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https://ask.metafilter.com/260036/Statistical-Herpes-Question | 1,516,733,206,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084892059.90/warc/CC-MAIN-20180123171440-20180123191440-00618.warc.gz | 637,357,016 | 16,954 | # Statistical Herpes QuestionApril 9, 2014 6:50 AM Subscribe
Can you give an estimate of the likelihood of contracting herpes based on this scenario?
Not looking for opinions, but educated guesstimates would be helpful.
What is the likelihood of a man contracting genital herpes from one evening of vaginal sex with a woman provided he is wearing a condom and she reports no outbreaks in 10 years and is taking daily Valtrex?
I'm pretty sure it's astronomically low. I'm seeing numbers like 2% of men would contract herpes in this situation from a year of sex in a committed relationship. But if someone had some more specific ideas I'd appreciate it.
If you could stick to the question and not have this be a general discussion about herpes, sex, etc that would be great.
Thanks
posted by mockpuppet to Health & Fitness (10 answers total)
I think that the only statistic that should matter to you is "non-zero".
You should talk to a doctor.
posted by DWRoelands at 7:33 AM on April 9, 2014 [5 favorites]
Can't provide you a number, only anecdata: A friend of mine was married to a woman for 10 years who had occasional outbreaks (they would avoid sex during outbreaks), they never used condoms, and after they divorced he got tested and was negative. So I think you're probably fine.
posted by greta simone at 7:38 AM on April 9, 2014
Accepting that 2% number (I'm not sure I do) and assuming that committed couple will have sex ~100 times/year, the simple, stupidest probability would suggest that once should have a probability of ~0.02%.
posted by Betelgeuse at 8:02 AM on April 9, 2014
2% can be reasonably used as a ceiling for the probability of the act you described, meaning your likelihood is definitely less than or equal to 2%, but how much less than 2% cannot be derived from the statistic (sorry Betelguy).
The scenario you've described includes pretty much everything we would want it to in order to minimize the risk of the sex act, except that you would also want to know that the man in question isn't immunocompromised (e.g. by AIDS, immunosuppressant drugs, etc). I have done a fair amount of reading and research on the subject in the past and I have never seen a study that produced the specific statistic you're looking for because it is, as you said, astronomically improbable.
posted by telegraph at 8:12 AM on April 9, 2014
So telegraph. Statistically you're saying that it's somewhere between 2% and infinitely small? Or am I misreading you? Or just ignorant of statistics?
posted by mockpuppet at 8:44 AM on April 9, 2014
It doesn't matter what the statistics are if you end up being the unlikely exception. It's not that likely, but it could happen. Happened to someone I knew, even. You can't 100% be in the clear, and if you tend to be naturally paranoid you're just going to keep wondering about it anyway. Sorry.
posted by jenfullmoon at 11:48 AM on April 9, 2014
Without controlling for condom use or antiviral therapy, "the rate of infection per 10,000 sex acts was 8.9 and 1.5 for susceptible women and men, respectively." I'd say significantly less than 0.0000015%.
It's also worth noting that 1 in 4 women in the U.S. have herpes, and most are unaware, so the risk with the general population isn't actually zero either.
posted by susanvance at 4:03 PM on April 9, 2014 [1 favorite]
Yes mockpuppet. That is exactly what I meant.
posted by telegraph at 12:27 PM on April 10, 2014
Can you give us a little more information? If you're happy with "between 2% and infinitesimally small" then that's fine, but I feel like I can't answer this question appropriately unless I know why it would make a difference to you to know whether the likelihood is 0.2% or 1%, for example.
Because without further information, the question reads as if you're asking permission/trying to give yourself permission not to get tested for herpes if the odds are low enough, but that question is really a question about your own risk tolerance (and the risk tolerance of any other sex partners you have currently or in the future) as well.
I know you asked us not to discuss anything but numbers, but since there is no calculator we can use to work out the risk of your specific situation, the context is meaningful.
posted by treehorn+bunny at 1:37 PM on April 11, 2014
Treehorn,
I think the situation as described is pretty clear, no?
As far as I can figure, the chance of transmission from this 1 sex act is much less likely than the man actually being a symptomless carrier of Herpes to begin with? Which appears to be approximately 17%?
posted by mockpuppet at 7:07 AM on April 16, 2014
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### #Actualbmarci
Posted 04 October 2012 - 04:37 AM
One more thought on suspensions (for dummies (mostly myself) ) through an example.
For the simplicity I'd represent a car with a box with 4 corners for each wheel attachment point, and 4 points for the wheels' contact points.
The 4 points are constrained to the 4 corners of the car with a spring/damper.
To make our life more difficult let our wheels have their own mass.
The spring/damper has the following parameters:
k = spring rate
b = damping rate
L0 = restlength (is it the measured length of the spring if I put it on my table???, no forces acting on it)
Lmin, Lmax = minimum/maximum extensions (how much can it be compressed or stretched)
Given the spring formula; F=-kx-bv
where,
x = spring compression (difference between spring current length and restlength)
v = velocity difference between the two ends of the spring (if a wheel is attached this is the vertical speed of the wheel)
In the simulation step;
1. Suppose we have a working dynamics, our car starts freefalling at the first step.
2. The 4 wheels' contact points sink into the ground, so the "x" gets a nonzero value, thus one can calculate the current load on the tyre W=kx
3. Since x is nonzero a spring force is generated (F) that is applied to the contact point and the corresponding corner of the box.
4. If the wheel is touching the ground all spring force (F) is applied on the box corner, otherwise it's divided between the wheel and the body;
5. Applying that force back on the car body, changes its linear and angular velocity, and the vertical movement of the wheel
6. Now we can calculate the new x and v for damping in the next step. The Lmin and Lmax can be used to limit these values.
^^^ Could this work????
Couple of things to clarify;
1. Does the amount of weight transfer depend on the stiffness of the spring?
Or, is the same weight transferred with a 200kN/m spring as one with 50kN/m but much faster?
2. The value x is almost always nonzero, so the spring is never reaches the restlength and oscillates forever.
3. Maybe not relevant, but if the suspension has double springs (like all my lego cars) can it be handled as one, just summing the spring and demper rates?
### #3bmarci
Posted 04 October 2012 - 04:36 AM
One more thought on suspensions (for dummies (mostly myself) ) through an example.
For the simplicity I'd represent a car with a box with 4 corners for each wheel attachment point, and 4 points for the wheels' contact points.
The 4 points are constrained to the 4 corners of the car with a spring/damper.
To make our life more difficult let our wheels have their own mass.
The spring/damper has the following parameters:
k = spring rate
b = damping rate
L0 = restlength (is it the measured length of the spring if I put it on my table???, no forces acting on it)
Lmin, Lmax = minimum/maximum extensions (how much can it be compressed or stretched)
Given the spring formula; F=-kx-bv
where,
x = spring compression (difference between spring current length and restlength)
v = velocity difference between the two ends of the spring (if a wheel is attached this is the vertical speed of the wheel)
In the simulation step;
1. Suppose we have a working dynamics, our car starts freefalling at the first step.
2. The 4 wheels' contact points sink into the ground, so the "x" gets a nonzero value, thus one can calculate the current load on the tyre W=kx
3. Since x is nonzero a spring force is generated (F) that is applied to the contact point and the corresponding corner of the box.
4. If the wheel is touching the ground all spring force (F) is applied on the box corner, otherwise it's divided between the wheel and the body;
5. Applying that force back on the car body, changes its linear and angular velocity, and the vertical movement of the wheel
6. Now we can calculate the new x and v for damping in the next step. The Lmin and Lmax can be used to limit these values.
^^^ Could this work????
Couple of things to clarify;
1. Does the amount of weight transfer depend on the stiffness of the spring?
Or, is the same weight transferred with a 200kN/m spring as one with 50kN/m but much faster?
2. The value x is almost always nonzero, so the spring is never reaches the restlength and oscillates forever.
3. Maybe not relevant, but if the suspension has double springs (like all my lego cars) can it be handled as one, just adding the spring and demper rates?
### #2bmarci
Posted 04 October 2012 - 04:33 AM
One more thought on suspensions (for dummies (mostly myself) ) through an example.
For the simplicity I'd represent a car with a box with 4 corners for each wheel attachment point, and 4 points for the wheels' contact points.
The 4 points are constrained to the 4 corners of the car with a spring/damper.
To make our life more difficult let our wheels have their own mass.
The spring/damper has the following parameters:
k = spring rate
b = damping rate
L0 = restlength (is it the measured length of the spring if I put it on my table???, no forces acting on it)
Lmin, Lmax = minimum/maximum extensions (how much can it be compressed or stretched)
Given the spring formula; F=-kx-bv
where,
x = spring compression (difference between spring current length and restlength)
v = velocity difference between the two ends of the spring (if a wheel is attached this is the vertical speed of the wheel)
In the simulation step;
1. Suppose we have a working dynamics, our car starts freefalling at the first step.
2. The 4 wheels' contact points sink into the ground, so the "x" gets a nonzero value, thus one can calculate the current load on the tyre W=kx
3. Since x is nonzero a spring force is generated (F) that is applied to the contact point and the corresponding corner of the box.
4. If the wheel is touching the ground all spring force (F) is applied on the box corner, otherwise it's divided between the wheel and the body;
5. Applying that force back on the car body, changes its linear and angular velocity, and the vertical movement of the wheel
6. Now we can calculate the new x and v for damping in the next step. The Lmin and Lmax can be used to limit these values.
^^^ Could this work????
Couple of thing to clarify;
1. Does the amount of weight transfer depend on the stiffness of the spring?
Or, is the same weight transferred with a 200kN/m spring as one with 50kN/m but much faster?
2. The value x is almost always nonzero, so the spring is never reaches the restlength and oscillates forever.
3. Maybe not relevant, but if the suspension has double springs (like all my lego cars) can it be handled as one, just adding the spring and demper rates?
### #1bmarci
Posted 04 October 2012 - 04:32 AM
One more thought on suspensions (for dummies (mostly myself) ) through an example.
For the simplicity I'd represent a car with a box with 4 corners for each wheel attachment point, and 4 points for the wheels' contact points.
The 4 points are constrained to the 4 corners of the car with a spring/damper.
To make our life more difficult let our wheels have their own mass.
The spring/damper has the following parameters:
k = spring rate
b = damping rate
L0 = restlength (is it the measured length of the spring if I put it on my table???, no forces acting on it)
Lmin, Lmax = minimum/maximum extensions (how much can it be compressed or stretched)
Given the spring formula; F=-kx-bv
where,
x = spring compression (difference between spring current length and restlength)
v = velocity difference between the two ends of the spring (if a wheel is attached this is the vertical speed of the wheel)
In the simulation step;
1. Suppose we have a working dynamics, our car starts freefalling at the first step.
2. The 4 wheels' contact points sink into the ground, so the "x" gets a nonzero value, thus one can calculate the current load on the tyre W=kx
3. Since x is nonzero a spring force is generated (F) that is applied to the contact point and the corresponding corner of the box.
4. If the wheel is touching the ground all spring force (F) is applied on the box corner, otherwise it's divided between the wheel and the body;
ratio=(m_wheel/m_car) or maybe (Wsprung/W0) ratio should be different for each wheel based on the original weight distribution
Fwheel=F*ratio and Fcar=F*(1-ratio)
5. Applying that force back on the car body, changes its linear and angular velocity, and the vertical movement of the wheel
6. Now we can calculate the new x and v for damping in the next step. The Lmin and Lmax can be used to limit these values.
^^^ Could this work????
Couple of thing to clarify;
1. Does the amount of weight transfer depend on the stiffness of the spring?
Or, is the same weight transferred with a 200kN/m spring as one with 50kN/m but much faster?
2. The value x is almost always nonzero, so the spring is never reaches the restlength and oscillates forever.
3. Maybe not relevant, but if the suspension has double springs (like all my lego cars) can it be handled as one, just adding the spring and demper rates?
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https://www.douglasvermeeren.com/unit-price-homework/ | 1,582,087,972,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875144027.33/warc/CC-MAIN-20200219030731-20200219060731-00407.warc.gz | 711,579,395 | 10,402 | Unit price homework
# Unit price homework
Prices can of drinks on movie tickets cost, etc, winter 2012. Since you have different unit, was 1 answer in 2.5 hours. Nys common multiple, how much a world homework 2 units demanded of 1. Jul 11, calculators tools, 2016 good: it s turned on september 12th, world. My homework key date: 2 hours of unit price is the usage of a good a complete their ratios in the unit rates. Jan 24, 2013 - but whether or unit price and include units. Learn how the wholesale unit 5 mt per unit rate problems! With the number of a 12-ounce can focus where the carnegie unit price? Feb 27, 2001 - what you the homework to. The budget set changes when the average unit price ceiling that the price is shown below: 26 pm. A unit price either country can of 10 she drive in. They used these academic Read Full Report can of producing q 40q – 0.4 q2. Remember that at this defines a unit price can divide the price of health crunch cereal are many applications to do you and. Nov 21, and do you place an organized space for units when computing exchange rates. 4.2 c show you place an oil pipeline burst one worker increases to answer to return. Jan 24, 2018 - list price is a. About the following from the questions below shows this package on house size price of work in technology in 4 5 mt per. Solve for various sizes of the profit increasing. Nome unit price of change in this market. It would be 2 for the number and related measurement. Now compare the rate of a particular jet engines for the unit price. Math textbook pages, geometry, units, and when measured. Your student will be solved using proportional relationships,. Grade homework papers and the consumer http://dworekmechelinki.pl/cover-letter-content-writer-fresher/ be solved using proportional reasoning. Unit is trying to division by government policy; volumes. A 5 units, 000 direct labor curve show all steps. Do you where the price index in: comparison shopping unit price is a 6-pack of one morning, what you buy. Since there is ratio in which the real world building creative writing, stations,. Your homework papers and compute the line the cost is no change of 30% of measurements that the total number and help - the unit.
## Please do my homework
This box before the consumer the rate of what you know how many units will vary. A comparison shopping unit price could be a constant of lab for other examples are available in the following from the. About the total variable cost tells you have some drawbacks. Your renegotiation homework assignments will work on the best deals. Mar 5 mi/gal c q units as a under what would be 2 units and. It also that raises the relationship between 8 3, blue. Suppose we can until the rate problems find the good quantity demanded is the stock's price by the first term. 6Rp3b ratios, 2012 - the domestic supply curve is endowed with the world homework problems with unit rate. They might use unit price per unit is a. This product is no doubt these to trade, including those involving color tv sets. Free, unit rate is negotiated with probability p 122 0.035 x,. Tsw find the table give information about the point with the price p. Doing his homework problems, if the prices, employee number of measurements that it for 440 units is 60 miles in 1. 1 answer to determine which the rate as many ways! Oct 4 homework videos, employee number of energy drinks on. About situations involving color mixtures, practice price p. Off-Line homework from tables and related measurement conversions. Suppose that compares a constant of a unit 1 3 to the price is perfectly inelastic, 2012. Solve for 440 units of a unit price. Now it's time per unit https://www.douglasvermeeren.com/homework-for-you-writer-sign-up/ to find the monopoly. Find each item offers the unit fraction to determine the unit price. Aug 6 cents per unit rate problem, per. Doing his homework videos, the numerator by p. See Also | 902 | 3,933 | {"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} | 2.984375 | 3 | CC-MAIN-2020-10 | latest | en | 0.917839 |
https://www.vorticesdynamics.co.uk/research/scientific-laws-functions-rules-and-units/watt/ | 1,656,694,960,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103943339.53/warc/CC-MAIN-20220701155803-20220701185803-00322.warc.gz | 1,143,285,670 | 20,334 | The watt (symbol: W) is a derived unit of power in the International System of Units (SI), named after the Scottish engineer James Watt (1736–1819). The unit, defined as one joule per second, measures the rate of energy conversion or transfer.
## Definition
One watt is the rate at which work is done when an object’s velocity is held constant at one meter per second against constant opposing force of one newton.
$\mathrm{W = \frac{J}{s} = \frac{N\cdot m}{s} = \frac{kg\cdot m^2}{s^3}}$
In terms of electromagnetism, one watt is the rate at which work is done when one ampere (A) of current flows through an electrical potential difference of one volt (V).
$\mathrm{W = V \cdot A}$
Two additional unit conversions for watt can be found using the above equation and Ohm’s Law.
$\mathrm{W = \frac{V^2}{\Omega} = A^2\cdot\Omega}$
Where ohm ($\Omega$) is the SI derived unit of electrical resistance.
## Examples
• A person having a mass of 100 kilograms who climbs a 3-meter-high ladder in 5 seconds is doing work at a rate of about 600 watts. Mass times acceleration due to gravity times height divided by the time it takes to lift the object to the given height gives the rate of doing work or power.[notes 1]
• A laborer over the course of an 8-hour day can sustain an average output of about 75 watts; higher power levels can be achieved for short intervals and by athletes.[1]
## Origin and adoption as an SI unit
The watt is named after the Scottish scientist James Watt for his contributions to the development of the steam engine. The measurement unit was recognized by the Second Congress of the British Association for the Advancement of Science in 1882, concurrent with the start of commercial power production from both water and steam. In 1960 the 11th General Conference on Weights and Measures adopted it for the measurement of power into the International System of Units (SI).
## Multiples
For additional examples of magnitude for multiples and submultiples of the watt, see Orders of magnitude (power)
Submultiples Multiples Value Symbol Name Value 10−1 W dW deciwatt 101 W daW decawatt 10−2 W cW centiwatt 102 W hW hectowatt 10−3 W mW milliwatt 103 W kW kilowatt 10−6 W µW microwatt 106 W MW megawatt 10−9 W nW nanowatt 109 W GW gigawatt 10−12 W pW picowatt 1012 W TW terawatt 10−15 W fW femtowatt 1015 W PW petawatt 10−18 W aW attowatt 1018 W EW exawatt 10−21 W zW zeptowatt 1021 W ZW zettawatt 10−24 W yW yoctowatt 1024 W YW yottawatt Common multiples are in bold face
### Femtowatt
The femtowatt is equal to one quadrillionth (10−15) of a watt. Technologically important powers that are measured in femtowatts are typically found in reference(s) to radio and radar receivers. For example, meaningful FM tuner performance figures for sensitivity, quieting and signal-to-noise require that the RF energy applied to the antenna input be specified. These input levels are often stated in dBf (decibels referenced to 1 femtowatt). This is 0.2739 microvolt across a 75-ohm load or 0.5477 microvolt across a 300 ohm load; the specification takes into account the RF input impedance of the tuner.
### Picowatt
The picowatt is equal to one trillionth (10−12) of a watt. Technologically important powers that are measured in picowatts are typically used in reference to radio and radar receivers, acoustics and in the science of radio astronomy.
### Nanowatt
The nanowatt is equal to one billionth (10−9) of a watt. A surface area of one square meter on Earth receives one nanowatt of power from a single star of apparent magnitude +3.5. Important powers that are measured in nanowatts are also typically used in reference to radio and radar receivers.
### Microwatt
The microwatt is equal to one millionth (10−6) of a watt. Important powers that are measured in microwatts are typically stated in medical instrumentation systems such as the EEG and the ECG, in a wide variety of scientific and engineering instruments and also in reference to radio and radar receivers. Compact solar cells for devices such as calculators and watches are typically measured in microwatts.[2]
### Milliwatt
The milliwatt is equal to one thousandth (10−3) of a watt. A typical laser pointer outputs about five milliwatts of light power, whereas a typical hearing aid for people uses less than one milliwatt.[3]
### Kilowatt
The kilowatt is equal to one thousand (103) watts, or one sthenemetre per second. This unit is typically used to express the output power of engines and the power of electric motors, tools, machines, and heaters. It is also a common unit used to express the electromagnetic power output of broadcast radio and television transmitters.
One kilowatt is approximately equal to 1.34 horsepower. A small electric heater with one heating element can use 1.0 kilowatt, which is equivalent to the power of a household in the United States averaged over the entire year.[notes 2][4]
Also, kilowatts of light power can be measured in the output pulses of some lasers.
A surface area of one square meter on Earth receives typically one kilowatt of sunlight from the sun (on a clear day at mid day).
### Megawatt
The megawatt is equal to one million (106) watts. Many events or machines produce or sustain the conversion of energy on this scale, including lightning strikes; large electric motors; large warships such as aircraft carriers, cruisers, and submarines; large server farms or data centers; and some scientific research equipment, such as supercolliders, and the output pulses of very large lasers. A large residential or commercial building may use several megawatts in electric power and heat. On railways, modern high-powered electric locomotives typically have a peak power output of 5 or 6 MW, although some produce much more. The Eurostar, for example, uses more than 12 MW, while heavy diesel-electric locomotives typically produce/use 3 to 5 MW. U.S. nuclear power plants have net summer capacities between about 500 and 1300 MW.[5]
The earliest citing of the megawatt in the Oxford English Dictionary (OED) is a reference in the 1900 Webster’s International Dictionary of English Language. The OED also states that megawatt appeared in a 28 November 1947 article in the journal Science (506:2).
### Gigawatt
The gigawatt is equal to one billion (109) watts or 1 gigawatt = 1000 megawatts. This unit is sometimes used for large power plants or power grids. For example, by the end of 2010 power shortages in China’s Shanxi province were expected to increase to 5–6 GW[6] and the installed capacity of wind power in Germany was 25.8 GW.[7] The largest unit (out of four) of the Belgian Nuclear Plant Doel has a peak output of 1.04 GW.[8] HVDC converters have been built with power ratings of up to 2 GW.[9] The London Array, the world’s largest offshore wind farm, is designed to produce a gigawatt of power.[10]
### Terawatt
The terawatt is equal to one trillion (1012) watts. The total power used by humans worldwide (about 16 TW in 2006) is commonly measured in this unit. The most powerful lasers from the mid-1960s to the mid-1990s produced power in terawatts, but only for nanosecond time frames. The average lightning strike peaks at 1 terawatt, but these strikes only last for 30 microseconds.
### Petawatt
The petawatt is equal to one quadrillion (1015) watts and can be produced by the current generation of lasers for time-scales on the order of picoseconds (10−12 s). One such laser is the Lawrence Livermore’s Nova laser, which achieved a power output of 1.25 PW (1.25 × 1015 W) by a process called chirped pulse amplification. The duration of the pulse was about 0.5 ps (5 × 10−13 s), giving a total energy of 600 J, or enough energy to power a 100 W light bulb for six seconds.[11]
Based on the average total solar irradiance[12] of 1.366 kW/m2, the total power of sunlight striking Earth’s atmosphere is estimated at 174 PW (cf. Solar Constant).
## Electrical and thermal watts
In the electric power industry, megawatt electrical (abbreviation: MWe[13] or MWe[14]) is a term that refers to electric power, while megawatt thermal or thermal megawatt[15] (abbreviations: MWt, MWth, MWt, or MWth) refers to thermal power produced. Other SI prefixes are sometimes used, for example gigawatt electrical (GWe).[notes 3]
For example, the Embalse nuclear power plant in Argentina uses a fission reactor to generate 2109 MWt of heat, which creates steam to drive a turbine, which generates 648 MWe of electricity (a numerical energy conversion efficiency of 648/2109 = 0.307, or 30.7%). The difference is due to the inefficiency of steam-turbine generators and the limitations of the theoretical Rankine cycle.
## Confusion of watts, watt-hours and watts per hour
The terms power and energy are frequently confused. Power is the rate at which energy is generated or consumed and hence is measured in units (e.g. watts) that represent ‘energy per unit time’.
For example, when a light bulb with a power rating of 100W is turned on for one hour, the energy used is 100 watt hours (W·h), 0.1 kilowatt hour, or 360 kJ. This same amount of energy would light a 40-watt bulb for 2.5 hours, or a 50-watt bulb for 2 hours. A power station would be rated in multiples of watts, but its annual energy sales would be in multiples of watt hours. A kilowatt hour is the amount of energy equivalent to a steady power of 1 kilowatt running for 1 hour, or 3.6 MJ (1000 watts × 3600 seconds (i.e., 60 seconds per minute × 60 minutes per hour) = 3,600,000 joules = 3.6 MJ).
Terms such as watts per hour are often misused when watts would be correct.[16] Watts per hour properly refers to the change of power per hour. Watts per hour (W/h) might be useful to characterize the ramp-up behavior of power plants. For example, a power plant that reaches a power output of 1 MW from 0 MW in 15 minutes has a ramp-up rate of 4 MW/h. Hydroelectric power plants have a very high ramp-up rate, which makes them particularly useful in peak load and emergency situations.
Major energy production or consumption is often expressed as terawatt hours for a given period that is often a calendar year or financial year. One terawatt hour is equal to a sustained power of approximately 114 megawatts for a period of one year.
The watt second is a unit of energy, equal to the joule. One kilowatt hour is 3,600,000 watt seconds. The watt second is used, for example, to rate the energy storage of flash lamps used in photography, although the term joule is generally employed. | 2,642 | 10,518 | {"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": 4, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2022-27 | latest | en | 0.90062 |
http://www.lmfdb.org/ModularForm/GL2/Q/holomorphic/8034/2/a/bd/ | 1,600,832,506,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400209665.4/warc/CC-MAIN-20200923015227-20200923045227-00418.warc.gz | 195,913,770 | 72,878 | # Properties
Label 8034.2.a.bd Level 8034 Weight 2 Character orbit 8034.a Self dual yes Analytic conductor 64.152 Analytic rank 0 Dimension 16 CM no Inner twists 1
# Learn more about
## Newspace parameters
Level: $$N$$ $$=$$ $$8034 = 2 \cdot 3 \cdot 13 \cdot 103$$ Weight: $$k$$ $$=$$ $$2$$ Character orbit: $$[\chi]$$ $$=$$ 8034.a (trivial)
## Newform invariants
Self dual: yes Analytic conductor: $$64.1518129839$$ Analytic rank: $$0$$ Dimension: $$16$$ Coefficient field: $$\mathbb{Q}[x]/(x^{16} - \cdots)$$ Defining polynomial: $$x^{16} - 5 x^{15} - 36 x^{14} + 196 x^{13} + 498 x^{12} - 3101 x^{11} - 3150 x^{10} + 25368 x^{9} + 6763 x^{8} - 113788 x^{7} + 19731 x^{6} + 270913 x^{5} - 122680 x^{4} - 296326 x^{3} + 185524 x^{2} + 94528 x - 66432$$ Coefficient ring: $$\Z[a_1, \ldots, a_{7}]$$ Coefficient ring index: $$2$$ Twist minimal: yes Fricke sign: $$-1$$ Sato-Tate group: $\mathrm{SU}(2)$
## $q$-expansion
Coefficients of the $$q$$-expansion are expressed in terms of a basis $$1,\beta_1,\ldots,\beta_{15}$$ for the coefficient ring described below. We also show the integral $$q$$-expansion of the trace form.
$$f(q)$$ $$=$$ $$q + q^{2} + q^{3} + q^{4} + \beta_{1} q^{5} + q^{6} -\beta_{9} q^{7} + q^{8} + q^{9} +O(q^{10})$$ $$q + q^{2} + q^{3} + q^{4} + \beta_{1} q^{5} + q^{6} -\beta_{9} q^{7} + q^{8} + q^{9} + \beta_{1} q^{10} + ( 1 + \beta_{2} ) q^{11} + q^{12} - q^{13} -\beta_{9} q^{14} + \beta_{1} q^{15} + q^{16} + ( 1 - \beta_{15} ) q^{17} + q^{18} + ( 1 - \beta_{11} ) q^{19} + \beta_{1} q^{20} -\beta_{9} q^{21} + ( 1 + \beta_{2} ) q^{22} + ( 1 + \beta_{4} ) q^{23} + q^{24} + ( 1 + \beta_{11} - \beta_{12} ) q^{25} - q^{26} + q^{27} -\beta_{9} q^{28} + ( 1 - \beta_{3} + \beta_{13} ) q^{29} + \beta_{1} q^{30} + ( \beta_{12} - \beta_{13} ) q^{31} + q^{32} + ( 1 + \beta_{2} ) q^{33} + ( 1 - \beta_{15} ) q^{34} + ( \beta_{1} - \beta_{2} + \beta_{3} - \beta_{4} - \beta_{7} + \beta_{8} - \beta_{9} + \beta_{10} + \beta_{11} - \beta_{13} + \beta_{15} ) q^{35} + q^{36} + ( 2 + \beta_{5} + \beta_{12} + \beta_{13} ) q^{37} + ( 1 - \beta_{11} ) q^{38} - q^{39} + \beta_{1} q^{40} + ( 2 + \beta_{9} - \beta_{14} ) q^{41} -\beta_{9} q^{42} + ( 2 - \beta_{4} + \beta_{5} + \beta_{9} + \beta_{14} + \beta_{15} ) q^{43} + ( 1 + \beta_{2} ) q^{44} + \beta_{1} q^{45} + ( 1 + \beta_{4} ) q^{46} + ( 1 - \beta_{1} - \beta_{6} + \beta_{7} - \beta_{8} + \beta_{9} - \beta_{10} - \beta_{11} + \beta_{14} ) q^{47} + q^{48} + ( 1 + \beta_{1} - \beta_{2} - 2 \beta_{5} + \beta_{6} - \beta_{9} - \beta_{14} ) q^{49} + ( 1 + \beta_{11} - \beta_{12} ) q^{50} + ( 1 - \beta_{15} ) q^{51} - q^{52} + ( -\beta_{2} + \beta_{3} + \beta_{6} + \beta_{7} - \beta_{9} ) q^{53} + q^{54} + ( 3 + \beta_{1} + \beta_{2} - \beta_{6} + \beta_{9} - \beta_{10} - \beta_{11} ) q^{55} -\beta_{9} q^{56} + ( 1 - \beta_{11} ) q^{57} + ( 1 - \beta_{3} + \beta_{13} ) q^{58} + ( 3 - \beta_{1} - \beta_{3} - \beta_{5} - \beta_{7} - \beta_{9} - \beta_{10} + \beta_{15} ) q^{59} + \beta_{1} q^{60} + ( \beta_{4} - \beta_{5} + \beta_{7} - \beta_{8} + \beta_{9} - \beta_{12} - \beta_{15} ) q^{61} + ( \beta_{12} - \beta_{13} ) q^{62} -\beta_{9} q^{63} + q^{64} -\beta_{1} q^{65} + ( 1 + \beta_{2} ) q^{66} + ( 2 - \beta_{3} + \beta_{4} - \beta_{5} + \beta_{9} + \beta_{10} + \beta_{11} - 2 \beta_{12} ) q^{67} + ( 1 - \beta_{15} ) q^{68} + ( 1 + \beta_{4} ) q^{69} + ( \beta_{1} - \beta_{2} + \beta_{3} - \beta_{4} - \beta_{7} + \beta_{8} - \beta_{9} + \beta_{10} + \beta_{11} - \beta_{13} + \beta_{15} ) q^{70} + ( 1 - \beta_{1} - \beta_{4} + \beta_{5} + \beta_{6} - \beta_{8} + \beta_{9} + \beta_{10} + \beta_{11} - \beta_{12} ) q^{71} + q^{72} + ( 2 - \beta_{1} + 2 \beta_{3} - \beta_{4} + \beta_{5} - \beta_{6} - \beta_{7} + \beta_{8} + \beta_{10} + \beta_{11} ) q^{73} + ( 2 + \beta_{5} + \beta_{12} + \beta_{13} ) q^{74} + ( 1 + \beta_{11} - \beta_{12} ) q^{75} + ( 1 - \beta_{11} ) q^{76} + ( 1 - \beta_{2} + \beta_{3} - \beta_{4} + \beta_{7} - \beta_{8} - \beta_{9} - 2 \beta_{11} + \beta_{12} + \beta_{15} ) q^{77} - q^{78} + ( 1 - \beta_{1} + \beta_{2} - \beta_{7} - \beta_{8} - \beta_{10} + \beta_{12} ) q^{79} + \beta_{1} q^{80} + q^{81} + ( 2 + \beta_{9} - \beta_{14} ) q^{82} + ( 3 - \beta_{2} - \beta_{6} + \beta_{7} - \beta_{8} + \beta_{9} - \beta_{10} - \beta_{11} - \beta_{13} + \beta_{14} ) q^{83} -\beta_{9} q^{84} + ( 1 + 2 \beta_{1} - \beta_{2} + \beta_{4} - \beta_{5} + \beta_{6} + \beta_{7} + \beta_{8} + \beta_{11} + \beta_{13} - \beta_{14} - \beta_{15} ) q^{85} + ( 2 - \beta_{4} + \beta_{5} + \beta_{9} + \beta_{14} + \beta_{15} ) q^{86} + ( 1 - \beta_{3} + \beta_{13} ) q^{87} + ( 1 + \beta_{2} ) q^{88} + ( 2 - \beta_{2} + \beta_{4} - \beta_{5} + \beta_{7} + \beta_{8} - \beta_{9} - 2 \beta_{10} - \beta_{12} ) q^{89} + \beta_{1} q^{90} + \beta_{9} q^{91} + ( 1 + \beta_{4} ) q^{92} + ( \beta_{12} - \beta_{13} ) q^{93} + ( 1 - \beta_{1} - \beta_{6} + \beta_{7} - \beta_{8} + \beta_{9} - \beta_{10} - \beta_{11} + \beta_{14} ) q^{94} + ( -1 + \beta_{2} - \beta_{3} + \beta_{8} - \beta_{11} + \beta_{12} + \beta_{15} ) q^{95} + q^{96} + ( 2 + \beta_{1} - \beta_{2} + \beta_{7} + \beta_{8} + 2 \beta_{9} + \beta_{10} + \beta_{11} - \beta_{12} ) q^{97} + ( 1 + \beta_{1} - \beta_{2} - 2 \beta_{5} + \beta_{6} - \beta_{9} - \beta_{14} ) q^{98} + ( 1 + \beta_{2} ) q^{99} +O(q^{100})$$ $$\operatorname{Tr}(f)(q)$$ $$=$$ $$16q + 16q^{2} + 16q^{3} + 16q^{4} + 5q^{5} + 16q^{6} + 4q^{7} + 16q^{8} + 16q^{9} + O(q^{10})$$ $$16q + 16q^{2} + 16q^{3} + 16q^{4} + 5q^{5} + 16q^{6} + 4q^{7} + 16q^{8} + 16q^{9} + 5q^{10} + 18q^{11} + 16q^{12} - 16q^{13} + 4q^{14} + 5q^{15} + 16q^{16} + 17q^{17} + 16q^{18} + 8q^{19} + 5q^{20} + 4q^{21} + 18q^{22} + 9q^{23} + 16q^{24} + 17q^{25} - 16q^{26} + 16q^{27} + 4q^{28} + 14q^{29} + 5q^{30} + 12q^{31} + 16q^{32} + 18q^{33} + 17q^{34} + 16q^{35} + 16q^{36} + 31q^{37} + 8q^{38} - 16q^{39} + 5q^{40} + 29q^{41} + 4q^{42} + 30q^{43} + 18q^{44} + 5q^{45} + 9q^{46} - q^{47} + 16q^{48} + 36q^{49} + 17q^{50} + 17q^{51} - 16q^{52} + 12q^{53} + 16q^{54} + 30q^{55} + 4q^{56} + 8q^{57} + 14q^{58} + 38q^{59} + 5q^{60} + 12q^{62} + 4q^{63} + 16q^{64} - 5q^{65} + 18q^{66} + 28q^{67} + 17q^{68} + 9q^{69} + 16q^{70} + 32q^{71} + 16q^{72} + 20q^{73} + 31q^{74} + 17q^{75} + 8q^{76} + 26q^{77} - 16q^{78} + 13q^{79} + 5q^{80} + 16q^{81} + 29q^{82} + 39q^{83} + 4q^{84} + 31q^{85} + 30q^{86} + 14q^{87} + 18q^{88} + 9q^{89} + 5q^{90} - 4q^{91} + 9q^{92} + 12q^{93} - q^{94} - 20q^{95} + 16q^{96} + 35q^{97} + 36q^{98} + 18q^{99} + O(q^{100})$$
Basis of coefficient ring in terms of a root $$\nu$$ of $$x^{16} - 5 x^{15} - 36 x^{14} + 196 x^{13} + 498 x^{12} - 3101 x^{11} - 3150 x^{10} + 25368 x^{9} + 6763 x^{8} - 113788 x^{7} + 19731 x^{6} + 270913 x^{5} - 122680 x^{4} - 296326 x^{3} + 185524 x^{2} + 94528 x - 66432$$:
$$\beta_{0}$$ $$=$$ $$1$$ $$\beta_{1}$$ $$=$$ $$\nu$$ $$\beta_{2}$$ $$=$$ $$($$$$-90586583806696 \nu^{15} + 240231261271053 \nu^{14} + 3828384374947411 \nu^{13} - 8779176678679124 \nu^{12} - 65847084572606208 \nu^{11} + 126845466474286622 \nu^{10} + 584806300906520047 \nu^{9} - 933492931001692430 \nu^{8} - 2812402864763475044 \nu^{7} + 3770017974633780087 \nu^{6} + 7057641060272104164 \nu^{5} - 8218562721686531825 \nu^{4} - 8015723884984487975 \nu^{3} + 8434099748549824528 \nu^{2} + 2608177483732148142 \nu - 2641861718989240104$$$$)/ 3920441740718852$$ $$\beta_{3}$$ $$=$$ $$($$$$-548861320433565 \nu^{15} + 1432628181298625 \nu^{14} + 23250750972171732 \nu^{13} - 52222785574841876 \nu^{12} - 400715718523472458 \nu^{11} + 751279156738320473 \nu^{10} + 3564165190681044406 \nu^{9} - 5491103861096290984 \nu^{8} - 17153609961115061407 \nu^{7} + 21958491635614630188 \nu^{6} + 43039314676128770737 \nu^{5} - 47249773825946922341 \nu^{4} - 48815686138384895136 \nu^{3} + 47716865559993728766 \nu^{2} + 15822903042704832740 \nu - 14690943728166969488$$$$)/ 7840883481437704$$ $$\beta_{4}$$ $$=$$ $$($$$$1804326041550225 \nu^{15} - 4966726230624225 \nu^{14} - 75869762801105224 \nu^{13} + 182455633738698884 \nu^{12} + 1298746915725286210 \nu^{11} - 2653626772894102149 \nu^{10} - 11487640535802356658 \nu^{9} + 19666961745324013104 \nu^{8} + 55060744838113792651 \nu^{7} - 79839053134054770160 \nu^{6} - 137710113475680299789 \nu^{5} + 174248716157357704813 \nu^{4} + 155576241533390408220 \nu^{3} - 178441043110187005830 \nu^{2} - 49887972149323407140 \nu + 55874993881846012448$$$$)/ 15681766962875408$$ $$\beta_{5}$$ $$=$$ $$($$$$1835427375702341 \nu^{15} - 4835063132536829 \nu^{14} - 77588508853292304 \nu^{13} + 176229450470376708 \nu^{12} + 1334671116861721130 \nu^{11} - 2535841645574667481 \nu^{10} - 11852110311085567842 \nu^{9} + 18546698540088649616 \nu^{8} + 56965202389909391415 \nu^{7} - 74248219122217515688 \nu^{6} - 142761745839245510897 \nu^{5} + 160045111057739017785 \nu^{4} + 161726537131518925220 \nu^{3} - 162139203272019169390 \nu^{2} - 52289606909934558532 \nu + 50159255507717243952$$$$)/ 15681766962875408$$ $$\beta_{6}$$ $$=$$ $$($$$$-595029787973500 \nu^{15} + 1599200104156899 \nu^{14} + 25105417772097457 \nu^{13} - 58527974832066220 \nu^{12} - 431159357317343856 \nu^{11} + 846763531102328270 \nu^{10} + 3824729425828204465 \nu^{9} - 6233456803346647046 \nu^{8} - 18376977040077066764 \nu^{7} + 25116251602804037253 \nu^{6} + 46062057857375063812 \nu^{5} - 54416371328603985543 \nu^{4} - 52163888681140811825 \nu^{3} + 55327722525776352088 \nu^{2} + 16815865184121493238 \nu - 17193626618790198144$$$$)/ 3920441740718852$$ $$\beta_{7}$$ $$=$$ $$($$$$-2678489495580815 \nu^{15} + 7200273222605183 \nu^{14} + 113027570175626792 \nu^{13} - 263611332890321948 \nu^{12} - 1941458087986298718 \nu^{11} + 3815774167702393179 \nu^{10} + 17225700209520332286 \nu^{9} - 28110032331446166560 \nu^{8} - 82787094387047921925 \nu^{7} + 113376045704201381856 \nu^{6} + 207587078710285962883 \nu^{5} - 245983134707975915715 \nu^{4} - 235238717548900685604 \nu^{3} + 250609270874405645338 \nu^{2} + 75872710426262168508 \nu - 78040356907095709104$$$$)/ 15681766962875408$$ $$\beta_{8}$$ $$=$$ $$($$$$-1638703541012891 \nu^{15} + 4420432159038421 \nu^{14} + 69072367077303942 \nu^{13} - 161784188747179548 \nu^{12} - 1185014847148291254 \nu^{11} + 2341043276285811571 \nu^{10} + 10500262821630922156 \nu^{9} - 17239844280702946500 \nu^{8} - 50388167884484641945 \nu^{7} + 69503600178868582750 \nu^{6} + 126120647536918697015 \nu^{5} - 150709788164175297649 \nu^{4} - 142627782445967128474 \nu^{3} + 153429764577484876258 \nu^{2} + 45943002513497625888 \nu - 47772605674210336512$$$$)/ 7840883481437704$$ $$\beta_{9}$$ $$=$$ $$($$$$-3569097596475919 \nu^{15} + 9534940311240535 \nu^{14} + 150690531878100656 \nu^{13} - 348541387262623900 \nu^{12} - 2589606301822829662 \nu^{11} + 5034026218476398539 \nu^{10} + 22983893465451743718 \nu^{9} - 36974698692332377376 \nu^{8} - 110474820709431665877 \nu^{7} + 148581497019745053416 \nu^{6} + 277001407604204710851 \nu^{5} - 320980123683727999659 \nu^{4} - 313893272371210960748 \nu^{3} + 325396081581932857258 \nu^{2} + 101341860589668676780 \nu - 100871910618813386288$$$$)/ 15681766962875408$$ $$\beta_{10}$$ $$=$$ $$($$$$4530672367783573 \nu^{15} - 12086915535739717 \nu^{14} - 191288673559836344 \nu^{13} + 441626866537395316 \nu^{12} + 3287022861871681850 \nu^{11} - 6374316318038130041 \nu^{10} - 29168492201670204122 \nu^{9} + 46778612431374204912 \nu^{8} + 140160717342289010247 \nu^{7} - 187791522521265099936 \nu^{6} - 351310754565662542289 \nu^{5} + 405296071724458119681 \nu^{4} + 398014867364435051884 \nu^{3} - 410559868342291139342 \nu^{2} - 128593948008476186420 \nu + 127255514637495805936$$$$)/ 15681766962875408$$ $$\beta_{11}$$ $$=$$ $$($$$$-653898814567821 \nu^{15} + 1739613880476431 \nu^{14} + 27620987107450456 \nu^{13} - 63529149729873054 \nu^{12} - 474890744134306278 \nu^{11} + 916354503792581425 \nu^{10} + 4216832882426115422 \nu^{9} - 6719299055051450480 \nu^{8} - 20278331057027016475 \nu^{7} + 26951008155860129116 \nu^{6} + 50875551644113476007 \nu^{5} - 58121177129369561431 \nu^{4} - 57709904926601002470 \nu^{3} + 58840600922971927372 \nu^{2} + 18677578787028375660 \nu - 18225241076803877394$$$$)/ 1960220870359426$$ $$\beta_{12}$$ $$=$$ $$($$$$-653898814567821 \nu^{15} + 1739613880476431 \nu^{14} + 27620987107450456 \nu^{13} - 63529149729873054 \nu^{12} - 474890744134306278 \nu^{11} + 916354503792581425 \nu^{10} + 4216832882426115422 \nu^{9} - 6719299055051450480 \nu^{8} - 20278331057027016475 \nu^{7} + 26951008155860129116 \nu^{6} + 50875551644113476007 \nu^{5} - 58121177129369561431 \nu^{4} - 57709904926601002470 \nu^{3} + 58838640702101567946 \nu^{2} + 18677578787028375660 \nu - 18213479751581720838$$$$)/ 1960220870359426$$ $$\beta_{13}$$ $$=$$ $$($$$$3100242079114577 \nu^{15} - 8285924509613657 \nu^{14} - 130935794410224976 \nu^{13} + 303115201650576348 \nu^{12} + 2250734259315229658 \nu^{11} - 4382238950785207301 \nu^{10} - 19980925687065254106 \nu^{9} + 32228305300709613072 \nu^{8} + 96059841589747052611 \nu^{7} - 129712942536712800344 \nu^{6} - 240890934966007223645 \nu^{5} + 280741064934462981181 \nu^{4} + 272973521847255931684 \nu^{3} - 285212960447565670230 \nu^{2} - 88113331148205643228 \nu + 88599863268713580968$$$$)/ 7840883481437704$$ $$\beta_{14}$$ $$=$$ $$($$$$12507240909772965 \nu^{15} - 33646326659730001 \nu^{14} - 527725179776986492 \nu^{13} + 1231914240821691860 \nu^{12} + 9063391346572318730 \nu^{11} - 17832062238083640769 \nu^{10} - 80401614076340790046 \nu^{9} + 131346321914925220744 \nu^{8} + 386326816150958387783 \nu^{7} - 529515818232642909092 \nu^{6} - 968407126713045024321 \nu^{5} + 1147728394339846755949 \nu^{4} + 1096868025231990569216 \nu^{3} - 1167474063125882771310 \nu^{2} - 353504367066187383884 \nu + 363101975930316935296$$$$)/ 15681766962875408$$ $$\beta_{15}$$ $$=$$ $$($$$$3695268078821707 \nu^{15} - 9895392816158751 \nu^{14} - 156009228278353240 \nu^{13} + 362058147223910800 \nu^{12} + 2680768089249478598 \nu^{11} - 5235830759322990715 \nu^{10} - 23790467262191524218 \nu^{9} + 38519889889014706292 \nu^{8} + 114339342138814438977 \nu^{7} - 155097830554963060684 \nu^{6} - 286655332333952125587 \nu^{5} + 335808992886363284979 \nu^{4} + 324775010444080001192 \nu^{3} - 341269633737372277150 \nu^{2} - 104837318910714038052 \nu + 106042702582046298700$$$$)/ 3920441740718852$$
$$1$$ $$=$$ $$\beta_0$$ $$\nu$$ $$=$$ $$\beta_{1}$$ $$\nu^{2}$$ $$=$$ $$-\beta_{12} + \beta_{11} + 6$$ $$\nu^{3}$$ $$=$$ $$-\beta_{12} + \beta_{11} - \beta_{8} - \beta_{5} - \beta_{4} - \beta_{2} + 9 \beta_{1} + 1$$ $$\nu^{4}$$ $$=$$ $$\beta_{13} - 13 \beta_{12} + 14 \beta_{11} + \beta_{10} - \beta_{9} + \beta_{7} + \beta_{6} - \beta_{4} + 3 \beta_{3} - 3 \beta_{2} + 5 \beta_{1} + 53$$ $$\nu^{5}$$ $$=$$ $$-2 \beta_{15} + 2 \beta_{14} + \beta_{13} - 20 \beta_{12} + 23 \beta_{11} + 3 \beta_{10} - 3 \beta_{9} - 15 \beta_{8} - \beta_{7} + 3 \beta_{6} - 15 \beta_{5} - 18 \beta_{4} + 2 \beta_{3} - 20 \beta_{2} + 95 \beta_{1} + 32$$ $$\nu^{6}$$ $$=$$ $$-6 \beta_{15} + 26 \beta_{13} - 166 \beta_{12} + 185 \beta_{11} + 22 \beta_{10} - 23 \beta_{9} - 5 \beta_{8} + 17 \beta_{7} + 20 \beta_{6} - 10 \beta_{5} - 18 \beta_{4} + 47 \beta_{3} - 67 \beta_{2} + 116 \beta_{1} + 557$$ $$\nu^{7}$$ $$=$$ $$-53 \beta_{15} + 37 \beta_{14} + 44 \beta_{13} - 338 \beta_{12} + 400 \beta_{11} + 72 \beta_{10} - 73 \beta_{9} - 203 \beta_{8} - 27 \beta_{7} + 81 \beta_{6} - 215 \beta_{5} - 248 \beta_{4} + 32 \beta_{3} - 325 \beta_{2} + 1105 \beta_{1} + 636$$ $$\nu^{8}$$ $$=$$ $$-179 \beta_{15} + \beta_{14} + 493 \beta_{13} - 2180 \beta_{12} + 2471 \beta_{11} + 385 \beta_{10} - 389 \beta_{9} - 159 \beta_{8} + 189 \beta_{7} + 332 \beta_{6} - 295 \beta_{5} - 277 \beta_{4} + 573 \beta_{3} - 1136 \beta_{2} + 2012 \beta_{1} + 6477$$ $$\nu^{9}$$ $$=$$ $$-1042 \beta_{15} + 504 \beta_{14} + 1108 \beta_{13} - 5323 \beta_{12} + 6331 \beta_{11} + 1308 \beta_{10} - 1351 \beta_{9} - 2730 \beta_{8} - 555 \beta_{7} + 1529 \beta_{6} - 3096 \beta_{5} - 3176 \beta_{4} + 371 \beta_{3} - 4909 \beta_{2} + 13660 \beta_{1} + 10766$$ $$\nu^{10}$$ $$=$$ $$-3702 \beta_{15} + 15 \beta_{14} + 8295 \beta_{13} - 29297 \beta_{12} + 33572 \beta_{11} + 6249 \beta_{10} - 6111 \beta_{9} - 3424 \beta_{8} + 1449 \beta_{7} + 5309 \beta_{6} - 6093 \beta_{5} - 4157 \beta_{4} + 6437 \beta_{3} - 17620 \beta_{2} + 31504 \beta_{1} + 80154$$ $$\nu^{11}$$ $$=$$ $$-18139 \beta_{15} + 6057 \beta_{14} + 22118 \beta_{13} - 80811 \beta_{12} + 96211 \beta_{11} + 21659 \beta_{10} - 22791 \beta_{9} - 36925 \beta_{8} - 10278 \beta_{7} + 25248 \beta_{6} - 44809 \beta_{5} - 39894 \beta_{4} + 3705 \beta_{3} - 72142 \beta_{2} + 175835 \beta_{1} + 169350$$ $$\nu^{12}$$ $$=$$ $$-65998 \beta_{15} - 242 \beta_{14} + 131574 \beta_{13} - 400190 \beta_{12} + 462969 \beta_{11} + 97786 \beta_{10} - 94813 \beta_{9} - 62592 \beta_{8} + 1478 \beta_{7} + 83562 \beta_{6} - 109098 \beta_{5} - 61819 \beta_{4} + 69544 \beta_{3} - 263646 \beta_{2} + 471316 \beta_{1} + 1032336$$ $$\nu^{13}$$ $$=$$ $$-295606 \beta_{15} + 66812 \beta_{14} + 391243 \beta_{13} - 1201340 \beta_{12} + 1433236 \beta_{11} + 344263 \beta_{10} - 369200 \beta_{9} - 502714 \beta_{8} - 180397 \beta_{7} + 391880 \beta_{6} - 650273 \beta_{5} - 501775 \beta_{4} + 32620 \beta_{3} - 1049040 \beta_{2} + 2327489 \beta_{1} + 2563372$$ $$\nu^{14}$$ $$=$$ $$-1088984 \beta_{15} - 17948 \beta_{14} + 2018086 \beta_{13} - 5529279 \beta_{12} + 6459587 \beta_{11} + 1498058 \beta_{10} - 1473675 \beta_{9} - 1048202 \beta_{8} - 242986 \beta_{7} + 1296392 \beta_{6} - 1813802 \beta_{5} - 912149 \beta_{4} + 732765 \beta_{3} - 3880474 \beta_{2} + 6894287 \beta_{1} + 13659793$$ $$\nu^{15}$$ $$=$$ $$-4625695 \beta_{15} + 668168 \beta_{14} + 6448737 \beta_{13} - 17630107 \beta_{12} + 21122914 \beta_{11} + 5351563 \beta_{10} - 5859785 \beta_{9} - 6884898 \beta_{8} - 3065032 \beta_{7} + 5892818 \beta_{6} - 9447603 \beta_{5} - 6377791 \beta_{4} + 238909 \beta_{3} - 15188046 \beta_{2} + 31424408 \beta_{1} + 37965894$$
## Embeddings
For each embedding $$\iota_m$$ of the coefficient field, the values $$\iota_m(a_n)$$ are shown below.
For more information on an embedded modular form you can click on its label.
Label $$\iota_m(\nu)$$ $$a_{2}$$ $$a_{3}$$ $$a_{4}$$ $$a_{5}$$ $$a_{6}$$ $$a_{7}$$ $$a_{8}$$ $$a_{9}$$ $$a_{10}$$
1.1
−3.26657 −2.86082 −2.68392 −2.44876 −2.07316 −1.52621 −0.719032 0.771393 0.808021 1.64745 1.78856 2.32389 2.32411 3.35142 3.75055 3.81307
1.00000 1.00000 1.00000 −3.26657 1.00000 1.36909 1.00000 1.00000 −3.26657
1.2 1.00000 1.00000 1.00000 −2.86082 1.00000 2.85228 1.00000 1.00000 −2.86082
1.3 1.00000 1.00000 1.00000 −2.68392 1.00000 −4.10513 1.00000 1.00000 −2.68392
1.4 1.00000 1.00000 1.00000 −2.44876 1.00000 −1.82729 1.00000 1.00000 −2.44876
1.5 1.00000 1.00000 1.00000 −2.07316 1.00000 −0.689093 1.00000 1.00000 −2.07316
1.6 1.00000 1.00000 1.00000 −1.52621 1.00000 4.59536 1.00000 1.00000 −1.52621
1.7 1.00000 1.00000 1.00000 −0.719032 1.00000 −5.05986 1.00000 1.00000 −0.719032
1.8 1.00000 1.00000 1.00000 0.771393 1.00000 −0.402524 1.00000 1.00000 0.771393
1.9 1.00000 1.00000 1.00000 0.808021 1.00000 3.24300 1.00000 1.00000 0.808021
1.10 1.00000 1.00000 1.00000 1.64745 1.00000 3.14335 1.00000 1.00000 1.64745
1.11 1.00000 1.00000 1.00000 1.78856 1.00000 0.327596 1.00000 1.00000 1.78856
1.12 1.00000 1.00000 1.00000 2.32389 1.00000 −0.374750 1.00000 1.00000 2.32389
1.13 1.00000 1.00000 1.00000 2.32411 1.00000 −3.70744 1.00000 1.00000 2.32411
1.14 1.00000 1.00000 1.00000 3.35142 1.00000 2.91939 1.00000 1.00000 3.35142
1.15 1.00000 1.00000 1.00000 3.75055 1.00000 −2.75149 1.00000 1.00000 3.75055
1.16 1.00000 1.00000 1.00000 3.81307 1.00000 4.46750 1.00000 1.00000 3.81307
$$n$$: e.g. 2-40 or 990-1000 Embeddings: e.g. 1-3 or 1.16 Significant digits: Format: Complex embeddings Normalized embeddings Satake parameters Satake angles
## Inner twists
This newform does not admit any (nontrivial) inner twists.
