). The solvent in this solution is water, H2O.Comment: have the right to an 8.77 M solution of one acid have actually a density of 0.853 g/mL? who cares? We"ll just solve the problem.Solution:There is a trick to addressing this type of problem: let united state assume 1.00 l (or 1000 mL) of the solution is present. (Another location where a comparable trick is work is in determining empirical formulas, wherein you assume 100 g the the problem is present.)1) some preliminary calculations:moles acid: 1.00 together x (8.77 moles / L) = 8.77 moles HX (used in molality and mole fraction)mass acid: 8.77 mole x 31.00 g / mole = 272 g HX (percent through mass)mass the solution: 1000 mL x 0.853 g / mL = 853 g systems (percent by mass)mass of solvent: 853 g minus 272 g = 581 g = 0.581 kg (molality)moles solvent: 581 g separated by 18.015 g) = 32.25 (mole fraction)2) Calculations come answer the questions:molality = 8.77 moles / 0.581 kg = 15.1mmole portion = 8.77 / (8.77 + 32.25) = 0.214% w/w = (272 g / 853 g) x 100 = 31.9%Problem #16: A systems of hydrogen peroxide, H2O2, is 30.0% by mass and has a thickness of 1.11 g/cm3. Calculation the (a) molality, (b) molarity, and (c) mole fractionSolution:1) fixed of 1 liter of solution:1.11 g/cm3 time (1000 cm3 / L) = 1110 g/L2) fixed of the two contents of the solution:mass of H2O2 ---> 30.0% that 1110 g = 333 gmass that H2O ---> 70.0% that 1110 g = 777 g3) mole of hydrogen peroxide:333 g / 34.0138 g/mol = 9.79 mol4) Molality:9.79 mol / 0.777 kg = 12.6 m5) Molarity9.79 mol / 1.00 l = 9.79 M6) Mole fractionmole the H2O2 = 9.79mole of H2O = 43.13total moles ---> 9.79 + 43.13 = 52.92mole portion ---> 9.79 / 52.92 = 0.185Problem #17: family hydrogen peroxide is an aqueous systems containing 3.0% hydrogen peroxide through mass. What is the molarity that this solution? (Assume a density of 1.01 g/mL.)Solution:1) permit us have 1000 g of systems on hand. For the volume of solution:1000 g split by 1.01 g/mL = 990.1 mL2) because H2O2 is 3% by mass, we understand that there are 30 grams the H2O2 current in the 1000 g of solution.MV = mass / molar mass(x) (0.9901 L) = 30 g / 34.0138 g/molx = 0.890814 MTwo sig figs appears reasonable, so 0.89 M.Problem #18: A 6.90 M KOH systems in water has 30% by load KOH. Calculation the density of the KOH solution.Solution:Let us assume we have actually 1.00 liter of solution present. This means we have 6.90 mol that KOH. Let us determine the massive of KOH:6.90 mol times 56.1049 g/mol = 387.124 g387.124 g represents 30% of the full weight of the solution. (The water provides up the other 70%.) To acquire the massive of the solution, perform this:387.124 g is to 0.3 as x is to 1x = 1290 gdensity the the equipment ---> 1290 g / 1000 mL = 1.29 g/mL (to three sig figs)Problem #19: one aqueous NaCl solution is made using 138 g of NaCl diluted to a total solution volume that 1.30 L.(A) calculate the molarity of the solution.(B) calculation the molality of the solution. (Assume a thickness of 1.08 g/mL because that the solution.)(C) calculation the fixed percent that the solution. (Assume a thickness of 1.08 g/mL for the solution.)Solution:Part A:MV = mass / molar mass(x) (1.30 L) = 138 g / 58.443 g/molx = 1.82 MPart B:molality is mole solute per kg that solvent. Ns will usage 2.3613 mol (keeping a couple of guard digits).Let united state assume 1000 mL that the solution is present. This tells us that 2.3613 mol of the solute is current (that"s the 138 g that NaCl).1.08 g/mL times 1000 mL = 1080 g 1080 g minus 138 g = 942 g 942 g = 0.942 kg2.3613 mol / 0.942 kg = 2.51 mPart C:138 g that solute was dissolved in 1080 total grams the solution(138 / 1080) times 100 = 12.8% 100% minus 12.8% = 87.2% 2OAnother kind of inquiry is this area is to ask you to identify the mole portion for every substance. For the you will require to know the mole of water:942 g / 18.015 g/mol = 52.29 molThe mole portion of NaCl is this:2.3613 mol / (2.3613 mol + 52.29 mol) = 0.0432The mole fraction of the water is this:1 − 0.0432 = 0.9568Problem #20: A equipment is all set by dissolve 28.0 g that glucose (C6H12O6) in 350 g that water. The last volume the the solution is 384 mL . Because that this solution, calculate each the the following:1) molarity; 2) molality; 3) percent through mass; 4) mole fraction; 5) mole percentSolution:1) Molarity is the number of moles separated by volume that solvent in liters.28.0 g / 180 g/mol = 0.156 mol0.156 mol / 0.384 together = 0.405 M2) Molality is the number of moles separated by fixed of solvent in kilograms.0.156 mol / 0.350 kg = 0.444 m3) Percent through mass, together the name implies, is the fixed of solute divided by full mass time 100%.28.0 g / (350 g + 28 g) time 100 = 7.41% C6H12O6 by mass4) Mole fraction is the variety of moles of solute split by full moles.350 g / 18.015 g/mol = 19.428 mol of water0.156 mol / (19.428 + 0.156) = 0.0080 mol fraction C6H12O6The mole portion of water is:1 − 0.0080 = 0.9925) Mole percent is mole portion times 100%.0.0080 x 100 = 0.80 % C6H12O6 through molesProblem #21: How plenty of grams of glucose is necessary to dissolve in 3 litres the water to obtain 40% solution?Solution:Let x gms of glucose dissolve in 3 liters that water to type a 40% solution.glucose weight = xwater load = 3 liters = 3000 g (take water thickness as 1.00 g/mL)so complete solution load = x + 3000 gglucose percent ---> x / (x + 3000) = 0.40 x = 0.4 (x + 3000)x = 0.4x + 1200x − 0.4x = 12000.6x = 1200x = 2000 gProblem #22: The vinegar offered in the grocery stores