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https://quant.stackexchange.com/questions/20879/need-help-understanding-basics-of-cash-flow-engineering | 1,624,269,813,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488269939.53/warc/CC-MAIN-20210621085922-20210621115922-00577.warc.gz | 429,108,864 | 36,026 | # Need help understanding basics of cash flow engineering
I'm studying Financial Engineering, a subject I'm completely new to.
I'm using Principles of Financial Engineering 3rd Edition and trying to solve the exercises of the chapter on Cash Flow Engineering.
The first question states:
You have a 4-year coupon bond that pays semiannual interest. The coupon rate is 8% and the par value is 100.
a. Can you construct a synthetic equivalent of this bond? Be explicit and show your cash flows.
Unfortunately, I don't know how to proceed with this. I'm not able to understand how the cash flows would be split.
Would it be 8 separate payments?
And if yes, what would the payment amount be at each ti, i=1,2,...8 and why?
• Try thinking about the following things. Do you understand exactly what the payouts are of your 4-year coupon bond (that pays semi annually)? How many payouts does your bond have? when are they? what is the amount at each period? Once you got that figured out, how does having zero coupons help you in replicating your original bond? – mbison Sep 23 '15 at 22:44
• "Annual coupon of x paid semiannually" means payments of x/2 each, spaced 6 months apart. – Alex C Sep 24 '15 at 2:21 | 286 | 1,211 | {"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 | 3 | CC-MAIN-2021-25 | latest | en | 0.949185 |
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### Probability question - chances [duplicate]
A game show offers contestants the following chance to win a car: There are three doors. A car is hidden behind one door, and goats are hidden behind each of the other doors. The contestant selects a ... | 789 | 3,606 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2016-30 | latest | en | 0.944205 |
https://www.supergb.com/cbt/assessment/c8239876-7dcf-409b-b488-19788212cabf/question/639bc556-8515-4cc9-9943-0e13fe0f2469 | 1,720,879,155,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514494.35/warc/CC-MAIN-20240713114822-20240713144822-00836.warc.gz | 825,339,099 | 5,453 | # A machine whose efficiency is 60% has a velocity ratio of 5. If a force of 500N is applied to lift a load P, what is the magnitude of P?
Question 1
A machine whose efficiency is 60% has a velocity ratio of 5. If a force of 500N is applied to lift a load P, what is the magnitude of P? | 82 | 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} | 2.546875 | 3 | CC-MAIN-2024-30 | latest | en | 0.962898 |
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### Go Spaceship Go
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Show that even a very powerful spaceship would eventually run out of overtaking power
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How fast would you have to throw a ball upwards so that it would never land?
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A ball whooshes down a slide and hits another ball which flies off the slide horizontally as a projectile. How far does it go?
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Gravity on the Moon is about 1/6th that on the Earth. A pole-vaulter 2 metres tall can clear a 5 metres pole on the Earth. How high a pole could he clear on the Moon?
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Consider the mechanics of pole vaulting
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Explore the energy of this incredibly energetic particle which struck Earth on October 15th 1991
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How high can a high jumper jump? How can a high jumper jump higher without jumping higher? Read on... | 359 | 1,526 | {"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-2019-22 | latest | en | 0.86698 |
https://measuringstuff.com/11-things-that-are-15-feet-long/ | 1,701,586,310,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100489.16/warc/CC-MAIN-20231203062445-20231203092445-00256.warc.gz | 451,870,161 | 48,745 | # 11 Things That Are 15 Feet Long
Many people use 15 feet to describe a certain length or distance, but this can be unhelpful if you don’t have a frame of reference.
Keeping a few examples in mind can help you get a much better feel for just how long or tall 15 feet is.
## One to Two Sunflowers
There are many varieties of sunflowers, and each variety grows to a different height.
Some miniature sunflowers grow to only 1 foot tall, while others can be much taller.
The average height of a sunflower is between 6 and 10 feet, so you would need about one and a half or two sunflower plants to equal 15 feet.
However, mammoth, titan, California Greystripe, and skyscraper sunflowers can grow up to 14 feet, and some might even reach a height of 20 feet.
The flower parts of these massive plants can be as large as 2 feet across.
## Six Footsteps
Everyone’s walking stride differs, and people take steps at different lengths depending on their height and other factors.
However, an average footstep length is about 2.5 feet, which means you would need to take about six steps to equal 15 feet.
A full stride is two steps and is therefore double the average step, so you would need to take about three strides to equal 15 feet.
## Two Artificial Christmas Trees
You can purchase an artificial Christmas tree in a variety of heights so that you can ensure it will fit in whatever room you want to use it in. One of the most common heights is 7.5 feet.
If you had two artificial Christmas trees of this height in boxes and you lined them up end to end, they would equal 15 feet.
## One Canoe
Canoes can be a few different sizes, and their length varies depending on what they will be used for and how many people will sit in them.
Most canoes, however, are between 14.5 and 17 feet long, and the most common canoe length is 16 feet. This means that the average canoe is just over 15 feet in length.
If you can picture a canoe, you can easily estimate what 15 feet looks like.
Kayaks are also close to 15 feet in length, with the longest kayak measuring around 16 feet.
Most kayaks, however, are closer to 10 feet long, so you would need about one and a half kayaks to equal 15 feet.
## One SUV
The average length of a vehicle is about 14 feet long, which is just shy of 15 feet. A small SUV is usually right around, if not exactly, 15 feet in length.
If you can picture any car, you can get a good feel for a 15-foot length estimate, but picturing an SUV will often give you a more exact estimate.
## A Giraffe
Most giraffes are between 14 and 19 feet tall. What the giraffe’s diet is like, what its genetics are, and what its adolescence was like can all impact how tall it will grow.
CHECK OUT Convert 208 MM Into Inches
Male giraffes often tend to be a bit taller than female giraffes as well. The average giraffe, however, is right around 15 feet tall, so they make a great estimate for that length or height.
A giraffe’s neck makes up almost half of its height. Most giraffes carry their necks at a 50 or 60-degree angle, but they can stretch to a 90-degree angle to reach food.
Most people think of giraffes as one species with several subspecies, but recent research has shown that there might be four or five distinct giraffe species.
Giraffes can weigh as much as 2,630 pounds, and they can live for up to 38 years.
## Five Baseball Bats
Baseball bats can vary somewhat in length, depending on how tall the person using them is. However, most adult baseball bats are right around, if not exactly, 3 feet long.
This means that if you lined up five bats end to end, they would equal 15 feet.
Picturing a baseball bat is an excellent way to break down 15 feet evenly.
## Three Park Benches
Most park benches are designed for two people to sit on, and they are relatively uniform in size.
A park bench is generally about 5 feet long, and many are exactly this length. This means that three park benches together equal 15 feet.
A bench is something most people are familiar with, so this makes for an excellent estimate for 15 feet.
One of the most common step ladder sizes is 15 feet. Many ladders are 15 feet and are designed to expand to a greater height, while other ladders might be 10 feet tall or long when they’ve been collapsed but can expand to 15 feet.
A 15-foot step ladder is a great choice for painting tall walls or reaching high ceilings. These ladders are also helpful if you need to reach something on your roof or if you need to trim a tall tree.
## Three Standing Lamps
Most standing lamps, such as what you might find in a living room office, or library, are about five feet tall, so picturing three of these lamps will bring you close to 15 feet.
Although this is the most common lamp height, as it’s perfect for shining light over the back of a chair or couch, other standing lamps can be much taller.
If the lamp is between 7 and 8 feet tall, you will only need two to equal close to 15 feet.
## An Extension Cord
Extension cords are extremely helpful, especially if you don’t have enough outlets or if you need to plug multiple items in.
One of the most common lengths for an extension cord is 15 feet. You can find regular extension cords with one plug at each end, but you can also purchase an extension cord with a power strip at one end, into which you can plug many items.
Extension cords for phones or that have USB ports are also commonly available in a length of 15 feet. | 1,268 | 5,453 | {"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-2023-50 | longest | en | 0.963232 |
https://www.ukessays.com/essays/general-studies/prime-numbers-divide.php | 1,607,158,579,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141747323.98/warc/CC-MAIN-20201205074417-20201205104417-00513.warc.gz | 891,539,696 | 15,922 | Prime Numbers: History, Facts and Examples
2565 words (10 pages) Essay
16th Jul 2019 General Studies Reference this
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Prime Numbers: An Introduction Prime number is the number, which is greater than 1 and cannot be divided by any number excluding itself and one. A prime number is a positive integer that has just two positive integer factors, including 1 and itself. Such as, if the factors of 28 are listed, there are 6 factors that are 1, 2, 4, 7, 14, and 28. Similarly, if the factors of 29 are listed, there are only two factors that are 1 and 29. Therefore, it can be inferred that 29 is a prime number, but 28 is not. Examples of prime numbers The first few prime numbers are as follows: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, etc. Identifying the primes The ancient Sieve of Eratosthenes is a simple way to work out all prime numbers up to a given limit by preparing a list of all integers and repetitively striking out multiples of already found primes. There is also a modern Sieve of Atkin, which is more complex when compared to that of Eratosthenes. A method to determine whether a number is prime or not, is to divide it by all primes less than or equal to the square root of that number. If the results of any of the divisions are an integer, the original number is not a prime and if not, it is a prime. One need not actually calculate the square root; once one sees that the quotient is less than the divisor, one can stop. This is called as the trial division, which is the simplest primality test but it is impractical for testing large integers because the number of possible factors grows exponentially as the number of digits in the number to be tested increases. Primality tests: A primality test algorithm is an algorithm that is used to test a number for primality, that is, whether the number is a prime number or not.
• AKS primality test
The AKS primality test is based upon the equivalence (x - a)n = (xn - a) (mod n) for a coprime to n, which is true if and only if n is prime. This is a generalization of Fermat's little theorem extended to polynomials and can easily be proven using the binomial theorem together with the fact that: for all 0 < k < n if n is prime. While this equivalence constitutes a primality test in itself, verifying it takes exponential time. Therefore AKS makes use of a related equivalence (x - a)n = (xn - a) (mod n, x r - 1), which can be checked in polynomial time.
• Fermat primality test
Fermat's little theorem asserts that if p is prime and 1 a < p, then a p -1≡ 1 (mod p) In order to test whether p is a prime number or not, one can pick random a's in the interval and check if there is an equality.
• Solovay-Strassen primality test
For a prime number p and any integer a, A (p -1)/2 ≡ (a/p) (mod p) Where (a/p) is the Legendre symbol. The Jacobi symbol is a generalisation of the Legendre symbol to (a/n); where n can be any odd integer. The Jacobi symbol can be computed in time O((log n)²) using Jacobi's generalization of law of quadratic reciprocity. It can be observed whether or not the congruence A (n -1)/2 ≡ (a/n) (mod n) holds for various values of a. This congruence is true for all a's if n is a prime number. (Solovay, Robert M. and Volker Strassen, 1977)
• Lucas-Lehmer test
This test is for a natural number n and in this test, it is also required that the prime factors of n − 1 should be already known. If for every prime factor (q) of n − 1, there exists an integer a less than n and greater than 1 such as a n -1 ≡1 (mod n) and then a n -1/q 1 (mod n) then n is prime. If no such number can be found, n is composite number.
• Miller-Rabin primality test
If we can find an a such that ad ≡ 1 (mod n), and a2nd -1 (mod n) for all 0 ≤ r ≤ s - 1 then ‘a' proves the compositeness of n. If not, ‘a' is called a strong liar, and n is a strong probable prime to the base a. “Strong liar” refers to the case where n is composite but yet the equations hold as they would for a prime number. There are several witnesses ‘a' for every odd composite n. But, a simple way to generate such an ‘a' is known. Making the test probabilistic is the solution: we choose randomly, and check whether it is a witness for the composite nature of n. If n is composite, majority of the ‘a's are witnesses, therefore the test will discover n as a composite number with high probability. (Rabin, 1980) A probable prime is an integer, which is considered to be probably prime by passing a certain test. Probable primes, which are actually composite (such as Carmichael numbers) are known as pseudoprimes. Besides these methods, there are other methods also. There is a set of Diophantine equations in 9 variables and one parameter in which the parameter is a prime number only if the resultant system of equations has a solution over the natural numbers. A single formula with the property of all the positive values being prime can be obtained with this method. There is another formula that is based on Wilson's theorem. The number ‘two' is generated several times and all other primes are generated exactly once. Also, there are other similar formulas that can generate primes. Some primes are categorized as per the properties of their digits in decimal or other bases. An example is that the numbers whose digits develop a palindromic sequence are palindromic primes, and if by consecutively removing the first digit at the left or the right generates only new prime numbers, a prime number is known as a truncatable prime. The first 5,000 prime numbers can be known very quickly by just looking at odd numbers and checking each new number (say 5) against every number above it (3); so if 5Mod3 = 0 then it's not a prime number. History of prime numbers The most ancient and acknowledged proof for the statement that “There are infinitely many prime numbers”, is given by Euclid in his Elements (Book IX, Proposition 20). The Sieve of Eratosthenes is a simple, ancient algorithm to identify all prime numbers up to a particular integer. After this, came the modern Sieve of Atkin, which is faster but more complex. The Sieve of Eratosthenes was created in the 3rd century BC by Eratosthenes. Some clues can be found in the surviving records of the ancient Egyptians regarding their knowledge of prime numbers: for example, the Egyptian fraction expansions in the Rhind papyrus have fairly different forms for primes and for composites. But, the first surviving records of the clear study of prime numbers come from the Ancient Greeks. Euclid's Elements (circa 300 BC) include key theorems about primes, counting the fundamental theorem of arithmetic and the infinitude of primes. Euclid also explained how a perfect number is constructed from a Mersenne prime. After the Greeks, nothing special happened with the study of prime numbers till the 17th century. In 1640, Pierre de Fermat affirmed Fermat's little theorem, which was later on proved by Leibniz and Euler. Chinese may have identified a special case of Fermat's theorem much earlier. Fermat assumed that all numbers of the form 22n + 1 are prime and he proved this up to n = 4. But, the subsequent Fermat number 232+1 is composite; whose one prime factor is 641). This was later on discovered by Euler and now no further Fermat numbers are recognized as prime numbers. A French monk, Marin Mersenne looked at primes of the form 2p - 1, with p as a prime number. They are known as Mersenne primes after his name. Euler showed that the infinite series 1/2 + 1/3 + 1/5 + 1/7 + 1/11 + … is divergent. In 1747, Euler demonstrated that even the perfect numbers are in particular the integers of the form 2p-1(2p-1), where the second factor is a Mersenne prime. It is supposed that there are no odd perfect numbers, but it is not proved yet. In the beginning of the 19th century, Legendre and Gauss independently assumed that because x tends to infinity, the number of primes up to x is asymptotic to x/log(x), where log(x) is the natural logarithm of x. Awards for finding primes A prize of US\$100,000 has been offered by the Electronic Frontier Foundation (EFF) to the first discoverers of a prime with a minimum 10 million digits. Also, \$150,000 for 100 million digits, and \$250,000 for 1 billion digits has been offered. In 2000, \$50,000 for 1 million digits were paid. Apart from this, prizes up to US\$200,000 for finding the prime factors of particular semi-primes of up to 2048 bits were offered by the RSA Factoring Challenge. Facts about prime numbers
• 73939133 is an amazing prime number. If the last or the digit at the units place is removed, every time you will get a prime number. It is the largest known prime with this property. Because, all the numbers which we get after removing the end digit of the number are also prime numbers. They are as follows: 7393913, 739391, 73939, 7393, 739, 73 and 7. All these numbers are prime numbers. This is a distinct quality of the number 73939133, which any other number does not have. (Amazing number facts, 2008)
• The only even prime number is 2. All other even numbers can be divided by 2. So, they are not prime numbers.
• Zero and 1 are not considered to be prime numbers.
• If the sum of the digits of a number is a multiple of 3, that number can be divided by 3.
• With the exception of 0 and 1, a number is either a prime number or a composite number. A composite number is identified as any number that is greater than 1 and that is not prime.
• The last digit of a prime number greater than 5 can never be 5. Any number greater than 5 whose last digit is 5 can be divided by 5. (Prime Numbers, 2008)
1/2 0.5 Terminates 1/3 0.33333... Repeating block: 1 digit 1/5 0.2 Terminates 1/7 0.1428571428... Repeating block: 6 digits 1/11 0.090909... Repeating block: 2 digits 1/13 0.0769230769... Repeating block: 6 digits 1/17 0.05882352941176470588... Repeating block: 16 digits 1/19 0.0526315789473684210526... Repeating block: 18 digits 1/23 0.04347826086956521739130434... Repeating block: 22 digits
For some of the prime numbers, the size of the repeating block is 1 less than the prime. These are known as Golden Primes. 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97 9 primes out of the 25 (less than 100) are golden primes; this forms 36% (9/25). (Amazing number facts, 2008) Examples of mathematicians specialized in prime numbers Arthur Wieferich, D. D. Wall, Zhi Hong Sun and Zhi Wei Sun, Joseph Wolstenholme, Joseph Wolstenholme, Euclid, Eratosthenes. Applications of prime numbers For a long time, the number theory and the study of prime numbers as well was seen as the canonical example of pure mathematics with no applications beyond the self-interest of studying the topic. But, in the 1970s, it was publicly announced that prime numbers could be used as a basis for creating the public key cryptography algorithms. They were also used for hash tables and pseudorandom number generators. A number of rotor machines were designed with a different number of pins on each rotor. The number of pins on any one rotor was either prime, or co-prime to the number of pins on any other rotor. With this, a full cycle of possible rotor positions (before repeating any position) was generated. Prime numbers in the arts and literature Also, prime numbers have had a significant influence on several artists and writers. The French composer Olivier Messiaen created ametrical music through "natural phenomena" with the use of prime numbers. In his works, La Nativité du Seigneur (1935) and Quatre études de rythme (1949-50), he has used motifs with lengths given by different prime numbers to create unpredictable rhythms: 41, 43, 47 and 53 are the primes that appear in one of the études. A scientist of NASA, Carl Sagan recommended (in his science fiction ‘Contact') that prime numbers could be used for communication with the aliens. The award-winning play ‘Arcadia' by Tom Stoppard was a willful attempt made to discuss mathematical ideas on the stage. In the very first scene, the 13 year old heroine baffles over the Fermat's last theorem (theorem that involves prime numbers). A popular fascination with the mysteries of prime numbers and cryptography has been seen in various films. References Amazing number facts, 2008. Retrieved April 28, 2008 from http://www.madras.fife.sch.uk/maths/amazingnofacts/fact018.html Prime Numbers, 2008. Retrieved April 28, 2008 from http://www.factmonster.com/ipka/A0876084.html Solovay, Robert M. & Strassen, V. (1977). "A fast Monte-Carlo test for primality". SIAM Journal on Computing 6 (1): 84-85. Rabin, M.O. (1980). Probabilistic algorithm for testing primality, Journal of Number Theory 12, no. 1, pp. 128-138.
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https://rdrr.io/bioc/aroma.light/man/normalizeAffine.html | 1,527,471,707,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794870604.65/warc/CC-MAIN-20180528004814-20180528024814-00430.warc.gz | 628,344,480 | 20,076 | # normalizeAffine: Weighted affine normalization between channels and arrays In aroma.light: Light-Weight Methods for Normalization and Visualization of Microarray Data using Only Basic R Data Types
## Description
Weighted affine normalization between channels and arrays.
This method will remove curvature in the M vs A plots that are due to an affine transformation of the data. In other words, if there are (small or large) biases in the different (red or green) channels, biases that can be equal too, you will get curvature in the M vs A plots and this type of curvature will be removed by this normalization method.
Moreover, if you normalize all slides at once, this method will also bring the signals on the same scale such that the log-ratios for different slides are comparable. Thus, do not normalize the scale of the log-ratios between slides afterward.
It is recommended to normalize as many slides as possible in one run. The result is that if creating log-ratios between any channels and any slides, they will contain as little curvature as possible.
Furthermore, since the relative scale between any two channels on any two slides will be one if one normalizes all slides (and channels) at once it is possible to add or multiply with the same constant to all channels/arrays without introducing curvature. Thus, it is easy to rescale the data afterwards as demonstrated in the example.
## Usage
```1 2 3``` ```## S3 method for class 'matrix' normalizeAffine(X, weights=NULL, typeOfWeights=c("datapoint"), method="L1", constraint=0.05, satSignal=2^16 - 1, ..., .fitOnly=FALSE) ```
## Arguments
`X` An NxK `matrix` (K>=2) where the columns represent the channels, to be normalized. `weights` If `NULL`, non-weighted normalization is done. If data-point weights are used, this should be a `vector` of length N of data point weights used when estimating the normalization function. `typeOfWeights` A `character` string specifying the type of weights given in argument `weights`. `method` A `character` string specifying how the estimates are robustified. See `iwpca`() for all accepted values. `constraint` Constraint making the bias parameters identifiable. See `fitIWPCA`() for more details. `satSignal` Signals equal to or above this threshold will not be used in the fitting. `...` Other arguments passed to `fitIWPCA`() and in turn `iwpca`(). For example, the weight argument of `iwpca`(). See also below. `.fitOnly` If `TRUE`, the data will not be back-transform.
## Details
A line is fitted robustly throught the (y_R,y_G) observations using an iterated re-weighted principal component analysis (IWPCA), which minimized the residuals that are orthogonal to the fitted line. Each observation is down-weighted by the inverse of the absolute residuals, i.e. the fit is done in L_1.
## Value
A NxK `matrix` of the normalized channels. The fitted model is returned as attribute `modelFit`.
## Negative, non-positive, and saturated values
Affine normalization applies equally well to negative values. Thus, contrary to normalization methods applied to log-ratios, such as curve-fit normalization methods, affine normalization, will not set these to `NA`.
Data points that are saturated in one or more channels are not used to estimate the normalization function, but they are normalized.
## Missing values
The estimation of the affine normalization function will only be made based on complete non-saturated observations, i.e. observations that contains no `NA` values nor saturated values as defined by `satSignal`.
## Weighted normalization
Each data point/observation, that is, each row in `X`, which is a vector of length K, can be assigned a weight in [0,1] specifying how much it should affect the fitting of the affine normalization function. Weights are given by argument `weights`, which should be a `numeric` `vector` of length N. Regardless of weights, all data points are normalized based on the fitted normalization function.
## Robustness
By default, the model fit of affine normalization is done in L_1 (`method="L1"`). This way, outliers affect the parameter estimates less than ordinary least-square methods.
For further robustness, downweight outliers such as saturated signals, if possible.
We do not use Tukey's biweight function for reasons similar to those outlined in `calibrateMultiscan`().
## Using known/previously estimated channel offsets
If the channel offsets can be assumed to be known, then it is possible to fit the affine model with no (zero) offset, which formally is a linear (proportional) model, by specifying argument `center=FALSE`. In order to do this, the channel offsets have to be subtracted from the signals manually before normalizing, e.g. `Xa <- t(t(X)-a)` where `e` is `vector` of length `ncol(X)`. Then normalize by `Xn <- normalizeAffine(Xa, center=FALSE)`. You can assert that the model is fitted without offset by `stopifnot(all(attr(Xn, "modelFit")\$adiag == 0))`.
Henrik Bengtsson
## References
[1] Henrik Bengtsson and Ola Hössjer, Methodological Study of Affine Transformations of Gene Expression Data, Methodological study of affine transformations of gene expression data with proposed robust non-parametric multi-dimensional normalization method, BMC Bioinformatics, 2006, 7:100.
`calibrateMultiscan`().
``` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104``` ```pathname <- system.file("data-ex", "PMT-RGData.dat", package="aroma.light") rg <- read.table(pathname, header=TRUE, sep="\t") nbrOfScans <- max(rg\$slide) rg <- as.list(rg) for (field in c("R", "G")) rg[[field]] <- matrix(as.double(rg[[field]]), ncol=nbrOfScans) rg\$slide <- rg\$spot <- NULL rg <- as.matrix(as.data.frame(rg)) colnames(rg) <- rep(c("R", "G"), each=nbrOfScans) layout(matrix(c(1,2,0,3,4,0,5,6,7), ncol=3, byrow=TRUE)) rgC <- rg for (channel in c("R", "G")) { sidx <- which(colnames(rg) == channel) channelColor <- switch(channel, R="red", G="green") # - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - # The raw data # - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - plotMvsAPairs(rg[,sidx]) title(main=paste("Observed", channel)) box(col=channelColor) # - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - # The calibrated data # - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - rgC[,sidx] <- calibrateMultiscan(rg[,sidx], average=NULL) plotMvsAPairs(rgC[,sidx]) title(main=paste("Calibrated", channel)) box(col=channelColor) } # for (channel ...) # - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - # The average calibrated data # # Note how the red signals are weaker than the green. The reason # for this can be that the scale factor in the green channel is # greater than in the red channel, but it can also be that there # is a remaining relative difference in bias between the green # and the red channel, a bias that precedes the scanning. # - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - rgCA <- rg for (channel in c("R", "G")) { sidx <- which(colnames(rg) == channel) rgCA[,sidx] <- calibrateMultiscan(rg[,sidx]) } rgCAavg <- matrix(NA_real_, nrow=nrow(rgCA), ncol=2) colnames(rgCAavg) <- c("R", "G") for (channel in c("R", "G")) { sidx <- which(colnames(rg) == channel) rgCAavg[,channel] <- apply(rgCA[,sidx], MARGIN=1, FUN=median, na.rm=TRUE) } # Add some "fake" outliers outliers <- 1:600 rgCAavg[outliers,"G"] <- 50000 plotMvsA(rgCAavg) title(main="Average calibrated (AC)") # - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - # Normalize data # - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - # Weight-down outliers when normalizing weights <- rep(1, nrow(rgCAavg)) weights[outliers] <- 0.001 # Affine normalization of channels rgCANa <- normalizeAffine(rgCAavg, weights=weights) # It is always ok to rescale the affine normalized data if its # done on (R,G); not on (A,M)! However, this is only needed for # esthetic purposes. rgCANa <- rgCANa *2^1.4 plotMvsA(rgCANa) title(main="Normalized AC") # Curve-fit (lowess) normalization rgCANlw <- normalizeLowess(rgCAavg, weights=weights) plotMvsA(rgCANlw, col="orange", add=TRUE) # Curve-fit (loess) normalization rgCANl <- normalizeLoess(rgCAavg, weights=weights) plotMvsA(rgCANl, col="red", add=TRUE) # Curve-fit (robust spline) normalization rgCANrs <- normalizeRobustSpline(rgCAavg, weights=weights) plotMvsA(rgCANrs, col="blue", add=TRUE) legend(x=0,y=16, legend=c("affine", "lowess", "loess", "r. spline"), pch=19, col=c("black", "orange", "red", "blue"), ncol=2, x.intersp=0.3, bty="n") plotMvsMPairs(cbind(rgCANa, rgCANlw), col="orange", xlab=expression(M[affine])) title(main="Normalized AC") plotMvsMPairs(cbind(rgCANa, rgCANl), col="red", add=TRUE) plotMvsMPairs(cbind(rgCANa, rgCANrs), col="blue", add=TRUE) abline(a=0, b=1, lty=2) legend(x=-6,y=6, legend=c("lowess", "loess", "r. spline"), pch=19, col=c("orange", "red", "blue"), ncol=2, x.intersp=0.3, bty="n") ``` | 2,674 | 9,204 | {"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-2018-22 | longest | en | 0.852081 |
https://hawthornecommunitycouncil.org/inspiration-complementary-and-supplementary-angles-worksheet-1-answers/ | 1,652,927,981,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662522741.25/warc/CC-MAIN-20220519010618-20220519040618-00443.warc.gz | 355,014,098 | 12,370 | Beautiful Complementary And Supplementary Angles Worksheet 1 Answers
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Complementary And Supplementary Angles Practice Activity Worksheet Fun Way To Review Complementary A Geometry Worksheets Practices Worksheets Angles Worksheet | 1,641 | 7,818 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2022-21 | latest | en | 0.824814 |
https://www.physicsforums.com/threads/calculating-albedo-and-eccentricity.810391/ | 1,526,998,520,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794864790.28/warc/CC-MAIN-20180522131652-20180522151652-00037.warc.gz | 802,033,175 | 15,200 | # Calculating albedo and eccentricity
Tags:
1. Apr 24, 2015
### Chief17
I have a three part question:
Background: For a planet on an orbit with semi-major axis a and eccentricity e, the distance of closest approach to the Sun is r = a(1 − e) and the farthest approach is r = a(1 + e).
(1) Assuming an albedo A = 0.2, estimate the temperature on Earth in equilibrium with irradiation from
the Sun. Estimate the correction factor necessary due to the greenhouse effect to bring us up to a balmy 300 K.
I don't know what to do.
(2) Assuming that this correction factor does not change, how large an eccentricity could the Earth have before the temperature extremes reach the point where the Earth reaches either boiling or freezing point?
Freezing:
Based off the equation T=T⊙((1-A)/4)^0.25(R⊙/r)^0.5,
273.15=5778((1-0.2)/4)^0.25(6.96*10^8/r)^0.5
0.005=(6.96/r)
r=1.39*10^11
Boiling:
373.15=5778((1-0.2)/4)^0.25(6.96*10^8/r)^0.5
r=7.46*10^10
These both seem reasonable to me except for one thing. The actual distance from the Sun to the Earth is 1.5*10^11. This means that the Earth is actually farther than the distance I calculated for the boiling part, which doesn't make sense.
Is this the equation I should use, and is the work (and answer) correct? Or did I do something wrong?
(3) If we define habitability as having a level of irradiation between these two extremes, consider the
habitable zone around a lower mass star. Assuming circular orbits again, and the same greenhouse correction factor as above, where is the habitable zone around a 0.5M⊙ star, which has radius 0.5R⊙ and effective temperature 3700 K?
Not sure where exactly to get started here.
Do I use the equations
L=4*pi*R⊙^2*stefan-boltzmann constant*T⊙^4
Labs=((R⊙^2*stefan-boltzmann constant*T⊙^4*pi*R^2)/r^2)(1-A)?
Not sure where the mass of the star fits in here.
2. Apr 25, 2015
### Staff: Mentor
(1): You have the formula in (2), you just have to plug in numbers for the first part. And then find out which prefactor you need in the equation to get the actual average temperature.
That comes from correction factor you have to include.
(3): right
You don't need it.
3. Apr 25, 2015
### Chief17
Not sure what you mean by prefactor. Would I just multiply the albedo times a value (i.e. x) and then solve for x? So:
300=5778((1-0.2x)/4)^0.25(6.96*10^8/(1.5*10^11))^0.5
----> x=-1.75
4. Apr 25, 2015
### Staff: Mentor
I'm sure you have a definition of the correction factor somewhere. | 760 | 2,488 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2018-22 | latest | en | 0.891351 |
http://www.docstoc.com/docs/137155305/Conditional-Averaging-a-New-Algorithm-for-Digital-Filter | 1,438,393,699,000,000,000 | text/html | crawl-data/CC-MAIN-2015-32/segments/1438042988399.65/warc/CC-MAIN-20150728002308-00126-ip-10-236-191-2.ec2.internal.warc.gz | 407,529,570 | 45,795 | # Conditional Averaging a New Algorithm for Digital Filter by ides.editor
VIEWS: 32 PAGES: 5
• pg 1
``` ACEEE Int. J. on Signal & Image Processing, Vol. 01, No. 03, Dec 2010
Conditional Averaging
a New Algorithm for Digital Filter
Sukesh Rao M, NMAMIT,
E&C Dept, Nitte, India.
Email: sukesh_muligar@yahoo.co.in
Dr. S. Narayana Iyer, NMAMIT, Nitte, India.
Email: iyer_narayana@yahoo.com
Abstract—This paper aims at designing a new algorithm for 1) For a fixed sampling frequency, the sample of the
digital filters. The traditional methods like FIR, IIR have been incoming signal is processed for three different levels of
improved in recent times with new approaches. However, the conditionality, resulting in three different types of cutoff
developments have used complex arithmetic calculation and frequencies or the signal bandwidth.
dedicated DSP processors. In this research project, effort has
been made to reduce such complexities using a procedure
2) A relation is formed between the conditional statements
based on the technique of Conditional Averaging. The entire and the cutoff frequencies which remain true for any
algorithm is developed using more of conditional statements number of remaining conditional statements and the
and less of arithmetic calculations. desired cutoff frequencies.
Digital signals are filtered at different stages of 3) A mixed harmonic signal is fed to the system to obtain
signal processing. However high speed processor is used for the system response for different cutoff frequencies.
different calculations associated with filtration process. An
averaging is one such scheme used in simple FIR filter, which II. CONDITIONAL AVERAGING- THE PROPOSED
performs low pass filtering operation. Conditional Averaging
TECHNIQUE
is a new technique, which is one of the improvements in
continuous time averaging. Conditional Averaging algorithm Conditional averaging is a new scheme proposed in the
is explained in this practice with different examples for the area of different averaging techniques [1]. Simple
design of low pass filter. This algorithm has been successfully averaging with N points will reduce any of the AC
tested using digital starter kit with TMS3206416v DSP
components corresponding to the amplitude variations. All
processor. Using code composer studio, the entire algorithm is
written in C/C++ language and compiled into an assembly averaging techniques act as low pass filter with the filter
language. Conditional averaging can be implemented with any coefficients equal to 1/N. conditional averaging will not
general purpose processor to arrive at other types of filters use any such kind of filter coefficients and any definite
with certain necessary modifications. mathematical relation as such.
The Fast Fourier Transform (FFT) is the method used in
Index Terms— FFT, DSP, LPF
the DSP (digital signal processing) to find out the
frequency spectrum of any given signal. It is a
I. INTRODUCTION
mathematical operation for obtaining the accurate
Filter is a system which selectively changes the wave frequency spectrum. It is also possible to predict the
shape, amplitude, frequency content and phase frequency spectrum of the signal with signal amplitude
characteristic of a signal as desired. In digital signal variations. The spectrum so obtained may not be accurate
processing, based on the design constraints digital filters of and it is approximated. If we know the sampling frequency
FIR, IIR or even other types are commonly in usage. In real or the timing interval of the incoming signal, it is possible
time signal processing, these filters use dedicated to predict the harmonics present in the signal. Figure 1
Microprocessor system to carry out complex floating point shows a random signal. We can determine the frequency
arithmetic operation. To acquire high accuracy and high plot accurately for the signal shown in figure 1 using the
precision of the system response, high frequency frequency transform techniques. Also possible to predict
supportive DSP processor is used. The scope of this work approximately the harmonics present in the signal by
namely Conditional Averaging is to minimize the complex knowing the sampling frequency. Let us assume that the
arithmetic operation and obtain better response for the signal is sampled at a rate of fs Hz and figure 2 (a) shows
required design using general purpose microprocessor. the resulting discrete samples.
From the sampled signal, one can conclude that, (1/ fs)
A. Problem Definition
is the time gap between two samples. If the signal varies
A specially devised technique of conditional averaging with more number of successive samples, then it
has been developed to design a low pass filter. The term corresponds to lower harmonics
‘conditional averaging’ has been assigned to this method as Fluctuation with lesser number of successive samples
it performs the averaging of a given data set based on corresponds to higher harmonics. All such samples
certain attributed conditions. This algorithm has been tested which contribute to the frequency component less than fs
for different cutoff frequencies maintaining the sampling can be easily predicted. These samples are marked and
frequency at a fixed rate. MATAB simulation technique shown in figure 2(a). The samples marked in red, blue and
has been used to confirm and validate the algorithm. The yellow color; corresponds to different frequency
algorithm is verified with the following criteria. components within the bandwidth fs. The higher harmonic
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is marked in red color and it contains approximately 7 samples considered for the averaging. Apart from this type
samples. This corresponds to a frequency component of averaging one more kind can be introduced, which
(1/7)fs. The next harmonic is marked in blue color. It performs the sample modification similar to the averaging.
Consider red colored samples from the figure 2 (b) for the
sample modification. To remove the fluctuation of this
signal, a linear path is predicted between first and the last
sample. The linear path is formed between time index t=1
and t=8. It is shown in figure 2 (b). If there are even
number of samples, then a value is obtained at the center
[t=4.5] of first and last sample. From this value at t=4.5, the
averaging with first and last sample leads to two more new
sample sets corresponding to t=3 and t=6. Same method of
averaging is performed to obtain the values at t= 2, 4, 5 and
Figure 1. An arbitrary Signal
consists of approximately 15 samples, which contributes to 7.
(1/15)fs frequency spectrum of the signal. Similarly the The newly obtained values will make the signal shown
yellow colored sample set belongs to lowest harmonic of in figure 2 (c). Same kind of averaging is performed to the
the signal and this frequency is less than (1/40)fs. yellow colored samples. Since we perform the averaging
To suppress the two harmonics with samples red and only for a set of samples with certain conditions on
blue colors and maintain only the harmonic lower than amplitude variations, this algorithm is named as
“conditional averaging”. It indicates that this proposed
algorithm results in reduced computations, which is
certainly an overwhelming advantage.
Signal averaging is performed on the basis of the
variation in the magnitude of the incoming samples, the
averaging takes place always between two samples.
Averaging of two samples is performed over N points to
determine the new possible sample of the signal. Not all the
(a) Samples set identified with different color incoming signal sample undergo for the averaging
(b) operation.
To test this algorithm, consider a signal, which is
sampled at a rate of fs with four different frequency
components f1, f2, f3 and f4 with different phase angle Φ1,
Φ2, Φ3 and Φ4.We express this signal as follows.
(b) Magnified view of samples with red color
By assuming suitable values of these frequencies and
phase angles, the simulation of the conditional averaging is
carried out. Consider f1, f2, f3 and f4 to be 600 Hz, 50 Hz ,
2000 Hz and 800 Hz respectively with phase angle of 2Π/3,
0, Π/5 and Π/2.
(c) Final reconstructed Signal
Figure 2
(1/40)fs, we can think of a technique which will process
only these samples. Continuous averaging will suppress the
higher frequency components depending on the number of
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bandwidth at
i.e.
After performing the conditional averaging, the resulting
sample is shown in figure 4 (b) with its FFT.
The frequency 1250 Hz corresponds to the m=25 and
all the values with m>25 are attenuated. Hence it is
Figure 3 Resulting wave for 100 samples after sampling
Conditional averaging is performed for the first 100
samples of the sampled input signals as shown in figure 4
(a) to get different cutoff frequency for a LPF. The
conditional averaging is carried out for three different cases
a) Four sample-conditional Averaging
b) Eight sample- conditional Averaging (a) Input FFT
c) Sixteen sample- conditional Averaging
A. Four sample-conditional Averaging
Four sample-conditional averaging uses only four sample
buffers. The samples in the buffer are used for the
comparison process to check the conditions of a relational
set. A relational set is an array, which is obtained by
comparing two samples each from the main sample set and
contains combination of only two values. If the signal
frequency of interest is f, then the sampling should be
carried out at the rate more than 4f. It is also possible that if
a signal is sampled at fs, then the signal can be band
limited to minimum of fs/4. Conditional averaging with
four sample is the minimum possible selection of the cutoff
(b) Output FFT
frequency for the design of LPF.
The Algorithm uses following steps functioning like a LPF for the designed frequency fs/4. The
1) Take new signal sample to the buffer of size four in first- process introduces a delay of 4 sample time between an
in last-out order. input and the output value at any instant of time. The
2) Check the relation between all 4 samples by getting a algorithm mentioned for the conditional averaging is
new set of relational array with 3 elements. implemented in the C language.
3) In the relational array search for the unwanted sequence
and eliminate that sequence in the buffer by performing B. Eight sample-conditional Averaging
Averaging of four samples. Four sample-conditional averaging is simpler and the
4) From the updated buffer, the last sample is taken as basic averaging. Whereas eight sample-conditional
output and entire process is repeated from step 1. averaging includes the signal constraints of 4 sample
In the figure 3 or figure 4 (a), samples are taken at conditional averaging also. An 8 sample buffer is
5000Hz sampling rate and a 100 point FFT is obtained for maintained for the comparison process to check for the
the frequency analysis. Here m varies from 0 to 99 and conditions. Eight sample conditional averaging gives a
m=100 correspond to the Nyquist rate, i.e. 5000 Hz in this signal bandwidth of fs/8. Conditional averaging with eight
case. sample uses the algorithm used similar to that of four
To find the frequency of analysis ‘f’ from the sample-conditional averaging to get the bandwidth for the
above FFT desired of LPF.
To find the frequency of analysis ‘f’ from the FFT
shown in figure 5.
Where N=100 an fs =5000 Hz
The four samples conditional averaging will result in a
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Where N=100 an fs =5000 Hz
The eight sample-Conditional Averaging will give cutoff
value at fs/8. i.e. 5000/8=625 Hz .This can be verified by
figure5.
The frequency 625 Hz corresponds to m=12.5 and all the
values m>12.5 are attenuated. Hence it is acting like a LPF
for the desired bandwidth. The whole process introduces a
delay of 8 sample time between an input and the output.
Figure 6 outputs FFT of 16 sample conditional averaging
III. RESULT
Conditional averaging algorithm is verified in real time
to compare the result of the theoretical simulation. To
perform this operation DSP Starter kit (DSK6416) is used
Figure 5. Output FFT of 8 sample conditional averaging for the programming and the debugging. TMS320C6416v
is the key element in DSK6416 system and it has got all the
other speech processing peripherals interfaced to it. Input
C. Sixteen sample -conditional Averaging signal are fed using audio codec and then processed by the
In this Averaging technique only sixteen sample buffers DSP processor.
are maintained for the comparison process to check the Audio codec performs analog to digital conversion and
conditions. Sixteen sample-conditional averaging includes vice versa. Analog signals are mixed using simple passive
the both the processing stages of four and eight sample components to get the mixture of all the input signals.
averaging. For a sampling frequency fs, the sixteen sample Signal selections are made such that, they will be within
conditional averaging given a bandwidth limit of fs /16 in the sampling frequency of the system. The sampling
the design of LPF. To find the frequency of analysis ‘f’ frequency of the DSK6416 is selected by the program
from the above FFT shown in figure 6. command to set at 8 KHz. All input signals are band
limited to less than 4 KHz to satisfy the Nyquist rate.
A real time testing has been carried out for sixteen sample
conditional averaging. Sixteen sample-conditional
where N=100 an fs =5000 Hz. The eight sample-
averaging gives a very small bandwidth of fs /16. More
conditional Averaging will give cutoff value at fs /8. i.e.
conditional statements has to be included in the processing
5000/16=312.5 Hz .This can be verified by figure 6.
of the signal, given that the number of samples considered
The frequency 312.5 Hz corresponds to m=6.25 and all
are more for the signal processing. The time domain input
the values m>6.25 are attenuated. Hence it is behaving like
and the output signal after the processing through
a LPF. The whole process introduces a delay of 16 sample
conditional averaging are shown in the figure 7 (a).
time between an input and the output.
A test is conducted with different input FFT, wherein a
FIR filtering obeys sinusoidal functional manipulation,
signal of higher frequency component within the input
hence the output samples are also the part of sine or cosine
band limit is fed into the system. The output FFT obtained
functions. The above work is still modified for the better
clearly shows the attenuation of the signal above 500Hz in
performance for different frequencies. When more samples
the spectrum. The signals above 4 KHz are also attenuated,
are taken for a particular conditionality, it is possible to
as the signal was sampled at 8 KHz rate and hence it can
modify the range of the bandwidth. Conditional Averaging
not reconstruct the signal beyond 4 KHz (refer figure 7
makes use of the deterministic signals like ECG, EMG and
(c)). Including more conditional statements for a signal of
EEG, which are the better sources of signals for the
higher frequency value, the output response of the system
analysis. To reiterate, this algorithm reduces the
we find is much better.
computation.
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processing, but the conditional averaging uses only limited
number of samples, which satisfy the conditional array set.
This results in reduced computational time and burden on
the processor.
Any type of general purpose processor can be used to
design the system for the signal processing using this
algorithm. Thus, unlike other algorithm, this method
requires only a general purpose processor. However, in the
(a) Input and Output signal
proposed algorithm, in order to verify all the sets of
possible combination within the short period of time,
before the next sample arrives, a high speed general
purpose processor is required. This processor should be
able to execute minimum of 100 to 1000 instructions
within one sampling time. For the lower bandwidth, system
undergoes more conditional statements and therefore it
requires more instruction for the processing.
Conditional algorithm can also be developed for the
analysis of ECG, EEG signals as well. Therefore,
conditional averaging becomes a very useful filtering
algorithm to analyze the biomedical signals. All such
biomedical signals are of low frequency and any high
frequency noise or the signal surge can be easily detected
(b)Input signal FFT using the conditional averaging.
The main drawback of the system is that, it may distort
the signal by adding many other lower harmonics within
the bandwidth. To alleviate this and get a desired
bandwidth, it is necessary to include more conditional
statements during processing.
This conditional averaging is possible to be applied to high
pass filtering operation also. It may not be just averaging as
performed in the case of the low pass filter design. Design
of narrow band pass filter using the conditional averaging
algorithm is another future work.
REFERENCES
(c) Out put FFT [1] Welch P. D “ The use of Fast Fourier Transform for the
estimation of power spectra: A method based on time
Fig 14 averaging over short, Modified periodgrams” IEEE
transaction on audio and electroacoust., Vol. AU-15 No.2
CONCLUSION June 1967
[2] Richard G Lyons, Understanding Digital signal processing
The desired bandwidth for the proposed low pass filter Pearson Education third edition.2001.
could be achieved after the simulation and debugging of [3] TMS320C6713 DSK Technical Reference, SPECTRUM
the algorithm using the conditional averaging, which works DIGITAL, INC
as a low pass filter. It may be noted that filter uses simple [4] TMS320 DSP datasheet by Texas instrumentation.
two values averaging at a time and gives the corrected [5] Emmanuel C. Ifeachor, Digital Signal Processing, Pearson
output by consuming less processing time. Normally time Education
domain averaging uses all the incoming samples for the
© 2010 ACEEE 32
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```
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### Orbits I (some basics)
"So... that's my man, Isaac Newton"
You can watch the video here (English subtitles available)
Newton (along with others) invented calculus, in part, to understand what’s going on behind Keppler’s laws – e.g. 'why are the orbits of the planets elliptical, with the sun at one focus'? He showed that this is the result of an inverse square behavior of gravity.
On earth, the strength of gravity seemed about the same everywhere. But, by studying the motion of the planets (including earth) it was possible to see that the force of gravity drops off as the square of the distance from the sun. Here is how this is expressed in both scalar and vector form:
F=G(m_1 m_2)/r^2 (scalar)
bb F=G(m_1 m_2)/|bb r_12|^2 hat bbr_12 (vector), where
bb r_12=bb r_2-bb r_1, and hat bbr_12=(bb r_2-bbr_1)/|bb r_2-bbr_1|.
Sometimes it's easier to code if you just write it this way:
bbF=-Gm_1m_2(bb r)/r^3
NOTE: In this post, we're assuming that the sun is fixed in space, the earth's mass is so small that it doesn't affect anything, and there are no other planets (or moons!). We're just getting started on this fishing expedition.
A very simple way to try to calculate orbits (with a computer) would be to just use finite differences.
(d bb v)/(d t) = bb a = (bb F)/(m_2) = - Gm_1(bb r)/r^3, (d bb x)/(d t) = bb v
Delta bb v = Gm(bb r)/(r^3) bb Delta t, Delta bb x = bb v bb Delta t
bb x_(i+1) = bb x + bb Delta bb x_i, bb v_(i+1) = bb v + bb Delta bb v_i
Note that Gm (the product of G and the mass of the sun) is known much more precisely than either G or m alone. Can you guess why? Imagine trying to measure one without the other!
Also,
r_(ellipse) = ( a (1 - e^2)) / (1 + e cos(theta))
r_(aphelion) = a ( 1 + e); theta = pi
r_(perihelion) = a ( 1 - e); theta = 0
and the vis-viva equation (which is really just conservation of energy):
v^2 = Gm( (2)/(r) - (1)/(a) )
So here's a simple python script to try: (no NumPy or SciPy yet)
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 import matplotlib.pyplot as plt import matplotlib.cm as cm import math Gm = 1.3271244002E+20 # m^3 s^-2 (Wikipedia) e = 0.01671123 # earth eccentricity (Wikipedia) a = 1.49598261E+11 # earth semi-major axis (Wikipedia) # r = a(1-e^2)/(1+e*cos(theta)) an equation for an ellipse r_aphelion = a * (1. + e) r_perihelion = a * (1. - e) # using vis-viva equation v^2 = Gm * (2./r - 1./a) (Wikipedia) # (this is really just conservation of energy: Kinetic + Potential = const) v_aphelion = math.sqrt(Gm * (2./(1.+e) - 1.) / a) t_year = 2. * math.pi * math.sqrt(a**3/Gm) nstep = 100 # CHANGE THIS TO SEE WHAT HAPPENS! Dt = t_year / float(nstep) # time step V = [(0.0, v_aphelion)] X = [(r_aphelion, 0.0 )] T = [0.0] x, y = X[0] vx, vy = V[0] t = T[0] for i in range(nstep): one_over_r3 = (x**2 + y**2)**-1.5 vx += -Dt * Gm * x * one_over_r3 vy += -Dt * Gm * y * one_over_r3 x += Dt * vx y += Dt * vy t += Dt V.append((vx, vy)) X.append(( x, y)) T.append(t) gap = math.sqrt((X[-1][0] - X[0][0])**2 + (X[-1][1] - X[0][1])**2) v_gap = math.sqrt((V[-1][0] - V[0][0])**2 + (V[-1][1] - V[0][1])**2) print nstep, gap, v_gap plot_it = True save_it = True save_name = 'nicefig' if nstep > 10**5: plot_it = False # don't be silly!
...and the next zillion lines just generate and formatting the plots.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 if plot_it: r = [math.sqrt(x**2 + y**2) for x,y in X] x, y = zip(*X) vx, vy = zip(*V) tmin, tmax = min(T), max(T) tcolor = [(tt-tmin)/(tmax-tmin) for tt in T] tc2 = [tt*0.6 + 0.2 for tt in tcolor] # avoid dark red/blue cmap = cm.jet c, cfirst, clast = cmap(tc2), cmap(min(tc2)), cmap(max(tc2)) dgrey = (0.5, 0.5, 0.5) # darker gray lgrey = (0.7, 0.7, 0.7) # lighter gray fsl = 16 # font size for axis labels fs = 14 # font size for tick labels params = ['lines.color', 'xtick.color', 'ytick.color', 'axes.labelcolor'] for param in params: plt.rcParams[param] = lgrey plt.rcParams['axes.edgecolor'] = dgrey # this one darker fig = plt.figure(figsize = (7.7,2.45)) # here is the figure ax1 = fig.add_axes([0.08, 0.15, 0.25, 0.8]) # axes for three plots ax2 = fig.add_axes([0.41, 0.15, 0.15, 0.8]) ax3 = fig.add_axes([0.64, 0.40, 0.30, 0.55]) fig.patch.set_alpha(0.0) # FIRST PLOT line = ax1.scatter(x, y, s=16) # scatter plot of all points line.set_facecolors(c) # go back and change color of each one line.set_edgecolors(c) l = ax1.scatter(x[0], y[0], s=50) # make a big blue one for starting point l.set_facecolors('b') l = ax1.scatter(0.0, 0.0, s=120) # the sun! l.set_facecolors((1.0, 1.0, 0.0)) # is yellow? l.set_edgecolors((1.0, 0.7, 0.0)) # with an orange photosphere? not really! ax1.arrow(x[0], y[0], 0.0, 0.25*a, head_width=0.08*a, head_length=0.16*a, fc='k', ec='k') # hwbox = 1.6E+11 ax1.set_xlim([0.0 - hwbox, 0.0 + hwbox]) ax1.set_ylim([0.0 - hwbox, 0.0 + hwbox]) ax1.set_xlabel('x(m)', fontsize = fsl, color=lgrey) ax1.set_ylabel('y(m)', fontsize = fsl, color=lgrey) ax1.tick_params(axis='both', which='major', labelsize = fs, colors=lgrey) # SECOND PLOT hwbox = 2E+10 ax2.plot(x, y, '-k') l = ax2.scatter(x[0], y[0], s=240) l.set_facecolors(c[0]) line = ax2.scatter(x, y, s=60) line.set_facecolors(c) line.set_edgecolors(c) ax2.arrow(x[0], y[0], 0.0, 0.25*hwbox, head_width=0.05*hwbox, head_length=0.1*hwbox, fc='k', ec='k') # ax2.set_xlim([x[0] - 0.5*hwbox, x[0] + 0.5*hwbox]) ax2.set_ylim([y[0] - hwbox, y[0] + hwbox]) ax2.set_xlabel('x (m)', fontsize = fsl, color=lgrey) ax2.set_ylabel('y (m)', fontsize = fsl, color=lgrey) ax2.tick_params(axis='both', which='major', labelsize = fs, colors=lgrey) # THIRD PLOT ax3.plot(T, r, '-b', linewidth=6) ax3.plot( [min(T), max(T)], [r_aphelion, r_aphelion], '-k') ax3.plot( [min(T), max(T)], [r_perihelion, r_perihelion], '-k') ax3.set_ylabel('r (m)', fontsize = fsl, color=lgrey) ax3.set_xlabel('time (sec)', fontsize = fsl, color=lgrey) ax3.tick_params(axis='both', which='major', labelsize = fs, colors=lgrey) for ax in [ax1, ax2, ax3]: ax.patch.set_facecolor(dgrey) ax.patch.set_alpha(1.0) line_1 = 'nstep = ' + str(nstep) line_2 = 'gap = ' + str(round(gap,1)) + ' m' line_3 = 'v_gap = ' + str(round(v_gap, 6)) + ' m/s' fig.text(0.62, 0.19, line_1, horizontalalignment = 'left', verticalalignment = 'center', fontsize = 16, color = lgrey) fig.text(0.62, 0.12, line_2, horizontalalignment = 'left', verticalalignment = 'center', fontsize = 16, color = lgrey) fig.text(0.62, 0.05, line_3, horizontalalignment = 'left', verticalalignment = 'center', fontsize = 16, color = lgrey) if save_it: plt.savefig(save_name) plt.show()
Just to note: the following values (Wikipedia):
Gm = 1.3271244002E+20 m^3 s^-2
e = 0.01671123
a = 1.49598261E+11
give the period of an orbit of 31558319.5 seconds, or about 365.25833 days. In reality, a year on earth is about 365.25636 days. This difference may be due to our simplifying assumption that the sun just sits in one place and the earth rotates around it. In reality the masses of the sun and earth are roughly 1.989E+30 and 5.972E+24 (Wikipedia) and if they were alone, the'd rotate around a common barycenter. You can see the difference in the year, and the ratio of the masses are both of the order 10E-06. But instead of looking at reduced mass yet, let's just see what happens with this approximation.