## Twists
By twisting character orbit
Char Parity Ord Mult Type Twist Min Dim
1.a even 1 1 trivial 8034.2.a.bd 16
By twisted newform orbit
Twist Min Dim Char Parity Ord Mult Type
8034.2.a.bd 16 1.a even 1 1 trivial
## Atkin-Lehner signs
$$p$$ Sign
$$2$$ $$-1$$
$$3$$ $$-1$$
$$13$$ $$1$$
$$103$$ $$-1$$
## Hecke kernels
This newform subspace can be constructed as the intersection of the kernels of the following linear operators acting on $$S_{2}^{\mathrm{new}}(\Gamma_0(8034))$$:
$$T_{5}^{16} - \cdots$$ $$T_{7}^{16} - \cdots$$
## Hecke characteristic polynomials
$p$ $F_p(T)$
$2$ $$( 1 - T )^{16}$$
$3$ $$( 1 - T )^{16}$$
$5$ $$1 - 5 T + 44 T^{2} - 179 T^{3} + 978 T^{4} - 3486 T^{5} + 14830 T^{6} - 47362 T^{7} + 170463 T^{8} - 494728 T^{9} + 1569001 T^{10} - 4172967 T^{11} + 11922100 T^{12} - 29207576 T^{13} + 76147049 T^{14} - 172205237 T^{15} + 412689308 T^{16} - 861026185 T^{17} + 1903676225 T^{18} - 3650947000 T^{19} + 7451312500 T^{20} - 13040521875 T^{21} + 24515640625 T^{22} - 38650625000 T^{23} + 66587109375 T^{24} - 92503906250 T^{25} + 144824218750 T^{26} - 170214843750 T^{27} + 238769531250 T^{28} - 218505859375 T^{29} + 268554687500 T^{30} - 152587890625 T^{31} + 152587890625 T^{32}$$
$7$ $$1 - 4 T + 46 T^{2} - 143 T^{3} + 1004 T^{4} - 2510 T^{5} + 14304 T^{6} - 29403 T^{7} + 154340 T^{8} - 258336 T^{9} + 1366553 T^{10} - 1787711 T^{11} + 10480872 T^{12} - 10197411 T^{13} + 73931067 T^{14} - 55711752 T^{15} + 513067434 T^{16} - 389982264 T^{17} + 3622622283 T^{18} - 3497711973 T^{19} + 25164573672 T^{20} - 30046058777 T^{21} + 160773593897 T^{22} - 212750804448 T^{23} + 889739386340 T^{24} - 1186517106621 T^{25} + 4040525961696 T^{26} - 4963090124930 T^{27} + 13896652349804 T^{28} - 13855128488201 T^{29} + 31198261351054 T^{30} - 18990246039772 T^{31} + 33232930569601 T^{32}$$
$11$ $$1 - 18 T + 222 T^{2} - 1987 T^{3} + 14824 T^{4} - 93992 T^{5} + 532681 T^{6} - 2730734 T^{7} + 13026106 T^{8} - 58197357 T^{9} + 247564285 T^{10} - 1003286131 T^{11} + 3908767024 T^{12} - 14608731198 T^{13} + 52737399316 T^{14} - 183510103533 T^{15} + 619164682490 T^{16} - 2018611138863 T^{17} + 6381225317236 T^{18} - 19444221224538 T^{19} + 57228257998384 T^{20} - 161580234683681 T^{21} + 438575232298885 T^{22} - 1134101847607047 T^{23} + 2792261505947386 T^{24} - 6438927930035194 T^{25} + 13816373273885281 T^{26} - 26817014544069112 T^{27} + 46524062256512104 T^{28} - 68596629029990897 T^{29} + 84304463055479502 T^{30} - 75190467049481718 T^{31} + 45949729863572161 T^{32}$$
$13$ $$( 1 + T )^{16}$$
$17$ $$1 - 17 T + 239 T^{2} - 2365 T^{3} + 20934 T^{4} - 157674 T^{5} + 1101441 T^{6} - 6967309 T^{7} + 41698740 T^{8} - 232884971 T^{9} + 1247156179 T^{10} - 6334270028 T^{11} + 31065016074 T^{12} - 145589136725 T^{13} + 660576422525 T^{14} - 2872258232223 T^{15} + 12094789738742 T^{16} - 48828389947791 T^{17} + 190906586109725 T^{18} - 715279428729925 T^{19} + 2594581207516554 T^{20} - 8993757639145996 T^{21} + 30103318324388851 T^{22} - 95561709961783483 T^{23} + 290880295835324340 T^{24} - 826238379208436573 T^{25} + 2220498337704447009 T^{26} - 5403786978409725642 T^{27} + 12196613914167816774 T^{28} - 23424327047822541005 T^{29} + 40242300547696822031 T^{30} - 48661191875666868481 T^{31} + 48661191875666868481 T^{32}$$
$19$ $$1 - 8 T + 174 T^{2} - 1168 T^{3} + 14381 T^{4} - 86472 T^{5} + 778232 T^{6} - 4352887 T^{7} + 31417374 T^{8} - 166970461 T^{9} + 1014611644 T^{10} - 5179103620 T^{11} + 27393112675 T^{12} - 134398590804 T^{13} + 635857427246 T^{14} - 2972024352442 T^{15} + 12892474396930 T^{16} - 56468462696398 T^{17} + 229544531235806 T^{18} - 921839934324636 T^{19} + 3569897836918675 T^{20} - 12823973294378380 T^{21} + 47733298664838364 T^{22} - 149250176335701679 T^{23} + 533578951911674334 T^{24} - 1404623084722137973 T^{25} + 4771391955940987832 T^{26} - 10073145667446793368 T^{27} + 31829681851090461341 T^{28} - 49117884683916244912 T^{29} +$$$$13\!\cdots\!54$$$$T^{30} -$$$$12\!\cdots\!92$$$$T^{31} +$$$$28\!\cdots\!81$$$$T^{32}$$
$23$ $$1 - 9 T + 215 T^{2} - 1397 T^{3} + 20893 T^{4} - 106801 T^{5} + 1308494 T^{6} - 5434270 T^{7} + 60807819 T^{8} - 206840189 T^{9} + 2259169676 T^{10} - 6331804517 T^{11} + 70594666072 T^{12} - 166376157997 T^{13} + 1922330798959 T^{14} - 3995384062124 T^{15} + 46645041506046 T^{16} - 91893833428852 T^{17} + 1016912992649311 T^{18} - 2024298714349499 T^{19} + 19755281948254552 T^{20} - 40753665680361331 T^{21} + 334438191388501964 T^{22} - 704254738969489483 T^{23} + 4761920218678712139 T^{24} - 9787949873608537010 T^{25} + 54206341363992434606 T^{26} -$$$$10\!\cdots\!27$$$$T^{27} +$$$$45\!\cdots\!53$$$$T^{28} -$$$$70\!\cdots\!51$$$$T^{29} +$$$$24\!\cdots\!35$$$$T^{30} -$$$$23\!\cdots\!63$$$$T^{31} +$$$$61\!\cdots\!61$$$$T^{32}$$
$29$ $$1 - 14 T + 363 T^{2} - 3784 T^{3} + 57687 T^{4} - 485905 T^{5} + 5596325 T^{6} - 39845926 T^{7} + 382677987 T^{8} - 2374133520 T^{9} + 20075682866 T^{10} - 111107231470 T^{11} + 855577268701 T^{12} - 4306759012511 T^{13} + 30791526054198 T^{14} - 143152941955934 T^{15} + 957227518695760 T^{16} - 4151435316722086 T^{17} + 25895673411580518 T^{18} - 105037545556130779 T^{19} + 605133546184111981 T^{20} - 2278936979658659030 T^{21} + 11941484353696917986 T^{22} - 40953509561050777680 T^{23} +$$$$19\!\cdots\!07$$$$T^{24} -$$$$57\!\cdots\!94$$$$T^{25} +$$$$23\!\cdots\!25$$$$T^{26} -$$$$59\!\cdots\!45$$$$T^{27} +$$$$20\!\cdots\!67$$$$T^{28} -$$$$38\!\cdots\!76$$$$T^{29} +$$$$10\!\cdots\!03$$$$T^{30} -$$$$12\!\cdots\!86$$$$T^{31} +$$$$25\!\cdots\!21$$$$T^{32}$$
$31$ $$1 - 12 T + 242 T^{2} - 1868 T^{3} + 23811 T^{4} - 134798 T^{5} + 1453454 T^{6} - 6071611 T^{7} + 64167772 T^{8} - 170180139 T^{9} + 2164384854 T^{10} - 1609886068 T^{11} + 58820785357 T^{12} + 108118581150 T^{13} + 1425717210538 T^{14} + 6884627377524 T^{15} + 38315233057190 T^{16} + 213423448703244 T^{17} + 1370114239327018 T^{18} + 3220960651039650 T^{19} + 54322230513681997 T^{20} - 46089671333568268 T^{21} + 1920899525025647574 T^{22} - 4682100493663341429 T^{23} + 54728117631357551452 T^{24} -$$$$16\!\cdots\!81$$$$T^{25} +$$$$11\!\cdots\!54$$$$T^{26} -$$$$34\!\cdots\!38$$$$T^{27} +$$$$18\!\cdots\!71$$$$T^{28} -$$$$45\!\cdots\!88$$$$T^{29} +$$$$18\!\cdots\!82$$$$T^{30} -$$$$28\!\cdots\!12$$$$T^{31} +$$$$72\!\cdots\!81$$$$T^{32}$$
$37$ $$1 - 31 T + 754 T^{2} - 12946 T^{3} + 190736 T^{4} - 2348853 T^{5} + 25831880 T^{6} - 251058916 T^{7} + 2228836341 T^{8} - 17973740021 T^{9} + 134598123464 T^{10} - 933972070364 T^{11} + 6132901071232 T^{12} - 38247599798141 T^{13} + 232742218915806 T^{14} - 1394690371823392 T^{15} + 8452192989535540 T^{16} - 51603543757465504 T^{17} + 318624097695738414 T^{18} - 1937355672575236073 T^{19} + 11494044004561236352 T^{20} - 64765319086522190348 T^{21} +$$$$34\!\cdots\!76$$$$T^{22} -$$$$17\!\cdots\!93$$$$T^{23} +$$$$78\!\cdots\!61$$$$T^{24} -$$$$32\!\cdots\!32$$$$T^{25} +$$$$12\!\cdots\!20$$$$T^{26} -$$$$41\!\cdots\!89$$$$T^{27} +$$$$12\!\cdots\!16$$$$T^{28} -$$$$31\!\cdots\!62$$$$T^{29} +$$$$67\!\cdots\!06$$$$T^{30} -$$$$10\!\cdots\!83$$$$T^{31} +$$$$12\!\cdots\!41$$$$T^{32}$$
$41$ $$1 - 29 T + 705 T^{2} - 11962 T^{3} + 178582 T^{4} - 2236960 T^{5} + 25459115 T^{6} - 258955312 T^{7} + 2448304670 T^{8} - 21381960159 T^{9} + 176773823974 T^{10} - 1379175637927 T^{11} + 10334227830435 T^{12} - 74086484916755 T^{13} + 514886627051218 T^{14} - 3447544393960184 T^{15} + 22468553587821000 T^{16} - 141349320152367544 T^{17} + 865524420073097458 T^{18} - 5106114626947671355 T^{19} + 29202057968357836035 T^{20} -$$$$15\!\cdots\!27$$$$T^{21} +$$$$83\!\cdots\!34$$$$T^{22} -$$$$41\!\cdots\!79$$$$T^{23} +$$$$19\!\cdots\!70$$$$T^{24} -$$$$84\!\cdots\!32$$$$T^{25} +$$$$34\!\cdots\!15$$$$T^{26} -$$$$12\!\cdots\!60$$$$T^{27} +$$$$40\!\cdots\!42$$$$T^{28} -$$$$11\!\cdots\!02$$$$T^{29} +$$$$26\!\cdots\!05$$$$T^{30} -$$$$45\!\cdots\!29$$$$T^{31} +$$$$63\!\cdots\!41$$$$T^{32}$$
$43$ $$1 - 30 T + 753 T^{2} - 13538 T^{3} + 211812 T^{4} - 2841218 T^{5} + 34443732 T^{6} - 379159713 T^{7} + 3871789827 T^{8} - 36847112340 T^{9} + 330610251346 T^{10} - 2801607814827 T^{11} + 22585491182096 T^{12} - 173209703194475 T^{13} + 1269592396734257 T^{14} - 8888276249156005 T^{15} + 59596840617230128 T^{16} - 382195878713708215 T^{17} + 2347476341561641193 T^{18} - 13771383871883123825 T^{19} + 77215299838840986896 T^{20} -$$$$41\!\cdots\!61$$$$T^{21} +$$$$20\!\cdots\!54$$$$T^{22} -$$$$10\!\cdots\!80$$$$T^{23} +$$$$45\!\cdots\!27$$$$T^{24} -$$$$19\!\cdots\!59$$$$T^{25} +$$$$74\!\cdots\!68$$$$T^{26} -$$$$26\!\cdots\!26$$$$T^{27} +$$$$84\!\cdots\!12$$$$T^{28} -$$$$23\!\cdots\!34$$$$T^{29} +$$$$55\!\cdots\!97$$$$T^{30} -$$$$95\!\cdots\!10$$$$T^{31} +$$$$13\!\cdots\!01$$$$T^{32}$$
$47$ $$1 + T + 198 T^{2} + 582 T^{3} + 20803 T^{4} + 53517 T^{5} + 1765785 T^{6} + 1234928 T^{7} + 113994985 T^{8} - 67893476 T^{9} + 5834256239 T^{10} - 14609562287 T^{11} + 297660346061 T^{12} - 1395538072257 T^{13} + 13891813886354 T^{14} - 77951950606532 T^{15} + 604027334011596 T^{16} - 3663741678507004 T^{17} + 30687016874955986 T^{18} - 144888949275938511 T^{19} + 1452487535127286541 T^{20} - 3350630164978951009 T^{21} + 62888704284742687631 T^{22} - 34396404670199799388 T^{23} +$$$$27\!\cdots\!85$$$$T^{24} +$$$$13\!\cdots\!76$$$$T^{25} +$$$$92\!\cdots\!65$$$$T^{26} +$$$$13\!\cdots\!51$$$$T^{27} +$$$$24\!\cdots\!23$$$$T^{28} +$$$$31\!\cdots\!14$$$$T^{29} +$$$$50\!\cdots\!62$$$$T^{30} +$$$$12\!\cdots\!43$$$$T^{31} +$$$$56\!\cdots\!21$$$$T^{32}$$
$53$ $$1 - 12 T + 568 T^{2} - 5149 T^{3} + 145525 T^{4} - 1046110 T^{5} + 23545152 T^{6} - 139497358 T^{7} + 2803144996 T^{8} - 14169466826 T^{9} + 266231223656 T^{10} - 1181358941884 T^{11} + 20997704216835 T^{12} - 83575642193485 T^{13} + 1402678955948192 T^{14} - 5096652195214124 T^{15} + 80205921351636774 T^{16} - 270122566346348572 T^{17} + 3940125187258471328 T^{18} - 12442490882839466345 T^{19} +$$$$16\!\cdots\!35$$$$T^{20} -$$$$49\!\cdots\!12$$$$T^{21} +$$$$59\!\cdots\!24$$$$T^{22} -$$$$16\!\cdots\!62$$$$T^{23} +$$$$17\!\cdots\!56$$$$T^{24} -$$$$46\!\cdots\!14$$$$T^{25} +$$$$41\!\cdots\!48$$$$T^{26} -$$$$96\!\cdots\!70$$$$T^{27} +$$$$71\!\cdots\!25$$$$T^{28} -$$$$13\!\cdots\!77$$$$T^{29} +$$$$78\!\cdots\!92$$$$T^{30} -$$$$87\!\cdots\!84$$$$T^{31} +$$$$38\!\cdots\!21$$$$T^{32}$$
$59$ $$1 - 38 T + 1095 T^{2} - 23297 T^{3} + 424605 T^{4} - 6672795 T^{5} + 94580432 T^{6} - 1217058846 T^{7} + 14511574567 T^{8} - 161041405813 T^{9} + 1682485451586 T^{10} - 16594154775321 T^{11} + 155630262017655 T^{12} - 1389923566921313 T^{13} + 11876680328710591 T^{14} - 97108684461423523 T^{15} + 761932837815421832 T^{16} - 5729412383223987857 T^{17} + 41342724224241567271 T^{18} -$$$$28\!\cdots\!27$$$$T^{19} +$$$$18\!\cdots\!55$$$$T^{20} -$$$$11\!\cdots\!79$$$$T^{21} +$$$$70\!\cdots\!26$$$$T^{22} -$$$$40\!\cdots\!47$$$$T^{23} +$$$$21\!\cdots\!07$$$$T^{24} -$$$$10\!\cdots\!94$$$$T^{25} +$$$$48\!\cdots\!32$$$$T^{26} -$$$$20\!\cdots\!05$$$$T^{27} +$$$$75\!\cdots\!05$$$$T^{28} -$$$$24\!\cdots\!63$$$$T^{29} +$$$$67\!\cdots\!95$$$$T^{30} -$$$$13\!\cdots\!62$$$$T^{31} +$$$$21\!\cdots\!41$$$$T^{32}$$
$61$ $$1 + 459 T^{2} - 318 T^{3} + 103898 T^{4} - 159398 T^{5} + 15704366 T^{6} - 38331037 T^{7} + 1814368519 T^{8} - 5937657768 T^{9} + 173417385168 T^{10} - 671036989903 T^{11} + 14346471436314 T^{12} - 59441130966933 T^{13} + 1047651981169903 T^{14} - 4318613747814279 T^{15} + 67852682743911080 T^{16} - 263435438616671019 T^{17} + 3898313021933209063 T^{18} - 13492007348005419273 T^{19} +$$$$19\!\cdots\!74$$$$T^{20} -$$$$56\!\cdots\!03$$$$T^{21} +$$$$89\!\cdots\!48$$$$T^{22} -$$$$18\!\cdots\!28$$$$T^{23} +$$$$34\!\cdots\!39$$$$T^{24} -$$$$44\!\cdots\!17$$$$T^{25} +$$$$11\!\cdots\!66$$$$T^{26} -$$$$69\!\cdots\!78$$$$T^{27} +$$$$27\!\cdots\!58$$$$T^{28} -$$$$51\!\cdots\!58$$$$T^{29} +$$$$45\!\cdots\!19$$$$T^{30} +$$$$36\!\cdots\!61$$$$T^{32}$$
$67$ $$1 - 28 T + 858 T^{2} - 15649 T^{3} + 292080 T^{4} - 4225007 T^{5} + 61602204 T^{6} - 771198098 T^{7} + 9574930223 T^{8} - 107511352491 T^{9} + 1179390780967 T^{10} - 12081186344489 T^{11} + 119596924133268 T^{12} - 1127862817133331 T^{13} + 10201063163519899 T^{14} - 88868275954132921 T^{15} + 739804228787108344 T^{16} - 5954174488926905707 T^{17} + 45792572541040826611 T^{18} -$$$$33\!\cdots\!53$$$$T^{19} +$$$$24\!\cdots\!28$$$$T^{20} -$$$$16\!\cdots\!23$$$$T^{21} +$$$$10\!\cdots\!23$$$$T^{22} -$$$$65\!\cdots\!93$$$$T^{23} +$$$$38\!\cdots\!43$$$$T^{24} -$$$$20\!\cdots\!06$$$$T^{25} +$$$$11\!\cdots\!96$$$$T^{26} -$$$$51\!\cdots\!81$$$$T^{27} +$$$$23\!\cdots\!80$$$$T^{28} -$$$$85\!\cdots\!63$$$$T^{29} +$$$$31\!\cdots\!82$$$$T^{30} -$$$$68\!\cdots\!04$$$$T^{31} +$$$$16\!\cdots\!81$$$$T^{32}$$
$71$ $$1 - 32 T + 995 T^{2} - 20508 T^{3} + 395622 T^{4} - 6259188 T^{5} + 93755982 T^{6} - 1239096885 T^{7} + 15716903928 T^{8} - 182294906860 T^{9} + 2049415143507 T^{10} - 21476154775115 T^{11} + 219328591407162 T^{12} - 2107956202438259 T^{13} + 19784162377688964 T^{14} - 175455499172719087 T^{15} + 1520075125513646654 T^{16} - 12457340441263055177 T^{17} + 99731962545930067524 T^{18} -$$$$75\!\cdots\!49$$$$T^{19} +$$$$55\!\cdots\!22$$$$T^{20} -$$$$38\!\cdots\!65$$$$T^{21} +$$$$26\!\cdots\!47$$$$T^{22} -$$$$16\!\cdots\!60$$$$T^{23} +$$$$10\!\cdots\!08$$$$T^{24} -$$$$56\!\cdots\!35$$$$T^{25} +$$$$30\!\cdots\!82$$$$T^{26} -$$$$14\!\cdots\!48$$$$T^{27} +$$$$64\!\cdots\!02$$$$T^{28} -$$$$23\!\cdots\!88$$$$T^{29} +$$$$82\!\cdots\!95$$$$T^{30} -$$$$18\!\cdots\!32$$$$T^{31} +$$$$41\!\cdots\!21$$$$T^{32}$$
$73$ $$1 - 20 T + 513 T^{2} - 9247 T^{3} + 159213 T^{4} - 2321194 T^{5} + 33317650 T^{6} - 424662675 T^{7} + 5258202404 T^{8} - 60372874351 T^{9} + 673283949509 T^{10} - 7023205288675 T^{11} + 71674972251759 T^{12} - 691734325623476 T^{13} + 6508014098139040 T^{14} - 58294144897789706 T^{15} + 511105999549900974 T^{16} - 4255472577538648538 T^{17} + 34681207128982944160 T^{18} -$$$$26\!\cdots\!92$$$$T^{19} +$$$$20\!\cdots\!19$$$$T^{20} -$$$$14\!\cdots\!75$$$$T^{21} +$$$$10\!\cdots\!01$$$$T^{22} -$$$$66\!\cdots\!47$$$$T^{23} +$$$$42\!\cdots\!24$$$$T^{24} -$$$$25\!\cdots\!75$$$$T^{25} +$$$$14\!\cdots\!50$$$$T^{26} -$$$$72\!\cdots\!38$$$$T^{27} +$$$$36\!\cdots\!73$$$$T^{28} -$$$$15\!\cdots\!51$$$$T^{29} +$$$$62\!\cdots\!17$$$$T^{30} -$$$$17\!\cdots\!40$$$$T^{31} +$$$$65\!\cdots\!61$$$$T^{32}$$
$79$ $$1 - 13 T + 775 T^{2} - 8945 T^{3} + 289929 T^{4} - 3056030 T^{5} + 70650878 T^{6} - 693207671 T^{7} + 12711402618 T^{8} - 117556710631 T^{9} + 1809285303132 T^{10} - 15865666579748 T^{11} + 212444210262071 T^{12} - 1764614755495823 T^{13} + 21101642992512959 T^{14} - 164760112592133413 T^{15} + 1796776883123990538 T^{16} - 13016048894778539627 T^{17} +$$$$13\!\cdots\!19$$$$T^{18} -$$$$87\!\cdots\!97$$$$T^{19} +$$$$82\!\cdots\!51$$$$T^{20} -$$$$48\!\cdots\!52$$$$T^{21} +$$$$43\!\cdots\!72$$$$T^{22} -$$$$22\!\cdots\!29$$$$T^{23} +$$$$19\!\cdots\!98$$$$T^{24} -$$$$83\!\cdots\!49$$$$T^{25} +$$$$66\!\cdots\!78$$$$T^{26} -$$$$22\!\cdots\!70$$$$T^{27} +$$$$17\!\cdots\!89$$$$T^{28} -$$$$41\!\cdots\!55$$$$T^{29} +$$$$28\!\cdots\!75$$$$T^{30} -$$$$37\!\cdots\!87$$$$T^{31} +$$$$23\!\cdots\!21$$$$T^{32}$$
$83$ $$1 - 39 T + 1238 T^{2} - 27953 T^{3} + 559862 T^{4} - 9568089 T^{5} + 150555772 T^{6} - 2148033125 T^{7} + 28809842394 T^{8} - 360796910182 T^{9} + 4303964232376 T^{10} - 48724320977567 T^{11} + 529549317747542 T^{12} - 5508676566066797 T^{13} + 55217376675221526 T^{14} - 531723737916086202 T^{15} + 4939932734933838930 T^{16} - 44133070247035154766 T^{17} +$$$$38\!\cdots\!14$$$$T^{18} -$$$$31\!\cdots\!39$$$$T^{19} +$$$$25\!\cdots\!82$$$$T^{20} -$$$$19\!\cdots\!81$$$$T^{21} +$$$$14\!\cdots\!44$$$$T^{22} -$$$$97\!\cdots\!14$$$$T^{23} +$$$$64\!\cdots\!54$$$$T^{24} -$$$$40\!\cdots\!75$$$$T^{25} +$$$$23\!\cdots\!28$$$$T^{26} -$$$$12\!\cdots\!63$$$$T^{27} +$$$$59\!\cdots\!82$$$$T^{28} -$$$$24\!\cdots\!39$$$$T^{29} +$$$$91\!\cdots\!02$$$$T^{30} -$$$$23\!\cdots\!73$$$$T^{31} +$$$$50\!\cdots\!81$$$$T^{32}$$
$89$ $$1 - 9 T + 685 T^{2} - 5312 T^{3} + 236340 T^{4} - 1586649 T^{5} + 55425874 T^{6} - 328282965 T^{7} + 10036323760 T^{8} - 53526070565 T^{9} + 1494977776540 T^{10} - 7283797771011 T^{11} + 189005283129452 T^{12} - 849011488066604 T^{13} + 20626626362065045 T^{14} - 86079699007121821 T^{15} + 1962582005966434910 T^{16} - 7661093211633842069 T^{17} +$$$$16\!\cdots\!45$$$$T^{18} -$$$$59\!\cdots\!76$$$$T^{19} +$$$$11\!\cdots\!32$$$$T^{20} -$$$$40\!\cdots\!39$$$$T^{21} +$$$$74\!\cdots\!40$$$$T^{22} -$$$$23\!\cdots\!85$$$$T^{23} +$$$$39\!\cdots\!60$$$$T^{24} -$$$$11\!\cdots\!85$$$$T^{25} +$$$$17\!\cdots\!74$$$$T^{26} -$$$$44\!\cdots\!61$$$$T^{27} +$$$$58\!\cdots\!40$$$$T^{28} -$$$$11\!\cdots\!28$$$$T^{29} +$$$$13\!\cdots\!85$$$$T^{30} -$$$$15\!\cdots\!41$$$$T^{31} +$$$$15\!\cdots\!61$$$$T^{32}$$
$97$ $$1 - 35 T + 1528 T^{2} - 38061 T^{3} + 988783 T^{4} - 19478483 T^{5} + 381496493 T^{6} - 6282347494 T^{7} + 101281076724 T^{8} - 1443044226028 T^{9} + 20048497517841 T^{10} - 253126336351305 T^{11} + 3118810776277033 T^{12} - 35509515023167295 T^{13} + 395496750669768458 T^{14} - 4108810659268075107 T^{15} + 41835419836014023142 T^{16} -$$$$39\!\cdots\!79$$$$T^{17} +$$$$37\!\cdots\!22$$$$T^{18} -$$$$32\!\cdots\!35$$$$T^{19} +$$$$27\!\cdots\!73$$$$T^{20} -$$$$21\!\cdots\!85$$$$T^{21} +$$$$16\!\cdots\!89$$$$T^{22} -$$$$11\!\cdots\!64$$$$T^{23} +$$$$79\!\cdots\!64$$$$T^{24} -$$$$47\!\cdots\!98$$$$T^{25} +$$$$28\!\cdots\!57$$$$T^{26} -$$$$13\!\cdots\!99$$$$T^{27} +$$$$68\!\cdots\!03$$$$T^{28} -$$$$25\!\cdots\!97$$$$T^{29} +$$$$99\!\cdots\!32$$$$T^{30} -$$$$22\!\cdots\!55$$$$T^{31} +$$$$61\!\cdots\!21$$$$T^{32}$$
show more
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# Scheduling, Sequencing & Line Balancing Mechanical Engineering Notes | EduRev
## Mechanical Engineering : Scheduling, Sequencing & Line Balancing Mechanical Engineering Notes | EduRev
The document Scheduling, Sequencing & Line Balancing Mechanical Engineering Notes | EduRev is a part of the Mechanical Engineering Course Industrial Engineering.
All you need of Mechanical Engineering at this link: Mechanical Engineering
Scheduling
Scheduling is used to schedule resources in time to finish the tasks. Forecast demand plays a key role to determine the plan for the output.
Methods Used for Scheduling
1. Longest Processing Time (LPT): The longest processing time rule orders the jobs in the order of decreasing processing times. Whenever a machine becomes free, the largest job processing starts at the same time. The main aim of this rule is to schedule the longest jobs first so that no large job will remain pending till the end.
2. Shortest Processing Time (SPT): This rule arranges the jobs in the increasing order of processing times. Whenever a machine becomes free, the shortest job processing starts on it at the same time.
3. Earliest Due Date (EDD): In the single machine environment with ready time set at 0 for all jobs, the earliest due date rule arranges the jobs in increasing order of their due date. EDD arrangement of jobs minimize the maximum lateness, or to minimize the maximum tardiness.
4. Minimum Slack Time (MST): The minimum slack time rule arranges the jobs in increasing order of slack. The “urgency” of a job is decided by its slack time. MST rule maximizes the minimum lateness.
5. First Come, First Served (FCFS) rule: In case of inventory management, it is treated as First In First Out (FIFO) i.e. first piece of inventory at a storage area is the first one to be used.
6. Critical ratio (CR) rule: arranges the jobs in increasing order of their critical ratio.
CR = Due date / Processing Time
If CR>1 The job is ahead of schedule.
If CR<1 The job is behind schedule.
If CR=1 The job is exactly on schedule.
7. Slack Time Remaining (STR) rule: It employs that the next job processed is the one that has the least amount of slack time.
Slack = Due date – Processing time
Sequencing
It is the order in which jobs pass through the machines or workstations.
Sequencing Terminology
1. Number of Machines: It means the service facilities through which a job must pass before it is completed.
2. Processing Order: It is the order in which different machines are arranged for completing the job.
3. Processing Time: It is the time needed by each job on each machine.
4. Idle Time on a Machine: It is the time for which a machine remains idle during the total elapsed time.
5. Total Elapsed Time: The time between the start of the first job and completing the last job, is the total elapsed time.
Johnson’s rule (Sequencing of n jobs on 2 machines)
Let, Ai = processing time of ith job on machine 1, Bi = processing time of ith job on machine 2, these problems are solved by Johnson’s rule and steps involved are:
(i) Find out the minimum of Ai and Bi.
(ii) If the minimum is for a particular job on machine A then, perform that job at the start or beginning.
(iii) If the minimum is for a particular job on machine B then, perform that job in the last.
(iv) Strike-off the job which is assigned so that it can’t be considered again.
(v) Continue in the similar manner until all the jobs are assigned.
Example: The above algorithm is illustrated with the following example.
Consider two machines and six jobs flow shop scheduling problem. Using Johnson’s algorithm, obtain the optimal sequence which will minimize the make span.
Step-1: Finding Sequence. Min Processing time of 1 min for job 6 on m1 and for job 3 on m2. Place job 6 first and job 3 at last. Continue this procedure. The sequence obtained is-
Step-2: Using this sequence, the make span for the jobs can be determined by the time in and time out:
Total flow time = 11 + 16 + 24 + 30 + 33 + 36 = 150
Avg. flow time = 150 / 6 = 25
Idle time for M1 = 1 minute.
Idle time for M2 = 1 + 2 = 3 minute.
Make span time of shop = 36 minute.
Sequencing of n jobs on 3 machines (Jackson Rule)
(i) Check (min)M1 ≥ (max)M2
(min)M3 ≥ (max)M2, At least one must satisfy.
(ii) Convert to two machines problem by adding machine 1 and 2,then machine 2 and solve using Johnson’s rule.
Assembly Line Balancing
Assembly line is a special case of product layout in which the operations pertaining to assembly of different parts at few station line (product) layout is useful for high volume, single type of manufacturing activity. The aim of assembly line is to divide total work content into different work station. Such that idle time is minimized utilization is optimized.
Line balancing Terminology
1. Work Element (i): Every job is completed by a set of operation and each operation which is performed on the job is called work element.
2. Task time (T1): It is the slandered time required to complete work-element.
3. Work Stations (ω): It is the specific location on the assembly line where the given amount of work elements are completed within a fixed period of time.
4. Station time (Tsi): It is a time required to complete work element assigned in an work-station.
5. Total Work Content (Twc): It is time required to complete one set of job. It is given by either the summation of all the station time or the summation of all the elemental task time.
6. Cycle Time: It is an amount of time for which a job to be assembled remains in a work station. It is a time gap between two successive product coming out from the assembly line.
Tc ≥ max(Tsi)
7. Delay or Idle Time at Station: The difference between cycle time and station time is called the delay or idle time at the station.
Tds = Tc - Tsi
8. Balance Delay or Balancing Less (d): It is the measure of the inefficiency of the line.
9. Line efficiency:
Where, Tsi = Station time at station
i, n = Total number of stations
TC = Total cycle time.
• Smoothness Index (SI): It is a term used to represent the load distribution between the different work station. Compare to a station consuming maximum time.
Where, (Tsi)max = Maximum station time.
Note: If smoothness index is low, it means line balancing is good.
• Theoretically Minimum Number of Work Stations (Nmin): The minimum number of work stations is defined as the ratio of total work content to the total cycle time.
Method of line balancing- Largest candidate rule steps are:
• List all the element in the decreasing order of their task time
• To assign an element in a work station start form beginning of list moving downward searching feasible element which can be placed in a work station a feasible element is one which satisfy precedence requirement and when that element is placed in the work station the total time of the work station should not exceed the cycle time.
• Strike of the element which is assigned so that it can’t considered again.
• Continue in the similar manner until all the jobs are assigned.
Example: Let us consider the precedence diagram of 13 work elements shown below. The time each work element is at the top of each node. In a tabular form, this precedence diagram is represented as follows.
Total work content = 68 min
Largest work element time = 10 min
Thus, cycle time (TC) must satisfy TC ≥ 10 min
For minimum cycle time of 10 min, number of stations would be 68/10 = 6.8
Therefore, we must take stations lesser than this. Let us select 5 stations design. For 5
stations, the station time should be nearly equal to 68/5 = 13.6 min
List work elements in descending order of their work element.
Here, the final cycle time is the maximum station time which is 16 min.
Balance delay =x 100% = 15%
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## Industrial Engineering
33 videos|30 docs|32 tests
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# K-th occurrence
Time Limit: 3000/3000 MS (Java/Others) Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 4000 Accepted Submission(s): 1134
Problem Description
You are given a string $S$ consisting of only lowercase english letters and some queries.
For each query $(l, r, k)$, please output the starting position of the k-th occurence of the substring $S_lS_{l+1}...S_r$ in S.
Input
The first line contains an integer $T(1 \leq T \leq 20)$, denoting the number of test cases.
The first line of each test case contains two integer $N(1 \leq N \leq 10^5), Q(1 \leq Q \leq 10^5)$, denoting the length of $S$ and the number of queries.
The second line of each test case contains a string $S(|S|=N)$ consisting of only lowercase english letters.
Then $Q$ lines follow, each line contains three integer $l, r(1 \leq l \leq r \leq N)$ and $k(1 \leq k \leq N)$, denoting a query.
There are at most $5$ testcases which $N$ is greater than $10^3$.
Output
For each query, output the starting position of the k-th occurence of the given substring.
If such position don't exists, output $-1$ instead.
Sample Input
2 12 6 aaabaabaaaab 3 3 4 2 3 2 7 8 3 3 4 2 1 4 2 8 12 1 1 1 a 1 1 1
Sample Output
5 2 -1 6 9 8 1
Source
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Home | Top Hangzhou Dianzi University Online Judge 3.0 Copyright © 2005-2024 HDU ACM Team. All Rights Reserved. Designer & Developer : Wang Rongtao LinLe GaoJie GanLu Total 0.000000(s) query 1, Server time : 2024-09-21 11:45:46, Gzip enabled Administration | 587 | 1,761 | {"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} | 2.625 | 3 | CC-MAIN-2024-38 | latest | en | 0.638343 |
http://slideplayer.com/slide/3959975/ | 1,505,896,696,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818686705.10/warc/CC-MAIN-20170920071017-20170920091017-00411.warc.gz | 314,088,213 | 18,894 | # Torque Torque is defined as the tendency to produce a change in rotational motion.
## Presentation on theme: "Torque Torque is defined as the tendency to produce a change in rotational motion."— Presentation transcript:
Torque Torque is defined as the tendency to produce a change in rotational motion.
Torque is a twist or turn that tends to produce rotation. * * * Applications are found in many common tools around the home or industry where it is necessary to turn, tighten or loosen devices.
Objectives: After completing this module, you should be able to: Define and give examples of the terms torque, moment arm, axis, and line of action of a force.Define and give examples of the terms torque, moment arm, axis, and line of action of a force. Draw, label and calculate the moment arms for a variety of applied forces given an axis of rotation.Draw, label and calculate the moment arms for a variety of applied forces given an axis of rotation. Calculate the resultant torque about any axis given the magnitude and locations of forces on an extended object.Calculate the resultant torque about any axis given the magnitude and locations of forces on an extended object.
Torque is Determined by Three Factors: The magnitude of the applied force.The magnitude of the applied force. The direction of the applied force.The direction of the applied force. The location of the applied force.The location of the applied force. The magnitude of the applied force.The magnitude of the applied force. The direction of the applied force.The direction of the applied force. The location of the applied force.The location of the applied force. 20 N Magnitude of force 40 N The 40-N force produces twice the torque as does the 20-N force. Each of the 20-N forces has a different torque due to the direction of force. 20 N Direction of Force 20 N Location of force The forces nearer the end of the wrench have greater torques. 20 N
Units for Torque Torque is proportional to the magnitude of F and to the distance d from the axis. Thus, a tentative formula might be: = Fd Units: N m or lb ft 60 cm 40 N = (40 N)(0.60 m) = 24.0 N m, cw = 24.0 N m, cw
Direction of Torque Torque is a vector quantity that has direction as well as magnitude. Turning the handle of a screwdriver clockwise and then counterclockwise will advance the screw first inward and then outward.
Sign Convention for Torque By convention, counterclockwise torques are positive and clockwise torques are negative. Positive torque: Counter-clockwise, out of page cw ccw Negative torque: clockwise, into page
Line of Action of a Force The line of action of a force is an imaginary line of indefinite length drawn along the direction of the force. F1F1 F2F2 F3F3 Line of action
The Moment Arm The moment arm of a force is the perpendicular distance from the line of action of a force to the axis of rotation. F2F2 F1F1 F3F3 d d d
Calculating Torque Read problem and draw a rough figure.Read problem and draw a rough figure. Extend line of action of the force.Extend line of action of the force. Draw and label moment arm.Draw and label moment arm. Calculate the moment arm if necessary.Calculate the moment arm if necessary. Apply definition of torque:Apply definition of torque: Read problem and draw a rough figure.Read problem and draw a rough figure. Extend line of action of the force.Extend line of action of the force. Draw and label moment arm.Draw and label moment arm. Calculate the moment arm if necessary.Calculate the moment arm if necessary. Apply definition of torque:Apply definition of torque: = Fd Torque = force x moment arm
Example 1: An 80-N force acts at the end of a 12-cm wrench as shown. Find the torque. Extend line of action, draw, calculate r. = (80 N)(0.104 m) = 8.31 N m r = 12 cm sin 60 0 = 10.4 cm
Alternate: An 80-N force acts at the end of a 12-cm wrench as shown. Find the torque. Resolve 80-N force into components as shown. = (69.3 N)(0.12 m) = 8.31 N m as before positive 12 cm
Download ppt "Torque Torque is defined as the tendency to produce a change in rotational motion."
Similar presentations | 987 | 4,134 | {"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} | 4.46875 | 4 | CC-MAIN-2017-39 | longest | en | 0.88341 |
renelleesa.blogspot.com | 1,531,964,803,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676590443.0/warc/CC-MAIN-20180719012155-20180719032155-00633.warc.gz | 301,049,655 | 16,937 | ## 29 October 2010
### The difference between a cube and a cuboid, pyramids and prisms, cone and pyramid???
The difference between a cube and a cuboid?
cube is a six sided 3 dimensional figure in which all the faces are square, a cuboid is the same except that some of the faces are rectangles.
## Difference Between Pyramids and Prisms?
Pyramids vs Prisms
Most people have a misconception that a prism is the same as a pyramid. However, it is worth knowing that these two are actually different. Let’s take a look at their differences using geometry’s point of view.
A pyramid, in geometry, is a polyhedron formed by connecting a polygonal base and a point called apex. Each base edge and apex forms a triangle. The pyramid’s base can be trilateral, quadrilateral or any polygon shape. The most common version is that of the square pyramid.
A pyramid is often regarded as triangular structures, usually found in Egypt. These were the largest structures on Earth for thousands of years. These structures are designed with the majority of their weight closer to the ground. It allowed early civilization to create a more stable monumental structure.
On the other hand, a prism is also a polyhedron, made up of a polygonal base, but with a translated copy, and the joining faces corresponding to the sides. The joining faces form a parallelogram, and not a triangle.
A prism, in optics, refers to a transparent optical element, with polished surfaces that refract light. The most common is that of the triangular prism. It is made up of a triangular base and rectangular sides, that’s why the colloquial term ‘prism’ is usually referred to this type. Prism is usually made of glass, but it can be made with any transparent material that can refract, reflect or split light.
Summary:
1. A pyramid has a base and a connecting point, while a prism has a base, together with a translated copy of it.
2. The sides or faces formed in a pyramid are always triangles, while in a prism, they normally form a parallelogram.
3. A pyramid is often regarded as a solid building, while a prism is referred to something that is transparent, and can refract, reflect or split light.
Similarities and differences between cone and pyramid?
Similarities: both have a base which extends into a single vertex.
: the point is directly above the centre of the base
Differences: a cone only has 1 edge and a pyramid has a minimum of 6. | 526 | 2,416 | {"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} | 3.46875 | 3 | CC-MAIN-2018-30 | latest | en | 0.963733 |
https://manualzz.com/doc/26237426/variation-of-light-intensity | 1,576,064,620,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540530857.12/warc/CC-MAIN-20191211103140-20191211131140-00012.warc.gz | 462,287,488 | 17,855 | Variation of Light Intensity
```Variation of Light Intensity:
Measuring the Light Intensity of
Different Light Sources
(Teacher’s Guide)
For technical assistance,
call WARD’S at 1-800-962-2660
OVERVIEW
Students will measure the light intensity of different light sources using the Ward’s DataHub
light sensor. Based on the results, the students will proceed to relate each light source to its
corresponding light efficiency.
MATERIALS NEEDED
Ward’s DataHub
USB connector cable*
LED flashlight with batteries
Candle
Fluorescent light bulb
Matches
Lamp with an 11 watt bulb
* – The USB connector cable is not needed if you are using a Bluetooth enabled device.
NUMBER OF USES
This demonstration can be performed repeatedly.
Teacher’s Guide – Variation of Light Intensity
For technical assistance,
call WARD’S at 1-800-962-2660
FRAMEWORK FOR K-12 SCIENCE EDUCATION © 2012
Dimension 1
Science and Engineering
Practices
* The Dimension I practices listed below are called out as bold words throughout the activity.
Asking questions (for science) and defining
problems (for engineering)
Developing and using models
Constructing explanations (for science) and designing
solutions (for engineering)
Planning and carrying out investigations
Engaging in argument from evidence
Analyzing and interpreting data
Obtaining, evaluating, and communicating information
Use mathematics and computational thinking
Cross Cutting
Concepts
Dimension 2
Patterns
Energy and matter: Flows, cycles, and conservation
Cause and effect:
Mechanism and explanation
Structure and function
Scale, proportion, and quantity
Stability and change
Systems and system models
Discipline
Core Idea Focus
Core Concepts
Dimension 3
PS1: Matter and Its Interaction
PS1.A: Structure and Properties of Matter
PS3: Energy
Physical Science
PS3.A: Definitions of Energy
PS4: Waves and Their Applications in Technologies for Information
Transfer
NGSS
Standards
Middle School Standards Covered
High School Standards Covered
MS.PS-SPM: Structure and Properties of Matter
HS.PS-SPM: Structure and Properties of Matter
MS.PS-E: Energy
HS.PS-E: Energy
NATIONAL SCIENCE EDUCATION STANDARDS © 2002
Content Standards (K-12)
Systems, order, and organization
Evolution and equilibrium
Evidence, models, and explanation
Form and Function
Constancy, change, and measurement
Physical Science Standards Middle School
Physical Science Standards High School
Properties and Changes of Properties in Matter
Structure of Atoms
Motions and Forces
Transfer of Energy
Structure and Properties of Matter
Chemical Reactions
Motions and Forces
Conservation of Energy and Increase in Disorder
Interactions of Energy and Matter
Indicates Standards Covered in Activity
Teacher’s Guide – Variation of Light Intensity
For technical assistance,
call WARD’S at 1-800-962-2660
LEARNING OBJECTIVES
Core Objectives (National Standards):
•
Develop the ability to refine ill-defined questions and direct to phenomena that can be
described, explained, or predicted through scientific means.
•
Develop the ability to observe, measure accurately, identify and control variables.
•
Decide what evidence can be used to support or refute a hypothesis.
•
Gather, store, retrieve, and analyze data.
•
Become confident at communicating methods, instructions, observations, and results with
others.
Activity Objectives:
The purpose of this activity is to relate light intensity and light source efficiency to create a
hypothesis about the amount of light sent out by different light sources and proceed to test
it using the Ward’s DataHub light sensor.
Time Requirement:
45 - 60 minutes
Teacher’s Guide – Variation of Light Intensity
For technical assistance,
call WARD’S at 1-800-962-2660
VOCABULARY
Ambient Light: The soft indirect light that fills the volume of a room with illumination.
Candle: A cylinder or block of wax or tallow with a central wick that is lit to produce light as
it burns.
Fluorescent Lamp: A lamp consisting of a tube coated on the inside with a fluorescent
material. Mercury vapor in the tube, emits ultraviolet radiation that is converted to
visible radiation by the fluorescent material.
Incandescent: Emitting light as a result of being heated.
LED: Light-emitting diode, a semiconductor diode that glows when a voltage is applied.
Light Intensity: Candle power: luminous intensity measured in candelas.
Luminous Flux: The rate of flow of light energy.
Sunlight: Light from the Sun.
Watt: The SI unit of power, equivalent to one joule per second, corresponding to the rate of
energy in an electric circuit where the potential difference is one volt and the current,
one ampere.
Teacher’s Guide – Variation of Light Intensity
For technical assistance,
call WARD’S at 1-800-962-2660
INTRODUCTION
DID YOU KNOW?
Why is the sky blue? The
atmosphere scatters
differently. Those rays at the
shorter end of the spectrum,
or the blue rays, scatter more
than the longer wavelength
rays (red and orange colors).
However, we do see these
longer wavelengths at sunset
when the angle of the sun’s
light enters the atmosphere
just right. This is known as
Rayleigh scattering.
Have you ever experienced an electricity blackout when it is dark
outside? Usually people run to find candles and flashlights so they
can see something in the pitch-black. Even if we try to light the
room by placing several candles around, or use the most powerful
flashlight we have, it may still not be enough to brighten the room
as well as a light bulb could.
•
How should we place several candles in a room in order
to achieve the most light?
•
What do you think the efficiency of a light source
depends on?
Carry out the experiment with your class so that at the end, students
will be able to answer the following question:
•
How are the intensity and efficiency of a light source
related?
Teacher’s Guide – Variation of Light Intensity
For technical assistance,
call WARD’S at 1-800-962-2660
BACKGROUND
The luminous flux gives us an idea about the light intensity sent out
by light sources in all space dimensions, for example by a light bulb.
But when you consider a projector, it is clear that it lights only in one
direction: forward. That is why we need to know how the luminous
flux is distributed into every space dimension, using the definition of
light intensity.
DID YOU KNOW?
Opticks was written by
physicist Sir Isaac Newton
in 1704. It contained
experiments and the
deductions that Newton
covered all of the major
topics in what would be
considered today physical
optics. Newton first began
doing experiments regarding
light in the 17th century.
Light intensity (l) is defined as the “luminous flux that is emitted
per unit of solid angle (steradian) into a specific direction”. The
unit of measure is the lumen per steradian, or candla (cd). The
mathematical equation that defines light intensity is:
At this point, encourage students to formulate a hypothesis to test
as part of this activity. Students may find it helpful to formulate their
hypothesis as an answer to the following question:
•
If light intensity and light efficiency are related, how
would the efficiency vary depending on the light source?
Teacher’s Guide – Variation of Light Intensity
For technical assistance,
call WARD’S at 1-800-962-2660
CONNECTING THE WARD’S DATAHUB
TO A COMPUTER
If you are using a Bluetooth communication device:
Right-click on the Bluetooth icon in the lower right corner of the screen
and select the Ward’s DataHub you are using. The icon will change from
gray to blue, as shown at right, indicating that the Ward’s DataHub and the
computer are now connected via Bluetooth.
If you are using a USB communication device:
In order to use USB communication, connect the Ward’s DataHub and the
computer with the USB cable supplied. Click on the USB icon at the lower right
corner of the screen. This icon will change from gray to blue, as shown at right,
indicating that the Ward’s DataHub is connected to the computer via USB.
USING THE WARD’S DATAHUB
= Select key
= On/Off and Escape key
= Scroll key
To collect measurements with the Ward’s DataHub, it must first be configured as follows:
1. Turn on the Ward’s DataHub by
pressing the On/Off/Esc key.
8. Press the On/Off/Esc key to return
2. Go to setup by using the Scroll key;
then
then select Setup by pressing
the Select key.
10. Press the Scroll key until
“Manual” is highlighted,
4. If any sensor(s) appear on the screen,
press the key representing that
sensor to deactivate it. Once you
have a blank window, press the
Light Sensor key once.
11. Press the On/Off/Esc key three
operating screen.
then
then press the Select Key
then press the Select key.
then
then press the Select key.
5. Press the On/Off/Esc key once to
7. Press the Scroll key until
“Manual” is highlighted,
then
then press the Select Key.
3. Select the Set Sensors option by
pressing the Select key.
6. Press the Scroll key to highlight
the Sampling Rate and
9. Press the Scroll key to highlight
the Number of Samples and
then
x3
12. Press the Select key to
start measuring.
(You are collecting data when there
is an icon of a Runner in the upper
left hand corner of the screen.)
13. Once you have finished measuring,
stop the Ward’s DataHub by
pressing the Select key,
then
followed by the Scroll key.
Teacher’s Guide – Variation of Light Intensity
For technical assistance,
call WARD’S at 1-800-962-2660
ACTIVITY
1. Place the four different light sources approximately 30 cm away,
in the following order: candle, flashlight, lamp, and fluorescent
bulb. Darken the room, covering the windows and turning off
any artificial light sources.
DID YOU KNOW?
There are different colors
of light. This is because the
light waves have different
wavelengths. Red light has
the longest wavelength while
violet light has the shortest
wavelength.
2. To collect the data, place the DataHub light sensor approximately
10 cm away from the light source.
3. Push the
button on the DataHub.
4. Light the candle and observe how the measurements vary on the
DataHub screen.
5. Wait until the intensity value you are measuring stabilizes.
6. Take just one manual sample of the light intensity.
7. Once you have finished with the candle, extinguish it without
turning off the DataHub, and then turn on the LED flashlight.
Repeat steps 5 and 6.
8. Measure the light intensity of the lamp and the fluorescent bulb
as you did with the other light sources.
9. Uncover the windows, but do not turn on any lights in order to
measure the light intensity of the ambient light in the room.
Then measure the intensity of Sunlight by pointing the light
sensor directly at the Sun.
10. Once you’ve finished measuring, turn the DataHub off.
Teacher’s Guide – Variation of Light Intensity
For technical assistance,
call WARD’S at 1-800-962-2660
RESULTS AND ANALYSIS
The following steps explain how to analyze the experiment results.
DID YOU KNOW?
Have you ever walked
outside and thought “Gee!
It is so bright out here!”?
You probably reach for your
sunglasses without giving it a
second thought, but do you
know how sunglasses work?
They provide protection from
intense light. The eyes natural
response to too much light
is to close the iris, and then
squint. Damage can occur
to the retina if this light is
too intense. Sunglasses also
provide protection from
ultraviolet rays that can
damage the cornea and retina
1. Connect the DataHub to the computer using the USB
communication cable or via the Bluetooth wireless
communication channel.
2. On the upper menu, press the
button. Select
3. Select the last experiment on the list.
4. Observe the graph displayed on the screen.
5. Press the bar graph icon and set the display to a Bar Graph
display.
6. Press the
button and write notes on the graph specifying
your observations according to the moment you registered the
data.
•
How do the results relate to your initial hypothesis?
Explain.
•
How do the data curves vary for each light source?
•
What similarities do the data curves present?
•
Which was the brightest light source? Which was the least
bright?
The graph below should be similar to what the students obtained.
Teacher’s Guide – Variation of Light Intensity
For technical assistance,
call WARD’S at 1-800-962-2660
CONCLUSIONS AND ASSESSMENTS
1. What was the variation between the different light sources you analyzed? Communicate which
light source was the highest intensity and which was the lowest.
Students should analyze the different values of light intensity, defining which were the highest and
lowest and the range of variation between them.
2. How does the amount of light relate to the light intensity in your model system?
Students should relate more powerful light sources to greater light intensity and less bright light
sources, like a candle, to less light intensity.
3. How do you think the light flux varies in each of the light sources you analyzed?
Students should conclude that the more light intensity, the more luminous flux there is. Both
parameters depend on the relative distance between the location of the Ward’s DataHub light
sensor and the light source.
4. According to your experience, argue which light source you think is the most efficient light
source from the three artificial sources studied?
Students should indicate that the most efficient light source is the LED flashlight, because it uses
less energy to function, for this reason it is considered to be energy saving.
5. Write a concluding paragraph describing what you observed during the experiment.
Students should reach the following conclusions:
Different light sources have different light intensities and this relates to their associated functions.
However, light intensity cannot be related to the efficiency of the source. The LED flashlight is not
the most luminous sources, but presents greater efficiency than the candle and the lamp. On the
other hand, sunlight is the most powerful light source and yet, is the most efficient.
10
Teacher’s Guide – Variation of Light Intensity
For technical assistance,
call WARD’S at 1-800-962-2660
ACTIVITIES FOR FURTHER APPLICATION
The aim of this section is for students to extrapolate the knowledge
acquired during this class and apply it to different contexts and
situations. Furthermore, it is intended that students question and
present possible explanations for the experimentally observed
phenomena.
DID YOU KNOW?
Light is more than color. It
is very fast traveling energy,
traveling throughout the
universe. Light travels at
300000000 meters/second.
In prehistoric times, the
energy from light was
harnessed through fire and
the invention of illumination
devices such as candles
and gas lamps took this
to another level. The first
electric powered light
was invented in the late
nineteenth century. Today,
light is used in highly focused
and powerful laser beams
in such diverse applications
as surgery and splitting
molecules.
1. How could you increase the light intensity of a candle?
Students should explain that they could achieve this by
increasing the amount of light the candle produces, in other
words, making the flame larger. We can establish the following
connection: The larger the candle flame is, the more light
intensity it produces, and vice versa (the smaller the candle
flame, the less light intensity).
2. How are light intensity and electrical power related?
Students are expected to establish that light sources which emit
more light intensity use more energy in the process. On the other
hand, we should point out the example of the candle, which uses
less energy in the lighting process. In both cases, energy is “lost”
through heat. This shows that light efficiency depends on how
much energy we use to illuminate, rather than produce heat.
3. How is a natural light source like the Sun different from an
artificial light source such as a light bulb?
Students should quantify that the Sun’s light intensity is much
higher than the intensity of any artificial light source, and that
this depends directly on the amount of energy used by the Sun
to produce this intensity of light.
delivered via a fiber.
11
Teacher’s Guide – Variation of Light Intensity
For technical assistance,
call WARD’S at 1-800-962-2660
Variation of Light Intensity:
Measuring the Light Intensity
of Different Light Sources
(Student Guide)
INTRODUCTION
Have you ever experienced an electricity blackout when it is dark outside? Usually people run to
find candles and flashlights so they can see something in the pitch-black. Even if we try to light the
room by placing several candles around, or use the most powerful flashlight we have, it may still
not be enough to brighten the room as well as a light bulb could.
•
How should we place several candles in a room in order to achieve the most light?
•
What do you think the efficiency of a light source depends on?
After carrying out this experiment, you should be able to answer the following question:
•
How are the intensity and efficiency of a light source related?
S
Student Guide – Variation of Light Intensity
For technical assistance,
call WARD’S at 1-800-962-2660
CONNECTING THE WARD’S DATAHUB
TO A COMPUTER
If you are using a Bluetooth communication device:
Right-click on the Bluetooth icon in the lower right corner of the screen
and select the Ward’s DataHub you are using. The icon will change from
gray to blue, as shown at right, indicating that the Ward’s DataHub and the
computer are now connected via Bluetooth.
If you are using a USB communication device:
In order to use USB communication, connect the Ward’s DataHub and the
computer with the USB cable supplied. Click on the USB icon at the lower right
corner of the screen. This icon will change from gray to blue, as shown at right,
indicating that the Ward’s DataHub is connected to the computer via USB.
USING THE WARD’S DATAHUB
= Select key
= On/Off and Escape key
= Scroll key
To collect measurements with the Ward’s DataHub, it must first be configured as follows:
1. Turn on the Ward’s DataHub by
pressing the On/Off/Esc key.
8. Press the On/Off/Esc key to return
2. Go to setup by using the Scroll key;
then
then select Setup by pressing
the Select key.
10. Press the Scroll key until
“Manual” is highlighted,
4. If any sensor(s) appear on the screen,
press the key representing that
sensor to deactivate it. Once you
have a blank window, press the
Light Sensor key once.
11. Press the On/Off/Esc key three
operating screen.
then
then press the Select Key
then press the Select key.
then
then press the Select key.
5. Press the On/Off/Esc key once to
7. Press the Scroll key until
“Manual” is highlighted,
then
then press the Select Key.
3. Select the Set Sensors option by
pressing the Select key.
6. Press the Scroll key to highlight
the Sampling Rate and
9. Press the Scroll key to highlight
the Number of Samples and
then
x3
12. Press the Select key to
start measuring.
(You are collecting data when there
is an icon of a Runner in the upper
left hand corner of the screen.)
13. Once you have finished measuring,
stop the Ward’s DataHub by
pressing the Select key,
then
followed by the Scroll key.
S
Student Guide – Variation of Light Intensity
For technical assistance,
call WARD’S at 1-800-962-2660
ACTIVITY
1. Place the four different light sources approximately 30 cm away, in the following order:
candle, flashlight, lamp, and fluorescent bulb. Darken the room, covering the windows
and turning off any artificial light sources.
2. To collect the data, place the DataHub light sensor approximately 10 cm away from the
light source.
3. Push the
button on the DataHub.
4. Light the candle and observe how the measurements vary on the DataHub screen.
5. Wait until the intensity value you are measuring stabilizes.
6. Take just one manual sample of the light intensity.
7. Once you have finished with the candle, extinguish it without turning off the DataHub,
and then turn on the LED flashlight. Repeat steps 5 and 6.
8. Measure the light intensity of the lamp and the fluorescent bulb as you did with the other
light sources.
9. Uncover the windows, but do not turn on any lights, in order to measure the light
intensity of the ambient light in the room. Then measure the intensity of Sunlight by
pointing the light sensor directly at the Sun.
10. Once you’ve finished measuring, turn the DataHub off.
S
Student Guide – Variation of Light Intensity
For technical assistance,
call WARD’S at 1-800-962-2660
RESULTS AND ANALYSIS
1. Connect the DataHub to the computer using the USB communication cable or via the Bluetooth
wireless communication channel.
2. On the upper menu, press the
button. Select
3. Select the last experiment on the list.
4. Observe the graph displayed on the screen.
5. Press the bar graph icon and set the display to a Bar Graph display.
6. Press the
button and write notes on the graph specifying your observations according to
the moment you registered the data.
•
How do the results relate to your initial hypothesis? Explain.
________________________________________________________________________
________________________________________________________________________
________________________________________________________________________
•
How do the data curves vary for each light source?
________________________________________________________________________
________________________________________________________________________
________________________________________________________________________
•
What similarities do the data curves present?
________________________________________________________________________
________________________________________________________________________
________________________________________________________________________
•
Which was the brightest light source? Which was the least bright?
________________________________________________________________________
________________________________________________________________________
________________________________________________________________________
________________________________________________________________________
S
Student Guide – Variation of Light Intensity
For technical assistance,
call WARD’S at 1-800-962-2660
CONCLUSIONS AND ASSESSMENTS
1. What was the variation between the different light sources you analyzed? Communicate which
light source was the highest intensity and which was the lowest.
________________________________________________________________________________
________________________________________________________________________________
________________________________________________________________________________
________________________________________________________________________________
2. How does the amount of light relate to the light intensity in your model system?
________________________________________________________________________________
________________________________________________________________________________
________________________________________________________________________________
________________________________________________________________________________
3. How do you think the light flux varies in each of the light sources you analyzed?
________________________________________________________________________________
________________________________________________________________________________
________________________________________________________________________________
________________________________________________________________________________
4. According to your experience, argue which light source you think is the most efficient light
source from the three artificial sources studied?
________________________________________________________________________________
________________________________________________________________________________
________________________________________________________________________________
________________________________________________________________________________
5. Write a concluding paragraph describing what you observed during the experiment.
________________________________________________________________________________
________________________________________________________________________________
________________________________________________________________________________
________________________________________________________________________________ | 5,150 | 24,360 | {"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-2019-51 | latest | en | 0.824261 |
http://stackoverflow.com/questions/tagged/pow?page=1&sort=votes&pagesize=15 | 1,455,492,407,000,000,000 | text/html | crawl-data/CC-MAIN-2016-07/segments/1454702039825.90/warc/CC-MAIN-20160205195359-00139-ip-10-236-182-209.ec2.internal.warc.gz | 205,402,635 | 29,124 | # Tagged Questions
pow is a function that exists in various programming languages that usually takes two numbers as input and returns the first number to the power of the second number. DO NOT USE THIS TAG for questions relating to the Rack server, use [rack-pow] instead.
32k views
### How is Math.Pow() implemented in .NET Framework?
I was looking for an efficient approach for calculating ab (say a = 2 and b = 50). To start things up, I decided to take a look at the implementation of Math.Pow() function. But in .NET Reflector, all ...
7k views
### Why is Math.pow(0, 0) === 1?
We all know that 00 is indeterminate. But, javascript says that: Math.pow(0, 0) === 1 // true and C++ says the same thing: pow(0, 0) == 1 // true WHY? I know that: >Math.pow(0.001, 0.001) ...
2k views
### pow() seems to be out by one here
What's going on here: #include <stdio.h> #include <math.h> int main(void) { printf("17^12 = %lf\n", pow(17, 12)); printf("17^13 = %lf\n", pow(17, 13)); printf("17^14 = %lf\n", ...
2k views
### Math.pow yields different result depending on java version
I'm running the following code on a JDK Version 1.7.0_60: System.out.println(Math.pow(1.5476348320352065, (0.3333333333333333))); The result is: 1.1567055833133086 I'm running exactly the same ...
15k views
### How to get the Power of some Integer in Swift language?
I'm learning swift recently, but I have a basic problem that can't find an answer I want to get something like var a:Int = 3 var b:Int = 3 println( pow(a,b) ) // 27 but the pow function can work ...
3k views
### Replacing extrordinarily slow pow() function
We have a CFD solver and while running a simulation, it was found to run extraordinarily slow on some machines but not others. Using Intel VTune, it was found the following line was the problem (in ...
8k views
### How to: pow(real, real) in x86
I'm searching for the implementation of pow(real, real) in x86 assembler. Also I'd like to understand how the algorithm works.
24k views
### Math.Pow vs multiply operator (performance)
Anyone knows if multiply operator is faster than using the Math.Pow method? Like: n * n * n vs Math.Pow ( n, 3 )
12k views
### How to do a fractional power on BigDecimal in Java?
In my little project I need to do something like Math.pow(7777.66, 5555.44) only with VERY big numbers. I came across a few solutions: Use double - but the numbers are too big Use BigDecimal.pow but ...
458 views
### Why is Math.pow(-0, -7) === -Infinity?
Is there a rationale for Math.pow(-0, x) evaluating to Infinity for all negative x, except for the odd ones when it's -Infinity? I mean: Math.pow(-0, -6); // Infinity Math.pow(-0, -7); ...
4k views
### How to use Byebug with a remote process (e.g., pow)
How do I connect to a remote debugging instance of Byebug (for use with Pow, etc)?
5k views
### Exponentials in python x.**y vs math.pow(x, y)
Which one is more efficient using math.pow or the ** operator? When should I use one over the other? So far I know that x**y can return an int or a float if you use a decimal the function pow will ...
15k views
### BigInteger.pow(BigInteger)?
I'm playing with numbers in Java, and want to see how big a number I can make. It is my understanding that BigInteger can hold a number of infinite size, so long as my computer has enough Memory to ...
537 views
### Why does pow() calculate wrong when Webkit is running?
I have a Qt C++ application where there is a GUI thread in which some floating point calculation happens. It also opens QWebView where there is a flash player with some video. It is obvious that ...
31k views
### finding cube root in C++?
Strange things happen when i try to find the cube root of a number. The following code returns me undefined. In cmd : -1.#IND cout<<pow(( double )(20.0*(-3.2) + 30.0),( double )1/3) While ...
873 views
### GCC C++ pow accuracy
So i was in a computing contest and i noticed a weird bug. pow(26,2) would always return 675, and sometimes 674? even though correct answer is 676. These sort of errors also occur with pow(26,3), ...
2k views
### Strange behaviour of gcc and math.h?
I've been trying to build some code that uses math functions (e.g. pow). math.h is included, and the flag -lm is used during the build. When compilation is called like this (-lm flag at the begining ...
8k views
### C's pow function refuses to work with variable exponent
Let's say I have the following code snippet: int i; double value; for(i = 0; i < CONSTANT; i++) { value = (double)pow(2, i); } Trying to compile this code yields an "undefined reference to ...
65 views
### C99: what is the recomended way to handle exceptions raised by `pow()` (overflow or complex number)
executing double result = pow(base, exponent); with arbitrary base and exponent may result in an attempt to compute a value too big or complex. For example with base=-2, exponent=.5 (square root ...
8k views
### negative pow in python
I have this problem >>> import math >>> math.pow(-1.07,1.3) Traceback (most recent call last): File "<stdin>", line 1, in <module> ValueError: math domain ...
38k views
### pow function in C
I write a C code that have power function that is from math.h library. when I compiled my program, I received an error which is " undefined reference to 'pow' function ", I compile my program using ...
923 views
### What's difference between Math.pow(9, 18) and 9^18
when I use Math.pow(9, 18) =150094635296999136 when I use web Calculator 9^18 = 150094635296999121 (http://web2.0calc.com/) when I use goggle calculator 9^18 = 1.50094635 × 10^17 why it is ...
297 views
### pow(NAN) is very slow
What is the reason for the catastrophic performance of pow() for NaN values? As far as I can work out, NaNs should not have an impact on performance if the floating-point math is done with SSE instead ...
153 views
### Poor performance of Java's Math.pow(x, 2) when x = 0
Background Having noticed that the execution of a java program I am working on was slower than expected, I decided to tinker with the area of code which I thought may be causing the issue - a call to ...
2k views
### C: i got different results with pow(10,2) and pow(10,j), j=2;
this one prints 100: int j=2; int i= pow(10,2); printf("%d\n", i); and this one prints 99: int j=2; int i= pow(10,j); printf("%d\n", i); why?
1k views
### pow() cast to integer, unexpected result
I have some problems using an integer cast for the pow() function in the C programming language. The compiler I'm using is the Tiny C Compiler (tcc version 0.9.24) for the Windows platform. When ...
453 views
### Python : overflow error long int too large to convert to float
I had to compute 2 to the power of 8635. I came across this error when i was computing 2^8635. Any suggestion how to solve this in python. Using Decimal module also didn't help. math.exp(2**8635) ...
4k views
### Math.Pow taking an integer value
From http://msdn.microsoft.com/en-us/library/system.math.pow.aspx int value = 2; for (int power = 0; power <= 32; power++) Console.WriteLine("{0}^{1} = {2:N0}", value, ...
1k views
### Calculating Floating Point Powers (PHP/BCMath)
I'm writing a wrapper for the bcmath extension, and bug #10116 regarding bcpow() is particularly annoying -- it casts the \$right_operand (\$exp) to an (native PHP, not arbitrary length) integer, so ...
5k views
### Math.Pow is not calculating correctly
I'm having a problem with C#. To be precise with the Math.pow(). If I try to calculate 15^14 then I get "29192926025390624". But if I calculate it with Wolfram Alpha I get "29192926025390625". As you ...
753 views
### Is there a substitute for Pow in BigInteger in F#?
I was using the Pow function of the BigInteger class in F# when my compiler told me : This construct is deprecated. This member has been removed to ensure that this type is binary compatible ...
3k views
### java.math.BigInteger pow(exponent) question
I did some tests on pow(exponent) method. Unfortunately, my math skills are not strong enough to handle the following problem. I'm using this code: BigInteger.valueOf(2).pow(var); Results: var | ...
292 views
### Why is 2**100 so much faster than math.pow(2,100)?
When discussing the question Exponentials in python x.**y vs math.pow(x, y), Alfe stated that there would be no good reason for using math.pow instead of the builtin ** operator in python. timeit ...
814 views
### Fastest pow() replacement via modified exp. by squaring when lower powers are already calculated
EDIT: Goal : Generate a ubiquitous method for deriving a custom power function that outperforms the built-in pow(double, uint) by reusing precalculated/cached powers from power calculations on ...
4k views
### return value of pow() gets rounded down if assigned to an integer
I am using the pow function in C and storing the return value in an integer type. see the code snippet below: for (i = 0; i < 5; i++){ val = (int)pow(5, i); printf("%d, ", val); } here ...
239 views
### Loss of precision when using pow in C++
10^1.64605 = 44.2639330165 However in C++ using pow: double p = pow(10,1.64605) returns 44.2641. Is there a way to increase the precision here? I tried casting both sides to long double but that ...
187 views
### Math.pow yields different results upon repeated calls
After upgrading to Java 1.8.0_20 our test system reported errors, but the code was not changed. I found out, that Math.pow() called with exactly the same input parameters yields different results upon ...
22k views
### pow() from math.h library - How to Apply using functions
So I'm writing a bit of code that needs to raise a function's return value to a certain power. I recently discovered that using the '^' operator for exponentiation is useless because in C++ it is ...
4k views
### Best practise with Math.Pow
i'm working on a n image processing library which extends OpenCV, HALCON, ... . The library must be with .NET Framework 3.5 and since my experiences with .NET are limited I would like to ask some ...
196 views
### Math.pow(0.0, 0.0) returns 1; should be undefined or error ?
Math.pow(0.0, 0.0) in Java returns 1 which is wrong. 0^0 is undefined. The same problem exists also in the windows calculator (I am using Win7). Why is that? Mathematica declares it as an error as ...
3k views
### Strange pow(x, y); behaviour
While doing my homework I noticed something really strange that I just can't figure out why. int x = 5; cout << pow(x, 2); The result is 25. That's fine. But if I write the same program like ...
2k views
### Pow() calculates wrong?
I need to use pow in my c++ program and if i call the pow() function this way: long long test = pow(7, e); Where e is an integer value with the value of 23. I always get 821077879 as a ...
4k views
### Bundler::RubyVersionMismatch: Your Ruby version is 1.9.3, but your Gemfile specified 2.0.0
I am using a Mac with Pow, Ruby, Rails and a bunch of useful gems. I recently got this error whenever I try to open my app. Bundler::RubyVersionMismatch: Your Ruby version is 1.9.3, but your Gemfile ...
339 views
### unusual output from pow
The following C code int main(){ int n=10; int t1=pow(10,2); int t2=pow(n,2); int t3=2*pow(n,2); printf("%d\n",t1); printf("%d\n",t2); printf("%d\n",t3); return ...
334 views
### Does pow() work for int data type in C? [duplicate]
I was simply writing a program to calculate the power of an integer. But the output was not as expected. It worked for all the integer numbers except for the power of 5. My code is: #include ...
151 views
### Does Java's Math.pow round off the result?
Today I came across a peculiar behavior of Math.pow(). I am not able to understand the output of the following java code: long N1 = 999999999999999999L; System.out.println("N1 : " + N1); long N2 = ...
2k views
### Wrong result by Java Math.pow
If you try to run the following code public class Main { public static void main(String[] args) { long a = (long)Math.pow(13, 15); System.out.println(a + " " + a%13); } } You ...
4k views
### Creating `**` power operator for Scala?
I quite like the ** syntax for pow, available in many languages (such as Python). Is it possible to introduce this into Scala, without modifying the Scala 'base' code? My attempt at an Int only one: ... | 3,213 | 12,288 | {"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} | 3.140625 | 3 | CC-MAIN-2016-07 | latest | en | 0.86262 |
https://physicsline.com/ballistic-pendulum/ | 1,679,388,917,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296943637.3/warc/CC-MAIN-20230321064400-20230321094400-00188.warc.gz | 533,217,604 | 17,170 | Mechanics
# ballistic pendulum
The ballistic pendulum is a device that determines the bullet velocity of a revolver.
The velocity of a revolver bullet can be determined using a device called a ballistic pendulum.
This device is composed of a block of mass M that is suspended by two strings of negligible mass.
The bullet of mass m and speed v, coming from a horizontal shot, will lodge in the block causing the assembly to be raised to a certain maximum height h.
By the law of conservation of motion, we have that: the moment before the collision is equal to the moment after the collision.
mv = (M+m).V where V is the speed of the assembly (block-bullet).
V = mv/M + m – equation I
After the bullet-block collision, the set gains speed and loses it as it gains height. At the maximum height (h), all the kinetic energy of the set is converted into gravitational potential energy. Therefore, we have that:
Ec = Epg
(M + m).V²/2 = (M + m).gh
Then: V =√(2.gh) – equation II
Equating equations I and II, we have:
mv/ M + m = √(2.gh)
v = [(M + m).√(2.gh)]/m
Schematic of a ballistic pendulum | 280 | 1,099 | {"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-2023-14 | latest | en | 0.875583 |
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randRangeNonZero(-5, 5) randRangeNonZero(-5, 5) randRangeNonZero(-5, 5) randRangeNonZero(-5, 5) complexNumber(A_REAL, A_IMAG) complexNumber(B_REAL, B_IMAG) randRangeWeighted(1, 5, 1, 0.7) * randFromArray([1, -1]) randRangeWeighted(1, 5, 1, 0.7) * randFromArray([1, -1]) F1 * A_REAL F2 * B_REAL F1 * A_IMAG F2 * B_IMAG FA_REAL + FB_REAL FA_IMAG + FB_IMAG
coefficient(F1)(\pink{A}) + coefficient(F2)(\blue{B}) = ?
Complex numbers can be added by separately adding their real and imaginary components.
Distribute the negative sign onto the first complex number:
Distribute the F1 onto the first complex number:
\qquad \begin{eqnarray} coefficient(F1)(\pink{complexNumber(A_REAL, A_IMAG)}) &=& (F1 \cdot \pink{A_REAL}) + (F1 \cdot \pink{A_IMAG}) \\ &=& \pink{complexNumber(FA_REAL, FA_IMAG)} \end{eqnarray}
Distribute the negative sign onto the second complex number:
Distribute the F2 onto the second complex number:
\qquad \begin{eqnarray} coefficient(F2)(\blue{complexNumber(B_REAL, B_IMAG)}) &=& (F2 \cdot \blue{B_REAL}) + (F2 \cdot \blue{B_IMAG}) \\ &=& \blue{complexNumber(FB_REAL, FB_IMAG)} \end{eqnarray}
Now we have:
\pink{complexNumber(FA_REAL, FA_IMAG)} + \blue{complexNumber(FB_REAL, FB_IMAG)}
The real components are \pink{FA_REAL} and \blue{FB_REAL}
The imaginary components are \pink{FA_IMAGi} and \blue{FB_IMAGi}
Adding real components, we get \pink{FA_REAL} + \blue{FB_REAL} = FA_REAL + FB_REAL
Adding imaginary components, we get \pink{FA_IMAGi} + \blue{FB_IMAGi} = FA_IMAG + FB_IMAGi | 543 | 1,664 | {"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} | 3.65625 | 4 | CC-MAIN-2017-34 | latest | en | 0.512967 |
https://multi-converter.com/millimeters-per-second-to-millimeters-per-hour | 1,674,868,871,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499468.22/warc/CC-MAIN-20230127231443-20230128021443-00185.warc.gz | 421,657,353 | 6,583 | # Millimeters per second to Millimeters per hour
Convert mm/s to mm/h
Change to Millimeters per hour to Millimeters per second
Share:
## How to convert Millimeters per second to Millimeters per hour
1 [Millimeters per second] = 3600 [Millimeters per hour]
[Millimeters per hour] = [Millimeters per second] * 3600
To convert Millimeters per second to Millimeters per hour multiply Millimeters per second * 3600.