is defined as 5% (v/v) acetic acid. What is the molarity the this solution (density the 100% acetic mountain is 1.05 g/mL)?Solution:1) 5% (v/v) method 5% that the volume is acetic acid. Therefore 1.00 together of vinegar includes 50 mL acetic acid:(0.05)(1000 mL) = 50 mL2) The fixed of this acetic mountain is:(1.05 g/mL)(50 mL) = 52.5 g3) because that the molarity, usage MV = massive / molar mass(x) (1.00 L) = 52.5 g / 60.0516 g/molx = 0.874 M (to three sig figs)Problem #23: by titration, the molarity of acetic acid in vinegar was established to be 0.870 M. Convert this come %(v/v). (The thickness of acetic mountain is 1.05 g/mL)Solution:1) i think 1.00 l of the systems to be present. Use MV = mass/molar fixed to identify mass the acetic acid present:(0.870 mol/L) (1.00 L) = x / 60.0516 g/molx = 52.245 g2) determine what volume the pure acetic acid this is:52.245 g divided by 1.05 g/mL = 49.757 mL3) identify %(v/v):(49.757 mL / 1000 mL) * 100 = 4.9757Rounded off, this is 4.98%(v/v), usually offered as 5%(v/v).Problem #24: In one aqueous systems of sulfuric acid, the acid concentration is 2.40 mole percent and the density of the equipment is 1.079 g/mL. Calculate (1) the molal concentration the the acid, (2) the weight percentage of the acid, and also (3) the molarity the the solutionSolution:2.40 mole percent that acid way 97.60 mole percent water.Let"s i think 100 moles of the solution is present. This means 2.40 mole the the systems is H2SO4 and also 97.60 mole is water.(1) for molality, we require to know kg of water ---> 97.60 mol time 18.015 g/mol = 1758.264 g = 1.758264 kgmolality ---> 2.40 mol / 1.758264 kg = 1.365 m (1.36 m to three sig figs)(2) for the load percent, we need the massive of H2SO4 ---> 2.40 mol times 98.0768 g/mol = 235.38432 gweight percent ---> 235.38432 g / (235.38432 + 1758.264 g) = 0.118067 = 11.8% (three sig figs)(3) because that molarity, we require to understand the volume the the systems ---> 1993.64832 g separated by 1.079 g/mL = 1847.68 mL = 1.84768 LNote: 1993.64832 g is the full mass that the solution.molarity ---> 2.40 mol / 1.84768 together = 1.2989 M (1.30 M to 3 sig figs) through the way, mole percent is mole fraction written as a percent. The mole fractions in the above problem room 0.0240 and also 0.9760.Problem #25: calculate the molality, molarity, and also mole fraction of FeCl3 in a 26.3% (w/w) equipment (density = 1.28 g/mL).Solution:Molarity:Assume 100. G of equipment present.26.3 g of FeCl3 is present.100. G divided by 1.28 g/mL = 78.125 mLUse MV = mass / molar mass(x) (0.078125 L) = 26.3 g / 162.204 g/molx = 2.08 M (to 3 sig figs)Molality100. G − 26.3 g = 73.7 g molality ---> (26.3 g / 162.204 g/mol) / 0.0737 kg = 2.20 m (to three sig figs)Mole portion of FeCl3:moles FeCl3 ---> 26.3 g / 162.204 g/mol = 0.1621415 molmoles water ---> 73.7 g / 18.015 g/mol = 4.091035 molmole fraction ---> <0.1621415 mol / (0.1621415 mol + 4.091035 mol)> = 0.0381 (to 3 sig figs)Problem #26: The mole portion in a solution of Na2S is 0.125. Calculate the fixed precent (w/w) of Na2S in this solution.Solution:The mole portion of Na2S is 0.125. Therefore, the mole fraction of water in the solution is 0.875.mass Na2S ---> 0.125 mol time 78.045 g/mol = 9.755625 gmass water ---> 0.875 mole 18.015 g/mol = 15.763125 g% (w/w) Na2S ---> <9.755625 / (9.755625 + 15.763125) * 100> = 38.2% (to 3 sig figs)Problem #27: In dilute nitric acid, the concentration the HNO3 is 6.00 M and the thickness of this equipment is 1.19 g/mL. Usage that information to calculate the mass percent and also mole portion of HNO3 in the solution.
See more: Are Optional Arguments Are Always Placed First In The Argument List ?
Solution #1:1) Let us assume 1000 mL of the systems is present. Some preliminary calculations:1000 mL x 1.19 g/mL = 1190 g (this is the mass of our 1000 mL)6.00 mol/L x 1.00 l = 6.00 mol (this is how countless moles of HNO3 room in ours 1000 mL)6.00 mol x 63.012 g/mol = 378.072 g (the fixed of HNO3 in our 1000 mL)1190 g minus 378.072 g = 811.928 g (the fixed of water in the 1000 mL that solution)2) calculate the massive percent the HNO3:378.072 g / 1190 g = 31.8% (to 3 sig figs)3) calculation the mole portion of HNO3:811.928 g / 18.015 g/mol = 45.07 mol of H2O6.00 mol / (6.00 mol + 45.07 mol) = 0.118 (to 3 sig figs)Solution #2:1) Let us assume 1000 g the the solution is present. Part preliminary calculations:1000 g / 1.19 g/mL = 840.336 mL (the volume of ours 1000 g that solution)6.00 mol/L x 0.840336 together = 5.042 mol (this is how countless moles the HNO3 space in our 840.336 mL)5.042 mol x 63.012 g/mol = 317.7065 g (the massive of HNO3 in our 840.336 mL)1000 g minus 317.7065 g = 682.2935 g (the massive of water in the 1000 g the solution)2) calculation the fixed percent of HNO3:317.7065 g / 1000 g = 31.8% (to 3 sig figs)3) calculate the mole fraction of HNO3:682.2935 g / 18.015 g/mol = 37.874 mol that H2O5.042 mol / (5.042 mol + 37.874 mol) = 0.117 (to 3 sig figs)Comment: In systems #2, I derived 0.117485 and that, technically, is 0.117 (not 0.118), as soon as rounded to three sig figs.Problem #28: A 0.100 M NaOH solution will be all set by dilution of a 50.0% (w/w) NaOH solution. This solution has a density of 1.53 g/mL. Compute the volume of this systems that is compelled to prepare 1.00 x 103 mL the 0.100 M NaOH.Solution:1) determine moles the NaOH in 1.0 x 103 mL the 0.10 M solution:MV = moles(0.100 mol/L) (1.00 L) = 0.100 mol2) recognize mass of 0.100 mol that NaOHmoles x molar massive = grams(0.100 mol) (40.00 