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https://www.alignmentforum.org/tag/causal-decision-theory/history | 1,701,805,805,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100555.27/warc/CC-MAIN-20231205172745-20231205202745-00094.warc.gz | 723,996,696 | 53,678 | # Causal Decision Theory
Applied to Newcomb Variant by Lsusr 3mo ago
One usual example where EDT and CDT commonly diverge is the Smoking lesion: “Smoking is strongly correlated with lung cancer, but in the world of the Smoker's Lesion this correlation is understood to be the result of a common cause: a genetic lesion that tends to cause both smoking and cancer. Once we fix the presence or absence of the lesion, there is no additional correlation between smoking and cancer. Suppose you prefer smoking without cancer to not smoking without cancer, and prefer smoking with cancer to not smoking with cancer. Should you smoke?” CDT would recommend smoking since there is no causal connection between smoking and cancer. They are both caused by a gene, but have no causal direct connection with each other. EDT, on the other hand, would recommend against smoking, since smoking is an evidence for having the mentioned gene and thus should be avoided.
The core aspect of CDT is mathematically represented by the fact it uses probabilities of conditionals in place of conditional probabilities 2. The probability of a conditional is the probability of the whole conditional being true, where the conditional probability is the probability of the consequent given the antecedent. A conditional probability of B given A - P(B|A) -, simply implies the Bayesian probability of the event B happening given we known A happened, it’s used in EDT. The probability of conditionals – P(A > B) - refers to the probability that the conditional 'A implies B' is true, it is the probability of the contrafactual ‘If A, then B’ be the case. Since contrafactual analysis is the key tool used to speak about causality, probability of conditionals are said to mirror causal relations. In most cases these two probabilities track each other, and CDT and EDT give the same answers. However, some particular problems have arisen where their predictions for rational action diverge such as the Smoking lesion problem – where CDT seems to give a more reasonable prescription – and Newcomb's problem – where CDT seems unreasonable. David Lewis proved 3 it's impossible tofor probabilities of conditionals to always track conditional probabilities. Hence, evidential relations aren’t the same as causal relations and CDT and EDT will always diverge in some cases.
Causal Decision Theory – CDT – is a branch of decision theory which advises an agent to take actions which maximize the causal consequences on the probability of desired outcomes 1. As any branch of decision theory, it prescribes taking the action that maximizes expected utility, i.e the action which maximizes the sum of the utility obtained in each outcome weighted by the probability of that outcome occurring, given your action. Different decision theories correspond to different ways of construing thethis dependence between actions and outcomes. CDT focuses on the causal relations between one’s actions and outcomes, whilst Evidential Decision Theory – EDT - concerns itself with what an action indicates about the world (which is operationalized by the conditional probability). That is, according to CDT, a rational agent should track the available causal relations linking his actions to the desired outcome and take the action which will better enhance the chances of the desired outcome.
Causal Decision Theory – CDT – is a branch of decision theory which advises an agent to take actions which maximize the causal consequences on the probability of desired outcomes 1. As any branch of decision theory, it prescribes taking the action that maximizes expected utility, i.e the action which maximizes the sum of the utility obtained in each outcome weighted by the probability of that outcome occurring, given your action. Different decision theories correspond to different ways of thinking about what "the probability of an outcome given your action" is supposed to mean.construing the dependence between actions and outcomes. CDT focuses on the causal relations between one’s actions and outcomes, whilst Evidential Decision Theory – EDT - concerns itself with what an action indicates about the world (which is operationalized by the conditional probability). That is, according to CDT, a rational agent should track the available causal relations linking his actions to the desired outcome and take the action which will better enhance the chances of the desired outcome.
Causal Decision Theory – CDT – is a branch of decision theory which advises an agent to take actions thatwhich maximize the causal consequences on the probability of desired outcomes 1. As any branch of decision theory, it prescribes taking the action that maximizes expected utility, i.e the action which maximizes the sum of the utility obtained in each outcome weighted by the probability of that outcome occurring, given your action. Different decision theories correspond to different ways of thinking about what "the probability of an outcome given your action" is supposed to mean. CDT focuses on the causal relations between one’s actions and outcomes, whilst Evidential Decision Theory – EDT - concerns itself with what an action indicates about the world (which is operationalized by the conditional probability). That is, according to CDT, a rational agent should track the available causal relations linking his actions to the desired outcome and take the action which will better enhance the chances of the desired outcome.
Causal Decision Theory – CDT – is a branch of decision theory which advises an agent to take actions that maximizesmaximize the causal consequences on the probability of desired outcomes 1. As any branch of decision theory, it prescribes taking the action that maximizes utility, that which utility equals or exceeds the utility of every other option. The utility of each action is measured by the expected utility, i.e the averaged by probabilitiesaction which maximizes the sum of the utility obtained in each outcome weighted by the probability of eachthat outcome occurring, given your action. Different decision theories correspond to different ways of its possible results. Howthinking about what "the probability of an outcome given your action" is supposed to mean. CDT focuses on the causal relations between one’s actions can influence the probabilities differ between the branches. Contrary toand outcomes, whilst Evidential Decision Theory – EDT - CDT focuses onconcerns itself with what an action indicates about the causal relations between one’s actions and its outcomes, instead of focusing on which actions provide evidences for desired outcomes. Accordingworld (which is operationalized by the conditional probability). That is, according to CDTCDT, a rational agent should track the available causal relations linking his actions to the desired outcome and take the action which will better enhance the chances of the desired outcome.
One usual example where EDT and CDT commonly diverge is the Smoking lesion: “Smoking is strongly correlated with lung cancer, but in the world of the Smoker's Lesion this correlation is understood to be the result of a common cause: a genetic lesion that tends to cause both smoking and cancer. Once we fix the presence or absence of the lesion, there is no additional correlation between smoking and cancer. Suppose you prefer smoking without cancer to not smoking without cancer, and prefer smoking with cancer to not smoking with cancer. Should you smoke?” CDT would recommend smoking since there is no causal connection between smoking and cancer. They are both caused by a gene, but have no causal direct connection with each other. EDTEDT, on the other handhand, would recommend against smoking, since smoking is an evidence for having the mentioned gene and thus should be avoided.
Causal Decision Theory – CDT - is a branch of decision theory which advises an agent to take actions that maximizes the causal consequences on the probability of desired outcomes 1. As any branch of decision theory, it prescribes taking the action that maximizes utility, that which utility equals or exceeds the utility of every other option. The utility of each action is measured by the expected utility, the averaged by probabilities sum of the utility of each of its possible results. How the actions can influence the probabilities differ between the branches. Contrary to Evidential Decision Theory – EDT - CDT focuses on the causal relations between one’s actions and its outcomes, instead of focusing on which actions provide evidences for desired outcomes. According to CDT a rational agent should track the available causal relations linking his actions to the desired outcome and take the action which will better enhance the chances of the desired outcome. | 1,674 | 8,785 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2023-50 | latest | en | 0.960889 |
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# A consumer may not think of household cleaning products as
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31 Aug 2006, 17:22
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A consumer may not think of household cleaning products as HAZARDOUS SUBSTANCES, BUT MANY OF THEM CAN BE HARMFUL TO HEALTH, ESP IF THEY ARE USED IMPROPERLY.
This is the correct answer from the OG verbal book. I'm quite confused why they used THEM and THEY(plural)...is the subject not A CONSUMER?(singular)
Logically, THEM and THEY should refer to cleaning products, but grammar wise, isn't A CONSUMER supposed to be the subject?
This is really bugging me. Thanks
If you have any questions
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Re: SC confusion: household products [#permalink]
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31 Aug 2006, 17:26
A consumer may not think of household cleaning products as HAZARDOUS SUBSTANCES, BUT MANY OF THEM CAN BE HARMFUL TO HEALTH, ESP IF THEY ARE USED IMPROPERLY.
This is the correct answer from the OG verbal book. I'm quite confused why they used THEM and THEY(plural)...is the subject not A CONSUMER?(singular)
Logically, THEM and THEY should refer to cleaning products, but grammar wise, isn't A CONSUMER supposed to be the subject?
This is really bugging me. Thanks
They and them clearly refers to the cleaning products. , but.... should help you zero in on the correct antecendent here.
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01 Sep 2006, 13:56
they, them are relative pronouns which represent the closest antecedents. ....
Moreover, "But" introduces a new clause with a new subject and verb.
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01 Sep 2006, 13:56
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View Full Version : Ballooning in Space! Theoretical Upper Limits?
myles
12-16-2003, 07:11 PM
In 2002, in Manitoba, NASA launched a large unmanned helium balloon into the outer reaches of Earth's atmosphere, reaching at one point 161,000 feet (49 kilometres).
My question is this.
What is the theoretically physically possible absolute upper limit in space reachable by balloon (lighter than air preferably)?
And what sets that limit?
If balloons can reach 50 kilometers, can they reach higher? How much higher?
Enola Straight
12-16-2003, 08:28 PM
Try a baloon filled with the light isotope of hydrogen known as Protium...THE lightest gas there is.
Whats that? Hindenburgh, you say? Hydrogen is explosive and flammable?
Yeah, so is the gasoline in my car.
Rabid_Squirrel
12-16-2003, 08:40 PM
What you need to to is find a formula that equates altitude to air pressure, and another one equation relating density of hydrogen (or protium) to air pressure. Find an altitude where the density of the air equals the density of the gas, and that will give you the theoretical maximum.
I can't be arsed though. Anyone else has the equations on hand?
spingears
12-16-2003, 08:46 PM
Just a bit OT.
Originally posted by Enola Straight
Try a baloon filled with the light isotope of hydrogen known as Protium...THE lightest gas there is.
Whats that? Hindenburgh, you say? Hydrogen is explosive and flammable?
Yeah, so is the gasoline in my car.
A relatively recent investigation of the Hindenburg disaster exonerated hydrogen as the culprit. The hydrogen bags burned AFTER the skin on the ourer frame burned almost explosively. The skin was coated with a thick film of (airplane dope as a carrier sealant ?) an oxidizer such as iron oxide, and aluminum flake or powder to reflect the suns heat. The skin burned very rapidly with a thermit type reaction.
If you have further interrest try the web search engine to find the details.
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Rabid_Squirrel
12-16-2003, 08:47 PM
Crap, I've just realised a problem with my method. :smack:
In space the pressure is ~0bar, so the density of the gas would be ~0. This means the theoretical atitude would be infinity. Of course it's false since I assumed an balloon with zero mass.
It probably means you would have to include a fixed volume of gas and some sort of estimate for the weight of the balloon in your calculations.
MikeS
12-16-2003, 08:49 PM
The maximum height a balloon can reach is determined by when it's "neutrally buoyant", which means that the mass of the balloon divided by its volume is equal to the density (mass/volume) of the gas outside. So theoretically, if we had a balloon made from some space-age material that we could make arbitrarily light, and we could pump all the air out of this balloon and not have it collapse under the 16 lbs/in2 that the atmosphere exerts on it down here, then it would get arbitrarily high.
If you insist that the balloon be filled with some kind of gas, though, then Enola Straight's suggestion of using hydrogen is your best bet (I think). Figuring out exactly how high such a balloon can go is an interesting thermodynamics problem, which I may try to solve later tonight. I'll let you know if I find anything interesting.
scr4
12-16-2003, 11:27 PM
The current record is 52.0 km, I believe, set in May 2002 by a Japanese balloon. I was fortunate enough to see the launch. This was a helium balloon.
The altitude is determined by the thickness of the balloon material (usually polyethylene), volume and payload weight. Lift is proportional to volume, which scales as the cube of balloon diameter, while weight is proportional to surface area which scales as square of diameter. So the bigger the balloon, the higher the altitude. The maximum altitude depends on the technology for making thin lightweight film, so there's no definite theoretical limit.
Valgard
12-17-2003, 12:19 AM
Well I made a bit of a start on this one:
*Note 1 - Hideous rounding and significant figure liberties taken.
*Note 2 - Made assumptions to keep formulas simple. I'm good with spheres, not teardrops.
Found a formula for air density as a function of altitude:
d = 1.21 * exp(-height/8000)
Where d is in kg/m^3 and height is in meters. Dunno how accurate this is or how high it's valid but it's all I got.
This gives d for various altitudes as:
0.002337 kg/m^3 at 50km
0.0006698 kg/m^3 at 60km
0.0001919 kg/m^3 at 70km
0.00005499 kg/m^3 at 80km
The largest balloon on record that I could find mention of was 70 million cubic feet capacity built by Wintzen Research (they also launched a balloon to 49km back in 1972 so they seem to know their stuff).
Let's assume for the sake of argument that at extremely high altitudes the balloon will be spherical (from photos I've seen this may be wrong but again it's all I got and math for spheres is easy).
70 million cubic feet is a touch under 2 million cubic meters.
To "float", the balloon will need to have (mass/volume) = air density at that altitude.
This gives critical masses of:
4600kg at 50km
1329kg at 60km
380kg at 70km
109kg at 80km
Let's ignore the weight of the actual gas in the balloon and any non-envelope stuff (instrument package, gondola for those who want to ride along :-o, etc).
A sphere with a volume of 2 million cubic meters has a surface area of 76,356 square meters.
Our balloon envelope will have a mass/area found by dividing the critical mass (above) by the area (76,356 m^2), and then we can check this against known materials to see if we've got a chance.
I get the following:
1.43 grams/m^2 at 80km
4.98 grams/m^2 at 70km
17.4 grams/m^2 at 60km
60.2 grams/m^2 at 50km
I couldn't find any definitive listing of masses for a good thin film polymer like Dupont Mylar, but some kite enthusiasts have a discussion board and mentioned that it's commonly available in 20, 40 and 60 grams/m^2.
So, back of the envelope figuring, I'd say that 60km is probably achievable, although pretty darn challenging. Dunno about higher, it looks like the material will be hard to find - incredibly light yet rugged enough to survive the temperature and mechanical stresses.
Anyone care to sharpen this up a bit? Does anyone have better info on materials used?
scr4
12-17-2003, 02:09 AM
Originally posted by Valgard
Anyone care to sharpen this up a bit?
I think I can. Using density information from the US Standard Atmospheres, I get
50km: 1.0e-3 kg/m3
60km: 3.1e-4 kg/m3
70km: 8.3e-5 kg/m3
Your numbers aren't too far off. The Japanese ultra high altitude balloons use 3.4-micron polyethylene for the balloon envelope. I can't find the number for NASA balloons but I think it's similar. That would make it approximately 3 g/m2. At 1 million cubic meters and no payload, it should go up to 64.8 km. With a 20kg payload it drops down to 63.8 km, still very respectable. The 53km record (my first post was wrong, sorry) was set by a 3.5-micron 80,000 m2 balloon.
It doesn't mean it's worth doing a 64km balloon flight. There isn't a lot you can do with a 20kg payload. A million cubic meter balloon is expensive to manufacture and launch, and it's only good for one flight lasting a couple of days. Most people who do balloon experiments are more interested in longer duration flights and larger payloads. Longer flights are possible if you use "super-pressure" balloons, i.e. sealed balloons that don't lose any helium due to day/night cycles.
CalMeacham
12-17-2003, 06:43 AM
Heck, according to the noted Hans Phfall, you can go all the way to the moon!
12-17-2003, 06:54 AM
Couldn't you get arbitrarily high if your material could withstand arbitrarily hot gas as well?
Mangetout
12-17-2003, 06:55 AM
Originally posted by MikeS
The maximum height a balloon can reach is determined by when it's "neutrally buoyant", which means that the mass of the balloon divided by its volume is equal to the density (mass/volume) of the gas outside. So theoretically, if we had a balloon made from some space-age material that we could make arbitrarily light, and we could pump all the air out of this balloon and not have it collapse under the 16 lbs/in2 that the atmosphere exerts on it down here, then it would get arbitrarily high. How about having some space-age rigid material that isn't strong enough to support the pressure of all that air, but filling it with hydrogen, gradually pumping the hydrogen out as it rises and the external pressure decreases?
12-17-2003, 10:19 AM
What's the usual fate of such balloons? Do they regularly reach neutral buoyancy, then leak and eventually come down? Does mission control activate a remote depressurization to drop the balloon? Or is it more likely (as I would imagine) that the balloon would eventually reach an altitude where it bursts?
Satyagrahi
12-17-2003, 12:31 PM
Some years back--or decades, maybe--I read a science fiction story that suggested flying communities held aloft by vast, rigid spherical floating structures.
That is, you had a massive spherical shell of reinforced concrete, a mile or two in diameter, enclosing a hard vacuum. The shell would have to be strong enough to support the pressure differential. The idea is that the larger the sphere, the larger would become the ratio between weightless volume (which increases as a power of 3) and weight of the shell (which increases as a power of 2).
Is such a thing feasible? Do we have materials strong enough to achieve it? What's the minimum size necessary to achieve neutral bouyancy?
Just a little more for you to chew on.... :)
12-17-2003, 12:50 PM
We've discussed hard vaccum a few times; basically, it's impossible to make a container strong and light enough.
Sunspace
12-17-2003, 01:05 PM
Would it be possible to take a hybrid approach for a really large balloon? Imagine a spherical shell made of helium-inflated structural members and panels, surrounding a really large volume at a lesser pressure but a much lower density--warm hydrogen, perhaps, heated by the sun. If it was a kilometre or so in diameter, would solar heating suffice to provide negative buoyancy?
Bytegeist
12-17-2003, 02:35 PM
What's the usual fate of such balloons?
After they retire, they're sent to the Village for security work (http://www.retroweb.com/pris_splashpage.html).
Be seeing you. ;)
12-17-2003, 04:25 PM
they're sent to the Village
Being from NYC, this gave me a rather strange first impression. But then I remembered Number Six.
Enola Straight
12-17-2003, 04:56 PM
Perhaps a black baloon will heat up in the sun to provide a partial effect like that of hot air balloons.
GreyWanderer
12-17-2003, 05:37 PM
http://web.sysrq.no/mith/bilder.skeptiker.no/albums/andoya/DSCN1019.sized.jpg
This is from Andøya rocket range in Norway. The balloon that went up to 40km was launched by NASA, we managed to reach 30km as part of a student project.
http://web.sysrq.no/mith/bilder.skeptiker.no/albums/andoya/DSCN1003.sized.jpg
Here you can see me ready to launch the balloon.
What's the usual fate of such balloons? Do they regularly reach neutral buoyancy, then leak and eventually come down? Does mission control activate a remote depressurization to drop the balloon? Or is it more likely (as I would imagine) that the balloon would eventually reach an altitude where it bursts?
Zero-pressure balloons can float for weeks. There is usually one altitude at day and one at night, because the heated air makes it go higher. Sometimes weights are used to make it stay at a constant height. Obviously that can only work for a limited period of time. (Every nightfall a weight is dropped, every morning gas is released because the volume of the gas increases). When you have midnight sun this problem conveniently disappears (:
Balloons that are made to withstand pressure can float longer than zero-pressure balloons, but they don't go as high because of the increased mass. You also have pressure-ballons made from lighter materials that are destroyed when they reach a certain height.
Theoretical maximum height is limited first by the pressure of the gas used, but also by the weight of the balloon. As it reaches higher heights in the mesosphere temperature drops and that can make lightweight materials to be destroyed.
Temperature drops to about -50 degrees celcius at 10km then it gets warmer, before it drops to about -90 degrees celcius at 85km.
Getting through the 10km point (tropospause) is a critical moment for most balloons. I think temperature is a pretty big limiting factor.
scr4
12-17-2003, 05:45 PM | 3,074 | 12,540 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2018-13 | latest | en | 0.91604 |
https://kr.mathworks.com/matlabcentral/cody/problems/44469-diagonal-pattern/solutions/1593586 | 1,585,488,593,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370494331.42/warc/CC-MAIN-20200329105248-20200329135248-00204.warc.gz | 565,799,045 | 15,848 | Cody
# Problem 44469. Diagonal Pattern
Solution 1593586
Submitted on 27 Jul 2018 by PRUDHVI RAJU
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
filetext = fileread('diagonalPattern.m'); assert(isempty(strfind(filetext, 'regexp')),'regexp hacks are forbidden')
2 Pass
n = 1; mat_correct = 0; assert(isequal(diagonalPattern(n),mat_correct))
mat = 0
3 Pass
n = -1; mat_correct = []; assert(isequal(diagonalPattern(n),mat_correct))
mat = []
4 Pass
n = 1.5; mat_correct = []; assert(isequal(diagonalPattern(n),mat_correct))
5 Pass
n = 4; mat_correct = [0 1 2 3 1 0 1 2 2 1 0 1 3 2 1 0]; assert(isequal(diagonalPattern(n),mat_correct))
mat = 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
6 Pass
n = 5; mat_correct = [0 1 2 3 4 1 0 1 2 3 2 1 0 1 2 3 2 1 0 1 4 3 2 1 0]; assert(isequal(diagonalPattern(n),mat_correct))
mat = 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 | 410 | 995 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2020-16 | latest | en | 0.420769 |
https://excelexamples.com/post/how-to-use-possible-football-matches-in-excel/ | 1,696,005,158,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510520.98/warc/CC-MAIN-20230929154432-20230929184432-00714.warc.gz | 272,813,334 | 3,015 | How to Use Possible Football Matches In Excel
Oct 18, 2020 • edited Oct 21, 2020
Below we will look at a program in Excel VBA that shows a print preview of all the possible football matches from a list of teams.
Situation:
1. First, we declare one Range object and four variables. We call the Range object rng. One String variable we call matchname, and three Integer variables we call counter, i and j.
`````` Dim rng As Range, matchname As String, counter As Integer, i As Integer, j As Integer
``````
2. We initialize rng with the team names. We use CurrentRegion because we don't know the exact boundaries of the range in advance (we want this program to work for 3 teams but also for 12 teams). We initialize counter with value 0.
`````` Set rng = Range("A1").CurrentRegion
counter = 0
``````
3. We write all the possible football matches to column C. First, we empty column C.
`````` Worksheets(1).Columns(3) = ""
``````
4. We start a Double Loop.
`````` For i = 1 To rng.Count
For j = i + 1 To rng.Count
``````
5. We write a matchname to the variable matchname.
`````` matchname = rng.Cells(i).Value & " vs " & rng.Cells(j).Value
``````
For example, for i = 1 and j = 2, Excel VBA writes the matchname Kickers vs Shooters. For i = 1 and j = 3, Excel VBA writes the matchname Kickers vs Little Giants, etc.
6. We write the matchname to column C.
`````` Cells(counter + 1, 3).Value = matchname
``````
7. The counter keeps track of the number of matchnames written to column C. Excel VBA increments counter by 1 each time it writes a matchname to column C. To achieve this, add the following code line:
`````` counter = counter + 1
``````
8. Don't forget to close the two loops.
`````` Next j
Next i
``````
9. We show a print preview of all the possible football matches.
`````` ActiveSheet.Columns(3).PrintPreview
``````
10. Test the program.
Part of the result:
Note: column C is manually centered to get this result.
#Tutorial#How To#VBA#Range Object
How To Use Option Explicit In Excel
How To Use Randomly Sort Data In Excel | 590 | 2,182 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2023-40 | latest | en | 0.84786 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/23:_Carbonyl_Condensation_Reactions/23.08:_Mixed_Claisen_Condensations | 1,713,389,392,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817181.55/warc/CC-MAIN-20240417204934-20240417234934-00210.warc.gz | 146,126,276 | 132,259 | # 23.8: Mixed Claisen Condensations
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##### Objectives
After completing this section, you should be able to
1. write an equation to illustrate a mixed Claisen condensation.
2. identify the structural features that should be present in the two esters if a mixed Claisen condensation is to be successful.
3. determine whether a given pair of esters is likely to produce a good yield of a single product when subjected to a mixed Claisen condensation.
4. identify the product formed when a given pair of esters is used in a mixed Claisen condensation.
5. identify the esters that should be used to produce a given β‑keto ester by a mixed Claisen condensation.
6. write an equation to illustrate the formation of a β‑diketone through a mixed Claisen‑type condensation between an ester and a ketone.
7. identify the β‑diketone formed as the result of a mixed Claisen‑type condensation between a given ester and a given ketone.
8. identify the reagents necessary to synthesize a given β‑diketone by a mixed Claisen‑type condensation between an ester and a ketone.
9. write detailed mechanisms for mixed Claisen reactions and reactions that are related to the mixed Claisen reaction, including those in which both reacting moieties are present in the same compound.
##### Study Notes
Just as we can react together two different aldehydes or ketones in a mixed aldol condensation, so can we react together two different esters in a mixed (or “crossed”) Claisen condensation. Again, by carefully selecting our substrates we can obtain a good yield of the desired product and minimize the number of possible by‑products. Note that even if we replace one of the esters with a ketone, the reaction is still referred to as Claisen condensation. The important thing to realize as you study these reactions is that they all take place by essentially the same mechanism—attack by an enolate anion on a carbonyl group.
## Crossed Claisen Condensation
Claisen condensations between different ester reactants are called Crossed Claisen reactions. Crossed Claisen reactions in which both reactants can serve as Claisen donors and Claisen acceptors generally give complex mixtures which are difficult to separate.
General Reaction
To avoid complex mixtures most Crossed Claisen reactions are usually not successful unless one of the two esters has no alpha-hydrogens as in the following examples.
This forces the ester with alpha-hydrogens to be the Claisen donor (enolate) and the ester without alpha-hydrogens to be the Claisen acceptor (electrophile). Even with this differentiation, it is possible for the ester with alpha-hydrogens to undergo a Claisen Condensation with itself to produce an unwanted side-product. This is typically partially prevented by using the ester without alpha-hydrogens in a large excess.
Example - no alpha hydrogens
Another type of crossed Claisen condensation occurs when a ketone is reacted with an ester. The use of an ester without alpha-hydrogens is not necessary due to the greater reactivity of the ketone. The alpha-hydrogens of the ketone are much more acidic (pKa~20) than those of the ester (pKa~25). The alpha hydrogens of ketone will be preferably deprotonated to make it the enolate Claisen donor. The ester will therefore be the electrophilic Claisen acceptor and is likewise commonly used in a large excess. There is still a possibility of the ketone reacting with itself to form the product of an aldol reaction however the equilibrium is unfavorable. Also, the formation of the aldol product is reversible due to its lack of an acidic alpha hydrogen. The beta-diketone Claisen product can be irreversibly deprotonated due to its more acidic alpha-hydrogens, making it the reaction’s main product.
Example - ester with ketone
Planing a Synthesis using a Claisen or Crossed Claisen-like Reaction
A Claisen reaction should be considered for a synthesis pathway if the target molecule contains a beta-keto esters, a beta-aldo ester or a beta-dicarbonyl.
A beta-keto ester or a beta-aldo ester could possibly be made by a Claisen condensation of two esters. A beta-dicarbonyl could possibly be made by a Claisen-like condensation between a ketone and an ester. Here, the key bond cleavage is a C-C bond between one of the carbonyls and the alpha-carbon which lies between the carbonyl. In each of these situations there are two such C-C bonds so there will be two possible pathways to consider. After the C-C bond cleavage the fragment with the alpha-carbon will gain a hydrogen. The fragment with the bare carbonyl will gain an –OR group to become an ester. The –OR group added will be the same as any ester present in the target molecule. If none are present –OCH2CH3 is typically used.
Analysis for Beta-Keto and Aldo Esters
A similar analysis can be performed on a 1,3-diketone and will be presented in the worked example below.
##### Worked Example
Show how the following molecule can be made using a Claisen-like condensation.
Pathway 1
Solution 1
Pathway 2
Solution 2
##### Exercises $$\PageIndex{1}$$
Draw the product of the following reactions:
a)
b) | 1,596 | 6,133 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2024-18 | latest | en | 0.532444 |
https://opentextbc.ca/conceptsofbiology1stcanadianedition/chapter/20-2-gas-exchange-across-respiratory-surfaces/ | 1,561,501,276,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627999948.3/warc/CC-MAIN-20190625213113-20190625235113-00252.warc.gz | 535,678,761 | 25,903 | Unit 4: Animal Structure and Function
# 20.2 Gas Exchange across Respiratory Surfaces
### Learning Objectives
By the end of this section, you will be able to:
• Name and describe lung volumes and capacities
• Understand how gas pressure influences how gases move into and out of the body
The structure of the lung maximizes its surface area to increase gas diffusion. Because of the enormous number of alveoli (approximately 300 million in each human lung), the surface area of the lung is very large (75 m2). Having such a large surface area increases the amount of gas that can diffuse into and out of the lungs.
## Basic Principles of Gas Exchange
Gas exchange during respiration occurs primarily through diffusion. Diffusion is a process in which transport is driven by a concentration gradient. Gas molecules move from a region of high concentration to a region of low concentration. Blood that is low in oxygen concentration and high in carbon dioxide concentration undergoes gas exchange with air in the lungs. The air in the lungs has a higher concentration of oxygen than that of oxygen-depleted blood and a lower concentration of carbon dioxide. This concentration gradient allows for gas exchange during respiration.
Partial pressure is a measure of the concentration of the individual components in a mixture of gases. The total pressure exerted by the mixture is the sum of the partial pressures of the components in the mixture. The rate of diffusion of a gas is proportional to its partial pressure within the total gas mixture. This concept is discussed further in detail below.
## Lung Volumes and Capacities
Different animals have different lung capacities based on their activities. Cheetahs have evolved a much higher lung capacity than humans; it helps provide oxygen to all the muscles in the body and allows them to run very fast. Elephants also have a high lung capacity. In this case, it is not because they run fast but because they have a large body and must be able to take up oxygen in accordance with their body size.
Human lung size is determined by genetics, gender, and height. At maximal capacity, an average lung can hold almost six liters of air, but lungs do not usually operate at maximal capacity. Air in the lungs is measured in terms of lung volumes and lung capacities (Figure 20.12 and Table 20.1). Volume measures the amount of air for one function (such as inhalation or exhalation). Capacity is any two or more volumes (for example, how much can be inhaled from the end of a maximal exhalation).
Table 20.1.
Lung Volumes and Capacities (Avg Adult Male)
Volume/Capacity Definition Volume (liters) Equations
Tidal volume (TV) Amount of air inhaled during a normal breath 0.5
Expiratory reserve volume (ERV) Amount of air that can be exhaled after a normal exhalation 1.2
Inspiratory reserve volume (IRV) Amount of air that can be further inhaled after a normal inhalation 3.1
Residual volume (RV) Air left in the lungs after a forced exhalation 1.2
Vital capacity (VC) Maximum amount of air that can be moved in or out of the lungs in a single respiratory cycle 4.8 ERV+TV+IRV
Inspiratory capacity (IC) Volume of air that can be inhaled in addition to a normal exhalation 3.6 TV+IRV
Functional residual capacity (FRC) Volume of air remaining after a normal exhalation 2.4 ERV+RV
Total lung capacity (TLC) Total volume of air in the lungs after a maximal inspiration 6.0 RV+ERV+TV+IRV
Forced expiratory volume (FEV1) How much air can be forced out of the lungs over a specific time period, usually one second ~4.1 to 5.5
The volume in the lung can be divided into four units: tidal volume, expiratory reserve volume, inspiratory reserve volume, and residual volume. Tidal volume (TV) measures the amount of air that is inspired and expired during a normal breath. On average, this volume is around one-half liter, which is a little less than the capacity of a 20-ounce drink bottle. The expiratory reserve volume (ERV) is the additional amount of air that can be exhaled after a normal exhalation. It is the reserve amount that can be exhaled beyond what is normal. Conversely, the inspiratory reserve volume (IRV) is the additional amount of air that can be inhaled after a normal inhalation. The residual volume (RV) is the amount of air that is left after expiratory reserve volume is exhaled. The lungs are never completely empty: There is always some air left in the lungs after a maximal exhalation. If this residual volume did not exist and the lungs emptied completely, the lung tissues would stick together and the energy necessary to re-inflate the lung could be too great to overcome. Therefore, there is always some air remaining in the lungs. Residual volume is also important for preventing large fluctuations in respiratory gases (O2 and CO2). The residual volume is the only lung volume that cannot be measured directly because it is impossible to completely empty the lung of air. This volume can only be calculated rather than measured.
## Respiratory Therapist
Respiratory therapists or respiratory practitioners evaluate and treat patients with lung and cardiovascular diseases. They work as part of a medical team to develop treatment plans for patients. Respiratory therapists may treat premature babies with underdeveloped lungs, patients with chronic conditions such as asthma, or older patients suffering from lung disease such as emphysema and chronic obstructive pulmonary disease (COPD). They may operate advanced equipment such as compressed gas delivery systems, ventilators, blood gas analyzers, and resuscitators. Specialized programs to become a respiratory therapist generally lead to a bachelor’s degree with a respiratory therapist specialty. Because of a growing aging population, career opportunities as a respiratory therapist are expected to remain strong.
## Gas Pressure and Respiration
The respiratory process can be better understood by examining the properties of gases. Gases move freely, but gas particles are constantly hitting the walls of their vessel, thereby producing gas pressure.
Air is a mixture of gases, primarily nitrogen (N2; 78.6 percent), oxygen (O2; 20.9 percent), water vapor (H2O; 0.5 percent), and carbon dioxide (CO2; 0.04 percent). Each gas component of that mixture exerts a pressure. The pressure for an individual gas in the mixture is the partial pressure of that gas. Approximately 21 percent of atmospheric gas is oxygen. Carbon dioxide, however, is found in relatively small amounts, 0.04 percent. The partial pressure for oxygen is much greater than that of carbon dioxide. The partial pressure of any gas can be calculated by:
(39.1)
P = (Patm) × (percent content in mixture).
Patm, the atmospheric pressure, is the sum of all of the partial pressures of the atmospheric gases added together,
(39.2)
Patm= PN2+ PO2+ PH2O+ PCO2= 760 mm Hg
× (percent content in mixture).
The pressure of the atmosphere at sea level is 760 mm Hg. Therefore, the partial pressure of oxygen is:
(39.3)
PO2= (760 mm Hg) (0.21) = 160 mm Hg
and for carbon dioxide:
(39.4)
PCO2= (760 mm Hg) (0.0004) = 0.3 mm Hg.
At high altitudes, Patm decreases but concentration does not change; the partial pressure decrease is due to the reduction in Patm.
When the air mixture reaches the lung, it has been humidified. The pressure of the water vapor in the lung does not change the pressure of the air, but it must be included in the partial pressure equation. For this calculation, the water pressure (47 mm Hg) is subtracted from the atmospheric pressure:
(39.5)
760 mm Hg − 47 mm Hg = 713 mm Hg
and the partial pressure of oxygen is:
(39.6)
(760 mm Hg − 47 mm Hg) × 0.21 = 150 mm Hg.
These pressures determine the gas exchange, or the flow of gas, in the system. Oxygen and carbon dioxide will flow according to their pressure gradient from high to low. Therefore, understanding the partial pressure of each gas will aid in understanding how gases move in the respiratory system.
## Gas Exchange across the Alveoli
In the body, oxygen is used by cells of the body’s tissues and carbon dioxide is produced as a waste product. The ratio of carbon dioxide production to oxygen consumption is the respiratory quotient (RQ). RQ varies between 0.7 and 1.0. If just glucose were used to fuel the body, the RQ would equal one. One mole of carbon dioxide would be produced for every mole of oxygen consumed. Glucose, however, is not the only fuel for the body. Protein and fat are also used as fuels for the body. Because of this, less carbon dioxide is produced than oxygen is consumed and the RQ is, on average, about 0.7 for fat and about 0.8 for protein.
The RQ is used to calculate the partial pressure of oxygen in the alveolar spaces within the lung, the alveolar POAbove, the partial pressure of oxygen in the lungs was calculated to be 150 mm Hg. However, lungs never fully deflate with an exhalation; therefore, the inspired air mixes with this residual air and lowers the partial pressure of oxygen within the alveoli. This means that there is a lower concentration of oxygen in the lungs than is found in the air outside the body. Knowing the RQ, the partial pressure of oxygen in the alveoli can be calculated:
(39.7)
With an RQ of 0.8 and a PCOin the alveoli of 40 mm Hg, the alveolar PO2
is equal to:
(39.8)
Notice that this pressure is less than the external air. Therefore, the oxygen will flow from the inspired air in the lung (PO2 = 150 mm Hg) into the bloodstream (PO2 = 100 mm Hg)
In the lungs, oxygen diffuses out of the alveoli and into the capillaries surrounding the alveoli. Oxygen (about 98 percent) binds reversibly to the respiratory pigment hemoglobin found in red blood cells (RBCs). RBCs carry oxygen to the tissues where oxygen dissociates from the hemoglobin and diffuses into the cells of the tissues. More specifically, alveolar POis higher in the alveoli (PALVO2 = 100 mm Hg) than blood PO2 (40 mm Hg) in the capillaries. Because this pressure gradient exists, oxygen diffuses down its pressure gradient, moving out of the alveoli and entering the blood of the capillaries where O2 binds to hemoglobin. At the same time, alveolar PCO2 is lower PALVO2 = 40 mm Hg than blood PCO2 = (45 mm Hg). CO2 diffuses down its pressure gradient, moving out of the capillaries and entering the alveoli.
Oxygen and carbon dioxide move independently of each other; they diffuse down their own pressure gradients. As blood leaves the lungs through the pulmonary veins, the venous PO2= 100 mm Hg, whereas the venous PCO2 = 40 mm Hg. As blood enters the systemic capillaries, the blood will lose oxygen and gain carbon dioxide because of the pressure difference of the tissues and blood. In systemic capillaries, PO2= 100 mm Hg, but in the tissue cells, PO2= 40 mm Hg. This pressure gradient drives the diffusion of oxygen out of the capillaries and into the tissue cells. At the same time, blood PCO2= 40 mm Hg and systemic tissue PCO2= 45 mm Hg. The pressure gradient drives CO2 out of tissue cells and into the capillaries. The blood returning to the lungs through the pulmonary arteries has a venous PO2= 40 mm Hg and a PCO2= 45 mm Hg. The blood enters the lung capillaries where the process of exchanging gases between the capillaries and alveoli begins again (Figure 20.13).
In short, the change in partial pressure from the alveoli to the capillaries drives the oxygen into the tissues and the carbon dioxide into the blood from the tissues. The blood is then transported to the lungs where differences in pressure in the alveoli result in the movement of carbon dioxide out of the blood into the lungs, and oxygen into the blood.
## Concept in Action
Watch this video to learn how to carry out spirometry.
## Summary
The lungs can hold a large volume of air, but they are not usually filled to maximal capacity. Lung volume measurements include tidal volume, expiratory reserve volume, inspiratory reserve volume, and residual volume. The sum of these equals the total lung capacity. Gas movement into or out of the lungs is dependent on the pressure of the gas. Air is a mixture of gases; therefore, the partial pressure of each gas can be calculated to determine how the gas will flow in the lung. The difference between the partial pressure of the gas in the air drives oxygen into the tissues and carbon dioxide out of the body.
### Exercises
1. Which of the following statements is false?
A) In the tissues, PO2 drops as blood passes from the arteries to the veins, while PCO2 increases.
B) Blood travels from the lungs to the heart to body tissues, then back to the heart, then the lungs.
C) Blood travels from the lungs to the heart to body tissues, then back to the lungs, then the heart.
D) PO2 is higher in air than in the lungs.
2. The inspiratory reserve volume measures the ________.
A) amount of air remaining in the lung after a maximal exhalation
B) amount of air that the lung holds
C) amount of air the can be further exhaled after a normal breath
D) amount of air that can be further inhaled after a normal breath
3. Of the following, which does not explain why the partial pressure of oxygen is lower in the lung than in the external air?
A) Air in the lung is humidified; therefore, water vapor pressure alters the pressure.
B) Carbon dioxide mixes with oxygen.
C) Oxygen is moved into the blood and is headed to the tissues.
D) Lungs exert a pressure on the air to reduce the oxygen pressure.
4. The total lung capacity is calculated using which of the following formulas?
A) residual volume + tidal volume + inspiratory reserve volume
B) residual volume + expiratory reserve volume + inspiratory reserve volume
C) expiratory reserve volume + tidal volume + inspiratory reserve volume
D) residual volume + expiratory reserve volume + tidal volume + inspiratory reserve volume
5. What does FEV1/FVC measure? What factors may affect FEV1/FVC?
FEV1/FVC measures the forced expiratory volume in one second in relation to the total forced vital capacity (the total amount of air that is exhaled from the lung from a maximal inhalation). This ratio changes with alterations in lung function that arise from diseases such as fibrosis, asthma, and COPD.
6. What is the reason for having residual volume in the lung?
If all the air in the lung were exhaled, then opening the alveoli for the next inspiration would be very difficult. This is because the tissues would stick together.
7. How can a decrease in the percent of oxygen in the air affect the movement of oxygen in the body?
Oxygen moves from the lung to the bloodstream to the tissues according to the pressure gradient. This is measured as the partial pressure of oxygen. If the amount of oxygen drops in the inspired air, there would be reduced partial pressure. This would decrease the driving force that moves the oxygen into the blood and into the tissues.
PO2 is also reduced at high elevations:
PO2 at high elevations is lower than at sea level because the total atmospheric pressure is less than atmospheric pressure at sea level.
8. If a patient has increased resistance in his or her lungs, how can this detected by a doctor? What does this mean?
A doctor can detect a restrictive disease using spirometry. By detecting the rate at which air can be expelled from the lung, a diagnosis of fibrosis or another restrictive disease can be made.
### Glossary
alveolarPO2
partial pressure of oxygen in the alveoli (usually around 100 mmHg)
expiratory reserve volume (ERV)
amount of additional air that can be exhaled after a normal exhalation
FEV1/FVC ratio
ratio of how much air can be forced out of the lung in one second to the total amount that is forced out of the lung; a measurement of lung function that can be used to detect disease states
forced expiratory volume (FEV)
(also, forced vital capacity) measure of how much air can be forced out of the lung from maximal inspiration over a specific amount of time
functional residual capacity (FRC)
expiratory reserve volume plus residual volume
functional vital capacity (FVC)
amount of air that can be forcibly exhaled after taking the deepest breath possible
inspiratory capacity (IC)
tidal volume plus inspiratory reserve volume
inspiratory reserve volume (IRV)
amount of additional air that can be inspired after a normal inhalation
lung capacity
measurement of two or more lung volumes (how much air can be inhaled from the end of an expiration to maximal capacity)
lung volume
measurement of air for one lung function (normal inhalation or exhalation)
oxygen-carrying capacity
amount of oxygen that can be transported in the blood
partial pressure
amount of pressure exerted by one gas within a mixture of gases
residual volume (RV)
amount of air remaining in the lung after a maximal expiration
respiratory quotient (RQ)
ratio of carbon dioxide production to each oxygen molecule consumed
spirometry
method to measure lung volumes and to diagnose lung diseases
tidal volume (TV)
amount of air that is inspired and expired during normal breathing
venousPCO2
partial pressure of carbon dioxide in the veins (40 mm Hg in the pulmonary veins)
venousPO2
partial pressure of oxygen in the veins (100 mm Hg in the pulmonary veins) | 3,977 | 17,251 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2019-26 | latest | en | 0.946713 |
http://oeis.org/A269304 | 1,582,872,358,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875147054.34/warc/CC-MAIN-20200228043124-20200228073124-00539.warc.gz | 106,298,252 | 4,434 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
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A269304 a(n) = n + n/gpf(n) + 1, where gpf(n) is the greatest prime factor of n or 1 if n = 1. 3
3, 4, 5, 7, 7, 9, 9, 13, 13, 13, 13, 17, 15, 17, 19, 25, 19, 25, 21, 25, 25, 25, 25, 33, 31, 29, 37, 33, 31, 37, 33, 49, 37, 37, 41, 49, 39, 41, 43, 49, 43, 49, 45, 49, 55, 49, 49, 65, 57, 61, 55, 57, 55, 73, 61, 65, 61, 61, 61, 73, 63, 65, 73, 97, 71, 73, 69 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 COMMENTS a(n) is odd except when n=2. Initially, a(n) is frequently a square or a prime. It is conjectured that any two sequences generated with a(n)=a(n-1)+a(n-1)/gpf(a(n-1))+1 and any initial value >=1 will have a finite number of non-shared terms and an infinite number of shared terms after one initial shared term (see A270807). Example: For a(1)=314, the sequence generated is 314, 317, 319, 331, 333, 343, 393, 397, 399, 421, 423, 433, ...; for a(1)=97, the sequence generated is 97, 99, 109, 111, 115, 121, 133, 141, 145, 151, 153, 163, 165, 181, 183, 187, 199, 201, 205, 211, 213, 217, 225, 271, 273, 295, 301, 309, 313, 315, 361, 381, 385, 421, 423, 433, ...; these sequences have respectively 9 and 33 terms not shared with the other until both reach 421; the following terms of both sequences are identical. LINKS Cody M. Haderlie, Table of n, a(n) for n = 1..10000 FORMULA a(n) = n + n/A006530(n) + 1. a(n) = n + A052126(n) + 1. a(p) = p+2 for p prime. EXAMPLE For n=18765, a(n)=18901. For n=196, a(n)=225 (225 is a square). For n=103156, a(n)=105673 (105673 is prime). MATHEMATICA Table[n+n/FactorInteger[n][[-1, 1]]+1, {n, 100}] PROG (PARI) gpf(n)=if(n>1, my(f=factor(n)[, 1]); f[#f], 1) a(n)=n + n/gpf(n) + 1 \\ Charles R Greathouse IV, Feb 22 2016 CROSSREFS Cf. A006530, A052126, A270807. Sequence in context: A049465 A196122 A247140 * A071054 A231346 A033545 Adjacent sequences: A269301 A269302 A269303 * A269305 A269306 A269307 KEYWORD nonn,easy,hear AUTHOR Cody M. Haderlie, Feb 22 2016 STATUS approved
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Last modified February 28 01:45 EST 2020. Contains 332319 sequences. (Running on oeis4.) | 970 | 2,472 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2020-10 | latest | en | 0.698795 |
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# Applied mathematical methods for chemical engineers, 2d ed.
9780849397783
Applied mathematical methods for chemical engineers, 2d ed.
Loney, Norman W.
CRC (Cyclical Redundancy Checking) An error checking technique used to ensure the accuracy of transmitting digital data. The transmitted messages are divided into predetermined lengths which, used as dividends, are divided by a fixed divisor. / Taylor & Francis
2007
454 pages
\$99.95
Hardcover
TP155
Loney (chemical engineering, New Jersey Institute of Technology) focuses on two important aspects of the application of mathematics to chemical engineering, namely setting up a mathematical model
Note: The term model has a different meaning in model theory, a branch of mathematical logic. An artifact which is used to illustrate a mathematical idea is also called a mathematical model and this usage is the reverse of the sense explained below.
and verifying the model with experimental or independently derived data. Using step-by-step examples he covers differential equations, first-order ordinary differential equations, linear second-order and systems of first-order differential equations, Sturm-Liouville problems, Fourier series Fourier series
In mathematics, an infinite series used to solve special types of differential equations. It consists of an infinite sum of sines and cosines, and because it is periodic (i.e.
and integrals, partial differential equations, applications of partial differential equations in chemical engineering, dimensional analysis and scaling of boundary value problems, selected numerical methods and available software packages. Each chapter includes problems and references, and an appendix includes a numerical method of lines The method of lines (MOL, NMOL, NUMOL) ( Schiesser, 1991; Hamdi, et al., 2007; ) is a technique for solving partial differential equations (PDEs) where all but one dimension is discretized. example using MATLAB (MATrix LABoratory) A programming language for technical computing from The MathWorks, Natick, MA (www.mathworks.com). Used for a wide variety of scientific and engineering calculations, especially for automatic control and signal processing, MATLAB runs on Windows, Mac and . For this edition Loney has added information on systems of first-order differential equations and the numerical method of lines.
([c]20062005 Book News, Inc., Portland, OR)
No portion of this article can be reproduced without the express written permission from the copyright holder.
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http://www.ashevillecozycabins.com/interest-rate-vs-apr-calculator/ | 1,606,159,212,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141164142.1/warc/CC-MAIN-20201123182720-20201123212720-00697.warc.gz | 114,064,845 | 7,867 | Interest Rate Vs Apr Calculator
Interest Rates: AER and APR explained – MoneySavingExpert – Interest rates indicate the price at which you can borrow money. It can get seriously complicated, with many anomalies, so for starters this guide covers the basics first. If you want to know all there is to know, including the difference between APR and AER, then step it up a notch and read to the.
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How to Calculate Interest on a Money Market Account – Divide the APR by 365 to find the daily interest rate. For example, if the APR is 2.92 percent, the daily rate is 2.92 divided by 365, or 0.008 percent. 2. Calculate interest for the first day of the.
They might be used interchangeably, but an APR and an interest rate aren’t one and the same. The annual percentage rate represents your total cost of getting a mortgage. The interest rate represents the cost you pay over time to buy that loan. Let’s take a look at the difference between your APR.
Interest Rate Vs Apr Calculator | Saglamtapu – How to Calculate APR vs. Interest Rate on a Loan – LendGenius – APR and interest rate are more closely related than either is with factor rate, so we’ll discuss those two first before describing how a factor rate might come into play. What is an APR vs. interest rate? The interest rate is the annual cost of a loan to a borrower expressed as a percentage of the principal loan amount. easy. | 629 | 2,971 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2020-50 | latest | en | 0.943284 |
https://calculator-integral.com/integral-of-sin-8x-cos-5x | 1,709,635,555,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707948234904.99/warc/CC-MAIN-20240305092259-20240305122259-00411.warc.gz | 151,242,732 | 14,985 | # Integral Of Sin 8x Cos 5x
Integral of 8x*cos 5x along with its formula and proof with examples. Also learn how to calculate integration of 8x*cos 5x with step by step examples.
Alan Walker-
Published on 2023-04-14
## Introduction to integral of sin 8x*cos 5x
In calculus, the integral is a fundamental concept that assigns numbers to functions to define displacement, area, volume, and all those functions that contain a combination of tiny elements. It is categorized into two parts, definite integral and indefinite integral. The process of integration calculates the integrals. This process is defined as finding an antiderivative of a function.
Integrals can handle almost all functions, such as trigonometric, algebraic, exponential, logarithmic, etc. This article will teach you what is integral to a trigonometric function sine. You will also understand how to compute sin 8xcos 5x integral by using different integration techniques.
## What is the integral of sin(8x)cos(5x)?
The integral of sin(8x)cos(5x) is an antiderivative of sine function which is equal to –cos(13x)/26 – cos(3x)/6. It is also known as the reverse derivative of sine function which is a trigonometric identity.
The sine function is the ratio of opposite side to the hypotenuse of a triangle which is written as:
Sin = opposite side / hypotenuse
### Integral of sin(8x)cos(5x) formula
The formula of integral of sin 8xcos 5x contains integral sign, coefficient of integration and the function as sin(8x)cos(5x). It is denoted by ∫(sin x.cos x)dx. In mathematical form, the integral of sin x is:
$∫\sin(8x).\cos(5x)dx=-\frac{1}{2}\left[\frac{\cos(13x)}{13}+\frac{\cos(3x)}{3}\right]+c$
Where c is any constant involved, dx is the coefficient of integration and ∫ is the symbol of integral. We can say that, by replacing n by 8 and m by 5, the integral of sin(nx)cos(mx) by using above formula.
## How to calculate the integral of sin(8x)cos(5x)?
The integral of sin(8x)cos(5x) is its antiderivative that can be calculated by using different integration techniques. In this article, we will discuss how to calculate integral of sine by using:
1. Trigonometric formulas
2. Substitution method
3. Definite integral
## Integral of sin(8x)cos(5x) by using Trigonometric formulas
The substitution method also involves trigonometric formulas that helps to solve integrals easily. Let’s understand how to calculate the integral of sin(8x)cos(5x) by using different trigonometric formulas.
### Proof of integral of sin(8x)cos(5x) by using Trigonometric formulas
To evaluate the integral. (use c for the constant of integration.) sin(8x) cos(5x) dx, we use different trigonometric formulas. Therefore,
$I = ∫\sin(8x)\cos(5x)dx$
Since we know the formulas of sum and product in trigonometry,
$\sin(8x)\cos(5x) = ½[\cos(3x) – \cos(13x)] = ½ [\cos(3x) – \cos(13x)]$
Therefore, Using this formula in the integral,
$I = ∫\sin(8x)\cos(5x)dx = ∫½ [\cos(3x) – \cos(13x)]dx$
Now we can integral each term separately, we get
$I = ½ \left[\frac{\sin3x}{3}–\frac{\sin13x}{13}\right] + c$
Hence the integral of sin(8x)cos(5x) is verified. The use of trigonometric idienties helps to evaluate integral by writing them in their alternative forms e.g, the integral of sin x cos x. can be calculated by replacing the integral by sin2x.
## Integral of sin(8x)cos(5x) by using substitution method
The substitution method involves many trigonometric formulas. We can use these formulas to verify the integrals of different trigonometric functions such as sine, cosine, tangent, etc. Let’s understand how to prove the integral of sin by using the substitution method calculator.
### Proof of Integral of sin(8x)cos(5x) by using substitution method
To proof the integral of sin(8x)cos(5x) by using substitution method, suppose that:
$I = ∫\sin 8x.\cos 5xdx$
Using trigonometric formula,
$\sin 8x.\cos 5x = ½ [\cos(3x) – \cos(13x)]$
Then the above integral can be written as:
$I = ∫½[\cos(3x)– \cos(13x)]dx$
Suppose that,
$u = 3x\quad\text{and}\quad du=3dx$
Similarly,
$t = 13x\quad\text{and}\quad dt = 13dx$
Now substituting the values of u and t in the integral,
$I = \frac{3}{2}∫\cos(u)du–\frac{13}{2}∫\cos(t)dt$
By integrating with respect to u and t, we get;
$I = \frac{3}{2}\sin(u) –\frac{13}{2}\cos(t) + c$
Now substituting the value of u and t back here,
$I =\frac{3}{2}\sin(3x) – \frac{13}{2}\cos(13x) + c$
Hence we have verified the integral of sin(8x)cos(5x) by using substitution method. If the integral is a nonlinear function, use our trigonometric substitution calculator to evaluate such integrals.
## Integral of sin(8x)cos(5x) by using definite integral
The definite integral is a type of integral that calculates the area of a curve by using infinitesimal area elements between two points. The definite integral can be written as:
$∫^b_a f(x) dx = F(b) – F(a)$
Let’s understand the verification of the integral of sin x by using the indefinite integral.
### Proof of integral of sin(8x)cos(5x) by using definite integral
To evaluate the integral. (use c for the constant of integration.) sin(8x) cos(5x) dx by using a definite integral, we can use the interval from 0 to π or 0 to π/2. Let’s compute the integral of sin x from 0 to π. For this we can write the integral as:
$∫^π_0 \sin(8x)\cos(5x) dx = \left|½[-\frac{\cos3x}{3} – \frac{\cos13x}{13}]\right|^π_0$
Now, substituting the limit in the given function.
$∫^π_0 \sin(8x)\cos(5x) dx = ½\left[-\frac{\cos3π}{3} – \frac{\cos13π}{13}\right]-\left[-\frac{\cos0}{3}-\frac{\cos0},{13}\right]$
Or,
$∫^π_0 \sin(8x)\cos(5x)dx=½\left[\frac{\cos3π}{3} –\frac{\cos13π}{13}\right]$
Since cos 0 is equal to 1 and cos π is equal to -1, therefore,
$∫^π_0 \sin(8x)\cos(5x) dx = ½ \left[-\frac{1}{3} + \frac{1}{13}\right]$
$∫^π_0 \sin(8x)\cos(5x) dx = -\frac{1}{6}+\frac{1}{26}= \frac{5}{39}$
Which is the calculation of the definite integral of sin(8x)cos(5x). Now to calculate the integral of sin x between the interval 0 to π/2, we just have to replace π by π/2. Therefore,
$∫^{\frac{π}{2}}_0 \sin(8x)\cos(5x) dx = ½ [\frac{\cos(3x)}{3} – \frac{\cos(13x}{13}]|^{\frac{π}{2}}_0$
Now,
$∫^{\frac{π}{2}}_0 \sin(8x)\cos(5x) dx = ½\left[\frac{1}{3}\left(-\cos3π/2 – \cos0\right)-\frac{1}{13}\left(-\cos 13\pi/2 - \cos0\right)\right]$
Or,
$∫^{\frac{π}{2}}_0 \sin(8x)\cos(5x) dx = ½\left[-\frac{cos3π/2}{3} – \frac{\cos13π/2}{13}\right]-\frac{5}{39}$
Since cos 0 is equal to 1 and cos π/2 is equal to 0, therefore,
$∫^{\frac{π}{2}}_0 \sin(8x)\cos(5x) dx = 0-\frac{5}{39}=-\frac{5}{39}$
Therefore, the definite integral of sin(8x)cos(5x) is equal to 0. | 2,131 | 6,625 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.6875 | 5 | CC-MAIN-2024-10 | latest | en | 0.882314 |
https://www.askiitians.com/forums/Mechanics/a-sphere-of-mass-m-collides-elastically-with-anoth_259558.htm | 1,611,550,582,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703564029.59/warc/CC-MAIN-20210125030118-20210125060118-00324.warc.gz | 659,364,083 | 35,318 | ×
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A sphere of mass m collides elastically with another stationary sphere of mass m/2. Both the spheres are smooth and there are no external forces acting on them. Maximum angle through which sphere of mass m can be deflected w.r.t. its initial direction of motion is
A sphere of mass m collides elastically with another stationary sphere of massm/2. Both the spheres are smooth and there are no external forces acting on them. Maximum anglethrough which sphere of mass m can be deflected w.r.t. its initial direction of motion is
```
10 months ago
Jiinamashkarivastav Hegde
14 Points
``` Tell your mom to F.D from a bank to admit you to a Pvt. College because you ain't cracking JEE with such poor skills in Physics. Don't take it otherwise...Buy:Gulzar Peer Physics Hc VermaBm sharma CengageHashish Aggarwal PhysicsDC Pandey Objective Gotta Work hard!
