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https://www.jiskha.com/questions/92725/write-3x-2-8x-25-in-the-form-a-x-h-2-where-a-h-and-k-are-real-numbers | 1,555,695,245,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578527865.32/warc/CC-MAIN-20190419161226-20190419183226-00182.warc.gz | 752,961,401 | 4,895 | # math
write 3x^2 +8x-25 in the form a(x+h)^2,where a,h and k are real numbers
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3. 👁 26
1. That expression is not a perfect square. You would have to add a constant. Maybe that is the "k" that you left out
3x^2 +8x -25 = 3(x^2 + 8x/3 -25/3)
= 3(x^2 + 8x/3 + 16/9) -25 - 48/9
= 3(x + 4/3)^2 - 123/9
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2. 👎 0
posted by drwls
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More Similar Questions | 845 | 2,556 | {"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} | 3.875 | 4 | CC-MAIN-2019-18 | latest | en | 0.927639 |
https://physics.stackexchange.com/questions/643807/hyperelastic-solids-why-cannot-the-work-in-a-closed-process-be-negative | 1,718,747,793,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861794.76/warc/CC-MAIN-20240618203026-20240618233026-00877.warc.gz | 401,582,307 | 39,953 | Hyperelastic solids: why cannot the work in a closed process be negative?
While introducing the theory of hyperelastic solids, our professor gave the following example.
"Suppose to have a rubber which is undeformed. Then stretch it, and then release it. The final configuration will be the same as the initial one. Then, the he overall work cannot be negative"
I really cannot understand the overall work can't be negative. While he was writing this, he was mentioning something like "if it were negative, then we have a contradiction". However, I really can't understand what was the idea.
• Commented Jun 7, 2021 at 0:34
The argument is probably this: it is assumed that the rubber body returns to the initial state; then its final energy has to be the same as the initial energy because internal energy of a body is function of state; same state means same energy. It is also assumed that no work was lost on heating the environment during this exercise. So if energy of both rubber and its environment did not change, then no total work was done.
In reality, when rubber is stretched, it heats up and some energy is transferred to the cooler environment. During contraction, rubber gets colder and part of this energy is absorbed back by the rubber, but not all of it, so the rubber won't really get back to the initial state and work got back won't be as large as work put in, so there will be some small amount of work lost.
I think that we could take the definition for a hyperelastic material. From Wikipedia
A hyperelastic or Green elastic material is a type of constitutive model for ideally elastic material for which the stress–strain relationship derives from a strain energy density function.
Then, you will have something as the following
$$S_{ij} = \frac{\partial U}{\partial E_{ij}}\, ,$$
where $$S$$ is a stress measure and $$E$$ is a strain measure, and these two are energy conjugate. This means that the product of the two of them represents a density of energy.
Now, you need to assume a sign convention for work.
Hyperelastic materials are a special case of Cauchy elastic materials and, as mentioned by @JánLalinský, you have a state variable and the final and initial configuration are the same. Then the work should be the difference in internal energy from these two states, but they are the same
• Thanks @nicoguaro. Why did someone downvote your answer? Commented Jun 7, 2021 at 7:04
• @bobinthebox, no idea. Commented Jun 7, 2021 at 15:22 | 559 | 2,482 | {"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": 3, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2024-26 | latest | en | 0.976071 |
https://www.studypool.com/discuss/264301/i-need-help-with-math-homework?free | 1,508,557,018,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187824543.20/warc/CC-MAIN-20171021024136-20171021044136-00194.warc.gz | 990,066,049 | 14,391 | Time remaining:
##### I need help with math homework.
label Algebra
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schedule 1 Day
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3u+z=15
u+2z+10
Oct 20th, 2017
3u+z=15
Multiplying by 2
6u+2z=30 1A
u+2z =10 2
Subtracting 2 from 1 A
5u =20
u =20/5 =4
12+z=15
z=15-12=3
u=4
z=3
Oct 28th, 2014
...
Oct 20th, 2017
...
Oct 20th, 2017
Oct 21st, 2017
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Mark as Final Answer
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check_circle | 191 | 470 | {"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-2017-43 | latest | en | 0.772403 |
https://xianblog.wordpress.com/tag/computational-budget/ | 1,632,594,635,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057733.53/warc/CC-MAIN-20210925172649-20210925202649-00313.warc.gz | 1,088,969,231 | 20,680 | ## approximations of Markov Chains [another garden of forking paths]
Posted in Books, Mountains, pictures, Statistics, University life with tags , , , , , , , , , , on March 15, 2016 by xi'an
James Johndrow and co-authors from Duke wrote a paper on approximate MCMC that was arXived last August and that I missed. David Dunson‘s talk at MCMski made me aware of it. The paper studies the impact of replacing a valid kernel with a close approximation. Which is a central issue for many usages of MCMC in complex models, as exemplified by the large number of talks on that topic at MCMski.
“All of our bounds improve with the MCMC sample path length at the expected rate in t.”
A major constraint in the paper is Doeblin’s condition, which implies uniform geometric ergodicity. Not only it is a constraint on the Markov kernel but it is also one for the Markov operator in that it may prove impossible to… prove. The second constraint is that the approximate Markov kernel is close enough to the original, which sounds reasonable. Even though one can always worry that the total variation norm is too weak a norm to mean much. For instance, I presume with some confidence that this does not prevent the approximate Markov kernel from not being ergodic, e.g., not irreducible, not absolutely continuous wrt the target, null recurrent or transient. Actually, the assumption is stronger in that there exists a collection of approximations for all small enough values ε of the total variation distance. (Small enough meaning ε is much smaller than the complement α to 1 of the one step distance between the Markov kernel and the target. With poor kernels, the approximation must thus be very good.) This is less realistic than assuming the availability of one single approximation associated with an existing but undetermined distance ε. (For instance, the three examples of Section 3 in the paper show the existence of approximations achieving a certain distance ε, without providing a constructive determination of such approximations.) Under those assumptions, the average of the sequence of Markov moves according to the approximate kernel converges to the target in total variation (and in expectation for bounded functions). With sharp bounds on those distances. I am still a bit worried at the absence of conditions for the approximation to be ergodic.
“…for relatively short path lengths, there should exist a range of values for which aMCMC offers better performance in the compminimax sense.”
The paper also includes computational cost into the picture. Introducing the notion of compminimax error, which is the smallest (total variation) distance among all approximations at a given computational budget. Quite an interesting, innovative, and relevant notion that may however end up being too formal for practical use. And that does not include the time required to construct and calibrate the approximations. | 599 | 2,919 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2021-39 | latest | en | 0.94182 |
http://www.algebra.com/algebra/homework/NumericFractions/Numeric_Fractions.faq.question.636862.html | 1,369,393,810,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368704590423/warc/CC-MAIN-20130516114310-00060-ip-10-60-113-184.ec2.internal.warc.gz | 310,571,988 | 4,635 | SOLUTION: Can I get help with these please. 4 4/5 / 8/15 5 6/7 + 4 5/8 2 3/8 + 3 3/4 4 5/9 - 7/8 8 2/7 - 3 1/3
Algebra -> Algebra -> Numeric Fractions Calculators, Lesson and Practice -> SOLUTION: Can I get help with these please. 4 4/5 / 8/15 5 6/7 + 4 5/8 2 3/8 + 3 3/4 4 5/9 - 7/8 8 2/7 - 3 1/3 Log On
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Algebra: Numeric Fractions Solvers Lessons Answers archive Quiz In Depth
Click here to see ALL problems on Numeric Fractions Question 636862: Can I get help with these please. 4 4/5 / 8/15 5 6/7 + 4 5/8 2 3/8 + 3 3/4 4 5/9 - 7/8 8 2/7 - 3 1/3Answer by stanbon(57361) (Show Source): You can put this solution on YOUR website!4 4/5 / 8/15 = (24/5)/(8/15) = (24/5)*(15/8) = 3*3 = 9 ------------------------------------ 5 6/7 + 4 5/8 = (41/7)+(37/8) = (8*41+7*37)/(7*8) = 587/56 ------------------------------------ 2 3/8 + 3 3/4 = (19/8) + (15/4) = (19 + 30)/8 = 49/8 ---------------------------------------- 4 5/9 - 7/8 = (41/9) - (7/8) = (8*41-9*7)/(9*8) = 262/75 --------------------- 8 2/7 - 3 1/3 = (58/7) - (10/3) = (3*58-7*10)/(7*3) = 104/21 ============== Cheers, Stan H. | 546 | 1,278 | {"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} | 3.96875 | 4 | CC-MAIN-2013-20 | latest | en | 0.698947 |
http://www.ehow.com/how_7296907_calculate-cost-living-increase.html | 1,429,926,808,000,000,000 | text/html | crawl-data/CC-MAIN-2015-18/segments/1429246645606.86/warc/CC-MAIN-20150417045725-00207-ip-10-235-10-82.ec2.internal.warc.gz | 450,589,635 | 13,016 | # How to Calculate the Cost of Living Increase
The cost of living is the amount of money it takes to attain a certain level of utility or a certain standard of living. A cost of living increase is the increase in the amount of money it takes to maintain a given standard of living. Cost of living increases are an important consideration for money managing and retirement planning, since it often takes more money to maintain a certain standard of living over time due to increases in prices.
Step 1:
Select the time period for which you want to compare the cost of living. For instance, you could find the cost of living increase between this year and last year.
Step 2:
Add up the total cost of expenses you paid during an average month in the first time period. For instance, if you were comparing this year to last year, add up your average monthly expenses from last year.
Step 3:
Add up the total cost of all the expenses you paid in an average month for the second time period. If you are comparing last year to this year, add up your current monthly expenses.
Step 4:
Subtract the amount you calculated in Step 2 from the amount you calculated in Step 3. This is the increase in your monthly cost of living between the two periods. If your standard of living or spending habits changed significantly between the two periods (for example, you have a bigger home, new car or go out to eat more often) the cost of living increase calculation may not be meaningful.
Related Searches
## Tips & Warnings
• Cost of living increases can only be calculated for time periods in the past up to the present.
• The U.S. Bureau of Labor Statistics releases new consumer price index (CPI) data on a regular basis. The consumer price index is a measure of the increase of prices in an economy over time. CPI data can be used as a rough estimate of the increase in the cost of living over time. | 395 | 1,897 | {"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-2015-18 | latest | en | 0.941861 |
https://gcbcbasketball.com/how-many-laps-around-each-court-makes-a-mile/ | 1,716,521,966,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058677.90/warc/CC-MAIN-20240524025815-20240524055815-00302.warc.gz | 234,524,365 | 38,727 | # How Many Laps Around Each Court Makes A Mile?
To measure this accurately, you need to know the court’s dimensions. On average, a regulation-size basketball court measures 94 feet long by 50 feet wide. That means one lap around the court equals 211 feet total (94 + 94 + 33). So if you divide 5280 feet (the number of feet in one mile) by 211, you get 24.836 – which, when rounded up, comes out as 18.
A standard high school basketball court is 84 feet long by 50 feet wide, making 19.7 laps necessary to complete one mile of running. If you’re looking to get in three miles in your workout, you’ll need 59 laps around the court—a daunting task!
### NCAA, NBA, And WNBA Basketball Courts:
Those playing on NCAA, NBA, and WNBA basketball courts would take 18.33 laps to complete one mile. This is because all these courts measure 94 feet by 50 feet and have four corners – which means that the total number of steps needed to do a lap around the entire court is 288 feet in length – equal to 18.33 laps around the perimeter of the court.
## How To Calculate The Number Of Laps Around A Basketball Court?
To accurately determine the number of laps around a basketball court that equal one mile, you must first find out the total length of one lap. To do this, use a measuring tape to measure all four sides of the court and add them to get the total length in feet. Once you have this figure, divide it by 5,280 (the number of feet in one mile) and round off your answer to getting your final result.
## How Many Laps Around A Basketball Court Makes A Half-Mile?
The length of the court from baseline to baseline is 94 feet. Due to this, one lap covers 94 feet times four sides for a total of 376 feet. A half-mile equals 2460 feet, so it would take 9.16 laps around the court—or three and one-quarter laps—to make up that distance. It may be helpful to count your footsteps as you go to keep track of your progress and not get lost in circles!
## How Many Laps Around A Basketball Court Makes A Quarter Mile?
The answer is 4.6 laps, So if you want to go for a quarter-mile run, you have to head out to your local basketball court or any other open field with marked lanes and begin running! A quarter-mile jog will take about 8-10 minutes, depending on your speed. For those more serious runners looking for a challenge, it’s possible to complete this distance in as little as four minutes! Once completed, take a break and stretch before continuing your workout routine!
## Dimensions Of Olympic Basketball Courts:
The Olympic basketball court is 91.86 feet long and 49.21 feet wide, with a total area of 4,493 square feet. The three-point line for men’s play is 19.69 feet away from the basket, while the women’s line is 17.72 feet away from their basket.
## Why Do Basketball Coaches Make Players Run Laps?
Stamina And Strength Training
Coaches use running laps as an effective way to help their players improve physical performance and conditioning on and off the court. For example, having players run shorter distances at higher speeds helps build agility, while longer runs can increase strength and endurance. Running also benefits basketball players by improving concentration levels during gameplay, leading to better performances.
Punishment
Running laps is one of the most commonly used disciplinary measures. Whether it’s to punish a player for missing practice or messing up a play, running laps can be an effective way to encourage better performance on the court.
## Conclusion:
Running laps around a basketball court is an effective way to measure miles. It can also be enjoyable and provide an opportunity to get outside and enjoy the fresh air. To find out how many laps you need to run, use the formula of approximately 1⅓ laps per one-tenth of a mile. With this formula, you can accurately measure your goals quickly and easily. | 859 | 3,868 | {"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 | 4 | CC-MAIN-2024-22 | longest | en | 0.932921 |
https://discourse.processing.org/t/creating-boundaries-for-my-ball-using-max-and-min/17367 | 1,656,123,022,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103033925.2/warc/CC-MAIN-20220625004242-20220625034242-00059.warc.gz | 253,507,882 | 5,220 | # Creating "boundaries" for my ball using max and min
So I was supposed to create this joystick like program in one of my questions but Im not sure how I can use functions max and min so the ball doesnt fly past the canvas when Im controlling it with my mouse.
float lineX = width/2;
float lineY = height/2;
float ballX = width/2;
float ballY = height/2;
final float DIAMETER_BALL = 20;
float moveX = mouseX-250;
float moveY = mouseY-250;
final float SPEED_OF_BALL = 1.2;
void setup() {
size(500,500);
}
void draw() {
background(255);
drawBall();
drawLine();
moveBall();
}
void drawBall() {
ballX = width/2;
ballY = height/2;
ballX = (ballX + moveX);
ballY = (ballY + moveY);
ellipse((ballX+moveX), (ballY+moveY), DIAMETER_BALL, DIAMETER_BALL);
}
void drawLine() {
lineX = width/2;
lineY = height/2;
line(lineX, lineY-150, lineX, lineY+150);
line(lineX-150, lineY, lineX+150, lineY);
}
void moveBall() {
moveX = (mouseX-250) / SPEED_OF_BALL;
moveY = (mouseY-250) / SPEED_OF_BALL;
}
Check the examples in the main page.This one would be helpful: Bounce / Examples / Processing.org
One point to watch:
The values of mouseX and mouseY are only available within `setup()` and `draw()`. You can define those variables globals but you need to assign their values in setup.
I recommend that instead of using 250 you sue `width/2` and `height/2`. this is easier to read and in case you change your sketch size, you don’t have to update all these “magic numbers”.
Finally, I like to say I do not fully understand what your function `drawBall()` does.
In general, when the ball is moving along the x axis and it reaches the left boundary (the width of you sketch) then you use something like `posX=min(width,posX)` which would return the min of the two values and assign it back to your posX. However, you will need to detect when this happens as you will need to reverse the velocity along the x axis as well so it “bounces back”. | 540 | 1,944 | {"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-2022-27 | latest | en | 0.81684 |
https://diy.stackexchange.com/questions/170685/above-ground-pole-support-for-sail-cloth/170693 | 1,571,575,078,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986707990.49/warc/CC-MAIN-20191020105426-20191020132926-00397.warc.gz | 443,854,227 | 31,816 | # above ground pole support for sail cloth
I need to create above ground, mobile "footings" for metal poles to hold sail cloth. I was thinking of a cardboard tubing, filled with concrete to hold a sleeve to insert the pole into that will hold a metal pole. the pole will hold a sail cloth. I am thinking of a 3-4' high, 16" wide concrete tube form. will this be sufficient to hold the poles for sail cloth?
• The shape is all wrong. You want something no taller in aspect than square. A short trapezoid would be good. I'm not sure how anyone can answer without a sail size and configuration, though. – isherwood Aug 6 at 14:26
It will be very hard to give guidance as to if a particular size of base will be adequate for your application. Here are some of the variables to consider:
1. Length of the metal pipe that holds the sail cloth. The longer (and presumed higher) this pole the more lateral leverage it can apply to the base.
2. Span of the sail cloth overhead. The longer span you have for the sail cloth the more tension it will place on the metal poles and in turn on the base.
3. Placing the base into the ground. I know that you desire a mobile application but be aware that putting the base assembly into the ground will give it a lot more stability that just letting it sit at ground level. It would also look a lot better than having a huge block of concrete sitting there. I can easily imagine a buried base that would only project enough from the ground level so that the dirt and debris does not easily get into the slide in hole for the metal poles.
Do also consider that even at the size you asked about that these bases would not be so "mobile". Concrete weighs approximately 150 pounds per cubic foot. A 16" (i.e. 1.333 feet) diameter concrete pillar 4 feet high is going to be:
V = 3.1415 x 0.666 feet x 0.666 feet x 4 feet
V = 5.57 cubic feet
Weight = 5.57 x 150 pounds
Weight = 836 pounds
That does not seem particularly mobile to me.
If you're building something that is hard to tip over, it needs to be wide and heavy. Your poles unfortunately have a huge lever arm compared to the base. The weight of your base multiplied by its radius needs to be more than the tension in the sail cloth multiplied by the height of the pole. If you have a 6' pole and a 16" wide base, the leverage works against you by a factor of 9 (because 6 * 12 / 8 = 9). I.e. you'd need at least 9 times more weight than tension!
Instead I think you'd be better with a simple pole and guy-ropes tied to weights. If the guy-ropes were really long, and the weights had a friction co-efficient of 1, the weights would need to be at least as heavy as the tension to avoid sliding. As the guys get shorter and steeper, the weights need to be heavier. At 45 degrees, about 150% of the tension. At 60 degrees, 200%.
Then your next problem is estimating how much tension the wind will cause. This is nearly impossible as the wind varies so much. I suggest estimating the force required to rip a corner off the sail cloth, as this is the most force that the sail can exert on the pole. | 737 | 3,089 | {"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-2019-43 | latest | en | 0.952932 |
https://studychacha.com/discuss/39530-industrial-automation-masters-degrees.html | 1,620,826,946,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243990929.24/warc/CC-MAIN-20210512131604-20210512161604-00035.warc.gz | 574,756,046 | 19,555 | 2021-2022 StudyChaCha
#1
October 10th, 2012, 12:02 PM
Super Moderator Join Date: Oct 2019
Industrial Automation Masters Degrees
I want to do Master of Business and Project Management in Industrial Automation course from Engineering Institute of Technology so can anyone provide me its course structure?
The syllabus of M.Tech Industrial Automation & Robotics Program offered by Department of Mechanical Engineering of National Institute of Engineering, Mysore is as follows:
APPLIED MATHEMATICS (4-0-0)
PART-A: Numerical Analysis
Unit I
Approximation & Errors. Significant figures, Accuracy & precision. Round off & truncation
errors. Numerical solution of equations – Fixed point iteration method false position method,
Secant method, Newton - Raphson method, multiple roots.
(SLE: Bisection, Graphical methods).
Unit II
Polynomials in Engineering & Science, Birge- Vieta method, Muller’s method, Horner’s
method, Graeffe’s roots squaring method. Numerical differentiation
(SLE: Applications to Engineering problems). 9 Hrs
Unit III
Numerical Integration – Newton cote’s quadrature formula. Trapezoidal rule (SLE:
Simpson’s one third and three eighth’s rule), Boole’s rule, Weddle’s rule. Romberg
integration.Numerical double integration.Gauss quadrature formulae -Gauss
Legendre.Numerical solution of ODE – Taylor series method, Adam-Bashforth- Moulton
method.
(SLE: Runge-Kutta method of order IV).
Unit IV
Solution of system of linear algebraic equations, Triangularization method, Cholesky’s
method, Partition method, Gauss Seidel iterative method.
(SLE: Gauss elimination method). 9 Hrs
Unit V
Eigen values & Eigen vectors,Bounds on eigen values. Given’s method, Jacobi’s method for
diagonalisation of symmetric matrices ,Rutishauser method for arbitrary matrices, Power
method , Inverse power method.
(SLE: Analytical method to obtain eigen values and eigen vectors) 9 Hrs
Unit VI
Vectors & vector spaces, Linear Transformations - Kernel, Range.Matrix of linear
transformation.Inverse linear transformation, Inner product, Length / Norm.Orthogonality,
orthogonal projections.Orthonormalbases.Gram-Schmidt process.Least square problems.
(SLE: Applications)
Books for Reference:
1.Introductory Methods of Numerical Analysis – S.S. Sastry, 5th edition.
2.Numerical Methods in engineering and science – B.S.Grewal ,Khanna Publications 8th
edition, 2009.
3.Higher Engineering Mathematics – Dr. B.V. Ramana,5th edition, Tata McGraw – Hil
publications.
4.Linear Algebra – Larson &Falvo (Cengage learning)
5.Numerical Methods – M.K. Jain S.R.K. Iyengar R.K. Jain, 2nd edition, New Age
International Pvt Ltd Publishers.
Syllabus M.Tech Industrial Automation & Robotics National Institute of Engineering, Mysore
#2
October 11th, 2012, 12:41 PM
Super Moderator Join Date: Dec 2011
Re: Industrial Automation Masters Degrees
You want course structure of Master of Business and Project Management in Industrial Automation course of Engineering Institute of Technology so following is its course structure:
Units 1 to 9 are presented by the Engineering Institute of Technology.
Unit 1: Project Scope, Time and Cost Management
Unit 2: Project Quality, Risk and Procurement Management
Unit 3: Industrial Process Instrumentation
Unit 4; Industrial Process Control Systems
Unit 5: Programmable Logic Controllers, SCADA and Distributed Control Systems
Unit 6: Industrial Data Communications
Unit 7: Management of Industrial Automation Projects
Unit 8: Processes, Tools and Templates for Management of Industrial Automation Projects
Unit 9: Integrated Management of Health, Safety and Environment (HSE)
Unit 10: Professional Development and Ethics
Unit 11: Project/Program Human Resources, Teams, Communication and Integration Management
Unit 12: Project/Program Strategic Intent, Business Case, Framework and Governance
Unit 13: Project and Program Information and Communication Systems
Unit 14: Venture/Project Economics and Finance
Unit 15: Research Project
__________________
#3
November 21st, 2019, 05:45 PM
Unregistered Guest
Re: Industrial Automation Masters Degrees
Can you provide me the syllabus of M.Tech. Industrial Automation & Robotics Program offered by Department of Mechanical Engineering of National Institute of Engineering, Mysore? | 985 | 4,258 | {"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-2021-21 | latest | en | 0.688284 |
https://www.neetprep.com/question/57677-two-tuning-forks-B-sounded-together-produce--beats-persecond-slightly-loaded-wax-produce--beats-when-soundedagain-frequency--frequency-B-will----/126-Physics--Waves/690-Waves | 1,575,646,639,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540488870.33/warc/CC-MAIN-20191206145958-20191206173958-00451.warc.gz | 782,460,680 | 68,975 | # NEET Physics Waves Questions Solved
If two tuning forks A and B are sounded together, they produce 4 beats per second. A is then slightly loaded with wax, they produce 2 beats when sounded again. The frequency of A is 256. The frequency of B will be
(1) 250
(2) 252
(3) 260
(4) 262 | 84 | 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} | 2.609375 | 3 | CC-MAIN-2019-51 | latest | en | 0.865915 |
https://www.pokerstrategy.com/forum/thread.php?threadid=349465 | 1,529,468,734,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267863411.67/warc/CC-MAIN-20180620031000-20180620051000-00493.warc.gz | 880,941,438 | 21,042 | # Article: Variance over the increasing hand sample
• Bronze
Joined: 10.04.2011
What makes poker the most thrilling and painful at the same time? Variance. Everyone can be a winner. But for how long?
A few months ago me and my poker buddies had the debate: ‘What happens with the variance over the increasing hand sample?’ The answer seems obvious to everybody, but surprisingly it was different. Everyone’s arguments, reasoning and logic seems legit. But statistics can be tricky and sometimes, just like in poker, your ‘feeling’ is not enough. So I, the nerdiest guy in the group, decided to show off.
Basics
“Half of the answer lies within a good constructed question,” wise man once said. If I’m first to say it, then I’m that wise man. So, let’s break it down. First, definition of variance: ‘Variance is a number which measures how far a set of numbers is spread out around mean.’ More useful term is standard deviation which is the square root of variance. Both represent how swingy your poker results are. But standard deviation can tell you something more. Let’s get right to the poker example. Our hero plays 0 EV (mean) game with standard deviation 100 big blinds per 100 hands1 (he will stay the same rake back grinder through the whole post). Because of this exact standard deviation, there is 68% chance2 his bankroll will change less than 100 big blinds after 100 hands.
But what would happen if hero played a different number of hands?
Variance of winnings
Let’s start with the obvious one, variance of winnings. It tells you how much swings you can expect in big blinds or moneywise. But, does it get smaller or bigger with more hands played? Think for yourself but be careful. Your peasant logic can get outplayed by the lady variance. And this time you can’t blame her.
Here are seven different simulated graphs of hero’s winnings trough 100k hand sample. To make those graphs more understandable you can think about them as graphs of players with the exact same skill level or it can be the same player3 who made 700k hands and those are parts4 of his graph.
It’s kind of hard to believe, those are all simulated graphs of 0 EV player. From one, you would think he plays like BalugaWhale, from another, like beluga whale. As you can see, lines are getting more and more dispersed. That means bigger variance and also standard deviation. If you still have thoughts such as “luck will cancel out at the end of the day”, let me show you another graph.
Let’s say hero plays 1k hands and at the end of a hand sample, he makes a dot of his winnings. Then he goes from scratch again and makes 2k hands. Dot again. Then 3k hands, 4k, 5k… So on, till the 1M hands. Here is what he gets:
One more time, it gets dispersed.
To make things clear, I even displayed a standard deviation line. It’s proportional to the square root of number of number of hands. Equation for standard deviation is:
(1)
std dev=K*sqrt(#hands)
In our case, K is equal to5 10. For example, if your standard deviation is 85bbs/100hands, the value of K would be 8.5. You can actually calculate a few interesting and useful information from equation (1). You can see that any fixed hand sample has the same standard deviation no matter where on the graph it is located.
Those results are somewhat hard to understand at first sight so let’s bring a few examples, which make ‘logical’ sense:
• You can’t win more than one stack in a single hand. If you play more hands, there is possibility to win more. That tiny portions add to dispersion.
• Let’s say you roll a ball in a straight line on the concrete which is not completely flat and has some sand on it. Ball will eventually go out of the line. The longer way it makes, there is a higher chance it will be farther away from the initial line. The last graph represents this situation from bird eye perspective.
Before you start playing as little as possible to lower your variance, read on.
Variance of a win rate
This case turns around the whole picture. Win rate is defined as big blinds won divided by number of hands6 played. Considering this we get a bit different standard deviation’s equation:
(2)
std dev=100*K / sqrt(#hands)
Actually, something dramatically changed. Standard deviation now gets smaller trough the increasing hand sample. Let’s just divide the winnings with a number of hands/100 and here is what we get from Graph 1:
Win rates are getting closer to 0 bb/100hands, which is hero's genuie win rate.
Same thing for Graph 2:
Obviously, the more hands you play, the more accurate your win rate should become. Using the (2) equation we can calculate how many hands you need to lower your standard deviation to 1 bb/100hands. That means, if your win rate is 1bb/100hands, you can be 67% sure your win rate is somewhere between 0 and 2 bb/100hands. And for that, you need 1 million hands. So grind on…
Conclusion
Understanding variance is one of the key factors of becoming a good poker players. Especially because of your mental game. You should expect sick things to happen and be capable of facing them. Don’t get too much into those poker software lines. They aren’t as good predictor of your skill as you may think, even EV line. Just remember – never underestimate variance and overestimate your poker skill. I wish you all the best at the poker tables.
wnbMG is a poker player who blames the variance for not making it as a pro. Nevertheless, he luckily had enough skill to become Slovenian poker champion (team event).
Notes:
1 This standard deviation is somewhat standard for no limit Texas hold' em shorthanded cash games.
2 Variables have to be normally distributed. In our case, they are.
3 We assume player and his opponent’s skill stayed the same for all 700k hands.
4 We will use that example later on.
5 Where do we get this 10? Our standard deviation should be 100 per 100 hands. So just insert this in equation you get. 100 = K*sqrt(100). K should be 10. It also means standard deviation for 1 hand.
6 We usually define win rate in big blinds won per 100 hands. That’s why number of hands should be divided by 100.
• 4 replies
• Bronze
Joined: 17.03.2008
So how lucky are you when you're a winning player at live cash games? Who can play 1 million hands at a live cash game?
• Super Moderator
Super Moderator
Joined: 02.09.2010
Good question, Maverick290784.
In live cash games there are personal skills that help that don't exist online.
Hiding ones emotions, and reading the emotions of the other players, for example.
Cheers,
VS
• Bronze
Joined: 10.04.2011
Originally posted by Maverick290784
So how lucky are you when you're a winning player at live cash games? Who can play 1 million hands at a live cash game?
You don't play poker to see your winrate, you play poker because you think you have an edge. But after a decent sample you can check it with specific statistical tools.
Btw, edges/winrates are ussually much bigger in live game, so you don't need as many hands as online to be certain if you are winner or not.
• Bronze
Joined: 25.01.2009
Originally posted by VorpalF2F
Good question, Maverick290784.
In live cash games there are personal skills that help that don't exist online.
Hiding ones emotions, and reading the emotions of the other players, for example.
Cheers,
VS
Here in my region there is a huge poker boom and a lot of people are starting to play at clubs, most of them play only live poker. I do agree with that argument, there are skills that you can use in a live table which are not available online, but on the other hand online you can get what - in my honest opinion - are the best tells about a player, that is, the stats and how he plays his hands.
Originally posted by wnbMG
Originally posted by Maverick290784
So how lucky are you when you're a winning player at live cash games? Who can play 1 million hands at a live cash game?
You don't play poker to see your winrate, you play poker because you think you have an edge. But after a decent sample you can check it with specific statistical tools.
Btw, edges/winrates are ussually much bigger in live game, so you don't need as many hands as online to be certain if you are winner or not.
I agree in parts, but even though winrates are usually much bigger, the variance can hit harder, specially depending on the situation you play. I usually play 5/10 and 10/20 games live and - well - sometimes it's really hard to be ahead, specially when you get those kind of players who don't really care about the money.
As for the article, I think it's a very good inside for those who don't really understand statistics, I really liked it | 1,999 | 8,607 | {"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} | 4.15625 | 4 | CC-MAIN-2018-26 | latest | en | 0.946823 |
https://alexanderpruss.blogspot.com/2021/04/decisions-in-heaven.html | 1,621,293,176,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991870.70/warc/CC-MAIN-20210517211550-20210518001550-00428.warc.gz | 119,564,528 | 39,902 | ## Monday, April 12, 2021
### Decisions in heaven
Suppose I will live forever in heaven, and I have two infinite decks of cards. Each card specifies the good things that will happen to me over the next day. Every card in the left deck provides a hundred units of goods. Every card in the right deck provides a thousand units of goods.
Each day I get to draw the top card from a deck I choose and then I get the specified goods.
Consider three of the strategies I could opt for:
1. Always draw from the left deck.
2. Always draw from the right deck.
3. Alternate between decks.
Clearly, strategy 1 is not a good idea, so let’s put that aside.
There is an obvious argument for preferring 2 to 3. If I opt for strategy 2, then every other day I will be much better off than on strategy 3, and on the other days I will be at least as well off as on strategy 3.
But there is also an argument for preferring 3 to 2: on option 3, over the course of eternity, I get all the goods from both decks.
Moreover, even if one does not buy the argument that option 3 is better than option 2, it seems no worse: for while on option 3, the greater goods of the right deck get delayed more, a good is no less valuable for being pushed further off into the future.
Walter Van den Acker said...
Of course you should always draw from the right deck. Drawing from the wrong deck would be a bit silly.
IanS said...
“… over the course of eternity, I will get all the goods …” put strictly, means that for each good, there will be a time at which you will get it. But there will be no time at which you will have got them all. In fact, there will be no time at which you will have got more than an infinitesimal part of them. So trying to get them all is not a reasonable aim.
“In the eyes of God” (who can presumably see eternity at a glance), if you choose 3, you will get all the goods in the left deck. With 2, you won’t. But your eyes are not godlike. You cannot experience eternity. You can only experience things day by day. On that basis, 2 beats 3.
SMatthewStolte said...
The hope for a good is itself a kind of good, so if I opt for alternating, then I benefit even in the present. | 530 | 2,183 | {"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-2021-21 | latest | en | 0.956682 |
https://mattcoaty.com/2012/06/23/1128/ | 1,685,936,494,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224650620.66/warc/CC-MAIN-20230605021141-20230605051141-00122.warc.gz | 432,344,856 | 35,588 | # Math Reasoning and Feedback
Image by: J. Creationz
Having math reasoning skills is important. Generally, math reasoning skills are taught and incorporated in early elementary school. In math, a problem is what a student is asked and expected to answer. If a student is unable to answer why their answer is correct, I believe that the student might not fully grasp the mathematical concept. The student might not be utilizing math reasoning skills.
For example, a student that measures area in linear feet might not completely have an understanding that area is measured in square units. The student could have the correct numerical answer, but include the wrong unit (centimeters compared to square centimeters).
How is mathematical reasoning taught? I’m going to be taking a proactive step next year to give opportunities for my students to utilize math reasoning. I’m deciding to use higher level questioning to enable students to think of the process in finding the solution. The learning process is key. I’ve found that math instruction isn’t always linear, just as mathematical reasoning isn’t rigid. By asking students why/how they arrived at a solution is vital in understanding their thinking.
As I’m planning for next school year, I’ve decided to ask students to explain their reasoning more frequently. By hearing their reasoning, I’m in a better position to give direct feedback. All math questions have some type of reasoning. I believe that multiple solution / open-ended questions can be used to display mathematical reasoning. Students need to be able to explain why they responded with a specific answer and what methods/connections were utilized to solve the problem. Based on the math Common Core, students are expected to reason abstractly and quantitatively. When students describe their mathematical process, teachers are better able to diagnose and assess a student’s current level of understanding. Math reasoning isn’t always quantifiable, but it can be documented via journaling and other communication methods. More importantly, teachers will be able to provide specific feedback to help a student understand concepts more clearly. I also feel that this questioning process develops self-confidence in students and prepares them to become more responsible for their own learning. See the chart below.
## Author: Matt Coaty
I've taught elementary students for the past 14 years. I enjoy reading educational research and learning from my PLN. Words on this blog are my own. | 521 | 2,527 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2023-23 | latest | en | 0.939702 |
https://www.vidyakul.com/ncert-solutions-class-12-physics-chapter-3-Current-Electricity/ | 1,624,473,003,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488539764.83/warc/CC-MAIN-20210623165014-20210623195014-00143.warc.gz | 971,079,506 | 30,901 | NCERT Solutions for Class 12 Physics Chapter 3 | Vidyakul
# NCERT Solutions for Class 12 Physics Chapter 3
## NCERT Solutions for Class 12 Physics Chapter 3 Current Electricity – PDF Download
Vidyakul presents NCERT Solutions for Class 12 Physics Chapter 3 for students who want to learn fast and score more in the examination. These NCERT Solutions for Class 12 Physics Chapter 3 answers all the questions mentioned in Chapter-Current Electricity of NCERT Textbook. The NCERT Solutions for Class 12 Physics Chapter 3 pdf also contains notes which would help you understand the basic concepts in a simple yet effective way.
Current Electricity is an essential chapter in class 12 Physics, both from the board’s perspective as well as competitive exam’s perspective. The NCERT Solutions for Class 12 Physics Chapter 3 PDF covers the entire CBSE Physics syllabus and solutions to in text questions as well. Thus students are advised to follow these solutions and notes as not only they will help them to achieve more marks, but could also be used as the answer key to complete an assignment or homework faster.
Given below is an overview of the topics covered in NCERT Solutions for Class 12 Physics Chapter 3 PDF:
• ELECTRIC CURRENT
• ELECTRIC CURRENTS IN CONDUCTORS
• OHM’S LAW
• DRIFT OF ELECTRONS AND THE ORIGIN OF RESISTIVITY
• LIMITATIONS OF OHM’S LAW
• RESISTIVITY OF VARIOUS MATERIALS
• TEMPERATURE DEPENDENCE OF RESISTIVITY
• ELECTRICAL ENERGY, POWER
• COMBINATION OF RESISTORS – SERIES AND PARALLEL
• CELLS, EMF, INTERNAL RESISTANCE
• CELLS IN SERIES AND IN PARALLEL
• KIRCHHOFF’S RULES
• WHEATSTONE BRIDGE
• METER BRIDGE
Electrical Current is the flow of charged particles. The flow of charges will be constant in current electricity. For the current to flow we need a circuit. The electrons in a current flow from negative to positive. Electrons flow in the direction which is opposite to the direction of electric current.Know More about this in NCERT Solutions Class 12 Physics Chapter 3 PDF.
Download the FREE PDF of Physics Class 12 NCERT Solutions Chapter 3-Current Electricity and start your preparation with Vidyakul! | 498 | 2,142 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2021-25 | latest | en | 0.859733 |
https://www.studysmarter.co.uk/explanations/math/calculus/tangent-lines/ | 1,721,214,528,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514759.39/warc/CC-MAIN-20240717090242-20240717120242-00622.warc.gz | 858,871,480 | 71,539 | # Tangent Lines
In Latin, the word tangent means "to touch." So then, a tangent line is a line that touches. Consider a bicycle moving along the flat pavement. The road is essentially tangent to the bicycle wheel as it touches the wheel at a point. In this article, we will further discuss the meaning of a tangent line, a tangent line's formula, and what the slope of a tangent line means.
#### Create learning materials about Tangent Lines with our free learning app!
• Flashcards, notes, mock-exams and more
• Everything you need to ace your exams
## Definition and formula of a Tangent Line
A tangent line is a line that "just touches" the point $P$. It can also be defined as a line joining 2 infinitely close points on a curve.
The tangent line to a curve$f\left(x\right)$ at a point $P$, with coordinates $\left(a,f\left(a\right)\right)$, is the line through $P$ with slope
$m=\underset{x\to a}{\mathrm{lim}}\frac{f\left(x\right)-f\left(a\right)}{x-a}$
if the limit exists.
## Equation of a Tangent Line
Once the slope $m$ is found, the equation of a tangent line is the same as any other line in point-slope form through a point $\left(a,f\left(a\right)\right)$:
$\left(y-f\left(a\right)\right)=m\left(x-a\right)$
## Tangent Lines on a Graph
In the graph below, we say that $y$ is a tangent line to the curve $f$ at point $P$. Or, we can say that $y$ is tangent to the curve $f$ at point $P$.
The tangent line in green merely touches the curve f at point P - StudySmarter Original
Notice how the tangent line "just touches" the curve at point P.
## The slope of a tangent line
### Geometry
The slope of the tangent line at a point on a curve is equal to the slope of the curve at that point. The assumption behind tangent lines is when looking at the graph of a curve, if you zoom in close enough to a segment of the curve, the curve will look indistinguishable from the tangent line.
For example, let's zoom in on the graph above.
Zooming in at the point where the tangent line touches the curve - StudySmarter Original
Zooming in a bit more...
Here we can see that the tangent line and the point of the curve where the tangent line touches are indistinguishable - StudySmarter Original
Notice how the tangent line comes from connecting two infinitely close points on a curve.
### Additional tangent line slope equation
There is another version of the equation for the slope of the tangent line that can be easier to work with. This equation says that the slope of the tangent line $m$ is
$m=\underset{h\to 0}{\mathrm{lim}}\frac{f\left(a+h\right)-f\left(a\right)}{h}$
This equation sets $h=x-a$ and $x=h+a$. As $x$ approaches $a$, $h$ approaches 0. Thus, the supplementary equation of the tangent line is formed.
This equation for the slope of the tangent line should look familiar to you.... It is the equation for the derivative of a function at a point (a, f(a)). So, we can say that the slope of the tangent line on a curve at point P is equal to the derivative of the curve at point P !
## Examples of finding the Tangent Line Equation
### Example 1
Find the equation for the tangent line to $f\left(x\right)={x}^{2}$ at the point (2, 4).
As we are given $f\left(x\right)$ and a point, all we need to form the equation of the tangent line is the slope. To find the slope, we will use the supplementary equation of the tangent line.
The slope of the tangent at (2, 4) is
$\begin{array}{rcl}m& =& \underset{h\to 0}{\mathrm{lim}}\frac{f\left(2+h\right)-f\left(2\right)}{h}\\ & =& \underset{h\to 0}{\mathrm{lim}}\frac{{\left(2+h\right)}^{2}-4}{h}\\ & =& \underset{h\to 0}{\mathrm{lim}}\frac{{h}^{2}+4h+4-4}{h}\\ & =& \underset{h\to 0}{\mathrm{lim}}\frac{{h}^{2}+4h}{h}\\ & =& \underset{h\to 0}{\mathrm{lim}}\frac{h\left(h+4\right)}{h}\\ & =& \underset{h\to 0}{\mathrm{lim}}h+4\\ & =& 4\end{array}$
So, the equation of the line tangent to f(x) at (2, 4) is $y=4\left(x-2\right)+4$.
Recall how the slope of the tangent line is the same of the derivative. So, we could also simply take the derivative of $f\left(x\right)$ and plug in $x=2$ to find the slope of the tangent line at the point $\left(2,4\right)$.
$f\text{'}\left(x\right)=2x\phantom{\rule{0ex}{0ex}}f\text{'}\left(2\right)=4$
The graph of f(x) and the line tangent to f(x) at (2, 4) - StudySmarter Original
### Example 2
Find the equation of the tangent line to the curve $f\left(x\right)=x-{x}^{3}$ at the point (1, 0).
Again, as we are given $f\left(x\right)$ and a point, all we need to form the equation of the tangent line is the slope. To find the slope, we will use the supplementary equation of the tangent line.
With , the slope of the tangent at (1, 0) is:
$\begin{array}{rcl}m& =& \underset{h\to 0}{\mathrm{lim}}\frac{f\left(1+h\right)-f\left(1\right)}{h}\\ & =& \underset{h\to 0}{\mathrm{lim}}\frac{\left[\left(1+h\right)-{\left(1+h\right)}^{3}\right]-0}{h}\\ & =& \underset{h\to 0}{\mathrm{lim}}\frac{\left(1+h\right)-\left({h}^{3}+3{h}^{2}+3h+1\right)}{h}\\ & =& \underset{h\to 0}{\mathrm{lim}}\frac{{h}^{3}-3{h}^{2}-2h}{h}\\ & =& \underset{h\to 0}{\mathrm{lim}}{h}^{2}-3h-2\\ & =& -2\end{array}$
So, the equation of the line tangent to f(x) at (1, 0) is $y=-2\left(x-1\right)$.
Again, recall how the slope of the tangent line is the same of the derivative. So, we could also simply take the derivative of $f\left(x\right)$ and plug in 1 to find the slope of the tangent line at the point $\left(1,0\right)$.
$f\text{'}\left(x\right)=1-3{x}^{2}\phantom{\rule{0ex}{0ex}}f\text{'}\left(1\right)=1-3\left(1\right)=-2$
The graph of f(x) and the line tangent to f(x) at (1, 0) - StudySmarter Original
## Tangent Lines in a Circle
A line is said to be tangent to a circle if it touches the circle at exactly one point. If a line is tangent to a circle at a point $P$, then the tangent line is perpendicular to the radius drawn to point $P$.
A line that is tangent to a circle at point P is perpendicular to the radius drawn to point P - StudySmarter Originals
For more information on tangent lines in a circle, check out our article on Tangent of a Circle!
## Tangent Lines - Key takeaways
• A tangent line is a line that touches a curve at a fixed point $P$
• The equation of a tangent line in point-slope form is $\left(y-f\left(a\right)\right)=m\left(x-a\right)$
• The slope of the tangent line at a point on a curve is equal to the slope of the curve at that point
• If you zoom in close enough to a segment of a curve, the tangent line at the segment and the curve will look indistinguishable
• In a circle, the tangent line drawn at any point is perpendicular to the radius at that point.
#### Flashcards in Tangent Lines 9
###### Learn with 9 Tangent Lines flashcards in the free StudySmarter app
We have 14,000 flashcards about Dynamic Landscapes.
How do you find tangent line?
To find a tangent line, use the equation for the slope of a tangent line at a point and plug in the slope and the point into point-slope form.
What is tangent line of a curve?
A tangent line of a curve is a line that touches the curve at a fixed point P.
What defines a tangent line?
A tangent line is a line that touches the curve at a fixed point P. The slope of the point at the curve is equal to the slope of the tangent line at point P.
How to find the equation of line tangent?
To find the equation of a tangent line, use the equation for the slope of a tangent line at a point and plug in the slope and the point into point-slope form.
How to find slope of a tangent line?
The slope of a tangent line can be found using the limit definition of the slope of a tangent line.
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• Checked by StudySmarter Editorial Team | 2,375 | 8,483 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 41, "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} | 4.5 | 4 | CC-MAIN-2024-30 | latest | en | 0.852859 |
https://www.physicsforums.com/threads/the-gradient-direction-and-rate-of-maximum-increase.550717/ | 1,585,903,864,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370510352.43/warc/CC-MAIN-20200403061648-20200403091648-00449.warc.gz | 1,042,943,416 | 15,058 | # The Gradient direction and rate of maximum increase
## Homework Statement
What is the direction and rate of maximum increase?
f(x,y) = x^2 + y^3, v = <4,3>, P = (1,2)
## The Attempt at a Solution
The direction should be as same as the gradient so < 2,12>
The rate of maximum increase is magnitude of the gradient so sqrt(4+144) = sqrt(148).
Is that right?
Related Calculus and Beyond Homework Help News on Phys.org
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Yes. | 127 | 462 | {"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-2020-16 | longest | en | 0.860583 |
https://ko.coursera.org/learn/linear-algebra-machine-learning/reviews?authMode=signup&page=83 | 1,642,402,221,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300343.4/warc/CC-MAIN-20220117061125-20220117091125-00027.warc.gz | 382,834,550 | 131,127 | Mathematics for Machine Learning: Linear Algebra(으)로 돌아가기
# 임페리얼 칼리지 런던의 Mathematics for Machine Learning: Linear Algebra 학습자 리뷰 및 피드백
4.7
별점
10,378개의 평가
2,077개의 리뷰
## 강좌 소개
In this course on Linear Algebra we look at what linear algebra is and how it relates to vectors and matrices. Then we look through what vectors and matrices are and how to work with them, including the knotty problem of eigenvalues and eigenvectors, and how to use these to solve problems. Finally we look at how to use these to do fun things with datasets - like how to rotate images of faces and how to extract eigenvectors to look at how the Pagerank algorithm works. Since we're aiming at data-driven applications, we'll be implementing some of these ideas in code, not just on pencil and paper. Towards the end of the course, you'll write code blocks and encounter Jupyter notebooks in Python, but don't worry, these will be quite short, focussed on the concepts, and will guide you through if you’ve not coded before. At the end of this course you will have an intuitive understanding of vectors and matrices that will help you bridge the gap into linear algebra problems, and how to apply these concepts to machine learning....
## 최상위 리뷰
CS
2018년 3월 31일
Amazing course, great instructors. The amount of working linear algebra knowledge you get from this single course is substantial. It has already helped solidify my learning in other ML and AI courses.
NS
2018년 12월 22일
Professors teaches in so much friendly manner. This is beginner level course. Don't expect you will dive deep inside the Linear Algebra. But the foundation will become solid if you attend this course.
필터링 기준:
## Mathematics for Machine Learning: Linear Algebra의 2,087개 리뷰 중 2051~2075
교육 기관: Roy M K
2021년 2월 3일
There are few or no worked out examples, making it hard to solve quizes.
교육 기관: Achal A
2021년 8월 23일
교육 기관: Rishabh J
2019년 2월 27일
Not as challenging. Too basic
교육 기관: Andrey L
2021년 3월 23일
Not for the beginner at all
교육 기관: Yash G
2019년 12월 14일
not clearly explained
교육 기관: Wei T L
2019년 7월 10일
it is too expensive
교육 기관: Sreeja D
2020년 11월 27일
Tough syllabus.
교육 기관: Petey C
2021년 8월 22일
This course is seriously lacking. I spent almost all of my time on Khan Academy due to the horribly high-level video lectures. The lectures in this course threw out strategies and ideas without explanation of the foundational mathematics behind them. Then the students are expected to apply these unexplained strategies to complex problems in the quizzes.
On the other hand, the programming assignments were a joke. They provided zero learning experience as everything needed to solve the assignments was provided in the already completed sections of code. Minimal thought was involved. This course needs a serious overhaul. These are complex topics that require more than a 4-10 minute video to explain. I understand that this is not an actual college course, and as such the professors cannot go as in depth comparatively. However, in its current state, this course has given me little understanding of linear algebra. I would suggest potentially increasing the price of the course so more in-depth lectures can be provided.
교육 기관: NG, S L
2021년 6월 1일
This is absolutely not a beginner's course. The professors did well at explaining but there is a huge gap between what I've heard and what I've done in the exercises. There is not enough information to complete the exercises. When they recommended Khan Academy as a reference source, I feel that it should be a complete whole alternative set of source instead of a supplementary source because Khan Academy explains much slower but in-depth. 5 weeks is not enough for me to go from scratch to an advanced learner. I couldn't retain what I've learnt too because all I heard were words that mean nothing to a novice like me.
교육 기관: Kim K
2020년 9월 18일
I wish I had read the most recent reviews for this course. I thought that by taking an online course, they would have produced a course that had a logical flow to it. I was trying to get caught up on some math that I missed taught by a TA that was all over the place. This was the wrong platform. I had to figure out what the first teacher was talking about by going online and learning it there. I was looking for a structured course that would teach me some basics. Not worth the money.
교육 기관: Chris D
2020년 2월 28일
The lectures don't cover the material in the quizzes. The only supplemental material is a single, incomplete formula sheet. In fact, it's hard to see what resources they are trying to provide other than a forum where other users fill in the many gaps and suggest alternate sources of information. Maybe it picks up significantly in weeks 4 and 5, but after going through weeks 1-3, spending most of my time trawling the forums and looking for study/learning materials on other sites, I'm done.
교육 기관: Lea S
2020년 7월 19일
While the course gave a rather high-level theoretical view of several concepts, the quizzes were based in applying those theories. Without examples of how to apply the theories, I don't feel I have a good understanding of the concepts or how they can be useful in real-world scenarios. I needed to use other resources to actually understand the concepts to be able to complete the course, which made the course rather frustrating and useless.
교육 기관: John D
2019년 6월 30일
Overall, rushed and poorly explained. It attempts to condense a semester-long course into laughably short videos (maybe 30 mins per "week"). If you had linear algebra before, probably won't be very challenging. But if you haven't... there is no help/explanation to the quizes and the message boards are filled with unanswered questions. Essentially, this course builds a very shaky foundation.
교육 기관: Tom D
2020년 5월 10일
The lectures will cover basic principles and then rather than build on basic principles to solidify the knowledge with practice and repetition, this goes straight to extremely challenging scenarios and me banging my head against a wall. Many comments within the forums are from other students like me who had to access external resources to bridge the gaps.
Not worth the time.
교육 기관: Gemma G
2021년 3월 8일
Do not bother with this course. There are some amazing free resources on youtube (3 blue 1 brown, and rootmath are two). This is poorly explained, the grading system is ridiculous and buggy, there is zero support from the tutors. Worst Coursera course I have ever taken.
교육 기관: Aerrow N
2021년 8월 28일
There is no support or interest from the creators to help you through and the automatic grading scripts are erroring out. You will never complete the course as the final assignment wont work at all and no one will help.
교육 기관: Rajeesh O T
2020년 10월 1일
to be honest it was not that understandable i had to go through other materials and videos to get some concepts. i think I speak for a few portion of people who too would have felt the same!
교육 기관: Rawitch M
2020년 4월 14일
Unclear explanation, week 3 final assignment is a huge discouraging no matter how you can do the previous exercise with ease or not. Maybe this course is not for me.
교육 기관: Ian G
2019년 12월 18일
Bad content and broken links run otherwise adequate lectures. Fortunately I was taking this course as a refresher--there are much better ways to learn linear algebra
교육 기관: Cindy C
2020년 7월 5일
Didn't thoroughly explain concepts. Tested on concepts not covered. I regularly needed to go to Khan Academy videos to fill in the gaps this course has.
교육 기관: Kimberely C
2019년 12월 19일
Not very good. Had to use YouTube as I had no knowledge of Python. Needs to have more examples and walk you through Python to be able to understand.
교육 기관: CHOI H J
2021년 12월 3일
It teaches simple math, very briefly,
and make test that needs Python and other Math basic knowledges.
It isn't kind for beginner!
교육 기관: Priyabrota S
2020년 6월 8일
No practical example. Like University hard mathematics lecture. This course content does not relate to the course title.
교육 기관: Alp S
2019년 7월 20일
It is disrespectful that programming assignments are not accessible for audit users. | 1,983 | 8,193 | {"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-2022-05 | longest | en | 0.819107 |
https://online.stat.psu.edu/stat200/book/export/html/479 | 1,721,291,157,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514826.9/warc/CC-MAIN-20240718064851-20240718094851-00548.warc.gz | 374,523,387 | 3,501 | # 1.4.2 - Causal Conclusions
1.4.2 - Causal Conclusions
In order to control for confounding variables, participants can be randomly assigned to different levels of the explanatory variable. This act of randomly assigning cases to different levels of the explanatory variable is known as randomization. An experiment that involves randomization may be referred to as a randomized experiment or randomized comparative experiment. By randomly assigning cases to different conditions, a causal conclusion can be made; in other words, we can say that differences in the response variable are caused by differences in the explanatory variable. Without randomization, an association can be noted, but a causal conclusion cannot be made.
Note that randomization and random sampling are different concepts. Randomization refers to the random assignment of experimental units to different conditions (e.g., different treatment groups). Random sampling refers to probability-based methods for selecting a sample from a population.
Randomization
The act of randomly assigning cases to different levels of the explanatory variable
Causation
Changes in one variable can be attributed to changes in a second variable
Association
A relationship between variables
## Example: Fitness Programs
Two teams have designed research studies to compare the weight loss of participants in two different fitness programs. Each team used a different research study design.
The first team surveyed people who already participate in each program. This is an observational study, which means there is no randomization. Each group is comprised of participants who made the personal decision to engaged in that fitness program. With this research study design, the researchers can only determine whether or not there is an association between the fitness program and participants' weight loss. A causal conclusion cannot be made because there may be confounding variables. The people in the two groups may be different in some key ways. For example, if the cost of the two programs is different, the two groups may differ in terms of their finances.
The second team of researchers obtained a sample of participants and randomly assigned half to participate in the first fitness program and half to participate in the second fitness program. They measured each participants' weight twice: both at the beginning and end of their study. This is a randomized experiment because the researchers randomly assigned each participant to one of the two programs. Because participants were randomly assigned to groups, the groups should be balanced in terms of any confounding variables and a causal conclusion may be drawn from this study.
[1] Link ↥ Has Tooltip/Popover Toggleable Visibility | 521 | 2,759 | {"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} | 3.5 | 4 | CC-MAIN-2024-30 | latest | en | 0.886001 |
http://mathhelpforum.com/differential-geometry/index221.html | 1,506,005,030,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818687820.59/warc/CC-MAIN-20170921134614-20170921154614-00058.warc.gz | 218,757,904 | 14,109 | # Differential Geometry Forum
Analysis, Topology and Differential Geometry Help Forum
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### nokio720 price
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Student[Calculus1]
NewtonQuotient
return the Newton quotient of an expression
Calling Sequence NewtonQuotient(f(x), x = a..b, opts) NewtonQuotient(f(x), x = c, a..b, opts) NewtonQuotient(f(x), a..b, opts) NewtonQuotient(f(x), c, a..b, opts)
Parameters
f(x) - algebraic expression in variable 'x' x - name; specify the independent variable a, b - algebraic expressions; specify the plot range c - algebraic expression; specify a point in a..b opts - equation(s) of the form option=value where option is one of central, derivativeoptions, functionoptions, h, iterations, output, pointoptions, quotientoptions, showderivative, showfunction, showpoints, showquotient, view, or Student plot options; specify output options
Description
• The NewtonQuotient(f(x), x) command returns the Newton quotient of the expression $f\left(x\right)$, that is, the slope of the secant line connecting $\left(c,f\left(c\right)\right)$ and $\left(c+h,f\left(c+h\right)\right)$.
• The NewtonQuotient(f(x), x=a..b) command returns a plot of $f\left(x\right)$ from a to b and the function defined by the Newton quotient for step size h at each point in the interval $\left(a,b\right)$.
• The NewtonQuotient(f(x), x=c) command returns a plot of $f\left(x\right)$ around c and shows the secant line connecting the points $\left(c,f\left(c\right)\right)$ and $\left(c+h,f\left(c+h\right)\right)$, that is, the secant line whose slope is the corresponding Newton quotient.
• The optional argument a..b gives the range of the plot. If no range is given, the interval $\left[c-2\left|h\right|,c+2\left|h\right|\right]$ is used.
• If the independent variable can be uniquely determined from the expression, the parameter x need not be included in the calling sequence.
• The opts argument can contain any of the Student plot options or any of the following equations that (excluding central, h, iterations, and output) set plot options.
central = true or false
Whether the secant line that approximates the tangent is drawn between $x=c+\frac{h}{2}$ and $x=c-\frac{h}{2}$ instead of $x=c$ and $x=c+h$. By default, the value is false, meaning the secant line is drawn using $x=c$ and $x=c+h$.
derivativeoptions = list
A list of options for the plot of the derivative of the expression $f\left(x\right)$. By default, the expression is plotted as a solid green line. For more information on plot options, see plot/options.
functionoptions = list
A list of options for the plot of the expression $f\left(x\right)$. By default, the expression is plotted as a solid red line. For more information on plot options, see plot/options.
h = positive, negative, list of positive or negative, or name
The value of the parameter h. The default value is $0.1$. If the option output is plot or animation, this option must be of type positive, negative, or a list of these types. (See the description of the central option for more information.)
iterations = posint
In an animation, by default $5$ iterations are shown with the value of h divided by $2$ each step. This option is ignored if the option h is a list.
output = value, line, plot, or animation
This option controls the return value of the function.
– output = value specifies that the value of the Newton quotient at the point (and corresponding value of h) are returned. Plotting options are ignored if output = value. This is the default.
– output = line specifies that the expression of the secant line passing through the points $\left(c,f\left(c\right)\right)$ and $\left(c+h,f\left(c+h\right)\right)$ is returned. Plotting options are ignored if output = line.
– output = plot specifies that a plot, which shows the expression and the secant through the points $\left(c,f\left(c\right)\right)$ and $\left(c+h,f\left(c+h\right)\right)$, is displayed.
– output = animation specifies that an animation, which shows the expression and a sequence of secants through the points $\left(c,f\left(c\right)\right)$ and $\left(c+\frac{h}{{2}^{n}},f\left(c+\frac{h}{{2}^{n}}\right)\right)$, $n=1,2,3,\dots$ , is displayed.
pointoptions = list
A list of options for any points in the plot. By default, all points are plotted as blue circles. If the second argument is a range, this option is ignored. For more information on plot options, see plot/options.
quotientoptions = list
A list of options for the plot of the Newton quotient or secant lines. By default, the Newton quotients on an interval or secant lines on a point are plotted in blue. For more information on plot options, see plot/options.
showderivative = true or false
Whether the tangent to the graph of f(x) at the point $\left(c,f\left(c\right)\right)$ is plotted. By default, the value is false.
showfunction = true or false
Whether the expression $f\left(x\right)$ is plotted. By default, the value is true.
showpoints = true or false
Whether the expression f(x) evaluated at the points $c$, $c+\frac{h}{2}$, or $c+h$, as appropriate, is marked. (See the description of the central option for more information.) By default, the value is true.
showquotient = true or false
Whether the quotient of the expression $f\left(x\right)$ is plotted. By default, the value is true.
caption = anything
A caption for the plot.
The default caption is constructed from the parameters and the command options. caption = "" disables the default caption. For more information about specifying a caption, see plot/typesetting.
title = anything
A title for the plot.
The default title is constructed from the parameters and the command options. title = "" disables the default title. For more information about specifying a title, see plot/typesetting.
Examples
> $\mathrm{with}\left({\mathrm{Student}}_{\mathrm{Calculus1}}\right):$
> $\mathrm{NewtonQuotient}\left(\mathrm{sin}\left(x\right),1.0,'h'=0.1\right)$
${0.4973637530}$ (1)
> $\mathrm{NewtonQuotient}\left(\mathrm{sin}\left(x\right),1.0,\mathrm{output}=\mathrm{line},'h'=0.1\right)$
${0.4973637530}{}{x}{+}{0.3441072318}$ (2)
> $\mathrm{NewtonQuotient}\left({x}^{3}-2{x}^{2}-x+1,x=-2..2,\mathrm{output}=\mathrm{plot}\right)$
To play the following animation in this help page, right-click (Control-click, on Mac) the plot to display the context menu. Select Animation > Play.
> $\mathrm{NewtonQuotient}\left({x}^{3}-2{x}^{2}-x+1,x=\frac{1}{2},-1..2,\mathrm{output}=\mathrm{animation},'h'=0.4\right)$
The command to create the plot from the Plotting Guide is
> $\mathrm{NewtonQuotient}\left({x}^{3}-2{x}^{2}-x+1,x=\frac{1}{2},\mathrm{output}=\mathrm{plot},'h'=1.0,\mathrm{view}=\left[-1..2,-2..1.5\right]\right)$ | 1,884 | 6,732 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 41, "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} | 3.890625 | 4 | CC-MAIN-2024-26 | latest | en | 0.46325 |
https://hstreasures.com/acceptable-pins-49391/ | 1,652,986,911,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662529658.48/warc/CC-MAIN-20220519172853-20220519202853-00128.warc.gz | 361,917,562 | 9,471 | # Acceptable Pins Essay
Case Problem 2:Acceptable pins Data Analysis The lengths of the pins made by the automatic lathe are normally distributed with a mean of 1. 012 inches and a standard deviation of 0. 018 inch. The customer will buy only those pins with lengths in the interval 1. 00 ( 0. 02 inch, that is, between 0. 98 inches to 1. 02 inches. The probability of a pin having a length between 0. 98 and 1. 02 is 0. 6339. Therefore, we can say that 63. 39% of the pins produced will be acceptable to the customers. This is a very low acceptance level.
In order to improve the percentage accepted, either the mean or the standard deviation needs to be adjusted. In order to maximize the acceptable percentage of pins keeping the standard deviation same, the mean needs to be adjusted to a value that is exactly between the tolerance limits. This will ensure that the tolerance limit will cover the maximum area under the standard normal curve which in turn means that the acceptance probability will be maximum. Thus, the mean needs to be set to 1. 00 inches.
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This will make the probability of creating a pin within the tolerance limit as 0. 7335. Thus, 73. 35% of the pins would be acceptable when the mean is set to 1. 00. If the mean cannot be adjusted, but the standard deviation can be reduced, the maximum value of standard deviation that would make 90% of the parts acceptable (probability = 0. 90) is 0. 00624 inches. For making 95% of the parts acceptable, the standard deviation needs to be reduced to 0. 00486. Similarly, for making 99% of the parts acceptable, the standard deviation needs to be reduced to 0. 0343. In practice it is easier to adjust the mean as compared to standard deviation. While adjusting the mean, the total cost involves only the engineer’s time and the cost of the production time lost. The cost of reducing the population standard deviation involves, in addition to these costs, the cost of overhauling the machine and reengineering the process. Assuming that it costs 150*(1000x)^2 to reduce the standard deviation by x inch, the cost of setting the standard deviation to 0. 00624 (90% acceptance) is Rs. 20736.
