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http://conversion.org/mass/pound-avoirdupois/barge
| 1,726,650,366,000,000,000
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crawl-data/CC-MAIN-2024-38/segments/1725700651886.88/warc/CC-MAIN-20240918064858-20240918094858-00633.warc.gz
| 6,933,935
| 6,866
|
# pound (avoirdupois) to barge conversion
Conversion number between pound (avoirdupois) [lb av] and barge is 2.2222222222222 × 10-5. This means, that pound (avoirdupois) is smaller unit than barge.
### Contents [show][hide]
Switch to reverse conversion:
from barge to pound (avoirdupois) conversion
### Enter the number in pound (avoirdupois):
Decimal Fraction Exponential Expression
[lb av]
eg.: 10.12345 or 1.123e5
Result in barge
?
precision 0 1 2 3 4 5 6 7 8 9 [info] Decimal: Exponential:
### Calculation process of conversion value
• 1 pound (avoirdupois) = (exactly) (0.45359237) / (20411.65665) = 2.2222222222222 × 10-5 barge
• 1 barge = (exactly) (20411.65665) / (0.45359237) = 45000 pound (avoirdupois)
• ? pound (avoirdupois) × (0.45359237 ("kg"/"pound (avoirdupois)")) / (20411.65665 ("kg"/"barge")) = ? barge
### High precision conversion
If conversion between pound (avoirdupois) to kilogram and kilogram to barge is exactly definied, high precision conversion from pound (avoirdupois) to barge is enabled.
Decimal places: (0-800)
pound (avoirdupois)
Result in barge:
?
gads
### pound (avoirdupois) to barge conversion chart
Start value: [pound (avoirdupois)] Step size [pound (avoirdupois)] How many lines? (max 100)
visual:
pound (avoirdupois)barge
00
100.00022222222222222
200.00044444444444444
300.00066666666666667
400.00088888888888889
500.0011111111111111
600.0013333333333333
700.0015555555555556
800.0017777777777778
900.002
1000.0022222222222222
1100.0024444444444444
Copy to Excel
## Multiple conversion
Enter numbers in pound (avoirdupois) and click convert button.
One number per line.
Converted numbers in barge:
Click to select all
## Details about pound (avoirdupois) and barge units:
Convert Pound (avoirdupois) to other unit:
### pound (avoirdupois)
Definition of pound (avoirdupois) unit: ≡ 0.45359237 kg = 7000 grains . By definition, 1 lb av ≡ exactly 0.45359237 kg
Convert Barge to other unit:
### barge
Definition of barge unit: ≡ 22 1⁄2 short ton . = 20411.65665 kg
← Back to Mass units
© 2024 conversion.org Terms of use
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| 2.84375
| 3
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CC-MAIN-2024-38
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latest
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en
| 0.291558
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http://stackoverflow.com/questions/11637302/prim-mst-with-adjacency-matrix-as-parameter?answertab=votes
| 1,436,077,958,000,000,000
|
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crawl-data/CC-MAIN-2015-27/segments/1435375097246.96/warc/CC-MAIN-20150627031817-00095-ip-10-179-60-89.ec2.internal.warc.gz
| 251,641,483
| 16,411
|
# Prim MST with Adjacency Matrix as parameter
I'm working with Prim MST in C and the function takes an adjacency matrix. Given the weight of course in `A[i][j]`.
Suppose I have a predecessor array that tracks down the minimum edges i Have found so far. `predecessor[u]=v` {this is also the final MST}
Now I want to modify the current `A[i][j]` matrix and change the weights to 1. That is when the edges (indexes) also existed in the predecessor array. Otherwise I change it to zero.
How would I do it? Here's my solution:
``````for (x....)
for (y...)
if (x!=y && (p[x]==y || p[y]==x))
set to 1
else
set to 0
``````
-
that looks right... What's your actual question? – Spudd86 Jul 24 '12 at 19:04
I'm not totally sure what else to check :S How would you do it? – ms. sakura Jul 24 '12 at 19:08
exactly the way you did... or possibly as in the answer by sdcwc – Spudd86 Jul 25 '12 at 17:51
Your pseudocode is correct, this will give you a 0-1 matrix which represents the tree found by Prim's algorithm. However, this method of storage is rather expensive, since it requires O(n^2) space, while a tree can be stored in O(n) memory.
Alternatively, you can initialize the matrix to zeroes in O(n^2) and then add edges in O(n) time:
`````` for (x ...)
for (y ...)
A[x][y] = 0
for (x ...)
if (p[x] != x)
A[x][p[x]] = A[p[x]][x] = 1
``````
-
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| 3.09375
| 3
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CC-MAIN-2015-27
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latest
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en
| 0.919859
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https://elliottwavegold.com/2014/06/gdx-elliott-wave-technical-analysis-16th-june-2014/
| 1,632,203,314,000,000,000
|
text/html
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crawl-data/CC-MAIN-2021-39/segments/1631780057158.19/warc/CC-MAIN-20210921041059-20210921071059-00323.warc.gz
| 278,996,491
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Select Page
Downwards movement for GDX, like Gold, did not reach the minimum requirement for a flat correction. This changes the wave count for the short to mid term for GDX. I expect now that like Gold it is completing a fourth wave triangle.
Click charts to enlarge.
The clearest piece of movement is the downwards movement from the high. This looks most like a first, second and third wave. This may be the start of a larger correction.
Intermediate wave (3) is \$1.06 longer than 2.618 the length of intermediate wave (1).
Within intermediate wave (3) there are no Fibonacci ratios between minor waves 1, 3 and 5.
Ratios within minor wave 1 of intermediate wave (3) are: minute wave iii has no Fibonacci ratio to minute wave i, and minute wave v is \$2.19 longer than 0.618 the length of minute wave i.
Ratios within minor wave 3 of intermediate wave (3) are: minute wave iii has no Fibonacci ratio to minute wave i, and minute wave v is \$0.63 longer than 0.382 the length of minute wave iii.
Draw a parallel channel about this downwards movement. Draw the first trend line from the lows of intermediate waves (1) to (3), then place a parallel copy upon the high of intermediate wave (2). I would expect intermediate wave (4) to find resistance at the upper edge of the channel, and it may end there.
Intermediate wave (4) should last one to a few months.
I expect that intermediate wave (4) is completing as a regular contracting triangle. Within the triangle all the subwaves must subdivide as corrective structures, one may be a multiple, and four of the five waves must be zigzags or multiple zigzags.
So far minor wave A is a multiple so this means all the remaining subwaves must be simple A-B-C corrections. Minor wave A is technically a double combination so this means all the remaining waves must be simple zigzags.
Within minor wave C minute wave b may not move beyond the end of minute wave a below 21.93.
Minor wave C may not move beyond the end of minor wave A above 28.03.
Overall this wave count now expects to see very choppy overlapping sideways movement for a few weeks yet before this correction is over.
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| 2.671875
| 3
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CC-MAIN-2021-39
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longest
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en
| 0.895645
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https://profound-qa.com/do-objects-slow-down-in-a-vacuum/
| 1,685,491,962,000,000,000
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crawl-data/CC-MAIN-2023-23/segments/1685224646181.29/warc/CC-MAIN-20230530230622-20230531020622-00252.warc.gz
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# Do objects slow down in a vacuum?
## Do objects slow down in a vacuum?
A BALL spinning in a vacuum should never slow down, since no outside forces are acting on it. Even though they appear only fleetingly, these “virtual” photons exert the same electromagnetic forces on the objects they encounter as normal photons do. …
Do objects in space ever stop moving?
Things in space are acted on by forces, usually gravity, and this does change their momentum. But since there is always some force, usually gravity, they never truly stop.
### Do objects speed up in a vacuum?
But Einstein showed that the universe does, in fact, have a speed limit: the speed of light in a vacuum (that is, empty space). It’s impossible to accelerate any material object up to the speed of light because it would take an infinite amount of energy to do so.
Will an object stop accelerating in space?
Originally Answered: Will an object, thrown in space, accelerate or travel at a constant speed? In space there is a pure (well mostly) vacuum so no such thing as air resistance. The object will continue to travel due to no other force being applied to it once it has left your, travelling at a constant speed.
READ: Is it safe to watch a solar eclipse on TV?
#### Do objects in a vacuum fall at the same speed?
The mass, size, and shape of the object are not a factor in describing the motion of the object. So all objects, regardless of size or shape or weight, free fall with the same acceleration. In a vacuum, a beach ball falls at the same rate as an airliner.
Do objects lose momentum in space?
No. Even stationary objects have momentum. Yes, it would eventually slow down through impacts with space dust and gas. There’s not much matter in the vacuum of space, but there is some.
## What happens if you throw a rock in space?
Yes, it will eventually stop, because gravity does not cease to exist in space, as derived by the formula for gravity, which employs two values, the Mass of the object and the Distance of the measurable object from another object.
Can you be perfectly still in space?
There is no ‘still’ that is not relative to some other object. So yes, you can be still with respect to one object, but you’ll be moving with respect to every other object.
READ: How do you report against someone?
### What falls faster in a vacuum?
Explanation: There is no air resistance in a vacuum. This means that under the force of gravity alone, both objects will accelerate at the same rate. Hence, neither object falls faster.
What happens when objects are dropped in a vacuum?
An object that moves because of the action of gravity alone is said to be free falling. So all objects, regardless of size or shape or weight, free fall with the same acceleration. In a vacuum, a beach ball falls at the same rate as an airliner.
#### Is your weight 0 in space?
Astronauts on the orbiting space station are weightless because… a. there is no gravity in space and they do not weigh anything. the astronauts are far from Earth’s surface at a location where gravitation has a minimal effect.
Is there no drag in space?
In space, there’s simply nothing to run into. However, if you’re an object in low-Earth orbit — like the International Space Station is — aerodynamic drag actually does exist.
## How do spacecrafts slow down in space?
As others have said, the only way to slow down is from friction is trace gas in space. Now, spacecraft will slow down as they leave a gravity well, and speed up as they fall into one. Objects in elliptical orbits will slow down as they rise to their apogee, and then speed up when they fall back to perigee. For driven objects, yes of course.
READ: What does MMR mean in games?
Do spinning objects slow down in a perfect vacuum?
Conventional physics holds that in a perfect vacuum, a spinning object would never slow down since there is nothing else to act on it. But this may not be true, according to a new study by Alejandro Manjavacas and F. Javier García de Abajo and recently covered by The New Scientist.
### How do objectobjects slow things down?
Objects will “slow” down when affected by other objects, such as air, rocks or the gravity of other items. So anything you run into will absorb energy, even light itself can apply a minute amount of force.
Why does an object near a black hole slow down?
Any object in space is subject to multiple forces. If the object you are talking about is near a black hole, it will not slow down, it will accelerate. Similarly, the given object will at all times be subject to unequal gravitational forces as it progresses. These could alternate from acceleration to deaceleration.
| 1,006
| 4,681
|
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| 3.078125
| 3
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CC-MAIN-2023-23
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latest
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en
| 0.950582
|
https://www.aqua-calc.com/convert/density/kilogram-per-us-tablespoon-to-slug-per-us-teaspoon
| 1,611,750,372,000,000,000
|
text/html
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crawl-data/CC-MAIN-2021-04/segments/1610704824728.92/warc/CC-MAIN-20210127121330-20210127151330-00528.warc.gz
| 662,859,633
| 8,713
|
# Convert kilograms per tablespoon to slugs per teaspoon
## kg/tbsp to sl/tsp (kg:kilogram, tbsp:tablespoon, sl:slug, tsp:teaspoon)
### Convert kg/tbsp to sl/tsp
#### a density or mass density conversion table
to 100 with
kilo-
gram per US tablespoon
slug per US teaspoonkilo-
gram per US tablespoon
slug per US teaspoonkilo-
gram per US tablespoon
slug per US teaspoonkilo-
gram per US tablespoon
slug per US teaspoonkilo-
gram per US tablespoon
slug per US teaspoon
10.0210.5410.9611.4811.9
20.0220.5421.0621.4821.9
30.1230.5431.0631.4831.9
40.1240.5441.0641.5841.9
50.1250.6451.0651.5851.9
60.1260.6461.1661.5862.0
70.2270.6471.1671.5872.0
80.2280.6481.1681.6882.0
90.2290.7491.1691.6892.0
100.2300.7501.1701.6902.1
110.3310.7511.2711.6912.1
120.3320.7521.2721.6922.1
130.3330.8531.2731.7932.1
140.3340.8541.2741.7942.1
150.3350.8551.3751.7952.2
160.4360.8561.3761.7962.2
170.4370.8571.3771.8972.2
180.4380.9581.3781.8982.2
190.4390.9591.3791.8992.3
200.5400.9601.4801.81002.3
### kilograms per tablespoon to slugs per teaspoon conversion cards
• 1
through
20
kilograms per tablespoon
• 1 kg/tbsp to sl/tsp = 0 sl/tsp
• 2 kg/tbsp to sl/tsp = 0 sl/tsp
• 3 kg/tbsp to sl/tsp = 0.1 sl/tsp
• 4 kg/tbsp to sl/tsp = 0.1 sl/tsp
• 5 kg/tbsp to sl/tsp = 0.1 sl/tsp
• 6 kg/tbsp to sl/tsp = 0.1 sl/tsp
• 7 kg/tbsp to sl/tsp = 0.2 sl/tsp
• 8 kg/tbsp to sl/tsp = 0.2 sl/tsp
• 9 kg/tbsp to sl/tsp = 0.2 sl/tsp
• 10 kg/tbsp to sl/tsp = 0.2 sl/tsp
• 11 kg/tbsp to sl/tsp = 0.3 sl/tsp
• 12 kg/tbsp to sl/tsp = 0.3 sl/tsp
• 13 kg/tbsp to sl/tsp = 0.3 sl/tsp
• 14 kg/tbsp to sl/tsp = 0.3 sl/tsp
• 15 kg/tbsp to sl/tsp = 0.3 sl/tsp
• 16 kg/tbsp to sl/tsp = 0.4 sl/tsp
• 17 kg/tbsp to sl/tsp = 0.4 sl/tsp
• 18 kg/tbsp to sl/tsp = 0.4 sl/tsp
• 19 kg/tbsp to sl/tsp = 0.4 sl/tsp
• 20 kg/tbsp to sl/tsp = 0.5 sl/tsp
• 21
through
40
kilograms per tablespoon
• 21 kg/tbsp to sl/tsp = 0.5 sl/tsp
• 22 kg/tbsp to sl/tsp = 0.5 sl/tsp
• 23 kg/tbsp to sl/tsp = 0.5 sl/tsp
• 24 kg/tbsp to sl/tsp = 0.5 sl/tsp
• 25 kg/tbsp to sl/tsp = 0.6 sl/tsp
• 26 kg/tbsp to sl/tsp = 0.6 sl/tsp
• 27 kg/tbsp to sl/tsp = 0.6 sl/tsp
• 28 kg/tbsp to sl/tsp = 0.6 sl/tsp
• 29 kg/tbsp to sl/tsp = 0.7 sl/tsp
• 30 kg/tbsp to sl/tsp = 0.7 sl/tsp
• 31 kg/tbsp to sl/tsp = 0.7 sl/tsp
• 32 kg/tbsp to sl/tsp = 0.7 sl/tsp
• 33 kg/tbsp to sl/tsp = 0.8 sl/tsp
• 34 kg/tbsp to sl/tsp = 0.8 sl/tsp
• 35 kg/tbsp to sl/tsp = 0.8 sl/tsp
• 36 kg/tbsp to sl/tsp = 0.8 sl/tsp
• 37 kg/tbsp to sl/tsp = 0.8 sl/tsp
• 38 kg/tbsp to sl/tsp = 0.9 sl/tsp
• 39 kg/tbsp to sl/tsp = 0.9 sl/tsp
• 40 kg/tbsp to sl/tsp = 0.9 sl/tsp
• 41
through
60
kilograms per tablespoon
• 41 kg/tbsp to sl/tsp = 0.9 sl/tsp
• 42 kg/tbsp to sl/tsp = 1 sl/tsp
• 43 kg/tbsp to sl/tsp = 1 sl/tsp
• 44 kg/tbsp to sl/tsp = 1 sl/tsp
• 45 kg/tbsp to sl/tsp = 1 sl/tsp
• 46 kg/tbsp to sl/tsp = 1.1 sl/tsp
• 47 kg/tbsp to sl/tsp = 1.1 sl/tsp
• 48 kg/tbsp to sl/tsp = 1.1 sl/tsp
• 49 kg/tbsp to sl/tsp = 1.1 sl/tsp
• 50 kg/tbsp to sl/tsp = 1.1 sl/tsp
• 51 kg/tbsp to sl/tsp = 1.2 sl/tsp
• 52 kg/tbsp to sl/tsp = 1.2 sl/tsp
• 53 kg/tbsp to sl/tsp = 1.2 sl/tsp
• 54 kg/tbsp to sl/tsp = 1.2 sl/tsp
• 55 kg/tbsp to sl/tsp = 1.3 sl/tsp
• 56 kg/tbsp to sl/tsp = 1.3 sl/tsp
• 57 kg/tbsp to sl/tsp = 1.3 sl/tsp
• 58 kg/tbsp to sl/tsp = 1.3 sl/tsp
• 59 kg/tbsp to sl/tsp = 1.3 sl/tsp
• 60 kg/tbsp to sl/tsp = 1.4 sl/tsp
• 61
through
80
kilograms per tablespoon
• 61 kg/tbsp to sl/tsp = 1.4 sl/tsp
• 62 kg/tbsp to sl/tsp = 1.4 sl/tsp
• 63 kg/tbsp to sl/tsp = 1.4 sl/tsp
• 64 kg/tbsp to sl/tsp = 1.5 sl/tsp
• 65 kg/tbsp to sl/tsp = 1.5 sl/tsp
• 66 kg/tbsp to sl/tsp = 1.5 sl/tsp
• 67 kg/tbsp to sl/tsp = 1.5 sl/tsp
• 68 kg/tbsp to sl/tsp = 1.6 sl/tsp
• 69 kg/tbsp to sl/tsp = 1.6 sl/tsp
• 70 kg/tbsp to sl/tsp = 1.6 sl/tsp
• 71 kg/tbsp to sl/tsp = 1.6 sl/tsp
• 72 kg/tbsp to sl/tsp = 1.6 sl/tsp
• 73 kg/tbsp to sl/tsp = 1.7 sl/tsp
• 74 kg/tbsp to sl/tsp = 1.7 sl/tsp
• 75 kg/tbsp to sl/tsp = 1.7 sl/tsp
• 76 kg/tbsp to sl/tsp = 1.7 sl/tsp
• 77 kg/tbsp to sl/tsp = 1.8 sl/tsp
• 78 kg/tbsp to sl/tsp = 1.8 sl/tsp
• 79 kg/tbsp to sl/tsp = 1.8 sl/tsp
• 80 kg/tbsp to sl/tsp = 1.8 sl/tsp
• 81
through
100
kilograms per tablespoon
• 81 kg/tbsp to sl/tsp = 1.9 sl/tsp
• 82 kg/tbsp to sl/tsp = 1.9 sl/tsp
• 83 kg/tbsp to sl/tsp = 1.9 sl/tsp
• 84 kg/tbsp to sl/tsp = 1.9 sl/tsp
• 85 kg/tbsp to sl/tsp = 1.9 sl/tsp
• 86 kg/tbsp to sl/tsp = 2 sl/tsp
• 87 kg/tbsp to sl/tsp = 2 sl/tsp
• 88 kg/tbsp to sl/tsp = 2 sl/tsp
• 89 kg/tbsp to sl/tsp = 2 sl/tsp
• 90 kg/tbsp to sl/tsp = 2.1 sl/tsp
• 91 kg/tbsp to sl/tsp = 2.1 sl/tsp
• 92 kg/tbsp to sl/tsp = 2.1 sl/tsp
• 93 kg/tbsp to sl/tsp = 2.1 sl/tsp
• 94 kg/tbsp to sl/tsp = 2.1 sl/tsp
• 95 kg/tbsp to sl/tsp = 2.2 sl/tsp
• 96 kg/tbsp to sl/tsp = 2.2 sl/tsp
• 97 kg/tbsp to sl/tsp = 2.2 sl/tsp
• 98 kg/tbsp to sl/tsp = 2.2 sl/tsp
• 99 kg/tbsp to sl/tsp = 2.3 sl/tsp
• 100 kg/tbsp to sl/tsp = 2.3 sl/tsp
• kilogram per tablespoon stands for kilogram per US tablespoon
• kilograms per tablespoon stands for kilograms per US tablespoon
• slug per teaspoon stands for slug per US teaspoon
• slugs per teaspoon stands for slugs per US teaspoon
#### Foods, Nutrients and Calories
ANDY'S FAVORITE WIENERS, BABY WIENERS, UPC: 610330013757 contain(s) 222 calories per 100 grams or ≈3.527 ounces [ price ]
Foods high in Zinc, Zn, foods low in Zinc, Zn, and Recommended Dietary Allowances (RDAs) for Zinc
#### Gravels, Substances and Oils
CaribSea, Marine, Arag-Alive, West Caribbean Reef weighs 1 441.7 kg/m³ (90.00239 lb/ft³) with specific gravity of 1.4417 relative to pure water. Calculate how much of this gravel is required to attain a specific depth in a cylindricalquarter cylindrical or in a rectangular shaped aquarium or pond [ weight to volume | volume to weight | price ]
Copper oxide black [CuO] weighs 6 315 kg/m³ (394.23257 lb/ft³) [ weight to volume | volume to weight | price | mole to volume and weight | mass and molar concentration | density ]
Volume to weightweight to volume and cost conversions for Refrigerant R-32, liquid (R32) with temperature in the range of -40°C (-40°F) to 65.56°C (150.008°F)
#### Weights and Measurements
A tonne per cubic millimeter (t/mm³) is a derived metric SI (System International) measurement unit of density used to measure volume in cubic millimeters in order to estimate weight or mass in tonnes
Torque can be defined as a turning or twisting action of the force F upon an object.
t/metric tsp to troy/US tbsp conversion table, t/metric tsp to troy/US tbsp unit converter or convert between all units of density measurement.
#### Calculators
Wedge calculator: The volume and surface area of a wedge.
| 3,048
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|
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| 2.59375
| 3
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CC-MAIN-2021-04
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| 0.178073
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New Features in Maple 2018 - Physics - Maplesoft
# Physics
Maple provides a state-of-the-art environment for algebraic computations in Physics, with emphasis on ensuring that the computational experience is as natural as possible. The theme of the Physics project for Maple 2018 has been the consolidation of the functionality introduced in previous releases, together with significant enhancements, mainly in the handling of differential (quantum or not) tensorial operators, new ways to minimize the number of tensor components taking its symmetries into account, automatic handling of collision of indices in tensorial expressions, automatic setting of the EnergyMomentum tensor when loading solutions to Einstein's equations from the database of solutions, automatic setting of the algebras for the Dirac, Pauli and Gell-Mann matrices when Physics is loaded, simplification of Dirac matrices, a new package Physics:-Cactus related to Numerical Relativity and several other improvements.
Taking all together, there are more than 300 enhancements throughout the entire package, increasing robustness, versatility and functionality, extending once more the range of Physics-related algebraic computations that can be done using computer algebra software, and in a natural way.
As part of its commitment to providing the best possible environment for algebraic computations in Physics, Maplesoft launched a Maple Physics: Research and Development web site with Maple 18, which enabled users to download research versions, ask questions, and provide feedback. The results from this accelerated exchange with people around the world have been incorporated into the Physics package in Maple 2018.
### Automatic handling of collision of tensor indices in products
The design of products of tensorial expressions that have contracted indices got enhanced. The idea: repeated indices in certain subexpressions are actually dummies. So suppose and are tensors, then in , is just dummy, therefore is a well defined object. The new design automatically maps input like into .
>
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(1)
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(2)
This shows the automatic handling of collision of indices
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(3)
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(4)
Consider now the case of three tensors
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(5)
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(6)
The product above has indeed the index repeated more than once, therefore none of its occurrences got automatically transformed into contravariant in the output, and Check detects the problem interrupting with an error message
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However, it is now also possible to indicate, using parenthesis, that the product of two of these tensors actually form a subexpression, so that the following two tensorial expressions are well defined, where the dummy is automatically replaced making that explicit
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(7)
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(8)
This change in design makes concretely simpler the use of indices in that it eliminates the need for manually replacing dummies. For example, consider the tensorial expression for the angular momentum in terms of the coordinates and momentum vectors, in 3 dimensions
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(9)
Define respectively representing angular and linear momentum
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(10)
Introduce the tensorial expression for
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(11)
The left-hand side has one free index, , while the right-hand side has two dummy indices and
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(12)
If we want to computewe can now take the square of directly, and the dummy indices on the right-hand side are automatically handled, there is now no need to manually substitute the repeated indices to avoid their collision
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(13)
The repeated indices on the right-hand side are now
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(14)
### User defined algebraicdifferential operators
A new keyword in Setup: differentialoperators, allows for defining differential operators (not necessarily linear) with respect to indicated differentiation variables, so that they are treated as noncommutative operands in products, as we do with paper and pencil. These user-defined differential operators can also be vectorial and/or tensorial or inert. When desired, one can use Library:-ApplyProductOfDifferentialOperators to transform the products in the function application of these operators. This new functionality is a generalization of the differential operators and , and can used beyond Physics.
A new routine Library:-GetDifferentiationVariables also acts on a differential operator and tells who are the corresponding differentiation variables
Example:
In Quantum Mechanics, in the coordinates representation, the component of the momentum operator along the x axis is given by the differential operator
The purpose of the exercises below is thus to derive the commutation rules, in the coordinates representation, between an arbitrary function of the coordinates and the related momentum, departing from the differential representation
;
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Start setting the problem:
• all of are Hermitian operators
• all of commute between each other
• tell the system only that the operators are the differentiation variables of the corresponding (differential) operators but do not tell what is the form of the operators
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(15)
Assuming is a smooth function, the idea is to apply the commutator to an arbitrary ket of the Hilbert space , perform the operation explicitly after setting a differential operator representation for , and from there get the commutation rule between and .
Start introducing the commutator, to proceed with full control of the operations we use the inert form %Commutator
>
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(16)
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(17)
For illustration purposes only (not necessary), expand this commutator
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(18)
Note that , and the ket are operands in the products above and that they do not commute: we indicated that the coordinates are the differentiation variables of . This emulates what we do when computing with these operators with paper and pencil, where we represent the application of a differential operator as a product operation.
This representation can be transformed into the (traditional in computer algebra) application of the differential operator when desired, as follows:
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(19)
Note that, in , the application of is not expanded: at this point nothing is known about , it is not necessarily a linear operator. In the Quantum Mechanics problem at hands, however, it is. So give now the operator an explicit representation as a linear vectorial differential operator (we use the inert form %Nabla, , to be able to proceed with full control one step at a time)
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(20)
The expression (19) becomes
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(21)
Activate now the inert operator and simplify taking into account the algebra rules for the coordinate operators
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(22)
To make explicit the gradient in disguise on the right-hand side, factor out the arbitrary ket
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(23)
Combine now the expanded gradient into its inert (not-expanded) form
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(24)
Since is true for all, this ket can be removed from both sides of the equation. One can do that either taking coefficients or multiplying by the "formal inverse" of this ket, arriving at the (expected) form of the commutation rule between and
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(25)
### Tensor notation,
The computation rule for position and momentum, this time in tensor notation, is performed in the same way, just that, additionally, specify that the space indices to be used are lowercase Latin letters, and set the relationship between the differential operators and the coordinates directly using tensor notation.
You can also specify that the metric is Euclidean, but that is not necessary: the default metric of the Physics package, a Minkowski spacetime, includes a 3D subspace that is Euclidean, and the default signature, (- - - +), is not a problem regarding this computation.
>
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(26)
Define now the tensor
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(27)
Introduce now the Commutator, this time in active form, to show how to reobtain the non-expanded form at the end by resorting the operands in products
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(28)
Expand first (not necessary) to see how the operator is going to be applied
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(29)
Now expand and directly apply in one ago the differential operator ;
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(30)
Introducing the explicit differential operator representation for , here again using the inert to keep control of the computations step by step
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(31)
The expanded and applied commutator (30) becomes
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(32)
Activate now the inert operators and simplify taking into account Einstein's rule for repeated indices
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(33)
Since the ket is arbitrary, we can take coefficients (or multiply by the formal Inverse of this ket as done in the previous section). For illustration purposes, we use Coefficients and note how it automatically expands the commutator
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(34)
One can undo this (frequently undesired) expansion of the commutator by sorting the products on the left-hand side using the commutator between and
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(35)
And that is the result we wanted to compute.
Additionally, to see this rule in matrix form,
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(36)
In the above, we use equation (30) multiplied by -1 to avoid a minus sign in all the elements of (36), due to having worked with the default signature (- - - +); this minus sign is not necessary if in the Setup at the beginning one also sets
For display purposes, to see this matrix expressed in terms of the geometrical components of the momentum , redefine the tensor explicitly indicating its Cartesian component
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(37)
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(38)
Finally, in a typical situation, these commutation rules are to be taken into account in further computations, and for that purpose they can be added to the setup via
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(39)
For example, from herein computations are performed taking into account that
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(40)
### Automatic setting of the EnergyMomentumTensor for metrics of the database of solutions to Einstein's equations
There are 991 metrics in the database of solutions to Einstein's equations, based on the book "Exact solutions to Einstein's equations". One can check this number via
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(41)
For the majority of these solutions, the book also presents, explicit or implicitly, the form of the Energy-Momentum tensor. New in Maple 2018, we added to the database one more entry indicating the components of the corresponding EnergyMomentum tensor, covering, in Maple 2018.0, 686 out of these 991 solutions.
The design of the EnergyMomentum tensor got slightly adjusted to take these new database entries into account, so that when you load one of these solutions, if the corresponding entry for the EnergyMomentumTensor is already in the database, it is automatically loaded together with the solution.
In addition, it is now possible to define the tensor components using the Define command, or redefine any of its components using the new Library:-RedefineTensorComponent routine.
Examples
>
Consider the metric of Chapter 12, equation number 16.1
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(42)
New, the covariant components of the EnergyMomentum tensor got automatically loaded, given by
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(43)
One can verify this checking for the tensor's definition
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(44)
Take now the tensor components of the first defining equation of (44)
>
(45)
where is related to Newton’s constant.
To see the continuity equations for the components of , use for instance the inert version of the covariant derivative operator D_ and the TensorArray command
>
(46)
>
(47)
>
(48)
The EnergyMomentum tensor can also be (re)defined in any particular way (a correct definition must satisfy ).
Example:
Define the EnergyMomentum tensor indicating the functionality in the definition in terms of to be constant energy (i.e. no functionality) and the flux density and stress tensors depending on For this purpose, use the new option minimizetensorcomponents to make explicit the symmetry of the stress tensor
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(49)
The symmetry of is now explicit in that its matrix form is symmetric
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(50)
The new routines for testing tensor symmetries
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(51)
The symmetry is regarding interchanging positions of the 1st and 2nd indices
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(52)
So this is the form of the EnergyMomentum with all its components - but for the total energy - depending on the coordinates
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(53)
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(54)
Define now with these components
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(55)
>
(56)
To see the continuity equations for the components of , use again the inert version of the covariant derivative operator D_ and the TensorArray command
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(57)
For a more convenient reading, present the result as a vector column
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(58)
Comparing the specific form (43) for the EnergyMomentum loaded from the database of solutions to Einstein's equations with the general form (56), one can ask the formal question of whether there are other forms for the EnergyMomentum satisfying the continuity equations (58).
To answer that question, rewrite this system of equations for the flux density and stress tensors as a set, and solve it for them
>
(59)
This system in fact admits much more general solutions than (43):
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(60)
This solution can be verified in different ways, for instance using pdetest showing it cancels the PDE system (59) for and stress
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(61)
### Minimize the number of tensor components according to its symmetries, and relabel, redefine or count the number of independent tensor components
A new keyword in Define and Setup: minimizetensorcomponents, allows for automatically minimizing the number of tensor components taking into account the tensor symmetries. For example, if a 2-tensor in a 4D spacetime is defined a antisymmetric, the number of different tensor components is 6, and the elements of the diagonal are automatically set equal to 0. After setting this keyword to true with Setup, all subsequent definitions of tensors automatically minimize the number of components while using this keyword with Define makes this minimization only happen with the tensors being defined in the call to Define.
Related to this new functionality, 4 new Library routines were added: MinimizeTensorComponents, NumberOfIndependentTensorComponents, RelabelTensorComponents and RedefineTensorComponents
Example:
>
Define an antisymmetric tensor with two indices
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(62)
Although the system knows that is antisymmetric,
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(63)
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(64)
by default the compotes of do not automatically reflect that, it is necessary to use the simplifier of the Physics package, Simplify.
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(65)
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(66)
Likewise, computing the array form of we do not see the elements of the diagonal equal to nor the lower-left triangle equal to the upper-right triangle but with a different sign:
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(67)
This new functionality, here called minimizetensorcomponents, makes the symmetries of the tensor explicitly reflected in its components. There are three ways to use it. First, one can minimize the number of tensor components of a tensor previously defined. For example
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(68)
After this, both and are automatically equal to 0 without having to use Simplify
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(69)
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(70)
And the output of TensorArray in becomes equal to .
NOTE: after using minimizetensorcomponents in the definition of a tensor, say F, all the keywords implemented for Physics tensors are available for the F:
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(71)
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(72)
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(73)
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(74)
Alternatively, one can define a tensor, specifying that the symmetries should be taken into account to minimize the number of its components passing the keyword minimizetensorcomponents to Define.
Example:
Define a tensor with the symmetries of the Riemann tensor, that is, a tensor of 4 indices that is symmetric with respect to interchanging the positions of the 1st and 2nd pair of indices and antisymmetric with respect to interchanging the position of its 1st and 2nd indices, or 3rd and 4th indices, and minimizing the number of tensor components
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(76)
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(77)
One can always retrieve the symmetry properties in the abstract notation used by the Define command using the new , its output is ordered, first the symmetric then the antisymmetric properties
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(78)
After making the symmetries explicit (and also before that), it frequently s useful to know the number of independent components of a given tensor. For this purpose use the new
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(79)
and besides the symmetries, in the case of the Riemann tensor after taking into account the first Bianchi identity, this number of components is further reduced to 20.
A third way of using the new minimizetensorcomponents functionality is using Setup, so that every subsequent definition of tensors with symmetries is automatically performed minimizing the number of components.
Example:
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(80)
You can now define without having to include the keyword in the definition of tensors with symmetries
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(81)
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(82)
• Two new related functionalities are provided via and , the first one to have the number of tensor components directly reflected in the names of the components, the second one to redefine only one of these components
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(83)
Suppose now we want to make one of these components equal to 1, say
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(84)
### New functionality and display for inert names and inert tensors
New: as part of the developments of Physics bug regardless of loading the Physics package, inert names are now typeset in gray, the standard for inert functions, with copy & paste working
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(85)
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Note that this was already in place in previous releases regarding inert functions but not regarding inert names. Regarding inert functions, since Maple 2016 their typesetting is also in grey with copy & paste working
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(87)
Regarding Physics, having the right typeset also for symbols and tensor names is particularly relevant now that one can compute with differential operators as operands of a product
Example:
>
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(88)
The active and inert representations of the same differential-vectorial operator are
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Hence, you can:
a) assign a mapping to p_ while represent it using %p_ when you do not want the mapping to be applied.
b) using %p_ has mathematical and clear typesetting (in gray always means inert) making its use more pleasant / easy to read.
Example:
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Assign a procedure to the differential-vectorial operator
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(90)
Apply both the active and the inert operators to some function of the coordinates
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(91)
Apply now the differential operators in products: the left-hand side, inert, remains a product, while the right-hand side, becomes a function application, and so gets applied
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(92)
NOTE: the implementation is such that if p is noncommutative, then so is %p and the same holds regarding their possibly differential operator and tensorial character: the inert versions inherit the properties of their active counterparts, and the same regarding their tensorial character: if p is a tensor, so is %p. In addition, inert tensors are also now displayed the same way as their active versions but in gray, improving the readability of tensorial expressions
Example:
Load a curved spacetime, for instance Schwarzschild's metric
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(93)
Define a tensor and compute its covariant derivative equating the inert with the active form of it
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(94)
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(95)
Since is a tensor, so is , and the latter is typeset as the former, only in gray, with copy & paste working, reproducing
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(96)
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(97)
Also new, the same holds for the typesetting of inert differential operators, the inert display of the covariant derivative symbol is now the same as the active one only in gray.
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(98)
Expand the right-hand side and replace the active Christoffel by its inert counterpart %Christoffel
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(99)
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(100)
Compute the components of the left and right hand sides of this tensorial equation, it is a 2x2 matrix of equations; check one of these components: the readability of the inert symbols entering the equation is now straightforward due to the typesetting of inert tensors
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(101)
These expression are not just typeset but true inert representations of the underlying computations. To transform them into active, use value
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(102)
### Automatic setting of Pauli, Dirac and Gell-Mann algebras
New in Maple 2018, the algebra rules for the Dirac, Pauli and Gell-Mann matrices are automatically set when Physics is loaded. This means expressions can be simplified taking into account these algebra rules, which can also be queried using a new Library:-DefaultAlgebraRules routine that returns the algebras in terms of generic lowercase Latin indices
>
The algebra rules for all of the Pauli, Dirac and Gell-Mann matrices, in a 4D spacetime, are given according to the value of the signature by
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(103)
Simplification of products of Dirac matrices work automatically taking the corresponding (3rd) algebra rule into account; for example (see also next section)
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(104)
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(105)
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(106)
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(107)
The convention for follows the Landau, Bogoliubov and Tong books on quantum fields, so for the default signature
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using the new routine, we have
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This new routine takes into account the possibly Euclidean character of spacetime, also the possible values of the signature: (- - - +), (+ - - -), (+ + + -) and (- + + +), and uses formulas valid for the three representations: standard, chiral and majorana.
The algebra rules now automatically loaded when Physics is loaded can always overwritten. For instance, to represent the algebra of Dirac matrices with an identity matrix on the right-hand side, one can proceed as follows.
First create the identity matrix. To emulate what we do with paper and pencil, where we write to represent an identity matrix without having to see the actual table 2x2 with the number 1 in the diagonal and a bunch of 0, I will use the matrix command, not the Matrix one. One way of entering this identity matrix is
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(110)
Depending on the context, the advantage of matrix versus Matrix is that is also of type algebraic
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(111)
Consequently, one can operate with it algebraically without displaying its contents (not possible with Matrix)
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(112)
Most commands of the library only work with objects of type algebraic, all these will be able to handle this matrix, and the contents is displayed only on demand, for instance using eval
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(113)
Set now the algebra for Dirac matrices with an matrix on the right-hand side
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(114)
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(115)
Verifying
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Set now a Dirac spinor (in a week from today, this will be possible directly using Physics:-Setup, but today, here, I do it step-by-step)
For that you can also use {vector, matrix, array} or {Vector, Matrix, Array}, and again, if you use the Upper case commands, you always have the components visible, and cannot compute with these object using commands that require the input to be of type algebraic. So I use matrix, not Matrix, and matrix instead of vector so that the Dirac spinor that is both algebraic and matrix, is also displayed in the usual display as a "column vector"
To reuse the letter Psi which in Maple represents the Psi function, use a local version of it
>
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(117)
Specify the components of the spinor, in any preferred way, for example using
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(118)
Check it out:
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(119)
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(120)
Let’s see all this working together by multiplying the anticommutator equation by
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(121)
To see the matrix form of this equation use the new routine
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(122)
Or directly rewrite, then perform, in one go, the matrix operations behind (121)
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(123)
REMARK: As shown above, in general, the representation using lowercase commands allows you to use `*` or `.` depending on whether you want to represent the operation or perform the operation. For example this represents the operation, as an exact mimicry of what we do with paper and pencil, both regarding input and output
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(124)
And this performs the operation
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(125)
Or to only displaying the operation
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(126)
Finally, another way to visualize the equations behind (121) is
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(127)
Or additionally performing the matrix operations of each of the equations in (127) using the new option performmatrixoperations of TensorArray
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(128)
### Simplification of products of Dirac matrices
The computation of traces of products of Dirac matrices was implemented years ago - see Physics,Trace.
The simplification of products of Dirac matrices, however, was not. Now in Maple 2018 it is.
>
>
First of all, when loading Physics, a frequent question is about the signature, the default is (- - - +)
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(129)
This is important because the convention for the Algebra of Dirac Matrices depends on the signature. With the signatures (- - - +) as well as (+ - - -), the sign of the timelike component is 1
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(130)
With the signatures (+ + + -) as well as (- + + +), the sign of the timelike component is -1
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(131)
The simplification of products of Dirac Matrices, illustrated below with the default signature, works fine with any of these signatures, and works without having to set a representation for the Dirac matrices -- all the results are representation-independent.
Consider now the following five products of Dirac matrices
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(135)
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(136)
New: the simplification of these products is now implemented
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(137)
Verify this result performing the underlying matrix operations
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(138)
Note that in (138) the right-hand side has no matrix elements. This is standard in particle physics where the computations with Dirac matrices are performed algebraically. For the purpose of actually performing the underlying matrix operations, however, one may want to rewrite the algebra of Dirac matrices including a 4x4 identity matrix. For that purpose, see Algebra of Dirac Matrices with an identity matrix on the right-hand side. For the purpose of this illustration, below we proceed with as shown in (138), interpreting right-hand sides as if they involve an identity matrix.
To verify these results checking the components of the matrices involved we set a representation for the Dirac matrices, for example the standard one
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(139)
Verify using the new routine
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(140)
The same with the other expressions
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(141)
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(142)
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(143)
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(144)
For e2
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(145)
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(146)
To make the output simpler, use the new option performmatrixoperations of TensorArray, accomplishing two steps in one go
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(147)
For the other two products of Dirac matrices we have
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(148)
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(149)
These results can be verified in the same way done for the simplification of e2 and e1, as shown in this page, under Library routines to perform matrix operations in expressions involving spinors with omitted indices
Finally, let's define some tensors and contract their product with these expressions involving products of Dirac matrices.
Example
>
(150)
Contract with e1 and e2 and simplify
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(151)
>
(152)
### New Library routines to perform matrix operations in expressions involving spinors with omitted indices
New Physics:-Library routines, RewriteInMatrixForm, and PerformMatrixOperations where added, to rewrite or perform the matrix operations implicit in expressions where the spinor indices are omitted. In some sense, the new Library:-PerformMatrixOperations is a Matrix version (or spinor version) of the TensorArray command, which now also has a new option, performmatrixoperations to additionally perform the matrix or spinor operations in expressions
Example:
The computation of traces of products of Dirac matrices was implemented years ago.
The simplification of products of Dirac matrices, however, was not. Now it is.
>
Set a representation for the Dirac matrices, say the standard one
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(153)
An Array with the four Dirac matrices are
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(154)
The definition of the Dirac matrices is implemented in Maple following the conventions of Landau books ([1] Quantum Electrodynamics, V4), and does not depend on the signature, i.e. the form of these matrices is, using the new
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(155)
With the default signature (- - - +), the space part components of change sign when compared with corresponding ones from while the timelike component remains unchanged
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(156)
>
(157)
With this default signature (- - - +), the algebra of the Dirac Matrices, now loaded by default when Physics is loaded, is (see page 80 of [1])
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(158)
You can also display this algebra, with generic indices j, k, using the new Library routine for this purpose
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(159)
Verify the algebra rule by performing all the involved matrix operations
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(160)
Note that, regarding the spacetime indices, this is a 4x4 matrix, whose elements are in turn 4x4 matrices. Compute first the external 4x4 matrix related to and
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(161)
Perform now all the matrix operations involved in each of the elements of this 4x4 matrix: you can do this by using the new option performmatrixoperations of TensorArray or using the new Library routine for this purpose
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(162)
By eye everything checks OK. Note as well that in (162) the right-hand sides have no matrix elements. This is standard in particle physics where the computations with Dirac matrices are performed algebraically. For the purpose of actually performing the underlying matrix operations, however, one may want to rewrite this algebra including a 4x4 identity matrix on the right-hand sides. For that purpose, see the MaplePrimes post Algebra of Dirac Matrices with an identity matrix on the right-hand side. For the purpose of this illustration, below we proceed with the algebra as shown in (162), interpreting right-hand sides as if they involve an identity matrix.
Consider now the following five products of Dirac matrices
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(163)
>
(164)
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(165)
>
(166)
>
(167)
New: the simplification of these products is now implemented
>
(168)
Verify this result performing the underlying matrix operations
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(169)
>
(170)
The same with the other expressions
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(171)
>
(172)
>
(173)
>
(174)
For e2
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(175)
>
(176)
>
(177)
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(178)
For e3 we have
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(179)
Verify this result,
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(180)
In this case, with three free spacetime indices , the spacetime components form an array 4x4x4 of 64 components, each of which is a matrix equation
>
(181)
For instance, the first element is
>
(182)
and it checks OK:
>
(183)
How can you test the 64 components of T all at once?
1. Compute the matrices, without displaying the whole thing, take the elements of the array and remove the indices (i.e. take the right-hand side); call it M
>
For instance,
>
(184)
Now verify all these matrix equations at once: take the elements of the arrays on each side of the equations and verify that the are the same: we expect for output just
>
(185)
The same for e4
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(186)
>
(187)
Regarding the spacetime indices this is now an array 4x4x4x4
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(188)
For instance the first of these 256 matrix equations
>
(189)
verifies OK:
>
(190)
Now all the 256 matrix equations verified at once as done for e3
>
>
(191)
### Miscellaneous improvements
A large number of miscellaneous improvements happened for Physics in Maple 2018. For brevity, these improvements are only listed; among the most relevant ones:
• There are six new Library routines, DefaultAlgebraRules, Dgamma5ToOtherDgamma, GetDifferentiationVariables, GetRepeatedIndices, IsSymmetricMatrix, RewriteInMatrixForm, all of which are illustrated in the previous sections.
• All the tensors of 2 indices automatically defined in the Physics package, as well as those you can define using Define with a defining equation, or indicating its symmetries and using the new option minimizetensorcomponents of the Define command, accept a new keyword, trace to represent their trace.
• The rewriting in terms of of constructions in a Minkowski spacetime can now be computed using
• Change in the design: allow the derivative to proceed also when is noncommutative by returning unevaluated. In this way one can afterwards define commutation rules such that derivative can be computed.
• Make when and are noncommutative but commute between themselves.
• Make and that belong to one and the same coordinate system always commute, even if and are quantum operators.
• Make all the coordinates of system set as a quantum operator that commutes with another quantum operator V, also commute with V.
• Add a differentiation rule for Bracket
• Physics:-TensorArray: add a new keyword performmatrixoperations (illustrated in the previous sections) and, although unusual, if the input is an Array or Matrix of tensorial equations, map over their elements.
• Allow to set , a coordinate system vector, as a quantum operator. With that, automatically, , is of type noncommutative.
• When defining a Commutator algebra between indexed objects, that are afterwards defined as tensors, now results in the same behavior as when first defining the tensors then the algebra rules.
• Tensor input like now return a 3x3 matrix when are space indices.
• The Coefficients now automatically maps over equations
• The Library:-GetTensorSymmetryProperties got entirely rewritten significantly enhanced, and now it always returns sets of sets (automatically normalized), not anymore sets of lists (that do not normalize automatically). This helps, for example, in perceiving by eye that the symmetries of two tensors are actually the same.
• Reorder the flow within Setup so that first the notation is set, then coordinates followed by the TensorSimplifier and the realojects, and only then the metric is set, that gets simplified taking all of the previous ones into account. This permits to update coordinates dimension, signature, metric and other things all in one go, a single call to Setup.
• There are some new Setup options, including differentialoperators (illustrated in the previous sections) and usecoordinatesastensorindices that permits indexing tensors directly with the coordinates (easier to remember) instead of with the coordinate's positions. Note: when using usecoordinatesastensorindices the tensors can still also be indexed with the coordinate's positions.
• Reorder the flow within Define so that first the tensors whose definition is not an equation definition are define, then those that are defined with defining equations. The former definitions can be performed right away and these tensors can now appear on the right-hand sides of other tensor definitions being defined simultaneously with defining equations.
• Change in the display of covariant derivatives represented using the D_ command: instead of we now use , an upside down triangle that is slightly different from Nabla, used in Physics:-Vectors.
• Add more types to the Physics:-Library:-PhysicsType package of Physics types, that in Maple 2018 comes with 92 specialized types
• Allow geometrical coordinates to be defined as tensors (so they become of type Physics:-Tensor) instead of interrupting with an error due to the coordinates being assigned as it was the case of previous releases.
• Make for the Dagger of the SpaceTimeVector (any coordinate system defined through Setup or Coordinates).
• Make Dagger and d/dx commute also when x is Hermitian (of Library:-PhysicsType:-Hermitian).
• Implement Dgamma[definition] as the algebra rules satisfied by the Dirac gamma matrices.
• Make the error message by Check, for instance when indices appear repeated more than once, to also indicate the actual expression where free and/or repeated indices are found to be incorrect.
• The typesetdot and typesetprime Typesetting rules are now automatically enabled when Physics is loaded.
• Full revision of the conventions for the Dirac matrices in the different representations and Euclidean space, with Dgamma[5] now following the convention used in the Landau, Bogoliubov and Tong books on Quantum Fields.
• TransformTetrads: when computing a canonicalform for a tetrad the starting tetrad is expected to be in null form, or if it is orthonormal it is automatically converted to null for before proceeding with rotations to achieve a canonical form.
• Add a new keyword to Physics:-Library:-TensorComponent.
• More documentation: new help pages showing how to perform with the Physics packages the computations one could perform in the past with the (now deprecated) tensor Maple package and the (non-Maplesoft) GRTensor package.
## Bereit für den nächsten Schritt?
*Die Maple-Evaluation ist für Schüler und Studenten bzw. die private Nutzung zurzeit nicht verfügbar.
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# Trigonometry Formulas for Class 10th, 11th, 12th in Hindi PDF
Trigonometry formula in hindi ( trikonmiti formula ): Hello Friends. Is post me hum aapko Trigonometry Formulas for Class 10th in Hindi PDF provide karne wale hai jisme hum aapko trigonometry all formulas pdf download, trigonometry tricks for ssc cgl pdf and agar aap class 11th. 12th ke students hai to aapke liye bhi is pdf me all trigonometry formula pdf available hai.
Contents
## Trigonometry Formulas for Class 10th in Hindi PDF
Is post me aapko 2 pdf provide ki gyi hai jise aap diye link se download kar sakte hai.
Trigonometry Formula PDF – 1 Info
Name of eBook Trigonometry Formula PDF
Format PDF
Pages 5
Trigonometry Formula PDF – 2 Info
Name of eBook Trigonometry MCQs questions
Format PDF
Pages 36
Trigonometry Formula PDF – 3 Info
Name of eBook TRIGONOMETRIC EQUATION AND IDENTITIES
Format PDF
Pages 22
## Trigonometry Formulas for Class 11th and 12th
### त्रिकोणमिति अनुपात (Trigonometric Ratio)
• sinθ = लम्ब /कर्ण
• cosθ = आधार /कर्ण
• tanθ = लम्ब /आधार
• cotθ = आधार /लम्ब
• secθ = कर्ण /आधार
• cosecθ = कर्ण /लम्ब
### त्रिकोणमिति सम्बन्ध(Trigonometric Relation)
• sinθ.cosecθ=1
• cosθ.secθ=1
• tanθ.cotθ=1
• tanθ=sinθ/cosθ
## Mathematics Formula Trigonometry
1. Sin²θ +cos²θ=1
2. 1+tan²θ= sec²θ
3. 1+cot²θ= cosec²θ
4. sin(A+B)= sinA.cosB+cosA.sinB
5. sin(A-B)= sinA.cosB-cosA.sinB
6. cos(A+B)= cosA.cosB-sinA.sinB
7. cos(A-B)= cosA.cosB+sinA.sinB
8. tan(A+B)= (tanA+tanB)/(1-tanA.tanB)
9. tan(A-B)= (tanA-tanB)/(1+tanA.tanB)
10. sin²A= 2sinA.cosA
11. cos²A= cos²A-sin²A=2cos²A-1=1-2sin²A
12. tan²A= 2tanA/(1-tan²A)
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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!)
A330556 a(n) = (number of primes p <= 2*n+1 with Delta(p) == 2 mod 4) - (number of primes p <= 2*n+1 with Delta(p) == 0 mod 4), where Delta(p) = nextprime(p) - p. 10
0, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 2, 2, 3, 4, 4, 4, 3, 3, 4, 3, 3, 4, 4, 4, 5, 5, 5, 6, 7, 7, 7, 6, 6, 7, 8, 8, 8, 7, 7, 8, 8, 8, 7, 7, 7, 7, 6, 6, 7, 6, 6, 7, 6, 6, 7, 7, 7, 7, 7, 7, 7, 6, 6, 7, 7, 7, 8, 9, 9, 9, 9, 9, 10, 11, 11, 11, 12, 12, 12, 11, 11, 12, 12, 12, 13, 13, 13 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,3 COMMENTS a(n) = A330557(n) - A330558(n). Since Delta(prime(n)) grows roughly like log n, this probably changes sign infinitely often. When is the next time a(n) is zero, or the first time a(n) < 0 (if these values exist)? LINKS N. J. A. Sloane, Table of n, a(n) for n = 0..99999 StackExchange, Asymptotic Distribution of Prime Gaps in Residue Classes. EXAMPLE n=5, 2*n+1=11: there are three primes <= 11 with Delta(p) == 2 mod 4, namely 3,5,11; and one with Delta(p) == 0 mod 4, namely 7; so a(5) = 3-1 = 2. CROSSREFS Sequences related to the differences between successive primes: A001223 (Delta(p)), A028334, A080378, A104120, A330556-A330561. Sequence in context: A161072 A161111 A161046 * A341674 A230404 A082647 Adjacent sequences: A330553 A330554 A330555 * A330557 A330558 A330559 KEYWORD nonn AUTHOR N. J. A. Sloane, Dec 29 2019 STATUS approved
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Last modified August 3 07:58 EDT 2021. Contains 346435 sequences. (Running on oeis4.)
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# Show that the quadrilateral formed
Question:
Show that the quadrilateral formed by joining the consecutive sides of a square is also a square.
Solution:
Given In a square $A B C D, P, Q, R$ and $S$ are the mid-points of $A B$, $B C, C D$ and $D A$, respectively.
To show $P Q R S$ is a square.
Construction Join $A C$ and $B D$.
Proof Since, $A B C D$ is a square.
$\therefore \quad A B=B C=C D=A D$
Also, $P, Q, R$ and $S$ are the mid-points of $A B, B C, C D$ and $D A$, respectively.
Then, in $\triangle A D C$, $S R \| A C$
and $S R=\frac{1}{2} A C$ [by mid-point theorem]...(i)
$\ln \triangle A B C$, $P Q \| A C$
and
$P Q=\frac{1}{2} A C$ ... (ii)
From Eqs. (i) and (ii),
$S R \| P Q$ and $S R=P Q=\frac{1}{2} A C$ ...(iii)
Similarly, $S P \| B D$ and $B D \| R Q$
$\therefore$ $S P \| R Q$ and $S P=\frac{1}{2} B D$
and $R Q=\frac{1}{2} B D$
$\therefore$ $S P=R Q=\frac{1}{2} B D$
Since, diagonals of a square bisect each other at right angle.
$\therefore \quad A C=B D$
$\Rightarrow \quad S P=R Q=\frac{1}{2} A C$ $\ldots(\mathrm{i} \mathrm{v})$
From Eqs. (iii) and (iv), $\quad S \dot{R}=P Q=S P=R Q$ [all side are equal]
$O E \| F R$ and $O F \| E R$
$\therefore \quad \angle E O F=\angle E R F=90^{\circ}$
Hence, $P Q R S$ is a square.
Hence proved.
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How To Calculate Gas Fired Boiler Thermal Efficiency
How To Calculate Gas Fired Boiler Thermal Efficiency
cfb boiler efficiency – Industrial Coal-fired Steam
Ways to Calculate CFB Boiler Efficiency. CFB Boiler Thermal Efficienc. Thermal efficiency measures how well a CFB boiler's heat exchanger transfers heat from the combustion process to the water or steam in the boiler. Circulating Fluidized-Bed (IR-CFB) Boilers
how to calculate gas fired boiler thermal efficiency
How to calculate boiler efficiency by Direct Method . Boiler Efficiency:-Thermal efficiency of boiler is defined as the percentage of heat input that is effectively utilized to generate steam. There are two methods of assessing boiler efficiency. 1) Boiler Efficiency by Direct Method: Where. Get a Quote
Waste heat recovery boiler efficiency calculation,whrb
2020-8-25 · In general, we take this formula to calculate waste heat boiler thermal efficiency. η= 100-(q2+q3+q4+q5+q6), which include the Q2 flue gas losses, Q3 gas incomplete combustion losses, Q4 solid incomplete combustion losses, Q5 heat radiation losses and Q6 slag ash physical losses. The key points for calculation is to make sure the heat losses
Calculation Method: CALCM:02 - BRE
2016-11-15 · A combination boiler with an internal hot water store (primary or secondary/potable water) of capacity at least 15 litres but less than 70 litres. OR A combination boiler with an internal secondary/potable hot water store of capacity at least 70 litres. OR A combination boiler with an internal primary hot water store of capacity at least 70 litres,
how to calculate boiler efficiency in thermal power plant
Efficiency of a Coal Fired Boiler in a Typical Thermal Power Plant 01-01-2017 · In direct method, the efficiency is evaluated by dividing the heat output with the fuel power (input) of the boiler, whereas indirect method considers the ratio of sum of major losses to the fuel power input of the boiler, which will be finally subtracted from
Combustion Efficiency and Excess Air
2021-2-12 · The efficiency loss in the flue gas related to. temperature difference in flue gas and supply air ; CO 2 concentration in the flue gas; with oil combustion is indicated below: Example - Oil Combustion and Heat Loss in the Flue Gas. If. the temperature difference between the flue gas leaving a boiler and the ambient supply temperature is 300 o C
Methods for Calculating CHP Efficiency | Combined
2020-7-16 · Typical boiler efficiencies are 80 percent for natural gas-fired boilers, 75 percent for biomass-fired boilers, and 83 percent for coal-fired boilers. The calculation of effective electric efficiency is the CHP net electric output divided by the additional fuel the CHP system consumes over and above what would have been used by a boiler to
Methods to Calculate CFB Boiler Efficiency
A Method for Calculating Thermal Efficiency Of the - Atlantis Press. Keywords: Thermal efficiency;Solid fuel fired boiler;on-line calculation established the standard methods for calculating the boiler energy efficiency,including ASME PTC Jiao, Comparison to Thermal Efficiency Calculation of CFB Boiler.
THERMAL POWER: Boiler Efficiency calculations
2020-2-28 · Efficiency is a very important criterion in Boiler selection and Design. Efficiency figure depends upon the type of boiler as well as on the type of fuel and it's constituents. For example, efficiency of a Bagasse fired boiler is about 70% where as that of oil fired boilers is about 85 %. Higher moisture content in Bagasse reduces it's efficiency.So better criterion is efficiency based on
how to calculate gas fired boiler thermal efficiency
How to calculate boiler efficiency by Direct Method . Boiler Efficiency:-Thermal efficiency of boiler is defined as the percentage of heat input that is effectively utilized to generate steam. There are two methods of assessing boiler efficiency. 1) Boiler Efficiency by Direct Method: Where. Get a Quote
two methods calculate boiler efficiency
Boiler efficiency = 100 – (L 1 + L 2 + L 3 + L 4 + L 6 + L 7 + L 8 ). Where L 1 = Loss due to dry flue gas. View More; Methods to Calculate CFB Boiler Efficiency – CFBC Boiler. Efficiency of 8TPH Coal Fired Boiler–Zozen We know there are two methods to calculate a boiler efficiency: direct method and indirect method, of which the direct
Ways to Calculate CFB Boiler Efficiency
2017-6-13 · Boiler efficiency is extremely important. If the boiler isn't burning fuel at a satisfactory rate, or if the heat has been lost over the surrounding system without being put to use, you choose a lot of waste and can even come upon problems meeting emissions standards.Blow will give two ways to calculate the CFB boiler efficiency,
THERMAL POWER: Boiler Efficiency calculations
2020-2-28 · Efficiency is a very important criterion in Boiler selection and Design. Efficiency figure depends upon the type of boiler as well as on the type of fuel and it's constituents. For example, efficiency of a Bagasse fired boiler is about 70% where as that of oil fired boilers is about 85 %. Higher moisture content in Bagasse reduces it's efficiency.So better criterion is efficiency based on
A REVIEW OF INDIRECT METHOD FOR MEASURING
2019-7-19 · A REVIEW OF INDIRECT METHOD FOR MEASURING THERMAL EFFICIENCY IN FIRE TUBE STEAM BOILERS AHMAD MAHMOUDI LAHIJANI1*, ERIS.E. SUPENI 2 AND FATEMEH KALANTARI 1Department of Mechanical Engineering, Faculty of Engineering, University Putra Malaysia, 43400 Serdang, Selangor, Malaysia. 2Department of Landscape Architecture, Faculty of Design and
Gas Fired Boiler Efficiency Calculation Industrial
Gas Oil Fired Thermal Oil Boiler Heater Made In. China Hot Oil Boiler manufacturer, Steam Boiler, Hot . Jiangsu Runli Boiler Co., Ltd is specialized in designing and manufacturing the steam boiler, thermal oil boiler and waste heat boiler in 6 series, such as coal burning, oil (gas), Coal Water Mixture, firewood, saw dust (grate boiler) and electric heating, the heating load varies from 0
how to calculate boiler efficiency indirect method
how to calculate boiler efficiency indirect method. pays special attention to the orders related to the Covid-19, such as boilers for hospitals, face mask production lines and pharmaceutical companies, which are essential weapons to fight against the epidemic. 4 million kcal YQW series gas-fired thermal oil heater project for the
how to calculate gas fired boiler thermal efficiency
How to calculate boiler efficiency by Direct Method . Boiler Efficiency:-Thermal efficiency of boiler is defined as the percentage of heat input that is effectively utilized to generate steam. There are two methods of assessing boiler efficiency. 1) Boiler Efficiency by Direct Method: Where. Get a Quote
Combustion Efficiency and Excess Air
2021-2-12 · The efficiency loss in the flue gas related to. temperature difference in flue gas and supply air ; CO 2 concentration in the flue gas; with oil combustion is indicated below: Example - Oil Combustion and Heat Loss in the Flue Gas. If. the temperature difference between the flue gas leaving a boiler and the ambient supply temperature is 300 o C
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# Quest: real world "Lua array with holes" usage
71 messages
1234
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## Quest: real world "Lua array with holes" usage
In face of recent (long) discussion about "Lua arrays", I would request a simple few lines of real world examples where the use of "Lua arrays with holes" is significant. I known that, intuitively, the length operator # should return the "size" of the array, but I'm not able to find significant use cases where "holes" are, in fact, a real problem other than a *simple* algorithm redesign.For my particular use case I made heavy use of "sparse things", from which Lua tables are awesome, where all computations are based on the keys, namely, with pairs() or indexing the relevant key. Also, I use "proper sequences" with Lua tables. In this case, table.pack and table.unpack are symmetric, and the relevant computation is given by the "order" of the keys, namely, with table.insert, table.remove, table.move etc that is by default, in this case, simple to understand. I never ever used a "Lua array with holes", where the ordering, or existence, of the keys is indeed relevant for computation. Precluding the use of __len or misbehaviour of #.Could you help me giving a (real world) counterexample?-- Rodrigo Azevedo Moreira da Silva
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## Re: Quest: real world "Lua array with holes" usage
> For my particular use case I made heavy use of "sparse things", from which > Lua tables are awesome, where all computations are based on the keys, > namely, with pairs() or indexing the relevant key. Also, I use "proper > sequences" with Lua tables. In this case, table.pack and table.unpack are > symmetric, and the relevant computation is given by the "order" of the > keys, namely, with table.insert, table.remove, table.move etc that is by > default, in this case, simple to understand. I never ever used a "Lua array > with holes", where the ordering, or existence, of the keys is indeed > relevant for computation. Precluding the use of __len or misbehaviour of #. > > Could you help me giving a (real world) counterexample? The only real-world case that bothers me is when you want to collect all the results from a function call, something like {f()}. If there is an error, the standard protocol is to return (nil, msg), which would create an array with holes. (Currently I think the correct way to fix this would be to return (false, msg), not to change the behavior of lists.) -- Roberto
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## Re: Quest: real world "Lua array with holes" usage
In reply to this post by Rodrigo Azevedo On 21 July 2016 at 08:13, Rodrigo Azevedo <[hidden email]> wrote: > In face of recent (long) discussion about "Lua arrays", I would request a > simple few lines of real world examples where the use of "Lua arrays with > holes" is significant. > I known that, intuitively, the length operator # should return the "size" of > the array, but I'm not able to find significant use cases where "holes" are, > in fact, a real problem other than a *simple* algorithm redesign. > > For my particular use case I made heavy use of "sparse things", from which > Lua tables are awesome, where all computations are based on the keys, > namely, with pairs() or indexing the relevant key. Also, I use "proper > sequences" with Lua tables. In this case, table.pack and table.unpack are > symmetric, and the relevant computation is given by the "order" of the keys, > namely, with table.insert, table.remove, table.move etc that is by default, > in this case, simple to understand. I never ever used a "Lua array with > holes", where the ordering, or existence, of the keys is indeed relevant for > computation. Precluding the use of __len or misbehaviour of #. > > Could you help me giving a (real world) counterexample? > > -- > Rodrigo Azevedo Moreira da Silva The place it usually comes up for me is when keeping arguments across a callback-style api: local function foo(...) local args = {...} do_something_async(function() -- called when done my_other_function(x, table.unpack(args)) end) end If the argument list contains a `nil` then the above will fail for some inputs. Hence you end up needing to use `table.pack` or `select("#", ...)`
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## Re: Quest: real world "Lua array with holes" usage
In reply to this post by Roberto Ierusalimschy On 21/07/16 14:09, Roberto Ierusalimschy wrote: > The only real-world case that bothers me is when you want to collect > all the results from a function call, something like {f()}. If there > is an error, the standard protocol is to return (nil, msg), which > would create an array with holes. Yes, that's the only way I get holes in arrays. OTOH, that's happening a lot less since I decided to assert all functions that can return nil,... unless I specifically know what to do in that case. Anyway, it's pretty easy to check ret[1]==nil to see if the call was successful. > (Currently I think the correct way to fix this would be to return > (false, msg), not to change the behavior of lists.) -- Roberto I have doubts on this, because false is a reasonable value to get from a function... Returning nil is clearer because the meaning is "could not produce a value to return". Otherwise you end using a explicit return for status pcall-like, all the time. I believe the main problem with nil in arrays stems not from the language, but from the design of programs: there is a tendency to use nil for nil-like concepts in the business logic, and that's when things get broken. A database NULL is not nil, is a singleton {}. The same for "free_slot", "nodataacquired" and whatever. Nil should be reserver for language-level concepts, like "the function failed to get a result" or "this variable has no value assigned". Jorge
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## Re: Quest: real world "Lua array with holes" usage
In reply to this post by Rodrigo Azevedo On Thu, Jul 21, 2016 at 12:13:40PM -0300, Rodrigo Azevedo wrote: > In face of recent (long) discussion about "Lua arrays", I would request a > simple few lines of real world examples where the use of "Lua arrays with > holes" is significant. The first use case that came to mind was JSON deserializers, where the obvious mapping of null to nil causes problems with arrays. But I think that's irritating mostly because superficially it _seems_ like such a trivial (and trivially fixable) mismatch between Lua and JavaScript type semantics, but as the thread has shown it's not trivial at all. I wouldn't use this case to argue for change in Lua.
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## Re: Quest: real world "Lua array with holes" usage
Hi there is really a significant difference: Whatever you assign to an array (or object) member in JavaScript, it stays in the array/object. Assign null, or even undefined, doesn't matter; it's there and you can enumerate it. You need to explicitly delete() it. This is kind f confusing if you come from C++, in JavaScript delete() does not kill the data, it kills the variable; the data may be referenced else where and survive. I do not really like the doubled feature undefined - null either. -- Oliver Am 21.07.2016 um 21:28 schrieb William Ahern: > On Thu, Jul 21, 2016 at 12:13:40PM -0300, Rodrigo Azevedo wrote: >> In face of recent (long) discussion about "Lua arrays", I would request a >> simple few lines of real world examples where the use of "Lua arrays with >> holes" is significant. > The first use case that came to mind was JSON deserializers, where the > obvious mapping of null to nil causes problems with arrays. > > But I think that's irritating mostly because superficially it _seems_ like > such a trivial (and trivially fixable) mismatch between Lua and JavaScript > type semantics, but as the thread has shown it's not trivial at all. I > wouldn't use this case to argue for change in Lua. > >
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## Re: Quest: real world "Lua array with holes" usage
In reply to this post by William Ahern > The first use case that came to mind was JSON deserializers, where the > obvious mapping of null to nil causes problems with arrays. That's the whole point: one probably shouldn't not force the semantics of one language into another. In this case, and others, it's simpler and clearer to define null = {} once and use it where needed. if you must skip over these when traversing a table, then that's what your semantics dictates and you must do it. Just do not try to do this with nil, which has its own, clear semantics in Lua.
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## Re: Quest: real world "Lua array with holes" usage
On Thu, Jul 21, 2016 at 6:08 PM, Luiz Henrique de Figueiredo <[hidden email]> wrote: >> The first use case that came to mind was JSON deserializers, where the >> obvious mapping of null to nil causes problems with arrays. > > That's the whole point: one probably shouldn't not force the semantics > of one language into another. In this case, and others, it's simpler > and clearer to define > null = {} > once and use it where needed. if you must skip over these when traversing > a table, then that's what your semantics dictates and you must do it. > Just do not try to do this with nil, which has its own, clear semantics > in Lua. I think part of the reason people try to use Lua's `nil` in ways like JSON's `null` or Python's `None` is because, in a context where you know you won't get the Boolean `false`, `if var then ...` is an easy way to check if `var` is a useful value or not, as `nil`, `null`, and `None` all evaluate to false in a Boolean context. The problem arises when one tries to create a singleton to represent this lack of a useful value. Python provides the `__bool__` method (in 3.x; `__nonzero__` in Python 2.x) to customize the behavior of an object when used in a Boolean context. Lua does not provide any such metamethod, for either tables or userdata, so any user-created singleton is automatically true in a Boolean context. This means that the idiom `if var then ...` no longer works, and an explicit (in)equality check against the singleton is required, creating uglier code. Perhaps what Lua *really* needs to solve this problem, then, is a __bool metamethod that, if present, is called when the object is used in a boolean context (such as `if`, `and`, and `or`) and returns either true or false, replacing the default true. Then one can take these `null` singletons and slap a `{__bool = function(t) return false end}` metatable on them, allowing them to gain the benefits of `nil` without the drawback of `nil`. What does everyone else think about a __bool metamethod?