Here are some results. Distances are in meters, time in seconds, etc. The total time is the theoretical period (one year) based on the values used for Gm, e, and a, assuming the earth's mass is very small, and no other planets. If the calculation were correct, then the earth should return to exactly the same position after a period of time given by t_year.
Color is used to represent time (from blue to red - indicated by arrow also). The plot on the left is the whole orbit (sun at one focus). The center plot is a zoom in on the starting position. On the right, the radial distance to the sun is plotted against time. The horizontal lines indicate perihelion and aphelion.
In this first simulation (above) where only 50 time steps were used for an entire orbit, the results are obviously extremely bad. The earth doesn’t even return to the same position after one orbit (center plot), and the variation of the distance to the sun (right plot) is right out! The "gap" values shown below the right plot are the scalar distances between initial and final values. Should be zero if we are orbiting correctly.
So let's try increasing the number of steps an see how quickly it improves.
With 200 and 1000 steps, it's not quite dead - it's getting better, and it doesn't want to go on the cart. We know how that story ends - it's all down hill. But if we plot the improvement, it's not really going down hill fast enough, especially if we want to start calculating more complicated things.
Orbits with large eccentricity are much more demanding because the forces are changing quickly. A quick way to make the orbit elliptical is to just cut the initial velocity in half, and let the planet fall towards the sun.
So clearly the orbits aren't going to come even close to returning to the same spot with a reasonable number of steps. Luckily there are better methods that can be extended to generalized multi-body problems.
Scene II: Enter Runge and Kutta
Newton did find some better ways to solve for the position of an object in an elliptical orbit as a function of time, but we’re going to do some more complicated calculations (multi-body, 3D) so let’s move on a few centuries.
Around 1900, Carl Runge and Martin Wilhelm Kutta developed a system for finding better algorithms for solving ODEs involving time-varying effects. Algorithms in this family are called Runge-Kutta algorithms. You can read about these everywhere.
The simplest or most common is called RK4, or “the” Runge-Kutta algorithm. Here it is implemented for simultaneous velocity and position expressions. The function a is acceleration (F/m). Be careful to note the x and v subscripts as they are written.
I've left off the terms for time dependence here.
Don't forget, these are all still vectors.
a() is a function returning the acceleration vector (dv)/(dt)
k_(v1) = a(x_i)
k_(x1) = v_i
k_(v2) = a(x_i + (h)/(2) k_(x1))
k_(x2) = v_i + (h)/(2) k_(v1)
k_(v3) = a (x_i + (h)/(2) k_(x2) )
k_(x3) = v_i + (h)/(2) k_(v2)
k_(v4) = a ( x_i + h k_(x3) )
k_(x4) = v_i + h k_(v3)
and finally to iterate:
v_(i+1) = v_i + (h)/(6) ( k_(v1) + 2 (k_(v2) + k_(v3)) + k_(v4) )
x_(i+1) = x_i + (h)/(6) ( k_(x1) + 2 (k_(x2) + k_(x3)) + k_(x4) )
A second, popular algorithm is the Runge-Kutta-Fehlberg or RK45. This method is also fourth order, but more accurate. It uses many more coefficients and floating point operations per step, but with a few more calculations it provides a fifth order solution as well. This can be compared to estimate the approximate size of the error. This comparison can be used to adjust the step size h dynamically, so time is not wasted “over-calculating”. But you have to think about how often and when you change the step size so that it makes sense for your calculation.
k_(v1) = h a( x_i )
k_(x1) = h v_i
k_(v2) = h a( x_i + b_(21) k_(x1) )
k_(x2) = h ( v_i + b_(21) k_(v1) )
k_(v3) = h a( x_i + b_(31) k_(x1) + b_(32) k_(x2) )
k_(x3) = h ( v_i + b_(31) k_(v1) + b_(32) k_(v2) )
k_(v4) = h a( x_i + b_(41) k_(x1) + b_(42) k_(x2) + b_(43) k_(x3) )
k_(x4) = h ( v_i + b_(41) k_(v1) + b_(42) k_(v2) + b_(43) k_(v3) )
k_(v5) = h a( x_i + b_(51) k_(x1) + b_(52) k_(x2) + b_(53) k_(x3) + b_(54) k_(x4) )
k_(x5) = h ( v_i + b_(51) k_(v1) + b_(52) k_(v2) + b_(53) k_(v3) + b_(54) k_(v4) )
k_(v6) = h a( x_i + b_(61) k_(x1) + b_(62) k_(x2) + b_(63) k_(x3) + b_(64) k_(x4) + b_(65) k_(x5) )
k_(x6) = h ( v_i + b_(61) k_(v1) + b_(62) k_(v2) + b_(63) k_(v3) + b_(64) k_(v4) + b_(65) k_(v5) )
and finally to iterate:
x_(i+1) = x_i + c_1 k_(x1) + c_3 k_(x3) + c_4 k_(x4) + c_5 k_(x5) (no c_2)
v_(i+1) = v_i + c_1 k_(v1) + c_3 k_(v3) + c_4 k_(v4) + c_5 k_(v5)
These methods are fourth order, so the error (per step) is O(5). Since the number of steps must increase as step size decreases, we can expect the total error to drop as step size to the fourth power.
Yep, for 100 steps, the position error after one orbit is 10^6 meters, or 1000 km, and for 1000 steps it drops to about 30 meters! Remember thought, this is just the numerical precision for this case. The input values have uncertainties, and the assumptions are unrealistic. We're just fishing with python here. Still it's nice to see what a few extra lines can do!
So we can see both methods are the same order of accuracy. Each step is O(5), and the total number of steps increases as the step size decreases, so the final error drops as step-size to the minus four power. The RK45 method is about a factor of ~30 more accurate in this particular example. You may wonder why you might want to use it. You can get the same accuracy (at least in this example) with RK4 by using 10^(1/4) or about twice the number of steps, which may be faster than using RK45 with roughly 5 times the number of floats and 50% more calls to the subroutine for acceleration.
With just a few more calculations, the RK45 method can also provide a more accurate fifth order approximation.
x_(5th) = x_i + d_1 k_(x1) + d_3 k_(x3) + d_4 k_(x4) + d_5 k_(x5) + d_6 k_(x6) (no d_2)
v_(5th) = v_i + d_1 k_(v1) + d_3 k_(v3) + d_4 k_(v4) + d_5 k_(v5) + d_6 k_(v6)
The best use of this is often to "check" the fourth order solution. If they agree to within a pre-determined tolerance, everything is OK. However if they disagree beyond the tolerance, the size of the disagreement can be used to choose a smaller step size. In the same way, if they agree much more closely than is necessary, a larger step size can be chosen. We'll see later that for some crazy orbits, most of the time you can use a large step size, but when something exciting happens (like a "swing-by") it's good to be able to dynamically (and automatically) reduce the step size during that short event.
s = ( (delta_x h)/(2 |x_(5th)-x| ) )^(1/4)
If you want to try it, the RK45 method (Runge-Kutta-Fehlberg) is very popular and you can read about it wherever ODEs are sold.
Here's the half-velocity elliptical orbit test again with both RK4 and RK45 methods. You can see the step size for the RK45 vary by roughly a factor of 50, and so it ends up at a different place. In a serious calculation, you would pay more attention to the algorithm that is actually changing step size. There are different scenarios out there to choose from.
Here is what they look like.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 181 182 183 184 185 186 187 188 189 190 191 192 193 194 195 196 197 198 199 200 201 202 203 def numburz(): """ possibly the coeficients for the RK45 (Runge-Kutta-Fehlberg) method, but you really better look these up for yourself!! """ global a11, a21, a31, a41, a51, a61 global b21 global b31, b32 global b41, b42, b43 global b51, b52, b53, b54 global b61, b62, b63, b64, b65 global c1, c3, c4, c5 # no c2 global d1, d3, d4, d5, d6 # no d2 a11 = 1./1. # not used a21 = 1./4. a31 = 3./8. a41 = 12./13. a51 = 1./1. a61 = 1./2. b21 = 1./4. b31 = 3./32. b32 = 9./32. b41 = 1932./2197. b42 = -7200./2197. b43 = 7296./2197. b51 = 439./216. b52 = -8./1. b53 = 3680./513. b54 = -845./4104. b61 = -8./27. b62 = 2./1. b63 = -3544./2565. b64 = 1859./4104. b65 = -11./40. c1 = 25./216. # no c2 c3 = 1408./2565. c4 = 2197./4104. c5 = -1./5. d1 = 16./135. #no d2 d3 = 6656./12825. d4 = 28561./56430. d5 = -9./50. d6 = 2./55. def RK45(x, v, t, n, h, F, tolx, varistep=False): sm = np.zeros_like(t) for i in range(n): # written for readability, not speed kv1 = h * F( x[i] ) kx1 = h * ( v[i] ) kv2 = h * F( x[i] + b21*kx1 ) kx2 = h * ( v[i] + b21*kv1 ) kv3 = h * F( x[i] + b31*kx1 + b32*kx2 ) kx3 = h * ( v[i] + b31*kv1 + b32*kv2 ) kv4 = h * F( x[i] + b41*kx1 + b42*kx2 + b43*kx3 ) kx4 = h * ( v[i] + b41*kv1 + b42*kv2 + b43*kv3 ) kv5 = h * F( x[i] + b51*kx1 + b52*kx2 + b53*kx3 + b54*kx4 ) kx5 = h * ( v[i] + b51*kv1 + b52*kv2 + b53*kv3 + b54*kv4 ) kv6 = h * F( x[i] + b61*kx1 + b62*kx2 + b63*kx3 + b64*kx4 + b65*kx5 ) kx6 = h * ( v[i] + b61*kv1 + b62*kv2 + b63*kv3 + b64*kv4 + b65*kv5 ) xx = x[i] + c1*kx1 + c3*kx3 + c4*kx4 + c5*kx5 # no c2 vv = v[i] + c1*kv1 + c3*kv3 + c4*kv4 + c5*kv5 xx5 = x[i] + d1*kx1 + d3*kx3 + d4*kx4 + d5*kx5 + d6*kx6 # no c2 vv5 = v[i] + d1*kv1 + d3*kv3 + d4*kv4 + d5*kv5 + d6*kv6 x[i+1] = xx v[i+1] = vv s = np.sqrt(np.sqrt(tolx * h / (2.*abs(xx5 - xx)))) # faster than **0.25 smin = s.min() sm[i] = smin t[i+1] = t[i] + h if varistep: h *= smin # usually you do this more carefully. return sm def RK4(x, v, t, n, h, F): h_over_two = 0.5 * h h_over_six = (1./6.) * h for i in range(n): # written for readability, not speed kv1 = F( x[i] ) kx1 = v[i] kv2 = F( x[i] + kx1 * h_over_two ) kx2 = v[i] + kv1 * h_over_two kv3 = F( x[i] + kx2 * h_over_two ) kx3 = v[i] + kv2 * h_over_two kv4 = F( x[i] + kx3 * h ) kx4 = v[i] + kv3 * h v[i+1] = v[i] + h_over_six * (kv1 + 2.*(kv2 + kv3) + kv4) x[i+1] = x[i] + h_over_six * (kx1 + 2.*(kx2 + kx3) + kx4) t[i+1] = t[i] + h def acc(x): """ acceleration due to the sun's gravity (NumPy version) """ return -Gm * x / (x**2).sum()**1.5 import matplotlib.pyplot as plt import matplotlib.cm as cm import numpy as np Gm = 1.3271244002E+20 # m^3 s^-2 (Wikipedia) e = 0.01671123 # earth eccentricity (Wikipedia) a = 1.49598261E+11 # earth semi-major axis (Wikipedia) # r = a(1-e^2)/(1+e*cos(theta)) an equation for an ellipse r_aphelion = a * (1. + e) r_perihelion = a * (1. - e) # using vis-viva equation v^2 = Gm * (2./r - 1./a) (Wikipedia) # (this is really just conservation of energy: Kinetic + Potential = const) v_aphelion = np.sqrt(Gm * (2./(1.+e) - 1.) / a) t_year = 2. * np.pi * np.sqrt(a**3/Gm) nstep = 500 # CHANGE THIS TO SEE WHAT HAPPENS! Dt = t_year / float(nstep) # time step X4 = np.zeros((1000,3)) V4 = np.zeros((1000,3)) T4 = np.zeros((1000)) V4[0] = 0.0, 0.5*v_aphelion, 0.0 X4[0] = r_aphelion, 0.0, 0.0 T4[0] = 0.0 X45 = X4.copy() V45 = V4.copy() T45 = T4.copy() RK4( X4, V4, T4, nstep, Dt, acc) vary = True # try both numburz() s = RK45(X45, V45, T45, nstep, Dt, acc, 1E-04, varistep=vary) plt.figure() plt.subplot(2, 3, 1) plt.plot(X4[:nstep,0], X4[:nstep,1]) plt.title("RK4 x, y orbit") plt.subplot(2, 3, 2) plt.plot(X45[:nstep,0], X45[:nstep,1]) plt.title("RK45 x, y orbit") if not vary: plt.subplot(2, 3, 4) plt.plot(X4[:nstep,0] - X45[:nstep,0]) plt.plot(X4[:nstep,1] - X45[:nstep,1]) plt.title("RK4-RK45 diff. x, y") plt.subplot(2, 3, 5) plt.plot(s[:nstep]) plt.title("RK45 s, vari = " + str(vary)) if vary: plt.subplot(2, 3, 6) plt.plot(T45[1:nstep]-T45[:nstep-1]) plt.title("RK45 dT, vari = " + str(vary)) plt.show() | 7,438 | 19,697 | {"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} | 3.90625 | 4 | CC-MAIN-2022-21 | longest | en | 0.897229 |
https://eudml.org/subject/MSC/65Qxx | 1,708,597,702,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473738.92/warc/CC-MAIN-20240222093910-20240222123910-00855.warc.gz | 257,082,496 | 8,422 | Page 1
## Displaying 1 – 11 of 11
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### A certain integral-recurrence equation with discrete-continuous auto-convolution
Archivum Mathematicum
Laplace transform and some of the author’s previous results about first order differential-recurrence equations with discrete auto-convolution are used to solve a new type of non-linear quadratic integral equation. This paper continues the author’s work from other articles in which are considered and solved new types of algebraic-differential or integral equations.
### Curvature and Flow in Digital Space
Actes des rencontres du CIRM
We first define the curvature indices of vertices of digital objects. Second, using these indices, we define the principal normal vectors of digital curves and surfaces. These definitions allow us to derive the Gauss-Bonnet theorem for digital objects. Third, we introduce curvature flow for isothetic polytopes defined in a digital space.
### Equation $f\left(p\left(x\right)\right)=q\left(f\left(x\right)\right)$ for given real functions $p$, $q$
Czechoslovak Mathematical Journal
We investigate functional equations $f\left(p\left(x\right)\right)=q\left(f\left(x\right)\right)$ where $p$ and $q$ are given real functions defined on the set $ℝ$ of all real numbers. For these investigations, we can use methods for constructions of homomorphisms of mono-unary algebras. Our considerations will be confined to functions $p,q$ which are strictly increasing and continuous on $ℝ$. In this case, there is a simple characterization for the existence of a solution of the above equation. First, we give such a characterization. Further, we present a construction...
### Implicit difference inequalities corresponding to first-order partial differential functional equations.
Journal of Applied Mathematics and Stochastic Analysis
### Nonuniqueness of implicit lattice Nagumo equation
Applications of Mathematics
We consider the implicit discretization of Nagumo equation on finite lattices and show that its variational formulation corresponds in various parameter settings to convex, mountain-pass or saddle-point geometries. Consequently, we are able to derive conditions under which the implicit discretization yields multiple solutions. Interestingly, for certain parameters we show nonuniqueness for arbitrarily small discretization steps. Finally, we provide a simple example showing that the nonuniqueness...
### On the rational recursive sequence ${x}_{n+1}=\frac{{\alpha }_{0}{x}_{n}+{\alpha }_{1}{x}_{n-l}+{\alpha }_{2}{x}_{n-k}}{{\beta }_{0}{x}_{n}+{\beta }_{1}{x}_{n-l}+{\beta }_{2}{x}_{n-k}}$
Mathematica Bohemica
The main objective of this paper is to study the boundedness character, the periodicity character, the convergence and the global stability of positive solutions of the difference equation ${x}_{n+1}=\frac{{\alpha }_{0}{x}_{n}+{\alpha }_{1}{x}_{n-l}+{\alpha }_{2}{x}_{n-k}}{{\beta }_{0}{x}_{n}+{\beta }_{1}{x}_{n-l}+{\beta }_{2}{x}_{n-k}},\phantom{\rule{1.0em}{0ex}}n=0,1,2,\cdots$ where the coefficients ${\alpha }_{i},{\beta }_{i}\in \left(0,\infty \right)$ for $i=0,1,2,$ and $l$, $k$ are positive integers. The initial conditions ${x}_{-k},\cdots ,{x}_{-l},\cdots ,{x}_{-1},{x}_{0}$ are arbitrary positive real numbers such that $l. Some numerical experiments are presented.
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Fixed Point Theory and Applications [electronic only]
### Volterra discrete inequalities of Bernoulli type.
Journal of Inequalities and Applications [electronic only]
Page 1 | 926 | 3,780 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 17, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2024-10 | latest | en | 0.816799 |
https://mail.scipy.org/pipermail/numpy-discussion/2007-March/026803.html | 1,495,750,660,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463608617.6/warc/CC-MAIN-20170525214603-20170525234603-00205.warc.gz | 775,416,412 | 2,236 | # [Numpy-discussion] fastest way to do multiplication with diagonal matrices from left or right
daniel.egloff@z... daniel.egloff@z...
Fri Mar 23 09:13:04 CDT 2007
```
Dear list
what is the fastet way to multiply with a diagonal matrix from left or
right and without to build a square matrix from the diagonal.
Here it what I am looking for:
import numpy as N
def diagmult(X, Y):
"""
Matrix multiplication X*Y where either X or Y is a diagonal matrix.
"""
if X.ndim == 1 and Y.ndim == 2:
R = Y.copy()
for i, d in enumerate(X):
R[i,:] *= d
return R
elif X.ndim == 2 and Y.ndim == 1:
R = X.copy()
for i, d in enumerate(Y):
R[:,i] *= d
return R
elif X.ndim == 1 and Y.ndim == 1:
return X*Y
else
raise ValueError('diagmult dimension mismatch X.ndim = %d, Y.ndim =
%d' % (X.ndim, Y.ndim))
Freundliche Grüsse
Daniel Egloff
Zürcher Kantonalbank
Leiter(in) Financial Computing, ZEF
Josefstrasse 222, 8005 Zürich
Telefon 044 292 45 33, Fax 044 292 45 95
Briefadresse: Postfach, 8010 Zürich, http://www.zkb.ch
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https://hackage.haskell.org/package/sbv-8.6/docs/Data-SBV-Trans.html | 1,582,952,499,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875148375.36/warc/CC-MAIN-20200229022458-20200229052458-00393.warc.gz | 394,466,065 | 61,811 | sbv-8.6: SMT Based Verification: Symbolic Haskell theorem prover using SMT solving.
Data.SBV.Trans
Description
More generalized alternative to Data.SBV for advanced client use
Synopsis
# Symbolic types
## Booleans
type SBool = SBV Bool Source #
A symbolic boolean/bit
### Boolean values and functions
Symbolic True
Symbolic False
Symbolic boolean negation
(.&&) :: SBool -> SBool -> SBool infixr 3 Source #
Symbolic conjunction
(.||) :: SBool -> SBool -> SBool infixr 2 Source #
Symbolic disjunction
(.<+>) :: SBool -> SBool -> SBool infixl 6 Source #
Symbolic logical xor
(.~&) :: SBool -> SBool -> SBool infixr 3 Source #
Symbolic nand
(.~|) :: SBool -> SBool -> SBool infixr 2 Source #
Symbolic nor
(.=>) :: SBool -> SBool -> SBool infixr 1 Source #
Symbolic implication
(.<=>) :: SBool -> SBool -> SBool infixr 1 Source #
Symbolic boolean equivalence
Conversion from Bool to SBool
oneIf :: (Ord a, Num a, SymVal a) => SBool -> SBV a Source #
Returns 1 if the boolean is sTrue, otherwise 0.
### Logical functions
sAnd :: [SBool] -> SBool Source #
Generalization of and
sOr :: [SBool] -> SBool Source #
Generalization of or
sAny :: (a -> SBool) -> [a] -> SBool Source #
Generalization of any
sAll :: (a -> SBool) -> [a] -> SBool Source #
Generalization of all
## Bit-vectors
### Unsigned bit-vectors
8-bit unsigned symbolic value
16-bit unsigned symbolic value
32-bit unsigned symbolic value
64-bit unsigned symbolic value
type SWord (n :: Nat) = SBV (WordN n) Source #
A symbolic unsigned bit-vector carrying its size info
data WordN (n :: Nat) Source #
An unsigned bit-vector carrying its size info
Instances
(KnownNat n, IsNonZero n) => Bounded (WordN n) Source # Bounded instance for WordN Instance detailsDefined in Data.SBV.Core.Sized Methods (KnownNat n, IsNonZero n) => Enum (WordN n) Source # Enum instance for WordN Instance detailsDefined in Data.SBV.Core.Sized Methodssucc :: WordN n -> WordN n #pred :: WordN n -> WordN n #toEnum :: Int -> WordN n #fromEnum :: WordN n -> Int #enumFrom :: WordN n -> [WordN n] #enumFromThen :: WordN n -> WordN n -> [WordN n] #enumFromTo :: WordN n -> WordN n -> [WordN n] #enumFromThenTo :: WordN n -> WordN n -> WordN n -> [WordN n] # Eq (WordN n) Source # Instance detailsDefined in Data.SBV.Core.Sized Methods(==) :: WordN n -> WordN n -> Bool #(/=) :: WordN n -> WordN n -> Bool # (KnownNat n, IsNonZero n) => Integral (WordN n) Source # Integral instance for WordN Instance detailsDefined in Data.SBV.Core.Sized Methodsquot :: WordN n -> WordN n -> WordN n #rem :: WordN n -> WordN n -> WordN n #div :: WordN n -> WordN n -> WordN n #mod :: WordN n -> WordN n -> WordN n #quotRem :: WordN n -> WordN n -> (WordN n, WordN n) #divMod :: WordN n -> WordN n -> (WordN n, WordN n) #toInteger :: WordN n -> Integer # (KnownNat n, IsNonZero n) => Num (WordN n) Source # Num instance for WordN Instance detailsDefined in Data.SBV.Core.Sized Methods(+) :: WordN n -> WordN n -> WordN n #(-) :: WordN n -> WordN n -> WordN n #(*) :: WordN n -> WordN n -> WordN n #negate :: WordN n -> WordN n #abs :: WordN n -> WordN n #signum :: WordN n -> WordN n # Ord (WordN n) Source # Instance detailsDefined in Data.SBV.Core.Sized Methodscompare :: WordN n -> WordN n -> Ordering #(<) :: WordN n -> WordN n -> Bool #(<=) :: WordN n -> WordN n -> Bool #(>) :: WordN n -> WordN n -> Bool #(>=) :: WordN n -> WordN n -> Bool #max :: WordN n -> WordN n -> WordN n #min :: WordN n -> WordN n -> WordN n # (KnownNat n, IsNonZero n) => Real (WordN n) Source # Real instance for WordN Instance detailsDefined in Data.SBV.Core.Sized MethodstoRational :: WordN n -> Rational # Show (WordN n) Source # Show instance for WordN Instance detailsDefined in Data.SBV.Core.Sized MethodsshowsPrec :: Int -> WordN n -> ShowS #show :: WordN n -> String #showList :: [WordN n] -> ShowS # (KnownNat n, IsNonZero n) => Bits (WordN n) Source # Instance detailsDefined in Data.SBV.Core.Sized Methods(.&.) :: WordN n -> WordN n -> WordN n #(.|.) :: WordN n -> WordN n -> WordN n #xor :: WordN n -> WordN n -> WordN n #complement :: WordN n -> WordN n #shift :: WordN n -> Int -> WordN n #rotate :: WordN n -> Int -> WordN n #bit :: Int -> WordN n #setBit :: WordN n -> Int -> WordN n #clearBit :: WordN n -> Int -> WordN n #complementBit :: WordN n -> Int -> WordN n #testBit :: WordN n -> Int -> Bool #bitSize :: WordN n -> Int #isSigned :: WordN n -> Bool #shiftL :: WordN n -> Int -> WordN n #unsafeShiftL :: WordN n -> Int -> WordN n #shiftR :: WordN n -> Int -> WordN n #unsafeShiftR :: WordN n -> Int -> WordN n #rotateL :: WordN n -> Int -> WordN n #rotateR :: WordN n -> Int -> WordN n #popCount :: WordN n -> Int # (KnownNat n, IsNonZero n) => HasKind (WordN n) Source # WordN has a kind Instance detailsDefined in Data.SBV.Core.Sized MethodskindOf :: WordN n -> Kind Source #hasSign :: WordN n -> Bool Source #intSizeOf :: WordN n -> Int Source #isReal :: WordN n -> Bool Source #isFloat :: WordN n -> Bool Source #isDouble :: WordN n -> Bool Source #isChar :: WordN n -> Bool Source #isString :: WordN n -> Bool Source #isList :: WordN n -> Bool Source #isSet :: WordN n -> Bool Source #isTuple :: WordN n -> Bool Source #isMaybe :: WordN n -> Bool Source #isEither :: WordN n -> Bool Source # (KnownNat n, IsNonZero n) => SymVal (WordN n) Source # SymVal instance for WordN Instance detailsDefined in Data.SBV.Core.Sized MethodsmkSymVal :: MonadSymbolic m => Maybe Quantifier -> Maybe String -> m (SBV (WordN n)) Source #literal :: WordN n -> SBV (WordN n) Source #fromCV :: CV -> WordN n Source #isConcretely :: SBV (WordN n) -> (WordN n -> Bool) -> Bool Source #forall :: MonadSymbolic m => String -> m (SBV (WordN n)) Source #forall_ :: MonadSymbolic m => m (SBV (WordN n)) Source #mkForallVars :: MonadSymbolic m => Int -> m [SBV (WordN n)] Source #exists :: MonadSymbolic m => String -> m (SBV (WordN n)) Source #exists_ :: MonadSymbolic m => m (SBV (WordN n)) Source #mkExistVars :: MonadSymbolic m => Int -> m [SBV (WordN n)] Source #free :: MonadSymbolic m => String -> m (SBV (WordN n)) Source #free_ :: MonadSymbolic m => m (SBV (WordN n)) Source #mkFreeVars :: MonadSymbolic m => Int -> m [SBV (WordN n)] Source #symbolic :: MonadSymbolic m => String -> m (SBV (WordN n)) Source #symbolics :: MonadSymbolic m => [String] -> m [SBV (WordN n)] Source #unliteral :: SBV (WordN n) -> Maybe (WordN n) Source #isConcrete :: SBV (WordN n) -> Bool Source #isSymbolic :: SBV (WordN n) -> Bool Source # (KnownNat n, IsNonZero n) => SatModel (WordN n) Source # Constructing models for WordN Instance detailsDefined in Data.SBV.Core.Sized MethodsparseCVs :: [CV] -> Maybe (WordN n, [CV]) Source #cvtModel :: (WordN n -> Maybe b) -> Maybe (WordN n, [CV]) -> Maybe (b, [CV]) Source # (KnownNat n, IsNonZero n) => SMTValue (WordN n) Source # Reading WordN values in queries. Instance detailsDefined in Data.SBV.Core.Sized MethodssexprToVal :: SExpr -> Maybe (WordN n) Source # (KnownNat n, IsNonZero n) => Metric (WordN n) Source # Optimizing WordN Instance detailsDefined in Data.SBV.Core.Sized Associated Typestype MetricSpace (WordN n) :: Type Source # MethodstoMetricSpace :: SBV (WordN n) -> SBV (MetricSpace (WordN n)) Source #fromMetricSpace :: SBV (MetricSpace (WordN n)) -> SBV (WordN n) Source #msMinimize :: (MonadSymbolic m, SolverContext m) => String -> SBV (WordN n) -> m () Source #msMaximize :: (MonadSymbolic m, SolverContext m) => String -> SBV (WordN n) -> m () Source # (KnownNat n, IsNonZero n) => SDivisible (SWord n) Source # SDivisible instance for SWord Instance detailsDefined in Data.SBV.Core.Sized MethodssQuotRem :: SWord n -> SWord n -> (SWord n, SWord n) Source #sDivMod :: SWord n -> SWord n -> (SWord n, SWord n) Source #sQuot :: SWord n -> SWord n -> SWord n Source #sRem :: SWord n -> SWord n -> SWord n Source #sDiv :: SWord n -> SWord n -> SWord n Source #sMod :: SWord n -> SWord n -> SWord n Source # (KnownNat n, IsNonZero n) => SDivisible (WordN n) Source # SDivisible instance for WordN Instance detailsDefined in Data.SBV.Core.Sized MethodssQuotRem :: WordN n -> WordN n -> (WordN n, WordN n) Source #sDivMod :: WordN n -> WordN n -> (WordN n, WordN n) Source #sQuot :: WordN n -> WordN n -> WordN n Source #sRem :: WordN n -> WordN n -> WordN n Source #sDiv :: WordN n -> WordN n -> WordN n Source #sMod :: WordN n -> WordN n -> WordN n Source # (KnownNat n, IsNonZero n) => SFiniteBits (WordN n) Source # SFiniteBits instance for WordN Instance detailsDefined in Data.SBV.Core.Sized MethodssFiniteBitSize :: SBV (WordN n) -> Int Source #lsb :: SBV (WordN n) -> SBool Source #msb :: SBV (WordN n) -> SBool Source #blastBE :: SBV (WordN n) -> [SBool] Source #blastLE :: SBV (WordN n) -> [SBool] Source #fromBitsBE :: [SBool] -> SBV (WordN n) Source #fromBitsLE :: [SBool] -> SBV (WordN n) Source #sTestBit :: SBV (WordN n) -> Int -> SBool Source #sExtractBits :: SBV (WordN n) -> [Int] -> [SBool] Source #sPopCount :: SBV (WordN n) -> SWord8 Source #setBitTo :: SBV (WordN n) -> Int -> SBool -> SBV (WordN n) Source #fullAdder :: SBV (WordN n) -> SBV (WordN n) -> (SBool, SBV (WordN n)) Source #fullMultiplier :: SBV (WordN n) -> SBV (WordN n) -> (SBV (WordN n), SBV (WordN n)) Source # (KnownNat n, IsNonZero n) => SIntegral (WordN n) Source # SIntegral instance for WordN Instance detailsDefined in Data.SBV.Core.Sized Source # SWord 8 instance for ByteConverter Instance detailsDefined in Data.SBV.Core.Sized MethodstoBytes :: SWord 8 -> [SWord 8] Source #fromBytes :: [SWord 8] -> SWord 8 Source # Source # SWord 16 instance for ByteConverter Instance detailsDefined in Data.SBV.Core.Sized MethodstoBytes :: SWord 16 -> [SWord 8] Source #fromBytes :: [SWord 8] -> SWord 16 Source # Source # SWord 32 instance for ByteConverter Instance detailsDefined in Data.SBV.Core.Sized MethodstoBytes :: SWord 32 -> [SWord 8] Source #fromBytes :: [SWord 8] -> SWord 32 Source # Source # SWord 64 instance for ByteConverter Instance detailsDefined in Data.SBV.Core.Sized MethodstoBytes :: SWord 64 -> [SWord 8] Source #fromBytes :: [SWord 8] -> SWord 64 Source # ByteConverter (SWord 128) Source # SWord 128 instance for ByteConverter Instance detailsDefined in Data.SBV.Core.Sized MethodstoBytes :: SWord 128 -> [SWord 8] Source #fromBytes :: [SWord 8] -> SWord 128 Source # ByteConverter (SWord 256) Source # SWord 256 instance for ByteConverter Instance detailsDefined in Data.SBV.Core.Sized MethodstoBytes :: SWord 256 -> [SWord 8] Source #fromBytes :: [SWord 8] -> SWord 256 Source # ByteConverter (SWord 512) Source # SWord 512 instance for ByteConverter Instance detailsDefined in Data.SBV.Core.Sized MethodstoBytes :: SWord 512 -> [SWord 8] Source #fromBytes :: [SWord 8] -> SWord 512 Source # ByteConverter (SWord 1024) Source # SWord 1024 instance for ByteConverter Instance detailsDefined in Data.SBV.Core.Sized MethodstoBytes :: SWord 1024 -> [SWord 8] Source #fromBytes :: [SWord 8] -> SWord 1024 Source # (KnownNat n, IsNonZero n) => Polynomial (SWord n) Source # Instance detailsDefined in Data.SBV.Tools.Polynomial Methodspolynomial :: [Int] -> SWord n Source #pAdd :: SWord n -> SWord n -> SWord n Source #pMult :: (SWord n, SWord n, [Int]) -> SWord n Source #pDiv :: SWord n -> SWord n -> SWord n Source #pMod :: SWord n -> SWord n -> SWord n Source #pDivMod :: SWord n -> SWord n -> (SWord n, SWord n) Source # type MetricSpace (WordN n) Source # Instance detailsDefined in Data.SBV.Core.Sized type MetricSpace (WordN n) = WordN n
### Signed bit-vectors
type SInt8 = SBV Int8 Source #
8-bit signed symbolic value, 2's complement representation
16-bit signed symbolic value, 2's complement representation
32-bit signed symbolic value, 2's complement representation
64-bit signed symbolic value, 2's complement representation
type SInt (n :: Nat) = SBV (IntN n) Source #
A symbolic signed bit-vector carrying its size info
data IntN (n :: Nat) Source #
A signed bit-vector carrying its size info
Instances
(KnownNat n, IsNonZero n) => Bounded (IntN n) Source # Bounded instance for IntN Instance detailsDefined in Data.SBV.Core.Sized Methods (KnownNat n, IsNonZero n) => Enum (IntN n) Source # Enum instance for IntN Instance detailsDefined in Data.SBV.Core.Sized Methodssucc :: IntN n -> IntN n #pred :: IntN n -> IntN n #toEnum :: Int -> IntN n #fromEnum :: IntN n -> Int #enumFrom :: IntN n -> [IntN n] #enumFromThen :: IntN n -> IntN n -> [IntN n] #enumFromTo :: IntN n -> IntN n -> [IntN n] #enumFromThenTo :: IntN n -> IntN n -> IntN n -> [IntN n] # Eq (IntN n) Source # Instance detailsDefined in Data.SBV.Core.Sized Methods(==) :: IntN n -> IntN n -> Bool #(/=) :: IntN n -> IntN n -> Bool # (KnownNat n, IsNonZero n) => Integral (IntN n) Source # Integral instance for IntN Instance detailsDefined in Data.SBV.Core.Sized Methodsquot :: IntN n -> IntN n -> IntN n #rem :: IntN n -> IntN n -> IntN n #div :: IntN n -> IntN n -> IntN n #mod :: IntN n -> IntN n -> IntN n #quotRem :: IntN n -> IntN n -> (IntN n, IntN n) #divMod :: IntN n -> IntN n -> (IntN n, IntN n) #toInteger :: IntN n -> Integer # (KnownNat n, IsNonZero n) => Num (IntN n) Source # Num instance for IntN Instance detailsDefined in Data.SBV.Core.Sized Methods(+) :: IntN n -> IntN n -> IntN n #(-) :: IntN n -> IntN n -> IntN n #(*) :: IntN n -> IntN n -> IntN n #negate :: IntN n -> IntN n #abs :: IntN n -> IntN n #signum :: IntN n -> IntN n # Ord (IntN n) Source # Instance detailsDefined in Data.SBV.Core.Sized Methodscompare :: IntN n -> IntN n -> Ordering #(<) :: IntN n -> IntN n -> Bool #(<=) :: IntN n -> IntN n -> Bool #(>) :: IntN n -> IntN n -> Bool #(>=) :: IntN n -> IntN n -> Bool #max :: IntN n -> IntN n -> IntN n #min :: IntN n -> IntN n -> IntN n # (KnownNat n, IsNonZero n) => Real (IntN n) Source # Real instance for IntN Instance detailsDefined in Data.SBV.Core.Sized MethodstoRational :: IntN n -> Rational # Show (IntN n) Source # Show instance for IntN Instance detailsDefined in Data.SBV.Core.Sized MethodsshowsPrec :: Int -> IntN n -> ShowS #show :: IntN n -> String #showList :: [IntN n] -> ShowS # (KnownNat n, IsNonZero n) => Bits (IntN n) Source # Instance detailsDefined in Data.SBV.Core.Sized Methods(.&.) :: IntN n -> IntN n -> IntN n #(.|.) :: IntN n -> IntN n -> IntN n #xor :: IntN n -> IntN n -> IntN n #complement :: IntN n -> IntN n #shift :: IntN n -> Int -> IntN n #rotate :: IntN n -> Int -> IntN n #bit :: Int -> IntN n #setBit :: IntN n -> Int -> IntN n #clearBit :: IntN n -> Int -> IntN n #complementBit :: IntN n -> Int -> IntN n #testBit :: IntN n -> Int -> Bool #bitSize :: IntN n -> Int #isSigned :: IntN n -> Bool #shiftL :: IntN n -> Int -> IntN n #unsafeShiftL :: IntN n -> Int -> IntN n #shiftR :: IntN n -> Int -> IntN n #unsafeShiftR :: IntN n -> Int -> IntN n #rotateL :: IntN n -> Int -> IntN n #rotateR :: IntN n -> Int -> IntN n #popCount :: IntN n -> Int # (KnownNat n, IsNonZero n) => HasKind (IntN n) Source # IntN has a kind Instance detailsDefined in Data.SBV.Core.Sized MethodskindOf :: IntN n -> Kind Source #hasSign :: IntN n -> Bool Source #intSizeOf :: IntN n -> Int Source #isBoolean :: IntN n -> Bool Source #isBounded :: IntN n -> Bool Source #isReal :: IntN n -> Bool Source #isFloat :: IntN n -> Bool Source #isDouble :: IntN n -> Bool Source #isChar :: IntN n -> Bool Source #isString :: IntN n -> Bool Source #isList :: IntN n -> Bool Source #isSet :: IntN n -> Bool Source #isTuple :: IntN n -> Bool Source #isMaybe :: IntN n -> Bool Source #isEither :: IntN n -> Bool Source # (KnownNat n, IsNonZero n) => SymVal (IntN n) Source # SymVal instance for IntN Instance detailsDefined in Data.SBV.Core.Sized MethodsmkSymVal :: MonadSymbolic m => Maybe Quantifier -> Maybe String -> m (SBV (IntN n)) Source #literal :: IntN n -> SBV (IntN n) Source #fromCV :: CV -> IntN n Source #isConcretely :: SBV (IntN n) -> (IntN n -> Bool) -> Bool Source #forall :: MonadSymbolic m => String -> m (SBV (IntN n)) Source #forall_ :: MonadSymbolic m => m (SBV (IntN n)) Source #mkForallVars :: MonadSymbolic m => Int -> m [SBV (IntN n)] Source #exists :: MonadSymbolic m => String -> m (SBV (IntN n)) Source #exists_ :: MonadSymbolic m => m (SBV (IntN n)) Source #mkExistVars :: MonadSymbolic m => Int -> m [SBV (IntN n)] Source #free :: MonadSymbolic m => String -> m (SBV (IntN n)) Source #free_ :: MonadSymbolic m => m (SBV (IntN n)) Source #mkFreeVars :: MonadSymbolic m => Int -> m [SBV (IntN n)] Source #symbolic :: MonadSymbolic m => String -> m (SBV (IntN n)) Source #symbolics :: MonadSymbolic m => [String] -> m [SBV (IntN n)] Source #unliteral :: SBV (IntN n) -> Maybe (IntN n) Source #isConcrete :: SBV (IntN n) -> Bool Source #isSymbolic :: SBV (IntN n) -> Bool Source # (KnownNat n, IsNonZero n) => SatModel (IntN n) Source # Constructing models for IntN Instance detailsDefined in Data.SBV.Core.Sized MethodsparseCVs :: [CV] -> Maybe (IntN n, [CV]) Source #cvtModel :: (IntN n -> Maybe b) -> Maybe (IntN n, [CV]) -> Maybe (b, [CV]) Source # (KnownNat n, IsNonZero n) => SMTValue (IntN n) Source # Reading IntN values in queries. Instance detailsDefined in Data.SBV.Core.Sized MethodssexprToVal :: SExpr -> Maybe (IntN n) Source # (KnownNat n, IsNonZero n) => Metric (IntN n) Source # Optimizing IntN Instance detailsDefined in Data.SBV.Core.Sized Associated Typestype MetricSpace (IntN n) :: Type Source # MethodstoMetricSpace :: SBV (IntN n) -> SBV (MetricSpace (IntN n)) Source #fromMetricSpace :: SBV (MetricSpace (IntN n)) -> SBV (IntN n) Source #msMinimize :: (MonadSymbolic m, SolverContext m) => String -> SBV (IntN n) -> m () Source #msMaximize :: (MonadSymbolic m, SolverContext m) => String -> SBV (IntN n) -> m () Source # (KnownNat n, IsNonZero n) => SDivisible (SInt n) Source # SDivisible instance for SInt Instance detailsDefined in Data.SBV.Core.Sized MethodssQuotRem :: SInt n -> SInt n -> (SInt n, SInt n) Source #sDivMod :: SInt n -> SInt n -> (SInt n, SInt n) Source #sQuot :: SInt n -> SInt n -> SInt n Source #sRem :: SInt n -> SInt n -> SInt n Source #sDiv :: SInt n -> SInt n -> SInt n Source #sMod :: SInt n -> SInt n -> SInt n Source # (KnownNat n, IsNonZero n) => SDivisible (IntN n) Source # SDivisible instance for IntN Instance detailsDefined in Data.SBV.Core.Sized MethodssQuotRem :: IntN n -> IntN n -> (IntN n, IntN n) Source #sDivMod :: IntN n -> IntN n -> (IntN n, IntN n) Source #sQuot :: IntN n -> IntN n -> IntN n Source #sRem :: IntN n -> IntN n -> IntN n Source #sDiv :: IntN n -> IntN n -> IntN n Source #sMod :: IntN n -> IntN n -> IntN n Source # (KnownNat n, IsNonZero n) => SFiniteBits (IntN n) Source # SFiniteBits instance for IntN Instance detailsDefined in Data.SBV.Core.Sized MethodssFiniteBitSize :: SBV (IntN n) -> Int Source #lsb :: SBV (IntN n) -> SBool Source #msb :: SBV (IntN n) -> SBool Source #blastBE :: SBV (IntN n) -> [SBool] Source #blastLE :: SBV (IntN n) -> [SBool] Source #fromBitsBE :: [SBool] -> SBV (IntN n) Source #fromBitsLE :: [SBool] -> SBV (IntN n) Source #sTestBit :: SBV (IntN n) -> Int -> SBool Source #sExtractBits :: SBV (IntN n) -> [Int] -> [SBool] Source #sPopCount :: SBV (IntN n) -> SWord8 Source #setBitTo :: SBV (IntN n) -> Int -> SBool -> SBV (IntN n) Source #fullAdder :: SBV (IntN n) -> SBV (IntN n) -> (SBool, SBV (IntN n)) Source #fullMultiplier :: SBV (IntN n) -> SBV (IntN n) -> (SBV (IntN n), SBV (IntN n)) Source # (KnownNat n, IsNonZero n) => SIntegral (IntN n) Source # SIntegral instance for IntN Instance detailsDefined in Data.SBV.Core.Sized type MetricSpace (IntN n) Source # Instance detailsDefined in Data.SBV.Core.Sized type MetricSpace (IntN n) = WordN n
### Converting between fixed-size and arbitrary bitvectors
type family IsNonZero (arg :: Nat) :: Constraint where ... Source #
Type family to create the appropriate non-zero constraint
Equations
IsNonZero 0 = TypeError ZeroWidth IsNonZero n = ()
type family FromSized (t :: Type) :: Type where ... Source #
Capture the correspondence between sized and fixed-sized BVs
Equations
FromSized (WordN 8) = Word8 FromSized (WordN 16) = Word16 FromSized (WordN 32) = Word32 FromSized (WordN 64) = Word64 FromSized (IntN 8) = Int8 FromSized (IntN 16) = Int16 FromSized (IntN 32) = Int32 FromSized (IntN 64) = Int64 FromSized (SWord 8) = SWord8 FromSized (SWord 16) = SWord16 FromSized (SWord 32) = SWord32 FromSized (SWord 64) = SWord64 FromSized (SInt 8) = SInt8 FromSized (SInt 16) = SInt16 FromSized (SInt 32) = SInt32 FromSized (SInt 64) = SInt64
type family ToSized (t :: Type) :: Type where ... Source #
Capture the correspondence between fixed-sized and sized BVs
Equations
ToSized Word8 = WordN 8 ToSized Word16 = WordN 16 ToSized Word32 = WordN 32 ToSized Word64 = WordN 64 ToSized Int8 = IntN 8 ToSized Int16 = IntN 16 ToSized Int32 = IntN 32 ToSized Int64 = IntN 64 ToSized SWord8 = SWord 8 ToSized SWord16 = SWord 16 ToSized SWord32 = SWord 32 ToSized SWord64 = SWord 64 ToSized SInt8 = SInt 8 ToSized SInt16 = SInt 16 ToSized SInt32 = SInt 32 ToSized SInt64 = SInt 64
fromSized :: FromSizedBV a => a -> FromSized a Source #
Convert a sized bit-vector to the corresponding fixed-sized bit-vector, for instance 'SWord 16' to SWord16. See also toSized.
toSized :: ToSizedBV a => a -> ToSized a Source #
Convert a fixed-sized bit-vector to the corresponding sized bit-vector, for instance SWord16 to 'SWord 16'. See also fromSized.
## Unbounded integers
Infinite precision signed symbolic value
## Floating point numbers
IEEE-754 single-precision floating point numbers
IEEE-754 double-precision floating point numbers
## Algebraic reals
Infinite precision symbolic algebraic real value
data AlgReal Source #
Algebraic reals. Note that the representation is left abstract. We represent rational results explicitly, while the roots-of-polynomials are represented implicitly by their defining equation
Instances
Source # Instance detailsDefined in Data.SBV.Core.AlgReals Methods(==) :: AlgReal -> AlgReal -> Bool #(/=) :: AlgReal -> AlgReal -> Bool # Source # NB: Following the other types we have, we require a/0 to be 0 for all a. Instance detailsDefined in Data.SBV.Core.AlgReals Methods Source # Instance detailsDefined in Data.SBV.Core.AlgReals Methods Source # Instance detailsDefined in Data.SBV.Core.AlgReals Methods(<) :: AlgReal -> AlgReal -> Bool #(<=) :: AlgReal -> AlgReal -> Bool #(>) :: AlgReal -> AlgReal -> Bool #(>=) :: AlgReal -> AlgReal -> Bool # Source # Instance detailsDefined in Data.SBV.Core.AlgReals Methods Source # Instance detailsDefined in Data.SBV.Core.AlgReals MethodsshowList :: [AlgReal] -> ShowS # Source # Instance detailsDefined in Data.SBV.Core.AlgReals Methodsshrink :: AlgReal -> [AlgReal] # Source # Instance detailsDefined in Data.SBV.Core.AlgReals MethodsrandomR :: RandomGen g => (AlgReal, AlgReal) -> g -> (AlgReal, g) #random :: RandomGen g => g -> (AlgReal, g) #randomRs :: RandomGen g => (AlgReal, AlgReal) -> g -> [AlgReal] #randoms :: RandomGen g => g -> [AlgReal] #randomRIO :: (AlgReal, AlgReal) -> IO AlgReal # Source # Instance detailsDefined in Data.SBV.Core.Kind Methods Source # Instance detailsDefined in Data.SBV.Core.Model MethodsisConcretely :: SBV AlgReal -> (AlgReal -> Bool) -> Bool Source #forall :: MonadSymbolic m => String -> m (SBV AlgReal) Source #forall_ :: MonadSymbolic m => m (SBV AlgReal) Source #mkForallVars :: MonadSymbolic m => Int -> m [SBV AlgReal] Source #exists :: MonadSymbolic m => String -> m (SBV AlgReal) Source #exists_ :: MonadSymbolic m => m (SBV AlgReal) Source #mkExistVars :: MonadSymbolic m => Int -> m [SBV AlgReal] Source #free :: MonadSymbolic m => String -> m (SBV AlgReal) Source #free_ :: MonadSymbolic m => m (SBV AlgReal) Source #mkFreeVars :: MonadSymbolic m => Int -> m [SBV AlgReal] Source #symbolic :: MonadSymbolic m => String -> m (SBV AlgReal) Source #symbolics :: MonadSymbolic m => [String] -> m [SBV AlgReal] Source # Source # AlgReal as extracted from a model Instance detailsDefined in Data.SBV.SMT.SMT MethodsparseCVs :: [CV] -> Maybe (AlgReal, [CV]) Source #cvtModel :: (AlgReal -> Maybe b) -> Maybe (AlgReal, [CV]) -> Maybe (b, [CV]) Source # Source # Instance detailsDefined in Data.SBV.Control.Utils MethodssexprToVal :: SExpr -> Maybe AlgReal Source # Source # Instance detailsDefined in Data.SBV.Core.Model Associated Types MethodsmsMinimize :: (MonadSymbolic m, SolverContext m) => String -> SBV AlgReal -> m () Source #msMaximize :: (MonadSymbolic m, SolverContext m) => String -> SBV AlgReal -> m () Source # Source # Instance detailsDefined in Data.SBV.Core.Floating Methods Source # Instance detailsDefined in Data.SBV.Core.Model
Convert an SReal to an SInteger. That is, it computes the largest integer n that satisfies sIntegerToSReal n <= r essentially giving us the floor.
For instance, 1.3 will be 1, but -1.3 will be -2.
## Characters, Strings and Regular Expressions
type SChar = SBV Char Source #
A symbolic character. Note that, as far as SBV's symbolic strings are concerned, a character is currently an 8-bit unsigned value, corresponding to the ISO-8859-1 (Latin-1) character set: http://en.wikipedia.org/wiki/ISO/IEC_8859-1. A Haskell Char, on the other hand, is based on unicode. Therefore, there isn't a 1-1 correspondence between a Haskell character and an SBV character for the time being. This limitation is due to the SMT-solvers only supporting this particular subset. However, there is a pending proposal to add support for unicode, and SBV will track these changes to have full unicode support as solvers become available. For details, see: http://smtlib.cs.uiowa.edu/theories-UnicodeStrings.shtml
A symbolic string. Note that a symbolic string is not a list of symbolic characters, that is, it is not the case that SString = [SChar], unlike what one might expect following Haskell strings. An SString is a symbolic value of its own, of possibly arbitrary but finite length, and internally processed as one unit as opposed to a fixed-length list of characters.
## Symbolic lists
type SList a = SBV [a] Source #
A symbolic list of items. Note that a symbolic list is not a list of symbolic items, that is, it is not the case that SList a = [a], unlike what one might expect following haskell lists/sequences. An SList is a symbolic value of its own, of possibly arbitrary but finite length, and internally processed as one unit as opposed to a fixed-length list of items. Note that lists can be nested, i.e., we do allow lists of lists of ... items.
# Arrays of symbolic values
class SymArray array where Source #
Flat arrays of symbolic values An array a b is an array indexed by the type SBV a, with elements of type SBV b.
If a default value is supplied, then all the array elements will be initialized to this value. Otherwise, they will be left unspecified, i.e., a read from an unwritten location will produce an uninterpreted constant.
While it's certainly possible for user to create instances of SymArray, the SArray and SFunArray instances already provided should cover most use cases in practice. Note that there are a few differences between these two models in terms of use models:
• SArray produces SMTLib arrays, and requires a solver that understands the array theory. SFunArray is internally handled, and thus can be used with any solver. (Note that all solvers except abc support arrays, so this isn't a big decision factor.)
• For both arrays, if a default value is supplied, then reading from uninitialized cell will return that value. If the default is not given, then reading from uninitialized cells is still OK for both arrays, and will produce an uninterpreted constant in both cases.
• Only SArray supports checking equality of arrays. (That is, checking if an entire array is equivalent to another.) SFunArrays cannot be checked for equality. In general, checking wholesale equality of arrays is a difficult decision problem and should be avoided if possible.
• Only SFunArray supports compilation to C. Programs using SArray will not be accepted by the C-code generator.
• You cannot use quickcheck on programs that contain these arrays. (Neither SArray nor SFunArray.)
• With SArray, SBV transfers all array-processing to the SMT-solver. So, it can generate programs more quickly, but they might end up being too hard for the solver to handle. With SFunArray, SBV only generates code for individual elements and the array itself never shows up in the resulting SMTLib program. This puts more onus on the SBV side and might have some performance impacts, but it might generate problems that are easier for the SMT solvers to handle.
As a rule of thumb, try SArray first. These should generate compact code. However, if the backend solver has hard time solving the generated problems, switch to SFunArray. If you still have issues, please report so we can see what the problem might be!
Minimal complete definition
Methods
newArray_ :: (MonadSymbolic m, HasKind a, HasKind b) => Maybe (SBV b) -> m (array a b) Source #
Generalization of newArray_
newArray :: (MonadSymbolic m, HasKind a, HasKind b) => String -> Maybe (SBV b) -> m (array a b) Source #
Generalization of newArray
readArray :: array a b -> SBV a -> SBV b Source #
Read the array element at a
writeArray :: SymVal b => array a b -> SBV a -> SBV b -> array a b Source #
Update the element at a to be b
mergeArrays :: SymVal b => SBV Bool -> array a b -> array a b -> array a b Source #
Merge two given arrays on the symbolic condition Intuitively: mergeArrays cond a b = if cond then a else b. Merging pushes the if-then-else choice down on to elements
Instances
data SArray a b Source #
Arrays implemented in terms of SMT-arrays: http://smtlib.cs.uiowa.edu/theories-ArraysEx.shtml
• Maps directly to SMT-lib arrays
• Reading from an unintialized value is OK. If the default value is given in newArray, it will be the result. Otherwise, the read yields an uninterpreted constant.