Similarly, the cost of setting the standard deviation to 0. 00486 (95% acceptance) is Rs. 25899 and to 0. 00343 (99% acceptance) is Rs. 31843. However, if the mean is set to 1. 00 inch (incurring cost Rs. 80) and the standard deviation is then reduced, the maximum value of standard deviation that would make 90% of the parts acceptable is 0. 0121 and the total cost incurred would be Rs. 5302. Similarly, for making 95% of the parts acceptable, the standard deviation needs to be reduced to 0. 0102 which would incur total cost of Rs. 206. For making 99% of the parts acceptable, the standard deviation needs to be reduced to 0. 0077 which would incur a total cost of Rs. 15994. Based on the above calculations of costs involved, we can conclude that the most cost effective way of increasing pins acceptability is to adjust the mean to 1. 00 first and then attempt to reduce the standard deviation. The standard deviation should be reduced to at least 0. 0102 so that 95% of the pins are acceptable. The total cost required for these adjustments would be Rs. 9206.
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Hi there, would you like to get such a paper? How about receiving a customized one? Check it out | 837 | 3,439 | {"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} | 3.84375 | 4 | CC-MAIN-2022-21 | latest | en | 0.91732 |
https://en.wikipedia.org/wiki/International_system_of_units | 1,696,000,305,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510516.56/warc/CC-MAIN-20230929122500-20230929152500-00502.warc.gz | 269,390,305 | 134,874 | # International System of Units
The International System of Units, internationally known by the abbreviation SI (for Système International),[a][1]: 125 [2]: iii [3] [b] is the modern form[1]: 117 [6][7] of the metric system[g] and the world's most widely used system of measurement.[1]: 123 [9][10] Established and maintained[11] by the General Conference on Weights and Measures[j] (CGPM[k]), it is the only system of measurement with an official status[m] in nearly every country in the world,[n] employed in science, technology, industry, and everyday commerce.
SI base units
Symbol Name Quantity
s second time
m metre length
kg kilogram mass
A ampere electric current
K kelvin thermodynamic temperature
mol mole amount of substance
cd candela luminous intensity
The SI comprises a coherent[o] system of units of measurement starting with seven base units, which are the second (symbol s, the unit of time), metre (m, length), kilogram (kg, mass), ampere (A, electric current), kelvin (K, thermodynamic temperature), mole (mol, amount of substance), and candela (cd, luminous intensity). The system can accommodate coherent units for an unlimited number of additional quantities. These are called coherent derived units, which can always be represented as products of powers of the base units.[p] Twenty-two coherent derived units have been provided with special names and symbols.[q]
The seven base units and the 22 coherent derived units with special names and symbols may be used in combination to express other coherent derived units.[r] Since the sizes of coherent units will be convenient for only some applications and not for others, the SI provides twenty-four prefixes which, when added to the name and symbol of a coherent unit[s] produce twenty-four additional (non-coherent) SI units for the same quantity; these non-coherent units are always decimal (i.e. power-of-ten) multiples and sub-multiples of the coherent unit.[t][u] The SI is intended to be an evolving system; units and prefixes are created and unit definitions are modified through international agreement as the technology of measurement progresses and the precision of measurements improves.
SI defining constants
Symbol Defining constant Exact value
ΔνCs hyperfine transition frequency of Cs 9192631770 Hz
c speed of light 299792458 m/s
h Planck constant 6.62607015×10−34 J⋅s
e elementary charge 1.602176634×10−19 C
k Boltzmann constant 1.380649×10−23 J/K
NA Avogadro constant 6.02214076×1023 mol−1
Kcd luminous efficacy of 540 THz radiation 683 lm/W
Since 2019, the magnitudes of all SI units have been defined by declaring that seven SI defining constants have certain exact numerical values when expressed in terms of their SI units. These defining constants are the speed of light in vacuum c, the hyperfine transition frequency of caesium ΔνCs, the Planck constant h, the elementary charge e, the Boltzmann constant k, the Avogadro constant NA, and the luminous efficacy Kcd. The nature of the defining constants ranges from fundamental constants of nature such as c to the purely technical constant Kcd. Prior to 2019, h, e, k, and NA were not defined a priori but were rather very precisely measured quantities. In 2019, their values were fixed by definition to their best estimates at the time, ensuring continuity with previous definitions of the base units.
The current way of defining the SI is a result of a decades-long move towards increasingly abstract and idealised formulation in which the realisations of the units are separated conceptually from the definitions. A consequence is that as science and technologies develop, new and superior realisations may be introduced without the need to redefine the unit. One problem with artefacts is that they can be lost, damaged, or changed; another is that they introduce uncertainties that cannot be reduced by advancements in science and technology. The last artefact used by the SI was the International Prototype of the Kilogram, a cylinder of platinum–iridium.
The original motivation for the development of the SI was the diversity of units that had sprung up within the centimetre–gram–second (CGS) systems (specifically the inconsistency between the systems of electrostatic units and electromagnetic units) and the lack of coordination between the various disciplines that used them. The General Conference on Weights and Measures (French: Conférence générale des poids et mesures – CGPM), which was established by the Metre Convention of 1875, brought together many international organisations to establish the definitions and standards of a new system and to standardise the rules for writing and presenting measurements. The system was published in 1960 as a result of an initiative that began in 1948, so it is based on the metre–kilogram–second system of units (MKS) rather than any variant of the CGS.
## Principles
The International System of Units, or SI,[1]: is a decimal[v] and metric[w] system of units established in 1960 and periodically updated since then. The SI has an official status in most countries,[x] including the United States,[y] Canada, and the United Kingdom, although these three countries are amongst a handful of nations that, to various degrees, also continue to use their customary systems. Nevertheless, with this nearly universal level of acceptance, the SI "has been used around the world as the preferred system of units, the basic language for science, technology, industry and trade."[1]: 123
The only other types of measurement system that still have widespread use across the world are the Imperial and US customary measurement systems,[z] and they are legally defined in terms of the SI.[aa] There are other, less widespread systems of measurement that are occasionally used in particular regions of the world. In addition, there are many individual non-SI units that do not belong to any comprehensive system of units, but that are nevertheless still regularly used in particular fields and regions. Both of these categories of unit are also typically defined legally in terms of SI units.[ab]
### International System of Quantities
The quantities and equations that provide the context in which the SI units are defined are now referred to as the International System of Quantities (ISQ). The ISQ is based on the quantities underlying each of the seven base units of the SI. Other quantities, such as area, pressure, and electrical resistance, are derived from these base quantities by clear, non-contradictory equations. The ISQ defines the quantities that are measured with the SI units.[16] The ISQ is formalised, in part, in the international standard ISO/IEC 80000, which was completed in 2009 with the publication of ISO 80000-1,[17] and has largely been revised in 2019–2020 with the remainder being under review.
### Controlling authority
The SI was established and is maintained by the General Conference on Weights and Measures (CGPM[k]).[11] In practice, the CGPM follows the recommendations of the Consultative Committee for Units (CCU), which is the actual body conducting technical deliberations concerning new scientific and technological developments related to the definition of units and the SI. The CCU reports to the International Committee for Weights and Measures (CIPM[ac]), which, in turn, reports to the CGPM.
The SI is regulated and continually developed by three international organisations that were established in 1875 under the terms of the Metre Convention. They are the General Conference on Weights and Measures (CGPM[k]), the International Committee for Weights and Measures (CIPM[ac]), and the International Bureau of Weights and Measures (BIPM[ad]). The ultimate authority rests with the CGPM, which is a plenary body through which its Member States[ae] act together on matters related to measurement science and measurement standards; it usually convenes every four years.[12] The CGPM elects the CIPM, which is an 18-person committee of eminent scientists. The CIPM operates based on the advice of a number of its Consultative Committees, which bring together the world's experts in their specified fields as advisers on scientific and technical matters.[18][af] One of these committees is the Consultative Committee for Units (CCU), which is responsible for "matters related to the development of the International System of Units (SI), preparation of successive editions of the SI brochure, and advice to the CIPM on matters concerning units of measurement."[19] It is the CCU which considers in detail all new scientific and technological developments related to the definition of units and the SI. In practice, when it comes to the definition of the SI, the CGPM simply formally approves the recommendations of the CIPM, which, in turn, follows the advice of the CCU.
The CCU has the following as members:[20][21] national laboratories of the Member States of the CGPM charged with establishing national standards;[ag] relevant intergovernmental organisations and international bodies;[ah] international commissions or committees;[ai] scientific unions;[aj] personal members;[ak] and, as an ex officio member of all Consultative Committees, the Director of the BIPM.
All the decisions and recommendations concerning units are collected in a brochure called The International System of Units (SI),[1][al] which is published in French and English by the BIPM and periodically updated. The writing and maintenance of the brochure is carried out by one of the committees of the CIPM. The definitions of the terms "quantity", "unit", "dimension" etc. that are used in the SI Brochure are those given in the International vocabulary of metrology.[24] The brochure leaves some scope for local variations, particularly regarding unit names and terms in different languages.[am][2]
## Units and prefixes
The International System of Units consists of a set of SI base units, SI derived units, and a set of decimal-based multipliers and submultipliers that are used as SI prefixes.[25]: 103–106 The units, excluding prefixed units,[an] form a coherent system of units, which is based on a system of quantities in such a way that the equations between the numerical values expressed in coherent units have exactly the same form, including numerical factors, as the corresponding equations between the quantities. For example, 1 N = 1 kg × 1 m/s2 says that one newton is the force required to accelerate a mass of one kilogram at one metre per second squared, as related through the principle of coherence to the equation relating the corresponding quantities: F = m × a.
Derived units apply to derived quantities, which may by definition be expressed in terms of base quantities, and thus are not independent; for example, electrical conductance is the inverse of electrical resistance, with the consequence that the siemens is the inverse of the ohm, and similarly, the ohm and siemens can be replaced with a ratio of an ampere and a volt, because those quantities bear a defined relationship to each other.[ao] Other useful derived quantities can be specified in terms of the SI base and derived units that have no named units in the SI, such as acceleration, which is defined in SI units as m/s2.
### SI base units
The SI selects seven units to serve as base units, corresponding to seven base physical quantities.[ap][aq] They are the second, with the symbol s, which is the SI unit of the physical quantity of time; the metre, symbol m, the SI unit of length; kilogram (kg, the unit of mass); ampere (A, electric current); kelvin (K, thermodynamic temperature); mole (mol, amount of substance); and candela (cd, luminous intensity).[1] All units in the SI can be expressed in terms of the base units, and the base units serve as a preferred set for expressing or analysing the relationships between units.
SI base units[2]: 6 [26][27]
Unit name Unit symbol Dimension symbol Quantity name Typical symbols Definition
second
[n 1]
s T time ${\displaystyle t}$ The duration of 9192631770 periods of the radiation corresponding to the transition between the two hyperfine levels of the ground state of the caesium-133 atom.
metre m L length ${\displaystyle l}$, ${\displaystyle h}$, ${\displaystyle a}$, ${\displaystyle b}$, ${\displaystyle x}$, ${\displaystyle y}$, ${\displaystyle r}$, etc.[n 2] The distance travelled by light in vacuum in 1/299792458 seconds.
kilogram
[n 3]
kg M mass ${\displaystyle m}$ The kilogram is defined by setting the Planck constant h exactly to 6.62607015×10−34 J⋅s (J = kg⋅m2⋅s−2), given the definitions of the metre and the second.[28]
ampere A I electric current ${\displaystyle I,\;i}$ The flow of exactly 1/1.602176634×10−19 times the elementary charge e per second.
Equalling approximately 6.2415090744×1018 elementary charges per second.
kelvin K Θ thermodynamic
temperature
${\displaystyle T}$ The kelvin is defined by setting the fixed numerical value of the Boltzmann constant k to 1.380649×10−23 J⋅K−1, (J = kg⋅m2⋅s−2), given the definition of the kilogram, the metre, and the second.
mole mol N amount of substance ${\displaystyle n}$ The amount of substance of exactly 6.02214076×1023 elementary entities.[n 4] This number is the fixed numerical value of the Avogadro constant, NA, when expressed in the unit mol−1.
candela cd J luminous intensity ${\displaystyle I_{v}}$ The luminous intensity, in a given direction, of a source that emits monochromatic radiation of frequency 5.4×1014 hertz and that has a radiant intensity in that direction of 1/683 watt per steradian.
Notes
1. ^ Within the context of the SI, the second is the coherent base unit of time, and is used in the definitions of derived units. The name "second" historically arose as being the 2nd-level sexagesimal division (1602) of some quantity, the hour in this case, which the SI classifies as an "accepted" unit along with its first-level sexagesimal division the minute.
2. ^ Symbols for length vary greatly with context. Problems involving intuitive three-dimensional quantities often use ${\displaystyle l}$, ${\displaystyle w}$, and ${\displaystyle h}$ for length, distance, and height, respectively. More generally, physicists tend to set up the coordinate system of a given problem so that one axis lies conveniently parallel to the length being measured. Length is then often denoted either by some constant (e.g. ${\displaystyle a}$, ${\displaystyle b}$) along said axis, or by the same symbol as the axis itself (e.g. ${\displaystyle x}$, ${\displaystyle y}$, or ${\displaystyle r}$ for horizontal, vertical, and radial axes, respectively).
3. ^ Despite the prefix "kilo-", the kilogram is the coherent base unit of mass, and is used in the definitions of derived units. Nonetheless, prefixes for the unit of mass are determined as if the gram were the base unit.
4. ^ When the mole is used, the elementary entities must be specified and may be atoms, molecules, ions, electrons, other particles, or specified groups of such particles.
### Derived units
The system allows for an unlimited number of additional units, called derived units, which can always be represented as products of powers of the base units, possibly with a nontrivial numeric multiplier. When that multiplier is one, the unit is called a coherent derived unit.[ar] The base and coherent derived units of the SI together form a coherent system of units (the set of coherent SI units).[as] Twenty-two coherent derived units have been provided with special names and symbols.[q] The seven base units and the 22 derived units with special names and symbols may be used in combination to express other derived units,[r] which are adopted to facilitate measurement of diverse quantities.
Prior to its redefinition in 2019, the SI was defined through the seven base units from which the derived units were constructed as products of powers of the base units. After the redefinition, the SI is defined by fixing the numerical values of seven defining constants. This has the effect that the distinction between the base units and derived units is, in principle, not needed, since all units, base as well as derived, may be constructed directly from the defining constants. Nevertheless, the distinction is retained because "it is useful and historically well established", and also because the ISO/IEC 80000 series of standards[at] specifies base and derived quantities that necessarily have the corresponding SI units.[1]: 129
The derived units in the SI are formed by powers, products, or quotients of the base units and are potentially unlimited in number.[25]: 103 [2]: 14, 16 Derived units are associated with derived quantities; for example, velocity is a quantity that is derived from the base quantities of time and length, and thus the SI derived unit is metre per second (symbol m/s). The dimensions of derived units can be expressed in terms of the dimensions of the base units.
Combinations of base and derived units may be used to express other derived units. For example, the SI unit of force is the newton (N), the SI unit of pressure is the pascal (Pa)—and the pascal can be defined as one newton per square metre (N/m2).[30]
SI derived units with special names and symbols[2]: 15
Name Symbol Quantity In SI base units In other SI units
radian[N 1] rad plane angle m/m 1
steradian[N 1] sr solid angle m2/m2 1
hertz Hz frequency s−1
newton N force, weight kg⋅m⋅s−2
pascal Pa pressure, stress kg⋅m−1⋅s−2 N/m2 = J/m3
joule J energy, work, heat kg⋅m2⋅s−2 N⋅m = Pa⋅m3
watt W power, radiant flux kg⋅m2⋅s−3 J/s
coulomb C electric charge s⋅A
volt V electric potential, voltage, emf kg⋅m2⋅s−3⋅A−1 W/A = J/C
farad F capacitance kg−1⋅m−2⋅s4⋅A2 C/V = C2/J
ohm Ω resistance, impedance, reactance kg⋅m2⋅s−3⋅A−2 V/A = J⋅s/C2
siemens S electrical conductance kg−1⋅m−2⋅s3⋅A2 Ω−1
weber Wb magnetic flux kg⋅m2⋅s−2⋅A−1 V⋅s
tesla T magnetic flux density kg⋅s−2⋅A−1 Wb/m2
henry H inductance kg⋅m2⋅s−2⋅A−2 Wb/A
degree Celsius °C temperature relative to 273.15 K K
lumen lm luminous flux cd⋅m2/m2 cd⋅sr
lux lx illuminance cd⋅m2/m4 lm/m2 = cd⋅sr⋅m−2
becquerel Bq activity referred to a radionuclide (decays per unit time) s−1
gray Gy absorbed dose (of ionising radiation) m2⋅s−2 J/kg
sievert Sv equivalent dose (of ionising radiation) m2⋅s−2 J/kg
katal kat catalytic activity mol⋅s−1
Notes
1. ^ a b The radian and steradian are defined as dimensionless derived units.
#### Coherent and non-coherent SI units
When prefixes are used with the coherent SI units, the resulting units are no longer coherent, because the prefix introduces a numerical factor other than one.[1]: 137 The one exception is the kilogram, the only coherent SI unit whose name and symbol, for historical reasons, include a prefix.[au]
The complete set of SI units consists of both the coherent set and the multiples and sub-multiples of coherent units formed by using the SI prefixes.[1]: 138 For example, the metre, kilometre, centimetre, nanometre, etc. are all SI units of length, though only the metre is a coherent SI unit. A similar statement holds for derived units: for example, kg/m3, g/dm3, g/cm3, Pg/km3, etc. are all SI units of density, but of these, only kg/m3 is a coherent SI unit.
Moreover, the metre is the only coherent SI unit of length. Every physical quantity has exactly one coherent SI unit, although this unit may be expressible in different forms by using some of the special names and symbols.[1]: 140 For example, the coherent SI unit of linear momentum may be written as either kg⋅m/s or as N⋅s, and both forms are in use (e.g. compare respectively here[31]:205 and here[32]:135).
On the other hand, several different quantities may share the same coherent SI unit. For example, the joule per kelvin (symbol J/K) is the coherent SI unit for two distinct quantities: heat capacity and entropy; another example is the ampere, which is the coherent SI unit for both electric current and magnetomotive force. This is why it is important not to use the unit alone to specify the quantity.[av]
Furthermore, the same coherent SI unit may be a base unit in one context, but a coherent derived unit in another. For example, the ampere is a base unit when it is a unit of electric current, but a coherent derived unit when it is a unit of magnetomotive force.[1]: 140 As perhaps a more familiar example, consider rainfall, defined as volume of rain (measured in m3) that fell per unit area (measured in m2). Since m3/m2 = m, it follows that the coherent derived SI unit of rainfall is the metre, even though the metre is also the base SI unit of length.[aw]
Examples of coherent derived units in terms of base units[2]: 17
Name Symbol Derived quantity Typical symbol
square metre m2 area A
cubic metre m3 volume V
metre per second m/s speed, velocity v
metre per second squared m/s2 acceleration a
reciprocal metre m−1 wavenumber σ,
vergence (optics) V, 1/f
kilogram per cubic metre kg/m3 density ρ
kilogram per square metre kg/m2 surface density ρA
cubic metre per kilogram m3/kg specific volume v
ampere per square metre A/m2 current density j
ampere per metre A/m magnetic field strength H
mole per cubic metre mol/m3 concentration c
kilogram per cubic metre kg/m3 mass concentration ρ, γ
candela per square metre cd/m2 luminance Lv
Examples of derived units that include units with special names[2]: 18
Name Symbol Quantity In SI base units
pascal-second Pa⋅s dynamic viscosity m−1⋅kg⋅s−1
newton-metre N⋅m moment of force m2⋅kg⋅s−2
newton per metre N/m surface tension kg⋅s−2
radian per second rad/s angular velocity, angular frequency s−1
radian per second squared rad/s2 angular acceleration s−2
watt per square metre W/m2 heat flux density, irradiance kg⋅s−3
joule per kelvin J/K entropy, heat capacity m2⋅kg⋅s−2⋅K−1
joule per kilogram-kelvin J/(kg⋅K) specific heat capacity, specific entropy m2⋅s−2⋅K−1
joule per kilogram J/kg specific energy m2⋅s−2
watt per metre-kelvin W/(m⋅K) thermal conductivity m⋅kg⋅s−3⋅K−1
joule per cubic metre J/m3 energy density m−1⋅kg⋅s−2
volt per metre V/m electric field strength m⋅kg⋅s−3⋅A−1
coulomb per cubic metre C/m3 electric charge density m−3⋅s⋅A
coulomb per square metre C/m2 surface charge density, electric flux density, electric displacement m−2⋅s⋅A
farad per metre F/m permittivity m−3⋅kg−1⋅s4⋅A2
henry per metre H/m permeability m⋅kg⋅s−2⋅A−2
joule per mole J/mol molar energy m2⋅kg⋅s−2⋅mol−1
joule per mole-kelvin J/(mol⋅K) molar entropy, molar heat capacity m2⋅kg⋅s−2⋅K−1⋅mol−1
coulomb per kilogram C/kg exposure (x- and γ-rays) kg−1⋅s⋅A
gray per second Gy/s absorbed dose rate m2⋅s−3
watt per steradian W/sr radiant intensity m2⋅kg⋅s−3
watt per square metre-steradian W/(m2⋅sr) radiance kg⋅s−3
katal per cubic metre kat/m3 catalytic activity concentration m−3⋅s−1⋅mol
### Dimensionless units
The unit of a dimensionless quantity is one (symbol 1), but it is rarely shown.[33] The radian and steradian are also dimensionless quantities, but use the symbols rad and sr respectively.[34]
### Prefixes
Like all metric systems, the SI uses metric prefixes to systematically construct, for the same physical quantity, a set of units that are decimal multiples of each other over a wide range.
For example, while the coherent unit of length is the metre,[ax] the SI provides a full range of smaller and larger units of length, any of which may be more convenient for any given application – for example, driving distances are normally given in kilometres (symbol km) rather than in metres. Here the metric prefix 'kilo-' (symbol 'k') stands for a factor of 1000; thus, 1 km = 1000 m.[ay]
The current version of the SI provides twenty-four metric prefixes that signify decimal powers ranging from 10−30 to 1030, the most recent being adopted in 2022.[1]: 143–144 [35][36][37] Most prefixes correspond to integer powers of 1000; the only ones that do not are those for 10, 1/10, 100, and 1/100.
In general, given any coherent unit with a separate name and symbol,[az] one forms a new unit by simply adding an appropriate metric prefix to the name of the coherent unit (and a corresponding prefix symbol to the coherent unit's symbol).[au] Since the metric prefix signifies a particular power of ten, the new unit is always a power-of-ten multiple or sub-multiple of the coherent unit. Thus, the conversion between different SI units for one and the same physical quantity is always through a power of ten.[ba] This is why the SI (and metric systems more generally) are called decimal systems of measurement units.[38][bb]
The grouping formed by a prefix symbol attached to a unit symbol (e.g. 'km', 'cm') constitutes a new inseparable unit symbol. This new symbol can be raised to a positive or negative power. It can also be combined with other unit symbols to form compound unit symbols.[1]: 143 For example, g/cm3 is an SI unit of density, where cm3 is to be interpreted as (cm)3.
Prefixes are added to unit names to produce multiples and submultiples of the original unit. All of these are integer powers of ten, and above a hundred or below a hundredth all are integer powers of a thousand. For example, kilo- denotes a multiple of a thousand and milli- denotes a multiple of a thousandth, so there are one thousand millimetres to the metre and one thousand metres to the kilometre. The prefixes are never combined, so for example a millionth of a metre is a micrometre, not a millimillimetre. Multiples of the kilogram are named as if the gram were the base unit, so a millionth of a kilogram is a milligram, not a microkilogram.[25]: 122 [39]: 14 When prefixes are used to form multiples and submultiples of SI base and derived units, the resulting units are no longer coherent.[25]: 7
The BIPM specifies 24 prefixes for the International System of Units (SI):
Prefix Base 10 Decimal Adoption
[nb 1]
Name Symbol
quetta Q 1030 1000000000000000000000000000000 2022[40]
ronna R 1027 1000000000000000000000000000
yotta Y 1024 1000000000000000000000000 1991
zetta Z 1021 1000000000000000000000
exa E 1018 1000000000000000000 1975[41]
peta P 1015 1000000000000000
tera T 1012 1000000000000 1960
giga G 109 1000000000
mega M 106 1000000 1873
kilo k 103 1000 1795
hecto h 102 100
deca da 101 10
100 1
deci d 10−1 0.1 1795
centi c 10−2 0.01
milli m 10−3 0.001
micro μ 10−6 0.000001 1873
nano n 10−9 0.000000001 1960
pico p 10−12 0.000000000001
femto f 10−15 0.000000000000001 1964
atto a 10−18 0.000000000000000001
zepto z 10−21 0.000000000000000000001 1991
yocto y 10−24 0.000000000000000000000001
ronto r 10−27 0.000000000000000000000000001 2022[40]
quecto q 10−30 0.000000000000000000000000000001
Notes
1. ^ Prefixes adopted before 1960 already existed before SI. The introduction of the CGS system was in 1873.
## Lexicographic conventions
### Unit names
According to the SI Brochure,[1]: 148 unit names should be treated as common nouns of the context language. This means that they should be typeset in the same character set as other common nouns (e.g. Latin alphabet in English, Cyrillic script in Russian, etc.), following the usual grammatical and orthographical rules of the context language. For example, in English and French, even when the unit is named after a person and its symbol begins with a capital letter, the unit name in running text should start with a lowercase letter (e.g., newton, hertz, pascal) and is capitalized only at the beginning of a sentence and in headings and publication titles. As a nontrivial application of this rule, the SI Brochure notes[1]: 148 that the name of the unit with the symbol °C is correctly spelled as 'degree Celsius': the first letter of the name of the unit, 'd', is in lowercase, while the modifier 'Celsius' is capitalized because it is a proper name.[bc][1]: 148
The English spelling and even names for certain SI units and metric prefixes depend on the variety of English used. US English uses the spelling deka-, meter, and liter, whilst International English uses deca-, metre, and litre. The name of the unit whose symbol is t and which is defined according to 1 t = 103 kg is 'metric ton' in US English and 'tonne' in International English.[2]: iii
### Unit symbols and the values of quantities
Symbols of SI units are intended to be unique and universal, independent of the context language.[25]: 130–35 The SI Brochure has specific rules for writing them.[25]: 130–35 The guideline produced by the National Institute of Standards and Technology (NIST)[42] clarifies language-specific details for American English that were left unclear by the SI Brochure, but is otherwise identical to the SI Brochure.[43]
#### General rules
General rules[bd] for writing SI units and quantities apply to text that is either handwritten or produced using an automated process:
• The value of a quantity is written as a number followed by a space (representing a multiplication sign) and a unit symbol; e.g., 2.21 kg, 7.3×102 m2, 22 K. This rule explicitly includes the percent sign (%)[25]: 134 and the symbol for degrees Celsius (°C).[25]: 133 Exceptions are the symbols for plane angular degrees, minutes, and seconds (°, ′, and ″, respectively), which are placed immediately after the number with no intervening space.
• Symbols are mathematical entities, not abbreviations, and as such do not have an appended period/full stop (.), unless the rules of grammar demand one for another reason, such as denoting the end of a sentence.
• A prefix is part of the unit, and its symbol is prepended to a unit symbol without a separator (e.g., k in km, M in MPa, G in GHz, μ in μg). Compound prefixes are not allowed. A prefixed unit is atomic in expressions (e.g., km2 is equivalent to (km)2).
• Unit symbols are written using roman (upright) type, regardless of the type used in the surrounding text.
• Symbols for derived units formed by multiplication are joined with a centre dot (⋅) or a non-breaking space; e.g., N⋅m or N m.
• Symbols for derived units formed by division are joined with a solidus (/), or given as a negative exponent. E.g., the "metre per second" can be written m/s, m s−1, m⋅s−1, or m/s. In cases where a solidus is followed by a centre dot (or space), or more than one solidus is present, parentheses must be used to avoid ambiguity; e.g., kg/(m⋅s2), kg⋅m−1⋅s−2, and (kg/m)/s2 are acceptable, but kg/m/s2 and kg/m⋅s2 are ambiguous and unacceptable.
• The first letter of symbols for units derived from the name of a person is written in upper case; otherwise, they are written in lower case. E.g., the unit of pressure is named after Blaise Pascal, so its symbol is written "Pa", but the symbol for mole is written "mol". Thus, "T" is the symbol for tesla, a measure of magnetic field strength, and "t" the symbol for tonne, a measure of mass. Since 1979, the litre may exceptionally be written using either an uppercase "L" or a lowercase "l", a decision prompted by the similarity of the lowercase letter "l" to the numeral "1", especially with certain typefaces or English-style handwriting. The American NIST recommends that within the United States "L" be used rather than "l".[39]
• Symbols do not have a plural form, e.g., 25 kg, not 25 kgs.
• Uppercase and lowercase prefixes are not interchangeable. E.g., the quantities 1 mW and 1 MW represent two different quantities (milliwatt and megawatt).
• The symbol for the decimal marker is either a point or comma on the line. In practice, the decimal point is used in most English-speaking countries and most of Asia, and the comma in most of Latin America and in continental European countries.[44]
• Thin spaces may be used as a thousands separator (1000000) in order to facilitate reading, but neither dots nor commas should be inserted between groups of three (1,000,000 or 1.000.000).[25]: 133 When there are only four digits, a space is ordinarily not used to isolate a single digit. NIST guideline has the same recommendations.[45]: 37
• Any line-break inside a number, inside a compound unit, or between number and unit should be avoided. Where this is not possible, line breaks should coincide with thousands separators.
• Because the value of "billion" and "trillion" varies between languages, the dimensionless terms "ppb" (parts per billion) and "ppt" (parts per trillion) should be avoided. The SI Brochure does not suggest alternatives.
#### Printing SI symbols
The rules covering printing of quantities and units are part of ISO 80000-1:2009.[46]
Further rules[bd] are specified in respect of production of text using printing presses, word processors, typewriters, and the like.
## Realisation of units
Metrologists carefully distinguish between the definition of a unit and its realisation. The definition of each base unit of the SI is drawn up so that it is unique and provides a sound theoretical basis on which the most accurate and reproducible measurements can be made. The realisation of the definition of a unit is the procedure by which the definition may be used to establish the value and associated uncertainty of a quantity of the same kind as the unit. A description of the mise en pratique[be] of the base units is given in an electronic appendix to the SI Brochure.[48][25]: 168–169
The published mise en pratique is not the only way in which a base unit can be determined: the SI Brochure states that "any method consistent with the laws of physics could be used to realise any SI unit."[25]: 111 Various consultative committees of the CIPM decided in 2016 that more than one mise en pratique would be developed for determining the value of each unit.[49] These methods include the following:
• At least three separate experiments be carried out yielding values having a relative standard uncertainty in the determination of the kilogram of no more than 5×10−8 and at least one of these values should be better than 2×10−8. Both the Kibble balance and the Avogadro project should be included in the experiments and any differences between these be reconciled.[50][51]
• The definition of the kelvin measured with a relative uncertainty of the Boltzmann constant derived from two fundamentally different methods such as acoustic gas thermometry and dielectric constant gas thermometry be better than one part in 10−6 and that these values be corroborated by other measurements.[52]
### Definition vs. realization of units
Since 2019, the magnitudes of all SI units have been defined in an abstract way, which is conceptually separated from any practical realisation of them.[1]: 126 [bf] Namely, the SI units are defined by declaring that seven defining constants[1]: 125–129 have certain exact numerical values when expressed in terms of their SI units. Probably the most widely known of these constants is the speed of light in vacuum, c, which in the SI by definition has the exact value of c = 299792458 m/s. The other six constants are ΔνCs, the hyperfine transition frequency of caesium; h, the Planck constant; e, the elementary charge; k, the Boltzmann constant; NA, the Avogadro constant; and Kcd, the luminous efficacy of monochromatic radiation of frequency 540×1012 Hz.[bg] The nature of the defining constants ranges from fundamental constants of nature such as c to the purely technical constant Kcd.[1]: 128–129 Prior to 2019, h, e, k, and NA were not defined a priori but were rather very precisely measured quantities. In 2019, their values were fixed by definition to their best estimates at the time, ensuring continuity with previous definitions of the base units.
As far as realisations, what are believed to be the current best practical realisations of units are described in the mises en pratique,[bh] which are also published by the BIPM.[55] The abstract nature of the definitions of units is what makes it possible to improve and change the mises en pratique as science and technology develop without having to change the actual definitions themselves.[bk]
In a sense, this way of defining the SI units is no more abstract than the way derived units are traditionally defined in terms of the base units. Consider a particular derived unit, for example, the joule, the unit of energy. Its definition in terms of the base units is kgm2/s2. Even if the practical realisations of the metre, kilogram, and second are available, a practical realisation of the joule would require some sort of reference to the underlying physical definition of work or energy—some actual physical procedure for realising the energy in the amount of one joule such that it can be compared to other instances of energy (such as the energy content of motor spirit put into a car or of electricity delivered to a household).
The situation with the defining constants and all of the SI units is analogous. In fact, purely mathematically speaking, the SI units are defined as if we declared that it is the defining constant's units that are now the base units, with all other SI units being derived units. To make this clearer, first note that each defining constant can be taken as determining the magnitude of that defining constant's unit of measurement;[1]: 128 for example, the definition of c defines the unit m/s as 1 m/s = c/299792458 ('the speed of one metre per second is equal to one 299792458th of the speed of light'). In this way, the defining constants directly define the following seven units:
Further, one can show, using dimensional analysis, that every coherent SI unit (whether base or derived) can be written as a unique product of powers of the units of the SI defining constants (in complete analogy to the fact that every coherent derived SI unit can be written as a unique product of powers of the base SI units). For example, the kilogram can be written as kg = (Hz)(J⋅s)/(m/s)2.[bl] Thus, the kilogram is defined in terms of the three defining constants ΔνCs, c, and h because, on the one hand, these three defining constants respectively define the units Hz, m/s, and J⋅s,[bm] while, on the other hand, the kilogram can be written in terms of these three units, namely, kg = (Hz)(J⋅s)/(m/s)2.[bn] While the question of how to actually realise the kilogram in practice would, at this point, still be open, that is not really different from the fact that the question of how to actually realise the joule in practice is still in principle open even once one has achieved the practical realisations of the metre, kilogram, and second.[improper synthesis?]
### Specifying fundamental constants vs. other methods of definition
The current way of defining the SI is the result of a decades-long move towards increasingly abstract and idealised formulation in which the realisations of the units are separated conceptually from the definitions.[1]: 126
The great advantage of doing it this way is that as science and technologies develop, new and superior realisations may be introduced without the need to redefine the units.[bi] Units can now be realised with an accuracy that is ultimately limited only by the quantum structure of nature and our technical abilities but not by the definitions themselves.[bj] Any valid equation of physics relating the defining constants to a unit can be used to realise the unit, thus creating opportunities for innovation... with increasing accuracy as technology proceeds.'[1]: 122 In practice, the CIPM Consultative Committees provide so-called "mises en pratique" (practical techniques),[55] which are the descriptions of what are currently believed to be best experimental realisations of the units.[59]
This system lacks the conceptual simplicity of using artefacts (referred to as prototypes) as realisations of units to define those units: with prototypes, the definition and the realisation are one and the same.[bo] However, using artefacts has two major disadvantages that, as soon as it is technologically and scientifically feasible, result in abandoning them as means for defining units.[bs] One major disadvantage is that artefacts can be lost, damaged,[bu] or changed.[bv] The other is that they largely cannot benefit from advancements in science and technology. The last artefact used by the SI was the International Prototype Kilogram (IPK), a particular cylinder of platinum-iridium; from 1889 to 2019, the kilogram was by definition equal to the mass of the IPK. Concerns regarding its stability on the one hand, and progress in precise measurements of the Planck constant and the Avogadro constant on the other, led to a revision of the definition of the base units, put into effect on 20 May 2019.[28] This was the biggest change in the SI since it was first formally defined and established in 1960, and it resulted in the definitions described above.[66]
In the past, there were also various other approaches to the definitions of some of the SI units. One made use of a specific physical state of a specific substance (the triple point of water, which was used in the definition of the kelvin[25]: 113–114 ); others referred to idealised experimental prescriptions[1]: 125 (as in the case of the former SI definition of the ampere[25]: 113 and the former SI definition (originally enacted in 1979) of the candela[25]: 115 ).
In the future, the set of defining constants used by the SI may be modified as more stable constants are found, or if it turns out that other constants can be more precisely measured.[bw]
## Evolution of the SI
### Changes to the SI
The International Bureau of Weights and Measures (BIPM) has described SI as "the modern form of metric system".[25]: 95 Changing technology has led to an evolution of the definitions and standards that has followed two principal strands – changes to SI itself, and clarification of how to use units of measure that are not part of SI but are still nevertheless used on a worldwide basis.
Since 1960 the CGPM has made a number of changes to the SI to meet the needs of specific fields, notably chemistry and radiometry. These are mostly additions to the list of named derived units, and include the mole (symbol mol) for an amount of substance, the pascal (symbol Pa) for pressure, the siemens (symbol S) for electrical conductance, the becquerel (symbol Bq) for "activity referred to a radionuclide", the gray (symbol Gy) for ionising radiation, the sievert (symbol Sv) as the unit of dose equivalent radiation, and the katal (symbol kat) for catalytic activity.[25]: 156 [67][25]: 156 [25]: 158 [25]: 159 [25]: 165
The range of defined prefixes pico- (10−12) to tera- (1012) was extended to quecto- (10−30) to quetta- (1030).[25]: 152 [25]: 158 [25]: 164
The 1960 definition of the standard metre in terms of wavelengths of a specific emission of the krypton-86 atom was replaced in 1983 with the distance that light travels in vacuum in exactly 1/299792458 second, so that the speed of light is now an exactly specified constant of nature.
A few changes to notation conventions have also been made to alleviate lexicographic ambiguities. An analysis under the aegis of CSIRO, published in 2009 by the Royal Society, has pointed out the opportunities to finish the realisation of that goal, to the point of universal zero-ambiguity machine readability.[68]
### 2019 redefinitions
After the metre was redefined in 1960, the International Prototype of the Kilogram (IPK) was the only physical artefact upon which base units (directly the kilogram and indirectly the ampere, mole and candela) depended for their definition, making these units subject to periodic comparisons of national standard kilograms with the IPK.[69] During the 2nd and 3rd Periodic Verification of National Prototypes of the Kilogram, a significant divergence had occurred between the mass of the IPK and all of its official copies stored around the world: the copies had all noticeably increased in mass with respect to the IPK. During extraordinary verifications carried out in 2014 preparatory to redefinition of metric standards, continuing divergence was not confirmed. Nonetheless, the residual and irreducible instability of a physical IPK undermined the reliability of the entire metric system to precision measurement from small (atomic) to large (astrophysical) scales.
A proposal was made that:[70]
• In addition to the speed of light, four constants of nature – the Planck constant, an elementary charge, the Boltzmann constant, and the Avogadro constant – be defined to have exact values
• The International Prototype of the Kilogram be retired
• The current definitions of the kilogram, ampere, kelvin, and mole be revised
• The wording of base unit definitions should change emphasis from explicit unit to explicit constant definitions.
The new definitions were adopted at the 26th CGPM on 16 November 2018, and came into effect on 20 May 2019.[71] The change was adopted by the European Union through Directive (EU) 2019/1258.[72]
## History
The original motivation for the development of the SI was the diversity of units that had sprung up within the centimetre–gram–second (CGS) systems (specifically the inconsistency between the systems of electrostatic units and electromagnetic units) and the lack of coordination between the various disciplines that used them. The General Conference on Weights and Measures (French: Conférence générale des poids et mesures – CGPM), which was established by the Metre Convention of 1875, brought together many international organisations to establish the definitions and standards of a new system and to standardise the rules for writing and presenting measurements.
Adopted in 1889, use of the MKS system of units succeeded the centimetre–gram–second system of units (CGS) in commerce and engineering. The metre and kilogram system served as the basis for the development of the International System of Units (abbreviated SI), which now serves as the international standard. Because of this, the standards of the CGS system were gradually replaced with metric standards incorporated from the MKS system.[74]
In 1901, Giovanni Giorgi proposed to the Associazione elettrotecnica italiana [it] (AEI) that this system, extended with a fourth unit to be taken from the units of electromagnetism, be used as an international system.[75] This system was strongly promoted by electrical engineer George A. Campbell.[76]
The International System was published in 1960, based on the MKS units, as a result of an initiative that began in 1948.
### The improvisation of units
The units and unit magnitudes of the metric system which became the SI were improvised piecemeal from everyday physical quantities starting in the mid-18th century. Only later were they moulded into an orthogonal coherent decimal system of measurement.
The degree centigrade as a unit of temperature resulted from the scale devised by Swedish astronomer Anders Celsius in 1742. His scale counter-intuitively designated 100 as the freezing point of water and 0 as the boiling point. Independently, in 1743, the French physicist Jean-Pierre Christin described a scale with 0 as the freezing point of water and 100 the boiling point. The scale became known as the centi-grade, or 100 gradations of temperature, scale.
The metric system was developed from 1791 onwards by a committee of the French Academy of Sciences, commissioned to create a unified and rational system of measures.[77] The group, which included preeminent French men of science,[78]: 89 used the same principles for relating length, volume, and mass that had been proposed by the English clergyman John Wilkins in 1668[79][80] and the concept of using the Earth's meridian as the basis of the definition of length, originally proposed in 1670 by the French abbot Mouton.[81][82]
In March 1791, the Assembly adopted the committee's proposed principles for the new decimal system of measure including the metre defined to be 1/10,000,000 of the length of the quadrant of Earth's meridian passing through Paris, and authorised a survey to precisely establish the length of the meridian. In July 1792, the committee proposed the names metre, are, litre and grave for the units of length, area, capacity, and mass, respectively. The committee also proposed that multiples and submultiples of these units were to be denoted by decimal-based prefixes such as centi for a hundredth and kilo for a thousand.[83]: 82
William Thomson (Lord Kelvin) and James Clerk Maxwell played a prominent role in the development of the principle of coherence and in the naming of many units of measure.[84][85][86][87][88]
Later, during the process of adoption of the metric system, the Latin gramme and kilogramme, replaced the former provincial terms gravet (1/1000 grave) and grave. In June 1799, based on the results of the meridian survey, the standard mètre des Archives and kilogramme des Archives were deposited in the French National Archives. Subsequently, that year, the metric system was adopted by law in France.[89][90] The French system was short-lived due to its unpopularity. Napoleon ridiculed it, and in 1812, introduced a replacement system, the mesures usuelles or "customary measures" which restored many of the old units, but redefined in terms of the metric system.
During the first half of the 19th century there was little consistency in the choice of preferred multiples of the base units: typically the myriametre (10000 metres) was in widespread use in both France and parts of Germany, while the kilogram (1000 grams) rather than the myriagram was used for mass.[73]
In 1832, the German mathematician Carl Friedrich Gauss, assisted by Wilhelm Weber, implicitly defined the second as a base unit when he quoted the Earth's magnetic field in terms of millimetres, grams, and seconds.[84] Prior to this, the strength of the Earth's magnetic field had only been described in relative terms. The technique used by Gauss was to equate the torque induced on a suspended magnet of known mass by the Earth's magnetic field with the torque induced on an equivalent system under gravity. The resultant calculations enabled him to assign dimensions based on mass, length and time to the magnetic field.[bx][91]
A candlepower as a unit of illuminance was originally defined by an 1860 English law as the light produced by a pure spermaceti candle weighing 16 pound (76 grams) and burning at a specified rate. Spermaceti, a waxy substance found in the heads of sperm whales, was once used to make high-quality candles. At this time the French standard of light was based upon the illumination from a Carcel oil lamp. The unit was defined as that illumination emanating from a lamp burning pure rapeseed oil at a defined rate. It was accepted that ten standard candles were about equal to one Carcel lamp.
### Metre Convention
A French-inspired initiative for international cooperation in metrology led to the signing in 1875 of the Metre Convention, also called Treaty of the Metre, by 17 nations.[by][78]: 353–354 Initially the convention only covered standards for the metre and the kilogram. In 1921, the Metre Convention was extended to include all physical units, including the ampere and others thereby enabling the CGPM to address inconsistencies in the way that the metric system had been used.[85][25]: 96
A set of 30 prototypes of the metre and 40 prototypes of the kilogram,[bz] in each case made of a 90% platinum-10% iridium alloy, were manufactured by British metallurgy specialty firm[who?] and accepted by the CGPM in 1889. One of each was selected at random to become the International prototype metre and International prototype kilogram that replaced the mètre des Archives and kilogramme des Archives respectively. Each member state was entitled to one of each of the remaining prototypes to serve as the national prototype for that country.[92]
The treaty also established a number of international organisations to oversee the keeping of international standards of measurement.[93][ca]
### The CGS and MKS systems
In the 1860s, James Clerk Maxwell, William Thomson (later Lord Kelvin) and others working under the auspices of the British Association for the Advancement of Science, built on Gauss's work and formalised the concept of a coherent system of units with base units and derived units christened the centimetre–gram–second system of units in 1874. The principle of coherence was successfully used to define a number of units of measure based on the CGS, including the erg for energy, the dyne for force, the barye for pressure, the poise for dynamic viscosity and the stokes for kinematic viscosity.[87]
In 1879, the CIPM published recommendations for writing the symbols for length, area, volume and mass, but it was outside its domain to publish recommendations for other quantities. Beginning in about 1900, physicists who had been using the symbol "μ" (mu) for "micrometre" or "micron", "λ" (lambda) for "microlitre", and "γ" (gamma) for "microgram" started to use the symbols "μm", "μL" and "μg".[94]
At the close of the 19th century three different systems of units of measure existed for electrical measurements: a CGS-based system for electrostatic units, also known as the Gaussian or ESU system, a CGS-based system for electromechanical units (EMU) and an International system based on units defined by the Metre Convention.[95] for electrical distribution systems. Attempts to resolve the electrical units in terms of length, mass, and time using dimensional analysis was beset with difficulties—the dimensions depended on whether one used the ESU or EMU systems.[88] This anomaly was resolved in 1901 when Giovanni Giorgi published a paper in which he advocated using a fourth base unit alongside the existing three base units. The fourth unit could be chosen to be electric current, voltage, or electrical resistance.[96] Electric current with named unit 'ampere' was chosen as the base unit, and the other electrical quantities derived from it according to the laws of physics. This became the foundation of the MKS system of units.
In the late 19th and early 20th centuries, a number of non-coherent units of measure based on the gram/kilogram, centimetre/metre, and second, such as the Pferdestärke (metric horsepower) for power,[97][cb] the darcy for permeability[98] and "millimetres of mercury" for barometric and blood pressure were developed or propagated, some of which incorporated standard gravity in their definitions.
At the end of the Second World War, a number of different systems of measurement were in use throughout the world. Some of these systems were metric system variations; others were based on customary systems of measure, like the US customary system and British Imperial system.
### The Practical system of units
In 1948, the 9th CGPM commissioned a study to assess the measurement needs of the scientific, technical, and educational communities and "to make recommendations for a single practical system of units of measurement, suitable for adoption by all countries adhering to the Metre Convention".[99] This working document was Practical system of units of measurement. Based on this study, the 10th CGPM in 1954 defined an international system derived from six base units including units of temperature and optical radiation in addition to those for the MKS system mass, length, and time units and Giorgi's current unit. Six base units were recommended: the metre, kilogram, second, ampere, degree Kelvin, and candela.
The 9th CGPM also approved the first formal recommendation for the writing of symbols in the metric system when the basis of the rules as they are now known was laid down.[100] These rules were subsequently extended and now cover unit symbols and names, prefix symbols and names, how quantity symbols should be written and used, and how the values of quantities should be expressed.[25]: 104, 130
### Birth of the SI
In 1960, the 11th CGPM synthesised the results of the 12-year study into a set of 16 resolutions. The system was named the International System of Units, abbreviated SI from the French name, Le Système International d'Unités.[25]: 110 [101]
### Historical definitions
When Maxwell first introduced the concept of a coherent system, he identified three quantities that could be used as base units: mass, length, and time. Giorgi later identified the need for an electrical base unit, for which the unit of electric current was chosen for SI. Another three base units (for temperature, amount of substance, and luminous intensity) were added later.
The early metric systems defined a unit of weight as a base unit, while the SI defines an analogous unit of mass. In everyday use, these are mostly interchangeable, but in scientific contexts the difference matters. Mass, strictly the inertial mass, represents a quantity of matter. It relates the acceleration of a body to the applied force via Newton's law, F = m × a: force equals mass times acceleration. A force of 1 N (newton) applied to a mass of 1 kg will accelerate it at 1 m/s2. This is true whether the object is floating in space or in a gravity field e.g. at the Earth's surface. Weight is the force exerted on a body by a gravitational field, and hence its weight depends on the strength of the gravitational field. Weight of a 1 kg mass at the Earth's surface is m × g; mass times the acceleration due to gravity, which is 9.81 newtons at the Earth's surface and is about 3.5 newtons at the surface of Mars. Since the acceleration due to gravity is local and varies by location and altitude on the Earth, weight is unsuitable for precision measurements of a property of a body, and this makes a unit of weight unsuitable as a base unit.
SI base units[2]: 6 [26][27]
Unit name Definition[n 1]
second
• Prior: (1675) 1/86400 of a day of 24 hours of 60 minutes of 60 seconds.TLB
• Interim (1956): 1/31556925.9747 of the tropical year for 1900 January 0 at 12 hours ephemeris time.
• Current (1967): The duration of 9192631770 periods of the radiation corresponding to the transition between the two hyperfine levels of the ground state of the caesium-133 atom.
metre
• Prior (1793): 1/10000000 of the meridian through Paris between the North Pole and the Equator.FG
• Interim (1889): The prototype of the metre chosen by the CIPM, at the temperature of melting ice, represents the metric unit of length.
• Interim (1960): 1650763.73 wavelengths in vacuum of the radiation corresponding to the transition between the 2p10 and 5d5 quantum levels of the krypton-86 atom.
• Current (1983): The distance travelled by light in vacuum in 1/299792458 second.
kilogram
• Prior (1793): The grave was defined as being the mass (then called weight) of one litre of pure water at its freezing point.FG
• Interim (1889): The mass of a small squat cylinder of ≈47 cubic centimetres of platinum-iridium alloy kept in the International Bureau of Weights and Measures (BIPM), Pavillon de Breteuil, France.[cc] Also, in practice, any of numerous official replicas of it.
• Current (2019): The kilogram is defined by setting the Planck constant h exactly to 6.62607015×10−34 J⋅s (J = kg⋅m2⋅s−2), given the definitions of the metre and the second.[28] Then the formula would be kg = h/6.62607015×10−34⋅m2⋅s−1
ampere
• Prior (1881): A tenth of the electromagnetic CGS unit of current. The [CGS] electromagnetic unit of current is that current, flowing in an arc 1 cm long of a circle 1 cm in radius, that creates a field of one oersted at the centre.[102] IEC
• Interim (1946): The constant current which, if maintained in two straight parallel conductors of infinite length, of negligible circular cross-section, and placed 1 m apart in vacuum, would produce between these conductors a force equal to 2×10−7 newtons per metre of length.
• Current (2019): The flow of 1/1.602176634×10−19 times the elementary charge e per second.
kelvin
• Prior (1743): The centigrade scale is obtained by assigning 0 °C to the freezing point of water and 100 °C to the boiling point of water.
• Interim (1954): The triple point of water (0.01 °C) defined to be exactly 273.16 K.[n 2]
• Previous (1967): 1/273.16 of the thermodynamic temperature of the triple point of water.
• Current (2019): The kelvin is defined by setting the fixed numerical value of the Boltzmann constant k to 1.380649×10−23 J⋅K−1, (J = kg⋅m2⋅s−2), given the definition of the kilogram, the metre, and the second.
mole
• Prior (1900): A stoichiometric quantity which is the equivalent mass in grams of Avogadro's number of molecules of a substance.ICAW
• Interim (1967): The amount of substance of a system which contains as many elementary entities as there are atoms in 0.012 kilogram of carbon-12.
• Current (2019): The amount of substance of exactly 6.02214076×1023 elementary entities. This number is the fixed numerical value of the Avogadro constant, NA, when expressed in the unit mol−1 and is called the Avogadro number.
candela
• Prior (1946): The value of the new candle (early name for the candela) is such that the brightness of the full radiator at the temperature of solidification of platinum is 60 new candles per square centimetre.
• Current (1979): The luminous intensity, in a given direction, of a source that emits monochromatic radiation of frequency 5.4×1014 hertz and that has a radiant intensity in that direction of 1/683 watt per steradian.
Note: both old and new definitions are approximately the luminous intensity of a spermaceti candle burning modestly bright, in the late 19th century called a "candlepower" or a "candle".
Notes
1. ^ Interim definitions are given here only when there has been a significant difference in the definition.
2. ^ In 1954 the unit of thermodynamic temperature was known as the "degree Kelvin" (symbol °K; "Kelvin" spelt with an upper-case "K"). It was renamed the "kelvin" (symbol "K"; "kelvin" spelt with a lower case "k") in 1967.
The Prior definitions of the various base units in the above table were made by the following authors and authorities:
All other definitions result from resolutions by either CGPM or the CIPM and are catalogued in the SI Brochure.
## Related units
### Non-SI units accepted for use with SI
Many non-SI units continue to be used in the scientific, technical, and commercial literature. Some units are deeply embedded in history and culture, and their use has not been entirely replaced by their SI alternatives. The CIPM recognised and acknowledged such traditions by compiling a list of non-SI units accepted for use with SI:[25]
Some units of time, angle, and legacy non-SI units have a long history of use. Most societies have used the solar day and its non-decimal subdivisions as a basis of time and, unlike the foot or the pound, these were the same regardless of where they were being measured. The radian, being 1/2π of a revolution, has mathematical advantages but is rarely used for navigation. Further, the units used in navigation around the world are similar. The tonne, litre, and hectare were adopted by the CGPM in 1879 and have been retained as units that may be used alongside SI units, having been given unique symbols. The catalogued units are given below.
Most of these, in order to be converted to the corresponding SI unit, require conversion factors that are not powers of ten. Some common examples of such units are the customary units of time, namely the minute (conversion factor of 60 s/min, since 1 min = 60 s), the hour (3600 s), and the day (86400 s); the degree (for measuring plane angles, = π/180 rad); and the electronvolt (a unit of energy, 1 eV = 1.602176634×10−19 J).
Non-SI units accepted for use with SI units
Quantity Name Symbol Value in SI units
time minute min 1 min = 60 s
hour h 1 h = 60 min = 3600 s
day d 1 d = 24 h = 86400 s
length astronomical unit au 1 au = 149597870700 m
plane and
phase angle
degree ° 1° = π/180 rad
arcminute 1′ = 1/60° = π/10800 rad
arcsecond 1″ = 1/60′ = π/648000 rad
area hectare ha 1 ha = 1 hm2 = 104 m2
volume litre l, L 1 l = 1 L = 1 dm3 = 103 cm3 = 10−3 m3
mass tonne (metric ton) t 1 t = 1 Mg = 103 kg
dalton Da 1 Da = 1.660539040(20)×10−27 kg
energy electronvolt eV 1 eV = 1.602176634×10−19 J
logarithmic
ratio quantities
neper Np In using these units it is important that the nature of the quantity be specified and that any reference value used be specified.
bel B
decibel dB
These units are used in combination with SI units in common units such as the kilowatt-hour (1 kW⋅h = 3.6 MJ).
### Metric units that are not recognised by the SI
Although the term metric system is often used as an informal alternative name for the International System of Units,[103] other metric systems exist, some of which were in widespread use in the past or are even still used in particular areas. There are also individual metric units such as the sverdrup and the darcy that exist outside of any system of units. Most of the units of the other metric systems are not recognised by the SI.[cd][cg]
Examples include the centimetre–gram–second (CGS) system, the dominant metric system in the physical sciences and electrical engineering from the 1860s until at least the 1960s, and still in use in some fields. It includes such SI-unrecognised units as the gal, dyne, erg, barye, etc. in its mechanical sector, as well as the poise and stokes in fluid dynamics. When it comes to the units for quantities in electricity and magnetism, there are several versions of the CGS system. Two of these are obsolete: the CGS electrostatic ('CGS-ESU', with the SI-unrecognised units of statcoulomb, statvolt, statampere, etc.) and the CGS electromagnetic system ('CGS-EMU', with abampere, abcoulomb, oersted, maxwell, abhenry, gilbert, etc.).[ch][cj] A 'blend' of these two systems is still popular and is known as the Gaussian system (which includes the gauss as a special name for the CGS-EMU unit maxwell per square centimetre).[ck]
In engineering (other than electrical engineering), there was formerly a long tradition of using the gravitational metric system, whose SI-unrecognised units include the kilogram-force (kilopond), technical atmosphere, metric horsepower, etc. The metre–tonne–second (mts) system, used in the Soviet Union from 1933 to 1955, had such SI-unrecognised units as the sthène, pièze, etc. Other groups of SI-unrecognised metric units are the various legacy and CGS units related to ionising radiation (rutherford, curie, roentgen, rad, rem, etc.), radiometry (langley, jansky), photometry (phot, nox, stilb, nit, metre-candle,[109]:17 lambert, apostilb, skot, brill, troland, talbot, candlepower, candle), thermodynamics (calorie), and spectroscopy (reciprocal centimetre).
Some other SI-unrecognised metric units that do not fit into any of the already mentioned categories include the are, bar, barn, fermi, gradian (gon, grad, or grade), metric carat, micron, millimetre of mercury, torr, millimetre (or centimetre, or metre) of water, millimicron, mho, stere, x unit, γ (unit of mass), γ (unit of magnetic flux density), and λ (unit of volume).[110]: In some cases, the SI-unrecognised metric units have equivalent SI units formed by combining a metric prefix with a coherent SI unit. For example, γ (unit of magnetic flux density) = 1 nT, 1 Gal = 1 cm⋅s−2, 1 barye = , etc. (a related group are the correspondences[ch] such as 1 abampere, 1 abhenry, etc.[cl]). Sometimes, it is not even a matter of a metric prefix: the SI-nonrecognised unit may be exactly the same as an SI coherent unit, except for the fact that the SI does not recognise the special name and symbol. For example, the nit is just an SI-unrecognised name for the SI unit candela per square metre and the talbot is an SI-unrecognised name for the SI unit lumen second. Frequently, a non-SI metric unit is related to an SI unit through a power-of-ten factor, but not one that has a metric prefix, e.g., 1 dyn = 10−5 newton, The angstrom (1 Å = 10−10 m), still used in various fields, etc. (and correspondences[ch] like 1 gauss10−4 tesla). Finally, there are metric units whose conversion factors to SI units are not powers of ten, e.g., 1 calorie = 4.184 joules and 1 kilogram-force = 9.806650 newtons. Some SI-unrecognised metric units are still frequently used, e.g., the calorie (in nutrition), the rem (in the US), the jansky (in radio astronomy), the gauss (in industry) and the CGS-Gaussian units[ck] more generally (in some subfields of physics), the metric horsepower (for engine power, in most of the non-English speaking world), the kilogram-force (for rocket engine thrust, in China and sometimes in Europe), etc. Others are now rarely used, such as the sthene and the rutherford.
### Unacceptable uses
Sometimes, SI unit name variations are introduced, mixing information about the corresponding physical quantity or the conditions of its measurement; however, this practice is unacceptable with the SI.[cm] Instances include: "watt-peak" and "watt RMS"; "geopotential metre" and "vertical metre"; "standard cubic metre"; "atomic second", "ephemeris second", and "sidereal second".
## See also
Organisations
Standards and conventions
## Notes
1. ^ 'SI' is an initialism of Système international, which is an abbreviated form of its full French name Système international d'unités,[1]: 165 which literally means 'International System of Units'. By Resolution 12 of the 11th CGPM (1960), the international abbreviation of the name of the system is: SI.[1]: 165
2. ^ Sometimes, it is referred to as the SI system, which is a pleonasm: 'SI' is the abbreviation of the French Système international, which translates to 'International System'; so, 'SI system' is functionally equivalent to 'International System system'. Nevertheless, examples of such usage exist.[4][5]
3. ^ In a decimal system, different units for a given kind of physical quantity are related by factors of 10, so that, within such a system, unit conversions involve the simple process of moving the decimal point to the right or to the left.[8] So instead of relations like 1 mile = 1760 yards, as in the imperial and US customary measurement systems (which are not decimal), in the SI (which is decimal) 1 kilometre = 1000 metres. Here the kilometre is comparable in size to the mile (1 km0.6 mi) and the metre to the yard (1 m1.1 yd).
4. ^ Or one of its decimal multiples or submultiples, like the centimetre.
5. ^ Or one of its decimal multiples or submultiples, like the gram.
6. ^ Or one of its decimal multiples or submultiples, like the gram-force.
7. ^ A metric system of units is any system of weights and measures that is decimal[c] and based on the metre[d] as the unit of length and either the kilogram[e] as the unit of mass or the kilogram-force[f] as the unit of force.
8. ^ As of 19 January 2021.
9. ^ a b The latter group includes economic unions such as the Caribbean Community (CARICOM).
10. ^ This is an international organization with[h] 63 member states and 39 Associate States and Economies of the General Conference.[i][12] It was established in 1875 under the terms of the Metre Convention.[11][13]
11. ^ a b c From French: Conférence générale des poids et mesures.
12. ^ a b It shall be lawful throughout the United States of America to employ the weights and measures of the metric system; and no contract or dealing, or pleading in any court, shall be deemed invalid or liable to objection because the weights or measures expressed or referred to therein are weights or measures of the metric system. (15 U.S.C. § 204)
13. ^ Here 'official status' means that the SI is recognized in some way by the laws and regulations of the country. In many countries, this means that using the SI units is mandatory for most commercial and administrative purposes (e.g. in the European Union). On the other hand, when it comes to the US, 'official status' means that federal law specifically allows, but does not require, the SI units to be used.[l] In fact, federal law even states that it is the declared policy of the United States to designate the metric system of measurement as the preferred system of weights and measures for United States trade and commerce (15 U.S.C. § 205b).
See metrication for more information.
14. ^ This includes the United States, Canada, and the United Kingdom, despite the fact these three countries also continue to use their customary systems to various degrees.
15. ^ Although the precise definition of coherence is complicated, the basic idea is that mathematical relations between the units for quantities should mirror the mathematical relations between the corresponding quantities themselves. For example, the coherent unit of volume is equal to the volume of a cube whose sides are one unit of length; the coherent unit of pressure is equal to the pressure exerted by a unit-magnitude force over a surface of unit area; etc. As an example of lack of coherence, consider how, in the US customary system, the units of fluid volume are related to the units of length. The principal units of length are inches, feet, yards, and miles; meanwhile, the principal units of fluid volume are based on the (US) gallon, which, at 231 cubic inches, is not a cubic inch, or a cubic foot, or a cubic yard, or a cubic mile (note that 231=3×7×11).
16. ^ For example, the SI unit of velocity is the metre per second, m⋅s−1; of acceleration is the metre per second squared, m⋅s−2; etc. These can also be written as m/s and m/s2, respectively.
17. ^ a b For example the newton (N), the unit of force, equivalent to kg⋅m⋅s−2; the joule (J), the unit of energy, equivalent to kg⋅m2⋅s−2, etc. The most recently named derived unit, the katal, was defined in 1999.
18. ^ a b For example, the recommended unit for the electric field strength is the volt per metre, V/m, where the volt is the derived unit for electric potential difference. The volt per metre is equal to kg⋅m⋅s−3⋅A−1 when expressed in terms of base units.
19. ^ This must be one of 29 coherent units with a separate name and symbol, i.e. either one of the seven base units or one of the 22 coherent derived units with special names and symbols.
