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http://www.chegg.com/homework-help/questions-and-answers/maximum-speed-1140-kg-car-round-turn-radius80-m-flat-road-coefficientof-static-friction-ti-q438007
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What is the maximum speed with which a 1140 kg car can round a turn of radius80 m on a flat road if the coefficientof static friction between tires and road is 0.80?
m/s
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# Tagged Questions
28 views
### Is mean pairwise distance a metric over subsets of a metric space.
Specifically, I am looking at finite subsets of a set that is a discrete metric space under Jaccard Distance. I'm having trouble proving the triangle inequality or coming up with a counterexample. ...
62 views
### Embedding finite (discrete) metric spaces to Eulidean space as isometrically as possible
Let $X = \{1, 2, 3, ..., k\}$ with the discrete metric (distance is 1 for every pair of points). How can this be embedded into $\mathbb{R}^n$ (with the usual metric) such that the embedding would be ...
82 views
### How to calculate number of lumps of a 1D discrete point distribution?
I would like to calculate the number of lumps of a given set of points. Defining "number of lumps" as "the number of groups with points at distance 1" Supose we have a discrete 1D space in this ...
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# Isosceles
A triangle in which two sides have the same length.
"I so love these (isosceles) socks" can be used to remind you that two sides or two lengths of socks are equal.
Examples
An isosceles triangle with two sides same length giving two angles at 50° and one at 80°.
An isosceles triangle with two sides same length giving two angles at 65° and one at 50°.
An isosceles triangle with two sides same length giving two angles at 59° and one at 62°.
A right angled triangle can be an isosceles as long as two of the sides are the same length.
A triangle with two angles the same will be an isosceles triangle.
NOTE:
Isosceles triangles have two internal angles that are the same and one which is different.
Remember to spell isosceles with an 'SOS' in the word.
Even when stranded on an island, she wore the socks and sent out an SOS for another pair.
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https://www.swirlzcupcakes.com/samples/what-does-the-nn-stand-for-in-youtube/
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What does the NN stand for in YouTube?
What does the NN stand for in YouTube?
Nnnn. Nnn. Nn. N nn n n. – YouTube Nnnnnn?nnnnnnnn. Nnnn. Nnn. Nn. N nn n n. If playback doesn’t begin shortly, try restarting your device. Videos you watch may be added to the TV’s watch history and influence TV recommendations. To avoid this, cancel and sign in to YouTube on your computer.
What does NNN mean in commercial real estate?
Wonder no more. In this post, we’ll demystify what the heck NNN means next to all those Office, Warehouse and Retail Space listings you’re seeing online. NNN stands for Triple Net rent. In this type of commercial real estate rent, you pay the amount listed and you also have pay additional costs (usually Operating Expenses) on top of that.
How much does NNN cost per square foot?
Using our example above, say the commercial real estate agent tells you the NNN charges/expenses are \$6.00 per square foot, per year. Then to find out your total monthly rent, you will have to add the \$24.00 base rent and add \$6.00, then divide by 12 months, then multiply by the total square feet you will be leasing.
What does NNN stand for in medical terms?
NNN is a derivative of nicotine that is produced in the Curing of tobacco, in the burning of tobacco (such as with cigarettes), and in the acidic conditions of the stomach.
Wonder no more. In this post, we’ll demystify what the heck NNN means next to all those Office, Warehouse and Retail Space listings you’re seeing online. NNN stands for Triple Net rent. In this type of commercial real estate rent, you pay the amount listed and you also have pay additional costs (usually Operating Expenses) on top of that.
Using our example above, say the commercial real estate agent tells you the NNN charges/expenses are \$6.00 per square foot, per year. Then to find out your total monthly rent, you will have to add the \$24.00 base rent and add \$6.00, then divide by 12 months, then multiply by the total square feet you will be leasing.
What kind of expenses are included in NNN?
The NNN expenses usually include your proportionate share of the exterior maintenance of the building, property insurance and property taxes. They are usually quoted on a per square foot basis. Are There Any Other Expenses to Consider? The short answer is, yes.
What does it mean to pay nnn for office space?
For example: say the Office Space listing you’re interested in says the rent is \$24.00 NNN per sqft/year. The \$24.00 number is your Base Rent. That’s the minimum amount that you pay in rent just to lease the space. In addition to Base Rent, you will pay have to pay additional money to cover the Operating Expenses (or NNN or Triple Net Expenses).
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https://en.wikipedia.org/wiki/Zn%C3%A1m's_problem
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# Znám's problem
Jump to navigation Jump to search
Graphical demonstration that 1 = 1/2 + 1/3 + 1/11 + 1/23 + 1/31 + 1/(2×3×11×23×31). Each row of k squares of side length 1/k has total area 1/k, and all the squares together exactly cover a larger square with area 1. The bottom row of 47058 squares with side length 1/47058 is too small to see in the figure and is not shown.
In number theory, Znám's problem asks which sets of k integers have the property that each integer in the set is a proper divisor of the product of the other integers in the set, plus 1. Znám's problem is named after the Slovak mathematician Štefan Znám, who suggested it in 1972, although other mathematicians had considered similar problems around the same time. One closely related problem drops the assumption of properness of the divisor, and will be called the improper Znám problem hereafter.
One solution to the improper Znám problem is easily provided for any k: the first k terms of Sylvester's sequence have the required property. Sun (1983) showed that there is at least one solution to the (proper) Znám problem for each k ≥ 5. Sun's solution is based on a recurrence similar to that for Sylvester's sequence, but with a different set of initial values.
The Znám problem is closely related to Egyptian fractions. It is known that there are only finitely many solutions for any fixed k. It is unknown whether there are any solutions to Znám's problem using only odd numbers, and there remain several other open questions.
## The problem
Znám's problem asks which sets of integers have the property that each integer in the set is a proper divisor of the product of the other integers in the set, plus 1. That is, given k, what sets of integers
${\displaystyle \{n_{1},\ldots ,n_{k}\}}$
are there, such that, for each i, ni divides but is not equal to
${\displaystyle {\Bigl (}\prod _{j\neq i}^{n}n_{j}{\Bigr )}+1\quad ?}$
A closely related problem concerns sets of integers in which each integer in the set is a divisor, but not necessarily a proper divisor, of one plus the product of the other integers in the set. This problem does not seem to have been named in the literature, and will be referred to as the improper Znám problem. Any solution to Znám's problem is also a solution to the improper Znám problem, but not necessarily vice versa.
## History
Znám's problem is named after the Slovak mathematician Štefan Znám, who suggested it in 1972. Barbeau (1971) had posed the improper Znám problem for k = 3, and Mordell (1973), independently of Znám, found all solutions to the improper problem for k ≤ 5. Skula (1975) showed that Znám's problem is unsolvable for k < 5, and credited J. Janák with finding the solution {2, 3, 11, 23, 31} for k = 5.
## Examples
One solution to k = 5 is {2, 3, 7, 47, 395}. A few calculations will show that
3 × 7 × 47 × 395 + 1 = 389866, which is divisible by but unequal to 2, 2 × 7 × 47 × 395 + 1 = 259911, which is divisible by but unequal to 3, 2 × 3 × 47 × 395 + 1 = 111391, which is divisible by but unequal to 7, 2 × 3 × 7 × 395 + 1 = 16591, which is divisible by but unequal to 47, and 2 × 3 × 7 × 47 + 1 = 1975, which is divisible by but unequal to 395.
An interesting "near miss" for k = 4 is the set {2, 3, 7, 43}, formed by taking the first four terms of Sylvester's sequence. It has the property that each integer in the set divides the product of the other integers in the set, plus 1, but the last member of this set is equal to the product of the first three members plus one, rather than being a proper divisor. Thus, it is a solution to the improper Znám problem, but not a solution to Znám's problem as it is usually defined.
## Connection to Egyptian fractions
Any solution to the improper Znám problem is equivalent (via division by the product of the xi's) to a solution to the equation
${\displaystyle \sum {\frac {1}{x_{i}}}+\prod {\frac {1}{x_{i}}}=y,}$
where y as well as each xi must be an integer, and conversely any such solution corresponds to a solution to the improper Znám problem. However, all known solutions have y = 1, so they satisfy the equation
${\displaystyle \sum {\frac {1}{x_{i}}}+\prod {\frac {1}{x_{i}}}=1.}$
That is, they lead to an Egyptian fraction representation of the number one as a sum of unit fractions. Several of the cited papers on Znám's problem study also the solutions to this equation. Brenton & Hill (1988) describe an application of the equation in topology, to the classification of singularities on surfaces, and Domaratzki et al. (2005) describe an application to the theory of nondeterministic finite automata.
## Number of solutions
As Janák & Skula (1978) showed, the number of solutions for any k is finite, so it makes sense to count the total number of solutions for each k.
Brenton and Vasiliu calculated that the number of solutions for small values of k, starting with k = 5, forms the sequence
2, 5, 18, 96 (sequence A075441 in the OEIS).
Presently, a few solutions are known for k = 9 and k = 10, but it is unclear how many solutions remain undiscovered for those values of k. However, there are infinitely many solutions if k is not fixed: Cao & Jing (1998) showed that there are at least 39 solutions for each k ≥ 12, improving earlier results proving the existence of fewer solutions (Cao, Liu & Zhang 1987, Sun & Cao 1988). Sun & Cao (1988) conjecture that the number of solutions for each value of k grows monotonically with k.
It is unknown whether there are any solutions to Znám's problem using only odd numbers. With one exception, all known solutions start with 2. If all numbers in a solution to Znám's problem or the improper Znám problem are prime, their product is a primary pseudoperfect number (Butske, Jaje & Mayernik 2000); it is unknown whether infinitely many solutions of this type exist.
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Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > cantnfclOLD Unicode version
Theorem cantnfclOLD 8137
Description: Basic properties of the order isomorphism used later. The support of an is a finite subset of , so it is well-ordered by and the order isomorphism has domain a finite ordinal. (Contributed by Mario Carneiro, 25-May-2015.) Obsolete version of cantnfcl 8107 as of 28-Jun-2019. (New usage is discouraged.) (Proof modification is discouraged.)
Hypotheses
Ref Expression
cantnfsOLD.1
cantnfsOLD.2
cantnfsOLD.3
cantnfvalOLD.3
cantnfvalOLD.4
Assertion
Ref Expression
cantnfclOLD
Proof of Theorem cantnfclOLD
StepHypRef Expression
1 cnvimass 5362 . . . . 5
2 cantnfvalOLD.4 . . . . . . . 8
3 cantnfsOLD.1 . . . . . . . . 9
4 cantnfsOLD.2 . . . . . . . . 9
5 cantnfsOLD.3 . . . . . . . . 9
63, 4, 5cantnfsOLD 8136 . . . . . . . 8
72, 6mpbid 210 . . . . . . 7
87simpld 459 . . . . . 6
9 fdm 5740 . . . . . 6
108, 9syl 16 . . . . 5
111, 10syl5sseq 3551 . . . 4
12 onss 6626 . . . . 5
135, 12syl 16 . . . 4
1411, 13sstrd 3513 . . 3
15 epweon 6619 . . 3
16 wess 4871 . . 3
1714, 15, 16mpisyl 18 . 2
18 cnvexg 6746 . . . . 5
19 imaexg 6737 . . . . 5
20 cantnfvalOLD.3 . . . . . 6
2120oion 7982 . . . . 5
222, 18, 19, 214syl 21 . . . 4
237simprd 463 . . . . 5
2420oien 7984 . . . . . 6
2523, 17, 24syl2anc 661 . . . . 5
26 enfii 7757 . . . . 5
2723, 25, 26syl2anc 661 . . . 4
2822, 27elind 3687 . . 3
29 onfin2 7729 . . 3
3028, 29syl6eleqr 2556 . 2
3117, 30jca 532 1
Colors of variables: wff setvar class Syntax hints: ->wi 4 /\wa 369 =wceq 1395 e.wcel 1818 cvv 3109 \cdif 3472 i^icin 3474 C_wss 3475 class class class wbr 4452 cep 4794 Wewwe 4842 con0 4883 'ccnv 5003 domcdm 5004 "cima 5007 -->wf 5589 (class class class)co 6296 com 6700 c1o 7142 cen 7533 cfn 7536 OrdIso`coi 7955 ccnf 8099 This theorem is referenced by: cantnfval2OLD 8139 cantnfleOLD 8141 cantnfltOLD 8142 cantnflt2OLD 8143 cantnfp1lem2OLD 8145 cantnfp1lem3OLD 8146 cantnflem1bOLD 8149 cantnflem1dOLD 8151 cantnflem1OLD 8152 cnfcomlemOLD 8172 cnfcomOLD 8173 cnfcom2lemOLD 8174 cnfcom3lemOLD 8176 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1618 ax-4 1631 ax-5 1704 ax-6 1747 ax-7 1790 ax-8 1820 ax-9 1822 ax-10 1837 ax-11 1842 ax-12 1854 ax-13 1999 ax-ext 2435 ax-rep 4563 ax-sep 4573 ax-nul 4581 ax-pow 4630 ax-pr 4691 ax-un 6592 This theorem depends on definitions: df-bi 185 df-or 370 df-an 371 df-3or 974 df-3an 975 df-tru 1398 df-fal 1401 df-ex 1613 df-nf 1617 df-sb 1740 df-eu 2286 df-mo 2287 df-clab 2443 df-cleq 2449 df-clel 2452 df-nfc 2607 df-ne 2654 df-ral 2812 df-rex 2813 df-reu 2814 df-rmo 2815 df-rab 2816 df-v 3111 df-sbc 3328 df-csb 3435 df-dif 3478 df-un 3480 df-in 3482 df-ss 3489 df-pss 3491 df-nul 3785 df-if 3942 df-pw 4014 df-sn 4030 df-pr 4032 df-tp 4034 df-op 4036 df-uni 4250 df-iun 4332 df-br 4453 df-opab 4511 df-mpt 4512 df-tr 4546 df-eprel 4796 df-id 4800 df-po 4805 df-so 4806 df-fr 4843 df-se 4844 df-we 4845 df-ord 4886 df-on 4887 df-lim 4888 df-suc 4889 df-xp 5010 df-rel 5011 df-cnv 5012 df-co 5013 df-dm 5014 df-rn 5015 df-res 5016 df-ima 5017 df-iota 5556 df-fun 5595 df-fn 5596 df-f 5597 df-f1 5598 df-fo 5599 df-f1o 5600 df-fv 5601 df-isom 5602 df-riota 6257 df-ov 6299 df-oprab 6300 df-mpt2 6301 df-om 6701 df-1st 6800 df-2nd 6801 df-supp 6919 df-recs 7061 df-rdg 7095 df-seqom 7132 df-1o 7149 df-er 7330 df-map 7441 df-en 7537 df-dom 7538 df-sdom 7539 df-fin 7540 df-fsupp 7850 df-oi 7956 df-cnf 8100
Copyright terms: Public domain W3C validator
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# A liter is equivalent to
## What litre is equal to?
one cubic decimetrelitre (l), also spelled liter, unit of volume in the metric system, equal to one cubic decimetre (0.001 cubic metre).
## What is a liter the same as?
To be exact, 1 liter is 0.264 gallon (a little more than a quart), and 4 liters is 1.06 gallons.
## How many 8 oz cups are in a liter?
A cup is 8 fluid ounces and a liter is 33.8 fluid ounces. What is this? Cups and liters both measure the volume of liquids, so whether you need to know how many cups are in a liter or water, oil or a bottle of soda, there will always be about 4.3 cups in a liter!
## How many oz are in a liter?
One liter equals 33.814 US fluid ounces.
## How many liters are in gallon?
3.785 litersThere are 3.785 liters in a gallon and a liter is 0.264 of a gallon.
## How many ml is in a L?
How many ml in a liter? 1 litre is equal to 1,000 milliliters, which is the conversion factor from liters to milliliters.
## Does 4 cups equal 1 liter?
Yes, there are 4 cups in a liter. A cup is equivalent to 250 mL, and there are 1,000 mL in a liter. Therefore, the number of cups in a liter is 1000 divided by 250, or 4 cups.
## Does 8 cups equal 2 liters?
Health experts commonly recommend eight 8-ounce glasses, which equals about 2 liters, or half a gallon a day.
## How many glasses of water is a liter?
four glassesA glass of water has an approximate volume of 8 ounces, while 1 litre has a capacity of 32 ounces. No one can answer this question with true accuracy since a glass does not have a standard size. ∴ four glasses of water are equal to 1 litre. Thus, one liter is equal to 4 glasses of water.
## What is an example of liter?
The definition of a liter is a unit for measuring volume in the metric system. An example of a liter is a bottle containing 33.76 ounces or 1.0567 quarters of soda. The basic unit of liquid volume or capacity in the metric system, equal to 1.06 quart or 2.12 pints.
## How many ml is in 24 oz?
If you’re using US customary fluid ounces, your 24 fl oz is equal to 709,76 ml when rounded off. If you’re putting this on a food label, you have to say that your 24 fl oz is the equivalent of 720 ml. If you’re using UK or imperial ounces, then 24 oz equals 681,91 ml rounded off.
## How many ounces are in 2 liters of liquid?
Note that 2 liters are equal to 67.628 fluid ounces. Notice that we are using fluid ounces instead of ounces because an ounce is a unit of weight whereas liters is a unit of volume.
## Does 4 cups equal 1 liter?
Yes, there are 4 cups in a liter. A cup is equivalent to 250 mL, and there are 1,000 mL in a liter. Therefore, the number of cups in a liter is 1000 divided by 250, or 4 cups.
## What is an example of 1 liter?
The definition of a liter is a unit for measuring volume in the metric system. An example of a liter is a bottle containing 33.76 ounces or 1.0567 quarters of soda. The basic unit of liquid volume or capacity in the metric system, equal to 1.06 quart or 2.12 pints.
## Is a quart the same size as a liter?
In the US, a liquid quart is equal to approximately 0.946353 liters and a dry quart is equal to approximately 1.101221 liters.
## What is the symbol for liter?
A liter is a unit of volume in the Metric System. The symbol for liter is L and the International spelling for this unit is litre.
## How many ounces are in a gallon?
Note: There is a difference between US Customary Units and the Imperial System for volume conversions. The US gallon contains 128 US fluid ounces, whereas the Imperial gallon contains 160 Imperial fluid ounces.
## What is a liter?
Liter. Definition: A liter (symbol: L) is a unit of volume that is accepted for use with the International System of Units (SI) but is technically not an SI unit. One liter is equal to 1 cubic decimeter (dm 3 ), 1,000 cubic centimeters (cm 3 ), or 1/1,000 cubic meters (m 3 ). History/origin: There was a point from 1901 to 1964 when …
## What is a liter of water?
History/origin: There was a point from 1901 to 1964 when a liter was defined as the volume of one kilogram of pure water under the conditions of maximum density at atmospheric pressure. However, due to the mass-volume relationship of water being based on a number of factors that can be cumbersome to control (temperature, pressure, purity, isotopic uniformity), as well as the discovery that the prototype of the kilogram was slightly too large (making the liter equal to 1.000028 dm 3 rather than 1 dm 3 ), the definition of the liter was reverted to its previous, and current definition.
## How many liters are in a gallon?
The US gallon is defined as 231 cubic inches (3.785 liters). In contrast, the imperial gallon, which is used in the United Kingdom, Canada, and some Caribbean nations, is defined as 4.54609 liters. In both systems, the gallon is divided into four quarts. Quarts are then divided into two pints and pints are divided into two cups.
## What is the purpose of liters?
Current use: The liter is used to measure many liquid volumes as well as to label containers containing said liquids. It is also used to measure certain non-liquid volumes such as the size of car trunks, backpacks and climbing packs, computer cases, microwaves, refrigerators, and recycling bins, as well as for expressing fuel volumes and prices in most countries around the world.
## What is a gallon used for?
Gallons are also widely used in fuel economy expression in the US, as well as some of its territories.
## What is the difference between liters and kilograms?
Liters and kilograms are approximately equivalent when the substance measured has a density of close to 1 kilogram per liter. A common substance with that density is water. Substances that float on water (like oil) are of lower density and have a greater volume for the same mass.
## How many pints are in a liter?
A liter is a metric unit of capacity, formerly defined as the volume of 1 kilogram of water under standard conditions, now equal to 1,000 cubic centimeters (about 1.75 pints).
## How many cups are in a quart?
Depends on how accurate you want to be. There are four cups in a quart. A quart is 32 ounces. A liter is 33.8 ounces in a liter. So four cups will leave you 1.8 ounces short of a liter.
## How many grams are in an ounce?
The international avoirdupois ounce (abbreviated oz) is defined as exactly 28.349523125 g under the international yard and pound agreement of 1959, signed by the United States and countries of the Commonwealth of Nations.
## How many glasses per liter?
This depends on the glass, but most glasses are about 250 ml so there are 4 glasses per liter.
## How to calculate the number of cups in 1.5L?
For US cups divide the previous quotient by 0.7789 and that quotient will be the number of US Cups in 1.5L
## What is the numerator of a fraction?
Definitions: The numerator of a fraction is the number on the top the denominator is the number on the bottom. Let’s get very basic; to add two fractions that have a common denominator (same number on the bottom) you add the numerators (the numbers on the top). Simplify 2/2 by dividing the numerator by the denominator.
## What is a liter?
Liter. Definition: A liter (symbol: L) is a unit of volume that is accepted for use with the International System of Units (SI) but is technically not an SI unit. One liter is equal to 1 cubic decimeter (dm 3 ), 1,000 cubic centimeters (cm 3 ), or 1/1,000 cubic meters (m 3 ). History/origin: There was a point from 1901 to 1964 when …
## What is a liter of water?
History/origin: There was a point from 1901 to 1964 when a liter was defined as the volume of one kilogram of pure water under the conditions of maximum density at atmospheric pressure. However, due to the mass-volume relationship of water being based on a number of factors that can be cumbersome to control (temperature, pressure, purity, isotopic uniformity), as well as the discovery that the prototype of the kilogram was slightly too large (making the liter equal to 1.000028 dm 3 rather than 1 dm 3 ), the definition of the liter was reverted to its previous, and current definition.
## What is the purpose of liters?
Current use: The liter is used to measure many liquid volumes as well as to label containers containing said liquids. It is also used to measure certain non-liquid volumes such as the size of car trunks, backpacks and climbing packs, computer cases, microwaves, refrigerators, and recycling bins, as well as for expressing fuel volumes and prices in most countries around the world.
## How many gallons are in a cup?
The metric cup is defined as 250 milliliters. One United States customary cup is equal to 236.5882365 milliliters as well as 1/16 U.S. customary gallons, 8 U.S. customary fluid ounces, 16 U.S. customary tablespoons, or 48 U.S. customary teaspoons.
## How many cups are in a liter?
We can say that one liters is approximately four point two two seven cups: 1 L ≅ 4.227 cup. An alternative is also that one cup is approximately zero point two three seven times one liters.
## How to convert liters to cups?
To convert 1 liters into cups we have to multiply 1 by the conversion factor in order to get the volume amount from liters to cups. We can also form a simple proportion to calculate the result: 1 L → 4.2267528198649 cup. 1 L → V (cup)
## How many liters is 1 cup?
We can also convert by utilizing the inverse value of the conversion factor. In this case 1 cup is equal to 0.2365882375 × 1 liters.
## What is the unit of volume of a cup?
Cup. The cup is an English unit of volume, most commonly associated with cooking and serving sizes. It is traditionally equal to half a liquid pint in either US customary units or the British imperial system but is now separately defined in terms of the metric system at values between 1⁄5 and 1⁄4 of a liter.
## What is a liter?
Liter. Definition: A liter (symbol: L) is a unit of volume that is accepted for use with the International System of Units (SI) but is technically not an SI unit. One liter is equal to 1 cubic decimeter (dm 3 ), 1,000 cubic centimeters (cm 3 ), or 1/1,000 cubic meters (m 3 ). History/origin: There was a point from 1901 to 1964 when …
## What is a liter of water?
History/origin: There was a point from 1901 to 1964 when a liter was defined as the volume of one kilogram of pure water under the conditions of maximum density at atmospheric pressure. However, due to the mass-volume relationship of water being based on a number of factors that can be cumbersome to control (temperature, pressure, purity, isotopic uniformity), as well as the discovery that the prototype of the kilogram was slightly too large (making the liter equal to 1.000028 dm 3 rather than 1 dm 3 ), the definition of the liter was reverted to its previous, and current definition.
## How many liters are in a quart?
Multiple definitions of the quart exist. In the US, a liquid quart is equal to approximately 0.946353 liters and a dry quart is equal to approximately 1.101221 liters. In the UK, the imperial quart is equal to 1.136523. In both the UK and the US, the quart is equal to ¼ of its respective gallon.
## What is the purpose of liters?
Current use: The liter is used to measure many liquid volumes as well as to label containers containing said liquids. It is also used to measure certain non-liquid volumes such as the size of car trunks, backpacks and climbing packs, computer cases, microwaves, refrigerators, and recycling bins, as well as for expressing fuel volumes and prices in most countries around the world.
## Where is the quart used?
Current use: The respective versions of the quart are used mainly in the United States and the United Kingdom, though in the UK, the use of the liter is now mandated as a result of metrication.
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# Correct line-breaking of long math expression
What is the correct way of breaking such unnumbered equation into multiple lines?
\begin{equation*}
\lim_{\bigtriangleup t \to 0^+}\int_{\bigtriangleup t}^{T} \! \int_{\Omega} \! D(t_1,x) \frac{\varphi(t_1-\bigtriangleup t,x)-\varphi(t_1,x)}{(-\bigtriangleup t)} \, \mathrm{d}x \, \mathrm{d}t_1
=\lim_{\bigtriangleup t \to 0^+} \int_{0}^{T} \! \int_{\Omega} \! D(t_1,x) \frac{\varphi(t_1-\bigtriangleup t,x)-\varphi(t_1,x)}{(-\bigtriangleup t)} \chi_{(\bigtriangleup t,T)}(t_1) \, \mathrm{d}x \, \mathrm{d}t_1
= \int_{0}^{T} \! \int_{\Omega} \! D(t_1,x) \frac{\partial \varphi}{\partial t_1} (t_1,x) \, \mathrm{d}x \, \mathrm{d}t_1 .
\end{equation*}
It's not an equation actually, I'm proving something in my thesis and I'm just modifying a math expression. So \multline is not this case I think. I used \split but then it can be aligned to the right side or the left side only because the expressions are too long and it would overfull.
-
Welcome to TeX.sx! Have a look at mathmode with a lot of examples: ctan.org/pkg/voss-mathmode – Marco Daniel Apr 21 '13 at 13:02
@MarcoDaniel Thank you, this could be really helpful. – vessel Apr 21 '13 at 13:18
This is a tough formula to typeset; there is no "right way", such complex formulas often need specific treatment which depend on their size and also on subtle semantic issues.
Here is my proposal: the initial term is set on a line by itself and the two developments below it with aligned equals, indented to the right.
I changed \bigtriangleup into the customary \Delta; I also added a definition for the differential symbol that spares you from explicit \, commands. Finally I changed the definition of \phi to produce \varphi, so you can change them all to the "closed" variant by simply commenting the redefinition.
Why aligned inside equation*? Because this is a single block, so if you decide to number it, you can simply remove the *. With align* the result would be identical, but numbering it would require more steps.
\documentclass{article}
\usepackage{amsmath}
\newcommand{\diff}{\mathop{}\!\mathrm{d}}
\renewcommand{\phi}{\varphi}
\begin{document}
\begin{equation*}
\begin{aligned}
\lim_{\Delta t \to 0^+}
& \int_{\Delta t}^{T} \int_{\Omega}
D(t_1,x)
\frac{\phi(t_1-\Delta t,x)-\phi(t_1,x)}{(-\Delta t)}
\diff x \diff t_1
\\
&=\lim_{\Delta t \to 0^+}
\int_{0}^{T} \! \int_{\Omega}
D(t_1,x)
\frac{\phi(t_1-\Delta t,x)-\phi(t_1,x)}{(-\Delta t)}
\chi_{(\Delta t,T)}(t_1)
\diff x \diff t_1
\\
&=\int_{0}^{T} \! \int_{\Omega}
D(t_1,x)
\frac{\partial\phi}{\partial t_1}(t_1,x)
\diff x \diff t_1 .
\end{aligned}
\end{equation*}
\end{document}
-
Oh, that's an exhaustive answer, thank you. I will use it. – vessel Apr 21 '13 at 13:22
+1 for paying such nice close attention to the semantics. If I could upvote twice I'd do it again for the time you spent. – Ethan Bolker Apr 21 '13 at 13:39
For example:
\documentclass{article}
\begin{document}
\begin{eqnarray*}
\lefteqn{
\lim_{\bigtriangleup t \to 0^+}\int_{\bigtriangleup t}^{T} \! \int_{\Omega} \! D(t_1,x) \frac{\varphi(t_1-\bigtriangleup t,x)-\varphi(t_1,x)}{(-\bigtriangleup t)} \, \mathrm{d}x \, \mathrm{d}t_1
}\\
&=&\lim_{\bigtriangleup t \to 0^+} \int_{0}^{T} \! \int_{\Omega} \! D(t_1,x) \frac{\varphi(t_1-\bigtriangleup t,x)-\varphi(t_1,x)}{(-\bigtriangleup t)} \chi_{(\bigtriangleup t,T)}(t_1) \, \mathrm{d}x \, \mathrm{d}t_1
\\
&= &\int_{0}^{T} \! \int_{\Omega} \! D(t_1,x) \frac{\partial \varphi}{\partial t_1} (t_1,x) \, \mathrm{d}x \, \mathrm{d}t_1 .
\end{eqnarray*}
But you'd rather write \Delta instead of \bigtriangleup.
-
eqnarray! Run for your lives! – Jubobs Apr 21 '13 at 13:10
@Jubobs I do know eqnarray's limitations. But I also know, how to adjust this environment. – Przemysław Scherwentke Apr 21 '13 at 13:17
@PrzemysławScherwentke Thank you, this is looking well. Is there a rule of how to line-break equations and math expressions? For example would it be considered wrong if I separated each line by and then it would be all centred? – vessel Apr 21 '13 at 13:17
@PrzemysławScherwentke Really, never ever use eqnarray. – egreg Apr 21 '13 at 13:20
@vessel Sugested eqnarray is the simplest one, making 3 columns. If you need more sophisticated breaking, please consider using of the amsmath package. – Przemysław Scherwentke Apr 21 '13 at 13:21
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# Coldness of Space
1. Nov 21, 2008
### Mentallic
From what I've been told, the temperature in space is approx. 3oK if not in direct contact with infrared rays from the sun/stars etc. This temperature is also decreasing as the universe expands. Space is also a vacuum, with very few particles per cubic unit.
My question is: If space is a vacuum and a point in space is completely blocked by objects so that no infrared radiation is in direct or indirect contact (reflections etc.) and there are no particles in the vicinity, why is it that the temperature is not absolute zero? Where does this tiny temperature come from? Also, if a vacuum were created here on Earth, would it also be the same temperature as in space?
2. Nov 21, 2008
### Jonathan Scott
Whatever is blocking off the sun/stars or surrounding the vacuum will be radiating some heat itself unless it too is near absolute zero. The effect may be reduced if its emissivity is designed to be much lower than that of a black body.
3. Nov 21, 2008
### Mentallic
Ahh so what I think you're implying is that all bodies of mass (even black holes?) radiate their own heat at the atomic level after being bombarded by heat from the other end of that body?
So the temperature in this point in space that I specified will actually be lower than 3oK which is the average of the universe?
What if, lets say we did attempt to bring that spherical body of mass that is isolating this point close to absolute zero also, by surrounding that mass with yet another large sphere. Would this mean that point in space will approach even closer to absolute zero, but never actually reach (apparently absolute zero can never be reached)?
4. Nov 21, 2008
### Jonathan Scott
Black holes have their own special rules relating to temperature and thermodynamics, so let's forget about them here.
Yes, all massive bodies radiate heat. The maximum rate is primarily determined (through Stefan's Law for "black bodies") by their temperature but the rate may be less than that depending on the emissivity of the surface.
I'm sorry, but I don't remember enough from my student days about how thermal equilibrium of radiation works to know whether passive shielding of a volume could actually make it colder than the shield in the long term.
If the inside surface of the shield had low emissivity, it would initially radiate less than a black body into the internal volume, but that energy would presumably last longer with reflection inside the shield before being reabsorbed, because lower emissivity means higher reflectivity.
I suspect that the long-term stable temperature inside a passive shield is likely to be not very different from the temperature of the shield itself. However, a highly reflective passive shield should prevent something already cool from warming up for a long time.
5. Nov 21, 2008
### D H
Staff Emeritus
This is somewhat of a misnomer. Space is not a pure vacuum. The space between stars and even galaxies has some stuff in it, called the interstellar and intergalactic medium. The interstellar and intergalactic medium intergalactic medium have all of the characteristics of an ionized gas, albeit a very, very tenuous one. Like any other gas, these media have a temperature, and this temperature can be extremely high, up to 108 Kelvins or more! So in one sense space is anything but "cold".
People say that space is "cold" because a macroscopic object such as a thermometer in deep space will never go into thermal equilibrium with the surrounding medium. Radiative heat transfer will complete dominate conductive heat transfer because the interstellar/intergalactic medium is so very tenuous.
It comes from the big bang, or more precisely, from the radiation released when the universe became transparent about 300,000 years after the big bang. This cosmic microwave background radiation pervades all of space. A macroscopic object well-shielded from or very far from any star will eventually come into thermal equilibrium with this background radiation: 2.725 Kelvin.
6. Nov 21, 2008
### Mentallic
And this microwave background radiation is everywhere in the universe? I don't see how this is possible. If the point where the big bang began released with it very high frequency radiation, wouldn't this radiation now be on the edges of the universe? Say for e.g. the point of the big bang is rather our sun. When both begin expanding (the sun explodes), all the radiation will move outwards and eventually our entire solar system will be starved of infrared, light, gamma rays etc. I highly doubt the point where the sun used to be, or even our Earth will still be bombarded by the sun's rays.
7. Nov 21, 2008
### D H
Staff Emeritus
That is correct.
You have a common misperception of the big bang: That it was an explosion in space. A much better way to look at the big bang is that it was an explosion of space.
Here are some introductory-level articles on the big bang and the cosmic microwave background radiation:
http://www.totse.com/en/technology/space_astronomy_nasa/300000.html [Broken]
http://www.astro.ubc.ca/people/scott/cmb_intro.html
http://nedwww.ipac.caltech.edu/level5/Guth/Guth_contents.html
Last edited by a moderator: May 3, 2017
8. Nov 21, 2008
### Integral
Staff Emeritus
You really cannot speak of the temperature of open space, you need something there to have a temperature. So lets assume a arbitrary object located in empty space. Initial temp = $T_0$ and emissivity 1.
Since the object is isolated in deep space the only path of energy exchange is radiation.
From Stephan Boltzmann law the Energy exchange between the surroundings and the body is :
$$E= \alpha (T_s^4 - T_0^4)$$
Where $T_s$ is the temperature of the suroundings. In deep space $T_s$ is 3.7K. The rate of heat loss will decrease as the temperature of the object decreases. So the equilbrium temperature approached over time will be 3.7K. That is when the energy exchanged is zero.
A simple fact is that there is no place in know space where you can see a uniform 3.7K in all directions. Any hot object, such as a star or planet, visible from the object will be exchanging energy at a rate determined by its temperature.
For a body in low earth orbit the side of the body facing the sun to be recieveing energy at one rate, the side faceing the earth will recieve energy at a different rate, while the side faceing deep space loses energy. This is one reason that virtually every satilte is rotating. The rotation ensures that the satilite maintains a uniform surface temperatue.
9. Nov 22, 2008
### D H
Staff Emeritus
There is something there, even in the space between galaxies. That something might be very tenuous, but it is not nothing at all. Astronomers regularly discuss the attributes of the intergalactic medium -- including its temperature. What you can't do is think that this interplanetary/interstellar/intergalactic medium temperature will have any bearing on the temperature of macroscopic objects imbedded in that medium.
10. Nov 22, 2008
### Vanir
Excuse me, just a tourist passing through.
Do you mean in the sense, of the duality of wavefunctions to be defined as a medium, in this case albeit quite tenuous? I've no idea for correct terminology...zero point energy field? Quantum field? Electromagnetic field? Gravitational field?
11. Nov 22, 2008
### D H
Staff Emeritus
No reason to get so exotic. The interplanetary/stellar/galactic medium is just plain old baryonic matter, mostly ionized hydrogen and helium. The intergalactic medium has a density of about one atom per cubic meter. This ionized gas has a random component to its kinetic energy, aka temperature.
12. Nov 22, 2008
### jambaugh
Actually you can. A given volume of space is a quantum system with excitations (particles/fields). You can directly talk about its entropy, energy and temperature.
If the volume is assumed to be truly empty you are asserting entropy=0 and energy =0 thus temperature = 0. (This is however impossible to actualize.)
13. Nov 22, 2008
### Integral
Staff Emeritus
DH has told us that the temperature is well defined because we may have 1 particle per cubic meter. Now you tell me that there is no such thing as empty space??? You guys need to get together and work this out. Seems that I could easily find a volume of deep space with NO particles. What is the temperature of that volume? O??? does it change in a step function when a stray particle drifts through? As I said you need something to have a temperature.
I think I will stick with my initial idea, slightly reworded to satisfy the nit pickers.
Temperature of deep space is a complicated issue and needs to be addressed carefully. You really need an "ensemble" of particles to define the temperature.
14. Nov 22, 2008
### marcus
Hello Integral, DH, Mentallic, and everybody.
In case it might help for me to butt in, simply to emphasize what DH already said, here goes:
DH is right. Deep intergalactic space is absolutely swarming with CMB photons. I forget how many per cubic meter but it is a huge number. That defines the temperature quite precisely. Mentallic please listen to what DH says. He is giving you the absolute straight dope.
An otherwise empty vacuum created on earth would be full of photons radiated off the walls of the box containing the vacuum. The temperature of a vacuum is the temperature of thermal radiation in the vacuum. It could be whatever, depending on the temp of the walls. If you use refrigeration to make the walls of the box 5 kelvin, then the temp inside will be 5 kelvin. If you make the walls 1 kelvin, the temp of the photons inside will be 1 kelvin.
It just happens that the temp of otherwise empty space (far enough out not to be affected by stars and stuff) is 2.728 kelvin. That's the temp of the swarm of CMB photons out there (which have a welldefined thermal energy bar-chart.)
So if you put a piece of metal out in deep space it will radiate off photons and absorb CMB photons until it settles into equilibrium 2.7 with the CMB photons and with the rest of space.
If you put a metal box around the piece of metal, to try to shield it from the CMB then the box itself will settle into equilibrium 2.7 and then it will be filled with its own 2.7 kelvin radiation. and then the original piece of metal will settle to the same 2.7 temp.
You can't shield against the 2.7 CMB temp unless you use some kind of refrigeration to cool the box.
Last edited: Nov 22, 2008
15. Nov 22, 2008
### marcus
Mentallic, here is something you might like to learn to do, since the
You can calculate for yourself how many CMB photons are in a cubic meter of the space out between the stars.
I just calculated it, using Google calculator, and it came to about 400 million photons per cubic meter.
Here is what I typed into the google window
0.24*(k*2.728 kelvin/(hbar*c))^3
then I pressed return and it said 411 million per cubic meter.
The main thing is if you want to know how many thermal photons per cubic meter are in any space, like your room, or an oven, or a skating rink---wherever, it's very simple.
You just take the temperature in kelvin, say T (in this case T was 2.728 kelvin), multiply it by the k constant (Google knows this and will do it if you say to) and divide that by hbar and c.
Then you cube whatever you got, and you are done! or nearly so.
All that's left is to multiply by a number that is close to 1/4 but which is actually more like 0.2436. If you forget to do that, it's OK because the answer will still be about right. (At least to get a rough order of magnitude estimate.)
So the main thing is simply take the temp, multiply it by the k constant, divide by Planck's hbar constant and by c, the speed of light.
Google knows all these universal constants like Boltzmann k, hbar, c. So it will take care of it.
So then you have kT/(hbar*c) and all you have to do is cube it.
Can you do this? Can you find the number of photons per cubic meter in your room?
Your room is probably around 293 kelvin. If you want, type this into Google and press return
(k*293 kelvin/(hbar*c))^3
this will say how many photons are with you right now, in the room (as a number per cubic meter)
and if you like, to make it more accurate divide by 4, or multiply by that auxilliary number I mentioned that is about 0.24 and doesnt affect the gross size very much.
The temperature in a space is revealed by the thermal light in that space no matter where it is. I'm telling you this because it's a very neat fact and today is Max Planck's birthday. Every day is.
Last edited: Nov 22, 2008
16. Nov 22, 2008
### Mentallic
Ahh no one has ever expressed it to me in such a way before. So concise, yet so understandable.
This same problem is what was bothering me when I first created this thread. But then Jonathan Scott said it is rather the radiation being emitted that heats up that point in space, not the particles passing through it.
But what if the universe were to be given enough time to expand enough such that the density of CMB photons per cubic metre is low enough so that there can be distinct points in space that are completely free of these photons. Will this assume 0 temperature?
Thanks for clarifying this dilemma for me
The photon density is about 1.2 million times more! At roughly 1.4 quadrillion / m3.
Just out of curiosity, I went scrounging and found the critical temperature required for hydrogen bombs to begin the fusion process is approx 40M K. This corresponds with 3.5x1030 photons per cubic metre? This sounds incredibly high, and makes me wonder if photons even have their own volume.
17. Nov 22, 2008
### marcus
Mentallic, I'm really glad to see that you took hold with that formula and calculated some interesting photon densities with various temperature. Then, looking back over my post I noticed I had left out a factor of 2.701 or approximately 3, that I should have divided by.
So I misled you, sorry. All our answers are too big by about a factor of 3. (But they still have the right order of magnitude so they give the right idea.)
I went back and corrected. the CMB is only 400 million photons per cubic meter.
18. Nov 22, 2008
### D H
Staff Emeritus
If the universe continues its expansion indefinitely (note well: the jury is still out on the ultimate fate of the universe), the universe will eventually cool toward absolute zero. Google "ultimate fate of the universe", "heat death", "big freeze".
It's rarely a good idea to talk about points. That is where demons such as singularities lie. What you should be talking about instead are volumes. Even now there are at any given instant of time volumes of space void of CMB photons. So that is not a particularly good way to look at it, either.
The concept of temperature starts to lose meaning when one looking at too small an ensemble of matter or too short a duration of time. When talking about temperature you need to look at a sufficiently large ensemble of matter and a sufficiently long duration of time.
As the universe ages and expands, the energy and flux density of the CMBR photons will decrease. Per some fixed time interval, a macroscopic object in deep space will absorb an ever decreasing number of photons of an ever decreasing frequency. The object in turn will emit an ever decreasing number of photons of an ever decreasing frequency as it cools toward absolute zero.
19. Nov 22, 2008
### Integral
Staff Emeritus
Thank you. I think that is what I have been trying to say.
20. Nov 22, 2008
### marcus
Mentallic,
(I concur with what DH and Integral just said* and have some side remarks)
Now that you know how many thermal photons there are in a cubic meter at some temperature, you might what to know the average energy of each photon.
If T is the temp (like T = 2.728 kelvin) then the average photon carries this energy:
2.701 kT which google probably likes to see written 2.701*k*2.728 kelvin
and if you want to know the wavelength of that average energy photon,
it is h*c/(2.701*k*2.728 kelvin)
that is h*c divided by the energy freight it carries. (the more energy, the shorter the wavelength).
So I'm inviting you to find, should you care to, the wavelength of an average-energy CMB photon
=========================
This is just for fun. You can also calculate the radiant energy in a cubic meter, by multiplying the number of photons by the average energy of each one. So you can find the radiant energy per cubic meter in your room, or in an oven etc. The energy density of the CMB, small as it is, is analogous to these more familiar thermal energy densities---just a lower temperature.
*including CMB photons along with other matter. I guess the point being that since you have 400 million CMB photons per cubic meter there is more of a statistical ensemble, to give a meaningful idea of temperature, than there is with other species of matter. Other species, like hydrogen atoms, can be rather sparse.
Last edited: Nov 22, 2008
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# Microsoft PowerPoint - Simulink 3
Document Sample
``` Introduction to Simulink
Design of Mechanical Systems
650:486
• Simulink is an extension to Matlab that
allows engineers to rapidly and accurately
build computer models of dynamical
systems using block diagram notation.
• Block diagram notation is a graphical
means to represent dynamical systems.
Block Diagram Vs Flowchart
• A flowchart describes a • A block diagram
sequence of operations, describes a set of
so only one block in the relationships that holds
flow chart is active at a simultaneously, all blocks
time. in a block diagram may
be active at once. So
block diagram can be
thought of being
represented by a set of
simultaneous equations.
• At the command prompt for matlab type
simulink. This will open up the window for
create a new model or
click the “new” button open an existing one
Continuous System
• Most physical systems are modeled as
continuous system since they can be
described by using differential equations.
• Simple models are linear and time
invariant.
Four fundamental blocks
• The four primitive blocks used to represent
continuous linear systems are
• Gain block
• Sum block
• Derivative block
• Integrator block.
Continuous system
Using source and sink blocks
Gain Block
• The simplest block
diagram element is the
gain block. The output of
the gain block is the input
multiplied by a constant.
• y(t)= kx(t) is represented
by the following block
diagram
Sum Block
• The sum block permit
us to add two or more
inputs.
• The expression
• c=a – b is represented
by the following
block diagram
Integrator Block
• The integrator block
computes the time
integral of its input
from the starting time
to the present.
Derivative block
• The derivative block
computes the time
rate of change of its
input.
y = dx/dt
Example 1
• Solving for a second order constant
coefficient linear differential equation
d2y/dt2 +c1dy/dt + c0y = b0f(t)
For a response to a ‘step’ command
Get an equivalent block diagram for the system
use mouse to drag blocks into
the model window and to
connect blocks with arrows
use integrators to get dy/dt and y
d2y/dt2 +c1dy/dt + c0y =
b0f(t)
Introducing the stepping function
Now, double click the blocks to open and set the block’s parameters
set gain value
set initial condition
set variable name
set output format to “array”
To set the simulation parameters….
select Simulation -> Simulation Parameters
set Start and Stop time (in seconds)
set numerical integration type
Time to run the simulation
click the “run” button to begin the simulation
when the simulation is complete, “Ready” appears at the bottom
Example
Simulink will automatically save a variable named “tout” to the
workspace.
This variable contains the time values used in the simulation, important
for variable time integration types
Simulink also will create the output variable(s) you specified
>>plot(tout,yoft)
Example #2
• The block diagram
denotes a cart of mass
m, on a frictionless
surface denoted by the
equation of motion :
d2x/dt2 = F/m
Cart Continued
• The block diagram of
the cart position
computation.
Simulation
• Simulating the cart: Using
a sine function as force
input, mass as 100 kg.
Problem #1
• Consider a spring-mass-
dashpot system
represented by the
equation of motion:
m(d2x/dt2) + c(dx/dt) + kx =
F
where m=100, c=10, k =5.
Simulate the position of the
mass.
Solution #1
Problem #2
• Represent the
differential equation
given by
dx/dt = bx – px2
where b =1 and p = 0.5
Hint: use product block
Solution #2
```
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TOPICS
# Complete Lattice
A partially ordered set (or ordered set or poset for short) is called a complete lattice if every subset of has a least upper bound (supremum, ) and a greatest lower bound (infimum, ) in .
Taking shows that every complete lattice has a greatest element (maximum, ) and a least element (minimum, ).
Of course, every complete lattice is a lattice. Moreover, every lattice with a finite set is a complete lattice.
Lattice, Partially Ordered Set, Tarski's Fixed Point Theorem
This entry contributed by Roland Uhl
## Explore with Wolfram|Alpha
More things to try:
## References
Birkhoff, G. Lattice Theory, 3rd ed. Providence, RI: Amer. Math. Soc., 1967.Grätzer, G. General Lattice Theory, 2nd ed. Boston, MA: Birkhäuser, 1998.
Complete Lattice
## Cite this as:
Uhl, Roland. "Complete Lattice." From MathWorld--A Wolfram Web Resource, created by Eric W. Weisstein. https://mathworld.wolfram.com/CompleteLattice.html
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# Questions on Algebra: Introduction to vectors, addition and scaling answered by real tutors!