• Can check for equality of these arrays
• Cannot be used in code-generation (i.e., compilation to C)
• Cannot quick-check theorems using SArray values
• Typically slower as it heavily relies on SMT-solving for the array theory
Instances
data SFunArray a b Source #
Arrays implemented internally, without translating to SMT-Lib functions:
• Internally handled by the library and not mapped to SMT-Lib, hence can be used with solvers that don't support arrays. (Such as abc.)
• Reading from an unintialized value is OK. If the default value is given in newArray, it will be the result. Otherwise, the read yields an uninterpreted constant.
• Cannot check for equality of arrays.
• Can be used in code-generation (i.e., compilation to C).
• Can not quick-check theorems using SFunArray values
• Typically faster as it gets compiled away during translation.
Instances
# Creating symbolic values
## Single value
sBool :: MonadSymbolic m => String -> m SBool Source #
Generalization of sBool
sWord8 :: MonadSymbolic m => String -> m SWord8 Source #
Generalization of sWord8
sWord16 :: MonadSymbolic m => String -> m SWord16 Source #
Generalization of sWord16
sWord32 :: MonadSymbolic m => String -> m SWord32 Source #
Generalization of sWord32
sWord64 :: MonadSymbolic m => String -> m SWord64 Source #
Generalization of sWord64
sWord :: (KnownNat n, IsNonZero n) => MonadSymbolic m => String -> m (SWord n) Source #
Generalization of sWord
sInt8 :: MonadSymbolic m => String -> m SInt8 Source #
Generalization of sInt8
sInt16 :: MonadSymbolic m => String -> m SInt16 Source #
Generalization of sInt16
sInt32 :: MonadSymbolic m => String -> m SInt32 Source #
Generalization of sInt32
sInt64 :: MonadSymbolic m => String -> m SInt64 Source #
Generalization of sInt64
sInt :: (KnownNat n, IsNonZero n) => MonadSymbolic m => String -> m (SInt n) Source #
Generalization of sInt
Generalization of sInteger
sReal :: MonadSymbolic m => String -> m SReal Source #
Generalization of sReal
sFloat :: MonadSymbolic m => String -> m SFloat Source #
Generalization of sFloat
sDouble :: MonadSymbolic m => String -> m SDouble Source #
Generalization of sDouble
sChar :: MonadSymbolic m => String -> m SChar Source #
Generalization of sChar
sString :: MonadSymbolic m => String -> m SString Source #
Generalization of sString
sList :: (SymVal a, MonadSymbolic m) => String -> m (SList a) Source #
Generalization of sList
## List of values
sBools :: MonadSymbolic m => [String] -> m [SBool] Source #
Generalization of sBools
sWord8s :: MonadSymbolic m => [String] -> m [SWord8] Source #
Generalization of sWord8s
sWord16s :: MonadSymbolic m => [String] -> m [SWord16] Source #
Generalization of sWord16s
sWord32s :: MonadSymbolic m => [String] -> m [SWord32] Source #
Generalization of sWord32s
sWord64s :: MonadSymbolic m => [String] -> m [SWord64] Source #
Generalization of sWord64s
sWords :: (KnownNat n, IsNonZero n) => MonadSymbolic m => [String] -> m [SWord n] Source #
Generalization of sWord64s
sInt8s :: MonadSymbolic m => [String] -> m [SInt8] Source #
Generalization of sInt8s
sInt16s :: MonadSymbolic m => [String] -> m [SInt16] Source #
Generalization of sInt16s
sInt32s :: MonadSymbolic m => [String] -> m [SInt32] Source #
Generalization of sInt32s
sInt64s :: MonadSymbolic m => [String] -> m [SInt64] Source #
Generalization of sInt64s
sInts :: (KnownNat n, IsNonZero n) => MonadSymbolic m => [String] -> m [SInt n] Source #
Generalization of sInts
sIntegers :: MonadSymbolic m => [String] -> m [SInteger] Source #
Generalization of sIntegers
sReals :: MonadSymbolic m => [String] -> m [SReal] Source #
Generalization of sReals
sFloats :: MonadSymbolic m => [String] -> m [SFloat] Source #
Generalization of sFloats
sDoubles :: MonadSymbolic m => [String] -> m [SDouble] Source #
Generalization of sDoubles
sChars :: MonadSymbolic m => [String] -> m [SChar] Source #
Generalization of sChars
sStrings :: MonadSymbolic m => [String] -> m [SString] Source #
Generalization of sStrings
sLists :: (SymVal a, MonadSymbolic m) => [String] -> m [SList a] Source #
Generalization of sLists
# Symbolic Equality and Comparisons
class EqSymbolic a where Source #
Symbolic Equality. Note that we can't use Haskell's Eq class since Haskell insists on returning Bool Comparing symbolic values will necessarily return a symbolic value.
Minimal complete definition
(.==)
Methods
(.==) :: a -> a -> SBool infix 4 Source #
Symbolic equality.
(./=) :: a -> a -> SBool infix 4 Source #
Symbolic inequality.
(.===) :: a -> a -> SBool infix 4 Source #
Strong equality. On floats ('SFloat'/'SDouble'), strong equality is object equality; that is NaN == NaN holds, but +0 == -0 doesn't. On other types, (.===) is simply (.==). Note that (.==) is the right notion of equality for floats per IEEE754 specs, since by definition +0 == -0 and NaN equals no other value including itself. But occasionally we want to be stronger and state NaN equals NaN and +0 and -0 are different from each other. In a context where your type is concrete, simply use fpIsEqualObject. But in a polymorphic context, use the strong equality instead.
NB. If you do not care about or work with floats, simply use (.==) and (./=).
(./==) :: a -> a -> SBool infix 4 Source #
Negation of strong equality. Equaivalent to negation of (.===) on all types.
distinct :: [a] -> SBool Source #
Returns (symbolic) sTrue if all the elements of the given list are different.
allEqual :: [a] -> SBool Source #
Returns (symbolic) sTrue if all the elements of the given list are the same.
sElem :: a -> [a] -> SBool Source #
Symbolic membership test.
Instances
Source # Instance detailsDefined in Data.SBV.Core.Model Methodsdistinct :: [Bool] -> SBool Source #allEqual :: [Bool] -> SBool Source #sElem :: Bool -> [Bool] -> SBool Source # EqSymbolic a => EqSymbolic [a] Source # Instance detailsDefined in Data.SBV.Core.Model Methods(.==) :: [a] -> [a] -> SBool Source #(./=) :: [a] -> [a] -> SBool Source #(.===) :: [a] -> [a] -> SBool Source #(./==) :: [a] -> [a] -> SBool Source #distinct :: [[a]] -> SBool Source #allEqual :: [[a]] -> SBool Source #sElem :: [a] -> [[a]] -> SBool Source # EqSymbolic a => EqSymbolic (Maybe a) Source # Instance detailsDefined in Data.SBV.Core.Model Methods(.==) :: Maybe a -> Maybe a -> SBool Source #(./=) :: Maybe a -> Maybe a -> SBool Source #(.===) :: Maybe a -> Maybe a -> SBool Source #(./==) :: Maybe a -> Maybe a -> SBool Source #distinct :: [Maybe a] -> SBool Source #allEqual :: [Maybe a] -> SBool Source #sElem :: Maybe a -> [Maybe a] -> SBool Source # Source # Instance detailsDefined in Data.SBV.Core.Model Methods(.==) :: SBV a -> SBV a -> SBool Source #(./=) :: SBV a -> SBV a -> SBool Source #(.===) :: SBV a -> SBV a -> SBool Source #(./==) :: SBV a -> SBV a -> SBool Source #distinct :: [SBV a] -> SBool Source #allEqual :: [SBV a] -> SBool Source #sElem :: SBV a -> [SBV a] -> SBool Source # EqSymbolic a => EqSymbolic (S a) Source # Symbolic equality for S. Instance details Methods(.==) :: S a -> S a -> SBool Source #(./=) :: S a -> S a -> SBool Source #(.===) :: S a -> S a -> SBool Source #(./==) :: S a -> S a -> SBool Source #distinct :: [S a] -> SBool Source #allEqual :: [S a] -> SBool Source #sElem :: S a -> [S a] -> SBool Source # (EqSymbolic a, EqSymbolic b) => EqSymbolic (Either a b) Source # Instance detailsDefined in Data.SBV.Core.Model Methods(.==) :: Either a b -> Either a b -> SBool Source #(./=) :: Either a b -> Either a b -> SBool Source #(.===) :: Either a b -> Either a b -> SBool Source #(./==) :: Either a b -> Either a b -> SBool Source #distinct :: [Either a b] -> SBool Source #allEqual :: [Either a b] -> SBool Source #sElem :: Either a b -> [Either a b] -> SBool Source # (EqSymbolic a, EqSymbolic b) => EqSymbolic (a, b) Source # Instance detailsDefined in Data.SBV.Core.Model Methods(.==) :: (a, b) -> (a, b) -> SBool Source #(./=) :: (a, b) -> (a, b) -> SBool Source #(.===) :: (a, b) -> (a, b) -> SBool Source #(./==) :: (a, b) -> (a, b) -> SBool Source #distinct :: [(a, b)] -> SBool Source #allEqual :: [(a, b)] -> SBool Source #sElem :: (a, b) -> [(a, b)] -> SBool Source # EqSymbolic (SArray a b) Source # Instance detailsDefined in Data.SBV.Core.Model Methods(.==) :: SArray a b -> SArray a b -> SBool Source #(./=) :: SArray a b -> SArray a b -> SBool Source #(.===) :: SArray a b -> SArray a b -> SBool Source #(./==) :: SArray a b -> SArray a b -> SBool Source #distinct :: [SArray a b] -> SBool Source #allEqual :: [SArray a b] -> SBool Source #sElem :: SArray a b -> [SArray a b] -> SBool Source # (EqSymbolic a, EqSymbolic b, EqSymbolic c) => EqSymbolic (a, b, c) Source # Instance detailsDefined in Data.SBV.Core.Model Methods(.==) :: (a, b, c) -> (a, b, c) -> SBool Source #(./=) :: (a, b, c) -> (a, b, c) -> SBool Source #(.===) :: (a, b, c) -> (a, b, c) -> SBool Source #(./==) :: (a, b, c) -> (a, b, c) -> SBool Source #distinct :: [(a, b, c)] -> SBool Source #allEqual :: [(a, b, c)] -> SBool Source #sElem :: (a, b, c) -> [(a, b, c)] -> SBool Source # (EqSymbolic a, EqSymbolic b, EqSymbolic c, EqSymbolic d) => EqSymbolic (a, b, c, d) Source # Instance detailsDefined in Data.SBV.Core.Model Methods(.==) :: (a, b, c, d) -> (a, b, c, d) -> SBool Source #(./=) :: (a, b, c, d) -> (a, b, c, d) -> SBool Source #(.===) :: (a, b, c, d) -> (a, b, c, d) -> SBool Source #(./==) :: (a, b, c, d) -> (a, b, c, d) -> SBool Source #distinct :: [(a, b, c, d)] -> SBool Source #allEqual :: [(a, b, c, d)] -> SBool Source #sElem :: (a, b, c, d) -> [(a, b, c, d)] -> SBool Source # (EqSymbolic a, EqSymbolic b, EqSymbolic c, EqSymbolic d, EqSymbolic e) => EqSymbolic (a, b, c, d, e) Source # Instance detailsDefined in Data.SBV.Core.Model Methods(.==) :: (a, b, c, d, e) -> (a, b, c, d, e) -> SBool Source #(./=) :: (a, b, c, d, e) -> (a, b, c, d, e) -> SBool Source #(.===) :: (a, b, c, d, e) -> (a, b, c, d, e) -> SBool Source #(./==) :: (a, b, c, d, e) -> (a, b, c, d, e) -> SBool Source #distinct :: [(a, b, c, d, e)] -> SBool Source #allEqual :: [(a, b, c, d, e)] -> SBool Source #sElem :: (a, b, c, d, e) -> [(a, b, c, d, e)] -> SBool Source # (EqSymbolic a, EqSymbolic b, EqSymbolic c, EqSymbolic d, EqSymbolic e, EqSymbolic f) => EqSymbolic (a, b, c, d, e, f) Source # Instance detailsDefined in Data.SBV.Core.Model Methods(.==) :: (a, b, c, d, e, f) -> (a, b, c, d, e, f) -> SBool Source #(./=) :: (a, b, c, d, e, f) -> (a, b, c, d, e, f) -> SBool Source #(.===) :: (a, b, c, d, e, f) -> (a, b, c, d, e, f) -> SBool Source #(./==) :: (a, b, c, d, e, f) -> (a, b, c, d, e, f) -> SBool Source #distinct :: [(a, b, c, d, e, f)] -> SBool Source #allEqual :: [(a, b, c, d, e, f)] -> SBool Source #sElem :: (a, b, c, d, e, f) -> [(a, b, c, d, e, f)] -> SBool Source # (EqSymbolic a, EqSymbolic b, EqSymbolic c, EqSymbolic d, EqSymbolic e, EqSymbolic f, EqSymbolic g) => EqSymbolic (a, b, c, d, e, f, g) Source # Instance detailsDefined in Data.SBV.Core.Model Methods(.==) :: (a, b, c, d, e, f, g) -> (a, b, c, d, e, f, g) -> SBool Source #(./=) :: (a, b, c, d, e, f, g) -> (a, b, c, d, e, f, g) -> SBool Source #(.===) :: (a, b, c, d, e, f, g) -> (a, b, c, d, e, f, g) -> SBool Source #(./==) :: (a, b, c, d, e, f, g) -> (a, b, c, d, e, f, g) -> SBool Source #distinct :: [(a, b, c, d, e, f, g)] -> SBool Source #allEqual :: [(a, b, c, d, e, f, g)] -> SBool Source #sElem :: (a, b, c, d, e, f, g) -> [(a, b, c, d, e, f, g)] -> SBool Source #
class (Mergeable a, EqSymbolic a) => OrdSymbolic a where Source #
Symbolic Comparisons. Similar to Eq, we cannot implement Haskell's Ord class since there is no way to return an Ordering value from a symbolic comparison. Furthermore, OrdSymbolic requires Mergeable to implement if-then-else, for the benefit of implementing symbolic versions of max and min functions.
Minimal complete definition
(.<)
Methods
(.<) :: a -> a -> SBool infix 4 Source #
Symbolic less than.
(.<=) :: a -> a -> SBool infix 4 Source #
Symbolic less than or equal to.
(.>) :: a -> a -> SBool infix 4 Source #
Symbolic greater than.
(.>=) :: a -> a -> SBool infix 4 Source #
Symbolic greater than or equal to.
smin :: a -> a -> a Source #
Symbolic minimum.
smax :: a -> a -> a Source #
Symbolic maximum.
inRange :: a -> (a, a) -> SBool Source #
Is the value withing the allowed inclusive range?
Instances
OrdSymbolic a => OrdSymbolic [a] Source # Instance detailsDefined in Data.SBV.Core.Model Methods(.<) :: [a] -> [a] -> SBool Source #(.<=) :: [a] -> [a] -> SBool Source #(.>) :: [a] -> [a] -> SBool Source #(.>=) :: [a] -> [a] -> SBool Source #smin :: [a] -> [a] -> [a] Source #smax :: [a] -> [a] -> [a] Source #inRange :: [a] -> ([a], [a]) -> SBool Source # OrdSymbolic a => OrdSymbolic (Maybe a) Source # Instance detailsDefined in Data.SBV.Core.Model Methods(.<) :: Maybe a -> Maybe a -> SBool Source #(.<=) :: Maybe a -> Maybe a -> SBool Source #(.>) :: Maybe a -> Maybe a -> SBool Source #(.>=) :: Maybe a -> Maybe a -> SBool Source #smin :: Maybe a -> Maybe a -> Maybe a Source #smax :: Maybe a -> Maybe a -> Maybe a Source #inRange :: Maybe a -> (Maybe a, Maybe a) -> SBool Source # (Ord a, SymVal a) => OrdSymbolic (SBV a) Source # If comparison is over something SMTLib can handle, just translate it. Otherwise desugar. Instance detailsDefined in Data.SBV.Core.Model Methods(.<) :: SBV a -> SBV a -> SBool Source #(.<=) :: SBV a -> SBV a -> SBool Source #(.>) :: SBV a -> SBV a -> SBool Source #(.>=) :: SBV a -> SBV a -> SBool Source #smin :: SBV a -> SBV a -> SBV a Source #smax :: SBV a -> SBV a -> SBV a Source #inRange :: SBV a -> (SBV a, SBV a) -> SBool Source # (OrdSymbolic a, OrdSymbolic b) => OrdSymbolic (Either a b) Source # Instance detailsDefined in Data.SBV.Core.Model Methods(.<) :: Either a b -> Either a b -> SBool Source #(.<=) :: Either a b -> Either a b -> SBool Source #(.>) :: Either a b -> Either a b -> SBool Source #(.>=) :: Either a b -> Either a b -> SBool Source #smin :: Either a b -> Either a b -> Either a b Source #smax :: Either a b -> Either a b -> Either a b Source #inRange :: Either a b -> (Either a b, Either a b) -> SBool Source # (OrdSymbolic a, OrdSymbolic b) => OrdSymbolic (a, b) Source # Instance detailsDefined in Data.SBV.Core.Model Methods(.<) :: (a, b) -> (a, b) -> SBool Source #(.<=) :: (a, b) -> (a, b) -> SBool Source #(.>) :: (a, b) -> (a, b) -> SBool Source #(.>=) :: (a, b) -> (a, b) -> SBool Source #smin :: (a, b) -> (a, b) -> (a, b) Source #smax :: (a, b) -> (a, b) -> (a, b) Source #inRange :: (a, b) -> ((a, b), (a, b)) -> SBool Source # (OrdSymbolic a, OrdSymbolic b, OrdSymbolic c) => OrdSymbolic (a, b, c) Source # Instance detailsDefined in Data.SBV.Core.Model Methods(.<) :: (a, b, c) -> (a, b, c) -> SBool Source #(.<=) :: (a, b, c) -> (a, b, c) -> SBool Source #(.>) :: (a, b, c) -> (a, b, c) -> SBool Source #(.>=) :: (a, b, c) -> (a, b, c) -> SBool Source #smin :: (a, b, c) -> (a, b, c) -> (a, b, c) Source #smax :: (a, b, c) -> (a, b, c) -> (a, b, c) Source #inRange :: (a, b, c) -> ((a, b, c), (a, b, c)) -> SBool Source # (OrdSymbolic a, OrdSymbolic b, OrdSymbolic c, OrdSymbolic d) => OrdSymbolic (a, b, c, d) Source # Instance detailsDefined in Data.SBV.Core.Model Methods(.<) :: (a, b, c, d) -> (a, b, c, d) -> SBool Source #(.<=) :: (a, b, c, d) -> (a, b, c, d) -> SBool Source #(.>) :: (a, b, c, d) -> (a, b, c, d) -> SBool Source #(.>=) :: (a, b, c, d) -> (a, b, c, d) -> SBool Source #smin :: (a, b, c, d) -> (a, b, c, d) -> (a, b, c, d) Source #smax :: (a, b, c, d) -> (a, b, c, d) -> (a, b, c, d) Source #inRange :: (a, b, c, d) -> ((a, b, c, d), (a, b, c, d)) -> SBool Source # (OrdSymbolic a, OrdSymbolic b, OrdSymbolic c, OrdSymbolic d, OrdSymbolic e) => OrdSymbolic (a, b, c, d, e) Source # Instance detailsDefined in Data.SBV.Core.Model Methods(.<) :: (a, b, c, d, e) -> (a, b, c, d, e) -> SBool Source #(.<=) :: (a, b, c, d, e) -> (a, b, c, d, e) -> SBool Source #(.>) :: (a, b, c, d, e) -> (a, b, c, d, e) -> SBool Source #(.>=) :: (a, b, c, d, e) -> (a, b, c, d, e) -> SBool Source #smin :: (a, b, c, d, e) -> (a, b, c, d, e) -> (a, b, c, d, e) Source #smax :: (a, b, c, d, e) -> (a, b, c, d, e) -> (a, b, c, d, e) Source #inRange :: (a, b, c, d, e) -> ((a, b, c, d, e), (a, b, c, d, e)) -> SBool Source # (OrdSymbolic a, OrdSymbolic b, OrdSymbolic c, OrdSymbolic d, OrdSymbolic e, OrdSymbolic f) => OrdSymbolic (a, b, c, d, e, f) Source # Instance detailsDefined in Data.SBV.Core.Model Methods(.<) :: (a, b, c, d, e, f) -> (a, b, c, d, e, f) -> SBool Source #(.<=) :: (a, b, c, d, e, f) -> (a, b, c, d, e, f) -> SBool Source #(.>) :: (a, b, c, d, e, f) -> (a, b, c, d, e, f) -> SBool Source #(.>=) :: (a, b, c, d, e, f) -> (a, b, c, d, e, f) -> SBool Source #smin :: (a, b, c, d, e, f) -> (a, b, c, d, e, f) -> (a, b, c, d, e, f) Source #smax :: (a, b, c, d, e, f) -> (a, b, c, d, e, f) -> (a, b, c, d, e, f) Source #inRange :: (a, b, c, d, e, f) -> ((a, b, c, d, e, f), (a, b, c, d, e, f)) -> SBool Source # (OrdSymbolic a, OrdSymbolic b, OrdSymbolic c, OrdSymbolic d, OrdSymbolic e, OrdSymbolic f, OrdSymbolic g) => OrdSymbolic (a, b, c, d, e, f, g) Source # Instance detailsDefined in Data.SBV.Core.Model Methods(.<) :: (a, b, c, d, e, f, g) -> (a, b, c, d, e, f, g) -> SBool Source #(.<=) :: (a, b, c, d, e, f, g) -> (a, b, c, d, e, f, g) -> SBool Source #(.>) :: (a, b, c, d, e, f, g) -> (a, b, c, d, e, f, g) -> SBool Source #(.>=) :: (a, b, c, d, e, f, g) -> (a, b, c, d, e, f, g) -> SBool Source #smin :: (a, b, c, d, e, f, g) -> (a, b, c, d, e, f, g) -> (a, b, c, d, e, f, g) Source #smax :: (a, b, c, d, e, f, g) -> (a, b, c, d, e, f, g) -> (a, b, c, d, e, f, g) Source #inRange :: (a, b, c, d, e, f, g) -> ((a, b, c, d, e, f, g), (a, b, c, d, e, f, g)) -> SBool Source #
class Equality a where Source #
Equality as a proof method. Allows for very concise construction of equivalence proofs, which is very typical in bit-precise proofs.
Methods
(===) :: a -> a -> IO ThmResult infix 4 Source #
Instances
(SymVal a, SymVal b, EqSymbolic z) => Equality ((SBV a, SBV b) -> z) Source # Instance detailsDefined in Data.SBV.Core.Model Methods(===) :: ((SBV a, SBV b) -> z) -> ((SBV a, SBV b) -> z) -> IO ThmResult Source # (SymVal a, SymVal b, SymVal c, EqSymbolic z) => Equality ((SBV a, SBV b, SBV c) -> z) Source # Instance detailsDefined in Data.SBV.Core.Model Methods(===) :: ((SBV a, SBV b, SBV c) -> z) -> ((SBV a, SBV b, SBV c) -> z) -> IO ThmResult Source # (SymVal a, SymVal b, SymVal c, SymVal d, EqSymbolic z) => Equality ((SBV a, SBV b, SBV c, SBV d) -> z) Source # Instance detailsDefined in Data.SBV.Core.Model Methods(===) :: ((SBV a, SBV b, SBV c, SBV d) -> z) -> ((SBV a, SBV b, SBV c, SBV d) -> z) -> IO ThmResult Source # (SymVal a, SymVal b, SymVal c, SymVal d, SymVal e, EqSymbolic z) => Equality ((SBV a, SBV b, SBV c, SBV d, SBV e) -> z) Source # Instance detailsDefined in Data.SBV.Core.Model Methods(===) :: ((SBV a, SBV b, SBV c, SBV d, SBV e) -> z) -> ((SBV a, SBV b, SBV c, SBV d, SBV e) -> z) -> IO ThmResult Source # (SymVal a, SymVal b, SymVal c, SymVal d, SymVal e, SymVal f, EqSymbolic z) => Equality ((SBV a, SBV b, SBV c, SBV d, SBV e, SBV f) -> z) Source # Instance detailsDefined in Data.SBV.Core.Model Methods(===) :: ((SBV a, SBV b, SBV c, SBV d, SBV e, SBV f) -> z) -> ((SBV a, SBV b, SBV c, SBV d, SBV e, SBV f) -> z) -> IO ThmResult Source # (SymVal a, SymVal b, SymVal c, SymVal d, SymVal e, SymVal f, SymVal g, EqSymbolic z) => Equality ((SBV a, SBV b, SBV c, SBV d, SBV e, SBV f, SBV g) -> z) Source # Instance detailsDefined in Data.SBV.Core.Model Methods(===) :: ((SBV a, SBV b, SBV c, SBV d, SBV e, SBV f, SBV g) -> z) -> ((SBV a, SBV b, SBV c, SBV d, SBV e, SBV f, SBV g) -> z) -> IO ThmResult Source # (SymVal a, SymVal b, SymVal c, SymVal d, SymVal e, SymVal f, SymVal g, EqSymbolic z) => Equality (SBV a -> SBV b -> SBV c -> SBV d -> SBV e -> SBV f -> SBV g -> z) Source # Instance detailsDefined in Data.SBV.Core.Model Methods(===) :: (SBV a -> SBV b -> SBV c -> SBV d -> SBV e -> SBV f -> SBV g -> z) -> (SBV a -> SBV b -> SBV c -> SBV d -> SBV e -> SBV f -> SBV g -> z) -> IO ThmResult Source # (SymVal a, SymVal b, SymVal c, SymVal d, SymVal e, SymVal f, EqSymbolic z) => Equality (SBV a -> SBV b -> SBV c -> SBV d -> SBV e -> SBV f -> z) Source # Instance detailsDefined in Data.SBV.Core.Model Methods(===) :: (SBV a -> SBV b -> SBV c -> SBV d -> SBV e -> SBV f -> z) -> (SBV a -> SBV b -> SBV c -> SBV d -> SBV e -> SBV f -> z) -> IO ThmResult Source # (SymVal a, SymVal b, SymVal c, SymVal d, SymVal e, EqSymbolic z) => Equality (SBV a -> SBV b -> SBV c -> SBV d -> SBV e -> z) Source # Instance detailsDefined in Data.SBV.Core.Model Methods(===) :: (SBV a -> SBV b -> SBV c -> SBV d -> SBV e -> z) -> (SBV a -> SBV b -> SBV c -> SBV d -> SBV e -> z) -> IO ThmResult Source # (SymVal a, SymVal b, SymVal c, SymVal d, EqSymbolic z) => Equality (SBV a -> SBV b -> SBV c -> SBV d -> z) Source # Instance detailsDefined in Data.SBV.Core.Model Methods(===) :: (SBV a -> SBV b -> SBV c -> SBV d -> z) -> (SBV a -> SBV b -> SBV c -> SBV d -> z) -> IO ThmResult Source # (SymVal a, SymVal b, SymVal c, EqSymbolic z) => Equality (SBV a -> SBV b -> SBV c -> z) Source # Instance detailsDefined in Data.SBV.Core.Model Methods(===) :: (SBV a -> SBV b -> SBV c -> z) -> (SBV a -> SBV b -> SBV c -> z) -> IO ThmResult Source # (SymVal a, SymVal b, EqSymbolic z) => Equality (SBV a -> SBV b -> z) Source # Instance detailsDefined in Data.SBV.Core.Model Methods(===) :: (SBV a -> SBV b -> z) -> (SBV a -> SBV b -> z) -> IO ThmResult Source # (SymVal a, EqSymbolic z) => Equality (SBV a -> z) Source # Instance detailsDefined in Data.SBV.Core.Model Methods(===) :: (SBV a -> z) -> (SBV a -> z) -> IO ThmResult Source #
# Conditionals: Mergeable values
class Mergeable a where Source #
Symbolic conditionals are modeled by the Mergeable class, describing how to merge the results of an if-then-else call with a symbolic test. SBV provides all basic types as instances of this class, so users only need to declare instances for custom data-types of their programs as needed.
A Mergeable instance may be automatically derived for a custom data-type with a single constructor where the type of each field is an instance of Mergeable, such as a record of symbolic values. Users only need to add Generic and Mergeable to the deriving clause for the data-type. See Status for an example and an illustration of what the instance would look like if written by hand.
The function select is a total-indexing function out of a list of choices with a default value, simulating array/list indexing. It's an n-way generalization of the ite function.
Minimal complete definition: None, if the type is instance of Generic. Otherwise symbolicMerge. Note that most types subject to merging are likely to be trivial instances of Generic.
Minimal complete definition
Nothing
Methods
symbolicMerge :: Bool -> SBool -> a -> a -> a Source #
Merge two values based on the condition. The first argument states whether we force the then-and-else branches before the merging, at the word level. This is an efficiency concern; one that we'd rather not make but unfortunately necessary for getting symbolic simulation working efficiently.
select :: (Ord b, SymVal b, Num b) => [a] -> a -> SBV b -> a Source #
Total indexing operation. select xs default index is intuitively the same as xs !! index, except it evaluates to default if index underflows/overflows.
symbolicMerge :: (Generic a, GMergeable (Rep a)) => Bool -> SBool -> a -> a -> a Source #
Merge two values based on the condition. The first argument states whether we force the then-and-else branches before the merging, at the word level. This is an efficiency concern; one that we'd rather not make but unfortunately necessary for getting symbolic simulation working efficiently.
Instances
Source # Instance detailsDefined in Data.SBV.Core.Model MethodssymbolicMerge :: Bool -> SBool -> () -> () -> () Source #select :: (Ord b, SymVal b, Num b) => [()] -> () -> SBV b -> () Source # Source # Instance details Methodsselect :: (Ord b, SymVal b, Num b) => [Mostek] -> Mostek -> SBV b -> Mostek Source # Source # Instance details Methodsselect :: (Ord b, SymVal b, Num b) => [Status] -> Status -> SBV b -> Status Source # Mergeable a => Mergeable [a] Source # Instance detailsDefined in Data.SBV.Core.Model MethodssymbolicMerge :: Bool -> SBool -> [a] -> [a] -> [a] Source #select :: (Ord b, SymVal b, Num b) => [[a]] -> [a] -> SBV b -> [a] Source # Mergeable a => Mergeable (Maybe a) Source # Instance detailsDefined in Data.SBV.Core.Model MethodssymbolicMerge :: Bool -> SBool -> Maybe a -> Maybe a -> Maybe a Source #select :: (Ord b, SymVal b, Num b) => [Maybe a] -> Maybe a -> SBV b -> Maybe a Source # Mergeable a => Mergeable (ZipList a) Source # Instance detailsDefined in Data.SBV.Core.Model MethodssymbolicMerge :: Bool -> SBool -> ZipList a -> ZipList a -> ZipList a Source #select :: (Ord b, SymVal b, Num b) => [ZipList a] -> ZipList a -> SBV b -> ZipList a Source # SymVal a => Mergeable (SBV a) Source # Instance detailsDefined in Data.SBV.Core.Model MethodssymbolicMerge :: Bool -> SBool -> SBV a -> SBV a -> SBV a Source #select :: (Ord b, SymVal b, Num b) => [SBV a] -> SBV a -> SBV b -> SBV a Source # Mergeable a => Mergeable (S a) Source # Instance details MethodssymbolicMerge :: Bool -> SBool -> S a -> S a -> S a Source #select :: (Ord b, SymVal b, Num b) => [S a] -> S a -> SBV b -> S a Source # Mergeable a => Mergeable (S a) Source # Instance details MethodssymbolicMerge :: Bool -> SBool -> S a -> S a -> S a Source #select :: (Ord b, SymVal b, Num b) => [S a] -> S a -> SBV b -> S a Source # Mergeable a => Mergeable (Move a) Source # Mergeable instance for Move simply pushes the merging the data after run of each branch starting from the same state. Instance details MethodssymbolicMerge :: Bool -> SBool -> Move a -> Move a -> Move a Source #select :: (Ord b, SymVal b, Num b) => [Move a] -> Move a -> SBV b -> Move a Source # SymVal a => Mergeable (AppS a) Source # Instance details MethodssymbolicMerge :: Bool -> SBool -> AppS a -> AppS a -> AppS a Source #select :: (Ord b, SymVal b, Num b) => [AppS a] -> AppS a -> SBV b -> AppS a Source # Mergeable a => Mergeable (FibS a) Source # Instance details MethodssymbolicMerge :: Bool -> SBool -> FibS a -> FibS a -> FibS a Source #select :: (Ord b, SymVal b, Num b) => [FibS a] -> FibS a -> SBV b -> FibS a Source # Mergeable a => Mergeable (GCDS a) Source # Instance details MethodssymbolicMerge :: Bool -> SBool -> GCDS a -> GCDS a -> GCDS a Source #select :: (Ord b, SymVal b, Num b) => [GCDS a] -> GCDS a -> SBV b -> GCDS a Source # Mergeable a => Mergeable (DivS a) Source # Instance details MethodssymbolicMerge :: Bool -> SBool -> DivS a -> DivS a -> DivS a Source #select :: (Ord b, SymVal b, Num b) => [DivS a] -> DivS a -> SBV b -> DivS a Source # Mergeable a => Mergeable (SqrtS a) Source # Instance details MethodssymbolicMerge :: Bool -> SBool -> SqrtS a -> SqrtS a -> SqrtS a Source #select :: (Ord b, SymVal b, Num b) => [SqrtS a] -> SqrtS a -> SBV b -> SqrtS a Source # SymVal a => Mergeable (LenS a) Source # Instance details MethodssymbolicMerge :: Bool -> SBool -> LenS a -> LenS a -> LenS a Source #select :: (Ord b, SymVal b, Num b) => [LenS a] -> LenS a -> SBV b -> LenS a Source # Mergeable a => Mergeable (SumS a) Source # Instance details MethodssymbolicMerge :: Bool -> SBool -> SumS a -> SumS a -> SumS a Source #select :: (Ord b, SymVal b, Num b) => [SumS a] -> SumS a -> SBV b -> SumS a Source # Mergeable b => Mergeable (a -> b) Source # Instance detailsDefined in Data.SBV.Core.Model MethodssymbolicMerge :: Bool -> SBool -> (a -> b) -> (a -> b) -> a -> b Source #select :: (Ord b0, SymVal b0, Num b0) => [a -> b] -> (a -> b) -> SBV b0 -> a -> b Source # (Mergeable a, Mergeable b) => Mergeable (Either a b) Source # Instance detailsDefined in Data.SBV.Core.Model MethodssymbolicMerge :: Bool -> SBool -> Either a b -> Either a b -> Either a b Source #select :: (Ord b0, SymVal b0, Num b0) => [Either a b] -> Either a b -> SBV b0 -> Either a b Source # (Mergeable a, Mergeable b) => Mergeable (a, b) Source # Instance detailsDefined in Data.SBV.Core.Model MethodssymbolicMerge :: Bool -> SBool -> (a, b) -> (a, b) -> (a, b) Source #select :: (Ord b0, SymVal b0, Num b0) => [(a, b)] -> (a, b) -> SBV b0 -> (a, b) Source # (Ix a, Mergeable b) => Mergeable (Array a b) Source # Instance detailsDefined in Data.SBV.Core.Model MethodssymbolicMerge :: Bool -> SBool -> Array a b -> Array a b -> Array a b Source #select :: (Ord b0, SymVal b0, Num b0) => [Array a b] -> Array a b -> SBV b0 -> Array a b Source # SymVal b => Mergeable (SFunArray a b) Source # Instance detailsDefined in Data.SBV.Core.Model MethodssymbolicMerge :: Bool -> SBool -> SFunArray a b -> SFunArray a b -> SFunArray a b Source #select :: (Ord b0, SymVal b0, Num b0) => [SFunArray a b] -> SFunArray a b -> SBV b0 -> SFunArray a b Source # SymVal b => Mergeable (SArray a b) Source # Instance detailsDefined in Data.SBV.Core.Model MethodssymbolicMerge :: Bool -> SBool -> SArray a b -> SArray a b -> SArray a b Source #select :: (Ord b0, SymVal b0, Num b0) => [SArray a b] -> SArray a b -> SBV b0 -> SArray a b Source # SymVal e => Mergeable (STree i e) Source # Instance detailsDefined in Data.SBV.Tools.STree MethodssymbolicMerge :: Bool -> SBool -> STree i e -> STree i e -> STree i e Source #select :: (Ord b, SymVal b, Num b) => [STree i e] -> STree i e -> SBV b -> STree i e Source # (Mergeable a, Mergeable b, Mergeable c) => Mergeable (a, b, c) Source # Instance detailsDefined in Data.SBV.Core.Model MethodssymbolicMerge :: Bool -> SBool -> (a, b, c) -> (a, b, c) -> (a, b, c) Source #select :: (Ord b0, SymVal b0, Num b0) => [(a, b, c)] -> (a, b, c) -> SBV b0 -> (a, b, c) Source # (Mergeable a, Mergeable b, Mergeable c, Mergeable d) => Mergeable (a, b, c, d) Source # Instance detailsDefined in Data.SBV.Core.Model MethodssymbolicMerge :: Bool -> SBool -> (a, b, c, d) -> (a, b, c, d) -> (a, b, c, d) Source #select :: (Ord b0, SymVal b0, Num b0) => [(a, b, c, d)] -> (a, b, c, d) -> SBV b0 -> (a, b, c, d) Source # (Mergeable a, Mergeable b, Mergeable c, Mergeable d, Mergeable e) => Mergeable (a, b, c, d, e) Source # Instance detailsDefined in Data.SBV.Core.Model MethodssymbolicMerge :: Bool -> SBool -> (a, b, c, d, e) -> (a, b, c, d, e) -> (a, b, c, d, e) Source #select :: (Ord b0, SymVal b0, Num b0) => [(a, b, c, d, e)] -> (a, b, c, d, e) -> SBV b0 -> (a, b, c, d, e) Source # (Mergeable a, Mergeable b, Mergeable c, Mergeable d, Mergeable e, Mergeable f) => Mergeable (a, b, c, d, e, f) Source # Instance detailsDefined in Data.SBV.Core.Model MethodssymbolicMerge :: Bool -> SBool -> (a, b, c, d, e, f) -> (a, b, c, d, e, f) -> (a, b, c, d, e, f) Source #select :: (Ord b0, SymVal b0, Num b0) => [(a, b, c, d, e, f)] -> (a, b, c, d, e, f) -> SBV b0 -> (a, b, c, d, e, f) Source # (Mergeable a, Mergeable b, Mergeable c, Mergeable d, Mergeable e, Mergeable f, Mergeable g) => Mergeable (a, b, c, d, e, f, g) Source # Instance detailsDefined in Data.SBV.Core.Model MethodssymbolicMerge :: Bool -> SBool -> (a, b, c, d, e, f, g) -> (a, b, c, d, e, f, g) -> (a, b, c, d, e, f, g) Source #select :: (Ord b0, SymVal b0, Num b0) => [(a, b, c, d, e, f, g)] -> (a, b, c, d, e, f, g) -> SBV b0 -> (a, b, c, d, e, f, g) Source #
ite :: Mergeable a => SBool -> a -> a -> a Source #
If-then-else. This is by definition symbolicMerge with both branches forced. This is typically the desired behavior, but also see iteLazy should you need more laziness.
iteLazy :: Mergeable a => SBool -> a -> a -> a Source #
A Lazy version of ite, which does not force its arguments. This might cause issues for symbolic simulation with large thunks around, so use with care.
# Symbolic integral numbers
class (SymVal a, Num a, Bits a, Integral a) => SIntegral a Source #
Symbolic Numbers. This is a simple class that simply incorporates all number like base types together, simplifying writing polymorphic type-signatures that work for all symbolic numbers, such as SWord8, SInt8 etc. For instance, we can write a generic list-minimum function as follows:
mm :: SIntegral a => [SBV a] -> SBV a
mm = foldr1 (a b -> ite (a .<= b) a b)
It is similar to the standard Integral class, except ranging over symbolic instances.
Instances
Source # Instance detailsDefined in Data.SBV.Core.Model Source # Instance detailsDefined in Data.SBV.Core.Model Source # Instance detailsDefined in Data.SBV.Core.Model Source # Instance detailsDefined in Data.SBV.Core.Model Source # Instance detailsDefined in Data.SBV.Core.Model Source # Instance detailsDefined in Data.SBV.Core.Model Source # Instance detailsDefined in Data.SBV.Core.Model Source # Instance detailsDefined in Data.SBV.Core.Model Source # Instance detailsDefined in Data.SBV.Core.Model (KnownNat n, IsNonZero n) => SIntegral (IntN n) Source # SIntegral instance for IntN Instance detailsDefined in Data.SBV.Core.Sized (KnownNat n, IsNonZero n) => SIntegral (WordN n) Source # SIntegral instance for WordN Instance detailsDefined in Data.SBV.Core.Sized
# Division and Modulus
class SDivisible a where Source #
The SDivisible class captures the essence of division. Unfortunately we cannot use Haskell's Integral class since the Real and Enum superclasses are not implementable for symbolic bit-vectors. However, quotRem and divMod both make perfect sense, and the SDivisible class captures this operation. One issue is how division by 0 behaves. The verification technology requires total functions, and there are several design choices here. We follow Isabelle/HOL approach of assigning the value 0 for division by 0. Therefore, we impose the following pair of laws:
x sQuotRem 0 = (0, x)
x sDivMod 0 = (0, x)
Note that our instances implement this law even when x is 0 itself.
NB. quot truncates toward zero, while div truncates toward negative infinity.
### C code generation of division operations
In the case of division or modulo of a minimal signed value (e.g. -128 for SInt8) by -1, SMTLIB and Haskell agree on what the result should be. Unfortunately the result in C code depends on CPU architecture and compiler settings, as this is undefined behaviour in C. **SBV does not guarantee** what will happen in generated C code in this corner case.
Minimal complete definition
Methods
sQuotRem :: a -> a -> (a, a) Source #
sDivMod :: a -> a -> (a, a) Source #
sQuot :: a -> a -> a Source #
sRem :: a -> a -> a Source #
sDiv :: a -> a -> a Source #
sMod :: a -> a -> a Source #
Instances
Source # Instance detailsDefined in Data.SBV.Core.Model MethodssQuotRem :: Int8 -> Int8 -> (Int8, Int8) Source #sDivMod :: Int8 -> Int8 -> (Int8, Int8) Source #sQuot :: Int8 -> Int8 -> Int8 Source #sRem :: Int8 -> Int8 -> Int8 Source #sDiv :: Int8 -> Int8 -> Int8 Source #sMod :: Int8 -> Int8 -> Int8 Source # Source # Instance detailsDefined in Data.SBV.Core.Model MethodssQuotRem :: Int16 -> Int16 -> (Int16, Int16) Source #sDivMod :: Int16 -> Int16 -> (Int16, Int16) Source # Source # Instance detailsDefined in Data.SBV.Core.Model MethodssQuotRem :: Int32 -> Int32 -> (Int32, Int32) Source #sDivMod :: Int32 -> Int32 -> (Int32, Int32) Source # Source # Instance detailsDefined in Data.SBV.Core.Model MethodssQuotRem :: Int64 -> Int64 -> (Int64, Int64) Source #sDivMod :: Int64 -> Int64 -> (Int64, Int64) Source # Source # Instance detailsDefined in Data.SBV.Core.Model Methods Source # Instance detailsDefined in Data.SBV.Core.Model MethodssQuotRem :: Word8 -> Word8 -> (Word8, Word8) Source #sDivMod :: Word8 -> Word8 -> (Word8, Word8) Source # Source # Instance detailsDefined in Data.SBV.Core.Model MethodssQuotRem :: Word16 -> Word16 -> (Word16, Word16) Source #sDivMod :: Word16 -> Word16 -> (Word16, Word16) Source # Source # Instance detailsDefined in Data.SBV.Core.Model MethodssQuotRem :: Word32 -> Word32 -> (Word32, Word32) Source #sDivMod :: Word32 -> Word32 -> (Word32, Word32) Source # Source # Instance detailsDefined in Data.SBV.Core.Model MethodssQuotRem :: Word64 -> Word64 -> (Word64, Word64) Source #sDivMod :: Word64 -> Word64 -> (Word64, Word64) Source # Source # Instance detailsDefined in Data.SBV.Core.Model MethodssQuotRem :: CV -> CV -> (CV, CV) Source #sDivMod :: CV -> CV -> (CV, CV) Source #sQuot :: CV -> CV -> CV Source #sRem :: CV -> CV -> CV Source #sDiv :: CV -> CV -> CV Source #sMod :: CV -> CV -> CV Source # Source # Instance detailsDefined in Data.SBV.Core.Model Methods Source # Instance detailsDefined in Data.SBV.Core.Model MethodssQuotRem :: SInt64 -> SInt64 -> (SInt64, SInt64) Source #sDivMod :: SInt64 -> SInt64 -> (SInt64, SInt64) Source # Source # Instance detailsDefined in Data.SBV.Core.Model MethodssQuotRem :: SInt32 -> SInt32 -> (SInt32, SInt32) Source #sDivMod :: SInt32 -> SInt32 -> (SInt32, SInt32) Source # Source # Instance detailsDefined in Data.SBV.Core.Model MethodssQuotRem :: SInt16 -> SInt16 -> (SInt16, SInt16) Source #sDivMod :: SInt16 -> SInt16 -> (SInt16, SInt16) Source # Source # Instance detailsDefined in Data.SBV.Core.Model MethodssQuotRem :: SInt8 -> SInt8 -> (SInt8, SInt8) Source #sDivMod :: SInt8 -> SInt8 -> (SInt8, SInt8) Source # Source # Instance detailsDefined in Data.SBV.Core.Model Methods Source # Instance detailsDefined in Data.SBV.Core.Model Methods Source # Instance detailsDefined in Data.SBV.Core.Model Methods Source # Instance detailsDefined in Data.SBV.Core.Model MethodssQuotRem :: SWord8 -> SWord8 -> (SWord8, SWord8) Source #sDivMod :: SWord8 -> SWord8 -> (SWord8, SWord8) Source # (KnownNat n, IsNonZero n) => SDivisible (SInt n) Source # SDivisible instance for SInt Instance detailsDefined in Data.SBV.Core.Sized MethodssQuotRem :: SInt n -> SInt n -> (SInt n, SInt n) Source #sDivMod :: SInt n -> SInt n -> (SInt n, SInt n) Source #sQuot :: SInt n -> SInt n -> SInt n Source #sRem :: SInt n -> SInt n -> SInt n Source #sDiv :: SInt n -> SInt n -> SInt n Source #sMod :: SInt n -> SInt n -> SInt n Source # (KnownNat n, IsNonZero n) => SDivisible (IntN n) Source # SDivisible instance for IntN Instance detailsDefined in Data.SBV.Core.Sized MethodssQuotRem :: IntN n -> IntN n -> (IntN n, IntN n) Source #sDivMod :: IntN n -> IntN n -> (IntN n, IntN n) Source #sQuot :: IntN n -> IntN n -> IntN n Source #sRem :: IntN n -> IntN n -> IntN n Source #sDiv :: IntN n -> IntN n -> IntN n Source #sMod :: IntN n -> IntN n -> IntN n Source # (KnownNat n, IsNonZero n) => SDivisible (SWord n) Source # SDivisible instance for SWord Instance detailsDefined in Data.SBV.Core.Sized MethodssQuotRem :: SWord n -> SWord n -> (SWord n, SWord n) Source #sDivMod :: SWord n -> SWord n -> (SWord n, SWord n) Source #sQuot :: SWord n -> SWord n -> SWord n Source #sRem :: SWord n -> SWord n -> SWord n Source #sDiv :: SWord n -> SWord n -> SWord n Source #sMod :: SWord n -> SWord n -> SWord n Source # (KnownNat n, IsNonZero n) => SDivisible (WordN n) Source # SDivisible instance for WordN Instance detailsDefined in Data.SBV.Core.Sized MethodssQuotRem :: WordN n -> WordN n -> (WordN n, WordN n) Source #sDivMod :: WordN n -> WordN n -> (WordN n, WordN n) Source #sQuot :: WordN n -> WordN n -> WordN n Source #sRem :: WordN n -> WordN n -> WordN n Source #sDiv :: WordN n -> WordN n -> WordN n Source #sMod :: WordN n -> WordN n -> WordN n Source #
# Bit-vector operations
## Conversions
sFromIntegral :: forall a b. (Integral a, HasKind a, Num a, SymVal a, HasKind b, Num b, SymVal b) => SBV a -> SBV b Source #
Conversion between integral-symbolic values, akin to Haskell's fromIntegral
## Shifts and rotates
sShiftLeft :: (SIntegral a, SIntegral b) => SBV a -> SBV b -> SBV a Source #
Generalization of shiftL, when the shift-amount is symbolic. Since Haskell's shiftL only takes an Int as the shift amount, it cannot be used when we have a symbolic amount to shift with.
sShiftRight :: (SIntegral a, SIntegral b) => SBV a -> SBV b -> SBV a Source #
Generalization of shiftR, when the shift-amount is symbolic. Since Haskell's shiftR only takes an Int as the shift amount, it cannot be used when we have a symbolic amount to shift with.
NB. If the shiftee is signed, then this is an arithmetic shift; otherwise it's logical, following the usual Haskell convention. See sSignedShiftArithRight for a variant that explicitly uses the msb as the sign bit, even for unsigned underlying types.
sRotateLeft :: (SIntegral a, SIntegral b) => SBV a -> SBV b -> SBV a Source #
Generalization of rotateL, when the shift-amount is symbolic. Since Haskell's rotateL only takes an Int as the shift amount, it cannot be used when we have a symbolic amount to shift with. The first argument should be a bounded quantity.
sBarrelRotateLeft :: (SFiniteBits a, SFiniteBits b) => SBV a -> SBV b -> SBV a Source #
An implementation of rotate-left, using a barrel shifter like design. Only works when both arguments are finite bitvectors, and furthermore when the second argument is unsigned. The first condition is enforced by the type, but the second is dynamically checked. We provide this implementation as an alternative to sRotateLeft since SMTLib logic does not support variable argument rotates (as opposed to shifts), and thus this implementation can produce better code for verification compared to sRotateLeft.
>>> prove $\x y -> (x sBarrelRotateLeft y) sBarrelRotateRight (y :: SWord32) .== (x :: SWord64) Q.E.D. sRotateRight :: (SIntegral a, SIntegral b) => SBV a -> SBV b -> SBV a Source # Generalization of rotateR, when the shift-amount is symbolic. Since Haskell's rotateR only takes an Int as the shift amount, it cannot be used when we have a symbolic amount to shift with. The first argument should be a bounded quantity. sBarrelRotateRight :: (SFiniteBits a, SFiniteBits b) => SBV a -> SBV b -> SBV a Source # An implementation of rotate-right, using a barrel shifter like design. See comments for sBarrelRotateLeft for details. >>> prove$ \x y -> (x sBarrelRotateRight y) sBarrelRotateLeft (y :: SWord32) .== (x :: SWord64)
Q.E.D.
sSignedShiftArithRight :: (SFiniteBits a, SIntegral b) => SBV a -> SBV b -> SBV a Source #
Arithmetic shift-right with a symbolic unsigned shift amount. This is equivalent to sShiftRight when the argument is signed. However, if the argument is unsigned, then it explicitly treats its msb as a sign-bit, and uses it as the bit that gets shifted in. Useful when using the underlying unsigned bit representation to implement custom signed operations. Note that there is no direct Haskell analogue of this function.
## Finite bit-vector operations
class (Ord a, SymVal a, Num a, Bits a) => SFiniteBits a where Source #
Finite bit-length symbolic values. Essentially the same as SIntegral, but further leaves out Integer. Loosely based on Haskell's FiniteBits class, but with more methods defined and structured differently to fit into the symbolic world view. Minimal complete definition: sFiniteBitSize.
Minimal complete definition
sFiniteBitSize
Methods
Bit size.
lsb :: SBV a -> SBool Source #
Least significant bit of a word, always stored at index 0.
msb :: SBV a -> SBool Source #
Most significant bit of a word, always stored at the last position.
blastBE :: SBV a -> [SBool] Source #
Big-endian blasting of a word into its bits.
blastLE :: SBV a -> [SBool] Source #
Little-endian blasting of a word into its bits.
fromBitsBE :: [SBool] -> SBV a Source #
Reconstruct from given bits, given in little-endian.
fromBitsLE :: [SBool] -> SBV a Source #
Reconstruct from given bits, given in little-endian.
sTestBit :: SBV a -> Int -> SBool Source #
Replacement for testBit, returning SBool instead of Bool.
sExtractBits :: SBV a -> [Int] -> [SBool] Source #
Variant of sTestBit, where we want to extract multiple bit positions.
Variant of popCount, returning a symbolic value.
setBitTo :: SBV a -> Int -> SBool -> SBV a Source #
A combo of setBit and clearBit, when the bit to be set is symbolic.
fullAdder :: SBV a -> SBV a -> (SBool, SBV a) Source #
fullMultiplier :: SBV a -> SBV a -> (SBV a, SBV a) Source #
Full multipler, returns both high and low-order bits. Only for unsigned quantities.
Count leading zeros in a word, big-endian interpretation.
Count trailing zeros in a word, big-endian interpretation.