20. ^ For example, the coherent SI unit of length is the metre, about the height of kitchen counter (just over 3 ft). But for driving distances, one would normally use kilometres, where one kilometre is 1000 metres; here the metric prefix 'kilo-' (symbol 'k') stands for a factor of 1000. On the other hand, for tailoring measurements, one would usually use centimetres, where one centimetre is 1/100 of a metre; here the metric prefix 'centi-' (symbol 'c') stands for a factor of 1/100.
21. ^ Non-coherent, customary systems have another tendency, well-illustrated by the US customary system. In that system, some liquid commodities are measured neither in the coherent units of volume (e.g. cubic inches) nor in gallons, but in barrels. Furthermore, the size of the barrel depends on the commodity: it means 31 US gallons for beer,[14] but 42 gallons for petroleum.[15] So different units for one and the same quantity (e.g. volume) are used depending on what is being measured, and these different units may not be related to each other in any obvious way—even if they have the same name.
22. ^ Meaning that different units for a given quantity, such as length, are related by factors of 10. Therefore, calculations involve the simple process of moving the decimal point to the right or to the left.[8]
23. ^ Although the terms the metric system and the SI are often used as synonyms, there are in fact many mutually incompatible metric systems. Moreover, there exist metric units that are not recognised by any larger metric system. See § Metric units that are not recognised by the SI, below.
24. ^ As of May 2020, only for the following countries is it uncertain whether the SI has any official status: Myanmar, Liberia, the Federated States of Micronesia, the Marshall Islands, Palau, and Samoa.
25. ^ In the United States, the history of legislation begins with the Metric Act of 1866, which legally protected use of the metric system in commerce. The first section is still part of US law (15 U.S.C. § 204).[l] In 1875, the US became one of the original signatories of the Metre Convention. In 1893, the Mendenhall Order stated that the Office of Weights and Measures ... will in the future regard the International Prototype Metre and Kilogramme as fundamental standards, and the customary units — the yard and the pound — will be derived therefrom in accordance with the Act of 28 July 1866. In 1954, the US adopted the International Nautical Mile, which is defined as exactly 1852 m, in lieu of the US Nautical Mile, defined as 6080.20 ft = 1853.248 m. In 1959, the US National Bureau of Standards officially adapted the International yard and pound, which are defined exactly in terms of the metre and the kilogram. In 1968, the Metric Study Act (Pub. L. 90-472, 9 August 1968, 82 Stat. 693) authorised a three-year study of systems of measurement in the US, with particular emphasis on the feasibility of adopting the SI. The Metric Conversion Act of 1975 followed, later amended by the Omnibus Trade and Competitiveness Act of 1988, the Savings in Construction Act of 1996, and the Department of Energy High-End Computing Revitalization Act of 2004. As a result of all these acts, the US current law (15 U.S.C. § 205b) states that
It is therefore the declared policy of the United States-
(1) to designate the metric system of measurement as the preferred system of weights and measures for United States trade and commerce;
(2) to require that each Federal agency, by a date certain and to the extent economically feasible by the end of the fiscal year 1992, use the metric system of measurement in its procurements, grants, and other business-related activities, except to the extent that such use is impractical or is likely to cause significant inefficiencies or loss of markets to United States firms, such as when foreign competitors are producing competing products in non-metric units;
(3) to seek out ways to increase understanding of the metric system of measurement through educational information and guidance and in Government publications; and
(4) to permit the continued use of traditional systems of weights and measures in non-business activities.
26. ^ There are differences between the United States customary system and the United Kingdom's imperial system. For example, the imperial gallon is about 20% larger than the US gallon.
27. ^ And have been defined in terms of the SI's metric predecessors since at least the 1890s.
28. ^ See e.g. here for the various definitions of the catty, a traditional Chinese unit of mass, in various places across East and Southeast Asia. Similarly, see this article on the traditional Japanese units of measurement, as well as this one on the traditional Indian units of measurement.
29. ^ a b from French: Comité international des poids et mesures
30. ^ from French: Bureau international des poids et mesures
31. ^ The official term is "States Parties to the Metre Convention"; the term "Member States" is its synonym and used for easy reference.[12] As of 13 January 2020,[12] there are 63 Member States and 39 Associate States and Economies of the General Conference.[i]
32. ^ "Among the tasks of these Consultative Committees are the detailed consideration of advances in physics that directly influence metrology, the preparation of Recommendations for discussion at the CIPM, the identification, planning and execution of key comparisons of national measurement standards, and the provision of advice to the CIPM on the scientific work in the laboratories of the BIPM."[18]
33. ^ As of April 2020, these include those from Spain (CEM), Russia (FATRiM), Switzerland (METAS), Italy (INRiM), South Korea (KRISS), France (LNE), China (NIM), US (NIST), Japan (AIST/NIMJ), UK (NPL), Canada (NRC), and Germany (PTB).
34. ^ As of April 2020, these include International Electrotechnical Commission (IEC), International Organization for Standardization (ISO), and International Organization of Legal Metrology (OIML).
35. ^ As of April 2020, these include International Commission on Illumination (CIE), CODATA Task Group on Fundamental Constants, International Commission on Radiation Units and Measurements (ICRU), and International Federation of Clinical Chemistry and Laboratory Medicine (IFCC).
36. ^ As of April 2020, these include International Astronomical Union (IAU), International Union of Pure and Applied Chemistry (IUPAC), and International Union of Pure and Applied Physics (IUPAP).
37. ^ These are "individuals with a long-term involvement in matters related to units, having actively contributed to publications on units, and having a global view and understanding of science as well as knowledge on the development and functioning of the International System of Units."[22] As of April 2020, these include[21][23] Marc Himbert and Terry Quinn.
38. ^ The SI Brochure for short. As of 2022, the latest edition is the ninth, originally published in 2019 and updated in 2022. It is Ref.[1] of this article.
39. ^ For example, the United States' National Institute of Standards and Technology (NIST) has produced a version of the CGPM document (NIST SP 330) which clarifies usage for English-language publications that use American English
40. ^ For historical reasons, the kilogram rather than the gram is treated as the coherent unit, making an exception to this characterisation.
41. ^ Ohm's law: 1 Ω = 1 V/A from the relationship E = I × R, where E is electromotive force or voltage (unit: volt), I is current (unit: ampere), and R is resistance (unit: ohm).
42. ^ The latter are formalised in the International System of Quantities (ISQ).[1]:
43. ^ The choice of which and even how many quantities to use as base quantities is not fundamental or even unique – it is a matter of convention.[1]: For example, four base quantities could have been chosen as velocity, angular momentum, electric charge and energy.
44. ^ Here are some examples of coherent derived SI units: the unit of velocity, which is the metre per second, with the symbol m/s; the unit of acceleration, which is the metre per second squared, with the symbol m/s2; etc.
45. ^ A useful property of a coherent system is that when the numerical values of physical quantities are expressed in terms of the units of the system, then the equations between the numerical values have exactly the same form, including numerical factors, as the corresponding equations between the physical quantities;[29]: 6 An example may be useful to clarify this. Suppose we are given an equation relating some physical quantities, e.g. T = 1/2{m}{v}2, expressing the kinetic energy T in terms of the mass m and the velocity v. Choose a system of units, and let {T}, {m}, and {v} be the numerical values of T, m, and v when expressed in that system of units. If the system is coherent, then the numerical values will obey the same equation (including numerical factors) as the physical quantities, i.e. we will have that T = 1/2{m}{v}2. Therefore, SI units can be converted without numerical factors: 1 J = 1 N·m = 1 C·V = 1 W·s.
On the other hand, if the chosen system of units is not coherent, this property may fail. For example, the following is not a coherent system: one where energy is measured in calories, while mass and velocity are measured in their SI units. After all, in that case, 1/2{m}{v}2 will give a numerical value whose meaning is the kinetic energy when expressed in joules, and that numerical value is different, by a factor of 4.184, from the numerical value when the kinetic energy is expressed in calories. Thus, in that system, the equation satisfied by the numerical values is instead {T} = 1/4.1841/2{m}{v}2.
46. ^ Which define the International System of Quantities (ISQ).
47. ^ a b For historical reasons, names and symbols for decimal multiples and sub-multiples of the unit of mass are formed as if it is the gram which is the base unit, i.e. by attaching prefix names and symbols, respectively, to the unit name "gram" and the unit symbol "g". For example, 10−6 kg is written as milligram, mg, not as microkilogram, μkg.[1]: 144
48. ^ As the SI Brochure states,[1]: 140 this applies not only to technical texts, but also, for example, to measuring instruments (i.e. the instrument read-out needs to indicate both the unit and the quantity measured).
49. ^ Customarily, however, rainfall is measured in non-coherent SI units such as millimetres in height collected on each square metre during a certain period, equivalent to litres per square metre.
50. ^ It is correct to say that an SI base unit (like the metre) is a coherent unit for its corresponding physical quantity. Recall that the set of coherent SI units consists of the base units and the coherent derived units. This usage is consistent with the definition of a coherent unit as one that is equal to 'a product of powers of the base units with a prefactor of 1'. After all, each base unit is obviously so representable—it is equal to itself to the power of 1 and with a prefactor of 1.
51. ^ One kilometre is about 0.62 miles, a length equal to about two and a half laps around a typical athletic track. Walking at a moderate pace for one hour, an adult human will cover about five kilometres (about three miles). The distance from London, UK, to Paris, France is about 350 km; from London to New York, 5600 km.
52. ^ In other words, given any base unit or any coherent derived unit with a special name and symbol.
53. ^ This last statement in fact applies to all SI units, not only those with special names and symbols. Consider the example of the SI units of torque. Because the SI does not have a unit with a special name and symbol for torque, its coherent SI unit is the newton-metre, N⋅m. The following are some examples of non-coherent SI units of torque: N⋅mm, kN⋅μm, mN⋅cm, etc. Note that these non-coherent units are obtained from the original coherent unit by replacing some (or all) of the units with special names and symbols that are present in the original coherent unit by their decimal multiples or submultiples. But then these different powers of ten combine into one overall power of ten. For example, kN⋅μm = (103 N)⋅(10−6 m) = 103–6 N⋅m = 10−3 N⋅m.
54. ^ Note, however, that there is a special group of units that are called non-SI units accepted for use with SI, most of which are not decimal multiples of the corresponding SI units; see below.
55. ^ The unit is named after Anders Celsius.
56. ^ a b Except where specifically noted, these rules are common to both the SI Brochure and the NIST brochure.
57. ^ This term is a translation of the official [French] text of the SI Brochure.
58. ^ See the next section for why this type of definition is considered advantageous.
59. ^ Their exactly defined values are as follows:[1]: 128
${\displaystyle \Delta \nu _{\text{Cs}}}$ = 9192631770 Hz
${\displaystyle c}$ = 299792458 m/s
${\displaystyle h}$ = 6.62607015×10−34 J⋅s
${\displaystyle e}$ = 1.602176634×10−19 C
${\displaystyle k}$ = 1.380649×10−23 J/K
${\displaystyle N_{\text{A}}}$ = 6.02214076×1023 mol−1
${\displaystyle K_{\text{cd}}}$ = 683 lm/W.
60. ^ A mise en pratique is French for 'putting into practice; implementation'.[53][54]
61. ^ a b The sole exception is the definition of the second, which is still given not in terms of fixed values of fundamental constants but in terms of a particular property of a particular naturally occurring object, the caesium atom. And indeed, it has been clear for some time that relatively soon, by using atoms other than caesium, it will be possible to have definitions of the second that are more precise than the current one. Taking advantage of these more precise methods will necessitate the change in the definition of the second, probably sometime around the year 2030.[56]: 196 [57]
62. ^ a b Again, except for the second, as explained in the previous note.
The second may eventually get fixed by defining an exact value for yet another fundamental constant (whose derived unit includes the second), for example the Rydberg constant. For this to happen, the uncertainty in the measurement of that constant must become so small as to be dominated by the uncertainty in the measurement of whatever clock transition frequency is being used to define the second at that point. Once that happens, the definitions will be reversed: the value of the constant will be fixed by definition to an exact value, namely its most recent best measured value, while the clock transition frequency will become a quantity whose value is no longer fixed by definition but which has to be measured. Unfortunately, it is unlikely that this will happen in the foreseeable future, because presently there are no promising strategies for measuring any additional fundamental constants with the necessary precision.[58]: 4112–4113
63. ^ The one exception being the definition of the second; see Notes [bi] and [bj] in the following section.
64. ^ To see this, recall that Hz = s−1 and J = kgm2s−2. Thus,
(Hz) (J⋅s) / (m/s)2
= (s−1) [(kgm2s−2)⋅s] (ms−1)−2
= s(−1−2+1+2)m(2−2)kg
= kg,
since all the powers of metres and seconds cancel out. It can further be shown that (Hz) (J⋅s) / (m/s)2 is the only combination of powers of the units of the defining constants (that is, the only combination of powers of Hz, m/s, J⋅s, C, J/K, mol−1, and lm/W) that results in the kilogram.
65. ^ Namely,
1 Hz = ΔνCs/9192631770
1 m/s = c/299792458 , and
1 J⋅s = h/6.62607015×10−34.
66. ^ The SI Brochure prefers to write the relationship between the kilogram and the defining constants directly, without going through the intermediary step of defining 1 Hz, 1 m/s, and 1 J⋅s, like this:[1]: 131 1 kg = (299792458)2/(6.62607015×10−34)(9192631770)hΔνCs/c2.
67. ^ For example, from 1889 until 1960, the metre was defined as the length of the International Prototype Metre, a particular bar made of platinum-iridium alloy that was (and still is) kept at the International Bureau of Weights and Measures, located in the Pavillon de Breteuil in Saint-Cloud, France, near Paris. The final artefact-based definition of the metre, which stood from 1927 to the redefinition of the metre in 1960, read as follows:[1]: 159
The unit of length is the metre, defined by the distance, at , between the axes of the two central lines marked on the bar of platinum-iridium kept at the Bureau International des Poids et Mesures and declared Prototype of the metre by the 1st Conférence Générale des Poids et Mesures, this bar being subject to standard atmospheric pressure and supported on two cylinders of at least one centimetre diameter, symmetrically placed in the same horizontal plane at a distance of 571 mm from each other.
The '' refers to the temperature of 0 °C. The support requirements represent the Airy points of the prototype—the points, separated by 4/7 of the total length of the bar, at which the bending or droop of the bar is minimised.[60]
68. ^ The latter was called the 'quadrant', the length of a meridian from the equator to the North Pole. The originally chosen meridian was the Paris meridian.
69. ^ At the time 'weight' and 'mass' were not always carefully distinguished.
70. ^ This volume is 1 cm3 = 1 mL, which is 1×10−6 m3. Thus, the original definition of mass used not the coherent unit of volume (which would be the m3) but a decimal submultiple of it.
71. ^ Indeed, the original idea of the metric system was to define all units using only natural and universally available measurable quantities. For example, the original definition of the unit of length, the metre, was a definite fraction (one ten-millionth) of the length of a quarter of the Earth's meridian.[bp] Once the metre was defined, one could define the unit of volume as the volume of a cube whose sides are one unit of length. And once the unit of volume was determined, the unit of mass could be defined as the mass of a unit of volume of some convenient substance at standard conditions. In fact, the original definition of the gram was 'the absolute weight[bq] of a volume of pure water equal to the cube of the hundredth part of a metre,[br] and at the temperature of melting ice.'
However, it soon became apparent that these particular 'natural' realisations of the units of length and mass simply could not, at that time, be as precise (and as convenient to access) as the needs of science, technology, and commerce demanded. Therefore, prototypes were adopted instead. Care was taken to manufacture the prototypes so that they would be as close as possible, given the available science and technology of the day, to the idealised 'natural' realisations. But once the prototypes were completed, the units of length and mass became equal by definition to these prototypes (see Mètre des Archives and Kilogramme des Archives).
Nevertheless, throughout the history of the SI, one keeps seeing expressions of hope that one day, one would be able to dispense with the prototypes and define all units in terms of standards found in nature. The first such standard was the second. It was never defined using a prototype, being originally defined as 1/86400 of the length of a day (since there are 60 s/min × 60 min/hr × 24 hr/day = 86400 s/day). As we mentioned, the vision of defining all units in terms of universally available natural standards was at last fulfilled in 2019, when the sole remaining prototype used by the SI, the one for the kilogram, was finally retired.
72. ^ The following references are useful for identifying the authors of the preceding reference: Ref.,,[62] Ref.,[63] and Ref.[64]
73. ^ a b As happened with British standards for length and mass in 1834, when they were lost or damaged beyond the point of useability in a great fire known as the burning of Parliament. A commission of eminent scientists was assembled to recommend the steps to be taken for the restoration of the standards, and in its report, it described the destruction caused by the fire as follows:[61][bt]
We shall in the first place describe the state of the Standards recovered from the ruins of the House of Commons, as ascertained in our inspection of them made on 1st June, 1838, at the Journal Office, where they are preserved under the care of Mr. James Gudge, Principal Clerk of the Journal Office. The following list, taken by ourselves from inspection, was compared with a list produced by Mr. Gudge, and stated by him to have been made by Mr. Charles Rowland, one of the Clerks of the Journal Office, immediately after the fire, and was found to agree with it. Mr. Gudge stated that no other Standards of Length or Weight were in his custody.
No. 1. A brass bar marked "Standard [G. II. crown emblem] Yard, 1758", which on examination was found to have its right hand stud perfect, with the point and line visible, but with its left hand stud completely melted out, a hole only remaining. The bar was somewhat bent, and discoloured in every part.
No. 2. A brass bar with a projecting cock at each end, forming a bed for the trial of yard-measures; discoloured.
No. 3. A brass bar marked "Standard [G. II. crown emblem] Yard, 1760", from which the left hand stud was completely melted out, and which in other respects was in the same condition as No. 1.
No. 4. A yard-bed similar to No. 2; discoloured.
No. 5. A weight of the form [drawing of a weight] marked [2 lb. T. 1758], apparently of brass or copper; much discoloured.
No. 6. A weight marked in the same manner for 4 lbs., in the same state.
No. 7. A weight similar to No. 6, with a hollow space at its base, which appeared at first sight to have been originally filled with some soft metal that had been now melted out, but which on a rough trial was found to have nearly the same weight as No. 6.
No. 8. A similar weight of 8 lbs., similarly marked (with the alteration of 8 lbs. for 4 lbs.), and in the same state.
No. 9. Another exactly like No. 8.
Nos. 10 and 11. Two weights of 16 lbs., similarly marked.
Nos. 12 and 13. Two weights of 32 lbs., similarly marked.
No. 14. A weight with a triangular ring-handle, marked "S.F. 1759 17 lbs. 8 dwts. Troy", apparently intended to represent the stone of 14 lbs. avoirdupois, allowing 7008 troy grains to each avoirdupois pound.
It appears from this list that the bar adopted in the Act 5th Geo. IV., cap. 74, sect. 1, for the legal standard of one yard, (No. 3 of the preceding list), is so far injured, that it is impossible to ascertain from it, with the most moderate accuracy, the statutable length of one yard. The legal standard of one troy pound is missing. We have therefore to report that it is absolutely necessary that steps be taken for the formation and legalising of new Standards of Length and Weight.
74. ^ Indeed, one of the motivations for the 2019 redefinition of the SI was the instability of the artefact that served as the definition of the kilogram.
Before that, one of the reasons the United States started defining the yard in terms of the metre in 1893 was that[65]: 381
[t]he bronze yard No. 11, which was an exact copy of the British imperial yard both in form and material, had shown changes when compared with the imperial yard in 1876 and 1888 which could not reasonably be said to be entirely due to changes in No. 11. Suspicion as to the constancy of the length of the British standard was therefore aroused.
In the above, the bronze yard No. 11 is one of two copies of the new British standard yard that were sent to the US in 1856, after Britain completed the manufacture of new imperial standards to replace those lost in the fire of 1834 (see [bu]). As standards of length, the new yards, especially bronze No. 11, were far superior to the standard the US had been using up to that point, the so-called Troughton scale. They were therefore accepted by the Office of Weights and Measures (a predecessor of NIST) as the standards of the United States. They were twice taken to England and recompared with the imperial yard, in 1876 and in 1888, and, as mentioned above, measurable discrepancies were found.[65]: 381
In 1890, as a signatory of the Metre Convention, the US received two copies of the International Prototype Metre, the construction of which represented the most advanced ideas of standards of the time. Therefore it seemed that US measures would have greater stability and higher accuracy by accepting the international metre as fundamental standard, which was formalised in 1893 by the Mendenhall Order.[65]: 379–381
75. ^ As mentioned above, it is all but certain that the defining constant ${\displaystyle \Delta \nu _{\text{Cs}}}$ will have to be replaced relatively soon, as it is becoming increasingly clear that atoms other than caesium can provide more precise time standards. However, it is not excluded that some of the other defining constants would eventually have to be replaced as well. For example, the elementary charge e corresponds to a coupling strength of the electromagnetic force via the fine-structure constant ${\displaystyle \alpha }$. Some theories predict that ${\displaystyle \alpha }$ can vary over time. The presently known experimental limits of the maximum possible variation of ${\displaystyle \alpha }$ are so low that 'any effect on foreseeable practical measurements can be excluded',[1]: 128 even if one of these theories turns out to be correct. Nevertheless, if the fine-structure constant turns out to slightly vary over time, science and technology may in the future advance to a point where such changes become measurable. At that point, one might consider replacing, for the purposes of defining the SI, the elementary charge with some other quantity, the choice of which will be informed by what we learn about the time variation of ${\displaystyle \alpha }$.
76. ^ The strength of the Earth's magnetic field was designated 1 G (gauss) at the surface (= 1 cm−1/2⋅g1/2⋅s−1).
77. ^ Argentina, Austria-Hungary, Belgium, Brazil, Denmark, France, German Empire, Italy, Peru, Portugal, Russia, Spain, Sweden and Norway, Switzerland, Ottoman Empire, United States, and Venezuela.
78. ^ The text "Des comparaisons périodiques des étalons nationaux avec les prototypes internationaux" (English: the periodic comparisons of national standards with the international prototypes) in article 6.3 of the Metre Convention distinguishes between the words "standard" (OED: "The legal magnitude of a unit of measure or weight") and "prototype" (OED: "an original on which something is modelled").
79. ^ These included:
80. ^ Pferd is German for "horse" and Stärke is German for "strength" or "power". The Pferdestärke is the power needed to raise 75 kg against gravity at the rate of one metre per second. (1 PS = 0.985 HP).
81. ^ It is known as the International Prototype of the Kilogram.
82. ^ Meaning, they are neither part of the SI nor one of the non-SI units accepted for use with that system.
83. ^ Almost invariably either the meter or the centimeter.
84. ^ All major systems of units in which force rather than mass is a base unit are of a type known as gravitational system (also known as technical or engineering system). In the most prominent metric example of such a system, the unit of force is taken to be the kilogram-force (kp), which is the weight of the standard kilogram under standard gravity, g = 9.80665 m/s2. The unit of mass is then a derived unit, defined as the mass that is accelerated at a rate of 1 m/s2 when acted upon by a net force of 1 kp; often called the hyl, it therefore has a value of 1 hyl = 9.80665 kg, so that it is not a decimal multiple of the gram.
85. ^ Having said that, some units are recognised by all metric systems. The second is a base unit in all of them. The metre is recognised in all of them, either as the base unit of length or as a decimal multiple or submultiple of the base unit of length. On the other hand, not every metric system recognises the gram as a unit (either the base unit or a decimal multiple of the base unit). In particular, in gravitational metric systems, the unit of force (gram-force or kilogram-force) replaces the unit of mass as a base unit. The unit of mass is then a derived unit, defined as the mass that, when acted upon by a net unit force, is accelerated at the unit rate (i.e. at a rate of 1 base unit of length[ce] per second squared).[cf]
86. ^ a b c Interconversion between different systems of units is usually straightforward; however, the units for electricity and magnetism are an exception, and a surprising amount of care is required. The problem is that, in general, the physical quantities that go by the same name and play the same role in the CGS-ESU, CGS-EMU, and SI—e.g. 'electric charge', 'electric field strength', etc.—do not merely have different units in the three systems; technically speaking, they are actually different physical quantities.[104]: [104]: Consider 'electric charge', which in each of the three systems can be identified as the quantity two instances of which enter in the numerator of Coulomb's law (as that law is written in each system). This identification produces three different physical quantities: the 'CGS-ESU charge', the 'CGS-EMU charge', and the 'SI charge'.[105]: 35 [104]: 423 They even have different dimensions when expressed in terms of the base dimensions: mass1/2 × length3/2 × time−1 for the CGS-ESU charge, mass1/2 × length1/2 for the CGS-EMU charge, and current × time for the SI charge (where, in the SI, the dimension of current is independent of those of mass, length, and time). On the other hand, these three quantities are clearly quantifying the same underlying physical phenomenon. Thus, we say not that 'one abcoulomb equals ten coulomb', but rather that 'one abcoulomb corresponds to ten coulomb',[104]: written as 1 abC10 C.[105]: 35 By that we mean, 'if the CGS-EMU electric charge is measured to have the magnitude of 1 abC, then the SI electric charge will have the magnitude of 10 C'.[105]: 35 [106]: 57–58
87. ^ Neither EMU nor ESU units had sizes that were convenient for practical work of electrical engineers. It was therefore decided to establish a 'practical' system of units, where each unit is an appropriate decimal multiple or submultiple of the corresponding EMU unit, so that the resulting units have convenient sizes and form a coherent system. These practical units were given names derived from the names of eminent scientists, and many of these units—both the names and the magnitudes—were later incorporated into the SI: the volt, the ampere, the ohm, etc.
88. ^ For several decades, the ESU and EMU units did not have special names; one would just say, for example, the ESU unit of resistance. In 1903, A. E. Kennelly suggested that the names of the EMU units be obtained by prefixing the name of the corresponding 'practical unit' by 'ab-' (short for 'absolute', giving the 'abohm', 'abvolt', the 'abampere', etc.), and that the names of the ESU units be analogously obtained by using the prefix 'abstat-', which was later shortened to 'stat-' (giving the 'statohm', 'statvolt', 'statampere', etc.).[107]: 534–535 In a sense, this brings us back full circle: historically, the magnitudes of the 'practical units' were first defined in terms of the magnitudes of the corresponding units of the EMU system;[ci] then, Kennelly proposed that the names of units in the ESU and EMU systems be derived from the corresponding names of practical units. Kennelly's naming system was widely used in the US, but, apparently, not in Europe.[108]
89. ^ a b The CGS-Gaussian units are a blend of the CGS-ESU and CGS-EMU, taking units related to magnetism from the latter and all the rest from the former. In addition, the system introduces the gauss as a special name for the CGS-EMU unit maxwell per square centimetre.
90. ^ Authors often abuse notation slightly and write these with an 'equals' sign ('=') rather than a 'corresponds to' sign ('≘').
91. ^ "Unacceptability of mixing information with units: When one gives the value of a quantity, any information concerning the quantity or its conditions of measurement must be presented in such a way as not to be associated with the unit."[25]
Attribution
[1] This article incorporates text from this source, which is available under the CC BY 3.0 license.
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70. ^ "Appendix 1. Decisions of the CGPM and the CIPM" (PDF). BIPM. p. 188. Retrieved 27 April 2021.
71. ^ Wood, B. (3–4 November 2014). "Report on the Meeting of the CODATA Task Group on Fundamental Constants" (PDF). BIPM. p. 7. [BIPM director Martin] Milton responded to a question about what would happen if ... the CIPM or the CGPM voted not to move forward with the redefinition of the SI. He responded that he felt that by that time the decision to move forward should be seen as a foregone conclusion.
72. ^
73. ^ a b "Amtliche Maßeinheiten in Europa 1842" [Official units of measure in Europe 1842] (in German). Retrieved 26 March 2011Text version of Malaisé's book:{{cite web}}: CS1 maint: postscript (link)Malaisé, Ferdinand von (1842). Theoretisch-practischer Unterricht im Rechnen [Theoretical and practical instruction in arithmetic] (in German). München: Verlag des Verf. pp. 307–322. Retrieved 7 January 2013.
74. ^ "Units: CGS and MKS". www.unc.edu. Archived from the original on 26 June 2018. Retrieved 22 January 2016.
75. ^ Giovanni Giorgi (1901), "Unità Razionali de Elettromagnetismo", in Atti dell' Associazione Elettrotecnica Italiana.
76. ^ Brainerd, John G. (1970). "Some Unanswered Questions". Technology and Culture. JSTOR. 11 (4): 601–603. doi:10.2307/3102695. ISSN 0040-165X. JSTOR 3102695. S2CID 112215565.
77. ^ "The name 'kilogram'". International Bureau of Weights and Measures. Archived from the original on 14 May 2011. Retrieved 25 July 2006.
78. ^ a b Alder, Ken (2002). The Measure of all Things—The Seven-Year-Odyssey that Transformed the World. London: Abacus. ISBN 978-0-349-11507-8.
79. ^ Quinn, Terry (2012). From artefacts to atoms: the BIPM and the search for ultimate measurement standards. Oxford University Press. p. xxvii. ISBN 978-0-19-530786-3. OCLC 705716998. he [Wilkins] proposed essentially what became ... the French decimal metric system
80. ^ Wilkins, John (1668). "VII". An Essay towards a Real Character and a Philosophical Language. The Royal Society. pp. 190–194.
"Reproduction (33 MB)" (PDF). Retrieved 6 March 2011.; "Transcription" (PDF). Retrieved 6 March 2011.
81. ^ "Mouton, Gabriel". Complete Dictionary of Scientific Biography. encyclopedia.com. 2008. Retrieved 30 December 2012.
82. ^ O'Connor, John J.; Robertson, Edmund F. (January 2004), "Gabriel Mouton", MacTutor History of Mathematics Archive, University of St Andrews
83. ^ Tavernor, Robert (2007). Smoot's Ear: The Measure of Humanity. Yale University Press. ISBN 978-0-300-12492-7.
84. ^ a b "Brief history of the SI". International Bureau of Weights and Measures. Retrieved 12 November 2012.
85. ^ a b Tunbridge, Paul (1992). Lord Kelvin, His Influence on Electrical Measurements and Units. Peter Pereginus Ltd. pp. 42–46. ISBN 978-0-86341-237-0.
86. ^ Everett, ed. (1874). "First Report of the Committee for the Selection and Nomenclature of Dynamical and Electrical Units". Report on the Forty-third Meeting of the British Association for the Advancement of Science Held at Bradford in September 1873. 43: 222–225. Retrieved 28 August 2013. Special names, if short and suitable, would ... be better than the provisional designation 'C.G.S. unit of ...'.
87. ^ a b Page, Chester H.; Vigoureux, Paul, eds. (20 May 1975). The International Bureau of Weights and Measures 1875–1975: NBS Special Publication 420. Washington, D.C.: National Bureau of Standards. p. 12.
88. ^ a b Maxwell, J. C. (1873). A treatise on electricity and magnetism. Vol. 2. Oxford: Clarendon Press. pp. 242–245. Retrieved 12 May 2011.
89. ^ Bigourdan, Guillaume (2012) [1901]. Le Système Métrique Des Poids Et Mesures: Son Établissement Et Sa Propagation Graduelle, Avec L'histoire Des Opérations Qui Ont Servi À Déterminer Le Mètre Et Le Kilogramme [The Metric System of Weights and Measures: Its Establishment and its Successive Introduction, with the History of the Operations Used to Determine the Metre and the Kilogram] (in French) (facsimile ed.). Ulan Press. p. 176. ASIN B009JT8UZU.
90. ^ Smeaton, William A. (2000). "The Foundation of the Metric System in France in the 1790s: The importance of Etienne Lenoir's platinum measuring instruments". Platinum Metals Rev. 44 (3): 125–134. Archived from the original on 29 October 2013. Retrieved 18 June 2013.
91. ^ "The intensity of the Earth's magnetic force reduced to absolute measurement" (PDF). {{cite journal}}: Cite journal requires |journal= (help)
92. ^ Nelson, Robert A. (1981). "Foundations of the international system of units (SI)" (PDF). Physics Teacher. 19 (9): 597. Bibcode:1981PhTea..19..596N. doi:10.1119/1.2340901.
93. ^ "The Metre Convention". Bureau International des Poids et Mesures. Retrieved 1 October 2012.
94. ^ McGreevy, Thomas (1997). Cunningham, Peter (ed.). The Basis of Measurement: Volume 2 – Metrication and Current Practice. Pitcon Publishing (Chippenham) Ltd. pp. 222–224. ISBN 978-0-948251-84-9.
95. ^ Fenna, Donald (2002). Weights, Measures and Units. Oxford University Press. International unit. ISBN 978-0-19-860522-5.
96. ^ "Historical figures: Giovanni Giorgi". International Electrotechnical Commission. 2011. Archived from the original on 15 May 2011. Retrieved 5 April 2011.
97. ^ "Die gesetzlichen Einheiten in Deutschland" [List of units of measure in Germany] (PDF) (in German). Physikalisch-Technische Bundesanstalt (PTB). p. 6. Retrieved 13 November 2012.
98. ^ "Porous materials: Permeability" (PDF). Module Descriptor, Material Science, Materials 3. Materials Science and Engineering, Division of Engineering, The University of Edinburgh. 2001. p. 3. Archived from the original (PDF) on 2 June 2013. Retrieved 13 November 2012.
99. ^ "BIPM – Resolution 6 of the 9th CGPM". Bipm.org. 1948. Retrieved 22 August 2017.
100. ^
101. ^ "BIPM – Resolution 12 of the 11th CGPM". Bipm.org. Retrieved 22 August 2017.
102. ^ McKenzie, A. E. E. (1961). Magnetism and Electricity. Cambridge University Press. p. 322.
103. ^ Olthoff, Jim (2018). "For All Times, For All Peoples: How Replacing the Kilogram Empowers Industry". NIST. Archived from the original on 16 March 2020. Retrieved 14 April 2020. ... the International System of Units (SI), popularly known as the metric system.
104. ^ a b c d Page, Chester H. (1970). "Relations among Systems of Electromagnetic Equations". Am. J. Phys. 38 (4): 421–424. Bibcode:1970AmJPh..38..421P. doi:10.1119/1.1976358.
105. ^ a b c
106. ^ Carron, Neal (2015). "Babel of Units. The Evolution of Units Systems in Classical Electromagnetism". arXiv:1506.01951 [physics.hist-ph].
107. ^ Kennelly, Arthur E. (July 1903). "Magnetic Units and Other Subjects that Might Occupy Attention at the Next International Electrical Congress". Transactions of the American Institute of Electrical Engineers. XXII: 529–536. doi:10.1109/T-AIEE.1903.4764390. S2CID 51634810. [p. 534] The expedient suggests itself of attaching the prefix ab or abs to a practical or Q. E. S. unit, in order to express the absolute or corresponding C. G. S. magnetic unit. … [p. 535] In a comprehensive system of electromagnetic terminology, the electric C. G. S. units should also be christened. They are sometimes referred to in electrical papers, but always in an apologetic, symbolical fashion, owing to the absence of names to cover their nakedness. They might be denoted by the prefix abstat.
108. ^ Silsbee, Francis (April–June 1962). "Systems of Electrical Units". Journal of Research of the National Bureau of Standards Section C. 66C (2): 137–183. doi:10.6028/jres.066C.014.
109. ^ Trotter, Alexander Pelham (1911). Illumination: Its Distribution and Measurement. London: Macmillan. OCLC 458398735.
110. ^ IEEE/ASTM SI 10 American National Standard for Use of the International System of Units (SI): The Modern Metric System. IEEE and ASTM. 2016. | 31,574 | 125,918 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 36, "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} | 2.78125 | 3 | CC-MAIN-2023-40 | latest | en | 0.88164 |
https://www.talkbass.com/threads/help-with-electronics-homework.38280/ | 1,511,496,620,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934807084.8/warc/CC-MAIN-20171124031941-20171124051941-00071.warc.gz | 878,562,543 | 23,730 | # help with electronics homework
Discussion in 'Amps and Cabs [BG]' started by Phat Ham, Feb 1, 2002.
1. ### Phat Ham
Feb 13, 2000
DC
Ok I'm doing my electronics homework and I'm too lazy to think (it's Friday) so I was wondering if someone could help me out. Here's the problem:
Design an inverting op amp (ideal) with a gain of 46dB that can deliver +/- 15V to a load resistance of 5kohms. The maximum current that can be delivered is 4mA. So, how should I go about doing this? I know the gain is -R2/R1 = Vo/Vs = 200. But does it matter what values I choose for R1 and R2?
2. ### throbbinnut
Man, I've already forgotten all that stuff. Do you mean design the op-amp external circuit? I'm assuming you do from the ideal part and the references to R1 and R2.
46dB = votlage gain of 39810.
So, R2/R1 = 39810, and like you say, you have to kind of just pick R1. If you pick 100 ohms for R1, then R2 = 4Meg. (4,000,000 / 100 = 40,000)
Since you said Ideal op-amp, the power supply doesn't matter, nor does the output load, nor does the output current capability. An ideal op amp has infinite input impedance, 0 output impedance, and infinite gain.
Chris
3. ### throbbinnut
I just realize the 15V supply, 5K load, and 4mA output limits help determine R2.
15V across 5K = 3 mA, so you have 1mA to send back around the feedback loop. Since this is an Ideal op amp with + input grounded, the - input will have 0V on it (infinite gain), and if the output is at +15V, and you want 1mA to flow, you need 15K minimum in the feedback loop. Then back calculate R2/R1 = 40,000 where R2 = 15K. R1 = 2.67 ohms.
Kind of freaky, but still works out. I'd rather use the 4Meg resistor and 100 ohm resistor myself.
Chris
4. ### Phat Ham
Feb 13, 2000
DC
I don't see where you got the voltage gain of 39810. We learned that Vout/Vsource = R2/R1 = Av. And 20*log(Av) = gain in dB. Working backwards from 46 dB I get Av = 200.
In the back of the book where all the answers are it says one possibility is R1 = 1k and R2 = 20k. But with those numbers R2/R1 = 20, not 200, so either I'm missing something or the book is wrong (which isn't all that uncommon).
I get what you're saying about how the feedback loop needs to get 1mA. That helps me out a lot. If you use R2/R1 = 20 like the book apparently did and arbitrarily choose 20k for R2 (which is > the minimum of 15k) you get 1k for R1.
If I use R2/R1 = 200, with R2 = 20k then R1 = 100 ohms. hmm. I wonder who's right?
5. ### throbbinnut
My bad on the dB calculation. I was doing power calculation which is (10)Log10 ratio instead of (20)Log10 ratio.
Now, using (20) Log10; 46/20 = 2.3, and 10**2.3 = 199.5 => 200. Cool. That's better.
Now, work it out like before, picking 20K for R2, gives 100 ohms for R1. I'd guess that the book is wrong. Go for 20K and 100R, or go for 200K and 1K.
I told you, 5 years after graduating you've forgotten all this stuff.
Chris
Chris
7. ### Rockin John
The input resistor is R1. You must choose that value so that it doesn't load the stage preceeding the op amp. The junction between the input and the feedback resistor is a virtual ground point in an inverting circuit: that means the Zin of the op amp stage will be virtually equal to R1. If the previous stage / signal generator or whatever can't drive 100 Ohms, you can't use 100 Ohms as an input resistor.
The actual values used will depend on what type of op amp you use. Really large values of feedback res. for bipolar op amps (eg 741) are a no-no. However, 2Meg Ohm shouldn't be a problem which means you could use a 10K for R1: no worries for the opamp or preceeding stage.
Calculate the load resistor from Ohms law. In reality a (bipolar) opamp will deliver o/p voltage to within approx. 2 Volts of either supply rail: that is, +- 13 Volts. So, 13/.004 = 3250 Ohms; 3K3 for the next prefered value. The amp should then sink or source slightly less than 4 milliamps with 3K3 so be within your design tollerances so driving 5K is no problem.
Does that make sense?
John
8. ### Rockin John
The real answer to your problem is, of course, that you should have taken up a profession that pays loads of \$\$\$ - lawyer, accountant, etc and leave electronics to the idiots like me.
RJ | 1,247 | 4,235 | {"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-2017-47 | longest | en | 0.920353 |
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# FP1 Complex numbers watch
1. Complete losing the plot.
2x^3 + ax^2 + bx - 10
has root (3 + i)
other root is therefore (3 - i)
or (x^2 - 6x + 10) together. Find the other root. Missing how I can find the real root and it blatantly obvious. Help?!
2. product of roots is .......
3. 2x^3 + ax^2 + bx - 10
= (x - 3+i)(x - 3-i)(px + q)
4. dude,
remember root1+root2+root3=-d/a?
6+rt3=-10/2=-5
rt3=-11
EDIT:
got it wrong its -1(i hope)
5. (Original post by mathz)
product of roots is .......
Errrrrr, I'm confused as to your question, the product of the complex conjugate roots is 10... how does that help me find the real root? Am I being dense?
6. (Original post by england_sucks)
dude,
remember root1+root2+root3=-d/a?
6+rt3=-10/2=-5
rt3=-11
EDIT:
got it wrong its -1(i hope)
not quite. -d/a is product of roots.
7. (Original post by AnonyMatt)
2x^3 + ax^2 + bx - 10
= (x - 3+i)(x - 3-i)(px + q)
Jaja, I get that but the cubic has two unknown variables so you can't use polynomial division, or well you can but you then have two unknowns, what's the method?
8. (Original post by Clarity Incognito)
Errrrrr, I'm confused as to your question, the product of the complex conjugate roots is 10... how does that help me find the real root? Am I being dense?
if i have 2 roots and i know their product, and also know the product of all 3 roots then i can work out the third root.
9. AAAAAAAAAAAAAAAAAAAAAAAAAH I've got it! Thank you all very much!!!!! I was being completely dense
10. (Original post by mathz)
if i have 2 roots and i know their product, and also know the product of all 3 roots then i can work out the third root.
Thanks, that oversight of mine was laughable.
11. (Original post by mathz)
not quite. -d/a is product of roots.
Grrr.
i'm a little rusty here and there. the sum is -b/a yeah?
just remembered...
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https://newpathworksheets.com/math/grade-2/odd-and-even-2/idaho-standards | 1,620,866,901,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991413.30/warc/CC-MAIN-20210512224016-20210513014016-00402.warc.gz | 472,260,281 | 7,535 | ## ◂Math Worksheets and Study Guides Second Grade. Odd and Even
### The resources above correspond to the standards listed below:
#### Idaho Content Standards
2.OA. Operations and Algebraic Thinking – 2.OA
Work with equal groups of objects to gain foundations for multiplication.
2.OA.3. Determine whether a group of objects (up to 20) has an odd or even number of members, e.g., by pairing objects or counting them by 2s; write an equation to express an even number as a sum of two equal addends.
Standards | 119 | 510 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2021-21 | latest | en | 0.879894 |
http://www.mathcs.emory.edu/~cheung/Courses/554/public/hw3/hw3.html | 1,521,766,472,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257648113.87/warc/CC-MAIN-20180323004957-20180323024957-00746.warc.gz | 418,572,626 | 3,022 | ### CS544, Homework 3
• Question 1 (40 pts)
• Exercise 13.3.1:
• Suppose we are scheduling I/O requests for a disk, and the requests arrives as follows:
``` Time of arrival Request for track ----------------------------------------------- 0 msec 8,000 1 msec 48,000 10 msec 4,000 20 msec 40,000 ```
The disk head initially at track 32,000.
The time it takes the disk head to move n tracks is 1 + 0.00025n msec.
The (average) latency and transfer time is total 4.3 msec.
Questions:
• At what time is each request serviced fully if we use the elevator algorithm (it is permissible to start moving in either direction at first).
Indicate the order in which the requests will be satisfied !!!
Answer must be in this form (with dervations !)
Correct solution: (seek is not interrupted by an arrival)
``` Request (for track#) Time of completion ------------------------------------------------------------------------ At time 0: serve track 8000 #1 track 8000 0 + (1+0.00025*24000) + 4.3 = 11.3 msec At time 11.3: 4000 has arrived, serve track 4000 next #2 track 4000 11.3 + (1+0.00025*4000) + 4.3 = 17.6 msec At time 17.6: only 48000 has arrived, serve track 48000 next #3 track 48000 17.6 + (1+0.00025*44000) + 4.3 = 33.9 msec At time 33.9: 40000 has arrived, serve track 40000 next #4 track 40000 33.9 + (1+0.00025*8000) + 4.3 = 41.2 msec ```
Acceptable solution: (arrival will interrupt the seek operation - harder to implement)
``` Request (for track#) Time of completion ------------------------------------------------------------------------ At time 0: serve track 8000 #1 track 8000 0 + (1+0.00025*24000) + 4.3 = 11.3 msec At time 11.3: 4000 has arrived, serve track 4000 before 48000 because it's "on the same way" (elevator is going down now) #2 track 4000 11.3 + (1+0.00025*4000) + 4.3 = 17.6 msec At time 17.6: only 48000 has arrived, go serve track 48000 next ** We pass track 40000 at time: 17.6 + (1+0.00025*36000) = 27.6 msec < 20.0 msec ** STOP at track 40000 on the way to 48000 - serve 40000 first !!! #3 track 40000 17.6 + (1+0.00025*36000) + 4.3 = 31.9 msec At time 31.9: go serve track 48000 #4 track 48000 31.9 + (1+0.00025*8000) + 4.3 = 39.2 msec ```
• At what time is each request serviced fully if we use the first come, first serve service discipline
Indicate the order in which the requests will be satisfied !!!
Answer must be in this form (with dervations !)
``` Request (for track#) Time of completion ------------------------------------------------------------------- #1 track 8000 0 + (1+0.00025*24000) + 4.3 = 11.3 msec #2 track 48000 11.3 + (1+0.00025*40000) + 4.3 = 26.6 msec #3 track 4000 26.6 + (1+0.00025*44000) + 4.3 = 42.9 msec #4 track 40000 42.9 + (1+0.00025*36000) + 4.3 = 57.2 msec ```
• Question 2 (Exercise 13.4.6-8) (30 pts)
• Suppose we are using a RAID level 4 scheme (using even parity) with four data disks and one redundant disk.
Aassume that blocks are a single byte.
Questions:
• Give the block of the redundant disk if the corresponding blocks of the data disks are:
1. 01010110, 11000000, 00111011, and 11111011.
``` 01010110 11000000 00111011 11111011 ------------- 01010110 ```
2. 11110000, 11111000, 00111111, and 00000001.
``` 11110000 11111000 00111111 00000001 -------------- 00110110 ```
• Suppose that data disk 1 has failed.
Recover the block of that disk under the following circumstances:
• The contents of disks 2 through 4 are 01010110, 11000000, and 00111011, while the redundant disk holds 11111011.
``` 01010110 11000000 00111011 11111011 -------------- 01010110 ```
• The contents of disks 2 through 4 are 11110000, 11111000, and 00111111, while the redundant disk holds 00000001.
``` 11110000 11111000 00111111 00000001 ------------- 00110110 ```
• Suppose the block on the first disk in part (1) is changed to 10101010.
What changes to the corresponding blocks on the other disks must be made?
1. Originally: 01010110, 11000000, 00111011, and 11111011.
Changed to: 10101010, 11000000, 00111011, and 11111011.
``` You also need to update the redundant disk to: 10101010 11000000 00111011 11111011 -------------- 10101010 ```
2. Originally: 11110000, 11111000, 00111111, and 00000001.
Changed to: 10101010, 11111000, 00111111, and 00000001.
``` You also need to update the redundant disk to: 10101010 11111000 00111111 00000001 -------------- 01101100 ```
• Question 3 (Exercise 13.7.1-3) (30 pts)
• A patient record consists of the following:
• 3 fixed-length fields: the patient's date of birth, social-security number, and patient ID, each field is 10 bytes long.
• It also has the following 3 variable-length fields: (1) name, (2) address, and (3) patient history.
• If pointers within a record require 4 bytes, and the record length is a 4-byte integer, how many bytes, exclusive of the space needed for the variable length fields, are needed for the record ?
You may assume that no alignment of fields is required.
``` We get 10+10+10 = 30 for the fixed-length fields, 4+4 = 8 for the pointers to varying-length Fields, and 4 for the record length. The total is 42 bytes. ```
• A patient record consists of the following:
• 3 fixed-length fields: the patient's date of birth, social-security number, and patient ID, each field is 10 bytes long.
• It also has the following 3 variable-length fields: (1) name, (2) address, and (3) patient history.
• A repeating field that represents cholesterol tests --- Each cholesterol test requires 16 bytes
• Show the layout of patient records if the repeating tests are kept with the record itself. | 1,715 | 5,575 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2018-13 | latest | en | 0.810649 |
http://www.squaring.net/sq/sr/spsr/spsr.html | 1,498,602,835,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128321938.75/warc/CC-MAIN-20170627221726-20170628001726-00183.warc.gz | 658,344,612 | 3,079 | # Catalogues of Simple Perfect Squared Rectangles (SPSR)
## Orders 9 to 17
Squared squares and squared rectangles are called simple if they do not contain a smaller squared square or rectangle. Simple perfect squared rectangles begin at order 9.
Squared squares and squared rectangles are called perfect if the squares in the tiling are all of different sizes.
Simple perfect squared rectangles have been have been catalogued from order 9 to 17 in javascript menus. Initially arranged by order and then by width and height, but further selected catalogues arranged by other characteristics are planned.
The main method of classifying Squared Rectangles is to organise them with a given number of elements, by order, that is, the number of constituent squares, then by width then by height.
Catalogues have been produced of simple perfect squared rectangles(SPSR) in the search for squared squares.
The On-Line Encyclopedia of Integer Sequences lists sequence A219766; the Number of nonsquare simple perfect squared rectangles of order n up to symmetry as; 0, 0, 0, 0, 0, 0, 0, 0, 2, 6, 22, 67, 213, 744, 2 609, 9 016, 31 426, 110 381, 390 223, 1 383 905, 4 931 307, 17 633 769 , 63 301 415, 228 130 900. In Ian Gambini's thesis he provided SPSR counts up to order 24, his counts for order 23 and 24 are confirmed despite being shown with question marks (indicating some uncertainty) in his thesis. Gambini's counts and Anderson's counts agree to order 24. It is not practical to display catalogues above order 17 due to the sheer number of tilings involved, however links to zipped bouwkampcodes (and tablecodes) up to order 21 are provided for connoisseurs of simple perfect squared rectangles.
A square is a rectangle, so the correct count for Simple Perfect Squared Rectangles includes Perfect Squared Squares from order 21 onwards. In the OEIS, this is listed as sequence A002839 ; Number of simple perfect squared rectangles of order n up to symmetry. ; 0, 0, 0, 0, 0, 0, 0, 0, 2, 6, 22, 67, 213, 744, 2609, 9016, 31426, 110381, 390223, 1383905, 4931308, 17633773, 63301427, 228130926. | 569 | 2,098 | {"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-2017-26 | longest | en | 0.895197 |
https://www.spiritualunite.com/articles/number-14-meaning-balance-freedom-with-control/ | 1,708,604,709,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473738.92/warc/CC-MAIN-20240222093910-20240222123910-00368.warc.gz | 1,027,551,519 | 72,993 | Through synchronicity, the universe often shows us clues, messages and instructions encoded within numbers.
The number 14 is one of the more common synchronous numbers, so we know more about its meaning.
We decode these messages using numerology, which can help us to understand more about ourselves and our path through this life.
So if you have experienced number 14 synchronicity, or it is significant in your numerological life chart, here is what you should know about its meaning:
## The Essence Of 1, 4 & 5
When we look at any number with more than one digit, we need to simplify the numbers to find their essence.
We do this by summing the digits. For the number 14, this is:
1 + 4 = 5
Therefore, the essence of the number 14 in numerology is a combination of the numbers 1, 4 and 5.
The number 5, which is the base essence of 14, is the most important part.
The essence of 1 and 4 play a smaller but still significant role in the meaning of the number 14.
So let’s look at what we have:
The number 5 represents personal freedom, independence, self-improvement, courage and other individualist ideals.
It strikes out on its own, treads the unbeaten path and hates to rely on anyone else.
The number 1 is all about beginnings, starting fresh, making decisions and generally getting going.
It makes sense that number 1 is all about origins as it is the first tentative step that every journey must begin with.
The number 4 represents the act of creation and creativity, building and maintaining, and the productive aspects of life. It revels in hard work and worthwhile endeavours.
By combining the essence of these three numbers, we can interpret the number 14 depending on the situation in which it arises.
Every situation is different, and context is everything – so we can’t tell you what your 14 means.
What we can do is show you how this might apply to a specific situation, so that you can apply it in the same way to your own.
You have been feeling stuck at work lately. You are overworked, underpaid and feeling under-appreciated as well.
You start to wonder if you should look for a new job. As you do, you will begin to notice number 14 synchronicity. 14 is showing up everywhere.
The essence of 5 is striking out on your own, gaining independence and taking the unconventional route.
The essence of 1 is to start on a new journey, to get a fresh start — the essence of 4 places emphasis on building and creation.
The number 14, in this situation, maybe suggesting that you start your own business, go freelance, or otherwise take control of your working life.
Taking control will be good for you spiritually, allowing some of your power to return to you where before it was in someone else’s hands.
Remember, if your interpretation feels right then it probably is right. Intuition is your number one tool when communicating with the universe.
But when the number 14 shows up over and over again, it is indeed time for you to take back control and start building something new that you can be proud of growing into something amazing. | 671 | 3,077 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2024-10 | latest | en | 0.943209 |
http://www.cakecentral.com/forum/t/675644/amount-of-ganache | 1,500,785,878,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549424247.30/warc/CC-MAIN-20170723042657-20170723062657-00117.warc.gz | 385,603,480 | 16,836 | ## Amount Of Ganache?
By lecrn Updated 14 Apr 2010 , 8:38pm by sillywabbitz
lecrn Posted 8 Apr 2010 , 8:55pm
post #1 of 11
Sorry if this has been asked before, but I can't find any info on the subject. For those of you that use ganache under fondant, how do you know how much to make? Do you go by the amount of buttercream that it would take to crumb coat (like in cups)? How many cups of ganache will 2pounds choc: 1 pound cream make?
I was planning on practicing on a one tier cake to see how easy it is to apply and smooth ganache before I try a tiered cake.
10 replies
Roxybc Posted 9 Apr 2010 , 12:06am
post #2 of 11
I'd also like to know this, as I think ganache might be easier to use on the base of the giant cupcake than buttercream will given all the "edges" of the base.
Rylan Posted 9 Apr 2010 , 9:36pm
post #3 of 11
It depends on how thick or thin you want to apply it. With my experience, 5 pounds of chocolate and 1 pound of cream can cover a 10", 8", and two 6" rounds plus a few left. Sometimes, it can go even more.
lecrn Posted 10 Apr 2010 , 1:04pm
post #4 of 11
Quote:
Originally Posted by Rylan
It depends on how thick or thin you want to apply it. With my experience, 5 pounds of chocolate and 1 pound of cream can cover a 10", 8", and two 6" rounds plus a few left. Sometimes, it can go even more.
I have a few more questions if you don't mind.
I know that the ratio is 2 parts choc to 1 part cream. Do you weigh the cream on a scale instead of by volume in a measuring cup?
So, 16oz of choc. : 8oz of weighed cream?
In you blog, you state that you refrigerate the ganache after it's made. About how long does it take to come to the right consistency?
You also put the cake in the fridge after the ganache is applied (before the fondant)? About how long does it take in the fridge before the fondant goes on? Have you ever had any problem with condensation when you apply the fondant to a cold cake?
Sorry for all the idiot questions, but I just re-watched a sugarshack dvd where she is putting ganache on a cake, and there is a lot of letting the ganache sit on the counter. The ganache is made, bowl sits on the counter for 12hrs until correct consistency. The cake is ganached, sits on the counter for 12hrs until hardened. It seems like it would be more time effective if the fridge could speed things along if that wouldn't mess up the ganache or cause condensation to the finished fondant cake.
Thanks for you time!
sillywabbitz Posted 10 Apr 2010 , 1:38pm
post #5 of 11
I've used setting ganache under Fondant which is the 2 to 1 choc to cream ratio. You weigh the choc and use fluid ounces for the liquid. So 16 oz of choc to 8 oz or 1 cup of cream. That recipe made approximately 3 cups of ganache for me. I used it for filling and coating and it seemed to follow the number of cups of buttercream required to ice a cake recommened on the Wilton website. I would make a litlle more ganache than you think you need just to give you some breathing room. I loved working with ganache under fondant.
I did sugar shacks approach of letting sit out. I see no problem with chilling the warm ganache. I don't know about putting the covered cake in the fridge just because I didn't do it.
Oh but here is an important trick, do not store all your ganache in one big bowl. Put it into smaller containers. This makes it easier to reheat and if one batch is too warm you can add a small amount of the hardened ganache to bring it to the right temp. I hope this all makes sense good luck.
lecrn Posted 10 Apr 2010 , 4:25pm
post #6 of 11
Thanks so much. That's so helpful! I just want to be careful since it will be my first time, & I don't want to waste a bunch of choc. & cream b/c it's so expensive.
ceshell Posted 11 Apr 2010 , 1:02am
post #7 of 11
I have had similar experience to what's been posted above, although I must say that I find that my volume of final product is about 75% less than the weight. Meaning, if I use 8oz chocolate and 8oz cream, the resultant 16oz of ganache does NOT yield me 2 cups. The weight and volume are different; I get more like 1.5c ganache.
Of course you can fatten this amount right back up by whipping some air into it before applying to the cake.
I see you all mention the 2:1 ratio - that is for a firmer ganache. Keep in mind you can do it at 1:1 and still yield a very firm ganache, at least if you use 70% chocolate.
Rylan Posted 12 Apr 2010 , 10:10am
post #8 of 11
Quote:
Originally Posted by lecrn
I know that the ratio is 2 parts choc to 1 part cream. Do you weigh the cream on a scale instead of by volume in a measuring cup?
So, 16oz of choc. : 8oz of weighed cream?
Yes, I always measure by weight on a scale. So, 2 pounds of chocolate : 1 pound cream.
Quote:
Originally Posted by lecrn
In you blog, you state that you refrigerate the ganache after it's made. About how long does it take to come to the right consistency?
I refrigerate it after it has cooled a bit. I let it stay in the fridge usually about 2 hours or so (don't cover it). If it does get too hard, I just microwave it in 10 second intervals, mixing it everytime it comes out.
Quote:
Originally Posted by lecrn
You also put the cake in the fridge after the ganache is applied (before the fondant)? About how long does it take in the fridge before the fondant goes on? Have you ever had any problem with condensation when you apply the fondant to a cold cake?
Yes I do. I just refrigerate the ganache covered cake until it becomes solid like a chocolate bar. That way, it is easier for me to achieve sharp edges (just like covering a dummy). It would usually set up hard in 10 minutes or so. Once the cake has come to room temperature, it will get softer. I've never had problems with condensation--maybe because I live in a desert.
I hope that clears everything.
Bel_Anne Posted 12 Apr 2010 , 11:03am
post #9 of 11
I've learnt soo much about ganache this year. And it caused me so much stress learning it! Haha. While you can absolutely pop your ganache in the fridge... for taste/texture purposes it really is best to leave it to set overnight at room temp. For some chemistry reason (haha), it makes the ganache melt in your mouth like a truffle when eaten, yet still set firm on a cake. I was told putting it in the fridge can seperate fat particles and ruin the texture of the chocolate.... or something (that sounds scientific, eh!)
Also.. in many, many books it says to 'scald' the cream and pour over chocolate then stir to desired consistency. I have had MANY failed attempts at this, where it gets an oily residue on the top and becomes grainy to the point it's virtually unusable (expensive experiences). Until I read this article... http://acselementsofchocolate.typepad.com/elements_of_chocolate/Chocolate.html about half way down the page titled "ganache". Heat the cream then add SHAVED chocolate to the pan the cream is in - never had a problem since. Chocolate is supposedly one of the hardest ingredients to use in the kitchen and it's better to understand it on a chemical level (not that I completely do yet, haha).
Another tip - if you're using cheaper chocolate that has a lower cocoa mass percentage (so not couverture) then increase the ratio of chocolate to cream. It just will not set hard at a 2:1 ratio with compound chocolate... If you're just experimenting with cheap stuff, I'd go 3:1
And don't rush the 'setting' process by popping it in the fridge... It WILL set on the bench, and that's the best way to do it.
Bunsen Posted 12 Apr 2010 , 11:04am
post #10 of 11
Quote:
Originally Posted by Rylan
I've never had problems with condensation--maybe because I live in a desert.
The desert would definitely help! I tried this on a humid day and had so much trouble with condensation - made it impossible to get the fondant on neatly and then lots of huge air bubbles formed, never again!!
sillywabbitz Posted 14 Apr 2010 , 8:38pm
post #11 of 11
Oh and I used bakers chocolate for one batch and choc chips for another. The chips make a very weak chocolate...so I would use baker's chocolate if you can. If not you will need a lot more chips than the normal amount. | 2,108 | 8,169 | {"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-2017-30 | latest | en | 0.955359 |
https://en.wikipedia.org/wiki/Minimum_mean_square_error | 1,726,384,141,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651616.56/warc/CC-MAIN-20240915052902-20240915082902-00019.warc.gz | 204,515,636 | 60,624 | # Minimum mean square error
In statistics and signal processing, a minimum mean square error (MMSE) estimator is an estimation method which minimizes the mean square error (MSE), which is a common measure of estimator quality, of the fitted values of a dependent variable. In the Bayesian setting, the term MMSE more specifically refers to estimation with quadratic loss function. In such case, the MMSE estimator is given by the posterior mean of the parameter to be estimated. Since the posterior mean is cumbersome to calculate, the form of the MMSE estimator is usually constrained to be within a certain class of functions. Linear MMSE estimators are a popular choice since they are easy to use, easy to calculate, and very versatile. It has given rise to many popular estimators such as the Wiener–Kolmogorov filter and Kalman filter.
## Motivation
The term MMSE more specifically refers to estimation in a Bayesian setting with quadratic cost function. The basic idea behind the Bayesian approach to estimation stems from practical situations where we often have some prior information about the parameter to be estimated. For instance, we may have prior information about the range that the parameter can assume; or we may have an old estimate of the parameter that we want to modify when a new observation is made available; or the statistics of an actual random signal such as speech. This is in contrast to the non-Bayesian approach like minimum-variance unbiased estimator (MVUE) where absolutely nothing is assumed to be known about the parameter in advance and which does not account for such situations. In the Bayesian approach, such prior information is captured by the prior probability density function of the parameters; and based directly on Bayes theorem, it allows us to make better posterior estimates as more observations become available. Thus unlike non-Bayesian approach where parameters of interest are assumed to be deterministic, but unknown constants, the Bayesian estimator seeks to estimate a parameter that is itself a random variable. Furthermore, Bayesian estimation can also deal with situations where the sequence of observations are not necessarily independent. Thus Bayesian estimation provides yet another alternative to the MVUE. This is useful when the MVUE does not exist or cannot be found.
## Definition
Let ${\displaystyle x}$ be a ${\displaystyle n\times 1}$ hidden random vector variable, and let ${\displaystyle y}$ be a ${\displaystyle m\times 1}$ known random vector variable (the measurement or observation), both of them not necessarily of the same dimension. An estimator ${\displaystyle {\hat {x}}(y)}$ of ${\displaystyle x}$ is any function of the measurement ${\displaystyle y}$. The estimation error vector is given by ${\displaystyle e={\hat {x}}-x}$ and its mean squared error (MSE) is given by the trace of error covariance matrix
${\displaystyle \operatorname {MSE} =\operatorname {tr} \left\{\operatorname {E} \{({\hat {x}}-x)({\hat {x}}-x)^{T}\}\right\}=\operatorname {E} \{({\hat {x}}-x)^{T}({\hat {x}}-x)\},}$
where the expectation ${\displaystyle \operatorname {E} }$ is taken over ${\displaystyle x}$ conditioned on ${\displaystyle y}$. When ${\displaystyle x}$ is a scalar variable, the MSE expression simplifies to ${\displaystyle \operatorname {E} \left\{({\hat {x}}-x)^{2}\right\}}$. Note that MSE can equivalently be defined in other ways, since
${\displaystyle \operatorname {tr} \left\{\operatorname {E} \{ee^{T}\}\right\}=\operatorname {E} \left\{\operatorname {tr} \{ee^{T}\}\right\}=\operatorname {E} \{e^{T}e\}=\sum _{i=1}^{n}\operatorname {E} \{e_{i}^{2}\}.}$
The MMSE estimator is then defined as the estimator achieving minimal MSE:
${\displaystyle {\hat {x}}_{\operatorname {MMSE} }(y)=\operatorname {argmin} _{\hat {x}}\operatorname {MSE} .}$
## Properties
• When the means and variances are finite, the MMSE estimator is uniquely defined[1] and is given by:
${\displaystyle {\hat {x}}_{\operatorname {MMSE} }(y)=\operatorname {E} \{x\mid y\}.}$
In other words, the MMSE estimator is the conditional expectation of ${\displaystyle x}$ given the known observed value of the measurements. Also, since ${\displaystyle {\hat {x}}_{\mathrm {MMSE} }}$ is the posterior mean, the error covariance matrix ${\displaystyle C_{e}=\operatorname {E} \{({\hat {x}}-x)({\hat {x}}-x)^{T}\}}$ is equal to the posterior covariance ${\displaystyle C_{X|Y}}$ matrix,
${\displaystyle C_{e}=C_{X|Y}}$.