Algebra -> -> Questions on Algebra: Introduction to vectors, addition and scaling answered by real tutors! Log On
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Question 582498: Let the plane V be defined by ax + by + cz + d = 0. Which is true? A) the vector ⟨a,b,c⟩ is perpendicular to V B) the distance between V and the origin is Click here to see answer by Edwin McCravy(9717)
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Question 616179: Please help me solve this question: An airplane flying at an altitude of 3600 feet is dropping aid supplies. The path of the plane is parallel to the ground at the time of the supplies are released and the plane is traveling at a speed of 330 mph. Note: 5280 feet is 1 mile. a) Write parametric equations that represent the path of the supplies b) How long will it take for the supplies to reach the ground? c) How far will the supplies travel horizontally before landing? Click here to see answer by josmiceli(10616)
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Question 633728: Can you help me with this one? Solve the ordered triple of point H. A is (5,0,0) E is (0,0,4) G (0,3,0) Q is (0,3,4) R is (5,3,0) and p is (5,0,4) Click here to see answer by Edwin McCravy(9717)
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Question 665255: Can someone please help me with this i really don't know how to do vectors, i have tried part a but i dont know if its right. Consider the two vectors a = î + ĵ and b = 2î – ĵ . (a) Calculate a + b and a – b . a + b = 2 î^2 + ĵ^2 a – b = 2 (b) On a Cartesian plane, illustrate the vector addition a + b and the vector subtraction a – b. Clearly label your diagrams with all its appropriate vectors. (c) Calculate the magnitude of a and the magnitude of b. (d) Find the scalar product a•b. (e) Find the orthogonal projection of vector a onto vector b. That is, find the vector projb a . Click here to see answer by lynnlo(4174)
Question 665255: Can someone please help me with this i really don't know how to do vectors, i have tried part a but i dont know if its right. Consider the two vectors a = î + ĵ and b = 2î – ĵ . (a) Calculate a + b and a – b . a + b = 2 î^2 + ĵ^2 a – b = 2 (b) On a Cartesian plane, illustrate the vector addition a + b and the vector subtraction a – b. Clearly label your diagrams with all its appropriate vectors. (c) Calculate the magnitude of a and the magnitude of b. (d) Find the scalar product a•b. (e) Find the orthogonal projection of vector a onto vector b. That is, find the vector projb a . Click here to see answer by Alan3354(34677)
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Question 717890: There is this homework problem I have for Pre-Algebra. It is on scale factors. This is the problem: What scale factor was used to build a 55 ft wide billboard from a 25 in. wide model? I tried to write a proportion (model to actual), and I ended up getting 5 in. to 11 ft. I remember my teacher saying that when doing a scale factor, and it is inches to feet, then it has to be a certain number of inches to 1 ft. He said the exception is that if the scale factor ends up like 1/6 in. to 1 ft because there is no 1/6 inch mark on a ruler. Now, I am confused as to how this problem should be done, and I would be very grateful if someone were to clearly explain how to do this problem. By the way, I am sorry if the category for this question is incorrect. Click here to see answer by stanbon(60771)
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Question 735088: Am I allowed to simple ask for assistance into understanding the concept of vectors in Mathematics? Does this site work like that? I kind of missed the whole stage of learning about vectors and I'm unable to do numerous questions due to lack of knowledge and understanding. What are vectors and what is a vector space? How would one, in a simple generalised form of instruction, go about trying to prove whether something is or isnt a vector space? So lost. Click here to see answer by rothauserc(643)
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Question 723360: Find a unit vector in the same direction as <-12,-5> Click here to see answer by tommyt3rd(3721)
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User reladenine vakalwe - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T08:18:01Z http://mathoverflow.net/feeds/user/23907 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/115359/computing-rational-cohomology-of-smooth-not-necessarily-compact-toric-varieties Computing rational cohomology of smooth (not necessarily compact) toric varieties Reladenine Vakalwe 2012-12-04T05:47:53Z 2013-01-15T08:22:00Z <p>The title pretty much says it all. I am looking for references (lecture notes, books, readable articles, suggestions), preferably example laden, that explain how to compute the rational cohomology of a smooth toric variety. I am particularly interested in methods for doing this when the variety is not compact. Implicit here is the assumption that this can be done in a practical way (please correct me if I am wrong, I do not know much about the subject).</p> <p><b>Some background:</b> I have a bunch of varieties whose cohomology I would really like to compute. I have reasonably explicit descriptions of these varieties as subsets of <code>$\mathbb{C}^n$</code>. These descriptions are along the lines of (but more involved) the variety in this question:</p> <p><a href="http://mathoverflow.net/questions/115141/a-cohomology-computation-request" rel="nofollow">http://mathoverflow.net/questions/115141/a-cohomology-computation-request</a></p> <p>I realized earlier today that my varieties are toric, and I am hoping this observation will be the answer to my prayers.</p> <p>If it helps, I can compute the equivariant cohomology of of my varieties (mainly because the Hodge structure on it is pure), but I am presuming this doesn't really completely determine the ordinary cohomology (apart from Hodge-Euler polynomials etc.).</p> http://mathoverflow.net/questions/118176/around-the-socle-filtration-of-a-verma-module Around the socle filtration of a Verma module Reladenine Vakalwe 2013-01-06T03:13:03Z 2013-01-06T23:39:05Z <p>Work in the context of a finite dimensional simple Lie algebra over $\mathbb{C}$. Write $W$ for the Weyl group and $\leq$ for the Bruhat order. For $w\in W$ let $\Delta_w$ denote the Verma module of highest weight $w^{-1}w_0\cdot 0$, where $w_0$ is the longest element in $W$ and $\cdot$ denotes the so-called dot action. So $\Delta_e$ is simple and there is a unique (up to scaling) inclusion $\Delta_v \hookrightarrow \Delta_w$ whenever $v\leq w$. In particular, there is an inclusion $\Delta_v \hookrightarrow \Delta_{w_0}$ for all $v\in W$.</p> <p>For a module $M$, let $0\subset soc^1M \subset soc^2 M \subset \cdots$ denote the socle filtration. Set $soc_iM = soc^i M/soc^{i-1}M$. For example, in the case of $\mathfrak{sl}_2$: <code>$W=\{e,s\}$</code> and $soc_1\Delta_s = \Delta_e = L_e$, $soc_2 \Delta_s = L_s$, where $L_w$ denotes the unique simple quotient of $\Delta_w$.</p> <p><strong>The basic question: </strong> Is it true that</p> <p>a) $soc_1(\Delta_{w_0}/\Delta_w) \subseteq soc_k(\Delta_{w_0})$,</p> <p>where $k$ is the smallest integer such that $soc^k(\Delta_{w_0})\not\subseteq \Delta_w$.</p> <p><strong>Why this has hope of being true: </strong> 1) It is true in type $A_1$ and $A_2$, I have not checked type $G_2$ yet out of sheer laziness (and hope that an expert will just point me to some place in the literature or tell me that I am overthinking matters). 2) It is true for $w=e$. 3) Miracles sometimes occur in Schubert varieties.</p> <p><strong>Why this has no hope of being true: </strong> 1) It has a sort of ridiculous feel to it (apologies for the cavalier attitude, but perhaps it will be justified by what follows): roughly the statement is saying that simples in Vermas can't move down in the layers of the socle filtration upon quotienting out by Verma submodules. This feels a bit nutty to me. 2) The examples of type $A_1$ and $A_2$ are both multiplicity one situations (ala occurences of simples in Vermas) and aren't really indicative of the general situation.</p> <p><strong>Some reformulations/related tidbits</strong> (that I am aware of but don't see how to leverage into a counterexample or proof):</p> <p>1') The radical filtration on a Verma coincides (up to shift) with the socle filtration.</p> <p>1) The statement is equivalent to the socle filtration on $\Delta_{w_0}/\Delta_w$ coinciding (up to shift) with the weight filtration (ala mixed sheaves or graded category $\mathcal{O}$), since the weight filtration on $\Delta_{w_0}$ coincides with the socle filtration (see "Proof of Jantzen conjectures" by Beilinson-Bernstein or "Koszul duality patterns in representation theory" by Beilinson-Ginzburg-Soergel). Note: the radical filtration on $\Delta_{w_0}/\Delta_w$ does coincide with the weight filtration.</p> <p>2) The statement implies the assertion obtained by replacing $w_0$ with any $x$ such that $w\leq x$, since in this situation $\Delta_w\hookrightarrow \Delta_{w_0}$ factors as $\Delta_w \hookrightarrow \Delta_x \hookrightarrow \Delta_{w_0}$.</p> <p>3) The question is motivated by trying to understand an analogous question for the anti-dominant projective in category $\mathcal{O}$ (principal block). Namely, let $P_e$ be the (canonical) indcomposable projective cover of $\Delta_e$ (note: $P_e$ is an amazing object, it is self-dual, injective, tilting). Are the following statements true:</p> <p>b) $soc_1(P_e/\Delta_{w_0}) \subseteq soc_k(P_e)$,</p> <p>where $k$ is the smallest integer such that $soc^k(P_e)\not\subseteq \Delta_{w_0}$.</p> <p>c) Same question as b) but replace $w_0$ by arbitrary $w\in W$. This is of course related to a).</p> <p><b>Added later: </b>c) is undoubtedly false, as indicated by Dag's counterexample in type $A_1\times A_1$ below.</p> <p><b>Added later: </b>c) is also false for type $A_2$.</p> <p>Here is <strong>why one might care</strong>: from the short exact sequence </p> <p>$0\to \Delta_{w_0} \to P_e \to P_e/\Delta_{w_0} \to 0$</p> <p>one deduces $Ext^1(\Delta_e, \Delta_{w_0}) = Hom(\Delta_e, P_e/\Delta_{w_0})$. Consequently, the purity (ala mixed sheaves/graded category $\mathcal{O}$) of this $Ext^1$ is (unless I am being screwy) equivalent to b). </p> <p>Unless I am completely misunderstanding things, V. Mazorchuk proves this latter purity statement (in slightly different language) in Theorem 32 of <a href="http://arxiv.org/abs/math/0607589" rel="nofollow">http://arxiv.org/abs/math/0607589</a>. </p> <p>In fact, Theorem 32 states that $Ext^1(\Delta_v, \Delta_{w_0})$ is pure for arbitrary $v$. Now for $v=e$ this translates to b) above. This is starting to smell like a proof/answer to my questions. However, the problem is that Mazorchuk's proof (which I don't understand very well) seems to be using statements along these lines.</p> <p>Related also is the fact that granted the purity of $Ext^1(\Delta_v,\Delta_{w_0})$ a downwards induction gives purity of $Ext^1(\Delta_v, \Delta_w)$. This in turn implies that the dimension of these $Ext^1$'s is given by the coefficient of $q$ (modulo sign) in the corresponding Kazhdan-Lusztig $R$-polynomial (these statements start getting me really worried, since they are certainly not true for all $Ext^i$ thanks to Boe's "Counterexample to the Gabber-Joseph conjecture"). </p> <p>Needless to say I am playing fast and loose with a number of things. So the assertions above should be treated with a healthy dose of suspicion (I would be grateful though to people pointing out the errors of my ways).</p> <p>This of course ties in with a number of toy questions that have been bugging me:</p> <p><a href="http://mathoverflow.net/questions/116348/morphisms-between-verma-modules" rel="nofollow">http://mathoverflow.net/questions/116348/morphisms-between-verma-modules</a></p> <p><a href="http://mathoverflow.net/questions/115141/a-cohomology-computation-request" rel="nofollow">http://mathoverflow.net/questions/115141/a-cohomology-computation-request</a></p> <p>Having typed all that, I really hope I didn't make a silly mistake right in the beginning!</p> http://mathoverflow.net/questions/117728/extensions-of-ic-sheaves/117742#117742 Answer by Reladenine Vakalwe for extensions of IC sheaves Reladenine Vakalwe 2012-12-31T17:12:21Z 2012-12-31T17:27:37Z <p>There is general convolution algebra type formalism that you can try and use, see Chriss and Ginzburg's "Complex geometry and representation theory". Chapter 8 in particular is very much in line with the your "source of examples".</p> <p>In general this can be hard. Heck, consider even the case that your local system is trivial and your IC-sheaf is the (shifted) constant sheaf on $X$. Then you are asking to compute the cohomology of the space. This may be a non-trivial endeavor depending on your space.</p> <p>An ideal example where things work out very nicely is that of flag varieties and the IC-sheaves are those corresponding to Schubert subvarieties. Then these Ext-computations can be carried out combinatorially in the Hecke algebra. Soergel's papers on this and related topics are particularly enlightening. A related point here is that in this situation considering hypercohomology as a functor to graded modules for the cohomology algebra of the flag variety is full and faithful. This result also generalizes to projective varieties with <code>$\mathbb{C}^*$</code>-actions. This is a result of Ginzburg "Perverse sheaves and $\mathbb{C}^*$-actions". Similar ideas are also worked out in some of Springer's papers on spherical varieties. Related are also the moment graph techniques that can be found in papers of Braden, MacPherson, etc.</p> <p>Regarding, "replace cohomology with equivariant cohomology". I am assuming you want to compute $Ext$ in the equivariant derived category. Then similar techniques as above can be tried. In the presence of suitable assumptions, often the equivariant calculation reduces to the non-equivariant one due to formality. Instead of trying to flesh this out let me just refer to Soergel's "Langlands philosophy and Koszul duality". A number of examples are worked out in there. </p> <p>In general though, there is no magic pill that I know of.</p> http://mathoverflow.net/questions/116348/morphisms-between-verma-modules Morphisms between Verma modules Reladenine Vakalwe 2012-12-14T04:59:57Z 2012-12-30T23:22:00Z <p>Let $\mathcal{O}_0$ be the principal block of the BGG category $\mathcal{O}$ for a finite dimensional simple Lie algebra over $\mathbb{C}$. For an element $w$ in the Weyl group $W$, let $\Delta_w$ denote the Verma module with highest weight $w_0w^{-1}\cdot 0$, where $w_0\in W$ is the longest element, and $\cdot$' denotes the dot-action.</p> <p>It is well known that</p> <p>$\mathrm{dim}(\mathrm{Hom}(\Delta_v, \Delta_w)) \leq 1$</p> <p>This fact is pretty straightforward to prove algebraically. However, I do not know how to see this topologically. Namely, I do not know how to prove this via the interpretation of Verma modules as perverse sheaves on the flag variety.</p> <p>I would be grateful if someone could explain how to see this fact topologically.</p> <p><strong>Added later: </strong>In response to Jim Humphreys comment let me add some motivation:</p> <ol> <li><p>In this regard I think of category $\mathcal{O}$ as a "toy example". I would like to know what sort of generality this fact holds for. For instance, is the corresponding statement true for perverse sheaves smooth along a stratification given by affine spaces? The latter is certainly a highest weight category, computations in it can be undertaken topologically, etc. So as a starting point I would like to understand the topological reason for its truth for the "toy example".</p></li> <li><p>In the same vein as 1) I would like to know whether this truly is a "geometric" fact, i.e., does it hold if I consider my sheaves with coefficients in a commutative ring say?</p></li> <li><p>Computing extension groups of Verma modules is an old problem. If there is any hope for doing this topologically, I would think a reasonable place to start would be to compute $\mathrm{Ext}^0$ topologically!</p></li> <li><p>In the same vein as 3). One can see that the extensions of Verma modules is given by (compactly supported) cohomology (appropriately shifted) of intersections of Schubert cells with opposite Schubert cells. This is related to my earlier questions:</p></li> </ol> <p><a href="http://mathoverflow.net/questions/111682/intersection-of-plus-minus-cells-in-bialynicki-birula-decomposition" rel="nofollow">http://mathoverflow.net/questions/111682/intersection-of-plus-minus-cells-in-bialynicki-birula-decomposition</a></p> <p><a href="http://mathoverflow.net/questions/115141/a-cohomology-computation-request" rel="nofollow">http://mathoverflow.net/questions/115141/a-cohomology-computation-request</a></p> <ul> <li>The above fact about homomorphisms between Vermas translates to the lowest non-vanishing cohomology (compactly supported) being one dimensional. These are smooth affine varieties, but (at least in low ranks) their Betti numbers satisfy a curious "Poincare duality"/palindromic type phenomenon. This phenomenon is even more starkly visible if one further looks at the Hodge numbers. Amusingly, since these varieties are smooth and irreducible, one immediately gets that the highest non-vanishing extension group (when it is possible to have morphisms between the Vermas) is one dimensional and concentrated in the "right" degree. This latter fact can also be shown algebraically, but requires a careful argument using translation functors (which can also be done geometrically without ever knowing anything about the intersections, but now I am digressing). Anyway, a topological reason as in my question may hopefully give some insight as to whether this palindromic phenomenon is a low rank coincidence or has any hope for holding in general.</li> </ul> <p>Apologies if any of the reasons above are too vague/ranting, I didn't want to throw in all of that in my original question in case the answer was something blatantly obvious that I had been overlooking.</p> http://mathoverflow.net/questions/117432/splitting-of-the-weight-filtration Splitting of the weight filtration Reladenine Vakalwe 2012-12-28T18:00:21Z 2012-12-29T16:08:43Z <p>All varieties are over $\mathbb{C}$. Notions related to weights etc. refer to mixed Hodge structures (say rational, but I would be grateful if the experts would point out any differences in the real setting).</p> <p>I am trying to get some intuition for the geometric meaning of/when to expect the weight filtration on the cohomology groups $H^i(X)$ of a variety to split. By the weight filtration splitting, I mean that the Hodge structure on each $H^i(X)$ is a direct sum of pure Hodge structures.</p> <p>The simplest situation in which this happens is the classical one of smooth projective varieties. The next simplest situation I think of is the "smooth" being weakened to "mild singularities" (for instance, rationally smooth). So at least in the projective case I think of the "mixed" as encoding singularities. This also gels well with the construction of these Hodge structures using resolution of singularities. Are there other helpful perspectives?</p> <p>Instead of fiddling with the "smooth" one can make the variety non-compact (but still smooth say). Examples: affine $n$-space, tori. Here I don't know how to think of or when to expect the weight filtration to split. Any intuition would be appreciated. The only rough picture I have is that this encodes information about the complement in a good compactification. But I don't find this particularly illuminating.</p> <p>Generalizing affine space and tori is the situation of toric varieties for which the weight filtration always splits (thanks to a lift of Frobenius to characteristic $0$). In general when should one expect a splitting of the weight filtration to be given "geometrically" by a morphism (or say correspondence in the context of Borel-Moore homology)?</p> <p>Related is the following: when should one expect the Hodge structure on each $H^i(X)$ to be pure (not necessarily of weight $i$). Here I am again more interested in weakening the "projective" rather than the "smooth".</p> <p>At the risk of being even more vague, let me add some motivation from left field. There are several situations in representation theory where one expects/knows that the weight filtration on some cohomology groups splits (and is even of Hodge-Tate type). For instance, the cohomology of intersections of Schubert cells with opposite Schubert cells. However, the reasoning/heuristic has, a priori, nothing to do with geometry but more with the philosophy of "graded representation theory" (ala Soergel, see for instance his ICM94 address) I would love to have a geometric reason/heuristic for this. Pertinent to this is also the question of when should one expect the canonical Hodge structure on extensions between perverse sheaves of geometric origin to be split Tate? The only examples I know come from representation theory (see Section 4 of Beilinson-Ginzburg-Soergel's "Koszul duality patterns in representation theory").</p> http://mathoverflow.net/questions/116855/geometric-interpretation-of-translation-through-the-wall/116883#116883 Answer by Reladenine Vakalwe for Geometric interpretation of translation through the wall Reladenine Vakalwe 2012-12-20T17:16:28Z 2012-12-20T20:20:24Z <p>I am guessing the following is well known to you/not what your are looking for, but nonetheless:</p> <p>Let $s$ be a simple reflection, $P_s$ the corresponding minimal parabolic, $\pi_s\colon G/B\to G/P_s$ the projection. Translation across the $s$-wall corresponds' to $\pi_s^*\pi_{s*}$. I use quotation marks because as stated this is clearly not true (translation across the wall is t-exact, $\pi_s^*\pi_{s*}$ is certainly not). However, $\pi_s^*\pi_{s*}$ does correspond to translation across the wall under Koszul duality. This is also the same as convolving with the $IC$-complex corresponding to $s$.</p> <p>Morally (as you point out), reflection across the wall should correspond to convolving with the corresponding tilting. But there is an annoying issue here: tiltings are not $B$-equivariant. Similar problem occurs if instead of convolution using equivariant derived categories you try to use the standard Fourier-Mukai formalism and try to use an object on $G/B\times G/B$ as a kernel. However, there is a fix that comes at some technical expense. Namely, Bezrukavnikov and Yun's free monodromic sheaves <a href="http://arxiv.org/abs/1101.1253" rel="nofollow">http://arxiv.org/abs/1101.1253</a>. The idea actually goes back to the paper of Beilinson and Ginzburg that you cite (look at Section 5).</p> http://mathoverflow.net/questions/115141/a-cohomology-computation-request A cohomology computation request. Reladenine Vakalwe 2012-12-02T02:53:12Z 2012-12-02T04:07:14Z <p><strong>The short: </strong>Let</p> <p><code>$X= \{(x,y,z) \in \mathbb{C}^* \times \mathbb{C} \times \mathbb{C} \, |\, yz-x\neq 0\}$</code></p> <p>Compute $H^*_c(X)$ (say with $\mathbb{C}$-coefficients).</p> <p><strong>The long: </strong>Unless I messed something up, the answer should be</p> <p>$H_c^3(X) = \mathbb{C}$, $H_c^4(X) = \mathbb{C}^2$, $H_c^5(X) = \mathbb{C}^2$, $H_c^6(X)=\mathbb{C}$, and $0$ otherwise.</p> <p>However, as will become clear I did this through an extremely convoluted argument and I am hoping someone can explain to me a simple way of doing this. </p> <p><b>Some context:</b> this question is closely related to</p> <p><a href="http://mathoverflow.net/questions/111682/intersection-of-plus-minus-cells-in-bialynicki-birula-decomposition" rel="nofollow">http://mathoverflow.net/questions/111682/intersection-of-plus-minus-cells-in-bialynicki-birula-decomposition</a> </p> <p>Namely, $X$ is the intersection of the big Bruhat cell and the big opposite Bruhat cell for $SL_3$.</p> <p>It is also closely related to </p> <p><a href="http://mathoverflow.net/questions/83877/are-kazhdan-lusztig-r-polynomials-the-poincare-polynomials-of-the-corresponding" rel="nofollow">http://mathoverflow.net/questions/83877/are-kazhdan-lusztig-r-polynomials-the-poincare-polynomials-of-the-corresponding</a> </p> <p>(the Hodge-Euler characteristic of $X$ is the $R$-polynomial corresponding to the identity and the longest element in type $A_2$; this is a special case of a general fact about $R$-polynomials).</p> <p><strong>My convoluted argument: </strong> Considering the alternatives $y\neq 0$ and $y= 0$, one obtains a decomposition of $X$ into</p> <p><code>$X = (\mathbb{C}^*)^3 \sqcup \mathbb{C}\times \mathbb{C}^*$</code>.</p> <p>This gives rise to a long exact sequence that puts several restrictions on <code>$H^*(X)$</code> (but doesn't fully determine it, namely $H_c^3(X), H_c^4(X), H_c^5(X)$ aren't fully determined). </p> <p>So far this is nice, but now the convoluted bit starts. It is not too hard to see that <code>$H^{*-3}(X)$</code> equals $Ext^*(\Delta_e, \Delta_{w_0})$ where $\Delta_e$ is the unique simple Verma and $\Delta_{w_0}$ is the unique projective Verma in the principal block of the BGG-category $\mathcal{O}$ of $\mathfrak{sl}_3$. </p> <p><b>Aside:</b> this a special case of a statement connecting extensions of Verma modules with cohomology of intersections of Bruhat cells and opposite Bruhat cells (and also why I am interested in the cohomology of these intersections).</p> <p>Now some standard representation theoretic facts about these $Hom$ spaces combined with the decomposition above yield what I claimed the answer to be.</p> <p>I would love a simpler/more geometric way of going about this computation!</p> http://mathoverflow.net/questions/115032/non-rigorous-reasoning-in-rigorous-mathematics/115045#115045 Answer by Reladenine Vakalwe for Non-rigorous reasoning in rigorous mathematics Reladenine Vakalwe 2012-12-01T02:04:29Z 2012-12-01T02:04:29Z <p>Unless I am misunderstanding, the Weil conjectures fit into this framework. I believe it took about two decades for the Grothendieck school to formalize Weil's heuristic that his conjectures follow from a Lefschetz fixed point formula for varieties over finite fields. Rather than try to flesh this post out, let me point to the Brian Osserman's article for the PCM: <a href="http://www.math.ucdavis.edu/~osserman/math/pcm.pdf" rel="nofollow">http://www.math.ucdavis.edu/~osserman/math/pcm.pdf</a>. The Wikipedia account of the history also seems to be not bad (but I haven't really read it in detail): <a href="http://en.wikipedia.org/wiki/Weil_conjectures" rel="nofollow">http://en.wikipedia.org/wiki/Weil_conjectures</a>. I also seem to remember learning about the history and some of the mathematics for the first time from an article by Steven Kleiman, but cannot remember the precise reference. </p> http://mathoverflow.net/questions/114020/how-to-understand-the-harish-chandra-isomorphism/114022#114022 Answer by Reladenine Vakalwe for How to understand the Harish-Chandra isomorphism? Reladenine Vakalwe 2012-11-21T05:33:52Z 2012-11-21T05:33:52Z <p>Here is one algebraic/representation theory perspective on why such a morphism might exist, although I don't think historically this is the way it went at all. Let's admit for a moment that one might be interested in Verma modules. It is easy to see that the center acts via scalars on these. Verma modules are parametrized by $\mathfrak{h}^*$ and in this way one gets an algebra morphism $Z(\mathfrak{g}) \to Sym(\mathfrak{h})$. Now it is also relatively straightforward to see (if I remember correctly) that for each simple reflection $s$, the Verma module $M(s\cdot \lambda)$ occurs as a submodule of the Verma module $M(\lambda)$, for $\lambda$ integral. It follows that the algebra morphism constructed earlier lands in $W$-invariants.</p> http://mathoverflow.net/questions/111682/intersection-of-plus-minus-cells-in-bialynicki-birula-decomposition Intersection of plus/minus cells in Bialynicki-Birula decomposition Reladenine Vakalwe 2012-11-06T20:55:09Z 2012-11-06T21:00:00Z <p>Let $X$ be a projective variety endowed with an algebraic $\mathbb{C}^*$-action. Assume the fixed point set $W$ is discrete. Then we have two cell decompositions of $X$:</p> <p>$X = \bigsqcup_{w\in W} C_w$ and $X = \bigsqcup_{w\in W} C^w$,</p> <p>where</p> <p><code>$C_w = \{x\in X | \lim_{t\to 0} t\cdot x = w \}$</code> and <code>$C^w=\{x\in X | \lim_{t\to \infty} t\cdot x = w\}$</code>,</p> <p>the so-called plus and minus Bialynicki-Birula decomposition. Assume that both of these decompositions are in fact stratifications (although this latter bit may be irrelevant to the question).</p> <p>My (slightly vague) question: Are there any general results about the structure of the intersections $C_w \cap C^v$? When non-empty are they smooth in general? My favorite example of the flag variety, these are smooth (I think). It would make me extremely happy if someone could point me to towards a general description of the cohomology groups of these intersections?</p> <p>Note: In the special situation of flag varieties I am familiar with Deodhar's and Curtis' results describing these intersections. But, unless I am missing something, their results don't help in describing the cohomology groups.</p> http://mathoverflow.net/questions/109901/do-mixed-hodge-modules-form-a-stack/109929#109929 Answer by Reladenine Vakalwe for Do mixed Hodge modules form a stack? Reladenine Vakalwe 2012-10-17T17:09:47Z 2012-10-17T17:09:47Z <p>This is an extremely partial (and sketchy) answer that I haven't fully thought through. But, I am essentially just running the proof for perverse sheaves. Let me first construct the required MHM if the cover just consists of two elements. In this case the required MHM can be define as the cone in the Mayer-Vietoris distinguished triangle. Now use induction to prove the statement for a countable cover. For arbitrary covers I am not quite sure how to proceed. On the other hand, all I am saying is that define the MHM as the cohomology of the Cech complex given by your data. Unless I am missing something, this (modulo some technicalities with finiteness assumptions) gives the construction. Now for uniqueness, probably one can just observe that the usual isomorphism that shows uniqueness for perverse sheaves is a morphism of MHM. Since it is an isomorphism on underlying perverse sheaves, it is an isomorphism on the MHMs. In the case of a 2-element cover this also follows from the fact that there are no negative Exts between MHMs (which leads to the cone in the Mayer-Vietoris triangle being unique).</p> http://mathoverflow.net/questions/109661/polarizable-variations-of-mixed-hodge-structures Polarizable variations of (mixed) Hodge structures Reladenine Vakalwe 2012-10-14T23:53:26Z 2012-10-15T16:02:44Z <p>I am trying to come to grips with Saito's theory of mixed Hodge modules (slightly) beyond just the basic axiomatic formalism. I will take my Hodge structures and sheaves to be rational, but I would be grateful if the experts would point out any subtleties with my questions if rational is replaced by real. Let me at the moment focus on questions related to polarizability:</p> <ol> <li><p>Let $\mathcal{L}$ be a local system underlying a polarizable variation of Hodge structure on a smooth variety. Does the polarizability imply $\mathcal{L}$ is self-dual? If yes, then does every direct summand of this local system also have to be self-dual (I am guessing no to the latter)?</p></li> <li><p>More generally, let $M$ be a polarizable mixed Hodge module on some variety (not necessarily smooth). Is polarizability equivalent to Verdier self-duality (up to Tate twist) in the derived category of mixed Hodge modules? If not equivalent, does it at least imply Verdier self-duality? </p></li> </ol> http://mathoverflow.net/questions/108984/mixed-structures-on-hom-spaces-induced-by-mixed-sheaves Mixed structures on Hom spaces induced by mixed sheaves Reladenine Vakalwe 2012-10-06T06:38:35Z 2012-10-08T16:06:52Z <p>Let $D^b_m(X)$ (resp $D^b(X)$) denote the derived category of mixed Hodge modules (resp. constructible sheaves) on a complex variety $X$. Let </p> <p>$rat\colon D^b_m(X)\to D^b(X)$</p> <p>be the forgetful' functor. This is t-exact for the perverse t-structure on the right. Write $MHM(X)$ for the abelian category of mixed Hodge modules on $X$. Then $MHM(pt)$ is the category of graded polarizable mixed Hodge structures, and $rat\colon MHM(pt) \to VectorSpaces$ is the evident forgetful functor. </p> <p>Now let $M,N\in D^b_m(X)$. Set </p> <p>$\mathcal{H}om(M,N) = \Delta^!(\mathbb{D}M \boxtimes N)$, </p> <p>where $\Delta\colon X\to X\times X$ is the diagonal map, and $\mathbb{D}$ is Verdier duality. </p> <p>Let $a\colon X \to pt$ be the evident map. Then</p> <p><code>$rat ( H^0(a_*\mathcal{H}om(M,N))) = H^0(rat(a_*\mathcal{H}om(M,N))) = Hom(rat(M), rat(N))$</code></p> <p>and in this way we get a Hodge structure on $Hom(rat(M),rat(N))$. All functors are derived.</p> <p><b>My question: </b> If $M,N$ are <s>pure</s> <b>pointwise pure (see Geordie Williamson's comment below)</b>, then is the induced structure on $Hom(rat(M), rat(N))$ pure?</p> <p>My gut answer is no (even if $X$ is complete, the $\Delta^!$ should be messing weights up), but it would make me happier if the answer is yes!</p> <p>If the answer is no, under what additional conditions (other than requiring $X$ to be smooth and complete plus $M,N$ being the constant' sheaf) can the answer be converted to yes? </p> <p>I guess one could also ask the same sort of question for mixed $\ell$-adic sheaves. But I am even less familiar with that setting.</p> http://mathoverflow.net/questions/1890/describe-a-topic-in-one-sentence/109102#109102 Answer by Reladenine Vakalwe for Describe a topic in one sentence. Reladenine Vakalwe 2012-10-07T22:00:36Z 2012-10-07T22:00:36Z <p>Geometric representation theory: keep translating the problem until you run into Hard Lefschetz, then you are done.</p> http://mathoverflow.net/questions/108491/about-an-argument-in-koszul-duality-patterns-in-representation-theory-by-beilin About an argument in Koszul duality patterns in representation theory' by Beilinson-Ginzburg-Soergel. Reladenine Vakalwe 2012-09-30T20:25:06Z 2012-09-30T21:14:44Z <p>I am trying to understand Proposition 3.4.2 in Koszul duality patterns in representation theory' by Beilinson-Ginzburg-Soergel [BGS]. A copy of the paper can be found at <a href="http://home.mathematik.uni-freiburg.de/soergel/" rel="nofollow">http://home.mathematik.uni-freiburg.de/soergel/</a></p> <p>I outline the setup below. Text in bold are my own commentary. I have taken the liberty to change some confusing notation in the paper. However, it is entirely possible that the reason I am confused is that one of my "changes of notation" isn't correct. I have tried to be careful in my changes, but apologies in advance for any additional confusion this may contribute to.</p> <p>My question is that indicated by the last bold face text below.</p> <p>Let $X$ be a complex variety with an algebraic stratification by <em>affine linear spaces</em> $X = \sqcup_{w\in W} X_w$. Let $IC_w$ denote the intersection cohomology complex on $X$ corresponding to the constant sheaf on $X_w$.</p> <p>Let $X = Y_0 \supset Y_1 \supset \cdots \supset Y_r = \emptyset$ be the corresponding filtration by closed subvarieties so that $Y_{p}- Y_{p+1} = X_p$ for some strata $X_p$.</p> <p>Let </p> <p>$j_w\colon X_w \to X$ </p> <p>be the inclusion of the strata. Assume parity vanishing, i.e., assume </p> <p>$H^ij_v^*IC_w = 0$ unless $i = dim(X_v) + dim(X_w) \mod 2$, for all $v,w\in W$. </p> <p>Here $H^*$ denotes perverse cohomology.</p> <p><strong>Proposition:</strong> Under the assumption of parity vanishing, hypercohomology induces an injection</p> <p>$Hom^{\bullet}_{D^b(X)}(IC_x, IC_y) \to Hom_{\mathbb{C}}(\mathbb{H}^{\bullet}IC_x, \mathbb{H}^{\bullet}IC_y)$.</p> <p><strong>Proof:</strong> By parity vanishing the spectral sequence $\mathbb{H}^{p+q}j_p^!IC_x \implies \mathbb{H}^n IC_x$ is degenerate (<strong>the spectral sequence is defined via the filtration by local hypercohomology along the strata, for details see Section 3.4 of [BGS]</strong>). So if $f\in Hom^{\bullet}_{D^b(X)}(IC_x, IC_y)$ is given such that $\mathbb{H}^{\bullet}f = 0$, then necessarily $0 = j_p^!f \in Hom^{\bullet}_{D^b(X)}(j_p^!IC_x, j_p^!IC_y)$ for all $p$. Let </p> <p>$a_p\colon Y_p \to X$ </p> <p>be the closed inclusion. We have a decomposition</p> <p>$u\colon X_p = Y_p - Y_{p+1} \to Y_p$, </p> <p>$i\colon Y_{p+1}\to Y_p$</p> <p>in an open and a closed subset and a distinguished triangle</p> <p><strong>Edit: the original distinguished triangle (as stated in [BGS]) wasn't correct, I have now made the fix</strong></p> <p>$i_*i^!a_p^! \to a_p^! \to u_*u^!a_p^!$</p> <p>(<b>so this distinguished triangle is the same as $i_*a_{p+1}^! \to a_p^! \to u_*j_p^!$ </b>)</p> <p>which shows that $a_{p+1}^!f = 0 = j_{p}^!f$ implies $a_p^!f = 0$ (<b>it is this implication that I don't understand, related to my confusion is an earlier question of mine <a href="http://mathoverflow.net/questions/108481/showing-morphism-of-sheaves-is-zero" rel="nofollow">http://mathoverflow.net/questions/108481/showing-morphism-of-sheaves-is-zero</a> </b>)</p> <p>Hence by induction $j_p^!f = 0$ for all $p$ implies $f = a_0^!f = 0$.</p> <p>Any comments that would clarify the above would be most appreciated!</p> http://mathoverflow.net/questions/108481/showing-morphism-of-sheaves-is-zero Showing morphism of sheaves is zero Reladenine Vakalwe 2012-09-30T16:16:38Z 2012-09-30T18:33:28Z <p>I work in derived category $D^b(X)$ of constructible sheaves on a reasonable space $X$. Let $j\colon U\to X$ be an open inclusion and $i\colon Y\to X$ the closed complement. Let $M,N\in D^b(X)$ and let $f\in Hom_{D^b(X)}(M,N)$. Suppose $i^*f = 0$ and $j^*f= 0$. </p> <p>Then is it true that $f=0$? </p> <p>My gut answer is yes, and I thought I would be able to lever the canonical distinguished triangle $j_! j^* \to id \to i_* i^* \to j_!j^*[1]$ into a proof. But I have failed so far.</p> http://mathoverflow.net/questions/106299/question-regarding-a-statement-in-a-proof-of-jantzen-conjectures Question regarding a statement in A proof of Jantzen conjectures' Reladenine Vakalwe 2012-09-04T04:00:38Z 2012-09-05T12:38:19Z <p>So I am trying to understand a statement in the proof of Corollary 5.2.3 in A proof of Jantzen conjectures' (a copy of the paper can be found at <a href="http://www.math.harvard.edu/~gaitsgde/grad_2009/" rel="nofollow">http://www.math.harvard.edu/~gaitsgde/grad_2009/</a>).</p> <p>The starting assumptions of the Corollary are:</p> <p>Let $M_1, M_2$ be pure perverse sheaves of weights $w_1, w_2$ that are both $*$- and $!$-pointwise pure. Suppose that $Ext^1_{mixed}(M_1, M_2)\neq 0$. Then ...'</p> <p>The first line of the proof says:</p> <p>Clearly either $Y_1 \subset Y_2$ or $Y_2 \subset Y_1$ (otherwise $Ext^1 = 0$)'</p> <p>Here $Y_i = Supp(M_i)$. This statement confuses me. The ordinary ( = non-mixed) $Ext^1$ group should be the extensions between the restrictions of $M_1$ and $M_2$ to the intersection $Y_1 \cap Y_2$. I don't see why this should vanish if either $Y_1$ isn't contained in $Y_2$ or vice versa. Anyway, even if the unmixed group does vanish, without vanishing of the unmixed $Hom$ group I don't see how I would get $Ext^1_{mixed}$ vanishes. </p> <p>Presumably I am making a stupid error here and both of the unmixed groups above do vanish? Any comments would be appreciated.</p> http://mathoverflow.net/questions/106299/question-regarding-a-statement-in-a-proof-of-jantzen-conjectures/106304#106304 Answer by Reladenine Vakalwe for Question regarding a statement in A proof of Jantzen conjectures' Reladenine Vakalwe 2012-09-04T06:18:10Z 2012-09-04T06:18:10Z <p>Too long to leave as a comment.</p> <p>Ben: Let $i_k$, $k=1,2$ be the closed inclusions $Y_i \to X$ (where $X$ is my ambient space). Then $M_k = i_{k*}i_k^*M_k$. So</p> <p>$Ext^1(M_1, M_2) = Ext^1(i_{1*}i_{1}^*M_1, i_{2*}i_2^*M_2) = Ext^1(i_2^*i_{1*}i_1^*M_1, i_2^*M_2)$</p> <p>Now let $r\colon Y_1\cap Y_2 \to X$ and $s\colon Y_1\cap Y_2 \to Y_2$ be the closed inclusions and we get:</p> <p>$Ext^1(i_2^*i_{1*}i_1^*M_1, i_2^*M_2) = Ext^1(s_*r^*M_1, i_2^*M_2) = Ext^1(r^*M_1, s^!i_2^*M_2)$</p> <p>Ah, so my initial error was to magically convert the $s^!$ to $s^*$, but it still reduces the computation of the Ext group to the intersection (or did I do something screwy again?). On the other hand, I don't see how to sanely deal with the $s^!i_2^*$. Regardless, I still don't see why the mixed Ext group in the original question is vanishing.</p> http://mathoverflow.net/questions/73430/on-the-definition-of-regularity/100838#100838 Answer by Reladenine Vakalwe for On the definition of regularity Reladenine Vakalwe 2012-06-28T02:55:52Z 2012-06-28T02:55:52Z <p>There are some comparison results in Chapter 5 of Bjork's Analytic D-modules and Applications'. Also see Chapter 8. In particular, I think Thm. 8.7.3 combined with Thm. 5.6.5 (almost) gives (1) iff (3). Further, I think Prop. 5.6.22 gives the equivalence with (2). There are also results in there comparing Deligne's description.</p> <p>I must admit though that I find Bjork quite notationally dense and am not particularly familiar with it, so I may be quite off with the references above. I am interested in this question also, so please comment/post if you find better references.</p> http://mathoverflow.net/questions/100795/characteristic-variety-of-a-d-module-along-smooth-pullback Characteristic variety of a D-module along smooth pullback Reladenine Vakalwe 2012-06-27T18:03:24Z 2012-06-27T18:03:24Z <p>All varieties are over the complex numbers. Given a smooth variety $X$, write $T^* X$ for its cotangent bundle. For a morphism of smooth varieties $f: X \to Y$ write $f_{\pi}: T^*Y \times_Y X \to T^*Y$ for the projection map, and let $f_d: T^*Y \times_Y X \to T^* X$ be the map dual to the derivative.</p> <p>Now let $f\colon X\to Y$ be a smooth morphism. Let $M$ be a coherent $D_Y$-module. I am trying to understand the proof of:</p> <p>$Ch(f^*M) \subseteq f_df_{\pi}^{-1}Ch(M)$,</p> <p>(where $Ch$ means characteristic variety) as presented in J.P. Schneider's notes <a href="http://www.analg.ulg.ac.be/jps/rec/idm.pdf" rel="nofollow">http://www.analg.ulg.ac.be/jps/rec/idm.pdf</a> (see pages 34-35).</p> <p>I need a bit more notation before I get to my exact question: we know $f^*M = O_X\otimes_{f^{-1}O_Y}f^{-1}M$ as an $O_X$-module. Endow $f^*M$ with the good filtration $F_i(f^*M) = O_X\otimes_{f^{-1}O_Y}f^{-1}F_iM$, where the $F_i$ on the right hand side is some good filtration of $M$. Then we have a surjection</p> <p>$O_X\otimes_{f^{-1}O_Y}f^{-1}gr(M) \twoheadrightarrow gr(f^*M)$</p> <p>of $gr(D_X)$-modules, where $gr$ denotes associated graded. Write $\tilde{gr}(-)$ for the sheaf on the cotangent bundle associated to $gr(-)$. Then the claim (last two lines of the proof on p35 of the aforementioned notes) is that:</p> <p>since $f_d$ is finite we obtain a surjection</p> <p>$f_{d*}f_{\pi}^*\tilde{gr}(M) \twoheadrightarrow \tilde{gr}(f^*M)$</p> <p>I don't understand this claim. Any clarifying thoughts would be most appreciated.</p> <p>Unless I made a mistake:</p> <p>$f_{d*}f_{\pi}^*\tilde{gr}(M) = f_{d*}(O_{T^*Y\times_Y X}\otimes_{f_{\pi}^{-1}\pi_Y^{-1}gr D_Y}f_{\pi}^{-1}\pi_Y^{-1}gr(M))$,</p> <p>where $\pi_Y\colon T^*Y\to Y$. What is required is that this equal</p> <p>$O_{T^*X}\otimes_{\pi_X^{-1}gr D_X}(O_X\otimes_{f^{-1}O_Y}f^{-1}gr(M))$</p> <p>I don't see how to proceed. Presumably I am making this way too complicated and there is a very simple explanation?</p> <p>Also, it is mentioned at the end of the proof that the estimate of the characteristic variety is in fact an equality. Does anyone know of a reference for this?</p> http://mathoverflow.net/questions/100264/c-equivariant-modules-on-a-vector-bundle-vs-graded-modules-on-the-pushforward C^*-equivariant modules on a vector bundle vs graded modules on the pushforward. Reladenine Vakalwe 2012-06-21T18:00:19Z 2012-06-22T09:14:54Z <p>All varieties are over $\mathbb{C}$.</p> <p>Let $X$ be a variety and $\pi \colon E \to X$ a geometric vector bundle. So $\pi$ is affine. Then certainly the assignment <code>$M \mapsto \pi_*M$</code> defines an equivalence between quasi-coherent <code>$\mathcal{O}_E$</code>-modules and quasi-coherent $\pi_*\mathcal{O}_E$-modules.</p> <p>Now $\mathbb{C}^{\times}$ acts on $E$ via dilation of the fibres of $\pi$. So <code>$\pi_* \mathcal{O}_E$</code> acquires a grading. Is it true that $M \mapsto \pi_* M$ gives an equivalence between $\mathbb{C}^{\times}$-equivariant quasi-coherent <code>$\mathcal{O}_E$</code>-modules and graded quasi-coherent $\pi_*\mathcal{O}_E$-modules?</p> <p><strike>If this is true, does it generalize to replacing $\pi$ being a vector bundle with $E$ just equipped with a $\mathbb{C}^{\times}$-action, $\pi$ affine and $\mathbb{C}^{\times}$-equivariant, $\mathbb{C}^{\times}$ acting on $X$ trivially? </strike></p> <p><strong>Added later</strong> (in response to a-fortiori's comment): Perhaps I hadn't done my homework as conscientiously as I thought. Regardless, here are some thoughts. As candidate for the quasi-inverse (is there a more sensible choice?) take</p> <p><code>$N\mapsto \mathcal{O}_E \otimes_{\pi^{-1}\pi_*\mathcal{O}_E}\pi^{-1}N$</code></p> <p>with $\mathbb{C}^{\times}$-equivariant structure given by</p> <p>$z \cdot (f(x,v) \otimes n_i) = f(x, z^{-1}v) \otimes z^{-i}n_i$,</p> <p>where $n_i$ is in the $i$-th component of $N$ and the rest of the notation is (I hope) self-explanatory. Hitting the structure sheaf $\mathcal{O}_E$ (with the trivial/evident equivariant structure) with these functors works fine, so this isn't completely ridiculous. But now I am not even sure whether there are other equivariant structures on $\mathcal{O}_E$ that would make this breakdown.</p> http://mathoverflow.net/questions/97777/non-characteristic-maps-ala-d-modules Non-characteristic maps (ala D-modules) Reladenine Vakalwe 2012-05-23T17:25:29Z 2012-05-23T17:25:29Z <p>I am trying to understand a <code>well known' fact (see Kashiwara's</code>Introduction to microlocal analysis page 63, remark 4.8) about non-characteristic morphisms. Here is the setup: </p> <p>All varieties are over the complex numbers. Given a smooth variety $X$, write $T^* X$ for its cotangent bundle, and $T^*_XX \subseteq T^*X$ for the zero section.</p> <p>Let $f: X \to Y$ be a morphism of smooth varieties. Write $f_{\pi}: T^*Y \times_Y X \to T^*Y$ for the projection map, and let $f_d: T^*Y \times_Y X \to T^* X$ be the map dual to the derivative. Let $\Lambda \subseteq T^* Y$ be a closed $\mathbb{C}^*$ stable subvariety ($\mathbb{C}^*$ acting on $T^*Y$ in the evident way). Then $f$ is called <em>non-characteristic</em> for $\Lambda$ if</p> <p>$f_{\pi}^{-1}(\Lambda) \cap f_d^{-1}(T^*_XX) \subseteq T^*_YY \times_Y X$</p> <p>The well known' fact: if $f$ is non-characteristic for $\Lambda$, then $f_d$ restricted to $f_{\pi}^{-1}(\Lambda)$ is finite.</p> <p>I would be grateful if someone could explain the truth of this to me.</p> <p>Some remarks:</p> <p>a) The statement is actually an if and only if, but the converse is straightforward, since the fibres of $f_d$ are $\mathbb{C}^*$ stable.</p> <p>b) I believe I understand how to show that $f_d$ restricted to $f_{\pi}^{-1}(\Lambda)$ is quasi-finite (using the the $\mathbb{C}^*$ stability and the upper semi-continuity of fibre dimension). But that's as far as I have got.</p> http://mathoverflow.net/questions/118176/around-the-socle-filtration-of-a-verma-module/118235#118235 Comment by Reladenine Vakalwe Reladenine Vakalwe 2013-01-07T00:26:39Z 2013-01-07T00:26:39Z Also, is Jantzen's Habilitationsschrift a reference to "Moduln mit einem hochsten Gewicht"? That text strikes fear in my heart! http://mathoverflow.net/questions/118176/around-the-socle-filtration-of-a-verma-module/118235#118235 Comment by Reladenine Vakalwe Reladenine Vakalwe 2013-01-07T00:22:44Z 2013-01-07T00:22:44Z To be honest, apart from being slightly open to "miracles sometimes occur", I am pretty skeptical about a). I do believe in b) though because it matches with some heuristics I have regarding higher extensions between Vermas (heuristics coming from some brute force computations, but I haven't managed to work out all the extensions in Boe's counterexample, so these may still be "too low rank"). I think I understand Mazorchuk's proof (regarding b)) enough to see that he defers the burden to a result of Backelin. But haven't got my hands on the Backelin paper yet. http://mathoverflow.net/questions/118176/around-the-socle-filtration-of-a-verma-module/118213#118213 Comment by Reladenine Vakalwe Reladenine Vakalwe 2013-01-06T20:14:36Z 2013-01-06T20:14:36Z I agree with the type $A_2$ check (although I don't think I could typeset the lattice/diagram). Stroppel's diagram is pretty impressive! I am reasonably sure that b) is true in general (since I believe Mazorchuk's result), but of course I don't understand why. http://mathoverflow.net/questions/118176/around-the-socle-filtration-of-a-verma-module/118213#118213 Comment by Reladenine Vakalwe Reladenine Vakalwe 2013-01-06T19:36:08Z 2013-01-06T19:36:08Z Inspired by your example, as long as I did it correctly, c) is false in type $A_2$ also, and I am reasonably sure is essentially always going to fail (with the exception of $\mathfrak{sl}_2$). http://mathoverflow.net/questions/118176/around-the-socle-filtration-of-a-verma-module/118213#118213 Comment by Reladenine Vakalwe Reladenine Vakalwe 2013-01-06T18:32:37Z 2013-01-06T18:32:37Z Yes! This is a nice counterexample. Not simple, but that's a minor quibble. Thank you! Any thoughts on b)? http://mathoverflow.net/questions/118176/around-the-socle-filtration-of-a-verma-module Comment by Reladenine Vakalwe Reladenine Vakalwe 2013-01-06T16:11:49Z 2013-01-06T16:11:49Z Dag: Could you elaborate on your second comment regarding the counterexample to c) (possibly as an answer)? http://mathoverflow.net/questions/118176/around-the-socle-filtration-of-a-verma-module Comment by Reladenine Vakalwe Reladenine Vakalwe 2013-01-06T16:10:07Z 2013-01-06T16:10:07Z Dag: Your first comment about the statement after "Note:" in a) being wrong is absolutely correct. I have removed it (sorry strikethrough wasn't showing up correctly). As to how I am choosing $k$. Basically I just want the first layer that isn't completely contained in $\Delta_w$. Does that clarify or am I being screwy? http://mathoverflow.net/questions/118176/around-the-socle-filtration-of-a-verma-module Comment by Reladenine Vakalwe Reladenine Vakalwe 2013-01-06T16:02:13Z 2013-01-06T16:02:13Z Jim: Thanks for pointing out the misprint. I have fixed it now. I agree about the risk of relying on rank 2 examples. But well, it's a start. Figuring out socle filtrations for higher rank is quickly going to start being a pain! Apologies for the length. I had hoped that getting a) out there quickly would alleviate some of the pain. My next step is to ask Mazorchuk, but am still holding out hope that I am missing something silly and MO will offer some instant gratification! http://mathoverflow.net/questions/117723/geometric-intuitive-interpretation-of-ext Comment by Reladenine Vakalwe Reladenine Vakalwe 2012-12-31T18:06:43Z 2012-12-31T18:06:43Z Not quite in line with your question. But if you were dealing with a reasonable topological space $X$, then $Ext$ groups of the constant sheaf with itself (in the category of constructible sheaves) are the cohomology groups of that space. More generally, extensions from the constant sheaf to any complex of sheaves is hypercohomology with coefficients in the complex. http://mathoverflow.net/questions/116348/morphisms-between-verma-modules Comment by Reladenine Vakalwe Reladenine Vakalwe 2012-12-31T02:25:39Z 2012-12-31T02:25:39Z As the Mathoverflow bot has so graciously pinged this question, I may as well add the following. The "curious Poincare duality" and "palindromic phenomenon" mentioned above has a high powered explanation. Namely: Koszul duality. This can in turn be used to show the dimension bound asked for. But this is using a blowtorch to light a candle. http://mathoverflow.net/questions/117432/splitting-of-the-weight-filtration/117509#117509 Comment by Reladenine Vakalwe Reladenine Vakalwe 2012-12-29T17:38:08Z 2012-12-29T17:38:08Z Dan: Thanks for the examples! They are helpful. I wasn't aware of Minhyong Kim's paper. It's really nice. http://mathoverflow.net/questions/117432/splitting-of-the-weight-filtration/117506#117506 Comment by Reladenine Vakalwe Reladenine Vakalwe 2012-12-29T17:31:37Z 2012-12-29T17:31:37Z Donu: Thanks! Your answer is helpful. At least how I am reading it is that the weight filtration has geometric as well as linear algebraic content to it (split over $\mathbb{R}$ vs. $\mathbb{Q}$). As an aside, related to your comment about varieties coming from linear algebra, most of my examples come from flag varieties and "miracles often happen in flag varieties"! http://mathoverflow.net/questions/116855/geometric-interpretation-of-translation-through-the-wall Comment by Reladenine Vakalwe Reladenine Vakalwe 2012-12-21T01:39:20Z 2012-12-21T01:39:20Z Jim: Perhaps I should make this a separate question. But don't Soergel's arguments (which I am implicitly using to justify my answer below) show this? Under Soergel's functor to combinatorics $\mathbb{V}$, translation across the wall corresponds to (roughly) restriction/induction for the coinvariant algebra. The latter depending only on the stabilizer (namely $s$) of $\lambda$. Or am I confused? Of course, $\mathbb{V}$ is not an equivalence but it is full and faithful on maps between projectives/tiltings, but this should be enough to show the desired independence? No? http://mathoverflow.net/questions/116830/non-characteristic-is-to-pullback-as-blank-is-to-pushforward Comment by Reladenine Vakalwe Reladenine Vakalwe 2012-12-20T00:52:15Z 2012-12-20T00:52:15Z Slightly confused about a certain point, why is pushforward along a proper map preserving t-structure? Regarding the question, there is a good estimate on singular support of $f_*M$ if $f_{\pi}\colon f_d^{-1} Ch(M)\to T^*Y$ is finite (I hope the notation is self-explanatory. See Kashiwara's D-Modules and microlocal calculus section 4.7. This condition is analogous to the non-characteristic condition. http://mathoverflow.net/questions/116348/morphisms-between-verma-modules Comment by Reladenine Vakalwe Reladenine Vakalwe 2012-12-17T01:56:20Z 2012-12-17T01:56:20Z Deleted a previous comment of mine where I thought I had an argument, since it was a pipe dream.
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Proof
Definitions occuring in Statement : radd-list: radd-list(L) real: list: List uall: [x:A]. B[x] member: t ∈ T
Definitions unfolded in proof : uall: [x:A]. B[x] member: t ∈ T radd-list: radd-list(L) callbyvalueall: callbyvalueall uimplies: supposing a has-value: (a)↓ has-valueall: has-valueall(a) nat: so_lambda: λ2x.t[x] so_apply: x[s] all: x:A. B[x] implies: Q bool: 𝔹 unit: Unit it: btrue: tt uiff: uiff(P;Q) and: P ∧ Q ifthenelse: if then else fi bfalse: ff exists: x:A. B[x] prop: or: P ∨ Q sq_type: SQType(T) guard: {T} bnot: ¬bb assert: b false: False nat_plus: + ge: i ≥ nequal: a ≠ b ∈ decidable: Dec(P) le: A ≤ B satisfiable_int_formula: satisfiable_int_formula(fmla) not: ¬A top: Top
Lemmas referenced : real-list-has-valueall evalall-reduce list_wf real_wf valueall-type-real-list value-type-has-value nat_wf set-value-type le_wf int-value-type length_wf_nat eq_int_wf length_wf bool_wf eqtt_to_assert assert_of_eq_int int-to-real_wf eqff_to_assert equal_wf bool_cases_sqequal subtype_base_sq bool_subtype_base assert-bnot neg_assert_of_eq_int accelerate_wf non_neg_length decidable__lt satisfiable-full-omega-tt intformand_wf intformnot_wf intformless_wf itermConstant_wf itermVar_wf intformle_wf intformeq_wf int_formula_prop_and_lemma int_formula_prop_not_lemma int_formula_prop_less_lemma int_term_value_constant_lemma int_term_value_var_lemma int_formula_prop_le_lemma int_formula_prop_eq_lemma int_formula_prop_wf less_than_wf reg-seq-list-add_wf
Rules used in proof : sqequalSubstitution sqequalTransitivity computationStep sqequalReflexivity isect_memberFormation introduction cut sqequalRule extract_by_obid sqequalHypSubstitution isectElimination thin hypothesisEquality hypothesis callbyvalueReduce independent_isectElimination intEquality lambdaEquality natural_numberEquality lambdaFormation unionElimination equalityElimination equalityTransitivity equalitySymmetry productElimination dependent_pairFormation promote_hyp dependent_functionElimination instantiate cumulativity independent_functionElimination because_Cache voidElimination dependent_set_memberEquality int_eqEquality isect_memberEquality voidEquality independent_pairFormation computeAll axiomEquality
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# Two sides of a triangle are 13 cm and 18 cm. The perimeter of a triangle is 48 cm. What is it third side?​
Given: Two sides of a triangle are 13 cm and 18 cm and assume third side is x.
The perimeter of a triangle = 48 cm => 13 + 18 + x = 48
x = 17
Hence the third side of triangle is 17 cm.​
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Intuitionistic Logic Explorer < Previous Next > Nearby theorems Mirrors > Home > ILE Home > Th. List > qbtwnzlemex Unicode version
Theorem qbtwnzlemex 9105
Description: Lemma for qbtwnz 9106. Existence of the integer. The proof starts by finding two integers which are less than and greater than the given rational number. Then this range can be shrunk by choosing an integer in between the endpoints of the range and then deciding which half of the range to keep based on rational number trichotomy, and iterating until the range consists of two consecutive integers. (Contributed by Jim Kingdon, 8-Oct-2021.)
Assertion
Ref Expression
qbtwnzlemex
Distinct variable group: ,
Proof of Theorem qbtwnzlemex
Dummy variables are mutually distinct and distinct from all other variables.
StepHypRef Expression
1 qre 8560 . . . 4
2 btwnz 8357 . . . 4
31, 2syl 14 . . 3
4 reeanv 2479 . . 3
53, 4sylibr 137 . 2
6 simpll 481 . . . . 5
7 simplrl 487 . . . . . . . 8
87zred 8360 . . . . . . 7
91ad2antrr 457 . . . . . . 7
10 simplrr 488 . . . . . . . 8
1110zred 8360 . . . . . . 7
12 simprl 483 . . . . . . 7
13 simprr 484 . . . . . . 7
148, 9, 11, 12, 13lttrd 7140 . . . . . 6
15 znnsub 8296 . . . . . . 7
1615ad2antlr 458 . . . . . 6
1714, 16mpbid 135 . . . . 5
188, 9, 12ltled 7135 . . . . . 6
197zcnd 8361 . . . . . . . 8
2010zcnd 8361 . . . . . . . 8
2119, 20pncan3d 7325 . . . . . . 7
2213, 21breqtrrd 3790 . . . . . 6
23 breq1 3767 . . . . . . . 8
24 oveq1 5519 . . . . . . . . 9
2524breq2d 3776 . . . . . . . 8
2623, 25anbi12d 442 . . . . . . 7
2726rspcev 2656 . . . . . 6
287, 18, 22, 27syl12anc 1133 . . . . 5
29 qbtwnzlemshrink 9104 . . . . 5
306, 17, 28, 29syl3anc 1135 . . . 4
3130ex 108 . . 3
3231rexlimdvva 2440 . 2
335, 32mpd 13 1
Colors of variables: wff set class Syntax hints: wi 4 wa 97 wb 98 wcel 1393 wrex 2307 class class class wbr 3764 (class class class)co 5512 cr 6888 c1 6890 caddc 6892 clt 7060 cle 7061 cmin 7182 cn 7914 cz 8245 cq 8554 This theorem was proved from axioms: ax-1 5 ax-2 6 ax-mp 7 ax-ia1 99 ax-ia2 100 ax-ia3 101 ax-in1 544 ax-in2 545 ax-io 630 ax-5 1336 ax-7 1337 ax-gen 1338 ax-ie1 1382 ax-ie2 1383 ax-8 1395 ax-10 1396 ax-11 1397 ax-i12 1398 ax-bndl 1399 ax-4 1400 ax-13 1404 ax-14 1405 ax-17 1419 ax-i9 1423 ax-ial 1427 ax-i5r 1428 ax-ext 2022 ax-coll 3872 ax-sep 3875 ax-nul 3883 ax-pow 3927 ax-pr 3944 ax-un 4170 ax-setind 4262 ax-iinf 4311 ax-cnex 6975 ax-resscn 6976 ax-1cn 6977 ax-1re 6978 ax-icn 6979 ax-addcl 6980 ax-addrcl 6981 ax-mulcl 6982 ax-mulrcl 6983 ax-addcom 6984 ax-mulcom 6985 ax-addass 6986 ax-mulass 6987 ax-distr 6988 ax-i2m1 6989 ax-1rid 6991 ax-0id 6992 ax-rnegex 6993 ax-precex 6994 ax-cnre 6995 ax-pre-ltirr 6996 ax-pre-ltwlin 6997 ax-pre-lttrn 6998 ax-pre-apti 6999 ax-pre-ltadd 7000 ax-pre-mulgt0 7001 ax-pre-mulext 7002 ax-arch 7003 This theorem depends on definitions: df-bi 110 df-dc 743 df-3or 886 df-3an 887 df-tru 1246 df-fal 1249 df-nf 1350 df-sb 1646 df-eu 1903 df-mo 1904 df-clab 2027 df-cleq 2033 df-clel 2036 df-nfc 2167 df-ne 2206 df-nel 2207 df-ral 2311 df-rex 2312 df-reu 2313 df-rmo 2314 df-rab 2315 df-v 2559 df-sbc 2765 df-csb 2853 df-dif 2920 df-un 2922 df-in 2924 df-ss 2931 df-nul 3225 df-pw 3361 df-sn 3381 df-pr 3382 df-op 3384 df-uni 3581 df-int 3616 df-iun 3659 df-br 3765 df-opab 3819 df-mpt 3820 df-tr 3855 df-eprel 4026 df-id 4030 df-po 4033 df-iso 4034 df-iord 4103 df-on 4105 df-suc 4108 df-iom 4314 df-xp 4351 df-rel 4352 df-cnv 4353 df-co 4354 df-dm 4355 df-rn 4356 df-res 4357 df-ima 4358 df-iota 4867 df-fun 4904 df-fn 4905 df-f 4906 df-f1 4907 df-fo 4908 df-f1o 4909 df-fv 4910 df-riota 5468 df-ov 5515 df-oprab 5516 df-mpt2 5517 df-1st 5767 df-2nd 5768 df-recs 5920 df-irdg 5957 df-1o 6001 df-2o 6002 df-oadd 6005 df-omul 6006 df-er 6106 df-ec 6108 df-qs 6112 df-ni 6402 df-pli 6403 df-mi 6404 df-lti 6405 df-plpq 6442 df-mpq 6443 df-enq 6445 df-nqqs 6446 df-plqqs 6447 df-mqqs 6448 df-1nqqs 6449 df-rq 6450 df-ltnqqs 6451 df-enq0 6522 df-nq0 6523 df-0nq0 6524 df-plq0 6525 df-mq0 6526 df-inp 6564 df-i1p 6565 df-iplp 6566 df-iltp 6568 df-enr 6811 df-nr 6812 df-ltr 6815 df-0r 6816 df-1r 6817 df-0 6896 df-1 6897 df-r 6899 df-lt 6902 df-pnf 7062 df-mnf 7063 df-xr 7064 df-ltxr 7065 df-le 7066 df-sub 7184 df-neg 7185 df-reap 7566 df-ap 7573 df-div 7652 df-inn 7915 df-n0 8182 df-z 8246 df-q 8555 df-rp 8584 This theorem is referenced by: qbtwnz 9106
Copyright terms: Public domain W3C validator
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# English to Samoan Meaning of tangent - tangent
Tangent :
tangent
uliuli
cotangentcotangentstangenttangentialtangents
Definitions of tangent in English
Noun(1) a straight line or plane that touches a curve or curved surface at a point but does not intersect it at that point(2) ratio of the opposite to the adjacent side of a right-angled triangle
Examples of tangent in English
(1) Note that this curvature is the inverse of the radius of a circle tangent to the neutral line at this point.(2) We can never fit a straight tangent line to the curve at the point.(3) Smooth infinitesimal analysis embodies a concept of intensive magnitude in the form of infinitesimal tangent vectors to curves.(4) However, Newton's approach, based on a decomposition of motion along the tangent and along the normal to the orbital curve, was missing an essential ingredient.(5) The discussion then went off on a tangent , to the question of how many spheres of equal size could rotate around a central ball of the same size.(6) The chair went one way, she went another the and keyboard headed off at a completely different tangent .(7) Therefore, we say the gradient of the tangent line is 2x, although in this case we knew that already.(8) To understand this notion fully requires understanding tangent spaces, computing with vector fields, and working with bracket products of vector fields.(9) The line of fall is the line tangent to the trajectory at the level point.(10) The midpoints of segments AM lie directly on the tangent lines.(11) Construct two tangent circles 1 and 2 and the line L through their centers.(12) Now the tangent line is much easier to visualize. Notice that the tangent line is at a right angle to the aiming line.(13) In addition, each circle is divided into 8 regions, by adding 3 tangent circles.(14) He breaks off, perhaps uncomfortable with this line of questioning, and scoots off at a tangent .(15) Water contact angle refers to the angle between the tangent plane of the liquid surface and the tangent plane of the solid surface at any point along the line of contact.(16) Instead, we use the radical plane, defined by the property that any point on it will have equal lengths of tangent line segments to the two atoms.