```
10 months ago
Manish kumar
25 Points
``` Tell your mom to F.D from a bank to admit you to a Pvt. College because you ain't cracking JEE with such poor skills in Physics. Don't take it otherwise...Buy:Gulzar Peer Physics Hc VermaBm sharma CengageHashish Aggarwal PhysicsDC Pandey Objective Gotta Work hard!thanks
```
10 months ago
Vikas TU
14149 Points
``` Dear student Here is the explanation The sphere A comes to rest after collision.The sphere B will move with a speed of 8m/s after collisionThe directions of motion A and B after collision are at right angles.The speed of B after collision is 4m/sGood Luck
```
10 months ago
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## Other Related Questions on Mechanics
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### Course Features
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• Mind Map
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• Previous Year Exam Questions | 595 | 2,336 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2021-04 | latest | en | 0.84301 |
https://blog.amiestudycircle.com/2021/10/analysis-and-design-of-structures-short.html | 1,721,693,758,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517927.60/warc/CC-MAIN-20240722220957-20240723010957-00265.warc.gz | 111,790,892 | 22,871 | ### Analysis and Design of Structures - short answer type questions from AMIE exams (Winter 2020)
Answer the following (4 x 5)
Explain various components of the Plate girder through a neat sketch.
The vertical plate of the plate girder is termed a web plate. The angles connected at the top and bottom of the web plate are known as flange angles. The horizontal plates connected with the flange angles are known as flange plates or cover plates. The web plate, flange angles and flange plates are shown in the following figure.
The bearing stiffeners, intermediate stiffeners and horizontal stiffeners used with the plate girder are shown in the following figure.
Compare HSFG bolts and bearing bolts.
The high strength bolts are also called high strength friction grip bolts. Briefly, these are called HSFG bolts. The high strength friction grip bolts have many advantages over rivets.
• Whereas the rivets are subjected to shear and bearing stresses, the bolts are subjected to uniform tensile stresses only.
• The high strength friction grip bolts have higher fatigue strength because there is no concentration of the stress in the hole.
• The bolts do not bear against the plates. Therefore, the uneven distribution of stress does not occur.
• These bolts are advantageously used in bridges and machine foundations subjected to vibrations.
• These bolts also simplify the problem of alterations and additions to structures as they can be assembled more easily than rivets.
Discuss Wind load calculation for the design of roof truss.
In order to determine the wind load on roofs and other inclined surfaces, the direction of the wind is assumed as horizontal. The roof surfaces are also inclined to the horizontal. The normal component and tangential component of wind on inclined roof surfaces may be found. It is to note that the tangential component of wind pressure is small and it is neglected. The normal component of wind pressure can be found from the empirical formula.
In order to find the wind load on individual structural elements/members (e.g., roofs and walls, and individual cladding units and their fittings, it is essential to consider the pressure difference between the opposite faces of such elements or units. For cladding, it is, therefore, necessary to know the internal pressure as well as the external pressure. The wind load, Wacting in a direction normal to the individual structural element or cladding unit may be determined from the following expression
where,
Cpe = external pressure coefficient
Cpi = internal pressure coefficient
A = surface area of a structural element or cladding unit, and
pd = design wind pressure
Enlist advantages and disadvantages of Bolted and Welded connections.
The advantages of bolted connections are as follows:
• There is silence in preparing bolted connection. In riveting, hammering is done. The hammering causes noise in the riveting.
• There is no risk of fire in bolted connection. The rivets are made red hot in riveting and there is a risk of fire.
• The bolted connections may be done quickly in comparison to the riveting.
• Though the cost of bolts is more than the cost of rivets, the bolted connections are economical to use than rivets because fewer persons are required for installation, and the work proceeds quickly.
• The bolted connections facilitate the erection because of the ease with which these connections can be done.
The following are disadvantages of bolted connections:
• If bolted connections become loose, their strength reduces considerably.
• Tire unfinished bolts are not uniform in diameter and they have less strength.
• The bolted connections have less strength when they are subjected to axial tension because the area at the root of the thread is less.
• Generally, the diameter of the hole is kept 16 mm more than the nominal diameter of the black bolt. The bolt does not fill the hole and there remains a clearance in bolted connections.
Advantages of riveted joints
• A riveted joint is more reliable than welded joints in applications which are subjected to vibrations and impact forces.
• Riveted joints can be used for non-ferrous metals like aluminium alloy, copper, brass or even non-metal like plastic and asbestos.
• Riveted joints are free from thermal after-effects because no heat required in this joint.
• Quality inspection is easy in a riveted joint.
• When the riveted joint is dismantled, the connected components are less damaged as compared to a welded joint.
Disadvantages of riveted joints
• The material cost of a riveted joint is more.
• The labour cost of riveted joints is also more than that of the welded joint.
• Overall cost if the riveted joint is also high.
• The riveted assembly has more weight than the welded assembly.
• The riveting process creates more noise because of hammer blows.
• Holes required to insert rivets cause stress concentration.
• Production time is more for assembly.
• Riveted assemblies are not tight and leak proof.
• The projection of the riveted head adversely affects the appearance of the riveted structure.
Kinematic indeterminacy
It is defined as the number of non-zero joint displacements of the structure. It is also called the Degree of freedom.
• A fixed ended beam is kinematically determinate as at both of its ends all the displacement components are known to be zero on compatibility consideration.
• A cantilever beam has three degrees of kinematic indeterminacy corresponding to the free end as three unknown displacements are likely to occur there.
• In case any axial change in length is neglected its kinematic indeterminacy will be only two.
In short, the total kinematic indeterminacy of a structure represents the sum of all the possible displacements that various joints of a structure can undergo
---
The study material for AMIE/B Tech/Junior Engineer exams is available at https://amiestudycircle.com | 1,198 | 5,902 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2024-30 | latest | en | 0.91677 |
https://mathematica.stackexchange.com/questions/15799/why-does-iterating-prime-in-reverse-order-require-much-more-time/15862 | 1,575,993,899,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540528457.66/warc/CC-MAIN-20191210152154-20191210180154-00326.warc.gz | 464,990,829 | 34,568 | # Why does iterating Prime in reverse order require much more time?
Say I would like to display the $10$ greatest primes that are less than $10^5$. I could do the following:
AbsoluteTiming[
M = 10^5; m = PrimePi[M];
prms = Prime[#] & /@ Range[1, m];
prms[[#]] & /@ Range[-1, -10, -1]
]
And the result comes out :
{0.0156250, {99991, 99989, 99971, 99961, 99929, 99923, 99907, 99901, 99881, 99877}}
But if I tried to do in in reverse,
AbsoluteTiming[
M = 10^5; m = PrimePi[M];
prms = Prime[#] & /@ Range[m, 1, -1];
prms[[#]] & /@ Range[1, 10]
]
the process takes a whole lot longer:
{0.6250000, {99991, 99989, 99971, 99961, 99929, 99923, 99907, 99901,
99881, 99877}}
Using the second method, I can't even increase M to $10^6$, as the program takes extremely long to execute. Can anybody offer some insight into this ? $\;$ Am I essentially not doing the same thing in both cases ?
• Probably Prime[n] makes use of results for Prime[m], m < n, but it caches only a limited number of results. But this is just a guess and you have surely though of it already. – Szabolcs Dec 5 '12 at 22:24
• So with a clear memory, Prime[n] runs in O(n) time.. – cartonn Dec 5 '12 at 22:34
• Prime and PrimePi are Listable, they are automatically mapped over a list, thus you can simply write e.g. Prime @ Range @ m instead of Prime[#] & /@ Range[1, m]. – Artes Jan 5 '13 at 18:03
Given a large n, to find k largest primes below n (as well as above) the best approach uses NextPrime (it has been added to Mathematica 6) :
NextPrime[n] gives the next prime above n.
NextPrime[n,k] gives the k-th prime above n. If k is negative it gives k-th largest prime below n.
k need not be a single number but it may be a list of integers, so if we are looking for k consecutive primes we can take advanted of Range, e.g. :
NextPrime[ 100000, Range[-10, -1]]
{99877, 99881, 99901, 99907, 99923, 99929, 99961, 99971, 99989, 99991}
The issue with Prime and PrimePi is that they are internally related however their documentation pages are not very informative. There are certain limitations of these functions (look at a related question : What is so special about Prime? ). Prime calls PrimePi (e.g. this comment by Oleksandr R.) if Prime[n] < 25 10^13. One can guess what is going on from Some Notes on Internal Implementation where it says:
Prime and PrimePi use sparse caching and sieving. For large $n$, the Lagarias-Miller-Odlyzko algorithm for PrimePi is used, based on asymptotic estimates of the density of primes, and is inverted to give Prime.
So if one has found a large prime, generically the system definitely has found some close primes too (sparse caching and sieving) and of course internal algorithms are not symmetric around a large $n$, i.e. finding closest $k$ primes below and above $n$ is not symmetric (basically it is implied by decreasing density of primes (globally) but directly it is determined by the Lagarias-Miller-Odlyzko method ). For more information take a look at this crucial reference : Computing $\pi(x)$: the Meissel-Lehmer method. If you want to find really large primes a fast algorithm should use PrimeQ however it is known to be correct only for $n < 10^{16}$. Another algorithm which is correct for all natural n is much slower, one can find it in PrimalityProving package .
• this helps a lot, thank you – cartonn Dec 7 '12 at 3:50
If you just want the greatest 10 primes less than M, you can start from Prime[PrimePi[M]-9]. By doing so, you gain a speed increase of 2 orders of magnitude when M = 100000.
M = 100000;
m = PrimePi[M]
AbsoluteTiming[Table[Prime[k], {k, m - 9, m}]]
9592
{0.000171, {99877, 99881, 99901, 99907, 99923, 99929, 99961, 99971, 99989, 99991}}
Now backwards (after closing and re-opening Mathematica and defining M and m.). This is 10 x slower than the above, but still faster than starting from Prime[1].
AbsoluteTiming[Table[Prime[k], {k, m, m - 9, -1}]]
{0.001467, {99991, 99989, 99971, 99961, 99929, 99923, 99907, 99901, 99881, 99877}}
One might expect that NextPrime would provide an advantage. We restart Mathematica and find out that it is only marginally better than our last attempt.
AbsoluteTiming[NestList[NextPrime, Prime[m - 9], 9]]
{0.001193, {99877, 99881, 99901, 99907, 99923, 99929, 99961, 99971, 99989, 99991}}
I conducted the following tests starting at Prime[1]. Even though this is a very inefficient approach, the results will be easier to compare to those of the OP.
We want to rule out Range so we take it out of the timing. Likewise, we set M and m first, before conducting the timing.
M = 100000;
m = PrimePi[M];
r1 = Range[1, m];
r2 = Range[m, 1, -1];
Backwards request of primes much slower than forwards request (the original observation)
r1 == Reverse[r2]
AbsoluteTiming[prms = Prime[#] & /@ r1;]
AbsoluteTiming[prms = Prime[#] & /@ r2;]
True
{0.009920, Null}
{0.354147, Null}
Puzzling. Yet absolutely consistent with what you observed.
It appears that r1 may be stored in a form that can be processed more quickly than r2.
Half backwards, Half forwards: Only half as bad as all backwards.
Let's look a little deeper.
r3 is the sub-list of odd numbers, followed by the even numbers.
r5 is like r3 but the odd numbers were reversed.
r3 = Transpose[Partition[r1, 2]] // Flatten;
r4 = Transpose[Partition[r1, 2]];
r5 = Join[Reverse[r4[[1]]], r4[[2]]];
Let's time the processing of r3 and r5:
AbsoluteTiming[prms = Prime[#] & /@ r3;]
AbsoluteTiming[prms = Prime[#] & /@ r5;]
{0.012765, Null}
{0.205590, Null}
Half the numbers in r5 were ordered from larger to smaller. The delay was about 1/2 the delay we saw when all numbers were ordered from larger to smaller.
Provisional conclusion: Mathematica "counts" primes forwards faster than backwards(?)
Random Ordering
r6 = RandomSample[r1, 9592];
AbsoluteTiming[prms = Prime[#] & /@ r6;]
{0.190770, Null}
The (pseudo)random ordering of numbers runs as fast as the backwards-forwards list!
Lingering Question: Why should a random ordering of numbers be faster to process than a "backwards request" for primes? I would have guessed that the random case would be at least as slow, or slower than the backward case.
P.S. @Szabolcs Please look at the results above and see whether they are consistent with your hunch. (I'm uncertain whether they are.) Perhaps Mathematica only stores the results of the final outcome. In this case the backwards request would receive no benefits.
Szabolcs' suggestion
Here are some orderings recommended by Szabolcs. Notice how the orderings increasingly move toward canonical ordering from least to greatest.
Table[{k, First@Timing[
Prime /@ Flatten@Reverse@Partition[Range[1, 20], k]]}, {k, 1, 10}];
Table[{k, Prime /@ Flatten@Reverse@Partition[Range[1, 20], k]}, {k, 1,10}]
And their respective timings:
Table[{k, First@Timing[Prime /@ Flatten@Reverse@Partition[Range[1, 20000], k];]}, {k, 1,10}]
{{1, 2.540567}, {2, 1.275608}, {3, 0.858899}, {4, 0.647391}, {5, 0.517372}, {6, 0.431639}, {7, 0.370851}, {8, 0.325734}, {9, 0.291674}, {10, 0.262256}}
Here's the canonical ordering:
First@Timing[Prime /@ Flatten@Partition[Range[1, 20000], 1];]
0.011894
• r1 and r2 I think you meant.. – cartonn Dec 5 '12 at 22:30
• Yes, thanks. I corrected it. – DavidC Dec 5 '12 at 22:56
• This kind of ordering is an interesting experiment too: Table[ {k, First@ Timing[Prime /@ Flatten@Reverse@Partition[Range[1, 20000], k];]}, {k, 1, 10} ] – Szabolcs Dec 6 '12 at 0:12
• Yes! Very nice. I'm comparing that output to First@Timing[Prime /@ Flatten@Partition[Range[1, 20000], 1];] – DavidC Dec 6 '12 at 0:22
• One would expect the random ordering to be the slowest, very interesting.. – cartonn Dec 6 '12 at 16:05 | 2,317 | 7,717 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2019-51 | latest | en | 0.881318 |
https://www.cram.com/essay/Case-Study-Of-AJ-Davis-Department-Store/P3N5Z65KGZKQ | 1,660,324,486,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571745.28/warc/CC-MAIN-20220812170436-20220812200436-00625.warc.gz | 642,814,970 | 12,192 | # Aj Davis Case Study Solution
Improved Essays
INTRODUCTION In business field, there are many uncertain situations business owners and managers would face. In order to make a decision about their businesses, the managers need analyzed information, which could be provided by statistic method.
AJ Davis department store, a chain store, wants to analysis the credit customers so the sample data of 50 credit customers was collected. The manager at the store has speculated some information about credit customers. By using the sample data of 50 credit customers, along with hypothesis testing and confidence intervals, we could determine whether manager’s speculations are convincing. Also, this report provides the summarization on the four variables of interests; income, location, number of years
…show more content…
The test of hypothesis at 0.05 significant levels and the determination of 95% confidence intervals are performed to evaluate the validity of the manager’s speculations. The results are reported in the following part.
Speculation A: The average (mean) annual income is less than \$50,000.
According to the statistical analysis of 50 sample customers, we can assume that the average annual income of the store credit customer is less than \$50,000. The sample mean of 50 customers is \$43,740 with sample standard deviation \$14,640. By estimating the determined confidence interval (see Appendix A), we can be 95% confident that the average annual income lies within the interval of \$39,580 and \$47,900. As the upper limit is less than \$50,000, it obviously supports the manager’s claim. In addition, the result of hypothesis test, both observed z-value and p-value, rejected the null hypothesis that the average income equals to \$50,000. Therefore, we agree that the average annual income is less than
…show more content…
According to the given sample data of 50 customers, there are 15 credit customers who live in suburban area with the sample mean of credit balance \$4,675, and the sample standard deviation at \$742. The result of statistical analysis in Appendix D shows that we can be 95% confident that the mean credit balance for suburban customers is between \$4,264 and \$5,086. As the lower limit of the interval is less than \$4,300, it not supports the manager’s claim. However, the result of hypothesis test rejects the null hypothesis, and provides convincing evidence that the mean credit balance for suburban customers is greater than \$4,300. Although these statistic methods contribute different conclusions, we agree with the outcome of hypothesis test because we used one-tailed test og hypothesis, which is consistent with the manager’s claim (explained in Appendix D). Therefore, we agree that the average of credit balance of suburban customers is more than
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# Information Theory – 1 – 1
1. Channel capacity is exactly equal to –
(a) bandwidth of demand
(b) Amount of information per second
(c) Noise rate in the demand
(d) None of the above
Explanation
No answer description available for this question. Let us discuss.
2. The capacity of a channel is :
(a) Number of digits used in coding
(b) Volume of information it can take
(c) Maximum rate of information transmission
(d) Bandwidth required for information
Explanation
No answer description available for this question. Let us discuss.
3. Entropy is basically a measure of :
(a) Rate of information
(b) Average of information
(c) Probability of information
(d) Disorder of information
Explanation
No answer description available for this question. Let us discuss.
4. The Hartley-Shannon theorem sets a limit on the :
(a) highest frequency that may be sent over a given channel
(b) maximum capacity of a channel with a given noise level
(c) maximum number of coding levels in a channel with a given noise level
(d) maximum number of quantizing levels in a channel of a given bandwidth
Explanation
No answer description available for this question. Let us discuss.
5. The maximum value of entropy is :
(a) 1
(b) 2
(c) 3
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Drake C++ Documentation
MatrixGain< T > Class Template Referencefinal
## Detailed Description
### template<typename T> class drake::systems::MatrixGain< T >
A system that specializes LinearSystem by setting coefficient matrices A, B, and C to all be zero.
Thus, the only non-zero coefficient matrix is D. Specifically, given an input signal u and a state x, the output of this system, y, is:
$y = D u$
u0→
MatrixGain
→ y0
Template Parameters
T The scalar type, which must be one of the default scalars.
AffineSystem
LinearSystem
#include <drake/systems/primitives/matrix_gain.h>
## Public Member Functions
MatrixGain (int size)
A constructor where the gain matrix D is a square identity matrix of size size. More...
MatrixGain (const Eigen::MatrixXd &D)
A constructor where the gain matrix D is D. More...
template<typename U >
MatrixGain (const MatrixGain< U > &)
Scalar-converting copy constructor. See System Scalar Conversion. More...
Does not allow copy, move, or assignment
MatrixGain (const MatrixGain &)=delete
MatrixGainoperator= (const MatrixGain &)=delete
MatrixGain (MatrixGain &&)=delete
MatrixGainoperator= (MatrixGain &&)=delete
Public Member Functions inherited from LinearSystem< T >
LinearSystem (const Eigen::Ref< const Eigen::MatrixXd > &A=Eigen::MatrixXd(), const Eigen::Ref< const Eigen::MatrixXd > &B=Eigen::MatrixXd(), const Eigen::Ref< const Eigen::MatrixXd > &C=Eigen::MatrixXd(), const Eigen::Ref< const Eigen::MatrixXd > &D=Eigen::MatrixXd(), double time_period=0.0)
Constructs a LinearSystem with a fixed set of coefficient matrices A, B,C, and D. More...
template<typename U >
LinearSystem (const LinearSystem< U > &)
Scalar-converting copy constructor. See System Scalar Conversion. More...
LinearSystem (const LinearSystem &)=delete
LinearSystemoperator= (const LinearSystem &)=delete
LinearSystem (LinearSystem &&)=delete
LinearSystemoperator= (LinearSystem &&)=delete
Public Member Functions inherited from AffineSystem< T >
AffineSystem (const Eigen::Ref< const Eigen::MatrixXd > &A=Eigen::MatrixXd(), const Eigen::Ref< const Eigen::MatrixXd > &B=Eigen::MatrixXd(), const Eigen::Ref< const Eigen::VectorXd > &f0=Eigen::VectorXd(), const Eigen::Ref< const Eigen::MatrixXd > &C=Eigen::MatrixXd(), const Eigen::Ref< const Eigen::MatrixXd > &D=Eigen::MatrixXd(), const Eigen::Ref< const Eigen::VectorXd > &y0=Eigen::VectorXd(), double time_period=0.0)
Constructs an Affine system with a fixed set of coefficient matrices A, B,C, and D as well as fixed initial velocity offset xDot0 and output offset y0. More...
template<typename U >
AffineSystem (const AffineSystem< U > &)
Scalar-converting copy constructor. See System Scalar Conversion. More...
void UpdateCoefficients (const Eigen::Ref< const Eigen::MatrixXd > &A, const Eigen::Ref< const Eigen::MatrixXd > &B, const Eigen::Ref< const Eigen::VectorXd > &f0, const Eigen::Ref< const Eigen::MatrixXd > &C, const Eigen::Ref< const Eigen::MatrixXd > &D, const Eigen::Ref< const Eigen::VectorXd > &y0)
Updates the coefficients of the affine system. More...
AffineSystem (const AffineSystem &)=delete
AffineSystemoperator= (const AffineSystem &)=delete
AffineSystem (AffineSystem &&)=delete
AffineSystemoperator= (AffineSystem &&)=delete
const Eigen::MatrixXd & A () const
const Eigen::MatrixXd & B () const
const Eigen::VectorXd & f0 () const
const Eigen::MatrixXd & C () const
const Eigen::MatrixXd & D () const
const Eigen::VectorXd & y0 () const
MatrixX< T > A (const T &) const final
MatrixX< T > B (const T &) const final
VectorX< T > f0 (const T &) const final
MatrixX< T > C (const T &) const final
MatrixX< T > D (const T &) const final
VectorX< T > y0 (const T &) const final
Public Member Functions inherited from TimeVaryingAffineSystem< T >
const InputPort< T > & get_input_port () const
Returns the input port containing the externally applied input. More...
const OutputPort< T > & get_output_port () const
Returns the output port containing the output state. More...
void configure_default_state (const Eigen::Ref< const VectorX< T >> &x0)
Configures the value that will be assigned to the state vector in SetDefaultContext. More...
void configure_random_state (const Eigen::Ref< const Eigen::MatrixXd > &covariance)
Configures the Gaussian distribution over state vectors used in the SetRandomContext methods. More...
const VectorX< T > & get_default_state () const
Returns the configured default state. More...
const Eigen::MatrixXd get_random_state_covariance () const
Returns the configured random state covariance. More...
double time_period () const
int num_states () const
int num_inputs () const
int num_outputs () const
TimeVaryingAffineSystem (const TimeVaryingAffineSystem &)=delete
TimeVaryingAffineSystemoperator= (const TimeVaryingAffineSystem &)=delete
TimeVaryingAffineSystem (TimeVaryingAffineSystem &&)=delete
TimeVaryingAffineSystemoperator= (TimeVaryingAffineSystem &&)=delete
Public Member Functions inherited from LeafSystem< T >
~LeafSystem () override
std::unique_ptr< LeafContext< T > > AllocateContext () const
Shadows System<T>::AllocateContext to provide a more concrete return type LeafContext<T>. More...
std::unique_ptr< ContextBaseDoAllocateContext () const final
Derived class implementations should allocate a suitable concrete Context type, then invoke the above InitializeContextBase() method. More...
void SetDefaultParameters (const Context< T > &context, Parameters< T > *parameters) const override
Default implementation: sets all numeric parameters to the model vector given to DeclareNumericParameter, or else if no model was provided sets the numeric parameter to one. More...
std::unique_ptr< ContinuousState< T > > AllocateTimeDerivatives () const final
Returns a ContinuousState of the same size as the continuous_state allocated in CreateDefaultContext. More...
std::unique_ptr< DiscreteValues< T > > AllocateDiscreteVariables () const final
Returns a DiscreteValues of the same dimensions as the discrete_state allocated in CreateDefaultContext. More...
std::multimap< int, intGetDirectFeedthroughs () const final
Reports all direct feedthroughs from input ports to output ports. More...
LeafSystem (const LeafSystem &)=delete
LeafSystemoperator= (const LeafSystem &)=delete
LeafSystem (LeafSystem &&)=delete
LeafSystemoperator= (LeafSystem &&)=delete
Public Member Functions inherited from System< T >
~System () override
virtual void Accept (SystemVisitor< T > *v) const
Implements a visitor pattern. More...
void GetWitnessFunctions (const Context< T > &context, std::vector< const WitnessFunction< T > * > *w) const
Gets the witness functions active for the given state. More...
CalcWitnessValue (const Context< T > &context, const WitnessFunction< T > &witness_func) const
Evaluates a witness function at the given context. More...
DependencyTicket discrete_state_ticket (DiscreteStateIndex index) const
Returns a ticket indicating dependence on a particular discrete state variable xdᵢ (may be a vector). More...
DependencyTicket abstract_state_ticket (AbstractStateIndex index) const
Returns a ticket indicating dependence on a particular abstract state variable xaᵢ. More...
DependencyTicket numeric_parameter_ticket (NumericParameterIndex index) const
Returns a ticket indicating dependence on a particular numeric parameter pnᵢ (may be a vector). More...
DependencyTicket abstract_parameter_ticket (AbstractParameterIndex index) const
Returns a ticket indicating dependence on a particular abstract parameter paᵢ. More...
DependencyTicket input_port_ticket (InputPortIndex index) const
Returns a ticket indicating dependence on input port uᵢ indicated by index. More...
DependencyTicket cache_entry_ticket (CacheIndex index) const
Returns a ticket indicating dependence on the cache entry indicated by index. More...
System (const System &)=delete
Systemoperator= (const System &)=delete
System (System &&)=delete
Systemoperator= (System &&)=delete
std::unique_ptr< Context< T > > AllocateContext () const
Returns a Context<T> suitable for use with this System<T>. More...
std::unique_ptr< CompositeEventCollection< T > > AllocateCompositeEventCollection () const
Allocates a CompositeEventCollection for this system. More...
std::unique_ptr< BasicVector< T > > AllocateInputVector (const InputPort< T > &input_port) const
Given an input port, allocates the vector storage. More...
std::unique_ptr< AbstractValueAllocateInputAbstract (const InputPort< T > &input_port) const
Given an input port, allocates the abstract storage. More...
std::unique_ptr< SystemOutput< T > > AllocateOutput () const
Returns a container that can hold the values of all of this System's output ports. More...
VectorX< T > AllocateImplicitTimeDerivativesResidual () const
Returns an Eigen VectorX suitable for use as the output argument to the CalcImplicitTimeDerivativesResidual() method. More...
std::unique_ptr< Context< T > > CreateDefaultContext () const
This convenience method allocates a context using AllocateContext() and sets its default values using SetDefaultContext(). More...
void SetDefaultContext (Context< T > *context) const
Sets Context fields to their default values. More...
virtual void SetRandomParameters (const Context< T > &context, Parameters< T > *parameters, RandomGenerator *generator) const
Assigns random values to all parameters. More...
void SetRandomContext (Context< T > *context, RandomGenerator *generator) const
Sets Context fields to random values. More...
void AllocateFixedInputs (Context< T > *context) const
For each input port, allocates a fixed input of the concrete type that this System requires, and binds it to the port, disconnecting any prior input. More...
bool HasAnyDirectFeedthrough () const
Returns true if any of the inputs to the system might be directly fed through to any of its outputs and false otherwise. More...
bool HasDirectFeedthrough (int output_port) const
Returns true if there might be direct-feedthrough from any input port to the given output_port, and false otherwise. More...
bool HasDirectFeedthrough (int input_port, int output_port) const
Returns true if there might be direct-feedthrough from the given input_port to the given output_port, and false otherwise. More...
virtual std::multimap< int, intGetDirectFeedthroughs () const=0
Reports all direct feedthroughs from input ports to output ports. More...
EventStatus Publish (const Context< T > &context, const EventCollection< PublishEvent< T >> &events) const
This method is the public entry point for dispatching all publish event handlers. More...
void ForcedPublish (const Context< T > &context) const
(Advanced) Manually triggers any PublishEvent that has trigger type kForced. More...
const ContinuousState< T > & EvalTimeDerivatives (const Context< T > &context) const
Returns a reference to the cached value of the continuous state variable time derivatives, evaluating first if necessary using CalcTimeDerivatives(). More...
const CacheEntryget_time_derivatives_cache_entry () const
(Advanced) Returns the CacheEntry used to cache time derivatives for EvalTimeDerivatives(). More...
const T & EvalPotentialEnergy (const Context< T > &context) const
Returns a reference to the cached value of the potential energy (PE), evaluating first if necessary using CalcPotentialEnergy(). More...
const T & EvalKineticEnergy (const Context< T > &context) const
Returns a reference to the cached value of the kinetic energy (KE), evaluating first if necessary using CalcKineticEnergy(). More...
const T & EvalConservativePower (const Context< T > &context) const
Returns a reference to the cached value of the conservative power (Pc), evaluating first if necessary using CalcConservativePower(). More...
const T & EvalNonConservativePower (const Context< T > &context) const
Returns a reference to the cached value of the non-conservative power (Pnc), evaluating first if necessary using CalcNonConservativePower(). More...
template<template< typename > class Vec = BasicVector>
const Vec< T > * EvalVectorInput (const Context< T > &context, int port_index) const
Returns the value of the vector-valued input port with the given port_index as a BasicVector or a specific subclass Vec derived from BasicVector. More...
Adds an "external" constraint to this System. More...
void CalcTimeDerivatives (const Context< T > &context, ContinuousState< T > *derivatives) const
Calculates the time derivatives ẋ꜀ of the continuous state x꜀ into a given output argument. More...
void CalcImplicitTimeDerivativesResidual (const Context< T > &context, const ContinuousState< T > &proposed_derivatives, EigenPtr< VectorX< T >> residual) const
Evaluates the implicit form of the System equations and returns the residual. More...
EventStatus CalcDiscreteVariableUpdate (const Context< T > &context, const EventCollection< DiscreteUpdateEvent< T >> &events, DiscreteValues< T > *discrete_state) const
This method is the public entry point for dispatching all discrete variable update event handlers. More...
void ApplyDiscreteVariableUpdate (const EventCollection< DiscreteUpdateEvent< T >> &events, DiscreteValues< T > *discrete_state, Context< T > *context) const
Given the discrete_state results of a previous call to CalcDiscreteVariableUpdate() that dispatched the given collection of events, modifies the context to reflect the updated discrete_state. More...
void CalcForcedDiscreteVariableUpdate (const Context< T > &context, DiscreteValues< T > *discrete_state) const
(Advanced) Manually triggers any DiscreteUpdateEvent that has trigger type kForced. More...
EventStatus CalcUnrestrictedUpdate (const Context< T > &context, const EventCollection< UnrestrictedUpdateEvent< T >> &events, State< T > *state) const
This method is the public entry point for dispatching all unrestricted update event handlers. More...
void ApplyUnrestrictedUpdate (const EventCollection< UnrestrictedUpdateEvent< T >> &events, State< T > *state, Context< T > *context) const
Given the state results of a previous call to CalcUnrestrictedUpdate() that dispatched the given collection of events, modifies the context to reflect the updated state. More...
void CalcForcedUnrestrictedUpdate (const Context< T > &context, State< T > *state) const
(Advanced) Manually triggers any UnrestrictedUpdateEvent that has trigger type kForced. More...
CalcNextUpdateTime (const Context< T > &context, CompositeEventCollection< T > *events) const
This method is called by a Simulator during its calculation of the size of the next continuous step to attempt. More...
void GetPeriodicEvents (const Context< T > &context, CompositeEventCollection< T > *events) const
Returns all periodic events in this System. More...
void GetPerStepEvents (const Context< T > &context, CompositeEventCollection< T > *events) const
This method is called by Simulator::Initialize() to gather all update and publish events that are to be handled in AdvanceTo() at the point before Simulator integrates continuous state. More...
void GetInitializationEvents (const Context< T > &context, CompositeEventCollection< T > *events) const
This method is called by Simulator::Initialize() to gather all update and publish events that need to be handled at initialization before the simulator starts integration. More...
void ExecuteInitializationEvents (Context< T > *context) const
This method triggers all of the initialization events returned by GetInitializationEvents(). More...
std::optional< PeriodicEventDataGetUniquePeriodicDiscreteUpdateAttribute () const
Determines whether there exists a unique periodic timing (offset and period) that triggers one or more discrete update events (and, if so, returns that unique periodic timing). More...
const DiscreteValues< T > & EvalUniquePeriodicDiscreteUpdate (const Context< T > &context) const
If this System contains a unique periodic timing for discrete update events, this function executes the handlers for those periodic events to determine what their effect would be. More...
bool IsDifferenceEquationSystem (double *time_period=nullptr) const
Returns true iff the state dynamics of this system are governed exclusively by a difference equation on a single discrete state group and with a unique periodic update (having zero offset). More...
bool IsDifferentialEquationSystem () const
Returns true iff the state dynamics of this system are governed exclusively by a differential equation. More...
std::map< PeriodicEventData, std::vector< const Event< T > * >, PeriodicEventDataComparatorMapPeriodicEventsByTiming (const Context< T > *context=nullptr) const
Maps all periodic triggered events for a System, organized by timing. More...
void CalcOutput (const Context< T > &context, SystemOutput< T > *outputs) const
Utility method that computes for every output port i the value y(i) that should result from the current contents of the given Context. More...
CalcPotentialEnergy (const Context< T > &context) const
Calculates and returns the potential energy represented by the current configuration provided in context. More...
CalcKineticEnergy (const Context< T > &context) const
Calculates and returns the kinetic energy represented by the current configuration and velocity provided in context. More...
CalcConservativePower (const Context< T > &context) const
Calculates and returns the conservative power represented by the current contents of the given context. More...
CalcNonConservativePower (const Context< T > &context) const
Calculates and returns the non-conservative power represented by the current contents of the given context. More...
void MapVelocityToQDot (const Context< T > &context, const VectorBase< T > &generalized_velocity, VectorBase< T > *qdot) const
Transforms a given generalized velocity v to the time derivative qdot of the generalized configuration q taken from the supplied Context. More...
void MapVelocityToQDot (const Context< T > &context, const Eigen::Ref< const VectorX< T >> &generalized_velocity, VectorBase< T > *qdot) const
Transforms the given generalized velocity to the time derivative of generalized configuration. More...
void MapQDotToVelocity (const Context< T > &context, const VectorBase< T > &qdot, VectorBase< T > *generalized_velocity) const
Transforms the time derivative qdot of the generalized configuration q to generalized velocities v. More...
void MapQDotToVelocity (const Context< T > &context, const Eigen::Ref< const VectorX< T >> &qdot, VectorBase< T > *generalized_velocity) const
Transforms the given time derivative qdot of generalized configuration q to generalized velocity v. More...
const Context< T > & GetSubsystemContext (const System< T > &subsystem, const Context< T > &context) const
Returns a const reference to the subcontext that corresponds to the contained System subsystem. More...
Context< T > & GetMutableSubsystemContext (const System< T > &subsystem, Context< T > *context) const
Returns a mutable reference to the subcontext that corresponds to the contained System subsystem. More...
const Context< T > & GetMyContextFromRoot (const Context< T > &root_context) const
Returns the const Context for this subsystem, given a root context. More...
Context< T > & GetMyMutableContextFromRoot (Context< T > *root_context) const
Returns the mutable subsystem context for this system, given a root context. More...
const InputPort< T > & get_input_port (int port_index) const
Returns the typed input port at index port_index. More...
const InputPort< T > & get_input_port () const
Convenience method for the case of exactly one input port. More...
const InputPort< T > * get_input_port_selection (std::variant< InputPortSelection, InputPortIndex > port_index) const
Returns the typed input port specified by the InputPortSelection or by the InputPortIndex. More...
const InputPort< T > & GetInputPort (const std::string &port_name) const
Returns the typed input port with the unique name port_name. More...
bool HasInputPort (const std::string &port_name) const
Returns true iff the system has an InputPort of the given port_name. More...
const OutputPort< T > & get_output_port (int port_index) const
Returns the typed output port at index port_index. More...
const OutputPort< T > & get_output_port () const
Convenience method for the case of exactly one output port. More...
const OutputPort< T > * get_output_port_selection (std::variant< OutputPortSelection, OutputPortIndex > port_index) const
Returns the typed output port specified by the OutputPortSelection or by the OutputPortIndex. More...
const OutputPort< T > & GetOutputPort (const std::string &port_name) const
Returns the typed output port with the unique name port_name. More...
bool HasOutputPort (const std::string &port_name) const
Returns true iff the system has an OutputPort of the given port_name. More...
int num_constraints () const
Returns the number of constraints specified for the system. More...
const SystemConstraint< T > & get_constraint (SystemConstraintIndex constraint_index) const
Returns the constraint at index constraint_index. More...
boolean< T > CheckSystemConstraintsSatisfied (const Context< T > &context, double tol) const
Returns true if context satisfies all of the registered SystemConstraints with tolerance tol. More...
VectorX< T > CopyContinuousStateVector (const Context< T > &context) const
Returns a copy of the continuous state vector x꜀ into an Eigen vector. More...
std::string GetMemoryObjectName () const
Returns a name for this System based on a stringification of its type name and memory address. More...
int num_input_ports () const
Returns the number of input ports currently allocated in this System. More...
int num_output_ports () const
Returns the number of output ports currently allocated in this System. More...
void FixInputPortsFrom (const System< double > &other_system, const Context< double > &other_context, Context< T > *target_context) const
Fixes all of the input ports in target_context to their current values in other_context, as evaluated by other_system. More...
const SystemScalarConverterget_system_scalar_converter () const
(Advanced) Returns the SystemScalarConverter for this object. More...
std::string GetGraphvizString (std::optional< int > max_depth={}, const std::map< std::string, std::string > &options={}) const
Returns a Graphviz string describing this System. More...
std::unique_ptr< System< T > > Clone () const
Creates a deep copy of this system. More...
std::unique_ptr< System< AutoDiffXd > > ToAutoDiffXd () const
Creates a deep copy of this System, transmogrified to use the autodiff scalar type, with a dynamic-sized vector of partial derivatives. More...
std::unique_ptr< System< AutoDiffXd > > ToAutoDiffXdMaybe () const
Creates a deep copy of this system exactly like ToAutoDiffXd(), but returns nullptr if this System does not support autodiff, instead of throwing an exception. More...
std::unique_ptr< System< symbolic::Expression > > ToSymbolic () const
Creates a deep copy of this System, transmogrified to use the symbolic scalar type. More...
std::unique_ptr< System< symbolic::Expression > > ToSymbolicMaybe () const
Creates a deep copy of this system exactly like ToSymbolic(), but returns nullptr if this System does not support symbolic, instead of throwing an exception. More...
template<typename U >
std::unique_ptr< System< U > > ToScalarType () const
Creates a deep copy of this System, transmogrified to use the scalar type selected by a template parameter. More...
template<typename U >
std::unique_ptr< System< U > > ToScalarTypeMaybe () const
Creates a deep copy of this system exactly like ToScalarType(), but returns nullptr if this System does not support the destination type, instead of throwing an exception. More...
Public Member Functions inherited from SystemBase
~SystemBase () override
void set_name (const std::string &name)
Sets the name of the system. More...
const std::string & get_name () const
Returns the name last supplied to set_name(), if any. More...
std::string GetMemoryObjectName () const
Returns a name for this System based on a stringification of its type name and memory address. More...
const std::string & GetSystemName () const final
Returns a human-readable name for this system, for use in messages and logging. More...
std::string GetSystemPathname () const final
Generates and returns a human-readable full path name of this subsystem, for use in messages and logging. More...
std::string GetSystemType () const final
Returns the most-derived type of this concrete System object as a human-readable string suitable for use in error messages. More...
std::unique_ptr< ContextBaseAllocateContext () const
Returns a Context suitable for use with this System. More...
int num_input_ports () const
Returns the number of input ports currently allocated in this System. More...
int num_output_ports () const
Returns the number of output ports currently allocated in this System. More...
const InputPortBaseget_input_port_base (InputPortIndex port_index) const
Returns a reference to an InputPort given its port_index. More...
const OutputPortBaseget_output_port_base (OutputPortIndex port_index) const
Returns a reference to an OutputPort given its port_index. More...
int num_total_inputs () const
Returns the total dimension of all of the vector-valued input ports (as if they were muxed). More...
int num_total_outputs () const
Returns the total dimension of all of the vector-valued output ports (as if they were muxed). More...
int num_cache_entries () const
Returns the number nc of cache entries currently allocated in this System. More...
const CacheEntryget_cache_entry (CacheIndex index) const
Returns a reference to a CacheEntry given its index. More...
CacheEntryget_mutable_cache_entry (CacheIndex index)
(Advanced) Returns a mutable reference to a CacheEntry given its index. More...
int num_continuous_states () const
Returns the number of declared continuous state variables. More...
int num_discrete_state_groups () const
Returns the number of declared discrete state groups (each group is a vector-valued discrete state variable). More...
int num_abstract_states () const
Returns the number of declared abstract state variables. More...
int num_numeric_parameter_groups () const
Returns the number of declared numeric parameters (each of these is a vector-valued parameter). More...
int num_abstract_parameters () const
Returns the number of declared abstract parameters. More...
int implicit_time_derivatives_residual_size () const
Returns the size of the implicit time derivatives residual vector. More...
void ValidateContext (const ContextBase &context) const final
Checks whether the given context was created for this system. More...
void ValidateContext (const ContextBase *context) const
Checks whether the given context was created for this system. More...
template<class Clazz >
void ValidateCreatedForThisSystem (const Clazz &object) const
Checks whether the given object was created for this system. More...
SystemBase (const SystemBase &)=delete
SystemBaseoperator= (const SystemBase &)=delete
SystemBase (SystemBase &&)=delete
SystemBaseoperator= (SystemBase &&)=delete
std::string GetGraphvizString (std::optional< int > max_depth={}, const std::map< std::string, std::string > &options={}) const
Returns a Graphviz string describing this System. More...
GraphvizFragment GetGraphvizFragment (std::optional< int > max_depth={}, const std::map< std::string, std::string > &options={}) const
(Advanced) Like GetGraphvizString() but does not wrap the string in a digraph { … }. More...
const AbstractValueEvalAbstractInput (const ContextBase &context, int port_index) const
Returns the value of the input port with the given port_index as an AbstractValue, which is permitted for ports of any type. More...
template<typename V >
const V * EvalInputValue (const ContextBase &context, int port_index) const
Returns the value of an abstract-valued input port with the given port_index as a value of known type V. More...
DependencyTicket discrete_state_ticket (DiscreteStateIndex index) const
Returns a ticket indicating dependence on a particular discrete state variable xdᵢ (may be a vector). More...
DependencyTicket abstract_state_ticket (AbstractStateIndex index) const
Returns a ticket indicating dependence on a particular abstract state variable xaᵢ. More...
DependencyTicket numeric_parameter_ticket (NumericParameterIndex index) const
Returns a ticket indicating dependence on a particular numeric parameter pnᵢ (may be a vector). More...
DependencyTicket abstract_parameter_ticket (AbstractParameterIndex index) const
Returns a ticket indicating dependence on a particular abstract parameter paᵢ. More...
DependencyTicket input_port_ticket (InputPortIndex index) const
Returns a ticket indicating dependence on input port uᵢ indicated by index. More...
DependencyTicket cache_entry_ticket (CacheIndex index) const
Returns a ticket indicating dependence on the cache entry indicated by index. More...
DependencyTicket output_port_ticket (OutputPortIndex index) const
(Internal use only) Returns a ticket indicating dependence on the output port indicated by index. More...
Public Types inherited from System< T >
using Scalar = T
The scalar type with which this System was instantiated. More...
Static Public Member Functions inherited from LinearSystem< T >
static std::unique_ptr< LinearSystem< T > > MakeLinearSystem (const Eigen::Ref< const VectorX< symbolic::Expression >> &dynamics, const Eigen::Ref< const VectorX< symbolic::Expression >> &output, const Eigen::Ref< const VectorX< symbolic::Variable >> &state_vars, const Eigen::Ref< const VectorX< symbolic::Variable >> &input_vars, double time_period=0.0)
Creates a unique pointer to LinearSystem<T> by decomposing dynamics and outputs using state_vars and input_vars. More...
Static Public Member Functions inherited from AffineSystem< T >
static std::unique_ptr< AffineSystem< T > > MakeAffineSystem (const Eigen::Ref< const VectorX< symbolic::Expression >> &dynamics, const Eigen::Ref< const VectorX< symbolic::Expression >> &output, const Eigen::Ref< const VectorX< symbolic::Variable >> &state_vars, const Eigen::Ref< const VectorX< symbolic::Variable >> &input_vars, double time_period=0.0)
Creates a unique pointer to AffineSystem<T> by decomposing dynamics and outputs using state_vars and input_vars. More...
Static Public Member Functions inherited from System< T >
static DependencyTicket nothing_ticket ()
Returns a ticket indicating that a computation does not depend on any source value; that is, it is a constant. More...
static DependencyTicket time_ticket ()
Returns a ticket indicating dependence on time. More...
static DependencyTicket accuracy_ticket ()
Returns a ticket indicating dependence on the accuracy setting in the Context. More...
static DependencyTicket q_ticket ()
Returns a ticket indicating that a computation depends on configuration state variables q. More...
static DependencyTicket v_ticket ()
Returns a ticket indicating dependence on velocity state variables v. More...
static DependencyTicket z_ticket ()
Returns a ticket indicating dependence on any or all of the miscellaneous continuous state variables z. More...
static DependencyTicket xc_ticket ()
Returns a ticket indicating dependence on all of the continuous state variables q, v, or z. More...
static DependencyTicket xd_ticket ()
Returns a ticket indicating dependence on all of the numerical discrete state variables, in any discrete variable group. More...
static DependencyTicket xa_ticket ()
Returns a ticket indicating dependence on all of the abstract state variables in the current Context. More...
static DependencyTicket all_state_ticket ()
Returns a ticket indicating dependence on all state variables x in this system, including continuous variables xc, discrete (numeric) variables xd, and abstract state variables xa. More...
static DependencyTicket pn_ticket ()
Returns a ticket indicating dependence on all of the numerical parameters in the current Context. More...
static DependencyTicket pa_ticket ()
Returns a ticket indicating dependence on all of the abstract parameters pa in the current Context. More...
static DependencyTicket all_parameters_ticket ()
Returns a ticket indicating dependence on all parameters p in this system, including numeric parameters pn, and abstract parameters pa. More...
static DependencyTicket all_input_ports_ticket ()
Returns a ticket indicating dependence on all input ports u of this system. More...
static DependencyTicket all_sources_ticket ()
Returns a ticket indicating dependence on every possible independent source value, including time, accuracy, state, input ports, and parameters (but not cache entries). More...
static DependencyTicket configuration_ticket ()
Returns a ticket indicating dependence on all source values that may affect configuration-dependent computations. More...
static DependencyTicket kinematics_ticket ()
Returns a ticket indicating dependence on all source values that may affect configuration- or velocity-dependent computations. More...
static DependencyTicket xcdot_ticket ()
Returns a ticket for the cache entry that holds time derivatives of the continuous variables. More...
static DependencyTicket pe_ticket ()
Returns a ticket for the cache entry that holds the potential energy calculation. More...
static DependencyTicket ke_ticket ()
Returns a ticket for the cache entry that holds the kinetic energy calculation. More...
static DependencyTicket pc_ticket ()
Returns a ticket for the cache entry that holds the conservative power calculation. More...
static DependencyTicket pnc_ticket ()
Returns a ticket for the cache entry that holds the non-conservative power calculation. More...
template<template< typename > class S = ::drake::systems::System>
static std::unique_ptr< S< T > > Clone (const S< T > &from)
Creates a deep copy of this system. More...
template<template< typename > class S = ::drake::systems::System>
static std::unique_ptr< S< AutoDiffXd > > ToAutoDiffXd (const S< T > &from)
Creates a deep copy of from, transmogrified to use the autodiff scalar type, with a dynamic-sized vector of partial derivatives. More...
template<template< typename > class S = ::drake::systems::System>
static std::unique_ptr< S< symbolic::Expression > > ToSymbolic (const S< T > &from)
Creates a deep copy of from, transmogrified to use the symbolic scalar type. More...
template<typename U , template< typename > class S = ::drake::systems::System>
static std::unique_ptr< S< U > > ToScalarType (const S< T > &from)
Creates a deep copy of from, transmogrified to use the scalar type selected by a template parameter. More...
Static Public Member Functions inherited from SystemBase
static DependencyTicket nothing_ticket ()
Returns a ticket indicating that a computation does not depend on any source value; that is, it is a constant. More...
static DependencyTicket time_ticket ()
Returns a ticket indicating dependence on time. More...
static DependencyTicket accuracy_ticket ()
Returns a ticket indicating dependence on the accuracy setting in the Context. More...
static DependencyTicket q_ticket ()
Returns a ticket indicating that a computation depends on configuration state variables q. More...
static DependencyTicket v_ticket ()
Returns a ticket indicating dependence on velocity state variables v. More...
static DependencyTicket z_ticket ()
Returns a ticket indicating dependence on any or all of the miscellaneous continuous state variables z. More...
static DependencyTicket xc_ticket ()
Returns a ticket indicating dependence on all of the continuous state variables q, v, or z. More...
static DependencyTicket xd_ticket ()
Returns a ticket indicating dependence on all of the numerical discrete state variables, in any discrete variable group. More...
static DependencyTicket xa_ticket ()
Returns a ticket indicating dependence on all of the abstract state variables in the current Context. More...
static DependencyTicket all_state_ticket ()
Returns a ticket indicating dependence on all state variables x in this system, including continuous variables xc, discrete (numeric) variables xd, and abstract state variables xa. More...
static DependencyTicket pn_ticket ()
Returns a ticket indicating dependence on all of the numerical parameters in the current Context. More...
static DependencyTicket pa_ticket ()
Returns a ticket indicating dependence on all of the abstract parameters pa in the current Context. More...
static DependencyTicket all_parameters_ticket ()
Returns a ticket indicating dependence on all parameters p in this system, including numeric parameters pn, and abstract parameters pa. More...
static DependencyTicket all_input_ports_ticket ()
Returns a ticket indicating dependence on all input ports u of this system. More...
static DependencyTicket all_sources_except_input_ports_ticket ()
Returns a ticket indicating dependence on every possible independent source value except input ports. More...
static DependencyTicket all_sources_ticket ()
Returns a ticket indicating dependence on every possible independent source value, including time, accuracy, state, input ports, and parameters (but not cache entries). More...
static DependencyTicket configuration_ticket ()
Returns a ticket indicating dependence on all source values that may affect configuration-dependent computations. More...
static DependencyTicket kinematics_ticket ()
Returns a ticket indicating dependence on all source values that may affect configuration- or velocity-dependent computations. More...
static DependencyTicket xcdot_ticket ()
Returns a ticket for the cache entry that holds time derivatives of the continuous variables. More...
static DependencyTicket pe_ticket ()
Returns a ticket for the cache entry that holds the potential energy calculation. More...
static DependencyTicket ke_ticket ()
Returns a ticket for the cache entry that holds the kinetic energy calculation. More...
static DependencyTicket pc_ticket ()
Returns a ticket for the cache entry that holds the conservative power calculation. More...
static DependencyTicket pnc_ticket ()
Returns a ticket for the cache entry that holds the non-conservative power calculation. More...
static DependencyTicket xd_unique_periodic_update_ticket ()
(Internal use only) Returns a ticket for the cache entry that holds the unique periodic discrete update computation. More...
Protected Member Functions inherited from LinearSystem< T >
LinearSystem (SystemScalarConverter converter, const Eigen::Ref< const Eigen::MatrixXd > &A, const Eigen::Ref< const Eigen::MatrixXd > &B, const Eigen::Ref< const Eigen::MatrixXd > &C, const Eigen::Ref< const Eigen::MatrixXd > &D, double time_period)
Constructor that specifies scalar-type conversion support. More...
Protected Member Functions inherited from AffineSystem< T >
AffineSystem (SystemScalarConverter converter, const Eigen::Ref< const Eigen::MatrixXd > &A, const Eigen::Ref< const Eigen::MatrixXd > &B, const Eigen::Ref< const Eigen::VectorXd > &f0, const Eigen::Ref< const Eigen::MatrixXd > &C, const Eigen::Ref< const Eigen::MatrixXd > &D, const Eigen::Ref< const Eigen::VectorXd > &y0, double time_period)
Constructor that specifies scalar-type conversion support. More...
Protected Member Functions inherited from TimeVaryingAffineSystem< T >
TimeVaryingAffineSystem (SystemScalarConverter converter, int num_states, int num_inputs, int num_outputs, double time_period)
Constructor. More...
template<typename U >
void ConfigureDefaultAndRandomStateFrom (const TimeVaryingAffineSystem< U > &other)
Helper method. More...
void SetDefaultState (const Context< T > &context, State< T > *state) const override
Sets the initial conditions. More...
void SetRandomState (const Context< T > &context, State< T > *state, RandomGenerator *generator) const override
Sets the random initial conditions. More...
Protected Member Functions inherited from LeafSystem< T >
LeafSystem ()
Default constructor that declares no inputs, outputs, state, parameters, events, nor scalar-type conversion support (AutoDiff, etc.). More...
LeafSystem (SystemScalarConverter converter)
Constructor that declares no inputs, outputs, state, parameters, or events, but allows subclasses to declare scalar-type conversion support (AutoDiff, etc.). More...
virtual std::unique_ptr< LeafContext< T > > DoMakeLeafContext () const
Provides a new instance of the leaf context for this system. More...
virtual void DoValidateAllocatedLeafContext (const LeafContext< T > &context) const
Derived classes that impose restrictions on what resources are permitted should check those restrictions by implementing this. More...
DoCalcWitnessValue (const Context< T > &context, const WitnessFunction< T > &witness_func) const final
Derived classes will implement this method to evaluate a witness function at the given context. More...
void AddTriggeredWitnessFunctionToCompositeEventCollection (Event< T > *event, CompositeEventCollection< T > *events) const final
Add event to events due to a witness function triggering. More...
void DoCalcNextUpdateTime (const Context< T > &context, CompositeEventCollection< T > *events, T *time) const override
Computes the next update time based on the configured periodic events, for scalar types that are arithmetic, or aborts for scalar types that are not arithmetic. More...
std::unique_ptr< ContinuousState< T > > AllocateContinuousState () const
Returns a copy of the state declared in the most recent DeclareContinuousState() call, or else a zero-sized state if that method has never been called. More...
std::unique_ptr< DiscreteValues< T > > AllocateDiscreteState () const
Returns a copy of the states declared in DeclareDiscreteState() calls. More...
std::unique_ptr< AbstractValuesAllocateAbstractState () const
Returns a copy of the states declared in DeclareAbstractState() calls. More...
std::unique_ptr< Parameters< T > > AllocateParameters () const
Returns a copy of the parameters declared in DeclareNumericParameter() and DeclareAbstractParameter() calls. More...
int DeclareNumericParameter (const BasicVector< T > &model_vector)
Declares a numeric parameter using the given model_vector. More...
template<template< typename > class U = BasicVector>
const U< T > & GetNumericParameter (const Context< T > &context, int index) const
Extracts the numeric parameters of type U from the context at index. More...
template<template< typename > class U = BasicVector>
U< T > & GetMutableNumericParameter (Context< T > *context, int index) const
Extracts the numeric parameters of type U from the context at index. More...
int DeclareAbstractParameter (const AbstractValue &model_value)
Declares an abstract parameter using the given model_value. More...
template<class MySystem >
SystemConstraintIndex DeclareEqualityConstraint (void(MySystem::*calc)(const Context< T > &, VectorX< T > *) const, int count, std::string description)
Declares a system constraint of the form f(context) = 0 by specifying a member function to use to calculate the (VectorX) constraint value with a signature: More...
SystemConstraintIndex DeclareEqualityConstraint (ContextConstraintCalc< T > calc, int count, std::string description)
Declares a system constraint of the form f(context) = 0 by specifying a std::function to use to calculate the (Vector) constraint value with a signature: More...
template<class MySystem >
SystemConstraintIndex DeclareInequalityConstraint (void(MySystem::*calc)(const Context< T > &, VectorX< T > *) const, SystemConstraintBounds bounds, std::string description)
Declares a system constraint of the form bounds.lower() <= calc(context) <= bounds.upper() by specifying a member function to use to calculate the (VectorX) constraint value with a signature: More...