• The MMSE estimator is unbiased (under the regularity assumptions mentioned above):
${\displaystyle \operatorname {E} \{{\hat {x}}_{\operatorname {MMSE} }(y)\}=\operatorname {E} \{\operatorname {E} \{x\mid y\}\}=\operatorname {E} \{x\}.}$
${\displaystyle {\sqrt {n}}({\hat {x}}_{\operatorname {MMSE} }-x)\xrightarrow {d} {\mathcal {N}}\left(0,I^{-1}(x)\right),}$
where ${\displaystyle I(x)}$ is the Fisher information of ${\displaystyle x}$. Thus, the MMSE estimator is asymptotically efficient.
• The orthogonality principle: When ${\displaystyle x}$ is a scalar, an estimator constrained to be of certain form ${\displaystyle {\hat {x}}=g(y)}$ is an optimal estimator, i.e. ${\displaystyle {\hat {x}}_{\operatorname {MMSE} }=g^{*}(y),}$ if and only if
${\displaystyle \operatorname {E} \{({\hat {x}}_{\operatorname {MMSE} }-x)g(y)\}=0}$
for all ${\displaystyle g(y)}$ in closed, linear subspace ${\displaystyle {\mathcal {V}}=\{g(y)\mid g:\mathbb {R} ^{m}\rightarrow \mathbb {R} ,\operatorname {E} \{g(y)^{2}\}<+\infty \}}$ of the measurements. For random vectors, since the MSE for estimation of a random vector is the sum of the MSEs of the coordinates, finding the MMSE estimator of a random vector decomposes into finding the MMSE estimators of the coordinates of X separately:
${\displaystyle \operatorname {E} \{(g_{i}^{*}(y)-x_{i})g_{j}(y)\}=0,}$
for all i and j. More succinctly put, the cross-correlation between the minimum estimation error ${\displaystyle {\hat {x}}_{\operatorname {MMSE} }-x}$ and the estimator ${\displaystyle {\hat {x}}}$ should be zero,
${\displaystyle \operatorname {E} \{({\hat {x}}_{\operatorname {MMSE} }-x){\hat {x}}^{T}\}=0.}$
• If ${\displaystyle x}$ and ${\displaystyle y}$ are jointly Gaussian, then the MMSE estimator is linear, i.e., it has the form ${\displaystyle Wy+b}$ for matrix ${\displaystyle W}$ and constant ${\displaystyle b}$. This can be directly shown using the Bayes theorem. As a consequence, to find the MMSE estimator, it is sufficient to find the linear MMSE estimator.
## Linear MMSE estimator
In many cases, it is not possible to determine the analytical expression of the MMSE estimator. Two basic numerical approaches to obtain the MMSE estimate depends on either finding the conditional expectation ${\displaystyle \operatorname {E} \{x\mid y\}}$ or finding the minima of MSE. Direct numerical evaluation of the conditional expectation is computationally expensive since it often requires multidimensional integration usually done via Monte Carlo methods. Another computational approach is to directly seek the minima of the MSE using techniques such as the stochastic gradient descent methods; but this method still requires the evaluation of expectation. While these numerical methods have been fruitful, a closed form expression for the MMSE estimator is nevertheless possible if we are willing to make some compromises.
One possibility is to abandon the full optimality requirements and seek a technique minimizing the MSE within a particular class of estimators, such as the class of linear estimators. Thus, we postulate that the conditional expectation of ${\displaystyle x}$ given ${\displaystyle y}$ is a simple linear function of ${\displaystyle y}$, ${\displaystyle \operatorname {E} \{x\mid y\}=Wy+b}$, where the measurement ${\displaystyle y}$ is a random vector, ${\displaystyle W}$ is a matrix and ${\displaystyle b}$ is a vector. This can be seen as the first order Taylor approximation of ${\displaystyle \operatorname {E} \{x\mid y\}}$. The linear MMSE estimator is the estimator achieving minimum MSE among all estimators of such form. That is, it solves the following optimization problem:
${\displaystyle \min _{W,b}\operatorname {MSE} \qquad {\text{s.t.}}\qquad {\hat {x}}=Wy+b.}$
One advantage of such linear MMSE estimator is that it is not necessary to explicitly calculate the posterior probability density function of ${\displaystyle x}$. Such linear estimator only depends on the first two moments of ${\displaystyle x}$ and ${\displaystyle y}$. So although it may be convenient to assume that ${\displaystyle x}$ and ${\displaystyle y}$ are jointly Gaussian, it is not necessary to make this assumption, so long as the assumed distribution has well defined first and second moments. The form of the linear estimator does not depend on the type of the assumed underlying distribution.
The expression for optimal ${\displaystyle b}$ and ${\displaystyle W}$ is given by:
${\displaystyle b={\bar {x}}-W{\bar {y}},}$
${\displaystyle W=C_{XY}C_{Y}^{-1}.}$
where ${\displaystyle {\bar {x}}=\operatorname {E} \{x\}}$, ${\displaystyle {\bar {y}}=\operatorname {E} \{y\},}$ the ${\displaystyle C_{XY}}$ is cross-covariance matrix between ${\displaystyle x}$ and ${\displaystyle y}$, the ${\displaystyle C_{Y}}$ is auto-covariance matrix of ${\displaystyle y}$.
Thus, the expression for linear MMSE estimator, its mean, and its auto-covariance is given by
${\displaystyle {\hat {x}}=C_{XY}C_{Y}^{-1}(y-{\bar {y}})+{\bar {x}},}$
${\displaystyle \operatorname {E} \{{\hat {x}}\}={\bar {x}},}$
${\displaystyle C_{\hat {X}}=C_{XY}C_{Y}^{-1}C_{YX},}$
where the ${\displaystyle C_{YX}}$ is cross-covariance matrix between ${\displaystyle y}$ and ${\displaystyle x}$.
Lastly, the error covariance and minimum mean square error achievable by such estimator is
${\displaystyle C_{e}=C_{X}-C_{\hat {X}}=C_{X}-C_{XY}C_{Y}^{-1}C_{YX},}$
${\displaystyle \operatorname {LMMSE} =\operatorname {tr} \{C_{e}\}.}$
Derivation using orthogonality principle
Let us have the optimal linear MMSE estimator given as ${\displaystyle {\hat {x}}=Wy+b}$, where we are required to find the expression for ${\displaystyle W}$ and ${\displaystyle b}$. It is required that the MMSE estimator be unbiased. This means,
${\displaystyle \operatorname {E} \{{\hat {x}}\}=\operatorname {E} \{x\}.}$
Plugging the expression for ${\displaystyle {\hat {x}}}$ in above, we get
${\displaystyle b={\bar {x}}-W{\bar {y}},}$
where ${\displaystyle {\bar {x}}=\operatorname {E} \{x\}}$ and ${\displaystyle {\bar {y}}=\operatorname {E} \{y\}}$. Thus we can re-write the estimator as
${\displaystyle {\hat {x}}=W(y-{\bar {y}})+{\bar {x}}}$
and the expression for estimation error becomes
${\displaystyle {\hat {x}}-x=W(y-{\bar {y}})-(x-{\bar {x}}).}$
From the orthogonality principle, we can have ${\displaystyle \operatorname {E} \{({\hat {x}}-x)(y-{\bar {y}})^{T}\}=0}$, where we take ${\displaystyle g(y)=y-{\bar {y}}}$. Here the left-hand-side term is
{\displaystyle {\begin{aligned}\operatorname {E} \{({\hat {x}}-x)(y-{\bar {y}})^{T}\}&=\operatorname {E} \{(W(y-{\bar {y}})-(x-{\bar {x}}))(y-{\bar {y}})^{T}\}\\&=W\operatorname {E} \{(y-{\bar {y}})(y-{\bar {y}})^{T}\}-\operatorname {E} \{(x-{\bar {x}})(y-{\bar {y}})^{T}\}\\&=WC_{Y}-C_{XY}.\end{aligned}}}
When equated to zero, we obtain the desired expression for ${\displaystyle W}$ as
${\displaystyle W=C_{XY}C_{Y}^{-1}.}$
The ${\displaystyle C_{XY}}$ is cross-covariance matrix between X and Y, and ${\displaystyle C_{Y}}$ is auto-covariance matrix of Y. Since ${\displaystyle C_{XY}=C_{YX}^{T}}$, the expression can also be re-written in terms of ${\displaystyle C_{YX}}$ as
${\displaystyle W^{T}=C_{Y}^{-1}C_{YX}.}$
Thus the full expression for the linear MMSE estimator is
${\displaystyle {\hat {x}}=C_{XY}C_{Y}^{-1}(y-{\bar {y}})+{\bar {x}}.}$
Since the estimate ${\displaystyle {\hat {x}}}$ is itself a random variable with ${\displaystyle \operatorname {E} \{{\hat {x}}\}={\bar {x}}}$, we can also obtain its auto-covariance as
{\displaystyle {\begin{aligned}C_{\hat {X}}&=\operatorname {E} \{({\hat {x}}-{\bar {x}})({\hat {x}}-{\bar {x}})^{T}\}\\&=W\operatorname {E} \{(y-{\bar {y}})(y-{\bar {y}})^{T}\}W^{T}\\&=WC_{Y}W^{T}.\\\end{aligned}}}
Putting the expression for ${\displaystyle W}$ and ${\displaystyle W^{T}}$, we get
${\displaystyle C_{\hat {X}}=C_{XY}C_{Y}^{-1}C_{YX}.}$
Lastly, the covariance of linear MMSE estimation error will then be given by
{\displaystyle {\begin{aligned}C_{e}&=\operatorname {E} \{({\hat {x}}-x)({\hat {x}}-x)^{T}\}\\&=\operatorname {E} \{({\hat {x}}-x)(W(y-{\bar {y}})-(x-{\bar {x}}))^{T}\}\\&=\underbrace {\operatorname {E} \{({\hat {x}}-x)(y-{\bar {y}})^{T}\}} _{0}W^{T}-\operatorname {E} \{({\hat {x}}-x)(x-{\bar {x}})^{T}\}\\&=-\operatorname {E} \{(W(y-{\bar {y}})-(x-{\bar {x}}))(x-{\bar {x}})^{T}\}\\&=\operatorname {E} \{(x-{\bar {x}})(x-{\bar {x}})^{T}\}-W\operatorname {E} \{(y-{\bar {y}})(x-{\bar {x}})^{T}\}\\&=C_{X}-WC_{YX}.\end{aligned}}}
The first term in the third line is zero due to the orthogonality principle. Since ${\displaystyle W=C_{XY}C_{Y}^{-1}}$, we can re-write ${\displaystyle C_{e}}$ in terms of covariance matrices as
${\displaystyle C_{e}=C_{X}-C_{XY}C_{Y}^{-1}C_{YX}.}$
This we can recognize to be the same as ${\displaystyle C_{e}=C_{X}-C_{\hat {X}}.}$ Thus the minimum mean square error achievable by such a linear estimator is
${\displaystyle \operatorname {LMMSE} =\operatorname {tr} \{C_{e}\}}$.
### Univariate case
For the special case when both ${\displaystyle x}$ and ${\displaystyle y}$ are scalars, the above relations simplify to
${\displaystyle {\hat {x}}={\frac {\sigma _{XY}}{\sigma _{Y}^{2}}}(y-{\bar {y}})+{\bar {x}}=\rho {\frac {\sigma _{X}}{\sigma _{Y}}}(y-{\bar {y}})+{\bar {x}},}$
${\displaystyle \sigma _{e}^{2}=\sigma _{X}^{2}-{\frac {\sigma _{XY}^{2}}{\sigma _{Y}^{2}}}=(1-\rho ^{2})\sigma _{X}^{2},}$
where ${\displaystyle \rho ={\frac {\sigma _{XY}}{\sigma _{X}\sigma _{Y}}}}$ is the Pearson's correlation coefficient between ${\displaystyle x}$ and ${\displaystyle y}$.
The above two equations allows us to interpret the correlation coefficient either as normalized slope of linear regression
${\displaystyle \left({\frac {{\hat {x}}-{\bar {x}}}{\sigma _{X}}}\right)=\rho \left({\frac {y-{\bar {y}}}{\sigma _{Y}}}\right)}$
or as square root of the ratio of two variances
${\displaystyle \rho ^{2}={\frac {\sigma _{X}^{2}-\sigma _{e}^{2}}{\sigma _{X}^{2}}}={\frac {\sigma _{\hat {X}}^{2}}{\sigma _{X}^{2}}}}$.
When ${\displaystyle \rho =0}$, we have ${\displaystyle {\hat {x}}={\bar {x}}}$ and ${\displaystyle \sigma _{e}^{2}=\sigma _{X}^{2}}$. In this case, no new information is gleaned from the measurement which can decrease the uncertainty in ${\displaystyle x}$. On the other hand, when ${\displaystyle \rho =\pm 1}$, we have ${\displaystyle {\hat {x}}={\frac {\sigma _{XY}}{\sigma _{Y}}}(y-{\bar {y}})+{\bar {x}}}$ and ${\displaystyle \sigma _{e}^{2}=0}$. Here ${\displaystyle x}$ is completely determined by ${\displaystyle y}$, as given by the equation of straight line.
### Computation
Standard method like Gauss elimination can be used to solve the matrix equation for ${\displaystyle W}$. A more numerically stable method is provided by QR decomposition method. Since the matrix ${\displaystyle C_{Y}}$ is a symmetric positive definite matrix, ${\displaystyle W}$ can be solved twice as fast with the Cholesky decomposition, while for large sparse systems conjugate gradient method is more effective. Levinson recursion is a fast method when ${\displaystyle C_{Y}}$ is also a Toeplitz matrix. This can happen when ${\displaystyle y}$ is a wide sense stationary process. In such stationary cases, these estimators are also referred to as Wiener–Kolmogorov filters.
## Linear MMSE estimator for linear observation process
Let us further model the underlying process of observation as a linear process: ${\displaystyle y=Ax+z}$, where ${\displaystyle A}$ is a known matrix and ${\displaystyle z}$ is random noise vector with the mean ${\displaystyle \operatorname {E} \{z\}=0}$ and cross-covariance ${\displaystyle C_{XZ}=0}$. Here the required mean and the covariance matrices will be
${\displaystyle \operatorname {E} \{y\}=A{\bar {x}},}$
${\displaystyle C_{Y}=AC_{X}A^{T}+C_{Z},}$
${\displaystyle C_{XY}=C_{X}A^{T}.}$
Thus the expression for the linear MMSE estimator matrix ${\displaystyle W}$ further modifies to
${\displaystyle W=C_{X}A^{T}(AC_{X}A^{T}+C_{Z})^{-1}.}$
Putting everything into the expression for ${\displaystyle {\hat {x}}}$, we get
${\displaystyle {\hat {x}}=C_{X}A^{T}(AC_{X}A^{T}+C_{Z})^{-1}(y-A{\bar {x}})+{\bar {x}}.}$
Lastly, the error covariance is
${\displaystyle C_{e}=C_{X}-C_{\hat {X}}=C_{X}-C_{X}A^{T}(AC_{X}A^{T}+C_{Z})^{-1}AC_{X}.}$
The significant difference between the estimation problem treated above and those of least squares and Gauss–Markov estimate is that the number of observations m, (i.e. the dimension of ${\displaystyle y}$) need not be at least as large as the number of unknowns, n, (i.e. the dimension of ${\displaystyle x}$). The estimate for the linear observation process exists so long as the m-by-m matrix ${\displaystyle (AC_{X}A^{T}+C_{Z})^{-1}}$ exists; this is the case for any m if, for instance, ${\displaystyle C_{Z}}$ is positive definite. Physically the reason for this property is that since ${\displaystyle x}$ is now a random variable, it is possible to form a meaningful estimate (namely its mean) even with no measurements. Every new measurement simply provides additional information which may modify our original estimate. Another feature of this estimate is that for m < n, there need be no measurement error. Thus, we may have ${\displaystyle C_{Z}=0}$, because as long as ${\displaystyle AC_{X}A^{T}}$ is positive definite, the estimate still exists. Lastly, this technique can handle cases where the noise is correlated.
### Alternative form
An alternative form of expression can be obtained by using the matrix identity
${\displaystyle C_{X}A^{T}(AC_{X}A^{T}+C_{Z})^{-1}=(A^{T}C_{Z}^{-1}A+C_{X}^{-1})^{-1}A^{T}C_{Z}^{-1},}$
which can be established by post-multiplying by ${\displaystyle (AC_{X}A^{T}+C_{Z})}$ and pre-multiplying by ${\displaystyle (A^{T}C_{Z}^{-1}A+C_{X}^{-1}),}$ to obtain
${\displaystyle W=(A^{T}C_{Z}^{-1}A+C_{X}^{-1})^{-1}A^{T}C_{Z}^{-1},}$
and
${\displaystyle C_{e}=(A^{T}C_{Z}^{-1}A+C_{X}^{-1})^{-1}.}$
Since ${\displaystyle W}$ can now be written in terms of ${\displaystyle C_{e}}$ as ${\displaystyle W=C_{e}A^{T}C_{Z}^{-1}}$, we get a simplified expression for ${\displaystyle {\hat {x}}}$ as
${\displaystyle {\hat {x}}=C_{e}A^{T}C_{Z}^{-1}(y-A{\bar {x}})+{\bar {x}}.}$
In this form the above expression can be easily compared with ridge regression, weighed least square and Gauss–Markov estimate. In particular, when ${\displaystyle C_{X}^{-1}=0}$, corresponding to infinite variance of the apriori information concerning ${\displaystyle x}$, the result ${\displaystyle W=(A^{T}C_{Z}^{-1}A)^{-1}A^{T}C_{Z}^{-1}}$ is identical to the weighed linear least square estimate with ${\displaystyle C_{Z}^{-1}}$ as the weight matrix. Moreover, if the components of ${\displaystyle z}$ are uncorrelated and have equal variance such that ${\displaystyle C_{Z}=\sigma ^{2}I,}$ where ${\displaystyle I}$ is an identity matrix, then ${\displaystyle W=(A^{T}A)^{-1}A^{T}}$ is identical to the ordinary least square estimate. When apriori information is available as ${\displaystyle C_{X}^{-1}=\lambda I}$ and the ${\displaystyle z}$ are uncorrelated and have equal variance, we have ${\displaystyle W=(A^{T}A+\lambda I)^{-1}A^{T}}$, which is identical to ridge regression solution.
## Sequential linear MMSE estimation
In many real-time applications, observational data is not available in a single batch. Instead the observations are made in a sequence. One possible approach is to use the sequential observations to update an old estimate as additional data becomes available, leading to finer estimates. One crucial difference between batch estimation and sequential estimation is that sequential estimation requires an additional Markov assumption.
In the Bayesian framework, such recursive estimation is easily facilitated using Bayes' rule. Given ${\displaystyle k}$ observations, ${\displaystyle y_{1},\ldots ,y_{k}}$, Bayes' rule gives us the posterior density of ${\displaystyle x_{k}}$ as
{\displaystyle {\begin{aligned}p(x_{k}|y_{1},\ldots ,y_{k})&\propto p(y_{k}|x,y_{1},\ldots ,y_{k-1})p(x_{k}|y_{1},\ldots ,y_{k-1})\\&=p(y_{k}|x_{k})p(x_{k}|y_{1},\ldots ,y_{k-1}).\end{aligned}}}
The ${\displaystyle p(x_{k}|y_{1},\ldots ,y_{k})}$ is called the posterior density, ${\displaystyle p(y_{k}|x_{k})}$ is called the likelihood function, and ${\displaystyle p(x_{k}|y_{1},\ldots ,y_{k-1})}$ is the prior density of k-th time step. Here we have assumed the conditional independence of ${\displaystyle y_{k}}$ from previous observations ${\displaystyle y_{1},\ldots ,y_{k-1}}$ given ${\displaystyle x}$ as
${\displaystyle p(y_{k}|x_{k},y_{1},\ldots ,y_{k-1})=p(y_{k}|x_{k}).}$
This is the Markov assumption.
The MMSE estimate ${\displaystyle {\hat {x}}_{k}}$ given the k-th observation is then the mean of the posterior density ${\displaystyle p(x_{k}|y_{1},\ldots ,y_{k})}$. With the lack of dynamical information on how the state ${\displaystyle x}$ changes with time, we will make a further stationarity assumption about the prior:
${\displaystyle p(x_{k}|y_{1},\ldots ,y_{k-1})=p(x_{k-1}|y_{1},\ldots ,y_{k-1}).}$
Thus, the prior density for k-th time step is the posterior density of (k-1)-th time step. This structure allows us to formulate a recursive approach to estimation.
In the context of linear MMSE estimator, the formula for the estimate will have the same form as before: ${\displaystyle {\hat {x}}=C_{XY}C_{Y}^{-1}(y-{\bar {y}})+{\bar {x}}.}$ However, the mean and covariance matrices of ${\displaystyle X}$ and ${\displaystyle Y}$ will need to be replaced by those of the prior density ${\displaystyle p(x_{k}|y_{1},\ldots ,y_{k-1})}$ and likelihood ${\displaystyle p(y_{k}|x_{k})}$, respectively.
For the prior density ${\displaystyle p(x_{k}|y_{1},\ldots ,y_{k-1})}$, its mean is given by the previous MMSE estimate,
${\displaystyle {\bar {x}}_{k}=\mathrm {E} [x_{k}|y_{1},\ldots ,y_{k-1}]=\mathrm {E} [x_{k-1}|y_{1},\ldots ,y_{k-1}]={\hat {x}}_{k-1}}$,
and its covariance matrix is given by the previous error covariance matrix,
${\displaystyle C_{X_{k}|Y_{1},\ldots ,Y_{k-1}}=C_{X_{k-1}|Y_{1},\ldots ,Y_{k-1}}=C_{e_{k-1}},}$
as per by the properties of MMSE estimators and the stationarity assumption.
Similarly, for the linear observation process, the mean of the likelihood ${\displaystyle p(y_{k}|x_{k})}$ is given by ${\displaystyle {\bar {y}}_{k}=A{\bar {x}}_{k}=A{\hat {x}}_{k-1}}$ and the covariance matrix is as before
{\displaystyle {\begin{aligned}C_{Y_{k}|X_{k}}&=AC_{X_{k}|Y_{1},\ldots ,Y_{k-1}}A^{T}+C_{Z}=AC_{e_{k-1}}A^{T}+C_{Z}.\end{aligned}}}.
The difference between the predicted value of ${\displaystyle Y_{k}}$, as given by ${\displaystyle {\bar {y}}_{k}=A{\hat {x}}_{k-1}}$, and its observed value ${\displaystyle y_{k}}$ gives the prediction error ${\displaystyle {\tilde {y}}_{k}=y_{k}-{\bar {y}}_{k}}$, which is also referred to as innovation or residual. It is more convenient to represent the linear MMSE in terms of the prediction error, whose mean and covariance are ${\displaystyle \mathrm {E} [{\tilde {y}}_{k}]=0}$ and ${\displaystyle C_{{\tilde {Y}}_{k}}=C_{Y_{k}|X_{k}}}$.
Hence, in the estimate update formula, we should replace ${\displaystyle {\bar {x}}}$ and ${\displaystyle C_{X}}$ by ${\displaystyle {\hat {x}}_{k-1}}$ and ${\displaystyle C_{e_{k-1}}}$, respectively. Also, we should replace ${\displaystyle {\bar {y}}}$ and ${\displaystyle C_{Y}}$ by ${\displaystyle {\bar {y}}_{k-1}}$ and ${\displaystyle C_{{\tilde {Y}}_{k}}}$. Lastly, we replace ${\displaystyle C_{XY}}$ by
{\displaystyle {\begin{aligned}C_{X_{k}Y_{k}|Y_{1},\ldots ,Y_{k-1}}&=C_{e_{k-1}{\tilde {Y}}_{k}}=C_{e_{k-1}}A^{T}.\end{aligned}}}
Thus, we have the new estimate as new observation ${\displaystyle y_{k}}$ arrives as
{\displaystyle {\begin{aligned}{\hat {x}}_{k}&={\hat {x}}_{k-1}+C_{e_{k-1}{\tilde {Y}}_{k}}C_{{\tilde {Y}}_{k}}^{-1}(y_{k}-{\bar {y}}_{k})\\&={\hat {x}}_{k-1}+C_{e_{k-1}}A^{T}(AC_{e_{k-1}}A^{T}+C_{Z})^{-1}(y_{k}-A{\hat {x}}_{k-1})\end{aligned}}}
and the new error covariance as
${\displaystyle C_{e_{k}}=C_{e_{k-1}}-C_{e_{k-1}}A^{T}(AC_{e_{k-1}}A^{T}+C_{Z})^{-1}AC_{e_{k-1}}.}$
From the point of view of linear algebra, for sequential estimation, if we have an estimate ${\displaystyle {\hat {x}}_{1}}$ based on measurements generating space ${\displaystyle Y_{1}}$, then after receiving another set of measurements, we should subtract out from these measurements that part that could be anticipated from the result of the first measurements. In other words, the updating must be based on that part of the new data which is orthogonal to the old data.
The repeated use of the above two equations as more observations become available lead to recursive estimation techniques. The expressions can be more compactly written as
${\displaystyle W_{k}=C_{e_{k-1}}A^{T}(AC_{e_{k-1}}A^{T}+C_{Z})^{-1},}$
${\displaystyle {\hat {x}}_{k}={\hat {x}}_{k-1}+W_{k}(y_{k}-A{\hat {x}}_{k-1}),}$
${\displaystyle C_{e_{k}}=(I-W_{k}A)C_{e_{k-1}}.}$
The matrix ${\displaystyle W_{k}}$ is often referred to as the Kalman gain factor. The alternative formulation of the above algorithm will give
${\displaystyle C_{e_{k}}^{-1}=C_{e_{k-1}}^{-1}+A^{T}C_{Z}^{-1}A,}$
${\displaystyle W_{k}=C_{e_{k}}A^{T}C_{Z}^{-1},}$
${\displaystyle {\hat {x}}_{k}={\hat {x}}_{k-1}+W_{k}(y_{k}-A{\hat {x}}_{k-1}),}$
The repetition of these three steps as more data becomes available leads to an iterative estimation algorithm. The generalization of this idea to non-stationary cases gives rise to the Kalman filter. The three update steps outlined above indeed form the update step of the Kalman filter.
### Special case: scalar observations
As an important special case, an easy to use recursive expression can be derived when at each k-th time instant the underlying linear observation process yields a scalar such that ${\displaystyle y_{k}=a_{k}^{T}x_{k}+z_{k}}$, where ${\displaystyle a_{k}}$ is n-by-1 known column vector whose values can change with time, ${\displaystyle x_{k}}$ is n-by-1 random column vector to be estimated, and ${\displaystyle z_{k}}$ is scalar noise term with variance ${\displaystyle \sigma _{k}^{2}}$. After (k+1)-th observation, the direct use of above recursive equations give the expression for the estimate ${\displaystyle {\hat {x}}_{k+1}}$ as:
${\displaystyle {\hat {x}}_{k+1}={\hat {x}}_{k}+w_{k+1}(y_{k+1}-a_{k+1}^{T}{\hat {x}}_{k})}$
where ${\displaystyle y_{k+1}}$ is the new scalar observation and the gain factor ${\displaystyle w_{k+1}}$ is n-by-1 column vector given by
${\displaystyle w_{k+1}={\frac {C_{e_{k}}a_{k+1}}{\sigma _{k+1}^{2}+a_{k+1}^{T}C_{e_{k}}a_{k+1}}}.}$
The ${\displaystyle C_{e_{k+1}}}$ is n-by-n error covariance matrix given by
${\displaystyle C_{e_{k+1}}=(I-w_{k+1}a_{k+1}^{T})C_{e_{k}}.}$
Here, no matrix inversion is required. Also, the gain factor, ${\displaystyle w_{k+1}}$, depends on our confidence in the new data sample, as measured by the noise variance, versus that in the previous data. The initial values of ${\displaystyle {\hat {x}}}$ and ${\displaystyle C_{e}}$ are taken to be the mean and covariance of the aprior probability density function of ${\displaystyle x}$.
Alternative approaches: This important special case has also given rise to many other iterative methods (or adaptive filters), such as the least mean squares filter and recursive least squares filter, that directly solves the original MSE optimization problem using stochastic gradient descents. However, since the estimation error ${\displaystyle e}$ cannot be directly observed, these methods try to minimize the mean squared prediction error ${\displaystyle \mathrm {E} \{{\tilde {y}}^{T}{\tilde {y}}\}}$. For instance, in the case of scalar observations, we have the gradient ${\displaystyle \nabla _{\hat {x}}\mathrm {E} \{{\tilde {y}}^{2}\}=-2\mathrm {E} \{{\tilde {y}}a\}.}$ Thus, the update equation for the least mean square filter is given by
${\displaystyle {\hat {x}}_{k+1}={\hat {x}}_{k}+\eta _{k}\mathrm {E} \{{\tilde {y}}_{k}a_{k}\},}$
where ${\displaystyle \eta _{k}}$ is the scalar step size and the expectation is approximated by the instantaneous value ${\displaystyle \mathrm {E} \{a_{k}{\tilde {y}}_{k}\}\approx a_{k}{\tilde {y}}_{k}}$. As we can see, these methods bypass the need for covariance matrices.
### Special Case: vector observation with uncorrelated noise
In many practical applications, the observation noise is uncorrelated. That is, ${\displaystyle C_{Z}}$ is a diagonal matrix. In such cases, it is advantageous to consider the components of ${\displaystyle y}$ as independent scalar measurements, rather than vector measurement. This allows us to reduce computation time by processing the ${\displaystyle m\times 1}$ measurement vector as ${\displaystyle m}$ scalar measurements. The use of scalar update formula avoids matrix inversion in the implementation of the covariance update equations, thus improving the numerical robustness against roundoff errors. The update can be implemented iteratively as:
${\displaystyle w_{k+1}^{(\ell )}={\frac {C_{e_{k}}^{(\ell )}A_{k+1}^{(\ell )T}}{C_{Z_{k+1}}^{(\ell )}+A_{k+1}^{(\ell )}C_{e_{k}}^{(\ell )}(A_{k+1}^{(\ell )T})}}}$
${\displaystyle C_{e_{k+1}}^{(\ell )}=(I-w_{k+1}^{(\ell )}A_{k+1}^{(\ell )})C_{e_{k}}^{(\ell )}}$
${\displaystyle {\hat {x}}_{k+1}^{(\ell )}={\hat {x}}_{k}^{(\ell -1)}+w_{k+1}^{(\ell )}(y_{k+1}^{(\ell )}-A_{k+1}^{(\ell )}{\hat {x}}_{k}^{(\ell -1)})}$
where ${\displaystyle \ell =1,2,\ldots ,m}$, using the initial values ${\displaystyle C_{e_{k+1}}^{(0)}=C_{e_{k}}}$ and ${\displaystyle {\hat {x}}_{k+1}^{(0)}={\hat {x}}_{k}}$. The intermediate variables ${\displaystyle C_{Z_{k+1}}^{(\ell )}}$ is the ${\displaystyle \ell }$-th diagonal element of the ${\displaystyle m\times m}$ diagonal matrix ${\displaystyle C_{Z_{k+1}}}$; while ${\displaystyle A_{k+1}^{(\ell )}}$ is the ${\displaystyle \ell }$-th row of ${\displaystyle m\times n}$ matrix ${\displaystyle A_{k+1}}$. The final values are ${\displaystyle C_{e_{k+1}}^{(m)}=C_{e_{k+1}}}$ and ${\displaystyle {\hat {x}}_{k+1}^{(m)}={\hat {x}}_{k+1}}$.
## Examples
### Example 1
We shall take a linear prediction problem as an example. Let a linear combination of observed scalar random variables ${\displaystyle z_{1},z_{2}}$ and ${\displaystyle z_{3}}$ be used to estimate another future scalar random variable ${\displaystyle z_{4}}$ such that ${\displaystyle {\hat {z}}_{4}=\sum _{i=1}^{3}w_{i}z_{i}}$. If the random variables ${\displaystyle z=[z_{1},z_{2},z_{3},z_{4}]^{T}}$ are real Gaussian random variables with zero mean and its covariance matrix given by
${\displaystyle \operatorname {cov} (Z)=\operatorname {E} [zz^{T}]=\left[{\begin{array}{cccc}1&2&3&4\\2&5&8&9\\3&8&6&10\\4&9&10&15\end{array}}\right],}$
then our task is to find the coefficients ${\displaystyle w_{i}}$ such that it will yield an optimal linear estimate ${\displaystyle {\hat {z}}_{4}}$.
In terms of the terminology developed in the previous sections, for this problem we have the observation vector ${\displaystyle y=[z_{1},z_{2},z_{3}]^{T}}$, the estimator matrix ${\displaystyle W=[w_{1},w_{2},w_{3}]}$ as a row vector, and the estimated variable ${\displaystyle x=z_{4}}$ as a scalar quantity. The autocorrelation matrix ${\displaystyle C_{Y}}$ is defined as
${\displaystyle C_{Y}=\left[{\begin{array}{ccc}E[z_{1},z_{1}]&E[z_{2},z_{1}]&E[z_{3},z_{1}]\\E[z_{1},z_{2}]&E[z_{2},z_{2}]&E[z_{3},z_{2}]\\E[z_{1},z_{3}]&E[z_{2},z_{3}]&E[z_{3},z_{3}]\end{array}}\right]=\left[{\begin{array}{ccc}1&2&3\\2&5&8\\3&8&6\end{array}}\right].}$
The cross correlation matrix ${\displaystyle C_{YX}}$ is defined as
${\displaystyle C_{YX}=\left[{\begin{array}{c}E[z_{4},z_{1}]\\E[z_{4},z_{2}]\\E[z_{4},z_{3}]\end{array}}\right]=\left[{\begin{array}{c}4\\9\\10\end{array}}\right].}$
We now solve the equation ${\displaystyle C_{Y}W^{T}=C_{YX}}$ by inverting ${\displaystyle C_{Y}}$ and pre-multiplying to get
${\displaystyle C_{Y}^{-1}C_{YX}=\left[{\begin{array}{ccc}4.85&-1.71&-0.142\\-1.71&0.428&0.2857\\-0.142&0.2857&-0.1429\end{array}}\right]\left[{\begin{array}{c}4\\9\\10\end{array}}\right]=\left[{\begin{array}{c}2.57\\-0.142\\0.5714\end{array}}\right]=W^{T}.}$
So we have ${\displaystyle w_{1}=2.57,}$ ${\displaystyle w_{2}=-0.142,}$ and ${\displaystyle w_{3}=.5714}$ as the optimal coefficients for ${\displaystyle {\hat {z}}_{4}}$. Computing the minimum mean square error then gives ${\displaystyle \left\Vert e\right\Vert _{\min }^{2}=\operatorname {E} [z_{4}z_{4}]-WC_{YX}=15-WC_{YX}=.2857}$.[2] Note that it is not necessary to obtain an explicit matrix inverse of ${\displaystyle C_{Y}}$ to compute the value of ${\displaystyle W}$. The matrix equation can be solved by well known methods such as Gauss elimination method. A shorter, non-numerical example can be found in orthogonality principle.
### Example 2
Consider a vector ${\displaystyle y}$ formed by taking ${\displaystyle N}$ observations of a fixed but unknown scalar parameter ${\displaystyle x}$ disturbed by white Gaussian noise. We can describe the process by a linear equation ${\displaystyle y=1x+z}$, where ${\displaystyle 1=[1,1,\ldots ,1]^{T}}$. Depending on context it will be clear if ${\displaystyle 1}$ represents a scalar or a vector. Suppose that we know ${\displaystyle [-x_{0},x_{0}]}$ to be the range within which the value of ${\displaystyle x}$ is going to fall in. We can model our uncertainty of ${\displaystyle x}$ by an aprior uniform distribution over an interval ${\displaystyle [-x_{0},x_{0}]}$, and thus ${\displaystyle x}$ will have variance of ${\displaystyle \sigma _{X}^{2}=x_{0}^{2}/3.}$. Let the noise vector ${\displaystyle z}$ be normally distributed as ${\displaystyle N(0,\sigma _{Z}^{2}I)}$ where ${\displaystyle I}$ is an identity matrix. Also ${\displaystyle x}$ and ${\displaystyle z}$ are independent and ${\displaystyle C_{XZ}=0}$. It is easy to see that
{\displaystyle {\begin{aligned}&\operatorname {E} \{y\}=0,\\&C_{Y}=\operatorname {E} \{yy^{T}\}=\sigma _{X}^{2}11^{T}+\sigma _{Z}^{2}I,\\&C_{XY}=\operatorname {E} \{xy^{T}\}=\sigma _{X}^{2}1^{T}.\end{aligned}}}
Thus, the linear MMSE estimator is given by
{\displaystyle {\begin{aligned}{\hat {x}}&=C_{XY}C_{Y}^{-1}y\\&=\sigma _{X}^{2}1^{T}(\sigma _{X}^{2}11^{T}+\sigma _{Z}^{2}I)^{-1}y.\end{aligned}}}
We can simplify the expression by using the alternative form for ${\displaystyle W}$ as
{\displaystyle {\begin{aligned}{\hat {x}}&=\left(1^{T}{\frac {1}{\sigma _{Z}^{2}}}I1+{\frac {1}{\sigma _{X}^{2}}}\right)^{-1}1^{T}{\frac {1}{\sigma _{Z}^{2}}}Iy\\&={\frac {1}{\sigma _{Z}^{2}}}\left({\frac {N}{\sigma _{Z}^{2}}}+{\frac {1}{\sigma _{X}^{2}}}\right)^{-1}1^{T}y\\&={\frac {\sigma _{X}^{2}}{\sigma _{X}^{2}+\sigma _{Z}^{2}/N}}{\bar {y}},\end{aligned}}}
where for ${\displaystyle y=[y_{1},y_{2},\ldots ,y_{N}]^{T}}$ we have ${\displaystyle {\bar {y}}={\frac {1^{T}y}{N}}={\frac {\sum _{i=1}^{N}y_{i}}{N}}.}$
Similarly, the variance of the estimator is
${\displaystyle \sigma _{\hat {X}}^{2}=C_{XY}C_{Y}^{-1}C_{YX}={\Big (}{\frac {\sigma _{X}^{2}}{\sigma _{X}^{2}+\sigma _{Z}^{2}/N}}{\Big )}\sigma _{X}^{2}.}$
Thus the MMSE of this linear estimator is
${\displaystyle \operatorname {LMMSE} =\sigma _{X}^{2}-\sigma _{\hat {X}}^{2}={\Big (}{\frac {\sigma _{Z}^{2}}{\sigma _{X}^{2}+\sigma _{Z}^{2}/N}}{\Big )}{\frac {\sigma _{X}^{2}}{N}}.}$
For very large ${\displaystyle N}$, we see that the MMSE estimator of a scalar with uniform aprior distribution can be approximated by the arithmetic average of all the observed data
${\displaystyle {\hat {x}}={\frac {1}{N}}\sum _{i=1}^{N}y_{i},}$
while the variance will be unaffected by data ${\displaystyle \sigma _{\hat {X}}^{2}=\sigma _{X}^{2},}$ and the LMMSE of the estimate will tend to zero.
However, the estimator is suboptimal since it is constrained to be linear. Had the random variable ${\displaystyle x}$ also been Gaussian, then the estimator would have been optimal. Notice, that the form of the estimator will remain unchanged, regardless of the apriori distribution of ${\displaystyle x}$, so long as the mean and variance of these distributions are the same.
### Example 3
Consider a variation of the above example: Two candidates are standing for an election. Let the fraction of votes that a candidate will receive on an election day be ${\displaystyle x\in [0,1].}$ Thus the fraction of votes the other candidate will receive will be ${\displaystyle 1-x.}$ We shall take ${\displaystyle x}$ as a random variable with a uniform prior distribution over ${\displaystyle [0,1]}$ so that its mean is ${\displaystyle {\bar {x}}=1/2}$ and variance is ${\displaystyle \sigma _{X}^{2}=1/12.}$ A few weeks before the election, two independent public opinion polls were conducted by two different pollsters. The first poll revealed that the candidate is likely to get ${\displaystyle y_{1}}$ fraction of votes. Since some error is always present due to finite sampling and the particular polling methodology adopted, the first pollster declares their estimate to have an error ${\displaystyle z_{1}}$ with zero mean and variance ${\displaystyle \sigma _{Z_{1}}^{2}.}$ Similarly, the second pollster declares their estimate to be ${\displaystyle y_{2}}$ with an error ${\displaystyle z_{2}}$ with zero mean and variance ${\displaystyle \sigma _{Z_{2}}^{2}.}$ Note that except for the mean and variance of the error, the error distribution is unspecified. How should the two polls be combined to obtain the voting prediction for the given candidate?
As with previous example, we have
{\displaystyle {\begin{aligned}y_{1}&=x+z_{1}\\y_{2}&=x+z_{2}.\end{aligned}}}
Here, both the ${\displaystyle \operatorname {E} \{y_{1}\}=\operatorname {E} \{y_{2}\}={\bar {x}}=1/2}$. Thus, we can obtain the LMMSE estimate as the linear combination of ${\displaystyle y_{1}}$ and ${\displaystyle y_{2}}$ as
${\displaystyle {\hat {x}}=w_{1}(y_{1}-{\bar {x}})+w_{2}(y_{2}-{\bar {x}})+{\bar {x}},}$
where the weights are given by
{\displaystyle {\begin{aligned}w_{1}&={\frac {1/\sigma _{Z_{1}}^{2}}{1/\sigma _{Z_{1}}^{2}+1/\sigma _{Z_{2}}^{2}+1/\sigma _{X}^{2}}},\\w_{2}&={\frac {1/\sigma _{Z_{2}}^{2}}{1/\sigma _{Z_{1}}^{2}+1/\sigma _{Z_{2}}^{2}+1/\sigma _{X}^{2}}}.\end{aligned}}}
Here, since the denominator term is constant, the poll with lower error is given higher weight in order to predict the election outcome. Lastly, the variance of ${\displaystyle {\hat {x}}}$ is given by
${\displaystyle \sigma _{\hat {X}}^{2}={\frac {1/\sigma _{Z_{1}}^{2}+1/\sigma _{Z_{2}}^{2}}{1/\sigma _{Z_{1}}^{2}+1/\sigma _{Z_{2}}^{2}+1/\sigma _{X}^{2}}}\sigma _{X}^{2},}$
which makes ${\displaystyle \sigma _{\hat {X}}^{2}}$ smaller than ${\displaystyle \sigma _{X}^{2}.}$ Thus, the LMMSE is given by
${\displaystyle \mathrm {LMMSE} =\sigma _{X}^{2}-\sigma _{\hat {X}}^{2}={\frac {1}{1/\sigma _{Z_{1}}^{2}+1/\sigma _{Z_{2}}^{2}+1/\sigma _{X}^{2}}}.}$
In general, if we have ${\displaystyle N}$ pollsters, then ${\displaystyle {\hat {x}}=\sum _{i=1}^{N}w_{i}(y_{i}-{\bar {x}})+{\bar {x}},}$ where the weight for i-th pollster is given by ${\displaystyle w_{i}={\frac {1/\sigma _{Z_{i}}^{2}}{\sum _{j=1}^{N}1/\sigma _{Z_{j}}^{2}+1/\sigma _{X}^{2}}}}$ and the LMMSE is given by ${\displaystyle \mathrm {LMMSE} ={\frac {1}{\sum _{j=1}^{N}1/\sigma _{Z_{j}}^{2}+1/\sigma _{X}^{2}}}.}$
### Example 4
Suppose that a musician is playing an instrument and that the sound is received by two microphones, each of them located at two different places. Let the attenuation of sound due to distance at each microphone be ${\displaystyle a_{1}}$ and ${\displaystyle a_{2}}$, which are assumed to be known constants. Similarly, let the noise at each microphone be ${\displaystyle z_{1}}$ and ${\displaystyle z_{2}}$, each with zero mean and variances ${\displaystyle \sigma _{Z_{1}}^{2}}$ and ${\displaystyle \sigma _{Z_{2}}^{2}}$ respectively. Let ${\displaystyle x}$ denote the sound produced by the musician, which is a random variable with zero mean and variance ${\displaystyle \sigma _{X}^{2}.}$ How should the recorded music from these two microphones be combined, after being synced with each other?
We can model the sound received by each microphone as
{\displaystyle {\begin{aligned}y_{1}&=a_{1}x+z_{1}\\y_{2}&=a_{2}x+z_{2}.\end{aligned}}}
Here both the ${\displaystyle \operatorname {E} \{y_{1}\}=\operatorname {E} \{y_{2}\}=0}$. Thus, we can combine the two sounds as
${\displaystyle y=w_{1}y_{1}+w_{2}y_{2}}$
where the i-th weight is given as
${\displaystyle w_{i}={\frac {a_{i}/\sigma _{Z_{i}}^{2}}{\sum _{j}a_{j}^{2}/\sigma _{Z_{j}}^{2}+1/\sigma _{X}^{2}}}.}$
## Notes
1. ^ "Mean Squared Error (MSE)". www.probabilitycourse.com. Retrieved 9 May 2017.
2. ^ Moon and Stirling. | 13,016 | 41,196 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 363, "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-2024-38 | latest | en | 0.886628 |
https://blog.csdn.net/ggn_2015/article/details/79154315 | 1,548,183,672,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583867214.54/warc/CC-MAIN-20190122182019-20190122204019-00321.warc.gz | 445,115,701 | 40,340 | # 证明
$\left[\alpha \cdot {x}_{1}^{n}+\beta \cdot {x}_{2}^{n}\right]=p\cdot \left[\alpha \cdot {x}_{1}^{n-1}+\beta \cdot {x}_{2}^{n-1}\right]+q\cdot \left[\alpha \cdot {x}_{1}^{n-2}+\beta \cdot {x}_{2}^{n-2}\right]$$[\alpha \cdot x_1^{n} +\beta \cdot x_2^{n}] = p\cdot [\alpha \cdot x_1^{n-1} +\beta \cdot x_2^{n-1}] + q\cdot [\alpha \cdot x_1^{n-2} +\beta \cdot x_2^{n-2}]$
$\left[\alpha \cdot {x}_{1}^{n}\right]+\beta \cdot {x}_{2}^{n}=\left[p\cdot \alpha \cdot {x}_{1}^{n-1}+q\cdot \alpha \cdot {x}_{1}^{n-2}\right]+p\cdot \beta \cdot {x}_{2}^{n-1}+q\cdot \beta \cdot {x}_{2}^{n-2}$$[\alpha\cdot x_1^{n}]+\beta\cdot x_2^{n}=[p\cdot \alpha \cdot x_1^{n-1}+q\cdot \alpha\cdot x_1^{n-2}]+p\cdot \beta\cdot x_2^{n-1}+q\cdot\beta \cdot x_2^{n-2}$
$\alpha \cdot {x}_{1}^{n}=p\cdot \alpha \cdot {x}_{1}^{n-1}+q\cdot \alpha \cdot {x}_{1}^{n-2}$$\alpha\cdot x_1^{n}=p\cdot \alpha \cdot x_1^{n-1}+q\cdot \alpha\cdot x_1^{n-2}$
$\beta \cdot {x}_{2}^{n}=p\cdot \beta \cdot {x}_{2}^{n-1}+q\cdot \beta \cdot {x}_{2}^{n-2}$$\beta\cdot x_2^{n}=p\cdot \beta\cdot x_2^{n-1}+q\cdot\beta \cdot x_2^{n-2}$
${x}_{1}^{n}=p\cdot {x}_{1}^{n-1}+q\cdot {x}_{1}^{n-2}$$x_1^{n}=p\cdot x_1^{n-1}+q\cdot x_1^{n-2}$
${x}_{2}^{n}=p\cdot {x}_{2}^{n-1}+q\cdot {x}_{2}^{n-2}$$x_2^{n}=p\cdot x_2^{n-1}+q\cdot x_2^{n-2}$
${x}_{1}^{2}=p\cdot {x}_{1}+q⇔{x}_{1}^{2}-p\cdot {x}_{1}-q=0$$x_1^2=p\cdot x_1+q \Leftrightarrow x_1^2-p\cdot x_1-q=0$
${x}_{2}^{2}=p\cdot {x}_{2}+q⇔{x}_{2}^{2}-p\cdot {x}_{2}-q=0$$x_2^2=p\cdot x_2+q \Leftrightarrow x_2^2-p\cdot x_2-q=0$
（不靠谱的证明到此结束）
# 用法案例：斐波那契数列
${F}_{0}={F}_{1}=1$$F_0 = F_1 = 1$
${F}_{n}={F}_{n-1}+{F}_{n-2}:n\ge 2$$F_n = F_{n-1}+F_{n-2}:n\geq2$
${x}^{2}-x-1=0$$x^2 - x -1=0$
${F}_{0},{F}_{1}$$F_0,F_1$带入通项公式。
${F}_{0}=\alpha +\beta =1$$F_0=\alpha+\beta=1$
${F}_{1}=\alpha \cdot \frac{1+\sqrt{5}}{2}+\beta \cdot \frac{1-\sqrt{5}}{2}=1$$F_1=\alpha\cdot\frac{1+\sqrt{5}}{2}+\beta\cdot \frac{1-\sqrt{5}}{2}=1$
$\alpha =\frac{\sqrt{5}+1}{2\sqrt{5}},\beta =\frac{\sqrt{5}-1}{2\sqrt{5}}$$\alpha=\frac{\sqrt{5}+1}{2\sqrt{5}},\beta=\frac{\sqrt{5}-1}{2\sqrt{5}}$
${F}_{1}=\alpha \cdot \frac{1+\sqrt{5}}{2}+\beta \cdot \frac{1-\sqrt{5}}{2}=1$$F_1=\alpha\cdot\frac{1+\sqrt{5}}{2}+\beta\cdot \frac{1-\sqrt{5}}{2}=1$
${F}_{2}=\alpha \cdot \left(\frac{1+\sqrt{5}}{2}{\right)}^{2}+\beta \cdot \left(\frac{1-\sqrt{5}}{2}{\right)}^{2}=1$$F_2=\alpha\cdot(\frac{1+\sqrt{5}}{2})^2+\beta\cdot (\frac{1-\sqrt{5}}{2})^2=1$
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120 | 1,355 | 2,481 | {"found_math": true, "script_math_tex": 33, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 33, "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-2019-04 | latest | en | 0.186483 |
http://www.orafaq.com/usenet/comp.databases.theory/2008/12/11/0066.htm | 1,477,494,822,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988720962.53/warc/CC-MAIN-20161020183840-00485-ip-10-171-6-4.ec2.internal.warc.gz | 630,339,551 | 3,552 | # Re: Date and McGoveran comments on view updating 'problem'
Date: Thu, 11 Dec 2008 14:24:48 -0800 (PST)
On Dec 11, 1:48 pm, vadim..._at_gmail.com wrote:
> On Dec 11, 1:28 pm, vadim..._at_gmail.com wrote:
>
>
>
> > On Dec 8, 4:13 pm, vadim..._at_gmail.com wrote:
>
> > > As for deletion here is my worksheet
> > ....
> > > Now, we are proving the following decomposition
>
> > > (SP v D')^(S v D') = (SP ^ S) v D'.
>
> > This assertion is wrong. If D is the set of tuples that we delete from
> > SP ^ S, then the condition is:
>
> > (SP ^ (D' v (SP ^ R00))) ^ (S ^ (D' v (S ^ R00))) = (SP ^ S) ^ D'
>
> > Informally, on the right side we have a relation which is the result
> > of deleting D tuples from SP ^ S. On the left side is a relation that
> > is a join of two base relations SP and S, each trimmed with some
> > projection of D.
>
> > I fail to derive it with any assumptions, except the obvious, like
> > demanding that D acted only on SP (or S). Therefore, there might be
> > some truth to people questioning the validity of deletion from join
> > view...
>
> 2. (D ^ (SP ^ S)') ^ R00 = R00
> "D is subset of SP ^ S"
Join instead of union typo again! These pendulum swings become entertaining...
There is little lesson here. The set containment expressed via empty intersection is a theorem in RL:
x ^ R00 = y ^ R00 ->
((x ^ y') v R00 = R00
<-> x ^ y = x). Received on Thu Dec 11 2008 - 23:24:48 CET
Original text of this message | 468 | 1,447 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2016-44 | longest | en | 0.885345 |
https://gmatclub.com/forum/calling-all-2012-columbia-gsb-applicants-113136-100.html | 1,510,995,307,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934804680.40/warc/CC-MAIN-20171118075712-20171118095712-00556.warc.gz | 607,726,120 | 49,779 | It is currently 18 Nov 2017, 01:55
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# Columbia 2012 - Calling All Applicants
Author Message
Forum Moderator
Status: mission completed!
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Re: Calling All 2012 Columbia GSB Applicants !!! [#permalink]
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26 Jul 2011, 13:19
tiggy wrote:
Sorry, saw that I had a couple of questions directed at me. I submitted for J-Term 6/29, got an interview invite on 7/12 and interviewed on 7/25.
Now the waiting game begins!
_________________
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Re: Calling All 2012 Columbia GSB Applicants !!! [#permalink]
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26 Jul 2011, 20:38
WTF??? Why are there so few posts in the CBS app thread this year? Last year this thread was BOOMING? Any ideas?
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Re: Calling All 2012 Columbia GSB Applicants !!! [#permalink]
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26 Jul 2011, 22:04
Applied for J term on 6/15, Interviewed on 7/20, been waiting for a week now.
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Re: Calling All 2012 Columbia GSB Applicants !!! [#permalink]
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26 Jul 2011, 22:50
Pkit wrote:
tiggy wrote:
Sorry, saw that I had a couple of questions directed at me. I submitted for J-Term 6/29, got an interview invite on 7/12 and interviewed on 7/25.
Now the waiting game begins!
I am pretty sure Tiggy is a woman --- a 'Tiggy' is a really cute girl
Any news for any of the others under review? I have been under review for the J term from the first week of July... No news yet.
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Re: Calling All 2012 Columbia GSB Applicants !!! [#permalink]
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27 Jul 2011, 01:13
1
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Moss wrote:
Hey guys I'll be matriculating this August for the Columbia Business School MBA class of 2013. If you have any questions feel free to send them my way.
Hey hi,
I am planning to apply to columbia this year, can you pls tell is there any difference in the way your application is viewed if you apply in ED?? What is the benefit of ED?? In case you are selected in ED & dont join columbia, whats the penalty you will have to bear??
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Re: Calling All 2012 Columbia GSB Applicants !!! [#permalink]
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27 Jul 2011, 06:40
2
KUDOS
prateekgiitd wrote:
Moss wrote:
Hey guys I'll be matriculating this August for the Columbia Business School MBA class of 2013. If you have any questions feel free to send them my way.
Hey hi,
I am planning to apply to columbia this year, can you pls tell is there any difference in the way your application is viewed if you apply in ED?? What is the benefit of ED?? In case you are selected in ED & dont join columbia, whats the penalty you will have to bear??
I want to know this too. I used to think you were locked in if applying ED, but I went to an information session this monday and the admissions rep said there are people every year that make the \$6K deposit after being accepted ED then go to another school, forfeiting the deposit, so it does happen. I think the ED decision is so early that its hard to get in AND wait for decisions from other schools without making the deposit.
Im thinking about ED, but i'm leaning toward RD. I've done much more research for Columbia then any other school, so i just don't feel like I know enough about the other schools to commit 100% to Columbia. But that could change in 2 months, I've wanted to attend Columbia in some form since high school. I will definitely be done with my application before the deadline.
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Re: Calling All 2012 Columbia GSB Applicants !!! [#permalink]
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27 Jul 2011, 10:23
Kudos [?]: 3 [0], given: 1
Forum Moderator
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Re: Calling All 2012 Columbia GSB Applicants !!! [#permalink]
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27 Jul 2011, 10:45
Guys,
no one will enroll you until you update your profile. Yes you may complete it and remain private.
Please update your profile and you will be rolled automatically to the roll-call list!
_________________
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Re: Calling All 2012 Columbia GSB Applicants !!! [#permalink]
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27 Jul 2011, 11:14
mbaer2012 wrote:
I want to know this too. I used to think you were locked in if applying ED, but I went to an information session this monday and the admissions rep said there are people every year that make the \$6K deposit after being accepted ED then go to another school, forfeiting the deposit, so it does happen. I think the ED decision is so early that its hard to get in AND wait for decisions from other schools without making the deposit.
Im thinking about ED, but i'm leaning toward RD. I've done much more research for Columbia then any other school, so i just don't feel like I know enough about the other schools to commit 100% to Columbia. But that could change in 2 months, I've wanted to attend Columbia in some form since high school. I will definitely be done with my application before the deadline.
With ED, once an admit has been made, you have only two weeks to make a decision. If you decide to accept the offer, you immediately owe a \$6,000 non-refundable deposit. While CBS would argue this, many people feel that CBS is more generous during ED than it is with regular decision.
Early Decision
• Available for August entry applicants only
• Application deadline in early October
• Early decision applications are reviewed, and decisions rendered, before regular decision applications
• Candidates have decided that Columbia is their first choice and must sign the following statement of commitment within their applications: I am committed to attending Columbia Business School and will withdraw all applications and decline all offers from other schools upon admission to Columbia Business School.
• Applicants must submit a nonrefundable \$6,000 tuition deposit within two weeks of admission
_________________
Last edited by MDF on 27 Jul 2011, 13:58, edited 1 time in total.
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Re: Calling All 2012 Columbia GSB Applicants !!! [#permalink]
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27 Jul 2011, 12:47
Applied for J-term on 6/28, Interviewed on 7/19, waiting for decision.
Last edited by wishnuv on 27 Jul 2011, 17:35, edited 1 time in total.
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Re: Calling All 2012 Columbia GSB Applicants !!! [#permalink]
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27 Jul 2011, 12:49
Pkit wrote:
Guys,
no one will enroll you until you update your profile. Yes you may complete it and remain private.
Please update your profile and you will be rolled automatically to the roll-call list!
J-Termers don't seem to be on the roll-call list. I updated my profile.
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Re: Calling All 2012 Columbia GSB Applicants !!! [#permalink]
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27 Jul 2011, 14:29
I know for RD the earlier you apply the better, but is the same true of ED? E.g. if i apply to ED by the deadline are my chances much less then if i had applied for ED in July? And if thats so, are you better off applying to RD in Mid October then applying to ED on the deadline?
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Re: Calling All 2012 Columbia GSB Applicants !!! [#permalink]
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27 Jul 2011, 16:38
I counted 6 people for the J term who went under review in May/June and have already had their interviews/got them scheduled:
Even9net, Tiggy, Nimabi, Miambaprosp, Mmbma, Wishnuv
Let us know how it goes! Good luck
Anyone went under review in July and has heard back?
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Re: Calling All 2012 Columbia GSB Applicants !!! [#permalink]
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27 Jul 2011, 21:06
MBAji wrote:
Pkit wrote:
tiggy wrote:
Sorry, saw that I had a couple of questions directed at me. I submitted for J-Term 6/29, got an interview invite on 7/12 and interviewed on 7/25.
Now the waiting game begins!
I am pretty sure Tiggy is a woman --- a 'Tiggy' is a really cute girl
Any news for any of the others under review? I have been under review for the J term from the first week of July... No news yet.
Thanks MBAji, you are correct! I am female.
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Re: Calling All 2012 Columbia GSB Applicants !!! [#permalink]
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27 Jul 2011, 21:49
1
KUDOS
No word so far. They say J term decisions should be made within 2 weeks after the completion of interview. Should be right around now.
Tiggy, that's soem crazy GPA you got there, the worst grade you got was probably a single B+
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Re: Calling All 2012 Columbia GSB Applicants !!! [#permalink]
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27 Jul 2011, 23:43
Count me in for ED...
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Re: Calling All 2012 Columbia GSB Applicants !!! [#permalink]
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28 Jul 2011, 08:20
mmbma wrote:
No word so far. They say J term decisions should be made within 2 weeks after the completion of interview.
Where did you read or hear this?
I sure hope you are right I am at 10 days since my status went complete. I feel like I'm seeing debt ceiling clock in my dreams, only when I look closer its a final decision clock...tick tick tick
Posted from my mobile device
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Re: Calling All 2012 Columbia GSB Applicants !!! [#permalink]
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28 Jul 2011, 09:06
I'm a reapplicant, going for J-Term this time. Submitted application on 7/15, under review 7/18, invited to interview today 7/28.
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28 Jul 2011, 11:57
Submitted today ED. First a vacation, then on to the next app.
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Re: Calling All 2012 Columbia GSB Applicants !!! [#permalink]
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28 Jul 2011, 14:31
I just got the call!!! Accepted!
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Re: Calling All 2012 Columbia GSB Applicants !!! [#permalink] 28 Jul 2011, 14:31
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Display posts from previous: Sort by | 3,912 | 13,327 | {"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} | 3.046875 | 3 | CC-MAIN-2017-47 | latest | en | 0.839684 |
https://www.k5learning.com/blog/rectangular-prisms-volume | 1,718,288,295,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861451.34/warc/CC-MAIN-20240613123217-20240613153217-00262.warc.gz | 774,972,815 | 18,008 | # How to Find the Volume of Rectangular Prisms
## Volume formula for rectangular prisms
The volume of a rectangular prism is given by the formula:
V = l x w x h
where V is the volume, l is the length, w is the width, and h is the height.
## How to calculate the volume of a rectangular prism
Here's an example:
Suppose we have a rectangular prism with the following dimensions:
Length (l) = 5 cm
Width (w) = 3 cm
Height (h) = 4 cm
To find the volume, we use the formula:
V = l x w x h = 5 cm x 3 cm x 4 cm = 60 cubic cm
Therefore, the volume of the rectangular prism is 60 cubic cm.
Note that the units for volume are cubic units (e.g. cubic cm, cubic inches, etc.).
## Grade 5 volume and surface area of rectangular prisms worksheets
In our worksheets, we combine the practice of calculating the volume and surface area of rectangular prisms. Here’s the set for grade 5.
## Volume and surface area of rectangular prisms practice for grade 6
This is the set of worksheets for grade 6 for finding the volume and surface areas of rectangular prisms.
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https://www.tutorialspoint.com/compute-the-logarithm-base-10-with-scimath-in-python | 1,675,829,640,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500671.13/warc/CC-MAIN-20230208024856-20230208054856-00471.warc.gz | 1,049,003,277 | 10,285 | # Compute the logarithm base 10 with scimath in Python
To compute the logarithm base 10 with scimath, use the scimath.log10() method in Python Numpy. The method returns the log base 10 of the x value(s). If x was a scalar, so is out, otherwise an array object is returned.
For a log10() that returns NAN when real x < 0, use numpy.log10 (note, however, that otherwise numpy.log10 and this log10 are identical, i.e., both return -inf for x = 0, inf for x = inf, and, notably, the complex principle value if x.imag != 0). The 1st parameter, x is the value(s) whose log base 10 is (are) required.