Instances
Source # Instance detailsDefined in Data.SBV.Core.Model MethodsblastBE :: SBV Int8 -> [SBool] Source #blastLE :: SBV Int8 -> [SBool] Source #fromBitsBE :: [SBool] -> SBV Int8 Source #fromBitsLE :: [SBool] -> SBV Int8 Source #sExtractBits :: SBV Int8 -> [Int] -> [SBool] Source # Source # Instance detailsDefined in Data.SBV.Core.Model MethodsblastBE :: SBV Int16 -> [SBool] Source #blastLE :: SBV Int16 -> [SBool] Source #fromBitsBE :: [SBool] -> SBV Int16 Source #fromBitsLE :: [SBool] -> SBV Int16 Source #sExtractBits :: SBV Int16 -> [Int] -> [SBool] Source # Source # Instance detailsDefined in Data.SBV.Core.Model MethodsblastBE :: SBV Int32 -> [SBool] Source #blastLE :: SBV Int32 -> [SBool] Source #fromBitsBE :: [SBool] -> SBV Int32 Source #fromBitsLE :: [SBool] -> SBV Int32 Source #sExtractBits :: SBV Int32 -> [Int] -> [SBool] Source # Source # Instance detailsDefined in Data.SBV.Core.Model MethodsblastBE :: SBV Int64 -> [SBool] Source #blastLE :: SBV Int64 -> [SBool] Source #fromBitsBE :: [SBool] -> SBV Int64 Source #fromBitsLE :: [SBool] -> SBV Int64 Source #sExtractBits :: SBV Int64 -> [Int] -> [SBool] Source # Source # Instance detailsDefined in Data.SBV.Core.Model MethodsblastBE :: SBV Word8 -> [SBool] Source #blastLE :: SBV Word8 -> [SBool] Source #fromBitsBE :: [SBool] -> SBV Word8 Source #fromBitsLE :: [SBool] -> SBV Word8 Source #sExtractBits :: SBV Word8 -> [Int] -> [SBool] Source # Source # Instance detailsDefined in Data.SBV.Core.Model MethodsblastBE :: SBV Word16 -> [SBool] Source #blastLE :: SBV Word16 -> [SBool] Source #fromBitsBE :: [SBool] -> SBV Word16 Source #fromBitsLE :: [SBool] -> SBV Word16 Source #sExtractBits :: SBV Word16 -> [Int] -> [SBool] Source # Source # Instance detailsDefined in Data.SBV.Core.Model MethodsblastBE :: SBV Word32 -> [SBool] Source #blastLE :: SBV Word32 -> [SBool] Source #fromBitsBE :: [SBool] -> SBV Word32 Source #fromBitsLE :: [SBool] -> SBV Word32 Source #sExtractBits :: SBV Word32 -> [Int] -> [SBool] Source # Source # Instance detailsDefined in Data.SBV.Core.Model MethodsblastBE :: SBV Word64 -> [SBool] Source #blastLE :: SBV Word64 -> [SBool] Source #fromBitsBE :: [SBool] -> SBV Word64 Source #fromBitsLE :: [SBool] -> SBV Word64 Source #sExtractBits :: SBV Word64 -> [Int] -> [SBool] Source # (KnownNat n, IsNonZero n) => SFiniteBits (IntN n) Source # SFiniteBits instance for IntN Instance detailsDefined in Data.SBV.Core.Sized MethodssFiniteBitSize :: SBV (IntN n) -> Int Source #lsb :: SBV (IntN n) -> SBool Source #msb :: SBV (IntN n) -> SBool Source #blastBE :: SBV (IntN n) -> [SBool] Source #blastLE :: SBV (IntN n) -> [SBool] Source #fromBitsBE :: [SBool] -> SBV (IntN n) Source #fromBitsLE :: [SBool] -> SBV (IntN n) Source #sTestBit :: SBV (IntN n) -> Int -> SBool Source #sExtractBits :: SBV (IntN n) -> [Int] -> [SBool] Source #sPopCount :: SBV (IntN n) -> SWord8 Source #setBitTo :: SBV (IntN n) -> Int -> SBool -> SBV (IntN n) Source #fullAdder :: SBV (IntN n) -> SBV (IntN n) -> (SBool, SBV (IntN n)) Source #fullMultiplier :: SBV (IntN n) -> SBV (IntN n) -> (SBV (IntN n), SBV (IntN n)) Source # (KnownNat n, IsNonZero n) => SFiniteBits (WordN n) Source # SFiniteBits instance for WordN Instance detailsDefined in Data.SBV.Core.Sized MethodssFiniteBitSize :: SBV (WordN n) -> Int Source #lsb :: SBV (WordN n) -> SBool Source #msb :: SBV (WordN n) -> SBool Source #blastBE :: SBV (WordN n) -> [SBool] Source #blastLE :: SBV (WordN n) -> [SBool] Source #fromBitsBE :: [SBool] -> SBV (WordN n) Source #fromBitsLE :: [SBool] -> SBV (WordN n) Source #sTestBit :: SBV (WordN n) -> Int -> SBool Source #sExtractBits :: SBV (WordN n) -> [Int] -> [SBool] Source #sPopCount :: SBV (WordN n) -> SWord8 Source #setBitTo :: SBV (WordN n) -> Int -> SBool -> SBV (WordN n) Source #fullAdder :: SBV (WordN n) -> SBV (WordN n) -> (SBool, SBV (WordN n)) Source #fullMultiplier :: SBV (WordN n) -> SBV (WordN n) -> (SBV (WordN n), SBV (WordN n)) Source #
## Splitting, joining, and extending bit-vectors
Arguments
:: (KnownNat n, IsNonZero n, SymVal (bv n), KnownNat i, KnownNat j, (i + 1) <= n, j <= i, IsNonZero ((i - j) + 1)) => proxy i i: Start position, numbered from n-1 to 0 -> proxy j j: End position, numbered from n-1 to 0, j <= i must hold -> SBV (bv n) Input bit vector of size n -> SBV (bv ((i - j) + 1)) Output is of size i - j + 1
Extract a portion of bits to form a smaller bit-vector.
>>> prove $\x -> bvExtract (Proxy @7) (Proxy @3) (x :: SWord 12) .== bvDrop (Proxy @4) (bvTake (Proxy @9) x) Q.E.D. (#) infixr 5 Source # Arguments :: (KnownNat n, IsNonZero n, SymVal (bv n), KnownNat m, IsNonZero m, SymVal (bv m)) => SBV (bv n) First input, of size n, becomes the left side -> SBV (bv m) Second input, of size m, becomes the right side -> SBV (bv (n + m)) Concatenation, of size n+m Join two bitvectors. >>> prove$ \x y -> x .== bvExtract (Proxy @79) (Proxy @71) ((x :: SWord 9) # (y :: SWord 71))
Q.E.D.
Arguments
:: (KnownNat n, IsNonZero n, SymVal (bv n), KnownNat m, IsNonZero m, SymVal (bv m), (n + 1) <= m, SIntegral (bv (m - n)), IsNonZero (m - n)) => SBV (bv n) Input, of size n -> SBV (bv m) Output, of size m. n < m must hold
Zero extend a bit-vector.
>>> prove $\x -> bvExtract (Proxy @20) (Proxy @12) (zeroExtend (x :: SInt 12) :: SInt 21) .== 0 Q.E.D. Arguments :: (KnownNat n, IsNonZero n, SymVal (bv n), KnownNat m, IsNonZero m, SymVal (bv m), (n + 1) <= m, SFiniteBits (bv n), SIntegral (bv (m - n)), IsNonZero (m - n)) => SBV (bv n) Input, of size n -> SBV (bv m) Output, of size m. n < m must hold Sign extend a bit-vector. >>> prove$ \x -> sNot (msb x) .=> bvExtract (Proxy @20) (Proxy @12) (signExtend (x :: SInt 12) :: SInt 21) .== 0
Q.E.D.
>>> prove $\x -> msb x .=> bvExtract (Proxy @20) (Proxy @12) (signExtend (x :: SInt 12) :: SInt 21) .== complement 0 Q.E.D. Arguments :: (KnownNat n, IsNonZero n, KnownNat i, (i + 1) <= n, ((i + m) - n) <= 0, IsNonZero (n - i)) => proxy i i: Number of bits to drop. i < n must hold. -> SBV (bv n) Input, of size n -> SBV (bv m) Output, of size m. m = n - i holds. Drop bits from the top of a bit-vector. >>> prove$ \x -> bvDrop (Proxy @0) (x :: SWord 43) .== x
Q.E.D.
>>> prove $\x -> bvDrop (Proxy @20) (x :: SWord 21) .== ite (lsb x) 1 0 Q.E.D. Arguments :: (KnownNat n, IsNonZero n, KnownNat i, IsNonZero i, i <= n) => proxy i i: Number of bits to take. 0 < i <= n must hold. -> SBV (bv n) Input, of size n -> SBV (bv i) Output, of size i Take bits from the top of a bit-vector. >>> prove$ \x -> bvTake (Proxy @13) (x :: SWord 13) .== x
Q.E.D.
>>> prove $\x -> bvTake (Proxy @1) (x :: SWord 13) .== ite (msb x) 1 0 Q.E.D. >>> prove$ \x -> bvTake (Proxy @4) x # bvDrop (Proxy @4) x .== (x :: SWord 23)
Q.E.D.
## Exponentiation
(.^) :: (Mergeable b, Num b, SIntegral e) => b -> SBV e -> b Source #
Symbolic exponentiation using bit blasting and repeated squaring.
N.B. The exponent must be unsigned/bounded if symbolic. Signed exponents will be rejected.
# IEEE-floating point numbers
class (SymVal a, RealFloat a) => IEEEFloating a where Source #
A class of floating-point (IEEE754) operations, some of which behave differently based on rounding modes. Note that unless the rounding mode is concretely RoundNearestTiesToEven, we will not concretely evaluate these, but rather pass down to the SMT solver.
Minimal complete definition
Nothing
Methods
fpAbs :: SBV a -> SBV a Source #
Compute the floating point absolute value.
fpNeg :: SBV a -> SBV a Source #
Compute the unary negation. Note that 0 - x is not equivalent to -x for floating-point, since -0 and 0 are different.
fpAdd :: SRoundingMode -> SBV a -> SBV a -> SBV a Source #
Add two floating point values, using the given rounding mode
fpSub :: SRoundingMode -> SBV a -> SBV a -> SBV a Source #
Subtract two floating point values, using the given rounding mode
fpMul :: SRoundingMode -> SBV a -> SBV a -> SBV a Source #
Multiply two floating point values, using the given rounding mode
fpDiv :: SRoundingMode -> SBV a -> SBV a -> SBV a Source #
Divide two floating point values, using the given rounding mode
fpFMA :: SRoundingMode -> SBV a -> SBV a -> SBV a -> SBV a Source #
Fused-multiply-add three floating point values, using the given rounding mode. fpFMA x y z = x*y+z but with only one rounding done for the whole operation; not two. Note that we will never concretely evaluate this function since Haskell lacks an FMA implementation.
fpSqrt :: SRoundingMode -> SBV a -> SBV a Source #
Compute the square-root of a float, using the given rounding mode
fpRem :: SBV a -> SBV a -> SBV a Source #
Compute the remainder: x - y * n, where n is the truncated integer nearest to x/y. The rounding mode is implicitly assumed to be RoundNearestTiesToEven.
Round to the nearest integral value, using the given rounding mode.
fpMin :: SBV a -> SBV a -> SBV a Source #
Compute the minimum of two floats, respects infinity and NaN values
fpMax :: SBV a -> SBV a -> SBV a Source #
Compute the maximum of two floats, respects infinity and NaN values
fpIsEqualObject :: SBV a -> SBV a -> SBool Source #
Are the two given floats exactly the same. That is, NaN will compare equal to itself, +0 will not compare equal to -0 etc. This is the object level equality, as opposed to the semantic equality. (For the latter, just use .==.)
Is the floating-point number a normal value. (i.e., not denormalized.)
Is the floating-point number a subnormal value. (Also known as denormal.)
fpIsZero :: SBV a -> SBool Source #
Is the floating-point number 0? (Note that both +0 and -0 will satisfy this predicate.)
Is the floating-point number infinity? (Note that both +oo and -oo will satisfy this predicate.)
fpIsNaN :: SBV a -> SBool Source #
Is the floating-point number a NaN value?
Is the floating-point number negative? Note that -0 satisfies this predicate but +0 does not.
Is the floating-point number positive? Note that +0 satisfies this predicate but -0 does not.
Is the floating point number -0?
Is the floating point number +0?
fpIsPoint :: SBV a -> SBool Source #
Is the floating-point number a regular floating point, i.e., not NaN, nor +oo, nor -oo. Normals or denormals are allowed.
Instances
Source # SDouble instance Instance detailsDefined in Data.SBV.Core.Floating Methods Source # SFloat instance Instance detailsDefined in Data.SBV.Core.Floating Methods
Rounding mode to be used for the IEEE floating-point operations. Note that Haskell's default is RoundNearestTiesToEven. If you use a different rounding mode, then the counter-examples you get may not match what you observe in Haskell.
Constructors
RoundNearestTiesToEven Round to nearest representable floating point value. If precisely at half-way, pick the even number. (In this context, even means the lowest-order bit is zero.) RoundNearestTiesToAway Round to nearest representable floating point value. If precisely at half-way, pick the number further away from 0. (That is, for positive values, pick the greater; for negative values, pick the smaller.) RoundTowardPositive Round towards positive infinity. (Also known as rounding-up or ceiling.) RoundTowardNegative Round towards negative infinity. (Also known as rounding-down or floor.) RoundTowardZero Round towards zero. (Also known as truncation.)
Instances
Source # Instance detailsDefined in Data.SBV.Core.Symbolic Methods Source # Instance detailsDefined in Data.SBV.Core.Symbolic Methods Source # Instance detailsDefined in Data.SBV.Core.Symbolic Methods Source # Instance detailsDefined in Data.SBV.Core.Symbolic Methodsgfoldl :: (forall d b. Data d => c (d -> b) -> d -> c b) -> (forall g. g -> c g) -> RoundingMode -> c RoundingMode #gunfold :: (forall b r. Data b => c (b -> r) -> c r) -> (forall r. r -> c r) -> Constr -> c RoundingMode #dataCast1 :: Typeable t => (forall d. Data d => c (t d)) -> Maybe (c RoundingMode) #dataCast2 :: Typeable t => (forall d e. (Data d, Data e) => c (t d e)) -> Maybe (c RoundingMode) #gmapT :: (forall b. Data b => b -> b) -> RoundingMode -> RoundingMode #gmapQl :: (r -> r' -> r) -> r -> (forall d. Data d => d -> r') -> RoundingMode -> r #gmapQr :: (r' -> r -> r) -> r -> (forall d. Data d => d -> r') -> RoundingMode -> r #gmapQ :: (forall d. Data d => d -> u) -> RoundingMode -> [u] #gmapQi :: Int -> (forall d. Data d => d -> u) -> RoundingMode -> u #gmapM :: Monad m => (forall d. Data d => d -> m d) -> RoundingMode -> m RoundingMode #gmapMp :: MonadPlus m => (forall d. Data d => d -> m d) -> RoundingMode -> m RoundingMode #gmapMo :: MonadPlus m => (forall d. Data d => d -> m d) -> RoundingMode -> m RoundingMode # Source # Instance detailsDefined in Data.SBV.Core.Symbolic Methods Source # Instance detailsDefined in Data.SBV.Core.Symbolic Methods Source # Instance detailsDefined in Data.SBV.Core.Symbolic MethodsshowList :: [RoundingMode] -> ShowS # Source # RoundingMode kind Instance detailsDefined in Data.SBV.Core.Symbolic Methods Source # RoundingMode can be used symbolically Instance detailsDefined in Data.SBV.Core.Data Methodsforall :: MonadSymbolic m => String -> m (SBV RoundingMode) Source #forall_ :: MonadSymbolic m => m (SBV RoundingMode) Source #mkForallVars :: MonadSymbolic m => Int -> m [SBV RoundingMode] Source #exists :: MonadSymbolic m => String -> m (SBV RoundingMode) Source #exists_ :: MonadSymbolic m => m (SBV RoundingMode) Source #mkExistVars :: MonadSymbolic m => Int -> m [SBV RoundingMode] Source #free :: MonadSymbolic m => String -> m (SBV RoundingMode) Source #free_ :: MonadSymbolic m => m (SBV RoundingMode) Source #mkFreeVars :: MonadSymbolic m => Int -> m [SBV RoundingMode] Source #symbolic :: MonadSymbolic m => String -> m (SBV RoundingMode) Source #symbolics :: MonadSymbolic m => [String] -> m [SBV RoundingMode] Source # Source # A rounding mode, extracted from a model. (Default definition suffices) Instance detailsDefined in Data.SBV.SMT.SMT MethodsparseCVs :: [CV] -> Maybe (RoundingMode, [CV]) Source #cvtModel :: (RoundingMode -> Maybe b) -> Maybe (RoundingMode, [CV]) -> Maybe (b, [CV]) Source #
The symbolic variant of RoundingMode
nan :: Floating a => a Source #
Not-A-Number for Double and Float. Surprisingly, Haskell Prelude doesn't have this value defined, so we provide it here.
infinity :: Floating a => a Source #
Infinity for Double and Float. Surprisingly, Haskell Prelude doesn't have this value defined, so we provide it here.
sNaN :: (Floating a, SymVal a) => SBV a Source #
Symbolic variant of Not-A-Number. This value will inhabit both SDouble and SFloat.
sInfinity :: (Floating a, SymVal a) => SBV a Source #
Symbolic variant of infinity. This value will inhabit both SDouble and SFloat.
## Rounding modes
Symbolic variant of RoundNearestTiesToEven
Symbolic variant of RoundNearestTiesToAway
Symbolic variant of RoundTowardPositive
Symbolic variant of RoundTowardNegative
Symbolic variant of RoundTowardZero
Alias for sRoundNearestTiesToEven
Alias for sRoundNearestTiesToAway
Alias for sRoundTowardPositive
Alias for sRoundTowardNegative
Alias for sRoundTowardZero
## Conversion to/from floats
class SymVal a => IEEEFloatConvertible a where Source #
Capture convertability from/to FloatingPoint representations.
Conversions to float: toSFloat and toSDouble simply return the nearest representable float from the given type based on the rounding mode provided.
Conversions from float: fromSFloat and fromSDouble functions do the reverse conversion. However some care is needed when given values that are not representable in the integral target domain. For instance, converting an SFloat to an SInt8 is problematic. The rules are as follows:
If the input value is a finite point and when rounded in the given rounding mode to an integral value lies within the target bounds, then that result is returned. (This is the regular interpretation of rounding in IEEE754.)
Otherwise (i.e., if the integral value in the float or double domain) doesn't fit into the target type, then the result is unspecified. Note that if the input is +oo, -oo, or NaN, then the result is unspecified.
Due to the unspecified nature of conversions, SBV will never constant fold conversions from floats to integral values. That is, you will always get a symbolic value as output. (Conversions from floats to other floats will be constant folded. Conversions from integral values to floats will also be constant folded.)
Note that unspecified really means unspecified: In particular, SBV makes no guarantees about matching the behavior between what you might get in Haskell, via SMT-Lib, or the C-translation. If the input value is out-of-bounds as defined above, or is NaN or oo or -oo, then all bets are off. In particular C and SMTLib are decidedly undefine this case, though that doesn't mean they do the same thing! Same goes for Haskell, which seems to convert via Int64, but we do not model that behavior in SBV as it doesn't seem to be intentional nor well documented.
You can check for NaN, oo and -oo, using the predicates fpIsNaN, fpIsInfinite, and fpIsPositive, fpIsNegative predicates, respectively; and do the proper conversion based on your needs. (0 is a good choice, as are min/max bounds of the target type.)
Currently, SBV provides no predicates to check if a value would lie within range for a particular conversion task, as this depends on the rounding mode and the types involved and can be rather tricky to determine. (See http://github.com/LeventErkok/sbv/issues/456 for a discussion of the issues involved.) In a future release, we hope to be able to provide underflow and overflow predicates for these conversions as well.
Minimal complete definition
Nothing
Methods
Convert from an IEEE74 single precision float.
Convert to an IEEE-754 Single-precision float.
>>> :{
roundTrip :: forall a. (Eq a, IEEEFloatConvertible a) => SRoundingMode -> SBV a -> SBool
roundTrip m x = fromSFloat m (toSFloat m x) .== x
:}
>>> prove $roundTrip @Int8 Q.E.D. >>> prove$ roundTrip @Word8
Q.E.D.
>>> prove $roundTrip @Int16 Q.E.D. >>> prove$ roundTrip @Word16
Q.E.D.
>>> prove $roundTrip @Int32 Falsifiable. Counter-example: s0 = RoundNearestTiesToEven :: RoundingMode s1 = 134280664 :: Int32 Note how we get a failure on Int32. The counter-example value is not representable exactly as a single precision float: >>> toRational (134280664 :: Float) 134280672 % 1 Note how the numerator is different, it is off by 8. This is hardly surprising, since floats become sparser as the magnitude increases to be able to cover all the integer values representable. toSFloat :: Integral a => SRoundingMode -> SBV a -> SFloat Source # Convert to an IEEE-754 Single-precision float. >>> :{ roundTrip :: forall a. (Eq a, IEEEFloatConvertible a) => SRoundingMode -> SBV a -> SBool roundTrip m x = fromSFloat m (toSFloat m x) .== x :} >>> prove$ roundTrip @Int8
Q.E.D.
>>> prove $roundTrip @Word8 Q.E.D. >>> prove$ roundTrip @Int16
Q.E.D.
>>> prove $roundTrip @Word16 Q.E.D. >>> prove$ roundTrip @Int32
Falsifiable. Counter-example:
s0 = RoundNearestTiesToEven :: RoundingMode
s1 = 134280664 :: Int32
Note how we get a failure on Int32. The counter-example value is not representable exactly as a single precision float:
>>> toRational (134280664 :: Float)
134280672 % 1
Note how the numerator is different, it is off by 8. This is hardly surprising, since floats become sparser as the magnitude increases to be able to cover all the integer values representable.
Convert from an IEEE74 double precision float.
Convert to an IEEE-754 Double-precision float.
>>> :{
roundTrip :: forall a. (Eq a, IEEEFloatConvertible a) => SRoundingMode -> SBV a -> SBool
roundTrip m x = fromSDouble m (toSDouble m x) .== x
:}
>>> prove $roundTrip @Int8 Q.E.D. >>> prove$ roundTrip @Word8
Q.E.D.
>>> prove $roundTrip @Int16 Q.E.D. >>> prove$ roundTrip @Word16
Q.E.D.
>>> prove $roundTrip @Int32 Q.E.D. >>> prove$ roundTrip @Word32
Q.E.D.
>>> prove $roundTrip @Int64 Falsifiable. Counter-example: s0 = RoundNearestTiesToEven :: RoundingMode s1 = 8005056270191255168 :: Int64 Just like in the SFloat case, once we reach 64-bits, we no longer can exactly represent the integer value for all possible values: >>> toRational (8005056270191255168 :: Double) 8005056270191255552 % 1 In this case the numerator is off by 384! toSDouble :: Integral a => SRoundingMode -> SBV a -> SDouble Source # Convert to an IEEE-754 Double-precision float. >>> :{ roundTrip :: forall a. (Eq a, IEEEFloatConvertible a) => SRoundingMode -> SBV a -> SBool roundTrip m x = fromSDouble m (toSDouble m x) .== x :} >>> prove$ roundTrip @Int8
Q.E.D.
>>> prove $roundTrip @Word8 Q.E.D. >>> prove$ roundTrip @Int16
Q.E.D.
>>> prove $roundTrip @Word16 Q.E.D. >>> prove$ roundTrip @Int32
Q.E.D.
>>> prove $roundTrip @Word32 Q.E.D. >>> prove$ roundTrip @Int64
Falsifiable. Counter-example:
s0 = RoundNearestTiesToEven :: RoundingMode
s1 = 8005056270191255168 :: Int64
Just like in the SFloat case, once we reach 64-bits, we no longer can exactly represent the integer value for all possible values:
>>> toRational (8005056270191255168 :: Double)
8005056270191255552 % 1
In this case the numerator is off by 384!
Instances
Source # Instance detailsDefined in Data.SBV.Core.Floating Methods Source # Instance detailsDefined in Data.SBV.Core.Floating Methods Source # Instance detailsDefined in Data.SBV.Core.Floating Methods Source # Instance detailsDefined in Data.SBV.Core.Floating Methods Source # Instance detailsDefined in Data.SBV.Core.Floating Methods Source # Instance detailsDefined in Data.SBV.Core.Floating Methods Source # Instance detailsDefined in Data.SBV.Core.Floating Methods Source # Instance detailsDefined in Data.SBV.Core.Floating Methods Source # Instance detailsDefined in Data.SBV.Core.Floating Methods Source # Instance detailsDefined in Data.SBV.Core.Floating Methods Source # Instance detailsDefined in Data.SBV.Core.Floating Methods Source # Instance detailsDefined in Data.SBV.Core.Floating Methods
## Bit-pattern conversions
Convert an SFloat to an SWord32, preserving the bit-correspondence. Note that since the representation for NaNs are not unique, this function will return a symbolic value when given a concrete NaN.
Implementation note: Since there's no corresponding function in SMTLib for conversion to bit-representation due to partiality, we use a translation trick by allocating a new word variable, converting it to float, and requiring it to be equivalent to the input. In code-generation mode, we simply map it to a simple conversion.
Reinterpret the bits in a 32-bit word as a single-precision floating point number
Convert an SDouble to an SWord64, preserving the bit-correspondence. Note that since the representation for NaNs are not unique, this function will return a symbolic value when given a concrete NaN.
See the implementation note for sFloatAsSWord32, as it applies here as well.
Reinterpret the bits in a 32-bit word as a single-precision floating point number
blastSFloat :: SFloat -> (SBool, [SBool], [SBool]) Source #
Extract the sign/exponent/mantissa of a single-precision float. The output will have 8 bits in the second argument for exponent, and 23 in the third for the mantissa.
blastSDouble :: SDouble -> (SBool, [SBool], [SBool]) Source #
Extract the sign/exponent/mantissa of a single-precision float. The output will have 11 bits in the second argument for exponent, and 52 in the third for the mantissa.
# Enumerations
Make an enumeration a symbolic type.
# Uninterpreted sorts, constants, and functions
class Uninterpreted a where Source #
Uninterpreted constants and functions. An uninterpreted constant is a value that is indexed by its name. The only property the prover assumes about these values are that they are equivalent to themselves; i.e., (for functions) they return the same results when applied to same arguments. We support uninterpreted-functions as a general means of black-box'ing operations that are irrelevant for the purposes of the proof; i.e., when the proofs can be performed without any knowledge about the function itself.
Minimal complete definition: sbvUninterpret. However, most instances in practice are already provided by SBV, so end-users should not need to define their own instances.
Minimal complete definition
sbvUninterpret
Methods
uninterpret :: String -> a Source # | 29,888 | 98,476 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2020-10 | latest | en | 0.678905 |
http://fer3.com/arc/m2.aspx/GMT-from-Lunar-Photo-Miseta-jan-2022-g51859 | 1,652,937,614,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662525507.54/warc/CC-MAIN-20220519042059-20220519072059-00194.warc.gz | 23,790,327 | 5,075 | # NavList:
## A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding
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Re: GMT from Lunar Photo
From: Tibor Miseta
Date: 2022 Jan 6, 12:49 -0800
I've tried to estimate by the illuminated disk. I measured the diameter of the Moon (between the ends of the horn) 732 pixs, and the shadowed disk from the terminator to the dark limb 672 pixels. The illuminated size is than: (732-672)/672 = 8.2%
A rough estimate from the Air Almanac: the illuminated disk for 01.04 18:00Z is approx 6% and for 01.05 06:00Z is approx 9%, so 12 hours change is 3% (almost linear in the neighbouring halfdays), a quick interpolation gives 01.05 02:48Z. But this is very inaccurate, because the almanac data was given as whole integers only.
A better estimation if we try to calculate the illuminated disk from GHA differences. By one of Meeus' approximate formula the illuminated fraction is k = (1-cos i)/2, where cos i is approx - cos(delta Lat) * cos (delta Lon), so an estimate is k = (1 - cos(delta GHA))/2. (I neglected the latitude differences, because they are so close, that the cosine of small angles are nearly 1.) So I got the following results:
01.04 18:00Z = 6.7%
01.05 06:00Z = 10.0%
A quick interpolation for 8.2% is 7:38 from 01.04 18:00Z, that is 01.05 01:38Z
The watch correction is approx +25 minutes (1:38 - 1:13)
This is still just an estimate of course, more accurate measurement and calculation could be performed.
Best regars:
Tibor
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Subject: Author: Start date: (yyyymm dd) End date: (yyyymm dd) | 565 | 2,087 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2022-21 | longest | en | 0.888154 |
https://mcqlearn.com/math/g9/proportions-in-math-mcqs.php | 1,708,488,623,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473370.18/warc/CC-MAIN-20240221034447-20240221064447-00620.warc.gz | 398,972,787 | 23,928 | Class 9 Online Courses
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Proportions in Math MCQ Quiz : Math MCQs
MCQ 1: If two quantities are related in such a way that increase in 1 quantity causes increase in other quantity, then this variation is said to be
A) joint proportion
B) extreme proportion
C) direct proportion
D) inverse proportion
MCQ 2: If 2 ratios a:b and c:d are equal then we can write it as
A) a : b ⁄ c : d
B) a : b = c : d
C) a + b = c + d
D) a : c = d : b
MCQ 3: A statement which is expressed as an equivalence of two ratios is known as
A) proportion
B) variation
C) ratio
D) probability
MCQ 4: 14 buffaloes consume 63kg hay in 18 days. The number of buffaloes to eat 770kg of hay in 28 days should be
A) 120 cows
B) 100 cows
C) 115 cows
D) 110 cows
MCQ 5: If two quantities are related in such a way that when 1 quantity increases, the other quantity decreases, then this variation is said to be
A) extreme proportion
B) joint proportion
C) direct proportion
D) inverse proportion
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7th Grade Math App (Android & iOS) | 640 | 2,561 | {"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-2024-10 | longest | en | 0.841277 |
https://www.keyword-suggest-tool.com/search/labor+mix+variance+formula/ | 1,610,990,526,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703515075.32/warc/CC-MAIN-20210118154332-20210118184332-00418.warc.gz | 864,488,151 | 8,937 | # Labor mix variance formula
Labor mix variance formula keyword after analyzing the system lists the list of keywords related and the list of websites with related content, in addition you can see which keywords most interested customers on the this website
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Labor mix variance formula
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### Labour/Labor - Mix/Gang-Composition Variance
Variance of Mix and Total Variance are the same. Variance Mix provides a method to find the total variance through calculations instead of by just adding up individual variances. Where there is only one labour/labor type being used, there is no meaning in thinking of the Labour/Labor Mix Variance. TLMV/TGCV = 0 as well as LMV/GCV Lab = 0 in ...
https://www.futureaccountant.com/standard-costing-variance-analysis/study-notes/labour-labor-mix-gang-composition-variance.php
DA: 24 PA: 50 MOZ Rank: 50
### Direct labor mix variance - Definition and Formula - Play ...
Variance used to analyze effect of changed in the mix of labor on erect labor cost. Computed as: [(Actual labor mix Standard labor mix percentage – Standard labor mix percentage) x (Actual total units of labor inputs used)] x (Standard individual price per unit of labor input – Standard average price per unit of labor input).
https://www.playaccounting.com/accounting-terms/d/direct-labor-mix-variance/
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### Labour Mix Variance | Accounting Assignment Help ...
Labour Mix Variance is one of the subject in which we provide homework and assignment help. Whether your problem is related to Managerial, Cost, Activity based or financial accounting, We provide a systematic way of looking at events, collecting data, analyzing information, and reporting the results. We have 24 / 7 live online tutors available to help you.
https://www.assignmenthelp.net/assignment_help/labour-mix-variance
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### Labour Mix | Revised Efficiency | Variance TutorsOnNet
Labour Mix Variance: Exactly analogous to material mix variance is the labour mix variance. It arises when due to shortage of a particular labour grade or some other particular reason; there is a difference between the composition of actual gang of labour & the composition of standard gang.
https://www.tutorsonnet.com/labour-mix-revised-efficiency-variance-homework-help.php
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### What is the Labor Efficiency Variance Formula? – Basic ...
Direct labor variance is a means to mathematically compare expected labor costs to actual labor costs. More specifically, the formula looks at the direct labor hours invested into an outcome (such as the number of units produced) and relates them to the projected labor hours. The formula assumes standard rates for both.
https://basicaccountinghelp.com/what-is-the-labor-efficiency-variance-formula/
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### Labor Mix Variance - YouTube
This video tutorial is the part of teaching of cost accounting topics. In this video, you will learn to calculate labor mix variance of two different company...
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### How to Calculate Direct Labor Variances - dummies
With these numbers in hand, you can apply the formula to compute the direct labor price variance: Direct labor price variance = (SR – AR) x AH = (\$12.00 – \$13.00) x 3,600 = –\$1.00 x 3,600 = –\$3,600 unfavorable According to the direct labor price variance, the increase in average wages from \$12 to \$13 cause costs to increase by \$3,600.
DA: 15 PA: 50 MOZ Rank: 87
### Labour/Labor Yield/Sub-Efficiency Variance
Any variation on account of varying the times of individual labour/labor types is revealed by the Labour/Labor Mix/Gang-Composition Variance. This can be identified from the fact that the calculation of the variance for individual labour/labor types is the equivalent of dividing the variance for the mix among the labour/labor types in the ...
https://www.futureaccountant.com/standard-costing-variance-analysis/study-notes/labour-labor-yield-variance.php
DA: 24 PA: 50 MOZ Rank: 50
### Mix and Yield Variances for Material and Labor - Budgeting ...
8.10. Mix and Yield Variances for Material and Labor. Mix refers to the relative proportion of various ingredients of input factors such as materials and labor. Yield is a measure of productivity.. Material and Labor Mix Variances. The material mix variance indicates the impact on material costs of the deviation from the standard mix.
https://www.oreilly.com/library/view/budgeting-basics-and/9780470389683/9780470389683_mix_and_yield_variances_for_material_and.html
DA: 15 PA: 50 MOZ Rank: 50 | 1,094 | 4,790 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2021-04 | latest | en | 0.841266 |
https://www.goconqr.com/mapamental/147792/estructural-equation-modeling-sem | 1,701,283,866,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100135.11/warc/CC-MAIN-20231129173017-20231129203017-00599.warc.gz | 890,201,482 | 11,546 | # Estructural Equation Modeling (SEM)
### Description
Mind Map on Estructural Equation Modeling (SEM), created by bamezcuan on 07/13/2013.
Mind Map by bamezcuan, updated more than 1 year ago
Created by bamezcuan over 10 years ago
99
1
## Resource summary
Estructural Equation Modeling (SEM)
1. Basic concepts
1. A tool for analyzing multivariate data that has been especially appropriate for theory testing (Bagozzi, 1980; Steenkamp and Baumgartner, 2000).
1. * Latent variables can not be observed directly (Chin, Peterson & Brown, 2008) * Path Analysis: is a special case of SEM that only involves observed (manifest) variables (Savalei & Bentler, 2010).
1. SEMs go beyond ordinary regression models to incorporate multiple independent and dependent variables as well as hypothetical latent constructs which may cluster several observed variables (Savalei & Bentler, 2010).
2. Related studies
1. SEM have become ubiquitous in all the social and behavioral sciences (e.g., MacCallum & Austin, 2000).
1. Anderson and Gerbin (1988) conducted a review and recommended a 2 step approach: 1) A confirmatory measurement, or factor analysis, model specifies the relations of the observed measures to their posited underlying constructs, with the constructs allowed to intercorrelate freely. 2) A confirmatory structural model then specifies the causal relations of the constructs to one another.
1. Baumgartner and Homburg (1996) conducted a review of use of SEM in marketing research.
2. Modeling Process (Savalei & Bentler, 2010)
1. 1) Model specification: Specify variables and constructs
1. 2) Model estimation: Iterative process to stimate parameters
1. 3) Model evaluation: There are two components to model fit: statistical fit and practical fit. Statistical fit is evaluated via a formal test of the hypothesis ( T). Practical fit is evaluated by examining various indices of fit: standardized root mean-square residual (SRMR) <.05. Goodness of Fit index (GFI) related to R2 in ordinary regression. AGFI (Adjusted GFI), is analogous to the adjusted R2 . Both GFI and AGFI take on values between 0 and 1, with values less than 0.90 often considered unacceptable. Comparative Fit Index (CFI) with values between 0 and 1. Preferable close to 0.
1. 4) Model modification : if the proposed model dont fit the data
1. 5) Parameter Testing: To interpret parameter estimates and to test for their statistical significance. Several tests used, in particular z-tests, Wald tests, and chi-square difference tests.
2. Rules of Thumb
1. Bentler & Bonett, 1980: at least three items per construct.
1. Bentler & Chau, 1987: Sample size of 5 to 10 subjects per item and up to 300
1. Baumgartner, 1996: Minimum 3 or 4 indicators per latent variable, large sample sizes 5:1, assumption of normality, model should be identified
1. Fornell & Larcker (1981): Standardized loading estimates should be .5 or higher, ideally .7 or higher. Average of variance Extracted (AVE) should be .5 or higher. Construct Reliability (CR) should be .7 or higher, between .6 and .7 is acceptable
2. The relationship between a manifest variable and a construct is expressed as being either formative or reflective. If the relationship is formative, the manifest variable (indicator) cause the construct, whereas if the relationship is reflective, the construct cause the manifest variable (Jarvis, Mckenzie, & podsakoff, 2003). Most structural equation models specify the manifest variable–construct relationship as reflective (Chin, Peterson & Brown, 2008)
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untitled 2 | 927 | 3,799 | {"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-2023-50 | latest | en | 0.8357 |
https://howtosearch.com/examine/how-much-does-1-yard-of-concrete-weigh-59221/ | 1,695,837,466,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510319.87/warc/CC-MAIN-20230927171156-20230927201156-00380.warc.gz | 342,994,894 | 11,893 | ## Is concrete heavier when set?
The difference in weight between the original weight of the fresh bag of cement and the final hardened bag is a measure of the amount of water which has reacted with the cement to cause it to set. For this reason, a hardened bag of cement will always be considerably heavier than a fresh bag.
## How many pounds of concrete is in a yard?
This standard concrete mix (used for most premix applications) weighs about 133 pounds per cubic foot of concrete, so a cubic yard (which contains 27 cubic feet) will weigh in at about 3,600 pounds.
## How much does concrete weigh after drying?
Standard reinforced concrete is 150 pounds per cubic foot when dry, but there are “light” versions that may be under 100 pounds per cubic foot, or as heavy as 300 pounds per cubic foot.
## How much does a 5 gallon bucket of hardened concrete weigh?
Regarding this, “how much does a 5 gallon bucket of concrete weight?”, generally used for fair estimate, on average, a 5 gallon bucket of concrete weighs around 100 pounds or 0.05 tons, which can hold volume about 0.66 cubic feet or 0.0247 cubic yards.
## How much does a 2 gallon bucket of concrete weigh?
Equals: 20.08 pounds (lb) in mass. Converting gallon to pounds value in the concrete units scale.
## Is water or concrete heavier?
The density of concrete varies with its exact composition, but averages around 2,400 kg per cubic meter, or 150lbs per cubic foot. … In fresh water, the apparent weight would be a little more, 87.6 pounds, because a cubic foot of fresh water weighs only 62.4 pounds so provides less upward force.
## Is concrete heavier when wet or dry?
In short, dry concrete and wet concrete mix each weigh about the same; roughly 2 tons per cubic yard.
## How many yards of concrete do I need for a 24×24 slab?
7.11 yards
For example, for a concrete slab that is 24′ X 24′ X 4”, simply enter 4 in the Thickness/Depth field, 24 in the Width field, and 24 in the Length field. Click “Calculate”. Your answer should be 7.11 yards. Note: The Concrete Volume Calculator can also be used to determine yardage for aggregate products.
## Is sand heavier than concrete?
Is concrete or sand heavier? The reference substance for liquids is nearly always water at its densest! The specific gravity of cement is around 3.15 while sand is 2.65 to 2.67! Scientifically, Cement is heavier than Sand!
## Whats heavier sand or concrete?
Since the specific gravity of diesel is less than water it floats on top of it. As specific gravity of sand is 2.6 – 2.7 and that of cement is 3.14 – 3.15, i.e. for the same volume occupied by cement and sand, cement is “3.15/2.7 = 1.16 times “heavier than sand.
## What is the heaviest concrete?
Coming in for a second round is the Three Gorges Dam from Hubei province, China, holding the title of the world’s heaviest concrete structure, weighing in at 144,309,356,753.51 pounds… of concrete that is!
## Which is heavier a gallon of milk or water?
A gallon is a measurement of volume and density is directly proportional to the mass of a fixed volume. Milk is about 87% water and contains other substances that are heavier than water, excluding fat. A gallon of milk is heavier than a gallon of water.
## How can you make concrete heavier?
Obviously, concrete can be made more dense by infusing it with material of greater density than the concrete itself. You could use iron ore instead of limestone aggregate. You could use uranium-bearing sand. You could dump lumps of any extremely dense material into the mix.
## Which is heavier concrete or lead?
Sea Sprite 23 #110 (20) concrete also has air trapped in it. So it is naturally lighter than lead which is very dense.
## What is the heaviest liquid on Earth?
What is the heaviest liquid substance on earth? Mercury is the densest liquid at standard conditions for temperature and pressure (STP). Also called quicksilver, mercury has been known for more than 3,500 years. It is an important metal in industry, but it is also toxic.
## Does gasoline weigh more than milk?
A gallon of water is almost eight and a half pounds, and milk is a few ounces more. A liter of gasoline is more than 1.03 liters of milk. …
## How heavy is a gallon of gas?
six pounds
According to the Science and Technology Desk Reference, the weight of a gallon of common fuel (such as gasoline) is six pounds. A gallon of water, on the other hand, weighs about 8.4 pounds.
## What is the lightest liquid on Earth?
The order of the liquids from heaviest to lightest will be syrup, glycerin, water, oil, and then alcohol will be on top.
## How dense is honey?
between 1.38 and 1.45 kg/l
The density of honey typically ranges between 1.38 and 1.45 kg/l at 20 °C.
## How much does a gallon of mercury weigh?
One gallon of mercury converted to pound equals to 112.95 lb. How many pounds of mercury are in 1 gallon? The answer is: The change of 1 gal ( gallon ) unit of a mercury amount equals = to 112.95 lb ( pound ) as the equivalent measure for the same mercury type.
## Which gas is very light?
Hydrogen is known to be the first element in the periodic table of elements. It has one proton in its nucleus and one outter electron. It is a very light gas and also flammable. Hydrogen, H, is the lightest of all gases and the most abundant element in the universe. | 1,267 | 5,337 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2023-40 | latest | en | 0.914804 |
https://ipv6.snipplr.com/view/92590/graphic-design--bresenhams-line-algorithm | 1,643,428,565,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320299927.25/warc/CC-MAIN-20220129032406-20220129062406-00387.warc.gz | 373,917,177 | 7,701 | Posted By
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Graphic Design | Bresenham's Line Algorithm
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This is an implementation of the Bresenham's Line Algorithm in Computer Graphics Design in the C programming language.
Copy this code and paste it in your HTML
`//Shubham Mehta, Write a program to implement Bresenham's Line Algorithm.#include<stdio.h>#include<conio.h>#include<graphics.h>void main(){ int gd, gm; float x1, x2, y1, y2, dx, dy, p, xk, yk, slope; clrscr(); printf("%s", "Please enter x1: "); scanf("%f", &x1); printf("%s", "Please enter y1: "); scanf("%f", &y1); printf("%s", "Please enter x2: "); scanf("%f", &x2); printf("%s", "Please enter y2: "); scanf("%f", &y2); gd=DETECT, gm; initgraph(&gd, &gm, "C:\\TC\\BGI"); putpixel(x1, y1, RED); slope=(y2-y1)/(x2-x1); dx=x2-x1; dy=y2-y1; xk=x1; yk=y1; if(slope<1) { p=(2*dy)-dx; for( ; x1<=x2; x1++) { if(p<0) { putpixel(++xk, yk, RED); p+=(2*dy); } else { putpixel(++xk, ++yk, RED); p=p+(2*dy)-(2*dx); } } } else { p=(2*dx)-dy; for( ; y1<=y2; y1++) { if(p<0) { putpixel(xk, ++yk, RED); p+=(2*dx); } else { putpixel(++xk, ++yk, RED); p=p+(2*dx)-(2*dy); } } } getch(); closegraph();}` | 508 | 1,273 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2022-05 | longest | en | 0.483599 |
https://liu.se/studieinfo/en/kurs/tadi31/ht-2018 | 1,638,440,371,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964361253.38/warc/CC-MAIN-20211202084644-20211202114644-00035.warc.gz | 426,180,486 | 8,338 | Diskret matematik, 6 hp
Main field of study
Mathematics Applied Mathematics
First cycle
Programme course
Daniel Carlsson
Director of studies or equivalent
Jesper Thorén
Course offered for Semester Period Timetable module Language Campus VOF
6IDAT Computer Engineering, B Sc in Engineering 3 (Autumn 2018) 2 1+3 Swedish Linköping o
6KKEM Chemistry, Bachelor´s Programme 5 (Autumn 2018) 2 1+3 Swedish Linköping v
Main field of study
Mathematics, Applied Mathematics
First cycle
G1X
Course offered for
• Computer Engineering, B Sc in Engineering
• Chemistry, Bachelor´s Programme
Entry requirements
Note: Admission requirements for non-programme students usually also include admission requirements for the programme and threshold requirements for progression within the programme, or corresponding.
Intended learning outcomes
To give the basic knowledge of discrete mathematics that is needed for further courses in mathematics, natural and computer science. After completing the course the student should be able to
• use the Euclidean algorithm to solve Diophantine equations
• use the principle of mathematical induction to solve recursive problems
• understand and use the terminology and laws of set theory
• formulate and solve combinatorial problems on combinations and permutations
• master the foundations of graph theory and use graphs as a tool to model real-life problems
• use the language of propositional logic, be familiar to logic operations and be able to evaluate the validity of logical conclusions.
Course content
Number theory; prime numbers, divisibility, Euclidean algorithm, Diophantine equations,
Mathematical induction and recursion.
Set theory, the laws of set theory and Venn diagrams.
Combinatorics with permutations and combinations.
Graphs: Euler paths, Hamilton cycles, trees and some applications in computer science
Logic; propositional logic, logic operations, truth tables and conclusions.
Teaching and working methods
Teaching is done through lectures and problem sessions
Examination
UPG1 Hand-in-assignment U, G 2 credits TEN1 A written examination U, 3, 4, 5 4 credits
Four-grade scale, LiU, U, 3, 4, 5
Department
Matematiska institutionen
Jesper Thorén
Examiner
Daniel Carlsson
http://courses.mai.liu.se/Lists/html/index-amne-tm.html
Education components
Preliminary scheduled hours: 50 h
Recommended self-study hours: 110 h
Course literature
Compendiums
Asratian, A., Björn A. och Turesson, B. O., Diskret matematik
Compendia
Asratian, A., Björn A. och Turesson, B. O., Diskret matematik | 589 | 2,566 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2021-49 | latest | en | 0.839821 |
https://esa.github.io/pygmo2/tutorials/coding_udp_simple.html | 1,709,505,389,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476399.55/warc/CC-MAIN-20240303210414-20240304000414-00582.warc.gz | 240,878,055 | 10,160 | # Coding a simple User Defined Problem#
While pagmo provides a number of UDPs (see List of problems) to help you test your own optimization strategy or user defined algorithm, the possibility to write your own UDP is at the core of pygmo’s use. In this tutorial we will show how to code a UDP. Remember that UDPs are classes that can be used to construct a problem which, in turn, is what an algorithm can solve. See Use of the class algorithm and Use of the class problem to learn the use of these core classes.
We encourage the user to read the documentation of the class problem to have a detailed list of methods that can be, or have to be, implemented in a UDP. To start simple we consider the simple problem of minimizing the two dimensional sphere function.
$\begin{split}\begin{array}{ll} \mbox{minimize: } & x_1^2+x_2^2 \\ \mbox{subject to:} & -1 \le x_i \le 1, i = 1..2 \end{array}\end{split}$
Note
In pagmo minimization is always assumed and should you need to maximize some objective function, just put a minus sign in front of that objective.
>>> class sphere_function:
... def fitness(self, x):
... return [sum(x*x)]
...
... def get_bounds(self):
... return ([-1,-1],[1,1])
The two mandatory methods you must implement in your class are fitness(self, x) and get_bounds(self). The problem dimension will be inferred by the return value of the second, while the actual fitness of decision vectors will be computed calling the first method and passing as argument x as a NumPy array. It is important to remember that x is a NumPy array, so that the NumPy array arithmetic applies in the body of fitness(). Note also how to define a UDP we do not need to inherit from some other pygmo related class. Since we do not define, in this case, any other method pygmo will assume a single objective, no constraints, no gradients etc…
Let’s now build a problem from our new UDP.
>>> import pygmo as pg
>>> prob = pg.problem(sphere_function())
That’s easy! To inspect what type of problem pygmo has detected from our UDP we may print on screen:
>>> print(prob)
Problem name: <class 'sphere_function'>
C++ class name: ...
Global dimension: 2
Integer dimension: 0
Fitness dimension: 1
Number of objectives: 1
Equality constraints dimension: 0
Inequality constraints dimension: 0
Lower bounds: [-1, -1]
Upper bounds: [1, 1]
Has batch fitness evaluation: false
Has hessians: false
User implemented hessians sparsity: false
Fitness evaluations: 0
Let’s now add some (mild) complexity. We want our UDP to be scalable:
$\begin{split}\begin{array}{ll} \mbox{minimize: } & \sum_i x_i^2 \\ \mbox{subject to:} & -1 \le x_i \le 1, i = 1..n \end{array}\end{split}$
and to have a human readable name.
>>> class sphere_function:
... def __init__(self, dim):
... self.dim = dim
...
... def fitness(self, x):
... return [sum(x*x)]
...
... def get_bounds(self):
... return ([-1] * self.dim, [1] * self.dim)
...
... def get_name(self):
... return "Sphere Function"
...
... def get_extra_info(self):
... return "\tDimensions: " + str(self.dim)
>>> prob = pg.problem(sphere_function(3))
>>> print(prob)
Problem name: Sphere Function
C++ class name: ...
Global dimension: 3
Integer dimension: 0
Fitness dimension: 1
Number of objectives: 1
Equality constraints dimension: 0
Inequality constraints dimension: 0
Lower bounds: [-1, -1, -1]
Upper bounds: [1, 1, 1]
Has batch fitness evaluation: false
Has hessians: false
User implemented hessians sparsity: false
Fitness evaluations: 0
Extra info:
Dimensions: 3
Well that was easy, but now have a problem to solve …
>>> algo = pg.algorithm(pg.bee_colony(gen = 20, limit = 20))
>>> pop = pg.population(prob,10)
>>> pop = algo.evolve(pop)
>>> print(pop.champion_f)
[ 3.75822114e-06]
Wow those bees!!
## Possible pitfalls#
Well that was nice as it worked like a charm. But the UDP can also be a rather complex class and the chances that it is somehow malformed are high. Let’s see some common mistakes.
>>> class sphere_function:
... def fitness(self, x):
... return [sum(x*x)]
...
>>> pg.problem(sphere_function())
NotImplementedError Traceback (most recent call last)
...
NotImplementedError: the mandatory 'get_bounds()' method has not been detected in the user-defined Python problem
'<sphere_function object at 0x1108cad68>' of type '<class 'sphere_function'>': the method is either not present or not callable
Oops, I forgot to implement one of the two mandatory methods. In this case it is not possible to construct a problem and, when we try, we then get a rather helpful error message.
In other cases while the UDP is still malformed, the construction of problem will succeed and the issue will be revealed only when calling the malformed method:
>>> class sphere_function:
... def fitness(self, x):
... return sum(x*x)
...
... def get_bounds(self):
... return ([-1,-1],[1,1])
>>> prob = pg.problem(sphere_function())
>>> prob.fitness([1,2])
AttributeError Traceback (most recent call last)
...
AttributeError: 'numpy.float64' object has no attribute '__iter__'
In this case, the issue is that the fitness() method returns a scalar instead of an array-like object (remember that pygmo is also able to solve multi-objective and constrained problems, thus the fitness value must be, in general, a vector). pygmo will complain about the wrong return type the first time the fitness() method is invoked.
## Notes on computational speed#
The most performant way to write a UDP is to code it in C++ and expose it to python. Most UDPs that are included in pygmo (see List of problems) are like that. When writing your own UDP, though, it is often quicker and less painful to code, as shown in this tutorial, directly in python. What effect does this have w.r.t. the ideal situation? Well, Let’s see, on a test machine, a simple example: the scalable Rosenbrock function:
$\begin{split}\begin{array}{ll} \mbox{minimize: } & \sum_{i=1}^{N-1} 100 (x_{i+1} - x_i^2 )^2 + (1-x_i)^2 \\ \mbox{subject to:} & -5 \le x_i \le 10, i = 1..N \end{array}\end{split}$
which in pygmo can be quickly written as:
>>> import numpy as np
>>> class py_rosenbrock:
... def __init__(self,dim):
... self.dim = dim
... def fitness(self,x):
... retval = np.zeros((1,))
... for i in range(len(x) - 1):
... retval[0] += 100.*(x[i + 1]-x[i]**2)**2+(1.-x[i])**2
... return retval
... def get_bounds(self):
... return (np.full((self.dim,),-5.),np.full((self.dim,),10.))
We now make a quick and dirty profiling instantiating a high dimensional instance of Rosenbrock: 2000 variables!!
>>> prob_python = pg.problem(py_rosenbrock(2000))
>>> prob_cpp = pg.problem(pg.rosenbrock(2000))
>>> dummy_x = np.full((2000,), 1.)
>>> import time
>>> start_time = time.time(); [prob_python.fitness(dummy_x) for i in range(1000)]; print(time.time() - start_time)
2.3352...
>>> start_time = time.time(); [prob_cpp.fitness(dummy_x) for i in range(1000)]; print(time.time() - start_time)
0.0114226...
wait a minute … really? Python is two orders of magnitude slower than cpp? Do not panic. This is a very large problem and that for loop is not going to be super optimized in python. Let’s see if we can do better in these cases …. Let us use the jit decorator from numba to compile our fitness method into C code.
>>> from numba import jit
>>> class jit_rosenbrock:
... def __init__(self,dim):
... self.dim = dim
... @jit
... def fitness(self,x):
... retval = np.zeros((1,))
... for i in range(len(x) - 1):
... retval[0] += 100.*(x[i + 1]-x[i]**2)**2+(1.-x[i])**2
... return retval
... def get_bounds(self):
... return (np.full((self.dim,),-5.),np.full((self.dim,),10.))
>>> prob_jit = pg.problem(jit_rosenbrock(2000))
>>> start_time = time.time(); [prob_jit.fitness(dummy_x) for i in range(1000)]; print(time.time() - start_time)
0.03771...