g/mol) = 4.00 g3) fixed of 50.0% (w/w) equipment that includes 4.00 g of NaOH:use a ratio and proportion50 g is come 100 g as 4 g is come xx = 8.00 mL4) recognize volume of equipment that includes 8.00 g the NaOH:Luke, uuuuuuse the density!8.00 g / 1.53 g/mL = 5.23 mLProblem #29: What is the molarity that a NaOH solution with a density of 1.33 g/mL the was made with 70.0 ml of water if the molality is 10.7 molal? Solution:1) use the molality come determine just how much NaOH was supplied with the 70.0 g of water:10.7 molal = 10.7 mol solute every kg solvent10.7 mol/kg = x / 0.0700 kgx = 0.749 mol of NaOH2) identify the grams the NaOH:(0.749 mol) (40.0 g/mol) = 29.96 g 3) full mass of the solution:70.0 g + 30.0 g = 100.0 g4) Volume the the solution:100.0 g / 1.33 g/mL = 75.2 mL5) Molarity:0.749 mol / 0.0752 together = 9.96 MProblem #30a: How numerous mL of an 3.78% (w/w) solution can be ready from 18.00 g that sucrose?Solution:3.78 is come 18 together 100 is to xx = 476.19 g 476.19 − 18.00 = 458.19 g i think 1 g/mL for the density and also no volume readjust when combine 458.19 g of water and also 18.00 g that sucrose.Answer = 458 mL (to 3 sig figs)Problem #30b: How countless mL of an 3.78% (w/v) solution have the right to be prepared from 18.00 g of sucrose?Solution:3.78%(w/v) method 3.78 g of sucrose every 100 mL of solution3.78 is to 100 together 18 is to xx = 476 mL (to three sig figs)Fifteen ExamplesProblems 1 - 10Return to services Menu
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# Gill (U.S.) to Cubic Decimeters Conversion
Convert Gill (U.S.) to Cubic Decimeters by entering the Gill (U.S.) (gi) value in the calculator form. gi to dm³ conversion.
gi to dm³ Converter
1 gi = 0.118294 dm³
Cubic Decimeters to Gill (U.S.) Conversion
Gill (U.S.) volume unit is equal to 0.118294 Cubic Decimeters.
## How Many Cubic Decimeters in a Gill (U.S.)
There are 0.118294 Cubic Decimeters in a Gill (U.S.).
### Conversion Factors for Cubic Decimeters and Gill (U.S.)
Volume UnitSymbolFactor
Gill (U.S.) gi 1.182941 × 10 -4
Cubic Decimeters dm³ 1 × 10 -3
### Gill (U.S.) to Cubic Decimeters Calculation
We calculate the base unit equivalent of Gill (U.S.) and Cubic Decimeters with the base unit factor of volume cubic meter.
```1 gi = 1.182941 * 10-4 m³
1 dm³ = 1 * 10-3 m³
1 dm³ = 0.001 m³
1 m³ = (1/0.001) dm³
1 m³ = 1000 dm³
1 gi = 1.182941 * 10-4 * 1000 dm³
1 gi = 0.1182941 dm³```
### Gill (U.S.) to Cubic Decimeters Conversion Table
Gill (U.S.)Cubic Decimeters
1 gi0.118294 dm³
10 gi1.18294 dm³
20 gi2.36588 dm³
30 gi3.54882 dm³
40 gi4.73176 dm³
50 gi5.9147 dm³
60 gi7.09764 dm³
70 gi8.28058 dm³
80 gi9.46352 dm³
90 gi10.64646 dm³
100 gi11.8294 dm³
110 gi13.01234 dm³
120 gi14.19528 dm³
130 gi15.37822 dm³
140 gi16.56116 dm³
150 gi17.7441 dm³
160 gi18.92704 dm³
170 gi20.10998 dm³
180 gi21.29292 dm³
190 gi22.47586 dm³
200 gi23.6588 dm³
#### Abbreviations
• gi : Gill (U.S.)
• dm³ : Cubic Decimeters
• m³ : Cubic Meter
Gill (U.S.) to Cubic Decimeters Conversion
### Related Volume Conversions
List all Gill (U.S.) Conversions »
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Why is null in JavaScript bigger than -1, less than 1, but not equal (==) to 0? What is it exactly then?
``````var x = null;
undefined
x > 0
false
x < 0
false
x > -1
true
x < 1
true
x == 1
false
x === 1
false
``````
• Not a number, so the assessment is inherently false? Commented Nov 15, 2012 at 22:44
• Note `null >= 0`, etc. is true Commented Nov 15, 2012 at 22:45
• Even more strangely, `null <= 0` is true, but both `null < 0` and `null == 0` are false Commented Apr 14, 2021 at 20:57
When you compare null for equality to 0, the result is false. If you force `null` to be interpreted in a numeric context then it is treated like 0 and the result becomes true.
You can force it to be numeric by putting `+` in front, or by using numeric operators like `<`, `<=`, `>`, and `>=`. Notice how `null >= 0` and `null <= 0` are both true.
``````> null == 0
false
> +null == 0
true
> null >= 0
true
> null <= 0
true
``````
The ECMAScript Language Specification defines when a so-called "ToNumber" conversion is performed. When it is, null and false are both converted to 0.
§9.1 Type Conversion and Testing:
Table 14 — To Number Conversions
```Argument Type Result
------------- ------
Undefined Return NaN
Null Return +0
Boolean Return 1 if argument is true. Return +0 if argument is false.
Number Return argument (no conversion).
String See grammar and note below.
```
Knowing when the ToNumber conversion is applied depends on the operator in question. For the relational operators `<`, `<=`, `>`, and `>=` see:
§11.8.5 The Abstract Relational Comparison Algorithm:
The comparison `x < y`, where x and y are values, produces true, false, or undefined (which indicates that at least one operand is NaN). Such a comparison is performed as follows:
1. Call ToPrimitive(x, hint Number).
2. Call ToPrimitive(y, hint Number).
3. If Type(Result(1)) is String and Type(Result(2)) is String, go to step 16. (Note that this step differs from step 7 in the algorithm for the addition operator + in using and instead of or.)
4. Call ToNumber(Result(1)).
5. Call ToNumber(Result(2)).
The `==` operator is different. Its type conversions are described below. Notice how null and false follow different rules.
§11.9.3 The Abstract Equality Comparison Algorithm
The comparison x == y, where x and y are values, produces true or false. Such a comparison is performed as follows:
1. If Type(x) is different from Type(y), go to step 14.
...