## Steps
At first, import the required libraries −
import numpy as np
Creating a numpy array using the array() method −
arr = np.array([10**1, -10**1, -10**2, 10**2, -10**3, 10**3])
Display the array −
print("Our Array...\n",arr)
Check the Dimensions −
print("\nDimensions of our Array...\n",arr.ndim)
Get the Datatype −
print("\nDatatype of our Array object...\n",arr.dtype)
Get the Shape −
print("\nShape of our Array object...\n",arr.shape)
To compute the logarithm base 10 with scimath, use the scimath.log10() method −
print("\nResult (log10)...\n",np.emath.log10(arr))
## Example
import numpy as np
# Creating a numpy array using the array() method
arr = np.array([10**1, -10**1, -10**2, 10**2, -10**3, 10**3])
# Display the array
print("Our Array...\n",arr)
# Check the Dimensions
print("\nDimensions of our Array...\n",arr.ndim)
# Get the Datatype
print("\nDatatype of our Array object...\n",arr.dtype)
# Get the Shape
print("\nShape of our Array object...\n",arr.shape)
# To compute the logarithm base 10 with scimath, use the scimath.log10() method in Python Numpy
print("\nResult (log10)...\n",np.emath.log10(arr))
## Output
Our Array...
[ 10 -10 -100 100 -1000 1000]
Dimensions of our Array...
1
Datatype of our Array object...
int64
Shape of our Array object...
(6,)
Result (log10)...
[1.+0.j 1.+1.36437635j 2.+1.36437635j 2.+0.j
3.+1.36437635j 3.+0.j ] | 612 | 1,994 | {"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} | 3.59375 | 4 | CC-MAIN-2023-06 | latest | en | 0.630002 |
https://puzzling.stackexchange.com/questions/55808/what-is-an-asymphonic-word | 1,701,651,992,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100518.73/warc/CC-MAIN-20231203225036-20231204015036-00519.warc.gz | 543,678,471 | 41,388 | # What is an Asymphonic Word™?
This is in the spirit of the What is a Word™/Phrase™ series started by JLee with a special brand of Phrase™ and Word™ puzzles.
If a word conforms to a certain rule, I call it a Asymphonic Word™. Use the following examples to find the rule:
Asymphonic Words™ Non-Asymphonic Words™
CACHE CRICKET
CIRCLE CROCODILE
CANCER COCK
GRUDGE GANG
GEOGRAPHY GEOLOGY
GRAPHOLOGY GEMOLOGY
GAUGE GAG
CROTCH CHARACTER
CSV version:
Asymphonic Words™, Non-Asymphonic Words™
CACHE, CRICKET
CIRCLE, CROCODILE
CANCER, COCK
GRUDGE, GANG
GEOGRAPHY, GEOLOGY
GRAPHOLOGY, GEMOLOGY
GAUGE, GAG
CROTCH, CHARACTER
Not only these but many such words exist.
• I just wanted to say that "Non-Asymphonic" is a double negative and should be replaced with "Symphonic". Oct 10, 2017 at 20:18
In a Non-asymphonic word, when
a letter appears more than once, it it pronounced the same way each time. C is a hard C, as in CROCODILE, if it appears that way at least once. GAG=hard g, hard g.
In an asymphonic word,
a letter that appears twice (or more?) has different pronunciations. CANCER= hard c, soft c. GAUGE=hard g, soft g.
Why you would call it non-asymphonic instead of symphonic is one of those deep mysteries.
• It is probably worth noting the O's in a few of the Non- words and A's in the last one. Either it is required for all pairs, or just consonants.
– Apep
Oct 10, 2017 at 15:56
• The rule seems to apply to repeated consonants only, rather than all repeated letters. Oct 10, 2017 at 16:05
• @Mike Q. You are correct. I overlooked the repeated vowels in CHARACTER, CROCODILE, and GRAPHOLOGY.
– user41265
Oct 10, 2017 at 16:13
• GANG probably needs more attention, since NG isn't same as G. (try to make them same, it sounds hilarious either way). +1 anyway.
– Bass
Oct 10, 2017 at 16:17
• @Bass Even if they don't sound quite the same, they're both a hard G sound, as opposed to being soft G sounds (Neither Jang or Ganj sound like Gang) Oct 10, 2017 at 16:21 | 606 | 1,977 | {"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} | 2.578125 | 3 | CC-MAIN-2023-50 | longest | en | 0.922644 |
https://math.stackexchange.com/questions/3649123/need-help-with-a-homogeneous-poisson-process-question | 1,721,582,205,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517747.98/warc/CC-MAIN-20240721152016-20240721182016-00134.warc.gz | 327,418,300 | 35,080 | # Need help with a Homogeneous Poisson Process Question
Customers arrive at a shop following a homogeneous Poisson process $$N(t), t ≥ 0$$, rate $$\lambda$$. Each customer spends some amount of time, $$t_i$$ in the shop, with mean $$E[t_i] = \mu_t$$. If there is a customer in the shop already and a new customer arrives, the new customer leaves. Suppose that each customer spends some amount of money, $$d_i$$, with mean $$E[d_i] = \mu_d$$. Let $$D(t)$$ be the total sales of the shop up to time $$t$$. Find $$\lim_{t \rightarrow \infty} \frac{D(t)}{t}$$. (Hint: Let $$r_i$$ be the time between customers being served).
I'm not sure where to start with this problem. Any hints would be appreciated. Thanks!
$$\lim_{t\to\infty}\frac{D(t)}t=\frac{\mu_d}{\mu_t+\frac1\lambda}\;.$$ | 228 | 781 | {"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": 11, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2024-30 | latest | en | 0.85421 |
http://read.pudn.com/downloads68/sourcecode/windows/network/245700/adHocSimtest/randomWalk.cc__.htm | 1,579,746,991,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250608062.57/warc/CC-MAIN-20200123011418-20200123040418-00533.warc.gz | 141,686,364 | 2,174 | www.pudn.com > adHocSimtest.rar > randomWalk.cc, change:2003-03-07,size:3209b
```
#include "h/randomWalk.h"
Define_Module_Like(RandomWalk,Mobility);
bool RandomWalk::rebound(int& x, int &y)
{
bool update=false;
d("rebound");
if( x < minX)
{
dX *= (-1); // change the sign
x = minX;
alfa = 3.14 - alfa;
update=true;
}
if( x > maxX)
{
x = maxX;
dX *= (-1);
update=true;
alfa = 3.14 - alfa;
}
if( y < minY)
{
dY *= -1;
y = minY;
alfa = 6.28 - alfa ;
update=true;
}
if( y > maxY)
{
dY *= -1;
y = maxY;
alfa = 6.28 - alfa;
update=true;
}
return update;
}
bool RandomWalk::torus(int &x, int &y)
{
bool update=false;
d("toru");
if( x < minX)
{
x = maxX;
update=true;
}
if( x > maxX)
{
x = minX;
update=true;
}
if( y < minY)
{
y = maxY;
update=true;
}
if( y > maxY)
{
y = minY;
update=true;
}
return update;
}
double RandomWalk::randomWalk(int& x, int& y)
{
double step;
//if the node has covered the defined distance chose a new direction
//and speed
if(steps == 0)
{
//choose the direction angle; 6.27 is because alfa is in [0,2PI[
alfa = genk_uniform(2,0,6.27);
//choose a random speed
speed = genk_uniform(1,minSpeed->doubleValue(),maxSpeed->doubleValue());
//compute a single step length
step = moveInterval->doubleValue() * speed;
steps = step > 0 ? (int )( distance->doubleValue() / step) : 1;
stepsNum += steps;
partial += steps * speed;
}
//copmpute a single step
step = moveInterval->doubleValue() * speed;
dX = (int) (step * cos(alfa));
dY = (int) (step * sin(alfa));
//do not go outside the map
if(0 == moveKind->boolValue())
{
//define new <x,y>
x = (x + dX);
y = (y + dY);
torus(x,y);
}
else
{
//define new <x,y>
x = (x + dX);
y = (y + dY);
rebound(x,y);
}
steps--;
return (double)moveInterval->doubleValue();
}
void RandomWalk::initialize()
{
d("Random Walk Module");
cGate *g = gate("out");
//pointer to the physic module that
//store tje actual position
physic =(Physic*) g->toGate()->ownerModule();
minX = 5;
maxX =(int) par("XRange") - 5;
minY = 0;
maxY = (int)par("YRange") - 5;
steps = 0;
moveInterval = &par("moveInterval");
moveKind = &par("movKind");
minSpeed = &par("minSpeed");
maxSpeed = &par("maxSpeed");
distance = &par("distance");
cMessage *moveMsg = new cMessage("Move");
//start miving
scheduleAt(simTime()+0.01, moveMsg);
//statistical variables
stepsNum =0;
partial = 0;
WATCH(speed);
WATCH(alfa);
}
void RandomWalk::handleMessage(cMessage *msg)
{
int x,y;
d("random Walk");
//get the current position from the physic module
physic->getPos(x, y);
//calcolate the new position
double time = randomWalk(x,y);
cMessage *moveMsg = new cMessage("Move",MOVE);
//inform to the physic module about
//the new position so it can be displayed
send(moveMsg,"out");
//tell to the physic module to move
scheduleAt(simTime()+time, msg);
}
void RandomWalk::finish()
{
//recordScalar("Speed avarage..........:",partial / stepsNum);
d("Walk random says bye");
FILE* fout = fopen("collcectedData.dat","a");
fprintf(fout,"\nSpeed avatage............... %.2f\n",partial/stepsNum);
fclose(fout);
}
``` | 962 | 3,045 | {"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-2020-05 | latest | en | 0.264545 |
https://snipplr.com/view/10460/project-euler-add-all-the-natural-numbers-below-one-thousand-that-are-multiples-of-3-or-5 | 1,618,317,058,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038072366.31/warc/CC-MAIN-20210413122252-20210413152252-00505.warc.gz | 611,519,102 | 7,883 | # Posted By
theobriscoe on 12/13/08
# Statistics
Viewed 297 times
Favorited by 0 user(s)
# Project Euler: Add all the natural numbers below one thousand that are multiples of 3 or 5.
/ Published in: Groovy
Add all the natural numbers below one thousand that are multiples of 3 or 5.
If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9.
The sum of these multiples is 23.
http://projecteuler.net/index.php?section=problems&id=1
Copy this code and paste it in your HTML
`def numberList = [] for (i in 1..999) { if (i.mod(3) == 0){ numberList.add(i) } else if (i.mod(5) == 0){ numberList.add(i) } } println "totoal = " + numberList.sum()` | 225 | 695 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2021-17 | longest | en | 0.598553 |
https://getrevising.co.uk/revision-cards/chemical-patters-c4-2 | 1,547,685,675,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583658662.31/warc/CC-MAIN-20190117000104-20190117022104-00117.warc.gz | 533,315,652 | 24,216 | # CHEMICAL PATTERS - C4
## Periodic Table - Proton Number
In the periodic table, the elements are arranged in order of proton number, also called atomic number. This is the number of positive protons in each atom.
1 of 27
## Periodic Table - Development .1
During, the period of the 18th century, actual masses could not be measured so they were compared against the mass of hydrogen (the lightest). This is called the relative atomic mass.
In 1817 Johann Döbereiner proposed his 'law of triads'. He realised the relative atomic mass of the middle element in a group of three elements (that had similar properties) was close to the average of the other two elements.
2 of 27
## Periodic Table - Development .2
The next significant stage in the development of the periodic table came from John Newlands. He arranged the known elements in order of their atomic masses. He proposed a 'law of octaves', meaning every eighth element had similar properties. This did not work for all the known elements, so was dismissed by the scientific community.
The most significant development of the periodic table was due to Dmitri Mendeleev. He put the known elements in order of relative atomic mass but left gaps for undiscovered elements. He also predicted what the properties of these undiscovered elements might be.
3 of 27
## Periodic Table - Relative Atomic Mass
As well as the proton number shown below each element, another number is shown above it. This is the relative atomic mass of the element. It is a comparative measurement of the mass of one atom of the element. You can use it to see how much heavier an atom of one element is compared with an atom of another element.
For example a magnesium atom has a relative atomic mass of 24. So we know it is twice as heavy as a carbon atom, which has a relative atomic mass of 12.
4 of 27
## Periodic Table - Groups
Each column in the table contains elements with similar properties, called a group. Each has a group number, shown across the top of the table. So group 1 contains the elements lithium (Li) to francium (Fr), and group 7 contains the elements fluorine (F) to astatine (At).
5 of 27
## Reactions of Elements - Group 1 Alkali Metals .1
Lithium, sodium and potassium are all soft metals that are easily cut with a scalpel or knife. The freshly cut surface is a shiny, silver colour, but this tarnishes quickly to a dull grey as the metal reacts with oxygen and water in the air. Pieces of such metals are stored in oil to prevent these reactions.
The shiny surface of sodium tarnishes more quickly than that of lithium. And potassium tarnishes more quickly than sodium. This shows the increasing reactivity of the metals as we go down the group.
6 of 27
## Reactions of Elements - Group 1 Alkali Metals .2
Because the alkali metals are so reactive, care has to be taken when using them. They must not be touched because they will react with the water in sweat on the skin. Gloves may be used, and goggles should be worn.
7 of 27
## Reactions of Elements - Alkali Metal Reactions
All the alkali metals react vigorously with cold water. In each reaction, hydrogen gas is given off and the metal hydroxide is produced. The speed and violence of the reaction increases as you go down the group. This shows that the reactivity of the alkali metals increases as you go down group 1.
Lithium + water → lithium hydroxide + hydrogen
2Li(s) + 2H2O(l) → 2LiOH(aq) + H2(g)
Sodium + water → sodium hydroxide + hydrogen
2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g)
Potassium + water → potassium hydroxide + hydrogen
2K(s) + 2H2O(l) → 2KOH(aq) + H2(g)
8 of 27
## Reactions of Elements - Alkali Metal Reactions .Ch
All of the alkali metals react vigorously with chlorine gas. Each reaction produces a white crystalline salt. The reaction gets more violent as you move down group 1, showing how reactivity increases down the group.
Lithium + chlorine → lithium chloride
2Li(s) + Cl2(g) → 2LiCl(s)
Sodium + chlorine → sodium chloride
2Na(s) + Cl2(g) → 2NaCl(s)
Potassium + chlorine → potassium chloride
2K(s) + Cl2(g) → 2KCl(s)
9 of 27
## Reactions of Elements - Group 7 Halogens
The halogens have low melting points and boiling points. This is a typical property of non-metals. Fluorine has the lowest melting point and boiling point. The melting points and boiling points then increase as you go down the group.
10 of 27
## Reactions of Elements - Group 7 Halogens .2
The halogens react with metals to make salts called metal halides.
Metal + halogen → metal halide
EG. Sodium + chlorine → sodium chloride
2Na(s) + Cl2(g) → 2NaCl(s)
The reaction between sodium and a halogen becomes less vigorous as we move down group 7. Fluorine reacts violently with sodium at room temperature. Chlorine reacts very vigorously when in contact with hot sodium. Iodine reacts slowly with hot sodium.
11 of 27
## Reactions of Elements - Uses of Halogens
Halogens are bleaching agents. They will remove the colour of dyes. Chlorine is used to bleach wood pulp to make white paper.
Halogens kill bacteria. Chlorine is added to drinking water at very low concentrations. This kills any harmful bacteria in the water, making it safe to drink. Chlorine is also added to the water in swimming pools.
12 of 27
## Reactions of Elements - Displacement
When chlorine (as a gas or dissolved in water) is added to sodium bromide solution the chlorine takes the place of the bromine. Because chlorine is more reactive than bromine, it displaces bromine from sodium bromide. The solution turns brown. This brown colour is the displaced bromine. The chlorine has gone to form sodium chloride.
If you look at the equation, you can see that the Cl and Br have swapped places.
Chlorine + sodium bromide → sodium chloride + bromine
Cl2(aq) + 2NaBr(aq) → 2NaCl(aq) + Br2(aq)
13 of 27
## Explaining Patterns - Protons, Neutrons and Electr
At the centre of every atom is a nucleus containing protons and neutrons. All atoms of the same element have the same number of protons. This number is used to arrange the elements in the periodic table, beginning with hydrogen, which has just one proton.
Electrons are contained in shells around the nucleus. In a neutral atom the total number of electrons is always the same as the number of protons in the nucleus.
14 of 27
## Explaining Patterns - Energy Levels
These shells are also called energy levels. The number of shells, and the number of electrons in the outer shell, varies from one element to another. For example, a lithium atom has two shells, with two electrons in the inner shell and one in the outer shell. A carbon atom also has two shells, but with two electrons in the inner shell and four in the outer shell.
15 of 27
## Explaining Patterns - Relative Masses and Charges
Protons and neutrons have the same mass, which is much larger than the mass of an electron.
Protons and electrons have an electrical charge. This electrical charge is the same size for both, but protons are positive and electrons are negative.
Neutrons have no electrical charge; they are neutral.
16 of 27
## Explaining Patterns - Line Sprectra
All atoms give off light when heated, although sometimes this light is not visible to the human eye. A prism can be used to split this light to form a spectrum, and each element has its own distinctive line spectrum. This technique is known as spectroscopy.
Scientists have used line spectra to discover new elements. In fact, the discovery of some elements, such as rubidium and caesium, was not possible until the development of spectroscopy. The element helium was discovered by studying line spectra emitted by the Sun.
17 of 27
## Explaining Patterns - Electron Arrangement
Electrons are arranged in shells at different distances around the nucleus. As we move across each row of the Periodic Table the proton number increases by one for each element. This means the number of electrons also increases by one for each element.Starting from the simplest element, hydrogen, and moving through the elements in order we can see how the electrons fill the shells. The innermost shell (or lowest energy level) of electrons is filled first. This shell can contain a maximum of two electrons.Next, the second shell fills with electrons. This can hold a maximum of eight electrons. When this is filled, electrons go into the third shell, which also holds a maximum of eight electrons. Then the fourth shell begins to fill.
18 of 27
## Explaining Patterns - Electronic Structure
The atomic number of an atom is the number of protons it has. This is the same as the number of electrons. If we know the atomic number we can work out the arrangement of the electrons. Fill the shells starting from the smallest and going outward.
For example silicon has atomic number 16. So we have to fill the shells with 16 electrons. That makes 2 in the first (to fill it), 8 in the second shell (to fill that) and 6 left to go into the third shell. So silicon has electronic structure 2.8.6
19 of 27
## Explaining Patterns - Group 1
The atoms of the elements in group 1 all have one electron in their highest occupied energy level (the outer shell). This is why their chemical properties are similar.
When you write the electronic structure of an alkali metal, the last number must be a 1. When you draw or complete a diagram showing the electronic structure of an alkali metal, there must only be one dot or cross in the outer circle.
In a reaction with a non-metal, each alkali metal atom loses its outer electron and becomes an ion with a single positive charge, +1.
20 of 27
## Explaining Patterns - Larger Atoms = More Reactive
As you go down group 1 the atoms become larger and the outer electron is further from the nucleus. The force of attraction between the positively-charged nucleus and the negatively-charged outer electron becomes weaker, which is why the outer electron is more easily lost.
So potassium is more reactive than lithium because the outer electron of a potassium atom is further from its nucleus than the outer electron of a lithium atom.
Francium atoms, with 7 shells, are the largest atoms in group 1. They are very reactive - now you know why.
21 of 27
## Explaining Patterns - Group 7
The atoms of the elements in group 7 (also called the halogens) have seven electrons in their highest occupied energy level (the outer shell). This is why their chemical properties are similar.
When you write the electronic structure of a halogen, the last number must be a 7. When you draw or complete a diagram showing the electronic structure of a halogen, there must be seven dots or crosses in the outer circle.
In a reaction with a metal, each halogen atom gains an outer electron and becomes an ion with a single negative charge, -1.
22 of 27
## Group 1&7 Elements - Metal Ions
When a metal reacts with a non-metal, each metal atom loses the electron, or electrons, from its outer shell. The atom loses negative electrons but still has the same number of positive protons, so it has an overall positive charge. It's not an atom now. Instead it is called an ion.
23 of 27
## Group 1&7 Elements - Non-Metal Ions
When a metal reacts with a non-metal, each non-metal atom gains the number of electrons needed to fill its outer shell. The atom gains negative electrons, but still has the same number of positive protons, so it becomes an ion with a negative charge
24 of 27
## Group 1&7 Elements - Ionic Compounds
When metals react with non-metals, electrons are transferred from the metal atoms to the non-metal atoms, forming ions. The resulting compound is called an ionic compound. EG.
sodium + chlorine → sodium chloride
magnesium + oxygen → magnesium oxide
calcium + chlorine → calcium chloride.
In each of these reactions, the metal atoms give electrons to the non-metal atoms, so that the metal atoms become positive ions and the non-metal atoms become negative ions. There is a strong electrostatic force of attraction between these oppositely-charged ions, called an ionic bond.
25 of 27
## Group 1&7 Elements - Properties of Ionic Compounds
• High melting and boiling points - ionic bonds are very strong and a lot of energy is needed to break them, so ionic compounds have high melting points and boiling points.
• Conductive when liquid - ions are charged particles, but ionic compounds can only conduct electricity if their ions are free to move. So ionic compounds do not conduct electricity when they are solid, but they do conduct electricity when they are dissolved in water or when they are melted.
26 of 27
## Group 1&7 Elements - Formulae and Charges
If you know the charges on the ions in an ionic compound, you can work out its formula.
For example, calcium has an ion with two positive charges, Ca2+, and chlorine has an ion with a single negative charge, Cl-. To balance the positive and negative charges, one calcium ion will need to be with two chloride ions, so the formula of calcium chloride is CaCl2.
If you have the formula of an ionic compound and the charge on one of the two ions, you can work out the charge on the other ion.
For example, sodium oxide has the formula Na2O, and the charge on a sodium ion, Na+, is +1. To balance up the charges from two sodium ions, the oxygen ion must have two negative charges, O2-.
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The Fibonacci Sequence is a common and frequently used series in Mathematics. It is shown below.
``0,1,1,2,3,5,8,13,21,34,55,89,144,229.... ``
The next number in the Fibonacci Sequence is the sum of the previous two numbers and can be shown mathematically as `Fn = Fn-1 + Fn-2`.
The first and second elements of the series are 0 and 1, respectively.
In this tutorial, we will discuss how to create such a sequence in Python.
## Use the Mathematical Formula to Create a Fibonacci Sequence in Python
Every element in a Fibonacci Sequence can be represented using the following mathematical formula.
We can implement this formula in Python to find the series till the required number and print the sequence. The following code shows how.
``from math import sqrt def F(n): return ((1+sqrt(5))**n-(1-sqrt(5))**n)/(2**n*sqrt(5)) def Fibonacci(startNumber, endNumber): n = 0 cur = F(n) while cur <= endNumber: if startNumber <= cur: print(cur) n += 1 cur = F(n) Fibonacci(1,100) ``
Output:
``1.0 1.0 2.0 3.0000000000000004 5.000000000000001 8.000000000000002 13.000000000000002 21.000000000000004 34.00000000000001 55.000000000000014 89.00000000000003 ``
The `Fibonacci()` function calculates the Fibonacci number at some position in a sequence specified by the start and end number.
## Use the `for` Loop to Create a Fibonacci Sequence in Python
We will create a function using the `for` loop to implement the required series. In this method, we will print a sequence of a required length. We will only use the `for` loop to iterate to the required length and alter the required variables every iteration. The following code explains how:
``def fibonacci_iter(n): a=1 b=1 if n==1: print('0') elif n==2: print('0','1') else: print('0') print(a) print(b) for i in range(n-3): total = a + b b=a a= total print(total) fibonacci_iter(8) ``
Output:
``0 1 1 2 3 5 8 13 ``
## Use a Recursive Function to Create a Fibonacci Sequence in Python
A recursive function is a function which calls itself, and such methods can reduce time complexity but use more memory. We can create such a function to return the Fibonacci Number and print the required series using a `for` loop.
For example,
``def rec_fib(n): if n > 1: return rec_fib(n-1) + rec_fib(n-2) return n for i in range(10): print(rec_fib(i)) ``
Output:
``0 1 1 2 3 5 8 13 21 34 ``
## Use Dynamic Programming Method to Create a Fibonacci Sequence in Python
Dynamic Programming is a method in which we divide problems into sub-problems and store the values of these sub-problems to find solutions. This method is usually used in optimizing problems and can be used to generate the Fibonacci Sequence as shown below:
``def fibonacci(num): arr = [0,1] if num==1: print('0') elif num==2: print('[0,','1]') else: while(len(arr)<num): arr.append(0) if(num==0 or num==1): return 1 else: arr[0]=0 arr[1]=1 for i in range(2,num): arr[i]=arr[i-1]+arr[i-2] print(arr) fibonacci(10) ``
Output:
``[0, 1, 1, 2, 3, 5, 8, 13, 21, 34] ``
Note that the sequence is stored in an array in this method. | 946 | 3,493 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2022-33 | latest | en | 0.730628 |
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posted by on .
A military helicopter on a training mission is flying horizontally at a speed of 60.0 m/s and accidentally drops a bomb (fortunately not armed) at an elevation of 350 m. You can ignore air resistance.
Find the horizontal and vertical components of its velocity just before it strikes the earth.
• Physics - ,
a) d = 1/2(a x t^2); rewrite to solve for t:
t = sqrt(2 x d / a) = sqrt( 2 x 350 / 9.8) = 8.45 seconds
b) horiz distance = v x t = 60 x 8.45 = 507 meters
c) bombs horizontal component of velocity doesn't change, neglecting air resistance = 60.0 m/s.
d) the helicopter is directly over the bomb if no air resistance
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"
Key Takeaway:
• The "Sum-of-the-Years" Digits Method is a depreciation method used in accounting that allocates a greater amount of depreciation expense to the earlier years of an asset's useful life and a lesser amount to the later years.
• To calculate the depreciation amount using this method, first determine the asset's useful life and then find the sum of the digits of the years of the asset's useful life. Then, for each year of the asset's useful life, divide the remaining useful life by the sum of the digits and multiply that by the depreciable cost of the asset.
• Advantages of using the "Sum-of-the-Years" Digits Method include front-loading the depreciation expense, which can be helpful for tax purposes, and recognizing the asset's decreasing usefulness over time. Disadvantages include complex calculations and potential overstatement of depreciation expenses in the earlier years of an asset's useful life.
Are you seeking help to understand the concept of Sum-of-the-Years' Digits and how to calculate it? This article will explain the definition of SOYD and provide examples to help you become an expert!
Definition of the "Sum-of-the-Years" Digits Method
The Sum-of-the-Years Digits Method is a depreciation method that allows you to allocate more significant depreciation expenses for the early years of an asset's life and fewer expenses for later years. It is calculated by adding up the digits of the asset's expected life and using this sum to determine the percentage of the asset's cost to be depreciated each year. For example, if an asset has a five-year life, the sum of the digits would be 15 (1+2+3+4+5), so the first-year depreciation expense would be 5/15 of the asset's cost, the second year would be 4/15, and so on.
The Sum-of-the-Years Digits Method is useful for assets that have a higher expected use in the early years, such as vehicles or computers. It front-loads the depreciation expenses, which helps companies to generate more significant tax savings in the earlier years of the asset's life. However, this also means that the asset's book value will be lower in the early years, which can affect financial ratios and complicate decision-making.
It is essential to note that using the Sum-of-the-Years Digits Method requires companies to maintain precise records of the asset's original cost, expected life, and salvage value to calculate accurate depreciation expenses each year. Improper calculations can lead to over- or under-stated financial statements, which can significantly affect investor confidence and stock valuation.
One construction company found itself in a difficult position after using the Sum-of-the-Years Digits Method to depreciate a new crane. The company did not accurately estimate the crane's expected life, and unforeseen repairs caused it to retire the crane after just three years. The inaccurate depreciation calculations overstated the crane's remaining value, leading to an understated loss on the asset's disposal, negative impact on the company's financial statements, and decreased investor confidence in the company's leadership. Thus, companies must be diligent in assessing reliable estimates to avoid similar scenarios.
How to Calculate the Depreciation Amount using "Sum-of-the-Years" Digits Method
To calculate the depreciation using the "Sum-of-the-Years" digits method, one can apply a 3-Step Guide:
1. First, find the sum of the years of the asset's useful life.
2. Next, determine the ratio of each year's useful life to the total sum of the years of useful life.
3. Lastly, multiply the asset's cost by the ratio, and repeat the process for each year. With this method, the asset's depreciation will be higher in the earlier years and decrease as it gets closer to the end of its useful life.
It is worth noting that the "Sum-of-the-Years" digits method compares favorably to other depreciation methods as it factors in the asset's diminishing value accurately. Additionally, this method can be used to maximize tax incentives and increase the amount of expenses claimable.
To ensure better accuracy, it is essential to use the appropriate method based on the asset's unique features and expected useful life. Further, it is advisable to avoid over or under-depreciating assets and ensure that the company's accounting records comply with legal and tax regulations.
By following these recommendations and using the "Sum-of-the-Years" digits method, an organization can improve its financial statements' accuracy and make informed decisions about how to manage the assets.
In depreciation accounting, using the "Sum-of-the-Years" Digits method has both advantages and disadvantages.
A table can be used to illustrate the pros and cons of this method. Advantages may include:
• faster depreciation of assets in their early years
• reduced tax liability
• higher depreciation expenses in later years
• a higher cost of replacing assets
It is important to note that this method may not be suitable for all types of assets, and businesses must consider their specific circumstances before choosing a depreciation method.
A software company once used this method for a computer system that was replaced every three years. However, as the company grew and the system became more essential, they found that using a straight-line method would have been more appropriate.
Some Facts About Sum-of-the-Years' Digits: Definition and How to Calculate:
• ✅ Sum-of-the-Years' Digits is a method of depreciation commonly used in accounting. (Source: AccountingVerse)
• ✅ It is an accelerated depreciation method that results in higher depreciation expenses in the earlier years of an asset's useful life. (Source: Corporate Finance Institute)
• ✅ The calculation involves adding the digits of an asset's useful life and then taking the ratio of the remaining useful life to that sum. (Source: My Accounting Course)
• ✅ Sum-of-the-Years' Digits can be advantageous for businesses that use assets more heavily in the early years of their useful lives. (Source: The Balance Small Business)
• ✅ This method is not allowed under International Financial Reporting Standards (IFRS) and is only permitted under Generally Accepted Accounting Principles (GAAP) in the United States. (Source: AccountingTools)
FAQs about Sum-Of-The-Years' Digits: Definition And How To Calculate
What is Sum-of-the-Years' Digits?
Sum-of-the-Years' Digits is a method of calculating depreciation for an asset where more depreciation is charged in the earlier years of the asset's useful life.
How is Sum-of-the-Years' Digits Calculated?
To calculate depreciation using Sum-of-the-Years' Digits, you first add the digits of the asset's useful life. Then, for each year of the asset's life, you divide the number of years left in the asset's useful life by the sum of years' digits and multiply it by the asset's original cost.
What are the Advantages of Using Sum-of-the-Years' Digits?
The main advantage of using Sum-of-the-Years' Digits is that it allocates a higher portion of the asset's value to the early years of its life, which better reflects the asset's declining effectiveness over time.
What are the Disadvantages of Using Sum-of-the-Years' Digits?
The disadvantage of Sum-of-the-Years' Digits is that it causes greater volatility in depreciation expenses than other methods, making it unsuitable for certain types of assets.
What Types of Assets are Appropriate for Sum-of-the-Years' Digits?
Sum-of-the-Years' digits is appropriate for assets that lose value rapidly in their early years, such as technology equipment or vehicles.
Is it Possible to Switch Depreciation Methods After Starting with Sum-of-the-Years' Digits?
Yes, it is possible to switch depreciation methods mid-stream. However, it is important to evaluate the impact of the switch on financial statements and tax filings before doing so.
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https://stats.stackexchange.com/questions/81659/mutual-information-versus-correlation | 1,721,455,756,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763515020.57/warc/CC-MAIN-20240720052626-20240720082626-00601.warc.gz | 476,112,314 | 47,633 | # Mutual information versus correlation
Why and when we should use Mutual Information over statistical correlation measurements such as "Pearson", "spearman", or "Kendall's tau" ?
Let's consider one fundamental concept of (linear) correlation, covariance (which is Pearson's correlation coefficient "un-standardized"). For two discrete random variables $X$ and $Y$ with probability mass functions $p(x)$, $p(y)$ and joint pmf $p(x,y)$ we have
$$\operatorname{Cov}(X,Y) = E(XY) - E(X)E(Y) = \sum_{x,y}p(x,y)xy - \left(\sum_xp(x)x\right)\cdot \left(\sum_yp(y)y\right)$$
$$\Rightarrow \operatorname{Cov}(X,Y) = \sum_{x,y}\left[p(x,y)-p(x)p(y)\right]xy$$
The Mutual Information between the two is defined as
$$I(X,Y) = E\left (\ln \frac{p(x,y)}{p(x)p(y)}\right)=\sum_{x,y}p(x,y)\left[\ln p(x,y)-\ln p(x)p(y)\right]$$
Compare the two: each contains a point-wise "measure" of "the distance of the two rv's from independence" as it is expressed by the distance of the joint pmf from the product of the marginal pmf's: the $\operatorname{Cov}(X,Y)$ has it as difference of levels, while $I(X,Y)$ has it as difference of logarithms.
And what do these measures do? In $\operatorname{Cov}(X,Y)$ they create a weighted sum of the product of the two random variables. In $I(X,Y)$ they create a weighted sum of their joint probabilities.
So with $\operatorname{Cov}(X,Y)$ we look at what non-independence does to their product, while in $I(X,Y)$ we look at what non-independence does to their joint probability distribution.
Reversely, $I(X,Y)$ is the average value of the logarithmic measure of distance from independence, while $\operatorname{Cov}(X,Y)$ is the weighted value of the levels-measure of distance from independence, weighted by the product of the two rv's.
So the two are not antagonistic—they are complementary, describing different aspects of the association between two random variables. One could comment that Mutual Information "is not concerned" whether the association is linear or not, while Covariance may be zero and the variables may still be stochastically dependent. On the other hand, Covariance can be calculated directly from a data sample without the need to actually know the probability distributions involved (since it is an expression involving moments of the distribution), while Mutual Information requires knowledge of the distributions, whose estimation, if unknown, is a much more delicate and uncertain work compared to the estimation of Covariance.
– SaZa
Commented Jan 9, 2014 at 9:25
• I was asking myself the same question but I have not completely understood the answer. @ Alecos Papadopoulos: I understood that the dependance measured is not the same, okay. So for what kind of relations beetween X and Y should we prefer mutual information I(X,Y) rather than Cov(X,Y)? I had a strange example recently where Y was almost linearly dependent on X (it was nearly a straight line in a scatter plot) and Corr(X,Y) was equal to 0.87 whereas I(X,Y) was equal to 0.45. So is there clearly some cases where one indicator should be choosen over the other one? Thanks for helping! Commented Mar 25, 2014 at 19:47
• This is a great and very clear answer. I was wondering if you have a readily available example where cov is 0, but pmi is not. Commented Jun 11, 2017 at 23:54
• @GENIVI-LEARNER In the context of my answer "point-wise" means that deviations are measured point by point (i.e. value per value that the rv takes). And no, it is not deviation from the mean here, this is clearly stated in my answer. It is deviation from the Independence relation. Commented Feb 18, 2020 at 19:39
• @Cybernetic "The mean is well defined only for the Gaussian"? Last time I checked, there is a really large number of distributions that have a finite mean. And as I already wrote, an otherwise inconsequential truncation allows us to have fat-tail/heavy-tail/long-tail/subexponential distributions together with finite moments (if they are missing in the first place). I understand your criticism for those statisticians that still cling myopically to the "Normal world", but this is a phenomenon of professional inertia. Finally, as my contribution to bold claims, nobody understands probability. Commented Feb 19, 2020 at 17:47
Here's an example.
In these two plots the correlation coefficient is zero. But we can get high shared mutual information even when the correlation is zero.
In the first, I see that if I have a high or low value of X then I'm likely to get a high value of Y. But if the value of X is moderate then I have a low value of Y. The first plot holds information about the mutual information shared by X and Y. In the second plot, X tells me nothing about Y.
Mutual information is a distance between two probability distributions. Correlation is a linear distance between two random variables.
You can have a mutual information between any two probabilities defined for a set of symbols, while you cannot have a correlation between symbols that cannot naturally be mapped into a R^N space.
On the other hand, the mutual information does not make assumptions about some properties of the variables... If you are working with variables that are smooth, correlation may tell you more about them; for instance if their relationship is monotonic.
If you have some prior information, then you may be able to switch from one to another; in medical records you can map the symbols "has genotype A" as 1 and "does not have genotype A" into 0 and 1 values and see if this has some form of correlation with one sickness or another. Similarly, you can take a variable that is continuous (ex: salary), convert it into discrete categories and compute the mutual information between those categories and another set of symbols.
• Correlation isn't a linear function. Should it say that correlation is a measure of the linear relationship between random variables? Commented Oct 20, 2016 at 9:33
• I think this: "You can have a mutual information between any two probabilities defined for a set of symbols, while you cannot have a correlation between symbols that cannot naturally be mapped into a R^N space" is probably the key. Corr doesn't make sense if you don't have a complete random variable; however, pmi makes sense even with just the pdf and sigma (the space). This is why in many applications where RVs don't make sense (e.g. NLP), pmi is used. Commented Jun 12, 2017 at 0:02
Although both of them are a measure of relationship between features, the MI is more general than correlation coefficient (CE) sine the CE is only able to takes into account linear relationships but the MI can also handle non-linear relationships.
• @meow Calculating Pearson's correlation does not assume linearity, but rather quantifies the tightness to a line. Spearman's correlation likewise quantifies monotonicity, it does not assume it. Second moments are assumed to exist, but calculating a Pearson's correlation does not assume a joint normal distribution. Commented Sep 8, 2022 at 14:18
Mutual Information (MI) uses the concept entropy to specify how much common certainty are there in two data samples $$X$$ and $$Y$$ with distribution functions $$p_{x}(x)$$ and $$p_y(y)$$. Considering this interpretation of MI: $$I(X:Y) = H(X) + H(Y) - H(X,Y)$$ we see that the last part says about the dependency of variables. In case of independence the MI is zero and in case of a consistency between $$X$$ and $$Y$$ the MI is equal with the entropy of $$X$$ or $$Y$$. Though, the covariance measures only the distance of every data sample $$(x,y)$$ from the average ($$\mu_X, \mu_Y)$$. Therefore, Cov is only one part of MI. Another difference is the extra information that Cov can deliver about the sign of Cov. This type of knowledge can not extracted from MI because of log-function.
Note that Correlation(Pearson, Spearman or Kendell) takes values in $$[-1,1]$$ while Mutual Information takes value in $$\mathbb{R^*}$$. This makes a big difference: a correlation score is a stronger description of the association between the two RVs than the mutual information. On the other hand, although mutual information is a weaker description it can capture a more general association between two RVs.
A non-zero correlation score not only tells you that the two RVs are related but also in which direction are they related in the value space. This could be very useful in some situations.
Consider the two RVs: the weight and height of a human. If I am told that they are positively correlated, then I can guess the height of a person based on his weight, i.e., a heavy person is likely to be tall(n.b. correlation is not equal to causation).
However, mutual information only tells you if there is some relationship between the two RVs, without precising how exactly the two RVs are related in values. In the same example as above, if I am told that weight and height has a non zero mutual information, then all I know is weight is somehow related to height but I don't know how exactly are they related. It could be negatively related or positively related. (Indeed, an example is a correlation of -1 and +1 both yield maximum mutual information, so the two scenarios are indistinguishable solely based on MI!)
However, the advantage of MI is that it could capture more general(or subtle) associations. Pearson, Spearman, and Kendall are all trying to capture a sort of an increasing or decreasing relationship between RVs in value space. But if the association between the two RVs is quadratic such as shown in the last figure of this reply: then neither of them would capture this relationship and gives 0 as a correlation.
The choice of MI versus a kind of correlation coefficient really depends on what you want to capture. If you want to investigate if there is any sort of association between the two RVs, then choose MI. If you want to find out if the values of two RVs are mutually indicative, then choose a correlation coefficient.
1. Complex Relationships: When you suspect that the relationship between variables is complex and not just linear or monotonic, MI can provide a more comprehensive measure of dependency.
2. Mixed Data Types: If you're dealing with a mix of continuous, ordinal, and nominal variables, MI can offer a consistent measure across variable types.
3. No Assumptions: If you want a measure that doesn't make underlying assumptions about data distributions, MI can be more suitable.
4. Feature Selection: In machine learning, MI is often used for feature selection since it can identify any statistical dependency between features and targets. | 2,419 | 10,614 | {"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": 13, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2024-30 | latest | en | 0.88375 |
coin.furuknap.net | 1,516,367,321,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084887981.42/warc/CC-MAIN-20180119125144-20180119145144-00559.warc.gz | 69,990,343 | 23,514 | # Why Investing in Mining is Always a Bet That Prices Will Drop
The one thing that really puzzles people when it comes to evaluating Bitcoin mining profitability is the fact that investing in mining equipment is always a bet that the prices of your chosen currency goes down.
I’ve tried numerous approaches to explaining why this is true, and I’m almost always met with either complete bewilderment (but, it’s free money, how can that not be profitable?) to rage (you simply don’t understanding mining, it’s all about difficulty/price/speed/etc) to weird math-related arguments (well, your calculations cannot be true because you didn’t prove it with my numbers).
As such, I’m going to explain this concept, hopefully in a way that makes it easy to understand.
However, to do so, I’m going to have to trick you.
I want to start with a completely unrelated topic, just to make sure we have something relatively easy to understand. The topic is going to be currency trading with three currencies.
These three currencies are not to be understood as the traditional currencies you usually handle, like US dollars, Euros, or Pounds. To accomplish this, I’m going to call them A, B, and C.
Because we’re going to do some trading with these currencies, we need to establish an initial exchange rate between the currencies. I’m going to start with the following exchange rates:
Currency Pair Exchange Rate A:B 1:2 B:C 1:50 A:C 1:100
This table should be fairly easy to understand. If you have 1 A, you can trade that for either 2 B or 100 C. If you have one B, you can trade that for either 0.1 A or 50 C. If you have 100 C you can trade that for either 2 B or 1 A.
In fact, let’s start with 2 B and see what we can do. To make this simple, we trade only once per day, using the final exchange rates for that day.
Initial Status: Our holdings initially is 2 B, the equivalent of 1 A or 100 C
On the first day of trading, we decide to buy 1 A for our 2 B.
Day 1 Status: Our holding before day 1 is thus 1 A, the equivalent of 2 B or 100 C.
After the first day of trading, the exchange rates have shifted, making C 2.5 times more valuable. Our exchange rates now look like this:
Currency Pair Exchange Rate A:B 1:2 B:C 1:20 A:C 1:40
Dang! Our value measured in C is now down to 40 C. Even if we still hold 1 A, the exact amount we started with, the increase in C value means our starting sum now translates to a much lower amount of C.
In other words, our value of 1 A means we can get 2.5 times less C today than initially.
Thinking that the C price surge on day one may be a flop, we decide to hold on to our A during day two.
Day 2 Status: Our holding before day 2 thus remains at 1 A, the equivalent of 2 B or 40 C.
After the second day of trading, it turns out we were right! The exchange rate of C drops down to only 10% to its original level, and our exchange rates thus look like this:
Currency Pair Exchange Rate A:B 1:2 B:C 1:500 A:C 1:1000
Luckily, we didn’t buy into the C hype. If we did, we would still hold 40 C, but measured in A, we would suddenly hold only 0.04 A. Measured in B, we would have had only 0.08 B.
In other words, our value of 1 A means we can now get 25 times more C than yesterday.
Final Status: Our holdings after day 2 is thus 1 A, the equivalent of 2 B or 1000 C.
Do we understand each other so far? Everything looks swell? Happy with what’s going on? Good! Because it’s time for me to spring my trap.
# The Trick
I mentioned earlier that I needed to trick you to explain how this all relates to investing in mining and how investing in mining is always a bet that the prices of Bitcoin or your favorite cryptocurrency will drop.
The trick here is that in the scenario above, there aren’t really three currencies. There are just two. One of the currencies, A, is actually a piece of mining equipment, for example a graphics card (GPU).
Hold on!” you say “That’s not fair! Mining equipment isn’t currency and you can’t trade it like that!
You’re right! However, you may notice that we didn’t trade A at all, we just bought it at the beginning. In the two days of trading, we used it only to measure how much of the other currencies we held.
Let’s see if my trickery goes further. What if we replaced currency B with US dollars, and currency C with Bitcoin or some other cryptocurrency? Let’s review our positions initially, after day 1 and after day 2.
Note: Remember that during day 1, the price of C, or our cryptocurrency, increased drastically while on day 2, the price of our cryptocurrency dropped like a rock.
Let’s just exchange our statuses with A being GPU, B being USD, and C being Bitcoin.
Initial Status: Our holdings initially is 2 USD, the equivalent of 1 GPU or 100 Bitcoin
Day 1 Status: Our holding before day 1 is thus 1 GPU, the equivalent of 2 USD or 100 Bitcoin.
Day 2 Status: Our holding before day 2 thus remains at 1 GPU the equivalent of 2 USD or 40 Bitcoin.
Final Status: Our holdings after day 2 is thus 1 GPU, the equivalent of 2 USD or 1000 Bitcoin.
See what happens here?
When the price of Bitcoin rises, our value denominated in Bitcoin drops drastically. When the price of Bitcoin crashes, the amount of Bitcoin our holdings represent goes through the roof!
In short, buying mining equipment yields far more reward in Bitcoins when the value of Bitcoins drop than if Bitcoins rise in value.
As always, I’m anticipating, partially because I’ve been trying to explain this to people many times, that you have some questions. Let me get ahead of you and answer some of them right now. If you have other questions, feel free to leave them as comments below.
### Q: You Forgot Mining, You Idiot!
Nope, I didn’t forget, I left it out because it would only add to your nightmare.
Go ahead, add mining into the equation. Let’s pick any number, say 10 Bitcoin per day. After day 1, you would have had 50 Bitcoins instead, an increase in Bitcoins of 25%! Amazing, increase, so mining must be profitable, right?
Well, after day two, you’d have 1020 Bitcoins, which represents an increase of Bitcoins of 1020% (yes, that’s one thousand and twenty percent) from our initial value. In other words, a price drop means you get 995% more Bitcoins than if you mine while the price goes up and manage to sell at the top. Clearly, a decreasing price yields far more Bitcoins than mining because a drop in price would add 900 Bitcoins, whereas mining would add 20 Bitcoins.
### Q: It’s All About Mining Difficulty, You Idiot!
Not really. If the mining difficulty goes up, you get fewer coins, but even a doubling of the difficulty would only reduce your mining revenue by half. The theory seems to be that increased difficulty leads to a higher price because the cost of mining one coin goes up.
Note: This theory is far from certain, and looking at how major difficulty shifts in other cryptocurrencies have affected prices recently, there doesn’t even seem to be a correlation, much less a dependency between difficulty and price.
In any case, difficulty increase or decrease does not affect profitability anywhere near enough to compensate for the changes in price of a coin.
An increase in difficulty means you get fewer coins, which if the price/difficulty theory holds true means the price will rise. Of course, with fewer coins, that also means less effect of that price increase. Conversely, if the difficulty drops and the price goes down, you have more coins affected by the price decrease.
In the end, it does balance out, but if you think difficulty affects the profitability like that, just run the numbers yourself and see.
### Q: You Forgot Equipment Depreciation, You Idiot!
OK, enough with the insults already!
Depreciation means that something loses value over time. For a GPU, you may expect a lifetime of 12 months, so you can on average expect the value of your GPU to depreciate 1/12 per month. The number of months may be different, but the idea is the same.
Let’s go back a couple of steps and look at the investment before my little text replacement trick. In our first example, depreciation would mean that the exchange value of our A would drop by, for example, 1/12 every month.
However, we would still have one A. Our value denominated in other currencies would drop over time, but our ability to mine with our A does not go down.
Our production from having a GPU increases over time when compared to the value of our GPU. For example, after one month, our A or GPU would be worth only 11/12 of the B/USD and C/Bitcoin value, and would give 10 C/Bitcoins. After 11 months, our GPU would be worth only 1/12 of its original value, but would still produce 10 Bitcoins, or whatever value you choose to use.
### Q: You’re Using Made-Up Numbers! Use My Numbers, You Idiot!
This is the counter-argument that ultimately demonstrates whether you understand math or are just being argumentative.
Look, replace the numbers with whatever makes you happy. It’s not about whether there is a 1:2 exchange rate or a 1:45, 2:31, or 86:15 exchange rate. It doesn’t matter whether a dollar currently is higher or lower than a Bitcoin.
Try it and see! It’s very easy. Just replace the A:B exchange rate with the price of your favorite mining equipment in USD (don’t forget to convert the value to USD, regardless of whether you buy it using USD or Bitcoins), the B:C ratio with the exchange rate of US dollars to Bitcoins, and A:C with the price of your favorite mining equipment in Bitcoins. Then, do the same experiment, using higher or lower decreases and increases if you like.
Don’t trust me, trust the math.
Oh, and if Bitcoins isn’t your chosen cryptocurrency, just swap Bitcoin in the previous paragraph with Whatevercoin.
### Q: Of Course I Want the Price of My Coins to be as High as Possible. Nobody Wants to Sell at a Low Price, You Idiot!
That is true, but tell me, would you rather have 1,000 coins or 100 coins to sell if the price was the same? You’re thinking right but ignore the acquisition of the coins completely.
Remember that when you sell your hardware, you are no longer a miner. You are a coin holder. The argument here is that a mining operation benefits from a falling price, but since your mining operation ceases the moment you sell your mining operation, the falling price no longer benefits you.
In fact, it’s the exact opposite when you just hold coins. You want the price to skyrocket! Until that happens, however, you want to gain as many coins as possible at the lowest price possible, and thus you gain more from a falling price than you do from a rising price.
In our simplified trading example, we stopped the analysis after the price dropped. Add one more day where you trade in your A for 1000 coins after day 2, and see what happens when the price of C or your chosen coin shoots up again on day 3 to the level it was after day 1:
Final Status: Our holdings after day 3 is thus 1020 Bitcoin, the equivalent of 25.5 GPUs or 510 US dollars.
If you were just mining at the rate of the 10 Bitcoins per day from the example in the first question, the results would be:
Final Status: Our holdings after day 3 is thus 1 GPU and 30 Bitcoin, the equivalent of 1.8 GPUs or 36 US dollars.
Mining yields a profit of 34 dollars while mining plus selling your hardware yields 508 dollars.
# The End?
I doubt it, because this is a topic that seems to bring rage to miners all over cryptocurrency land. However, the short version of this article is this:
Mining is always most profitable when the price of Bitcoin goes down. If you invest in mining equipment your highest profit comes when the price of Bitcoin crashes.
That doesn’t mean that mining isn’t profitable when the price rises, only that you’re missing out on a lot of coins when that happens.
Still disagree? Leave your comment below and I’ll try to answer any question you have. Perhaps you know better? Heck, I might even update this article to include your question, and you’ll be famous for setting me straight!
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.b
## 21 thoughts on “Why Investing in Mining is Always a Bet That Prices Will Drop”
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2. Christophe says:
I’m don’t think this is correct. I’ll try to poke holes in it, interested in your response.
You effectively have three options in this model:
1. Hold USD
2. Hold Bitcoins
3. Hold (and use) mining equipment.
The value of the mining equipment (especially GPUs) can be approximated to fall slowly against the USD, due to depreciation. In short timespans, it’s probably negligible, but let’s just assume we lose 0.01 USD’s worth per day.
Then, given your data, the most profitable course of action is to hold US dollars. If we hold the USD from start to finish, we maximise our return (regardless of which currency you measure it in). So why would anyone mine when the currency falls?
Instead of tracing out a half dozen possible price scenarios by hand let’s set up the exact model of mining/holding usd/holding btc.
I’m using a single time period for this, but the same reasoning can be extended to handle multiple periods. I’m also valuing everything against the USD, but it’s trivial to show that the logic remains the same regardless of which unit you use to measure.
Let B0 be the value of 1 BTC in USD at the start of the investment period.
Let B1 be the value of 1 BTC in USD at the end of the investment period.
Let M0 be the purchase price of the mining rig (in USD).
Let M1 be the resale value of the mining rig at the end of the investment period (in USD).
Let E be the cost of running a mining rig (in USD)
Let R be the amount of bitcoins mined during the investment period.
Then the return of holding 1.00 USD is 1.00 USD, since we’re not doing anything.
The return on holding 1.00 USD’s worth of Bitcoins is (B1/B0) USD.
This is the trivial part: As long as Bitcoins appreciate against USD, they’re a better pick.
Now the return on 1.00 USD’s worth of mining rig is:
The return on the physical hardware: M1/M0
Plus the value of the generated bitcoins: B1*R
Minus the value of the electricity spent: E
Giving us: (M1/M0) + B1*R – E
Now it’s entirely reasonable to assume that (M1/M0) < 1 and E > 0, so the only source of profits against the USD comes from B1*R. So higher bitcoin prices, and more bitcoins generated are what leads to mining being profitable. In the extreme case of bitcoins going to zero, it’s obvious that all the mining did is waste electricity and depreciate a piece of equipment.
I’m sorry for digging all the way into the gory details, but this is the best way of taking all factors into account without fixating on what numbers are being used.
1. Thanks for your response, and what you’re saying is correct but doesn’t change the math. Your calculations, however, are exactly the same as mine with the exception that you introduce more complexity through depreciation and ming efficiency.
You further assume that the miner will keep all coins to the end of the period and sell for a given exchange rate at that time. This introduces currency price speculation, which again just complicates things. I could counter-argue that a massive profit can be made by selling the card for coins at the low price and selling the coins at the high price.
In the end, you’re just repeating the same argument, that total profit must the the result of mining revenue less depreciation and cost.
In my calculations, mining doesn’t come into play because it is a minute factor and would balance out by a lower price with more coins being the same as a higher price with fewer coins. If you introduce mining, like you say, you need to introduce cost of electricity, depreciation, cost of operation (and for a \$1000 rig, flipping burgers for minimum wage is far more profitable), and so on.
I also don’t assume the end result denominted in a certain currency, which is exactly why I ‘purchased’ one A at the beginning and held on to it. In the end, you hold exactly the same A as you had when you started. The point isn’t that one A is the best deal you could get, but to take trading and currency speculation out of the picture and show that when prices decline, you get far more coins than you do if you had held on to fiat or bought coins when the prices are higher.
I go into far more details about in the book I’m writing (http://coin.furuknap.net/bitcoin-and-litecoin-mining-book/). You can download the preview chapter for free and you’ll see 8 scenarios with various permutations of mining, depreciation, selling low/high, and so on, but I’m afraid the results remain the same.
Mining is most profitable when prices go down.
3. kyl says:
You left out one (albeit unlikely) possibility. What if Bitcoin exchange rates remain unchanged or become pegged to the USD? In that scenario, mining may be more profitable than just holding USD or Bitcoin.
1. You’re right, both in your analysis and that this won’t ever happen 🙂 For one, who would pick the fiat currency to which Bitcoin would be pegged? I’m fairly certain the US isn’t on the top of the list…
That’s another show, though.
Hi, I think this argument is incorrect for a different fundamental reason.
The numbers remain exactly the same if we had kept our \$2 as dollars throughout, instead of investing it in one GPU. Namely our \$2 would be worth first 100BTC, then 50BTC, and finally 1000BTC.
Thus we can say with equal validity, “investing in \$ is always a bet that the BTC price will drop”. But this is obvious. To sell a dollar gives more BTC if the BTC rate drops relative to dollar. Similarly to sell a gpu gives more BTC if the BTC rate drops relative to gpu.
This is enough to show that you are not actually analysing mining profitability – there is no difference between investing in mining and simply holding the money in \$. You have simply taken the value of the gpu (assumed constant in \$) in btc.
To analyse mining profitability it’s necessary to include depreciation, and also the coins mined which is the return or income from the mining investment.
Cheers
(I’m only commenting because I’ve learnt quite a lot from your other posts – thanks.)
You’re right, except…
The reason for simplifying the exchange rate between A and B is to account for the fact that a GPU is relatively stable in USD price but fluctuates wildly in BTC.
If you want to go down that path, however, the equation becomes a bit different, but not much more complex, because the mining operation output in BTC needs to counter the deprecation of the GPU in USD plus electricity and other operational expenses. This is a fairly simple proposition which may or may not yield a profit, but in the end, a falling BTC price will blow this profit or loss completely out of the water.
Remember, and I’ll actually add this as a question at the end, that the moment you sell your GPU for tons of Bitcoin, you are no longer a miner, you are a coin holder. The argument that investing in mining is a bet that the price goes down no longer applies to your profitability; now you are a coin holder and you want the price to go up.
.b
“…GPU is relatively stable in USD price but fluctuates wildly in BTC”
-> Agreed, this is a reasonable assumption.
On including btc income, & expenses: “… but in the end, a falling BTC price will blow this profit or loss completely out of the water”.
-> Actually it won’t necessarily. This applies only when residual gpu value is large compared to possible net income. To see this take the (reasonable?) case where the gpu value depreciates right down to zero at the end of its useful lifetime. Now this net income you acquired over the period is all you are left with, and can’t be swamped by the residual gpu value.
And the claim “Investing in Mining is Always a Bet That [btc to \$] Prices Will Drop” is not true. Indeed the opposite is the case now: a strong btc increases profitability.
“…you are a coin holder…” etc
-> Agreed, but the decision whether to hold cash as btc or \$ is a seperate decision from whether to invest cash in mining.
This is not to say mining profitability is unaffected by btc/\$ rate, as you need the btc mined income to exceed the \$ expenses, as you say.
Interestingly in mining you have the valuable option to switch off your gpu while the btc/\$ rate is too unfavouable and btc income doesn’t cover \$ expenses.
Cheers A
You assumption on the full depreciation of value is true only in ASIC investing where the reuse value is zero. For calculating profitability then, one must time the end of the operation at a point where the decline in resale value of the card is optimal.
However, during any time of your mining operation where the price of BTC drops, you will get more coins (and become a coin holder) if you sell your hardware for BTC (even via USD) than you will by mining.
The only reason why mining can be more profitable than holding USD is if the earned BTC exceeds the depreciation value. Because depreciation is much higher in the beginning and trending towards zero over time (a card wil never have zero value) in the long run, mining versus depreciation favors mining.
However, in the long run, and this would be at least a couple of years, the chances that the BTC value drops increases, thus favoring selling the card and become a coin holder, in which case you’d better hope the price goes through the roof.
I’ve actually made a far more elaborate scenario setup in my upcoming book, and the free preview chapter available in the top menu details the various permutations of selling, holding cuirrencies, mining, and so on.
.b
5. fluttershy77x says:
Sorry, I feel confussed right now and I’m not sure, if I understood correctly.
1. Lets begin: we can choose between A: Hardware, B: USD, C: BTC
2. Without depreciation and mining profitability and for the sake of simplicity, we can say A (Hardware) has always the same exchange value as B (USD), thus A = B.
3. My choice now is a) A = B, b) B = A, c) BTC – because (2).
4. Because B (USD) = A (Hardware) and vice versa, we can reduce the choice to A: Hardware, B: BTC or A: USD, B: BTC. Doesn’t matter at this point. To give it a name, I’ll call it A: HardwareDollar and B: BTC now.
1 HardwareDollar = 0.01 BTC
or 1 BTC = 100 HardwareDollar
7. We have two possible outcomes (three tbh, but unrelevant):
– BTC rises:
1 HardwareDollar = 0.005 BTC
or 1 BTC = 200 HardwareDollar
– BTC falls:
1 HardwareDollar = 0.02 BTC
or 1 BTC = 50 HardwareDollar
8. Conclusion: if you expect the BTC price to rise, you want to invest in BTC, if you expect the BTC price to fall, you want to invest in USD.
This is quite obvious, so let’s include mining profitability now:
Mining becomes profitable if you can generate at least as much BTC as if you could have brought _at the time of hardware purchase. If this is true, it doesn’t matter, if BTC price falls or rises.
You excluded the part of mining profit completely, even though it is the only relevant variable we have to consider in the first place.
Are you out of your mind, good sir, or is it me? 🙂
Have a nice day.
1. This is quite obvious, so let’s include mining profitability now:
Mining becomes profitable if you can generate at least as much BTC as if you could have brought _at the time of hardware purchase. If this is true, it doesn’t matter, if BTC price falls or rises.
You excluded the part of mining profit completely, even though it is the only relevant variable we have to consider in the first place.
Are you out of your mind, good sir, or is it me?
Your simplification isn’t very simple when you ignore the most profitable aspect of the operation, namely a falling coin price.
I skipped the mining operation because it makes matters worse for profitability. The mining operation itself must compete with the depreciation of the hardware, which is large in the beginning and slows down. Mining cannot usually keep up with this so any mining is usually at a loss in the beginning. In a longer term, the chance of volatility cancelling out any mining revenue increases.
For BTC and particularly ASIC mining, this effect is permanent because the hardware depreciation is permanent and tends towards zero or even negative profitability.
Remember that when you sit on coins as opposed to mining, you want the price to increase, but not until then. If the price increases while you are mining, you lose out on a large portion of your profit potential.
A more detailed analysis of these scenarios are available in the free chapter of Bitcoin and Litecoin Mining.
.b
6. Most people today would not consider buying cryptocurrency mining equipment in the hope that they can exchange it for ANYTHING in the future. They just want to use it to mine coins right now. Consider it a way to get into the market. People who started mining when it started could NOT have just “bought” coins and hung on to them instead of mining, because the coins did not EXIST yet!! My argument is that the exchange rate to fiat currencies will one day become irrelevant because everyone will be using a cryptocurrency as trade, thus mining the coins AND holding coins is the thing to do since ANY mining of cryptocurrency is new money. The whole point of a cryptocurrency is to have a currency that is non-corruptable and have people using it, not trading it for something like USD. People who are willing to trade cryptocurrencies for fiat currency are desperate, they have invested all this fiat money in their mining equipment and they need a way to pay the rent and bills. They would much rather have a way to pay the bills in cryptocurrency. Don’t forget, the only way to GENERATE cryptocurrency is to mine it. One day either the cryptocurrency exchanges will be “illegal” in every country, or everyone will just be using cryptocurrency, or both.
1. 1. What people consider as their motivation is really irrelevant to whether it makes financial sense. Whether someone considers 2+2 to be 4 or something else is just as irrelevant.
2. For most coins, they exist the moment they are announced becuase the genesis block will already have been mined. That, however, is vastly more irrelevant because people don’t buy mining equipment in the hopes that somewhere in the future there will come a coin that will be valuable. They buy mining equipment to mine established coins, thus your argument is also irrelevant.
3. The only way to generate fiat money is to be a central bank. That does not mean the only way to get fiat money is to be a central bank. Your argument, I’m afraid, is again false or irrelevant.
People’s feelings and opinions are irrelevant to math and the math in this article shows that investing in mining equipment yields far more profit if prices go down.
.b
7. tom says:
Actually you can reach this conclusion with a simpler analysis, but you should state your assumptions:
1) you are investing bitcoins
2) the mining equipment’s value at the end of the investment will vastly outweigh accumulated net mining income.
If those are true, your investment is a sale of bitcoins and a purchase of dollar denominated assets (mining equipment), therefore you are short bitcoins and long the dollar.
If you replace the second assumption, your investment is:
– Short bitcoins in the amount of your investment
– Long whatever official currency your mining equipment is priced in, in the amount of your investment less depreciation
– Short electric power, more or less predictably priced in said currency
– Long bitcoins in the amount of mining output
That doesn’t look to me like it always must add up to a short bitcoin long some official currency investment. Usually but not absolutely always.
1. Tom, thanks for your comment,
I had intended to include a second component of the scenario, that while trading, as a holder of A you get a job that pays a fixed amount of C per day, based on a world-wide total of C available each day. Because the output of C would be fixed (regardless of whether you say that it is really C+(Electricity-Electricity)) it would correlate to the mining income.
The fact remains that in order to counter the C-denominated value increase, you would have to have a massive salary. A theoretical situation can exist where that is true, but of course, if all A stock holders got a job that paid in C, then what each A holder could get would diminish.
This is a very nice balance in the Bitcoin system; the more profitable mining becomes, the more people will invest, and the less profitable mining becomes, and fewer people will invest, making mining more profitable again. Over a sufficiently long time, this will reach a balance, so the trick is to invest in the periods where profitability declines (to have the investments when profitability rises) and sell during rises (to have money for when profitability declines again).
.b
8. Hector Sanchez says:
Hi B,
Thanks a lot for this very well explained exercise. I’m starting with Bitcoin/LiteCoin and your analysis helped me to better understand the economics of exchanging cryptocurrencies.