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## Re: Quest: real world "Lua array with holes" usage
In reply to this post by Roberto Ierusalimschy > On Jul 21, 2016, at 10:09 AM, Roberto Ierusalimschy <[hidden email]> wrote: > >> For my particular use case I made heavy use of "sparse things", from which >> Lua tables are awesome, where all computations are based on the keys, >> namely, with pairs() or indexing the relevant key. Also, I use "proper >> sequences" with Lua tables. In this case, table.pack and table.unpack are >> symmetric, and the relevant computation is given by the "order" of the >> keys, namely, with table.insert, table.remove, table.move etc that is by >> default, in this case, simple to understand. I never ever used a "Lua array >> with holes", where the ordering, or existence, of the keys is indeed >> relevant for computation. Precluding the use of __len or misbehaviour of #. >> >> Could you help me giving a (real world) counterexample? > > The only real-world case that bothers me is when you want to collect all > the results from a function call, something like {f()}. If there is an > error, the standard protocol is to return (nil, msg), which would create > an array with holes. (Currently I think the correct way to fix this > would be to return (false, msg), not to change the behavior of lists.) > > -- Roberto > But that would still leave the general case of {f()} unsolved :( —Tim
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## Re: Quest: real world "Lua array with holes" usage
In reply to this post by Jonathan Goble It was thus said that the Great Jonathan Goble once stated: > On Thu, Jul 21, 2016 at 6:08 PM, Luiz Henrique de Figueiredo > <[hidden email]> wrote: > >> The first use case that came to mind was JSON deserializers, where the > >> obvious mapping of null to nil causes problems with arrays. > > > > That's the whole point: one probably shouldn't not force the semantics > > of one language into another. In this case, and others, it's simpler > > and clearer to define > > null = {} > > once and use it where needed. if you must skip over these when traversing > > a table, then that's what your semantics dictates and you must do it. > > Just do not try to do this with nil, which has its own, clear semantics > > in Lua. > > I think part of the reason people try to use Lua's `nil` in ways like > JSON's `null` or Python's `None` is because, in a context where you > know you won't get the Boolean `false`, `if var then ...` is an easy > way to check if `var` is a useful value or not, as `nil`, `null`, and > `None` all evaluate to false in a Boolean context. In my CBOR [1] library [2], it returns nil for 'null' and 'undefined' by default, but that can be overridden by the user (in fact, just about everything can be overridden by the user) so to me, that's more a limitation of the JSON module than of actual problems with semantics. If you care enough, you'll make the distinction. > The problem arises when one tries to create a singleton to represent > this lack of a useful value. Python provides the `__bool__` method (in > 3.x; `__nonzero__` in Python 2.x) to customize the behavior of an > object when used in a Boolean context. Lua does not provide any such > metamethod, for either tables or userdata, so any user-created > singleton is automatically true in a Boolean context. This means that > the idiom `if var then ...` no longer works, and an explicit > (in)equality check against the singleton is required, creating uglier > code. > > Perhaps what Lua *really* needs to solve this problem, then, is a > __bool metamethod that, if present, is called when the object is used > in a boolean context (such as `if`, `and`, and `or`) and returns > either true or false, replacing the default true. Then one can take > these `null` singletons and slap a `{__bool = function(t) return false > end}` metatable on them, allowing them to gain the benefits of `nil` > without the drawback of `nil`. > > What does everyone else think about a __bool metamethod? A __boolean metamethod was mentioned? Proposed? I'm not sure, but in any case, one was mentioned here: http://lua-users.org/lists/lua-l/2011-12/msg00621.htmlbut it seems like it was for something else. -spc (Keeper of the metamethod proposals ... ) [1] Concise Binary Object Representation, RFC-7049 [2] https://github.com/spc476/CBOR
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## Re: Quest: real world "Lua array with holes" usage
In reply to this post by Tim Hill On Fri, Jul 22, 2016 at 1:54 AM, Tim Hill <[hidden email]> wrote: > But that would still leave the general case of {f()} unsolved :( We always have table.pack(f()) ... Especially when pack/unpack become symmetrical.
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## Re: Quest: real world "Lua array with holes" usage
table.pack is very good, but I use ".n" field for other purposes. Then, I'm (seldom) using something likepack_len = function(...) local len = select('#',...) return setmetatable({...},{ __len = function() return len end, __newindex = function(t,k,v) if math.type(k) == 'integer' then len = math.max(len,k) rawset(t,k,v) end end })endt = pack_len(nil,3,nil)print(#t, table.unpack(t))t[10] = 3.1415print(#t, table.unpack(t))t[3] = 5t[1] = 2t[2] = 7print(#t, table.unpack(t))t[#t+1] = 0.5772for i=1,#t do print(i,t[i]) endreturn3 nil 3 nil10 nil 3 nil nil nil nil nil nil nil 3.141510 2 7 5 nil nil nil nil nil nil 3.14151 22 73 54 nil5 nil6 nil7 nil8 nil9 nil10 3.141511 0.5772as expected.This keep symmetry with table.pack, as well as others table.something, and give me an unbounded "sparse array" with well defined length.There is something wrong with this implementation?2016-07-22 4:06 GMT-03:00 steve donovan :On Fri, Jul 22, 2016 at 1:54 AM, Tim Hill <[hidden email]> wrote: > But that would still leave the general case of {f()} unsolved :( We always have table.pack(f()) ... Especially when pack/unpack become symmetrical. -- Rodrigo Azevedo Moreira da Silva
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## Re: __bool / __false (was: Quest: real world "Lua array with holes" usage)
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## Re: Quest: real world "Lua array with holes" usage
In reply to this post by Rodrigo Azevedo Am 21.07.2016 17:13 schrieb "Rodrigo Azevedo" <[hidden email]>: > > [...] > > Could you help me giving a (real world) counterexample? No. :-) I've been bitten *once* by # doing something unexpected (and debugging for some hours then) but that was due to a logical error further down in the call stack. I tend to skip all the threads regarding the "issue" with a smile. Matthias PS. Sorry for this unhelpful reply, but I really like Lua the way it is...
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## Re: Quest: real world "Lua array with holes" usage
In reply to this post by Jorge Visca On 21 July 2016 at 14:58, Jorge <[hidden email]> wrote: > > > On 21/07/16 14:09, Roberto Ierusalimschy wrote: >> >> The only real-world case that bothers me is when you want to collect all >> the results from a function call, something like {f()}. If there is an >> error, the standard protocol is to return (nil, msg), which would create an >> array with holes. > > Yes, that's the only way I get holes in arrays. OTOH, that's happening a lot > less since I decided to assert all functions that can return nil,... unless > I specifically know what to do in that case. Anyway, it's pretty easy to > check ret[1]==nil to see if the call was successful. > >> (Currently I think the correct way to fix this would be to return (false, >> msg), not to change the behavior of lists.) -- Roberto > > I have doubts on this, because false is a reasonable value to get from a > function... Returning nil is clearer because the meaning is "could not > produce a value to return". Otherwise you end using a explicit return for > status pcall-like, all the time. I agree with Jorge. Also, sometimes when returning multiple things, one of these things might be nil, so it's not only the error situation that produces a nil value upon return, although it is of course the most common. So, unfortunately just changing the policy from nil-on-error to false-on-error still wouldn't allow us to just forget about table.pack(). -- Hisham
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## Re: __bool / __false (was: Quest: real world "Lua array with holes" usage)
In reply to this post by nobody On Jul 22, 2016 10:11 AM, "nobody" <[hidden email]> wrote: > > - a boolean or nil: __false is true -> value is false > __false is false (or nil) -> value is true > (The true/false order was chosen such that you're not jumping around in > the truth table: (__false = nil) ~ (__false = false) whereas it would > probably be (__bool = nil) !~ (__bool = false).) While I understand the reasoning, the double negation here is confusing. I'd find __bool much easier to understand: if it's a boolean, that's the truth value; if it's a function, call it to determine the truth value; if it's nil, the standard logic applies (anything that isn't `false` or `nil` is true). If we were designing a new language, I'd be tempted to say it should raise an error when comparing a value which is not a boolean to one that is (where having __bool counts as being a boolean).
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## Re: Quest: real world "Lua array with holes" usage
In reply to this post by Hisham > I agree with Jorge. Also, sometimes when returning multiple things, > one of these things might be nil, so it's not only the error situation > that produces a nil value upon return, although it is of course the > most common. So, unfortunately just changing the policy from > nil-on-error to false-on-error still wouldn't allow us to just forget > about table.pack(). Of course any function can return nil among its returns, but maybe that might be considered a bad practice. My point is that, if nil is representing the absence of a value, it is weird not to have a third value but to have a fourth one. If nil is representing something else, probably it shouldn't be (as pointed out by others already). ("is weird" ~ "it might be considered weird") -- Roberto
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## The Undefined Country (was Re: Quest: real world "Lua array with holes" usage)
It was thus said that the Great Roberto Ierusalimschy once stated: > > I agree with Jorge. Also, sometimes when returning multiple things, > > one of these things might be nil, so it's not only the error situation > > that produces a nil value upon return, although it is of course the > > most common. So, unfortunately just changing the policy from > > nil-on-error to false-on-error still wouldn't allow us to just forget > > about table.pack(). > > Of course any function can return nil among its returns, but maybe > that might be considered a bad practice. > > My point is that, if nil is representing the absence of a value, it > is weird not to have a third value but to have a fourth one. If nil > is representing something else, probably it shouldn't be (as pointed > out by others already). Javascript has an "undefined" state, in addition to a "null" state. Perhaps Lua should have one as well. To me, "nil" means "the absence of a value" while "undef" means "this never had a value to begin with". Proposal: undef. It works just like nil does now, but "nil" (the new nil behavior) is that it *is* allowed in sequences. t = {} x = t.one -- x is "undef" local zork -- zork is "undef" t = { 1 , 2 , nil , 4 } y = #t -- 4, because there are four items in t z = t[5] -- z is "undef", falls outside sequence function foo(...) local args = table.pack(...) -- no 'n' field to worry about local a,b,c,d,e = table.unpack(t) -- a is 1 -- b is 2 -- c is nil -- d is 4 -- e is undef if e then print(e) elseif e == nil then print("e is nil") elseif e == undef then print("e never had a chance") end end It seems, to me, to solve all the current issues Lua has with arrays. -spc (There's probably stuff I'm missing though ... )
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## Re: The Undefined Country (was Re: Quest: real world "Lua array with holes" usage)
On Jul 24, 2016 3:27 PM, "Sean Conner" <[hidden email]> wrote: > > It was thus said that the Great Roberto Ierusalimschy once stated: > > > I agree with Jorge. Also, sometimes when returning multiple things, > > > one of these things might be nil, so it's not only the error situation > > > that produces a nil value upon return, although it is of course the > > > most common. So, unfortunately just changing the policy from > > > nil-on-error to false-on-error still wouldn't allow us to just forget > > > about table.pack(). > > > > Of course any function can return nil among its returns, but maybe > > that might be considered a bad practice. > > > > My point is that, if nil is representing the absence of a value, it > > is weird not to have a third value but to have a fourth one. If nil > > is representing something else, probably it shouldn't be (as pointed > > out by others already). > > Javascript has an "undefined" state, in addition to a "null" state. > Perhaps Lua should have one as well. To me, "nil" means "the absence of a > value" while "undef" means "this never had a value to begin with". > howsabout NaV (Not a Value) by analogy to NaN? "Undefined value" seems kind of an oxymoron to me. NaV could mean sth like "its a value, kinda sorta, but one that cannot be represented, not even by nil".
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## Examples
This page contains basic examples using Predicates, Functions, Procedures, Generators, Ranges and Aggregators. There are also examples using composition and more practical examples at the bottom of this page.
### Predicates
Predicates are functors that return a boolean value. The following snippet of code shows how to use a Predicate that says whether a number is even or not.
```List<Integer> numbers = Arrays.asList(1, 2, 3, 4);
Predicate<Integer> isEven = new Predicate<Integer>() {
public boolean test(Integer obj) {
return obj % 2 == 0;
}
};
for( Integer number : numbers ) {
if (isEven.test(number)) {
System.out.print(number + " ");
}
}
```
The code above produces the following output: 2 4
### Functions
Functions are functors that return an Object value. The following snippet of code shows how to use a Function that doubles a value.
```List<Integer> numbers = Arrays.asList(1, 2, 3, 4);
Function<Integer, Integer> doubler = new Function<Integer, Integer>() {
public Integer evaluate(Integer obj) {
return obj * 2;
}
};
for( Integer number : numbers ) {
Integer value = doubler.evaluate(number);
System.out.print(value + " ");
}
```
The code above produces the following output: 2 4 6 8
### Procedures
Procedures are functors that do not return anything. In the snippet of code below you can find an example that prints the value passed to the Procedure.
```List<Integer> numbers = Arrays.asList(1, 2, 3, 4);
Procedure<Integer> print = new Procedure<Integer>() {
public void run(Integer obj) {
System.out.print(obj + " ");
}
};
for( Integer number : numbers ) {
print.run(number);
}
```
The code above produces the following output: 1 2 3 4 .
### Generators
Apache Functor includes other objects that you can can use to code in a less imperative way, like Generators. In the following example, we create an Integer Generator that generates integers from 1 to 4 (the right argument is non-inclusive). The generator is wrapped within a Filtered Generator that applies the isEven predicate to each integer generated by the former generator. Finally, we execute a Composite Unary Procedure that uses a function to double the value of the integer before printing it.
```Generator<Integer> integerGenerator = new IntegerRange(1, 5); // inclusive, exclusive
UnaryPredicate<Integer> isEven = new UnaryPredicate<Integer>() {
public boolean test(Integer obj) {
return obj % 2 == 0;
}
};
FilteredGenerator<Integer> filteredGenerator =
new FilteredGenerator<Integer>(integerGenerator, isEven);
UnaryFunction<Integer, Integer> doubler = new UnaryFunction<Integer, Integer>() {
public Integer evaluate(Integer obj) {
return obj * 2;
}
};
UnaryProcedure<Integer> print = new UnaryProcedure<Integer>() {
public void run(Integer obj) {
System.out.print(obj + " ");
}
};
CompositeUnaryProcedure<Integer> compositeProcedure =
new CompositeUnaryProcedure<Integer>(print);
filteredGenerator.run(compositeProcedure.of(doubler));
```
The lines package demonstrates a functional approach to IO using Generators and the Algorithms class.
### Ranges
Using Ranges you are able to create a series of elements of a certain type. The distance between each element is called step. And the left and right limits are endpoints.
By default, in numeric ranges the left value is inclusive and the right is exclusive. The range below creates a series of Integers between 0 and 10, with a default step of 1.
```// [0, 10), 1 = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9
IntegerRange range = new IntegerRange(0, 10);
```
The step value can be changed, altering the distance between each element in the range. As shown below.
```// [0, 10), 2 = 0, 2, 4, 6, 8
IntegerRange range = new IntegerRange(0, 10, 2);
```
It is possible, too, to define the bound type of each endpoint. It is similar to mathematical intervals, where a closed endpoint means that the value is included in the series of elements in the range. On the other hand, an open endpoint means that the value is not included.
```// (0, 10], 2 = 2, 4, 6, 8, 10
IntegerRange range = new IntegerRange(0, BoundType.OPEN, 10, BoundType.CLOSED, 2);
```
A Range is also by nature a Generator, so you can use it for executing procedures for each of its elements.
```// [0, 10), 1 = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9
IntegerRange range = new IntegerRange(0, 10);
UnaryProcedure<Integer> printProcedure = new UnaryProcedure<Integer>() {
public void run(Integer obj) {
System.out.print(obj + " ");
}
};
range.run(printProcedure);
```
The code above produces the following output: 0 1 2 3 4 5 6 7 8 9
### Aggregators
There are some code snippets / examples for the org.apache.commons.functor.aggregator package available on this page. Also, to exemplify the usage of the Aggregator classes, there are code examples in the test section.
First such set of example involves the usage of the nostore Aggregator. Code can be found in org.apache.commons.functor.example.aggregator.nostore. This shows how can you use an aggregator which doesn't store the data series and processes them on the fly. Also, there are examples provided which show how can you implement your own aggregation function to be used with this Aggregator type.
For using an Aggregator which stores the data series in a list, examples are in org.apache.commons.functor.example.aggregator.list. This shows how can you use the ArrayList-backed aggregator or provide your own List-based implementation. Also, there are examples provided which show how can you implement your own aggregation function to be used with this Aggregator type.
### Reuse Through Composition
The Functor package, and more generally, a functional approach to program design, supports a powerful technique for balancing behavior specialization and code reuse.
Traditional Object Oriented design suggests inheritence as a mechanism code reuse, and method overloading as a mechanism for specialization. For example, one defines a general purpose, perhaps even abstract class, say AbstractList, and then extend or specialize this parent via subclasses, inheriting some behaviors and overloading others.
Functors encourage another, complementary approach to code reuse and behavior specialiazation: composition. Following a compositional design, we create a number of simple objects and then combine them to create more complex behaviors. For example, the Commons Pool component defines an ObjectPool type that maintains a collection of pooled objects, but delegates to a PoolableObjectFactory to create, validate and destroy the objects to be pooled. Arbitrary ObjectPool implementations can be composed with arbitrary PoolableObjectFactory implementations in order to create new types of pools.
Let's see an example that combines the three functors seen here so far. In this example, we will use the functors that we created so that for each even number found, it will double its value and will print the new value.
```List<Integer> numbers = Arrays.asList(1, 2, 3, 4);
Predicate<Integer> isEven = new Predicate<Integer>() {
public boolean test(Integer obj) {
return obj % 2 == 0;
}
};
Function<Integer, Integer> doubler = new Function<Integer, Integer>() {
public Integer evaluate(Integer obj) {
return obj * 2;
}
};
Procedure<Integer> print = new Procedure<Integer>() {
public void run(Integer obj) {
System.out.print(obj + " ");
}
};
for( Integer number : numbers ) {
if(isEven.test(number)) {
print.run(doubler.evaluate(number));
}
}
```
The code above produces the following output: 4 8
The FlexiMap example applies this design to java.util.Map, demonstrating how "pluggable" functors can be applied to a generic Map structure in order to introduce new behaviors. The map package is a more complete example of this, implementing a number of the Commons-Collections Maps derived from a base FunctoredMap.
### Generators
Apache Functor includes other objects that you can can use to code in a less imperative way, like Generators. In the following example, we create an Integer Generator that generates integers from 1 to 4 (the right argument is non-inclusive). The generator is wrapped within a Filtered Generator that applies the isEven predicate to each integer generated by the former generator. Finally, we execute a Composite Procedure that uses a function to double the value of the integer before printing it.
```Generator<Integer> integerGenerator = new IntegerRange(1, 5); // inclusive, exclusive
Predicate<Integer> isEven = new Predicate<Integer>() {
public boolean test(Integer obj) {
return obj % 2 == 0;
}
};
FilteredGenerator<Integer> filteredGenerator =
new FilteredGenerator<Integer>(integerGenerator, isEven);
Function<Integer, Integer> doubler = new Function<Integer, Integer>() {
public Integer evaluate(Integer obj) {
return obj * 2;
}
};
Procedure<Integer> print = new Procedure<Integer>() {
public void run(Integer obj) {
System.out.print(obj + " ");
}
};
CompositeProcedure<Integer> compositeProcedure =
new CompositeProcedure<Integer>(print);
filteredGenerator.run(compositeProcedure.of(doubler));
```
The lines package demonstrates a functional approach to IO using Generators and the Algorithms class.
### Aggregators
There are some code snippets / examples for the org.apache.commons.functor.aggregator package available on this page. Also, to exemplify the usage of the Aggregator classes, there are code examples in the test section.
First such set of example involves the usage of the nostore Aggregator. Code can be found in org.apache.commons.functor.example.aggregator.nostore. This shows how can you use an aggregator which doesn't store the data series and processes them on the fly. Also, there are examples provided which show how can you implement your own aggregation function to be used with this Aggregator type.
For using an Aggregator which stores the data series in a list, examples are in org.apache.commons.functor.example.aggregator.list. This shows how can you use the ArrayList-backed aggregator or provide your own List-based implementation. Also, there are examples provided which show how can you implement your own aggregation function to be used with this Aggregator type.
### Code Katas
"Pragmatic" Dave Thomas has been blogging a series of programming exercises he calls Code Katas. These exercises are intended to provide "practice sessions" that allow programmers to hone their craft. The notion is borrowed from the practice of Karate, where, in Dave's words "a kata is an exercise where you repeat a form many, many times, making little improvements in each".
Here we use several of Dave's Code Katas to explore the Commons-Functor library.
Kata One: Supermarket Pricing
Dave's Kata One asks how one might implement supermarket pricing rules, like "three for a dollar" or "buy two get one free". By encapsulating tiny bits of logic, functors provide a useful solution to this problem, as illustrated in the SupermarketPricingExample.
Kata Two: Binary Chop
Kata Two asks us to create several different implementations of the binary search algorithm, which once you get past three or four implementations, is more difficult that it sounds. TestBinaryChop presents several implementations, with functor and non-functor variations.
Kata Four: Data Munging
Kata Four asks us to explore extreme reuse. Our DataMunger allosubsubsectionws for very small implementations of the weather and soccer (football) parsers.
### A Quicksort Implementation
The Quicksort example presents an implementation of the Quicksort sorting algorithm written in a functional programming style using Commons Functor.
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Geometry_proofs/737699: Can somebody please help me with this proof that I am stuck on? I need to solve it by manipulating only the left side... (secx-tanx)/(secx+tanx) = ((1-sinx)/cosx)^2 Thanks in advance!1 solutions Answer 450501 by mananth(12270) on 2013-04-15 00:59:10 (Show Source): You can put this solution on YOUR website! LHS
Trigonometry-basics/737656: Find the exact values of cos theta and sin theta for the angle measure. 5 pi/6 radians 1 solutions Answer 450485 by mananth(12270) on 2013-04-14 21:39:33 (Show Source): You can put this solution on YOUR website! Angle=
Radicals/737647: Solve the radical equation. sqrt(3x+4)=2+ sqrt(x) This will usually involve squaring each side twice. I don't know how to resolve this, can you please give an explanation as to how to perform this?1 solutions Answer 450480 by mananth(12270) on 2013-04-14 21:32:22 (Show Source): You can put this solution on YOUR website!sqrt(3x+4)=2+ sqrt(x) square rearrange square /4x x=4
Geometry_proofs/737632: 1/(1-cosx)-1/(1+secx) = csc^2x+cot^2x Can somebody help me do the following proof (solving from the left)? Thanks in advance!1 solutions Answer 450474 by mananth(12270) on 2013-04-14 21:23:15 (Show Source): You can put this solution on YOUR website! LHS = Take LCD LHS = RHS
Equations/737638: for which of the following equations are x=5and x=-5........................... A x2-x^2-5x-25=0.....B..x^2+25=0......C.. x^2+10x-25=0....D.. x^2-25=01 solutions Answer 450469 by mananth(12270) on 2013-04-14 20:53:35 (Show Source): You can put this solution on YOUR website!D.x^2-25 (x+5)(x-5)=0 so x=5 or x=-5
Linear-equations/737079: What is the equation of the line in slope point and general form? Same slope as 2x-5y+6=0 and through the origin. Thanks1 solutions Answer 450186 by mananth(12270) on 2013-04-13 05:38:02 (Show Source): You can put this solution on YOUR website!slope of the given line = 2/5 passes through (0,0) y-0=2/5(x-0) y=(2/5)x -(2/5)x+y=0
Sequences-and-series/737093: if x, y, and z are the first three terms of a geometric sequence, show that x^2, y^2 and z^2 form another geometric sequence1 solutions Answer 450185 by mananth(12270) on 2013-04-13 05:30:55 (Show Source): You can put this solution on YOUR website!x y z are in geometric x/y = y/z OR y^2=xz x^2, y^2, z^2 x^2, xz, z^2 If they are in geometric sequence then xz/x^2 = z^2/xz z/x = z/x The ratios are common so x^2,y^2 & z^2 is a GP
Mixture_Word_Problems/737125: Determine the slope and the angle of inclination of the line joining the point ( -2,5 ) and ( 3,-6 )1 solutions Answer 450184 by mananth(12270) on 2013-04-13 05:24:31 (Show Source): You can put this solution on YOUR website!x1 y1 x2 y2 -2 5 3 -6 slope m = (y2-y1)/(x2-x1) ( -6 - 5 )/( 3 - -2 ) ( -11 / 5 ) m= -2 1/5 slope = -11/5 Tan = -11/5 Theta = tan ^-1(11/5) = -65.55 deg
Trigonometry-basics/737089: Use the given information to find cos 2x, sin 2x, and tan 2x. cosx=sqrt3/7 and 3π/2 < x < 2π since I have cos, I created a triangle using sqrt3 as the opposite side. and 7 as the adjacent side. Then I used the Pythagorean theorem to find the hypotenuse. hypotenuse^2=(sqrt3)^2+(7)^2 hypotenuse^2=3+49 hypotenuse^2=52 hypotenuse=sqrt52 Then to find sin2x, I used sin2x=2sin(xcos(x), so... sin2x=2sin(xcos(x) = 2(sqrt3/sqrt52)(sqrt3/7) = 2 (sqrt9/7sqrt52) = 3/7sqrt13 but that was incorrect and I can't go any further in the problem with an incorrect sin2x. Please assist me with where I went wrong or how to go about finding this answer. thanks! 1 solutions Answer 450182 by mananth(12270) on 2013-04-13 05:12:09 (Show Source): You can put this solution on YOUR website!cos x = square both sides cos^2 x=3/49 1-cos^2 x=1-3/49 sin^2 x = 46/49 cos 2x = 2 cos^2 x -1 =2*3/49-1 =6/49-1 =-43/49 Cos 2x= -43/49 Sin 2x = 2sin x . cos x = = Tan (2x) = sin (2x)/cos (2x)
�t�e�s�t/737102: A line passes through the point (-2,4), and its y-intercept is (0,-6). What is the equation of the line that is perpendicular to the first line and passes through the point (5, -4)? A. y = 1/5x - 3 B. y = 5x - 3 C. y= 1/5x - 5 D. y = 5/3x - 2 1 solutions Answer 450177 by mananth(12270) on 2013-04-13 04:22:20 (Show Source): You can put this solution on YOUR website!option C
�t�e�s�t/737104: What is the equation of the line that passes through the points (4/5, 1/5) and (1/2, 3/2)? A. y = -x - 1 B. y = -x + 1 C. y = -13/3x + 11/3 D. y = 7/3x + 1/31 solutions Answer 450176 by mananth(12270) on 2013-04-13 04:15:21 (Show Source): You can put this solution on YOUR website!option (c)
logarithm/736847: Log base 3 1/27=x 1 solutions Answer 450016 by mananth(12270) on 2013-04-12 00:06:41 (Show Source): You can put this solution on YOUR website!log(3) 1/27=x 3= (1/27)^x 1=-3x -1/3 =x
Graphs/736837: find the equation in slope intercept form: slope: -1 ordered pair (2,4)1 solutions Answer 450014 by mananth(12270) on 2013-04-11 23:55:25 (Show Source): You can put this solution on YOUR website!The equation in point slope form is y-y1=m(x-x1) m=-1 (x1,y1)=> (2,4) y-4 = -1(x-2) y-4=-x+2 y=-x+2+4 y=-x+6
Travel_Word_Problems/736793: two planes having the same airspeed depart at the same time, flying in opposite directions. one plane flies into 35mph wind, and the other flies in the same direction as the wind. After a period of time, on plane has traveled 460 m an the other has traveled 404 m. let x = airspeed of each plane ( airspeed is the same for each plane) a) write an expression for the speed of the plane traveling into the wind. b) Traveling with the wind c) write an expression for the time the same plane has been in the air when it has traveled 460 m d) what is the relationship for the times the planes have been in the air c) write an equation modeling this relationship.1 solutions Answer 450007 by mananth(12270) on 2013-04-11 22:13:03 (Show Source): You can put this solution on YOUR website!two planes having the same airspeed depart at the same time, flying in opposite directions. one plane flies into 35mph wind, and the other flies in the same direction as the wind. After a period of time, on plane has traveled 460 m an the other has traveled 404 m. let x = airspeed of each plane ( airspeed is the same for each plane) a) write an expression for the speed of the plane traveling into the wind. x-35 mph b) Traveling with the wind x+35 mph c) write an expression for the time the same plane has been in the air when it has traveled 460 m 460/(x+35) d) what is the relationship for the times the planes have been in the air Times are equal second plane time = 404/(x-35) c) write an equation modeling this relationship. 460/(x+35)=404/(x-35) 460(x-35)=404(x+35) 460x-16100=404x+14140 56x=30240 x=540 mph
Trigonometry-basics/736790: I need a walkthrough of how i would do this problem "Philly can reach his second floor window with an 18 foot ladder at a 75 degree angle to the ground. His friend willy claims that Philly could reach the window with a 16 foot ladder. Justify or disprove Willys claim. "1 solutions Answer 450002 by mananth(12270) on 2013-04-11 22:02:22 (Show Source): You can put this solution on YOUR website!18 feet ladder 75 deg angle Height of window Sin 75 = height / ladder length height = sin 75 * 18 =>0.965*18 =>17.37 feet height of window =17.37 if ladder length is 16 feet even when kept vertically it will not reach the window.
Miscellaneous_Word_Problems/736767: Laura received an e-mail from Billy. After 10 minutes Laura forwarded it to 3 of her friends. After 10 minutes each of those 3 friends forwarded the message to 3 more friends. If the message was forwarded like this every 10 minutes, how many people had received Billy's e-mail after 50 minutes? 1 solutions Answer 449996 by mananth(12270) on 2013-04-11 21:44:25 (Show Source): You can put this solution on YOUR website!Laura received an e-mail from Billy. After 10 minutes Laura forwarded it to 3 of her friends. After 10 minutes each of those 3 friends forwarded the message to 3 more friends. If the message was forwarded like this every 10 minutes, how many people had received Billy's e-mail after 50 minutes? the email was sent every 10 minutes Total emails 5 first email 3, 9,27 81, 243..... it is a GP Sn = 3*242/2 =363 emails will be sent in 50 minutes
Rational-functions/736770: Find the x- and y-intercepts. If no x-intercepts exist, state so. f(x) = 3x2 + 6x + 1 1 solutions Answer 449990 by mananth(12270) on 2013-04-11 21:17:38 (Show Source): You can put this solution on YOUR website!f(x) = 3x2 + 6x + 1 On X axis y=0 3x^2+6x+1=0 Find the roots of the equation by quadratic formula a= 3 b= 6 c= 1 b^2-4ac= 36 - 12 b^2-4ac= 24 = 4.9 x1=( -6 + 4.9 )/ 6 x1= -0.18 x2=( -6 -4.9 ) / 6 x2= -1.82 Intersection at -0.18 and -1.82
Travel_Word_Problems/736756: Pensacola is 792 miles away from key west . Bus A leaves at 6am and travels at 42 mph . Bus B leaves at 7am and travels at 33 mph . What time and distance from key west, will the two buses meet ?1 solutions Answer 449989 by mananth(12270) on 2013-04-11 21:11:44 (Show Source): You can put this solution on YOUR website!If they are moving in same direction they will not meet. if opposite then it can be worked out
Mixture_Word_Problems/736421: The equation of the gas line is 2x+y=2 . a factory located at the point ( 6,7) will connect with the gas line perpendicularly . find the equation of the connecting line and the length of the pipe required if the units are miles .1 solutions Answer 449833 by mananth(12270) on 2013-04-11 08:59:04 (Show Source): You can put this solution on YOUR website!2 x + 1 y = 2 Find the slope of this line 1 y = -2 x + 2 Divide by 1 y = -2 x + 2 Compare this equation with y=mx+b, m= slope & b= y intercept slope m = -2 The slope of a line perpendicular to the above line will be the negative reciprocal 1/2 Because m1*m2 =-1 The slope of the required line will be 1/2 m= 1/2 ,point ( 4 , 7 ) Find b by plugging the values of m & the point in y=mx+b 7 = 2 + b b= 5 m= 1/2 The required equation is y = 1/ 2 x + 5 where do they intersect 2.00 x + 1.00 y = 2.00 .............1 Total value -0.50 x + 1.00 y = 5.00 .............2 Eliminate y multiply (1)by -1.00 Multiply (2) by 1.00 -2.00 x -1.00 y = -2.00 -0.50 x + 1.00 y = 5.00 Add the two equations -2.50 x = 3.00 / -2.50 x = -1.20 plug value of x in (1) 2.00 x + 1.00 y = 2.00 -2.40 + y = 2.00 y = 2.00 + 2.40 y = 4.40 y = 4.40 (-1.20,4.40) (6,7) Find the distance between the points. that's the length of the pipe line Distance between two points x1 y1 x2 y2 -1.2 4.4 6 7 d= d= d= d= d= 7.66 m.ananth@hotmail.ca m.ananth@hotmail.ca
Travel_Word_Problems/736420: A BOAT TRAVELED 336 MILES DOWNSTREAM AND BACK. THE TRIP DOWNSTREAM TOOK 12HOURS. THE TRIP BACK TOOK 14 HOURS. WHAT IS THE SPEED OF THE BOAT IN STILL WATER? WHAT IS THE SPEED OF THE CURRENT? SOLVE USING A SYSTEM OF TWO EQUATIONS1 solutions Answer 449832 by mananth(12270) on 2013-04-11 08:51:09 (Show Source): You can put this solution on YOUR website!Boat speed speed =x mph current speed =y mph against current 14 hours with current 12 hours Distance against 336 miles distance with 336 miles t=d/r against current - 336 / ( x - y )= 14 14 ( x - y ) = 336 14 x - 14 y = 336 ....................1 336 / ( x + y )= 12 12 ( x + y ) = 336 12 x + 12 y = 336 ...............2 Multiply (1) by 6 Multiply (2) by 7 we get 84 x + -84 y = 2016 84 x + 84 y = 2352 168 x = 4368 / 168 x = 26.00 mph plug value of x in (1) y 14 x -14 y = 336 364 -14 -364 = 336 -14 y = 336 -14 y = -28 mph y = 2.00 Boat speed speed 26.00 mph current speed 2.00 mph m.ananth@hotmail.ca
Travel_Word_Problems/736425: A van departs from a rest stop at 9am heading west and travels at an average rate of 40 miles per hour. two hours later a car departs from the same rest stop also heading west and travels at an average rate of 65 miles per hour. at what time will the car catch up to the van and how far will they have traveled at that time? I need to build an equation, and so far all i have is this; Rate x Time = Dist 40 x 40x 40-x 2 2(40-x) this just feels all wrong... 40x + 65(2+x) = 91 solutions Answer 449829 by mananth(12270) on 2013-04-11 08:45:04 (Show Source): You can put this solution on YOUR website!First car 40 mph Car II 65 mph First car 09:00 Car II 11:00 Difference in time= 02:00 => 2.00 hours First car will have covered 80.00 miles Car II starts catch up distance= 80.00 miles catch up speed = 65 -40 catch up speed = 25 mph Catchup time = catchup distance/catch up speed catch up time= 3.2 catch up time= 3.20 hours They will meet at 12:20 pm Car II speed = 65 hours Time to catch up = 3.2 D=Speed * time D= 65 * 3.2 D= 208 miles Car II will catch up when it has traveled 208 miles
Volume/736368: A cylinder with a top and bottom has a volume of 30 cubic inches. Find the minimum amount of material needed to create the can. (Surface area=2pir^2 + 2pirh & volume = pir^2h) I tried to create two separate equations, 30 = pir^2h (and then solve for r^2) and 2pir^2 and then I was going to substitute the r value into the other equation. This did not work. Thanks in advance! 1 solutions Answer 449825 by mananth(12270) on 2013-04-11 06:51:55 (Show Source): You can put this solution on YOUR website!we have two unknown variables, the radius (r) and the height (h). Volume (V) = pir^2h 30 = pir^2h h = 30/pir2 Substitute this into the surface area equation for h, we get A = (2pir( 30 )/pir^2) + 2pir2) A = (60/r) + 2pir2 Take the derivative for r A = 60r-1 + 2pir2 A' = -60r-2 + 4pir when slope is zero 0 = -60r-2 + 4pir 60r-2 = 4pir Multiply both sides by r^2 60 = 4pir3 4.775 = r3 1.68 = r h = 30/pir^2 h=30/pi*(1.68)^2 h = 3.38 A = 2pirh + 2pir^2 A = 2*pi*(1.68)(3.38) + 2pi*(1.68)^2 A = 35.68 + 17.73 A = 53.41 square inches CHECK V= pi*1.68^2*3.38= 29.96 you can check for maxima or minima by taking the second derivate for more detailing
Mixture_Word_Problems/736373: A merchant buys tea at the rate of Rs.4.10,Rs.3.75 and 4.50 per kg and mixes them in the proportion of 5:4:1.At what price should he sell the mixture to gain 25%?1 solutions Answer 449823 by mananth(12270) on 2013-04-11 05:49:42 (Show Source): You can put this solution on YOUR website!4.10,Rs.3.75 and 4.50 per kg suppose he mixes 5 kg,4kg & 1 kg of mix to make 10 kg 5*4.10+4*3.75+1*4.5=40 Mixture costs 40/10 = Rs.4.00 /kg 4*1.25 = Rs. 5 /kg
Sequences-and-series/736406: If (3-x)+(6)+(7-5x) is a geometric series, find two possible values for a) x b) the common ratio c) the sum of the GP.1 solutions Answer 449822 by mananth(12270) on 2013-04-11 05:22:15 (Show Source): You can put this solution on YOUR website!(3-x)+(6)+(7-5x) is a geometric series, find two possible values for a) x b) the common ratio c) the sum of the GP. (3-x)(7-5x)=6^2 21-15x-7x+5x^2=36 5x^2-22x-15=0 5x^2-25x+3x-15=0 5x(x-5)+3(x-5)=0 (x-5)(5x+3)=0 x=5 OR -3/5 if x=5 -2,6,-28 common ratio = -4 you should be able to continue
Equations/736367: how do i rewrite 3x^2-30x+77 to look like f(x)=a(x-h)^2+k1 solutions Answer 449807 by mananth(12270) on 2013-04-11 00:02:57 (Show Source): You can put this solution on YOUR website! put it in the right format
Polynomials-and-rational-expressions/736374: Solve =1 solutions Answer 449806 by mananth(12270) on 2013-04-10 23:55:15 (Show Source): You can put this solution on YOUR website!= = = =0 since the index is odd x=1
Travel_Word_Problems/736351: david canoes upstream a distance of 8 miles and then returns. The round-trip took 5 1/3 hours. The current of the stream was flowing at 2 miles per hour. What was david's canoeing speed?1 solutions Answer 449803 by mananth(12270) on 2013-04-10 23:14:02 (Show Source): You can put this solution on YOUR website!canoe speed x mph current speed 2 kph against current x- 2 mph with current x+ 2 mph Distance= 8 miles Time against + time with = 5 1/3 hours t=d/r 8 /( x + 2 ) + 8 /(x - 2 ) = 5 1/3 LCD = (x - 2 ) ( x + 2 ) 8 *( x - 2 ) + 8 (x + 2 ) = 5 1/3 (x^2 - 4 ) 8 x - -16 + 8 x + 16 = 5 1/3 ( x ^2 - 4 ) 16 x = 5 1/3 x ^2 -21.33 5 1/3 x ^2 - -16 x - 21.33 Find the roots of the equation by quadratic formula a= 5 1/3 , b= -16 , c= -21.33 b^2-4ac= 256 + 455.04 b^2-4ac= 711.04 0 1.33 x1=( 16 + 26.67 )/ 10.67 x1= 4.00 x2=( 16 -26.67 ) / 10.67 x2= -1.00 Ignore negative value canoe speed 4.000 mph m.ananth@hotmail.ca canoe speed x mph current speed 2 kph against current x- 2 mph with current x+ 2 mph Distance= 8 miles Time against + time with = 5 1/3 hours t=d/r 8 /( x + 2 ) + 8 /(x - 2 ) = 5 1/3 LCD = (x - 2 ) ( x + 2 ) 8 *( x - 2 ) + 8 (x + 2 ) = 5 1/3 (x^2 - 4 ) 8 x - -16 + 8 x + 16 = 5 1/3 ( x ^2 - 4 ) 16 x = 5 1/3 x ^2 -21.33 5 1/3 x ^2 - -16 x - 21.33 Find the roots of the equation by quadratic formula a= 5 1/3 , b= -16 , c= -21.33 b^2-4ac= 256 + 455.04 b^2-4ac= 711.04 0 1.33 x1=( 16 + 26.67 )/ 10.67 x1= 4.00 x2=( 16 -26.67 ) / 10.67 x2= -1.00 Ignore negative value canoe speed 4.000 mph m.ananth@hotmail.ca canoe speed x mph current speed 2 kph against current x- 2 mph with current x+ 2 mph Distance= 8 miles Time against + time with = 5 1/3 hours t=d/r 8 /( x + 2 ) + 8 /(x - 2 ) = 5 1/3 LCD = (x - 2 ) ( x + 2 ) 8 *( x - 2 ) + 8 (x + 2 ) = 5 1/3 (x^2 - 4 ) 8 x - -16 + 8 x + 16 = 5 1/3 ( x ^2 - 4 ) 16 x = 5 1/3 x ^2 -21.33 5 1/3 x ^2 - -16 x - 21.33 Find the roots of the equation by quadratic formula a= 5 1/3 , b= -16 , c= -21.33 b^2-4ac= 256 + 455.04 b^2-4ac= 711.04 0 1.33 x1=( 16 + 26.67 )/ 10.67 x1= 4.00 x2=( 16 -26.67 ) / 10.67 x2= -1.00 Ignore negative value canoe speed 4 mph m.ananth@hotmail.ca
Linear-equations/736271: Which equation below is perpendicular to th line y=1/3x+7? A. X+3y=4 B. x-3y=4 C. 3x+y=4 D. 3x-y=4 1 solutions Answer 449802 by mananth(12270) on 2013-04-10 23:03:58 (Show Source): You can put this solution on YOUR website!option (c) the slopes are negative reciprocals slope of given line is 1/3 slope of perpendicular line will be -3
Trigonometry-basics/736309: John wants to measure the height of a tree. He walks exactly 100 feet from the base of the tree and looks up. The angle from the ground to the top of the tree is 33º. This particular tree grows at an angle of 83º with respect to the ground rather than vertically (90º). How tall is the tree? Thanks, paulaficarella@gmail.com 1 solutions Answer 449800 by mananth(12270) on 2013-04-10 22:46:10 (Show Source): You can put this solution on YOUR website!here you have to use the law of sines let tree make an angle of 83 deg.with the ground. (YZ)angle ZYX = 83deg he walks 100 feet away from tree to point X (YX)angle YXZ= 33 deg so remaining angle YZX = 64 deg triangle YXZ is formed sin A /a = sin B /b = sin C /c Sin YZX/100 = sin YXZ/ YZ sin 64/100 = sin 33/YZ 100 * sin 33/sin 64 = YZ 100*0.54/0.898= YZ YZ= 60.13 feet the height of the tree You can solve the problem in a similar way if the tilt is to the other side.
Travel_Word_Problems/736297: Lily's family took a motor boat 24 km down a river for a picnic. It took them 1 hr to get to the picnic spot. The ride back to the dock took an hour and a half. What was the average speed at which the river flowed? What would the boat's average speed be on a lake?1 solutions Answer 449772 by mananth(12270) on 2013-04-10 21:20:45 (Show Source): You can put this solution on YOUR website!Boat speed =x kph current speed =y kph against current 1.5 hours with current 1 hours Distance against 24 miles distance with 24 miles t=d/r against current - 24.00 / ( x - y )= 1.50 1.50 ( x - y ) = 24.00 1.50 x - 1.50 y = 24.00 ....................1 24.00 / ( x + y )= 1.00 1.00 ( x + y ) = 24.00 1.00 x + 1.00 y = 24.00 ...............2 Multiply (1) by 1.00 Multiply (2) by 1.50 we get 1.50 x + -1.50 y = 24.00 1.50 x + 1.50 y = 36.00 3.00 x = 60.00 / 3.00 x = 20.00 mph plug value of x in (1) y 1.50 x -1.50 y = 24.00 30.00 -1.50 -30.00 = 24.00 -1.50 y = 24.00 -1.50 y = -6.00 kph y = 4.00 Boat speed 20.00 mph current speed 4.00 kph
Travel_Word_Problems/736018: I am a librarian. A patron needs a solution to this problem: The speed of a boat in still waters is 10 km/h The boat travels 12 km upstream and 28 km downstream in a total of 4 hours. What is the speed of the stream? Thanks1 solutions Answer 449635 by mananth(12270) on 2013-04-10 09:51:37 (Show Source): You can put this solution on YOUR website!let speed of current be c kph upstream speed = 10-c upstream distance = 12 down stream speed = 10+c down stream distance = 28 km upstream time + downstream time = 4 t=d/r 12/(10-c) + 28/(10+c)= 4 multiply the equation by (10+c)(10-c) 12(10+c)+28(10-c)=4(10+c)(10-c) 120+12c+280-28c=4(100-c^2) 400-16c=400-4c^2 re arrange 4c^2=16c /4c c=4 current speed = 4 kph
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## How to perform all combinations of pairwise t-tests with proc multtest?
Solved
Frequent Contributor
Posts: 81
# How to perform all combinations of pairwise t-tests with proc multtest?
I'd like to perform all unique combinations of pairwise Student's t-test comparisons of average length of stay between a family of 31 hospitals, and then output the p-values for each test (with the goal of plotting the ordered p-values to help estimate the # of true null hypotheses according to Schweder and Spjotvoll (1982)).
Is there a simple way to do this with proc multtest without specifiying all possible contrasts with multiple CONTRAST statements? I want to avoid this:
proc multtest data=hospital;
class hospital_name;
test mean(los);
contrast 'A vs B' (-1 1 0 0 ... 0);
contrast 'A vs C' (-1 0 1 0 ... 0);
contrast 'B vs C' (0 -1 1 0 ... 0);
...
contrast 'A vs Z' (1 0 0 0 ... -1);
run;
I've checked the online manual for proc multtest and it isn't terribly helpful.
Accepted Solutions
Solution
12-12-2014 10:57 PM
Posts: 5,543
## Re: How to perform all combinations of pairwise t-tests with proc multtest?
Proc GLM has the ability to do all pairwise comparisons and correct p-values for test multiplicity. You will have to do a bit of reformatting of the results as in the following example::
proc glm data=sashelp.baseball plots=none;
class team;
model nRuns = team;
*lsmeans team / pdiff adjust=simulate; /* Corrected p-values */
lsmeans team / pdiff adjust=t; /* Uncorrected p-values */
ods output Diff=teamDiffs LSMeans=teamLSMeans;
run; quit;
proc transpose data=teamDiffs out=teamDiffsLong;
by rowName;
var _:;
run;
proc sql;
create table teamDiffsPList as
select
b.team as team,
c.team as withTeam,
a.COL1 as tProb
from
teamDiffsLong as a inner join
teamLSMeans as b on left(a.rowName) = cats(b.LSMeanNumber) inner join
teamLSMeans as c on a._NAME_ = cats("_", c.LSMeanNumber)
where input(a.rowName, best.) > input(substr(a._NAME_,2), best.)
order by tProb;
quit;
PG
PG
All Replies
Solution
12-12-2014 10:57 PM
Posts: 5,543
## Re: How to perform all combinations of pairwise t-tests with proc multtest?
Proc GLM has the ability to do all pairwise comparisons and correct p-values for test multiplicity. You will have to do a bit of reformatting of the results as in the following example::
proc glm data=sashelp.baseball plots=none;
class team;
model nRuns = team;
*lsmeans team / pdiff adjust=simulate; /* Corrected p-values */
lsmeans team / pdiff adjust=t; /* Uncorrected p-values */
ods output Diff=teamDiffs LSMeans=teamLSMeans;
run; quit;
proc transpose data=teamDiffs out=teamDiffsLong;
by rowName;
var _:;
run;
proc sql;
create table teamDiffsPList as
select
b.team as team,
c.team as withTeam,
a.COL1 as tProb
from
teamDiffsLong as a inner join
teamLSMeans as b on left(a.rowName) = cats(b.LSMeanNumber) inner join
teamLSMeans as c on a._NAME_ = cats("_", c.LSMeanNumber)
where input(a.rowName, best.) > input(substr(a._NAME_,2), best.)
order by tProb;
quit;
PG
PG
Frequent Contributor
Posts: 81
## Re: How to perform all combinations of pairwise t-tests with proc multtest?
Fantastic, thank you PGStats!
BTW I also edited the format of the p-values from PVALUEw.d format to 12.8 numerical format to avoid values of "<.001".
🔒 This topic is solved and locked.
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CBSE Class 9 Science NCERT Solutions: Chapter 8, Motion
Get accurate NCERT Solutions for Class 9 Science Chapter 8 - Motion. All the solutions are explained by subject experts and are quite helpful to clear all the concepts. Students may download these NCERT solutions in PDF format.
CBSE Class 9 Science NCERT Solutions
Students searching for Class 9 Science NCERT solutions can now get all the solutions at jagranjosh.com/cbse and find a better approach to the questions that follow in each and every chapter in NCERT textbook.
NCERT books are written in the most lucid and clear manner to help students break the complex problems in most efficient way. Central Board of Secondary Education (CBSE) has made it mandatory to make use of the books that have been prepared by the researchers that will help the students build a strong foundation in a subject.
NCERT books contain a variety of questions given at the end of each chapter which are based on every conceptual topic. Students must solve all these NCERT questions to keep a track of their understanding. Practicing the NCERT questions will assure good results as most of the questions in annual examinations are asked from NCERT textbooks.
In this article we are providing NCERT Solutions for Class 9 Science Chapter 8 - Motion. Prepared by the subject matter experts, these solutions will help you get the right approach to all the NCERT questions and clear your concepts.
Students may download the NCERT solutions for Class 9 Science chapter - Motion, in the form of PDF.
Main topics discussed in Class 9 Science chapter- Motion are:
• Motion along a straight line
• Uniform motion and non-uniform motion
• Measuring the rate of motion
• Speed with direction
• Rate of change of velocity
• Graphical representation of motion
• Equations of motion by graphical method
• Equation for velocity-time relation
• Equation for position-time relation
• Equation for position–velocity relation
• Uniform circular motion.
Some of the questions and their solutions from NCERT Solutions for Class 9: Motion, are as follows:
Q. An object has moved through a distance. Can it have zero displacement? If yes, support your answer with an example.
Sol.
Yes, an object which has moved through a distance can have zero displacement, if it returns to its initial position after covering the desired distance.
For example: If a boy starts from his home goes to the market and then returns back to home, then his displacement is zero as the initial and the final position are the same which in this case is the home.
Q. Under what condition(s) is the magnitude of average velocity of an object equal to its average speed?
Sol.
The magnitude of average velocity is equal to the average speed when a body moves in a straight line.
Q. When will you say a body is in:
(i) Uniform acceleration?
(ii) Non-uniform acceleration?
Sol.
(i) A body has uniform acceleration, if its velocity changes by equal amount in equal interval of time.
(ii) A body has non-uniform acceleration, if its velocity changes by unequal amount in equal interval of time.
Q. What is the nature of the distance-time graphs for uniform and non-uniform motion of an object?
Sol.
(i) For uniform motion of an object, its distance-time graph is a straight line with constant slope.
(ii) For non-uniform motion of an object, its distance-time graph is a curved line with increasing or decreasing slope.
Q. State which of the following situations are possible and give an example for each of these:
(a) An object with a constant acceleration but with zero velocity.
(b) An object moving in a certain direction with acceleration in the perpendicular direction.
Sol.
(a) When an object is thrown vertically upward, then at highest point its velocity is zero but it has constant acceleration of 9.8 ms–2 which is termed as acceleration due to gravity.
(b) An aeroplane flies in horizontal direction but the acceleration due to gravity acts on it in vertically downward direction i.e., along the direction perpendicular to the direction of motion.
To get the complete solution click on the following link:
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## Influence of quantum disjunctive normal form formulae
This is the second post in a series of posts aimed at developing lower bounds for a natural class of quantum circuits. The first couple of posts are aimed at working out how this “natural” class, tentatively called qAC0, should be defined; essentially I’ll just try and argue by analogy with the classical definition of AC0, as far as I can.
At the current time we have defined a representation of QBFs called quantum disjunctive normal form (qDNF). However, we haven’t actually related this to quantum circuits yet (it’s a representation, but there’s no guarantee that it has anything to do with natural quantum circuits), I intend to tackle this problem a little later. For the moment I think it might be interesting to develop a better understanding of qDNF formulae before moving on. To this end I want to discuss, in this post, a couple of (conjectured) results concerning the influence of qDNF formulae.
### Influence of boolean variables
The classical definition of the influence of variable $j$ on a boolean function $f$ is the probability that $f$’s value is undefined if the value of $j$ is unknown, formally defined as
$I_j(f) = \mathbb{P}_x[f(x)\not=f(x\oplus e_j)]$,
where $x \oplus e_j$ flips the $j$th bit of $x$. (We are putting the uniform measure on the space of all $2^n$ bit strings.) The total influence $\mathbb{I}(f)$ is just the sum of the $n$ influences $I_j(f)$.
The influence plays an important role in the analysis of boolean functions, providing a useful tool to understand the behaviour of classes of boolean functions. There are many good sources about the influence, including, Gil Kalai’s posts on influence, and this, and this. I wouldn’t be able to do it justice here.
In this post I want to focus on one particular classical result, namely, the influence of DNF formulae, and to see if it can be generalised somehow. The reason for this is that certain boolean functions like PARITY have a huge total influence, and if you can show that some class of circuits cannot have a large influence you automatically learn that this class cannot compute PARITY. The particular result I want to generalise here is the following
Proposition 1. If a function $f:\{-1,+1\}^{\times n}\rightarrow \{-1,+1\}$ has a width $w$ DNF, then $\mathbb{I}(f) \le 2w$.
Proof. This is left as an exercise in O’Donnell’s notes, but, irritatingly, I can’t see how to immediately prove it…
### Influence of quantum variables
Now I want to set up the main conjecture of this post. To do this I’ll need to introduce the quantum generalisation of the influence of a variable. As we argue in our paper, a natural generalisation of the influence can be cooked up as follows. We define the derivative operator on the $j$th subsystem to be
$d_j(f) = f-\mbox{tr}_j(f)\otimes \frac{\mathbb{I}_j}{2}$,
where $f$ is any operator, $\mbox{tr}_j$ denotes the partial trace over subsystem $j$, and $\mathbb{I}_j$ is the $2\times 2$ identity matrix. Using this the quantum influence of variable $j$ is defined as
$I_j(f) = \frac{1}{2^n}\mbox{tr}(d_j(f)^2)$.
There are a couple of nice things to note about the quantum influence. Firstly, it reduces to exactly the classical influence for classical boolean functions embedded as diagonal operators. Secondly, in terms of our noncommutative fourier expansion, the quantum influence has a rather compelling expression. Thirdly, there are several alternative expressions (see lemma 67 in our paper) for $d_j(f)$ that might be useful. (I haven’t seen these in the classical literature, and maybe they might be useful in some contexts?)
So now we come to the main conjecture of this post
Conjecture 1. Let $f$ be defined by a width-$w$ qDNF. (Should we worry about the maxrank?) Then the total influence of $f$ satisfies $\mathbb{I}(f) \le 2w$.
Proof ideas. Seeing as how I don’t even know how to prove this result classically I don’t have much of a clue how to generalise the classical proof to the quantum setting. I’m guessing the classical proof is combinatorial in nature? If so, it’ll probably be impossible to quantise. However, it might be more fruitful to approach this problem directly from a quantum perspective.
I’ll just briefly describe one reformulation of the conjecture which might be easier to deal with. To do this I’ll write our qDNF as
$f = -\mathbb{I} + 2R$
where $R$ is the rejecting subspace given by
$R = \bigwedge_{j=1}^m (\mathbb{I}-c_j)$
where $c_j$ are the conjunctive clauses and we’ve used the fact for projectors that $P\vee Q = \mathbb{I} - (\mathbb{I}-P)\wedge(\mathbb{I}-Q)$. This gets rid of the nasty subspace joins and puts everything in terms of subspace intersections which are nicer.
We now rewrite the influence of $f$ as follows. The derivative operator becomes
$d_j(f) = 2(R-\mbox{tr}_j(R)\otimes \frac{\mathbb{I}_j}{2})$
so that
$I_j(f) = \frac{1}{2^n}\mbox{tr}(d_j(f)^2) = \frac{4}{2^n}\mbox{tr}(R) - \frac{2}{2^n}\mbox{tr}(\mbox{tr}_j(R)^2)$.
Thus the total influence can be written
$\mathbb{I}(f) = \frac{4n}{2^n}\mbox{tr}(R) - \frac{2}{2^n}\sum_{j=1}^m\mbox{tr}(\mbox{tr}_j(R)^2)$.
So the conjecture concerns how the projector $R$ onto the rejecting space changes when we trace out a qubit… At this point I’m out of ideas…
#### Implications of conjecture 1.
One nice thing about conjecture 1, if true, would be the corollary that width-$w$ qDNF cannot compute, eg., PARITY, because PARITY has total influence $n$. This is a pretty fundamental starting point for this whole research program! Thus I think conjecture 1 is a great place to start this whole program: it is a small enough conjecture that it looks provable (or disprovable!) and it would go a long way toward provide the basic building block of the larger quantum circuit lower bound approach.
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suivant: ibpu monter: Integration by parts : précédent: Integration by parts : Table des matières Index
### ibpdv
ibpdv is used to search the primitive of an expression written as u(x).v'(x).
ibpdv takes two arguments :
• an expression u(x).v'(x) and v(x) (or a list of two expressions [F(x), u(x)*v'(x)] and v(x)),
• or an expression g(x) and 0 (or a list of two expressions [F(x), g(x)] and 0).
ibpdv returns :
• if v(x) 0, the list [u(x).v(x), - v(x).u'(x)] (or [F(x) + u(x).v(x), - v(x).u'(x)]),
• if the second argument is zero, a primitive of the first argument g(x) (or F(x)+a primitive of g(x)) :
hence, ibpdv(g(x),0) returns a primitive G(x) of g(x) or
ibpdv([F(x),g(x)],0) returns F(x)+G(x) where diff(G(x))=g(x).
Hence, ibpdv returns the terms computed in an integration by parts, with the possibility of doing several ibpdv succesively.
When the answer of ibpdv(u(x)*v'(x),v(x)) is computed, to obtain a primitive of u(x).v'(x), it remains to compute the integral of the second term of this answer and then, to sum this integral with the first term of this answer : to do this, just use ibpdv command with the answer as first argument and a new v(x) (or 0 to terminate the integration) as second argument.
Input :
ibpdv(ln(x),x)
Output :
[x.ln(x),-1]
then
ibpdv([x.ln(x),-1],0)
Output :
-x+x.ln(x)
Remark
When the first argument of ibpdv is a list of two elements, ibpdv works only on the last element of this list and adds the integrated term to the first element of this list. (therefore it is possible to do several ibpdv successively).
For example :
ibpdv((log(x))`^`2,x) = [x*(log(x))`^`2,-(2*log(x))]
it remains to integrate -(2*log(x)), the input :
ibpdv(ans(),x) or input :
ibpdv([x*(log(x))`^`2,-(2*log(x))],x)
Output :
[x*(log(x))`^`2+x*(-(2*log(x))),2]
and it remains to integrate 2, hence input ibpdv(ans(),0) or
ibpdv([x*(log(x))`^`2+x*(-(2*log(x))),2],0).
Output : x*(log(x))`^`2+x*(-(2*log(x)))+2*x
suivant: ibpu monter: Integration by parts : précédent: Integration by parts : Table des matières Index
giac documentation written by Renée De Graeve and Bernard Parisse
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Partner with ConvertIt.com
New Online Book! Handbook of Mathematical Functions (AMS55)
Conversion & Calculation Home >> Measurement Conversion
Measurement Converter
Convert From: (required) Click here to Convert To: (optional) Examples: 5 kilometers, 12 feet/sec^2, 1/5 gallon, 9.5 Joules, or 0 dF. Help, Frequently Asked Questions, Use Currencies in Conversions, Measurements & Currencies Recognized Examples: miles, meters/s^2, liters, kilowatt*hours, or dC.
Conversion Result: Roman as = 0.3265865064 mass (mass) Related Measurements: Try converting from "as" to Babylonian shekel, cental (British cental), drachma2 (Greek drachma), dram (avoirdupois dram), dram troy (troy dram), electron mass (electron rest mass), funt (Russian funt), gram, Greek obolos, hyl, kwan (Japanese kwan), long hundredweight (avoirdupois long hundredweight), long ton (avoirdupois long ton), Mexican libra, oz ap (apothecary ounce), oz troy (troy ounce), pennyweight troy (troy pennyweight), short hundredweight (avoirdupois short hundredweight), short quarter, UK quintal (British quintal), or any combination of units which equate to "mass" and represent mass. Sample Conversions: as = .54 catty (Chinese catty), .0072 cental (British cental), .00096 cotton bale Egypt, 3,604.71 crith, .00326587 doppelzentner, .8 funt (Russian funt), 326.59 gram, .32658651 kg (kilogram), .54421769 kin (Japanese kin), .66650307 livre (French livre), .00642857 long hundredweight (avoirdupois long hundredweight), .00032143 long ton (avoirdupois long ton), .70935961 Mexican libra, .76085808 mina (Greek mina), 100,800 mite (English mite), 450 obol (Greek obol), 11.52 oz (ounce), .01993908 pood (Russian pood), .008 Roman talent, .00710059 Spanish quintal.
Feedback, suggestions, or additional measurement definitions?
Please read our Help Page and FAQ Page then post a message or send e-mail. Thanks!
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https://community.qlik.com/thread/30450
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3 Replies Latest reply: Jun 24, 2011 2:55 AM by jan.luchterhand
# Delivery on Time - Noobie formula question
Dear Community & Experts,
as I am still a noobie with QlikView I hope you can help me to solve the following task.
I habe a list of our suppliers and I would like to know, how their Delivery on Time accuracy is. I have attached a short excel file, which you
can use.
The way we want to calculate the DOT is like this:
We take all lines of one (the same) supplier in which the order status (column L) is 5. In column S we see, if this supplier has delivered in time.
If there is a 1 in column S, then he was on time. If there is a 0, he was not.
In my example list the company "Schulzenschnulze AG" has 21 lines with order status 5. 20 of these lines have a 1 in column S, 1 line has a 0.
That means 20 out of 21 deliveries were on time, so the DOT is 95,2%.
What are the right formulas to:
- count the number of lines of each supplier, where the order status in column L is 5
- count the 1's in column S for these lines
- divide number of 1's of column s by the number of lines for supplier X with status 5 in column L
I hope you get what I mean.
The data source will be an excel file like the one I've attached.
• ###### Re: Delivery on Time - Noobie formula question
Maybe I have to simplify my question... in Excel the formula I am looking for is most likely "countif"...
Let's say I have the following table:
Supplier Order Status DOT
A 5 1
A 5 1
A 5 0
A 5 1
B 5 1
B 3
B 5 1
a) I want to count the number of lines of Supplier A, only if the status in the second column is 5.
Should be something like: if(Order Status="5", count(Supplier), "N/A") but that doesn't work
The result should be: A count 4; B count 2
b) I want to count the number of 1's in column DOT for the above calculated number of rows for each supplier with order status 5. No idea about the formula, the result should be for Supplier A: 3, for B: 2.
c) Now I want to divide result b) by a)... so for
Supplier A: 3 times 1 / 4 rows with status 5: DOT accuracy 75% (3 out of 4 deliveries on time)
Supplier B: 2 times 1 / 2 rows with status 5: DOT accuracy 100% (2 out of 2)
• ###### Re: Delivery on Time - Noobie formula question
Hi Jan,
Here's an example using Set Analysis. Please see attached file.
Expressions are:
1. For counting the number of 5 occurences per supplier:
Count({<[Order Status]={5}>} [Order Status])
2. For counting the number of 1 occurences in DOT per supplier:
Sum({<[DOT]={1}>} [DOT])
3. For creating the percentage of deliveries on time:
Sum({<[DOT]={1}>} [DOT])
/
Count({<[Order Status]={5}>} [Order Status])
• ###### Re: Delivery on Time - Noobie formula question
Thank you Johannes!
That's exactly what I needed!
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# Derivation of Holomorphic Ward Identities in Franceso's CFT
In equation 5.37 of francesco's CFT he writes the Ward Identities for traslation symmetry in the language of holomorphic functions. He goes from
$$$$\frac{\partial}{\partial x^\mu} \langle T^\mu_{\ \ \ \nu}(x) X\rangle =-\sum_i^N\delta(x-x_i)\frac{\partial}{\partial x_i^\nu}\langle X\rangle \tag{5.32a}$$$$
to
\begin{align} 2\pi\partial_\bar{z}\langle T_{zz}X\rangle+2\pi\partial_z\langle T_{\bar{z}z}X\rangle=-\sum_i^N\partial_{\bar{z}}\frac{1}{z-\omega_i}\partial_{\omega_i}\langle X\rangle \tag{5.37a} \\ 2\pi\partial_\bar{z}\langle T_{z\bar{z}}X\rangle+2\pi\partial_z\langle T_{\bar{z}\bar{z}}X\rangle=-\sum_i^N\partial_{z}\frac{1}{\bar{z}-\bar{\omega}_i}\partial_{\bar{\omega}_i}\langle X\rangle. \tag{5.37b} \end{align}
I understand how to write the delta function as
$$$$\delta(x-x_i)=\frac{1}{\pi}\partial_z\frac{1}{\bar{z}-\bar{\omega}_i}=\frac{1}{\pi}\partial_\bar{z}\frac{1}{z-\omega_i}\tag{5.33}$$$$
but the index manipulation on the left hand side is very obscure to me. When I try to perform the change of variables explicitly I get
\begin{align} \frac{\partial}{\partial x^\mu}T^\mu_{\ \ \ \nu}&=\frac{\partial}{\partial x^1} T^1_{\ \ \ \nu}+\frac{\partial}{\partial x^2} T^2_{\ \ \ \nu} \\ &= \Big\{\frac{\partial z}{\partial x^1}\partial_z +\frac{\partial \bar{z}}{\partial x^1}\partial_\bar{z} \Big\}T^1_{\ \ \ \nu}+\Big\{\frac{\partial z}{\partial x^2}\partial_z +\frac{\partial \bar{z}}{\partial x^2}\partial_\bar{z} \Big\} T^2_{\ \ \ \nu} \\ &=\partial_z\Big\{T^1_{\ \ \ \nu}+iT^2_{\ \ \ \nu}\Big\}+\partial_\bar{z}\Big\{T^1_{\ \ \ \nu}-iT^2_{\ \ \ \nu}\Big\} \end{align}
Since we are working in Euclidean metric I'm assuming indeces $$\mu=1,2$$ can be changed from upper to lower indistinctively (we can't do that if the indeces represent a $$z$$ or $$\bar{z}$$ since the metric is different in those coordinates). Using the energy-momentum tensor in complex coordinates
\begin{align} T_{zz}=\frac{1}{2}\big(T_{11}-iT_{12}\big) \\ T_{\bar{z}\bar{z}}=\frac{1}{2}\big(T_{11}+iT_{12}\big) \end{align}
Then I get $$\partial_zT_{\bar{z}\bar{z}}+\partial_\bar{z}T_{zz}$$ for $$\nu=1$$ and $$-i\partial_zT_{\bar{z}\bar{z}}-i\partial_\bar{z}T_{zz}$$ for $$\nu=2$$ (I'm using that the energy-momentum tensor is traceless symmetric to get these results). This two equations clearly don't resemble the two equations that Francesco gets, specially since he gets crossed terms in each equation. It seems like he is simply replacing the old indeces with complex indeces but I'd like to see how the equation is actually derived.
If anyone can point where I'm messing up the calculation it would be really helpful.
• FWIW, the metric conventions are given in eqs. (2.89) & (5.7). Commented Apr 8, 2019 at 9:05
You have assumed that $$T_{z\bar{z}}=0$$ (i.e. $$T_{11}+T_{22}=0$$) in writing down the last pair of displayed equations. More generally, \begin{aligned} T_{zz}&=\frac{1}{4}\big(T_{11}-T_{22}-2iT_{12}\big)\\ T_{\bar{z}\bar{z}}&=\frac{1}{4}\big(T_{11}-T_{22}+2iT_{12}\big)\\ T_{z\bar{z}}&=\frac{1}{4}\big(T_{11}+T_{22}\big) \end{aligned} Also, if you use $$ds^2=dzd\bar{z}$$ (rather than a general conformal metric $$ds^2=2g_{z\bar{z}}dzd\bar{z}$$) then you get factors of 2 when raising indices, e.g., $$T^z_{\phantom{a}z}=g^{z\bar{z}}T_{\bar{z}z}=2T_{\bar{z}z}$$. Finally, when you go through it in detail you'll see it is effectively correct to let $$\mu$$ range over $$z,\bar{z}$$ directly, so that provides a fast route to the Di Francesco result.
It's easiest to just use the $$z,\bar{z}$$ basis directly. The $$\mu,\nu$$ indices don't need to start out in the $$1,2$$ basis.
$$\frac{\partial}{\partial x^\mu} T^\mu_{\ \ \ \nu}=g^{\mu\lambda}\frac{\partial}{\partial x^\mu} T_{\lambda\nu}=2\frac{\partial}{\partial z} T_{\bar{z}\nu}+2\frac{\partial}{\partial \bar{z}} T_{z\nu},$$ since the inverse metric only has the $$z\bar{z}$$ components. Then you can choose $$\nu$$ as either $$z$$ (53.7a) or $$\bar{z}$$ (53.7b). Note on the RHS the $$\nu$$ becomes a $$\omega_i,\bar{\omega}_i$$, whereas the $$\partial_z,\partial_\bar{z}$$ are coming from the delta function.
Now I suspect the problem in comparing the way you did it to the RHS of (53.7a) and (53.7b) was that you substituted in $$\nu=1,2$$ which mixes together the two equations. I'll show that the last equation you wrote (trivially lowering indices as you mention) agrees with mine. \begin{align} \frac{\partial}{\partial x^\mu}T^\mu_{\ \ \ \nu}&=\partial_z\Big\{T_{1 \nu}+iT_{2 \nu}\Big\}+\partial_\bar{z}\Big\{T_{1 \nu}-iT_{2 \nu}\Big\} \end{align} Now rather than substituting in more $$1,2$$ for $$\nu$$, I'll transform the first index to the $$z,\bar{z}$$ basis. I'll just show the first term. $$\partial_z\Big\{T_{1 \nu}+iT_{2 \nu}\Big\}=\partial_z\Big\{\frac{\partial z}{\partial x^1}T_{z \nu}+\frac{\partial \bar{z}}{\partial x^1}T_{\bar{z} \nu}+i\left(\frac{\partial z}{\partial x^2}T_{z \nu}+\frac{\partial \bar{z}}{\partial x^2}T_{\bar{z} \nu}\right)\Big\}=2\partial_z T_{\bar{z} \nu}$$ This is exactly what I got for the first term
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# physics
posted by .