Related Phrases of tangent
(1) tangent plane ::
vaalele tangent
(2) tangent line ::
laina tangent
(3) tangent point ::
tulaga tangent
Synonyms
Noun
1. tan ::
uliuli
Different Forms
cotangent, cotangents, tangent, tangential, tangents
Word Example from TV Shows
I feel like an inverse TANGENT function that's approaching an asymptote.
The Big Bang Theory Season 2, Episode 13
- Stuck? - What part of "inverse TANGENT function...
The Big Bang Theory Season 2, Episode 13
English to Samoan Dictionary: tangent
Meaning and definitions of tangent, translation in Samoan language for tangent with similar and opposite words. Also find spoken pronunciation of tangent in Samoan and in English language.
Tags for the entry 'tangent'
What tangent means in Samoan, tangent meaning in Samoan, tangent definition, examples and pronunciation of tangent in Samoan language.
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# How to simplify an improper fractions
1
by s422494
25/5 is a improper fraction so if 5 goes into 25-5 times than ur answer would be 5whole
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https://www.lawnsite.com/threads/question-on-measuring-mulch.69711/
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# question on measuring mulch!
Discussion in 'Starting a Lawn Care Business' started by zmak, Apr 25, 2004.
1. ### zmakLawnSite Memberfrom DelawareMessages: 58
I have about 5300 square feet to mulch at about 2 in thick.
How many yards of mulch do you think I will need.
and how do you come up with your numbers.
Thanks.
2. ### gwwilsonLawnSite Memberfrom coopersburg, paMessages: 89
31.55 yards of mulch
2" = 168 sq ft
3" = 126 sq ft
4" = 84 sq ft
3. ### glenndLawnSite Memberfrom Roxboro, North CarolinaMessages: 18
This is the formula I use and it works pretty good. 1 Cubic yard mulch, 2 inches thick will cover an area 144 sq ft. 1 Cubic yard of mulch, 3 inches thick will cover an area 100 sq ft. So with you having an area 5300 sq ft, divide 144 sq ft into 5300 sq ft = 36.8 yards. So you'd be safe with 36-37 yards. Just my .02
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Hits: 242
# Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 2 (P2-9709/02) | Year 2008 | May-Jun | (P2-9709/02) | Q#7
Hits: 242 Question The equation of the curve is . i. Show that ii. Find the coordinates of each of the points on the curve where the tangent is parallel to the x- axis. Solution i. We are given; We are required to find . To find from […]
# Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 2 (P2-9709/02) | Year 2008 | May-Jun | (P2-9709/02) | Q#6
Hits: 206 Question It is given that the curve has one stationary point. i. Find the x-coordinates of this point. ii. Determine whether this point is a maximum or a minimum point. Solution i. We are required to find the coordinates of stationary point of the curve; […]
# Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 2 (P2-9709/02) | Year 2008 | Oct-Nov | (P2-9709/02) | Q#3
Hits: 1709 Question The variables x and y satisfy the equation y = A(b-x), where A and b are constants. The graph of ln y against ln x is a straight line passing through the points (0, 1.3) and (1.6, 0.9), as shown in the diagram. Find the values of A and b, correct […]
# Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 2 (P2-9709/02) | Year 2008 | Oct-Nov | (P2-9709/02) | Q#1
Hits: 335 Question Solve the inequality . Solution SOLVING INEQUALITY: PIECEWISE Let, . We can write it as; We have to consider both moduli separately and it leads to following cases; When If then above four intervals translate to following with their corresponding inequality; When When When If then above four intervals translate […]
# Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 2 (P2-9709/02) | Year 2008 | May-Jun | (P2-9709/02) | Q#1
Hits: 195 Question Solve the inequality . Solution SOLVING INEQUALITY: PIECEWISE Let and , then; We have to consider two separate cases; When When We have the inequality; It can be written as; We have to consider two separate cases; When When Therefore the inequality will hold for ; SOLVING INEQUALITY: ALGEBRAICALLY Let, […]
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Welcome! On this website you can find information about each number.
# Number 324470102
## Number 324470102 basic info
Number 324470102 has 9 digits. Number 324470102 can be formatted as 324,470,102 or 324.470.102 or 324 470 102 or in case this was a phone number 324-470-102 or 32-447-0102 to be easier to read. Number 324470102 in English words is "three hundred and twenty-four million, four hundred and seventy thousand, one hundred and two". Number 324470102 can be read by triplets (groups of 3 digits) as "three hundred and twenty-four, four hundred and seventy, one hundred and two". Number 324470102 can be read digit by digit as "three two four four seven zero one zero two". Number 324470102 is even. Number 324470102 is divisible by: two. Number 324470102 is a composite number (non-prime number).
## Number 324470102 conversions
Number 324470102 in binary code is 10011010101110000010101010110. Number 324470102 in octal code is: 2325602526. Number 324470102 in hexadecimal (hexa): 13570556.
The sum of all digits of this number is 23. The digital root (repeated digital sum until you get single-digit number) is 5. Number 324470102 divided by two (halved) equals 162235051. Number 324470102 multiplied by two (doubled) equals 648940204. Number 324470102 multiplied by ten equals 3244701020. Number 324470102 raised to the power of 2 equals 1.0528084709189E+17. Number 324470102 raised to the power of 3 equals 3.4160487194552E+25. The square root (sqrt) of 324470102 is 18013.053655613. The sine (sin) of 324470102 degree is -0.84804809615145. The cosine (cos) of 324470102 degree is 0.52991926424117. The base-10 logarithm of 324470102 equals 8.5111746853132. The natural logarithm of 324470102 equals 19.59770395427. The number 324470102 can be encoded to characters as CBDDGJAJB. The number 324470102 can be encrypted to chemical element names as lithium, helium, beryllium, beryllium, nitrogen, neon, hydrogen, neon, helium.
## Numbers simmilar to 324470102
Numbers simmilar to number 324470102 (one digit altered): 224470102424470102314470102334470102323470102325470102324370102324570102324460102324480102324471102324470002324470202324470112324470101324470103
Possible variations of 324470102 with a digit pair swapped: 234470102342470102324740102324407102324471002324470012324470120
Number 324470102 typographic errors with one digit missing: 244701023447010232470102324701023244010232447102324470023244701232447010
Number 324470102 typographic errors with one digit doubled: 332447010232244701023244470102324447010232447701023244700102324470110232447010023244701022
Previous number: 324470101
Next number: 324470103
## Several randomly selected numbers:
3653040608249239288309026798110726591977892276347391197170174149628583294977369787120123440661527585614563528919245938552271071031188625981451165605380406978100139187497546276408284885969987492127374772623431394380417617756463275485219017102446536308731715294811849300828976968189991325104427452399463819271938632648705226414797109753682274365201937722691408137828316279279164177389.
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# How to hire a Calculus expert for assignment assistance?
How to hire a Calculus expert for assignment assistance? Calculate the following 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 (1) Find the first ten most common mistakes in Calculus. Write down each error of Calculus. 1. I have three types of problems with the third equation. You can find this rule, but you should write down the last ten errors of Calculus. 2. As you can see, you made exactly the same mistake, but when you do this, you are getting different kinds of errors of Calculus. Write down the next ten errors of Calculus. 3. These errors are “little” errors. If you only use “little” errors, you will get same error results as you tried. This is because when you use big errors, it will cause problems in your Calculus calculations. When you use small errors, you will get the same error results as you tried. 4. The last ten questions are very obvious: 1. I know most people believe I made the right mistake of the first nine numbers. But I can give you some steps by focusing on the first 10 mistakes I made. Just calculate each one of these 10-bit numbers from: 4 to 6, from 1 to 10. 2. At least once you find out what number you would expect to find yourself: 1.
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As you know, vegetables not only taste good, but they are good for you. Many vegetables are a great source of vitamin C. Vitamin C is a water-soluble antioxidant that plays an important role in protecting the body from infection and disease. Humans do not make vitamin C on their own, so we must get it from dietary sources. Potatoes, like the ones shown in Figure 1, below, are one good source of vitamin C. Does cooking them affect how much vitamin C they have? In other words, if you boil a… Read more
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Science Fair Project Idea
Vitamin C is a water-soluble vitamin that has many functions in the body. Vitamin C is needed to bolster the immune system. It is an antioxidant that protects LDL cholesterol from oxidative damage, and it is needed to make collagen, a substance that strengthens many parts of the body, such as muscles and blood vessels. Our bodies do not make vitamin C, so we must get it from dietary sources. Citrus fruits, carrots, avocados, and spinach all have vitamin C. Bell peppers, like the ones shown in… Read more
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Time Required Average (6-10 days) Prerequisites Material Availability Titration equipment and supplies are needed. A kit is available from the . Cost High (\$100 - \$150) Safety Adult supervision required. Iodine solution is poisonous. Avoid skin and eye contact. Wear chemical safety goggles and rubber gloves when handling the concentrated solution. For more tips, consult the
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# Help with a few questions
• Jan 22nd 2007, 07:05 PM
dbcavs23
Help with a few questions
I would really appreciate any help on these math questions. They are giving me tons of trouble
Simplify- and tell which conic section it is: 2x^2 + 4x +y^2 +10y = 25
Ellipse: x^2 +4y^2 + 8y + 6x = 23
Standard Equation:
Center:
Vertices
Co-certices
Foci
Circle: x^2 + 4y +y^2 = 12
Translated 6 units left and 4 units up
Standard Form:
Translated Form:
Thanks for any help,
Dan
• Jan 23rd 2007, 09:07 AM
CaptainBlack
Quote:
Originally Posted by dbcavs23
I would really appreciate any help on these math questions. They are giving me tons of trouble
Simplify- and tell which conic section it is: 2x^2 + 4x +y^2 +10y = 25
I don't know what you are looking for but:
$2x^2 + 4x +y^2 +10y = 25$
becomes:
$2(x^2 + 2x) +(y^2 +10y) = 25$
or:
$2(x + 1)^2 - 2 +(y + 5)^2 -25 = 25$
which simplifies to:
$(x + 1)^2 +\frac{(y + 5)^2}{2} = 26$
Which is an ellipse with centre at (-1,-5).
RonL
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https://create.roblox.com/docs/fr-fr/reference/engine/classes/IntValue
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# IntValue
Afficher les obsolètes
An IntValue stores a single signed 64-bit integer. The highest allowed value is 2^63-1 or around 9.2 quintillion (9.2^18); attempting to store larger numbers will cause integer overflow. The lowest allowed value is -2^63 or about -9.2 quintillion. Practically, however, working with integers larger than 2^53 (9.0^15) will cause loss of precision since Luau uses double-precision floating-point to store numbers.
Note that it's possible to store values between 2^53 and 2^63-1 via the Properties window since it uses strings to pass data to the engine, but manipulating large values via Luau scripts will result in loss of precision and rounding as mentioned above.
The main advantage of using IntValue lies in its rounding of values to the nearest integer, with halfway cases rounded away from 0. For values outside of this range, use a NumberValue instead. Like all ValueBase objects, this single value is stored in the Value property.
The Changed event for this (and other objects like it) will run with the new value being stored in the object, instead of a string representing the property being changed.
## Résumé
### Propriétés
• Lecture parallèle
Used to hold an integer.
## Propriétés
### Value
Lecture parallèle
Used to hold an integer.
## Évènements
### Changed
This event fires whenever the IntValue.Value is changed. It will run with the new value being stored in the argument object, instead of a string representing the property being changed.
Equivalent change events exist for similar objects such as NumberValue and StringValue, depending on what object type best suits the need.
#### Paramètres
value: number
The new value after the change.
#### Échantillons de code
How to Use IntValue.Changed
``````local value = Instance.new("IntValue")
value.Parent = workspace
local function onValueChanged(newValue)
print(newValue)
end
value.Changed:Connect(onValueChanged)
value.Value = 20``````
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http://www.formuladirectory.com/user/formula/89
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HOSTING A TOTAL OF 318 FORMULAS WITH CALCULATORS
## Electric Potential Due to a Point Charge
This formula calculates the value of the electric potential due to a point charge, as measured in volts. The values needed for the calculation of this value are the charge, in coulombs and distance in meters.
## $\frac{8.99*{10}^{9}*q}{r}$
The values needed for this formula are : q = charge and r = distance
ENTER THE VARIABLES TO BE USED IN THE FORMULA
Similar formulas which you may find interesting.
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https://askthetask.com/25/what-is-a-negative-coefficient-and-how-do-i-work-it-out
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0 like 0 dislike
what is a negative coefficient and how do i work it out
For example $$5^{-2}= \frac{1}{5^{2} }$$
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https://www.lmfdb.org/ModularForm/GL2/Q/holomorphic/3150/2/d/d/
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# Properties
Label 3150.2.d.d Level 3150 Weight 2 Character orbit 3150.d Analytic conductor 25.153 Analytic rank 0 Dimension 8 CM no Inner twists 2
# Related objects
## Newspace parameters
Level: $$N$$ = $$3150 = 2 \cdot 3^{2} \cdot 5^{2} \cdot 7$$ Weight: $$k$$ = $$2$$ Character orbit: $$[\chi]$$ = 3150.d (of order $$2$$, degree $$1$$, not minimal)
## Newform invariants
Self dual: no Analytic conductor: $$25.1528766367$$ Analytic rank: $$0$$ Dimension: $$8$$ Coefficient field: 8.0.7442857984.4 Coefficient ring: $$\Z[a_1, \ldots, a_{17}]$$ Coefficient ring index: $$2^{4}\cdot 3^{2}$$ Twist minimal: no (minimal twist has level 630) Sato-Tate group: $\mathrm{SU}(2)[C_{2}]$
## $q$-expansion
Coefficients of the $$q$$-expansion are expressed in terms of a basis $$1,\beta_1,\ldots,\beta_{7}$$ for the coefficient ring described below. We also show the integral $$q$$-expansion of the trace form.
$$f(q)$$ $$=$$ $$q + q^{2} + q^{4} -\beta_{2} q^{7} + q^{8} +O(q^{10})$$ $$q + q^{2} + q^{4} -\beta_{2} q^{7} + q^{8} -\beta_{3} q^{11} + ( -1 + \beta_{7} ) q^{13} -\beta_{2} q^{14} + q^{16} + ( \beta_{3} - \beta_{6} ) q^{17} + ( \beta_{1} - \beta_{2} + \beta_{5} + \beta_{6} ) q^{19} -\beta_{3} q^{22} + ( -1 - \beta_{1} - \beta_{2} + \beta_{7} ) q^{23} + ( -1 + \beta_{7} ) q^{26} -\beta_{2} q^{28} + ( \beta_{1} - \beta_{2} + \beta_{3} + \beta_{5} ) q^{29} + ( -\beta_{1} + \beta_{2} + \beta_{3} + 2 \beta_{5} ) q^{31} + q^{32} + ( \beta_{3} - \beta_{6} ) q^{34} + ( \beta_{1} - \beta_{2} - \beta_{3} + \beta_{6} ) q^{37} + ( \beta_{1} - \beta_{2} + \beta_{5} + \beta_{6} ) q^{38} + ( -1 + \beta_{7} ) q^{41} + ( -\beta_{3} + \beta_{5} + \beta_{6} ) q^{43} -\beta_{3} q^{44} + ( -1 - \beta_{1} - \beta_{2} + \beta_{7} ) q^{46} + ( \beta_{1} - \beta_{2} + 2 \beta_{3} - 2 \beta_{5} + \beta_{6} ) q^{47} + ( \beta_{1} + \beta_{3} - \beta_{4} + \beta_{5} - \beta_{6} ) q^{49} + ( -1 + \beta_{7} ) q^{52} + ( -1 + \beta_{1} + \beta_{2} + \beta_{7} ) q^{53} -\beta_{2} q^{56} + ( \beta_{1} - \beta_{2} + \beta_{3} + \beta_{5} ) q^{58} + ( 1 + \beta_{1} + \beta_{2} + \beta_{3} - 2 \beta_{4} + \beta_{5} + \beta_{7} ) q^{59} + ( \beta_{1} - \beta_{2} - 2 \beta_{5} - \beta_{6} ) q^{61} + ( -\beta_{1} + \beta_{2} + \beta_{3} + 2 \beta_{5} ) q^{62} + q^{64} + ( 2 \beta_{1} - 2 \beta_{2} - \beta_{3} + 3 \beta_{5} + \beta_{6} ) q^{67} + ( \beta_{3} - \beta_{6} ) q^{68} + ( \beta_{1} - \beta_{2} + \beta_{5} + \beta_{6} ) q^{71} + ( 6 - 2 \beta_{1} - 2 \beta_{2} ) q^{73} + ( \beta_{1} - \beta_{2} - \beta_{3} + \beta_{6} ) q^{74} + ( \beta_{1} - \beta_{2} + \beta_{5} + \beta_{6} ) q^{76} + ( -\beta_{2} + \beta_{3} - \beta_{4} + 4 \beta_{5} + \beta_{6} + \beta_{7} ) q^{77} + ( -1 - 3 \beta_{1} - 3 \beta_{2} - \beta_{7} ) q^{79} + ( -1 + \beta_{7} ) q^{82} + ( -2 \beta_{1} + 2 \beta_{2} - 2 \beta_{5} ) q^{83} + ( -\beta_{3} + \beta_{5} + \beta_{6} ) q^{86} -\beta_{3} q^{88} + ( 2 \beta_{1} + 2 \beta_{2} - \beta_{3} + 2 \beta_{4} - \beta_{5} - 2 \beta_{7} ) q^{89} + ( \beta_{2} + \beta_{3} - \beta_{4} - 3 \beta_{5} + \beta_{6} + \beta_{7} ) q^{91} + ( -1 - \beta_{1} - \beta_{2} + \beta_{7} ) q^{92} + ( \beta_{1} - \beta_{2} + 2 \beta_{3} - 2 \beta_{5} + \beta_{6} ) q^{94} + ( -3 - \beta_{1} - \beta_{2} + 3 \beta_{7} ) q^{97} + ( \beta_{1} + \beta_{3} - \beta_{4} + \beta_{5} - \beta_{6} ) q^{98} +O(q^{100})$$ $$\operatorname{Tr}(f)(q)$$ $$=$$ $$8q + 8q^{2} + 8q^{4} + 8q^{8} + O(q^{10})$$ $$8q + 8q^{2} + 8q^{4} + 8q^{8} - 8q^{13} + 8q^{16} - 8q^{23} - 8q^{26} + 8q^{32} - 8q^{41} - 8q^{46} - 4q^{49} - 8q^{52} - 8q^{53} + 8q^{64} + 48q^{73} - 4q^{77} - 8q^{79} - 8q^{82} + 8q^{89} - 4q^{91} - 8q^{92} - 24q^{97} - 4q^{98} + O(q^{100})$$
Basis of coefficient ring in terms of a root $$\nu$$ of $$x^{8} + 26 x^{6} + 205 x^{4} + 540 x^{2} + 324$$:
$$\beta_{0}$$ $$=$$ $$1$$ $$\beta_{1}$$ $$=$$ $$($$$$\nu^{7} + 26 \nu^{5} + 18 \nu^{4} + 223 \nu^{3} + 234 \nu^{2} + 774 \nu + 324$$$$)/216$$ $$\beta_{2}$$ $$=$$ $$($$$$-\nu^{7} - 26 \nu^{5} + 18 \nu^{4} - 223 \nu^{3} + 234 \nu^{2} - 774 \nu + 324$$$$)/216$$ $$\beta_{3}$$ $$=$$ $$($$$$\nu^{7} + 20 \nu^{5} + 73 \nu^{3} - 54 \nu$$$$)/72$$ $$\beta_{4}$$ $$=$$ $$($$$$-\nu^{7} - 26 \nu^{5} + 18 \nu^{4} - 223 \nu^{3} + 450 \nu^{2} - 558 \nu + 1836$$$$)/216$$ $$\beta_{5}$$ $$=$$ $$($$$$-5 \nu^{7} - 112 \nu^{5} - 665 \nu^{3} - 954 \nu$$$$)/216$$ $$\beta_{6}$$ $$=$$ $$($$$$\nu^{7} + 26 \nu^{5} + 187 \nu^{3} + 306 \nu$$$$)/36$$ $$\beta_{7}$$ $$=$$ $$($$$$\nu^{6} + 22 \nu^{4} + 123 \nu^{2} + 138$$$$)/12$$
$$1$$ $$=$$ $$\beta_0$$ $$\nu$$ $$=$$ $$($$$$\beta_{5} + \beta_{3} - \beta_{2} + \beta_{1}$$$$)/2$$ $$\nu^{2}$$ $$=$$ $$($$$$-\beta_{5} + 2 \beta_{4} - \beta_{3} - \beta_{2} - \beta_{1} - 14$$$$)/2$$ $$\nu^{3}$$ $$=$$ $$($$$$-2 \beta_{6} - 13 \beta_{5} - 13 \beta_{3} + 7 \beta_{2} - 7 \beta_{1}$$$$)/2$$ $$\nu^{4}$$ $$=$$ $$($$$$13 \beta_{5} - 26 \beta_{4} + 13 \beta_{3} + 25 \beta_{2} + 25 \beta_{1} + 146$$$$)/2$$ $$\nu^{5}$$ $$=$$ $$($$$$50 \beta_{6} + 187 \beta_{5} + 163 \beta_{3} - 73 \beta_{2} + 73 \beta_{1}$$$$)/2$$ $$\nu^{6}$$ $$=$$ $$($$$$24 \beta_{7} - 163 \beta_{5} + 326 \beta_{4} - 163 \beta_{3} - 427 \beta_{2} - 427 \beta_{1} - 1766$$$$)/2$$ $$\nu^{7}$$ $$=$$ $$($$$$-854 \beta_{6} - 2737 \beta_{5} - 2113 \beta_{3} + 895 \beta_{2} - 895 \beta_{1}$$$$)/2$$
## Character values
We give the values of $$\chi$$ on generators for $$\left(\mathbb{Z}/3150\mathbb{Z}\right)^\times$$.
$$n$$ $$127$$ $$451$$ $$2801$$ $$\chi(n)$$ $$-1$$ $$-1$$ $$-1$$
## Embeddings
For each embedding $$\iota_m$$ of the coefficient field, the values $$\iota_m(a_n)$$ are shown below.
For more information on an embedded modular form you can click on its label.
Label $$\iota_m(\nu)$$ $$a_{2}$$ $$a_{3}$$ $$a_{4}$$ $$a_{5}$$ $$a_{6}$$ $$a_{7}$$ $$a_{8}$$ $$a_{9}$$ $$a_{10}$$
3149.1
− 3.73923i 3.73923i − 0.916813i 0.916813i − 1.91681i 1.91681i − 2.73923i 2.73923i
1.00000 0 1.00000 0 0 −2.64404 0.0951965i 1.00000 0 0
3149.2 1.00000 0 1.00000 0 0 −2.64404 + 0.0951965i 1.00000 0 0
3149.3 1.00000 0 1.00000 0 0 −0.648285 2.56510i 1.00000 0 0
3149.4 1.00000 0 1.00000 0 0 −0.648285 + 2.56510i 1.00000 0 0
3149.5 1.00000 0 1.00000 0 0 1.35539 2.27220i 1.00000 0 0
3149.6 1.00000 0 1.00000 0 0 1.35539 + 2.27220i 1.00000 0 0
3149.7 1.00000 0 1.00000 0 0 1.93693 1.80230i 1.00000 0 0
3149.8 1.00000 0 1.00000 0 0 1.93693 + 1.80230i 1.00000 0 0
$$n$$: e.g. 2-40 or 990-1000 Embeddings: e.g. 1-3 or 3149.8 Significant digits: Format: Complex embeddings Normalized embeddings Satake parameters Satake angles
## Inner twists
Char Parity Ord Mult Type
1.a even 1 1 trivial
105.g even 2 1 inner
## Twists
By twisting character orbit
Char Parity Ord Mult Type Twist Min Dim
1.a even 1 1 trivial 3150.2.d.d 8
3.b odd 2 1 3150.2.d.a 8
5.b even 2 1 3150.2.d.c 8
5.c odd 4 1 630.2.b.a 8
5.c odd 4 1 3150.2.b.e 8
7.b odd 2 1 3150.2.d.f 8
15.d odd 2 1 3150.2.d.f 8
15.e even 4 1 630.2.b.b yes 8
15.e even 4 1 3150.2.b.f 8
20.e even 4 1 5040.2.f.f 8
21.c even 2 1 3150.2.d.c 8
35.c odd 2 1 3150.2.d.a 8
35.f even 4 1 630.2.b.b yes 8
35.f even 4 1 3150.2.b.f 8
60.l odd 4 1 5040.2.f.i 8
105.g even 2 1 inner 3150.2.d.d 8
105.k odd 4 1 630.2.b.a 8
105.k odd 4 1 3150.2.b.e 8
140.j odd 4 1 5040.2.f.i 8
420.w even 4 1 5040.2.f.f 8
By twisted newform orbit
Twist Min Dim Char Parity Ord Mult Type
630.2.b.a 8 5.c odd 4 1
630.2.b.a 8 105.k odd 4 1
630.2.b.b yes 8 15.e even 4 1
630.2.b.b yes 8 35.f even 4 1
3150.2.b.e 8 5.c odd 4 1
3150.2.b.e 8 105.k odd 4 1
3150.2.b.f 8 15.e even 4 1
3150.2.b.f 8 35.f even 4 1
3150.2.d.a 8 3.b odd 2 1
3150.2.d.a 8 35.c odd 2 1
3150.2.d.c 8 5.b even 2 1
3150.2.d.c 8 21.c even 2 1
3150.2.d.d 8 1.a even 1 1 trivial
3150.2.d.d 8 105.g even 2 1 inner
3150.2.d.f 8 7.b odd 2 1
3150.2.d.f 8 15.d odd 2 1
5040.2.f.f 8 20.e even 4 1
5040.2.f.f 8 420.w even 4 1
5040.2.f.i 8 60.l odd 4 1
5040.2.f.i 8 140.j odd 4 1
## Hecke kernels
This newform subspace can be constructed as the intersection of the kernels of the following linear operators acting on $$S_{2}^{\mathrm{new}}(3150, [\chi])$$:
$$T_{11}^{8} + 52 T_{11}^{6} + 820 T_{11}^{4} + 4320 T_{11}^{2} + 5184$$ $$T_{13}^{4} + 4 T_{13}^{3} - 22 T_{13}^{2} - 24 T_{13} + 72$$ $$T_{23}^{4} + 4 T_{23}^{3} - 44 T_{23}^{2} - 96 T_{23} + 288$$
## Hecke Characteristic Polynomials
$p$ $F_p(T)$
$2$ $$( 1 - T )^{8}$$
$3$ 1
$5$ 1
$7$ $$1 + 2 T^{2} + 24 T^{3} + 2 T^{4} + 168 T^{5} + 98 T^{6} + 2401 T^{8}$$
$11$ $$1 - 36 T^{2} + 776 T^{4} - 11916 T^{6} + 146094 T^{8} - 1441836 T^{10} + 11361416 T^{12} - 63776196 T^{14} + 214358881 T^{16}$$
$13$ $$( 1 + 4 T + 30 T^{2} + 132 T^{3} + 514 T^{4} + 1716 T^{5} + 5070 T^{6} + 8788 T^{7} + 28561 T^{8} )^{2}$$
$17$ $$1 - 36 T^{2} + 248 T^{4} + 5364 T^{6} - 133842 T^{8} + 1550196 T^{10} + 20713208 T^{12} - 868952484 T^{14} + 6975757441 T^{16}$$
$19$ $$1 - 44 T^{2} + 1256 T^{4} - 31140 T^{6} + 702382 T^{8} - 11241540 T^{10} + 163683176 T^{12} - 2070018764 T^{14} + 16983563041 T^{16}$$
$23$ $$( 1 + 4 T + 48 T^{2} + 180 T^{3} + 1438 T^{4} + 4140 T^{5} + 25392 T^{6} + 48668 T^{7} + 279841 T^{8} )^{2}$$
$29$ $$1 - 128 T^{2} + 8732 T^{4} - 399744 T^{6} + 13409894 T^{8} - 336184704 T^{10} + 6175977692 T^{12} - 76137385088 T^{14} + 500246412961 T^{16}$$
$31$ $$1 - 32 T^{2} + 572 T^{4} - 21984 T^{6} + 1650310 T^{8} - 21126624 T^{10} + 528254012 T^{12} - 28400117792 T^{14} + 852891037441 T^{16}$$
$37$ $$1 - 152 T^{2} + 10004 T^{4} - 394632 T^{6} + 13477382 T^{8} - 540251208 T^{10} + 18749106644 T^{12} - 389990414168 T^{14} + 3512479453921 T^{16}$$
$41$ $$( 1 + 4 T + 142 T^{2} + 468 T^{3} + 8354 T^{4} + 19188 T^{5} + 238702 T^{6} + 275684 T^{7} + 2825761 T^{8} )^{2}$$
$43$ $$1 - 228 T^{2} + 23256 T^{4} - 1476524 T^{6} + 70398222 T^{8} - 2730092876 T^{10} + 79507636056 T^{12} - 1441270775172 T^{14} + 11688200277601 T^{16}$$
$47$ $$1 + 68 T^{2} + 10040 T^{4} + 446124 T^{6} + 34404910 T^{8} + 985487916 T^{10} + 48991997240 T^{12} + 732986642372 T^{14} + 23811286661761 T^{16}$$
$53$ $$( 1 + 4 T + 160 T^{2} + 636 T^{3} + 11630 T^{4} + 33708 T^{5} + 449440 T^{6} + 595508 T^{7} + 7890481 T^{8} )^{2}$$
$59$ $$( 1 + 42 T^{2} - 312 T^{3} + 3106 T^{4} - 18408 T^{5} + 146202 T^{6} + 12117361 T^{8} )^{2}$$
$61$ $$1 - 236 T^{2} + 31208 T^{4} - 2804772 T^{6} + 194357422 T^{8} - 10436556612 T^{10} + 432101005928 T^{12} - 12158808349196 T^{14} + 191707312997281 T^{16}$$
$67$ $$1 - 180 T^{2} + 22904 T^{4} - 1953756 T^{6} + 148114446 T^{8} - 8770410684 T^{10} + 461541275384 T^{12} - 16282508790420 T^{14} + 406067677556641 T^{16}$$
$71$ $$1 - 460 T^{2} + 98600 T^{4} - 12828132 T^{6} + 1106047246 T^{8} - 64666613412 T^{10} + 2505591746600 T^{12} - 58926130603660 T^{14} + 645753531245761 T^{16}$$
$73$ $$( 1 - 24 T + 404 T^{2} - 4680 T^{3} + 45878 T^{4} - 341640 T^{5} + 2152916 T^{6} - 9336408 T^{7} + 28398241 T^{8} )^{2}$$
$79$ $$( 1 + 4 T + 48 T^{2} + 532 T^{3} + 9470 T^{4} + 42028 T^{5} + 299568 T^{6} + 1972156 T^{7} + 38950081 T^{8} )^{2}$$
$83$ $$1 - 424 T^{2} + 86492 T^{4} - 11335896 T^{6} + 1081357798 T^{8} - 78092987544 T^{10} + 4104765099932 T^{12} - 138622718308456 T^{14} + 2252292232139041 T^{16}$$
$89$ $$( 1 - 4 T + 118 T^{2} + 1308 T^{3} - 670 T^{4} + 116412 T^{5} + 934678 T^{6} - 2819876 T^{7} + 62742241 T^{8} )^{2}$$
$97$ $$( 1 + 12 T + 176 T^{2} + 2148 T^{3} + 29438 T^{4} + 208356 T^{5} + 1655984 T^{6} + 10952076 T^{7} + 88529281 T^{8} )^{2}$$
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## ››Convert kilometre to heer
kilometer heer
How many kilometer in 1 heer? The answer is 0.073152.
We assume you are converting between kilometre and heer.
You can view more details on each measurement unit:
kilometer or heer
The SI base unit for length is the metre.
1 metre is equal to 0.001 kilometer, or 0.013670166229221 heer.
Note that rounding errors may occur, so always check the results.
Use this page to learn how to convert between kilometres and heer.
Type in your own numbers in the form to convert the units!
## ››Quick conversion chart of kilometer to heer
1 kilometer to heer = 13.67017 heer
2 kilometer to heer = 27.34033 heer
3 kilometer to heer = 41.0105 heer
4 kilometer to heer = 54.68066 heer
5 kilometer to heer = 68.35083 heer
6 kilometer to heer = 82.021 heer
7 kilometer to heer = 95.69116 heer
8 kilometer to heer = 109.36133 heer
9 kilometer to heer = 123.0315 heer
10 kilometer to heer = 136.70166 heer
## ››Want other units?
You can do the reverse unit conversion from heer to kilometer, or enter any two units below:
## Enter two units to convert
From: To:
## ››Definition: Kilometer
A kilometre (American spelling: kilometer, symbol: km) is a unit of length equal to 1000 metres (from the Greek words khilia = thousand and metro = count/measure). It is approximately equal to 0.621 miles, 1094 yards or 3281 feet.
## ››Metric conversions and more
ConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 100 kg, US fluid ounce, 6'3", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more!
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From one of the premier authors in higher education comes a new linear algebra textbook that fosters thinking, problem-solving abilities, and exposure to real-world applications.
Without sacrificing mathematical precision, and Busby focus on the aspects of linear algebra that are most likely to have practical value to the while not compromising the intrinsic mathematical form of the subject. Throughout Contemporary Linear Algebra, students are encouraged to look at ideas and problems from multiple points of view.
CHAPTER 1 Vectors 1
1.1 Vectors and Matrices in and ; n-Space 1
1.2 Dot Product and Orthogonality 15
1.3 Vector Equations of Lines and Planes 29
CHAPTER 2 Systems of Linear Equations 39
2.1 Introduction to Systems of Linear Equations 39
2.2 Solving Linear Systems by Row Reduction 48
2.3 Applications of Linear Systems 63
CHAPTER 3 Matrices and Matrix Algebra 79
3.1 Operations on Matrices 79
3.2 Inverses; Algebraic Properties of Matrices 94
3.3 Elementary Matrices; A Method for Finding A−1 109
3.4 Subspaces and Linear Independence 123
3.5 The Geometry of Linear Systems 135
3.6 Matrices with Special Forms 143
3.7 Matrix Factorizations; LU-Decomposition 154
3.8 Partitioned Matrices and Parallel Processing 166
CHAPTER 4 175
4.1 Determinants; Cofactor Expansion 175
4.2 Properties of Determinants 184
4.3 Cramer’s Rule; Formula for A −1; Applications of Determinants 196
4.4 A First Look at Eigenvalues and Eigenvectors 210
CHAPTER 5 Matrix Models 225
5.1 Dynamical Systems and Markov Chains 225
5.2 Leontief Input-Output Models 235
5.3 Gauss–Seidel and Jacobi Iteration; Sparse Linear Systems 241
5.4 The Power Method; to Internet Search Engines 249
CHAPTER 6 Linear Transformations 265
6.1 Matrices as Transformations 265
6.2 Geometry of Linear Operators 280
6.3 Kernel and Range 296
6.4 Composition and Invertibility of Linear Transformations 305
6.5 Computer 318
CHAPTER 7 Dimension and Structure 329
7.1 Basis and Dimension 329
7.2 Properties of Bases 335
7.3 The Spaces of a Matrix 342
7.4 The Dimension Theorem and Its Implications 352
7.5 The Rank Theorem and Its Implications 360
7.6 The Pivot Theorem and Its Implications 370
7.7 The Projection Theorem and Its Implications 379
7.8 Best Approximation and Least Squares 393
7.9 Orthonormal Bases and the Gram–Schmidt Process 406
7.10 QR-Decomposition; Householder Transformations 417
7.11 Coordinates with Respect to a Basis 428
CHAPTER 8 Diagonalization 443
8.1 Matrix Representations of Linear Transformations 443
8.2 Similarity and Diagonalizability 456
8.3 Orthogonal Diagonalizability; of a Matrix 468
8.5 Application of Quadratic Forms to Optimization 495
8.6 Singular Value Decomposition 502
8.7 The Pseudoinverse 518
8.8 Complex Eigenvalues and Eigenvectors 525
8.9 Hermitian, Unitary, and Normal Matrices 535
8.10 Systems of Differential Equations 542
CHAPTER 9 General Vector Spaces 555
9.1 Vector Space Axioms 555
9.2 Inner Product Spaces; Fourier Series 569
9.3 General Linear Transformations; Isomorphism 582
APPENDIX A How to Read Theorems A1
APPENDIX B Complex Numbers A3
PHOTO CREDITS C1
INDEX I-1
Title: Contemporary Linear Algebra
Author: Howard Anton / Robert C. Busby
Edition: 1st Edition
ISBN: 978-0471450016
Type: Solution Manual
Language: English
Linear Algebra
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Seconds to Weeks conversion
Enter the time in seconds and press the Convert button:
How to convert Seconds to Weeks:1 second = 1/604800 week = 1.65344 * 10-6 weekor1 week = 604800 secondsSeconds to Weeks formulaThe time t in months is equal to the time t in seconds divided by 604800:t(w) = t(s) / 604800ort(s) = t(w) * 604800
Example:Convert 7257600 seconds to weeks:t(w) = 7257600(s) / 604800 = 12(w)
Seconds to Weeks conversion table
Seconds (s) Weeks (w)
1 s 0.0000016534 w
2 s 0.0000033069 w
3 s 0.0000049603 w
4 s 0.0000066138 w
5 s 0.0000082672 w
6 s 0.0000099206 w
7 s 0.000011574 w
8 s 0.000013228 w
9 s 0.000014881 w
10 s 0.000016534 w
100 s 0.000165 w
1000 s 0.001653 w
10000 s 0.016534 w
100000 s 0.165344 w
1000000 s 1.6534 w
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# RsLogix 500 Training - Counters - CTU and CTD
## In this lesson you will learn about the CTU and CTD counter instructions.
Rung 0
Each false to true transition of the CTU will increment C5:0.ACC by "1".
-CU-If the CTU is false, it will write a "0" to the CU (Count Up Enable) bit. If the CTU is true, it will write a "1" to the CU (Count Up Enable) bit.
-CD-The CTU does not write to the CD (Count Down Enable) bit.
-DN-If the ACC (Accumulated) value is less than the PRE (Preset) then the CTU will write a "0" to the DN (Done) bit. If the ACC (Accumulated) value is greater than or equal to the PRE (Preset) then the CTU will write a "1" to the DN (Done) bit.
-OV-If when the CTU is incrementing, it "rolls over" (goes from +32767 to -32768), it will write a "1" to the OV (Count Up Overflow) bit.
-UN-The CTU instruction does no write to the UN (Count Down Underflow) bit.
-UA-The UA (Update Accumulator) is only used by the HSC (High Speed Counter) of the Fixed SLC Controller.
Rung 1
If the CTU is false, it will write a "0" to the CU (Count Up Enable) bit. If the CTU is true, it will write a "1" to the CU (Count Up Enable) bit.
Rung 2
The CTU does not write to the CD (Count Down Enable) bit.
Rung 3
If the ACC (Accumulated) value is less than the PRE (Preset) then the CTU will write a "0" to the DN (Done) bit.
If the ACC (Accumulated) value is greater than or equal to the PRE (Preset) then the CTU will write a "1" to the DN (Done) bit.
To see this toggle the B3:0/0 bit until the ACC is equal to the PRE.
Rung 4
If when the CTU is incrementing, it "rolls over" (goes from +32767 to -32768), it will write a "1" to the OV (Count Up Overflow) bit.
To see this set the ACC (Accumulated) value to "32767" then toggle B3:0/0 allowing it to "rollover" to -32768.
Rung 5
The CTU instruction does no write to the UN (Count Down Underflow) bit.
Rung 6
The UA (Update Accumulator) is only used by the HSC (High Speed Counter) of the Fixed SLC Controller.
Rung 7
Each false to true transition of the CTD will decrement C5:1.ACC by "1".
-CU-The CTD does not write to the CU (Count Up Enable) bit.
-CD-If the CTD is false, it will write a "0" to the CD (Count Down Enable) bit. If the CTD is true, it will write a "1" to the CD (Count Down Enable) bit.
-DN-If the ACC (Accumulated) value is less than the PRE (Preset) then the CTD will write a "0" to the DN (Done) bit. If the ACC (Accumulated) value is greater than or equal to the PRE (Preset) then the CTD will write a "1" to the DN (Done) bit.
-OV-The CTD instruction does no write to the OV (Count Up Overflow) bit.
-UN-If when the CTD is decrementing, it "rolls over" (goes from -32768 to +32767), it will write a "1" to the UN (Count Down Underflow) bit.
-UA-The UA (Update Accumulator) is only used by the HSC (High Speed Counter) of the Fixed SLC Controller.
Rung 8
The CTD does not write to the CU (Count Up Enable) bit.
Rung 9
If the CTD is false, it will write a "0" to the CD (Count Down Enable) bit. If the CTD is true, it will write a "1" to the CD (Count Down Enable) bit.
Rung 10
If the ACC (Accumulated) value is less than the PRE (Preset) then the CTD will write a "0" to the DN (Done) bit.
If the ACC (Accumulated) value is greater than or equal to the PRE (Preset) then the CTD will write a "1" to the DN (Done) bit.
THIS IS IMPORTANT! Even though the CTD counts down, its DN bit operates the same as the CTU.
Rung 11
The CTD instruction does no write to the UV (Count Up Overflow) bit.
Rung 12
If when the CTD is decrementing, it "rolls over" (goes from -32768 to +32767), it will write a "1" to the ON (Count Down Underflow) bit.
To see this set the ACC (Accumulated) value to "-32768" then toggle B3:0/1 allowing it to "rollover" to +32767.
Rung 13
The UA (Update Accumulator) is only used by the HSC (High Speed Counter) of the Fixed SLC Controller.
Rung 14
The next two rungs show that two different counter instructions can address the same address (C5:2). There is nothing "illegal" about this. A common example of a use of this would be a part counter using a CTU instruction and there being a CTD for rejected parts.
Toggling B3:0/2 will increment C5:0.ACC and B3:0/3 will decrement the same C5:0.ACC.
Rung 15
Rung 16
You can do more with counters than just look at the status bits. You can compare their ACC (Accumulated) values as well.
The next two rungs show examples of this.
Rung 17
Next you will do some basic PLC fault understanding in the RsLogix 500 - The Big RED Fault - Finding, Understanding, and Clearing lesson.
## Next Steps
Go to the Allen Bradley RsLogix 500 PLC Training PLC Training Getting Started Lesson series to select your next lesson. There are also many other Lesson Series on PLC Programming and Industrial Automation.
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Home » How Many Times Can 30 Go Into 100? New Update
# How Many Times Can 30 Go Into 100? New Update
Let’s discuss the question: how many times can 30 go into 100. We summarize all relevant answers in section Q&A of website Countrymusicstop.com in category: MMO. See more related questions in the comments below.
## What will go into 100?
The factors of 100 are 100, 50, 25, 20, 10, 5, 4, 2, and 1.
## How many times does 30 go in to 120?
Ans . 4 Times does 30 go into 120.
### Dividing by Decimals 2
Dividing by Decimals 2
Dividing by Decimals 2
## How many times does 15% go into 100%?
100 divided by 15 is equal to 6 with a remainder of 10. You can also write this as the fraction 6 2/3. When you divide 100 by 15, you are essentially…
## What number can go into 100 4 times?
4 times 25 equals 100. This can also be expressed as 4 x 25 = 100. The best way to figure out what 4 x ‘unknown’ = 100 is, is to divide 100 by 4….
## How many 2s are in a 100?
Digit 2 appears 20 times in first 100 natural numbers.
## What can go into 30?
Frequently Asked Questions on Factors of 30
The factors of 30 are 1, 2, 3, 5, 6, 10, 15 and 30.
## How many times can 16 reach 100?
100 divided by 16 is 6 1/4, or 6.25.
## How do you solve 11 divided by 100?
Extra calculations for you
Using a calculator, if you typed in 100 divided by 11, you’d get 9.0909.
## How do you work out 60 divided by 100?
60 divided by 100 is 3/5, an answer that can also be expressed in decimal form as 0.6.
## How many sevens is 100?
Answer. Just count the 7s now, so 20.
### How To Butterfly Click 20 CPS
How To Butterfly Click 20 CPS
How To Butterfly Click 20 CPS
## How many 5’s are there in 100?
5, 15, 25, … , 95: There are ten 5s at the unit’s place. 50, 51, …, 59: There are ten 5s at the ten’s place. So, the total number of times digit 5 appears between 1 to 100 = 20 times. Was this answer helpful?
## What times does 6 give you 100?
16 2/3 times 6 equals 100.
## How many times can 9 Enter 100?
There are total 20 nine’s between 1 and 100. 9, 19, 29, 39, 49, 59, 69, 79, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99. you can count total 9 in these series.
## How do you do 100 divided by 2?
Using a calculator, if you typed in 100 divided by 2, you’d get 50. You could also express 100/2 as a mixed fraction: 50 0/2.
## How many times does alphabet A appear from 0 100?
If we write number from 0 to 100 then alphabet “a” will not appear because our fingers will not touch on this alphabet and hence we can say that answer of this question is Zero “A”. When you start writing 0 to 100 then zero will not appear becasue it is a number form not a alphabet form.
## What is the multiple of 30?
Multiples of 30: 30, 60, 90, 120, 150, 180, 210, 240, 270, 300, 330… Common multiples of 25 and 30 include 150 and 300.
## What can go into 33?
There are 4 factors of 33, which are 1, 3, 11, and 33.
## How do you factor 30?
Step 5: All these numbers when multiplied make up the factors of 30. That is 30 = 1 × 30, 2 × 15, 3 × 10 and 5 × 6.
### How this trader turned \$200 into \$190,000 in 4 hours
How this trader turned \$200 into \$190,000 in 4 hours
How this trader turned \$200 into \$190,000 in 4 hours
## How many 16s are there in 80?
There are 5 times 16 in 80. The answer you divide 80 by 16 which would get you 5.
## How many multiples of 8 are there in the numbers from 1 to 100?
What are the multiples of 8 through 100? The multiples of 8 until 100 are 8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96.
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0
# Why do I have to force my car into gear?
Updated: 4/28/2022
Wiki User
13y ago
Could depend on the age of the vehicle, but most likely a problem with your clutch, but it's more likely that it's your transmission, have everything checked out to make sure everything's working properly.
Wiki User
13y ago
Earn +20 pts
Q: Why do I have to force my car into gear?
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Still have questions?
Related questions
### What is the function of a gear?
a gear is used to transport force.
### What keeps the car in gear?
A car is in gear when the clutch is engaged on the transmission
### Why is my car slipping out of 1st gear?
my car is not working in 1st gear
### How force is changed when machines increases force?
It depends on the working principle of the machine. Like for example in a gear mechanism if you replace the bigger gear with an even bigger one then a smaller amount of force would rotate the smaller gear in the same angular velocity as the initial gear did but with a greater force. But now if you increases the force in the bigger gear the smaller gear would tend to rotate more fast.
### When machines increases force how is force changed?
It depends on the working principle of the machine. Like for example in a gear mechanism if you replace the bigger gear with an even bigger one then a smaller amount of force would rotate the smaller gear in the same angular velocity as the initial gear did but with a greater force. But now if you increases the force in the bigger gear the smaller gear would tend to rotate more fast.
### Is gear a force multiplier or a speed multiplier?
Both. A small driving gear and a large driven gear is a force multiplier. Whilst a large driving gear and a small driven gear is a speed multiplier
### Can you give me a sentence for gear?
the gear is on the car
### Why doesn't car go into gear at proper speeds?
on a 2000 intrigue; car does not shift into 3rd gear or 4th gear. 1st 2nd gear ok
absolutely!
### If the car starts but when you put it in gear it cuts off?
car starts but when i put it in gear it stop
### If you're in neutral at an intersection is it ok to skip first gear and go straight to second when the light changes?
If you are sitting at a light and you start in second, it won't usually stall your car. However it is not a good idea to always do this because if this is how you always shift, your transmission will not last very long. When you do shift back to first gear, be sure that your car has stopped completely-it's not good to force the car into first gear.
### When do you use drive in a standard transmission car?
There is no drive in a standard transmission car. In most cases you will have reverse, neutral, 1st gear, 2nd gear, 3rd gear, 4th gear, and 5th gear.
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# Worksheet: Subsets and Venn Diagrams
Q1:
Use or to fill in the gap: The Venn diagram shows that .
• A
• B
Q2:
Use or to fill in the gap: The Venn diagram shows that .
• A
• B
Q3:
Use or to fill in the gap: The Venn diagram shows that .
• A
• B
Q4:
If , , and are all subsets of a universal set , which Venn diagram represents these sets?
• A(b)
• B(a)
• C(c)
• D(d)
Q5:
Use , , or to fill in the gap: The Venn Diagram shows .
• A
• B
• C
• D
Q6:
Use , , or to fill in the gap: The Venn Diagram shows .
• A
• B
• C
• D
Q7:
Use or to fill in the gap: The Venn diagram shows that .
• A
• B
Q8:
Using the given Venn diagram, complete the following with or : .
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Q9:
Using the given Venn diagram, complete the following with or : .
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Q10:
Use or to fill in the gap: The Venn Diagram shows .
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Q11:
Use or to fill in the gap: The Venn Diagram shows .
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Q12:
Use or to fill in the gap: The Venn Diagram shows .
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Q13:
Use or to fill in the gap: The Venn Diagram shows .
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Q14:
Use or to fill in the gap: The Venn Diagram shows .
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Q15:
Use or to fill in the gap: The Venn Diagram shows .
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• B
Q16:
Use or to fill in the gap: The Venn Diagram shows .
• A
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Q17:
Use or to fill in the gap: The Venn Diagram shows .
• A
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1
Hi all!
I am putting some time in improving my code and make my tiny game smooth and decently playable.
The result I obtained is really far from this (bugs everywhere, the game is slow and boring, it accelerates and slows down without any apparent reason).
Clearly there is a lot to do and a long path to walk but I'd love if you guys can share any suggestion like changes I need to implement, maths / physics / trigonometry I need to study.
I am sorry if this question is very broad but I'd be incredibly happy to get even the tiniest hint.
This is my current code:
https://github.com/ltpitt/lua-pico-8-pong
A version of the game can be found online:
http://www.davidenastri.it/pong
Thanks for your time and patience :)
P#39138 2017-04-06 16:06 ( Edited 2017-04-09 16:55)
Hi!
Why don't you just post a cart on BBS so everyone can play and check the code without going anywhere?
P#39141 2017-04-07 04:19 ( Edited 2017-04-07 08:19)
You're doing it the right way, I mean, recognize the boringness of your game and asking for help, and only for that, I will spend some time to give you my feedback.
I think, if the game is boring, it's because of the slow speed of the ball. So first, I suggest increasing it. Also, every short amount of hit with any pad, you need to increase it again, in a way that the player will notice it, plus a special sound to tell the players.
Second cause, you decide to implement a blockout-style control: the direction the ball will take after a hit with a pad is relative to the distance of the ball with the center of the pad. The player and the CPU can throw the ball to each other in a straight line, and can never lose. Remove the feature and do it like pong. 45° forever.
There is also a problem with your actual implementation of ball speed. The ball is faster if it goes n diagonal than if it goes in a straight line. If you want to keep the blockout-style control you have to fix that, using two information: a number for the speed, and a normalized vector for the direction.
Third reason. The CPU always try to catch the ball using the center of its pad, making the ball it hit, always goes in a straight line. I can write an entire post about how to code the CPU to make it more fun. Maybe later if you ask for it.
Fourth, if you increase the ball of the speed you will notice that you can't catch the ball anymore because of the slow speed of the pad. Guess what you have to increase also the pad speed.
Now, because the ball will gain speed faster it should be harder to catch it after a couple of seconds. The player has no other choice to guess where the ball will hit when it will return to predict the exact position where he should move his pad to catch it again. Right? (make sure you understand this). But, because the ball will always go 45°, it is possible now. And that's part of the game-play. That's making the game fun, like pong. Newbies players will chase the ball and good players will predict the next hit.
Regarding the pad control, there is tones of way to create a good feeling. I will explain a simple solution, easy to implement and good enough for a pico-8 game. Use a variable for the pad acceleration. Increase it by one every frame when player press the down button. Combine it with a max value, something like 3 or 4. And add this value to the position of the pad. (y axis). When the down key is not pressed anymore, decrease the acceleration until it reachs zero. Do the same thing for the up key. Test it, it should have a weirld effect if player rapidly switch between up and down key. To fix that, you can invert the sign of the acceleration value when, e.g.: if down key and acceleration lesser than zero.
To finish :
• the green color is ugly :p
• the sfx when a ball is missed should not sound like success, but failure.
• pad and ball are too big.
• win message burn my eye.
• the game need a lot of polishes.
Have fun.
P#39144 2017-04-07 07:06 ( Edited 2017-04-07 11:06)
To be fair, your title music is pretty solid. Why not keep it around during the actual game?
You could even increase the speed of the music as the ball gains speed. Would require some poke()'ing into the RAM though, I guess. Has anyone ever tried this?
P#39160 2017-04-07 16:19 ( Edited 2017-04-07 20:19)
@TriBar
Interesting indeed!
How can I push my cart to the bbs?
I will check pico-8's manual.
@moechofe
Wow, thanks for such a long and detailed feedback!
Love it!
Are you sure that pong just does 45 degrees forever?
All the implementation I tried allow the player to change the degrees depending on position / speed...
"There is also a problem with your actual implementation of ball speed. The ball is faster if it goes n diagonal than if it goes in a straight line. If you want to keep the blockout-style control you have to fix that, using two information: a number for the speed, and a normalized vector for the direction."
This is not really clear to me...
Would you mind explaining with a small example?
"Third reason. The CPU always try to catch the ball using the center of its pad, making the ball it hit, always goes in a straight line. I can write an entire post about how to code the CPU to make it more fun. Maybe later if you ask for it."
I would be VERY happy to read such a post (and I think I would not be the only one) :)
Other suggestions are more or less clear, really thanks for your time and patience.
@gyfe
I'll find something to play during the game, good idea...
Awesome quality of my music is just thanks to Gruber's genius:
https://www.lexaloffle.com/bbs/?pid=38646#p38646
P#39180 2017-04-08 12:51 ( Edited 2017-04-08 16:51)
ltpitt: I have been thinking about vector movement recently, so I made a demo cart that shows what (I think) moechofe means about using a number for the speed and a vector for the direction.
https://www.lexaloffle.com/bbs/?tid=29127
Basically, your ball is moving in a certain direction (the angle) at a certain speed. If you store these values in their own variables, then you can change them very easily. When your ball actually moves, you can use sin() and cos() to calculate the x and y distance using the angle and speed.
P#39429 2017-04-08 17:26 ( Edited 2017-04-08 21:27)
Simply WOW, LRP!
Amazing implementation.
I will study your code ASAP :)
Really thanks, great share.
P#39450 2017-04-09 12:55 ( Edited 2017-04-09 16:55)
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# Quillen–Suslin theorem
Quillen–Suslin theorem "Serre's problem" redireciona aqui. Para outros usos, see Serre's conjecture (desambiguação). Quillen–Suslin theorem Field Commutative algebra Conjectured by Jean-Pierre Serre Conjectured in 1955 First proof by Daniel Quillen Andrei Suslin First proof in 1976 The Quillen–Suslin theorem, also known as Serre's problem or Serre's conjecture, is a theorem in commutative algebra concerning the relationship between free modules and projective modules over polynomial rings. In the geometric setting it is a statement about the triviality of vector bundles on affine space.
The theorem states that every finitely generated projective module over a polynomial ring is free.
Conteúdo 1 História 1.1 Fundo 2 Generalização 3 Notas 4 References History Background Geometrically, finitely generated projective modules over the ring {estilo de exibição R[x_{1},pontos ,x_{n}]} correspond to vector bundles over affine space {estilo de exibição mathbb {UMA} _{R}^{n}} , where free modules correspond to trivial vector bundles. This correspondence (from modules to (algebraic) vector bundles) is given by the 'globalisation' or 'twiddlification' functor, sending {displaystyle Mto {widetilde {M}}} (cite Hartshorne II.5, página 110). Affine space is topologically contractible, so it admits no non-trivial topological vector bundles. A simple argument using the exponential exact sequence and the d-bar Poincaré lemma shows that it also admits no non-trivial holomorphic vector bundles.
Jean-Pierre Serre, in his 1955 paper Faisceaux algébriques cohérents, remarked that the corresponding question was not known for algebraic vector bundles: "It is not known whether there exist projective A-modules of finite type which are not free."[1] Aqui {estilo de exibição A} is a polynomial ring over a field, isso é, {estilo de exibição A} = {estilo de exibição k[x_{1},pontos ,x_{n}]} .