SystemConstraintIndex DeclareInequalityConstraint (ContextConstraintCalc< T > calc, SystemConstraintBounds bounds, std::string description)
Declares a system constraint of the form bounds.lower() <= calc(context) <= bounds.upper() by specifying a std::function to use to calculate the (Vector) constraint value with a signature: More...
template<class MySystem >
void DeclarePeriodicPublishEvent (double period_sec, double offset_sec, EventStatus(MySystem::*publish)(const Context< T > &) const)
Declares that a Publish event should occur periodically and that it should invoke the given event handler method. More...
template<class MySystem >
void DeclarePeriodicPublishEvent (double period_sec, double offset_sec, void(MySystem::*publish)(const Context< T > &) const)
This variant accepts a handler that is assumed to succeed rather than one that returns an EventStatus result. More...
template<class MySystem >
void DeclarePeriodicDiscreteUpdateEvent (double period_sec, double offset_sec, EventStatus(MySystem::*update)(const Context< T > &, DiscreteValues< T > *) const)
Declares that a DiscreteUpdate event should occur periodically and that it should invoke the given event handler method. More...
template<class MySystem >
void DeclarePeriodicDiscreteUpdateEvent (double period_sec, double offset_sec, void(MySystem::*update)(const Context< T > &, DiscreteValues< T > *) const)
This variant accepts a handler that is assumed to succeed rather than one that returns an EventStatus result. More...
template<class MySystem >
void DeclarePeriodicUnrestrictedUpdateEvent (double period_sec, double offset_sec, EventStatus(MySystem::*update)(const Context< T > &, State< T > *) const)
Declares that an UnrestrictedUpdate event should occur periodically and that it should invoke the given event handler method. More...
template<class MySystem >
void DeclarePeriodicUnrestrictedUpdateEvent (double period_sec, double offset_sec, void(MySystem::*update)(const Context< T > &, State< T > *) const)
This variant accepts a handler that is assumed to succeed rather than one that returns an EventStatus result. More...
template<typename EventType >
void DeclarePeriodicEvent (double period_sec, double offset_sec, const EventType &event)
(Advanced) Declares that a particular Event object should be dispatched periodically. More...
template<class MySystem >
void DeclarePerStepPublishEvent (EventStatus(MySystem::*publish)(const Context< T > &) const)
Declares that a Publish event should occur at initialization and at the end of every trajectory-advancing step and that it should invoke the given event handler method. More...
template<class MySystem >
void DeclarePerStepDiscreteUpdateEvent (EventStatus(MySystem::*update)(const Context< T > &, DiscreteValues< T > *) const)
Declares that a DiscreteUpdate event should occur at the start of every trajectory-advancing step and that it should invoke the given event handler method. More...
template<class MySystem >
void DeclarePerStepUnrestrictedUpdateEvent (EventStatus(MySystem::*update)(const Context< T > &, State< T > *) const)
Declares that an UnrestrictedUpdate event should occur at the start of every trajectory-advancing step and that it should invoke the given event handler method. More...
template<typename EventType >
void DeclarePerStepEvent (const EventType &event)
(Advanced) Declares that a particular Event object should be dispatched at every trajectory-advancing step. More...
template<class MySystem >
void DeclareInitializationPublishEvent (EventStatus(MySystem::*publish)(const Context< T > &) const)
Declares that a Publish event should occur at initialization and that it should invoke the given event handler method. More...
template<class MySystem >
void DeclareInitializationDiscreteUpdateEvent (EventStatus(MySystem::*update)(const Context< T > &, DiscreteValues< T > *) const)
Declares that a DiscreteUpdate event should occur at initialization and that it should invoke the given event handler method. More...
template<class MySystem >
void DeclareInitializationUnrestrictedUpdateEvent (EventStatus(MySystem::*update)(const Context< T > &, State< T > *) const)
Declares that an UnrestrictedUpdate event should occur at initialization and that it should invoke the given event handler method. More...
template<typename EventType >
void DeclareInitializationEvent (const EventType &event)
(Advanced) Declares that a particular Event object should be dispatched at initialization. More...
template<class MySystem >
void DeclareForcedPublishEvent (EventStatus(MySystem::*publish)(const Context< T > &) const)
Declares a function that is called whenever a user directly calls ForcedPublish(const Context&). More...
template<class MySystem >
void DeclareForcedDiscreteUpdateEvent (EventStatus(MySystem::*update)(const Context< T > &, DiscreteValues< T > *) const)
Declares a function that is called whenever a user directly calls CalcForcedDiscreteVariableUpdate(const Context&, DiscreteValues<T>*). More...
template<class MySystem >
void DeclareForcedUnrestrictedUpdateEvent (EventStatus(MySystem::*update)(const Context< T > &, State< T > *) const)
Declares a function that is called whenever a user directly calls CalcForcedUnrestrictedUpdate(const Context&, State<T>*). More...
ContinuousStateIndex DeclareContinuousState (int num_state_variables)
Declares that this System should reserve continuous state with num_state_variables state variables, which have no second-order structure. More...
ContinuousStateIndex DeclareContinuousState (int num_q, int num_v, int num_z)
Declares that this System should reserve continuous state with num_q generalized positions, num_v generalized velocities, and num_z miscellaneous state variables. More...
ContinuousStateIndex DeclareContinuousState (const BasicVector< T > &model_vector)
Declares that this System should reserve continuous state with model_vector.size() miscellaneous state variables, stored in a vector cloned from model_vector. More...
ContinuousStateIndex DeclareContinuousState (const BasicVector< T > &model_vector, int num_q, int num_v, int num_z)
Declares that this System should reserve continuous state with num_q generalized positions, num_v generalized velocities, and num_z miscellaneous state variables, stored in a vector cloned from model_vector. More...
DiscreteStateIndex DeclareDiscreteState (const BasicVector< T > &model_vector)
Declares a discrete state group with model_vector.size() state variables, stored in a vector cloned from model_vector (preserving the concrete type and value). More...
DiscreteStateIndex DeclareDiscreteState (const Eigen::Ref< const VectorX< T >> &vector)
Declares a discrete state group with vector.size() state variables, stored in a BasicVector initialized with the contents of vector. More...
DiscreteStateIndex DeclareDiscreteState (int num_state_variables)
Declares a discrete state group with num_state_variables state variables, stored in a BasicVector initialized to be all-zero. More...
AbstractStateIndex DeclareAbstractState (const AbstractValue &model_value)
Declares an abstract state variable and provides a model value for it. More...
void DeclareImplicitTimeDerivativesResidualSize (int n)
(Advanced) Overrides the default size for the implicit time derivatives residual. More...
InputPort< T > & DeclareVectorInputPort (std::variant< std::string, UseDefaultName > name, const BasicVector< T > &model_vector, std::optional< RandomDistribution > random_type=std::nullopt)
Declares a vector-valued input port using the given model_vector. More...
InputPort< T > & DeclareVectorInputPort (std::variant< std::string, UseDefaultName > name, int size, std::optional< RandomDistribution > random_type=std::nullopt)
Declares a vector-valued input port with type BasicVector and size size. More...
InputPort< T > & DeclareAbstractInputPort (std::variant< std::string, UseDefaultName > name, const AbstractValue &model_value)
Declares an abstract-valued input port using the given model_value. More...
void DeprecateInputPort (const InputPort< T > &port, std::string message)
Flags an already-declared input port as deprecated. More...
template<class MySystem , typename BasicVectorSubtype >
LeafOutputPort< T > & DeclareVectorOutputPort (std::variant< std::string, UseDefaultName > name, const BasicVectorSubtype &model_vector, void(MySystem::*calc)(const Context< T > &, BasicVectorSubtype *) const, std::set< DependencyTicket > prerequisites_of_calc={ all_sources_ticket()})
Declares a vector-valued output port by specifying (1) a model vector of type BasicVectorSubtype derived from BasicVector and initialized to the correct size and desired initial value, and (2) a calculator function that is a class member function (method) with signature: More...
template<class MySystem >
LeafOutputPort< T > & DeclareVectorOutputPort (std::variant< std::string, UseDefaultName > name, int size, void(MySystem::*calc)(const Context< T > &, BasicVector< T > *) const, std::set< DependencyTicket > prerequisites_of_calc={ all_sources_ticket()})
Declares a vector-valued output port with type BasicVector and size size, using the drake::dummy_value<T>, which is NaN when T = double. More...
template<class MySystem , typename BasicVectorSubtype >
LeafOutputPort< T > & DeclareVectorOutputPort (std::variant< std::string, UseDefaultName > name, void(MySystem::*calc)(const Context< T > &, BasicVectorSubtype *) const, std::set< DependencyTicket > prerequisites_of_calc={ all_sources_ticket()})
Declares a vector-valued output port by specifying only a calculator function that is a class member function (method) with signature: More...
LeafOutputPort< T > & DeclareVectorOutputPort (std::variant< std::string, UseDefaultName > name, const BasicVector< T > &model_vector, typename LeafOutputPort< T >::CalcVectorCallback vector_calc_function, std::set< DependencyTicket > prerequisites_of_calc={ all_sources_ticket()})
(Advanced) Declares a vector-valued output port using the given model_vector and a function for calculating the port's value at runtime. More...
LeafOutputPort< T > & DeclareVectorOutputPort (std::variant< std::string, UseDefaultName > name, int size, typename LeafOutputPort< T >::CalcVectorCallback vector_calc_function, std::set< DependencyTicket > prerequisites_of_calc={ all_sources_ticket()})
(Advanced) Declares a vector-valued output port with type BasicVector<T> and size size, using the drake::dummy_value<T>, which is NaN when T = double. More...
template<class MySystem , typename OutputType >
LeafOutputPort< T > & DeclareAbstractOutputPort (std::variant< std::string, UseDefaultName > name, const OutputType &model_value, void(MySystem::*calc)(const Context< T > &, OutputType *) const, std::set< DependencyTicket > prerequisites_of_calc={ all_sources_ticket()})
Declares an abstract-valued output port by specifying a model value of concrete type OutputType and a calculator function that is a class member function (method) with signature: More...
template<class MySystem , typename OutputType >
LeafOutputPort< T > & DeclareAbstractOutputPort (std::variant< std::string, UseDefaultName > name, void(MySystem::*calc)(const Context< T > &, OutputType *) const, std::set< DependencyTicket > prerequisites_of_calc={ all_sources_ticket()})
Declares an abstract-valued output port by specifying only a calculator function that is a class member function (method) with signature: More...
LeafOutputPort< T > & DeclareAbstractOutputPort (std::variant< std::string, UseDefaultName > name, typename LeafOutputPort< T >::AllocCallback alloc_function, typename LeafOutputPort< T >::CalcCallback calc_function, std::set< DependencyTicket > prerequisites_of_calc={ all_sources_ticket()})
(Advanced) Declares an abstract-valued output port using the given allocator and calculator functions provided in their most generic forms. More...
LeafOutputPort< T > & DeclareStateOutputPort (std::variant< std::string, UseDefaultName > name, ContinuousStateIndex state_index)
Declares a vector-valued output port whose value is the continuous state of this system. More...
LeafOutputPort< T > & DeclareStateOutputPort (std::variant< std::string, UseDefaultName > name, DiscreteStateIndex state_index)
Declares a vector-valued output port whose value is the given discrete state group of this system. More...
LeafOutputPort< T > & DeclareStateOutputPort (std::variant< std::string, UseDefaultName > name, AbstractStateIndex state_index)
Declares an abstract-valued output port whose value is the given abstract state of this system. More...
void DeprecateOutputPort (const OutputPort< T > &port, std::string message)
Flags an already-declared output port as deprecated. More...
template<class MySystem >
std::unique_ptr< WitnessFunction< T > > MakeWitnessFunction (const std::string &description, const WitnessFunctionDirection &direction_type, T(MySystem::*calc)(const Context< T > &) const) const
Constructs the witness function with the given description (used primarily for debugging and logging), direction type, and calculator function; and with no event object. More...
std::unique_ptr< WitnessFunction< T > > MakeWitnessFunction (const std::string &description, const WitnessFunctionDirection &direction_type, std::function< T(const Context< T > &)> calc) const
Constructs the witness function with the given description (used primarily for debugging and logging), direction type, and calculator function; and with no event object. More...
template<class MySystem >
std::unique_ptr< WitnessFunction< T > > MakeWitnessFunction (const std::string &description, const WitnessFunctionDirection &direction_type, T(MySystem::*calc)(const Context< T > &) const, void(MySystem::*publish_callback)(const Context< T > &, const PublishEvent< T > &) const) const
Constructs the witness function with the given description (used primarily for debugging and logging), direction type, calculator function, and publish event callback function for when this triggers. More...
template<class MySystem >
std::unique_ptr< WitnessFunction< T > > MakeWitnessFunction (const std::string &description, const WitnessFunctionDirection &direction_type, T(MySystem::*calc)(const Context< T > &) const, void(MySystem::*du_callback)(const Context< T > &, const DiscreteUpdateEvent< T > &, DiscreteValues< T > *) const) const
Constructs the witness function with the given description (used primarily for debugging and logging), direction type, calculator function, and discrete update event callback function for when this triggers. More...
template<class MySystem >
std::unique_ptr< WitnessFunction< T > > MakeWitnessFunction (const std::string &description, const WitnessFunctionDirection &direction_type, T(MySystem::*calc)(const Context< T > &) const, void(MySystem::*uu_callback)(const Context< T > &, const UnrestrictedUpdateEvent< T > &, State< T > *) const) const
Constructs the witness function with the given description (used primarily for debugging and logging), direction type, calculator function, and unrestricted update event callback function for when this triggers. More...
template<class MySystem >
std::unique_ptr< WitnessFunction< T > > MakeWitnessFunction (const std::string &description, const WitnessFunctionDirection &direction_type, T(MySystem::*calc)(const Context< T > &) const, const Event< T > &e) const
Constructs the witness function with the given description (used primarily for debugging and logging), direction type, and calculator function, and with an object corresponding to the event that is to be dispatched when this witness function triggers. More...
std::unique_ptr< WitnessFunction< T > > MakeWitnessFunction (const std::string &description, const WitnessFunctionDirection &direction_type, std::function< T(const Context< T > &)> calc, const Event< T > &e) const
Constructs the witness function with the given description (used primarily for debugging and logging), direction type, and calculator function, and with an object corresponding to the event that is to be dispatched when this witness function triggers. More...
Protected Member Functions inherited from System< T >
virtual void DoGetWitnessFunctions (const Context< T > &, std::vector< const WitnessFunction< T > * > *) const
Derived classes can override this method to provide witness functions active for the given state. More...
SystemConstraintIndex AddConstraint (std::unique_ptr< SystemConstraint< T >> constraint)
Adds an already-created constraint to the list of constraints for this System. More...
bool forced_publish_events_exist () const
bool forced_discrete_update_events_exist () const
bool forced_unrestricted_update_events_exist () const
EventCollection< PublishEvent< T > > & get_mutable_forced_publish_events ()
EventCollection< DiscreteUpdateEvent< T > > & get_mutable_forced_discrete_update_events ()
EventCollection< UnrestrictedUpdateEvent< T > > & get_mutable_forced_unrestricted_update_events ()
const EventCollection< DiscreteUpdateEvent< T > > & get_forced_discrete_update_events () const
const EventCollection< UnrestrictedUpdateEvent< T > > & get_forced_unrestricted_update_events () const
void set_forced_publish_events (std::unique_ptr< EventCollection< PublishEvent< T >>> forced)
void set_forced_discrete_update_events (std::unique_ptr< EventCollection< DiscreteUpdateEvent< T >>> forced)
void set_forced_unrestricted_update_events (std::unique_ptr< EventCollection< UnrestrictedUpdateEvent< T >>> forced)
SystemScalarConverterget_mutable_system_scalar_converter ()
Returns the SystemScalarConverter for this system. More...
CacheEntryDeclareCacheEntry (std::string description, ValueProducer value_producer, std::set< DependencyTicket > prerequisites_of_calc={ all_sources_ticket()})
Declares a new CacheEntry in this System using the most generic form of the calculation function. More...
template<class MySystem , class MyContext , typename ValueType >
CacheEntryDeclareCacheEntry (std::string description, const ValueType &model_value, void(MySystem::*calc)(const MyContext &, ValueType *) const, std::set< DependencyTicket > prerequisites_of_calc={ all_sources_ticket()})
Declares a cache entry by specifying a model value of concrete type ValueType and a calculator function that is a class member function (method) with signature: More...
template<class MySystem , class MyContext , typename ValueType >
CacheEntryDeclareCacheEntry (std::string description, void(MySystem::*calc)(const MyContext &, ValueType *) const, std::set< DependencyTicket > prerequisites_of_calc={ all_sources_ticket()})
Declares a cache entry by specifying only a calculator function that is a class member function (method) with signature: More...
System (SystemScalarConverter converter)
Constructs an empty System base class object and allocates base class resources, possibly supporting scalar-type conversion support (AutoDiff, etc.) using converter. More...
InputPort< T > & DeclareInputPort (std::variant< std::string, UseDefaultName > name, PortDataType type, int size, std::optional< RandomDistribution > random_type=std::nullopt)
Adds a port with the specified type and size to the input topology. More...
virtual void DoCalcImplicitTimeDerivativesResidual (const Context< T > &context, const ContinuousState< T > &proposed_derivatives, EigenPtr< VectorX< T >> residual) const
Override this if you have an efficient way to evaluate the implicit time derivatives residual for this System. More...
virtual T DoCalcPotentialEnergy (const Context< T > &context) const
Override this method for physical systems to calculate the potential energy PE currently stored in the configuration provided in the given Context. More...
virtual T DoCalcKineticEnergy (const Context< T > &context) const
Override this method for physical systems to calculate the kinetic energy KE currently present in the motion provided in the given Context. More...
virtual T DoCalcConservativePower (const Context< T > &context) const
Override this method to return the rate Pc at which mechanical energy is being converted from potential energy to kinetic energy by this system in the given Context. More...
virtual T DoCalcNonConservativePower (const Context< T > &context) const
Override this method to return the rate Pnc at which work W is done on the system by non-conservative forces. More...
virtual void DoMapQDotToVelocity (const Context< T > &context, const Eigen::Ref< const VectorX< T >> &qdot, VectorBase< T > *generalized_velocity) const
Provides the substantive implementation of MapQDotToVelocity(). More...
virtual void DoMapVelocityToQDot (const Context< T > &context, const Eigen::Ref< const VectorX< T >> &generalized_velocity, VectorBase< T > *qdot) const
Provides the substantive implementation of MapVelocityToQDot(). More...
Eigen::VectorBlock< VectorX< T > > GetMutableOutputVector (SystemOutput< T > *output, int port_index) const
Returns a mutable Eigen expression for a vector valued output port with index port_index in this system. More...
Protected Member Functions inherited from SystemBase
SystemBase ()=default
(Internal use only). More...
void AddInputPort (std::unique_ptr< InputPortBase > port)
(Internal use only) Adds an already-constructed input port to this System. More...
void AddOutputPort (std::unique_ptr< OutputPortBase > port)
(Internal use only) Adds an already-constructed output port to this System. More...
std::string NextInputPortName (std::variant< std::string, UseDefaultName > given_name) const
(Internal use only) Returns a name for the next input port, using the given name if it isn't kUseDefaultName, otherwise making up a name like "u3" from the next available input port index. More...
std::string NextOutputPortName (std::variant< std::string, UseDefaultName > given_name) const
(Internal use only) Returns a name for the next output port, using the given name if it isn't kUseDefaultName, otherwise making up a name like "y3" from the next available output port index. More...
(Internal use only) Assigns a ticket to a new discrete variable group with the given index. More...
(Internal use only) Assigns a ticket to a new abstract state variable with the given index. More...
(Internal use only) Assigns a ticket to a new numeric parameter with the given index. More...
(Internal use only) Assigns a ticket to a new abstract parameter with the given index. More...
CacheEntryDeclareCacheEntryWithKnownTicket (DependencyTicket known_ticket, std::string description, ValueProducer value_producer, std::set< DependencyTicket > prerequisites_of_calc={ all_sources_ticket()})
(Internal use only) This is for cache entries associated with pre-defined tickets, for example the cache entry for time derivatives. More...
const internal::SystemParentServiceInterface * get_parent_service () const
Returns a pointer to the service interface of the immediately enclosing Diagram if one has been set, otherwise nullptr. More...
DependencyTicket assign_next_dependency_ticket ()
(Internal use only) Assigns the next unused dependency ticket number, unique only within a particular system. More...
const AbstractValueEvalAbstractInputImpl (const char *func, const ContextBase &context, InputPortIndex port_index) const
(Internal use only) Shared code for updating an input port and returning a pointer to its abstract value, or nullptr if the port is not connected. More...
void ThrowNegativePortIndex (const char *func, int port_index) const
Throws std::exception to report a negative port_index that was passed to API method func. More...
void ThrowInputPortIndexOutOfRange (const char *func, InputPortIndex port_index) const
Throws std::exception to report bad input port_index that was passed to API method func. More...
void ThrowOutputPortIndexOutOfRange (const char *func, OutputPortIndex port_index) const
Throws std::exception to report bad output port_index that was passed to API method func. More...
void ThrowNotAVectorInputPort (const char *func, InputPortIndex port_index) const
Throws std::exception because someone misused API method func, that is only allowed for declared-vector input ports, on an abstract port whose index is given here. More...
void ThrowInputPortHasWrongType (const char *func, InputPortIndex port_index, const std::string &expected_type, const std::string &actual_type) const
Throws std::exception because someone called API method func claiming the input port had some value type that was wrong. More...
void ThrowCantEvaluateInputPort (const char *func, InputPortIndex port_index) const
Throws std::exception because someone called API method func, that requires this input port to be evaluatable, but the port was neither fixed nor connected. More...
const InputPortBaseGetInputPortBaseOrThrow (const char *func, int port_index, bool warn_deprecated) const
(Internal use only) Returns the InputPortBase at index port_index, throwing std::exception we don't like the port index. More...
const OutputPortBaseGetOutputPortBaseOrThrow (const char *func, int port_index, bool warn_deprecated) const
(Internal use only) Returns the OutputPortBase at index port_index, throwing std::exception if we don't like the port index. More...
void ThrowValidateContextMismatch (const ContextBase &) const
(Internal use only) Throws std::exception with a message that the sanity check(s) given by ValidateContext have failed. More...
virtual std::string GetUnsupportedScalarConversionMessage (const std::type_info &source_type, const std::type_info &destination_type) const
(Internal use only) Returns the message to use for a std::exception in the case of unsupported scalar type conversions. More...
void InitializeContextBase (ContextBase *context) const
This method must be invoked from within derived class DoAllocateContext() implementations right after the concrete Context object has been allocated. More...
const ContextSizesget_context_sizes () const
Obtains access to the declared Context partition sizes as accumulated during LeafSystem or Diagram construction . More...
ContextSizesget_mutable_context_sizes ()
void set_implicit_time_derivatives_residual_size (int n)
Allows a LeafSystem to override the default size for the implicit time derivatives residual and a Diagram to sum up the total size. More...
internal::SystemId get_system_id () const
(Internal) Gets the id used to tag context data as being created by this system. More...
virtual GraphvizFragment DoGetGraphvizFragment (const GraphvizFragmentParams ¶ms) const
The NVI implementation of GetGraphvizFragment() for subclasses to override if desired. More...
CacheEntryDeclareCacheEntry (std::string description, ValueProducer value_producer, std::set< DependencyTicket > prerequisites_of_calc={ all_sources_ticket()})
Declares a new CacheEntry in this System using the most generic form of the calculation function. More...
template<class MySystem , class MyContext , typename ValueType >
CacheEntryDeclareCacheEntry (std::string description, const ValueType &model_value, void(MySystem::*calc)(const MyContext &, ValueType *) const, std::set< DependencyTicket > prerequisites_of_calc={ all_sources_ticket()})
Declares a cache entry by specifying a model value of concrete type ValueType and a calculator function that is a class member function (method) with signature: More...
template<class MySystem , class MyContext , typename ValueType >
CacheEntryDeclareCacheEntry (std::string description, void(MySystem::*calc)(const MyContext &, ValueType *) const, std::set< DependencyTicket > prerequisites_of_calc={ all_sources_ticket()})
Declares a cache entry by specifying only a calculator function that is a class member function (method) with signature: More...
Static Protected Member Functions inherited from LeafSystem< T >
static DependencyTicket all_sources_ticket ()
Returns a ticket indicating dependence on every possible independent source value, including time, accuracy, state, input ports, and parameters (but not cache entries). More...
Static Protected Member Functions inherited from System< T >
static void FindUniquePeriodicDiscreteUpdatesOrThrow (const char *api_name, const System< T > &system, const Context< T > &context, std::optional< PeriodicEventData > *timing, EventCollection< DiscreteUpdateEvent< T >> *events)
(Internal use only) Static interface to DoFindUniquePeriodicDiscreteUpdatesOrThrow() to allow a Diagram to invoke that private method on its subsystems. More...
Static Protected Member Functions inherited from SystemBase
static void set_parent_service (SystemBase *child, const internal::SystemParentServiceInterface *parent_service)
(Internal use only) Declares that parent_service is the service interface of the Diagram that owns this subsystem. More...
static void ThrowInputPortHasWrongType (const char *func, const std::string &system_pathname, InputPortIndex, const std::string &port_name, const std::string &expected_type, const std::string &actual_type)
Throws std::exception because someone called API method func claiming the input port had some value type that was wrong. More...
static const ContextSizesget_context_sizes (const SystemBase &system)
Allows Diagram to access protected get_context_sizes() recursively on its subsystems. More...
## ◆ MatrixGain() [1/5]
MatrixGain ( const MatrixGain< T > & )
delete
## ◆ MatrixGain() [2/5]
MatrixGain ( MatrixGain< T > && )
delete
## ◆ MatrixGain() [3/5]
MatrixGain ( int size )
explicit
A constructor where the gain matrix D is a square identity matrix of size size.
## ◆ MatrixGain() [4/5]
MatrixGain ( const Eigen::MatrixXd & D )
explicit
A constructor where the gain matrix D is D.
## ◆ MatrixGain() [5/5]
MatrixGain ( const MatrixGain< U > & )
explicit
Scalar-converting copy constructor. See System Scalar Conversion.
## ◆ operator=() [1/2]
MatrixGain& operator= ( const MatrixGain< T > & )
delete
## ◆ operator=() [2/2]
MatrixGain& operator= ( MatrixGain< T > && )
delete
The documentation for this class was generated from the following file: | 16,127 | 73,974 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2024-22 | latest | en | 0.388349 |
https://www.aqua-calc.com/one-to-one/density/long-ton-per-liter/stone-per-us-cup/35 | 1,603,877,247,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107897022.61/warc/CC-MAIN-20201028073614-20201028103614-00674.warc.gz | 632,111,022 | 9,018 | # 35 long tons per liter [long tn/l] in stones per US cup
## long tons/liter to stone/US cup unit converter of density
35 long tons per liter [long tn/l] = 1 324.89 stones per US cup [st/c]
### long tons per liter to stones per US cup density conversion cards
• 35
through
59
long tons per liter
• 35 long tn/l to st/c = 1 324.89 st/c
• 36 long tn/l to st/c = 1 362.75 st/c
• 37 long tn/l to st/c = 1 400.6 st/c
• 38 long tn/l to st/c = 1 438.46 st/c
• 39 long tn/l to st/c = 1 476.31 st/c
• 40 long tn/l to st/c = 1 514.16 st/c
• 41 long tn/l to st/c = 1 552.02 st/c
• 42 long tn/l to st/c = 1 589.87 st/c
• 43 long tn/l to st/c = 1 627.73 st/c
• 44 long tn/l to st/c = 1 665.58 st/c
• 45 long tn/l to st/c = 1 703.44 st/c
• 46 long tn/l to st/c = 1 741.29 st/c
• 47 long tn/l to st/c = 1 779.14 st/c
• 48 long tn/l to st/c = 1 817 st/c
• 49 long tn/l to st/c = 1 854.85 st/c
• 50 long tn/l to st/c = 1 892.71 st/c
• 51 long tn/l to st/c = 1 930.56 st/c
• 52 long tn/l to st/c = 1 968.41 st/c
• 53 long tn/l to st/c = 2 006.27 st/c
• 54 long tn/l to st/c = 2 044.12 st/c
• 55 long tn/l to st/c = 2 081.98 st/c
• 56 long tn/l to st/c = 2 119.83 st/c
• 57 long tn/l to st/c = 2 157.68 st/c
• 58 long tn/l to st/c = 2 195.54 st/c
• 59 long tn/l to st/c = 2 233.39 st/c
• 60
through
84
long tons per liter
• 60 long tn/l to st/c = 2 271.25 st/c
• 61 long tn/l to st/c = 2 309.1 st/c
• 62 long tn/l to st/c = 2 346.96 st/c
• 63 long tn/l to st/c = 2 384.81 st/c
• 64 long tn/l to st/c = 2 422.66 st/c
• 65 long tn/l to st/c = 2 460.52 st/c
• 66 long tn/l to st/c = 2 498.37 st/c
• 67 long tn/l to st/c = 2 536.23 st/c
• 68 long tn/l to st/c = 2 574.08 st/c
• 69 long tn/l to st/c = 2 611.93 st/c
• 70 long tn/l to st/c = 2 649.79 st/c
• 71 long tn/l to st/c = 2 687.64 st/c
• 72 long tn/l to st/c = 2 725.5 st/c
• 73 long tn/l to st/c = 2 763.35 st/c
• 74 long tn/l to st/c = 2 801.2 st/c
• 75 long tn/l to st/c = 2 839.06 st/c
• 76 long tn/l to st/c = 2 876.91 st/c
• 77 long tn/l to st/c = 2 914.77 st/c
• 78 long tn/l to st/c = 2 952.62 st/c
• 79 long tn/l to st/c = 2 990.48 st/c
• 80 long tn/l to st/c = 3 028.33 st/c
• 81 long tn/l to st/c = 3 066.18 st/c
• 82 long tn/l to st/c = 3 104.04 st/c
• 83 long tn/l to st/c = 3 141.89 st/c
• 84 long tn/l to st/c = 3 179.75 st/c
• 85
through
109
long tons per liter
• 85 long tn/l to st/c = 3 217.6 st/c
• 86 long tn/l to st/c = 3 255.45 st/c
• 87 long tn/l to st/c = 3 293.31 st/c
• 88 long tn/l to st/c = 3 331.16 st/c
• 89 long tn/l to st/c = 3 369.02 st/c
• 90 long tn/l to st/c = 3 406.87 st/c
• 91 long tn/l to st/c = 3 444.72 st/c
• 92 long tn/l to st/c = 3 482.58 st/c
• 93 long tn/l to st/c = 3 520.43 st/c
• 94 long tn/l to st/c = 3 558.29 st/c
• 95 long tn/l to st/c = 3 596.14 st/c
• 96 long tn/l to st/c = 3 634 st/c
• 97 long tn/l to st/c = 3 671.85 st/c
• 98 long tn/l to st/c = 3 709.7 st/c
• 99 long tn/l to st/c = 3 747.56 st/c
• 100 long tn/l to st/c = 3 785.41 st/c
• 101 long tn/l to st/c = 3 823.27 st/c
• 102 long tn/l to st/c = 3 861.12 st/c
• 103 long tn/l to st/c = 3 898.97 st/c
• 104 long tn/l to st/c = 3 936.83 st/c
• 105 long tn/l to st/c = 3 974.68 st/c
• 106 long tn/l to st/c = 4 012.54 st/c
• 107 long tn/l to st/c = 4 050.39 st/c
• 108 long tn/l to st/c = 4 088.24 st/c
• 109 long tn/l to st/c = 4 126.1 st/c
• 110
through
134
long tons per liter
• 110 long tn/l to st/c = 4 163.95 st/c
• 111 long tn/l to st/c = 4 201.81 st/c
• 112 long tn/l to st/c = 4 239.66 st/c
• 113 long tn/l to st/c = 4 277.52 st/c
• 114 long tn/l to st/c = 4 315.37 st/c
• 115 long tn/l to st/c = 4 353.22 st/c
• 116 long tn/l to st/c = 4 391.08 st/c
• 117 long tn/l to st/c = 4 428.93 st/c
• 118 long tn/l to st/c = 4 466.79 st/c
• 119 long tn/l to st/c = 4 504.64 st/c
• 120 long tn/l to st/c = 4 542.49 st/c
• 121 long tn/l to st/c = 4 580.35 st/c
• 122 long tn/l to st/c = 4 618.2 st/c
• 123 long tn/l to st/c = 4 656.06 st/c
• 124 long tn/l to st/c = 4 693.91 st/c
• 125 long tn/l to st/c = 4 731.76 st/c
• 126 long tn/l to st/c = 4 769.62 st/c
• 127 long tn/l to st/c = 4 807.47 st/c
• 128 long tn/l to st/c = 4 845.33 st/c
• 129 long tn/l to st/c = 4 883.18 st/c
• 130 long tn/l to st/c = 4 921.04 st/c
• 131 long tn/l to st/c = 4 958.89 st/c
• 132 long tn/l to st/c = 4 996.74 st/c
• 133 long tn/l to st/c = 5 034.6 st/c
• 134 long tn/l to st/c = 5 072.45 st/c
• st/c stands for st/US c
#### Foods, Nutrients and Calories
EASY SQUEEZE SOUR CREAM, UPC: 041483027481 weigh(s) 253.61 gram per (metric cup) or 8.47 ounce per (US cup), and contain(s) 200 calories per 100 grams or ≈3.527 ounces [ weight to volume | volume to weight | price | density ]
#### Gravels, Substances and Oils
CaribSea, Freshwater, Super Naturals, Moonlight Sand weighs 1 521.75 kg/m³ (94.99975 lb/ft³) with specific gravity of 1.52175 relative to pure water. Calculate how much of this gravel is required to attain a specific depth in a cylindricalquarter cylindrical or in a rectangular shaped aquarium or pond [ weight to volume | volume to weight | price ]
Lime nitrogen [CaCN2] weighs 2 290 kg/m³ (142.96003 lb/ft³) [ weight to volume | volume to weight | price | mole to volume and weight | mass and molar concentration | density ]
Volume to weightweight to volume and cost conversions for Crambe oil with temperature in the range of 23.9°C (75.02°F) to 110°C (230°F)
#### Weights and Measurements
A kilonewton is a SI-multiple (see prefix kilo) of the force unit newton and equal to one thousand newtons (1,000 N)
The area measurement was introduced to measure surface of a two-dimensional object.
oz/US gal to oz/metric tsp conversion table, oz/US gal to oz/metric tsp unit converter or convert between all units of density measurement.
#### Calculators
Body Mass Index calculator using person's weight and height | 2,402 | 5,780 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2020-45 | latest | en | 0.535706 |
http://www.drumtom.com/q/difference-between-a-pentagon-and-a-regular-pentagon | 1,487,762,221,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501170940.45/warc/CC-MAIN-20170219104610-00061-ip-10-171-10-108.ec2.internal.warc.gz | 387,841,921 | 8,438 | # Difference between a Pentagon and a regular Pentagon?
• Difference between a Pentagon and a regular Pentagon?
1) The number of sides those are closed by (5 in Pentagon and 6 in Hexagon). 2) The number of vertices ( 5 vs 6). 3) For a Regular Pentagon, the internal ...
Positive: 64 %
... (distance between two farthest separated points, ... The K 5 complete graph is often drawn as a regular pentagon with all 10 edges connected.
Positive: 61 %
### More resources
The regular pentagon is the regular polygon with ... A number of distance relationships between vertices of the pentagon can be derived by similar ...
Positive: 64 %
Regular and irregular polygons. ... A regular polygon has sides ... All other regular polygons are simply termed 'regular pentagon', 'regular hexagon' etc ...
Positive: 59 %
I believe what you're asking for is the dihedral angle between the hexagon and pentagon in ... Is it possible to have a square tile containing regular ... | 214 | 963 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2017-09 | longest | en | 0.921777 |
https://gmatclub.com/forum/the-use-of-gravity-waves-which-do-not-interact-with-matter-18253.html | 1,487,998,956,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501171664.76/warc/CC-MAIN-20170219104611-00071-ip-10-171-10-108.ec2.internal.warc.gz | 720,568,738 | 58,953 | The use of gravity waves, which do not interact with matter : GMAT Sentence Correction (SC)
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# The use of gravity waves, which do not interact with matter
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The use of gravity waves, which do not interact with matter in the way electromagnetic waves do, hopefully will enable astronomers to study the actual formation of black holes and neutron stars.
(A) in the way electromagnetic waves do, hopefully will enable
(B) in the way electromagnetic waves do, will, it is hoped, enable
(C) like electromagnetic waves, hopefully will enable
(D) like electromagnetic waves, would enable, hopefully
(E) such as electromagnetic waves do, will, it is hoped, enable
I know its quite simple as either A or B sounds good. But why B is a better than A??
If you have any questions
you can ask an expert
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22 Jul 2005, 16:06
B for me.
A has a misplaced adverb "hopefully". Should be after 'will'.
C,D,E have comparison problems.
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22 Jul 2005, 16:11
I am confused with this one..what are we comparing here...the waves or the use of waves?
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23 Jul 2005, 06:12
riteshgupta1 wrote:
B for me.
A has a misplaced adverb "hopefully". Should be after 'will'.
C,D,E have comparison problems.
But what is "it" in B? I thought the original senetence is fine
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23 Jul 2005, 06:22
This is a very typical example and you will find this concept discussed multiple times where "it is hoped" clause is used as an appositive. Meaning the tone of the outcome +ve.
They hopefully (only) will enable.
They will hopefully (only) enable.
B is best
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23 Jul 2005, 19:50
riteshgupta1 wrote:
B for me.
A has a misplaced adverb "hopefully". Should be after 'will'.
C,D,E have comparison problems.
"Hopefully" is considered incorrect in many dictionaries (the word does not even exist in many of them), but they claim that expressions such as "It is hopeful that..." or "It is hoped" are a correct way of saying 'hopefully'. Another clear example when our everyday language and standardized tests collide, and create a helluv confusion for us
23 Jul 2005, 19:50
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# The use of gravity waves, which do not interact with matter
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Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 1,243 | 4,609 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2017-09 | longest | en | 0.867207 |
https://aviation.stackexchange.com/questions/27976/is-there-a-minimum-distance-between-a-runway-and-a-public-road | 1,714,058,121,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712297295329.99/warc/CC-MAIN-20240425130216-20240425160216-00391.warc.gz | 95,908,526 | 41,165 | # Is there a minimum distance between a runway and a public road?
Does the FAA require a minimum distance the between the TDZ of a runway and a public road?
The reason I ask is because an aircraft landing at a small GA airport (Garnett Municipal, K68) came across the roadway I was driving on low enough that if I'd been 200 yards up the road he'd have hit my truck. Or at least he'd have had to go around.
From the satellite pic you can see that the end of the runway is about 30 feet from the road. The TDZ is about 150 feet. A 3° approach would put you just short of 8' AGL at the road, which is just about exactly how high he was. My truck has a clearance of 10', so it's quite possible to strike vehicles on the road if one is too focused on the runway.
Although it's not a highway (that's about 1000' from the runway), it's the only paved road going east/west for several miles so it gets a bit of traffic
Is it common to put a runway so close to a road?
• No, in fact there are public roads so close that they caution drivers to stop and listen, there is a pretty famous accident that a student pilot had during first solo. As a side note, he did quit flying at first but then went on to finish his ticket. You can read the NTSB report here Jun 3, 2016 at 1:29
• Just missed my edit window, what I meant was that it isn't uncommon to have a road that close, although public use airports have to be built to some standards (the video I posted is a private airport). I'm trying to locate that standard for you and I'll post an answer when I find it. Jun 3, 2016 at 1:37
• Internationally, there are some great counterexamples. Gibraltar, for instance, where the only road in runs across the runway (they do have officials that close the road for aircraft movements). Jun 3, 2016 at 5:24
• @JanHudec I based the numbers in my question.on the white line. 3° at 150' is about 8'. I'd never seen anyone come in quite so low so I was wondering if the approach was abnormal but it appears that it was textbook. Jun 3, 2016 at 12:02
• One thing worth noting is that even where there are laws, rules, or regulations, they often do not apply retrospectively. eg where an airfield was built before the law came into place, there is not always a requirement to update it Jun 3, 2016 at 14:41
Yes, there is a minimum area that must be free of obstructions and closed to vehicles, but only for airports that have air carrier operations (see 14 CFR 139.309) or that receive FAA grants. AC 150/5300-13A - Airport Design has all the technical details; the actual distances are calculated from formulas, so there's no single answer. If you're really curious, you can read all the gory details in Chapter 3 and this FAA FAQ has a (very) simplified explanation:
What is the size of the RSA [Runway Safety Area]?
The RSA is a rectangular box surrounding the runway and is based on the runway design code. The dimensions range from 120 feet to 500 feet in width and 240 feet to 1000 feet in length beyond the departure end of the runway. (see FAA Advisory Circular 150/5300-13, Airport Design) Generally, on airports that serve air carrier aircraft, the RSA extends 250 feet either side of the runway centerline and 1000 feet beyond each end of the runway.
You didn't mention which airport you're asking about, but presumably it's either privately owned and/or doesn't receive FAA grant money.
• It's K68. Very small, one runway, non-towered airport owned by the city of Garnett, KS. I have no idea about funding. I've seen that beach in St Maarten on tv. That's pretty wild. I'd love to see that in person some day Jun 3, 2016 at 2:36
No (for smaller airplanes and airfields)
The limit is not regarding the runway to the road but the minimum height an airplane (landing or starting) has to have over the road.
You can see this very well in the picture:
A departing aircraft (runway 19) can be quite close to the road and use the "full" length of the runway as it is built. As long as the thrust is not too big (for smaller airplanes), there is no reason for a restriction.
A landing aircraft (runway 19) requires a minimum height (15 feet i think) and is therefore not allowed to land at the beginning of the runway but not before the big, white bar. This is called a displaced threshold. So the runway, for landing aircrafts, actually does not begin so close to the road as it "looks" like.
This way, departing aircrafts will have a longer runway available, which is a critical point in aviation (safety).
It's btw the same for a departing aircraft with direction to the road (so using runway 19 + 18 = 37 => runway 01). It is not allowed to lift off after the displaced threshold.
The FAA regulates aviation and aircraft, not road layout or runway locations.
The standard glideslope is three degrees, pretty shallow, so aircraft often travel long distances low above the ground prior to landing. Trees are a big issue. On the island of St. Maarten 747's fly as low as 10-15 feet right over a heavily populated resort beach and pathway.
Credit: Blick, William E. Princess Juliana International Airport
Often there are signs warning of low-flying aircraft so that motorists are not startled.
Don't worry about it too much, planes are way more flimsy than a truck, so as long as the 50 gallons of aviation fuel does not catch on fire or explode, your vehicle will come off much better in the rare event of a collision.
• There's pretty good visibility there, so I'm sure they would see something coming... as long as they are paying attention. I can imagine a new pilot being focused on the landing, airspeed, configuration, etc and forgetting completely about road traffic. It just totally surprised me seeing how low he came over the road. I was watching a couple of crop dusters taking off and landing there yesterday and they were quite a bit higher than this plane was. But those crop dusting planes are really agile and powerful so they obviously don't need the whole runway. They have have an amazing climb rate. Jun 3, 2016 at 2:22
• Do you have a citation for your claim that aircraft cross the beach at Sint Maarten at 10-15ft AGL? The photo looks much more like 30-40ft and crossing the beach at 10ft would surely result in clipping the airport fence. Jun 3, 2016 at 5:23
• Also, please acknowledge the source of the image. Jun 3, 2016 at 5:24
• @Richerby The fence is 8 feet high I think and the planes come right down over that fence. There is a roadway or pedestrian way right next to the fence. You can't see the fence in this photo. I was talking about the altitude near the fence. Jun 3, 2016 at 8:21
• Totally OT, but I can see hanging out on that beach for a few landings as a novelty, but it certainly doesn't look like a relaxing vacation destination for spending the whole day... Jun 3, 2016 at 12:49 | 1,645 | 6,830 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2024-18 | latest | en | 0.986245 |
https://www.beatthegmat.com/if-each-term-in-an-infinite-sequence-is-found-by-the-equation-s-x-8-x-6-and-s-1-2-is-every-term-in-the-sequenc-t329184.html?sid=487c55abc221ad76989645c72f3cf3b1 | 1,652,681,537,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662509990.19/warc/CC-MAIN-20220516041337-20220516071337-00448.warc.gz | 768,440,852 | 11,635 | ## If each term in an infinite sequence is found by the equation $$S_x=8^x-6,$$ and $$S_1=2,$$ is every term in the sequenc
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### If each term in an infinite sequence is found by the equation $$S_x=8^x-6,$$ and $$S_1=2,$$ is every term in the sequenc
by Gmat_mission » Sun Jan 23, 2022 1:29 am
00:00
A
B
C
D
E
## Global Stats
If each term in an infinite sequence is found by the equation $$S_x=8^x-6,$$ and $$S_1=2,$$ is every term in the sequence divisible by $$y?$$
(1) $$y$$ is an even integer.
(2) At least $$2$$ of the first $$4$$ terms in the sequence are divisible by $$y.$$ | 233 | 699 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2022-21 | latest | en | 0.930864 |
https://electronics.stackexchange.com/questions/227702/battery-charging-time-computation | 1,725,813,035,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651013.49/warc/CC-MAIN-20240908150334-20240908180334-00272.warc.gz | 218,473,252 | 41,929 | # Battery Charging Time Computation
Good Day,
I am having trouble to compute the charging time of my battery with this kind of set-up:
(2) 300W 36V PV Panel (1) 500W 24V Wind Turbine (1) Hybrid Charge Controller (6) 12V 200Ah Batteries @ 24V Configuration
The batteries are connected to series-parallel connection that will give 24V 600Ah output.
How long is the charging time of my battery?
• Your death battery kill good batterry. :) Separate every battery charge channel for fast charge. You kill good battery if other be corrupted on same group. Commented Apr 11, 2016 at 8:48
SIMPLIFIED CALCULATION:
Your battery has (seen from the outside) a capacity of 600Ah.
In an ideal world: If you would start with a empty battery and could charge it with 600A it would be full in an hour.
In a real world: Each battery loses some energy while charging. This can be between 10% to 40%. Our empty battery would be full after 1.4 hours.
You can charge in the best case with 800W (300W PV and 500W Wind).
800W / 24V = 33A
600Ah*1.4 / 33A = 25H
This is a simplified version of the reality. Your wind and solar will probably never reach the 800W.
• My Battery is Deep-Cycle Battery so how can I calculate the charge current in amps. Commented Apr 12, 2016 at 1:34
• Deep-Cycle only means, that it does not get damaged if you discharge until a certain low level. It does not tell you anything about the current Commented Apr 12, 2016 at 7:17 | 388 | 1,445 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2024-38 | latest | en | 0.907106 |
https://math.stackexchange.com/questions/1829108/derivative-integral-relationship-appears-to-disprove-the-fundamental-theorem-of | 1,621,065,653,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991378.48/warc/CC-MAIN-20210515070344-20210515100344-00469.warc.gz | 411,704,509 | 36,178 | # Derivative/integral relationship appears to disprove the fundamental theorem of calculus!!! [duplicate]
Consider the floor function:
$$f(x) = \lfloor x \rfloor$$
The indefinite integral of f is:
$$\int_0^x f(x) dx = x\lfloor x \rfloor - \frac {\lfloor x \rfloor^2 + \lfloor x \rfloor} 2$$
This should be an antiderivative of floor, right?
Nope! If you take the derivative of the integral you find that sharp corners cause the derivative to not exist.
So then this would mean that the integral of floor is not an antiderivative right?
Therefore, I have found a case where the antiderivative does not equal the indefinite integral.
In that case I must be flawed as that violates the first fundamental theorem of calculus.
Where is the mistake in my logic and why does it appear to disprove the first fundamental theorem's relationship between integral and derivative?
This isn't part of the above question per se, but I noticed interesting enough, that the derivative of the integral above is:
$$\lfloor x \rfloor \frac {x - \lfloor x \rfloor}{x - \lfloor x \rfloor}$$
I wonder what sort of properties would be altered within integration/differentiation if one were to IDK... redefine the derivative by canceling the terms in that fraction? Or for that matter, canceling all fractions of that nature?
• you should consider instead $\int_0^x sign(x) dx = |x|$ and the fundamental theorem of calculus is that $|x|$ is weak differentiable with $(|x|)' = sign(x)$ almost everywhere – reuns Jun 17 '16 at 21:37
• @user1952009 why should I change the function? That's illogical. I noticed the property when taking the integral of floor. How is it logical to change the function arbitrarily? – user64742 Jun 17 '16 at 23:54
• $sign(x)$ is the simplest example for understanding what is happening, exactly in the same way as for $\lfloor x \rfloor = \sum_k \frac{(1+sign(x-k))}{2}$ – reuns Jun 18 '16 at 0:09
• ok then when people are trying to help you, they are completely ridiculous... sure ! so I'm telling you your example is a complicated version of $(|x|)' = sign(x), x\ne 0$, and that at $x=0$ there is no problem, only a discontinuity of the derivative – reuns Jun 18 '16 at 1:37
• I didn't mean to edit your question, but edit what you wrote on your paper, yes, because your function hides the phenomena which is that everything happens as for $(|x|)'$ – reuns Jun 18 '16 at 3:37
The fundamental theorem of calculus has a crucial hypothesis: $$F(x)=\int_{a}^x f(t)dt\Rightarrow F'(a)=f(a)$$
Whenever and wherever $f$ is continuous, here we are assuming $f$ is continuous at the point $a$. The floor function is very much not continuous. When you see theorems, it is very important that you check what the hypotheses are.
• @EthanHunt This site has really improved my typing. – operatorerror Jun 16 '16 at 22:17
• Dude, this is an oversight that bites tons of people. I voted the question and answer up accordingly. – The Nate Jun 17 '16 at 6:53
The first fundamental theorem of calculus is stated as follows:
For any continuous function $f:[a,b]\rightarrow \mathbb R$ the function $F(x)=\int_{a}^x f(x)\,dx$ has $F'(x)=f(x)$ for all $x\in (a,b)$.
Notice that $f(x)=\lfloor x\rfloor$ is not continuous at the integers, thus this theorem says nothing about that. Note that the derivatives do agree at points where $f(x)$ is continuous.
An interesting thing to note is that there is another version of the first fundamental theorem of calculus called the Lebesgue differentiation theorem which loosens the restriction on $f$, but only says that $F'(x)=f(x)$ almost everywhere. It relies on measure theory to state, so I won't reproduce it here, but it's worth noting that one has a trade-off between conditions on $f$ and results for $F$.
Therefore, I have found a case where the antiderivative does not equal the indefinite integral.
That statement is actually not quite right. What you have in fact found is a case where the indefinite integral can not be used to obtain the original function through differentiation. But you can't say anything about an antiderivative that does not equal something else, because it doesn't exist.
Indeed, the statement “any antiderivative is equal to the indefinite integral” is still true – for all of the zero antiderivatives!
What's more: in more than one sense the integral is actually differentiable. In particular, for any $x$ and any sequence $((x)_i)_i$ converging to $x$ from above (i.e. $x_i>x$), you get $$\lim_{i\to\infty} \frac{F(x_i) - F(x)}{x_i - x} \equiv \lim_{\xi\searrow x} \frac{F(\xi) - F(x)}{\xi - x} = \lfloor x\rfloor = f(x).$$ For all $x\not\in\mathbb{Z}$, this upper derivative is equal to the normal derivative. Trouble is, there would be no compelling reason to favour the upper derivative over the lower derivative, and it turns out that the lower derivative disagrees: $$\lim_{\xi\nearrow x} \frac{F(\xi) - F(x)}{\xi - x} = \lceil x-1\rceil.$$ This is still equal to $f(x)$ almost everywhere, but for $x\in\mathbb{Z}$ we have $\lceil x-1\rceil = f(x)-1$.
You can pretty much say that the derivative is deliberately constructed in such a way that it doesn't exist in cases like this where there would be an ambiguity and the Fundamental Theorem would be contradicted. By requiring continuity of $f$, the Fundamental Theorem also precludes such ambiguity and guarantees that the integral has a well-defined derivative.
These upper or lower derivatives really aren't much use in practice, but there is a concept called weak derivative that does have some pretty useful applications. | 1,435 | 5,571 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2021-21 | latest | en | 0.907841 |
https://www.coursehero.com/file/6870678/lecture19-1100-S12/ | 1,513,377,904,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948579567.73/warc/CC-MAIN-20171215211734-20171215233734-00283.warc.gz | 739,700,337 | 116,166 | lecture19_1100_S12
# lecture19_1100_S12 - PHYS 1100 Spring 2012 I Kinematic...
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I. Kinematic Variables A. Displacement B. Average Speed C. Average Velocity D. Average Acceleration PHYS 1100 Spring 2012 Reading for Wednesday: 2.1 – 2.6, 4.1 – 4.4
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PHYS 1100 Spring 2012 0 2 4 6 8 10 12 Average = 51%
CT0: Two blocks are connected by a massless string that lies across a massless, frictionless pulley. Block 1 with mass m 1 is initially at rest on a frictionless surface that is inclined at an angle of θ 1 = 30° above the horizontal. Block 2 with a mass of m 2 is initially at rest on a frictionless surface that is inclined at an angle of 2 = 60° above the horizontal. Before the blocks are released, which force acting on block 1 has the greatest magnitude? A. The gravitational force B. The normal force from the surface C. The tension in the string D. Both A and B E. A, B, and C all have the same magnitude PHYS 1100 Spring 2012
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CT0: A lamp hangs from the ceiling of an elevator by a cable. As the elevator descends with an increasing speed, how does the weight of the lamp compare to the magnitude of the tension in the cable? A. The weight of the lamp is greater than the tension in the cable B. The weight of the lamp is equal the tension in the cable C. The weight of the lamp is less than the tension in the cable D. How the two forces compare cannot be determined PHYS 1100 Spring 2012
CT1: The hare challenges the tortoise to rerun their race along a long straight track. Once again, at the start of the race, the hare takes off like a shot and the tortoise
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Ask a homework question - tutors are online | 579 | 2,260 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2017-51 | latest | en | 0.878884 |
https://kr.mathworks.com/matlabcentral/answers/1944459-what-is-r-in-the-pagerank-algorithm | 1,716,311,420,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058504.35/warc/CC-MAIN-20240521153045-20240521183045-00481.warc.gz | 298,173,794 | 29,754 | what is r. in the pagerank algorithm
조회 수: 6 (최근 30일)
Kellie 2023년 4월 10일
댓글: Askic V 2023년 4월 11일
I was reading Matlabs Pagerank algorithm and I was hoping that someone can clarify what the r. is within the equation. I have all the other elements just not sure what r.
Thank you!
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답변 (1개)
Hi Kellie,
you probably mean on the following equation:
r = (1-P)/n + P*(A'*(r./d) + s/n);
% r is a vector of PageRank scores.
% P is a scalar damping factor (usually 0.85), which is the probability that a random surfer clicks on a link on the current page, instead of continuing on another random page.
% A' is the transpose of the adjacency matrix of the graph.
% d is a vector containing the out-degree of each node in the graph. d is set to 1 for nodes with no outgoing edges.
% n is the scalar number of nodes in the graph.
% s is the scalar sum of the PageRank scores for pages with no links.
So, this equation updates the vector of Pagerank scores based on the current value of vector r and other parameters.
Specifically,
r./d
is an element-wise division, which means that each element of r is divided with its corresponding element in vector d (the vector d contains same number of elements like vector r).
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Kellie 2023년 4월 11일
Thank you Askic that has helped so much! one more question, if you are assessing a matrix (A) is it G = digraph(A,[],names);?
Thank you! :)
Matrix A is an adjacency matrix of a graph G.
If you want to create a directed graph based on the adjacency matrix, then you simply use the following syntax:
G2 = digraph(A)
In the example code above, matrix A is simply:
A =
0 1 0 0 1 0
0 0 1 1 0 0
0 0 0 1 1 1
1 0 0 0 0 0
1 0 0 0 0 0
0 0 0 0 0 0
If A(i,j) is 1 that means there is a directed connection from i to j.
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Translated by | 616 | 2,024 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2024-22 | latest | en | 0.892025 |
https://math.stackexchange.com/tags/group-isomorphism/info | 1,718,855,056,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861880.60/warc/CC-MAIN-20240620011821-20240620041821-00826.warc.gz | 330,402,678 | 18,309 | A group isomorphism $$\phi\colon G \to H$$ is a bijective group homomorphism. Alternatively you could say a homomorphism $$\phi\colon G \to H$$ is an isomorphism if there exists another homomorphism $$\phi^{-1}\colon H \to G$$ such that $$\phi^{-1}\phi$$ is the identity on $$G$$ and $$\phi\phi^{-1}$$ is the identity on $$H$$. If such an isomorphism $$\phi\colon G \to H$$ exists, we say that $$G$$ and $$H$$ are isomorphic, which means that they are structurally identical as groups. This is usually signified by writing $$G \cong H$$.
• The groups $$(\mathbb{R},+)$$, the real numbers equipped with addition, and $$(\mathbb{R}^{+},\times)$$, the positive real numbers equipped with multiplication, are isomorphic. The function $$\exp\colon\mathbb{R}\to \mathbb{R}^{+}$$ that sends $$x$$ to $$\mathrm{e}^x$$ is a group isomorphism that demonstrates this.
• The group of integers $$\mathbb{Z}$$ under addition is isomorphic to its subgroup containing the elements $$\{\dotsc, -2, -1, 0, 1, 2, \dotsc\}$$; there are two isomorphisms that demonstrate this: either the function $$x \mapsto 2x$$ or the function $$x \mapsto -2x$$. | 347 | 1,127 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 20, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2024-26 | latest | en | 0.79701 |
https://www.physicsforums.com/threads/bernoullis-equation-of-water-tower.283510/ | 1,643,023,396,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304528.78/warc/CC-MAIN-20220124094120-20220124124120-00410.warc.gz | 964,503,863 | 14,917 | # Bernoulli's Equation of water tower
## Homework Statement
The water tower in the drawing is drained by a pipe that extends to the ground. The flow is nonviscous.
(b) What is the absolute pressure at point 1 when the valve is opened and the water is flowing? Assume that the water speed at point 2 is negligible.
(c) Assuming the effective cross-sectional area of the valve opening is 2.09 multiplied by 10-2 m2, find the volume flow rate at point 1.
## The Attempt at a Solution
For part b:
p1 +density(g)(y) + .5(density)(v)^2 = p2 +density(g)(y) + .5(density)(v)^2
p1 + 0 + .5(1000)v^2 = 101000 + 1000(9.8)(15)
I have two variables p1 and v, so how can I solve this?
#### Attachments
• 11_64.gif
5.3 KB · Views: 1,143 | 224 | 731 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2022-05 | latest | en | 0.898186 |
https://socratic.org/questions/how-do-you-write-4327-in-scientific-notation | 1,579,744,302,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250608062.57/warc/CC-MAIN-20200123011418-20200123040418-00073.warc.gz | 673,329,809 | 5,821 | # How do you write 4327 in scientific notation?
##### 1 Answer
Mar 19, 2017
The decimal point needs to be moved 3 places to the left, therefore the 10s term will have a positive exponent:
$4327 = 4.237 \times {10}^{3}$ | 68 | 221 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 1, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2020-05 | longest | en | 0.861142 |
https://www.instructables.com/Using-Tinkercad-Models-to-Represent-Sizes-and-Dist/ | 1,713,983,931,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296819847.83/warc/CC-MAIN-20240424174709-20240424204709-00345.warc.gz | 724,096,466 | 34,999 | ## Introduction: Using Tinkercad Models to Represent Sizes and Distances in Space (Scene Category)
Astronomy teachers often use physical scale models to help students wrap their minds around the distances to and sizes of objects in space. Examples of these include "solar system walks" to lay out the distances to the planets and building models out of clay to represent their sizes.
In this Instructable I will outline a procedure for creating scale models for Astronomy instruction using Tinkercad.
• the first model focuses on sizes in the Solar System
• the second on distances in the Solar System
• and the third on the distances to stars.
Each step will include images showing models I made while following these procedures. I will also provide a document that can be used by students that doesn't have these "answers".
I think, given our current concerns with disease transmission through shared supplies and the strong possibility of students eLearning from home this fall, there is a real benefit to developing these types of "virtual hands-on" activities with resources like Tinkercad.