The Space Shuttle travels at a speed of about 5.45 x 103 m/s. The blink of an astronaut's eye lasts about 111 ms. How many football fields (length = 91.4 m) does the Space Shuttle cover in the blink of an eye?
• physics -
vt/L=5.45•10³•111•10⁻³/91.4 =6.62
### Related Questions
More Related Questions
Post a New Question
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https://web.cs.swarthmore.edu/~knerr/teaching/f12/weeks/week11W.html
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```
WEEK10: recursion
---------------------------------------------------------------
W: recursion in graphics, binary search
REVIEW: concentric circles solution
""" draw recursive circles """
c.setOutline(color)
c.draw(w)
RECURSIVE GRAPHICS:
- how can we use recursion to draw something like this??
```
Here are some other interesting examples of recursion in graphics:
```
BINARY SEARCH with RECURSION:
- the binary search algorithm is a divide-and-conquer algorithm, and these
are usually easy to express using recursion:
[ 2, 6, 9, 20, 21, 33, 49, 72, 74, 99, 106]
^ ^ ^
low mid high
if what we're looking for is at the mid position, we're done
elif what we're looking for is less than what's at the mid position
do the binary search again with high=mid-1
else (what we're looking for is greater than what's at the mid position)
do the binary search again with low=mid+1
and we keep doing this until either we find what we want or
high ends up being less than low
- here are two ways to express this in python:
def recbinsearch(x, y, low, high):
""" return True if x is in y """
if high < low:
return False
else:
mid = (low + high) / 2
if x == y[mid]:
return True
elif x < y[mid]:
return recbinsearch(x,y,low,mid-1)
else:
return recbinsearch(x,y,mid+1,high)
####################################################
def recbinsearch(x, y):
""" return True if x is in y """
if len(y) <= 0:
return False
else:
low = 0
high = len(y)-1
mid = (low + high) / 2
if x == y[mid]:
return True
elif x < y[mid]:
return recbinsearch(x,y[:mid])
else:
return recbinsearch(x,y[(mid+1):])
- what is the difference between these two functions?
- do they both work?
- why might one be slower than the other??
```
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# Math
posted by on .
21. Find the square root ±√121
Is it 11?
28. Use the distributive property to simplify x(4x^2+x+4)
38. 4/5 + 2/10
Is it 3/5 ?
• Math - ,
21. Find the square root ±√121
Is it 11?
yes, 11^2 = 121
28. Use the distributive property to simplify x(4x^2+x+4)
4 x^3 + 4 x^2 + 4 x
38. 4/5 + 2/10
Is it 3/5 ?
No 4/5 + 1/5 = 5/5 = 1
### Answer This Question
First Name: School Subject: Answer:
### Related Questions
More Related Questions
Post a New Question
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# GSEB Solutions Class 9 Maths Chapter 7 Triangles Ex 7.3
Gujarat Board GSEB Solutions Class 9 Maths Chapter 7 Triangles Ex 7.3 Textbook Questions and Answers.
## Gujarat Board Textbook Solutions Class 9 Maths Chapter 7 Triangles Ex 7.3
Question 1.
ΔABC and ΔDBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see figure). If AD is extended to intersect BC at P, show that
1. ΔABD ≅ ΔACD
2. ΔABP ≅ ΔACP
3. AP bisects ∠A as well as ∠D
4. AP is the perpendicular bisector of BC.
Solution:
1. In ΔABD and ΔACD,
AB = AC (given) (ΔABC is an isosceles triangle)
BD = CD (given) (ΔDBC is an isosceles triangle)
∴ ΔABD ≅ ΔACD (by SSS congruency)
2. In ΔABP and ΔACP,
AB = AC (given)
∠BAP = ∠CAP (CPCT as AABD ≅ ∠ACD)
AP = AP (common)
∴ ΔABP ≅ ΔACP (by SAS congruency)
3. ΔABP ≅ ΔACP (Proved in part (ii))
BP = CP ………(1) (byCPCT)
and ∠BAP = ∠CAP (by CPCT)
∴ AP bisect ∠A.
In ΔBDP and ΔCDP,
BD = CD (given ΔBDC is an isosceles triangle)
BP = CP (Proved above in part (ii))
DP = DF (Common)
∠BDP = ∠CDP (by SSS congruency)
Hence ∠BDP = ∠CDP (by CPCT)
∴ AP bisect ∠D.
4. Since ΔBDP≅ΔCDP (Proved above)
BP = CP (by CPCT)
∠BPD = ∠CPD (by CPCT)
∠BPD + ∠CPD = 180° (linear pair)
∴ ∠BPD + ∠BPD = 180°
2∠BPD = 180°
∠BPD = 90°
∴ AP is the perpendicular bisector of BC.
Question 2.
AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that
solution:
AB = AC (ΔABC is an isosceles triangle)
∴ ΔABD ≅ ΔACD (by RHS congruency)
BD = CD (by CPCT)
Question 3.
Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of åPQR (see figure). Show that
(i) ΔABM ≅ ΔPQN
(ii) ΔABC ≅ ΔPQR.
Solution:
Given In ΔABC and ΔPQR,
AB = PQ
BC = QR
and AM = PN
(i) In ΔABM and ΔPQN,
AB = PQ (given)
BC = QR (given)
∴ $$\frac {1}{2}$$BC = $$\frac {1}{2}$$QR (Dividing by 2)
Hence BM = QN (median bisect opposite sides)
AM = PN
∴ ΔABM ≅ ΔPQN (by SSS congruency)
∠ABM = ∠PQN (by CPCT)
⇒ ∠ABC = ∠PQR
(ii) In ΔABC and ΔPQR,
AB = PQ (given)
∠ABC = ∠PQR (proved above in part (i))
BC = QR (Given)
∴ ΔABC ≅ ΔPQR (by SAS congruency)
Question 4.
BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that the triangle ABC is isosceles.
Solution:
Given: ΔABC in which BE = CF.
To Prove: ΔABC is an isosceles triangle
Proof: In ABEC and ΔCFB,
∠BEC = ∠CFB (each 900)
BC = CB (Common)
BE = CF (given)
∴ ∠BEC = ∠CFB (by RHS congruency)
∴ ∠CBF = ∠BCE (by CPCT)
Hence AB = AC (Sides opposite to equal angle are equal)
Therefore, ABC is an isosceles triangle.
Question 5.
ABC is an isoscele8 triangle with AB = AC. Draw AP ⊥ BC to show that ∠B = ∠C.
Solution:
Given: ABC is an isosceles triangle in which
AB = AC.
To Prove: ∠B = ∠C
Construction: Draw AP ⊥ BC
Proof: In right triangles APB and APC,
∠APB = ∠APC (each 90°)
AB = AC (given)
AP = AP (common)
∴ ΔAPB ≅ ΔAPC (By RHS congruency)
∠ABP = ∠ACP (by CPCT)
⇒ ∠B = ∠C
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lecture5_small
# lecture5_small - Sample and hold systems Sample and Hold...
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Unformatted text preview: Sample and hold systems Sample and Hold Systems The continuous-time plant, P ( s ), is preceeding by a zero-order hold and followed by a sampler. P ( s ) ZOH C ( z ) a64 a64 T a14a13 a15a12 + a117 a27 a27 a27 a27 a27 a27 a54 r ( k ) u ( k ) u ( t ) y ( t ) y ( k ) Performance and stability is specified in terms of the digital domain signals, r ( k ), y ( k ), u ( k ), etc. Analog/Digital (A/D) board: sampler. Digital/Analog (D/A) board: zero-order hold. Other options are possible but the above are by far the most common. Roy Smith: ECE 147b 5 : 2 Modeling sampled systems Modeling P ( z ) P ( s ) C ( s ) C ( z ) P ( z ) Approximation of C ( s ) with C ( z ) Model P ( s ), and sample/hold as P ( z ) Continuous-time design Discrete-time design Develop a model of the discrete-time behavior of the plant. Allows digital designs to be performed directly. Evaluating the stability of the discrete-time system ( C ( z ) and P ( z ) in feedback). Roy Smith: ECE 147b 5 : 1 Sample and hold systems Zero-order hold equivalence This is a reasonable model of a typical digital to analog (D/A) converter. At the sample-time, t = kT , the discrete input, u ( k ), is put on the output, u ( t ). This value is held constant for the entire sample period. So, u ( t ) = u ( k ) , for kT t &lt; kT + T. T 2T 3T 4T 5T 6T-2-1 1 2 3 4 Time: t u ( k ) u ( t ) Roy Smith: ECE 147b 5 : 4 Sample and hold systems Sample and Hold Systems Model the system from the ZOH block to the sampler: P ( s ) ZOH a64 a64 T a27 a27 a27 a27 u ( k ) u ( t ) y ( t ) y ( k ) P ( s ) is an LTI system = the system from u ( k ) to y ( k ) is LSI. It has an equivalent Z-transform, P ( z ). P ( z ) a102 a102 u ( k ) y ( k ) Zero-order hold equivalence: The closed-loop combination of P ( z ) and C ( z ) exactly models P ( s ) in closed-loop at the sample times....
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## This note was uploaded on 04/06/2010 for the course ECE 145 taught by Professor Rodwell during the Spring '07 term at UCSB.
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## DEV Community
Mercedes Bernard
Posted on • Updated on • Originally published at mercedesbernard.com
# Big O notation and cupcakes
This post is the first of a series adapted from my conference talk, Fun, Friendly Computer Science.
I am a terrible cook. Actually, that’s probably not true, I just really dislike cooking so never put any effort into it. But if you’re not like me and you enjoy being in the kitchen, you may be skilled enough to cook with ratios instead of recipes.
Cooking with ratios is when you don't memorize recipes but instead memorize the ratios of ingredients. The amount of ingredients you need changes in proportion to how much of the food you want to make. For example, cupcakes follow a 4:3:2:1 ratio. 4 eggs, 3 cups of flour, 2 cups of sugar, and one cup of butter.
So how is big O notation related to cupcakes? Big O notation measures the relative complexity of a function or algorithm. It measures the complexity in proportion to the size of the function’s input.
Most often complexity refers to running time but it could also be used to measure memory consumption, stack depth, and other resources. We’re going to focus on running time since in a coding interview that’s what they’re usually asking about. And because we’re talking about relative complexity, the actual input size is unimportant because we want to measure the proportional complexity of the code. This will become clearer in our examples.
At its core, big O notation is a mathematical representation of complexity. It is a functional notation of this measurement so all “O”s can be represented on a graph, which can be helpful for those more visually or graphically minded.
If visuals aren’t your thing, this post will also have code samples. For this post, the samples are in Javascript and can be found here, but I also have samples in Ruby and Python if you’re more comfortable with either of those languages.
## O(1)
The first “O” that you may encounter is O(1) (spoken as O of 1). O(1) is a constant running time. As the input size increases, the time to execute the code remains the same.
An example of this might be if you’re making cupcakes and you need to beat your butter and sugar together for 3 minutes. So regardless of if you’re making a single batch or double batch of cupcakes, it will take 3 minutes to mix together your butter and sugar.
``````combineButterAndSugar(batches) {
const steps = [];
const butter = {
ingredient: this.recipeRatios.butter,
amount: batches * this.recipeRatios.butter.number
};
const sugar = {
ingredient: this.recipeRatios.sugar,
amount: batches * this.recipeRatios.sugar.number
};
steps.push(this.beatWithMixer([butter, sugar], 3));
steps.push("Combined butter and sugar: O(1)");
return steps.join("<br/>");
}
``````
## O(n)
Next up, we have O(n). The running time of the function increases in direct proportion to the input size. In other words, as the input size increases the time to execute increases linearly.
A single for loop or while loop is a great indicator that a function is O(n). For example, when adding eggs to our cupcake batter, we need to add 1 egg at a time and then beat the batter with an electric mixer for 1 minute. So if we need 4 eggs per batch of cupcakes, it would look like 1) add an egg, beat for a minute, 2) add an egg, beat for a minute, 3) add an egg, beat for a minute, 4) add an egg, beat for a minute. You may notice that this running time is actually 4n but with big O notation, coefficients are unimportant. Again, we’re measuring relative complexity and degrees of complexity in terms of n are specific enough, we don’t need to increase specificity with the coefficients.
``````addEggs(batches) {
const steps = [];
const oneEgg = { ingredient: this.recipeRatios.eggs, amount: 1 };
const butterMixture = { ingredient: "butter mixture" };
const amount = batches * this.recipeRatios.eggs.number;
for (let egg = 0; egg < amount; egg++) {
steps.push(this.beatWithMixer([oneEgg, butterMixture], 1));
}
return steps.join("<br/>");
}
``````
## O(n^2)
Our next “O” is getting a little less performant. O(n^2) means that the running time increases in proportion to the square of the input size.
Increasing in proportion to the square of the input size may seem uncommon, but it's actually very common. Any time you see nested loops, you are most likely dealing with O(n^2) or its siblings. As you increase the depth of the nested loops, the exponent increases. So loops nested to 2 levels deep are O(n^2), 3 levels deep are O(n^3), 4 levels deep are O(n^4) and so on and so forth.
``````combineFlourMixtureAndMilkAndButterMixture(batches) {
const steps = [];
const butterMixture = { ingredient: "butter mixture" };
const flourMixture = { ingredient: "flour mixture" };
const milk = {
ingredient: this.recipeRatios.milk,
amount:
(batches * this.recipeRatios.milk.number) / (batches * batches)
};
steps.push(this.beatWithMixer([butterMixture, flourMixture], 1));
for (let batch = 0; batch < batches; batch++) {
for (let portion = 0; portion < batches; portion++) {
steps.push(this.beatWithMixer([butterMixture, milk], 1));
steps.push(this.beatWithMixer([butterMixture, flourMixture], 1));
}
}
steps.push("Slowly combined milk, flour mixture, and butter mixture: O(n^2)");
return steps.join("<br/>");
}
``````
## O(2^n)
When the time to execute code grows exponentially with the size of the input, you’re looking at O(2^n). This means that very small increases in input size result in very large increases in running time.
Calculating Fibonacci numbers recursively is an example of an O(2^n) algorithm. If we were throwing a birthday party for Fibonacci and wanted to frost Fibonacci numbers on our cupcakes, we would be looking at exponential running time as we frosted more and more cupcakes. (We’ll be covering recursion soon so don’t worry if that’s new to you.)
``````fibonacciFrosting(batches) {
const numberToFrost = this.calculateFibonacciNumber(batches);
return `Iced the fibonacci number \${numberToFrost} to all of the cupcakes: O(2^n)`;
}
calculateFibonacciNumber(number) {
if (number <= 1) {
console.log("Fibonacci base case!");
return number;
}
return this.calculateFibonacciNumber(number - 1) + this.calculateFibonacciNumber(number - 2);
}
``````
## O(log n)
Finally, we have logarithmic running time. I think of O(log n) as the inverse of O(2^n). Where exponential growth sees larger and larger increases in running time as the input size grows, logarithmic growth sees smaller and smaller increases in running time as the input size grows until it approaches a constant running time for large input sizes.
I don’t have a code sample for this since my cupcake metaphor is getting a little stretched 😬, but logarithmic running time is common in divide and conquer algorithms like binary search. If we wanted to hand out our cupcakes using an algorithm with logarithmic running time, we might cut our batch of cupcakes in half and hand to the 2 people nearest to us and ask them to cut their collection of cupcakes in half and hand to the 2 people nearest to them and continue that pattern until everyone has 1 cupcake.
## Big O notation uses
Big O notation can be intimidating since it’s very hard to piece together context clues of what it means if you never had the chance to learn it. In the 10 years I’ve been coding, I’ve never used big O notation except in my college CS classes and interviews. So before you interview for a new opportunity, it might be worth refreshing your memory, just in case you have an interviewer who values CS fundamentals over other, more valuable skills (in my opinion).
Big O notation can also be useful for having a shared language to talk about relative code performance with your peers. But don’t feel inadequate if you don’t remember the different “O”s and what they mean. A shared language is nice, but you can also point to nested loops and talk about the negative impact that has on performance without remembering which “O” that is 🤗
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Small and large extra dimension(s) of the physical space
Trying to make sense of small and large extra dimension(s) of phyiscal space in a simple intuitive example.
Consider a two dimensional manifold like $$\mathbb{R}^2$$ and we are trying to add a small and a large extra dimension.
Do we mean by small extra dimension in this case something like $$(0,1) \times \mathbb{R}$$ (the flat case) or $$S^1 \times \mathbb{R}$$ (the curved case)?
Do we mean by large extra dimension something like $$\mathbb{R^2} \times \mathbb{R}=\mathbb{R}^3$$?
Do we mean in the case of our three dimensional space that basically we have a base space of our phyiscal three dimensional space with a total space built by adding a fiber and thus creating a fiber bundle or a even more general an arbitrary total space?
Does the extra dimension need to be real or can we even consider the complex manifolds, in the case of adding extra dimension to the phyiscal space, for example $$\mathbb{C} \times \mathbb{R^3}$$ or (Riemann surface) $$\times \mathbb{R^3}$$
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Type emmean SE df lower.
# Reporting Two Way Repeated Measures Anova
Key Topics and Links to Files Data Analysis Guide Which Therapy is Best for Treating Eating Disorders? Ccetsw In Social.
For example, Marie; Verbeke, or whether other analyses may have provided more insight. One way anova assumptions pdf Think Nature in Cogne. First we must define our two repeated-measures variables Figure 1. Construct a profile plot.
Imagine a set of three numbers, red, these packages have really simplified these procedures. Stata Interpreting Two-way ANOVA CrunchEconometrix. Drug classes are two way anova, measures anova in measurement as used. ANOVA table should be ignored.
Does mean reaction tasks, reporting allows readers to this way we have two or pattern in. One measurement were measured as well as a report. As time period as the measurement of multiply comparison procedures for each snake. We can analysis data using a repeated measures ANOVA for two types of. Images are still loading. Mixed designs are research designs in which there are both between subjects and within subjects variables being compared.
The previous chapters, or gender and confidence interval level of this dataset.
The grey hair makes you look distinguished and the hemorrhoids make you look concerned. In measurement were not implemented in advance! Thus making the reporting of overall treatment effects appear rather confusing. We use a one-way repeated measures ANOVA in two specific situations. Read my post of them to learn more!
Since there are only two levels for each of our variables, you may obtain invalid results. Repeated-measures ANOVA STATA Support ULibraries. Here, gender is a categorical factor that has the two levels of male and female. Reporting Partial Eta Squared instead of Generalized Eta Squared. Please correct mse for two way repeated measures anova family.
Way anova is repeated measures anova assumption of two way to report effect size datasets. In essence, and statistics in APA style, no interactions can be calculated. We appreciate your patience.
The reason is that I am not happy with any of the traditional multiple comparison procedures. How to conduct repeated measures ANOVA SAS Support. Then will inflate fwer you will focus on a repeated measured over from the approach. However, as the measurement variable is measured repeatedly on each snake. Press J to jump to the feed. However things are not as complicated as you might think.
By default this includes the title author date and the format of the report votes.
We deal with complicated as grip strength of gender attitudes scores are equal variances may i could point, you for this is measured.
This two repeated measures anova sheet for reporting partial eta and value to a best guess. Error Sum of Squares, it has many other applications. Present computation of the A x B x S totally within-subjects two-way design. An interaction effect size calculations into groups need a known as well. Repeated-Measures ANOVA Basicmedical Key.
Lecture 3- Two-Way Repeated Measures ANOVA & Mixed. Advanced ANOVA Procedures.
But perhaps you want a more formal way to do it, the multivariate tests are easier to justify. The example provided in class was aimed at addressing two distinct tasks 1 to give. Transparent reporting is essential for the critical evaluation of studies.
Jk combinations of accounting for our anova when possible values shown in sas will vary. This is robust with two way will elicit faster to. No use, Abbie Van Nuland, and there are multiple items for each condition. Factorial ANOVA Amazon AWS. Thank you measure the repeated measures.
Beverly Hills, some research hypotheses require repeated measures.
Therefore if you have a repeated-measures variable that has only two levels then sphericity. There is measured over the anova is a substitute for? But now read it again and still feel very useful and interesting. Interaction Effects in ANOVA. Which Effect Size Measure is Appropriate for One-Way and.
In so that again the correlation, in general linear or pattern of two way anova performed in. For example, Kand σwhen the other parameters in the design are held constant. Repeated measures ANOVA is also known as 'within-subjects' ANOVA.
Many of the random sample between measures anova table at the above for the alternative to. Lesson 9 Mixed Factorial ANOVA Furman University. Just be sure you're reporting the df for the correct error term for all those. Click to see our collection of resources to help you on your path. Chapter 15 Mixed design ANOVA SAGE edge.
Disagreements were measured over restudying. Banking In the presence of imbalanced data, as used throughout the current article.
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## Removable discontinuities
Determine whether the following functions can be made be continuous by defining a value at $$x=0$$. The functions are defined on $$\mathbb R\setminus\{0\}$$ by the following formulas:
• #### Variant 1
$$f(x)=e^{|x|}$$
• #### Variant 2
$$f(x)=\arctan\frac1{x^2}$$
• #### Variant 3
$$f(x)=\frac{1-\cos x}{x^3}$$
• #### Variant 4
$$f(x)=x^3\sin^2\frac1x$$
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## C # summary of all the classic sorting algorithms
` 1.Selection and SortingSelection and SortingClass SelectionSorter {private int min; public void Sort (int [] arr) {for (int i = 0; i <arr.Length-1; ++ i) {min = i; for (int j = i + 1; j <arr.Length; ++ j) {if (arr [j] <arr [min]) min = j;} int t = arr [min]; arr [min] = arr [i ]; arr [i] = t;}}} 2. EbullitionSorter {public void Sort (int [] arr) {int i, j, temp; bool done = false; j = 1; while ((j <arr.Length) && (! done)) // Judgment length {done = true; for (i = 0; i <arr.Length-j; i ++) {if (arr [i]> arr [i + 1]) {done = false; temp = arr [i]; arr [i] = arr [i + 1]; // exchange data arr [i + 1] = temp;}} j ++;}}} 3. Quick Sort Quick Sort class QuickSorter {private void swap (ref int l, ref int r) {int temp; temp = l; l = r; r = temp;} public void Sort (int [] list, int low, int high) {int pivot; // storage branch Point int l, r; int mid; if (high <= low) return; else if (high == low + 1) {if (list [low]> list [high]) swap (ref list [low], ref list [high]); return;} mid = (low + high) >> 1; pivot = list [mid]; swap (ref list [low], ref list [mid]); l = low + 1; r = high; do { while (l <= r && list [l] <pivot) l ++; while (list [r]> = pivot) r--; if (l <r) swap (ref list [l], ref list [r]) ;} while (l <r); list [low] = list [r]; list [r] = pivot; if (low + 1 <r) Sort (list, low, r-1); if (r + 1 <high) Sort (list, r + 1, high);}} 4, insert sort sort public class InsertionSorter {public void Sort (int [] arr) {for (int i = 1; i <arr.Length; i ++ ) {int t = arr [i]; int j = i; while ((j> 0) && (arr [j-1]> t)) {arr [j] = arr [j-1]; // Exchange Order --j;} arr [j] = t;}}} 5. Hill sort public sort ShellSorter {public void Sort (int [] arr) {int inc; for (inc = 1; inc <= arr.Length / 9; inc = 3 * inc + 1); for (; inc> 0; inc / = 3) {for (int i = inc + 1; i <= arr.Length; i + = inc) {int t = arr [i-1]; int j = i; while ((j> inc) && (arr [j-inc-1 ]> t)) {arr [j-1] = arr [j-inc-1]; // exchange data j-= inc;} arr [j-1] = t;}}}} 6. Merge Sort Merge Sort // // <summary> /// merge sort sort: merge sort entry // // </ summary> /// <param name = "data"> unordered array </ param> /// <returns> Ordered arrays </ returns> /// <author> Lihua (www.zivsoft.com) </ author> int [] Sort (int [] data) {// Take the middle index of the array int middle = data.Lengt h / 2; // Initialize temporary array let, right, and define result as the final ordered array int [] left = new int [middle], right = new int [middle], result = new int [data.Length]; if (data.Length% 2! = 0) // If the array has an odd number of elements, re-initialize the right temporary array {right = new int [middle + 1];} if (data.Length <= 1) // only left 1 or 0 quaternions, return without sorting {return data;} int i = 0, j = 0; foreach (int x in data) // start sorting {if (i <middle) // fill the left array {left [i] = x; i ++;} else // fill the right array {right [j] = x; j ++;}} left = Sort (left); // recursive left array right = Sort (right); // recursive right Array result = Merge (left, right); // Begin sorting ///this.Write(result);// Output sorting, test (lihua debug) return result;} /// <summary> /// Merge and sort and merge: sort in This step // </ summary> /// <param name = "a"> left array </ param> /// <param name = "b"> right array </ param> /// <returns> merging Sort the left and right arrays and return </ returns> int [] Merge (int [] a, int [] b) {// Define the result array to store the final result int [] result = new int [a.Length + b. Length]; int i = 0, j = 0, k = 0; while (i <a.Length && j <b.Length) {if (a [i] <b [j]) // The elements in the left array are less than Elements in the right array {result [k ++] = a [i ++]; // Place the smaller one in the result array} else /// The elements in the left array are greater than the elements in the right array {result [k ++] = b [j ++]; / / Put the smaller one in the result array }} while (i <a.Length) // There is actually a left element but no right element {result [k ++] = a [i ++];} while (j <b.Length) // right and right elements, No left element {result [k ++] = b [j ++];} return result; // return result array} Note: This algorithm is provided by Zhou Lihua (http://www.cnblogs.com/architect/archive/2009/05/ 06 / 1450489.html) 7. Cardinality sorting Cardinality sorting // Cardinality sorting public int [] RadixSort (int [] ArrayToSort, int digit) {// low to high digit for (int k = 1; k <= digit; k ++ ) {// temp array to store the sort result inside digit int [] tmpArray = new int [ArrayToSort.Length]; // temp array for countingsort int [] tmpCountingSortArray = new int [10] {0,0,0,0 , 0,0,0,0,0,0}; // CountingSort for (int i = 0; i <ArrayToSort.Length; i ++) {// split the specified digit from the element int tmpSplitDigit = ArrayToSort [i] / (int) Math.Pow (10, k-1)-(ArrayToSort [ i] / (int) Math.Pow (10, k)) * 10; tmpCountingSortArray [tmpSplitDigit] + = 1;} for (int m = 1; m <10; m ++) {tmpCountingSortArray [m] + = tmpCountingSortArray [m -1];} // output the value to result for (int n = ArrayToSort.Length-1; n> = 0; n--) {int tmpSplitDigit = ArrayToSort [n] / (int) Math.Pow (10, k-1)-(ArrayToSort [n] / (int) Math.Pow (10, k)) * 10; tmpArray [tmpCountingSortArray [tmpSplitDigit] -1] = ArrayToSort [n]; tmpCountingSortArray [tmpSplitDigit]-= 1; } // copy the digit-inside sort result to source array for (int p = 0; p <ArrayToSort.Length; p ++) {ArrayToSort [p] = tmpArray [p];}} return ArrayToSort;} 8. Sorting // counting sorting /// <summary> /// counting sort /// </ summary> /// <param name = "arrayA"> input array </ param> /// <param name = "arrange" > the value arrange in input array </ param> /// <returns> </ returns> public int [] CountingSort (int [] arrayA, int arrange) {// array to store the sorted result, // size is the same with input array. int [] arrayResult = new int [arrayA.Length]; // array to store the direct value in sorting process // include index 0; // size is arrange + 1; int [] ar rayTemp = new int [arrange + 1]; // clear up the temp array for (int i = 0; i <= arrange; i ++) {arrayTemp [i] = 0;} // now temp array stores the count of value equal for (int j = 0; j <arrayA.Length; j ++) {arrayTemp [arrayA [j]] + = 1;} // now temp array stores the count of value lower and equal for (int k = 1; k <= arrange; k ++) {arrayTemp [k] + = arrayTemp [k-1];} // output the value to result for (int m = arrayA.Length-1; m> = 0; m--) {arrayResult [arrayTemp [arrayA [m]]-1] = arrayA [m]; arrayTemp [arrayA [m]]-= 1;} return arrayResult;} 9. Small root heap sort Small root heap sort // // <summary> / // small root heap sort /// < / summary> /// <param name = "dblArray"> </ param> /// <param name = "StartIndex"> </ param> /// <returns> </ returns> private void HeapSort (ref double [ ] dblArray) {for (int i = dblArray.Length-1; i> = 0; i--) {if (2 * i + 1 <dblArray.Length) {int MinChildrenIndex = 2 * i + 1; // compare Left subtree and right subtree, record the index of the minimum if (2 * i + 2 <dblArray.Length) {if (dblArray [2 * i + 1]> dblArray [2 * i + 2]) MinChildrenIndex = 2 * i + 2;} if (dblArray [i]> dblArray [MinChildrenIndex]) {ExchageValue (ref dblArray [i], ref dblArray [MinChildrenIndex]); NodeSort (ref dblArray, MinChildrenIndex);}}}} /// <summary> /// Node sorting /// </ summary> /// <param name = "dblArray"> </ param> /// <param name = "StartIndex"> </ param> private void NodeSort (ref double [] dblArray, int StartIndex) {while (2 * StartIndex + 1 <dblArray.Length) {int MinChildrenIndex = 2 * StartIndex + 1; if (2 * StartIndex + 2 <dblArray.Length) {if (dblArray [2 * StartIndex + 1]> dblArray [2 * StartIndex + 2]) {MinChildrenIndex = 2 * StartIndex + 2;}} if (dblArray [StartIndex]> dblArray [MinChildrenIndex]) {ExchageValue (ref dblArray [StartIndex], ref dblArray [MinChildrenIndex]); StartIndex = MinChildrenIndex;}}} /// <summary> /// exchange value // // </ summary> // / <param name = "A"> </ param> /// <param name = "B"> </ param> private void ExchageValue (ref double A, ref double B) {double Temp = A; A = B; B = Temp;} `
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# A machine gun fires a bullet of mass 40 g with a velocity 1200 m/s The man holding it, can exert . a maximum force of 144 N on the gun. How many bullets can he fire per second at the most?
upender surepally
126 Points
10 years ago
The force exerted by machine gun on man''s hand in firing a bullet = change in momentum per second on a bullet or rate of change of momentum =(40/1000*)1200= 48 N .
The force exerted by man on machine gun = 144 N
Hence, number of bullets fired = 3
Shivam Bhagat
22 Points
10 years ago
According to newtons 2nd law..... F=m(v-u)/t
144=0.04*n*1200/1 (per second...so t=1s)
n=3 bullets / sec
ALL THE BEST..!
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# Chapter 2 - Set Theory - 2.3 Venn Diagrams and Set Operations - Exercise Set 2.3 - Page 81: 36
C ∩ ∅ = {} = ∅
#### Work Step by Step
C∩∅ U = {1,2,3,4,5,6,7} C = {2,3,4,5,6} The set C ∩ ∅ represents the elements which are common in both C and ∅ (∅ represent empty set). Therefore the common elements of C and ∅ is an empty set . Therefore C ∩ ∅ = {} = ∅
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.
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# Finding Fibonacci In Golden Trees
Aug 22, 2017 · 15 min read
Learning new things always brings the opportunity to have your mind completely blown. Mind you, this always doesn’t happen — at least when it comes to the realm of computer science. However, every once in awhile, you learn something that truly awes you, something that is actually awesome in the actual dictionary definition sense of the word; something that really feels like a lightbulb going off, making you feel very grateful to have set about trying to understand it in the first place.
I’m speaking from personal experience, of course, because this was exactly what happened to me when I learned about the wonders of AVL trees. It turns out that computer science has elements of mathematics under the surface, some of which we have already seen in the course of this series. But mathematics also has another hidden layer to it, as well: a connection to the natural world around us.
Oftentimes, when we learn about super theoretical stuff (like a lot of core computer science concepts) it’s hard to connect the dots between where these concepts were created, why we use them, and why we should care. Understanding the history of why something was invented is just as important as comprehending its applications and usage. It’s a bit rare for us to be able to see how something in computer science directly connects to other fields that have nothing to do with computing!
In today’s post, however, mathematics, computer science, and nature all come together in the most magical of ways. So, let’s get to it!
# Looking for tree patterns
Last week, we learned about a unique data structure called an AVL tree, which is a type of self-balancing binary search tree. Hopefully you remember that an AVL tree is a height-balanced tree, which means that it follows a golden rule: no single leaf in the tree should have a significantly longer path from the root node than any other leaf on the tree.
The important thing to keep in mind about this golden rule is that as an AVL tree grows in height, it must have a minimum number of nodes at each level before another one can be added.
The illustration shown here exemplifies this. In this drawing, we can see five different AVL trees, each with a varying heights.
In the AVL tree with a height of 0, there are no nodes, so n is equal to `0`. In the AVL tree with a height of 1, there is exactly `1` node, while in an AVL tree with a height of 2, there are exactly `2` nodes. Remember, we’re only considering the minimium number of nodes. Technically, we could have added a third node to our AVL tree with a height of 2, and that would still be fine; but, we need at least two nodes in order to create a height-balanced AVL tree with a height of 2.
What happens when we need to create an AVL tree with a height of 3? Well, we need `4` nodes at minimum to do that, since we need to add a right subtree before we can go about adding another level. It’s a similar situation for the tree with a height of 4: we need `7` nodes in order to create a height-balanced right subtree before adding another node to the left subtree.
Let’s take it one step further: what is the minimum number of nodes that we need in order to create an AVL tree with a height of 5?
Well, based on what we learned last week about AVL trees, we know that the difference between the left and the right subtrees cannot be more than one level in height. This means that we’ll need to add to our right subtree before we can grow our left subtree.
The example shown here illustrates what our minimum-node number for an AVL tree with a height of 5 would be: we need, at a minimum, 12 nodes in order to create an AVL tree with a height of 5. If we break this down further, we can see that we need 7 of those nodes in the left subtree, and 4 nodes for the right subtree. And, of course, we need a node for the root.
If we look at the left subtree and the right subtree for long enough, we’ll notice that they seem familiar. That’s because we just saw them a minute ago when we were looking at our AVL trees with a height of 4 and 3! Those two trees just happen to be the two structures that came before the AVL tree with a minimum node count and a height of 5.
If you feel like there’s an interesting pattern here, you’re be right.
The height of an AVL tree with the minimum number of nodes is a combination of the two minimum-height AVL trees preceding it.
This is not unique to the AVL tree with a height of 5; in fact, if we look back at the earlier illustration and close closely at the AVL tree with a height of 4, we’ll notice the exact same pattern. The left subtree of the an AVL tree with a height of 4 is the AVL tree with a height of 3, while the right subtree is the the AVL tree with a height of 2.
We could even continue to draw larger height-balanced trees and see this pattern continue, even as the height of a tree gets immensely large.
So, how can transform this pattern out into something more abstract? We derive this pattern to the following formula: “the height of a tree `n` is equivalent to the sum of the heights of trees `n-1` plus `n-2` plus `1`.
Another way to think about this is that the minimum number of nodes to create a tree of height n is by combining the two trees that come before it, and adding another node for the root node. In other words, in order to determine the minimum number of nodes necessary to create a tree with a height of 10, you’d need the number of nodes from the tree with a height of 9 and with a height of 8, plus one additional node for the root node.
If we sum the two AVL trees before the one we’re looking for, we can programmatically figure out the minimum number of nodes we’ll need to create any given height-balanced AVL tree.
Voilà! We’ve discovered the Fibonacci sequence right under our noses!
# Finding Fibonacci and the golden ratio
The pattern of “summing two numbers to get the third number” is known as the Fibonacci sequence, named after Leonardo Fibonacci. What young Leonardo stumbled upon could be summarized as a mathematical sequence that follows a single rule, which is exactly what we know to be the Fibonacci sequence:
every number after the first two numbers is the sum of the two numbers that preceded it.
Once we have the first two numbers of the Fibonacci sequence, we have enough to build the entire thing (but it goes on forever and forever, so we’ll just skip that for now)! And it just so happens that the first two numbers are pretty easy to remember: they’re 0 and 1.
We already know that there’s some kind of correlation between the Fibonacci sequence and the left and right subtree of a height-balanced AVL tree. Namely, we know that the minimum number of nodes for a tree with a height of H will be the two minimum node trees that come before it, plus one more node for the root node.
Since we derived a formula in the previous section to determine the height of the tree based on the number of nodes, can we use the Fibonacci sequence to determine the reverse? Can we abstract out a formula to determine the minimum number of nodes for a tree based on it’s height?
Of course we can!
Let’s break down the math that’s working behind the scenes here and see if we can figure out what’s actually going on.
In the illustration shown here, we’re going to do the reverse of the formula we deduced earlier, which means that we’ll continue working with a tree with a height H of `5`. If we’re looking for n, the minimum number of nodes for a tree of height H, we can find the Fibonacci number located at the index of `Fibonacci[H+2]`, and subtract one from it in order to determine the minimum number of nodes to create a tree of height H.
So, in the case of a tree with a height H of `5`, we’ll need to find the Fibonacci number at the index of `5 + 2`, or `7`. Recall that we indexed our Fibonacci sequence beginning with `0` as the first index, so the element at `Fibonacci[7]` will give us `13`. Since `13 - 1` is `12`, we know that we’ll need a minimum of `12` nodes to create a tree with a height of 5.
When we drew out a balanced tree with a height of 5 in the previous section, that’s exactly how many nodes we had!
Okay, what about a tree with a height H of `6`? Well, `6 + 2` is `8`, and the element located at the index of `Fibonacci[8]` would be the number `21`. Remember, even if we don’t know the element at the index of `Fibonacci[8]`, we know that `Fibonacci[6] = 8` and `Fibonacci[7] = 13`, which means we can sum these two in order to get the next element in the sequence. Since `Fibonacci[8] = 21`, we can subtract one from it and know that we need, at a minimum, `20` nodes in order to create a height-balanced AVL tree with a height of 6.
Rad! We just reversed our first formula, which used the Fibonacci pattern, into its opposite…which also uses the Fibonacci pattern!
Okay, that’s pretty cool, but it’s about to get even cooler.
The sequence that has long been attributed to (and named for!) Fibonacci was actually discovered hundreds of years earlier by ancient Indian mathematicians, way back in the 6th century. It was only with Fibonacci’s discovery of it 600 years later that the western world came to know of this sequence, and were thus able to study it in even greater depth.
As more mathematicians, scientists, and artists started studying this sequence, they began to understand its deep connections to the world around us. One of the things that they discovered was that the Fibonacci sequence was very closely linked to another special property in geometry and mathematics: the golden spiral and the golden ratio.
It’s unclear when the golden ratio was discovered; although it was Greek mathematician Euclid who first mentioned it in his famous, well-studied and well-cited text, “Elements”, it’s also very possible that the ancient Egyptians used it when they built the pyramids, thousands of years ago. The golden ratio is often referred to as “phi” (φ or ϕ), and is also known as DaVinci’s “golden proportion”. In mathematical terms, two quantities have a golden ratio if the ratio between the two quantities is the same as the ratio of the two elements summed when compared to the larger of the two quantities.
The golden ratio is usually illustrated by using the example of a rectangle made up of a square and a rectangle, respectively.
In the example shown above, we can see the golden ratio at work. When we divide a line into two parts such that the whole length divided by the larger part is equal to the larger part divided by the smaller part, we end up with the golden ratio. In this particular illustration, the ratio of the longer side of the rectangle, the sum of a + b, to the side a will be the same as the ratio of the length of a to the side b.
But what exactly is the golden ratio? We’re comparing a’s and b’s, but what does that mean, really? Well, let’s look at the golden ratio using actual numbers, and see if we can get to a more concrete answer.
We know that the golden ratio is often described as follows:
a + b is to a, as a is to b
Let’s replace a and b here with integers: `a = 5` and `b = 3`.
If a + b is to a, as a is to b, then we should be able to prove that `5 + 3` to `5` has the same ratio as `5` does to `3`.
Well, it’s time to do the math and find out if that’s really the case!
Since `5 + 3 = 8`, we determine the ratio here by dividing `8/3`, which is `1.6`. If we divide `5/3`, we get 1.6, repeating.
In both of these instances, we’re actually coming very close to the value of phi (φ), which is an irrational number that is the sum of 1 and the square root of 5, divided by 2. This irrational number roughly approximates to `1.6180339887`, often just referred to as `1.618`, for short.
So how does the golden ratio tie into the golden spiral? Well, the golden spiral is a geometric, logarithmic spiral whose growth factor actually happens to be — surprise, surprise! — the golden ratio. And guess what? The Fibonacci sequence comes into play here again, too!
If we construct squares with the width of each consecutive element in the Fibonacci sequence, we can start to build up rectangles. We’ll notice that each one of these rectangles has the golden ratio within it! And, as these rectangles grow, a spiral shape begins to emerge.
We’ll notice that each square’s width is made up of two parts: another square and a rectangle, each of whose widths are the sums of the two squares (the two Fibonacci numbers) that precede it. For example, the width of the square 8, drawn in pink, is the sum of the widths of the two squares that came before it: 5 and 3. Similarly, the width of the red square 13 is the sum of the two squares that came before it: 8 and 5.
The Fibonacci sequence appears once again! The ratio between each of these squares that we used to build up our golden spiral is tied directly to the golden ratio. If we divide each Fibonacci number by the number that preceded it, this becomes more obvious.
Notice that we get closer and closer to the actual value of phi (φ) as our Fibonacci numbers grow in size. We won’t ever hit the golden ratio exactly, since it’s an irrational number; but the larger our Fibonacci number, the closer we’ll get to it as we compare one ratio to another.
The Fibonacci sequence and the Golden Spiral exist all around us — in the arms of a starfish, the patterns of a seashell, the branching of a tree — and even within us — inside the human ear, within the proportions of our fingers, and even the ratio of the human arm! My personal favorite example of this sequence in nature comes from my favorite flower: the sunflower.
According to research from The American Association for the Advancement of Science,
The telltale sign is the number of different seed spirals on the sunflower’s face. Count the clockwise and counterclockwise spirals that reach the outer edge, and you’ll usually find a pair of numbers from the sequence: 34 and 55, or 55 and 89, or — with very large sunflowers — 89 and 144. Although the math may be beautiful, plant biologists have not worked out a mechanistic model that fully explains how the sunflower seed patterns arise.
And guess what? There’s yet another way that this pattern is connected to AVL trees, which is perhaps what makes them so damn golden.
Time to find out!
# The golden (tree height) ratio
What is the final petal in this metaphorical flower of Fibonacci?
In order to bring it all home, we’re going to need to go back to where we started: determining the relationship between the minimum number of nodes and the height of an AVL tree. We’ve looked at two different formulas: first, we figured out how we could determine the height based on the minimum number of nodes we needed; then, we determined at the number of nodes we’d need based on the height that knew a tree had.
Well, it turns out that the Fibonacci sequence and the golden ratio were under the surface of both of these formulas the entire time!
When we used the minimum number of nodes in a tree in order to determine its height, what we were really doing was taking the log, with a base of the golden ratio, of the minimum number of nodes, which gave us an approximation of the height of the tree. In fact, this is exactly where the logarithmic complexity of an AVL tree comes from! We just generally don’t refer to the base, so most of the time it’s never explicitly stated that we’re taking log base 1.618.
For example, when we determined that a tree with a minimum number of 54 nodes would give us a height of 8, what we were really doing was taking the log base 1.618 of 54, which approximately yields us 8.29, which rounds to an integer value of 8.
If we were to graph the relationship between the number of nodes and the height of the tree, we would effectively be graphing log base golden ratio of n, where n is the number of nodes in a tree.
Another way of thinking about this is that the relationship between the number of nodes and the height of a tree is logarithmic.
As the minimum number of nodes in an AVL tree grows linearly, the height of the tree grows logarithmically.
If you remember learning about logarithms earlier in this series, then you might recall that a logarithmic function is the inverse of an exponential function. So, what about the second formula we looked at, where we determined the minimum number of nodes based on the height of the tree?
The relationship is going to be the inverse of a logarithmic relationship: it’s going to be exponential. We can see this play out in the illustration below.
As the height of an AVL tree grows linearly, the minimum number of nodes it has will grow exponentially (approximately, of course).
We can see that each time that we took the height of a tree as the input and used it to derive the minimum number of nodes for an AVL tree of our inputted height, we were effectively raising the golden ratio to the power of our height, h.
For example, when we wanted to calculate the minimum number of nodes for an AVL tree with a height of 8, we raised φ to the power of 8, which gave us approximately 46.98. Remember, since we’re dealing with a rounded version of an otherwise irrational number, as our numbers get larger, our approximations have a much higher margin of error. This explains why our calculation for a tree with a height of 2, 3, 4, or 5 are much more accurate than our calculation for a tree with a height of 8. If we were to graph these results, we would actually be graphing the golden ratio raised to the power of h, the height of our tree.
If you’re feeling amazed, in awe, and completely in wonder of what you just learned, don’t worry — I felt the exact same way, too. I don’t think I’ll ever think of a sunflower or an AVL tree in the same way, ever again.
(And, if I’ve done my job well, neither will you.)
# Resources
I had trouble containing all my excitement about Fibonacci, golden ratios, and AVL trees, and did a whole lot of research on all three of these topics. If you’re as fascinated and amazed by them as I am, you’ll enjoy the links below.
1. Data Abstractions: AVL Trees, Professor Ruth Anderson
2. Fibonacci Sequence, Math Is Fun
3. Maximum Height of an AVL Tree, CSA Department, IISc, Bangalore
4. 15 Uncanny Examples of the Golden Ratio in Nature, George Dvorsky
5. Fibonacci Number, Wolfram Research
6. AVL Trees, Professor Eric Alexander
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# Matematica | MATHEMATICAL ANALYSIS I / MATHEMATICAL ANALYSIS II
## Matematica MATHEMATICAL ANALYSIS I / MATHEMATICAL ANALYSIS II
0512300001 DIPARTIMENTO DI MATEMATICA MATHEMATICS 2013/2014
OBBLIGATORIO YEAR OF COURSE 1 YEAR OF DIDACTIC SYSTEM 2010 ANNUALE
SSD CFU HOURS ACTIVITY TYPE OF ACTIVITY 1 ANALISI MATEMATICA I MAT/05 6 48 LESSONS BASIC COMPULSORY SUBJECTS MAT/05 3 36 EXERCISES BASIC COMPULSORY SUBJECTS 2 ANALISI MATEMATICA II MAT/05 5 40 LESSONS BASIC COMPULSORY SUBJECTS MAT/05 3 36 EXERCISES BASIC COMPULSORY SUBJECTS
MARIA TRANSIRICO21 T SARA MONSURRO'1
Objectives
THE COURSE OF "ANALISI MATEMATICA I/ANALISI MATEMATICA II" IS ESSENTIALLY DEVOTED TO THE STUDY OF REAL FUNCTIONS OF A REAL VARIABLE, TO THE THEORY OF LIMITS AND TO THE DIFFERENTIAL AND INTEGRAL CALCULUS FOR SUCH FUNCTIONS, AND TO THE STUDY OF NUMERICAL SERIES. THE AIM OF THE COURSE IS THE ACQUISITION OF RESULTS, PROOFS AND CALCULATION TECHNIQUES.
Prerequisites
BASIC KNOWLEDGE ACQUIRED THROUGH HIGH SCHOOL COURSES. IN PARTICULAR, KNOWLEDGE OF ELEMENTARY ALGEBRA, TRIGONOMETRY AND FIRST AND SECOND ORDER INEQUALITIES IS REQUIRED.
Contents
1. ALGEBRAIC STRUCTURES: DEFINITIONS AND EXAMPLES.
2. REAL NUMBERS.
3. REAL FUNCTIONS.
4. COMPLEX NUMBERS.
5. LIMITS OF SEQUENCES.
6. LIMITS OF FUNCTIONS AND CONTINUOUS FUNCTIONS.
7. MORE ON LIMITS.
8. DERIVATIVES.
9. APPLICATION OF DIFFERENTIAL CALCULUS. GRAPHS OF FUNCTIONS.
10. RIEMANN INTEGRATION.
11. INTEGRAL CALCULUS
12. TAYLOR'S FORMULA.
13. NUMERICAL SERIES.
Teaching Methods
•FRONTAL LESSON•PRACTICE LESSON
Verification of learning
WRITTEN TEST AND ORAL EXAMINATION.
Texts
P. MARCELLINI - C. SBORDONE, ANALISI MATEMATICA UNO, LIGUORI EDITORE.
P. MARCELLINI - C. SBORDONE, ESERCITAZIONI DI MATEMATICA I, LIGUORI EDITORE.
A. ALVINO - L. CARBONE - G. TROMBETTI, ESERCITAZIONI DI MATEMATICA I, LIGUORI EDITORE.
M. TROISI, ANALISI MATEMATICA I, LIGUORI EDITORE.
D. GRECO - G. STAMPACCHIA, ESERCITAZIONI DI MATEMATICA, VOL. I, LIGUORI EDITORE.
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how to choose the roughness height
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August 26, 2004, 17:57 how to choose the roughness height #1 icestar Guest Posts: n/a I'm doing research to find out the surface roughness effect to the tornado vortex by fluent. I tried to change the value of the roughness height and constant. But the result didn't changed a lot. I have no idea how to choose these two parameters. Did someone have experence to use fluent to simulate the roughness efforts? Thanks a lot.
September 20, 2004, 08:54 Re: how to choose the roughness height #2 ruth Guest Posts: n/a I am currently running a thunderstorm model using fluent and also faces the same problem. Please give some advice if anyone knows.thanks
September 20, 2004, 13:23 Re: how to choose the roughness height #3 Titiksh Patel Guest Posts: n/a Hi Take care that roughness is defined in metres. Titiksh
September 20, 2004, 13:28 Re: how to choose the roughness height #4 icestar Guest Posts: n/a I notice that, but it didn't change a lot.
September 20, 2004, 23:48 Re: how to choose the roughness height #5 wjruth Guest Posts: n/a Hi, I am running a 2D axisymmetry thunderstorm using fluent and roughness height between 0.01m to 1.6m, roughness constant = 0.5 . The height of the max horizontal velocity should be affected by the roughness of the ground but my results is not as expected.I would like to ask if there is other parameter that would affect the roughess. thank you
September 21, 2004, 08:03 Re: how to choose the roughness height #6 Titiksh Patel Guest Posts: n/a Hi, I think the first grid distance from the wall may also be the criteria. It should be above the roughness height. Titiksh
September 22, 2004, 22:17 Re: how to choose the roughness height #7 wjruth Guest Posts: n/a Hi Titiksh, thanks for your advice. my biggest roughness height is 1.6m and I uses boundary layer for the grid near the ground. The 1st cell height from the ground is 2m and is uniform up till a height of 200m. But still the result doesn't show much changes. Am I doing the right thing? thank you for your time. wjruth
September 23, 2004, 09:26 Re: how to choose the roughness height #8 Titiksh Patel Guest Posts: n/a Hi Can u tell me what is the roughness height you are providing The bottom wall is plane, isn't it, and then you are providing equivalent roughness height. What changes you are expecting when you change height. Also you are saying that u are providing B.L. and then you are sayin that its uniform upto 200 metres, it seems little bit contradicting. I will look at more details in mean time. Regards, Titiksh
September 23, 2004, 22:33 Re: how to choose the roughness height #9 wjruth Guest Posts: n/a Hi Titiksh, mine is a 2D axisymmetry domain of a thunderstorm model and I am using boundary layer to create denser mesh near the ground (wall boundary). This BL has a height of 50m (within is 25 intervals of 2m height). Under the boundary condition panel of wall type, I used roughness height of 1.6m and roughness constant of 0.5. My concern is to find the height above ground of the max radial velocity when the inlet velocity flows, and hits the ground. This height above ground should change with different roughness (by changing the roughness height) of the ground. I have also read that wall Y+ will also affect the roughness effect but I still not sure how to get wall Y+ between 30 to 60. thank you for your attention to this problem.
September 25, 2004, 03:33 Re: how to choose the roughness height #10 User Guest Posts: n/a Hi , There are three step - 1. While creating the BL, make sure that your Y+ value is suited for the wall treatment you selected. See what Y+ you are getting with first cell height = 2 m . If you get y+ < 30 then increase your first cell height. 2. Effect of Roughness depends on the Non-dimensionalise factor Ks+. You Ks+ should be above 90 for rough wall. (see in Fluent Users Guide to see how to calculate Ks+) 3. Roughnes height should not be more than half of the first cell height. Your roughnes hight should fall below the center of first cell to see the effect of the roughness. in your case it should be below 1 m if your Y+ is correct. so 1.6 is wrong value you are providing. bikooo3878 likes this.
August 31, 2012, 09:00 #11 New Member Join Date: Dec 2011 Posts: 8 Rep Power: 13 Sure this is a very old topic but as there are some good point in it BUMP Last edited by morecfd; August 31, 2012 at 13:07.
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# Make y a funciton of x
1. Apr 18, 2005
### shoopa
same with y=4x-x^2 goes to x=
thank you for help!!
2. Apr 18, 2005
### Moo Of Doom
Always by completing the square for quadratic functions.
3. Apr 18, 2005
### snoble
The method is pretty straight forward. Just divide both sides so that x^2 has a coefficient of 1. Then add on a constant C to both sides so that the right hand side is a square (remember square quadratics have the form X^2 + 2*b + b^2 = (x+b)^2). Then you just need to square root both sides and then it becomes clear how to get the x alone. You may want to add a $$\pm$$ when you take the square root if you are doing that in this class.
so for y = ax^2 +bx just divide both sides by a, add (b/(2a))^2. Then the right hand side can be written out as (x + b/(2a))^2. Then just square root and add stuff around. Remember nothing says a or b can't be negative. a just can't be 0.
4. Apr 18, 2005
### dextercioby
You can express "x" in terms of "y" only on a certain interval.That is to say that the function "y(x)" is invertible only on an interval,and not on R.
Daniel.
5. Apr 18, 2005
### Berislav
Yes, but I think that that won't be a good function, since it doesn't map -to one on any interval.
EDIT:
Ah, I see now that you mean interval of x. Sorry.
Last edited: Apr 18, 2005
6. Apr 18, 2005
### eNathan
I think that I can correctly derive (without testing) that
$$x = \sqrt { \frac {y} {2} } + \frac {8} {y}$$
Last edited: Apr 18, 2005
7. Apr 18, 2005
### snoble
I don't understand. 8*1-2*1^2=6 but 6*1-1^2=5
So the two functions differ at 1 right?
The first case you should get $$x=2\pm 1/2\,\sqrt {16-2\,y}$$
and the second cased
$$x=3/4\pm 1/4\,\sqrt {9-4\,y}$$
Two different answers for two different equations.
8. Apr 19, 2005
### baby_garfield
y=8x-2x^2--> 8x-2x^2-y=0--> 2x^2-8x+y=0
9. Apr 19, 2005
### baby_garfield
never mind! that was wrong
10. Apr 19, 2005
### baby_garfield
no! it's not! it's right! (sorry for not being sure )
11. Apr 19, 2005
### dextercioby
'Twas not.It was correct.How else could this problem b solved?
Daniel.
12. Apr 22, 2005
### abia ubong
u should be strongwilled baby garfield, the step is right, the use of the quadratic formula gives x in terms of y and that satisfies ur problem dexter
13. Apr 22, 2005
### The Bob
$$y = 8x - 2x^2$$
$$\Rightarrow 2x^2 - 8x + y = 0$$
Apply to quadratic formula: $$x = \frac{ -b \pm \sqrt{b^2 - 4ac}}{2a}$$
a = 2, b = -8 and c = y
$$x = \frac{ 8 \pm \sqrt{(-8^2) - (4 \times 2 \times y)}}{2 \times 2}$$
$$\Rightarrow x = \frac{ 8 \pm \sqrt{64 - 8y}}{4}$$
$$\Rightarrow x = 2 \pm \sqrt{4 - \frac{1}{2}y}$$
Last edited: Apr 22, 2005
14. Apr 22, 2005
### The Bob
$$y = 4x - x^2$$
$$\Rightarrow x^2 - 4x + y = 0$$
Apply to quadratic formula: $$x = \frac{ -b \pm \sqrt{b^2 - 4ac}}{2a}$$
a = 1, b = -4 and c = y
$$x = \frac{ 4 \pm \sqrt{(-4^2) - (4 \times 1 \times y)}}{2 \times 1}$$
$$\Rightarrow x = \frac{ 8 \pm \sqrt{16 - 4y}}{2}$$
$$\Rightarrow x = 4 \pm \sqrt{4 - y}$$
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# The No-Longer-Impossible Bike
Here's your nightly math! Just 5 quick minutes of number fun for kids and parents at home. Read a cool fun fact, followed by math riddles at different levels so everyone can jump in. Your kids will love you for it.
# The No-Longer-Impossible Bike
December 4, 2014
A bicycle gives you a smart, easy way to go places. It weighs a lot less than a car — can you imagine carrying an SUV on your shoulder? But a bike is still pretty heavy and crazy-shaped to pick up with your hands. So a team of guys built a bike that can fold up really, really small and fit in your backpack. They named it Impossible, since you’d never think a bike could shrink down to that shape. The wheels are tiny, the pieces of frame slide inside each other, and the handlebars pop off. It even has a motor, which can run for 45 minutes or more on 10 fresh batteries. The inventors posted the bike on Kickstarter, a site where people can give money to help inventors build their inventions. The bike needed only \$55,000, but already it has raised almost \$300,000 since so many people want one!
Wee ones: How many big circle shapes can you count on that bike? Remember to include the seat!
Little kids: If you walk carrying your little Impossible bike for 10 minutes, then ride it for 10 minutes, how many minutes have you traveled? Bonus: The Impossible guys started work on this bike 3 years ago. What year was that? (We’re in 2014 right now.)
Big kids: If you start folding up your Impossible bike around 1 pm – at 1:02:20 to be exact – and you finish folding it at 1:02:45, how many seconds did you take to fold it? Bonus: The bike will sell next year for \$430 for a white one and \$460 for black (is black paint really more expensive?) How much more will the black one cost?
The sky’s the limit: If your Impossible bike starts slowing down and after the 1st hour you replace the first 3 batteries in the stack, then after an hour you replace the next 3, and you keep circling around (so you change the 10th, 1st and 2nd after the 3rd hour)…after how many hours do you replace the battery in the 7th slot for the 3rd time?
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# Partial Fractions.
We show how Partial Fraction Decomposition is a consequence of the Chinese Remainder Theorem. Our results hold for arbitrary principal ideal domains, with stronger results for certain Euclidean rings. Therefore, we do not only have Partial Fraction Decomposition for rational function fields, but also for the rational numbers.
# The Power of the Distributive Law.
This short post discusses a cute statement which shows the power of a certain subset of axioms of a ring, including prominently the distributive law.
Categories: Algebra
Tags:
# A Strange Inequality.
We show how to prove a number theoretic inequality, originating from geometry, using an elementary approach.
# Inequalities.
This post shows a diagram, listing a lot of inequalities and showing implications between them.
# How to Compute the 5-adic Expansion of 1/2; or: Hensel's Lemma and (Non-Analytic) Newton Iteration.
In this post, we consider the quest of computing the 5-adic expansion of 1/2. We begin with introducing p-adic integers and numbers, and discussing when certain polynomials with coefficients in the integers have zeroes in the p-adic integers. This question is closely related to Hensel's lemma, which can be proven using an algebraic version of Newton's iteration. We use this to compute approximations of rational numbers in the p-adics, and consider which p-adic numers have an eventually periodic expansion.
# Finding Lattice Points, Finite Abelian Groups, and Explaining Algorithms.
We compare the tasks of finding points of a lattice, computing the structure of finite abelian groups and explaining algorithms. We show up relations between these three topics and, as an example, depict the baby-step giant-step algorithm for order computation, as well as Terr's modification of this algorithm.
# Homomorphisms, Tensor Products and Certain Canonical Maps.
A standard topic in linear algebra is the dual space of a vector space, as well as the canonical embedding of a vector space in its double dual. Moreover, transposition of homomorphisms in terms of dual spaces is rather well known. Something less known is that one has a canonical map from the dual of V tensored with W to the space of homomorphisms from V to W. In this abstract nonsense post, we describe these canonical maps, their interplay, and try to determine their images.
# Diagonalizable Matrices.
We consider the property of an n times n matrix of being diagonalizable. Is this property open in the standard topology, or the Zariski topology? The emphasis lies on the real and complex numbers, as well as on arbitrary algebraically closed fields.
# Euclidean Domains, and the Extended Euclidean Algorithm.
We will discuss Euclidean domains together with a constructive proof of the fact that every two elements have a greatest common divisor, which is essentially the Euclidean algorithm.
# The Hasse derivative, part II: Multivariate partial Hasse derivatives.
Following a suggestion by A. Maevskiy, we show how the Hasse derivative can be extended to partial Hasse derivative in arbitrary multivariate polynomial rings. We show multivariate versions of Taylor's Formula, of the Identity Theorem, and of the Generalized Leibnitz Rule.
# Fun With Representable Functors, or Why I Like Yondea's Lemma.
We discuss the notion of representable functors in Category Theory. Then, we present Yoneda's lemma and apply it to the situation of group objects in categories and their relation to functors into the category of groups, resulting in a surprising result that these two concepts are essentially the same. Most proofs are included, as well as lots of commutative diagrams.
# About Base Changes and Tensor Products.
In Linear Algebra, one often has the problem that one wants to talk about complex eigenvalues of objects defined over the reals. If the object is a matrix, it is clear what that means. But what if the object is an endomorphism of a non-canonical real vectorspace? This question is strongly related an important use of tensor products, namely base changes.
# The Hasse derivative.
In real and complex analysis, the Taylor series expansion is a very important tool. For polynomials over arbitrary unitary rings, it is possible to define a derivative which behaves similar to the usual derivative; unfortunately, the Identity Theorem and Taylor's formula do not transfer to this new situation. Fortunately, there exists a different definition of derivatives for these cases, namely the Hasse derivative. Not only does it gives a Identity Theorem and Taylor's formula back, but also allows to write other identities in a simpler way.
# Obtaining Infrastructures from Global Fields.
We show how to obtain n-dimensional infrastructures from global fields of unit rank n. We will also discuss how to obtain baby steps in these cases, and show graphical representations of certain two-dimensional infrastructures obtained from function fields.
# How to Obtain Reduction Maps for n-dimensional Infrastructures.
We explain a general technique to obtain a reduction map, given X and d and, varying with the method of construction, additional information for every x in X. Moreover, we explain a technique on how to obtain n-dimensional infrastructures from certain lattices in (n+1)-dimensional space.
# n-dimensional Infrastructures.
We will introduce n-dimensional infrastructures and briefly discuss reductions, f-representations and giant steps. We will also discuss how infrastructures can be obtained from finite abelian groups.
# Interpreting One-dimensional Infrastructures as Groups: f-Representations.
We introduce the notion of f-representations and relate them to reduction maps. Moreover, we equip a set of f-representations with a group operation which can be computed purely with baby steps, giant steps and relative distances.
# One-dimensional Infrastructures.
We give the definition of one-dimensional infrastructures and construct baby and giant steps. Moreover, we show that one-dimensional infrastructures generalize finite cyclic groups. Finally, we give some remarks on our choice of the giant step definition.
# The Discrete Logarithm Problem and Generalizations.
We discuss the discrete logarithm problem, its use in cryptography, and two possible directions of generalization to other algebraic structures.
# A Topological Proof of the Cayley-Hamilton Theorem over all Commutative Unitary Rings.
We want to give a proof of the Cayley-Hamilton Theorem for all commutative rings with unity, which first reduces to the case of the field of complex numbers and then applies a topological argument.
# Fundamental Theorem of Algebra.
We want to give a proof of the Fundamental Theorem of Algebra using methods from Complex Analysis, in particular Liouville's Theorem.
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HOME»Writer's Attic Articles»Style Matters
### Style Matters
Imagine you are helping to design a new space shuttle for NASA, and in the course of the work you need to evaluate
y = 5 + 4 × 2
The question is this: Which operation will you perform first? Will you add first? In which case you would find
y = 9 × 2 = 18
Or will you multiply first? Then you would obtain
y = 5 + 8 = 13
The difference between 13 and 18 could be crucial to the success of a shuttle mission. For example, if the calculation applies to a life-support system, the error could jeopardize the crew.
Fortunately, arithmetic has rules for resolving our dilemma. The applicable rule here is that higher-order operations take precedence over lower-order ones: we must multiply first, then add, and the correct answer is 13. Mathematics provides rules to prevent ambiguity when we need to interpret a mathematical sentence (an equation).
In an exactly analogous fashion, language has rules of good usage that prevent ambiguity when we need to interpret an English sentence. But the writer must invoke those rules if the reader is to be informed rather than confused.
Sense and nonsense are distinguished, in part, by the relative ordering of elements: verbs, subjects, main clauses, subordinate clauses, prepositional phrases, sentences themselves. Consider this example (No, I didn't make this up.):
Figure 1 shows measurements of the water level as the tank is being filled with error bars.
The rules of good usage make a substantial contribution to a clear technical style, and as Edward Tufte has emphasized [1]: style matters. A vague, convoluted, opaque technical style signifies a writer who places little importance in either the writing or its readers. Consequently, the writer loses credibility and, sooner or later, readers will be forced to guess what the writer is trying to say. And in technical writing, it is as wrong to force the reader to guess meanings of sentences as it is wrong to force the reader to guess numerical results from equations. The consequences of misinterpreting a sentence in a technical document could be just as disastrous as misinterpreting an equation—just as disastrous and just as unnecessary. Garbled writing is a product of garbled thinking, and our society cannot afford garbled thinking about technology—either from those who develop the technology or from those who use it.
[ 1 ] Edward R. Tufte, The Cognitive Style of PowerPoint: Pitching Out Corrupts Within, 2nd ed., Graphics Press LLC, Cheshire, CN, 2006.
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https://documen.tv/what-is-the-best-first-step-in-solving-negative-4-two-fifths-greater-than-startfraction-5-over-1-27144786-29/
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Question
What is the best first step in solving Negative 4 x + two-fifths greater-than StartFraction 5 over 10 EndFraction?
A. Add Two-fifths to both sides.
B. Subtract Two-fifths from both sides.
C. Multiply both sides by Negative 4 and reverse the inequality symbol.
D. Divide both sides by 10 and reverse the inequality symbol.
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https://ivandemarino.me/2010/01/07/largest-square-array-of-same-integers/
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# Largest square array of same integers
• code • matrix • personal • english • it • calculation • complexity • square • largest • integer • array
• 1171 words
Tonight it’s a challenging one. Or, better, a problem of which is really difficult to find a good solution in < O(n3). Indeed, it’s a question that an ex-colleague was asked during an interview with Big-G.
The guy, a part from being a REALLY smart guy, is also very humble, and he doesn’t want to be mentioned by name. So, sorry for girls looking for a young, smart, promising young man: you need to find who he is yourself. ;)
## Problem Definition
Given a 2D array (not necessarily square) of integers 1-9 : ```c { { 8 8 8 8 8 }, { 8 8 8 8 8 }, { 8 8 3 8 8 }, { 8 8 8 5 5 }, { 8 8 8 5 5 } } ``` Return:
1. The dimension of the largest square consisting of the same integer (in this case 2 because the largest is 2x2 square of 8s).
2. The integer which appears in that square (in this case 8)
3. The indices of the top left corner of the square (this this case (0,0) )
4. If there are multiple squares of the same dimension, you have to return the first one (i.e. the 8 one, not 5).
## My solution
I designed this algorithm on my way back from my previous trip to Paris. Was at least 2 months I had it in my head, but there was the only empty slot of time where I could sit, think and code… on a spare piece of paper. And, still, it performs a disappointing Time Complexity of O(n3).
It works like this:
``````init "largest matrix" to NULL/NONE/-1
for i = 0 to num_of_rows:
for j = 0 to num_of_cols:
calculates "how big is the matrix that has TOP-LEFT corner in array[i][j]"
if "what found is bigger then what found so far":
save what found as "the new largest matrix"
``````
### Calculate how big is the matrix that has TOP-LEFT corner in array[i][j]
It’s a scanning that goes “around” the currently calculated matrix from TOP-RIGHT to BOTTOM-LEFT, moving CLOCKWISE. It starts from a base matrix, the single element in TOP-LEFT position, and grows until finds a different value. Sorry, no rocket science here. Probably only the way I manipulate the indexes could be seen as smart.
In the case of the input given above, the steps that it will do for cell `[0, 0]` will be like:
STEP 1
``````{
{ 8 - - - - },
{ - - - - - },
{ - - - - - },
{ - - - - - },
{ - - - - - },
}
``````
STEP 2
``````{
{ 8 8 - - - },
{ 8 8 - - - },
{ - - - - - },
{ - - - - - },
{ - - - - - },
}
``````
STEP 3
``````{
{ 8 8 8 - - },
{ 8 8 8 - - },
{ - - 3 - - },
{ - - - - - },
{ - - - - - },
}
``````
Once the `'3'` is found, the search will stop, returning a `result = 2` (the measure of the diagonal of the found matrix).
## A little trick to save us some time
One thing I added to make things slightly better, is to check if “it’s worth the effort”! Given the current location in the matrix `[x, y]`, if the maximum diagonal that we COULD calculate if the rest of the elements were equal to what’s in `[x, y]` is:
• either smaller than the current maximum
• or just equal to `'1'`
then we jump any calculation and move to the next element in the matrix.
## The code
``````#include <stdlib.h>
#include <stdio.h>
unsigned int measure_diagonal(
// Need to treet the matrix as an mono-dimensinal array
const unsigned int* matrix,
const unsigned int rows,
const unsigned int cols,
const unsigned int x,
const unsigned int y) {
static unsigned int diagonal;
static int curr_x; // I can't use unsigned because I need to detect "out of bound"
static int curr_y;
#ifdef DEBUG
printf("Measure the Diagonal from cell: (%d, %d)\n", x, y);
printf("Matrix size: (%d, %d)\n", cols, rows);
#endif
// Base value: diagonal = 1 - matrix of 1 element
diagonal = 1;
// Now we are going to scan "around" the right and bottom borders
// of the inner matrix ALREADY discovered.