To Serre's dismay, this problem quickly became known as Serre's conjecture. (Serre wrote, "I objected as often as I could [to the name]."[2]) The statement does not immediately follow from the proofs given in the topological or holomorphic case. These cases only guarantee that there is a continuous or holomorphic trivialization, not an algebraic trivialization.
Serre made some progress towards a solution in 1957 when he proved that every finitely generated projective module over a polynomial ring over a field was stably free, meaning that after forming its direct sum with a finitely generated free module, it became free. The problem remained open until 1976, when Daniel Quillen and Andrei Suslin independently proved the result. Quillen was awarded the Fields Medal in 1978 in part for his proof of the Serre conjecture. Leonid Vaseršteĭn later gave a simpler and much shorter proof of the theorem which can be found in Serge Lang's Algebra.
Generalization A generalization relating projective modules over regular Noetherian rings A and their polynomial rings is known as the Bass–Quillen conjecture.
Note that although {displaystyle GL_{n}} -bundles on affine space are all trivial, this is not true for G-bundles where G is a general reductive algebraic group.
Notes ^ "On ignore s'il existe des A-modules projectifs de type fini qui ne soient pas libres." Apertado, FAC, p. 243. ^ Lam, p. 1 References Serre, Jean Pierre (Marchar 1955), "Feixes algébricos coerentes", Anais da Matemática, Segunda Série, 61 (2): 197-278, doi:10.2307/1969915, JSTOR 1969915, SENHOR 0068874 Apertado, Jean Pierre (1958), "Modules projectifs et espaces fibrés à fibre vectorielle", Séminaire P. Dubreil, M.-L. Dubreil-Jacotin et C. Pisot, 1957/58, Fasc. 2, Exposé 23 (em francês), SENHOR 0177011 Quillen, Daniel (1976), "Projective modules over polynomial rings", Descobertas matemáticas, 36 (1): 167-171, doi:10.1007/BF01390008, SENHOR 0427303 Suslin, Andrei A. (1976), Проективные модули над кольцами многочленов свободны [Projective modules over polynomial rings are free], Doklady Akademii Nauk SSSR (em russo), 229 (5): 1063–1066, SENHOR 0469905. Translated in "Projective modules over polynomial rings are free", Soviet Mathematics, 17 (4): 1160–1164, 1976. Lang, Sarja (2002), Álgebra, Textos de Graduação em Matemática, volume. 211 (Revised third ed.), Nova york: Springer-Verlag, ISBN 978-0-387-95385-4, SENHOR 1878556 An account of this topic is provided by: Lam, T. S. (2006), Serre's problem on projective modules, Monografias Springer em Matemática, Berlim; Nova york: Springer Science+Business Media, pp. 300pp., ISBN 978-3-540-23317-6, SENHOR 2235330 Categorias: Commutative algebraTheorems in abstract algebra
Se você quiser conhecer outros artigos semelhantes a Quillen–Suslin theorem você pode visitar a categoria Commutative algebra.
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You are Here: Home >< Oxbridge
# The Big TSA 2009 Thread watch
1. the CIE 2009 specimen paper 1 is worth looking at if you havent already for anyone uneasy about the multiple choice - the mark scheme actually explains how they to get to the right answer..here is the link: http://www.cie.org.uk/qualifications...def_id=765_804
2. [QUOTE=*shooting*star*]can anyone help me with this question (question 5 from the CIE May/June 2008 paper):
what CIE may/june 2008 paper??
3. Merk that Sike of a Mike- thank you lovie! Do you know how to work out the silage question?? Once I see the algebra/workings for a question I understand it, but I just don't ever think about that method when trying to answer it!!
4. p.s- you hertford applicants....did you guys by any chance make an open application :P also...has anyone received emails from two colleges?
5. (Original post by *shooting*star*)
are you a hertford E&M applicant too?! i guess so then lol
Yup. E&M
6. (Original post by amy123123)
p.s- you hertford applicants....did you guys by any chance make an open application :P also...has anyone received emails from two colleges?
I applied direct to Hertford and yes, I have received emails detailing general information about the application process for E&M incl. provisional interview dates and another just acknowledging my application.
7. Coolio! Has anyone got the mark scheme for the CIE Nov 2006 papers?
8. (Original post by *shooting*star*)
can anyone help me with this question (question 5 from the CIE May/June 2008 paper):
"In a certain family, the father gives out pocket money to his children in the following way. Every month he chooses a number between 1 and 100, and then shares that amount of money equally between the seven children, giving then whole-number sums of money, and giving anything left over to charity.
For instance, if he chose the number 30, he would give \$4 to each child (\$4 x 7 = \$28) and \$2 to charity.
If the father realises after 100 months that every number between 1 and 100 (inclusive) has been used once, how much has he given to charity?
A \$276
B \$294
C \$295
D \$297
E \$300"
the correct answer is D but i am not sure how they get that! thanks
Are u sure it's D? I get A:
alright it goes like this:
100-98 (he gets 2+1+0), 97-91 (he gets 6+5+4+3+2+1+0), 91-84 (gets 6+5+4+3+2+1+0) and so on u get the idea.
If u take the lower boundaries that makes a total of 13 (91, 84, ..., 7).
Each time u get 6+5+4+3+2+1+0=21, so 13*21= 273.
Then u add 3 (for no. 100-98) so u get 276.
edit:
alright just realized my mistake. 0 is the last lower boundary so it's another 21 which makes 297
9. (Original post by Buffyboy)
Btw take all of my opinions with a pinch of salt, I don't really know what iI'm doing! But its helpful for me too, to assess your essay, so heregoes;
Spoiler:
Show
"Freedom of speech doesn't mean that you can say anything you want"
The freedom of speech is a key part of any modern constitution. To be precise it is generally defined (I laughed geekily at this!! Cannot generally precisily define.) as the freedom of opinion rather than just the freedom to engage in the act of speaking which does not necessarily mean that opinions are stated. (I'd say this was good as you open with define the terms of the question. Personally I might briefly outline my line or argument - but perhaps not due to time constraints, I will however definatly have trigger sentences at the start of each paragraph to show what I am arguing - just makes things seem clearer)
This freedom is based on the concept that to begin with every opinion is as valid as any other and should initially always be allowed to be expressed. It is a fact that opinions on different issues often vary greatly. I would perhaps argue a bit more for the merits of free speech before showing cases it is limited, be careful to show that you haven't assumed free speech is desirable, and if you have assumed it, say you have assumed it. Whilst most differences in opinion on factual matters can usually be resolved by discussion there are certain cases in which there is a call for an opinion to be forbidden. When this claim is made the other parts of a constitution upon which the freedom of speech rests can be consulted.
As the freedom of speech is initially considered as an absolute, as described above, all other basic rights defined in the constitution must be as well. If a person using his freedom of speech to hurt another person's basic rights as defined in the constitution then the offender may be sanctioned on the basis of acting against what is defined as an absolute in the constitution. I like this, I'd say "rights" rather than "absolute rights", yes a lot of constitutions seem to embrace deontological rights but I'd still stick with "rights". Perhaps define which rights rank above freedom of speech, your essay is perhaps too vauge on this... Personally I'd take a Millian route, that freedom of speech is justified always unless it causes harm to others. Then I'd argue what circumstances constitute harm (Is offence harm? If I indirectly cause harm is that a problem?)
Cases such as the Mohammad caricatures which provoked a large part of the muslim world in 2007 illustrate how this definition can and should be applied. Really good example, I probably couldn't think of such a relevant one. The author of the caricatures claimed to be exercising his freedom of speech. One might suspect that he was hurting muslim believers' right to exercise their religion but upon closer consideration one must conclude that he was not trying to inhibit anyone's religious beliefs but was expressing his opinion on a certain matter. This is because there is a difference between trying to take direct influence on a matter and expressing one's opinion. This sentence is clumbsy, its not agaisnt freedom of speech (or your definition of it!) to take direct influnce on a matter, I may directly influnce you to do your homework but the state wouldn't declare my actions illegal. Would be better if it more clearly explained how freedom of speech isn't a violation of religious rights...
Freedom of speech consequently does not mean the right to say anything one wants. It can be marked by the condition that it may not be abused to infringe another fundamental right accepted in our society. Good clear conclusion/defintion (Again I'd have just prefered a more clear discussion of which fundamental rights are infringed, and which are unacceptable to infringe (unacceptable to infringe someones right to health, however perhaps acceptable to infringe someones right to practice religion unchallenged if rights to freedom of speech clash with it... (if that makes sense...)
This can be extended to define a way of dealing with laws that suppress free speech. From the standpoint that the fundamental rights in the constitution of the western world are correct, laws that go against such a right can not legitimately suppress it. Laws suppressing the freedom of speech such as exist many countries such as China are hence illigitimate.
I think it was a good essay. Very eloquent, which is something i never manage in timed conditions, had examples and definitions - *tick, tick*. Perhaps could be slightly more clear... Plan it a little better maybe...
But yeah seems good to me! Anyone else please criticise what I've said.
My essay plan for this would be;
Defintion of "freedom of speech"
Maybe a brief statement of why freedom of speech is important... (However I'm really unsure on this, someone please give opinions, Its not explicitly asked for in question - so probably not in fact...)
"Freedom of speech very important, as Mill argued for individuality, to challenge existing views, if the incorrect view is tolerated it forces the defendants of the correct view to evaluate their own positions, furthermore humanity is not infallible and cannot assume the truth of its arguments. (Example). However in some cases freedom of speech may other rights, most pertinetly in the case of harm. "
Assess harm principle. Show cases it is relevant etc...
Conclude.
My plans pretty much what you did anyway (only probably worse!)... One thing I would say is just focus on paragraph clarity and header sentences.
Your right: trying to be precise on something generally being defined as X doesn't really fit. Writing "is widely" defined would probably have been better.
Right indeed on requesting a more precise description of why freedom of speech is actually so important...but here's the time dilemma again
Nevertheless I would surely have had enough time to add an example of a right that might be infringed by someone expressing his opinion e.g. by questioning someone's human dignity.
Anyway thanks for the good ideas. Think it's quite helpful discussing these issues here and also reading what the criticism other people's work recieves.
Cheers
EDIT: I must find how to hide something here...Using more than a page here
10. (Original post by Stashup)
Your right: trying to be precise on something generally being defined as X doesn't really fit. Writing "is widely" defined would probably have been better.
Right indeed on requesting a more precise description of why freedom of speech is actually so important...but here's the time dilemma again
Nevertheless I would surely have had enough time to add an example of a right that might be infringed by someone expressing his opinion e.g. by questioning someone's human dignity.
Anyway thanks for the good ideas. Think it's quite helpful discussing these issues here and also reading what the criticism other people's work recieves.
Cheers
EDIT: I must find how to hide something here...Using more than a page here
lol put write SPOILER with square brackets around it [ ] and then /SPOILER at the end with squared brackets around it.
Quote this post if to see
Spoiler:
Show
11. [QUOTE=matija_v]5 more days!! just want to get over it..
ok, first here are my answers to the 2007 paper
Spoiler:
Show
1 B
2 B
3 A
4 C
5 C
6 A
7 B
8 D
9 D
10 C
11 C
12 D
13 A
14 D
15 B
16 E
17 Pig=5 Dog=12 Cat=7 Horse=9
18 C
19 A
20 B
22 A
23 B
24 A oldest, c youngest
25 B
26 E
27 E
28 A
29 B
30 B
31 D
32 C
33 C
34 A
35 A
36 C
37 B
38 E
39 A
40 C
41 B
42 A
43 D
44 C
45 C
46 E
47 D
48 D
49 D
50 B
Spoiler:
Show
3.c
9.b
19.a
25.e
26.c
27.d
29.d
30.e
31.c
35.d
36.b
41.a
50.a
Where we differ you are likely the one in the right as my answers arn't close to any of the other suggestions posted. That said I'm averaging 84% on the other papers including some daft mistakes which I hopefully won't make in the real thing. I thought this was good until I saw the many "90% in an hour" claims on here
The 2007 paper was posted on here somewhere in the last 15 pages, in word document form, for free. Although some c**t posting as "Mr.Diamonds" also tried to sell it me and probably others for £50 last month using this very thread.
12. Okay, here's my essay- everyone please be nice (im hopeless at writing essays ) , and tell me how to improve it etc. I did use the phrase, 'to use fear as a tool'- i couldn't find another way to describe it!
To be a successful leader is it better to be loved or feared?
The qualities required for a successful leader varies, depending upon the situation. For example the characteristic's for a successful political leader would inevitably be different to that of a teacher. The emotions love and fear are extreme, and alone cannot encourage or catalyse tasks to be completed, as impediments will occur.
It is not appropriate or ethical for leaders to be feared, and I would question why leaders require to use fear as tool to create a successful leadership. Leaders are usually feared in totalitarian regimes, where a population have to conform and even obey inhumane orders. As a result, an individuals human rights are subconsciously impaired. A historical example of this is Hilter's autocracy, many Germans and Jews were denied freedom of speech and were both unable to stop the Nazi regime. If a leader is feared, the attentions and outcomes are not democratic. Therefore this should be avoided as all leaders ought be virtuous.
In conclusion, a leader requires many qualities. In the example of a teacher, they should be loved, as this would create a comfortable learning environment and respected, as this will enable them to receive homework by the deadlines set. The most common and important qualities for successful leaders are empathy, obedience and respect. A leader's attentions should be made clear, so everyone can understand why they want a certain task to be completed or why co-co-operation is required. Hence why it is not necessary for a leader to be either loved or feared.
(Original post by matija_v)
5 more days!! just want to get over it..
ok, first here are my answers to the 2007 paper
Spoiler:
Show
1 B
2 B
3 A
4 C
5 C
6 A
7 B
8 D
9 D
10 C
11 C
12 D
13 A
14 D
15 B
16 E
17 Pig=5 Dog=12 Cat=7 Horse=9
18 C
19 A
20 B
22 A
23 B
24 A oldest, c youngest
25 B
26 E
27 E
28 A
29 B
30 B
31 D
32 C
33 C
34 A
35 A
36 C
37 B
38 E
39 A
40 C
41 B
42 A
43 D
44 C
45 C
46 E
47 D
48 D
49 D
50 B
Spoiler:
Show
3.c
9.b
19.a
25.e
26.c
27.d
29.d
30.e
31.c
35.d
36.b
41.a
50.a
Where we differ you are likely the one in the right as my answers arn't close to any of the other suggestions posted. That said I'm averaging 84% on the other papers including some daft mistakes which I hopefully won't make in the real thing. I thought this was good until I saw the many "90% in an hour" claims on here
The 2007 paper was posted on here somewhere in the last 15 pages, in word document form, for free. Although some c**t posting as "Mr.Diamonds" also tried to sell it me and probably others for £50 last month using this very thread.
Yeah i was considering buying the mark scheme from him as well =/
i hate it how people try and make money off the system and off people who don't know any better.
14. I thought I'd join in with the essay posting. First one I have done so far in 30mins and reading it back does not fill me with confidence lol. Any comments on how to improve are very welcome!!
Is ethical consumerism a solution to poverty or a dangerous distraction?
Over recent years we have seen a growing shift towards the relatively new phenomenon of fair-trade and ethical consumerism. The market is now flooded with such products and consumers are responding favourably.
Claiming ethical consumerism is a solution to poverty is a bold claim to make, but I argue it is definitely beneficial to the cause. Products such as fair-trade items increase the awareness of the public to some of the problems we face on a global scale. This increased awareness on a national scale can lead to pressure on government and consequently more foreign aid. Due to our electoral system the parties are essentially bound by the demands of the majority (if their aim is to win votes). If this more informed public now view poverty as a prominent issue, government must respond in kind in order to remain in office.
Whichever side of the argument one supports, it is undeniable that the direct benefits of ethical consumerism (e.g. suppliers are paid a fair price) can only be desirable. This will never solve poverty in itself but if the principle was applied globally then perhaps the result would be different. The profit motivation behind big businesses would prevent them from ever voluntarily paying suppliers more than necessary but consumer pressure for fair-trade products has backed them into a corner. The prices paid now are a step in the right direction and this can only be of benefit to those in poverty which it directly affects.
Ethical consumerism can be criticised, although not to the extent the title suggests. It has been viewed as a marketing gimmick deployed by businesses who do the bare minimum required to achieve fair trade status. This can be a problem if people feel the only charity they need to give is to buy with ethical considerations, as the effects of this are not substantial enough to address the problem sufficiently.
Ethical consumerism is neither a solution to poverty nor a dangerous distraction. But I argue it increases the likelihood of solving poverty. The increased awareness of poverty and how this relates to trade and employment results from ethical consumerism. Whether the direct benefits of ethical consumerism (ie fair trading prices) actually works to solving poverty is questionable but a society who is more aware of the problem is in a better place to do so – ethical consumerism brings this about. After all, how can we solve something when those in control (the electorate) do not understand the extent of the problem?
16. (Original post by matija_v)
5 more days!! just want to get over it..
ok, first here are my answers to the 2007 paper
Spoiler:
Show
1 B
2 B
3 A
4 C
5 C
6 A
7 B
8 D
9 D
10 C
11 C
12 D
13 A
14 D
15 B
16 E
17 Pig=5 Dog=12 Cat=7 Horse=9
18 C
19 A
20 B
22 A
23 B
24 A oldest, c youngest
25 B
26 E
27 E
28 A
29 B
30 B
31 D
32 C
33 C
34 A
35 A
36 C
37 B
38 E
39 A
40 C
41 B
42 A
43 D
44 C
45 C
46 E
47 D
48 D
49 D
50 B
And, because Stashup asked for it, I wrote an essay to the leader-question a while ago...
Here
Spoiler:
Show
To be a successful leader is it better to be loved or feared?
The need for leadership and authority is a natural characteristic not only of animals but also of human beings. The characteristics of leaders are hard to define, but emotions like love or fear from followers are often distracting and damaging a productive relationship. I believe that a successful leader earns respect and obedience, rather than feelings like love or fear.
In the case of love, a very positive emotion, the leader is admired and cared for, thus tasks are completed out of love. For example if a teacher is loved, his or her students will do their homework and listen in class because they like the teacher. However, love can cause a lack of discipline, and furthermore if the leader imposes stricter behavior, this leads to less love and unsuccessful leadership. To balance the positive emotions amongst the group of people one is leading is a very difficult task, and it often backfires at some point.
Fear, presented as the contrary to love, implies an association with hate, another strongly negative feeling. People acting out of fear or despair results in many problematic situations, and fear was often the source of wars, violence and destructive behavior. Therefore using fear as a tool to establish successful leadership is very questionable. Nevertheless, it is certainly true that the levels of discipline and obedience are very high in societies based on fear and control, as indicated by many totalitarian regimes. However, unhappy and controlled minds should not be the goal of a leader if he is not only aiming for success. But if success is the only factor a leader wants to achieve, then fear represents a very effective but unethical option.
To sum up, a loved leader makes those who are being led satisfied but success is hard to achieve in such a friendly environment. A feared leader is able to be successful but this in an unethical way, respectively. In my opinion, a successful leader should be more concerned about respect towards him or her than to aim for feelings like love or fear. Moreover these feelings should be balanced and neutral, and extremes of emotions should better be avoided.
any feedback on that would be appreciated.
btw, @stashup: bist du aus deutschland, weil ich hab irgendwo gelesen dass du abitur gemacht hast?! was war dein durchschnitt, 1,0?
Hi. I think you did rather a good job here. Haven't written anything on this myself yet and was having a few problems which is why I asked for someone to post his example.
In the first sentence I would maybe explicitly say "some animals" or use a sentece like "Much like many animals humans have an inherent tendency to organise themselves into social and group structures often lead by a single individual" ... but this is not sooooo important.
In paragraph one maybe explain how love can cause a lack of discipline.
I think paragraph number 2 is very good and understandable! You might want to replace "unhappy" with "fearful". Although it seems quite obvious you haven't assumed that fear leads to being unhappy.
Further I think somewhere in your essay you should distinguish between personal success and success for the group which will make it easier for you in paragraph 2.
You have given me a bit of inspiration in showing that it is probably quite smart to answer "either A or B" questions with saying: neither
In this way you can get around having to describe both A and B in the greatest detail possible cointaining all pros and cons you can think of and then still having to compare them in your conclusion.
Now for you personally in German: Ja, wohne in der Nähe von Stuttgart. Hab 1,0 aber ich glaub hier fühlt sich jeder relativ doof^^ Meine multiple choice ergebnisse sind bis jetzt auch immer unter 70%
Naja, mal sehen. Kannst mich gerne im ICQ adden wenn du möchtest. 230774494
Woch machst du den TSA? Ich in Salem
17. (Original post by spaceman657)
I thought I'd join in with the essay posting. First one I have done so far in 30mins and reading it back does not fill me with confidence lol. Any comments on how to improve are very welcome!!
Is ethical consumerism a solution to poverty or a dangerous distraction?
Over recent years we have seen a growing shift towards the relatively new phenomenon of fair-trade and ethical consumerism. The market is now flooded with such products and consumers are responding favourably.
Claiming ethical consumerism is a solution to poverty is a bold claim to make, but I argue it is definitely beneficial to the cause. Products such as fair-trade items increase the awareness of the public to some of the problems we face on a global scale. This increased awareness on a national scale can lead to pressure on government and consequently more foreign aid. Due to our electoral system the parties are essentially bound by the demands of the majority (if their aim is to win votes). If this more informed public now view poverty as a prominent issue, government must respond in kind in order to remain in office.
Whichever side of the argument one supports, it is undeniable that the direct benefits of ethical consumerism (e.g. suppliers are paid a fair price) can only be desirable. This will never solve poverty in itself but if the principle was applied globally then perhaps the result would be different. The profit motivation behind big businesses would prevent them from ever voluntarily paying suppliers more than necessary but consumer pressure for fair-trade products has backed them into a corner. The prices paid now are a step in the right direction and this can only be of benefit to those in poverty which it directly affects.
Ethical consumerism can be criticised, although not to the extent the title suggests. It has been viewed as a marketing gimmick deployed by businesses who do the bare minimum required to achieve fair trade status. This can be a problem if people feel the only charity they need to give is to buy with ethical considerations, as the effects of this are not substantial enough to address the problem sufficiently.
Ethical consumerism is neither a solution to poverty nor a dangerous distraction. But I argue it increases the likelihood of solving poverty. The increased awareness of poverty and how this relates to trade and employment results from ethical consumerism. Whether the direct benefits of ethical consumerism (ie fair trading prices) actually works to solving poverty is questionable but a society who is more aware of the problem is in a better place to do so – ethical consumerism brings this about. After all, how can we solve something when those in control (the electorate) do not understand the extent of the problem?
Nice.
18. (Original post by amy123123)
p.s- you hertford applicants....did you guys by any chance make an open application :P also...has anyone received emails from two colleges?
nope, i didn't send an open application, i applied directly to hertford...and was e-mailed an acknowledgment letter, admissions booklet and test/written work booklet...
19. (Original post by bachmts)
Are u sure it's D? I get A:
alright it goes like this:
100-98 (he gets 2+1+0), 97-91 (he gets 6+5+4+3+2+1+0), 91-84 (gets 6+5+4+3+2+1+0) and so on u get the idea.
If u take the lower boundaries that makes a total of 13 (91, 84, ..., 7).
Each time u get 6+5+4+3+2+1+0=21, so 13*21= 273.
Then u add 3 (for no. 100-98) so u get 276.
edit:
alright just realized my mistake. 0 is the last lower boundary so it's another 21 which makes 297
thank you so much ...i made the same mistake you made when i did it! thanks again
20. Shooting star- are you applying for PPE?
### Related university courses
Updated: October 8, 2010
Today on TSR
### He broke up with me because of long distance
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# Prove that : cot^(-1)7+cot^(-1)8+cot^(-1)18=cot^(-1)3
Updated On: 17-04-2022
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Step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams.
Transcript
hello plants in this question we have to prove that Cos inverse 7 + cot inverse 8 + cot inverse 18 is equal to COT inverse 3 ok so with the left-hand side of the question we have gotten worse 7 + cot inverse 8 + cot inverse a team and you know that the formula of cot inverse x is equal to tan inverse 1 by X so this formula we can I coronavirus 7s 10 1.7 and we can write cot inverse 8 as tan inverse 1 upon it and similarly we can write quarter was 18 as tan inverse 1 upon a time now we know that 10 inverse X + 10 inverse y is equal to tan inverse
X + Y upon 1 minus x into I so applying this formula for this week and it says inverse X + Y that is 1.7 + 1 upon 1 upon 1 minus x into that is 1.7 into 1.8 ok plus this is additive inverse 1 upon 18 now on for resolving this week and addresses tan inverse this will be 56 der se Mili 56 and in the moment we get 15 / this will be 56 - 1 that is 55 upon 50 and this is tested was 1.18 ok so this 56 and get cancelled and we get tan inverse
15 upon 55 + 10 inverse 1 upon 18 now adding this y83 and it will be 11 hence we get inverse 3 by 11 + 10 inverse 1 by 8 Naagin episode 10 3 by 11 + 10 inverse 1 by 18 this year will apply again this formula OK it is equal to 10 inverse X + Y that is 3 upon 1 + 1 upon 1 3 divided by 1 minus 3 upon 11 into 1.10 ok on this year
10 inverse their LCM of 11 and 28 and it will be 3 into 3 + 1 into 11 good night bhai Ye Dil Se Mil v11 into it and it will be 11 into 18 - 3121 ok handsome father for doing this we get here tan inverse this will be a 54 + 11 Dec 11 2011 in 1898 and this is minus 3 upon 11 into a 338 and this will be cancelled and get 10 inverse 54 + 11 that is 65 upon 198 -3 that is 195 aunty wearing this time
65 we get air 1 and this is 3 and forget tan inverse 1 by 3 that is equal to COT inverse 3 and this is equal to Rs of the question ok we can write cot inverse 3 is equal to this question has proved hence proved thank you
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http://www.rotaryforum.com/forum/archive/index.php/t-1576.html
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PDA
View Full Version : Belt Slippage
KenSandyEggo
07-09-2004, 06:37 PM
I've been wondering if the prerotator belt could slip under high loads on the RAF or Sparrowhawk and how to determine this. If I make a paint mark on the small pulley and a mark on the belt, both at the same locations, should those marks always line up after running the engine if there is no slippage? I know that this is basic physics, but it's been a long time. I don't know if the reduction ratio changes that. :confused:
Mike Hook
07-09-2004, 06:55 PM
Yes Ken they will not line up you have a ratio there with two different sizes of pulleys.
Example if you have a 2 in pulley the circumference = (3.14)Pi.x dia. so if the dia was 2 the cur would be 6.28 inches
if you have a belt of say 20 inch dia. cir is dia x (3.14) PI. =cir of 62.8 in you have a ratio of 10 to 1 so the belt would line up only every ten revolutions of the pulley. Now the pulley is measured at the point of pitch and the belt is measured on the inside so you see it is a hard thing to get to work accurate.
Mike
KenSandyEggo
07-09-2004, 10:07 PM
Thanks, Mike. Guess I'll have to figure some other way to see if there's any slippage. When I stop prerotating after a 1/2 hour or so of testing, the belt is not hot at all. I'd think that if it were slipping, it would be hot or quickly shred, but I'd like to find out for sure.
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http://www.java2s.com/Code/Cpp/Set-Multiset/Createthesymmetricdifferencebetweenlist1andlist2.htm
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# Create the symmetric difference between list1 and list2 : difference « Set Multiset « C++
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C++ » Set Multiset » difference Screenshots
Create the symmetric difference between list1 and list2
``` #include #include #include using namespace std; template void show_range(const char *msg, InIter start, InIter end); int main() { list list1, list2, result(15); list::iterator res_end; for(int i=0; i < 5; i++) list1.push_back('A'+i); for(int i=3; i < 10; i++) list2.push_back('A'+i); show_range("Contents of list1: ", list1.begin(), list1.end()); show_range("Contents of list2: ", list2.begin(), list2.end()); res_end = set_symmetric_difference(list1.begin(), list1.end(),list2.begin(), list2.end(),result.begin()); show_range("Symmetric difference of list1 and list2: ",result.begin(), res_end); return 0; } template void show_range(const char *msg, InIter start, InIter end) { InIter itr; cout << msg << endl; for(itr = start; itr != end; ++itr) cout << *itr << endl; } ```
Related examples in the same category
1 Get the difference between two sets 2 Get the symmetric difference between two sets 3 set_difference and back_inserter 4 Use set_difference to get values from only one list 5 Use set_symmetric_difference() to get the difference between two lists
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http://www.chegg.com/homework-help/hubbard-s-leaky-bucket-given-empty-bucket-like-one-fig-8-hol-chapter-1.5-problem-24p-solution-9780131860612-exc
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View more editions
# TEXTBOOK SOLUTIONS FOR Differential Equations and Linear Algebra 2nd Edition
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Chapter: Problem:
Hubbard’s Leaky Bucket If you are given an empty bucket like the one in Fig. 8, having a hole in the bottom from which all the water has leaked, can you tell how long ago the bucket was full? Of course not, and the answer can be related to nonunique solutions. The equation describing the height h of the water level comes from Torricelli’s Law:
where the positive constant k depends on the size and shape of the bucket and of the hole.
(a) Even before setting a time scale, we can see that Picard’s Theorem will not apply. Explain why.
(b) Formulate a general initial condition and solve the DE by separation of variables. Why is it impossible to tell how much time has elapsed? Draw several possible solution curves.
(c) Show that the total emptying time is given by [a-z]here h0 is the initial height of the water in the bucket.
Figure 8 Hubbard’s leaky bucket.
STEP-BY-STEP SOLUTION:
Chapter: Problem:
Corresponding Textbook
Differential Equations and Linear Algebra | 2nd Edition
9780131860612ISBN-13: 0131860615ISBN:
Alternate ISBN: 9780131860636
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# Kevin You're Fat? Then Understand Weight and Balance of an Airplane
Silent Bob actor, Kevin Smith, has tweeted about his embarrassment after being kicked off a Southwest Airline flight for being too fat. Let's talk about physics. Airplanes fly in the air because there are certain indelible laws of gravity and physics. For example: the airplane is going to be able to get into the air because the weight and balance of the aircraft is congruent with the ability of the engines and the wings. This is something pilots are taught and expect to repeat much like a mantra or meditation: "Weight and balance, take off and landing."
I became a pilot years ago at John Wayne airport, it was right after the singer Aaliyah died because she forced her pilot to carry more baggage than the plane could physically handle. Her Louis Vuitton survived, she did not. The NTSB report stated that the total gross weight of her airplane was "substantially exceeded," which caused the center of gravity to be pushed too far forward. In other words, there was too much weight on the plane and the engines and wings could not lift the front of the airplane high enough to get the aircraft balanced and stable for a safe flight. The rest is tragic history.
So, back to Southwest, pilots are taught how much weight his or her aircraft can handle. And if the plane crashes, the pilot is the first person who will be questioned, so therefore a pilot has a right to be proactive and notify a passenger if their size is putting the weight and balance in question. Especially when, like Kevin Smith, the passenger's weight was taken into consideration at the time of purchasing an extra ticket to compensate the passenger's size. Mr. Smith knew that he needed to buy two seats, he was taking a chance (and putting other passengers in danger by getting his big body on that earlier flight). Remember, the engines and wings are able to get off the ground because the weight in the airplane is not exceeding its ability to fly safely.
If it the airplane surpasses its weigh restriction, the plane won't be able to take off. Or even worse, the airplane will crash going at a very fast speed. The pilot had a right to ask Mr. Smith to get off the plane. We all know the saying, "The straw that broke the camel's back." Kevin, you could have been that straw. Honestly, if it was your family on that plane and it crashed, who would you blame?
The math for each aircraft has been determined by the FAA and is available online. (Just google the airplane you are flying, FAA, and "weight and balance.") For example the calculated average weight for a passenger on a 737 is 170 pounds. That breaks down to:
• Forward compartment ............................................... 18 people
• Aft compartment ........................................................ 95 people
How about the average folks who are flying, how much do they weigh? The Journal of the American Medical Association states that the prevalence of obesity in the United States continues to be high, exceeding 30% in most sex and age groups. If that is the case, then is it safe to say that 30% of passengers are overweight?
Knowing all this, do overweight people have a right to get mad when they need to purchase an extra ticket or get kicked off fully-booked airplane? I think not. And, Dear Mr. Kevin Smith, I don't know how much you weigh, but based on your photos, you weigh more than 170 pounds. Let's suppose that I am on a fully-booked airplane with 20 overweight people who weigh, let's say 240 pounds -- that is 1400 more pounds than the aircraft should be carrying. Now, this will not be a problem if those overweight passengers have purchased an additional ticket to compensate for their extra weight, but if not, then everyone on the aircraft is in danger.
Yes, flying has become a tiresome drag and the airlines are not doing much to spruce up the friendly skies. But as passengers, it is our responsibility to understand the gravitational and physical laws that govern the sky. Even though this is physics, it ain't rocket science. You put too much weight on an aircraft and you will compromise the airplane's ability to do what is built to do.
Maybe obese passengers don't have shoe bombs or box cutters on them, but it seems to me that what could happen on board an aircraft filled with too many overweight people could be just as deadly.
My hat goes off to the captain of the Southwest flight who asked Kevin to get off the aircraft. Instead of a \$100 flight voucher, perhaps Southwest should have given Kevin a \$100 membership to the gym.
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# Introduction to Human Body Model
Introduction to Skinned Multi-Person Linear Model (SMPL), which is a popular models used in both academia and industry computer graphics
In the previous post, we have introduced the concept of Linear Blend Skinning (LBS), which is a highly important technique in graphics and animation. However, LBS is just so simple that it has many problems such as volume collapsing and candy wrapper (see Figure 4 in this post), which can lead to unrealistic body deformation. Furthermore, in many practical problems, we may want a realistic body model that has the ability to represent any human body in the real world and the model should not be too complicated to use. In this article, we will go into a human body model called Skinned Multi-Person Linear Model (SMPL) [1], which utilize a very effective data-driven strategy to address all of these concerns.
## #Linear Blend Skinning Recall
To recap, linear blend skinning parameters contain:
• Mesh vertices in rest pose: $\mathbf{T} \in \mathbb{R}^{3N}$
• Joint locations: $\mathbf{J} \in \mathbb{R}^{3K}$.
• Joint rotations (pose parameters): $\mathbf{\theta} \in \mathbb{R}^{3K}$
• Blend skinning weights: $\mathbf{W} \in \mathbb{R}^{N \times K}$
where $N$ is the number of vertices, and $K$ is the number of articulated body joints. The local bone transformation matrix $\mathbf{M}_k$ at joint $k$ depends on the joint location and the joint location, which can be computed as:
$\mathbf{M}_k = \begin{bmatrix} Rot(\theta_k) & \mathbf{j}_k \\ \mathbf{0}^T & 1 \end{bmatrix}$
The function $Rot(\theta_k)$ transforms the joint rotation vector $\theta_k$ into a $3\times 3$ rotation matrix and $\mathbf{j}_k$ is the offsets of the current joint $k$ from its parent ($\mathbf{J}_k - \mathbf{J}_{p(k)}$). To obtain the world transformation $\mathbf{G}_k(\theta,\mathbf{J})$ at joint $k$, we simply accumulate the local transformation matrix along the kinematic tree up to the root joint: $\mathbf{G}_k(\theta,\mathbf{J}) = \Pi_{j\in A(k)}\mathbf{M}_j$, where $A(k)$ is the ordered list of joints in the path from the root to joint $k$. Finally, the position of the vertex $i$ after LBS is calculated as
$\mathbf{t}_i' = \sum_{k=1}^K w_{ik}\mathbf{G'}_k(\theta,\mathbf{J})\mathbf{t}_i, \tag{1} \\ \mathbf{G'}_k(\theta,\mathbf{J}) = \mathbf{G}_k(\theta,\mathbf{J})\mathbf{G}_k(\theta^0,\mathbf{J})^{-1}$
where $\mathbf{G}_k(\theta^0,\mathbf{J})$ is the world transformation of joint $k$ according to the rest pose $\theta^0$; $\mathbf{t}_i$ is the vertex position in the rest pose; $\mathbf{t}'_i$ is the transformed vertex corresponding to the desired pose $\theta$.
## #Pose corrective blendshapes
To deal with the errors of linear blend skinning, in practice, people often artistically sculpt the blendshape, and then add it into the template mesh to correct the posed mesh. A blendshape is a vector of vertex displacements to the original template mesh. As shown in Figure 1, after adding the blendshape to the original mesh, the artifacts around the elbow and the hip go away, and the posed mesh is correct.
Figure 1. Skinning results with and without added corrective blendshapes.
However, it is quite a tedious and expensive process. Inspired by this, the SMPL's author resolved the problem of LBS by learning the corrective blendshape from a large amount of scanned data of real humans. Specifically, in the SMPL formulation, the pose corrective blendshapes are added to the original LBS (Equation 1) as follows:
$\mathbf{t}_i' = \sum_{k=1}^K w_{ik}\mathbf{G'}_k(\theta,\mathbf{J})(\mathbf{t}_i + \mathbf{P}_i(\theta)), \tag{2}$
where $\mathbf{P}(\theta) \in \mathbb{R}^{3N}$ is the pose correctives. Let $f(\theta)$ be some function of $\theta$ and $f_n(\theta)$ denote the $n$-th element. The pose correctives depend on the current pose $\theta$ and can be calculated as the combination of the learned blendshapes:
$\mathbf{P}(\theta) = \sum_{n=1}^{\vert f(\theta)\vert} f_n(\theta) \mathbf{P}_n \tag{3}$
The authors of SMPL designed $f(\theta)$ to be the function converting $\theta$ into a vectorized version of the concatenated joint rotation matrices $f: \mathbb{R}^{3K} \mapsto \mathbb{R}^{9K}$. Note that although $\mathbf{P}(\theta)$ may be non-linear in pose $\theta$, it is linear in elements of the rotation matrices $f(\theta)$ (the authors also tried to make the model linear, i.e., $f(\theta) = \theta$, but it did not work). Therefore, SMPL is an additive and simple yet very effective model. These blendshape vectors $\mathbf{P}_n \in \mathbb{R}^{3N}$ are the model's parameters and are learned from real-world data.
### #Training of pose-related parameters
Figure 2. The joint location could be estimated as an weighted average of surrounding vertices. The vertex weights for each joint are indicated by the colored lines on the body surface. The joint regressor (matrix containing vertex weights) is sparse since the joint is only influenced by nearby vertices (Source).
The pose-related parameters of the model are trained using a 3D scan dataset when people are in different poses (we called multi-pose dataset). SMPL pose-related parameters in this stage include:
• $\mathcal{J} \in \mathbb{R}^{K\times N}$: Joint regressor matrix, which is used to regress the joint location based of the surrounding vertices (Figure 2): $\mathbf{J} = \mathcal{J}\mathbf{t}$
• $\mathbf{W} \in \mathbb{R}^{N\times K}$: blend skinning weights matrix.
• $\mathbf{P} \in \mathbb{R}^{3N\times 9K}$: pose corrective blendshapes.
To train these parameters, we minimize the surface reconstruction error (squared euclidean distance) between the ground-truth mesh vertices (template aligned from the scanned human) and the output vertices of the corrective skinning function (Equation 2):
$\Vert\mathbf{V}_i - \sum_{k=1}^K w_{ik}\mathbf{G'}_k(\theta,\mathbf{J})(\mathbf{t}_i + \mathbf{P}_i(\theta))\Vert^2 \tag{4}$
where $\mathbf{V}$ is the ground-truth aligned mesh, $i$ denotes the $i$-th vertex position. In addition, there are many subjects in the dataset, each human subject has a different body configuration (e.g., fat, skinny, tall,...). Therefore, we also need to compute the rest template mesh $\mathbf{t}$ and the joint location $\mathbf{J}$ for each subject. This result in an alternating optimization scheme where we alternate updating between the pose parameter $\theta$, the subject-specific parameters $\{\mathbf{t}$, $\mathbf{J}\}$, and the global parameters $\{\mathbf{P}$, $\mathbf{W}\}$
Since the model consists of a large number of parameters, we also apply several regularizations to prevent overfitting:
• $\mathbf{P}$: regularized the blendshapes towards zeros.
• $\mathcal{J}$: regularized towards the predicting joints near the boundaries between the body parts and to be sparse.
• $\mathbf{W}$: regularized towards the initial artist-design LBS blend weights.
## #Identity-dependent (Shape) blendshapes
Next, we want the body model to be able represent a wide variety of human body shape in the population. To do so, the authors also built a dataset for shape training, called multi-shape dataset. This dataset contains several template meshes with high variation of body shape of real human (mostly from US and Europe) . In order to learn the shape space properly (the shape must be completely independent with the pose), we also need to factor out the pose (pose normalization) so that every people in this dataset are in the same rest pose.
Figure 3. Given the shape training dataset with multiple human meshes in the default pose, we first calculate the mean body shape and then subtract it to obtain the data matrix for PCA.
Our goal is to learn a statistical model from a body shape space. Note that each body can be considered as a vector in a $3N$-dimensional space (3D positions of $N$ vertices). However, this space is infeasible to deal with since $N$ is typically large, and not all vectors in this space correspond to a valid body shape (they only take the place of a tiny subspace). Therefore, we need to reduce the dimensionality to compress the shape space into a low dimensional space.
We first start by computing the mean body shape from the training data. We subtract the mean mesh from each of all the meshes in the training dataset (Figure 3). After that, we stack these body mesh vertices into a matrix and perform principal component analysis (PCA). This results in a set of eigenvalues and eigenvectors that describes the major directions of variation in the body shape space. The top ordered eigenvectors give us a low-dimensional linear subspace (typically 10-300D) that captures most of the variance in the $3N$-D space. Consequently, the body shapes of different people are approximated by a linear combination of the shape parameters $\beta$ and the shape-dependent blendshapes $\mathbf{S}$ plus a mean body $\mu$:
$\mathbf{T} = \sum_{n=1}^{\vert\beta\vert} \beta_n\mathbf{S}_n + \mathbf{\mu} = \mathbf{S}\beta + \mathbf{\mu} \tag{5}$
Figure 4. An arbitrary body shape can be represented by the sum of the mean body and a combination of the shape-dependent blendshapes weighted by some shape coefficient parameters.
Figure 5 below shows the variation in the first principal components of the shape PCA subspace. We also note that this procedure works because the non-linearities of the pose are factored out so that the body meshes live in a linear euclidean space of point. Moreover, the joint locations $\mathbf{J}$ is now a function depending on the shape parameters $\beta$ since it is regressed from the body mesh that may vary with the shape.
Figure 5. The first principal component of the trained shape space (Source).
Putting it all together, we obtained the final skinning equation of SMPL:
$\mathbf{t}_i' = \sum_{k=1}^K w_{ik}\mathbf{G'}_k(\theta,\mathbf{J}(\beta))(\mathbf{t}_i + \mathbf{S}_i(\beta) + \mathbf{P}_i(\theta)) \tag{6}$
where
• $t = \mu$ is the mean body template mesh (in the rest pose).
• $\mathbf{S}(\beta) = \mathbf{S}\beta$ is the shape blendshapes (Equation 5).
• $\mathbf{P}(\theta)$ is the pose blendshapes (Equation 3).
The SMPL model can represent "any" human body in different poses in an effective way thanks to the disentanglement of the shape and the pose. This means we can easily transfer the pose from a person with this body shape to another one with different body shape by just directly copying the pose parameters. We can also see that the SMPL is a simple additive model and is built upon the Linear Blend Skinning. Therefore, it is compatible with existing graphics engines. Another advantage is that it is trained from real data so that it is accurate and can resolve many problems of the traditional skinning method.
This model cannot just only be used to synthesize artificial characters but can also be used to estimate the pose and the shape of humans precisely even from just a single image, thanks to the differentiability and linearity in the formulation. We can arbitrarily generate a new body shape by randomly sampling the shape vector $\beta$ around a Gaussian distribution with zero means. Moreover, the emergence of this model pushes forward many interesting research directions such as modeling clothing/garments, human reconstruction and motion dynamics, virtual avatars, scene interaction... Since SMPL is simple, it may not fully describe all features of human. There are several extensions of SMPL that can be useful in some specific tasks: DMPL [1] to model body and dynamic soft tissue movements, SMPL-H [2] to model body and hand/finger movement, SMPL-X [3] to model body, hand, and facial expression.
## #References
[1] Loper, M., Mahmood, N., Romero, J., Pons-Moll, G., & Black, M. J. SMPL: A skinned multi-person linear model. ACM transactions on graphics (TOG).
[2] Romero, J., Tzionas, D., & Black, M. J. Embodied hands: Modeling and capturing hands and bodies together. ACM transactions on graphics (TOG).
[3] Pavlakos, G., Choutas, V., Ghorbani, N., Bolkart, T., Osman, A. A., Tzionas, D., & Black, M. J. Expressive body capture: 3d hands, face, and body from a single image. In CVPR 2019.
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## Graduate Theses, Dissertations, and Problem Reports
#### Title
Cycles in graph theory and matroids
Spring
2008
Dissertation
PhD
#### College
Eberly College of Arts and Sciences
Mathematics
Hong-Jian Lai.
#### Abstract
A circuit is a connected 2-regular graph. A cycle is a graph such that the degree of each vertex is even. A graph G is Hamiltonian if it has a spanning circuit, and Hamiltonian-connected if for every pair of distinct vertices u, v ∈ V( G), G has a spanning (u, v)-path. A graph G is s-Hamiltonian if for any S ⊆ V (G) of order at most s, G -- S has a Hamiltonian-circuit, and s-Hamiltonian connected if for any S ⊆ V( G) of order at most s, G -- S is Hamiltonian-connected. In this dissertation, we investigated sufficient conditions for Hamiltonian and Hamiltonian related properties in a graph or in a line graph. In particular, we obtained sufficient conditions in terms of connectivity only for a line graph to be Hamiltonian, and sufficient conditions in terms of degree for a graph to be s-Hamiltonian and s-Hamiltonian connected.;A cycle C of G is a spanning eulerian subgraph of G if C is connected and spanning. A graph G is supereulerian if G contains a spanning eulerian subgraph. If G has vertices v1, v2, &cdots; ,vn, the sequence (d( v1),d(v2), &cdots; ,d(vn)) is called a degree sequence of G. A sequence d = ( d1,d2, &cdots; ,dn) is graphic if there is a simple graph G with degree sequence d. Furthermore, G is called a realization of d. A sequence d ∈ G is line-hamiltonian if d has a realization G such that L(G) is hamiltonian. In this dissertation, we obtained sufficient conditions for a graphic degree sequence to have a supereulerian realization or to be line hamiltonian.;In 1960, Erdos and Posa characterized the graphs G which do not have two edge-disjoint circuits. In this dissertation, we successfully extended the results to regular matroids and characterized the regular matroids which do not have two disjoint circuits.
COinS
#### DOI
https://doi.org/10.33915/etd.2720
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• Brainly User
2014-07-19T13:42:40-04:00
### This Is a Certified Answer
Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
A=lW divide both sides by L
W=A / L
----------------------------------------------------------------------
A=8m^2; L=4m
W=(8m^2) / (4m) = 8 / 4 * m^2 / m =2m
ok i will
And??
still nothing.. I'll figure something out..
look now
I got it!!! =) Thanks so much! It was correct!
2014-07-19T13:56:48-04:00
### This Is a Certified Answer
Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
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# FORECAST Function
Calculates a future value using existing values
## What is the FORECAST Function?
The FORECAST Function[1] is categorized under Excel Statistical functions. It will calculate or predict a future value using existing values.
In financial modeling, the FORECAST function can be useful in calculating the statistical value of a forecast made. For example, if we know the past earnings and expenses, we can forecast the future amounts using the function.
### Formula
=FORECAST(x, known_y’s, known_x’s)
The FORECAST function uses the following arguments:
1. X (required argument) – This is a numeric x-value for which we want to forecast a new y-value.
2. Known_y’s (required argument) – The dependent array or range of data.
3. Known_x’s (required argument) – This is the independent array or range of data that is known to us.
### How to use the FORECAST Function in Excel?
As a worksheet function, FORECAST can be entered as part of a formula in a cell of a worksheet. To understand the uses of the function, let’s consider an example:
#### Example
Suppose we are given earnings data, which are the known x’s, and expenses, which are the known y’s. We can use the FORECAST function to predict an additional point along the straight line of best fit through a set of known x- and y-values. Using the data below:
Using earnings data for January 2019, we can predict the expenses for the same month using the FORECAST function.
The formula to use is:
We get the results below:
The FORECAST function will calculate a new y-value using the simple straight-line equation:
Where:
and:
The values of x and y are the sample means (the averages) of the known x- and the known y-values.
### A few notes about the function:
1. The length of the known_x’s array should be the same length as the known_y’s, and the variance of the known_x’s must not be zero.
2. #N/A! error – Occurs if:
1. The supplied values known_x’s and the supplied known_y’s arrays have different lengths.
2. One or both of the known_x’s or the known_y’s arrays are empty.
1. #DIV/0! error – Occurs if the variance of the supplied known_x’s is equal to zero.
2. #VALUE! error – Occurs if the given future value of x is non-numeric.
Thanks for reading CFI’s guide to this important Excel function. By taking the time to learn and master these functions, you’ll significantly speed up your financial modeling and valuation analysis. To learn more, check out these additional CFI resources:
### Free Excel Tutorial
To master the art of Excel, check out CFI’s FREE Excel Crash Course, which teaches you how to become an Excel power user. Learn the most important formulas, functions, and shortcuts to become confident in your financial analysis.
Launch CFI’s Free Excel Course now
to take your career to the next level and move up the ladder!
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# reading structured arrays, changing their structure
• 05-15-2003
_kevin007
reading structured arrays, changing their structure
hi,
i am trying to read in two different structured arrays, place them in a different structure so they are both the same and then compare them.
the first array, 'primary', has the structure..
Code:
``` struct primary_plot { double bearing; /*degrees */ int range; /*metres */ double elevation; /*degrees */ };```
the second array, 'secondary', has a similar structure..
Code:
``` struct secondary_plot { double bearing; /*degrees */ int range; /*metres */ int altitude; /*metres */ int flight_number; };```
i need to read these arrays into a file, and store them in seperate arrays with this structure..
Code:
```struct aircraft_information { double bearing; /*degrees */ double range; /*kilometres */ int altitude; /*metres */ int flight_number; int match; /*array index of matching plot */```
obviously i will need to convert the range's from both the primary and secondary into kilometres which isnt a problem, im just unsure of how to carry out this task.
any help will be appreciated!
• 05-15-2003
poccil
Given your information alone, it's difficult to determine an answer. Maybe multiply the two values and divide by 1000?
• 05-15-2003
_kevin007
sorry, i should of explained better.
i have the formula to convert metres into kilometres, i was unsure about how to read the arrays in and put them into the different structure shown.
thanks again
• 05-15-2003
stumon
Did you create the file you are going to be reading from, or was it already created for you? If so, is it a random access, or sequential file? Depending on which type, you could use fwrite() or fgets() to get that information. Then break it up and assign each struct member its values.
• 05-15-2003
_kevin007
i created the file myself, using the following code..
Code:
```#include <stdio.h> int main() { struct primary_plot { double bearing; int range; double elevation; }; typedef struct primary_plot PLOT; PLOT plots[25]; struct primary_location { double bearing; double km_range; double altitude; }; typedef struct primary_location LOCATION; LOCATION loc[25]; int i; FILE *fp; loc[0].km_range = 140.1 ; loc[0].bearing = 260.17 ; loc[0].altitude = 12063 ; loc[1].km_range = 20.3; loc[1].bearing = 281.31; loc[1].altitude = 4312; loc[2].km_range = 20.10; loc[2].bearing = 301.52; loc[2].altitude = 3310; loc[3].km_range = 120.35; loc[3].bearing = 301.51; loc[3].altitude = 10078; loc[4].km_range = 97.15; loc[4].bearing = 328.29; loc[4].altitude = 7320; loc[5].km_range = 170.12; loc[5].bearing = 353.20; loc[5].altitude = 6340; loc[6].km_range = 273.16; loc[6].bearing = 349.17; loc[6].altitude = 12750; loc[7].km_range = 52.73; loc[7].bearing = 0.02; loc[7].altitude = 2160; loc[8].km_range = 81.18; loc[8].bearing = 28.31; loc[8].altitude = 3720; loc[9].km_range = 221.31; loc[9].bearing = 42.11; loc[9].altitude = 6815; loc[10].km_range = 42.31; loc[10].bearing = 83.97; loc[10].altitude = 2170; loc[11].km_range = 42.47; loc[11].bearing = 84.11; loc[11].altitude = 2130; loc[12].km_range = 16.15; loc[12].bearing = 86.10; loc[12].altitude = 753; loc[13].km_range = 73.42; loc[13].bearing = 103.97; loc[13].altitude = 7300; loc[14].km_range = 73.15; loc[14].bearing = 102.73; loc[14].altitude = 8020; loc[15].km_range = 259.63; loc[15].bearing = 151.96; loc[15].altitude = 12630; loc[16].km_range = 278.10; loc[16].bearing = 146.39; loc[16].altitude = 13150; loc[17].km_range = 103.15; loc[17].bearing = 198.21; loc[17].altitude = 8420; loc[18].km_range = 45.18; loc[18].bearing = 207.52; loc[18].altitude = 5190; loc[19].km_range = 310.15; loc[19].bearing = 248.68; loc[19].altitude = 13150; loc[20].km_range = 10.28; loc[20].bearing = 252.13; loc[20].altitude = 853; for (i=0; i<21; i++) { plots[i].bearing = loc[i].bearing; plots[i].range = (int) (loc[i].km_range * 40.0); plots[i].elevation = (loc[i].altitude/(loc[i].km_range*1000) - loc[i].km_range/15200)*180/3.14159; }; fp = fopen("primary", "wb"); fwrite(&plots, sizeof(PLOT), 21, fp); fclose(fp); return 0; }```
how would i go about using fgets() in my situation?
• 05-15-2003
stumon
Here, i have edited your code in some places, you will notice, and i have added a fuction to read the file. There are some bad practices in here, but you can use this to get an idea.
Code:
```#include <stdio.h> #include <stdlib.h> void blah(void); struct primary_plot { double bearing; double range; double elevation; }; typedef struct primary_plot PLOT; struct primary_location { double bearing; double km_range; double altitude; }; typedef struct primary_location LOCATION; int main() { int i; FILE *fp; LOCATION loc[25]; PLOT plots[25]; loc[0].km_range = 140.1 ; loc[0].bearing = 260.17 ; loc[0].altitude = 12063 ; loc[1].km_range = 20.3; loc[1].bearing = 281.31; loc[1].altitude = 4312; loc[2].km_range = 20.10; loc[2].bearing = 301.52; loc[2].altitude = 3310; loc[3].km_range = 120.35; loc[3].bearing = 301.51; loc[3].altitude = 10078; loc[4].km_range = 97.15; loc[4].bearing = 328.29; loc[4].altitude = 7320; loc[5].km_range = 170.12; loc[5].bearing = 353.20; loc[5].altitude = 6340; loc[6].km_range = 273.16; loc[6].bearing = 349.17; loc[6].altitude = 12750; loc[7].km_range = 52.73; loc[7].bearing = 0.02; loc[7].altitude = 2160; loc[8].km_range = 81.18; loc[8].bearing = 28.31; loc[8].altitude = 3720; loc[9].km_range = 221.31; loc[9].bearing = 42.11; loc[9].altitude = 6815; loc[10].km_range = 42.31; loc[10].bearing = 83.97; loc[10].altitude = 2170; loc[11].km_range = 42.47; loc[11].bearing = 84.11; loc[11].altitude = 2130; loc[12].km_range = 16.15; loc[12].bearing = 86.10; loc[12].altitude = 753; loc[13].km_range = 73.42; loc[13].bearing = 103.97; loc[13].altitude = 7300; loc[14].km_range = 73.15; loc[14].bearing = 102.73; loc[14].altitude = 8020; loc[15].km_range = 259.63; loc[15].bearing = 151.96; loc[15].altitude = 12630; loc[16].km_range = 278.10; loc[16].bearing = 146.39; loc[16].altitude = 13150; loc[17].km_range = 103.15; loc[17].bearing = 198.21; loc[17].altitude = 8420; loc[18].km_range = 45.18; loc[18].bearing = 207.52; loc[18].altitude = 5190; loc[19].km_range = 310.15; loc[19].bearing = 248.68; loc[19].altitude = 13150; loc[20].km_range = 10.28; loc[20].bearing = 252.13; loc[20].altitude = 853; for (i=0; i<21; i++) { plots[i].bearing = loc[i].bearing; plots[i].range = (loc[i].km_range * 40.0); plots[i].elevation = (loc[i].altitude/(loc[i].km_range*1000) - loc[i].km_range/15200)*180/3.14159; }; fp = fopen("primary", "wb"); if (fp == NULL) { printf("Blah. Error."); exit(0); }; fwrite(&plots, sizeof(PLOT), 21, fp); fclose(fp); blah(); return 0; } void blah(void) { FILE *fp; int i; PLOT tempstruct[25]; fp = fopen("primary", "r"); if (fp == NULL) { printf("Blah Error again!!!"); exit(0); }; for (i=0; i<20; i++) { fread(&tempstruct[i], sizeof(PLOT), 1, fp); printf("%lf %lf %lf\n", tempstruct[i].range, tempstruct[i].bearing, tempstruct[i].elevation); }; fclose(fp); }```
• 05-15-2003
quzah
Code:
```loc[0].km_range = 140.1 ; loc[0].bearing = 260.17 ; loc[0].altitude = 12063 ;```
Ack. What a nightmare. You should just initialize your arrays when you declare them:
Code:
```struct foo bar[SIZE] = { { 1.2, 3.4, 5.6 }, { 7.8, 9.0, 1.2 }, ... lather, rinse, repeat... { 3.4, 5.6, 7.8 } };```
So much cleaner.