14. If x is null and y is undefined, return true.
15. If x is undefined and y is null, return true.
16. If Type(x) is Number and Type(y) is String, return the result of the comparison x == ToNumber(y).
17. If Type(x) is String and Type(y) is Number, return the result of the comparison ToNumber(x) == y.
18. If Type(x) is Boolean, return the result of the comparison ToNumber(x) == y.
19. If Type(y) is Boolean, return the result of the comparison x == ToNumber(y).
20. If Type(x) is either String or Number and Type(y) is Object, return the result of the comparison x == ToPrimitive(y).
21. If Type(x) is Object and Type(y) is either String or Number, return the result of the comparison ToPrimitive(x) == y.
22. Return false.
If you read carefully you can see why `false == 0` is true but `null == 0` is false.
• For `false == 0`, Type(x) is Boolean. This means Step 18's type conversion is applied, and false is converted to a number. ToNumber(false) is 0, and `0 == 0` is true, so the comparison succeeds.
• For `null == 0`, Type(x) is Null. None of the type checks match so the comparison falls through to Step 22, which returns false. The comparison fails.
• If someone does not know `null >=0`; that could lead to some very troubling bugs. Suppose one allowing a user to do something when their balance is greater than zero; but got null for some reason from API; & they did not check for null (` x != null`) Commented Jan 26, 2022 at 14:42
null casts to 0 as a number: `(+null)` is 0. > and < cast null to this value, so when compared to numbers it acts as zero. `==` doesn't cast null to a number, so `null == 0` is false.
• Yes - + can be used to quickly cast things to numbers: `(+new Date())` is a common trick to get millisecond time. For instance, `4 + '4' = 44`, while `4 + (+'4') = 8`
– tmcw
Commented Nov 16, 2012 at 15:37
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Tricky SC Series- #9 : GMAT Sentence Correction (SC)
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# Tricky SC Series- #9
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31 Oct 2009, 22:10
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Hi All !
Sharing some SCs from a recent mock test (thought that u may or may not get the same q when you take the same mock and that it'd be a good practice for all... ). Please post some explanation with the answers.
Others in this series:
1. tricky-sc-series-86098.html
2. tricky-sc-series-86099.html
3. tricky-sc-series-86100.html
4. tricky-sc-series-86101.html
5. tricky-sc-series-86102.html
6. tricky-sc-series-86104.html
7. tricky-sc-series-86105.html
8. tricky-sc-series-86106.html
10. tricky-sc-series-86108.html
11. tricky-sc-series-86110.html
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Re: Tricky SC Series- #9 [#permalink]
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01 Nov 2009, 12:13
Go for A..
Known to have + past participles.
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02 Nov 2009, 21:35
thanks GMAT TIGER... if no other inputs, OA in a day or two
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Re: Tricky SC Series- #9 [#permalink]
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03 Nov 2009, 07:53
+1 for A.
to have driven it looks the appropriate usage to me.
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Re: Tricky SC Series- #9 [#permalink]
### Show Tags
03 Nov 2009, 19:12
I'd chosen A as well.
Unfortunately it is not the OA:
[Reveal] Spoiler:
C
Reason given is that A is unidiomatic. "to drive it" is generally preferred over "to have driven it"
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Re: Tricky SC Series- #9 [#permalink]
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19 Jun 2010, 04:03
I was also with A.
Could anybody elaborate on the explanation?
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Re: Tricky SC Series- #9 [#permalink]
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02 Jul 2010, 06:11
my pick is A too.
can someone explain why C is correct and why A is incorrect?
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Re: Tricky SC Series- #9 [#permalink]
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11 Jul 2010, 07:30
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Americans' love affair with the Mustang is readily apparent: few Americans have been knon to have driven it for the first time without taking it out for a another spin.
A. few Americans have been known to have driven it
B. few having been known to drive it
C. few Americans have been known to drive it
D. it has been driven by few Americans
E. few Americans having driven it
Hi Guys,
This is a good opportunity to look at what the real difference between "to drive" and "to have driven" is.
to+base verb (V1) is not a verb, but (as many of you know) an infinitive. Because it is not a verb, it has no time.
to+have+V3 is the perfect infinitive. This is a structure that indicates the past in an infinitive.
So...
Joe is known to be a good student = We know (in the present) that Joe is a good student (in the present).
Joe is known to have been a good student = We know (in the present) that Joe was a good student (in the past).
So let's look at the answer choices one by one:
A. few Americans have been known to have driven it
"To have driven" indicates that Americans drove the Mustang in the past. But the sentences is talking about a present-day phenomenon.
B. few having been known to drive it
Error in Ellipses! "Few" indicates "few Americans," but the word "Americans" appears nowhere in the sentence! At the beginning of the sentence, we get Americans' -- the possessive, not the noun!
C. few Americans have been known to drive it
"To drive" correctly indicates "to drive (in the present)."
D. it has been driven by few Americans
"...driven by Americans" is passive voice-- as long as the doer is present, active voice should be used.
E. few Americans having driven it
We have lost the intended meaning. The sentences wanted to indicate what is "known".
(More on comparisons in SC Lesson 6 and on ellipses in Lesson 9 at gmaxonline!)
Best,
Sarai
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Re: Tricky SC Series- #9 [#permalink]
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04 Sep 2010, 11:23
The explanation is still not clear to me...
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Re: Tricky SC Series- #9 [#permalink]
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08 Mar 2014, 17:36
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Hello from the GMAT Club VerbalBot!
Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
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Re: Tricky SC Series- #9 [#permalink] 08 Mar 2014, 17:36
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# What is parabolic path in physics?
Parabolic path is defined as the angle of trajectory of a projectile.
## What is a parabolic path called?
Projectile motion is a form of motion where an object moves in parabolic path; the path that the object follows is called its trajectory.
## What is the equation of parabolic path?
y=xtanθ−(2u2cos2θg)x2.
## Why the path is parabolic?
If these two motions are combined – vertical free fall motion and constant horizontal motion – then the trajectory will be that of a parabola. An object which begins with an initial horizontal velocity and is acted upon only by the force of gravity will follow the path of the blue sphere.
## What is an example of parabolic motion?
Projectile motion here’s a ball I fire it into the air, boom and it travels in a parabola hence the name parabolic motion right. so it’s basically any object that’s fired into the air and when it moves there’s a lot of different forces acting on it. It has got a vertical velocity, it has got a horizontal velocity.