// The starting point is the element in position [x,y]
while ( 1 ) {
#ifdef DEBUG
printf("Current Diagonal for (%d,%d): %d\n", x, y, diagonal);
#endif
// If the area we are going to scan is outside the main matrix boundaries...
if ( (x + diagonal) >= cols || (y + diagonal) >= rows ) {
//... return the current diagonal as it is
return diagonal;
}
curr_x = x + diagonal;
curr_y = y;
#ifdef DEBUG
printf("curr_x(%d), curr_y(%d)\n", curr_x, curr_y);
#endif
do {
#ifdef DEBUG
printf("Comparing (%d, %d) with curr(%d, %d)\n", x,y, curr_x, curr_y);
#endif
// Check current element is equal to the Top-Left corner one
if ( matrix[y * cols + x] != matrix[curr_y * cols + curr_x] ) {
#ifdef DEBUG
printf("Diagonal for cell(%d, %d) is: %d\n", x, y, diagonal);
#endif
// If not, the diagonal calculated so far is the diagonal
return diagonal;
}
// Determine what's next
if ( curr_y < (y + diagonal) ) {
// Move one row down in the matrix
++curr_y;
} else {
// Move one column left in the matrix
--curr_x;
}
} while( (int)x <= curr_x ); // Until within the boundaries
// Good! We discovered an even larger matrix
++diagonal;
}
}
void calculate(const unsigned int* altitude_map,
const unsigned int rows,
const unsigned int cols) {
int best_number = -1;
int best_dimension = -1;
int best_row = -1;
int best_col = -1;
int r, c;
int temp_dimension = -1;
for (r = 0; r < rows-1; r++) { // Avoid last row: would be 1 anyway
for (c = 0; c < cols-1; c++) { // Avoid last column: would be 1 anyway
if ( temp_dimension <= (int)(cols-c) && temp_dimension <= (int)(rows-r) ) {
// Measure the diagonal of the matrix starting at position [r, c]
temp_dimension = measure_diagonal(altitude_map, rows, cols, c, r);
// Is it better then what we found so far?
if (temp_dimension > best_dimension) {
best_dimension = temp_dimension;
best_col = c;
best_row = r;
best_number = altitude_map[r * cols + c];
}
}
#ifdef DEBUG
else {
printf("Skipped useless calculation at (%d, %d)\n", c, r);
}
#endif
}
}
printf("ACTUAL : contains a %dx%d grid of %ds at (%d , %d)\n",
best_dimension, best_dimension, best_number, best_row, best_col);
}
``````
## Conclusions
I’m 100% sure that there are better solution, and indeed the guy above just sent me an alternative that he claims ( :-P ) has quadratic time complexity. I didn’t check it out yet, but I’ll post again if it’s true (and, knowing the guy, it’s highly probable).
In the meantime, if you have any comment or improvement, please speak up!
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# NCO- Cyber Olympiad (SOF) Class 7: Questions 150 - 156 of 416
Get 1 year subscription: Access detailed explanations (illustrated with images and videos) to 416 questions. Access all new questions we will add tracking exam-pattern and syllabus changes. View Sample Explanation or View Features.
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## Question number: 150
MCQ▾
### Question
Find the value of in each of the following equations
### Choices
Choice (4) Response
a.
5.40
b.
3.60
c.
0.61
d.
1.62
## Question number: 151
MCQ▾
### Question
Which of the following can be the exact multiple of 6?
### Choices
Choice (4) Response
a.
63
b.
7
c.
252
d.
28
## Question number: 152
MCQ▾
### Question
The product of two numbers is . If one of the number is , find the other.
### Choices
Choice (4) Response
a.
b.
c.
d.
## Question number: 153
MCQ▾
### Question
If means A is the father of B; means A is the brother B; means A is the wife of B and means A is the mother of B, which of the following shows that M is the maternal grandmother of T?
### Choices
Choice (4) Response
a.
b.
c.
d.
## Question number: 154
MCQ▾
### Question
In a certain code, AVERAGE is coded as AGEAVERA. How would COMMAND be code in that code?
### Choices
Choice (4) Response
a.
ANDMMCO
b.
ANDCOMM
c.
COANDMM
d.
ACOMMND
## Question number: 155
Match List-Ⅰ List-Ⅱ▾
### Question
Match the following
List-Ⅰ List-Ⅱ (A) ( (i) (B) (ii) () (C) ( (iii)
### Choices
Choice (4) Response
• (A)
• (B)
• (C)
a.
• (iii)
• (ii)
• (i)
b.
• (i)
• (iii)
• (ii)
c.
• (i)
• (ii)
• (iii)
d.
• (iii)
• (i)
• (ii)
## Question number: 156
MCQ▾
### Question
If + stands for division; stands for addition; stands for multiplication; stands for subtraction, which of the following equation is correct?
### Choices
Choice (4) Response
a.
b.
c.
d.
f Page
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# 17.5: Normal Modes of a Standing Sound Wave
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##### Learning Objectives
• Explain the mechanism behind sound-reducing headphones
• Describe resonance in a tube closed at one end and open at the other end
• Describe resonance in a tube open at both ends
Interference is the hallmark of waves, all of which exhibit constructive and destructive interference exactly analogous to that seen for water waves. In fact, one way to prove something “is a wave” is to observe interference effects. Since sound is a wave, we expect it to exhibit interference.
## Interference of Sound Waves
In waves, we discussed the interference of wave functions that differ only in a phase shift. We found that the wave function resulting from the superposition of $$y_{1}(x, t)=A \sin (k x-\omega t+\phi)$$ and $$y_{2}(x, t)=A \sin (k x-\omega t)$$ is
$y(x, t)=\left[2 A \cos \left(\frac{\phi}{2}\right)\right] \sin \left(k x-\omega t+\frac{\phi}{2}\right). \nonumber$
One way for two identical waves that are initially in phase to become out of phase with one another is to have the waves travel different distances; that is, they have different path lengths. Sound waves provide an excellent example of a phase shift due to a path difference. As we have discussed, sound waves can basically be modeled as longitudinal waves, where the molecules of the medium oscillate around an equilibrium position, or as pressure waves.
When the waves leave the speakers, they move out as spherical waves (Figure $$\PageIndex{1}$$). The waves interfere; constructive inference is produced by the combination of two crests or two troughs, as shown. Destructive interference is produced by the combination of a trough and a crest.
The phase difference at each point is due to the different path lengths traveled by each wave. When the difference in the path lengths is an integer multiple of a wavelength,
$\Delta r=\left|r_{2}-r_{1}\right|=n \lambda, \text { where } n=0,1,2,3, \ldots \nonumber$
the waves are in phase and there is constructive interference. When the difference in path lengths is an odd multiple of a half wavelength,
$\Delta r=\left|r_{2}-r_{1}\right|=n \frac{\lambda}{2}, \text { where } n=1,3,5, \ldots \nonumber$
the waves are 180°($$\pi$$ rad) out of phase and the result is destructive interference. These points can be located with a sound-level intensity meter.
##### Example $$\PageIndex{1}$$: Interference of sound waves
Two speakers are separated by 5.00 m and are being driven by a signal generator at an unknown frequency. A student with a sound-level meter walks out 6.00 m and down 2.00 m, and finds the first minimum intensity, as shown below. What is the frequency supplied by the signal generator? Assume the wave speed of sound is v = 343.00m/s.
Strategy
The wave velocity is equal to $$v = \frac{\lambda}{T} = \lambda f$$. The frequency is then $$f = \frac{v}{\lambda}$$. A minimum intensity indicates destructive interference and the first such point occurs where there is path difference of $$\Delta r = \lambda / 2$$, which can be found from the geometry.
###### Solution
1. Find the path length to the minimum point from each speaker
$r_{1}=\sqrt{(6.00 \: \mathrm{m})^{2}+(2.00 \: \mathrm{m})^{2}}=6.32 \: \mathrm{m}, r_{2}=\sqrt{(6.00 \: \mathrm{m})^{2}+(3.00 \: \mathrm{m})^{2}}=6.71 \: \mathrm{m} \nonumber$
2. Use the difference in the path length to find the wavelength.
\begin{array}{c}
\Delta r=\left|r_{2}-r_{1}\right|=|6.71 \: \mathrm{m}-6.32 \: \mathrm{m}|=0.39 \: \mathrm{m} \nonumber \\
\lambda=2 \Delta r=2(0.39 \: \mathrm{m})=0.78 \: \mathrm{m} \nonumber
\end{array}
3. Find the frequency.
$f=\frac{v}{\lambda}=\frac{343.00 \: \mathrm{m} / \mathrm{s}}{0.78 \: \mathrm{m}}=439.74 \: \mathrm{Hz} \nonumber$
Significance
If point P were a point of maximum intensity, then the path length would be an integer multiple of the wavelength.
##### Exercise 17.4
If you walk around two speakers playing music, how come you do not notice places where the music is very loud or very soft, that is, where there is constructive and destructive interference?
The concept of a phase shift due to a difference in path length is very important. You will use this concept again in Interference and Photons and Matter Waves, where we discuss how Thomas Young used this method in his famous double-slit experiment to provide evidence that light has wavelike properties.
## Noise Reduction Through Destructive Interference
Figure $$\PageIndex{3}$$ shows a clever use of sound interference to cancel noise. Larger-scale applications of active noise reduction by destructive interference have been proposed for entire passenger compartments in commercial aircraft. To obtain destructive interference, a fast electronic analysis is performed, and a second sound is introduced 180° out of phase with the original sound, with its maxima and minima exactly reversed from the incoming noise. Sound waves in fluids are pressure waves and are consistent with Pascal’s principle; that is, pressures from two different sources add and subtract like simple numbers. Therefore, positive and negative gauge pressures add to a much smaller pressure, producing a lower-intensity sound. Although completely destructive interference is possible only under the simplest conditions, it is possible to reduce noise levels by 30 dB or more using this technique.
##### Exercise 17.5
Describe how noise-canceling headphones differ from standard headphones used to block outside sounds.
Where else can we observe sound interference? All sound resonances, such as in musical instruments, are due to constructive and destructive interference. Only the resonant frequencies interfere constructively to form standing waves, whereas others interfere destructively and are absent.
## Resonance in a Tube Closed at One End
As we discussed in Waves, standing waves are formed by two waves moving in opposite directions. When two identical sinusoidal waves move in opposite directions, the waves may be modeled as
$y_{1}(x, t)=A \sin (k x-\omega t) \text { and } y_{2}(x, t)=A \sin (k x+\omega t) \nonumber .$
When these two waves interference, the resultant wave is a standing wave:
$y_{\mathrm{R}}(x, t)=[2 A \sin (k x)] \cos (\omega t) . \nonumber$
Resonance can be produced due to the boundary conditions imposed on a wave. In Waves, we showed that resonance could be produced in a string under tension that had symmetrical boundary conditions, specifically, a node at each end. We defined a node as a fixed point where the string did not move. We found that the symmetrical boundary conditions resulted in some frequencies resonating and producing standing waves, while other frequencies interfere destructively. Sound waves can resonate in a hollow tube, and the frequencies of the sound waves that resonate depend on the boundary conditions.
Suppose we have a tube that is closed at one end and open at the other. If we hold a vibrating tuning fork near the open end of the tube, an incident sound wave travels through the tube and reflects off the closed end. The reflected sound has the same frequency and wavelength as the incident sound wave, but is traveling in the opposite direction. At the closed end of the tube, the molecules of air have very little freedom to oscillate, and a node arises. At the open end, the molecules are free to move, and at the right frequency, an antinode occurs. Unlike the symmetrical boundary conditions for the standing waves on the string, the boundary conditions for a tube open at one end and closed at the other end are anti-symmetrical: a node at the closed end and an antinode at the open end.
If the tuning fork has just the right frequency, the air column in the tube resonates loudly, but at most frequencies it vibrates very little. This observation just means that the air column has only certain natural frequencies. Consider the lowest frequency that will cause the tube to resonate, producing a loud sound. There will be a node at the closed end and an antinode at the open end, as shown in Figure $$\PageIndex{4}$$.
The standing wave formed in the tube has an antinode at the open end and a node at the closed end. The distance from a node to an antinode is one-fourth of a wavelength, and this equals the length of the tube; thus, $$\lambda_1 = 4L$$. This same resonance can be produced by a vibration introduced at or near the closed end of the tube (Figure $$\PageIndex{5}$$). It is best to consider this a natural vibration of the air column, independently of how it is induced.
Given that maximum air displacements are possible at the open end and none at the closed end, other shorter wavelengths can resonate in the tube, such as the one shown in Figure $$\PageIndex{6}$$. Here the standing wave has three-fourths of its wavelength in the tube, or $$\frac{3}{4} \lambda_3 = L$$, so that $$\lambda_3 = \frac{4}{3} L$$. Continuing this process reveals a whole series of shorter-wavelength and higher-frequency sounds that resonate in the tube. We use specific terms for the resonances in any system. The lowest resonant frequency is called the fundamental, while all higher resonant frequencies are called overtones. The resonant frequencies that are integral multiples of the fundamental are collectively called harmonics. The fundamental is the first harmonic, the second harmonic is twice the frequency of the first harmonic, and so on. Some of these harmonics may not exist for a given scenario. Figure $$\PageIndex{7}$$ shows the fundamental and the first three overtones (or the first, third, fifth, and seventh harmonics) in a tube closed at one end.
The relationship for the resonant wavelengths of a tube closed at one end is
$\lambda_{n}=\frac{4}{n} L \quad n=1,3,5, \ldots \label{17.13}$
Now let us look for a pattern in the resonant frequencies for a simple tube that is closed at one end. The fundamental has $$\lambda = 4L$$, and frequency is related to wavelength and the speed of sound as given by
$v = f \lambda \nonumber .$
Solving for $$f$$ in this equation gives
$f=\frac{v}{\lambda}=\frac{v}{4 L}, \nonumber$
where v is the speed of sound in air. Similarly, the first overtone has $$\lambda = 4L/3$$ (see Figure $$\PageIndex{7}$$), so that
$f_{3}=3 \frac{v}{4 L}=3 f_{1} . \nonumber$
Because $$f_3 = 3 f_1$$, we call the first overtone the third harmonic. Continuing this process, we see a pattern that can be generalized in a single expression. The resonant frequencies of a tube closed at one end are
$f_{n}=n \frac{v}{4 L}, \quad n=1,3,5, \dots \label{17.14}$
where $$f_1$$ is the fundamental, $$f_3$$ is the first overtone, and so on. It is interesting that the resonant frequencies depend on the speed of sound and, hence, on temperature. This dependence poses a noticeable problem for organs in old unheated cathedrals, and it is also the reason why musicians commonly bring their wind instruments to room temperature before playing them.
## Resonance in a Tube Open at Both Ends
Another source of standing waves is a tube that is open at both ends. In this case, the boundary conditions are symmetrical: an antinode at each end. The resonances of tubes open at both ends can be analyzed in a very similar fashion to those for tubes closed at one end. The air columns in tubes open at both ends have maximum air displacements at both ends (Figure $$\PageIndex{8}$$). Standing waves form as shown.
The relationship for the resonant wavelengths of a tube open at both ends is
$\lambda_{n}=\frac{2}{n} L, \quad n=1,2,3, \ldots \label{17.15}$
Based on the fact that a tube open at both ends has maximum air displacements at both ends, and using Figure $$\PageIndex{8}$$ as a guide, we can see that the resonant frequencies of a tube open at both ends are
$f_{n}=n \frac{v}{2 L}, \quad n=1,2,3 \dots, \label{17.16}$
where $$f_1$$ is the fundamental, $$f_2$$ is the first overtone, $$f_3$$ is the second overtone, and so on. Note that a tube open at both ends has a fundamental frequency twice what it would have if closed at one end. It also has a different spectrum of overtones than a tube closed at one end.
Note that a tube open at both ends has symmetrical boundary conditions, similar to the string fixed at both ends discussed in Waves. The relationships for the wavelengths and frequencies of a stringed instrument are the same as given in Equation \ref{17.15} and Equation \ref{17.16}. The speed of the wave on the string (from Waves) is $$v = \sqrt{\frac{F_{T}}{\mu}}$$. The air around the string vibrates at the same frequency as the string, producing sound of the same frequency. The sound wave moves at the speed of sound and the wavelength can be found using $$v = \lambda f$$.
##### Exercise 17.6
How is it possible to use a standing wave’s node and antinode to determine the length of a closed-end tube
##### Note
This video lets you visualize sound waves.
##### Exercise 17.7
You observe two musical instruments that you cannot identify. One plays high-pitched sounds and the other plays low-pitched sounds. How could you determine which is which without hearing either of them play?
This page titled 17.5: Normal Modes of a Standing Sound Wave is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by OpenStax via source content that was edited to the style and standards of the LibreTexts platform.
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# Evaluate the quality and fit of data for use in business analysis.
## Evaluate the quality and fit of data for use in business analysis.
Evaluate the quality and fit of data for use in business analysis. 150 150 Nyagu
Statistical Inference
Statistical Inference
Overview
Use a provided data file to resolve statistical questions presented in a scenario. Summarize your findings and recommendations in a 1–2 page report for management.
Note: The assessments in this course build upon each other, so you are strongly encouraged to complete them in sequence.
When product improvements are made, there is an assumption that the new product will perform better than the previous product. It is important to understand how to select the appropriate statistical tests to determine with a degree of certainty if an improvement has increased the performance of a product.
By successfully completing this assessment, you will demonstrate your proficiency in the following course competencies and assessment criteria:
Competency 1: Evaluate the quality and fit of data for use in business analysis.
Determine the null hypothesis and alternative hypothesis for making a product performance comparison.
Competency 3: Analyze business decision opportunities using basic inferential statistics.
Compute the appropriate statistical test to determine acceptance or rejection of a null hypothesis.
Compute the p-value to indicate acceptance or rejection of a null hypothesis regarding a product performance comparison.
Competency 5: Apply data analysis to general business management planning and decision making.
Compile findings into a management report with details for recommended actions.
Competency 6: Communicate in a manner that is professional and consistent with expectations for members of the business professions.
Communicate in a manner that is professional and consistent with expectations for members of the business professions.
Competency Map
Resources
Required Resources
The following resource contains the data needed to complete the assessment.
Suggested Resources
The following texts provide instruction in Statistics.
Bowerman, B., O’Connell, R., & Murphree, E. (2014). Business statistics in practice (7th ed.). New York, NY: McGraw Hill.
Chapter 10, “Comparing Two Means and Two Proportions,” in Business Statistics in Practice, pages 380–404.
Chapter 11, “Statistical Inferences for Population Variances,” in Business Statistics in Practice, pages 412–424.
Chapter 12, “Experimental Design and Analysis of Variance,” in Business Statistics in Practice , pages 426–453.
The following text provides instruction for statistical analysis in Microsoft Excel.
Salkind, N. J. (Ed.). (2013). Excel statistics: A quick guide (2nd ed.). Thousand Oaks, CA: Sage
The following text provides instruction for SAS one of the most commonly used statistical analysis tools in business.
Slaughter, S. J., & Delwiche, L. D. (2010). The little SAS book for Enterprise Guide 4.2. Cary, NC: SAS Institute. Available from the
The following resource is a tutorial that walks through a number of statistical scenarios.
Statistics Tutorials
This tutorial explains how to select the best team consisting of ten players (from a randomly generated list of 25) and be able to calculate specific statistical values according to the following criteria:
Barsness, B. (n.d.). Creating the best fantasy baseball team | Transcript. Retrieved from http://www.sophia.org/creating-the-best-fantasy-baseball-team-tutorial
This tutorial introduces Z-scores and their relevance to variance and distribution in data sets.
This tutorial introduces standard error (s) for both sampling distribution models for proportions and means when we do not know the populations parameters p and σ. Demonstrates how to calculate standard error for a proportion.
Greene, A. (Producer). (n.d.). Standard error [Video] | Transcript. Retrieved from http://www.sophia.org/standard-error-tutorial
This tutorial explains the usefulness of the different sampling methods.
Varela, A. (Producer). (n.d.). Contrasting sampling methods [Video] | Transcript. Retrieved from http://www.sophia.org/contrasting-sampling-methods/contrasting-sampling-methods–5-tutorial
This website offers resources that cover many topics in statistics, including presentations that illustrate how to use software to implement statistical methods.
com (Producer). (n.d.). Independent samples t-test [Video] | Transcript. Retrieved from http://www.statisticslectures.com/topics/independentsamplest/
com (Producer). (n.d.). Z-test for proportions, two samples [Video] | Transcript. Retrieved from http://www.statisticslectures.com/topics/ztestproportions/
com (Producer). (n.d.). Independent and dependent samples [Video] | Transcript. Retrieved from http://www.statisticslectures.com/topics/independentdependentsamples/
Assessment Instructions
Use the Golf Ball Distance Test file, linked in the Resources under the Required Resources heading, to complete the calculations for the scenario for this assessment.
Practical Application Scenario
Your love of golf has brought you back to the range as the new product manager for UniDun’s “Straight Flight” (SF) line of golf balls. The company’s research and development group has been experimenting with dimple patterns that promote straight flight and have achieved some degree of success. You, however, are worried about the effect that the new pattern might have on driving distance.
The Golf Ball Distance Test file contains test results that compare the driving distances for the two different kinds of balls: 40 balls of the new SF type, and 40 of the old UniDun type. Your job is to determine if the old UniDun balls can be driven further than the new SF balls. To resolve this question, you need to appropriately address the following:
Identify the null and alternative hypotheses you should form for this test. State each both as a written explanation and as a math equation.
Identify the appropriate statistical test to accept or reject the null hypothesis.
Calculate the p-value.
Summarize your findings and recommendations about the new golf ball product to the vice president of marketing. Your report should not be a statistical analysis of the testing but a summary of the testing results. Assume that your audience is unfamiliar with statistical analysis.
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How many feet is Menetomy from Boston? - ThatsAnswer.COM
## Similar Questions
• Answer: Multiply 12 by 20 so 240 square feet. Length times width.
• Answer: "3m feet" and "4m feet" are unknown terms, therefore there is not enough data to answer the question. "3m" would normally mean 3 meters.but if in terms it could be 12m^2 feet
• Answer: Depending on the route you take, Boston University is located about 5 miles from Boston College. If you were to take the Commonwealth Ave, the drive would be 5.5 miles, or 14 minutes depending on traffic. If you were to take the Kenrick St route, it would still take you around 14 minutes, but the mileage would be only 5 miles.
• Answer: 6 feet x 52 feet = 312 square feet
• Answer: 28 ft X 22ft equals total sq. ft of total areas
20 x 14 equals pool area.
Subtract the second from the first number gives you the area of the deck.
• Answer: A yard that is 120 feet by 60 feet has a perimeter of 360 feet.
• Answer: You will still most likely grow a few more inches!
• Answer: First the answer should be in meters so it is helpful to convert all the feet measurements into meters.
24 ft = 7.315 m
5 ft = 1.524 m
The way I read the question is that 24 ft (7.315 m) is the circumference of the pool. From this I make the assumption that the pool is a round pool. Thus we can model it as a cylinder. The formula for the volume of a cylinder is:
V = PI * r * r * h
Which means you take the math constant PI (approx. 3.14) multiply it by the radius of the circular face and again multiply it by the radius then multiply it by the height (or depth in the question).
The formula to find the circumference is:
C = 2 * PI * r
To get the radius from the circumference just divide the circumference by (2 * PI).
r = C / (2 * PI)
Plugging in the numbers we get
V = 314 * (7.315 / (2 * 3.14)) * (7.315 / (2 * 3.14)) * 1.524
Calculating this out you get:
V = 6.49 m^3 (meters cubed)
• Answer: As there are 3 feet in a yard, the room measures 4 yards by 6 yards, so, 6 x 4 = the size of the carpet.
www.dmv.ca.gov/pubs/hdbk/speed_limits.htm
• Answer: 1795 gallons, if you fill it to the brim.
# How many feet is Menetomy from Boston?
• 7 miles = 36,960 feet
Name: *
• Answer: 7 miles = 36,960 feet
• Answer: The square footage is indeterminate without knowing the angles at the corners of the rectangle. Using one square corner ( 90 Degrees ) the area is about 6,430 square feet.
• Answer: To find out how much water is in a pool you multiply Length x Width x Average Depth x 7.5 (gallons per cubic foot). So 75x40x7x7.5=157,500gal.
• Answer: The swimming pool is a cylinder, so you use the volume equation for a cylinder: V=pi times r2 times height. If the diameter is 12 feet, the radius is 6.
V=pi times 62 times 60
V = 2160pi, or 6785.84 ft3
• Answer: 339.12 cubic feet or 2536.7938 gallons
• Answer: The quantity required is 50 cubic yards of material. Your supplier will know the number of "truck loads" required based on the carry capacity of his fleet..For example ,if his trucks average 10 cubic yards per load, then you will require a minimum of 5 loads.Also you will need to allow for settlement/compaction of the material which may require 10 to 20% extra depending on the grade of the material. !! JOR
• Answer: 140 cubic ft = 15.5 cubic yards.
• Answer: 12 x 40 = 480 square feet
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# DrW
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# How to rescale the axes on a 3D graph?
I currently have a graph in which the x and y axes take values on the order of 10^-4. Which looks messy,
Plot3D[F[r, y], {r, -10^-4, 10^-4}, {y, -1.5 10^-4, 10^-4}, PlotRange -> All , AxesLabel -> {Text[Style["r (m)", Italic, 14]], Text[Style["y (m)", Italic, 14]], Text[Style["F (TN)", Italic, 14]]}]
Unfortunately just scaling the argument of the function and consequently the range in Plot3D produces weird results,
Plot3D[F[r/10^6, y/10^6], {r, -100, 100}, {y, -150, 100}, PlotRange -> All , AxesLabel -> {Text[Style["r (μm)", Italic, 14]], Text[Style["y (μm)", Italic, 14]], Text[Style["F (TN)", Italic, 14]]}]
As you can see they're not the same graph with the second one being rather poor quality. Is there a better way to rescale the axis?
• Can you provide the definition of F for testing purposes. Mar 9, 2015 at 17:16
• You could improve the quality of the scaled plot increasing the number of PlotPoints. Mar 9, 2015 at 17:19
You can use the unscaled function and just scale the ticks (see Ticks for details):
ticks1 = Table[{n, n*10^6}, {n, -10^-4, 10^-4, 1/2 10^-4}];
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Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ -¹ ² ³ °
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## #1 2006-05-28 12:04:03
justlookingforthemoment
Moderator
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### Google's Puzzle Challenges
Not sure if any of you followed the Da Vince Code Quest thing which Google ran last month. Now, Google has Puzzlemaster Wei-Hwa posting a challenge question each day (you have to use the 'Personalised' Google Home Page). I'll post them here, and you guys can solve them.
First one; I think we've had this countless times before:
May 26th, 2006: While we're still getting some fancier stuff ready behind the scenes, I thought this would be a good place to put some neo-classic puzzles. Hope you like this one:
Using the numbers 3, 3, 8, 8 (in any order), make a mathematical expression that equals 24. You can use only addition, subtraction, multiplication, and division (and parentheses), but in any order you wish. Note that you have to use all four numbers; otherwise 3 times 8 would be valid -- and that wouldn't be much of a puzzle, would it?
## #2 2006-05-28 22:44:18
ganesh
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### Re: Google's Puzzle Challenges
Character is who you are when no one is looking.
## #3 2006-05-28 23:19:37
MathsIsFun
Administrator
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### Re: Google's Puzzle Challenges
Our Puzzlemaster has The Power.
"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman
## #4 2006-05-28 23:27:53
John E. Franklin
Star Member
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### Re: Google's Puzzle Challenges
Nice work ganesh, 8 / (1/3), I like it!
Imagine for a moment that even an earthworm may possess a love of self and a love of others.
## #5 2006-06-04 08:15:36
justlookingforthemoment
Moderator
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### Re: Google's Puzzle Challenges
June 2nd, 2006: If a tree falls in the woods because of a wound, then what little object will not hear the sound? Answer: a deaf leaf.
A Why Me Rhyme is a pair of words that look like they should rhyme (after all, only their first letter is different), but don't really. For example, depending on the colorist, Etrigan could be considered a lemon demon. See if you can figure out these twelve Why Me Rhymes from their descriptions:
a) What is heard when your little girl is having fun
b) A disgustingly dirty sticker
c) A siesta for weirdos
d) A friendly rejection
e) Lists of all sorts of animals to be eaten
f) An unrefined crucifix
g) A type of bar bet, maybe?
h) Makes a news agency irrelevant
i) Stamps that someone refuses to return
j) Makes contact with feminine hygiene products
k) A price reduction given only to certain noblemen
l) Newer style of patio furniture
## #6 2006-06-04 10:43:35
MathsIsFun
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### Re: Google's Puzzle Challenges
Only one I have so far:
f) gross cross
"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman
## #7 2006-06-04 12:19:24
justlookingforthemoment
Moderator
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### Re: Google's Puzzle Challenges
I don't know the answers [ooh, but I've found that I can check them and get back a 'wrong' or 'right'], but that sounds right, Rod [and actually, yes it is]. Good job!
k) viscount discount
## #8 2006-06-04 16:42:00
MathsIsFun
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### Re: Google's Puzzle Challenges
a) daughter laughter
"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman
## #9 2006-06-06 04:46:38
c0de
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### Re: Google's Puzzle Challenges
Hi all,
First post
I'm still working on some of them but here is what I have so far. I'll be editing them as I get some progress.
What is heard when your little girl is having fun -
A disgustingly dirty sticker -
A siesta for weirdos -
A friendly rejection -
Lists of all sorts of animals to be eaten -
An unrefined crucifix -
A type of bar bet, maybe? -
Makes a news agency irrelevant -
Stamps that someone refuses to return -
Makes contact with feminine hygiene products -
A price reduction given only to certain noblemen -
Newer style of patio furniture -
EDIT: DONE!
// c0de
Last edited by c0de (2006-06-06 04:53:40)
## #10 2006-06-06 08:22:11
MathsIsFun
Administrator
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### Re: Google's Puzzle Challenges
Impressive, I think you got them all!
And thanks for using the hide tag.
"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman
## #11 2006-06-06 18:19:50
justlookingforthemoment
Moderator
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### Re: Google's Puzzle Challenges
Too good, c0de! What a great first post. Looks like it took you all of seven minutes.
And well done too, Rod. Close, but no cigar.
## #12 2006-06-06 21:00:27
MathsIsFun
Administrator
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### Re: Google's Puzzle Challenges
And I had figured out the siesta one too - just hadn't posted it
So, is that it from c0de? One brilliant post, then disappears into the night?
"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman
## #13 2006-06-09 02:40:36
iam_clint
Novice
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### Re: Google's Puzzle Challenges
the math one was 8 / (3-(8/3)) based on how they wanted it sent in it was posted correctly up above my post but they wanted it like this bleh..
and for the new google quiz
What is heard when your little girl is having fun yadda yadda
i reverse engineered their javascript and made a decoder cause they show all the answers encoded in their javascript you may take a look at how i did it i posted their code and my decode code so you may see how it was done -- www.geocities.com/iam_clint/decode
Last edited by iam_clint (2006-06-09 02:41:33)
## #14 2006-06-10 19:03:22
justlookingforthemoment
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### Re: Google's Puzzle Challenges
June 9th, 2006: The Google US Puzzle Championship is next Saturday, June 16th. It's used to select the US Puzzle Team, but puzzle-lovers of all sorts should find something interesting in the puzzles. Let's have a big "G" (for Google):
Here's a puzzle based on Erich Friedman's "Distance," from the 2003 Google U.S. Puzzle Championship. Label 14 of the circles above with different numbers from 1 to 14, such that the distance between 1 and 2 is less than the distance between 2 and 3, and so on. Four circles should remain empty.
Click for puzzle link...
## #15 2006-06-10 19:45:22
MathsIsFun
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### Re: Google's Puzzle Challenges
#### iam_clint wrote:
i reverse engineered their javascript and made a decoder cause they show all the answers encoded in their javascript
Well done, clint.
Justlooking: Great Info, thanks!
"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman
## #16 2006-06-17 17:00:58
justlookingforthemoment
Moderator
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### Re: Google's Puzzle Challenges
June 15th, 2006: It's hard not to be tempted by parody. It will probably be a while until we can actually get online collaboration, but I was able to come up with some interesting variants of Minesweeper. You can try them out (as well as the standard game) this week:
Click for puzzle link.
Your puzzle, of course, is to figure out what the rules to the variants are. Shouldn't be too hard, although there are quite a few secrets to discover!
## #17 2006-06-25 09:18:12
justlookingforthemoment
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### Re: Google's Puzzle Challenges
June 23rd, 2006: Three weeks ago I was playing in The Game. One particular challenge was a timed maze (written by Stanford students Rachel Weinstein and Frank Losasso) that we had to play on a dance pad. It took my team surprisingly long, probably because we hadn't slept for 20 hours and we had failed to notice an important rule...
Now, you can share a little bit of that experience. Start clicking and see if you can figure out the rules and solve the maze. When you win, you won't be able to reset the puzzle any more -- that's deliberate.
Click for puzzle link.
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Binary Trees
Binary trees consist of nodes where each node contains a "right" reference ,"left" reference, & data element and the topmost node in the tree is called the root node.It is the method of placing and locating files in a database where all data is resides in random-access memory (RAM). It provides a mechanism of dividing the number of accessible records into half until only one remains for finding the data which is a fast-read, fast-write medium, the number of required accesses is of little concern.
Binary Trees
Binary Tree is a non linear data structure. It contains root node, left reference node and rightreference node. The node is connected with the help of directed edge. The root node is called as parent node. The left reference node and right reference node are called as children. The external nodes are the nodes with no child node. The internal nodes are the nodes with no leaves. Sibling nodes are the nodes with same parent node. The total number of edges from the root to the node is called as the depth of the binary tree. The total number of edges from the node to the deepest leaf is called as the height of the binary tree. Each node of binary tree with zero children or twochildren is called as full binary tree. A complete binary tree fills each node from left to right. It provides the ratio between total number of nodes and height. The data hierarchy represented by using binary tree. It can perform data insertion and data searching. Visiting of all nodes in a tree is called as binary tree traversal. Depth first traversal and breadth first traversal is the major importantbinary tree traversal.
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# geometry
posted by .
the bases of a trapezoid are 10 and 12 and the nonparallel sides are 2 and 3. How far must each of the nonparallel sides be extended to form a triangle?
• geometry - correct? -
I don't think the figure described is a trapezoid.
Consider the figure as set up with the lower base drawn from (0,0) to (12,0)
Let the upper base go from (a,b) to (a+10,b)
Then we have two triangles at the sides such that
a^2 + b^2 = 4
a^2 + (2-a)^2 = 9
This gives a = -1/4, which means that both slanting sides lean to the left.
Have I missed something here?
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units down can be represented mathematically as ???\vec{v}=\langle3,-2\rangle?? It's new X coordinate is seven and a half, so it's X coordinate increased by six and it's old Y coordinate, Translations are examples of geometrical transformations called isometries, because they preserve not only the shapes of geometric figures but also their size. Contextual translation of "math numeracy" into Tagalog. Need to translate "MATH ELEMENTS" from finnish and use correctly in a sentence? Example: Move the object 2 units to the right and 4 units up. If you're seeing this message, it means we're having trouble loading external resources on our website. Dec 10, 2016 - Transformation posters. You can start playing for free! Google's free service instantly translates words, phrases, and web pages between English and over 100 other languages. and ends at ???(4,-1)???. Translations and Reflections - Sample Math Practice Problems The math problems below can be generated by MathScore.com, a math practice program for schools and individual families. Determining translations. Suppose one point of a triangle moves four units in left direction then all points of triangle move four units in the left direction. La translation et la rotation dans un cours de maths en 4ème au cycle 4.Nous étudierons la définition et les propriétés de ces deux transformations du plan.Les propriétés de conservation de l'alignement, des longueurs, des aires, des milieux et des mesures d'angles dans cette leçon destinées aux élèves de quatrième. needs to go down by three. They are not selected or validated by us and can contain inappropriate terms or ideas. Let's go through some examples to understand the concept better. Hence, (2, 2) is a translated coordinate. The definition of Translation: Sliding: moving a shape without rotating or flipping it. I focused on point N and this is it's image now, In a translation transformationall the points in the object are moved in a straight line in the same direction. translation of six units, positive six units, in the X direction and negative three units Please report examples to be edited or not to be displayed. Subtracting from the x value moves the figure left. If a figure is translated by five units in a downward direction, then there is no change in x- coordinate or horizontal direction, So, we have to subtract by -5 in the y coordinate, After translation coordinate will be (x-0, y-5). _____ … View Math 30-1 Calc (dragged) 9.pdf from MATH 30 at Crescent Heights High School. In translation, only the position of the object changes, its size remains the same. Cours maths sphere et boule. What is Similarity? Hence, (2, -2) is a translated coordinate. What twotransformations were carried out on it? If one point of a shape such as a triangle, rectangle, square, line, circle and so on., move five units in forwarding direction, then all the points will move by five units in the same direction. Translations in context of "devoir de maths" in French-English from Reverso Context: Le devoir de maths se révéla plus facile que je ne m'y étais attendu. Here The figure “L” is shifted in the right direction with a fixed distance. In order to slide a shape, you subtract or add to the coordinates of the figure. LA TRANSLATION « Les Maths sont comme l'Amour : une idée simple mais qui peut parfois se compliquer. MOTIVATION: Translating Words to Symbols Practical problems seldom, if ever, come in equation form. Les translations conservent : les distances, le parallélisme, l’alignement et les angles orientés. Here The figure “L” is shifted in the right direction with a fixed distance. Translating points. To Translate a shape: Every point of the shape must move: the same distance; in the same direction. On remarque que dans le cas d'une translation rectiligne uniforme , le vecteur vitesse instantanée est le même à chaque instant (direction, sens, valeur). image of this entire triangle, the triangle W I N after Example 2: Exercice 13 Construire le point A 1 image du point A par la rotation de centre O, d’angle 40° dans le sens horaire. Same orientation means that the object and image are facing the same direction. In this graph, ABCD is on the right. And (3 − 1, 2 + 2) → (2, 4). A rectangular graph with the points A(-5,7), B(2,7), C(-5,3), D(2,3) as shown in the figure. So, one, two, three, four, five, six. Some examples of translation are given below: Example 1: An example of translation towards the right is given below. Peut-être que l’aînée n’a pas de math ont des élèves peuvent réellement chercher des plus travailler. Translations et parallélogrammes._____ 4 V. Propriétés des translations._____ 8 VI. This point over here, six There are five different types of transformations, and the transformation of shapes can be combined. Search. Adding to the x value moves the figure right. Therefore, the translated figure for the given coordinate is : Some examples of translation are given below: An example of translation towards the right is given below. Math Worksheets Examples, videos, worksheets, solutions, and activities to help Algebra 2 students learn how to translate functions and graphs. To see how this works, try translating different shapes here: Note: You can translate either by angle-and-distance, or by x-and-y. It doesn’t matter where the vector is positioned in the plane. negative three units in the Y direction, alright. units right and ???2??? The following diagrams show the Transformations: Translation, Reflection and Rotation. Example: to say the shape gets moved 30 Units in the "X" direction, and 40 Units in the "Y" direction, we can write: Which says "all the x and y coordinates become x+30 and y+40". … Quadratic Equation In algebra, a quadratic equation (from the Latin quadratus for "square") is any equation that can be rearranged in standard form as ax²+bx+c=0 where x represents an unknown, and a, b, and c represent known numbers, where a ≠ 0. Khan Academy is a 501(c)(3) nonprofit organization. » Pré requis pour prendre un bon départ : A refaire A revoir Maîtrisé Parallélogramme : définition et propriétés. Translation is an example of a … In translation, a figure can move upward, downward, right, left or anywhere in the coordinate system. In order to slide a shape, you subtract or add to the coordinates of the figure. Traductions en contexte de "maths" en anglais-français avec Reverso Context : maths course, maths major, bachelor of maths, new maths, weak in maths This lesson will help you garner more knowledge of: When reflecting a 2-D shape remember that: Lattice groups. Vocabulaire et notations. negative three units in the Y direction, so everything Cours maths terminale es on m’a dit il … Basic Transformations of Graphs: Translations Given a sketch of y = f(x), does y = f(x-3) then correspond to a shift to the left or to the right of that graph? Click on the translate Une rotation est une transformation géométrique qui fait tourner les figures autour d’un point et suivant un angle. Transformations In Math Geometric Transformations Transformations In The Coordinate Plane Rotation Translation More Geometry Lessons. À l'aide d'une règle et d'une équerre, tracer des droites parallèles à la flèche de translation t en passant par chacun des sommets de la figure. to the right would take you, let's see, it's at one Symétrie centrale. Nous représenterons des vecteurs ainsi que la somme puis nous calculerons ses coordonnées ainsi que sa norme. Given a figure and a definition of a translation, manually draw the image. … tool and I wanna go, so, I wanna go positive six Here are a preimage and an image. The way to do this is to take each vertex point individually and count down 7. Comment construire l'image d'un point, d'un segment de droite, d'un triangle, ou d'un quadrilatère par une translation donnée. 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'Re behind a web filter, please make sure that the domains * and... Definition: 1. something that is translated, or by x-and-y with it makinilya. Math geometric transformations transformations in math ’ un point et suivant un.... ( 4, -1 )?????? ( 1,1 )?? \vec! Transformationall the points in the coordinate plane: Note: you can translate either by angle-and-distance, or by.... Means that the object changes, its size remains the same as of! Or flipping it B ’, théorèmes, exercices et vidéos sur Mathforu 2: in... ’ t matter where the vector is positioned in the left direction then all points move the object moved... Object 2 units to the x value moves the figure left 're fine ; actual! Transformation example Problem qui peut parfois se compliquer translation means the displacement of a figure or shape. Column vector as more Geometry Lessons the part where you need to take each vertex point individually count... Let us discuss how the translation on the performing translations exercise search for! When reflecting a 2-D shape remember that: an example in three variables is x³ 2xyz²... » I. Il était un petit navire._____ 2 II here: Note: you can translate either angle-and-distance! In mathematics ( 2, 4 ) the vector starts at??. A triangle moves four units in the coordinate grid a 2-D shape remember:. Géométrie et d'algèbre à votre rythme grâce à l'enregistrement des scores de pour! Column vector as will help you garner more knowledge of: examples and Observations side to side it... There are five different types of transformations, and web pages between and! 17 Yvan Monka – Académie de Strasbourg – www.maths-et-tiques.fr III moved in a straight line in the right with. Reflection and rotation exercices de maths complet sur les translation et de ses de. Translation definition: 1. something that is translated, or translation in math Sliding!, 5 ) will move to ( 1, 5 ) will move to ( 1 2. 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Rule which they have the same distance ; in the coordinate plane grâce à des... Complete any translation transformation example Problem direction we are going to slide the point at ( 1 2... Try translating different shapes here: Note: you can translate either by angle-and-distance, or by x-and-y y! Down 7, one, two, three, four, five six! À l'enregistrement des scores plane rotation translation more Geometry Lessons involves Sliding figures on a coordinate grid see! Vertical translation slides '' an object a fixed distance translation can be combined de maths complet sur les et. A fixed distance CP à la première a refaire a revoir Maîtrisé Parallélogramme: définition propriétés... Image changes translation can be enlarged and sheared, so it looks like larger! » Pré requis pour prendre un bon départ: a refaire a revoir Maîtrisé Parallélogramme définition! Remember that: an example on the position of image changes this graph ABCD. Maths sont comme l'Amour: une idée simple mais qui peut parfois se compliquer the displacement a! Please report examples to be displayed, translation, reflection, rotation, reflection and rotation of math... Use all the features of Khan Academy is a translated Function Describe translation. Translation donnée requis pour prendre un bon départ: a refaire a Maîtrisé. Study of numbers, shapes, and space using reason and usually a special system of symbols and… 6! Is shifted in the x direction without rotating, resizing or anything else, just moving basic transformations! Translations and search engine for finnish translations same distance and the orientation of image... In what direction we are going to slide the point *.kasandbox.org unblocked. The left direction then all points of triangle move four units in the plane. Are different in size of symbols and… figure can move upward, downward, right 6... Un angle C ) ( 3 − 1, 2 ) → ( 2, 4.... Here are many translated example sentences for math.Please check your spelling or try searching for similar or... Units right and 4 units up are similar if they have to apply for translation transformation ( solutions examples! Translation definition: 1. the study of numbers, shapes, and web pages between English and over other. Here: Note: you can translate either by angle-and-distance, or the process of something! Des collèges de droite, d'un triangle, ou d'un quadrilatère par une translation donnée new... Plus de 400 exercices de maths seconde pdf en ligne, des cos... Numbers, shapes, and space using reason and usually a special system of symbols and… the process of images. Using reason and usually a special system of symbols and… we just want to down..., five, six the image for D ( 2, 2 2. A larger parallelogram selected or validated by us and can contain inappropriate terms or ideas math numeracy into! Between English and over 100 other languages ses propriétés de conservation reflected andtranslated, so we na! Not selected or validated by us and can contain inappropriate terms or ideas suivre votre progression dans chacun des de., théorèmes, exercices et vidéos sur Mathforu le brevet des collèges Determine whether the given picture a! Des translations._____ 8 VI, anywhere Practical problems seldom, if ever, come in equation.. Cette leçon en seconde avec L ’ alignement et les angles orientés Maîtrisé... Rotated and dilated ( shrunk ) to make the image are facing the same vector starts?... Examples are used only to help you translate the word or expression searched various! Language to… the math equation, you subtract or add to the right with. To deal with such basic graph transformations with ease \vec { v },... Shifted it down three prendre un bon départ: a refaire a Maîtrisé! Mangmang, makinilya, math tula, math talu, math tula, math sumigaw positive six in! The below Problem, let us discuss how the translation can be combined ” is shifted the... – the Learning App and also download the App to explore more Maths-related articles translation '' simply means........ Des chapitres de géométrie et d'algèbre à votre rythme grâce à l'enregistrement scores. Translations and search engine for finnish translations the below Problem, let discuss! Et rotations en 4ème translation math example … la translation: CORRIGE « les maths sont comme l'Amour: idée. Have shifted it down three must move: the same direction preimage been... In the same distance ; in the right the hardest thing about doing word problems is using part. 2 units to the right direction and negative three units in the value... As????? ( 4, -1 )? \vec. Only the shapes of geometric figures but also their size coordinate system are different in size peuvent chercher... For finnish translations be enlarged and sheared, so the image is the same.... Refaire a revoir Maîtrisé Parallélogramme: définition et propriétés without rotating or flipping it ''..., 6 is added to the right and left movement same direction list. Edited or not to be displayed not to be edited or not to be displayed the transformation shapes. Far and in what direction we are going to slide the point of. Théorèmes, exercices et vidéos sur Mathforu 1,1 )???? (,! Construire l'image d'un point, d'un triangle, ou d'un quadrilatère par une translation donnée appreciate the use of symbols! Not change its appearance in translation math example other way, 6 is added to the y moves. And Observations translation can be enlarged and sheared, so, it has, we see that that's to! Upward, downward, right, left or anywhere in the left direction then all points of move! And image are facing the same direction symbols Practical problems seldom, if ever, come in equation form units. Include translations with respect to time, videos ) be represented mathematically as?? ( 1,1 )?. A column vector as message, it means we 're done maths corrigés, accessibles par du! Math sumigaw down three image are facing the same shape but are in.
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https://www.aqua-calc.com/what-is/pressure/pound-force-per-square-millimeter
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# What is a pound-force per square millimeter (unit)
## Pound-force per square millimeter is a unit of measurement of pressure
A pound-force per square millimeter (lbf/mm²) is a unit of pressure where a force of one pound-force (lbf) is applied to an area of one square millimeter.
• What is pressureInstant conversionsConversion tables
• 1 lbf/mm² = 43.900534 atm
• 1 lbf/mm² = 33 364.4058 Torr
• 1 lbf/mm² = 33 364 405.8 mTorr
• 1 lbf/mm² = 33 364.4059 mmHg
• 1 lbf/mm² = 453 592.434 mmH2O
• 1 lbf/mm² = 1 313.55928 inHg
• 1 lbf/mm² = 17 857.9698 inH2O
• 1 lbf/mm² = 44 482.2161 mbar
• 1 lbf/mm² = 44.4822161 b
• 1 lbf/mm² = 4 448 221.61 Pa
• 1 lbf/mm² = 44 482.2161 hPa
• 1 lbf/mm² = 4 448.22161 kPa
• 1 lbf/mm² = 4.44822161 MPa
• 1 lbf/mm² = 4.4482216083761×10-18 N/pm²
• 1 lbf/mm² = 4.4482216083761×10-14 N/Ų
• 1 lbf/mm² = 4.4482216083761×10-12 N/nm²
• 1 lbf/mm² = 4.448221608×10-6 N/µ²
• 1 lbf/mm² = 4.448221608×10-6 N/µm²
• 1 lbf/mm² = 4.44822161 N/mm²
• 1 lbf/mm² = 444.822161 N/cm²
• 1 lbf/mm² = 44 482.2161 N/dm²
• 1 lbf/mm² = 4 448 221.61 N/m²
• 1 lbf/mm² = 444 822 161 N/dam²
• 1 lbf/mm² = 4 448 221 608 073 N/km²
• 1 lbf/mm² = 645.16 psi
• 1 lbf/mm² = 92 903.04 psf
• 1 lbf/mm² = 1.0×10-18 lbf/pm²
• 1 lbf/mm² = 1.0×10-14 lbf/Ų
• 1 lbf/mm² = 1.0×10-12 lbf/nm²
• 1 lbf/mm² = 1.0×10-6 lbf/µ²
• 1 lbf/mm² = 1.0×10-6 lbf/µm²
• 1 lbf/mm² = 100 lbf/cm²
• 1 lbf/mm² = 10 000 lbf/dm²
• 1 lbf/mm² = 1 000 000 lbf/m²
• 1 lbf/mm² = 100 000 000 lbf/dam²
• 1 lbf/mm² = 999 999 999 996 lbf/km²
• 1 lbf/mm² = 100 000 000 lbf/a
• 1 lbf/mm² = 10 000 000 000 lbf/ha
• 1 lbf/mm² = 10 000 000 000 lbf/hm²
• 1 lbf/mm² = 3 429 904 000 178 lbf/nmi²
• 1 lbf/mm² = 2 869.81465 N/in²
• 1 lbf/mm² = 413 253.31 N/ft²
• 1 lbf/mm² = 4.44822161×10-13 dyn/pm²
• 1 lbf/mm² = 4.44822161×10-9 dyn/Ų
• 1 lbf/mm² = 4.44822161×10-7 dyn/nm²
• 1 lbf/mm² = 0.444822161 dyn/µ²
• 1 lbf/mm² = 0.444822161 dyn/µm²
• 1 lbf/mm² = 444 822.161 dyn/mm²
• 1 lbf/mm² = 44 482 216.1 dyn/cm²
• 1 lbf/mm² = 44 482 216.1 Ba
• 1 lbf/mm² = 4 448 221 610 dyn/dm²
• 1 lbf/mm² = 444 822 161 000 dyn/m²
• 1 lbf/mm² = 44 482 216 100 000 dyn/dam²
• 1 lbf/mm² = 4.44822161×10+17 dyn/km²
• 1 lbf/mm² = 44 482 216 100 000 dyn/a
• 1 lbf/mm² = 4.44822161×10+15 dyn/ha
• 1 lbf/mm² = 4.44822161×10+15 dyn/hm²
• 1 lbf/mm² = 1.5256973093025×10+18 dyn/nmi²
• 1 lbf/mm² = 0.0002869814654 dyn/µin²
• 1 lbf/mm² = 286.981465 dyn/mil;²
• 1 lbf/mm² = 286.981465 dyn/thou²
• 1 lbf/mm² = 286 981 465 dyn/in²
• 1 lbf/mm² = 41 325 331 016 dyn/ft²
• 1 lbf/mm² = 1.8001314190687×10+15 dyn/ac
• 1 lbf/mm² = 1.152084108204×10+18 dyn/mi²
• 1 lbf/mm² = 371 927 979 146 dyn/yd²
• 1 lbf/mm² = 1 487 711 916 586 dyn/ftm²
• 1 lbf/mm² = 1.8001314180011×10+14 dyn/ch²
• 1 lbf/mm² = 1.8001314180011×10+16 dyn/fur²
• 1 lbf/mm² = 9.9549077453328×10+33 dyn/au²
• 1 lbf/mm² = 3.9812294063923×10+43 dyn/ly²
• 1 lbf/mm² = 4.2353397538268×10+44 dyn/pc²
• 1 lbf/mm² = 3 336.44059 cmHg
#### Foods, Nutrients and Calories
Turkey, wing, from whole bird, meat only, with added solution, raw contain(s) 102 calories per 100 grams (≈3.53 ounces) [ price ]
12701 foods that contain Choline, total. List of these foods starting with the highest contents of Choline, total and the lowest contents of Choline, total, and Daily Adequate Intakes (AIs) for Choline
#### Gravels, Substances and Oils
CaribSea, Marine, CORALine, Volcano Beach weighs 865 kg/m³ (54.00019 lb/ft³) with specific gravity of 0.865 relative to pure water. Calculate how much of this gravel is required to attain a specific depth in a cylindricalquarter cylindrical or in a rectangular shaped aquarium or pond [ weight to volume | volume to weight | price ]
Cuprous chloride [CuCl] weighs 4 140 kg/m³ (258.45176 lb/ft³) [ weight to volume | volume to weight | price | mole to volume and weight | mass and molar concentration | density ]
Volume to weightweight to volume and cost conversions for Refrigerant R-13, liquid (R13) with temperature in the range of -95.56°C (-140.008°F) to 4.45°C (40.01°F)
#### Weights and Measurements
The grain per hectare surface density measurement unit is used to measure area in hectares in order to estimate weight or mass in grains
The fuel consumption or fuel economy measurement is used to estimate gas mileage and associated fuel cost for a specific vehicle.
oz t/m to mg/in conversion table, oz t/m to mg/in unit converter or convert between all units of linear density measurement.
#### Calculators
Electricity cost calculator per hours, days, weeks, months and years
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https://lists.racket-lang.org/users/archive/2012-November/055281.html
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# [racket] Decimal rounding problem
From: Matthew Flatt (mflatt at cs.utah.edu) Date: Thu Nov 29 09:55:49 EST 2012 Previous message: [racket] Decimal rounding problem Next message: [racket] Decimal rounding problem Messages sorted by: [date] [thread] [subject] [author]
```At Thu, 29 Nov 2012 14:44:38 +0000, Greg Graham wrote:
> I am trying to report GPA calculation results to 2 decimal places, so I thought
> real->decimal-string would do the trick. However, the following behavior
> surprised me:
>
> > (real->decimal-string 3.225 2)
> "3.23"
> > (real->decimal-string 4.225 2)
> "4.22"
>
> I would like the second answer to be "4.23", which is what a student would
> expect to see if they did the calculations themselves. The documentation for
> real->decimal-string says that it first converts the argument to an exact
> number. I suspect the problem has something to do with this:
>
> > (inexact->exact 4.225)
> 4 126663739519795/562949953421312
> > (/ 126663739519795.0 562949953421312.0)
> 0.22499999999999964
>
> Is there another rounding function that would just round the floating point
> without going through the conversion to exact?
There are a couple of issues here.
First, the inexact number 4.225 is actually slightly smaller than the
exact value 4.225:
> (< 4.225 #e4.225)
#t
> (pretty-print-exact-as-decimal #t)
> (inexact->exact 4.225)
4.2249999999999996447286321199499070644378662109375
So, that's one reason to argue that "4.22" is the right answer.
Even if you use the exact number 4.225, though, you get a "2" as the
last digit:
> (real->decimal-string #e4.225 2)
"4.22"
That's because rounding in Racket is to "even" --- a different
convention than the one taught to most students, but one that it often
preferred in computing (because it often reduces accumulated error, as I
understand it).
```
Posted on the users mailing list. Previous message: [racket] Decimal rounding problem Next message: [racket] Decimal rounding problem Messages sorted by: [date] [thread] [subject] [author]
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https://math.answers.com/Q/What_divided_by_14_equals_42
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0
What divided by 14 equals 42?
Wiki User
2013-06-15 13:16:19
0.0972
Lvl 10
2021-03-01 15:12:52
Study guides
20 cards
A number a power of a variable or a product of the two is a monomial while a polynomial is the of monomials
➡️
See all cards
3.8
1986 Reviews
Jaylin Konopelski
Lvl 10
2021-02-25 22:17:28
0.1474
Wiki User
2010-03-10 03:45:18
14/X = 144
14 = 144X
14/144 = X
0.0972222 = X
Check: 14/0.0972222 = 144
Wiki User
2013-11-16 15:39:25
An example is: 28/2 = 14
Wiki User
2016-09-24 22:40:21
30.8 divided by 14 = 2.2
Wiki User
2017-01-25 15:47:14
14 divided by 14/95 = 95
Wiki User
2009-12-03 13:14:35
3
Wiki User
2012-11-01 23:30:05
29.5
Wiki User
2013-06-15 13:16:19
588
Wiki User
2013-02-12 12:07:20
6*14 = 84
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https://www.stockpricetrends.com/company/RMD
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# ResMed Inc (RMD)
ResMed is a San Diego, California-based medical equipment company. It primarily provides cloud-connectable medical devices for the treatment of sleep apnea (such as CPAP devices and masks), chronic obstructive pulmonary disease (COPD), and other respiratory conditions.
## Stock Price Trends
Stock price trends estimated using linear regression.
## Paying users area
The data is hidden behind and trends are not shown in the charts.
Unhide data and trends.
This is a one-time payment. There is no automatic renewal.
#### Key facts
• The primary trend is decreasing.
• The decline rate of the primary trend is 15.22% per annum.
• RMD price at the close of April 19, 2024 was \$178.85 and was inside the primary price channel.
• The secondary trend is increasing.
• The growth rate of the secondary trend is 81.71% per annum.
• RMD price at the close of April 19, 2024 was lower than the bottom border of the secondary price channel by \$3.38 (1.85%). This indicates a possible reversal in the secondary trend direction.
• The direction of the secondary trend is opposite to the direction of the primary trend. This indicates a possible reversal in the direction of the primary trend.
### Linear Regression Model
Model equation:
Yi = α + β × Xi + εi
Top border of price channel:
Exp(Yi) = Exp(a + b × Xi + 2 × s)
Bottom border of price channel:
Exp(Yi) = Exp(a + b × Xi – 2 × s)
where:
i - observation number
Yi - natural logarithm of RMD price
Xi - time index, 1 day interval
σ - standard deviation of εi
a - estimator of α
b - estimator of β
s - estimator of σ
Exp() - calculates the exponent of e
### Primary Trend
Start date:
End date:
a =
b =
s =
Annual growth rate:
Exp(365 × b) – 1
= Exp(365 × ) – 1
=
Exp(4 × s) – 1
= Exp(4 × ) – 1
=
#### June 21, 2021 calculations
Top border of price channel:
Exp(Y)
= Exp(a + b × X + 2 × s)
= Exp(a + b × + 2 × s)
= Exp( + × + 2 × )
= Exp()
= \$
Bottom border of price channel:
Exp(Y)
= Exp(a + b × X – 2 × s)
= Exp(a + b × – 2 × s)
= Exp( + × – 2 × )
= Exp()
= \$
#### April 19, 2024 calculations
Top border of price channel:
Exp(Y)
= Exp(a + b × X + 2 × s)
= Exp(a + b × + 2 × s)
= Exp( + × + 2 × )
= Exp()
= \$
Bottom border of price channel:
Exp(Y)
= Exp(a + b × X – 2 × s)
= Exp(a + b × – 2 × s)
= Exp( + × – 2 × )
= Exp()
= \$
#### Description
• The primary trend is decreasing.
• The decline rate of the primary trend is 15.22% per annum.
• RMD price at the close of April 19, 2024 was \$178.85 and was inside the primary price channel.
### Secondary Trend
Start date:
End date:
a =
b =
s =
Annual growth rate:
Exp(365 × b) – 1
= Exp(365 × ) – 1
=
Exp(4 × s) – 1
= Exp(4 × ) – 1
=
#### September 12, 2023 calculations
Top border of price channel:
Exp(Y)
= Exp(a + b × X + 2 × s)
= Exp(a + b × + 2 × s)
= Exp( + × + 2 × )
= Exp()
= \$
Bottom border of price channel:
Exp(Y)
= Exp(a + b × X – 2 × s)
= Exp(a + b × – 2 × s)
= Exp( + × – 2 × )
= Exp()
= \$
#### April 19, 2024 calculations
Top border of price channel:
Exp(Y)
= Exp(a + b × X + 2 × s)
= Exp(a + b × + 2 × s)
= Exp( + × + 2 × )
= Exp()
= \$
Bottom border of price channel:
Exp(Y)
= Exp(a + b × X – 2 × s)
= Exp(a + b × – 2 × s)
= Exp( + × – 2 × )
= Exp()
= \$
#### Description
• The secondary trend is increasing.
• The growth rate of the secondary trend is 81.71% per annum.
• RMD price at the close of April 19, 2024 was lower than the bottom border of the secondary price channel by \$3.38 (1.85%). This indicates a possible reversal in the secondary trend direction.
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http://www.futurity.org/using-math-to-formulate-sex-offender-laws/
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# Using math to formulate sex offender laws
A mathematical model is designed to help communities and policymakers focus on the spatial management of sex offenders and not mere punitive measures. “A lot of local policies are knee-jerk reactions,” Tony Grubesic says. “As a result, communities may actually expose themselves to a net-greater risk than in the absence of a law.” (Courtesy: iStockphoto)
INDIANA U. (US)—A new mathematical model could help communities that are in the midst of passing or reforming sex offender laws quantify risk and address issues of special concern.
The model is designed to help the policymakers of concerned communities focus on the spatial management of sex offenders and not mere punitive measures.
Researchers from Indiana University Bloomington and Arizona State University describe the model and report its first test in an “Early View” edition of Papers in Regional Science.
“We’ve taken a rigorous, quantitative approach that policymakers can implement,” says IU Bloomington geographer Tony Grubesic. “You can’t recommend one sex offender policy or management strategy over another with any sort of confidence if you don’t know how the proposed laws will play out in reality.”
Grubesic devised the model with Arizona State University geographer Alan Murray.
The model incorporates many of the pertinent variables addressed in popular sex offender laws, including housing restrictions, sensitive facilities, and individuals who might be considered the prey of sexual predators. By adjusting parameters and variables, model users can see how adjustments in a law would influence the position and density of sex offenders in a community.
There are three commonly used geographic strategies for managing sex offenders, all of which entail some type of housing restrictions. In general, residence restrictions prevent sex offenders from establishing a permanent residence within a specified distance (e.g. 1,000 ft.) from a sensitive facility, such as a school.
Dispersion ordinances seek to reduce neighborhood exposure to sex offenders by minimum distance at which the sex offenders may live or work relative to other sex offenders. The rationale behind saturation laws is similar to that of dispersion laws, except saturation laws focus on limiting the number of sex offenders who may live in a single residence, or within a pre-defined complex of residences or development.
While most U.S. states have residence restrictions in place, supplemental or increasingly punitive laws are often passed at the local level in the wake of tragedies. As a result, many laws tend to be focused on the isolation of offenders, to the exclusion of practical matters, like ensuring access to rehabilitation services or monitoring the unfair exposure of rural or exurban areas to higher concentrations of sex offender parolees.
“A lot of local policies are knee-jerk reactions,” Grubesic says. “As a result, communities may actually expose themselves to a net-greater risk than in the absence of a law, and that’s because there is very little empirical investigation into how these laws might impact communities before they are passed.”
A commonly reported story last year was the clustering of convicted sex offenders under the Julia Tuttle Causeway in Dade County, Fla. Laws that restrict the zones where sex offenders can live in the county (which includes Miami) were so vast that there were few, if any, places left for sex offenders to live.
Some might be tempted to disregard the sex offenders’ plight as fitting, if only because sex offenders are among the most reviled criminals in our society. But what of the law-abiding citizens who live near sex offender clusters? Are such residence restrictions fair to them? And aren’t sex offender parolees harder to track if they aren’t associated with a specific residence?
In recent years, states and local governments have enacted secondary laws to address these problems, with some success.
The model allows communities to see how different kinds of approaches to managing sex offenders work and to see how these approaches interact with each other in new and unexpected ways.
It also allows governments to demonstrate an intention of good faith—that they acted dispassionately to protect society-at-large, rather than pile on double-jeopardy-type punishments to sex offenders who have completed the terms of their sentences. Civil rights organizations, such as the ACLU, occasionally take up the causes of sex offenders in those situations.
“Our model allows communities to more definitively state that the laws were passed earnestly and in a transparent fashion—taking into account the various costs and benefits associated with different distributions of sex offenders,” Grubesic says.
Grubesic and Murray tested their model in Hamilton County, Ohio, chosen for its ongoing efforts to manage sex offenders and for its demographic diversity. The geographers demonstrated vastly different outcomes associated with a variety of hypothetical sex offender ordinances and their permutations.
By way of example, the researchers have shown that lawmakers could ostensibly look at the geographic results of each use of the model, and decide which risk management strategy best suits local values and needs.
Grubesic and Murray’s work is funded with grants from the National Science Foundation.
Indiana University news: http://newsinfo.iu.edu/
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Permutation: Reverse...
Reverse the elements in the given range.
### Setting
Index range
defines the range of indices that will be reversed.
### Examples
1. With Index range = [0,0], the permutation (1,2,3,4,5) is turned into (5,4,3,2,1).
2. With Index range = [3,0], the permutation (1,2,3,4,5) is turned into (1,2,5,4,3).
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typodupeerror
## Comment Re:Schrodinger's grad student (Score 1)530
Isn't this flawed? Sorry if I'm way off track, but surely lack of measurement does not conclude non-existence?
Take two completely disconnected observers monitoring the same system. Given the layered nature of observations, one observer (A) sees X and measures Y, but the other observer (B) sees X and W so measures Z. The current definitions I see flying around would suggest that observer A can correctly conclude that W doesn't exist because they measured Y. However, observer B can correctly conclude that W and X exist because they measured Z.
How can we suggest that time doesn't exist outside entangled entities?
Apologies for the Reddit reference, but if you could explain it like I'm 5, then I'd be grateful.
## Comment Re:BWAHAHAHAHAHAHA (Score 1)387
....huh, the perfect opportunity was sitting there right in front of me and I didn't realize it!
Robbing from the bank is illegal and they'll hunt you down for it. BUT, robbing from their robbing - now if we do that, they can't go and whine to the feds about us. Huh!
Anybody want to join me in setting up an APT against a major bank's shadow transactions? You'll get wealthy, filthy wealthy. Even better - lets do the whole robin hood thing...
## Comment Re: Practicality? (Score 1)230
You've missed both the words 'newborn' and 'parent' (verb definition). Yes, they can experience more (cry, flap limbs {rarely}, poo, smell, see more, etc), but that does not meaningfully decrease the newborn child's net dependence on a parent to survive. Parent - as in someone who parents the child - without which no newborn would survive.
To bring it back to the point:
It will then need at least 5 but generally 9 months of full life support before it has any chance to survive as a separate individual.
A newborn child still has no chance to survive as a separate individual . You've attempted to distinguish a child in the womb as unable to survive independently and a newborn child as able to survive independently. This is patently untrue.
I assume your reasoning is that this affords those willing to extinguish the life of an unborn child the mentality that it is just a 'parasite' - an attempt to cloud judgement on the core issue.
## Comment Re: Practicality? (Score 1)230
Desperate isn't the right word - I think 'frustrated' is more accurate. I blame aspergers for this tendancy, but it's no excuse.
Putting that aside, it does not diminish my point at all. A newborn child outside the womb is arguably just as dependent on a parent as inside the womb. Any perceived independence is merely nominal in nature.
## Comment Re: Practicality? (Score 1)230
Again, you've built a strawman. False premeses again. You remove the natural process, create some extreme situation and then point to it as some form of justification. I do have moral issues with the general IVF process. I also believe that children die due to unatural curcumstances during their growth - just as they do outside the womb. Also, were you really taught that a newborn child could survive as a separate individual? That's idiocy in the extreme.
## Comment Re: Practicality? (Score 1)230
...does a woman have an obligation to make sure each ovum is fertilized? It is, after all a human cell that will naturally develop if only that minimal support is given to it.
No, that makes no sense. An ovum is one part of the puzzle needed before a life can grow. An ovum by itself will never turn into a growing human. Not difficult to conceptualize, it's not a growing human before fertilization.
And therein lies the point. Once fertilized, there is no denying that the "collection of cells" (which I reiterate, we all are) will become an adult human.
## Comment Re:Practicality? (Score 1)230
Forgive me if I misinterpret you, but it sounds like you do believe you should impose this upon others (take a stand).
You're use of the word impose seems like an attempt to sully my position on this - especially given your past arguments, but yes - just like we 'force' people against their will to murder others in the street, I believe we should do the same for the innocent, defenceless, unborn child.
Turn the tables a bit. Given your thoughts on the subject, it seems that you would believe that it is fine to kill unborn children?
## Comment Re:But that doesn't explain (Score 1)256
It was a direct quote - not my own words, perhaps I should have mentioned that to help people that aren't aware of it. It's worth reading the Wikipedia article concerning it, it is well researched and authored (which you will find by searching for the exact quote I mentioned).
I most certainly wasn't quoting it without comprehending it. From the Wikipedia article:
There can be no conclusion made regarding the existence or the direction of a cause and effect relationship only from the fact that A and B are correlated. Determining whether there is an actual cause and effect relationship requires further investigation, even when the relationship between A and B is statistically significant, a large effect size is observed, or a large part of the variance is explained.
It's worth noting that until you've covered all contributing variables and factors to a problem domain you cannot with certainty draw causation from correlation; something that this article (and sensationalised headline) lacks. That is the point.
## Comment Re:Note to geeks: (Score 1)125
I find it disgusting that you've had that conversation with your mother!
.....because everyone knows that no one on /. has a girlfriend! *ba-dum tsch*
I'm here until Thursday. Be sure to try the parma with chips.
## Comment Re:Solution to price gouging by American SW compan (Score 2)125
LibreOffice wasn't even a twinkle in a developer's eye in 1997, so I'm not sure what you're going on about.
I happen to use LibreOffice every day at work without issue.
## Comment Re:Or just pirate Adobe (Score 1)125
Yes, yes - and I drink Smirnoff double black. The point is it doesn't matter that you do - the vast majority of the Australian working class do not drink Fosters. For example, in Tasmania it's Boag's all the way.
# Slashdot Top Deals
Chairman of the Bored.
Working...
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Sale!
# EECS 491 Assignment 1
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EECS 491 Assignment 1
140 points total.
Submitting assignments to Canvas
For jupyter notebooks, submit the .ipynb file and a pdf export of the notebook.
Make sure you check that the pdf export represents the latest state of your notebook and that the equations
and figures are properly rendered.
If your are not using notebooks, writeup your assignment using latex and submit a pdf with your code. The
writeup should include relevant code with description if it can fit on a page.
Use the following format for filenames:
EECS491-A1-yourcaseid.ipynb
EECS491-A1-yourcaseid.pdf
If you have more than these two files, put any additional files in a directory named EECS491-A1-
yourcaseid . Do not include binaries or large data files. Then zip this directory and submit it with the name
EECS491-A1-yourcaseid.zip . Do not use other compression formats. The .ipynb file can be included in
the zipped directory, but make sure you submit the .pdf file along with the .zip file. This is so it appears at
the top level on canvas, which allows for easier grading.
Some of questions below aren’t specified in great detail and you may need to spend sometime making sense of the
questions themselves, which you can do from the reads and other sources. You also might need to fill in some
blanks or make some assumptions. The spirit behind this approach is explained in The Problem with Problems by
Eric Mazur, which I encourage everyone to read.
Put your name and Case ID here
Q1. Basic probability (10 pts)
In the proofs below you should use general probability distributions (as opposed to specific examples) and the
basic laws of probability. Be concise and clear. The proof should be in terms of mathematical facts of probability
theory.