Quzah.
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Mega 6 Math Quizzes 11-20
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These Mega 6 quizzes 11-20 provide students a good six-problem morning (or anytime) review over Math Expressions topics we’ve covered thus far (from Math Expressions Common Core Grade Five 2013). I will continue to write these throughout the year and add them to TpT in groups of 10. Because of my shortened afternoon math block, I have found these morning quizzes very valuable to my math instruction. These 10 quizzes cover the following topics: multiplication, division, rounding decimals, mixed number subtraction and addition, work with properties (associative and commutative), simplifying fractions, word problems, adding decimals, subtraction, multiplying fractions, equation chains, solving for an unknown (quizzes 17 and 20 are heavy on solving for unknowns using variables and fractions), writing numbers in expanded and word form, G.C.F. (greatest common factor), L.C.M. (lowest common multiple), and listing factors. These mini quizzes have six problems, but there are two quizzes per page to save on paper (except for quizzes 17 and 20 that require a full page since a grid is needed to solve the problems). Feel free to modify these to fit your curriculum. Enjoy quiz #17 and the answer key as a preview.
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https://www.openlb.net/forum/topic/implement-gravity-in-shen-chen-model/
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OpenLB – Open Source Lattice Boltzmann Code Forums on OpenLB General Topics Implement Gravity in Shen Chen model
Viewing 5 posts - 1 through 5 (of 5 total)
• Author
Posts
• #4504
zshi6193
Participant
Hello OpenLB,
I am trying to use Shen-chen multicomponent model to simulate the gravity-driven drainage process. To apply a gravity force, I am current using “slatticeOne.defineExteneralField ( geometry, 1, externalforcebeginat, sizeofexternalforce, f) ”
I am not sure how to choose the value of f to link to the gravity. As f=rho*g, f is not a unique value and it is related to the density in a specific cell. Could anyone help me with the implementation of gravity?
Many thanks.
Regards.
#4509
mgaedtke
Keymaster
Hi zshi,
using `slatticeOne.defineExteneralField ( geometry, 1, externalforcebeginat, sizeofexternalforce, f)` to set additional forces when using Shan-Chen-Model is exactly right. Note that our field externalforce is actually F/rho (acceleration), so that you don’t have any problems with variing densities. Just set the externalforce to g in lattice units and you should be fine.
Cheers,
Max
#4514
zshi6193
Participant
Dear Max,
I am sorry that I have one more question regarding the pressure boundary used in Shan-Chen two compartment multiphase model. To set the pressure inlet and pressure outlet, I used the following methods:
slatticeWater.defineRho(supergeometry,3(inlet),rhowater)
slatticeOil.defineRho(supergeometry,3(inlet),zero)
slatticeWater.defineRho(supergeometry,4(outlet),rhowater)
slatticeOil.defineRho(supergeometry,4(outlet),zero).
(I applied above four codes every iteraction to keep the constant pressure)
Alternatively, I also try:
slatticeOil.defineRho(supergeometry,3(inlet),zero).
slatticeOil.defineRho(supergeometry,4(outlet),zero).
The middle part of the domain is filled with oil and the rest is filled with water. But under above set up, the equilibrium state can not be achieved. The pre-filled oil layer moves downwards.
I am not sure that I set up the boundary conditions correctly or not? .Could you help me to check it? Ideally, at the same pressure at the top and bottom (no pressure gradient), the oil in the middle part should be stable.
Thank you very much.
Simon
#4515
mgaedtke
Keymaster
Hi Simon,
that sounds like you have implemented a buoyancy force already and the densities of oil and water differ in your initial conditions.
Best,
Max
#4519
zshi6193
Participant
Dear Max,
Sorry that I didn’t express myself clearly. I would like to test whether the system can reach equilibrium without gravity. So the middle of the domain is filled with oil and the rest part of the domain is filled with water. I set the same density for oil and water. If I use the following code, the system should reach the equilibrium but unfortunately, it didn’t. Would you think that the following codes are correct for the constant pressure boundary at both inlet and outlet?
Also, a very small question, do we need add ” sLatticeWater.addLatticeCoupling(superGeometrytest, 3(inlet), couplingtest, sLatticeOil ) for the inlet and outlet.
Thank you.
Code:
slatticeWater.defineRho(supergeometry,3(inlet),rhowater)
slatticeOil.defineRho(supergeometry,3(inlet),zero) (not sure whether we need to set the pressure for oil is zero)
slatticeWater.defineRho(supergeometry,4(outlet),rhowater)
slatticeOil.defineRho(supergeometry,4(outlet),zero).
(applied in each step)
Viewing 5 posts - 1 through 5 (of 5 total)
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Welcome to 90% as a decimal. In this post you can discover everything around 90 percent together a decimal, including the calculation.
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Right triangle - 9th grade (14y) - examples - page 14
1. Cube wall
The perimeter of one cube wall is 120 meters. Calculate the surface area and the body diagonal of this cube.
2. Isosceles trapezoid
Calculate the content of an isosceles trapezoid whose bases are at ratio 5:3, the arm is 6cm long and it is 4cm high.
3. Circle and rectangle
A rectangle with sides of 11.7 cm and 175 mm is described by circle. What is its length? Calculate the content area of the circle described by this circle.
4. Diagonals
Draw a square ABCD whose diagonals have a length of 6 cm
5. Is right?
Determine whether the triangle with legs (catheti) 19.5 cm and 26 cm and length of the hypotenuse 32.5 cm is rectangular?
6. Inscribed circle
XYZ is right triangle with right angle at the vertex X that has inscribed circle with a radius 5 cm. Determine area of the triangle XYZ if XZ = 14 cm.
7. Lift
The largest angle at which the lift rises is 16°31'. Give climb angle in permille.
8. Roof 7
The roof has the shape of a regular quadrangular pyramid with a base edge of 12 m and a height of 4 m. How many percent is folds and waste if in construction was consumed 181.4m2 of plate?
9. Rhumbline
Find circumference and area of the rhumbline ABCD if the short side AD of which has a length of 5 cm, and the heel of the height from D leading to the AB side divides the AB side into two sections of 3 cm and 4 cm.
10. Concrete block
Determine the volume of concrete block whose one edge of the base has a length 3 meters, body diagonal is 13 meters and its height is 12 meters.
11. Acceleration 2
if a car traveling at a velocity of 80 m/s/south accelerated to a velocity of 100 m/s east in 5 seconds, what is the cars acceleration? using Pythagorean theorem
12. Isosceles right triangle
Contents of an isosceles right triangle is 18 dm2. Calculate the length of its base.
13. Chord 2
Point A has distance 13 cm from the center of the circle with radius r = 5 cm. Calculate the length of the chord connecting the points T1 and T2 of contact of tangents led from point A to the circle.
14. Pyramid 4sides
Calculate the volume and the surface of a regular quadrangular pyramid when the edge of the base is 4 cm long and the height of the pyramid is 7 cm.
15. Right triangle
It is given a right triangle angle alpha of 90 degrees beta angle of 55 degrees c = 10 cm use Pythagorean theorem to calculate sides a and b
16. Cube diagonals
Determine the volume and surface area of the cube if you know the length of the body diagonal u = 216 cm.
17. Diagonal - simple
Calculate the length of the diagonal of a rectangle with dimensions 5 cm and 12 cm.
Show that the quadrilateral with vertices P1(0,1), P2(4,2) P3(3,6) P4(-5,4) has two right triangles.
19. Broken tree
The tree was 35 meters high. The tree broke at a height of 10 m above the ground. Top but does not fall off it refuted on the ground. How far from the base of the tree lay its peak?
20. Cube
Calculate the surface of the cube ABCDA'B'C'D' if the area of rectangle ACC'A' = 344 mm2.
Do you have an interesting mathematical example that you can't solve it? Enter it, and we can try to solve it.
To this e-mail address, we will reply solution; solved examples are also published here. Please enter e-mail correctly and check whether you don't have a full mailbox.
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+0
# help
0
30
1
The length of the longer leg of a right triangle is 19 ft more than five times the length of the shorter leg. The length of the hypotenuse is 20 ft more than five times the length of the shorter leg. Find the side lengths of the triangle.
Feb 9, 2020
#1
+20217
+1
It's a right triangle , so Pythagorean theorem applies
s = short leg
l = long leg = 5s + 19
h= hypotenuse= 5s +20
s^2 + (5s+19)^2 = (5s+20)^2 solve for s then 5s + 19 = other leg and 5s+20 = hypot
can you take it frome here?
Feb 9, 2020
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# Sankarsan
Members
8
1. ## Importing Problem in Acusolve.
Hello experts, I am trying to import a parasolid file, to Acusolve. The parasolid file was generated after deleting the solid and enclosing the fluid volume in Hypermesh. But, while importing to acusolve... It is showing error as attached picture.. Kindly tell me, why it is showing error and how to get rid of it.
2. ## AcuProbe. Sign convection error in Mass flux
And, while using Massflux BC at outlet, can I uncheck Simple B.C?
3. ## AcuProbe. Sign convection error in Mass flux
Hello acupro, While giving mass flux, should I give -8.88kg/s at outlet? To correct the sign convection, should I do this?
4. ## AcuProbe. Sign convection error in Mass flux
Hello acupro, I performed the simulation according to your suggestion. At inlet I gave 8.88 kg/s and at both the outlet I gave 16kp of pressure. After solving.. My Pressure at outlet is also changing. Which itself is a BC. May I know, what is this mean?
5. ## AcuProbe. Sign convection error in Mass flux
Ok, The pressure value in Outlet will be a stagnation pressure or Normal.. As Acusolve gives 2 options for pressure BC. I am bit confused. Can you explain a bit about these 2 different types and in which applications we should use them?!
6. ## AcuProbe. Sign convection error in Mass flux
So, For my problem, Which set of BC will be best from the given known values? -Flow rate from each 3 side and Pressure at 2 Outlets. **Objective is to determine, pressure loss through the Manifold.
7. ## AcuProbe. Sign convection error in Mass flux
Ok, Thanks for the info. 1. So, If this is to be true, is My way of giving BC wrong? and If wrong, How to correct it. 2. Can we give mass flow rate at both inlet and outlet as Bc? 3. Similarly Can we give Pressure as Inlet and outlet Bc?
8. ## AcuProbe. Sign convection error in Mass flux
Hello Experts, I am simulating fluidflow through 2 outlet one inlet manifold. My inlet BC is 8.88 Kg/s with Bc type Inflow and other two as outflow. After running simulation, I plotted graph of mass flux at inlet using acu probe, surface output. Surprisingly, It was showing -8.88 Kg/s throughout the time. But the outlet mass flow rate was 4.44 kg/s. Which is expected. My Q, Why is the sign convection is negative at inlet. From my understanding It should be Positive , ad I have given it as inflow with 8.88 Kg/s. Kindly, give me your views. Thank you in advance.
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# 24. The equation 12x² + 4kx + 3 = 0 has real and equal roots, if (a) k = ±3 (b) k = ±9 (c) k = 4 (d) k = ±2
Grade:12th pass
## 1 Answers
Pawan Prajapati
askIITians Faculty 60785 Points
one year ago
Answer: a Explaination:Reason: Here a = 12, b = 4k, c = 3 Since the given equation has real and equal roots ∴ b² – 4ac = 0 ⇒ (4k)² – 4 × 12 × 3 = 0 ⇒ 16k² – 144 = 0 ⇒ k² = 9 ⇒ k = ±3
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At present, a manufacturer has 2500 units of product in stock
Math
At present, a manufacturer has 2500 units of product in stock. The product is now selling at RM4 per unit. Next month the unit price will increase by RM0.50. The manufacturer wants the total revenue received from the sale of 2500 units to be no less than RM10750. If all 2500 units were to be sold in the two months, what is the maximum number of units that can be sold this month?
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In order for an organism to evolve or survive any length of : GMAT Sentence Correction (SC)
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# In order for an organism to evolve or survive any length of
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In order for an organism to evolve or survive any length of [#permalink]
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In order for an organism to evolve or survive any length of time, it must learn to somehow deal with the issue of UV rays, which implies that the very oldest sorts of organisms that exist on Earth today all have mechanisms such as the production of sunscreen pigments like melanin, also known as tanning, to protect themselves.
A) which implies that the very oldest sorts of organisms that exist on Earth today all have mechanisms such as the production of sunscreen pigments like melanin, also known as tanning, to protect themselves.
B) which infers the fact that the very oldest sorts of organisms that exist on Earth today all have mechanisms such as the production of sunscreen pigments like melanin, also known as tanning, to protect themselves.
C) which is inferred from the fact that the very oldest sorts of organisms that exist on Earth today all have mechanisms such as the production of sunscreen pigments like melanin, also known as tanning, to protect themselves.
D) which can be interpreted from the fact that the very oldest sorts of organisms that exist on Earth today all have mechanisms such as the production of sunscreen pigments like melanin, also known as tanning, to protect themselves.
E) which can be implied from the fact that the very oldest sorts of organisms that exist on Earth today all have mechanisms such as the production of sunscreen pigments like melanin, also known as tanning, to protect themselves.
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21 Dec 2003, 07:35
I would go with C.
"organism need to deal with the issue of UV rays" (X)
can be inferred from the fact that
"the very oldest sorts of organisms have protection mechanism". (Y)
i.e X can be inferred from Y.
A confuses X and Y saying that,
X implies Y
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07 Jul 2011, 08:39
mbamantra wrote:
In order for an organism to evolve or survive any length of time, it must learn to somehow deal with the issue of UV rays, which implies that the very oldest sorts of organisms that exist on Earth today all have mechanisms such as the production of sunscreen pigments like melanin, also known as tanning, to protect themselves.
A) which implies that the very oldest sorts of organisms that exist on Earth today all have mechanisms such as the production of sunscreen pigments like melanin, also known as tanning, to protect themselves.
B) which infers the fact that the very oldest sorts of organisms that exist on Earth today all have mechanisms such as the production of sunscreen pigments like melanin, also known as tanning, to protect themselves.
C) which is inferred from the fact that the very oldest sorts of organisms that exist on Earth today all have mechanisms such as the production of sunscreen pigments like melanin, also known as tanning, to protect themselves.
D) which can be interpreted from the fact that the very oldest sorts of organisms that exist on Earth today all have mechanisms such as the production of sunscreen pigments like melanin, also known as tanning, to protect themselves.
E) which can be implied from the fact that the very oldest sorts of organisms that exist on Earth today all have mechanisms such as the production of sunscreen pigments like melanin, also known as tanning, to protect themselves.
the question here is imply vs infer.
I go with C here .
A little bit of CR skills help here. the first line' In order for an organism to evolve or survive any length of time, it must learn to somehow deal with the issue of UV rays' is the conclusion . and rest of the passage i simple fact.
The first sentence conclusion is based on the rest of the passage.
thus a conclusion drawn can only 'infer' from rest of the passage and not imply.
Re: SC: organisms [#permalink] 07 Jul 2011, 08:39
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# 4.3 Introduction to Probability
Kanya Shah
Jed Quiaoit
A
Aly Moosa
Kanya Shah
Jed Quiaoit
A
Aly Moosa
💡 The big idea of this section and the rest of the math-y part of probability: The likelihood of a random event can be quantified!
That being said, we'll learn how to (1) calculate probabilities for events and their and (2) interpret probabilities for events. Not too daunting for an introductory dive into the basics of probability, huh? 🔐
## Basic Probability Rules
A is a mathematical representation of a chance process that is used to describe the likelihood of different outcomes occurring. It consists of two main components: a list of all possible outcomes and the probability of each outcome occurring.
The sample space is a list of all the possible outcomes of a chance process, such as flipping a coin or rolling a die. For example, the sample space for flipping a coin might be {heads, tails}, while the sample space for rolling a die might be {1, 2, 3, 4, 5, 6}.
Using this model, probabilities are usually expressed as fractions or decimals. For example, the probability of flipping heads when flipping a coin is 0.5, and the probability of rolling a 6 when rolling a die is 1/6.
A is a description of some chance process that is made up of two portions: a list of all possible outcomes and the probability for each outcome. The list of all possible outcomes is called a sample space.
### Rule 1: Equally Likely Probability
If all outcomes in a sample space are equally likely, the probability that event A occurs can be calculated by dividing the number of outcomes in event A by the total number of outcomes in the sample space. This is known as the classical definition of probability. 🟰
P(A) = number of outcomes in event A / total number of outcomes in the sample space.
For example, if we have a sample space of 6 outcomes (such as rolling a die), and event A consists of 3 of those outcomes (such as rolling a 1, 2, or 3), the probability of event A occurring would be 3/6, or 0.5. This is because there are 3 outcomes in event A, and there are a total of 6 outcomes in the sample space, so the probability of event A occurring is 3/6.
### Rule 2: 0 Probability 1
Probabilities are always expressed as numbers between 0 and 1, inclusive, with a probability of 0 indicating that an event is impossible, a probability of 1 indicating that an event is certain to occur, and probabilities between 0 and 1 indicating that an event is possible but not certain.
For example, the probability of flipping heads when flipping a coin is 0.5, the probability of rolling a 6 when rolling a die is 1/6, and the probability of drawing an ace from a deck of cards is 4/52. These probabilities all fall between 0 and 1, indicating that these events are possible but not certain to occur.
When you do your math, getting probability values like -0.82 and 1.63 is an instant red flag! The good thing with this rule is that you can immediately sense something's wrong should you get such answers and immediately troubleshoot. 🚩
### Rule 3: P(all outcomes) 1!
All possible outcomes MUST add up to 1. That's right! The probabilities of all possible outcomes in a sample space add up to 1. This is because the probability of an event occurring is a measure of the likelihood of that event occurring, and the probability of all possible events occurring must be 1 (since one of those events must occur).
For example, consider the probability of rolling a die. There are six possible outcomes (rolling a 1, 2, 3, 4, 5, or 6), and the probability of each outcome occurring is 1/6. If we add up the probabilities of all six outcomes, we get:
1/6 + 1/6 + 1/6 + 1/6 + 1/6 + 1/6 = 6/6 = 1
This demonstrates that the probabilities of all possible outcomes (rolling a 1, 2, 3, 4, 5, or 6) add up to 1, as expected. ➕
### Rule 4: Complements
The probability of the complement of an event (also known as the inverse or opposite of an event) is equal to 1 minus the probability of the event occurring. This is because the probability of an event occurring and the probability of the event not occurring must add up to 1, since one of those two events must occur. ❌
Source: eCampus Ontario
For example, consider the probability of flipping heads when flipping a coin. The probability of flipping heads is 0.5, so the probability of flipping tails (which is the complement of flipping heads) is 1 - 0.5 = 0.5. Similarly, if the probability of rolling a 6 when rolling a die is 1/6, the probability of rolling any number other than a 6 (which is the complement of rolling a 6) is 1 - 1/6 = 5/6.
The College Board sometimes refers to the complement as E' or E^c (i.e. not E) = 1 - P(E). You'll most likely have to use this rule when problems ask for probability values mentioninig "at least ___" or "at most ___."
As you go through more sections, we'll add onto this growing list of probability rules. 😉
## Probabilities and Context
Be sure to know how to interpret the probability you calculate. Remember to use words involving context!
### Example
A company that manufactures car parts wants to understand the quality of their production process. To do this, they decide to sample a of 100 car parts and test them for defects. The results of the testing show that 20 parts are defective and 80 parts are non-defective.
Based on the results of the testing, the company makes the following conclusions:
• The probability that a randomly selected car part is defective is 20%.
• The probability that a randomly selected car part is non-defective is 80%.
• There is a 20% chance that a randomly selected car part will be defective.
• There is an 80% chance that a randomly selected car part will be non-defective.
Write a brief summary explaining the company's conclusions and the statistical evidence that supports them.
The company concludes that the probability that a randomly selected car part is defective is 20%, and that the probability that a randomly selected car part is non-defective is 80%, based on the statistical evidence provided in the results of the testing. The company also concludes that there is a 20% chance that a randomly selected car part will be defective, and an 80% chance that a randomly selected car part will be non-defective.
The statistical evidence that supports these conclusions is the fact that 20 out of 100 car parts were defective, and 80 out of 100 car parts were non-defective. This means that the probability of a randomly selected car part being defective is 20/100, or 20%, and the probability of a randomly selected car part being non-defective is 80/100, or 80%.
🎥Watch: AP Stats - Probability Rules and Random Variables
Courtesy of Make-a-Meme
# Key Terms to Review (4)
Complements
: Complements are events that consist of all outcomes not included in a given event.
Equally Likely Probability
: Equally likely probability refers to the situation where all possible outcomes of an event have the same chance of occurring.
Probability Model
: A probability model is a mathematical representation of a random phenomenon. It describes all possible outcomes and assigns probabilities to each outcome.
Random Selection
: Random selection refers to the process of selecting individuals from a population in such a way that every individual has an equal chance of being chosen.
# 4.3 Introduction to Probability
Kanya Shah
Jed Quiaoit
A
Aly Moosa
Kanya Shah
Jed Quiaoit
A
Aly Moosa
💡 The big idea of this section and the rest of the math-y part of probability: The likelihood of a random event can be quantified!
That being said, we'll learn how to (1) calculate probabilities for events and their and (2) interpret probabilities for events. Not too daunting for an introductory dive into the basics of probability, huh? 🔐
## Basic Probability Rules
A is a mathematical representation of a chance process that is used to describe the likelihood of different outcomes occurring. It consists of two main components: a list of all possible outcomes and the probability of each outcome occurring.
The sample space is a list of all the possible outcomes of a chance process, such as flipping a coin or rolling a die. For example, the sample space for flipping a coin might be {heads, tails}, while the sample space for rolling a die might be {1, 2, 3, 4, 5, 6}.
Using this model, probabilities are usually expressed as fractions or decimals. For example, the probability of flipping heads when flipping a coin is 0.5, and the probability of rolling a 6 when rolling a die is 1/6.
A is a description of some chance process that is made up of two portions: a list of all possible outcomes and the probability for each outcome. The list of all possible outcomes is called a sample space.
### Rule 1: Equally Likely Probability
If all outcomes in a sample space are equally likely, the probability that event A occurs can be calculated by dividing the number of outcomes in event A by the total number of outcomes in the sample space. This is known as the classical definition of probability. 🟰
P(A) = number of outcomes in event A / total number of outcomes in the sample space.
For example, if we have a sample space of 6 outcomes (such as rolling a die), and event A consists of 3 of those outcomes (such as rolling a 1, 2, or 3), the probability of event A occurring would be 3/6, or 0.5. This is because there are 3 outcomes in event A, and there are a total of 6 outcomes in the sample space, so the probability of event A occurring is 3/6.
### Rule 2: 0 Probability 1
Probabilities are always expressed as numbers between 0 and 1, inclusive, with a probability of 0 indicating that an event is impossible, a probability of 1 indicating that an event is certain to occur, and probabilities between 0 and 1 indicating that an event is possible but not certain.
For example, the probability of flipping heads when flipping a coin is 0.5, the probability of rolling a 6 when rolling a die is 1/6, and the probability of drawing an ace from a deck of cards is 4/52. These probabilities all fall between 0 and 1, indicating that these events are possible but not certain to occur.
When you do your math, getting probability values like -0.82 and 1.63 is an instant red flag! The good thing with this rule is that you can immediately sense something's wrong should you get such answers and immediately troubleshoot. 🚩
### Rule 3: P(all outcomes) 1!
All possible outcomes MUST add up to 1. That's right! The probabilities of all possible outcomes in a sample space add up to 1. This is because the probability of an event occurring is a measure of the likelihood of that event occurring, and the probability of all possible events occurring must be 1 (since one of those events must occur).
For example, consider the probability of rolling a die. There are six possible outcomes (rolling a 1, 2, 3, 4, 5, or 6), and the probability of each outcome occurring is 1/6. If we add up the probabilities of all six outcomes, we get:
1/6 + 1/6 + 1/6 + 1/6 + 1/6 + 1/6 = 6/6 = 1
This demonstrates that the probabilities of all possible outcomes (rolling a 1, 2, 3, 4, 5, or 6) add up to 1, as expected. ➕
### Rule 4: Complements
The probability of the complement of an event (also known as the inverse or opposite of an event) is equal to 1 minus the probability of the event occurring. This is because the probability of an event occurring and the probability of the event not occurring must add up to 1, since one of those two events must occur. ❌
Source: eCampus Ontario
For example, consider the probability of flipping heads when flipping a coin. The probability of flipping heads is 0.5, so the probability of flipping tails (which is the complement of flipping heads) is 1 - 0.5 = 0.5. Similarly, if the probability of rolling a 6 when rolling a die is 1/6, the probability of rolling any number other than a 6 (which is the complement of rolling a 6) is 1 - 1/6 = 5/6.
The College Board sometimes refers to the complement as E' or E^c (i.e. not E) = 1 - P(E). You'll most likely have to use this rule when problems ask for probability values mentioninig "at least ___" or "at most ___."
As you go through more sections, we'll add onto this growing list of probability rules. 😉
## Probabilities and Context
Be sure to know how to interpret the probability you calculate. Remember to use words involving context!
### Example
A company that manufactures car parts wants to understand the quality of their production process. To do this, they decide to sample a of 100 car parts and test them for defects. The results of the testing show that 20 parts are defective and 80 parts are non-defective.
Based on the results of the testing, the company makes the following conclusions:
• The probability that a randomly selected car part is defective is 20%.
• The probability that a randomly selected car part is non-defective is 80%.
• There is a 20% chance that a randomly selected car part will be defective.
• There is an 80% chance that a randomly selected car part will be non-defective.
Write a brief summary explaining the company's conclusions and the statistical evidence that supports them.
The company concludes that the probability that a randomly selected car part is defective is 20%, and that the probability that a randomly selected car part is non-defective is 80%, based on the statistical evidence provided in the results of the testing. The company also concludes that there is a 20% chance that a randomly selected car part will be defective, and an 80% chance that a randomly selected car part will be non-defective.
The statistical evidence that supports these conclusions is the fact that 20 out of 100 car parts were defective, and 80 out of 100 car parts were non-defective. This means that the probability of a randomly selected car part being defective is 20/100, or 20%, and the probability of a randomly selected car part being non-defective is 80/100, or 80%.
🎥Watch: AP Stats - Probability Rules and Random Variables
Courtesy of Make-a-Meme
# Key Terms to Review (4)
Complements
: Complements are events that consist of all outcomes not included in a given event.
Equally Likely Probability
: Equally likely probability refers to the situation where all possible outcomes of an event have the same chance of occurring.
Probability Model
: A probability model is a mathematical representation of a random phenomenon. It describes all possible outcomes and assigns probabilities to each outcome.
Random Selection
: Random selection refers to the process of selecting individuals from a population in such a way that every individual has an equal chance of being chosen.
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## Sunday, January 29, 2012
### Problem 723: Squares, Circumscribed Circles, Collinearity
Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.
Click the figure below to see the problem 723 details.
1. First we observe that diagonal AOC is a diameter of circle (O) since <ADC is a rt angle.
<AHC = <ADC = 90 deg
Similarly <EHG = < EFG = 90 deg
By Problem 722,
A, E, H are collinear and so <AHG is the same as <EHG = 90 deg
Hence <AHC + <AHG = 180 deg and
C, H, G are collinear.
2. Solution 2:
<DHG = <DEG = 45 deg
<DHC = supplement of <DBC = 180deg - 45deg
Sum = 180 deg
So G, H, C are collinear.
[Note: DB passes thro'O and bisects the rt <ABC]
3. We can prove this problem using result of problem 722 or
Note that DF and DB are diameters of circles O and O’
So (BHD)=(DHF)=90 >> B,H, F are collinear
(GHF)=(CHB)=45 ----- (angles face 90 arc)
So (CHG)=(BHF)+45-45= 180 >> C, H, G are collinear
4. Reference 722 < AHC = 90 = < EHG. Hence CHG are collinear
5. Let the interecting point of CG & EF be Y
Let <ECY=x
<EYC=90-x=<FYG
<CYF=180-<EYC=90+x
<CYF+<FYG=90+x+90-x=180
CHG is a st. line
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# math
posted by .
A manufacturer has a maximum of 240, 360, and 180 kilograms of wood, plastic and steel available. The company produces two products, A and B. Each unit of A requires 1, 3 and 2 kilograms of wood, plastic and steel respectively; each unit of B requires 3, 4 and 1 kilograms of wood, plastic and steel respectively, and each unit of B requires 3, 4 and 1 kilograms of wood, plastic and steel respectively. The profit per unit of A and B is \$4.00 and \$6.00 respectively.
Identify all constraints.
Identify all applicable corner points of the feasibility region.
How many units of A and B should be manufactured in order to maximize profits? What would the maximum profit be?
• math -
I did two of these linear programming problems. You do one.
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# How many DRB's in LTE? can some one explain their Functions?
1,820 views
How many DRB's in LTE? can some one explain their Functions?
posted Sep 16, 2013
Answer is 8. As per Table B.1-1 of 36.331 following DRB combinations are supported, based on featureGroupIndicators, received in UE capability information message .
Case 1 ( Bit 7 and bit 20 are set 1 ) :
``````SRB1 and SRB2 for DCCH + 8x AM DRB OR SRB1 and SRB2 for DCCH + 5x AM DRB + 3x UM DRB
``````
Case 2 ( BIT 7 is set 0 and BIT 20 is set to 1) :
``````SRB1 and SRB2 for DCCH + 8x AM DRB
``````
Case 3 ( BIT 7 and BIT 20 are set to any value ie 0/1) :
``````SRB1 and SRB2 for DCCH + 4x AM DRB
``````
Case 4 ( BIT 7 is set to 1 and BIT 20 is set to any value 0/1) :
``````SRB1 and SRB2 for DCCH + 4x AM DRB + 1 UM DRB
``````
I hope this should clarify ur doubt.
Thank You for your earliest Response.
can u please explain little more about the BIT 7 and BIT 20? why depends on this value only RLC mode is changing?
Not sure what u are trying to ask, let me explain bit 20 and Bit 7 -
Regardless of what bit number 20 is set to, if bit number 7 is set to ‘1’, UE shall support at least SRB1 and SRB2 for DCCH + 4x AM DRB + 1x UM DRB
Similar Questions
+1 vote
Hi All,
It is mentioned in different websites that in LTE max. of 8 DRB can be established per UE. But as per page 652 of 36.331 V13.2.0, the max no. of DRB is 11.
"maxDRB INTEGER ::= 11 -- Maximum number of Data Radio Bearers"
Based on what source are we saying that max. DRB in LTE is 8 and not 11.
It was mentioned in one of the old threads the following:
If EPS bearer ID is = x +4 ;
Then
DRB ID = x;
Logical Channel ID = x + 2
I found the above relationship of DRB ID & EPS Bearer ID to be true for one of the eNB vendors but not for another one.
Example:
EPS ID is 5 and DRB ID is 4 instead of 1. Can somebody comment in this?
NOTE : All these ranges is with respect to DRB establishment
EPS Bearer ID Range is INTEGER (0..15) (But 0-4 is reserved so we use starting from 5 )
DRB ID Range is INTEGER (1..32) (As in LTE max 8 DRB can be established per UE so we use from 1-8)
Logical Channel ID range for DRBs is INTEGER (3..10) ( For for DRB1 we use 3 as logical channed ID)
Thanks
+1 vote
Why the numbers of SRB’s and DRB’s are fixed as 3 and 8 in LTE?
+1 vote
I have very limited insight in mutlimedia multicast/broadcast services. An article says one LTE cell can have maximum of 8 MCCH channels and each MCCH configuration corresponds to configuration of one MBSFN area.
If this is correct then it concludes that one LTE cell can be part of max 8 MBSFN areas.
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# Chapter 4 Basic Geometrical Ideas
Posted on: October 16, 2021 Posted by: user Comments: 0
## Exercise 4.1
Question 1.
Use the figure to name:
(a) Five points
(b) A line
(c) Four rays
(d) Five line segments.
Solution :
(a) O, B, C. D, E
Question 2.
Name the line given in all possible (twelve) ways, choosing only two letters at a time from the four given.
Solution :
Question 3.
Use the figure to name:
(a) The line containing point E.
(b) The line passing through A.
(c) The line on which O lies
(d) Two pairs of intersecting lines.
Solution :
(a)AE, etc.
(b) AE, etc.
(c) CO or OC
(d) CO, AE; AE, EF.
Question 4.
How many lines can pass through
(a) one given point?
(b) two given points?
Solution :
(a) Countless lines can pass through one given point.
(b) One and only one line can pass through two given points.
Question 5.
Draw a rough figure and label suitably in each of the following cases :
(a) Point P lies on AB.
(b) XY and PQ intersect at M.
(c) Line contains E and F but not D.
(d) Op and OQ meet at O.
Solution :
(a)
(b)
(c)
(d)
Question 6.
Consider the following figure of line MN. Say whether following statements are true or false in context of the given figure.
(a) Q, M, O, N, P are points on the line MN
(b) M, O, N are points on a line segment MN.
(c) M and N are end points of line segment MN.
(d) O and N are end points of line segment OP.
(e) M is one of the end points of line segment QO.
(f) M is point on ray OP.
(g) Ray OP is different from ray QP.
(h) Ray OP is same as ray OM.
(i) Ray OM is not opposite to ray OP.
(j) O is not an initial point of OP.
(k) N is the initial point of NP and NM.
Solution :
(a) True
(b) True
(c) True
(d) False
(e) False
(f) False
(g) True
(h) False
(i) False
(j) False
(k) True.
## Exercise 4.2
Question 1.
Classify the following curves as
(i) open
(ii) closed.
Solution :
(i) a, c
(ii) b,d,e.
Question 2.
Draw rough diagrams to illustrate the following :
(a) Open curve
(b) Closed curve.
Solution :
(a)
(b)
Question 3.
Draw any polygon and shade its interior.
Solution :
Question 4.
Consider the given figure and answer the questions:
(a) Is it a curve?
(b) Is it closed?
Solution :
(a) Yes! It is a curve
(b) Yes! It is closed.
Question 5.
Illustrate, if possible, each one of the following with a rough diagram :
(a) A closed curve that is not a polygon
(b) An open curve made up entirely of line segments.
(c) A polygon with two sides.
Solution :
(a)
(b)
(c) Not possible.
## Exercise 4.3
Question 1.
Name the angles in the given figure.
Solution :
∠A or ∠DAB; ∠B or ∠ABC; ∠C or ∠BCD; ∠D or ∠CDA.
Question 2.
In the given diagram, name the point (s)
(a) in the interior of ∠DOE
(b) in the exterior of∠EOF
(c) on ∠EOF.
Solution :
(a) A
(b) C, A, D
(c) E, B,0, F.
Question 3.
Draw rough diagrams of two angles such that they have
(a) One point in common
(b) Two points in common
(c) Three points in common
(d) Four points in common
(e) One ray in common.
Solution :
(a)
∠AOB and ∠BOC have one point O in common.
(b) ∠AOB and ∠OBC have two points O and B in common.
(c) Not possible
(d) Not possible
(e)
∠AOB and ∠BOC have one ray OB in common
## Exercise 4.4
Question 1.
Draw a rough sketch of a triangle ABC. Mark a point P in its interior and a point Q in its exterior. Is the point A in its exterior or in its exterior?
Solution :
The point A is neither in the exterior nor in the interior of triangle ABC. It is on the triangle ABC.
Question 2.
(a) Identify three triangles in the figure.
(b) Write the names of seven angles,
(c) Write the names of the six line segments,
(d) Which two triangles have ∠B as common?
Solution :
(a) Three triangles
Triangle ABC,
Triangle ABD,
(b) Seven Angles
(c) Six line segments
AB, AC, BC , AD, BD, DC
(d) ΔABC. ΔABD.
## Exercise 4.5
Question 1.
Draw a rough sketch of a quadrilateral PQRS. Draw its diagonals. Name them. Is the meeting point of the diagonal in the interior or exterior of the quadrilateral?
Solution :
The meeting point O of the diagonals PR and QS of the quadrilateral PQRS is in the interior of the quadrilateral PQRS.
Question 2.
Draw a rough sketch of a quadrilateral KLMN. State :
(a) two pairs of opposite sides.
(b) two pairs of opposite angles.
(c) two pairs of adjacent sides.
(d) two pairs of adjacent angles.
Solution :
(a) KL, NM and KN, ML
(b) ∠K, ∠M and ∠N, ∠L
(c) KL, KN and NM, ML or KL , LM and NM, ML
(d) ∠K, ∠L, and ∠M. ∠N or ∠K. ∠L and ∠L, ∠M etc.
Question 3.
Investigate:
Use strips and fasteners to make a triangle and a quadrilateral. Try to push inward at any one vertex of the triangle. Do the same to the quadrilateral. Is the triangle distorted? Is the quadrilateral distorted? Is the triangle rigid? Why is it that structures like electric towers make use of triangular shapes and not quadrilaterals?
Solution :
On pushing inward at any one vertex of the triangle, the triangle is not distorted, However, the quadrilateral is distorted. Hence, a triangle is a rigid figure. This is why structures like electric towers make use of triangular shapes and not quadrilaterals.
## Exercise 4.6
Question 1.
From the figure identify
(a) the center of the circle
(c) a diameter
(d) a chord
(e) two points in the interior
(f) a point in the exterior
(g) a sector
(h) a segment.
Solution :
(a) O is the centre of the circle.
(b) OA, OB, OC are three radii of the circle.
(c) AC is a diameter of the circle.
(d) ED is a chord of the circle,
(e) O and P are two points in the interior.
(f) Q is a point in the exterior.
(e) OAB (shaped portion) is a sector of the circle.
(f) Shaded portion of the circular region enclosed by line segment ED and the corresponding arc.
Question 2.
(a) Is every diameter of a circle also a chord?
(b) Is every chord of a circle also a diameter?
Solution :
(a) Yes! every diameter of a circle is also a chord.
(b) No ! every chord of a circle is not also a diametre.
Question 3.
Draw any circle and mark:
(a) it’s center
(c) a diameter
(d) a sector
(e) a segment
(f) a point in its interior
(g) a point in its exterior
(h) an arc.
Solution :
(a) O is the centre.
(c) AB is a diameter.
(d) OBC is a sector.
(e) AGD is a segment.
(f) P is a point in its interior.
(g) Q is a point in its exterior.
(h) EF is an arc.
Question 4.
Say true or false:
(a) Two diameters of a circle will necessarily intersect.
(b) The center of a circle is always in its interior.
Solution :
(a) True
(b) True.
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# Getting critical values - clarify solution [duplicate]
Two brands of coffee were compared. Two independent random samples of 50 people each were asked to taste either Brand A or Brand B coffee, and indicate whether they liked it or not. Eighty four percent of the people who tasted Brand A liked it; the analogous sample proportion for Brand B was ninety percent.
(A) At α = 0.01, is there a significant difference in the proportions of individuals who like the two coffees? Use the p-value approach.
(B) What is the critical value(s) for the test in Part(A)?
So I understand my teacher's solution for part A but I don't really understand how they got the values for B. Can someone explain? I tried looking up tables but I couldn't find where 2.575 came from. I feel like I might be looking at the wrong table or reading them wrong.
• Aug 12, 2019 at 18:04
Exact p values cannot always be found in tables. 2.575 comes from being the halfway point between 2.57 and 2.58. Please look at the following table:
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# SQL - ABS() Function
The SQL ABS() function accepts a single numeric value as an argument and returns the corresponding absolute value for this numeric value.
The absolute value is defined as the distance of a particular point on a number line from zero, irrespective of its direction. Since, the absolute value does not take the direction into consideration, it is never negative. Therefore, for this method, if the argument passed is not negative, the original argument is returned. But if the argument passed is negative, the result will be the negation of this argument.
### Syntax
Following is the syntax of SQL ABS() function −
```ABS(number)
```
where, number is the value for which we need to find the absolute value.
### Example
If we pass a positive value as an argument to the SQL ABS() function, it returns the same value as shown below −
```SELECT ABS(0.8)
AS Aboslute_Value
```
When we run above program, it produces following result −
```+----------------+
| Aboslute_Value |
+----------------+
| 0.8 |
+----------------+
```
### Example
If we pass a negative value as an argument to this function, it returns the same value without the negative sign as shown below −
```SELECT ABS(-55787)
AS Aboslute_Value
```
While executing the above code we get the following output −
```+----------------+
| Aboslute_Value |
+----------------+
| 55787 |
+----------------+
```
### Example
If we pass the mathematical constant PI as an argument, this function returns its equivalent constant value as shown below −
```SELECT ABS(PI())
AS Aboslute_Value
```
Following is an output of the above code −
```+------------------+
| Aboslute_Value |
+------------------+
| 3.14159265358979 |
+------------------+
```
### Example
If the value passed is NULL or, in case of an error, this function returns NULL.
```SELECT ABS(NULL)
AS Aboslute_Value
```
Output of the above code is as follows −
```+----------------+
| Aboslute_Value |
+----------------+
| NULL |
+----------------+
```
### Example
If we pass the value to this function in the form of a string, it returns the same value without negative sign(if given).
```SELECT ABS('-225')
AS Aboslute_Value
```
Following is the output of the above code −
```+----------------+
| Aboslute_Value |
+----------------+
| 225 |
+----------------+
```
### Example
If the value passed to this function is neither null nor a numeric value, it raises an error.
```SELECT ABS('string')
AS Aboslute_Value
```
We get the following output while executing the above code −
```Msg 8114, Level 16, State 5, Line 1
Error converting data type varchar to float.
```
### Example
Assume we have created a table with name CUSTOMERS as shown below −
```create table CUSTOMERS(ID INT NOT NULL,
NAME VARCHAR(20) NOT NULL,
AGE INT NOT NULL,
SALARY DECIMAL(18, 2),
PRIMARY KEY(ID));
Commands completed successfully.
```
Let us insert r values into it −
```insert INTO CUSTOMERS VALUES(1, 'Ramesh', 32, 'Ahmedabad', 2000.00);
insert INTO CUSTOMERS VALUES(2, 'Khilan', 25, 'Delhi', 1500.00);
insert INTO CUSTOMERS VALUES(3, 'kaushik', 23, 'Kota', 2000.00);
insert INTO CUSTOMERS VALUES(4, 'Chaitali', 25, 'Mumbai', 6500.00);
insert INTO CUSTOMERS VALUES(5, 'Hardik', 27, 'Bhopal', 8500.00);
insert INTO CUSTOMERS VALUES(6, 'Komal', 22, 'MP', 4500.00);
insert INTO CUSTOMERS VALUES(7, 'Muffy', 24, 'Indore', 10000.00);
()
()
()
()
()
()
()
```
Following query calculates the EPF (equal to 12% of the basic salary) of all the customers that is deducted from the salary per month −
```SELECT SALARY,
abs(SALARY*0.12)
as EPF
FROM customers;
```
The result produced is as follows −
```+----------+-----------+
| SALARY | EPF |
+----------+-----------+
| 2000.00 | 240.0000 |
| 1500.00 | 180.0000 |
| 2000.00 | 240.0000 |
| 6500.00 | 780.0000 |
| 8500.00 | 1020.0000 |
| 4500.00 | 540.0000 |
| 10000.00 | 1200.0000 |
+----------+-----------+
```
sql-numeric-functions.htm
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Cody
# Problem 167. Pizza!
Solution 1689180
Submitted on 10 Dec 2018 by Qiling Zhao
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
z = 1; a = 1; v_correct = pi; assert(isequal(pizza(z,a),v_correct))
2 Pass
z = 2; a = 1; v_correct = 4*pi; assert(isequal(pizza(z,a),v_correct))
3 Pass
z = 1; a = 2; v_correct = 2*pi; assert(isequal(pizza(z,a),v_correct))
4 Pass
z = 1; a = 2; v_correct = 2*pi; assert(isequal(pizza(z,a),v_correct))
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Question:1. Deduce the combined law of Charles & Boyle’s law.
Or, Find out the relationship among temperature, pressure & volume of gases
Or, Show that, the product of pressure and volume is directly proportional to the absolute temp.
Ans.: According to Boyle’s law: “At constant temperature, the volume of a definite mass of a given gas is inversely proportional to the pressure exerted on it.”
According to Charles’ law: “At constant pressure, the volume of a definite mass of a given gas is directly proportional to the absolute temperature.
Combining above two laws, (equations i & ii) –
That is, the product of pressure & volume of a fixed mass of gas is directly proportional to its absolute temperature. This is called the combined law of Charles’ & Bovle’s
Question:2. Deduce combined law of Boyle’s, Charles’ & Avogadro’s law.
Or, Deduce ideal gas equation. Or, Show that PV = nRT.
Ans.: According to Boyle’s law: “At constant temp., the volume of a definite mass of a given gas is inversely proportional to the pressure exerted on the gas.”
According to Charles’ law: At constant pressure, the volume of a definite mass of a given gas is directly proportional to its absolute temperature.”
According to Avogadro’s law: Amadeo Avogadro put a relationship between volume of a gas to the number of molecules at constant temperature and pressure. This has now been accepted as a law and is known as Avogadro’s law. The law states that, “Equal volumes of all gases, under similar conditions of temperature and pressure contain equal number of molecules”. Then, from Avogadro’s law one can write,
Combining above three laws, (equations i, ii & iii) –
Where R is a constant of proportionality and is known as universal gas constant. The value of R is same for all gases. However, the numerical value of R varies with the units in which pressure, volume and temp. are expressed. Finally we can express above equation as
PV = nRT.
The above equation (4) is called ideal gas equation.
So, the product of the volume of a given gas and the pressure exerted on it, is equal to the product of mole number of gas, molar gas constant & absolute temperature.
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# Gall (Gall Stereographic)¶
The Gall stereographic projection, presented by James Gall in 1855, is a cylindrical projection. It is neither equal-area nor conformal but instead tries to balance the distortion inherent in any projection.
Classification Transverse and oblique cylindrical Available forms Forward and inverse, Spherical Defined area Global Implemented by Gerald I. Evenden Options No special options for this projection
## Usage¶
The need for a world map which avoids some of the scale exaggeration of the Mercator projection has led to some commonly used cylindrical modifications, as well as to other modifications which are not cylindrical. The earliest common cylindrical example was developed by James Gall of Edinburgh about 1855 (Gall, 1885, p. 119-123). His meridians are equally spaced, but the parallels are spaced at increasing intervals away from the Equator. The parallels of latitude are actually projected onto a cylinder wrapped about the sphere, but cutting it at lats. 45° N. and S., the point of perspective being a point on the Equator opposite the meridian being projected. It is used in several British atlases, but seldom in the United States. The Gall projection is neither conformal nor equal-area, but has a blend of various features. Unlike the Mercator, the Gall shows the poles as lines running across the top and bottom of the map.
Example using Gall Stereographical
$echo 9 51 | proj +proj=gall +lon_0=0 +x_0=0 +y_0=0 +ellps=WGS84 +datum=WGS84 +units=m +no_defs 708432.90 5193386.36 Example using Gall Stereographical (Central meridian 90°W)$ echo 9 51 | proj +proj=gall +lon_0=90w +x_0=0 +y_0=0 +ellps=WGS84 +datum=WGS84 +units=m +no_defs
7792761.91 5193386.36
## Mathematical definition¶
The formulas describing the Gall Stereographical are all taken from Snyder’s [Snyder1993].
### Spherical form¶
#### Forward projection¶
$x = \frac{\lambda}{\sqrt{2}}$
$y = (1+\frac{\sqrt{2}}{2}) \tan(\phi/2)$
#### Inverse projection¶
$\phi = 2 \arctan( \frac{y}{1+\frac{\sqrt{2}}{2}} )$
$\lambda = \sqrt{2} x$
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Jun 22, 2024
College Catalog 2022-2023
Select a Catalog College Catalog 2024-2025 Workforce and Continuing Education Catalog 2023-2024 Archived Course Descriptions College Catalog 2022-2023 [ARCHIVED CATALOG]
HELP College Catalog 2022-2023 [ARCHIVED CATALOG] Print-Friendly Page (opens a new window) Add to Favorites (opens a new window)
# MATH 1000 - Intermediate Algebra
Credit Hours: 4.00
Prerequisites: MATH 0070 with grade C or better; or an equivalent college course; or an acceptable score on a placement exam or prerequisite exam
(3 credit hrs prior to Fall 1990)
MATH 1000 is an additional course in algebra and includes systems of linear equations in three variables; expressions and equations containing quadratic, rational, radical, exponential, and logarithmic terms; rational and quadratic inequalities; complex numbers; graphs of lines, parabolas, and circles; and an introduction to functions and functional notation.
Billable Contact Hours: 4
Search for Sections
Transfer Possibilities
Michigan Transfer Network (MiTransfer) - Utilize this website to easily search how your credits transfer to colleges and universities.
OUTCOMES AND OBJECTIVES
Outcome 1: Upon completion of the course, the student will be able to solve systems of linear equations in three variables.
Objectives: Students will perform the following without the use of a calculator:
1. Solve systems of 3 equations with 3 unknowns (with elimination)
2. Applications.
Outcome 2: Upon completion of the course, the student will be able to solve rational equations.
Objectives: Students will perform the following without the use of a calculator:
1. Simplify rational expressions.
2. Add and subtract rational expressions with like and unlike denominators.
3. Multiply and divide rational expressions.
4. Simplify complex fractions.
5. Solve rational equations and check for extraneous solutions.
6. Solve applications with rational equations.
Outcome 3: Upon completion of the course, the student will be able to evaluate functions.
Objectives: Students will perform the following without the use of a calculator:
1. Determine when a relation is a function.
2. Evaluate functions.
3. Find domain and range of a function.
Outcome 4: Upon completion of the course, the student will be able to solve absolute value equations and inequalities.
Objectives: Students will perform the following without the use of a calculator:
1. Solve absolute value equations.
2. Solve absolute value inequalities.
3. Determine when an absolute value or inequality has no solution or infinite solutions.
Outcome 5: Upon completion of the course, the student will be able to solve radical equations.
Objectives: Students will perform the following without the use of a calculator:
1. Use properties of rational exponents.
2. Convert rational exponents to radical form and vice versa.
3. Simplify and perform basic operations on radicals.
4. Solve equations with rational exponents and radicals.
5. Perform basic operations on complex numbers.
Outcome 6: Upon completion of the course, the student will be able to solve quadratic equations.
Objectives: Students will perform the following without the use of a calculator:
1. Solve quadratic equations by factoring, completing the square, square root property, and quadratic formula.
2. Solve equations in quadratic form.
3. Solve applications with quadratic equations.
4. Find the vertex of a parabola and graph.
5. Solve quadratic and rational inequalities.
Outcome 7: Upon completion of the course, the student will be able to graph circles.
Objectives: Students will perform the following without the use of a calculator:
1. Find the distance between two points using the distance formula.
2. Identify the center and radius of a circle.
3. Graph a circle.
Outcome 8: Upon completion of the course, the student will be able to solve logarithmic and exponential equations.
Objectives: Students will perform the following without the use of a calculator:
1. Convert expressions from exponential form to logarithmic form and vice versa.
2. Evaluate logarithms.
3. Use properties of logarithms to expand and condense logarithmic expressions.
4. Solve exponential and logarithmic equations.
5. Solve applications using exponential and logarithmic equations.
COMMON DEGREE OUTCOMES (CDO)
• Communication: The graduate can communicate effectively for the intended purpose and audience.
• Critical Thinking: The graduate can make informed decisions after analyzing information or evidence related to the issue.
• Global Literacy: The graduate can analyze human behavior or experiences through cultural, social, political, or economic perspectives.
• Information Literacy: The graduate can responsibly use information gathered from a variety of formats in order to complete a task.
• Quantitative Reasoning: The graduate can apply quantitative methods or evidence to solve problems or make judgments.
• Scientific Literacy: The graduate can produce or interpret scientific information presented in a variety of formats.
CDO marked YES apply to this course:
Critical Thinking: YES
Quantitative Reasoning: YES
COURSE CONTENT OUTLINE
1. Review
1. Systems of equations in two variables
2. Graph lines using slope-intercept form
3. Factor polynomials
2. Systems of Linear Equations in Three Variables
1. Solve systems of equations in three variables
2. Applications
3. Rational Expressions and Equations
1. Simplify rational expressions
2. Add, subtract, multiply, and divide
3. Simplify complex fractions
4. Rational equations
5. Applications
4. Functions
1. Relations
2. Vertical Line Test
3. Evaluate functions
4. Domain and range
5. Absolute Value Equations and Inequalities
1. Equations
2. Inequalities
6. Exponents, Radicals and Complex Numbers
1. Rational exponents
3. Add, subtract, multiply, and divide radicals
4. Solve equations with radicals
5. Complex numbers
7. Quadratic Equations and Functions
1. Solve quadratic equations by
1. Factoring
2. Square Root Property
3. Completing the square
2. Solve equations in quadratic form
3. Applications
4. Solve quadratic and rational inequalities
5. Graph parabolas
8. Circles
1. Use distance formula
2. Find center and radius of a circle and graph
9. Exponents and Logarithms
1. Convert between exponential and logarithmic forms
2. Evaluate logarithms
3. Expand and condense logarithmic expressions
4. Solve exponential and logarithmic equations
5. Applications
Primary Faculty
Friday, David
Secondary Faculty
Chapman, Lori
Associate Dean
McMillen, Lisa
Dean
Pritchett, Marie
Official Course Syllabus - Macomb Community College, 14500 E 12 Mile Road, Warren, MI 48088
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Instruction
1
The first is to understand how this formed the sample average. For example, given some set of numeric values that consists of n units. All these units form the so-called sample. The sum of all these numbers is a formula to be expressed as ΣXi (Xi is any of the values of this sample, where i = 1,2,3...i-1,i. That is, i is the number of values in the sample). Then, in order to find the sample mean, you add together all the values from the sample and divided by their number n.
2
On top of all the recorded data can be expressed by only one formula, which is listed above. The sample mean is the simplest of the characteristics that reveal the nature of the sampling. It is widely used in mathematical statistics, probability theory, and also in many other fields of knowledge.
3
In the school curriculum is not given any formulas for finding average, although it becomes immediately clear that when children in math class in 5th grade are asked to find the average value of any numbers the children already know in order to find the average of these numbers, they will need to fold them all and then divided by their number. In fact, they also find the sample mean.
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# Questions tagged [computational-physics]
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# Find all roots of a function with parabolic cylinder functions in a range of the variable
I want to find all roots of a function involving Parabolic Cylinder Functions. In what follows, I define 2 variables $\xi1$ and $\xi2$, which in turn depend on $\omega$. My function is then defined as f. I go on defining g and h (where I take specific values for my parameters $M$ and $\lambda$ which are real and positive. I then plot the real and imaginary part of h to locate the roots. I would like, however, to be able to find all roots of $Im[h]$ (The real part is essentially 0) in a range of $\omega$, say from 0 to 50.
ξ1[ ω_] := (-1 + I) (ω/Sqrt[λ] - Sqrt[λ]/2)
ξ2[ ω_] := (-1 + I) (ω/Sqrt[λ] + Sqrt[λ]/2)
f := (I ParabolicCylinderD[I M/(2 λ), I ξ1[ ω]] - Sqrt[M/λ]*(-I - 1)/2*
ParabolicCylinderD[I M/(2 λ) - 1, I ξ1[ ω]])*(
ParabolicCylinderD[-I M/(2 λ), ξ2[ ω]] + I *Sqrt[M/λ]*(I - 1)/2*
ParabolicCylinderD[-I M/(2 λ) - 1, ξ2[ ω]]) + (I ParabolicCylinderD[
I M/(2 λ), I ξ2[ ω]] + Sqrt[M/λ]*(-I - 1)/2*ParabolicCylinderD[I M/(2 λ) - 1,
I ξ2[ ω]])*(ParabolicCylinderD[-I M/(2 λ), ξ1[ ω]] - I *Sqrt[M/λ]*(I - 1)/2*
ParabolicCylinderD[-I M/(2 λ) - 1, ξ1[ ω]])
g:=FullSimplify[f, {ω>0&&M>0&&λ>0}]
h:=FullSimplify[g/.{M->2, λ->100}]
Plot[{ Re[h], Im[h]}, {ω, 0, 20}, PlotPoints -> 50, MaxRecursion -> 0]
FindRoot[Im[h],{ω,5}]
I have searched through some posts with the keyword "find all roots in a range"; however, most of the solutions are for simpler functions than this special parabolic cylinder functions, c.f. About multi-root search in Mathematica for transcendental equations and Find all roots of an interpolating function (solution to a differential equation).
I would appreciate any help. Thank you in advance.
• To help the reader, please provide values for all your constants. Also, do you mean f [ω_] rather than f? You could improve your question by providing the code to calculate at least one root. Commented Jun 25, 2015 at 18:30
• Thanks, I forgot some codes. I have updated. I mean f [ω_] as a function of ω, after assigning values for M and λ. Since I am not too familiar with Mathematica, so there is some anomaly in writing the codes. Commented Jun 25, 2015 at 18:55
...most of the solutions are for simpler functions...
I'm not quite sure what gave OP that impression; certainly, FindAllCrossings[] is quite capable of handling transcendental equations, as long as all the roots being sought are simple.