With a bit more elbow grease, we can further improve performance:
>>> from numba import jit, float64
>>> class jit_rosenbrock2:
... def fitness(self,x):
... return jit_rosenbrock2._fitness(x)
... @jit(float64[:](float64[:]),nopython=True)
... def _fitness(x):
... retval = np.zeros((1,))
... for i in range(len(x) - 1):
... tmp1 = (x[i + 1]-x[i]*x[i])
... tmp2 = (1.-x[i])
... retval[0] += 100.*tmp1*tmp1+tmp2*tmp2
... return retval
... def get_bounds(self):
... return (np.full((self.dim,),-5.),np.full((self.dim,),10.))
... def __init__(self,dim):
... self.dim = dim
>>> prob_jit2 = pg.problem(jit_rosenbrock2(2000))
>>> start_time = time.time(); [prob_jit2.fitness(dummy_x) for i in range(1000)]; print(time.time() - start_time)
0.01687...
Much better, right?
Note
For more information on using Numba to speed up your python code see the Numba documentation pages. In particular, note that only a limited part of NumPy and the python language in general is supported by this use. | 2,416 | 9,368 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2024-10 | latest | en | 0.853992 |
https://numberworld.info/31500 | 1,582,676,900,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875146176.73/warc/CC-MAIN-20200225233214-20200226023214-00011.warc.gz | 485,877,999 | 4,197 | # Number 31500
### Properties of number 31500
Cross Sum:
Factorization:
2 * 2 * 3 * 3 * 5 * 5 * 5 * 7
Divisors:
1, 2, 3, 4, 5, 6, 7, 9, 10, 12, 14, 15, 18, 20, 21, 25, 28, 30, 35, 36, 42, 45, 50, 60, 63, 70, 75, 84, 90, 100, 105, 125, 126, 140, 150, 175, 180, 210, 225, 250, 252, 300, 315, 350, 375, 420, 450, 500, 525, 630, 700, 750, 875, 900, 1050, 1125, 1260, 1500, 1575, 1750, 2100, 2250, 2625, 3150, 3500, 4500, 5250, 6300, 7875, 10500, 15750, 31500
Count of divisors:
Sum of divisors:
Prime number?
No
Fibonacci number?
No
Bell Number?
No
Catalan Number?
No
Base 2 (Binary):
Base 3 (Ternary):
Base 4 (Quaternary):
Base 5 (Quintal):
Base 8 (Octal):
7b0c
Base 32:
uoc
sin(31500)
0.68130031386551
cos(31500)
-0.73200401797173
tan(31500)
-0.93073302487231
ln(31500)
10.357742824814
lg(31500)
4.4983105537896
sqrt(31500)
177.48239349299
Square(31500)
### Number Look Up
Look Up
31500 which is pronounced (thirty-one thousand five hundred) is a very great number. The cross sum of 31500 is 9. If you factorisate 31500 you will get these result 2 * 2 * 3 * 3 * 5 * 5 * 5 * 7. 31500 has 72 divisors ( 1, 2, 3, 4, 5, 6, 7, 9, 10, 12, 14, 15, 18, 20, 21, 25, 28, 30, 35, 36, 42, 45, 50, 60, 63, 70, 75, 84, 90, 100, 105, 125, 126, 140, 150, 175, 180, 210, 225, 250, 252, 300, 315, 350, 375, 420, 450, 500, 525, 630, 700, 750, 875, 900, 1050, 1125, 1260, 1500, 1575, 1750, 2100, 2250, 2625, 3150, 3500, 4500, 5250, 6300, 7875, 10500, 15750, 31500 ) whith a sum of 113568. The number 31500 is not a prime number. The figure 31500 is not a fibonacci number. The figure 31500 is not a Bell Number. 31500 is not a Catalan Number. The convertion of 31500 to base 2 (Binary) is 111101100001100. The convertion of 31500 to base 3 (Ternary) is 1121012200. The convertion of 31500 to base 4 (Quaternary) is 13230030. The convertion of 31500 to base 5 (Quintal) is 2002000. The convertion of 31500 to base 8 (Octal) is 75414. The convertion of 31500 to base 16 (Hexadecimal) is 7b0c. The convertion of 31500 to base 32 is uoc. The sine of 31500 is 0.68130031386551. The cosine of the figure 31500 is -0.73200401797173. The tangent of 31500 is -0.93073302487231. The root of 31500 is 177.48239349299.
If you square 31500 you will get the following result 992250000. The natural logarithm of 31500 is 10.357742824814 and the decimal logarithm is 4.4983105537896. You should now know that 31500 is very great figure! | 1,087 | 2,403 | {"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-2020-10 | latest | en | 0.673493 |
http://www.thenakedscientists.com/forum/index.php?topic=19840.0 | 1,481,429,743,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698544097.11/warc/CC-MAIN-20161202170904-00028-ip-10-31-129-80.ec2.internal.warc.gz | 754,381,322 | 18,243 | # The Naked Scientists Forum
### Author Topic: Can one unify gravity and temperature? (Read 11457 times)
#### A Davis
• Sr. Member
• Posts: 106
##### Can one unify gravity and temperature?
« on: 27/01/2009 15:33:14 »
The following equation can be used to unify Newtons gravity equation with Maxwells equation. The equation was determined numerically.
g = α2μoμr
Alpha is the Fine Structure constant and μo the permeability of free space, μr = 1.
The fine structure constant has four main parts, the most significant being the Intrinsic Spin of the Electron S.
α = π S/8 ( appoximately)
From the Curie Weiss law.
g = π2 S2(1 + C/(T-θf))/64
Testing this equation on the Sun at it's current temperatue of above 1 million deg μr = 1, and assuming it has an iron core then at the Curie temperature of 1090degK μr = 5000, as the Sun cools it's spin reduces, so at the Curie temperature it's spin should be 0.9 revolutions per day (11years/5000). It's not possible, the increased gravitational force causes the Sun to implode and produce a smaller solution rotating faster.
There is a second example and that is Binary Star Systems when a star cools down it's gravitational force increases, but it can't increase it's spin, nature has a solution the cooling star must find a companion and it's increased gravitational force helps it to do so. This is why Binary Star Systems are so common, the rotational spin of the pair brings gravity back to it's normal value. The above equation predicts
that the square of the period of rotation is proportional to the temperature of the coldest star. (The spin value is universal)
A.Davis. B.Sc. 27.01.2009
Mod edit - formatted the subject as a question - please do this to help keep the forum tidy and easy to navigate - thanks!
Mod edit - formatted the subject as a question again - please do not undo this change
« Last Edit: 14/02/2009 17:01:03 by BenV »
#### Vern
• Neilep Level Member
• Posts: 2072
##### Re: Can one unify gravity and temperature?
« Reply #1 on: 27/01/2009 16:38:49 »
Do you not need permittivity or is that the 1.
#### A Davis
• Sr. Member
• Posts: 106
##### Re: Can one unify gravity and temperature?
« Reply #2 on: 28/01/2009 00:24:22 »
Permittivity is in Maxwells equation for the velocity of light, if one rearranges the equation I have shown and puts it in Maxwells equation then the gravitational constant is in the equation, Unification.
#### Vern
• Neilep Level Member
• Posts: 2072
##### Re: Can one unify gravity and temperature?
« Reply #3 on: 28/01/2009 12:22:14 »
Permittivity is in Maxwells equation for the velocity of light, if one rearranges the equation I have shown and puts it in Maxwells equation then the gravitational constant is in the equation, Unification.
The way I understood Maxwell is that permeability was the property of a material that allowed a magnetic field and permittivity was the property that allowed the electric field. Maybe "allowed" is not the right word.
I'm not all that swift when it comes to manipulating maths; but it just seemed that permittivity might be unaccounted for in your equation since it is in the Maxwell equations. If you could show your derivation I might could follow it even though I can't do it from scratch.
« Last Edit: 28/01/2009 16:12:47 by Vern »
#### A Davis
• Sr. Member
• Posts: 106
##### Re: Can one unify gravity and temperature?
« Reply #4 on: 28/01/2009 23:46:40 »
The equation I have shown is all magnetic Vern there is no electric field or charge in the equation. If one rearranges it to give.
μo = g/(α2μr) and substitute it into c2=1/(ε0.μ0) Maxwell then one gets
c2 = α2.μr /(εo.g)
There is no derivation it was determined numerically it's a new equation to Science, Science has to test it numerically and then dimensionally when they have a result they have to form a conclusion, is it correct or not!
« Last Edit: 28/01/2009 23:48:33 by A Davis »
#### Vern
• Neilep Level Member
• Posts: 2072
##### Re: Can one unify gravity and temperature?
« Reply #5 on: 28/01/2009 23:57:43 »
The equation I have shown is all magnetic Vern there is no electric field or charge in the equation. If one rearranges it to give.
μo = g/(α2μr) and substitute it into c2=1/(ε0.μ0) Maxwell then one gets
c2 = α2.μr /(εo.g)
There is no derivation it was determined numerically it's a new equation to Science, Science has to test it numerically and then dimensionally when they have a result they have to form a conclusion, is it correct or not!
Well; good luck It is a little bit over my head to put it to the test. Maybe someone will offer some insight.
#### LeeE
• Neilep Level Member
• Posts: 3382
##### Re: Can one unify gravity and temperature?
« Reply #6 on: 30/01/2009 12:31:04 »
The following equation can be used to unify Newtons gravity equation with Maxwells equation. The equation was determined numerically.
g = α2μoμr
Alpha is the Fine Structure constant and μo the permeability of free space, μr = 1.
A.Davis. B.Sc. 27.01.2009
This doesn't seem to work for me. What values and units were you working in?
If I use the following values:
α = 7.29735257x10-3
μ0 = 1.2566x10-6
then:
α2 = 5.325135x10-5
and:
5.325135x10-5 x 1.2566x10-6 = 6.69156464x10-11
but:
g = 6.67428x10-11
The values for the constants were obtained from wikipedia, with the value for α being the corrected value. However, even the older uncorrected value doesn't give g.
#### A Davis
• Sr. Member
• Posts: 106
##### Re: Can one unify gravity and temperature?
« Reply #7 on: 30/01/2009 23:19:32 »
Nice to see somebody doing some maths, your values agree with mine there is a small error. Life isn't simple in Science there are always errors which lead to corrections, if you want a correction look at the Anomalous Magnetic Moment of the Electron square the correction value and divide it into the value you have calculated from my equation, there's no point in going any further the measured value isn't accurate enough. You have to do the dimensional analysis yourself and come to a conclusion, it's not easy.
#### Bored chemist
• Neilep Level Member
• Posts: 8676
• Thanked: 42 times
##### Re: Can one unify gravity and temperature?
« Reply #8 on: 01/02/2009 10:00:00 »
"There is no derivation it was determined numerically it's a new equation to Science, Science has to test it numerically and then dimensionally when they have a result they have to form a conclusion, is it correct or not!"
And Science has tested it, and it's not correct because it gives the wrong answer for G.
A small error as you call it is bigger than the errors on the numbers you are working with. The least acurately known is G which is known to 6 significant figures so the agreement should be that good. It isn't, it's only good to 2 figures and that's far to imprecise to be experimental error.
You have just come up with an amusing (near) coincidence.
Heres' a joke based on a much better coincidence (5 sig fig).
http://xkcd.com/217/
#### A Davis
• Sr. Member
• Posts: 106
##### Re: Can one unify gravity and temperature?
« Reply #9 on: 08/02/2009 21:32:43 »
The error between the two values is 1.00258973
Squaring the anomalous magnetic moment gives 1.00232054 and removing it gives, error 1.00026856
Taking the squareroot of the reduced mass correction gives 1.0002722 and removing it gives, error 1.000004.
The way to attack the theory is to prove that the predicted square law in the Binary Star System, is wrong.
If the period of rotation squared is not proportional to the temperature of the coldest star, then I will
concede defeat.
#### LeeE
• Neilep Level Member
• Posts: 3382
##### Re: Can one unify gravity and temperature?
« Reply #10 on: 08/02/2009 21:44:52 »
Hang on a minute - you've changed the title of the thread to relate to temperatures now, which had nothing to do with your original postings. What are you up to?
#### A Davis
• Sr. Member
• Posts: 106
##### Re: Can one unify gravity and temperature?
« Reply #11 on: 09/02/2009 23:19:50 »
Unifying both of them. Put the new equations at the beginning. More to come.
#### A Davis
• Sr. Member
• Posts: 106
##### Re: Can one unify gravity and temperature?
« Reply #12 on: 11/02/2009 00:01:24 »
The next question is what happens when both the stars in a binary star system reach their curie temperatures. First they will implode and then they will explode.
The result is the same as eta carinae.
« Last Edit: 11/02/2009 00:05:29 by A Davis »
#### lyner
• Guest
##### Re: Can one unify gravity and temperature?
« Reply #13 on: 11/02/2009 10:23:58 »
A Davis
Quote
There is a second example and that is Binary Star Systems when a star cools down it's gravitational force increases, but it can't increase it's spin, nature has a solution the cooling star must find a companion and it's increased gravitational force helps it to do so. This is why Binary Star Systems are so common, the rotational spin of the pair brings gravity back to it's normal value.
So, we've got this star which cools down to this limit of yours. It's been doing an essentially Newtonian thing and following its lonely (minimum energy) path through the Galaxy. Suddenly it breaks out of this behaviour - getting incredible amounts of ENERGY (from where, exactly?) and locates a suitable chum to visit. They then set up home together in a mutual orbit of low eccentricity - involving a load more energy to make the change in trajerctory. This corresponds to a major cosmological event. Why haven't any stars observed doing this (you claim it is a common phenomenon)?
Should I believe your bit of numerology or the evidence, or rather, non-evidence produced from Astronomy?
#### A Davis
• Sr. Member
• Posts: 106
##### Re: Can one unify gravity and temperature?
« Reply #14 on: 11/02/2009 23:51:26 »
January edition Astronomy now page 68, quote Estimates suggest that two-thirds of stars in our local
neighbourhood are doubles. Binary star systems are a big part of nature. One can also get triplets and other multiples of stars. I don't understand where you are getting all this energy from, most of it is locked up inside the Stars.
#### Vern
• Neilep Level Member
• Posts: 2072
##### Re: Can one unify gravity and temperature?
« Reply #15 on: 12/02/2009 18:27:45 »
Hi A. Davis; do you have any candidates for stars that might become Sol's companion?
#### BenV
• Neilep Level Member
• Posts: 1503
##### Re: Can one unify gravity and temperature?
« Reply #16 on: 12/02/2009 18:48:25 »
A. Davis - The subject line of this thread was changed, by me, to be a question, as this is in line with forum policy. I put a note in your post to explain as much, and now I find that you have altered it back without explanation.
Kindly edit your original post so that the subject line is a question. I would also appreciate it in future if you did not revert changes that moderators have made.
#### A Davis
• Sr. Member
• Posts: 106
##### Re: Can one unify gravity and temperature?
« Reply #17 on: 13/02/2009 22:52:56 »
I don't understand your reasoning. There's a lot of new Science to do , I am in no position to do it, only work on the Maths, to me it's a challenge to young graduate science students to take it forwards could get a P.hd. in the following areas.
1. The half integer solution is interesting.
2. Put the equation into precession theory, very interesting.
3. The maximum value for c is in the maths.
4. Relativistic change in spin.
5. Relativistic change in charge.
6. Our Sun has a companion, worth looking at.
I will continue doing calculations, if I come up with something new I'll post it.
#### Bored chemist
• Neilep Level Member
• Posts: 8676
• Thanked: 42 times
##### Re: Can one unify gravity and temperature?
« Reply #18 on: 14/02/2009 16:57:12 »
I think this says it all.
"There's a lot of new Science to do , I am in no position to do it,"
#### lyner
• Guest
##### Can one unify gravity and temperature?
« Reply #19 on: 14/02/2009 23:47:58 »
January edition Astronomy now page 68, quote Estimates suggest that two-thirds of stars in our local
neighbourhood are doubles. Binary star systems are a big part of nature. One can also get triplets and other multiples of stars. I don't understand where you are getting all this energy from, most of it is locked up inside the Stars.
I have no problem with the statistics - just your explanation of them.
What proof do you have that there is any causal relationship?
#### A Davis
• Sr. Member
• Posts: 106
##### Can one unify gravity and temperature?
« Reply #20 on: 17/02/2009 23:27:31 »
To Bored chemist, I am not in a position to do experiments, it would require funding. My pension wouldn't cover it. To SC none, do you have a better explanation. Will post possible Pulsar theory tomorrow.
#### lyner
• Guest
##### Can one unify gravity and temperature?
« Reply #21 on: 18/02/2009 11:31:06 »
To Bored chemist, I am not in a position to do experiments, it would require funding. My pension wouldn't cover it. To SC none, do you have a better explanation. Will post possible Pulsar theory tomorrow.
My "better explanation" is the conventional one which is, at least backed up by observations and has some logical consistency. Have you actually spotted any major flaws in it?
From the very start of this thread - its title,even - you have used scientific inconsistency. If we can't use dimensional analysis then the whole of our present Science is very shaky. If you want to be as revolutionary as that, then you can't really quote any of the formulae from established Science - which were based upon the idea of dimensional consistency. You are on your own from the very start. You have leaped in, half way through and cherry picked the bits that fit your fancy. As I said before, it's no more than numerology.
#### A Davis
• Sr. Member
• Posts: 106
##### Can one unify gravity and temperature?
« Reply #22 on: 20/02/2009 00:11:02 »
I agree, I am on my own. Newtons Laws do describe planetary motion, if the universe is made from elecromagnetic radiation then what electrical unit replaces mass. There are a number of electrical problems with our solar system, the suns magnetic field should decrease as the cube of the distance from the sun it doesn't. Why. Mars has a very small magnetic field. Why. I am trying to prove that the numbers work first, if they didn't I would put the theory into the dustbin.
#### lyner
• Guest
##### Can one unify gravity and temperature?
« Reply #23 on: 20/02/2009 09:57:38 »
Quote
the suns magnetic field should decrease as the cube of the distance from the sun
why?
#### A Davis
• Sr. Member
• Posts: 106
##### Can one unify gravity and temperature?
« Reply #24 on: 23/02/2009 23:57:47 »
I can't answer the question fully, not done enough Maths on the problem. It's interesting to note that Keplers equation can be seen in Stratton i.e. r 3/2
#### The Naked Scientists Forum
##### Can one unify gravity and temperature?
« Reply #24 on: 23/02/2009 23:57:47 » | 3,923 | 14,983 | {"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-2016-50 | latest | en | 0.875175 |
https://homework.cpm.org/category/CC/textbook/cc1/chapter/2/lesson/2.3.3/problem/2-76 | 1,603,730,984,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107891428.74/warc/CC-MAIN-20201026145305-20201026175305-00579.warc.gz | 356,327,946 | 14,693 | ### Home > CC1 > Chapter 2 > Lesson 2.3.3 > Problem2-76
2-76.
James is painting his $10$-by-$8$-foot bedroom wall that contains a $2$-by-$3$-foot window.
1. Draw a diagram of his wall and the window.
There should be one rectangle with a smaller rectangle inside it. Remember to include labels and units..
2. How many square feet of wall does he need to paint?
James does not want to paint his window, so the answer is NOT $80$ square feet. | 124 | 445 | {"found_math": true, "script_math_tex": 5, "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-2020-45 | longest | en | 0.963125 |
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# MAE5010-B Control Engineering
## Objectives:
• In Part I of this assignment you will use the Root Locus method to design a compensator for a plant to satisfy a set performance specifications. You will use Matlab to verify your calculations.
• In Part II of this assignment you will draw the straight-line approximation of the Bode plots and calculate the phase and gain margins
## Submission:
Submit a typed Word report that documents your answers to each section and provides the appropriate plots and put the Matlab code you generated at the end of the report. Do not use screenshots but export your graphs using ‘copy figure’ then paste into your report.
### Compensator Design via the Root Locus
The root locus plot displays the system closed-loop pole locations as a parameter (usually the controller gain) varies from zero to infinity. For a given plant, it may not be possible to achieve the desired closed-loop performance by adjusting the controller gain (proportional control). In this case, a compensator that introduces additional poles and zeros is needed to change the shape of the root locus in a desired manner.
#### Problem Statement:
The constants a, b, and c are the last three digits of your UoB number, for instance if your UoB number is 12610906 then select a = 9, b = 0, c = 6.
#### Part A: Root Locus Plot – 25%
Let C(s) = K, where K {`>`}0
1) Use Word drawing tool to sketch manually the root locus and determine:
• The number of branches, the asymptotes and their angles and intersection with the real axis. Use arrows to show direction of increasing gain on each branch of the root locus
• The breakaway point and the value of K at this point
• The range of K that yield a stable closed-loop system
Discuss how the closed-loop system behaves when the gain K is increased from zero to infinity. Explain closed-loop behaviour in terms of type of response, stability, overshoot/damping, settling time, dominant time constant, etc.
2. Write down the expression for the position constant and hence determine the theoretical value of K that gives a steady-state error of 10% to a unit step input. Is this value of K within the stability range? Is there a value of K that reduces the error to zero? Justify your answer.
### Part B: Compensator Design C(s) – 45%
Use the Root Locus approach to design the simplest compensator possible that achieves the following requirements
1. Zero steady-state error for a step input
2. Overshoot of 20% or less is acceptable.
3. 5% Settling time of 2 seconds or less if possible
Show all the intermediate steps and calculations of your design procedure and explain the reason(s) you chose the compensator. Once your compensator has been designed check with Matlab to see if all the performance specifications have been met.
If the compensated system does not meet all the requirements you will need to adjust the compensator zero(s) until the specifications are achieved. Discuss your approach. Use Matlab to check your results.
Calculate the steady-state error for a unit step input. Use Matlab to plot the closed-loop unit step response of the compensated system and compute the closed-loop poles.
Bode Plot - 30%
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Tip/Trick
# Stardates in C#
, 5 Aug 2010 CPOL
Rate this:
Calculate a *Proper* Star Trek style Stardate
Just a little bit of boredom here, and I happened to be watching an episode of Star Trek: TNG, so I decided I would cobble up a little function to calculate a valid stardate.
Fairly novel function, but who knows...someone writing a video game could use it.
```public double calculateStardate()
{
DateTime calenderStarTrek = new DateTime(2323, 1, 1, 0, 0, 0);
// you can replace the DateTime.Now with year values running all
// the way back to January 1, 1946 at 0:00:00 and still maintain
// a positive stardate by virtue of the way the calculation is
// done, but this is contingent upon application of a 377 year
// offset which puts us into the current Trek year. Follow the
// code for a little bit clearer understanding...
DateTime presentLocalDate = DateTime.Now;
// derive the total offset between present date and trek date
// if we don't do the year offset, we will end up with a date
// that is in the negative, which while technically correct
// it's probably not what we want so we adjust the year value
// of the current date to bring us into the proper Trek year
TimeSpan timeOffset = presentLocalDate - calenderStarTrek;
// we divide into a highly granular value to get the most
// accurate value possible for our calculation. What we are
// actually figuring out the average number of seconds in a
// 4 year leap/non-leap cycle and dividing the total number of
// milliseconds in our time offset to arrive at our raw stardate
// value.
//
// we further round this value to 2 decimal places and miliply it
// by 100 in rounded form, and then divide by 100 to get our two
// decimal places back. 2.7 stardate units account for 1 earth day
// so we do the rounding and multiply / divide operations to get
// granularity back into the final date value.
//
// it makes sense when you look at it :-) trust me.
double yearValue = timeOffset.TotalMilliseconds / (60 * 60 * 24 * 365.2422);
double stardate = Math.Floor(yearValue * 100);
stardate = stardate / 100;
return stardate;
}```
Comments, suggestions and code rewrites appreciated. I handle criticism well Cheers.
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http://easy-ciphers.com/hobnobbing | 1,591,222,286,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347436466.95/warc/CC-MAIN-20200603210112-20200604000112-00196.warc.gz | 40,064,213 | 18,516 | Easy Ciphers Tools:
Caesar cipher
Caesar cipher, is one of the simplest and most widely known encryption techniques. The transformation can be represented by aligning two alphabets, the cipher alphabet is the plain alphabet rotated left or right by some number of positions.
When encrypting, a person looks up each letter of the message in the 'plain' line and writes down the corresponding letter in the 'cipher' line. Deciphering is done in reverse.
The encryption can also be represented using modular arithmetic by first transforming the letters into numbers, according to the scheme, A = 0, B = 1,..., Z = 25. Encryption of a letter x by a shift n can be described mathematically as
Plaintext: hobnobbing
cipher variations: ipcopccjoh jqdpqddkpi kreqreelqj lsfrsffmrk mtgstggnsl nuhtuhhotm oviuviipun pwjvwjjqvo qxkwxkkrwp rylxyllsxq szmyzmmtyr tanzannuzs uboaboovat vcpbcppwbu wdqcdqqxcv xerderrydw yfsefsszex zgtfgttafy ahughuubgz bivhivvcha cjwijwwdib dkxjkxxejc elyklyyfkd fmzlmzzgle gnamnaahmf
Decryption is performed similarly,
(There are different definitions for the modulo operation. In the above, the result is in the range 0...25. I.e., if x+n or x-n are not in the range 0...25, we have to subtract or add 26.)
Atbash Cipher
Atbash is an ancient encryption system created in the Middle East. It was originally used in the Hebrew language.
The Atbash cipher is a simple substitution cipher that relies on transposing all the letters in the alphabet such that the resulting alphabet is backwards.
The first letter is replaced with the last letter, the second with the second-last, and so on.
An example plaintext to ciphertext using Atbash:
Plain: hobnobbing Cipher: slymlyyrmt
Baconian Cipher
To encode a message, each letter of the plaintext is replaced by a group of five of the letters 'A' or 'B'. This replacement is done according to the alphabet of the Baconian cipher, shown below.
```a AAAAA g AABBA m ABABB s BAAAB y BABBA
b AAAAB h AABBB n ABBAA t BAABA z BABBB
c AAABA i ABAAA o ABBAB u BAABB
d AAABB j BBBAA p ABBBA v BBBAB
e AABAA k ABAAB q ABBBB w BABAA
f AABAB l ABABA r BAAAA x BABAB
```
Plain: hobnobbing Cipher: AABBB ABBAB AAAAB ABBAA ABBAB AAAAB AAAAB ABAAA ABBAA AABBA
Affine Cipher
In the affine cipher the letters of an alphabet of size m are first mapped to the integers in the range 0..m - 1. It then uses modular arithmetic to transform the integer that each plaintext letter corresponds to into another integer that correspond to a ciphertext letter. The encryption function for a single letter is
where modulus m is the size of the alphabet and a and b are the key of the cipher. The value a must be chosen such that a and m are coprime.
Considering the specific case of encrypting messages in English (i.e. m = 26), there are a total of 286 non-trivial affine ciphers, not counting the 26 trivial Caesar ciphers. This number comes from the fact there are 12 numbers that are coprime with 26 that are less than 26 (these are the possible values of a). Each value of a can have 26 different addition shifts (the b value) ; therefore, there are 12*26 or 312 possible keys.
Plaintext: hobnobbing
cipher variations:
ipcopccjoh
wreoreezot
ktgotggpof
yvioviifor
mxkoxkkvod
azmozmmlop
cdqodqqron
qfsofsshoz
ehuohuuxol
sjwojwwnox
glyolyydoj
unaonaatov
jqdpqddkpi
xsfpsffapu
luhpuhhqpg
zwjpwjjgps
nylpyllwpe
banpannmpq
derperrspo
rgtpgttipa
fivpivvypm
tkxpkxxopy
hmzpmzzepk
vobpobbupw
kreqreelqj
ytgqtggbqv
mviqviirqh
axkqxkkhqt
ozmqzmmxqf
cboqboonqr
efsqfsstqp
shuqhuujqb
gjwqjwwzqn
ulyqlyypqz
inaqnaafql
wpcqpccvqx
lsfrsffmrk
zuhruhhcrw
nwjrwjjsri
bylrylliru
panrannyrg
dcprcppors
fgtrgtturq
tivrivvkrc
hkxrkxxaro
vmzrmzzqra
jobrobbgrm
xqdrqddwry
mtgstggnsl
avisviidsx
oxksxkktsj
czmszmmjsv
qbosboozsh
edqsdqqpst
ghushuuvsr
ujwsjwwlsd
ilyslyybsp
wnasnaarsb
kpcspcchsn
yresreexsz
nuhtuhhotm
bwjtwjjety
pyltyllutk
dantannktw
rcptcppati
ferterrqtu
hivtivvwts
vkxtkxxmte
jmztmzzctq
xobtobbstc
lqdtqddito
zsftsffyta
oviuviipun
cxkuxkkfuz
qzmuzmmvul
ebouboolux
sdqudqqbuj
gfsufssruv
ijwujwwxut
wlyulyynuf
ypcupcctud
mreureejup
atgutggzub
pwjvwjjqvo
dylvyllgva
ranvannwvm
fcpvcppmvy
terverrcvk
hgtvgttsvw
jkxvkxxyvu
xmzvmzzovg
lobvobbevs
zqdvqdduve
nsfvsffkvq
buhvuhhavc
qxkwxkkrwp
ezmwzmmhwb
sbowbooxwn
gdqwdqqnwz
ufswfssdwl
ihuwhuutwx
klywlyyzwv
ynawnaapwh
mpcwpccfwt
arewreevwf
otgwtgglwr
cviwviibwd
rylxyllsxq
fanxannixc
tcpxcppyxo
herxerroxa
vgtxgttexm
jivxivvuxy
lmzxmzzaxw
zobxobbqxi
nqdxqddgxu
bsfxsffwxg
puhxuhhmxs
dwjxwjjcxe
szmyzmmtyr
gboyboojyd
udqydqqzyp
ifsyfsspyb
whuyhuufyn
kjwyjwwvyz
mnaynaabyx
apcypccryj
oreyreehyv
ctgytggxyh
qviyviinyt
exkyxkkdyf
tanzannuzs
hcpzcppkze
verzerrazq
jgtzgttqzc
xivzivvgzo
lkxzkxxwza
nobzobbczy
bqdzqddszk
psfzsffizw
duhzuhhyzi
rwjzwjjozu
fylzyllezg
uboaboovat
wfsafssbar
yjwajwwhap
mlyalyyxab
opcapccdaz
creareetal
qtgatggjax
eviaviizaj
sxkaxkkpav
gzmazmmfah
vcpbcppwbu
jerberrmbg
xgtbgttcbs
livbivvsbe
zkxbkxxibq
nmzbmzzybc
pqdbqddeba
dsfbsffubm
ruhbuhhkby
fwjbwjjabk
tylbyllqbw
hanbanngbi
wdqcdqqxcv
kfscfssnch
yhuchuudct
mjwcjwwtcf
alyclyyjcr
onacnaazcd
qrecreefcb
etgctggvcn
svicviilcz
gxkcxkkbcl
uzmczmmrcx
ibocboohcj
xerderrydw
lgtdgttodi
zivdivvedu
nkxdkxxudg
bmzdmzzkds
rsfdsffgdc
fuhduhhwdo
twjdwjjmda
hyldyllcdm
vandannsdy
jcpdcppidk
yfsefsszex
mhuehuupej
ajwejwwfev
olyelyyveh
cnaenaalet
qpcepccbef
stgetgghed
gvieviixep
uxkexkkneb
izmezmmden
wboebootez
kdqedqqjel
zgtfgttafy
nivfivvqfk
bkxfkxxgfw
pmzfmzzwfi
dobfobbmfu
rqdfqddcfg
tuhfuhhife
hwjfwjjyfq
vylfyllofc
janfannefo
xcpfcppufa
lerferrkfm
ahughuubgz
ojwgjwwrgl
clyglyyhgx
qnagnaaxgj
epcgpccngv
sregreedgh
uvigviijgf
ixkgxkkzgr
wzmgzmmpgd
kbogboofgp
ydqgdqqvgb
mfsgfsslgn
bivhivvcha
pkxhkxxshm
dmzhmzzihy
robhobbyhk
fqdhqddohw
tsfhsffehi
vwjhwjjkhg
jylhyllahs
xanhannqhe
lcphcppghq
zerherrwhc
ngthgttmho
cjwijwwdib
qlyilyytin
enainaajiz
spcipcczil
greireepix
utgitggfij
wxkixkklih
kzmizmmbit
yboiboorif
mdqidqqhir
afsifssxid
ohuihuunip
dkxjkxxejc
rmzjmzzujo
fobjobbkja
tqdjqddajm
hsfjsffqjy
vuhjuhhgjk
xyljyllmji
lanjanncju
zcpjcppsjg
nerjerrijs
bgtjgttyje
pivjivvojq
elyklyyfkd
snaknaavkp
gpckpcclkb
urekreebkn
itgktggrkz
wvikviihkl
yzmkzmmnkj
mbokboodkv
ofskfssjkt
chukhuuzkf
qjwkjwwpkr
fmzlmzzgle
toblobbwlq
hqdlqddmlc
vsflsffclo
juhluhhsla
xwjlwjjilm
zanlannolk
ncplcppelw
berlerruli
pgtlgttklu
divlivvalg
rkxlkxxqls
gnamnaahmf
upcmpccxmr
iremreenmd
wtgmtggdmp
kvimviitmb
yxkmxkkjmn
abomboopml
odqmdqqfmx
cfsmfssvmj
qhumhuulmv
ejwmjwwbmh
slymlyyrmt
hobnobbing
vqdnqddyns
jsfnsffone
xuhnuhhenq
lwjnwjjunc
zylnyllkno
bcpncppqnm
pernerrgny
dgtngttwnk
rivnivvmnw
fkxnkxxcni
tmznmzzsnu
The decryption function is
where a - 1 is the modular multiplicative inverse of a modulo m. I.e., it satisfies the equation
The multiplicative inverse of a only exists if a and m are coprime. Hence without the restriction on a decryption might not be possible. It can be shown as follows that decryption function is the inverse of the encryption function,
ROT13 Cipher
Applying ROT13 to a piece of text merely requires examining its alphabetic characters and replacing each one by the letter 13 places further along in the alphabet, wrapping back to the beginning if necessary. A becomes N, B becomes O, and so on up to M, which becomes Z, then the sequence continues at the beginning of the alphabet: N becomes A, O becomes B, and so on to Z, which becomes M. Only those letters which occur in the English alphabet are affected; numbers, symbols, whitespace, and all other characters are left unchanged. Because there are 26 letters in the English alphabet and 26 = 2 * 13, the ROT13 function is its own inverse:
ROT13(ROT13(x)) = x for any basic Latin-alphabet text x
An example plaintext to ciphertext using ROT13:
Plain: hobnobbing Cipher: uboaboovat
Polybius Square
A Polybius Square is a table that allows someone to translate letters into numbers. To give a small level of encryption, this table can be randomized and shared with the recipient. In order to fit the 26 letters of the alphabet into the 25 spots created by the table, the letters i and j are usually combined.
1 2 3 4 5
1 A B C D E
2 F G H I/J K
3 L M N O P
4 Q R S T U
5 V W X Y Z
Basic Form:
Plain: hobnobbing Cipher: 32432133432121423322
Extended Methods:
Method #1
Plaintext: hobnobbing
method variations: ntgstggosm symxymmtxr xdrcdrrycw ciwhiwwdhb
Method #2
Bifid cipher
The message is converted to its coordinates in the usual manner, but they are written vertically beneath:
```h o b n o b b i n g
3 4 2 3 4 2 2 4 3 2
2 3 1 3 3 1 1 2 3 2 ```
They are then read out in rows:
34234224322313311232
Then divided up into pairs again, and the pairs turned back into letters using the square:
Plain: hobnobbing Cipher: smirhmlcfh
Method #3
Plaintext: hobnobbing
method variations: rhlshfqmhm hlshfqmhmr lshfqmhmrh shfqmhmrhl hfqmhmrhls fqmhmrhlsh qmhmrhlshf mhmrhlshfq hmrhlshfqm mrhlshfqmh
Permutation Cipher
In classical cryptography, a permutation cipher is a transposition cipher in which the key is a permutation. To apply a cipher, a random permutation of size E is generated (the larger the value of E the more secure the cipher). The plaintext is then broken into segments of size E and the letters within that segment are permuted according to this key.
In theory, any transposition cipher can be viewed as a permutation cipher where E is equal to the length of the plaintext; this is too cumbersome a generalisation to use in actual practice, however.
The idea behind a permutation cipher is to keep the plaintext characters unchanged, butalter their positions by rearrangement using a permutation
This cipher is defined as:
Let m be a positive integer, and K consist of all permutations of {1,...,m}
For a key (permutation) , define:
The encryption function
The decryption function
A small example, assuming m = 6, and the key is the permutation :
The first row is the value of i, and the second row is the corresponding value of (i)
The inverse permutation, is constructed by interchanging the two rows, andrearranging the columns so that the first row is in increasing order, Therefore, is:
Total variation formula:
e = 2,718281828 , n - plaintext length
Plaintext: hobnobbing
first 5040 cipher variations(3628800 total)
hobnobbing
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Read more ...[1] , [2] , [3] | 36,084 | 65,778 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2020-24 | latest | en | 0.803347 |
https://brilliant.org/problems/electric-mushroom-2/ | 1,596,721,401,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439736962.52/warc/CC-MAIN-20200806121241-20200806151241-00083.warc.gz | 240,207,275 | 9,340 | # Electric Mushroom !
Let a solid uniformly charged cone having total charge $Q$, radius $R$ and height $H$ is kept in contact with uniformly charged wire having charge per unit length $\lambda$ as shown. Then calculate the electric interaction force between them and report your answer to nearest integer.
Details and Assumptions
\begin{aligned} Q & = & 1 C \\ R & = & 1m \\ H & = & 1m \\ \lambda & = & 10^{-9} Cm^{-1} \\ \frac {1}{4\pi \epsilon} & = & 9 \times 10^9 Nm C^{-2} \\ \end{aligned}
× | 147 | 500 | {"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": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2020-34 | latest | en | 0.846485 |
http://www.nurulamal.com/math-drills-worksheet-7th-grade-2/ | 1,590,704,304,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347400101.39/warc/CC-MAIN-20200528201823-20200528231823-00055.warc.gz | 185,363,274 | 12,859 | # Math Drills Worksheet 7th Grade
Math Drills Worksheet 7th Grade– Why use free math worksheets? Easy. These worksheets can help you save plenty of time and money and when you’re an active homeschool mom teaching several children, this means a lot.
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### Save Money
It’s obvious how free worksheets can help you save money. If you want, you can skip buying math books and just use worksheets you will get free of charge on the internet. All you have to to do is make use of a “scope and sequence” book that informs you what your son or daughter must be doing in math by age and grade. This book is essential whenever you homeschool.
7th Grade Math and Division Worksheets from Math Drills Worksheet 7th Grade , source: pinterest.com
I would suggest getting one of these simple books when you begin homeschooling and utilize it as a guide through your homeschool journey. It doesn’t matter how long you homeschool, you’ll always have doubts and questions about how your son or daughter is performing.A scope and sequence book can put your brain at ease.
When you have a scope and sequence book, make a listing of each area in math that he needs to focus on for the school year.
Reading Worskheets Kids Worksheet e Worksheets Division from Math Drills Worksheet 7th Grade , source: asucartstudio.org
For instance for grades three and four, by the finish of the entire year in subtraction, your child should manage to:
• Solve vertical and horizontal computation problems
• Review subtraction of 2 numbers whose sums will be 18 or less
• Subtract 1- or 2-digit number from a 2-digit number with/without renaming
• Subtract 1-, 2-, or 3-digit numbers from 3- and 4-digit number with/without renaming
• Subtract 1-, 2-, 3-, 4-, or 5-digit number from the 5-digit number
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If you have this list, begin searching online free of charge math worksheets that fit your child’s scope and sequence for the year and the goals you have set for the child.
### Save Time
Free worksheets not just help you save money, they could also save time. If you choose that it’s best for your son or daughter to do worksheets especially tailored for his needs, by doing a little research for printable math worksheets found online, you do not have to really make the worksheets yourself. This will save plenty of time. Worksheets aren’t that difficult to produce, but it can be time consuming. | 531 | 2,413 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2020-24 | latest | en | 0.905616 |
https://electronics.stackexchange.com/questions/109635/digitally-controlled-smoothly-variable-gain | 1,558,616,318,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232257244.16/warc/CC-MAIN-20190523123835-20190523145835-00478.warc.gz | 480,741,694 | 36,126 | # Digitally controlled smoothly variable gain?
I have a circuit that outputs about 600mVpk wave shape. I need to be able to output between 10mVpk to 10Vpk accurately, preferably with no calibration but a known error. The whole process should be supervised by digital means.
The general scheme I was thinking about is amplifying the signal to 10Vpk and some attenuation device to set different gains. This could be a series of resistive dividers that can be inserted in line. Each one is attenuating the signal by a different amount like -2dB,-4dB,-8dB,-16dB and they can be combined to create more attenuation points like -6dB or -30dB. This can be done using 4 SPDT relays but this is not enough. What additional element can I use to make up the spaces. I would like a high resolution. Possibly 10mV if possible.
• Use a multiplying DAC? – Dave Tweed May 8 '14 at 15:33
• @Dave: You should make that a answer. I'd upvote it. – Olin Lathrop May 8 '14 at 16:17
• What's the highest frequency component of the signal? If it's audio you could use a digital pot, higher frequency an MDAC, higher again and some other method such as switched attenuators will have to be used. – Spehro Pefhany May 8 '14 at 17:26
As Dave Tweed commented, a multiplying DAC (MDAC) seems the way to go. You don't mention your digital interface, frequency range or your output power requirements, so some details you'll have to work out for yourself. Analog Devices makes a wide variety of MDACs, which go all the way up to 16 bits resolution, which would give you amplitude steps of 150 microvolts. See, for instance, http://www.analog.com/static/imported-files/circuit_notes/CN0055.pdf
As to your "no calibration" requirement, I doubt if that's going to happen. Assuming you want accuracy equal to your resolution, this implies an overall gain of up to about 16, with a gain accuracy of 0.1% (10 volt max / 10 mV resolution). This in turn suggests that your external gain components will need at least 0.05% accuracy, and this will not come cheap. You'd do better to allow some adjustment. This assumes that by "calibration" you mean "adjustment". If you use lower-tolerance parts and accept the resulting inaccuracies, you'll still need a calibration cycle in the sense of putting in various input levels and measuring the resulting outputs.
• Wow I totally forgot about these! This app note seems almost perfect for my needs. The output of this system in 50 Ohm terminated so it should have enough juice to put 10Vppk on that load. Freq. range is under 1Mhz. – user34920 May 8 '14 at 17:24
• Eeek! 10 volts into 50 ohms? Not with that op amp, you won't. Assuming your 10Vppk is +/- 5 V (not 0 - 10 V), you can series terminate with 50 ohms, and your actual output (at the amplifier) will be +/- 10V @ 100 mA. This will require a serious output stage. If you don't terminate the output as suggested, an accidental short on the output will quite possibly kill your amp. – WhatRoughBeast May 8 '14 at 17:56
• I did not think of using the op amp shown at the application note, but the idea is still the same. Off course a series fuse + protection diodes will also be placed at the output. – user34920 May 8 '14 at 18:47
I'm assuming you have a stable "reference" wave-shape that you can feed into a precision rectifier circuit. From this, you can obtain the peak value (or rectified mean value if that is more suitable) of that wave-shape. If you then use a linear attenuator such as from a JFET or analogue multiplier you can "pot" the AC amplitude down by a suitable amount.
This new output will be shaped like the reference waveform but not have a very precise amplitude ratio to the reference - there will be an amplitude error BUT, using the same design precision rectifier circuit, you can measure the attenuated signal and calculate the imprecise ratio.
The next thing to do is to make a correction to the JFET or analogue multiplier. This would be based on making the ratio error as small as possible. Now, you have a self calibrating system that relies on: -
• the control an analogue multiplier/attenuator with a DC value
• the accuracy/linearity of a precision rectifier
• the accuracy of being able to differentiate DC levels representing the reference and attenuated waveform.
Because you are dealing with DC values you can choose to go to 24 bit ADC technology to get very high resolution or just stick with analogue control.
Just a thought. | 1,086 | 4,441 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2019-22 | longest | en | 0.948375 |
https://iconvert.org/convert/area/9788-square-feet-to-hectares | 1,686,284,460,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224655247.75/warc/CC-MAIN-20230609032325-20230609062325-00616.warc.gz | 347,317,244 | 12,627 | # 9788 Square Feet to Hectares
Do you want to know how much is 9788 square feet converted to hectares? With our free square feet to hectares conversion tool, you can determine the value in hectares of 9788 square feet.
Convert square feet to hectares
9788 square feet = 0.0909334564 hectares
Convert 9788 hectares to square feet
## How to convert 9788 square feet to hectares?
Note: ft2 is the abbreviation of square feet and ha is the abbreviation of hectares.
1 square feet is equal to 0.0000092903 hectares:
1 ft2 = 0.0000092903 ha
In order to convert 9788 ft2 to ha you have to multiply 9788 by 0.0000092903:
9788 ft2 x (0.0000092903 ha / 1 ft2) = 9788 x 0.0000092903 ha = 0.0909334564 ha
So use this simple rule to calculate how many hectares is 9788 square feet.
## 9788 square feet in other area units
Want to convert 9788 square feet to other area units? Use these links below: | 256 | 902 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2023-23 | longest | en | 0.825359 |
https://www.adellejenike.com/how-to-use-fibonacci-technical-analysis/ | 1,582,696,306,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875146187.93/warc/CC-MAIN-20200226054316-20200226084316-00323.warc.gz | 603,938,242 | 7,816 | # How To Use Fibonacci Technical Analysis
One of the best ways to use the Fibonacci retracement tool is to spot potential support and resistance levels and see if they line up with Fibonacci retracement levels. If Fibonacci levels are already support and resistance levels, and you combine them with other price areas that a lot of other traders are watching, then the chances of price bouncing from those areas are much higher.
Hello, and welcome to another episode of technical analysis show for trading binary options. Today I will show you detailed examples of trading while using one of the most basic things – Fibonacci lines. We will learn how to draw the lines, how to use it to trade, but also a.
Fibonacci sequences have long secured their place as one of the most reliable technical analysis tools. Using them properly is an art, and we explain how to use them briefly with TradingView.com’s free charting software.
Where Was Thomas Edison Raised Many have characterized Tesla and inventor Thomas Edison as enemies (see this and this,) but Carlson says this relationship has been misrepresented. Early in his career, Tesla worked for Edison. Rage, rage against the dying of the light. His parents’ first language was Welsh, but Thomas was raised speaking English so that he wouldn’t sound
Fibonacci was an Italian mathematician who came up with the Fibonacci numbers. They are extremely popular with technical analysts who trade the financial markets, since they can be applied to any timeframe. The most common kinds of Fibonacci levels are retracement levels and extension levels.
How To Use Fibonacci Retracements for Technical Analysis. December 15, 2013. By Vlad Karpel. Fibonacci Retracement is a popular tool used by many investors around the world in predicting the reversal levels of stocks. Since stock and option.
A beginners guide to Fibonacci retracement. What is Fibonacci retracement? – "In technical analysis, Fibonacci retracement is created by taking two extreme points (usually a major peak and trough) on a stock chart and dividing the vertical distance by the key Fibonacci ratios of 23.6%, 38.2%, 50%, 61.8% and 100%." – Investopedia In english?
Einstein Brothers Bagels Honey Blue Cheese 1 Bagel. 107. 280. 15. 2. 0. 0. 0. 480. 57. 2. 5. 10. X X. Honey Whole Wheat. 1 Bagel. 102. 260. 25. 3. 0. 0. 0. 550. 49. 7. 7. 6. 10. X X. Sourdough*. 1 bagel. 106. 270. 35. 4.0. 0.5. 0. 0. 440. 49. 2. 3. 10. X1 X. Asiago Cheese.
Story Points Agile Fibonacci Rather than trying to build the fastest or farthest-running electric car, the brand focused on making the machine feel as. The job in today’s spread game, especially against good run-oriented teams like the 49ers and Baltimore Ravens, requires a. Einstein Brothers Bagels Honey Blue Cheese 1 Bagel. 107. 280. 15. 2. 0. 0. 0. 480.
To help better understand how to use Fibonacci Spirals when trading, I have decided to dedicate a topic about them. My quest here, is to consistently generate predictions using Fibonacci Spirals. Polar Equations Applied To Technical Analysis – Spirals. EURUSD, 60.
Peer Reviewed Studies Questioning E=mc2 After formulating the clinical question with the format of PICO, the next step is to search for the relevant evidence that will help answer the clinical questions. Unfortunately, studies have shown that, even in countries where hospitals have facilities for internet access allowing healthcare personnel access to a number of electronic databases. In a new
Fibonacci was an Italian mathematician who came up with the Fibonacci numbers. They are extremely popular with technical analysts who trade the financial markets, since they can be applied to any timeframe. The most common kinds of Fibonacci levels are retracement levels and extension levels.
The use of Fibonacci levels in trading is perhaps one the best examples of the core philosophy of Technical Analysis and the belief of many, that trading decisions can be made purely from studying.
Fibonacci was an Italian mathematician who came up with the Fibonacci numbers. They are extremely popular with technical analysts who trade the financial markets, since they can be applied to any timeframe. The most common kinds of Fibonacci levels are retracement levels and extension levels.
Technical Analysis – Fibonacci Levels Retracements A retracement is a pullback within the context of a trend. Dip After a rise from 0 to 1, short term market participants start to take profit. This drives the price lower until such a point that the bulls, sensing the price is better value, enter
28-1-2020 · Jan 28, 2020 NZD/USD Technical Analysis: Kiwi Dollar May Bounce Before Drop. Jan 28, 2020 Japanese Yen Gains On Wuhan Virus, Can You Use Fibonacci As A Leading Indicator? | 1,081 | 4,755 | {"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-2020-10 | latest | en | 0.949993 |
http://slidegur.com/doc/309176/8.4--ellipses | 1,568,534,892,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514570830.42/warc/CC-MAIN-20190915072355-20190915094355-00231.warc.gz | 178,734,156 | 8,142 | ### 8.4: Ellipses
```8.4: Ellipses
• Write equations of ellipses
• Graph ellipses
Ellipse
• An ellipse is like an oval.
• Every ellipse has two axes of symmetry
• Called the major axis and the minor axis
• The axes intersect at the center of the ellipse
• The major axis is bigger than the minor axis
• We use c2 = a2 – b2 to find c
• a is always greater b
• The equation is always equal to 1
Ellipses Chart (pg 434)
Standard Form
of Equation
Center
Direction of
Major Axis
Foci
Length of Major
Axis
Length of Minor
Axis
x
- h
a
2
2
+
y
- k
b
2
2
=1
y
- k
a
2
2
+
x
- h
b
2
2
=1
(h,k)
(h,k)
Horizontal
Vertical
(h + c, k), (h – c, k)
(h, k + c), (h, k – c)
2a units
2a units
2b units
2b units
Example One:
Graph the ellipse
x
- 2
4
2
+
y
+ 5
1
2
=1
Graph the ellipse
x
+ 2
81
2
+
y
- 5
16
2
=1
Example Two:
Graph the ellipse
y
- 4
64
2
+
x
- 2
4
2
=1
Graph the ellipse
y
- 2
36
2
+
x
- 4
9
2
=1
Example Three:
Write the equation of the ellipse in the
graph:
Write the equation of the ellipse in the
graph:
Example Four:
Write the equation of the ellipse in the
graph:
Write the equation of
the ellipse in the graph:
Standard Form
Find the coordinates of the center and
foci and the lengths of the major and
minor axes of the ellipse with equation:
x2 + 4y2 + 24y = -32
Standard Form
Find the coordinates of the center and
foci and the lengths of the major and
minor axes of the ellipse with equation:
9x2 + 6y2 – 36x + 12y = 12
``` | 569 | 1,456 | {"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} | 4.71875 | 5 | CC-MAIN-2019-39 | latest | en | 0.793743 |
https://www.bhavinionline.com/2018/08/logical-riddle-equal-distribution-of-the-stolen-diamonds/ | 1,601,516,168,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600402130531.89/warc/CC-MAIN-20200930235415-20201001025415-00769.warc.gz | 665,215,805 | 20,060 | # Logical Riddle: Equal Distribution of the Stolen Diamonds
Use your logical and mathematical skills to solve this riddle.
7 thieves rob a diamond merchant of some diamonds.
When everybody was sleeping, two of them woke up and decided to divide the diamonds equally among themselves.
But when they divided the diamonds equally, one diamond is left.
So they woke up the 3rd thief and tried to divide the diamonds equally again but still one diamond was left.
Then they woke up the 4th thief to divide the diamonds equally again, and again one diamond was left.
This happened with the 5th and 6th thief – one diamond was still left.
Finally, they woke up the 7th thief and this time the diamonds were divided equally.
How many diamonds did they steal in total?
So were you able to solve the riddle? Leave your answers in the comment section below.
## 5 Replies to “Logical Riddle: Equal Distribution of the Stolen Diamonds”
1. samarth says:
they stole 7 diamonds
2. Sanjiv says:
Should be 49 as it looks 5th & 6th we’re woken together. Otherwise 301 is the right answer
3. MS PRIME says: | 254 | 1,101 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2020-40 | longest | en | 0.975216 |
https://www.teachstarter.com/us/teaching-resource/number-talks-mental-division-task-cards/ | 1,702,075,407,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100779.51/warc/CC-MAIN-20231208212357-20231209002357-00461.warc.gz | 1,108,795,270 | 76,265 | teaching resource
# Number Talks - Mental Division Task Cards
• Updated: 03 Aug 2022
Build number sense skills with this set of 24 mental math task cards.
• Non-Editable: PDF
• Pages: 24 Pages
Tag #TeachStarter on Instagram for a chance to be featured!
teaching resource
# Number Talks - Mental Division Task Cards
• Updated: 03 Aug 2022
Build number sense skills with this set of 24 mental math task cards.
• Non-Editable: PDF
• Pages: 24 Pages
Build number sense skills with this set of 24 mental math task cards.
Use this set of task cards to easily implement number talks into your classroom.
Number talks are meant to be short, daily, math activities that allow students to have meaningful and highly engaging conversations about math. Simply show students the front of the card, and ask the prompts on the back. These exchanges will lead to the development of more accurate, efficient, and flexible strategies for students.
This card set is a great teaching resource for helping students to mentally divide numbers by one and two-digit divisors using strategies based on:
## Tips for Differentiation + Scaffolding
A team of dedicated, experienced educators created this resource to support your math lessons.
In addition to individual student work time, use this number talks activity to enhance learning through guided math groups or whole class lessons.