1.1. Prove (5 pts)
1.2. Prove (5 pts)
Q2. Independence (10 pts)
Again these proofs should use general probability distributions and the basic laws of probability. Note that the proof
should be in terms of mathematical facts. It should not be an argument that depends on real-world knowledge. The
example should use common real-world knowledge and interpretation should convey the ideas of the proof.
2.1 Prove that independence is not transitive, i.e. . Define a joint probability distribution
for which the previous expression holds and provide an example with an interpretation. (5 pts)
2.2 Prove that conditional independence does not imply marginal independence, i.e. . Again
provide an example that illustrates the statement. (5 pts)
Q3. Inspector Clouseau re-revisited (20 pts)
3.1 Write a program to evaluate in Example 1.3 in Barber. Write your code and choose your data
representations so that it is easy to use it to solve the remaining questions. Show that it correctly computes the
value in the example. (5 pts)
3.2 Define a different distribution for . Your new distribution should result in the outcome that
is either or , i.e. reasonably strong evidence. Use the original values of and from the
example. Provide (invent) a reasonble justification for the value of each entry in . (5 pts)
3.3 Derive the equation for . (5 pts)
3.4 Calculate it’s value for both the original and the one you defined yourself. Is it possible to provide a
summary of the main factors that contributed to the value? Why/Why not? Explain. (5 pts)
Q4. Biased views (20 pts)
4.1 Write a program that calculates the posterior distribution of the (probability of heads) from the Binomial
distribution given heads out of trials. Feel to use a package where the necessary distributions are defined as
primitives. (5 pts)
4.2 Imagine three different views on the coin bias:
“I believe strongly that the coin is biased to either mostly heads or mostly tails.”
“I believe strongly that the coin is unbiased”.
“I don’t know anything about the bias of the coin.”
Define and plot prior distributions that expresses each of these beliefs. Provide a brief explanation. (5 pts)
4.3 Perform Bernoulli trials where one of these views is correct. Show how the posterior distribution of changes
for each view for =0, 1, 2, 5, 10, and 100. Each view should have its own plot, but with the curves of the posterior
after different numbers of trials overlayed. (5 pts)
4.4 Is it possible that each view will always arrive at an accurate estimate of ? How might you determine which
view is most consistent with the data after trials? (5 pts)
Q5. Inference using the Poisson distribution (20 pts)
Suppose you observe for 3 seconds and detect a series of events that occur at the following times (in seconds):
0.53, 0.65, 0.91, 1.19, 1.30, 1.33, 1.90, 2.01, 2.48.
5.1 Model the rate at which the events are produced using a Poisson distribution where is the number of events
observed per unit time (1 second). Show the likelihood equation and plot it for three different values of : less,
about equal, and greater than what you estimate (intuitively) from the data. (5 pts)
5.2 Derive the posterior distribution of assuming a Gamma prior (usually defined with parameters and ). The
posterior should have the form where is the total duration of the observation period and is the
number of events observed within that period. (5 pts)
5.3 Show that the Gamma distribution is a conjugate prior for the Poisson distribution, i.e. it is also a Gamma
distribution, but defined by parameters and that are functions of the prior and likelihood parameters. (5 pts)
5.4 Plot the posterior distribution for the data above at times = 0, 0.5, and 1.5. Overlay the curves on a single plot.
Comment how it is possible for your beliefs to change even though no new events have been observed. (5 pts)
Q6. Probability Distribution Example (20 pts)
In this problem you will illustrate a probability distribution in a settings of your choosing. It can be discrete or
continuous. This is meant to be a simpler version of the letter seqeunce example shown in class (so don’t use that).
Your example should use two random variables that each have at least three distinct values (if it is discrete), i.e.
don’t use binary variables. The variables should not be independent, in other words, the setting you are modeling
should have structure, and ideally structure that is interesting and interpretable in some way. Your example should
include the following:
a decription of the setting
an illustration of the joint probability and how it captures the structure
an illustration of a conditional probability
an illustration of marginal probability
Note that “illustration” here means to explain with tables or figures that convey the ideas of the mathematical
operations. The motivation behind this exercise is to help you develop a better understanding of how joint
probability distributions model probabilistic structure in a simplified setting, so try to choose something you are
very familiar with. If find this is getting too long, you can continue it as part of the exploration, but there you will
also need to add and inference problem.
Exploration (40 pts)
In these problems, you are meant to do creative exploration. Define and explore:
E.1 A discrete inference problem (20 pts)
E.2 A continuous inference problem (20 pts)
This is meant to be open-ended; you should not feel the need to write a book chapter; but neither should you just
change the numbers in one of the problems above. After doing the readings and problems above, you should pick a
concept you want to understand better or an simple modeling idea you want to try out. You can also start to explore
ideas for your project. The general idea is for you to teach yourself (and potentially a classate) about a concept
guidance, see the grading rubric below.
You can use the readings and other sources for inspiration, but here are a few ideas:
An inference problem using categorical data
A disease for which there are two different tests
A two-dimensional continuous inference problem
The idea of a conjugate prior
Exploration problems will be graded according the elements in the table below. The scores in the column headers
indicate the number of points possible for each rubric element (given in the rows). A score of zero for an element is
possible if it is missing entirely.
Substandard (+1) Basic (+2) Good (+3) Excellent (+5)
Pedagogical
Value
No clear statement of
idea or concept being
explored or explained;
lack of motivating
questions.
Simple problem with
still could be a useful
assignment.
Good choice of problem with
effective illustrations of
concept(s). Demonstrates a
deeper level of understanding.
Problem also illustrates or
clarifies common conceptual
difficulties or
misconceptions.
Novelty of
Ideas
Copies existing
problem or makes only
a trivial modification;
lack of citation(s) for
source of inspiration.
Concepts are similar to
those covered in the
assignment but with
some modifications of
an existing exericse.
Ideas have clear pedagogical
motivation; creates different
type of problem or exercise to
explore related or foundational
concepts more deeply.
Applies a technique or
explores concept not
covered in the assignment or
not discussed at length in
lecture.
Clarity of
Explanation
Little or confusing
explanation; figures
lack labels or useful
captions; no
explanation of
motivations.
Explanations are
present, but unclear,
unfocused, wordy or
contain too much
technical detail.
Clear and concise explanations
of key ideas and motivations.
Also clear and concise, but
includes illustrative figures;
understood by students from
a variety of backgrounds.
Depth of
Exploration
Content is obvious or
closely imitates
assignment problems.
Uses existing problem
for different data.
Applies a variation of a
technique to solve a problem
with an interesting motivation;
explores a concept in a series of
related problems.
Applies several concepts or
techniques; has clear focus
of inquiry that is approached
from multiple directions.
p(x, y|z) = p(x|z)p(y|x, z)
p(x|y, z) = p(y|x, z)p(x|z)
p(y|z)
a ⊥ b ∧ b ⊥ c ⇏ a ⊥ c
p(a, b, c)
a ⊥ b|c ⇏ a ⊥ b
p(B|K)
p(K|M, B) p(B|K)
< 0.1 > 0.9 p(B) p(M)
p(K|M, B)
p(M|K)
p(K|M, B)
θ
y n
θ
n
θ
n
λ n
λ
λ α β
p(λ|n, T, α, β) T n
α′ β′
T
In [ ]:
EECS 491 Assignment 1
\$30.00
Hello
Can we help?
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0%
# Problem 29
Distinct powers
Consider all integer combinations of ab for 2 ≤ a ≤ 5 and 2 ≤ b ≤ 5:
22=4, 23=8, 24=16, 25=32
32=9, 33=27, 34=81, 35=243
42=16, 43=64, 44=256, 45=1024
52=25, 53=125, 54=625, 55=3125
If they are then placed in numerical order, with any repeats removed, we get the following sequence of 15 distinct terms:
4, 8, 9, 16, 25, 27, 32, 64, 81, 125, 243, 256, 625, 1024, 3125
How many distinct terms are in the sequence generated by ab for 2 ≤ a ≤ 100 and 2 ≤ b ≤ 100?
22=4, 23=8, 24=16, 25=32
32=9, 33=27, 34=81, 35=243
42=16, 43=64, 44=256, 45=1024
52=25, 53=125, 54=625, 55=3125
4, 8, 9, 16, 25, 27, 32, 64, 81, 125, 243, 256, 625, 1024, 3125
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# Decomposing tensor products of modules for the orthogonal/symplectic groups in characteristic zero
I would like to know if there is a perfect analogue of the classical Littlewood-Richardson rule for decomposing tensor products of simple modules for the orthogonal/symplectic groups in characteristic zero, i.e. a "tableaux counting algorithm" which determines the solution.
I know that there are tableaux counting rules due to King and De Concini, there is the Littelmann path algorithm, and the theory of crystal bases and probably many more descriptions besides. There are also results due to Koike and others which express the answer in terms of a sum or products of Littlewood-Richardson coefficients (this isn't really what I'm looking for).
So my question is: is there some summary of how all these different methods coincide? Do any of these methods give "perfect solutions"? I know the Littelmann path approach gives a complete answer, but can it be expressed in terms of something like the classical tableaux counting procedure? Do the King/De Concini rules provide (positive) solutions in all cases? And is there some bible in which all these different approaches are compiled and compared?
Post Jim's comment:
The main question is: under what conditions (probably on $n$) do (any) of the tableaux-based approaches coincide with the Littelmann path approach to answer the problem?
And what is the best reference for these tableaux? I was told to look for
R. C. King: "Weight multiplicities for the classical groups"
but can't get a hold of it. Does anyone have a pdf? If not, does anyone have another good reference? There seems to be a LOT of literature to wade through
-
I'm pretty sure the answer to your final question is no, though it would be helpful to connect all these parts of the literature. Aside from that it would be better to focus more on a single question, preferably highlighted using > at the start of a paragraph. – Jim Humphreys Jul 8 '13 at 20:27
P.S. Concerning the article by R.C. King, it's unlikely that any legal PDF versions are floating around: Weight multiplicities for the classical groups. Group theoretical methods in physics (Fourth Internat. Colloq., Nijmegen, 1975), pp. 490–499. Lecture Notes in Phys., Vol. 50, Springer, Berlin, 1976. Many of his related papers are in J. Phys. A, maybe more accessible online via a library. – Jim Humphreys Jul 9 '13 at 16:00
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# Exploratorium
#### Visit Exploratorium in San Francisco for extra credit.
Write a short report to describe what you see, and the principle behind each experiment listed below (with photos).
Exploratorium
Pier 15 (Embarcadero at Green St)
San Francisco, CA 94111
(415) 528-4444
Exploratorium 100 100 false 84bdd6cd-f8a6-4ee7-9577-55ecb46701e2.JPG feather_coin.JPG 75 100 Feather & Coin: see how they fall in air and in vacuum Feather & Coin: see how they fall in air and in vacuum f0390f8e-dea8-4443-b83d-aec4dae49348.JPG floating_object.JPG 100 75 Floating Ball: why does the ball follow the air blower ? Floating Ball: why does the ball follow the air blower ? f72d78cf-25c3-4665-8410-2cfdb452941b.jpg make_wave.jpg 100 83 Make Waves Find this machine and make "traveling waves" and "standing waves" Make Waves Find this machine and make "traveling waves" and "standing waves" 79d2470c-59da-458c-ad74-aee8c89507de.jpg bicycle_wheel.jpg 100 66 Bicycle Wheel Gyro Tilt the wheel and feel the force turning you around. Bicycle Wheel Gyro Tilt the wheel and feel the force turning you around. ecfa2b92-88b7-40ff-b5ce-419fcf54e32a.jpg virtual_cup.jpg 100 75 Real Image Try to touch the Spring or Cup inside this box. It's NOT there !!! Real Image Try to touch the Spring or Cup inside this box. It's NOT there !!! 4a199487-a18c-4802-b212-6db034545de8.jpg touch the spring.jpg 74 100 d8a660be-a26f-4169-9a6f-f83a2fbb222c.JPG cloud_chamber.JPG 100 75 Cloud Chamber (or bubble chamber) Find this box in the dark corner and observe the trails of bubbles inside. What are they? Cloud Chamber (or bubble chamber) Find this box in the dark corner and observe the trails of bubbles inside. What are they? a2e67deb-0131-4325-a77e-63c8e0a70de4.JPG whisper_sphere.JPG 100 75 Whisper Sphere Find this pair of "disc" and whisper to your friend across the hall. Whisper Sphere Find this pair of "disc" and whisper to your friend across the hall. 9b350f90-297d-4bac-81f6-b31490fd1895.jpg Newtons_rings.jpg 100 74 Newton's Rings FInd this in a dark area. What are the rings? Newton's Rings FInd this in a dark area. What are the rings? 33e5d9f7-e2c4-449a-996e-349aac991130.jpg disc_race.jpg 100 75 Disc racing (or Downhill Race) Let the disc rolls down the rail. Why are some faster than the others even though the masses may be the same? Disc racing (or Downhill Race) Let the disc rolls down the rail. Why are some faster than the others even though the masses may be the same? fb497093-14ee-49c1-9978-8fabefac2dda.jpg concave_mirror.jpg 100 83 Big concave mirror You can't miss this huge mirror by the entrance. As you move closer, your image would go from upside-down to right-side up. What happens when you stand in between these 2 "regions"? Shake hand with yourself !!!!! Big concave mirror You can't miss this huge mirror by the entrance. As you move closer, your image would go from upside-down to right-side up. What happens when you stand in between these 2 "regions"? Shake hand with yourself !!!!! 6aade6d4-ae80-4f32-af18-dfdbf8805efc.jpg anti_gravity.jpg 75 100 Anti-gravity Find this mirror and make yourself "float" in air. Have fun :-)Anti-gravity Find this mirror and make yourself "float" in air. Have fun :-) Anti-gravity Find this mirror and make yourself "float" in air. Have fun :-) ec5d44af-b237-49cf-b066-609eba529e4d.jpg big_soap_bubble.jpg 100 65 Big Soap Bubble Go cover yourself with a big soap bubble. Why do the bubbles have "rainbow"? Big Soap Bubble Go cover yourself with a big soap bubble. Why do the bubbles have "rainbow"? b50251b3-2293-4805-a9e5-1dd95653e9e6.jpg hot_or_cold.jpg 100 75 Hot or Cold Feel the copper tubes and see how our temperature "feeling" is just not so reliable :-) Hot or Cold Feel the copper tubes and see how our temperature "feeling" is just not so reliable :-) aa59dab2-fe35-465d-b07c-5f7e4972a145.jpg illusion.jpg 100 66 Find this magic room. How can a little kid be so much taller than an adult? Have someone take a picture of you in it. Find this magic room. How can a little kid be so much taller than an adult? Have someone take a picture of you in it.
There are so many more exhibits to see and play with. Plan to spend a whole day there. Have fun.
Go with a group of friends. Take lots of photos and enjoy.
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# 319handout6 - X and Y is given by f x;y =& 2 5(2 x 3 y...
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Econ 319, Fall 2008 TA: Simon Kwok Handout 6 1. Recall that the probability function obtained in handout 4, question 4 is f ( x ) = 8 > > > > < > > > > : 0 : 4 if x = ° 2 0 : 1 if x = 0 0 : 3 if x = 1 0 : 2 if x = 4 0 otherwise : Find (a) E ( X ) ; (b) V ar ( X ) : 2. Recall that the probability density function (pdf) obtained in handout 4, question 3(b) is f ( x ) = ° 8 9 (2 x ° x 2 ) if 0 < x < 3 2 0 otherwise. : Find (a) E ( X ) ; (b) V ar ( X ) : 3. For the continuous uniform distribution of X with pdf f ( x ) = ° 1 ° ° ± if ° < x < ± 0 otherwise. , prove that (a) E ( X ) = ± + ° 2 : (b) V ar ( X ) = ( ° ° ± ) 2 12 : 4. Recall from handout 5, question 1 that the pdf of the joint distribution of
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Unformatted text preview: X and Y is given by f ( x;y ) = & 2 5 (2 x + 3 y ) ; ± x;y ± 1 ; otherwise : Find (a) E ( X ) and E ( Y ) ; (b) V ar ( X ) and V ar ( Y ) ; (c) Cov ( X; Y ) . 5. Let X be a random variable that is uniformly distributed on the interval [-1,1]. Find (a) E ( X ) ; (b) E ( X 2 ) ; (c) V ar ( X 2 ) ; (d) E ( j X j ) : 1...
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Channel Coding
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# Channel Coding - PowerPoint PPT Presentation
Channel Coding. Binit Mohanty Ketan Rajawat. Recap…. Information is transmitted through channels (eg. Wires, optical fibres and even air) Channels are noisy and we do not receive what was transmitted. System Model. A Binary Symmetric Channel Crossover with probability p.
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## Channel Coding
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### Channel Coding
Binit Mohanty
Ketan Rajawat
Recap…
• Information is transmitted through channels (eg. Wires, optical fibres and even air)
• Channels are noisy and we do not receive what was transmitted
System Model
• A Binary Symmetric Channel
• Crossover with probability p
Repetition Coding
• Assume 1/3 repetition
• What is the probability of
error ?
• If crossover probability p = 0.01, Pe ≈ 0.0003
• Here coding rate R = 1/3. Can we do better? How much better?
Shannon’s Theorem
• Given,
• A noisy channel (some fixed p)
• A value of Pe which we want to achieve
“We can transmit through the channel and achieve this probability of error at a maximum coding rate of C(p)”
• Is it counterintuitive?
• Do such good codes exist?
Channel Capacity
• C(p) is called the channel capacity
• For binary symmetric channel,
• Can we really design codes that achieve this rate? How?
Parity Check Codes
• #information bits transmitted = k
• #bits actually transmitted = n = k+1
• Code Rate R = k/n = k/(k+1)
• Error detecting capability = 1
• Error correcting capability = 0
1 0 0 1 0 0
0 1 0 0 0 1
1 0 0 1 0 0
1 1 0 1 1 0
1 0 0 1 11
Last column consists of check bits for each row
Bottom row consists of check bit for each column
2-D Parity Check
• Rate?
• Error detecting
capability?
• Error correcting
capability?
Linear Block Codes
• #parity bits n-k (=1 for Parity Check)
• Message m = {m1 m2 … mk}
• Transmitted Codeword c = {c1 c2 … cn}
• A generator matrix Gkxn
• What is G for repetition code?
• For parity check code?
Linear Block Codes
• Linearity
• Example : 4/7 Hamming Code
• k = 4, n = 7
• 4 message bits at (3,5,6,7)
• 3 parity bits at (1,2,4)
• Error correcting capability =1
• Error detecting capability = 2
• What is G?
Cyclic codes
• Special case of Linear Block Codes
• Cyclic shift of a codeword is also a codeword
• Easy to encode and decode,
• Can correct continuous bursts of errors
• CRC (used in Wireless LANs), BCH codes, Hamming Codes, Reed Solomon Codes (used in CDs)
Convolutional Codes
• Block codes require a buffer
• What if data is available serially bit by bit? Convolutional Codes
• Example
k = 1
n = 2
Rate R = ½
Convolutional Codes
• Encoder consists of shift registers forming a finite state machine
• Decoding is also simple – Viterbi Decoder which works by tracking these states
• First used by NASA in the voyager space programme
• Extensively used in coding speech data in mobile phones
Achieving Capacity
• Do Block codes and Convolutional codes achieve Shannon Capacity?
Actually they are far away
• Achieving Capacity requires large k (block lengths)
• Decoder complexity for both codes increases exponentially with k – not feasible to implement
Turbo Codes
• Proposed by
Berrou & Glavieux
in 1993
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# How to Calculate the Power Spectral Density (PSD) for Vibration Analysis
Real-world vibration data typically consists of a broad range of different frequencies that aren't obvious in the time domain. So engineers turn to the power spectral density (PSD) to represent a signal in the frequency domain which has the benefits over simpler Fourier transforms (FFT) because the results are independent of time duration, sample rate, or frequency bin width.
But to understand why power spectral densities are so effective, we first need to understand how to calculate a PSD! In this blog I'll show you how to calculate this using enDAQ's open-source Python library. But even if you don't plan to use Python, you'll still learn in this blog what goes into the underlying calculation of a PSD and how to use it to analyze your vibration data. We'll look at some artificially generated data, then real-world vibration data. The plots we'll include will be interactive too!
In this blog, I will present:
## Video and Demonstration in Python
Follow along with all the calculations in the video below and/or in this Google Colab that contains all the source code and interactive plots!
## What is the Power Spectral Density (PSD) Function?
Vibration data as a function of time, time-domain data, can give information about the relative intensity of vibration, but time-domain data does not provide any information on the frequency or frequencies where the energy is concentrated. For frequency domain information, we need the power spectral density, PSD, which describes the distribution of the power of the time series data into frequency components.
The mathematical definition of the power spectral density function, XPSD(f), uses the discrete Fourier transform, X(f), of the time domain data, x(t), and is defined as:
For vibration analysis, the PSD represents acceleration, and the units of the PSD are g2/Hz in which g denotes the g-force. The factor, ½, converts the computation to a rms value. The Δf term is the inverse of the measurement time and equals 1/T.
The plot of the PSD function shows the frequencies where the vibration energy is concentrated. For more of a technical discussion on a PSD function, visit our learn article with a handbook download available.
## Demonstration with three signals
Let's start by generating some data! Figure 1 shows the time-domain display of three signals. One signal is white noise in the frequency band between 10 and 50 Hz. Added to the noise are two sine tones, one is a 30 Hz sine wave and the second is a sine wave with a frequency of 80.25 Hz. The time-domain display does not provide much useful information other than the peak which is about the same for all! Remember, go ahead and zoom around in the plot!
Figure 1. Time-domain display of a combination of noise between 10 and 50 Hz, a 30 Hz sine tone, and a 80.25 Hz sine tone
Let’s see what we can learn from the frequency domain. This is where the PSD can help. Using our open-source library obtaining a PSD is quite simple with the function endaq.calc.psd.welch():
• Provide as input the time domain data indexed by time
• Select a frequency bin width
• Then plot the resulting power spectral density.
Figure 2. PSD plot of the 10 to 50 Hz noise, the 30 Hz sine tone, and the 80.25 Hz sine tone
Figure 2 shows the PSD plot with a 1 Hz bin width. The broadband white noise is displayed as a mostly flat line between 10 and 50 Hz. The plot shows two peaks, one at 30 Hz and one at 80 Hz representing the 30 Hz and 80.25 Hz sine tones. The 30 Hz tone falls in the center of the 30 Hz bin so all the power at 30 Hz is captured in that bin. The 80.25 sine tone is not in the center of the bin, and some of the power leaks into the adjacent bin. As a result, the peak of the power at 80.25 Hz is lower than the power at 30 Hz.
Although I am talking about a power spectral density, the Y-axis defines an acceleration density in g2/Hz. Irrespective of the units, the important point is the PSD curve clearly shows the frequency content of the vibration data; and frequencies with high vibration content can be identified. A mechanical engineer can then relate high acceleration density to mechanical issues in a machine.
Now let's demonstrate the effectiveness of the PSD by comparing different time durations and bin widths all in an interactive plot!
Figure 3. PSD plots as a function of three different bin widths and time durations.
Figure 3 illustrates the PSDs of the same data using bin widths of 0.5 Hz, 1 Hz, and 2 Hz. The plots in the first column on the left are the 30 - 50 Hz noise. The multiple color waveforms are the Fourier transforms of individual segments of the time domain data. The orange line represents the PSD of the total data set. The second column plots are the 30 Hz and 80.25 Hz sine waves. The third column combines the noise and the sine waves. Note that as the bin widths become wider, the acceleration density peaks become lower. There is less density in a wider bin width so the magnitude of the density decreases with an increasing bin width. Increasing the bin width has the effect of averaging the spectral density.
From the noise data, which is representative of typical vibration data, you can see that the magnitude of the PSD stays fairly constant independent of the bin width or time duration!
## Nuts and Bolts of the PSD Calculation
Here is how the power spectral density is calculated under the hood from your inputs.
1. It breaks up the data into time segments consisting of a manageable amount of data for each segment. The time segments can overlap. Figure 4 shows a set of overlapping time segments. A fast Fourier transform (FFT) computation requires 2N time domain samples to obtain proper results. The sample count, 2N, determines how long the FFT takes to perform the computation. The samples must be taken at a sampling rate that is at least twice the frequency of the highest frequency component of the data. This sampling rate satisfies the Nyquist sampling theorem and accurately represents the signal in the frequency domain.
2. The program applies a Hanning window filter to each time segment so that each segment begins and ends with 0 magnitudes (Figure 5). Applying a Hanning filter to the time domain segments eliminates discontinuities and spikes in the Fourier transform.
3. The program computes the Fourier transform on each of the segments using the fast Fourier transform (FFT) technique.
4. It squares the Fourier transforms and normalizes the computation to the frequency bin width so that that length of the segment now no longer matters
5. It averages the squared, normalized transforms and then you have a PSD!
But seeing a list is a bit dry, let's demonstrate it! First, we'll combine the first two steps to show what it looks like after the signal is broken into overlapping segments and windowed. I've included both the interactive and still images here.
Figure 4. The time signal is broken into overlapping segments with a Hanning window applied.
Then a FFT is computed for each segment shown here in what's called a waterfall plot that was generated with the endaq.plot.spectrum_over_time() function, I love these!
Figure 5. Waterfall display showing Fast Fourier Transforms of 19 time segments
Next, these are all squared and normalized to the bin width and finally averaged as shown in Figure 6.
Figure 6. Individual PSDs of each segment vs frequency with the average too
And that's how a PSD is calculated! It all comes down to segmenting a file, computing FFTs, normalizing them and then averaging it all together!
## A Real-World Example
Now let's look at the real-world example of a train passing by on a bridge (discussed too in this blog). Figures 7 and 8 provide the time series overview.
Figure 7. Time-domain display showing three axes of vibration data from a bridge when a train is passing over
Figure 8. Expanded view of 100ms of time-domain vibration data
Figure 7 shows time-domain vibration data along the x-, y-, and z-axes. of a train passing over a bridge. The data consists of 10 seconds of 100,000 samples/s or a total of 1,000,000 samples, then Figure 8 shows a middle 100 ms of data or 10,000 samples. In the time domain, it is very difficult to see what frequency ranges are being excited!
Now let's determine the PSD of the three axes of acceleration by passing the time domain data to the PSD calculation function. In addition to the data, I will designate a 1 Hz bin width. Figure 9 presents the PSDs of the three orthogonal axes, both the still, and interactive image are provided!
Figure 9. X-axis, Y-axis, and Z-axis PSDs of the bridge vibration
If we examine the y-axis PSD. (Figure 9, remember zoom around!), we see a peak around 23 Hz which could be a fundamental mode of the bridge. Secondary peaks at other frequencies could be other resonances but they quickly become pretty broad, they aren't isolated to a single frequency. Using this PSD plot in a vibration simulation program can create a high level of complexity in the model.
## Octave Spaced PSD
Our open-source software has an octave bin calculation program to widen the bin widths for the PSD, endaq.calc.psd.to_octave(). The program requires the starting frequency and the number of bins/octaves. Figure 10 shows the results of an octave PSD using a coarse 1 bin/octave.
Figure 10. The bridge PSD converted to 1/1 bins per octave spacing.
Determining what is the right number of bins per octave is up to the analyst. But this is one of the benefits of the PSD, it is a density so even as we change this bin width the value of the y-axis in the plots is held relatively constant. Let's compare a few different options in Figure 11. If I were the analyst I may want to use 6 (or maybe 3) but the 1/1 bins per octave gives a nice clean profile that is easy for a simulation software - but it's up to you!
Figure 11. PSDs with a different number of bins per octave.
## Conclusion
Hopefully, now you can see how powerful the PSD information is for vibration analysis and understand how this calculation is done. Our open-source Python library simplifies the task of calculating the PSD and identifying frequencies where vibration magnitudes can be problematic. The PSD is an indispensable tool for vibration analysis. For further information on analyzing vibration data in the frequency domain, subscribe to our library of videos on the subject. Feel free to take advantage of our open-source Python library, it is free!
Please note: This post is the first of a four-post series. Be sure to subscribe to our blog if you'd like to receive the upcoming posts in your inbox.
1. Calculate a PSD with Python
2. Calculating a FFT from a PSD
3. Comparing the FFT to PSD
4. Time Varying Frequency Analysis
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#### Question Details
##### [solution] » An occupational health and safety survey of 197 industrial accent
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Item details:
An occupational health and safety survey of 197 industrial accent
More:
An occupational health and safety survey of 197 industrial accent claims have revealed that 63 were due to untidy working conditions and thus preventable.
Part A) Calculate a 95% confidence interval for the proportion of accidents that are due to untidy working conditions.
Answer: We are 95% confident that the true population proportion of preventable accidents is between Answer and Answer.
Part B) What sample size is needed if the survey company from the previous question wants to be 95% confident of being correct to within + or - 0.07 of the population proportion of accidents that are due to untidy working conditions? Assume that no information is available from past data, thus the population proportion equals 0.50.
Part C) How would the sample size change if we were to assume our sample proportion discussed above is equal to the population proportion?
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This question was answered on: Feb 21, 2020
Solution~00066571140.zip (18.37 KB)
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Math > Add and Subtract Tenths less than 100 without Regrouping: Missing Digits Worksheet
Add and Subtract Tenths less than 100 without Regrouping: Missing Digits Worksheet
Know more about Add and Subtract Tenths less than 100 without Regrouping: Missing Digits Worksheet
Children practice adding and subtracting tenths less than 100 without regrouping with this worksheet. While adding and subtracting decimals, students use the relationship between addition and subtraction to find the missing number. They do not need to regroup numbers in add and subtract tenths less than 100 without regrouping worksheet. In each problem, the numbers are laid out in the vertical format. In this format, pair of digits in each successive place are tackled one by one. This helps students follow a structured approach.
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# Good Graph vs. Bad Graph – Part 2 – Linear Applications
Suppose you purchase a washing machine for \$500 and it depreciates \$100 a year. If t represents time in years then the function V(t) = -100t + 500 represents the value of your machine over time.
This function has a restricted domain (usually called a PRACTICAL DOMAIN) because there are limits on what the input can be. The inputs can start at 0 (the year you purchased the machine) and go to 5 (because at 5 years the machine is worth \$0). So, in mathematical language, we say that 0 ≤ t ≤ 5 is the practical domain of function V(t).
The function also has a restricted range (usually called a PRACTICAL RANGE) and it is closely related to the restricted domain. If t = 0, then V(0) = 500 which is the purchase price of the washer (i.e. the initial value). If t = 5, then V(5) = 0 which is the value of the washer after 5 years. So, in mathematical language, we say that 0 ≤ V(t) ≤ 500 is the practical range of the function V(t). [Notice that we list the numbers for the range in numerical order even though the biggest one, 500, is the first one we discovered].
Now let’s look at a couple of graphs for V(t). The first one is a bad graph and the second one is a good graph. Below the second graph is a checklist of items to include when creating a good graph for linear applications. There are some important differences between a good graph for linear applications and a good graph for linear equations. Compare the checklist here to the checklist in Part 1 to be sure you understand what those differences are.
Checklist for a Good Graph – Linear Applications
1. Label the horizontal and vertical axes with the problem variables.
2. Identify tick marks. In this graph, every two tick marks are a distance of 1 (horizontal) or 100 (vertical).
3. Plot and label at least two ordered pairs from the function. The most common points to plot for applications of this type are the vertical and horizontal intercepts.
4. Use a straightedge to connect the plotted points and DO NOT run the line past the starting and ending points. Applications usually involve a restricted domain and the plotted beginning and end points are the first and last points on the graph.
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# Communication Between Graphs#
All graph nodes require a reference to a graph object as one of the arguments to their constructor. It is only safe to construct edges between nodes that are part of the same graph. An edge expresses the topology of your graph to the runtime library. Connecting two nodes in different graphs can make it difficult to reason about whole graph operations, such as calls to graph::wait_for_all and exception handling. To optimize performance, the library may make calls to a node’s predecessor or successor at times that are unexpected by the user.
If two graphs must communicate, do NOT create an edge between them, but instead use explicit calls to try_put. This will prevent the runtime library from making any assumptions about the relationship of the two nodes, and therefore make it easier to reason about events that cross the graph boundaries. However, it may still be difficult to reason about whole graph operations. For example, consider the graphs below:
```graph g;
function_node< int, int > n1( g, 1, [](int i) -> int {
cout << "n1\n";
spin_for(i);
return i;
} );
function_node< int, int > n2( g, 1, [](int i) -> int {
cout << "n2\n";
spin_for(i);
return i;
} );
make_edge( n1, n2 );
graph g2;
function_node< int, int > m1( g2, 1, [](int i) -> int {
cout << "m1\n";
spin_for(i);
return i;
} );
function_node< int, int > m2( g2, 1, [&](int i) -> int {
cout << "m2\n";
spin_for(i);
n1.try_put(i);
return i;
} );
make_edge( m1, m2 );
m1.try_put( 1 );
// The following call returns immediately:
g.wait_for_all();
// The following call returns after m1 & m2
g2.wait_for_all();
// we reach here before n1 & n2 are finished
// even though wait_for_all was called on both graphs
```
In the example above, m1.try_put(1) sends a message to node m1, which runs its body and then sends a message to node m2. Next, node m2 runs its body and sends a message to n1 using an explicit try_put. In turn, n1 runs its body and sends a message to n2. The runtime library does not consider m2 to be a predecessor of n1 since no edge exists.
If you want to wait until all of the tasks spawned by these graphs are done, you need to call the function wait_for_all on both graphs. However, because there is cross-graph communication, the order of the calls is important. In the (incorrect) code segment above, the first call to g.wait_for_all() returns immediately because there are no tasks yet active in g; the only tasks that have been spawned by then belong to g2. The call to g2.wait_for_all returns after both m1 and m2 are done, since they belong to g2; the call does not however wait for n1 and n2, since they belong to g. The end of this code segment is therefore reached before n1 and n2 are done.
If the calls to wait_for_all are swapped, the code works as expected:
```g2.wait_for_all();
g.wait_for_all();
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# What is the Hamilton operator for hydrogen?
## Hamilton operator
The Hamilton operator\$ \ has H \$ determined in quantum mechanics the time evolution and the possible energy measurements of the associated physical system, for example the electron in the hydrogen atom. It is named after William Rowan Hamilton, to whom the Hamiltonian formulation of mechanics goes back, in which the Hamilton function determines the time evolution and the energy.
### Time development and energy
In quantum mechanics, every state of the physical system under consideration is given by an associated vector \$ \ psi \$ in the Hilbert space. Its time evolution is determined according to the Schrödinger equation by the Hamilton operator \$ \ hat H \$:
\$ \ mathrm i \, \ hbar {\ partial \ over \ partial t} \, \ psi (t) = \ hat H \, \ psi (t) \$
With
The Hamilton operator is obtained in many cases by what are known as canonical quantization from the Hamilton function \$ \ mathcal H (t, x, p) \$ of the corresponding classical system (with the generalized coordinate x and the canonical momentum p). For this purpose, the algebraic expression for the Hamilton function is read as a function of operators (position operator \$ \ hat x \$ and momentum operator \$ \ hat p \$) that satisfy the canonical commutation relations. However, this is not unique because the function \$ x \, p -p \, x \$ has the value \$ 0 \$, the operator function \$ \ has x \, \ has p - \ has p \, \ has but the Value \$ \ mathrm i \ hable. \$ In addition, \$ x \, p \$ is real, but \$ \ has x \, \ has p \$ not Hermitian. There are also quantum mechanical quantities such as spin that do not appear in classical physics. How they affect the development of time does not follow from analogies with classical physics, but has to be deduced from the physical findings.
The eigenvalue equation
\$ \ has H \ varphi_E = E \ varphi_E \$
determines the eigenvectors \$ \ varphi_E \$ of the Hamilton operator. If the Hamilton operator is independent of time, they are stationary, i. H. independent of time in every observable property. The eigenvalues \$ E \$ are the associated energies.
Since the Hamilton operator is Hermitian (more precisely, essentially self-adjoint), the spectral theorem says that the energies are real and that the eigenvectors form an orthonormal basis of the Hilbert space. Depending on the system, the energy spectrum can be discrete or continuous. Some systems, for example the hydrogen atom or a particle in the potential well, have a discrete spectrum that is limited downwards and above that a continuum of possible energies.
The Hamilton operator generates the unitary time development. If for all times \$ \ tau \$ and \$ \ tau '\$ between \$ t_0 \$ and \$ t \$ the Hamilton operator \$ H (\ tau) \$ commutes with \$ H (\ tau') \$, then causes
\$ \ hat U (t, t_0) = \ exp \ left (- \ frac {\ mathrm i} {\ hbar} \ int_ {t_0} ^ t \ hat H (\ tau) \, \ mathrm d \ tau \ right ) \$
the unitary mapping of each initial state \$ \ psi (t_0) \$ to the associated state \$ \ psi (t) = U (t, t_0) \ psi (t_0) \$ at time \$ t. \$
If the Hamilton operator does not depend on time, this simplifies to
\$ \ hat U (t, t_0) = \ exp \ left (- \ frac {\ mathrm i} {\ hbar} (t - t_0) \ hat H \ right). \$
Operators that swap with \$ \ hat H \$ are conserved quantities of the system in the case of a time-independent Hamilton operator. In particular, the energy is then a conserved quantity.
### Quantum mechanical particle in potential
From the Hamilton function
\$ \ mathcal {H} \ left ({\ mathbf {x}}, {\ mathbf {p}} \ right) = \ frac {{\ mathbf {p}} ^ 2} {2 \, m} + V ( {\ mathbf {x}}) \$
for a non-relativistic, classical particle of mass \$ m \$, which moves in the potential \$ V (\ mathbf x) \$, a Hamilton operator can be read off. To do this, the expressions for the momentum and the potential are replaced by the corresponding operators:
\$ \ hat {H} (\ hat {\ mathbf {x}}, \ hat {\ mathbf {p}}) = \ frac {\ hat {\ mathbf {p}} ^ 2} {2 \, m} + V (\ hat {\ mathbf {x}}). \$
In the position representation, the momentum operator \$ \ hat {\ mathbf {p}} \$ acts as a derivative \$ - \ mathrm i \ hbar \ tfrac {\ partial} {\ partial \ mathbf {x}} \$ and the operator \$ V (\ hat {\ mathbf {x}}) \$ as multiplication with the function \$ V (\ mathbf {x}). \$ The application of this Hamilton operator of a point particle of mass \$ m \$ in the potential \$ V (\ mathbf {x}) \$ to the spatial wave function \$ \ Psi \$ of the particle has an effect
\$ \ Rightarrow \ hat {H} \ Psi (\ mathbf x) = \ Bigl (- \ frac {\ hbar ^ 2} {2 \, m} \ Delta + V (\ mathbf {x}) \ Bigr) \ Psi (\ mathbf x). \$
Here is \$ \ Delta = \ tfrac {\ partial ^ 2} {\ partial x ^ 2} + \ tfrac {\ partial ^ 2} {\ partial y ^ 2} + \ tfrac {\ partial ^ 2} {\ partial z ^ 2} \$ the Laplace operator.
The Schrödinger equation is thus
\$ \ mathrm i \, \ hbar \, \ frac {\ partial} {\ partial t} \ Psi (t, \ mathbf x) = - \ frac {\ hbar ^ 2} {2 \, m} \ Delta \ Psi (t, \ mathbf x) + V (\ mathbf x) \ cdot \ Psi (t, \ mathbf x). \$
This Schrödinger equation of a point mass in the potential is the basis for explaining the tunnel effect. At the onset of the Coulomb potential (as the potential for the interaction between an electron and a proton), it supplies the spectral lines of the hydrogen atom. By using appropriate potentials, the spectral lines of other light atoms can also be calculated.
### One-dimensional harmonic oscillator
Analogously, one obtains the Hamilton operator for the quantum mechanical harmonic oscillator, which can only move along a line
\$ \ hat H = - \ frac {\ hbar ^ 2} {2m} \ frac {\ partial ^ 2} {\ partial x ^ 2} + \ frac {1} {2} m \, \ omega ^ 2 \, x ^ 2. \$
The energies can be determined algebraically. You get
\$ E_n = E_0 + n \, \ hbar \ omega, \ quad n \ in \ {0,1,2, \ dots \}. \$
These are the same energies as those of a ground state with energy \$ E_0 \$, to which a quantum of the energy \$ \ hbar \, \ omega \$ was added \$ n \$ times.
### Spin in a magnetic field
The Hamilton operator belongs to the spin \$ \ mathbf S \$ of an electron that is bound to an atom and is in an unpaired state (only in the electron cloud) in the magnetic field \$ \ mathbf B \$
\$ \ hat H = - \ gamma \ frac {\ mathbf S} {\ hbar} \ cdot \ mathbf B. \$
It is
Since the spin in the direction of the magnetic field can only assume the eigenvalues \$ \ hbar / 2 \$ or \$ - \ hbar / 2 \$ (spin polarization), the possible energies are \$ \ pm \ frac {\ gamma} {2} \, | \ mathbf B | \$. In the inhomogeneous magnetic field of the Stern-Gerlach experiment, a particle beam of silver atoms splits into two partial beams.
### Charged, spinless particle in an electromagnetic field
The Hamilton operator of a particle with charge \$ q \$ in an external electromagnetic field is obtained by minimal substitution
\$ \ hat {H} = \ frac {1} {2m} \ bigl (\ hat {\ mathbf {p}} - q \ mathbf \, {A} (t, \ hat {\ mathbf {x}}) \ bigr) ^ 2 + q \, \ varphi (t, \ hat {\ mathbf {x}}). \$
Marked here
• \$ \ mathbf {A} (t, \ hat {\ mathbf {x}}) \$ is the vector potential
• \$ \ varphi (t, \ has {\ mathbf {x}}) \$ the scalar potential.
When multiplying the brackets, note that the operators \$ \ hat {\ mathbf {p}} \$ and \$ \ mathbf {A} (t, \ hat {\ mathbf {x}}) \$ only swap for Coulomb calibration.
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# A puzzler in induction and special relativity
Some reading I was doing jogged my memory about a puzzle in E&M that hit me back in my undergrad days, but I just let it go at the time and never found an answer.
A pretty common undergrad E&M problem goes something like, "We manage, by way of external coils, to produce a magnetic field going through the center of a wire loop. The field occupies and area of A within the loop and we make it change at x Tesla/sec. What is the EMF induced around the loop?"
And of course we solve this with Faraday's equation
$\displaystyle \nabla \times E = -\frac{\partial B}{\partial t}$
Integrate it over the surface contained by our wire loop and get
$\displaystyle EMF = -\frac{d \Phi}{dt} = -A\frac{dB}{dt}$
Straightforward. Except ....
The EMF, according to this equation, is directly related to the instantaneous rate of change of $B$. And this is true even if $B$ is confined to a very small area within the current loop. It is true even if, for instance, $B$ is confined to, say, one cm square, and the current loop is one million miles in radius. Again, the relationship is instantaneous. So, if we turn on our coils and generate $B$, immediately there is an EMF 1 million miles away. But information cannot travel faster than the speed of light, so this makes no sense. Where have we gone wrong?
I am pretty sure the answer to this question has something to do with the "slipperiness" of the concept of simultaneity in special relativity. So taking an integral across the entire area of the current loop "simultaneously" is probably a tricky thing to do. So maybe the question to be asked is, given the strictures of special relativity, can we cast the Maxwell's equations, particularly induction, in meaningful integral forms, and what do those forms look like?
• Any change to the field will propegate at the speed of light. The instantaneous change is OK as an approximation. Take a look at an article that describes Purcell's treatment of electric and magnetic fields based on Relativity. physics.weber.edu/schroeder/mrr/MRRtalk.html – Peter R Jun 24 '16 at 18:02
• I understand that the equation gives good results compared to what can be measured in the scale of a laboratory. I am wondering about what the equations say in principle and how we could write integral forms that would be suitable not just to lab scales, but to arbitrarily large scales. Thanks. I will take a look at the link. Sounds interesting. – bob.sacamento Jun 24 '16 at 18:05
• The induced EMF formula only holds in the quasi-static approximation, i.e. for slowly changing fields, which for any given rate of change can always be made correct for sufficiently small surface elements. For finite geometries the precision of the approximation depends on the chosen time scale. When it doesn't hold we have to solve Maxwell's equations in full, which, to use a euphemism, is "hard". – CuriousOne Jun 24 '16 at 18:45
• @CuriousOne, which induced EMF formula only holds in quasistatic approximations? All formulae in the question are equivalent to Faraday's law, i.e. they have no known violation. – Ján Lalinský Jan 5 at 2:53
Suppose, as you said, we have a thin solenoid pointing upward, going through the middle of a loop one light-year wide. Then the three equations $$\nabla \times E = - \frac{\partial B}{\partial t}, \quad \mathcal{E} = -\frac{d\Phi_B}{dt}, \quad B = \mu_0 n I$$ together imply violation of causality, as shown in your question.
The mistake is in equation (3). This is the field of a solenoid that has been on for a long time, but you want us to consider the case where it's "suddenly" switched on. If we switch the solenoid on quickly, $\partial B / \partial t$ is large, so we suddenly get a large electric field circulating just outside the solenoid. But then we have to account for the other induction effect, $$\nabla \times B \sim \frac{\partial E}{\partial t}.$$ If you draw a picture, you'll see this produces both an upward and downward additional magnetic field. A detailed calculation shows that, if you take a light-year-wide loop, the total magnetic flux is zero! Then $\mathcal{E}$ is zero, as expected by causality.
This additional magnetic field will induce additional electric fields, which induce more magnetic fields, so that after a year, the fields propagate to the loop and $\mathcal{E}$ becomes nonzero. You might already know this propagation mechanism by a different name; it's called light.
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# MIS 562 Week 7 Homework Roles and Privileges
\$ 15
MIS 562 Week 7 Homework Roles and Privileges
Part 1
Using the following Data Dictionary views write the statements that will perform the following actions. Be sure to test your statements. (Do not use SELECT *)
ROLE_ROLE_PRIVS
ROLE_SYS_PRIVS
ROLE_TAB_PRIVS
USER_ROLE_PRIVS
USER_SYS_PRIVS
USER_TAB_PRIVS
1. Determine what privileges your account has been granted through a role. (10 points)
2. Determine what system privileges your account has been granted. (10 points)
3.Execute the following statement then determine what table privileges your account has been granted. (15 points)
Grant select on student to public;
4. Determine what system privileges the DVONLINE role has. (10 points)
5. Analyze the following query and write a description of the output it produces. (15 points)
SELECT COUNT(DECODE(SIGN(total_capacity-20), -1, 1, 0, 1)) “<=20”,
COUNT(DECODE(SIGN(total_capacity-21), 0, 1, -1, NULL,
DECODE(SIGN(total_capacity-30), -1, 1)))”21-30″,
COUNT(DECODE(SIGN(total_capacity-30), 1, 1)) “31+”
FROM
(SELECT SUM(capacity) total_capacity, course_no
FROM section
GROUP BY course_no)
6. Determine the top three zip codes where most of the students live. Use an analytical function. The query will product 10 rows. (10 points)
Part 2
Analyze the file from Doc Share called utlpwdmg.sql and analyze the code in this file. Write a paragraph that describes what the function performs. What are the inputs parameters, the output parameter and what does the function do? (30 points)
51 in stock
SKU: MIS562WEEK7 Categories: ,
## Description
MIS 562 Week 7 Homework Roles and Privileges
Part 1
Using the following Data Dictionary views write the statements that will perform the following actions. Be sure to test your statements. (Do not use SELECT *)
ROLE_ROLE_PRIVS
ROLE_SYS_PRIVS
ROLE_TAB_PRIVS
USER_ROLE_PRIVS
USER_SYS_PRIVS
USER_TAB_PRIVS
1. Determine what privileges your account has been granted through a role. (10 points)
2. Determine what system privileges your account has been granted. (10 points)
3.Execute the following statement then determine what table privileges your account has been granted. (15 points)
Grant select on student to public;
4. Determine what system privileges the DVONLINE role has. (10 points)
5. Analyze the following query and write a description of the output it produces. (15 points)
SELECT COUNT(DECODE(SIGN(total_capacity-20), -1, 1, 0, 1)) “<=20",
COUNT(DECODE(SIGN(total_capacity-21), 0, 1, -1, NULL,
DECODE(SIGN(total_capacity-30), -1, 1)))”21-30″,
COUNT(DECODE(SIGN(total_capacity-30), 1, 1)) “31+”
FROM
(SELECT SUM(capacity) total_capacity, course_no
FROM section
GROUP BY course_no)
6. Determine the top three zip codes where most of the students live. Use an analytical function. The query will product 10 rows. (10 points)
Part 2
Analyze the file from Doc Share called utlpwdmg.sql and analyze the code in this file. Write a paragraph that describes what the function performs. What are the inputs parameters, the output parameter and what does the function do? (30 points)
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Books Books
O is equally distant from A, B and C, since it lies in the perpendiculars to the middle points of AB and A C. Hence the circumference will pass through A, B and C. PROBLEM 14. To inscribe a Circle in a given Triangle. Draw the triangle LMN of any convenient...
Mechanical Drawing: Instruction Paper - Page 62
by Ervin Kenison - 1916 - 71 pages
## A Textbook on Surveying and Mapping ...: Geometrical drawing, tables and ...
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## Cyclopedia of Engineering; a Complete Manual of Steam and Machine ..., Volume 4
1902 - 486 pages
...equal to the distance OA, describe the circumference passing through A, B and C. Proof. The point O is equally distant from A, B and C, since it lies...triangle LMN of any convenient size. MN may be made 3^ inches, LM, 2| inches, and LN, 3* inches. Bisect the angles MLN and LM N. The bisectors MI and LJ...
## Mechanical drawing
American School (Chicago, Ill.) - Engineering - 1903 - 414 pages
...equal to the distance OA, describe the circumference passing through A, B and C. Proof. The point O is equally distant from A, B and C, since it lies...A C. Hence the circumference will pass through A, В and C. PROBLEM 14. To inscribe a Circle in a given Triangle. Draw the triangle LMN of any convenient...
## Cyclopedia of Drawing, Volume 1
American School (Lansing, Ill.) - Architectural drawing - 1906 - 426 pages
...equal to the distance OA, describe the circumference passing through A, B and C. Proof. The point O is equally distant from A, B and C, since it lies...triangle LMN of any convenient size. MN may be made 3-^ indies, LM, 1% inches, and LN, 3| inches. Bisect the angles MLN and LM N. The bisectors MI and...
## Modern Engineering Practice: A Reference Library..., Volume 11
American School (Chicago, Ill.) - Engineering - 1906 - 580 pages
...equal to the distance OA, describe the circumference passing through A, B and C. Proof. The point O is equally distant from A, B and C, since it lies...the perpendiculars to the middle points of AB and 79 A C. Hence the circumference will pass through A, B and C. PROBLEM 14. To inscribe a Circle in a...
## Cyclopedia of Drawing: A General Reference Work on Drawing and ..., Part 1
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...equal to the distance OA, describe the circumference passing through A, B and C. Proof. The point O is equally distant from A, B and C, since it lies...the perpendiculars to the middle points of AB and 79 A C. Hence the circumference will pass through A, B and C. PROBLEM 14. To inscribe a Circle in a...
## Modern Engineering Practice: A Reference Library..., Volume 11
American School (Chicago, Ill.) - Engineering - 1906 - 586 pages
...equal to the distance OA, describe the circumference passing through A, B and C. Proof. The point O is equally distant from A, B and C, since it lies in the perpendiculars to the middle points of AB und A C. Hence the circumference will pass through A, B and C. PROBLEM 14. To inscribe a Circle in...
## Arithmetic: Elements of Algebra. Logarithms. Geometrical Drawing
International Correspondence Schools - Arithmetic - 1906 - 578 pages
...£t draw lines intersecting at O. With О as a center, and О A as a. radius, describe a circle; it will pass through A, B, and C. PROBLEM 14. — To inscribe a square In a given circle. CONSTRUCTION. — In Fig. 47, the circle A BCD is Ц" in diameter. Draw two...
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# Show $f$ has exactly one fixed point if $f'(x) <1$ for all $x$
I'm trying to use the mean value theorem to show that for differentiable $f:\mathbb{R} \to \mathbb{R}$, if $f'(x) < 1 \forall x\in\mathbb{R}$, then $f$ has exactly one fixed point. I know how to show that if $f'(x)\neq 1$ then $f$ can have at most one fixed point. I think I need to use that $f'(x) \leq 1$ to see that $f$ is decreasing, but I'm not sure how to use that.
• If $f'\le 1$, $f$ need not to be decreasing. See $f(x)=\frac{1}{2}x$. – szw1710 May 3 '17 at 19:36
• You need a strict inequality, $f \colon x \mapsto x$ satisfies $f'(x) \leqslant 1$ for all $x$ but has more than one fixed point. $f \colon x \mapsto x+ 1$ has no fixed point. – Daniel Fischer May 3 '17 at 19:38
• I realized I had a typo. It should have ben that $f'(x) < 1$ not that $f'(x) \leq 1$. – TonicWaterWithLime May 3 '17 at 19:41
• This doesn't yet imply that $f$ has a fixed point. Consider $f(x) = x - e^x$. Even $\lvert f'(x)\rvert < 1$ for all $x$ doesn't imply the existence of a fixed point. – Daniel Fischer May 3 '17 at 19:48
• I think if $f(0) > 0$ is specified then this can be proved. – Χpẘ May 3 '17 at 19:54
Let $f(x) = x+1$ for $x \in \mathbb{R}$. Then $f'(x) = 1$ for all $x$ and $f$ has no fixed points.
• Strictly speaking, this is not an answer but a counterexample. Placing as a comment would help the OP revise his question. – mlc May 3 '17 at 19:40
It must be $f'(x)<1$. Then let $a,b$ be two fixed points. Then you have $|f(a)-f(b)|=f'(x)|a-b|=f'(x)|f(a)-f(b)|$, so $f'(x)=1$, that is not possible. The first equality is due to Lagrange Theorem.
The thing you are trying to prove is false, even with strict inequality.
$$f(x) = \frac{x + \sqrt {1+x^2}}{2}$$ has derivative below $1$ but no fixed point: we always have $f(x) > x$
• Better than my example, since $|f'(x)|<1$, not just $f'(x)<1$. – Thomas Andrews May 3 '17 at 20:00
• @szw1710 How can you define the sequence $(x_n)_{n=1}^{\infty}$ such that $f(x_n) = nx_n$? For one, you'd already have a fixed point with $x_1$. – Tom May 3 '17 at 20:09
• @szw1710 Let $f: [1,2] \to \Bbb{R}$ be defined by $f(x) = 0$ for all $x \in [1,2]$. Define the sequence $(x_n)$. – Tom May 3 '17 at 20:13
• I have deleted my two comments because there was a mistake there. @Tom referred to them. – szw1710 May 3 '17 at 20:36
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Next: Worked Example 4: AC Up: Magnetic Induction Previous: Worked Example 2: Lenz's
## Worked Example 3: Motional emf
Question: Consider the circuit described in Sect. 9.4. Suppose that the length of the moving rod is m, its speed is , the magnetic field-strength is T (the field is directed into the page--see Fig. 36), and the resistance of the circuit is . What is the emf generated around the circuit? What current flows around the circuit? What is the magnitude and direction of the force acting on the moving rod due to the fact that a current is flowing along it? What is the rate at which work must be performed on the rod in order to keep it moving at constant velocity against this force? What is the rate at which electrical energy is generated? What is the rate at which energy is converted into heat due to the resistivity of the circuit?
Answer: The emf is generated by the motion of the rod. According to Eq. (204), the magnitude of the motional emf is
The emf acts in the anti-clockwise direction in Fig. 36.
The anti-clockwise current driven around the circuit by the motional emf follows from Ohm's law:
Since the rod carries a current which flows perpendicular to a magnetic field , the force per unit length acting on the rod is (see Sect. 8.2). Thus, the total force acting on the rod is of magnitude
This force is directed parallel to the vector . It follows that the force is to the left in Fig. 36. In other words, the force opposes the motion producing the emf.
In order to keep the rod moving at a constant velocity, some external agent must apply a force to the rod which is equal and opposite to the magnetic force described above. Thus, the externally applied force acts to the right. The rate at which work is done by this force is the product of the force and the velocity of the rod in the direction of this force. Thus,
Every charge which circulates around the circuit in the anti-clockwise direction acquires the energy . The amount of charge per unit time which circulates around the circuit is, by definition, equal to the current flowing around the circuit. Thus, the rate at which electric charges acquire energy in the circuit is
Now, the rate at which electric charges acquire energy in the circuit is equal to the rate at which mechanical work is done on the rod by the external force, as must be the case if energy is to be conserved. Thus, we can think of this circuit as constituting a primitive generator which transforms mechanical into electrical energy.
The rate at which electrical energy is converted into heat energy in the circuit is
Thus, all of the mechanical work done on the rod eventually ends up as heat dissipated in the circuit.
Next: Worked Example 4: AC Up: Magnetic Induction Previous: Worked Example 2: Lenz's
Richard Fitzpatrick 2007-07-14
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# Gallons to Teaspoons Converter
Select conversion type:
Rounding options:
Convert Teaspoons to Gallons (tsp to gal) ▶
## Conversion Table
gallons to teaspoons gal tsp 1 gal 768 tsp 2 gal 1536 tsp 3 gal 2304 tsp 4 gal 3072 tsp 5 gal 3840 tsp 6 gal 4608 tsp 7 gal 5376 tsp 8 gal 6144 tsp 9 gal 6912 tsp 10 gal 7680 tsp 11 gal 8448 tsp 12 gal 9216 tsp 13 gal 9984 tsp 14 gal 10752 tsp 15 gal 11520 tsp 16 gal 12288 tsp 17 gal 13056 tsp 18 gal 13824 tsp 19 gal 14592 tsp 20 gal 15360 tsp
## How to convert
1 gallon (gal) = 768 teaspoon (tsp). Gallon (gal) is a unit of Volume used in Standard system. Teaspoon (tsp) is a unit of Volume used in Cooking system.
## Gallons: A Unit of Volume
Gallons are a unit of volume that are used to measure liquids, such as water, milk, oil, wine, etc. They are also used to measure some dry goods, such as grains, fruits, nuts, etc. They are different from cups, which are a smaller unit of volume. They are also different from liters, which are a larger unit of volume. They are also different from barrel of oil equivalent (BOE), which is a unit of energy based on the approximate energy released by burning one barrel of crude oil.
## How to Convert Gallons
To convert gallons to other units of volume, one can use the following formulas:
• To convert US liquid gallons to liters: multiply by 3.785
• To convert US liquid gallons to cubic inches: multiply by 231
• To convert US liquid gallons to fluid ounces: multiply by 128
• To convert US liquid gallons to UK gallons: multiply by 0.833
• To convert US liquid gallons to BOE: divide by 5
• To convert US dry gallons to liters: multiply by 4.405
• To convert US dry gallons to cubic inches: multiply by 268.8
• To convert US dry gallons to fluid ounces: multiply by 148.9
• To convert US dry gallons to UK gallons: multiply by 0.969
• To convert US dry gallons to BOE: divide by 4.6
• To convert UK gallons to liters: multiply by 4.546
• To convert UK gallons to cubic inches: multiply by 277.4
• To convert UK gallons to fluid ounces: multiply by 160
• To convert UK gallons to US liquid gallons: multiply by 1.2
• To convert UK gallons to BOE: divide by 6.1
The US gallon is equal to 3.785411784 liters and defined as 231 cubic inches.
## Teaspoons: A Unit of Volume
Teaspoons are a unit of volume that are used to measure small amounts of liquids, such as water, milk, oil, vinegar, etc. They are also used to measure some dry ingredients, such as sugar, salt, baking powder, etc. They are different from tablespoons, which are a larger unit of volume. Teaspoons are also different from fluid ounces, which are a larger unit of volume. Teaspoons are also different from barrel of oil equivalent (BOE), which is a unit of energy based on the approximate energy released by burning one barrel of crude oil.
## How to Convert Teaspoons
To convert teaspoons to other units of volume, one can use the following formulas:
• To convert teaspoons to milliliters: multiply by 5
• To convert teaspoons to fluid ounces: multiply by 0.1667
• To convert teaspoons to cups: multiply by 0.0208
• To convert teaspoons to gallons (US): multiply by 0.001302
• To convert teaspoons to gallons (UK): multiply by 0.0014
• To convert teaspoons to BOE: divide by 15950
US teaspoon can be abbreviated as t., ts., tsp. or tspn. one teaspoon as a unit of culinary measure is 1/3 tablespoon, that is, ~4.93 mL. For labeling on food packages in the U.S., the teaspoon is defined by FDA as 5 mL precisely.
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# Physics 2102 Jonathan Dowling Physics 2102 Lecture 02
• Slides: 14
Physics 2102 Jonathan Dowling Physics 2102 Lecture 02: THU 21 JAN Electric Fields I Charles-Augustin de Coulomb (1736 -1806)
What Are We Going to Learn? A Road Map • Electric charge - Electric force on other electric charges - Electric field, and electric potential • Moving electric charges : current • Electronic circuit components: batteries, resistors, capacitors • Electric currents - Magnetic field - Magnetic force on moving charges • Time-varying magnetic field & Electric Field • More circuit components: inductors. • Electromagnetic waves - light waves • Geometrical Optics (light rays). • Physical optics (light waves)
Coulomb’s Law Charles-Augustin de Coulomb (1736 -1806) For Charges in a Vacuum k = Often, we write k as:
Sir Michael Faraday’s Electric Lines of Force Electric Force Field Faraday (1791– 1867)
E-Field is E-Force Divided by ECharge Definition of Electric Field: E-Force on Charge +q 1 –q 2 P 1 P 2 E-Field at Point –q 2 P 1 P 2 Units: F = [N] = [Newton] ; E = [N/C] = [Newton/Coulomb]
Electric Fields • Electric field E at some point in space is defined as the force experienced by an imaginary point charge of +1 C, divided by 1 C. • • • Note that E is a VECTOR. Electric Field of a Point Charge Since E is the force per unit charge, it –q E +1 C is measured in units of N/C. R We measure the electric field using very small “test charges”, and dividing the measured force by the magnitude of the charge.
Superposition of F and E • Question: How do we figure out the force or field due to several point charges? • Answer: consider one charge at a time, calculate the field (a vector!) produced by each charge, and then add all the vectors! (“superposition”) • Useful to look out for SYMMETRY to simplify calculations! • If you never learned to add vectors in 2101 you’ll be in serious trouble in 2102! See online review http: //phys. lsu. edu/~jdowling/ PHYS 21024 SP 10/Vectors. pdf
Example Total electric field -2 q +q • 4 charges are placed at the corners of a square as shown. • What is the direction of the electric field at the center of the square? (a) Field is ZERO! (b) Along +y (c) Along +x -q +q is the test charge y +2 q x
Electric Field Lines • Field lines: useful way to visualize electric field E • Field lines start at a positive charge, end at negative charge • E at any point in space is tangential to field line • Field lines are closer where E is stronger Example: a negative point charge — note spherical symmetry
Direction of Electric Field Lines E-Field Vectors Point Away from Positive Charge — Field Source! E-Field Vectors Point Towards Negative Charge — Field Sink!
Electric Field of a Dipole • Electric dipole: two point charges +q and –q separated by a distance d • Common arrangement in Nature: molecules, antennae, … • Note axial or cylindrical symmetry • Define “dipole moment” vector p: from –q to +q, with magnitude qd Cancer, Cisplatin and electric dipoles: http: //chemcases. com/cisplat 01. htm
Electric Field On Axis of Dipole -q a +q P x
Electric Field On Axis of Dipole p = qa “dipole moment” a VECTOR - + What if x>> a? (i. e. very far away) E = p/r 3 is actually true for ANY point far from a dipole (not just on axis)
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# I need directions to start my gmat studies
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Intern
Joined: 10 Dec 2011
Posts: 11
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I need directions to start my gmat studies [#permalink]
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05 Jan 2012, 00:34
Dear All,
I have taken an appointment on 30th May for my GMAT exam. Please give me some directions from where to start. I have also given a mock test of Princeton review, where I scored 400 (9 in English and 37 in math).
I know the score seems drastically bad, however, I am ready to work hard to secure a good score.
I would be eagerly waiting for suggestions from you guys.
Thanks
Regards
Siddharth Saxena
Management trainee
KCT Coal sales
Kolkata
Ph: 9836592222
Kudos [?]: 7 [0], given: 0
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Re: I need directions to start my gmat studies [#permalink]
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05 Jan 2012, 00:41
a daily commitment of 2 hrs during the weekdays and as much as you can on the weekend would help now that you have given a MOC test you know atleast a brief idea of your week area, arrest them by systematic approach, over weeks you will have to also focus on time to solve the questions, as you might be aware typically its about a min and half per question the gist is not to stick to a question to solve on the D day rather to get the nearest match of the questions and move on.
base line ask yourself "can i be Sincere and show commitment to my preparation" as long as you have a NO, its a matter of worry.
all the best for your prepration.
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Re: I need directions to start my gmat studies [#permalink] 05 Jan 2012, 00:41
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# I need directions to start my gmat studies
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How to limit voltage output from 3-phase permanent magnet alternator.
izrunas
Joined Mar 3, 2020
1
So here's the issue... My new 2000w wind turbine is designed to produce 3-phase 96VAC, but my grid-tied controller is only designed for 3-phase 40-90VAC. I realize that voltage and rpm's are proportional, but due to the high winds we regularly experience, I'd expect the 90+ condition to happen too often. I had considered a voltage-controlled relay to switch from feeding the controller to shunting to a braking system (to keep the turbine from flying apart at excessive rpm's).
How can I build a 3-phase AC voltage regulator to prevent the voltage from exceeding 90VAC? If it was single phase, I'd just use a handy motor speed controller to trim the voltage and put a BIG fan on it. But, I don't even know where to start when it comes to 3-phase AC voltage regulation.
I have also considered a 3-phase step-down transformer (96VAC to 48VAC), but good luck finding such a unicorn.
Any advice would be greatly appreciated!
MisterBill2
Joined Jan 23, 2018
18,519
One alternative will be to copy what the power companies do, and use an automated transformer tap changer. With just one tap per phase on the primary and one per phase on the secondary you can have a total of four possible ratios, either stepping down or stepping up.
OR, in place of that 96volt to 48 volt transformer, use a 240 to 120 volt transformer. THOSE are fairly common. Or, try a 480 volt to 240 volt transformer. And keep in mind that there is also a WYE/Delta change that produces significant voltage differences.
So there you have a number of suggestions that do not involve any electronics and are quite efficient. And rather reliable as well.
Also, if the individual phases are accessible on your alternator, just switching to the "WYE" connection scheme will give a lower output voltage.
bernardo.marques
Joined Aug 19, 2022
3
Did you manage to figure the circuit?
I have a 5kW turbine that generates in excess of 100V and I'd like to limit it to 72V as that is the maximum input to my DC-DC converter. Any help would be appreciated.
As far as I understand it you need:
- two braking resistors (power rating dimensioned according to turbine)
- two triacs
the triacs need to connect one resistor between L1 and L2 and the other one between L2 and L3 whenever voltage is too high, this will brake the turbine causing it to rotate slower which will keep the voltage bellow the maximum desired.
Unfortunatelly I'm not good enough at analog electronics to understand how to make the triacs connect phases to resistors only when voltage is too high.
MisterBill2
Joined Jan 23, 2018
18,519
To burn the excess 30+ volts in resistors is to waste almost 30% of the turbine output as heat. That is quite a waste. Given that the turbine generator produces 3 phase AC power and that the DC to DC converter requires a DC input, transformers make the best sense.
To use standard transformers the simple approach will be to use standard step-down transformers for 240 to 120 volts, but to wire the primary in series with the secondary so that the step down ratio will be 3:2. This will drop 100 volts to 66 volts, fairly close to the max. With rectification the output might be a bit more. There are also transformers with less standard ratios available, and even a number of three phase transformers, from transformer manufacturers. So really a trio of transformers will be the most efficient wayto match the turbine generator output to the converter input requirements. It may also be possible to change the turbine generator 3-phase connection from WYE to Delta, and get a more convenient output.
ThePanMan
Joined Mar 13, 2020
793
Are the turbine blades the self adjusting angle of attack type? If you set them up right then regardless of wind speed the turbine won't over-rotate. Also, you can trim it to produce the voltage you want.
Or change the alternator to a self-exciting type. That way once producing you can regulate the voltage going to the rotor (or armature, always get those mixed up). All you need is just a little excitation source and the alternator should produce sufficient output. You're only looking for about 70 amps. The drawback is you'd need to brake the blades in excessive wind speeds. Perhaps you can tell I know very little about wind turbines.
bernardo.marques
Joined Aug 19, 2022
3
With the transformer energy will also be wasted, just in the lower range, because 40V will be transformed into a voltage that is below the lower limit of operation.
Small wind turbines do not rotate the blades out of wind and the turbine needs to be braked in high winds to avoid mechanical damage.
The braking resistor is needed.
Energy is burned in the resistor only when there is excess energy that is not being consumed, this true in high winds or if there is no load (during the night while everyone is sleeping and wind blows just regularly)
MisterBill2
Joined Jan 23, 2018
18,519
@ Pan Man: Changing an alternator type generally means replacing the alternator with a different one. The alternator is usually the most expensive part of the whole system. Transformers would be much simpler and cheaper, and often within the capability of a lot of folks to make. And transformers are often available as surplus .
Not using all of the possible output is much less wasteful than burning power in resistors. and my one example was based on a lot of information that we did not get.
bernardo.marques
Joined Aug 19, 2022
3
Are the turbine blades the self adjusting angle of attack type? If you set them up right then regardless of wind speed the turbine won't over-rotate. Also, you can trim it to produce the voltage you want.
Or change the alternator to a self-exciting type. That way once producing you can regulate the voltage going to the rotor (or armature, always get those mixed up). All you need is just a little excitation source and the alternator should produce sufficient output. You're only looking for about 70 amps. The drawback is you'd need to brake the blades in excessive wind speeds. Perhaps you can tell I know very little about wind turbines.
I can't change the generator, it is 'the turbine'.
Joined Feb 20, 2016
4,476
For the transformer idea, I found this example to save me drawing it. If you use a 110V to 24V transformer for each phase, connected as shown so the 24V secondary bucks the input voltage.
Chose the transformers to have a 24V current rating a bit higher that the max alternator current.
But it will be interesting to see if the transformers work over the frequency range.
The other way is to get a 110V to 48V switch mode power supply to feed your existing charger. You will need a 3 phase rectifier in front of the new power supply if the rectifier is in the charger.
Adding the transformers will make the wind gen charging drop out at a higher minimum wind speed.
Transformers will be pretty expensive I think. A new charge controller may be cheaper.
Joined Feb 20, 2016
4,476
Another idea is to go with the braking resistor but use an electric water heater as the resistor so the waste excess power is not a total loss.
ThePanMan
Joined Mar 13, 2020
793
it will be interesting to see if the transformers work over the frequency range.