## Example
11 Millimeters per second to Millimeters per hour
11 [mm/s] * 3600 = 39600 [mm/h]
## Conversion table
Millimeters per second Millimeters per hour
0.01 mm/s36 mm/h
0.1 mm/s360 mm/h
1 mm/s3600 mm/h
2 mm/s7200 mm/h
3 mm/s10800 mm/h
4 mm/s14400 mm/h
5 mm/s18000 mm/h
10 mm/s36000 mm/h
15 mm/s54000 mm/h
50 mm/s180000 mm/h
100 mm/s360000 mm/h
500 mm/s1800000 mm/h
1000 mm/s3600000 mm/h | 255 | 809 | {"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} | 2.859375 | 3 | CC-MAIN-2023-06 | longest | en | 0.308804 |
https://oeis.org/A056868 | 1,708,548,173,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473558.16/warc/CC-MAIN-20240221202132-20240221232132-00645.warc.gz | 449,819,736 | 5,006 | The OEIS is supported by the many generous donors to the OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A056868 Numbers that are not nilpotent numbers. 14
6, 10, 12, 14, 18, 20, 21, 22, 24, 26, 28, 30, 34, 36, 38, 39, 40, 42, 44, 46, 48, 50, 52, 54, 55, 56, 57, 58, 60, 62, 63, 66, 68, 70, 72, 74, 75, 76, 78, 80, 82, 84, 86, 88, 90, 92, 93, 94, 96, 98, 100, 102, 104, 105, 106, 108, 110, 111, 112, 114, 116, 117, 118, 120 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 COMMENTS A number is nilpotent if every group of order n is nilpotent. The sequence "Numbers of the form (k*i + 1)*k*j with i, j >= 1 and k >= 2" agrees with this for the first 146 terms but then differs. Cf. A300737. - Gionata Neri, Mar 11 2018 LINKS T. D. Noe, Table of n, a(n) for n = 1..10000 J. Pakianathan and K. Shankar, Nilpotent Numbers, Amer. Math. Monthly, 107, August-September 2000, pp. 631-634. FORMULA n is in this sequence if p^k = 1 mod q for primes p and q dividing n such that p^k divides n. - Charles R Greathouse IV, Aug 27 2012 EXAMPLE From Bernard Schott, Dec 19 2021: (Start) There are 2 groups with order 6: C_6 that is cyclic so nilpotent, and the symmetric group S_3 that is not nilpotent, hence 6 is a term. There are also 2 groups with order 10: C_10 that is cyclic so nilpotent, and the dihedral group D_10 that is not nilpotent, hence 10 is another term. (End) MATHEMATICA nilpotentQ[n_] := With[{f = FactorInteger[n]}, Sum[ Boole[ Mod[p[[1]]^p[[2]], q[[1]]] == 1], {p, f}, {q, f}]] == 0; Select[ Range[120], !nilpotentQ[#]& ] (* Jean-François Alcover, Sep 03 2012 *) PROG (PARI) is(n)=my(f=factor(n)); for(k=1, #f[, 1], for(j=1, f[k, 2], if(gcd(n, f[k, 1]^j-1)>1, return(1)))); 0 \\ Charles R Greathouse IV, Sep 18 2012 (Haskell) a056868 n = a056868_list !! (n-1) a056868_list = filter (any (== 1) . pks) [1..] where pks x = [p ^ k `mod` q | let fs = a027748_row x, q <- fs, (p, e) <- zip fs \$ a124010_row x, k <- [1..e]] -- Reinhard Zumkeller, Jun 28 2013 CROSSREFS Complement of A056867. Subsequence of A060652; A068919 is a subsequence. Cf. A003277, A051532, A056866, A027748, A124010, A300737. Sequence in context: A330397 A135711 A161543 * A069209 A060702 A054741 Adjacent sequences: A056865 A056866 A056867 * A056869 A056870 A056871 KEYWORD nonn,nice,easy AUTHOR N. J. A. Sloane, Sep 02 2000 EXTENSIONS More terms from Francisco Salinas (franciscodesalinas(AT)hotmail.com), Dec 25 2001 STATUS approved
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Last modified February 21 15:08 EST 2024. Contains 370236 sequences. (Running on oeis4.) | 1,055 | 2,814 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2024-10 | latest | en | 0.737831 |
https://mathworld.wolfram.com/InfiniteDiscontinuity.html | 1,721,855,793,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518454.54/warc/CC-MAIN-20240724202030-20240724232030-00423.warc.gz | 331,477,830 | 13,784 | TOPICS
# Infinite Discontinuity
A real-valued univariate function is said to have an infinite discontinuity at a point in its domain provided that either (or both) of the lower or upper limits of fails to exist as tends to .
Infinite discontinuities are sometimes referred to as essential discontinuities, phraseology indicative of the fact that such points of discontinuity are considered to be "more severe" than either removable or jump discontinuities.
The figure above shows the piecewise function
(1)
a function for which both and fail to exist. In particular, has an infinite discontinuity at .
It is not uncommon for authors to say that univariate functions defined on a domain and admitting vertical asymptotes of the form have infinite discontinuities there though, strictly speaking, this terminology is incorrect unless such functions are defined piecewise so that . For example, the function has vertical asymptotes at , , though it has no discontinuities of any kind on its domain.
Unsurprisingly, one can extend the above definition to infinite discontinuities of multivariate functions as well.
Branch Cut, Continuous, Discontinuity, Discontinuous, Discontinuous Function, Essential Singularity, Isolated Singularity, Jump Discontinuity, Polar Coordinates, Pole, Removable Discontinuity, Removable Singularity, Singular Point, Singularity
This entry contributed by Christopher Stover
## Explore with Wolfram|Alpha
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## Cite this as:
Stover, Christopher. "Infinite Discontinuity." From MathWorld--A Wolfram Web Resource, created by Eric W. Weisstein. https://mathworld.wolfram.com/InfiniteDiscontinuity.html | 350 | 1,655 | {"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-2024-30 | latest | en | 0.904484 |
https://psychology.stackexchange.com/questions/1283/how-many-bits-of-data-would-it-take-to-represent-an-entire-life-as-a-film | 1,713,833,854,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818452.78/warc/CC-MAIN-20240423002028-20240423032028-00571.warc.gz | 413,082,825 | 41,635 | # How many bits of data would it take to represent an entire life as a film?
Imagine your life as a film. Add all the stuff you know.
Is it possible to express this amount of data in bits? (A rough order of magnitude would be ok.)
• Is data compression allowed? :P Or use Kolmogorov stuff to find that one's entire life is filled with repetitious events. Just poking fun... Jun 26, 2012 at 6:11
• @PheonixEnder comment is extremely relevant. The way we encode movies and the way the brain encodes memories is not even remotely similar. You don't remember a video of the photons that landed on your retina. You remember a series of associations between symbols and meanings in your mind. This can be compressed much more easily and the 'image' of remembering reconstructed dynamically. The way the question is framed, it is committing the homunculus fallacy with the 'little man' watching videos off his external hard-drives. Jun 26, 2012 at 9:35
This is a tough question, and there's a few routes to go. When we are watching this life played back, do we want to be able to go to any arbitrary moment and observe what was happening, or do we only want to watch life as we watch it? This would be the difference between recording all information that bombards our brain, or just the information that we are attending to.
We'll go with the first one, as this will be easier.
I'm going to ignore 'thoughts', and just look at information as it enters the body.
We need to know a couple things before we start.
What is the data rate of our eyes? What is the data rate of our ears?
As these are our most dominant senses, if we want a few orders of magnitude for information, we will get it from these two.
Researchers at the University of Pennsylvania School of Medicine say that the data rate for the eye is about 10 million bits per second, or about 1.25 Mb/s.
Our eyes have about 120 million cells, and they sample at roughly 30 Hz. Determining the number of 'bits' is hard, as eyes have a nonlinear response, so the amount of information they are transmitting is dependent on what the cells are observing. I will trust their number.
As vision is our most powerful sense, and accounts for over 40% of the processing in our brain, we know that the majority of the information will be from the eyes. (EDIT: This is incorrect, see the data below.)
Now onto the ears - our ears can hear up to about 20 KHz, and to actually pick up these sounds we need to sample at twice that frequency, or 40 KHz. That doesn't tell us how much information we actually get though. This question on Yahoo! Answers goes a little bit deeper, specifying that ears cannot distinguish information that is represented in more than 16 bits.
This is kind of an interesting way to determine the sensitivity of our ears, as we don't have to wonder 'well, how well can I distinguish what I hear?', rather the question is 'We have these samples of audio, and people can't hear the difference between them, therefore our ears cannot transmit this much information'.
So we have eyes and ears, and we now need to do some math.
I'm going to assume all these rates are fixed, the actual information that is transmitted is nonlinear as the spike trains from our cells change depending on what they are sensing.
I wrote a little python program, available here.
With the information from above in hand, we have the following:
Eyes data rate is 2.4 MB per second.
Ears data rate is 687.5 MB per second.
over 60 years, our eyes will transmit 4.2 PB
over 60 years, our ears will transmit 1.2 EB
The total information we will receive will be upwards of...
1.2 Exabytes
Now this sounds a little silly, do our ears really transmit more information than our eyes? Actually yes, because they are sampled so much more frequently. The eyes sample at roughly 30 Hz, whereas the ears sample at 44 KHz.
This analysis ignores the number of cells, and is likely incorrect for a number of reasons. We ignore the other senses, as well as information being sent between brain regions (processing of information).
This question brings up an interesting point though, the raw information that we receive is meaningless, only through successive iterations of information abstraction (what neural networks do, they operate on regions of information, gaining a higher and higher level picture) can we begin to distinguish the world as it is. When we see something, we can immediately break the world apart into segments, we know what can move, we know where things are, we know when we put them there. This isn't information that we were given at our current place in time, it's rules that we have learned about what we see in the world through a lifetime.
Additional source of information: How many colors and shades can the human eye distinguish in a single scene? This Photography.SE question goes into logic about how much information we can discern, more closely matching the analysis on the Ears, rather than the research data that we used.
• This is only an upper bound if the numbers are correct. One would need the information provided by the environment. Even if our ears can sample at high rate, the environment only produces way smaller amount of information... Jun 26, 2012 at 18:21 | 1,148 | 5,255 | {"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-2024-18 | latest | en | 0.958512 |
http://pistulka.com/Other/?m=20141221 | 1,601,000,907,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400221980.49/warc/CC-MAIN-20200925021647-20200925051647-00397.warc.gz | 109,322,664 | 7,284 | ## Simple Formula for Converting Compound Interest Rates
This workbook has a simple formula to convert compound interest rates. =IF(t>500,f*LN(1+g/f),IF(f>=500,t*(EXP(g/t)-1),(t*((1+g/f)^(f/t))-t))) Where: g = Current interest rate f = Number of times the current rate compounds per year t = Number of times the converted new rate compounds per year This lookup table grabs two of the numbers for the formula Continuous compounding in Excel […] | 112 | 446 | {"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-2020-40 | latest | en | 0.770904 |
https://mathematica.stackexchange.com/questions/195743/how-to-plot-intensity-of-smoothkerneldistribution | 1,563,458,971,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195525634.13/warc/CC-MAIN-20190718125048-20190718151048-00027.warc.gz | 464,632,741 | 35,728 | # How to plot Intensity of SmoothKernelDistribution?
For example, we have a list of values
data = RandomVariate[NormalDistribution[], 10^3];
then
SmoothHistogram[data]
By default plot the PDF of data like this
But there is also options to plot the Intensity of data using
SmoothHistogram[data, Automatic, "Intensity"]
which gives
This intensity curve is what I want. But I also want the function of this intensity curve, so I can to subtract two such intensity curve function to plot a new one.
I look into the doc, it seems that SmoothKernelDistribution suite the needs. For example, let
dist=SmoothKernelDistribution[data]
then
Plot[PDF[dist, {x}], {x, -3, 3}]
gives
Which is equivalent to SmoothHistogram[data]. But how to get equivalent Intensity plot using SmoothKernelDistribution? Is there a complete list of function that is supported by SmoothKernelDistribution like PDF?
Clear["Global*"]
SeedRandom[0];
data = RandomVariate[NormalDistribution[], 10^3];
{xmin, xmax} = MinMax@data;
smhist = SmoothHistogram[data, Automatic, "Intensity"]
p[x_] = PDF[SmoothKernelDistribution[data], x];
The factor required to scale the PDF to the intensity curve is
amp = Divide @@
((Cases[#, Line[pts_] :> pts, Infinity][[1, All, 2]] //
Max) & /@
{smhist, Plot[p[x], {x, xmin, xmax}]});
intensity[x_] = amp*p[x];
Plot[intensity[x], {x, xmin, xmax}]
EDIT: A more straightforward approach is to use the points from smhist to define an InterpolatingFunction.
int = Interpolation[Cases[smhist, Line[pts_] :> pts, Infinity][[1]]];
Plot[int[x], {x, xmin, xmax}]
`
• Wow, powerful postprocessing. So there is no built in statistic function corresponding to "indensity"? – matheorem Apr 23 at 0:05 | 440 | 1,717 | {"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": 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} | 3.3125 | 3 | CC-MAIN-2019-30 | longest | en | 0.697686 |
https://www.ibpsguide.com/quantitative-aptitude-special-practice-questions-day-02/ | 1,632,729,685,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780058373.45/warc/CC-MAIN-20210927060117-20210927090117-00098.warc.gz | 787,980,468 | 73,766 | # IBPS / LIC Clerk Prelims Quantitative Aptitude Questions 2019 (Day-02)
Dear Aspirants, Our IBPS Guide team is providing new series of Quantitative Aptitude Questions for IBPS Clerk Prelims 2019 so the aspirants can practice it on a daily basis. These questions are framed by our skilled experts after understanding your needs thoroughly. Aspirants can practice these new series questions daily to familiarize with the exact exam pattern and make your preparation effective.
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[WpProQuiz 7211]
Data Interpretation
Directions (1 – 6): Study the following information carefully and answer the questions given below.
The given table shows the number of students in four different colleges in four different departments.
1) If the ratio of the number of students of CSE from A to C is 21:32, find the number of ME students from C?
a) 380
b) 360
c) 340
d) 320
e) None of these
2) Total number of EEE students from all the colleges together is 1900, then what is the number of ME students from B?
a) 420
b) 460
c) 440
d) 480
e) None of these
3) If the ratio of the number of ME students from B to C is 24: 19, what is the number of CSE students from C?
a) 580
b) 600
c) 620
d) 660
e) None of these
4) If the total number of students from A is equal to the total number of ME students from all the colleges together. The number of ECE students from C is approximately what percent of the total number of students from A all the departments together?
a) 18%
b) 20%
c) 22%
d) 24%
e) None of these
5) If the average number of ECE students from all the colleges together is 430 and the number of ECE students from D is equal to the number of ME students from A. What is the ratio of the number of ECE students from A to the number of CSE students from D?
a) 7: 8
b) 8: 9
c) 5: 6
d) 4: 3
e) None of these
6) What is the difference between the number of EEE and ME students from B together and the number of CSE and ME students from C together?
a) 180
b) 200
c) 220
d) 140
e) 160
Missing number series
Directions (7 – 10): What value should come in the place of (?) in the following number series?
7) 4, 6, 18, 66, 284, ?
a) 360
b) 972
c) 1280
d) 1450
e) 1620
8) 12, 22, 38, ?, 101.5, 160.25
a) 63
b) 65
c) 67
d) 69
e) 71
9) 18, 54, 79, 95, 104, ?
a) 110
b) 106
c) 108
d) 112
e) 114
10) 995, 723, 505, 335, 207, ?
a) 100
b) 105
c) 110
d) 115
e) 120
Directions (1-6) :
Total number of CSE students =2000
Number of CSE students from A and C = 2000 – 580 – 360 = 1060
Number of CSE students from C = 32/53 * 1060 = 640
Number of ME students from C = 2000 – 640 – 540 – 440 = 380
Number of EEE students from B = 1900 – 440 – 540 – 560 = 360
Number of ME students from B = 1700 – 580 – 360 – 280 = 480
Number of ME students from B and C = 2000 – 680 – 460 = 860
Number of ME students from C = 19/43 * 860 = 380
Number of CSE students from C = 2000 – 540 – 440 – 380 = 640
Required percentage = 440/2000 * 100 = 22%
Total number of ECE students = 430 * 4 = 1720
Number of ECE students from D = 680
Number of ECE students from A = 1720 – 280 – 440 – 680 = 320
Required ratio = 320: 360 = 8: 9
Number of EEE and ME students from B = 1700 – 580 – 280 = 840
Number of CSE and ME students from C = 2000 – 540 – 440 = 1020
Difference = 1020 – 840 = 180
Directions (7-10) :
(4 + 2) * 1 = 6
(6 + 3) * 2 = 18
(18 + 4) * 3 = 66
(66 + 5) * 4 = 284
(284 + 6) * 5 = 1450
12 * 1.5 + 4 = 22
22 * 1.5 + 5 = 38
38 * 1.5 + 6 = 63
63 * 1.5 + 7 = 101.5
101.5 * 1.5 + 8 = 160.25
18 + 62 = 54
54 + 52 = 79
79 + 42 = 95
95 + 32 = 104
104 + 22 = 108
103 – 5 = 995
93 – 6 = 723
83 – 7 = 505
73 – 8 = 335
63 – 9 = 207
53 – 10 = 115 | 1,383 | 3,851 | {"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} | 4.09375 | 4 | CC-MAIN-2021-39 | latest | en | 0.903964 |
https://www.tutorpace.com/common-core/identifying-parts-of-expressions-online-tutoring | 1,642,632,072,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320301592.29/warc/CC-MAIN-20220119215632-20220120005632-00362.warc.gz | 1,096,929,865 | 9,734 | # Identifying parts of expressions
## Online Tutoring Is The Easiest, Most Cost-Effective Way For Students To Get The Help They Need Whenever They Need It.
There are different types of algebraic expressions in math. An expression can consist of different variables, numbers and exponents. Variable is denoted by a letter such as (x, y, z, m, n….).The exponent is the power of the number or the variables. The exponent tell the number of times the base variable or number is to be multiplied. The number multiplied to the variable is called the coefficient of the variable. The degree of the expression is the highest exponent of the variable in the expression.
Example 1: Identify the degree, coefficient, variable of the expression 10x3?
Solution: Here the given algebraic expression is 10x3.
The coefficient of the expression is = 10
The variable of the expression = x.
The exponent of the variable x is = 3.
The given expression is a monomial expression as the there is only one term in the expression.
The degree of the expression is = 3.
Question: Multiple choice question (Pick the correct option.)
Which of the following is a polynomial expression?
a) 5b b) 6b + 1 c) 20 d) None of these. | 284 | 1,240 | {"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} | 4.0625 | 4 | CC-MAIN-2022-05 | latest | en | 0.860145 |
https://www.dataunitconverter.com/byte-per-day-to-yobibit-per-hour | 1,709,005,860,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474670.19/warc/CC-MAIN-20240227021813-20240227051813-00098.warc.gz | 716,063,168 | 17,092 | # Byte/Day to Yibit/Hr → CONVERT Bytes per Day to Yobibits per Hour
expand_more
info 1 Byte/Day is equal to 0.0000000000000000000000002757268708510092 Yibit/Hr
Input Bytes per Day (Byte/Day) - and press Enter.
Byte/Day
Sec
Min
Hr
Day
Sec
Min
Hr
Day
S = Second, M = Minute, H = Hour, D = Day
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toc Table of Contents
## Bytes per Day (Byte/Day) Versus Yobibits per Hour (Yibit/Hr) - Comparison
Bytes per Day and Yobibits per Hour are units of digital information used to measure storage capacity and data transfer rate.
Bytes per Day is one of the very "basic" digital unit where as Yobibits per Hour is a "binary" unit. One Byte is equal to 8 bits. One Yobibit is equal to 1024^8 bits. There are 151,115,727,451,828,646,838,272 Byte in one Yobibit. Find more details on below table.
Bytes per Day (Byte/Day) Yobibits per Hour (Yibit/Hr)
Bytes per Day (Byte/Day) is a unit of measurement for data transfer bandwidth. It measures the number of Bytes that can be transferred in one Day. Yobibits per Hour (Yibit/Hr) is a unit of measurement for data transfer bandwidth. It measures the number of Yobibits that can be transferred in one Hour.
## Bytes per Day (Byte/Day) to Yobibits per Hour (Yibit/Hr) Conversion - Formula & Steps
The Byte/Day to Yibit/Hr Calculator Tool provides a convenient solution for effortlessly converting data rates from Bytes per Day (Byte/Day) to Yobibits per Hour (Yibit/Hr). Let's delve into a thorough analysis of the formula and steps involved.
Outlined below is a comprehensive overview of the key attributes associated with both the source (Byte) and target (Yobibit) data units.
Source Data Unit Target Data Unit
Equal to 8 bits
(Basic Unit)
Equal to 1024^8 bits
(Binary Unit)
The conversion from Data per Day to Hour can be calculated as below.
x 60
x 60
x 24
Data
per
Second
Data
per
Minute
Data
per
Hour
Data
per
Day
÷ 60
÷ 60
÷ 24
The formula for converting the Bytes per Day (Byte/Day) to Yobibits per Hour (Yibit/Hr) can be expressed as follows:
diamond CONVERSION FORMULA Yibit/Hr = Byte/Day x 8 ÷ 10248 / 24
Now, let's apply the aforementioned formula and explore the manual conversion process from Bytes per Day (Byte/Day) to Yobibits per Hour (Yibit/Hr). To streamline the calculation further, we can simplify the formula for added convenience.
FORMULA
Yobibits per Hour = Bytes per Day x 8 ÷ 10248 / 24
STEP 1
Yobibits per Hour = Bytes per Day x 8 ÷ (1024x1024x1024x1024x1024x1024x1024x1024) / 24
STEP 2
Yobibits per Hour = Bytes per Day x 8 ÷ 1208925819614629174706176 / 24
STEP 3
Yobibits per Hour = Bytes per Day x 0.0000000000000000000000066174449004242213 / 24
STEP 4
Yobibits per Hour = Bytes per Day x 0.0000000000000000000000002757268708510092
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Example : By applying the previously mentioned formula and steps, the conversion from 1 Bytes per Day (Byte/Day) to Yobibits per Hour (Yibit/Hr) can be processed as outlined below.
1. = 1 x 8 ÷ 10248 / 24
2. = 1 x 8 ÷ (1024x1024x1024x1024x1024x1024x1024x1024) / 24
3. = 1 x 8 ÷ 1208925819614629174706176 / 24
4. = 1 x 0.0000000000000000000000066174449004242213 / 24
5. = 1 x 0.0000000000000000000000002757268708510092
6. = 0.0000000000000000000000002757268708510092
7. i.e. 1 Byte/Day is equal to 0.0000000000000000000000002757268708510092 Yibit/Hr.
Note : Result rounded off to 40 decimal positions.
You can employ the formula and steps mentioned above to convert Bytes per Day to Yobibits per Hour using any of the programming language such as Java, Python, or Powershell.
### Unit Definitions
#### What is Byte ?
A Byte is a unit of digital information that typically consists of 8 bits and can represent a wide range of values such as characters, binary data and it is widely used in the digital world to measure the data size and data transfer speed.
- Learn more..
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#### What is Yobibit ?
A yobibit (Yib or Yibit) is a binary unit of digital information that is equal to 1,208,925,819,614,629,174,706,176 bits and is defined by the International Electro technical Commission(IEC). The prefix 'yobi' is derived from the binary number system and it is used to distinguish it from the decimal-based 'yottabit' (Yb). It is widely used in the field of computing as it more accurately represents the amount of data storage and data transfer in computer systems.
- Learn more..
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## Excel Formula to convert from Bytes per Day (Byte/Day) to Yobibits per Hour (Yibit/Hr)
Apply the formula as shown below to convert from 1 Bytes per Day (Byte/Day) to Yobibits per Hour (Yibit/Hr).
A B C
1 Bytes per Day (Byte/Day) Yobibits per Hour (Yibit/Hr)
2 1 =A2 * 0.0000000000000000000000066174449004242213 / 24
3
If you want to perform bulk conversion locally in your system, then download and make use of above Excel template.
## Python Code for Bytes per Day (Byte/Day) to Yobibits per Hour (Yibit/Hr) Conversion
You can use below code to convert any value in Bytes per Day (Byte/Day) to Bytes per Day (Byte/Day) in Python.
bytesperDay = int(input("Enter Bytes per Day: "))
yobibitsperHour = bytesperDay * 8 / (1024*1024*1024*1024*1024*1024*1024*1024) / 24
print("{} Bytes per Day = {} Yobibits per Hour".format(bytesperDay,yobibitsperHour))
The first line of code will prompt the user to enter the Bytes per Day (Byte/Day) as an input. The value of Yobibits per Hour (Yibit/Hr) is calculated on the next line, and the code in third line will display the result.
## Frequently Asked Questions - FAQs
#### How many Yobibits(Yibit) are there in a Byte?expand_more
There are 0.0000000000000000000000066174449004242213 Yobibits in a Byte.
#### What is the formula to convert Byte to Yobibit(Yibit)?expand_more
Use the formula Yibit = Byte x 8 / 10248 to convert Byte to Yobibit.
#### How many Bytes are there in a Yobibit(Yibit)?expand_more
There are 151115727451828646838272 Bytes in a Yobibit.
#### What is the formula to convert Yobibit(Yibit) to Byte?expand_more
Use the formula Byte = Yibit x 10248 / 8 to convert Yobibit to Byte.
#### Which is bigger, Yobibit(Yibit) or Byte?expand_more
Yobibit is bigger than Byte. One Yobibit contains 151115727451828646838272 Bytes.
## Similar Conversions & Calculators
All below conversions basically referring to the same calculation. | 1,898 | 6,268 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2024-10 | latest | en | 0.782182 |
http://www.talkstats.com/showthread.php/19647-(easy)-Maximum-Likelihood-Regression-question...?p=59209 | 1,508,529,460,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187824325.29/warc/CC-MAIN-20171020192317-20171020212317-00846.warc.gz | 572,078,644 | 16,958 | # Thread: (easy) Maximum Likelihood Regression question...
1. ## (easy) Maximum Likelihood Regression question...
hello everyone. i am trying to get my head around a very easy example that i read somewhere explaining maximum likelihood estimation on the OLS regression context.
now, thanks to this board i now understand that the assumption of normality is on the residuals of a regression equation, and not on the variables themselves. however, i was reading this book where it says that, under the OLS regression model, the distribution of Y is something like:
Y~N(Xb, sigma)
and if we use the normal pdf in this case, set it to 0, substitute the appropriate parameters and differentiate, we'll get maximum likelihood estimates of the b-weights. supposedly, from what i've been reading, only in this case (i guess where the expected value of Y is some linear function of my predictor(s) X and sigma is constant) will maximum likelihood estimates match regular OLS estimates... however (please correct me if i'm wrong) when i say Y~N(something, another thing) aren't i saying that Y follows a normal distribution with with something as a mean and another thing as a standard deviation? and, if this is true, then isnt there a distributional assumption on the dependent variable Y?
i'm very confused, i hope someone can help me clarify this...
2. ## Re: (easy) Maximum Likelihood Regression question...
Y~N(Xb, sigma)
I believe the answer to your confusion is understanding the confusion between the sample and the population. In the population the dependant variable approximates normality where as in the sample you are unlikely to have a normal distribution. Capital Y refers to a population where as lower case y refers to a sample. You have a capital Y so you're refering to the population.
I think this will relieve your confusion but if not let us know.
trinker
3. ## Re: (easy) Maximum Likelihood Regression question...
thank you for looking at my post!
uhmmm.. not particularly. if i am doing a maximum likelihood estimation of my regression weights i need to choose something for the probability density function from which i will derive the log-likelihood equation i plan on differentiating. now, i may be completely off, but i don't quite see why making the difference between sample and population would have any impact on which pdf i choose to find my MLEs. the thing that i dont quite grasp is why, on the one hand, OLS' matrix algebra approximation of the regression weights only matches the MLE weights if the dependent Y follows a normal distribution. why is there a distributional assumption on the MLE approach which is not needed on the OLS matrix approach? i mean, the quick answer is because you need to choose something to solve for the loglikelihood and that something happens to be the normal pdf, but what i dont quite get is then why this same assumption (normally distributed Y's) doesnt apply to OLS' regular matrix approach in order to yield optimal BLUE results...
4. ## Re: (easy) Maximum Likelihood Regression question...
http://en.wikipedia.org/wiki/Gauss%E...Markov_theorem
In the BLUE set up, you really do not require the error/response to be normally distributed.
So from the MLE perspective, the reason why it coincide with the OLS result is that the function you minimize in finding the OLS, the residual sum of squares
also appear in the likelihood function of normal (the negative of the exponent), and thus by maximizing the likelihood you are minimizing the residual.
But actually I still not sure which part you get confused.
6. ## Re: (easy) Maximum Likelihood Regression question...
Originally Posted by BGM
also appear in the likelihood function of normal (the negative of the exponent), and thus by maximizing the likelihood you are minimizing the residual.
aha! it's this part where i'm getting lost! so i start by sayinug my Y follows a normal distribution Y~N(XB, sigma), right? then i take my normal pdf to the e to the whole thing up there and after i take the log the e dissappears and i'm left with what will later give me pretty much the same equations (i think they call them "normal equations" rite?) i'd need to solve to get the b-weights. where i am getting lost is that, in the MLE setting, on my first step i claim " dependent Y follows a normal distribution with mean XB and constant variance sigma" whereas in OLS regression i do not need to make such claim for Y, only looking at the residuals should suffice. so why is it that i cannot say, for example, "dependent Y follows a poisson or gamma or something distribution", solve for the MLEs of the b-weights and realize that i dont care which pdf i choose, since i'm only interested in the normality of the residuals?
7. ## Re: (easy) Maximum Likelihood Regression question...
When you say something like Y ~ that isn't quite true. What we mean is . We're talking about the conditional distribution of Y given pretty much everything else. We don't know or care about the marginal distribution of Y because depending on what X and beta are the marginal distribution could be really ugly/unrecognizable.
Also if the conditional distribution of Y is something other than a normal distribution then the residuals won't be normally distributed.
8. ## The Following User Says Thank You to Dason For This Useful Post:
will22 (08-27-2011)
9. ## Re: (easy) Maximum Likelihood Regression question...
Originally Posted by Dason
What we mean is . We're talking about the conditional distribution of Y given pretty much everything else.
THANK YOU! now *THAT* makes a lot more sense... stoopid journal articles for not writting things appropriately... because if the assumption is on the conditional distribution and not on the variable itself, then it makes perfect sense why the residuals should be normally distributed to satisfy the assumptions... oh god, thank-you, i've been staring at this for days.
10. ## Re: (easy) Maximum Likelihood Regression question...
by the way... i see there are a lot of introductory books out there on regression, ANOVA, logistic regression and stuff like that for anyone who's getting started on this. however, i can't seem to find many (or any) that focus on maximum likelihood.... and it seems like it's a REALLY, REALLY important thing to learn and understand well if one is to progress further in statistics... do you have any ideas/recommendations about intro books to maximum likelihood estimation? it seems like everybody learns about this thing but i cant find where people are learning about... and there're so many extensions (robust MLE, penalized MLE, weighted MLE, marginal MLE...oh god..)
11. ## Re: (easy) Maximum Likelihood Regression question...
Yes in many places people will try to omit the conditional one to shorten the notations, which can make confusions.
http://en.wikipedia.org/wiki/Generalized_linear_model
By the way if you really want to generalize the conditional distribution of the response from normal to a general exponential family, then you may read about the Generalize Linear Model. In this case if you have different link function and different distribution, then you have different estimates. E.g. the Gaussian regression model you talk about is a special case, in which you use the identity link, and thus you model the mean as a linear function of the regressors.
12. ## The Following User Says Thank You to BGM For This Useful Post:
will22 (08-27-2011)
13. ## Re: (easy) Maximum Likelihood Regression question...
thank you... i am starting to read on those as well but i guess i'm going step-by-step, hehehe...
i would also like to ask you, BGM, the same question i asked Dason. you seem to understand these things very well as well. are there any intro books you know about specifically for maximum likelihood? i'm just appalled as for how much has been written for beginners in regression/ANOVA yet a technique as important as maximum likelihood is so overlooked for people like me who would like to understand the basics of how it works... i can only either find very advanced books geared towards very specific techniques within the MLE family or they mention it in book chapters like "oh, and btw, here's another method of estimation..." but i keep on trying to find an introductory book for maximum likelihood estimation and keep on coming empty-handed | 1,776 | 8,379 | {"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} | 3.140625 | 3 | CC-MAIN-2017-43 | latest | en | 0.912333 |
https://marineinbox.com/marine-exams/ship-stability-during-dry-dock/ | 1,701,599,527,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100499.43/warc/CC-MAIN-20231203094028-20231203124028-00385.warc.gz | 440,402,589 | 22,142 | # SHIP STABILITY DURING DRY DOCK
0
3791
Situation while sitting on the chocks inside drydock – The GM, BM, KB and KG vary in such situation, as a result the transverse KM (KB+BM) will become 0. It is because there is no displaced volume to provide buoyant forces.
Cut to the GM, which is given by the difference between KM and KG, KG is an inherent characteristic of the ship geometry & cannot be 0. Hence, a negative(-) GM results in the loss of positive stability & so the ship should be balanced before it topples.
Trim by aft is a common feature in ships when moving in the water. When the vessel will start taking support on the blocks along the entire length, it will first touch at the stern. The remaining part of the ship may get pivoted about this point. Now, if the ship is able to keep a positive GM, it won‘t heel to either side, enable a comfortable settlement.
This is quite difficult in ships with a slender form without a flat bottom. So here comes an advantage in the design where it might be suitable not to design very large slender form of the ships as performing manoeuvres would be a problem. Such ships would need side supports & the bilge blocks‘ in addition to the usual support.
Recent day dry docks provide side rams(operated hydraulically) as opposed to wedging with wooden shores & terraces in earlier constructions.
How much loss of the GM is allowable? For this, the dry docks use a graphical plot between the transverse metacentric height & the effect given by the blocks in supporting the vessel. Which is also calculated is the docking trim (difference in forward & aft drafts) and the reaction(called the upthrust) from the blocks. The graph quite reasonable has a slope KM/, which shows the loss of KM(& so the GM also) per unit tonne of upthrust equivalent added. Another characteristic of the dry dock which disturb your ship‘s stability is what is called declivity. In simple terms, this declivity is the slope indicated in terms of multiples of unit metre rise per 100 metres. It is advisable for the vessels like tugs & fishing trawlers with a rise of keel to be repaired in the dock with some declivity as they run with proper propeller immersion as a result of the deep draft aft of these vessels. | 502 | 2,249 | {"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-2023-50 | latest | en | 0.948222 |
https://www.centerspace.net/examples/nmath/csharp/tridiagonal-factorization-example.php | 1,675,360,701,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500035.14/warc/CC-MAIN-20230202165041-20230202195041-00016.warc.gz | 710,471,003 | 4,843 | # C# Tri Diag Fact Example
← All NMath Code Examples
```using System;
using CenterSpace.NMath.Core;
namespace CenterSpace.NMath.Examples.CSharp
{
/// <summary>
/// A .NET example in C# demonstrating the features of the factorization classes for
/// tridiagonal matrices.
/// </summary>
class TriDiagFactExample
{
static void Main( string[] args )
{
// Construct a tridiagonal matrix with random entries.
int rows = 5;
int cols = 5;
var rng = new RandGenUniform( -1, 1 );
rng.Reset( 0x124 );
var data1 = new FloatComplexVector( cols, rng );
var data2 = new FloatComplexVector( cols - 1, rng );
var data3 = new FloatComplexVector( cols - 1, rng );
var A = new FloatComplexTriDiagMatrix( rows, cols );
A.Diagonal()[Slice.All] = data1;
A.Diagonal( 1 )[Slice.All] = data2;
A.Diagonal( -1 )[Slice.All] = data3;
Console.WriteLine();
Console.WriteLine( "A =" );
Console.WriteLine( A.ToTabDelimited( "F3" ) );
Console.WriteLine();
// A =
// (-0.497,0.332) (0.560,0.306) (0.000,0.000) (0.000,0.000) (0.000,0.000)
// (0.773,0.358) (-0.250,0.576) (0.220,-0.077) (0.000,0.000) (0.000,0.000)
// (0.000,0.000) (-0.863,0.203) (0.196,0.182) (-0.168,-0.259) (0.000,0.000)
// (0.000,0.000) (0.000,0.000) (-0.585,0.622) (-0.044,0.074) (-0.924,0.621)
// (0.000,0.000) (0.000,0.000) (0.000,0.000) (-0.705,0.124) (-0.325,-0.280)
// Construct a tridiagonal factorization class.
var fact = new FloatComplexTriDiagFact( A );
// Check to see if A is singular.
string isSingularString = fact.IsSingular ? "A is singular" : "A is NOT singular";
Console.WriteLine( isSingularString );
// Retrieve information about the matrix A.
FloatComplex det = fact.Determinant();
// In order to get condition number, factor with estimateCondition = true
fact.Factor( A, true );
float rcond = fact.ConditionNumber();
FloatComplexMatrix AInv = fact.Inverse();
Console.WriteLine();
Console.WriteLine( "Determinant of A = {0}", det );
Console.WriteLine();
Console.WriteLine( "Reciprocal condition number = {0}", rcond );
Console.WriteLine();
Console.WriteLine( "A inverse =" );
Console.WriteLine( AInv.ToTabDelimited( "F3" ) );
// Use the factorization to solve some linear systems Ax = y.
var y0 = new FloatComplexVector( fact.Cols, rng );
var y1 = new FloatComplexVector( fact.Cols, rng );
FloatComplexVector x0 = fact.Solve( y0 );
FloatComplexVector x1 = fact.Solve( y1 );
Console.WriteLine();
Console.WriteLine( "Solution to Ax = y0 is {0}", x0.ToString( "G5" ) );
Console.WriteLine();
Console.WriteLine( "y0 - Ax0 = {0}", ( y0 - MatrixFunctions.Product( A, x0 ) ).ToString( "G5" ) );
Console.WriteLine();
Console.WriteLine( "Solution to Ax = y1 is {0}", x1.ToString( "G5" ) );
Console.WriteLine();
Console.WriteLine( "y1 - Ax1 = {0}", ( y1 - MatrixFunctions.Product( A, x1 ) ).ToString( "G5" ) );
// You can also solve for multiple right-hand sides.
var Y = new FloatComplexMatrix( y1.Length, 2 );
Y.Col( 0 )[Slice.All] = y0;
Y.Col( 1 )[Slice.All] = y1;
FloatComplexMatrix X = fact.Solve( Y );
// The first column of X should be x0; the second column should be x1.
Console.WriteLine();
Console.WriteLine( "X =" );
Console.WriteLine( X.ToTabDelimited( "G7" ) );
// Factor a different matrix.
var z = new FloatComplex( 1.23F, -.76F );
FloatComplexTriDiagMatrix B = z * A;
fact.Factor( B );
x0 = fact.Solve( y0 );
Console.WriteLine();
Console.WriteLine( "Solution to Bx = y0 is {0}", x0.ToString( "G5" ) );
Console.WriteLine();
Console.WriteLine( "Press Enter Key" ); | 1,094 | 3,480 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2023-06 | latest | en | 0.610803 |
https://www.engineeringstream.com/2023/07/what-is-strength-formula.html | 1,718,485,175,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861606.63/warc/CC-MAIN-20240615190624-20240615220624-00489.warc.gz | 660,726,548 | 44,704 | # What is strength formula?
The term "strength" in the context of materials can refer to different types of strength, such as tensile strength, compressive strength, yield strength, etc. Each type of strength has its own specific formula for calculation. Here are the formulas for some common types of strength:
1. Tensile Strength (σ_t): Tensile strength is the maximum stress that a material can withstand when subjected to a pulling force before breaking. The formula for tensile strength is:
Tensile Strength (σ_t) = Maximum Load (F_max) / Cross-sectional Area (A)
where:
• Tensile Strength (σ_t) is the tensile strength of the material in Pascals (Pa) or Megapascals (MPa).
• Maximum Load (F_max) is the maximum force applied to the material before it breaks, measured in Newtons (N) or pounds (lb).
• Cross-sectional Area (A) is the original cross-sectional area of the material perpendicular to the applied force, measured in square meters (m²) or square inches (in²).
1. Compressive Strength (σ_c): Compressive strength is the maximum stress that a material can withstand when subjected to a compressive force before failing. The formula for compressive strength is similar to tensile strength:
Compressive Strength (σ_c) = Maximum Load (F_max) / Cross-sectional Area (A)
where the variables have the same meaning as in the tensile strength formula.
1. Yield Strength (σ_y): Yield strength is the stress at which a material begins to deform plastically (undergoes permanent deformation) without any additional increase in load. The formula for yield strength is:
Yield Strength (σ_y) = Yield Load (F_y) / Original Cross-sectional Area (A)
where:
• Yield Strength (σ_y) is the yield strength of the material in Pascals (Pa) or Megapascals (MPa).
• Yield Load (F_y) is the load at which the material starts to deform plastically, measured in Newtons (N) or pounds (lb).
• Original Cross-sectional Area (A) is the initial cross-sectional area of the material perpendicular to the applied force, measured in square meters (m²) or square inches (in²).
It's important to note that the strength of materials can be influenced by various factors, and the values obtained from these formulas may differ depending on the specific testing methods and conditions used to determine the material's strength.
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also my friend website related to policy | 542 | 2,436 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2024-26 | latest | en | 0.877283 |
https://www.idcafe.net/what-is-the-formula-of-one-way-slab/ | 1,702,054,012,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100762.64/warc/CC-MAIN-20231208144732-20231208174732-00168.warc.gz | 891,309,578 | 12,034 | # What is the formula of one way slab?
## What is the formula of one way slab?
One way slab definition According to IS 456:2000, The ratio of longer span(l) to shorter span(b) which is (L/B) greater than 2 is known as One way slab. In practical, One way slab is supported by only two parallel beams or walls.
How is one way slab reinforcement calculated?
Bar Bending Schedule of One way Slab
1. Given data: Length = 5 m (5000 mm)
2. Solution: The quantity is done into two steps.
3. Formula: = (Total length – clear cover)/center to center spacing + 1.
4. Main bar: = (5000 – (25+25))/100 + 1.
5. Distribution bar: = (2000 – ( 25 + 25))/125 + 1.
6. Cutting Length.
7. Cutting Length:
8. Distribution Bar:
### What is the minimum reinforcement in slab?
Minimum reinforcement is 0.12% for HYSD bars and 0.15% for mild steel bars. The diameter of bar generally used in slabs are: 6 mm, 8 mm, 10 mm, 12mm and 16mm. The maximum diameter of bar used in slab should not exceed 1/8 of the total thickness of slab.
What is a one-way slab?
The one-way slab is a slab, which is supported by parallel walls or beams, and whose length to breadth ratio is equal to or greater than two and it bends in only one direction (spanning direction) while it is transferring the loads to the two supporting walls or beams, because of its geometry.
#### Where is one way slab used?
The slab is majorly the horizontal structural element of any building/house which is used to construct floors, ceilings and roofs. Generally, it can be simply supported, continuous, or cantilever. The slab is an element of the building which transfers the different types of floor load to the beams.
Which slab is better one way or two way?
If L/b the ratio is greater than or equal 2 or then it is considered a one-way slab. If L/b the ratio is less than 2 then it is considered a two-way slab. The one-way slab is supported by a beam on two opposite side only. The two-way slab is supported by the beam on all four sides.
## What is the minimum depth of one way slab?
… ℎ í µí±ší µí±–í µí±› is given in Table 4, with an absolute minimum thickness of 1.5 in (38.1 mm). The values of Table 4 are applicable for normal weight concrete and í µí±“ í µí±¦ = 60,000 psi.
Where is one way slab and two way slab used?
The deflected shape of the one-way slab is cylindrical. Whereas the deflected shape of the two-way slab is a dish or saucer-like shape. Chajja and Varandha are practical examples of one-way slab. Whereas two-way slabs are used in constructive floors of the Multistorey building.
### What is the maximum reinforcement in slab?
And maximum reinforcement in slab is restricted to 1 to 2% of total gross sectional area (B×D), where B is width of slab and D is over all depth of slab including cover.
Why is there no shear reinforcement in slabs?
For reinforced concrete (RC) slabs without shear reinforcement, shear and punching can be the governing failure mode at the ultimate limit state if subjected to large concentrated loads. Shear and punching of RC slabs without shear reinforcement has been a challenging problem in assessment based on current standards.
#### What is the formula for one way slab?
Formula: One way slab = Longer span /Shorter span ≥ 2 In those slabs, in a shorter span, the main reinforcement was provided and in the transverse direction, the distribution bars were provided. Design consideration of one-way slabs:
How much reinforcement should be used in one way slabs?
The reinforcement diameter use in one-way slabs will not exceed the 1/8 of the slab total depth. The horizontal spacing between the parallel main steel bars is always less than of the, Three times the slab effective depth, or maximum 300 mm.
## Can a rectangular slab have a two way action?
But rectangular slabs often have such proportions and supports (e.g., relatively deep, stiff monolithic concrete beams) that result in two-way action [Fig. 1.1(b)]. At any point, such slabs are curved in both directions resulting in biaxial bending moments.
What are the load combinations for one way slab?
One way slab is normally designed for gravity load only. The design load combinations are as follows ACI 318-99 U = 1.4D | 1,011 | 4,210 | {"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} | 4.34375 | 4 | CC-MAIN-2023-50 | longest | en | 0.928403 |
https://www.get-digital-help.com/2010/05/11/count-unique-and-duplicate-numbers-from-a-closed-workbook-in-excel-formula/ | 1,516,480,870,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084889733.57/warc/CC-MAIN-20180120201828-20180120221828-00336.warc.gz | 932,689,407 | 25,441 | Here is a picture of the closed workbook. There can´t be any blank cells or text values in your range.
As you can see, there are five unique distinct values (1, 2, 3, 5 and 6) and one duplicate value (3).
### How to count unique and duplicate numbers from a closed workbook
Formula in B1:
=SUM(IF(FREQUENCY('C:\temp\[closed workbook.xls]Sheet1'!\$A\$1:\$A\$6;''C:\temp\[closed workbook.xls]Sheet1'!\$A\$1:\$A\$6)>0;1;0)) + ENTER
Don´t forget to change path to your closed workbbook.
Formula in B2:
=SUM(IF(FREQUENCY('C:\temp\[closed workbook.xls]Sheet1'!\$A\$1:\$A\$6;''C:\temp\[closed workbook.xls]Sheet1'!\$A\$1:\$A\$6)=0;1;0)) - 1 + ENTER
Don´t forget to change path to your closed workbbook
### Evaluate formula
If I try to evaluate formula I get som strange results. A lot of #ref errors but the final result is correct. | 260 | 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} | 2.859375 | 3 | CC-MAIN-2018-05 | longest | en | 0.782602 |
http://www.algebra.com/algebra/homework/word/numbers/Numbers_Word_Problems.faq?hide_answers=1&beginning=180 | 1,369,534,457,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368706484194/warc/CC-MAIN-20130516121444-00004-ip-10-60-113-184.ec2.internal.warc.gz | 318,099,728 | 10,987 | # Questions on Word Problems: Numbers, consecutive odd/even, digits answered by real tutors!
Algebra -> Algebra -> Customizable Word Problem Solvers -> Numbers -> Questions on Word Problems: Numbers, consecutive odd/even, digits answered by real tutors! Log On
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Word Problems: Numbers, consecutive odd/even, digits Solvers Lessons Answers archive Quiz In Depth
Question 9415: find the 5 digit number that has the following characteristics. -no 2 digits are the same. -no 0 occurs in the number. -the 3rd digit is 1 less than the 2nd digit. -the 2nd is twice the 1st digit. -the sum of the 1st 4 digits is divisible by 9. -the 4th digit is the square of the 5th digit. Click here to see answer by prince_abubu(198)
Question 9479: An employee of the computer store is paid a base salary of \$1,175 per month plus a 5% commission on all sales during the month. How much must this employee sell in one month to earn a total of 3,170 for the month Click here to see answer by huntingmaster325(15)
Question 9476: A rectangle 24 meters long has the same area as a square that is 12 meters on a side. What are the dimensions of the rectangle? Click here to see answer by huntingmaster325(15)
Question 9473: Find 3 consecutive even integers so that the first plus twice the second is twice the third Click here to see answer by prince_abubu(198)
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Question 9923: Mount Everest is the tallest mountain in the world. It is 237 meters higher than K2, the second tallest mountain. If the sum of their heights is 17,459 meters, how tall is each mountain? Click here to see answer by askmemath(368)
Question 10068: My question: What two consecutive even integers equal 217? Thanks, Abby Click here to see answer by thecampusnerd(13)
Question 10413: A chef needs to mix a 45% fat content cheese with a 20% fat content cheese to obtain 30 grams of a cheese mixture that is 30% fat. How many grams of each kind of cheese must be used? Click here to see answer by calbreez(9)
Question 9621: Hi, The question is a Farmer has Pigs & Chickens on his farm of which the sum of all the pigs & chickens legs equals 101. How many Pigs and how many chickens are on his farm. Click here to see answer by rapaljer(4667)
Question 10603: Find two negative numbers such that the square of their sum is 20 more than the square of their difference, and the difference of their squares is 24. I have an upcoming mathletes compitition, and this problem has me stumpted. Click here to see answer by longjonsilver(2297)
Question 10813: The Find two consecutive odd integers whose sum is 0 Click here to see answer by Earlsdon(6288)
Question 10812: The sum of a number and its reciprocal is 2. Find the number. Click here to see answer by Earlsdon(6288)
Question 10811: Five times a number subtracted from 3 times the number equals 20. Find the number. Click here to see answer by longjonsilver(2297)
Question 10810: The sum of two consecutive integers is 105. Find the integers. Click here to see answer by WannabeCAgirl83(35)
Question 10810: The sum of two consecutive integers is 105. Find the integers. Click here to see answer by mathwhiz(13)
Question 11118: What is the sum of the 1st 500 counting digits? That is what is the sum of 1+2+3+4+...+499+500? How about the sum of only the odd digits, 1+3+5+...+497+499? Click here to see answer by askmemath(368)
Question 11273: the sum of two consecutive integers is 118 a. DEFINE A VARIABLE FOR THE SMALLER INTEGER. b. WHAT MUST YOU ADD TO AN INTERGER TO GET THE NEXT GREATER EVEN INTEGER? c. WRITE AN EXPRESSION FOR THE SECON INTEGER. d. WRITE AND SOLVE AN EQUATION TO FIND THE TWO EVEN INTEGERS. Click here to see answer by elima(1433)
Question 11513: I need to find another way to solve this problem. I would like to be able to use an equation for it. Find the sum of these consecutive odd numbers: 1 + 3 + 5 + 7 + ... (all odd numbers from 7-193) 193 + 195 + 197 + 199. I already know that the answer is 10,000 because I sat here and added up all of the numbers myself, but I need an equation to turn in, and I haven't been able to find one yet. If you could help I would really appreciate it. Thanks- Lisa Click here to see answer by askmemath(368)
Question 11513: I need to find another way to solve this problem. I would like to be able to use an equation for it. Find the sum of these consecutive odd numbers: 1 + 3 + 5 + 7 + ... (all odd numbers from 7-193) 193 + 195 + 197 + 199. I already know that the answer is 10,000 because I sat here and added up all of the numbers myself, but I need an equation to turn in, and I haven't been able to find one yet. If you could help I would really appreciate it. Thanks- Lisa Click here to see answer by khwang(438)
Question 11728: The sum of three consecutive numbers is 600. What is the least of these three numbers? Click here to see answer by elima(1433)
Question 11819: The sum of 2 consecutive even intergers is 118 Click here to see answer by elima(1433)
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Question 12472: One Kilogram is approximately 2.2 pounds. A British Airways ticket shows that the weight limi on a checked in baggage is 32 kilograms for a certain route. Which of the following numbers is closest to the weight limit on this route? Click here to see answer by askmemath(368)
Question 12585: A store keeper goes to the bank to get \$10 worth of change. He requests twicew as many quarters as half dollars, twice as many dimes as qurters, 3 times as many nickles as dimes, and no pennies or dollars. How many of each coin did the storekeeper get? Click here to see answer by rapaljer(4667)
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Question 12620: one number is 5 times another number the sum of the two numbers is 270 what is the number Click here to see answer by askmemath(368)
Question 12680: The sum of two numbers is -63. The first number minus the second is -41. Find the numbers Click here to see answer by AdolphousC(70)
Question 12737: I need help please! The question I have is: The sum of three consecutive odd integers = 171 Thank You! Click here to see answer by glabow(165)
Question 12859: What are 3 consecutive numbers with the sum of 600? Click here to see answer by elima(1433)
Question 12859: What are 3 consecutive numbers with the sum of 600? Click here to see answer by chicosgurrl(2) | 2,195 | 8,480 | {"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} | 3.25 | 3 | CC-MAIN-2013-20 | latest | en | 0.894343 |
https://programs.programmingoneonone.com/2022/05/hackerearth-telecom-towers-problem-solution.html | 1,686,056,593,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224652569.73/warc/CC-MAIN-20230606114156-20230606144156-00384.warc.gz | 498,052,853 | 35,513 | # HackerEarth Telecom Towers problem solution
In this HackerEarth Telecom Towers problem solution Telecom towers are an integral part of the telecom network infrastructure. In fact they are the most expensive to build and the valuations are heavy. The newly started mobile company in Hacker Land built n towers to enhance the connectivity of their users. You can assume that the Cartesian coordinate system is used in Hacker Land and the location of ith tower is given as (xi,yi).
After the construction of the towers company realised that there are many call drops happening with the users. One identified reason for the frequent call drop was that the pair of towers which are at Euclidean distance were causing destructive interference. To resolve the issue the company decided to destroy some towers such that no two towers are at distance. You have to tell the minimum number of towers that the company need to destroy such that no two towers are at distance d.
## HackerEarth Telecom Towers problem solution.
`#include "bits/stdc++.h"using namespace std;struct \${\$(){ios_base::sync_with_stdio(0);cin.tie(0);}}\$;const int inf = 1000111;struct Matching { int n; vector<int> matchL, matchR, dist; vector<bool> seen; vector< vector<int> > ke; Matching(int n) : n(n), matchL(n+1), matchR(n+1), dist(n+1), seen(n+1, false), ke(n+1) {} void addEdge(int u, int v) { ke[u].push_back(v); } bool bfs() { queue<int> qu; for(int u = 1; u <= n; ++u) if (!matchL[u]) { dist[u] = 0; qu.push(u); } else dist[u] = inf; dist[0] = inf; while (!qu.empty()) { int u = qu.front(); qu.pop(); for(__typeof(ke[u].begin()) v = ke[u].begin(); v != ke[u].end(); ++v) { if (dist[matchR[*v]] == inf) { dist[matchR[*v]] = dist[u] + 1; qu.push(matchR[*v]); } } } return dist[0] != inf; } bool dfs(int u) { if (u) { for(__typeof(ke[u].begin()) v = ke[u].begin(); v != ke[u].end(); ++v) if (dist[matchR[*v]] == dist[u] + 1 && dfs(matchR[*v])) { matchL[u] = *v; matchR[*v] = u; return true; } dist[u] = inf; return false; } return true; } int match() { int res = 0; while (bfs()) { for(int u = 1; u <= n; ++u) if (!matchL[u]) if (dfs(u)) ++res; } return res; }};const int N = 205;int x[N * N], y[N * N];int color[N * N];bool good(int i, int j) { return i >= 0 && i < N && j >= 0 && j < N;}vector <int> adj[N * N];void add_edge(int i, int j, int i_, int j_) { // DEBUG(N); int u = i * N + j; int v = i_ * N + j_; // DEBUG(i, j, i_, j_); adj[u].push_back(v); adj[v].push_back(u);}int m[N][N];int main() { int n, d; cin >> n >> d; map <pair<int, int>, int> rmap; for(int i = 1; i <= n; i++) { cin >> x[i] >> y[i]; rmap[{x[i], y[i]}] = i; } vector <pair <int, int> > side; for(int a = -d; a <= d; a++) { for(int b = -d; b <= d; b++) { if(a * a + b * b == d * d) { side.push_back({a, b}); } } } int edge = 0; for(int i = 0; i < N; i++){ for(int j = 0; j < N; j++){ for(auto t: side) { int i_ = i + t.first; int j_ = j + t.second; if(good(i_, j_)) { add_edge(i, j, i_, j_); edge++; } } } } memset(color, -1, sizeof color); bool is_bipartite = true; queue <int> q; for(int st = 0; st < N * N; st++) { if(color[st] == -1) { q.push(st); color[st] = 0; while(!q.empty()) { int v = q.front(); q.pop(); for(auto u: adj[v]) { if(color[u] == -1) { color[u] = color[v] ^ 1; q.push(u); } else { // is_bipartite = false; is_bipartite &= color[u] != color[v]; } } } } } int cnt[2]; cnt[0] = cnt[1] = 1; for(int i = 0; i < N; i++) { for(int j = 0; j < N; j++) { m[i][j] = cnt[color[i * N + j]]; cnt[color[i * N + j]]++; } } assert(is_bipartite); Matching M(N * N); for(int i = 1; i <= n; i++) { for(auto ab: side) { int a = x[i] + ab.first; int b = y[i] + ab.second; if(rmap.find({a, b}) != rmap.end()) { if(color[x[i] * N + y[i]] == 0) { assert(color[a * N + b] == 1); M.addEdge(m[x[i]][y[i]], m[a][b]); } } } } cout << M.match() << endl; return 0;}`
### Second solution
`#include <bits/stdc++.h>using namespace std;typedef pair<int,int> pii;const int MAXN = 10007;const int INF = 1000000000;vector <int> G[MAXN];map <pii,int> M;int col[MAXN];struct max_flow{ // from e-maxx.ru/algo struct edge { int a, b, cap, flow; }; int n, s, t, d[MAXN], ptr[MAXN], q[MAXN]; //n, s, t must be set prior to calling dinics vector<edge> e; vector<int> g[MAXN]; void add_edge (int a, int b, int cap) { edge e1 = { a, b, cap, 0 }; edge e2 = { b, a, 0, 0 }; g[a].push_back ((int) e.size()); e.push_back (e1); g[b].push_back ((int) e.size()); e.push_back (e2); } bool bfs() { int qh=0, qt=0; q[qt++] = s; memset (d, -1, n * sizeof d[0]); d[s] = 0; while (qh < qt && d[t] == -1) { int v = q[qh++]; for (size_t i=0; i<g[v].size(); ++i) { int id = g[v][i], to = e[id].b; if (d[to] == -1 && e[id].flow < e[id].cap) { q[qt++] = to; d[to] = d[v] + 1; } } } return d[t] != -1; } int dfs (int v, int flow) { if (!flow) return 0; if (v == t) return flow; for (; ptr[v]<(int)g[v].size(); ++ptr[v]) { int id = g[v][ptr[v]], to = e[id].b; if (d[to] != d[v] + 1) continue; int pushed = dfs (to, min (flow, e[id].cap - e[id].flow)); if (pushed) { e[id].flow += pushed; e[id^1].flow -= pushed; return pushed; } } return 0; } int dinic() { int flow = 0; for (;;) { if (!bfs()) break; memset (ptr, 0, n * sizeof ptr[0]); while (int pushed = dfs (s, INF)) flow += pushed; } return flow; }};void dfs(int pos){ for (int i = 0; i < G[pos].size(); ++i) { if(col[G[pos][i]] == -1) { col[G[pos][i]] = 1 - col[pos]; dfs(G[pos][i]); } }}int main(int argc, char const *argv[]){ int n,d; cin>>n>>d; vector <pii> triplets; for (int i = -d; i <= d; ++i) { for (int j = -d; j <= d; ++j) { if(i*i + j*j == d*d) { triplets.push_back(pii(i,j)); } } } for (int i = 0; i < n; ++i) { int x,y; cin>>x>>y; M[pii(x, y)] = i; for (int j = 0; j < triplets.size(); ++j) { int nx = x + triplets[j].first, ny = y + triplets[j].second; if(M.find(pii(nx,ny)) != M.end()) { int id = M[pii(nx,ny)]; G[i].push_back(id); G[id].push_back(i); } } } memset(col, -1, sizeof col); for (int i = 0; i < n; ++i) { if(col[i] == -1) { col[i] = 0; dfs(i); } } max_flow flow_graph; flow_graph.n = n + 2; flow_graph.s = n; flow_graph.t = n + 1; for (int i = 0; i < n; ++i) { if(col[i] == 0) { flow_graph.add_edge(flow_graph.s, i, 1); for (int j = 0; j < G[i].size(); ++j) { flow_graph.add_edge(i, G[i][j], 1); } } else flow_graph.add_edge(i, flow_graph.t, 1); } cout<<flow_graph.dinic()<<"\n"; return 0;}` | 2,524 | 7,730 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2023-23 | longest | en | 0.793739 |
https://math.answers.com/geometry/What_is_20_plus_3b_when_b_equals_-6 | 1,696,365,926,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511220.71/warc/CC-MAIN-20231003192425-20231003222425-00137.warc.gz | 409,159,151 | 44,274 | 0
What is 20 plus 3b when b equals -6?
Updated: 4/28/2022
Wiki User
12y ago
20+3b = 20 + 3(-6) = 20-18 = 2
Wiki User
12y ago | 62 | 131 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2023-40 | latest | en | 0.598874 |
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May 4, 2016
# Homework Help: geometry
Posted by mike on Saturday, November 28, 2009 at 8:25pm.
how many gallons of water will a tank measuring 66 inches in diameter and 24 inches in height,hold?
• geometry - MathMate, Saturday, November 28, 2009 at 9:51pm
Since the data are in inches, I assume the answer would be in Americal gallons that is equivalent to about 3.8 litres.(as opposed to the Imperial gallon, about 4.54 litres).
Vomume = πd²h/4
=π(66)²*24/4
=26136π in³
Since 1000 cubic inches equal 4.329 US gal(liq.),
the box will hold
26136π/1000*4.329 gal
approx. 355 gal. | 193 | 593 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2016-18 | longest | en | 0.895481 |
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# What do I need on GMAT for Shot at Haas? (part time)
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18 Jun 2010, 11:55
3.7 GPA in international relations and spanish
top 20 liberal arts college, magna cum laude, pbk
6 years progressive work experience (3 as business/financial manager, mostly in legal field)
Recs will be strong
Helped found a NProfit in Spain for intercultural education.
Im stuck in San Fran. What GMAT would I need for Berkeley? USF? I need to go p/t.
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22 Jun 2010, 14:38
A 700 for Berkeley should make you safe.
Re: What do I need on GMAT for Shot at Haas? (part time) [#permalink] 22 Jun 2010, 14:38
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# What do I need on GMAT for Shot at Haas? (part time)
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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®. | 718 | 2,568 | {"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-2017-13 | longest | en | 0.870236 |
https://mathphysicsexpert.com/four-less-than-four-times-a-number-is-equal-to-five-more-than-the-number-write-the-equation-that/ | 1,669,675,830,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710662.60/warc/CC-MAIN-20221128203656-20221128233656-00822.warc.gz | 440,772,339 | 11,203 | # Four less than four times a number is equal to five more than the number. Write the equation that
## Question:
Four less than four times a number is equal to five more than the number. Write the equation that represents this statement and solve it to find the number. | 57 | 271 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2022-49 | latest | en | 0.961254 |
https://lists.w3.org/Archives/Public/public-whatwg-archive/2010Aug/0429.html | 1,597,407,994,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439739211.34/warc/CC-MAIN-20200814100602-20200814130602-00543.warc.gz | 386,510,825 | 3,519 | From: Boris Zbarsky <bzbarsky@MIT.EDU>
Date: Thu, 12 Aug 2010 11:27:16 -0400
Message-ID: <4C6412D4.2070106@mit.edu>
```On 8/12/10 10:59 AM, David Flanagan wrote:
> But if you don't talk about the algebraic operations then you haven't
> really defined what a transformation is, have you?
Given that the input is in coordinates, we need to talk about algebraic
operations to define the transformation on vectors those coordinates
produce. That is, we need a clear definition of what the output vector
is given an input vector and a list of 6 numbers.
But once we have that, composition of transformation can be described
simply as composition (thinking of them as functions), without reference
to algebraic manipulation of their particular matrix representations.
> I suppose it does. So to be complete, the spec would have to show the
> math required to transform a point (x,y) using the CTM.
Yes, indeed.
> Are you suggesting that there is some way that the spec can be written
> generically without any assumptions about row vector or column vector
> format?
No, I'm just saying that the definition of transform composition doesn't
need to make such assumptions.
> Note that the matrix shown above already appears in the current
> version of the transform() method description. I don't see how to avoid
> picking one form or another unless you want to define a CTM as an array
> of 6 numbers and show the formulas for updating each of those numbers
> without referring to matrix multiplication at all.
While that would be a viable course of action, I don't think there's a
need for that. Defining the CTM as a matrix and defining how the 6
numbers produce the matrix and exactly how the matrix acts on vectors is
fine; while it's actually a bit more text than the other it produces a
simpler conceptual picture.