## What is difference between parabola and trajectory?
This type of curved motion is called motion of projectile and the path is called trajectory. It is a parabola. This type of motion is two dimensional motion. If we ignore air-resistance motion of a projectile in due gravity only path of projectile is always parabolic.
## What is a parabola in simple terms?
1 : a plane curve generated by a point moving so that its distance from a fixed point is equal to its distance from a fixed line : the intersection of a right circular cone with a plane parallel to an element of the cone. 2 : something bowl-shaped (such as an antenna or microphone reflector)
## How do you prove the path of a projectile is parabolic?
If we put tan θ = A and g/2u2cos2θ = B then equation (5) can be written as y = Ax – Bx2 where A and B are constants. This is the equation of a parabola. Hence, the path of the projectile is a parabola.
## Is every trajectory a parabola?
No, most trajectories are ellipses. If an object has less than escape velocity, it is an ellipse. If the object just has escape velocity, it has a parabolic trajectory.
## What is difference between hyperbola and parabola?
A parabola is defined as a set of points in a plane which are equidistant from a straight line or directrix and focus. The hyperbola can be defined as the difference of distances between a set of points, which are present in a plane to two fixed points is a positive constant.
## Can an orbit be parabolic?
The parabolic orbit is the Borderline case between open and closed orbits and therefore identifies the border line condition between space vehicles that are tied to paths (elliptical) in the general vicinity of their parent planet and those that can take up paths (hyperbolic) extending to regions remote from their …
## What is the shape of the path of any projectile?
Parabola: in projectile motion, it is the shape of the path traced by a projectile.
## What is projectile motion and show that its path is parabola?
Put equation (1) here, y = usinθ × x/ucosθ – 1/2g × x²/u²cos²θ y = tanθx – 1/2gx²/u²cos²θ This equation is similar to Standard equation of parabola y = ax² + bx + c her, a, b and c are constant. So, A projectile motion is a parabolic motion.
## What is projectile called?
Projectile motion : Projectile motion is a form of motion in which object or particle ( called a projectile , is thrown near earth’s surface and it moves along a curved path under the action of gravity only.
## Why do projectiles move in a curve path?
projectile motion, the motion of a falling object (projectile) after it is given an initial forward velocity. Air resistance and gravity are the only forces acting on a projectile. The combination of an initial forward velocity and the downward vertical force of gravity causes the ball to follow a curved path.
## What type of motion is parabolic motion?
Projectile motion, also known as parabolic motion, consists in launching a body with a velocity that form an angle α with the horizontal. In the following figure, you can see a representation of the situation. This motion is characteristic of projectiles, moving objects being affected only by gravity.
## What are the 3 types of projectile motion?
• Oblique projectile motion.
• Horizontal projectile motion.
• Projectile motion on an inclined plane.
## What are 2 examples of projectile motion?
• Firing a Canon.
• Sneezing.
• Javelin Throw.
• Archery.
• Water Escaping a Hose.
• Car and Bike Stunts.
• Golf Ball.
## Why is gravity a parabola?
When an object is thrown, it moves forward but is constantly affected by gravity as well. Consequently, it will move downward toward gravity’s pull, creating a parabolic movement for the object.
## What is projectile is the trajectory of a projectile is a parabola?
a=tanθ,b=2u2cos2θg, are constants. for given values of θ, u and g. Equation (3) represents a parabola. Therefore trajectory of a projectile is parabola.
## Why is it called a parabola?
The name “parabola” is due to Apollonius, who discovered many properties of conic sections. It means “application”, referring to “application of areas” concept, that has a connection with this curve, as Apollonius had proved. The focus–directrix property of the parabola and other conic sections is due to Pappus.
## Why is a curve called a parabola?
A parabola is a curve that looks like the one shown above. Its open end can point up, down, left or right. A curve of this shape is called ‘parabolic’, meaning ‘like a parabola’.
## What is another name for a parabola?
Parabola synonyms In this page you can discover 8 synonyms, antonyms, idiomatic expressions, and related words for parabola, like: eccentricity, ellipse, parallelogram, paraboloid, hyperbola, polar-coordinates, ellipsoid and vector-field.
## What forces is acting on the projectile?
The only force acting upon a projectile is gravity!
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# Cartesian tree: Part 1. Description, operations, applications
Part 1. Description, operations, applications.
Part 2. Valuable information in the tree and multiple operations with it.
Part 3. Cartesian tree by implicit key.
To be continued ...
Cartesian tree (treap) is a beautiful and easily implemented data structure that with minimal effort will allow you to perform many high-speed operations on your data arrays. What is characteristic, on Habrahabr I found his only mention in the review post of the esteemed winger , but then the continuation of that cycle did not follow. It's a shame, by the way.
I will try to cover everything that I know on the subject - despite the fact that I know relatively little, there will be enough material for two or even three posts. All the algorithms are illustrated by C # sources (and since I am a lover of functional programming, somewhere in the afterword we will also talk about F # - but this is not necessary to read :). So let's get started.
#### Introduction
As an introduction, I recommend reading a post about binary search trees of the same winger , because without understanding what a tree is, a search tree, as well as without knowing the complexity of the algorithm, much of the material in this article will remain for you a Chinese letter. It's a shame, right?
The next item in our compulsory program is heap . I think that many people also know the data structure, but I will give a brief overview.
Imagine a binary tree with some data (keys) at the vertices. And for each vertex, we without fail require the following: its key is strictly greater than the keys of its immediate sons. Here is a small example of a valid heap:
I’ll say right away that it’s not necessary to think of the heap solely as a structure in which the parent has more than his descendants. Nobody forbids to take the opposite option and assume that the parent is smaller than the descendants - most importantly, choose one for the whole tree. For the needs of this article, it will be much more convenient to use the option with the “more” sign.
Now behind the scenes the question remains, how can you add and remove elements from the heap. Firstly, these algorithms require a separate place for inspection, and secondly, we still do not need them.