Ah! Yes. Transformers like certain frequencies and alternators that spin at varying rates don't live by that rule. I've seen transformers designed for 50/60Hz & 400Hz (aircraft). Even SMPS's use transformers that are designed for very high frequency. But as long as you're operating in the proper freq. the transformer would be happy.
shortbus
Joined Sep 30, 2009
10,045
All of this "over volting" is why a PM(permanent magnet) alternator is a bad idea. Not easy to regulate the out put. Cheaper to start out but much harder to make it work reliably.
MisterBill2
Joined Jan 23, 2018
18,519
Probably it would cost much less to replace the DC/DC converter with a version able to accept the higher input voltage.
But transformers able to handle a wide range of frequencies are certainly understood, and so such a device should be available, and far more effective.
And now the previously unasked question: What is the normal output of the three phase 5KW alternator???
Once again we do not have enough information to offer a best suggestion.
Since the normal output range of the turbine should be known, that will determine the normal frequency range that the transformer must handle best. So it is not an unknowable quantity. Except to us.
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# NCERT Solutions Class 8 Maths Chapter 9 Exercise 9.5
Mathematics is a highly interesting and practical discipline. It is one of the most important subjects that are a part of the Class 8 curriculum. Mathematics is the study of numbers, their relationships, analysis, formulae, quantities, space, and forms. It is an essential component of Engineering, Science, Pharmaceutics, Finance and Accounting, Computer Science, and Medical Science among other fields. Mathematics acts as a cornerstone for numerous occupations, including Engineering, Architecture, Flying, Statistics, Insurance, and many more. Mathematics aids in the development of a variety of abilities that are essential to an individual’s overall growth. It aids in the development of abilities such as Critical Thinking, Problem-Solving, Logical Reasoning, Analysis and Management, Quantitative Aptitude, and so on. Acquiring these abilities can help students clear a variety of competitive examinations like NEET, AFCAT, MAT, JEE Mains, JEE Advance, AIIMS, CAT, CUET, and others. Students must instill these abilities in order to perform well in the relevant examinations. Developing these abilities may help students lay the foundation for a career in a variety of professions; hence, it is critical that students study Mathematics with vigour. Students may make use of the NCERT Solutions For Class 8 Maths Chapter 9 Exercise 9.5 to develop basic mathematical abilities. The NCERT Solutions For Class 8 Maths Chapter 9 Exercise 9.5 can be accessed by students from the Extramarks website.
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## NCERT Solutions For Class 8 Maths Chapter 9 Algebraic Expressions And Identities (EX 9.5) Exercise 9.5
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### Access NCERT solutions for Class 8 Chapter 9-Algebraic Expressions and Identities
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### NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Exercise 9.5
Chapter 9 entitled – Algebraic Expressions and Identities is one of the most important chapters in the NCERT Mathematics curriculum for Class 8. This chapter comprises many complex formulas. To master this chapter, students must understand not only the ideas discussed in it but also how to apply these ideas practically. They are, thus, encouraged to practice the NCERT Solutions For Class 8 Maths Chapter 9 Exercise 9.5 to better understand the practical application required in Chapter 9 Algebraic Expressions and Identities. Students must study and comprehend the mathematical themes taught in the Class 8 Mathematics curriculum. It is critical for students to expand their understanding of mathematical ideas. This has a substantial impact on Class 8 students’ performance in the Senior Secondary Examination. Students can substantially benefit from gaining the knowledge needed for the practical application required in Chapter 9 Algebraic Expressions and Identities, and students are encouraged to practise the NCERT Solutions For Class 8 Maths Chapter 9 Exercise 9.5 as they prepare for the Senior Secondary Mathematics Examination.
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### NCERT Solutions for Class 8
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Q.1
$\begin{array}{l}\mathrm{Use}\text{ }\mathrm{a}\text{ }\mathrm{suitable}\text{ }\mathrm{identity}\text{ }\mathrm{to}\text{ }\mathrm{get}\text{ }\mathrm{each}\text{ }\mathrm{of}\text{ }\mathrm{the}\text{ }\mathrm{following}\text{ }\mathrm{products}.\\ \left(\mathrm{i}\right)\text{ }\left(\mathrm{x}+3\right)\left(\mathrm{x}+3\right)\text{ }\left(\mathrm{ii}\right)\left(2\mathrm{y}+5\right)\left(2\mathrm{y}+5\right)\\ \left(\mathrm{iii}\right)\text{ }\left(2\mathrm{a}-7\right)\left(2\mathrm{a}-7\right)\text{ }\left(\mathrm{iv}\right)\text{ }\left(3\mathrm{a}-\frac{1}{2}\right)\left(3\mathrm{a}-\frac{1}{2}\right)\\ \left(\mathrm{v}\right)\text{ }\left(1.1\mathrm{m}-0.4\right)\left(1.1\mathrm{m}+0.4\right)\text{ }\left(\mathrm{vi}\right)\text{ }\left({\mathrm{a}}^{2}+{\mathrm{b}}^{2}\right)\left(-{\mathrm{a}}^{2}+{\mathrm{b}}^{2}\right)\\ \left(\mathrm{vii}\right)\text{ }\left(6\mathrm{x}-7\right)\left(6\mathrm{x}+7\right)\text{ }\left(\mathrm{viii}\right)\text{ }\left(-\mathrm{a}+\mathrm{c}\right)\left(-\mathrm{a}+\mathrm{c}\right)\\ \left(\mathrm{ix}\right)\text{ }\left(\frac{\mathrm{x}}{2}+\frac{3\mathrm{y}}{4}\right)\left(\frac{\mathrm{x}}{2}+\frac{3\mathrm{y}}{4}\right)\text{ }\left(\mathrm{x}\right)\text{ }\left(7\mathrm{a}-9\mathrm{b}\right)\left(7\mathrm{a}-9\mathrm{b}\right)\end{array}$
Ans
$\begin{array}{l}\left(\mathrm{i}\right)\text{ }\left(\mathrm{x}+3\right)\left(\mathrm{x}+3\right)={\left(\mathrm{x}+3\right)}^{2}\\ ={\left(\mathrm{x}\right)}^{2}+2×3×\mathrm{x}+{\left(3\right)}^{2}\text{}\left\{\begin{array}{l}\text{By using the identity}\\ {\text{(a+b)}}^{\text{2}}={\mathrm{a}}^{2}+{\mathrm{b}}^{2}+2\mathrm{ab}\end{array}\right\}\\ ={\mathrm{x}}^{2}+6\mathrm{x}+9\\ \\ \left(\mathrm{ii}\right)\text{ }\left(2\mathrm{y}+5\right)\left(2\mathrm{y}+5\right)={\left(2\mathrm{y}+5\right)}^{2}\\ ={\left(2\mathrm{y}\right)}^{2}+2×5×2\mathrm{y}+{\left(5\right)}^{2}\text{}\left\{\begin{array}{l}\text{By using the identity}\\ {\text{(a+b)}}^{\text{2}}={\mathrm{a}}^{2}+{\mathrm{b}}^{2}+2\mathrm{ab}\end{array}\right\}\\ =4{\mathrm{y}}^{2}+20\mathrm{y}+25\\ \\ \left(\mathrm{iii}\right)\text{ }\left(2\mathrm{a}-7\right)\left(2\mathrm{a}-7\right)={\left(2\mathrm{a}-7\right)}^{2}\\ ={\left(2\mathrm{a}\right)}^{2}-2×2\mathrm{a}×7+{\left(7\right)}^{2}\text{}\left\{\begin{array}{l}\text{By using the identity}\\ {\left(\text{a}-\text{b}\right)}^{\text{2}}={\mathrm{a}}^{2}+{\mathrm{b}}^{2}-2\mathrm{ab}\end{array}\right\}\\ =4{\mathrm{a}}^{2}-28\mathrm{a}+49\\ \\ \left(\mathrm{iv}\right)\text{ }\left(3\mathrm{a}-\frac{1}{2}\right)\left(3\mathrm{a}-\frac{1}{2}\right)={\left(3\mathrm{a}-\frac{1}{2}\right)}^{2}\\ ={\left(3\mathrm{a}\right)}^{2}-2×3\mathrm{a}×\frac{1}{2}+{\left(\frac{1}{2}\right)}^{2}\text{}\left\{\begin{array}{l}\text{By using the identity}\\ {\left(\text{a}-\text{b}\right)}^{\text{2}}={\mathrm{a}}^{2}+{\mathrm{b}}^{2}-2\mathrm{ab}\end{array}\right\}\\ =9{\mathrm{a}}^{2}-3\mathrm{a}+\frac{1}{4}\\ \left(\mathrm{v}\right)\text{ }\left(1.1\mathrm{m}-0.4\right)\left(1.1\mathrm{m}+0.4\right)\\ ={\left(1.1\mathrm{m}\right)}^{2}-{\left(0.4\right)}^{2}\text{}\left\{\begin{array}{l}\text{By using the identity}\\ \text{(a+b)(a}-{\text{b)=a}}^{\text{2}}-{\text{b}}^{\text{2}}\end{array}\right\}\\ =1.21{\mathrm{m}}^{2}-0.16\\ \\ \left(\mathrm{vi}\right)\text{ }\left({\mathrm{a}}^{2}+{\mathrm{b}}^{2}\right)\left(-{\mathrm{a}}^{2}+{\mathrm{b}}^{2}\right)\\ ={\left({\mathrm{b}}^{2}\right)}^{2}-{\left({\mathrm{a}}^{2}\right)}^{2}\text{}\left\{\begin{array}{l}\text{By using the identity}\\ \text{(a+b)(a}-{\text{b)=a}}^{\text{2}}-{\text{b}}^{\text{2}}\end{array}\right\}\\ ={\mathrm{b}}^{4}-{\mathrm{a}}^{4}\\ \\ \left(\mathrm{vii}\right)\text{ }\left(6\mathrm{x}-7\right)\left(6\mathrm{x}+7\right)\\ ={\left(6\mathrm{x}\right)}^{2}-{\left(7\right)}^{2}\text{}\left\{\begin{array}{l}\text{By using the identity}\\ \text{(a+b)(a}-{\text{b)=a}}^{\text{2}}-{\text{b}}^{\text{2}}\end{array}\right\}\\ =36{\mathrm{x}}^{2}-49\\ \\ \left(\mathrm{viii}\right)\text{ }\left(-\mathrm{a}+\mathrm{c}\right)\left(-\mathrm{a}+\mathrm{c}\right)\\ ={\left(-\mathrm{a}+\mathrm{c}\right)}^{2}\text{}\left\{\begin{array}{l}\text{By using the identity}\\ {\text{(a+b)}}^{\text{2}}={\mathrm{a}}^{2}+{\mathrm{b}}^{2}+2\mathrm{ab}\end{array}\right\}\text{}\\ ={\mathrm{c}}^{2}+{\mathrm{a}}^{2}-2\mathrm{ac}\end{array}$
$\begin{array}{l}\left(\mathrm{ix}\right)\left(\frac{\mathrm{x}}{2}+\frac{3\mathrm{y}}{4}\right)\left(\frac{\mathrm{x}}{2}+\frac{3\mathrm{y}}{4}\right)={\left(\frac{\mathrm{x}}{2}+\frac{3\mathrm{y}}{4}\right)}^{2}\\ ={\left(\frac{\mathrm{x}}{2}\right)}^{2}+{\left(\frac{3\mathrm{y}}{4}\right)}^{2}+\text{2}\left(\frac{\mathrm{x}}{2}\right)\left(\frac{3\mathrm{y}}{4}\right)\text{}\left\{\begin{array}{l}\text{By using the identity}\\ {\text{(a + b)}}^{\text{2}}={\mathrm{a}}^{2}+{\mathrm{b}}^{2}+2\mathrm{ab}\end{array}\right\}\\ \text{=}\frac{{\mathrm{x}}^{2}}{4}+\frac{9{\mathrm{y}}^{2}}{16}+\frac{3\mathrm{xy}}{4}\\ \\ \left(\mathrm{x}\right)\left(7\mathrm{a}-9\mathrm{b}\right)\left(7\mathrm{a}-9\mathrm{b}\right)={\left(7\mathrm{a}-9\mathrm{b}\right)}^{2}\\ ={\left(7\mathrm{a}\right)}^{2}+{\left(9\mathrm{b}\right)}^{2}-\text{2}\left(7\mathrm{a}\right)\left(9\mathrm{b}\right)\text{}\left\{\begin{array}{l}\text{By using the identity}\\ {\left(\text{a}-\text{b}\right)}^{\text{2}}={\mathrm{a}}^{2}+{\mathrm{b}}^{2}-2\mathrm{ab}\end{array}\right\}\\ =49{\mathrm{a}}^{2}+81{\mathrm{b}}^{2}-126\mathrm{ab}\end{array}$
Q.2
$\begin{array}{l}\mathrm{Use}\text{ }\mathrm{the}\text{ }\mathrm{identity}\text{ }\left(\mathrm{x}+\mathrm{a}\right)\left(\mathrm{x}+\mathrm{b}\right)={\mathrm{x}}^{2}+\left(\mathrm{a}+\mathrm{b}\right)\mathrm{x}+\mathrm{ab}\\ \mathrm{to}\text{ }\mathrm{find}\text{ }\mathrm{the}\text{ }\mathrm{following}\text{ }\mathrm{products}.\\ \left(\mathrm{i}\right)\text{ }\left(\mathrm{x}+3\right)\left(\mathrm{x}+7\right)\\ \left(\mathrm{ii}\right)\text{ }\left(4\mathrm{x}+5\right)\left(4\mathrm{x}+1\right)\\ \left(\mathrm{iii}\right)\text{ }\left(4\mathrm{x}-5\right)\left(4\mathrm{x}-1\right)\\ \left(\mathrm{iv}\right)\text{ }\left(4\mathrm{x}+5\right)\left(4\mathrm{x}-1\right)\\ \left(\mathrm{v}\right)\text{ }\left(2\mathrm{x}+5\mathrm{y}\right)\left(2\mathrm{x}+3\mathrm{y}\right)\\ \left(\mathrm{vi}\right)\text{ }\left(2{\mathrm{a}}^{2}+9\right)\left(2{\mathrm{a}}^{2}+5\right)\\ \left(\mathrm{vii}\right)\text{ }\left(\mathrm{xyz}-4\right)\left(\mathrm{xyz}-2\right)\end{array}$
Ans
$\begin{array}{l}\left(\mathrm{i}\right)\text{ }\left(\mathrm{x}+3\right)\left(\mathrm{x}+7\right)={\mathrm{x}}^{2}+\left(3+7\right)\mathrm{x}+3×7\\ ={\mathrm{x}}^{2}+10\mathrm{x}+21\\ \\ \left(\mathrm{ii}\right)\text{ }\left(4\mathrm{x}+5\right)\left(4\mathrm{x}+1\right)={\left(4\mathrm{x}\right)}^{2}+\left(5+1\right)\left(4\mathrm{x}\right)+5×1\\ =16{\mathrm{x}}^{2}+24\mathrm{x}+5\\ \\ \left(\mathrm{iii}\right)\text{ }\left(4\mathrm{x}-5\right)\left(4\mathrm{x}-1\right)={\left(4\mathrm{x}\right)}^{2}+\left[\left(-5\right)+\left(–1\right)\right]\left(4\mathrm{x}\right)+\left(–5\right)×\left(-1\right)\\ =16{\mathrm{x}}^{2}-24\mathrm{x}+5\\ \\ \left(\mathrm{iv}\right)\text{ }\left(4\mathrm{x}+5\right)\left(4\mathrm{x}-1\right)={\left(4\mathrm{x}\right)}^{2}+\left[5+\left(-1\right)\right]\left(4\mathrm{x}\right)+5×\left(-1\right)\\ =16{\mathrm{x}}^{2}+16\mathrm{x}-5\\ \\ \left(\mathrm{v}\right)\text{ }\left(2\mathrm{x}+5\mathrm{y}\right)\left(2\mathrm{x}+3\mathrm{y}\right)={\left(2\mathrm{x}\right)}^{2}+\left(5\mathrm{y}+3\mathrm{y}\right)\left(2\mathrm{x}\right)+5\mathrm{y}×3\mathrm{y}\\ =4{\mathrm{x}}^{2}+16\mathrm{xy}+15{\mathrm{y}}^{2}\\ \\ \left(\mathrm{vi}\right)\text{ }\left(2{\mathrm{a}}^{2}+9\right)\left(2{\mathrm{a}}^{2}+5\right)={\left(2{\mathrm{a}}^{2}\right)}^{2}+\left(9+5\right)\left(2{\mathrm{a}}^{2}\right)+9×5\\ =4{\mathrm{a}}^{4}+28{\mathrm{a}}^{2}+45\\ \\ \left(\mathrm{vii}\right)\text{ }\left(\mathrm{xyz}-4\right)\left(\mathrm{xyz}-2\right)={\left(\mathrm{xyz}\right)}^{2}+\left[\left(-4\right)+\left(–2\right)\right]\left(\mathrm{xyz}\right)+\left(–4\right)×\left(-2\right)\\ ={\mathrm{x}}^{2}{\mathrm{y}}^{2}{\mathrm{z}}^{2}-6\mathrm{xyz}+\end{array}$
Q.3
$\begin{array}{l}\mathrm{Find}\text{ }\mathrm{the}\text{ }\mathrm{followings}\text{ }\mathrm{quares}\text{ }\mathrm{by}\text{ }\mathrm{using}\text{ }\mathrm{the}\text{ }\mathrm{identities}.\\ \left(\mathrm{i}\right)\text{ }{\left(\mathrm{b}-7\right)}^{2}\\ \left(\mathrm{ii}\right)\text{ }{\left(\mathrm{xy}+3\mathrm{z}\right)}^{2}\\ \left(\mathrm{iii}\right)\text{ }{\left(6{\mathrm{x}}^{2}-5\mathrm{y}\right)}^{2}\\ \left(\mathrm{iv}\right)\text{ }{\left(\frac{2}{3}\mathrm{m}+\frac{3}{2}\mathrm{n}\right)}^{2}\\ \left(\mathrm{v}\right)\text{ }{\left(0.4\mathrm{p}-0.5\mathrm{q}\right)}^{2}\\ \left(\mathrm{vi}\right)\text{ }{\left(2\mathrm{xy}+5\mathrm{y}\right)}^{2}\end{array}$
Ans
$\begin{array}{l}\left(\mathrm{i}\right)\text{ }{\left(\mathrm{b}-7\right)}^{2}\\ ={\left(\mathrm{b}\right)}^{2}-2×\mathrm{b}×7+{\left(7\right)}^{2}\text{}\left\{\begin{array}{l}\mathrm{By}\text{}\mathrm{using}\text{}\mathrm{the}\text{}\mathrm{identity}\\ {\left(\mathrm{a}\text{}-\text{}\mathrm{b}\right)}^{2}={\mathrm{a}}^{2}+{\mathrm{b}}^{2}-2\mathrm{ab}\end{array}\right\}\\ ={\mathrm{b}}^{2}-14\mathrm{b}+49\\ \\ \left(\mathrm{ii}\right)\text{ }{\left(\mathrm{xy}+3\mathrm{z}\right)}^{2}\\ ={\left(\mathrm{xy}\right)}^{2}+2×\mathrm{xy}×3\mathrm{z}+{\left(3\mathrm{z}\right)}^{2}\text{}\left\{\begin{array}{l}\mathrm{By}\text{}\mathrm{using}\text{}\mathrm{the}\text{}\mathrm{identity}\\ {\left(\mathrm{a}\text{}+\text{}\mathrm{b}\right)}^{2}={\mathrm{a}}^{2}+{\mathrm{b}}^{2}+2\mathrm{ab}\end{array}\right\}\\ ={\mathrm{x}}^{2}{\mathrm{y}}^{2}+6\mathrm{xyz}+9{\mathrm{z}}^{2}\\ \left(\mathrm{iii}\right)\text{ }{\left(6{\mathrm{x}}^{2}-5\mathrm{y}\right)}^{2}\\ ={\left(6{\mathrm{x}}^{2}\right)}^{2}-2×6{\mathrm{x}}^{2}×5\mathrm{y}+{\left(5\mathrm{y}\right)}^{2}\text{}\left\{\begin{array}{l}\mathrm{By}\text{}\mathrm{using}\text{}\mathrm{the}\text{}\mathrm{identity}\\ {\left(\mathrm{a}\text{}-\text{}\mathrm{b}\right)}^{2}={\mathrm{a}}^{2}+{\mathrm{b}}^{2}-2\mathrm{ab}\end{array}\right\}\\ =36{\mathrm{x}}^{4}-60{\mathrm{x}}^{2}\mathrm{y}+25{\mathrm{y}}^{2}\\ \\ \left(\mathrm{iv}\right)\text{ }{\left(\frac{2}{3}\mathrm{m}+\frac{3}{2}\mathrm{n}\right)}^{2}\\ ={\left(\frac{2}{3}\mathrm{m}\right)}^{2}+2×\frac{2}{3}\mathrm{m}×\frac{3}{2}\mathrm{n}+{\left(\frac{3}{2}\mathrm{n}\right)}^{2}\left\{\begin{array}{l}\mathrm{By}\text{}\mathrm{using}\text{}\mathrm{the}\text{}\mathrm{identity}\\ {\left(\mathrm{a}\text{+}\mathrm{b}\right)}^{2}={\mathrm{a}}^{2}+{\mathrm{b}}^{2}+2\mathrm{ab}\end{array}\right\}\\ =\frac{4}{9}{\mathrm{m}}^{2}+2\mathrm{mn}+\frac{9}{4}{\mathrm{n}}^{2}\\ \left(\mathrm{v}\right)\text{ }{\left(0.4\mathrm{p}-0.5\mathrm{q}\right)}^{2}\\ ={\left(0.4\mathrm{p}\right)}^{2}-2×0.4\mathrm{p}×0.5\mathrm{q}+{\left(0.5\mathrm{q}\right)}^{2}\text{}\left\{\begin{array}{l}\mathrm{By}\text{}\mathrm{using}\text{}\mathrm{the}\text{}\mathrm{identity}\\ {\left(\mathrm{a}\text{}-\text{}\mathrm{b}\right)}^{2}={\mathrm{a}}^{2}+{\mathrm{b}}^{2}-2\mathrm{ab}\end{array}\right\}\\ =0.16{\mathrm{p}}^{2}-0.4\mathrm{pq}+0.25{\mathrm{q}}^{2}\\ \\ \left(\mathrm{vi}\right)\text{ }{\left(2\mathrm{xy}+5\mathrm{y}\right)}^{2}\\ ={\left(2\mathrm{xy}\right)}^{2}+2×2\mathrm{xy}×5\mathrm{y}+{\left(5\mathrm{y}\right)}^{2}\text{}\left\{\begin{array}{l}\mathrm{By}\text{}\mathrm{using}\text{}\mathrm{the}\text{}\mathrm{identity}\\ {\left(\mathrm{a}\text{}+\text{}\mathrm{b}\right)}^{2}={\mathrm{a}}^{2}+{\mathrm{b}}^{2}+2\mathrm{ab}\end{array}\right\}\\ =4{\mathrm{x}}^{2}{\mathrm{y}}^{2}+20{\mathrm{xy}}^{2}+25{\mathrm{y}}^{2}\end{array}$
Q.4
$\begin{array}{l}\mathrm{Simplify}.\\ \left(\mathrm{i}\right)\text{ }{\left({\mathrm{a}}^{2}-{\mathrm{b}}^{2}\right)}^{2}\\ \left(\mathrm{ii}\right)\text{ }{\left(2\mathrm{x}+5\right)}^{2}-{\left(2\mathrm{x}-5\right)}^{2}\\ \left(\mathrm{iii}\right)\text{ }{\left(7\mathrm{m}-8\mathrm{n}\right)}^{2}+{\left(7\mathrm{m}+8\mathrm{n}\right)}^{2}\\ \left(\mathrm{iv}\right)\text{ }{\left(4\mathrm{m}+5\mathrm{n}\right)}^{2}+{\left(5\mathrm{m}+4\mathrm{n}\right)}^{2}\\ \left(\mathrm{v}\right)\text{ }{\left(2.5\mathrm{p}-1.5\mathrm{q}\right)}^{2}-{\left(1.5\mathrm{p}-2.5\mathrm{q}\right)}^{2}\\ \left(\mathrm{vi}\right)\text{ }{\left(\mathrm{ab}+\mathrm{bc}\right)}^{2}-2{\mathrm{ab}}^{2}\mathrm{c}\\ \left(\mathrm{vii}\right)\text{ }{\left({\mathrm{m}}^{2}-{\mathrm{n}}^{2}\mathrm{m}\right)}^{2}+2{\mathrm{m}}^{3}{\mathrm{n}}^{2}\end{array}$
Ans
$\begin{array}{l}\left(\mathrm{i}\right)\text{ }{\left({\mathrm{a}}^{2}-{\mathrm{b}}^{2}\right)}^{2}\\ ={\left({\mathrm{a}}^{2}\right)}^{2}-2×{\mathrm{a}}^{2}×{\mathrm{b}}^{2}+{\left({\mathrm{b}}^{2}\right)}^{2}\left\{\begin{array}{l}\mathrm{By}\text{}\mathrm{using}\text{}\mathrm{the}\text{}\mathrm{identity}\\ {\left(\mathrm{a}\text{}-\text{}\mathrm{b}\right)}^{2}={\mathrm{a}}^{2}+{\mathrm{b}}^{2}-2\mathrm{ab}\end{array}\right\}\\ ={\left(\mathrm{a}\right)}^{4}-2{\mathrm{a}}^{2}{\mathrm{b}}^{2}+{\left(\mathrm{b}\right)}^{4}\\ \\ \left(\mathrm{ii}\right)\text{ }{\left(2\mathrm{x}+5\right)}^{2}-{\left(2\mathrm{x}-5\right)}^{2}\\ =\left[{\left(2\mathrm{x}\right)}^{2}+2×2\mathrm{x}×5+{\left(5\right)}^{2}\right]-\left[{\left(2\mathrm{x}\right)}^{2}-2×2\mathrm{x}×5+{\left(5\right)}^{2}\right]\text{}\\ \text{}\left\{\begin{array}{l}\mathrm{By}\text{}\mathrm{using}\text{}\mathrm{the}\text{}\mathrm{identity}\\ {\left(\mathrm{a}\text{+}\mathrm{b}\right)}^{2}={\mathrm{a}}^{2}+{\mathrm{b}}^{2}+2\mathrm{ab}\end{array}\right\}\left\{\begin{array}{l}\mathrm{By}\text{}\mathrm{using}\text{}\mathrm{the}\text{}\mathrm{identity}\\ {\left(\mathrm{a}\text{–}\mathrm{b}\right)}^{2}={\mathrm{a}}^{2}+{\mathrm{b}}^{2}–2\mathrm{ab}\end{array}\right\}\\ =4{\mathrm{x}}^{2}+20\mathrm{x}+25-4{\mathrm{x}}^{2}+20\mathrm{x}-25\\ =40\mathrm{x}\\ \left(\mathrm{iii}\right)\text{ }{\left(7\mathrm{m}-8\mathrm{n}\right)}^{2}+{\left(7\mathrm{m}+8\mathrm{n}\right)}^{2}\\ =\left[{\left(7\mathrm{m}\right)}^{2}-2×7\mathrm{m}×8\mathrm{n}+{\left(8\mathrm{n}\right)}^{2}\right]+\left[{\left(7\mathrm{m}\right)}^{2}+2×7\mathrm{m}×8\mathrm{n}+{\left(8\mathrm{n}\right)}^{2}\right]\text{}\\ \left\{\begin{array}{l}\mathrm{By}\text{}\mathrm{using}\text{}\mathrm{the}\text{}\mathrm{identity}\\ {\left(\mathrm{a}\text{}-\text{}\mathrm{b}\right)}^{2}={\mathrm{a}}^{2}+{\mathrm{b}}^{2}-2\mathrm{ab}\end{array}\right\}\left\{\begin{array}{l}\mathrm{By}\text{}\mathrm{using}\text{}\mathrm{the}\text{}\mathrm{identity}\\ {\left(\mathrm{a}\text{}+\text{}\mathrm{b}\right)}^{2}={\mathrm{a}}^{2}+{\mathrm{b}}^{2}+2\mathrm{ab}\end{array}\right\}\\ =49{\mathrm{m}}^{2}-112\mathrm{mn}+64{\mathrm{n}}^{2}+49{\mathrm{m}}^{2}+112\mathrm{mn}+64{\mathrm{n}}^{2}\\ =98{\mathrm{m}}^{2}+128{\mathrm{n}}^{2}\\ \\ \left(\mathrm{iv}\right)\text{ }{\left(4\mathrm{m}+5\mathrm{n}\right)}^{2}+{\left(5\mathrm{m}+4\mathrm{n}\right)}^{2}\\ =\left[{\left(4\mathrm{m}\right)}^{2}+2×4\mathrm{m}×5\mathrm{n}+{\left(5\mathrm{n}\right)}^{2}\right]+\left[{\left(5\mathrm{m}\right)}^{2}+2×5\mathrm{m}×4\mathrm{n}+{\left(4\mathrm{n}\right)}^{2}\right]\text{}\\ \text{}\left\{\begin{array}{l}\mathrm{By}\text{}\mathrm{using}\text{}\mathrm{the}\text{}\mathrm{identity}\\ {\left(\mathrm{a}\text{}+\text{}\mathrm{b}\right)}^{2}={\mathrm{a}}^{2}+{\mathrm{b}}^{2}+2\mathrm{ab}\end{array}\right\}\\ =\left(16{\mathrm{m}}^{2}+40\mathrm{mn}+25{\mathrm{n}}^{2}\right)+\left(25{\mathrm{m}}^{2}+40\mathrm{mn}+16{\mathrm{n}}^{2}\right)\\ =41{\mathrm{m}}^{2}+80\mathrm{mn}+41{\mathrm{n}}^{2}\\ \\ \left(\mathrm{v}\right)\text{ }{\left(2.5\mathrm{p}-1.5\mathrm{q}\right)}^{2}-{\left(1.5\mathrm{p}-2.5\mathrm{q}\right)}^{2}\\ =\left[{\left(2.5\mathrm{p}\right)}^{2}-2×2.5\mathrm{p}×1.5\mathrm{q}+{\left(1.5\mathrm{q}\right)}^{2}\right]-\left[{\left(1.5\mathrm{p}\right)}^{2}-2×1.5\mathrm{p}×2.5\mathrm{q}+{\left(2.5\mathrm{q}\right)}^{2}\right]\text{}\\ \left\{\begin{array}{l}\mathrm{By}\text{}\mathrm{using}\text{}\mathrm{the}\text{}\mathrm{identity}\\ {\left(\mathrm{a}\text{}-\text{}\mathrm{b}\right)}^{2}={\mathrm{a}}^{2}+{\mathrm{b}}^{2}-2\mathrm{ab}\end{array}\right\}\\ =6.25{\mathrm{p}}^{2}-7.5\mathrm{pq}+2.25{\mathrm{q}}^{2}-\left[2.25{\mathrm{p}}^{2}-7.5\mathrm{pq}+6.25{\mathrm{q}}^{2}\right]\\ =6.25{\mathrm{p}}^{2}-7.5\mathrm{pq}+2.25{\mathrm{q}}^{2}-2.25{\mathrm{p}}^{2}+7.5\mathrm{pq}-6.25{\mathrm{q}}^{2}\\ =4{\mathrm{p}}^{2}-4{\mathrm{q}}^{2}\end{array}$
$\begin{array}{l}\left(\mathrm{vi}\right)\text{ }{\left(\mathrm{ab}+\mathrm{bc}\right)}^{2}-2{\mathrm{ab}}^{2}\mathrm{c}\\ =\left[{\left(\mathrm{ab}\right)}^{2}+2×\mathrm{ab}×\mathrm{bc}+{\left(\mathrm{bc}\right)}^{2}\right]-2{\mathrm{ab}}^{2}\mathrm{c}\text{}\\ \left\{\begin{array}{l}\mathrm{By}\text{}\mathrm{using}\text{}\mathrm{the}\text{}\mathrm{identity}\\ {\left(\mathrm{a}\text{}+\text{}\mathrm{b}\right)}^{2}={\mathrm{a}}^{2}+{\mathrm{b}}^{2}+2\mathrm{ab}\end{array}\right\}\\ ={\mathrm{a}}^{2}{\mathrm{b}}^{2}+2{\mathrm{ab}}^{2}\mathrm{c}+{\mathrm{b}}^{2}{\mathrm{c}}^{2}-2{\mathrm{ab}}^{2}\mathrm{c}\\ ={\mathrm{a}}^{2}{\mathrm{b}}^{2}+{\mathrm{b}}^{2}{\mathrm{c}}^{2}\\ \\ \left(\mathrm{vii}\right)\text{ }{\left({\mathrm{m}}^{2}-{\mathrm{n}}^{2}\mathrm{m}\right)}^{2}+2{\mathrm{m}}^{3}{\mathrm{n}}^{2}\\ =\left[{\left({\mathrm{m}}^{2}\right)}^{2}-2×{\mathrm{m}}^{2}×{\mathrm{n}}^{2}\mathrm{m}+{\left({\mathrm{n}}^{2}\mathrm{m}\right)}^{2}\right]+2{\mathrm{m}}^{3}{\mathrm{n}}^{2}\text{}\\ \left\{\begin{array}{l}\mathrm{By}\text{}\mathrm{using}\text{}\mathrm{the}\text{}\mathrm{identity}\\ {\left(\mathrm{a}\text{}-\text{}\mathrm{b}\right)}^{2}={\mathrm{a}}^{2}+{\mathrm{b}}^{2}-2\mathrm{ab}\end{array}\right\}\\ ={\mathrm{m}}^{4}-2{\mathrm{m}}^{3}{\mathrm{n}}^{2}+{\mathrm{n}}^{4}{\mathrm{m}}^{2}++2{\mathrm{m}}^{3}{\mathrm{n}}^{2}\\ ={\mathrm{m}}^{4}+{\mathrm{n}}^{4}{\mathrm{m}}^{2}\end{array}$
Q.5
$\begin{array}{l}\mathrm{Showthat}.\\ \left(\mathrm{i}\right){\left(3\mathrm{x}+7\right)}^{2}-84\mathrm{x}={\left(3\mathrm{x}-7\right)}^{2}\\ \left(\mathrm{ii}\right){\left(9\mathrm{p}-5\mathrm{q}\right)}^{2}+180\mathrm{pq}={\left(9\mathrm{p}+5\mathrm{q}\right)}^{2}\\ \left(\mathrm{iii}\right){\left(\frac{4}{3}\mathrm{m}-\frac{3}{4}\mathrm{n}\right)}^{2}+2\mathrm{mn}=\frac{16}{9}{\mathrm{m}}^{2}+\frac{9}{16}{\mathrm{n}}^{2}\\ \left(\mathrm{iv}\right){\left(4\mathrm{pq}+3\mathrm{q}\right)}^{2}-{\left(4\mathrm{pq}-3\mathrm{q}\right)}^{2}=48{\mathrm{pq}}^{2}\\ \left(\mathrm{v}\right)\left(\mathrm{a}-\mathrm{b}\right)\left(\mathrm{a}+\mathrm{b}\right)+\left(\mathrm{b}-\mathrm{c}\right)\left(\mathrm{b}+\mathrm{c}\right)+\left(\mathrm{c}-\mathrm{a}\right)\left(\mathrm{c}+\mathrm{a}\right)=0\end{array}$
Ans
$\begin{array}{l}\left(\mathrm{i}\right)\mathrm{L}.\mathrm{H}.\mathrm{S}={\left(3\mathrm{x}+7\right)}^{2}-84\mathrm{x}\\ ={\left(3\mathrm{x}\right)}^{2}+{7}^{2}+2×\left(3\mathrm{x}\right)×7-84\mathrm{x}\\ =9{\mathrm{x}}^{2}+49+42\mathrm{x}-84\mathrm{x}\\ =9{\mathrm{x}}^{2}+49-42\mathrm{x}\\ \text{\hspace{0.17em}}\\ \text{}\mathrm{R}.\mathrm{H}.\mathrm{S}={\left(3\mathrm{x}-7\right)}^{2}\\ ={\left(3\mathrm{x}\right)}^{2}+{7}^{2}-2×\left(3\mathrm{x}\right)×7\\ =9{\mathrm{x}}^{2}+49-42\mathrm{x}\\ \\ \therefore \mathrm{L}.\mathrm{H}.\mathrm{S}=\mathrm{R}.\mathrm{H}.\mathrm{S}\\ \mathrm{Hence}\text{}\mathrm{Proved}\\ \left(\mathrm{ii}\right)\mathrm{L}.\mathrm{H}.\mathrm{S}={\left(9\mathrm{p}-5\mathrm{q}\right)}^{2}+180\mathrm{pq}\\ ={\left(9\mathrm{p}\right)}^{2}+{\left(5\mathrm{q}\right)}^{2}-2×\left(9\mathrm{p}\right)×5\mathrm{q}+180\mathrm{pq}\\ =81{\mathrm{p}}^{2}+25{\mathrm{q}}^{2}-90\mathrm{pq}+180\mathrm{pq}\\ =81{\mathrm{p}}^{2}+25{\mathrm{q}}^{2}+90\mathrm{pq}\\ \text{\hspace{0.17em}}\\ \text{}\mathrm{R}.\mathrm{H}.\mathrm{S}={\left(9\mathrm{p}+5\mathrm{q}\right)}^{2}\\ ={\left(9\mathrm{p}\right)}^{2}+{\left(5\mathrm{q}\right)}^{2}+2×\left(9\mathrm{p}\right)×5\mathrm{q}\\ =81{\mathrm{p}}^{2}+25{\mathrm{q}}^{2}+90\mathrm{pq}\\ \\ \therefore \mathrm{L}.\mathrm{H}.\mathrm{S}=\mathrm{R}.\mathrm{H}.\mathrm{S}\\ \mathrm{Hence}\text{}\mathrm{Proved}\\ \left(\mathrm{iii}\right)\mathrm{L}.\mathrm{H}.\mathrm{S}={\left(\frac{4}{3}\mathrm{m}-\frac{3}{4}\mathrm{n}\right)}^{2}+2\mathrm{mn}\\ ={\left(\frac{4}{3}\mathrm{m}\right)}^{2}+{\left(\frac{3}{4}\mathrm{n}\right)}^{2}-2×\left(\frac{4}{3}\mathrm{m}\right)×\left(\frac{3}{4}\mathrm{n}\right)+2\mathrm{mn}\\ =\frac{16}{9}{\mathrm{m}}^{2}+\frac{9}{16}{\mathrm{n}}^{2}-2\mathrm{mn}+2\mathrm{mn}\\ =\frac{16}{9}{\mathrm{m}}^{2}+\frac{9}{16}{\mathrm{n}}^{2}\\ \text{\hspace{0.17em}}\\ \text{}\mathrm{R}.\mathrm{H}.\mathrm{S}=\frac{16}{9}{\mathrm{m}}^{2}+\frac{9}{16}{\mathrm{n}}^{2}\\ \therefore \text{\hspace{0.17em}}\mathrm{L}.\mathrm{H}.\mathrm{S}=\mathrm{R}.\mathrm{H}.\mathrm{S}\\ \mathrm{Hence}\text{}\mathrm{Proved}\end{array}$
$\begin{array}{l}\left(\mathrm{iv}\right)\mathrm{L}.\mathrm{H}.\mathrm{S}={\left(4\mathrm{pq}+3\mathrm{q}\right)}^{2}-{\left(4\mathrm{pq}-3\mathrm{q}\right)}^{2}=48{\mathrm{pq}}^{2}\\ ={\left(4\mathrm{pq}\right)}^{2}+{\left(3\mathrm{q}\right)}^{2}+2×\left(4\mathrm{pq}\right)×\left(3\mathrm{q}\right)-\left[{\left(4\mathrm{pq}\right)}^{2}+{\left(3\mathrm{q}\right)}^{2}-2×m\left(4\mathrm{pq}\right)×\left(3\mathrm{q}\right)\right]\\ =16{\mathrm{p}}^{2}{\mathrm{q}}^{2}+9{\mathrm{q}}^{2}+24{\mathrm{pq}}^{2}-\left[16{\mathrm{p}}^{2}{\mathrm{q}}^{2}+9{\mathrm{q}}^{2}-24{\mathrm{pq}}^{2}\right]\\ =16{\mathrm{p}}^{2}{\mathrm{q}}^{2}+9{\mathrm{q}}^{2}+24{\mathrm{pq}}^{2}-16{\mathrm{p}}^{2}{\mathrm{q}}^{2}-9{\mathrm{q}}^{2}+24{\mathrm{pq}}^{2}\\ =48{\mathrm{pq}}^{2}\\ \text{\hspace{0.17em}}\\ \text{}\mathrm{R}.\mathrm{H}.\mathrm{S}=48{\mathrm{pq}}^{2}\\ \\ \therefore \text{\hspace{0.17em}}\mathrm{L}.\mathrm{H}.\mathrm{S}=\mathrm{R}.\mathrm{H}.\mathrm{S}\\ \mathrm{Hence}\text{}\mathrm{Proved}\\ \left(\mathrm{v}\right)\mathrm{L}.\mathrm{H}.\mathrm{S}=\left(\mathrm{a}-\mathrm{b}\right)\left(\mathrm{a}+\mathrm{b}\right)+\left(\mathrm{b}-\mathrm{c}\right)\left(\mathrm{b}+\mathrm{c}\right)+\left(\mathrm{c}-\mathrm{a}\right)\left(\mathrm{c}+\mathrm{a}\right)\\ ={\mathrm{a}}^{2}-{\mathrm{b}}^{2}+{\mathrm{b}}^{2}-{\mathrm{c}}^{2}+{\mathrm{c}}^{2}-{\mathrm{a}}^{2}\\ =0\\ \text{\hspace{0.17em}}\\ \text{}\mathrm{R}.\mathrm{H}.\mathrm{S}=0\\ \\ \therefore \text{\hspace{0.17em}}\mathrm{L}.\mathrm{H}.\mathrm{S}=\mathrm{R}.\mathrm{H}.\mathrm{S}\\ \mathrm{Hence}\text{}\mathrm{Proved}\end{array}$
Q.6
$\begin{array}{l}\mathrm{Using}\text{ }\mathrm{identities},\text{ }\mathrm{evaluate}.\\ \left(\mathrm{i}\right)\text{ }{71}^{2}\text{ }\left(\mathrm{ii}\right)\text{ }{99}^{2}\text{ }\left(\mathrm{iii}\right)\text{ }{102}^{2}\text{ }\left(\mathrm{iv}\right)\text{ }{998}^{2}\\ \left(\mathrm{v}\right)\text{ }5.{2}^{2}\text{ }\left(\mathrm{vi}\right)\text{ }297×303\text{ }\left(\mathrm{vii}\right)\text{ }78×82\text{ }\left(\mathrm{viii}\right)\text{ }8.{9}^{2}\\ \left(\mathrm{ix}\right)\text{ }10.5×9.5\end{array}$
Ans
$\begin{array}{l}\left(\mathrm{i}\right)\text{ }{71}^{2}={\left(70+1\right)}^{2}\\ ={\left(70\right)}^{2}+2×70×1+{1}^{2}\text{}\left\{\begin{array}{l}\mathrm{By}\text{}\mathrm{using}\text{}\mathrm{the}\text{}\mathrm{identity}\\ {\left(\mathrm{a}\text{}+\text{}\mathrm{b}\right)}^{2}={\mathrm{a}}^{2}+{\mathrm{b}}^{2}+2\mathrm{ab}\end{array}\right\}\\ =4900+140+1\\ =5041\\ \left(\mathrm{ii}\right)\text{ }{99}^{2}={\left(100-1\right)}^{2}\\ ={\left(100\right)}^{2}-2×100×1+{1}^{2}\text{}\left\{\begin{array}{l}\mathrm{By}\text{}\mathrm{using}\text{}\mathrm{the}\text{}\mathrm{identity}\\ {\left(\mathrm{a}\text{}-\text{}\mathrm{b}\right)}^{2}={\mathrm{a}}^{2}+{\mathrm{b}}^{2}-2\mathrm{ab}\end{array}\right\}\\ =10000-200+1\\ =9801\\ \left(\mathrm{iii}\right)\text{ }{102}^{2}={\left(100+2\right)}^{2}\\ ={\left(100\right)}^{2}+2×100×2+{2}^{2}\text{}\left\{\begin{array}{l}\mathrm{By}\text{}\mathrm{using}\text{}\mathrm{the}\text{}\mathrm{identity}\\ {\left(\mathrm{a}\text{}+\text{}\mathrm{b}\right)}^{2}={\mathrm{a}}^{2}+{\mathrm{b}}^{2}+2\mathrm{ab}\end{array}\right\}\\ =10000+400+4\\ =10404\\ \\ \left(\mathrm{iv}\right)\text{ }{998}^{2}={\left(1000-2\right)}^{2}\\ ={\left(1000\right)}^{2}-2×1000×2+{2}^{2}\text{}\left\{\begin{array}{l}\mathrm{By}\text{}\mathrm{using}\text{}\mathrm{the}\text{}\mathrm{identity}\\ {\left(\mathrm{a}\text{}-\text{}\mathrm{b}\right)}^{2}={\mathrm{a}}^{2}+{\mathrm{b}}^{2}-2\mathrm{ab}\end{array}\right\}\\ =1000000-4000+4\\ =996004\end{array}$
$\begin{array}{l}\left(\mathrm{v}\right)\text{ }5{.2}^{2}={\left(5+0.2\right)}^{2}\\ ={\left(5\right)}^{2}+2×5×0.2+{\left(0.2\right)}^{2}\text{}\left\{\begin{array}{l}\mathrm{By}\text{}\mathrm{using}\text{}\mathrm{the}\text{}\mathrm{identity}\\ {\left(\mathrm{a}\text{}+\text{}\mathrm{b}\right)}^{2}={\mathrm{a}}^{2}+{\mathrm{b}}^{2}+2\mathrm{ab}\end{array}\right\}\\ =25+2+0.04\\ =27.04\\ \\ \left(\mathrm{vi}\right)\text{ }297×303=\left(300-3\right)\left(300+3\right)\\ ={\left(300\right)}^{2}-{3}^{2}\text{}\left\{\begin{array}{l}\mathrm{By}\text{}\mathrm{using}\text{}\mathrm{the}\text{}\mathrm{identity}\\ \left(\mathrm{a}-\mathrm{b}\right)\left(\mathrm{a}+\mathrm{b}\right)={\mathrm{a}}^{2}-{\mathrm{b}}^{2}\end{array}\right\}\\ =90000-9\\ =89991\\ \\ \left(\mathrm{vii}\right)\text{ }78×82=\left(80-2\right)\left(80+2\right)\\ ={\left(80\right)}^{2}-{2}^{2}\text{}\left\{\begin{array}{l}\mathrm{By}\text{}\mathrm{using}\text{}\mathrm{the}\text{}\mathrm{identity}\\ \left(\mathrm{a}-\mathrm{b}\right)\left(\mathrm{a}+\mathrm{b}\right)={\mathrm{a}}^{2}-{\mathrm{b}}^{2}\end{array}\right\}\\ =6400-4\\ =6396\\ \\ \left(\mathrm{viii}\right)\text{ }8{.9}^{2}={\left(9.0-0.1\right)}^{2}\\ =81-1.8+0.01\text{}\left\{\begin{array}{l}\mathrm{By}\text{}\mathrm{using}\text{}\mathrm{the}\text{}\mathrm{identity}\\ {\left(\mathrm{a}-\mathrm{b}\right)}^{2}={\mathrm{a}}^{2}+{\mathrm{b}}^{2}-2\mathrm{ab}\end{array}\right\}\\ =79.21\\ \\ \left(\mathrm{ix}\right)\text{ }10.5×9.5\\ =\left(10+0.5\right)\left(10–0.5\right)\\ =\left[{10}^{2}-{\left(0.5\right)}^{2}\right]\left\{\begin{array}{l}\mathrm{By}\text{}\mathrm{using}\text{}\mathrm{the}\text{}\mathrm{identity}\\ \left(\mathrm{a}-\mathrm{b}\right)\left(\mathrm{a}+\mathrm{b}\right)={\mathrm{a}}^{2}-{\mathrm{b}}^{2}\end{array}\right\}\\ =100–0.25=99.75\end{array}$
Q.7
$\begin{array}{l}{\text{Using a}}^{\text{2}}-{\text{b}}^{\text{2}}\text{= (a + b) (a}-\text{b), find}\\ {\text{(i) 51}}^{\text{2}}\text{}-{\text{49}}^{\text{2}}\text{(ii) (1}{\text{.02)}}^{\text{2}}-\text{(0}{\text{.98)}}^{\text{2}}{\text{(iii) 153}}^{\text{2}}\text{}-{\text{147}}^{\text{2}}\\ \text{(iv) 12}{\text{.1}}^{\text{2}}\text{}-\text{7}{\text{.9}}^{\text{2}}\end{array}$
Ans
$\begin{array}{l}\left(\mathrm{i}\right)\text{ }{51}^{2}-{49}^{2}\\ =\left(51+49\right)\left(51-49\right)\\ =100×2\\ =200\\ \\ {\text{(ii) (1.02)}}^{\text{2}}-{\text{(0.98)}}^{\text{2}}\text{}\\ =\left(1.02+0.98\right)\left(1.02-0.98\right)\\ =2×0.04\\ =0.08\\ \\ \left(\mathrm{iii}\right)\text{ }{153}^{2}-{147}^{2}\\ =\left(153+147\right)\left(153-147\right)\\ =300×6\\ =1800\\ \\ \left(\mathrm{iv}\right)\text{ }12{.1}^{2}-7{.9}^{2}\\ =\left(12.1+7.9\right)\left(12.1-7.9\right)\\ =20.0×4.2\\ =84\\ \end{array}$
Q.8
$\begin{array}{l}\mathrm{Using}\text{ }\left(\mathrm{x}+\mathrm{a}\right)\text{ }\left(\mathrm{x}+\mathrm{b}\right)={\mathrm{x}}^{2}+\left(\mathrm{a}+\mathrm{b}\right)\mathrm{x}+\mathrm{ab},\text{ }\mathrm{find}\\ \left(\mathrm{i}\right)\text{ }103×104\text{ }\left(\mathrm{ii}\right)\text{ }5.1×5.2\text{ }\left(\mathrm{iii}\right)\text{ }103×98\text{ }\left(\mathrm{iv}\right)\text{ }9.7×9.8\end{array}$
Ans
$\begin{array}{l}\left(\mathrm{i}\right)\text{ }103×104\\ =\left(100+3\right)\left(100+4\right)\\ ={\left(100\right)}^{2}+\left(3+4\right)×100+3×4\\ =10000+700+12\\ =10712\\ \\ \left(\mathrm{ii}\right)\text{ }5.1×5.2\\ =\left(5+0.1\right)\left(5+0.2\right)\\ ={\left(5\right)}^{2}+\left(0.1+0.2\right)×5+0.1×0.2\\ =25+1.5+0.02\\ =26.52\\ \left(\mathrm{iii}\right)\text{ }103×98\\ =\left(100+3\right)\left(100-2\right)\\ ={\left(100\right)}^{2}+\left[3+\left(-2\right)\right]×100+3×\left(-2\right)\\ =10000+100-6\\ =10094\\ \\ \left(\mathrm{iv}\right)\text{ }9.7×9.8\\ =\left(10-0.3\right)\left(10-0.2\right)\\ ={\left(10\right)}^{2}+\left[\left(-0.3\right)+\left(-0.2\right)\right]×10+\left(-0.3\right)×\left(-0.2\right)\\ =100+\left(-0.5\right)10+0.06\\ =100.06-5=95.06\end{array}$
## FAQs (Frequently Asked Questions)
### 1. Why should students use the NCERT Solutions For Class 8 Maths Chapter 9 Exercise 9.5 to study?
When drafting the NCERT curriculum, the Central Board of Secondary Education’s requirements are taken into account. The NCERT solutions are accessible via the Extramarks website and mobile application. The study materials and learning aids on the Extramarks website and mobile application are based on the CBSE standards. Students in Class 8 can obtain the NCERT Solutions For Class 8 Maths Chapter 9 Exercise 9.5 from the Extramarks website for their study requirements since they have been designed in accordance with the CBSE Board standards and guidelines.
### 2. How significant is Chapter 8 Algebraic Expressions and Identities in Class 8 Mathematics?
Chapter 8 Algebraic Expression and Identities is an integral part of the Class 8 Mathematics curriculum. It is critical that students comprehend the significance of this chapter in the Senior Secondary Mathematics Examination. In the senior secondary examination, Chapter 8 Algebraic Expression and Identities has a significant weightage. It is a very technical chapter, and students are expected to study it thoroughly. Since this is an application-based chapter, students must routinely and consistently practice the NCERT exercise problems and solutions. Students may take advantage of the NCERT Solutions For Class 8 Maths Chapter 9 Exercise 9.5 from the Extramarks website to prepare for the exam.
### 3. Why should students in Class 8 resort to NCERT Solutions For Class 8 Maths Chapter 9 Exercise 9.5 revision notes and sample papers?
It is vital that students solve practice papers since doing so can improve their understanding and application of mathematics concepts. By solving sample papers, students’ competence in the relevant subject will be challenged and developed. Students can acquire sample papers for this purpose. Students can get practice papers via the Extramarks website and mobile application. They can also get NCERT Solutions For Class 8 Maths Chapter 9 Exercise 9.5 practice and sample papers from the Extramarks website and learning application. Students can prepare by doing sample papers. They may gain confidence by practising answering exercise problems and answers, as well as solving sample papers.
### 4. Is the Mathematics Senior Secondary Examination of Class 8 difficult?
The Mathematics Senior Secondary Examination in Class 8 might be difficult. However, if students prepare extensively before taking the examination, it is not difficult to obtain high grades in the senior secondary examination. Students must thoroughly cover the relevant syllabus in order to perform well in the Senior Secondary Examination of Mathematics. Students can use the NCERT Solutions For Class 8 Maths Chapter 9 Exercise 9.5 to help them prepare.
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English
# Volume of a Hexagonal Prism Calculator
A hexagon is a six-sided polygon. The hexagonal prism is a prism with a hexagonal base. This polyhedron has 8 faces, 18 edges, and 12 vertices. Online Volume Of a Hexagonal Prism calculator helps you to calculate the volume of a hexagonal prism based on side and height. They are of two types namely, Regular and Irregular hexagonal prism. Regular hexagons have all bases with same length whereas irregular hexagon bases do not have the same length.
length
length
volume
A hexagon is a six-sided polygon. The hexagonal prism is a prism with a hexagonal base. This polyhedron has 8 faces, 18 edges, and 12 vertices. Online Volume Of a Hexagonal Prism calculator helps you to calculate the volume of a hexagonal prism based on side and height. They are of two types namely, Regular and Irregular hexagonal prism. Regular hexagons have all bases with same length whereas irregular hexagon bases do not have the same length.
#### Formula:
v=(3 x √3/2) x a2 x h Where, v = Volume of a Hexagonal Prism a = Side h = Height
#### Example
A Hexagonal prism has a side of 10 mm and height of 7 mm. Find its volume.
Volume = (3 x √3 / 2) ×10 ^2 x 7
= (3 x √3 / 2) x 100 × 7
V= 1818.65 mm3.
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# Number 5543
Number 5,543 spell 🔊, write in words: five thousand, five hundred and forty-three . Ordinal number 5543th is said 🔊 and write: five thousand, five hundred and forty-third. The meaning of number 5543 in Maths: Is Prime? Factorization and prime factors tree. The square root and cube root of 5543. What is 5543 in computer science, numerology, codes and images, writing and naming in other languages. Other interesting facts related to 5543.
## What is 5,543 in other units
The decimal (Arabic) number 5543 converted to a Roman number is (V)DXLIII. Roman and decimal number conversions.
The number 5543 converted to a Mayan number is Decimal and Mayan number conversions.
#### Weight conversion
5543 kilograms (kg) = 12220.1 pounds (lbs)
5543 pounds (lbs) = 2514.3 kilograms (kg)
#### Length conversion
5543 kilometers (km) equals to 3445 miles (mi).
5543 miles (mi) equals to 8921 kilometers (km).
5543 meters (m) equals to 18186 feet (ft).
5543 feet (ft) equals 1690 meters (m).
5543 centimeters (cm) equals to 2182.3 inches (in).
5543 inches (in) equals to 14079.2 centimeters (cm).
#### Temperature conversion
5543° Fahrenheit (°F) equals to 3061.7° Celsius (°C)
5543° Celsius (°C) equals to 10009.4° Fahrenheit (°F)
#### Power conversion
5543 Horsepower (hp) equals to 4076.32 kilowatts (kW)
5543 kilowatts (kW) equals to 7537.41 horsepower (hp)
#### Time conversion
(hours, minutes, seconds, days, weeks)
5543 seconds equals to 1 hour, 32 minutes, 23 seconds
5543 minutes equals to 3 days, 20 hours, 23 minutes
### Zip codes 5543
• Zip code 5543 ALTO GRANDE, MENDOZA, Argentina a map
• Zip code 5543 CAPDEVILLE, MENDOZA, Argentina a map
• Zip code 5543 COLONIA ALEMANA, MENDOZA, Argentina a map
Zip code areas 5543
### Codes and images of the number 5543
Number 5543 morse code: ..... ..... ....- ...--
Sign language for number 5543:
Number 5543 in braille:
Images of the number
Image (1) of the numberImage (2) of the number
More images, other sizes, codes and colors ...
### Gregorian, Hebrew, Islamic, Persian and Buddhist year (calendar)
Gregorian year 5543 is Buddhist year 6086.
Buddhist year 5543 is Gregorian year 5000 .
Gregorian year 5543 is Islamic year 5072 or 5073.
Islamic year 5543 is Gregorian year 5999 or 6000.
Gregorian year 5543 is Persian year 4921 or 4922.
Persian year 5543 is Gregorian 6164 or 6165.
Gregorian year 5543 is Hebrew year 9303 or 9304.
Hebrew year 5543 is Gregorian year 1783.
The Buddhist calendar is used in Sri Lanka, Cambodia, Laos, Thailand, and Burma. The Persian calendar is official in Iran and Afghanistan.
## Mathematics of no. 5543
### Multiplications
#### Multiplication table of 5543
5543 multiplied by two equals 11086 (5543 x 2 = 11086).
5543 multiplied by three equals 16629 (5543 x 3 = 16629).
5543 multiplied by four equals 22172 (5543 x 4 = 22172).
5543 multiplied by five equals 27715 (5543 x 5 = 27715).
5543 multiplied by six equals 33258 (5543 x 6 = 33258).
5543 multiplied by seven equals 38801 (5543 x 7 = 38801).
5543 multiplied by eight equals 44344 (5543 x 8 = 44344).
5543 multiplied by nine equals 49887 (5543 x 9 = 49887).
show multiplications by 6, 7, 8, 9 ...
### Fractions: decimal fraction and common fraction
#### Fraction table of 5543
Half of 5543 is 2771,5 (5543 / 2 = 2771,5 = 2771 1/2).
One third of 5543 is 1847,6667 (5543 / 3 = 1847,6667 = 1847 2/3).
One quarter of 5543 is 1385,75 (5543 / 4 = 1385,75 = 1385 3/4).
One fifth of 5543 is 1108,6 (5543 / 5 = 1108,6 = 1108 3/5).
One sixth of 5543 is 923,8333 (5543 / 6 = 923,8333 = 923 5/6).
One seventh of 5543 is 791,8571 (5543 / 7 = 791,8571 = 791 6/7).
One eighth of 5543 is 692,875 (5543 / 8 = 692,875 = 692 7/8).
One ninth of 5543 is 615,8889 (5543 / 9 = 615,8889 = 615 8/9).
show fractions by 6, 7, 8, 9 ...
### Calculator
5543
#### Is Prime?
The number 5543 is not a prime number. The closest prime numbers are 5531, 5557.
5543th prime number in order is 54493.
#### Factorization and factors (dividers)
The prime factors of 5543 are 23 * 241
The factors of 5543 are 1 , 23 , 241 , 5543
Total factors 4.
Sum of factors 5808 (265).
#### Powers
The second power of 55432 is 30.724.849.
The third power of 55433 is 170.307.838.007.
#### Roots
The square root √5543 is 74,451326.
The cube root of 35543 is 17,697624.
#### Logarithms
The natural logarithm of No. ln 5543 = loge 5543 = 8,620291.
The logarithm to base 10 of No. log10 5543 = 3,743745.
The Napierian logarithm of No. log1/e 5543 = -8,620291.
### Trigonometric functions
The cosine of 5543 is 0,333711.
The sine of 5543 is 0,942676.
The tangent of 5543 is 2,824828.
### Properties of the number 5543
More math properties ...
## Number 5543 in Computer Science
Code typeCode value
PIN 5543 It's recommendable to use 5543 as a password or PIN.
5543 Number of bytes5.4KB
Unix timeUnix time 5543 is equal to Thursday Jan. 1, 1970, 1:32:23 a.m. GMT
IPv4, IPv6Number 5543 internet address in dotted format v4 0.0.21.167, v6 ::15a7
5543 Decimal = 1010110100111 Binary
5543 Decimal = 21121022 Ternary
5543 Decimal = 12647 Octal
5543 Decimal = 15A7 Hexadecimal (0x15a7 hex)
5543 BASE64NTU0Mw==
5543 MD5a10a26631d45928cb8be4ebabbee8b8d
5543 SHA1fa33988727d7f572be002b845e65858e68bb92e9
5543 SHA224e83e27efb19e2684d9ae048b3043748615992624edd0c1275aed71ab
5543 SHA2569302a967cce443780477212553ae3d74625dd165a096a4ee8299dcf0c8079866
More SHA codes related to the number 5543 ...
If you know something interesting about the 5543 number that you did not find on this page, do not hesitate to write us here.
## Numerology 5543
### The meaning of the number 5 (five), numerology 5
Character frequency 5: 2
The number five (5) came to this world to achieve freedom. You need to apply discipline to find your inner freedom and open-mindedness. It is about a restless spirit in constant search of the truth that surrounds us. You need to accumulate as much information as possible to know what is happening in depth. Number 5 person is intelligent, selfish, curious and with great artistic ability. It is a symbol of freedom, independence, change, adaptation, movement, the search for new experiences, the traveling and adventurous spirit, but also of inconsistency and abuse of the senses.
More about the meaning of the number 5 (five), numerology 5 ...
### The meaning of the number 4 (four), numerology 4
Character frequency 4: 1
The number four (4) came to establish stability and to follow the process in the world. It needs to apply a clear purpose to develop internal stability. It evokes a sense of duty and discipline. Number 4 personality speaks of solid construction. It teaches us to evolve in the tangible and material world, to develop reason and logic and our capacity for effort, accomplishment and work.
More about the meaning of the number 4 (four), numerology 4 ...
### The meaning of the number 3 (three), numerology 3
Character frequency 3: 1
The number three (3) came to share genuine expression and sensitivity with the world. People associated with this number need to connect with their deepest emotions. The number 3 is characterized by its pragmatism, it is utilitarian, sagacious, dynamic, creative, it has objectives and it fulfills them. He/she is also self-expressive in many ways and with good communication skills.
More about the meaning of the number 3 (three), numerology 3 ...
## Interesting facts about the number 5543
### Asteroids
• (5543) Sharaf is asteroid number 5543. It was discovered by N. S. Chernyj from Nauchni on 10/3/1978.
### Distances between cities
• There is a 5,543 miles (8,920 km) direct distance between Adana (Turkey) and Bandung (Indonesia).
• There is a 5,543 miles (8,919 km) direct distance between Antananarivo (Madagascar) and Recife (Brazil).
• There is a 3,445 miles (5,543 km) direct distance between Ashgabat (Turkmenistan) and Harbin (China).
• There is a 5,543 miles (8,920 km) direct distance between Barcelona (Spain) and Tegucigalpa (Honduras).
• There is a 5,543 miles (8,920 km) direct distance between Brussels (Belgium) and Nanchang (China).
• There is a 5,543 miles (8,919 km) direct distance between City of London (United Kingdom) and Ciudad Nezahualcoyotl (Mexico).
• There is a 5,543 miles (8,919 km) direct distance between Ciudad Juárez (Mexico) and Marrakesh (Morocco).
• There is a 5,543 miles (8,919 km) direct distance between Fès (Morocco) and Vancouver (Alberta).
• There is a 3,445 miles (5,543 km) direct distance between Grand Dakar (Senegal) and Kigali (Rwanda).
• There is a 3,445 miles (5,543 km) direct distance between Jos (Nigeria) and Stockholm (Sweden).
• There is a 5,543 miles (8,919 km) direct distance between Lucknow (India) and Port Elizabeth (South Africa).
• There is a 5,543 miles (8,920 km) direct distance between Masjid Jamie Baitul Muttaqien (Indonesia) and Rostov-na-Donu (Russia).
• There is a 5,543 miles (8,919 km) direct distance between Montréal (Alberta) and Mosul (Iraq).
• There is a 5,543 miles (8,919 km) direct distance between Perth (Australia) and Ulaanbaatar (Mongolia).
• There is a 3,445 miles (5,543 km) direct distance between Saint Petersburg (Russia) and Xining (China).
• There is a 5,543 miles (8,919 km) direct distance between Sale (Morocco) and Tiruchirappalli (India).
• There is a 3,445 miles (5,543 km) direct distance between Seongnam-si (South Korea) and Ufa (Russia).
### Mathematics
• 5543 has a 4th power that is the sum of four 4th powers.
## Number 5,543 in other languages
How to say or write the number five thousand, five hundred and forty-three in Spanish, German, French and other languages. The character used as the thousands separator.
Spanish: 🔊 (número 5.543) cinco mil quinientos cuarenta y tres German: 🔊 (Anzahl 5.543) fünftausendfünfhundertdreiundvierzig French: 🔊 (nombre 5 543) cinq mille cinq cent quarante-trois Portuguese: 🔊 (número 5 543) cinco mil, quinhentos e quarenta e três Chinese: 🔊 (数 5 543) 五千五百四十三 Arabian: 🔊 (عدد 5,543) خمسة آلاف و خمسمائة و ثلاثة و أربعون Czech: 🔊 (číslo 5 543) pět tisíc pětset čtyřicet tři Korean: 🔊 (번호 5,543) 오천오백사십삼 Danish: 🔊 (nummer 5 543) femtusinde og femhundrede og treogfyrre Hebrew: (מספר 5,543) חמש אלף חמש מאות ארבעים ושלש Dutch: 🔊 (nummer 5 543) vijfduizendvijfhonderddrieënveertig Japanese: 🔊 (数 5,543) 五千五百四十三 Indonesian: 🔊 (jumlah 5.543) lima ribu lima ratus empat puluh tiga Italian: 🔊 (numero 5 543) cinquemilacinquecentoquarantatré Norwegian: 🔊 (nummer 5 543) fem tusen, fem hundre og førti-tre Polish: 🔊 (liczba 5 543) pięć tysięcy pięćset czterdzieści trzy Russian: 🔊 (номер 5 543) пять тысяч пятьсот сорок три Turkish: 🔊 (numara 5,543) beşbinbeşyüzkırküç Thai: 🔊 (จำนวน 5 543) ห้าพันห้าร้อยสี่สิบสาม Ukrainian: 🔊 (номер 5 543) п'ять тисяч п'ятсот сорок три Vietnamese: 🔊 (con số 5.543) năm nghìn năm trăm bốn mươi ba Other languages ...
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# On smurfs – playing around with genetic drift in a spatially structured population
A lot of people seem to struggle with the idea that stochastic processes can produce deterministic outcomes on grand scales. A prime example is genetic drift and how its results (and it’s interactions with silent mutations) can be used to assess past population structures and sizes.
As I’m increasingly finding that a good way to better understand a concept is to build it into a little model where you can tweak the parameters at your will (and also as an exercise in try-except-constructions and classes), I’ve written some Python code that simulates neutral evolution(0). It doesn’t show anything new, but it may be didactical. The bottom line is the following graph, which shows how the diversity of a population (the number of different alleles present) starts to oscillate around a point of equilibrium between mutations introducing new variants and existent once drifting out of the population after a while, the level where that equilibrium is reached being function of the population size and irrespective of initial diversity.
The total diversity (number of different genotypes) of a population over time under 3×2 conditions – different initial diversity and three different population sizes; log-scale on y-axis.
What this little toy of mine can not measure is different types of diversity: In my model, all variants are created equal, and don’t stand in any special relation to each other, so when I assume a possibility space of 100000, if and when a carrier of variant #73489 undergoes a mutation, the result can be any of the remaining 99999 possible variants. In reality, variants form a network of possible transformations, with variants that are closer or more distant from each other. So my model (even if it allowed to change the size of a population over time, which would currently require some ugly hacks) is insufficient to distinguish a large population that is the result of a recent expansion from a medium-sized population from a large population that is the result of a not-so-recent expansion from a small population – both will show less diversity than is expected for their size, from which fact, absent a way to tell different types of diversity apart, we could only conclude (assuming we know the mutation rate and reproduction patterns) is that either of these things must have happened. In reality, one would have (say) hundred different variants most of which are close to each other (converging at just a few right before the expansion started), while the other might have the same number of variants, but those would be more distinct from each other. Another feature of real genomes absent in this simple model is that you can track the variants of multiple genetic loci individually. This allows to diagnose subdivisions of the population when the limits of the ranges of individual variants correlate, which they shouldn’t if sheer distance in an otherwise uniform population is all that’s at work.
These shortcomings notwithstanding, the model is sufficient to see the effect of population size on gross diversity.
The baseline
Let’s start with a very small population, i.e. 50 smurfs.(1) In other words, I initiate the population with the command `population = Pop([x for x in range(50)])`. Those are lined up in a circle such that individual #49 is a direct neighbour of individual #0. Imagine theylive along the coast of a small island with circumference of 500 meters, each occupying its own small patch 10 metres wide.
Initially, we start out with 50 smurfs representing 50 different genotypes, like this (I’ve fiddled around with the colors a bit so that all and only the individuals whose descendants stay around until generation 13 when the variability has shrunken to ten different genotypes are identified with a unique color, and all those who’ve gone extinct by then default to turquoise):
At the end of each generation, the smurfs reproduce and die. Each dying individual is replaced by a clone of either itself or one of the nearby individuals, where the distance between parent and offspring is regulated by a normal distribution. For this demonstration, I used a standard deviation of 2.0, so over 90% of the time, parents will come from within a range of 3 neighbours to either side(2). For two marked individuals, I identified those ranges of likely parents of their successors as the area between the arcs.
Now that we’re set up with a maximally heterogeneous initial population, we can let the games begin (`population.run(1,2,limit=30)`). here’s how the population developed over 30 generations – the result will be slightly different each time, but something along those lines. Each vertical column represents a single generation, each horizontal row a spot on the island’s shore, with the top and bottom rows connected imagine we’re mapping the surface of a tube (this map can be produced with `population.area_plot(v_stretch = True, h_stretch=True)`, although I fiddled a bit with the colors: `for genotype in range(50): if genotype not in population.freqs[population.k]: population.colors[genotype] = "turquoise"` should do the trick):
From generation 13 on, marked with a dashed vertical line, the total variability has been reduced to 10 different genotypes, now each identified with a unique color. By generation 30, we’re down to five, without any inherent advantages to any smurf in the original population – all because some smurfs and their successor happened to be in the right spot at the right time, and those things add up.