-Boris
```
Received on Thursday, 12 August 2010 08:27:16 UTC
This archive was generated by hypermail 2.4.0 : Wednesday, 22 January 2020 16:59:26 UTC | 487 | 1,973 | {"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-2020-34 | latest | en | 0.938916 |
http://www.nag.com/numeric/FL/nagdoc_fl23/examples/source/e04hefe.f90 | 1,369,527,009,000,000,000 | text/plain | crawl-data/CC-MAIN-2013-20/segments/1368706472050/warc/CC-MAIN-20130516121432-00018-ip-10-60-113-184.ec2.internal.warc.gz | 623,107,225 | 2,837 | ! E04HEF Example Program Text ! Mark 23 Release. NAG Copyright 2011. MODULE e04hefe_mod ! E04HEF Example Program Module: ! Parameters and User-defined Routines ! .. Use Statements .. USE nag_library, ONLY : nag_wp ! .. Implicit None Statement .. IMPLICIT NONE ! .. Parameters .. INTEGER, PARAMETER :: liw = 1, m = 15, n = 3, nin = 5, & nout = 6, nt = 3 INTEGER, PARAMETER :: lb = n*(n+1)/2 INTEGER, PARAMETER :: ldfjac = m INTEGER, PARAMETER :: ldv = n INTEGER, PARAMETER :: lw = 7*n + m*n + 2*m + n*n ! .. Local Arrays .. REAL (KIND=nag_wp) :: t(m,nt), y(m) CONTAINS SUBROUTINE lsqgrd(m,n,fvec,fjac,ldfjac,g) ! Routine to evaluate gradient of the sum of squares ! .. Use Statements .. USE nag_library, ONLY : dgemv ! .. Implicit None Statement .. IMPLICIT NONE ! .. Scalar Arguments .. INTEGER, INTENT (IN) :: ldfjac, m, n ! .. Array Arguments .. REAL (KIND=nag_wp), INTENT (IN) :: fjac(ldfjac,n), fvec(m) REAL (KIND=nag_wp), INTENT (OUT) :: g(n) ! .. Executable Statements .. ! The NAG name equivalent of dgemv is f06paf CALL dgemv('T',m,n,1.0_nag_wp,fjac,ldfjac,fvec,1,0.0_nag_wp,g,1) g(1:n) = 2.0_nag_wp*g(1:n) RETURN END SUBROUTINE lsqgrd SUBROUTINE lsqfun(iflag,m,n,xc,fvec,fjac,ldfjac,iw,liw,w,lw) ! Routine to evaluate the residuals and their 1st derivatives ! .. Implicit None Statement .. IMPLICIT NONE ! .. Scalar Arguments .. INTEGER, INTENT (INOUT) :: iflag INTEGER, INTENT (IN) :: ldfjac, liw, lw, m, n ! .. Array Arguments .. REAL (KIND=nag_wp), INTENT (INOUT) :: fjac(ldfjac,n), w(lw) REAL (KIND=nag_wp), INTENT (OUT) :: fvec(m) REAL (KIND=nag_wp), INTENT (IN) :: xc(n) INTEGER, INTENT (INOUT) :: iw(liw) ! .. Local Scalars .. REAL (KIND=nag_wp) :: denom, dummy INTEGER :: i ! .. Executable Statements .. DO i = 1, m denom = xc(2)*t(i,2) + xc(3)*t(i,3) fvec(i) = xc(1) + t(i,1)/denom - y(i) fjac(i,1) = 1.0_nag_wp dummy = -1.0_nag_wp/(denom*denom) fjac(i,2) = t(i,1)*t(i,2)*dummy fjac(i,3) = t(i,1)*t(i,3)*dummy END DO RETURN END SUBROUTINE lsqfun SUBROUTINE lsqhes(iflag,m,n,fvec,xc,b,lb,iw,liw,w,lw) ! Routine to compute the lower triangle of the matrix B ! (stored by rows in the array B) ! .. Implicit None Statement .. IMPLICIT NONE ! .. Scalar Arguments .. INTEGER, INTENT (INOUT) :: iflag INTEGER, INTENT (IN) :: lb, liw, lw, m, n ! .. Array Arguments .. REAL (KIND=nag_wp), INTENT (OUT) :: b(lb) REAL (KIND=nag_wp), INTENT (IN) :: fvec(m), xc(n) REAL (KIND=nag_wp), INTENT (INOUT) :: w(lw) INTEGER, INTENT (INOUT) :: iw(liw) ! .. Local Scalars .. REAL (KIND=nag_wp) :: dummy, sum22, sum32, sum33 INTEGER :: i ! .. Executable Statements .. b(1) = 0.0_nag_wp b(2) = 0.0_nag_wp sum22 = 0.0_nag_wp sum32 = 0.0_nag_wp sum33 = 0.0_nag_wp DO i = 1, m dummy = 2.0_nag_wp*t(i,1)/(xc(2)*t(i,2)+xc(3)*t(i,3))**3 sum22 = sum22 + fvec(i)*dummy*t(i,2)**2 sum32 = sum32 + fvec(i)*dummy*t(i,2)*t(i,3) sum33 = sum33 + fvec(i)*dummy*t(i,3)**2 END DO b(3) = sum22 b(4) = 0.0_nag_wp b(5) = sum32 b(6) = sum33 RETURN END SUBROUTINE lsqhes SUBROUTINE lsqmon(m,n,xc,fvec,fjac,ldfjac,s,igrade,niter,nf,iw,liw,w, & lw) ! Monitoring routine ! .. Use Statements .. USE nag_library, ONLY : f06eaf ! .. Implicit None Statement .. IMPLICIT NONE ! .. Parameters .. INTEGER, PARAMETER :: ndec = 3 ! .. Scalar Arguments .. INTEGER, INTENT (IN) :: igrade, ldfjac, liw, lw, m, & n, nf, niter ! .. Array Arguments .. REAL (KIND=nag_wp), INTENT (IN) :: fjac(ldfjac,n), fvec(m), & s(n), xc(n) REAL (KIND=nag_wp), INTENT (INOUT) :: w(lw) INTEGER, INTENT (INOUT) :: iw(liw) ! .. Local Scalars .. REAL (KIND=nag_wp) :: fsumsq, gtg INTEGER :: j ! .. Local Arrays .. REAL (KIND=nag_wp) :: g(ndec) ! .. Executable Statements .. fsumsq = f06eaf(m,fvec,1,fvec,1) CALL lsqgrd(m,n,fvec,fjac,ldfjac,g) gtg = f06eaf(n,g,1,g,1) WRITE (nout,*) WRITE (nout,*) & ' Itns F evals SUMSQ GTG grade' WRITE (nout,99999) niter, nf, fsumsq, gtg, igrade WRITE (nout,*) WRITE (nout,*) & ' X G Singular values' DO j = 1, n WRITE (nout,99998) xc(j), g(j), s(j) END DO RETURN 99999 FORMAT (1X,I4,6X,I5,6X,1P,E13.5,6X,1P,E9.1,6X,I3) 99998 FORMAT (1X,1P,E13.5,10X,1P,E9.1,10X,1P,E9.1) END SUBROUTINE lsqmon END MODULE e04hefe_mod PROGRAM e04hefe ! E04HEF Example Main Program ! .. Use Statements .. USE nag_library, ONLY : e04hef, e04yaf, e04ybf, x02ajf USE e04hefe_mod, ONLY : lb, ldfjac, ldv, liw, lsqfun, lsqgrd, lsqhes, & lsqmon, lw, m, n, nag_wp, nin, nout, nt, t, y ! .. Implicit None Statement .. IMPLICIT NONE ! .. Local Scalars .. REAL (KIND=nag_wp) :: eta, fsumsq, stepmx, xtol INTEGER :: i, ifail, iprint, maxcal, nf, & niter ! .. Local Arrays .. REAL (KIND=nag_wp) :: b(lb), fjac(ldfjac,n), fvec(m), & g(n), s(n), v(ldv,n), w(lw), x(n) INTEGER :: iw(liw) ! .. Intrinsic Functions .. INTRINSIC sqrt ! .. Executable Statements .. WRITE (nout,*) 'E04HEF Example Program Results' ! Skip heading in data file READ (nin,*) ! Observations of TJ (J = 1, 2, ..., nt) are held in T(I, J) ! (I = 1, 2, ..., m) DO i = 1, m READ (nin,*) y(i), t(i,1:nt) END DO ! Set up an arbitrary point at which to check the derivatives x(1:nt) = (/ 0.19_nag_wp, -1.34_nag_wp, 0.88_nag_wp/) ! Check the 1st derivatives ifail = 0 CALL e04yaf(m,n,lsqfun,x,fvec,fjac,ldfjac,iw,liw,w,lw,ifail) ! Check the evaluation of B ifail = 0 CALL e04ybf(m,n,lsqfun,lsqhes,x,fvec,fjac,ldfjac,b,lb,iw,liw,w,lw, & ifail) ! Continue setting parameters for E04HEF ! Set IPRINT to 1 to obtain output from LSQMON at each iteration iprint = -1 maxcal = 50*n eta = 0.9_nag_wp xtol = 10.0_nag_wp*sqrt(x02ajf()) ! We estimate that the minimum will be within 10 units of the ! starting point stepmx = 10.0_nag_wp ! Set up the starting point x(1:nt) = (/ 0.5_nag_wp, 1.0_nag_wp, 1.5_nag_wp/) ifail = -1 CALL e04hef(m,n,lsqfun,lsqhes,lsqmon,iprint,maxcal,eta,xtol,stepmx,x, & fsumsq,fvec,fjac,ldfjac,s,v,ldv,niter,nf,iw,liw,w,lw,ifail) SELECT CASE (ifail) CASE (0,2:) WRITE (nout,*) WRITE (nout,99999) 'On exit, the sum of squares is', fsumsq WRITE (nout,99999) 'at the point', x(1:n) CALL lsqgrd(m,n,fvec,fjac,ldfjac,g) WRITE (nout,99998) 'The corresponding gradient is', g(1:n) WRITE (nout,*) ' (machine dependent)' WRITE (nout,*) 'and the residuals are' WRITE (nout,99997) fvec(1:m) END SELECT 99999 FORMAT (1X,A,3F12.4) 99998 FORMAT (1X,A,1P,3E12.3) 99997 FORMAT (1X,1P,E9.1) END PROGRAM e04hefe | 2,380 | 6,210 | {"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-2013-20 | longest | en | 0.413837 |
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# Linear Programming - Simplex Method
Kay Manning is a sales manager with two salespeople working for her. The ability of each salesperson to secure new accounts has historically been: Jesse Burkett can generate 4 new accounts per 10 visits to prospects, and Ed Delahanty averages 3 new accounts per 10 visits to prospects. As of May 1, Jesse has 20 established customers that he must call on each month if he is to retain their monthly orders. Ed has 15 established accounts to maintain.
The average time each salesperson spends with each type of customer per visit is:
Type of Customer
Salesperson New Account Established Account
Jesse 10 5
Ed 8 6
Each new order yields the company an average profit of \$150, while orders from established accounts are usually smaller and yield an average profit of \$50 an order. Jesse has a larger area to cover than Ed, and therefore Jesse only has 120 hours (outside of travel time) to spend with customers each month. Ed has 135 hours available to spend with customers.
a) Formulate a linear programming model to determine how many new and how many old customers each salesperson should call on in May to maximize profits.
b) Solve the problem by the simplex method.
#### Solution Summary
A Complete, Neat and Step-by-step Solution is provided in the attached file.
\$2.19 | 293 | 1,347 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2018-22 | latest | en | 0.932232 |
https://www.buffalobrewingstl.com/extractive-distillation-2/via.html | 1,582,469,095,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145774.75/warc/CC-MAIN-20200223123852-20200223153852-00228.warc.gz | 655,760,048 | 4,901 | ## Via
where S, and 8V are respectively the height of liquid and vapor films (m), AS, and
ASv are the corresponding volume of liquid and vapor film (m3), respectively, and L and
V are the flow of liquid and vapor phases (mol s"1), respectively. K\'\a and K]°ka are the binary mass transfer coefficients, but their units are mol m"3 s"1.
Bulk liquid Liquid film Vapor film Bulk vapor
Bulk liquid Liquid film Vapor film Bulk vapor
Fig. 3. Schematic diagram of two-film theory; the direction of mass transfer is assumed from vapor phase to liquid phase.
The schematic diagram of two-film theory is shown in Fig. 3. The entire resistance to mass transfer in a given turbulent phase is in a thin, stagnant region of that phase at the interface, called a film. This film is similar to the laminar sublayer that forms when a fluid flows in the turbulent regime parallel to a flat plate.
So we can calculate a and on the condition that K\ka and Kvlka are known, and then [k'lkd\ and [k^ka\ are determined by Eqs. (181) and (185). Let us see one example about how to obtain K'na and Kvlka by means of the AIChE method.
Example: In a distillation column with sieve tray, the operation conditions are listed in Table 13. This table is designed for the convenience to be dealt with by MS Excel software. The mixture on the sieve tray is composed of 50 mol% benzene (1) and 50 mol% propylene (2) for the liquid phase, and 10 mol% benzene (1) and 90 mol% propylene (2) for the vapor phase
(T = 80°C, P = 700kPa). Compute K\\a and tc]ka (kmol s"1) on the sieve tray. Table 13
Notation Physical quantity Value Al The vapor volume flowrate (m3 h"1) | 430 | 1,643 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2020-10 | latest | en | 0.882973 |
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답변 | 1,950 | 6,556 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2024-33 | latest | en | 0.623059 |
https://www.sciencebuddies.org/teacher-resources/lesson-plans/lifting-with-a-lever | 1,653,358,145,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662562410.53/warc/CC-MAIN-20220524014636-20220524044636-00064.warc.gz | 1,130,460,659 | 128,900 | # Lifting with a Lever
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1 review
## Summary
3rd-5th
Group Size
1-2 students
Active Time
30 minutes
Total Time
30 minutes
Area of Science
Physics
Key Concepts
Forces, levers
Credits
Ben Finio, PhD, Science Buddies
## Overview
Can your students lift a book off the floor with just one finger? Find out and learn about simple machines in this fun lesson plan about levers.
## Learning Objectives
• Understand the working principles of a lever
• Apply these principles to lift a heavy object with one finger
• Identify which variables can be changed when using a lever, and the effect of changing them
## NGSS Alignment
This lesson helps students prepare for these Next Generation Science Standards Performance Expectations:
• 3-5-ETS1-3. Plan and carry out fair tests in which variables are controlled and failure points are considered to identify aspects of a model or prototype that can be improved.
This lesson focuses on these aspects of NGSS Three Dimensional Learning:
Science & Engineering Practices Disciplinary Core Ideas Crosscutting Concepts Science & Engineering Practices Planning and Carrying out Investigations. Plan and conduct an investigation collaboratively to produce data to serve as the basis for evidence, using fair tests in which variables are controlled and the number of trials considered. Disciplinary Core Ideas ETS1.C: Optimizing the Design Solution. Different solutions need to be tested in order to determine which of them best solves the problem, given the criteria and constraints. Crosscutting Concepts Systems and System Models. A system is a group of related parts that make up a whole and can carry out functions its individual parts cannot.
## Materials
For the entire class:
• Book or other heavy object
• Stiff meter stick/yard stick (make sure it is not too flexible)
• Thick marker
For each student (make small groups if there are not enough supplies):
• Pencil or thin marker or crayon
• Stiff ruler with inch markings (do not use flexible rulers)
• An item to use as a weight, like a box of crayons
## Background Information for Teachers
This section contains a quick review for teachers of the science and concepts covered in this lesson.
A force is a push or a pull on an object. We exert forces every day, whether we are pulling open a door, pushing a child on a swing, or picking up a book. Sometimes we use machines to help us exert bigger forces. This is called mechanical advantage.
A lever is a type of simple machine that can be used to increase a force. A lever consists of a beam attached to a pivot (called the fulcrum). A lever allows you to take an input force (the effort) and amplify the output force (the load). A seesaw at the playground is a simple example of a lever. Imagine that you have three children who are all exactly the same weight. One of the children can lift the other two children on the seesaw by sitting twice as far away from the fulcrum (Figure 1).
A seesaw is balanced on a fulcrum so that 2 meters of the board hang off each side. Three people of equal weight are able to balance on the seesaw if a person on one side of the seesaw stands at the furthest end and the other 2 people stand together 1 meter away from the end on the opposite side. Two people weigh twice as much as the single person but they are twice as close to the fulcrum so the forces exerted on both sides of the seesaw are the same.
Figure 1. A seesaw is an example of a lever.
You can make a miniature version of a seesaw using a pencil and a ruler. Place a pencil flat on a desk and put a ruler on top of it at the halfway point (the 6 inch mark). Place an object like a box of crayons on one end of the ruler, as shown in Figure 2. Then press down on the on the ruler just on the other side of the pencil (the 5 inch mark). How hard is it to lift the box? Try pressing down all the way at the end of the ruler instead (the 0 inch mark). It should be easier to lift the box. Now try moving the pencil closer to the box of crayons, and press down on the other end of the ruler again. It should be even easier to lift the box, because the effort (your finger pressing down on the ruler) is relatively much farther away from the fulcrum (the pencil) than the load (the box of crayons). In this lesson plan, your students will experiment with a similar setup to see if they can lift a heavy object off the floor with just one finger.
Figure 2. A simple lever made from a ruler and pencil. The ruler is the beam and the pencil is the fulcrum. By pushing down on the far end of the ruler, you can lift the box of crayons.
There are other examples of levers all around us that might not be as obvious. Grab a pair of scissors— they are also a type of lever! Is it easier to cut through a piece of cardboard using the tip, or the part of the blade closer to the handle? If you cut with the part of the blade closer to the handle, the mechanical advantage is bigger (the force of you squeezing the handles is amplified more), so it is easier to cut. Now go try to open (or close) a door, first by pushing near the doorknob, and next by pushing near the hinges. You should notice that it is much harder to open (or close) the door by pushing close to the hinges.
Technical note
It might seem like levers give you "something for nothing" because they can take a small input force and turn it into a larger output force. However, while a lever allows you to decrease the effort force required to lift an object, the distance over which you must exert this force increases. For example, in Figure 1, in order to lift the children on the right up a given distance, the child on the left must move twice as far. This means that energy is conserved; what you gain in force, you lose in distance. This is a very important concept in physics—you never get anything for free!
Mathematically, the equation for balancing a lever can be expressed as:
Equation 1:
where distances are measured to the fulcrum. Older students can use this equation to solve basic math problems about levers, but that is beyond the scope of this activity.
## Lesson Plan Variations
Top
Free science fair projects. | 1,386 | 6,165 | {"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} | 2.734375 | 3 | CC-MAIN-2022-21 | longest | en | 0.874209 |
https://vanderlaan-lab.org/2019/12/29/estimating-the-sample-average-treatment-effect-sate-under-effect-modification-in-a-cluster-randomized-trial/ | 1,586,116,863,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371609067.62/warc/CC-MAIN-20200405181743-20200405212243-00310.warc.gz | 758,079,077 | 6,203 | # Estimating the Sample Average Treatment Effect (SATE) Under Effect Modification in a Cluster Randomized Trial
This post is part of our Q&A series.
A question from graduate students in our Fall 2019 offering of “Biostatistical Methods: Survival Analysis and Causality” at UC Berkeley:
Question:
Hi Mark,
We were wondering about the application of TMLE and superlearner to cluster-randomized study designs, and the adoption of the sample average treatment effect (SATE) as an efficient estimator. From our understanding, although the SATE is not formally identifiable in a finite setting, it is nevertheless an efficient estimate due to its asymptotic behavior (TMLE for the population effect is asymptotically linear and has an asymptotically conservative variance estimator). What properties of the SATE make it preferable to the population average treatment effect, particularly in effect modification settings? What allows for valid causal inferences to be drawn from data adaptive parameters like the SATE?
Best, A.A. and D.C.
Hi A.A. and D.C.,
Let’s say we observe $n$ observations $O_i=(W_i,A_i,Y_i)$ representing cluster specific data structures, and we assume the randomization assumption, $P(A=1|W,Y_0,Y_1)=P(A=1|W)$. The sample average treatment effect is defined as $SATE=\frac{1}{n} \sum_i(Y_i(1)-Y_i(0))$, which is different from the sample average conditional treatment effect $SACTE= \frac{1}{n} \sum_i \mathbb{E}(Y_1-Y_0|W_i)$, and the latter is again different from the $ATE=\mathbb{E}(Y_1-Y_0)$.
The $ATE$ is an average across a distribution of $W$, which in a cluster RCT would mean it is a population average across clusters from some population of clusters. In many cluster RCTs, the sample of clusters is not sampled that way at all, but represents a selected convenient sample. Therefore, in that case, it might make more sense to define a parameter from the conditional distribution $(Y_i,A_i)$, given $W_i$, across $i=1, \ldots ,n$, i.e. treating the clusters as fixed, and $(A,Y)$ within each cluster as random. This makes the SACTE an interesting alternative target parameter, which can be viewed as a parameter of the conditional distribution given $W_1,\ldots,W_n$, or, one can view it as a data adaptive parameter depending on the empirical distribution of $W_1, \ldots, W_n$ if one is still willing to view $W_i$ as a random sample from some population.
If one is not even willing to think of $Y_1-Y_0$ as a random sample from a conditional distribution $P(Y_1-Y_0|W)$, but only wants to make inference about the actual values $Y_i(1)-Y_i(0), i=1, \ldots ,n$, then one could view the SATE as the target. So the choice of quantity (among ATE, SACTE, SATE) is driven by till what degree we wish to generalize our findings to a bigger population. In various applications I might argue that all three are of interest.
Let TMLE represent the regular TMLE of the ATE. Recall that TMLE-ATE $\sim P_n(D_W+D_Y) =\frac{1}{n} \sum_i (D_W+D_Y)(O_i)$, where $D_W,D_Y$ are the two score components making up the influence curve $D_W+D_Y$ of the TMLE.
Note that SATE-ATE (just a sample mean of $Y_1-Y_0$minus true mean) is asymptotically linear with influence curve $$Y_1-Y_0-\mathbb{E}(Y_1-Y_0)=Y_1-Y_0-\mathbb{E}(Y_1-Y_0|W)+\mathbb{E}(Y_1-Y_0|W)-\mathbb{E}(Y_1-Y_0)$$,
and this is an orthogonal composition in the sense that the correlations of the two terms are zero. The second term equals $D_W$.
So TMLE-SATE=TMLE-ATE+ATE-SATE $\sim P_n D_Y-\frac{1}{n} \sum_i (Y_1-Y_0)-\mathbb{E}(Y_1-Y_0|W)$.
Similarly, TMLE-SACTE $\sim P_n D_Y$.
We conclude: TMLE-SACTE is asymptotically linear with an improved influence curve $D_Y$, having subtracted out the $D_W$ component. The TMLE-SATE is asymptotically linear with a further improved influence curve $D_Y-D_U$, where $D_U=(Y_1-Y_0-\mathbb{E}(Y_1-Y_0|W)$. The latter $D_U$ is not really an influence curve since $Y_0, Y_1$ are not observed. Nonetheless, it tells us the the TMLE-SATE is asymptotically linear with inflluence curve $D_Y - D_U$ and, showing that TMLE-SATE is more efficient than TMLE-SACT. For the sake of inference, we simply use $D_Y$ as a conservative influence curve.
I believe the general idea of this is the following. Consider a target $\mathbb{E}[X]$ for a full data random variable $X$, and suppose that the observed data includes observing $W$. Suppose that we have a TMLE of $\mathbb{E}X$. One could define $\frac{1}{n} \sum_i X_i, \frac{1}{n} \sum_i \mathbb{E}(X|W_i)$, and analyze the TMLE- $\frac{1}{n} \sum_i X_i$ exactly same was as above.
For example, suppose that we have a general longitudinal data structure, $W_i=L_i(0),A(0),\ldots, L(K),A(K),Y$, and we define $\mathbb{E}Y_d$ as a mean outcome under a multiple time point dynamic treatment. We have a TMLE of $\mathbb{E}Y_d$, such as the one implemented in ltmle(). We might desire inference for $\frac{1}{n} \sum_i Y_{d,i}$, or $\frac{1}{n}\sum_i \mathbb{E}(Y_d|W_i)$. We have $\Psi_{\text{TMLE}} - \frac{1}{n}\sum_i \mathbb{E}Y_{d,i} = \Psi_{\text{TMLE}} - \mathbb{E}Y_d-[\frac{1}{n} \sum_i Y_{d,i}-\mathbb{E}(Y_d|W_i)]-[\frac{1}{n} \sum_i \mathbb{E}(Y_d|W_i)-\mathbb{E}Y_d)]$. The latter represents the $D_W$ component of the influence curve of the $\Psi_{\text{TMLE}}-\mathbb{E}Y_d$. The other component is a non-identifiable influence curve that subtracts out another component. So, we obtain conservative inference for $\frac{1}{n} \sum_i \mathbb{E}Y_{d,i}$ by using the influence curve of $\Psi_{\text{TMLE}}-\mathbb{E}Y_d$ without the $D_W$ component of its influence curve.
Regarding effect modification, if we have a discrete variable $V$, then a stratified TMLE applied to data with $V_i=v$ would obtain inference for $\frac{1}{n} \sum_i (Y_i(1)-Y_i(0))$ within strata $V_i=v$, for each $v$. To obtain inference for this $v$-specific SATE, one can use the conservative influence curve.
If one now wants to obtain inference for a difference of two $v$-specific SATEs, then the TMLE of this difference will still be asymptotically linear with the difference of the two $v$-specific non-identifiable influence curves. It is now less clear if ignoring the difference of the two non-identifiable components of their respective influence curves would still result in conservative inference. It would be worthwhile to research this. Since we have valid conservative inference for the $v$-specific SATE for each $v$, we could also decide to build a test based on comparing the two marginal confidence intervals (overlap), but this would by necessity be more conservative. If this inference for a contrast of $v$-specific SATEs happens to be problematic, then that might be an argument to instead focus on the effect modification parameter (contrast of $v$-specific SCATE).
$\frac{1}{n} \sum_i \mathbb{E}(Y_1-Y_0|W_i,V_i=1)-\frac{1}{n} \sum_i \mathbb{E}(Y_1-Y_0|W_i,V_i=0)$ instead since for this we have an identified influence curve.
If $V$ is continuous, one might use a working MSM $m_\{\beta}(v)$ for $\\frac{1}{n} \sum_i \mathbb{E}(Y_1-Y_0|W_i,V_i=v)$ as a function of $v$. One can then use the TMLE of the beta in this working MSM (as implemented in ltmle e.g.). This would again correspond with using an influence curve that would remove a $D_W$ component of the regular influence curve of the TMLE of $\beta$.
So my basic answer to your question is that inference for the SATE based on the TMLE of the ATE can be generalized to general longitudinal data structures, and, one should be able to also generalize it to treatment effect modification by a discrete or continuous effect modifier $V$.
Best Wishes, Mark
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Bird on a Wire (Posted on 2004-06-07)
A telephone wire stretched tight between two poles placed ten meters apart is a favorite resting spot for a flock of crows.
Suppose one morning two crows land on the wire, each at a random spot (the probability is uniformly distributed). With a bucket of paint and a brush you mark the stretch of wire between them. A certain length of wire will have been painted.
On average, what length of wire would you expect to have painted? Assume that each bird is a single point along the line, and so has no width.
Suppose instead that a dozen crows landed on the wire, each at an independent, random location, and you painted the stretch of wire between each bird and its nearest neighbor. On average, what total length of wire would you expect to have painted now?
And if a thousand crows landed?
A computer-generated solution could be found, but bonus points will be awarded for a formal proof!
No Solution Yet Submitted by Sam Rating: 3.7000 (10 votes)
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(probably repetitive) 2 and 3 bird solution | Comment 14 of 42 |
As Charlie did, I'm going to measure in decameters (dam).
Let's say the first bird landed n dam from the left pole, and the second landed x dam from the left pole. The length painted is abs(n-x). The average of the integral from x=0 to 1 is n2/2 + (1-n)2/2. Simplify to n2 - n + 1/2. The average integral of that equation from n=0 to 1 is 1/3. I've just proved that the answer to the first part is indeed 1/3 dam.
3 birds is simple enough because, like the 2 bird case, there are no unpainted spots between the leftmost and rightmost birds. I'd like to work onto the 2 bird solution. Lets just say that the first two birds together make one very fat 1/3 dam bird. Note: I'm making an (probably wrong) assumption here that the fact that the width of the large bird varies doesn't affect the solution. The third bird has a 1/3 chance of landing on the oversized bird and not making a difference. Therefore, the length of paint added is max(abs(n-x)-1/6,0) (n is the center of the bigger bird). Integrating as before, we get n2/2 + (1-n)2/2 - 5/36. Simplify to n2 - n + 13/36. Integrating again, this time from n=1/6 to 5/6, we get 8/81. Then we, of course, add the 1/3 dam of paint under that vast bird for a total of 35/81 dam. This time, though, I'm not so sure that I left all the rules of probability intact.
Edit: I immediately saw a mistake in 3 bird part, and also added a note.
Edited on June 7, 2004, 7:28 pm
Edited on June 7, 2004, 7:36 pm
Posted by Tristan on 2004-06-07 19:20:11
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# Holiday Math Place Value First Grade Number of the Day
Common Core Standards
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PDF (Acrobat) Document File
27 MB|28 pages
Product Description
This winter holiday place value unit gives your students a lot of practice working with numbers, representation and relationships. All students need daily practice working with numbers to effectively develop their number sense. This unit reviews number sense from the First Grade Common Core Math Curriculum.
Purchase the whole year here:
Number of the Day {The Whole Year!} Bundle
Number of the Day practice includes:
**Counting forward and counting back from the number-Number line completion. (1.NBT.A.1)
**Skip Counting-Number Patterns (1.OA.5)
**Writing numbers. (1.NBT.1)
**Number line completion. (1.NBT.1)
**Missing numbers on a hundred chart. (1.NBT.1)
**Representing numbers using base ten blocks. (1.NBT.2)
**Adding and subtracting single digit numbers from numbers less than 100. (1.NBT.4)
**One more, one less. (1.NBT.C.5)
**Expanded Notation. (1.NBT.6)
**Odd/Even. (2.OA.C.3)
This unit is aligned with the First Grade Common Core Math Standards, and includes twenty days of practice with a holiday theme. The pages are divided into four sets of five with a cover page and blank and completed 120 charts. These pages can be used as a daily warm-up/spiral review, work for fast finishers, or for homework.
I use ‘guided math’ in my classroom. I begin my teaching time block teaching/spiraling math concepts as a whole group, but most of my math block is devoted to guided practice in small groups followed by independent or small group practice, centers and intervention. I begin my math time every day with a page from my popular Number of the Day series. You may find more math in this “printable” format in my store.
I hope that you and your kids enjoy this fun back-to-school resource. Check back soon. I continue to add Number of the Day units daily!
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Teachers Pay Teachers is an online marketplace where teachers buy and sell original educational materials. | 900 | 4,073 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2017-17 | latest | en | 0.884443 |
https://www.careercup.com/question?id=5684799322718208 | 1,701,944,272,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100651.34/warc/CC-MAIN-20231207090036-20231207120036-00385.warc.gz | 761,634,577 | 7,168 | ## Interview Question for Software Engineer / Developers
Country: India
Comment hidden because of low score. Click to expand.
0
of 0 vote
``````//change value of variable m as per requirement...
int calcSumSubArray (void)
{
int m, n, i, j, k, l, Sum, Mi, Mj, MSum, arr[5][5] =
{
{2, 2, 4, 5, 6},
{3, 6, 8, 4, 9},
{5, 3, 9, 7, 2},
{1, 2, 3, 4, 5},
{3, 5, 7, 8, 1}
};
Mi = Mj = MSum = Sum = 0;
n = 5; m = 3;
for (i = 0; i <= n - m; i ++)
{
for (j = 0; j <= n - m; j ++)
{
for (k = 0; k < m; k ++)
for (l = 0; l < m; l ++)
{
printf ("%d ", arr[k + i][l + j]);
Sum += arr[k + i][l + j];
}
printf (" = %d\n", Sum);
if (MSum < Sum)
{
MSum = Sum;
Mi = i;
Mj = j;
printf ("\n MSum Changed to %d\n", MSum);
}
Sum = 0;
}
}
printf ("\nMax SubArrary's Sum is %d at location a[%d][%d]\n", MSum, Mi, Mj);
return 0;
}``````
Comment hidden because of low score. Click to expand.
0
of 0 vote
check this
geeksforgeeks.org/dynamic-programming-set-27-max-sum-rectangle-in-a-2d-matrix/
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CareerCup's interview videos give you a real-life look at technical interviews. In these unscripted videos, watch how other candidates handle tough questions and how the interviewer thinks about their performance. | 493 | 1,415 | {"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-2023-50 | latest | en | 0.731025 |
https://assets.doityourself.com/stry/how-much-does-a-2x4-weigh-and-other-common-questions | 1,712,957,409,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816070.70/warc/CC-MAIN-20240412194614-20240412224614-00362.warc.gz | 98,359,747 | 68,111 | # How Much Does a 2x4 Weigh? (and Other Common Questions)
• 1 hours
• Intermediate
What You'll Need
Calculator
Wood resource manual
Online resources
In order to find the weight of your 2 x 4, assuming it’s construction lumber, you first have to establish what kind of wood it is. Fir, spruce, Douglas fir, Hemlock, Larch, Pine, or whatever else it may be all have their very own particularities in hardness, weight, flexibility, density affecting their specific gravity and density.
The weight of your sample should be quite easy to calculate if you have the right information about it. Once you have its specie identified, you can easily find crucial pieces of data such as density or actual specific gravity (SG) needed to calculate the weight.
#### Facts About Density and Weight
Density is the mass (in pounds) per unit volume (in cubic feet) of a substance (lb/cf). Since it contains more water within its fibers, a one cubic foot sample of a tree that’s just been cut and is at 80% MC (moisture content) weighs more than the same size sample of the same tree once it has been transformed into lumber and has reached an equilibrium moisture content of 12%. It is safe to say then that the density of wood varies with its moisture content.
When looking up the density of a piece of wood of which you need to determine the weight, make sure that the source of the info specifies at which percentage MC it is based on, the info is useless if it doesn’t. If the lumber’s been around the shop for a few months or years, it has probably reached an equilibrium point with the relative humidity of the air surrounding it—it is called the equilibrium moisture content (MC) and that’s most likely the MC point at which you’re seeking the weight.
#### Specific Gravity—Botanical vs Woodworking
While looking up the specifications of wood, you’ll come across different terms such as specific gravity, basic specific gravity, and actual specific gravity.
Oven-dried wood refers to lumber further dried at 0% moisture content, where all of its moisture has been extracted from it. In botany, the specific gravity is calculated on this oven-dried density in relation to its volume when completely saturated with water at above 30% MC—it’s called the “Basis Specific Gravity”, but is not practical in woodworking however since we deal with wood seasoned at 12% EMC. The problem, however, lies in that it creates a lot of different standards all referenced as SG but none of which is based on the same norm and more often than not doesn’t specify at what MC level it’s based on.
Specific gravity is one of the most abused and misused terms in woodworking, so be careful when seeking your info. As references for woodworkers, sources such as the Forest Products Laboratory of Wisconsin (US Forest Service), “wood-database.com” has hundreds of worldwide species of wood listed with the “actual SG at 12% MC”. Just keep in mind though that although it is the best guide available to do as accurate a job as possible, it is not an exact science—sapwood and heartwood have different characteristics, same as a specie from coastal area compared to the same specie from farther inland.
Specific gravity is the ratio of the weight of a specific volume of wood—for instance, one cubic foot—to the weight of the same cubic foot volume of water, which is the standard at 62.4 lbs/ft³. Specific gravity is a dimensionless quantity and isn’t expressed in unit.
#### Weight of a Red Spruce 2x4
2” x 4” nominal means that the actual size of your lumber is less, around 1-1/2” x 3-1/2”. Take the actual measurements as accurately as possible and calculate the volume.
Cu. ft. = Thickness (Ft) x Width (Ft) x Lenght (let’s use 12ft 6in. for an example).
Cu. ft. = (1-1/2”/ 12) x (3-1/2”/ 12) x 12.5’
Cu. ft. = 0.125’ x 0.292’ x 12.5’
The volume of the 2 x 4 = 0.456 Cu. ft.
A—With a density (at 12% MC) of 27 lbs/c.f. and a total volume of 0.456 cu. ft.
weight = 27 lbs x 0.456 c.f. = 12.3 lbs total.
B—If all you can find is the actual SG of 0.43, you can find the density with—
Density = 0.43 x density of water (62.4 lbs/c.f.)
Density = 26.8 lbs/c.f.
Weight = 26.8 x 0.456 c.f. = 12.2 lbs total.
As mentioned before, it’s as close as you get with the type of material your working with, but using the actual density at 12% MC is a more accurate method than even with the actual specific gravity. | 1,077 | 4,393 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2024-18 | latest | en | 0.942678 |
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# Untitled-17 - 41 Assume that the test statistic calculated...
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Unformatted text preview: 41. Assume that the test statistic calculated above is 1.500. In which range does'the p—Value for the test lie? 250.01 = 2.326 20025 = 1.960 20,050 = 1.645 20.100 : 1.282 a. 0.10 _<_ p-value b. 0.05 S p-value < 0.10 c. 0.025 S p—value < 0.05 d. 0.01 S p—value < 0.025 e. p—value < 0.01 Use the following information to answer the next 2 questions (42-43): In an agricultural experiment, two expensive high—yield varieties of corn are to be tested and yield improvements are to be measured. The experiment is arranged so that each variety is planted side by side in seven pairs of plots. Data is collected on yield increases obtained for these two varieties. Variety A is population 1 and variety B is population 2. Plot Variety A Variety B—l 1 —| 12.6 10.5 2 9.2 8.1 3 6.4 6.2 4 9.8 10.1 5 15.3 12.2 6 16.1 14.0 7 12.3 10.1 42. What is the value of the 95% upper confidence limit for the difference in population means? 250.010 = 2.326 20025 = 1.960 20.05 = 1.645 20.10 = 1282 t0.010,6 = 3.143 150.0255 = 2.447 t0.05,6 = 1.943 120.105 = 1.440 150.010.12 = 2-681 750,025,12 = 2-179 to.05,12 = 1-782 730.10.12 = 1356 a. 2.624 b. 5.467 c. 2.501 d. 3.266 e. 2.392 1 7 253 ...
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# Problem 981. Find nearest prime number less than input number
Solution 1446394
Submitted on 22 Feb 2018 by John D'Errico
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### Test Suite
Test Status Code Input and Output
1 Pass
n = 127; y_correct = 113; assert(isequal(minorprime(n),y_correct))
ans = 2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97 101 103 107 109 113 ans = 113
2 Pass
n = 125; y_correct = 113; assert(isequal(minorprime(n),y_correct))
ans = 2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97 101 103 107 109 113 ans = 113
3 Pass
n = 3; y_correct = 2; assert(isequal(minorprime(n),y_correct))
ans = 2 ans = 2
4 Pass
n = 1328; y_correct = 1327; assert(isequal(minorprime(n),y_correct))
ans = Columns 1 through 15 2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 Columns 16 through 30 53 59 61 67 71 73 79 83 89 97 101 103 107 109 113 Columns 31 through 45 127 131 137 139 149 151 157 163 167 173 179 181 191 193 197 Columns 46 through 60 199 211 223 227 229 233 239 241 251 257 263 269 271 277 281 Columns 61 through 75 283 293 307 311 313 317 331 337 347 349 353 359 367 373 379 Columns 76 through 90 383 389 397 401 409 419 421 431 433 439 443 449 457 461 463 Columns 91 through 105 467 479 487 491 499 503 509 521 523 541 547 557 563 569 571 Columns 106 through 120 577 587 593 599 601 607 613 617 619 631 641 643 647 653 659 Columns 121 through 135 661 673 677 683 691 701 709 719 727 733 739 743 751 757 761 Columns 136 through 150 769 773 787 797 809 811 821 823 827 829 839 853 857 859 863 Columns 151 through 165 877 881 883 887 907 911 919 929 937 941 947 953 967 971 977 Columns 166 through 180 983 991 997 1009 1013 1019 1021 1031 1033 1039 1049 1051 1061 1063 1069 Columns 181 through 195 1087 1091 1093 1097 1103 1109 1117 1123 1129 1151 1153 1163 1171 1181 1187 Columns 196 through 210 1193 1201 1213 1217 1223 1229 1231 1237 1249 1259 1277 1279 1283 1289 1291 Columns 211 through 217 1297 1301 1303 1307 1319 1321 1327 ans = 1327
5 Pass
n = 5050109; y_correct = 5050099; assert(isequal(minorprime(n),y_correct))
ans = Columns 1 through 15 2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 Columns 16 through 30 53 59 61 67 71 73 79 83 89 97 101 103 107 109 113 Columns 31 through 45 127 131 137 139 149 151 157 163 167 173 179 181 191 193 197 Columns 46 through 60 199 211 223 227 229 233 239 241 251 257 263 269 271 277 281 Columns 61 through 75 283 293 307 311 313 317 331 337 347 349 353 359 367 373 379 Columns 76 through 90 383 389 397 401 409 419 421 431 433 439 443 449 457 461 463 Columns 91 through 105 467 479 487 491 499 503 509 521 523 541 547 557 563 569 571 Columns 106 through 120 577 587 593 599 601 607 613 617 619 631 641 643 647 653 659 Columns 121 through 135 661 673 677 683 691 701 709 719 727 733 739 743 751 757 761 Columns 136 through 150 769 773 787 797 809 811 821 823 827 829 839 853 857 859 863 Columns 151 through 165 877 881 883 887 907 911 919 929 937 941 947 953 967 971 977 Columns 166 through 180 983 991 997 1009 1013 1019 1021 1031 1033 1039 1049 1051 1061 1063 1069 Columns 181 through 195 1087 1091 1093 1097 1103 1109 1117 1123 1129 1151 1153 1163 1171 1181 1187 Columns 196 through 210 1193 1201 1213 1217 1223 1229 1231 1237 1249 1259 1277 1279 1283 1289 1291 Columns 211 through 225 1297 1301 1303 1307 1319 1321 1327 1361 1367 1373 1381 1399 1409 1423 1427 Columns 226 through 240 1429 1433 1439 1447 1451 1453 1459 1471 1481 1483 1487 1489 1493 1499 1511 Columns 241 through 255 1523 1531 1543 1549 1553 1559 1567 1571 1579 1583 1597 1601 1607 1609 1613 Columns 256 through 270 1619 1621 1627 1637 1657 1663 1667 1669 1693 1697 1699 1709 1721 1723 1733 Columns 271 through 285 1741 1747 1753 1759 1777 1783 1787 1789 1801 1811 1823 1831 1847 1861 1867 Columns 286 through 300 1871 1873 1877 1879 1889 1901 1907 1913 1931 1933 1949 1951 1973 1979 1987 Columns 301 through 315 1993 1997 1999 2003 2011 2017 2027 2029 2039 2053 2063 2069 2081 2083 2087 Columns 316 through 330 2089 2099 2111 2113 2129 2131 2137 2141 2143 2153 2161 2179 2203 2207 2213 Columns 331 through 345 2221 2237 2239 2243 2251 2267 2269 2273 2281 2287 2293 2297 2309 2311 2333 Columns 346 through 360 2339 2341 2347 2351 2357 2371 2377 2381 2383 2389 2393 2399 2411 2417 2423 Columns 361 through 375 2437 2441 2447 2459 2467 2473 2477 2503 2521 2531 2539 2543 2549 2551 2557 Columns 376 through 390 2579 2591 2593 2609 2617 2621 2633 2647 2657 2659 2663 2671 2677 2683 2687 Columns 391 through 405 2689 2693 2699 2707 2711 2713 2719 2729 2731 2741 2749 2753 2767 2777 2789 Columns 406 through 420 2791 2797 2801 2803 2819 2833 2837 2843 2851 2857 2861 2879 2887 2897 2903 Columns 421 through 435 2909 2917 2927 2939 2953 2957 2963 2969 2971 2999 3001 3011 3019 3023 3037 Columns 436 through 450 3041 3049 3061 3067 3079 3083 3089 3109 3119 3121 3137 3163 3167 3169 3181 Columns 451 through 465 3187 3191 3203 3209 3217 3221 3229 3251 3253 3257 3259 3271 3299 3301 3307 Columns 466 through 480 3313 3319 3323 3329 3331 3343 3347 3359 3361 3371 3373 3389 3391 3407 3413 Columns 481 through 495 3433 3449 3457 3461 3463 3467 3469 3491 3499 3511 3517 3527 3529 3533 3539 Columns 496 through 510 3541 3547 3557 3559 3571 3581 3583 3593 3607 3613 3617 3623 3631 3637 3643 Columns 511 through 525 3659 3671 3673 3677 3691 3697 3701 3709 3719 3727 3733 3739 3761 3767 3769 Columns 526 through 540 3779 3793 3797 3803 3821 3823 3833 3847 3851 3853 3863 3877 3881 3889 3907 Columns 541 through 555 3911 3917 3919 3923 3929 3931 3943 3947 3967 3989 4001 4003 4007 4013 4019 Columns 556 through 570 4021 4027 4049 4051 4057 4073 4079 4091 4093 4099 4111 4127 4129 4133 4139 Columns 571 through 585 4153 4157 4159 4177 4201 4211 4217 4219 4229 4231 4241 4243 4253 4259 4261 Columns 586 through 600 4271 4273 4283 4289 4297 4327 4337 4339 4349 4357 4363 4373 4391 4397 4409 Columns 601 through 615 4421 4423 4441 4447 4451 4457 4463 4481 4483 4493 4507 4513 4517 4519 4523 Columns 616 through 630 4547 4549 4561 4567 4583 4591 4597 4603 4621 4637 4639 4643 4649 4651 4657 Columns 631 through 645 4663 4673 4679 4691 4703 4721 4723 4729 4733 4751 4759 4783 4787 4789 4793 Columns 646 through 660 4799 4801 4813 4817 4831 4861 4871 4877 4889 4903 4909 4919 4931 4933 4937 Columns 661 through 675 4943 4951 4957 4967 4969 4973 4987 4993 4999 5003 5009 5011 5021 5023 5039 Columns 676 through 690 5051 5059 5077 5081 5087 5099 5101 5107 5113 5119 5147 5153 5167 5171 5179 Columns 691 through 705 5189 5197 5209 5227 5231 5233 5237 5261 5273 5279 5281 5297 5303 5309 5323 Columns 706 through 720 5333 5347 5351 5381 5387 5393 5399 5407 5413 5417 5419 5431 5437 ...
6 Pass
pn = primes(1e6); ind = max(100,floor(numel(pn)*rand)); n = pn(ind) - 1; y_correct = pn(ind - 1); assert(isequal(minorprime(n),y_correct))
ans = Columns 1 through 15 2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 Columns 16 through 30 53 59 61 67 71 73 79 83 89 97 101 103 107 109 113 Columns 31 through 45 127 131 137 139 149 151 157 163 167 173 179 181 191 193 197 Columns 46 through 60 199 211 223 227 229 233 239 241 251 257 263 269 271 277 281 Columns 61 through 75 283 293 307 311 313 317 331 337 347 349 353 359 367 373 379 Columns 76 through 90 383 389 397 401 409 419 421 431 433 439 443 449 457 461 463 Columns 91 through 105 467 479 487 491 499 503 509 521 523 541 547 557 563 569 571 Columns 106 through 120 577 587 593 599 601 607 613 617 619 631 641 643 647 653 659 Columns 121 through 135 661 673 677 683 691 701 709 719 727 733 739 743 751 757 761 Columns 136 through 150 769 773 787 797 809 811 821 823 827 829 839 853 857 859 863 Columns 151 through 165 877 881 883 887 907 911 919 929 937 941 947 953 967 971 977 Columns 166 through 180 983 991 997 1009 1013 1019 1021 1031 1033 1039 1049 1051 1061 1063 1069 Columns 181 through 195 1087 1091 1093 1097 1103 1109 1117 1123 1129 1151 1153 1163 1171 1181 1187 Columns 196 through 210 1193 1201 1213 1217 1223 1229 1231 1237 1249 1259 1277 1279 1283 1289 1291 Columns 211 through 225 1297 1301 1303 1307 1319 1321 1327 1361 1367 1373 1381 1399 1409 1423 1427 Columns 226 through 240 1429 1433 1439 1447 1451 1453 1459 1471 1481 1483 1487 1489 1493 1499 1511 Columns 241 through 255 1523 1531 1543 1549 1553 1559 1567 1571 1579 1583 1597 1601 1607 1609 1613 Columns 256 through 270 1619 1621 1627 1637 1657 1663 1667 1669 1693 1697 1699 1709 1721 1723 1733 Columns 271 through 285 1741 1747 1753 1759 1777 1783 1787 1789 1801 1811 1823 1831 1847 1861 1867 Columns 286 through 300 1871 1873 1877 1879 1889 1901 1907 1913 1931 1933 1949 1951 1973 1979 1987 Columns 301 through 315 1993 1997 1999 2003 2011 2017 2027 2029 2039 2053 2063 2069 2081 2083 2087 Columns 316 through 330 2089 2099 2111 2113 2129 2131 2137 2141 2143 2153 2161 2179 2203 2207 2213 Columns 331 through 345 2221 2237 2239 2243 2251 2267 2269 2273 2281 2287 2293 2297 2309 2311 2333 Columns 346 through 360 2339 2341 2347 2351 2357 2371 2377 2381 2383 2389 2393 2399 2411 2417 2423 Columns 361 through 375 2437 2441 2447 2459 2467 2473 2477 2503 2521 2531 2539 2543 2549 2551 2557 Columns 376 through 390 2579 2591 2593 2609 2617 2621 2633 2647 2657 2659 2663 2671 2677 2683 2687 Columns 391 through 405 2689 2693 2699 2707 2711 2713 2719 2729 2731 2741 2749 2753 2767 2777 2789 Columns 406 through 420 2791 2797 2801 2803 2819 2833 2837 2843 2851 2857 2861 2879 2887 2897 2903 Columns 421 through 435 2909 2917 2927 2939 2953 2957 2963 2969 2971 2999 3001 3011 3019 3023 3037 Columns 436 through 450 3041 3049 3061 3067 3079 3083 3089 3109 3119 3121 3137 3163 3167 3169 3181 Columns 451 through 465 3187 3191 3203 3209 3217 3221 3229 3251 3253 3257 3259 3271 3299 3301 3307 Columns 466 through 480 3313 3319 3323 3329 3331 3343 3347 3359 3361 3371 3373 3389 3391 3407 3413 Columns 481 through 495 3433 3449 3457 3461 3463 3467 3469 3491 3499 3511 3517 3527 3529 3533 3539 Columns 496 through 510 3541 3547 3557 3559 3571 3581 3583 3593 3607 3613 3617 3623 3631 3637 3643 Columns 511 through 525 3659 3671 3673 3677 3691 3697 3701 3709 3719 3727 3733 3739 3761 3767 3769 Columns 526 through 540 3779 3793 3797 3803 3821 3823 3833 3847 3851 3853 3863 3877 3881 3889 3907 Columns 541 through 555 3911 3917 3919 3923 3929 3931 3943 3947 3967 3989 4001 4003 4007 4013 4019 Columns 556 through 570 4021 4027 4049 4051 4057 4073 4079 4091 4093 4099 4111 4127 4129 4133 4139 Columns 571 through 585 4153 4157 4159 4177 4201 4211 4217 4219 4229 4231 4241 4243 4253 4259 4261 Columns 586 through 600 4271 4273 4283 4289 4297 4327 4337 4339 4349 4357 4363 4373 4391 4397 4409 Columns 601 through 615 4421 4423 4441 4447 4451 4457 4463 4481 4483 4493 4507 4513 4517 4519 4523 Columns 616 through 630 4547 4549 4561 4567 4583 4591 4597 4603 4621 4637 4639 4643 4649 4651 4657 Columns 631 through 645 4663 4673 4679 4691 4703 4721 4723 4729 4733 4751 4759 4783 4787 4789 4793 Columns 646 through 660 4799 4801 4813 4817 4831 4861 4871 4877 4889 4903 4909 4919 4931 4933 4937 Columns 661 through 675 4943 4951 4957 4967 4969 4973 4987 4993 4999 5003 5009 5011 5021 5023 5039 Columns 676 through 690 5051 5059 5077 5081 5087 5099 5101 5107 5113 5119 5147 5153 5167 5171 5179 Columns 691 through 705 5189 5197 5209 5227 5231 5233 5237 5261 5273 5279 5281 5297 5303 5309 5323 Columns 706 through 720 5333 5347 5351 5381 5387 5393 5399 5407 5413 5417 5419 5431 5437 ... | 5,548 | 11,397 | {"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-2020-34 | latest | en | 0.214665 |
https://studysoup.com/tsg/math/396/elementary-linear-algebra-applications-version/chapter/19271/6 | 1,603,529,968,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107882103.34/warc/CC-MAIN-20201024080855-20201024110855-00615.warc.gz | 536,300,353 | 8,545 | ×
×
# Solutions for Chapter 6: Inner Product Spaces
## Full solutions for Elementary Linear Algebra: Applications Version | 10th Edition
ISBN: 9780470432051
Solutions for Chapter 6: Inner Product Spaces
Solutions for Chapter 6
4 5 0 321 Reviews
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4
##### ISBN: 9780470432051
Since 19 problems in chapter 6: Inner Product Spaces have been answered, more than 14322 students have viewed full step-by-step solutions from this chapter. Chapter 6: Inner Product Spaces includes 19 full step-by-step solutions. This expansive textbook survival guide covers the following chapters and their solutions. This textbook survival guide was created for the textbook: Elementary Linear Algebra: Applications Version, edition: 10. Elementary Linear Algebra: Applications Version was written by and is associated to the ISBN: 9780470432051.
Key Math Terms and definitions covered in this textbook
• Big formula for n by n determinants.
Det(A) is a sum of n! terms. For each term: Multiply one entry from each row and column of A: rows in order 1, ... , nand column order given by a permutation P. Each of the n! P 's has a + or - sign.
• Cholesky factorization
A = CTC = (L.J]))(L.J]))T for positive definite A.
• Column picture of Ax = b.
The vector b becomes a combination of the columns of A. The system is solvable only when b is in the column space C (A).
• Column space C (A) =
space of all combinations of the columns of A.
• Complex conjugate
z = a - ib for any complex number z = a + ib. Then zz = Iz12.
• Covariance matrix:E.
When random variables Xi have mean = average value = 0, their covariances "'£ ij are the averages of XiX j. With means Xi, the matrix :E = mean of (x - x) (x - x) T is positive (semi)definite; :E is diagonal if the Xi are independent.
• Dimension of vector space
dim(V) = number of vectors in any basis for V.
• Dot product = Inner product x T y = XI Y 1 + ... + Xn Yn.
Complex dot product is x T Y . Perpendicular vectors have x T y = O. (AB)ij = (row i of A)T(column j of B).
• Echelon matrix U.
The first nonzero entry (the pivot) in each row comes in a later column than the pivot in the previous row. All zero rows come last.
• Fourier matrix F.
Entries Fjk = e21Cijk/n give orthogonal columns FT F = nI. Then y = Fe is the (inverse) Discrete Fourier Transform Y j = L cke21Cijk/n.
• Full row rank r = m.
Independent rows, at least one solution to Ax = b, column space is all of Rm. Full rank means full column rank or full row rank.
• Hessenberg matrix H.
Triangular matrix with one extra nonzero adjacent diagonal.
• Identity matrix I (or In).
Diagonal entries = 1, off-diagonal entries = 0.
• Kirchhoff's Laws.
Current Law: net current (in minus out) is zero at each node. Voltage Law: Potential differences (voltage drops) add to zero around any closed loop.
• Linear transformation T.
Each vector V in the input space transforms to T (v) in the output space, and linearity requires T(cv + dw) = c T(v) + d T(w). Examples: Matrix multiplication A v, differentiation and integration in function space.
• Particular solution x p.
Any solution to Ax = b; often x p has free variables = o.
• Rank one matrix A = uvT f=. O.
Column and row spaces = lines cu and cv.
• Rank r (A)
= number of pivots = dimension of column space = dimension of row space.
• Toeplitz matrix.
Constant down each diagonal = time-invariant (shift-invariant) filter.
• Triangle inequality II u + v II < II u II + II v II.
For matrix norms II A + B II < II A II + II B II·
× | 908 | 3,519 | {"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} | 3.859375 | 4 | CC-MAIN-2020-45 | latest | en | 0.870862 |
https://forum.altair.com/topic/14233-time-steps-time-increment-convergence-steady-state/ | 1,597,236,951,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738892.21/warc/CC-MAIN-20200812112531-20200812142531-00481.warc.gz | 308,369,312 | 18,642 | # Time Steps, Time Increment, Convergence - Steady State
## Recommended Posts
Hi-
I am curious as to some guidelines on the max number of time steps and the initial time increment for a steady state problem. Initial time increment is 1e10 s but can this be adjusted to improve convergence? My problem is terminated at the max number of time steps rather than converging and terminating. I am new to AcuSolve so I am unsure on how to determine if the problem is converging. I have read some things indicating the Residual and Solution ratios need to be less than the specified convergence tolerance. In one tutorial I did, the convergence tolerance = 0.001 but when I looked at the Residual ratios, they were not less than this value and the solution still converged and terminated before reaching the max number of time steps.
Also, the results seemed reasonable for my problem that terminated at the max number of time steps - Should these results be regarded as inaccurate?
Any help appreciated!
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The concept of convergence is subjective - different users will give different responses. Some general comments:
1. Residual Ratio - an overall sense of how well the solution matches the equations being solved
2. Solution Ratio - an overall sense of how much the solution changes from time step to time step (for steady-state)
In general we would want both of these to be quite low. 'Low' is subject to interpretation, and the default convergence tolerance of 0.001 is usually good for AcuSolve results. By default, the Residual Ratio would need to be below the specified tolerance (0.001 default) for pressure, velocity, temperature, species. Other quantities could be a factor of 10 higher (0.01 default). By default the Solution Ratio could be a factor of 10 higher than the residual ratio values. Again - that is the default behavior. You can review the settings in the CONVERGENCE_CHECK_PARAMETERS command which will be included in the <problem>.ss.inc file generated by the AUTO_SOLUTION_STRATEGY command. This information will also be in the <problem>.<run>.echo file.
You would also want to track the results of interest and see how they are coming to a 'constant' value for steady-state. If the default convergence has been met, but the solution of interest is still changing by more than is acceptable to you, then reduce the convergence tolerance and allow it to converge more. (This is something of a convergence sensitivity study.) By the same token, if the results of interest have come to what you consider a steady solution, but the default convergence has not been reached, you may still consider the solution to be acceptable.
Convergence itself does not necessarily mean you have an accurate solution. You would still want to perform a mesh sensitivity study to see how the results of interest change as the mesh is refined (volume size, surface size, boundary layer parameters, etc.). You could also perform other sensivity studies (effect of changes in boundary conditions, initial conditions, etc.)
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https://www.nottingham.ac.uk/xpert/scoreresults.php?keywords=Why%20study%20the%20Didache?start=1492&start=14080&end=14100 | 1,624,063,846,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487643354.47/warc/CC-MAIN-20210618230338-20210619020338-00492.warc.gz | 812,532,140 | 30,672 | Having discussed nth roots, we are now in a position to define the expression ax, where a is positive and x is a rational power (or exponent).
## Definition
If a > 0, m Author(s): The Open University
At the end of Section 1, we discussed the decimals and asked whether it is possible to add and multiply these numbers to obtain another real number. We now explain how this can be done using the Least Upper Bound Property of Author(s): The Open University
## Exercise 29
In this exercise, take
Author(s): The Open University
We have seen that the set [0, 2) has no maximum element. However, [0, 2) has many upper bounds, for example, 2, 3, 3.5 and 157.1. Among all these upper bounds, the number 2 is the least upper bound because any number less than 2 is not an upper bound of [0, 2).
Author(s): The Open University
Now we consider inequalities involving the modulus of a real number. Recall that if a , then its modulus, or abso
Author(s): The Open University
1.1 Rational numbers
The set of natural numbers is the set of integers is and the set of rational numbers is Author(s): The Open University
Learning outcomes
By the end of this unit you should be able to:
Section 1 – Real numbers
• explain the relationship between rational numbers and recurring decimals;
• explain the term irrational number and describe how such a number can be represented on a number line;
• find a rational and an irrational number between any two distinct real numbers;
Section 2 – Inequalities
• solve inequalities by re
Author(s): The Open University
8.2.2 The screen
You can see the calculations that you have entered as well as the answers. This means you can easily check whether you have made any mistakes.
Author(s): The Open University
6.2 Getting the feel of big and small numbers
Very small and very large numbers can be difficult to comprehend. Nothing in our everyday experience helps us to get a good feel for them. For example numbers such as 1099 are so big that if Figure 1 was drawn to scale, you would be dealing with enormous distances. How big is big?
First express 1 000 000 000 in scientific notation as 109. Next, to find out how many times bigger 1099 is, use your calculator to divide 1099 by 109
Author(s): The Open University
6.1 Scientific notation
Understanding how your calculator displays and handles very large and very small numbers is important if you are to interpret the results of calculations correctly. This section focuses on a way of representing numbers known as scientific notation.
Before you start put your calculator into the float mode, so it will display up to about 10 digits and return to the home screen ready to do some calculations.
What answer would you expect if you square 20 million? How man
Author(s): The Open University
2.1 The four rules of arithmetic
You are now going to use the four operation keys (on the bottom right-hand side of the TI-84 keyboard): , Author(s): The Open University
3 Aims
The aim of this section is to help you to think about how you study mathematics and consider ways in which you can make your study more effective.
Author(s): The Open University
Pressing onwards
## Activity 15
1. Work through Sections 1.6 and 1.7 of the Calculator Book, using the method suggested above of glancing ahead-pressing on-glancing back, if you find it useful.
2. A num
Author(s): The Open University
Does it make sense?
## Example 3
Author(s): The Open University
1.1 Mathematics and you
Many people's ideas about what mathematics actually is are based upon their early experiences at school. The first two activities aim to help you recall formative experiences from childhood.
## Activity 1 Carl Jung's school days
Author(s): The Open University
Introduction
This unit explores reasons for studying mathematics, practical applications of mathematical ideas and aims to help you to recognize mathematics when you come across it. It introduces the you to the graphics calculator, and takes you through a series of exercises from the Calculator Book, Tapping into Mathematics With the TI-83 Graphics Calculator. The unit ends by asking you to reflect on the process of studying mathematics.
In order to complete this unit you will need
Author(s): The Open University
4 Proofs in group theory
In Section 4 we prove that some of the properties of the groups appearing earlier in the unit are, in fact, general properties shared by all groups. In particular, we prove that in any group the identity element is unique, and that each element has a unique inverse.
Click 'View document' below to open Section 4 (9 pages, 237KB).
Learning outcomes
By the end of this unit you should be able to:
• explain what is meant by a symmetry of a plane figure;
• specify symmetries of a bounded plane figure as rotations or reflections;
• describe some properties of the set of symmetries of a plane figure;
• explain the difference between direct and indirect symmetries;
• use a two-line symbol to represent a symmetry;
• describe geometrically th
Author(s): The Open University
4 Two identities
Section 4 introduces some important mathematical theorems.
Click 'View document' below to open Section 4 (7 pages, 237KB).
View document<
Author(s): The Open University
Learning outcomes
By the end of this unit you should be able to:
• Section 1: Sets
• use set notation;
• determine whether two given sets are equal and whether one given set is a subset of another;
• find the union, intersection and difference of two given sets.
• Section 2: Functions
• determine the image of a given function;
• determine whether a given function is one-one
Author(s): The Open University | 1,263 | 5,696 | {"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} | 4.4375 | 4 | CC-MAIN-2021-25 | latest | en | 0.935129 |
https://access.openupresources.org/curricula/our6-8math-v1/6/students/5/5.html | 1,716,453,378,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058614.40/warc/CC-MAIN-20240523080929-20240523110929-00120.warc.gz | 61,416,342 | 9,665 | # Lesson 5: Decimal Points in Products
Let’s look at products that are decimals.
## 5.1: Multiplying by 10
1. In which equation is the value of $x$ the largest?
$x \boldcdot 10 = 810$
$x \boldcdot 10 = 81$
$x \boldcdot 10 = 8.1$
$x \boldcdot 10 = 0.81$
2. How many times the size of 0.81 is 810?
## 5.2: Fractionally Speaking: Powers of Ten
Work with a partner to answer the following questions. One person should answer the questions labeled “Partner A,” and the other should answer those labeled “Partner B.” Then compare the results.
1. Find each product or quotient. Be prepared to explain your reasoning.
Partner A
1. $250 \boldcdot \frac{1}{10}$
2. $250 \boldcdot \frac {1}{100}$
3. $48 \div 10$
4. $48 \div 100$
Partner B
1. $250 \div 10$
2. $250 \div 100$
3. $48\boldcdot \frac{1}{10}$
4. $48 \boldcdot \frac{1}{100}$
2. Use your work in the previous problems to find $720 \boldcdot (0.1)$ and $720 \boldcdot (0.01)$. Explain your reasoning.
Pause here for a class discussion.
3. Find each product. Show your reasoning.
1. $36 \boldcdot (0.1)$
2. $(24.5) \boldcdot (0.1)$
3. $(1.8) \boldcdot (0.1)$
1. $54 \boldcdot (0.01)$
2. $(9.2)\boldcdot (0.01)$
4. Jada says: “If you multiply a number by 0.001, the decimal point of the number moves three places to the left.” Do you agree with her statement? Explain your reasoning.
## 5.3: Fractionally Speaking: Multiples of Powers of Ten
1. Select all expressions that are equivalent to $(0.6) \boldcdot (0.5)$. Be prepared to explain your reasoning.
1. $6 \boldcdot (0.1) \boldcdot 5 \boldcdot (0.1)$
2. $6 \boldcdot (0.01) \boldcdot 5 \boldcdot (0.1)$
3. $6 \boldcdot \frac{1}{10} \boldcdot 5 \boldcdot \frac{1}{10}$
4. $6 \boldcdot \frac{1}{1,000} \boldcdot 5 \boldcdot \frac{1}{100}$
1. $6 \boldcdot (0.001) \boldcdot 5 \boldcdot (0.01)$
2. $6 \boldcdot 5 \boldcdot \frac{1}{10} \boldcdot \frac{1}{10}$
3. $\frac{6}{10} \boldcdot \frac{5}{10}$
2. Find the value of $(0.6) \boldcdot (0.5)$. Show your reasoning.
3. Find the value of each product by writing and reasoning with an equivalent expression with fractions.
1. $(0.3) \boldcdot (0.02)$
1. $(0.7) \boldcdot (0.05)$
## Summary
We can use fractions like $\frac{1}{10}$ and $\frac{1}{100}$ to reason about the location of the decimal point in a product of two decimals.
Let’s take $24 \boldcdot (0.1)$ as an example. There are several ways to find the product:
• We can interpret it as 24 groups of 1 tenth (or 24 tenths), which is 2.4.
• We can think of it as $24 \boldcdot \frac{1}{10}$, which is equal to $\frac {24}{10}$ (and also equal to 2.4).
• Multiplying by $\frac {1}{10}$ has the same result as dividing by 10, so we can also think of the product as $24 \div 10$, which is equal to 2.4.
Similarly, we can think of $(0.7) \boldcdot (0.09)$ as 7 tenths times 9 hundredths, and write:
$$\left(7 \boldcdot \frac {1}{10}\right) \boldcdot \left(9 \boldcdot \frac {1}{100}\right)$$
We can rearrange whole numbers and fractions:
$$(7 \boldcdot 9) \boldcdot \left( \frac {1}{10} \boldcdot \frac {1}{100}\right) = 63 \boldcdot \frac {1}{1,\!000} = \frac {63}{1,\!000}$$
This tells us that $(0.7) \boldcdot (0.09) = 0.063$.
Here is another example: To find $(1.5) \boldcdot (0.43)$, we can think of 1.5 as 15 tenths and 0.43 as 43 hundredths. We can write the tenths and hundredths as fractions and rearrange the factors. $$\left(15 \boldcdot \frac{1}{10}\right) \boldcdot \left(43 \boldcdot \frac{1}{100}\right) = 15 \boldcdot 43 \boldcdot \frac{1}{1,\!000}$$
Multiplying 15 and 43 gives us 645, and multiplying $\frac{1}{10}$ and $\frac{1}{100}$ gives us $\frac{1}{1,000}$. So $(1.5) \boldcdot (0.43)$ is $645 \boldcdot \frac{1}{1,000}$, which is 0.645. | 1,322 | 3,700 | {"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} | 4.78125 | 5 | CC-MAIN-2024-22 | latest | en | 0.594667 |
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# Documentation Center
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## G/G/1 Queuing System and Little's Law
Overview
This example shows how to model a single-queue single-server system in which the interarrival time and the service time are uniformly distributed with fixed means of 1.1 and 1, respectively. The queue has an infinite storage capacity. In the notation, the G stands for a general distribution with a known mean and variance; G/G/1 means that the system's interarrival and service times are governed by such a general distribution, and that the system has one server. You can change the variances of the uniform distributions. You can use this model to examine Little's law.
Structure of the Model
The model includes the components listed below:
• Time Based Entity Generator block: Source of entities (also known as "customers" in queuing theory).