#### Problems
When it comes to search trees (you already read the recommended article, right?), The main question posed to the structure is the speed of operations, regardless of the data stored in it, and the sequence of their arrival. So, the binary search tree gives a guarantee that the search for a specific key in this tree will be performed in O (H), where H is the height of the tree. But what height can be - the devil knows him. Under adverse circumstances, the height of the tree can easily become N(the number of elements in it), and then the search tree degenerates into a regular list - and why then is it needed? To achieve this situation, it is enough to add elements from 1 to N in the ascending queue in the search tree - with the standard algorithm for adding to the tree, we get the following picture:
A huge number of so-called balanced search trees were invented - roughly speaking, those in which as the tree exists at Each operation on it maintains the optimality of the maximum depth of the tree. The optimal depth is of the order of O (log 2N) - then the same order has the execution time of each search in the tree. There are many data structures supporting this depth, the most famous are red-black wood or AVL-tree. Their distinguishing feature in most cases is a difficult implementation, based on the size of a hell of a heap of cases in which you can get confused. Conversely, the Cartesian tree compares favorably with its simplicity and beauty, and even gives us in some way the very coveted logarithmic time, but only with a rather high probability ... however, about such details and subtleties later.
#### Definition
So we have the data of the tree - the keys x (hereinafter it is assumed that the key is the very information that we store in the tree; when later it will be necessary to separate the user information from the keys in meaning, I will say it separately). Let's add another parameter to them in a pair - y , and call it priority . Now we’ll build such a magic tree that stores two parameters at each vertex, and at the same time it ’s a search tree by its keys and a bunch of priorities . Such a tree will also be called Cartesian.
By the way: the name treap is very popular in English literature , which clearly shows the essence of the structure: tree + heap. In the Russian-speaking, one can sometimes find those compiled according to the same principle: deramide (tree + pyramid) or ducha (tree + heap).
Why is a tree called cartesian? This will immediately become clear as soon as we try to draw it. Take some set of key-priority pairs and arrange the corresponding points (x, y) on the grid. And then we connect the corresponding vertices with lines, forming a tree. Thus, the Cartesian tree fits perfectly on the plane due to its limitations, and its two main parameters - the key and priority - in a sense, the coordinates. The result of the construction is shown in the figure: on the left in the standard tree notation, on the right on the Cartesian plane.
So far, it’s not very clear why this is necessary. And the answer is simple, and it lies in the following statements. First, let many keys be given: many different trees, including list-like ones, can be built from the correct search trees. But after adding priorities to them, a tree of these keys can be built only one, regardless of the order of receipt of the keys. This is pretty obvious.
And secondly, let's now make our priorities random. That is, we simply associate a random number from a sufficiently large range with each key, and it will serve as the corresponding player. Then the resulting Cartesian tree with a very high probability of 100% will have a height not exceeding 4 log 2N. (I will leave this fact without proof here.) So, although it may not be perfectly balanced, the search time for a key in such a tree will still be of the order of O (log 2 N), which we, in fact, sought.
Another interesting approach is not to make priorities random, but to remember that we have a huge amount of some additional user information, which, as a rule, has to be stored at the tops of the tree. If there is reason to believe that this information is essentially random enough (the user's birthday, for example), then you can try to use it for personal gain. Take either information directly or the result of some function from it as a priority (only then should the function be reversible in order to restore information from priorities if necessary). However, it is necessary to act here at your own peril and risk - if after some time the tree is very unbalanced and the whole program starts to slow down significantly, you will have to urgently invent something to save the situation.
Further, for simplicity, we assume that all keys and all priorities in the trees are different. In fact, the possibility of equality of keys does not create any special problems, you just need to clearly determine where the elements equal to this x will be located - either only in its left subtree, or only in the right one. Equality of priorities in theory also does not constitute a particular problem, except for contamination of evidence and reasoning by special cases, but in practice it is better to avoid it. The random generation of whole priorities is quite suitable in most cases, real between 0 and 1 - in almost all cases.
Before starting the story about operations, I will give you a blank of the C # class, which will implement our Cartesian tree.
``````public class Treap
{
public int x;
public int y;
public Treap Left;
public Treap Right;
private Treap(int x, int y, Treap left = null, Treap right = null)
{
this.x = x;
this.y = y;
this.Left = left;
this.Right = right;
}
// здесь будут операции...
}``````
For simplicity, I made an exposition of the x and y types `int`, but it is clear that in their place there could be any type whose instances we can compare with each other - that is, any that implements `IComparable `either in C # terms. In Haskell, it could be any type from the class , in F # any with restrictions on the comparison operator, in Java it implements an interface , and so on.`IComparable``Ord``Comparable`
#### The magic of glue and scissors
The pressing issues on the agenda are how to work with the Cartesian tree. I’ll postpone the question of how to build it from a crude set of keys altogether, but for now suppose that some kind soul of the initial tree has already been built for us, and now we need to change it if necessary.
All the ins and outs of working with the Cartesian tree consists of two main operations: Merge and Split . With the help of them, all other popular operations are elementarily expressed, so let's start with the basics.
The Merge operation accepts two Cartesian trees L and R as input . From it is required to merge them into one, also correct, Cartesian tree T. It should be noted that the Merge operation can work not with any pairs of trees, but only with those for which all the keys of one tree (L) do not exceed the keys of the second (R). (Pay special attention to this condition - it will come in handy more than once in the future!)
The algorithm of Merge is very simple. Which element will be the root of the future tree? Obviously with the highest priority. We have two candidates for maximum priority - only the roots of two source trees. Compare their priorities; for uniqueness, let the priority y of the left root be greater, and the key in it is equal to x. A new root is defined, now it’s worth considering which elements will appear in its right subtree and which in the left one.
It is easy to understand that the whole tree R will be in the right subtree of the new root, because it has keys greater than x by condition. In the same way, the left subtree of the old L.Left root has all keys smaller than x and should remain the left subtree, and the right subtree L.Right ... but the right subtree should be on the right for the same reasons, however, it’s not clear where to put its elements, and where are the elements of the tree R?
Stop, why is it unclear? We have two trees, the keys in one are less than the keys in the other, and we need to somehow combine them and attach the result to the new root as the right subtree. Just recursively call Merge for L.Right and the R tree, and use the tree returned by it as the new right subtree. The result is obvious.
The figure in blue shows the right subtree of the resulting tree after the Merge operation and the connection from the new root to this subtree.
The symmetric case - when the priority at the root of the tree R is higher - is disassembled similarly. And, of course, we must not forget about the basis of recursion, which in our case occurs if one of the trees L and R, or both, are empty.
``````public static Treap Merge(Treap L, Treap R)
{
if (L == null) return R;
if (R == null) return L;
if (L.y > R.y)
{
var newR = Merge(L.Right, R);
return new Treap(L.x, L.y, L.Left, newR);
}
else
{
var newL = Merge(L, R.Left);
return new Treap(R.x, R.y, newL, R.Right);
}
}
``````
Now about Split operation. At the input, she receives the correct Cartesian tree T and a certain key x 0 . The task of the operation is to divide the tree into two so that in one of them (L) there are all elements of the source tree with keys less than x 0 , and in the other (R) with large ones. There are no special restrictions on the tree.