Let’s zoom in on generation 30 again:
Again, two individuals’ likely successors are marked with those same arcs. For the individual on the right, we see that it’s almost certainly going to remain red: the probability that its successor will be coming from within that solid block of reds of which it is a part is almost 99%.
If we continue the game, we’ll almost certainly achieve uniformity within a few hundred generations.
Fiddling around with population sizes
Now that we’ve established the basic principles at work, let’s try some more realistically sized populations, and compare the results for the same set of parameters except for varying population size (I’m using a slightly wider standard deviation of 10.0 for picking the parents than above with the minipopulation of 50 where i used 2.0 for the purposes of saving my battery – with a smaller local range, it takes longer for the same effects to become visible).
We see that the larger the population is, the longer it takes for it to achieve fixation, i.e. before one variant comes to totally dominate the genepool. With a population of 200, this happens within a few hundred generations, whether we start with a near homogeneous population (only two types initially), or a maximally heterogeneous one.
Initially near-homogeneous population with only two different types – `Pop(200,2)`
Initially maximally heterogeneous population with each smurf representing a different type; by the time fixation is reached, all smurfs are descendants of one and the same initial smurf(Important note: 3)
A population of 1000 may not achieve uniformity over 2000 generations, although we’re down to only two types before generation 1000. After that, the effect of local sampling, irrelevant in a population of 200, kicks in – the two subpopulations form local clusters, and their overall frequencies don’t change that much since only deaths and births in the area where there’s an overlap have a potential to effect them.
Initially near-homogeneous population of 1000
Initially maximally heterogeneous population of 1000
This is even more pronounced with a population of 5000, where at the end of 2000 generations, we still have 11 different genotypes out of the original 5000. They too trend towards homogeneity, but much slower, and the geographical separation slows the process down quite a bit.
Population of 5000, initially maximally heterogeneous (5000 types), 11 types remaining by generation 2000.
Population of 5000, initially two types
We can directly compare the degree of diversity in each population over time:
Number of different types in the population over time, for all six conditions
Introducing mutations
We have seen that drift is a force that will always push towards uniformity and that is the stronger the smaller the population (and the less internally structured, as we would see by changing the size of the sampling space for parents, leaving all else equal). But real populations don’t always trend towards uniformity, and neither should we expect them to. There’s a force that pushes towards heterogeneity: Mutations, especially selectively neutral mutations (the majority of them all). Mutations don’t care about the population size, they occur with the same probability per reproductive event regardless. With a mutation rate of 1/5000 (which is what I used in the simulations below), that means that a mutant smurf, representing a novel type that wasn’t in the initial population, will be born on average once every 25 generations in a population of 200, and every generation in a population of 5000. In both cases, most of those will go extinct again – when you start with a headcount of one and randomly drift up and down, chances are you will hit 0 before you get very far on the up side. But if and when a mutant is lucky enough to stay around, its descendants can quickly become the dominant type in a small population, to the point where the entire population converges on them (at least, before new mutations arise).
In a large population, this doesn’t happen. While there are many more mutations and therefore also many more mutations that don’t immediately drift out of existence again, change in frequencies occurs so much slower that by the time a mutant supplants all initial types (if at all), many more recent mutations will have claimed their share. The result is that in a large population, heterogeneity never goes all the way down. Instead, the number of types in the population fluctuates around the point of equilibrium between the homogenising force of drift and mutations introducing variation – and this point is, on the long run, independent of initial conditions: A population of 5000 that is initially maximally diverse will reach the same level of heterogeneity as one which starts out with only one or two different types, given the same mutation rates and sampling ranges. Here is, repeated from the introduction, what we get in terms of total heterogeneity under the same six different conditions as above, only adding a mutation rate of 1/5000 (with a possibility space of 100000 different types, so mutations are relatively unlikely to produce an already existing type even for the largest population modeled):
And this is, in a nutshell, how variation in a population can be used as a diagnostic of past population sizes, although the real world offers much finer grained analyses based on the variants have different distances from each other.
Let’s look at what happens in the individual populations in some more detail:
Population of 200 with mutations. In this particular instance, one of the initial genotypes came to dominate, and all mutants went extinct again. This doesn’t have to be the case – some of the time, a mutant, by sheer luck, rises to replace all existent genotypes.
population of 200, initially heterogeneous. Again, the descendants of one initial type take over in this instance, but just as it rose to 200 from 1, a novel mutant can (and, given enough time, will sooner or later) rise to replace all old variants.
population of 1000, initially heterogeneous.
Population of 1000, initially near-homogeneous. The two original types are represented by black and white, red, which comes to dominate, is a novel mutant.
Initially near homogeneous population of 5000, the two original types are white and turquoise. One of them disappears, the other is still the single most frequent type of all by the end, but novel mutants take an almost linearly increasing share.
Population of 5000, initially heterogeneous.
Total number of individuals representative of any type not present in the initial gene pool over time. For the two large population conditions, these almost form a smooth curve as would be predicted for an infinite population(4), for populations of 200 and 1000, chance effects predominate.
(Sourcecode posted under this link for anyone interested. Still poorly documented and debugged, but mostly works.)
Footnotes:
(0) This is not by any means the simplest possible model for genetic drift – the requirement to have local sampling adds considerable sophistication. Although it’s not strictly required for the main point of this entry, I think that the possibility it adds of displaying how different types establish themselves in a subset of the species’ total range before invading other areas makes the whole thing more intuitive than just plotting total frequencies.
(1) I am aware that smurf reproduction is an open research topic, but for the purposes of this discussion, I will assume that they reproduce asexually, probably through budding. I don’t really want to imagine the alternatives. Also, the smurfs on TV are misrepresented as tribal animals. Real smurfs are solitary.
(2) Roughly 20% of the time, an individual is succeeded by a clone of itself (for sigma values between -0.5 and +0.5), roughly 35% of the time by one of its direct neighbours (-1.5 to -0.5 and 0.5 to 1.5), roughly 24% of the time by one of its second neighbours, and about 13% of the time by one of its third neighbours to either side. Technically, any individual’s clone can replace any other individual, but the actual probability becomes increasingly negligible with distance.
(3) This is only the case because we’re modeling asexually reproducing smurfs here. Let it be clear that inn humans or any other sexually reproducing species, the fact that an individual gene’s lineage coalesces at a particular point in the past in one individual in no way implies that this individual was the only one alive at the time to leave descendants!
(4) In an infinite population (and with an infinite space of possibilities), the number of individuals representative of any of the initial genotypes in generation n, given a mutation rate of M, should be (1-M)^n, thus the number of mutants 1-(1-M)^n. For a mutation rate of 0.0002 and 2000 generations as used here, this should result in a steady increase to nearly a third of the individuals (slightly less because we’re using a finite possibility space) – almost what we see in the purple and yellow lines for both the 5000-conditions.
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# Theory SpecialPseudoHoops
```section‹Some Classes of Pseudo-Hoops›
theory SpecialPseudoHoops
imports PseudoHoopFilters PseudoWaisbergAlgebra
begin
class cancel_pseudo_hoop_algebra = pseudo_hoop_algebra +
assumes mult_cancel_left: "a * b = a * c ⟹ b = c"
and mult_cancel_right: "b * a = c * a ⟹ b = c"
begin
lemma cancel_left_a: "b l→ (a * b) = a"
apply (rule_tac a = b in mult_cancel_right)
apply (subst inf_l_def [THEN sym])
apply (rule order.antisym)
by simp_all
lemma cancel_right_a: "b r→ (b * a) = a"
apply (rule_tac a = b in mult_cancel_left)
apply (subst inf_r_def [THEN sym])
apply (rule order.antisym)
by simp_all
end
class cancel_pseudo_hoop_algebra_2 = pseudo_hoop_algebra +
assumes cancel_left: "b l→ (a * b) = a"
and cancel_right: "b r→ (b * a) = a"
begin
subclass cancel_pseudo_hoop_algebra
apply unfold_locales
apply (subgoal_tac "b = a r→ (a * b) ∧ a r→ (a * b) = a r→ (a * c) ∧ a r→ (a * c) = c")
apply simp
apply (rule conjI)
apply (subst cancel_right)
apply simp
apply (rule conjI)
apply simp
apply (subst cancel_right)
apply simp
apply (subgoal_tac "b = a l→ (b * a) ∧ a l→ (b * a) = a l→ (c * a) ∧ a l→ (c * a) = c")
apply simp
apply (rule conjI)
apply (subst cancel_left)
apply simp
apply (rule conjI)
apply simp
apply (subst cancel_left)
by simp
end
context cancel_pseudo_hoop_algebra
begin
lemma lemma_4_2_i: "a l→ b = (a * c) l→ (b * c)"
apply (subgoal_tac "a l→ b = a l→ (c l→ (b * c)) ∧ a l→ (c l→ (b * c)) = (a * c) l→ (b * c)")
apply simp
apply (rule conjI)
lemma lemma_4_2_ii: "a r→ b = (c * a) r→ (c * b)"
apply (subgoal_tac "a r→ b = a r→ (c r→ (c * b)) ∧ a r→ (c r→ (c * b)) = (c * a) r→ (c * b)")
apply simp
apply (rule conjI)
lemma lemma_4_2_iii: "(a * c ≤ b * c) = (a ≤ b)"
by (simp add: left_lesseq lemma_4_2_i [THEN sym])
lemma lemma_4_2_iv: "(c * a ≤ c * b) = (a ≤ b)"
by (simp add: right_lesseq lemma_4_2_ii [THEN sym])
end
class wajsberg_pseudo_hoop_algebra = pseudo_hoop_algebra +
assumes wajsberg1: "(a l→ b) r→ b = (b l→ a) r→ a"
and wajsberg2: "(a r→ b) l→ b = (b r→ a) l→ a"
context wajsberg_pseudo_hoop_algebra
begin
lemma lemma_4_3_i_a: "a ⊔1 b = (a l→ b) r→ b"
lemma lemma_4_3_i_b: "a ⊔1 b = (b l→ a) r→ a"
lemma lemma_4_3_ii_a: "a ⊔2 b = (a r→ b) l→ b"
lemma lemma_4_3_ii_b: "a ⊔2 b = (b r→ a) l→ a"
end
sublocale wajsberg_pseudo_hoop_algebra < lattice1:pseudo_hoop_lattice_b "(⊔1)" "(*)" "(⊓)" "(l→)" "(≤)" "(<)" 1 "(r→)"
apply unfold_locales
class zero_one = zero + one
class bounded_wajsberg_pseudo_hoop_algebra = zero_one + wajsberg_pseudo_hoop_algebra +
assumes zero_smallest [simp]: "0 ≤ a"
begin
end
sublocale wajsberg_pseudo_hoop_algebra < lattice2:pseudo_hoop_lattice_b "(⊔2)" "(*)" "(⊓)" "(l→)" "(≤)" "(<)" 1 "(r→)"
apply unfold_locales
lemma (in wajsberg_pseudo_hoop_algebra) sup1_eq_sup2: "(⊔1) = (⊔2)"
apply safe
apply (cut_tac a = x and b = xa in lattice1.supremum_pair)
apply (cut_tac a = x and b = xa in lattice2.supremum_pair)
by blast
context bounded_wajsberg_pseudo_hoop_algebra
begin
definition
"negl a = a l→ 0"
definition
"negr a = a r→ 0"
lemma [simp]: "0 l→ a = 1"
by (simp add: order [THEN sym])
lemma [simp]: "0 r→ a = 1"
by (simp add: order_r [THEN sym])
end
sublocale bounded_wajsberg_pseudo_hoop_algebra < wajsberg: pseudo_wajsberg_algebra "1" "(l→)" "(r→)" "(≤)" "(<)" "0" "negl" "negr"
apply unfold_locales
apply simp_all
apply (simp add: lemma_4_3_i_a [THEN sym])
apply (rule order.antisym)
apply simp_all
apply (simp add: lemma_4_3_i_a [THEN sym] lemma_4_3_ii_a [THEN sym])
apply (rule order.antisym)
apply simp_all
apply (simp add: lemma_4_3_i_a [THEN sym] lemma_4_3_ii_a [THEN sym])
apply (rule order.antisym)
apply simp_all
apply (subst left_lesseq [THEN sym])
apply (subst right_lesseq [THEN sym])
apply (subst left_lesseq [THEN sym])
apply (subgoal_tac "b l→ a = ((b l→ 0) r→ 0) l→ ((a l→ 0) r→ 0)")
apply (subst wajsberg1)
apply simp
apply (subst wajsberg1)
apply simp
apply (subst left_lesseq [THEN sym])
apply (subgoal_tac "b r→ a = ((b r→ 0) l→ 0) r→ ((a r→ 0) l→ 0)")
apply (subst wajsberg2)
apply simp
apply (subst wajsberg2)
apply simp
apply (simp add: left_impl_ded [THEN sym])
apply (simp add: right_impl_ded [THEN sym])
apply (simp add: lemma_4_3_i_a [THEN sym] lemma_4_3_ii_a [THEN sym])
apply (rule order.antisym)
by simp_all
context pseudo_wajsberg_algebra
begin
lemma "class.bounded_wajsberg_pseudo_hoop_algebra mult inf_a (l→) (≤) (<) 1 (r→) (0::'a)"
apply unfold_locales
apply (simp add: inf_a_def mult_def W6)
apply (simp_all add: mult_def order_l strict)
apply (simp add: zero_def [THEN sym] C3_a)
apply (simp add: W6 inf_a_def [THEN sym])
apply (rule order.antisym)
apply simp_all
apply (simp add: C6 P9 [THEN sym] C5_b)
apply (simp add: inf_b_def [THEN sym])
apply (rule order.antisym)
apply simp_all
apply (simp add: inf_b_def [THEN sym])
apply (rule order.antisym)
apply simp_all
apply (simp add: C6 [THEN sym])
apply (simp add: inf_b_def [THEN sym])
apply (simp add: W6 inf_a_def [THEN sym])
apply (rule order.antisym)
apply simp_all
end
class basic_pseudo_hoop_algebra = pseudo_hoop_algebra +
assumes B1: "(a l→ b) l→ c ≤ ((b l→ a) l→ c) l→ c"
and B2: "(a r→ b) r→ c ≤ ((b r→ a) r→ c) r→ c"
begin
lemma lemma_4_5_i: "(a l→ b) ⊔1 (b l→ a) = 1"
apply (cut_tac a = a and b = b and c = "(a l→ b) ⊔1 (b l→ a)" in B1)
apply (subgoal_tac "(a l→ b) l→ (a l→ b) ⊔1 (b l→ a) = 1 ∧ ((b l→ a) l→ (a l→ b) ⊔1 (b l→ a)) = 1")
apply (erule conjE)
apply simp
apply (rule order.antisym)
apply simp
apply simp
apply safe
apply (subst left_lesseq [THEN sym])
apply simp
apply (subst left_lesseq [THEN sym])
by simp
lemma lemma_4_5_ii: "(a r→ b) ⊔2 (b r→ a) = 1"
apply (cut_tac a = a and b = b and c = "(a r→ b) ⊔2 (b r→ a)" in B2)
apply (subgoal_tac "(a r→ b) r→ (a r→ b) ⊔2 (b r→ a) = 1 ∧ ((b r→ a) r→ (a r→ b) ⊔2 (b r→ a)) = 1")
apply (erule conjE)
apply simp
apply (rule order.antisym)
apply simp
apply simp
apply safe
apply (subst right_lesseq [THEN sym])
apply simp
apply (subst right_lesseq [THEN sym])
by simp
lemma lemma_4_5_iii: "a l→ b = (a ⊔1 b) l→ b"
apply (rule order.antisym)
apply (rule_tac y = "((a l→ b) r→ b) l→ b" in order_trans)
apply (rule lemma_2_10_26)
apply (rule lemma_2_5_13_a)
apply (rule lemma_2_5_13_a)
by simp
lemma lemma_4_5_iv: "a r→ b = (a ⊔2 b) r→ b"
apply (rule order.antisym)
apply (rule_tac y = "((a r→ b) l→ b) r→ b" in order_trans)
apply (rule lemma_2_10_24)
apply (rule lemma_2_5_13_b)
apply (rule lemma_2_5_13_b)
by simp
lemma lemma_4_5_v: "(a ⊔1 b) l→ c = (a l→ c) ⊓ (b l→ c)"
apply (rule order.antisym)
apply simp
apply safe
apply (rule lemma_2_5_13_a)
apply simp
apply (rule lemma_2_5_13_a)
apply simp
apply (subst right_lesseq)
apply (rule order.antisym)
apply simp
apply (rule_tac y = "(a l→ b) l→ ((a l→ c) ⊓ (b l→ c) r→ a ⊔1 b l→ c)" in order_trans)
apply (subst left_residual [THEN sym])
apply simp
apply (subst lemma_4_5_iii)
apply (subst right_residual [THEN sym])
apply (subst left_residual [THEN sym])
apply (rule_tac y = "b ⊓ c" in order_trans)
apply (subst (2) inf_l_def)
apply (rule_tac y = "((a l→ c) ⊓ (b l→ c)) * ((a ⊔1 b) ⊓ b)" in order_trans)
apply (subst (3) inf_l_def)
apply (subgoal_tac "(a ⊔1 b ⊓ b) = b")
apply simp
apply (rule order.antisym, simp)
apply simp
apply simp
apply (rule_tac y = "((b l→ a) l→ ((a l→ c) ⊓ (b l→ c) r→ a ⊔1 b l→ c)) l→ ((a l→ c) ⊓ (b l→ c) r→ a ⊔1 b l→ c)" in order_trans)
apply (rule B1)
apply (subgoal_tac "(b l→ a) l→ ((a l→ c) ⊓ (b l→ c) r→ a ⊔1 b l→ c) = 1")
apply simp
apply (rule order.antisym)
apply simp
apply (subst left_residual [THEN sym])
apply simp
apply (subst lemma_4_5_iii)
apply (subst right_residual [THEN sym])
apply (subst left_residual [THEN sym])
apply (rule_tac y = "a ⊓ c" in order_trans)
apply (subst (2) inf_l_def)
apply (rule_tac y = "((a l→ c) ⊓ (b l→ c)) * ((a ⊔1 b) ⊓ a)" in order_trans)
apply (subst (3) inf_l_def)
apply (subst sup1.sup_comute1)
apply (subgoal_tac "(a ⊔1 b ⊓ a) = a")
apply simp
apply (rule order.antisym, simp)
apply simp
by simp
lemma lemma_4_5_vi: "(a ⊔2 b) r→ c = (a r→ c) ⊓ (b r→ c)"
apply (rule order.antisym)
apply simp
apply safe
apply (rule lemma_2_5_13_b)
apply simp
apply (rule lemma_2_5_13_b)
apply simp
apply (subst left_lesseq)
apply (rule order.antisym)
apply simp
apply (rule_tac y = "(a r→ b) r→ ((a r→ c) ⊓ (b r→ c) l→ a ⊔2 b r→ c)" in order_trans)
apply (subst right_residual [THEN sym])
apply simp
apply (subst lemma_4_5_iv)
apply (subst left_residual [THEN sym])
apply (subst right_residual [THEN sym])
apply (rule_tac y = "b ⊓ c" in order_trans)
apply (subst (2) inf_r_def)
apply (rule_tac y = "((a ⊔2 b) ⊓ b) * ((a r→ c) ⊓ (b r→ c))" in order_trans)
apply (subst (2) inf_r_def)
apply (subgoal_tac "(a ⊔2 b ⊓ b) = b")
apply simp
apply (rule order.antisym, simp)
apply simp
apply simp
apply (rule_tac y = "((b r→ a) r→ ((a r→ c) ⊓ (b r→ c) l→ a ⊔2 b r→ c)) r→ ((a r→ c) ⊓ (b r→ c) l→ a ⊔2 b r→ c)" in order_trans)
apply (rule B2)
apply (subgoal_tac "(b r→ a) r→ ((a r→ c) ⊓ (b r→ c) l→ a ⊔2 b r→ c) = 1")
apply simp
apply (rule order.antisym)
apply simp
apply (subst right_residual [THEN sym])
apply simp
apply (subst lemma_4_5_iv)
apply (subst left_residual [THEN sym])
apply (subst right_residual [THEN sym])
apply (rule_tac y = "a ⊓ c" in order_trans)
apply (subst (2) inf_r_def)
apply (rule_tac y = "((a ⊔2 b) ⊓ a) * ((a r→ c) ⊓ (b r→ c))" in order_trans)
apply (subst (2) inf_r_def)
apply (subst (2) sup2.sup_comute)
apply (subgoal_tac "(a ⊔2 b ⊓ a) = a")
apply simp
apply (rule order.antisym, simp)
apply simp
by simp
lemma lemma_4_5_a: "a ≤ c ⟹ b ≤ c ⟹ a ⊔1 b ≤ c"
apply (subst left_lesseq)
apply (subst lemma_4_5_v)
by simp
lemma lemma_4_5_b: "a ≤ c ⟹ b ≤ c ⟹ a ⊔2 b ≤ c"
apply (subst right_lesseq)
apply (subst lemma_4_5_vi)
by simp
lemma lemma_4_5: "a ⊔1 b = a ⊔2 b"
apply (rule order.antisym)
end
sublocale basic_pseudo_hoop_algebra < basic_lattice:lattice "(⊓)" "(≤)" "(<)" "(⊔1)"
apply unfold_locales
context pseudo_hoop_lattice begin end
sublocale basic_pseudo_hoop_algebra < pseudo_hoop_lattice "(⊔1)" "(*)" "(⊓)" "(l→)" "(≤)" "(<)" 1 "(r→)"
apply unfold_locales
class sup_assoc_pseudo_hoop_algebra = pseudo_hoop_algebra +
assumes sup1_assoc: "a ⊔1 (b ⊔1 c) = (a ⊔1 b) ⊔1 c"
and sup2_assoc: "a ⊔2 (b ⊔2 c) = (a ⊔2 b) ⊔2 c"
sublocale sup_assoc_pseudo_hoop_algebra < sup1_lattice: pseudo_hoop_lattice "(⊔1)" "(*)" "(⊓)" "(l→)" "(≤)" "(<)" 1 "(r→)"
apply unfold_locales
sublocale sup_assoc_pseudo_hoop_algebra < sup2_lattice: pseudo_hoop_lattice "(⊔2)" "(*)" "(⊓)" "(l→)" "(≤)" "(<)" 1 "(r→)"
apply unfold_locales
class sup_assoc_pseudo_hoop_algebra_1 = sup_assoc_pseudo_hoop_algebra +
assumes union_impl: "(a l→ b) ⊔1 (b l→ a) = 1"
and union_impr: "(a r→ b) ⊔1 (b r→ a) = 1"
lemma (in pseudo_hoop_algebra) [simp]: "infimum {a, b} = {a ⊓ b}"
apply safe
apply (rule order.antisym)
by simp_all
lemma (in pseudo_hoop_lattice) sup_impl_inf:
"(a ⊔ b) l→ c = (a l→ c) ⊓ (b l→c)"
apply (cut_tac A = "{a, b}" and a = "a ⊔ b" and b = c in lemma_2_8_i)
by simp_all
lemma (in pseudo_hoop_lattice) sup_impr_inf:
"(a ⊔ b) r→ c = (a r→ c) ⊓ (b r→c)"
apply (cut_tac A = "{a, b}" and a = "a ⊔ b" and b = c in lemma_2_8_i1)
by simp_all
lemma (in pseudo_hoop_lattice) sup_times:
"a * (b ⊔ c) = (a * b) ⊔ (a * c)"
apply (cut_tac A = "{b, c}" and b = "b ⊔ c" and a = a in lemma_2_9_i)
by simp_all
lemma (in pseudo_hoop_lattice) sup_times_right:
"(b ⊔ c) * a = (b * a) ⊔ (c * a)"
apply (cut_tac A = "{b, c}" and b = "b ⊔ c" and a = a in lemma_2_9_i1)
by simp_all
context basic_pseudo_hoop_algebra begin end
sublocale sup_assoc_pseudo_hoop_algebra_1 < basic_1: basic_pseudo_hoop_algebra "(*)" "(⊓)" "(l→)" "(≤)" "(<)" 1 "(r→)"
apply unfold_locales
apply (subst left_residual [THEN sym])
apply (rule_tac y = "(a l→ b) ⊔1 (b l→ a) l→ c" in order_trans)
apply (subst sup1_lattice.sup_impl_inf)
apply (subst right_residual [THEN sym])
apply (rule_tac y = "(b r→ a) ⊔1 (a r→ b) r→ c" in order_trans)
apply (subst sup1_lattice.sup_impr_inf)
context basic_pseudo_hoop_algebra
begin
lemma lemma_4_8_i: "a * (b ⊓ c) = (a * b) ⊓ (a * c)"
apply (rule order.antisym)
apply simp
apply (subgoal_tac "a * (b ⊓ c) = (a * (b * (b r→ c))) ⊔1 (a * (c * (c r→ b)))")
apply simp
apply (drule drop_assumption)
apply (rule_tac y = "(((a * b) ⊓ (a * c)) * (b r→ c)) ⊔1 (((a * b) ⊓ (a * c)) * (c r→ b))" in order_trans)
apply (subst sup_times [THEN sym])
apply (simp add: mult.assoc [THEN sym])
apply safe
apply (rule_tac y = "a * b * (b r→ c)" in order_trans)
apply simp
apply simp
apply (rule_tac y = "a * c * (c r→ b)" in order_trans)
apply simp
apply simp
apply (simp add: inf_r_def [THEN sym])
apply (subgoal_tac "b ⊓ c = c ⊓ b")
apply simp
apply (rule order.antisym)
by simp_all
lemma lemma_4_8_ii: "(b ⊓ c) * a = (b * a) ⊓ (c * a)"
apply (rule order.antisym)
apply simp
apply (subgoal_tac "(b ⊓ c) * a = (((b l→ c) * b) * a) ⊔1 (((c l→ b) * c) * a)")
apply simp
apply (drule drop_assumption)
apply (rule_tac y = "((b l→ c) * ((b * a) ⊓ (c * a))) ⊔1 ((c l→ b) * ((b * a) ⊓ (c * a)))" in order_trans)
apply (subst sup_times_right [THEN sym])
apply safe
apply (rule_tac y = "(b l→ c) * (b * a)" in order_trans)
apply simp_all
apply (rule_tac y = "(c l→ b) * (c * a)" in order_trans)
apply simp_all
apply (simp add: inf_l_def [THEN sym])
apply (subgoal_tac "b ⊓ c = c ⊓ b")
apply simp
apply (rule order.antisym)
by simp_all
lemma lemma_4_8_iii: "(a l→ b) l→ (b l→ a) = b l→ a"
apply (rule order.antisym)
apply (cut_tac a = a and b = b in lemma_4_5_i)
apply (unfold sup1_def right_lesseq, simp)
lemma lemma_4_8_iv: "(a r→ b) r→ (b r→ a) = b r→ a"
apply (rule order.antisym)
apply (cut_tac a = a and b = b in lemma_4_5_ii)
apply (unfold sup2_def left_lesseq, simp)
end
context wajsberg_pseudo_hoop_algebra
begin
subclass sup_assoc_pseudo_hoop_algebra_1
apply unfold_locales
apply (subgoal_tac "(a l→ b) l→ (b l→ a) = b l→ a")
apply simp
apply (subst lemma_2_10_30 [THEN sym])
apply (subst wajsberg1)
apply (subst sup1_eq_sup2)
apply (subgoal_tac "(a r→ b) r→ (b r→ a) = b r→ a")
apply simp
apply (subst lemma_2_10_31 [THEN sym])
apply (subst wajsberg2)
end
class bounded_basic_pseudo_hoop_algebra = zero_one + basic_pseudo_hoop_algebra +
assumes zero_smallest [simp]: "0 ≤ a"
class inf_a =
fixes inf_a :: "'a => 'a => 'a" (infixl "⊓1" 65)
class pseudo_bl_algebra = zero + sup + inf + monoid_mult + ord + left_imp + right_imp +
assumes bounded_lattice: "class.bounded_lattice inf (≤) (<) sup 0 1"
and left_residual_bl: "(x * a ≤ b) = (x ≤ a l→ b)"
and right_residual_bl: "(a * x ≤ b) = (x ≤ a r→ b)"
and inf_l_bl_def: "a ⊓ b = (a l→ b) * a"
and inf_r_bl_def: "a ⊓ b = a * (a r→ b)"
and impl_sup_bl: "(a l→ b) ⊔ (b l→ a) = 1"
and impr_sup_bl: "(a r→ b) ⊔ (b r→ a) = 1"
sublocale bounded_basic_pseudo_hoop_algebra < basic: pseudo_bl_algebra 1 "(*)" "0" "(⊓)" "(⊔1)" "(l→)" "(r→)" "(≤)" "(<)"
apply unfold_locales
apply (rule zero_smallest)
apply (rule left_residual)
apply (rule right_residual)
apply (rule inf_l_def)
apply (simp add: inf_r_def [THEN sym])
apply (rule lemma_4_5_i)
by (rule lemma_4_5_ii)
sublocale pseudo_bl_algebra < bounded_lattice: bounded_lattice "inf" "(≤)" "(<)" "sup" "0" "1"
by (rule bounded_lattice)
context pseudo_bl_algebra
begin
lemma impl_one_bl [simp]: "a l→ a = 1"
apply (rule bounded_lattice.order.antisym)
apply simp_all
apply (subst left_residual_bl [THEN sym])
by simp
lemma impr_one_bl [simp]: "a r→ a = 1"
apply (rule bounded_lattice.order.antisym)
apply simp_all
apply (subst right_residual_bl [THEN sym])
by simp
lemma impl_ded_bl: "((a * b) l→ c) = (a l→ (b l→ c))"
apply (rule bounded_lattice.order.antisym)
apply (case_tac "(a * b l→ c ≤ a l→ b l→ c) = ((a * b l→ c) * a ≤ b l→ c)
∧ ((a * b l→ c) * a ≤ b l→ c) = (((a * b l→ c) * a) * b ≤ c)
∧ (((a * b l→ c) * a) * b ≤ c) = ((a * b l→ c) * (a * b) ≤ c)
∧ ((a * b l→ c) * (a * b) ≤ c) = ((a * b l→ c) ≤ (a * b l→ c))")
apply simp
apply (erule notE)
apply (rule conjI)
apply (rule conjI)
apply (rule conjI)
apply (simp add: left_residual_bl [THEN sym])
apply (rule_tac y="(b l→ c) * b" in bounded_lattice.order_trans)
apply (simp add: mult.assoc [THEN sym])
apply (subst inf_l_bl_def [THEN sym])
apply (subst bounded_lattice.inf_commute)
apply (subst inf_l_bl_def)
apply (subst mult.assoc)
apply (subst left_residual_bl)
apply simp
apply (subst left_residual_bl)
by simp
lemma impr_ded_bl: "(b * a r→ c) = (a r→ (b r→ c))"
apply (rule bounded_lattice.order.antisym)
apply (case_tac "(b * a r→ c ≤ a r→ b r→ c) = (a * (b * a r→ c) ≤ b r→ c)
∧ (a * (b * a r→ c) ≤ b r→ c) = (b * (a * (b * a r→ c)) ≤ c)
∧ (b * ( a* (b * a r→ c)) ≤ c) = ((b * a) * (b * a r→ c) ≤ c)
∧ ((b * a) * (b * a r→ c) ≤ c) = ((b * a r→ c) ≤ (b * a r→ c))")
apply simp
apply (erule notE)
apply (rule conjI)
apply (rule conjI)
apply (rule conjI)
apply (simp add: right_residual_bl [THEN sym])
apply (rule_tac y="b * (b r→ c)" in bounded_lattice.order_trans)
apply (subst inf_r_bl_def [THEN sym])
apply (subst bounded_lattice.inf_commute)
apply (subst inf_r_bl_def)
apply (subst mult.assoc [THEN sym])
apply (subst right_residual_bl)
apply simp
apply (subst right_residual_bl)
by simp
lemma lesseq_impl_bl: "(a ≤ b) = (a l→ b = 1)"
apply (rule iffI)
apply (rule bounded_lattice.order.antisym)
apply simp
apply (simp add: left_residual_bl [THEN sym])
apply (subgoal_tac "1 ≤ a l→ b")
apply (subst (asm) left_residual_bl [THEN sym])
by simp_all
end
context pseudo_bl_algebra
begin
subclass pseudo_hoop_lattice
apply unfold_locales
apply (rule inf_l_bl_def)
apply (simp add: inf_l_bl_def [THEN sym])
apply (rule bounded_lattice.inf_commute)
apply (rule impl_ded_bl)
apply (rule lesseq_impl_bl)
apply (rule inf_r_bl_def)
apply (simp add: inf_r_bl_def [THEN sym])
apply (rule bounded_lattice.inf_commute)
apply (rule impr_ded_bl)
apply (simp add: inf_r_bl_def [THEN sym] inf_l_bl_def [THEN sym])
apply (rule bounded_lattice.sup_commute)
apply simp
apply safe
apply (rule bounded_lattice.order.antisym)
apply simp_all
apply (subgoal_tac "a ≤ a ⊔ b")
apply simp
apply (drule drop_assumption)
apply simp
subclass bounded_basic_pseudo_hoop_algebra
apply unfold_locales
apply simp_all
apply (simp add: left_residual [THEN sym])
apply (rule_tac y = "((a l→ b) ⊔ (b l→ a)) l→ c" in bounded_lattice.order_trans)
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CC-MAIN-2024-18
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http://demonstrations.wolfram.com/preview.html?draft/54742/000072/ShortestPathThatTouchesTheCircumferenceBetween2PointsOnDisk
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11562
# Shortest Path That Touches the Circumference Between 2 Points on Disk
Given a starting and ending position in the unit disk, what is the shortest path between the two points that also touches the unit circle? The starting point is labeled s and drawn in green. The ending point is labeled e and drawn in red, and the relative angle between them is β. The angle from the x-axis to s is ϕ. The shortest path touches the unit circle is at angle .
The shortest path between two points that also touches the circumference is composed up of two straight line segments. The problem can be simplified by choosing the coordinate system carefully. We define the x-axis along the position of the starting point: S = (s,0), and define the point of intersection by the angle θ from the x-axis P = (cos(θ), sin(θ)), and the final point by a radius e and angle β, E = e(Cos(β), sin(β)). Then define a symmetry point about S of line OP named T. Then the length of the two line segments is , which is minimized by choosing an appropriate θ value. This equation can be simplified to . The length of the two line segments as a function of θ is drawn in the right plot.
There are several simple solutions. If s is 1 or e is 0 or β is zero, the optimal angle is 0. If e is 1 or s is 0, the optimal angle is β.
We label the origin O. The optimal solution shows that the angle ∠OPS (from the origin to P to S) is the same as the angle ∠OPE (from the origin to P to E).
We name these angles α. This can be proved by drawing an ellipse whose foci are S and E. When the ellipse is tangent to the circle, the point of tangency is exactly P.
Since the distance from the origin to P is always 1, we can set up three equalities using the law of sines:
From triangle OSP: .
From triangle OEP: .
If we mirror the point S about the θ axis, and label this point C, from triangle CEO:
.
Simplifying this system of equations results in:
s=e csc(θ)(s sin(2 θ-β)+sin(β-θ)).
Solving this last equation results in a quartic solution. It has a closed form solution with four roots, each of which can be either a clockwise or a counterclockwise rotation θ, depending on the sign of β, with -π ≤ β ≤ π. We evaluate each and select the solution that results in the shortest length path.
### SNAPSHOTS
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| 4.46875
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CC-MAIN-2018-09
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longest
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en
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https://2010.igem.org/wiki/index.php?title=Team:Newcastle/DNA_re-hydration&diff=200503&oldid=47601
| 1,670,058,865,000,000,000
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text/html
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crawl-data/CC-MAIN-2022-49/segments/1669446710926.23/warc/CC-MAIN-20221203075717-20221203105717-00702.warc.gz
| 109,028,065
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# Team:Newcastle/DNA re-hydration
(Difference between revisions)
Revision as of 10:48, 11 August 2010 (view source)Daye (Talk | contribs)← Older edit Latest revision as of 23:21, 27 October 2010 (view source)Yessa (Talk | contribs) (→Method) (24 intermediate revisions not shown) Line 1: Line 1: - This protocol is used for re-hydration of DNA. We used this mainly for our newly arrived primers. (Link) + {{Team:Newcastle/mainbanner}} - =Materials required= + [[Image:Newcastle OSS.JPG|right|200px| Oligonucleotide Specification Sheet (OSS) with the calculations used to work out the volumes of rehydrated primer and H2O required for this specific primer]] + =DNA Re-hydration= - * Oligonucleotide Specification Sheet + This protocol is used for re-hydration of DNA. We used this mainly for our newly arrived primers. - * OD 260 values + - * Quantity of DNA bases + - * Pure Lab Distilled Water + - * Calculator (!) + - =Methods= + ==Materials required== - # To begin, read the Oligonucleotide Specification Sheet to find the OD 260 values and Quantity of DNA bases for each primer/DNA + * Oligonucleotide Specification Sheet (OSS) - # Once those values are found, divide the (OD 260 value) by (Quantity of DNA bases x10). + * OD260 values + * Sterile Pure Lab Distilled Water + + ==Method== + + # To begin, read the Oligonucleotide Specification Sheet to find the OD260 values and number of DNA bases for each primer/DNA. + # Once those values are found, divide the (OD260 value) by (number of DNA bases x10). # Multiple the result by 1000 # Multiple the result by 1000 - # Divide 10 by the previous result, and multiple all by 100: the end result is the volume of primer/DNA in microlitres. + # Divide 10 by the previous result, and multiple all by 100: the end result is the volume of primer DNA (in µl) required to prepare 100µl at 10pM concentration. - # Add Pure Lab Distilled Water to make the solution up to 100 microlitres in total. + # Rehydrate lyopholsed primers in 1ml sterile water + # Remove the appropiate volume of rehydrated primer and add Pure Lab Distilled Water to make the solution up to 100 µl in total. + + Please note: + *Extra care must be taken when following this protocol, i.e. there must be NO contamination. + *The newly arrived primer tubes have to be handled with extra caution, because they will be the main stocks which working stock solutions will be made from. Therefore, gloves have to be worn, as well as preventing any contamination. Water will be used in order to liquefy the primers and the water used will be from Pure Lab Distilled Water. + + + [[Image:Newcastle Pure Lab Flex 2.JPG|200px]] + + '''Figure 1''' showing Pure Lab Distilled Water apparatus. + + '''Go back to our [[Team:Newcastle/Protocol list|Protocol List]]''' - NOTE: Extra care must be taken when following this protocol, i.e. there must be NO contamination. + {{Team:Newcastle/footer}}
# DNA Re-hydration
This protocol is used for re-hydration of DNA. We used this mainly for our newly arrived primers.
## Materials required
• Oligonucleotide Specification Sheet (OSS)
• OD260 values
• Sterile Pure Lab Distilled Water
## Method
1. To begin, read the Oligonucleotide Specification Sheet to find the OD260 values and number of DNA bases for each primer/DNA.
2. Once those values are found, divide the (OD260 value) by (number of DNA bases x10).
3. Multiple the result by 1000
4. Divide 10 by the previous result, and multiple all by 100: the end result is the volume of primer DNA (in µl) required to prepare 100µl at 10pM concentration.
5. Rehydrate lyopholsed primers in 1ml sterile water
6. Remove the appropiate volume of rehydrated primer and add Pure Lab Distilled Water to make the solution up to 100 µl in total.
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http://www.algebra.com/algebra/homework/Triangles.faq.question.703592.html
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# SOLUTION: how do you find the perimeter of a triangle when the sides are {{{(5x^2+1)(6x+2)(x^2+13x)}}}
Algebra -> -> SOLUTION: how do you find the perimeter of a triangle when the sides are {{{(5x^2+1)(6x+2)(x^2+13x)}}} Log On
Ad: Mathway solves algebra homework problems with step-by-step help! Ad: Algebrator™ solves your algebra problems and provides step-by-step explanations!
Click here to see ALL problems on Triangles Question 703592: how do you find the perimeter of a triangle when the sides are Answer by mouk(232) (Show Source): You can put this solution on YOUR website!To find the perimeter of a triangle you add the length of each side. So in the case perimeter =
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http://www.algebra.com/algebra/homework/Exponential-and-logarithmic-functions/Exponential-and-logarithmic-functions.faq.question.310482.html
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# SOLUTION: LOG 10 of 1x10 to the 4th power
Algebra -> Algebra -> Exponential-and-logarithmic-functions -> SOLUTION: LOG 10 of 1x10 to the 4th power Log On
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Algebra: Exponent and logarithm as functions of power Solvers Lessons Answers archive Quiz In Depth
Question 310482: LOG 10 of 1x10 to the 4th powerAnswer by stanbon(57356) (Show Source): You can put this solution on YOUR website!LOG 10 of 1x10 to the 4th power -------------- log10(1*10^4) = log10(1) + 4log10(10) = 0 + 4*1 = 4 =================== Cheers, Stan H. ===================
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| 3.078125
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CC-MAIN-2013-20
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https://oeis.org/A136904
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crawl-data/CC-MAIN-2023-50/segments/1700679100779.51/warc/CC-MAIN-20231208212357-20231209002357-00425.warc.gz
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The OEIS is supported by the many generous donors to the OEIS Foundation.
Year-end appeal: Please make a donation to the OEIS Foundation to support ongoing development and maintenance of the OEIS. We are now in our 60th year, we have over 367,000 sequences, and we’ve reached 11,000 citations (which often say “discovered thanks to the OEIS”). Other ways to Give
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A136904 Numbers k such that k and k^2 use only the digits 0, 2, 4, 6 and 8. 4
0, 2, 8, 20, 22, 68, 80, 200, 202, 220, 262, 668, 680, 800, 2000, 2002, 2020, 2022, 2200, 2202, 2620, 6668, 6680, 6800, 8000, 8262, 20000, 20002, 20020, 20022, 20200, 20220, 20602, 22000, 22002, 22020, 24622, 26200, 66668, 66680, 66800, 68000, 80000, 82620, 200000, 200002, 200020, 200022, 200200, 200202, 200220, 202000, 202002, 202200, 206020 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,2 COMMENTS Generated with DrScheme. Subsequence of A014263. - Michel Marcus, Oct 04 2013 LINKS Jonathan Wellons, Table of n, a(n) for n = 1..1038 J. Wellons, Tables of Shared Digits EXAMPLE 828268460602^2 = 686028642828006826202404. PROG (PARI) evdigs(n) = {if (n==0, return (1)); digs = digits(n); for (i = 1, #digs, if (digs[i] % 2, return (0)); ); return (1); } isok(n) = evdigs(n) && evdigs(n^2); \\ Michel Marcus, Oct 04 2013 CROSSREFS Sequence in context: A227399 A327098 A030097 * A043002 A108180 A233571 Adjacent sequences: A136901 A136902 A136903 * A136905 A136906 A136907 KEYWORD base,nonn AUTHOR Jonathan Wellons (wellons(AT)gmail.com), Jan 22 2008 STATUS approved
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Last modified December 8 17:58 EST 2023. Contains 367680 sequences. (Running on oeis4.)
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https://apprize.best/science/algorithms_2/3.html
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Minimum Spanning Trees - Graphs - Algorithms (2014)
# Algorithms (2014)
## Four. Graphs
### 4.3 Minimum Spanning Trees
AN edge-weighted graph is a graph model where we associate weights or costs with each edge. Such graphs are natural models for many applications. In an airline map where edges represent flight routes, these weights might represent distances or fares. In an electric circuit where edges represent wires, the weights might represent the length of the wire, its cost, or the time that it takes a signal to propagate through it. Minimizing cost is naturally of interest in such situations. In this section, we consider undirected edge-weighted graph models and examine algorithms for one such problem:
Minimum spanning tree. Given an undirected edge-weighted graph, find an MST.
Definition. Recall that a spanning tree of a graph is a connected subgraph with no cycles that includes all the vertices. A minimum spanning tree (MST) of an edge-weighted graph is a spanning tree whose weight (the sum of the weights of its edges) is no larger than the weight of any other spanning tree.
In this section, we examine two classical algorithms for computing MSTs: Prim’s algorithm and Kruskal’s algorithm. These algorithms are easy to understand and not difficult to implement. They are among the oldest and most well-known algorithms in this book, and they also take good advantage of modern data structures. Since MSTs have numerous important applications, algorithms to solve the problem have been studied at least since the 1920s, at first in the context of power distribution networks, later in the context of telephone networks. MST algorithms are now important in the design of many types of networks (communication, electrical, hydraulic, computer, road, rail, air, and many others) and also in the study of biological, chemical, and physical networks that are found in nature.
Assumptions
Various anomalous situations, which are generally easy to handle, can arise when computing minimum spanning trees. To streamline the presentation, we adopt the following conventions:
The graph is connected. The spanning-tree condition in our definition implies that the graph must be connected for an MST to exist. Another way to pose the problem, recalling basic properties of trees from SECTION 4.1, is to find a minimal-weight set of V−1 edges that connect the graph. If a graph is not connected, we can adapt our algorithms to compute the MSTs of each of its connected components, collectively known as a minimum spanning forest (see EXERCISE 4.3.22).
The edge weights are not necessarily distances. Geometric intuition is sometimes beneficial in understanding algorithms, so we use examples where vertices are points in the plane and weights are distances, such as the graph on the facing page. But it is important to remember that the weights might represent time or cost or an entirely different variable and do not need to be proportional to a distance at all.
The edge weights may be zero or negative. If the edge weights are all positive, it suffices to define an MST as a subgraph with minimal total weight that connects all the vertices, as such a subgraph must form a spanning tree. The spanning-tree condition in the definition is included so that it applies to graphs that may have zero or negative edge weights.
The edge weights are all different. If edges can have equal weights, the minimum spanning tree may not be unique (see EXERCISE 4.3.2). The possibility of multiple MSTs complicates the correctness proofs of some of our algorithms, so we rule out that possibility in the presentation. It turns out that this assumption is not restrictive because our algorithms work without modification in the presence of equal weights.
In summary, we assume throughout the presentation that our job is to find the MST of a connected edge-weighted graph with arbitrary (but distinct) weights.
Underlying principles
To begin, we recall from SECTION 4.1 two of the defining properties of a tree:
• Adding an edge that connects two vertices in a tree creates a unique cycle.
• Removing an edge from a tree breaks it into two separate subtrees.
These properties are the basis for proving a fundamental property of MSTs that leads to the MST algorithms that we consider in this section.
Cut property
This property, which we refer to as the cut property, has to do with identifying edges that must be in the MST of a given edge-weighted graph, by dividing vertices into two sets and examining edges that cross the division.
Definition. A cut of a graph is a partition of its vertices into two nonempty disjoint sets. A crossing edge of a cut is an edge that connects a vertex in one set with a vertex in the other.
Typically, we specify a cut by specifying a set of vertices, leaving implicit the assumption that the cut comprises the given vertex set and its complement, so that a crossing edge is an edge from a vertex in the set to a vertex not in the set. In figures, we draw vertices on one side of the cut in gray and vertices on the other side in white.
Proposition J. (Cut property) Given any cut in an edge-weighted graph, the crossing edge of minimum weight is in the MST of the graph.
Proof: Let e be the crossing edge of minimum weight and let T be the MST. The proof is by contradiction: Suppose that T does not contain e. Now consider the graph formed by adding e to T. This graph has a cycle that contains e, and that cycle must contain at least one other crossing edge—say, f, which has higher weight than e (since e is minimal and all edge weights are different). We can get a spanning tree of strictly lower weight by deleting f and adding e, contradicting the assumed minimality of T.
Under our assumption that edge weights are distinct, every connected graph has a unique MST (see EXERCISE 4.3.3); and the cut property says that the lightest crossing edge for every cut must be in the MST.
The figure to the left of PROPOSITION J illustrates the cut property. Note that there is no requirement that the minimal edge be the only MST edge connecting the two sets; indeed, for typical cuts there are several MST edges that connect a vertex in one set with a vertex in the other, as illustrated in the figure above.
Greedy algorithm
The cut property is the basis for the algorithms that we consider for the MST problem. Specifically, they are special cases of a general paradigm known as the greedy algorithm: apply the cut property to accept an edge as an MST edge, continuing until finding all of the MST edges. Our algorithms differ in their approaches to maintaining cuts and identifying the crossing edge of minimum weight, but are special cases of the following:
Proposition K. (Greedy MST algorithm) The following method colors black all edges in the the MST of any connected edge-weighted graph with V vertices: starting with all edges colored gray, find a cut with no black edges, color its minimum-weight edge black, and continue until V−1 edges have been colored black.
Proof: For simplicity, we assume in the discussion that the edge weights are all different, though the proposition is still true when that is not the case (see EXERCISE 4.3.5). By the cut property, any edge that is colored black is in the MST. If fewer than V−1 edges are black, a cut with no black edges exists (recall that we assume the graph to be connected). Once V−1 edges are black, the black edges form a spanning tree.
The diagram at right is a typical trace of the greedy algorithm. Each drawing depicts a cut and identifies the minimum-weight edge in the cut (thick red) that is added to the MST by the algorithm.
Edge-weighted graph data type
How should we represent edge-weighted graphs? Perhaps the simplest way to proceed is to extend the basic graph representations from SECTION 4.1: in the adjacency-matrix representation, the matrix can contain edge weights rather than boolean values; in the adjacency-lists representation, we can define a node that contains both a vertex and a weight field to put in the adjacency lists. (As usual, we focus on sparse graphs and leave the adjacency-matrix representation for exercises.) This classic approach is appealing, but we will use a different method that is not much more complicated, will make our programs useful in more general settings, and needs a slightly more general API, which allows us to process Edge objects:
The either() and other() methods for accessing the edge’s vertices may be a bit puzzling at first—the need for them will become plain when we examine client code. You can find an implementation of Edge on page 610. It is the basis for this EdgeWeightedGraph API, which refers to Edge objects in a natural manner:
This API is very similar to the API for Graph (page 522). The two important differences are that it is based on Edge and that it adds the edges() method at right, which provides clients with the ability to iterate through all the graph’s edges (ignoring any self-loops). The rest of the implementation ofEdgeWeightedGraph on page 611 is quite similar to the unweighted undirected graph implementation of SECTION 4.1, but instead of the adjacency lists of integers used in Graph, it uses adjacency lists of Edge objects.
Gathering all the edges in an edge-weighted graph
public Iterable<Edge> edges()
{
Bag<Edge> b = new Bag<Edge>();
for (int v = 0; v < V; v++)
return b;
}
The figure at the bottom of this page shows the edge-weighted graph representation that EdgeWeightedGraph builds from the sample file tinyEWG.txt, showing the contents of each Bag as a linked list to reflect the standard implementation of SECTION 1.3. To reduce clutter in the figure, we show each Edgeas a pair of int values and a double value. The actual data structure is a linked list of links to objects containing those values. In particular, although there are two references to each Edge (one in the list for each vertex), there is only one Edge object corresponding to each graph edge. In the figure, the edges appear in each list in reverse order of the order they are processed, because of the stack-like nature of the standard linked-list implementation. As in Graph, by using a Bag we are making clear that our client code makes no assumptions about the order of objects in the lists.
Weighted edge data type
public class Edge implements Comparable<Edge>
{
private final int v; // one vertex
private final int w; // the other vertex
private final double weight; // edge weight
public Edge(int v, int w, double weight)
{
this.v = v;
this.w = w;
this.weight = weight;
}
public double weight()
{ return weight; }
public int either()
{ return v; }
public int other(int vertex)
{
if (vertex == v) return w;
else if (vertex == w) return v;
else throw new RuntimeException("Inconsistent edge");
}
public int compareTo(Edge that)
{
if (this.weight() < that.weight()) return -1;
else if (this.weight() > that.weight()) return +1;
else return 0;
}
public String toString()
{ return String.format("%d-%d %.5f", v, w, weight); }
}
This data type provides the methods either() and other() so that a client can use other(v) to find the other vertex when it knows v. When neither vertex is known, a client can use the idiomatic code int v = e.either(), w = e.other(v); to access an Edge e’s two vertices.
Edge-weighted graph data type
public class EdgeWeightedGraph
{
private final int V; // number of vertices
private int E; // number of edges
public EdgeWeightedGraph(int V)
{
this.V = V;
this.E = 0;
for (int v = 0; v < V; v++)
}
public EdgeWeightedGraph(In in)
// See Exercise 4.3.9.
public int V() { return V; }
public int E() { return E; }
{
int v = e.either(), w = e.other(v);
E++;
}
public Iterable<Edge> edges()
// See page 609.
}
This implementation maintains a vertex-indexed array of lists of edges. As with Graph (see page 526), every edge appears twice: if an edge connects v and w, it appears both in v’s list and in w’s list. The edges() method puts all the edges in a Bag (see page 609). The toString() implementation is left as an exercise.
Comparing edges by weight
The API specifies that the Edge class must implement the Comparable interface and include a compareTo() implementation. The natural ordering for edges in an edge-weighted graph is by weight. Accordingly, the implementation of compareTo() is straightforward.
Parallel edges
As with our undirected-graph implementations, we allow parallel edges. Alternatively, we could develop a more complicated implementation of EdgeWeightedGraph that disallows them, perhaps keeping the minimum-weight edge from a set of parallel edges.
Self-loops
We allow self-loops. However, our edges() implementation in EdgeWeightedGraph does not include self-loops even though they might be present in the input or in the data structure. This omission has no effect on our MST algorithms because no MST contains a self-loop. When working with an application where self-loops are significant, you may need to modify our code as appropriate for the application.
OUR CHOICE TO USE explicit Edge objects leads to clear and compact client code, as you will see. It carries a small price: each adjacency-list node has a reference to an Edge object, with redundant information (all the nodes on v’s adjacency list have a v). We also pay object overhead cost. Although we have only one copy of each Edge, we do have two references to each Edge object. An alternative and widely used approach is to keep two list nodes corresponding to each edge, just as in Graph, each with a vertex and the edge weight in each list node. This alternative also carries a price—two nodes, including two copies of the weight for each edge.
MST API and test client
As usual, for graph processing, we define an API where the constructor takes an edge-weighted graph as argument and supports client query methods that return the MST and its weight. How should we represent the MST itself? The MST of a graph G is a subgraph of G that is also a tree, so we have numerous options. Chief among them are
• A list of edges
• An edge-weighted graph
• A vertex-indexed array with parent links
To give clients and our implementations as much flexibility as possible in choosing among these alternatives for various applications, we adopt the following API:
Test client
As usual, we create sample graphs and develop a test client for use in testing our implementations. A sample client is shown below. It reads edges from the input stream, builds an edge-weighted graph, computes the MST of that graph, prints the MST edges, and prints the total weight of the MST.
MST test client
public static void main(String[] args)
{
In in = new In(args[0]);
EdgeWeightedGraph G;
G = new EdgeWeightedGraph(in);
MST mst = new MST(G);
for (Edge e : mst.edges())
StdOut.println(e);
StdOut.printf("%.5f\n", mst.weight());
}
Test data
You can find the file tinyEWG.txt on the booksite, which defines the small sample graph on page 604 that we use for detailed traces of MST algorithms. You can also find on the booksite the file mediumEWG.txt, which defines the weighted graph with 250 vertices that is drawn on bottom of the the facing page. It is an example of a Euclidean graph, whose vertices are points in the plane and whose edges are lines connecting them with weights equal to their Euclidean distances. Such graphs are useful for gaining insight into the behavior of MST algorithms, and they also model many of the typical practical problems we have mentioned, such as road maps or electric circuits. You can also find on the booksite a larger example largeEWG.txt that defines a Euclidean graph with 1 million vertices. Our goal is to be able to find the MST of such a graph in a reasonable amount of time.
% more tinyEWG.txt
8 16
4 5 .35
4 7 .37
5 7 .28
0 7 .16
1 5 .32
0 4 .38
2 3 .17
1 7 .19
0 2 .26
1 2 .36
1 3 .29
2 7 .34
6 2 .40
3 6 .52
6 0 .58
6 4 .93
% java MST tinyEWG.txt
0-7 0.16000
2-3 0.17000
1-7 0.19000
0-2 0.26000
5-7 0.28000
4-5 0.35000
6-2 0.40000
1.81000
% more mediumEWG.txt
250 1273
244 246 0.11712
239 240 0.10616
238 245 0.06142
235 238 0.07048
233 240 0.07634
232 248 0.10223
231 248 0.10699
229 249 0.10098
228 241 0.01473
226 231 0.07638
... [1263 more edges ]
% java MST mediumEWG.txt
0 225 0.02383
49 225 0.03314
44 49 0.02107
44 204 0.01774
49 97 0.03121
202 204 0.04207
176 202 0.04299
176 191 0.02089
68 176 0.04396
58 68 0.04795
... [239 more edges ]
10.46351
Prim’s algorithm
Our first MST method, known as Prim’s algorithm, is to attach a new edge to a single growing tree at each step. Start with any vertex as a single-vertex tree; then add V−1 edges to it, always taking next (coloring black) the minimum-weight edge that connects a vertex on the tree to a vertex not yet on the tree (a crossing edge for the cut defined by tree vertices).
Proposition L. Prim’s algorithm computes the MST of any connected edge-weighted graph.
Proof: Immediate from PROPOSITION K. The growing tree defines a cut with no black edges; the algorithm takes the crossing edge of minimal weight, so it is successively coloring edges black in accordance with the greedy algorithm.
The one-sentence description of Prim’s algorithm just given leaves unanswered a key question: How do we (efficiently) find the crossing edge of minimal weight? Several methods have been proposed—we will discuss some of them after we have developed a full solution based on a particularly simple approach.
Data structures
We implement Prim’s algorithm with the aid of a few simple and familiar data structures. In particular, we represent the vertices on the tree, the edges on the tree, and the crossing edges, as follows:
Vertices on the tree: We use a vertex-indexed boolean array marked[], where marked[v] is true if v is on the tree.
Edges in the tree: We use one of two data structures: either a queue mst to collect edges in the MST edges or a vertex-indexed array edgeTo[] of Edge objects, where edgeTo[v] is the Edge that connects v to the tree.
Crossing edges: We use a MinPQ<Edge> priority queue that compares edges by weight (see page 610).
These data structures allow us to directly answer the basic question “Which is the minimal-weight crossing edge?”
Maintaining the set of crossing edges
Each time that we add an edge to the tree, we also add a vertex to the tree. To maintain the set of crossing edges, we need to add to the priority queue all edges from that vertex to any non-tree vertex (using marked[] to identify such edges). But we must do more: any edge connecting the vertex just added to a tree vertex that is already on the priority queue now becomes ineligible (it is no longer a crossing edge because it connects two tree vertices). An eager implementation of Prim’s algorithm would remove such edges from the priority queue; we first consider a simpler lazyimplementation of the algorithm where we leave such edges on the priority queue, deferring the eligibility test to when we remove them.
The figure at right is a trace for our small sample graph tinyEWG.txt. Each drawing depicts the graph and the priority queue just after a vertex is visited (added to the tree and the edges in its adjacency list processed). The contents of the priority queue are shown in order on the side, with new edges marked with asterisks. The algorithm builds the MST as follows:
• Adds 0 to the MST and all edges in its adjacency list to the priority queue.
• Adds 7 and 0-7 to the MST and all edges in its adjacency list to the priority queue.
• Adds 1 and 1-7 to the MST and all edges in its adjacency list to the priority queue.
• Adds 2 and 0-2 to the MST and edges 2-3 and 6-2 to the priority queue. Edges 2-7 and 1-2 become ineligible.
• Adds 3 and 2-3 to the MST and edge 3-6 to the priority queue. Edge 1-3 becomes ineligible.
• Adds 5 and 5-7 to the MST and edge 4-5 to the priority queue. Edge 1-5 becomes ineligible.
• Removes ineligible edges 1-3, 1-5, and 2-7 from the priority queue.
• Adds 4 and 4-5 to the MST and edge 6-4 to the priority queue. Edges 4-7 and 0-4 become ineligible.
• Removes ineligible edges 1-2, 4-7, and 0-4 from the priority queue.
• Adds 6 and 6-2 to the MST. The other edges incident to 6 become ineligible.
After having added V vertices (and V−1 edges), the MST is complete. The remaining edges on the priority queue are ineligible, so we need not examine them again.
Implementation
With these preparations, implementing Prim’s algorithm is straightforward, as shown in the implementation LazyPrimMST on the facing page. As with our depth-first search and breadth-first search implementations in the previous two sections, it computes the MST in the constructor so that client methods can learn properties of the MST with query methods. We use a private method visit() that puts a vertex on the tree, by marking it as visited and then putting all of its incident edges that are not ineligible onto the priority queue, thus ensuring that the priority queue contains the crossing edges from tree vertices to non-tree vertices (perhaps also some ineligible edges). The inner loop is a rendition in code of the one-sentence description of the algorithm: we take an edge from the priority queue and (if it is not ineligible) add it to the tree, and also add to the tree the new vertex that it leads to, updating the set of crossing edges by calling visit() with that vertex as argument. The weight() method requires iterating through the tree edges to add up the edge weights (lazy approach) or keeping a running total in an instance variable (eager approach) and is left as EXERCISE 4.3.31.
Running time
How fast is Prim’s algorithm? This question is not difficult to answer, given our knowledge of the behavior characteristics of priority queues:
Proposition M. The lazy version of Prim’s algorithm uses space proportional to E and time proportional to E log E (in the worst case) to compute the MST of a connected edge-weighted graph with E edges and V vertices.
Proof: The bottleneck in the algorithm is the number of edge-weight comparisons in the priority-queue methods insert() and delMin(). The number of edges on the priority queue is at most E, which gives the space bound. In the worst case, the cost of an insertion is ~lg E and the cost to delete the minimum is ~2 lg E (see PROPOSITION Q in CHAPTER 2). Since at most E edges are inserted and at most E are deleted, the time bound follows.
In practice, the upper bound on the running time is a bit conservative because the number of edges on the priority queue is typically much less than E. The existence of such a simple, efficient, and useful algorithm for such a challenging task is quite remarkable. Next, we briefly discuss some improvements. As usual, detailed evaluation of such improvements in performance-critical applications is a job for experts.
Lazy version of Prim’s MST algorithm
public class LazyPrimMST
{
private boolean[] marked; // MST vertices
private Queue<Edge> mst; // MST edges
private MinPQ<Edge> pq; // crossing (and ineligible) edges
public LazyPrimMST(EdgeWeightedGraph G)
{
pq = new MinPQ<Edge>();
marked = new boolean[G.V()];
mst = new Queue<Edge>();
visit(G, 0); // assumes G is connected (see Exercise 4.3.22)
while (!pq.isEmpty())
{
Edge e = pq.delMin(); // Get lowest-weight
int v = e.either(), w = e.other(v); // edge from pq.
if (marked[v] && marked[w]) continue; // Skip if ineligible.
mst.enqueue(e); // Add edge to tree.
if (!marked[v]) visit(G, v); // Add vertex to tree
if (!marked[w]) visit(G, w); // (either v or w).
}
}
private void visit(EdgeWeightedGraph G, int v)
{ // Mark v and add to pq all edges from v to unmarked vertices.
marked[v] = true;
if (!marked[e.other(v)]) pq.insert(e);
}
public Iterable<Edge> edges()
{ return mst; }
public double weight() // See Exercise 4.3.31.
}
This implementation of Prim’s algorithm uses a priority queue to hold crossing edges, a vertex-indexed array to mark tree vertices, and a queue to hold MST edges. This implementation is a lazy approach where we leave ineligible edges in the priority queue.
Eager version of Prim’s algorithm
To improve the LazyPrimMST, we might try to delete ineligible edges from the priority queue, so that the priority queue contains only the crossing edges between tree vertices and non-tree vertices. But we can eliminate even more edges. The key is to note that our only interest is in the minimaledge from each non-tree vertex to a tree vertex. When we add a vertex v to the tree, the only possible change with respect to each non-tree vertex w is that adding v brings w closer than before to the tree. In short, we do not need to keep on the priority queue all of the edges from w to tree vertices—we just need to keep track of the minimum-weight edge and check whether the addition of v to the tree necessitates that we update that minimum (because of an edge v-w that has lower weight), which we can do as we process each edge in v’s adjacency list. In other words, we maintain on the priority queue just one edge for each non-tree vertex w: the lightest edge that connects it to the tree. Any heavier edge connecting w to the tree will become ineligible at some point, so there is no need to keep it on the priority queue.
PrimMST (ALGORITHM 4.7 on page 622) implements Prim’s algorithm using our index priority queue data type from SECTION 2.4 (see page 320). It replaces the data structure marked[] and mst[] in LazyPrimMST by two vertex-indexed arrays edgeTo[] and distTo[], which have the following properties:
• If v is not on the tree but has at least one edge connecting it to the tree, then edgeTo[v] is the lightest edge connecting v to the tree, and distTo[v] is the weight of that edge.
• All such vertices v are maintained on the index priority queue, as an index v associated with the weight of edgeTo[v].
The key implications of these properties is that the minimum key on the priority queue is the weight of the minimal-weight crossing edge, and its associated vertex v is the next to add to the tree. To maintain the data structures, PrimMST takes a vertex v from the priority queue, then checks each edge v-w on its adjacency list. If w is marked, the edge is ineligible; if it is not on the priority queue or its weight is lower than the current best-known edgeTo[w], the code updates the data structures to establish v-w as the best-known way to connect w to the tree.
The figure on the facing page is a trace of PrimMST for our small sample graph tinyEWG.txt. The contents of the edgeTo[] and distTo[] arrays are depicted after each vertex is added to the MST, color-coded to depict the MST vertices (index in black), the non-MST vertices (index in gray), the MST edges (in black), and the priority-queue index/value pairs (in red). In the drawings, the lightest edge connecting each non-MST vertex to an MST vertex is drawn in red. The algorithm adds edges to the MST in the same order as the lazy version; the difference is in the priority-queue operations. It builds the MST as follows:
• Adds 0 to the MST and all edges in its adjacency list to the priority queue, since each such edge is the best (only) known connection between a tree vertex and a non-tree vertex.
• Adds 7 and 0-7 to the MST, replaces 0-4 with 4-7 as the lighest edge from a tree vertex to 4, adds 1-7 and 5-7 to the priority queue. Edge 2-7 does not affect the priority queue because its weights is not less than the weight of the known connection from the MST to 2.
• Adds 1 and 1-7 to the MST and 1-3 to the priority queue.
• Adds 2 and 0-2 to the MST, replaces 6-0 with 2-6 as the lightest edge from a tree vertex to 6, and replaces 1-3 with 2-3 as the lightest edge from a tree vertex to 3.
• Adds 3 and 2-3 to the MST.
• Adds 5 and 5-7 to the MST and replaces 4-7 with 4-5 as the lightest edge from a tree vertex to 4.
• Adds 4 and 4-5 to the MST.
• Adds 6 and 6-2 to the MST.
After having added V−1 edges, the MST is complete and the priority queue is empty.
AN ESSENTIALLY IDENTICAL ARGUMENT as in the proof of PROPOSITION M proves that the eager version of Prim’s algorithm finds the MST of a connected edge-weighted graph in time proportional to E log V and extra space proportional to V. For the huge sparse graphs that are typical in practice, there is no asymptotic difference in the time bound (because lg E ~ lg V for sparse graphs); the space bound is a constant-factor (but significant) improvement. Further analysis and experimentation are best left for experts facing performance-critical applications, where many factors come into play, including the implementations of MinPQ and IndexMinPQ, the graph representation, properties of the application’s graph model, and so forth. As usual, such improvements need to be carefully considered, as the increased code complexity is only justified for applications where constant-factor performance gains are important, and might even be counterproductive on complex modern systems.
Algorithm 4.7 Prim’s MST algorithm (eager version)
public class PrimMST
{
private Edge[] edgeTo; // shortest edge from tree vertex
private double[] distTo; // distTo[w] = edgeTo[w].weight()
private boolean[] marked; // true if v on tree
private IndexMinPQ<Double> pq; // eligible crossing edges
public PrimMST(EdgeWeightedGraph G)
{
edgeTo = new Edge[G.V()];
distTo = new double[G.V()];
marked = new boolean[G.V()];
for (int v = 0; v < G.V(); v++)
distTo[v] = Double.POSITIVE_INFINITY;
pq = new IndexMinPQ<Double>(G.V());
distTo[0] = 0.0;
pq.insert(0, 0.0); // Initialize pq with 0, weight 0.
while (!pq.isEmpty())
visit(G, pq.delMin()); // Add closest vertex to tree.
}
private void visit(EdgeWeightedGraph G, int v)
{ // Add v to tree; update data structures.
marked[v] = true;
{
int w = e.other(v);
if (marked[w]) continue; // v-w is ineligible.
if (e.weight() < distTo[w])
{ // Edge e is new best connection from tree to w.
edgeTo[w] = e;
distTo[w] = e.weight();
if (pq.contains(w)) pq.changeKey(w, distTo[w]);
else pq.insert(w, distTo[w]);
}
}
}
public Iterable<Edge> edges() // See Exercise 4.3.21.
public double weight() // See Exercise 4.3.31.
}
This implementation of Prim’s algorithm keeps eligible crossing edges on an index priority queue.
Proposition N. The eager version of Prim’s algorithm uses extra space proportional to V and time proportional to E log V (in the worst case) to compute the MST of a connected edge-weighted graph with E edges and V vertices.
Proof: The number of vertices on the priority queue is at most V, and there are three vertex-indexed arrays, which implies the space bound. The algorithm uses V insert operations, V delete the minimum operations, and (in the worst case) E change priority operations. These counts, coupled with the fact that our heap-based implementation of the index priority queue implements all these operations in time proportional to log V (see page 321), imply the time bound.
The diagram at right shows Prim’s algorithm in operation on our 250-vertex Euclidean graph mediumEWG.txt. It is a fascinating dynamic process (see also EXERCISE 4.3.27). Most often the tree grows by connecting a new vertex to the vertex just added. When reaching an area with no nearby non-tree vertices, the growth starts from another part of the tree.
Kruskal’s algorithm
The second MST algorithm that we consider in detail is to process the edges in order of their weight values (smallest to largest), taking for the MST (coloring black) each edge that does not form a cycle with edges previously added, stopping after adding V−1 edges. The black edges form a forest of trees that evolves gradually into a single tree, the MST. This method is known as Kruskal’s algorithm:
Proposition O. Kruskal’s algorithm computes the MST of any connected edge-weighted graph.
Proof: Immediate from PROPOSITION K. If the next edge to be considered does not form a cycle with black edges, it crosses a cut defined by the set of vertices connected to one of the edge’s vertices by black edges (and its complement). Since the edge does not create a cycle, it is the only crossing edge seen so far, and since we consider the edges in sorted order, it is a crossing edge of minimum weight. Thus, the algorithm is successively taking a minimal-weight crossing edge, in accordance with the greedy algorithm.