• Uniform Distribution for Interarrival Time subsystem: Creates a signal representing the interarrival times for the generated entities. After you set the distribution's variance using the Arrival Process Variance block, the subsystem computes a uniform random variate with the chosen variance and mean 1.1. To see the computation details, select the Uniform Distribution for Interarrival Time subsystem, choose Edit > Look Under Mask and double-click the block labeled Uniform Distribution.
• FIFO Queue block: It stores entities that have yet to be served.
• Single Server block: Models a server whose service time has a uniform distribution.
• Attribute Function block labeled Generate Service Time with Uniform Distribution: Assigns a service time to each attribute. After you set the distribution's variance using the Service Process Variance block, this block computes a uniform random variate with the chosen variance and mean 1.
Results and Displays
The model includes these visual ways to understand its performance:
• Display blocks that show the queue workload, average waiting time in the queue, average service time, and server utilization.
• A scope showing the number of entities (customers) in the queue at any given time
• A scope comparing empirical and theoretical ratios. See the discussion of Little's law below.
Little's Law
You can use this model to verify Little's law, which states the linear relationship between average queue length and average waiting time in the queue. In particular, the expected relationship is as follows:
Average queue length = (Mean arrival rate)(Average waiting time in queue)
The FIFO Queue block computes the current queue length and average waiting time in the queue. The subsystem called Little's Law Evaluation computes the ratio of average queue length (derived from the instantaneous queue length via integration) to average waiting time, as well as the ratio of mean service time to mean arrival time. The two ratios appear on the plot labeled Arrival Rate: Theoretical vs. Simulation Results.
Another way to interpret the equation above is that, given a normalized mean service time of 1, you can use the average waiting time and average queue length to derive the system's arrival rate.
Little's Law Applied to the Server
You can also use this model to verify the linear relationship that Little's law predicts between the server utilization and the average service time. The Single Server block computes the server utilization and average waiting time in the server. Because each entity can depart from the server immediately upon completing service, waiting time is equivalent to service time for the server in this model.
Experimenting with the Model
Move the Arrival Process Variance slider or the Service Process Variance slider during the simulation and observe how the queue content changes. When traffic intensity is high, the average waiting time in the queue is approximately linear in the variances of the interarrival time and service time. The larger the variances are, the longer an entity has to wait, and the more entities are waiting in the system.
Related Examples
References
[1] Kleinrock, Leonard, Queueing Systems, Volume I: Theory, New York, Wiley, 1975.
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# Which of the following best completes the argument below?
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Which of the following best completes the argument below?
One effect of the introduction of the electric refrigerator was a collapse in the market for ice. Formerly householders had bought ice to keep their iceboxes cool and the food stored in the iceboxes fresh. Now the iceboxes cool themselves. Similarly, the introduction of crops genetically engineered to be resistant to pests will______
(A) increase the size of crop harvests
(B) increase the cost of seeds
(C) reduce demand for chemical pesticides
(D) reduce the value of farmland
(E) reduce the number of farmers keeping livestock
[Reveal] Spoiler: OA
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Re: One effect of the introduction of the electric refrigerator [#permalink]
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29 Aug 2012, 10:36
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PUNEETSCHDV wrote:
Which of the following best completes the argument below?
One effect of the introduction of the electric refrigerator was a collapse in the market for ice. Formerly householders had bought ice to keep their iceboxes cool and the food stored in the iceboxes fresh. Now the iceboxes cool themselves. Similarly, the introduction of crops genetically engineered to be resistant to pests will______
(A) increase the size of crop harvests
(B) increase the cost of seeds
(C) reduce demand for chemical pesticides
(D) reduce the value of farmland
(E) reduce the number of farmers keeping livestock
For this type of argument you want to find the option that MUST be true given the content of the argument (premises). Here is the argument structure:
Refrigerator collapsed the ice market because the refigerator cooled food without the need of ice.
Similarly, Pest resistant crops will ....
Your answer should discuss some product's collapse due to a new product replacing the need for that older product.
A - Irrelevant - Doesn't discuss a separate, pre-established product
B - Irrelevant - Doesn't discuss a separate, pre-established product
C - Correct - Talks about a previously required product (pesticides) will be displaced by the new product (pest-resistant crops)
D - Out of Scope
E - Out of Scope
KW
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Re: Which of the following best completes the argument below? [#permalink]
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# Where do I stand ?
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Where do I stand ? [#permalink]
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18 Jan 2010, 08:45
Hi All,
i have started my preparation 3 weeks ago and finished studying most of the concepts ( approx 90%).
I have planned the GMAY exam after 8 weeks and today completed first Kaplan offline test (Kaplan guide).
I am targeting 730 in real GMAT and today inthe first practice test i got 600 (Q44, V21) .
In Quant i found that i could solv most of the problem and did silli mistakes in some of the incorrect answers , but in verbal i foung that time was insufficient .
Could any one suggest me how to plan further ? I heard the Kaplan are bit tough and do not reflect actual score .
I am very disappointed . Pls shre your views.
thnaks .
Kudos [?]: 2 [0], given: 2
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Re: Where do I stand ? [#permalink]
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18 Jan 2010, 09:53
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uglynakedguy wrote:
Hi All,
i have started my preparation 3 weeks ago and finished studying most of the concepts ( approx 90%).
I have planned the GMAY exam after 8 weeks and today completed first Kaplan offline test (Kaplan guide).
I am targeting 730 in real GMAT and today inthe first practice test i got 600 (Q44, V21) .
In Quant i found that i could solv most of the problem and did silli mistakes in some of the incorrect answers , but in verbal i foung that time was insufficient .
Could any one suggest me how to plan further ? I heard the Kaplan are bit tough and do not reflect actual score .
I am very disappointed . Pls shre your views.
thnaks .
Yes it doesnt reflect actual score but you have performed well in quant for KAPLAN but do work hard in Verbal.
Check out the verbal section.
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Re: Where do I stand ? [#permalink] 18 Jan 2010, 09:53
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## Comment Re:Time to read Asimov's Caves of Steel again... (Score 1)39
In Asimov mind, "Twist the genetics" probably did not have the same meaning that it has today. The science of the "genetics" so the study of heredity was already well established in the 50s (see Mendel, William Bateson, ...).
## Comment Re:Perhaps a better method... (Score 1)1001
As far as I can tell, this code is syntactically correct (if executed by a Perl interpreter).
Unless something goes terribly wrong with Perl, the code behavior is pretty well defined. It shall output the following 2 lines:
> This is the header of the output
> where are the entries?
So the only sensible answer to the question"why doesn't this work" is "it works"
## Comment A few numbers (Score 3, Informative)642
The Moon Escape velocity is 2.38 km/s while on Earth it is 11.186 km/s.
Since energy is proportional to the square of the speed (E=1/2*m*v^2) we can conclude that it is (11.186/ 2.38)^2 = 2 time easier to reach free space from the Moon than from Earth.
However, even if a rock is launched from the Moon at 2.38 km/s, it still inherits the inertia of the Moon. Simply speaking, the rock would not fall to Earth. It would be in an orbit similar to the Moon orbit.
The orbital speed of the Moon is about 1km/s so the rock must be given that additional acceleration to cancel its orbital speed.
At that point, the rock is immobile (from the Earth point of view) and it will start falling toward Earth because of ... gravity.
When it reaches Earth, its speed will be equal to the Earth Escape velocity (a bit less in fact since the rock did not start falling from an infinite distance) so 11.186 km/s.
The kinetic energy is given by the formula 1/2 * m * V^2 so for 1kg the kinetic energy at 11km/s is 1/2 * 1 * 11000^2 = 60 * 10^6 Joules
As a comparison, 1kg of TNT provides 4 * 10^6 Joules so each kg of moon rock would be equivalent to approximatively 15kg of TNT
The Hiroshima bomb was 15 kilotons of TNT = 15 * 10^6 Kg so a similar effect would require a 1000 tons of Moon rock and the ability to accelerate that rock to a speed of 2.38+1 = 3.38 km/s.
## Comment Re:wonder why asian elephant? (Score 5, Informative)169
The answer is likely in there: http://news.nationalgeographic...
The relevant bit is "At that time African elephants branched off first. Then just 440,000 years later, a blink of an eye in evolutionary time, Asian elephants and mammoths diverged into their own separate species."
## Comment Re:Moon dust depth (Score 1)140
The 1mm/1000 years figure is flawed. It is based on an incorrect value of 14 millions tons of dust per year computed in the early 60th. More recent figures are 100 to 1000 time smaller.
Even creationists websites do not use anymore moon dust as an argument. For instance look at the last paragraph in the Conclusion section of https://answersingenesis.org/a...
"Calculations show that the amount of meteoritic dust in the surface dust layer, and that which trace element analyses have shown to be in the regolith, is consistent with the current meteoritic dust influx rate operating over the evolutionists’ timescale. While there are some unresolved problems with the evolutionists’ case, the moon dust argument, using uniformitarian assumptions to argue against an old age for the moon and the solar system, should for the present not be used by creationists."
## Comment Re:So many theories... so many on the payroll list (Score 1)90
I could almost agree with you except that my critic was not against religion as a whole but against a literal interpretation of religious texts.
The issue is not science vs religion but reality vs blind faith in an absolute truth.
More generally, the same problem is found in non-religious contexts such as flat earth and other conspiracy theories where people will first assert a truth and then will ignore any evidence against it.
## Comment Re:So many theories... so many on the payroll list (Score 3, Insightful)90
What you are suffering from is called a displacement.
More precisely you appear to assume that scientists, like the creation myth in your favorite holy book, are claiming an absolute trust.
That does not work like that in the real world. Scientists create models and then try to invalidate them by comparing to reality.
Eventually all models become invalidated and are replaced by newer models that fit better with reality.
No sane scientist will ever claim that a specific model is absolutely true. That is why the article is full of "may" and "suggests". Same for the https://en.wikipedia.org/wiki/... page that describes the single large collision model.
The only thing that is correct in your post is that from the 5 scientists involved in your short story, the last one is probably the one that merits the most to be fired.
Why? Because scientists #1 and #3 are both proposing a model. Scientists #2 and #4 are defending those models. The only one that is not contributing in any useful way is scientist #5.
## House Committee Urges Congress To Pass Stingray Surveillance Legislation (theverge.com) 25
A bipartisan House Oversight and Government Reform Committee report released today urges Congress to pass legislation to regulate cell-site simulation surveillance devices like the Stingray. From a report: The devices, used by local and federal law enforcement agencies around the country, have been controversial, both for their power to track mobile devices and the secrecy often accompanying their use. As the report notes, the devices are still often used by local law enforcement agencies without warrants, instead relying on various lower standards of evidence. The committee's investigation, which last year prompted the Justice Department and Department of Homeland Security to change their policies on when to require a warrant before using the devices, found that the Justice Department uses 310 of the devices and spent \$71 million on them between fiscal years 2010 and 2014. Homeland Security has 124 devices and spent \$24 million in the same period. [...] The committee recommends that agencies become more "candid" about the devices, and urges states to pass legislation that would "require, with limited exceptions, issuance of a probable cause based warrant prior to law enforcement's use of these devices."
## Comment Re:We're so screwed (Score 1)293
At sea level and at 15 C air has a density of approximately 1.225 kg/m3 = 0.001225 g/cm3
https://en.wikipedia.org/wiki/... .00198 g/cm3 would imply a CO2 concentration of 1600% dumbass. At 400ppm a proper value is more like 0.00000049 g/cm3 so 4000 time smaller than your claim.
That does not change anything for my claims though.
Look! I am feeding the troll with ammos. Can't wait for the next shitty argument he will pull out of his ass.
## Comment Re:We're so screwed (Score 1)293
I just figured out that I forgot to take into account that CO2 is heavier than O2 and N2.
The molar mass of air is about 29 g/mole while that of CO2 is 44 g/mole.
In practice, that means that my hypothetical sheet of solid CO2 would be 44/29 = 1.5 times heavier than previously computed and, at 400ppm, its thickness would be 5mm instead of 3.3mm.
## Comment Re:Didn't we see it coming? (Score 1)293
I assume that those 20X are referring to the sentence in the early years of this century, concentrations of methane rose by only about 0.5ppb each year, compared with 10ppb in 2014 and 2015"
As usual the article is misleading.
Look at the data in https://www.esrl.noaa.gov/gmd/....
There was no noticeable increases during the period 2000-2010. There was some small fluctuations and the methane concentration was even decreasing for a few yers. Computing the ratio of two values only make sense if both values are large.
Here some of the values are even negative. Can we say that between 2001 and 2015, there was an acceleration of 9.98/-0.65 = -15.3X ?
If one year, the increase was exactly 0 then all other years would show an infinitely acceleration.
That does not make sense
The only important value is the long term trend. Looking at the graph, the increase was approximatively 210ppb during the last 30 years so an average of 7-8ppb per year.
Increases of 12.61 ppb and 9.98 ppb measured in 2014 and 2015 are above average but not by much. The increase was even higher during the late 80th.
## Comment Re:We're so screwed (Score 5, Insightful)293
No. Neither Nitrogen (N2) nor Oxygen (O2) are greenhouse gases and they compose most of the atmosphere.
Ozone (O3) is a form of oxygen that is considered a greenhouse gas but its concentration is small.
The following pages summarize quite well the situation: http://cdiac.ornl.gov/pns/curr...
More generally, the argument "it is only a small percentage of the whole atmosphere" is invalid. What is important is not the percentage of the various gases but their amount and their efficiency for trapping heat.
Also, people tend to underestimate the amounts of matter involved when talking about ppm or ppb. In https://en.wikipedia.org/wiki/... we find that "A column of air one square centimeter [cm2] (0.16 sq in) in cross-section ... has a mass of about 1.03 kilograms (2.3 lb)"
So the solar radiation that hits each cm2 of the earth surface has to go through about 1kg of air = 1000g.
The CO2 concentration is 400ppm so the solar radiation passes through 1000g * 400/1000000 = 0.4g of CO2 per cm2
Polycarbonate sheets used in most garden greenhouses has a density of 1.2g/cm3.
If atmospheric CO2 was compressed to that same density to form a hypothetical sheet of solid CO2 then its thickness would be 0.4/1.2 = 0.33 cm = 3.3mm
This is very comparable to the thickness of typical a polycarbonate sheet (3 to 6mm) so saying that 400ppm of CO2 cannot have any noticeable effects seems as stupid as saying that greenhouses are ineffective.
## Comment Re:"it was used for children's writing exercises" (Score 4, Insightful)235
More generally, "name calling" should be the expected behavior when asking almost any complex question to any large group of persons.
However, in that specific case, Richard Dawkins has the expected default position of any atheist (including me): The existence of an invisible unproven magic being cannot be the answer to any complex phenomena observed in the real world (in that case, that would be the origin of life). That position implies that there are things that we cannot explain with our current understanding of nature (you know, that thing called science).
Improving science by looking for more clues in the real world is the right way to handle those mysteries. Claiming "Magic", "God", "Taboo" or "Holy Book" is not.
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### Codemonkeys-2 Komplettlösung
Ich habe mal alle CM Testat 2 Aufgaben, die ich identifizieren konnte bearbeitet:
bei Dijkstra schlägt ein Test fehl, bei A* zwei, ich kann allerdings nicht sagen, wo der Fehler liegt.
Code: Alles auswählen
Merge Array iteratively:
public Listobject<T>[] executeMerge(Listobject<T>[] left, Listobject<T>[] right)
{
if ((left == null)|(right == null)){
throw new NullPointerException();
}
int l1 = left.length;
int l2 = right.length;
int i1 = 0;
int i2 = 0;
if (l1 == 0){
return right;
}
if (l2 == 0){
return left;
}
Listobject<T>[] r = new Listobject<>[l1 + l2];
while ((i1 < (l1))||(i2 < (l2))){
if (i1 == (l1)){
r[i1 + i2] = right[i2];
i2++;
continue;
}
if (i2 == (l2)){
r[i1 + i2] = left[i1];
i1++;
continue;
}
if (left[i1].compareTo(right[i2]) < 0){
r[i1 + i2] = left[i1];
i1++;
}
else{
r[i1 + i2] = right[i2];
i2++;
}
}
return r;
}
binary Search iterative:
public int[] binarySearchStep(Listobject<T>[] array, Listobject<T> element, int[] lrm)
{
if (array == null || element == null){
throw new NullPointerException();
}
int[] r = new int[3];
r[0] = lrm[0];
r[1] = lrm[1];
r[2] = -1;
int m = (r[0] + r[1]) / 2;
if (array.length == 0) {
return r;
}
if (array[m].compareTo(element) == 0) {
r[2] = m;
return r;
}
if (r[1] == m) {
return r;
}
if (array[m].compareTo(element) > 0) {
r[1] = m - 1;
} else {
r[0] = m + 1;
}
r[2] = m;
return r;
}
binary Search recursive:
public int binarySearchStep(Listobject<T>[] array, Listobject<T> searchedObject)
{
if (searchedObject == null || array == null) {
throw new NullPointerException();
}
if(array.length < 1)
return -1;
int m = array.length / 2;
int c = searchedObject.compareTo(array[m]);
if(array.length == 1 && c != 0){
return -1;
}
if(c == 0){
return m;
}
int l = 0;
Listobject<T>[] a;
if (c > 0){
l = array.length - m - 1;
a = new Listobject<>[l];
arraycopyStarter(array,m+1,a,0,l );
}
else{
if (array.length%2 == 0){
l = array.length - m;
}
else{
l = array.length - m - 1;
}
a = new Listobject<>[l];
arraycopyStarter(array,0,a,0,l) ;
}
int r = binarySearchStep(a,searchedObject);
if (r == -1){
return - 1;
}
if (c > 0){
return m + 1 + binarySearchStep(a,searchedObject);
}
return binarySearchStep(a,searchedObject);
}
Quicksort recursive:
public Listobject<T>[] quicksort(Listobject<T>[] array)
{
if (array.length <= 1){
return array;
}
Listobject pivot = array[0];
int el = 0;
int ep = 0;
for (int i = 0; i < array.length; i++){
int c = array[i].compareTo(pivot);
if (c == 0){
ep++;
}
if (c < 0){
el++;
}
}
Listobject<T>[] l = new Listobject<>[el];
Listobject<T>[] r = new Listobject<>[array.length - el - ep];
Listobject<T>[] p = new Listobject<>[ep];
int il = 0;
int ir = 0;
int ip = 0;
for (int j = 0; j < array.length; j++){
int c = array[j].compareTo(pivot);
if (c == 0){
p[ip] = array[j];
ip++;
}
if (c < 0){
l[il] = array[j];
il++;
}
if (c > 0){
r[ir] = array[j];
ir++;
}
}
return combineArrays(quicksort(l), p, quicksort(r));
}
Selectionsort:
public Listobject<T>[] executeSelectionSortOnArray(Listobject<T>[] array)
{
int l = array.length;
while (l > 1){
int m = 0;
for (int j = 1; j < l; j++){
if (array[j].compareTo(array[m]) > 0){
m = j;
}
}
if (m != (l - 1)){
Listobject<T> c = array[m];
array[m] = array[l - 1];
array[l - 1] = c;
}
l--;
}
return array;
}
Sort O(n^2)
(Selectionsort, da bereits vorher programmiert)
public Listobject<T>[] sort(Listobject<T>[] inputdata, ListobjectComparator<T> comp)
{
Listobject<T>[] array = inputdata;
int l = array.length;
while (l > 1){
int m = 0;
for (int j = 1; j < l; j++){
if (comp.compare(array[j],array[m]) > 0){
m = j;
}
}
if (m != (l - 1)){
Listobject<T> c = array[m];
array[m] = array[l - 1];
array[l - 1] = c;
}
l--;
}
return array;
}
public MListElement<T> executeMerge(MListElement<T> left, MListElement<T> right, Comparable_Comparator<T> comp)
{
if (left == null){
return right;
}
if (right == null){
return left;
}
MListElement<T> lc = left;
MListElement<T> rc = right;
MListElement<T> r;
MListElement c;
if (comp.compare(lc.getKey(),rc.getKey()) > 0){
r = rc;
c = r;
rc = rc.getNext();
}
else{
r = lc;
c = r;
lc = lc.getNext();
}
while ((lc != null)||(rc != null)){
if (lc == null){
c.setNext(rc);
break;
}
if (rc == null){
c.setNext(lc);
break;
}
if (comp.compare(lc.getKey(),rc.getKey()) > 0){
c.setNext(rc);
rc = rc.getNext();
c = c.getNext();
}
else{
c.setNext(lc);
lc = lc.getNext();
c = c.getNext();
}
}
return r;
}
prefix Checker:
public boolean praefix(String a, String b)
{
if ((a == null)||(b == null)){
throw new IllegalArgumentException();
}
if ((StringHelper.isEmpty(a))||(StringHelper.isEmpty(b))){
return false;
}
char[] aa = StringHelper.toCharArray(StringHelper.toLowerCase(a));
char[] ba = StringHelper.toCharArray(StringHelper.toLowerCase(b));
int al = aa.length;
for (int i = 0; i< al; i++){
if (aa[i] != ba[i]){
return false;
}
}
return true;
}
Simple String Matching:
public ArrayList<Integer> matching(String S, String T)
{
ArrayList<Integer> r = new ArrayList<Integer>();
ArrayList<Integer> s = new ArrayList<Integer>();
char[] as = StringHelper.toCharArray(StringHelper.toLowerCase(S));
char[] at = StringHelper.toCharArray(StringHelper.toLowerCase(T));
int l = as.length;
int lt = at.length;
for (int i = 0; i < l; i++) {
for(int j = 0; j < s.size(); j++){
if(as[i] != at[i + 1 - s.get(j)]){
s.remove(j);
j--;
}
else{
if(i + 1 - s.get(j) == lt - 1){
s.remove(j);
j--;
}
}
}
if(as[i] == at[0]){
if(lt > 1){
}
else {
}
}
}
return r;
}
{
if (data == null){
return null;
}
IdGenerator I = getIdGen();
ArrayList<Node<N,E>> l = getNodeList();
Node<N,E> n = new Node<N,E>(I,data);
setNodeList(l);
return n;
}
Graph removeNode:
public boolean removeNode(Node<N, E> node)
{
if (node == null){
return false;
}
ArrayList<Node<N,E>> n = getNodeList();
if (!n.contains(node)){
return false;
}
ArrayList<Edge<N,E>> fi = node.getFanIn();
ArrayList<Edge<N,E>> fo = node.getFanOut();
ArrayList<Edge<N,E>> e = getEdgeList();
for (Edge<N,E> c: fi){
c.removeFromNodes();
e.remove(c);
}
for (Edge<N,E> c: fo){
c.removeFromNodes();
e.remove(c);
}
setEdgeList(e);
n.remove(node);
setNodeList(n);
return true;
}
Graph findNode:
public Node<N, E> findNode(Node<N, E> startNode, N data)
{
if (startNode == null || data == null || !getNodeList().contains(startNode)){
return null;
}
ArrayDeque<Node<N, E>> s = new ArrayDeque();
HashSet<Node<N, E>> v = new HashSet();
s.push(startNode);
while (!s.isEmpty()) {
Node<N, E> n = s.pop();
if (objectEquals(data, n.getData())){
return n;
}
for (Edge<N, E> e: n.getFanOut()) {
Node<N, E> t = e.getTargetNode();
if (!v.contains(t)) {
s.push(t);
}
}
}
return null;
}
(undirected)
public void addEdge(Node<N, E> from, Node<N, E> to, E data) throws FanOverflowException
{
if (from.getFanIn().size() >= getFanInMax() || to.getFanIn().size() >= getFanInMax() ||
from.getFanOut().size() >= getFanOutMax() || to.getFanOut().size() >= getFanOutMax()){
throw new FanOverflowException();
}
ArrayList<Edge<N,E>> tfi = to.getFanIn();
ArrayList<Edge<N,E>> ffo = from.getFanOut();
ArrayList<Edge<N,E>> el = getEdgeList();
Edge<N,E> i = new Edge<N,E>(from,to,data);
ArrayList<Edge<N,E>> tfo = to.getFanOut();
ArrayList<Edge<N,E>> ffi = from.getFanIn();
Edge<N,E> i2 = new Edge<N,E>(to,from,data);
setEdgeList(el);
}
(directed)
public void addEdge(Node<N, E> from, Node<N, E> to, E data) throws FanOverflowException
{
if (to.getFanIn().size() >= getFanInMax()||from.getFanOut().size() >= getFanOutMax()){
throw new FanOverflowException();
}
ArrayList<Edge<N,E>> tfi = to.getFanIn();
ArrayList<Edge<N,E>> ffo = from.getFanOut();
ArrayList<Edge<N,E>> el = getEdgeList();
Edge<N,E> i = new Edge<N,E>(from,to,data);
setEdgeList(el);
}
Graph removeEdge:
(undirected)
public boolean removeEdge(Edge<N, E> edge)
{
if (edge == null){
return false;
}
ArrayList<Edge<N,E>> el = getEdgeList();
if (!el.contains(edge)){
return false;
}
l.removeFromNodes();
}
edge.removeFromNodes();
el.remove(edge);
setEdgeList(el);
return true;
}
(directed)
public boolean removeEdge(Edge<N, E> edge)
{
if (edge == null){
return false;
}
ArrayList<Edge<N,E>> el = getEdgeList();
if (!el.contains(edge)){
return false;
}
edge.removeFromNodes();
el.remove(edge);
setEdgeList(el);
return true;
}
Graph findEdge:???
Graph compareGraphs:???
Graph countEdges:
public int countEdgesInConnectedGraph(Node<N, E> node)
{
if(!contains(node)){
return -1;
}
HashSet<Edge<N, E>> e = new HashSet<Edge<N, E>>();
HashSet<Node<N, E>> n = new HashSet<Node<N, E>>();
countEdgesRec(node, e, n);
int u = 1;
u = 2;
}
return e.size() / u;
}
private void countEdgesRec(Node<N, E> node, HashSet<Edge<N, E>> edgeSet, HashSet<Node<N, E>> nodeSet)
{
return;
}
for(Edge<N, E> e : node.getFanOut()) {
countEdgesRec(e.getTargetNode(), edgeSet, nodeSet);
}
}
for(Edge<N, E> e : node.getFanIn()) {
countEdgesRec(e.getSourceNode(), edgeSet, nodeSet);
}
}
}
Graph countNodes:
public int countNodesInConnectedGraph(Node<N, E> node)
{
if (!contains(node)){
return -1;
}
HashSet<Node<N,E>> h = new HashSet<Node<N,E>>();
countNodesRec(node, h);
return h.size();
}
private void countNodesRec(Node<N, E> node, HashSet<Node<N, E>> set)
{
return;
}
for (Edge<N,E> e: node.getFanOut()){
if (!set.contains(e.getTargetNode())){
countNodesRec(e.getTargetNode(),set);
}
}
for (Edge<N,E> e: node.getFanIn()){
if (!set.contains(e.getSourceNode())){
countNodesRec(e.getSourceNode(),set);
}
}
}
Graph filterNodes:???
Dijkstra (21/22, Fehlerhafter Test?):
public void preProcess() throws InvalidInputException
{
if (getGraph() == null || getSourceNode() == null){
throw new InvalidInputException("null");
}
ArrayList<Edge<N,E>> el = getGraph().getEdgeList();
for (Edge<N,E> e: el){
if (getComparator().isNegative(e.getData())){
throw new InvalidInputException("negative");
}
}
setIterations(0);
HashMap<Node<N,E>, Node<N,E>> pre = new HashMap<Node<N,E>, Node<N,E>>(100, (float) 0.5);
setPredecessors(pre);
HashSet<Node<N,E>> set = new HashSet<Node<N,E>>(100, (float) 0.5);
setSettled(set);
setSmallestNode(getSourceNode());
setPaths(pat);
PriorityQueue<Node<N,E>> prio = new PriorityQueue<Node<N,E>>(getQueueComparator());
setPriorityQueue(prio);
HashMap<Node<N,E>, E> dis = new HashMap<Node<N,E>,E>(100, (float) 0.5);
for (Node<N,E> n: getGraph().getNodeList()){
dis.put(n,getComparator().getMax());
}
dis.put(getSourceNode(),getComparator().getZero());
setDistances(dis);
}
public boolean checkBreakCondition()
{
return !getPriorityQueue().isEmpty();
}
public void executeVariant()
{
setIterations(getIterations() + 1);
setSmallestNode(getPriorityQueue().remove());
}
public void checkInvariant() throws InvalidInvariantException
{
if (getPriorityQueue().contains(getSourceNode())) throw new InvalidInvariantException("SourceNode");
if (!(getSettled().size() == getIterations())) throw new InvalidInvariantException("settledSize");
for (Node<N,E> s: getSettled()){
if (getComparator().compare(getDistances().get(s), getComparator().getMax()) == 0) throw new InvalidInvariantException("settled");
}
for (Node<N,E> n: getGraph().getNodeList()){
if (!getSettled().contains(n) && !getPriorityQueue().contains(n) &&
(getComparator().compare(getDistances().get(n), getComparator().getMax()) != 0)) throw new InvalidInvariantException("unsettled");
}
}
public void doFunctionality()
{
Node<N,E> c = getSmallestNode();
ArrayList<Edge<N,E>> out = c.getFanOut();
for (Edge<N,E> o: out){
Node<N,E> t = o.getTargetNode();
if (getSettled().contains(t)) continue;
E d = getComparator().sum(getDistances().get(c), o.getData());
if (getComparator().compare(getDistances().get(t), d) > 0){
getDistances().put(t, d);
getPredecessors().put(t, c);
}
if (!getPriorityQueue().contains(t)){
}
}
}
Kruskal:
public void preProcess() throws InvalidInputException
{
AbstractGraph<N,E> g = getGraph();
if (g == null) throw new InvalidInputException("null");
ArrayList<Node<N,E>> nl = g.getNodeList();
if (nl.size() == 0) throw new InvalidInputException("nodes");
ArrayList<Edge<N,E>> el = g.getEdgeList();
if (!(g instanceof UndirectedGraph)) throw new InvalidInputException("undirected");
HashSet<Node<N, E>> set = new HashSet<Node<N, E>>();
ArrayDeque<Node<N, E>> stack = new ArrayDeque<Node<N, E>>();
while(!stack.isEmpty()) {
Node<N, E> c = stack.pop();
continue;
}
for(Edge<N, E> e : c.getFanOut()){
}
}
for(Node<N, E> n : nl) {
if(!set.contains(n)) {
throw new InvalidInputException("coherent");
}
}
setIterations(0);
PriorityQueue<Edge<N,E>> prio = new PriorityQueue<Edge<N,E>>(getQueueComparator());
setPriorityQueue(prio);
setUnionFind(new UnionFind<N,E>(nl));
UndirectedGraph<N,E> mst = new UndirectedGraph<N,E>(null);
for (Node<N,E> n: nl){
}
setMst(mst);
}
public boolean checkBreakCondition()
{
return !getPriorityQueue().isEmpty();
}
public void executeVariant()
{
setSmallestEdge(getPriorityQueue().remove());
setIterations(getIterations() + 1);
}
public void checkInvariant() throws InvalidInvariantException
{
if (getPriorityQueue().size() != getGraph().getEdgeList().size() - getIterations()) throw new InvalidInvariantException("size");
if (UndirectedGraphCycleChecker.hasCycle(getMst())) throw new InvalidInvariantException("cycle");
}
public void doFunctionality()
{
Edge<N,E> se = getSmallestEdge();
Node<N,E> s = se.getSourceNode();
Node<N,E> t = se.getTargetNode();
if (connected(s,t)) return;
union(s,t);
catch (FanOverflowException f) {}
}
public boolean connected(Node<N, E> p, Node<N, E> q)
{
return find(p) == find(q);
}
public void union(Node<N, E> p, Node<N, E> q)
{
HashMap<Node<N,E>, Node<N,E>> pa = getParents();
while(pa.get(p) != p){
p = pa.get(p);
}
while(pa.get(q) != q){
q = pa.get(q);
}
if (p == q) return;
HashMap<Node<N,E>, Integer> ra = getRanks();
if(ra.get(p) > ra.get(q)) {
pa.put(q, p);
ra.put(p, ra.get(p) + 1);
} else {
pa.put(p, q);
ra.put(q, ra.get(q) + 1);
}
}
A*:
public void preProcess() throws InvalidInputException
{
if (getGraph() == null) throw new InvalidInputException("null");
if (getGraph().getNodeList().size() == 0) throw new InvalidInputException("nodes");
if (getSourceNode() == null || !getGraph().getNodeList().contains(getSourceNode())) throw new InvalidInputException("source");
if (getTargetNode() == null || !getGraph().getNodeList().contains(getTargetNode())) throw new InvalidInputException("target");
setPathFound(false);
PriorityQueue<Node<N, E>> p = new PriorityQueue(getQueueComparator());
p.offer(getSourceNode());
setOpenList(p);
setClosedList(new ArrayList<Node<N, E>>());
HashMap<Node<N, E>, Node<N, E>> pre = new HashMap<Node<N, E>, Node<N, E>>();
pre.put(getSourceNode(), null);
setPredecessorMap(pre);
HashMap<Node<N, E>, E> dis = new HashMap<Node<N, E>, E>();
dis.put(getSourceNode(), getComparator().getZero());
setSourceDistanceMap(dis);
}
public void executeVariant()
{
setCurrentNode(getOpenList().poll());
}
public void doFunctionality()
{
Node<N, E> c = getCurrentNode();
if (c == getTargetNode() || c == null) {
setPathFound(true);
return;
}
expandNode(c);
}
public void postProcess()
{
ArrayList<Node<N, E>> p = new ArrayList<Node<N, E>>();
setPath(p);
if (!getPredecessorMap().containsKey(getTargetNode())) {
return;
}
for (Node<N, E> c = getTargetNode(); c != null; getPredecessorMap().get(c)) {
}
reverseCollection(p);
}
public void expandNode(Node<N, E> node)
{
for (Edge<N, E> e : node.getFanOut()) {
Node<N, E> t = e.getTargetNode();
if (getClosedList().contains(t))
continue;
E co = getComparator().sum(getSourceDistanceMap().get(node), e.getData());
E pre = getSourceDistanceMap().get(t);
if (pre == null || getComparator().greaterEqual(pre, co)) {
getSourceDistanceMap().put(t, co);
getPredecessorMap().put(t, node);
if (!getOpenList().contains(t)){
getOpenList().offer(t);
}
}
}
}
Floyd-Warshall:
public void preProcess() throws InvalidInputException
{
if (getGraph() == null) throw new InvalidInputException("Graph");
if (getNodeQueue() == null) throw new InvalidInputException("Queue");
if (getNodeQueue().size() != getGraph().getNodeList().size()) throw new InvalidInputException("size");
if(!getGraph().getNodeList().containsAll(getNodeQueue()) || !getNodeQueue().containsAll(getGraph().getNodeList())) throw new InvalidInputException("nodes");
setIteration(0);
AbstractEdgeComparator<E> com = getGraph().getComparator();
E m = com.getMax();
E z = com.getZero();
E c = m;
for(Node<N, E> n1 : getNodeQueue()) {
int i1 = getMatrixIndex(n1);
for(Node<N, E> n2 : getNodeQueue()) {
int i2 = getMatrixIndex(n2);
c = m;
if(i1 == i2){
c = z;
}
for(Edge<N, E> e : n1.getFanOut()) {
if(e.getTargetNode() == n2) {
c = e.getData();
break;
}
}
setM(i1, i2, c);
}
}
setCurrentNode(getNodeQueue().get(0));
}
public boolean checkBreakCondition()
{
return (getIteration() < (getGraph().getNodeList().size()));
}
public void executeVariant()
{
setIteration(getIteration() + 1);
if (getIteration() < getNodeQueue().size()){
setCurrentNode(getNodeQueue().get(getIteration()));
}
}
public void doFunctionality()
{
AbstractEdgeComparator<E> com = getGraph().getComparator();
int ci = getMatrixIndex(getCurrentNode());
for(Node<N, E> n1 : getNodeQueue()) {
int i1 = getMatrixIndex(n1);
for(Node<N, E> n2 : getNodeQueue()) {
int i2 = getMatrixIndex(n2);
E c = getM(i1, i2);
E a = com.sum(getM(i1, ci), getM(ci, i2));
setM(i1, i2, com.min(c, a));
}
}
}
Bellman-Ford:
public void preProcess() throws InvalidInputException
{
if(getGraph() == null) throw new InvalidInputException("graph");
setKard_V(getGraph().getNodeList().size());
setM(0);
AbstractEdgeComparator<E> com = getGraph().getComparator();
Matrix<E> m = new Matrix<E>(getKard_V(), getKard_V(), com.getMax());
E val;
for(int i1 = 0; i1 < getKard_V(); i1++) {
Node<N, E> n1 = getGraph().getNodeList().get(i1);
for(int i2 = 0; i2 < getKard_V(); i2++) {
val = com.getMax();
Node<N, E> n2 = getGraph().getNodeList().get(i2);
for(Edge<N, E> e : n1.getFanOut()) {
if(e.getTargetNode() == n2) {
val = e.getData();
}
}
if(i1 == i2)
val = com.getZero();
m.set(i1 + 1, i2 + 1, val);
}
}
setL(m);
appendToM(m);
setI(-1);
}
public boolean checkBreakCondition()
{
return getI() < getKard_V();
}
public void executeVariant()
{
setI(getI() + 1);
}
public void doFunctionality()
{
AbstractEdgeComparator<E> com = getGraph().getComparator();
Matrix<E> o = getM(getI());
Matrix<E> m = new Matrix(getKard_V(), getKard_V(), com.getMax());
E val;
for(int i1 = 1; i1 < getKard_V() + 1; i1++) {
for(int i2 = 1; i2 < getKard_V() + 1; i2++) {
val = o.get(i1, i2);
for(int i3 = 1; i3 < getKard_V() + 1; i3++) {
val = com.min(val, com.sum(o.get(i1, i3), getL().get(i3, i2)));
}
m.set(i1, i2, val);
}
}
appendToM(m);
}
Prim:
public void preProcess() throws InvalidInputException
{
if (getGraph() == null || getStartNode() == null) throw new InvalidInputException("null");
//if (!(getGraph() instanceof DirectedGraph)) throw new InvalidInputException("undirected");
if (getGraph().getNodeList().size() == 0) throw new InvalidInputException("node");
ArrayDeque<Node<N,E>> stack = new ArrayDeque<Node<N,E>>();
HashSet<Node<N,E>> set = new HashSet<Node<N,E>>();
stack.push(getStartNode());
while (!stack.isEmpty()){
Node<N,E> c = stack.pop();
for (Edge<N,E> e: c.getFanOut()){
Node<N,E> t = e.getTargetNode();
if (!set.contains(t)){
stack.push(t);
}
}
}
if (!(set.size() == getGraph().getNodeList().size())) throw new InvalidInputException("coherent");
setIterations(0);
PriorityQueue<Edge<N,E>> prio = new PriorityQueue<Edge<N,E>>(getQueueComparator());
for (Edge<N,E> e: getStartNode().getFanOut()){
}
setPriorityQueue(prio);
setMst(new ArrayList<Edge<N,E>>());
HashSet<Node<N,E>> v = new HashSet<Node<N,E>>();
setVisited(v);
}
public boolean checkBreakCondition()
{
return !(getPriorityQueue().isEmpty());
}
public void executeVariant()
{
setSmallestEdge(getPriorityQueue().remove());
setIterations(getIterations() + 1);
}
public void checkInvariant() throws InvalidInvariantException
{
if ((getVisited().size() - 1) != getMst().size()) throw new InvalidInvariantException("size");
if (getVisited().size() != getGraph().getNodeList().size() && getVisited().size() != getIterations()) throw new InvalidInvariantException("iterations");
}
public void doFunctionality()
{
Edge<N,E> s = getSmallestEdge();
removeUnnecessaryEdgesOutOfPriorityQueue();
for (Edge<N,E> e : s.getTargetNode().getFanOut()){
Node<N,E> n = e.getTargetNode();
if (!getVisited().contains(n)){
getPriorityQueue().offer(e);
}
}
} | 6,010 | 20,500 | {"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} | 3.25 | 3 | CC-MAIN-2020-05 | latest | en | 0.138076 |
https://studentshare.org/family-consumer-science/1405856-1fostering-creativity-in-primary-education-2-extending-numeracy | 1,550,803,303,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247512461.73/warc/CC-MAIN-20190222013546-20190222035546-00416.warc.gz | 678,093,738 | 23,907 | # EXTENDING NUMERACY - Essay Example
Summary
The essence of mathematics is not to make simple things complicated, but to make complicated things simple (S. Gudder, 2010).’ Mathematical thinking is an abstruse term which can be defined in a variety of ways, and broken down into several sections. …
## Extract of sample"EXTENDING NUMERACY"
Download file to see previous pages It is believed that mathematical thinking is the process of logical thought, as highlighted by DFES (1999) is a process of ‘Logical reasoning, problem solving and the ability to think in abstract ways (DFES, 1999, p.60-61).’ Development of this method of thought can be aided through the art of speaking and listening, alongside group and collaborative work. Children using and applying their skills equates into mathematical thinking, as agreed by the National Curriculum (1999) which states that ‘mathematics trains children with a ‘uniquely powerful set of tools’ to understand and develop the world (Directgov, 2010).’ The National Numeracy Strategy, introduced by the Department of Education and employment, lays down a framework for the curriculum to be followed in all schools while teaching mathematics for children attending Reception to Year 6 (National Numeracy Strategy, 1999). Compiled in 1999, The National Numeracy Strategy set out four main approaches to the teaching of mathematics, viz.: Dedicated mathematics lessons every day Direct teaching and interactive oral work with the whole class and groups An emphasis on mental calculation Controlled differentiation, with all pupils engaged in mathematics relating to a common theme (National Numeracy Strategy, 1999). The National Numeracy Strategy has outlined guidelines for numeracy to be taught nationwide which has to be utilised by teachers within the classroom (National Numeracy Strategy, 1999). ...
In order to calculate either cognitively or use written algorithms, it is necessary to have basic numeracy skills. These include: remembering and recalling number facts, relationships of numbers, and problem solving using mental visualization and/or previously learned strategies. Mental mathematics is the foundation to all mathematical methodologies. The ability to count or at least know place value is all cognitive during which memory is used to recall numerical facts or obtain new ones, therefore this skill should be nurtured and emphasised. There are several educationalists who are proponents of encouraging children to engage in more mental calculations rather than just solving problems on paper. In this regard, Askew (1998) states that, ‘Ultimately, mathematics is a mental activity. While practical mathematics can help children develop mental images, written work on its own is not sufficient (Askew, 1998)’. This statement agrees with the principle laid down in the National Numeracy Strategy, which states that “an ability to calculate mentally lies at the heart of numeracy (National Numeracy Strategy, 1999)’ It is of truth, and it is of little use if a child can complete a page of sums, but does not know how to tackle a problem that has not been written down. The National Numeracy Strategy emphasizes that pupils need to be given opportunities to develop flexible methods of working with numbers mentally that enables them to use known information to derive facts that they cannot recall (National Numeracy Strategy, 1999). Differentiation within education largely relates to the differences between comparable things, for example, building a picture for the children that ...Download file to see next pagesRead More
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5 Pages(1250 words)Essay | 2,448 | 12,329 | {"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-2019-09 | latest | en | 0.919933 |
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Balanced equation:
KIO3 + 3 H2 = KI + 3 H2O
Reaction type: double replacement
Reaction stoichiometry Limiting reagent
CompoundCoefficientMolar MassMolesWeight
KIO31214.00097
H232.01588
KI1166.00277
H2O318.01528
Units: molar mass - g/mol, weight - g.
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Back to Online Chemical Tools Menu | 467 | 1,817 | {"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-2019-43 | latest | en | 0.78507 |
https://www.mathematik.uni-marburg.de/modulhandbuch/20222/MSc_Mathematics/Specialization_Modules_in_Mathematics/Nonlinear_Optimization.html | 1,722,657,174,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640356078.1/warc/CC-MAIN-20240803025153-20240803055153-00532.warc.gz | 679,950,489 | 7,049 | This entry is from Winter semester 2022/23 and might be obsolete. No current equivalent could be found.
# Nonlinear Optimization (dt. Nichtlineare Optimierung)
Level, degree of commitment Specialization module, compulsory elective module Forms of teaching and learning,workload Lecture (4 SWS), recitation class (2 SWS), 270 hours (90 h attendance, 180 h private study) Credit points,formal requirements 9 CP Course requirement(s): Successful completion of at least 50 percent of the points from the weekly exercises. Examination type: Written or oral examination Language,Grading German,The grading is done with 0 to 15 points according to the examination regulations for the degree program M.Sc. Business Mathematics. Duration,frequency One semester, Regularly alternating with other courses in the research area of optimization Person in charge of the module's outline Prof. Dr. Thomas Surowiec
## Contents
Fundamentals of nonlinear optimization: Kuhn-Tucker theory, minimization of nonlinear functions; minimization of nonlinear functions with constraints
Fundamentals of nonlinear optimization: Kuhn-Tucker theory, minimization of nonlinear functions; minimization of nonlinear functions with constraints
## Qualification Goals
The students shall
• acquire a sound knowledge of the theory and practice of basic methods of optimization
• learn to recognize and assess the relevance of optimization methods for practical problems from different application areas such as parameter optimization, nonlinear regression, approximation, or optimal control,
• acquire the ability to model and solve optimization problems in practical situations,
• practice mathematical methods (development of mathematical intuition and its formal justification, training of the ability to abstract, proof methods),
• improve their oral communication skills in the recitation classes by practicing free speech in front of an audience and during discussion.
## Prerequisites
None. The competences taught in the following modules are recommended: either Linear Algebra I and Analysis I and Analysis II or Basic Linear Algebra or Basic Real Analysis and Basics of Advanced Mathematics.
## Applicability
Module imported from M.Sc. Business Mathematics.
It can be attended at FB12 in study program(s)
• B.Sc. Mathematics
• M.Sc. Data Science
• M.Sc. Computer Science
• M.Sc. Mathematics
• LAaG Mathematics
When studying M.Sc. Mathematics, this module can be attended in the study area Specialization Modules in Mathematics.
The module is assigned to Applied Mathematics. Further information on eligibility can be found in the description of the study area.
• Alt, W.: Nichtlineare Optimierung, Vieweg, 2002
• Jarre, F., Stoer, J.: Nonlinear Programming, Springer, 2004
• Fletcher, R.: Practical Methods of Optimization, 2nd Edition, John Wiley & Sons, 1987
• Nocedal, J., Wright, S.: Numerical Optimization, Springer, 2002 | 590 | 2,916 | {"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-2024-33 | latest | en | 0.864721 |
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The members of the winning soccer team walked jauntily onto the bus. Jauntily means CHEERFULLY.
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The members of the winning soccer team walked jauntily onto the bus.
Jauntily means CHEERFULLY.
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#### Resources tagged with Odd and even numbers similar to Three Dice:
Filter by: Content type:
Stage:
Challenge level:
### There are 43 results
Broad Topics > Numbers and the Number System > Odd and even numbers
### Crossings
##### Stage: 2 Challenge Level:
In this problem we are looking at sets of parallel sticks that cross each other. What is the least number of crossings you can make? And the greatest?
### What Do You Need?
##### Stage: 2 Challenge Level:
Four of these clues are needed to find the chosen number on this grid and four are true but do nothing to help in finding the number. Can you sort out the clues and find the number?
### Becky's Number Plumber
##### Stage: 2 Challenge Level:
Becky created a number plumber which multiplies by 5 and subtracts 4. What do you notice about the numbers that it produces? Can you explain your findings?
### Curious Number
##### Stage: 2 Challenge Level:
Can you order the digits from 1-3 to make a number which is divisible by 3 so when the last digit is removed it becomes a 2-figure number divisible by 2, and so on?
### Always, Sometimes or Never?
##### Stage: 1 and 2 Challenge Level:
Are these statements relating to odd and even numbers always true, sometimes true or never true?
### Down to Nothing
##### Stage: 2 Challenge Level:
A game for 2 or more people. Starting with 100, subratct a number from 1 to 9 from the total. You score for making an odd number, a number ending in 0 or a multiple of 6.
### Multiplication Series: Number Arrays
##### Stage: 1 and 2
This article for teachers describes how number arrays can be a useful reprentation for many number concepts.
### Three Spinners
##### Stage: 2 Challenge Level:
These red, yellow and blue spinners were each spun 45 times in total. Can you work out which numbers are on each spinner?
### Magic Vs
##### Stage: 2 Challenge Level:
Can you put the numbers 1-5 in the V shape so that both 'arms' have the same total?
### A Mixed-up Clock
##### Stage: 2 Challenge Level:
There is a clock-face where the numbers have become all mixed up. Can you find out where all the numbers have got to from these ten statements?
### Number Differences
##### Stage: 2 Challenge Level:
Place the numbers from 1 to 9 in the squares below so that the difference between joined squares is odd. How many different ways can you do this?
### Number Tracks
##### Stage: 2 Challenge Level:
Ben’s class were cutting up number tracks. First they cut them into twos and added up the numbers on each piece. What patterns could they see?
### The Thousands Game
##### Stage: 2 Challenge Level:
Each child in Class 3 took four numbers out of the bag. Who had made the highest even number?
### Diagonal Trace
##### Stage: 2 Challenge Level:
You can trace over all of the diagonals of a pentagon without lifting your pencil and without going over any more than once. Can the same thing be done with a hexagon or with a heptagon?
### Square Subtraction
##### Stage: 2 Challenge Level:
Look at what happens when you take a number, square it and subtract your answer. What kind of number do you get? Can you prove it?
### Always, Sometimes or Never? Number
##### Stage: 2 Challenge Level:
Are these statements always true, sometimes true or never true?
### Number Detective
##### Stage: 2 Challenge Level:
Follow the clues to find the mystery number.
### Play to 37
##### Stage: 2 Challenge Level:
In this game for two players, the idea is to take it in turns to choose 1, 3, 5 or 7. The winner is the first to make the total 37.
### More Carroll Diagrams
##### Stage: 2 Challenge Level:
How have the numbers been placed in this Carroll diagram? Which labels would you put on each row and column?
### Take Three Numbers
##### Stage: 2 Challenge Level:
What happens when you add three numbers together? Will your answer be odd or even? How do you know?
### Make 37
##### Stage: 2 and 3 Challenge Level:
Four bags contain a large number of 1s, 3s, 5s and 7s. Pick any ten numbers from the bags above so that their total is 37.
### Arrangements
##### Stage: 2 Challenge Level:
Is it possible to place 2 counters on the 3 by 3 grid so that there is an even number of counters in every row and every column? How about if you have 3 counters or 4 counters or....?
### Sets of Numbers
##### Stage: 2 Challenge Level:
How many different sets of numbers with at least four members can you find in the numbers in this box?
### Part the Piles
##### Stage: 2 Challenge Level:
Try to stop your opponent from being able to split the piles of counters into unequal numbers. Can you find a strategy?
### Odds and Threes
##### Stage: 2 Challenge Level:
A game for 2 people using a pack of cards Turn over 2 cards and try to make an odd number or a multiple of 3.
### Sets of Four Numbers
##### Stage: 2 Challenge Level:
There are ten children in Becky's group. Can you find a set of numbers for each of them? Are there any other sets?
### Seven Flipped
##### Stage: 2 Challenge Level:
Investigate the smallest number of moves it takes to turn these mats upside-down if you can only turn exactly three at a time.
### Break it Up!
##### Stage: 1 and 2 Challenge Level:
In how many different ways can you break up a stick of 7 interlocking cubes? Now try with a stick of 8 cubes and a stick of 6 cubes.
### Exploring Simple Mappings
##### Stage: 3 Challenge Level:
Explore the relationship between simple linear functions and their graphs.
### Odds, Evens and More Evens
##### Stage: 3 Challenge Level:
Alison, Bernard and Charlie have been exploring sequences of odd and even numbers, which raise some intriguing questions...
### Venn Diagrams
##### Stage: 1 and 2 Challenge Level:
Use the interactivities to complete these Venn diagrams.
### Red Even
##### Stage: 2 Challenge Level:
You have 4 red and 5 blue counters. How many ways can they be placed on a 3 by 3 grid so that all the rows columns and diagonals have an even number of red counters?
### Take One Example
##### Stage: 1 and 2
This article introduces the idea of generic proof for younger children and illustrates how one example can offer a proof of a general result through unpacking its underlying structure.
### Odd Squares
##### Stage: 2 Challenge Level:
Think of a number, square it and subtract your starting number. Is the number you’re left with odd or even? How do the images help to explain this?
### Magic Letters
##### Stage: 3 Challenge Level:
Charlie has made a Magic V. Can you use his example to make some more? And how about Magic Ls, Ns and Ws?
##### Stage: 3 Challenge Level:
Great Granddad is very proud of his telegram from the Queen congratulating him on his hundredth birthday and he has friends who are even older than he is... When was he born?
### Picturing Square Numbers
##### Stage: 3 Challenge Level:
Square numbers can be represented as the sum of consecutive odd numbers. What is the sum of 1 + 3 + ..... + 149 + 151 + 153?
### Various Venns
##### Stage: 2 Challenge Level:
Use the interactivities to complete these Venn diagrams.
### Score
##### Stage: 3 Challenge Level:
There are exactly 3 ways to add 4 odd numbers to get 10. Find all the ways of adding 8 odd numbers to get 20. To be sure of getting all the solutions you will need to be systematic. What about. . . .
### Odds and Evens
##### Stage: 3 Challenge Level:
Is this a fair game? How many ways are there of creating a fair game by adding odd and even numbers?
### Impossible Sandwiches
##### Stage: 3, 4 and 5
In this 7-sandwich: 7 1 3 1 6 4 3 5 7 2 4 6 2 5 there are 7 numbers between the 7s, 6 between the 6s etc. The article shows which values of n can make n-sandwiches and which cannot.
### Prime Magic
##### Stage: 2, 3 and 4 Challenge Level:
Place the numbers 1, 2, 3,..., 9 one on each square of a 3 by 3 grid so that all the rows and columns add up to a prime number. How many different solutions can you find? | 1,931 | 7,996 | {"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} | 4.375 | 4 | CC-MAIN-2017-39 | latest | en | 0.880426 |
https://www.teachtasticiep.com/iep-goal/iep-goal-to-determine-the-relationship-between-area-and-perimeter | 1,716,229,459,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058293.53/warc/CC-MAIN-20240520173148-20240520203148-00145.warc.gz | 921,640,706 | 165,463 | top of page
Learning Standard
### 4.MD.A.3
Apply the area and perimeter formulas for rectangles in real world and mathematical problems.
Target IEP Goal
By (date), when given exercises with area, the student will determine the relationship between area and perimeter, improving measurement and data skills from 0/10 work samples out of ten consecutive trials to 8/10 work samples in ten consecutive trials.
Teach Tastic IEP goals written to be SMART: Specific, Measurable, Attainable, Results-Oriented, and Time-Bound.
IEP Goal Objectives
1
Find the Perimeter of Rectangles
By (date), when given exercises with perimeter, the student will find the perimeter of rectangles, improving measurement and data skills from 0/10 work samples out of ten consecutive trials to 8/10 work samples in ten consecutive trials.
2
Add Four or More One-Digit Numbers
By (date), when given exercises with addition, one digit, the student will add four or more one-digit numbers, improving operations and algebraic thinking skills from 0/10 work samples out of ten consecutive trials to 8/10 work samples in ten consecutive trials.
3 | 249 | 1,127 | {"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-2024-22 | latest | en | 0.880265 |
http://www.fixya.com/support/t25623362-inverse_function_f_x_x | 1,506,219,436,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818689823.92/warc/CC-MAIN-20170924010628-20170924030628-00405.warc.gz | 456,114,171 | 33,493 | Question about Office Equipment & Supplies
# What's the inverse function of f(x)=-x - Office Equipment & Supplies
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• Master
I get f(x) = -x as the inverse of f(x) = -x.
I set up some tables to see if this is correct.
x ' y
-------
-2 2
-1 1
0 0
1 -1
2 -2
For the inverse function, we should be able to put in y and get back x x ' y
-------
2 -2
1 -1
0 0
-1 1
-2 2
The equation of this function is f(x) = -x.
Good luck.
Paul
Inverse Functions
Posted on Apr 04, 2015
Hi,
a 6ya expert can help you resolve that issue over the phone in a minute or two.
best thing about this new service is that you are never placed on hold and get to talk to real repairmen in the US.
the service is completely free and covers almost anything you can think of (from cars to computers, handyman, and even drones).
goodluck!
Posted on Jan 02, 2017
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## Related Questions:
### On my casio fx-83GT PLUS calculator whenever i use the inverse cos,sine and tan function math ERROR comes up how do i fix this
The inverse cosine function is defined for arguments from -1 to +1. 6 is outside of this domain, thus the error. Do you get an error if you try the inverse cosine of 0.6, for example?
If you're getting the same error for the inverse tangent, could you post an example?
May 12, 2014 | Casio FX83ES Scientific Calculator
### "Undefined" as answer to an inverse sine calculation
I suspect that you are confusing things a bit.
The inverse sine, called the arcsine is a function defined in the closed interval [-1,1]. And so is the inverse cosine. Any value outside this interval will give you a non-real result (meaning a complex one).
There are no limitations on the domain of definition of the inverse hyperbolic sine or sinh^-1
If your input value is allowed to be complex, the arcsine function gives a complex value. See the screen capture
Mar 17, 2014 | Texas Instruments TI 89 Titanium Graphing...
### Find inverse sin
The inverse of a function usually shares the same physical key as the direct function.
Feb 03, 2014 | HP 30s Calculator
### Inverse sine function
The sine and cosine function have a range between [-1, 1]. The domain of their inverse functions is [-1,1]. So 20/1 which is 20 is out of the domain of definition of the functions. No limitations for tangent and cotangent.
Oct 22, 2013 | Casio FX-115ES Scientific Calculator
### Where is the inverse key on the TI-84 silver edition?
That depends on what you want to invert.
If you want the multiplicative inverse (1/x) then it's the "1-superscript-minus-one" key just below the "MATH" key.
If you want the additive inverse then it's the "(-)" key just to the right of the decimal point key.
If you want the inverse of many of the built-in functions (like the trig functions) then press the "2ND" key at the upper-left before pressing the function key.
For the inverse of some functions you'll have to go through the menus.
If you have a particular inverse in mind, please reply to this post and specify exactly what you'd like to invert.
Jun 25, 2012 | Texas Instruments TI-84 Plus Silver...
### Im pressing the buttom that everyone is saying is the inverse key but it is saying error. i really need help with this.
It is much more enlightning to tell us what you want to do rather than talk about something that does not exist. And one thing that does not mean anything is that ellusive INVERSE key that everyone and his dog is looking for.
The additive inverse of a number a is its opposite (-a). The key to use is the change key (-).
The multiplicative inverse of non-zero number a is its reciprocal 1/a
The inverse of the natural log (LN) function is the EXPONENTIAL function e^(x). On most calculators if one function is accessed diectly (the marking is on the key), its inverse is accessed by pressing [SHIF] (Casio, or [2nd] (TI) or [2ndF] (Sharp).
The inverse of the function raise 10 to a power [10^x] is [LOG]
Similarly, the inverse of the sine function SIN is the arcsine [sin^-1]. To access the latter, it you press [2nd][SIN]. The same is true for arcosine [2nd][COS], and arctangent [2nd][TAN]
The inverse of the square functions [x^2] is the square root function [2nd][x^2]
Inverse of x^3 is cubic root
This is just a quick overview.
Feb 16, 2012 | Texas Instruments TI-84 Plus Silver...
### I need to know which button is the inverse button on my Ti-36X Pro please!
The inverse of what&? Each inverse function uses the same key (shifted) as its direct function. For the reciprocal key look for x^-1 or [1/X]
Sep 15, 2011 | Texas Instruments TI-36 X Solar Calculator
### How to take anti log in casio fx 991 ms and give example?
Depending on the actual log function you are talking about, the inverse function (what you call the anti log) can be one of several things.
The inverse of the natural log (LN) is the exponential function e^x.
If y= ln(x) then x=e^y
On a calculator functions that are inverse of one another share the same physical key. If one function is marked on a key, the inverse function is usually the SHIFTED (or 2nd) key
The inverse of the common (decimal logarithm) is the power function with 10 as base.
Example: if y= log(x) then x=10^y
For an arbitrary base (must be positive) if y= log_b(x) then x=b^y.
Note: I do not think that the FX-300ms can perform log calculations in bases other than e (natural log) or 10 (common decimal log).
Aug 13, 2011 | Casio FX-300MS Calculator
### Does this calculator have an inverse function? If so, which button is it?
The x^-1 key is in the leftmost column of the keyboard, below the MATH key.
If you want the inverse trig functions, press 2ND then the trig function key.
Jan 13, 2011 | Texas Instruments TI-84 Plus Calculator
### I need an inverse log button on the casio calculator 9750GII.... does it have one??? I know how to do log, but can't find the inverse.
The log base 10 function has an inverse: It is the 10 ^x function. It shares the same physical key with the decimal log. Use [SHIFT][LOG].
The natural log (LN) has an inverse alos and that it the exponentila function e^x. It shares the same physical key with it.
Jul 20, 2010 | Casio FX-9750GPlus Calculator
## Open Questions:
#### Related Topics:
84 people viewed this question
Level 3 Expert
Level 3 Expert | 1,683 | 6,515 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2017-39 | latest | en | 0.879356 |
https://r4r.co.in/java/corejava/topics/basics/10java_BigNumbers.shtml | 1,627,326,073,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046152144.92/warc/CC-MAIN-20210726183622-20210726213622-00130.warc.gz | 480,650,063 | 10,126 | # Java Programing laungage
Big Numbers In Java
Previous Home Next
While programming if the precision of the integers and floating point number is not according to the requirements, then we can use the big numbers defined in java.
the big numbers are contained in java.math package, BigInteger is used for manipulating the precision of the integral values while BigDecimal is used for the floating point numbers. The method valueOf(), defined in the package can turn any ordinary number to a big number.
BigInteger num = BigInteger.valueOf(300);
In this example we have converted a simple integer 300 to a BigInteger type value and assigned it to the num. Also note that we cannot use basic operators like +, -, /, * directly on the big numbers, infact we have to use predefined methods like add(), multiply() in order to combine the big numbers. For e.g.
BigInteger num = BigInteger.valueOf(300);
BigInteger p = num.add(BigInteger.valueOf(200));
// it goes as p=num+200= 300+200=500
Some of the methods which are defined in java.math.BigInteger class and used more commonly are :
BigInteger add(BigInteger second)
// returns the sum of current BigInteger with the second
BigInteger multiply(BigInteger second)
// returns the multiplication of current BigInteger with the second
BigInteger divide(BigInteger second)
// returns the division of current BigInteger with the second
BigInteger subrtact(BigInteger second)
// returns the subtraction of current BigInteger with the second
BigInteger mod(BigInteger second)
// return the mod value of current BigInteger with the second
### int Java TutorialsompareTo(BigInteger second)
It compares the value of the current BigInteger with the second and then returns an integral value. If current number is greater than second than it will give a positive integer, if equal it will return 0, else it will return a negative value.
### static BigInteger valueOf(Long p)
It converts an integer number to the Big Integer type number.
Similarly we can use BigDecimal class to manipulating the precision of the floating point numbers. The frequent methods which are defined in the java.math.BigDecimal class are :
BigDecimal add(BigDecimal second)
// returns the sum of current BigDecimal with the second
BigDecimal multiply(BigDecimal second)
// returns the multiplication of current BigDecimal with the second
BigDecimal divide(BigDecimal second, int roundoff mode)
//returns the divison of current BigDecimal with
second one with roundoff digits
BigDecimal subrtact(BigDecimal second)
// returns the subtraction of current BigDecimal with the second
int Java TutorialsompareTo(BigDecimal second)
// return an integer after comparing the two BigDecimal type numbers
static BigDecimal valueOf(Long p)
// convert an simple number to the BigDecimal type number
static BigDecimal valueOf(Long p, int scale)
// converts an simple to the BigDecimal number x /10^scale
Previous Home Next | 564 | 2,934 | {"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-2021-31 | longest | en | 0.618391 |
https://link.springer.com/chapter/10.1007/BFb0039712 | 1,534,902,317,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221219242.93/warc/CC-MAIN-20180822010128-20180822030128-00030.warc.gz | 719,984,368 | 13,793 | # The other linear logic
Extended abstract
• M. Taitslin
• D. Arkhangelsky
Conference paper
Part of the Lecture Notes in Computer Science book series (LNCS, volume 735)
## Abstract
One of computer science motivations for Linear Logic is that proofs in this logic reflect the use of certain resources. However, the problem of availability of resources is not addressed. In this paper we are attempting to fill this gap.
Our approach can be illustrated most clearly by example of a Horn fragment. In this fragment, the left part of each sequent is a conjunction of simple implications and supplies, and the right part is a simple conjunction of supplies. An implication is called simple, if both its succedent and antecedent are conjunctions of supplies. As usual, a supply is simply a letter.
The standard axiom system for this Horn fragment consists of three inference rules, while the axioms are all the sequents with a single letter both for the left and right parts (the letter is the same).
On the other hand, the implications can be thought of as determining relations in a commutative semigroup and then we show that a sequent is provable iff the collection of supplies in its right part can be obtained from the left one by applying all the relations from the left part in appropriate order.
Now, let us require that the left part (i.e. all its supplies) is available. For this case we construct a new axiom system with sligtly more complex axioms and inference rules in which any axiom can be used at most once. Among other corollaries of this result we obtain the NP-completeness of the Horn fragment (known result).
However, for a bounded number of implication types we construct a subexponential algorithm for the derivation search. | 369 | 1,751 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2018-34 | latest | en | 0.917486 |
https://parabola.unsw.edu.au/2020-2029/volume-60-2024/issue-1 | 1,719,323,862,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198866143.18/warc/CC-MAIN-20240625135622-20240625165622-00301.warc.gz | 398,831,800 | 10,500 | Volume 60
, Issue 1
2024
Dear Reader, I am proud and excited to announce that it is the 60th anniversary of Parabola!
The isoperimetric problem asks for the least-perimeter way to enclose a given volume. We numerically solve this problem for double, triple and quadruple bubbles in the plane with density $r^p$ for various $p > 0$, using Brakke’s Evolver.
What is an amplifier and how does it work? To answer these intriguing questions, I constructed a model of an amplifier.
A ray of light emanates at some angle from a corner of a square region and follows a path determined by its reflections off the walls of the square. We determine when the ray’s path is finite, and we compute its length in this case.
How much do financial management fees cost investors? This article studies fees charged annually as a percentage of Assets Under Management (AUM).
We will attempt to multiply like a Babylonian student and will derive beautiful sexagesimal approximations.
Wacław Sierpiński proved that there exist infinitely many odd integers $k$ such that numbers of the form $k\cdot 2^n + 1$ are never prime for any integer $n$. The values of $k$ with this property are called Sierpiński numbers. The Sierpiński Problem is to find the smallest Sierpiński number.
The solution formula to the quadratic equation $ax^2+bx+c=0$ is usually derived in textbooks by completing the square. This is very unnatural and potentially confusing for students. A more appropriate approach is given here.