But first: I slightly tidied up the definition of f[] (e.g. by using auxiliary variables for common subexpressions), as the original version brought tears to my sensitive eyes:
f[M_?NumericQ, λ_?NumericQ, ω_?NumericQ] := Module[{c, k, ξ1, ξ2},
c = Sqrt[M/λ]; k = I M/(2 λ);
ξ1 = (I - 1) (ω/Sqrt[λ] - Sqrt[λ]/2); ξ2 = (I - 1) (ω/Sqrt[λ] + Sqrt[λ]/2);
({1 + I, -I c}.ParabolicCylinderD[{-k, -k - 1}, ξ2]
{c, 1 + I}.ParabolicCylinderD[{k - 1, k}, I ξ1] -
{I c, 1 + I}.ParabolicCylinderD[{-k - 1, -k}, ξ1]
{c, -1 - I}.ParabolicCylinderD[{k - 1, k}, I ξ2])/2]
...and with that,
roots = FindAllCrossings[Im[f[2, 100, ω]], {ω, 0, 50}, WorkingPrecision -> 20]
{1.4210217375131208861, 4.9080677718060732317, 7.6276760758692264160,
11.242328271551264279, 14.025220377481373494, 17.188413671355074367,
20.686743750589305061, 23.568643080603343806, 26.490531437543067517,
29.849368653509459477, 33.222900929283185978, 36.429230282166527794,
39.466210718845193558, 42.459671861175573218, 45.512697669849073416,
48.625869297148536333}
As a graphical verification:
Plot[Im[f[2, 100, ω]], {ω, 0, 50},
Epilog -> {Red, AbsolutePointSize[4], Point[Thread[{roots, 0}]]},
Frame -> True, PlotStyle -> RGBColor[59/67, 11/18, 1/7]]
• Hi, thank you very much! I have made an extension to the problem. Please have a look. Commented Jun 26, 2015 at 14:40
It is enormously faster to use
x = f /. {M -> 2, λ -> 100} // Simplify;
FindRoot[Im[x], {ω, 5}]
Then, given the space of roots for Im[x]
DeleteCases[Union[Table[ω /. FindRoot[Im[x], {ω, i}], {i, 50}],
SameTest -> (Abs[#1 - #2] < 10^-8 &)], z_ /; z < 0 || z > 50, Infinity]
finds all positive roots less than 50
(* {1.42102, 4.90807, 7.62768, 11.2423, 14.0252, 17.1884, 20.6867, 23.5686, 26.4905,
29.8494, 33.2229, 36.4292, 39.4662, 42.4597, 45.5127, 48.6259} *)
These results are consistent with a Plot of x.
Plot[{Re[x], Im[x]}, {ω, 0, 50}]
### Introduction
For many transcendental functions, NSolve can solve for the roots, but not in this case. Since the roots are real we can apply the "Chebyshev-proxy rootfinder" method (CPR) which is based on the "colleague matrix" of a truncated Chebyshev series approximation to the function (see this answer by J.M. and the book by Boyd (2014)).
The first step is to form the Chebyshev series $f(\omega) = \sum a_n T_n(\zeta)$, where $\zeta = \frac{1}{2} (\omega (b-a)+(a+b))$ ranges over $-1 \le \zeta \le 1$ while $\omega$ ranges over $a \le \omega \le b$. We will take $[a,b]$ to be OP's example $[0,20]$. The series is truncated when the desired precision is achieved. By default, this is set to $10^{-14}$ in adaptiveChebSeries[]. For a function that is analytic in a complex neighborhood of $[a,b]$, the series will converge at least geometrically (or exponentially with an exponential rate of $1$). One can expect to get nearly machine precision for such a function. See Boyd (2014) for further discussion.
The next step is to get the eigenvalues of the colleague matrix. The roots will be the rescaled values $\omega_k$ in interval $[a,b]$ of the eigenvalues $\zeta_k$ in the interval $[-1,1]$. This is done by chebRoots[].
The success of this method depends primarily on the success of approximating $f(x)$. This has not been automated in adaptiveChebSeries[], so I'll show some hand checking at the end. One could also use this method to get a rough single-precision approximation of the roots and polish them off with FindRoot. (The option adaptiveChebSeries[..., PrecisionGoal->6] would accomplish this.)
### Finding the roots
Code for the functions is given at the end of this answer and can also be found here.
h = f /. {M -> 2, λ -> 100} // Simplify; (* Simplify seems sufficient *)
obj = Function @@ {ω, Im@h}; (* function equivalent of f[w] *)
dobj = Function @@ {ω, Im@D[h, ω]}; (* to be used in analyzing the roots *)
(a0 = 0; b0 = 20; (* specify interval *)
coeff = adaptiveChebSeries[obj, a0, b0]; (* approximate f(w) === obj[w] *)
roots = chebRoots[coeff, {a0, b0}] (* get roots of approximant *)
) // AbsoluteTiming
Length[roots] (* check how many roots found *)
(*
{1.57934,
{1.42102, 4.90807, 7.62768, 11.2423, 14.0252, 17.1884}}
6
*)
### Checking the roots
We can see in the table below that the roots are calculated to roughly machine precision. (We approximate the error by a single Newton-Raphson step, which a fussier person could check with FindRoot at a higher precision than machine precision.)
The relative error shows the roots are almost all accurate to machine precision.
TableForm[Transpose@
{roots,
obj /@ roots,
(obj /@ roots/dobj /@ roots)/roots // Abs},
TableHeadings -> {None, {"root", "residual", "rel. err. in rt."}}
]
We can visualize the roots:
Plot showing the OP's function and roots (indicated by the red lines).
objPlot = Plot[obj[ω], {ω, 0, 20}];
Show[objPlot, GridLines -> {roots, None},
GridLinesStyle -> Directive[Red, Thickness[0.002]]]
### Checking the Chebyshev approximant
We can see graphically that the approximant is close to the function.
Plot showing the OP's function (blue) and the Chebyshev approximant (light gray).
Show[
objPlot /. _AbsoluteThickness -> AbsoluteThickness[4.4],
Plot[obj2[ω], {ω, 0, 20}, PlotStyle -> LightGray]
]
Comparing the difference, the relative error shows the approximant is not quite down to machine precision. Except for near the roots where one expects the relative error to grow, the error stays below $10^{-14}$.
Plot showing the OP's function (blue) and the Chebyshev approximant (light gray).
LogPlot[(obj[ω] - obj2[ω])/obj2[ω] // Abs,
{ω, 0, 20},
PlotPoints -> 100, MaxRecursion -> 2,
Frame -> True, Axes -> False,
FrameLabel -> {HoldForm@ω, "relative error"},
GridLines -> {roots, None},
GridLinesStyle -> Directive[Red, Thickness[0.002]]]
Inspecting the Chebyshev coefficients, we can see that once $n >12$, the coefficients decrease roughly geometrically until $n=37$ when machine precision $\approx 10^{-16}$ is reached. After that, they bounce around at that level. (The function adaptiveChebSeries[] actually computed 65 coefficients and discarded about twenty using Boyd's heuristic with a cutoff of $10^{-14}$. Perhaps a few more should have been discarded.) One could recompute the coefficients at a higher working precision and see the geometric convergence continue.
Plot showing the the Chebyshev coefficients $a_n$. Once the degree of the approximation reaches roughly twice the number of roots (6), the series converges geometrically until it reaches machine precision. At that point rounding error dominates the coefficients.
ListLogPlot[Abs@coeff,
Frame -> True, Axes -> False,
FrameLabel -> {Row[{"Degree ", HoldForm@n, " of coefficient"}],
HoldForm@Abs[Subscript[a, n]]}]
Code dump
See my answer here for a discussion of the code.
(* Chebyshev extreme points *)
chx[n_, prec_: MachinePrecision] := N[Cos[Range[0, n]/n Pi], prec];
(* Chebyshev series approximation to f *)
Clear[chebSeries];
chebSeries[f_, a_, b_, n_, prec_: MachinePrecision] := Module[{x, y, cc},
x = Rescale[chx[n, prec], {-1, 1}, {a, b}];
y = f /@ x; (* function values at Chebyshev points *)
cc = Sqrt[2/n] FourierDCT[y, 1]; (* get coeffs from values *)
cc[[{1, -1}]] /= 2; (* adjust first & last coeffs *)
cc
];
(* recursively double the Chebyshev points
* until the PrecisionGoal is met
* The function values are memoized in f0
* *)
Options[adaptiveChebSeries] = {PrecisionGoal -> 14, "Points" -> 32,
WorkingPrecision -> MachinePrecision, MaxIterations -> 5};
Module[{cc, f0, max, len = 0, sum},
f0[x_] := f0[x] = f[x]; (* values reused in subsequent series *)
NestWhile[
(cc = chebSeries[f0, a, b, #, OptionValue[WorkingPrecision]];
(* check error estimate *)
max = Max@Abs@cc; (* sum the tail of the series *)
sum = 0; (* relative to the max coefficient *)
len = LengthWhile[
Reverse@cc, (sum += Abs@#) < 10^-OptionValue[PrecisionGoal]*max &];
2 #) &, (* double the number of points *)
OptionValue["Points"],
len < 3 && # <= 2^OptionValue[MaxIterations] OptionValue["Points"] &
(* at least two coefficients dropped *)
];
If[len < 3,
If[TrueQ[len > 1], Drop[cc, 1 - len], cc]
]
(* "Chebyshev companion matrix" (Boyd, 2014) /
"Colleague matrix" (Good, 1961) *)
colleagueMatrix[cc_] := With[{n = Length[cc] - 1},
SparseArray[{{i_, j_} /; i == j + 1 :> 1/2,
{i_, j_} /; i + 1 == j :> 1/(2 - Boole[i == 1])}, {n, n}] -
SparseArray[{{n, i_} :> cc[[i]]/(2 cc[[n + 1]])}, {n, n}]
];
(* Find the real roots of a truncated Chebyshev series
representing a function over an interval [a,b] *)
Options[chebRoots] = {(* TBD *)};
chebRoots::usage =
"chebRoots[c,{a,b}], c = {c0, c1,..., cn} Chebyshev coefficients, over the interval {a,b}
computes the (real) roots in the interval {a,b}";
chebRoots[coeff_, dom_: {-1, 1}, OptionsPattern[]] := Module[{eigs},
eigs = Eigenvalues@colleagueMatrix[coeff];
roots = Sort@Rescale[
Re@Select[
eigs,
Abs[Im[#]] < 1*^-15 && (* Im error tolerance *)
-1.0001 < Re[#] < 1.0001 &], (* Re error tolerance *)
{-1, 1}, dom]
]
• This is great! A note: the type 1 DCT is conventionally done for lengths of the form $2^n +1$ (corresponding to a power-of-two FFT), so the traditional Clenshaw-Curtis hierarchy proceeds as $3, 5, 9, 17, 33, \dots$. Thankfully, Mathematica is not constrained to these lengths, so this works fine. Commented May 6, 2016 at 0:23
• @J.M. Probably I should rename "Points" -> n to "Order -> n", because it actually uses n+1 points. It iteratively doubles n until the desired precision is reached. So if you start with n = 2^k, it will always use 2^(k+m} + 1 points for m = 0, 1,...`. Commented May 6, 2016 at 19:23
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simplify (4x)2 + 42x
(4x)2 + 42x = 8x + 42x = 50x
Question
Updated 8/15/2014 12:09:37 AM
Rating
3
(4x)2 + 42x = 8x + 42x = 50x
Confirmed by yumdrea [8/15/2014 12:24:36 AM]
Questions asked by the same visitor
Simplify. (4x)2 + 42x
Weegy: 2(2x+1) User: simplify x^4 x^2 x^4 Weegy: Simplified x^4 x^2 x^4 is x ^10. (More)
Question
Updated 1/9/2014 3:40:01 PM
Simplified (4x)2 + 42x is 50x.
Determine the value of x(9 - x) + 1 when x = –2.
Question
Updated 10 days ago|9/7/2018 10:42:06 PM
x(9 - x) + 1 when x = –2
-2(9 + 2) + 1;
= -2(11) + 1;
= -22 + 1;
= -21
Added 10 days ago|9/7/2018 10:41:55 PM
Simplified expression for 4(x - 8) - 2(x - 3) + 3x.
Question
Updated 119 days ago|5/22/2018 6:50:43 AM
4(x - 8) - 2(x - 3) + 3x
= 4x - 32 - 2x + 6 + 3x
= 2x - 26 + 3x
= 5x - 26
Added 119 days ago|5/22/2018 6:50:43 AM
solve the inequality 0.2(0.1x + 0.9) < -0.8
Question
Updated 1/11/2014 12:48:54 AM
0.2(0.1x + 0.9) < -0.8
0.02x + 0.18 < -0.8
0.02x < -0.8 - 0.18
0.02x < -0.98
x < -49
solve 6x+9=8
Question
Updated 1/12/2014 2:34:21 PM
6x+9=8
6x = 8 - 9
6x = -1
x = -1/6
27,344,316
*
Get answers from Weegy and a team of really smart live experts.
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# Segment2D
@Serializable
data class Segment2D(val start: Vector2, val control: List<Vector2>, val end: Vector2, val corner: Boolean = false) : BezierSegment<Vector2> , ShapeContourProvider(source)
Creates a new Segment2D, which specifies a linear or a Bézier curve path between two anchor points (and up to two control points for curvature).
## Constructors
constructor(start: Vector2, control: List<Vector2>, end: Vector2, corner: Boolean = false)
## Functions
open fun adaptivePositions(distanceTolerance: Double = 0.5): List<Vector2>
Recursively subdivides Segment2D to approximate Bézier curve.
open fun adaptivePositionsWithT(distanceTolerance: Double = 0.5): List<Pair<Vector2, Double>>
open override fun derivative(t: Double): Vector2
open fun direction(): Vector2
Returns the direction T of between the Segment2D anchor points.
open fun direction(t: Double): Vector2
operator fun div(scale: Double): Segment2D
open operator override fun equals(other: Any?): Boolean
open fun equidistantPositions(pointCount: Int, distanceTolerance: Double = 0.5): List<Vector2>
Samples specified amount of points on the Segment2D.
open fun equidistantPositionsWithT(pointCount: Int, distanceTolerance: Double = 0.5): List<Pair<Vector2, Double>>
Returns the t values of the extrema for the current Segment2D.
Returns the extrema points as Vector2s for current Segment2D
open override fun hashCode(): Int
Calculates a List of all points where two Segment2Ds intersect.
Calculates a List of all points of where a Segment2D and a Shape intersect.
Calculates a List of all points of where a Segment2D and a ShapeContour intersect.
fun isStraight(tolerance: Double = 0.01): Boolean
Determines if the Segment2D forms a straight line.
fun lut(size: Int = 100): List<Vector2>
operator fun minus(right: Segment2D): Segment2D
Find point on segment nearest to given point.
fun normal(ut: Double, polarity: YPolarity = YPolarity.CW_NEGATIVE_Y): Vector2
Returns a normal Vector2 at given value of t in the range of `0.0` to `1.0`.
fun on(point: Vector2, error: Double = 5.0): Double?
operator fun plus(right: Segment2D): Segment2D
open fun pointAtLength(length: Double, distanceTolerance: Double): Vector2
Calculates the point at a given distance along this Segment2D.
fun pose(t: Double, polarity: YPolarity = YPolarity.CW_NEGATIVE_Y): Matrix44
Calculates the pose Matrix44 (i.e. translation and rotation) that describes an orthonormal basis formed by normal and tangent of the contour at t.
open override fun position(ut: Double): Vector2
Returns a point on the segment.
open override fun split(t: Double): Array<Segment2D>
Splits the path into one or two parts, depending on if the cut was successful.
open override fun sub(t0: Double, t1: Double): Segment2D
Samples a new Segment2D from the current Segment2D starting at t0 and ending at t1.
open override fun tForLength(length: Double): Double
Estimate t value for a given length
operator fun times(scale: Double): Segment2D
open override fun toString(): String
fun transform(transform: Matrix44): Segment2D
Applies given linear transformation.
## Properties
Returns the bounding box.
open override val contour: ShapeContour
the provided contour
open override val control: List<Vector2>
val corner: Boolean = false
open override val cubic: Segment2D
Converts the Segment2D to a cubic Bézier curve.
open override val end: Vector2
open override val length: Double
Calculates approximate Euclidean length of the Segment2D.
@Transient
open val linear: Boolean
Indicates whether the Segment2D is linear.
Converts the Segment2D to a quadratic Bézier curve.
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# RE: [R] creating a formula on-the-fly inside a function
From: Berton Gunter <gunter.berton_at_gene.com>
Date: Fri 04 Mar 2005 - 03:41:07 EST
If these are nested models, see ?drop.terms.
• Bert Gunter Genentech Non-Clinical Statistics South San Francisco, CA
"The business of the statistician is to catalyze the scientific learning process." - George E. P. Box
> -----Original Message-----
> From: r-help-bounces@stat.math.ethz.ch
> [mailto:r-help-bounces@stat.math.ethz.ch] On Behalf Of Marc Schwartz
> Sent: Thursday, March 03, 2005 7:52 AM
> To: Dr Carbon
> Cc: r-help@r-project.org
> Subject: Re: [R] creating a formula on-the-fly inside a function
>
> On Thu, 2005-03-03 at 10:28 -0500, Dr Carbon wrote:
> > I have a function that, among other things, runs a linear model and
> > returns r2. But, the number of predictor variables passed to the
> > function changes from 1 to 3. How can I change the formula
> inside the
> > function depending on the number of variables passed in?
> >
> > An example:
> >
> > get.model.fit <- function(response.dat, pred1.dat, pred2.dat = NULL,
> > pred3.dat = NULL)
> > {
> > res <- lm(response.dat ~ pred1.dat + pred2.dat + pred3.dat)
> > summary(res)\$r.squared
> > # other stuff happens here...
> > }
> >
> > y <- rnorm(10)
> > x1 <- y + runif(10)
> > x2 <- y + runif(10)
> > x3 <- y + runif(10)
> > get.model.fit(y, x1, x2, x3)
> > get.model.fit(y, x1, x2)
> > get.model.fit(y, x1)
>
>
> Consider using as.formula() to take a character vector that
> you pass as
> an argument instead of specifying each IV separately:
>
> get.model.fit <- function(my.form)
> {
> res <- lm(as.formula(my.form))
> summary(res)\$r.squared
> # other stuff happens here...
> }
>
>
> Then call it with:
>
> get.model.fit("y ~ x1 + x2 + x3")
>
> Internally, the vector will be converted to:
>
> > as.formula("y ~ x1 + x2 + x3")
> y ~ x1 + x2 + x3
>
> Doing it this way provides for greater flexibility if you
> want to use a
> more complicated formula construct.
>
> including the
> use of paste() if you want to separate the DV from the IVs for an
> additional approach for a long set of similarly named IV's
> (ie x1:x25).
>
> HTH,
>
> Marc Schwartz
>
> ______________________________________________
> R-help@stat.math.ethz.ch mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
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# Presentation1.pptx
software effort prediction using machine learning
software effort prediction using machine learning
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## Weitere Verwandte Inhalte
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### Presentation1.pptx
1. 1. Software Effort Prediction using Statistical and Machine Learning Methods Presented by: EMAN AMJAD 70077643 NARMEEN KAZMI 70078283 KASHFA MEHMOOD 70075058
2. 2. ABSTRACT • Software project managers have to make estimates of how much a software development is going to cost. Effort is measured in terms of person months and duration. There are various effort estimation models, but it is difficult to determine which model gives more accurate estimation on which dataset. The results show that decision tree method is better than all the other compared method.
3. 3. Introduction: Accurate estimation of effort is crucial for successful management and control of software project. Effort estimation techniques fall under following categories: Expert judgment, Algorithmic estimation, Machine Learning, Empirical techniques, Regression techniques, and Theory-based techniques. We have analysed these methods on large datasets collected from 499 projects.
4. 4. Related work • Software effort estimation is a key consideration to software cost estimation. There are numerous Software Effort Estimation Methods such as Algorithmic effort estimation, machine learning, empirical techniques, regression techniques and theory based techniques. An important task in software project management is to understand and control critical variables that influence software effort. Another method of improving estimation accuracy is proposed by Tosun. In the traditional formula for Euclidean distance, the features are either unweighted or same weight is assigned to each of the features. A lot of research has also been done in Machine learning techniques of estimation.
5. 5. RESEARCH METHODOLOGY 1. Linear Regression 2. Support Vector Machine 3. Artificial Neural network 4. Decision Tree 5. Bagging
6. 6. Linear Regression • Linear regression is a method of estimating the conditional expected value of a variable x given the values of some other variable or. One variable is the dependent variable and the other is the independent variable. For doing this, it finds a line which minimizes the sum of the squares of the vertical distances of the points from the line.
7. 7. Support Vector Machine • Support Vector Machine is a learning technique used for classifying unseen data correctly. SVM builds a hyperplane which separates the data into into different categories. The dataset may or may not be linearly separable, i.e. the cases with one category on one side of the hyperplane and the other on the other side.
8. 8. VECTORMACHINE
9. 9. Artificial Neural network • Artificial Neural network comprises a network of interconnected units called ―neurons‖ or processing units‖. The ANN has three layers, i.e. the input layer, hiddenlayer and the output layer. More complex systems have more than one hidden layer. Error back-propagation learning consists of two passes: a forward pass and a backward pass.
10. 10. Decision Tree • Decision tree is a methodology used for classification and regression. • Decision tree algorithm is a data mining induction technique that recursively partitions a data set of records using depth-first greedy approach or breadth-first approach until all the data items belong to a particular class. • The tree structure is used in classifying unknown data records. • This method generated M5 model rules and trees
11. 11. Bagging • E Bagging which is also known as bootstrap aggregating is a technique that repeatedly samples (with replacement) from a data set according to a uniform probability distribution. • Each bootstrap sample has the same size as the original data. • Because the sampling is done with replacement, some instances may appear several times in the same training set, while others may be omitted from the training set.
12. 12. ANALYSIS RESULTS Model Prediction Results • China Dataset was used to carry out the prediction of effort estimation model.A holdout technique of cross validation was used to estimate the accuracy of the effort estimation model.The dataset was divided into two parts i.e. training and validation set. • Four machine learning methods and one regression method was used to analyze the results. • Hence the decision tree method is found to be effective in predicting effort. Also, the results of the decision tree are competent with the traditional linear regression model.
13. 13. OUR RESULTS
14. 14. RESULTSUSINGLINEARREGRESSION
15. 15. RESULTUSING SUPPORT VECTOR MACHINE
16. 16. RESULT USING ARTIFICIAL NURAL NETWORK
17. 17. RESULTS USING DECISION TREE
18. 18. RESULT USING BAGGING
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CheckPoint Week 5
# CheckPoint Week 5 - Define count_below to store the number...
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CheckPoint Week 5 Define n to store number of employees Output “Enter the number of employees:” Input n Define namearray[n] Define salaryarray[n] Define sum_salary to store running sum of salaries Set sum_salary = 0 Define loop_integer Set loop_integer = 1 While (loop_integer – 1 < n) Output “Enter the name of employee” loop_integer Input namearray[loop_integer – 1] Output “Enter the salary of employee” loop_integer Input salaryarray[loop_integer – 1] Add salaryarray[loop_integer – 1] to sum_salary Increase loop_integer by 1 End while Define mean_salary to store mean salary of employees mean_salary = sum_salary / n Output “The mean salary is” mean_salary
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Unformatted text preview: Define count_below to store the number of salaries below the mean salary Set count_below = 0 Define count_above to store the number of salaries above the mean salary Set count_above = 0 Set loop_integer = 1 While (loop_integer – 1 < n) If (salaryarray[loop_integer – 1] < mean_salary) Increase count_below by 1 Else if (salaryarray[loop_integer – 1] > mean_salary) Increase count_above by 1 End if End while Output “The number of salaries below the mean salary is” count_below Output “The number of salaries above the mean salary is” count_above...
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View the step-by-step solution to:
# Future value of an annuity. For the case below answer the questions that follow:
Future value of an annuity. For the case below answer the questions that follow:
amount of annuity \$3500
interest rate 10%
Show all work including formula and answer
deposit period 8 years
calculate the future value of the annuity assuming that it is
1. an ordinary annuity
2. an annuity due
Amount of annuity
\$3500
Interest rate
10% i.e..1
No. of Years
8
Calculate the future value of the annuity assuming that it is
1. An ordinary annuity
Future Value = C *
i
= \$3,500
.1
= \$3,500...
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# nLab convenient category of topological spaces
### Context
#### Topology
topology
algebraic topology
# Contents
## Idea
The term convenient category of topological spaces is used for a category of topological spaces nice enough to address many of the needs of working topologists, notably including the condition of being a cartesian closed category. As such, they are examples of nice categories of spaces.
A primary example is the category of compactly generated spaces.
## Definition
While the authors of this article don’t know whether there exists in the literature a widely accepted definition of “convenient category of topological spaces”, we propose the following definition as reasonable and convenient (see also the discussion below on the distinction between “nice” and “convenient”):
###### Definition
A convenient category of topological spaces is a full replete subcategory $C$ of the category of all topological spaces Top such that the following conditions 1-3 hold:
1. Every CW complex is an object of $C$;
2. $C$ is cartesian closed;
3. $C$ is complete and cocomplete.
Frequently it is also felt desirable to add closure under certain types of subspaces. For instance, in the well-known examples one has
• $C$ is closed under closed subspaces in Top, i.e., if $X$ belongs to $C$ and $A \subseteq X$ is a closed subspace (in $Top$), then $A$ also belongs to $C$.
At times one might hope that $C$ is closed under open subspaces as well, but this does not hold for all objects in some of the well-known examples of convenient categories.
It may be well to note that colimits and limits in $C$ need not agree with the corresponding colimits and limits in $Top$, except under certain conditions. Some convenient categories are reflective or coreflective in $Top$, in which case they are closed under limits or colimits respectively. Moreover, because $C$ is a full subcategory of $Top$ which contains all CW complexes, the usual sorts of colimits used to present CW complexes are the same whether interpreted in $Top$ or in $C$. Also, if $C$ is closed under closed subspaces, then an equalizer of a pair of maps between Hausdorff spaces in $C$ (being a closed subspace) is the same whether computed in $Top$ or in $C$.
On the other hand, products of $C$-objects in $Top$ need not land in $C$, so in that situation the product in $Top$ and the product in $C$ do not agree. This is in particular the case for compactly generated spaces. In fact, the “compactly generated product?” is sometimes preferable to the $Top$-product for more explicit reasons: for instance, if $X$ and $Y$ are CW complexes, then $X \times Y$ need not be a CW complex in the usual product topology, but it is in the compactly generated topology.
## Examples
The original example of a convenient category of topological spaces is described at
and there are closely related convenient categories, including for example the category of compactly generated weakly Hausdorff spaces. The article by Neil Strickland (ref), written with the needs of algebraic topology in mind, covers the latter example in thorough detail.
A reasonably large class of examples, including the examples of compactly generated spaces and sequential spaces, is given in the article by Escardó, Lawson, and Simpson (ref). These may be outlined as follows. An exponentiable space in $Top$ is a space $X$ such that $X \times -: Top \to Top$ has a right adjoint. These may be described concretely as core-compact spaces (spaces whose topology is a continuous lattice). Suppose given a collection $\mathcal{C}$ of core-compact spaces, with the property that the product of any two spaces in $\mathcal{C}$ is a colimit in $Top$ of spaces in $\mathcal{C}$. Such a collection $\mathcal{C}$ is called productive. Spaces which are $Top$-colimits of spaces in $\mathcal{C}$ are called $\mathcal{C}$-generated.
###### Theorem
(Escardó, Lawson, Simpson)
If $\mathcal{C}$ is a productive class, then the full subcategory of $Top$ whose objects are $\mathcal{C}$-generated is a coreflective subcategory of $Top$ (hence complete and cocomplete) that is cartesian closed.
The other convenience conditions listed in this article (inclusion of CW-complexes, closure under closed subspaces) are in practice usually satisfied as well. For example, if closed subspaces of objects of $\mathcal{C}$ are $\mathcal{C}$-generated, then closed subspaces of $\mathcal{C}$-generated spaces are also $\mathcal{C}$-generated. If the unit interval $I$ is $\mathcal{C}$-generated, then so are all CW-complexes.
## Counterexamples
A naive approach to the problem of constructing “convenient categories” usually runs into problems. For example, one could try to work with the full subcategory of $Top$ consisting of exponentiable spaces; the problem is that even if $X$ and $Y$ are exponentiable, the exponential $Y^X$ may not be:
• The category of exponentiable spaces is not cartesian closed. (It is however cartesian: the product of two exponentiable spaces under the usual product topology is exponentiable.)
To see this, we recall that a Hausdorff space is exponentiable if and only if it is locally compact, and that an exponential $Y^X$ (provided it exists) is Hausdorff if $Y$ is. Thus, it is enough to exhibit two locally compact Hausdorff spaces $X$, $Y$ whose exponential is not locally compact.
Take $X = \mathbb{R}$ with its usual topology and $Y = \mathbb{N}$ (the set of natural numbers) with the discrete topology. Suppose that an exponential $\mathbb{R}^\mathbb{N}$ exists in the category of locally compact Hausdorff spaces. Then it must be a countable product of copies of $\mathbb{R}$ by the following calculation:
$LCH(X \times \mathbb{N}, \mathbb{R}) \cong LCH(\sum_{\mathbb{N}} X, \mathbb{R}) \cong \prod_{\mathbb{N}} LCH(X, \mathbb{R})$
where the last functor in $X$, in order to be representable, would have to be represented by a product $\prod_{\mathbb{N}} \mathbb{R}$. Using the universal property of products, one may easily exhibit a scalar multiplication
$\mathbb{R} \times \mathbb{R}^{\mathbb{N}} \to \mathbb{R}^{\mathbb{N}}$
rendering $\mathbb{R}^{\mathbb{N}}$ a topological vector space (a TVS) over the real numbers. But it is well-known that locally compact TVS are finite-dimensional, and we have reached an absurdity.
Similarly, the topological product $\mathbb{N}^{\mathbb{N}}$ is homeomorphic to Baire space, and this is not locally compact either.
## “Nice” versus “convenient” categories of spaces
A related entry is nice category of spaces. Here we explain the difference between “convenient” and “nice” categories of spaces.
“A convenient category of topological spaces” is the title of a well-known paper by Norman Steenrod, who emphasized particularly function space constructions (i.e., cartesian closure) as a great convenience for working algebraic topologists, on top of (generally less problematic) completeness and cocompleteness assumptions. (See also the historical remarks that follow.)
It is well-known that $Top$ is not cartesian closed. One can characterize the exponentiable spaces, which include all locally compact Hausdorff spaces, but as we saw above, the naive idea of simply cutting down to some of these does not give a good cartesian closed category either, since firstly it need not be complete and cocomplete, and secondly even if $X$ and $Y$ are exponentiable, the exponential $Y^X$ need not be.
However, one can cut down to some full subcategory of spaces which does admit function spaces. This typically involves the subtle and delicate interplay between compactness conditions and openness conditions. It comes at a cost – that limits and/or colimits in the subcategory might not be computed as they are in $Top$ – but this is generally considered a very small price to pay in exchange for the great convenience of these assumptions.
That a convenient category of topological spaces contains all the CW complexes was not explicitly declared by Steenrod, but we feel certain that algebraic topologists want these as part of their convenient category. In practice this is a mild assumption, because the usual examples certainly contain all finite topological products of the unit interval $I$, and are closed under those $Top$-colimits used to present CW complexes, as built inductively from the basic spaces $I^n \cong D^n$.
“Nice categories of spaces” should be thought of as a wider and vaguer term; it really means the category of spaces has “nice” categorical properties for some mathematical purpose at hand. Certainly any convenient category of spaces should be considered a nice category of spaces for the general purposes of algebraic topology. On the other hand, there are categories of spaces which are not “convenient” in the technical sense described above, but which may be very nice for certain purposes.
For example, the category of compact Hausdorff spaces can be considered as being very nice for certain purposes, because it is monadic over $Set$ (!). It is, however, not “convenient”, as it very far from being cartesian closed. In a different direction, there is the (complete, cocomplete) cartesian closed category of equilogical spaces, but this is not a full subcategory of $Top$, and the core concerns of mathematicians working with equilogical spaces are somewhat different from those of algebraic topologists.
It should also be noted that “space” itself has a wider meaning than the technical notion of “topological space”, even if topological intuitions come into play. For example, in domain theory, one often considers certain types of posets (for example, dcpos) as certain types of “spaces”. In this direction, we have that the category of dcpos and Scott-continuous map?s between them forms a complete, cocomplete, cartesian closed category. However, these types of spaces are quite far removed from the traditional concerns of topology, hence are not “convenient” in the sense given above (despite the manifold relations between Scott continuity and the point-set topology which underlies the classical results on compactly generated spaces). In another direction, the category of locales is in some respects “nice”, but these are not topological spaces either.
Along with the entry on nice category of spaces and the examples above, see Johnstone's topological topos and Spanier’s quasitopological spaces (ref). None of these is convenient in the precise sense above. However there are advantages in having a category which is not only cartesian closed but also locally cartesian closed. In algebraic topology, this has led to Peter Booth’s work on “fibred mapping spaces” (ref), (ref).
There are possible advantages in homotopy theory of using a “topological topos”, (Johnstone ref), Harasani ref).
## Historical remarks
The phrase “convenient category of topological spaces” predates Steenrod’s paper by a number of years. Indeed, Ronnie Brown $[1963]$ had written the following in the Introduction to his paper “Ten topologies for X x Y “:
• “It may be that the category of Hausdorff k-spaces is adequate and convenient for all purposes of topology.”
The requirements for convenience were spelled out in the sequel.
In fact, Brown (not Steenrod, as has sometimes been assumed) was the first to prove that Hausdorff k-spaces? (with the “kelleyfied” weakening of the compact--open topology on function spaces) formed what is now called a cartesian closed category (see his 1961 thesis); this work was influenced by the paper of Cohen listed below. Note Brown’s work predates the formal introduction of cartesian closed categories in Bill Lawvere’s thesis by a couple of years; in fact there is clear anticipation in Brown’s thesis of the notions of monoidal closed and cartesian closed categories, which was to attract much attention throughout the sixties. Further, Brown’s results on the topological case are more precise than Steenrod’s, since the 1964 paper deals with functions continuous on compact subsets, and for these obtains a homeomorphism without kellyfication. An account of this may be found in the book Topology and Groupoids.
Of course, as has often been emphasized by Lawvere, the need for and convenience of considering function spaces is a very old idea in geometry (going back to the roots of the calculus of variations, for example). The constructions of exponentials of topological spaces via the compact–open topology had been known for a long time; see for example John Kelley’s General Topology (1955). The relevance of what are today called Kelley space?s (or k-spaces1) had also been recognized; for example, Kelley’s book indicates the completeness of function spaces (wrt the compact–open topology) when the base is a complete uniform space and the exponent is a k-space. However, the problem of obtaining a class of spaces closed under function spaces wasn’t solved prior to Brown’s thesis. Brown also observed the relevance of k-spaces to studies of how products interact with quotient spaces. The general appreciation of the connection between cartesian closure and preservation of quotients under products came with the appreciation of the conceptual simplicity of categorical adjunctions, namely the point that the functor $- \times X$ preserves colimits if it has an exponential right adjoint $(-)^X$. The question also solved in Brown’s 1964 paper was to give, in the Hausdorff case, a left adjoint to the functor $(-)^Y$ when the function spaces have the compact-open topology, and so to give a monoidal closed structure on the category of Hausdorff spaces; the generalisation to the non-Hausdorff case is given in the paper of Booth and Tillotson listed below.
Appreciation of the role of convenient categories was in full force by the early seventies (for a sample, see Peter May’s Geometry of Iterated Loop Spaces, where the category of Hausdorff k-spaces plays a foundational role). The notion of a “convenient category” is recognized elsewhere too (and not just within the categorical and algebraic topology communities); see for example the book by Kriegl and Michor.
## References
• Booth, Peter I. The exponential law of maps. II. Math. Z. 121 (1971), 311–319. (#Booth)
• Ronnie Brown, Some problems of algebraic topology: a study of function spaces, function complexes, and FD-complexes, DPhil thesis (part A), Oxford University, 1961 (pdf)
• Ronnie Brown, Ten topologies for $X\times Y$, Quart. J.Math. (2) 14 (1963), 303–319. (pdf)
• Ronnie Brown, Function spaces and product topologies, Quart. J. Math. (2) 15 (1964), 238–250. (pdf)
• Ronnie Brown Topology and Groupoids, Booksurge (2006), Section 5.9: Spaces of functions and the compact-open topology. (#Brown06)
• Cohen, D. E. Spaces with weak topology, Quart. J. Math., (2) 5 (1954) 77–-80.
• Crabb, Michael; James, Ioan, Fibrewise homotopy theory.
Springer Monographs in Mathematics. Springer-Verlag London, Ltd., London, 1998. viii+341 pp. ISBN: 1-85233-014-7 (#CrabbJames)
• Norman Steenrod, A convenient category of topological spaces, Michigan Math. J. 14 (1967) 133–152, project euclid
• Booth, P.; Tillotson, J., Monoidal closed, Cartesian closed and convenient categories of topological spaces. Pacific J. Math. 88 (1980), no. 1, 35–53.
• Peter T. Johnstone, On a topological topos, Proc. London Math. Soc. (3) 38 (1979) 237–271.
• Harasani, Hamed A. “Topological methods in general topology”, PhD Thesis, University of Wales, Bangor, (1988) (link to pdf files). (#Harasani)
• Edwin Spanier, “Quasi-topologies”, Duke Mathematical Journal 30 (1) (1963), 1–14.
• Kriegl, A.; Michor, P.W., The convenient setting of global analysis, Mathematical Surveys and Monographs, Volume 53. American Mathematical Society, Providence, RI (1997).
• M. Escardó, J.Lawson, A. Simpson, Comparing Cartesian closed categories of (core) compactly generated spaces, Topology and its Applications, 143 (2004), 105–145. (link to ps file)
1. The ‘k’ certainly refers to ‘kompakt’ rather than Kelley’s initial.
Revised on July 16, 2013 13:43:11 by Ronnie Brown (109.149.198.118)
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# Difference between revisions of "Knot Theory"
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Bronze broach from the Iron Age, found on Gottland, Sweden
Relevant examples from Escher's work:
## Introduction
Knots have been used in decorations for centuries. Ornamental knotwork borders were used by the Egyptians, Greeks, Romans and many other civilizations.
Knotwork Border - Time of Tutankhamen (CA 1350 BCE)
Knotwork Border - Time of Tutankhamen (CA 1350 BCE)
These two examples of knotted borders from staffs belonging to the Pharaoh Tutankhamen (ca.1340 BCE). These decorations are not much different than some found on a highland dirk handle, Byzantine borders, etc.
Knotted figures appear in Celtic and Nordic art and these images date back at least two thousand years. Sometimes they created intricately knotted figures of varying shapes, and at other times the knotted figures were combined with animals. One might find for instance a lion with its body made of of some intricately knotted figure.
One can think of a knot as a string that is knotted up, and then has the ends tied together to prevent it from becoming undone. Mathematically we would say that a knot is an embedding of the circle in 3-space.
The knot as an unending loop which is twisted up is sometimes seen as a symbol of the infinite. In Ireland the Celtic crosses are often shown as intricate knotted figures. These Celtic crosses are a combination of some pagan imagery and Christian beliefs. The Irish believed that the Celtic cross was brought to Ireland by Saint Patrick. The intricate knot work was also used to decorate the gospels of the day, and the monks produces for instance the Lindisfarne gospels, the Book of Kells, the Echternach gospels, etc.
In the 19th century the interest in knots saw a resurgence. Lord Kelvin, a British scientist, conjectured that atoms were knotted vortices in the ether. The idea was that the universe was filled with ether (an invisible fluid) and that the building blocks of the molecules could be thought of as knots in this ether. Scientists attempted to create tables of knots depending on how many crossings there were in a drawing (diagram) of the knot. This resulted in the first known knot tables. The theory of Kelvin was later replaced by other theories about the nature of our basic matter and the theory of knots was put on the shelf.
In the latter part of the 20th century there was a renewed interest in knot theory. Strands of DNA can sometimes form knots, and some mathematical techniques that play a role in quantum mechanics are related to knot theory for instance.
This section is unfinished.
## Knot Tables
In the 19th century Tait, Kirkman and Little started tabulating the knots in so called knot tables. The knots are listed in order grouped by the number of crossings. Below you see a copy of the knot table for knots with 3 to 7 crossings.
The knot in the top left corner is special. It is often called the unknot. It is a simple circle with no knotting. As you see there is exactly one knot with 3 crossongs. This knot is often called the trefoil knot. There is also a unique knot with four crossings: the figure-eight-knot. In some pictures part of the knot will resemble a figure eight.
The original image can be found at The knot Atlas page.[1]. On this site you can click on the knots to obtain more information.
An important question is: Can we recognize the unknot?
Knots with 8 crossings.
Knots with 9 crossings.
This section is unfinished.
## References
• Bain, George, Celtic Art: The Methods of Construction, Dover (1973) ISBN 0-486-22923-8
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## surjective proof
Using the following definition:
Definition: Let X and Y be any subsets of Real Numbers. A function f: X to Y from a set X to a set Y is called bounded above if and only if there exists M in Reals such that for all x in X, f(x) less than or equal to M.
Prove or disprove: Let A, B and C be any sets. For any functions g:A to B, h:A to B, and f:B to C, if fog=foh and f is surjective, then g=h.
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# Fin 534 week 2 homework set #1
Don't use plagiarized sources. Get Your Assignment on
Fin 534 week 2 homework set #1
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FIN 534 – Homework Set #1
© 2014 Strayer University. All Rights Reserved. This document contains Strayer University Confidential and Proprietary information and may not be copied, further distributed, or otherwise disclosed in whole or in part, without the expressed written permission of Strayer University. FIN 534 Homework Set #1 Page 1 of 4
Directions: Answer the following questions on a separate document. Explain how you reached the answer or show your work if a mathematical calculation is needed, or both. Submit your assignment using the assignment link in the course shell. This homework assignment is worth 100 points.
Use the following information for Questions 1 through 8:
Assume that you recently graduated and have just reported to work as an investment advisor at the one of the firms on Wall Street. You have been presented and asked to review the following Income Statement and Balance Sheets of one of the firm’s clients. Your boss has developed the following set of questions you must answer.
Income Statements and Balance Sheet Balance Sheet 2012 2013 Cash \$9,000 \$7,282 Short-term investments 48,600 20,000 Accounts receivable 351,200 632,160 Inventories 715,200 1,287,360 Total current assets \$1,124,000 \$1,946,802 Gross fixed assets 491,000 1,202,950 Less: Accumulated depreciation 146,200 263,160 Net fixed assets \$344,800 \$939,790 Total assets \$1,468,800 \$2,886,592 Liabilities and Equity Accounts payable \$145,600 \$324,000 Notes payable 200,000 720,000 Accruals 136,000 284,960 Total current liabilities \$481,600 \$1,328,960 Long-term debt 323,432 1,000,000 Common stock (100,000 shares) 460,000 460,000 Retained earnings 203,768 97,632 Total equity \$663,768 \$557,632 Total liabilities and equity \$1,468,800 \$2,886,592
Statements 2012 2013 Sales \$3,432,000 \$5,834,400 Cost of goods sold except depr. 2,864,000 4,980,000 Depreciation and amortization 18,900 116,960 Other expenses 340,000 720,000 Total operating costs \$3,222,900 \$5,816,960 EBIT \$209,100 \$17,440 Interest expense 62,500 176,000 EBT \$146,600 (\$158,560) Taxes (40%) 58,640 -63,424 Net income \$87,960 (\$95,136)
Other Data 2012 2013 Stock price \$8.50 \$6.00 Shares outstanding 100,000 100,000 EPS \$0.88 (\$0.95) DPS \$0.22 0.11 Tax rate 40% 40% Book value per share \$6.64 \$5.58 Lease payments \$40,000 \$40,000
Ratio Analysis 2012 2013 Current 2.3 1.5 Quick 0.8 0.5 Inventory turnover 4 4 Days sales outstanding 37.3 39.6 Fixed assets turnover 10 6.2 Total assets turnover 2.3 2 Debt ratio 35.60% 59.60% Liabilities-to-assets ratio 54.80% 80.70% TIE 3.3 0.1 EBITDA coverage 2.6 0.8 Profit margin 2.60% −1.6% Basic earning power 14.20% 0.60% ROA 6.00% −3.3% ROE 13.30% −17.1% Price/Earnings (P/E) 9.7 −6.3 Price/Cash flow 8 27.5 Market/Book 1.3 1.1
What is the free cash flow for 2013?
2. Suppose Congress changed the tax laws so that Berndt’s depreciation expenses doubled. No changes in operations occurred. What would happen to reported profit and to net cash flow?
3. Calculate the 2013 current and quick ratios based on the projected balance sheet and income statement data. What can you say about the company’s liquidity position in 2013?
4. Calculate the 2013 inventory turnover, days sales outstanding (DSO), fixed assets turnover, and total assets turnover.
5. Calculate the 2013 debt ratio, liabilities-to-assets ratio, times-interest-earned, and EBITDA coverage ratios. What can you conclude from these ratios?
6. Calculate the 2013 profit margin, basic earning power (BEP), return on assets (ROA), and return on equity (ROE). What can you say about these ratios?
7. Calculate the 2013 price / earnings ratio, price / cash flow ratio, and market / book ratio.
8. Use the extended DuPont equation to provide a summary and overview of company’s financial condition as projected for 2013. What are the firm’s major strengths and weaknesses?
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MAE2_Lecture15
# MAE2_Lecture15 - Structures and Materials Turning Flight...
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Structures and Materials Turning Flight V-n Diagrams Lift Distribution Stress and Strain Materials Weight Estimation MAE 2
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2 MAE 2 Turning Flight L cos = W L = lift W = weight V = velocity = bank angle = heading angle mV d dt = L sin A level turn implies that the airplane flight path changes, but is still restricted to a horizontal plane. By banking the wings, the lift vector is tilted toward the center of the turn. Vertical: Horizontal: W L Airplane is flying into page during a right turn (right wing down) Equations of Motion
3 MAE 2 Load Factor Define load factor as the ratio of lift to weight. L cos = W Vertical: cos = 1 n n L W n = load factor sin = 1 cos 2 = 1 1 n 2 = n 2 1 n 2
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4 MAE 2 Turn Rate and Radius Turn rate is the rate at which the airplane changes (compass) heading. ≡ d dt mV d dt = L sin Horizontal: mV = mg L W sin = turn rate R = turn radius = g V n sin = g V n n 2 1 n 2 = g V n 2 1 V = R R = V 2 g n 2 1
5 MAE 2 Maximum Turn Rate
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# Why protons and neutrons don't have less mass than their constituents?
A system of gravitational attracted objects weight less than the sum of their individual masses because it needs energy to move them apart and overcome the gravitational attraction. Same is true for electromagnetic force, where the attraction between opposite charges must be overcome with additional energy to keep the objects move apart, therefore the system has less energy and less mass.
But why this isn't the case for three quarks bound together to form a proton/neutron?. The color force is (in simple terms) also attractive between the three quarks in a bound system and therefore energy must be applied to move them apart. As a consequence, the bounded system of quarks should have less energy (and less mass) than unbound quarks. Why this isn't the case?
Imagine free quarks being brought together to make a baryon. Your intuition is that, since the end product is bound, (i) the total mass-energy has declined and, though you may not have thought of this implication, (ii) putting the difference back in could break the quarks out. Neither is correct.
Tackling (ii) first, if you try pulling quarks out, your energy is spent on the creation of quark-antiquark pairs, and so what's released is new mesons, not the original valence quarks. This is deeply weird when you first learn of it: it's as if we couldn't create cations, because the attempt to remove an electron from a neutral atom simply released positronium.
The difference comes from gluons carrying their own charge; photons, by contrast, are electrically neutral. So not only do quarks emit gluons, the gluons do too. A baryon's three valence quarks are joined by flux tubes of gluons. These tubes can absorb energy, then turn it into mesons, if you try to pull out the original valence quarks.
A gluon has a color and anticolor charge, and can turn into a quark-antiquark pair. This doesn't just give baryons a meson-making party trick; it also means "sea quarks" populate the forcefields between the valence quarks in a baryon. (Gluons are also believed to clump into glueballs.) So if we turn now to (i), what mass do we reduce by a binding energy's worth to get a baryon's mass? Answer: that of many more than just three particles. Another way to put it is to assign multiple notions of mass to a quark: its "current mass" is what we imagined before the baryon formed, but its being in the baryon means it really has a larger constituent mass.
It's a good question and has a subtle answer, as answer from J.G. has already tried to explain. I think it may help to point out a related fact about something more familiar: the electron.
The mass of an electron is generally stated to be about $$9.10938 \times 10^{-31}\,$$kg. Now let's consider the electric field around a non-moving electron. This field has an energy density $$(1/2) \epsilon_0 E^2$$, and a magnitude $$E = \frac{e}{4 \pi \epsilon_0 r^2}.$$ Suppose we could imagine the electron as a point charge. In that case the electric field would tend to infinity at locations approaching the point charge and so would the energy density. So that suggests the point charge model is questionable. If instead we model an electron as a sphere of some very small but non-zero radius, then we get a finite electric field and a finite energy density. By integrating the field energy density over volume, we get an energy. It is the total energy of the electric field of our electron. This energy has a mass associated with it, and this mass contributes to the $$9.10938 \times 10^{-31}\,$$kg mentioned above. In fact, if you choose the radius of the electron appropriately, then the whole mass of the electron is accounted for as coming from the field!
Please note, this 'classical charged sphere' model of an electron is not the correct model in the end---one needs quantum theory for that---but it makes the point that an electron has more mass overall than the mass which you might assign to the particle in the absence of its surrounding field.
Similar statements apply to quarks, only more so. In view of the field energy contribution, the term 'the mass of a quark' is rather ambiguous. What mass is it referring to? If it is referring to some notion of what the mass would be if there were no field (no gluons etc) then it could be quite unrelated to the mass which is relevant to things like protons and neutrons. It could even be zero, and then the whole mass of the proton comes from the gluons. Indeed that idea is a pretty good first approximation to what happens.
In view of the above, my answer to the question "why is the proton mass more than the sum of the quark masses, which one does not expect for a bound system?" is "it's because you need to understand more clearly what mass you are referring to when you quote some given value for a quark mass".
A bound system can have greater mass than sum of masses of the components. This does not happen in case of gravitational and electric forces, because the corresponding potential energies are monotonic functions of distance. This is because the forces are attractive irrespective of distance.
However, if there was a conservative force acting on the two bodies that was repulsive for distances greater than some distance $$r_0$$, and attractive for smaller distances, potential energy would have a maximum for some distance "near" $$r_0$$. If the system was in such a state or close to it, it would have greater mass than the system in decayed state (where the components are infinitely far from each other).
Simply put, a bound system can be like a "compressed locked spring", storing more energy than system whose parts are far from each other.
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# math
posted by on .
Subtract simplify by collecting like radical terms 2square root symbol 50 -2 squareroot symbol 2.
• math - ,
2√50 - 2√2
2√25√2 - 2√2
10√2 - 2√2
8√2
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# Math
posted by .
4s-(3-2s)-6=9
Please check my work:
4s-3+2s-6=9
6s-9=9
3s=18
s=3
• Math -
Your answer is right, but you have an error in this line:
3s = 18
I think you mean 6s = 18
• Math -
Yes that's what I meant.
Thank you
• Math -
You are welcome.
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# [closed] Strange score values for Feb19 Challenge
Hi,
It seems there is some error in calculating absolute scores for Feb19 challenge problem CHORCKIT.
In the problem statements we have:
“The length of the string S must not exceed 10^6 at any time.”, “There are M melodies … M <= 100”, “Ci <= 1000”. So even if every pretty melody occurs at every position in the resulting string and every melody has the maximal possible pretty value, we can get at most V = 10^6 * 100 * 1000 = 10^11 for each test. Now, the absolute score is V/100, i.e. no more than 10^9 for each test or 5*10^9 for 5 tests.
However currently the best solution shows some astronomic numbers like 22271556114636.004. And what is .004 while we are dividing integer value by 100? Is this just a float-precision issue?
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# How Much Electricity Does A 40 Gallon Water Heater Use?
factor of energy
factor
#### With a 0.95 energy efficiency, a 50-gallon electric tank uses 4622 kWh.
factor
• A 50-gallon gas tank with a 0.62 energy efficiency produces 242 therms each year.
factor
factor
## For a month, how much does it cost to run a hot water heater?
The number of watts an electric water heater consumes is determined by a variety of factors, including the unit’s age and size, whether it’s a tank or on-demand model, the temperature you set it to, how much hot water you use in a day, and other considerations. You may calculate the cost by multiplying the number of watts your heater consumes by the price per kWh multiplied by the number of hours the heater is turned on every day, then dividing by 1,000. A tank-style hot water heater will typically run for three to five hours every day. So, at \$.10 per kWh, a 4,000-watt heater operated for three hours each day will cost \$1.20 per day, \$36.50 per month, or \$438 per year.
## Is it true that water heaters consume a lot of electricity?
However, your hot water heater consumes a significant amount of electricity. The second largest consumer of electricity in your home is water heating systems. According to the US Department of Energy, this accounts for 18% of your total electricity bills.
## What is the daily energy consumption of a water heater?
Water heaters have an average thermal dispersion of 1 to 2 kWh/24 hr. This means that an unattended electric water heater uses between 1 and 2 kWh per hour.
## What is the most cost-effective method of heating hot water?
Simply said, if you have a central heating gas boiler, heating your water with it is the cheapest and most efficient choice. Of course, many individuals do not have access to gas, so it is not always possible.
### >> Find out whether a biomass boiler is a viable option for your home by clicking here
If you live off the grid, you can heat your water using a biomass burner or an oil boiler.
## Is the difference between a 40-gallon and a 50-gallon water heater significant?
With a larger tank, you won’t have to worry about running out of hot water. Water heaters with a capacity of 50 gallons have a longer lifespan. 40-gallon heaters typically last eight years, but 50-gallon heaters can last up to 12 years. Because 50-gallon heaters have thicker tanks, this is the case. As a result, they have a lower risk of developing leaks.
A 50-gallon water heater has a lifespan of 8 to 20 years, which is significantly longer than the average water heater.
#### They are less expensive than water heaters with a capacity of 40 gallons.
An extra ten gallons may not appear to make much of a difference. However, going with a 50-gallon water heater implies going with a more cost-effective water heater.
A 50-gallon water heater, according to experts, is less expensive to operate than a 40-gallon water heater. Furthermore, a water heater with a capacity of 50 gallons is more efficient than one with a capacity of 40 gallons.
## What in a house consumes the most electricity?
The breakdown of energy use in a typical home is depicted in today’s infographic from Connect4Climate.
It displays the average annual cost of various appliances as well as the appliances that consume the most energy over the course of the year.
Modern convenience comes at a cost, and keeping all those air conditioners, freezers, chargers, and water heaters running is the third-largest energy demand in the US.
Here are the things in your house that consume the most energy:
• Cooling and heating account for 47% of total energy consumption.
• Water heater consumes 14% of total energy.
• 13 percent of energy is used by the washer and dryer.
• Lighting accounts for 12% of total energy use.
• Refrigerator: 4% of total energy consumption
• Electric oven: 34% energy consumption
• TV, DVD, and cable box: 3% of total energy consumption
• Dishwasher: 2% of total energy consumption
• Computer: 1% of total energy consumption
One of the simplest ways to save energy and money is to eliminate waste. Turn off “vampire electronics,” or devices that continue to draw power even when switched off. DVRs, laptop computers, printers, DVD players, central heating furnaces, routers and modems, phones, gaming consoles, televisions, and microwaves are all examples.
A penny saved is a cent earned, and being more energy efficient is excellent for your wallet and the environment, as Warren Buffett would undoubtedly agree.
## How much does a 50 gallon water heater cost to run?
Your water heater is the second-largest energy consumer in your home, accounting for roughly 17% of overall energy use, according to the Department of Energy. With this in mind, knowing how much energy your electric hot water heater uses and how much it costs you each month is beneficial. You can make the most efficient use of your water heater and lower your electric bill with this knowledge.
### Are electric water heaters expensive to run?
The average American household spends between \$400 and \$600 per year on water heating, according to the Department of Energy. This cost will vary based on the cost of power in your area and the efficiency of your water heater.
The quantity of energy consumed by an electric water heater is determined by a number of factors, including:
• electric water heater type
• the amount of water used on a daily basis
When it comes to operating costs, the type of electric heater is extremely essential. The following is a list of electric hot water heating system alternatives.
### Storage Water Heaters
When it comes to the system itself, storage heaters are often the least expensive, with low operating expenses.
One issue with this method is that heat energy is lost when the tank struggles to keep the water warm even when it isn’t being utilized, so you are paying for wasted energy. If you decide to go this route, make sure you choose an insulated model that will keep this from happening.
### Tankless Water Heaters
Tankless water heaters are slightly more expensive to buy and install, but they are between 8% and 34% more energy efficient than storage water heaters, depending on the volume of water consumed, according to the Department of Energy. This means that upgrading to this type of water heater could save you \$100 or more per year.
The disadvantage is that, depending on the type and your home’s electrics, the initial purchase and installation costs may not be worth the energy savings until much later.