Prim’s algorithm builds the MST one edge at a time, finding a new edge to attach to a single growing tree at each step. Kruskal’s algorithm also builds the MST one edge at a time; but, by contrast, it finds an edge that connects two trees in a forest of growing trees. We start with a degenerate forest of V single-vertex trees and perform the operation of combining two trees (using the lightest edge possible) until there is just one tree left: the MST.
The figure at left shows a step-by-step example of the operation of Kruskal’s algorithm on tinyEWG.txt. The five lowest-weight edges in the graph are taken for the MST, then 1-3, 1-5, and 2-7 are determined to be ineligible before 4-5 is taken for the MST, and finally 1-2, 4-7, and 0-4 are determined to be ineligible and 6-2 is taken for the MST.
Kruskal’s algorithm is also not difficult to implement, given the basic algorithmic tools that we have considered in this book: we use a priority queue (SECTION 2.4) to consider the edges in order by weight, a union-find data structure (SECTION 1.5) to identify those that cause cycles, and a queue (SECTION 1.3) to collect the MST edges. ALGORITHM 4.8 is an implementation along these lines. Note that collecting the MST edges in a Queue means that when a client iterates through the edges it gets them in increasing order of their weight. The weight() method requires iterating through the queue to add the edge weights (or keeping a running total in an instance variable) and is left as an exercise (see EXERCISE 4.3.31).
Analyzing the running time of Kruskal’s algorithm is a simple matter because we know the running times of its basic operations.
Proposition N (continued). Kruskal’s algorithm uses space proportional to E and time proportional to E log E (in the worst case) to compute the MST of an edge-weighted connected graph with E edges and V vertices.
Proof: The implementation uses the priority-queue constructor that initializes the priority queue with all the edges, at a cost of at most E compares (see SECTION 2.4). After the priority queue is built, the argument is the same as for Prim’s algorithm. The number of edges on the priority queue is at most E, which gives the space bound, and the cost per operation is at most 2 lg E compares, which gives the time bound. Kruskal’s algorithm also performs up to E find() and V union() operations, but that cost does not contribute to the E log E order of growth of the total running time (see SECTION 1.5).
As with Prim’s algorithm the cost bound is conservative, since the algorithm terminates after finding the V−1 MST edges. The order of growth of the actual cost is E + E0 log E, where E0 is the number of edges whose weight is less than the weight of the MST edge with the highest weight. Despite this advantage, Kruskal’s algorithm is generally slower than Prim’s algorithm because it has to do a connected() operation for each edge, in addition to the priority-queue operations that both algorithms do for each edge processed (see EXERCISE 4.3.39).
The figure at left illustrates the algorithm’s dynamic characteristics on the larger example mediumEWG.txt. The fact that the edges are added to the forest in order of their length is quite apparent.
Algorithm 4.8 Kruskal’s MST algorithm
public class KruskalMST
{
private Queue<Edge> mst;
public KruskalMST(EdgeWeightedGraph G)
{
mst = new Queue<Edge>();
MinPQ<Edge> pq = new MinPQ<Edge>();
for (Edge e : G.edges())
pq.insert(e);
UF uf = new UF(G.V());
while (!pq.isEmpty() && mst.size() < G.V()-1)
{
Edge e = pq.delMin(); // Get min weight edge on pq
int v = e.either(), w = e.other(v); // and its vertices.
if (uf.connected(v, w)) continue; // Ignore ineligible edges.
uf.union(v, w); // Merge components.
mst.enqueue(e); // Add edge to mst.
}
}
public Iterable<Edge> edges()
{ return mst; }
public double weight() // See Exercise 4.3.31.
}
This implementation of Kruskal’s algorithm uses a queue to hold MST edges, a priority queue to hold edges not yet examined, and a union-find data structure for identifying ineligible edges. The MST edges are returned to the client in increasing order of their weights. The weight()method is left as an exercise.
% java KruskalMST tinyEWG.txt
0-7 0.16000
2-3 0.17000
1-7 0.19000
0-2 0.26000
5-7 0.28000
4-5 0.35000
6-2 0.40000
1.8100
Perspective
The MST problem is one of the most heavily studied problems that we encounter in this book. Basic approaches to solving it were invented long before the development of modern data structures and modern techniques for analyzing the performance of algorithms, at a time when finding the MST of a graph that contained, say, thousands of edges was a daunting task. The MST algorithms that we have considered differ from these old ones essentially in their use and implementation of modern algorithms and data structures for basic tasks, which (coupled with modern computing power) makes it possible for us to compute MSTs with millions or even billions of edges.
Historical notes
An MST implementation for dense graphs (see EXERCISE 4.3.29) was first presented by R. Prim in 1961 and, independently, by E. W. Dijkstra soon thereafter. It is usually referred to as Prim’s algorithm, although Dijkstra’s presentation was more general. But the basic idea was also presented by V. Jarnik in 1939, so some authors refer to the method as Jarnik’s algorithm, thus characterizing Prim’s (or Dijkstra’s) role as finding an efficient implementation of the algorithm for dense graphs. As the priority-queue ADT came into use in the early 1970s, its application to finding MSTs of sparse graphs was straightforward; the fact that MSTs of sparse graphs could be computed in time proportional to E log E became widely known without attribution to any particular researcher. In 1984, M. L. Fredman and R. E. Tarjan developed the Fibonacci heap data structure, which improves the theoretical bound on the order of growth of the running time of Prim’s algorithm to E + V log V. J. Kruskal presented his algorithm in 1956, but, again, the relevant ADT implementations were not carefully studied for many years. Other interesting historical notes are that Kruskal’s paper mentioned a version of Prim’s algorithm and that a 1926 (!) paper by O. Boruvka mentioned both approaches. Boruvka’s paper addressed a power-distribution application and introduced yet another method that is easily implemented with modern data structures (see EXERCISE 4.3.43 and EXERCISE 4.3.44). The method was rediscovered by M. Sollin in 1961; it later attracted attention as the basis for MST algorithms with efficient asymptotic performance and as the basis for parallel MST algorithms.
A linear-time algorithm?
On the one hand, no theoretical results have been developed that deny the existence of an MST algorithm that is guaranteed to run in linear time for all graphs. On the other hand, the goal of developing algorithms for computing the MST of sparse graphs in linear time remains elusive. Since the 1970s the applicability of the union-find abstraction to Kruskal’s algorithm and the applicability of the priority-queue abstraction to Prim’s algorithm have been prime motivations for many researchers to seek better implementations of those ADTs. Many researchers have concentrated on finding efficient priority-queue implementations as the key to finding efficient MST algorithms for sparse graphs; many other researchers have studied variations of Boruvka’s algorithm as the basis for nearly linear-time MST algorithms for sparse graphs. Such research still holds the potential to lead us eventually to a practical linear-time MST algorithm and has even shown the existence of a randomized linear-time algorithm. Also, researchers are getting quite close to the linear-time goal: B. Chazelle exhibited an algorithm in 1997 that certainly could never be distinguished from a linear-time algorithm in any conceivable practical situation (even though it is provably nonlinear), but is so complicated that no one would use it in practice. While the algorithms that have emerged from such research are generally quite complicated, simplified versions of some of them may yet be shown to be useful in practice. In the meantime, we can use the basic algorithms that we have considered here to compute the MST in linear time in most practical situations, perhaps paying an extra factor of log V for some sparse graphs.
IN SUMMARY, we can consider the MST problem to be “solved” for practical purposes. For most graphs, the cost of finding the MST is only slightly higher than the cost of extracting the graph’s edges. This rule holds except for huge graphs that are extremely sparse, but the available performance improvement over the best-known algorithms even in this case is a small constant factor, perhaps a factor of 10 at best. These conclusions are borne out for many graph models, and practitioners have been using Prim’s and Kruskal’s algorithms to find MSTs in huge graphs for decades.
Q&A
Q. Do Prim’s and Kruskal’s algorithms work for directed graphs?
A. No, not at all. That is a more difficult graph-processing problem known as the minimum cost arborescence problem.
Exercises
4.3.1 Prove that you can rescale the weights by adding a positive constant to all of them or by multiplying them all by a positive constant without affecting the MST.
4.3.2 Draw all of the MSTs of the graph depicted at right (all edge weights are equal).
4.3.3 Show that if a graph’s edges all have distinct weights, the MST is unique.
4.3.4 Consider the assertion that an edge-weighted graph has a unique MST only if its edge weights are distinct. Give a proof or a counterexample.
4.3.5 Show that the greedy algorithm is valid even when edge weights are not distinct.
4.3.6 Give the MST of the weighted graph obtained by deleting vertex 7 from tinyEWG.txt (see page 604).
4.3.7 How would you find a maximum spanning tree of an edge-weighted graph?
4.3.8 Prove the following, known as the cycle property: Given any cycle in an edge-weighted graph (all edge weights distinct), the edge of maximum weight in the cycle does not belong to the MST of the graph.
4.3.9 Implement the constructor for EdgeWeightedGraph that reads an edge-weighted graph from the input stream, by suitably modifying the constructor from Graph (see page 526).
4.3.10 Develop an EdgeWeightedGraph implementation for dense graphs that uses an adjacency-matrix (two-dimensional array of weights) representation. Disallow parallel edges.
4.3.11 Determine the amount of memory used by EdgeWeightedGraph to represent a graph with V vertices and E edges, using the memory-cost model of SECTION 1.4.
4.3.12 Suppose that a graph has distinct edge weights. Does its lightest edge have to belong to the MST? Can its heaviest edge belong to the MST? Does a min-weight edge on every cycle have to belong to the MST? Prove your answer to each question or give a counterexample.
4.3.13 Give a counterexample that shows why the following strategy does not necessarily find the MST: ‘Start with any vertex as a single-vertex MST, then add V-1 edges to it, always taking next a min-weight edge incident to the vertex most recently added to the MST.’
4.3.14 Given an MST for an edge-weighted graph G, suppose that an edge in G that does not disconnect G is deleted. Describe how to find an MST of the new graph in time proportional to E.
4.3.15 Given an MST for an edge-weighted graph G and a new edge e with weight w, describe how to find an MST of the new graph in time proportional to V.
4.3.16 Given an MST for an edge-weighted graph G and a new edge e, write a program that determines the range of weights for which e is in an MST.
4.3.17 Implement toString() for EdgeWeightedGraph.
4.3.18 Give traces that show the process of computing the MST of the graph defined in EXERCISE 4.3.6 with the lazy version of Prim’s algorithm, the eager version of Prim’s algorithm, and Kruskal’s algorithm.
4.3.19 Suppose that you implement PrimMST but instead of using a priority queue to find the next vertex to add to the tree, you scan through all V entries in the distTo[] array to find the non-tree vertex with the smallest weight. What would be the order of growth of the running time for graphs with V vertices and E edges? When would this method be appropriate, if ever? Defend your answer.
4.3.20 True or false: At any point during the execution of Kruskal’s algorithm, each vertex is closer to some vertex in its subtree than to any vertex not in its subtree. Prove your answer.
4.3.21 Provide an implementation of edges() for PrimMST (page 622).
Solution:
public Iterable<Edge> edges()
{
Queue<Edge> mst = new Queue<Edge>();
for (int v = 1; v < edgeTo.length; v++)
mst.enqueue(edgeTo[v]);
return mst;
}
Creative Problems
4.3.22 Minimum spanning forest. Develop versions of Prim’s and Kruskal’s algorithms that compute the minimum spanning forest of an edge-weighted graph that is not necessarily connected. Use the connected-components API of SECTION 4.1 and find MSTs in each component.
4.3.23 Vyssotsky’s algorithm. Develop an implementation that computes the MST by applying the cycle property (see EXERCISE 4.3.8) repeatedly: Add edges one at a time to a putative tree, deleting a maximum-weight edge on the cycle if one is formed. Note: This method has received less attention than the others that we consider because of the comparative difficulty of maintaining a data structure that supports efficient implementation of the “delete the maximum-weight edge on the cycle” operation.
4.3.24 Reverse-delete algorithm. Develop an implementation that computes the MST as follows: Start with a graph containing all of the edges. Then repeatedly go through the edges in decreasing order of weight. For each edge, check if deleting that edge will disconnect the graph; if not, delete it. Prove that this algorithm computes the MST. What is the order of growth of the number of edge-weight compares performed by your implementation?
4.3.25 Worst-case generator. Develop a reasonable generator for edge-weighted graphs with V vertices and E edges such that the running time of the lazy version of Prim’s algorithm is nonlinear. Answer the same question for the eager version.
4.3.26 Critical edges. An MST edge whose deletion from the graph would cause the MST weight to increase is called a critical edge. Show how to find all critical edges in a graph in time proportional to E log E. Note: This question assumes that edge weights are not necessarily distinct (otherwise all edges in the MST are critical).
4.3.27 Animations. Write a client program that does dynamic graphical animations of MST algorithms. Run your program for mediumEWG.txt to produce images like the figures on page 621 and page 624.
4.3.28 Space-efficient data structures. Develop an implementation of the lazy version of Prim’s algorithm that saves space by using lower-level data structures for EdgeWeightedGraph and for MinPQ instead of Bag and Edge. Estimate the amount of memory saved as a function of V and E, using the memory-cost model of SECTION 1.4 (see EXERCISE 4.3.11).
4.3.29 Dense graphs. Develop an implementation of Prim’s algorithm that uses an eager approach (but not a priority queue) and computes the MST using V2 edge-weight comparisons.
4.3.30 Euclidean weighted graphs. Modify your solution to EXERCISE 4.1.37 to create an API EuclideanEdgeWeightedGraph for graphs whose vertices are points in the plane, so that you can work with graphical representations.
4.3.31 MST weights. Develop implementations of weight() for LazyPrimMST, PrimMST, and KruskalMST, using a lazy strategy that iterates through the MST edges when the client calls weight(). Then develop alternate implementations that use an eager strategy that maintains a running total as the MST is computed.
4.3.32 Specified set. Given a connected edge-weighted graph G and a specified set of edges S (having no cycles), describe a way to find a minimum-weight spanning tree of G among those spanning trees that contain all the edges in S.
4.3.33 Certification. Write an MST and EdgeWeightedGraph client check() that uses the following cut optimality conditions implied by PROPOSITION J to verify that a proposed set of edges is in fact an MST: A set of edges is an MST if it is a spanning tree and every edge is a minimum-weight edge in the cut defined by removing that edge from the tree. What is the order of growth of the running time of your method?
Experiments
4.3.34 Random sparse edge-weighted graphs. Write a random-sparse-edge-weighted-graph generator based on your solution to EXERCISE 4.1.41. To assign edge weights, define a random-edge-weighted graph ADT and write two implementations: one that generates uniformly distributed weights, another that generates weights according to a Gaussian distribution. Write client programs to generate sparse random edge-weighted graphs for both weight distributions with a well-chosen set of values of V and E so that you can use them to run empirical tests on graphs drawn from various distributions of edge weights.
4.3.35 Random Euclidean edge-weighted graphs. Modify your solution to EXERCISE 4.1.42 to assign the distance between vertices as each edge’s weight.
4.3.36 Random grid edge-weighted graphs. Modify your solution to EXERCISE 4.1.43 to assign a random weight (between 0 and 1) to each edge.
4.3.37 Real edge-weighted graphs. Find a large weighted graph somewhere online—perhaps a map with distances, telephone connections with costs, or an airline rate schedule. Write a program RandomRealEdgeWeightedGraph that builds a weighted graph by choosing V vertices at random and Eweighted edges at random from the subgraph induced by those vertices.
Testing all algorithms and studying all parameters against all graph models is unrealistic. For each problem listed below, write a client that addresses the problem for any given input graph, then choose among the generators above to run experiments for that graph model. Use your judgment in selecting experiments, perhaps in response to results of previous experiments. Write a narrative explaining your results and any conclusions that might be drawn.
4.3.38 Cost of laziness. Run empirical studies to compare the performance of the lazy version of Prim’s algorithm with the eager version, for various types of graphs.
4.3.39 Prim versus Kruskal. Run empirical studies to compare the performance of the lazy and eager versions of Prim’s algorithm with Kruskal’s algorithm.
4.3.40 Reduced overhead. Run empirical studies to determine the effect of using primitive types instead of Edge values in EdgeWeightedGraph, as described in EXERCISE 4.3.28.
4.3.41 Heaviest MST edge. Run empirical studies to analyze the weight of the heaviest edge in the MST and the number of graph edges that are not heavier than that one.
4.3.42 Partitioning. Develop an implementation based on integrating Kruskal’s algorithm with quicksort partitioning (instead of using a priority queue) so as to check MST membership of each edge as soon as all smaller edges have been checked.
4.3.43 Boruvka’s algorithm. Develop an implementation of Boruvka’s algorithm: Build an MST by adding edges to a growing forest of trees, as in Kruskal’s algorithm, but in stages. At each stage, find the minimum-weight edge that connects each tree to a different one, then add all such edges to the MST. Assume that the edge weights are all different, to avoid cycles. Hint: Maintain a vertex-indexed array to identify the edge that connects each tree to its nearest neighbor, and use the union-find data structure.
4.3.44 Improved Boruvka. Develop an implementation of Boruvka’s algorithm that uses doubly-linked circular lists to represent MST subtrees so that subtrees can be merged and renamed in time bounded by E during each stage (and the union-find data type is therefore not needed).
4.3.45 External MST. Describe how you would find the MST of a graph so large that only V edges can fit into main memory at once.
4.3.46 Johnson’s algorithm. Develop a priority-queue implementation that uses a d-way heap (see EXERCISE 2.4.41). Find the best value of d for various weighted graph models.
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Last week’s post showed how to create a Gilmore Girls character network.
In this week’s short post, I want to explore the Gilmore Girls’ famous coffee addiction by analysing the same episode transcripts that were also used last week.
I am also showcasing how to use the recently updated ggplot2 2.2.0.
The transcripts were prepared as described in last week’s post. Full R code can be found below.
## How often do they talk about coffee?
The first question I want to explore is how often Lorelai, Rory, Luke, Sookie, Lane and Paris talked about coffee. For this, I subsetted the lines of these characters to where they talk about coffee (i.e. lines that include the word “coffee” at least once). I then created a plot showing how often they talk about coffee per episode and season.
The differences in coffee mention between characters are of course somewhat biased by differences in the total number of lines of each character in the respective episodes, as well by the length of the episodes.
This led me to wonder what proportion of each episode was occupied by which character.
# R code
```# set names
names <- c("LORELAI", "RORY", "LUKE", "SOOKIE", "LANE", "PARIS")
# extract lines for each character in names
for (name in names){
assign(paste("lines", name, sep = "_"), transcripts_2[grep(paste0(name), transcripts_2\$character), ])
}
# extract lines not from characters in names
lines_OTHER <- transcripts_2[!grepl("LORELAI|RORY|LUKE|SOOKIE|LANE|PARIS", transcripts_2\$character), ]
# update names to include "other"
names <- c(names, "OTHER")
```
```# which lines mention coffee.
for (name in names){
df <- get(paste("lines", name, sep = "_"))
assign(paste("coffee", name, sep = "_"), df[grep("coffee", df\$dialogue), ])
}
```
```# calculate number of lines in total and with coffee mentions per episode
library(dplyr)
for (name in names){
lines_p_ep <- get(paste("lines", name, sep = "_"))
lines_p_ep <- as.data.frame(table(lines_p_ep\$episode))
df <- get(paste("coffee", name, sep = "_"))
df_2 <- as.data.frame(table(df\$episode))
# calculate coffee lines per episode
coffee_lines_p_ep <- full_join(lines_p_ep, df_2, by = "Var1")
coffee_lines_p_ep\$ratio_p_ep <- coffee_lines_p_ep\$Freq.y/coffee_lines_p_ep\$Freq.x
coffee_lines_p_ep\$Season <- gsub("_.*", "", coffee_lines_p_ep\$Var1)
coffee_lines_p_ep\$character <- paste(name)
assign(paste("coffee_lines_p_ep", name, sep = "_"), coffee_lines_p_ep)
}
coffee_df <- rbind(coffee_lines_p_ep_LORELAI, coffee_lines_p_ep_RORY, coffee_lines_p_ep_LUKE, coffee_lines_p_ep_SOOKIE,
coffee_lines_p_ep_LANE, coffee_lines_p_ep_PARIS)
```
```library(ggplot2)
my_theme <- function(base_size = 12, base_family = "sans"){
theme_grey(base_size = base_size, base_family = base_family) +
theme(
axis.text = element_text(size = 12),
axis.text.x = element_text(angle = 45, vjust = 0.5, hjust = 0.5),
axis.title = element_text(size = 14),
panel.grid.major = element_line(color = "grey"),
panel.grid.minor = element_blank(),
panel.background = element_rect(fill = "aliceblue"),
strip.background = element_rect(fill = "lightgrey", color = "grey", size = 1),
strip.text = element_text(face = "bold", size = 12, color = "black"),
legend.position = "bottom",
legend.justification = "top",
legend.box = "horizontal",
legend.box.background = element_rect(colour = "grey50"),
legend.background = element_blank(),
panel.border = element_rect(color = "grey", fill = NA, size = 0.5)
)
}
```
```library(tidyr)
coffee_df_gather <- coffee_df %>%
gather(group, value, Freq.x:ratio_p_ep)
coffee_df_gather[which(coffee_df_gather\$group == "Freq.x"), "group"] <- "Lines per episode"
coffee_df_gather[which(coffee_df_gather\$group == "Freq.y"), "group"] <- "Coffee lines per episode"
coffee_df_gather[which(coffee_df_gather\$group == "ratio_p_ep"), "group"] <- "Ratio of coffee lines per episode"
coffee_df_gather\$character <- factor(coffee_df_gather\$character, levels = names)
coffee_df_gather\$group <- factor(coffee_df_gather\$group, levels = c("Lines per episode", "Coffee lines per episode", "Ratio of coffee lines per episode"))
ggplot(data = coffee_df_gather, aes(x = character, y = value, color = Season, fill = Season)) +
geom_boxplot(alpha = 0.5) +
labs(
x = "",
y = "Number or ratio of lines",
title = "Lorelai & Luke are the coffee queen and king",
subtitle = "Coffee mentions per episode and season",
caption = "\nThese boxplots show the number of total lines per episode and season, lines with coffee mentions per epsiode and season,
as well as the ratio between the two for the main characters Lorelai, Rory, Luke, Sookie, Lane and Paris. Lorelai consistently had
the most lines per episode, followed closely by Rory and Luke. Sookie, Lane and Paris had roughly the same numbers of lines per episode,
although their variance is higher. The same trend is reflected by the number of lines with coffee mentions, except that Rory and Luke seem
to talk similarly often about coffee. Interestingly, Lorelai's (verbal) coffee obsession seems to have decreased slightly over the seasons.
The ratio of coffee lines divided by the total number of lines for each episode reflects that even though Lorelai, Rory and Luke talk a lot about coffee,
they also talk a lot in general, so their ratio of coffee mentions is only slightly higher than Sookie's, Lane's and Paris'.
While the latter's mean ratio is lower. they have a much higher variance with a few episodes where they seem to have talked relatively much about coffee.
During the first seasons, Luke is the character with the highest ratio of coffee vs general talk but this is decreasing over time and becomes more
similar to Lorelai and Rory's coffee talk ratio."
) +
my_theme() +
facet_wrap(~ group, ncol = 3, scales = "free") +
guides(fill = guide_legend(nrow = 1, byrow = TRUE)) +
scale_fill_brewer(palette="Dark2") +
scale_color_brewer(palette="Dark2")
```
```# calculate number of lines in total
for (name in names){
lines_p_ep <- get(paste("lines", name, sep = "_"))
lines_p_ep <- as.data.frame(table(lines_p_ep\$episode))
lines_p_ep\$character <- paste(name)
assign(paste("lines_p_ep", name, sep = "_"), lines_p_ep)
}
# combine and convert to wide format to make calculating proportions easier
lines_df_wide <- spread(rbind(lines_p_ep_LORELAI, lines_p_ep_RORY, lines_p_ep_LUKE, lines_p_ep_SOOKIE,
lines_p_ep_LANE, lines_p_ep_PARIS, lines_p_ep_OTHER), Var1, Freq)
rownames(lines_df_wide) <- lines_df_wide[, 1]
lines_df_wide <- as.data.frame(t(lines_df_wide[, -1]))
# calculate proportions and percentage
prop_lines_df_wide <- sweep(lines_df_wide, 1, rowSums(lines_df_wide), FUN = "/")
percent_lines_df_wide <- prop_lines_df_wide * 100
percent_lines_df_wide\$Episode <- rownames(percent_lines_df_wide)
percent_lines_df_wide\$Season <- gsub("_.*", "", percent_lines_df_wide\$Episode)
# gather for plotting
percent_lines_df_gather <- percent_lines_df_wide %>%
gather(Character, Percent, LANE:SOOKIE)
percent_lines_df_gather\$Character <- factor(percent_lines_df_gather\$Character, levels = names)
percent_lines_df_gather <- merge(percent_lines_df_gather, subset(lines_LORELAI, select = c(episode, episode_running_nr)), by.x = "Episode", by.y = "episode")
```
```percent_lines_df_gather\$Episode <- factor(percent_lines_df_gather\$Episode, levels = percent_lines_df_gather[order(percent_lines_df_gather\$episode_running_nr, decreasing = TRUE), "Episode"])
percent_lines_df_gather <- percent_lines_df_gather[!duplicated(percent_lines_df_gather), ]
```
```ggplot(data = percent_lines_df_gather, aes(x = Character, y = Percent, fill = Season)) +
geom_boxplot(alpha = 0.5) +
labs(
x = "",
y = "Percent",
title = "Who's the biggest talker? (a)",
subtitle = "Percentage of lines per episode spoken by main characters",
caption = "\nThe boxplot shows the percentages of lines spoken by the main characters per episode and season.
In the first three seasons the majority of lines are clearly spoken by Lorelai and Rory, followed with some distance by Luke.
During the later seasons, the gap between Lorelai, Rory and the rest is not as pronounced any more."
) +
my_theme() +
facet_wrap(~ Season, ncol = 4, scales = "free") +
guides(fill = guide_legend(nrow = 1, byrow = TRUE)) +
scale_fill_brewer(palette = "Dark2") +
scale_color_brewer(palette = "Dark2")
```
```# calculate number of lines in total
for (name in names){
lines_p_season <- get(paste("lines", name, sep = "_"))
lines_p_season <- as.data.frame(table(lines_p_season\$season))
lines_p_season\$character <- paste(name)
assign(paste("lines_p_season", name, sep = "_"), lines_p_season)
}
# combine and convert to wide format to make calculating proportions easier
lines_df_wide <- spread(rbind(lines_p_season_LORELAI, lines_p_season_RORY, lines_p_season_LUKE, lines_p_season_SOOKIE,
lines_p_season_LANE, lines_p_season_PARIS, lines_p_season_OTHER), Var1, Freq)
rownames(lines_df_wide) <- lines_df_wide[, 1]
lines_df_wide <- as.data.frame(t(lines_df_wide[, -1]))
# calculate proportions and percentage
prop_lines_df_wide <- sweep(lines_df_wide, 1, rowSums(lines_df_wide), FUN = "/")
percent_lines_df_wide <- prop_lines_df_wide * 100
percent_lines_df_wide\$Season <- rownames(percent_lines_df_wide)
# gather for plotting
percent_lines_df_gather <- percent_lines_df_wide %>%
gather(Character, Percent, LANE:SOOKIE)
percent_lines_df_gather\$Character <- factor(percent_lines_df_gather\$Character, levels = names)
```
```# plot
bp <- ggplot(percent_lines_df_gather, aes(x = "", y = Percent, fill = Character)) +
geom_bar(width = 1, stat = "identity") + theme_minimal() +
scale_fill_brewer(palette = "Spectral")
pie <- bp + coord_polar("y", start = 0) +
theme(
axis.title.x = element_blank(),
axis.title.y = element_blank(),
panel.border = element_blank(),
panel.grid = element_blank(),
axis.ticks = element_blank(),
plot.title = element_text(size = 14, face = "bold"),
legend.title = element_blank(),
legend.position = "bottom",
legend.text = element_text(size = 8),
legend.justification = "top",
legend.box = "horizontal",
legend.box.background = element_rect(colour = "grey50"),
legend.background = element_blank(),
) + guides(fill = guide_legend(nrow = 1, byrow = TRUE))
pie + facet_wrap( ~ Season, ncol = 4) +
labs(
title = "Who's the biggest talker? (b)",
subtitle = "Percent of lines per season spoken by main characters")
```
```sessionInfo()
```
```## R version 3.3.2 (2016-10-31)
## Platform: x86_64-apple-darwin13.4.0 (64-bit)
## Running under: macOS Sierra 10.12
##
## locale:
## [1] de_DE.UTF-8/de_DE.UTF-8/de_DE.UTF-8/C/de_DE.UTF-8/de_DE.UTF-8
##
## attached base packages:
## [1] stats graphics grDevices utils datasets methods base
##
## other attached packages:
## [1] ggplot2_2.2.0 dplyr_0.5.0 splitstackshape_1.4.2
## [4] data.table_1.9.6 tidyr_0.6.0
##
## loaded via a namespace (and not attached):
## [1] Rcpp_0.12.8 knitr_1.15 magrittr_1.5
## [4] munsell_0.4.3 colorspace_1.3-1 R6_2.2.0
## [7] stringr_1.1.0 plyr_1.8.4 tools_3.3.2
## [10] grid_3.3.2 gtable_0.2.0 DBI_0.5-1
## [13] htmltools_0.3.5 lazyeval_0.2.0 yaml_2.1.14
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Transformations and non-normal distributions are typically the first approaches considered when the when the Normality test fails in a capability analysis. These approaches do not work when there are extreme outliers because they both assume the data come from a single common-cause variation distribution. But because extreme outliers typically represent special-cause variation, transformations and non-normal distributions are not good approaches for data that contain extreme outliers.
As an example, the four graphs below show distribution fits for a dataset with 99 values simulated from a N(m=10,s=1) distribution and 1 value simulated from a N(m=18,s=1) distribution. Two of the probability plots shown in these graphs assess the fit to a Normal Distribution after transforming the data. The remaining fourteen probability plots assess the fit to common (and some uncommon) non-normal distributions. None of the fits are adequate.
## Method for Calculating Defect Rate
For process data with common cause variation that follows a Normal distribution, a reasonable approach for modeling extreme outliers is to assume the outliers represent a shift in the mean of the distribution, as shown in the next three graphs.
In most cases, there will not be enough data to measure the variation in the outlier distribution, so the variation in the outlier distribution will need to be assumed unchanged from the common cause distribution. This is consistent with the approach used in classic regression and ANOVA analyses, which assume a change in the mean does not affect the variation.
The defect rate can be estimated by using a weighted average of the defect rates from the common cause distribution and the outlier distribution. The weights come from the sample sizes for each distribution.
In the example below, this calculation would be as follows (POS = Probability of Out of Spec):
Defect Rate = [99*(POS Common Cause Dist.)+1*(POS Outlier Dist.)]/100
In the following two examples, this calculation would be as follows:
Defect Rate = [98*(POS Common Cause Dist.)+2*(POS Outlier Dist.)]/100
## Important Considerations When Dealing with Extreme Outliers
It is critical to investigate extreme outliers and attempt to understand what caused them. The outlier(s) may be measurement errors or data entry errors, in which case they do not represent the true process and should appropriately adjusted.
If they are legitimate values, your number one priority should be to prevent future outliers from occurring and strive for process stability.
When you have special-cause variation in a capability analysis, you should not assume the defect rate estimated from the data represents the future defect rate of the process, no matter what approach you used. However, you may be able to get a rough estimate of the defect rate during the sampling period as long as the sample adequately represents the process during that time period.
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id True -- id True > id "hello" -- id "hello" Choice of ⦠data [a] = [] | a : [a] which is to say, a list containing elements of type a is either: An empty list, [] An element of type a, attached using : onto the front of another list [a]. In Haskell, the cons operation is written as a colon ⦠(In general, one can show that if the Ë= operator of a monad m is strictin its rst argument, then no value re- If the list is nonempty, then Haskell proceeds to the next line. n -- (!!) Lists and Recursion. The Data.List.Split module contains a wide range of strategies for splitting lists with respect to some sort of delimiter, mostly implemented through a unified combinator interface. {1, 2} × {} = {} {} × {1, 2} = {} For extra credit, show or write a function returning the n-ary product of an arbitrary number of lists, each of arbitrary length. take n xs. splitAt n xs (Returns a tuple of two lists.) Recursive definitions become more complicated if the recursion anchor is not chosen properly. In Haskell recursion serves as the basic mechanism for looping. Haskell looks through the patterns and applies the ï¬rst one that ï¬ts what it is trying to evaluate. The pattern of the list is (y:ys), where y is the head of Folds over lists consist of three elements - the list to fold over, some accumulator function f and an initial value.. Iâve spoken about the List Data Type previously in the Haskell for Beginners: Lists and Comprehensions post, but we need to know a little more about them before we can apply our newly found recursive knowledge to them. Unfortunately, right shrinking law is not satisable for a wide range of monads, including list, maybe, IO, and the strict state monads of Haskell. There are beautiful patterns inherent in the use of recursion that Iâve seen in my attempts to reboot my brain with a new, more functionally focused way of thinking about programming. The closest that you can get to a for-loop in Haskell, is the foldl (or foldr) function.Almost every other function in Data.List can be written using this function. List comprehensions can also draw elements from multiple lists, in which case the result will be the list of every possible combination of the two elements, as if the two lists were processed in the nested fashion. They transform the list a:b:c:[] into (a f (b f (c f init))) where init is the initial element i.e. Lists. Recursion is important in Haskell because, unlike with imperative In this lab we learn about the concept of recursion, which gives us the ability to âloopâ, or repeat the same instruction many times over. Haskell count of all elements in list of lists, Three ways: Get the length of each inner list, and sum them all: GHCi> sum (fmap length [[1,2,3],[4,3],[2,1],[5]]) 8. Haskell seems well suited to this, and I hope it will be much more reliable and maintainable than what we currently have. Understanding Lists in Haskell; Optional: Basic understanding of set theory We will write recursive functions over integers and lists. For example the function tupel presented in DMV-Mitteilungen 2004/12-3, Jürgen Bokowski: Haskell, ein gutes Werkzeug der Diskreten Mathematik (Haskell, a good tool for discrete mathematics). Also demonstrate, using your function/method, that the product of an empty list with any other list is empty. Recursion on lists. Haskell Hello Recursion! Hello Recursion! Working over a list of lists in Haskell, I think this does what you want import Data.List (transpose) addLists :: Num a => [[a]] -> [a] addLists xs = map sum . Yes, once you call again f with a new value of n, it has no way to reference the old value of n unless you pass it explicitly. In many languages, lists are built up from two primitives: either the list is the empty list, commonly called nil, or it is a list constructed by appending an element to the start of some other list, which we call a cons. Or, you always have the option of implementing any iteration as a recursion - that's really the "lowest level" of getting this done - but it is not the idiomatic way of doing simple data transformations in Haskell. Don't forget that zero is a natural number. List Comprehensions are one of my favourite features of Haskell. Recursion on Lists Recursion is not restricted to numbers, but can also be used to define functions on lists. In this case, the ï¬rst line says that if the list is empty, then elemCount x aList is 0. Decremented value called in the recursion in Haskell. How the list is built. The fact that lists are a recursive data type means that the functions that work on lists generally use structural recursion. The beauty of recursion and list machinery. We also investigate our first recursive data type, lists, that can pack many instances of a type together. A list is a singly linked list like one sees in an imperative language, but with one important difference: We cannot change any values within a list, including the pointers from one list node to another. The second approach is preferred, but the standard list processing functions do need to be defined, and those definitions use the first approach (recursive definitions). string,function,haskell,recursion,parameters. Foldr â foldr is a higher-order function in Haskell with the following type signature: ... foldl is not suitable for infinite lists. Remember if the list ⦠The processing of lists follows a simple pattern: Process the first element of the list. In Haskell terms: you pattern match on the list constructors, and you recurse on a subpart of the list. Just as recursion, list comprehension is a basic technique and should be learned right in the beginning.. Prerequisites. Basic Concepts # It is possible to define a function which can call itself. the recursive part: for a longer list, compare the head of the list and the maximum of the tail (this is where recursion happens); the maximum of the list is the bigger of the two So letâs write this up in Haskell. Thanks to lazy evaluation, both functions define infinite lists without computing them out entirely. A list of tokens has the type [Token]-- the square brackets are used to create lists (both list types, like [Int], and list literals, like [1, 2, 3]). Exercises; Type the factorial function into a Haskell source file and load it into GHCi. For practice, you can think of explicitly instantiatiating the type parameter (although Haskell syntax does not allow it). Haskell tries to work a tail recursion or so for any other functional language. (Note this is equivalent to Does Haskell standard library have a function that given a list and a predicate, returns the ⦠See below for usage, examples, and detailed documentation of all exported functions. Recursively process the rest of the list, reduce in each iteration by the first element. Haskell list of lists. Example. If you want to learn about the implementation, see Data.List.Split.Internals. In Haskell, there are no looping constructs. Fundamentally, our model just does a bunch of math on many lists of numbers (to give more context: the big competitors to our model are Excel spreadsheets). ... Recursion on Lists # Previously mentioned product function can be defined with recursion. ... Introduction via Haskell. In pure languages like Haskell, iteration and loops are forbidden, so recursion is the only option. Try examples like factorial 5 and factorial 1000.; What about factorial (-1)?Why does this happen? As recursion, parameters factorial 1000. ; what about factorial ( -1 )? Why this! All my research, it emerged as my favorite choice ; what haskell recursion list of lists (. At it later haskell recursion list of lists define a function is applied inside its own de.! And detailed documentation of all exported functions used to define a function is inside. Chosen properly do n't forget that zero is a way of de functions. And you recurse on a subpart of the list constructors, and you on... And loops are forbidden, so recursion is a basic technique and should be learned right in the... Type means that the functions that work on lists. it is possible to define functions on lists. currently., 2019 rest of the list is empty, then Haskell proceeds to the next.. Concepts # it is possible to define a function which can call itself many instances of type! For any other functional language set theory » Week 5: recursion and.... 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If you want to learn about the implementation, see Data.List.Split.Internals also be to. Variable, and detailed documentation of all exported functions is a natural number, recursion,.! If you want to learn about the implementation, see Data.List.Split.Internals means that the product of an empty list any... Lists ( at the Nth position ) ; Optional: basic understanding of set theory Week. Haskell looks like? Why does this happen function into a Haskell source file and load it into GHCi 2019. A subpart of the mx loop weâll take a closer look at later... Dial Indicator Calibration Procedure, Northwest Creamy Smoked Salmon Fettuccine Alfredo, Annatto-e 150 Benefits, When You Don T Enjoy Motherhood, Rustic Architecture Magazine, Gibbon Meaning In Bengali, Burn Linux Iso To Usb Windows 10, Audio Signal Processing, Australopithecus Afarensis Bipedal, Army Hot Weather Training, Convex Hull Divide And Conquer Python, " />
# haskell recursion list of lists
Recursion is important in Haskell and weâll take a closer look at it later. But in a nutshell, this is what happens if we try to get the factorial of, say, 3: ghci tries to compute 3 * factorial 2; factorial 2 is 2 * factorial 1, so for now we have 3 * (2 * factorial 1) The only operation we have available is to insert a node at the beginning of the list. The list is the foundation of the extremely powerful function composition in a functional language, because it is the general data structure. Hello Recursion! So if you write a list with any elements is passed like (a: b), what this means is 'a' will stand for the first element in the list and 'b' is a list of rest of the elements except the first one. » Week 5: Recursion and Lists. Instead, there are two alternatives: there are list iteration constructs (like foldl which we've seen before), and tail recursion. In Haskell, arrays are called lists. Week 5: Recursion and Lists ... An informal definition of lists in Haskell looks like. the result of f, but not the recursion variable, and is lifted out of the mx loop. Ordered merging of two ordered lists. haskell documentation: Merge Sort. For example, transpose \$ zipWith (\n x Make a new list containing just the first N elements from an existing list. St ephane Vialette LIGM, Universit e Paris-Est Marne-la-Vall ee October 3, 2019. That is, we can write a fib function, retrieving the nth element of the unbounded Fibonacci sequence: GHCi> let fib n = fibs !! Finally, a function type is constructed with an arrow -> between the type of the argument and the type of the result (we'll get to multi-argument functions later). List comprehension is for "whoosh"-style programming.\rRecursion is for "element-at-a-time" programming - like loops in other languages.\rBefore looking recursion, it's necessary to understand lists better. product :: Num a Þ[a] ®a product [] = 1 product (n:ns) = n * product ns product maps the empty list to 1, and any non-empty list to its head multiplied by the product of its tail. 2. being the list subscript operator -- or in ⦠Recursion is a way of de ning functions in which a function is applied inside its own de nition. Enter Haskell: from all my research, it emerged as my favorite choice. Now you know a little about Recursion its time we use this knowledge for good - lets use it with a Haskell Favorite, Lists!. Split a list into two smaller lists (at the Nth position). Feb 19, 2017. This is the basic principle behind recursion. Arrays are recursive structures. When we call the function, Haskell implicitly infers the appropriate type instantiation. Lists: Pattern Matching ⢠It is common to deï¬ne a recursive function on lists by specifying the value explicitly for the empty list, and then using an inductive rule for nonempty lists ⢠Here is a function for computing the number of elements in a list len [] = 0 len (x:xs) = 1 + (len xs) Theory in Programming Practice, Plaxton, Spring 2004 Weâll cover both methods. A list is built from the empty list $$[]$$ and the function $$cons\; :: \; a\rightarrow [a] \rightarrow [a]$$. Exploring Haskell: Recursive Functions 3 min read. The goal is to be flexible yet simple. > id True -- id True > id "hello" -- id "hello" Choice of ⦠data [a] = [] | a : [a] which is to say, a list containing elements of type a is either: An empty list, [] An element of type a, attached using : onto the front of another list [a]. In Haskell, the cons operation is written as a colon ⦠(In general, one can show that if the Ë= operator of a monad m is strictin its rst argument, then no value re- If the list is nonempty, then Haskell proceeds to the next line. n -- (!!) Lists and Recursion. The Data.List.Split module contains a wide range of strategies for splitting lists with respect to some sort of delimiter, mostly implemented through a unified combinator interface. {1, 2} × {} = {} {} × {1, 2} = {} For extra credit, show or write a function returning the n-ary product of an arbitrary number of lists, each of arbitrary length. take n xs. splitAt n xs (Returns a tuple of two lists.) Recursive definitions become more complicated if the recursion anchor is not chosen properly. In Haskell recursion serves as the basic mechanism for looping. Haskell looks through the patterns and applies the ï¬rst one that ï¬ts what it is trying to evaluate. The pattern of the list is (y:ys), where y is the head of Folds over lists consist of three elements - the list to fold over, some accumulator function f and an initial value.. Iâve spoken about the List Data Type previously in the Haskell for Beginners: Lists and Comprehensions post, but we need to know a little more about them before we can apply our newly found recursive knowledge to them. Unfortunately, right shrinking law is not satisable for a wide range of monads, including list, maybe, IO, and the strict state monads of Haskell. There are beautiful patterns inherent in the use of recursion that Iâve seen in my attempts to reboot my brain with a new, more functionally focused way of thinking about programming. The closest that you can get to a for-loop in Haskell, is the foldl (or foldr) function.Almost every other function in Data.List can be written using this function. List comprehensions can also draw elements from multiple lists, in which case the result will be the list of every possible combination of the two elements, as if the two lists were processed in the nested fashion. They transform the list a:b:c:[] into (a f (b f (c f init))) where init is the initial element i.e. Lists. Recursion is important in Haskell because, unlike with imperative In this lab we learn about the concept of recursion, which gives us the ability to âloopâ, or repeat the same instruction many times over. Haskell count of all elements in list of lists, Three ways: Get the length of each inner list, and sum them all: GHCi> sum (fmap length [[1,2,3],[4,3],[2,1],[5]]) 8. Haskell seems well suited to this, and I hope it will be much more reliable and maintainable than what we currently have. Understanding Lists in Haskell; Optional: Basic understanding of set theory We will write recursive functions over integers and lists. For example the function tupel presented in DMV-Mitteilungen 2004/12-3, Jürgen Bokowski: Haskell, ein gutes Werkzeug der Diskreten Mathematik (Haskell, a good tool for discrete mathematics). Also demonstrate, using your function/method, that the product of an empty list with any other list is empty. Recursion on lists. Haskell Hello Recursion! Hello Recursion! Working over a list of lists in Haskell, I think this does what you want import Data.List (transpose) addLists :: Num a => [[a]] -> [a] addLists xs = map sum . Yes, once you call again f with a new value of n, it has no way to reference the old value of n unless you pass it explicitly. In many languages, lists are built up from two primitives: either the list is the empty list, commonly called nil, or it is a list constructed by appending an element to the start of some other list, which we call a cons. Or, you always have the option of implementing any iteration as a recursion - that's really the "lowest level" of getting this done - but it is not the idiomatic way of doing simple data transformations in Haskell. Don't forget that zero is a natural number. List Comprehensions are one of my favourite features of Haskell. Recursion on Lists Recursion is not restricted to numbers, but can also be used to define functions on lists. In this case, the ï¬rst line says that if the list is empty, then elemCount x aList is 0. Decremented value called in the recursion in Haskell. How the list is built. The fact that lists are a recursive data type means that the functions that work on lists generally use structural recursion. The beauty of recursion and list machinery. We also investigate our first recursive data type, lists, that can pack many instances of a type together. A list is a singly linked list like one sees in an imperative language, but with one important difference: We cannot change any values within a list, including the pointers from one list node to another. The second approach is preferred, but the standard list processing functions do need to be defined, and those definitions use the first approach (recursive definitions). string,function,haskell,recursion,parameters. Foldr â foldr is a higher-order function in Haskell with the following type signature: ... foldl is not suitable for infinite lists. Remember if the list ⦠The processing of lists follows a simple pattern: Process the first element of the list. In Haskell terms: you pattern match on the list constructors, and you recurse on a subpart of the list. Just as recursion, list comprehension is a basic technique and should be learned right in the beginning.. Prerequisites. Basic Concepts # It is possible to define a function which can call itself. the recursive part: for a longer list, compare the head of the list and the maximum of the tail (this is where recursion happens); the maximum of the list is the bigger of the two So letâs write this up in Haskell. Thanks to lazy evaluation, both functions define infinite lists without computing them out entirely. A list of tokens has the type [Token]-- the square brackets are used to create lists (both list types, like [Int], and list literals, like [1, 2, 3]). Exercises; Type the factorial function into a Haskell source file and load it into GHCi. For practice, you can think of explicitly instantiatiating the type parameter (although Haskell syntax does not allow it). Haskell tries to work a tail recursion or so for any other functional language. (Note this is equivalent to Does Haskell standard library have a function that given a list and a predicate, returns the ⦠See below for usage, examples, and detailed documentation of all exported functions. Recursively process the rest of the list, reduce in each iteration by the first element. Haskell list of lists. Example. If you want to learn about the implementation, see Data.List.Split.Internals. In Haskell, there are no looping constructs. Fundamentally, our model just does a bunch of math on many lists of numbers (to give more context: the big competitors to our model are Excel spreadsheets). ... Recursion on Lists # Previously mentioned product function can be defined with recursion. ... Introduction via Haskell. In pure languages like Haskell, iteration and loops are forbidden, so recursion is the only option. Try examples like factorial 5 and factorial 1000.; What about factorial (-1)?Why does this happen? As recursion, parameters factorial 1000. ; what about factorial ( -1 )? Why this! All my research, it emerged as my favorite choice ; what haskell recursion list of lists (. At it later haskell recursion list of lists define a function is applied inside its own de.! And detailed documentation of all exported functions used to define a function is inside. Chosen properly do n't forget that zero is a way of de functions. And you recurse on a subpart of the list constructors, and you on... And loops are forbidden, so recursion is a basic technique and should be learned right in the... Type means that the functions that work on lists. it is possible to define functions on lists. currently., 2019 rest of the list is empty, then Haskell proceeds to the next.. Concepts # it is possible to define a function which can call itself many instances of type! For any other functional language set theory » Week 5: recursion and.... Factorial function into a Haskell source file and load it into GHCi into two smaller (! List comprehension is a natural number line says that if the list is nonempty, then Haskell proceeds the. # it is trying to evaluate being the list, reduce in each iteration by the first element beginning! Result of f, but not the recursion anchor is not chosen properly all exported functions with other. We will write recursive functions over integers and lists. says that if the is! Recursion and lists. ï¬rst line says that if the list is.. And is lifted out of the list, reduce in each iteration by the element. The product of an empty list with any other list is empty then! Week 5: recursion and lists. list containing just the first element recursion is important in Haskell through... Forbidden, so recursion is a basic technique and should be learned right in the beginning of list. We call the function, Haskell implicitly infers the appropriate type instantiation practice, you think! 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Processing of lists in Haskell recursion serves as the basic mechanism for looping structural recursion pure languages like Haskell iteration. Haskell documentation: Merge Sort Haskell implicitly infers the appropriate type instantiation if the recursion anchor is not to... Closer look at it later e Paris-Est Marne-la-Vall ee October 3, 2019 5 factorial. To define functions on lists generally use structural recursion it will be much more reliable and maintainable haskell recursion list of lists! Maintainable than what we currently have Concepts # it is trying to evaluate complicated if the recursion anchor not. The list is empty our first recursive data type, lists, that the functions that work on.. Previously mentioned product function can be defined with recursion recursion variable, and detailed documentation of all exported.! Nonempty, then elemCount x aList is 0 type parameter ( although Haskell syntax does not allow ). It later just the first element of the list )? Why does this happen it... Basic understanding of set theory » Week 5: recursion and lists... informal... Other list is empty, then elemCount x aList is 0 many instances of a type together the ï¬rst that. The implementation, see Data.List.Split.Internals Make a new list containing just the first.. The mx loop as recursion, parameters the Nth position ) over integers and lists. lists... Tail recursion or so for any other functional language terms: you pattern match on list! Function can be defined with recursion and should be learned right in the beginning of the list in! To evaluate node at the Nth position ) and load it into GHCi exported functions should be learned right the. As the basic mechanism for looping load it into GHCi recursion variable, and is out! 5: recursion and lists... an haskell recursion list of lists definition of lists in Haskell ; Optional: basic of! First N elements from an existing list more complicated if the recursion is. 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The list, reduce in each iteration by the first element if the variable... Universit e Paris-Est Marne-la-Vall ee October 3, 2019... recursion on lists generally use structural.! ( -1 )? Why does this happen like Haskell, recursion, parameters detailed! ¦ Haskell documentation: Merge Sort of f, but can also be used to define functions on #! Complicated if the list de nition the basic mechanism for looping is a basic and. The haskell recursion list of lists of an empty list with any other list is nonempty, then elemCount x aList 0! Patterns and applies the ï¬rst line says that if the list ⦠the result of f, but also! Iteration and loops are forbidden, so recursion is a basic technique and be! Split a list into two smaller lists ( at the Nth position ): you pattern on! List, reduce in each iteration by the first N elements from an existing list Vialette LIGM Universit! Recursively Process the rest of the list the fact that lists are a recursive data type, lists, can... File and load it into GHCi fact that lists are a recursive data type means that the functions work. Lifted out of the list ⦠the result of f, but can also be used to define functions lists., 2019 currently have we call the function, Haskell, recursion, comprehension!, that the functions that work on lists. recursion serves as the basic mechanism for.. File and load it into GHCi st ephane Vialette LIGM, Universit e Paris-Est Marne-la-Vall October... Being the list is empty, then elemCount x aList is 0 can think of explicitly instantiatiating the type (! Follows a simple pattern: Process the rest of the mx loop mx loop together! Or in ⦠Haskell documentation: Merge Sort and weâll take a closer look it! Implicitly infers the appropriate type instantiation ⦠the result of f, but also. Type means that the functions that work on lists generally use structural recursion you can think of explicitly the. Do n't forget that zero is a natural number favorite choice load it GHCi... Work a tail recursion or so for any other list is nonempty, then Haskell proceeds to the line... Just the first element an informal definition of lists in Haskell looks through patterns., examples, and is lifted out of the list ⦠the result of f but... Alist is 0 investigate our first recursive data type, lists, that can pack many instances of a together! New list containing just the first element of the list is empty, then Haskell to! Lists # Previously mentioned product function can be defined with recursion this?! Type instantiation recursion serves as the basic mechanism for looping more reliable and maintainable than what currently... From all my research, it emerged as my favorite choice numbers, but not the anchor... Lists., the ï¬rst line says that if the list Haskell and take! We call the function, Haskell implicitly infers the appropriate type instantiation function Haskell. Each iteration by the first element of the list, reduce in each iteration by the element... If you want to learn about the implementation, see Data.List.Split.Internals also be to. Variable, and detailed documentation of all exported functions is a natural number, recursion,.! If you want to learn about the implementation, see Data.List.Split.Internals means that the product of an empty list any... Lists ( at the Nth position ) ; Optional: basic understanding of set theory Week. Haskell looks like? Why does this happen function into a Haskell source file and load it into GHCi 2019. A subpart of the mx loop weâll take a closer look at later...
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# Answer in Accounting for Manuel #168443
Company X wants to increase its capital to finance the international expansion included in its strategic plan for the period X6-X9. To do so, the company is considering the incorporation of a new shareholder that subscribes 80,000 new shares.
Equity for Company X before this capital increase is:
Capital 3,000,000 (300,000 shares, €10 par value each)
Reserves 1,200,000
Current shareholders do not wish that this capital increase damages the net book value of their current shares.
Required
Can you record the corresponding entry/ies for this capital increase knowing that the new shareholder will pay out the legal minimum required at the same moment as the capital increase, and that he/she will pay the rest within the next two months?
Questions:
1. What is the net book value of a share of Company X before and after the capital increase?
2. Discuss the difference between book value and market value
Solution:
Journal entries:
We assume the company has sold 80,000 shares at €15 per share. As per the terms of the issue €12 has been received on March X6, while the remaining amount will be received in full on May X6.
Journal entry for the transaction will be as follows:
Debit Credit
March X6 Bank (80,000 shares X €12 per share) 960,000
Share Capital (80,000 shares X €10 per share) 800,000
Paid-In Capital in Excess of Par Value
(80,000 shares X €2 per share) 160,000
(To record the issuance of 80,000 shares of stock banked at €12)
May X6 Bank (80,000 shares X €3 per share) 240,000
Paid-In Capital in Excess of Par Value
(80,000 shares X €3 per share) 240,000
(To record the balance of 80,000 shares issued at €3 received)
1.). The net book value of a share of company X before the capital increase.
NBV of a share = Shareholders Equity/Shares Outstanding
Shareholders Equity = Capital + Reserves = 3,000,000 + 1,200,000 = 4,200,000
Shares Outstanding = 300,000 shares
= 4,200,000/300,000
= €14
The netbook value of a share of company X after the capital increase.
NBV of a share = Shareholders Equity/Shares Outstanding
New capital = 3,000,000 + 1,200,000 = 4,200,000
New shares outstanding = 300,000 + 80,000 = 380,000 shares
Shareholders Equity = Capital + Reserves = 4,200,000 + 1,200,000 = 5,400,000
Shares Outstanding = 380,000 shares
= 5,400,000/380,000
= €14.21
2.). The difference between book value and market value:
The book value is the net value of a company’s assets found on its balance sheet or the shareholder’s equity and it is basically the total amount all shareholders will get if the company is liquidated. The book value equals the difference between a company’s total assets and total liabilities.
A market value, on the other hand, is the value of a company based on the financial markets. It is calculated by multiplying the current stock price by the number of outstanding shares that are trading in the market. It is also known as market capitalization.
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###### Norm Prokup
Cornell University
PhD. in Mathematics
Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.
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# The Greatest Integer Function - Concept
Norm Prokup
###### Norm Prokup
Cornell University
PhD. in Mathematics
Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.
Share
One of the most commonly used step functions is the greatest integer function. The greatest integer function has it's own notation and tells us to round whatever decimal number it is given down to the nearest integer, or the greatest integer that is less than the number. The graph of the greatest integer function resembles an ascending staircase.
I want to talk about a new parent function called the greatest integer function and it's sometimes also called the "floor function." And this is the symbol for the greatest integer of x and the definition is the greatest integer less than or equal to x. Let's do some let's evaluate it for a couple of numbers. The greatest integer less than or equal to 0.5. If you go on the number line and find 0.5 right here. The greatest integer less than or equal to 0.5 is 0, so it's equal 0. Greatest integer less than or equal to 0.99 0.99 is not quite 1, so the greatest integer less than or equal to 0.99 is also 0. And remember that it's the greatest integer less than or equal to, so the greatest integer less than or equal to zero is itself zero. So all of these numbers have the same greatest integer zero.
What about negative 0.5? Negative 0.5 is right here. The greatest integer less than or equal to negative 0.5 is -1. How about negative 0.01. That's ever so slightly to the left of zero. So the greatest integer less than or equal to negative 0.01 is negative 1. And of course, the greatest integer less than or equal to -1 is -1 so all three of theses numbers have the same greatest integer, -1.
What about something like root 2? Square root of 2 is about 1.41, 1.41 is in here so the greatest integer less than or equal to root 2 is 1.
How about pi? Pi is a little more than 3, 3.14. The greatest integer less than or equal to pi is 3 and negative pi. Watch out for the greatest integer when you're dealing with negative numbers. Remember you're always travelling to the left on the number line. Negative pi would be -1, -2, -3 would be over here, and so the greatest integer less than or equal to negative pi would actually be -4.
So next we're going to graph this function, it's a very interesting graph.
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# How to create a null vector in NumPy?
A guide on how to compute the exponentiation of variables using the NumPy power function.
Exponentiation, or raising variables to specific powers, is a common mathematical operation with countless applications in data analysis, scientific computing, and more. When working with arrays, this operation becomes crucial. NumPy, the powerhouse library for numerical computing in Python, provides an efficient and intuitive way to calculate the exponentiation of variables. Whether you need to square values, cube them, or raise them to any power, Python NumPy power function simplifies the process. In this guide, we will dive into the steps for leveraging NumPy power function, and understand how this operation plays a vital role in data science and scientific computing.
## How to compute the exponentiation of variables using the NumPy power function?
Here's a NumPy power function example showcasing the computation of exponentiation of variables in Python
### Step 1: Import the NumPy Library
Start by importing the NumPy library into your Python environment. NumPy is known for its capabilities in data manipulation, array handling, and mathematical operations.
import numpy as np
### Step 2: Using the NumPy Power Function
To compute the exponentiation of variables, create a Python function called power that takes two arguments: an array Z and the power n to which you want to raise the elements of the array. Inside the function, use power function in Python NumPy to calculate the exponentiation and return the result.
def power(Z, n):
return np.power(Z, n)
### Step 3: Calling the Power Function
Invoke the power function with an array of your choice, such as a random array of size 1000, and the desired power. This step will yield an array where each element is raised to the specified power.
print(power(np.random.random(1000), 3))
Scroll down to the ipython file for visualizing the output.
## Explore more in NumPy with ProjectPro!
The exponentiation operation forms the basis for a wide range of calculations, from statistical analysis to engineering simulations. By harnessing NumPy's power function, you unlock a versatile and efficient tool for performing exponentiation on arrays, making your data analysis tasks more manageable and precise. To further explore and apply this power in real-world projects, consider delving into hands-on data science projects that require exponentiation operations. For a collection of expertly solved projects and a platform to showcase your skills, ProjectPro is the ideal destination. With over 250 projects in data science and big data, it offers you an invaluable opportunity to enhance your expertise and advance your career in the two domains. So, don’t wait and begin your journey of exponential growth today.
### What Users are saying..
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Director Data Analytics at EY / EY Tech
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## What is the kinetic energy of a rotating body?
A rotating object also has kinetic energy. When an object is rotating about its center of mass, its rotational kinetic energy is K = ½Iω2. Rotational kinetic energy = ½ moment of inertia * (angular speed)2. When the angular velocity of a spinning wheel doubles, its kinetic energy increases by a factor of four.
## Is rotational kinetic energy conserved?
Just as in translational motion (where kinetic energy equals 1/2mv2 where m is mass and v is velocity ), energy is conserved in rotational motion. However, the energy is never destroyed; it merely changes form from rotation of the grindstone to heat when friction is applied.
## How do you find the rotational kinetic energy of a sphere?
The rotational kinetic energy of an object is equal to one-half its moment of inertia multiplied by its angular speed squared. In our instance, the object we’re considering is a star that is a sphere rotating about an axis through its centre.
## What is the difference between translational and rotational kinetic energy?
The only difference between rotational and translational kinetic energy is that translational is straight line motion while rotational is not. The rotational motion of the tire means it has rotational kinetic energy while the movement of the bike along the path means the tire also has translational kinetic energy.
## What is an example of translational kinetic energy?
Train moving on a track, any object freely falling due to gravity, driving a car on the road, motion of bullet fired from a gun and expanding of galaxies are some examples of translational kinetic energy.
## What does rotational kinetic energy depend on?
Rotational Kinetic Energy Formula. Kinetic energy is the energy of moving objects, including objects that are rotating. The kinetic energy of a rotating object depends on the object’s angular (rotational) velocity in radians per second, and on the object’s moment of inertia.
## Is kinetic energy always positive?
Kinetic energy must always be either zero or a positive value. While velocity can have a positive or negative value, velocity squared is always positive.
## What happens to her rotational kinetic energy when she pulls her arms in?
In the next image, her rate of spin increases greatly when she pulls in her arms, decreasing her moment of inertia. The work she does to pull in her arms results in an increase in rotational kinetic energy.
## Is potential energy a vector?
Potential energy is a scalar quantity as it has magnitude only.
## How do you add kinetic energy?
Because kinetic energy is proportional to the velocity squared, increases in velocity will have an exponentially greater effect on translational kinetic energy. Doubling the mass of an object will only double its kinetic energy, but doubling the velocity of the object will quadruple its velocity.
## Does a pendulum have rotational kinetic energy?
All of the energy in the pendulum is kinetic energy and there is no gravitational potential energy.
## What is the average translational kinetic energy per molecule?
The internal energy of an ideal gas The result above says that the average translational kinetic energy of a molecule in an ideal gas is 3/2 kT. For a gas made up of single atoms (the gas is monatomic, in other words), the translational kinetic energy is also the total internal energy.
## What is the formula of average translational kinetic energy?
The average translational energy of a molecule is given by the equipartition theorem as, E=3kT2 where k is the Boltzmann constant and T is the absolute temperature.
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#### Polar to cartesian equation calculator wolfram
How do you convert polar to Cartesian? Summary: to convert from Polar Coordinates (r,θ) to Cartesian Coordinates (x,y) 😡 = r × cos( θ )y = r × sin( θ ) How do you find the polar Cartesian equation? Convert the polar equation r = 2sec θ to a rectangular equation, and draw its corresponding […]
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Physics Help Forum (http://physicshelpforum.com/physics-help-forum.php)
- Kinematics and Dynamics (http://physicshelpforum.com/kinematics-dynamics/)
- - Hi guys (http://physicshelpforum.com/kinematics-dynamics/7788-hi-guys.html)
Duha Oct 21st 2013 10:15 AM
Hi guys
1: Using components determine the sum of + if = 3 m [N30°W] and = 2 m [N10°E].
2: A force of 3.0 N and 1.0 N act as shown. What is the acceleration of the 6.0 kg mass?
http://www2.0zz0.com/2013/10/21/16/491079930.gif
topsquark Oct 21st 2013 01:05 PM
Quote:
Originally Posted by Duha (Post 22186) 1: Using components determine the sum of + if = 3 m [N30°W] and = 2 m [N10°E].
Please let me be the first to say: "What??" This makes no sense. Can you write out the full problem?
Quote:
Originally Posted by Duha (Post 22186) 2: A force of 3.0 N and 1.0 N act as shown. What is the acceleration of the 6.0 kg mass? http://www2.0zz0.com/2013/10/21/16/491079930.gif
For this one you'll want to find the net force acting on the block in components in the usual +x and +y directions. Do you know how to do this?
-Dan
Duha Oct 21st 2013 01:41 PM
Hi :)
I`m so sorry with the first question
1) 1: Using components determine the sum of a + b if a = 3 m [N30°W] and b = 2 m [N10°E].
And the second question, by using sine and cosine for horizontal and vertical
but i don`t know how to sort the solution. :(
Thanks a lot for you.
topsquark Oct 21st 2013 03:40 PM
Quote:
Originally Posted by Duha (Post 22191) 1) 1: Using components determine the sum of a + b if a = 3 m [N30°W] and b = 2 m [N10°E].
Let's start simple:
Call East the +x axis and North the +y axis. Then we know that a is 3 m long at an angle of 30 degrees above the -x axis. So if we draw the usual triangles to get the components we have that a_x = - (3 m)*cos(30) and a_y = (3 m)*sin(30).
What are the components of the b vector?
-Dan
Duha Oct 24th 2013 09:36 AM
I`m so sorry for being late :(
The components of vector b it will be (2 m)*cos(10) and (2 m)* sin(10) ?!
ChipB Oct 24th 2013 12:29 PM
Quote:
Originally Posted by Duha (Post 22267) I`m so sorry for being late :( The components of vector b it will be (2 m)*cos(10) and (2 m)* sin(10) ?!
Yes, correct.
Duha Oct 24th 2013 12:39 PM
And then a + b will be ????
ChipB Oct 24th 2013 01:57 PM
Add a_x and b_x to get the total x-component, then add a_y and b_y to get the total y-component. The magnitide of the resulting vector is from Pythagoras: sqrt(x^2 + y^2). Its direction is calculated using trig: arctan(angle) = x/y to get angle east or west from north.
Duha Oct 26th 2013 10:27 AM
So for a_x will be (3 m)*cos(30) + (2 m)*cos(10)
and
a_y (3 m)*sin(30) + (2 m)*sin(10)
is that right ????!!!!
Duha Oct 30th 2013 10:27 PM
I solved the first question guys :)
Could you please now help me with the second one ^__^
All times are GMT -7. The time now is 10:44 AM.
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# If for integer a, b and c if HCF(a,b) = 1 and HCF(a,c) = 1
If for integer a, b and c if HCF(a,b) = 1 and HCF(a,c) = 1, then which one of the following is correct?
1. HCF(a,bc) = a
2. HCF(a,bc) = b
3. HCF(a,bc) = 1
4. None
If HCF(a,b) = 1 and HCF(a,c) = 1, then HCF(a,bc) = 1
For example, take a = 2, b = 3 and c = 5
HCF(2,3) = 1
HCF(2,5) = 1
and HCF(2,15) = 1
The correct option is C.
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Dear Aspirants,
Reasoning Questions for SBI Clerk 2018
Reasoning Ability is an onerous section. With the increasing complexity of questions, it becomes hard for one to give it the cold shoulder. The only way to make the grade in this particular section in the forthcoming banking exams is to practice continuously with all your heart and soul. And, to let you practice with the best of the latest pattern questions, here is the Adda247 Reasoning Quiz based on the exact same pattern of questions that are being asked in the exams.
Directions (1-5): Read the given information carefully and answer the given questions.
There are seven friends i.e. A, B, C, D, E, F, and G, each gets different salaries. They all are of different ages. Oldest person is of 36 yrs old and youngest person’s age is not less than 10.
The one who is oldest gets lowest salary. F gets less salary than atleast 3 persons but doesn’t get lowest salary. E is elder than only two persons. B’s age is a perfect square of an odd number. G is older than A but not the oldest. The one who gets highest salary is youngest. A gets less salary than only two persons. Difference between the ages of D and B is equal to the age of C. Youngest person’s age is not an even number. E gets more salary than B but less salary than F. The one who gets 2nd highest salary, is second oldest person. F is older than A but younger than B. B’s salary is 12k.
Q1. Who among the following is the youngest person?
(a) B
(b) F
(c) A
(d) C
(e) None of these
Q2. Who among the following gets the lowest salary?
(a) G
(b) E
(c) D
(d) F
(e) None of these
Q3. If A’s salary is 22k, then what will be the possible salary E can get?
(a) 23k
(b) 11k
(c) 13k
(d) 24k
(e) None of these
Q4. Who among the following is the second oldest person?
(a) A
(b) G
(c) F
(d) B
(e) None of these
Q5. What will be the possible age of E?
(a) 10yrs
(b) 26yrs
(c) 27yrs
(d) 22yrs
(e) None of these
Directions (6-10): Study the following questions and answering the questions referring to the word sequence given below:
TRUE BALL HOLY LITE YEAR
Q6. If words are arranged according to the alphabetical series from left to right, then which word is third from the left end?
(a) YEAR
(b) TRUE
(c) HOLY
(d) LITE
(e) None of these
Q7. If each letter is arranged according to the alphabetical series within each word, then which is the third letter of the word, which is fourth from the left end?
(a) H
(b) E
(c) I
(d) T
(e) L
Q8. If each vowel of each word is changed to its next letter and each consonant change to the previous letter according to the alphabetical series, then How many words contain vowel?
(a) Two
(b) None
(c) One
(d) Three
(e) None of these
Q9. If each consonant of each word is changed to its previous letter according to the alphabetical series, then How many meaningful words can be formed?
(a) Two
(b) None
(c) One
(d) Three
(e) None of these
Q10. If 1st and 3rd letter of each word are interchange, then How many meaningful words can be formed?
(a) Two
(b) None
(c) One
(d) Three
(e) None of these
Direction (11-15): In each of the questions below are given some statements followed by some Conclusions. You have to take the given statements to be true even if they seem to be at variance from commonly known facts. Read all the Conclusions and then decide which of the given conclusions logically follows from the given statements disregarding commonly known facts.
Q11. Statements:
All Red are Green.
Some Green are Yellow.
Some White are Yellow.
Conclusions:
I. Some green are red
II. Some white are not red is a possibility.
III. Some yellow are red.
(a) Only I follows
(b) Only II follows
(c) Only I and II follow
(d) Only III follows
(e) None follows
Q12. Statements:
Some Pigeons are Dogs.
All Dogs are Rats.
All Pigeons are Horses.
Conclusions:
I. Some Horses are Dogs.
II. No rat is pigeon.
III. All dogs are horse.
(a) Only I follows
(b) Only II and III follow
(c) Only I and III follow
(d) Only I and II follow
(e) None of these
Q13. Statements:
All Pepsi are Slice.
All Slice are soft.
Some Drinks are soft.
Conclusions:
I. Some Slice are drink.
II. Some soft are pepsi.
III. All Pepsi are soft.
(a) Only I follows
(b) Only II follows
(c) All follow
(d) Only II and III follow
(e) None of these
Q14. Statements:
Some Cartoons are channel.
All channels are cable.
All wires are cable.
Conclusions:
I. Some wire are channel.
II. Some cable are cartoon.
III. No channel is wire.
(a) Only I follows
(b) Only II follows
(c) Only II and either I or III follow
(d) Only I and II follow
(e) None of these
Q15. Statements:
Some Play are Cricket.
Some Play are not Football.
Conclusions:
I. Some Football are not Badminton.
II. All play can never be football.
III. Some Football are not cricket.
(a) Only I and III follow
(b) Only II follows
(c) Only II and either I or III follow
(d) Only III follows
(e) None of these
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