I run your numbers into my calculation and I agree to your conclusions. Here is my spreadsheet in case you want to take a look. I think it helps to illustrate your point.
http://www.hectorsanchez.net/BitCoin_Calculator.xlsx
Regards
Hector
9. First, it’s crude and abrasive to use the word “idiot” dozens of times in your article. You will convince people more easily by omitting this, in my opinion.
Second, it is easy to construct a counterexample to the statement “investing in mining is always a bet that prices will drop”.
Counterexample 1:
You have \$100.
Assume BTC price is fixed at \$100.
Spend \$100 on a mining machine.
Mine on said machine that produces 1 BTC per day for 10 days.
You now have 10 BTC, worth \$1000.
Holding BTC would be less profitable (you would only have 1 BTC now). Holding dollars would be less profitable (you would only have \$100 now).
QED.
Counterexample 2:
Make all the assumptions of counterexample 1, except that the value of a BTC increases by \$10 per day.
At the end of 10 days of mining, you have 10 BTC, worth \$2000.
Holding BTC would be less profitable (you would have 1 BTC now). Holding dollars would be less profitable (you would have \$100 now).
QED.
Many other counterexamples are possible, so your thesis is shown to be false.
1. How many BTC would you have if you held on to your USD and the price dropped to \$1 per BTC then? You would have 100BTC. What something is worth in USD is somewhat irrelevant because as an earner, you want the price to drop so you get more of whatever you earn. As an investor, you want the price to rise when you hold them and drop when you don’t.
I see you don’t understand this, which would make you… Which term would you prefer instead of an idiot? Logically challenged?
Which term would you prefer instead of an idiot? Logically challenged?
If you continue to feel the need to resort to insult in place of civil argument, then you need more practice arguing logic. After all, I am not the one who seems to fail to understand the concept of a counterexample to a universal.
If we can show that sometimes holding mining equipment is optimal when the price is not falling, then we have shown that your statement that “holding mining equipment is only optimal when the price drops” (reworded, but logically equivalent) is false.
10. Yes, but your statement was “investing in mining is always a bet that prices will drop”.
As an investor in mining equipment, you benefit from the rise in the price of a bitcoin. As shown in my counterexample #2, it is possible for the price to rise while holding mining equipment, so that holding mining equipment was the optimal strategy (vs. holding USD or BTC). This is the case as long as the price didn’t rise too quickly.
In counterexample #1, we saw that holding mining equipment was optimal when the price did not fall (nor did it rise).
If we can construct a case where the price doesn’t fall, and holding mining equipment instead of USD or BTC gives the optimum returns, then we know that investing in mining is not always a bet that the price will fall. How could it be, when it gave the best possible return without a fall in BTC price? | 7,300 | 32,599 | {"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} | 3.421875 | 3 | CC-MAIN-2018-05 | latest | en | 0.932187 |
https://thedatatalks.in/education/uniform-distribution/ | 1,716,510,242,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058675.22/warc/CC-MAIN-20240523235012-20240524025012-00183.warc.gz | 486,798,446 | 8,527 | # Uniform distribution
by theDataTalks . 26 Apr 2020
## Uniform-distribution:
A distribution, where each outcome has probability of equally, likely chances and they are impossible beyond a range.
## These are discrete uniform distribution:
Consider a draw of cards(suit) from a deck of playing cards, each suit (clubs, hearts, diamonds & spades) will have a probability of equally, likely chances of 1/4.
The number of values are finite for each suit. We can not get other than 2 to 10, ace, jack, queen & king.
Consider a roll of a die having 6 faces (1,2,3,4,5 & 6), each face will have a probability of equally, likely chances of 1/6.
The number of values are finite viz., 1,2,3,4,5 & 6. We can not get 1.2, 8, 2.6 or some other value.
Consider a roll of a die having 6 faces (1,2,3,4,5 & 6), each face will have a probability of equally, likely chances of 1/6.
The number of values are finite viz., 1,2,3,4,5 & 6. We can not get 1.2, 8, 2.6 or some other value.
The probability mass function(PMF), which gives the probability of x
$\begin{array}{c} p(x) \geq 0 AND \sum_{i=1}^n p(x) =1\\\ p(x) = p(X=x) \end{array}$
The below images are explaining the PMF of single fair dice & two fair dice.
Here the dice are fair and they are not biased.
Notice the change in the shape of the graph. When the number of dice increases, the curve get closer to bell shape and the curve widens.
It always approximates to normal distribution.
## Examples:
1. Single fair dice - Probability to get 3 is 1/6
2. Single fair dice - Probablity to get greater than or equal to 3 & less than are equal to 5 is 1/6+1/6+1/6 = 3/6 = 1/2
3. Two fair dice - Probability to get 2 is 1/36
4. Two fair dice - Probablity to get greater than or equal to 3 & less than are equal to 5 is 3/36+6/36+10/36 = 19/36
The cumulative distribution function(CDF), which gives the probability that the variable less than or equal to the value x
$\begin{array}{c} F(x) = p(X \leq x)\\\\ = \sum_{x_i \leq x} p(x_i)\\\\ = p(x_1)+p(x_2)+p(x_3)+...+p(x) \end{array}$
The below images are explaining the CDF of single fair dice & two fair dice.
Here the dices are fair & they are not biased.
## Examples:
1. Single fair dice - Probability to get less than 3 is 1/6+1/6+1/6 = 3/6 = 1/2
2. Two fair dice - Probablity to get less 3 is 1/36+3/36 = 4/36 = 1/9
If X is a discrete random variable then the Population Mean
$\begin{array}{c} \mu = E[X] = \sum {x * p(x)} \end{array}$
If X is a discrete random variable then the Population variance
$\begin{array}{c} \sigma^2 = E[(X-\mu)^2] = \sum {(x-\mu)^2 * p(x)}\\\\ \text{The same can be written as }\\\\ \sigma^2 = E[X^2]-\mu^2 = \sum{x^2*p(x)-\mu^2} \end{array}$
## Examples:
Mean & variance of a single dice are
Mean = (1 * 1/6) + (2 * 1/6) + (3 * 1/6) + (4 * 1/6) + (5 * 1/6) + (6 * 1/6) = 21/6 = 3.5
The same can be calculated using below formula
Mean = (n + n * 6)/2
Here, 6 - is the number of faces on a dice & n - is the number of dice
So, Mean = (1+1*6)/2 = 7/2 = 3.5
Variance = [(1-3.5)^2 * 1/6 + (2-3.5)^2 * 1/6 + (3-3.5)^2 * 1/6 + (4-3.5)^2 * 1/6 + (5-3.5)^2 * 1/6 + (6-3.5)^2 * 1/6
= 70/24 = 35/12 = 2.91
The same can be calculated using below formula
Mean = n * 35/12
n - is the number of dice
So, Variance = 1* 35/12 = 2.91
Mean & variance of a 2 & 3 dice are
For, 2 Dice
Mean = (2+2*6)/ = 14/2 = 7
Variance = 2 * 35/12 = 5.83
For, 3 Dice
Mean = (3+3*6)/ = 21/2 = 10.5
Variance = 3 * 35/12 = 8.75
## These are continious uniform distribution:
Consider a random variable(a random number generation) between any two numbers. It can take any real numbers with in the range.
The number of values are finite with in the range and they are continuous.
These variables will have probability of equally likely chances with in a range at the same time the chances are impossible beyond the specified range.
The below image represents a continuous uniform distribution and it is generated using the below specified R-command
randomVariable <- runif(1000, min = -2, max = 2.5)
hist(randomVariable, density = 50, freq = FALSE, xlim = c(-3,3), ylim = c(0, 0.3))
curve(dunif(x, min = -2, max =2.5), from = -3, to = 3, n=1000, add = TRUE, col = "red", lwd=2)
1000 random numbers were generated between -2 & 2.5.
Minimum value ‘a’ = -2 & maximum value ‘b’ = 2.5.
Here,
$\begin{array}{c} Density = f(x) = {1 \over (b-a)} \\\\ Mean = \mu = {(b+a) \over 2} \\\\ Variance = \sigma^2 = {(b-a)^2 \over 12} \end{array}$
Don't be in delusion, understand the reality from historical data | Copyright © 2020 theDataTalks | 1,568 | 4,548 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 5, "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} | 4.46875 | 4 | CC-MAIN-2024-22 | latest | en | 0.881287 |
https://www.physicsforums.com/threads/au-196-decay-help-reading-tables.348168/ | 1,511,346,612,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806543.24/warc/CC-MAIN-20171122084446-20171122104446-00252.warc.gz | 840,119,232 | 16,883 | # Au-196 decay: help reading tables
1. Oct 22, 2009
### JustinLevy
I'm sorry, this is probably a pretty stupid question as I probably am just not understanding something basic about nuclear reactions. Can someone please help?
I am trying to understand what happens when Au-196 goes to Hg-196.
I am assuming it is something like:
$Au^{196}_{79} \rightarrow Hg^{196}_{80} + e^- + \bar{\nu}_e + 0.686 MeV$
according to this table:
http://atom.kaeri.re.kr/cgi-bin/nuclide?nuc=Au196
What is confusing me is that it says the Au isotope is spin 2, and the table for Hg says the isotope is spin 0.
http://atom.kaeri.re.kr/cgi-bin/nuclide?nuc=Hg196
How is this possible?
Somehow the electron and neutrino must carry away some specific orbital angular momentum?
Also, what is the +/- they use in front of the spin numbers. How would a +2 be different than a -2?
If I was to look at an electron capture:
$Hg^{196}_{80} + e^- + energy \rightarrow Au^{196}_{79} + \nu_e$
Are there certain configurations/polarizations I can rule out due to the spin?
Does the negative beta decay energy it lists in the table mean this capture would require the electron have have at least ~ 4.4MeV of energy?
Any help understanding these tables would be great.
2. Oct 24, 2009
### arivero
Ah, you are looking for the Stone :tongue:. Ok, trick is first to do alpha from Hg to Pt, then beta from Pt to Au stable. In this way the energy balance is favorable, but as you say you could still have some issue with angular momentum, plus a very very slow rate (so you need the Stone, of course... and it is not easy to find in the market Joachimsthal black, nowadays. Of course you could do a "particular".).
About reading the tables, I have always some doubt about the electron binding energy. I believe that the tables are always for neutral atoms, so it is accounted. But no sure.
3. Oct 24, 2009
### JustinLevy
What do they mean by spin in those tables?
I'm starting to realize it clearly doesn't mean what I thought (I was taking it to be the total angular momentum of the nucleus. But if that was the case, it wouldn't mean anything to give a sign in front of it ... yet there are some nuclei in the full table it lists as spin=+2 and some as spin=-2).
http://atom.kaeri.re.kr/cgi-bin/nuclide?nuc=Au184 ... shows a state with spin = 2+
http://atom.kaeri.re.kr/cgi-bin/nuclide?nuc=Au196 ... shows a state with spin = 2-
Just understanding what they mean by spin here, and what the sign means, would be quite useful.
4. Oct 24, 2009
### OmCheeto
The http://www.nndc.bnl.gov/chart/reColor.jsp?newColor=dm" refers to your charts "spin" numbers as "Jπ"
Last edited by a moderator: Apr 24, 2017
5. Nov 9, 2009
### bcrowell
Staff Emeritus
This looks correct to me.
Right. The maximum angular momentum they could take away in the form of spin would be 1/2+1/2=1. But they also carry orbital angular momentum, which brings the total up to 2 in this case.
This is a notation for the parity of the state. The usual notation in nuclear physics is to write the parity to the right, as a superscript.
Ordinarily when we talk about electron capture, we're talking about the capture of an electron in an inner shell of the atom. Such an electron would have an energy in the low keV range, so there's no way you'd get one with 4.4 MeV of energy. The reaction you've written is still possible in theory, but it's not obvious to me that there's any situation where it would actually happen. In an accelerator, I think the cross-section would be way too low to detect, since it's a weak-force process. Conceivably this could happen in a white dwarf or something, assuming you had some heavy nuclei around. I don't think there's any easy way to rule out any possibilities according to spin. The minimum angular momentum for the final state is 3/2, so the incoming electron has to have at least 1 hbar of orbital angular momentum, which is perfectly possible. I don't think the parity of the incoming electron is constrained, since the outgoing neutrino's parity could be + or -. | 1,046 | 4,052 | {"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} | 2.71875 | 3 | CC-MAIN-2017-47 | longest | en | 0.928954 |
https://socratic.org/questions/592f1c2111ef6b1a91f66a9c#432666 | 1,652,729,622,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662512229.26/warc/CC-MAIN-20220516172745-20220516202745-00112.warc.gz | 615,343,150 | 6,467 | How many formula units constitute a mass of 450*g of "calcium carbonate"?
2 Answers
May 31, 2017
Approx. $2.7 \times {10}^{24}$ $\text{formula units.....}$
Explanation:
First we work out the molar quantity:
$\text{Moles"="Mass"/"Molar mass} = \frac{450 \cdot g}{100.09 \cdot g \cdot m o {l}^{-} 1} = 4.50 \cdot m o l$.
Now, by definition, $1 \cdot m o l$ of stuff specifies $6.022 \times {10}^{23}$ individual items of stuff; we can also use the abbreviation ${N}_{A} = 6.022 \times {10}^{23} \cdot m o {l}^{-} 1$.
Now calcium carbonate is NOT MOLECULAR. And thus we say that there are $4.50 \cdot m o l \times {N}_{A}$ $\text{formula units}$ of calcium carbonate.
May 31, 2017
$2.71 \times {10}^{24}$ ions
Explanation:
The number of particles $N -$ or in this case molecules $-$ of a substance is given by the equation $N = n L$; where $n$ is the number of mole and $L$ is Avogrado's constant.
Also, the definition of $n$ is given by the equation $n = \frac{m}{M}$; where $m$ is the mass and $M$ is the molar mass.
Let's substitute the definition of $n$ into the equation for $N$:
$R i g h t a r r o w N = \frac{m}{M} \times L$
Then, let's substitute the value of $m$ and $L$ into the equation:
$R i g h t a r r o w N = \frac{450 \text{ g}}{M} \times 6.022 \times {10}^{23}$ ${\text{mol}}^{- 1}$
$R i g h t a r r o w N = \frac{2.7099 \times {10}^{26} {\text{ g mol}}^{- 1}}{M}$
We now need to calculate the molar mass $M$ of ${\text{CaCO}}_{3}$:
$R i g h t a r r o w M \left({\text{CaCO}}_{3}\right) = \left(40.078 + 12.011 + 3 \times 15.999\right)$ ${\text{g mol}}^{- 1}$
$R i g h t a r r o w M \left({\text{CaC)}}_{3}\right) = 100.086$ ${\text{g mol}}^{- 1}$
Now, let's substitute this value into the equation:
Rightarrow N = frac(2.7099 times 10^(26) " g mol"^(- 1))(100.086 " g mol"^(- 1))
$R i g h t a r r o w N \approx 2.71 \times {10}^{24}$
Therefore, there are around $2.71 \times {10}^{24}$ ions of ${\text{CaCO}}_{3}$. | 725 | 1,953 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 36, "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} | 4.25 | 4 | CC-MAIN-2022-21 | latest | en | 0.649503 |
https://convertit.com/Go/ConvertIt/Measurement/Converter.ASP?From=arpentlin&To=height | 1,685,708,969,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224648635.78/warc/CC-MAIN-20230602104352-20230602134352-00449.warc.gz | 210,305,333 | 3,877 | Partner with ConvertIt.com
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Please read our Help Page and FAQ Page then post a message or send e-mail. Thanks! | 465 | 1,657 | {"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} | 2.90625 | 3 | CC-MAIN-2023-23 | latest | en | 0.685692 |
https://nolanbconaway.github.io/blog/2019/a-single-sample-poisson-test.html | 1,600,676,355,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400201601.26/warc/CC-MAIN-20200921081428-20200921111428-00679.warc.gz | 563,876,057 | 4,936 | # A single sample poisson test.
May 2019
I am working with a lot of Poisson data recently. The more I encounter these data, the more I realize that data scientists often incorrectly treat Poisson data as normally or linearly distributed variables.
I recently added a new tool to my Poisson kit, which I think of as a "Single Sample Poisson test". The question being asked is:
What is the probability of drawing $$N$$ samples with mean $$\bar{x}=\sum_i{N_i} \div N$$ from a Poisson distribution with mean $$\mu$$?
An example to make things more concrete. Search platforms like google/bing/yahoo allow you to bid a certain amount to show an ad when someone searches for a given keyword. You only pay the search engine if the ad is clicked on. They usually won't charge you the full price of the bid, but some unknown lower amount that we refer to as the cost-per-click (CPC).
Imagine we are hoping to pay a 100 cent (\$1 USD) CPC. We can measure how much we actually paid for each click, but as clicks come in they have variable prices:
Time Price (cents)
9:01am 120
9:01am 95
9:01am 99
9:02am 101
9:02am 98
Above are 5 samples with an average CPC of 102.6 cents. How likely are we to obtain 5 samples with an average of 102.6, assuming that the true CPC is 100? If $$N=5$$ samples with $$\bar{x}=102.6$$ is unlikely given a distribution with $$\mu=100$$, then clearly we are not correctly targeting the CPC value.
## The Test
A traditional approach to this question might treat the samples as normally distributed and use a Student's T-Test or a Sign test. This works a lot of the time due to the fact that Poisson data are approximately normal when $$\mu > 20$$.
If you want a "pure-Poisson" approach, you need to think about it another way. I took part in a very interesting SO discussion in which a Very Smart Person basically gave me the answer. I will relate that answer here.
In short, drawing $$N$$ samples with an average of $$\bar{x}$$ is like drawing a single sample from a higher distribution, which has the value $$\sum_i{x_i} = N\bar{x}$$. We can think about that value within the context of a distribution $$\mathsf{Poisson}(N \mu)$$, which would tell us how unlikely our samples are:
$$N \bar x \sim \mathsf{Poisson}(N \mu).$$
For example, the CDF of $$\mathsf{Poisson}(N \mu)$$ would tell us the proportion of values which are greater than or less than $$N\bar{x}$$.
If only a small proportion of values from the CPC distribution $$\mathsf{Poisson}(5 \cdot 100)$$ are greater than $$5 \cdot 102.6$$, this would indicate that it is unlikely to have obtained $$\bar{x}=102.6$$ by chance and thus maybe there's something wrong with the CPC targeting.
### In Python
>>> from scipy.stats import poisson
>>>
>>> N = 5
>>> xbar = 102.6
>>> mu = 100
>>> p = poisson.cdf(N * xbar, N * mu)
0.7285633495908114
>>>
That shows that 72.9% of $$N=5$$ samples from a distribution with $$\mu=100$$ would have an average value of less than $$\bar{x}=102.6$$. Conversely, 27.1% of samples would have a greater value. So our samples aren't very unlikely given a CPC of 100.
As you increase the number of samples you'd expect $$\bar{x}=102.6$$ to become more and more unlikely.
>>> for n in [10, 20, 30, 40, 50]:
... print(n, poisson.cdf(n * xbar, n * mu))
...
10 0.7994310510662929
20 0.8795150478303122
30 0.9236492476438214
40 0.9503028216143891
50 0.9671124727069668
>>>
When $$N=50$$, 96.7% of samples from the distribution would have a lower average value (conversely, only 3.3% of samples would have a greater value). We're much less likely to obtain $$\bar{x}=102.6$$ by chance, and I might suggest looking into how CPCs are being targeted. | 1,023 | 3,669 | {"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": 2, "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} | 3.46875 | 3 | CC-MAIN-2020-40 | latest | en | 0.943232 |
https://plainmath.net/pre-algebra/79757-let-f-mi-mathvariant-norma | 1,686,359,156,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224656869.87/warc/CC-MAIN-20230609233952-20230610023952-00418.warc.gz | 495,458,729 | 22,792 | hawatajwizp
2022-06-24
Let $f:\mathrm{\Omega }\to B$ be Bochner-measurable, i.e. the point-wise limit of a sequence of simple (i.e. countably-valued measurable) functions $\left({s}_{n}\right)$. I know that if
${\int }_{\mathrm{\Omega }}||f\left(\omega \right)||\phantom{\rule{thickmathspace}{0ex}}d\mu \left(\omega \right)<\mathrm{\infty }\phantom{\rule{1em}{0ex}}\left(\ast \right)$
then one can show that f is Bochner-integrable, i.e. there is even a Bochner-integrable sequence of simple functions $\left({\stackrel{~}{s}}_{n}\right)$ such that
$\underset{n\to \mathrm{\infty }}{lim}\phantom{\rule{thickmathspace}{0ex}}{\int }_{\mathrm{\Omega }}||f\left(\omega \right)-{\stackrel{~}{s}}_{n}\left(\omega \right)||\phantom{\rule{thickmathspace}{0ex}}d\mu \left(\omega \right)\phantom{\rule{mediummathspace}{0ex}}=\phantom{\rule{mediummathspace}{0ex}}0\phantom{\rule{1em}{0ex}}\left(\ast \ast \right)$
I am unsure, however, how to derive this sequence when the measure space $\mathrm{\Omega }$ is not finite. Apparently one is supposed to make use of the fact that due to $\left(\ast \right)$, the set
$A:=\left\{f\ne 0\right\}=\bigcup _{n=1}^{\mathrm{\infty }}\phantom{\rule{thickmathspace}{0ex}}\underset{=:{A}_{n}}{\underset{⏟}{\left\{||f||>\frac{1}{n}\right\}}}$
is $\sigma$-finite, as each ${A}_{n}$ has finite measure. But how to proceed? Setting ${\stackrel{~}{s}}_{n}={1}_{A}{s}_{n}$ would not be enough to make the functions Bochner-integrable, as the entire set $A$ might still have infinite measure. Setting ${\stackrel{~}{s}}_{n}={1}_{{A}_{n}}{s}_{n}$ would do it and maintain pointwise convergence, but then I don't know how to show the convergence in $\left(\ast \ast \right)$ anymore...any tips are much appreciated.
Schetterai
Think of ${A}_{n}$ as a measure space with the restriction of the $\sigma -$−algebra on $\mathrm{\Omega }$ and the restriction of the measure $\mu$. Since you already know the result for finite measure space you can find a simple function ${t}_{n}$ on this space such that ${\int }_{{A}_{n}}‖f{\chi }_{{A}_{n}}-{t}_{n}‖d\mu <\frac{1}{n}$. Let ${s}_{n}={t}_{n}$ on An and 0 outside. Then ${s}_{n}$ is a simple function on $\mathrm{\Omega }$ and $\int ‖f-{s}_{n}‖d\mu \le {\int }_{{A}_{n}}‖f{\chi }_{{A}_{n}}-{t}_{n}‖+{\int }_{\mathrm{\Omega }\setminus {A}_{n}}‖f‖d\mu$. Now ${\int }_{\mathrm{\Omega }\setminus {A}_{n}}‖f‖d\mu \to 0$ because $\underset{n}{lim}{\int }_{{A}_{n}}‖f‖d\mu =\int ‖f{\chi }_{\left\{f\ne 0\right\}}‖d\mu \left(\equiv \int ‖f‖d\mu \right)$ by Monotone ConvegenceTheorem. | 915 | 2,541 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 36, "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} | 3.0625 | 3 | CC-MAIN-2023-23 | latest | en | 0.731587 |
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## Computer Science
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problem 1:
describe, in detail, an algorithm to clip the straight line against an axis-aligned rectangle.
problem 2:
Describe why homogeneous coordinates are employed for handling the geometric transformations.
problem 3:
Give a matrix or a product of matrices that will transform the square ABCD to the square A’B’C’D’.
problem 4:
Show what happens if similar transformation is applied to the square A’B’C’D’?
Computer Graphics, Computer Science
• Category:- Computer Graphics
• Reference No.:- M98979
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### World of Tan 1 - Granma T
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Can you fit the tangram pieces into the outline of Granma T? | 1,925 | 8,433 | {"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} | 3.921875 | 4 | CC-MAIN-2018-17 | latest | en | 0.855152 |
https://forum.altair.com/profile/49653-diana-m/reputation/?type=forums_topic_post&change_section=1 | 1,601,292,843,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600401600771.78/warc/CC-MAIN-20200928104328-20200928134328-00587.warc.gz | 353,273,076 | 11,711 | # Diana M.
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## Reputation Activity
1. Simon Križnik liked a post in a topic by Diana M. in Info about the GRSM algorithm?
Hi Rhodel,
Thank you for your interest in HyperStudy and GRSM.
I thought you may be interested in looking at this white paper dealing with HyperStudy optimization algorithms benchmark:
https://www.altair.com/resource/benchmark-of-hyperstudy-optimization-algorithms
Kind regards,
Diana
2. Amasker liked a post in a topic by Diana M. in Hyperstudy & Dyna optimization
To smooth the curve in HyperStudy, you can use saefilter or polyfit functions available in the Expression builder. You would need to create data sources for each vector Time and Velocity and then create a Response defined with the filter expression.
Hope it helps. Let us know if further questions.
Diana
3. Rahul R liked a post in a topic by Diana M. in Hyperstudy & Dyna optimization
To smooth the curve in HyperStudy, you can use saefilter or polyfit functions available in the Expression builder. You would need to create data sources for each vector Time and Velocity and then create a Response defined with the filter expression.
Hope it helps. Let us know if further questions.
Diana
4. Rahul R liked a post in a topic by Diana M. in Create the response from fuction
Hello,
I confirm you that you can create a response of f(x1,x2) as (var_2-(5*var_1^2)/(4*pi^2)+var_1/pi-6)^2+10*(1-1/8*pi)*cos(var_1)+10, where var_1 is x1 and var_2 is x2
It works: f(1,1)=36.4
What was not working in your case?
Best regards
5. Sriramaero liked a post in a topic by Diana M. in Design variable decimals
Hello, | 416 | 1,610 | {"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-2020-40 | latest | en | 0.867884 |
https://www.reddit.com/r/Art/comments/dh2oj/tiltshift_photography_brings_van_gogh_to_life/ | 1,477,286,096,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988719465.22/warc/CC-MAIN-20161020183839-00157-ip-10-171-6-4.ec2.internal.warc.gz | 987,454,335 | 23,137 | This is an archived post. You won't be able to vote or comment.
[–] 27 points28 points (6 children)
This is just blurring with photoshop: no cameras involved.
[–] 8 points9 points (4 children)
Yup. These paintings are all over the world. She would've had to spent many weeks and thousands of dollars traveling to execute all these shots. And I know for a fact that many of the museums that house these painting don't allow tripods or photography at all.
[–] 1 point2 points * (3 children)
super edit: i'm totally wrong, and davvblack is totally right.
tl;dr -- tilt shift photography wouldn't work on a painting.
Tilt-shift relies on exaggerating depth of field by making things in front of and behind your subject very blurry. There's no depth of field when taking a photograph of a painting from straight in front of it (shooting from a point that lies on a line perpendicular to the plane of the painting).
The only thing that I can think of that might work is shooting a picture of the painting at an angle, rather than from directly ahead. From the camera's perspective, there would be depth that could be used with a tilt-shift lens to get this effect. This would, however, cause the image to be distorted. You'd have to paint the image with the opposite transformation built-in, to negate the transformation due to tilt so that your final image is not warped.
[–] 4 points5 points (2 children)
You do not understand how tilt shift photography works. It does not actually enact depth of field, but it SIMULATES it based on the uv coordinates of the image, so, for example, take this picture:
There is a diagonal band across the picture of focus, but it is not by distance from camera, but by the location on the lens that was not moved by the pivot. Everything above and below the pivot is blurred.
tl;dr tilt shift WOULD work (as well as photography of paintings normally works) but it is not what we are seeing here (though it would have worked for the first 5 paintings, it's clearly not used after and I can't imagine she switched techniques).
[–] 3 points4 points (1 child)
Bah! You're totally right. I forgot about the mechanics of tilt-shift and made them up again in my head. Very good point.
[–] 0 points1 point (0 children)
Thank you for this discussion. I learned something new and i went from "OMFG THIS IS COOL" to "oh god damnit, not so cool" in a very short span. This is why Reddit rocks.
[–] 5 points6 points (0 children)
Yeah, and it looks like shit. Especially Starry Night.
[–] 9 points10 points (1 child)
I actually thought it was cool. It may need some work, but I enjoyed it
[–] 2 points3 points (0 children)
Agreed. No matter how the work was created, it's a fine idea to play with. I love his paintings already, and am happy to see an interesting permutation of them.
[–] 5 points6 points (0 children)
Blog spam with no link to the original source material. Fuck the Chive. Ebams incarnate.
[–] 4 points5 points (3 children)
These look much worse than the originals.
However, an animated film in the Van Gogh style could be pretty awesome.
[–] 2 points3 points (0 children)
The film Frieda about Frieda Khalo incorporated excellent animated versions of her paintings. Highly recommend the movie, even if you're not a big fan of her work.
[–] 0 points1 point (1 child)
[–] 1 point2 points (0 children)
Completely unsure why you were downvoted. Kurosawa's Dreams is one of the best films ever.
[–] 2 points3 points (0 children)
I don't need a gimmicky photo technique to bring Van Gogh to life.
[–] 0 points1 point (1 child)
sacrilege.
[–] 0 points1 point (0 children)
Doesn't even rise to the level of sacrilege. To be sacrilegous you have to understand what you're defacing. This kid doesn't understand what Van Gogh was trying to do.
[–] 0 points1 point (0 children)
You could get an effect very similar to this by using a large format camera with an independent lens board and film plane... But you would still have the problem of the position of the painting itself. It WOULD have to be tilted in some way to achieve these effects, at which point more distortion would occur.
[–] 0 points1 point (0 children)
Way to ruin all the paintings! Holy fuck some people are lazy!
[–] 0 points1 point (0 children)
Totally clueless student uses modern technology to make a mess of Van Gogh's paintings. This is exactly what Van Gogh (and many modern painters after him) was not trying to do. He was purposely trying to flatten the pictorial space in search of a new graphic language, one very much influenced by Japanese prints, for example. This bogus depth runs totally counter to his intentions. | 1,152 | 4,686 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2016-44 | latest | en | 0.974194 |
https://www.cbserankers.com/2018/03/ncert-solutions-for-class-10th-ch-7_93.html | 1,652,912,229,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662522556.18/warc/CC-MAIN-20220518215138-20220519005138-00614.warc.gz | 798,901,277 | 32,881 | NCERT Solutions for Class 10th: Ch 7 Coordinate Geometry Maths
NCERT Solutions for Class 10th: Ch 7 Coordinate Geometry Maths
Page No: 161
Exercise 7.1
1. Find the distance between the following pairs of points:
(i) (2, 3), (4, 1) (ii) (−5, 7), (−1, 3) (iii) (ab), (− a, − b)
(i) Distance between the points is given by
(ii) Distance between (−5, 7) and (−1, 3) is given by
(iii) Distance between (ab) and (− a, − b) is given by
2. Find the distance between the points (0, 0) and (36, 15). Can you now find the distance between the two towns A and B discussed in Section 7.2.
Distance between points (0, 0) and (36, 15)
Yes, Assume town A at origin point (0, 0).
Therefore, town B will be at point (36, 15) with respect to town A.
And hence, as calculated above, the distance between town A and B will be 39 km.
3. Determine if the points (1, 5), (2, 3) and (-2, -11) are collinear.
Let the points (1, 5), (2, 3), and (- 2,-11) be representing the vertices A, B, and C of the given triangle respectively.Let A = (1, 5), B = (2, 3) and C = (- 2,-11)
Since AB + BC ≠ CA
Therefore, the points (1, 5), (2, 3), and ( - 2, - 11) are not collinear.
4. Check whether (5, - 2), (6, 4) and (7, - 2) are the vertices of an isosceles triangle.
Let the points (5, - 2), (6, 4), and (7, - 2) are representing the vertices A, B, and C of the given triangle respectively.
Therefore, AB = BC
As two sides are equal in length, therefore, ABC is an isosceles triangle.
5. In a classroom, 4 friends are seated at the points A, B, C and D as shown in the following figure. Champa and Chameli walk into the class and after observing for a few minutes Champa asks Chameli, “Don't you think ABCD is a square?” Chameli disagrees.
Using distance formula, find which of them is correct.
Clearly from the figure, the coordinates of points A, B, C and D are (3, 4), (6, 7), (9, 4) and (6, 1).
By using distance formula, we get
It can be observed that all sides of this quadrilateral ABCD are of the same length and also the diagonals are of the same length.
Therefore, ABCD is a square and hence, Champa was correct
6. Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer:
(i) (- 1, - 2), (1, 0), (- 1, 2), (- 3, 0)
(ii) (- 3, 5), (3, 1), (0, 3), (- 1, - 4)
(iii) (4, 5), (7, 6), (4, 3), (1, 2)
Let the points ( - 1, - 2), (1, 0), ( - 1, 2), and ( - 3, 0) be representing the vertices A, B, C, and D of the given quadrilateral respectively.
It can be observed that all sides of this quadrilateral are of the same length and also, the diagonals are of the same length. Therefore, the given points are the vertices of a square.
(ii) Let the points ( - 3, 5), (3, 1), (0, 3), and ( - 1, - 4) be representing the vertices A, B, C, and D of the given quadrilateral respectively.
It can be observed that all sides of this quadrilateral are of different lengths. Therefore, it can be said that it is only a general quadrilateral, and not specific such as square, rectangle, etc.
(iii) Let the points (4, 5), (7, 6), (4, 3), and (1, 2) be representing the vertices A, B, C, and D of the given quadrilateral respectively.
It can be observed that opposite sides of this quadrilateral are of the same length. However, the diagonals are of different lengths. Therefore, the given points are the vertices of a parallelogram.
7. Find the point on the x-axis which is equidistant from (2, - 5) and (- 2, 9).
We have to find a point on x-axis. Therefore, its y-coordinate will be 0.
Let the point on x-axis be (x,0)
(- 2)2 + 25 = (x - 2)2 + 81
x2 + 4 - 4x + 25 = x2 + 4 + 4x + 81
8x = 25 -81
8x = -56
x = -7
Therefore, the point is ( - 7, 0).
8. Find the values of y for which the distance between the points P (2, - 3) and Q (10, y) is 10 units.
It is given that the distance between (2, - 3) and (10, y) is 10.
64 + (y + 3)2 = 100
(y +3)2 = 36
y + 3 = ±6
y + 3 = +6 or y + 3 = -6
Therefore, y = 3 or -9
9. If Q (0, 1) is equidistant from P (5, - 3) and R (x, 6), find the values of x. Also find the distance QR and PR.
PQ = QR
41 = x2 + 25
16 = x2
x = ±4
Therefore, point R is (4, 6) or ( - 4, 6).
When point R is (4, 6),
10. Find a relation between x and y such that the point (x, y) is equidistant from the point (3, 6) and (- 3, 4).
Point (xy) is equidistant from (3, 6) and ( - 3, 4).
x2 + 9 - 6x + y2 + 36 - 12y = x2 + 9 + 6x + y2 + 16 - 8y
36 - 16 = 6x + 6x + 12y - 8y
20 = 12x + 4y
3x + y = 5
3x + y - 5 = 0
Exercise 7.2
1. Find the coordinates of the point which divides the join of (- 1, 7) and (4, - 3) in the ratio 2:3.
Let P(xy) be the required point. Using the section formula, we get
Therefore, the point is (1, 3).
2. Find the coordinates of the points of trisection of the line segment joining (4, -1) and (-2, -3).
Let P (x1y1) and Q (x2y2) are the points of trisection of the line segment joining the given points i.e., AP = PQ = QB
Therefore, point P divides AB internally in the ratio 1:2.
3. To conduct Sports Day activities, in your rectangular shaped school ground ABCD, lines have been drawn with chalk powder at a distance of 1 m each. 100 flower pots have been placed at a distance of 1 m from each other along AD, as shown in the following figure. Niharika runs 1/4th the distance AD on the 2nd line and posts a green flag. Preet runs 1/5th the distance AD on the eighth line and posts a red flag. What is the distance between both the flags? If Rashmi has to post a blue flagexactly halfway between the line segment joining the two flags, where should she post her flag?
It can be observed that Niharika posted the green flag at 1/4th of the distance AD i.e., (1×100/4)m = 25m from the starting point of 2nd line. Therefore, the coordinates of this point G is (2, 25).
Similarly, Preet posted red flag at 1/5 of the distance AD i.e., (1×100/5) m = 20m from the starting point of 8th line. Therefore, the coordinates of this point R are (8, 20).
Distance between these flags by using distance formula = GR
Therefore, Rashmi should post her blue flag at 22.5m on 5th line.
4. Find the ratio in which the line segment joining the points (-3, 10) and (6, - 8) is divided by (-1, 6).
Let the ratio in which the line segment joining ( -3, 10) and (6, -8) is divided by point ( -1, 6) be k : 1.Therefore, -1 = 6k-3/k+1
-k - 1 = 6k -3
7k = 2
k = 2/7
Therefore, the required ratio is 2:7.
5. Find the ratio in which the line segment joining A (1, - 5) and B (- 4, 5) is divided by the x-axis. Also find the coordinates of the point of division.
Let the ratio in which the line segment joining A (1, - 5) and B ( - 4, 5) is divided by x-axisbe.
Therefore, the coordinates of the point of division is (-4k+1/k+1, 5k-5/k+1).
We know that y-coordinate of any point on x-axis is 0.
∴ 5k-5/k+1 = 0
Therefore, x-axis divides it in the ratio 1:1.
6. If (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find and y.
Let A,B,C and D be the points (1,2) (4,y), (x,6) and (3,5) respectively.
Mid point of diagonal AC is
and Mid point of Diagonal BD is
Since the diagonals of a parallelogram bisect each other, the mid point of AC and BD are same.
∴ x+1/2 = 7/2 and 4 = 5+y/2
⇒ x + 1 = 7 and 5 + y = 8
⇒ x = 6 and y = 3
7. Find the coordinates of a point A, where AB is the diameter of circle whose centre is (2, - 3) and B is (1, 4).
Let the coordinates of point A be (xy).
Mid-point of AB is (2, - 3), which is the center of the circle.
⇒ x + 1 = 4 and y + 4 = -6
⇒ x = 3 and y = -10
Therefore, the coordinates of A are (3,-10).
8. If A and B are (–2, –2) and (2, –4), respectively, find the coordinates of P such that AP = 3/7 AB and P lies on the line segment AB.
The coordinates of point A and B are (-2,-2) and (2,-4) respectively.
Since AP = 3/7 AB
Therefore, AP:PB = 3:4
Point P divides the line segment AB in the ratio 3:4.
9. Find the coordinates of the points which divide the line segment joining A (- 2, 2) and B (2, 8) into four equal parts.
From the figure, it can be observed that points X, Y, Z are dividing the line segment in a ratio 1:3, 1:1, 3:1 respectively.
10. Find the area of a rhombus if its vertices are (3, 0), (4, 5), (-1, 4) and (-2,-1) taken in order. [Hint: Area of a rhombus = 1/2(product of its diagonals)]
Let (3, 0), (4, 5), ( - 1, 4) and ( - 2, - 1) are the vertices A, B, C, D of a rhombus ABCD.
Page No: 170
Exercises 7.3
1. Find the area of the triangle whose vertices are:
(i) (2, 3), (-1, 0), (2, -4)
(ii) (-5, -1), (3, -5), (5, 2)
(i) Area of a triangle is given by
Area of triangle = 1/2 {x1 (y2 - y3)+ x2 (y3 - y1)+ x3 (y1 - y2)}
Area of the given triangle = 1/2 [2 { 0- (-4)} + (-1) {(-4) - (3)} + 2 (3 - 0)]
= 1/2 {8 + 7 + 6}
= 21/2 square units.
(ii) Area of the given triangle = 1/2 [-5 { (-5)- (4)} + 3(2-(-1)) + 5{-1 - (-5)}]
= 1/2{35 + 9 + 20}
= 32 square units
2. In each of the following find the value of 'k', for which the points are collinear.
(i) (7, -2), (5, 1), (3, -k
(ii) (8, 1), (k, -4), (2, -5)
(i) For collinear points, area of triangle formed by them is zero.
Therefore, for points (7, -2) (5, 1), and (3, k), area = 0
1/2 [7 { 1- k} + 5(k-(-2)) + 3{(-2) + 1}] = 0
7 - 7k + 5k +10 -9 = 0
-2k + 8 = 0
k = 4
(ii) For collinear points, area of triangle formed by them is zero.
Therefore, for points (8, 1), (k, - 4), and (2, - 5), area = 0
1/2 [8 { -4- (-5)} + k{(-5)-(1)} + 2{1 -(-4)}] = 0
8 - 6k + 10 = 0
6k = 18
k = 3
3. Find the area of the triangle formed by joining the mid-points of the sides of the triangle whose vertices are (0, -1), (2, 1) and (0, 3). Find the ratio of this area to the area of the given triangle.
Let the vertices of the triangle be A (0, -1), B (2, 1), C (0, 3).
Let D, E, F be the mid-points of the sides of this triangle. Coordinates of D, E, and F are given by
D = (0+2/2 , -1+1/2) = (1,0)
E = (0+0/2 , -3-1/2) = (0,1)
F = (2+0/2 , 1+3/2) = (1,2)
Area of a triangle = 1/2 {x1 (y2 - y3) + x2 (y3 - y1) + x3 (y1 - y2)}
Area of ΔDEF = 1/2 {1(2-1) + 1(1-0) + 0(0-2)}
= 1/2 (1+1) = 1 square units
Area of ΔABC = 1/2 [0(1-3) + 2{3-(-1)} + 0(-1-1)]
= 1/2 {8} = 4 square units
Therefore, the required ratio is 1:4.
4. Find the area of the quadrilateral whose vertices, taken in order, are (-4, -2), (-3, -5), (3, -2) and (2, 3).
Let the vertices of the quadrilateral be A ( - 4, - 2), B ( - 3, - 5), C (3, - 2), and D (2, 3). Join AC to form two triangles ΔABC and ΔACD.
Area of a triangle = 1/2 {x1 (y2 - y3) + x2 (y3 - y1) + x3 (y1 - y2)}
Area of ΔABC = 1/2 [(-4) {(-5) - (-2)} + (-3) {(-2) - (-2)} + 3 {(-2) - (-5)}]
= 1/2 (12+0+9)
= 21/2 square units
Area of ΔACD = 1/2 [(-4) {(-2) - (3)} + 3{(3) - (-2)} + 2 {(-2) - (-2)}]
= 1/2 (20+15+0)
= 35/2 square units
Area of ☐ABCD = Area of ΔABC + Area of ΔACD
= (21/2 + 35/2) square units = 28 square units
5. You have studied in Class IX that a median of a triangle divides it into two triangles of equal areas. Verify this result for ΔABC whose vertices are A (4, - 6), B (3, - 2) and C (5, 2). | 4,194 | 10,953 | {"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} | 4.59375 | 5 | CC-MAIN-2022-21 | latest | en | 0.913863 |
https://atomparticles.com/how-many-atoms-are-there-in-salt/ | 1,702,230,734,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679102612.80/warc/CC-MAIN-20231210155147-20231210185147-00831.warc.gz | 133,320,110 | 12,155 | # How Many Atoms Are There In Salt
## How many atoms are there in salt?
Sodium (Na) and chlorine (Cl) are the two components that make sodium chloride or NaCl. Each sodium and chlorine atom makes up one molecule of sodium chloride or NaCl. Thus, there are two total atoms in each NaCl molecule.
## How much salt is a grain?
In the old-fashioned English units of weight, a grain weighs approximately 65 mg, which is about how much table salt a person might pick up between the fingers as a pinch.
## How many atoms is a grain of sand?
Originally Answered: How many atoms are there in a grain of sand ? Short answer: ~ 50 000 000 000 000 000 000 atoms. 50 quintillion (short scale ) in the US, Canada or UK.
## How many moles is a grain of salt?
Originally Answered: how many atoms are in a grain of salt? The molar mass of salt is 58.49 g/mol (about 22.99 for Sodium, 35.5 for Chlorine). 70e-6\58.49 = 1.197e-6 mol.
## What is the atoms of salt?
It is also a basic building block of table salt (a molecule of which consists of one atom of sodium and one atom of chlorine: NaCl).
## How many atoms are in a grain of sugar?
What is sugar? The white stuff we know as sugar is sucrose, a molecule composed of 12 atoms of carbon, 22 atoms of hydrogen, and 11 atoms of oxygen (C12H22O11). Like all compounds made from these three elements, sugar is a carbohydrate.
## What is a grain of salt made of?
The salt we eat is a compound made up of one atom of the element sodium (Na) and one atom of the element chlorine (Cl). Its chemical name is sodium chloride.
## What is an example of grain of salt?
The phrase ‘take with a grain of salt’ means that the listener should to take the source of the information as prone to be unreliable or exaggerated. Example of use: Yolanda tells some great stories, but we take what she says with a grain of salt because she has quite a vivid imagination and tends to exaggerate.
## Is a grain of salt a cube?
Sodium chloride crystals are cubic in form. Table salt consists of tiny cubes tightly bound together through ionic bonding of the sodium and chloride ions. The salt crystal is often used as an example of crystalline structure. The size and shape of salt crystals can be modified by temperature.
## How many atoms are in a human?
In summary, for a typical human of 70 kg, there are almost 7*1027 atoms (that’s a 7 followed by 27 zeros!) Another way of saying this is seven billion billion billion. Of this, almost 2/3 is hydrogen, 1/4 is oxygen, and about 1/10 is carbon. These three atoms add up to 99% of the total!
## What is a grain in atoms?
These individual crystals are calledgrains. In any one grain, all atoms are arranged with one particular orientation and one particular pattern. The juncture between adjacent grains is called a grain boundary. The grain boundary is a transition region in which some atoms are not exactly aligned with either grain.
## How many atoms are in Earth?
The number of atoms in the world is around 1.3 x 1050. The number is an approximation because its calculation requires making estimates about the mass and composition of the Earth and also because the number of atoms is always changing. The Earth loses atoms as they escape the atmosphere.
## How many atoms are in water?
A water molecule has three atoms: two hydrogen (H) atoms and one oxygen (O) atom. That’s why water is sometimes referred to as H2O.
## How many molecules are in salt?
Based on Avagadro’s number, there are 6.022 x 1023 particles in every one mole of substance. So I multiplied the moles of NaCl (5.133… x 10-6 moles) by Avagadro’s number and got the answer: one grain of salt ≈ 3.09 x 1018 molecules.
## How many electrons are in salt?
An atom of sodium has 11 protons, 11 electrons, and 12 neutrons.
## What atoms are in sea salt?
The six most abundant ions of seawater are chloride (Cl−), sodium (Na+), sulfate (SO24−), magnesium (Mg2+), calcium (Ca2+), and potassium (K+). By weight these ions make up about 99 percent of all sea salts. | 994 | 4,027 | {"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-2023-50 | longest | en | 0.938845 |
http://quizlet.com/1666376/business-statistics-chapter-1-flash-cards/ | 1,427,825,892,000,000,000 | text/html | crawl-data/CC-MAIN-2015-14/segments/1427131300905.36/warc/CC-MAIN-20150323172140-00171-ip-10-168-14-71.ec2.internal.warc.gz | 224,485,659 | 18,728 | ### 22 terms by dhenrich
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### Define Statistics
math that transforms numbers into useful information for decision makers
### Define descriptive statistics
focuses on collecting, summarizing, presenting, and analyzing a set of data( tables, charts)
### define inferential statistics
data that have been collected from a small group to draw conclusions about a larger group. To make decisions about which investment might lead to a higher return and what marketing strategy might lead to increased sales
### operational definitions
universally accepted meanings that are clear to all associated with an analysis
### Population
consists of all the items or individuals about which you want to draw a conclusion
### Sample
a portion of population for analysis
### Parameter
numerical measure that describes a characteristic of a population
### Statistic
numerical measure that describes a characteristic of a sample
### data collection
collecting the values for those variables
### Primary sources
When the data collector is the one using the date for analysis
### Secondary sources
Person forming the statistical analysis is not the data collector
### Categorical variables(qualitative)
Values that can only be placed into categories. Ex. Yes or No
### Numerical variables(quantitative)
Values that represent quantities. Ex. How many or How much?
### Discrete variables
numerical values that arise from a counting process. Ex. number magazines subscribed to
### Continuous variables
numerical responses that arise from a measuring process. Ex. Time to wait
### Nominal scale
classifies data into distinct categories in which no ranking is implied.Ex. buy a product in 12 months
### Ordinal scale
classifies data into distinct categories in which ranking is implied. Ex. Rate satisfaction
### Interval scale
ordered scale in which the difference between measurements is a meaningful quantity but does not involve a true zero point
### Ratio scale
ordered scale in which the difference between the measurements involves a true zero point such as height or weight.
### frequency distribution
a summary table in which the data re arranged into numerical ordered classes
### width class interval
divide the range(highest value- lowest value) by the number of classes desired.
### Relative Frequency
Frequency in each class/total # of values
Example: | 484 | 2,485 | {"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} | 3.40625 | 3 | CC-MAIN-2015-14 | latest | en | 0.864882 |
https://www.expertsmind.com/library/find-the-final-speed-of-the-space-craft-5261451.aspx | 1,725,920,691,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651157.15/warc/CC-MAIN-20240909201932-20240909231932-00765.warc.gz | 724,090,863 | 13,035 | ### Find the final speed of the space craft
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A 20t space craft leaving the dock of a space stationmakes a 12s engine burn. If the thrust force from the engine is 58KN, find the final speed of the space craft with respect to the station?
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All rights reserved! Copyrights ©2019-2020 ExpertsMind IT Educational Pvt Ltd | 916 | 4,181 | {"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} | 3.234375 | 3 | CC-MAIN-2024-38 | latest | en | 0.91003 |
http://stackoverflow.com/questions/12095484/fastest-language-for-computationally-intensive-functions-in-oracle?answertab=votes | 1,440,705,161,000,000,000 | text/html | crawl-data/CC-MAIN-2015-35/segments/1440644059455.0/warc/CC-MAIN-20150827025419-00058-ip-10-171-96-226.ec2.internal.warc.gz | 191,673,587 | 22,624 | # Fastest language for computationally intensive functions in Oracle
Currently we have quite a few functions (normal CDF, inverse CDF, Vasicek and all kinds of derivatives) coded out in PL/SQL, but they are very slow.
I can get much better performance by streaming the data over a workstation where I have coded out things in C# and then bulk insert the results back. This appproach however leaves the network as the bottleneck, it would be much better if I could 'put the mill where the wood is' by having faster functions inside the Oracle DB.
I want to see how I can speed that up by coding it out in either c(++) or Java (or any other alternative you may have). Does anyone here have any experience with this? Hopefully one of you has tried all approaches and can explain wich has worked best overall.
Extra complication here is that IT is busy as it is, so if I want a waiver to use some feature on the DB I need to make a solid case. I don;t get to play around much on that box, else I would do that.
We're on Oracle Database 11g Enterprise Edition Release 11.2.0.2.0 - 64bit Production
Gert-Jan
EDIT
Here's an example of what a function, wich is the Normal CDF by Cody.
The difference between this and the `cume_dist` is that `cume_dist` finds the distribution within a set of rows. I just need to convert a probability into standard deviations and back (a lot of times), like the `NORMDIST` and `NORMINV` functions in Excel.
`````` function stdnormal_cdf(u number) return number is
z number;
y Number;
begin
y:=abs(u);
if y <= 0.6629126073623883041257915894732959743297 then
z:=y * y;
y:=u * ((((1.161110663653770e-002 * z + 3.951404679838207e-001) * z + 2.846603853776254e + 001) * z + 1.887426188426510e + 002) * z + 3.209377589138469e + 003)/((((1.767766952966369e-001 * z + 8.344316438579620) * z + 1.725514762600375e + 002) * z + 1.813893686502485e + 003) * z + .044716608901563e + 003);
return 0.5 + y ;
else
z:=exp(-y * y/2)/2;
if y <= 5.65685424949238019520675489683879231428 then
y:=y/1.41421356237309504880168872420969807857;
y:=((((((((2.15311535474403846e-8 * y + 5.64188496988670089e-1) * y + 8.88314979438837594) * y + 6.61191906371416295e01) * y + 2.98635138197400131e02) * y + 8.81952221241769090e02) * y + 1.71204761263407058e03) * y + 2.05107837782607147e03) * y + 1.23033935479799725e03)/((((((((1.00000000000000000e00 * y + 1.57449261107098347e01) * y + 1.17693950891312499e02) * y + 5.37181101862009858e02) * y + 1.62138957456669019e03) * y + 3.29079923573345963e03) * y + 4.36261909014324716e03) * y + 3.43936767414372164e03) * + 1.23033935480374942e03);
y:=z * y;
else
z:=z * 1.41421356237309504880168872420969807857/y;
y:=2/(y * y);
y:=y * (((((1.63153871373020978e-2 * y + 3.05326634961232344e-1) * y + 3.60344899949804439e-1) * y + 1.25781726111229246e-1) * y + 1.60837851487422766e-2) * y + 6.58749161529837803e-4)/(((((y + 2.56852019228982242) * y + 1.87295284992346047) * y + 5.27905102951428412e-1) * y + 6.05183413124413191e-2) * y + 2.33520497626869185e-3);
y:=z * (1/1.77245385102123321827450760252310431421-y);
end if;
if u < 0 then
return y;
else
return 1-y;
end if;
end if;
end;
``````
EDIT 2
Ok so here are the benchmarks. Test table with 100k rows. The functions between Oracle and F# are pretty straight translations of each other and give the same result.
The qeury: `select sum(get_rwa(approach, exposure_class_code, pd_r, lgd_r, ead_r, maturity_r, net_sale, rwf_r)) from functest`
Interpreted: 12.8 sec
Native: 13.2 sec
.Net (F#): 0.04 sec.
This would make the .Net function 320x (!) faster than the Oracle implementation, I really don't understand where this difference could come from. Anything up to 3-10x would seem reasonable. I really think I'm missing something here. Anyone?
In F# I loaded the 100k rows into a List first. (seemed fair, just summing up any other column in Oracle cost 0.06 seconds, so it seemed fair to exclude the data acces time in both cases. It takes about 3 sec to load the data into a list, so even if I include the time it takes to open up the connection, execute and stream over the networks etc, then still it's 4x faster.)
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Have you profiled the pl/sql? If so what did it point at? – EvilTeach Aug 23 '12 at 15:50
Actually no, didn't know that existed. Hope I have priviliges for that. I'll get back on this. – gjvdkamp Aug 23 '12 at 15:55
Google DBMS_PROFILER as a start – EvilTeach Aug 23 '12 at 16:00
I'd be interested to see a stored proc (with easy access to the data) that performs slower than some C# program. NOt saying it can't happen, but most likely the pl/sql code could be improved dramatically. CAn you post a simple example of the type of pl/sql you tried? – tbone Aug 23 '12 at 23:35
Thanks for example, native compilation as Justin notes can help, but I doubt it will be dramatically faster. Are you sure this function is the performance bottleneck? How are you calling it? (i assume looping through some cursor of data). If you say that C# pgm gives much better performance, I doubt its because of a .NET implementation of this one function. – tbone Aug 24 '12 at 12:56
Gert-Jan,
Likely, the time difference is due to context switching between the SQL engine and the PL/SQL engine. Each of the 100,000 rows in functest is put through the PL/SQL routine get_rwa (and/or stdnormal_cdf). A context switch involves saving state and restoring state which you probably won't notice when done once. But doing it 100,000 times adds up.
So I'd suggest loading up the 100,000 rows in a nested table with 100,000 rows and pass this nested table only once to a PL/SQL-routine that does a simple "for i in 1 .. [nested_table_variable].count loop ... end loop;", while summing up the individual outcomes.
Another alternative is doing it all in SQL without resorting to PL/SQL.
Regards,
Rob.
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Do you know if using any of the external procedures (c, java, .Net) would not incur this overhead? Will give doing the whole thing from PL/SQL a try. – gjvdkamp Aug 25 '12 at 9:14
It depends on how you use those procedures. Inside SQL (if possible, I don't know for sure) context switching will occur once for each row as well. When called from PL/SQL, after loading all rows in a collection upfront, you won't have this overhead. – Rob van Wijk Aug 25 '12 at 9:45
Ok thanks for that.. so if I can chop up my work in smaller batches then I could run them from within the PL/SQL engine and be faster. Need to do some 30M rows and each of them with many different scenarios etc.. Was hoping for a lean as possible user experience but maybe I can wrap this in some kind of procedure. Thanks! – gjvdkamp Aug 25 '12 at 9:58
Oracle supports the ability to define and call external procedures. Assuming you can compile your C/ C++/ C# application into a DLL/ .so and move that library to the database server, you could then expose the DLL's functions as external procedures and then call the DLL's functions from within the database. Since everything would be running on the same machine, the network wouldn't be a bottleneck. Of course, that would mean that your C/ C++/ C# code would be using the server's processing resources-- that may or may not be a good thing depending on how beefy the server's CPUs are compared to the workstation's and what else the server is doing.
Depending on exactly how you have coded the logic in PL/SQL, you might also want to look into either leveraging Oracle's built-in analytic functions like cume_dist for cumulative distributions (I'm assuming that's what you mean by "normal CDF") or writing your own analytic functions. Since your code is computationally intensive, it's also reasonably likely that you could benefit from native compilation. Of course, this assumes that you have profiled the code and that there aren't obvious places/ approaches to tuning the PL/SQL as it sits.
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hi, I updated te question, cume_dist finds the cum dostribution of a set of rows, I need to convert probablities (0-100%) into standard deviations (roughly -5-+5) and back again a lot of times, like the NORMDIST and NORMINV functions in Excel. They don;t seem to exist in Oracle so we wrote our own. But that seems to be quite slow and I don;t know yet why. – gjvdkamp Aug 24 '12 at 10:34
Theexternal procedures is what I was thinking of, I was hoping someone would have had good results with one approach. There might be pitfalls or issues that might make one approach work better than the other, but I couldn;t find anything on the perf of these approaches. – gjvdkamp Aug 24 '12 at 10:36
Voted answer for native compilation. If all this stuff was indeed inetrpreted that would easiliy explain the poor perf. Wow, bnever expected it could be that simple. Haven;t tested it yet but this seems very likely to fix it. – gjvdkamp Aug 24 '12 at 12:00
Hmmm, so far for the native theory, the native functions are actually a litle bit slower than the interpreted ones. Not getting this at all. – gjvdkamp Aug 24 '12 at 16:11
Sorry, native compilation turned out not to speed things up after all. I updated the example with benchmarks. – gjvdkamp Aug 24 '12 at 16:49 | 2,640 | 9,106 | {"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-2015-35 | latest | en | 0.873232 |
https://math.stackexchange.com/questions/2852399/minimum-value-of-fracb1ab-2 | 1,638,895,604,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363400.19/warc/CC-MAIN-20211207140255-20211207170255-00118.warc.gz | 435,056,409 | 38,198 | # Minimum value of $\frac{b+1}{a+b-2}$
If $a^2 + b^2= 1$ and $u$ is the minimum value of the $\dfrac{b+1}{a+b-2}$, then find the value of $u^2$.
Attempt:
Then I tried this way: Let $a= bk$ for some real $k$.
Then I got $f(b)$ in terms of b and k which is minmum when $b = \dfrac{2-k}{2(k+1)}$ ... then again I got an equation in $k$ which didn't simplify.
Please suggest an efficient way to solve it.
• Just an idea: you can interpret the question as having the points $(a,b)$ on the unit circle. Perhaps you can try creating a bijection by expressing $a$ and $b$ in terms of the angle $\theta$. Jul 15 '18 at 11:27
Try with $b=\cos 2x$ and $a= \sin 2x$.
\begin{eqnarray}{b+1\over a+b-2}&=& {2\cos^2 x\over -\cos^2x+2\sin x \cos x -3\sin^2x}\\ &=& {2\over -1+2\tan x -3\tan^2x}\\ &=& {2\over -1+2t -3t^2} \end{eqnarray} where $t= \tan x$. So the expression will take a minimum when quadratic function $g(t)=-3t^2+2t-1$ will take a maximum. Note that $g(t)<0$ for all $t \in \mathbb{R}$
So $$u= {2\over -{2\over 3}} = -3\implies ....$$
• This was the first thing that came to my mind. Jul 15 '18 at 11:33
Note that $$u=\frac{b+1}{\sqrt{1-b^2}+b-2}$$ so $$\frac{du}{db}=\frac{1\big(\sqrt{1-b^2}+b-2\big)-(b+1)\left(-\frac{2b}{\sqrt{1-b^2}}+1\right)}{\big(\sqrt{1-b^2}+b-2\big)^2}$$ and setting to zero gives $$-3\sqrt{1-b^2}+b+1=0\implies 1-b^2=\frac{b^2+2b+1}9\implies 5b^2+b-4=0$$ and we see that $b=4/5,-1$ are roots.
Checking second derivatives, we have that $4/5$ is a minimum.
Hence $$u^2=\left(\frac{\frac45+1}{\frac35+\frac45-2}\right)^2=9.$$
Note that the negative root ($-3/5$) is also possible, but that yields a lower value of $u^2$ since $$\bigg|-\frac35+\frac45-2\bigg|>\bigg|\frac35+\frac45-2\bigg|$$
• You may need to consider $v=\frac{b+1}{-\sqrt{1-b^2}+b-2}$ as well Jul 15 '18 at 11:24
Let $\displaystyle u=\frac{b+1}{a+b-2}\Rightarrow ua+ub-2u=b+1$
So $ua+(u-1)b=1+2u$
Now Using Cauchy Schwarz Inequality
$\displaystyle \bigg[u^2+(u-1)^2\bigg]\cdot \bigg[a^2+b^2\bigg]\geq \bigg[ua+(u-1)b\bigg]^2$
So $\displaystyle 2u^2-2u+1\geq (1+2u)^2\Rightarrow 2u^2+6u\leq 0$
So $\displaystyle 2u(u+3)\leq 0\Rightarrow -3 \leq u\leq 0$
So $\displaystyle u^2 \geq 9.$
A bit geometry;
1)$x^2+y^2 = 1$, a circle, centre $(0,0)$, $r=1$.
2) Minimum of $C$:
$C:=\dfrac{y+1}{x+y-2}$
(Note: $x+y-2 \not =0$).
$C(x+y-2) = y+1$, or
$Cx +y(C-1) -(2C+1)= 0$, a straight line.
The line touches or intersects the circle 1)
if the distance line-to-origin $\le 1$ (radius).
Distance to $(0,0):$
$d =\dfrac{|2C+1|}{\sqrt{C^2+(C-1)^2}} \le 1.$
$(2C+1)^2 \le C^2 + (C-1)^2;$
$4C^2 +4C +1 \le 2C^2 -2C+1;$
$2C(C+3) \le 0.$
Hence: $-3 \le C \le 0$.
Minimum at $C =-3$.
Used: Line to point distance formula: http://mathworld.wolfram.com/Point-LineDistance2-Dimensional.html
• What's the geometrical interpretation of minimum at C= -3 ? Jul 15 '18 at 14:17
• C=-3 ; Line: -3x -4y +5=0 , y intercept 5/4 , slope: -3/4, distance from (0,0) : =1, hence a tangent to the circle. You can sketch it. Jul 15 '18 at 16:10
Just for a variation, using Lagrange’s method: $$f(a,b,t)=\frac{b+1}{a+b-2}-t(a^2+b^2-1)$$ Then \begin{align} \frac{\partial f}{\partial a}&=-\frac{b+1}{(a+b-2)^2}-2at \\[6px] \frac{\partial f}{\partial b}&=\frac{a-3}{(a+b-2)^2}-2bt \end{align} If these equal $0$, then $$-\frac{b+1}{a(a+b-2)^2}=\frac{a-3}{b(a+b-2)^2}$$ so that $-b^2-b=a^2-3a$, that gives $3a-b=1$ and from $a^2+b^2=1$ we derive $a=0$ or $a=3/5$.
The critical points are thus $(0,-1)$ and $(3/5,4/5)$. We have $$f(0,-1,0)=\frac{2}{3},\quad f(3/5,4/5,0)=-3,\quad$$ This also shows the maximum.