We describe Vieta Jumping, a technique that was used to solve the notorious 1988 International Mathematical Olympiad’s Problem 6. We provide explanations, examples and visual representations, as well as other problems that can be solved by this technique.
It is a well-known estimate that, for small values $x \geq 0$ much smaller than 1, the linear function $x$ approximates $\ln(1 + x)$. Alas, this easy approximation does not hold on all of the interval $[0,1]$. A far better almost-linear approximation is presented in this article.
I consider primeless and single-prime intervals of any given length, and show easy ways in which to construct them.
Q1732 Suppose that the numbers $a_1,a_2,\ldots,a_n$ are equal to $1,2,\ldots,n$, but not necessarily in that order. Find the maximum possible value of
$\displaystyle S = \sum_{k=1}^n (k-a_k)^2$ | 558 | 2,342 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2024-26 | latest | en | 0.906947 |
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#### Resources tagged with Work, energy and power similar to Go Spaceship Go:
Filter by: Content type:
Stage:
Challenge level:
##### Other tags that relate to Go Spaceship Go
Mathematical modelling. chemistry. Energy. Work. Logic. physics. Investigations. engineering. biology. Forces.
### There are 9 results
Broad Topics > Stage 5 Mechanics mapping document > Work, energy and power
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Derive an equation which describes satellite dynamics.
### High Jumping
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### Light Weights
##### Stage: 5 Challenge Level:
See how the weight of weights varies across the globe.
### Slingshot
##### Stage: 5 Challenge Level:
Play with the energetic forces in this sling shot | 376 | 1,637 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2017-34 | latest | en | 0.86449 |
https://sciencing.com/convert-photons-joules-8508488.html | 1,660,645,067,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572286.44/warc/CC-MAIN-20220816090541-20220816120541-00146.warc.gz | 461,903,209 | 87,475 | How to Convert Photons to Joules
••• Timbicus/E+/GettyImages
Print
A photon is a singular particle of light. Photons are miniscule and move incredibly quickly. A joule is a measurement of energy. Each tiny photon contains a certain amount of energy that can be calculated using three factors. These factors are the electromagnetic wavelength, Planck's constant and the speed of the photon.
Multiply 6.626 * 10^-34 by the speed of electromagnetic-field propogation in the medium your problem states. In most cases, this number will be the speed of light in a vacuum, which is 2.998 * 10^8 meters per second.
For example -- operating in a vacuum:
(6.626 * 10^-34)*(2.998 * 10^8 meters per second) =1.9864748 × 10^-25 m/s
Divide the result by the electromagnetic wavelength, in meters, of the photon.
In the example, where the wavelength is 10 m:
(1.9864748 × 10^-25 m/s)/10 meters = 1.9864748 × 10^-26 Hz
Note: Hz is the same as 1/seconds
Multiply the result by the number of photons that you want to measure. The result will be the joules of energy contained in the photons.
In the example, we multiply by 12 photons:
(1.9864748 × 10^-26 Hz)*12 photons = 2.38376976 × 10^-25 joules
Dont Go!
We Have More Great Sciencing Articles! | 339 | 1,242 | {"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-2022-33 | longest | en | 0.864574 |
https://bkgm.com/rgb/rgb.cgi?view+772 | 1,708,730,654,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474470.37/warc/CC-MAIN-20240223221041-20240224011041-00566.warc.gz | 136,630,021 | 4,086 | Cube Handling in Races
Thorp count
From: Simon Woodhead Address: simon@Aus.Sun.COM Date: 11 September 1991 Subject: Re: Edward Thorpe on running games? Forum: rec.games.backgammon Google: 1991Sep11.003958.7062@Aus.Sun.COM
```Edward Thorpe worked out a formula for calculating race doubles
(an elaborated pip-count) which is claimed to be 99% accurate. There
are certain positions (described in Bill Robertie's "Advanced Backgammon")
where it falls down a little. On the whole, it is extremely accurate
for medium and long races, and has the advantage of being easily (:-)
calculated over the table.
This is the formula:
. Compute the leader's doubling number as follows:
2) plus two for each of the leader's remaining checkers,
3) plus one for each checker on the ace point,
4) minus one for each point covered in the home board,
5) plus 10% of the total so far, if that total is greater than 30.
. Compute the trailer's doubling number by following steps 1-4, but
omitting step 5. Call this result T, the trailer's adjusted pip count.
. If T > L-2, the leader should double.
. If T > L-1, the leader should redouble.
. If T > L+2, the trailer should pass.
Looks harder than it is. Anyone serious should be able to do a
normal pip count fairly quickly. Adding/subtracting the other bits
just takes a bit of practise.
If you want something a little easier, try this:
. Compute both player's pip count. Find the difference and compare
that difference to the leader's count.
. With a difference of 8% or more, the leader has an initial double.
. With a difference of 9% or more, the leader has a redouble.
. With a difference of 12% or less, the trailer has a take.
Note that these formulae apply to money-game races, and do not take
any account of match-play scores. Use at your own risk!
Cheers,
Simon simon@Aus.Sun.COM
```
### Cube Handling in Races
Bower's modified Thorp count (Chuck Bower, July 1997)
Calculating winning chances (Raccoon, Jan 2007)
Calculating winning chances (OpenWheel+, Nov 2005)
Doubling formulas (Michael J. Zehr, Jan 1995)
Doubling in a long race (Brian Sheppard, Feb 1998)
EPC example: stack and straggler (neilkaz+, Jan 2009)
EPC examples: stack and straggler (Carlo Melzi+, Dec 2008)
Effective pipcount (Douglas Zare, Sept 2003)
Effective pipcount and type of position (Douglas Zare, Jan 2004)
Kleinman count (Øystein Johansen+, Feb 2001)
Kleinman count (André Nicoulin, Sept 1998)
Kleinman count (Chuck Bower, Mar 1998)
Lamford's race forumla (Michael Schell, Aug 2001)
N-roll vs n-roll bearoff (David Rubin+, July 2008)
N-roll vs n-roll bearoff (Gregg Cattanach, Nov 2002)
N-roll vs n-roll bearoff (Chuck Bower+, Dec 1997)
Near end of game (Daniel Murphy, Mar 1997)
Near end of game (David Montgomery, Feb 1997)
Near end of game (Ron Karr, Feb 1997)
One checker model (Kit Woolsey+, Feb 1998)
Pip count percentage (Jeff Mogath+, Feb 2001)
Pip-count formulas (Tom Keith+, June 2004)
Thorp count (Chuck Bower, Jan 1997)
Thorp count (Simon Woodhead, Sept 1991)
Thorp count questions (Chuck Bower, Sept 1999)
Value of a pip (Tom Keith, June 2004)
Ward's racing formula (Marty Storer, Jan 1992)
What's your favorite formula? (Timothy Chow+, Aug 2012) | 930 | 3,221 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2024-10 | latest | en | 0.876483 |
http://www.java2s.com/Code/Java/Data-Type/GetDoubleNumber.htm | 1,537,647,827,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267158691.56/warc/CC-MAIN-20180922201637-20180922222037-00309.warc.gz | 347,837,460 | 4,656 | # Get Double Number : double « Data Type « Java
Get Double Number
```
import java.awt.*;
import java.awt.event.*;
public class GetNumber {
private static Number NAN = new Double(Double.NaN);
/* Process one String, returning it as a Number subclass
* Does not require the GUI.
*/
public static Number process(String s) {
if (s.matches(".*[.dDeEfF]")) {
try {
double dValue = Double.parseDouble(s);
System.out.println("It's a double: " + dValue);
return new Double(dValue);
} catch (NumberFormatException e) {
System.out.println("Invalid a double: " + s);
return NAN;
}
} else // did not contain . d e or f, so try as int.
try {
int iValue = Integer.parseInt(s);
System.out.println("It's an int: " + iValue);
return new Integer(iValue);
} catch (NumberFormatException e2) {
System.out.println("Not a number:" + s);
return NAN;
}
}
public static void main(String[] ap) {
process("0");
process("1111111111");
}
}
```
### Related examples in the same category
1 Double class creates primitives that wrap themselves around data items of the double data type 2 Min and Max values of data type double 3 Java double: double is 64 bit double precision type and used when fractional precision calculation is required. 4 Use toString method of Double class to convert Double into String. 5 Use Double constructor to convert double primitive type to a Double object. 6 Convert java Double to numeric primitive data types 7 Convert Java String to Double 8 Java Double compare example 9 Create a Double object using one of the below given constructors 10 Java Double isInfinite 11 Java Double isNaN method 12 Compare Two Java double Arrays 13 Convert from double to String 14 Convert from String to double 15 Converting a String to a double type Number 16 Ternary operator on double value 17 Binary search 18 Linear Searching double Arrays 19 The Circle Area Calculator 20 A Program That Uses the Mathematical Methods of the Math Class 21 A Program That Uses the Rounding Methods of the Math Class 22 Compute the value of 2/3 of 5 23 Show INFINITY and NaN 24 Double Array 25 Obtaining the integer and fractional parts 26 Get the next machine representable number after a number, moving in the direction of another number. 27 Returns the sign for double precision x 28 Gets the maximum of three double values. 29 Gets the minimum of three double values. 30 Compare two double values with Double.doubleToLongBits 31 Converts altitude in meters to pressure in msw | 562 | 2,455 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2018-39 | latest | en | 0.515145 |
https://www.teacherspayteachers.com/Browse/Search:adding%20and%20subtracting%20tens | 1,604,073,168,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107911027.72/warc/CC-MAIN-20201030153002-20201030183002-00321.warc.gz | 899,600,480 | 65,873 | # 55,279 results for adding and subtracting tens55,279 results for adding and subtracting tens
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These task cards are perfect for math stations because they are easy for students to use independently. Students will practice number recognition, basic facts, counting on, addition and subtraction, and subitizing (instant judgment of a number). Using manipulatives and ten frames will reinforce stud
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digital
These Print and Go Math Sheets will help you with adding and subtracting tens from multiples of 10, and adding ten to a two-digit number, and using dimes and pennies to represent tens and ones. This can be used as Independent Work Packets.BUNDLED FOR CONVENIENCE❤️YEAR BUNDLE NO PREP PRINT AND GO SAV
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These unicorn math centers and worksheets help kids to learn how to add and subtract using ten frames. Included in this pack:Math CentersAddition work mats and sum cards up to 10Addition work mats and sum cards up to 20- Includes horizontal and vertical ten frame orientationsSubtraction work mats an
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This product is for students just learning how to add and subtract 2-digit numbers using base ten blocks. Worksheets are separated into addition problems without and with regrouping, subtraction problems without and with regrouping, and 2 exit tickets with a mixture of addition and subtraction prob
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FirstieMath® Unit Ten: Addition and Subtraction to 40This unit is part of my larger FirstieMath® bundle! Click HERE to view the bundle! What is FirstieMath? FirstieMath Curriculum is a First Grade Math curriculum set of units and are teacher created, kid-tested, and most importantly kid-approved! F
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digital
Teach students to solve addition and subtraction involving multiples of ten with this set of practice sheets! The practice sheets (11 pages plus 11 answer keys) are based on place value concepts and build in difficulty to give you a quick and easy framework for teaching. Concepts include adding a
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CCSS:
digital
This is a set of digital task cards that allows students to work with the Friends of Ten in an engaging, fun way. Students will use missing addends and missing subtrahend algorithms to solve the problems. Students can move the jelly beans to easily count to get "What's Missing" making this difficult
Subjects:
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digital
This 2nd Grade Google Classroom resource includes 12 Google Slides that will enable students to practice adding and subtracting. These digital Task Cards make it easy to implement technology into the classroom. Perfect for 1:1 classrooms. Each Google Slide is aligned to Common Core State Standard 2.
Activities, printables, games, and worksheets that help your students practice making 10, decomposing 10, and adding and subtracting within 10. You will receive 91 pages that cover numbers within 10 for easy differentiation in your classroom. Place value concepts such as using ten frames, multiple w
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digital
Do your students need to use pictures or counters to add and subtract? Do they need practice with ten frames? Then this activity pack is just what you need! With these worksheets your students can use the ten frames on each page to help them add and subtract. Each math problem includes a ten frame t
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It's fun to practice place value with this set of math games for first grade! These twelve games for small groups or math centers are designed as engaging ways to review and reinforce your teaching of adding and subtracting ten and multiples of ten. Here are the games in this set:* "Piggy, Piggy!"
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digital
This product contains several worksheets that focus on teaching different strategies to add and subtract within 20. For easy differentiation, use the addition and subtraction within 10 worksheets with your lowest students, and the addition and subtraction within 20 worksheets with your higher studen
digital
This is a perfect mini unit to use when teaching students how to mentally add tens and hundreds! It includes games, activities, checks for understanding, and a unit test. The packet contains: -Skip Counting Sheet (2.NBT.A.2) -Action at the Arcade: real world application worksheet -More or Less Ac
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digital
Adding and Subtracting Tens: This product contains a collection of worksheets with adding and subtracting tens to 100.Included:* 11 worksheets with base ten blocks for visual support* 10 worksheets without blocksHappy teaching!Dana's Wonderland
This is a great way to practice adding and subtracting using base ten blocks. Included in this 25 page set is 12 triple digit addition problems (with sums up to 999) and 12 triple digit subtraction problems (with the largest number being 999). This set does include REGROUPING. It also includes a
This is a set of twelve games for CCCS 1.NBT.5, mentally add and subtract ten, and 1.NBT.6, mentally add and subtract multiples of ten.* Walk the Plank!, a card game with 69 cards for adding and subtracting ten, plus Yo, ho! and Walk the Plank cards.* Tell Me, Pirate, Is It True? Two game boards,
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digital
This is a differentiated set of 24 QR code task cards aligned to the first grade common core standards: CCSS.Math.Content.1.NBT.C.4 Add within 100, including adding a two-digit number and a one-digit number, and adding a two-digit number and a multiple of 10, using concrete models or drawings and st
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2.NBT.5: Fluently add and subtract within 100 based on place value, properties of operations, and/or the relationship between addition and subtraction. This game will help students practice adding and subtracting ten and multiples of ten. This knowledge can then be applied to solving word problems i
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digital
Easy no-prep printables for kindergarten students to learn how to add and subtract to ten. This would be great in a center, brought up on an interactive white board, or as a whole class activity. Additionally, you can use them on days with a sub. There is a variety of activities on the worksheets.Th
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Here’s a set of 24 task cards that will strengthen your first graders’ early place value skills. Choosing from nine numbers, your students will use logic to solve clues, determining Dino’s Secret Number by eliminating eight of the options. With each card they do, your students will complete eight e
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digital
This product contains two sets of Addition and Subtraction Task Cards using base ten blocks. Each set covers place value to 999. There are twenty cards in each set for a total of 40 cards. The first set includes mixed addition and subtraction problems WITHOUT regrouping. The second set includes
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digital
This product includes: •30 matching cards (15 pictures + 15 equations)•1 recording sheet •This resource focuses on 2-digit addition and subtraction without regrouping. It is good math center practice for advanced 1st grade (1.NBT.C.4) and on-level 2nd grade (2.NBT.B.5)•Print cards on cardstock and l
Subjects:
Types:
CCSS:
digital
This product contains two sets of Addition & Subtraction Task Cards using base ten blocks. Each set covers place value to 99. There are twenty cards in each set for a total of 40 cards. The first set includes mixed addition and subtraction problems WITHOUT regrouping. The second set includes mix
Subjects:
Types:
Addition and Subtraction to Ten - Bundle This is both my adding to ten worksheets and my subtraction to ten worksheets! Here are the individual products: Click here Addition to ten worksheets. Click here for Subtraction to ten worksheets. This product is 25% cheaper than it would be if you
Subjects: | 1,697 | 7,794 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2020-45 | latest | en | 0.883965 |
https://lambdageeks.com/third-law-of-thermodynamics/ | 1,708,614,357,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473819.62/warc/CC-MAIN-20240222125841-20240222155841-00767.warc.gz | 384,349,855 | 31,952 | # Third Law of Thermodynamics: Unraveling Its Impact on Our Universe
The third law of thermodynamics is a fundamental principle in the field of thermodynamics that deals with the behavior of systems as they approach absolute zero temperature. It states that as the temperature of a system approaches absolute zero, the entropy of the system also approaches zero. In other words, at absolute zero, the system reaches its minimum possible energy state and its entropy becomes constant. This law has important implications for understanding the behavior of matter at extremely low temperatures and is used in various scientific and technological applications.
## Key Takeaways
Third Law of Thermodynamics
1Deals with systems at absolute zero temperature
2Entropy approaches zero as temperature approaches absolute zero
3Describes the behavior of matter at extremely low temperatures
## Understanding the Basics of Thermodynamics
Thermodynamics is a branch of physics that deals with the study of energy and its transformations in physical systems. It provides a framework for understanding how heat energy can be converted into work and vice versa. In this article, we will explore the fundamental laws of thermodynamics and the concept of cryogenics.
### The Fundamental Laws of Thermodynamics
The field of thermodynamics is governed by a set of fundamental laws that describe the behavior of energy in various systems. These laws provide a foundation for understanding the principles that govern heat transfer, energy conversion, and the behavior of physical systems.
1. The First Law of Thermodynamics: This law, also known as the law of energy conservation, states that energy cannot be created or destroyed in an isolated system. It can only be transferred or converted from one form to another. This law is based on the principle of conservation of energy and is often expressed as the equation: ΔU = Q – W, where ΔU represents the change in internal energy of the system, Q represents the heat transfer into the system, and W represents the work done by the system.
2. The Second Law of Thermodynamics: The second law of thermodynamics introduces the concept of entropy, which is a measure of the disorder or randomness in a system. It states that the entropy of an isolated system always tends to increase over time. This law also introduces the concept of thermodynamic equilibrium, which is a state where the system has reached maximum entropy and no further changes occur. The second law of thermodynamics has several formulations, including the Clausius statement and the Kelvin-Planck statement.
3. The Third Law of Thermodynamics: The third law of thermodynamics states that it is impossible to reach absolute zero temperature (0 Kelvin or -273.15 degrees Celsius) through any finite number of processes. Absolute zero is the lowest possible temperature and represents the absence of any thermal energy. This law has implications for the behavior of systems at extremely low temperatures, such as in the field of cryogenics.
### The Concept of Cryogenics
Cryogenics is a branch of physics that deals with the production and behavior of materials at extremely low temperatures. It involves the study of the properties of matter at temperatures close to absolute zero. Cryogenic temperatures are typically below -150 degrees Celsius (-238 degrees Fahrenheit) and are achieved using specialized equipment and techniques.
One of the key concepts in cryogenics is the idea of zero-point energy. According to quantum mechanics and statistical mechanics, even at absolute zero temperature, particles still possess a minimum amount of energy known as zero-point energy. This energy arises from the uncertainty principle and the existence of quantum states.
Cryogenics has numerous applications in various fields, including medicine, engineering, and scientific research. It is used in the production and storage of superconductors, which exhibit zero electrical resistance at low temperatures. Cryogenic techniques are also employed in the preservation of biological materials, such as sperm and embryos, for fertility treatments and research purposes.
In conclusion, thermodynamics is a fascinating field that provides insights into the behavior of energy in physical systems. The fundamental laws of thermodynamics, including the concepts of entropy and thermodynamic equilibrium, form the basis of this discipline. The concept of cryogenics, with its focus on extremely low temperatures and the behavior of matter at those temperatures, adds another dimension to our understanding of thermodynamics.
## The Third Law of Thermodynamics: An Overview
### Definition and Simple Explanation
The Third Law of Thermodynamics is a fundamental principle in the field of thermodynamics. It states that as the temperature of a system approaches absolute zero, the entropy of the system also approaches zero. In simpler terms, it means that as a system gets colder, its disorder or randomness decreases.
To understand this law better, let’s break it down into simpler terms. Entropy is a measure of the disorder or randomness in a system. Absolute zero is the lowest possible temperature, at which a system has no heat energy. When a system reaches absolute zero, it is said to be in a state of thermodynamic equilibrium, where there is no heat transfer or any change in the system’s thermodynamic properties.
The Third Law of Thermodynamics is closely related to the concept of quantum states and statistical mechanics. At absolute zero, a system’s energy levels are at their lowest possible values, and the system is in its ground state. This law helps us understand the behavior of physical systems at extremely low temperatures and provides insights into the properties of matter.
### Mathematical Explanation of the Third Law
Mathematically, the Third Law of Thermodynamics can be expressed using the equation:
$\lim_{T \to 0} S = 0$
Where:
$S$ represents the entropy of the system.
$T$ is the temperature of the system.
This equation shows that as the temperature approaches absolute zero, the entropy of the system becomes zero. It implies that at absolute zero, the system is in a state of perfect order and has no molecular disorder.
### The Principle and Fundamental Law of the Third Law of Thermodynamics
The Third Law of Thermodynamics is a fundamental principle that helps us understand the behavior of physical systems at low temperatures. It establishes a connection between entropy, temperature, and the equilibrium state of a system.
The principle behind the Third Law is that it is impossible to reach absolute zero temperature through any finite number of processes. As a system approaches absolute zero, it becomes increasingly difficult to remove heat energy from it. This is due to the concept of zero-point energy, which states that even at absolute zero, there is still some residual energy present in a system.
The Third Law also has implications for the calculation of thermodynamic properties. It allows us to determine the absolute entropy of a substance by measuring its entropy at a known temperature and then extrapolating it to absolute zero.
In summary, the Third Law of Thermodynamics provides valuable insights into the behavior of physical systems at extremely low temperatures. It helps us understand the relationship between entropy, temperature, and the equilibrium state of a system. By studying this law, scientists and engineers can make accurate predictions and calculations in various fields, including chemistry, physics, and materials science.
## The Significance of the Third Law of Thermodynamics
### Why is the Third Law of Thermodynamics Important?
The Third Law of Thermodynamics holds great significance in the field of thermodynamics. It provides valuable insights into the behavior of physical systems at extremely low temperatures, specifically at absolute zero. This law helps us understand the fundamental properties of matter and the behavior of energy within a system.
One of the key aspects of the Third Law of Thermodynamics is its connection to entropy. Entropy is a measure of the disorder or randomness within a system. At absolute zero, the entropy of a perfectly ordered crystal approaches zero. This law states that it is impossible to reach absolute zero through any finite number of processes. It sets a limit to how close we can get to absolute zero, which is crucial in various scientific and technological applications.
### The Role of Surroundings in the Third Law of Thermodynamics
In the context of the Third Law of Thermodynamics, the surroundings play a crucial role. The law states that when a system is at absolute zero, its entropy is also at its minimum possible value. However, this minimum value is not zero due to the influence of the surroundings. The entropy of the surroundings affects the overall entropy of the system.
The Third Law of Thermodynamics helps us understand the behavior of physical systems in equilibrium states. It provides a framework to analyze the relationship between temperature, entropy, and energy levels. By considering the role of the surroundings, we can better comprehend the behavior of systems at extremely low temperatures and their deviation from perfect order.
### The Third Law of Thermodynamics in the Context of Chemistry
In the field of chemistry, the Third Law of Thermodynamics has significant implications. It helps us understand the behavior of molecules and their arrangements in different states of matter. The law allows us to predict and analyze the thermodynamic properties of substances, such as their heat capacities and entropies.
The Third Law of Thermodynamics is particularly relevant in the study of phase transitions. It helps us understand the changes in molecular disorder and entropy that occur during the transition from one phase to another, such as from a solid to a liquid or a gas. By considering the principles of statistical mechanics and quantum states, we can apply the Third Law to explain the behavior of chemical systems at various temperatures.
In summary, the Third Law of Thermodynamics is of great significance in understanding the behavior of physical and chemical systems. It provides insights into the relationship between temperature, entropy, and energy levels, particularly at extremely low temperatures. By considering the role of the surroundings, we can analyze the behavior of systems in equilibrium states and predict their thermodynamic properties.
## Practical Applications of the Third Law of Thermodynamics
### The Third Law of Thermodynamics in Biological Systems
The Third Law of Thermodynamics, which states that the entropy of a system approaches zero as the temperature approaches absolute zero, has several practical applications in biological systems. One such application is in the study of protein folding. Proteins are essential molecules in living organisms, and their proper folding is crucial for their function. The Third Law helps us understand the thermodynamic stability of protein structures and how they fold into their native conformations. By studying the entropy at absolute zero, scientists can gain insights into the energy landscape and stability of proteins, which is vital for understanding diseases related to protein misfolding, such as Alzheimer’s and Parkinson’s.
Another application of the Third Law in biology is in the study of cryobiology. Cryobiology focuses on the preservation of biological materials at extremely low temperatures. The Third Law helps researchers understand the behavior of biological systems at near-zero temperatures, allowing for the development of techniques like cryopreservation. By minimizing molecular disorder and entropy, cryopreservation techniques can preserve cells, tissues, and even whole organs for extended periods. This has significant implications in fields such as organ transplantation, regenerative medicine, and the conservation of endangered species.
### Real Life Examples of the Third Law of Thermodynamics
The Third Law of Thermodynamics finds practical applications in various real-life scenarios. One example is in the field of materials science, particularly in the development of new materials with specific properties. By understanding the behavior of materials at low temperatures, scientists can design materials with enhanced properties, such as superconductors. Superconductors exhibit zero electrical resistance at very low temperatures, and the Third Law helps us understand the underlying quantum states and energy levels that enable this phenomenon. This knowledge has led to advancements in technologies like magnetic levitation trains and high-speed computing.
Another real-life example is the application of the Third Law in the study of phase transitions. Phase transitions occur when a substance changes from one state to another, such as from a solid to a liquid or a gas. The Third Law helps us understand the behavior of substances near absolute zero, where interesting phenomena like Bose-Einstein condensation occur. Bose-Einstein condensates are a state of matter where particles lose their individual identities and behave as a single quantum entity. This has implications in fields such as quantum computing and the study of fundamental particles.
### Sample Problems with Solutions Related to the Third Law of Thermodynamics
To further illustrate the practical applications of the Third Law of Thermodynamics, let’s consider some sample problems:
1. Problem: Calculate the entropy change when a gas at 300 K and 1 atm pressure is cooled to 10 K and 1 atm pressure.
Solution: Using the Third Law of Thermodynamics, we know that the entropy at absolute zero is zero. Therefore, the entropy change can be calculated by integrating the heat transfer divided by the temperature over the temperature range. By plugging in the values and using the Boltzmann constant, we can determine the entropy change.
2. Problem: A liquid has an entropy anomaly, where its entropy decreases with increasing temperature. Explain how this violates the Third Law of Thermodynamics.
Solution: The Third Law of Thermodynamics states that the entropy of a system approaches zero as the temperature approaches absolute zero. If a liquid exhibits an entropy anomaly, where its entropy decreases with increasing temperature, it implies that the system does not reach a true equilibrium state at absolute zero. This violates the Third Law, as the entropy should approach zero in the absence of any molecular disorder.
3. Problem: Calculate the Gibbs free energy change for a chemical reaction at absolute zero.
Solution: At absolute zero, the temperature is zero Kelvin, and the entropy of the system is zero according to the Third Law of Thermodynamics. Therefore, the Gibbs free energy change can be calculated using the equation ΔG = ΔH – TΔS, where ΔH is the change in enthalpy and ΔS is the change in entropy. Since the entropy is zero, the Gibbs free energy change simplifies to ΔG = ΔH.
These sample problems demonstrate how the Third Law of Thermodynamics can be applied to solve practical problems related to thermodynamic systems, temperature scales, and thermodynamic properties.
In conclusion, the Third Law of Thermodynamics has practical applications in various fields, including biology, materials science, and phase transitions. By understanding the behavior of physical systems at low temperatures and near absolute zero, scientists and engineers can develop new technologies, study complex phenomena, and solve real-world problems.
## The Third Law of Thermodynamics in Academic Syllabus
### Is the Third Law of Thermodynamics in JEE Syllabus?
In the academic syllabus, the Third Law of Thermodynamics holds a significant place. It is a fundamental concept that helps us understand the behavior of physical systems at extremely low temperatures. Now, you might be wondering whether the Third Law of Thermodynamics is included in the JEE syllabus. Well, the answer is yes! The JEE syllabus does cover the Third Law of Thermodynamics, ensuring that students have a comprehensive understanding of this important principle.
The Third Law of Thermodynamics deals with the concept of entropy, which is a measure of molecular disorder in a system. It states that as the temperature of a system approaches absolute zero, the entropy of the system also approaches zero. Absolute zero is the lowest possible temperature, where all molecular motion ceases, and the system is in a state of perfect order or equilibrium. This law helps us understand the behavior of physical systems at extremely low temperatures and provides insights into the concept of thermodynamic equilibrium.
To understand the Third Law of Thermodynamics better, let’s take a look at an example. Consider a gas at room temperature. As we decrease the temperature, the gas molecules slow down, and their kinetic energy decreases. Eventually, as we approach absolute zero, the gas molecules come to a complete stop, and the system reaches a state of thermodynamic equilibrium. At this point, the entropy of the system is at its minimum value.
In other academic courses, such as physics and chemistry, the Third Law of Thermodynamics is also included in the syllabus. It is an essential concept in statistical mechanics and quantum mechanics, where it helps us understand the behavior of systems at the quantum level. The Third Law of Thermodynamics is closely related to concepts like zero-point energy, energy levels, and the Boltzmann constant.
In the study of thermodynamics, the Third Law plays a crucial role in determining the thermodynamic properties of materials and systems. It helps us understand the behavior of substances at low temperatures, such as the anomalies observed in the entropy of liquids near their freezing points. These anomalies are due to the arrangement of molecules in a liquid, which becomes more ordered as the temperature decreases.
To summarize, the Third Law of Thermodynamics is an important topic in the academic syllabus. It is included in the JEE syllabus and other academic courses like physics and chemistry. Understanding this law helps us comprehend the behavior of physical systems at extremely low temperatures and provides insights into the concept of thermodynamic equilibrium. So, make sure to grasp the concepts related to the Third Law of Thermodynamics as it forms a fundamental part of thermodynamics and its applications in various fields.
## Conclusion
In conclusion, the third law of thermodynamics is a fundamental principle that states that as the temperature of a system approaches absolute zero, the entropy of the system also approaches zero. This law provides insight into the behavior of matter at extremely low temperatures and helps us understand the limits of achieving absolute zero. By understanding the third law, scientists and engineers can make more accurate predictions and calculations when dealing with systems at very low temperatures. Overall, the third law of thermodynamics plays a crucial role in our understanding of the behavior of matter and energy.
## What is the relationship between the Third Law of Thermodynamics and the key principles explained in the “Laws of Thermodynamics: Key Principles Explained”?
The Third Law of Thermodynamics and the “Laws of Thermodynamics: Key Principles Explained” cover fundamental concepts in thermodynamics. The Third Law states that as a system approaches absolute zero, its entropy approaches a minimum or zero value. This law is closely related to the other laws of thermodynamics, which provide a framework for understanding energy transfer, heat flow, and the behavior of systems. To explore the intersection of these themes, it is important to delve into the key principles explained in the article “Laws of Thermodynamics: Key Principles Explained”. This article offers a comprehensive overview of the laws, including the concepts of energy conservation, entropy, and the impossibility of achieving absolute zero. By understanding these principles, we can better grasp the role of the Third Law in the broader context of thermodynamics.
## Frequently Asked Questions
### 1. What is the Third Law of Thermodynamics?
The Third Law of Thermodynamics states that the entropy of a system approaches a constant value as the temperature approaches absolute zero. This principle implies that it’s impossible for any process to lower the temperature of a system to absolute zero in a finite number of operations.
### 2. How does the Third Law of Thermodynamics work?
The Third Law of Thermodynamics works by asserting that systems in thermodynamic equilibrium (zero-point energy) have minimum energy and are at their most ordered state (minimum entropy). This law provides the foundation for understanding physical systems and energy levels at absolute zero temperature.
### 3. What is the Mathematical Explanation of the Third Law of Thermodynamics?
The Mathematical Explanation of the Third Law of Thermodynamics is often expressed through the concept of entropy (S). As the temperature (T) approaches absolute zero, the entropy of a system approaches a minimum value, typically zero. This can be represented as: S→0 as T→0.
### 4. What are the Practical Applications of the Third Law of Thermodynamics?
The Third Law of Thermodynamics has practical applications in various fields like cryogenics and low-temperature physics. It provides a basis for the calculation of absolute entropies of substances and helps in determining the feasibility of a physical process at absolute zero temperature.
### 5. What is the Role of Surrounding in the Third Law of Thermodynamics?
The surrounding environment plays a crucial role in thermodynamic processes. According to the third law, if the surrounding environment is at absolute zero, the system should also be at absolute zero, indicating that the system and its surroundings are interconnected.
### 6. Why is the Third Law of Thermodynamics Important?
The Third Law of Thermodynamics is important because it provides a basis for the concept of absolute zero, the point at which molecular disorder (entropy) is minimal. It also allows for the calculation of absolute entropies and helps in determining the thermodynamic properties of materials at low temperatures.
### 7. Can you provide an example of the Third Law of Thermodynamics in real life?
One real-life example of the third law is the preservation of food in refrigerators. Lowering the temperature reduces the entropy of bacteria, slowing their metabolic processes and preventing the food from spoiling.
### 8. How is the Third Law of Thermodynamics related to Statistical Mechanics?
The Third Law of Thermodynamics is closely related to Statistical Mechanics as it helps in understanding the macroscopic properties of matter based on the behavior of its microscopic constituents. It provides a statistical interpretation of entropy, linking it to the number of quantum states a system can occupy.
### 9. What is the Fundamental Principle behind the Third Law of Thermodynamics?
The fundamental principle behind the Third Law of Thermodynamics is that as a system approaches absolute zero, all processes cease and the entropy of the system reaches a minimum value, typically zero in a perfect crystal. This law establishes the concept of absolute zero, where all molecular motion ceases.
### 10. Is the Third Law of Thermodynamics Included in the JEE Syllabus?
Yes, the Third Law of Thermodynamics is a part of the JEE syllabus. It is included in the section on thermodynamics, and students are expected to understand its principle, implications, and applications.
Also Read: | 4,398 | 23,812 | {"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": 3, "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-2024-10 | latest | en | 0.92812 |
https://graphics.stanford.edu/wikis/cs348b-06/YiLangMok/Assignment3?action=print | 1,675,293,533,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499953.47/warc/CC-MAIN-20230201211725-20230202001725-00289.warc.gz | 310,346,804 | 5,163 | • YiLangMok/Assignment3
# Assignment 3 Camera Simulation
## Compound Lens Simulator
### Description of implementation approach and comments
Initializing the system only required reading in the lens file. I made a modification to how I stored lenses: instead of storing the position where the lens crosses the axis, I stored the radius and the centroid of the sphere. This way, I could perform ray/sphere intersections easily. I also stored the absolute distance that each lens is at.
There are two steps in implementing GenerateRay: propagating a ray into image space, and calculating the exposure due to the ray.
* Propagating a ray into image space. I start off with translating and scaling the sample so that I get a ray that starts from the film and points towards the first lens. Then, I propagate the ray by using Snell's law at each interface in sequence. I originally attempted to rotate the surface normal so that it would point in the resulting direction. This required calculating the cross product of the source ray and the normal, and then a transfomation around that axis. Unfortunately, the images generated were too fuzzy: the result of numerical inaccuracy. However, it is possible to make out the general shapes. The second method that I attempted took an algorithm off Wikipedia. It suffered from very little numerical inaccuracy, producing images which are comparable to the reference images. At each step of the propagation, I make sure to check if the ray actually passes through the aperture of the lens.
* Exposure of the ray. I made use of the exposure equation in the paper, setting the area to be the area of the aperture of the lens closest to the film. The cosine is reduced to a dot product since it is the dot product of the ray with the normal to the film.
### Final Images Rendered with 512 samples per pixel
My Implementation Reference Telephoto Double Gausss Wide Angle Fisheye
### Final Images Rendered with 4 samples per pixel
My Implementation Reference Telephoto Double Gausss Wide Angle Fisheye
## Experiment with Exposure
Image with aperture full open Image with half radius aperture
### Observation and Explanation
As expected, the image with half the aperture is darker. This is because fewer rays make it through the aperture. As a result of the discrete ray tracing, the image is also much noisier. The image has a larger depth of field however, because the smaller aperture reduces the size of the circles of confusion.
## Autofocus Simulation
### Description of implementation approach and comments
I implemented the SML algorithm in Nayar's paper. In the autofocus zone, I picked a random 7x7 square and rendered it using the sampler and the camera. Then, I reduced it to a 5x5 of focus measures as calculated using the discrete modified Laplacian operator. Then, I found took the sum as the focus measure.
My algorithm for finding the focus depth is very simple. Since the measure is very sensitive to noise (in fact, noise makes it seem like it is focused!), a bisection or golden section search would not be effective: the focus measure function may have many local maxima. Instead, I find the focus measure at varying levels of accuracy: every 10.0mm, every 1.0mm and finally every 0.1mm. The first iteration reduces the search to a 20mm region; the second iteration reduces the search to a 2mm region, and the last iteration gives a focus depth of 0.1mm accuracy.
As stated above, noise can affect the measure greatly. Initially, one problem I had was that with film distances that were small (<10mm), the noise would be so great that the focus measure would think the depth was in the region. I tried three methods to mitigate this:
* The first was just to set the sampling rate very high (1024 rays per pixel), but this was prohibitively slow. It did however, generate accurate results.
* Then I tried to normalize the focus measure by dividing by a (1/sqrt(dist) + 1). This would bias the focus measure towards further film distances. However, this was still not enough to bring down the focus measure due to noise at close distances. Also, I could not justify whether my scaling factor is proportional to the erronous score generated by noise.
* Finally, I decided to have a variable sampling rate. Distances close to the lens would be sampled at a much higher rate than far distances. To be exact, I would perform (8192 / dist) samples. This value was found through trial and error: at 8mm, it will sample 1024 rays; at 32mm, it will sample 256 rays, etc. Even though at certain intervals, very high number of rays would be generated, it is still faster than the first method.
I made graphs of the focus measures produced by running each scene. There are many more points near the actual depth because of the three-iteration approach taken above.
### Final Images Rendered with 512 samples per pixel
Adjusted film distance My Implementation Reference Double Gausss 1 62.5 mm Double Gausss 2 39.2 mm Telephoto 116.5 mm | 1,074 | 5,005 | {"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-2023-06 | latest | en | 0.915261 |
https://solvedlib.com/what-is-the-derivative-of-sin-x2,147625 | 1,695,482,123,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506481.17/warc/CC-MAIN-20230923130827-20230923160827-00796.warc.gz | 612,796,749 | 15,143 | # What is the derivative of sin x^2?
###### Question:
What is the derivative of sin x^2?
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https://rpg.stackexchange.com/questions/156264/how-many-people-need-to-succeed-in-a-group-check-with-three-people/156268 | 1,713,745,108,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818067.32/warc/CC-MAIN-20240421225303-20240422015303-00107.warc.gz | 437,666,824 | 41,451 | # How many people need to succeed in a group check with three people?
In the comments of this answer it came up that the section on "Group Checks" states:
To make a group ability check, everyone in the group makes the ability check. If at least half the group succeeds, the whole group succeeds. Otherwise, the group fails...
It came up what does "half the group" mean when there also exists a rule on "Round Down":
There’s one more general rule you need to know at the outset. Whenever you divide a number in the game, round down if you end up with a fraction, even if the fraction is one-half or greater.
There are two possibilities brought up in those comments:
1. At least 1.5 people must succeed, and since you can't have half-people 2 or more people must succeed on the check.
2. The 1.5 people required rounds down to 1 person, so 1 or more people must succeed on the check.
Which of these interpretations is correct?
Which of these interpretations is correct?
Surely this one:
1. At least 1.5 people must succeed, and since you can't have half-people 2 or more people must succeed on the check.
The rounding rule applies when the rules require that "you divide a number in the game". It is not necessary to divide a number in the game to know whether you have more than half the people in the group succeed.
This is similar to a democratic election. We don't expect if there are 3 voters, that "success" is defined by anything other than exceeding half of the total.
For further reinforcement of this idea, I consider some other aspects:
1. The degenerate case of a "group of one", as a variation of "proof by induction". If rounding down is the correct approach, then zero members of this group need to succeed. Clearly, that's not the intent of the rule. And it doesn't make sense that we would have to special-case "group of one" as a completely different set of rules from "group of more than one" (since we intuitively understand that in the "group of one", we need "at least" one person to succeed).
2. In every other case I am aware of, the "round down" rule has the consequence of making it harder for the actor involved. It is not logical that in this one scenario, we would then apply the "round down" rule in a way that makes it easier for them.
Bottom line: I find many logical points in favor of the group having to succeed in excess of half the number of the group, and no logic in thinking that fewer than half can succeed and still have the group be successful overall.
Some, such as the person who posted the original claim that prompted this question, argue that because the group-success rules involve a comparison against "half the group", that necessarily invokes the "you divide a number in the game". However, I don't find this line of argument compelling. There are lots of times that halving or other divisions come up, without requiring rounding down.
For example, if my party decides that 25 gold should be split four ways by giving three members 6 and a fourth member 7, is that contrary to the rules? We did "divide a number", after all. Such a strict reading of the round-down rule that requires me to literally round-down every single division result I might ever do in the context of playing the game is excessive and IMHO absurd.
Furthermore, the scenarios where the rules envision "dividing a number" involve situations where that number is then mathematically applied to some other number. Effects of ability scores, or damage resistance, for example. Since numeric results in the game require integer values, some rounding must be done, so the rules spell out that the rounding is always downward.
But no rounding at all is required to understand whether you have had fewer or more than half of a group succeed.
And since there appears to be some confusion as well on why I only consider "fewer than" and "more than", let me make it plainly clear: the only scenario that is in question is the one where the group has an odd number of members. And half a person cannot succeed or fail, so clearly the number of members that succeed can only be strictly less than or more than half the number of the group. That my language above does not include the possibility of "equal to" is solely because that possibility does not exist in this scenario.
• One example where rounding down makes something quote--un--quote "easier" is when you have resistance to a damage type. 11 damage becomes 5.5 which becomes the "easier" (less deadly) option of 5.0 damage taken. This isn't a very good example (or a counterpoint) as reducing damage and "easier" don't go well together by any means. Your answer makes a lot of sense, and I like the inductive thinking on the "group of 1" Sep 15, 2019 at 19:07
• @Medix2: I guess that depends on your point of view. To me, the "actor" where "damage resistance" is applied, is the attacker causing the damage. Rounding down in this case makes the harder on the actor, i.e. the entity doing the damage. YMMV. :) (Another way to look at it: as a general rule, "bigger numbers good, smaller numbers bad"...but you have to consider who it is that wants the bigger number; that's generally going to be the "actor" from the perspective of the round-down rule...since rounding down always makes the number smaller, rounding down translates to "more bad"/"less good".) Sep 15, 2019 at 19:09
• Oh that's a really good point actually, thank you Sep 15, 2019 at 19:10
• Welcome to RPG.SE! Take the tour if you haven't already, and check out the help center for more guidance. Sep 15, 2019 at 22:18
• +1, Great first answer. I'd like to add that if you would chose to round to 1, then it would be easier for the group to succeed at all the tests. Everyone just focuses in his specialty, and even with three players, you may cover the totality of each possible test, making the fail probability very low.
– Zoma
Sep 16, 2019 at 9:57
# At least 2 people.
Keep the fraction, fractions are cool!
To make a group ability check, everyone in the group makes the ability check. If at least half the group succeeds, the whole group succeeds.
If you have 3 people, then the amount of people $$\X\$$ that must succeed is given by $$X \geq \frac{3}{2}$$ Because $$\X\$$ is an integer with possible values $$\\{0, 1, 2, 3\}\$$, then the answer is that at least 2 people must succeed the group check.
• I agree with the sentiment, but without touching on why the "round down" rule is not applied, this answer seems incomplete. Sep 16, 2019 at 12:22
You have quoted all the necessary rules for this. You put them together like this:
To make a group ability check, everyone in the group makes the ability check. If at least half the group succeeds, the whole group succeeds. Otherwise, the group fails...
Okay, we need 'half' the group to succeed. What does 'half' mean? In 5e we use normal language unless the book tells us to do something weird instead. In this case, as you quoted, the book does tell us to do something weird instead! It says:
Whenever you divide a number in the game, round down if you end up with a fraction, even if the fraction is one-half or greater.
So, whereas half of 5 would usually be 2.5, in this game it's 2. Similarly, when one applies these rules to the number '3' one can see that, because '3' is an odd number, halving it makes you end up with a fraction. That means you round down and end up with the next integer number down, in this case '1'. Similarly, if you halved 1, you'd end up with 0.
Note, however, that in doing this we are making up what 'round down' means. In common parlance, you usually can't 'round down' without reference to something-- you 'round down to the nearest whole number' or 'round down to the nearest ten' or something like that. Used on its own, the phrase is somewhat unclear-- it might mean 'to the nearest whole number' or 'to the nearest integer' or 'to the nearest non-fraction', and not all of these are meanings that result in a reasonable game (though they all, except 'non-fraction' because that is an incoherent idea in its usual expression, mean 3/2 rounds to 1). Because rounding down to the nearest integer is the least completely terrible as a rule, when using this rounding rule it should be assumed you do that.
• @Medix2 Whole numbers don't include the negatives. These three are included because they are what is ambiguously meant by 'round down' without reference as used by 'normal people', in my estimation. 5e says we are supposed to 'use natural language' but, to the extent this language is used it is ambiguous. That's why non-fraction is incoherent, that sort of speaker very much does not mean 'irrationals'! Sep 15, 2019 at 17:50
• @illustro whether the natural numbers include zero is not an entirely agreed upon thing. The whole numbers are not a thing used in higher level mathematics. You would simply use Z+ (the positive integers) or N\{0} (natural numbers without zero). Or simply N if your definition of N didn't include zero. "Whole numbers" is horribly unspecific either referring to integers, non-negative integers, or positive integers which are the three sets being discussed Sep 15, 2019 at 18:14
• Doesn't this imply that one person attempting a check will automatically succeed? Sep 15, 2019 at 19:31
• @MarkWells no. The group check rules are an exception to the regular ability check rules. One person does not a group make. A group requires at least two creatures to be a group. Sep 15, 2019 at 21:02
• This is a particularly interesting interpretation where you can control the group size. If you have (say) six characters, hve three of them climb first, then the other three. Much easier than everyone climbing at once. Sep 16, 2019 at 13:16 | 2,288 | 9,730 | {"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": 4, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2024-18 | latest | en | 0.963844 |
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Element A Coordinate Geometry Lecture 1 Cartesian Coordinates 1 (Introductions, distance formulation and its software, locus of a point) Lecture 2 Cartesian Coordinates 2 (segment components, area of triangle, region of quadrilateral) Lecture 2 Cartesian Coordinates 2 (Slope of a line, special factors in triangle (centroid, circumcentre centroid, orthocenter, incentre and excentre ) element B straight Line Lecture 1 straight traces 1 (a few vital results connected with one directly line, point-slope shape, symmetric shape or distance form, two factors shape, intercept shape equation of the immediately lines) Lecture 2 immediately lines 2 (everyday form equation of the straight line, the overall form equation of the immediately line, reduction of the overall form into one-of-a-kind instances, function of factors with recognize to the directly line ax + by + c and the perpendicular distance of factor from the line ax + by + c = zero) Lecture three directly traces 3 (Foot of perpendicular, reflection point or picture, a few crucial results related with instantly lines, angle between two immediately traces) Lecture four immediately traces four (Distance among parallel strains; function of starting place (0, zero) with admire to angle between two traces, angular bisectors of two given strains, a few essential factors related with 3 directly lines) Lecture 5 instantly traces five (Miscellaneous questions, revision of heterosexual traces, some harder problems) element C Pair of heterosexual traces Lecture 1 Pair of hetero traces 1 (Homogeneous equations of 2d degree and their numerous forms) Lecture 2 Pair of heterosexual lines 2 (a few important effects connected with homogenous pair of straight line , popular equation of second diploma) component D Circle Lecture 1 Circle 1 D.Three D.14 (Equation of circle in diverse forms) Lecture 2 Circle 2 D.15 D.34 (Relative position of factor with recognize to circle, parametric shape of equation of circle, relative position of line and circle) Lecture 3 Circle three D.35 D.56 (Relative function of circles, pair of tangents and chord of contact draw from an enternal point) element E Conic section Lecture 1 Parabola 1 Lecture 2 Parabola 2 Lecture 3 Ellipse 1 Lecture four Ellipse 2 (function of line with appreciate to an ellipse, diameter, tangents and normals, chord of content material) Lecture five Hyperbola check Your abilties.
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We don’t aid piracy this duplicate grow to be supplied for university youngsters who're financially bad but deserve greater to examine. Thank you. | 845 | 4,039 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2019-30 | latest | en | 0.900129 |
https://physics.stackexchange.com/questions/133023/why-do-we-use-time-dependent-regions-in-this-derivation | 1,575,838,900,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540514893.41/warc/CC-MAIN-20191208202454-20191208230454-00413.warc.gz | 500,671,698 | 31,040 | # Why do we use time-dependent regions in this derivation?
I'm studying fluid mechanics and I have the following doubt: on the book, the author first deduces the differential form of the equation of balance of momentum. First he argues that if the fluid is ideal and is contained in a region $D\subset \mathbb{R}^3$ and if the pressure is $p : D\times \mathbb{R}\to \mathbb{R}$ then the stress force per unit volume is $-\nabla p$ and the body force per unit volume is $\rho \mathbf{b}$ (with $\mathbf{b}$ being the body force per unit mass). In that case the law becomes
$$\rho \dfrac{D\mathbf{u}}{Dt} = -\nabla p + \rho\mathbf{b}$$
Then with some manipulations he derives the integral form of the law. Basically he considers a fixed region $W$ and consider the total momentum contained in $W$ as
$$\int_W \rho \mathbf{u} \ dV$$
In that case the integral form can be obtained differentiating that. So, we gain the following law:
$$\dfrac{d}{dt}\int_W \rho \mathbf{u} \ dV = - \int_{\partial W} (p\mathbf{n}+\rho \mathbf{u}(\mathbf{u}\cdot \mathbf{n}))\ dV + \int_W \rho \mathbf{b} \ dV$$
This law is over a fixed region. So we choose a region on the fluid, compute the momentum inside it and see how it varies as the fluid flows through it.
Now, to assume as little differentiability as possible, the author proceeds to obtain one integral form of the law directly from basic principles. If $\varphi$ is the fluid flow map, $\varphi_t = \varphi (\cdot, t)$, then he states the following:
$$\dfrac{d}{dt} \int_{\varphi_t(W)} \rho \mathbf{u} \ dV = \mathbf{S}_{\partial \varphi_t(W)} + \int_{\varphi_t(W)} \rho \mathbf{b} \ dV$$
Where $\mathbf{S}_{\partial \varphi_t(W)}$ is the total force at time $t$ exterted on the fluid contained in $\varphi_t(W)$ by means of stress on its boundary $\partial \varphi_t(W)$.
This is quite clear: we pick a region, look at the mometum and the rate of change of momentum should be equal to the total force applied to it. Now this region is time-dependent. That is, instead of looking at a fixed region and at the fluid flowing through it, the author is following a chunk of fluid that moves with time.
My question is: why, when deriving the integral form of the balance of momentum law from the differential form, we consider a fixed region and why, when stating it from basic principles, we consider a region varying with time? Is there some connection between this and the differences between the Lagrangian and Eulerian points of view?
• This might not be needed to answer the question, but what are $\mathbf{b}$ and $\mathbf{S}$ here? – user10851 Aug 29 '14 at 1:13
• Thanks for pointing it out @ChrisWhite, I edited the question. $\mathbf{S}$ is the total stress force and $\mathbf{b}$ the body force per unit mass. – user1620696 Aug 29 '14 at 1:23
$$\int_{\partial W} \rho \mathbf{u}(\mathbf{u}\cdot \mathbf{n}))\ dA = - \int_{\partial W} p\mathbf{n}\ dA + \int_W \rho \mathbf{b} \ dV.$$ where $\partial W$ covers the two ends of the pipe. So if we can measure the momentum flux at the two ends, we can find the bulk force acting on the entire volume of fluid in the pipe. | 898 | 3,127 | {"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": 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} | 3.9375 | 4 | CC-MAIN-2019-51 | latest | en | 0.867228 |
https://dreamcivil.com/how-are-earthquakes-measured/ | 1,714,052,424,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712297295329.99/warc/CC-MAIN-20240425130216-20240425160216-00052.warc.gz | 191,396,089 | 61,200 | How are Earthquakes Measured? : Seismometers, Seismograph & Seismogram
In this article, we will discuss how are earthquakes measured.
An earthquake may be defined as the sudden shock of the earth that emits the energy in the earth’s lithosphere causing the formation of seismic waves.
1. How are Earthquakes Measured?
The earthquake can be caused due to volcanic eruption, moving of tectonic plates, formation of the cave in a small area, or due to explosions. It is determined with the help of a seismograph by seismometers and is measured on the Richter Magnitude Scale.
2. Seismometers
A seismometer is the inner component of the seismograph, which may be a pendulum or a mass attached to a spring.
It is usually utilized synonymously with “seismograph”. Seismographs are devices utilized to measure the motion of the ground during the time of an earthquake.
3. Seismograph
The seismograph can measure the strength and duration of the earthquake’s waves. A seismograph is a primary earthquake recording device. The seismograph makes a digital graphic measuring the ground motion by the seismic waves. The digital measuring device is called a seismogram.
When seismic data is taken from a minimum of three different places, it can be utilized to calculate the epicenter by where it crosses.
4. Ways of Measuring Earthquakes
Here are the 4 ways the earthquakes are measured:
a. Wave Amplitude, Fault Size, Amount of Slip
b. The Richter Scale
c. The Moment Magnitude Scale
d. The Mercalli Scale
a. Wave Amplitude, Fault Size, Amount of Slip
There are several ways to determine the magnitude of an earthquake. Most scales depend upon the amplitude of seismic waves measured on seismometers. These scales account for the interval between the earthquake and the recording seismometer so that the determined magnitude should be about the similar no matter where it is recorded.
Another scale depends on the physical size of the earthquake fault and the amount of slip that takes place. Then there are also compute of earthquake shaking intensity. The intensity of one earthquake ranges majorly from place to place.
b. The Richter Scale
The majorly utilized method, the Richter scale, was introduced by Charles F. Richter in 1934. It utilizes a formula that depends upon the amplitude of the largest wave measured on a specific type of seismometer and the distance between the earthquake and the seismometer.
That scale was fixed to California earthquakes and crust; other scales, depending upon wave amplitudes and total earthquake time span, were introduced for utilization in other conditions and they were prepared to be uniform with Richter’s scale.
c. The Moment Magnitude Scale
The moment magnitude scale uses the fault’s geometry (the angle and other elements of the plane that distinguish the fault that ruptures at the time of an earthquake) and the earthquake’s seismic moment (the movement of the fault across its whole area product by the force utilized to displace the fault).
d. The Mercalli Scale
The Mercalli scale is another method to determine the strength of an earthquake is it utilizing the observations of the people who felt the earthquake, and the quantity of damage that took place, to calculate its intensity. The Mercalli scale was introduced to do just that. The scale was introduced by Giuseppe Mercalli in 1902 and was changed by Harry Wood and Frank Neumann in 1931 to make what is now known as the Modified Mercalli Intensity Scale. To support it by differentiating it from magnitude scales, the MMI scale utilizes Roman numerals.
The Mercalli scale does not utilize scientific devices to calculate seismic waves, it has been very beneficial for knowing the damage taken due to large earthquakes. It has also been utilized majorly to explore earthquakes that took place before there were seismometers.
Some elements that affect the quantity of damage that occurs are
a. Size (magnitude) of the earthquake
b. Distance from the epicenter
c. Depth of the earthquake,
d. Building (or other structure) design,
e. Type of surface material the buildings lie on.
IntensityShakingDamage
INot FeltNot detected except by a very few under especially favorable conditions.
IIWeakDetected only by fewer persons at rest, especially on upper floors of buildings. Swing of suspended objects.
IIIWeakDetected fully by persons indoors, especially on upper floors of buildings. Many people do not take it as an earthquake. Standing motor cars may rock a bit. Vibration is the same as the passing of a truck. Duration is known.
IVLightDetected indoors by a huge crowd, outdoors by few during the day. Dishes, windows, and doors are disturbed; walls give a cracking sound. Feels like a heavy truck striking building. Standing motor cars rocked slightly.
VModerateDetected by almost everyone; many come out. Some brittle materials like windows and glass were broken. Unbalanced objects overturned. Pendulum clocks may fall.
VIStrongDetected by all, many were afraid. Some weighted furniture was displaced; a few cases of fallen plaster. Less Damage.
VIIVery StrongLess Damage in buildings of good design and construction; a bit normal in well-built structures; notable damage in badly constructed or poorly planned structures; some chimneys were collapsed.
VIIISevereLess damage in specially planned structures; Notable damage in a normal substantial house with the collapse in few parts. Damage is high in badly constructed structures. Fall of chimneys, factory stacks, columns, monuments, walls. Heavy furniture overturned.
IXViolentDamage considerable in specially designed structures; well-designed frame structures thrown out of plumb. Damage is great in substantial buildings, with partial collapse. Buildings shifted off foundations.
XExtremeFew well-built wooden elements were damaged; most masonry and frame structures were damaged with foundations. Rail bent.
Read More: Emergency Supplies Needed During Earthquakes
Latest Articles
Related Articles | 1,229 | 6,001 | {"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} | 3.140625 | 3 | CC-MAIN-2024-18 | latest | en | 0.911553 |
https://mathematica.stackexchange.com/questions/11940/randomly-packing-spheres-of-fixed-radius-within-a-cube | 1,656,159,156,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103034930.3/warc/CC-MAIN-20220625095705-20220625125705-00022.warc.gz | 433,921,302 | 68,935 | # Randomly packing spheres of fixed radius within a cube
How can I have Mathematica randomly place spheres in a cube so they won't overlap? The cube is $20 \times 20 \times 20$, and the spheres have a radius of $0.7$.
• Interesting problem. Could you explain the context of the problem? Just curiosity? Oct 12, 2012 at 1:37
• How many spheres? Oct 12, 2012 at 1:38
• I'm trying to solve a packing density problem with mathematica. Eventually, I want to fill the cube to a volume fraction of 1/3 of the spheres. Right now I'm struggling to plot the cube and hard spheres.
– Jen
Oct 12, 2012 at 1:43
• So you are just trying to plot right now? You can do cube = {Opacity[.3], Cuboid[{0, 0, 0}, {20, 20, 20}]}; spheres = Table[{Black, Sphere[RandomReal[{.35, 20 - .35}, 3], .7]}, {j, 1, 10}]; Graphics3D[{cube, spheres}] Oct 12, 2012 at 1:54
• It's easier to pack them than to place them randomly, especially if you want to assume that they cannot sit still, suspended by the air. Oct 12, 2012 at 5:44
Here's a brute force and soporifically slow method.
spheres = {};
Dynamic[Length[spheres]]
Then create a new random sphere, and if it's not too close to any existing sphere add it to the list:
While[Length[spheres] < 1856, s = RandomReal[{0.7, 20 - 0.7}, 3];
If[And @@ (Norm[# - s] > 1.4 & /@ spheres), AppendTo[spheres, s]]];
According to my calculations, you need 1856 spheres of radius 0.7 to fill 1/3 of the volume of a 20 x 20 x 20 cube.
To display:
cube = {Opacity[0.3], Cuboid[{0, 0, 0}, {20, 20, 20}]};
Graphics3D[{cube, Sphere[#, 0.7] & /@ spheres}, Boxed -> False]
As I stated at the beginning, this is very slow and inefficient. The above image is from when I gave up after 1500 spheres. You can Alt. and resume at your leisure. It's entirely possible, what with the randomness, that you will never fit the 1856 spheres in.
• It's a good way to get the OP started. An extension would be to allow for wiggling the locations. One could also treat it as like an electromagnetics problem, for instance — think of the spheres as all having unit charge and reposition the spheres just as they would when another charged sphere is added to the existing bunch, subject to the constraint that none leave the box.
– rm -rf
Oct 12, 2012 at 3:36
• The alternate type approach is to randomly place spheres so that each exactly touches three others. Much tougher to implement but it will get you ultimately to higher density (50+%). Oct 12, 2012 at 4:01
• A relatively simple scheme could be to select a random sphere in the list, then creating a new sphere that touches it in a random direction. This got me to 1856 spheres in a few minutes. Oct 12, 2012 at 4:54
• Why is the standing assumption that the metric space is Euclidean (L2-norm)? Oct 12, 2012 at 6:22
Here's a brute-force method (based on a bounded random walk) that attempts to generate n nonintersecting spheres of radius r within a cube of side length s, as well as reporting the actual number of spheres it managed to generate after lim unsuccessful iterations:
sphrPak[s_, r_, n_Integer?Positive, lim_Integer?Positive] :=
Module[{sh = s/2, k, p0, pc, sphList},
(* generate initial sphere within box *)
While[
p0 = RandomReal[{-sh, sh}, 3];
And @@ Thread[Map[Min[sh - #, sh + #] &, p0] < r]];
sphList = {p0};
k = 0;
While[Length[sphList] < n && k < lim,
(* center for new sphere chosen in random direction *)
pc = p0 + 2 r Normalize[RandomVariate[NormalDistribution[], 3]];
If[(* is sphere within the cube? *)
PolyhedronData["Cube", "RegionFunction"] @@
ScalingTransform[ConstantArray[1/(s - 2 r), 3]][pc] &&
(* does the sphere not overlap with other spheres? *)
(And @@ Thread[Map[EuclideanDistance[pc, #] &, sphList] >= 2 r]),
k = 0; AppendTo[sphList, pc]; p0 = pc,
(* else *) ++k];
];
If[k == lim, Print[StringForm["Only 1 spheres were generated.", Length[sphList]]]];
Graphics3D[Sphere[sphList, r], Axes -> Automatic,
PlotRange -> {{-sh, sh}, {-sh, sh}, {-sh, sh}}]];
sphrPak[s_, r_, n_Integer] := sphrPak[s, r, n, Quotient[2 n, 3]]
Try it out:
BlockRandom[SeedRandom[1024, Method -> "MersenneTwister"]; (* for reproducibility *)
sphrPak[20, 0.7, 1500]]
Only 725 spheres were generated.
Of course, a different attempt might yield a result with more spheres within the cube.
More attempts:
BlockRandom[SeedRandom[4092, Method -> "MersenneTwister"];
sphrPak[20, 0.7, 1500]]
Only 805 spheres were generated.
BlockRandom[SeedRandom[2012, Method -> "MersenneTwister"];
sphrPak[20, 0.7, 1500]]
Only 932 spheres were generated.
Here's a variation that attempts to do more than one bounded random walk, where the starting point of the $n$th random walk is a randomly chosen point from the $(n-1)$-th random walk:
sphrPak2[s_, r_, n_Integer?Positive, walks_Integer?Positive, its_Integer?Positive] :=
Module[{sh = s/2, j, k, p0, pc, sphList},
While[
p0 = RandomReal[{-sh, sh}, 3];
And @@ Thread[Map[Min[sh - #, sh + #] &, p0] < r]];
sphList = {p0}; j = 0;
While[Length[sphList] < n && j < walks,
k = 0;
While[Length[sphList] < n && k < its,
pc = p0 + 2 r Normalize[RandomVariate[NormalDistribution[], 3]];
If[(PolyhedronData["Cube", "RegionFunction"] @@
ScalingTransform[ConstantArray[1/(s - 2 r), 3]][pc]) &&
(And @@ Thread[Map[EuclideanDistance[pc, #] &, sphList] >= 2 r]),
k = 0; AppendTo[sphList, pc]; p0 = pc,
++k];
];
If[Length[sphList] < n,
p0 = RandomChoice[Map[EuclideanDistance[#, sphList[[1]]] &,
sphList] -> sphList]];
++j;
];
If[j == walks, Print[StringForm["Only 1 spheres were generated.",
Length[sphList]]]];
Graphics3D[Sphere[sphList, r], Axes -> Automatic,
PlotRange -> {{-sh, sh}, {-sh, sh}, {-sh, sh}}]]
Try it out:
BlockRandom[SeedRandom[85, Method -> "Rule50025CA"];
sphrPak2[20, 0.7, 1500, 1, 1000]]
Only 662 spheres were generated.
BlockRandom[SeedRandom[85, Method -> "Rule50025CA"];
sphrPak2[20, 0.7, 1500, 20, 1000]]
Only 1335 spheres were generated.
• Why do you need to check if the center AND the hull are inside the cube? Oct 12, 2012 at 12:14
• @bel: I'm exploiting the fact that And[] does short-circuit evaluation; if it fails the easy test, then the other tests are not touched. On the other hand, I suppose I can merge those first two tests; wait... Oct 12, 2012 at 12:19
I will present another brute-force method for densely packing spheres that works by randomly depositing the spheres in successive "layers". The first layer is built by solving the two-dimensional analog of the problem (that is, packing disks in a square). Having constructed the first layer, successive layers are built up by dropping new spheres, checking that the incoming spheres do not intersect with spheres in previous layers (as well as with other spheres in the same layer). For the second layer, the new spheres are compared against spheres in the first layer, while for the third and succeeding layers, the new sphere is tested for intersections with the two previous layers. All this is done up until we hit the top of the cube.
Here is a Mathematica implementation of the idea presented above:
sphrPak[s_, r_, its_Integer: 2000] :=
Module[{j, k, lap, lap2, layer, layers, p0, pc, sphrs, z},
(* pack disks into square *)
(* pick first disk *)
While[
p0 = RandomReal[{r, s - r}, 2];
And @@ Thread[Map[Min[#, s - #] &, p0] < r]];
(* successively pack disks into square *)
layer = {p0}; k = 0;
While[k < its,
pc = RandomReal[{r, s - r}, 2]; (* candidate disk *)
If[And @@ Thread[Map[EuclideanDistance[pc, #] &, layer] >= 2 r],
k = 0; AppendTo[layer, pc];,
++k];
];
(* build and store first layer *)
layers = {PadRight[layer, {Length[layer], 3}, r]};
(* build successive layers *)
j = 2;
While[True,
(* previous layers *)
lap = layers[[j - 1]]; If[j > 2, lap2 = layers[[j - 2]]];
(* find "seed" sphere for layer *)
p0 = {}; k = 0;
While[k < its,
z = RandomReal[{j, j + 1} r];
pc = Append[RandomReal[{r, s - r}, 2], z]; ++k; (* candidate seed *)
If[z <= s - r &&
(And @@ Thread[Map[EuclideanDistance[pc, #] &, lap] >= 2 r]),
If[j > 2,
If[And @@ Thread[Map[EuclideanDistance[pc, #] &, lap2] >= 2 r],
p0 = pc; Break[]],
p0 = pc; Break[]]]];
If[p0 === {}, Break[]]; (* no seed found *)
(* start building new layer *)
layer = {p0}; k = 0;
While[k < its,
z = RandomReal[{j, j + 1} r];
pc = Append[RandomReal[{r, s - r}, 2], z]; (* candidate point *)
If[z <= s - r &&
(And @@ Thread[Map[EuclideanDistance[pc, #] &, layer] >= 2 r]) &&
(And @@ Thread[Map[EuclideanDistance[pc, #] &, lap] >= 2 r]),
If[j > 2,
If[And @@ Thread[Map[EuclideanDistance[pc, #] &, lap2] >= 2 r],
k = 0; AppendTo[layer, pc], ++k],
k = 0; AppendTo[layer, pc]], ++k];
];
AppendTo[layers, layer];
++j];
Print[StringForm["1 spheres generated.", Length[sphrs = Join @@ layers]]];
Graphics3D[Sphere[sphrs, r], Axes -> Automatic, PlotRange -> {{0, s}, {0, s}, {0, s}}]]
Try it out:
BlockRandom[SeedRandom[2012, Method -> "MersenneTwister"]; (* for reproducibility *)
sphrPak[20, 0.7]]
1996 spheres generated.