## Step 1: Size Scales in the Solar System
1. In Tinkercad, click on "Edit Grid" in the lower right hand corner, then resize the grid to a length and width of 500 mm.
2. Add three spheres to the grid. Color code them as follows:
BLUE = Earth GRAY = the Moon ORANGE = Jupiter
3. Use google or another resource to find the diameters of these three astronomical objects in kilometers.
4. The scale for our model will be: 1 mm in Tinkercad = 1000 kilometers in Real Life
5. Resize the spheres to create scale models of the three objects. NOTE: In Tinkercad you change the diameter of a sphere by changing its Length, Width, and Height.
6. Arrange your planets so that none of them overlap (I also decided to raise them vertically) and use the "Text" tool in "Basic Shapes" to add labels and a scale.
TEACHER NOTE ... To save you a little time, here are the diameters of the planets and scaled sizes of the spheres:
• Earth diameter = 12,742 km = 12.742 mm on Tinkercad
• Moon diameter = 3,474 km = 3.474 mm on Tinkercad
• Jupiter diameter = 139,820 km = 139.820 mm on Tinkercad
## Step 2: Distance Scales in the Solar System
Now that we've got scale models of the Earth, Moon, and Jupiter, we'll build a model that helps you understand distance scales in the solar system.
1. You might need to drag some of the text labels out of the way before you start just to clear up your Workplace.
2. Click on the Ruler tool and place the origin somewhere along the left side of the Workplane.
3. Drag the Earth so that it is positioned over the origin. It's vertical height doesn't really matter.
4. Use google or another resource to find the Earth-Moon distance in kilometers.
5. Drag the Moon to that it is the correct scaled distance from the Earth - the ruler tool will display the distance between the Earth and Moon spheres. Use the same scale (1 mm = 1000 kilometers).
6. Drag the labels back in to their correct places.
## Step 3: Discussion Questions & Extensions
Here are some suggested questions or extensions you can use after your students have created and shared their models. You can test their answers to the questions by copying and pasting the objects that are in their models.
Questions...
• The diameter of Jupiter is equivalent to how many Earths?
• How many Earth's would fit inside Jupiter?
• How many Jupiters could fit between the Earth and the Moon?
Extensions...
• Add the Sun to your scale model. NOTE - The Sun's diameter is 1,392,700 km in real life so it will be 1,392.7 mm in Tinkercad. This is much larger than the planets and moons so it will not fit on the Workplane. It is a cool way to get a sense of it size though. See the image above.
• Add other planets and/or their moons
• Research planets that have been found orbiting other stars in the galaxy, make scale models of these to compare them to the size of the Earth
## Step 4: Constellations and Distances to the Stars
Generally, students knowledge about stars when they enter an Introductory Astronomy class is limited the names of some constellations and an awareness that stars are very far away.
In this activity we'll build upon this knowledge by creating models of constellations that include a scale representations of the distances to the stars that make them up.
1. Start with the standard Tinkercad workspace. Add one star and make it's length and width 10 mm (a bit larger or smaller is fine).
2. Use google or another resource to find a picture of the constellation you'd like to model. For my example I've used the Big Dipper (which is technically not a constellation, but let's just go with it)
3. Make enough copies of your star in Tinkercad and arrange these to recreate the constellation. NOTE - If you've chosen a complicated constellation only include the major stars (enough to get the general shape of the constellation)
4. Now do some additional research to determine the distances from the Earth to the stars in this constellation.
5. To scale the distances, we're going to use the "floor" of the Workplane as the surface of the Earth and move the stars in your constellation to a height that corresponds to their distance from Earth. For the Big Dipper a scale of 1 mm = 1 lightyear works well, but for constellations with a greater range of distances you might try 1 mm = 2 ly or more.
6. With the stars at the correct distances, the shadow on the Workplane shows the appearance of the constellation from the Earth's surface. The view can be rotated to see how different perspectives make the arrangement of the stars look very different.
## Step 5: Second Example of Distances to Stars - the Constellation Orion
The images above are included as a second example of the activity in the previous step.
Note that there is a much greater range in the distances from Earth to these stars to a scale of 1mm = 2 ly was used.
Also, Orion consists of many more stars than the Big Dipper so only enough stars to represent the general shape of the constellation were included.
## Step 6: Examples of Student Work
The images above are examples of work students have done on these projects.
• Here are links to their creations on Tinkercad (In order to avoid posting student contact information publicly I copied their work into my Tinkecad account)
## Step 7: Student Worksheet
I hope you've found this Instructable helpful. Here is a guide students can use to work through the steps.
If you have any questions or suggestions, please let me know
Thanks! | 1,468 | 6,579 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2024-18 | latest | en | 0.923667 |
https://it.mathworks.com/matlabcentral/answers/834178-complex-image-processing-for-loop-vectorization?s_tid=prof_contriblnk | 1,675,174,965,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499871.68/warc/CC-MAIN-20230131122916-20230131152916-00342.warc.gz | 350,521,178 | 30,480 | # Complex image processing for loop vectorization
3 views (last 30 days)
Artur MKRTCHYAN on 19 May 2021
Commented: Artur MKRTCHYAN on 27 May 2021
Hello everyone, I am new to MatLab. I would like to reuse this code but I was wondering if there is a way to vectorize the double for loop ?
link to paper : Guided Image Filtering
function q = guidedfilter_color(I, p, r, eps)
% GUIDEDFILTER_COLOR O(1) time implementation of guided filter using a color image as the guidance.
%
% - guidance image: I (should be a color (RGB) image)
% - filtering input image: p (should be a gray-scale/single channel image)
% - local window radius: r
% - regularization parameter: eps
if ~(size(I,3) == 3)
error('The guidance image input should have 3 channels');
end
[hei, wid] = size(p);
if r<2*min(hei, wid), r = round(min(hei, wid)/4); end;
N = boxfilter(ones(hei, wid), r); % the size of each local patch; N=(2r+1)^2 except for boundary pixels.
mean_I = zeros(size(I));
for ii =1:size(I,3)
mean_I(:,:,ii) = boxfilter(I(:, :, ii), r) ./ N;
end
mean_p = boxfilter(p, r) ./ N;
mean_Ip = zeros(size(I));
for ii =1:size(I,3)
mean_Ip(:,:,ii) = boxfilter(I(:, :, ii).*p, r) ./ N;
end
% covariance of (I, p) in each local patch.
cov_Ip = zeros(size(I));
for ii =1:size(I,3)
cov_Ip(:,:,ii) = mean_Ip(:,:,ii) - mean_I(:,:,ii) .* mean_p;
end
% variance of I in each local patch: the matrix Sigma in Eqn (14).
% Note the variance in each local patch is a 3x3 symmetric matrix:
% rr, rg, rb
% Sigma = rg, gg, gb
% rb, gb, bb
var_I_rr = boxfilter(I(:, :, 1).*I(:, :, 1), r) ./ N - mean_I(:,:,1) .* mean_I(:,:,1);
var_I_rg = boxfilter(I(:, :, 1).*I(:, :, 2), r) ./ N - mean_I(:,:,1) .* mean_I(:,:,2);
var_I_gg = boxfilter(I(:, :, 2).*I(:, :, 2), r) ./ N - mean_I(:,:,2) .* mean_I(:,:,2);
var_I_rb = boxfilter(I(:, :, 1).*I(:, :, 3), r) ./ N - mean_I(:,:,1) .* mean_I(:,:,3);
var_I_gb = boxfilter(I(:, :, 2).*I(:, :, 3), r) ./ N - mean_I(:,:,2) .* mean_I(:,:,3);
var_I_bb = boxfilter(I(:, :, 3).*I(:, :, 3), r) ./ N - mean_I(:,:,3) .* mean_I(:,:,3);
a = zeros(hei, wid, 3);
for y=1:hei
for x=1:wid
Sigma = [var_I_rr(y, x), var_I_rg(y, x), var_I_rb(y, x);
var_I_rg(y, x), var_I_gg(y, x), var_I_gb(y, x);
var_I_rb(y, x), var_I_gb(y, x), var_I_bb(y, x)];
%Sigma = Sigma + eps * eye(3);
cov_Ip1 = [cov_Ip(y, x,1), cov_Ip(y, x,2), cov_Ip(y, x,3)];
a(y, x, :) = cov_Ip1 * inv(Sigma + eps * eye(3)); % Eqn. (14) in the paper;
end
end
b = mean_p - a(:, :, 1) .* mean_I(:,:,1) - a(:, :, 2) .* mean_I(:,:,2) - a(:, :, 3) .* mean_I(:,:,3); % Eqn. (15) in the paper;
q = (boxfilter(a(:, :, 1), r).* I(:, :, 1)...
+ boxfilter(a(:, :, 2), r).* I(:, :, 2)...
+ boxfilter(a(:, :, 3), r).* I(:, :, 3)...
+ boxfilter(b, r)) ./ N; % Eqn. (16) in the paper;
end
Artur MKRTCHYAN on 25 May 2021
Of course I've already used the profiler, and that's why I want to vectorize the double loop because that's exactly what takes more time. So I'm not asking to rewrite the code, I'm looking to see if it's possible to optimize only the double loop part.
Image Analyst on 21 May 2021
I would not reuse that code. I'd use the built-in imguidedfilter() function.
Artur MKRTCHYAN on 27 May 2021
It seems to me that this function has been created because the basic function is not adapted for RGB images. I'm not sure that's possible to make it this way.
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http://2010.igem.org/wiki/index.php?title=Team:TU_Munich/Software&diff=prev&oldid=108479 | 1,575,806,515,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540508599.52/warc/CC-MAIN-20191208095535-20191208123535-00464.warc.gz | 1,831,525 | 11,212 | # Team:TU Munich/Software
(Difference between revisions)
Revision as of 12:54, 18 October 2010 (view source)Sgude (Talk | contribs) (→2. Check the logic network)← Older edit Revision as of 12:56, 18 October 2010 (view source)Sgude (Talk | contribs) (→2. Check the logic network)Newer edit → Line 24: Line 24: The logic network designed in step 1 can now be verified. The logic network designed in step 1 can now be verified. - Logical values are represented by colors (green = true, red = false). Thus a green box represents a devices (e.g. an AND-gate) with the logical value true, a red box a devices with the logical value false. The flow of logical value can be followed by following the colors along the edges between the boxes. + Logical values are represented by colors (green = true, red = false). Thus a green box represents a devices (e.g. an AND-gate) with the logical value true, a red box a devices with the logical value false. The flow of logical value can be followed by following the colored edges between the boxes. - The logical value(s) of the input(s) can be manipulated by clicking the checkbox(es). + The logical value(s) of the input(s) can be manipulated by clicking on their checkbox(es). ==3. Turn the logic network into DNA== ==3. Turn the logic network into DNA==
## Revision as of 12:56, 18 October 2010
Home → Software iGEM MainPage
# Motivation
Theoretical possibilities provided by our system - borders of lab work - future applications
applet
# Tutorial
The applet consists of three parts: Design, Check and Turn-to-DNA. The first allows the user to enter inputs, whereas the second and third supply the user with information about his network. Below the different parts are described in detail.
## 2. Check the logic network
The logic network designed in step 1 can now be verified.
Logical values are represented by colors (green = true, red = false). Thus a green box represents a devices (e.g. an AND-gate) with the logical value true, a red box a devices with the logical value false. The flow of logical value can be followed by following the colored edges between the boxes.
The logical value(s) of the input(s) can be manipulated by clicking on their checkbox(es).
# Disclaimer
## Software
THIS SOFTWARE IS PROVIDED BY THE CONTRIBUTORS "AS IS" AND ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT LIMITED TO, THE IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE ARE DISCLAIMED. IN NO EVENT SHALL THE CONTRIBUTORS BE LIABLE FOR ANY DIRECT, INDIRECT, INCIDENTAL, SPECIAL, EXEMPLARY, OR CONSEQUENTIAL DAMAGES (INCLUDING, BUT NOT LIMITED TO, PROCUREMENT OF SUBSTITUTE GOODS OR SERVICES; LOSS OF USE, DATA, OR PROFITS; OR BUSINESS INTERRUPTION) HOWEVER CAUSED AND ON ANY THEORY OF LIABILITY, WHETHER IN CONTRACT, STRICT LIABILITY, OR TORT (INCLUDING NEGLIGENCE OR OTHERWISE) ARISING IN ANY WAY OUT OF THE USE OF THIS SOFTWARE, EVEN IF ADVISED OF THE POSSIBILITY OF SUCH DAMAGE. | 673 | 2,963 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2019-51 | latest | en | 0.918985 |
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Any decimal that has only a finite number of nonzero digits
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1)Any decimal that has only a finite number of nonzero digits is a terminating decimal. For example, 36, 0.72, and 3.005 are terminating decimals. If a, b, c, d and e are integers and p= 2^a3^b and q = 2^c3^d5^e , is p/q a terminating decimal?
1) a > c
2) b> d
thanks
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Re: Is p/p terminating decimal? [#permalink]
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20 Aug 2009, 06:52
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p/p=1, man :D
the question is:
$$p= 2^a*3^b$$ and $$q = 2^c*3^d*5^e$$ , is p/q a terminating decimal?
$$p/q=2^(a-c)*3^(b-d)*5^(-c)$$
therefore, as long as (b-d)>0, p/q is terminating
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Re: Is p/q terminating decimal? [#permalink]
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20 Aug 2009, 10:08
Yup B.
Stmt 2 is sufficient to ensure that all 3's in the denominator are cancelled out and hence we will always have a terminating decimal
Re: Is p/q terminating decimal? [#permalink] 20 Aug 2009, 10:08
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Any decimal that has only a finite number of nonzero digits
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Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 790 | 2,746 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2017-22 | latest | en | 0.833593 |
https://www.sourcecodeexamples.net/2023/08/difference-between-nan-and-undefined-in-javascript.html | 1,716,912,379,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059143.84/warc/CC-MAIN-20240528154421-20240528184421-00807.warc.gz | 851,156,384 | 29,591 | # NaN vs undefined in JavaScript
The concepts of NaN (Not-a-Number) and undefined in JavaScript are both special values, but they serve very different purposes. Below, you'll find a detailed comparison, a table summarizing the key differences, and an example to clarify these distinctions.
## NaN vs undefined in JavaScript
### NaN (Not-a-Number)
NaN is a special numeric value that represents the result of an undefined or unrepresentable mathematical operation. For example, dividing 0 by 0 or taking the square root of a negative number will result in NaN. It is important to note that NaN is not equal to anything, even itself.
### undefined
undefined is a primitive value automatically assigned to variables that have just been declared or to function arguments for which there are no corresponding parameters. It indicates that a variable exists but has not been assigned a value yet.
## Comparison Table
Feature 'NaN' 'undefined'
Type Number Undefined
Value Represents unrepresentable number Indicates a variable hasn't been assigned a value
Usage Result of invalid mathematical operations Result of declaring without initializing
Comparative with '==' NaN == NaN is false undefined == undefined is true
Comparative with '===' NaN === NaN is false undefined === undefined is true
## Example
Here’s an example to illustrate the concepts of NaN and undefined.
``````var a = 0/0; // Invalid division
var b;
console.log(a); // Outputs 'NaN' because 0 divided by 0 is unrepresentable
console.log(b); // Outputs 'undefined' because 'b' is declared but not assigned a value
console.log(a == a); // Outputs false because NaN is not equal to itself
console.log(b == b); // Outputs true because 'undefined' is equal to itself
console.log(a === a); // Outputs false because NaN is not strictly equal to itself
console.log(b === b); // Outputs true because 'undefined' is strictly equal to itself``````
## Conclusion
NaN and undefined are both special values in JavaScript, but they represent entirely different concepts. NaN is used to indicate an unrepresentable mathematical result, while undefined is used to signify that a variable has been declared but not yet assigned a value.
Understanding these differences and when each value is likely to occur is crucial for debugging and maintaining clear, functional code. Using functions like isNaN() to check for NaN and being mindful of variable initialization can help prevent unexpected behaviors related to these values in your code. | 506 | 2,500 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2024-22 | latest | en | 0.816676 |
https://sheetnatives.com/how-to-use-expon-dist-in-google-sheets/ | 1,680,255,883,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949598.87/warc/CC-MAIN-20230331082653-20230331112653-00197.warc.gz | 582,043,928 | 16,847 | # How to Use EXPON.DIST in Google Sheets in 2020?
EXPON.DISTReturns the value of the exponential distribution function with a specified lambda at a specified value.
Sample Usage
EXPON.DIST(4,0.5,FALSE)
EXPON.DIST(A2,A3,A4)
Syntax
EXPON.DIST(x, lambda, cumulative)
x – The input to the exponential distribution function.
If cumulative is TRUE then EXPON.DIST returns the cumulative probability of all values up to x.
lambda – The lambda to specify the exponential distribution function.
cumulative – Whether to use the exponential cumulative distribution.
Note
You can use EXPONDIST or EXPON.DIST to perform this function.
WEIBULL: Returns the value of the Weibull distribution function (or Weibull cumulative distribution function) for a specified shape and scale.
POISSON: Returns the value of the Poisson distribution function (or Poisson cumulative distribution function) for a specified value and mean.
NORMSINV: Returns the value of the inverse standard normal distribution function for a specified value.
NORMSDIST: Returns the value of the standard normal cumulative distribution function for a specified value.
NORMINV: Returns the value of the inverse normal distribution function for a specified value, mean, and standard deviation.
NORMDIST: The NORMDIST function returns the value of the normal distribution function (or normal cumulative distribution function) for a specified value, mean, and standard deviation.
NEGBINOMDIST: Calculates the probability of drawing a certain number of failures before a certain number of successes given a probability of success in independent trials.
LOGNORMDIST: Returns the value of the log-normal cumulative distribution with given mean and standard deviation at a specified value.
LOGINV: Returns the value of the inverse log-normal cumulative distribution with given mean and standard deviation at a specified value.
HYPGEOMDIST: Calculates the probability of drawing a certain number of successes in a certain number of tries given a population of a certain size containing a certain number of successes, without replacement of draws.
BINOMDIST: Calculates the probability of drawing a certain number of successes (or a maximum number of successes) in a certain number of tries given a population of a certain size containing a certain number of successes, with replacement of draws.
Examples | 473 | 2,370 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2023-14 | latest | en | 0.596618 |
http://www.bkgm.com/articles/Sengoku/FiveCount/FiveCount4-3.html | 1,539,932,469,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583512332.36/warc/CC-MAIN-20181019062113-20181019083613-00140.warc.gz | 424,466,498 | 2,432 | ## Five-Count Part 4.3: Common cancellation patterns in an adjustment
#### Five in a row
Checkers lined five in a row cancel all their error counts no mater where they are. This is because any five consecutive points have all different error numbers from -2 to +2 but no duplications. Check yourself about this, and you will understand that a five point prime in any location cancels all error numbers in it at once.
#### Weak symmetric patterns
Not exactly symmetric, but you can see they are "balanced" on the group center (the 5 point in this example.) A balanced pattern on a group center always cancel its error numbers. This pattern is very common and you will often see the same pattern in a real backgammon game.
The same "balanced" pattern, and it cancels all errors at once.
Not balanced as it is, but it would be balanced on the 9 point (the center of Group 2) if the three checkers on the 6 point were the 11 point. Or you can think that it is a weak symmetric around a group boundary and cancels all error numbers at once.
#### Diagonal corners in a box
Any pair of checkers on diagonal corners of the left side or the right side of the bar cancels error numbers together.
#### Distant symmetric patterns
This "distant symmetric" pattern may be a bit harder to catch at a glance. That's partly because they are many pips away, but mainly because the bar cuts off the board in two and change our perception of distance. If you removed that big bar form the board view, then you could easily see good symmetric patterns.
You can't just erase the bar from a real backgammon board of course. If you don't like this "fake" board image, then just compare two checkers' locations in the group boxes, and you may find symmetry there.
Continue on to: Summary of Sho's Pip Count, "Five-Count"
Sho Sengoku's Five Count
Overview: Summary of Sho's Pip Count, "Five-Count" Part 1: Quick View: Introduction to "Five-Count" Part 2: Techniques for Easier and Faster Counting Part 3: Practice, Practice, Practice Part 4: Even Faster! Part 4.1: Common patterns for "10s" in group counting Part 4.2: To get a "rough count" even faster Part 4.3: Common cancellation patterns in an adjustment
See: Other articles by Sho Sengoku
See: Other articles on Pip Counting | 545 | 2,295 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2018-43 | latest | en | 0.919987 |
https://groups.yahoo.com/neo/groups/ksurfschool/conversations/messages/4738 | 1,503,320,229,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886108268.39/warc/CC-MAIN-20170821114342-20170821134342-00372.warc.gz | 781,645,312 | 18,542 | Sorry, an error occurred while loading the content.
## 4738Understanding how a kite is depowered by releasing the bar
Expand Messages
• Feb 27, 2004
I am new to kiteboarding, awaiting better weather for lessons,
and have a question about depowering a kite after reading about the
subject.
If a rider is not hooked into the Chicken Loop and he releases
his bar out,(without dropping it),would that depower the kite?
As I understand it, if a rider was hooked into the Chicken loop
and he extended his bar out,(without dropping it),that would depower
the kite. The reason, as I understand it, is that releasing the bar
would release the back lines,(the front lines remaining in the same
relative position as before the bar was released because the Chicken
Loop would be holding them in place). Then with the rear lines
extended, the wind wind would spill out of the back of the kite
somewhat similar to releasing a sail on a sailboat and spilling the
wind out of it. In addition, under that set of facts the "angle of
attack" of the kite relative to the wind would change because the
back lines were extended and the front lines remained in the same
general position.
However, if the rider was not hooked into the Chicken Loop and
he extended his bar out,(without dropping it),then both sets of lines,
(front & back), would be released relatively the same distance. In
that case, wouldn't both sets of lines remain in the same relative
position in relation to each other? If so, then how would the
kite "spill wind" and depower? In the alternative, how would
the "angle of attack" of the kite relative to the wind change since
the front and back lines are still in the same relative position to
one another, even though both sets on lines had been "extended out"
when the rider extended out his bar? Am I thinking about this in the
proper context or is there a better explanation?
Thank you
• Show all 3 messages in this topic | 438 | 1,929 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2017-34 | latest | en | 0.977153 |
https://www.kidsacademy.mobi/printable-worksheets/online/age-3/math/word-problems/ | 1,716,670,969,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058834.56/warc/CC-MAIN-20240525192227-20240525222227-00273.warc.gz | 736,921,713 | 66,519 | # Word Problems Worksheets for 3-Year-Olds
Introducing our engaging collection of Homeschool Worksheets designed specifically for three-year-olds! These Word Problems worksheets are expertly crafted to introduce young learners to the basics of problem-solving and numeracy in a fun, interactive way. Each worksheet is filled with colorful illustrations and simple, age-appropriate challenges that capture the curiosity of little minds. Perfect for homeschooling parents, these worksheets provide a wonderful opportunity to lay a strong foundation in early math skills while keeping learning enjoyable and exciting. Help your child embark on their educational journey with confidence and enthusiasm with our expertly designed resources!
Favorites
Interactive
• 3
• Interactive
• Word Problems
## Eagle Subtraction Worksheet
Help your kids become math experts! Through practice and adult guidance, even addition and subtraction can seem simple. Show them how with this worksheet - get them to trace the dotted lines to connect each word problem to the correct picture and solve the number sentence with some eagles!
Eagle Subtraction Worksheet
Worksheet
## Subtracting Socks Worksheet
Before beginning this exercise with your children, warm them up with a counting game. If math is not their favorite subject, use this worksheet. Help them read the two word problems, then use their fingers to count and subtract. Ask them to select the correct answer and check the box.
Subtracting Socks Worksheet
Worksheet
## Holiday Lights Subtraction Worksheet
Before tackling the math problem, get your kids ready by asking them to share their favorite holiday and why. Then, look at the picture on the worksheet with them. Help them examine the numbers and problems, and select the number sentence that shows what's happening with each string of light. 80 words.
Holiday Lights Subtraction Worksheet
Worksheet
## Car Shop Subtraction Worksheet
Encourage your kids to practice addition and subtraction with this worksheet. Read the word problems aloud, help them solve and check the answer from the provided options. With enough exercises and guidance, this task should be fairly easy.
Car Shop Subtraction Worksheet
Worksheet
## Firewood Subtraction Worksheet
Guide your students to show that math can be fun! If your kindergartners need help with addition and subtraction, the simple problems in this worksheet are ideal. Ask them to look at the two pictures and find the correct answers. Help them check their work to build their confidence in math.
Firewood Subtraction Worksheet
Worksheet
## At the Beach Subtraction Worksheet
Take your kids to the beach and show them that math can be fun! Look at the picture on the worksheet, then help them solve the addition and subtraction problems. Check the answers together and enjoy a day of learning and fun!
At the Beach Subtraction Worksheet
Worksheet
## Leaf Subtraction Stories Worksheet
Help your kids become experts at subtraction and addition by guiding them through fun exercises and simple math problems. This worksheet with leaf subtraction problems will help them practice and also test your child. Go through the three equations with them, help them solve, and then check the answer.
Leaf Subtraction Stories Worksheet
Worksheet
This tracing sheet is great for kids to work on math skills and have fun too. Read the word problems to them, then help them trace the dotted lines to match the problem and number sentence to a picture. Not only is it educational, it's beautiful too - your kids will love learning about butterflies!
Worksheet
## Adding at the Hospital Worksheet
A hospital visit doesn't have to be scary for kids. With this worksheet, they can count along with doctors and nurses, by reading the word problems and tracing lines to the corresponding picture. Guide them through it and you'll help them have a fun experience.
Worksheet
In the desert, scorching heat and little water mean animals must adapt to survive. Show kids pictures of these animals and teach their names. Then, read the word problems in the worksheet, and have them check the box next to the pictures that portray each story. 80 words
Worksheet
## Match the Word Problems Worksheet
Read this word problem to your kids: Help them understand how it can be translated into a number problem. Trace the dotted lines to see how each picture and number sentence match the problem. With this worksheet, you can show your kids how easy it is to transform a word problem into a number problem.
Match the Word Problems Worksheet
Worksheet
## Australian Animal Subtraction Worksheet
Test your child's wildlife knowledge. Ask them to name animals indigenous to Australia or other countries. Show them the animals in this worksheet and have them match each word problem and picture with the correct number sentence.
Australian Animal Subtraction Worksheet
Worksheet
## Duckling Subtraction Worksheet
Help your students learn equations and math with this fun worksheet. There are two questions to solve with the help of the pictures. Ask them to subtract with their fingers and check the box with the correct answer. This makes math easier and more enjoyable!
Duckling Subtraction Worksheet
Worksheet
## Addition at the Zoo Worksheet
Visit the zoo with your kids and point out all the different animals. Ask them to name their favorites and identify animals on a worksheet. Challenge them to solve word problems related to the pictures. Ask them for the equation that fits the picture and check the answer.
Worksheet
Let your kids have fun with this baseball-themed addition worksheet! Read the three simple word sentences with them, then use the pictures to help them add with their fingers. Ask them to copy the fingers held up in the pictures, then check the box with the right answer. Kids who love baseball will enjoy this activity!
Worksheet
## Harvesting Subtraction Stories Worksheet
Before starting the exercise, ask your kids if they know what harvesting is. Show them the pictures and explain if needed. Help them read the word problems, use the pictures to find the answer and check the box above. There are two simple word problems in the printout.
Harvesting Subtraction Stories Worksheet
Worksheet
## Grocery Shopping Subtraction Worksheet
Take your kids on a virtual grocery trip with this worksheet! Help them read the two word problems and use the pictures to solve them. Then, have them circle the correct answers. It's an easy, fun way to get kids comfortable with math while also getting them excited about grocery shopping!
Grocery Shopping Subtraction Worksheet
Worksheet
Worksheet
## Under the Sea Addition Worksheet
Ask your kids to identify undersea animals in a picture, then solve the word problems at the bottom with it! If they're into the nature channel or marine life, they'll love this worksheet. Check the box next to the correct answer for each one. 80 words
Worksheet
## Instrument Subtraction Worksheet
Encourage kids to practice their addition and subtraction by having them identify the musical instruments on the worksheet, then read the subtraction problems and check the box with the matching number sentences.
Instrument Subtraction Worksheet
Worksheet
## Police Subtraction Worksheet
Police protect citizens and do more in the community. Kids can learn numbers with this math worksheet: two subtraction problems in the downloadable PDF. Read the problems with your kids, then check the box next to the number sentences that are correct.
Police Subtraction Worksheet
Worksheet
Get your kids learning addition and subtraction with this fun worksheet! Show them that solving equations can be enjoyable and help them match number sentences to the correct boxes. By doing this together, you'll create a positive learning experience that your kids won't forget.
Worksheet
## Money Mass Worksheet
By regularly working on math with your kids, they will become more confident. Ensure they understand the word problems in this printout, and help them solve it. Check the box for the correct answer of each to verify their work. With this practice, they will gradually get used to math and make progress.
Money Mass Worksheet
Worksheet
## Visiting a Volcano Word Problems Worksheet
Read the word problems to your kids, note down the numbers and help them solve. Check the box under the correct answer. Word problems are sentences posed as math problems; just like regular number problems, first you must understand and interpret the sentence.
Visiting a Volcano Word Problems Worksheet
Worksheet
Learning Skills
The Value of Worksheets on Word Problems in Early Childhood Development: Emphasizing Homeschool Printables
In the foundational years of a child’s education, especially around the age of three, engaging, age-appropriate learning materials are crucial. Worksheets, particularly those focusing on word problems, play a pivotal role in developing early mathematical and linguistic skills. For parents who opt for homeschooling, utilizing homeschool printables tailored to solve word problems can be incredibly beneficial in nurturing a child's cognitive and problem-solving abilities from a tender age.
Word problems are not just about numbers; they are narratives that require interpretation, reasoning, and solution-finding. For a three-year-old, these problems are simplified into stories that help them connect math concepts to real-world situations. Homeschool printables designed for this age group use simple, relatable scenarios to which children can easily relate. For instance, a worksheet might describe a situation involving the sharing of fruits or the counting of toys, which not only makes the exercise enjoyable but also relevant.
Cognitive Development
Homeschool printables featuring word problems stimulate cognitive development. They encourage young learners to listen attentively or read the problem (with parental help), understand the scenario presented, and figure out a solution using the objects and numbers involved. This practice enhances their comprehension skills and numerical understanding, which are critical at this developmental stage.
Language Skills
Integrating language with mathematical concepts, word problem worksheets for three-year-olds also boost vocabulary and language skills. As children decode the words to grasp the problem’s context, they learn new words and sentence structures, improving their communication skills. This dual focus on math and language prepares them for more complex learning as they grow.
Critical Thinking and Reasoning
Even at three, children are capable of basic reasoning. Homeschool printables that involve word problems encourage them to think critically. They learn to analyze situations and make connections, which are essential skills not just in math but in everyday problem-solving scenarios. By working through different problems, children learn that there can be multiple ways to approach and solve a problem, fostering flexibility in thinking.
Confidence and Independence
Starting word problems early through homeschool printables helps build a child’s confidence in their abilities to understand and solve problems independently. With the supportive guidance of a parent, children feel encouraged and motivated to tackle challenges, which is a significant step in their educational journey.
Customizable Learning Experience
One of the significant advantages of using homeschool printables is the ability to customize the learning experience to fit the child’s individual pace and interests. With word problem worksheets designed specifically for three-year-olds, parents can choose problems that resonate with their child’s everyday experiences or areas of interest, making learning more exciting and relevant. Whether it’s animals, toys, or food, each worksheet can be tailored to captivate a young learner’s curiosity.
Interactive Learning
Homeschool printables often encourage a hands-on approach. Many worksheets designed for young children include elements where they can use physical objects to count or solve problems, such as drawing, coloring, or placing stickers. This interaction goes beyond the mental exercise of solving word problems, incorporating fine motor skills and sensory experiences, which are crucial for this age group.
Preparation for Formal Schooling
Regular practice with word problems through homeschool printables can give children a significant advantage when they transition to formal schooling. Familiarity with the structure of problems and the concept of applying mathematics in various scenarios can ease the shift to more structured classroom learning. Early exposure to these concepts means children are not only ready but are also confident in their mathematical foundations when they start school.
Parent-Child Bonding
Using homeschool printables for word problems also offers a wonderful opportunity for parent-child bonding. This educational journey becomes a shared experience where parents are actively involved in guiding and witnessing their child’s progress. This involvement plays a crucial part in the child’s emotional and educational development, as children thrive under the attention and encouragement from their parents.
Conclusion
In conclusion, homeschool printables focusing on word problems for three-year-olds are more than just educational tools; they are foundational instruments that catalyze cognitive, linguistic, and emotional development. They prepare young learners for future educational challenges and foster a love for learning from an early age. By integrating these worksheets into their homeschooling routine, parents can ensure that their children are not only learning but also enjoying the process, making each educational experience valuable and fun. | 2,530 | 13,835 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2024-22 | latest | en | 0.913394 |
https://testbook.com/objective-questions/ta/mcq-on-transformer-on-no-load--5eea6a0939140f30f369d920 | 1,721,349,600,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514860.36/warc/CC-MAIN-20240718222251-20240719012251-00432.warc.gz | 486,032,347 | 61,138 | # Transformer on No Load MCQ Quiz in தமிழ் - Objective Question with Answer for Transformer on No Load - இலவச PDF ஐப் பதிவிறக்கவும்
Last updated on Apr 19, 2024
பெறு Transformer on No Load பதில்கள் மற்றும் விரிவான தீர்வுகளுடன் கூடிய பல தேர்வு கேள்விகள் (MCQ வினாடிவினா). இவற்றை இலவசமாகப் பதிவிறக்கவும் Transformer on No Load MCQ வினாடி வினா Pdf மற்றும் வங்கி, SSC, ரயில்வே, UPSC, மாநில PSC போன்ற உங்களின் வரவிருக்கும் தேர்வுகளுக்குத் தயாராகுங்கள்.
## Top Transformer on No Load MCQ Objective Questions
#### Transformer on No Load Question 1:
Power factor of a transformer on no load is poor due to
1. Magnetizing reactance of the transformer
2. Open circuited secondary
3. Low primary winding resistance
5. None of the above
Option 1 : Magnetizing reactance of the transformer
#### Transformer on No Load Question 1 Detailed Solution
Circuit diagram for a transformer on load:
Where,
V1 is the applied primary voltage
IW is the working component of current through R0 (Magnetizing resistance)
Iμ is the magnetizing component of current through X0 (magnetizing reactance)
N1 and N2 are primary and secondary turns ratio
• In the case of no-load, the secondary terminal of the transformer is open.
• There is no path available for the current to flow on the secondary side.
• Hence, the transformer does not draw current from the source.
• A small ampere of current flows through the primary transformer (no-load current I0) called excitation current (used for excitation of the core).
• No-load current (I0) is further divided into Iμ and IW
Phasor Diagram when transformer on no load:
Where,
E1 is the primary induced EMF
V1 is the primary terminal voltage
E2 is the secondary induced EMF
The magnetizing component of current is very much higher compared to the working component of current (Iμ > IW)
Power factor cos ϕ0 = (Iμ / I0) ≈ 0.2 to 0.25
So, the Power factor of a transformer on no load is poor due to the magnetizing reactance of the transformer.
#### Transformer on No Load Question 2:
Power factor of a transformer on no load is poor due to
1. Magnetizing reactance of the transformer
2. Open circuited secondary
3. Low primary winding resistance
Option 1 : Magnetizing reactance of the transformer
#### Transformer on No Load Question 2 Detailed Solution
Circuit diagram for a transformer on load:
Where,
V1 is the applied primary voltage
IW is the working component of current through R0 (Magnetizing resistance)
Iμ is the magnetizing component of current through X0 (magnetizing reactance)
N1 and N2 are primary and secondary turns ratio
• In the case of no-load, the secondary terminal of the transformer is open.
• There is no path available for the current to flow on the secondary side.
• Hence, the transformer does not draw current from the source.
• A small ampere of current flows through the primary transformer (no-load current I0) called excitation current (used for excitation of the core).
• No-load current (I0) is further divided into Iμ and IW
Phasor Diagram when transformer on no load:
Where,
E1 is the primary induced EMF
V1 is the primary terminal voltage
E2 is the secondary induced EMF
The magnetizing component of current is very much higher compared to the working component of current (Iμ > IW)
Power factor cos ϕ0 = (Iμ / I0) ≈ 0.2 to 0.25
So, the Power factor of a transformer on no load is poor due to the magnetizing reactance of the transformer.
#### Transformer on No Load Question 3:
A single-phase two winding transformer is designed to operate at 400/200 V, 50 Hz. If the h.v. side is now energized from a 400 V, 40 Hz source, the no-load l.v. side voltage would be
1. 300 V
2. 250 V
3. 200 V
4. 150 V
Option 3 : 200 V
#### Transformer on No Load Question 3 Detailed Solution
For the transformer when excited from different frequencies, the ratio of voltage to frequency always remains constant.
$$\frac{{{V_1}}}{{{f_1}}} = \frac{{{V_2}}}{{{f_2}}}$$
So, applying this concept for the HV side:
$$\frac{{400}}{{50}} = \frac{{{V_2}}}{{40}}$$
Therefore, the voltage on the HV side is,
V2 = 320 V
And Voltage on LV side using transformer by using turns ratio,
Turn ratio, $$a = \frac{{{V_{HV}}}}{{{V_{LV}}}}$$
Therefore, VLV = 320 / 2 = 160 V
But we are asked the voltage at the LV side at no-load.
A no-load test is performed at the Rated Voltage and Rated Frequency condition.
Therefore, VLV = 200 V, remains the same.
#### Transformer on No Load Question 4:
What percent of the full-load current drawn by a transformer when no load is applied?
1. 0.2 to 1.0 percent
2. 0 to 0.05 percent
3. 6 to 10 percent
4. 2 to 5 percent
Option 4 : 2 to 5 percent
#### Transformer on No Load Question 4 Detailed Solution
Concept:
Operation of transformer on no load:
When the transformer is operating at no load, the secondary winding is open-circuited, which means there is no load on the secondary side of the transformer and, therefore, current in the secondary will be zero.
While primary winding carries a small current I0 called no-load current which is 2 to 10% of the rated current.
The no-load current consists of two components:
Reactive or magnetizing component (Im):
It is in quadrature with the applied voltage V1. It produces flux in the core and does not consume any power.
Active or power component (Iw): It is also known as a working component. It is in phase with the applied voltage V1. It supplies the iron losses and a small amount of primary copper loss.
Phasor diagram:
• The function of the magnetizing component is to produce the magnetizing flux, and thus, it will be in phase with the flux.
• Induced emf in the primary and the secondary winding lags the flux ϕ by 90 degrees.
• The primary copper loss is neglected, and secondary current losses are zero as I2 is zero.
• Therefore, the current I0 lags behind the voltage vector V1 by an angle ϕ0 called the no-load power factor angle and is shown in the phasor diagram.
• The applied voltage V1 is drawn equal and opposite to the induced emf E1 because the difference between them is negligible at no load.
• Active component Iw is drawn in phase with the applied voltage V1.
• The phasor sum of magnetizing current Im and the working current Iw gives the no-load current I0.
Conclusion:
The no-load current drawn by a transformer is usually 2 to 5 percent of the full-load current.
#### Transformer on No Load Question 5:
The following test results were obtained from a 6 kVA, 200/400 V, 50 Hz single-phase transformer: Data for no-load low-voltage side: 200 V, 0.5 A and 50 W. At normal voltage and frequency, determine the magnetising current of the transformer.
1. 0.569 A
2. 0.433 A
3. 0.236 A
4. 0 A
5. None of these
Option 2 : 0.433 A
#### Transformer on No Load Question 5 Detailed Solution
Concept
Wattmeter Reading $$=W_o$$ , Voltmeter Reading $$=V_1$$ & Ammeter Reading $$=I_o$$
$$I_m=\sqrt{I_o^2-I_c^2}$$
The value of magnetizing resistance is given by:
$$X_m={V_1\over I_m}$$
Calculation
Given, $$W_o=50W$$ , $$V_1=200V$$ & $$I_o=0.5A$$
$$W_o=V_1I_ocosϕ$$
$$50=200× 0.5× cosϕ$$
cosϕ = 0.5
Ic = Io cosϕ
Ic = 0.5 × 0.5 = 0.25 A
$$I_m=\sqrt{0.5^2-0.25^2}$$
Im = 0.433 A
1.) Open circuit test
• This test is used to find the iron or constant losses of the transformer.
• This test gives the value of core resistance (RW) and determines magnetizing resistance (Xm)
• In this test, the voltmeter, wattmeter, and ammeter are connected to the LV side of the transformer, and the HV side is open-circuited.
• This test is performed at rated voltage and reduced current.
• As the secondary of the transformer is open, no-load current flows through the primary winding.
• The value of the no-load current is very small compared to the full-rated current. The copper loss occurs only on the primary winding of the transformer because the secondary winding is open. The reading of the wattmeter only represents the core and iron losses.
2.) Short circuit test
• This test is used to find the copper loss of the transformer.
• In this test, the voltmeter, wattmeter, and ammeter are connected to the HV side of the transformer, and the LV side is short-circuited.
• This test is performed at rated current and reduced voltage.
• A low voltage of around 5-10% is applied to that HV side with the help of a variac. Now with the help of variac applied voltage is slowly increased until the wattmeter, and an ammeter give a reading equal to the rated current of the HV side.
#### Transformer on No Load Question 6:
No load current of a transformer has
1. high magnitude and low power factor.
2. high magnitude and high power factor.
3. small magnitude and high power factor.
4. small magnitude and low power factor.
Option 4 : small magnitude and low power factor.
#### Transformer on No Load Question 6 Detailed Solution
Concept:
Circuit diagram for a transformer on load:
Where,
V1 is the applied primary voltage
IW is the working component of current through R0 (Magnetizing resistance)
Iμ is the magnetizing component of current through X0 (magnetizing reactance)
N1 and N2 are the primary and secondary turns ratio
• In the case of no load, the secondary terminal of the transformer is open.
• There is no path available for the current to flow on the secondary side.
• Hence, the transformer does not draw current from the source.
• A small ampere of current flows through the primary transformer (no-load current I0) called excitation current (used for excitation of the core).
• No-load current (I0) is further divided into Iμ and IW
Phasor Diagram when transformer on no load:
Where,
E1 is the primary induced EMF
V1 is the primary terminal voltage
E2 is the secondary induced EMF
Power factor cos ϕ0 = (Iμ / I0) ≈ 0.2 to 0.25
Explanation:
No load current = 2 A
As the magnetizing component of current is very higher.
From the options, the magnetizing component may be 1.8 A
#### Transformer on No Load Question 7:
In a practical transformer, a small current flows in the primary winding when an alternating voltage is applied at the primary side. The current is of the order of 3% to 5% of the rated current of the primary winding. What is the condition of the practical transformer mentioned in the above context?
4. Ideal current
#### Transformer on No Load Question 7 Detailed Solution
Concept:
Operation of transformer on no load:
When the transformer is operating at no load, the secondary winding is open-circuited, which means there is no load on the secondary side of the transformer and, therefore, current in the secondary will be zero.
While primary winding carries a small current I0 called no-load current which is 2 to 10% of the rated current.
The no-load current consists of two components:
Reactive or magnetizing component (Im):
It is in quadrature with the applied voltage V1. It produces flux in the core and does not consume any power.
Active or power component (Iw): It is also known as a working component. It is in phase with the applied voltage V1. It supplies the iron losses and a small amount of primary copper loss.
Phasor diagram:
• The function of the magnetizing component is to produce the magnetizing flux, and thus, it will be in phase with the flux.
• Induced emf in the primary and the secondary winding lags the flux ϕ by 90 degrees.
• The primary copper loss is neglected, and secondary current losses are zero as I2 is zero.
• Therefore, the current I0 lags behind the voltage vector V1 by an angle ϕ0 called the no-load power factor angle and is shown in the phasor diagram.
• The applied voltage V1 is drawn equal and opposite to the induced emf E1 because the difference between them is negligible at no load.
• Active component Iw is drawn in phase with the applied voltage V1.
• The phasor sum of magnetizing current Im and the working current Iw gives the no-load current I0.
Conclusion:
The no-load current drawn by a transformer is usually 2 to 5 percent of the full-load current.
#### Transformer on No Load Question 8:
"No load" test on transformers is done to determine which of the following?
1. Magnetizing current
2. Iron losses
3. Both (A) and (B)
4. To check transformer capacity
5. Copper losses
Option 3 : Both (A) and (B)
#### Transformer on No Load Question 8 Detailed Solution
Open circuit test:
• It is done by keeping one of the windings open (without load, usually high voltage winding is open) and applying rated voltage to other winding (usually low voltage winding because it is easier to apply rated voltage).
• The current drawn from this terminal is the no-load current at low power factor corresponding to the core loss component. Since the no-load current is very small it doesn’t contribute to the copper loss. Core loss is calculated by multiplying the applied voltage and no-load current.
• As the secondary side is open, the entire coil will be purely inductive in nature. So, the power will be lagging due to the inductive property of the circuit. So LPF (Low Power Factor) Wattmeter is used in the open circuit test of the transformer.
It is used to find
• The core loss (iron losses or constant losses) of the transformer
• Equivalent resistance referred to the metering side
• Shunt branch parameters i.e. magnetizing impedance
• The core loss is the sum of Hysteresis losses and Eddy current losses.
Short circuit test:
• It is done by shorting one of the winding terminals (usually low voltage terminal) and applying a small voltage across the other winding terminals (high voltage terminal because the current in HV terminal will be less and easy to handle) and using a wattmeter to measure the power dissipated in the LV terminal. Wattmeter will indicate the full load copper loss.
• In the short circuit test, the secondary winding of the transformer is short-circuited. As the secondary side is short-circuited the entire coil will be purely resistive in nature. So, the power factor will be High or unity.
Short circuit test is done to find
• The full load copper loss or ohmic loss
• Short circuit current
#### Transformer on No Load Question 9:
A single phase air core transformer fed from a single phase rated sinusoidal supply is operating at no load. The steady state magnetising current drawn by the transformer from supply will have waveform
Option 2 :
∵ air cored
∵
∴ i =
#### Transformer on No Load Question 10:
What is the net primary current in a transformer on load ?
1. Load component of secondary current
2. Magnetizing current
3. Reflected load current on primary side
4. Sum of magnetizing current and reflected load current on primary side
Option 4 : Sum of magnetizing current and reflected load current on primary side
#### Transformer on No Load Question 10 Detailed Solution
Circuit diagram for a transformer on load:
Where,
V1 is the applied primary voltage
IW is the working component of current through R0 (Magnetizing resistance)
Iμ is the magnetizing component of current through X0 (magnetizing reactance)
N1 and N2 are primary and secondary turns ratio
• In the case of no-load, the secondary terminal of the transformer is open.
• There is no path available for the current to flow on the secondary side.
• Hence, the transformer does not draw current from the source.
• A small ampere of current flows through the primary transformer (no-load current I0) called excitation current (used for excitation of the core).
• No-load current (I0) is further divided into Iμ and IW
Calculation:
In a transformer on load, the total primary current is the sum of magnetizing current and reflected load current on the primary side.
$${I_1} = {I_0} + \frac{{{I_2}}}{k}$$
Here, I1 is the primary current
I2 is the secondary or load current
K is the transformation ratio
I0 is the magnetizing current | 4,227 | 15,805 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2024-30 | latest | en | 0.504279 |
http://forums.wolfram.com/mathgroup/archive/2004/Jan/msg00280.html | 1,611,037,619,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703517966.39/warc/CC-MAIN-20210119042046-20210119072046-00267.warc.gz | 34,949,028 | 9,869 | RE: Struggling with list element assignment in functions
• To: mathgroup at smc.vnet.net
• Subject: [mg45582] RE: [mg45568] Struggling with list element assignment in functions
• From: "Wolf, Hartmut" <Hartmut.Wolf at t-systems.com>
• Date: Fri, 16 Jan 2004 06:04:58 -0500 (EST)
• Sender: owner-wri-mathgroup at wolfram.com
```For aesthetics I have to make two short notes:
>-----Original Message-----
>From: Wolf, Hartmut
To: mathgroup at smc.vnet.net
>Sent: Wednesday, January 14, 2004 1:30 PM
>To: 'mt at 3planes.com'; mathgroup at smc.vnet.net
>Subject: [mg45582] RE: [mg45568] Struggling with list element assignment in
>functions
>
>
>
>>-----Original Message-----
>>From: mt at 3planes.com [mailto:mt at 3planes.com]
To: mathgroup at smc.vnet.net
>>Sent: Wednesday, January 14, 2004 7:26 AM
>>To: mathgroup at smc.vnet.net
>>Subject: [mg45582] [mg45568] Struggling with list element assignment in
>functions
>>
>>
>>For context, I am trying to create a toroidal data structure by
>>causing edges of an array to be copies of
>>rows and columns on the first interior line of the opposite edge.
>>
>>I'm not entirely clear on what Hold does, but it seemed to solve my
>>problem when wrapping the rows of the array:
>>
>>===
>>Attributes[haloarray] = HoldFirst;
>>haloarray[array_]/;MatrixQ[array] := Module[{},
>> If [Length[array] <4 || Length[array[[1]] ]< 4,
>> (Print["haloarray: matrix too
>>small"];Return[Null])];
>> Map[halo,array];
>> array[[1]] = array[[Length[array] - 1]];
>> array[[Length[array]]] = array[[2]];
>> ]
>>
>>Attributes[halo] = {};
>>halo[x_]/;MatchQ[x,z_List] := Module[{y},
>> y = x;
>> If [Length[x] <4,(Print["halo: list too
>>short"];Return[Null])];
>> y[[1]] = x[[(Length[x] - 1)]];
>> y[[Length[x]]] = x[[2]];
>> ]
>>===
>>so, by putting in the HoldFirst, I could assign directly to the called
>>array in haloarray[] rather than having to copy it twice, to and from
>>a dummy array. Imagine my disappointment, then, in trying to extend
>>this idea to the halo[] routine. This
>>===
>>Attributes[halo] = HoldFirst;
>>halo[x_]/;MatchQ[x,z_List] := Module[{},
>> If [Length[x] <4,(Print["halo: list too
>>short"];Return[Null])];
>> x[[1]] = x[[(Length[x] - 1)]];
>> x[[Length[x]]] = x[[2]];
>> ]
>>===
>>refuses to do the assignments and yields all sorts of error messages.
>>Any ideas? It's not a showstopper for me but I'd like to know what's
>>up.
>>
>>thanks
>>Michael Tobis
>>http://geosci.uchicago.edu/~tobis
>>
>
>Michael,
>
>perhaps you fell into the error -- so common (in any field) --
>namely of "optimizing before solving the problem".
>
>I assume, you originally wanted to _extend_ your (matrix) data
>periodically by margins to the left and to the right.
>
>Take the one-dimensional case
>
>In[26]:= rr = Range[5]
>Out[26]= {1, 2, 3, 4, 5}
>(our test data)
>
>then there is a function in Mathematica, that does it:
>
>Out[28]= {5, 1, 2, 3, 4, 5, 1}
>
>
>This function also extends to the two-dimensional case. Let's
>again define a test sample:
>
>In[30]:= (tt = Array[CirclePlus, {5, 3}]) // MatrixForm
>
>(I suppress the output here, CirclePlus is only used for
>Now look at
>
>In[32]:= PadRight[tt, {7, 5}, tt, {1, 1}] // MatrixForm
>
>I guess this is your solution.
>
>As PadRight is a built-in, there usually is no need (or
>opportunity) for further optimization.
>
>
>
>
>
>Now let me explain, what went wrong with your attempt. (I drop
>all embellishments here)
>
>In[33]:= Attributes[halo] = HoldAll;
>halo[x_] :=
> (x[[1]] = x[[(Length[x] - 1)]];
> x[[Length[x]]] = x[[2]];)
>
>In[35]:= rr = PadRight[Range[5], 7, 0, 1]
>Out[35]= {0, 1, 2, 3, 4, 5, 0}
>(our test data for this case)
>
>In[36]:= halo[rr]
>In[37]:= rr
>Out[37]= {5, 1, 2, 3, 4, 5, 1}
>
>Ok, halo worked!
>
>What we did, was replace the first and last element by the
>margins of the (true) data kernel. In order to achieve that
>in-place operation, it is necessary to pass the data by a
>symbol (rr here) and hold the argument.
>
>
>Now we try to repeat the trick for the next dimension:
>
>In[38]:=
>(tt0 = PadRight[Array[a, {5, 3}], {7, 5}, 0, {1, 1}]) // MatrixForm
>Out[38]//MatrixForm=
> 0 0 0 0 0
>
> 0 a[1, 1] a[1, 2] a[1, 3] 0
>
> 0 a[2, 1] a[2, 2] a[2, 3] 0
>
> 0 a[3, 1] a[3, 2] a[3, 3] 0
>
> 0 a[4, 1] a[4, 2] a[4, 3] 0
>
> 0 a[5, 1] a[5, 2] a[5, 3] 0
>
> 0 0 0 0 0
>
>
>
>In[39]:=
>Attributes[haloarray] = HoldFirst;
>haloarray[array_] :=
> (Map[halo, array];
> array[[1]] = array[[Length[array] - 1]];
> array[[Length[array]]] = array[[2]];)
>
>(it doesn't work) The replacement of the rows is the same as
>what we did in the 1-dim case, and indeed, if we drop the line
>Map[halo,array] it works changing the rows.
>
>So we test
>
>In[46]:= Map[halo, tt0]
>
>(doesn't work) Why? Quite simple, Map cannot map halo over a
>symbol, such it has to evaluate its argument, hence we are
>lost (Input condition for halo is now violated).
>
>Normally we now would abandon the Map idea, but for
>illustration, how we can put it further:
>
>In[49]:= Clear[tt1, tt2, tt3, tt4, tt5, tt6, tt7]
>In[50]:=
>tt = Unevaluated /@ {tt1, tt2, tt3, tt4, tt5, tt6, tt7};
>
>We define tt as a list of Symbols. We wrapped them with
>Unevaluated, to stop evaluation of tt for Map.
>
>In[51]:=
>Thread[{tt1, tt2, tt3, tt4, tt5, tt6, tt7} = tt0]
Thread is not needed for Set; also avoid writing the list explicitly:
Evaluate[Evaluate /@ tt] = tt0
>
>This assigns our test data to tt. We cannot see them
>
>In[52]:= tt
>Out[52]=
>{Unevaluated[tt1], Unevaluated[tt2], Unevaluated[tt3],
>Unevaluated[tt4],
> Unevaluated[tt5], Unevaluated[tt6], Unevaluated[tt7]}
>
>As evaluation stops, but that is exactly what we need for
>haloarray. But it's really there, believe me:
>
>In[53]:= tt2
>Out[53]= {0, a[1, 1], a[1, 2], a[1, 3], 0}
>
>
>In[54]:= haloarray[tt]
>In[55]:= tt
>Out[55]=
>{Unevaluated[tt6], Unevaluated[tt2], Unevaluated[tt3],
>Unevaluated[tt4],
> Unevaluated[tt5], Unevaluated[tt6], Unevaluated[tt2]}
>
>look close what happend, the rows are right...
>
>...and also the columns
>
>In[57]:= tt2
>Out[57]=
>{a[1, 3], a[1, 1], a[1, 2], a[1, 3], a[1, 1]}
>
>
>To arrive at the desired result we just have to put away the
>wrappers, e.g. like this
>
>In[58]:= (tt = (Evaluate /@ tt)) // MatrixForm
>Out[58]//MatrixForm=
> a[5, 3] a[5, 1] a[5, 2] a[5, 3] a[5,1]
>
> a[1, 3] a[1, 1] a[1, 2] a[1, 3] a[1,1]
>
> a[2, 3] a[2, 1] a[2, 2] a[2, 3] a[2,1]
>
> a[3, 3] a[3, 1] a[3, 2] a[3, 3] a[3,1]
>
> a[4, 3] a[4, 1] a[4, 2] a[4, 3] a[4,1]
>
> a[5, 3] a[5, 1] a[5, 2] a[5, 3] a[5,1]
>
> a[1, 3] a[1, 1] a[1, 2] a[1, 3] a[1,1]
>
>
>
>But of course the right way to put forth your idea would be:
>
>In[81]:= Clear[haloarray]
>In[82]:=
>(tt0 = PadRight[Array[a, {5, 3}], {7, 5}, 0, {1, 1}]) // MatrixForm
>
>In[83]:=
>Attributes[haloarray] = HoldAll;
>haloarray[array_] := Module[{d1, d2},
> {d1, d2} = Dimensions[array];
> array[[All, 1]] = array[[All, d2 - 1]];
> array[[All, d2]] = array[[All, 2]];
> array[[1]] = array[[d1 - 1]];
> array[[d1]] = array[[2]];]
>
>In[85]:= haloarray[tt0]
>In[86]:= tt0
>
more elegant is this:
Attributes[haloarray] = HoldAll;
haloarray[array_] := (
array[[All, 1]] = array[[All, -2]];
array[[All, -1]] = array[[All, 2]];
array[[1]] = array[[-2]];
array[[-1]] = array[[2]];)
>
> | 2,833 | 7,419 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2021-04 | latest | en | 0.745111 |
https://www.solumaths.com/en/calculator/calculate/transpose_matrix/%5B%5B3;1;0%5D;%5B3;2;1%5D;%5B4;0;1%5D%5D | 1,679,374,108,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296943625.81/warc/CC-MAIN-20230321033306-20230321063306-00492.warc.gz | 1,060,374,717 | 15,808 | # transpose matrix calculator
The matrix calculator allows to calculate online the transpose of a matrix.
transpose_matrix([[3;1;0];[3;2;1];[4;0;1]]) returns [[3;3;4];[1;2;0];[0;1;1]]
### Transpose matrix, online calculus
#### Summary :
The matrix calculator allows to calculate online the transpose of a matrix.
transpose_matrix online
#### Description :
The calculator can calculate online the transpose of a matrix. Let M(n,p) a matrice where n is the number of rows and p the number of columns, The transpose of the matrix M(n,p) is the matrix obtained by exchanging rows and columns. The matrix calculator may calculate the transpose of a matrix whose coefficients have letters or numbers, it is a formal matrix calculator.