We reason in a similar way. Where will the root of the tree T be? If its key is less than x 0 , then in L, otherwise in R. Again, for the sake of clarity, suppose that the root key is less than x 0 .
Then we can immediately say that all the elements of the left subtree T will also be in L - their keys, after all, will also all be less than x 0. Moreover, the root of T will also be the root of L, since its priority is greatest in the whole tree. The left subtree of the root will be completely preserved unchanged, but the right subtree will decrease - you will have to remove elements with keys greater than x 0 from it and put it into the tree R. And save the rest of the keys as the new right subtree L. Again we see the identical task, recursion again begs!
Take the right subtree and cut it recursively using the same key x 0 into two trees L 'and R'. Then it becomes clear that L 'will become the new right subtree of the tree L, and R' is the tree R itself - it consists of those and only those elements that are greater than x 0 .
Symmetric case in which the root key is greater than x 0, also completely identical. The basis of recursion here is when some of the subtrees is empty. Well, the source code of the function:
``````public void Split(int x, out Treap L, out Treap R)
{
Treap newTree = null;
if (this.x <= x)
{
if (Right == null)
R = null;
else
Right.Split(x, out newTree, out R);
L = new Treap(this.x, y, Left, newTree);
}
else
{
if (Left == null)
L = null;
else
Left.Split(x, out L, out newTree);
R = new Treap(this.x, y, newTree, Right);
}
}
``````
By the way, note: the trees issued by the Split operation are suitable as input for the Merge operation: all the keys of the left tree do not exceed the keys in the right. This valuable circumstance will come in handy after a few paragraphs.
The final question is the uptime of Merge and Split. It can be seen from the description of the algorithm that Merge for each iteration of recursion reduces the total height of two merged trees by at least one, so that the total operating time does not exceed 2H, that is, O (H). And with Split, everything is very simple - we work with a single tree, its height decreases with each iteration by at least one as well, and the asymptotic behavior of the operation is also O (H). And since a Cartesian tree with random priorities, as already mentioned, is very likely to have a logarithmic height, Merge and Split work for the desired O (log 2 N), and this gives us tremendous scope for their application.
#### Tree operations
Now that you and I have perfect knowledge of glue and scissors, it’s not at all difficult to just use them to implement the most necessary actions with the Cartesian tree: adding an element to the tree and deleting it. I will give the simplest version of their implementation, based entirely on Merge and Split. It will work for all the same logarithmic time, however it will differ, as ACM-Olympians say, by a larger constant: that is, the order of the time depends on the size of the tree will still be O (log 2 N), but the exact time will differ several times - a constant time. Say 4 log 2 N vs just log 2 N. In practice, this difference is almost not felt until the size of the tree reaches a truly galactic size.
There are also optimal implementations of additions with deletion, as well as other necessary operations of deramide, whose constant is much smaller. I will definitely present these implementations in one of the following parts of the cycle, and also talk about the pitfalls associated with using the faster option. Pitfalls are primarily associated with the need to support additional queries to the tree and store special information in it ... however, I will not bother the reader with this now, before multiple operations with the tree (an extremely important feature!) Time will come.
So, let us be given a Cartesian tree and some element x that you want to insert into it (as we recall, in the context of the article it is assumed that all elements are different and X is not in the tree yet). I would like to apply the approach from the binary search tree: go down the keys, each time choosing the path left or right, until we find a place where we can insert our x, and add it. But this decision is wrong, because we forgot about priorities. The place where the search tree algorithm wants to add a new vertex uniquely satisfies the restrictions of the search tree by x, however, it may violate the heap restriction by y. So, you have to act a little higher and more abstract.
The second solution is to present the new key as a tree from a single vertex (with a random priority y), and merge it with the original one using Merge. This is again wrong: in the source tree there may be vertices with keys greater than x, and then we break the promise given to the Merge function regarding the relationship between its input trees.
The problem can be fixed. Remembering the versatility of Split / Merge operations, the solution suggests itself almost immediately:
1. We split the tree by the key x into the tree L, with the keys smaller than X, and the tree R, with large ones.
2. Create from this key a tree M from a single vertex (x, y), where y is the newly generated random priority.
3. Combine (merge) in turn L with M, what happened - with R.
All steps of the algorithm can be illustrated.
We have 1 Split application here, and 2 Merge applications - the total operating time of O (log 2 N). Short source code attached.
``````public Treap Add(int x)
{
Treap l, r;
Split(x, out l, out r);
Treap m = new Treap(x, rand.Next());
return Merge(Merge(l, m), r);
}
``````
With the removal, there are also no questions. Let us be asked to remove an element with the key x from the Cartesian tree. Now I assume that you figured out the equality of keys, giving preference to the left side: in the right subtree of the vertex with the key x, other elements with the same key do not occur, but in the left one they can. Then we perform the following sequence of actions:
1. First, divide the tree by the key x-1. All elements less than or equal to x-1 went to the left result, which means that the desired element is in the right.
2. We divide the right result by the key x (here it is worth being careful with equality!). All the elements with keys greater than x went to the new right result, and to the "middle" (left from the right) - all less than or equal to x. But since everyone who was strictly smaller after the first step was eliminated, the middle tree is the desired element.
3. Now just combine the left tree with the right one again, without the middle one, and the deramide is left without the x keys.
Now it’s clear why I constantly focused on how all the same it is necessary to take into account the equality of keys. Say, if your comparator thought that elements with equal keys should be sent to the right subtree, then in the first step you would have to divide by the key x, and in the second - by x + 1. But if there were no specifics in this matter at all, then the deletion procedure in this option would not have fulfilled the desired at all - after the second step, the middle tree will remain empty, and the desired element will slip somewhere, either left or right, and look for him now.
The operation time is still O (log 2 N), since we applied 2 times Split and 1 time Merge.
Source:
``````public Treap Remove(int x)
{
Treap l, m, r;
Split(x - 1, out l, out r);
r.Split(x, out m, out r);
return Merge(l, r);
}
``````
#### Act of creation
Now, knowing the algorithm for adding an element to the finished Cartesian tree, we can give the simplest way to build a tree from an incoming set of keys: simply add them in turn using the standard algorithm, starting from a tree from one vertex - the first key. Remembering that the add operation is performed in logarithmic time, we get the total execution time of the complete tree construction - O (N log 2 N).