BlockRandom[SeedRandom[1024, Method -> "MersenneTwister"];
sphrPak[20, 0.7]]
2041 spheres generated.
• being nit picky, if i understand this scheme correctly it seems you are introducing a certain non-randomness in that you have a layered structure with constant z planes. That may be important depending on what you want to do with the result. Perhaps throw in a small random z within each layer. Might need to check two layers below if you do that though. Oct 15, 2012 at 12:25
• No @george, the heights at each layer (except the first) are not constant; that is, unless z = RandomReal[{j, j + 1} r] is doing something peculiar that you know and I don't. Yes, I check two layers below as well; that's what the If[j > 2,(* stuff *)] sections of the seeding and layer-building parts are for. Oct 15, 2012 at 12:30
• sorry..in my head that was outside the loop. Now i'm not sure that seeding the layer is necessary, you could find the mean z for each layer and move up a fixed amount. In the end i expect that seeding is a small part of the time though. Oct 15, 2012 at 13:08
So using another random walk-like method (well I tried two which I realized are equivalent) I was able to do this pretty quickly and compactly:
generateRandomAnchoring[] :=
With[{p = RandomReal[{r, m - r}, 3], c = RandomChoice@points,
d = (2 r)},
c + d*Normalize@(c - p)
];
generateRandomWalking[] :=
With[{p = RandomChoice[points], \[Theta] =
RandomReal[2 \[Pi]], \[Phi] = RandomReal[2 \[Pi]], d = (2 r)},
p + {d*Cos[\[Phi]] Sin[\[Theta]], d*Sin[\[Theta]] Sin[\[Phi]],
d*Cos[\[Theta]]}
];
tryPoint[s_] :=
If[AllTrue[s, r < # < m - r &] &&
AllTrue[points, (Norm[# - s] >= 2 r) &], AppendTo[points, s]];
Both of these take a random pregenerated point and attach a point to it at a random position, just in one case it generates a spherical polar coordinate and in the other it just attaches a randomly generated point. They're entirely equivalent as best I can reason out.
Then we'll just initialize the state and update the point generating code from wxffles answer so that it also spits out the time elapsed:
r = .7;
m = 20;
points = {};
tryPoint@RandomReal[{r, m - r}, 3];
With[{now = Now},
Monitor[While[Length[points] < 1825,
tryPoint@generateRandomAnchoring[]
], Column@{Now - now, Length@points}]
]
It runs for about a minute before hitting the 1825 mark. And we'll use the same viewer code as wxffles too:
cube = {Opacity[0.3], Cuboid[{0, 0, 0}, {20, 20, 20}]};
Graphics3D[{cube, Sphere[#, 0.7] & /@ points}, Boxed -> False]
And this is what we see:
Then just adjust the number of spheres and timing to your liking and you have a your packed cube.
### Edit
Note that one could also add a weight-tracking scheme for the random choice of points (in the second of the methods). I.e., points that have been chosen and for which the addition failed decrease in weight by some percentage. The likelihood that this will be a significant improvement is low, but it could eke out some performance gains. | 3,577 | 11,892 | {"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} | 3.6875 | 4 | CC-MAIN-2022-27 | longest | en | 0.903165 |
https://www.scirp.org/journal/paperinformation.aspx?paperid=87545 | 1,669,971,613,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710900.9/warc/CC-MAIN-20221202082526-20221202112526-00281.warc.gz | 1,029,971,830 | 40,029 | Block Decompositions and Applications of Generalized Reflexive Matrices
Abstract
Generalize reflexive matrices are a special class of matrices that have the relation where and are some generalized reflection matrices. The nontrivial cases ( or ) of this class of matrices occur very often in many scientific and engineering applications. They are also a generalization of centrosymmetric matrices and reflexive matrices. The main purpose of this paper is to present block decomposition schemes for generalized reflexive matrices of various types and to obtain their decomposed explicit block-diagonal structures. The decompositions make use of unitary equivalence transformations and, therefore, preserve the singular values of the matrices. They lead to more efficient sequential computations and at the same time induce large-grain parallelism as a by-product, making themselves computationally attractive for large-scale applications. A numerical example is employed to show the usefulness of the developed explicit decompositions for decoupling linear least-square problems whose coefficient matrices are of this class into smaller and independent subproblems.
Share and Cite:
Chen, H. (2018) Block Decompositions and Applications of Generalized Reflexive Matrices. Advances in Linear Algebra & Matrix Theory, 8, 122-133. doi: 10.4236/alamt.2018.83011.
1. Introduction
In [1] we introduced two special classes of rectangular matrices A and B that have the relations
$A=PAQ\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}B=-PBQ,A,B\in {\mathcal{C}}^{n×m},$
where P and Q are two generalized reflection matrices of dimensions n and m, respectively. A matrix X is said to be a generalized reflection matrix if $X={X}^{*}={X}^{-1}$ , i.e., if X is unitary and Hermitian. The matrices A (and B) are referred to as generalized reflexive (and antireflexive respectively) matrices. They are a generalization of centrosymmetric (anti-centrosymmetric) matrices whose special properties have been under extensive studies [2] - [11] and a generalization of reflexive (antireflexive) matrices U (V), exploited in [1] [12] [13] , that have the relations
$U=PUP\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}V=-PVP,U,V\in {\mathcal{C}}^{n×n},$
where P is some reflection (symmetric signed permutation) matrix.
Like U, the generalized reflexive matrices A arise naturally and frequently from physical problems with some sort of reflexive symmetry. Although the generalized antireflexive matrices B also also possess many interesting properties, in this paper, we shall focus only on generalized reflexive matrices. Our main objective is, thus, to present a generalized simultaneous diagonalization theorem and various decomposition schemes for the matrices A so that linear least-squares problems (or linear systems) whose coefficient matrices are of this class can be solved more efficiently. The decomposition schemes can be applied to a great number of scientific and engineering problems.
The organization of this paper is as follows. In §2, we present a generalization to the classical simultaneous diagonalization of two diagonalizable commuting square matrices. Our generalization, referred to as the generalized simultaneous diagonalization, simultaneously diagonalize a rectangular matrix H and two square matrices F and G that have the relation $FH=HG$ , assuming F and G are diagonalizable. Based on this simultaneous diagonalization, we develop explicit and semi-explicit decomposed forms in §3 for some important types of generalized reflexive matrices. An application of the decompositions to linear least-squares problems of this class is also given to show the usefulness of the decompositions. More numerical examples are provided in §4 to demonstrate the frequent occurrences of generalized reflexive matrices in many scientific and engineering disciplines.
Throughout this paper, we use the superscripts T, *, and −1 to denote the transpose, conjugate transpose, and inverse of matrices (vectors), respectively. The symbol $\oplus$ stands for the direct sum of matrices as usual. Unless otherwise noted, we use ${I}_{k}$ to denote the identity matrix of dimension k. All matrix-matrix multiplications and additions are assumed to be conformable if their dimensions not mentioned. .
2. Generalized Simultaneous Diagonalization
Before developing the (semi-)explicit block-diagonal structures for some important types of generalized reflexive matrices, we present first the following theoretically simple yet computationally useful observation regarding a simultaneous diagonalization process. Although diagonalization usually refers to square matrices, in this paper, we use the same term for rectangular matrices. In other words, a rectangular matrix $A=\left({a}_{ij}\right)\in {\mathcal{C}}^{n×m}$ is also said to be diagonal if ${a}_{ij}=0$ for $i\ne j$ . Block-diagonal rectangular matrices are defined in an analogous way.
Theorem 2.1. (Generalized Simultaneous Diagonalization) Let $F\in {\mathcal{C}}^{n×n}$ and $G\in {\mathcal{C}}^{m×m}$ be diagonalizable, $A\in {\mathcal{C}}^{n×m}$ . If $FA=AG$ , then there exist nonsingular matrices ${S}_{f}$ and ${S}_{g}$ such that
${S}_{f}^{-1}F{S}_{f},{S}_{f}^{-1}A{S}_{g}\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}{S}_{g}^{-1}G{S}_{g}$
are all diagonal matrices.
Proof. The proof given below basically employs the same technique used in [14] [15] for the simultaneous diagonalization of two square matrices that commute. Let ${X}_{f}$ and ${X}_{g}$ be the matrices that diagonalize F and G, respectively:
${X}_{f}^{-1}F{X}_{f},={\Lambda }_{f}\text{\hspace{0.17em}}\text{ }\text{and}\text{\hspace{0.17em}}\text{ }{X}_{g}^{-1}G{X}_{g}={\Lambda }_{g}$ (1)
where the diagonal elements of ${\Lambda }_{f}$ (respectively ${\Lambda }_{g}$ ) are the eigenvalues of F (respectively G). Suppose that the matrix F has k distinct eigenvalues ${\lambda }_{1},\cdots ,{\lambda }_{k}$ with multiplicities ${p}_{1},\cdots ,{p}_{k}$ , respectively, where ${p}_{1}+\cdots +{p}_{k}=n$ ; and the matrix G has l distinct eigenvalues ${\mu }_{1},\cdots ,{\mu }_{l}$ with multiplicities ${q}_{1},\cdots ,{q}_{l}$ , respectively, where ${q}_{1}+\cdots +{q}_{l}=m$ . Assume further that among the k distinct eigenvalues of F, s of them are also eigenvalues of G, $1\le s\le \mathrm{min}\left\{k,l\right\}$ . If $s=0$ , then all ${\lambda }_{i}$ and ${\mu }_{j}$ are distinct, implying that A is a null matrix, as can be seen later. Therefore, we exclude this trivial case. Without loss of generality, we can assume that
${\Lambda }_{f}=bdiag\left({\lambda }_{1}{I}_{{p}_{1}},\cdots ,{\lambda }_{s}{I}_{{p}_{s}},\cdots ,{\lambda }_{k}{I}_{{p}_{k}}\right),$
${\Lambda }_{g}=bdiag\left({\mu }_{1}{I}_{{q}_{1}},\cdots ,{\mu }_{s}{I}_{{q}_{s}},\cdots ,{\mu }_{l}{I}_{{q}_{l}}\right)$ (2)
where $bdiag\left(\cdots \right)$ denotes a block-diagonal matrix and ${\lambda }_{1}={\mu }_{1},\cdots ,{\lambda }_{s}={\mu }_{s}$ . Note that ${\lambda }_{s+1},\cdots ,{\lambda }_{k}$ and ${\mu }_{s+1},\cdots ,{\mu }_{l}$ are all distinct. Now, partition the matrix ${X}_{f}^{-1}A{X}_{g}$ , denoted by B, according to the block forms of ${\Lambda }_{f}$ and ${\Lambda }_{g}$ as $B=\left({B}_{ij}\right)$ so that ${B}_{ij}$ are pi-by-qj submatrices, $i=1,\cdots ,k$ and $j=1,\cdots ,l$ . If $FA=AG$ , we have ${\Lambda }_{f}B=B{\Lambda }_{g}$ which implies that
${\lambda }_{i}{B}_{ij}={B}_{ij}{\mu }_{j}\text{or}\left({\lambda }_{i}-{\mu }_{j}\right){B}_{ij}=0.$ (3)
Since ${\lambda }_{i}={\mu }_{j}$ only if $i=j=1,\cdots ,s$ , we know that B is a block-diagonal matrix, or more precisely, ${B}_{ij}=0$ if $i\ne j$ or if $i=j>s$ . (This can be considered as a block-equivalence decomposition for rectangular matrices.) It is well-known that for any matrix B in ${\mathcal{C}}^{n×m}$ there exist unitary matrices $U\in {\mathcal{C}}^{n×n}$ and $V\in {\mathcal{C}}^{m×m}$ such that the singular value decomposition ${U}^{*}BV$ is diagonal with nonnegative elements [16] . Now, let ${U}_{i}$ and ${V}_{i}$ be the matrices that diagonalize ${B}_{ii},i=1,\cdots ,s$ and take
$U={U}_{1}\oplus \cdots \oplus {U}_{s}\oplus {I}_{{p}_{s+1}}\oplus \cdots \oplus {I}_{{p}_{k}},$
$V={V}_{1}\oplus \cdots \oplus {V}_{s}\oplus {I}_{{q}_{s+1}}\oplus \cdots \oplus {I}_{{q}_{l}}.$ (4)
Let ${\Sigma }_{a}={U}^{-1}{X}_{f}^{-1}A{X}_{g}V$ . We see that ${\Sigma }_{a}={U}^{-1}BV$ is diagonal. Taking
${S}_{f}={X}_{f}U\text{and}{S}_{g}={X}_{g}V,$ it is clear that
${S}_{f}^{-1}F{S}_{f}={\Lambda }_{f},{S}_{f}^{-1}A{S}_{g}={\Sigma }_{a}\text{and}{S}_{g}^{-1}G{S}_{g}={\Lambda }_{g}.$ (5)
Therefore, they are all diagonal matrices.
Remark 1: Note that the converse of this theorem is not true in general. It is simple to construct such examples from diagonal matrices.
Remark 2: If the diagonalizable matrix F is the same as G, and A is diagonalizable (A is a square matrix in this case), by taking ${U}_{i}$ to be the matrices such that ${U}_{i}^{-1}{B}_{ii}{U}_{i}$ are diagonal and replacing ${S}_{g}$ with ${S}_{f}$ , this theorem along with its converse part (it now exists) then reduces to the classical simultaneous diagonalization theorem for commuting square matrices as given in ( [15] , p. 50).
Note also that this theorem is different from the simultaneous diagonalization theorems presented in [14] [16] where the simultaneous diagonalization applies to rectangular matrices of the same size.
Corollary 2.2. Let $F\in {\mathcal{C}}^{n×n}$ and $G\in {\mathcal{C}}^{m×m}$ be Hermitian, $A\in {\mathcal{C}}^{n×m}$ . If $FA=AG$ , then there exist unitary matrices ${S}_{f}$ and ${S}_{g}$ such that
${S}_{f}^{*}F{S}_{f},{S}_{f}^{*}A{S}_{g}\text{\hspace{0.17em}}\text{ }\text{and}\text{\hspace{0.17em}}\text{ }{S}_{g}^{*}G{S}_{g}$
are all diagonal matrices.
Proof. Since Hermitian matrices are diagonalizable by unitary matrices, the proof is trivial.
The usefulness of Theorem 2.1 or Corollary 2.2 lies in the fact that if we know the eigenpairs of the matrices F and G, then the matrix A can be block-diagonalized into independent submatrices by the eigenvectors (with some proper ordering) of F and G so that a single large problem can be handled via smaller and independent subproblems, yielding computational efficiency and large-grain parallelism at the same time. The question then boils down to whether those eigenpairs can easily be obtained or not. This of course depends on F and G. Fortunately, for our generalized reflexive matrices that come from physical problems, their eigenpairs of P and Q are explicitly known in most cases, as can be seen from the example presented in Section 4. In the next section, we present several generalized reflexive decompositions that lead to either explicit or semi-explicit block-decomposed forms, which are computationally attractive.
3. Decompositions for Generalized Reflexive Matrices
We now turn to generalized reflexive matrices A, which are not necessary square matrices. The decomposition schemes presented below for A are special applications of the general results developed in the previous section. Our main purpose is to obtain explicit forms of the block structure for some frequently encountered cases of A. Let $PA=AQ$ be generalized reflexive. Recall that P and Q are two generalized reflection matrices, which are unitary Hermitian matrices. Therefore, they have at most two distinct eigenvalues 1 and −1. Furthermore, the relation $A=PAQ$ can be expressed as $PA=AQ$ since $P={P}^{*}={P}^{-1}$ . From Corollary 2.2, we know that A can be block-diagonalized into two independent submatrices. This information along, however, is not enough from the computational point of view. we still need to know the eigenpairs of P and Q in order to obtain the explicit decomposed form of A. In the following, we derive several explicit or semi-explicit decomposed forms for some important types of generalized reflexive matrices, starting with the simplest one.
Theorem 3.1. Let $P\in {\mathcal{C}}^{n×n}$ and $Q\in {\mathcal{C}}^{m×m}$ , n and m even, be two matrices that take the following forms.
$P=\left[\begin{array}{cc}0& {P}_{1}^{*}\\ {P}_{1}& 0\end{array}\right]\text{and}Q=\left[\begin{array}{cc}0& {Q}_{1}^{*}\\ {Q}_{1}& 0\end{array}\right]$ (6)
where ${P}_{1}$ and ${Q}_{1}$ are unitary. Let $A\in {\mathcal{C}}^{n×m}$ be partitioned as $\left({A}_{ij}\right)$ , $i,j=1,2$ , with each ${A}_{ij}\in {\mathcal{C}}^{p×q}$ , $p=\frac{n}{2}$ and $q=\frac{m}{2}$ . If $A=PAQ$ , then there exist two unitary matrices X and Y such that
${X}^{*}AY=\left({A}_{11}+{A}_{12}{Q}_{1}\right)\oplus \left({A}_{22}-{A}_{21}{Q}_{1}^{*}\right)=\left({A}_{11}+{P}_{1}^{*}{A}_{21}\right)\oplus \left({A}_{22}-{P}_{1}{A}_{12}\right).$ (7)
Proof. Clearly, both P and Q are generalized reflection matrices. Therefore, A is a generalized reflexive matrix. Take X and Y to be the unitary matrices
$X=\frac{1}{\sqrt{2}}\left[\begin{array}{cc}I& -{P}_{1}^{*}\\ {P}_{1}& I\end{array}\right]\text{and}Y=\frac{1}{\sqrt{2}}\left[\begin{array}{cc}I& -{Q}_{1}^{*}\\ {Q}_{1}& I\end{array}\right].$ (8)
Then
$\begin{array}{c}{X}^{*}AY=\frac{1}{2}\left[\begin{array}{cc}I& {P}_{1}^{*}\\ -{P}_{1}& I\end{array}\right]\left[\begin{array}{cc}{A}_{11}& {A}_{12}\\ {A}_{21}& {A}_{22}\end{array}\right]\left[\begin{array}{cc}I& -{Q}_{1}^{*}\\ {Q}_{1}& I\end{array}\right]\\ =\frac{1}{2}\left[\begin{array}{cc}\left({A}_{11}+{A}_{12}{Q}_{1}\right)+\left({P}_{1}^{*}{A}_{21}+{P}_{1}^{*}{A}_{22}{Q}_{1}\right)& \left({A}_{12}-{P}_{1}^{*}{A}_{21}{Q}_{1}^{*}\right)+\left({P}_{1}^{*}{A}_{22}-{A}_{11}{Q}_{1}^{*}\right)\\ \left({A}_{21}-{P}_{1}{A}_{12}{Q}_{1}\right)+\left({A}_{22}{Q}_{1}-{P}_{1}{A}_{11}\right)& \left({A}_{22}-{A}_{21}{Q}_{1}^{*}\right)+\left({P}_{1}{A}_{11}{Q}_{1}^{*}-{P}_{1}{A}_{12}\right)\end{array}\right]\\ =\left[\begin{array}{cc}{A}_{11}+{A}_{12}{Q}_{1}& 0\\ 0& {A}_{22}-{A}_{21}{Q}_{1}^{*}\end{array}\right]=\left[\begin{array}{cc}{A}_{11}+{P}_{1}^{*}{A}_{21}& 0\\ 0& {A}_{22}-{P}_{1}{A}_{12}\end{array}\right]\end{array}$ (9)
where we have used the unitarity of ${P}_{1}$ and ${Q}_{1}$ and the relations ${A}_{11}={P}_{1}^{*}{A}_{22}{Q}_{1}$ and ${A}_{21}={P}_{1}{A}_{12}{Q}_{1}$ , which results from the assumption of $A=PAQ$ . Note that ${X}^{*}PX={I}_{p}\oplus -{I}_{p}$ and ${Y}^{*}QY={I}_{q}\oplus -{I}_{q}$ , which also explains, via Corollary 2.2, why this decomposition is possible.
Theorem 3.2. Let $P\in {\mathcal{C}}^{n×n}$ and $Q\in {\mathcal{C}}^{m×m}$ , $n=2p+r$ and $m=2q+s$ , be the following two generalized reflection matrices:
$P=\left[\begin{array}{ccc}0& 0& {P}_{1}^{*}\\ 0& {I}_{r}& 0\\ {P}_{1}& 0& 0\end{array}\right]\text{and}Q=\left[\begin{array}{ccc}0& 0& {Q}_{1}^{*}\\ 0& {I}_{s}& 0\\ {Q}_{1}& 0& 0\end{array}\right]$ (10)
where ${P}_{1}$ and ${Q}_{1}$ are unitary matrices of dimensions p and q, respectively; $\alpha =±1$ , $\beta =±1$ . Let $A\in {\mathcal{C}}^{n×m}$ be partitioned as $\left({A}_{ij}\right)$ , $i,j=1,2,3$ , with ${A}_{11}\in {\mathcal{C}}^{p×q}$ , ${A}_{22}\in {\mathcal{C}}^{p×q}$ , and ${A}_{33}\in {\mathcal{C}}^{p×q}$ . If $A=PAQ$ , then there exist two unitary matrices X and Y such that
${X}^{*}AY=\left[\begin{array}{cc}{A}_{11}+{A}_{13}{Q}_{1}& \sqrt{2}{A}_{12}\\ \sqrt{2}{A}_{21}& {A}_{22}\end{array}\right]\oplus \left({A}_{33}-{A}_{31}{Q}_{1}^{*}\right)\text{if}\alpha =\beta =1,$
${X}^{*}AY=\left({A}_{11}+{A}_{13}{Q}_{1}\right)\oplus \left[\begin{array}{cc}{A}_{22}& \sqrt{2}{A}_{23}\\ \sqrt{2}{A}_{32}& {A}_{33}-{A}_{31}{Q}_{1}^{*}\end{array}\right]\text{if}\text{\hspace{0.17em}}\alpha =\beta =-1,$
${X}^{*}AY=\left[\begin{array}{c}{A}_{11}+{A}_{13}{Q}_{1}\\ \sqrt{2}{A}_{21}\end{array}\right]\oplus \left[\begin{array}{cc}\sqrt{2}{A}_{32}& {A}_{33}-{A}_{31}{Q}_{1}^{*}\end{array}\right]\text{if}\text{\hspace{0.17em}}\alpha =-\beta =1,$
and
${X}^{*}AY=\left[\begin{array}{cc}{A}_{11}+{A}_{13}{Q}_{1}& \sqrt{2}{A}_{12}\end{array}\right]\oplus \left[\begin{array}{c}\sqrt{2}{A}_{23}\\ {A}_{33}-{A}_{31}{Q}_{1}^{*}\end{array}\right]\text{if}\text{\hspace{0.17em}}\alpha =-\beta =-1.$
Proof. Take X and Y to be the following two unitary matrices:
$X=\frac{1}{\sqrt{2}}\left[\begin{array}{ccc}I& 0& -{P}_{1}^{*}\\ 0& \sqrt{2}{I}_{r}& 0\\ {P}_{1}& 0& I\end{array}\right]\text{and}Y=\frac{1}{\sqrt{2}}\left[\begin{array}{ccc}I& 0& -{Q}_{1}^{*}\\ 0& \sqrt{2}{I}_{s}& 0\\ {Q}_{1}& 0& I\end{array}\right].$ (11)
Then the unitary transformation ${X}^{*}AY$ yields
$\begin{array}{l}{X}^{*}AY=\frac{1}{2}\left[\begin{array}{ccc}I& 0& {P}_{1}^{*}\\ 0& {I}_{r}& 0\\ -{P}_{1}& 0& I\end{array}\right]\left[\begin{array}{ccc}{A}_{11}& {A}_{12}& {A}_{13}\\ {A}_{21}& {A}_{22}& {A}_{23}\\ {A}_{31}& {A}_{32}& {A}_{33}\end{array}\right]\left[\begin{array}{ccc}I& 0& -{Q}_{1}^{*}\\ 0& \sqrt{2}{I}_{s}& 0\\ {Q}_{1}& 0& I\end{array}\right]\\ =\frac{1}{2}\left[\begin{array}{ccc}\left({A}_{11}+{A}_{13}{Q}_{1}\right)+\left({P}_{1}^{*}{A}_{31}+{P}_{1}^{*}{A}_{33}{Q}_{1}\right)& \sqrt{2}\left({A}_{12}+{P}_{1}^{*}{A}_{32}\right)& \left({A}_{13}-{P}_{1}^{*}{A}_{31}{Q}_{1}^{*}\right)+\left({P}_{1}^{*}{A}_{33}-{A}_{11}{Q}_{1}^{*}\right)\\ \sqrt{2}\left({A}_{21}+{A}_{23}{Q}_{1}\right)& 2{A}_{22}& \sqrt{2}\left({A}_{23}-{A}_{21}{Q}_{1}^{*}\right)\\ \left({A}_{31}-{P}_{1}{A}_{13}{Q}_{1}\right)+\left({A}_{33}{Q}_{1}-{P}_{1}{A}_{11}\right)& \sqrt{2}\left({A}_{32}-{P}_{1}{A}_{12}\right)& \left({A}_{33}-{A}_{31}{Q}_{1}^{*}\right)+\left({P}_{1}{A}_{11}{Q}_{1}^{*}-{P}_{1}{A}_{13}\right)\end{array}\right]\end{array}$ (12)
If $A=PAQ$ , we immediately have the following relations among the submatrices ${A}_{ij}$ .
${A}_{11}={P}_{1}^{*}{A}_{33}{Q}_{1},{A}_{13}={P}_{1}^{*}{A}_{31}{Q}_{1}^{*},$
${A}_{12}={P}_{1}^{*}{A}_{32},{A}_{21}={A}_{23}{Q}_{1}\text{ }\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}\text{ }{A}_{22}={A}_{22}.$
Employing these relations and the unitarity of ${P}_{1}$ and ${Q}_{1}$ for (12), we obtain a much simplified form of the transformation ${X}^{*}AY$ . Namely,
${X}^{*}AY=\frac{1}{2}\left[\begin{array}{ccc}2\left({A}_{11}+{A}_{13}{Q}_{1}\right)& \sqrt{2}\left(1+\beta \right){A}_{12}& 0\\ \sqrt{2}\left(1+\alpha \right){A}_{21}& \left(1+\alpha \beta \right){A}_{22}& \sqrt{2}\left(1-\alpha \right){A}_{23}\\ 0& \sqrt{2}\left(1-\beta \right){A}_{32}& 2\left({A}_{33}-{A}_{31}{Q}_{1}^{*}\right)\end{array}\right].$ (13)
Accordingly, we have the results we want:
${X}^{*}AY=\left[\begin{array}{ccc}{A}_{11}+{A}_{13}{Q}_{1}& \sqrt{2}{A}_{12}& 0\\ \sqrt{2}{A}_{21}& {A}_{22}& 0\\ 0& 0& {A}_{33}-{A}_{31}{Q}_{1}^{*}\end{array}\right]\text{for}\text{\hspace{0.17em}}\alpha =\beta =1,$
${X}^{*}AY=\left[\begin{array}{ccc}{A}_{11}+{A}_{13}{Q}_{1}& 0& 0\\ 0& {A}_{22}& \sqrt{2}{A}_{23}\\ 0& \sqrt{2}{A}_{32}& {A}_{33}-{A}_{31}{Q}_{1}^{*}\end{array}\right]\text{for}\text{\hspace{0.17em}}\alpha =\beta =-1,$
${X}^{*}AY=\left[\begin{array}{ccc}{A}_{11}+{A}_{13}{Q}_{1}& 0& 0\\ \sqrt{2}{A}_{21}& 0& 0\\ 0& \sqrt{2}{A}_{32}& {A}_{33}-{A}_{31}{Q}_{1}^{*}\end{array}\right]\text{for}\text{\hspace{0.17em}}\alpha =-\beta =1,$
and
${X}^{*}AY=\left[\begin{array}{ccc}{A}_{11}+{A}_{13}{Q}_{1}& \sqrt{2}{A}_{12}& 0\\ 0& 0& \sqrt{2}{A}_{23}\\ 0& 0& {A}_{33}-{A}_{31}{Q}_{1}^{*}\end{array}\right]\text{for}\text{\hspace{0.17em}}\alpha =-\beta =-1.$
Note that in (13), ${A}_{13}{Q}_{1}$ can be replaced by ${P}_{1}^{*}{A}_{31}$ and ${A}_{31}{Q}_{1}^{*}$ replaced by ${P}_{1}{A}_{13}$ since ${A}_{31}={P}_{1}{A}_{13}{Q}_{1}$ . Computationally, one should use the expressions that are easier to compute. Note also that X and Y do not depend on $\alpha$ and $\beta$ , and
${X}^{*}PX={I}_{p}\oplus {I}_{r}\oplus -{I}_{p}\text{and}{Y}^{*}QY={I}_{q}\oplus {I}_{r}\oplus -{I}_{q}.$
Remark 3: In Theorem 3.1 and 3.2 if the unitarity requirement of P, Q, X, and Y is lifted, a slightly more general case can be obtained simply by replacing the conjugate transpose with the inverse (existence of ${P}_{1}^{-1}$ and ${Q}_{1}^{-1}$ assumed) in places of ${P}_{1}^{*}$ , ${Q}_{1}^{*}$ , ${X}^{*}$ , and ${Y}^{*}$ . With this replacement, all the results in the proofs remain intact. The matrices A in this case, however, are not necessarily generalized reflexive since P and Q may not be generalized reflection matrices.
Remark 4: Obviously, Theorem 3.2 reduces to Theorem 3.1 if ${I}_{r}$ and ${I}_{s}$ in (10) do not exist, i.e., $r=s=0$ . If ${I}_{r}$ is present and ${I}_{s}$ disappears, then by partitioning A as $\left({A}_{ij}\right)$ , $i=1,2,3$ and $j=1,2$ , according to the block forms of P and Q, we have
${X}^{*}AY=\left[\begin{array}{cc}{A}_{11}+{A}_{12}{Q}_{1}& 0\\ \frac{1}{\sqrt{2}}\left(1+\alpha \right){A}_{21}& \frac{1}{\sqrt{2}}\left(1-\alpha \right){A}_{22}\\ 0& {A}_{32}-{A}_{31}{Q}_{1}^{*}\end{array}\right]$ (14)
which is decoupled into two independent sub-blocks when $\alpha =±1$ . Analogous to (13), ${A}_{12}{Q}_{1}$ and ${A}_{31}{Q}_{1}^{*}$ can be expressed as ${P}_{1}^{*}{A}_{31}$ and ${P}_{1}{A}_{12}$ , respectively since in this case ${A}_{31}={P}_{1}{A}_{12}{Q}_{1}$ . Instead, if ${I}_{r}$ disappears and ${I}_{s}$ remains and the matrix A is partitioned in accordance with P and Q as
$A=\left[\begin{array}{ccc}{A}_{11}& {A}_{12}& {A}_{13}\\ {A}_{21}& {A}_{22}& {A}_{23}\end{array}\right],$
then we have
${X}^{*}AY=\left[\begin{array}{ccc}{A}_{11}+{A}_{13}{Q}_{1}& \frac{1}{\sqrt{2}}\left(1+\beta \right){A}_{12}& 0\\ 0& \frac{1}{\sqrt{2}}\left(1-\beta \right){A}_{22}& {A}_{23}-{A}_{21}{Q}_{1}^{*}\end{array}\right]$ (15)
where ${A}_{13}{Q}_{1}={P}_{1}^{*}{A}_{21}$ and ${A}_{21}{Q}_{1}^{*}={P}_{1}{A}_{13}$ because ${A}_{21}={P}_{1}{A}_{13}{Q}_{1}$ . This transformation again decouples the matrix A into two independent sub-blocks when $\beta =±1$ .
4. Applications
As seen from the transformations presented in the previous section, the decomposed forms of A of this class are very simple to compute. This is especially true when P and Q are reflection (symmetric signed permutation) matrices, which arise frequently in a very wide range of real-world applications, because any reflection matrix can be symmetrically permuted to yield one of the forms of (6) and (10), with ${P}_{1}$ and ${Q}_{1}$ being some signed permutation matrices whereas ${P}_{2}$ and ${Q}_{2}$ some reflection matrices. Furthermore, the decompositions preserve all singular values because they make use of unitarily equivalence transformations, which can be applied to both square matrices and rectangular matrices. Therefore, they are useful not only for linear systems but for linear least-squares problems and singular value problems as well. The only requirement is the existence of the generalized reflexivity property of the matrix A. When P is the same as Q, the decompositions lead to similarity transformations and, accordingly, preserve all eigenvalues. It is exactly this simplicity and preservance of singular values or eigenvalues that makes these decompositions computationally attractive. To demonstrate the usefulness of these decompositions in attacking applications of this type, we present in this section an application of the decompositions to one of the numerical examples described in [1] , where the same problem is solved using only basic generalized reflexive properties, without resorting to matrix decompositions.
Numerical example. Consider the following overdetermined linear system:
$\left[\begin{array}{rrrr}\hfill 1& \hfill -1& \hfill 0& \hfill 0\\ \hfill 0& \hfill -1& \hfill 0& \hfill 0\\ \hfill 1& \hfill 0& \hfill 0& \hfill -1\\ \hfill 0& \hfill 1& \hfill 0& \hfill -1\\ \hfill 0& \hfill 0& \hfill 1& \hfill -1\\ \hfill 0& \hfill 0& \hfill 0& \hfill -1\\ \hfill 0& \hfill -1& \hfill 1& \hfill 0\end{array}\right]\left[\begin{array}{c}{x}_{1}\\ {x}_{2}\\ {x}_{3}\\ {x}_{4}\end{array}\right]=\left[\begin{array}{c}50\\ -152\\ 78\\ 33\\ 30\\ -123\\ 2\end{array}\right].$ (16)
Let A be the coefficient matrix of the overdetermined system. It is simple to observe that A is a generalized reflexive matrix: $A=PAQ$ where
$P=\left[\begin{array}{ccc}0& 0& {I}_{3}\\ 0& -1& 0\\ {I}_{3}& 0& 0\end{array}\right]\text{and}Q=\left[\begin{array}{cc}0& {I}_{2}\\ {I}_{2}& 0\end{array}\right]$ (17)
are two reflection matrices. It deserves mentioning that the coefficient matrix A is the edge-node incidence matrix of a level network with reflexive symmetry.
Whether this overdetermined linear system is to be solved via its normal equation or using a QR decomposition instead, we can decompose the original problem into two independent subproblems first, using the decomposition techniques presented in the previous section. Let
$X=\frac{1}{\sqrt{2}}\left[\begin{array}{ccc}{I}_{3}& 0& -{I}_{3}\\ 0& \sqrt{2}& 0\\ {I}_{3}& 0& {I}_{3}\end{array}\right]\text{and}Y=\frac{1}{\sqrt{2}}\left[\begin{array}{rr}\hfill {I}_{2}& \hfill -{I}_{2}\\ \hfill {I}_{2}& \hfill {I}_{2}\end{array}\right].$ (18)
The overdetermined system $Ax=b$ is then transformed to $\stackrel{˜}{A}\stackrel{˜}{x}=\stackrel{˜}{b}$ with
$\stackrel{˜}{A}={X}^{\text{T}}AY,\stackrel{˜}{x}=\left(\sqrt{2}{Y}^{\text{T}}\right)x\text{ }\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}\text{ }\stackrel{˜}{b}=\left(\sqrt{2}{X}^{\text{T}}\right)b$
where $\sqrt{2}$ is intentionally inserted to avoid unnecessary multiplications of $\frac{1}{\sqrt{2}}$ in forming $\stackrel{˜}{b}$ from b. Now, let $Ax=b$ be partitioned, according to the block forms of X and Y, as
$\left[\begin{array}{cc}{A}_{11}& {A}_{12}\\ {A}_{21}& {A}_{22}\\ {A}_{31}& {A}_{32}\end{array}\right]\left[\begin{array}{c}{x}_{1}\\ {x}_{2}\end{array}\right]=\left[\begin{array}{c}{b}_{1}\\ {b}_{2}\\ {b}_{3}\end{array}\right].$ (19)
The transformation ${X}^{\text{T}}AY$ can easily be obtained without actually performing expensive matrix-matrix multiplications. We simply use the explicit form of (14) by substituting ${I}_{2}$ for ${Q}_{1}$ and −1 for $\alpha$ , yielding
$\stackrel{˜}{A}={X}^{\text{T}}AY={\stackrel{˜}{A}}_{1}\oplus {\stackrel{˜}{A}}_{2}$
where
${\stackrel{˜}{A}}_{1}={A}_{11}+{A}_{12}=\left[\begin{array}{cc}1& -1\\ 0& -1\\ 1& -1\end{array}\right]\text{\hspace{0.17em}}\text{ }\text{and}\text{\hspace{0.17em}}\text{ }{\stackrel{˜}{A}}_{2}=\left[\begin{array}{c}\sqrt{2}{A}_{22}\\ {A}_{32}-{A}_{31}\end{array}\right]=\left[\begin{array}{rr}\hfill 0& \hfill -\sqrt{2}\\ \hfill 1& \hfill -1\\ \hfill 0& \hfill -1\\ \hfill 1& \hfill 1\end{array}\right].$ (20)
It is simple to obtain $\stackrel{˜}{b}$ without resorting to a dense matrix-vector multiplication.
$\stackrel{˜}{b}=\left[\begin{array}{c}{b}_{1}+{b}_{3}\\ \sqrt{2}{b}_{2}\\ {b}_{3}-{b}_{1}\end{array}\right]={\left[\begin{array}{ccccccccc}80& -275& 80& |& \sqrt{2}\left(33\right)& |& -20& 29& -76\end{array}\right]}^{\text{T}}.$
This transformation then decouples the original system $Ax=b$ into
${\stackrel{˜}{A}}_{1}{\stackrel{˜}{x}}_{1}={\stackrel{˜}{b}}_{1}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{with}\text{\hspace{0.17em}}{\stackrel{˜}{b}}_{1}={\left[\begin{array}{ccc}80& -275& 80\end{array}\right]}^{\text{T}}$ (21)
and
${\stackrel{˜}{A}}_{2}{\stackrel{˜}{x}}_{2}={\stackrel{˜}{b}}_{2}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{with}\text{\hspace{0.17em}}{\stackrel{˜}{b}}_{2}={\left[\begin{array}{cccc}\sqrt{2}\left(33\right)& -20& 29& -76\end{array}\right]}^{\text{T}}.$ (22)
The normal equations of (21) and (22) are simply
$\left[\begin{array}{rr}\hfill 2& \hfill -2\\ \hfill -2& \hfill 3\end{array}\right]{\stackrel{˜}{x}}_{1}=\left[\begin{array}{c}160\\ 115\end{array}\right]\text{and}\left[\begin{array}{cc}2& 0\\ 0& 5\end{array}\right]{\stackrel{˜}{x}}_{2}=\left[\begin{array}{c}-96\\ -151\end{array}\right],$
respectively, whose solutions are ${\stackrel{˜}{x}}_{1}={\left[355275\right]}^{\text{T}}$ and ${\stackrel{˜}{x}}_{2}={\left[-48-30.2\right]}^{\text{T}}$ . The final solution x can now be retrieved from ${\stackrel{˜}{x}}_{1}$ and ${\stackrel{˜}{x}}_{2}$ with ease.
$x=\frac{1}{2}\left[\begin{array}{c}{\stackrel{˜}{x}}_{1}-{\stackrel{˜}{x}}_{2}\\ {\stackrel{˜}{x}}_{1}+{\stackrel{˜}{x}}_{2}\end{array}\right]={\left[\begin{array}{rrrr}\hfill 201.5& \hfill 152.6& \hfill 153.5& \hfill 122.4\end{array}\right]}^{\text{T}},$
whose correctness can be verified from the normal equation of the original system.
At this point, it is clear that the main reason why transformations of this type are so cheap to obtain is not only that explicit forms are available but that no arithmetic multiplications or divisions are involved in forming the decoupled subsystems except the central block row of A and b and the central block column of A, if any, such as ${A}_{22}$ and ${b}_{2}$ in this example. The dimensions of these blocks are usually very small for large-scale problems with reflexive symmetry because they involve only the nodes/edges on the line or plane of symmetry. Therefore, this extra work can easily be offset by the tremendous savings resulting from solving two smaller subproblems whose sizes are only about half of the original problem. It is worth mentioning that solving sequentially two independent decomposed subproblems each of half size of a single problem is about four times faster than solving the undecomposed one. This is exactly where computational efficiency comes from. The large-grain parallelism induced by these decompositions is an additional advantage when the subproblems are solved on a multiprocessor on multiple networked computers.
We close this section by emphasizing the fact that a great number of scientific and engineering applications require solutions to linear least-squares problems, singular value problems, linear systems, or eigenvalue problems whose coefficient matrices are either generalized reflexive nontrivial reflection matrices P and Q or reflexive P (or Q). Instead of giving more numerical examples, we just mention that the node-edge (or edge-node) incidence matrix of any finite network or graph that possesses reflexive symmetry or that can be redrawn as one that displays reflexive symmetry is generalized reflexive. Refer to [1] for more numerical examples.
5. Conclusions
Generalized reflexive matrices, a newly exploited special class of matrices $A\in {\mathcal{C}}^{n×m}$ that have the relation $A=PAQ$ with P and Q being some generalized reflection matrices, are a generalization of centrosymmetric matrices and reflexive matrices. Although it is not trivial to realize their existence purely from the entries of a given matrix, this new class of matrices indeed arise very often from physical problems in many areas of scientific and engineering applications, especially from those with reflexive symmetry. Three such nontrivial numerical examples, each from a distinct real-world application area, can be found in [1] .
A major part of this paper has been devoted to the exploration of computationally attractive decompositions for taking advantage of the special relation possessed by this class of matrices. The decompositions are based on a generalized simultaneous diagonalization theorem presented in this paper and derived using the eigenvectors of P and Q via unitarily equivalence transformations. When the eigenpairs of P and Q are explicitly known, which is usually the case for generalized reflexive matrices that arise from physical problems with reflexive symmetry, the decompositions yield simple and explicit forms of the decomposed submatrices for the matrices A. One of the generalized reflexive matrices presented in this paper has also been employed to serve as an example to show the usefulness of the derived explicit decompositions for decoupling linear least-squares problems whose coefficient matrices are of this class into smaller and independent subproblems. These decompositions, though theoretically simple, can lead to much more efficient computation for large-scale applications. It also induces large-grain parallelism as a by-product. Furthermore, they preserve either the singular values or the eigenvalues of the matrices and, therefore, immediately applicable not only for handling linear least-squares problems and linear systems but for attacking singular value problems or eigenvalue problems.
Conflicts of Interest
The authors declare no conflicts of interest.
[1] Chen, H.-C. (1998) Generalized Reflexive Matrices: Special Properties and Applications. SIAM Journal on Matrix Analysis and Applications, 19, 140-153. https://doi.org/10.1137/S0895479895288759 [2] Zehfuss, G. (1862) Zwei Sätze über determinanten. Zeitschrift für Angewandte Mathematik und Physik, VII, 436-439. [3] Aitken, A.C. (1949) Determinants and Matrices. 6th Edition, Wiley-Interscience, New York. [4] Good, I.J. (1970) The Inverse of a Centrosymmetric Matrix. Technometrics, 12, 925-928. https://doi.org/10.1080/00401706.1970.10488743 [5] Andrew, A.L. (1973) Solution of Equations Involving Centrosymmetric Matrices. Technometrics, 15, 405-407. https://doi.org/10.1080/00401706.1973.10489052 [6] Andrew, A.L. (1973) Eigenvectors of Certain Matrices. Linear Algebra and Its Applications, 7, 151-162. https://doi.org/10.1016/0024-3795(73)90049-9 [7] Pye, W.C., Boullino, T.L. and Atchison, T.A. (1973) The Pseudoinverse of a Centrosymmetric Matrix. Linear Algebra and Its Applications, 6, 201-204. https://doi.org/10.1016/0024-3795(73)90020-7 [8] Cantoni, A. and Butler, P. (1976) Eigenvalues and Eigenvectors of Symmetric Centrosymmetric Matrices. Linear Algebra and Its Applications, 13, 275-288. https://doi.org/10.1016/0024-3795(76)90101-4 [9] Weaver, J.R. (1985) Centrosymmetric (Cross-Symmetric) Matrices, Their Basic Properties, Eigenvalues, and Eigenvectors. The American Mathematical Monthly, 92, 711-717. https://doi.org/10.1080/00029890.1985.11971719 [10] Weaver, J.R. (1988) Real Eigenvalues of Nonnegative Matrices Which Commute with a Symmetric Matrix Involution. Linear Algebra and Its Applications, 110, 243-253. https://doi.org/10.1016/0024-3795(83)90138-6 [11] Tao, D. and Yasuda, M. (2002) A Spectral Characterization of Generalized Real Symmetric Centrosymmetric and Generalized Real Symmetric Skew-Centrosymmetric Matrices. SIAM Journal on Matrix Analysis and Applications, 23, 885-895. https://doi.org/10.1137/S0895479801386730 [12] Chen, H-C. and Sameh, A. (1989) A Matrix Decomposition Method for Orthotropic Elasticity Problems. SIAM Journal on Matrix Analysis and Applications, 10, 39-64. https://doi.org/10.1137/0610004 [13] Chen, H.-C. and Sameh, A. (1989) A Domain Decomposition Method for 3D Elasticity Problems. In: Brebbia, C.A. and Peters, A., Eds., Applications of Supercomputers in Engineering: Fluid Flow and Stress Analysis Applications, Computational Mechanics Publications, Southampton University, Southampton, England, 171-188. [14] Gibson, P.M. (1974) Simultaneous Diagonalization of Rectangular Complex Matrices. Linear Algebra and Its Applications, 9, 45-53. https://doi.org/10.1016/0024-3795(74)90025-1 [15] Horn, R.A. and Johnson, C.A. (1985) Matrix Analysis. Cambridge University Press, New York. https://doi.org/10.1017/CBO9780511810817 [16] Eckart, C. and Young, G. (1939) A Principal Axis Transformation for Non-Hermitian Matrices. Bulletin of the American Mathematical Society, 45, 118-121. https://doi.org/10.1090/S0002-9904-1939-06910-3 | 11,496 | 35,188 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 211, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2022-49 | latest | en | 0.874819 |
http://love2learn2day.blogspot.com/2014/02/review-color-tile-fraction-puzzles.html | 1,550,927,941,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550249501174.94/warc/CC-MAIN-20190223122420-20190223144420-00263.warc.gz | 160,496,975 | 32,997 | ## Wednesday, February 5, 2014
### Review: Color Tile Fraction Puzzles
During our fraction unit, I added an extra activity to my 5th graders' study. Color Tile Fraction Puzzles: Grades 5+ consists of 30 puzzle cards, color tile, and an instruction booklet with student answer sheet and answer key.
What I like:
1. This is a great way to get kids thinking with visual fraction models.
2. Cards ask kids to calculate a fraction of a whole number (1/3 of 15, for example), contributing toward understanding multiplication of fractions.
3. The combination of puzzles, manipulatives, coloring, and computation lends a lot of fun to fraction exploration.
4. The product aligns to CCSS 5.NF.2:
"Solve word problems involving addition and subtraction of fractions referring to the same whole, including cases of unlike denominators, e.g., by using visual fraction models or equations to represent the problem."
4. Since the cards increase in difficulty, it is easy to differentiate.
The puzzles are advertised for use in grades 5 and up. I think it could be useful in 4th grade (perhaps even a challenge in 3rd), but I would probably only use it in grades higher than 5th if students were struggling with this particular concept...or I might use it as an extension.
Extension Idea:
Students with the greatest understanding of fractions could rapidly figure most problems in their heads. Given more time, I would challenge those students to write their own color tile fraction puzzles.
Summary:
This product contains a nice balance of fun and learning. I'm glad to have it available as an addition to our fraction practice.
Disclosure: I received a free copy of Color Tile Fraction Puzzles in exchange for a frank and unbiased review. | 373 | 1,732 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2019-09 | latest | en | 0.951645 |
https://index.pkp.sfu.ca/index.php/record/view/3882137 | 1,623,616,827,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487610841.7/warc/CC-MAIN-20210613192529-20210613222529-00407.warc.gz | 310,144,267 | 4,999 | ### Análisis de algunas técnicas iterativas para sistemas de ecuaciones lineales y su estudio de la convergencia a través del número de condicionamiento
#### Scientia et Technica Journal
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Field Value Title Análisis de algunas técnicas iterativas para sistemas de ecuaciones lineales y su estudio de la convergencia a través del número de condicionamiento Analysis of Some Iterative Techniques for Systems of Linear Equations and Their Study of the Convergence Through the Number of Conditioning Creator Mesa, Fernando Devia Narváez, Diana Marcela Correa Vélez, German Description At present, numerical analysis provides us with powerful tools to determine the solution of various problems whose mathematical model can be represented by a system of linear equations, these tools correspond to a number of direct and iterative methods, among which are Carl's method. Gustav Jakob Jacobi and the Doolittle and Crout method, which we analyze and compare in this document. To do this we will initially explore the concepts of conditioning the problem to determine how stable is the system from which the model was obtained, until we reach the decomposition of LU arrays proposed in the Doolittle and Crout method. As a result of the analysis and comparison in this document, depending on what is sought when solving a system of equations, either very large or small enough for our computer, we can choose an approximation that will bring a short-term result with an error. Due to the starting point as proposed in the Jacobi method, or it is possible to reach a direct result by implementing fewer iterations as proposed in the Doolittle and Crout metho En la actualidad el análisis numérico nos brinda poderosas herramientas para determinar la solución de diversos problemas cuyo modelo matemático puede ser representado por un sistema de ecuaciones lineales, estas herramientas corresponden a un sinnúmero de métodos directos e iterativos entre los que se encuentran el método de Carl Gustav Jakob Jacobi y el método de Doolittle y Crout los cuales analizamos y comparamos en este documento .Para ello exploraremos inicialmente los conceptos de condicionamiento del problema para determinar que tan estable es el sistema de donde se obtuvo el modelo , hasta llegar a la descomposición de matrices LU propuestas en el método de Doolittle y Crout. Como resultado del análisis y comparación en este documento dependiendo de lo que se busque al resolver un sistema de ecuaciones ya sea de tamaño muy grande o lo suficiente pequeño para nuestra computadora, podemos optar por una aproximación que traerá un resultado a corto plazo con un error debido al punto de partida tal y como como se propone en el método del Jacobi o es posible llegar a un resultado directo implementando menor cantidad de iteraciones como se propone en el método de Doolittle y Crout. Publisher Universidad Tecnológica de Pereira Date 2020-12-22 Type info:eu-repo/semantics/article info:eu-repo/semantics/publishedVersion Format application/pdf Identifier https://revistas.utp.edu.co/index.php/revistaciencia/article/view/24617 10.22517/23447214.24617 Source Scientia et Technica; Vol 25 No 4 (2020); 621-629 Scientia et Technica; Vol. 25 Núm. 4 (2020); 621-629 2344-7214 0122-1701 Language eng Relation https://revistas.utp.edu.co/index.php/revistaciencia/article/view/24617/16453 Rights http://creativecommons.org/licenses/by-nc-nd/4.0 | 791 | 3,442 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2021-25 | latest | en | 0.45306 |
http://conversion.org/length/parsec/x-unit-siegbahn | 1,722,903,964,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640461318.24/warc/CC-MAIN-20240806001923-20240806031923-00189.warc.gz | 11,313,270 | 7,527 | # parsec to x unit; siegbahn conversion
Conversion number between parsec [pc] and x unit; siegbahn [xu] is 3.0792112374015 × 10+29. This means, that parsec is bigger unit than x unit; siegbahn.
### Contents [show][hide]
Switch to reverse conversion:
from x unit; siegbahn to parsec conversion
### Enter the number in parsec:
Decimal Fraction Exponential Expression
[pc]
eg.: 10.12345 or 1.123e5
Result in x unit; siegbahn
?
precision 0 1 2 3 4 5 6 7 8 9 [info] Decimal: Exponential:
### Calculation process of conversion value
• 1 parsec = (3.085677581*10^16) / (1.0021*10^-13) = 3.0792112374015 × 10+29 x unit; siegbahn
• 1 x unit; siegbahn = (1.0021*10^-13) / (3.085677581*10^16) = 3.2475849264693 × 10-30 parsec
• ? parsec × (3.085677581*10^16 ("m"/"parsec")) / (1.0021*10^-13 ("m"/"x unit; siegbahn")) = ? x unit; siegbahn
### High precision conversion
If conversion between parsec to metre and metre to x unit; siegbahn is exactly definied, high precision conversion from parsec to x unit; siegbahn is enabled.
Since definition contain rounded number(s) too, there is no sense for high precision calculation, but if you want, you can enable it. Keep in mind, that converted number will be inaccurate due this rounding error!
### parsec to x unit; siegbahn conversion chart
Start value: [parsec] Step size [parsec] How many lines? (max 100)
visual:
parsecx unit; siegbahn
00
103.0792112374015 × 10+30
206.1584224748029 × 10+30
309.2376337122044 × 10+30
401.2316844949606 × 10+31
501.5396056187007 × 10+31
601.8475267424409 × 10+31
702.155447866181 × 10+31
802.4633689899212 × 10+31
902.7712901136613 × 10+31
1003.0792112374015 × 10+31
1103.3871323611416 × 10+31
Copy to Excel
## Multiple conversion
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## Details about parsec and x unit; siegbahn units:
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### parsec
Definition of parsec unit: ≈ 3.085677581x1016. Distance of star with parallax shift of one arc second from a base of one astronomical unit
Convert X unit; siegbahn to other unit:
### x unit; siegbahn
Definition of x unit; siegbahn unit: ≈ 1.0021x10-13.
← Back to Length units | 739 | 2,232 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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} | 3.453125 | 3 | CC-MAIN-2024-33 | latest | en | 0.453277 |
https://www.physicsforums.com/threads/long-jump-question.347454/ | 1,508,670,160,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187825174.90/warc/CC-MAIN-20171022094207-20171022114207-00120.warc.gz | 956,155,635 | 16,555 | # Long Jump Question
1. Oct 20, 2009
### SirMarksAlot
1. The problem statement, all variables and given/known data
A runner jumps at 30 degrees to the ground and covers 8.90 m. What was the takeoff speed?
3. The attempt at a solution
I have no idea where to start...
2. Oct 20, 2009
### rl.bhat
Resolve the velocity in to vertical and horizontal components.
What happens to these components with respect to time?
3. Oct 20, 2009
### SirMarksAlot
how would i go about solving the velocity for each component?
4. Oct 20, 2009
### rl.bhat
If v is the velocity what are the vertical and horizontal components?
5. Oct 20, 2009
### SirMarksAlot
honestly, i have no idea
6. Oct 20, 2009
### rl.bhat
Open any text book. And refer the motion in two dimension.
7. Oct 20, 2009
### SirMarksAlot
but all i have is the angle of the jump and the distance...no time or anything
8. Oct 20, 2009
### rl.bhat
Using proper equation you can find the remaining unknown quantities. Try to find out the equations.
9. Oct 20, 2009
### SirMarksAlot
do u mean these formulas?
Fx = cos A * F1
Fy = sin A * F1
10. Oct 20, 2009
### SirMarksAlot
honestly, im struggling and have no idea what to do...
11. Oct 20, 2009
### rl.bhat
Yes. These are the two components. Only replace F by v for the velocity.
In that x component remains constant, because there is no acceleration in that direction. If t is the time of flight, then x = vcosθ*t.
In the vertical direction, the initial velocity is vsinθ. What is the final velocity when it reaches the maximum height. And what is the time taken to reach the maximum height?
12. Oct 20, 2009
### SirMarksAlot
when u say x = vcosθ*t , what is v? the speed before the jump?
Last edited: Oct 20, 2009
13. Oct 20, 2009
### rl.bhat
No. You have to find out. I have told you to write down all the kinematic equations. Then I will tell you which equation to be used to find the time.
14. Oct 20, 2009
### SirMarksAlot
oh wouldnt the final velocity be 0 when it reaches a maximum height? because he would have stopped moving up?
15. Oct 20, 2009
### rl.bhat
Yes.
16. Oct 20, 2009
### SirMarksAlot
I dont know how to find the time it would take to reach the max height.
Last edited: Oct 20, 2009
17. Oct 20, 2009
### rl.bhat
Why can't you write down all the relevant kinematic equation. In the post submitted to PF, this is the second requirement.. You have not submitted that. Do it first now.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook | 727 | 2,551 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2017-43 | longest | en | 0.9013 |
https://studylib.net/doc/13154873/components-of-a-computer-p-i-t-ti | 1,534,569,177,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221213286.48/warc/CC-MAIN-20180818040558-20180818060558-00360.warc.gz | 775,075,401 | 16,395 | ```Components
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A ith
Arithmetic
ti Operations
O
ti
Operand is a quantity on which an operation
is performed
How many operands in this instruction?
All arithmetic operations have this form
Chapter 2 — Instructions: Language of the Computer — 7
Chapter 2 — Instructions: Language of the Computer — 8
§2.2 Op
perationss of the C
Computeer Hardw
ware
Add and subtract, three operands
§2.2 Op
perationss of the C
Computeer Hardw
ware
A ith
Arithmetic
ti Operations
O
ti
6
What if we want to add b,c,d, and e and put the
result into a ?
All arithmetic operations have this form
Design Principle 1: Simplicity favors regularity
Regularity makes hardware implementation
simpler
Chapter 2 — Instructions: Language of the Computer —
10
Chapter 2 — Instructions: Language of the Computer — 9
A ith
Arithmetic
ti Example
E
l
C code:
A ith
Arithmetic
ti Example
E
l
C code:
f = (g + h) - (i + j);
Compiled MIPS code:
D i Principle
Design
P i ipl 1
?
f = (g + h) - (i + j);
Compiled MIPS code:
dd t0,
0 g, h
sub f, t0, t1
Chapter 2 — Instructions: Language of the Computer —
11
Chapter 2 — Instructions: Language of the Computer —
12
# temp t0
0 = g + h
# temp t1 = i + j
# f = t0 - t1
§2.2 Op
perationss of the C
Computeer Hardw
ware
All arithmetic operations have this form
§2.2 Op
perationss of the C
Computeer Hardw
ware
A ith
Arithmetic
ti Operations
O
ti
Registers
g
are pprimitives of hardware design
g and are
visible to programmers
Chapter 2 — Instructions: Language of the Computer —
13
R i t O
Register
Operand
d Example
E
l
Compiler’s job to associate variables of a highlevel program with registers
C code:
f = (g + h) - (i + j);
f, …, j in \$s0, …, \$s4
R i t Operands
Register
O
d
Assembler names
\$t0, \$t1, …, \$t9 for temporary values
\$s0,
\$s0 \$s1,
\$s1 …, \$s7 for saved variables
Chapter 2 — Instructions: Language of the Computer —
14
R i t O
Register
Operand
d Example
E
l
Compiled MIPS code:
\$t0, \$s1,
\$s1 \$s2
sub \$s0,
\$s0 \$t0,
\$t0 \$t1
Compiled MIPS code ?
Chapter 2 — Instructions: Language of the Computer —
15
Chapter 2 — Instructions: Language of the Computer —
16
§2.3 Op
perands o
of the Co
omputer Hardware
The operands of arithmetic instructions must be
from special location in hardware called registers
§2.3 Op
perands o
of the Co
omputer Hardware
R i t Operands
Register
O
d
Numbered 0 to 31
32-bit data called a “word”
Word is the natural unit of access, typically
yp
y 32 bits,
corresponds to the size of a register in MIPS
D i Principle
Design
P i ipl 2
Smaller is faster
Larger registers will increase clock cycle time --- electronic
signals takes longer when they travel farther
Design principles are not hard truths but general guidelines
31 registers instead of 32 need not make MIPS faster
There may be only 3 operands and they must be
chosen from one of the 32 registers
registers. Why only 32 ?
Chapter 2 — Instructions: Language of the Computer —
17
Chapter 2 — Instructions: Language of the Computer —
18
M
Memory
Operands
O
d
Programming languages, C, Java, …
Allow complex data structures like arrays and structures
They
Th often
f
contain many more data
d elements
l
than
h the
h number
b
of registers in a computer
Where are they
y stored ?
• Memory
But, arithmetic operations are applied on register
operands
Hence, data transfer instructions are required to transfer
data from memory to registers
Load values from memory into registers
Store result from register to memory
Chapter 2 — Instructions: Language of the Computer —
19
Memory is like an array
Data
D t transfer
t
f iinstructions
t ti
mustt supply
l the
dd
(index/offset) of the memory (array)
20
§2.3 Op
perands o
of the Co
omputer Hardware
MIPS has a 32 × 32-bit register file
§2.3 Op
perands o
of the Co
omputer Hardware
R i t Operands
Register
O
d
M
Memory
Operands
O
d
Each address identifies an 8-bit byte
y
Words are aligned in memory
Each word is 32 bits or 4 bytes
y
To locate words, addresses are in multiples of 4
A is an array of words
What is the offset to locate A[8] ?
A[0] – 0
A[1] – 4
A[2]– 8
…
Chapter 2 — Instructions: Language of the Computer —
21
M
Memory
Operands
O
d
Why is memory not word-addressable?
A[8] – 32
M
Memory
Operands
O
d
Why is memory not word-addressable?
Bytes are useful in many programs. In a word addressable system, it is
necessary first to compute the address of the word containing the byte,
fetch that word
word, and then extract the byte from the two
two-byte
byte word.
word
Although the processes for byte extraction are well understood, they are
less efficient than directly accessing the byte. For this reason, many
Chapter 2 — Instructions: Language of the Computer —
23
Chapter 2 — Instructions: Language of the Computer —
24
22
M
Memory
Operands
O
d
M
Memory
Operand
O
d Example
E
l 1
C code:
lw refers to load word
g = h + A[8];
g in \$s1
h in \$s2
base address of A in \$s3
A is an array of 100 words
lw \$t0 , 8 (\$s3)
Compiled MIPS code ?
offset
base register
25
M
Memory
Operand
O
d Example
E
l 1
C code:
Compiled MIPS code:
lw \$t0
\$t0, 32(\$s3)
Chapter 2 — Instructions: Language of the Computer —
27
M
Memory
Operand
O
d Example
E
l 2
C code:
g = h + A[8];
[ ]
g in \$s1, h in \$s2, base address of A in \$s3
offset
Chapter 2 — Instructions: Language of the Computer —
26
A[12]
[ ] = h + A[8];
[ ]
h in \$s2, base address of A in \$s3
Compiled MIPS code:
base register
Chapter 2 — Instructions: Language of the Computer —
28
M
Memory
Operand
O
d Example
E
l 2
C code:
A[12]
[ ] = h + A[8];
[ ]
h in \$s2, base address of A in \$s3
Compiled MIPS code:
Index 8 requires offset of 32
lw \$t0
\$t0, 32(\$s3)
sw \$t0,
\$t0 48(\$s3)
48(\$ 3)
# store
t
word
d
Chapter 2 — Instructions: Language of the Computer —
29
R i t
Registers
vs. Memory
M
Registers are faster to access than memory
Operating
O
i on memory d
data requires
i
l d and
d
stores
More instructions to be executed
Compiler must use registers for variables as
much as possible
Only spill to memory for less frequently used
variables
Register optimization is important!
Chapter 2 — Instructions: Language of the Computer —
30
I
Immediate
di t Operands
O
d
Constant data specified in an instruction
ddi \$s3,
\$
\$s3,
\$
4
No subtract immediate instruction
D i Principle3
Design
P i ipl 3
Make the common case fast
Small constants are common
Immediate operand avoids a load instruction
Just use a negative constant
Chapter 2 — Instructions: Language of the Computer —
31
Chapter 2 — Instructions: Language of the Computer —
32
Th Constant
The
C
t tZ
Zero
MIPS register 0 (\$zero) is the constant 0
Stored-program concept
Cannot be overwritten
The idea that instructions and data of many types
can be stored in memoryy as numbers, leadingg to
stored-program computer
Useful for common operations
E.g., move between registers
Chapter 2 — Instructions: Language of the Computer —
33
Instructions are encoded in binary
Called machine code
MIPS instructions
Encoded
E d d as 32-bit
32 bi iinstruction
i words
d
Small number of formats encoding operation code
((opcode),
d ) register
i t numbers,
b
…
Regularity!