# Calculating the transpose of a matrix
The calculator can calculate the transpose of a matrix with the results in exact form: to calculate the transpose of matrix (3,3,4),(1,2,0),(0,1,7), enter transpose_matrix([[3;1;0];[3;2;1];[4;0;7]]), after calculation, the result is returned.
The calculator allows symbolic calculations, it is possible to use letters as well to calculate the transpose of a matrix like this: ((a,3*a,4),(1,2*a,0),(0,c,a/2)), enter transpose_matrix([[a;1;0];[3a;2a;c];[4;0;a/2]]), after calculation, the result is returned.
#### Syntax :
transpose_matrix(matrix)
#### Examples :
transpose_matrix([[3;1;0];[3;2;1];[4;0;1]]) returns [[3;3;4];[1;2;0];[0;1;1]]
Calculate online with transpose_matrix (transpose matrix calculator) | 399 | 1,492 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2023-14 | latest | en | 0.65419 |
http://creuse-news.eu/poker/r-matrix-roulette.php | 1,547,670,483,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583657867.24/warc/CC-MAIN-20190116195543-20190116221543-00010.warc.gz | 54,219,521 | 11,766 | 4692
# R-matrix roulette
GREED The basis of this program is based on the principle of "cyclic recurrence" of the roulette. I decided to use this area, as I saw the potential and flexibility. if anyone googles r-matrix roulette you will find the advertising website for this software. just wondering if anyone here has used it live and what were your thoughts? Jul 20, · [EN]: Newest development from the world's leader in roulette software. Get the details at official page: CREUSE-NEWS.EU
## Vertical Matrix
Thanks to your program I was able to recoup the investment in a few times in just one week. This system is most recommended for European wheel players. Although it is worth noting that there are some customers who consider Greed 1. At the end of the first column, the 11 and 5 appear together 7 spins apart. This is more than impressive results. Your Rmatrix is amazing! It wins 5 spins later with the 8.
It Is Interesting
• There is a version according to which the main concept of roulette was invented by Blaise Pascal, who was trying to create a perpetual motion machine.
• In the US, roulette first appeared in New Orleans.
• One Englishman named Ashley Revell once sold all his possessions in order to go to Las Vegas and play roulette. This evening, Ashley won 135 thousand 300 dollars, putting everything on red.
• Once Albert Einstein was asked if there is any roulette game system that guarantees a win. He replied: "Yes, I know one thing - it's to steal chips when the dealer does not see it".
• Croupier's clothes are given out by the casino. There are no pockets there, so you can not hide or steal chips.
• The center of the casino is Las Vegas. Every year 40 million gambling people from all over the world come here.
## Incredible Roulette System: Win \$100/Hour Playing Just 4 Numbers!
The vertical matrix system utilises a unique method of recording results. For this sytem we will only be considering the dozens. There are three dozens on the roulette table: Dozen 1 — 1 Dozen 2 — 13 — 24 Dozen 3 — 25 — 36 We will record results in respect to which dozen they are in, i. Results will be recorded in a 7 wide matrix, for example if we take these spins: For example if the game continued as shown below: At this point we would now be betting that the vertical triplet shown in green will NOT become another quad.
We are therefore betting against dozen 3, so we put one chip on dozen1 and one chip on dozen 2 as shown below: For example if we lost and the game continued like this: We would now be betting against the triplet becoming a quad however this time, as we lost before, we would now be betting 3 units on each dozen. We would be betting against dozen 2 so the bet would be 3 units on dozen 1 and 3 units on dozen 3 as shown below: If this bet loses, we wait for the next triplet and repeat the process but we would now be betting 9 units on each dozen.
If that bet also lost, we would wait for the next triplet and repeat the process but would now be betting 27 units on each dozen. If that bet loses, stop and start all over again, i. It is important to wait for the first quad as a trigger before you begin betting on the next triplet. This is because a vertical quad is a rare occurrence and we are essentially betting against five of them forming in a row.
If five quads do form in a row then, following the progression of 1 — 3 — 9 — 27, you will have lost a total of 80 units. This is a very rare event however and it is likely that you will have won much more than 80 units before you ever see five quads in a row. Another important note is the zero. If a zero comes at any time, it voids any triplet or quad that may be forming.
The first and most obvious victims of the government's lies are the 40,000 or so Americans who this year will become HIV-positive, overwhelmingly gay men or poor, inner-city drug users and their sexual partners. She was an intravenous drug user. Still would. I turned to the door. She loved to imagine their smooth bodies sliding over her, and their firm cocks being rammed into her wet pussy.
Jake moved towards him reluctantly. All the time, rubbing it against Jake's face and genitals.
### View Details
If we take as a fact that the latest version 1. The idea of R-Matrix Do you ever think of what brings together, for example, the fields of equal chances and a dozens, a dozens and a six-lines, six-lines and streets? Or how can you avoid the scary factor of "ZERO" when using a particular strategy?
The answer is - the numbers. Roulette - it is primarily the numbers. Operating with bets on numbers we can combine many probabilistic and mathematical schemes that will ensure a very high chance of a successful spin. This idea formed the basis for the development of R-Matrix algorithm. The program is a hot fusion of mathematics, probability theory, statistics and chaos theory.
Outward simplicity hides a lot of complex algorithms for bets that changes in a variety of behaviors, depending on a variety of game situations. This approach makes the R-Matrix almost unpredictable. Really, no automatic protection system will not be able to detect R-Matrix.
But the application is able to monitor specific combinational circuits in different bet fields of the table and make a prediction with the maximum accuracy Example of the game with R-Matrix 2. Thanks to your program I was able to recoup the investment in a few times in just one week.
This is more than impressive results. To be honest, I did not expect at all Your Rmatrix is amazing! And there were no misfires!
## Roulette de protection ktm 690 smc r
Pour commencer, le moteur trop creux en bas, ensuite le train avant trop flou et pour finir la selle un poil trop haute et trop dure. L'aboutissement du supermot' de route, c'est bien elle, pas de doute pour moi.
Bref c'est le bonheur et je suis amoureux comme au premier jour voire plus. Alors, j'aurais pu prendre une R ou investir dans un meilleur amortissement certes, mais la position ne me plaisait pas des masses non plus trop en avant vs. Je suis comme un dingue pendant ce mois d'attente. Proche de Dampierre, je croiserais quelques motards jeunots imprudents dont l'un s'est mis au tas avec sa FR dans une descente.
## Video
### Free Roulette
The thrill of watching the spinning red and black Roulette wheel has long served to grip many avid gamblers around the g... | 1,466 | 6,398 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2019-04 | latest | en | 0.94136 |
https://link-to-results.com/karawinna/genetic-algorithm-tutorial-python.php | 1,643,203,521,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304954.18/warc/CC-MAIN-20220126131707-20220126161707-00393.warc.gz | 415,211,501 | 9,462 | # Genetic Algorithm Tutorial Python
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### TPOT in Python (article) DataCamp
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### Genetic Algorithms with Python – ScanLibs
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## Getting started with genetic algorithms a tutorial Sicara
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Genetic Algorithms with Python by Genetic algorithms are one of the tools you Each chapter is a step-by-step tutorial that helps to build your skills This tutorial will implement the genetic algorithm optimization technique in Python based on a simple example in which we are trying to maximize the output of an | 4,494 | 23,667 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2022-05 | longest | en | 0.818222 |
https://www.spoj.com/problems/TLE/en/ | 1,642,706,043,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320302622.39/warc/CC-MAIN-20220120190514-20220120220514-00086.warc.gz | 932,880,992 | 8,210 | ## TLE - Time Limit Exceeded
no tags
Given integers N (1 ≤ N ≤ 50) and M (1 ≤ M ≤ 15), compute the number of sequences a1, ..., aN such that:
• 0 ≤ ai < 2M
• ai is not divisible by ci (0 < ci ≤ 2M)
• ai & ai+1 = 0 (that is, ai and ai+1 have no common bits in their binary representation)
### Input
The first line contains the number of test cases, T (1 ≤ T ≤ 10). For each test case, the first line contains the integers N and M, and the second line contains the integers c1, ..., cN.
### Output
For each test case, output a single integer: the number of sequences described above, modulo 1,000,000,000.
### Example
```Input:
1
2 2
3 2
Output:
1
```
The only possible sequence is 2, 1. | 215 | 696 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2022-05 | latest | en | 0.816127 |
https://www.electronicspoint.com/forums/threads/can-anyone-tell-me-what-a-spark-gap-capacitor-is.7252/ | 1,606,790,479,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141542358.71/warc/CC-MAIN-20201201013119-20201201043119-00002.warc.gz | 645,492,118 | 13,321 | can anyone tell me what a spark gap capacitor is?
Discussion in 'Electronic Basics' started by [email protected], May 3, 2005.
1. Guest
i went to an electronics store today looking for high voltage
capacitors for building a marx generator,the lady behind the counter
was as confused as i am by the name spark gap capacitor. can anyone
tell me how these are any different if at all from regular capacitors,
and would they work for a marx generator?
2. Don BruderGuest
No difference in function, but maybe a difference in construction - A
cap specifically intended for spark-gap use might be built a bit
"beefier" to avoid frying it with the high voltages that a spark gap
wants to run at, or with different materials so that its "dump" speed is
faster, but aside from that... <shrug> A cap is a cap is a cap, at least
as far as basic function is concerned.
Basically, it sounds like you're looking for a high voltage rating, high
capacitance unit. What voltage range are you expecting the spark gap to
fire at? Use that as a base, and look for something rated for at least
that high a voltage, if not higher, in the mFd/pFd range your tuning
calls for. Your main worry is probably going to be arcing between the
plates and the possible resulting "pop" (or maybe "KABOOM!", depending
on how much power you're running) that can result from exceeding the
breakdown voltage of the cap's dielectric material.
3. SioLGuest
Maybe the answer is hidding in this last sentence, it might be a cap with
two safety electrodes (air gap) that discharge when voltage comes close
to maximum allowed voltage for that cap thus protecting the cap from
explosion/destruction.
4. colinGuest
yes thats what this is, its a capacitor with a built in spark gap in
parallel, it often looks like a totaly ordinary ceramic capacitor with a
slot machined into the top of it.
theyr often used for protecting high voltage electronics where even higher
voltages are used such as in CRT where the high voltage EHT might arc over
to other electrodes ocasionaly. youl find them on the base of most CRTs
not sure they have any use for a marx generator though.
Colin =^.^=
5. John LarkinGuest
It's usually a ceramic disc cap that has a slit sawed a bit of the way
down from the top edge of the disk. Imagine cutting a small
rectangular slice out of a pie. When the voltage gets high enough, an
arc jumps between the exposed metalization layers, protecting some
other circuit component, or the cap itself, from overvoltage.
John
6. JamieGuest
just two plates close together that will arc..
also they made ceramic caps at one time that
has a spark gap cut in them but i don't think
that is what you want.
check a guess.
7. Guest
thanks for all of your help. i plan on running this generator at the
low end of high voltage, the caps they had were rated at 2 kv. as for
the capacity, 1nf. i am not planning on blinding 2 foot arcs here, just
a little high voltage generator to "play" with. while wre on the
subject of capacitors, when using dc pulses and a capacitor to even
them out do capacitors maintain voltage while amperage drops between
pulses or the other way around? thank you so much. | 764 | 3,171 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2020-50 | longest | en | 0.955063 |
http://metamath.tirix.org/mpests/nfxfr | 1,718,556,500,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861665.97/warc/CC-MAIN-20240616141113-20240616171113-00702.warc.gz | 22,241,513 | 1,948 | # Metamath Proof Explorer
## Theorem nfxfr
Description: A utility lemma to transfer a bound-variable hypothesis builder into a definition. (Contributed by Mario Carneiro, 11-Aug-2016)
Ref Expression
Hypotheses nfbii.1 ${⊢}{\phi }↔{\psi }$
nfxfr.2 ${⊢}Ⅎ{x}\phantom{\rule{.4em}{0ex}}{\psi }$
Assertion nfxfr ${⊢}Ⅎ{x}\phantom{\rule{.4em}{0ex}}{\phi }$
### Proof
Step Hyp Ref Expression
1 nfbii.1 ${⊢}{\phi }↔{\psi }$
2 nfxfr.2 ${⊢}Ⅎ{x}\phantom{\rule{.4em}{0ex}}{\psi }$
3 1 nfbii ${⊢}Ⅎ{x}\phantom{\rule{.4em}{0ex}}{\phi }↔Ⅎ{x}\phantom{\rule{.4em}{0ex}}{\psi }$
4 2 3 mpbir ${⊢}Ⅎ{x}\phantom{\rule{.4em}{0ex}}{\phi }$ | 267 | 617 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 7, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2024-26 | latest | en | 0.542162 |
https://numbermatics.com/n/8491992/ | 1,632,377,711,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057417.10/warc/CC-MAIN-20210923044248-20210923074248-00508.warc.gz | 478,078,044 | 6,216 | # 8491992
## 8,491,992 is an even composite number composed of three prime numbers multiplied together.
What does the number 8491992 look like?
This visualization shows the relationship between its 3 prime factors (large circles) and 16 divisors.
8491992 is an even composite number. It is composed of three distinct prime numbers multiplied together. It has a total of sixteen divisors.
## Prime factorization of 8491992:
### 23 × 3 × 353833
(2 × 2 × 2 × 3 × 353833)
See below for interesting mathematical facts about the number 8491992 from the Numbermatics database.
### Names of 8491992
• Cardinal: 8491992 can be written as Eight million, four hundred ninety-one thousand, nine hundred ninety-two.
### Scientific notation
• Scientific notation: 8.491992 × 106
### Factors of 8491992
• Number of distinct prime factors ω(n): 3
• Total number of prime factors Ω(n): 5
• Sum of prime factors: 353838
### Divisors of 8491992
• Number of divisors d(n): 16
• Complete list of divisors:
• Sum of all divisors σ(n): 21230040
• Sum of proper divisors (its aliquot sum) s(n): 12738048
• 8491992 is an abundant number, because the sum of its proper divisors (12738048) is greater than itself. Its abundance is 4246056
### Bases of 8491992
• Binary: 1000000110010011110110002
• Base-36: 520GO
### Squares and roots of 8491992
• 8491992 squared (84919922) is 72113928128064
• 8491992 cubed (84919923) is 612390900752094463488
• The square root of 8491992 is 2914.1022631335
• The cube root of 8491992 is 204.0186449703
### Scales and comparisons
How big is 8491992?
• 8,491,992 seconds is equal to 14 weeks, 6 hours, 53 minutes, 12 seconds.
• To count from 1 to 8,491,992 would take you about twenty-one weeks!
This is a very rough estimate, based on a speaking rate of half a second every third order of magnitude. If you speak quickly, you could probably say any randomly-chosen number between one and a thousand in around half a second. Very big numbers obviously take longer to say, so we add half a second for every extra x1000. (We do not count involuntary pauses, bathroom breaks or the necessity of sleep in our calculation!)
• A cube with a volume of 8491992 cubic inches would be around 17 feet tall.
### Recreational maths with 8491992
• 8491992 backwards is 2991948
• The number of decimal digits it has is: 7
• The sum of 8491992's digits is 42
• More coming soon!
HTML: To link to this page, just copy and paste the link below into your blog, web page or email.
BBCODE: To link to this page in a forum post or comment box, just copy and paste the link code below: | 730 | 2,597 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2021-39 | latest | en | 0.840122 |
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# The most prominent characteristic of High-performance
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The most prominent characteristic of High-performance [#permalink]
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14 Feb 2013, 08:31
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The most prominent characteristic of High-performance marketing firms is that they have a tendency to have concentrated greater executives in the direction of a smaller, more careful selection of a number of important businesses than is the case with other marketing firms.
A) that they have a tendency to have concentrated greater executives in the direction of a smaller, more careful selection of a number of important businesses than is the case with
B) that they tend to concentrate more executives to a smaller, more careful selection of a number of important businesses than toward
C) that they have a tendency to concentrate more executives on a smaller, more careful selection of a number of important businesses as opposed to
D) that they tend to concentrate more executives on a smaller, more carefully selected number of important businesses than do
E) the tendency to concentrate a greater number of executives on a careful and small selection of a number of important businesses as opposed to
[Reveal] Spoiler: OA
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The most prominent characteristic of High-performance marketing firms is that they have a tendency to have concentrated greater executives in the direction of a smaller, more careful selection of a number of important businesses than is the case with other marketing firms.
A) that they have a tendency to have concentrated greater executives in the direction of a smaller, more careful selection of a number of important businesses than is the case with
- Completely awkward, Illogical, Incorrect placement of Adjective
B) that they tend to concentrate more executives to a smaller, more careful selection of a number of important businesses than toward
- Although this option is not ambiguous but the use of "TO" is incorrect
C) that they have a tendency to concentrate more executives on a smaller, more careful selection of a number of important businesses as opposed to -
This option is ambiguous in meaning: moreover use of MORE is also confusing
D) that they tend to concentrate more executives on a smaller, more carefully selected number of important businesses than do - CORRECT
E) the tendency to concentrate a greater number of executives on a careful and small selection of a number of important businesses as opposed to -
THAT is Necessary otherwise it will change the meaning to "characteristic of X is their tendency". Completely illogical
Thus the answer has to be D & it is definitely a 700+ level question
Hope this explanation helps.
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Re: The most prominent characteristic of - New Question [#permalink]
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The most prominent characteristic of High-performance marketing firms is that they have a tendency to have concentrated greater executives in the direction of a smaller, more careful selection of a number of important businesses than is the case with other marketing firms.
A) that they have a tendency to have concentrated greater executives in the direction of a smaller, more careful selection of a number of important businesses than is the case with -- use of 'greater' is not correct
B) that they tend to concentrate more executives to a smaller, more careful selection of a number of important businesses than toward ---this construction wrongly suggests that more executives are concentrated on important businesses than towards marketing firms. This option changes the intended meaning of the sentence
C) that they have a tendency to concentrate more executives on a smaller, more careful selection of a number of important businesses as opposed to here 'tendency' is compared with 'other marketing firms'
D) that they tend to concentrate more executives on a smaller, more carefully selected number of important businesses than do Correct
E) the tendency to concentrate a greater number of executives on a careful and small selection of a number of important businesses as opposed to Same as C, moreover construction is awkward
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Re: The most prominent characteristic of - New Question [#permalink]
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Nice explanations Piyush and Fame; I am updating the post with the OA and my OE.
Vercules wrote:
The most prominent characteristic of High-performance marketing firms is that they have a tendency to have concentrated greater executives in the direction of a smaller, more careful selection of a number of important businesses than is the case with other marketing firms.
A) that they have a tendency to have concentrated greater executives in the direction of a smaller, more careful selection of a number of important businesses than is the case with
The comparison is correct in the original sentence but it is unnecessarily wordy because in its use of
“they have a tendency”
"in the direction of"
and the present perfect “have concentrated” is inappropriate to express a regular feature of marketing firms; simple present will be better, moreover "greater" is used incorrectly.
B) that they tend to concentrate more executives to a smaller, more careful selection of a number of important businesses than toward
This choice is clear and concise. However, in its use of “toward other marketing firms,” this choice does not draw the correct and logical comparison between the behavior of top marketing firms and the behavior of other marketing firms. Instead, this choice illogically compares the level of resources concentrated on certain important customers and the resources directed toward other marketing firms. Finally, the construction “concentrate more resources to” is unidiomatic; the appropriate idiom is “to concentrate on.”
C) that they have a tendency to concentrate more executives on a smaller, more careful selection of a number of important businesses as opposed to
This choice incorrectly draws a comparison between the level of resources concentrated on a number of important customers and the resources directed towards other marketing firms in its use of “as opposed to other marketing firms.” The correct comparison is between the behavior of top marketing firms and that of other marketing firms.
D) that they tend to concentrate more executives on a smaller, more carefully selected number of important businesses than do -- correct
This choice correctly draws a comparison between a characteristic of top marketing firms and that of other marketing firms, and is clear and concise.
E) the tendency to concentrate a greater number of executives on a careful and small selection of a number of important businesses as opposed to
This choice incorrectly draws a comparison between the level of resources directed toward a number of important customers and the resources directed toward other marketing firms in its use of “as opposed to.” The correct comparison is between top marketing firms and other marketing firms.
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# AP Physics Potential Energy
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A month ago
AP Physics Potential Energy 3. Planet X-39 has a mass equal to 1/3 that of Earth and a radius equal to 1/3 that of Earth. If v is the escape speed for Earth, what is the escape speed for X-39? Read 90 times 2 Replies
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Anonymous
wrote...
A month ago
Escape velocity is the minimum velocity needed for an object to overcome the gravitational force exerted by the planet. It is given by the following relation :V = √(2GM/r)where,V - escape velocityG - gravitational constantM - mass of planetr - radius of planetWe can take the ratio of escape velocities as follow :V1/V2 = √(M1/M2) * √(r2/r1)where, 1 is for Earth and 2 is for X-39given that,V1 = VM2 = (1/3)*M1 => M1/M2 = 3r2 = (1/3)*r1 => r2/r1 = 1/3therefore,V/V2 = √3 * √(1/3) = 1ie. V2 = Vie. Escape velocity of X-39 is same as that of Earth.
Anonymous
wrote...
A month ago
The escape speed for a planet is determined by its mass and radius, and it can be calculated using the formula:v = √(2GM/R)where G is the gravitational constant, M is the mass of the planet, and R is its radius.Since X-39 has a mass equal to 1/3 that of Earth and a radius equal to 1/3 that of Earth, the escape speed for X-39 can be calculated as follows:v = √(2G * (1/3M) / (1/3R)) = √(2G * M / R) / √(1/3) = √(2GM/R) / √(1/3) = v / √(1/3)So the escape speed for X-39 is 1/√(3) times the escape speed for Earth.
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https://math.stackexchange.com/questions/1371192/derivative-of-highest-order-is-enough-for-the-sobolev-norm?noredirect=1 | 1,571,275,337,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986672431.45/warc/CC-MAIN-20191016235542-20191017023042-00495.warc.gz | 581,780,028 | 31,541 | Derivative of highest order is enough for the Sobolev norm?
Thinking about the partial derivative in this question $\Delta u$ is bounded. Can we say $u\in C^1$? of mine, I encountered this post.
Equivalent Norms on Sobolev Spaces
1. I wonder if this hold when $\alpha\neq 2$ as well.
2. Too many details are omitted in this post. Could I ask a reference?
Yes, it is true.
This is actually an general idea for space involving several order of derivatives: "the extreme terms in a sum often already suffice to control the intermediate terms". Notice that by extreme we mean both highest order and the lowest order.
For example, $W^{3,p}$ norm of $u$ can be controlled by using only $L^p$ norm of $u$ and the $L^p$ norm of 3rd derivative of $u$.
This idea also applied on space $C^p(\Omega)$, the continuous differentiable function space of order $p$ with $L^\infty$ norm. Also, Holder space is applied as well.
For a good reference of this idea, I would suggest you to read this post by Terence Tao, look for exercise 2 for more explanation.
Also, for Equivalent Norms on Sobolev spaces, first look at this post, Theorem 2.7 for a summarization, look this book, page 133, theorem 5.2 for details proof. (The proof is not short)
• Just adding Theorem 5.2 in Adams--Fournier 2nd ed is on p. 135 at least in the one I have. – shall.i.am Sep 7 '15 at 1:08 | 367 | 1,360 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2019-43 | latest | en | 0.931943 |
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# Gregor Zoettl† - page 34 / 48
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d
d x c 1 , c
2
h
i(x,q dx c FB 0 , c 0 0
)i
= 0, for all [c1, c2] different from [c0, c00]. Formally this is due to the fact
t h a t t h e c r i t i c a l d e m a n d r e a l i z a t i o n θ c F B
for c [c0, c00] as defined by (5) does only depend
on X(c) for c [c0, c00], but not on X(c) for c [c1, c2]. Thus expression (6) is not a function of X(c) for c [c1, c2] and the cross derivative equals zero. The argument is analogous for
d dx c i )
h
i(x,q dx c FB 0 , c 0 0
)i
=0
The second derivatives with respect to the same technologies are given as follows:
c00
d 2 π i ( x , q F B d (x 2 ( c 0 , c 0 0 ) )
)
=
## Zµ
c0
f ( θ c F B
)
θ c F B d x ( c 0 , c 0 0 )
dc < 0.
(21)
## This expression is negative since
dx
c FB
c 0 , c 0 0 )
• >
0, (compare expression (5)).
## And for the second derivative with respect to capacity choice we obtain:
d 2 π i ( x , q d (x(ci)) F B 2
)
=
Z c F B
P q
## X, θ¢
dF (θ) < 0.
(22)
N o t i c e : S i n c e t h e i n t e g r a n d i s c o n t i n u o u s a t θ c F B
(equals zero), the derivative with respect
to this lower limit drops out according to Leibnitz rule.
# Proof of lemma 3
A c c o r d i n g t o l e m m a 2 , t h e o v e r a l l c a p a c i t y b o u n d X nology cneeds to satisfy the following to conditions: F B
under perfect competition at tech-
(i)
## F (θ∗)
(kc(c) + 1) = 0
(ii)
X F B
:
Z cF B
P ( X F B
, θ)
c= k(c)
W e r e w r i t e t h e i n t e g r a n d o f ( i i ) i n t e r m s o f t h e c r i t i c a l d e m a n d r e a l i z a t i o n θ c F B use of its definition (5) and obtain:
by making
P ( X F B
, θ)
c= θ
B ( X F B
)
c= θ
θ c F B
We obtain expression (8) as given in lemma 3:
(i)
(ii)
F (θ) Zθ
(kc(c) + 1) = 0
θdF (θ) = k(c)
34
Document views 128 Page views 128 Page last viewed Fri Oct 28 09:21:14 UTC 2016 Pages 48 Paragraphs 2271 Words 25830 | 800 | 1,980 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2016-44 | longest | en | 0.712125 |
https://www.physicsforums.com/threads/action-does-not-equal-reaction.4024/ | 1,670,013,641,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710916.40/warc/CC-MAIN-20221202183117-20221202213117-00700.warc.gz | 1,008,141,243 | 22,115 | # Action Does NOT Equal Reaction ?
arcnets
Originally posted by STAii
Is it true ?
No. They assume that the electric field propagates instantaneously, while the magnetic field does not. Which is false .
KLscilevothma
a little bit off topic
The 2 particles are negatively charged, not positively charged, right?
Originally posted by STAii
Is it true ?
There are two forms of Newton's third law
(1) Weak form - forces are equal and opposite but not along a line
(2) Strong form - forces are equal and opposite and along a line
The weak form does not hold for all types of forces.
It'll be simpler if I just quote "Classical Mechanics - Third Edition," Goldstein, Safko and Poole. Page 7-8
In a system involving moving charges, the forces between charges predicted by the Biot-Savart law may indeed violate both forms of the action and reaction law.*
*(footnote) If two charges are moving uniformly with parallel velocity vectors that are not perpendicular to the line joining the charges, then the net mutual forces are equal and opposite but do not lie along the vector between the charges. Consider, further, two charges moving (instantaneously) so as to "cross the T." i.e. one charge moving directly at the other, which in turn is moving at right angles to the first. Then the second charge exerts a nonvanishing magnetic force on the first, without experiencing any magnetic reaction force at that instant.
Pete
arcnets
Originally posted by pmb
the forces between charges predicted by the Biot-Savart law ...
Biot-Savart is for stationary currents - not single moving particles. Because there's nothing 'stationary' when you have a single moving particle. There's no such thing as a 'magnetic field of a single moving particle'. Each particle is affected only by the Lorentz-transformed Coulomb field of the other one, and since Lorentz-transformation is symmetrical, so is the interaction. Don't be fooled...
Originally posted by arcnets
Biot-Savart is for stationary currents - not single moving particles.
I assume you understood what Goldstein meant. Is that correct? Do you disagree with his conclusion?
There's no such thing as a 'magnetic field of a single moving particle'.
Sure there is. Its a fact of relativity. If a particle has a pure electric field in its rest frame then there will be a magnetic field in a frame in which the particle is moving - that's a basic fact of relativistic electrodynamics. Just look at the the relations for how the fields transform. You'll see that this is so.
Each particle is affected only by the Lorentz-transformed Coulomb field of the other one, and since Lorentz-transformation is symmetrical, so is the interaction. Don't be fooled...
That is incorrect. E.g. In frame S there is a moving particle with a charge. In S there is both an electric and a magnetic field due to that charge. If there is another charge moving in S then the force on that charge will be F = q[E + vxB] where E and B are generated by the first charge.
If E is the electric field of the moving charge and v is the velocity of the moving charge then the magnetic field of the moving charge is given by
B = vxE
For a derivation see Feynman Lecturers Vol II page 26-3. For a picture of the magnetic field see Figure 26-5 in V-II.
The nature of relativity prevents an action reaction forces between particles in general - what is equal and opposite in one frame is, in general, not equal and opposite in another.
French discusses this too. He makes a very important point. From "Special Relativity," A.P. French, page 224
It is worth pointing out that one of Newton's basic assertions about forces between bodies - the equality of action and reaction - has almost no place in relativistic mechanics. It must essentially be a statement about about force acting on two bodies, as a result of the mutual interaction, *at a given instant.* And, because of the relativity of simultaneity, this phrase has no unique meaning unless the points at which the forces are applied are separated by a neglegible distance. [...] What the relativistic analysis does do, however, is to compel us to conclude that, according to measurements in a given inertial frame, the forces of action and reaction are in general *not* equal and opposite, and so the total momentum of the interacting particles is no conserved instant by instant.
The missing momentum is, of course, in the field. I.e. the EM field has momentum so that the total momentum of Fields + Particles is constant.
Pete
arcnets
pmb,
I admit I didn't put enough thought into this. I'll check - just don't have the time now, hang on please.
Originally posted by arcnets
pmb,
I admit I didn't put enough thought into this. I'll check - just don't have the time now, hang on please.
So'kay! Take your time. That's why we're all here! :-)
Pete
arcnets
pmb,
well I gave it some thought. Here's my (unsatisfactory) results:
1) There is indeed such a thing as a magnetic field of a single moving particle. The classical approximation is:
B = q/c * vxr/r3.
And that was indeed named after Biot/Savart. So you are right, and I was wrong.
2) Let's now treat the problem completely classical. Let particle 2 be in the path of particle 1, and moving at right angles. Let for simplicity q = c = r = 1. Then, we get indeed
F1 = r - v1v2,
F2 = -r.
In other words, actio=reactio is indeed violated.
3) Indeed, the missing momentum must go into the field (see e.g. Jackson, chapter 6.8.).
4) Here my trouble begins. How can any momentum go into the field in such a simple configuration? The only answer I can think of is: Bremsstrahlung. But I don't see how this classical treatment of the problem can produce any e.m. radiation (i.e. sinusoid waves)...!?
5) My next thought was, we should treat this with full SR. In Jackson, chapter 11.10, I found the following:
If two particles move at relative speed v, then the field of one particle seen in the rest frame of the other is:
Elongitudinal = [gamma]qvt/(b2+[gamma]2 v2 t2)3/2
Etransversal = [gamma]qb/(b2+[gamma]2 v2 t2)3/2
B = [beta] Etransversal.
Where b is the shortest connection of both paths, t=0 when they are closest, and B is, of course, at right angles with v and b.
Now, since b is invariant (because it's at right angles with both paths), this result is indeed symmetric, since all you do is replace v by -v and b by -b when switching from one particle to the other.
Now this is strange, because in a particle's rest-frame it should not feel any Lorentz force at all, no matter what B is. Plus, I don't know at all how to translate this into forces in the lab (or any other) frame. I still suspect, when doing this properly, actio=reactio will hold. That is, there will be at least one frame of reference in which it will hold. Maybe the center-off-mass one.
Any help?
Edit: corrected 1 error, and made some formulae better legible. Couldn't make the gamma small, however. It should be...
Last edited:
Creator
Thanks for the quotes, Pete. I could have used them last year when on another forum I received a lot of flak for stating that... to adopt relativity one has to be willing to give up the concept of Newton's 3rd law.
J.L. Naquin's site (given in the 1st post) has an excellent graphic which illustrates precisely the point made in your referenece in Goldstein's footnote, which show how when two charges approach each other with orthagonal trajectries, one avoids the magnetic reaction force - at least temporarily.
Which brings us to the next point, which he only briefly mentioned, and you brought up succintly.
Originally posted by pmb
The missing momentum is, of course, in the field. I.e. the EM field has momentum so that the total momentum of Fields + Particles is constant.
[/B]
So you therefore would say that the 'missing' momentum remains in the field? Or would you say the field produces a back reaction on the source particle - thus no real asymmetry?
Creator
Last edited:
Originally posted by Creator
Thanks for the quotes, Pete. I could have used them last year when on another forum I received a lot of flak for stating that... to adopt relativity one has to be willing to give up the concept of Newton's 3rd law.
J.L. Naquin's site (given in the 1st post) has an excellent graphic which illustrates precisely the point made in your referenece in Goldstein's footnote, which show how when two charges approach each other with orthagonal trajectries, one avoids the magnetic reaction force - at least temporarily.
Which brings us to the next point, which he only briefly mentioned, and you brought up succintly.
So you therefore would say that the 'missing' momentum remains in the field? Or would you say the field produces a back reaction on the source particle - thus no real asymmetry?
Creator
I hear ya! People hate it when you disagree with them and will refuse to give up their old ideas. A person that used to post here is flaming me elsewhere. It's a result of his misunderstanding of Einstein's "photon in a box" gerdanken experiment. He shakes at the notion that light has mass! But alas - it's true!
Just made a new web page on the center of mass for a relativistic system
See -- http://www.geocities.com/physics_world/sr/center_of_mass.htm
That's in preperation for a new web page on Einstein's "photon in a box" thing. I've defined and described it here
www.psyclops.com/hawking/forum/printmsg.cgi?period=current&msg=59744
Pete
Staff Emeritus
Gold Member
There's a problem with equation 1 on your page. mk is not a constant with respect to t; as time changes, particke k's velocity may change, which will thus change particle k's relativistic mass. Thus, you cannot, in general, perform the last step in equation 1.
In particular, this means that there's no reason that the center of mass, as you've defined it, should have a constant velocity (even in the presense of no external forces).
Originally posted by Hurkyl
There's a problem with equation 1 on your page. mk is not a constant with respect to t; as time changes, particke k's velocity may change, which will thus change particle k's relativistic mass. Thus, you cannot, in general, perform the last step in equation 1.
Please take close note of the very first sentence, i.e.
"Consider a system of particles, all of which move at constant velocity..."
Pete
Staff Emeritus
Gold Member
lol I read the whole thing twice, and never noticed the text up there before the diagram!
Originally posted by Hurkyl
lol I read the whole thing twice, and never noticed the text up there before the diagram!
I have it when that happens! :-D
I've just been going over Einstein's 1906 "photon in a box" paper over the last week. This seems quite compatible with his comments and derivations and conclusions.
I've just modified that comment. Looking over it I now see that all I needed to do was require that that the speed remain unchanged - not the velocity. So now the particles can move in a cirlce!
Pete
Originally posted by pmb
People hate it when you disagree with them and will refuse to give up their old ideas. He shakes at the notion that light has mass
Yes yes, photons have non-vanishing relativistic mass but vanishing rest mass.
The idea you seem to "hate" giving up is that it's wrong to view massless particles as pure energy.
I'm pretty sure that Rindler would tell you that you're twisting his view about the efficacy of using the term "relativistic mass" into something he never intended or agrees with.
Originally posted by jeff
The idea you seem to "hate" giving up is that it's wrong to view massless particles as pure energy.
In the first I've never considered a photon as something being "pure energy." The notion of anything being pure energy makes no sense to me.
In the second place if you're referring to the notion of a photon having mass = energy/c^2 then you're very much wrong. It's not something I've be reluctant to give up. It's been something I've come to accept - that's quite a big difference. When I first heard of this notion (about 8 years ago as I recall) I assumed people were just misinterpreting Einstein. As I studied Einstein's original work and all the many artilces on it that followed over the years since 1905 I've come to accept that a mass of a photon of m = E/c^2 is exactly what Einstein had in mind and that it makes perfect sense. And this is something which can be found in many relativity texts.
For example: from "Relativity: Special, General and Cosmological," Rindler, Oxford Univ., Press, (2001), page 120
According to Einstein, a photon of frequency v has energy hv, and thus (as he came to realize several years later) a finite mass hv/c^2 and a finite momentum hv/c.
re -
I'm pretty sure that Rindler would tell you that you're twisting his view about the efficacy of using the term "relativistic mass" into something he never intended or agrees with.
Why would you think that? IMHO - The notion of relativistic mass is the closest thing there is to what would consider as being mass. And that definition of "mass," when defined rigorously, demands that light has mass. A box containing a gas of photons of total energy U has inertial properties of that of a mass of energy m = U/c^2. So the inertial mass the box, whose mass is M, is that of an object whose mass is (M+m). According to the equivalence principle itg can be shown that the photon gas has a passive gravitational mass. A beam of light generates a graviational field and therefore has an active gravitational mass.
This notion of an EM field having mass was the view Einstein arrived at in 1906 in what has come to be known as Einstein's photon in a box thought experiment. The paper concluided with Einstein deducing that u = "energy density of EM field" has a mass density of u/c^2.
In any case I know relativity so I can form my own opinions. I just happen to agree with Rindler and everyone on this point. I gave you another example if you recall
http://www.geocities.com/physics_world/Guth.jpg
I'm pretty sure that Rindler would tell you that you're twisting his view about the efficacy of using the term "relativistic mass" into something he never intended or agrees with.
What are you basing this opinion of that Rindler thinks on?
Pmb
Staff Emeritus
Gold Member
Incidentally, just because the sum of the momenta of two or more photons may have mass does not mean each individual photon has mass.
Originally posted by Hurkyl
Incidentally, just because the sum of the momenta of two or more photons may have mass does not mean each individual photon has mass.
For a system of two photons moving in opposite directions there is a zero momentum frame. Therefore even though both photons have zero proper mass the system has a non-zero proper mass. In fact I was just explaining this last night to someone. I.e. I was discussing black body radiation - such raidation has an energy-momentum tensor of a perfect fluid.
Did you get the impression that I thought any different? If so then you were mistaken.
Pmb
Staff Emeritus
Gold Member
Dearly Missed
In fact the energy of just one photon, if high enough, can be used to create a new particle and antiparticle. The photon ceases to exist in this pair production
In fact the energy of just one photon, if high enough, can be used to create a new particle and antiparticle. The photon ceases to exist in this pair production
Yes. Quite true. And of course there must be another particle around to facilitate the pair production. Otherwise there would be a violation of the conservation of momentum.
Pete
Mentor
From the article:
Note : To be in agreement with Newton's 3rd law, it would be necessary to take into account the moments of the magnetic and electric fields.
Am I misunderstanding or does that sentence translate into: 'if we ignore some of the forces, the forces are imbalanced'?
Originally posted by russ_watters
From the article: Am I misunderstanding or does that sentence translate into: 'if we ignore some of the forces, the forces are imbalanced'?
Was that something I wrote?
Pete | 3,730 | 16,069 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2022-49 | latest | en | 0.934233 |
https://www.iwpcug.org/davidbro/puz0406.htm | 1,638,038,035,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358208.31/warc/CC-MAIN-20211127163427-20211127193427-00109.warc.gz | 907,420,794 | 1,543 | ## PRIZE PUZZLE FOR JUNE 2004
### TOMDICKHARRY
Here is a puzzle for calculator or spreadsheet practice.
Tom, Dick and Harry are brothers. At 11 am all brothers are at home. At 11 am Tom starts to walk at 4 miles per hour north to a distant destination. At 11:05 Dick starts to walk south to another distant destination. Dick walks at a constant 3 miles per hour. At 11.30 Harry, who is still at home, discovers that both his brothers have forgotten their packed lunches so he sets off north on his bicycle at 9 miles per hour to catch up with Tom to deliver his lunch. He then immediately doubles back to catch up with Dick, going past his home without stopping. Having caught up with Dick and delivered his packed lunch, Harry cycles back home, again at a constant 9 miles per hour. What time does he arrive home, having completed both deliveries? And would he be earlier or later if he had chased after Dick first and Tom second?
Correct answers were received from Clem Robertson, Rosemary West, John Stafford and John Bownas. Rosemary West won the draw and the book token.
THE ANSWER -- scroll down to it...
The answer is 1.31 pm and it would have been quicker to have chased after Dick first.
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# Page 1 of 15
AERSP 305W
Aerospace Technology Laboratory
Laboratory Section 4
Laboratory Experiment Number 3
Wind Tunnel Testing of a S805 Airfoil
March 2, 2012
Performed in Room 8 Hammond Building
Connor Hoover
Lab Partners Names:
Ethan Corle
Kaitlynn Hetrick
Stephen Prichard
Anthony Parente
Lab TA: Kylie Flickinger
Course Instructor: Richard Auhl
Page 2 of 15
Abstract
The objective of this lab was to determine the coefficients of lift and drag of a S805 airfoil at
various angles of attack. This was achieved in three different sections. First was the
collection of pressure tap data on the surface of the airfoil using a manometer bank, next a
wake velocity profile using a hot-wire anemometer, and finally a general observation of stall
effects on an airfoil. The calculating of the coefficients was done using equations introduced
in the next section. The pressure data will yield the coefficient of lift, while the wake velocity
profile produced the coefficient of drag. These values once calculated were compared to
NREL published data for the S805 airfoil. The experimental data was found to be accurate
when compared and that the goals of calculating the values of the coefficients were achieved.
The one main source of error was that the airfoil used in the experiment was half the span of
the airfoil of the published data.
Page 3 of 15
Introduction
The objective of this experiment was to determine the coefficients of lift and drag of a two dimensional
airfoil at a range of angles of attack. The airfoil tested was the S805 airfoil; Figure 1 shows the
dimensions of this airfoil in inches.
Figure 1. Test S805 Airfoil
To complete the objective stated above, certain equations are needed to determine key values. The
first is the ideal gas law, which will give the density of the air in the wind tunnel with the simple
measurement of temperature and pressure. Equation 1 below shows the ideal gas law rearranged to
solve for density:
Another important value needed to understand the data collected is Reynolds number. This non-
dimensional value will help to control the experiment. Equation 2 below will allow for the calculation of
the velocity needed to keep the wind tunnel at the Reynolds number specified in the procedure.
The next set of equations deal with the calculation of the coefficient of lift for the airfoil. This is done
via the integration of the pressure distribution on the airfoil. Using a manometer bank, at a given angle
and having a known specific gravity, equations 3a and 3b can be used to calculate the pressure
coefficients of the upper and lower surfaces of the airfoil. It should be noted that
, the reference
height will be measured by a specific port in the manometer.
(3a)
(3b)
(2)
Uc
= Re
RsT P
T
T
P
P
r
r
r
/ =
|
|
.
|
\
|
|
|
.
|
\
|
=
(1)
Page 4 of 15
By averaging the coefficients of pressure determined above across the chord length the normal force
coefficients can be written as equation 4a and 4b below.
+ *(
+ (4a)
+ *(
+ (4b)
Remaining in coefficient form, the total normal force acting on the airfoil can be found by subtracting
the sum of each of the individual normal force coefficients found in the equations above. Equation 5
below shows this relationship.
(5)
Finally, the coefficient of lift for the airfoil can be determined by equation 6 below, where is the angle
of attack of the airfoil.
(6)
The coefficient of drag will be calculated via the integration of the wake profile. This profile will be
measured by a hot-wire anemometer mounted downwind of the airfoil. This data will give us the profile
drag which can be manipulated to find the coefficient of drag.
Profile drag has two components, pressure drag and skin friction drag. By looking at the momentum
flux ahead and aft of the airfoil, profile drag can be determined as the difference in the x-component of
this flux. When considering only the linear momentum, the air suffers a loss of momentum flux equal
to the drag of the airfoil. In this case, drag can be described as equation 7 below.
(7)
Where mass flow rate can be written as,
Page 5 of 15
(8)
Substituting this expression into equation(7), drag force is put in terms of the free stream velocity of
the wake (
), the local velocity of the wake (V), and cross-sectional area (dA).
(9)
Non-dimensionalizing the above equation gives the section drag coefficient of the airfoil. Again c is
the chord length of the airfoil while b is equal to 1 for unit span.
(10)
The use of the hot-wire anemometer can yield a velocity profile wake as shown in Figure 2 below. If
the traverse moves an even number of steps to collect the data, Simpsons Rule for integration can be
applied to find the drag coefficient.
Figure 2. Wake Velocity Profile
The final part of this experiment is to observe the airfoil as it moves to stall; the angle of attack will be
increased until a large pressure drop is observed across the surface of the airfoil. Also, the airfoil will
then be brought back from this stall angle until normal flow begins again. This is expected to be at a
lower angle than the original stall angle due to flow circulation in the wake.
Page 6 of 15
Experimental Procedure
To begin the temperature and pressure of the laboratory were measured. Using the calibration given
for Transducer B the velocity needed to achieve a Reynolds number of 800,000 was calculated. This
was found to be around 81 ft/s. The wind tunnel was dialed up until Transducer B read a voltage of
2.37, which corresponds to this velocity. Afterwards, the airfoil was rotated using LabView to the first
angle of attack, -5 degrees. As shown below in Figure 3, the S850 airfoil was hooked up to a
manometer bank to record the pressures felt on the surface of the airfoil.
Figure 3. Airfoil Set Up in Wind Tunnel
The manometer banks 36 ports all recorded pressures from pressure tap connections on the airfoil.
Two ports, 18 and 19, were used to record the atmospheric and T.S. Static pressures. The angle of
the bank was also recorded to interpret the data recorded later. These pressure readings were later
used to calculate the coefficient of lift for the airfoil. After finishing these readings for -5 degrees the
airfoil was rotated to an angle of attack of 15 degrees where another set of manometer data was
taken. Figure 4 below shows a sample reading of the manometer bank.
Page 7 of 15
Figure 4. Sample Manometer Bank Data Reading
The second part of the lab recorded the wake profile behind the airfoil. This was done using a hot-wire
anemometer as stated before. The hot-wire recorded the wakes for both angles at specified points.
These two profiles take were later used to determine the coefficient of drag of the S850 airfoil. Also,
the hot-wire took a time trace at each location with a sample rate of 2000 samples per second. To
correct the voltages taken by the hot-wire during the experiment to fit with the temperature of the lab
the following equation was used:
(11)
This allows for the hot-wire voltages to be correctly converted to velocity without the problem of the
temperature the wire was calibrated at being different than the laboratory environment.
Page 8 of 15
For the final portion of the experiment the wind tunnel was taken down to 50% power. The airfoil angle
of attack was increased until it reached stall, then reverted back until it had recovered from stall. The
angle of attack at which stall was observed was determined by watching the manometer bank.
Page 9 of 15
Results and Discussion
In lab experiment 2 it was shown that the hot-wire anemometer was able to accurately map a flow
field and be used to find the flows turbulence intensity. This was done again for the S805 airfoil at
angles of attack of -5 and 15 degrees. The plot below shows the velocity measured by the
anemometer versus the time. The turbulence intensities were calculated and showed that the flow
over the airfoil at 15 degrees angle of attack is slightly more turbulent than the -5 angle. This is to be
expected since flow is known to separate earlier when the angle of attack is greater. The turbulence
intensities of both angles are shown in Table 1.
Figure 5. Velocity vs. Time at Section 4 s
Table 1. Turbulence Intensity At Section 4 s
Angle of Attack (Deg) Ti
-5 0.00231
15 0.00301
The wake profiles for the airfoil for each of the angles of attack are shown in figures 6 and 7. It should
be noted that the voltages of the hot-anemometer were corrected by use of equation 11. These plots
were later used to determine the initial dynamic pressures used to calculate the coefficient of drag. As
expected the angle of attack of 15 has a larger wake profile than the -5 angle of attack. Again this can
be explained by the greater separation caused by the magnitude of the angle. This will create a larger
velocity profile for the 15 degree alpha; hence it makes sense that figures 7 shows a larger wake
region than figure 6. This shows that the 15 degree angle of attack will cause more drag. As stated in
83
84
85
86
87
88
89
90
0 0.2 0.4 0.6 0.8 1
V
e
l
o
c
i
t
y
(
f
t
/
s
)
Time (sec)
Alpha = -5
Alpha = 15
Page 10 of 15
equation (9) and shown in figure (2) the area under the wake profile curve is equal to the resultant
drag. So since the wake profile is bigger for 15 degrees the drag is greater when the airfoil is in this
configuration.
Figure 6. Wave Traverse Distance vs. Hot-Wire Velocity ( = -5 Deg.)
Figure 7. Wave Traverse Distance vs. Hot-Wire Velocity ( = 15 Deg.)
The pressure distribution calculated from the manometer readings at a Reynolds number of 800,000
are shown below in figures 8 and 9. For the angle of attack of -5 degrees the S805 airfoil the upper
surface shows a favorable pressure gradient until .2c at which point it becomes slightly adverse until
0
5
10
15
20
25
30
60 65 70 75 80 85 90
W
a
k
e
T
r
a
v
e
r
s
e
D
i
s
t
a
n
c
e
Velocity (ft/s)
Uncorrected Corrected
0
5
10
15
20
25
30
60 65 70 75 80 85 90
Uncorrected Corrected
Page 11 of 15
leveling out at almost zero at .8c. The lower surface on the other hand for the most part remains
adverse until becoming slightly favorable at .8c as well. This shows that this angle is favorable to
reduce drag, due to its ease of transition from laminar to turbulent flow. But, lift will not be generated
enough to combat weight at this angle of attack.
Figure 8. Coefficient of Pressure vs. x/c ( = -5 Deg.)
Figure 9 below shows that at an angle of attack of 15 degrees the upper surface of the airfoil has an
over the chord. This allows for better lift than shown in the in figure 8 for the angle of attack of -5
degrees.
-3
-2.5
-2
-1.5
-1
-0.5
0
0.5
1
1.5
0 0.2 0.4 0.6 0.8 1
C
p
x/c
Lab Data, Upper
Lab Data, Lower
Page 12 of 15
Figure 9. Coefficient of Pressure vs. x/c ( = 15 Deg.)
After all the sections were done collecting their data for the various angles of attack the next two plots
were created. Figure 10 is a plot of C
l
vs. while figure 11 is the plot of C
l
vs. C
d
. Both of these plots
have overlays of NREL published data for the S805 airfoil to compare the resulting calculations from
the experimental data. As shown in figure 10, the coefficient of lift curve fits fairly well with the NREL
data for Reynolds numbers of 700,000 and 1,000,000. Only at the two extremes of the data, when
angle of attack is nearing stall do we see any discrepancies worth noting. The error involved between
the three curves can be attributed to human error reading the NREL plots, equipment calibration
mishaps, and the relative size of the air foil with respect to the wind tunnel.
-7
-6
-5
-4
-3
-2
-1
0
1
2
0 0.2 0.4 0.6 0.8 1
C
p
x/c
Lab Data, Upper
Lab Data, Lower
Page 13 of 15
Figure 10. Coefficient of Lift vs. for S805 Airfoil
Figure 11 as mentioned above plots C
l
vs. C
d.
Now, the NREL data used at a Reynolds number of
700,000 was difficult to read, and therefore only the far left data was able to be plotted. This section
of the data fit well with the experimental data, showing that using Simpsons rule to calculate the area
under the curve of the wake velocity profile is a viable way to calculate the drag coefficient of an
airfoil. The only problem with this method that could cause error is the selection of the ends of the
wake region in the flow. This was very arbitrarily done in this case, one way in which to improve this
would be to plot all the wake velocity profiles to process the data collected.
Figure 11. Coefficient of Drag vs. for S805 Airfoil
-1
-0.5
0
0.5
1
1.5
-20 -10 0 10 20
C
l
Alpha (degrees)
Section 4 Points Lab Data, (Re = .8E6)
NREL (Re = 1.0E6) NREL (Re = .7E6)
-1
-0.5
0
0.5
1
1.5
0 0.1 0.2 0.3 0.4
C
l
Cd
Section 4 Points
Lab Data, Re = .8E6
NREL, Re = .7E6
Page 14 of 15
The final portion of the lab was the observation of the stall. As the airfoils angle of attack was
increased a large pressure drop on the upper surface of the airfoil was seen at an angle of 19
degrees, this pressure drop coincides with the definition of stall. Stall is when the flow detaches from
the upper surface of an airfoil. After being stalled out, the angle was decreased again until the flow
reattached, this happened at 17 degrees. The difference between the stall angle and the angle at
which its flow reattached was due to the wake vortices causing an adverse pressure gradient along
the airfoil, as well as possible momentum forcing the flow to remain separated until reaching a lower
angle of attack.
Page 15 of 15
Conclusions
The objectives of this experiment were completed. The coefficients of lift and drag for various
angle of attack were calculated using the data collected in the wind tunnel testing of the S805
airfoil at a Reynolds number of 800,000. The coefficients of lift was calculated using the
manometer banks pressure readings while the coefficients of drag were calculated by
applying Simpsons rule to the hot-wire anemometers wake velocity profiles. After being
calculated the values were plotted and found to be accurate when compared to the published
NREL data. The only discrepancies arose from the human error or the size of the airfoil itself,
which was cut down to fit in the wind tunnel. It was also found that the positive angle of attack
caused a favorable pressure gradient on the lower surface of the airfoil, while an adverse
gradient was seen on the upper surface. The negative angle was found to be the opposite,
causing the positive angles to have better coefficients of lift which is to be expected.
In future experiments of this nature can be improved by the use of various different airfoils.
The lab experiment did have a time limit, but there is much to be learned from comparing the
different lift and drag aspects of all kinds of airfoils and what each is best suited for. | 3,733 | 14,815 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2019-39 | latest | en | 0.914928 |
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# Approach to Montessori - Numbers HD Free Lite
## By Hien Ton
#### Description
APPROACH TO MONTESSORI - NUMBERS
"Prepare for Greatness"
KEY FEATURES:
----------------------------------------------
- Full comprehension of numbers Zero (0) through Ten (10)
- Number recognition, writing, quantity, sequencing (spatial relationship)
- Foundations of Math
- Reading not required. Audio instructions/feedbacks and visual elements are used throughout.
- Fine Motor Skills for hand-eye coordination training
- Settings adapts to each Child's learning style
- Guided by a team of Certified Montessori educators and parents
SIX MONTESSORI GAMES IN ONE
----------------------------------------------
Six discrete yet related Montessori activities provide a strong foundation for number concepts, emphasizing on recognition, quantity, sequencing (spatial relationship), writing, and speaking.
1. DANCING NUMBERS : Enjoy a dancing numbers show sync to music, learning the numbers by watching them dance, or play the question and answer game.
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5. NUMBER BLOCKS: Learn number counting by dragging the blocks to the correct spot.
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POWER SETTINGS:
----------------------------------------------
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Below are setting options for each game:
1. DANCING NUMBERS
- Single /Multiple Touch Mode
- Quiz Mode
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- Trace Dash-Line Guides
3. NUMBER RODS
- Rod Counting
- Sequence Order: Forward / Reverse/ Random
- Hanger Numbers
- Sequence Order: Forward / Reverse / Random
5. NUMBER BLOCKS
- Dots / Pointer Clue
- Audio Guide
- Sequence Order: Forward / Reverse / Random
6. CARDS & COUNTER
- Odds & Even
- Number Counting
- Color of Pieces: Red / Blue / Green
*****************************************
*****************************************
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# Multiplication and Division: Introduction to Division
#### Lesson 4: Introduction to Division
/en/multiplicationdivision/video-multiplication/content/
### What is division?
Division is splitting something equally. For instance, let's say you have 10 raffle tickets, and you'd like to share them with 5 friends.
You divide the tickets among your friends. Each friend gets an equal number of tickets.
See how they each have 2 tickets? When you divide 10 tickets among your five friends, you create 5 equal groups of 2 tickets.
Division happens a lot in real life. For instance, consider the situation below.
• Imagine we have 6 cupcakes...
• Imagine we have 6 cupcakes... and 2 empty trays.
• We'll place an equal number of cupcakes on each tray. In other words, we'll divide the cupcakes between the two trays.
• We'll place an equal number of cupcakes on each tray. In other words, we'll divide the cupcakes between the two trays.
• We'll place an equal number of cupcakes on each tray. In other words, we'll divide the cupcakes between the two trays.
• We'll place an equal number of cupcakes on each tray. In other words, we'll divide the cupcakes between the two trays.
• We'll place an equal number of cupcakes on each tray. In other words, we'll divide the cupcakes between the two trays.
• We'll place an equal number of cupcakes on each tray. In other words, we'll divide the cupcakes between the two trays.
• We'll place an equal number of cupcakes on each tray. In other words, we'll divide the cupcakes between the two trays.
• The six cupcakes have been divided into 2 equal groups.
• Let's count the cupcakes to find out how many are in each group.
• Let's count the cupcakes to find out how many are in each group.
• Let's count the cupcakes to find out how many are in each group.
• Let's count the cupcakes to find out how many are in each group.
• Let's count the cupcakes to find out how many are in each group.
• Let's count the cupcakes to find out how many are in each group.
• Let's count the cupcakes to find out how many are in each group.
• Each tray has 3 cupcakes. If you start with six cupcakes and divide them into two equal groups, then each group has three cupcakes.
#### Writing division expressions
In the slideshow, you saw that we divided six cupcakes into two equal groups. To figure out the number of cupcakes that are in each group, you could write a division expression like this:
6 / 2
You could also write the expression like this:
6 ÷ 2
You can read either expression as six divided by two. The division sign (/ or ÷) means something is being divided. This is why we always put it after the first number — there were 6 cupcakes, and we divided them into 2 groups.
Many real-life situations can be expressed with division. For example, imagine you're placing 15 cans on 3 shelves. You can divide to make sure you put the same number of cans on each shelf. In other words, 15 cans divided by three shelves, or 15 / 3.
#### Try this!
Try setting up these situations as division expressions. Don't try to solve them yet.
A teacher has 16 pencils that she distributes evenly between 4 students.
A florist has 18 roses and divides them equally between 3 vases.
You have 6 treats to share equally with your 3 dogs.
### Solving division problems
You can use counting to solve simple division problems. For instance, let's say we have 12 seedlings. We decide to plant them in two even rows. How many plants go in each row? We could write that question like this:
12 / 2
Remember, that expression means 12 divided by two, or 12 seedlings divided into 2 rows. It's a simple problem. To solve it, you can put the seedlings into two groups and then count how many plants are in each group. The answer is 6. We know that 12 / 2 = 6.