Interestingly, but can it be faster?
As it turned out, in some cases, yes. Let's imagine that the keys to our input come in ascending order. This, in principle, may well happen if these are some freshly created identifiers with auto increment. So, in this case, there is a simple algorithm for constructing a tree in O (N). True, we will have to pay for this with a temporary overhead from memory: to store for each vertex of the tree under construction a link to its ancestor (in fact, it’s not even necessary for each, but this is subtlety).
We will store the link to the last added vertex in the tree. By compatibility, it will be the rightmost one in it - because it has the largest key of all the keys of the tree currently built. Now suppose that the next key x comes in with some priority y. Where to put it?
The last peak is the most right, therefore, she has no right son. If its priority is greater than the y added, then you can simply ascribe a new vertex to the right son and, with a clear conscience, move on to the next key at the entrance. Otherwise, you need to think. The new vertex still has the largest key, so in the end it will definitely become the rightmost vertex of the tree. Therefore, it makes no sense to look for a place to insert it anywhere, except on the rightmost branch. What we need to find is just a place where the priority of the vertex is greater than y. So, we will rise from the rightmost one up the branch, each time checking the priority of the currently viewed vertex. In the end, we either come to the root, or stop somewhere in the middle of the branch.
Suppose we come to the root. Then it turns out that y is greater than all the priorities in the tree. We have no choice but to make (x, y) a new root and hang an old tree on it with our left son.
If we did not come to the root, the situation is similar. At some vertex of the right branch (x 0 , y 0 ) we have y 0 > y . And her immediate right descendant has priority less than y. To preserve the structure of the key search tree and make (x, y) the most right one in the tree, we suspend it as the new right son of (x 0 , y 0 ) , and the entire old right subtree becomes the left subtree (x, y).
To make it clearer, I will illustrate with examples. Take some Cartesian tree and show what happens if you try to add some vertex to it. The key is the same everywhere (22), and the priority is variable.
y = 4:
y = 16:
y = 11:
Why does this algorithm work for O (N)? Note that you will visit a maximum of two times in each life of a tree:
• perhaps adding some other while it remains in the right branch. Immediately after this, she will leave the right branch and will not be visited anymore.
Thus, the total number of transitions does not exceed 2N, and the asymptotic behavior of the construction is O (N).
For starters - the source code for building on a given array of keys and priorities. Here it is assumed that each vertex of the tree has a property `Parent`, and also that the already used private vertex constructor (the fifth in a row) has the same parameter with the same name.
``````public static Treap Build(int[] xs, int[] ys)
{
Debug.Assert(xs.Length == ys.Length);
var tree = new Treap(xs[0], ys[0]);
var last = tree;
for (int i = 1; i < xs.Length; ++i)
{
if (last.y > ys[i])
{
last.Right = new Treap(xs[i], ys[i], parent: last);
last = last.Right;
}
else
{
Treap cur = last;
while (cur.Parent != null && cur.y <= ys[i])
cur = cur.Parent;
if (cur.y <= ys[i])
last = new Treap(xs[i], ys[i], cur);
else
{
last = new Treap(xs[i], ys[i], cur.Right, null, cur);
cur.Right = last;
}
}
}
while (last.Parent != null)
last = last.Parent;
return last;
}
``````
#### Summary
We have built a tree-like data structure with the following properties:
• possesses almost guaranteed logarithmic height with respect to the number of its vertices;
• allows for logarithmic time to search for any key in the tree, add it and delete it;
• the source code of all its methods does not exceed 20 lines, they are easily understood and it is extremely difficult to make mistakes in them
• contains some overhead from memory, in comparison with truly self-balancing trees, for storing priorities.
In principle, the result is quite powerful. However, it may still be unclear to some whether it was worthwhile to make such a big garden with such a ton of text for it. It was worth it. The trick is that the capabilities of the Cartesian tree and the potential for its application are far from being limited to the functions described in this article. This is just a preface.
In the following parts:
1. Multiple operations on a Cartesian tree (looking for O (log 2 N) sum, maximum, etc.)
2. Cartesian tree by implicit key (or how to improve a regular array)
3. Accelerated implementations of Cartesian tree functions (and their problems)
4. Functional implementation of the Cartesian tree in F #.
#### Sources
I indicate the sources once and for all, they are the same for all planned articles.
First of all, when writing these articles, I am based on a lecture by Vitaliy Goldshtein, given at the Kharkov winter school for programming ACM ICPC in 2010. It can be downloaded from the school’s video gallery (year 2010, day 2), as soon as it starts working again, because in recent days the server is catastrophically junk. Maxim’s e-maxx
website by Ivanov is a rich storehouse of information on various algorithms and data structures used in sports programming. In particular, there is an article on the Cartesian tree on it .
In the famous book by Cormen, Leiserson, Rivest, Stein, “Algorithms: Construction and Analysis,” one can find evidence that the expectation of the height of a random binary search tree is O (log 2 N), although its definition of a random search tree is different from what we used here.
Deramides were first proposed in an article by Seidel, Raimund; Aragon, Cecilia R. (1996), Randomized Search Trees . In principle, there you can find the full amount of information on the topic.
That's all for now. I hope you were interested :)
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# I Induced charge density -- non-zero potential case
1. Apr 30, 2017
### mertcan
Hi, Let's think 2 arbitrary shape conductors with non zero charged. If these 2 conductors are closed, there will be induced charge density over surfaces of these conductors. I have not seen such an example, instead there are lots of problems which involve zero(grounded) potential case and method of images theorem is applied. So, I am asking how can we mathematically find induced charge density over conductors' surfaces including non zero potential case(not grounded)?
Thanks....
Last edited by a moderator: Apr 30, 2017
2. Apr 30, 2017
### Henryk
Finding charge density distribution involves solving Poisson's equation $\nabla ^2 U = - \frac {\rho}{\epsilon \epsilon_0}$.
This equation can be solved analytically only in some simple cases when the geometry is rather regular (e.g. planar, cylindrical, spherical, etc.) and these you find in textbooks. In other cases, you can get a power (or other) series expansion. But if you want to solve the Poisson's equation for a completely arbitrary shapes (conductors, dielectrics) you have to use Finite Element Analysis or similar methods.
3. Apr 30, 2017
### mertcan
Then, please let me ask the following question: I am really looking for that kind of examples, could you tell me how I can find some examples and solutions related to calculating charge density with finite element analysis ?
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