### Heat Pump Heating Systems
Heat pump systems use two to three times less energy than storage heaters, resulting in cheaper running expenses. Energy Star’s most efficient models will save you up to \$300 per year on your energy expenses.
However, because the system must maintain a constant temperature throughout the year, performance may vary depending on where it is situated in your home. They also require a large amount of room for both the pump and the surrounding air.
The most significant disadvantage is the initial cost, which starts at roughly \$1,000 for the purchase price alone, plus installation and maintenance charges.
### Can a water heater cause a higher electric bill?
To begin with, if you set your electric hot water heater to a higher temperature than is required, your heater will use more energy, resulting in higher energy costs. Below, we’ll look at how to fix this.
Another issue could be that your water heater is either too big or too little for your home. If your water heater is too big, it will just heat water that will never be utilized all at once, costing you more money than necessary.
If your water heater is not large enough, you will not only not have enough hot water for your entire home, but it will also be overworked since it will have to work more frequently throughout the day. Even if this does not immediately increase your electric bill, you will wind up paying for maintenance and repairs more frequently.
As a general guideline, 30- to 40-gallon water heaters are best for one or two people, 40- to 50-gallon heaters are best for two or three people, 50-to-60-gallon heaters are best for three or four people, and 60-to-80-gallon heaters are best for five or more people.
If your water heater is too old, you may also notice an increase in your electric bill. Electric water heaters are typically built to last 10 to 15 years, after which their efficiency drastically diminishes, resulting in higher energy expenses.
### What is the average cost of an electric hot water heater?
The average cost of an electric hot water heater varies depending on the kind, ranging from \$300 to \$700, with additional installation fees ranging from \$700 to \$1,000.
This implies that depending on the size of the system and the type of water heater you choose, you may expect to pay anywhere from \$1,000 to \$1,700.
### Are new electric water heaters more efficient?
Electric water heaters are no exception to the rule that newer appliances are more efficient than older counterparts.
Yes, they will cost more up front in terms of purchase price and installation fees, but they will save you money on your energy bills in the long run because they need less energy to function.
If you’re in the market for a new electric water heater, search for one that has the Energy Star designation because you’ll know it’ll be as energy-efficient as possible.
### Should I turn off my water heater at night?
You don’t have to worry about turning off your tankless on-demand heater at night because they are intended to only heat the water when it is actually needed.
If you have a tank water heater, try shutting it off not only at night, but whenever it is not in use for an extended length of time to save money on your power bill.
This type of water heater just heats the entire tank of water and keeps it at the desired temperature until it is needed again. Even with insulation, roughly 10% of the heat generated is lost to the atmosphere. By turning off your water heater, you can stop the heat from fleeing, as well as the few dollars that are escaping each day!
### Does turning down the water heater save money?
The majority of water heaters have a preset temperature of 140 degrees Fahrenheit, which is similar to the temperature required for dishwashers to function and clean dishes properly in the past.
In fact, most other home uses only require a hot water temperature of 120 degrees Fahrenheit, and dishwashers are now equipped with heating boosters that allow them to raise the hot water up to the required temperature on their own.
If you set your water heater at 120 degrees Fahrenheit, you will receive the most efficient temperature and save money by not heating to a higher temperature than is required.
### How much does it cost to run a 50-gallon electric water heater?
Based on the premise that an electric water heater is used for around three hours per day, a 50-gallon water heater running at 5,500 watts with an electricity rate of \$0.16 per kWh will cost \$781 per year to operate, according to the Department of Energy.
This cost will vary depending on the water heater’s efficiency rating and local electricity costs.
### Are electric water heaters worth it?
Electric water heaters are a popular choice in many households since their upfront costs are lower than those of other options, while installation prices vary based on the type of water heater.
They’re also better for the environment than other water heating options because they’re one of the more efficient forms of water heaters and, unlike gas heaters, can be powered by renewable energy sources.
Electric water heaters are also readily available to all persons who are linked to the electric grid, even if other sources, such as gas, are not. This is only a concern if there is a significant power outage, in which case you will be without hot water until the situation is resolved.
If you can pay a little extra on a newer, more energy-efficient model, electric water heaters are a terrific option. This will help you save money on your electricity costs on a daily basis, especially if you make an effort to use less hot water at a lower temperature.
## A 50 gallon electric water heater consumes how many watts?
A 50 gallon electric water heater uses 4500 watts of power. 4500 watts is equal to 18.75 amps in a 240 volt electric circuit. This means that a 50 gallon water heater will not flip a 20 amp circuit breaker, but it will push it to work at 95 percent of its capacity, which most circuit breakers are not designed to handle.
## Is it true that shutting off the water heater saves money?
Like many other electrical appliances, turning off your water heater can help you save money on your monthly payment. Even though the water heater is adequately insulated, a little quantity of heat escapes when it is turned on. This energy waste accounts for around 10% of your total expenditure. You may save ten percent by turning off your water heater and preventing additional energy loss.
## Is it true that turning down the water heater saves money?
Homeowners will save between 6 to 10% by lowering the temperature of their hot water tank by roughly 20 degrees to 120 degrees Fahrenheit or 48 degrees Celsius.
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# Convert kilometer/hour to centimeter/hour
## How to Convert kilometer/hour to centimeter/hour
To convert kilometer/hour to centimeter/hour , the formula is used,
$\mathrm{cm/h}=\mathrm{km/h}×100000$
where the km,h to cm,h value is substituted to get the answer from Speed Converter.
1 km,h
=
10000e+1 cm,h
1 cm,h
=
0 km,h
Example: convert 15 km,h to cm,h:
15 km,h
=
15
x
10000e+1 cm,h
=
15000e+2 cm,h
## kilometer/hour to centimeter/hour Conversion Table
kilometer/hour (km,h) centimeter/hour (cm,h)
0.01 km,h 1000 cm,h
0.1 km,h 10000 cm,h
1 km,h 100000 cm,h
2 km,h 200000 cm,h
3 km,h 300000 cm,h
5 km,h 500000 cm,h
10 km,h 1000000 cm,h
20 km,h 2000000 cm,h
50 km,h 5000000 cm,h
100 km,h 10000000 cm,h
1000 km,h 100000000 cm,h
### Popular Unit Conversions Speed
The most used and popular units of speed conversions are presented for quick and free access.
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incremental sort
closed account (oyA74iN6)
Given a stack, the task is to sort it such that the top of the stack has the greatest element.
Input:
The first line of input will contains an integer T denoting the no of test cases . Then T test cases follow. Each test case contains an integer N denoting the size of the stack. Then in the next line are N space separated values which are pushed to the the stack.
Output:
For each test case output will be the popped elements from the sorted stack.
Constraints:
1<=T<=100
1<=N<=100
Example(To be used only for expected output):
Input:
2
3
3 2 1
5
11 2 32 3 41
Output:
3 2 1
41 32 11 3 2
Explanation:
For first test case stack will be
1
2
3
After sorting
3
2
1
When elements popped : 3 2 1
Last edited on
when writing your comparison for the std sort routine, you can make it complicated...
in pseudo code:
a is less than b if uppercase(a.city) < uppercase(b.city)
if the above is equal, the
a is less than b if (a.time) < (b.time) //figure out how to compare the timestamps
etc. At the end of it all, your code should have prioritized the city comparison first, and when those are equal, figured out the time comparison instead.
Then just feed that to the sort routine, or add the comparison operator overload to your class and it should work.
Topic archived. No new replies allowed.
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# Computation with no memory, and rearrangeable multicast networks.
1 Introduction
The mathematical definition of a mapping E : [S.sup.n] [right arrow] [S.sup.n] can be thought of as the parallel computation of n assignment mappings [S.sup.n] [right arrow] S performing the mapping E, either by modifying at the same time the n component variables, or mapping the n input component variables onto n separate output component variables. If one wants to compute sequentially the mapping E by modifying the components one by one and using no other memory than the input variables whose modified values overwrite the initial values, one necessarily needs to transform the n mappings [S.sup.n] [right arrow] S in a suitable way. We call in situ computation this way of computing a mapping, and it turns out that it is always possible with a number of assignments linear with respect to n and a small factor depending on the mapping type.
The idea of developing in situ computation came from a natural computation viewpoint: transformation of an entity or structure is made inside this entity or structure, meaning with no extra space. As a preliminary example, consider the mapping E : [S.sup.2] [right arrow] [S.sup.2] defined by E([x.sub.1], [x.sub.2]) = ([x.sub.2], [x.sub.1]) consisting in the exchange of two variables for a group S. A basic program computing E is: x' := [x.sub.1]; [x.sub.1] := [x.sub.2]; [x.sub.2] := x'. An in situ program for E avoids the use of the extra variable x', with the successive assignments [x.sub.1] := [x.sub.1] + [x.sub.2]; [x.sub.2] := [x.sub.1]--[x.sub.2]; [x.sub.1] := [x.sub.1]--[x.sub.2]. In situ computation can be seen as a far reaching generalization of this classical computational trick. See Section 2 for a formal definition.
The problem of building in situ programs has already been introduced and considered under equivalent terms in [4, 5, 6, 7, 8, 9]. In the first papers, it had been proved that in situ computations are always possible [4], that three types of assignments are sufficient to perform this kind of computations [5], that the length of in situ computations of mappings on [{0,1}.sup.n] is bounded by [n.sup.2] [6], and that any linear mapping on [{0,1}.sup.n] is computed with 2n--1 linear assignments [7]. It turned out that, though this had not been noticed in those papers, this problem has close relations with the problem of finding rearrangeable (non-blocking) multicast routing methods for multistage interconnection networks obtained by concatenations of butterfly networks (see Section 3 for a formal definition). Also, several existence results on in situ programs can be deduced from results in the network field. This relation has been partially presented in [9], which proposed also improved bounds for mappings of various types on more general sets than the boolean set, and which can be considered as a preliminary conference version of the present paper (with weaker results, fewer detailed constructions, fewer references and fewer illustrations). Let us also mention [8] which presented the subject to an electronics oriented audience for the sake of possible applications, [12] which presented the subject to non-specialists and general public, and [13] which gives some results in the continuation of ours.
In the present paper, we survey the relation between in situ computations and multicast rearrangeable networks, through precise results and historical references. Also, we recall that a bijective mapping can be computed by a program of length 2n--1; we give, for a general arbitrary mapping, two methods to build a program with maximal length 4n--3, one of which is equivalent to a known method in network theory (Section 4), one of which available on the boolean set is new and more flexible (Section 5); and we build, for a linear mapping of a rather general kind, a program with maximal length 2n--1 (Section 6). Moreover, we end each main section with an open problem. Our techniques use combinatorics and modular arithmetic. Finally, our aim is to give a precise and illustrated survey on this subject, for both viewpoints (computations and networks), so as to sum up old and new available results in a unified appropriate framework.
Let us first detail some links with references from the network viewpoint. Multistage interconnection networks have been an active research area over the past forty years. We refer the reader to [14][16] for background on this field. Here, an assignment, which is a mapping [S.sup.n] [right arrow] S, is regarded as a set of edges in a bipartite graph between [S.sup.n] (input) and [S.sup.n] (output) where an edge corresponds to the modification of the concerned component. See Section 3 for details. All the results of the paper can be translated in this context. More precisely, making successive modifications of all consecutive components of X [member of] [S.sup.n] is equivalent to routing a butterfly network (i.e. a suitably ordered hypercube, sometimes called indirect binary cube) when S = {0,1}, or an s-ary butterfly network for an arbitrary finite set S with [absolute value of S] = s. Butterfly networks are a classical tool in network theory, and we mention that butterfly-type structures also appear naturally in recursive computation, for example in the implementation of the well-known FFT algorithm [11], see [16].
In the boolean case, the existence of an in situ program with 2n--1 assignments for a bijective mapping is equivalent to the well known [2] rearrangeability of the Benes network (i.e. of the concatenation of two reversed butterflies), that is: routing a Benes network can perform any permutation of the input vertices to the output vertices. Such a rearrangeability result can be extended to an arbitrary finite set S and an s-ary Benes network by means of the rearrangeability properties of the Clos network [14].
The problem of routing a general arbitrary mapping instead of a permutation, where several inputs may have the same output, is equivalent, up to reversing the direction of the network, to the rearrangeable multicast routing problem: one input may have several outputs, each output must be reachable from the associated input, and the different trees joining inputs to their outputs must be edge-disjoint. This general problem is a classical one in network theory, where sometimes rearrangeable is called rearrangeable non-blocking, and a huge number of routing methods have been developed for various networks, whose aims are to minimize the number of connections and to maximize the flexibility of the routings. As an instance of network derived from the butterfly network, an efficient construction consists in stacking butterfly networks [15]. Other examples and further references can be found for instance in [14].
In this paper, we are interested in networks obtained by concatenation of butterfly networks (a construction sometimes called cascading). A rearrangeable multicast routing method for such (boolean) networks was proposed in [18], involving five copies of the butterfly network, with possible reversions. It was noticed in [20] that one can remove one of these copies preserving the same rearrangeability property, yielding four copies only. In [17], a similar construction has been given, based on (boolean) baseline networks instead of butterfly networks (yielding an equivalent result since those two [log.sub.2]N networks are known to be topologically equivalent, see [3] for instance).
In Section 4, we investigate this problem under the setting of in situ programs, and we provide similar results than those cited above, with slight variants and complementary results (arbitrary finite sets, inversion of bijections...). Also, this provides a practical framework that unifies those network results from the literature. This connection is not always clear from the way those references were written, and we feel that this survey work is interesting on its own. And this framework will serve again for the next section. This yields finally an in situ program of length 4n--3 for a general mapping. The common general idea of those constructions involving four butterfly copies is the following: first group the vectors having same image, using two copies that provide a (unicast) rearrangeable network; then use one copy to give all those vectors a common image; and lastly use one copy to bring those images to the final required output. The efficiency of the two last steps relies on the capability of the butterfly network to map arbitrary inputs onto consecutive outputs in same order (a sorting property called infra-concentrator property in [14], or packing problem property in [16]). So, the limitation of this type of multicast routing strategy is that the groups formed at the middle stage have to be exactly in the same order than the final images, thus this middle stage is (almost) totally determined, yielding a poor flexibility.
In Section 5, we provide a new and more sophisticated construction, relying on the same framework. It involves arithmetical properties because of which we assume that [absolute value of S] is a power of 2. We mention that those propeties, and so the whole construction, can be extended to arbitrary [absolute value of S] as noticed in [13]. It yields new results on butterfly routing properties refining its classical sorting property (Proposition 23), and a more flexible multicast routing method for general mappings, with the same number of stages (four copies of the butterfly network), i.e. same in situ program length (Theorem 25). The improvement is to allow a huge number of possible orderings of the groups at the middle stage (see Remark 19 for details).
Let us now give some details from the algorithmic viewpoint. Building assignments whose number is linear in n to perform a mapping of [S.sup.n] to itself is satisfying in the following sense. If the input data is an arbitrary mapping E : [S.sup.n] [right arrow] [S.sup.n] with [absolute value of S] = s, given as a table of n x [s.sup.n] values, then the output data is a linear number of mappings [S.sup.n] [right arrow] S whose total size is a constant times the size of the input data. This means that the in situ program of E has the same size as the definition of E by its components, up to a multiplicative constant. This complexity bound is essentially of theoretical interest, since in terms of effective technological applications, it may be difficult to deal with tables of n x [s.sup.n] values for large n. Alternatively, assignments can be defined with algebraic expressions, for instance mappings [{0,1}.sup.n] [right arrow] {0,1} are exactly multivariate polynomials of degree at most n on n variables on the binary field {0,1}. Hence, it is interesting to deal with an input data given by algebraic expressions of restricted size, like polynomials of bounded degree for instance, and compare the complexity of the assignments in the output data with the input one, for instance using polynomial assignments of a related bounded degree. This general question (also related to the number of gates in a chip design) can motivate further research (examples are given in [8], see also Open problems 3 at the end of the paper).
Here, in Section 6, we prove that, in the linear case, i.e. if the input is given by polynomials with degree at most 1, with respect to any suitable algebraic structure for S (e.g. any field, or Z/sZ), then the assignments are in number 2n-1 and overall are also linear. Hence, we still obtain a program whose size is proportional to the restricted size of the input mapping. We mention that this decomposition method takes O([n.sup.3]) steps to build the program, and that if the mapping is invertible, then we get naturally a program for the inverse. This result generalizes to a large extent the result in [7] obtained for linear mappings on the binary field. In terms of multistage interconnection networks, a similar result is given in [19], also for the binary field only. Here, we get rearrangeable non-blocking multicast routing methods for the s-ary Benes network as soon as the input/outputs are related through a linear mapping on any suitable more general algebraic structure. Let us insist on the fact that this result is way more general than its restriction to the boolean field. First, there is a generalization from the boolean field Z/2Z to any field, such as Z/pZ for p prime. Secondly, there is a generalization to general rings such as Z/nZ for any integer n (which are not necessarily fields, i.e. elements are not necessarily invertible). Linear mappings on such general rings are fundamental and much used in mathematics and computer science (e.g. in cryptography). Those two generalizations should be considered as non-trivial theoretical jumps. Also, from the algebraic viewpoint, this provides a new result on matrix decompositions.
Finally, let us mention that some of the original motivation for this research was in terms of technological applications. A permanent challenge in computer science consists in increasing the performances of computations and the speed of processors. A computer decomposes a computation in elementary operations on elementary objects. For instance, a 64 bits processor can only perform operations on 64 bits, and any transformation of a data structure must be decomposed in successive operations on 64 bits. Then, as shown in the above example on the exchange of the contents of two registers [x.sub.1] and [x.sub.2], the usual solution to ensure the completeness of the computation is to make copies from the initial data. But this solution can generate some memory errors when the structures are too large, or at least decrease the performances of the computations. Indeed, such operations involving several registers in a micro-processor, through either a compiler or an electronic circuit, will have either to make copies of some registers in the cache memory or in RAM, with a loss of speed, or to duplicate signals in the chip design itself, with an extra power consumption. On the contrary, the theoretical solution provided by in situ computation would possibly avoid the technological problems alluded to, and hence increase the performance.
2 In situ programs
For the ease of the exposition, we fix for the whole paper a finite set S of cardinality s = [absolute value of S], a strictly positive integer n and a mapping E : [S.sup.n] [right arrow] [S.sup.n]. This paper strongly relies on the following definition.
Definition 1 An in situ program [product] of a mapping E : [S.sup.n] [right arrow] [S.sup.n] is a finite sequence
([[psi].sub.1], [i.sub.1]), ([[psi].sub.2], [i.sub.2]), ..., ([[psi].sub.m], [i.sub.m])
of assignments such that:
-for k = 1,2, ..., m, we have [[psi].sub.k] : [S.sup.n] [right arrow] S and [i.sub.k] [member of] {1, n};
--every transformation X = ([x.sub.1], ..., [x.sub.n]) [right arrow] E(X) is computed through the sequence
X = [X.sub.0], [X.sub.1], ..., [X.sub.m--1], [X.sub.m] = E(X)
where, for k = 1, 2, ..., m, the vector [X.sub.k] has the same components as [X.sub.k--1] except component [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] which is equal to [[psi].sub.k] ([X.sub.k--1]).
In other words, [[psi].sub.k] modifies only the [i.sub.k]-th component of the current vector, that is: every assignment ([[psi].sub.k], [i.sub.k]) of an in situ program performs the elementary operation
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII].
The length of [product] is the number m. The signature of [product] is the sequence
[i.sub.1], [i.sub.2], ..., [i.sub.m].
All in situ programs considered throughout this paper operate on consecutive components, traversing the list of all indices, possibly several times in forward or backward order. Thus program signatures will all be of type:
1, 2, ..., n--1, n, n--1, ..., 2, 1, 2, ... n--1, n, ...
For ease of exposition, the mappings [S.sup.n] [right arrow] S in the corresponding sequence of assignments will be simply distinguished by different letters, e.g. [f.sub.i] denotes the mapping affecting the variable [x.sub.i] on the first traversal, [g.sub.i] the one affecting [x.sub.i] on the second traversal, and so on, providing a sequence of assignment mappings denoted
[f.sub.1], [f.sub.2], [f.sub.n--1], [f.sub.n],[g.sub.n--1], ..., [g.sub.2], [g.sub.1], ...
where each index gives the index of the component modified by the mapping. For instance, a program [f.sub.1], [f.sub.2], [g.sub.1] on [S.sup.2] represents the sequence of operations: [x.sub.1] := [f.sub.1]([x.sub.1], [x.sub.2]); [x.sub.2] := [f.sub.2]([x.sub.1], [x.sub.2]); [x.sub.1] := [g.sub.1]([x.sub.1], [x.sub.2]).
As an example, it is easy to see that a mapping consisting in a cyclic permutation of k variables in a group S can be computed in k + 1 steps using the k variables only. This is an extension of the case of the exchange of two variables. Precisely ([x.sub.1], ..., [x.sub.k]) [right arrow] ([x.sub.2], ..., [x.sub.k], [x.sub.1]) is computed by the in situ program: [x.sub.1] := [x.sub.1] + [x.sub.2] +... + [x.sub.k];[x.sub.k] := [x.sub.1]--[x.sub.2]--...--[x.sub.k]; ... ; [x.sub.2] := [x.sub.1]--[x.sub.2]--...--[x.sub.k]. This length turns out to be a minimal bound for this type of mapping, as shown in the next proposition, which we state as an example of an in situ computation property whose proof is not so trivial. Let us mention that this proposition has been suggested by [12], and that [13] provides a similar but slightly more general result authorizing overwriting of variables.
Proposition 2 If S is a finite set, and the mapping E : [S.sup.n] [right arrow] [S.sup.n] consists in a permutation of the n variables, then an in situ program for E has a length greater than n--f + c, where c is the number of cycles of E non-reduced to one element, and f is the number of invariant elements of E.
Proof: In what follows, we assume that f = 0, then the proof can be extended directly to the case where f > 0 by applying it to the restriction of E to [S.sup.n--f]. Assume there exists an in situ program [product] for E of length strictly smaller than n + c.
Then there exists a cycle (non-reduced to a single variable) whose variables are modified once and only once each in the program (since each non-invariant variable is modified at least once). Assume the sequence of assignments modifying the variables of that cycle transforms the variables [x.sub.1], ..., [x.sub.k] into [x.sub.2], ..., [x.sub.k], [x.sub.1] respectively. Let us consider the first variable modified by the program amongst those variables, say it is [x.sub.1]. Then it is necessarily modified by the assignment [x.sub.1] := [x.sub.2].
Let us consider the program [product]' formed by all assignments of the program [product] from the first one to the assignment [x.sub.1] := [x.sub.2], included. The variables which are modified by those assignments are [x.sub.1] and some variables [y.sub.1], ... [y.sub.i]. Let us consider every other variable from [product] as a constant for [product]'.
The vector ([y.sub.1], [y.sub.i], [x.sub.1]) can have at the beginning every possible value in [S.sup.i + 1]. Since the permutation of variables is a bijection, this program [product]' has to compute a bijection from [S.sup.i + 1] into [S.sup.i + 1]. But this is impossible since the image of the mapping computed by [product]' has a size bounded by [[absolute value of S].sup.i], because the assignment [x.sub.1] := [x.sub.2] ends the program [product]' where [x.sub.2] is a constant. Note that we use that S is finite.
3 Multistage interconnection networks
Among formalism and terminology variants in the network theory field, we will remain close to that of [14] and [16]. Also, we prefer to define a network as a directed graph, whose routing consists in choosing edges to define directed paths, rather than considering vertices as switches with several routing positions to choose. Those two formal options are obviously equivalent.
A multistage interconnection network, or MIN for short, is a directed graph whose set of vertices is a finite number of copies [S.sup.n.sub.1], [S.sup.n.sub.2], ..., [S.sup.n.sub.k] of [S.sup.n], called stages, and whose edges join elements of [S.sup.n.sub.i] towards some elements of [S.sup.n.sub.i + 1] for 1 [less than or equal to] i < k. Then routing a MIN is specifying one outgoing edge from each vertex of [S.sup.n.sub.i] for 1 [less than or equal to] i < k. A mapping E of [S.sup.n] is performed by a routing of a MIN if for each element X [member of] [S.sup.n.sub.1] there is a directed path using specified edges from X to E(X) [member of] [S.sup.n.sub.k]. The fact that a MIN performs a mapping E can be seen as the reverse of a multicast communication pattern where one input may lead to several outputs. So, a MIN is called rearrangeable non-blocking multicast if every mapping of [S.sup.n] can be performed by this MIN. The concatenation of two MINs M, M' is the MIN M|M' obtained by identifying the last stage of M and the first stage of M'.
The assignment network [A.sub.i] is the MIN with two stages whose edges join ([x.sub.1], ..., [x.sub.n]) to ([x.sub.1], ..., [x.sub.i--1], e, [x.sub.i + 1], ..., [x.sub.n]) for an arbitrary e [member of] S. Hence each vertex has degree s = [absolute value of S]. With notations of Definition 1, given an assignment ([[psi].sub.k], [i.sub.k]) in an in situ program, we naturally define a routing of [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] by specifying the edge between X = ([x.sub.1], ..., [x.sub.n]) and [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII].
The s-ary butterfly network, or simply butterfly, denoted [B.sub.s,n], or B for short, is the MIN [A.sub.n]| ... [absolute value of [A.sub.2]][A.sub.1]. Then [B.sup.-1] is the MIN [A.sub.1][absolute value of [A.sub.2]] ... |[A.sub.n]. The usual 2-ary butterfly, also called indirect binary cube, stands here as [B.sub.2, n]. The Benes network is the network obtained from [B.sup.-1]|B by replacing the two consecutive assignment networks [A.sub.n] by a single one. Note that this last reduction is not part of the usual definition, however it is more convenient here since two successive assignments on a same component can always be replaced with a single one. Note also that the historical definition of a Benes network [2] is not in terms of butterflies, but that ours is topologically equivalent thanks to classical results (see [1] and [3] for instance) implying that they are equivalent in terms of mappings performed.
The point is that the signature of an in situ program defines a MIN, and the set of assignments that realize a given mapping define the routing of the MIN by specifying some routing edges between stages. From the above definitions, an in situ program of signature n, ..., 1, or 1, ..., n, or 1, ..., n ..., 1 corresponds to a routing in B, or [B.sup.-1], or the Benes network, respectively. Figure 1 gives an example for the Benes network, with corresponding in situ program [f.sub.1], [f.sub.2], [f.sub.3], [g.sub.2], [g.sub.1] (where indices show the modified components). As explained above, routing this network is exactly specifying these mappings.
4 A formalization and survey of results for both viewpoints
In this section, we provide reformulations, variations, complements, or extensions for known network theory results, in terms of in situ programs. We point out that deriving these constructions from existing literature is not straightforward and is interesting on its own, notably because of various approaches and formalisms used. Useful references are recalled. The formalism and preliminary constructions introduced here will also serve as a base for the next section.
The classical property of the Benes network is that it is rearrangeable (see [2][14]), that is: for any permutation of [S.sup.n], there exists a routing performing the permutation (note that a routing performs a permutation when it defines edge-disjoint directed paths). Theorem 3 below reformulates this result. A short proof in terms of in situ programs and using graph colouration is given in [9], yielding a construction of the in situ program in DTIME(t.log(t)), where t = n.[2.sup.n] is the size of the table defining the mapping.
Theorem 3 Let E be a bijective mapping on [S.sup.n]. There exists an in situ program for E of length 2n--1 and signature 1 ... n... 1. Equivalently, [B.sup.-1]|B has a routing performing E.
From a routing for a bijection E, one immediately gets a routing for [E.sup.-1] by reversing the network. Hence, the mappings corresponding to the reversed arcs in the reserved network define an in situ program of [E.sup.-1]. In the boolean case, we even obtain more: one just has to use exactly the same assignments but in the reserved way, as stated in Corollary 4.
Corollary 4 If [product] is an in situ program of a bijection E on [{0,1}.sup.n], then the reversed sequence of assignments is an in situ program of the inverse bijection [E.sup.-1].
Proof: First, we show that operations in the program [product] are necessarily of the form
[x.sub.i] := [x.sub.i] + h ([x.sub.1], [x.sub.i--1], ..., [x.sub.n]).
One can assume without loss of generality that i = 1. Let [x.sub.1] := f([x.sub.1], ..., [x.sub.n]) be an operation of [product]. Denote
h([x.sub.2], ..., [x.sub.n]) = f(0, [x.sub.2], ..., [x.sub.n]).
We necessarily have f(1, [x.sub.2], ..., [x.sub.n]) = [x.sub.1] + h([x.sub.2], ..., [x.sub.n]). Otherwise two different vectors would map to the same image. This yields f([x.sub.1], ..., [x.sub.n]) = [x.sub.1] + h([x.sub.2], ..., [x.sub.n]). As a straightforward consequence, performing the operations in reverse order will compute the inverse bijection [E.sup.-1].
Now, in order to build a program for a general mapping E on [S.sup.n], for which different vectors may have same images, we will use a special kind of mappings on [S.sup.n], that can be computed with n assignments.
Definition 5 It is assumed that S = {0,1, ..., s--1}. Denote [[s.sup.n]] the interval of integers [0, ..., [s.sup.n]-- 1]. The index of a vector ([x.sub.1], [x.sub.2], ..., [x.sub.n]) is the integer [x.sub.1] + s.[x.sub.2] +... + [s.sup.n-- 1].[x.sub.n] of [[s.sup.n]]. For every i [member of] [[s.sup.n]], denote by [X.sub.i] the vector of index i. The distance of two vectors [X.sub.a], [X.sub.b] is the integer [DELTA]([X.sub.a], [X.sub.b]) = [absolute value of b--a].
A mapping I on [S.sup.n] is distance-compatible if for every x, y [member of] [S.sup.n], we have [DELTA](I(x), I(y)) [less than or equal to] [DELTA](x, y), which is equivalent to [DELTA](I([X.sub.a]), I([X.sub.a + 1])) [less than or equal to] 1 for every a with 0 [less than or equal to] a < [s.sup.n]--1.
Proposition 6 and Corollary 7 below provide an extension of a well-known property of the butterfly network in terms of in situ programs: it can be used to map the first k consecutive inputs onto any set of k outputs in the same order. In [18], this property is used in a similar way than ours, as recalled in [14] (see notably Theorem 4.3.7, where this network is shown to be a multicast infra-concentrator, and see also [16], Section 3.4.3., where this property is used to solve the packing routing problem, with a proof similar to ours).
Proposition 6 Every distance-compatible mapping I on [S.sup.n] is computed by an in situ program with signature 1, ..., n This program [p.sub.1], [p.sub.2], ..., [p.sub.n] satisfies, for I ([x.sub.1], ..., [x.sub.n]) = ([y.sub.1], ..., [y.sub.n]) and for each i = 1, 2, ..., n:
[p.sub.i]([y.sub.1], .... [y.sub.i--1],[x.sub.i], ..., [x.sub.n]) = [y.sub.i].
Proof: Since each component is modified exactly one time in a program with signature 1, ..., n, necessarily each function [p.sub.i] must give its correct final value to each component [x.sub.i]. It remains to prove that this unique possible method is correct, that is the mappings [p.sub.i] are well defined by the above formula, that is, for each [p.sub.i], a same vector cannot have two different images according to the definition. Note that the given definition is partial, but sufficient for computing the image of any x.
Assume that [p.sub.1], ...,[p.sub.i] are well defined. Assume that, after step i, two different vectors x, x' are given the same image by the process whereas their final expected images I(x) and I(x') were different. The components [x.sub.j], j > i, of x and x' have not been modified yet. Hence, they are equal and we deduce [DELTA](x,x') < [s.sup.i]. On the other hand, the components [y.sub.j], j [less than or equal to] i, of I(x) and I(x') are equal but I(x) [not equal to] I(x'). Hence [DELTA](I(x), I(x')) [greater than or equal to] [s.sup.i]: a contradiction. So [p.sub.i + 1] is also well defined by the given formula.
Corollary 7 Let I be a mapping on [S.sup.n] preserving the strict ordering of a set of consecutive vectors [X.sub.i], ..., [X.sub.j],for 0 [less than or equal to] i < j < [s.sup.n]--1. Then the restriction of I to the set of vectors{[X.sub.i], ..., [X.sub.j]} can be computed by an in situ program with signature n, ..., 1.
Proof: By assumption, the restriction of I to {[X.sub.i], ..., [X.sub.j]} is injective. Let [I.sup.-1] be a mapping of Sn whose restriction to I({[X.sub.i], ..., [X.sub.j]}) is the inverse of I, and completed so that [I.sup.-1] is distance-compatible. Applying Proposition 6 to [I.sup.-1] provides a sequence of assignments performing [I.sup.-1] with signature 1, ..., n. Since I is injective on {[X.sub.i], ..., [X.sub.j]}, those assignments can be reversed to provide a sequence of assignments computing the restriction of I to this set of vectors. Observe that this result can be seen more simply in terms of networks: the in situ program of the mapping [I.sup.-1] corresponds to a routing of the reversed butterfly, reversing this routing provides directly a routing of the butterfly performing the required restriction of I.
Example 8 For S = {0,1} and n = 3, consider the mapping I defined by I([X.sub.0]) = I([X.sub.1]) = [X.sub.0], I([X.sub.2]) = [X.sub.1], I([X.sub.3]) = I([X.sub.4]) = I([X.sub.5]) = [X.sub.2] and I([X.sub.6]) = I([X.sub.7]) = [X.sub.3], as shown on the following left tables. The mapping I is computed by the in situ program [p.sub.1], [p.sub.2], [p.sub.3] as defined in Proposition 6 and as illustrated in the following right tables. For consistency with Figure 1, and for better readability of the index of a vector, the vector ([x.sub.1], [x.sub.n]) is written in reversed order in columns of the tables: from [x.sub.3] to [x.sub.1].
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]
Definition 9 We call partition-sequence of [S.sup.n] a sequence
P = ([P.sub.0], [P.sub.1], ..., [P.sub.k])
of subsets of [S.sup.n], for some integer k [greater than or equal to] 0, such that the non-empty subsets in the sequence form a partition of [S.sup.n]. Then, we denote by [I.sub.P] the mapping on [S.sup.n] which maps [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] respectively to
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII].
Observe that [I.sub.P] is well defined since the sum of sizes of the subsets equals [s.sup.n], and that [I.sub.P] depends only on the sizes of the subsets and their ordering. Observe also that if no subset is empty, then [I.sub.P] is distance-compatible since, by construction, [DELTA](I([X.sub.a]), I([X.sub.a + 1])) [less than or equal to] 1 for every a.
Example 10 The mapping I from Example 8 equals [I.sub.[??]] for the partition-sequence [??] = ([[??].sub.0], [[??].sub.1], [[??].sub.2], [[??].sub.3]) of [{0,1}.sup.3] such that [[absolute value of [[??].sub.0]], [absolute value of [[??].sub.1]], [absolute value of [[??].sub.2]], [absolute value of [[??].sub.3]]] = [2,1,3,2].
Definition 11 Let E be a mapping on [S.sup.n], and P = ([P.sub.0], [P.sub.1], ..., [P.sub.k]) be a partition-sequence of [S.sup.n] whose underlying partition of [S.sup.n] is given by the inverse images of E, that is precisely: for every 0 [less than or equal to] i [less than or equal to] k, if [P.sub.i] [not equal to] 0 then there exists (a unique) [y.sub.i] [member of] [S.sup.n] such that [P.sub.i] = [E.sup.- 1]([y.sub.i]). Then, we call P-factorisation of E a triple of mappings (F, I, G) on [S.sup.n] such that:
* I is the mapping [I.sub.P];
* G is bijective and maps the set [P.sub.i] onto the set [I.sup.-1]([X.sub.i]), for every 0 [less than or equal to] i [less than or equal to] k (it is arbitrary within each set [P.sub.i]);
* F is bijective and maps [X.sub.i] to [y.sub.i], for every 0 [less than or equal to] i [less than or equal to] k such that [P.sub.i] [not equal to] 0 (it is arbitrary for other values [X.sub.i]).
By construction, we have
E = F [??] I [??] G.
Using this construction with no empty subset in the sequence P, we obtain Theorem 12 below, which significantly improves the result of [6] where boolean mappings on [{0,1}.sup.n] are computed in [n.sup.2] steps. This result is similar, in terms of in situ programs, to the result of [18] for boolean mappings, as presented in [14].
Theorem 12 For every finite set S, every mapping E on [S.sup.n] can be computed by an in situ program of signature 1 ... n ... 1 ... n ... 1 ... n and length 5n-4 the following way:
* Consider any P-factorisation (F, I, G) of E with no empty subset in the sequence P
* Use Theorem 3 to compute G (resp. F) by a program of signature 1 ... n ... 1 (resp. n ... 1 ... n).
* Use Proposition 6, to compute I by a program of signature 1...n.
* Reduce into one assignment the consecutive assignments operating on the same component.
In terms of MIN, we get a routing of [B.sup.-1][absolute value of B][B.sup.-1][absolute value of B][B.sup.-1] performing E, and a multicast routing of B[absolute value of [B.sup.-1]]B[absolute value of [B.sup.-1]]B performing [E.sup.-1].
Proof: Consider any P-factorisation (F, I, G) of E with no empty subset in the sequence P. Then the mapping I is distance compatible, as already observed. By Theorem 3, G (resp. F) can be computed by a program of signature 1 ... n ... 1 (resp. n ... 1 ... n). By Proposition 6, I is computed by a program of signature 1 ... n. By composition and by reducing two successive assignments of the same variable in one, E is computed by a sequence of 5n-4 assignments of signature 1 ...n... 1 ...n... 1 ...n.
Now we can refine Theorem 12 to get Theorem 13 below. This modification is similar to that noticed in [20] about [18], and is similar to the construction of [17] in terms of boolean baseline networks, yielding equivalent results. Observe that this refinement is an improvement in terms of number of assignments (number of routing edges in the MIN setting), but not in terms of flexibility, since the ordering of subsets in the sequence P of the P-factorisation of E has now to be the same than the ordering of the outputs of E.
Theorem 13 For every finite set S, every mapping E on Sn can be computed by an in situ program of signature 1 ... n ... 1 ... n ... 1 and length 4n-3 the following way:
* Consider a P-factorisation (F, I, G) of E with no empty subset in the sequence P = ([P.sub.0], ..., [P.sub.k]) and such that [P.sub.i] = [E.sup.-1]([y.sub.i]) where [y.sub.0], ...,[y.sub.k] are the images of E in increasing ordering
* Use Theorem 3 to compute G by a program of signature 1... n... 1.
* Use Proposition 6, to compute I by a program ofsignature 1 ... n.
* Use Corollary 7, to compute the restriction of F to the image of I * G by a program of signature n ... 1.
* Reduce into one assignment the consecutive assignments operating on the same component.
In terms of MIN, we get a routing of [B.sup.-1]|B|[B.sup.-1]B performing E, and a multicast routing of B|[B.sup.- 1]B|[B.sup.-1] performing [E.sub.-1].
Proof: The proof is similar to that of Theorem 12 except for the bijection F. Here, by the choice of P, the restriction of F to the image of I o G maps consecutive vectors onto a set of vectors preserving the ordering. Hence it satisfies the hypothesis of Corollary 7.
Example 14 Figure 2 gives an example for the construction of Theorem 13. The elements of [{0,1}.sup.3] are grouped by the bijection G at stage 6, in the same ordering than the images of E. Then, at stage 9 all elements with same final image have been given a same image by I, again in the same ordering than the images of E. Hence, at last, the restriction of bijection F can finalize the mapping E in 3 stages only.
Remark 15 To end this section, let us notice that, due to the fact that successive assignments operate on consecutive components, successive assignments of type [S.sup.mn] [right arrow] S can be grouped in assignments of fewer variables on a larger base set [S.sup.m] defining successive mappings [S.sup.mn] [right arrow] [S.sup.m]:
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]
Hence, for instance, the case S = [{0,1}.sup.m] can be reduced to the case S = {0,1}. This is a particular case of the register integrability property described in [8]: the signatures of type 1, ..., n, ..., 1, ... allow to adapt in situ programs to memory registers with non-constant sizes.
Open problem 1 Up to our knowledge, no better bound than 4n--3 is known for the case of arbitrary mappings. It would be interesting to improve this bound, and to find the best possible general bound. Experiments on a computer make us think that the factor 4 could be replaced with a factor 3. Note that the in situ program may not have a signature with consecutive indices, or, in other words, that other combinations of assignment networks than the butterfly network may be used.
5 A more flexible new method
The more involved method given here is a refinement of the method given by Theorem 12. It is completely new with respect to network literature, and had been presented in the preliminary conference paper [9]. It provides the same number of assignments than Theorem 13 and a better flexibility.
We still use a P-factorisation (F, I, G) but the sequence P will possibly contain empty sets, and will be suitably ordered with respect to the sizes of its elements, in order to satisfy some boolean arithmetic properties. So doing, the intermediate mapping I = [I.sub.P] will have a so-called suffix-compatibility property. We show that the composition of a mapping having this property with any in situ program with signature 1, ..., n can be also computed in n steps. Hence the composition of I with the first n steps of the in situ program of the bijection F can also be computed with n assignments, performing the computation of F o I in 2n-1 steps instead of 3n-2.
Each intermediate result in this section provides a new result on in situ programs with signature 1, ..., n, or equivalently on routing properties of the butterfly network. In terms of multicast routing strategy, the flexibility of this method comes from the freedom one has in building a suitable ordering for the sequence P, as detailed in Remark 19.
In the whole section, we will fix S = {0,1}. The method is given here when S = {0,1}, hence it is directly available, by extension, when S = [{0,1}.sup.m] (cf. Remark 15). However it can be extended to a set S of aribtrary size. While this paper was under publication process, it has been noticed in [13] that Definition 16 and Lemma 17 below could be formulated using any integer q = [absolute value of S] instead of 2, by means of an arithmetical property. Then the rest of the construction can be adapted directly.
Definition 16 A block-sequence [[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]] is a sequence of [2.sup.n] non- negative integers such that, for every i = 0 ... n, the sum of values in each of the consecutive blocks of size [2.sup.i] is a multiple of [2.sup.i], that is, for all 0 [less than or equal to] j < [2.sup.n-i]:
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]
Lemma 17 Every sequence of [2.sup.n] non-negative integers whose sum equals [2.sup.n] can be reordered in a block-sequence.
Proof: The ordering is built inductively. Begin at level i = 0 with [2.sup.n] blocks of size 1 having each value in the sequence. At level i + 1, form consecutive pairs of blocks [B, B'] that have values v, v' of same parity and define the value of this new block to be (v + v')/2. Each new level doubles the size of blocks and divides their number by 2. The construction is valid since the sum of values of blocks at level i is [2.sup.n-i].
Example 18 We illustrate below the process described in the proof of Lemma 17 (n = 4 and each block has its value as an exponent):
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]
Remark 19 Let us anticipate the sequel of the detailed construction and already explain roughly in what respect the use of block-sequences will allow a more flexible routing strategy than the classical construction recalled in Section 4. As shown before, this construction consists in grouping and sorting pre-images of the mapping at some middle stage of the in situ program. The ordering of these pre-images is determined and thus the sequence of assignments is completely constrained at this middle stage. There is essentially one available sorting to get the 4n--3 length (up to a few possible shifts along the ordering).
In the more involved construction given in the present section, we will apply Lemma 17 to the sequence of integers given by the cardinalities of the pre-images of the mapping to compute. We will prove later that any ordering of the pre-images whose cardinalities satisfy the block-sequence property can be used at this middle stage. The point is that, given a sequence of integers, there is a number of such possible orderings given by block-sequences, built as in the proof of Lemma 17, and also a number of possible associations between the pre-images and those cardinality integers.
For instance, a given a block-sequence ordering can be represented as a bracket system, forming a binary tree, the following way, continuing Example 18:
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]
Then, one can always permute any sons of a node in the tree and still get a block-sequence. So, from one block-sequence, one can obtain potentially a number of available block-sequences (there are [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] permutations of the leaves obtained by this way for such a tree with 2n leaves). For instance, making such permutations in Example 18 leads to the following possible block-sequences (brackets have been added to identify blocks that have been permuted):
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII].
Of course, various possible block-sequences may be obtained independently from such permutations. Moreover, given a block-sequence, pre-images having same cardinality may be associated with any occurrence of the corresponding integer in the sequence, leading to a number of possibilities for building the in situ program. For instance, if every pre-image has size 2, then the sequence of integers to consider is [2,2, ..., 2,2,0,0, ..., 0,0], which is already a block-sequence and is invariant under permutation of the non-zero integers. In this case, any ordering of the pre-images can be used at the middle stage to provide finally a 4n--3 length program.
All those "any" in this new construction, compared with the "one" in the known construction, witness how the method is more flexible.
Definition 20 For a vector ([x.sub.1], ..., [x.sub.n]), we call prefix of order k, resp. suffix of order k, the vector ([x.sub.i], ..., [x.sub.k]),resp. ([x.sub.k], ..., [x.sub.n])."
A mapping I of [{0,1}.sup.n] is called suffix-compatible if, for every 1 [less than or equal to] k [less than or equal to] n, if two vectors X, X' have same suffixes of order k, then their images I(X), I(X') also have same suffixes of order k.
Lemma 21 Let [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] be a partition-sequence of [{0,1}.sup.n] such that [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] is a block-sequence. Then the mapping [I.sub.P] on [{0,1}.sup.n] is suffix-compatible.
Proof: The sketch of the proof is the following. First, define the j-th block of level i of [{0,1}.sup.n] as the set of vectors whose part has index j[2.sup.i] [less than or equal to] l < (j + 1)[2.sup.i]. Observe that the inverse image by [I.sub.P] of a block is a union of consecutive blocks of same level. The result follows.
Let us now detail the proof. For 0 [less than or equal to] i [less than or equal to] n and j [member of] [[2.sup.n-i]], define the j-th block at level i of [{0,1}.sup.n] as
[V.sub.i,j] = {[X.sub.l] : l [member of] [j[2.sup.i] (j + 1)[2.sup.i]--1]}.
(i) First, we prove that, for every i, j as above, there exists k, k' [member of] [[2.sup.n-i]], such that
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]
Let us call interval of [{0,1}.sup.n] the set of vectors [X.sub.l] for l belonging to an interval of [[2.sup.n]]. First, notice that the inverse image by [I.sub.P] of an interval of [{0,1}.sup.n] is an interval of [{0,1}.sup.n]. By definition of [I.sub.P], we have [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. Remark that [I.sup.-1.sub.P]([V.sub.i,j]) may be empty, when [v.sub.l] = 0 for all l [member of] [j[2.sup.i], (j + 1)[2.sup.i]]. Since [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] is a block sequence, we have [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] mod [2.sup.i]. Hence, [absolute value of ([I.sup.-1.sub.P]([V.sub.i,j]))] = 0 mod [2.sup.i].
For a fixed i, we prove the result by induction on j. If j = 0 then [absolute value of ([I.sup.-1.sub.P]([V.sub.i,0]))] = k.[2.sup.i] for some k [member of] [[2.sup.n-i]]. If [I.sup.-1.sub.P]([V.sub.i,0]) is not empty, then it is an interval of [{0,1}.sup.n] containing (0, ... 0) by definition of [I.sub.P]. Since this interval has a size k.[2.sup.i] multiple of [2.sup.i], it is of the form [[union].sub.0 [less than or equal to] l [less than or equal to] k] [V.sub.i,l];.
If the property is true for all l with 0 [less than or equal to] l < j, then [I.sup.-1.sub.P]([[union].sub.0 [less than or equal to] l [less than or equal to] j] [V.sub.i,l]) = [[union].sub.0 [less than or equal to] l [less than or equal to] j'] [V.sub.i,l]. Since [absolute value of ([I.sup.-1.sub.P]([V.sub.i,j]))] = k.[2.sup.i] for some k [member of] [[2.sup.n-i]], we must have [I.sup.-1.sub.P]([[union].sub.0 [less than or equal to] l [less than or equal to] k] [V.sub.i,l]) = [[union].sub.0 [less than or equal to] l [less than or equal to] j'] [V.sub.i,l], hence [I.sup.-1.sub.P]([V.sub.i,j]) = [[union].sub.j'< l [less than or equal to] j' + '] [V.sub.i,l].
(ii) Now, we prove the lemma. Assume a = ([a.sub.1], [a.sub.n]) and b = ([b.sub.1], ... [b.sub.n]) have same suffix of order i. For all l [greater than or equal to] i we have a; = b;. Let c [member of] [S.sup.n] be defined by [c.sub.n] = [a.sub.n] = [b.sub.n], ...,[c.sub.i] = [a.sub.i] = [b.sub.i], [c.sub.k-1] = 0, ..., [c.sub.1] =0. Let [phi](x) denote the index of vector x. We have [phi](c) = 0 mod [2.sup.i-1], that is [phi] (c) = j.[2.sup.i-1] for some j [member of] [[2.sup.n--i + 1]]. And [phi](a) and [phi](b) belong to the same interval [j.[2.sup.i--1], (j + 1).[2.sup.i--1]--1] whose elements have same components for l [greater than or equal to] i. That is a and b belong to [V.sub.i--1, j]. By (i), the inverse images of intervals of type [V.sub.i--1, k] by [I.sub.P] are unions of such consecutive intervals. Hence the image of an interval [V.sub.i--1, j] by [I.sub.p] is an interval contained in an interval [V.sub.i--1, k] for some k [member of] [[2.sup.n- i+1]]. Hence [I.sub.P](a) and [I.sub.P](b) have same components l [greater than or equal to] i.
Example 22 First consider again the mapping [I.sub.[??]] from Examples 8 and 10 obtained from the partition-sequence [??] = ([[??].sub.0], [[??].sub.1], [[??].sub.2], [[??].sub.s]) of [{0,1}.sup.3] such that [absolute value of [[??].sub.0]], [absolute value of [[??].sub.1]], [absolute value of [[??].sub.2]], [absolute value of [[??].sub.3]] = [2,1, 3,2], which is not a block-sequence. Observe that this mapping is not suffix-compatible, since ([x.sub.1],[x.sub.2],[x.sub.s]) = (0,1,0) and ([x'.sub.1],[x'.sub.2], [x'.sub.3]) = (1,1,0) have same suffix of order 2 equal to (1,0), but I([x.sub.1],[x.sub.2],[x.sub.3]) = (1,0,0) and I([x'.sub.1], [x'.sub.2], [x'.sub.3]) = (0,1,0) have not same suffix of order 2.
Now consider the mapping I shown on the next tables, obtained from the partition-sequence P = ([P.sub.0], [P.sub.1], [P.sub.2], [P.sub.s]) such that [[absolute value of [P.sub.0]], [absolute value of [P.sub.1]], [absolute value of [P.sub.2]], [absolute value of [P.sub.3]]] = [1,3,2,2], which is a block-sequence. Then one can check that [I.sub.P] is suffix-compatible, as claimed by Lemma 21.
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]
Proposition 12 Let I be a suffix-compatible mapping on [{0; 1}.sup.n] and let B be a mapping on [{0; 1}.sup.n] computed by an in situ program [b.sub.1], ..., [b.sub.n]. The mapping B * I is computed by an in situ program with signature 1, ..., n, namely [p.sub.1]; [p.sub.2], ..., [p.sub.n], with, for B * I([x.sub.1], ..., [x.sub.n]) = ([y.sub.1], ..., [y.sub.n]):
[p.sub.i]([y.sub.1], ..., [y.sub.i--1], [x.sub.i], ..., [x.sub.n]) = [y.sub.i]
Observe that Proposition 23 provides anew property of the butterfly network, which refines the classical sorting property of this network recalled in Proposition 6. We mention also that Proposition 23 is stated and proved for a general mapping B, but we will use it in what follows only when B is bijective.
Proof: Just as for Proposition 6, assume that [p.sub.1], ..., [p.sub.i] are well defined by the necessary above formula, and that, after step i, two different vectors x, x' are given the same image by the process whereas their final expected images y = B * I(x) and y' = B * I(x') were different (hence I(x) [not equal to] I(x')). By construction, y, y' have a same prefix P of order i and x, x' have a same suffix Q of order i + 1. Moreover, since I is suffix-compatible, the vectors I(x), I(x') also have a same suffix R of order i + 1. Hence one has some relation F(I(x)) = F(uR) = y = Pv and F(I(x')) = F(u'R) = y' = Pv', where u,u' are some prefix, and v, v' are some suffix. Let [F.sub.i] be the mapping defined by the first i assignments defining F, that is dealing with components 1, ..., i. Since F(uR) = Pv, we have [F.sub.i](uR) = PR. And since F(u'R) = Pv' we have [F.sub.i](u'R) = PR. Hence [F.sub.i](uR) = [F.sub.i](u'R). Since F is computed by computing [F.sub.i] first, we get that F(uR) = F(u'R), that is y = y', a contradiction with our assumption.
Example 24 Consider the bijective mapping B computed by the program [b.sub.1], [b.sub.2], [b.sub.3] shown on the next left tables (each step modifies one column). And consider the suffix-compatible mapping [I.sub.P] from Example 22 above. Then, by Proposition 23, the composition B * [I.sub.P] is computed by the program [p.sub.1],[p.sub.2],[p.sub.3] as shown on the next right tables. One can check that, at each step, two vectors which have same image through 1 [less than or equal to] k [less than or equal to] 3 assignments will have eventually same images.
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]
Now, given a mapping E of [S.sup.n], using a P-factorisation of E for a sequence P whose sequence of cardinalities is a block-sequence, we can improve the result of Section 3 in terms of flexibility. Indeed, the only constraint is now to have that the sets of vectors having a same final image are grouped after the first bijection according to any block-sequence representing the sequence of cardinalities of those sets. And there is a number of such possible block-sequences (cf. construction of Lemma 17).
Theorem 25 Every mapping E on [{0,1}.sup.n] is computed by an in situ program of length 4n--3 and signature 1 ... n ... 1 ... n ... 1 the following way:
* Consider a P -factorisation (F, I, G) of E with [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] is a block-sequence (built by Lemma 17)
* Use Theorem 3 to compute G and F by programs with signature 1...n... 1.
* Call B the in situ program formed by the n first assignments of the program of F, and use Proposition 23 to compute B o I by a program with signature 1, ..., n.
* Reduce into one assignment the consecutive assignments operating on the same component.
In terms of MIN, we get a routing of [B.sup.-1]|B|[B.sup.-1]|B performing E, and a multicast routing of B|[B.sup.- 1]|B|[B.sup.-1] performing [E.sup.-1].
Proof: Let (F, I, G) be a P-factorisation of E for a sequence [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] is a block-sequence (it exists thanks to Lemma 17). By Theorem 3, G (resp. F) can be computed by a program of signature 1 ... n ... 1 (resp. 1 ... n ... 1). By Lemma 21, the mapping I = [I.sub.P] on [{0,1}.sup.n] is suffix-compatible. Call B the mapping computed by the n first assignments [b.sub.1], ...,[b.sub.n] of the program of F. By Proposition 23, B * I is also computed by a program of signature 1...n. Then, by composition and by reducing two successive assignments of the same variable in one, the mapping E = F * I * G is computed by a sequence of 4n--3 assignments of signature 1 ... n ... 1 ... n ... 1.
Example 26 Figure 3 gives an example for the construction of this section (on the same mapping as in Figure 2). The elements of [{0,1}.sup.3] are grouped by the bijection G at stage 6, accordingly with the blocksequence [[1, 3], [2,2]] induced by [E.sup.-1]. Then, at stage 9 all elements with same final image have been given a same image by B o I (which is the mapping detailed in Example 24), where B is the first part of the bijection F. At last, the second part of the bijection F allows to finalize the mapping E. Observe that we could have chosen other block-sequences, such as [[3,1], [2, 2]] or [[2, 2], [1,3]] for instance, and we would have obtained other in situ programs and routing patterns for the same mapping (see Remark 19).
Open problem 2 A first natural question was to extend the notion of block-sequence and Lemma 17 to a set S of arbitrary size, in such a way that the rest of the construction remains valid. An efficient answer has been given recently in [13], as mentioned in the introduction of this Section. Another general and probably demanding question is to study the reach of the flexibility provided by the present construction: either to get general in situ programs of shorter length, as requested by Open problem 1; or, in terms of networks, to get efficient (wide-sense) non-blocking routing methods, i.e. methods to update dynamically the routing with respect to updates of the computed mapping. Finding such dynamic non-blocking routing strategies is a main concern of the network field (see [14] for a general background).
6 Linear mappings on suitable ring powers
In this section, we assume that S is given with an algebraic structure: S is a (non-necessarily finite) quotient of an Euclidean domain R by an ideal I. Classical examples for S are: any field (the result of this section for S being a field is easier, since most technicalities can be skipped), the rings Z/sZ, or K[x]/(P) for some polynomial P with coefficients in a field K.
Given S and an integer n, we consider a linear mapping [S.sup.n] [right arrow] [S.sup.n], that is an application from [S.sup.n] to [S.sup.n] which is a linear application with respect to the canonical structure of S-module of [S.sup.n]. The results of Section 3 show that O(n) assignments are sufficient to compute such a mapping. Here, we achieve a stronger result: the number of required mappings is bounded by 2n--1, and all intermediary assignments are linear.