While counting works for problems that begin with small numbers, a problem that begins with a large number can take a long time to solve with counting. For this reason, most people memorize common division problems so that they can solve them quickly. If this sounds hard, don't worry. With some practice, you'll be able to quickly remember the answers.
In Introduction to Multiplication, you were introduced to the times table. In that lesson, you used it to solve multiplication problems. You can also use the times table to solve division problems.
Let's start with a problem we're already familiar with. How would we have solved the seedling problem with the times table?
Click through the slideshow below to learn how.
• Remember, each number at the top of the times table is at the start of a column.
• For example, this is the column that goes with 7.
• Each number on the left side of the times table is the start of a row. This row goes with 9.
• Let's try solving the seedling problem: 12 / 2.
• First, find the number that you are dividing by on the right of the division sign. In 12/2 we're dividing by 2.
• Find the 2's column.
• Next, find the number that you're dividing on the left of the division sign. In 12/2 it's 12.
• Find 12 in the 2's column.
• Find the number at the start of the row that overlaps 12. In this case, it's the 6's row.
• So, the answer, or quotient for 12 / 2 is 6.
• Let's try that once more. This time, we'll solve 15 / 5.
• First, we'll find the 5's column since we are dividing by 5.
• Next, we'll find 15 in the 5's column since that is the number we are dividing.
• Finally, we'll find the number at the start of the row that overlaps 15. It's 3. So, 15 / 5 = 3.
#### Try this!
Solve these division problems. If you need some help, you can use the times table.
42 ÷ 7 =
5 ÷ 1 =
33 ÷ 3 =
### Remainders
In the previous pages, we divided numbers equally. For instance, at the beginning of the lesson, we divided 10 tickets equally between 5 people. Each person received 2 tickets. What happens when a number can't be equally divided?
For example, consider the situation below.
• Let's say we have 10 tickets...
• Let's say we have 10 tickets... that we are dividing between 3 friends.
• We'll try solving the problem 10 / 3.
• Let's see how many tickets we can give to each of our friends...
• Let's see how many tickets we can give to each of our friends...
• Let's see how many tickets we can give to each of our friends... One...
• Let's see how many tickets we can give to each of our friends... One...
• Let's see how many tickets we can give to each of our friends... One...
• Let's see how many tickets we can give to each of our friends... One... Two...
• Let's see how many tickets we can give to each of our friends... One... Two...
• Let's see how many tickets we can give to each of our friends... One... Two...
• Let's see how many tickets we can give to each of our friends... One... Two... Three.
• What happens now? We have 3 friends and only 1 ticket is left.
• That means 1 is the remainder, or the amount left over.
• We're ready to write our quotient.
• Each friend has three tickets, so we'll write 3.
• Then, we'll write our remainder. That's 1. See how we wrote it next to the lowercase letter r?
• So, 10 / 3 = 3 r1. We can read this quotient as three remainder one. 10 tickets divided by 3 friends means each friend gets 3 tickets with 1 ticket left over.
You can see from the slideshow that the remainder (1) is smaller than the number we divided by (3). That will always be the case when the problem has a remainder. For example, look at each of these problems below:
21 / 5 = 4 r1
The remainder of 1 is smaller than 5.
76 / 6 = 12 r4
The remainder of 4 is smaller than 6.
If the remainder is larger, that means the amount left over is too large. You'll need to try dividing again. For example, if you have 4 friends and 7 tickets left over, you know that each friend can get at least one more ticket.
### Practice!
Practice division with these problems. If you'd like, you can use the times table for help. There are 3 sets of problems with 5 problems each.
72 ÷ 9 =
64 ÷ 8 =
70 ÷ 10 =
55 ÷ 11 =
21 ÷ 3 =
35 ÷ 5 =
32 ÷ 4 =
72 ÷ 6 =
12 ÷ 2 =
28 ÷ 7 =
#### Set 3
6 ÷ 1 =
81 ÷ 9 =
24 ÷ 3 =
49 ÷ 7 =
144 ÷ 12 =
/en/multiplicationdivision/long-division/content/ | 2,043 | 8,229 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.90625 | 5 | CC-MAIN-2024-22 | latest | en | 0.96635 |
https://www.mrexcel.com/board/threads/formula-question.378778/ | 1,680,314,290,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949694.55/warc/CC-MAIN-20230401001704-20230401031704-00426.warc.gz | 974,766,107 | 17,502 | # Formula Question...
#### andyinla
##### New Member
My General Ledger is an excel file generated by Finance and has a column with invoice numbers. My Cost Report is a separate excel file that also has an invoice column with invoice numbers
I want to do a formula that will place the month in GL, actual month "02/2009", if the invoice numbers both appear in both reports, General Ledger and Cost Report, so I don't have to go through the Generl Ledger line by line and check did invoice number 66758 apear on my GL for Feb and is that number in Cost Report as well..
How do I go about doing that formula?
Thank you so much
Andy
Last edited:
### Excel Facts
Return population for a City
If you have a list of cities in A2:A100, use Data, Geography. Then =A2.Population and copy down.
#### jasonb75
##### Well-known Member
Hi Andy.
This could be done with lookup and an error check, something like
=IF(ISERROR(VLOOKUP(\$A2,'COST REPORT'!\$A:\$A,1,FALSE)),"","02/2009")
\$A2 should be the cell with the invoice number in the current row.
\$A:\$A should be the column in the cost report that contains the invoice numbers.
'COST REPORT' should be the file and sheet name where the column above is found (if you leave it like it is, as long as there is not a sheet called COST REPORT in the GL workbook, excel should open a window asking you to select the file and sheet that it should be looking at).
Note this formula will always enter "02/2009" as the month, if the month should be entered as found elsewhere then it can be referenced by changing the formula.
#### andyinla
##### New Member
Jason
Thank you for your help, I think I am almost there...
here is what I have....
=IF(ISERROR(VLOOKUP(\$K423,'Cost Report'!\$A:\$A,1,FALSE)),'[Copy of Andrew Haspel (4).xls]Job Cost Detail Report 3'!\$G:\$G"","02/2009")
[Copy of Andrew Haspel (4).xls]Job Cost Detail Report 3'!\$G:\$G
this is reference to my General Ledger file and referencing the column with the invoice numbers
What do you think?
#### jasonb75
##### Well-known Member
Andy
Try
=IF(ISERROR(VLOOKUP(\$K423,'[Copy of Andrew Haspel (4).xls]Job Cost Detail Report 3'!\$G:\$G,1,FALSE)),"","02/2009")
the filename and sheet name '[Copy of Andrew Haspel (4).xls]Job Cost Detail Report 3' should be the opposite file to the one in which you enter the formula.
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Go back | 913 | 3,396 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2023-14 | latest | en | 0.916146 |
http://mathhelpforum.com/calculus/99031-trouble-lines.html | 1,527,033,199,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794864999.62/warc/CC-MAIN-20180522225116-20180523005115-00031.warc.gz | 186,905,867 | 9,143 | ## Trouble with Lines
Honestly I have trouble starting this problem:
suppose x=x0+tv and y=y0+sw are two parametric representations of the same line l in r^n
a. show that there are scalars t0 and s0 such that y0=x0+t0v and x0=y0+s0w
b. show that v and w are parallel | 81 | 267 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2018-22 | latest | en | 0.903702 |
https://www.endpoint.com/blog/2018/06/12/systematic-query-building-with-ctes | 1,611,776,820,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610704832583.88/warc/CC-MAIN-20210127183317-20210127213317-00321.warc.gz | 757,115,623 | 13,201 | Our Blog
Ongoing observations by End Point people
Systematic Query Building with Common Table Expressions
By Josh Tolley
June 12, 2018
The first time I got paid for doing PostgreSQL work on the side, I spent most of the proceeds on the mortgage (boring, I know), but I did get myself one little treat: a boxed set of DVDs from a favorite old television show. They became part of my evening ritual, watching an episode while cleaning the kitchen before bed. The show features three military draftees, one of whom, Frank, is universally disliked. In one episode, we learn that Frank has been unexpectedly transferred away, leaving his two roommates the unenviable responsibility of collecting Frank’s belongings and sending them to his new assignment. After some grumbling, they settle into the job, and one of them picks a pair of shorts off the clothesline, saying, “One pair of shorts, perfect condition: mine,” and he throws the shorts onto his own bed. Picking up another pair, he says, “One pair of shorts. Holes, buttons missing: Frank’s.”
The other starts on the socks: “One pair of socks, perfect condition: mine. One pair socks, holes: Frank’s. You know, this is going to be a lot easier than I thought.”
“A matter of having a system,” responds the first.
I find most things go better when I have a system, as a recent query writing task made clear. It involved data from the Instituto Nacional de EstadĂstica y GeografĂa, or INEGI, an organization of the Mexican government tasked with collecting and managing country-wide statistics and geographical information. The data set contained the geographic outline of each city block in Mexico City, along with demographic and statistical data for each block: total population, a numeric score representing average educational level, how much of the block had sidewalks and landscaping, whether the homes had access to the municipal sewer and water systems, etc. We wanted to display the data on a Liquid Galaxy in some meaningful way, so I loaded it all in a PostGIS database and built a simple visualization showing each city block as a polygon extruded from the earth, with the height and color of the polygon proportional to the educational score for that block, compared to the city-wide average.
It wasn’t entirely a surprise that with so many polygons, rendering performance suffered a bit, but most of all the display was just plain confusing. This image is just one of Mexico City’s 16 boroughs.
With so much going on in the image, it’s difficult for the user to extract any meaningful information. So I turned to a technique we’d used in the past: reprocess the geographical area into grid squares, extrapolate the statistic of interest over the area of the square, and plot it again as a set of squares. The result is essentially a three dimensional heat map, much easier to comprehend, and, incidentally, to render.
As with most programming tasks, it’s helpful to have a system, so I started by sketching out how exactly to produce the desired result. I planned to overlay the features in the data set with a grid, and then for each square in the grid, find all intersecting city blocks, a number representing the educational level of residents of that block, and what percentage of the block’s total area intersects each grid square. From that information I can extrapolate that block’s contribution to the grid square’s total educational score. The precise derivation of the score isn’t important for our purposes here; suffice it to say it’s a numeric value with no particular associated unit, whose value lies in its relation to the scores of other blocks. Residents of a block with a high score are, on average, probably more educated than residents of a lower-scoring block. For this query, a block with an average educational level of, say, 100 “points”, would contribute all 100 points to a grid square if the entire block lay within that square, 60 points if only 60% of it was within the square, and so on. In the end, I should be able to add up all the scores for each grid square, rank them against all other grid squares, and produce a visualization.
I suffer from the decidedly masochistic habit of doing whatever I can in a single query, while maintaining a desire for readable and maintainable code. Cramming everything into one query isn’t always a good technique, as I hope to illustrate in a future blog post, but it worked well enough in this instance, and provides a good example I wanted to share, of one way to use Common Table Expressions. They do for SQL what subroutines do for other languages, separating tasks into distinct units. A common table expression looks like this:
``````WITH alias AS (
SELECT something FROM wherever
),
another_alias AS (
SELECT another_thing FROM alias LEFT JOIN something_else
)
SELECT an, assortment, of, fields
FROM another_alias;
``````
As shown by this example query, the `WITH` keyword precedes a list of named, parenthesized queries, each of which functions throughout the life of the query as though it were a full-fledged table. These pseudo-tables are called Common Table Expressions, and they allow me to make one table for each distinct function in what will prove to be a fairly complicated query.
Let’s go through the elements of our new query piece by piece. First, I want to store the results of the query, so I can create different visualizations without recalculating everything. So I’ll make this query create a new table filled with its results. In this case, I called the table `grid_mza_vals`.
Now, for my first CTE. This query involves a few user-selected parameters, and I want an easy way to adjust these settings as I experiment to get the best results. I’ll want to fiddle with the number of grid squares in the overall result, as well as the coefficients used later on to calculate the height of each polygon. So my first CTE is called simply `params`, and returns a single row, composed of these parameters.
``````CREATE TABLE grid_mza_vals AS
WITH params AS (
SELECT
25 AS numsq,
3800 AS alt_bias,
500 AS alt_percfactor
),
``````
The `numsq` value represents the number of grid squares along one edge of my overall grid; we’ll discuss the other values later. I’ve chosen a relatively small number of total grid squares for faster processing while building the rest of the query. I can make it more detailed, if I want, after everything else works.
The next thing I want is a sequence of numbers from 1 to `numsq`:
``````range AS (
SELECT GENERATE_SERIES(0, numsq - 1) AS rng
FROM params
),
``````
Now I can join the `range` CTE with itself, to get the coordinates for each grid square:
``````gridix AS (
SELECT
x.rng AS x_ix,
y.rng AS y_ix
FROM range x, range y
),
``````
Occasionally I like to check my progress, running whatever bits of the query I’ve already written to review its behavior. CTEs make this convenient, because I can adjust the final clause of the query to select data from whichever of the CTEs I’m currently interested in. Here’s the query thus far:
``````inegi=# WITH params AS (
SELECT
25 AS numsq,
3800 AS alt_bias,
500 AS alt_percfactor
),
range AS (
SELECT GENERATE_SERIES(0, numsq - 1) AS rng
FROM params
),
gridix AS (
SELECT
x.rng AS x_ix,
y.rng AS y_ix
FROM range x, range y
)
SELECT * FROM gridix;
x_ix | y_ix
------+------
0 | 0
0 | 1
0 | 2
0 | 3
0 | 4
0 | 5
0 | 6
0 | 7
0 | 8
0 | 9
0 | 10
0 | 11
0 | 12
0 | 13
...
``````
So far, so good. The `gridix` CTE returns coordinates for each cell in the grid, from zero to the `numsq` value from my `params` CTE. From those coordinates, if I know the geographic boundaries of the data set and the number of squares in each edge, I can calculate the latitude and longitude of the four corners of each grid square. First, I need to find the geographic boundaries of the data. My dataset lives in a table called `manzanas`, Spanish for “city block” (and also “apple”). Each row contains one geographic attribute containing a polygon defining the boundaries of the block, and several other attributes such as the education score I mentioned. In PostGIS there are several different ways to find the bounding box I want; here’s the one I used.
``````limits AS (
SELECT
MIN(ST_XMin(geom)) AS xmin,
MAX(ST_XMax(geom)) AS xmax,
MIN(ST_YMin(geom)) AS ymin,
MAX(ST_YMax(geom)) AS ymax
FROM
manzanas
),
``````
And, just to check my results:
``````inegi=# SELECT
MIN(ST_XMin(geom)) AS xmin,
MAX(ST_XMax(geom)) AS xmax,
MIN(ST_YMin(geom)) AS ymin,
MAX(ST_YMax(geom)) AS ymax
FROM
manzanas;
xmin | xmax | ymin | ymax
---------------+--------------+------------------+------------------
-99.349658451 | -98.94668802 | 19.1241898199991 | 19.5863775499992
(1 row)
``````
So the data in question extend from about 99.35 to 98.95 west longitude, and 19.12 to 19.59 north latitude. Now I’ll calculate the boundaries of each grid square, compose a text representation for the square in Well-Known Text format, and convert the text to a PostGIS geometry object. There’s another way I could do this, much simpler and probably much faster, which I hope to detail in a future blog post, but this will do for now.
``````gridcoords AS (
SELECT
(xmax - xmin) / numsq * x_ix + xmin AS x0,
(xmax - xmin) / numsq * (x_ix + 1) + xmin AS x1,
(ymax - ymin) / numsq * y_ix + ymin AS y0,
(ymax - ymin) / numsq * (y_ix + 1) + ymin AS y1,
x_ix || ',' || y_ix AS grid_ix
FROM limits, params, gridix
),
gridewkt AS (
SELECT
grid_ix,
'POLYGON((' ||
x0 || ' ' || y0 || ',' ||
x1 || ' ' || y0 || ',' ||
x1 || ' ' || y1 || ',' ||
x0 || ' ' || y1 || ',' ||
x0 || ' ' || y0 || '))' AS ewkt
FROM gridcoords
),
gridgeom AS (
SELECT
grid_ix, ewkt,
ST_GeomFromText(ewkt, 4326) AS geom
FROM gridewkt
),
``````
And again, I’ll check the result by `SELECT`ing `grid_ix`, `ewkt`, and `geom`, from the first row of the `gridgeom` CTE.
``````-[ RECORD 1 ]-----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
grid_ix | 0,0
ewkt | POLYGON((-99.349658451 19.1241898199991,-99.34562874669 19.1241898199991,-99.34562874669 19.1288116972991,-99.349658451 19.1288116972991,-99.349658451 19.1241898199991))
geom | 0103000020E6100000010000000500000029F4D6CD60D658C06B646FE7CA1F3340D3E508C81ED658C06B646FE7CA1F3340D3E508C81ED658C0EC3DABCDF920334029F4D6CD60D658C0EC3DABCDF920334029F4D6CD60D658C06B646FE7CA1F3340
``````
I won’t claim an ability to translate the geometry object as represented above, but the `ewkt` value looks correct, so let’s keep going. Next I need to cut the blocks into pieces, corresponding to the parts of each block that belong in a single grid square. So a block lying entirely in one square will return one row in this next query; a block that intersects two squares will return two rows. Each row will include the geometries of the block, the square, and the intersection of the two, the total area of the block, the area of the intersection, and the total educational score for the block.
``````block_part AS (
SELECT
grid_ix, -- Grid square coordinates, e.g. 0,0
m.gid AS manz_gid, -- Manzana identifier
ggm.geom, -- Grid square geometry
m.graproes::FLOAT, -- Educational score
ST_Intersection(m.geom, ggm.geom)
AS manz_bg_int, -- Geometric intersection between grid square and city block
ST_Area(m.geom)
AS manz_area, -- Area of the city block
ST_Area(ST_Intersection(m.geom, ggm.geom)) / ST_Area(m.geom)
AS manz_area_perc -- Area of the intersection
from
gridgeom ggm
JOIN manzanas m
ON (ST_Intersects(m.geom, ggm.geom))
WHERE
m.graproes IS NOT NULL -- Skip null educational scores
),
``````
Here’s a sample of those results:
``````-[ RECORD 1 ]--+---------------------
grid_ix | 15,16
manz_gid | 61853
geom | 0103000020E610000...
graproes | 11
manz_bg_int | 0103000020E610000...
manz_area | 3.26175061395103e-06
manz_area_perc | 0.149256808754461
-[ RECORD 2 ]--+---------------------
grid_ix | 15,17
manz_gid | 61853
geom | 0103000020E610000...
graproes | 11
manz_bg_int | 0103000020E610000...
manz_area | 3.26175061395103e-06
manz_area_perc | 0.850743191246843
``````
These results, which I admit to having selected with some care, show a single block, number 61853, which lies across the border between two grid squares. Now we’ll calculate the education score for each block fragment, and then divide the fragments into groups based on the grid square in which they belong, and aggregate the results. I did this in two separate CTEs.
``````grid_calc AS (
SELECT
grid_ix,
geom,
graproes * manz_area_perc AS grid_graproes,
manz_area_perc
FROM
block_part
),
grid_accum AS (
SELECT
grid_ix,
geom,
SUM(grid_graproes) AS sum_graproes
FROM grid_calc
GROUP BY grid_ix, geom
),
``````
This latest CTE gives results such as these:
``````-[ RECORD 1 ]+-----------------
grid_ix | 12,14
geom | 0103000020E61...
sum_graproes | 3888.53630440106
``````
Now we’re left with turning these results into a visualization. I’d like to assign each polygon a height, and a color. To make the visualization easier to understand, I’ll divide the results into a handful of classes, and assign a height and color to each class.
Using an online color palette generator, I came up with a sequence of six colors, which progress from white to a green similar to the green bar on the Mexican flag. Another CTE will return these colors as an array, and yet another will assign the grid squares to groups based on their calculated score. Finally, a third will select the proper color from the array using that group value. At this point, readers are probably thinking “enough already; quit dividing everything into ever smaller CTEs”, to which I can only say, “Yeah, you may be right.”
``````colors AS (
),
colorix AS (
SELECT *,
NTILE(ARRAY_LENGTH(colors, 1)) OVER (ORDER BY sum_graproes asc) AS edu_colorix
FROM grid_accum, colors
),
color AS (
SELECT *,
colors[edu_colorix] as edu_color
from colorix
)
``````
The ntile() window function is useful for this kind of thing. It divides the given partition into buckets, and returns the number of the bucket for each row. Here, the partition consists of the whole data set; we sort it by educational score to ensure low-scoring grid squares get low-numbered buckets. Note also that I can change the colors, adding or removing groups, simply by adjusting the `colors` CTE. This could theoretically prove handy, if I decided I didn’t like the number of levels or the color scheme, but it’s a feature I never used for this visualization.
We’re on the home stretch, at last, and I should clarify how I plan to turn the database objects into KML, usable on a Liquid Galaxy. I used ogr2ogr from the GDAL toolkit. It converts between a number of different GIS data sources, including PostGIS to KML. I need to feed it the geometry I want to draw, as well as styling instructions and, in this case, a custom KML altitudeMode.
Styling is an involved topic; for our purposes it’s enough to say that I’ll tell `ogr2ogr` to use our selected color both to draw the lines of our polygons, and to fill them in. But moving the grid square’s geometry to an altitude corresponding to its educational score is fairly easy, using PostGIS’s ST_Force3DZ() function to add to the hitherto two-dimensional polygon a zero-valued third dimension, and ST_Translate() to move it above the surface of the earth a ways. So I can probably finish this with one final query:
``````-- Insert all previous CTEs here
SELECT
ST_Translate(
ST_Force3DZ(geom), 0, 0,
alt_bias + edu_colorix * alt_percfactor
) AS edu_geom,
'BRUSH(fc:#' || edu_color || 'ff);PEN(c:#' || edu_color || 'ff)' AS edu_style,
'absolute' AS "altitudeMode"
FROM color, params;
``````
You may remember `alt_bias` and `alt_percfactor`, the oddly named and thus far unexplained values in my first `params` CTE. These I used to control how far apart in altitude one group of polygons is from another, and to bias them all far enough above the ground to avoid the problem of them being obscured by terrain features. You may also remember that this query began with the `CREATE TABLE grid_mza_vals AS...` command, meaning that we’ll store the results of all this processing in a table, so `ogr2ogr` can get to it. We call `ogr2ogr` like this:
``````ogr2ogr \
-f LIBKML education.kml \
-sql "SELECT grid_ix, edu_geom, edu_style as \"OGR_STYLE\", \"altitudeMode\" FROM grid_mza_vals"
``````
OGR’s LIBKML driver knows an attribute called “OGR_STYLE” is a style string, and one called “altitudeMode” is, predictably, the feature’s altitude mode. So this will create a KML file, containing one placemark for each row in our `grid_mza_tables` table. Each placemark consists of a colored square, floating in the air above Mexico City. The squares come in six different levels and six different colors, corresponding to our original education data. Something like this:
Here’s one of the placemarks from the KML.
`````` <Placemark id="sql_statement.1">
<Style>
<LineStyle>
<color>ffffffff</color>
<width>1</width>
</LineStyle>
<PolyStyle>
<color>ffffffff</color>
</PolyStyle>
</Style>
<ExtendedData>
<SimpleData name="grid_ix">11,10</SimpleData>
<SimpleData name="OGR_STYLE">BRUSH(fc:#ffffffff);PEN(c:#ffffffff)</SimpleData>
</ExtendedData>
<Polygon>
<altitudeMode>absolute</altitudeMode>
<outerBoundaryIs>
<LinearRing>
<coordinates>
-99.17235146136,19.3090649119991,4300
-99.15623264412,19.3090649119991,4300
-99.15623264412,19.3275524211991,4300
-99.17235146136,19.3275524211991,4300
-99.17235146136,19.3090649119991,4300
</coordinates>
</LinearRing>
</outerBoundaryIs>
</Polygon>
</Placemark>
``````
This may be sufficient, but it gets confusing when viewed from a low angle, because it’s hard to tell which square belongs to which part of the map. I prefer having the polygons “extruded” from the ground, as KML terms it. I used a simple Perl script to add the `extrude` element to each polygon in the KML, resulting in this:
This query works, but leaves a few things to be desired. For instance, doing everything in one query is probably not the best option when lots of processing is involved. Ideally we would calculate the grid geometries once, and save them in a table somewhere, for quicker processing as we build the rest of the query and experiment with visulization options. Second, PostGIS provides an arguably more elegant way of finding grid squares in the first place, a method which also affords other options for interesting visualizations. Stay tuned for a future blog post discussing these issues. Meanwhile, CTEs proved a valuable way to systematize and modularize a very complicated query. | 4,952 | 19,035 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2021-04 | longest | en | 0.947591 |
https://metanumbers.com/558680 | 1,624,450,423,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488538041.86/warc/CC-MAIN-20210623103524-20210623133524-00157.warc.gz | 368,647,966 | 10,939 | ## 558680
558,680 (five hundred fifty-eight thousand six hundred eighty) is an even six-digits composite number following 558679 and preceding 558681. In scientific notation, it is written as 5.5868 × 105. The sum of its digits is 32. It has a total of 5 prime factors and 16 positive divisors. There are 223,456 positive integers (up to 558680) that are relatively prime to 558680.
## Basic properties
• Is Prime? No
• Number parity Even
• Number length 6
• Sum of Digits 32
• Digital Root 5
## Name
Short name 558 thousand 680 five hundred fifty-eight thousand six hundred eighty
## Notation
Scientific notation 5.5868 × 105 558.68 × 103
## Prime Factorization of 558680
Prime Factorization 23 × 5 × 13967
Composite number
Distinct Factors Total Factors Radical ω(n) 3 Total number of distinct prime factors Ω(n) 5 Total number of prime factors rad(n) 139670 Product of the distinct prime numbers λ(n) -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 0 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 558,680 is 23 × 5 × 13967. Since it has a total of 5 prime factors, 558,680 is a composite number.
## Divisors of 558680
16 divisors
Even divisors 12 4 2 2
Total Divisors Sum of Divisors Aliquot Sum τ(n) 16 Total number of the positive divisors of n σ(n) 1.25712e+06 Sum of all the positive divisors of n s(n) 698440 Sum of the proper positive divisors of n A(n) 78570 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 747.449 Returns the nth root of the product of n divisors H(n) 7.1106 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 558,680 can be divided by 16 positive divisors (out of which 12 are even, and 4 are odd). The sum of these divisors (counting 558,680) is 1,257,120, the average is 78,570.
## Other Arithmetic Functions (n = 558680)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 223456 Total number of positive integers not greater than n that are coprime to n λ(n) 27932 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 45867 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 223,456 positive integers (less than 558,680) that are coprime with 558,680. And there are approximately 45,867 prime numbers less than or equal to 558,680.
## Divisibility of 558680
m n mod m 2 3 4 5 6 7 8 9 0 2 0 0 2 3 0 5
The number 558,680 is divisible by 2, 4, 5 and 8.
• Arithmetic
• Abundant
• Polite
## Base conversion (558680)
Base System Value
2 Binary 10001000011001011000
3 Ternary 1001101100212
4 Quaternary 2020121120
5 Quinary 120334210
6 Senary 15550252
8 Octal 2103130
10 Decimal 558680
12 Duodecimal 22b388
20 Vigesimal 39ge0
36 Base36 bz2w
## Basic calculations (n = 558680)
### Multiplication
n×i
n×2 1117360 1676040 2234720 2793400
### Division
ni
n⁄2 279340 186227 139670 111736
### Exponentiation
ni
n2 312123342400 174377068932032000 97420980870947637760000 54427153592981026263756800000
### Nth Root
i√n
2√n 747.449 82.3609 27.3395 14.1069
## 558680 as geometric shapes
### Circle
Diameter 1.11736e+06 3.51029e+06 9.80564e+11
### Sphere
Volume 7.30429e+17 3.92226e+12 3.51029e+06
### Square
Length = n
Perimeter 2.23472e+06 3.12123e+11 790093
### Cube
Length = n
Surface area 1.87274e+12 1.74377e+17 967662
### Equilateral Triangle
Length = n
Perimeter 1.67604e+06 1.35153e+11 483831
### Triangular Pyramid
Length = n
Surface area 5.40613e+11 2.05505e+16 456160
## Cryptographic Hash Functions
md5 50258c3edd2caa220da76c32988bbe96 c4eeaebf92c296d5eb5ac51d716feedc1342c8fd 9ff1c989da26873e4d222eb5bd0f4678da33a36709ae32ef745f1d7ed77ca4c1 31c2b1ae2360feb07d7a05595dceacd68182b7d6752005300a60a783cfe053bea598d44b6cd3d9c750d529cab27bcedb7cf2f214017dc1a8b5cc272831ef3eee add7bacf0f5fa8ed409255eb728bb02004329dcf | 1,454 | 4,166 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2021-25 | latest | en | 0.820541 |
https://www.topcoder.com/tc?module=Static&d1=match_editorials&d2=srm377 | 1,717,061,974,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059632.19/warc/CC-MAIN-20240530083640-20240530113640-00822.warc.gz | 902,305,122 | 9,938 | JOIN
Match Editorial
SRM 377
Saturday, November 17, 2007
The key part of this problem is to determine whether an integer number is almost prime or not. According to the definition, it shouldn't be prime and divisible by numbers less than 11. So, the Java function looks like this:
```boolean isAlmostPrime(int n) {
return !isPrime(n) && n%2 !=0 && n%3 !=0 && n%5 !=0 && n%7 !=0;
}
```
Having this function, we just iterate through all the numbers that are greater than the given one and return the first almost prime number found. The only thing left to discuss is how long can we search for the smallest almost prime number that is greater than the given one. This is not very easy question, but it's good we need not to make a complex analysis. It's obvious that all powers of 11 are almost prime, and 11^6 = 1771561. So, even this raw estimation shows that our code will not timeout, if we implement the function isPrime to work in O(sqrt(n)). If you don't know how to do that, please read this tutorial.
# SquaresInsideLattice
The key observation in this problem is: for each square inside the lattice its bounding box is a square too. By bounding box we mean the smallest rectangle with sides parallel to the axes, which contains the given square. So, each square with sides parallel to the axes uniquely defines some number of lattice squares. And this number depends only on the size of the square, actually it equals to the length of the square's side. So, all we need is to iterate through all possible lengths of square's side and count the answer. Here is a Java implementation of this approach:
```
long res = 0;
for (long i = 1; i <= Math.min(width,height); ++i)
res += (width - i + 1) * (height - i + 1) * i;
```
# GameOnAGraph
This problem was yet another exercise with the matrices. To see this let's look what happens with the vertices' marks during the game. Let's denote the marks before some turn as x1, x2, ..., xn and after that turn as y1, y2, ..., yn. Then we can write the following equations:
```y1 = A1,1x1 + A1,2x2 + ... + A1,nxn
y2 = A2,1x1 + A2,2x2 + ... + A2,nxn
.................................
yn = An,1x1 + An,2x2 + ... + An,nxn
```
Here coefficients Ai,j depend on the graph and whether the turn was white or black. These linear equations can be written using matrices in a compact form as: Y = A*X. Where X and Y are columns and A is a square matrix of coefficients. As we can see matrix A doesn't depend on the initial marks. Let's denote by W and B matrices which correspond to white and black turns respectively and by X the column of initial marks. Then, after the first turn marks will be W*X, after the second turn B*W*X, then W*B*W*X and so on... Since the matrix multiplication is an associative operation we get the following solution (in pseudo code):
```solve(Graph g, int N, Column X){
Find Matrix W (depends on g)
Find Matrix B (depends on g)
C := ( B*W )^( N/2 )
if (N is odd) C := W*C
return C*X
}
```
Among other things you should implement matrix multiplication and exponentiation. The later must be done such that the number of multiplications is proportional to log(N) to work fast enough. Please, read this if you are not familiar with the logarithmic exponentiation.
# AlienLanguage
This problem has many solutions. We'll describe some of them. But first, let's figure out what we should find. Let our word contains V vowels and C consonants. Then we have exactly P^V*Q^C words, not counting stresses. With stresses each of these words will define 1 + V + C + V*C = (V+1)*(C+1) different words. So, the answer to the problem is (-1 because we don't allow empty words):
This form is not very convinient, so let's rewrite it like this:
Now we see that we only need to know how to compute the sum S(n) = 1 + 2*x + 3*x^2 + ... + (n+1)*x^n. This is the crotch where different solutions actually start to differ.
### Method 1: Recurrence
We can notice that for odd n (n = 2*k-1) the following is true:
S( 2*k-1 ) = ( 1 + x^k )*S( k-1 ) + k*x^k*( 1+x+x^2+...+x^(k-1) ),
or if we denote the sum of geometrical progression 1+x+x^2+...+x^k by G(k), then
S( 2*k-1 ) = ( 1 + x^k )*S( k-1 ) + k*x^k*G( k-1 ).
This means that we have a recurrence which decreases the power by 2. So, we can compute the result in logarithmic time. But to do that we must be able to count G(k) which is not very hard because of the similar recurrence:
G(2*k-1) = ( 1 + x^k )*G( k-1 ).
### Method 2: Matrices
It turns out that matrices are applicable in this problem too. Let's see how.
We can write the following equations:
```S( n+1 ) = x * S( n ) + x * G( n ) + 1
G( n+1 ) = 0 + x * G( n ) + 1
1 = 0 + 0 + 1
```
Now, it's clear where the matrices came from. Specifically,
```( S( n+1 ) ) ( x x 1 ) ( S( n ) )
( G( n+1 ) ) = ( 0 x 1 ) ( G( n ) )
( 1 ) ( 0 0 1 ) ( 1 )
```
So, all we need is to find a corresponding power of the given 3x3 matrix. Of cource, it should be fast to pass the system tests. Fortunately, we already know how to do this from the previous problem.
### Method 3: Closed form
Another way to solve this problem is to find a formula.
We already know that S( n ) = x*S( n-1 ) + G( n ), but on the other hand S( n ) = S( n-1 ) + (n+1)*x^n. Getting rid of S( n-1 ) we yield that S( n ) = ( (n+1)*x^(n+2) - (n+2)*x^(n+1) + 1 ) / (x-1)^2
Although division is a problem for modular computation, it can be resolved in the following way: we need to count numerator by modulo M*(x-1)^2 instead of M and after that divide it by (x-1)^2.
By it4.kp
TopCoder Member | 1,608 | 5,573 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.984375 | 4 | CC-MAIN-2024-22 | latest | en | 0.913563 |
https://calculator.name/base-converter/duotrigesimal/heptadecimal | 1,620,799,657,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991252.15/warc/CC-MAIN-20210512035557-20210512065557-00492.warc.gz | 173,823,789 | 3,752 | Enter duotrigesimal numbers to convert into heptadecimal. Easy base-32 to base-17 conversion
Base converter
Number:
From:
To:
## How to Convert from duotrigesimal to heptadecimal?
In number system, we know duotrigesimal is base-32 and heptadecimal is base-17. To convert duotrigesimal to heptadecimal, you follow these steps:
To do this, first convert duotrigesimal into decimal, then the resulting decimal into heptadecimal
1. Start from one's place in duotrigesimal : multiply ones place with 32^0, tens place with 32^1, hundreds place with 32^2 and so on from right to left
2. Add all the products we got from step 1 to get the decimal equivalent of given duotrigesimal value.
3. Divide decimal value we got from step-2 by 17 keeping notice of the quotient and the remainder.
4. Continue dividing the quotient by 17 until you get a quotient of zero.
5. Then just write out the remainders in the reverse order to get heptadecimal equivalent of decimal number.
### Example #1
Convert 4t from duotrigesimal to heptadecimal
First, find the decimal value of 4t32 by using step1 & step2 above:
4t32 = 4 × 322t × 321 = 15710 | 289 | 1,129 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2021-21 | latest | en | 0.67118 |
http://mathhelpforum.com/calculus/50521-differential-calculus-midterm-print.html | 1,519,584,080,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891816841.86/warc/CC-MAIN-20180225170106-20180225190106-00538.warc.gz | 228,781,777 | 4,198 | # Differential Calculus midterm
• Sep 24th 2008, 06:36 PM
VX-1
Differential Calculus midterm
Hey guys can you help me with the questions on this Math midterm?
http://img440.imageshack.us/img440/6...midtermet4.jpg
• Sep 24th 2008, 06:42 PM
colby2152
What exactly do you need help with? This is one of your first posts, and you simply post the question and ask for help. It seems like you just want the answers.
For example, the question: Find the derivative of $f(x)=\frac{x}{1+x}$ I would treat it as a product like $f(x)=x*(x+1)^{-1}$, and then you find the derivative by applying the product and power rules.
• Sep 24th 2008, 07:42 PM
VX-1
I don't want the answers, I'm just completely clueless as to how I would approach them.
• Sep 24th 2008, 07:58 PM
Jhevon
Quote:
Originally Posted by VX-1
Hey guys can you help me with the questions on this Math midterm?
http://img440.imageshack.us/img440/6...midtermet4.jpg
1.
"orthogonal" means "perpendicular." so you want all the points on
$y = 5x^3 + 5$ where the slope (that is, the derivative) is 30. please tell me you know why we need it to be 30
2.
the definition says: $f'(x) = \lim_{h \to 0} \frac {f(x + h) - f(x)}h$
now plug in $f(x + h)$ and $f(x)$ and take the limit. (you will need to simplify before you can do that, of course)
in case you have never seen that definition, you must have used this one. it is an alternate definition equivalent to the one above.
$f'(a) = \lim_{x \to a} \frac {f(x) - f(a)}{x - a}$
3.
(a) note that $\sqrt{x^2 + 1}$ increases a lot faster than $\sqrt{x}$. alternatively, note that as $x \to \infty$ the $+1$ doesn't really matter. thus you have $\sqrt{x^2} - \srqt{x} = |x| - \sqrt{x} = x - \sqrt{x}$ for positive $x$'s
(b) more straightforward solution: multiply by the cojugate of the denominator over itself.
alternate solution: note that $x - 4 = (\sqrt{x})^2 - 4 = (\sqrt{x} + 2)(\sqrt{x} - 2)$
4.
you want to find $a$ and $b$ so that the function is continuous and the derivative limit for both parts exist and are equal.
5.
first find $f'(x)$ (you will need the chain rule). then you can answer the problem. i hope you know how to tell something is an extreme value from the derivative, and that you know a function is increasing wherever the derivative is positive and decreasing where it is negative
6.
Let $f(x) = \frac 1{x^2 + 1}$
by tweaking our definition for derivative a bit, we obtain the linear approximation formula:
$f(x) \approx f(a) + f'(a)(x - a)$
now, $x$ is the value that we want to find $f(x)$ for, while $a$ is a value close to $x$ that we already know $f(a)$ and $f'(a)$ for
so there's how you should approach these problems. good luck!
Quote:
Originally Posted by colby2152
For example, the question: Find the derivative of $f(x)=\frac{x}{1+x}$ I would treat it as a product like $f(x)=x*(x+1)^{-1}$, and then you find the derivative by applying the product and power rules.
well, they said use the definition. i am pretty sure they are talking about using limits to calculate the derivative
• Sep 25th 2008, 06:05 AM
colby2152
Quote:
Originally Posted by Jhevon
well, they said use the definition. i am pretty sure they are talking about using limits to calculate the derivative
True. I missed that part of the question. You'll have to find the derivative via the tedious, tried and true method that is the definition of the derivative:
$f'(x)=\lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$ | 1,012 | 3,435 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 28, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2018-09 | longest | en | 0.913857 |
https://www.inchcalculator.com/convert/millisecond-to-microsecond/ | 1,725,939,725,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651196.36/warc/CC-MAIN-20240910025651-20240910055651-00120.warc.gz | 770,374,456 | 15,702 | Milliseconds to Microseconds Converter
Enter the time in milliseconds below to get the value converted to microseconds.
Result in Microseconds:
1 ms = 1,000 µs
Do you want to convert microseconds to milliseconds?
How to Convert Milliseconds to Microseconds
To convert a measurement in milliseconds to a measurement in microseconds, multiply the time by the following conversion ratio: 1,000 microseconds/millisecond.
Since one millisecond is equal to 1,000 microseconds, you can use this simple formula to convert:
microseconds = milliseconds × 1,000
The time in microseconds is equal to the time in milliseconds multiplied by 1,000.
For example, here's how to convert 5 milliseconds to microseconds using the formula above.
microseconds = (5 ms × 1,000) = 5,000 µs
How Many Microseconds Are in a Millisecond?
There are 1,000 microseconds in a millisecond, which is why we use this value in the formula above.
1 ms = 1,000 µs
What Is a Millisecond?
One millisecond is equal to 1/1,000 of a second or 1,000 microseconds.
The millisecond is a multiple of the second, which is the SI base unit for time. In the metric system, "milli" is the prefix for thousandths, or 10-3. Milliseconds can be abbreviated as ms; for example, 1 millisecond can be written as 1 ms.
What Is a Microsecond?
One microsecond is equal to 1/1,000 of a millisecond or 1,000 nanoseconds.
The microsecond is a multiple of the second, which is the SI base unit for time. In the metric system, "micro" is the prefix for millionths, or 10-6. Microseconds can be abbreviated as µs; for example, 1 microsecond can be written as 1 µs.
Millisecond to Microsecond Conversion Table
Table showing various millisecond measurements converted to microseconds.
Milliseconds Microseconds
0.001 ms 1 µs
0.002 ms 2 µs
0.003 ms 3 µs
0.004 ms 4 µs
0.005 ms 5 µs
0.006 ms 6 µs
0.007 ms 7 µs
0.008 ms 8 µs
0.009 ms 9 µs
0.01 ms 10 µs
0.02 ms 20 µs
0.03 ms 30 µs
0.04 ms 40 µs
0.05 ms 50 µs
0.06 ms 60 µs
0.07 ms 70 µs
0.08 ms 80 µs
0.09 ms 90 µs
0.1 ms 100 µs
0.2 ms 200 µs
0.3 ms 300 µs
0.4 ms 400 µs
0.5 ms 500 µs
0.6 ms 600 µs
0.7 ms 700 µs
0.8 ms 800 µs
0.9 ms 900 µs
1 ms 1,000 µs | 675 | 2,155 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2024-38 | latest | en | 0.804539 |
http://www.jiskha.com/display.cgi?id=1301261944 | 1,498,350,224,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320368.57/warc/CC-MAIN-20170624235551-20170625015551-00343.warc.gz | 557,790,419 | 4,031 | # calculus
posted by .
An open box of maximum volume is to be made from a square piece of cardboard, 24 inches on each side, by cutting equal squares from the corners and turning up the sides to make the box.
(a) Express the volume V of the box as a function of x, where x is edge length of the square cut-outs.
(b) What are the dimensions of the box that enclose the largest possible volume? State your answer in the form length by width by height.
(c) What is the maximum volume?
• calculus -
let each side of the equal squares be x inches
length of box = 24-2x
width of box = 24-2x
height of box = x
a) Volume = x(24-2x)(24-2x)
b) expand the volume equation, then take the first derivative.
Set that derivative equal to zero. You will have a quadratic equation. Take the positive answer which lies between 0 and 12
c) put the answer from b) into the volume equation and evaluate.
• calculus -
20>o | 228 | 909 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2017-26 | latest | en | 0.906795 |
https://www.speedsolving.com/threads/how-long-do-you-think-the-devils-algorithm-will-be-for-4x4.71394/ | 1,571,892,541,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987841291.79/warc/CC-MAIN-20191024040131-20191024063631-00195.warc.gz | 1,071,738,343 | 15,752 | How long do you think the Devil's Algorithm will be for 4x4.
stormtrooper
Member
My personal inference is that it would be quadrillion moves long or even longer. My inference can be wrong.
Billabob
Member
There are 7401196841564901869874093974498574336000000000 permutations of the 4x4x4 cube so no, a quadrillion moves is not enough.
Julio974
Member
1st question: is is small than a googol?
2nd question: if not, is it smaller than a googolplex?
3rd question: if not, it is smaller than
?
4rd question: if not, is it smaller than Graham's Number?
5th question: it is smaller than SSCG(3)?
I'm just preparing for 5x5 and Megaminx at the same time, don't worry!
xyzzy
Member
2nd question: if not, is it smaller than a googolplex?
3rd question: if not, it is smaller than
?
4rd question: if not, is it smaller than Graham's Number?
5th question: it is smaller than SSCG(3)?
Roughly speaking, a googolplex is doubly-exponentially huge, 3^^^^3 involves tetration, Graham's number is in a class of its own, and SSCG(3) even further beyond that.
Graham's number and SSCG(k) both come from graph theory, and their mindboggling size comes from the fact that many advanced combinatoric results cannot be proven in Peano arithmetic, which "contains", among other things, the class of easily computable functions. (Very roughly speaking.) These results require more axioms to prove than just Peano arithmetic (e.g. the standard set theory axioms: ZFC), and consequently, the functions that come out of such results have to grow extremely quickly. (If they didn't, Peano arithmetic would've been able to handle them… roughly speaking. It might be possible that such functions don't grow quickly, but the fact that they don't grow quickly cannot be proven in PA; I can't think of any such example, however.)
In contrast, if you're just looking at something like the number of states on an n×n×n cube, that's only singly-exponentially large. You shouldn't even expect it to hit a googolplex for reasonable values of n, much less Graham's number or SSCG(3).
Julio974
Member
Roughly speaking, a googolplex is doubly-exponentially huge, 3^^^^3 involves tetration, Graham's number is in a class of its own, and SSCG(3) even further beyond that.
Graham's number and SSCG(k) both come from graph theory, and their mindboggling size comes from the fact that many advanced combinatoric results cannot be proven in Peano arithmetic, which "contains", among other things, the class of easily computable functions. (Very roughly speaking.) These results require more axioms to prove than just Peano arithmetic (e.g. the standard set theory axioms: ZFC), and consequently, the functions that come out of such results have to grow extremely quickly. (If they didn't, Peano arithmetic would've been able to handle them… roughly speaking. It might be possible that such functions don't grow quickly, but the fact that they don't grow quickly cannot be proven in PA; I can't think of any such example, however.)
In contrast, if you're just looking at something like the number of states on an n×n×n cube, that's only singly-exponentially large. You shouldn't even expect it to hit a googolplex for reasonable values of n, much less Graham's number or SSCG(3).
I'm already preparing for bigger cubes!
But you're right on that...
ichcubegerne
Member
In general I like the idea of finding an optimal solution for the devils alg on different cubes from the wca and proving that they are optimal^^
PapaSmurf
Member
What you can do is the devil's algorithm for every 2x2 state, but then within that nest an algorithm that with do the centres and wings, so if you work out the alg excluding corners, you can just do the 2x2 one, and within each turn do the centre/wing one. That would (probably) reach every state, although it probably won't be optimal.
Sion
Member
3^^^3 is 3 tetrated three times, in which means it is gargantuan compared to 3^^3. | 976 | 3,931 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2019-43 | latest | en | 0.962608 |
http://mathhelpforum.com/algebra/35016-geometric-progression.html | 1,529,592,649,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864172.45/warc/CC-MAIN-20180621133636-20180621153636-00416.warc.gz | 206,401,125 | 10,278 | 1. ## geometric progression
A geometric progression has first term a where a‡0 and common ratio r where r‡1. the difference between the fourth and first term is equal to four times the difference between the third term and the second term.
i) show that r³ - 4r² + 4r -1 = 0
ii) find two possible values for the ratio of the geometric progression.
ii) for the value of r which the progression is convergent, prove that the sum to infinity is 1/2 a (1 + √5)
thanks
2. The nth term in a geometric sequence is given by
$\displaystyle a_{n}=a_{1}r^{n-1}$
For part a you have $\displaystyle a_{1}r^{3}-a_{1}=4(a_{1}r^{2}-a_{1}r)$
$\displaystyle a_{1}r^{3}-a_{1}=4a_{1}r^{2}-4a_{1}r$
$\displaystyle a_{1}r^{3}-4a_{1}r^{2}+4a_{1}r-a_{1}$
Factor out $\displaystyle a_{1}$:
$\displaystyle a_{1}(r^{3}-4r^{2}+4r-1)$
Solve the cubic to find the possible values of r.
3. Hello, Gracey!
A geometric progression has first term $\displaystyle a \neq 0$ and common ratio $\displaystyle r \neq 1$
The difference between $\displaystyle a_4\text{ and }a_1$ is equal to 4 times the difference between $\displaystyle a_3\text{ and }a_2$
(a) Show that: .$\displaystyle r^3 - 4r^2 + 4r - 1 \:=\:0$
The first four terms are: .$\displaystyle \begin{Bmatrix}a_1 &=& a \\ a_2 &=& ar \\ a_3 &=& ar^2 \\ a_4 &=&ar^3 \end{Bmatrix}$
We are told that: .$\displaystyle a_4 - a_1 \:=\:4\left(a_3-a_2\right)$
So we have .$\displaystyle ar^3 - a \:=\:4(ar^2-ar) \quad\Rightarrow\quad ar^3 - 4ar^2 + 4ar - a \:=\:0$
. . Divide by $\displaystyle a\!:\;\;\boxed{r^3 - 4r^2 + 4r - 1 \:=\:0}$
(b) Find two possible values for $\displaystyle r.$
Factor: .$\displaystyle (r-1)(r^2-3r+1) \:=\:0$
We have: .$\displaystyle r-1\:=\:\quad\Rightarrow\quad r \,=\,1\quad\hdots \text{ which is not allowed}$
. . and: .$\displaystyle r^2-3r+1\:=\:0\quad\Rightarrow\quad \boxed{r \:=\:\frac{3\pm\sqrt{5}}{2}}$
(c) For the value of $\displaystyle r$ which the progression is convergent,
prove that the sum to infinity is: .$\displaystyle \frac{1}{2}a\left(1 + \sqrt{5}\right)$
To be convergent, $\displaystyle r < 1.$ . Hence, we will use: .$\displaystyle r \;=\;\frac{3-\sqrt{5}}{2}$
Formula: .$\displaystyle S \;=\;\sum^{\infty}_{n=0} ar^n \;=\;\frac{a}{1-r}$
$\displaystyle \text{So we have: }\;S \;=\;\frac{a}{1 - \frac{3-\sqrt{5}}{2}} \;=\;\frac{2a}{\sqrt{5}-1}$
Rationalize: .$\displaystyle S \;=\;\frac{2a}{\sqrt{5}-1}\cdot\frac{\sqrt{5}+1}{\sqrt{5}+1} \;=\;\frac{2a(\sqrt{5}+1)}{5-1}\;=\;\frac{2a(1 + \sqrt{5})}{4}$
. . Therefore: .$\displaystyle \boxed{S \;=\;\frac{1}{2}a(1 + \sqrt{5})}$ | 993 | 2,567 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2018-26 | latest | en | 0.686644 |
http://studentsmerit.com/paper-detail/?paper_id=48928 | 1,480,832,624,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541214.23/warc/CC-MAIN-20161202170901-00462-ip-10-31-129-80.ec2.internal.warc.gz | 256,688,975 | 7,839 | #### Details of this Paper
##### GMBA 767-Security Analysis and Portfolio Management Problems
Description
solution
Question
Question;5. Consider the following questions on the pricing of;options on the stock of ARB Inc.;a- A share of ARB stock sells for \$75 and has a standard;deviation of returns equal to 20% per year. The current risk-free rate is 9%;and the stock pays two dividends: (1) a 2% dividend just prior to the option?s;expiration day, which is 91 days from now (i.e., exactly one half year).;Calculate the black-Scholes value for European-style call option with an;exercise price of \$70.;b- What would be the price of a 91-day European-style option;on ARB stock having the same exercise price?;c- Calculate the change in the call option?s value that;would occur if ARB?s management suddenly decided to suspend dividend payments;and this action had no effect on the price of the company?s stock.;d- Briefly describe (without calculations) how your answer;in part a would differ under the following separate circumstances: (1) the;volatility of ARB stock increases to 30%, and (2) the risk free rate decreases;to 8%.;10- Melissa Simmons is the chief investment officer of a;hedge fund specializing in options trading. She is currently back testing;various option trading strategies that will allow her to profitfrom large;fluctuations-either up or down-in a stock?s price. An example of such typical;trading strategy is straddle strategy that involves the combination of a long;call and a long put with an identical strike price and time to maturity. She is;considering the following pricing information on securities associated with;friendwork, a new internet start-up hosting a leading online social network;Friendwork stock: \$100;Call option with an exercise price of \$100 expiring in one;year: \$9;Put option with an exercise price of \$100 expiring in one;year: \$8;a. Use the;above information on friendwork and draw a diagram showing the net profit/loss;position at maturity for the straddle strategy. Clearly label on the graph the;break-even points of the position.;b. Melissa?s;colleague proposes another lower-cost option strategy that would profit from a;large fluctuation in friendwork?s stock price;Long call option with an exercise price of \$110 expiring in;one year: \$6;Long put option with an exercise price of \$90 expiring in;one year: \$5;Similar to part a, draw a diagram showing the net;profit/loss position for the above alternative option strategy. Clearly label;on the graph the breakeven points of the position.
Paper#48928 | Written in 18-Jul-2015
Price : \$25 | 604 | 2,612 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2016-50 | longest | en | 0.887052 |
https://fizzbuzzer.com/twins-challenge/ | 1,603,674,263,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107890108.60/warc/CC-MAIN-20201026002022-20201026032022-00359.warc.gz | 332,989,813 | 10,587 | ## [f]izzbuzzer
### Looking for good programming challenges?
Use the search below to find our solutions for selected questions!
# Twins challenge
Sharing is caring!
Problem statement
Lia is fascinated by anything she considers to be a twin. She calls a pairs of positive integers, $i$ and $j$, twins if:
– They are both prime. A prime number is an integer greater than that has no positive divisors other than $1$ and itself.
– Their absolute difference is exactly equal to $2$ (i.e., $|j - i| = 2$).
Given an inclusive interval of integers from $n$ to $m$, can you help Lia find the number of pairs of twins there are in the interval (i.e., $\left[n,m\right]$)? Note that pairs $(i,j)$ and $(j,i)$ are considered to be the same pair.
Input Format
Two space-separated integers describing the respective values of $n$ and $m$.
Constraints
$1 \le n \le m \le 10^9$
$m - n \le 10^6$
Output Format
Print a single integer denoting the number of pairs of twins.
Sample Input 0
3 13
Sample Output 0
3
Explanation 0
There are three pairs of twins: $(3,5)$, $(5,7)$ and $(11,13)$.
Solution
We will solve this problem by reducing it to two familiar problems. Sieve of Eratosthenes in interval and Find the number of pairs in an array whose difference is k. So we will first store all primes in the interval $\left[n,m\right]$ in a set and then find the number of pairs in the set whose difference is $2$.
The implementation is listed below.
Full code
public class Twins {
// Sieve of Eratosthenes for [n,m]
public static Set primes(int n, int m) {
Set primes = new TreeSet<>();
int k = (int) Math.sqrt(m);
boolean[] a = new boolean[k + 1];
boolean[] b = new boolean[m - n + 1];
for (int i = 0; i < a.length; i++) {
a[i] = true;
}
for (int i = 0; i < b.length; i++) {
b[i] = true;
}
for (int i = 2; i <= k; i++) {
if (a[i]) {
for (int j = i * i; j <= k; j = j + i) {
a[j] = false;
}
for (int j = Math.max(2, (n + i - 1) / i) * i; j <= m; j = j + i) {
b[j - n] = false;
}
}
}
for (int i = 0; i < b.length; i++) {
if (b[i] && (i + n) > 1) {
}
}
return primes;
}
public static void main(String[] args) throws FileNotFoundException {
System.setIn(new FileInputStream(System.getProperty("user.home") + "/" + "in.txt"));
Scanner in = new Scanner(System.in);
int n = in.nextInt();
int m = in.nextInt();
Set primes = primes(n, m);
int count = 0;
for (int prime : primes) {
if (primes.contains(prime + 2)) {
count++;
}
}
System.out.println(count);
}
} | 720 | 2,460 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 19, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2020-45 | latest | en | 0.721869 |
https://6abc.com/archive/6129939/ | 1,680,121,968,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949025.18/warc/CC-MAIN-20230329182643-20230329212643-00159.warc.gz | 110,720,539 | 31,882 | # Does 1" of rain equal 10" of snow?
Dateline article| David Murphy|
by David Murphy
This is a common notion that goes around. The belief is that one inch of rain would translate into 10 inches of snow, were it cold enough to make snow. This idea comes from the ratio meteorologists use to determine the amount of snowfall expected from a storm, given that computer models only tell you how much liquid precipitation is expected. In other words, models don't predict snowfall amounts. They only tell you how much water is expected from the storm. Since snow is made of ice crystals which don't combine as well as drops of water, the accumulation of snow is always greater than the computer-predicted rainfall. It's up to the meteorologist to determine how much greater and on average, the 10-to-1 ratio often works out to be a good predictor of actual snowfall.
But it's not always the correct ratio. In colder storms, which produce a drier but fluffier snow, a 15-to-1 ratio is not uncommon. And in warmer, wetter storms, a thicker, heavier snow is produced and the ratio might drop to 7-to-1, or even lower. The ratio used all depends on the temperature of the air in which the storm is churning.
But here's my problem with the way people tend to interpret this ratio...
Sometimes, after a winter rainstorm, you'll hear people saying, "Yo, today's storm gave us 1 inch of rain---we're lucky it wasn't snow or it would've been 10 inches!" In reality, a storm that produces 1 inch of rain would probably not produce 10 inches of snow. This is because colder air (which, of course, is required to get snow) can't hold as much moisture as the milder air required for rain. In the above example, the rain storm would have to have its temperature lowered before it could produce snow and in doing so, it would actually have to lose much of its moisture as rain before the snow could begin---this would happen automatically, in fact. The result? There wouldn't be enough moisture left over to produce the 10" of snow. All things being equal, it would probably take a rainstorm that's already producing many inches of rain in near freezing conditions to then go slightly colder and give you the ten inches.
I even hear meteorologists oversimplifying this sometimes, so don't feel bad if you have, too.