If you have a mixture of above and below-level learners, check out these suggestions for keeping students on track with the concepts:
### 🆘 Support Struggling Students
Help students who need help understanding the concepts by focusing on the single digit divisor cards first. Additionally, if students are struggling to do these mentally, focus on using visual models before solving them mentally.
### ➕ Challenge Fast Finishers
For students who need a bit of a challenge, encourage them to create their own division problems to answer the prompts with.
### 👋 Exit Ticket
Use these cards as a formative assessment after your lesson. Pick a random assortment of cards and project them on the board for the whole class to see. Students can record their answers on a sheet of paper, sticky note, or their notebook.
Plan lessons for all ability levels with our 10 Best Scaffolding Strategies!
## Easily Prepare This Resource for Your Students
Use the dropdown icon on the Download button to choose between the color PDF or black and white PDF version of this resource.
Print on cardstock for added durability and longevity. Place all pieces in a folder or large envelope for easy access.
Print out the task cards front and back so that the prompts are displayed on the back of each card. The cards can also be put on a ring for added convenience.
Don’t stop there! We’ve got more activities and resources that cut down on lesson planning time:
[resource:4704990] [resource:2661098] [resource:2665722] | 604 | 2,901 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2023-50 | latest | en | 0.904377 |
http://www.perlmonks.org/?node_id=340521 | 1,496,047,743,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463612036.99/warc/CC-MAIN-20170529072631-20170529092631-00077.warc.gz | 768,203,962 | 17,605 | The stupid question is the question not asked PerlMonks
### Estimating continuous functions
by zdog (Priest)
on Mar 29, 2004 at 06:50 UTC Need Help??
I'm presently working on a programming project (of which the programming language shall remain nameless :-), and I've reached a point where my education and/or my creativity does not suffice to come to a satisfactory solution.
Dear friends! Say you are given a table with 4 columns. Each column represents a variable in a continuous mathematical function. So, say the function takes 3 independent variables and its result is the dependent variable. The function is one-to-one-to-one-to-one, or three-to-one, or something like that, so if you are given any three of the variables, you should be able to solve for a specific value for the 4th variable. Unfortunately, this function is not known to us. Instead, we are given a table containing a limited set of input and results of the function.
In case that description is not clear, hopefully this example will clarify. Say the table looks something like this:
```x | y
-----
3 | 4
4 | 5
We do not know the function, but we do know that for those values of x, we get those values of y. However, we do not know what we get for y if x were to equal 3.5. Not necessarily an accurate estimate, but surely not an unreasonable one, of the resulting value is 4.5 because the value of x was directly in between 3 and 4 so we take the value in the middle of 4 and 5 for the y value. Unfortunately, as we add more independent variables, the relationship between them becomes increasingly obfuscated given such a data set.
So I think you can guess where I am going with this ...
Does anyone know of a method for coming up with such reasonable estimates for those more complex functions and data tables, if such a method that would be reasonably implementable in a language like Perl exists? The value 1 can be an estimate for any set of values, but certainly something that tends to approach reality is favorable. This problem busted my brain for a few hours before I decided to ask someone else.
One thing I tried is if one of the given points (given by a set of inputed variables) from the table is not close enough to the required point, the midpoints between the points are calculated and those become treated as "given" points. This is repeated until something close arises. This wouldn't be bad if the number of midpoints didn't multiply so quickly. I tried limiting the number of "given" midpoints for each loop to those that are closest, but then issues with the domain arose so that the desired point was no longer within the "given" set.
Basically, this comes down to drawing or estimating the curve of a continuous function given a set of points. The example I gave is a reasonable solution to the problem in two dimensions. Is there any good or decent way to expand this to more dimensions? 3, 4, 5, and so on? Visually, it doesn't seem like too difficult of a problem, but mathematically, especially in the realm of programming, it may be ..
Thanks for any thoughts, ideas, or insights!
Zenon Zabinski | zdog | zdog@perlmonk.org
Replies are listed 'Best First'.
Re: Estimating continuous functions
by Itatsumaki (Friar) on Mar 29, 2004 at 07:33 UTC
There are two general approaches to this kind of problem. You can either try to estimate the underlying function from the data that you are given, or you can simply try to generate a smooth curve between the data-points. There are (naturally) advantages and disadvantages to each approach. :)
Predicting the function is a more complex business, and usually requires some simplifying assumptions about the type of relationship between the dependent variable (the output) and the various independent variables (inputs). For instance, you can assume that the relationship is linear, or quadratic, or logarithmic, or... just about anything else. However that assumption generally has to be made up front.
You then perform a statistical operation called a "regression" on the data, using the relationships that you defined above. The regression attempts to estimate the "importance" of each parameter (as measured by a set of constant coefficients) by minimizing the deviations between the predicted and actual values. Actually you typically minimize the squared-deviations. To get an estimate of the "goodness-of-fit" you would take a look at the "residuals": how much of the actual data does your model explain?
This is really only a rough overview of regression. You have multiple dependent variables, so you are going to want a "multiple regression". The biggest problem with regression is that it takes a lot of data to accurately estimate a sophisticated model. So, if you have n independent variables (dimensions), and you assume that: i) the variables are completely orthogonal (e.g. the independent variables are totally unrelated) and ii) there is a linear relationship between each independent variable and the dependent variable then you will need at a minimum n independent observations to be able to estimate the model.
Overall, regression thus works really well when you have a lot of data, and can make reasonable estimates about the relationships between the dependent and independent variables. One nice thing about regression techniques is that they can (relatively) seamlessly handle categorical (non-numeric) data like {TRUE/FALSE} or {RED/GREEN/YELLOW}.
Your other option is to use a interpolating method. These methods do not attempt to directly model the interactions between dependent and independent models. Instead, they simply attempt to find the best smooth-fitting curve for the data. The two most important methods in use today are cubic-spline-based interpolation and fourier-approximations.
Cubic splines essentially fit cubic functions to subsets of your data. That is, a separate cubic function is fitted (regressed, actually) between each pair of data-points. This regression uses some (but not all!) of the surrounding data points. So if you have a continuous set of say 100 observations, you might estimate a cubic function between points 50 and 51 using only the data-points from 45 to 55. It turns out that this local interpolation handles discontinuities and large abrupt changes much better than global interpolations. Of course this (can) come at the price of much more computational overhead.
Fourier approximation attempts to estimate your data with a set of sinuisoidal functions (usually just sine and cosine). You can essentially approximate or fit any function (even whacked out engineering things like the heavy-side or weirdo physics stuff like the Dirac delta) with an infinite series of sinuisoidal functions. The theory behind Fourier approximations is a bit complex, but at some level you can simply think of it as regressing a large set of sine/cosine functions.
So... how are you supposed to chooose between these alternatives? The first two questions you should ask yourself are:
1. Do you need to understand the dependency structure between your dependent and independent variables? Do you already have some understanding of it that you would like to incorporate into your estimation?
2. Do you simply need to interpolate amongst the data-points that you already have, or will you need to extrapolate to new and untested values of the dependent and/or independent variables?
If you wish to understand or exploit the dependency structure amongst variables, then you will want to use a regression approach. On the other hand, if you are primarily interested in interpolation, you will almost certainly just need a curve-fitting method. Your choice gets a little trickier if you have no desire/ability to understand the relationships amongst variables, but need to extrapolate: both approaches can be used in that case, but with serious concerns about accuracy.
If you are using a curve-fitting approach, cubic splines are much easier to implement yourself and much more intuitive than Fourier transforms. If you data is inherently oscillatory, however, you might get a lot of value out of learning the details behind Fourier approximation.
If that sounds like a whole bunch of "maybes" and "possiblies" that's what it is. There are tons of numerical methods, and there are lots of application-specific trade-offs that can be made on complexity, computational efficiency, generality, and so forth.
Finally, here are a bunch of links on this stuff you might find helpful. When I first learned numerical methods I used Chapra and Canale, which was a good general book that suited people w/o a lot of mathematical background. You might want to check that out.
Good luck!
-Tats
Update: s/durve/curve/; thanks to PodMaster
Re: Estimating continuous functions
by hossman (Prior) on Mar 29, 2004 at 07:38 UTC
As I understand it, this is generally considered a "doozy" of a problem. You have to accept the fact, that for any sets of points, there are an infinte number of curves that fit them, and except in simple cases, there's usually no way to determine that one curve is "more correct" then another -- and in the really simple cases, it's even worse (consider the points where y=cos(x) and y=sin(x) intersect, then consider how man even simpler curves all go through the same points).
There do however seem to be a few papers online that discuss how you can approximate it if you're willing to make ceratin assumptions. The math is way over my head, but knock yourself out...
Re: Estimating continuous functions
by dragonchild (Archbishop) on Mar 29, 2004 at 12:33 UTC
It is a doozy, but it's also a very interesting doozy. There are many, many programs that have been written that will do this. One of the best, imho, is Mathematica. (At least, it was in '93 when I was an aspiring math major.) If you can use it, I'd suggest looking at Math::ematica. (If you don't have it and you're being paid for this work, indicate to your paychecksource that they can shell \$ for Mathematica or \$\$\$ for you to replicate it.)
------
We are the carpenters and bricklayers of the Information Age.
Then there are Damian modules.... *sigh* ... that's not about being less-lazy -- that's about being on some really good drugs -- you know, there is no spoon. - flyingmoose
Re: Estimating continuous functions
by tilly (Archbishop) on Mar 30, 2004 at 03:16 UTC
Several people have pointed out that this is a very hard problem to get right. There is no perfect answer, and tremendous amounts of energy has been spent on finding pretty good ones.
But nobody has given you a simple to implement "OK" answer. Not great. Just something that can readily be implemented which gives an answer that you can defend as somewhat reasonable. Depending on the application, this is often enough.
Here is an outline of one.
Let's say that the last variable a function of the rest, call it f. Make f(x_1, x_2, ... , x_(n-1)) into the weighted average of the known values of f at the points that you have. A reasonable weighting is that a point is weighted with weight 1/(square of distance from that point to the spot you're estimating). Then you can do it something like this (with some error checking added, of course...):
```sub make_estimator {
return sub {0} unless @_; # You might want to die?
my @points = @_;
return sub {
my \$weight_total;
my \$weighted_sum;
foreach my \$point (@points) {
my \$dist_squared = 0;
foreach my \$coord (0..\$#_) {
\$dist_squared += (\$point->[\$coord] - \$_[\$coord])**2;
}
if (0 == \$dist_squared) {
return \$point->[-1];
}
else {
my \$weight = 1/\$dist_squared;
\$weight_total += \$weight;
\$weighted_sum += \$weight * \$point->[-1];
}
}
return \$weighted_sum/\$weight_total;
};
}
my \$estimator = make_estimator(
[3, 4],
[4, 5],
);
print \$estimator->(3.5);
This works perfectly well. The resulting function is trivially smooth everywhere except at the estimating points, and I'm reasonably surepositive that it is smooth there as well. It isn't linear or particularly simple though.
Feel free to play with the weighting function.
Update: I verified it. The function is also smooth at the estimating points. Its derivative at all estimating points is 0 (ie each is a maxima, minima, or inflection point).
Cool.
I guess that's the essence of that two dimensional example that I gave. The difference being that the program does not have to determine which points it takes into account. It just uses all of the data points, and the weighting allows it to do so. That's also its weakness in some ways .. Correct? .. that it considers all data points to an extent no matter how removed they are... or for some other reason?
The fact that it isn't linear isn't necessarily a bad thing, though, is it? Depends on the type of function, I guess. I suppose another aspect that adds to its simplicity is that it doesn't require you to select whether the function is linear, quadratic, logarithmic, etc, before approximating it.
When (if?) I get a chance, I'll generate some graphs using the data set (which I actually don't have yet :-), varying the weighting, to see how the estimation looks visually. Perhaps I'll post a sample of them up here later on.
Thanks!
Zenon Zabinski | zdog | zdog@perlmonk.org
I'd say that its biggest weakness is probably the opposite. It tends to overuse the nearest available data point resulting in flat sections and fairly sharp transitions from one point dominating to the next. This results in functions that don't look very reasonable.
Playing around with the weighting should let you achieve an acceptable trade-off between one point dominating and distant points having too much of an impact. The choice is going to be empirical however, there isn't a "best answer" to this problem.
Furthermore if you know anything about what your underlying function looks like, you would probably do a lot better to use more traditional estimation techniques. In particular if you can get samples on some kind of useful grid, one of the usual curve-fitting algorithms will be easy to calculate and should give excellent results. Two standard kinds of often-used curve-fitting algorithms are polynomial (eg cubic splines) and wavelets.
That's pretty cool. That seems almost like a loess smoother, isn't it?
Re: Estimating continuous functions
by zentara (Archbishop) on Mar 29, 2004 at 16:18 UTC
I'm trying to "pull a rabbit out of my hat" here, so read this with that in mind.
My first thoughts were Runge-Kutta methods. Do a google search for Runge-Kutta. I don't know how advanced your math skills are, (and mine are a bit rusty).
But what I would try to do is try and develope a set of differential equations which describe the differences between adjacent table entries. These can be as long as you need, and can involve "higher order differentials" with linear constants. Then try to combine the equations set and maybe integrate the whole thing to get a function out of it.
It may only give you a segmented function like it's "this if between 3 and 4", and that if between "4 and 5", etc.
I'm just guessing from your limited description, but you could setup equations with elements like
```w1 = w0 + a(dx/dw) + b(dy/dw) + c(dz/dw)
+ d(dx2/dw2) + e(dy2/dw2) + f(dz2/dw2)
w2 = w1 + a(dx/dw) + b(dy/dw) + c(dz/dw)
+ d(dx2/dw2) + e(dy2/dw2) + f(dz2/dw2)
..etc..
where w1 and w2 etc are known values from the table. Then try to solve for a,b,d,e,f,g and integrating the whole mess. I've only shown the second order differential, but you can go as high as you need to get accuracy. ( I think the space shots use a "4th order Runge-Kutta"). That is they evaluate the velocity, rate of change of velocity(acceleration), rate of change of acceleration, and rate of rate of change of acceleration. They do this for all x,y,z. They have to account for time, and rotation of the planet too.
The Runge-Kutta method assumes you know the function to get the results. What I'm suggesting is to "reverse engineer" the equation. But that's what make computers so powerful...you can make a big sloppy list of equations to run thru to calculate a value, and the computer dosn't care.
Not that it will solve your problem, but there is a perl module for Runge-Kutta, Math::RungeKutta, here is the README The examples in the module are very cool, one is an ascii animation of plantary orbits.
The problem with solving these types of problems is the "time factor", if you think long enough on it you will eventually come to the best answer. The trouble is it may take "years of contemplation", which dosn't cut it for homework assignments. :-)
I'm not really a human, but I play one on earth. flash japh
Runge-Kutta techniques are useful for solving a known set of differential equations numerically. This does not give you a convenient way to figure out which set of differential equations will give you a known set of points. Worse yet, many sets of equations will give you those points, and Runge-Kutta offers you no way to decide between them.
Thus while the similarity of certain key-words might make you think that there is a useful connection, there really isn't one.
Sorry...
Isn't Runge-Kutta a method for numeric integration of known curves? I think essentially you're suggesting to regress a high-order Runge-Kutta and then minimize the differential area. That's an interesting approach, but I suspect pretty computationally slow compared to directly fitting splines or something like that. I suspect the OP would be better off doing direct interpolation instead, especially because Runge-Kutta won't allow for extrapolation beyond predicted values.
One other note: remember that if you try to estimate a high-order equation without a lot of input data, you can run into over-fitting problems. Most numerical techniques get much more accurate (and slower) with lots of data
Runge-Kutta is a method for solving differential equations numerically. However, there is a mapping from integrals to differential equations, so in that respect Runge-Kutta is also a method for solving integrals numerically.
thor
Re: Estimating continuous functions
by hardburn (Abbot) on Mar 29, 2004 at 14:22 UTC
Not having as strong a math background as I would like, I jump straight to a (possibly) practical solution: Neural Nets. It'll probably either get close to the real function, or it'll generate a confused mass of code much like the one you say you have now :) If nothing else, it might be a fun diversion.
----
: () { :|:& };:
Note: All code is untested, unless otherwise stated
Or write a Genetic Algorithm (using AI::Genetic for example). Use a variety of functions as genes from which to build the genome of each 'individual'. The fitness test could use least squares.
Problem with Neural Nets is they tend to be black boxes, though they might approximate your function they can't easily tell you what it is. Unless you implement a *really* big one that can parse and generate natural language as well as your function ;)
Re: Estimating continuous functions
by husker (Chaplain) on Mar 29, 2004 at 17:13 UTC
Two "lectures" on curve-fitting:
and
Note that both of these use matrix algebra, in which case Math::Matrix may ride to your rescue.
Re: Estimating continuous functions
by flyingmoose (Priest) on Mar 29, 2004 at 16:06 UTC
What kind of function (or curve, whatever) you fit depends on your data. Can you give us a sample data set so that we can suggest appropriate regressions? If it does not follow a set order, you might have to fit a very large polynomial to it and even then it may not be accurate outside the range of the data set -- since a piecewise function is still technically a function, fitting cubic splines may be enough. It depends on the data model and what you are trying to model. Either way, there are a lot of good web references on the requisite math, so with a better understanding of the problem space, we can give you better tips.
Re: Estimating continuous functions
by johndageek (Hermit) on Mar 30, 2004 at 02:14 UTC
Love the problem!
Just a couple of twisted thoughts.
Given a set of data, our first move would be to apply a set of operators to the independent variables, such that the result equals the dependent variable for one line of input data.
Then we apply the formula to the following lines of input, one after another until we get a set of data that fails.
Having successfully negotiated all of the given data, do we assume (yes yes I know) the formula isthe correct one? Or do we need to attempt to find all (sigh) possible formulas that will work out when applied to the data set?
Either way we are left with the sticky problem of trying to know all possible fomulae that will solve a given set of data for a given set of answers. This begins to smack of cryptography.
Some series may not play fair such as:
1 1 2 3 5 8 13 21
o t t f f s s e
would the following fit in as a possible data set?
x y z D
1 4 1 3
4 1 5 4
1 5 9 5
5 9 2 6
"formula" to get x,y and z needs to be blown out for each value.
int( (\$pi*10**(\$d-3)-(int(\$pi*10**(\$d-3)))) *10**3)
Thanks for a fun problem and headache! Good luck.
Enjoy!
Dageek
Re: Estimating continuous functions
by zentara (Archbishop) on Mar 30, 2004 at 13:54 UTC
"Hey Rockie, watch me pull a rabbit out of my hat...oops wrong hat". :-)
If the simultaneous differential equations approach didn't fly well, how about using "fourier transformations?". Since no one has mentioned it yet.
I had a teacher once who claimed you could use fourier analysis to get an "equation of your handwritten signature". He never demonstrated it, but it would seem that it would be done from taking datapoints of the signature.
I just did a groups.google.com search for "fourier equation from data points" and it seems to be a well discussed problem. One of the solutions said to use "Legendre polynomials."
I'm not really a human, but I play one on earth. flash japh
Hmm, thats kind of interesting. Pretty cool.
Re: Estimating continuous functions
by tsee (Curate) on Apr 02, 2004 at 17:31 UTC
If you're trying to approximate a function f: R -> R with a polynomial, which is a pretty limited approach but usually a good start, you could have a look at Math::Approx and Math::Approx::Symbolic.
Steffen
Re: Estimating continuous functions
by Dr.Altaica (Scribe) on Mar 31, 2004 at 22:02 UTC
This sounds perfect for Bayesian networks. While it is fairly easy to find implimentations of Beyesian Networks most are discrete not continuous like you want.
As hossman said "..that for any sets of points, there are an infinte number of curves that fit them, and except in simple cases, there's usually no way to determine that one curve is "more correct" then another..."
With Bayesian dependence modeling you can compare the probabilities of two models. if you know the probability of your data you could get the actual probability of the model. One of the best places for reading more is the B-Course library at http://b-course.hiit.fi/library.html
```|\_/|
/o o\
(>_<)
`-'
```
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Results (192 votes). Check out past polls. | 5,592 | 24,248 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2017-22 | longest | en | 0.937643 |
https://tradingucdt.web.app/mangieri68769syd/yield-vs-rate-function-excel-1459.html | 1,643,362,114,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320305423.58/warc/CC-MAIN-20220128074016-20220128104016-00134.warc.gz | 618,105,430 | 6,292 | ## Yield vs rate function excel
Feb 6, 2020 Understanding a bond's yield to maturity (YTM) is an essential task for fixed income investors. The rate used to discount these cash flows and principal is called the "required rate The formula to price a traditional bond is: flow you receive over the life of the bond (the 5% vs. the required return of 2.5%). Jun 27, 2019 Yield is prospective and should not be confused with the rate of All dates should be entered using the DATE function in Excel rather than as In this section we will see how to calculate the rate of return on a bond investment . If you are comfortable using the built-in time value functions, then this will be a
The yield to maturity formula looks at the effective yield of a bond based on To calculate the actual yield to maturity requires trial and error by putting rates into the Excel is helpful for the trial and error method by setting the spreadsheet so As these calculations show, two bonds with the same maturity will usually have different yields to maturity if the coupons differ. 1The quadratic formula may be Jan 15, 2016 I've had numerous requests to show how the constant yield rate for debt cost amortization is computed in the sample Excel effective interest method calculations. The idea is pretty Before the Goal Seek function is run: Before. Jul 11, 2019 Learn how to calculate the Compound Annual Growth Rate in Excel, The CAGR formula is a way of calculating the Annual Percentage Yield,
## Jun 27, 2019 Yield is prospective and should not be confused with the rate of All dates should be entered using the DATE function in Excel rather than as
Learn how to use Excel's YIELD function for both Mac and PC. Includes numerous formula 1. =YIELD(settlement,maturity,rate,pr,redemption,frequency, basis) Also called annual percentage rate (APR) and annual percentage yield (APY), Excel makes it Effective vs. Read on to learn how to use Excel's EFFECT formula to calculate an effective interest rate (APY) from a nominal interest rate ( APR). The APY (Annual Percentage Yield), or AER (Annual Equivalent Rate) is 7.76%. The compounding effect results in a slightly higher rate than you're quoted by the Jul 16, 2019 The Excel YIELD function calculates the yield to maturity on a bond, Its syntax is YIELD (Settlement, Maturity, Rate, Pr, Redemption, Frequency,
### The par yield is the coupon rate required to produce a bond price I noticed you used Excel's Rate function to calculate YTM to be 6.71% on the Par Yield tab. That bring us to the bond-equivalent yield of 6.88% versus your
The APY (Annual Percentage Yield), or AER (Annual Equivalent Rate) is 7.76%. The compounding effect results in a slightly higher rate than you're quoted by the Jul 16, 2019 The Excel YIELD function calculates the yield to maturity on a bond, Its syntax is YIELD (Settlement, Maturity, Rate, Pr, Redemption, Frequency, Jun 8, 2015 Because yield is a function of price, changes in price result in bond yields moving in the opposite direction. There are two ways of looking at The yield to maturity formula looks at the effective yield of a bond based on To calculate the actual yield to maturity requires trial and error by putting rates into the Excel is helpful for the trial and error method by setting the spreadsheet so As these calculations show, two bonds with the same maturity will usually have different yields to maturity if the coupons differ. 1The quadratic formula may be Jan 15, 2016 I've had numerous requests to show how the constant yield rate for debt cost amortization is computed in the sample Excel effective interest method calculations. The idea is pretty Before the Goal Seek function is run: Before.
### Calculating Yield in Excel. To calculate the YTM of a bond in Excel, you need the following information: Settlement Date: The date when you purchased the security. All dates should be entered using the DATE function in Excel rather than as text. Maturity Date: This is the date when the security will expire.
Calculate the effective annual interest rate or APY (annual percentage yield) calculation for effective rate is similar to Excel function EFFECT(nominal_rate Feb 26, 2019 Yield Curve Building in Excel using Bond Prices (QuantLibXL vs Deriscope) bonds issued by a specific company rather than Libor or swap rates. It exports to Excel hundreds of functions that call internal QuantLib routines. The par yield is the coupon rate required to produce a bond price I noticed you used Excel's Rate function to calculate YTM to be 6.71% on the Par Yield tab. That bring us to the bond-equivalent yield of 6.88% versus your
## YIELD is an Excel function that returns the yield to maturity of a bond given its coupon rate, current price, principal amount and coupon payment frequency per year.. In the context of debt securities, yield is the return that a debt-holder earns by investing in a security at its current price.
How to use the yield function in excel. How to use the yield function in excel. Skip navigation Sign in. Search. Use the RATE Function - Duration: 6:25. Doug H 41,463 views. 6:25. The RATE function is an Excel Financial function that is used to calculate the interest rate charged on a loan or the rate of return needed to reach a specified amount on an investment over a given period. For a financial analyst, the RATE function can be useful to calculate the interest rate on zero coupon bonds. The Excel RATE function is a financial function that returns the interest rate per period of an annuity. You can use RATE to calculate the periodic interest rate, then multiply as required to derive the annual interest rate. The RATE function calculates by iteration. The Yield function is helpful for tracking interest income on bonds. Whereas IRR simply calculates interest rate gains, Yield is best suited for calculating bond yield over a set period of maturity. YIELD is an Excel function that returns the yield to maturity of a bond given its coupon rate, current price, principal amount and coupon payment frequency per year.. In the context of debt securities, yield is the return that a debt-holder earns by investing in a security at its current price. The rate that normalizes this difference is the yield to maturity. Calculating the Yield to Maturity in Excel The above examples break out each cash flow stream by year.
For example, Microsoft Excel and Google Sheets have built-in functions to calculate IRR for both fixed and variable time-intervals; "=IRR()" and "=XIRR()" . The Bond Pricing Calculator Based on Current Market Price and Yield Annual Coupon Rate – The annual coupon rate is the posted interest rate on Then you should use the 'PV' formula (use ';' to separate inputs in OpenOffice, use ',' in Excel). Learn how to use Excel's YIELD function for both Mac and PC. Includes numerous formula 1. =YIELD(settlement,maturity,rate,pr,redemption,frequency, basis) Also called annual percentage rate (APR) and annual percentage yield (APY), Excel makes it Effective vs. Read on to learn how to use Excel's EFFECT formula to calculate an effective interest rate (APY) from a nominal interest rate ( APR). The APY (Annual Percentage Yield), or AER (Annual Equivalent Rate) is 7.76%. The compounding effect results in a slightly higher rate than you're quoted by the Jul 16, 2019 The Excel YIELD function calculates the yield to maturity on a bond, Its syntax is YIELD (Settlement, Maturity, Rate, Pr, Redemption, Frequency, | 1,627 | 7,491 | {"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} | 3.78125 | 4 | CC-MAIN-2022-05 | latest | en | 0.914787 |
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I am designing a binary coded watch with an attiny1616 MCU and an LED watchface. The matrix I designed is displayed below along with the rest of the schematic. The LEDs is 2.1V 30mA and I plan to drive the anode with PWM. The supply voltage is 3V from a CR2032 coin cell. What resistor values do the LEDs need, and will resistors degrade PWM performance? Do I need resistors at all?
• What is the actual current you expect to drive through the LEDs? You don't expect to sink 120mA with a GPIO pin, do you? Or to source 120mA from your coin cell? Jul 24, 2022 at 15:00
• It’s impossible to supply each LED with 30mA average from a CR2032, that would be 600mA, maybe 200x what is available. Even if only one LED is ever intended to appear on at a time, it’s still 5-10x too much. Jul 24, 2022 at 15:02
• Those are the specifications of the LED, the actual current per LED will be much closer to 10mA, would it be better to do this with PWM or with a resistor? The maximum number of LEDs on at once is 14 due to the time encoding. For power savings the watch face will not be on at all times. Jul 24, 2022 at 15:07
• Would it be better to use a li-ion cell for this? Jul 24, 2022 at 15:09
According to the attiny1616 datasheet the absolute max source/sink current of any IO pin is +/-40mA and the total source or the total sink of all IO pins is limited to 100mA.
I am assuming you shall be driving R1 to R4 high, one at a time. With your advertised resistor placement, the brightness of a particular LED on a given channel (eg. on R1) will depend on how many other LEDs are also illuminated on that same channel. This is because they all share the same current limiting resistor. To avoid this problem, you should put the current limiting resistors on the C1-C6 lines instead.
If you do this then you must choose a resistance so that even when all LEDs are on, you don't exceed the maximum source current (40mA) on the R1-R4 pins. Dividing 40mA/6channels = ~6mA allocation to each C1-C6. Looking at the LED datasheet the forward voltage drop at 6mA is about 1.9V. So 180ohm resistors would work well.
Assuming each of the R1-R4 channels are on for 25% of the time. The average LED current would be 6mA x 25% = 1.5mA. That's still enough current to have a bright enough LED for you purposes.
Ensure you run at a high enough frequency to have no flicker (>90 Hz). Perhaps add a 100uF decoupling capacitor to smoothen and reduce the peak currents on you small coin cell battery.
• Yes, modern LEDs are very bright at low currents. Jul 24, 2022 at 15:41
• Thank you for the help, I had a 100uF capacitor that was just a typo, good catch Jul 25, 2022 at 16:16
This is not something you can calculate because it depends on how bright the LEDs look to you and what is acceptable. You can't get anywhere near the rated current.
I suggest you try playing around with a breadboard with a few K and see if that's usable. If you don't have a breadboard to test, multiply the resistor value by 4 (your matrix when properly driven will have a 1:4 duty cycle) and test that brightness. For example, if a 5K resistor was to be deemed suitable for the final unit try a 20K resistor with a CR2032 + LED and see how it looks to you. The time to do this is before you finalize circuit boards and send them off for fabrication and PCBA.
If that's too dim you may have to look at some other power source such as a Li-ion battery, but of course that raises a whole bunch of other considerations (charging and whether you have enough energy for the device to be useable for long enough). Smart watches are pretty clever with how they switch their LEDs (backlight or OLED) to maximize time between charges.
$$R = \frac{V-V_{f}}{I_{f}} = \frac{3-2.1}{0.03} = 30 \Omega$$=
• This can't be a correct answer. While calculation is correct in theory, the battery can't provide 30mA per LED and the MCU absoute maximum ratings would be exceeded if more than one LED is turned on at a time. Jul 24, 2022 at 15:28
• This could work if the firmware ensures that only a single LED is enabled at any given time. Jul 24, 2022 at 17:25
You do not need any resistors, the internal resistance of the battery will be the limiting factor on how much current gets to the LEDs. You can then PWM to control the brightness of the LEDs.
In my experience it is also usually easier to scan the LEDs individually rather than by full row or column. If you try to light more than one LED at once then you will have to compensate for the decreased brightness becuase both LEDs are drawing current at the same time. That chip should be more than fast enough to do this without visual flashing.
Finally, if you do not want to display to get dimmer as the battery voltage drops, you can compensate by measuring the battery voltage and then increasing the LED duty cycle as it goes down.
• This can't be a correct answer. Without a resistor the current can easily exceed the absolute maximum ratings, both for the IO pin and the LED. The battery output voltage would also get clamped to the LED forward voltage and if your MCU needs more than 2.1V to run at the speed you want it will not run. A CR2032 battery can't output the current needed for multiple LEDs at the same time. Jul 24, 2022 at 15:31
• @Justme It definitely works, I've done it on dozens of products that have shipped 10's of thousands of units and never had a problem - try it and see! It not only works - it is better since it gives you a wider dynamic range of possible LED brightnesses and driving an LED at higher current but lower duty cycle is more power efficient. The ATTINY can run down to 1.8V. Jul 25, 2022 at 3:43
• It depends what products those are. For toys it does not matter. For any real product, where reliability matters, you would not leave out the resistors, as you are stressing the components. So even if you have never had an issue in your scenario, please don't suggest other people to use bad practices, as it may not work in their scenario. Jul 25, 2022 at 8:30
• @Justme Can you refer to any literature or empirical data showing that this "stresses" the components or effects reliability? Which components do you think are being stressed and how? What failure modes do you predict? Jul 26, 2022 at 13:37
• A CR2032 battery has about 10-20 ohms internal resistance, and MCU IO driver also might have some 20-30 ohms of driver impedance. It is likely that safe operating area and absolute maximum ratings are exceeded and localized heating might be too much for the IO driver and the LED. Sure, some LEDs are rated for nominal current and allow higher pulsed current given the PWM duty, period and current are within specs. But if the LED is not rated for it, you are overheating it too. So the LED or IO pin can degrade. If the LED Vf is lower than what MCU needs to operate it may reset/hang. Jul 26, 2022 at 14:00 | 1,726 | 6,901 | {"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": 1, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2023-50 | latest | en | 0.930827 |
https://www.coursehero.com/file/7446377/lecture33/ | 1,516,154,175,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084886792.7/warc/CC-MAIN-20180117003801-20180117023801-00543.warc.gz | 872,664,487 | 213,035 | lecture+3.3
# lecture+3.3 - Analytical Solutions to Linear Programming(LP...
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Artificial Variables Analytical Solutions to Linear Programming (LP) Problems
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Plastic Design of a Frame Structure Example: Simplex Solution - Artificial Variables 20’ 15’ 15’ 2 kips 1 kip A B C D E
Model: Augmented Form subject to: and non-negativity conditions: 1 2 3 4 5 6 4 30 15 10 5 2 25 2 25 B B C B C C B C B C P P P P P P P P P P M S M M S M M S M S M M S M M S - = + - = + - = - = + - = + - = Min 30 40 B C P P Z M M = + 4 30 15 10 5 2 25 2 25 B B C B C C B C B C P P P P P P P P P P M M M M M M M M M M + + + + , 0 B C P P M M 1 2 3 4 5 6 Min 30 40 0 0 0 0 0 0 B C P P Z M M S S S S S S = + + + + + + + 1 2 3 4 5 6 , , , , , , , 0 B C P P M M S S S S S S
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Model: Fully Augmented Form Artificial Variable “Large” Coefficient Not a valid initial feasible solution 1 2 3 4 5 6 4 30 15 10 5 2 25 2 25 B B C B C C B C B C P P P P P P P P P P M S M M S M M S M S M M S M M S - = + - = + - = - = + - = + - = 1 2 3 4 5 6 Min 30 40 0 0 0 0 0 0 B C P P Z M M S S S S S S = + + + + + + + 1 2 3 4 5 6 , , , , , , , 0 B C P P M M S S S S S S 1 2 3 4 5 6 Min 30 40 0 0 0 0 0 0 100 B C P P Z M M S S S S S S A = + + + + + + + + 1 2 3 4 5 6 4 30 15 10 5 2 25 2 25 B B C B C C B C B C P P P P P P P P P P M S A M M S A M M S A M S A M M S A M M S A - + = + - + = + - + = - + = + - + = + - + = 1 2 3 4 5 6 , , , , , , , , 0 B C P P M M S S S S S S A 1 2 3 4 5 6 30 15 10 5 25 25 S S S S S S = - = - = - = - = - = - , 0 B C P P M M Big Problem: setting
Model: Fully Augmented Form [ ] 1 2 3 4 5 6 Min 30 40 0 0 0 0 0 0 100 B C P P M M S S Z S S S S A =
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Model: Fully Augmented Form subject to: 1 2 3 4 5 6 4 0 1 0 0 0 0 0 1 30 1 1 0 1 0 0 0 0 1 15 1 1 0 0 1 0 0 0 1 10 0 1 0 0 0 1 0 0 1 5 2 1 0 0 0 0 1 0 1 25 1 2 0 0 0 0 0 1 1 25 B C P P M M S S S S S S A - - - = - - -
Model: Fully Augmented Form and, non-negativity constraints 1 2 3 4 5 6 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 B C P P M M S S S S S S A
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Model: Fully Augmented Form 1 2 3 4 5 6 1 2 3 4 5 6 30 40 0 0 0 0 0 0 100 0 0 0 0 0 0 4 30 15 10 5 2 25 0 2 25 B C B B C B C C B C B C P P P P P P P P P P P P Z M M S S S S S S A M S A M M S A M M S A M S A M M S A M M Z Z Z Z Z Z S A - - - - - - - - - = + - + = + + - + = + + - + = + - + = + + - + = + + - + = 1 2 3 4 5 6 , , , , , , , , 0 B C P P M M S S S S S S A Min Z
Model: Fully Augmented Form 1 2 3 4 5 6 1 2 3 4 5 6 30 40 0 0 0 0 0 0 100 0 0 0 0 0 0 4 30 15 10 5 2 25 0 2 25 B C B B C B C C B C B C P P P P P P P P P P P P Z M M S S S S S S A M S A M M S A M M S A M S A M M S A M M Z Z Z Z Z Z S A - - - - - - - - - = + - + = + + - + = + + - + = + - + = + + - + = + + - + = 1 2 3 4 5 6 1 30 40 0 0 0 0 0 0 100 0 0 4 0 1 0 0 0 0 0 1 30 0 1 1 0 1 0 0 0 0 1 15 0 1 1 0 0 1 0 0 0 1 10 0 0 1 0 0 0 1 0 0 1 5 0 2 1 0 0 0 0 1 0 1 25 0 1 2 0 0 0 0 0 1 1 25 B C P P Z M M S S S S S S A
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lecture+3.3 - Analytical Solutions to Linear Programming(LP...
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Ask a homework question - tutors are online | 2,045 | 3,970 | {"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-2018-05 | longest | en | 0.755062 |
http://stackoverflow.com/questions/11249436/for-statement-not-executing/11249460 | 1,435,637,138,000,000,000 | text/html | crawl-data/CC-MAIN-2015-27/segments/1435375091587.3/warc/CC-MAIN-20150627031811-00100-ip-10-179-60-89.ec2.internal.warc.gz | 246,068,544 | 17,289 | # for statement not executing [closed]
I am writing a program, for homework, that will add 2 8-bit binary numbers. I am to use arrays for storage for the read binary numbers. In the function that would actually do the "addition", it will not execute the for loop. When I step through, it shows the initialization of the variable n for the loop, then it goes straight to the end of loop and exits. Here is my code:
``````for ( int n = 7; n < 0 ; n-- )
{
if ( carry == 0 )
{
if ( bin1[n] == 0 )
{
if ( bin2[n] == 0 )
{
sum[n] = 0;
carry = 0;
}
else
{
sum[n] = 1;
carry = 0;
}
}
else
{
if ( bin2[n] == 0 )
{
sum[n] = 1;
carry = 0;
}
else
{
sum[n] = 0;
carry = 1;
}
}
}
else
{
if ( bin1[n] == 0 )
{
if ( bin2[n] == 0 )
{
sum[n] = 1;
carry = 0;
}
else
{
sum[n] = 0;
carry = 1;
}
}
else
{
if ( bin2[n] == 0 )
{
sum[n] = 0;
carry = 1;
}
else
{
sum[n] = 1;
carry = 1;
}
}
}
}
``````
}
I know this may not be the most efficient way to write this so please avoid those answers.
-
## closed as too localized by Mat, Andrey, dmckee, Kris, hochlOct 18 '12 at 15:00
This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, visit the help center. If this question can be reworded to fit the rules in the help center, please edit the question.
Hmmm...`int n = 7; n < 0` It is doing exactly what you asked. – dmckee Oct 18 '12 at 14:42
Your condition is false from the start, so the loop exits immediately:
``````for ( int n = 7; n < 0 ; n-- )
``````
You probably meant:
``````for ( int n = 7; n >= 0 ; n-- )
``````
-
Attempted the change, though I didn't see the condition being wrong as max = 8 ( i know i didn't post that ), and it still jumps the body. he loop starts reading the array at subscript 7 and goes to 0. – Bizkutz Jun 28 '12 at 17:03
@Bizkutz: The for loop starts with `n` at the initial value 7 and then keeps executing the body while the condition is true. If you leave the initial condition it's clearly not correct, as `n` is initially 7 so n < 0 is false. Isn't going from index 7 to 0 what you wanted? – Tudor Jun 28 '12 at 17:05
my apologies, as I was obviously confusing myself with the condition of the for statement. The change solved my problem, Thank you. – Bizkutz Jun 28 '12 at 17:10
@Bizkutz: No problem. Don't forget to mark the answer as accepted if it solved your problem. :) – Tudor Jun 28 '12 at 17:11
If `n` is initialized to 7, it will never be `< 0`... Your condition is wrong. Should have been `>` or `>=`.
-
Your loop condition is wrong. `n` will never be less than 0. Try this instead:
``````for (int n = 7; n >= 0 ; n--)
``````
-
( `int n = 7; n < 0 ; n-- )`
the < is turned the wrong way. Either do:
``````( int n = 7; n > 0 ; n-- )
``````
or `( int n = 0; n < 7 ; n++ )`
-
If n=7, it is already > 0, so wouldn't that exit the loop immediately? Notice I am using the -- operator to decrement my counter. – Bizkutz Jun 28 '12 at 17:06
The loop executes `while(condition) is true` not `until(condition)`. – Bo Persson Jun 28 '12 at 17:08
I was confusing the condition of the for statement. Thank you. – Bizkutz Jun 28 '12 at 17:11
< means less then. so the statement you orginally had reads: keep looping as long as n is less then 0. And obviously since N starts at 7 it is greater then 0 and the statement will be false. If you turn the sign it will read: keep looping as long as n is greater then 0. And when N starts at 7 it is greater then 0 and the loop will enter. – John Snow Jun 28 '12 at 17:14
Your loop never executes, as 7 is not less than 0. To fix it, just write:
``````for(int n = 7; n >= 0; --n) // note: I use --n, because n--
{ // creates temporary objects
..... // and makes the program slower
}
``````
If doesn't work, write some `printf`s (or `cout`s if you use the iostream library) to see where is the problem
- | 1,256 | 4,079 | {"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-2015-27 | latest | en | 0.889837 |
https://www.thestudentroom.co.uk/showthread.php?t=3935627 | 1,534,518,707,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221212598.67/warc/CC-MAIN-20180817143416-20180817163416-00373.warc.gz | 986,827,633 | 43,775 | You are Here: Home >< Maths
# Algebra Expand and Simplify watch
1. Expand and Simplify
Does anyone know how to work out, this Algebra question out?
If you do it would be much appreciated .
(x+2) (x+3)
2. (Original post by mizliljaz)
Expand and Simplify
Does anyone know how to work out, this Algebra question out?
If you do it would be much appreciated .
(x+2) (x+3)
Can you simplify that?
Image hint
3. Moved to maths.
4. What have you moved to maths and did you get my Algebra sum simplify?
5. I'll explain in detail once I'm at home, give me a minute.
6. (Original post by mizliljaz)
What have you moved to maths and did you get my Algebra sum simplify?
The thread has been moved to Maths since it is maths-related.
7. I like to call this the 'Smiley face method'.
Multiply the lines out - so whichever value the coloured line starts at, multiply it by the value it ends at. These lines will in the same position for any two brackets that are next to each other.
Once you've multiplied each coloured line, add the products together and this will form a quadratic.
Would you like to give it a shot and show me what you get?
8. I got x2 + 5x + 6
Thanks for the "Smiley face method" it was very helpful.
9. (Original post by mizliljaz)
I got x2 + 5x + 6
Thanks for the "Smiley face method" it was very helpful.
Correct!
10. Thanks!
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Top tips from students who have already aced their exams | 553 | 2,154 | {"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-2018-34 | latest | en | 0.939983 |
https://www.aqua-calc.com/one-to-one/density/long-ton-per-liter/milligram-per-us-cup/1 | 1,606,904,601,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141706569.64/warc/CC-MAIN-20201202083021-20201202113021-00146.warc.gz | 576,596,600 | 9,731 | # 1 long ton per liter [long tn/l] in milligrams per US cup
## long tons/liter to milligram/US cup unit converter of density
1 long ton per liter [long tn/l] = 240 384 747 milligrams per US cup [mg/c]
### long tons per liter to milligrams per US cup density conversion cards
• 1
through
25
long tons per liter
• 1 long tn/l to mg/c = 240 384 747 mg/c
• 2 long tn/l to mg/c = 480 769 494 mg/c
• 3 long tn/l to mg/c = 721 154 241 mg/c
• 4 long tn/l to mg/c = 961 538 988 mg/c
• 5 long tn/l to mg/c = 1 201 923 735 mg/c
• 6 long tn/l to mg/c = 1 442 308 482 mg/c
• 7 long tn/l to mg/c = 1 682 693 229 mg/c
• 8 long tn/l to mg/c = 1 923 077 976 mg/c
• 9 long tn/l to mg/c = 2 163 462 723 mg/c
• 10 long tn/l to mg/c = 2 403 847 470 mg/c
• 11 long tn/l to mg/c = 2 644 232 217 mg/c
• 12 long tn/l to mg/c = 2 884 616 964 mg/c
• 13 long tn/l to mg/c = 3 125 001 711 mg/c
• 14 long tn/l to mg/c = 3 365 386 458 mg/c
• 15 long tn/l to mg/c = 3 605 771 205 mg/c
• 16 long tn/l to mg/c = 3 846 155 952 mg/c
• 17 long tn/l to mg/c = 4 086 540 699 mg/c
• 18 long tn/l to mg/c = 4 326 925 446 mg/c
• 19 long tn/l to mg/c = 4 567 310 193 mg/c
• 20 long tn/l to mg/c = 4 807 694 940 mg/c
• 21 long tn/l to mg/c = 5 048 079 687 mg/c
• 22 long tn/l to mg/c = 5 288 464 434 mg/c
• 23 long tn/l to mg/c = 5 528 849 181 mg/c
• 24 long tn/l to mg/c = 5 769 233 928 mg/c
• 25 long tn/l to mg/c = 6 009 618 675 mg/c
• 26
through
50
long tons per liter
• 26 long tn/l to mg/c = 6 250 003 422 mg/c
• 27 long tn/l to mg/c = 6 490 388 169 mg/c
• 28 long tn/l to mg/c = 6 730 772 916 mg/c
• 29 long tn/l to mg/c = 6 971 157 663 mg/c
• 30 long tn/l to mg/c = 7 211 542 410 mg/c
• 31 long tn/l to mg/c = 7 451 927 157 mg/c
• 32 long tn/l to mg/c = 7 692 311 904 mg/c
• 33 long tn/l to mg/c = 7 932 696 651 mg/c
• 34 long tn/l to mg/c = 8 173 081 398 mg/c
• 35 long tn/l to mg/c = 8 413 466 145 mg/c
• 36 long tn/l to mg/c = 8 653 850 892 mg/c
• 37 long tn/l to mg/c = 8 894 235 639 mg/c
• 38 long tn/l to mg/c = 9 134 620 386 mg/c
• 39 long tn/l to mg/c = 9 375 005 133 mg/c
• 40 long tn/l to mg/c = 9 615 389 880 mg/c
• 41 long tn/l to mg/c = 9 855 774 627 mg/c
• 42 long tn/l to mg/c = 10 096 159 374 mg/c
• 43 long tn/l to mg/c = 10 336 544 121 mg/c
• 44 long tn/l to mg/c = 10 576 928 868 mg/c
• 45 long tn/l to mg/c = 10 817 313 615 mg/c
• 46 long tn/l to mg/c = 11 057 698 362 mg/c
• 47 long tn/l to mg/c = 11 298 083 109 mg/c
• 48 long tn/l to mg/c = 11 538 467 856 mg/c
• 49 long tn/l to mg/c = 11 778 852 603 mg/c
• 50 long tn/l to mg/c = 12 019 237 350 mg/c
• 51
through
75
long tons per liter
• 51 long tn/l to mg/c = 12 259 622 097 mg/c
• 52 long tn/l to mg/c = 12 500 006 844 mg/c
• 53 long tn/l to mg/c = 12 740 391 591 mg/c
• 54 long tn/l to mg/c = 12 980 776 338 mg/c
• 55 long tn/l to mg/c = 13 221 161 085 mg/c
• 56 long tn/l to mg/c = 13 461 545 832 mg/c
• 57 long tn/l to mg/c = 13 701 930 579 mg/c
• 58 long tn/l to mg/c = 13 942 315 326 mg/c
• 59 long tn/l to mg/c = 14 182 700 073 mg/c
• 60 long tn/l to mg/c = 14 423 084 820 mg/c
• 61 long tn/l to mg/c = 14 663 469 567 mg/c
• 62 long tn/l to mg/c = 14 903 854 314 mg/c
• 63 long tn/l to mg/c = 15 144 239 061 mg/c
• 64 long tn/l to mg/c = 15 384 623 808 mg/c
• 65 long tn/l to mg/c = 15 625 008 555 mg/c
• 66 long tn/l to mg/c = 15 865 393 302 mg/c
• 67 long tn/l to mg/c = 16 105 778 049 mg/c
• 68 long tn/l to mg/c = 16 346 162 796 mg/c
• 69 long tn/l to mg/c = 16 586 547 543 mg/c
• 70 long tn/l to mg/c = 16 826 932 290 mg/c
• 71 long tn/l to mg/c = 17 067 317 037 mg/c
• 72 long tn/l to mg/c = 17 307 701 784 mg/c
• 73 long tn/l to mg/c = 17 548 086 531 mg/c
• 74 long tn/l to mg/c = 17 788 471 278 mg/c
• 75 long tn/l to mg/c = 18 028 856 025 mg/c
• 76
through
100
long tons per liter
• 76 long tn/l to mg/c = 18 269 240 772 mg/c
• 77 long tn/l to mg/c = 18 509 625 519 mg/c
• 78 long tn/l to mg/c = 18 750 010 266 mg/c
• 79 long tn/l to mg/c = 18 990 395 013 mg/c
• 80 long tn/l to mg/c = 19 230 779 760 mg/c
• 81 long tn/l to mg/c = 19 471 164 507 mg/c
• 82 long tn/l to mg/c = 19 711 549 254 mg/c
• 83 long tn/l to mg/c = 19 951 934 001 mg/c
• 84 long tn/l to mg/c = 20 192 318 748 mg/c
• 85 long tn/l to mg/c = 20 432 703 495 mg/c
• 86 long tn/l to mg/c = 20 673 088 242 mg/c
• 87 long tn/l to mg/c = 20 913 472 989 mg/c
• 88 long tn/l to mg/c = 21 153 857 736 mg/c
• 89 long tn/l to mg/c = 21 394 242 483 mg/c
• 90 long tn/l to mg/c = 21 634 627 230 mg/c
• 91 long tn/l to mg/c = 21 875 011 977 mg/c
• 92 long tn/l to mg/c = 22 115 396 724 mg/c
• 93 long tn/l to mg/c = 22 355 781 471 mg/c
• 94 long tn/l to mg/c = 22 596 166 218 mg/c
• 95 long tn/l to mg/c = 22 836 550 965 mg/c
• 96 long tn/l to mg/c = 23 076 935 712 mg/c
• 97 long tn/l to mg/c = 23 317 320 459 mg/c
• 98 long tn/l to mg/c = 23 557 705 206 mg/c
• 99 long tn/l to mg/c = 23 798 089 953 mg/c
• 100 long tn/l to mg/c = 24 038 474 700 mg/c
• mg/c stands for mg/US c
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JUST LETTUCE, UPC: 071430009543 contain(s) 12 calories per 100 grams or ≈3.527 ounces [ price ]
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CaribSea, Freshwater, Eco-Complete Cichlid, White Sand weighs 1 169.35 kg/m³ (73.00014 lb/ft³) with specific gravity of 1.16935 relative to pure water. Calculate how much of this gravel is required to attain a specific depth in a cylindricalquarter cylindrical or in a rectangular shaped aquarium or pond [ weight to volume | volume to weight | price ]
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#### Weights and Measurements
The sievert fission fragments [Sv fission] is a derived unit of ionizing radiation dose in the International System of Units (SI) and is a measure of the effective biological damage of low levels of ionizing radiation on the human body caused by exposure to nuclear fission fragments.
The speed measurement was introduced to measure distance traveled by an object per unit of time,
troy/yd to oz t/in conversion table, troy/yd to oz t/in unit converter or convert between all units of linear density measurement.
#### Calculators
Calculate volume of a spherical segment and its surface area | 2,587 | 6,358 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2020-50 | latest | en | 0.325253 |
https://www.convertunits.com/from/kilogram/cubic+centimeter/to/ounce/gallon+%5BU.S.%5D | 1,621,383,105,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243989874.84/warc/CC-MAIN-20210518222121-20210519012121-00450.warc.gz | 730,725,709 | 13,038 | ## ››Convert kilogram/cubic centimetre to ounce/gallon [U.S.]
kilogram/cubic centimeter ounce/gallon [U.S.]
Did you mean to convert kilogram/cubic centimeter to ounce/gallon [troy/U.S.] ounce/gallon [troy/U.K.] ounce/gallon [U.S.] ounce/gallon [U.K.]
How many kilogram/cubic centimeter in 1 ounce/gallon [U.S.]? The answer is 7.4891517E-6.
We assume you are converting between kilogram/cubic centimetre and ounce/gallon [U.S.].
You can view more details on each measurement unit:
kilogram/cubic centimeter or ounce/gallon [U.S.]
The SI derived unit for density is the kilogram/cubic meter.
1 kilogram/cubic meter is equal to 1.0E-6 kilogram/cubic centimeter, or 0.13352647136257 ounce/gallon [U.S.].
Note that rounding errors may occur, so always check the results.
Use this page to learn how to convert between kilograms/cubic centimeter and ounces/gallon.
Type in your own numbers in the form to convert the units!
## ››Quick conversion chart of kilogram/cubic centimeter to ounce/gallon [U.S.]
1 kilogram/cubic centimeter to ounce/gallon [U.S.] = 133526.47136 ounce/gallon [U.S.]
2 kilogram/cubic centimeter to ounce/gallon [U.S.] = 267052.94273 ounce/gallon [U.S.]
3 kilogram/cubic centimeter to ounce/gallon [U.S.] = 400579.41409 ounce/gallon [U.S.]
4 kilogram/cubic centimeter to ounce/gallon [U.S.] = 534105.88545 ounce/gallon [U.S.]
5 kilogram/cubic centimeter to ounce/gallon [U.S.] = 667632.35681 ounce/gallon [U.S.]
6 kilogram/cubic centimeter to ounce/gallon [U.S.] = 801158.82818 ounce/gallon [U.S.]
7 kilogram/cubic centimeter to ounce/gallon [U.S.] = 934685.29954 ounce/gallon [U.S.]
8 kilogram/cubic centimeter to ounce/gallon [U.S.] = 1068211.7709 ounce/gallon [U.S.]
9 kilogram/cubic centimeter to ounce/gallon [U.S.] = 1201738.24226 ounce/gallon [U.S.]
10 kilogram/cubic centimeter to ounce/gallon [U.S.] = 1335264.71363 ounce/gallon [U.S.]