In [7], a similar result is obtained in the particular case of linear boolean mappings. The paper [19] achieves this result with an on-the-fly self routing strategy, again restricted to the linear boolean case. We note that the more general result we obtain here is not of this efficient nature, since the actual computation of the decomposition we obtain has a complexity which is of the order of O([n.sup.3]). We insist however on the fact that the in situ decomposition we provide can be proven to be of length at most 2n--1, and is available for much more general useful algebraic structures.
Also, finding an in situ program of a linear mapping using linear assignments is equivalent to rewriting a matrix as a product of assignment matrices (matrices equal to the identity matrix except on one row). Theorem 29 below is proven using this alternate formalism.
We denote S* the set of invertible elements of S. For convenience we also define the Kronecker symbol [[delta].sup.j.sub.i], which is defined for two integers i and j as being 1 for i = j, and 0 otherwise.
Lemma 27 Let [x.sub.1], ...,[x.sub.n] be coprime elements of R. Let [i.sub.0] [member of] [1... n]. There exists multipliers [[lambda].sub.1], ..., [[lambda].sub.n] such that [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], and [[summation].sub.i] [[lambda].sub.i] [x.sub.i] [member of] S*.
Proof: By assumption, the index [i.sub.0] is fixed. Without loss of generality we may safely assume that [i.sub.0] = 1. In virtue of the Chinese Remainder Theorem, it suffices to define the multipliers [[lambda].sub.1], ... ,[[lambda].sub.n] separately modulo each prime power [p.sup.v] dividing I, thus it is also valid to restrict to the case where the ideal I is generated by a prime power [p.sup.v].
Now we distinguish two cases. In the first case, x1 is coprime to p, and thus coprime to I. We thus choose [[lambda].sub.1] = 1, and [[lambda].sub.i] = 0 for all i > 1. Then [[lambda].sub.1][x.sub.1] = [x.sub.1] is in (R/I)*. In the second case, [x.sub.1] is divisible by p. Since by assumption, the [x.sub.i]'s are coprime, there exists an integer [i.sub.1] such that [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] is coprime to p. Therefore [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] is coprime to p, hence we may set [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], and [[lambda].sub.i] = 0 for all other indices i (in other words, we may write [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]). We thus have [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], which is coprime to p, whence in (R/I)* as well.
Corollary 28 Let [x.sub.1], ..., [x.sub.n] be elements of R, and g = gcd([x.sub.1], ..., xn). Let [i.sub.0] [member of] [1 ... n]. There exists multipliers [[lambda].sub.1], ...,[[lambda].sub.n] such that [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], and [[summation].sub.i] [[lambda].sub.i][x.sub.i] [member of] gS*.
Proof: This is a trivial application of Lemma 27 to ([x.sub.1] /g, ..., [x.sub.n]/g).
Theorem 29 Every linear mapping E on [S.sup.n] is computed by an in situ program of length 2n--1 and signature 1,2,n,n--1,1 made of linear assignments.
Furthermore, if E is bijective, then the inverse mapping [E.sup.-1] is computed by the in situ program defined by the same sequence of assignments in reversed order together with the following transformation:
[[x.sub.i] := a x [x.sub.i] + f ([x.sub.1], ..., [x.sub.i--1], [x.sub.i + 1], ..., [x.sub.n])] [right arrow] [[x.sub.i] := [a.sup.-1].([x.sub.i]--f ([x.sub.1], ..., [x.sub.i--1], [x.sub.i + 1], ..., [x.sub.n]))].
Proof: The proof proceeds by induction. Let k be an integer, and let M be a matrix representing a linear mapping E on [S.sup.n] which leaves the first k--1 variables unchanged. In other words, the first k--1 rows of M equal those of the identity matrix. The matrix which defines the input linear mapping E satisfies this property with k = 1, thus our induction initiates at k = 1 with the matrix defining E. We explore the possibility of rewriting M as a product [L.sub.k] M'[R.sub.k], where the first k rows of M' match those of the identity matrix.
Let g be the greatest common divisor of (arbitrary representatives in R of) the coefficients of column k in M. A favourable situation is when mkkk is in gS*. Should this not be the case, let us see how we can transform the matrix to reach this situation unconditionally. Assume then for a moment that [m.sub.k, k] [not member of] gS*. Lemma 27 gives multipliers [[lambda].sub.1], ..., [[lambda].sub.n] such that [[summation].sub.l][[lambda].sub.l] [m.sub.l ,k] [member of] gS*, with the additional constraint that [[lambda].sub.k] = 1. Let us now denote by T the n x n matrix which differs from the identity matrix only at row k, and whose coefficients in row number k are defined by [t.sub.k, j] = [[lambda].sub.j]. Clearly T is an invertible assignment matrix, and the product TM has a coefficient at position (k, k) which is in gS*.
Now assume [m.sub.k,k] G gS*. Let G be the diagonal matrix having Gkkk = g as the only diagonal entry not equal to 1. Let M" = [MG.sup.-1] (M" has coefficients in R because g is the g.c.d. of column k). We have [m.sub.".sub.k, k] [member of] S*. We form an assignment matrix U which differs from the identity matrix only at row k, and whose coefficients in row number k are exactly the coefficients of the k-th row of M". The matrix U is then an invertible assignment matrix (its determinant is [m".sub.k, k]). The k first rows of the matrix M' = M"[U.sup.-1] match the k first rows of In, and we have M = [T.sup.-1] x M' x (UG). Our goal is therefore reached with [L.sub.k] = [T.sup.-1] and [R.sub.k] = UG.
Repeating the procedure, our input matrix is rewritten as a product
[L.sub.1][L.sub.2] ... [L.sub.n--1][R.sub.n] ... [R.sub.1],
where all matrices are assignment matrices. No left multiplier [L.sub.n] is needed for the last step, since the g.c.d. of one single element is equal to the element itself. Finally, the determinant of M is invertible if and only if all the matrices [R.sub.k] are invertible, hence the reversibility for bijective mappings.
Corollary 30 Every square matrix of size n on S (quotient of an Euclidean domain by an ideal) is the product of 2n--1 assignment matrices (equal to the identity matrix except on one row).
Remark 31 We digress briefly on the computational complexity of building the in situ programs for the linear mappings considered here. The matrix operations performed here all have complexity O([n.sup.2]) because of the special shape of the assignment matrices. Therefore, the overall computational complexity of the decomposition is O([n.sup.3]).
Example 32 The procedure in the proof of Theorem 29 can be illustrated by a small example. Assume we want to decompose the mapping in Z/12Z given by the matrix
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]
Let M be this matrix. The g.c.d of column 1 in M is gcd(4, 6) = 2, which is not invertible modulo 12. We therefore firstly use Corollary 28 to make gcd(4, 6) = 2 appear in the top left coefficient. We use the relation
1 * (4/2) + 1 * (6/2) = 5 [member of] (Z/12Z)*, 1 * 4 + 1 * 6 = 10 [member of] 2(Z/12Z)*.
We take therefore [[lambda].sub.1] = [[lambda].sub.2] = 1 and the matrix [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]
The common divisor 2 of column 1 can then be set aside. Let [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. We have [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] reproduce the first row of M". We have [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], exploiting the fact that because 5 is invertible modulo 12, U is an invertible matrix modulo 12. This eventually unfolds as the following factorization of M:
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]
This corresponds to the following sequence of assignments:
[x.sub.1] : = 10[x.sub.1] + 9[x.sub.2]; [x.sub.2] : = 3[x.sub.1] + [x.sub.2]; [x.sub.1] : = [x.sub.1]--[x.sub.2].
Open problem 3 As mentioned in the introduction, natural subsequent questions are the following. Let S = {0,1} be the binary field, and E : [S.sup.n] [right arrow] [S.sup.n] be a mapping the components of which are polynomials of degree at most k.
1. Does there always exist an in situ program of E the assignments of which are polynomials of degree at most k?
2. If so, what is the maximum (over all such E) of the minimum (over all such in situ programs of E) number of assignments in the program?
The following example tested on a computer shows that, in contrast to the linear case or the boolean bijective case developed earlier in the paper, this bound has to be strictly larger than 2n--1. For the mapping
E : ([x.sub.1], [x.sub.2], [x.sub.3]) [right arrow] ([x.sub.2][x.sub.3], [x.sub.1][x.sub.3], [x.sub.1][x.sub.2]),
the following in situ program of E has the smallest possible number of assignments using degree-2 polynomials:
[x.sub.1] := [x.sub.2] + [x.sub.2] [x.sub.3] + [x.sub.1]; [x.sub.2] := [x.sub.3] + [x.sub.1] + [x.sub.2]; [x.sub.3] := [x.sub.3] + [x.sub.2] + [x.sub.1][x.sub.2]; [x.sub.1] := [x.sub.3] + [x.sub.2] [x.sub.3] + [x.sub.1][x.sub.3]; [x.sub.2] := [x.sub.3] + [x.sub.2][x.sub.3] + [x.sub.1][x.sub.3]; [x.sub.3] := [x.sub.3] + [x.sub.2][x.sub.3] + [x.sub.1][x.sub.3].
Conclusion To conclude, further work can consist in applications: for instance the context of computations modulo e.g. Z/2[sup.64]Z is close to the concern of integer arithmetic with machine words. Theorem 29 shows that we can obtain a short sequence for computing linear mappings on such data. About the in situ approach of this computation, a question may arise, though, as to whether the constants appearing in the computation defeat the claim that the computation avoids temporaries. In fact, such constants can be directly embodied in the code, so that they contribute exclusively to the code size and not to its requested variable data size (or number of registers). Further work can also consist in improving bounds: for instance, it has been claimed very recently in [10] that the tight bound is [3.n/2j] linear assignments to compute linear mappings when S is the field Z/qZ for a prime power q (see also Open problem 1). Further work can also consist in dynamic routing for a network approcah (see [14] and Open problem 2), or in algebraic generalizations (such as proposed by Open problem 3)...
Acknowledgments
We are grateful to Jean-Paul Delahaye for presenting our in situ approach of computation in [12] and giving to us the idea of Proposition 2. Also, we are grateful to several anonymous referees for useful comments, suggestions and references, that helped notably searching the network literature.
References
[1] Agrawal, D.P.: Graph theoretical analysis and design of multistage interconnection networks. IEEE Trans. Computers C32 (1983), 637-648.
[2] Benes, V.E.: Optimal Rearrangeable Multistage Connecting Networks. Bell System Technical J. 43 (1964), 1641-1656.
[3] Bermond J.C., Fourneau J.M. and Jean-Marie A.: A graph theoretical approach to equivalence of multi-stage interconnection networks. Discrete Applied Maths 22 (1988), 201-214.
[4] Burckel, S.: Closed Iterative Calculus. Theoretical Computer Science 158 (1996), 371-378.
[5] Burckel, S., Morillon, M.: Three Generators for Minimal Writing-Space Computations. Theoretical Informatics and Applications 34 (2000) 131-138
[6] Burckel, S., Morillon, M.: Quadratic Sequential Computations. Theory of Computing Systems 37(4) (2004), 519-525
[7] Burckel, S., Morillon, M.: Sequential Computation of Linear Boolean Mappings. Theoretical Computer Science serie A 314 (2004), 287-292
[8] Burckel, S., Gioan, E.: In situ design of register operations. Proceedings of IEEE Computer Society Annual Symposium on Very Large Scale Integration ISVLSI 08 (2008), 4p.
[9] Burckel S., Gioan E., Thome E.: Mapping Computation with No Memory. Proc. 8th International Conference on Unconventional Computation UC09. LNCS 5715 (2009) 85-97.
[10] Cameron, P., Fairbairn, B., Gadouleau, M.: Computing in matrix groups without memory. ArXiv: 1310.6009 (2013).
[11] Cooley, J. W., Tukey, J. W.: An algorithm for the machine calculation of complex Fourier series. Math. Comput. 19 (1965), 297-301
[12] Delahaye, J.-P.: Le calculateur amnesique. Pour La Science 404 (2011), 88-92. Available at: http://www2.lifl.fr/~delahaye/pls/2011/208
[13] Gadouleau, M., Riis, S.: Memoryless computation: new results, constructions, and extensions. ArXiv: 1111.6026(2011).
[14] Hwang, F.K.: The mathematical theory of nonblocking switching networks. Series on Applied Mathematics vol. 11. World Scientific Publishing
Co., River Edge, NJ, USA, 2004 (2nd ed.).
[15] Lea, C.-T.: Multi-Log2N Networks and Their Applications in High-speed Electronic and Photonic Switching Systems IEEE Transactions on Communications 38(10) (1990) 1740-1749
[16] Leighton, F.T.: Introduction to Parallel Algorithms and Architectures: Arrays, Trees, Hypercubes. Morgan Kaufmann Publishers, San Francisco, CA, USA, 1992
[17] Lin, W., Chen, W.-S. E.: Efficient nonblocking multicast communications on baseline networks. Computer Communications 22 (1999), 556-567
[18] Ofman, Y. P.: A universal automaton. Trans. Moscow Math. Soc. 14 (1965), 200-215.
[19] Raghavendra, C. S. and Boppana, R. V. On self-routing in Benes and s shuffle-exchange networks. IEEE Trans. on Computers, 40(9):1057-1064, 1991.
[20] Thompson, C. D.: Generalized connection network for parallel processor interconnection. IEEE Trans. Comput. 27 (1978), 1119-1124
Serge Burckel (1) Emeric Gioan (2) Emmanuel Thom'e (3)
(1) ERMIT, Universit'e de La R'eunion, France
(2) LIRMM, CNRS, France
(3) LORIA, INRIA, France
Corresponding author. Emeric.Gioan@lirmm.fr
received 23rd Aug. 2013, revised 20th Feb. 2014, accepted 20th Feb. 2014.
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# How do we measure things in science?
## Presentation on theme: "How do we measure things in science?"— Presentation transcript:
How do we measure things in science?
The Metric System How do we measure things in science?
Length (measures the distance from one point to another)
Base Unit Meter Tool Meter stick/ruler
Meter sticks
This lizard is 3 meters long!
(a meter is about 39 inches)
Volume (measures the amount of space an object takes up )
Base Unit Liter (liquids) Cubic centimeters (solids) Tool Graduated Cylinder (liquids) Ruler (solids)
Graduated Cylinders Measure the volume of liquids
Place the graduated cylinder on level surface Read at eye level Read from the bottom of the meniscus
Measuring volume of a solid
This solid is three cubic meters in volume. L x W x H = volume
Formula for finding the volume of a rectangular solid
Length x Width x Height L x W x H
Finding the volume of an irregular object
1. Place enough water in a graduated cylinder to cover the object. Record the volume. 2. Add the object and record the new volume. 3. Find the difference between the two volumes and that is the volume of the irregular object.
Mass (measures the amount of matter in an object)
Base Unit gram Tool Electric balance
Electronic balance used to measure mass
Steps before using a balance
Clean the weigh pan Tare the balance (zero it out)
This stuff is supposed to give you more muscle mass
Temperature (measures heat and chill)
Base Unit Degrees celcius Tool Thermometer
Both Celsius and Farenheit scales are seen here
Both Celsius and Farenheit scales are seen here. In science we do not use degrees Farenheit.
This man is in a freezing cold environment as can be seen by his frozen hair and beard. He is sitting in a natural hot spring that is very warm at the same time!
Density How closely the molecules of an object are packed together.
Density can be calculated if the mass and volume are known Density = mass divided by volume
Measuring density using an electric balance for mass and a graduated cylinder for volume.
Less dense liquids float on more dense liquids in this density column
Weight of an object The mass of an object multiplied by the force of gravity. Weight changes if gravity changes. (like on the moon) Mass does not change on the moon.
Your weight will be less on the moon, but your mass does not change
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# Strange results by integrating Abs[Sin[a - t]]
I have a (simple) integral with a unknown parameter a: $$\int_0^{2\pi}|\sin(a-t)|dt$$ One can observe that it is just two times the area under sin curve from $0$ to $\pi$ and the result is $4$, Integrate agrees with it when given exact values of a.
However, when I try to evaluate this integral directly, Integrate[Abs@Sin[a - t], {t, 0, 2 Pi}] won't finish running. When I try to give it assumptions, interesting results appeared:
Table[Integrate[Abs@Sin[a - t], {t, 0, 2 Pi}, Assumptions -> a > i], {i, -8, 8, 2}]
Table[Integrate[Abs@Sin[a - t], {t, 0, 2 Pi}, Assumptions -> a < i], {i, -8, 8, 2}]
It just gives $0$, and generates the condition with the nearest feasible $k\pi,k\in \mathbb Z$.
I am using version 11.1 on OSX, is there any problem with my usage? How can I get the result $4$?
• A very limited solution: Assuming[0 <= a < 2 π, FullSimplify[Integrate[PiecewiseExpand[Abs[Sin[a - t]]], {t, 0, 2 π}]]] Mar 25, 2017 at 17:37
• @J.M. Nice observation for the assumption intervals. Now I found Integrate[Abs[Sin[a - t]], {t, 0, 2 \[Pi]}, Assumptions -> 0 < a < 23 Pi] generates Boole expressions that can be simplified to correct result, while parameters larger than 24, Integrate[Abs[Sin[a - t]], {t, 0, 2 \[Pi]}, Assumptions -> 0 < a < 24 Pi] directly gives wrong result. It could be that Mathematica gives up because the increasing size of the Boole expressions. Mar 25, 2017 at 17:58
• This looks like a bug. Wrapping the result in an unnecessary ConditionalExpression is excusable, but giving an incorrect answer is not. Mar 25, 2017 at 20:49
• @mikado I agree, I have reported this issue to support: case 3869442 Mar 26, 2017 at 6:26
Extending the answer of @J. M.♦ you can write the a as a == b + k * 2 * π and you will get the right general solution:
Assuming[0 <= b < 2 π && k ∈ Integers,
• Hah, manually taking a Mod[]... Mar 26, 2017 at 3:21
• Actually given the parameter a short interval assumption, e.g. Integrate[Abs[Sin[a - t]], {t, 0, 2 \[Pi]}, Assumptions -> 0 < a < 2 Pi] // FullSimplify is enough to get the correct answer. Mar 26, 2017 at 6:21
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# Posts by Samantha
Total # Posts: 1,702
Math
1808
Math
Jefferson Davis, President of the Confederate States of America, lived a square number of years. he died in 1889. The digits of the year in which he was born sum to 17. When was Jefferson Davis Born.
Science
Javier lives in an area where there is a lot of igneous rock. What is probably near by.
Science
I do not have shape or volume. You cannot see me, but i fill any container in which i am placed. I have weight, but you cannot hold me.What am I? Is it air?
Math
By how many degrees Fahrenheit do the frezzing point and the boiling point of water differ?
Science
I can live in both fresh water and salt water. When full grown, I migrate long distaces to my place of birth. What am I?
PreCalculus
Thanks! I didn't realize it could be so easy!
PreCalculus
I'm so confused... I've always been good at math but this year I'm just not getting it. But I know i can I always get help and a REAL explanation from the people on here so... A steel drum in the shape of a right circular cylinder required to have a volume of 100 ...
Science
Cells make tissue and tissue makes what?
math
Without actually graphing it, show that the graph of the equation x^3+y=x is symmetric with respect to the origin
Pre Calc
Without actually graphing it, show that the graph of the equation x^3+y=x is symmetric with respect to the origin.
Pre Calc
simplify the expression x-4 when x is less than 4
Science
Whats an example and non-example for cardiac muscle? Thanks again
Science
Whats a characterisic for axial skeleton? Thanks
Science
Whats an example and non-example of an axial skeleton?
supposed to be math
how many u.s. patents did Thomas Edison have? The number is 93 more than the number of pounds in half a ton. so basically i just need to know how many pounds are in half a ton
where can light travel, but not sound?
thanks
how many grams in a lb.
science
thanks
science
a cup of coffee, a 10 gallon aquarium, and a bottle of ginger ale were on the same table. on which container was there the greatest amount of air pressure
Science
thanks so much
Science
what is the largest organ human beings have
English
thank u greatly apreciated
English
what is a guide word and an entry word on a dictionary page. the one that i'm stuck on is what is a guide word
math
ok ya that could work. thanks ms sue!:)
math
im sorry that's what it says in my book its my last question. i know it doesn't make sense but tom im gonna get in trouble for not doing it, and there's no answer in the back of the book either for this one so im dead. that's exactly what it says i read it over.
math
sorry i wrote it wrong: Can you use a proportion to solve any ratio problem? if your answer is yes,which ratio problem would you not use a proportion to solve? if your answer in no, which ratio problem could you not use a proportion to solve?
math
Can you use a proportion to solve any ratio problem? If your answer is yes, which ratio problem would you not use a proportion to solve? if your answer is no, which ratio problem would you be able to solve?
math
thank u. :)
math
how would u do 60% of 65?
Math
its just at school for questions i can't ask for help, im fine with homework and all because ur here to explain but i don't have anyone right by my side for the tests. On the math tests the questions are really weird and stuff that wasn't even in our textbooks.
Math
lol ur so awesome!! im not so sure i did so well on my test last week though. i depended on u for all my homework, and for the test i had no help and i was frustrated. :(
Math
thanks Ms sue ur the best teacher ever! im for sure getting great in math this term! :D
anatomy
C- left common cartid artery It may be helpful to think of the aorta as a big hose coming out of the heart to supply the different parts of the body with oxygenated blood, that is fresh and high in oxygen. As that hose (Aorta) leaves the left ventricle it first gives a branch ...
language arts
I have to put together and present a strategy that is effective in the teaching of language arts at a specific grade level. What would a good idea be for around the grade 2 level? and why is it important?
health care
WATCH THE MOVIE BY MICHEAL MOORE! ITS GOOD but then again biased- but you can take your own views out of it!
math
n=30
english
nothing
Science
I grew in a cave, but i'm not alive. i began as water and minerals, but now i am solid. i grow upwards, but i never see the sky. What am i?
math
you take 10x10=100x10=1,000 then u take 1,000x500=500,000
Descriptive Writing
Colby gave her Romeo a lecture for the mean comment he made about her in front of his friends. Is this a simile, metaphor, personification or hyperbole?
science
What happens when you rub and ebonite stick with a wool cloth?
Algebra
I'm having trouble with inequalities. Basically my problem is about which way the sign goes. Here's an example.. X(X+1)>2 so then it becomes X^2 + X>2 Correct? Then X^2 + X - 2>0 then factor so it becomes (X+2)(X-1)>0 And so I end up with X>-2 and X>1...
Science
What is an example of aerobic organisms? Thank You
math
How do i find the area of a triangle? For example: A squared + b squared= c squared 8.5 squared + 6.4 squared= c squared 72.25 + 40.96= c squared 113.21=c squared 10.6400188=c even tho i thought this was the answer it is not because it does not solve the code on the worksheet...
geometry
find the circumference of each circle. use 3.14 for
science
How are a living thing and an automobile alike in carrying out life processes and how are they different?
english
i had an essay prompt, write about what Machiavelli would think if he encountered mencius' novel, mencius. I was looking for ideas to get started. Ways I thought he would disagree with Mencius is the importance of profit (Mencius- for the people, Machiavelli- for the ...
geometry
waht can I do to get help with my graph question ?
english
hi could anyone help me i need pun words for sucker and lollipop
biology
I was wondering if anyone could help me to better understand the Dichotomous Key and how to fill one in.
ethnics cultural diversity
I have to write a 700-1,050 word essay about my ethnic group but I don't know where to begin or where to search, do you have any advice?
math
IT CANT BE REDUCED
Vocabulary
i don't understand an homework it is to do with findind a diffrent word otherwase than walked.
algebra
6. Find the equation of each parabola described below. a) parabola with vertex (0,0) and the focus (0,7) b) parabola with focus (-3,0) and directrix x=3 c) parabola with vertex (3,3) and directrix x=-1 d) parabola with focus (-2,-1) and directrix y=5 e) ysquared=-6x translated...
chem
Water---> H2O (bi7) Fl2
social studies
Its china
Science ( the sun)
Is the sun a high mass star or a low mass star?
Science
Is the sun a high mass star or a low mass?
Literature-Poetry
All of Ashley's answers are right.
geography
for my homework i need to find out about broughton airbus factory.i need to write a report and a paragraph about these questions:- where it is what alterations have been made to the factory how it helps the airbus industry how the wings are transported to other countries when ...
GEOGRAPHY
for the question about how the wings are transported to toulouse it says about the A380 wings does this apply to all airbus. thanks ms sue for your help i appreciate it.
Science
What is another name for atmospheric conditions?
Math
i like to do maths
Algebra II
Oh, I see. So there would be two solutions for both B and C. I would just solve it like a regular triangle, and B' and C' would be 180 minus the values of B and C. Thanks!
Algebra II
How do you solve a triangle with two solutions? The problem says that A=40 degrees, b=10, and a=14. So I did this: 14>(10)(sin 40) And this means that there are two solutions according to the equations my teacher gave me. What do I do now?
Algebra
Sorry for asking another question, but I don't know how to set this problem up. Ship A is due west of a lighthouse. Ship B is 12 km south of ship A. From ship B the bearing to the lighthouse is N63E. How far is ship A from the lighthouse?
Algebra
I'm fine with angle of elevation problem, but angle of depression problems are tricky for me. Is this right? From a balloon 835 m high, a command post is seen with an angle of depression of 7 degrees. How far is it from the point on the ground directly below the balloon to...
Health
Oops! I'm sorry, there are 11 letters
Health
What is the process by which alcohol leaves the body? It's the only one on my crossword puzzle that I can't figure out and it's supposed to start with an E
Health
I have to make an advertisement to convince people to stop smoking that looks like a cigarette box, but I can't think of a title for my fake cigarettes. It's supposed to be a creative name that makes cigarettes look bad. For example, "Virginia Slimes" instead...
pre-calculus
Hi I have this question regarding ellipses. the equation is.. x^2/25 + y^2/169= 1 it asks for major and minor axis (i got 10 and 26) x-intercept y-intercept ((5,0)(0,13)) foci (-12,0) points of intersection with line y= 1-2x graph graph the equation of the ellipses after it is...
Informed Consent
It would be Substituted consent she is a minor and the parents have to give consent until she is of legal majority or becomes emancipated in the eyes of the law.
chemistry
A volume of 100 of 1.00 solution is titrated with 1.00 solution. You added the following quantities of 1.00 to the reaction flask. Classify the following conditions based on whether they are before the equivalence point, at the equivalence point, or after the equivalence point.
chemistry
sorry i didn't post before.. Lab: Electrolytic Cells Purpose: The purpose of this experiment is to test the method of stoichiometry in cells. Materials: Balance Steel can Tin electrode Power Source Wire lead (x2) tin(II) chloride solution (3.25 M) Timer Large beaker ...
chemistry
well im not all that worried about rounding off my numbers. When it comes to actually handing in the assignment i use it the way it's given to me from the book. My question was if i am doing the equation correctly, and if you could help me get started with the following ...
chemistry
yeah sorry about the 3600..in my lab i used 360 im not sure why i put 3600. And your right about the SnCl2 instead of SnCl3. and as far as the half cell, since im using tin II chloride the half cell should be Sn^2+ + 2e- --> Sn^+ Right? this is the full lab.. purpose: to ...
chemistry
TO DR.BOB222 hi i have a question. It's regarding a electrolytic cells. This lab has already been posted but i wanted to show you what i have so maybe you can check it for me. question: what is the mass of tin produced? what is the theoretical mass of tin that should have ...
chemistry
A weak base with a concentration of 1.3 mol/L has a percent ionization of 0.72%. What is the Kb of this weak base? could someone help me with this question or how to get started?
Science
Could you please help me. When water is electrolyzed, two gases ar produced. As indicated in the chemical equation, twice the volume of one gas is produced compared to the other. a) Which gas is produced in larger quantities? Explain * I think it's hydrogen, but i am not ...
Debate
I am new at debate and i need help on the new CX debate cause i have a 6 weeks grade and the teacher wont help me so i need help writing an affermative CX case on RESOLVED: the united states federal government should substantually increase alternative energy incentives to the ...
Spelin!!! plse chek my spellin!
imoshinal did i spell it right
English
How do we do this
Pre-Algebra
( copy/ orginal) = 10/3 (10/3)=(copy perimeter)/(orig.perimeter) (10/3)=(36/X) cross multiply and you get 10x=108 then x= 10 4/5 =10.8 x=perimeter of org. triangle Do the same with part b but this time it's area. (10/3)=(48/x) ===> x= 14 2/5 =14.4
geometry
I'm thinking that since the octagon is made up of eight equal size triangles with one of the sides being 1 cm. Then find the area of one triangle which has all sides equal to 1 cm. then multiply answer by 8 and that should give you total area of octagon. hope this helps.
physics
At a certain distance from a point charge, the potential due to the charge is |5.93 V, and the magnitude of the electric field is 13.5 V/m. (Take the potential to be zero at infinity.) Is the electric field directed toward or away from the charge? Here is my attempt. ...
physics
I saw that electric field has unit of V/m, so also tried doing (9m X 53V/m)+1200V but still wrong!
physics
The figure shows two points in an E-field: Point 1 is at (X1,Y1) = (3,4), and point 2 is at (X2,Y2) = (12,9), with coordinates in meters. The electric field is constant, with a magnitude of 53 V/m, and is directed parallel to the +x axis. The potential at point 1 is 1200 V. ...
physics
A 10000 car comes to a bridge during a storm and finds the bridge washed out. The 700 driver must get to the other side, so he decides to try leaping it with his car. The side the car is on is 24.0 above the river, while the opposite side is a mere 6.80 m above the river. The ...
Physics
0.2
us history
What is the "Hawley-Smoot Tariff"?
Spelling/Analogy
educated is to trained as spoken is to
Geography
I do not understand grid co-ordinates I need help!
math
The length of a rectangle is the perimeter of the. The length of a rectangle with the perimeter of the rectangle is 126 feet
Cultural Diversity & Ethics
Post your response to this question: Consider racial imbalances in education, the economy, family life, housing, criminal justice, health care, and politics. Of these societal challenges facing modern African Americans, which do you think are most difficult to overcome, and ...
Predication
Thanks.
Predication
Predication in logic, the attributing of characteristics to a subject to produce a meaningful statement combining verbal and nominal elements.
Predication
I do I correct the prediaction sentences below? Our resolutions should begin at midnight on January 1. The reason for the cat’s illness is because she was allergic to the dog. A book that has an exciting introduction and a surprising ending would be the best story for me ...
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# Lorentz transformation for change in coordinates | Physics | Khan Academy | Summary and Q&A
35.6K views
January 30, 2016
by
Lorentz transformation for change in coordinates | Physics | Khan Academy
## TL;DR
The video explains how to calculate the change in X prime and C T prime using algebraic manipulation of the Lorentz transformations.
## Install to Summarize YouTube Videos and Get Transcripts
### Q: How can we calculate the change in X prime in terms of change in X and C T?
To calculate the change in X prime, we can use the formula gamma times change in X minus beta times change in C T. This formula allows us to relate the changes in X and C T to the change in X prime.
### Q: What is the significance of the Lorentz factor gamma in the calculations?
The Lorentz factor gamma represents the factor by which lengths and time intervals change in relativistic situations. It is crucial in the calculations as it accounts for time dilation and length contraction effects.
### Q: Can the change in C T prime be calculated using the same algebraic manipulation?
Yes, the change in C T prime can be calculated using the formula gamma times change in C T minus beta times change in X. This formula relates the changes in C T and X to the change in C T prime.
### Q: How does the algebraic manipulation of the Lorentz transformations help understand velocities in different frames of reference?
By understanding the change in coordinates (X prime and C T prime), we can determine the velocities in different frames of reference. This allows us to analyze how objects move relative to each other in special relativity.
## Summary & Key Takeaways
• The video discusses the concept of change in X prime and C T prime in terms of change in X and C T in the Lorentz transformations.
• It demonstrates how to calculate X prime final and X prime initial using the formula gamma times X final minus beta times C T final.
• The video also explains how to calculate the change in X prime using algebraic manipulation and provides the formula gamma times change in X minus beta times change in C T.
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# Agda: Pair of vectors that have the same length
In Agda, how can I define a pair of vectors that must have the same length?
``````-- my first try, but need to have 'n' implicitly
b : ∀ ( n : ℕ ) → Σ (Vec ℕ n) (λ _ → Vec ℕ n)
b 2 = (1 ∷ 2 ∷ []) , (3 ∷ 4 ∷ [])
b 3 = (1 ∷ 2 ∷ 3 ∷ []) , (4 ∷ 5 ∷ 6 ∷ [])
b _ = _
-- how can I define VecSameLength correctly?
VecSameLength : Set
VecSameLength = _
c : VecSameLength
c = (1 ∷ 2 ∷ []) , (1 ∷ 2 ∷ [])
d : VecSameLength
d = (1 ∷ 2 ∷ 3 ∷ []) , (4 ∷ 5 ∷ 6 ∷ [])
-- another try
VecSameLength : Set
VecSameLength = Σ (Vec ℕ ?) (λ v → Vec ℕ (len v))
``````
-
I realize this is probably an attempt to learn how to work with dependent types, but you can get guaranteed equal length "pairs" of vectors by simply making a vector of pairs and unzipping it. I enjoy dependent types, but it's important to realize that you can get a lot of guarantees by clever manipulation of much simpler types. – copumpkin Mar 16 '13 at 20:37
If you want to keep the length as a part of the type, you just need to pack up two vectors with the same size index. Necessary imports first:
``````open import Data.Nat
open import Data.Product
open import Data.Vec
``````
Nothing extra fancy: just as you would write your ordinary vector of size `n`, you can do this:
``````2Vec : ∀ {a} → Set a → ℕ → Set a
2Vec A n = Vec A n × Vec A n
``````
That is, `2Vec A n` is the type of pairs of vectors of `A`s, both with `n` elements. Note that I took the opportunity to generalize it to arbitary universe level - which means you can have vectors of `Set`s, for example.
The second useful thing to note is that I used `_×_`, which is an ordinary non-dependent pair. It is defined in terms of `Σ` as a special case where the second component doesn't depend on the value of first.
And before I move to the example where we'd like to keep the size hidden, here's an example of a value of this type:
``````test₁ : 2Vec ℕ 3
-- We can also infer the size index just from the term:
-- test₁ : 2Vec ℕ _
test₁ = 0 ∷ 1 ∷ 2 ∷ [] , 3 ∷ 4 ∷ 5 ∷ []
``````
You can check that Agda rightfully complains when you attempt to stuff two vectors of uneven size into this pair.
Hiding indices is job exactly suited for dependent pair. As a starter, here's how you'd hide the length of one vector:
``````data SomeVec {a} (A : Set a) : Set a where
some : ∀ n → Vec A n → SomeVec A
someVec : SomeVec ℕ
someVec = some _ (0 ∷ 1 ∷ [])
``````
The size index is kept outside of the type signature, so we only know that the vector inside has some unknown size (effectively giving you a list). Of course, writing a new data type everytime we needed to hide an index would be tiresome, so standard library gives us `Σ`.
``````someVec : Σ ℕ λ n → Vec ℕ n
-- If you have newer version of standard library, you can also write:
-- someVec : Σ[ n ∈ ℕ ] Vec ℕ n
-- Older version used unicode colon instead of ∈
someVec = _ , 0 ∷ 1 ∷ []
``````
Now, we can easily apply this to the type `2Vec` given above:
``````∃2Vec : ∀ {a} → Set a → Set a
∃2Vec A = Σ[ n ∈ ℕ ] 2Vec A n
test₂ : ∃2Vec ℕ
test₂ = _ , 0 ∷ 1 ∷ 2 ∷ [] , 3 ∷ 4 ∷ 5 ∷ []
``````
copumpkin raises an excellent point: you can get the same guarantee just by using a list of pairs. These two representations encode exactly the same information, let's take a look.
Here, we'll use a different import list to prevent name clashes:
``````open import Data.List
open import Data.Nat
open import Data.Product as P
open import Data.Vec as V
open import Function
open import Relation.Binary.PropositionalEquality
``````
Going from two vectors to one list is a matter of zipping the two vectors together:
``````vec⟶list : ∀ {a} {A : Set a} → ∃2Vec A → List (A × A)
vec⟶list (zero , [] , []) = []
vec⟶list (suc n , x ∷ xs , y ∷ ys) = (x , y) ∷ vec⟶list (n , xs , ys)
-- Alternatively:
vec⟶list = toList ∘ uncurry V.zip ∘ proj₂
``````
Going back is just unzipping - taking a list of pairs and producing a pair of lists:
``````list⟶vec : ∀ {a} {A : Set a} → List (A × A) → ∃2Vec A
list⟶vec [] = 0 , [] , []
list⟶vec ((x , y) ∷ xys) with list⟶vec xys
... | n , xs , ys = suc n , x ∷ xs , y ∷ ys
-- Alternatively:
list⟶vec = ,_ ∘ unzip ∘ fromList
``````
Now, we know how to get from one representation to the other, but we still have to show that these two representations give us the same information.
Firstly, we show that if we take a list, convert it to vector (via `list⟶vec`) and then back to list (via `vec⟶list`), then we get the same list back.
``````pf₁ : ∀ {a} {A : Set a} (xs : List (A × A)) → vec⟶list (list⟶vec xs) ≡ xs
pf₁ [] = refl
pf₁ (x ∷ xs) = cong (_∷_ x) (pf₁ xs)
``````
And then the other way around: first vector to list and then list to vector:
``````pf₂ : ∀ {a} {A : Set a} (xs : ∃2Vec A) → list⟶vec (vec⟶list xs) ≡ xs
pf₂ (zero , [] , []) = refl
pf₂ (suc n , x ∷ xs , y ∷ ys) =
cong (P.map suc (P.map (_∷_ x) (_∷_ y))) (pf₂ (n , xs , ys))
``````
In case you are wondering what `cong` does:
``````cong : ∀ {a b} {A : Set a} {B : Set b}
(f : A → B) {x y} → x ≡ y → f x ≡ f y
cong f refl = refl
``````
We've shown that `list⟶vec` together with `vec⟶list` form an isomorphism between `List (A × A)` and `∃2Vec A`, which means that these two representations are isomorphic.
-
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http://www.ehow.co.uk/how_6468859_create-math-probability-board-game.html
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DISCOVER
# How to Create a Math Probability Board Game
Updated July 20, 2017
Probability is sometimes a difficult concept for students to grasp. Probability is the mathematical term for expressing the likelihood of an event's occurrence. For example, if you spin a spinner divided into four equal sections (red, green, blue, yellow), the probability of landing on a yellow section would be one-fourth. A great way to teach elementary children about probability is through playing games. While there are many versions of probability games on the market, creating your own is another option worth exploring and leaves an opportunity for tweaking if needed. This simple board game can be adapted for older students by changing the questions on the cards.
Cut a piece of plywood or cardboard to the desired size of the game board using scissors or a saw. Cardboard is more flexible and able to be bent for storage. Remember to keep the size reasonable for storage purposes.
Create a design for the board, either drawn by hand or computer created. Most game boards have a path to follow. There should be a start space, a finish space and equally divided spaces along the path. These spaces will be where game pieces will be placed during game play and will mark a player's current position on the board. The design should leave an empty space or a rectangular space on the board to mark where game cards will be placed. Clipart, photos from magazines or drawn pictures may be added to the design to make the board more attractive and interesting to the players. Another feature that may be added are spaces marked Lose a Turn, Go Back 3 Spaces or Move Forward 2 Spaces to make the game more interesting. Use a ruler to make sure all lines and spaces are equally divided if drawing by hand. Glue the finished design onto the board.
To create game pieces, find desirable clipart using a word processor. Print the clipart and glue onto cardstock or thicker paper such as the cardboard found on cereal or cracker boxes. Make sure that the pieces you select are a fit for the spaces on the game board design. If not, resize and reprint. Game cards can be made from index cards cut in half or by computer-generated boxes. Write or type a probability problem onto each card. For example, "Johnny woke up with no electricity but had to get ready for school. His sock drawer contained three red socks, five blue socks, and eight white socks. What is the probability that Johnny would pull out two of the same coloured socks in the first two draws?" Number the game cards in the lower right hand corner for matching with the answer sheet.
Using the numbers on the bottom right hand corner of the game cards, create a corresponding answer sheet that the other players can use to check the answers during game play. Laminate the game pieces, answer sheet and game cards for durability. Gather two die, which will be used to determine the number of spaces a player will move along the path if he answers correctly, and place all of the materials in a baggie for safekeeping.
As with any other game, there needs to be a clear set of rules. Create a direction sheet that includes information on how to set up the board for game play, how to move along the spaces, what the special spaces (e.g., Lose a Turn) mean and what to do in the event that a player answers a question incorrectly. For example, if answering correctly means you get another turn or if answering incorrectly means that you do not go forward along the path, then the direction sheet should make that clear to the player. Once the direction sheet is completed, play the game with a friend and make sure that the directions are clear so that any needed changes can be made before presenting it to students for game play.
#### Things You'll Need
• Cardboard or plywood
• Markers, paper, pencils and other drawing tools
• Ruler
• Scissors or saw
• Glue
• Index cards
• Laminating paper
• Baggies
• Dice
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https://www.studysmarter.us/textbooks/physics/physics-for-scientists-and-engineers-a-strategic-approach-with-modern-physics-4th/the-electric-potential/q47-the-electron-gun-in-an-old-tv-picture-tube-accelerates-e/
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Q.47
Expert-verified
Found in: Page 711
### Physics for Scientists and Engineers: A Strategic Approach with Modern Physics
Book edition 4th
Author(s) Randall D. Knight
Pages 1240 pages
ISBN 9780133942651
# The electron gun in an old TV picture tube accelerates electrons between two parallel plates 1.2 cm apart with a 25 kV potential difference between them. The electrons enter through a small hole in the negative plate, accelerate, then exit through a small hole in the positive plate. Assume that the holes are small enough not to affect the electric field or potential.a. What is the electric field strength between the plates?b. With what speed does an electron exit the electron gun if its entry speed is close to zero?Note: The exit speed is so fast that we really need to use the theory of relativity to compute an accurate value. Your answer to part b is in the right range but a little too big.
a. The electric field strength between the plates
b. At speed an electron exit the electron gun if its entry speed is close to zero
See the step by step solution
## Step 1: Definition of Electric Potential and Electric field strength
the equation of electric potential inside an parallel plate capacitor is ;
Here in this equation E is the electric field strength, and d is the distance between two plates;
## Step 2: Given Data
Here the given data is
## Step 3: Substitution of data within Equation
However this can be Written as;
As;
a. The electric field strength between the plates
In this scenario the initial speed of electron can be considered as Zero. The more the electron gets accelerated the more change has seen in the kinetic energy.
b) hence the calculated speed of electron will be;
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https://www.digitalengineering247.com/article/how-to-cook-a-turkey-like-an-engineer/
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Thanksgiving Throwback: How to Cook a Turkey Like an Engineer
Editor's Note: As Thanksgiving approaches, we take a long look back at one engineer's quest to cook the perfect turkey with the help of computational fluid analysis (CFD). Have a happy Thanksgiving.
Editor's Note: As Thanksgiving approaches, we take a long look back at one engineer's quest in 2013 to cook the perfect turkey with the help of computational fluid analysis (CFD). Have a happy Thanksgiving.
Figure 8: Airspeed near the turkey surface, opposite the fan.
By Travis Mikjaniec, Hardware Engineer, Juniper Networks; and former Application Engineer, Mechanical Analysis Division, Mentor Graphics
Editor's Note: As Thanksgiving approaches, we're happy to share one engineer's quest to cook the perfect turkey with the help of computational fluid analysis (CFD). Have a happy Thanksgiving.
As the annual Thanksgiving feast approached at our house (and my wife isn’t, let’s say, experienced at cooking American food), taking the engineer’s perspective, I decided I would prefer not to have hit or miss Thanksgiving holiday dinners for the next 10 years while figuring out what is important and what isn’t when it comes to roasting a turkey. I had some empirical (experimental) results from the last two Thanksgiving dinners, but not enough to make my conclusions, so I built a computational fluid dynamics model in the Mentor Graphics FloEFD software and ran a number of airflow analyses to answer some of the questions that were plaguing me. Just exactly what is the best way to cook a large turkey, what effect does the type of oven have, and what type of rack and roasting pan produces the best results? In this article, I discuss the details of the model and my analyses, providing answers to these age-old mysterious questions once and for all.
Modeling the Turkey
The turkey model I found wasn’t quite correct because it was just one solid block. It didn’t have a cavity for stuffing or a neck cavity, so I just eyeballed those and made some cuts. I wasn’t going to buy a turkey to take measurements! Now, because one of our most important questions revolves around airflow amounts through various spaces (into the cavities, under the turkey), I created some objects so we could track this data (Figure 1). The oven was setup as a convection oven, in that there is a fan at the back pushing air horizontally over the roasting area. Lastly, the turkey was placed on a rack inside the roaster, with a height of 0.4 inches, to provide some space for air to circulate underneath the bird.
Figure 2 shows the boundary conditions for cooking the turkey. The turkey model was run as a snapshot in time, meaning I set the turkey temperature to some not-quite-fully-cooked yet temperature (120°F). The oven temperature was set to 375°F, with the heating elements being slightly hotter at 400°F at the bottom of the oven.
Once I had a good turkey model in the oven (Figure 3), the model was solved in FloEFD and I looked at some streamlines (Figure 4). Streamlines are similar to smoke streams shown in car commercials when the car is in a wind tunnel. The streamlines show where the air is going. I set the fan as the starting point of these streamlines. Although the streamlines are quite chaotic, there is still information that can be obtained for the analysis.
By sectioning the model, I could see into the roaster and into the cavity of the turkey. Compared to the streamlines outside the turkey, not a lot of air is going into the roaster and under the turkey, and even less is going through the turkey (Figure 5). But this is what I expected, especially for the cavity airflow. The fan is pushing the air across the width of the turkey, not along the length of the bird. The air would have to flow around the turkey and then make a 180 degree turn to flow into the cavity, and there is no reason for it to do so. Because turkeys are so big, they can’t be oriented in the direction of the airflow from the fan.
Calculating Airspeed
Looking at a contour plot of velocity (Figure 6), going through the middle of the turkey and oven, the airspeed through the turkey and under the turkey is very low, while the airspeed above the turkey and under the roasting pan is more significant. What constitutes very slow air? Well if we didn’t pay for a fan and had a natural convection oven, the typical airspeed for would be about 0.2 m/s. So the fan isn’t providing any improved heat transfer because the majority of the surface area of the turkey is experiencing airspeed less than 0.2 m/s.
A surface plot of the velocity near the surface of the turkey illustrates the airspeed (Figures 7, 8, and 9). I split this into three images: one shows at the turkey surface close to the fan, another image shows the opposite side, and the last shows underneath. There is a clear difference between the fan side and the opposite side, and the result is that one side of the turkey will cook much faster than the other (or dry out faster), if the turkey isn’t rotated periodically. Faster air equals faster convection, which is why we blow on soup to cool it quicker. Underneath, it’s the same as was illustrated before, that the air is not moving underneath the turkey.
Looking at temperature shown in the contour plot, the air is a lot colder inside the turkey (Figure 10). This is because the air is stagnant, that is, no hot air is making its way into the turkey. Also below the turkey, in the 0.4 inch space provided by the rack, the air temperature is cooler than the rest of the oven—again, because not a lot of fresh hot air is making its way into this space.
Why is air having a hard time making it into the space between the turkey and the roaster bottom? The streamlines in the image below show that essentially it’s the same reason that the air isn’t going into the turkey. The air coming out of the fan follows the path of least resistance. To go under the turkey, it has to travel around the wall of the roaster, down between the turkey and roaster wall, and then make a 90° turn to flow under the turkey. All the while, the air slows down, and cools down. Both are important because cold air sinks, right? And because the air speed is lower than that of a natural convection current, hot fresh air isn’t going to displace this slower, cooler air. That is why it seems like the air makes it down into the roaster, but then can’t move further under the bird and just recirculates by the roaster wall (Figure 11).
In Summary
What did I learn? Well, stuffing doesn’t impede airflow through the turkey because there is very little airflow to speak of. But the added thermal mass of stuffing inside the turkey doesn’t necessarily lead to longer cooking times and dryer turkey meat. A lot of airflow under the turkey won’t make a full crispy skin bird. A taller rack might improve this or a roaster with shallower walls, but I have my doubts. It seems to me that a rack or carrots/celery/potatoes would do the same thing: get the turkey out of the drippings. Any claim about allowing air to circulate all around the bird is false. And it’s very important to rotate the turkey so it cooks evenly.
All this thinking about Thanksgiving, and airflow analysis, makes me pretty hungry. But there are more possibilities to keep me going on this analysis, such as what would be the difference with a non-convection oven, taller rack height/roaster wall height, different roaster materials, or stuffing cooking time versus unstuffed? And what about cooking the turkey in a smoker BBQ? Clearly, much progress has been made on solving the mystery of cooking the perfect Thanksgiving turkey, but more research needs to be done.
Travis Mikjaniec achieved his Masters Degree in Aerospace Engineering from Carleton University in 2006, with a research focus on rotorcraft aerodynamics and aeroelasticity. Travis has spent the past five years working for Mentor Graphics Corporation, Mechanical Analysis Division, specializing in the application of CFD to the design of electronics equipment. Before this, he worked for the National Research Council of Canada’s (NRC) Institute for Aerospace Research (IAR) Flight Research Laboratory in a role that focused on flight tests. Before that, he worked at the NRC's IAR Aeroacoustic and Structural Dynamics Laboratory in a role that focused on experimental testing of smart structures. Currently, Travis is a Thermal Engineer in the engineering team of the Mechanical Analysis Division, training and supporting the user base to enhance the accessibility of thermal analysis to engineering designer. Over this time, he has gained experience, not only in the use of software, but also of the general design processes and the best way to apply the CFD techniques to the real world problems encountered by engineers every day.
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#19901
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2014
03-13
# Equal Sum Partitions
An equal sum partition of a sequence of numbers is a grouping of the numbers (in the same order as the original sequence) in such a way that each group has the same sum. For example, the sequence:
2 5 1 3 3 7
may be grouped as:
(2 5) (1 3 3) (7)
to yield an equal sum of 7.
Note: The partition that puts all the numbers in a single group is an equal sum partition with the sum equal to the sum of all the numbers in the sequence.
For this problem, you will write a program that takes as input a sequence of positive integers and returns the smallest sum for an equal sum partition of the sequence.
The first line of input contains a single integer P, (1 ≤ P ≤ 1000), which is the number of data sets that follow. The first line of each data set contains the data set number, followed by a space, followed by a decimal integer M, (1 ≤ M ≤ 10000), giving the total number of integers in the sequence. The remaining line(s) in the dataset consist of the values, 10 per line, separated by a single space. The last line in the dataset may contain less than 10 values.
The first line of input contains a single integer P, (1 ≤ P ≤ 1000), which is the number of data sets that follow. The first line of each data set contains the data set number, followed by a space, followed by a decimal integer M, (1 ≤ M ≤ 10000), giving the total number of integers in the sequence. The remaining line(s) in the dataset consist of the values, 10 per line, separated by a single space. The last line in the dataset may contain less than 10 values.
3
1 6
2 5 1 3 3 7
2 6
1 2 3 4 5 6
3 20
1 1 2 1 1 2 1 1 2 1
1 2 1 1 2 1 1 2 1 1
1 7
2 21
3 2
/*
用了线段树+二分搜索，树状数组+二分搜索也行。
2012-10-03
*/
#include"stdio.h"
#include"string.h"
#include"stdlib.h"
struct seg
{
__int64 l,r,mid;
__int64 sum;
}T[30333];
__int64 set[10011];
void build(__int64 l,__int64 r,__int64 k)
{
T[k].l=l;
T[k].r=r;
T[k].mid=(l+r)>>1;
if(l==r) {T[k].sum=set[l];return ;}
build(l,T[k].mid,2*k);
build(T[k].mid+1,r,2*k+1);
T[k].sum=T[2*k].sum+T[2*k+1].sum;
}
__int64 find(__int64 l,__int64 r,__int64 k)
{
__int64 ans_t=0;
if(T[k].l==l && T[k].r==r) return T[k].sum;
if(r<=T[k].mid) ans_t+=find(l,r,2*k);
else if(l>T[k].mid) ans_t+=find(l,r,2*k+1);
else
{
ans_t+=find(l,T[k].mid,2*k);
ans_t+=find(T[k].mid+1,r,2*k+1);
}
return ans_t;
}
__int64 main()
{
__int64 T,Case;
__int64 n;
__int64 i;
__int64 left,up,mid,low;
__int64 count;
__int64 sum,base;
__int64 ans,ans_t;
scanf("%I64d",&T);
for(Case=1;Case<=T;Case++)
{
scanf("%I64d%I64d",&n,&n);
sum=0;
for(i=1;i<=n;i++) {scanf("%I64d",&set[i]);sum+=set[i];}
build(1,n,1);
ans=0;
for(i=sum;i>1;i--)
{
if(ans) break;
if(sum%i==0)
{
base=sum/i;
count=0;
left=1;
while(count<i)
{
up=left;
low=n;
mid=(up+low)>>1;
while(up<=low)
{
ans_t=find(left,mid,1);
if(ans_t==base) break;
if(ans_t>base) low=mid-1;
else up=mid+1;
mid=(low+up)>>1;
}
if(ans_t==base)
{
count++;
left=mid+1;
}
else break;
}
if(count==i) ans=base;
}
}
if(ans==0) ans=sum;
printf("%I64d %I64d\n",Case,ans);
}
return 0;
}
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|
# Convolutions
To perform linear convolutions on images, use image.convolve(). The only argument to convolve is an ee.Kernel which is specified by a shape and the weights in the kernel. Each pixel of the image output by convolve() is the linear combination of the kernel values and the input image pixels covered by the kernel. The kernels are applied to each band individually. For example, you might want to use a low-pass (smoothing) kernel to remove high-frequency information. The following illustrates a 15x15 low-pass kernel applied to a Landsat 8 image:
### Code Editor (JavaScript)
// Load and display an image.
var image = ee.Image('LANDSAT/LC08/C02/T1_TOA/LC08_044034_20140318');
Map.setCenter(-121.9785, 37.8694, 11);
Map.addLayer(image, {bands: ['B5', 'B4', 'B3'], max: 0.5}, 'input image');
// Define a boxcar or low-pass kernel.
var boxcar = ee.Kernel.square({
radius: 7, units: 'pixels', normalize: true
});
// Smooth the image by convolving with the boxcar kernel.
var smooth = image.convolve(boxcar);
Map.addLayer(smooth, {bands: ['B5', 'B4', 'B3'], max: 0.5}, 'smoothed');
The output of convolution with the low-pass filter should look something like Figure 1. Observe that the arguments to the kernel determine its size and coefficients. Specifically, with the units parameter set to pixels, the radius parameter specifies the number of pixels from the center that the kernel will cover. If normalize is set to true, the kernel coefficients will sum to one. If the magnitude parameter is set, the kernel coefficients will be multiplied by the magnitude (if normalize is also true, the coefficients will sum to magnitude). If there is a negative value in any of the kernel coefficients, setting normalize to true will make the coefficients sum to zero.
Use other kernels to achieve the desired image processing effect. This example uses a Laplacian kernel for isotropic edge detection:
### Code Editor (JavaScript)
// Define a Laplacian, or edge-detection kernel.
var laplacian = ee.Kernel.laplacian8({ normalize: false });
// Apply the edge-detection kernel.
var edgy = image.convolve(laplacian);
{bands: ['B5', 'B4', 'B3'], max: 0.5, format: 'png'},
'edges');
Note the format specifier in the visualization parameters. Earth Engine sends display tiles to the Code Editor in JPEG format for efficiency, however edge tiles are sent in PNG format to handle transparency of pixels outside the image boundary. When a visual discontinuity results, setting the format to PNG results in a consistent display. The result of convolving with the Laplacian edge detection kernel should look something like Figure 2.
There are also anisotropic edge detection kernels (e.g. Sobel, Prewitt, Roberts), the direction of which can be changed with kernel.rotate(). Other low pass kernels include a Gaussian kernel and kernels of various shape with uniform weights. To create kernels with arbitrarily defined weights and shape, use ee.Kernel.fixed(). For example, this code creates a 9x9 kernel of 1’s with a zero in the middle:
### Code Editor (JavaScript)
// Create a list of weights for a 9x9 kernel.
var row = [1, 1, 1, 1, 1, 1, 1, 1, 1];
// The center of the kernel is zero.
var centerRow = [1, 1, 1, 1, 0, 1, 1, 1, 1];
// Assemble a list of lists: the 9x9 kernel weights as a 2-D matrix.
var rows = [row, row, row, row, centerRow, row, row, row, row];
// Create the kernel from the weights.
var kernel = ee.Kernel.fixed(9, 9, rows, -4, -4, false);
print(kernel);
[{ "type": "thumb-down", "id": "missingTheInformationINeed", "label":"Missing the information I need" },{ "type": "thumb-down", "id": "tooComplicatedTooManySteps", "label":"Too complicated / too many steps" },{ "type": "thumb-down", "id": "outOfDate", "label":"Out of date" },{ "type": "thumb-down", "id": "samplesCodeIssue", "label":"Samples / code issue" },{ "type": "thumb-down", "id": "otherDown", "label":"Other" }]
[{ "type": "thumb-up", "id": "easyToUnderstand", "label":"Easy to understand" },{ "type": "thumb-up", "id": "solvedMyProblem", "label":"Solved my problem" },{ "type": "thumb-up", "id": "otherUp", "label":"Other" }]
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CC-MAIN-2024-22
|
latest
|
en
| 0.684871
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