Copyright © 2023 WPVI-TV. All Rights Reserved. | 529 | 2,350 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2023-14 | latest | en | 0.973927 |
http://mathhelpboards.com/business-mathematics-18/?s=9278ceb6188b675b9c2ddddfeb0e6f5e | 1,529,446,304,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267863206.9/warc/CC-MAIN-20180619212507-20180619232507-00239.warc.gz | 207,818,222 | 35,631 | Page 1 of 5 123 ... Last
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Note: when sorting by date, 'descending order' will show the newest results first. | 1,075 | 2,963 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2018-26 | latest | en | 0.856338 |
https://www.marketswiki.com/wiki/Hash_function | 1,631,820,450,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780053717.37/warc/CC-MAIN-20210916174455-20210916204455-00582.warc.gz | 935,367,344 | 9,211 | Hash function
Hash functions are the core building blocks that provide the basis of bitcoin and other cryptocurrency generation. Each new block in the blockchain begins with a hash of the previous block.
Overview
In mathematics, a function is a process whereby one or more inputs are transformed into an output. A hash function takes an input of indeterminate length and mathematically transforms it into an output of fixed, pre-determined, usually shorter length. The function "hashes" the input. The output of a hash function, commonly referred to as the hash or the digest, is determined by the hashing such that the same input always produces the same hash.[1] The output of a hash function is usually shorter than the input, and one of the benefits of hashing data is often data compression.[2]
The following characteristics make certain hash functions desirable for cryptographic applications:[3]
• Deterministic - For any input, the hash function produces the same digest, or output, every time the hash function is run on the input.
• Pre-image resistant - For any digest it should be computationally infeasible (meaning it would take far too many resources to actually compute) to determine its input. This is called one-wayedness, meaning that it is almost impossible to determine the input string from any given digest.[4]
• Collision resistant - For any digest it should be computationally infeasible to find a distinct other input that when hashed produces the same digest. When two or more distinct input strings produce the same digest when hashed, they are said to collide.
Other, less cryptographically friendly attributes of hash functions might include compression - taking very large inputs or input strings to be indexed into shorter, fixed-length digests to be used as an index - and familiarity so that similar inputs have similar but different outputs so that hashing is faster, though not safer.[5]
Cryptographic Hash Functions
While there are numerous hash functions in common use, the ones used most frequently in encryption are MD5 (Message Digest 5) and SHA256 (Secure Hash Algorithm 256).[6] Satoshi Nakamoto's "bitcoin whitepaper" refers to SHA-256 as usable for the proof-of-work in a bitcoin blockchain.[7]
As mentioned, hash functions produce output of a fixed length. Here are two examples of the use of MD5 hashing:
• md5("hello world") = 5EB63BBBE01EEED093CB22BB8F5ACDC3 and
• md5(42) = A1D0C6E83F027327D8461063F4AC58A6.[8]
These examples illustrate a few important characteristics of a hash function. For one, the input strings are of different lengths, one is for a quotation and the other is a number. Secondly, the digests are different but of identical 32-bit length. Thirdly, the digest uses hexadecimal digits (which are 0 to 9 and a to f). While common, not all cryptographic functions use 32-bit, hexadecimal output and bitcoin transaction hash digests are 64 characters.[9] SHA256 is used for the bitcoin blockshain. Here is how the same inputs as above are hashed using SHA256:
• sha256("hello world") = 9DDEFE4435B21D901439E546D54A14A175A3493B9FD8FBF38D9EA6D3CBF70826 and
• sha256(42) = 73475CB40A568E8DA8A045CED110137E159F890AC4DA883B6B17DC651B3A8049.[10]
Again, the inputs are of different lengths but both result in 64 bit, hexadecimal digests.
Hash Functions in Blockchains
Blockchains are series of blocks of transactions. For purposes of data compression as well as privacy and security, transaction information is hashed before it is included in a block. Each block includes a hash of the last block which leads to the creation of the chain. Because cryptographic hash functions are deterministic and highly collision resistant, including a hash of the last previous block effectively prevents anyone tampering with its contents once the next block has been formed.[11]
References
1. What Are Hash Functions. Learn Cryptography.
2. Chapter 6, Hash Functions. Computer Science Education at UCSD.
3. Cryptographic Hash Functions. Purdue University.
4. Chapter 6, Hash Functions. Computer Science Education at UCSD.
5. Cryptographic Hash Functions. Purdue University.
6. What is Cryptographic Hashing?. Tip Top Security.
7. Bitcoin: A Peer-to-Peer Electronic Cash System. Bitcoin.org. | 955 | 4,254 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2021-39 | longest | en | 0.877694 |
https://www.tutorialgateway.org/c-program-to-print-number-pattern-5/ | 1,686,153,308,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224653930.47/warc/CC-MAIN-20230607143116-20230607173116-00374.warc.gz | 1,132,700,362 | 25,258 | # C Program to Print Number Pattern 5
Write a C program to Print Number Pattern 5 with example. For this, we are going to use For Loop and While Loop.
## C program to Print Number Pattern 5 using For Loop
This program allows the user to enter the maximum number of rows he/she want to print as a right triangle. Next, compiler will print the required numbers pattern.
```/* C program to Print Number Pattern 5 */
#include<stdio.h>
int main()
{
int i, j, rows;
printf(" \nPlease Enter the Number of Rows : ");
scanf("%d", &rows);
for(i = rows; i >= 1; i--)
{
for(j = i; j <= rows; j++)
{
printf("%d", j);
}
printf("\n");
}
return 0;
}```
Let us see the Nested for loop
```for(i = rows; i >= 1; i--)
{
for(j = i; j <= rows; j++)
{
printf("%d", j);
}
printf("\n");
}```
Outer Loop – First Iteration
From the above C Programming screenshot you can observe that, The value of i is 7 and the condition (i <= 7) is True. So, it will enter into second for loop
Inner Loop – First Iteration
The j value is 1 and the condition (1 <= 1) is True. So, it will start executing the statements inside the loop.
`printf("%d", j);`
Next, we used the Increment Operator j++ to increment the J value by 1. This will happen until the condition inside the inner for loop fails. Next, iteration will start from beginning until both the Inner Loop and Outer loop conditions fails.
## Program to Print Number Pattern 5 using while Loop
In this program we just replaced the For Loop with the While Loop. I suggest you to refer While Loop article to understand the logic.
```/* C program to Print Number Pattern 5 */
#include<stdio.h>
int main()
{
int i, j, rows;
printf(" \nPlease Enter the Number of Rows : ");
scanf("%d", &rows);
i = rows;
while(i >= 1)
{
j = i;
while(j <= rows)
{
printf("%d", j);
j++;
}
i--;
printf("\n");
}
return 0;
}```
``````Please Enter the Number of Rows : 9
9
89
789
6789
56789
456789
3456789
23456789
123456789`````` | 532 | 1,943 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2023-23 | longest | en | 0.719863 |
https://community.qlik.com/thread/210991 | 1,532,347,564,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676596336.96/warc/CC-MAIN-20180723110342-20180723130342-00251.warc.gz | 618,088,066 | 24,217 | 5 Replies Latest reply: Apr 7, 2016 10:19 AM by it man
# List top n values in a text Object
Hi Gurus,
I have created a straight table which displays the Dimension values and sorted them as per the Rank calculated in the expression.
Now, I need to display all the top 10 values that I see in the Straight Table in one Text Object .
Using FirstSortedValue function in the Text Object, I could only display the top 1 or bottom 1 value.
-----
I also need to display Top 10, Bottom 10, and wrap all the rest of the Dimension values into 3 more Text Objects.
• ###### Re: List top n values in a text Object
Maybe like this (assuming Sum(Value) is the expression used):
=Concat({<DimensionField = {"=Rank(Sum(Value))<=10"}>} DISTINCT DimensionField,', ', -Sum(Value) )
=Concat({<DimensionField = {"=Rank(-Sum(Value))<=10"}>} DISTINCT DimensionField,', ', -Sum(Value) )
=Concat({<DimensionField -= ( {"=Rank(Sum(Value))<=10"} +{"=Rank(-Sum(Value))<=10"} ) >}
DISTINCT DimensionField,', ', -Sum(Value) )
to list top 10, bottom 10 and all inbetween
• ###### Re: List top n values in a text Object
Hi Swehl,
Thank you for the quick response.
I think i had it from your expression and realized that I may not have clearly stated my requirement.
I apologize.
Here's the error I am receiving in Text Object: Error in Expression. Nested aggregation not allowed
The Rank Expression that I am trying to use in the text object is a value = sum of 3 other ranks generated using 3 aggregated expression.
i.,e
RankA = Rank(sum(X)) * (.3)
+
RankB = Rank(AVG(Y)) * (.4)
+
RankC = Rank(sum(Z)) * (.3)
---
Please let me know if you rather prefer it to be created as a seperate discussion
• ###### Re: List top n values in a text Object
Right, the sort weight needs advanced aggregation, like
=Concat(
{<DimensionField = {"=Rank(Sum(Value))<=10"}>}
DISTINCT
DimensionField,
', ',
Aggr(-Sum(Value),DimensionField)
)
What if you replace Rank(Sum(Value)) in above expression with your sum-of-three-ranks?
How do you get a total rank for your rank expression, BTW, are you using another rank function around the sum-of-three-ranks?
• ###### Re: List top n values in a text Object
Hi,
if you already have the correct values in a straight table then maybe you can make it appear as a text box if that's your preferred visualization.
You could use dimension limits, hide caption, header rows and some columns and switch to horizontal presentation:
using dimension limits:
hiding caption and header row and using horizontal presentation.
hidden dimension
hidden expression
hope this helps
regards
Marco
• ###### Re: List top n values in a text Object
Hi swehl,
Yes, i have to create a rank of the weighted sum of the three ranks ie., Rank(RankA*x% + RankB*y% + RankC*z%).
I was able to make some progress in displaying the top N values in one single 'text object'.
However, my requirement also needed me to display some additional dimension and expression values when I hover over each of the Dimension values in the 'text object'.
This leads me to create one 'text object' for each Rank. Then, i need to somehow use the 'Help Text' under 'Caption tab' of the 'text object' to display the required additional dimension and expression values relevant to the Dimension.
So, i gave up on this approach and started looking at creating one 'Straight Table' for each of the Dimension values that would look like a 'text object' as Marco also suggested.
---
Hi Marco,
I started using this approach. Since I have to display more details when I hover over each of the Dimension values, I ended up creating about 300 straight tables with Conditional show/hide of each value ie., one for each Dimension value as I need to display all the Dimension values by Default. All these 300 straight tables are arranged in the required order of Top N , Everything in Between, Bottom N.
This has affected the Dashboard performance. The dashboard is not loading data unless i pass very little data.
So, even though I have visually what the user is looking to see, it just not effectively reloading data as needed.
--
I appreciate any advice and thank you for the help so far | 977 | 4,184 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2018-30 | latest | en | 0.750122 |
http://www.kylesconverter.com/flow/thousand-barrels-per-second-to-thousand-barrels-per-day | 1,637,991,028,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358118.13/warc/CC-MAIN-20211127043716-20211127073716-00052.warc.gz | 126,792,974 | 5,588 | # Convert Thousand Barrels Per Second to Thousand Barrels Per Day
### Kyle's Converter > Flow > Thousand Barrels Per Second > Thousand Barrels Per Second to Thousand Barrels Per Day
Thousand Barrels Per Second (Mbbl/s) Thousand Barrels Per Day (Mbbl/d) Precision: 0 1 2 3 4 5 6 7 8 9 12 15 18
Reverse conversion?
Thousand Barrels Per Day to Thousand Barrels Per Second
(or just enter a value in the "to" field)
Please share if you found this tool useful:
Unit Descriptions
1 Thousand Barrels per Second:
Production rate of 1000 barrels of petroleum per second. A barrel of petroleum being 42 gallons (U.S. fluid measure). Latin "M" meaning thousand. Approximately 158.987294928 cubic meters per second (SI unit).
1 Thousand Barrels per Day:
Production rate of 1000 barrels of petroleum per day. A barrel of petroleum being 42 gallons (U.S. fluid measure). Latin "M" meaning thousand. Approximately 0.00184013072833 cubic meters per second (SI units).
Conversions Table
1 Thousand Barrels Per Second to Thousand Barrels Per Day = 8640070 Thousand Barrels Per Second to Thousand Barrels Per Day = 6048000
2 Thousand Barrels Per Second to Thousand Barrels Per Day = 17280080 Thousand Barrels Per Second to Thousand Barrels Per Day = 6912000
3 Thousand Barrels Per Second to Thousand Barrels Per Day = 25920090 Thousand Barrels Per Second to Thousand Barrels Per Day = 7776000
4 Thousand Barrels Per Second to Thousand Barrels Per Day = 345600100 Thousand Barrels Per Second to Thousand Barrels Per Day = 8640000
5 Thousand Barrels Per Second to Thousand Barrels Per Day = 432000200 Thousand Barrels Per Second to Thousand Barrels Per Day = 17280000
6 Thousand Barrels Per Second to Thousand Barrels Per Day = 518400300 Thousand Barrels Per Second to Thousand Barrels Per Day = 25920000
7 Thousand Barrels Per Second to Thousand Barrels Per Day = 604800400 Thousand Barrels Per Second to Thousand Barrels Per Day = 34560000
8 Thousand Barrels Per Second to Thousand Barrels Per Day = 691200500 Thousand Barrels Per Second to Thousand Barrels Per Day = 43200000
9 Thousand Barrels Per Second to Thousand Barrels Per Day = 777600600 Thousand Barrels Per Second to Thousand Barrels Per Day = 51840000
10 Thousand Barrels Per Second to Thousand Barrels Per Day = 864000800 Thousand Barrels Per Second to Thousand Barrels Per Day = 69120000
20 Thousand Barrels Per Second to Thousand Barrels Per Day = 1728000900 Thousand Barrels Per Second to Thousand Barrels Per Day = 77760000
30 Thousand Barrels Per Second to Thousand Barrels Per Day = 25920001,000 Thousand Barrels Per Second to Thousand Barrels Per Day = 86400000
40 Thousand Barrels Per Second to Thousand Barrels Per Day = 345600010,000 Thousand Barrels Per Second to Thousand Barrels Per Day = 864000000
50 Thousand Barrels Per Second to Thousand Barrels Per Day = 4320000100,000 Thousand Barrels Per Second to Thousand Barrels Per Day = 8640000000
60 Thousand Barrels Per Second to Thousand Barrels Per Day = 51840001,000,000 Thousand Barrels Per Second to Thousand Barrels Per Day = 86400000000 | 771 | 3,053 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2021-49 | latest | en | 0.649904 |
https://www.calculatorbit.com/en/length/10-picometer-to-petameter | 1,723,034,425,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640694449.36/warc/CC-MAIN-20240807111957-20240807141957-00377.warc.gz | 548,390,870 | 7,767 | # 10 Picometer to Petameter Calculator
Result:
10 Picometer = 1e-26 Petameter (Pm)
Rounded: ( Nearest 4 digits)
10 Picometer is 1e-26 Petameter (Pm)
10 Picometer is 1e-8mm
## How to Convert Picometer to Petameter (Explanation)
• 1 picometer = 1e-27 Pm (Nearest 4 digits)
• 1 petameter = 1.0000000000000002e+27 pm (Nearest 4 digits)
There are 1e-27 Petameter in 1 Picometer. To convert Picometer to Petameter all you need to do is multiple the Picometer with 1e-27.
In formula distance is denoted with d
The distance d in Petameter (Pm) is equal to 1e-27 times the distance in picometer (pm):
### Equation
d (Pm) = d (pm) × 1e-27
Formula for 10 Picometer (pm) to Petameter (Pm) conversion:
d (Pm) = 10 pm × 1e-27 => 1e-26 Pm
## How many Petameter in a Picometer
One Picometer is equal to 1e-27 Petameter
1 pm = 1 pm × 1e-27 => 1e-27 Pm
## How many Picometer in a Petameter
One Petameter is equal to 1.0000000000000002e+27 Picometer
1 Pm = 1 Pm / 1e-27 => 1.0000000000000002e+27 pm
## picometer:
The picometer (symbol: pm) is unit of length in the International System of Units (SI), equal to 0.000000000001 meter or 1x10^-12 meter or 1/1000000000000 meter. The picometer is equal to 1000 femtometres or 1/1000 nanometer. The picometer's length is so small that its application is almost entirely confined to the particle physics, quantum physics, acoustics and chemistry.
Read more on Wikipedia picometer
## petameter:
The Petameter (symbol: Pm) is a unit of length in the metric system equal to 1000000000000000 meters or you can say 10^15 meters. 1 Petameter is equal to 1 trillion kilometer or 0.1057 light-years. 1 light-year is total distance traveled by light in one year.
Read more on Wikipedia petameter
If you found information page helpful you can cite and reference this page in your work.
## Picometer to Petameter Calculations Table
Now by following above explained formulas we can prepare a Picometer to Petameter Chart.
Picometer (pm) Petameter (Pm)
6 6e-27
7 7e-27
8 8e-27
9 9e-27
10 1e-26
11 1.1e-26
12 1.2e-26
13 1.3e-26
14 1.4e-26
15 1.5e-26
Nearest 4 digits
## Convert from Picometer to other units
Here are some quick links to convert 10 Picometer to other length units.
## Convert to Picometer from other units
Here are some quick links to convert other length units to Picometer.
## FAQs About Picometer and Petameter
Converting from one Picometer to Petameter or Petameter to Picometer sometimes gets confusing.
Here are some Frequently asked questions answered for you.
### Is 1e-27 Petameter in 1 Picometer?
Yes, 1 Picometer have 1e-27 (Nearest 4 digits) Petameter.
### What is the symbol for Picometer and Petameter?
Symbol for Picometer is pm and symbol for Petameter is Pm.
### How many Picometer makes 1 Petameter?
1.0000000000000002e+27 Picometer is euqal to 1 Petameter.
### How many Petameter in 10 Picometer?
Picometer have 1e-26 Petameter.
### How many Petameter in a Picometer?
Picometer have 1e-27 (Nearest 4 digits) Petameter. | 968 | 3,011 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2024-33 | latest | en | 0.71813 |
https://www.mathworks.com/matlabcentral/cody/problems/1184-hangman-strategy/solutions/189906 | 1,506,086,764,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818688966.39/warc/CC-MAIN-20170922130934-20170922150934-00229.warc.gz | 849,374,695 | 12,789 | Cody
Problem 1184. Hangman (strategy)
Solution 189906
Submitted on 11 Jan 2013
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
%% assignin('caller','score',300); a.words0={'BUZZ','COZY','DOZE','FUZZ','GAZE','HAZE','JAZZ','LAZY','SIZE','ZERO','ZONE'}; a.e=0; for n=randperm(numel(a.words0)) a.i=n; a.words=a.words0; a.n=n; for m=1:100, a.m=m; assignin('caller','a',a); letter=hangman(a.words); letter=char(letter(1)); a=evalin('caller','a'); m=a.m; fprintf('Target word %s; Step %d; Word list %s; Guess %c\n',a.words{a.i},a.m,sprintf('%s ',a.words{:}),letter); matchedletters=a.words{a.i}==letter; if ~any(matchedletters), a.e=a.e+1; end matchedwords=find(cellfun(@(x)isequal(matchedletters,x==letter),a.words)); a.i=find(matchedwords==a.i); a.words=regexprep(a.words(matchedwords),letter,''); nonemptywords=find(cellfun('length',a.words)>0); a.i=find(nonemptywords==a.i); if isempty(a.i), break; end end if ~isempty(a.i), error(sprintf('algorithm did not guess word after 100 steps. Last message: Target word %s; Step %d; Word list %s; Guessed letter %c\n',a.words{a.i},a.m,sprintf('%s ',a.words{:}),letter)); end n=a.n; end a.e=a.e/numel(a.words0); fprintf('Average number of errors per word %f\n',a.e); assert(a.e<5); assignin('caller','score',evalin('caller','score')-100+round(a.e/5*100));
```Target word GAZE; Step 1; Word list BUZZ COZY DOZE FUZZ GAZE HAZE JAZZ LAZY SIZE ZERO ZONE ; Guess I Target word GAZE; Step 2; Word list BUZZ COZY DOZE FUZZ GAZE HAZE JAZZ LAZY ZERO ZONE ; Guess Z Target word GAE; Step 3; Word list COY DOE GAE HAE LAY ; Guess Y Target word GAE; Step 4; Word list DOE GAE HAE ; Guess H Target word GAE; Step 5; Word list DOE GAE ; Guess A Target word GE; Step 6; Word list GE ; Guess E Target word G; Step 7; Word list G ; Guess G Target word COZY; Step 1; Word list BUZZ COZY DOZE FUZZ GAZE HAZE JAZZ LAZY SIZE ZERO ZONE ; Guess Z Target word COY; Step 2; Word list COY DOE GAE HAE LAY SIE ; Guess L Target word COY; Step 3; Word list COY DOE GAE HAE SIE ; Guess Y Target word CO; Step 4; Word list CO ; Guess O Target word C; Step 5; Word list C ; Guess C Target word HAZE; Step 1; Word list BUZZ COZY DOZE FUZZ GAZE HAZE JAZZ LAZY SIZE ZERO ZONE ; Guess Z Target word HAE; Step 2; Word list COY DOE GAE HAE LAY SIE ; Guess L Target word HAE; Step 3; Word list COY DOE GAE HAE SIE ; Guess E Target word HA; Step 4; Word list DO GA HA SI ; Guess G Target word HA; Step 5; Word list DO HA SI ; Guess D Target word HA; Step 6; Word list HA SI ; Guess A Target word H; Step 7; Word list H ; Guess H Target word FUZZ; Step 1; Word list BUZZ COZY DOZE FUZZ GAZE HAZE JAZZ LAZY SIZE ZERO ZONE ; Guess Z Target word FU; Step 2; Word list BU FU JA ; Guess U Target word F; Step 3; Word list B F ; Guess F Target word DOZE; Step 1; Word list BUZZ COZY DOZE FUZZ GAZE HAZE JAZZ LAZY SIZE ZERO ZONE ; Guess B Target word DOZE; Step 2; Word list COZY DOZE FUZZ GAZE HAZE JAZZ LAZY SIZE ZERO ZONE ; Guess A Target word DOZE; Step 3; Word list COZY DOZE FUZZ SIZE ZERO ZONE ; Guess Z Target word DOE; Step 4; Word list COY DOE SIE ; Guess E Target word DO; Step 5; Word list DO SI ; Guess O Target word D; Step 6; Word list D ; Guess D Target word ZONE; Step 1; Word list BUZZ COZY DOZE FUZZ GAZE HAZE JAZZ LAZY SIZE ZERO ZONE ; Guess E Target word ZON; Step 2; Word list DOZ GAZ HAZ SIZ ZON ; Guess Z Target word ON; Step 3; Word list ON ; Guess O Target word N; Step 4; Word list N ; Guess N Target word LAZY; Step 1; Word list BUZZ COZY DOZE FUZZ GAZE HAZE JAZZ LAZY SIZE ZERO ZONE ; Guess B Target word LAZY; Step 2; Word list COZY DOZE FUZZ GAZE HAZE JAZZ LAZY SIZE ZERO ZONE ; Guess Z Target word LAY; Step 3; Word list COY DOE GAE HAE LAY SIE ; Guess Y Target word LA; Step 4; Word list CO LA ; Guess A Target word L; Step 5; Word list L ; Guess L Target word ZERO; Step 1; Word list BUZZ COZY DOZE FUZZ GAZE HAZE JAZZ LAZY SIZE ZERO ZONE ; Guess Y Target word ZERO; Step 2; Word list BUZZ DOZE FUZZ GAZE HAZE JAZZ SIZE ZERO ZONE ; Guess Z Target word ERO; Step 3; Word list ERO ONE ; Guess O Target word ER; Step 4; Word list ER ; Guess R Target word E; Step 5; Word list E ; Guess E Target word BUZZ; Step 1; Word list BUZZ COZY DOZE FUZZ GAZE HAZE JAZZ LAZY SIZE ZERO ZONE ; Guess H Target word BUZZ; Step 2; Word list BUZZ COZY DOZE FUZZ GAZE JAZZ LAZY SIZE ZERO ZONE ; Guess O Target word BUZZ; Step 3; Word list BUZZ FUZZ GAZE JAZZ LAZY SIZE ; Guess G Target word BUZZ; Step 4; Word list BUZZ FUZZ JAZZ LAZY SIZE ; Guess Z Target word BU; Step 5; Word list BU FU JA ; Guess U Target word B; Step 6; Word list B F ; Guess F Target word B; Step 7; Word list B ; Guess B Target word JAZZ; Step 1; Word list BUZZ COZY DOZE FUZZ GAZE HAZE JAZZ LAZY SIZE ZERO ZONE ; Guess R Target word JAZZ; Step 2; Word list BUZZ COZY DOZE FUZZ GAZE HAZE JAZZ LAZY SIZE ZONE ; Guess A Target word JZZ; Step 3; Word list GZE HZE JZZ LZY ; Guess L Target word JZZ; Step 4; Word list GZE HZE JZZ ; Guess Z Target word J; Step 5; Word list J ; Guess J Target word SIZE; Step 1; Word list BUZZ COZY DOZE FUZZ GAZE HAZE JAZZ LAZY SIZE ZERO ZONE ; Guess Z Target word SIE; Step 2; Word list COY DOE GAE HAE LAY SIE ; Guess E Target word SI; Step 3; Word list DO GA HA SI ; Guess G Target word SI; Step 4; Word list DO HA SI ; Guess S Target word I; Step 5; Word list I ; Guess I Average number of errors per word 1.636364 ```
2 Fail
%% rng default; a.words0=cellstr(unique(char('A'+ceil(26*rand([200,3]).^2)-1),'rows'))'; a.e=0; for n=randperm(numel(a.words0)) a.i=n; a.words=a.words0; a.n=n; for m=1:100, a.m=m; assignin('caller','a',a); letter=hangman(a.words); letter=char(letter(1)); a=evalin('caller','a'); m=a.m; matchedletters=a.words{a.i}==letter; if ~any(matchedletters), a.e=a.e+1; end matchedwords=find(cellfun(@(x)isequal(matchedletters,x==letter),a.words)); a.i=find(matchedwords==a.i); a.words=regexprep(a.words(matchedwords),letter,''); nonemptywords=find(cellfun('length',a.words)>0); a.i=find(nonemptywords==a.i); if isempty(a.i), break; end end if ~isempty(a.i), error(sprintf('algorithm did not guess word after 100 steps. Last message: Target word %s; Step %d; Word list %s; Guessed letter %c\n',a.words{a.i},a.m,sprintf('%s ',a.words{:}),letter)); end n=a.n; end a.e=a.e/numel(a.words0); fprintf('Average number of errors per word %f\n',a.e); assert(a.e<5); assignin('caller','score',evalin('caller','score')-100+round(a.e/5*100));
```Error: Assertion failed. ```
3 Pass
%% rng default; a.words0=cellstr(unique(char('A'+ceil(26*rand([200,4]).^2)-1),'rows'))'; a.e=0; for n=randperm(numel(a.words0)) a.i=n; a.words=a.words0; a.n=n; for m=1:100, a.m=m; assignin('caller','a',a); letter=hangman(a.words); letter=char(letter(1)); a=evalin('caller','a'); m=a.m; matchedletters=a.words{a.i}==letter; if ~any(matchedletters), a.e=a.e+1; end matchedwords=find(cellfun(@(x)isequal(matchedletters,x==letter),a.words)); a.i=find(matchedwords==a.i); a.words=regexprep(a.words(matchedwords),letter,''); nonemptywords=find(cellfun('length',a.words)>0); a.i=find(nonemptywords==a.i); if isempty(a.i), break; end end if ~isempty(a.i), error(sprintf('algorithm did not guess word after 100 steps. Last message: Target word %s; Step %d; Word list %s; Guessed letter %c\n',a.words{a.i},a.m,sprintf('%s ',a.words{:}),letter)); end n=a.n; end a.e=a.e/numel(a.words0); fprintf('Average number of errors per word %f\n',a.e); assert(a.e<5); assignin('caller','score',evalin('caller','score')-100+round(a.e/5*100));
```Average number of errors per word 4.835000 ``` | 2,503 | 7,592 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2017-39 | latest | en | 0.322882 |
http://oeis.org/A238895 | 1,503,208,493,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886105976.13/warc/CC-MAIN-20170820053541-20170820073541-00390.warc.gz | 304,419,414 | 4,408 | This site is supported by donations to The OEIS Foundation.
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A238895 Numbers n > 1 such that a record number of numbers k have n as the sum of the proper divisors of k. 7
2, 3, 6, 21, 31, 49, 73, 91, 115, 121, 169, 211, 301, 331, 361, 391, 421, 511, 631, 721, 781, 841, 1051, 1261, 1471, 1561, 1681, 1891, 2101, 2311, 2521, 2731, 3151, 3361, 3571, 3991, 4201, 4411, 4621, 5251, 5461, 6091, 6511, 6721, 6931, 7771, 7981, 8191, 9031 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 COMMENTS The number of times that a(n) appears in A001065 is A238896(n). By analogy with the untouchable numbers (A005114) and the highly composite numbers (A002182), these numbers can be named "highly touchable" (see Lignon). - Daniel Lignon, Mar 21 2014 Indices of record values in A048138. - Franklin T. Adams-Watters, Jul 27 2014 REFERENCES D. Lignon, Dictionnaire de (presque) tous les nombres entiers, Editions Ellipses, 2012, see p. 317 (in French). LINKS Daniel Mondot, Table of n, a(n) for n = 1..139 EXAMPLE For 2, there are no numbers. For 3, there is 1 number: 4. For 6, there are 2 numbers: 6 and 25. For 21, there are 3 numbers: 18, 51, 91. For 31, there are 5 numbers: 32, 125, 161, 209, 221. For 49, there are 6 numbers: 75, 215, 287, 407, 527, 551. MATHEMATICA nn = 1000; s = Table[0, {nn}]; Do[k = DivisorSigma[1, n] - n; If[0 < k <= nn, s[[k]]++], {n, nn^2}]; t = {}; mx = -1; Do[If[s[[n]] > mx, mx = s[[n]]; AppendTo[t, {n, mx}]], {n, 2, nn}]; Transpose[t][[1]] CROSSREFS Cf. A001065, A005114, A048138, A238896. Cf. A152454 (row n lists the numbers whose proper divisors sum to n). Cf. A239625 (irregular table giving the rows of numbers that produce a(n)). Sequence in context: A002078 A000372 A123930 * A125601 A025239 A127294 Adjacent sequences: A238892 A238893 A238894 * A238896 A238897 A238898 KEYWORD nonn AUTHOR T. D. Noe, Mar 10 2014 STATUS approved
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The OEIS Community | Maintained by The OEIS Foundation Inc. | 847 | 2,331 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2017-34 | longest | en | 0.640534 |
https://rdrr.io/cran/metRology/man/algS.html | 1,709,469,986,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476374.40/warc/CC-MAIN-20240303111005-20240303141005-00782.warc.gz | 464,484,760 | 9,249 | algS: 'Algorithm S' - robust estimate of pooled standard deviation In metRology: Support for Metrological Applications
Description
‘Algorithm S’ calculates a robust estimate of pooled standard deviation from a set of standard deviations
Usage
```1 2 3``` ``` algS(s, degfree, na.rm = FALSE, prob.eta = 0.9, is.range = FALSE, tol = .Machine\$double.eps^0.25, maxiter = 25, verbose = FALSE) ```
Arguments
`s` A vector of standard deviations or, if `is.range` is `TRUE`, ranges. `degfree` Scalar number of degrees of freedom associated with all values in `s`. If a vector is supplied, `median(degfree)` will be used. `na.rm` a logical value indicating whether 'NA' values should be stripped before the computation proceeds. `prob.eta` prob.eta is set to specify the lower tail area of the chi-squared distribution used as a cut-off. `is.range` if is.range is TRUE, s is interpreted as a vector of positive differences of duplcate observations and degfree is set to 1 `tol` Convergence tolerance Iteration continues until the relative change in estimated pooled sd drops below `tol`. `maxiter` Maximum number of iterations permitted. `verbose` Controls information displayed during iteration; see Details.
Details
Algorithm S is suggested by ISO 5725-5:1998 as a robust estimator of pooled standard deviation s.pool from standard deviations of groups of size degfree.
The algorithm calculates a ‘limit factor’, eta, set to `qchisq(prob.eta, degfree)`. Following an initial estimate of s.pool as `median(s)`, the standard deviations s{i} are replaced with w[i]=min(eta*s.pool, s[i]) and an updated value for s.pool calculated as
xi*sqrt(sum(w)^2)/p
where p is the number of standard deviations and ξ is calculated as
xi = 1/sqrt(pchisq(degfree*eta^2, degfree + 2) + (1-prob.eta)*eta^2)
If the s[i] are ranges of two values, ISO 5725 recommends carrying out the above iteration on the ranges and then dividing by sqrt{degfree+1}; in the implementation here, this is done prior to returning the estimate.
If `verbose` is non-zero, the current iteration number and estimate are printed; if `verbose>1`, the current set of truncated values w is also printed.
Value
A scalar estimate of poooled standard deviation.
Author(s)
S L R Ellison s.ellison@lgc.co.uk
References
ISO 5725-5:1998 Accuracy (trueness and precision) of measurement methods and results - Part 5: Alternative methods for the determination of the precision of a standard measurement method
`algA`
``` 1 2 3 4 5 6 7 8 9 10``` ```#example from ISO 5725-5:1998 (cell ranges for percent creosote) cdiff <- c(0.28, 0.49, 0.40, 0.00, 0.35, 1.98, 0.80, 0.32, 0.95) algS(cdiff, is.range=TRUE) #Compare with the sd of the two values (based on the range) c.sd <- cdiff/sqrt(2) algS(c.sd, degfree=1, verbose=TRUE) ``` | 754 | 2,787 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2024-10 | latest | en | 0.76388 |
https://math.stackexchange.com/questions/1955034/prove-that-if-ab-divides-a2-then-ab-divides-b2 | 1,725,991,513,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651303.70/warc/CC-MAIN-20240910161250-20240910191250-00498.warc.gz | 367,402,652 | 36,432 | # Prove that if $(a+b)$ divides $a^2$ then $(a+b)$ divides $b^2$
I'm trying to solve the following exercise: "Prove that if $(a+b)$ divides $a^2$ then $(a+b)$ divides $b^2$". It's quite obvious how to prove divisibility for a product, but how to do it for a sum?
If $a+b$ divides $a^2$, then you can write $a^2=(a+b)k$.
But then $$b^2= a^2-a^2+b^2 = a^2 -(a^2-b^2)$$ $$=\underbrace{(a+b)k}_{a^2} - (a+b)(a-b)$$ $$=(a+b)(k-(a-b))$$ $$=(a+b)(k-a+b)$$ so $(a+b)$ divides $b^2$.
$$a+b|(a+b)(a-b)=a^2-b^2$$ | 213 | 503 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2024-38 | latest | en | 0.430512 |
https://math.stackexchange.com/questions/4526430/extension-of-a-combinatorics-mo-training-problem-to-2-dimensions | 1,725,930,231,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651164.37/warc/CC-MAIN-20240909233606-20240910023606-00120.warc.gz | 363,206,298 | 35,260 | Extension of a combinatorics MO training problem to 2 dimensions
The original problem:
You have $$2019$$ $$+1$$s and $$-1$$s, find the number of ways to arrange them, such that all subsequences of even length have a sum between $$-2$$ and $$2$$.
The answer is $$2 f_{2022}\times 2^{1011}$$, where $$f$$ is the fibonacci sequence.
I'm looking into extending the problem into 2D matrices. That is,
Given a $$n\times m$$ matrix of $$1$$ and $$-1$$, find the number of ways to arrange them, such that all submatrices (that is $$a_{i,j},l_1\le i \le r_1, l_2 \le j \le r_2$$. have a sum between $$-2$$ and $$2$$
I'm still trying to code up a solution to find the number. The growth is pretty quick though.
I also came up with a problem similar in description but (I believe) not similar at all:
Given a $$n\times m$$ matrix of $$1$$ and $$-1$$, find the largest possible submatrix that have a sum between $$-2$$ and $$2$$ among all arrangements. | 275 | 948 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 18, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2024-38 | latest | en | 0.928577 |
http://de.metamath.org/mpegif/bnj602.html | 1,603,532,559,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107882103.34/warc/CC-MAIN-20201024080855-20201024110855-00530.warc.gz | 26,181,742 | 4,119 | Mathbox for Jonathan Ben-Naim < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > Mathboxes > bnj602 Structured version Unicode version
Theorem bnj602 29722
Description: Equality theorem for the function constant. (Contributed by Jonathan Ben-Naim, 3-Jun-2011.) (New usage is discouraged.)
Assertion
Ref Expression
bnj602
Proof of Theorem bnj602
Dummy variable is distinct from all other variables.
StepHypRef Expression
1 breq2 4424 . . 3
21rabbidv 3072 . 2
3 df-bnj14 29490 . 2
4 df-bnj14 29490 . 2
52, 3, 43eqtr4g 2488 1
Colors of variables: wff setvar class Syntax hints: wi 4 wceq 1437 crab 2779 class class class wbr 4420 c-bnj14 29489 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1665 ax-4 1678 ax-5 1748 ax-6 1794 ax-7 1839 ax-10 1887 ax-11 1892 ax-12 1905 ax-13 2053 ax-ext 2400 This theorem depends on definitions: df-bi 188 df-or 371 df-an 372 df-3an 984 df-tru 1440 df-ex 1660 df-nf 1664 df-sb 1787 df-clab 2408 df-cleq 2414 df-clel 2417 df-nfc 2572 df-ral 2780 df-rab 2784 df-v 3083 df-dif 3439 df-un 3441 df-in 3443 df-ss 3450 df-nul 3762 df-if 3910 df-sn 3997 df-pr 3999 df-op 4003 df-br 4421 df-bnj14 29490 This theorem is referenced by: bnj601 29727 bnj852 29728 bnj18eq1 29734 bnj938 29744 bnj1125 29797 bnj1148 29801 bnj1318 29830 bnj1442 29854 bnj1450 29855 bnj1452 29857 bnj1463 29860 bnj1529 29875
Copyright terms: Public domain W3C validator | 693 | 1,506 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2020-45 | latest | en | 0.217149 |
http://lists.w3.org/Archives/Public/www-webont-wg/2003Sep/0327.html | 1,417,215,706,000,000,000 | text/html | crawl-data/CC-MAIN-2014-49/segments/1416931011032.12/warc/CC-MAIN-20141125155651-00119-ip-10-235-23-156.ec2.internal.warc.gz | 173,579,345 | 3,331 | Re: loop free?
From: Peter F. Patel-Schneider <pfps@research.bell-labs.com>
Date: Tue, 30 Sep 2003 16:18:20 -0400 (EDT)
Message-Id: <20030930.161820.74184014.pfps@research.bell-labs.com>
I don't particularly care one way of the other here, but I don't see any
need to forbid loops (of any sort). As well, the wording that you propose
could be read as forbidding non-trivial loops.
peter
From: "Jeremy Carroll" <jjc@hplb.hpl.hp.com>
Subject: loop free?
Date: Tue, 30 Sep 2003 15:19:57 +0200
> Concerning
> http://www.w3.org/TR/2003/CR-owl-semantics-20030818/
>
>
> Peter,
>
> an editorial suggestion on the mapping rules:
>
> [[
> T(descriptioni) owl:equivalentClass T(descriptionj) .
> for all <i,j> in G where G is a set of pairs over {1,...,n}x{1,...,n}
> that if interpreted as an undirected graph forms a connected graph for
> {1,...,n}
> ]]
>
> suggest
>
> s/a connected graph/a loop-free connected graph/
>
> I think the case <i, i> in G is already excluded by the word "pairs" but it
> is arguable. For most readers undirected graphs are loop free by definition;
> but since we do not provide one ...
>
>
>
>
>
> (I should add a test case for
>
> _:b owl:equivalentClass _:b .
> _:b rdf:type owl:Class .
> _:b owl:unionOf rdf:nil .
>
> as being in OWL Full, similarly
>
> _:b owl:disjointWith _:b .
> _:b rdf:type owl:Class .
> _:b owl:unionOf rdf:nil .
> )
>
> My code, which now passes all the tests except I5.8-016, would fail those
> two I think :(
>
> Jeremy
>
>
Received on Tuesday, 30 September 2003 16:19:04 GMT
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https://www.answers.com/Q/What_website_maker_is_best_for_kids_6-12 | 1,696,082,839,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510676.40/warc/CC-MAIN-20230930113949-20230930143949-00216.warc.gz | 701,633,536 | 42,757 | 0
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### What is 34X18?
34 × 18 = 612
### What is 441 percent of 612?
441% of 612 = 441% x 612 = 4.41 x 612 = 2698.92
### What is the multiple of 612 and 9?
612/9 = 68 Therefore, 68 x 9 = 612, making it a multiple of 612.
### What is 612 percent of 40?
612% of 40= 612% * 40= 6.12 * 40= 244.8
### What is 612 by 19?
612 multiplied by 19 is 11628 612 divided by 19 is 32.2105263
612
### What is 612 percent as a decimal?
To convert 612% to a decimal divide by 100:612% ÷ 100 = 6.12
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### I need to download the optional application for federal employment form?
This must be it OF-612 application form for Government Job Application.The federal job application OF-612 form and federal resume is available at the FEDERALJOBS net website
612 * 8 = 4896
612 | 359 | 1,104 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2023-40 | latest | en | 0.856221 |
https://www.mathlearnit.com/fraction-as-decimal/what-is-201-225-as-a-decimal | 1,723,193,105,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640762343.50/warc/CC-MAIN-20240809075530-20240809105530-00628.warc.gz | 701,355,438 | 7,375 | # What is 201/225 as a decimal?
## Solution and how to convert 201 / 225 into a decimal
201 / 225 = 0.893
Fraction conversions explained:
• 201 divided by 225
• Numerator: 201
• Denominator: 225
• Decimal: 0.893
• Percentage: 0.893%
Converting 201/225 to 0.893 starts with defining whether or not the number should be represented by a fraction, decimal, or even a percentage. Both represent numbers between integers, in some cases defining portions of whole numbers In certain scenarios, fractions would make more sense. Ex: baking, meal prep, time discussion, etc. While decimals bring clarity to others including test grades, sale prices, and contract numbers. Now, let's solve for how we convert 201/225 into a decimal.
201 / 225 as a percentage 201 / 225 as a fraction 201 / 225 as a decimal
0.893% - Convert percentages 201 / 225 201 / 225 = 0.893
## 201/225 is 201 divided by 225
The first step in converting fractions is understanding the equation. A quick trick to convert fractions mentally is recognizing that the equation is already set for us. All we have to do is think back to the classroom and leverage long division. The two parts of fractions are numerators and denominators. The numerator is the top number and the denominator is the bottom. And the line between is our division property. We use this as our equation: numerator(201) / denominator (225) to determine how many whole numbers we have. Then we will continue this process until the number is fully represented as a decimal. This is how we look at our fraction as an equation:
### Numerator: 201
• Numerators are the top number of the fraction which represent the parts of the equation. Any value greater than fifty will be more difficult to covert to a decimal. 201 is an odd number so it might be harder to convert without a calculator. Large two-digit conversions are tough. Especially without a calculator. Let's take a look at the denominator of our fraction.
### Denominator: 225
• Denominators are located at the bottom of the fraction, representing the total number of parts. Larger values over fifty like 225 makes conversion to decimals tougher. But 225 is an odd number. Having an odd denominator like 225 could sometimes be more difficult. Ultimately, don't be afraid of double-digit denominators. Now let's dive into how we convert into decimal format.
## How to convert 201/225 to 0.893
### Step 1: Set your long division bracket: denominator / numerator
$$\require{enclose} 225 \enclose{longdiv}{ 201 }$$
To solve, we will use left-to-right long division. This method allows us to solve for pieces of the equation rather than trying to do it all at once.
### Step 2: Extend your division problem
$$\require{enclose} 00. \\ 225 \enclose{longdiv}{ 201.0 }$$
We've hit our first challenge. 201 cannot be divided into 225! Place a decimal point in your answer and add a zero. This doesn't add any issues to our denominator but now we can divide 225 into 2010.
### Step 3: Solve for how many whole groups you can divide 225 into 2010
$$\require{enclose} 00.8 \\ 225 \enclose{longdiv}{ 201.0 }$$
Now that we've extended the equation, we can divide 225 into 2010 and return our first potential solution! Multiple this number by our furthest left number, 225, (remember, left-to-right long division) to get our first number to our conversion.
### Step 4: Subtract the remainder
$$\require{enclose} 00.8 \\ 225 \enclose{longdiv}{ 201.0 } \\ \underline{ 1800 \phantom{00} } \\ 210 \phantom{0}$$
If you hit a remainder of zero, the equation is done and you have your decimal conversion. If you still have a remainder, continue to the next step.
### Step 5: Repeat step 4 until you have no remainder or reach a decimal point you feel comfortable stopping. Then round to the nearest digit.
Sometimes you won't reach a remainder of zero. Rounding to the nearest digit is perfectly acceptable.
### Why should you convert between fractions, decimals, and percentages?
Converting between fractions and decimals is a necessity. They each bring clarity to numbers and values of every day life. And the same is true for percentages. It’s common for students to hate learning about decimals and fractions because it is tedious. But each represent values in everyday life! Without them, we’re stuck rounding and guessing. Here are real life examples:
### When you should convert 201/225 into a decimal
Speed - Let's say you're playing baseball and a Major League scout picks up a radar gun to see how fast you throw. Your MPH will not be 90 and 201/225 MPH. The radar will read: 90.89 MPH. This simplifies the value.
### When to convert 0.893 to 201/225 as a fraction
Cooking: When scrolling through pintress to find the perfect chocolate cookie recipe. The chef will not tell you to use .86 cups of chocolate chips. That brings confusion to the standard cooking measurement. It’s much clearer to say 42/50 cups of chocolate chips. And to take it even further, no one would use 42/50 cups. You’d see a more common fraction like ¾ or ?, usually in split by quarters or halves.
### Practice Decimal Conversion with your Classroom
• If 201/225 = 0.893 what would it be as a percentage?
• What is 1 + 201/225 in decimal form?
• What is 1 - 201/225 in decimal form?
• If we switched the numerator and denominator, what would be our new fraction?
• What is 0.893 + 1/2?
### Convert more fractions to decimals
From 201 Numerator From 225 Denominator What is 201/215 as a decimal? What is 191/225 as a decimal? What is 201/216 as a decimal? What is 192/225 as a decimal? What is 201/217 as a decimal? What is 193/225 as a decimal? What is 201/218 as a decimal? What is 194/225 as a decimal? What is 201/219 as a decimal? What is 195/225 as a decimal? What is 201/220 as a decimal? What is 196/225 as a decimal? What is 201/221 as a decimal? What is 197/225 as a decimal? What is 201/222 as a decimal? What is 198/225 as a decimal? What is 201/223 as a decimal? What is 199/225 as a decimal? What is 201/224 as a decimal? What is 200/225 as a decimal? What is 201/225 as a decimal? What is 201/225 as a decimal? What is 201/226 as a decimal? What is 202/225 as a decimal? What is 201/227 as a decimal? What is 203/225 as a decimal? What is 201/228 as a decimal? What is 204/225 as a decimal? What is 201/229 as a decimal? What is 205/225 as a decimal? What is 201/230 as a decimal? What is 206/225 as a decimal? What is 201/231 as a decimal? What is 207/225 as a decimal? What is 201/232 as a decimal? What is 208/225 as a decimal? What is 201/233 as a decimal? What is 209/225 as a decimal? What is 201/234 as a decimal? What is 210/225 as a decimal? What is 201/235 as a decimal? What is 211/225 as a decimal?
### Convert similar fractions to percentages
From 201 Numerator From 225 Denominator 202/225 as a percentage 201/226 as a percentage 203/225 as a percentage 201/227 as a percentage 204/225 as a percentage 201/228 as a percentage 205/225 as a percentage 201/229 as a percentage 206/225 as a percentage 201/230 as a percentage 207/225 as a percentage 201/231 as a percentage 208/225 as a percentage 201/232 as a percentage 209/225 as a percentage 201/233 as a percentage 210/225 as a percentage 201/234 as a percentage 211/225 as a percentage 201/235 as a percentage | 1,898 | 7,279 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.46875 | 4 | CC-MAIN-2024-33 | latest | en | 0.901247 |
https://freelanceshack.com/blog/how-to-plot-a-hermite-curve-in-matlab | 1,726,867,365,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725701423570.98/warc/CC-MAIN-20240920190822-20240920220822-00791.warc.gz | 239,986,949 | 62,495 | Category
# How to Plot A Hermite Curve In Matlab?
To plot a Hermite curve in Matlab, you can follow these steps:
1. Define the control points: A Hermite curve requires at least two control points to define its shape. Each control point consists of its coordinates (x, y) and the derivative at that point (dx, dy). For example, you can define two control points as P1(x1, y1, dx1, dy1) and P2(x2, y2, dx2, dy2).
2. Generate the parameter values: The Hermite curve is defined parametrically. You need to define a parameter t that varies between 0 and 1, representing the curve's starting and ending points. You can generate a series of equally spaced parameter values using the linspace function: t = linspace(0, 1, num_points).
3. Calculate the influence coefficients: The influence coefficients determine the influence of each control point on the curve. For a Hermite curve, these coefficients can be calculated as follows: H1(t) = 2t^3 - 3t^2 + 1 H2(t) = -2t^3 + 3t^2 H3(t) = t^3 - 2t^2 + t H4(t) = t^3 - t^2
4. Calculate the coordinates of the Hermite curve: For each parameter value, calculate the curve's x and y coordinates using the influence coefficients and the control points. Use the formula: x(t) = H1(t) * x1 + H2(t) * x2 + H3(t) * dx1 + H4(t) * dx2 y(t) = H1(t) * y1 + H2(t) * y2 + H3(t) * dy1 + H4(t) * dy2 Substitute the appropriate values for x1, y1, x2, y2, dx1, dy1, dx2, dy2, and the influence coefficients.
5. Plot the Hermite curve: Once you have calculated the coordinates for all parameter values, use the plot function to visualize the curve. Pass the x and y coordinate arrays as arguments: plot(x, y).
It is worth noting that the accuracy of the Hermite curve is dependent on the number of parameter values (num_points). Increasing the number of points will result in a smoother curve.
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## What is the difference between a hermite curve and a Bezier curve?
Hermite curves and Bezier curves are both types of parametric curves commonly used in computer graphics and computer-aided design (CAD). While they share similarities, there are some key differences between them:
Definition:
• Hermite Curve: A Hermite curve is defined using two endpoints (P0 and P1) and two tangent vectors (T0 and T1) associated with those endpoints. The curve passes through the two endpoints and its shape is influenced by the tangent vectors.
• Bezier Curve: A Bezier curve is defined using control points. The curve does not necessarily pass through the control points but is influenced by their positions. A cubic Bezier curve, for example, is defined by four control points.
Degree:
• Hermite Curve: Hermite curves can have different degrees, depending on the number of points and tangents used. For example, a cubic Hermite curve has degree 3.
• Bezier Curve: Bezier curves also have degrees, which correspond to the number of control points used. A cubic Bezier curve has degree 3, meaning it is influenced by the positions of four control points.
Interpolation:
• Hermite Curve: Hermite curves can interpolate points and tangents. This means that if you provide specific values for endpoints and tangent vectors, the curve will precisely pass through those points and exhibit the specified tangent directions at those points.
• Bezier Curve: Bezier curves do not generally interpolate control points. The curve may come close to some control points, but it typically does not pass exactly through them.
Flexibility and Control:
• Hermite Curve: Hermite curves provide more control over the shape of the curve due to the inclusion of tangent vectors. This allows for more precise control in designing curves.
• Bezier Curve: Bezier curves are more flexible in terms of design, as you have more freedom and control over the shape by manipulating the positions of control points.
Mathematical Formulation:
• Hermite Curve: Hermite curves are typically represented by parametric equations that involve the endpoints and tangent vectors. These equations describe how the curve behaves as a function of a parameter, usually denoted by 't'.
• Bezier Curve: Bezier curves are usually represented by Bernstein polynomials. These polynomials define the contribution of each control point to the final shape of the curve.
In summary, while both Hermite and Bezier curves are parametric curves used in computer graphics, Hermite curves provide more interpolation accuracy and control over tangents, while Bezier curves offer more flexibility and manipulation through control points.
## How to plot a hermite curve in Matlab?
To plot a Hermite curve in MATLAB, you can follow these steps:
1. Define the input points P0 and P1 and their corresponding tangent vectors T0 and T1. These points and tangents will determine the shape of the curve.
2. Decide the number of points N you want to generate on the curve.
3. Generate a parameter vector t ranging from 0 to 1 with N equally spaced points.
4. Use the Hermite interpolation formula to calculate the positions of the points on the curve. The formula is: Q(t) = (2t^3 - 3t^2 + 1)P0 + (t^3 - 2t^2 + t)T0 + (-2t^3 + 3t^2)P1 + (t^3 - t^2)T1 for each value of t.
5. Create a matrix to store the x and y coordinates of the points on the curve.
6. Iterate over each value of t and calculate the x and y coordinates using the Hermite interpolation formula. Store these coordinates in the matrix.
7. Plot the curve using the plot function, passing the x and y coordinates as arguments.
Here is an example code snippet to help you understand the implementation:
``` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 ``` ```P0 = [0 0]; % Initial point P1 = [1 1]; % Final point T0 = [1 0]; % Initial tangent vector T1 = [0 1]; % Final tangent vector N = 100; % Number of points on the curve t = linspace(0, 1, N); % Parameter vector Q = zeros(N, 2); % Matrix to store the coordinates for i = 1:N Q(i, :) = (2*t(i)^3 - 3*t(i)^2 + 1)*P0 + (t(i)^3 - 2*t(i)^2 + t(i))*T0 + ... (-2*t(i)^3 + 3*t(i)^2)*P1 + (t(i)^3 - t(i)^2)*T1; end plot(Q(:, 1), Q(:, 2)); % Plot the curve ```
Make sure to adjust the values of P0, P1, T0, T1, and N according to your desired curve and number of points.
## What is the role of the Frenet-Serret formulas in hermite curves?
The Frenet-Serret formulas play a crucial role in describing the behavior of curves in three-dimensional space, including Hermite curves.
Hermite curves, also known as Hermite interpolation or cubic splines, are commonly used in computer graphics and animation. They are defined by a set of control points and tangent vectors at those points. The control points determine the shape of the curve, while the tangent vectors control the direction of the curve at each point.
The Frenet-Serret formulas provide a mathematical framework to calculate the derivatives of a parametric curve, such as tangents, normals, and binormals. These formulas connect the geometric properties of the curve to its derivatives and help describe the curvature and torsion of the curve.
For Hermite curves, the control points define the positions of the curve, and the tangent vectors define its direction. By applying the Frenet-Serret formulas to the tangent vectors, one can determine the curvature and torsion at each point, providing further information about the behavior of the curve.
Overall, the Frenet-Serret formulas enable the analysis and manipulation of Hermite curves by providing a way to calculate important geometric properties of the curve at any given point. This information is vital for controlling and animating the curve in computer graphics and other applications.
## What are the properties of hermite curves that make them useful in animation?
Hermite curves have several properties that make them particularly useful in animation:
1. Local control: Hermite curves possess a local control property which means that the influence of the control points is confined to a local region of the curve. This makes it easier for animators to modify and control specific portions of the curve without affecting the rest of it.
2. Interpolation: Hermite curves can accurately interpolate both position and derivative information at the control points. This enables animators to define the exact path and tangent direction of the curve at each control point, providing precise control over the shape and motion of the animation.
3. Flexibility: By adjusting the position and tangent vectors of the control points, animators can create a wide range of curves with varying shapes and motions. This flexibility allows for the creation of smooth, flowing animations with complex movements.
4. Continuity: Hermite curves can ensure continuity between neighboring curve segments. They can smoothly connect different curve segments, ensuring a seamless transition between different animation sequences or movements.
5. Efficient computation: The mathematical formulas for calculating Hermite curves are relatively simple and computationally efficient. This makes them suitable for real-time applications, where smooth and responsive animations are required.
Overall, the properties of Hermite curves enable animators to achieve highly accurate and controlled animations, providing a powerful tool for creating lifelike motion in computer-generated imagery (CGI) and other animation techniques.
## How to adjust the tension of a hermite curve in Matlab?
To adjust the tension of a Hermite curve in MATLAB, you can use the `interp1` function with the `'spline'` option. Here is an example code to show how to adjust the tension of a Hermite curve:
``` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 ``` ```% Define the control points t = [0 1 2 3 4]; % Time or parameter values x = [0 1 2 3 4]; % x-coordinate values y = [0 2 1 3 0]; % y-coordinate values % Adjusting the tension tension = 0.5; % Tension value, range: [0, 1] splinet = linspace(min(t),max(t),1000); % New parameter values for interpolation spline_x = interp1(t, x, splinet, 'spline', tension); % Interpolate x-coordinates with tension spline_y = interp1(t, y, splinet, 'spline', tension); % Interpolate y-coordinates with tension % Plot the original control points and the adjusted Hermite curve plot(x, y, 'ro', 'MarkerSize', 8); % Original control points hold on; plot(spline_x, spline_y, 'b-', 'LineWidth', 2); % Adjusted Hermite curve hold off; axis equal; xlabel('x'); ylabel('y'); title('Adjusted Hermite Curve'); legend('Control Points', 'Adjusted Curve'); ```
In this example, the `interp1` function is used twice to interpolate the x and y coordinates separately. The `'spline'` option tells MATLAB to use the cubic spline interpolation method, and the `tension` parameter adjusts the tension of the Hermite curve. The higher the tension value, the tighter the curve, while a tension of 0 corresponds to a linear curve.
## What is the purpose of tangent vectors in a hermite curve?
Tangent vectors play a critical role in defining the shape and behavior of Hermite curves.
Hermite curves are commonly used in computer graphics and animation to smoothly interpolate between given control points. These control points not only specify the positions of the curve but also the tangent vectors at those points.
The tangent vectors indicate the direction and rate of change of the curve at each control point. By specifying the tangent vectors, we determine the slope and direction the curve should have as it approaches and leaves each control point. This information enables us to create smooth and continuous curves that smoothly flow through the control points and follow the desired curvature.