## ››Want other units?
You can do the reverse unit conversion from ounce/gallon [U.S.] to kilogram/cubic centimeter, or enter any two units below:
## Enter two units to convert
From: To:
## ››Metric conversions and more
ConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 100 kg, US fluid ounce, 6'3", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more! | 732 | 2,552 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2021-21 | latest | en | 0.595272 |
https://www.physicsoverflow.org/39325/gravitational-waves-of-object-in-superposition | 1,718,395,379,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861568.20/warc/CC-MAIN-20240614173313-20240614203313-00257.warc.gz | 825,781,150 | 22,504 | # Gravitational waves of object in superposition?
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Background
---
I recently read a blog post on measuring gravity waves of objects in superposition:http://backreaction.blogspot.in/2016/03/researchers-propose-experiment-to.html
And did some calculations regarding the same.
Calculations
---
We go into the heisenberg picture to define velocity:
$$\hat v = \frac{dU^\dagger x U}{dt} = U^\dagger\frac{[H,x]}{- i \hbar}U$$
where $\hat v$ is the velocity operator $U^\dagger$ is the unitary operator and $H$ is the Hamiltonian.
Now we can again differentiate to get acceleration $\hat a$:
$$\hat a = \hat U^\dagger\frac{[[\hat H, \hat x], \hat x]}{-i \hbar} \hat U = \hat U^\dagger\frac{(\hat H^2 \hat x + \hat x \hat H^2 - 2\hat H \hat x \hat H)}{\hbar^2} \hat U$$
We can simplify the calculation by splitting the Hamiltonian into potential $\hat V$ and kinetic energy $\hat T$: $\hat H = \hat T + \hat V$
By noticing (one can also calculate this) that the acceleration of an object in a constant potential is $0$:
$$\hat 0 = \hat T^2 x + x \hat T^2 - 2 \hat T \hat x \hat T$$
We also know $[\hat V, \hat x] = 0$ as potential is a function of position. Thus, we can simplify acceleration as:
$$\hat a = \hat V \hat T \hat x + \hat x \hat T \hat V - \hat V \hat x \hat T - \hat T \hat x \hat V$$
Note this acceleration operator also commutes with position:
$$[\hat a , \hat x ]=0$$
Implications
---
Now by the equivalence principle: "acceleration of the object is indistinguishable from gravity." Hence, I argue if the quantum version of the Ricci curvature tensor is measured then the position operator will also collapse and the superposition will exist no more.
Questions
---
Is this line of reasoning correct and also existent in the literature? What do other theories of quantum gravity predict for the gravitational waves for objects in superposition?
Note: I used a similar line of reasoning here (for something else). However, this is a different question.
https://www.physicsoverflow.org/38402/there-any-astrophysical-situation-where-this-idea-testable
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https://search.r-project.org/CRAN/refmans/DescTools/html/StdCoef.html | 1,701,369,172,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100229.44/warc/CC-MAIN-20231130161920-20231130191920-00707.warc.gz | 566,735,796 | 3,199 | StdCoef {DescTools} R Documentation
## Standardized Model Coefficients
### Description
Standardize model coefficients by Standard Deviation or Partial Standard Deviation.
### Usage
StdCoef(x, partial.sd = FALSE, ...)
PartialSD(x)
### Arguments
x a fitted model object. partial.sd logical, if set to TRUE, model coefficients are multiplied by partial SD, otherwise they are multiplied by the ratio of the standard deviations of the independent variable and dependent variable. ... additional arguments passed to coefTable, e.g. dispersion.
### Details
The standardized coefficients are meant to allow for a comparison of the importance of explanatory variables that have different variances. Each of them shows the effect on the response of increasing its predictor X(j) by one standard deviation, as a multiple of the response's standard deviation. This is often a more meaningful comparison of the relevance of the input variables.
Note, however, that increasing one X(j) without also changing others may not be possible in a given application, and therefore, interpretation of coefficients can always be tricky. Furthermore, for binary input variables, increasing the variable by one standard deviation is impossible, since an increase can only occur from 0 to 1, and therefore, the standardized coeffient is somewhat counterintuitive in this case.
Standardizing model coefficients has the same effect as centring and scaling the input variables.
“Classical” standardized coefficients are calculated as \betaᵢ* = \betaᵢ (sₓᵢ / Sᵧ) , where \beta is the unstandardized coefficient, sₓᵢ is the standard deviation of associated depenent variable Xᵢ and Sᵧ is SD of the response variable.
If the variables are intercorrelated, the standard deviation of Xᵢ used in computing the standardized coefficients \betaᵢ* should be replaced by a partial standard deviation of Xᵢ which is adjusted for the multiple correlation of Xᵢ with the other X variables included in the regression equation. The partial standard deviation is calculated as s*ₓᵢ = sₓᵢ √(VIFₓᵢ⁻¹) √((n-1)/(n-p)) , where VIF is the variance inflation factor, n is the number of observations and p number of predictors in the model. Coefficient is then transformed as \betaᵢ* = \betaᵢ s*ₓᵢ .
### Value
A matrix with at least two columns for standardized coefficient estimate and its standard error. Optionally, third column holds degrees of freedom associated with the coefficients.
Kamil Bartoń
### References
Cade, B.S. (2015) Model averaging and muddled multimodel inferences. Ecology 96, 2370-2382.
Afifi A., May S., Clark V.A. (2011) Practical Multivariate Analysis, Fifth Edition. CRC Press.
Bring, J. (1994). How to standardize regression coefficients. The American Statistician 48, 209-213.
coef
### Examples
# Fit model to original data:
fm <- lm(Fertility ~ Agriculture + Examination + Education + Catholic,
data = swiss)
# Partial SD for the default formula:
psd <- PartialSD(lm(data = swiss))[-1] # remove first element for intercept
# Standardize data:
zswiss <- scale(swiss, scale = c(NA, psd), center = TRUE)
# Note: first element of 'scale' is set to NA to ignore the first column 'y'
# Coefficients of a model fitted to standardized data:
# zapsmall(coefTable(stdizeFit(fm, data = zGPA)))
# Standardized coefficients of a model fitted to original data:
# zapsmall(StdCoef(fm, partial = TRUE))
# Standardizing nonlinear models:
fam <- Gamma("inverse")
fmg <- glm(log(Fertility) ~ Agriculture + Examination + Education + Catholic,
data = swiss, family = fam)
psdg <- PartialSD(fmg)
# zGPA <- stdize(GPA, scale = c(NA, psdg[-1]), center = FALSE)
# fmgz <- glm(log(y) ~ z.x1 + z.x2 + z.x3 + z.x4, zGPA, family = fam)
# Coefficients using standardized data:
# coef(fmgz) # (intercept is unchanged because the variables haven't been
# centred)
# Standardized coefficients:
# coef(fmg) * psdg
[Package DescTools version 0.99.51 Index] | 973 | 3,939 | {"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} | 2.765625 | 3 | CC-MAIN-2023-50 | latest | en | 0.897116 |
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# Yale SOM 2013 - Calling All Applicants
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Re: Yale SOM 2013 - Calling All Applicants [#permalink]
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16 May 2013, 13:45
Ding - my status was just updated.
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Re: Yale SOM 2013 - Calling All Applicants [#permalink]
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16 May 2013, 14:21
For whom the bell tolls? It tolls for me. Ding.
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Re: Yale SOM 2013 - Calling All Applicants [#permalink]
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16 May 2013, 15:09
I'm sad there's no good news today.
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Re: Yale SOM 2013 - Calling All Applicants [#permalink]
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16 May 2013, 16:48
hailmary620 wrote:
Got waitlisted at Yale. I was expecting a DING. Super Excited.
I got dinged tt
Congratulations for staying alive.
Were you waitlisted after a real interview or just the wowzer video interview?
I wonder if anybody got in after just taking the video interview.
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Re: Yale SOM 2013 - Calling All Applicants [#permalink]
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16 May 2013, 17:16
I'm in UK and I haven't got any calls or emails about the result of my application. I did not receive a real person interview after the wowzer video one...I guess I must be rejected.........T T
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Re: Yale SOM 2013 - Calling All Applicants [#permalink]
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16 May 2013, 17:21
turbo83 wrote:
hailmary620 wrote:
Got waitlisted at Yale. I was expecting a DING. Super Excited.
I got dinged tt
Congratulations for staying alive.
Were you waitlisted after a real interview or just the wowzer video interview?
I wonder if anybody got in after just taking the video interview.
I only did the video interview; I was actually 99.9% positive that I will get dinged, and was surprised to get waitlisted. Seems like there are a lot of people on the wl, so I don't fancy my chances.
Posted from my mobile device
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Re: Yale SOM 2013 - Calling All Applicants [#permalink]
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16 May 2013, 19:32
llxmh1226 wrote:
I'm in UK and I haven't got any calls or emails about the result of my application. I did not receive a real person interview after the wowzer video one...I guess I must be rejected.........T T
You should check your status on-line. They're updated.
Also, I realize that all the people who got in or got wailisted on round 3 come from a finance background.
Additionally, I don't think it's a coincidence that all the round 2 accepted people (at least 6 people) that I know happened to be finance people.
This is somewhat in line with the fact that the new Dean of SOM just came from Booth (UChicago), which is a hardcore finance school.
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Re: Yale SOM 2013 - Calling All Applicants [#permalink]
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16 May 2013, 20:50
turbo83 wrote:
llxmh1226 wrote:
I'm in UK and I haven't got any calls or emails about the result of my application. I did not receive a real person interview after the wowzer video one...I guess I must be rejected.........T T
You should check your status on-line. They're updated.
Also, I realize that all the people who got in or got wailisted on round 3 come from a finance background.
Additionally, I don't think it's a coincidence that all the round 2 accepted people (at least 6 people) that I know happened to be finance people.
This is somewhat in line with the fact that the new Dean of SOM just came from Booth (UChicago), which is a hardcore finance school.
Well it may not be a coincidence, but I think it's more indicative of who applies to Yale. Yale tends to draw people with an interest in non-profit, finance, and consulting. I don't think we can assume that they are admitting a disproportionate number of finance candidates. Do you yourself work in finance? Perhaps you just have been connected with those types? GMATclub just isn't that representative of the pool at large.
That said, I agree that Yale is probably moving to increase its class primarily in the for-profit arena. Basically they can improve recruiter perspectives about Yale by increasing the number of hires that recruiters make and by ensuring that there are enough bodies at Yale to make it worth their while to send a representative.
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http://www.mybreakaway.com/ Recent post: September 20, "Transitions"
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Re: Yale SOM 2013 - Calling All Applicants [#permalink]
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16 May 2013, 22:46
turbo83 wrote:
llxmh1226 wrote:
I'm in UK and I haven't got any calls or emails about the result of my application. I did not receive a real person interview after the wowzer video one...I guess I must be rejected.........T T
You should check your status on-line. They're updated.
Also, I realize that all the people who got in or got wailisted on round 3 come from a finance background.
Additionally, I don't think it's a coincidence that all the round 2 accepted people (at least 6 people) that I know happened to be finance people.
This is somewhat in line with the fact that the new Dean of SOM just came from Booth (UChicago), which is a hardcore finance school.
I got waitlisted and I don't work in finance. I think Yale is probably one of the most well rounded schools - when I visited Yale, I met people from all sorts of backgrounds (even met a full time ballet dancer and a pro. golfer).
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Re: Yale SOM 2013 - Calling All Applicants [#permalink]
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16 May 2013, 23:21
hailmary620 wrote:
turbo83 wrote:
llxmh1226 wrote:
I'm in UK and I haven't got any calls or emails about the result of my application. I did not receive a real person interview after the wowzer video one...I guess I must be rejected.........T T
You should check your status on-line. They're updated.
Also, I realize that all the people who got in or got wailisted on round 3 come from a finance background.
Additionally, I don't think it's a coincidence that all the round 2 accepted people (at least 6 people) that I know happened to be finance people.
This is somewhat in line with the fact that the new Dean of SOM just came from Booth (UChicago), which is a hardcore finance school.
I got waitlisted and I don't work in finance. I think Yale is probably one of the most well rounded schools - when I visited Yale, I met people from all sorts of backgrounds (even met a full time ballet dancer and a pro. golfer).
Agreed with hm. Sure, I met a good number of folks with finance backgrounds while interviewing, but I don't feel like I met "too many" when meeting admitted/matriculating students. I feel like in every group I met, everyone had really varied interests and any combination of us would lead to a diverse learning team or study group. I do know people though who didn't even consider Yale because they felt it was just a finance school, so positions vary widely...
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Re: Yale SOM 2013 - Calling All Applicants [#permalink]
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17 May 2013, 04:16
turbo83 wrote:
llxmh1226 wrote:
I'm in UK and I haven't got any calls or emails about the result of my application. I did not receive a real person interview after the wowzer video one...I guess I must be rejected.........T T
You should check your status on-line. They're updated.
Also, I realize that all the people who got in or got wailisted on round 3 come from a finance background.
Additionally, I don't think it's a coincidence that all the round 2 accepted people (at least 6 people) that I know happened to be finance people.
This is somewhat in line with the fact that the new Dean of SOM just came from Booth (UChicago), which is a hardcore finance school.
I'm a Chinese student running my own business in UK so actually I don't know what does "ding" mean....sorry for this silly question but could you please tell me what the "ding" is?
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Re: Yale SOM 2013 - Calling All Applicants [#permalink]
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17 May 2013, 11:09
turbo83 wrote:
Also, I realize that all the people who got in or got wailisted on round 3 come from a finance background.
Additionally, I don't think it's a coincidence that all the round 2 accepted people (at least 6 people) that I know happened to be finance people.
This is somewhat in line with the fact that the new Dean of SOM just came from Booth (UChicago), which is a hardcore finance school.
I'm matriculating from R2 and wasn't from a finance background; in fact, most of the people I met during welcome weekend were not from a finance background. I would agree with aerien in that there was a lot of diverse backgrounds in the incoming class as well as the current students I met.
llxmh1226 wrote:
I'm a Chinese student running my own business in UK so actually I don't know what does "ding" mean....sorry for this silly question but could you please tell me what the "ding" is?
ding means they didn't get in and were not on the waitlist either
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Re: Yale SOM 2013 - Calling All Applicants [#permalink]
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17 May 2013, 15:38
After months of waiting, I finally have a working email address (yes, machichi, it took 3 more months after you got yours lol)!! So happy
Next up, actually leaving work...
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Re: Yale SOM 2013 - Calling All Applicants [#permalink]
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17 May 2013, 18:37
aerien wrote:
After months of waiting, I finally have a working email address (yes, machichi, it took 3 more months after you got yours lol)!! So happy
Next up, actually leaving work...
That's awesome! I remember how great I felt when I got my (Sloan) email address!
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Re: Yale SOM 2013 - Calling All Applicants [#permalink]
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22 May 2013, 00:25
kingfalcon wrote:
aerien wrote:
After months of waiting, I finally have a working email address (yes, machichi, it took 3 more months after you got yours lol)!! So happy
Next up, actually leaving work...
That's awesome! I remember how great I felt when I got my (Sloan) email address!
Right?! I feel like there's an appropriate YouTube video of some sort I could post, but it's 3am and why am I awake? Haha
My crazy antics aside, has anyone heard of yield results this year? R3 results on GC were miserable at best (sorry guys ) so not much insight into what movement the WL will see if the same pattern continues. Maybe more movement will come once R3 deposits are due?
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Re: Yale SOM 2013 - Calling All Applicants [#permalink]
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22 May 2013, 06:37
In case anyone was wondering, I emailed Yale to ask for feedback on my application and their reply was:
"After all application rounds are complete, we will begin to field feedback requests. This process will begin in June. Please expect to receive feedback at that time."
Posted from my mobile device
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Re: Yale SOM 2013 - Calling All Applicants [#permalink]
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22 May 2013, 06:57
kingfalcon wrote:
aerien wrote:
After months of waiting, I finally have a working email address (yes, machichi, it took 3 more months after you got yours lol)!! So happy
Next up, actually leaving work...
That's awesome! I remember how great I felt when I got my (Sloan) email address!
Ha, agreed here with my Cornell email address. Once I got it I emailed my mom
It's a pretty exciting feeling... now it's time to sign up for a student amazon prime account IMO
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Re: Yale SOM 2013 - Calling All Applicants [#permalink]
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22 May 2013, 07:51
Does anyone knows when the last desposit deadline for Yale is?
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Re: Yale SOM 2013 - Calling All Applicants [#permalink]
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22 May 2013, 09:36
highwyre237 wrote:
kingfalcon wrote:
aerien wrote:
After months of waiting, I finally have a working email address (yes, machichi, it took 3 more months after you got yours lol)!! So happy
Next up, actually leaving work...
That's awesome! I remember how great I felt when I got my (Sloan) email address!
Ha, agreed here with my Cornell email address. Once I got it I emailed my mom
It's a pretty exciting feeling... now it's time to sign up for a student amazon prime account IMO
Yes! Student discounts FTW!!
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Re: Yale SOM 2013 - Calling All Applicants [#permalink]
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22 May 2013, 12:21
highwyre237 wrote:
kingfalcon wrote:
aerien wrote:
After months of waiting, I finally have a working email address (yes, machichi, it took 3 more months after you got yours lol)!! So happy
Next up, actually leaving work...
That's awesome! I remember how great I felt when I got my (Sloan) email address!
Ha, agreed here with my Cornell email address. Once I got it I emailed my mom
It's a pretty exciting feeling... now it's time to sign up for a student amazon prime account IMO
Ditto to both of those!
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Re: Yale SOM 2013 - Calling All Applicants [#permalink] 22 May 2013, 12:21
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https://helpseeker.net/find-the-next-number-formula-for-2-3-4-6-learn-your-kua-number-for-feng-shui-success/ | 1,674,838,677,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764495001.99/warc/CC-MAIN-20230127164242-20230127194242-00412.warc.gz | 290,566,962 | 11,008 | # Find The Next Number Formula For 2 3 4 6 Learn Your Kua Number For Feng Shui Success
You are searching about Find The Next Number Formula For 2 3 4 6, today we will share with you article about Find The Next Number Formula For 2 3 4 6 was compiled and edited by our team from many sources on the internet. Hope this article on the topic Find The Next Number Formula For 2 3 4 6 is useful to you.
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## Learn Your Kua Number For Feng Shui Success
Many feng shui practitioners use what is called “8 mansion feng shui” or “directional feng shui”. It helps determine your lucky and inauspicious directions based on your Kua number.
How to figure your Kua number
First, add the last two digits of your birth year. If this results in a double-digit number, add again until you get a single-digit number.
Example:
Year of Birth: 1950 Add: 5 + 0 = 5
Year of Birth: 1985 Add: 8 + 5 = 13, 1+3 = 4
Next, if you are male, subtract this number from 10, the result is your cow number.
If you are female, add 5 to this number, and reduce it to one digit again.
Please note that if you were born early in the year (January of February), you need to consult the Chinese lunar calendar, which is based on the formula, to see if you need to subtract a year from your birth.
So now you have your number, what does it mean?
Your personal number will tell you whether you are from the East or the West.
Eastern people have the following numbers: 1, 3, 4 and 9
Westerners have 2, 5, 6, 7 or 8
Lucky directions east
If you are the former, here are your lucky directions. If you live in an East group house (one of these lucky directions), it is good that either its back yard faces the East direction or its front door faces the East group direction: East, Southeast, North, South.
Lucky directions west
If you are a West group person, if you live in a West group house (one of these lucky directions), it is better that either its back yard faces the Lucky West direction or its front door faces the Lucky West group direction: West. , southwest, northwest, northeast
Inauspicious directions
If it is not listed as a lucky direction, then consider it unlucky for you. Try not to place your bedroom in one of your inauspicious directions, and avoid sleeping with your head pointing in that direction. It is also best not to face those directions whenever possible, especially for important business meetings or when balancing your checkbook, paying bills, facing that direction at work, etc.
## Question about Find The Next Number Formula For 2 3 4 6
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## Search keywords Find The Next Number Formula For 2 3 4 6
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Source: https://ezinearticles.com/?Learn-Your-Kua-Number-For-Feng-Shui-Success&id=2473840 | 878 | 3,389 | {"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} | 3.890625 | 4 | CC-MAIN-2023-06 | latest | en | 0.929812 |
https://www.in2013dollars.com/australia/inflation/2010?amount=1&endYear=2006 | 1,611,785,508,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610704833804.93/warc/CC-MAIN-20210127214413-20210128004413-00390.warc.gz | 834,328,779 | 11,376 | # \$1 in 2010 is worth \$0.89 in 2006
\$
## Value of \$1 from 2010 to 2006
\$1 in 2010 is equivalent in purchasing power to about \$0.89 in 2006, an increase of \$-0.11 over 4 years. The dollar had an average inflation rate of 2.84% per year between 2006 and 2010, producing a cumulative price increase of -10.61%.
This means that prices in 2006 are 10.61% lower than average prices since 2010, according to the Bureau of Statistics consumer price index.
The 2006 inflation rate was 3.49%. The inflation rate in 2010 was 2.89%. The 2010 inflation rate is higher compared to the average inflation rate of 1.98% per year between 2010 and 2020.
Cumulative price change -10.61% Average inflation rate 2.84% Converted amount (\$1 base) \$0.89 Price difference (\$1 base) \$-0.11 CPI in 2010 96.100 CPI in 2006 85.900 Inflation in 2006 3.49% Inflation in 2010 2.89% \$1 in 2010 \$0.89 in 2006
AUD Inflation since 1922
Annual Rate, the Bureau of Statistics CPI
## Buying power of \$1 in 2006
This chart shows a calculation of buying power equivalence for \$1 in 2006 (price index tracking began in 1922).
For example, if you started with \$1, you would need to end with \$0.89 in order to "adjust" for inflation (sometimes refered to as "beating inflation").
When \$1 is equivalent to \$0.89 over time, that means that the "real value" of a single Australian dollar decreases over time. In other words, a dollar will pay for fewer items at the store.
This effect explains how inflation erodes the value of a dollar over time. By calculating the value in 2006 dollars, the chart below shows how \$1 is worth less over 4 years.
According to the Bureau of Statistics, each of these AUD amounts below is equal in terms of what it could buy at the time:
Dollar inflation: 2006-2010
Year Dollar Value Inflation Rate
2006 \$1.00 3.49%
2007 \$1.02 2.33%
2008 \$1.07 4.44%
2009 \$1.09 1.74%
2010 \$1.12 2.89%
2011 \$1.16 3.33%
2012 \$1.18 1.71%
2013 \$1.20 2.48%
2014 \$1.24 2.51%
2015 \$1.25 1.51%
2016 \$1.27 1.30%
2017 \$1.29 1.92%
2018 \$1.31 1.26%
2019 \$1.34 1.90%
2020 \$1.36 1.90%*
* Compared to previous annual rate. Not final. See inflation summary for latest 12-month trailing value.
This conversion table shows various other 2006 amounts in 2010 dollars, based on the -10.61% change in prices:
Conversion: 2006 dollars in 2010
Initial value Equivalent value
\$1 dollar in 2006 \$1.12 dollars in 2010
\$5 dollars in 2006 \$5.59 dollars in 2010
\$10 dollars in 2006 \$11.19 dollars in 2010
\$50 dollars in 2006 \$55.94 dollars in 2010
\$100 dollars in 2006 \$111.87 dollars in 2010
\$500 dollars in 2006 \$559.37 dollars in 2010
\$1,000 dollars in 2006 \$1,118.74 dollars in 2010
\$5,000 dollars in 2006 \$5,593.71 dollars in 2010
\$10,000 dollars in 2006 \$11,187.43 dollars in 2010
\$50,000 dollars in 2006 \$55,937.14 dollars in 2010
\$100,000 dollars in 2006 \$111,874.27 dollars in 2010
\$500,000 dollars in 2006 \$559,371.36 dollars in 2010
\$1,000,000 dollars in 2006 \$1,118,742.72 dollars in 2010
## How to Calculate Inflation Rate for \$1, 2006 to 2010
Our calculations use the following inflation rate formula to calculate the change in value between 2006 and 2010:
CPI in 2006 CPI in 2010
×
2010 AUD value
=
2006 AUD value
Then plug in historical CPI values. The Australian CPI was 96.1 in the year 2010 and 85.9 in 2006:
85.996.1
×
\$1
=
\$0.89
\$1 in 2010 has the same "purchasing power" or "buying power" as \$0.89 in 2006.
To get the total inflation rate for the 4 years between 2006 and 2010, we use the following formula:
CPI in 2006 - CPI in 2010CPI in 2010
×
100
=
Cumulative inflation rate (4 years)
Plugging in the values to this equation, we get:
85.9 - 96.196.1
×
100
=
-11%
Politics and news often influence economic performance. Here's what was happening at the time:
• The Copiapo mining accident in Chile ends, after 33 miners resurface having spent 69 days trapped in the ruins.
• Big Haiti earthquake kills 230,000 people and leaves most of Port-au-Prince, its capital, in ruins.
• An explosion on the Deepwater Horizon (a drilling rig), kills 11 people and spills a massive amount of oil into the Gulf of Mexico.
• The US army abolishes the "Don't Ask Don't Tell" policy, which had banned homosexuals from openly serving in the US military.
## Data Source & Citation
Raw data for these calculations comes from the government of Australia's annual (CPI) as provided by the Reserve Bank of Australia. The consumer price index was established in 1922 and is tracked by Australian Bureau of Statistics (ABS).
You may use the following MLA citation for this page: “\$1 in 2010 → 2006 | Australia Inflation Calculator.” Official Inflation Data, Alioth Finance, 27 Jan. 2021, https://www.officialdata.org/australia/inflation/2010?amount=1&endYear=2006.
Special thanks to QuickChart for their chart image API, which is used for chart downloads.
in2013dollars.com is a reference website maintained by the Official Data Foundation. | 1,518 | 4,980 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2021-04 | latest | en | 0.895396 |
https://convertoctopus.com/56-miles-to-decimeters | 1,659,896,499,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882570692.22/warc/CC-MAIN-20220807181008-20220807211008-00072.warc.gz | 190,241,105 | 7,525 | ## Conversion formula
The conversion factor from miles to decimeters is 16093.44, which means that 1 mile is equal to 16093.44 decimeters:
1 mi = 16093.44 dm
To convert 56 miles into decimeters we have to multiply 56 by the conversion factor in order to get the length amount from miles to decimeters. We can also form a simple proportion to calculate the result:
1 mi → 16093.44 dm
56 mi → L(dm)
Solve the above proportion to obtain the length L in decimeters:
L(dm) = 56 mi × 16093.44 dm
L(dm) = 901232.64 dm
The final result is:
56 mi → 901232.64 dm
We conclude that 56 miles is equivalent to 901232.64 decimeters:
56 miles = 901232.64 decimeters
## Alternative conversion
We can also convert by utilizing the inverse value of the conversion factor. In this case 1 decimeter is equal to 1.1095914147095E-6 × 56 miles.
Another way is saying that 56 miles is equal to 1 ÷ 1.1095914147095E-6 decimeters.
## Approximate result
For practical purposes we can round our final result to an approximate numerical value. We can say that fifty-six miles is approximately nine hundred one thousand two hundred thirty-two point six four decimeters:
56 mi ≅ 901232.64 dm
An alternative is also that one decimeter is approximately zero times fifty-six miles.
## Conversion table
### miles to decimeters chart
For quick reference purposes, below is the conversion table you can use to convert from miles to decimeters
miles (mi) decimeters (dm)
57 miles 917326.08 decimeters
58 miles 933419.52 decimeters
59 miles 949512.96 decimeters
60 miles 965606.4 decimeters
61 miles 981699.84 decimeters
62 miles 997793.28 decimeters
63 miles 1013886.72 decimeters
64 miles 1029980.16 decimeters
65 miles 1046073.6 decimeters
66 miles 1062167.04 decimeters | 477 | 1,756 | {"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} | 4.1875 | 4 | CC-MAIN-2022-33 | latest | en | 0.778825 |
http://english.stackexchange.com/questions/39461/higher-greater-or-bigger-distance | 1,469,717,039,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257828283.6/warc/CC-MAIN-20160723071028-00238-ip-10-185-27-174.ec2.internal.warc.gz | 83,556,914 | 20,131 | # Higher, greater or bigger distance?
Which of the following is most correct?
The distance of the shortest path must not be higher than 10km.
The distance of the shortest path must not be greater than 10km.
The distance of the shortest path must not be bigger than 10km.
Or is there an even better word?
-
'Distance' should be replaced by 'length' for idiomatic English, but that doesn't change any of the answers. – TimLymington Aug 26 '11 at 11:09
As height is not being mentioned here, but rather distance on a horizontal scale, "higher" would be inappropriate.
"Bigger" refers to size, not magnitude, and therefore, in this case, is also inappropriate. It's inappropriate because "distance" cannot be measured in size, but in magnitude. "Great length" not "big length".
That leaves, "greater" which is correct.
A better phrasing could be:
The distance of the shortest path must not be more than 10km.
-
what is size if not magnitude? – Matt E. Эллен Aug 26 '11 at 10:35
Size, as in amount. Length is not an amount, but a measure of dimensions – Thursagen Aug 26 '11 at 10:38
I disagree with your definitions. However I agree that greater is the better word to use. – Matt E. Эллен Aug 26 '11 at 10:46
According to this NGram, "greater distance" is the most common:
Next after this is "further", as in "I ran further than he did". You could also say:
The distance of the shortest path must not be further than 10km.
However, I would suggest using "greater" as it seems to fit the context better.
-
"greater distance" suggests that something is farther away. I don't think the path is far away, it is long or short. The end of the path might be far away. So I wouldn't say "The distance of X must not be further than....". "Distance of" isn't idomatic, IMO. – Mr. Shiny and New 安宇 Aug 26 '11 at 12:33
While I agree that "greater than" is correct, I'm not sure about the rest of the sentence. Two places have a distance between them, but does a path have a distance? I don't think it does.
The length of the shortest path must not be greater than 10km
or even
The shortest path must not be longer than 10km
-
+1 For "longer". – Gurzo Aug 26 '11 at 11:25
+1 For "The shortest path must not be longer than 10km." I feel that this is the most accurate, appropriate way to express the idea in the question. Or alternatively, "The shortest path must be no longer than 10km." – Kit Z. Fox Aug 26 '11 at 11:34
+1 for "paths don't have distance". – Mr. Shiny and New 安宇 Aug 26 '11 at 12:32
"must not be over 10km" even simpler – z7sg Ѫ Sep 16 '11 at 12:35
The distance of the shortest path must not be greater than 10km.
This one is correct.
-
Shouldn't you go for longer distance? E.g. the distance is much longer than 2 miles.
- | 739 | 2,748 | {"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-2016-30 | latest | en | 0.966743 |
https://bizfluent.com/how-7464663-calculate-food-cost-restaurant.html | 1,603,946,382,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107902745.75/warc/CC-MAIN-20201029040021-20201029070021-00277.warc.gz | 241,827,795 | 31,195 | # How to Calculate Food Cost in a Restaurant
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Share It
If you own a restaurant, you know the importance of calculating food cost for such things as proper pricing and inventory. Proper pricing allows you to price your dishes at profitable levels, and accurate inventory allows you to assess the value of current food levels for such things as business valuation. Additionally, your food costs represent more than just the cost of food. The cost of your food includes the cost of everything it takes to sell a plate of food, so you must factor employee costs and overhead costs into your food costs since without them, you wouldn't have any food to sell. Not knowing how to calculate food costs could threaten your restaurant's viability or throw your business value off.
Make a checklist of each menu item.
Add together all the single-item foods such as steaks, hamburger patties or buns by dividing the total cost of the the case amount by the number of units per case. For instance, restaurants usually buy food items by the case, so if you purchase a case of 24 burgers for \$18, then your per-serving food cost for burgers equals .75 because \$18 divided by 24 equals .75. The per-serving cost for single item foods represents the minimum amount you must charge for that item to break even.
Calculate per-serving food cost for bulk ingredients like condiments or vegetables by dividing the volume cost of your food by the number of servings per volume. For instance, if the total volume of a bottle of mustard offers 300 servings and the total cost for the volume equals \$10.95 then your per-serving cost for mustard equals 3.65 cents because \$10.95 divided by 300 servings equals .0365 per serving. The per-serving cost for bulk ingredients represents the amount you must add to each menu item containing that bulk item to break even.
Calculate total drink costs by adding together the amounts you paid for individual "drink units" such as bottles or cans of soda, juice, milk or beer. The cost of each drink item represents the minimum amount you must charge to break even.
Calculate bulk drink costs by adding together the amounts you paid for individual case unit such as bottles or cans of soda, juice, milk or beer. Divide the cost of the case by the number of units in the case. For instance, if you purchased 24 cans of tomato juice for \$4.00, then your bulk drink costs for tomato juice equals .16 because \$4.00 divided by 24 equals .16. The cost of each item represents the minimum amount you must charge to break even.
Calculate the per-serving cost for drink ingredients like soda syrup used in fountain drinks by dividing the total price for the syrup by the average number of servings yielded per syrup canister. For example, if your syrup canisters costs \$35, and the canister rates approximately 2,000 12-ounce cups of soda, your per-serving cost for soda syrup equals .175 because \$35 divided by 2,000 equals .175. The per-serving cost for your drink ingredients represents the minimum amount of money you must charge per drink to break even.
#### Tips
• To make a profit, multiply all your per-serving food and drink costs by a factor of four or six. Such markup allows you to cover overhead costs like labor or building rent. | 684 | 3,301 | {"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} | 3.546875 | 4 | CC-MAIN-2020-45 | longest | en | 0.923035 |
https://brainmass.com/math/ring-theory/division-subrings-division-ring-subset-37699 | 1,708,590,490,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473735.7/warc/CC-MAIN-20240222061937-20240222091937-00296.warc.gz | 148,545,956 | 7,302 | Purchase Solution
# Division subring S = division ring D or S is a subset of D
Not what you're looking for?
Modern Algebra
Division Ring
Let D be a division ring, Z its center and let S be a division subring of D which is stabilized by every map
x→dxd-1, d ≠ 0 in D. Show that either S = D or S is a subset of D.
See the attached file.
##### Solution Summary
This solution is comprised of a detailed explanation of the problems of division rings. It contains step-by-step explanation for the problem.
##### Solution Preview
Modern Algebra
Division Ring
...
Solution provided by:
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Some questions on probability | 309 | 1,442 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2024-10 | latest | en | 0.889177 |
http://www.gurufocus.com/term/deb2equity/PPL/Debt%2Bto%2BEquity/PPL%2BCorp | 1,490,814,921,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218191353.5/warc/CC-MAIN-20170322212951-00274-ip-10-233-31-227.ec2.internal.warc.gz | 536,593,029 | 27,661 | Switch to:
GuruFocus has detected 6 Warning Signs with PPL Corp \$PPL.
More than 500,000 people have already joined GuruFocus to track the stocks they follow and exchange investment ideas.
PPL Corp (NYSE:PPL)
Debt-to-Equity
1.95 (As of Dec. 2016)
PPL Corp's current portion of long-term debt for the quarter that ended in Dec. 2016 was \$1,441 Mil. PPL Corp's long-term debt for the quarter that ended in Dec. 2016 was \$17,808 Mil. PPL Corp's total equity for the quarter that ended in Dec. 2016 was \$9,899 Mil. PPL Corp's debt to equity for the quarter that ended in Dec. 2016 was 1.94.
A high debt to equity ratio generally means that a company has been aggressive in financing its growth with debt. This can result in volatile earnings as a result of the additional interest expense.
Definition
Debt to Equity measures the financial leverage a company has.
PPL Corp's Debt to Equity Ratio for the fiscal year that ended in Dec. 2016 is calculated as
Debt to Equity = Total Debt / Total Equity = (Current Portion of Long-Term Debt + Long-Term Debt) / Total Equity = (1441 + 17808) / 9899 = 1.94
PPL Corp's Debt to Equity Ratio for the quarter that ended in Dec. 2016 is calculated as
Debt to Equity = Total Debt / Total Equity = (Current Portion of Long-Term Debt + Long-Term Debt) / Total Equity = (1441 + 17808) / 9899 = 1.94
* All numbers are in millions except for per share data and ratio. All numbers are in their local exchange's currency.
Explanation
In the calculation of Debt to Equity, we use the total of Current Portion of Long-Term Debt and Long-Term Debt divided by Total Equity. In some calculations, Total Liabilities is used to for calculation.
Be Aware
Because a company can increase its Return on Equity by having more financial leverage, it is important to watch the leverage ratio when investing in high Return on Equity companies.
Related Terms
Historical Data
* All numbers are in millions except for per share data and ratio. All numbers are in their local exchange's currency.
PPL Corp Annual Data
Dec07 Dec08 Dec09 Dec10 Dec11 Dec12 Dec13 Dec14 Dec15 Dec16 deb2equity 1.24 1.68 1.42 1.63 1.72 1.92 1.73 1.39 2.01 1.95
PPL Corp Quarterly Data
Sep14 Dec14 Mar15 Jun15 Sep15 Dec15 Mar16 Jun16 Sep16 Dec16 deb2equity 1.56 1.39 1.56 1.93 1.93 2.01 2.03 1.94 1.92 1.95
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https://mattermodeling.stackexchange.com/tags/supercell/hot | 1,722,861,561,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640447331.19/warc/CC-MAIN-20240805114033-20240805144033-00218.warc.gz | 304,615,721 | 26,106 | # Tag Info
## Hot answers tagged supercell
### Using LAMMPS for a small system
There are a couple of potential problems. First, the assumption of periodic simulations is that your box size is large enough that there is no effect on an atom from its nearest periodic replica. I ...
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### Extracting the coordinates of a super cell in Vesta or Avogadro
One option is to use the Atomic Simulation Environment's sort function. The function ase.build.sort() takes a required Atoms ...
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### ASE: why do I get warnings about "monoclinic" and "orthorhombic" not being interpreted?
As far as I understand, this warning can be safely ignored. While parsing a cif structure, the get_spacegroup() function in the cif parser of ASE, checks whether ...
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### Indirect to direct band gap transition in SUPERCELLS?
This is allowed, in fact you can almost always construct a supercell where this has to happen. Consider the band eigenvalue at k-point $\vec{k}=(0,0,\frac{1}{p})$ in a given unit cell; if I now make a ...
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### For a single unit cell calculation, is it necessary to add a vacuum?
In a single unit cell calculation, the inclusion of vacuum space depends on the specific requirements and characteristics of your system. Here are a few situations that you should consider about ...
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### How to get rotation matrix elements from root integer notation?
$\sqrt{3}$ Case $\left( \sqrt{3} \times \sqrt{3}\right)R30$ is a short way of representing a supercell in Woods notation (1, 2), which essentially means that each of the basis vectors of the ...
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### How to sort a supercell for a molecular crystal?
I don't know if this is exactly what you want to do, but pymatgen may be able to help (disclosure, I am a developer of this package). See this example code below ...
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### Energy per unit cell in NWChem differs for different supercells despite the same k-points density
It turns out, that NWchem do not support K-points with Gaussian basis. The nwpw section in my input files, that contains the K-point grid specification, actually ...
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### What is the meaning of the "PRIMCELL.vasp" file generated by VASPKIT tool during bandstructure inputs generation?
I have never used VASPKIT, so my answer is of a general nature. If you have a $4\times4\times1$ supercell of a perfect crystal, you have redundant information ...
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### Supercell model for a disordered system in phonon calculations using the harmonic approximation
If you have a system without periodicity like your disordered solid solution, then you should use a $1\times1\times1$ $\mathbf{q}$-point grid for a phonon calculation (equivalent to a $1\times1\times1$...
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### Different band structure in supercell, but not due to band folding
Answer: band folding is a little more subtle than I originally thought. I'd only really come across band folding in the textbook example of 1D and rather naively assumed that in higher dimensions if ...
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### Density of states does not scale properly with supercell
When the size of the supercell increases, you can reduce the number of k-points in an inverse manner to get the same accuracy as the one with the smaller cell. This is because the relation between ...
### Help understanding square root notation of supercell
If you want to achieve this using VESTA you can follow what Shahid said in his answer but you have to keep in mind how lattice vectors are defined in VESTA which in turn will change the rotation ...
### How to make left and right handed material interface using python
CIF2Cell (only about generating supercells) CIF2Cell is a tool to generate the geometrical setup for various electronic structure codes from a CIF (Crystallographic Information Framework) file. The ...
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### What is the meaning of the "PRIMCELL.vasp" file generated by VASPKIT tool during bandstructure inputs generation?
I assume that you are considering 2D materials. In short, you can use the undoped primitive cell to generate a k-path to plot the band structure of your $4\times4\times1$ supercell with doping. Why ...
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### How to construct thin film structure of desired thickness in quantum espresso?
You can't construct the thin film itself with Quantum Espresso, but only calculate it's properties. You first have to create the unit cell of the thin film with some other software. As an example here ...
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### Transform Supercell to Unitcell
ASE You could use ASE (Atomic Simulation Environment) to achieve this. Here is a sample code. I have tried the below code on three CIF files which can be found on the Crystallography Open Database ...
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### Errors in CASTEP supercell geometry optimization
The error message says that the computer did not have enough memory to perform the calculation (line 2 of the log you photographed). You don't give any details about your calculation, but in general ...
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### Kpt boundaries for cubic supercell
Ciao Marco! Once you make the supercell the reciprocal lattice vectors are divided by 2 (or whatever factor) automatically. The kptbounds are in reduced coordinates so no need to change the bound. ...
### VESTA using command line
In line with the comments, I'm not sure if you can achieve this in VESTA, but a simple ASE or Pymatgen based script can easily do it. This is the Pymatgen script I use. ...
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### What happens to the position coordinates of the nearest neighbors in supercell studies?
DFT calculations based on a supercell method utilize the periodicity of a perfect, infinite lattice by imposing periodic boundary conditions onto the wavefunction of the system (Bloch's theorem). In ...
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### Extracting the coordinates of a super cell in Vesta or Avogadro
This is in Avogadro 1.2.0. I will check behavior on Avogadro 2 later: File -> Open -> Choose your cif file Build -> Super Cell Builder -> Choose the number of repeats File -> Save As -&...
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• 1. Heat and Temperature THERMODYNAMICS
• 2. HEAT HEAT is energy (thermal) in transit, energy that transfers from one body to another because of temperature difference. Heat flows from an area of high temperature to an area of low temperature (unless an external work is done). Heat is measured with the following SI units: joules (J); calorie (cal); kilocalorie (kcal or Cal) where: 1 cal = 4.184J and 1 Cal = 1000cal
• 3. SOURCES OF HEAT A. NATURAL SOURCES 1. the Sun 2. interior of the Earth B. ARTIFICIAL SOURCES 1. chemical action 2. mechanical energy 3. electrical energy 4. nuclear energy
• 4. THERMAL ENERGY AND TEMPERATURE THERMAL ENERGY - also called Internal Energy - the total energy in a system or body; the sum of kinetic and potential energy of the atoms or molecules of a body TEMPERATURE - measure of the degree of hotness or coldness of a body - the average kinetic energy of the molecules of a body
• 5. Touching - Very easy to use - However, it is unreliable and has limitations
• 6. *Liquid-in-glass *thermocouple *Rotary *Liquid crystal thermometer
• 7. - A liquid inside a glass tube expands and contracts with a change in temperature
• 8. - Uses circuits to measure changes in electric current due to changes in temperature
• 9. - Uses a coiled bimetallic strip
• 10. - Uses a strip of liquid crystals that changes color with temperature
• 11. *Fahrenheit, F *Celsius, C *Kelvin, K
• 12. METHODS OF HEAT TRANSFER 1. CONDUCTION – heat transfer through solids *Thermal conductivity – ability of a material to allow heat to pass through * Kinds of material based on thermal conductivity: a. conductors – allow heat to pass through b. insulators – resist heat transfer 2. CONVECTION – heat transfer through fluids (liquids and gases), which involves currents/flow 3. RADIATION – heat transfer through empty space, through electromagnetic radiation, such as light, microwave, etc.
• 13. HEAT TRANSFER | 590 | 2,208 | {"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-2014-52 | latest | en | 0.818347 |
http://www.pokerlistings.com/omaha-poker-betting-rules | 1,516,194,962,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084886939.10/warc/CC-MAIN-20180117122304-20180117142304-00168.warc.gz | 514,500,645 | 14,162 | # Omaha Poker Betting Rules: No-Limit, Limit, Pot-Limit
Thanks to the action and excitement of Pot-Limit Omaha, Omaha has become the second most popular form of poker in the world.
Omaha can be played with a variety of betting structures, but only two are used with any regularity:
• Pot-Limit
• Fixed Limit
Omaha and No-Limit
Although it's technically possible to play Omaha as a No-Limit game, this combination is extremely rare. The rules and game-play of Omaha are such that the game requires rules for betting that have some semblance of structure.
In other words, the game is so draw- and action-laden that it doesn't function particularly well as No-Limit.
For this reason, players wishing to play a high-action Omaha game turn to Pot-Limit as their go-to betting structure.
### (Video) How to Play Pot-Limit Omaha
Pot-Limit Omaha
The popularity of Pot-Limit Omaha has surged recently, to the point that this particular Omaha variation is now the second most played poker variant both online and live. In fact, it's not uncommon for 100% of the night's online high-stakes action to take place over Omaha tables.
1. Betting proceeds clockwise from the button. The player to the left of the button is the small blind and the player on his left is the big blind. The player on his left is under the gun, and acts first.
2. His options are to call the big blind, raise or fold.
3. Your minimum bet is equal to the size of the big blind (this is assuming no players have bet before you on this betting round).
4. To determine the maximum bet, count all the money in the pot and all the bets on the table, including any call you would make before raising. (It sounds more complicated than it really is.) Two examples for you:
• You're first to act on the flop with a pot of \$15. You have the option to check or bet. You can bet anywhere from as little as the amount of the big blind, to the full amount of the pot (\$15). Any bet in between is a "legal bet."
• You're second to act on the flop with a pot of \$15. The first player bets \$10. You now have the option to fold, call (\$10) or raise.
5. Your minimum raise is equal to the amount of the previous bet. In this hand your minimum raise is \$10 (\$10 + \$10 for a total bet of \$20).
6. Your maximum raise is the amount of the pot. To figure this out, add up the pot + the bet + your call (\$15 + \$10 + \$10 = \$35). You are allowed to bet that total amount in addition to your call, meaning your total bet is \$45 (\$10 for the call + \$35 for the size of the pot).
7. You can raise any amount in between the minimum and the maximum raise amount.
8. The size of the game is determined by the blind size. The buy-in is usually minimum 20 big blinds and maximum 100 big blinds.
Fixed-Limit Omaha
1. In Limit Omaha the betting limits are fixed.
2. The size of the game is determined by the bet size. For example, in a \$4/\$8 game the small bet is \$4 and the big bet is \$8. The blinds would be \$2 and \$4.
3. Play proceeds as it does in any community card game, with the blinds to the left of the button and the play proceeding clockwise.
4. Betting and raising are done in increments.
5. Before the flop betting works in increments of the small bet; \$4 in our example. A bet would be equal to \$4, a raise would be to a total of \$8.
6. On the turn and river betting works in increments of the big bet; \$8 in our example. A bet would be equal to \$8, a raise would be equal to \$16.
The Limit betting structure puts a cap on the number of raises. Most venues allow a maximum of a bet and three raises, although some rooms have a cap of four raises.
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kris 2012-06-25 17:11:57
\$1-\$2 plo, if there's \$3 in the pot pre flop how can someone first to act come in and raise to \$15. Could that player have bet \$50 instead. I'm confused on the betting.
John 2012-02-22 06:55:28
pot limit 2-5 game, 8 players, if the all call what would pot be to the 7th player if every one calls, to the 8th, player, to the small blind, and finally to the big blind.
m klians 2011-10-30 23:18:38
please send the betting pattern for pot limit omaha
mel klians 2011-10-30 23:16:01
please send the rules for pot limit omaha & how to figure it out
thanks
Rick 2011-01-06 18:10:40
In Omaha no limit game. If a player calls raise, but does not say the amount, the next to act says All In. Is the player calling all in bound to that call. Even though the player who called Raise didn't call the amount.
Nabeel 2010-02-20 22:26:18
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https://extranet.education.unimelb.edu.au/SME/TNMY/Decimals/Decimals/teaching/models/lab.htm | 1,708,464,887,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473347.0/warc/CC-MAIN-20240220211055-20240221001055-00064.warc.gz | 268,472,226 | 6,134 | Linear Arithmetic Blocks (LAB)
Our preferred model for decimals
Description of LAB Features of LAB How to make LAB Using LAB as a model of the numberline Using the organiser Why we prefer LAB to MAB Sample lesson
### Description of LAB
Linear Arithmetic Blocks - our concrete model of choice for teaching decimals!
Numbers are represented by length of pieces of plastic pipe. The longest piece, representing "one" is just over a metre long. This piece is shown to the students first. Discuss cutting into 10 equal pieces. Ask students to use their hands to indicate how long the new piece will be, and what name it should have (a tenth).
Now show the students a tenth piece and discuss cutting it into 10 equal pieces. Again, ask students to predict the length of the new piece and a name (a hundredth).
Now show the students the hundredth piece and discuss cutting it into 10 equal pieces. Again, ask students to predict the length of the new piece and a name (a thousandth).
Now show the students the thousandth piece. We use washers for thousandths because it is too hard to cut such thin pieces of plastic pipe. Then discuss cutting the washer into 10 equal pieces, to make ten-thousandths and beyond to smaller and smaller pieces so that students understand the process that creates the endless base ten chain. In practice, we make nothing smaller than thousandths.
LAB pieces can be arranged randomly in piles or in 2 systematic ways. On the right, they are shown on an organiser. The organiser is a wooden stand with three dowel rods to hold nine tenths, hundredths and thousandths. This is very useful to demonstrate trading in the addition and subtraction algorithms. In the 3 photos below, the pieces are laid end to end to confirm that:
10 tenths have the same length as the one piece 10 hundredths have the same length as the tenth piece 10 thousandths have the same length as the hundredth piece
### Features of LAB as a model
Uses the physical quantity of length to represent the size of a number. Represents size of number from the digits in the numeral very well. Represents base 10 properties (bundling and column overflow) very well. Useful from ones to thousandths and can be mentally extended easily in both directions. (This is an important discussion to have with students!) The "endless base ten chain" multiplicative relations between the values of places are shown reasonably well. Demonstrates addition and subtraction algorithms well. Demonstrates multiplication and division of a decimal by a small whole number or by a power of ten well. Only some divisions by a decimal can be shown well (selected quotitions). This is a general limitation of concrete models. Multiplication by a decimal cannot be easily demonstrated with LAB. LAB represents numbers by the quantity of length, not by units of length such as millimetres. This is an important distinction because using units of length may perpetuate misconceptions that the decimal point simply separates one whole-number quantity (the number of metres) from another (the number of millimetres). It happens that the unit piece is approximately one metre long, but this is only a consequence of the size of suitable materials.
### How to make LAB
Linear Arithmetic Blocks (LAB) can be made at home or at school from ordinary washers and PVC pipe of a similar diameter. To make LAB, first purchase the washers (thousandths). The next pieces (hundredths) are then made by cutting small lengths of plastic tubing to exactly to the length of 10 washers. Then make tenths (medium size lengths of tubing ) cut to 10 hundredths. Lastly, the one (whole) is cut to 10 tenths and is a rather long piece of tube, probably over a metre long (as the washer may well be more than 1 mm thick). A good set for classroom demonstration requires about 7m of 25 mm diameter plastic pipe and contains about 40 thousandths about 30 tenths and hundredths at least two ones. Children's sets can have fewer pieces. The wooden organiser is optional. It is made of 3 rods (tenths, hundredths and thousandths) set into a base of wood. A decimal point can be drawn on the base with texta. The heights of the rods on the organiser are such that only 9 of the pieces can be placed on the relevant rod. It is obviously not practical to make a rod for the ones, as it would be over 9 metres high! (Involve students in this discussion) Approximately 1.5 m of dowel is required and a base measuring approximately 30cm x 15cm x 10cm.
### Using LAB as a model of the numberline
In the pictures below, the LAB pieces representing the decimals have been laid linearly instead of on the organiser. This allows us to compare the total length of the pieces and hence the size of the decimals they represent. LAB clearly shows that 0.2 is larger than 0.13 and not the other way around! Many longer-is-larger children think that 0.13 is larger than 0.2 (as 13 is larger than 2). It also shows other properties clearly including: equivalence of 0.2 and 0.20 (2 tenths and 20 hundredths) equivalence of 0.13 (13 hundredths) with 0.1 + 0.03 (one tenth and 3 hundredths) density of decimal numbers (that there are other decimals between 0.24 and 0.25 or between 0.247 and 0.248 etc) LAB can be used to round decimal numbers. For example, to round 0.27 to the nearest tenth, make several numbers using only the tenths pieces (1 tenth, 2 tenths, 3 tenths, 4 tenths...) and compare. Children can see that the number 0.27 is between 2 tenths and 3 tenths and closer to 0.3. Many children and even some adults are confused by numberlines. Building a number line is the ideal way to help students understand the ideas involved.
### Using the organiser
The organiser serves a similar purpose to the place value chart often used with MAB, holding pieces of the same size together and representing the left-right spatial arrangement of decimal numeration. (Beware - you must be looking from the front!) It also if there are ten or more pieces of any one size, they will not fit onto the appropriate rod and so ten of them must be exchanged for a single piece of the next highest value. This forces the trading required in the algorithms.
Overuse of the organiser or place value chart may reduce the LAB or MAB model (in the children's minds) to that of an abacus; where it is the position of the beads only which represents the size of the number. To ensure that students do not become reliant on the organiser (or place value chart), make sure that students:
appreciate that the size of the number represented does depend on the physical length of the blocks and not their arrangement on a chart, and are able to move the blocks onto the organiser (by themselves) to assist in reading off the symbolic representation.
### Why we prefer LAB to MAB
Our research has demonstrated that LAB has a number of advantages over MAB.
Our published research paper (Stacey et al, 2001a) compares the two materials on the basis of epistemic fidelity (how true the model is to the mathematical principles involved) and accessibility for students. Two teaching experiments involving 30 matched students indicated that LAB is considerably more accessible for students. There are three reasons for this:
students get confused with MAB simply because it has been used before with the "mini" representing one (see above); LAB models number with length whereas MAB models number with volume and many students in upper primary do not yet have a strong grasp of volume; the various pieces of MAB seem to be of different dimensions (1-D, 2-D, 3-D ) and this makes generalizing to more place value columns difficult.