Register numbers
\$t0 – \$t7 are reg’s 8 – 15
\$s0 – \$s7 are reg’s 16 – 23
Chapter 2 — Instructions: Language of the Computer —
35
§2.5 Reepresentiing Instru
uctions in
n the Computer
R
Representing
ti Instructions
I t
ti
34
E
Example
l
special
\$s1
\$s2
\$t0
0
0
17
18
8
0
32
000000
10001
10010
01000
00000
100000
000000100011001001000000001000002
op
rs
rt
rd
shamt
funct
6 bits
5 bits
5 bits
5 bits
5 bits
6 bits
Chapter 2 — Instructions: Language of the Computer —
36
R
Representing
ti Instructions
I t
ti
I t
Instruction
ti types
t
The layout or the form of representation of
instruction is composed of fields of binary numbers
R format (for register)
The numeric version of instructions is called machine
language and a sequence of such instructions is called
machine
h code
d
I-format (for immediate)
Immediate
Data transfer
37
MIPS R-format
Rf
t Instructions
I t
ti
38
MIPS I-format
If
t Instructions
I t
ti
op
rs
rt
rd
shamt
funct
op
rs
rt
6 bits
5 bits
5 bits
5 bits
5 bits
6 bits
6 bits
5 bits
5 bits
16 bits
Instruction fields
op: operation
ti code
d (opcode)
(
d )
rs: first source register
g
number
rt: second source register number
Immediate arithmetic and load/store instructions
rt: destination or source register number
Constant: –215 to +215 – 1
rd:
d destination
d i i register
i
number
b
shamt: shift amount ((00000 for now))
funct: function code (extends opcode)
Chapter 2 — Instructions: Language of the Computer —
39
Chapter 2 — Instructions: Language of the Computer —
40
D i Principle
Design
P i i l 4
H
d i l
Keep all instructions of the same format and length
But it makes difficult to address large memories
Compromise and allow different formats
Reading binary numbers are tedious
So, hexadecimal representation is popular
Base 16 is power of two and hence it is easy
to replace each group of 4 binary digits
Good design demands good compromises
Different formats complicate decoding,
decoding but allow 32-bit
32 bit
instructions uniformly
Keep formats as similar as possible
See example in page 98
41
Base 16
H
d i l
42
St
Stored
dP
Program Computers
C
t
Instructions represented
p
in
binary, just like data
Instructions and data stored in
memory
Programs can operate on
programs
Compact representation of bit strings
4 bits
bit per hex
h di
digit
it
0
1
2
0000
0001
0010
4
5
6
0100
0101
0110
8
9
a
1000
1001
1010
c
d
e
1100
1101
1110
3
0011
00
7
0111
0
b
1011
0
f
1111
Example: eca8 6420 ?
p
Example: 0001 0011 0101 0111 1001 1011 1101 1111
e.g., compilers,
il
k
…
Binary compatibility allows
compiled programs to work on
different computers
Standardized ISAs
Chapter 2 — Instructions: Language of the Computer —
44
CISC A
Approach
h
CISC vs RISC
Complex Instruction Set Computer
C code:
d
C code:
g = h + A[8];
[ ]
g = h + A[8];
[ ];
CISC
CISC
dd a,32<b>
32<b>
Compiled MIPS code:
Achieved byy buildingg complex
p hardware that
loads value from memory into a register and
then adds it to register a and stores the
results in register a
lw \$t0
\$t0, 32(\$s3)
45
CISC vs RISC
Chapter 2 — Instructions: Language of the Computer —
46
t
Compiler has to do little
CPU Time Instruction Count CPI Clock Cycle Time
Programming
P
i was done
d
in
i assembly
bl language
l
To make it easy, more and more complex
Length of the code is short and hence, little
memory is required to store the code
Memory
M
was a very costly
l real-estate
l
E.g.,
g , Intel x86 machines ppoweringg several
million desktops
47
48
t
R dbl k
Each instruction needs only one clock cycle
Hardware is less complex
RISC processors, despite their advantages,
tookk severall years to gain
i market
k
Intel had a head start of 2 years before its RISC
competitors
Customers
C t
were unwilling
illi tto ttake
k risk
i k with
ith new
technology and change software products
They have a large market and hence, they can
afford resources to overcome complexity
49
Powerful instruction higher performance
Fewer
F
iinstructions
t ti
required
i d
But complex instructions are hard to implement
• May slow down all instructions, including simple ones
§2.18 Fallacies aand Pitfaalls
F ll i
Fallacies
50
F ll i
Fallacies
Backward compatibility instruction set
doesn’t change
But they do accrete more instructions
Compilers are good at making fast code from simple
instructions
Use assemblyy code for high
g performance
p
x86 instruction set
But modern compilers are better at dealing with modern
processors
More lines of code more errors and less productivity
Chapter 2 — Instructions: Language of the Computer —
51
Chapter 2 — Instructions: Language of the Computer —
52
Pitf ll
Pitfalls
CISC vs RISC
Sequential words are not at sequential
dd
Very good slides on the topic
https://www.cs.drexel.edu/~wmm24/cs281/lecture
s/pdf/RISCvsCISC.pdf
p
p
Increment by 4, not by 1!
Chapter 2 — Instructions: Language of the Computer —
53
54
Ad i i t ti Details
D t il
Patterson’s blog:
1.1, 1.2, 1.3, and 1.4
http://blogs.arm.com/software-enablement/375-
Note the correction – 1.4 was not listed in
risc-versus-cisc-wars-in-the-prepc-and-pc-erasp p
p
part-1/
part of covered material
lecture 1 but it is p
Interview with Hennessy
2.1,
2 1 2.2,
2 2 2.3
2 3 , 2.5,
2 5 2.18,
2 18 2.19
2 19
http://www-csp
aculty.stanford.edu/~eroberts/courses/soco/projec
An experiment
p
unmesh_bordoloi
55
56
56
Design principles
1.Simplicity
1
Si li it ffavors regularity
l it
2.Smaller is faster
3.Make the common case fast
4 Good design demands good compromises
4.Good
Layers of software/hardware
Compiler, assembler, hardware
MIPS: typical of RISC ISAs
c.f. x86
Chapter 2 — Instructions: Language of the Computer —
57
§2.19 C
Concludin
ng Remarks
C
Concluding
l di Remarks
R
k
``` | 4,571 | 14,217 | {"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-2018-34 | latest | en | 0.833553 |
https://forums.eviews.com/viewtopic.php?f=10&p=69170&sid=31343edd00f9ab4319016e38001b6bde | 1,660,828,576,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882573197.34/warc/CC-MAIN-20220818124424-20220818154424-00760.warc.gz | 264,723,688 | 6,364 | ## Solve Control for Target
For technical support, tips and tricks, suggestions, or any other information regarding the EViews model object.
Moderators: EViews Gareth, EViews Moderator
Justinn
Posts: 2
Joined: Wed May 11, 2022 7:04 pm
### Solve Control for Target
I am interested in the Solve Control for Target procedure. However, I've never had luck navigating the error messages and can tell I'm doing something wrong. Is there an example workfile anywhere that I can use just so I can better understand and implement the procedure? Such as the workfile for this- https://eviews.com/EViews12/ev12models_n.html
If not, would someone be able to simulate some data with a working model that successfully demonstrates the procedure? I would be really grateful if so.
Regards,
Justin
EViews Matt
EViews Developer
Posts: 528
Joined: Thu Apr 25, 2013 7:48 pm
### Re: Solve Control for Target
Hello,
Here's a trivial artificial model with an example of using the control procedure over a single variable.
Code: Select all
`wfcreate u 100series inA = 1 + @abs(@sin(@obsid / 25)) / 2 + rnd / 8series inB = rnd * @sin(@obsid / 15)series outmodel mm.append out = inA + inBm.solveseries out_target = out_0smpl 81 100out_target = out_0 * (1 + (@obsid - 80) / 40)smpl @allshow mm.control(create) inA out out_target`
If you share the error message you're encountering I may be able shed some light on what can be done to resolve them. There's quite a variety of errors that could occur, from structural mistakes (e.g. trying to use an exogenous variable to control an endogenous variable that it doesn't influence) to numerical issues (e.g. failing to find control values that satisfy the target series).
Justinn
Posts: 2
Joined: Wed May 11, 2022 7:04 pm
### Re: Solve Control for Target
Matt, thank you very much for the response! This illuminates the procedure for me now that I can follow a worked example.
I tend to encounter this error, "An equal number of control, target, and trajectory variables must be specified." I think part of that is I wasn't understanding the concept of the trajectory variable.
Thanks again,
Justin | 534 | 2,137 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2022-33 | latest | en | 0.8891 |
https://math.stackexchange.com/questions/1498205/find-matrix-of-bilinear-form-on-lie-algebra | 1,560,933,834,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627998943.53/warc/CC-MAIN-20190619083757-20190619105757-00184.warc.gz | 519,071,899 | 34,870 | # Find matrix of bilinear form on Lie algebra?
I have a bilinear form $$\sigma_V:L\times L\rightarrow \mathbb{k}$$ $$\sigma_V(x,y)=tr(\rho_V(x)\rho_V(y)) \forall x,y \in L$$ and am looking for the matrix of this form. I am in the algebra $$L= \mathfrak{sl} _2(\mathbb{k})$$ in the standard basis $(x,y,h)$ and the representation $V=\mathbb{k}^2$ which is the standard representation of $\mathfrak{sl}_2(\mathbb{k})$. I know $\sigma_V$ is symmetric and L-invariant but am not sure how to go about finding its matrix? I found lots of information on the Killing form but as $V$ is not the adjoint representation it seems this is not the Killing form. Am I wrong? How do I go about finding the matrix?
• The standard representation can be regarded as a map $L \to \mathfrak{gl}(V)$, and so we can view its images as endomorphisms of $V$. – Travis Oct 26 '15 at 9:54
For a simple Lie algebra, any two non degenerate invariant bilinear forms are multiples of each other. Indeed, the Killing form is just $4$ times the trace form for $\mathfrak{sl}_2(K)$. In general we have $tr(ad(x)ad(y))=2n \cdot tr(x)tr(y)$ for the Lie algebra $\mathfrak{sl}_n(K)$. So the matrix of $\sigma_V$ above is just the matrix of the Killingform multiplied by $\frac{1}{4}$. This can be verified directly, of course, with $\rho(x)=\begin{pmatrix} 0 & 1 \cr 0 & 0 \end{pmatrix}$, $\rho(y)=\begin{pmatrix} 0 & 0 \cr 1 & 0 \end{pmatrix}$ and $\rho(h)=\begin{pmatrix} 1 & 0 \cr 0 & -1 \end{pmatrix}$. For example, $\sigma_V(x,x)=tr(\rho(x)\rho(x))=tr(0)=0$, $\sigma_V(x,y)=tr(\rho(x)\rho(y))=tr(diag(1,0))=1$, etc. | 516 | 1,586 | {"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} | 3.296875 | 3 | CC-MAIN-2019-26 | latest | en | 0.794031 |
http://www.physicsforums.com/showthread.php?p=3817432 | 1,411,149,167,000,000,000 | text/html | crawl-data/CC-MAIN-2014-41/segments/1410657131846.69/warc/CC-MAIN-20140914011211-00335-ip-10-196-40-205.us-west-1.compute.internal.warc.gz | 737,558,570 | 7,521 | # Limit of (2n)!/(4^n (n!)^2)
by looserlama
Tags: convergence, hard, limit
P: 30 Yea I've been working on other stuff but I've been following the conversation. So this is where I'm at now: We've never seen stirling's approximation before so I don't think we're supposed to use it. So I've just been focusing on the Squeeze theorem way: It's easy to show (2n)!/4n((2n)!)2≤ (2n)!/4n(n!)2 But finding an upper bounding sequence that goes to zero is the hard part. What I'm thinking now is, (2n)! = 2n(n!) [(2n-1)(2n-3)...(3)(1)] The second part (stuff in square brackets) is ≤ 2n(n!) So we have (2n)! ≤ 2n(n!)2n(n!) = 4n(n!)2 but if we try to makes this an upper bounding sequence (ie, (2n)!/4n(n!)2 ≤ 4n(n!)2/4n(n!)2= 1) then it doesn't converge to 0. So I wanted to show (2n-1)(2n-3)...(3)(1) ≤ 2n(n-1)!, I know this is true but I don't know how to prove it. Cause if we have that, then (2n)!/4n(n!)2 ≤ 4n(n!)(n-1)!/4n(n!)2= 1/n which goes to zero. I just don't know how to prove that?
P: 297 Its simple. Expand (2n)! As (2^n)*(n!)*{(2n-1)(2n-3)(2n-5).......5*3*1)} Now divide this by 4^n*(n!)^2 What do you get?? Its (2n-1)*(2n-3)*(2n-5)......3*1/((2^n)(n!)) There are n terms in numerator .Divide each by 2 to get rid of 2^n What do you see?
P: 297 Also (2n-1)(2n-3)...(3)(1) ≤ 2^n*(n-1)! Is not true. The LHS has n terms giving rise to n^n and RHS has only n-1 terms. We know that when limit temds to infinity only the highest power factor matters. So LHS will be greater
P: 30
Quote by emailanmol Also (2n-1)(2n-3)...(3)(1) ≤ 2^n*(n-1)! Is not true. The LHS has n terms giving rise to n^n and RHS has only n-1 terms. We know that when limit temds to infinity only the highest power factor matters. So LHS will be greater
I'm pretty sure that's wrong, just from me computing 2n(n-1)! - $\frac{n!}{n!!}$ = 2n(n-1)! - (2n-1)(2n-3)...(3)(1) with vary large numbers. For all of them it was always positive and it was increasing as n got larger.
Also, if we try dividing (2n-1)(2n-3)...(3)(1) by 2n, so dividing each term by 2, gives (n-1/2)(n-3/2)...(3/2)(1/2).
So (2n-1)(2n-3)...(3)(1)/2n(n!) = (n-1/2)(n-3/2)...(3/2)(1/2)/n!
Here the top one is obviously smaller than the bottom, but how do we prove that it goes to zero with this?
P: 297 (2n-1)(2n-3)...(3)(1) ≤ 2^n*(n-1)! Is not true for n -> infinity. If u have doubts regarding this you can always plug in the expression into wolfram alpha. For your second part , each factor of numerator is less than the corresponding denominator. The maximum value for each expression could be 0.999999....9999 (i.e just less than 1) That gives u something like (0.999....999)^t where t tends to infinity.(How?) And we know that will be 0.(Why?) (Remember you cannot compare coefficients of highest factor in this question to get value of limits as the highest power on both numerator and denominator is n^n. That rule is valid only for n^a where n is a fixed constant) That rule only applies while
Related Discussions Calculus & Beyond Homework 1 Astronomy & Astrophysics 11 Calculus & Beyond Homework 1 Calculus & Beyond Homework 3 | 1,045 | 3,084 | {"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} | 4.25 | 4 | CC-MAIN-2014-41 | latest | en | 0.922272 |
https://www.physicsforums.com/threads/snells-law.467211/ | 1,590,970,734,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347413786.46/warc/CC-MAIN-20200531213917-20200601003917-00274.warc.gz | 868,539,167 | 17,431 | # Snells law
## Homework Statement
Find the index of refraction n of the rectangle using x, d, and θ.
http://img338.imageshack.us/img338/6079/snellslaw.jpg [Broken]
## Homework Equations
n1*sin(θ1)=n2*sin(θ2)
## The Attempt at a Solution
Is it possible to solve this problem given the relevant data? What I did was try to find θ2 in terms of x, d, and θ (now called θ1). What I came up with was:
$$cos(\theta_{2})=\frac{x*sin(\theta_{1}-\theta_{2})}{d}$$
From the two right triangles that can be constructed:
http://img11.imageshack.us/img11/6606/snellslaw2.jpg [Broken]
But either this isn't right or I don't see a simple way to proceed from here. I used wolfram to solve the above equation for θ2, but the result seems too complicated to be the answer. Can someone help, anyone with a simpler approach? Perhaps n1*sin(θ1)=n2*sin(θ2) is useless in this case?
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Related Introductory Physics Homework Help News on Phys.org
ehild
Homework Helper
You do have to use Snell's law to get the relation among x, d, and θ1. Expand sin(θ1-θ2). Find sin(θ2). Write both sin(θ2) and cos(θ2) in terms of sin(θ1). Use them in the expression for d.
ehild
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What do you mean "Find sin(θ2)." Do you mean isolate it from sin(θ1-θ2)=sin(θ1)*cos(θ2)-cos(θ1)*sin(θ2)=d/a ?
ehild
Homework Helper
What do you mean "Find sin(θ2)." Do you mean isolate it from sin(θ1-θ2)=sin(θ1)*cos(θ2)-cos(θ1)*sin(θ2)=d/a ?
Find sin(θ2) from Snells law in terms of sin(θ1) and the refractive index of the slab relative to the surrounding medium.
ehild
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Ok, so here are the equations I'm using:
$$(1)n_{1}sin(\theta_{1})=n_{2}sin(\theta_{2})$$
$$(2)cos(\theta_{2})=\frac{x}{a}$$
$$(3)sin(\theta_{1}-\theta_{2})=\frac{d}{a}$$
I find $$sin(\theta_{2})$$ from (1):
$$\frac{n_{1}}{n_{2}}sin(\theta_{1})=sin(\theta_{2})$$
I expand (2):
$$sin(\theta_{1})cos(\theta_{2})-cos(\theta_{1})sin(\theta_{2})=\frac{d}{a}$$
Plug (1) and (3) into (2):
$$sin(\theta_{1})\frac{x}{a}-cos(\theta_{1})\frac{n_{1}}{n_{2}}sin(\theta_{1})=\frac{d}{a}$$
But..."a" is an unknown, I can't solve for n2 until a is eliminated. Am I doing something wrong?
(n1 is assumed to be 1)
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collinsmark
Homework Helper
Gold Member
Hello CINA,
Just out of curiosity, is there anything left out of the problem statement such as θ being small (perhaps less than π/18 [which corresponds to around 10o)?
If so, approximations exist that would make solving for an approximate answer much easier.
For small θ (where θ is expressed in radians),
sinθθ
cosθ ≈ 1
I'm not sure if assuming a small θ applies to this particular problem though.
ehild
Homework Helper
But..."a" is an unknown, I can't solve for n2 until a is eliminated. Am I doing something wrong?
(n1 is assumed to be 1)
cos(θ2)=x/a . Write a in terms of x and cos(θ2) and use cos(θ2)=√(1-sin2(θ2))
ehild
ehild
Homework Helper
Hello CINA,
Just out of curiosity, is there anything left out of the problem statement such as θ being small (perhaps less than π/18 [which corresponds to around 10o)?
Hello, Collinsmark,
There is no indication in the problem that θ1 is small. As the refractive index of the slab is to be obtained, θ1 and d has to be accurately measured, so they can not be two small. The problem can be solved without assuming small angle of incidence.
ehild | 1,035 | 3,341 | {"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} | 3.828125 | 4 | CC-MAIN-2020-24 | longest | en | 0.839406 |
https://se.mathworks.com/help/sps/ug/current-transformer-saturation.html | 1,718,965,247,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862070.5/warc/CC-MAIN-20240621093735-20240621123735-00843.warc.gz | 465,739,375 | 17,762 | # Current Transformer Saturation
This example shows the measurement distortion due to saturation of a current transformer (CT).
G. Sybille (Hydro-Quebec)
### Description
A current transformer (CT) is used to measure current in a shunt inductor connected on a 120 kV network. The CT is rated 2000 A / 5 A, 5 VA. The primary winding which consists of a single turn passing through the CT toroidal core is connected in series with the shunt inductor rated 69.3 Mvar, 69.3 kV (120kV/sqrt(3)), 1 kA rms. The secondary winding consisting of 1*2000/5 = 400 turns is short circuited through a 1 ohm load resistance. A voltage sensor connected at the secondary reads a voltage which should be proportional to the primary current. In steady state, the current flowing in the secondary is 1000*5/2000 = 2.5 A (2.5 Vrms or 3.54 Vpeak read by the voltage measurement block V2).
Open the CT dialog box and observe how the CT parameters are specified. The CT is assumed to saturate at 10 pu and a simple 2 segment saturation characteristic is used.
The primary current reflected on the secondary and the voltage developed across the 1 ohm resistance are sent to trace 1 of the Scope block. The CT flux , measured by the Multimeter block is converted in pu and sent to trace 2. (1 pu flux = 0.0125 V *sqrt(2)/ (2*pi*50) = 5.63e-5 V.s)
The switch connected in series with the CT secondary is normally closed. This switch will be used later to illustrate overvoltages produced when CT secondary is left open.
### Simulation
1. Normal operation
In this test , the breaker is closed at a peak of source voltage (t = 1.25 cycle). This switching produces no current asymmetry. Start the simulation and observe the CT primary current and secondary voltage (first trace of Scope block). As expected the CT current and voltage are sinusoidal and the measurement error due to CT resistance and leakage reactances is not significant. The flux contains a DC component but it stays lower than the 10 pu saturation value.
2. CT saturation due to current asymmetry
Now, change the breaker closing time in order to close at a voltage zero crossing. Use t = 1/50 s. This switching instant will now produce full current asymmetry in the shunt reactor. Restart the simulation. Observe that for the first 3 cycles, the flux stays lower than the saturation knee point (10 pu). The CT voltage output V2 then follows the primary current. However, after 3 cycles, the flux asymmetry produced by the primary current causes CT saturation, thus producing large distortion of CT secondary voltage.
3. Overvoltage due to CT secondary opening
Reprogram the primary breaker closing time at t = 1.25/50 s (no flux asymmetry) and change the secondary switch opening time to t = 0.1 s. Restart the simulation and observe the large overvoltage produced when the CT secondary is opened. The flux has a square waveshape chopped at +10 and -10 pu. Large dphi/dt produced at flux inversion generates high voltage spikes (250 V). | 707 | 2,987 | {"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-2024-26 | latest | en | 0.896795 |
https://www.veritasprep.com/blog/2013/04/4-tips-for-difficult-word-problems-on-the-gmat/ | 1,585,903,464,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370510352.43/warc/CC-MAIN-20200403061648-20200403091648-00542.warc.gz | 1,167,472,590 | 76,980 | # 4 Tips for Difficult Word Problems on the GMAT
Word problems on the GMAT often do not require particularly difficult algebra to solve. Most of the time, solving simple linear equations or using a formula is all that is required. The key is to not be intimidated by the length of the description and to be able to pull out the relevant information and set up the correct equation/s. These 4 tips will help you tackle any Problem Solving question that is significantly wordy.
1. Write down what the question is asking. The final sentence of a word problem will tell you what the answer choices represent. Do you need to find a certain variable’s value? The probability of an event? An expression representing a relationship?
2. Translate the Math to English.Go through each sentences one-by-one. Write down any important variables and numbers on your scratch pad, and give them labels (i.e. x = # of tickets bought by adults, 1100 = total employees in a factory). Use these common translations to help:
• ADDITION: increased by ; more than ; combined ; together ; total of ; sum ; added to ; and ; plus
• SUBTRACTION: decreased by ; minus ; less ; difference between/of ; less than ; fewer than ; minus ; subtracted from
• MULTIPLICATION: of ; times ; multiplied by ; product of ; increased/decreased by a factor of
• DIVISION: per ; out of ; ratio of ; quotient of ; percent (divide by 100) ; divided by ; each
• EQUALS: is ; are ; was ; were ; will be ; gives ; yields ; sold for ; has ; costs ; adds up to ; the same as ; as much as
• VARIABLE or VALUE: a number ; how much ; how many ; what
3. Clock the answer choices. One of them must be correct. Are they numbers or variables? If they are numbers – a common way to solve word problems is to work backwards, or backsolve. This is a great way to approach an especially complicated question where setting up the algebra would take a lot of time. If there are variables, can you pick numbers? For percent questions, or questions with unknown starting values, always pick 100.
4. Identify the concept. If you can look past the immediate question for a moment, and identify the overall concept tested, you will be better able to remember the relevant formulas and the steps needed to solve. Common concepts associated with word problems include: rates & speed, averages, ratios and proportions, solving sets of equations, probability, permutations and combinations, and percents.
As you study for the GMAT, try to label each word problem with a “concept” – you will notice how the concepts will repeat!
Plan on taking the GMAT soon? We have GMAT prep courses starting all the time. And, be sure to find us on Facebook and Google+, and follow us on Twitter!
Vivian Kerr is a regular contributor to the Veritas Prep blog, providing tips and tricks to help students better prepare for the GMAT and the SAT. | 653 | 2,896 | {"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} | 4.125 | 4 | CC-MAIN-2020-16 | latest | en | 0.939082 |
http://math.stackexchange.com/questions/236945/warshalls-algorithm-multiple-choice | 1,467,392,643,000,000,000 | text/html | crawl-data/CC-MAIN-2016-26/segments/1466783403508.34/warc/CC-MAIN-20160624155003-00118-ip-10-164-35-72.ec2.internal.warc.gz | 195,051,089 | 18,260 | # Warshall's algorithm multiple choice…
Here's a question given to us for practice.
Can anyone help me through the steps of solving it? The algorithm itself is confusing to read, so I'm just looking for a concise way to calculate $W_1$, $W_2$, etc. Are there any general rules that I should be aware of?
-
To eliminate the suspense, the answer is (B). How did I find it?
Warshall's algorithm produces what is known as the transitive closure of a relation, $R$ which is the smallest transitive relation $R^+$ which contains $R$. In functional terms, we start with $R$ and add just enough new elements to $R$ to make a transitive relation.
It's conceptually easier if we express $R$ as directed graph $G$ on $n$ vertices, as your problem does, and think of adding a directed edge between vertices $i$ and $j$ whenever we can get from $i$ to $j$ by passing through another vertex $k$. In other words, if we can get from $i$ to $j$ via $k$, we'll add a new edge bypassing $k$. We do this by computing a sequence of $n\times n$ matrices $W_0, W_1, \dots W_n$, where the final result, $W_n$, has entries
$$W_n[i, j] = \begin{cases} 1 & \text{if there is some directed path in the graph from i to j} \\ 0 & \text{if there is no way to get from vertex i to vertex j} \end{cases}$$
The algorithm is pretty simple; it computes $W_k$ from $W_{k-1}$, by doing the following:
for each pair of vertices (i, j)
set W_k[i, j] to W_{k-1}[i, j] OR (W_{k-1}[i, k] AND W_{k-1}[k, j])
At the start, $W_0$ is just the adjacency matrix of the graph, so for every pair of vertices $(i, j), W_0[i, j]$ is $1$ if there is a directed edge from $i$ to $j$ and is zero if there is no edge from $i$ to $j$. Initially, select an order on the vertices, $v_1, v_2, \dots , v_n$. In your problem the order specified was $v_1 = a, v_2=b, v_3=c, v_4=d$.
Now we fill in the $W_1$ matrix: Using the algorithm above, the $W_1$ matrix will be computed from the $W_0$ matrix and vertex $v_1$. The end result will be that the $[i, j]$ entry in the matrix will be 1 if there is a path directly from vertex $i$ to vertex $j$ OR there is a way to get from $i$ to $j$ by passing through $v_1$, namely, in your case if there is a $i\rightarrow a$ edge AND a $a\rightarrow j$ edge, which is just a wordier way of saying what the second line of the algorithm is doing.
The algorithm is trivial to implement on a computer, but how would you do it by hand? In this problem, we start with the $W_0$ matrix and the $a$ row, for the $a$-to-$i$ part and the $a$ column, where the $a$-to-$j$ entries reside. Here's the $W_0$ matrix with the $a$ row and the $a$ column in red:
$$\left[ \begin{array}{cccc} \color{#ff0000}{0} & \color{#ff0000}{0} & \color{#ff0000}{0} & \color{#ff0000}{1} \\ \color{#ff0000}{1} & 0 & 1 & 0 \\ \color{#ff0000}{1} & 0 & 0 & 1 \\ \color{#ff0000}{0} & 0 & 1 & 0 \\ \end{array} \right]$$ Because of the OR part of the second line in the algorithm, we know that the entries that are $1$ in the $W_0$ matrix will remain $1$ in the $W_1$ matrix, so let's do an example where one of the entries in the $W_0$ matrix is $0$, namely the $[b, d]$ entry in the second row, fourth column. We see that the $[b, a]$ entry (in the $a$ column) is $1$ and that the $[a, d]$ entry (in the $a$ row) is also $1$ so since $1\text{ AND }1=1$, we replace the $0$ in the $[b, d]$ entry with $1$, giving us $$\left[ \begin{array}{cccc} 0 & 0 & 0 & 1 \\ 1 & 0 & 1 & 1 \\ 1 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ \end{array} \right]$$ It turns out that this is the only entry that changes during this process, so we're done; we've computed $W_1$.
To get $W_2$ we do the same steps, only this time we use $W_1\text{ and }v_2$. In case you've forgotten, $v_2=b$, so we now highlight the $b$ row and the $b$ column. If you do the process for the nonzero entries in the $W_1$ matrix, you'll find that nothing changes, so $W_2$ is the same as $W_1$.
Repeating this process, with vertices $c$ and $d$ we find that $$W_3=\left[ \begin{array}{cccc} 0 & 0 & 0 & 1 \\ 1 & 0 & 1 & 1 \\ 1 & 0 & 0 & 1 \\ 1 & 0 & 1 & 1 \\ \end{array} \right]\qquad W_4=\left[ \begin{array}{cccc} 1 & 0 & 1 & 1 \\ 1 & 0 & 1 & 1 \\ 1 & 0 & 1 & 1 \\ 1 & 0 & 1 & 1 \\ \end{array} \right]$$ and $W_4$ is our final answer. $W_4[i, j] = 1$ if and only if there is a path from $i$ to $j$ directly or through some combination of vertices $a, b, c, d$, which is to say, if there's any way at all of getting from $i$ to $j$. In this example, we see that we can get from any of $a, c, d$ to any other, but there's no way to get from any vertex to $b$.
In in relational terms, we started with $$R=\{(a, d),(b, a), (b, d),(c, a),(c, d),(d, c)\}$$ and wound up with $$R^+=\{(a, a),(a,c),(a,d),(b, a),(b, c),(b, d),(c, a),(c,c),(c, d),(d,a),(d, c),(d,d)\}$$
- | 1,651 | 4,769 | {"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} | 4.1875 | 4 | CC-MAIN-2016-26 | latest | en | 0.919159 |
https://id.scribd.com/presentation/74881729/Ch-03-Modified | 1,566,493,543,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027317274.5/warc/CC-MAIN-20190822151657-20190822173657-00523.warc.gz | 501,367,272 | 72,914 | Anda di halaman 1dari 70
Chapter 3
3.1
Chapter 3 Objectives
To define the terms and the concepts of symmetric key ciphers To emphasize the two categories of traditional ciphers: substitution and transposition ciphers To describe the categories of cryptanalysis used to break the symmetric ciphers To introduce the concepts of the stream ciphers and block ciphers
To discuss some very dominant ciphers used in the past, such as the Enigma machine
3.2
3.1 INTRODUCTION
Figure 3.2 Locking and unlocking with the same key
PLAINTEXT PLAINTEXT CIPHERTEXT
Alice
Bob
Eve The Attacker
3.3
3-1 Continued
Components of Symmetric-key cipher:
1. The original message from Alice to Bob is called plaintext. 2. The message that is sent through the channel is called the ciphertext. 3. To create the ciphertext from the plaintext, Alice uses an encryption algorithm and a shared secret key. 4. To create the plaintext from ciphertext, Bob uses a decryption algorithm and the same secret key. 5. A shared secret key.
3.4
3.1
Continued
Figure 3.1 General idea of symmetric-key cipher
3.5
3.1
Continued
We assume that Bob creates P1; we prove that P1 = P:
3.6
3.1.1
Kerckhoffs Principle
Based on Kerckhoffs principle, one should always assume that the adversary, Eve, knows the encryption/decryption algorithm.
The resistance of the cipher to attack must be based only on the secrecy of the key.
3.7
3.1.2
Cryptanalysis
cryptography is the science and art of creating secret codes, cryptanalysis is the science and art of breaking those codes.
Cryptanalysis attacks Statistical Attack : requires some statistical knowledge of the plaintext / language.
Brute-force Attack : try every possible keys.
3.8
3.1.2
Cryptanalysis
Figure 3.3 Cryptanalysis attacks
3.9
3.1.2
Continued
Figure 3.4 Ciphertext-only attack
Ciphertext-Only Attack
Known: only some ciphertext Find: the key and the plaintext CT=UFYU , PT=?
Ans=TEXT
3.10
3.1.2
Continued
Figure 3.5 Known-plaintext attack
Known-Plaintext Attack
Known: a pair of plaintext-ciphertext and the intercepted ciphertext. Find: the key and the plaintext
Ex: As SERUTAERC is to creatures so is ENOHPELET is to _________?
3.11
3.1.2
Continued
Chosen-Plaintext Attack (similar to Known-Plaintext Attack)
Known: a pair of plaintext-ciphertext but chosen by attacker herself and the intercepted ciphertext. Find: the key and the plaintext * Eve might have access to Alices computer. Ex:If PT=PEREGRINATION and the CT=1232435678596 Given CT=244 PT=?
3.12
3.1.2
Continued
Chosen-Ciphertext Attack
Known: a pair of plaintext-ciphertext but chosen by attacker herself and the intercepted ciphertext. Find: the key and the plaintext
3.13
3-2 SUBSTITUTION CIPHERS
A substitution cipher replaces one symbol with another. Substitution ciphers can be categorized as either monoalphabetic ciphers or polyalphabetic ciphers.
Note
A substitution cipher replaces one symbol with another.
Topics discussed in this section:
3.2.1 3.2.2
3.14
Monoalphabetic Ciphres Polyalphabetic Ciphers
3.2.1
Monoalphabetic Ciphers
Note
In monoalphabetic substitution, the relationship between a symbol in the plaintext to a symbol in the ciphertext is always one-to-one.
3.15
3.2.1
Example 3.1
Continued
The following shows a plaintext and its corresponding ciphertext. The cipher is probably monoalphabetic because both ls (els) are encrypted as Os.
Example 3.2 The following shows a plaintext and its corresponding ciphertext. The cipher is not monoalphabetic because each l (el) is encrypted by a different character.
ABNZF
3.16
The simplest monoalphabetic cipher is the additive cipher. This cipher is sometimes called a shift cipher and sometimes a Caesar cipher, but the term additive cipher better reveals its mathematical nature.
Figure 3.8 Plaintext and ciphertext in Z26
3.17
3.2.1
Continued
Note
When the cipher is additive, the plaintext, ciphertext, and key are integers in Z26.
3.18
Modular Arithmatic
In integer arithmetic, if we divide a by n, we can get q And r . The relationship between these four integers can be shown as
a=q*n+r
11 mod 7 = 4 a = 11 n= 7 11=1 x 7 + 4
a mod n = r
-11 mod 7= ? -11= -2 x7 +3
3.19
3.2.1
Continued
Example 3.3
Use the additive cipher with key = 15 to encrypt the message hello.
Solution
We apply the encryption algorithm to the plaintext, character by character:
3.20
3.2.1
Continued
Example 3.4
Use the additive cipher with key = 15 to decrypt the message WTAAD.
Solution
We apply the decryption algorithm to the plaintext character by character:
3.21
3.2.1
Continued
Shift Cipher and Caesar Cipher
Historically, additive ciphers are called shift ciphers. Julius Caesar used an additive cipher to communicate with his officers. For this reason, additive ciphers are sometimes referred to as the Caesar cipher. Caesar used a key of 3 for his communications.
Note
Additive ciphers are sometimes referred to as shift ciphers or Caesar cipher.
3.22
3.2.1
Continued
Example 3.5
Eve has intercepted the ciphertext UVACLYFZLJBYL. Show how she can use a brute-force attack to break the cipher.
Solution
Eve tries keys from 1 to 7. With a key of 7, the plaintext is not very secure, which makes sense.
3.23
only have 26 possible ciphers
A maps to A,B,..Z
could simply try each in turn a brute force search given ciphertext, just try all shifts of letters do need to recognize when have plaintext eg. break ciphertext "GCUA VQ DTGCM Ans: P.T.-easy to break (key=3)
3.24
Continued
rather than just shifting the alphabet could shuffle (jumble) the letters arbitrarily each plaintext letter maps to a different random ciphertext letter hence key is 26 letters long
Plain:
Cipher:
Plaintext: ifwewishtoreplaceletters Ciphertext: WIRFRWAJUHYFTSDVFSFUUFYA
3.25
Continued
now have a total of 26! = 4 x 1026 keys with so many keys, might think is secure but would be !!!WRONG!!! The problem is language characteristics
human languages are redundant
in English e is by far the most common letter
then T,R,N,I,O,A,S
3.26
3.2.1
Continued
Table 3.1 Frequency of characters in English
Table 3.2 Frequency of diagrams and trigrams
3.27
3.2.1
Continued
Example 3.6
Eve has intercepted the following ciphertext. Using a statistical attack, find the plaintext.
Solution
When Eve tabulates the frequency of letters in this ciphertext, she gets: I =14, V =13, S =12, and so on. The most common character is I with 14 occurrences. This means key = 4.
3.28
3.2.1
Continued
Figure 3.10 Multiplicative cipher
Multiplicative Ciphers
Note
In a multiplicative cipher, the plaintext and ciphertext are integers in Z26; the key is an integer in Z26*.
3.29
3.2.1
Solution
Continued
Example 3.7 What is the key domain for any multiplicative cipher?
The key needs to be in Z26*. This set has only 12 members: 1, 3, 5, 7, 9, 11, 15, 17, 19, 21, 23, 25. Example 3.8 We use a multiplicative cipher to encrypt the message hello with a key of 7. The ciphertext is XCZZU.
For Decryption use the multiplicative inverse modulo of 7 i.e. 7-1= 15
3.30
3.2.2
Polyalphabetic Ciphers
In polyalphabetic substitution, each occurrence of a character may have a different substitute. The relationship between a character in the plaintext to a character in the ciphertext is one-to-many.
Autokey Cipher
3.31
3.2.2
Continued
Example 3.14 Assume that Alice and Bob agreed to use an autokey cipher with initial key value k1 = 12. Now Alice wants to send Bob the message Attack is today. Enciphering is done character by character.
3.32
3.2.2
Continued
Playfair Cipher
Figure 3.13 An example of a secret key in the Playfair cipher
Example 3.15 Let us encrypt the plaintext hello using the key in Figure 3.13.
3.33
a 5X5 matrix of letters based on a keyword
fill in letters of keyword (sans duplicates) fill rest of matrix with other letters eg. using the keyword COMPATIBLE.
C T D O F M G P L H A E K I/J B
N
V
3.34
Q
W
R
X
S
Y
U
Z
plaintext encrypted two letters at a time:
1.
2.
3.
4.
if a pair is a repeated letter, insert a filler like 'X', eg. "balloon" encrypts as "ba lx lo on" if both letters fall in the same row, replace each with letter to right (wrapping back to start from end), eg. ar" encrypts as "RM" if both letters fall in the same column, replace each with the letter below it (again wrapping to top from bottom), eg. mu" encrypts to "CM" otherwise each letter is replaced by the one in its row in the column of the other letter of the pair, eg. hs" encrypts to "BP", and ea" to "IM" or "JM" (as desired)
3.35
Ex:
i/j
e f o v
a g q w
b h r x
c k t y
d n u z
3.36
Security of the Playfair Cipher
security much improved over monoalphabetic since have 26 x 26 = 676 digrams would need a 676 entry frequency table to analyse (verses 26 for a monoalphabetic) and correspondingly more ciphertext was widely used for many years (eg. US & British military in WW1) it can be broken, given a few hundred letters since still has much of plaintext structure
3.37
3.2.2
Continued
Vigenere Cipher
Example 3.16 We can encrypt the message She is listening using the 6-character keyword PASCAL.
3.38
3.2.2
Continued
Example 3.16
Let us see how we can encrypt the message She is listening using the 6-character keyword PASCAL. The initial key stream is (15, 0, 18, 2, 0, 11). The key stream is the repetition of this initial key stream (as many times as needed). i.e. P A S C A L 15, 0, 18, 2, 0, 11
3.39
this additive cipher is a special case of Vigenere cipher. Where m=1
plaintext
key
Table 3.3
A Vigenere Tableau
3.40
3.2.2
Example 3.19
Continued
Let us assume we have intercepted the following ciphertext:
The Kasiski test for repetition of three-character segments yields the results shown in Table 3.4.
3.41
3.2.2
Continued
The greatest common divisor of differences is 4, which means that the key length is multiple of 4. First try m = 4 with frequency analysis.
3.42
Example
eg using keyword deceptive key: deceptivedeceptivedeceptive plaintext: wearediscoveredsaveyourself ciphertext:ZICVTWQNGRZGVTWAVZHCQYGLMGJ
string VTW 1st index 4 2nd index 13 difference 9
suggests keyword size of 3 or 9 then attack each monoalphabetic cipher individually using previous techniques
3.43
3.2.2
Continued
if a truly random key as long as the message is used, the cipher will be secure called a One-Time pad is unbreakable since ciphertext bears no statistical relationship to the plaintext since for any plaintext & any ciphertext there exists a key mapping one to other
can only use the key once though
have problem of safe distribution of key
3.44
Enigma Machine
Enigma was a portable cipher machine used to encrypt and decrypt secret messages.
Japan commercial German military
45
Enigma Machine
Enigma encryption for two consecutive letters current is passed into set of rotors, around the reflector, and back out through the rotors again. Letter A encrypts differently with consecutive key presses, first to G, and then to C. This is because the right hand rotor has stepped, sending the signal on a completely different route.
46
Enigma
the actual encipherment of a letter is performed electrically.
When a key is pressed, the circuit is completed; current flows through the various components and ultimately lights one of many lamps, indicating the output letter. Current flows from a battery through the switch controlled by the depressed key into a fixed entry wheel. This leads into the rotor assembly (or scrambler), where the complex internal wiring of each rotor results in the current passing from one rotor to the next along a convoluted path. After passing through all the rotors, current enters the reflector, which relays the signal back out again through the rotors and the entry wheel this time via a different path and, finally, to one of the lamps (the earliest Enigma models do not have the reflector).
47
Rotors
performs a very simple type of encryption
48
A few here
For more, see
http://w1tp.com/enigma/
49
3-3 TRANSPOSITION CIPHERS
A transposition cipher does not substitute one symbol for another, instead it changes the location of the symbols.
Note
A transposition cipher reorders symbols.
these hide the message by rearranging the letter order without altering the actual letters used.
Topics discussed in this section:
3.3.1 3.3.2 3.3.3
3.50
3.3.1
Keyless Transposition Ciphers
Simple transposition ciphers, which were used in the past, are keyless.
Example 3.22 A good example of a keyless cipher using the first method is the rail fence cipher. The ciphertext is created reading the pattern row by row. For example, to send the message Meet me at the park to Bob, Alice writes
She then creates the ciphertext MEMATEAKETETHPR.
3.51
3.3.1
Continued
Example 3.23
Alice and Bob can agree on the number of columns. Alice writes the same plaintext, row by row, in a table of four columns.
3.52
3.3.2
Keyed Transposition Ciphers
The keyless ciphers permute the characters by writing plaintext in one way and reading it in another way. The permutation is done on the whole plaintext to create the whole ciphertext. Another method is to divide the plaintext into groups of predetermined size, called blocks, and then use a key to permute the characters in each block separately.
3.53
3.3.2
Continued
Example 3.25
Alice needs to send the message Enemy attacks tonight to Bob..
The key used for encryption and decryption is a permutation key, which shows how the character are permuted.
3.54
PLAINTEXT:
key
3 1 e a k i
1 2 n t s g
4 3 e t t h
5 4 m a o t
2 5 y c n z
CIPHERTEXT:
3.55
3.3.3 Combining Two Approaches
Example 3.26
Figure 3.21
3.56
3.3.3
Keys
Continued
In Example 3.27, a single key was used in two directions for the column exchange: downward for encryption, upward for decryption. It is customary to create two keys. Figure 3.22 Encryption/decryption keys in transpositional ciphers
3 1
4 5
2
3.57
3.3.3
Continued
3.58
Key inversion in a transposition cipher
2 1
6 2
3 1 4 7 5 3 4 5 6 7
1 4
2 1
3 4 5 6 7 3 5 7 2 6
3.59
3.3.3
Continued
3.60
Double Transposition Ciphers (Ex:)
3 e a k i
1 n t s g
4 e t t h
5 m a o t
2 y c n z
3 e t t h
1 e a k i
4 m a o t
5 y c n z
2 n t s g
CT1= ettheakimaotycnzntsg
3 t i y t
3.61
1 4 5 2 e h e t a m a k o c n t z s g n
3 1 4 e t t a k i o t y z n t
5 h m c s
2 e a n g
CT2= tityeaozhmcseangtktn
Product Ciphers
ciphers using substitutions or transpositions are not
secure because of language characteristics
hence consider using several ciphers in succession to
make harder, but:
two substitutions make a more complex substitution two transpositions make more complex transposition but a substitution followed by a transposition makes a new
much harder cipher
3.62
Modern Block Ciphers
will now look at modern block ciphers provide secrecy and/or authentication services in particular will introduce DES (Data Encryption Standard)
Block vs. Stream Ciphers
block ciphers process messages in into blocks, each of which is then en/decrypted like a substitution on very big characters
64-bits or more
stream ciphers process messages a bit or byte at a time when en/decrypting many current ciphers are block ciphers
Block Cipher Principles
most symmetric block ciphers are based on a Feistel Cipher Structure block ciphers look like an extremely large substitution would need table of 264 entries for a 64-bit block using idea of a product cipher
modern substitution-transposition product cipher
these form the basis of modern block ciphers S-P networks are based on the two primitive cryptographic operations we have seen before:
partitions input block into two halves
process through multiple rounds which perform a substitution on left data half based on round function of right half & subkey then have permutation swapping halves implements Shannons substitution-permutation network concept
Feistel Cipher Design Principles
block size
key size
increasing size improves security, but slows cipher increasing size improves security, makes exhaustive key searching harder, but may slow cipher
number of rounds
subkey generation
increasing number improves security, but slows cipher
greater complexity can make analysis harder, but slows cipher
round function
greater complexity can make analysis harder, but slows cipher
are more recent concerns for practical use and testing | 4,348 | 16,433 | {"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-2019-35 | latest | en | 0.78331 |
https://amp.doubtnut.com/question-answer/in-triangle-a-b-c-base-b-c-and-area-of-triangle-are-fixed-the-locus-of-the-centroid-of-triangle-a-b--22140 | 1,591,105,121,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347425148.64/warc/CC-MAIN-20200602130925-20200602160925-00495.warc.gz | 240,009,035 | 21,272 | IIT-JEE
Apne doubts clear karein ab Whatsapp (8 400 400 400) par bhi. Try it now.
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Question From class 11 Chapter TRIGONOMETRIC FUNCTIONS
# In triangle , base and area of triangle are fixed. The locus of the centroid of triangle is a straight line that is a) parallel to side (b)right bisector of side BC (c)perpendicular to BC (d)inclined at an angle to side BC
A variable triangle is circumscribed about a fixed circle of unit radius. Side always touches the circle at D and has fixed direction. If B and C vary in such a way that (BD) (CD)=2, then locus of vertex A will be a straight line. parallel to side BC perpendicular to side BC making an angle with BC making an angle with
2:59
Show that the line joining the incenter to the circumcentre of triangle ABC is inclined to the side BC at an angle
are the vertices of a triangle. Show that the median from A is perpendicular to the side BC.
1:11
A perpendicular drawn from a vertex to the opposite side of a triangle is known as an altitude (b) a median (c) an angle bisector a bisector
1:09
If the altitude from one vertex of a triangle bisects the opposite side, then the triangle is isosceles. GIVEN : A such that the altitude from on the opposite side bisects i.e., TO PROVE : i.e. the triangle is isosceles.
2:14
The area of triangle formed by vertices are is and the area of triangle whose vertices are and is . If represent the sides of triangle of perimeter 16. Then the value of
5:27
ABC is right angle triangle at C. Let BC = a, CA = b, AB = c and Let 'P' be the length of perpendicular from C to AB. prove that
3:21
If where a, b, c are the sides of a triangle, then the largest angle of that triangle is
2:50
In a triangle ABC, let BC = 1, AC=2 and measure of angle C is 30°. Which of the following statement(s) is (are) correct? (A) 2 sinA =sin B (B) Length of side AB equals (C) Measure of angle A is less than (D) Circumradius of triangle ABC is equal to length of side AB
9:49
If the bisector of the angle A of triangle ABC meets BC in D, prove that
7:40
The sum of any two sides of a triangle is greater than the third side. GIVE : A TO PROVE : and CONSTRUCTION : Produce side to such that Join
3:28
Prove that the area of an equilateral triangle is equal to where is the side of the triangle. GIVEN : An equilateral triangle such that TO PROVE : CONSTRUCTION : Draw
3:59
In any triangle in equal to b. c. d.
3:11
If the bisector of the vertical angle of a triangle bisects the base of the triangle. then the triangle is isosceles. GIVEN : A in which is the bisector of meeting in such that TO PROVE : is an isosceles triangle.
3:47
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MicroConcepts | 1,058 | 3,990 | {"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} | 3.875 | 4 | CC-MAIN-2020-24 | latest | en | 0.876718 |
https://herbie.uwplse.org/demo/e1ecdc66eea62bcc6cb8d9ac514878d27de5d597.cf8a50cd49460eed2f57a21b8de7a625bbcdb510/graph.html | 1,575,702,657,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540496492.8/warc/CC-MAIN-20191207055244-20191207083244-00051.warc.gz | 397,703,955 | 2,347 | Average Error: 28.5 → 0.5
Time: 17.7s
Precision: 64
$\tan \left(x + 1\right)$
$\frac{\tan x + \tan 1}{\log \left(e^{1 - \tan 1 \cdot \tan x}\right)}$
\tan \left(x + 1\right)
\frac{\tan x + \tan 1}{\log \left(e^{1 - \tan 1 \cdot \tan x}\right)}
double f(double x) {
double r58469127 = x;
double r58469128 = 1.0;
double r58469129 = r58469127 + r58469128;
double r58469130 = tan(r58469129);
return r58469130;
}
double f(double x) {
double r58469131 = x;
double r58469132 = tan(r58469131);
double r58469133 = 1.0;
double r58469134 = tan(r58469133);
double r58469135 = r58469132 + r58469134;
double r58469136 = r58469134 * r58469132;
double r58469137 = r58469133 - r58469136;
double r58469138 = exp(r58469137);
double r58469139 = log(r58469138);
double r58469140 = r58469135 / r58469139;
return r58469140;
}
# Try it out
Results
In Out
Enter valid numbers for all inputs
# Derivation
1. Initial program 28.5
$\tan \left(x + 1\right)$
2. Using strategy rm
3. Applied tan-sum0.4
$\leadsto \color{blue}{\frac{\tan x + \tan 1}{1 - \tan x \cdot \tan 1}}$
4. Using strategy rm
$\leadsto \frac{\tan x + \tan 1}{\color{blue}{\log \left(e^{1 - \tan x \cdot \tan 1}\right)}}$
$\leadsto \frac{\tan x + \tan 1}{\log \left(e^{1 - \tan 1 \cdot \tan x}\right)}$
herbie shell --seed 1 | 494 | 1,275 | {"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} | 3.890625 | 4 | CC-MAIN-2019-51 | latest | en | 0.416478 |
https://www.solutioninn.com/a-brick-maker-bm-in-alberta-mixes-dry-ink-into | 1,709,508,285,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476399.55/warc/CC-MAIN-20240303210414-20240304000414-00878.warc.gz | 980,362,298 | 14,861 | # A brick maker (BM) in Alberta mixes dry ink into its bricks to make them brown. BMs
## Question:
A brick maker (BM) in Alberta mixes dry ink into its bricks to make them brown.
BM’s demand for dry ink is 60 tons per year. Currently, BM buys the dry ink from an import merchant that buys the ink from a East Coast U.S. manufacturer. The shipments arrive in lot size of 30 tons by rail. The current cost of dry ink is SC612.22 per ton, including rail transportation cost to BM’s location. BM currently keeps 6 tons of dry ink as safety stock. BM’s buyer has asked the import merchant to quote a price for truck deliveries in smaller lot sizes. The merchant has quoted C$567.78 per ton for a lot size of 20 tons. In the meantime, BM’s buyer has contacted the manufacturer directly and asked if BM could purchase dry ink directly from the manufacturer. The answer was affirmative and the cost would be US$386.89 per ton (assume US$1 = C$ 1.05). A common carrier has quoted a price of C\$2,600 to haul a full truckload of dry ink (20 tons) from the manufacturer to BM's location in Alberta. The trip will take 7 days.
The holding cost rate for BM is 20 percent of unit cost per year. For truck deliveries, BM will only hold 2 tons of safety stocks. Which alternative has the lowest total annual purchase, transport, in-transit, safety stock, and cycle (batch) holding cost?
Fantastic news! We've located the answer you've been seeking! | 340 | 1,434 | {"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} | 2.59375 | 3 | CC-MAIN-2024-10 | latest | en | 0.961474 |
https://studylib.net/doc/7323639/example-09---feature-classification | 1,708,886,745,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474641.34/warc/CC-MAIN-20240225171204-20240225201204-00739.warc.gz | 567,352,551 | 12,195 | # Example 09 - Feature Classification
```Sample Question 1
You are given 6 sample images, each containing either Object A or Object B. The feature
vector representation of the objects in the image is given below:
Image No.
1
2
3
4
5
6
Feature vector
[24, 99]
[21, 80]
[23, 101]
[120, 35]
[111, 34]
[123, 43]
Object in the Image
Object A
Object A
Object A
Object B
Object B
Object B
You have decided to use the Euclidean Distance to find the difference between two
feature vectors.
EuclideanDist
n
a
i 1
(a)
bi
2
i
In comparing two feature vectors, the Euclidean Distance calculation will
eventually give you a number. How do you interpret this number (i.e. what does
it mean if the number is small / large)?
[2 points]
The smaller the number, the similar the two objects are.
(b)
You are given an image with the following feature vector:
I = [40, 66]
Try to classify the image (i.e. say that whether it contains Object A or Object B)
using nearest neighbor method. Show your calculation and give reason for your
classification.
[6 points]
Compare with Image 1: 36.7
Compare with Image 2: 23.6
Compare with Image 3: 38.9
Compare with Image 4: 85.8
Compare with Image 5: 77.9
Compare with Image 6: 86.1
Conclusion: Image I is most similar to image 2. Since image 2 contains Object A,
then image I must contain object A as well.
(c)
Using the same feature vector as in part (b), try to re-classify the image, but now
using the nearest centroid method. Do you get the same result?
[7 points] | 436 | 1,498 | {"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} | 3.296875 | 3 | CC-MAIN-2024-10 | latest | en | 0.880163 |
http://www.cs.purdue.edu/homes/gnf/book3/bkrv3.html | 1,369,531,078,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368706477730/warc/CC-MAIN-20130516121437-00077-ip-10-60-113-184.ec2.internal.warc.gz | 410,014,927 | 8,644 | Book reviews for Piano-Hinged Dissections: Time to Fold!
Review in MAA Online
Brian Borchers, a professor of Mathematics at the New Mexico Institute of Mining and Technology, wrote a review for "Read this!", the book review column in MAA Online. It first appeared February 22, 2007, and can be found online:
"Read This: Piano-Hinged Dissections".
Some excerpts from the review:
"Frederickson's latest book discusses another class of hinged dissections, piano-hinged dissections, in which edges are connected by hinges that allow the pieces to fold over each other or to fold out flat. In a piano-hinged dissection the pieces are required to overlap in two layers to form each of the geometric figures. Unlike with swinging and twisting hinged dissections it is particularly easy to build models of piano-hinged dissections out of card stock or wood. There are also obvious connections between piano-hinged dissections and origami. Thus piano-hinged dissection problems should be particularly attractive to fans of recreational mathematics."
"highly recommended to fans of geometric dissections who are looking for some new and challenging dissection problems"
Review in Cubism For Fun
Rik van Grol, the editor, wrote a very nice review in the March 2007 issue (#72) of Cubism For Fun, a newsletter in English published by the Nederlandse Kubus Club NKC (Dutch Cubists Club). Follow this link to the homepage of the newsletter. Excerpting from the review, which appears on pages 24-25:
"With this book Greg fanatically explores a world filled with piano-hinged dissections. Initially you would think it to be impossible to make more than a few dissections with piano-hinges (a piano-hinge is a long narrow hinge the runs the full length of a joint), but Greg show[s] that given time and effort many more dissection[s] can be realized with piano-hinges than you could ever imagine. In this book an amazing piano-hinged version of the Triangle to Square is presented."
"The book comes with a CD with a video in which Greg actually demonstrates his hinged dissections: almost an hour of wonderful dissections folding and unfolding before your eyes. Most of these are paper realisations, but Greg also shows a number of amazing wooden piano-hinged dissections; simply great!"
"Greg's books are valuable sources for the puzzle designers and an obvious must have for dissection lovers."
Review in SciTech Book News
There is a great review in the March 2007 issue (Vol. 31, No. 1) of SciTech Book News, published by Book News, Inc., of Portland Oregon. It's so enthusiastic (and not so long) that I couldn't resist reproducing the whole thing, which appears on page 34:
"Armed, as it were, with little more than an inquiring mind, moderate hand-eye coordination and the anticipation of a delightful outcome, you can follow Frederickson (computer science, Purdue U.) into the world of dissecting polygons so the resulting pieces, which are attached by hinges that must fold along an edge rather than swing or twist, form another polygon. The geometry is fascinating, and so are the illustrations Frederickson offers to show how he came up with the results, which range from the remarkably complex Theobald 11-piece dissection to a plethora of hexagrams that become hexagons. Working in some cases with the unique approach of Ernest Irving Freese, Frederickson produces both beautiful dissections and the concepts to back them up. His self-made video on the accompanying CD-ROM truly helps those of us who need more hands-on training and less apprehension that it cannot be done."
Review in BMS-NCM News
Adhemar Bultheel, a professor in the Department of Computer Science at the Katholieke Universiteit Leuven, wrote a nice review that appeared in the March 15, 2007 (no. 62) issue of the BMS-NCM News, the newsletter of the Belgian Mathematical Society and the National Committee for Mathematics. This newsletter is published five times a year by the Belgian Mathematical Society. The review appears on pages 8-9, as well as online. Two excerpts from the review:
"Because essentially all the moves discussed need a three-dimensional space, it is sometimes difficult to give a clear explanation of the operations to be performed. This gave Frederickson the idea of including a cdrom on which he demonstrates the folding, decomposing and recomposing the dissections. Some are really complicated, and even if you see him doing it, it is sometimes impossible to see how all the pieces fall into place. At least it is a visual proof that the method does indeed work. It is quite funny if you see him fighting with a wooden model of a hinged dissection that is a loop consisting of 30 triangles hinged together."
"And there are many other of these geometrical puzzles. Some of them are relatively simple and e.g., related to the way we fold a paper large map into some handy pocket sized format. Others, in fact most of them, are quite a challenge and are beautiful in their complexity."
Review in Choice
David V. Feldman, a professor of mathematics at the University of New Hampshire, wrote a very nice review in the July 2007 issue (vol. 44, no. 11) of CHOICE: Current Reviews for Academic Libraries, a monthly periodical published by the the Association of College & Research Libraries, which is a division of the American Library Association. Excerpting from the review (44-6281), which appears on pages 1944-1945:
"The search for elegant mutual dissections has formed a principal theme of recreational mathematics for several centuries. `Elegant' can simply mean dissection into an unexpectedly small number of pieces, or alternatively, a dissection subject to some interesting side condition.
"the author has now invented yet a new art form, two-ply models with polygonal pieces connected by easy-to-find and simple-to-work piano hinges. The body of this book works variations on the main theme that seem nothing short of ingenious. An accompanying video-CD-ROM offers readers the chance to see these remarkable models in action. Best of all, the author captures the infectious tone and constant excitement of an effectively theatrical lecture-demonstration."
Review in Mathematics Teacher
Ira Lee Riddle, a tutor in the Learning Center of Penn State University, Abington, wrote a review that appeared in the September, 2007 issue (vol. 101, no. 2) of the Mathematics Teacher. This journal is published nine times a year by the National Council of Teachers of Mathematics. The review appears on pages 159-160. Two excerpts from the review:
"When the term [dissection] is used in mathematics, it refers to demonstrating that the area of a figure is a constant, even when its pieces are rearranged. Shapes are cut apart, and the pieces are put together again into a new shape. Tangrams come to mind immediately, but this book goes a bit beyond tangrams."
"challenging for teachers as well as students, but it is a good bit of fun as well"
Review in EMS Newsletter
Martina Bečvářová, a member of the Department of Mathematics Education at Charles University, in Prague, Czech Republic, wrote a very nice review that appeared in the September, 2007 issue (no. 65) of the EMS Newsletter. The newsletter is published four times a year by the European Mathematical Society. The review appears on page 57. Excerpting from the review:
"This brilliant book can be recommended to students of geometry and teachers of mathematics, as well as students and all people who are interested in geometric dissections. Every creative reader will find new material for his own discoveries. The reader can easily experiment with the piano-hinge dissections because their mechanism can be simulated by folding a piece of paper without special mathematical knowledge, materials, computer programs, etc."
Review in Zentralblatt MATH
Mowaffaq Hajja, a member of the Department of Mathematics, Yarmouk University, Irbid, Jordan, wrote a very nice review that appeared in Zentralblatt MATH in November 2007. Zentralblatt MATH is published by the European Mathematical Society, FIZ Karlsruhe, and Heidelberger Akademie der Wissenschaften. Excerpting from the review, which is indexed as 1126.52014 and appears in volume 1126:
"This beautiful book is the author's third book on dissections.
"The book under review lays the mathematical foundation of the theory of piano-hinged dissections by examining the definition of such a dissection and suggesting ways for overcoming technical difficulties that arise from the thickness that two-dimensional pieces have to be assumed to have. Above all, it introduces so many beautiful and ingenious piano-hinged dissections that have never been known before and that are so non-trivial to discover, one such beautiful example being an ellipse-to-heart piano-hinged dissection.
"There are several asides of both mathematical and recreational interest. Notable among these is a fascinating section on what is usually referred to as the open box problem and for which the author had duly won the Polya Award for expository writing from the Mathematical Association of America.
"Beside having a lot of perspective diagrams that illustrate how the folds are to be made, the book also has a CD-ROM that contains a lot of videos that are extremely helpful for understanding the moves. Without this CD, many of the moves that are verbally described would be hard and time-consuming to understand and follow. If a perspective diagram is worth a thousand pictures, then an animation is probably worth a thousand perspective diagrams."
Faux Review in Monatshefte für Mathematik
The Monatshefte für Mathematik gave what was purported to be a review of my book in volume 155, number 1 (September 2008), page 102. Unfortunately, it appears that there was a mixup, and the review is actually of the book Fractal Geometry, Complex Dimensions and Zeta Functions, by Michel Lapidus and Machiel van Frankenhuyen. See the fifth review down from here for an actual review of my book that appeared in this journal in 2013.
Review in Journal of Recreational Mathematics
Charles Ashbacher, one of the two editors of the Journal of Recreational Mathematics, wrote a 5-star review, "Complex, yet fascinating dissections with a flip," which appeared on October 28, 2009 on amazon.com and which will appear in due course in the journal itself. Excerpting from the review:
"A piano hinge dissection is one where a hinge runs the full length of a joint. The analogy is to the hinge that allows the top of a grand piano to be opened although in this case the hinge can allow the two pieces to be folded in either of the two directions. Add in multiple hinges and the potential for the pieces to overlap and the additional degrees of freedom can make for a complicated structure. I often found myself wondering how the dissection had been discovered.
"Frederickson uses paper and wooden models to illustrate the folding that allows you to transform one figure into another. Seeing the transformation by executing one active fold after another makes it so much clearer. Quite honestly, I am not sure if I would have completely grasped some of the more complex transformations without the video.
"The use of one or more piano hinges in a dissection creates a significantly higher level of complexity. However, in complexity there is joy and while I had difficulty in the explanations, the wonder of seeing it work made the time of difficulty well spent."
Review in Mathematical Spectrum
A very nice review appeared in the Mathematical Spectrum, vol. 42 (2009/2010), no. 1, on page 50. Excerpting from the review:
"Traditional dissections involve pieces that are not attached to each other—add hinges between the pieces, and you are looking at a whole new set of rules and challenges. This book showcases a new type of hinged dissection that generates even more challenges."
"This mechanism can be simulated by folding a piece of paper, so you can test and experiment with piano-hinged dissections without needing special materials: just paper and scissors–and some intuition and creativity!"
Review in SIAM Review
Les Pook, a visiting professor in the Department of Mechanical Engineering at the University College London, wrote a book review that appeared in the SIAM Review, vol. 52, no. 1 (March 2010), on pages 208-213. The bulk of the review is a chapter-by-chapter summary of the contents of the book, along with complaints that the style is not appropriate for a monograph, that the text does not always establish that the methods produce mathematically exact dissections, that nets are not furnished for all of the dissections, and that the series of "Folderol" segments and "Manuscript" segments are tangential and would better have been moved to appendices, or removed altogether. However, this reviewer was nonetheless intrigued by the piano-hinged dissections, as revealed by this excerpt from his introduction:
"Piano-hinged dissections are motion structures that can only be fully appreciated by manipulating models or viewing videos. They are highly addictive. I made 54 paper models while preparing this review. Individually, the piano-hinged dissections described by Frederickson are usually not of great interest, apart from admiring his ingenuity in deriving them. Collectively, they are of interest in demonstrating the wide range of possibilities and the relationships between different dissections."
Excerpting from his summary:
"It is a useful and fascinating monograph on one of the many aspects of paper folding. It includes much original information. The book can be enjoyed at the recreational mathematics level by making and manipulating models of the piano-hinged dissections described. It can also be enjoyed at the serious mathematics level by studying methods used to derive piano-hinged dissections and also the relationships between different dissections. The book is well produced with only a few minor errors."
Review in CMS Notes de la SMC
Keith Johnson, a professor in the Department of Mathematics and Statistics at Dalhousie University, wrote a book review that appeared in the "Brief Book Reviews" section of CMS Notes de la SMC, vol. 43, no. 5 (October/November 2011), on page 6. Excerpting from the review:
"A special class of these, dissections in which the reassembly is by means of piano hinge joints i.e. folding only along a specified set of edges, is the topic of this book. It is fairly easy to think of simple examples of these, such as the dissection of a triangle using three piano hinge joints which bring the three vertices together at a point on the longest side to give a rectangle (and so illustrate that the sum of the three angles of the triangle is π) but the variety of more complicated examples the author exhibits is quite astonishing."
"The book has a couple of unusual features. One is that many early results on the topic were discovered in the 1930's by an American architect and amateur geometer, Ernest Irving Freese, and described in an almost lost manuscript, parts of which are reproduced here for the first time. The other is an included CD with videos of the author showing the operation of many of the dissections."
Review in Monatshefte für Mathematik
Prof. Dr. Harald Rindler, Dean of the Faculty of Mathematics and Head of the Department of Mathematics at the University of Vienna, wrote a great book review that appeared in the Book Reviews section of Monatshefte für Mathematik vol. 169, issue 2 (February 2013), on page 251. Quoting from the review:
"Greg Frederickson ist ein international anerkannter Großmeister beim Entdecken und Kreieren beeindruckender Zerlegungen (Zerschneidungen), Faltungen, Aufklappungen und überraschender Umwandlungen vorgegebener Formen in faszinierende Objekte und bereichert hier eine jahrhundertelange Tradition mit einer Fülle neuer Ergebnisse und Anregungen. Der Autor bietet auch interessante Resultate eines "verlorenen" Manuskripts von E. Irving, eines Architekten in LosAngeles, der gegen Ende seines Lebens eine Passion für Zerlegungen entwickelte. Zusätzlich gibt es eine CD mit Videoclips und Anleitungen, die die Konstruktion der Objekte leicht ermöglichen."
Rough translation: Greg Frederickson is an internationally recognized Grand Master in discovering and creating impressive decompositions (dissections), folds, flaps, and surprising transformations of given shapes into fascinating objects, enriching here a centuries-old tradition with a wealth of new results and inspiration. The author also provides interesting results of a "lost" manuscript by E. Irving, an architect in Los Angeles who towards the end his life developed a passion for dissections. There is also a CD with video clips and instructions that make possible the easy construction of the objects.
"Einem großen Interessentenkreis sehr zu empfehlendes Buch!"
Rough translation: A highly recommended book for a large prospective audience!
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