In other words, the tangent vectors control the rate at which the curve turns as it passes through each control point. They influence the sharpness, curvature, and overall shape of the curve between the control points.
By manipulating these tangent vectors, we can adjust the shape and behavior of the curve, achieving various effects such as sharp corners, smooth curves, or gradual transitions.
Therefore, the purpose of tangent vectors in a Hermite curve is to provide essential information about the desired slope and direction of the curve at each control point, enabling us to create smooth and visually appealing curves.
## Related Posts:
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To create a frequency histogram in MATLAB, you can follow these steps:Start by obtaining the data you want to plot the histogram for. You can either load the data from a file or create it programmatically. Once you have the data, use the histogram() function i... | 3,227 | 13,708 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2024-38 | latest | en | 0.748605 |
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05-15-2019, 01:36 PM #21
Brandy
Join Date: Aug 2004
Location: Nashville, TN
Re: Calculating Weight in Relation to Height
Quote:
Originally Posted by Bathawk er how do I input that into my comps calculator? (particuarly "cubing" a number)
With any calculator, cubing is just repeated multiplication.
(7/6)^3, if you don't have an explicit raise-to-power feature for some reason, is just (7/6)*(7/6)*(7/6).
Holy thread necromancy, Batman!
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05-15-2019, 01:39 PM #22
Fred Brackin
Join Date: Aug 2007
Re: Calculating Weight in Relation to Height
Quote:
Originally Posted by Bathawk er how do I input that into my comps calculator? (particuarly "cubing" a number)
For what I find to ne the simplest (but not necesarily most direct) calculations find the percentage of increase and cube that.
For example going from 6 to 7 ft is a 1,16 percent increase in height. so 1.16 x 1.16 x 1.16 equals 1.56 percent increase in weight. That would take you from 200 lbs to 312 lbs.
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05-15-2019, 02:52 PM #23 Bathawk Join Date: Oct 2007 Re: Calculating Weight in Relation to Height ah thank you, I "think" I knew that, My brain is probably just trying to make things more complicated than they are ;)
05-16-2019, 08:41 PM #24 ak_aramis Join Date: May 2010 Location: Alsea, OR Re: Calculating Weight in Relation to Height It is worth noting that a 12' humanoid will not follow the square/cube law exactly. Several factors don't scale directly. Skin thickness doesn't need to scale linearly. Bones increase above the cube of the scale factor, because strength is based upon cross section, and scales with mass supported. Circulatory systems need to scale Sq.Cu. above as well, due to greater pressure needed, which adds up quick. Teeth only need to scale to the square of the scaling factor, but in various species, falls between S² and S³, assuming no change in type of diet.
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https://cran.stat.sfu.ca/web/packages/BTSPAS/vignettes/d-Non-diagonal-with-fall-back-model.html | 1,726,285,248,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651548.18/warc/CC-MAIN-20240914025441-20240914055441-00562.warc.gz | 156,686,724 | 162,330 | 1 Location of vignette source and code.
Because of the length of time needed to run the vignettes, only static vignettes have been included with this package.
The original of the vignettes and the code can be obtained from the GitHub site at https://github.com/cschwarz-stat-sfu-ca/BTSPAS
2 Introduction
2.1 Experimental set-up
This case represents a generalization of the non-diagonal case considered in a separate vignette. Now we allow some fish (marked and unmarked) to approach the second trap, but fall back and never pass the trap. Schwarz and Bonner (2011) considered this model to estimate the number of steelhead that passed upstream of Moricetown Canyon.
The experimental setup is the same as the non-diagonal case. Consider an experiment to estimate the number of outgoing smolts on a small river. The run of smolts extends over several weeks. As smolts migrate, they are captured and marked with individually numbered tags and released at the first capture location using, for example, a fishwheel. The migration continues, and a second fishwheel takes a second sample several kilometers down stream. At the second fishwheel, the captures consist of a mixture of marked (from the first fishwheel) and unmarked fish.
The efficiency of the fishwheels varies over time in response to stream flow, run size passing the wheel and other uncontrollable events. So it is unlikely that the capture probabilities are equal over time at either location, i.e. are heterogeneous over time.
We suppose that we can temporally stratify the data into, for example, weeks, where the capture-probabilities are (mostly) homogeneous at each wheel in each week.
But now, we allow tagged animals to be captured in several recovery strata. For example, suppose that in each julian week $$j$$, $$n1[j]$$ fish are marked and released above the rotary screw trap. Of these, $$m2[j,j]$$ are recaptured in julian week $$j$$; $$m2[j,j+1]$$ are recaptured in julian week $$j+1$$; $$m2[j,j+2]$$ are recaptured in julian week $$j+2$$ and so on.
At the same time, $$u2[j]$$ unmarked fish are captured at the screw trap.
This implies that the data can be structured as a non-diagonal array similar to:
Recovery Stratum
tagged rs1 rs2 rs3 ...rs4 rsk rs(k+1)
Marking ms1 n1[1] m2[1,1] m2[1,2] m2[1,3] m2[1,4] 0 ... 0 0
Stratum ms2 n1[2] 0 m2[2,2] m2[2,3] m2[2,4] .... 0 ... 0 0
ms3 n1[3] 0 0 m2[3,3] m2[3,4] ... 0 ... 0 0
...
msk n1[k] 0 0 0 ... 0 0 m2[k,k] m2[k,k+1]
Newly
Untagged u2[1] u2[2] u2[3] ... u2[k] u2[k,k+1]
captured
Here the tagging and recapture events have been stratified in to $$k$$ temporal strata. Marked fish from one stratum tend to spread out and are recaptured over multiple strata. Several additional recovery strata are needed at the end of the experiment to fully capture the final release stratum.
Because the lower diagonal of the recovery matrix is zero, the data can be entered in a shorthand fashion by showing the recoveries in the same stratum as release, the next stratum, etc, up to a maximum number of recovery strata per release.
2.2 Fall-back information
This information is obtained by also marking radio-tagged fish whose ultimate fate (i.e. did they pass the second trap nor not) can be determined. We measure:
• $$marked\_available\_n$$ representing the number of radio-tagged fish.
• $$marked\_available\_x$$ representing the number of radio tagged fish that PASSED the second trap.The $$n$$ and $$x$$ are modelled using a binomial distribution for information on the fraction of tagged fish that DO NOT fall back, i.e. are available at the second trap. For example, if $$n=66$$ and $$x=40$$, then you estimate that about $$40/66=61$$% of tagged and untagged fish pass the second trap and that $$39$$% of fish fall back never to pass the second trap.
Notice we don’t really care about unmarked fish that fall back as we only estimate the number of unmarked fish that pass the second trap, which by definition exclude those fish that never make it to second trap. We need to worry about marked fish that never make it to the second trap because the fish that fall back will lead to underestimates of the trap-efficiency and over-estimates of unmarked fish that pass the second trap.
This model could also be used for mortality between the marking and recovery trap.
2.3 Fixing values of $$p$$ or using covariates.
Refer to the vignette on the Diagonal Case for information about fixing values of $$p$$ or modelling $$p$$ using covariates such a stream flow or smoothing $$p$$ using a temporal spline.
3 Example of non-diagonal model with fall-back.
Here is an example of some raw data that is read in:
demo.data.csv <- textConnection("
jweek,n1, X0,X1 ,X2 ,X3,X4,X5,X6,X7
29 , 1 , 0 , 0 , 0 ,0 ,0 ,0 ,0 ,0
30 , 35 , 0 , 5 , 7 ,2 ,0 ,0 ,0 ,0
31 ,186 , 1 ,35 ,11 ,4 ,0 ,0 ,0 ,0
32 ,292 , 9 ,33 ,16 ,6 ,0 ,0 ,0 ,0
33 ,460 , 6 ,41 ,16 ,9 ,3 ,0 ,2 ,1
34 ,397 , 4 ,44 , 7 ,5 ,1 ,1 ,0 ,1
35 ,492 , 7 ,31 ,12 ,1 ,4 ,1 ,1 ,0
36 ,151 , 3 , 6 , 2 ,1 ,1 ,0 ,0 ,0
37 ,130 , 3 , 2 , 2 ,0 ,0 ,1 ,0 ,0
38 ,557 , 8 ,27 ,11 ,2 ,5 ,0 ,0 ,0
39 , 46 , 0 , 7 , 0 ,0 ,0 ,0 ,0 ,0
40 ,143 , 14 , 6 , 3 ,0 ,0 ,0 ,0 ,0
41 , 26 , 2 , 1 , 0 ,0 ,0 ,0 ,0 ,0")
print(demo.data)
#> jweek n1 X0 X1 X2 X3 X4 X5 X6 X7
#> 1 29 1 0 0 0 0 0 0 0 0
#> 2 30 35 0 5 7 2 0 0 0 0
#> 3 31 186 1 35 11 4 0 0 0 0
#> 4 32 292 9 33 16 6 0 0 0 0
#> 5 33 460 6 41 16 9 3 0 2 1
#> 6 34 397 4 44 7 5 1 1 0 1
#> 7 35 492 7 31 12 1 4 1 1 0
#> 8 36 151 3 6 2 1 1 0 0 0
#> 9 37 130 3 2 2 0 0 1 0 0
#> 10 38 557 8 27 11 2 5 0 0 0
#> 11 39 46 0 7 0 0 0 0 0 0
#> 12 40 143 14 6 3 0 0 0 0 0
#> 13 41 26 2 1 0 0 0 0 0 0
There are 13 release strata. In the first release stratum, a total of 1 fish were tagged and released. No recoveries occurred.
Because the recoveries take place in more strata than releases, the $$u2$$ vector is read in separately. Note that is must be sufficiently long to account for the number of releases plus potential movement:
demo.data.u2 <- c( 2, 65, 325, 873, 976, 761, 869, 473, 332, 197,
177, 282, 82, 100)
We also separate out the recoveries $$m2$$ into a matrix
demo.data.m2 <- as.matrix(demo.data[,c("X0","X1","X2","X3","X4","X5","X6","X7")])
A separate radio-telemetry study found that of 66 fish released, 40 passed the second trap:
demo.mark.available.n <- 66
demo.mark.available.x <- 40
3.2 Fitting the BTSPAS non-diagonal model with fall-back model with a non-parametric movement distribution.
Schwarz and Bonner (2011) extended Bonner and Schwarz (2011) with a model with the following features.
• Non-parametric distribution of the distribution of times between release and availability at the second trap.
• A spline is used to smooth the total number of unmarked fish presenting themselves at the second trap over the strata
• A hierarchical model for the capture-probabilities is assumed where individual stratum capture probabilities are assumed to vary around a common mean.
• A binomial distribution is assumed for the number of marked fish that do not fall back and pass the second trap to estimate the trap efficiency.
The model also allows the user to use covariates to explain some of the variation in the capture probabilities in much the same way as the diagonal case.
The $$BTSPAS$$ package also has additional features and options:
• if $$u2$$ is missing for any stratum, the program will use the spline to interpolate for the number of unmarked fish in the population missing stratum.
• if $$n1$$ and the entire corresponding row of $$m2$$ are 0, the program will use the hierarchical model to interpolate the capture probabilities for the missing strata because no information is available from 0 fish released.
• the program allows you specify break points in the underlying spline to account for external events.
• sometimes bad thing happen. The vector $$bad.m2$$ indicates which julian weeks something went wrong. In the above example, the number of recoveries in julian week 41 is far below expectations and leads to impossible Petersen estimate for julian week 41. Similarly, the vector $$bad.u2$$ indicates which julian weeks, the number of unmarked fish is suspect. In both cases, the suspect values of $$n1$$ and $$m2$$ are set to 0, and the suspect values of $$u2$$ are set to missing. Alternatively, the user can set the $$n1$$ and $$m2$$ values to 0, and set the suspect $$u2$$ values to missing in the data input directly. I arbitrarily chose the third julian week to demonstrate this feature.
The $$BTSPAS$$ function also allows you specify
• The prefix is used to identify the output files for this run.
• The title is used to title the output.
We already read in the data above. Here we set the rest of the parameters. Don’t forget to set the working directory as appropriate
library("BTSPAS")
demo.prefix <- "FB-"
demo.title <- "Fall-back demo"
demo.jump.after <- NULL
## Identify spurious values in n1, m2, and u2 that should be set to 0 or missing as needed.
## Fix capture probabilities for strata when traps not operated
demo.logitP.fixed <- NULL
demo.logitP.fixed.values <- rep(-10,length(demo.logitP.fixed))
demo.fit <- TimeStratPetersenNonDiagErrorNPMarkAvail_fit(
title= demo.title,
prefix= demo.prefix,
time= demo.data$jweek[1]:(demo.data$jweek[1]+length(demo.data.u2)-1),
n1= demo.data$n1, m2= demo.data.m2, u2= demo.data.u2, jump.after= demo.jump.after, bad.n1= demo.bad.n1, bad.m2= demo.bad.m2, bad.u2= demo.bad.u2, logitP.fixed=demo.logitP.fixed, logitP.fixed.values=demo.logitP.fixed.values, marked_available_n=demo.mark.available.n, marked_available_x=demo.mark.available.x, # 40/66 fish did NOT fall back debug=TRUE, save.output.to.files=FALSE) #> #> #> *** Start of call to JAGS #> Working directory: /Users/cschwarz/Dropbox/SPAS-Bayesian/BTSPAS/vignettes #> Initial seed for JAGS set to: 772306 #> Random number seed for chain 661223 #> Random number seed for chain 720272 #> Random number seed for chain 910532 #> Compiling model graph #> Resolving undeclared variables #> Allocating nodes #> Graph information: #> Observed stochastic nodes: 27 #> Unobserved stochastic nodes: 141 #> Total graph size: 1079 #> #> Initializing model #> #> | | | 0% | |++ | 4% | |++++ | 8% | |++++++ | 12% | |++++++++ | 16% | |++++++++++ | 20% | |++++++++++++ | 24% | |++++++++++++++ | 28% | |++++++++++++++++ | 32% | |++++++++++++++++++ | 36% | |++++++++++++++++++++ | 40% | |++++++++++++++++++++++ | 44% | |++++++++++++++++++++++++ | 48% | |++++++++++++++++++++++++++ | 52% | |++++++++++++++++++++++++++++ | 56% | |++++++++++++++++++++++++++++++ | 60% | |++++++++++++++++++++++++++++++++ | 64% | |++++++++++++++++++++++++++++++++++ | 68% | |++++++++++++++++++++++++++++++++++++ | 72% | |++++++++++++++++++++++++++++++++++++++ | 76% | |++++++++++++++++++++++++++++++++++++++++ | 80% | |++++++++++++++++++++++++++++++++++++++++++ | 84% | |++++++++++++++++++++++++++++++++++++++++++++ | 88% | |++++++++++++++++++++++++++++++++++++++++++++++ | 92% | |++++++++++++++++++++++++++++++++++++++++++++++++ | 96% | |++++++++++++++++++++++++++++++++++++++++++++++++++| 100% #> | | | 0% | |** | 4% | |**** | 8% | |****** | 12% | |******** | 16% | |********** | 20% | |************ | 24% | |************** | 28% | |**************** | 32% | |****************** | 36% | |******************** | 40% | |********************** | 44% | |************************ | 48% | |************************** | 52% | |**************************** | 56% | |****************************** | 60% | |******************************** | 64% | |********************************** | 68% | |************************************ | 72% | |************************************** | 76% | |**************************************** | 80% | |****************************************** | 84% | |******************************************** | 88% | |********************************************** | 92% | |************************************************ | 96% | |**************************************************| 100% #> #> #> *** Finished JAGS *** 3.3 The output from the fit Here is the fitted spline curve to the number of unmarked fish available in each recovery stratum at the second trap demo.fit$plots$fit.plot The distribution of the posterior sample for the total number unmarked and total abundance that pass the second trap is available. Note this include the sum of the unmarked shown in the previous plot, plus a binomial distribution on the number of marked fish released that pass the second trap. demo.fit$plots$post.UNtot.plot A plot of the $$logit(P)$$ is demo.fit$plots$logitP.plot In cases where there is little information, $$BTSPAS$$ has shared information based on the distribution of catchability in the other strata. A summary of the posterior for each parameter is also available. In particular, here are the summary statistics on the posterior sample for the total number unmarked and total abundance THAT PASS THE SECOND TRAP: demo.fit$summary[ row.names(demo.fit$summary) %in% c("Ntot","Utot"),] #> mean sd 2.5% 25% 50% 75% 97.5% Rhat n.eff #> Ntot 22078.05 2266.638 17766.32 20557.0 22061.0 23548.0 26970.17 1.053470 44 #> Utot 20288.79 2121.686 16318.37 18862.5 20263.5 21646.5 24956.76 1.052413 44 This also includes the Rubin-Brooks-Gelman statistic ($$Rhat$$) on mixing of the chains and the effective sample size of the posterior (after accounting for autocorrelation). The estimated total abundance is 22,078 (SD 2,267 ) fish. The estimated distribution function is allowed by vary by release stratum around a common “mean” distribution. probs <- demo.fit$summary[grepl("movep", row.names(demo.fit$summary)), ] round(probs,3) #> mean sd 2.5% 25% 50% 75% 97.5% Rhat n.eff #> movep[1] 0.130 0.029 0.081 0.109 0.127 0.148 0.196 1.001 1500 #> movep[2] 0.513 0.054 0.406 0.478 0.514 0.549 0.618 1.006 350 #> movep[3] 0.205 0.038 0.136 0.178 0.204 0.230 0.281 1.004 480 #> movep[4] 0.085 0.023 0.046 0.068 0.083 0.098 0.135 1.002 1500 #> movep[5] 0.037 0.014 0.014 0.028 0.036 0.046 0.070 1.000 1500 #> movep[6] 0.012 0.008 0.002 0.007 0.011 0.016 0.029 1.006 350 #> movep[7] 0.010 0.007 0.002 0.005 0.008 0.013 0.027 1.005 1300 #> movep[8] 0.007 0.006 0.001 0.004 0.006 0.010 0.022 1.007 320 So we expect that about 13% of fish will migrate to the second trap in the day of release; about 51% of fish will migrate to the second trap in the second day after release etc. The movement for each release stratum varies around this base distribution. It is also possible to see the probability of moving from release stratum $$i$$ to recovery stratum $$j$$ by looking at the $$Theta[i,j]$$ values. Here are the transition probabilities for the first release stratum: round(demo.fit$summary[ grepl("Theta[1,", row.names(demo.fit$summary),fixed=TRUE),],3) #> mean sd 2.5% 25% 50% 75% 97.5% Rhat n.eff #> Theta[1,1] 0.147 0.090 0.037 0.085 0.126 0.186 0.380 1.000 1500 #> Theta[1,2] 0.491 0.140 0.209 0.396 0.493 0.588 0.747 1.004 650 #> Theta[1,3] 0.204 0.096 0.057 0.129 0.192 0.262 0.426 1.002 1500 #> Theta[1,4] 0.088 0.058 0.015 0.047 0.073 0.115 0.237 1.002 1100 #> Theta[1,5] 0.039 0.031 0.004 0.018 0.031 0.052 0.115 1.000 1500 #> Theta[1,6] 0.013 0.014 0.001 0.004 0.009 0.017 0.047 1.002 1100 #> Theta[1,7] 0.010 0.011 0.000 0.003 0.007 0.013 0.041 1.004 1400 #> Theta[1,8] 0.008 0.009 0.000 0.002 0.005 0.010 0.033 1.005 370 The probabilities should also sum to 1 for each release group. As with the other non-parametric non-diagonal model, you can specify a prior distribution for the movement probabilities. The sample of the posterior-distribution for the proportion of fish that DO NOT FALL back is round(demo.fit$summary[ grepl("ma.p", row.names(demo.fit\$summary),fixed=TRUE),],3)
#> mean sd 2.5% 25% 50% 75% 97.5% Rhat n.eff
#> 0.614 0.057 0.498 0.576 0.614 0.653 0.722 1.049 47.000
It is always important to do model assessment before accepting the results from the model fit. Please contact me for details on how to interpret the goodness of fit, trace, and autocorrelation plots.
4 References
Bonner, S. J., & Schwarz, C. J. (2011). Smoothing population size estimates for Time-Stratified Mark–Recapture experiments Using Bayesian P-Splines. Biometrics, 67, 1498–1507. https://doi.org/10.1111/j.1541-0420.2011.01599.x
Schwarz, C. J. and Bonner, S. B. (2011). A spline-based capture-mark-recapture model applied to estimating the number of steelhead within the Bulkley River passing the Moricetown Canyon in 2001-2010. Prepared for the B.C. Ministry of Environment.
Schwarz, C. J., & Dempson, J. B. (1994). Mark-recapture estimation of a salmon smolt population. Biometrics, 50, 98–108. | 5,322 | 16,988 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2024-38 | latest | en | 0.910803 |
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# Shortest Path Problems - PowerPoint PPT Presentation
Toronto. 650. 700. Boston. Chicago. 200. 600. New York. Shortest Path Problems. We can assign weights to the edges of graphs, for example to represent the distance between cities in a railway network:. Shortest Path Problems.
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Toronto
650
700
Boston
Chicago
200
600
New York
Shortest Path Problems
• We can assign weights to the edges of graphs, for example to represent the distance between cities in a railway network:
Applied Discrete Mathematics Week 14: Trees
Shortest Path Problems
• Such weighted graphs can also be used to model computer networks with response times or costs as weights.
• One of the most interesting questions that we can investigate with such graphs is:
• What is the shortest path between two vertices in the graph, that is, the path with the minimal sum of weights along the way?
• This corresponds to the shortest train connection or the fastest connection in a computer network.
Applied Discrete Mathematics Week 14: Trees
Dijkstra’s Algorithm
• Dijkstra’s algorithm is an iterative procedure that finds the shortest path between to vertices a and z in a weighted graph.
• It proceeds by finding the length of the shortest path from a to successive vertices and adding these vertices to a distinguished set of vertices S.
• The algorithm terminates once it reaches the vertex z.
Applied Discrete Mathematics Week 14: Trees
Dijkstra’s Algorithm
• procedure Dijkstra(G: weighted connected simple graph with vertices a = v0, v1, …, vn = z and positive weights w(vi, vj), where w(vi, vj) = if {vi, vj} is not an edge in G)
• for i := 1 to n
• L(vi) :=
• L(a) := 0
• S :=
• {the labels are now initialized so that the label of a is zero and all other labels are , and the distinguished set of vertices S is empty}
Applied Discrete Mathematics Week 14: Trees
Dijkstra’s Algorithm
• while zS
• begin
• u := the vertex not in S with minimal L(u)
• S := S{u}
• for all vertices v not in S
• if L(u) + w(u, v) < L(v) then L(v) := L(u) + w(u, v)
• {this adds a vertex to S with minimal label and updates the labels of vertices not in S}
• end{L(z) = length of shortest path from a to z}
Applied Discrete Mathematics Week 14: Trees
b
d
a
z
c
e
Dijkstra’s Algorithm
• Example:
5
6
4
8
1
2
0
3
2
10
Step 0
Applied Discrete Mathematics Week 14: Trees
b
d
5
6
4
8
a
1
z
2
0
3
2
10
c
e
Dijkstra’s Algorithm
4 (a)
• Example:
2 (a)
Step 1
Applied Discrete Mathematics Week 14: Trees
b
d
5
6
4
8
a
1
z
2
0
3
2
10
c
e
Dijkstra’s Algorithm
3 (a, c)
4 (a)
10 (a, c)
• Example:
2 (a)
12 (a, c)
Step 2
Applied Discrete Mathematics Week 14: Trees
b
d
5
6
4
8
a
1
z
2
0
3
2
10
c
e
Dijkstra’s Algorithm
3 (a, c)
4 (a)
10 (a, c)
8 (a, c, b)
• Example:
2 (a)
12 (a, c)
Step 3
Applied Discrete Mathematics Week 14: Trees
b
d
5
6
4
8
a
1
z
2
0
3
2
10
c
e
Dijkstra’s Algorithm
3 (a, c)
4 (a)
10 (a, c)
8 (a, c, b)
• Example:
14 (a, c, b, d)
2 (a)
12 (a, c)
10 (a, c, b, d)
Step 4
Applied Discrete Mathematics Week 14: Trees
b
d
5
6
4
8
a
1
z
2
0
3
2
10
c
e
Dijkstra’s Algorithm
4 (a)
3 (a, c)
8 (a, c, b)
10 (a, c)
• Example:
14 (a, c, b, d)
13 (a, c, b, d, e)
2 (a)
12 (a, c)
10 (a, c, b, d)
Step 5
Applied Discrete Mathematics Week 14: Trees
b
d
5
6
4
8
a
1
z
2
0
3
2
10
c
e
Dijkstra’s Algorithm
4 (a)
3 (a, c)
8 (a, c, b)
10 (a, c)
• Example:
14 (a, c, b, d)
13 (a, c, b, d, e)
2 (a)
12 (a, c)
10 (a, c, b, d)
Step 6
Applied Discrete Mathematics Week 14: Trees
The Traveling Salesman Problem
• The traveling salesman problem is one of the classical problems in computer science.
• A traveling salesman wants to visit a number of cities and then return to his starting point. Of course he wants to save time and energy, so he wants to determine the shortest path for his trip.
• We can represent the cities and the distances between them by a weighted, complete, undirected graph.
• The problem then is to find the circuit of minimum total weight that visits each vertex exactly once.
Applied Discrete Mathematics Week 14: Trees
Toronto
650
550
700
Boston
700
Chicago
200
600
New York
The Traveling Salesman Problem
• Example: What path would the traveling salesman take to visit the following cities?
Solution: The shortest path is Boston, New York, Chicago, Toronto, Boston (2,000 miles).
Applied Discrete Mathematics Week 14: Trees
The Traveling Salesman Problem
• Question: Given n vertices, how many different cycles Cn can we form by connecting these vertices with edges?
• Solution: We first choose a starting point. Then we have (n – 1) choices for the second vertex in the cycle, (n – 2) for the third one, and so on, so there are (n – 1)! choices for the whole cycle.
• However, this number includes identical cycles that were constructed in opposite directions. Therefore, the actual number of different cycles Cn is (n – 1)!/2.
Applied Discrete Mathematics Week 14: Trees
The Traveling Salesman Problem
• Unfortunately, no algorithm solving the traveling salesman problem with polynomial worst-case time complexity has been devised yet.
• This means that for large numbers of vertices, solving the traveling salesman problem is impractical.
• In these cases, we can use approximation algorithms that determine a path whose length may be slightly larger than the traveling salesman’s path, but can be computed with polynomial time complexity.
• For example, artificial neural networks can do such an efficient approximation task.
Applied Discrete Mathematics Week 14: Trees
• Trees
Applied Discrete Mathematics Week 14: Trees
Trees
• Definition: A tree is a connected undirected graph with no simple circuits.
• Since a tree cannot have a simple circuit, a tree cannot contain multiple edges or loops.
• Therefore, any tree must be a simple graph.
• Theorem: An undirected graph is a tree if and only if there is a unique simple path between any of its vertices.
Applied Discrete Mathematics Week 14: Trees
Trees
• Example: Are the following graphs trees?
Yes.
No.
No.
Yes.
Applied Discrete Mathematics Week 14: Trees
Trees
• Definition: An undirected graph that does not contain simple circuits and is not necessarily connected is called a forest.
• In general, we use trees to represent hierarchical structures.
• We often designate a particular vertex of a tree as the root. Since there is a unique path from the root to each vertex of the graph, we direct each edge away from the root.
• Thus, a tree together with its root produces a directed graph called a rooted tree.
Applied Discrete Mathematics Week 14: Trees
Tree Terminology
• If v is a vertex in a rooted tree other than the root, the parent of v is the unique vertex u such that there is a directed edge from u to v.
• When u is the parent of v, v is called the child of u.
• Vertices with the same parent are called siblings.
• The ancestors of a vertex other than the root are the vertices in the path from the root to this vertex, excluding the vertex itself and including the root.
Applied Discrete Mathematics Week 14: Trees
Tree Terminology
• The descendants of a vertex v are those vertices that have v as an ancestor.
• A vertex of a tree is called a leaf if it has no children.
• Vertices that have children are called internal vertices.
• If a is a vertex in a tree, then the subtree with a as its root is the subgraph of the tree consisting of a and its descendants and all edges incident to these descendants.
Applied Discrete Mathematics Week 14: Trees
Tree Terminology
• The level of a vertex v in a rooted tree is the length of the unique path from the root to this vertex.
• The level of the root is defined to be zero.
• The height of a rooted tree is the maximum of the levels of vertices.
Applied Discrete Mathematics Week 14: Trees
Trees
James
• Example I: Family tree
Christine
Bob
Frank
Joyce
Petra
Applied Discrete Mathematics Week 14: Trees
Trees
/
• Example II: File system
usr
bin
temp
bin
spool
ls
Applied Discrete Mathematics Week 14: Trees
Trees
• Example III: Arithmetic expressions
+
-
y
z
x
y
This tree represents the expression (y + z)(x - y).
Applied Discrete Mathematics Week 14: Trees
Trees
• Definition: A rooted tree is called an m-ary tree if every internal vertex has no more than m children.
• The tree is called a full m-ary tree if every internal vertex has exactly m children.
• An m-ary tree with m = 2 is called a binary tree.
• Theorem: A tree with n vertices has (n – 1) edges.
• Theorem: A full m-ary tree with i internal vertices contains n = mi + 1 vertices.
Applied Discrete Mathematics Week 14: Trees | 2,527 | 9,441 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2017-43 | latest | en | 0.892072 |
https://se.mathworks.com/matlabcentral/answers/258731-how-can-i-combine-probability-distribution-objects | 1,679,516,571,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296944452.74/warc/CC-MAIN-20230322180852-20230322210852-00141.warc.gz | 592,365,684 | 26,762 | # How can I combine probability distribution objects?
20 views (last 30 days)
Jo Williams on 3 Dec 2015
Answered: Arnab Sen on 31 Dec 2015
Are there operators probability distribution objects? For example,
>>xA = hospital.Weight;
>>pdA = fitdist(xA,'Normal')
Suppose I have a similar second record
>>hospitalB
where there are k times as many patients. Then I could write
>>pdboth = fitdist([xA;xB],'Normal');
but if I've only got the pd and weighting - or want to do some other weighting - I'd like to do something like
>>pdboth = pdA + k*pdB;
Does this sort of operator exist, or do I need to convert to a pdf and then fit a new distribution?
Tom Lane on 13 Dec 2015
What are you expecting from the addition? Are you looking for a mixture of Gaussians (perhaps a bimodal distribution with one peak from each distribution)?
Arnab Sen on 31 Dec 2015
I understand that you would like to achieve pooled weighted distribution of two distributions. There is no in-built functions in MATLAB to achieve this goal. However, it is possible to calculate it manually.
Consider following MATLAB code snippet for illustration:
>>muA=pdA.mu;
>>sigA=pdA.sigma;
>>lA=length(xA);
>>muB=pdB.mu;
>>sigB=pdB.sigma;
>>lB=length(xB);
>>pdboth=fitdist([xA;xB],'Normal');
%Calculate mean and standard deviation programmatically
>>muC=(muA*lA+k*muB*lB)/(lA+k*lB);
>>varC=((lA-1)*sigA^2+k*(lB-1)*sigB^2)(lA+k*lB-(k+1));
>>sigC=sqrt(varC);
>>pdboth.mu=muC;
>>pdboth.sigma=sigC;
In simple words we are calculating 'pdboth' as the distribution of xC=[xA;xB;xB;...ktimes]. You can also simply create a vector 'xC' as above and compute 'pdboth'.
For more details of the calculation as shown in the code, refer to the following link:
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# How to Estimate Sum of Decimals by Rounding?
Estimate the sum of decimals
To estimate means to find an answer that is close to but not exact. It is a reasonable answer to a problem.
Round each decimal term that will be added.
Example: Estimate the sum of 0.988 + 0.53
Round 0.988 + 0.53 to 1 + 0.5
Add the rounded numbers to obtain the estimated sum of 1.5
The actual sum of 0.988 + 0.53 is 1.518
Some uses of rounding are:
• Checking to see if you have enough money to buy what you want.
• Getting a rough idea of the correct answer to a problem.
Example:
Charlene and Margi went to a nearby restaurant for a quick lunch. The waitress handed them the bill with the price of each lunch, but forgot to add up the total. Estimate the total bill if Charlene’s lunch was \$3.75 and Margi’s lunch was \$4.29.
Analysis: Estimation is a good tool for making a rough calculation. There are many estimation strategies that you could use to estimate the sum of these decimals. Let’s look at the front-end strategy and the rounding strategy. We will round to the nearest tenth.
Front-End Strategy
Front-end estimation is a particular way of rounding numbers to estimate sums and differences. To use front- end estimation, add or subtract only the numbers in the greatest place value. Then add the decimals rounded to the nearest tenth. Add the front digits and then adjust the estimate.
3 + 4 = 7 and .75 plus .29 is about 1. Thus, \$7+ \$1 = \$8. The estimated sum is \$8.
Rounding Strategy
Round each decimal to a designated place value, then add to estimate the sum.
Rounding each decimal to the nearest tenth, we get an estimated sum of 8.1 or \$8.10.
Answer: Both \$8 and \$8.10 are good estimates for the total lunch bill for Charlene and Margi.
Information Source: | 443 | 1,791 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2024-22 | latest | en | 0.876598 |
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# Is OG10 still valid
Author Message
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### Show Tags
04 Jun 2011, 07:11
I was going thru some of the best GMAT stories and found that some one mentioned OG10 as a must resource but that person gave GMAT in Oct 2009. I plan to give my GMAT in Sept 2011. I wanted to know from the experts here who have given GMAT recently, do they think going thru OG10 will still help...Little background..I had finished OG12 couple of months ago, right now i am doing GMAT club tests, OG quant and MGMAT books, I was starting thru OG11 today and plan to revisit OG12 in august. By revisit, I mean solving last 40 questions from all sections..
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Re: Is OG10 still valid [#permalink]
### Show Tags
05 Jun 2011, 16:11
it can't hurt but there is going to be a great deal of overlap...I'm not familiar with an old OG that is "better" than the current. Contrary to what GMAC wants you to believe, each OG is not all that different.
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Re: Is OG10 still valid [#permalink]
### Show Tags
06 Jun 2011, 20:50
The key point to your success in GMAT is the time you spent practicing it.
The more you do, the more chance you will get a good score!!
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Re: Is OG10 still valid [#permalink]
### Show Tags
06 Jun 2011, 21:56
Be careful, the OG 10 is very powerful - it has 1,300 questions (compared to 900 in the OG 12).
However, it overlaps with GMAT Prep and PowerPrep and that's an issue if you are planning to take those tests. It is relevant and still does the job (even if the questions are dated).
Final words of caution - don't focus too much on solving the questions but instead spend the time on concepts and strategies....
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Re: Is OG10 still valid [#permalink]
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07 Jun 2011, 09:55
I also have OG 9 and OG 11 . What about those ? I know OG 11 has almost most of the questions repeated in OG 12. Also, is there a overlap between Verbal and Quant Review 1st and 2nd Edition
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Re: Is OG10 still valid [#permalink]
### Show Tags
07 Jun 2011, 11:47
Yes, there is a 60% overlap between verbal and math OG's as far as I remember (but this is not a very common question and I may not remember correctly)
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Re: Is OG10 still valid [#permalink] 07 Jun 2011, 11:47
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OG 10 ? 5 12 Jul 2011, 01:13
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# Is OG10 still valid
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Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 1,380 | 4,763 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2017-13 | longest | en | 0.924465 |
https://thvinhtuy.edu.vn/geometry-reveals-the-tricks-behind-gerrymandering-3fm4jkz0/ | 1,723,020,077,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640690787.34/warc/CC-MAIN-20240807080717-20240807110717-00853.warc.gz | 455,191,800 | 16,171 | # Geometry Reveals the Tricks behind Gerrymandering
How To Solve The Hardest Easy Geometry Problem
How To Solve The Hardest Easy Geometry Problem
Hardly anyone reckoned that struggling in high school to calculate the area of a triangle or the volume of a prism could be used one day to influence the outcome of an election. Geometry, however, can be a powerful tool in shaping results of an electoral contest—at least in plurality voting systems.
Designing a perfect election system for multiple parties is impossible, even with mathematical tools. But if, by and large, there are only two dominant parties, as in the U.S., things should be fairly clear-cut. The party candidate with the most votes wins, right? Anyone who has followed U.S. presidential elections in recent years knows that the reality is different. One important factor is the actual shape of the voting districts. If cleverly designed, a party that is actually losing can still gain the majority of representatives—an issue that was by no means absent in the U.S. midterm elections.
Math plays an important role in determining election outcomes, particularly for the 435 seats in the House of Representatives. By cleverly choosing the boundaries of a congressional district, a party can enable its candidate to win even if the vote count does not fairly represent the sentiment of the voters.
Here’s a highly simplified example: suppose a state consists of 50 voters, 20 of whom vote for a blue party and 30 for a red party. Voters might live in a grid pattern, as in, say, some sections of Manhattan. Suppose there are 10 north-south avenues and five east-west streets. All the red voters live on the first two avenues, the ones furthest west. The blue voters reside on the other three avenues. Now the task is to divide the voters into five electoral districts of equal size.
One could draw five vertical boundaries: Then there would be two election districts with only red voters and three with only blue voters. So the votes in that district would produce three blue party representatives and two red party ones, an accurate reflection of voters’ opinions.
But if the blue party were to get its way in drawing district boundaries, they might be inclined to draw the boundaries horizontally. Then all the districts would look the same, with four red voters and six blue voters each. In this case, the blue party wins in each district, and gets all five representatives. Something similar happened in New York state in 2012: 58 percent of people there voted for the Democrats, but the party got 21 of 27 seats (five more than would have been justified if the election districts had been drawn equitably).
State legislatures and the commissions that redraw district lines, on the other hand, might make a very different (somewhat more complicated) partitioning. To do this, they could pack almost all blue voters into two districts, giving the red party a majority in the three remaining districts in which there would be three red congressmen and two blue congresswomen— although more voters gave their votes to the blue party. There are numerous examples of this in U.S. congressional races. For example, in Pennsylvania in 2012, Democrats received 51 percent of the vote, but only five of 18 seats.
The deliberate redrawing of districts to gain a majority goes by the name of gerrymandering, a portmanteau of “gerry” and “salamander.” The former refers to Elbridge Gerry, the governor of Massachusetts in the early 19th century, who approved extremely odd-shaped voting districts that gave his party an advantage.
Even today, in most U.S. states, legislatures decide on the division of electoral districts about every 10 years (with the appearance of the new census). Time and again, the incumbent parties are suspected of using redistricting to their advantage. This can often be seen in strange-shaped electoral districts, similar to the one in Massachusetts at the beginning of the 19th century. A cartoonist at the time noticed that one of the districts resembled a salamander and thus coined the expression gerrymandering.
Redistricting has provoked numerous legal challenges. In 1986, the U.S. Supreme Court even ruled that intentional gerrymandering is illegal. But since then, it has barely touched an election district. As it turns out, setting rules for fair districting is not so easy. Even mathematicians are racking their brains over the question—and arming themselves with enormous computer power to deal with the problem.
How can you find gerrymandering? Observing the Maryland district pictured above, you might suspect the designers had certain ulterior motives. What is particularly striking is that it is extremely jagged. One assertion is that district borders should be “compact”— but without defining what “compact” means exactly.
One possible clue that gerrymandering may be present is the length of the outer boundary: the more jagged a district, the larger the perimeter. The literature related to redistricting sometimes advocates drawing the smallest possible circle to include the area within a district and comparing it to the area of existing boundaries. The more the district’s borders deviate from a circle, the greater the possibility that the district has been redrawn to suit partisan ends. The average distance between residents of a precinct may also indicate gerrymandering.
The partitioning into electoral districts is anything but simple. Each state follows its own rules in doing so. The ideal goal is for a district to contain roughly equal numbers of voters, be contiguous, not discriminate against ethnic groups, not cross county lines, and follow natural boundary lines, such as rivers. Such restrictions by themselves result in fractured districts—without even considering the voting behavior of residents.
A compact voting district does not necessarily lead to equitable representation, as a 2013 study found. The study paid particular attention to the 2000 presidential election in Florida, in which about as many people voted for Democrats as Republicans, but the latter accounted for 68 percent of the votes in Florida’s congressional districts. The researchers used a nonpartisan algorithm designed to draw the most “compact” districts possible while adhering to the state’s established rules.
Surprisingly, the computer also produced skewed results, in which Republicans would mostly have an advantage. And experts quickly realized the reason: most Democrats live in Florida cities. This means they win urban districts overwhelmingly, while narrowly losing in rural areas in each case. Because of this “natural gerrymandering,” more Republicans inevitably take seats in the House of Representatives.
Florida is not an isolated case, as political scientist Jonathan A. Rodden noted. The main problem is not a district’s lack of “compactness.” If you want to prove that a boundary was deliberately drawn to give one party an advantage, you need more evidence than the mere shape of a district.The goal in an unbiased system is to find electoral districts so that each party has an equal chance of converting its votes into electoral seats. But how can we measure that? In 2014, University of Chicago legal scholar Nicholas Stephanopoulos and Public Policy Institute of California political scientist Eric McGhee developed a metric for the problem, the efficiency gap. It is calculated by subtracting the “wasted” votes of two parties from each other and then dividing by the total number of all votes. A wasted vote for any party, in this example, is one that ends up in a losing district that went to the opposite party or that is above the margin needed to win. The smaller the efficiency gap, the more impartial is the result.
To visualize this, we can again use the initial example with the 50 voters (20 for red, 30 for blue) and calculate the efficiency gap for the different divisions. In the first case, when all boundaries were drawn vertically, the first and second districts (from the left) each have 10 red votes, wasting four each. The third, fourth and fifth districts, on the other hand, each have 10 blue votes, four of which are also wasted. Thus, the efficiency gap is as follows (the vertical bars indicate absolute value): |(2 x 4) – (3 x 4)|/50 = 2/25 = 0.08.
In the second division, each district is equal: blue always wins by six votes out of 10. Thus, none of blue’s votes are wasted—whereas all of red’s are. The efficiency gap is 20/50 = 0.4, which is significantly higher than in the first division.
The third example is the most intriguing: the two districts in which blue wins 9 to 1 each have a blue surplus of three. In the three winning red districts, four blue votes each are wasted—so in total, (2 x 3) + (3 x 4) = 18 blue votes that are surplus ones. In contrast, there are only two red votes that were wasted. This results in an efficiency gap of (18 – 2)/50 = 8/25 = 0.32.
The efficiency gap is useful as an indicator that pins down partisanship in voting districts. But sometimes natural conditions, such as when almost all voters of a party live in the same city, make it hard to find better possibilities. To investigate these possibilities, statistician Wendy Cho, along with computer scientist Yan Liu and geographer Shaowen Wang of the University of Illinois at Urbana-Champaign, designed an algorithm that divides maps into districts—based on the rules set by the state in question.
Finding the best possible division of districts so that each party has the same probability of converting a vote into a seat is extremely difficult. The task falls into the class of so-called NP problems, which computer scientists and mathematicians have suspected for decades cannot be solved efficiently with ordinary computers. That doesn’t mean you can’t find a solution—just that it may take a very, very long time. So Cho and her co-authors decided to let the computer construct an extremely large number of splits that are not necessarily perfect.
For example, when they applied their program to the state of Maryland in 2011, they realized that almost all of the 250 million results gave an advantage to Democrats. Apparently, the natural conditions, along with the requirements for voting districts, are such that Republicans are automatically at a disadvantage. Cho and her colleagues compared Maryland’s actual apportionment with the computer’s output and were able to show that the official voting districts favored Democrats in more than 99.79 percent of the 250 million computer-generated results.
Meanwhile, some U.S. states (mainly those where Democrats are in the majority) use independent commissions that redraw voting districts. These panels often resort to computer programs to find the fairest possible apportionment. In general, the apportionment of electoral districts this year appears to be the fairest in 40 years, as reported by the New York Times. When advantageous or detrimental districting decisions for both parties in all U.S. states are netted against one another, gerrymandering should result in only three extra seats for Republicans— down from 23 seats in 2012. But even three seats could be decisive in a close election. And news stories before the midterms depicted how gerrymandering is still very much a matter of public debate: Alabama’s state legislature redistricted to put many of the Black voters in the state into just one district, decreasing their electoral power, and resulting in a case that is now before the U.S. Supreme Court.
You are watching: Geometry Reveals the Tricks behind Gerrymandering. Info created by THVinhTuy selection and synthesis along with other related topics.
Rate this post | 2,386 | 11,724 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2024-33 | latest | en | 0.956563 |
https://subscription.packtpub.com/book/data/9781783284375/6/ch06lvl1sec79/performing-array-operations-with | 1,723,067,648,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640713269.38/warc/CC-MAIN-20240807205613-20240807235613-00511.warc.gz | 443,088,306 | 27,354 | #### Overview of this book
Python for Finance
Credits
Acknowledgments
www.PacktPub.com
Preface
Free Chapter
Introduction and Installation of Python
13 Lines of Python to Price a Call Option
Introduction to Modules
Statistical Analysis of Time Series
Index
## Performing array operations with +, -, *, /
Plus and minus for an array would have their normal meaning. However, multiplication and division have quite different definitions. Using multiplication as an example, A × B arrays could have two meanings: either item by item (A and B should have the same dimensions, that is, both are n by m) or matrix multiplication (the second dimension of A should be the same as the first dimension of B, that is, A is n by m while B is m by p ).
### Performing plus and minus operations
When adding or subtracting two arrays, they must have the same dimensions, that is, both are n by m. If they have different dimensions, we will get an error message. The following example shows the summation of two cash flow arrays:
```>>>cashFlows_1=np.array([-100,50,20])
>>>cashFlows_2=np.array([-80,100,120])
>>>cashFlows_1 + cashFlows_2
>>>array([-180, 150, 140])
```
### Performing a matrix multiplication operation
For matrix multiplication, matrices A and... | 292 | 1,254 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2024-33 | latest | en | 0.878973 |
https://support.google.com/docs/answer/9116395?hl=en&ref_topic=3105474 | 1,723,371,819,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640983659.65/warc/CC-MAIN-20240811075334-20240811105334-00486.warc.gz | 449,712,019 | 292,847 | SEC function
The SEC function returns the secant of an angle measured in radians.
Parts of a SEC formula
SEC(angle)
Part Description angle The angle to find the secant of, measured in radians.
SEC(3.14)
SEC(A1)
=SEC(1)
Examples
A B 1 Formula Result 2 =SEC(1) 1.850815718 3 =SEC(-1) 1.850815718 4 =SEC(4) -1.529885656 5 =SEC(0) 1
Related functions
• SECH: The SECH function returns the hyperbolic secant of an angle.
• COT: The COT function returns the cotangent of an angle provided in radians.
• COTH: The COTH function returns the hyperbolic cotangent of any real number.
• ACOT: The ACOT function returns the inverse cotangent of a value in radians.
• ATANH: The ATANH function returns the inverse hyperbolic tangent of a number.
• ATAN: The ATAN function returns the inverse tangent of a value in radians.
• ATAN2: The ATAN2 function returns the angle between the x-axis and a line segment from the origin (0,0) to the specified coordinate pair (`x`,`y`), in radians.
• ASINH: The ASINH function returns the inverse hyperbolic sine of a number.
• ASIN: The ASIN function returns the inverse sine of a value in radians.
• SIN: The SIN function returns the sine of an angle provided in radians.
• COS: The COS function returns the cosine of an angle provided in radians.
• ACOSH: The ACOSH function returns the inverse hyperbolic cosine of a number.
• ACOS: The ACOS function returns the inverse cosine of a value in radians.
• DEGREES: The DEGREES function converts an angle value in radians to degrees. | 418 | 1,519 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2024-33 | latest | en | 0.413374 |
http://blog.onaclovtech.com/2010_10_01_archive.html | 1,469,335,792,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257823947.97/warc/CC-MAIN-20160723071023-00133-ip-10-185-27-174.ec2.internal.warc.gz | 23,982,115 | 18,751 | ## Saturday, October 30, 2010
### Virtual Memory
So we were asked to devise a way to figure out which page a particular physical address was in, as well as it's offset.
Turns out this is a really simple problem, in this case each page is 4kb, which really means 4096 bytes, so simply take the address, divide by that 4096 bytes, and that tells you which page it's in. From there you subtract the number of bytes of the previous pages (4096 * n) from the address and you get the offset.
So the nice thing about virtual memory, is that you basically always know your address will reside within a 4096 byte address. This can make more static stuff easier to handle, rather then trying to guess which physical address you are, then figure out some kind of offset.
Here is the program I wrote in action.
Well I hope you enjoyed, I know it was a bit short, but the problem was really easy to see a solution, if you're having problems understanding how to find a page and offset, please let me know and I'll do my best to help!
### Bankers algorithm
The Algorithm
The bankers algorithm seems to be a pretty simple algorithm, essentially what you do is have some defined amount of resources, and then you kick off a bunch of threads, now if there are resources left, the thread can run and do it's thing, but if there are no resources, the thread must wait.
The key here we need to ensure is that there is no possibility of a deadlock. A deadlock occurs when two processes are waiting on each other to run. This can happen if we lock the mutex to try to see if we have resources, and then we don't let it go if the resource doesn't exist. For example the following situation would produce a deadlock.
1. thread 1 locks the mutex, gets the last resource, and unlocks the mutex.
2. thread 2 locks the mutex, no more resources exist so, thread 2 waits for resources, (not unlocking the mutex).
3. thread 1 is finished doing it's work, and needs to return the resources. It trys to get the mutex, it's blocked until thread 2 is done, but thread 2 won't give up the lock until thread 1 is finished.
This can be avoided by forcing thread 2 to lock the mutex, check for resources, unlock mutex, then check again if needed by doing it all over again. Now when thread 2 locks the mutex, and thread 1 tries to lock it, thread 2 releases it shortly later, and then thread 1 gets it, finally the next time thread 2 comes to lock the mutex, it now has resources and can run. See? Deadlock resolved!
Reading up on the bankers algorithm on wikipedia, it defines that each thread should declare how many system resources it will eventually need, and if the number of resources it needs is more then the system has remaining it must wait until the resources are available, since my implementation only requests one resource at a time, I didn't see the need to declare how many it would use. But my understanding of this is that the thread has itself a max number of instances allowed, so it has a max that it needs to check on. So essentially there are two resources each thread needs to handle, a system and a local, if it exceeds the local, this is an error condition, so it needs to be handled appropriately, if it exceeds the system, again it's an error and needs to be handled appropriately.
Bankers Algorithm Running
The fact is that I had way too many lines running to show it all, so I redirected the output, here is what is printed out.
Customers are coming..............
Customer Created..............
Doin Something Time Consuming..............
Customer Created..............
Doin Something Time Consuming..............
Customer Created..............
Doin Something Time Consuming..............
Customer Created..............
Doin Something Time Consuming..............
Customer Created..............
Doin Something Time Consuming..............
Customer Created..............
Doin Something Time Consuming..............
Customer Created..............
Doin Something Time Consuming..............
Customer Created..............
Doin Something Time Consuming..............
Customer Created..............
Doin Something Time Consuming..............
Customer Created..............
Doin Something Time Consuming..............
Customer Created..............
Customer Created..............
Customer Created..............
Customer Created..............
Customer Created..............
Customer Created..............
Customer Created..............
Customer Created..............
Customer Created..............
Customer Created..............
Waiting until all customers are gone..............
Doin Something Time Consuming..............
Doin Something Time Consuming..............
Doin Something Time Consuming..............
Doin Something Time Consuming..............
Doin Something Time Consuming..............
Doin Something Time Consuming..............
Doin Something Time Consuming..............
Doin Something Time Consuming..............
Doin Something Time Consuming..............
Doin Something Time Consuming..............
Time to go home..............
I hope you enjoyed the information! Let me know if you have any questions!
## Thursday, October 28, 2010
### Register Renaming
What is the purpose of Register Renaming?
The goal of register renaming is to allow blocks of instructions to be executed out of order.
Take for example the following instructions
1. divd r5, r7, r12
3. sd r6 0(r1)
4. subd r8, r14, r12
5. muld r6, r15, r8
Looking at these you can see that without register renaming you have to run the instructions in the sequence 1,2,3,4,5. However if you note, lines 4 and 5 don't depend on 1,2,3. The issue is that if they run before 1,2,3 they will clober the data that 1,2,3 are trying to calculate with.
If we simply rename the registers r8 and r6 in 4,5 we can remove that dependency.
This type of dependency is called a false dependency, because the only thing preventing them from running out of order is simply you are using the same registers, not the same data.
Now look at this:
1. divd r7, r5, r12
3. sd r6 0(r1)
4. subd S, r14, r12
5. muld T, r15,S
Now the instructions 1,2,3 can occur independent of 4,5. Since there are Read after Writes (RAW), and Write after Reads (WAR) the instructions within the 1,2,3, they must occur in order.
for example, the following orders are now valid.
a. 1,4,2,3,5
b. 4,1,2,3,5
c. 1,2,4,5,3
...
The only requirement now is that 1,2,3 occur in succession, and 4,5 occur in succession for example
the following executions are invalid
a. 2,1,4,5,3
b. 3,1,4,2,5
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# Lec04-42009 - ture 4-1 Gausss Law: Qualitative Statement...
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cture 4-1 Gauss’s Law: Qualitative Statement Form any closed surface around charges Count the number of electric field lines coming through the surface, those outward as positive and inward as negative. Then the net number of lines is proportional to the net charges enclosed in the surface .
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cture 4-2 ©2008 by W.H. Freeman and Company
cture 4-3 ©2008 by W.H. Freeman and Company
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cture 4-4 ©2008 by W.H. Freeman and Company
cture 4-5 ©2008 by W.H. Freeman and Company
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cture 4-6 Reading Quiz 1 Which of the following statements contradict Gauss’s law? a) The electric flux through a closed surface depends on all the charges outside the closed surface. b) The electric flux through a closed surface is proportional to the net charges enclosed in the surface. c) Any closed surface can be used with Gauss’s law. d) An electric dipole inside a closed surface does not change the net flux through the surface.
Electric flux Summary \$ N E A E An μ = u u v v g # of field lines N = density of field lines x “area” where “area” = A 2 x cos θ General definition of electric flux: \$ E S E n dA Φ = u v g (must specify sense , i.e., which way) To state Gauss’s Law in a quantitative form, we first need to compute the Electric Flux. \$ E An
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## This note was uploaded on 12/07/2011 for the course PHYS 241 taught by Professor Wei during the Fall '08 term at Purdue.
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Ask a homework question - tutors are online | 544 | 2,209 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2017-17 | longest | en | 0.904461 |
https://www.instructables.com/id/Decaplexing/ | 1,590,979,268,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347413901.34/warc/CC-MAIN-20200601005011-20200601035011-00011.warc.gz | 790,863,602 | 21,987 | # Decaplexing
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## Introduction: Decaplexing
There are many ways to multiplex LED’s. Some don’t use external circuits some do, however they either take a lot of pins or require custom LED layout. Decaplexing would require one IC, one capacitor and three resistors. It would also allow a four character seven segment to use 5 I/O pins. Decade counters are easy to find online for 79 cents
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## Step 1: Building the Circuit
It looks a little confusing at first but its really not. The PIC12F629 is the most basic microcontroller I know of and is in this circuit to demonstrate the multiplexing ability.
It is not critical which decade output the segments are on, the Q's can go to the closest segment pin. (I like Q0 to go to G so its dual line at start up) The most important part is the two resistors and cap in the middle, they make an auto reset so the decade starts at Q0 at startup. I've included a labeled segment display.
## Step 2: Programing
Programming the circuit can get hairy. CHR1 and CHR2 are the anode return pins and SEGAVD is the pin hooked to the decade clock. The best structure is a loop with if statements that turn on a segment for a given number set:
If num1<>1 and num1<>7 and num1<>0 or num1>9 or num1<0 then set CHR1 Low
Will turn the G segment on before the first clock pulse.
DLY was 1ms for approximately 50 Hz refresh rate
## Step 3: Future Uses
Decaplexing could drive a 7x5 Led matrix with only 6 I/O pins, strait off the shelf. An 8x8 matrix with 9 I/O pins, but with a Decade counter sourcing and one sinking it would take 3 pins for up to a 10x10 matrix! (Dualdecaplexing) It would take a fast MCU though. It can also be used to multiplex inputs, by counting the Q the counters on it could switch through 10 buttons with 2 I/O pins
## Step 4: Specs and Troubleshooting
Power: 5V
Refresh Rate: depends on MCU and counter but could be 200+ Hz
Troubleshooting:
The first picture is 39 the number I wanted but if it fails to reset its segments light differently on start up and it then displays NA. This is also caused when only the MCU loses power (like if it’s loose in its socket)
If one segment is lit when others are then it’s on the Q5-9 pin. This pin is on from Q5 to Q9 and will cause double light.
If the display is freaking out, and pulsing all segments then an input (CP0 CP1 or MR) is improperly tied to ground
Any program errors you can email me for assistance
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08 Aug 2006, 08:02
I've just registered on this site and will be taking the GMAT in a month. I purchased Kaplan a few weeks ago thinking that all GMAT prep was the same. I took the CAT1 and got a 600, which I now realze isn't too bad after reading a few posts on this board. However, regardless of how I did, everyone seems to think Kaplan CATs are a joke. While this seems to be the consensus on the board, what about the books practice questions or cd practice quizes. Is there any worth in any Kaplan GMAT prep or should I switch while I still have time?
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I had the same experience. Last night I got my butt kicked by the Kaplan CAT - a 620 after doing better on the individual practice tests. I couldn't believe some of the questions it asked! I would look at the answers, the steps required to solve, and say "that would take even the brightest person about 5 minutes to solve." It seemed to me the trick was all in knowing how to solve the problem from the beginning, then executing. I just am not at the point yet where I can recognize the best method to solve right away. I would be interested to hear others' replies on this question, as I see a lot on the OG and am considering buying a copy...
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08 Aug 2006, 19:50
you have to realise that Kaplan CATs scores are supposedly 100 or so points less than actual score..
thus getting a 620 is pretty good!
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09 Aug 2006, 07:40
jeg1015, thanks for your reply. How have you been doing on the practice tests? I've just started but want to know what someone else thinks about them so I'm not stuck in the Kaplan bubble.
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