• Why should the partial derivatives equal 0? Jul 18 '18 at 7:52
• @Abcd Because we are looking for critical points. Jul 18 '18 at 8:10
• With respect to the maximum part, can't maximum be at the endpoints of the definition interval? How do we know if it is or not? Jul 20 '18 at 2:10
• @Abcd There's no “definition interval”: the function is restricted to the circle $a^2+b^2=1$, which is compact and has no “endpoints”. Jul 20 '18 at 8:32
Pulling a small rabbit from a hat, consider
$$f(a,b)=3+{b+1\over a+b-2}={3a+4b-5\over a+b-2}$$
It's clear that $a^2+b^2=1$ implies $a+b-2\lt0$. By Cauchy-Schwartz, we have
$$(3a+4b)^2\le(a^2+b^2)(3^2+4^2)=25=5^2$$
and therefore $3a+4b-5\le0$ if $a^2+b^2=1$. Thus $f(a,b)\ge0$ for all $a$ and $b$ for which $a^2+b^2=1$. But also $f({3\over5},{4\over5})=0$, so the minimum value of $(b+1)/(a+b-2)=f(a,b)-3$ with $a^2+b^2=1$ is $-3$, the square of which is $9$.
From
$$\frac{b+1}{a+b-2}= u\Rightarrow L\to a = 2-b\frac{b+1}{u}$$
So $L$ should be tangent to $a^2+b^2=1\;$ then substituting we have the condition
$$(2b^2-4b+3)u^2+(4+2b-2b^2)u+(b+1)^2 = 0$$
and solving for $u$
$$u = \frac{(b+1)^2}{b^2-b\pm\sqrt{(1-b) (b+1)^3}-2}$$
but tangency implies on $\sqrt{(1-b) (b+1)^3}=0\;$ hence the solutions for tangency are $b = \pm 1$ etc.
• Didn't understand why L should be a tangent and why it should tend to a Jul 15 '18 at 11:40
• Interesting method, though please explain the concepts used Jul 15 '18 at 11:40
• @Abcd The problem $\min\max u=f(a,b)$ s. t. $a^2+b^2-1=0$ when $f(a,b)$ is regular, is defined at the tangency points for $u=f(a,b)$ and $a^2+b^2-1=0$ At those points we have $u = \{u_{\min},\cdots,u_{\max}\}$. This is the idea of Lagrange multipliers method. Jul 15 '18 at 13:43 | 2,281 | 5,345 | {"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": 1, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2021-49 | latest | en | 0.725455 |
https://www.physicsforums.com/threads/sr-thought-spacetime-interval-w-bounds-of-0-alpha-1.981955/ | 1,718,402,309,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861575.66/warc/CC-MAIN-20240614204739-20240614234739-00539.warc.gz | 835,266,133 | 18,440 | # SR Thought: Spacetime Interval w/ Bounds of ##0<\alpha<1##
• B
• benorin
I still wouldn't call it a time dilation factor - that's usually reserved for ##\gamma##. That's not to say it's wrong, just that I've not seen it used that way.I think I see what you're getting at. You're thinking of the classic light-clock in a train moving at constant velocity problem. In that case you do indeed get the time dilation equation because the time it takes for the light to go from one mirror to the other is longer for an observer in the lab than for an observer in the train. Take a look at the time dilation equation - it has the same form as your final equation, except with ##c## replaced by ##v##. But that's because we're dealing with a
benorin
Homework Helper
TL;DR Summary
Just a note, wonder where the rabbit hole leads
Suppose the time interval in the Lab frame is a multiple of the time interval in the Rocket frame ##\alpha \Delta t_L = \Delta t_R##, where ##0 < \alpha < 1## without loss of generality. Then the spacetime interval is
##\left( \Delta t_L\right) ^2-\left( \Delta x_L\right) ^2 = \left( \Delta t_R\right) ^2-\left( \Delta x_R\right) ^2\Rightarrow \left( \Delta t_L\right) ^2=\frac{\left( \Delta x_L\right) ^2 -\left( \Delta x_R\right) ^2}{1-\alpha^2}##
The thought ended abruptly there. I guess I just wanted to jot this down before I forgot, and hey, if you've got some good idea how to proceed from here: post it! Thanks!
Edit: something is off, like a minor error... but I can't put my finger on it. Should the bounds for be ##1< \alpha < \infty##?
Last edited:
The spacetime interval between which pair of events? You haven’t specified that, and until you do there’s no way of checking your calculation.
Remember that an event is a point in spacetime, not time. “The moment when the lab clock reads ##T## and the rocket clock reads ##T/\gamma##” is not an event. “The lab clock at this point in space reads ##T##” is an event. “The rocket clock at that point in space reads ##T/\gamma##” is a different event even if it happens at the same time.
So the first step is for you to write down the ##x## and ##t## coordinates of two events. You can use either the lab frame or the rocket frame to assign these coordinates, as long as you’re consistent and only use one frame or the other.
Once you’ve done that and you can calculate the spacetime interval between the two events, use the Lorentz transformations to find the coordinates of the two events in the other frame, and then check that you get the same interval using those coordinates.
Orodruin
As Nugatory says, this depends on the pair of events you are considering. You can make ##\alpha## anything between ##\pm\infty## depending on that choice.
I would guess that you are thinking of ##\alpha## as the time dilation factor, which would imply ##\alpha=\sqrt{1-v^2}## (i.e. your ##\alpha## is the reciprocal of Lorentz' ##\gamma##). If I've guessed correctly, then for your initial equation to hold you need ##\Delta x_R=0##. In that case your final result can be rearranged to read ##(\Delta x_L)^2=(v\Delta t_L)^2##, a restatement of the fact that the rocket moves at ##v## in the lab frame.
Last edited:
I was indeed as you suspected Ibix thinking along the lines of a rocket going with constant velocity through the Lab frame so that ##\Delta x_R =0 \Rightarrow \alpha = \sqrt{1-\left( \frac{\Delta x_L}{\Delta t_L}\right) ^2}## is what I derive from the OP. Now my question is doesn't ##\frac{\Delta x_L}{\Delta t_L}:=v##? But then I also worked out the moving light-clock problem and suspect this should rather be ##\frac{v}{c}## to give the the expected Lorentz equation (or time dilation equation, whatever?). What am I missing?
It's usually easier to work in units where ##c=1##, since then you can drop factors of ##c## that pop up all over the place. I did that without commenting - apologies. Your ##\alpha## is ##\sqrt{1-v^2/c^2}## if you don't work in such units, yes.
## What is the concept of "SR Thought: Spacetime Interval w/ Bounds of ##0<\alpha<1##"?
The concept of "SR Thought: Spacetime Interval w/ Bounds of ##0<\alpha<1##" is a mathematical expression used in special relativity to describe the distance between two events in spacetime. The ##0<\alpha<1## represents the bounds of the interval, which indicates that the interval is bounded and finite.
## How is the "SR Thought: Spacetime Interval w/ Bounds of ##0<\alpha<1##" calculated?
The "SR Thought: Spacetime Interval w/ Bounds of ##0<\alpha<1##" is calculated using the Minkowski metric, which is a mathematical formula that takes into account the dimensions of time and space. It is calculated by taking the square root of the difference between the squares of the time and space intervals.
## What is the significance of the bounds in the "SR Thought: Spacetime Interval w/ Bounds of ##0<\alpha<1##"?
The bounds in the "SR Thought: Spacetime Interval w/ Bounds of ##0<\alpha<1##" indicate that the interval is finite and cannot be negative. This is important in special relativity because it helps to define the causality of events and ensures that the interval is consistent with the principles of relativity.
## How does the "SR Thought: Spacetime Interval w/ Bounds of ##0<\alpha<1##" relate to the concept of spacetime?
The "SR Thought: Spacetime Interval w/ Bounds of ##0<\alpha<1##" is a fundamental concept in special relativity that helps to define the geometry of spacetime. It is used to measure the distance between two events in spacetime and is essential in understanding the relationship between time and space in the theory of relativity.
## Can the "SR Thought: Spacetime Interval w/ Bounds of ##0<\alpha<1##" have a value of zero?
No, the "SR Thought: Spacetime Interval w/ Bounds of ##0<\alpha<1##" cannot have a value of zero. This is because the interval is bounded and cannot be negative, so the smallest possible value is ##\alpha##, which is greater than zero. A value of zero would indicate that the two events are coincident, which is not possible in special relativity.
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2K | 1,714 | 6,620 | {"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} | 3.5625 | 4 | CC-MAIN-2024-26 | latest | en | 0.903851 |
https://community.glideapps.com/t/multi-user-calculator/14452/12 | 1,675,503,039,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500095.4/warc/CC-MAIN-20230204075436-20230204105436-00422.warc.gz | 202,187,891 | 6,814 | # Multi User Calculator
My goal is to have a simple gSheets based calculator that is publicly available and usable by multiple users simultaneously. I am open to different ways of doing this, but my approach so far is like this:
-Have 3 unique numerical input fields
-Have 1 numerical output field
The issue I am having is that:
1. Input from one user overrides the input from another user, so it can only be used by one user at a time.
2. The output field is very slow to update. It takes about 6 seconds to multiply 3 triple-digit values together on the Glide portal. Note that the output field in gSheets updates instantaneously.
Hi,
Take a look at this demo https://multicalc.glideapp.io and use it
Saludos
1 Like
Is there a way to do this without generating a new row each time the calculation is performed? The app may have several thousand users.
Yes, of course!
Sorry, I didn’t remember that this APP was using the old way/technique.
Now using the user-specific columns you can do it in the way you want.
Saludos
In the Basic Table component, how were you able to create those math functions? I would need to create a Trig function.
Using a Math column you can create basic math operations (addition, subtraction, multiplication and division).
Something like this
Create 2 new user-specific columns to save the values there and later, use a Math column to get your operation by using them.
Chao @Koenner
1 Like
2 Likes
Is there a way for the calculations to be done in gSheets without a ~6 second delay for the results to show on Glide. It seems like it would be preferable to do it that way, if you’re limited by what math functions are available. My calculator requires inverse tangent and a multi-variable equation, which is quite easy to do in gSheets.
There must be a sync back and forth to update it, so I don’t think there’s a way.
1 Like
Any thoughts on how to truncate a math value from 4 significant figures to 3 sig. figs? The value must be 3 sig. figs. and not just displaying 3 sig. figs. The function library really needs to grow!
Use 3 significant figures for your math column, then use a template column referencing that math column to lock it in.
Brilliant! Thank you
1 Like | 511 | 2,225 | {"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-2023-06 | latest | en | 0.891489 |
https://sonichours.com/how-much-is-121-pounds-in-kg/ | 1,721,161,897,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514789.44/warc/CC-MAIN-20240716183855-20240716213855-00167.warc.gz | 472,580,070 | 20,569 | General
How Much Is 121 Pounds in Kilograms?
If you’re wondering how much is 121 pounds in kilograms, you’ve come to the right place. Here you can find out how much 121 pounds is in kilograms and other common measurements. Read on to find out how to convert lb to kg, and how to find out how much is 121 kilograms in imperial units.
121 pounds in kilograms
If you’re interested in converting 121 pounds to kilograms, you’ve come to the right place. There are many ways to convert pounds to kilograms. Whether you want to use imperial measurements or historical mass-pounds, our conversion calculator can help. It’s not hard to convert pounds to kilograms and vice versa.
First, you’ll need to figure out how many kilograms 121 pounds are. A kilogram equals roughly 10cm3. Luckily, you can use a simple calculator or formula to find out how much 121 pounds weigh in kilograms. Just use the search form on the sidebar to find the answer to your question.
Another way to figure out how many kilograms 121 pounds weigh is to look up the conversion factor for pounds and kilograms. For example, 121 pounds equals 121kg. A conversion factor of four can be used to figure out a specific number. Once you’ve calculated the factor, you can then use the formula to find out how much 121 pounds is in kilograms.
To get the weight of a kilogram, multiply it by its mass. A kilogram is approximately the same as the same number of grams. To convert 121 pounds to kilograms, multiply its mass by two. For example, if you’re 5’10, 121 pounds will equal approximately 5.4 grams of mass.
Converting lb to kg
If you’re trying to figure out how to convert 121 pounds into kilograms, the first step is to figure out how many kilograms 121 pounds is. To do so, simply use a calculator or a formula. You can also use an online calculator. If you’d rather work with a formula, you can type in 121 lbs into the search field on the sidebar of this website. Once you’ve completed the search, a list of relevant posts will appear.
Using the formula below, you can determine how many kilograms 121 pounds is. You’ll find that 121 pounds is equivalent to 54 kg, which is 884 grams. The same formula applies to the other conversion factors. When using this formula, remember that some values may be rounded. The author of the website cannot be held responsible for inaccuracies and the information provided is for educational purposes only. It’s not a substitute for professional medical advice, nor is it intended for use in dangerous or potentially hazardous situations.
Kilograms are the basic unit of mass in the International System of Units (SI). It’s also known as the International Avoirdupois Pound and is most widely used in the United States. One Kilogram is equal to 16 avoirdupois ounces. In SI base units, one kilogram is equal to 0.453592 pounds. The unit’s symbol is “kg.” In addition to the metric system, kilograms are the only base unit with an SI prefix. This is the reason that 121 pounds are written as 121 kg.
Finding 121 pounds in kilograms
You might be wondering how to convert 121 pounds to kilograms. Well, luckily there is a quick and easy way to find out. One way is to use an online calculator. A simple calculator can be found at sites like Google. There, you can type in a search term such as “finding 121 pounds in kilograms”. You will then see a list of relevant posts.
There are many online calculators that will let you convert 121 pounds to kilograms. These calculators are easy to use and can be extremely accurate. The conversion factor is based on a formula. Be aware that some values may be rounded. Also, keep in mind that these calculators are not intended for risky uses. | 819 | 3,699 | {"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-2024-30 | latest | en | 0.886211 |
https://hubpages.com/technology/Know-Some-Facts-Before-Purchasing-a-Wind-Turbine-for-Your-Home | 1,516,457,781,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084889617.56/warc/CC-MAIN-20180120122736-20180120142736-00418.warc.gz | 679,386,711 | 18,380 | • »
• Technology»
• Renewable & Alternative Energy
# Know Some FAQs Before Purchasing a Wind Turbine for Your Home
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## How much cost of wind energy per kilowatt hour?
With the innovation of new technology, the price of wind energy has much lowere than ever before. In fact, in USA wind energy has cut down its price more than 90% relative to 1980. Since, wind energy is totally dependable on wind resources, the cost efficiency is calculated against how fast and how frequently the turbine is rotated. Higher turbine speed means more power and low price cost. Now, how to calculate the “average capacity” of wind energy? It’s simple, “average capacity” means the percentage of turbine power divided by the power of turbine if it blows all the time.
For home purpose the average installation cost of wind energy varies from \$30,000 to \$70,000 depending on your demand. However, if you need very tiny off grid wind turbine for example 1 KW that will cost you around \$4,000 to \$10,000. Many federal state government encourages people for installing wind turbines under a leasing program.
## How long turbine should I install for producing electricity?
Wind turbine comes with 20 watt to 100 KW power depending on your application. For residential purpose you should calculate the power consumption in your home before using it. Generally, The resident of the USA needs 5-KW power in their home for operating all the equipments. For that amount of electricity 18 foot diameter turbine will be alright for you. However, the size of the turbine depends on the total power you actually need. Typically, for residential purposes that range varies between 2 KW to 10 KW (12 feet to 25 feet diameter), the average speed of wind blow, the height of the turbine from the ground and so on. The manufacturer will provide you enough information that actually you need for generating exact power.
## What will happen if the wind stops blowing?
Wind turbines can be variable as the wind will not continually blowing to your system. Off-grid turbines can store power in the battery and can supply electricity at the critical moment. However, if you are within the national grid connection system, you will not notice the effect of wind on your turbine. Your utility will supply you the required electricity and, if any extra electricity is generated it will send back to the utility for supplying power to the other user.
## Quiz 1
view quiz statistics
## Is it necessary to take measurement of wind?
No, it is not necessary to take measurement of the wind, your supplier will measure wind blowing and will decide the type of turbine you require. If the wind blowing rate is average below than 12 mph, then you should think twice before purchasing it. How in this case the direction of wind is not mandatory. It is required if you live a busy city or irregular terrain. Request your supplier to take wind measurement as long as possible not less than three months because the more you take wind measurement the more you will get confident.
## How much land is required for installing wind turbines?
Industrial wind turbines are much bigger relative to the home turbines. Typically a 2-MW wind turbine requires around 116 feet long blades and that will cover 80 acre land. However, that does not mean you need 80 acres in total. You need only 1% of the total land and the rest of the land will be used for different purposes, for example farming, cattle ranching and so on. A wind turbine need as little as 1 acre per megawatt power.
## How can I maintain my wind turbine power system?
The wind turbine should be performed as little as 20 years. An expert professional will inspect the total system of your system throughout the whole year, whereas other small turbine can be maintained remotely from a computer.
## Why seashore is suitable for placing wind turbines?
Wind blowing from the seashore is more reliable than offshore. Offshore turbine are still not developed that mark and still they are expensive. If we install an offshore wind turbine than your total project might be failed completely. However, in coming years the offshore turbine will be available and the cost will deduced at a significant price.
## Quiz 2
view quiz statistics
Cast your vote for Wind energy
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5 | 881 | 4,307 | {"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-2018-05 | latest | en | 0.930159 |
https://max.berger.name/teaching/si04/project.html | 1,713,861,528,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818468.34/warc/CC-MAIN-20240423064231-20240423094231-00248.warc.gz | 345,946,909 | 3,212 | # CS3364 - Design and analysis of Algorithms, Summer I / 2004
## Final Project
Please send me which problem you picked by Friday, June 25, 2004, 10 am, or tell me in class.
This is due Wednesday, June 30, 2004, 10 am
### Background
You should now be proficent enough to solve any problem with the help of extra documentation.
### Problem
Pick any problem from the ACM Problemset Archive. Please note that the problems have a percent number, which gives a good indication of the difficulty level of this problem. The Problem you pick should be between 10% and 50% and should be online-judgable. Please pick a problem that is appropriate for you personal skill level (=current grade). If you want an A+ in this class, pick a problem that has at most 20%. To avoid possible cheating no two students should pick the same problem.
### Programming
Solve the problem. If possible, use C/C++/Java/Pascal, so that you can test your solution online. This time you may use all features of your language, etc., as long as they come with the base install of your compiler (NO extra libraries). If you uses C++ please see the notes on C++
Please send me your solution, along with any test-data you have used.
### Presentation
The presentations are going to be on wednesday and thursday in class. The order will be based on volunteers or randomized, so you should be ready on wednesday! You presentation should be about 10, but no more than 15 minutes long. You should:
• Shortly describe the problem (unix ls: given a list of names, output them alphabectically and top-down left-right)
• Describe which part caused the most trouble (Not the sorting, but the output on unix ls)
• Describe which algorithm you used to solve this problem. If time permits, give a short overview over the algorithm, if not, just something like "I used merge sort, which is a O(n log n) sort, I used Dijkstra, which searches for shortest path, etc."
For the presentation you may use transparencies, slides on the computer or the board.
Both parts (programming + presentation) are going to count about the same (8/10 or 9/9). Since the grades will be fixed by thursday, there will be no late arrangements.
### Submission
Submit the programming part eletronically, either by putting them on the unix accounts in the cs department in a folder that I can read, or by sending them via email to max@berger.name | 547 | 2,385 | {"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-2024-18 | latest | en | 0.912196 |
http://mss3.libraries.rutgers.edu/dlr/showfed.php?pid=rutgers-lib:27189 | 1,369,247,760,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368702298845/warc/CC-MAIN-20130516110458-00064-ip-10-60-113-184.ec2.internal.warc.gz | 179,469,272 | 6,343 | RUcore Resource Object
RUcore Resource Object
TitleTracing students' growing understanding of rational numbers
NameSchmeelk, Suzanna E. (author), Maher, Carolyn A. (chair), Alston, Alice (internal member), Powell, Arthur B (internal member), Steencken, Elena P (internal member), Gjone, Gunnar (outside member), Rutgers University, Graduate School of Education,
Degree Date2010-01
Date Created2010
SubjectMathematics Education, Mathematics--Study and teaching, Numbers, Rational, Fractions, Learning--Mathematical models
DescriptionThis research, a component of a year long National Science Foundation funded study, traces and documents how rational number ideas are built by students as they move from placing fractions on a line segment (finite concept) to placing fractions on an infinite number line (infinite concept). The evidence is supported by representations used by students to express their ideas, explanations given by students and student justifications about their reasoning. The study was guided by the following research questions: 1. What evidence, if any, is there of the students' understanding of the idea of fraction as number? 2. How do students extend their understanding of fraction ideas to rational numbers? 3. What representations do students use to express their fraction ideas and extend these ideas to rational numbers? The subjects consisted of a heterogeneous class of twenty-five, fourth grade (nine and early ten year old) students. Digitized videos, transcripts, student work, observation notes, and student overhead transparencies comprised the data from extended classroom sessions, videotaped with three cameras. The study gives evidence that the students built understanding of fraction ideas such as equivalence and extended these ideas to negative fractions and improper fractions. It also showed that students successfully ordered fractions on line segments, then number lines, after working out distinctions between operator and number ideas. Student ideas revealed in these sessions showed that they were comfortable and successful with basic fraction operations. Lively classroom discussions and arguments worked out obstacles in the placement of fractions on a number line. Engagement in discussions about fraction ideas and negative fractions extended to rational numbers to include improper fractions as students identified equivalent number names for fractions. In the active student-centered environment the students worked together on tasks and shared their personal representations of rational number ideas and density of the rationals. This study provides detailed evidence that students can build understanding of fraction as number and successfully make connections to extend their understanding of number, generating and interest and understanding of fraction ideas that generally are not made accessible to students of this age.
NoteEd.D.
NoteIncludes abstract
NoteIncludes bibliographical references
Noteby Suzanna E. Schmeelk
Genretheses
Persistent URLhttp://hdl.rutgers.edu/1782.2/rucore10001500001.ETD.000052898
LanguageEnglish
CollectionGraduate School of Education Electronic Theses and Dissertations
Organization NameRutgers, The State University of New Jersey
RightsThe author owns the copyright to this work.
Version 7.1 | 621 | 3,287 | {"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-2013-20 | latest | en | 0.911765 |
https://www.media4math.com/TEKS-3-2A | 1,680,194,772,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949355.52/warc/CC-MAIN-20230330163823-20230330193823-00608.warc.gz | 981,558,634 | 13,546 | ## TEKS-3-2A: Compose and decompose numbers up to 100,000 as a sum of so many ten thousands, so many thousands, so many hundreds, so many tens, and so many ones using objects, pictorial models, and numbers, including expanded notation as appropriate.
There are 174 resources.
Title Description Thumbnail Image Curriculum Nodes
## Math Examples Collection: Writing Decimals in Expanded Form
This collection aggregates all the math examples around the topic of Writing Decimals in Expanded Form. There are a total of 14 images. Numerical Expressions
## Math Examples Collection: Models of Numbers in Expanded Form
This collection aggregates all the math examples around the topic of Models of Numbers in Expanded Form. There are a total of 10 Math Examples. Place Value
## Math Examples Collection: Four-Digit Numbers in Expanded Form
This collection aggregates all the math examples around the topic of Four-Digit Numbers in Expanded Form. There are a total of 9 Math Examples. Numerical Expressions
## Math Examples Collection: Decomposing Numbers
This collection aggregates all the math examples around the topic of Decomposing Numbers. There are a total of 14 Math Examples. Numerical Expressions
## Math Examples Collection: Decomposing Fractions
This collection aggregates all the math examples around the topic of Decomposing Fractions. There are a total of 18 Math Examples. Fractions
## Closed Captioned Video: Place Value: Reading and Writing Decimals in Expanded Form
Closed Captioned Video: Place Value: Reading and Writing Decimals in Expanded Form
In this tutorial, students learn how to write decimals in expanded and standard form using place value techniques.
Place Value
## Closed Captioned Video: Place Value: Reading and Writing Whole Numbers in Expanded Form
Closed Captioned Video: Place Value: Reading and Writing Whole Numbers in Expanded Form
In this tutorial, students learn how to write whole numbers in expanded and standard form using place value techniques.<
Place Value
## Math Example--Numbers--Decomposing Numbers--Example 6
Math Example--Numbers--Decomposing Numbers--Example 6
This is part of a collection of math examples that focus on numbers and their properties.
Counting
## Math Example--Numbers--Decomposing Numbers--Example 7
Math Example--Numbers--Decomposing Numbers--Example 7
This is part of a collection of math examples that focus on numbers and their properties.
Counting
## Math Example--Numbers--Decomposing Numbers--Example 8
Math Example--Numbers--Decomposing Numbers--Example 8
This is part of a collection of math examples that focus on numbers and their properties.
Counting
## Math Example--Numbers--Decomposing Numbers--Example 9
Math Example--Numbers--Decomposing Numbers--Example 9
This is part of a collection of math examples that focus on numbers and their properties.
Counting
## Math Example--Numbers--Decomposing Numbers--Example 10
Math Example--Numbers--Decomposing Numbers--Example 10
This is part of a collection of math examples that focus on numbers and their properties.
Counting
## Math Example--Numbers--Decomposing Numbers--Example 11
Math Example--Numbers--Decomposing Numbers--Example 11
This is part of a collection of math examples that focus on numbers and their properties.
Counting
## Math Example--Numbers--Decomposing Numbers--Example 12
Math Example--Numbers--Decomposing Numbers--Example 12
This is part of a collection of math examples that focus on numbers and their properties.
Counting
## Math Example--Numbers--Decomposing Numbers--Example 13
Math Example--Numbers--Decomposing Numbers--Example 13
This is part of a collection of math examples that focus on numbers and their properties.
Counting
## Math Example--Numbers--Decomposing Numbers--Example 14
Math Example--Numbers--Decomposing Numbers--Example 14
This is part of a collection of math examples that focus on numbers and their properties.
Counting
## Math Example--Numbers--Writing Four-Digit Numbers in Expanded Form--Example 1
Math Example--Numbers--Writing Four-Digit Numbers in Expanded Form--Example 1
This is part of a collection of math examples that focus on numbers and their properties.
Place Value
## Math Example--Numbers--Writing Four-Digit Numbers in Expanded Form--Example 2
Math Example--Numbers--Writing Four-Digit Numbers in Expanded Form--Example 2
This is part of a collection of math examples that focus on numbers and their properties.
Place Value
## Math Example--Numbers--Writing Four-Digit Numbers in Expanded Form--Example 3
Math Example--Numbers--Writing Four-Digit Numbers in Expanded Form--Example 3
This is part of a collection of math examples that focus on numbers and their properties.
Place Value
## Math Example--Numbers--Writing Four-Digit Numbers in Expanded Form--Example 4
Math Example--Numbers--Writing Four-Digit Numbers in Expanded Form--Example 4
This is part of a collection of math examples that focus on numbers and their properties.
Place Value
## Math Example--Numbers--Writing Four-Digit Numbers in Expanded Form--Example 5
Math Example--Numbers--Writing Four-Digit Numbers in Expanded Form--Example 5
This is part of a collection of math examples that focus on numbers and their properties.
Place Value
## Math Example--Numbers--Writing Four-Digit Numbers in Expanded Form--Example 6
Math Example--Numbers--Writing Four-Digit Numbers in Expanded Form--Example 6
This is part of a collection of math examples that focus on numbers and their properties.
Place Value
## Math Example--Numbers--Writing Four-Digit Numbers in Expanded Form--Example 7
Math Example--Numbers--Writing Four-Digit Numbers in Expanded Form--Example 7
This is part of a collection of math examples that focus on numbers and their properties.
Place Value
## Math Example--Numbers--Writing Four-Digit Numbers in Expanded Form--Example 8
Math Example--Numbers--Writing Four-Digit Numbers in Expanded Form--Example 8
This is part of a collection of math examples that focus on numbers and their properties.
Place Value
## Math Example--Numbers--Writing Four-Digit Numbers in Expanded Form--Example 9
Math Example--Numbers--Writing Four-Digit Numbers in Expanded Form--Example 9
This is part of a collection of math examples that focus on numbers and their properties.
Place Value | 1,359 | 6,338 | {"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-2023-14 | latest | en | 0.818173 |
https://virtualpsychcentre.com/question-how-much-should-i-pay-for-mortgage/ | 1,708,673,900,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474361.75/warc/CC-MAIN-20240223053503-20240223083503-00121.warc.gz | 621,038,037 | 12,812 | ## How much of your income should you spend on a mortgage?
The 28% rule states that you should spend 28% or less of your monthly gross income on your mortgage payment (e.g. principal, interest, taxes and insurance). To determine how much you can afford using this rule, multiply your monthly gross income by 28%.
## What is the 28 36 rule?
One way to decide how much of your income should go toward your mortgage is to use the 28/36 rule. According to this rule, your mortgage payment shouldn’t be more than 28% of your monthly pre-tax income and 36% of your total debt. This is also known as the debt-to-income (DTI) ratio.
## How much of a mortgage can I afford based on my salary?
The general rule is that you can afford a mortgage that is 2x to 2.5x your gross income. Total monthly mortgage payments are typically made up of four components: principal, interest, taxes, and insurance (collectively known as PITI).
## How much mortgage can I get for 90000 salary?
For instance, if your net salary is Rs. 55,000, you will be eligible for a loan of approximately Rs 33 lakhs.
How to calculate your home loan eligibility?
Net Monthly Income (Rs.)Home Loan Amount (Rs.)
60,00046,43,370
70,00054,81,756
80,00063,20,142
90,00071,58,529
## How much house can I afford if I make \$40000 a year?
3. The 36% Rule
Gross Income28% of Monthly Gross Income36% of Monthly Gross Income
\$20,000\$467\$600
\$30,000\$700\$900
\$40,000\$933\$1,200
\$50,000\$1,167\$1,500
•
Mar 17, 2022
## Can I buy a house if I make 45000 a year?
It’s definitely possible to buy a house on a \$50K salary. For many borrowers, low-down-payment loans and down payment assistance programs are putting homeownership within reach.
## What house can I afford on 70k a year?
Personal finance experts recommend spending between 25% and 33% of your gross monthly income on housing. Someone who earns \$70,000 a year will make about \$5,800 a month before taxes.
## How much house can I afford 300k salary?
Before you get into determining if you can afford monthly payments, figure out how much money you have available now for up-front costs of a home purchase. These include: A down payment: You should have a down payment equal to 20% of your home’s value. This means that to afford a \$300,000 house, you’d need \$60,000.
## What credit score is needed to buy a \$250000 house?
But because credit scores estimate the risk that you won’t repay the loan, lenders will reward a higher score with more choices and lower interest rates. For most loan types, the credit score needed to buy a house is at least 620.
## Can I buy a house with 780 credit score?
Your FICO® Score falls within a range, from 740 to 799, that may be considered Very Good. A 780 FICO® Score is above the average credit score. Borrowers with scores in the Very Good range typically qualify for lenders’ better interest rates and product offers.
## What house can I afford on 80k a year?
So, if you make \$80,000 a year, you should be looking at homes priced between \$240,000 to \$320,000. You can further limit this range by figuring out a comfortable monthly mortgage payment. To do this, take your monthly after-tax income, subtract all current debt payments and then multiply that number by 25%.
## How much can I borrow with a 800 credit score?
The average mortgage loan amount for consumers with Exceptional credit scores is \$208,977. People with FICO® Scores of 800 have an average auto-loan debt of \$18,764.
## Is 760 a good credit score?
Your score falls within the range of scores, from 740 to 799, that is considered Very Good. A 760 FICO® Score is above the average credit score. Consumers in this range may qualify for better interest rates from lenders. 25% of all consumers have FICO® Scores in the Very Good range.
## How much can I borrow with a 680 credit score?
Based on Bankrate’s national interest rate survey, a consumer with a FICO score between 680 and 699 trying to borrow \$300,000 in early April would have qualified for a 3.709 percent rate on a 30-year fixed mortgage, resulting in a \$1,382 monthly payment.
## Is a 900 credit score good?
The best-known range of FICO scores is 300 to 850. Anything above 670 is generally considered to be good. FICO also offers industry-specific FICO scores, such as for credit cards or auto loans, which can range from 250 to 900.
## Is a 900 credit score possible?
A credit score of 900 is either not possible or not very relevant. The number you should really focus on is 800. On the standard 300-850 range used by FICO and VantageScore, a credit score of 800+ is considered “perfect.” That’s because higher scores won’t really save you any money.
## Is 720 a good credit score?
A 720 FICO® Score is Good, but by raising your score into the Very Good range, you could qualify for lower interest rates and better borrowing terms. A great way to get started is to get your free credit report from Experian and check your credit score to find out the specific factors that impact your score the most.
## Is 862 a good credit score?
An 862 credit score is excellent.
## Is 874 a good credit score?
We provide a score from between 0-999 and consider a ‘good’ score to be anywhere between 881 and 960, with ‘fair’ or average between 721 and 880.
## What is the average credit score by age?
And for the first time, the average FICO® Score of Generation X (ages 41 through 56 in 2021) is in the 700s.
The Average FICO® Score Increased Among All Generations.
Average FICO® Score by Generation
Generation20202021
Baby boomers (57-75)736740
Generation X (41-56)698705
Millennials (25-40)679686
•
Feb 22, 2022
## What is a perfect credit score?
A perfect credit score of 850 is hard to get, but an excellent credit score is more achievable. If you want to get the best credit cards, mortgages and competitive loan rates — which can save you money over time — excellent credit can help you qualify. “Excellent” is the highest tier of credit scores you can have.
## How can I lift my credit score?
Steps to Improve Your Credit Scores
1. Build Your Credit File. …
2. Don’t Miss Payments. …
3. Catch Up On Past-Due Accounts. …
4. Pay Down Revolving Account Balances. …
5. Limit How Often You Apply for New Accounts. | 1,578 | 6,270 | {"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} | 2.84375 | 3 | CC-MAIN-2024-10 | latest | en | 0.946986 |
https://en.wikipedia.org/wiki/Pr%C3%BCfer_group | 1,550,879,462,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550249406966.99/warc/CC-MAIN-20190222220601-20190223002601-00536.warc.gz | 551,227,492 | 15,727 | # Prüfer group
The Prüfer 2-group with presentation <gn: gn+12 = gn, g12 = e>, illustrated as a subgroup of the unit circle in the complex plane.
In mathematics, specifically in group theory, the Prüfer p-group or the p-quasicyclic group or p-group, Z(p), for a prime number p is the unique p-group in which every element has p different p-th roots.
The Prüfer p-groups are countable abelian groups which are important in the classification of infinite abelian groups: they (along with the group of rational numbers) form the smallest building blocks of all divisible groups.
The groups are named after Heinz Prüfer, a German mathematician of the early 20th century.
## Constructions of Z(p∞)
The Prüfer p-group may be identified with the subgroup of the circle group, U(1), consisting of all pn-th roots of unity as n ranges over all non-negative integers:
${\displaystyle \mathbf {Z} (p^{\infty })=\{\exp(2\pi im/p^{n})\mid 0\leq m
The group operation here is the multiplication of complex numbers.
There is a presentation
${\displaystyle \mathbf {Z} (p^{\infty })=\langle \,g_{1},g_{2},g_{3},\ldots \mid g_{1}^{p}=1,g_{2}^{p}=g_{1},g_{3}^{p}=g_{2},\dots \,\rangle .}$
Here, the group operation in Z(p) is written as multiplication.
Alternatively and equivalently, the Prüfer p-group may be defined as the Sylow p-subgroup of the quotient group Q/Z, consisting of those elements whose order is a power of p:
${\displaystyle \mathbf {Z} (p^{\infty })=\mathbf {Z} [1/p]/\mathbf {Z} }$
(where Z[1/p] denotes the group of all rational numbers whose denominator is a power of p, using addition of rational numbers as group operation).
For each natural number n, consider the quotient group Z/pnZ and the homomorphism Z/pnZZ/pn+1Z induced by multiplication by p. The direct limit of this system is Z(p):
${\displaystyle \mathbf {Z} (p^{\infty })=\varinjlim \mathbf {Z} /p^{n}\mathbf {Z} .}$
We can also write
${\displaystyle \mathbf {Z} (p^{\infty })=\mathbf {Q} _{p}/\mathbf {Z} _{p}}$
where Qp denotes the additive group of p-adic numbers and Zp is the subgroup of p-adic integers.
## Properties
The complete list of subgroups of the Prüfer p-group Z(p) = Z[1/p]/Z is:
${\displaystyle 0\subset \left({1 \over p}\mathbf {Z} \right)/\mathbf {Z} \subset \left({1 \over p^{2}}\mathbf {Z} \right)/\mathbf {Z} \subset \left({1 \over p^{3}}\mathbf {Z} \right)/\mathbf {Z} \subset \cdots \subset \mathbf {Z} (p^{\infty })}$
(Here ${\displaystyle \left({1 \over p^{n}}\mathbf {Z} \right)/\mathbf {Z} }$ is a cyclic subgroup of Z(p) with pn elements; it contains precisely those elements of Z(p) whose order divides pn and corresponds to the set of pn-th roots of unity.) The Prüfer p-groups are the only infinite groups whose subgroups are totally ordered by inclusion. This sequence of inclusions expresses the Prüfer p-group as the direct limit of its finite subgroups. As there is no maximal subgroup of a Prüfer p-group, it is its own Frattini subgroup.
Given this list of subgroups, it is clear that the Prüfer p-groups are indecomposable (cannot be written as a direct sum of proper subgroups). More is true: the Prüfer p-groups are subdirectly irreducible. An abelian group is subdirectly irreducible if and only if it is isomorphic to a finite cyclic p-group or to a Prüfer group.
The Prüfer p-group is the unique infinite p-group which is locally cyclic (every finite set of elements generates a cyclic group). As seen above, all proper subgroups of Z(p) are finite. The Prüfer p-groups are the only infinite abelian groups with this property.[1]
The Prüfer p-groups are divisible. They play an important role in the classification of divisible groups; along with the rational numbers they are the simplest divisible groups. More precisely: an abelian group is divisible if and only if it is the direct sum of a (possibly infinite) number of copies of Q and (possibly infinite) numbers of copies of Z(p) for every prime p. The (cardinal) numbers of copies of Q and Z(p) that are used in this direct sum determine the divisible group up to isomorphism.[2]
As an abelian group (that is, as a Z-module), Z(p) is Artinian but not Noetherian.[3] It can thus be used as a counterexample against the idea that every Artinian module is Noetherian (whereas every Artinian ring is Noetherian).
The endomorphism ring of Z(p) is isomorphic to the ring of p-adic integers Zp.[4]
In the theory of locally compact topological groups the Prüfer p-group (endowed with the discrete topology) is the Pontryagin dual of the compact group of p-adic integers, and the group of p-adic integers is the Pontryagin dual of the Prüfer p-group.[5]
• p-adic integers, which can be defined as the inverse limit of the finite subgroups of the Prüfer p-group.
• Dyadic rational, rational numbers of the form a/2b. The Prüfer 2-group can be viewed as the dyadic rationals modulo 1.
## Notes
1. ^ See Vil'yams (2001)
2. ^ See Kaplansky (1965) | 1,375 | 4,938 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 7, "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} | 3.890625 | 4 | CC-MAIN-2019-09 | latest | en | 0.862738 |
https://aprove.informatik.rwth-aachen.de/eval/JAR06/JAR_TERM/TRS/HofWald/2.trs.Thm26:POLO_7172_DP:OLD.html.lzma | 1,718,896,460,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861957.99/warc/CC-MAIN-20240620141245-20240620171245-00854.warc.gz | 86,151,201 | 1,812 | Term Rewriting System R:
[x]
f(a, f(f(a, a), x)) -> f(f(a, a), f(a, f(a, x)))
Termination of R to be shown.
` R`
` ↳Dependency Pair Analysis`
R contains the following Dependency Pairs:
F(a, f(f(a, a), x)) -> F(f(a, a), f(a, f(a, x)))
F(a, f(f(a, a), x)) -> F(a, f(a, x))
F(a, f(f(a, a), x)) -> F(a, x)
Furthermore, R contains one SCC.
` R`
` ↳DPs`
` →DP Problem 1`
` ↳Polynomial Ordering`
Dependency Pairs:
F(a, f(f(a, a), x)) -> F(a, x)
F(a, f(f(a, a), x)) -> F(a, f(a, x))
F(a, f(f(a, a), x)) -> F(f(a, a), f(a, f(a, x)))
Rule:
f(a, f(f(a, a), x)) -> f(f(a, a), f(a, f(a, x)))
The following dependency pair can be strictly oriented:
F(a, f(f(a, a), x)) -> F(f(a, a), f(a, f(a, x)))
Additionally, the following usable rule w.r.t. to the implicit AFS can be oriented:
f(a, f(f(a, a), x)) -> f(f(a, a), f(a, f(a, x)))
Used ordering: Polynomial ordering with Polynomial interpretation:
POL(a) = 1 POL(f(x1, x2)) = 0 POL(F(x1, x2)) = x1
resulting in one new DP problem.
` R`
` ↳DPs`
` →DP Problem 1`
` ↳Polo`
` →DP Problem 2`
` ↳Narrowing Transformation`
Dependency Pairs:
F(a, f(f(a, a), x)) -> F(a, x)
F(a, f(f(a, a), x)) -> F(a, f(a, x))
Rule:
f(a, f(f(a, a), x)) -> f(f(a, a), f(a, f(a, x)))
On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule
F(a, f(f(a, a), x)) -> F(a, f(a, x))
one new Dependency Pair is created:
F(a, f(f(a, a), f(f(a, a), x''))) -> F(a, f(f(a, a), f(a, f(a, x''))))
The transformation is resulting in one new DP problem:
` R`
` ↳DPs`
` →DP Problem 1`
` ↳Polo`
` →DP Problem 2`
` ↳Nar`
` ...`
` →DP Problem 3`
` ↳Remaining Obligation(s)`
The following remains to be proven:
Dependency Pairs:
F(a, f(f(a, a), f(f(a, a), x''))) -> F(a, f(f(a, a), f(a, f(a, x''))))
F(a, f(f(a, a), x)) -> F(a, x)
Rule:
f(a, f(f(a, a), x)) -> f(f(a, a), f(a, f(a, x)))
Termination of R could not be shown.
Duration:
0:00 minutes | 766 | 2,071 | {"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} | 3.5625 | 4 | CC-MAIN-2024-26 | latest | en | 0.660574 |
vvella.blogspot.com | 1,532,353,296,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676596463.91/warc/CC-MAIN-20180723125929-20180723145929-00355.warc.gz | 380,514,154 | 21,422 | ### Swapping two variables without using a third
I came across an interesting algorithm ... the normal way how programmers swap values between two variables is by using a temporary third variable ... temp = x, x = y, y = temp. An interesting trick can actually do the same process without having to use a third variable, and this is done by using the XOR operator. Below I am showing how this can be done in java:
public static void main(String[] args) {
int x = 15;
int y = 20;
System.out.println(x+" "+y);
x ^= y;
y ^= x;
x ^= y;
System.out.println(x+" "+y);
}
Try it out !
Andre Genovese said…
Aw Vince,
just stumbled upon your blog - the variable-swapping algorithm is interesting!
Have been thinking about converting to bloggism myself...I just need a little nudge :)
See you, Andre
### Interfacing C# .Net and R - Integrating the best of both worlds
In specific software areas like in quantitative finance or else in other mathematical domains, data centric programming typically requires a good balance between three requirements - (1) a solid platform with rich mathematical/statistical functionality (2) having an easy to use, contemporary, programming environment which permits easy and flexible front end code development and (3) an easy to use interface between the two environments.
In this artcile I am going to explain how such a balance can be attained by using two of the best products in their specific worlds - using the rich R library as the mathematical/statistical component but then interfacing with C# for the front end application design. As an interfacing option I banked on using R (D)COM which provides an easy to use interfacing method which keeps you away from spending hours identifying interfacing problems.
The software required for this tutorial is the following: 1. R software (download from here) 2. R (D)COM Interface (dow…
### Interfacing C# .Net and R - Integrating the best of both worlds (Part 3) - Trader Desk Example
In this third and final part of the series (Part 1, Part 2), I am going to continue shaping the examples in the previous posts by quickly building a small application - a simple Trading Desk application. In the example I will use the same C# and R interface method together with a specific Quantitative Financial Modelling R library called Quantmod.
Example Objective In the example the objective is to build a C# Web Form which acts as the main application controller that captures various user option parameters with the main functions being:
1. To automatically kick off an R routine to download data from Yahoo Finance. 2. To select specific dates of interest 3. To add or udpate different charts 4. To add different chart indicators
5. To calculate Period Return Statistics
Most of these functions are provided through the set of R functions exposed by the Quantmod R library.
Additions under the hood
Below I am showing some of the salient additions done to the code over and above the code in the pr…
### Interfacing C# .Net and R - Integrating the best of both worlds (Part 2)
This post is a continuation from the previous post (Part I) focusing on interfacing C# with R using the R (D)COM. In this post I am going to enhance my previous exercise by creating a Facade .Net Class which facilitates access to specific functions in R.
Creating the R Facade Class
Creating a Facade Class (or a set of .Net classes) which acts as a .Net wrapper to R functions greatly facilitate the use of R functions and their integration within the .Net programming environment. Below I am showing an excerpt from the class RFacade that I have created in this example.
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using StatConnectorCommonLib;
using STATCONNECTORSRVLib;
using System.Runtime.InteropServices;
namespace R
{
class RFacade : IDisposable
{
private StatConnector rconn;
private bool disposed = false; | 849 | 3,919 | {"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-2018-30 | latest | en | 0.878392 |
https://onlinejudge.org/board/viewtopic.php?f=23&t=6425&p=7146 | 1,591,041,719,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347419593.76/warc/CC-MAIN-20200601180335-20200601210335-00168.warc.gz | 479,852,132 | 10,891 | ## 10407 - Simple division
All about problems in Volume 104. If there is a thread about your problem, please use it. If not, create one with its number in the subject.
Moderator: Board moderators
yahoo
Learning poster
Posts: 93
Joined: Tue Apr 23, 2002 9:55 am
### 10407 - Simple division
I can't understand why my this code got runtime error. Can anyobody explain where i am doing wrong.Here is my code:
#include<stdio.h>
#define N 3000
main()
{
int tmp,a[N],i,j,r,hi,lo,n,m,b[N];
while(1)
{
n=0;
while(1)
{
scanf("%d",&a[n]);
if(a[n]==0) break;
n++;
}
if(!n) break;
for(i=0;i<n-1;i++)
{
m=i;
for(j=i+1;j<n;j++)
if(a[m]>a[j])
m=j;
tmp=a[m];
a[m]=a[i];
a[i]=tmp;
}
for(i=0;i<n-1;i++)
b[i]=a[i+1]-a[i];
lo=b[0];
for(i=1;i<n-1;i++)
if(b[i]<lo)
lo=b[i];
hi=b[0];
for(i=1;i<n-1;i++)
if(b[i]>hi)
hi=b[i];
int gcd=b[0];
for(i=1;i<n-1;i++)
{
if(b[i]>gcd) { hi=b[i];lo=gcd;}
else{hi=gcd;lo=b[i];}
while (1)
{
r = hi%lo;
if (r!=0)
{
hi = lo;
lo = r;
}
else {gcd=lo;break;}
}
}
printf("%d\n",lo);
}
return 0;
}
:oops:
abc
New poster
Posts: 15
Joined: Sun Dec 15, 2002 2:51 pm
How do you solve this? I don't get it
yahoo
Learning poster
Posts: 93
Joined: Tue Apr 23, 2002 9:55 am
Thanks for your reply. I got it accepted after recoding it. My algorithm was wrong here. Thanks again.
anupam
A great helper
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Joined: Wed Aug 28, 2002 6:45 pm
Contact:
actually yahoo,
your method donot work for repeated input. try it.
--anupam
"Everything should be made simple, but not always simpler"
anupam
A great helper
Posts: 405
Joined: Wed Aug 28, 2002 6:45 pm
Contact:
again,
there may be problem about the array size.
check it.
--thanks.
"Everything should be made simple, but not always simpler"
b3yours3lf
New poster
Posts: 44
Joined: Wed Aug 14, 2002 3:02 am
### 10407-Simple Division
I always get TLE.
Can I use long int for this problem or I must use string?
saiqbal
New poster
Posts: 36
Joined: Wed Aug 07, 2002 4:52 pm
Location: Dhaka, Bangladesh
Contact:
i used int for this problem.
-sohel
..
A great helper
Posts: 454
Joined: Thu Oct 18, 2001 2:00 am
Location: Hong Kong
### 10407
I solved it by 0.082s, but I see there are many solutions of 0.002s.
Now I solve it by finding some possible values of d, and then test each value for d.....Is there some way to solve it in O(nlgn) or even O(n)?
My signature:
• Please make discussion about the algorithm BRFORE posting source code.
We can learn much more in discussion than reading source code.
• I HATE testing account.
• Don't send me source code for debug.
makbet
New poster
Posts: 44
Joined: Tue Nov 26, 2002 11:01 pm
Location: Wejherowo
It's very easy, once you know it :
if a mod d = b mod d <=> d divides (a-b). Now the problem gets simple
..
A great helper
Posts: 454
Joined: Thu Oct 18, 2001 2:00 am
Location: Hong Kong
Yes, I know that,
so what I do is, find the minimum (but positive) pairwise difference (diff), then test each factor of diff to check if all the input values get the same remainder r.
But it is slower than others, what do I miss??
My signature:
• Please make discussion about the algorithm BRFORE posting source code.
We can learn much more in discussion than reading source code.
• I HATE testing account.
• Don't send me source code for debug.
cytse
Learning poster
Posts: 67
Joined: Mon Sep 16, 2002 2:47 pm
Location: Hong Kong
Contact:
why do you consider the minimum one only?
the most important point here is that d divides ALL pairwise-differences.
..
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Joined: Thu Oct 18, 2001 2:00 am
Location: Hong Kong
cytse wrote: the most important point here is that d divides ALL pairwise-differences.
As you say, d divideds ALL pairwise-difference, so d is a factor of any pairwise-difference. Then d is also a factor of the min. pairwise-difference, so d <= min. pairwise-difference. It is safe to only consider the factors of min. pairwise difference.
My signature:
• Please make discussion about the algorithm BRFORE posting source code.
We can learn much more in discussion than reading source code.
• I HATE testing account.
• Don't send me source code for debug.
makbet
New poster
Posts: 44
Joined: Tue Nov 26, 2002 11:01 pm
Location: Wejherowo
I don't know what you guys are talking about: you calculate differences of consecutive numbers and then calculate gcd of those. For examle:
17 5 9 3 99 55
differences:
12 4 6 96 44
nwd of those: 2, which is the correct anwser.
..
A great helper
Posts: 454
Joined: Thu Oct 18, 2001 2:00 am
Location: Hong Kong
Oh~~~
I get it, thanks a lot!
My signature:
• Please make discussion about the algorithm BRFORE posting source code.
We can learn much more in discussion than reading source code.
• I HATE testing account.
• Don't send me source code for debug.
Riyad
Experienced poster
Posts: 131
Joined: Thu Aug 14, 2003 10:23 pm
Location: BUET
Contact:
### 10407,WA why ?????????????????
Hi,
My Algorithm is very simple for this problem . i calculated the differences between all the successive numbers and stored them in an array of integer . than calculated the GCD of all the number . which should be the answer.
lets take an example , the sequence of number be
a b c d e f g h i j
i calculated abs(a-b),abs(b-c),abs(c-d),abs(d-e),........... and stored them in an array .
than i calculated the GCD s of them like this ..
a= it is the int array where i stored the diffences
x=GCD(a[0],a[1])
for(i=2;i<index;i++){
x=GCD(x,a);
}
x is the ans .
What is wrong with my algorithm and code .
plizzzzzzzzzz help.
Riyad
HOLD ME NOW ,, I AM 6 FEET FROM THE EDGE AND I AM THINKIN.. MAY BE SIX FEET IS SO FAR DOWN | 1,700 | 5,583 | {"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} | 3.109375 | 3 | CC-MAIN-2020-24 | latest | en | 0.762828 |
http://www.ehow.co.uk/how_5021957_calculate-electric-motor-torque.html | 1,490,257,746,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218186841.66/warc/CC-MAIN-20170322212946-00064-ip-10-233-31-227.ec2.internal.warc.gz | 518,806,287 | 17,006 | # How to calculate electric motor torque
Written by g.k. bayne
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Calculating electric motor torque can be a useful skill for estimating the potential of work a motor can perform. Although not listed on the motor's nameplate data tag, the torque value can be figured out with the information provided. By applying some basic equations you can calculate the electric motor torque for most electrical devices.
Skill level:
Easy
### Things you need
• Equations:
• Watts = volts * amps
• 1HP (horsepower) = 746 watts
• Torque (T) = ((HP(horsepower) / RPM (rotations per minute))) * 63,025 (constant)
## Instructions
1. 1
Find the wattage from a motor that has the following nameplate data on its metal tag; 120 Volts at 10 amperes with a 3600-RPM. From the formula above we see that watts is equal to volts times amperes. Plugging in the numbers, we have 120 volts times 10 amperes is equal to 1200 watts.
2. 2
Calculate the horsepower that a 1200-watt motor can deliver. 1 HP is equal to 746 watts. We can then divide 1200 watts by 746 watts per HP and the answer is 1.6 HP.
3. 3
Understand that torque is described as the amount of “force” from any given distance from that force's rotational centre point. In this case the centre point of force will be the exact centre of the round motor shaft. Torque values are given in either inch-ounces (in-ozs) or foot-pounds (ft-lb). In the formula described above, the torque value formula will be described in ft-lbs. You can think of 1 ft-lb as the amount of “force” that is placed on an object that has an “arm” 1 foot long and a 1-pound weight hanging from the end of that 1-foot arm.
4. 4
Find the amount of Torque that the 1.6 Hp motor delivers at 3600 RPM. Taking the torque formula one step at a time, divide the HP of 1.6 by the RPM of 3600. The answer will be 0.00044. This is a very small number. Next multiply 0.00044 times the constant of 63,025. The final Torque value of the motor is 28.16 ft-lbs of Torque. This is equivalent to a 28 and 1/8-pound weight hanging on the end of a 1-foot long arm or a 1-pound weight hanging on the end of a 28 foot 2 inch long arm.
#### Tips and warnings
• Remember that these calculations are theoretical only and do not take into consideration any form of friction or real world applications.
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• Most recent | 652 | 2,464 | {"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} | 4.28125 | 4 | CC-MAIN-2017-13 | longest | en | 0.873495 |
http://www.cypress.com/forum/proc-ble/proc-ble-power-calculation | 1,477,181,726,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988719079.39/warc/CC-MAIN-20161020183839-00029-ip-10-171-6-4.ec2.internal.warc.gz | 390,708,628 | 25,310 | # Cypress Developer CommunityTM
## PRoC BLE Power calculation
Summary: 4 Replies, Latest post by roit on 12 Jul 2015 11:40 PM PDT
User
8 posts
I want to calculate minimum typical power needed for PRoC to be operate. For this I can list application wise needed power. i.e. power needed for BLE working, power needed for CPU in active mode. In this way I can plan for optimal use of battery.
For this I have searched datasheet and AN92584 app note.
I have following doubts/quarries :-
1. On CYBL10X6X Family Datasheet :-
a. On page 1 it is mentioned that "TX current: 15.6 mA at 0 dBm ■ RX current: 16.4 mA" , What is meant by that? , It is at what voltage? Is this amount of current needed for BLE Tx and Rx? Then why power calculator tool mentioned in app note display different current?
b. table 5 :- Active 850 μA + 260 μA per MHz[1], At what vdd? Is it needed for CPU computation only without BLE?
2. Power calculator tool gives typical current 27uA and 17uA for advertisement and connection respectively? What is the meaning of that current? In this case what is Vdd?
Generally I asked for Vdd, because current needed is depends on Vin.
User
7646 posts
1. On CYBL10X6X Family Datasheet :-
a. On page 1 it is mentioned that "TX current: 15.6 mA at 0 dBm ■ RX current: 16.4 mA" , What is meant by that? , It is at what voltage? Is this amount of current needed for BLE Tx and Rx? Then why power calculator tool mentioned in app note display different current?
b. table 5 :- Active 850 μA + 260 μA per MHz[1], At what vdd? Is it needed for CPU computation only without BLE?
There is an extensive table of Vdd and Power consumption in the PRoC datasheet. It shows
in pinout the seperate pins for the Radio (Vddr) and the general Digital (Vddd).
2. Power calculator tool gives typical current 27uA and 17uA for advertisement and connection respectively? What is the meaning of that current? In this case what is Vdd?
Vdd is the operating voltage applied to PROC. That spec is in its datasheet.
Regards, Dana.
User
8 posts
@ Dana,
User
7646 posts
The datasheet is missing qualifier notes as you noticed. I would file a CASE on this to get
To create a technical or issue case at Cypress -
www.cypress.com
“Support”
“Technical Support”
“Create a Case”
You have to be registered on Cypress web site first.
Then post results back here for benefit of all.
Regards, Dana.
Cypress Employee
221 posts
As Dana suggested, please raise a CyLink case for any issue that you see in the datasheets,
As for your questions, I would like to reply to some of them to best of my extent.
TX current of 15.6 mA and RX current of 16.4 mA (at 0 dBm) are the peak currents of the BLE radio in PRoC device. These peak currents will be observed whenever there is an BLE radio activity. For the rest of the time, these currents are not present. Also, as these are momentary currents, they are responsible for only fraction of the overall current requirements, depending upon the interval you choose.
As for the voltage, the device has an internal translator that drops the voltage from VDDR to 1.9V. Thus, the current values (only for radio) is at 1.9V. The other part of the device operates at the input voltage.
The current you see as part of power calculator calculations are averaged currents. As you know, all the BLE activities happen over some interval. If it is advertising, it happens every advertising interval set. If it is in connection, the interval == connection interval. The current values are averaged over 1 second interval, which is common for various low power BLE devices. In this 1 second, the device is active for only few milliseconds and in deep-sleep for the rest of the time. As the deep sleep current is ~1.3 uA, the averaged current (including the BLE TX and RX) comes about 17 uA for connection and 27 uA for advertisement (advertisement happens over three BLE channels, that is why it is more). | 973 | 3,928 | {"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-2016-44 | latest | en | 0.906724 |
https://www.educationquizzes.com/ks3/maths/level-7-8-algebra---simultaneous-equations/ | 1,718,936,452,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862032.71/warc/CC-MAIN-20240620235751-20240621025751-00819.warc.gz | 638,776,119 | 10,340 | UK USIndia
Every Question Helps You Learn
Can you get ten out of ten in this maths quiz?
# Level 7-8 Algebra - Simultaneous Equations
With all the work you've done on algebra in KS3 Maths, you will by now be familiar with equations. One type you will have come across are the simultaneous equations.
Simultaneous equations have the same values. You should be able to recognise a problem that is based on simultaneous equations – it will have two equations, and include two different variables. When a variable is the subject of one of the equations (it will start with y =… or x =...) substitute that expression into the second equation. This will produce one equation with one unknown, which can be solved. Substitute this value back into the first equation to find the second variable.
Simultaneous equations sound a bit daunting! The subject is not as fearsome as the title suggests. Play the following quiz a few times and read the helpful comments - you will soon find simultaneous equations are a doddle!
1.
Which of the following equations means exactly the same as 2x + 3y = 13?
4x + 6y = 26
6x + 4y = 26
6x + 6y = 26
6x + 6y = 13
The correct answer is double the original expression. Each term in the expression has been multiplied by two and therefore 2x becomes 4x, 3y becomes 6y and 13 becomes 26
2.
Which of the following equations means exactly the same as 7x - 5y = 17?
12x - 12y = 51
21x - 15y = 51
21x + 15y = 51
2x - 5y = 51
Here we took the original expression and multiplied each term by three. As long as we multiply each term by the SAME number, the letters in the resulting expression will always have the same value as those in the original expression
3.
Equation 1 is 2x + 3y = 19. Equation 2 is 3x + 6y = 30. How would we 'balance' one of the terms?
Add the terms in Equation 1 to the terms in Equation 2
Deduct the terms in Equation 1 from Equation 2
Multiply the terms in Equation 1 by 2
Multiply the terms in Equation 2 by 2
We would then have 4x + 6y = 38. The y's in each equation would then match and we would then say that the y's are 'balanced'
4.
Equation 1 is 8x - 3y = 17. Equation 2 is 4x + y = 21. How would we 'balance' one of the terms?
Add the terms in Equation 1 to the terms in Equation 2
Deduct the terms in Equation 1 from Equation 2
Multiply the terms in Equation 1 by 2
Multiply the terms in Equation 2 by 2
We would then have 8x + 2y = 42. The x's in each equation would be balanced
5.
We have balanced the y terms in two equations as follows: Equation 1 is 3x + 6y = 30 and Equation 2 is 4x + 6y = 38. What do we now do with them?
Add the terms in Equation 1 to the terms in Equation 2
Take the terms in Equation 1 from the terms in Equation 2
Divide Equation 1 by Equation 2
Multiply Equation 1 by Equation 2
Because the balanced terms (6y) are both positive, we take one from the other to cancel them out and we are then left with a value for x
6.
We have balanced the x terms in two equations as follows: Equation 1 is 8x - 3y = 17 and Equation 2 is 8x + 2y = 42. What do we now do with them?
Take the terms in Equation 1 from the terms in Equation 2
Add the terms in Equation 1 to the terms in Equation 2
Divide Equation 1 by Equation 2
Multiply Equation 1 by Equation 2
This time we have balanced the x terms and when we take one equation from the other we are left with a value for y. If one value for x had been positive and the other value for x had been negative we would ADD the equations together
7.
By adding (or subtracting) one equation from another we have concluded that x has a value of 8 in the equation 3x + 6y = 30. What is the value for y?
1
2
3
4
3x = 24
30 - 24 = 6
6 ÷ 6 = 1 therefore y = 1
8.
By adding (or subtracting) one equation from another we have concluded that y has a value of 5 in the equation 8x - 3y = 17. What is the value for x?
2
4
6
8
3 x 5 = 15
17 + 15 = 32
32 ÷ 8 = 4 therefore x = 4
9.
What are the values of x and y that can be derived from the following simultaneous equations: 8x - y = 13 and 12x - 3y = 15?
x = 2 and y = 2
x = 2 and y = 3
x = 3 and y = 2
x = 3 and y = 3 | 1,202 | 4,075 | {"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} | 4.625 | 5 | CC-MAIN-2024-26 | latest | en | 0.957268 |
https://www.physicsforums.com/threads/dimension-of-subspace-of-v-n-with-orthogonal-vectors.370347/ | 1,521,931,894,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257651007.67/warc/CC-MAIN-20180324210433-20180324230433-00246.warc.gz | 855,000,032 | 14,315 | # Dimension of subspace of V^n with orthogonal vectors
1. Jan 17, 2010
### raphael3d
in a space V^n, prove that the set of all vectors {v1,v2,..}, orthogonal to any v≠0, form a subspace V^(n-1).
i know that a subspace of V^n must be at least one dimension less and the set of vector v1,v2,... build a orthogonal basis, but how can one show with this preconditions that the subspace has to be of dimension (n-1)?
2. Jan 17, 2010
### rasmhop
For a fixed non-zero vector v extend (v) to a basis $(v,b_2,b_3,\ldots,b_n)$ of V^n. Now what is the dimension of $\text{Span}(b_2,\ldots,b_n)$ and how this space relate to all vectors orthogonal to v?
3. Jan 17, 2010
### raphael3d
so would v1+ span(b2,...,bn) be a subspace of v^n ?
but why should (v,b2,b3,...,bn) be a basis, when v isnt normalized to a unit vector?
I'm not talking about an orthonormal basis or anything like that, just a basis. In general in a vector space V of dimension n, for m linearly independent vectors $b_1,\ldots,b_m$ we can find vectors $b_{m+1},\ldots,b_n$ such that $b_1,\ldots,b_n$ is a basis for V.
The idea of my first reply was that we let $W = \text{Span}(b_2,\ldots,b_n)$ and let W' be the vector space of vectors orthogonal to v. Now it's possible to show W=W' by showing that if a vector is in W', then it must be in W; and if a vector is in W, then it is in W'. | 424 | 1,354 | {"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} | 3.625 | 4 | CC-MAIN-2018-13 | longest | en | 0.876175 |
http://www.jiskha.com/display.cgi?id=1321502370 | 1,462,493,276,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461861700245.92/warc/CC-MAIN-20160428164140-00140-ip-10-239-7-51.ec2.internal.warc.gz | 605,662,945 | 3,720 | Thursday
May 5, 2016
# Homework Help: Math
Posted by Bryana on Wednesday, November 16, 2011 at 10:59pm.
find the missing value that will make ratios 51 to 18, ? to 6
• Math - Ms. Sue, Wednesday, November 16, 2011 at 11:03pm
18/3 = 6
51/3 = ?
Or you can set up a proportion and cross multiply.
## Answer This Question
First Name: School Subject: Answer:
## Related Questions
More Related Questions | 123 | 407 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2016-18 | longest | en | 0.919348 |
https://www.coursehero.com/file/6329464/ex7-9/ | 1,490,310,312,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218187225.79/warc/CC-MAIN-20170322212947-00117-ip-10-233-31-227.ec2.internal.warc.gz | 882,074,171 | 40,115 | # ex7_9 - 1.1; -2.00; 0; 0];;Q = [0; 0; -0.70; 0; 0];%...
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% Example 7.9 bus system data % Number of generator busesng = 2;% Number of load busesnl = 3;% Number of busesn = ng+nl; % Ybus (from the last homework)A=zeros(8,5);y15=.9323-j*12.4301; A(1,1)=1; A(1,5)=-1;y25=.4464-j*4.9598; A(2,2)=1; A(2,5)=-1; y45=.8928-j*9.9197; A(3,4)=1; A(3,5)=-1; y24=.2232-j*2.4799; A(4,2)=1; A(4,4)=-1; y34=1.8645-j*24.8602; A(5,3)=1; A(5,4)=-1; ys25=j*.22/2; ys24=j*.43/2; ys45=j*.11/2; % Form Y y20=ys24+ys25; A(6,2)=1;y40=ys24+ys45; A(7,4)=1;y50=ys25+ys45; A(8,5)=1; Yl=diag([y15; y25; y45; y24; y34; y20; y40; y50]);Y=A'*Yl*A% Power injections (unknown values set to 0)P = [0;
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Unformatted text preview: 1.1; -2.00; 0; 0];;Q = [0; 0; -0.70; 0; 0];% Specified voltages (unknown magnitudes set to 1 unknown angles set to 0) V=[1.0; 1.05; 1.0; 1.0; 1.0];delta=zeros(5,1); % Relabel buses so bus 1 is slack and bus 2 is the generator busYbus=[Y(1,1) Y(1,3) Y(1,2) Y(1,4:5); Y(3,1) Y(3,3) Y(3,2) Y(3,4:5); Y(2,1) Y(2,3) Y(2,2) Y(2,4:5); Y(4:5,1) Y(4:5,3) Y(4:5,2) Y(4:5,4:5)] % B1 and B2 for calculating angle and voltage updates respectivelyBp=-imag(Ybus-diag(sum(Ybus))); %B' with neglecting any shunt capacitance Bpp = -imag(Ybus); % Neglect phase-shifting transformers but none here N...
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## This note was uploaded on 07/14/2011 for the course ECE 522 taught by Professor Tomsovic during the Summer '10 term at University of Florida.
Ask a homework question - tutors are online | 736 | 1,623 | {"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-2017-13 | longest | en | 0.685445 |
http://www.averest.org/examples/Quartz/Algorithms/Arithmetic/HardwareCircuits/IntMulBooth.html | 1,656,885,451,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104249664.70/warc/CC-MAIN-20220703195118-20220703225118-00272.warc.gz | 58,610,681 | 2,258 | ```// ************************************************************************** //
// //
// eses eses //
// eses eses //
// eses eseses esesese eses Embedded Systems Group //
// ese ese ese ese ese //
// ese eseseses eseseses ese Department of Computer Science //
// eses eses ese eses //
// eses eseses eseseses eses University of Kaiserslautern //
// eses eses //
// //
// ************************************************************************** //
// The following module implements a multiplication of 2-complement numbers //
// by Booth recoding. In addition to Booth recoding, we also further optimize //
// the Booth recoding by bit-pair recoding so that the number of zeros is //
// increased and as many as possible additions can be skipped. //
// Note that the Booth multiplier has depth O(M+N) and size O(M*N) and is thus//
// not optimal and concerning the worst case complexity comparable to the //
// paper-and-pencil methods. In practice, it is reported to behave better due //
// to the high probability of numbers with small absolute values. Such numbers//
// have 2-complement representations with large blocks of 1's or 0's so that //
// the Booth multiplier can simply skip these blocks. //
// ************************************************************************** //
macro M = 5; // number of digits used
macro N = 3; // number of digits used
macro dval(x,i,k) = (i==k-1 ? -(x[i]?1:0) : (x[i]?1:0));
macro intval(x,k) = sum(i=0..k-1) (dval(x,i,k) * exp(2,i));
module IntMulBooth([M]bool ?x,[N]bool ?y,[M+N]bool p) {
event [M]int{2} bx; // Booth recoding of x
event [M]int{2} bpx; // bit-pair optimization of bx
event [M-1][N]bool pp; // digits of partial products
event [M][N-1]bool cp; // carries for summation
event sM;
// Booth encoding of x
bx[0] = (x[0] ? -1 : 0);
for(i=1..M-1)
bx[i] = (x[i] <-> x[i-1] ? 0 : (x[i] ? -1 : +1));
// bit-pair optimization of Booth encoding
for(i=0..(M/2)-1)
case
(bx[2*i+1] == +1 & bx[2*i] == -1) do {
bpx[2*i+1] = 0;
bpx[2*i] = +1;
}
(bx[2*i+1] == -1 & bx[2*i] == +1) do {
bpx[2*i+1] = 0;
bpx[2*i] = -1;
}
default {
bpx[2*i+1] = bx[2*i+1];
bpx[2*i] = bx[2*i];
}
if(2*(M/2) != M)
bpx[M-1] = bx[M-1];
// construct M x N multiplier array
for(i=0..M-1) {
// IntAdd/IntSub of pp[i-1][N-1..0] and y[N-1..0] depending on bpx[i]
for(j=0..N-1) {
let(doSkp = bpx[i] == 0)
let(doSub = bpx[i] == -1)
let(yj = !doSkp & (doSub xor y[j]))
let(xyin = (j==N-1 ? !yj : yj))
let(pin = (i==0 ? (j==N-1) : pp[i-1][j]))
let(cin = (j==0 ? doSub : cp[i][j-1]))
let(pout = (j==0 | i==M-1 ? p[i+j] : pp[i][j-1]))
let(cout = (j==N-1 ? (i==M-1 ? sM : pp[i][N-1]) : cp[i][j]))
FullAdd(xyin,pin,cin,cout,pout);
}
}
p[M+N-1] = !sM;
// check the result
assert(intval(p,M+N) == intval(x,M) * intval(y,N));
}
``` | 992 | 3,345 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2022-27 | latest | en | 0.265835 |
http://geekological.com/c/pairs-challenge-in-interview-street/83/ | 1,542,347,907,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039742978.60/warc/CC-MAIN-20181116045735-20181116071735-00165.warc.gz | 126,868,562 | 33,374 | Problem Statement
Given N numbers , [N<=10^5] we need to count the total pairs of numbers that have a difference of K. [K>0 and K<1e9]
Input Format:
1st line contains N & K (integers).
2nd line contains N numbers of the set. All the N numbers are assured
to be distinct.
Output Format:
One integer saying the no of pairs of numbers that have a diff K.
Sample Input #00:
5 2
1 5 3 4 2
Sample Output #00:
3
Sample Input #01:
10 1
363374326 364147530 61825163 1073065718 1281246024 1399469912 428047635
491595254 879792181 1069262793
Sample Output #01:
0
Solution:
```#include<iostream>
#include<algorithm>
using namespace std;
int compare (const void * a, const void * b)
{
return ( *(int*)a - *(int*)b );
}
int main()
{
int n,k;
cin>>n>>k;
int a[n];
for(int i=0;i<n;i++)
cin>>a[i];
qsort(a,n,sizeof(int),compare);
int seq=0;
for(int i=0;i<n;i++)
{
int key=a[i]+k;
if(((int*)bsearch(&key,a+i+1,n-i-1,sizeof(int),compare))!=NULL)
seq++;
}
cout<<seq;
}```
Time Complexity : O(nlogn)+O(n) | 330 | 1,001 | {"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} | 2.984375 | 3 | CC-MAIN-2018-47 | longest | en | 0.461858 |
https://mathradical.com/radical-mathematics/interval-notation/consolidation-debt-delaware-en.html | 1,643,420,639,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320299894.32/warc/CC-MAIN-20220129002459-20220129032459-00468.warc.gz | 434,896,243 | 10,963 | Algebra Tutorials!
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Depdendent Variable
Number of equations to solve: 23456789
Equ. #1:
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Equ. #4:
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Equ. #6:
Equ. #7:
Equ. #8:
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Solve for:
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Number of inequalities to solve: 23456789
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• implicit differentiation calculator
• ti-83 factoring program | 1,024 | 4,396 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2022-05 | latest | en | 0.931368 |
https://teachingbd24.com/jsc-math-note-10th-chapter-circle/ | 1,656,130,182,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103034170.1/warc/CC-MAIN-20220625034751-20220625064751-00457.warc.gz | 613,836,397 | 19,748 | # JSC Math Note 10th Chapter Circle
JSC Math Note 10th Chapter Circle. Every day we see and use some things which are round: for example, car wheels, bracelets, watches, buttons, dishes, coins Etc. We see that the nozzle of the clock’s thorns rotates round. Second-hand thorn nozzle The path that marks it as a circle. We use round objects in many ways. With a one-rupee coin, put a white paper on the white paper and press down the middle of the currency and press down. In this situation, with a narrow pencil on the right hand, around the currency of the currency. If you move the currency, you will see a circular bound curve on the paper. This is a circle
## JSC Math Note 10th Chapter Circle
The perfect pencil compass is used to draw the circle. Connecting the thorns to the other side is attached to the paper A circle is drawn when the pencil turns around the paper, as shown in the figure. Then the time of drawing the circle Draws distant points from a particular point this. The point of the circle is the point The radius of any distance from the center is called the radius.
The length of the circle’s circomeference cannot be measured with the ruler, not the straight line. There is an easy way to measure the circomeference. Cut the picture shapes on a circle with a circle curved circle. Mark a point on the circomeference. Now draw a segment on the paper and keep the round card vertically on the paper so that the circomeference mark is matched to one end of the segment. Now move the card to the segment until the radius is identified. The point is to touch the segment again. Identify the tangent and measure the length of its line from the endpoint of the segment. This measure is the length of the circomeference. Notice that the small circle diameter is small, the circomeference is too small; On the other hand, the diameter of the big circle is larger, the circomeference is larger.
teachingbd24.com is such a website where you would get all kinds of necessary information regarding educational notes, suggestions and questions’ patterns of school, college, and madrasahs. Particularly you will get here special notes of physics that will be immensely useful to both students and teachers. The builder of the website is Mr. Md. Shah Jamal Who has been serving for 30 years as an Asst. Professor of BAF Shaheen College Dhaka. He expects that this website will meet up all the needs of Bengali version learners /students. He has requested both concerned students and teachers to spread this website home and abroad.
গণিত বিষয়ের অন্যান্য অধ্যায়ের নোটের জন্য এখানে ক্লিক করুন। | 622 | 2,604 | {"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} | 4.65625 | 5 | CC-MAIN-2022-27 | latest | en | 0.927092 |
http://drhuang.com/science/mathematics/math%20word/math/l/l407.htm | 1,718,669,755,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861741.14/warc/CC-MAIN-20240617215859-20240618005859-00756.warc.gz | 10,148,464 | 3,718 | ## Look and Say Sequence
The Integer Sequence beginning with a single digit in which the next term is obtained by describing the previous term. Starting with 1, the sequence would be defined by one 1, two 1s, one 2 two 1s,'' etc., and the result is 1, 11, 21, 1211, 111221, 312211, 13112221, 1113213211, ... (Sloane's A005150).
Starting the sequence instead with the digit for gives , 1, 111, 311, 13211, 111312211, 31131122211, 1321132132211, ... The sequences for and 3 are Sloane's A006751 and A006715. The number of Digits in the th term of both the sequences for is asymptotic to , where is a constant and
(Sloane's A014715) is Conway's Constant. is given by the largest Root of the Polynomial
In fact, the constant is even more general than this, applying to all starting sequences (i.e., even those starting with arbitrary starting digits), with the exception of 22, a result which follows from the Cosmological Theorem. Conway discovered that strings sometimes factor as a concatenation of two strings whose descendants never interfere with one another. A string with no nontrivial splittings is called an element,'' and other strings are called compounds.'' Every string of 1s, 2s, and 3s eventually decays'' into a compound of 92 special elements, named after the chemical elements.
References
Conway, J. H. The Weird and Wonderful Chemistry of Audioactive Decay.'' Eureka 45, 5-18, 1985.
Conway, J. H. The Weird and Wonderful Chemistry of Audioactive Decay.'' §5.11 in Open Problems in Communications and Computation. (Ed. T. M. Cover and B. Gopinath). New York: Springer-Verlag, pp. 173-188, 1987.
Conway, J. H. and Guy, R. K. The Look and Say Sequence.'' In The Book of Numbers. New York: Springer-Verlag, pp. 208-209, 1996.
Sloane, N. J. A. Sequences A005150/M4780, A006715/M2965, and A006751/M2052 in An On-Line Version of the Encyclopedia of Integer Sequences.'' http://www.research.att.com/~njas/sequences/eisonline.html and Sloane, N. J. A. and Plouffe, S. The Encyclopedia of Integer Sequences. San Diego: Academic Press, 1995.
Vardi, I. Computational Recreations in Mathematica. Reading, MA: Addison-Wesley, pp. 13-14, 1991. | 611 | 2,154 | {"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} | 2.53125 | 3 | CC-MAIN-2024-26 | latest | en | 0.812734 |
https://forum.kerbalspaceprogram.com/index.php?/tags/satellites/&_nodeSelectName=forums_topic_node&_noJs=1 | 1,579,423,136,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250594333.5/warc/CC-MAIN-20200119064802-20200119092802-00556.warc.gz | 458,983,370 | 24,683 | # Search the Community
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Found 11 results
3. ## Deep Space Relay Networks
This tutorial, Setting Up A Commnet System, suggests placing 6 satellites into a 4Gm Kerbolar orbit, 12 into 25Gm and, optionally, 24 into a 50Gm orbit. To do this in a finite amount of time requires launching them all individually at specific time intervals. I've calculated the following: To launch 6 Comm Sats to a 4Gm orbit, launch each with 13.48 day separation. To launch 12 Comm Sats to a 25Gm orbit, launch each with a 59.30 day separation. To launch 24 Comm Sats to a 50Gm orbit (optional), launch each with a 20.69 day separation. The 3 formulae used in this computation are: v = sqrt(G * M / R) p = 2 * PI * R / v / (6 * 3600) dt = 1 / (N * (1 / p1 - 1 / p2)) where: v is orbital speed G is grav:6.67408e-11 M is mass of Kerbol: 1.7565459E28 R is orbital altitude p is orbital period (days) PI is pi dt is launch separation (days) N is number of satellites to occupy an orbital altitude p1 is orbital period of Kerbin (days) p2 is orbital period of the target orbit (days) Note: dt for inner orbits will be negative which merely indicates the satellites will arrive in counter-revolutionary order 1/p is angular speed expressed as radians/day and subtracting the speed of the target frame of reference form the launch frame of reference is the insight that makes this achieve a full orbit with even spacing (your mileage may vary depending on how timely your launches are) I plan once arriving at apoapsis to not worry too much about the orbital parameters or spacing but to have the exact same orbital period for all sats in the orbit. This will keep them locked in their relative positions over a very long duration. If you're more picky about getting exact spacing, this tutorial, https://wiki.kerbalspaceprogram.com/wiki/Tutorial:Ideal_Orbits_for_Communication_Satellites, particularly the treatment of the Law of Cosines, is quite fascinating. It's most relevant to low altitude orbits, e.g. spacing N space stations around a body. [All of the above may have been covered elsewhere but I haven't seen it/didn't find it and I've just had to create all this from scratch; so I hope it's useful to someone else. Please let me know if you find any errors, as I have not executed this yet.]
4. ## How many satellites?!?! *reboot*
This is a simple challenge, get as many satellites into lko as possible in one launch. 2 categories: modded unmodded rules: No cheats Proof All satellites must have regeneratI've power source, the ability to orient itself in a specific angle, and have an external antenna. Please say modded or unmodded, mechjeb counts as mod, but if it doesn't affect the craft itself, like a clouds mod, then it's considered not modded. My attempt, It all exploded in the end, but it usually works.
5. ## NO Kerbnet access
So, i have 22 Satellites and non of them have Kerbnet access, if says always this http://imgur.com/a/h5LWb. I can not really understand this because it works in the past.
6. ## Add-on for Satellite Contracts
I am a bit bored with the basic "put a satellite here" contracts, and wanted to spice it up. I use Contract Configurator for Clever Sats but even still it hasn't quite scratched the itch. I thought that being given a randomly generated satellite (sizes/parts/layout) and told where to place it might be a fun spin on the existing satellite contract model. It could be displayed as a popup when entering the VAB (similar to the passenger UI when on the launch pad) where you would select the contract and it would insert the part for you to start attaching items to. Alternatively, the contracts could specify satellite requirements themselves (similar to the space station check lists that already exist) Must generate power Must have x units of electricity Must have x science experiments on board Specific number of satellites per rocket (if even possible?) Could also name specific parts to be installed on the satellite, but might be problematic with multiple part mods installed. Does this exist as its own add-on or as an enhancement to an existing add-on?
7. ## How to appreciate satellites again...
Step 1: put a satellite in Kerb-o-synchronous (geostationary/geosynchronous) equatorial orbit. Step 2: once in desired orbit, push 'V' until in "chase camera". Step 3: play your favorite space themed ambience music. Step 4: push F2. Step 5: profit. Working on putting a little video together to share here shortly. It's times like these that remind me how much I appreciate this game and its developers. 400+ hours deep and it's the little things.
8. ## Solar panel problems
hey I'm pretty new to KSP but watched a lot of Scott Manley's uploads and managed to get a satellite into orbit. It has 5 solar panels and 4 batteries which was working fine as I transmitted science back home but after the first I think 1 or 2 transmissions it ran out of power and I figured "ok, the panels need a minute to recharge" but they aren't. Is there a certain place on the craft I need to place them or do they maybe require more time to recharge? (OX-STAT Photovoltaic Panels) thanks a lot for anyone who takes time to help me out:)
9. ## Old contract satellites
Is there any reason to keep old satellites in orbit? Ive done a few sat contracts and they always say something along the lines of "this satellite shall henceforth belong to insert company name here". I already have the money and rewards for the contract and the space around kerbin is getting cluttered, so does it even matter if i just destroy them all? | 1,552 | 6,517 | {"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} | 3.203125 | 3 | CC-MAIN-2020-05 | latest | en | 0.655098 |
https://functions.wolfram.com/ElementaryFunctions/ArcSech/27/02/04/09/02/0002/ | 1,713,839,749,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818452.78/warc/CC-MAIN-20240423002028-20240423032028-00779.warc.gz | 246,533,885 | 7,511 | html, body, form { margin: 0; padding: 0; width: 100%; } #calculate { position: relative; width: 177px; height: 110px; background: transparent url(/images/alphabox/embed_functions_inside.gif) no-repeat scroll 0 0; } #i { position: relative; left: 18px; top: 44px; width: 133px; border: 0 none; outline: 0; font-size: 11px; } #eq { width: 9px; height: 10px; background: transparent; position: absolute; top: 47px; right: 18px; cursor: pointer; }
ArcSech
http://functions.wolfram.com/01.30.27.0683.01
Input Form
ArcSech[Sqrt[z + 1]] == (Pi I)/2 - I ArcTan[1/Sqrt[z]] /; Inequality[-Pi, Less, Arg[z], LessEqual, 0]
Standard Form
Cell[BoxData[RowBox[List[RowBox[List[RowBox[List["ArcSech", "[", SqrtBox[RowBox[List["z", "+", "1"]]], "]"]], "\[Equal]", RowBox[List[FractionBox[RowBox[List["\[Pi]", " ", "\[ImaginaryI]"]], "2"], "-", RowBox[List["\[ImaginaryI]", " ", RowBox[List["ArcTan", "[", FractionBox["1", SqrtBox["z"]], "]"]]]]]]]], "/;", RowBox[List[RowBox[List["-", "\[Pi]"]], "<", RowBox[List["Arg", "[", "z", "]"]], "\[LessEqual]", "0"]]]]]]
MathML Form
sech - 1 ( z + 1 ) π 2 - tan - 1 ( 1 z ) /; - π < arg ( z ) 0 Condition z 1 1 2 2 -1 -1 1 z 1 2 -1 Inequality -1 z 0 [/itex]
Rule Form
Cell[BoxData[RowBox[List[RowBox[List["HoldPattern", "[", RowBox[List["ArcSech", "[", SqrtBox[RowBox[List["z_", "+", "1"]]], "]"]], "]"]], "\[RuleDelayed]", RowBox[List[RowBox[List[FractionBox[RowBox[List["\[Pi]", " ", "\[ImaginaryI]"]], "2"], "-", RowBox[List["\[ImaginaryI]", " ", RowBox[List["ArcTan", "[", FractionBox["1", SqrtBox["z"]], "]"]]]]]], "/;", RowBox[List[RowBox[List["-", "\[Pi]"]], "<", RowBox[List["Arg", "[", "z", "]"]], "\[LessEqual]", "0"]]]]]]]]
Date Added to functions.wolfram.com (modification date)
2003-08-21 | 635 | 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} | 2.6875 | 3 | CC-MAIN-2024-18 | latest | en | 0.269059 |
https://nivent.github.io/blog/understanding-math/ | 1,716,901,203,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059139.85/warc/CC-MAIN-20240528123152-20240528153152-00033.warc.gz | 371,695,321 | 14,676 | Calculating isn’t enough
I remember one day in middle school when I was complaining about my math homework 1. Fed up with my complaints, and wanting to put me in my place, my sister decided to show me her homwork. It made no sense. It was asking strange questions about even stranger concepts.
Question
What’s the derivative of $\hspace{2pt}3x^4-7x^2+x-8$?
My sister quickly explained to me that to take one of these derivatives, you multiply the number in front by the exponent, then subtract one from it. Simple enough. I now knew about derivatives and how to compute them, except I didn’t. For years, all a derivaitve was to me was a process of transforming polynomials. It had no connection to slopes, areas, linear transformations, etc.;2 it was just symbolic manipulation. I saw examples where people were asked to take the derivative of strange things like $\sin(x)$, but although I knew of $\sin$, I had no idea what mysterious black magic was required to differentiate something like it.
Locality isn’t enough
Fast forward a number of years, and I am a junior in high school. I am no longer that ignorant middle schooler; my mathematical knowledge is leaps and bounds ahead of what it was before. At this time I was taking an “Advance Logic” course at a local university, and I was loving it. A more descriptive title for the course would have been Metalogic. In it, we considered a few different logic systems, and proved soundness and completeness of them. We did other stuff, but that was the big idea.
Aside3
When studying something reasoning system, you can study it in terms of semantics or syntax. Semantics is about assigning meaning to the system. You define models for it which can be thought of as examples of things that follow its rules. Syntax is all about manipulating symbols. You say your system has some ground truths (axioms) and deduction rules for getting new truths, and then see what true things result from this (this is how you might want a computer to view your system). A system is sound if syntactic manipulation always results in semantically true conclusions. A system is complete if every true statement can be proven syntactically.
We started with propositional logic, moved onto modal logic, and then did a little excursion into three-valued logics. The whole journey was amazing, made sense, and really gave me a newfound appreciation and interest for and in logic. Then, in the last part of the class, we moved on to first order logic. I had been looking forward to this all semester.
First order logic was a different beast altogether. Everything before made sense; it had clicked; FOL did not. The further along we went, the less I understood. Don’t get me wrong. I followed what was going on. Every single step made sense and was internally verified, but I could not grasp the big picture. In the words of the topologist inside of me, I had local understanding, but still lacked understanding. I could see what was going on in class, and I could reproduce and modify proofs for homework and tests, but almost nothing that happened4 felt like it made sense.
Passing isn’t enough
This brings us to now. I just finished up a quarter at school, and among the classes in my schedule was one called Algebraic Topology. In this class, something happened that I had convinced myself would never happen 5. I failed - I failed a math class. I didn’t fail in the obvious interpretation of that word; my grade in the class is fine 6. I failed in a more personal sense. Like derivatives and FOL metatheory before it, I do not understand algebraic topology, but this case is worse than those.
When I did not understand derivatives, I had no reason to. My only formative experience with them was a quick demonstration of how they’re calculated on a specific type of function. I had no need or incentive to know them any better at the time. When I did not understand FOL, it was only one part of a much larger course. Everything else felt more intuitive, and even FOL itself locally made sense. When it comes to algebraic topology, I just don’t feel like I understand it. There were moments of me appreciating things7, but I’d be hard pressed to say there was even local understanding of the material.
I still did well on the problem sets. I would work together with a friend, and every week we managed to push out 10 (mostly) correct proofs. What little reasoning power I had that was trying to approach understanding was cultivated in these sessions, but there was no shift. Often8 when taking a math class, the first couple of problem sets are the most difficult. In the beginning, you only have your default mathematical prowess available to you to use. You have yet to build up a feeling for the field; you do not yet know how to approach a proof. Somewhere during the course, however, there is a shift. You start thinking like a topologist, geometer, algebraist or whatever; when seeing a proposed claim, you also see techniques to prove it or at the very least, you see an idea of its placement in your internal map of the field. You see other results its related to, other theorems it could have similar proofs to, ideas it depends on, etc.
This was a change I had become accustomed to. When I took a Graph Theory class, every problem seemed novel and individual for the first half of the class. It felt like I had to reinvent the wheel everytime I had to prove some claim, and it was not easy. By the second half of the class, though, I had begun thinking like a graph theorist. I had a more standard set of tricks I could employ, I had better examples to refer to, and I felt like I was working on a theory and not a collection of claims. The same thing was true on a smaller scale in my math class my first quarter here, which was a sort of linear algebra for combinatorics course. There were two techniques we learned in the class that had become my gotos by the end, but were very mystical for quite some time: dimension arguments and the related polynomial method9. Again, in Game Theory I had an extremely hard time doing the first problem set, but when I finished the course, I returned to it, and everything seemed so obvious and simple. The arguments were nothing special, and applying them was mentally quick. In all these cases, I had built up an intution and appreciation for the subject.
I cannot say the same for algebraic topology. This really hit me when, the day before the final, a friend asked me what class I was studying for, and I drew a blank on trying to explain the course to him; I not only drew a blank in how to verbally descibe the class, but I also drew an internal blank in my own understanding of the course. It is normal for me to not know how to tell someone about math10, but I at least always have some idea of what the math is to me. Here, there was nothing. I had no internalized overarching picture of algebraic topology, and this was a first.
My suspisions were confirmed when I took the final. It was brutal; I am legitimately scared to see my grade on it because (almost) nothing I did felt convincing. The whole class seems to have similarly struggled with the final, so there is an argument to be made that it was just a particularly hard test, but I don’t think that’s the case. The test was not trying to trick you. Every question felt unfair while I was taking it, but in hindsight, they were questions that I believe you could have answered if you understood the material. To someone who really absorbed and built up a good intuition for the material, the test would have been appropriate. Sadly, that was not me.
Final Thoughts
Question
What is enough?
That’s the million dollar question, and one I do not have the answer to, except to say you may never know. I thought I understood derivatives, until some weeks ago I learned that a derivative is not just a number. It’s a linear transformation; specifically, it is the linear transformation that best locally approximates the function 11. I am now in a position similar to my first derivative experience. I know enough to say that, but I have not looked into things enough to understand what derivatives really are. After my logic course, I thought I understood my logic was about, then I learned that there were way more logics out there than the few we studied, that logic had interplays with some surprisign fields 12, and that this stuff gets really deep. In the end, I’m not sure there’s anything I do understand.
When I first decided to write a post talking about my experiences in AT, I wanted to do so in the context of revisitng mathematics in order to learn it. That clearly didn’t happen, but in order to shine some light in the no so hopeful picture of understanding mathematics that I painted, I will say that there is benefit to serious revisiting of mathematics. I have learned linear algebra in 3 different settings now 13, and each time I revealed something I had previously missed. Not necessarily some piece of knowledge I did not have before, but some insight I had yet to see. With that in mind, understanding math just takes time, and I might try to revisit AT at some point…14
1. It was either too hard or too easy. I can’t remember which.
2. What’s the proper grammatical syntax for this
3. warning, the following is simplified and so erroneous
4. whether I did it or the professor did it, whether it was correct or not
5. specifically, would never happen to me
6. technically to soon to say this because I don’t have my grade yet, but with high probability, it’s fine
7. a good proof, a nice example, the one time it felt like communtative diagrams actually had some use
8. for me at least
9. to this day whenever I see a bound, my first instinct is to form polynomials
10. describing math in an understandable, engaging, nonsuperficial way is no easy task
11. That is, the derivative operator does not take in a function and a point, and spit out a number (or vector). It takes a function and a point, and spits out a function. If you want to learn more about this, look into differential geometry. That’s what I was looking at when I first came accross it.
12. e.g. Game Theory, Measure Theory, Graph Theory, etc.
13. never feeling like my understanding was lacking after any of them
14. Alternate titles for this post that I considered include “Learning math”, “(Failing at) Learning math”, and “Math is Hard” | 2,270 | 10,411 | {"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} | 3.890625 | 4 | CC-MAIN-2024-22 | latest | en | 0.974851 |
http://www.maplesoft.com/support/help/category.aspx?CID=1604 | 1,454,904,094,000,000,000 | text/html | crawl-data/CC-MAIN-2016-07/segments/1454701152130.53/warc/CC-MAIN-20160205193912-00011-ip-10-236-182-209.ec2.internal.warc.gz | 520,390,233 | 13,084 | FastArithmeticTools Subpackage - RegularChains - Factorization and Solving Equations - Mathematics - Maple Help - Maplesoft
Home : Support : Online Help : Mathematics : Factorization and Solving Equations : RegularChains : FastArithmeticTools Subpackage
# FastArithmeticTools Subpackage
## Category Documents
Page 1 of 1
RegularChains/FastArithmeticTools/BivariateModularTriangularize
triangular decomposition of a bivariate square system by a modular method
RegularChains/FastArithmeticTools/IteratedResultantDim0
iterated resultant of a polynomial w.r.t a 0-dim regular chain
RegularChains/FastArithmeticTools/IteratedResultantDim1
iterated resultant of a polynomial w.r.t a one-dim regular chain
RegularChains/FastArithmeticTools/NormalFormDim0
normal form of a polynomial w.r.t. a 0-dim regular chain
RegularChains/FastArithmeticTools/NormalizePolynomialDim0
normalize a polynomial w.r.t a 0-dim regular chain
RegularChains/FastArithmeticTools/NormalizeRegularChainDim0
normalize a zero-dimensional regular chain
RegularChains/FastArithmeticTools/RandomRegularChainDim0
generate a random 0-dim regular chain
RegularChains/FastArithmeticTools/RandomRegularChainDim1
generate a random one-dim regular chain
RegularChains/FastArithmeticTools/ReduceCoefficientsDim0
reduce the coefficients of a polynomial w.r.t a 0-dim regular chain
RegularChains/FastArithmeticTools/RegularGcdBySpecializationCube
regular GCD of two polynomials modulo a regular chain
RegularChains/FastArithmeticTools/RegularizeDim0
Test the regularity of a polynomial w.r.t. a 0-dim regular chain
RegularChains/FastArithmeticTools/SubresultantChainSpecializationCube
evaluation of the subresultant chain of two polynomials
RegularChains/FastArithmeticTools
Overview of the RegularChains[FastArithmeticTools] Subpackage of RegularChains
Page 1 of 1 | 456 | 1,836 | {"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-2016-07 | longest | en | 0.705425 |
https://gamedev.stackexchange.com/questions/68612/how-to-compute-tangent-and-bitangent-vectors/68617 | 1,716,836,431,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059045.25/warc/CC-MAIN-20240527175049-20240527205049-00039.warc.gz | 230,617,063 | 39,746 | How to compute tangent and bitangent vectors
I have a texture loaded in three.js, then passed to the shaders. In the vertex shader I compute the normal, and I save into a variable the uv vector.
<script id="vertexShader" type="x-shader/x-vertex">
varying vec3 N,P;
varying vec2 UV;
void main() {
gl_Position= projectionMatrix * modelViewMatrix * vec4(position,1.0);
P= position;
N= normalMatrix * vec3(normal);
UV= uv;
}
</script>
varying vec3 N,P;
varying vec2 UV;
uniform sampler2D texture;
void main() {
gl_FragColor= texture2D(texture,UV);
}
</script>
How do I compute the T and B vectors?
• do you want the algorithm in general or specifically for your library of choice ? Jan 9, 2014 at 12:27
• If I can compute it with three.js it would be better. Jan 9, 2014 at 13:07
• You may want to look up the MikkTSpace algorithm, which is a standardised way of generating a tangent basis that's designed to inter-operate well between different software packages. Aug 11, 2022 at 9:17
First of all, for every 3D vertex there is infinite tangent and bi-tangent vectors. The below image explains why there is an infinite number of tangent spaces for each vertex, the tangent and bitangent can have any direction in the shown plane.
So in order to properly calculate the most useful1 tangent space, we want our tangent space to be aligned such that the x axis (the tangent) corresponds to the u direction in the bump map and the y axis (bitangent) corresponds to the v direction in the bump map, we should already have normal of the vertex which already corresponds to the Z direction in tangent space.
(1) most useful because in the end we want normal vectors to be sampled from the texture
That best be explained with pictures, we want our tangent space to be aligned like (u, v) shown below.
Source of the image though not strictly related to computer graphics
In computer graphics developers usually use (u,v) also known as texture coordinates. We will assume T is the tangent and B is the bitangent, and P0 is our target vertex, that is part of the triangle (P0,P1,P2).
First remember what we wanted to do, is to calculate tangent and bitanget that:
1. T aligned with u and B aligned with v.
2. T and B lays in the plane with the vertex normal (the plane shown in the above image).
The point is we already assumed that T and B lays in the same plane and corresponds to U and V now if we can know their values we can cross product and the third vector to construct a transformation matrix from world to tangent space.
Given that we know that any 2D vector can be written as a linear combination of two independent vectors2 and since we already have the triangle points (edges), shown in the above image. We can write:
E1 = (u1-u0)T + (v1-v0)B
E2 = (u2-u0)T + (v2-v0)B
(2) actually that's is how basis matrix is derived
The above equation can be written in a matrix form,
| E1x E1y E1z | | deltaU1 deltaV1 | * | Tx Ty Tz |
| E2x E2y E2z | = | deltaU2 deltaV2 | | Bx By Bz |
By solving the matrixs equation we can determine T and B values we can construct a transformation matrix.
The full source code in C++
#include "Vector4D.h"
struct Triangle
{
unsigned short index[3];
};
void CalculateTangentArray(long vertexCount, const Point3D *vertex, const Vector3D *normal,
const Point2D *texcoord, long triangleCount, const Triangle *triangle, Vector4D *tangent)
{
Vector3D *tan1 = new Vector3D[vertexCount * 2];
Vector3D *tan2 = tan1 + vertexCount;
ZeroMemory(tan1, vertexCount * sizeof(Vector3D) * 2);
for (long a = 0; a < triangleCount; a++)
{
long i1 = triangle->index[0];
long i2 = triangle->index[1];
long i3 = triangle->index[2];
const Point3D& v1 = vertex[i1];
const Point3D& v2 = vertex[i2];
const Point3D& v3 = vertex[i3];
const Point2D& w1 = texcoord[i1];
const Point2D& w2 = texcoord[i2];
const Point2D& w3 = texcoord[i3];
float x1 = v2.x - v1.x;
float x2 = v3.x - v1.x;
float y1 = v2.y - v1.y;
float y2 = v3.y - v1.y;
float z1 = v2.z - v1.z;
float z2 = v3.z - v1.z;
float s1 = w2.x - w1.x;
float s2 = w3.x - w1.x;
float t1 = w2.y - w1.y;
float t2 = w3.y - w1.y;
float r = 1.0F / (s1 * t2 - s2 * t1);
Vector3D sdir((t2 * x1 - t1 * x2) * r, (t2 * y1 - t1 * y2) * r,
(t2 * z1 - t1 * z2) * r);
Vector3D tdir((s1 * x2 - s2 * x1) * r, (s1 * y2 - s2 * y1) * r,
(s1 * z2 - s2 * z1) * r);
tan1[i1] += sdir;
tan1[i2] += sdir;
tan1[i3] += sdir;
tan2[i1] += tdir;
tan2[i2] += tdir;
tan2[i3] += tdir;
triangle++;
}
for (long a = 0; a < vertexCount; a++)
{
const Vector3D& n = normal[a];
const Vector3D& t = tan1[a];
// Gram-Schmidt orthogonalize
tangent[a] = (t - n * Dot(n, t)).Normalize();
// Calculate handedness
tangent[a].w = (Dot(Cross(n, t), tan2[a]) < 0.0F) ? -1.0F : 1.0F;
}
delete[] tan1;
}
Full source code and derivation can be found here.
• What if I don't have a triangle? In my case I have a texture that should be applied on a sphere. How to adapt it to this case? Jan 10, 2014 at 12:05
• @RamyAlZuhouri isn't the sphere build from triangles? You just loop over the vertices like in the code. If your sphere isn't triangle based that's a totally different story. Jan 10, 2014 at 12:26
• I'm using three.js SphereGeometry (in javascript). Maybe I should pass the face property to the shaders? The sphere I draw has 1089 vertices and 1084 faces. Jan 10, 2014 at 12:33
• you calculate tangent space and then pass the tangents to the shaders. And you should have access to face/vertices in order to calculate tangent space. Jan 10, 2014 at 13:45
• In my case I'll have 1084 tangents, how do I map the tangents with the vertices? Jan 11, 2014 at 11:35 | 1,685 | 5,606 | {"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} | 3.4375 | 3 | CC-MAIN-2024-22 | latest | en | 0.854476 |
https://mersenneforum.org/showthread.php?s=678cc3fb77156573080fbf23bcefb475&t=25275 | 1,695,529,304,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506559.11/warc/CC-MAIN-20230924023050-20230924053050-00725.warc.gz | 435,664,911 | 8,082 | mersenneforum.org Why primes are either of the form 8n+3 or 8n+7
Register FAQ Search Today's Posts Mark Forums Read
2020-02-19, 18:55 #1 enzocreti Mar 2018 10000111102 Posts Why primes are either of the form 8n+3 or 8n+7 2, 3, 7, 19, 67, 79, 359, 2131, 3371, 331259 are the k's (k is prime) such that pg(k) is also prime First I note that the primes have either the form 8n+3 or 8n+7 or 8n+2 (why?) The primes that are not of the form 8n+3, that is 2, 7 and 359 are of the form s^2-2 where s is a prime... Infact 2=2^2-2, 7=3^2-2 and 359=19^2-2 Last fiddled with by enzocreti on 2020-02-19 at 19:06
2020-02-20, 04:00 #2 jwaltos 3×29×83 Posts One of the best ways to answer this question (without posting) is to try and find the answer in a text book or peer reviewed paper. If you haven't found it after a quick search then you haven't looked hard enough.
2020-02-20, 08:46 #3 enzocreti Mar 2018 2·271 Posts .... if a prime p has the form 8s+1 or 8s+5 then 10^d*(2^p-1)+2^(p-1)-1 is divisible by 5
Similar Threads Thread Thread Starter Forum Replies Last Post sweety439 sweety439 179 2023-08-08 19:54 enzocreti enzocreti 0 2020-02-17 16:28 enzocreti enzocreti 7 2019-05-05 13:19 a1call Miscellaneous Math 6 2018-12-11 03:34 carpetpool carpetpool 3 2017-01-26 01:29
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Sun Sep 24 04:21:44 UTC 2023 up 11 days, 2:04, 0 users, load averages: 2.24, 2.02, 1.81 | 567 | 1,409 | {"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-2023-40 | latest | en | 0.842893 |
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I was wondering about this for quite a while now, I hope i get something new by throwing question out to you guys.
How can we better the 21st century transport scheduling system? Because can it be made more smarter?
• Thank you in advance
Note by Mathews James
5 years, 10 months ago
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## Comments
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A revolution in transport technology is really what we need.
- 5 years, 10 months ago
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Sorry guys been working away from my computer. Basically i was thinking of these -
1) positon of transport vehicles is determined by automated positioning technology eg. GPS .. 2)Routes of vehicles are available 3)vehicles( i am talking about bus or trains) can tag locations that they anticipate will change the schedule of the other vehicles in the vicinity. Anticipated schedule change. 4)vehicles be re-routed and re-scheduled dynamically based on "Schedule change" locations. 5) If the capacity of the bus/train does not meet the number of people waiting on route to board the same. There are a few more but just working on this for now. In my opinion a system architecture that makes use of distributed computing ,iHCI, and a using a smart environment ( maybe embedded devices). Not really sure if @Raghav Vaidyanathan Kochi uses a smart environment for transport or just a historic pre-set route. if the traffic lights could communicate,anticipate traffic that can be used in real time by bus drivers and travelers?
Well I am stuck as well like you guys, and need to come up with an architecture. Regardless a good topic for discussion cheers.
- 5 years, 10 months ago
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Kochi uses a historic pre-set route. And yes, what you are saying definitely makes sense. I've thought about it. Almost makes it seem like a living organism. But both of us know that it will be very challenging to implement. I'm no expert at all this, but what you are talking about is very similar to the "internet of things" where everything can communicate between each other. I think 3) leaves a lot of room for human error? Not all anticipation leads to the real thing. Also, I would like to add onto 5), the converse may also happen, when number of people are much lesser than anticipated, you wouldn't need multiple vehicles for 2 or 3 people.
- 5 years, 10 months ago
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and @Shashwat Shukla , I am working under the assumption that no one has choppers. 8-)
- 5 years, 10 months ago
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Makes things that much harder then(P.S: I was just fooling about, sorry if I might have gone off topic)
- 5 years, 10 months ago
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Hmm.. that is a question of great practical importance. Well, I have faced problems in the past with the public transport system in my city(kochi, kerala, India).
The problem here is that the transport system is not well equipped for rush hour traffic. Get out to the highway around 6.30 pm, you'll reach your destination at 9, get out at 8:45, you'll still reach at 9. Being largely unplanned, kochi never had a well planned public transport system. And since we have a metro rail project in process, there are even more detours and blockades. So what I want to know is whether transport scheduling systems could adapt to these changes. Rush hours are predictable, but when it comes to maintenance works and things of that sort, can a scheduling system accommodate these as well?
- 5 years, 10 months ago
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Well , I am not the traveling type so I guess I can't comment here . But I'd very much like to know about other people's views !
What are your ideas sir ?
- 5 years, 10 months ago
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Set Loading... | 1,358 | 5,501 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 8, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2021-04 | longest | en | 0.879205 |
https://www.authorea.com/users/13114/articles/13404/_show_article | 1,527,429,744,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794868316.29/warc/CC-MAIN-20180527131037-20180527151037-00250.warc.gz | 680,491,837 | 269,191 | # DTMF Decoding
## Simple Bandpass Filter Design
function hh = dtmfdesign(fb, L, fs)
%DTMFDESIGN
% hh = dtmfdesign(fb, L, fs)
% returns a matrix (L by length(fb)) where each column contains
% the impulse response of a BPF, one for each frequency in fb
% fb = vector of center frequencies
% L = length of FIR bandpass filters
% fs = sampling freq
%
% Each BPF must be scaled so that its frequency response has a
% maximum magnitude equal to one.
hh = zeros(L, length(fb)); % Create a matrix of length L times length(fb)
BB = zeros(1, length(fb)); % Create a vector that'll be used for scaling
ww = 0:pi/10000:pi;
% This loop will calculate the normalized freq response for all fb
for ii = 1:length(fb)
bb = cos(2 * pi * fb(ii) * (0:L-1) / fs); % Coefficients
HH = freqz(bb, 1, ww); % Calculate the impulse response
max_value = max(abs(HH)); % Get the max value of response
BB(ii) = 1 / max_value; % Insert into BB for scaling
HH2 = bb * BB(ii); % Scale the impulse response to one
hh(:, ii) = HH2; % Store normalized impulse response into hh matrix
end
ww_d = 0:pi/10000:pi;
L_min = 0; % Default
% This loop determines the smallest L that can be used
for L = 40:100
hh = dtmfdesign([697], L, 8000); % Returns a single vector (impulse response)
HH = freqz(hh, 1, ww_d); % Generates the frequency response
stopband = find(abs(HH) >= 0.25); % Determines the indexes in the stopband
stopband_right = ww_d(stopband(length(stopband))); % Determines the right index
if (2 * pi * 770 / 8000 >= stopband_right) % Checks if 770 Hz is in the stopband
L_min = L; % Successfully determined L_min -> Leave loop
break
end
end
all_frequencies = [697 770 852 941 1209 1336 1477 1633];
ww = 0:pi/10000:pi;
fs = 8000; % Sampling rate is still set at 8000
L = 40; % Should result in a passing bandwidth that is too wide
A = dtmfdesign(all_frequencies, L, fs); % Generate a matrix of impulse responses
HH = zeros(length(ww), 8); % Instantiating a matrix of frequency responses
for ii = 1:8
HH(:, ii) = freqz(A(:, ii), 1, ww); % Store frequency responses into HH
end
% Plot
figure(1);
title('Eight Bandpass Filters at Different Frequencies with L=40');
ylabel('Amplitude');
xlabel('Normalized Frequency');
plot(ww, abs(HH)); | 674 | 2,235 | {"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} | 2.8125 | 3 | CC-MAIN-2018-22 | latest | en | 0.656452 |
https://cbsenews.com/important-questions-for-class-12-chapter-2-inverse-trigonometric-function/ | 1,571,266,309,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986670928.29/warc/CC-MAIN-20191016213112-20191017000612-00246.warc.gz | 421,526,616 | 4,886 | # Important Questions for Class 12 Chapter 2 - Inverse Trigonometric Function
CBSE Board has laid down specific guidelines for the pattern of questions that a student of class 12th should be acquainted with. Adhering to these guidelines, we have come up with class 12th Important questions which carries questions related to all topics in Mathematics.
Important questions based on NCERT syllabus for Class 12 Chapter 2 - Inverse Trigonometric Function:
Question 1: if sin(sin^{-1} 1/5 + cos^{-1} x) = 1, Then find the valuee of x.
Solution: Given sin(sin^{-1} 1/5 + cos^{-1} x) = 1
=> sin^{-1} 1/5 + cos^{-1} x = sin^{-1}(1) [since sinθ = x => θ = sin^{-1}x ]
=> sin^{-1} 1/5 + cos^{-1} x = sin^{-1}(sin π/2) [since sin π/2 = 1]
=> sin^{-1} 1/5 + cos^{-1} x = pi/2
But we know that sin^{-1} x + cos^{-1} x = π/2
x belongs to [-1, 1]
sin^{-1} 1/5 = sin^{-1}x => x = 1/5
Question 2: Write the value of tan(2 tan^{-1} 1/5)
Solution: tan(2 tan^{-1} 1/5) = tan[tan^{-1}[(2 x 1/5)/(1-(1/5)^2)]
since 2 tan^{-1}x = tan^{-1} [ 2x/(1-x^2)]
tan[tan^{-1} (2x5/24)] = tan[tan^{-1}(5/12)] = 5/12
Question 3: Find the value of tan^{-1}[2 sin(2cos^{-1}√3/2)]
Solution: Consider tan^{-1}[2 sin(2cos^{-1}√3/2)]
tan^{-1}[2 sin(cos^{-1}(2 . 3/4 - 1))]
Since 2cos^{-1} x = cos^{-1}(2x^2 - 1)]
= tan^{-1}[2 sin{cos^{-1}(3/2 - 1)}]
= tan^{-1}[2 sin{cos^{-1}(1/2)}]
= tan^{-1}[2 sin{cos^{-1}(cos π/3)}]
= tan^{-1}[2sin π/3] [ since cos^{-1}(cos θ) = θ]
= tan^{-1}(2 . √3/2)
= tan^{-1}(√3) = tan^{-1}(tan π/3) = π/3
since tan^{-1}(tan θ) = θ | 615 | 1,542 | {"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-2019-43 | latest | en | 0.723818 |
https://blog.csdn.net/qq_29695087/article/details/51542979 | 1,527,047,749,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794865411.56/warc/CC-MAIN-20180523024534-20180523044534-00416.warc.gz | 504,294,236 | 12,534 | just hold on！
###### 堆的简单应用
#include<iostream>
#include<algorithm>
using namespace std;
//1.在N个数据当中找出最大的K个数
const int N = 10000;
const int K = 100;
void AdjustDown1(int a[], int size, int parent) //建一个小堆
{
int child = parent * 2 + 1;
while (child < K)
{
if ((child + 1 < K) && (a[child + 1] < a[child]))
{
child++;
}
if (a[child] < a[parent])
{
swap(a[child], a[parent]);
parent = child;
child = parent * 2 + 1;
}
else
{
break;
}
}
}
void GetTopK(int a[],int TopK[])
{
assert(K < N);
int i = 0;
int j = 0;
int m = 0;
int n = 0;
for (i = 0; i < K; i++)
{
TopK[i] = a[i]; //取出a中的前k个数字放到topk[]里面
}
//建堆
for (j = (K - 2) / 2; j >0; --j)
{
}
for (int m = 0; m < N; ++m)
{
if (a[m]>TopK[0])
{
TopK[0] = a[m];
}
}
for (int n = 0; n < K; ++n) //一次输出K个最大数
{
cout << TopK[n] << " ";
}
cout << endl;
}
#include"BIgData.h"
void TestTopK()
{
int a[N];
int TopK[K];
for (int i = 0; i < N; ++i)
{
a[i] = i;
}
GetTopK(a, TopK);
}
int main()
{
TestTopK();
system("pause");
return 0;
}
//2.堆排序：建大堆，每次找到最大的数据交换到数组末尾，将剩下的数据AdjustDown，再进行交换
void AdjustDown2(int a[],int size,size_t parent)
{
int child = parent * 2 + 1;
while (child<size)
{
if ((child + 1 < size)&&a[child] < a[child + 1])
{
++child;
}
if (a[child] > a[parent])
{
swap(a[child], a[parent]);
parent = child;
child = parent * 2 + 1;
}
else
{
break;
}
}
}
void Heap_Sort(int a[], size_t n)
{
for (int i = (n - 2) / 2; i >= 0; i--) //注意边界条件
{
}
for (int i = 0; i < n; ++i)
{
swap(a[0], a[n - 1-i]);
AdjustDown2(a, n - 1 - i, 0);
}
for (int i = 0; i < n; ++i)
{
cout << a[i] << " ";
}
cout << endl;
}
void TestHeap_Sort()
{
int a[] = { 10, 12, 9, 15, 13, 17, 16, 18, 20,14 };
Heap_Sort(a, 10);
}
int main()
{
TestHeap_Sort();
system("pause");
return 0;
}
#### 优先队列——斐波那契堆(without source code)
2016-01-24 10:24:39
#### 数据结构::堆及堆的应用~
2016-12-31 13:47:35
#### [大、小根堆应用总结一]堆排序的应用场景
2016-05-10 09:53:55
#### 堆和堆的应用：堆排序和优先队列
2017-11-19 11:57:02
#### 数据结构(Java)——堆的应用
2015-12-01 20:14:42
#### 算法导论之二项堆
2016-08-01 14:52:23
#### 堆原理及其基本应用
2016-12-02 21:38:22
#### 算法导论之斐波那契堆
2016-08-05 11:28:12
#### 堆的一些简单应用
2016-06-01 19:01:46
#### 堆的简单应用——TopK
2018-05-14 00:47:41 | 984 | 2,158 | {"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} | 3.234375 | 3 | CC-MAIN-2018-22 | latest | en | 0.155094 |
http://www.docstoc.com/docs/106263189/u11_h_titration_practice | 1,411,498,390,000,000,000 | text/html | crawl-data/CC-MAIN-2014-41/segments/1410657139395.59/warc/CC-MAIN-20140914011219-00325-ip-10-234-18-248.ec2.internal.warc.gz | 463,893,985 | 14,741 | # u11_h_titration_practice
Document Sample
Titration
Acid-base titration involves the gradual addition of acids and bases to one another, generally for
the purpose of determining reacting volumes or concentrations. In a laboratory titration, 1M HCl was
gradually added to 50 mL of 1M NaOH. The pH was monitored throughout the analysis, and the following
VOLUME OF HCl pH OF
0 14.0
5 13.9
10 13.8
15 13.7
20 13.6
30 13.4
40 13.0
45 12.4
49 11.7
50 7.0
51 2.3
55 1.6
60 1.0
75 0.7
100 0.64
Plot these data on the grid above, and then answer the following questions.
1. What volume of 1.0 M HCl was required to reach the equivalent point?
2. How would the graph differ if the concentration of titrating HCl were increased to 2.0 M?
3. List an indicator that could be used to recognize the equivalence point of this titration. Describe
the color changes that the solution would undergo with the addition of HCl.
Titration Practice Problems: solve the following problems on a separate piece of paper or the back of
this worksheet.
1. What volume of 0.275 M hydrochloric acid solution reacts with 36.4 mL of 0.150 M sodium
hydroxide solution? HCl + NaOH NaCl + H2O
2. What volume of 0.112 M NaOH solution would be required to react with 25.3 mL of 0.400 M
H2SO4 solution? H2SO4 + 2NaOH Na2SO4 + 2H2O
3. A 0.08964 M solution of NaOH was used to titrate a solution of unknown concentration of HCl. A
30.00 mL sample of the HCl solution required 24.21 mL of the NaOH solution for complete
reaction. What is the molarity of the HCl solution? (see #1 for equation)
4. A 34.53 mL sample of a solution of sulfuric acid, H 2SO4, reacts with 27.86 mL of 0.08964 M
NaOH solution. Calculate the molarity and normality of the sulfuric acid solution. (see #2 for
equation)
DOCUMENT INFO
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This page features video tutorials to help you do political analysis with Microsoft Excel. We created this page, and related pages, to supplement our Excel Companion to Political Analysis. Where possible, we leverage existing videos, but have created a number of custom tutorial videos for our textbook.
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# By a vote of 9 to 0, the Supreme Court awarded the Central
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By a vote of 9 to 0, the Supreme Court awarded the Central [#permalink]
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186. By a vote of 9 to 0, the Supreme Court awarded the Central Intelligence Agency broad discretionary powers enabling it to withhold from the public the identities of its sources of intelligence information.
(A) enabling it to withhold from the public
(B) for it to withhold from the public
(C) for withholding disclosure to the public of
(D) that enable them to withhold from public disclosure
(E) that they can withhold public disclosure of
[Reveal] Spoiler: OA
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Re: 186/1000 - Supreme court awarded CIA some powers [#permalink]
### Show Tags
08 Mar 2011, 23:00
There is missing comma before "enabling". A is correct.
fanatico wrote:
186. By a vote of 9 to 0, the Supreme Court awarded the Central Intelligence Agency broad discretionary powers enabling it to withhold from the public the identities of its sources of intelligence information.
(A) enabling it to withhold from the public
(B) for it to withhold from the public
(C) for withholding disclosure to the public of
(D) that enable them to withhold from public disclosure
(E) that they can withhold public disclosure of
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Re: 186/1000 - Supreme court awarded CIA some powers [#permalink]
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08 Mar 2011, 23:02
isn't "it" ambiguous in A and B?
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Re: 186/1000 - Supreme court awarded CIA some powers [#permalink]
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08 Mar 2011, 23:16
Actually there's more to it than meets the eye
The modifier "ing" in A modifies the previous clause and sentence was written with that intention.
In D and E- they and them are spoilers.
B and C - are unidiomatic. - for withhold/withholding
The idiom is "enable to withhold"
fanatico wrote:
isn't "it" ambiguous in A and B?
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Re: 186/1000 - Supreme court awarded CIA some powers [#permalink]
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15 Apr 2011, 01:16
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When there is a pronoun ambiguity in GMAT, one of the best ways of solving the muddle is to apply logic rather than mere structure.
In the given context, there are two eligible contenders for the pronoun’ it’ i.e., Supreme Court and the CIA, but the question is why will Supreme Court award the CIA something in order to enable itself? Isn’t logic missing then? That is the reason to conclude without further doubt that the pronoun refers to the CIA.
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Re: 186/1000 - Supreme court awarded CIA some powers [#permalink]
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27 Apr 2011, 06:50
The Answer is "A" but one must say that it has messy choices .I mean "It" in the underline portion may either refer to supreme court or to CIA!
fanatico wrote:
186. By a vote of 9 to 0, the Supreme Court awarded the Central Intelligence Agency broad discretionary powers enabling it to withhold from the public the identities of its sources of intelligence information.
(A) enabling it to withhold from the public
(B) for it to withhold from the public
(C) for withholding disclosure to the public of
(D) that enable them to withhold from public disclosure
(E) that they can withhold public disclosure of
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Re: By a vote of 9 to 0, the Supreme Court awarded the Central [#permalink]
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01 Jan 2015, 03:37
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IMO - A
By a vote of 9 to 0, the Supreme Court awarded the Central Intelligence Agency broad discretionary powers enabling it to withhold from the public the identities of its sources of intelligence information.
Let's re-write this sentence for more clarity : The Central Intelligence Agency broad was awarded discretionary powers by the Supreme Court, enabling it to withhold from the public the identities of its sources of intelligence information.. Here 'it' refers back to 'The Central Intelligence Agency', not to anyone else.
(A) enabling it to withhold from the public >> this is a cause & effect scenario: V-ing (enabling) is correctly producing the effect of the cause (awarded the Central Intelligence Agency broad discretionary powers). And 'it' is referring back to the CIA (closest noun indeed)
(B) for it to withhold from the public >> this is changing the meaning. Now, it is sounding as if SC awarded the DP to CIAB just to enable it to withhold.....so, nonsensical
(C) for withholding disclosure to the public of >> As per VAN rule (Verb>Adj>Noun), using 'withhold' (verb) is preferred over 'withholding' (noun). Also, this is changing the meaning as explained in B.
(D) that enable them to withhold from public disclosure >> them is incorrect pronoun. Referring to whom??
(E) that they can withhold public disclosure of >> they is incorrect pronoun. Referring to whom??
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Re: By a vote of 9 to 0, the Supreme Court awarded the Central [#permalink]
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20 Apr 2015, 01:43
fanatico wrote:
186. By a vote of 9 to 0, the Supreme Court awarded the Central Intelligence Agency broad discretionary powers enabling it to withhold from the public the identities of its sources of intelligence information.
(A) enabling it to withhold from the public
(B) for it to withhold from the public
(C) for withholding disclosure to the public of
(D) that enable them to withhold from public disclosure
(E) that they can withhold public disclosure of
regarding choice b,
i realy do not know why it is wrong. this meaning is logic though can be considered distorted meaning.
for it to withhold modifies the preceding clause and is good. but if we use "enabling" modifying "power", the meaning relation is tighter.
we need more discussion on this point.
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Re: By a vote of 9 to 0, the Supreme Court awarded the Central [#permalink]
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02 Jul 2015, 07:10
I was confused between A and B.
Chose A over B because B looked awkward to me...
for it to withhold...
the act enabled the intelligence agency to withhold seems more correct
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Posts: 1412
Re: By a vote of 9 to 0, the Supreme Court awarded the Central [#permalink]
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02 Jul 2015, 18:45
fanatico wrote:
186. By a vote of 9 to 0, the Supreme Court awarded the Central Intelligence Agency broad discretionary powers enabling it to withhold from the public the identities of its sources of intelligence information.
(A) enabling it to withhold from the public
(B) for it to withhold from the public
(C) for withholding disclosure to the public of
(D) that enable them to withhold from public disclosure
(E) that they can withhold public disclosure of
"for it to withold" modify the main clause, not "power". this is why B is wrong
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Re: By a vote of 9 to 0, the Supreme Court awarded the Central [#permalink] 02 Jul 2015, 18:45
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Display posts from previous: Sort by | 2,353 | 9,410 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2017-26 | longest | en | 0.908913 |
https://www.begalileo.com/math/math-calculators/decimal-to-percentage-converter/ | 1,679,407,824,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296943698.79/warc/CC-MAIN-20230321131205-20230321161205-00463.warc.gz | 748,581,147 | 35,447 | # Decimal to Percentage Converter
This decimal to percentage converter helps in converting any decimal value to a percentage. Both decimal value and Percentage value are common ways of representing a fraction.
## Uses of decimal to percentage converter
This converter can be used when a single expression has both forms i.e. decimal and percentage. To solve such an expression, either all numbers must be in decimal form or all should be in percentage form. The converter can be used to bring this uniformity by converting the numbers in decimal form to percentage form.
## How to use the calculator
Step 1: Enter the decimal value
Step 2: Click on the Convert button
The percentage form of the entered decimal value will be displayed.
You can click on the checkbox beside the text “Solution” to view how the conversion was done.
You can click on the Reset button to find the percentage value of another decimal value.
## Examples to try
Decimal Value: 0.56
Percentage Value: 56%
Solution explained:
Multiply the decimal value by 100:
0.56 * 100 = 56%
Decimal Value: 0.89
Percentage value: 89%
Solution:
0.89 * 100 = 89%
Decimal Value: 6.99
Percentage Value: 699%
Solution:
6.99 * 100 = 699%
Converting decimals to percentages is an important skill for a variety of applications, from academic work to personal finance. In order to convert a decimal to a percentage, you need to multiply the decimal by 100 and add the “%” symbol. A decimal to percentage converter makes this process quick and easy.
Decimals are used to represent values that are not necessarily whole numbers, and are written with a decimal point followed by one or more digits. For example, the decimal 0.5 represents half of a whole, while the decimal 0.75 represents three-quarters of a whole.
Percentages are used to represent values as a portion of 100, and are written with the “%” symbol after the number. For example, the percentage 50% represents half of a whole, while the percentage 75% represents three-quarters of a whole.
Converting a decimal to a percentage is a simple process that involves multiplying the decimal by 100 and adding the “%” symbol. For example, to convert the decimal 0.5 to a percentage, you would multiply it by 100 to get 50, and then add the “%” symbol to get 50%. Similarly, to convert the decimal 0.75 to a percentage, you would multiply it by 100 to get 75, and then add the “%” symbol to get 75%.
A decimal to percentage converter makes this process even easier by allowing you to enter the decimal value and automatically converting it to a percentage. Some converters allow you to specify the number of decimal places to include in the output, while others automatically round the output to a certain number of decimal places.
In addition to converting decimals to percentages, some converters can also convert percentages to decimals, and some can convert fractions to percentages or vice versa.
There are many different applications for decimal to percentage conversion in various fields. In mathematics, decimal to percentage conversion is used to express ratios and proportions as percentages, and to calculate percentages of quantities. In science, decimal to percentage conversion is used to express measurements and data in a more meaningful way, and to compare different measurements or data sets. In business and finance, decimal to percentage conversion is used to calculate interest rates, investment returns, and other financial metrics.
One important concept related to decimal to percentage conversion is the idea of significant figures. When converting a decimal to a percentage, it is important to consider the number of significant figures in the original decimal and the desired level of precision in the output percentage. For example, if the original decimal value is 0.123456 and the desired output precision is two significant figures, then the percentage equivalent would be 12.%, which represents an approximation of the original value. However, if the original decimal value is 0.123450 and the desired output precision is also two significant figures, then the percentage equivalent would be 12.%, which represents a more precise approximation of the original value.
Another important concept related to decimal to percentage conversion is the idea of rounding. When converting a decimal to a percentage, it is important to consider the precision of the original decimal and the desired precision of the output. For example, if the original decimal value is 0.123456 and the desired output precision is two decimal places, then the percentage equivalent would be 12.3456%, which can be rounded to 12.35%. However, if the original decimal value is 0.123450 and the desired output precision is also two decimal places, then the percentage equivalent would be 12.3450%, which can be rounded to 12.35%.
Overall, a decimal to percentage converter is a useful tool that can help you quickly and accurately convert decimals to percentages for a variety of applications.
per-sc-ce3d2z-0 LzDqj"> | 1,041 | 5,041 | {"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} | 3.9375 | 4 | CC-MAIN-2023-14 | longest | en | 0.843219 |
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