Use of LAB was associated with more active engagement by students and deeper discussion. Epistemic fidelity is critical to facilitate teaching with the models, but we attribute the enhanced classroom environment to the greater accessibility of the LAB material.
Both models have excellent epistemic fidelity, so that they both show how the size of numbers depends on the digits and the place value columns and they both can be used to demonstrate the operations. However, a significant difference between the two models is that LAB, with pieces laid end to end, has structural similarity to the number line, and is thus better able than MAB to model number density (the property that between any two decimals, a third decimal can always be inserted). LAB is therefore better able to demonstrate the principles of rounding. The limitations of MAB in regard to the continuous properties of decimals were noted by Hiebert, Wearne and Taber (1991). Following instruction with MAB, Year 4 students' performance on discrete-context tasks (e.g. writing the number represented by a picture of MAB; choosing the larger of two decimals) improved. However their performance on continuous-representation tasks (e.g. shading 2.6 of a continuous quantity; finding a number between two decimals) did not. Thus MAB appeared to support understandings of decimals as discrete quantities but not as continuous quantities.
Top
### Sample lesson plan
LAB was first shown to us by Heather McCarthy, a Melbourne teacher. Heather had seen it at an in-service day, but we cannot trace the source. Heather's worksheets.
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Assignment: Globalization of World Food Markets
Analyze the impact food-borne illnesses, genetic engineering and the organic food movements have on the global food markets. Instructions: Choose a topic from the list below to investigate further. Write a 2-3 page paper (APA format) on this topic and include a separate reference page. You may also submit a topic related to globalization to your instructor for approval.
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• The impact of globalization on food choices and health.
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https://housemetric.co.uk/house-price-analysis/BN3-5/Hove | 1,723,262,271,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640789586.56/warc/CC-MAIN-20240810030800-20240810060800-00003.warc.gz | 239,034,507 | 17,261 | House prices in BN3 5 (Hove)
This article reveals price per square metre data and various charts to help you understand current housing market in 'BN3 5' (Hove) - statistics were last calculated on 31 July 2024.
Defining 'BN3 5'
This analysis is limited to properties whose postcode starts with "BN3 5", this is also called the postcode sector. It is shown in red on the map above. There are no official names for postcode sectors so I've just labelled it Hove.
You can click on the map labels to change to a neighbouring sector, or you can enter a different postcode sector (e.g. CM23 4) below.
FYI, a postcode sector is the full postcode without the last two letters.
Price per square metre
Knowing the average house price in Hove is not much use. However, knowing average price per square metre can be quite useful. Price per sqm allows some comparison between properties of different size. We define price per square metre as the sold price divided by the internal area of a property:
£ per sqm = price ÷ internal area
E.g. 32, Linton Road, Hove, sold for £730,000 on Jun-2024. Given the internal area of 118 square metres, the price per sqm is £6,186.
England & Wales have been officially metric since 1965. However house price per square foot is prefered by some estate agents and those of sufficiently advanced age ;-) They may want to convert square meters on this page to square feet.
The chart below is called a histogram, it helps you see the distribution of this house price per sqm data. To make this chart we put the sales data into a series of £ per sqm 'buckets' (e.g. £5,600-£6,000, £6,000-£6,400, £6,400-£6,800 etc...) we then count the number of sales with within in each bucket and plot the results. The chart is based on 227 sales in Hove (BN3 5) that took place in the last two years.
Distribution of £ per sqm for Hove
Distribution of £ per sqm house prices in Hove
You can see the spread of prices above. This is because although internal area is a key factor in determining valuation, it is not the only factor. Many factors other than size affect desirability; these factors could be condition, aspect, garden size, negotiating power of the vendor etc.
The spread of prices will give you a feel of the typical range to expect in Hove (BN3 5). Of the 227 transactions, half were sold for between £5,580 and £6,840 per square metre. The median, or 'middle', price per square metre in 'BN3 5' is £6,180. Notably, only 25% of properties that sold recently were valued at more than £6,840 sqm. For anything to be valued more than this means it has to be more desireable than the clear majority of homes.
Price map for Hove
Do have a look at the interactive price map I created. I find it useful and I am sure it will help you in exploring Hove. You can zoom in all the way to individual properties and then all the way back out to see the whole country. The colours show the current estimated property values.
House price heatmap for Hove
Comparison with neighbouring postcode sectors
The table below shows how 'BN3 5' compares to neighbouring postcode sectors.
Postcode sector Lower quartile Middle quartile Upper quartile
BN3 1 Hove £5,000 sqm £5,640 sqm £6,380 sqm
BN3 7 Hove £4,690 sqm £5,490 sqm £6,280 sqm
BN3 8 Hove £4,760 sqm £5,450 sqm £6,310 sqm
BN3 6 Hove £5,520 sqm £6,220 sqm £7,420 sqm
BN3 5 Hove £5,580 sqm £6,180 sqm £6,840 sqm
BN3 3 Hove £4,900 sqm £5,540 sqm £6,330 sqm
BN3 4 Hove £5,070 sqm £5,860 sqm £7,020 sqm
BN3 2 Hove £4,990 sqm £5,800 sqm £6,580 sqm
Will Hove house prices drop in 2024?
I cannot tell the future and don't believe anyone who says they can. I can however plot price trends - I have done this in the chart below for BN3 5 (Hove) compared with both the wider area BN3 and inflation (CPIH from the Office of National Statistics). The dashed trend lines in the chart show the average over time.
Historic price per square metre in Hove,Hove
House price trends for Hove
For the most recent sales activity, rather than a summarized average, it is better to see the underlying data. This is shown in the chart below, where blue dots represent individual sales, click on them to see details. If there is an obvious trend you should be able to spot it here amid the noise from outliers.
Most recent BN3 5 sales
Recent trends for Hove
Data from Land Registry comes in gradually over time. I update it every month but it takes about 5 months for the majority of sales for Hove to be recorded. Disclaimer: I do not verify and cannot guarantee the accuracy of any data shown. Outliers exist in the data, typically these are where the EPC registry records the internal area incorrectly, sometimes although very rarely the Land Registry price paid data can be wrong. The data provided throughout this website about Hove and any other area, is not financial advice. Any information provided does not and cannot ever take in to account the particular financial situation, objectives or property needs of either you or anyone reading this information.
Street level data
Street Avg size Avg £sqm Recent sales
Wordsworth Street, Hove, BN3 5B 86 sqm £6,226 27
Coleridge Street, Hove, BN3 5A 86 sqm £5,641 26
Montgomery Street, Hove, BN3 5B 82 sqm £6,283 24
Bolsover Road, Hove, BN3 5H 74 sqm £6,205 21
Portland Road, Hove, BN3 5D 86 sqm £4,866 20
Byron Street, Hove, BN3 5B 87 sqm £6,342 20
Grange Road, Hove, BN3 5H 67 sqm £6,098 20
Cowper Street, Hove, BN3 5B 82 sqm £6,604 19
Raw data
Our analysis of Hove is derived from what is essentially a big table of sold prices from Land Registry with added property size information. Below are three rows from this table to give you an idea. | 1,516 | 5,655 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2024-33 | latest | en | 0.938326 |
https://studysoup.com/tsg/647531/elementary-linear-algebra-with-applications-9-edition-chapter-5-problem-5 | 1,653,470,627,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662584398.89/warc/CC-MAIN-20220525085552-20220525115552-00379.warc.gz | 608,064,381 | 13,713 | ×
Get Full Access to Elementary Linear Algebra With Applications - 9 Edition - Chapter 5 - Problem 5
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# The Cauchy- Schwarz inequality implies that the inner product of a pair of \"ectors is
ISBN: 9780132296540 301
## Solution for problem 5 Chapter 5
Elementary Linear Algebra with Applications | 9th Edition
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Problem 5
The Cauchy- Schwarz inequality implies that the inner product of a pair of \"ectors is less than or equal to theproduct of their lengths.
Step-by-Step Solution:
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L19 - 6 Derivatives of General Exponential Functions d x Recall (e )= dx d (b )= or b> 0 dx Writef(x)= b =x −1 ex. Find the slope of the tangent line to f(x)=2 sec(x) √ at x = 2.
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## Discover and learn what students are asking
Calculus: Early Transcendental Functions : Basic Differentiation Rules and Rates of Change
?Finding a Derivative In Exercises 3–24, use the rules of differentiation to find the derivative of the function. $$y=x^{7}$$
Calculus: Early Transcendental Functions : Introduction to Functions of Several Variables
?In Exercises 7-18, find and simplify the function values. $$g(x, y)=\int_{x}^{y} \frac{1}{t} d t$$ (a) (4,1) (b) (6,3)
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?A test is conducted at the = 0.05 level of significance. What is the probability of a Type I error?
#### Related chapters
Unlock Textbook Solution | 471 | 1,750 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2022-21 | latest | en | 0.819111 |
https://nobeltextiles.com/qa/question-what-is-the-difference-between-gain-and-directivity-of-an-antenna.html | 1,603,213,238,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107874026.22/warc/CC-MAIN-20201020162922-20201020192922-00696.warc.gz | 468,933,662 | 9,312 | # Question: What Is The Difference Between Gain And Directivity Of An Antenna?
## What is the highest DBI Antenna?
This powerful 2.4 GHz dipole antenna has a whopping 22 dB of gain.
This is one of the most powerful omni-directional dipole antenna available..
## What is antenna gain and directivity?
Directivity is the measure of the concentration of an antennas’s radiation pattern in a particular direction. Directivity is expressed in dB. The higher the directivity, the more concentrated or focussed is the beam radiated by an antenna. … Gain is the product of directivity and efficiency.
## Is higher antenna gain better?
The higher the dBi number of the antenna, the higher the gain, but less of a broad field pattern, meaning that the signal strength will go further but in a narrower direction, as illustrated in the diagram below.
## Does antenna gain affect reception?
When transmitting, a high-gain antenna allows more of the transmitted power to be sent in the direction of the receiver, increasing the received signal strength. … As a consequence of their directivity, directional antennas also send less (and receive less) signal from directions other than the main beam.
## How is antenna directivity calculated?
Directivity is the ratio of the radiation intensity in a given direction from the antenna to the radiation intensity averaged over all directions (IEEE 1993, p. 362). Notes: (1) The average radiation intensity is equal to the total power radiated by the antenna divided by 4p (area of sphere in steradians).
## What is VSWR in antenna?
VSWR (Voltage Standing Wave Ratio), is a measure of how efficiently radio-frequency power is transmitted from a power source, through a transmission line, into a load (for example, from a power amplifier through a transmission line, to an antenna).
## What is dB in antenna?
Antenna Definitions. Antenna Theory Home. Many parameters related to antennas are measured in decibels; for instance, gain is often specified in decibels, written as 10 dB. Or maybe the minimum received power for an antenna system to work is specified as -70 dBm (decibels relative to a milliWatt).
## What is the gain of a dipole antenna?
When mounted horizontally, the radiation peaks at right angles (90°) to the conductor, with nulls in the direction of the dipole. Neglecting electrical inefficiency, the antenna gain is equal to the directive gain, which is 1.5 (1.76 dBi) for a short dipole, increasing to 1.64 (2.15 dBi) for a half-wave dipole.
## What does the beamwidth of an antenna tell us?
Definition of Half Power Beamwidth The 3 dB, or half power, beamwidth of the antenna is defined as the angular width of the radiation pattern, including beam peak maximum, between points 3 dB down from maximum beam level (beam peak).
## What is 3dB gain of antenna?
A 2dB or 3dB gain antenna is the compromise in suburban and general settings. A 5dB gain antenna radiates more energy toward the horizon (compared to the 0, 2, and 3dB antennas) to reach radio communication sites that are further apart and less obstructed.
## Where is horn antenna used?
Horn antennas are commonly used as the active element in a dish antenna. The horn is pointed toward the center of the dish reflector. The use of a horn, rather than a dipole antenna or any other type of antenna, at the focal point of the dish minimizes loss of energy (leakage) around the edges of the dish reflector.
## What is antenna gain formula?
Antenna gain is usually defined as the ratio of the power produced by the antenna from a far-field source on the antenna’s beam axis to the power produced by a hypothetical lossless isotropic antenna, which is equally sensitive to signals from all directions.
## What is a good antenna gain?
The gain of a real antenna can be as high as 40-50 dB for very large dish antennas (although this is rare). Directivity can be as low as 1.76 dB for a real antenna (example: short dipole antenna), but can never theoretically be less than 0 dB.
## How far will a 9dBi antenna reach?
Antenna BasicsAntennaTypeMax Range9dbiOmni directional1200ft9dBi PanelDirectional.25 miles11dBiOmni directional.25 miles14dbiOmni directional.4 miles10 more rows
## What is the directivity of antenna?
In electromagnetics, directivity is a parameter of an antenna or optical system which measures the degree to which the radiation emitted is concentrated in a single direction.
## What is the value of directivity of an isotropic antenna?
The directivity of an antenna is defined as the power density of the antenna in its direction of maximum radiation in three-dimensional space divided by its average power density. The directivity of the hypothetical isotropic radiator is 1 or 0 dB. The directivity of a half-wave dipole antenna is 1.64 or 2.15 dB.
## How will you increase the gain of an antenna?
Therefore, antennas primarily increase their gain by concentrating the signal over a smaller area. For example, an omnidirectional antenna that transmits and receives signal in all directions will generally have a smaller gain then a directional antenna that transmits and receives a signal in only one direction.
## What is the highest gain TV antenna?
TV Antenna UFO X 12V – Extremely High Gain Omni Directional TV Antenna. The UFO X is the latest in digital technology and high performance for TV Reception in any mobile environment. It has the highest gain of any TV antenna for its tiny 24cm Diameter. | 1,212 | 5,466 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2020-45 | latest | en | 0.922869 |
https://www.studypool.com/discuss/504867/the-longer-leg-of-a-30-60-90-triangle-is-18-what-is-the-length-of-the-other?free | 1,495,488,546,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463607120.76/warc/CC-MAIN-20170522211031-20170522231031-00490.warc.gz | 931,033,028 | 14,220 | ##### The longer leg of a 30°-60°-90° triangle is 18. What is the length of the other
Mathematics Tutor: None Selected Time limit: 1 Day
The longer leg of a 30°-60°-90° triangle is 18. What is the length of the other leg?
a.6√3
b.9√3
c.9
d.12√3
Apr 30th, 2015
Let the length of the shorter leg be 'x'.
The hypotenuse of the 30°-60°-90° is 2x
--> --> --> -->
Since the length of a segment cannot be a negative number, the only solution is
Please let me know if you have any other questions and best me if you are satisfactory.
Apr 30th, 2015
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Apr 30th, 2015
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Apr 30th, 2015
May 22nd, 2017
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check_circle | 231 | 733 | {"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-2017-22 | longest | en | 0.828937 |
https://ncatlab.org/nlab/show/%28n+%C3%97+k%29-category | 1,723,196,703,000,000,000 | application/xhtml+xml | crawl-data/CC-MAIN-2024-33/segments/1722640762343.50/warc/CC-MAIN-20240809075530-20240809105530-00235.warc.gz | 338,409,857 | 9,063 | # nLab (n × k)-category
Contents
### Context
#### Higher category theory
higher category theory
# Contents
## Idea
An $(n \times k)$-category (read “n-by-k category”) is an n-category internal to the $(k+1)$-category of $k$-categories. The term is “generic” in that it does not specify the level of strictness of the $n$-category and the $k$-category.
For example:
• A $(1 \times 0)$-category, as well as a $(0 \times 1)$-category, is precisely a category. More generally, $(n\times 0)$-categories and $(0\times n)$-categories are precisely $n$-categories.
• A $(1 \times 1)$-category is precisely a double category (either strict or weak).
• Generalizing to a 3rd axis, a $(1 \times 1 \times 1)$-category is precisely a triple category, that is, a category internal to (categories internal to categories), i.e. a catgory internal to double categories, or a double category internal to categories — which again could be strict or weak.
• An $(n \times 1)$-category is what Batanin calls a monoidal n-globular category?.
An $(n \times k)$-category has $(n + 1)(k + 1)$ kinds of cells.
Under suitable fibrancy conditions, a $(n \times k)$-category will have an underlying $(n + k)$-category (where here, $n + k$ is to be read arithmetically, rather than simply as notation). Fibrant $(1 \times 1)$-categories are known as framed bicategories.
## Examples
• Commutative rings, algebras and modules form a symmetric monoidal $(2 \times 1)$-category.
• Conformal nets form a symmetric monoidal $(2 \times 1)$-category.
## Relationships
At least in some cases, if the structure is sufficiently strict or sufficiently fibrant, we can shift cells from $k$ to $n$. For instance:
• A sufficiently strict $(1 \times 2)$-category canonically gives rise to a $(2 \times 1)$-category. (Cor. 3.11 in DH10)
• Any double category (i.e. a $(1\times 1)$-category) has an underlying 2-category.
• A sufficiantly fibrant $(2\times 1)$-category has an underlying tricategory (i.e. $(3\times 0)$-category).
## References
• Mike Shulman, Constructing symmetric monoidal bicategories, arXiv preprint arXiv:1004.0993 (2010)
• Michael Batanin, Monoidal globular categories as a natural environment for the theory of weak $n$-categories , Advances in Mathematics 136 (1998), no. 1, 39–103.
The following paper contains some discussion on the relationship between various (weak) $(n \times k)$-categories for $n, k \leq 3$.
There is some discussion on this n-Category Café post as well as this one.
Last revised on March 29, 2024 at 23:26:36. See the history of this page for a list of all contributions to it. | 728 | 2,605 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 33, "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} | 3.0625 | 3 | CC-MAIN-2024-33 | latest | en | 0.763938 |
https://daxos.org/how-many-4-digit-combinations-are-there-using-0-9/ | 1,669,612,631,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710473.38/warc/CC-MAIN-20221128034307-20221128064307-00407.warc.gz | 236,120,545 | 13,543 | # How many 4 digit combinations are there using 0-9?
(*4*)
## How many 4 digit combinations are there using 0-9?
10,000 imaginable combinations
There are 10,000 conceivable combinations that the digits 0-9 can also be arranged into to form a four-digit code.
## How many ways can 0-9 numbers be organized?
How many quantity combinations are there? There are 10,000 possible combinations that the digits 0-9 can be arranged into to shape a four-digit code.
How many different 3 digit PIN codes using best the digits 0 9 are possible?
By comparison, this 3-dial lock (three wheels, each and every with digits 0-9) has 10 × 10 × 10 = 1, 000 imaginable combinations. The total selection of combinations isn’t very different, but the Simplex Lock is way harder to pick as a result of it is harder to systematically check each and every possible combination.
### How many different combinations of Nine numbers are there?
3265920
Therefore, the whole selection of probabilities is given by means of 9×9! =3265920. Hence, 9 digit numbers of different digits will also be formed in 3265920 tactics.
### How many 5 digit combinations are there using 0-9?
a hundred and five ways
Now, there are one hundred and five tactics during which the digits 0-9 can be selected for the 5 puts of a 5 digit quantity. Out of these, 104 start with 0 (once we get started with 0, there are simplest 4 slots to fill, where we now have 10 alternatives each and every).
How many imaginable 4Digit combinations are there between 0 and 9?
The 1st digit has 10 imaginable alternatives (0–9), the 2d digit has 9 possible choices (because one number has already been used for the first digit), the 3rd digit has Eight alternatives and the 4th digit has 7 choices.
#### How to calculate the choice of possible combinations?
To calculate the number of possible combinations of n non-repeating elements from a set of r forms of elements, the components is: If the elements can repeat in the mixture, the components is: In each formulas “!” denotes the factorial operation: multiplying the series of integers from 1 up to that number.
#### How many conceivable combinations are there in NCR?
It will checklist all imaginable combinations, too! However, be aware that 792 other combinations are already slightly numerous to turn. To keep away from a situation the place there are too many generated combinations, we limited this combination generator to a definite, most number of combinations (2000 by default).
How to find the collection of diversifications in a mix generator?
In fact, if you already know the selection of combinations, you’ll easily calculate the choice of diversifications: P (n,r) = C (n,r) * r!. If you switch on the complex mode of this mix calculator, you will be able to find the collection of diversifications. You would possibly wonder when you should utilize permutation instead of a combination.
in | 661 | 2,912 | {"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-2022-49 | latest | en | 0.91013 |
https://www.vedantu.com/question-answer/the-element-chromium-exists-in-a-bcc-lattice-class-12-chemistry-cbse-5fd6f2d7aef1aa270cdd876c | 1,726,699,578,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651941.8/warc/CC-MAIN-20240918201359-20240918231359-00322.warc.gz | 970,695,578 | 29,556 | Courses
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# The element Chromium exists in a bcc lattice with unit cell edge $2.88 \times {10^{ - 10}}m$. The density of Chromium is $7.2 \times {10^3}Kg\;{m^{ - 3}}$. How many atoms does $52 \times {10^{ - 3}}kg$ of chromium contain?
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Hint: As we know that by knowing the dimensions of a unit cell, we can calculate the volume of the unit cell, similarly by knowing the density of a unit cell we can calculate the mass and number of atoms in the unit cell. We also know that the bcc structure of a unit cell possesses eight lattice points at eight corners and one additional lattice point at the centre of the body.
As we know that each unit cell is adjacent to another and most of the atoms are shared by neighbouring unit cells due to which only some portion of each atom belongs to a particular unit cell.
So, by knowing the dimensions of a unit cell, we can calculate the volume of the unit cell, similarly by knowing the density of a unit cell we can calculate the mass and number of atoms in the unit cell.
Now, we are given that the edge length of bcc lattice of chromium elements is $2.88 \times {10^{ - 10}}m$.
And the density of the chromium is given as $7.2 \times {10^3}Kg\;{m^{ - 3}}$.
Let us recall the formula to calculate the density of the given unit cell which is $d = \dfrac{{Z \times M}}{{{a^3} \times N}}$ where $N$, is the Avogadro ’s number is in general but here we need to calculate the number of atoms which will be given by the $N$corresponding to the molecular mass of the chromium. And mass of chromium is given $52 \times {10^{ - 3}}kg$.
And we know that for a body centred cubic unit cell the number of atoms per unit cell or the value of $Z$ is $2$.
So, Put all the given values in the above formula we will get:
$N = \dfrac{{Z \times M}}{{{a^3} \times d}}$
$\Rightarrow N = \dfrac{{2 \times 52 \times {{10}^{ - 3}}}}{{{{(2.88 \times {{10}^{ - 10}})}^3} \times 7.2 \times {{10}^3}}}$
$\Rightarrow N = 6.04 \times {10^{23}}atoms$
Therefore the correct answer is $6.04 \times {10^{23}}atoms$.
Note: Remember that the body centred cubic arrangement contains $8$ lattice point at the corners and each one of these is shared by $8$cells and another lattice point is completely inside the unit cell or present at the centre of the body thus the number of points or atoms per unit cell of bcc is $8 \times \dfrac{1}{8} + 1 = 2$, therefore the value of $Z = 2$. Similarly, the value of ccp arrangement can be calculated which comes out to be $Z = 4$. | 741 | 2,635 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2024-38 | latest | en | 0.847362 |
https://engineering.electrical-equipment.org/forum/users/steven-mill/replies | 1,627,908,771,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154320.56/warc/CC-MAIN-20210802110046-20210802140046-00429.warc.gz | 248,466,652 | 15,544 | # Steven Mill
## Forum Replies Created
Viewing 3 posts - 1 through 3 (of 3 total)
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Posts
• in reply to: Power System Operation and Control #13100
Steven Mill
Participant
@guest said:
very interesting article…bt tell me abt fuzzy logic and lfc
Thanks. I’ll write an article about LFC, you could read it at the beginning of next week.
in reply to: Power System Operation and Control #12995
Steven Mill
Participant
Hello, you’re welcome!
Jeff Williams, sure next week I’ll post an article about AGC first.
in reply to: Stability Factor: K-Factor #13000
Steven Mill
Participant
Spir Georges GHALI, sorry for the late reply and thank you for your interesting remarks.
@Spir Georges GHALI said:
Dear;
It’s very good and important to mention the effects of Harmonics on the Transformers, but I have some remarks that are :
– I think the percentages of Harmonics currents mentioned in this topic mean the percentage of the “ THD-I ” ( Total Harmonic Distortion of Current ). If not, please clarify.
Yes these percentages are of THD. Sorry If that confused you.
– It’s known that the effect of K-Factor is to over-size the Transformer’s power, but what is the formula between the K-Factor and the Transformer’s power ?
Normally in order to calculate the K-Factor you keep into account two things the KVA & load Harmonic Current. So accordingly the K-Factor would be:
K – Factor = KVA * IL
As IL is the load harmonic content so you can easily calculate the K-Factor value to be used.
– You mention that “ the inductive loads like Motors are known as harmonic generating loads ”, but most of motors especially “ Squirrel Cage Motors ”, and after the transient running case are always “ Linear Loads ”, but for these loads the Harmonic currents can be generated or not depending on the kind of running equipment used to run these motors. ( for exp. Contactors with Thermal relays doesn’t generate any harmonic, VFD equipment generates harmonics ).
I was specifically referring to induction motors which are widely used in industry as compared to squirrel cage motors which are most suited for small tasks. Very rare have I seen squirrel cage motors used in industry. Induction motors generate harmonics and come in the domain of non-linear loads. Also in industries the Switch mode power supplies & DC rectifiers also add to the non-linearity of the system.
– It’s mentioned that “ the working principal of K-rated Transformers involves the use of a double sized neutral conductor ”, but in general, depending on the Harmonics Currents value or percentage, we over-size “ double or even more ” the cables’ sections of Phases & Neutral “, and also depending on the value or percentage of the “ Third multiple Harmonics ” and the “ Unbalance current’s value ” we decide the final section of neutral conductor.
Yes you are right about conductor size. It depends on the harmonic currents value. I also wrote in the article that Some K-rated transformers use more than one conductor. Normally you don’t need a transformer with K-factor rating more than K-4 or K-13 in industries which are coherent to the design specs presented in the article.
– As mentioned, the K-Factor Value’s range is from “ 1 ” to “ 50 ”, and also, the rules mentioned the K-Factor Transformers that should be used depending on the Harmonic Current, ( exp. for Harmonics Currents more than 75% the “ K-20 Transformers ” should be used ) but, at which levels or where the values “ K-30 or K-35 Transformers ” or more should be used ?
Transformer size increase as you increases the value of K-Factor & so does the heat tolerance due to harmonic currents. Normally transformers of K-Factor 30, 40 & 50 are used where harmonic currents are somewhere b/w 125-150%. These are worst harmonic conditions. As mentioned before the size of transformer increases with the increase in K-Factor so the optimal K-Factor should be chosen which would be a tradeoff between transformer size & its heat tolerance. Normally the K-50 transformers are very expensive & much larger in size so manufacturers recommend a transformer of K-30 with modified designs as a replacement.
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# Inventions in Everything: Drilling Square Holes
Friday, November 18, 2016 2:34
Have you ever needed to cut a square or rectangular hole into drywall? Such as you might need if you needed to install the electrical box for a light switch?
For most people, that’s something that would take drilling at least two round holes at the opposite corners of the square or rectangle you intend to cut, large enough to insert the blade of a saw, which you would then proceed to use very carefully and very slowly to cut out a square or rectangular plug from the drywall. When you’re done cutting the hole, in addition to a hole in your wall in the rough shape of a square or rectangle and a drywall plug that you might have accidentally dropped into the wall, you might also have tears or broken drywall to repair in addition to a coating of gypsum dust to clean up from yourself, the wall and the floor.
Wouldn’t it be cool if you could have just cut a perfectly square or rectangular hole in the wall to do the job more efficiently, more effectively and, let’s be honest, better?
Now you can! Inventor Teklemichael Sebhatu has a U.S. patent pending tool to do the job. (Via Core77):
The future has gotten to be a lot cooler than it used to be! And by future, we mean 2017:
The QuadSaw was patented and developed by Genius IP and is set for release in the UK in the summer of 2017. The company is said to have models sized for the electrical boxes used in the US. The tool is expected to sell for £199 (about \$220 USD). That’s a lot to spend for a hole cutting attachment but could be worth it to the electrician who regularly installs old work (retrofit) boxes in drywall.
Just in case you’re looking for ideas of what to get such a person for their birthday or for Christmas next year!
Source: http://politicalcalculations.blogspot.com/2016/11/inventions-in-everything-drilling.html
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## What does pro-rata mean?
(20 Posts)
Fri 14-Sep-12 12:40:00
As in when applying for a part time role the salary is £15,000 pro-rata?
Fri 14-Sep-12 12:42:03
if you did it full time you'd get £15K but you'll get whatever proportion of that you work.
So if FT was 40 hours and you work 20 you'd get £7.5K
WinklyFriedChicken Fri 14-Sep-12 12:42:34
If a person working full time (say, 40 hours a week), then a PT person gets the amount of that salary proportionate to the hours worked.
For example - full time - 40 hours = £15k
PT - 20 hours - 50% of the hours so 50% of the salary so £7.5k
Bluegrass Fri 14-Sep-12 12:42:43
It means that if the salary for full time is 15k and you worked for, say 3 out of 5 days a week you would get 3/5 ths of the full time salary.
It means that if you were working fulltime (whatever THAT means!) you would earn £15k, but it's a part-time job so you earn less. Eg, if fulltime = 40 hours a week and you get £15k, if the job is actually 20 hours you'll earn £7.5k.
What always bugs me about those job ads is that they don't usually tell you what "full time" actually means, 40 hours, 37.5, 35....
Fri 14-Sep-12 12:43:15
Probably that the salary would be £15k for a full-time job and that you would be paid the right proportion for the proportion of hours you worked - eg if you worked 3 out of the 5 days you would get 3/5 of the money (9k) but would need to see the exact words and also need to be sure that they meant what the ad actually said.
Bluegrass Fri 14-Sep-12 12:43:26
Beaten to it!
Lol, lots of x posts, why did so many of us pick 40 hours?
Fri 14-Sep-12 12:44:38
It means if you were working full-time (39hrs?) that's what you'd earn. If the the actual role was, for example, 13hrs pw what you'd actually earn would be £5000 - third of the hours, third of the money.
Fri 14-Sep-12 12:44:56
I thought it was something like that. Why can't they just say what they're actually paying for the advertised hours? Grrr. I'm not getting frustrated trying to find a job no not at all
WinklyFriedChicken Fri 14-Sep-12 12:45:27
Cos it's easy to divide into halves and quarters - 35 is a pest for dividing
Also in theory I do a 40 hour week. (HA.)
Fri 14-Sep-12 12:46:07
I was so slow! Lots of x posts there
Fri 14-Sep-12 12:47:59
Is there an easy way to work out what the actual pay would be? My brain is not working properly today.
WinklyFriedChicken Fri 14-Sep-12 12:52:23
What are the full time & part time hours? (part time/full time) x full time salary should do it.
Fri 14-Sep-12 12:52:31
According to Yahoo answers, divide actual hours by full-time hours then multiply by salary.
Rockchick1984 Fri 14-Sep-12 13:05:01
Or divide salary by full time hours, then multiply by part time hours
Fri 14-Sep-12 13:38:37
Thanks all
Juliehar Mon 26-Sep-16 21:01:55
If a job is advertised as 40 weeks a year at a pro rata salary of £15,000 can someone tell me how to work out what the weekly and monthly wage would be please?????
Mon 26-Sep-16 21:23:45
40/52*15000 = £11538 yearly
Divided over 12 months £961.50 per month
Juliehar Mon 26-Sep-16 21:49:31
Thank you so much X X
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http://learnmathblog.com/can-you-solve-it-pythagorass-best-puzzles.html/ | 1,513,353,048,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948575124.51/warc/CC-MAIN-20171215153355-20171215175355-00702.warc.gz | 159,215,382 | 11,857 | # Can you solve it? Pythagoras’s best puzzles
Three teasers from the vaults
Hi guzzlers,
The most famous theorem in maths is named after the Greek thinker Pythagoras. So is the most famous recreational mathematics publication in the Netherlands.
Pythagoras Magazine was founded in 1961, and to celebrate its half century it recently published a selection of its best brainteasers in English. Ive selected three of them here, in increasing order of difficulty.
1) Dollar bills. In a bag are 26 bills. If you take out 20 bills from the bag at random, you have at least one 1-dollar bill, two 2-dollar bills, and five 5-dollar bills. How much money was in the bag?
2) Yin and Yang. The Yin-Yang symbol is based on the figure below, bordered by three semi-circles. How can you divide this shape into two identical shapes?
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Thanks
# Accumulation of double precision vector
dattorro 1 year ago updated by Pavel Holoborodko 1 year ago
Quadruple precision floating-point accumulation of a double precision vector p is emulated numerically in Matlab:
```function t = orosumvec(p, recurs)
x = cumsum(p);
z = x - p;
u = (z - x) + p;
v = [0
x(1:end-1)] - z;
if ~recurs
t = sum([u
v
x(end)]);
else
t = orosumvec([u
v
x(end)], 0);
end
end```
This is verified by Advanpix Multiprecision Toolbox.
Matlab's built-in variable precision arithmetic, vpa() from Mathworks Symbolic Math Toolbox,
produces erroneous results in 2018One recursive call is necessary and sufficient.
Thanks
Thank you very much for the report!
Could you please provide full example, so that other people can easily copy and reproduce the situation?
If you want, we can add this example to the VPA vs. MCT comparison here:
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# 1) A quantity of 7.480g of an organic compound is dissolved in water to make 300.0 mL of solution.
Please see the attached file for the fully formatted problems.
1) A quantity of 7.480g of an organic compound is dissolved in water to make 300.0 mL of solution. The solution has an osmotic pressure of 1.43 atm at 27 degrees Celsius. The analysis of this compound shows that it contains 41.8 percent C, 4.7 percent H, 37.3 percent O and 16.3 percent N. Calculate the molecular formula of the compound.
2) The Ka for benzoic acid is 6.5 x 10^-5. Calculate the pH of a 0.10 M benzoic acid solution.
3) If 20.0 mL of 0.10 M Ba(NO3)2 are added to 50.0 mL of 0.10 M Na2CO3, will BaCO3 precipitate?
4) Compare the molar solubility of Mg(OH)2 in water and in a solution buffered at a pH of 9.0.
5) Calculate the pH of the 0.20M NH3/0.20M NH4Cl buffer. What is the pH of the buffer after the addition of 10.0 mL of 0.10M HCl to 65.0 mL of the buffer?
6) Given the following absolute entropies, determine So for the reaction
SO3(g) + H2O(l) H2SO4(l)
So (J/K mol)
SO3 256.2
H2O 69.9
H2SO4 156.9
7) Given the following free energies of formation, calculate Go for the reaction
3NO2(g) + H2O(l) 2HNO3(l) + NO(g)
G (kJ/mol)
H2O(l) -237.2
HNO3(l) -79.9
NO(g) 86.7
NO2(g) 51.8
8) Nitrosyl chloride (NOCl) decomposes at elevated temperatures according to the equation
2NOCL(g) 2NO(g) + Cl2(g)
Use the following information to calculate KP for this reaction at 227oC:
Ho = 81.2 kJ
So = 128 J/K
9) Given the following free energies of formation, calculate KP for the reaction below at 298 K.
SO2(g) + NO2(g) SO3(g) + NO(g)
G (kJ/mol)
SO2(g) -300.4 kJ/mol
SO3(g) -370.4 kJ/mol
NO(g) 86.7 kJ/mol
NO2(g) 51.8 kJ/mol
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Browse Documents | 720 | 2,133 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2019-04 | latest | en | 0.786236 |
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1. May 29, 2006
### choey
A circular loop with radius a = 0.25 m and N = 17 turns lies in the plane of the page (x-y plane). The wire used in constructing the loop has a resistance per unit length of dR/dl = 0.11 W/m.
A spatially uniform magnetic field points in the -z direction (into the page). In the interval between t = 0 and 15 s, the strength of this field varies according to the expression B(t) = 0.01 t^3 T/s3.
Calculate the current in the windings at t = 8 s. (Give the magnitude and algebraic sign - let a current that is clockwise in the view shown in the figure defined to be positive.)
---
First, I calculated the B-field @ t=8, which turned out to be 6.408849013 V.
Now, to find the current, I'm advised to write the equivalent Kirchoff's loop equation. I'm having a hard time doing this, because I'm given dR/dI, instead of just R. Since it's "resistance per unit length", I multiplied by 2pi*RN. What happens then?
2. May 29, 2006
### arunbg
What is the general expression for finding out induced emf wrt flux change ?
What is the value of induced emf at t = 8sec
Multiplying by 2pi*r*N would give you total resistance R.
Now you have induced emf and resistance. How will you find the current ?
Arun
3. May 29, 2006
### choey
Huh. I swear I tried that to find R and therefore I. But this time it worked, actually. :)
Now I'm asked, "In the time interval between 0 and 15 s, how much electrical charge passes any given point in the windings? (Give magnitude only.)"
I have...
emf @ t=8 : dPhi/dt = 6.408849013 V
I @ t=8 : dQ/dt = -2.181818 A
R = 2.937389131 Ohms
Q = -2.181818t, right?
So, Q(15) - Q(0) = -2.181818 * 15 = 32.72727 C, but that was too easy and wrong. Where should I be headed?
EDIT:
I also tried this:
Since I know that Phi = BA = 0.01t^3 * 2pi * 0.25^2 * 17 (the # of windings) = 0.0333794219t^3
Then I have this equation:
d(0.1963495408t^3)/dt = 2.937389131 * dQ/dt.
I integrate both sides w/ respect to t, and I get
0.1963495408t^3 = 2.937389131*Q(t)
Then Q(t) = 0.0668449198t^3
Evaluating Q(t) for t=0..15, I get
0.0668449198(15)^3 = 225.6016042 C, which is still wrong.
EDIT:
I got it :)
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• ## Related Books
1 answerLast reply by: Richard GregoryThu Nov 28, 2013 1:44 AMPost by Joel Fredin on November 12, 2013In your last example, i don't really understand how you could multiply with (2x-1) where did you got your "-1" from? :) Other than that, your making really good tutorials, keep it up!
### Integrals of Exponential Functions
• All other integration rules still apply
• Look for lots of substitution or chain rules applied with exponential functions
### Integrals of Exponential Functions
∫ex dx
∫ex dx = ex + c
∫e5x dx
• u = 5x
• du = 5 dx
• ∫e5x dx = [1/5] ∫eu du
• ∫e5x dx = [1/5] eu + c
∫e5x dx = [1/5] e5x + c
∫x ex2 dx
• u = x2
• du = 2x dx
• ∫x ex2 dx = ∫ex2 [1/2] 2x dx
• ∫x ex2 dx = [1/2] ∫eu du
• ∫x ex2 dx = [1/2] eu + c
∫x ex2 dx = [1/2] ex2 + c
∫9x dx
• ∫9x dx = ∫e(ln9) x
• u = ln9 x
• du = ln9 dx
• ∫9x dx = [1/ln9] ∫eln9 x ln9 dx
• ∫9x dx = [1/ln9] ∫eu du
• ∫9x dx = [(eu)/ln9] + c
• ∫9x dx = [(eln9, x)/ln9] + c
∫9x dx = [(9x)/ln9] + c
∫57x dx
• u = 7x
• du = 7 dx
• ∫57x dx = [1/7] ∫57x 7 dx
• ∫57x dx = [1/7] ∫5u du
• ∫57x dx = [1/7] [(5u)/ln5] + c
∫57x dx = [(57x)/7 ln5] + c
∫e2x + 1 dx
• u = 2x + 1
• du = 2 dx
• ∫e2x + 1 dx = [1/2] ∫e2x + 1 2 dx
• ∫e2x + 1 dx = [1/2] ∫eu du
• ∫e2x + 1 dx = [1/2] eu + c
∫e2x + 1 dx = [(e2x + 1)/2] + c
∫cos(x)esinx dx
• u = sinx
• du = cosx dx
• ∫cos(x)esinx dx = ∫esinx cosx dx
• ∫cos(x)esinx dx = ∫eu du
• ∫cos(x)esinx dx = eu + c
∫cos(x)esinx dx = esinx + c
∫[(e√x)/(√x)] dx
• u = √x
• du = [1/(2√x)] dx
• ∫[(e√x)/(√x)] dx = 2 ∫e√x [1/(2√x)] dx
• ∫[(e√x)/(√x)] dx = 2 ∫eu du
• ∫[(e√x)/(√x)] dx = 2 eu + c
∫[(e√x)/(√x)] dx = 2 e√x + c
∫[(e√{7x})/(√x)] dx
• u = √{7x}
• du = [1/2] [1/(√{7x})] [d/dx] 7x
• du = [7/2] [1/(√{7x})] dx
• du = [7/2] [1/((√7)√x)] dx
• du = [(√7)/2] [1/(√x)] dx
• ∫[(e√{7x})/(√x)] dx = [2/(√7)] ∫e√{7x} [(√7)/(2√x)] dx
• ∫[(e√{7x})/(√x)] dx = [2/(√7)] ∫eu du
• ∫[(e√{7x})/(√x)] dx = [2/(√7)] eu + c
∫[(e√{7x})/(√x)] dx = [2/(√7)] e√{7x} + c
∫sec2(5x) 4tan5x dx
• u = tan5x
• du = 5 sec2 (5x) dx
• ∫sec2(5x) 4tan5x dx = [1/5] ∫4tan5x 5 sec2 (5x) dx
• ∫sec2(5x) 4tan5x dx = [1/5] ∫4u du
• ∫sec2(5x) 4tan5x dx = [1/5] [(4u)/ln4] + c
∫sec2(5x) 4tan5x dx = [(4tan5x)/5 ln4] + c
*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.
### Integrals of Exponential Functions
Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.
• Intro 0:00
• Types of Integrals: Exponential Functions 0:09
• Rule 1
• Rule 2
• Example 1
• Example 2 2:54
• Example 3 4:19
• Example 4 5:19
• Example 5 7:37
• Example 6 9:04 | 1,458 | 2,920 | {"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-2017-43 | latest | en | 0.614905 |
https://tefdysplipo.firebaseapp.com/624.html | 1,669,755,091,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710711.7/warc/CC-MAIN-20221129200438-20221129230438-00590.warc.gz | 554,439,713 | 6,035 | # Nmarginal cost formula pdf merger
Marginal cost formula definition, examples, calculate marginal cost. Submarginal definition of submarginal by merriamwebster. Moreover, the change in marginal cost is used as the measure of efficiency gains in the calculation of. You can learn how to find marginal cost by using a formula. Subtract operating costs from gross profit and then divide by sales. So once youve figured out the change in total cost and the change in quantity. When marginal cost is plotted through graph, it results in u shaped curve at its minimum and for maximum, average cost increases as when quantity.
Understanding the relationship between marginal cost and average variable cost. The prediction of successful corporate takeovers the purpose of this section is to use the information implicit in the arbitrageurs decisionmaking process to estimate the likelihood that a cash tender or ex change offer will eventually be accepted. Here are total cost formulas, average variable, marginal cost, and more, work out your own algebra to. This situation usually arises in either of the following circumstances. Marginal cost is a key concept to be aware of in the field of business, and this quizworksheet will help you test your understanding of its calculation and use. In economics, marginal cost is the change in the total cost that arises when the quantity produced is incremented by one unit. Marginal cost mc is the cost of producing an extra unit of output. Marginal cost is an increase in total cost that results from a one unit increase in output.
In simple words we can say, marginal cost is cost of producing an additional unit. The marginal social cost of skiers msc is equal to the sum of both the marginal private cost and marginal external cost. Similarly, marginal revenue is the revenue earned by the sale of an additional unit. Any costs related to turning out a companys product or service are first combined and subtracted from the resulting revenues. Study of the effects of mergers and acquisitions in the.
Average variable cost avc is the cost of labor per unit of output produced. It is the difference between the total cost of the 6th unit and the total cost of the, 5th unit and so forth. This approach typically relates to shortterm price setting situations. Marginal cost financial definition of marginal cost. These numbers can be used to determine the markup percent. Because some of these expenses, such as labor and administrative costs, are more fixed than variable, theres another version of the gross margin ratio that takes this fact into. In other words, the marginal rate of technical substitution of labor l for capital k is the slope of an isoquant multiplied by 1. Marginal cost of capital definition of marginal cost of. It refers to the change in the total cost a business will incur by producing one additional unit of an item.
Upward pricing pressure screens in the new merger guidelines. Home category finance mcq questions and answers management accounting previous. Risk arbitrage and the prediction of successful corporate. Marginal cost pricing is the practice of setting the price of a product at or slightly above the variable cost to produce it. Marginal cost is defined by cima as the cost of one unit of a product or service which would be avoided if that unit were not provided or produced. Citeseerx document details isaac councill, lee giles, pradeep teregowda. The formula to calculate marginal cost is the change in cost divided by the change in quantity. Mc indicates the rate at which the total cost of a product changes as the production increases by one unit. Rearranging this identity, equilibrium price equals variable cost plus equilibrium contribution margin per unit the latter term being the xed cost plus equilibrium income, all divided by quantity. Marginal costing equation, profit volume ratio, break even point, margin of safety, cost break even point,finding the selling price, finding the profit.
The marginal cost mc at q items is the cost of producing the next item. You can also choose from calendar, alarm marginal cost calculator, as well as from tax calculator, general purpose calculator, and scientific marginal cost calculator, and. Upp taxes, are related to the compensating marginal cost reductions14 as follows. Aggregate social surplus before the merger is given by the area abca. Horizontal merger guidelines competition economics llc. Treatments include the use of simulated or human buyers, seller consolidations and mergerinduced fixed cost and unit. He is a very good basketball player with an nba career in his future. Choosing among tools for assessing unilateral merger effects biicl. How to calculate contribution per unit accountingtools.
The formula to obtain the marginal cost is change in costschange in quantity. Intuitively condition 1 makes sense, as the merged firm will try to reallocate resources among and so as to minimize. Marginal cost is computed as published on 08 sep 15. A user to the site will want to create a booklet from a subset of these. Subtract all other costs associated with making a profit from the operating profit. Kalin is deciding whether to enter the 2010 nba draft or play another year at msu and then enter the 2011 nba draft. It is usually computed to find at which point the company meets its economic growth. This is the marginal cost of capital, measured in cents. The average cost ac for q items is the total cost divided by q, or tcq. Fixed cost are costs that remain same in total in each period.
In evaluating how a merger will likely change a firms behavior, the agencies focus primarily on how the merger affects conduct that would be most profitable for the firm. Mixed costs can be separated into a variable cost per unit and a fixed cost per period. This is the equilibrium quantity substitute this number to the marginal social cost function. You can also talk about the average fixed cost, fcq, or the average variable cost, tvcq.
In this scenario, the formula for the markup percent is. Marginal cost is a production and economics calculation that tells you the cost of producing additional items. However, because fixed costs do not change based on the number of products produced, the marginal cost is. The experiment consists of 40 postedoffer quadropolies. It is the rate of change of the total cost of production that arises when the quantity produced is incremented by one unit. Price effects from differentiated products mergers are largely determined by diversion. Fixed cost pv ratio in value or fixed cost sales value per unit 1. Marginal cost definition, explanation and example formula.
Marginal costs are a key input to merger price simulations. Marginal revenue is the increase in revenue that results from the sale of one additional unit of output. So the calculation of the marginal cost will be 25. A wide variety of marginal cost calculator options are available to you, such as battery, solar. The marginal cost formula change in costs change in quantity. While marginal revenue can remain constant over a certain level of. They find evidence of decreasing marginal cost for the large. Remember when youre using these formulas there are a variety of assumptions, namely, that the the firm is profitmaximizing making as much money as they can.
After the merger, the marginal cost falls to c0 and the price rises to p0. It is calculated in the situations when a company meets its breakeven point. Its the sum of the fixed cost and the total variable cost for producing q items. When only one product is being sold, the concept can also be used to estimate the number of units that must be sold so that a business as a whole can break even.
Marginal cost of capital synonyms, marginal cost of capital pronunciation, marginal cost of capital translation, english dictionary definition of marginal cost of capital. The marginal cost formula represents the incremental costs incurred when producing additional units of a good or service. In general, the cost function of a merged firm is given by 1 where and are potentially different cost functions for the premerger firms. The cost that results from a one unit change in the production rate. At what output level does the marginal cost curve cross. This paper reports an experiment designed to evaluate interrelationships between strategic buyers, market power and mergerinduced synergies.
Divide the change in cost by the change in quantity. In this paper, we estimate the effects of hospital mergers on marginal costs. Marginal benefit cost ratio how is marginal benefit cost. Sales variable cost per unit margin of safety mop 1. The two ipr calculation done by the oft were premised on different.
I have a large collection of halfpage sized pdf cutsheets that are held in a folder on my linux server. Marginal cost which is really an incremental cost can be expressed in symbols. The assumption of the formula is that in steady state, this. We define the reaction functions of the firms in the standard way. We nd that the average merger target in our sample can expect to save 1. Suppose a firms average cost curve is described by the equation ac. Add margin to pdf file when merging using pdftk or similar. But as the formulas above show, once you have both upps.
Brewed in north america vancouver school of economics. Farrell and shapiro 1990 provide a more precise formula for the. The additional cost needed to produce or purchase one more unit of a good or service. Marginal cost of capital financial definition of marginal. Understanding the relationship between marginal cost and. The booklet will be bound therefore the even pages of the collection will want more margin on the right side and the odd pages will want more margin on the left side. The nba draft occurs every year at the end of june. Of these elements tax constitute 37 % of the sales price, product cost 34%, gross margin 9% and vat 20% of the sales price for gasoline 95 octane in december 2012. You must know several production variables, such as fixed costs and variable costs in order to find it. Intuitively, marginal cost at each level of production includes the cost of any additional inputs required to produce the next unit.
How to calculate the marginal cost of capital pocketsense. Microeconomics cost formulas here is a list of some of basic microeconomics formulas pertaining to revenues and costs of a firm. Marginal cost is the increase or decrease in total production cost if output is increased by one more unit. The variable costs included in the calculation are labor and materials, plus increases in fixed costs, administration, overhead. The formula to calculate marginal cost is then applied. Submarginal definition is adjacent to a margin or a marginal part or structure. Learn vocabulary, terms, and more with flashcards, games, and other study tools. Marginal cost is governed only by variable cost which changes with changes in output. Marginal profit is the profit earned by a firm or individual when one additional unit is produced and sold. Marginal cost formula definition, examples, calculate. Kalin lucas is the point guard on the msu basketball team.
685 968 1132 460 83 1003 1495 1352 260 962 1373 888 594 643 314 944 1163 844 620 191 1233 23 1007 1444 999 78 1447 1337 1376 378 1104 301 586 1089 628 593 639 437 164 110 28 571 1143 465 5 803 | 2,343 | 11,402 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2022-49 | latest | en | 0.928918 |
http://dictionnaire.sensagent.leparisien.fr/Hausdorff%20space/en-en/ | 1,610,965,548,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703514495.52/warc/CC-MAIN-20210118092350-20210118122350-00697.warc.gz | 28,188,846 | 20,436 | Hausdorff space : définition de Hausdorff space et synonymes de Hausdorff space (anglais)
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## définition - Hausdorff space
voir la définition de Wikipedia
Wikipedia
# Hausdorff space
Kolmogorov (T0) version Separation Axioms in Topological Spaces T0 | T1 | T2 | T2½ | completely T2 T3 | T3½ | T4 | T5 | T6
In topology and related branches of mathematics, a Hausdorff space, separated space or T2 space is a topological space in which distinct points have disjoint neighbourhoods. Of the many separation axioms that can be imposed on a topological space, the "Hausdorff condition" (T2) is the most frequently used and discussed. It implies the uniqueness of limits of sequences, nets, and filters. Intuitively, the condition is illustrated by the pun that a space is Hausdorff if any two points can be "housed off" from each other by open sets.[1]
Hausdorff spaces are named for Felix Hausdorff, one of the founders of topology. Hausdorff's original definition of a topological space (in 1914) included the Hausdorff condition as an axiom.
## Definitions
The points x and y, separated by their respective neighbourhoods U and V.
Points x and y in a topological space X can be separated by neighbourhoods if there exists a neighbourhood U of x and a neighbourhood V of y such that U and V are disjoint (UV = ∅). X is a Hausdorff space if any two distinct points of X can be separated by neighborhoods. This condition is the third separation axiom (after T0 and T1), which is why Hausdorff spaces are also called T2 spaces. The name separated space is also used.
A related, but weaker, notion is that of a preregular space. X is a preregular space if any two topologically distinguishable points can be separated by neighbourhoods. Preregular spaces are also called R1 spaces.
The relationship between these two conditions is as follows. A topological space is Hausdorff if and only if it is both preregular (i.e. topologically distinguishable points are separated) and Kolmogorov (i.e. distinct points are topologically distinguishable). A topological space is preregular if and only if its Kolmogorov quotient is Hausdorff.
## Equivalences
For a topological space X, the following are equivalent:
## Examples and counterexamples
Almost all spaces encountered in analysis are Hausdorff; most importantly, the real numbers (under the standard metric topology on real numbers) are a Hausdorff space. More generally, all metric spaces are Hausdorff. In fact, many spaces of use in analysis, such as topological groups and topological manifolds, have the Hausdorff condition explicitly stated in their definitions.
A simple example of a topology that is T1 but is not Hausdorff is the cofinite topology defined on an infinite set.
Pseudometric spaces typically are not Hausdorff, but they are preregular, and their use in analysis is usually only in the construction of Hausdorff gauge spaces. Indeed, when analysts run across a non-Hausdorff space, it is still probably at least preregular, and then they simply replace it with its Kolmogorov quotient, which is Hausdorff.
In contrast, non-preregular spaces are encountered much more frequently in abstract algebra and algebraic geometry, in particular as the Zariski topology on an algebraic variety or the spectrum of a ring. They also arise in the model theory of intuitionistic logic: every complete Heyting algebra is the algebra of open sets of some topological space, but this space need not be preregular, much less Hausdorff.
While the existence of unique limits for convergent nets and filters imply that a space is Hausdorff, there are non-Hausdorff T1 spaces in which every convergent sequence has a unique limit.[5]
## Properties
Subspaces and products of Hausdorff spaces are Hausdorff,[6] but quotient spaces of Hausdorff spaces need not be Hausdorff. In fact, every topological space can be realized as the quotient of some Hausdorff space.[7]
Hausdorff spaces are T1, meaning that all singletons are closed. Similarly, preregular spaces are R0.
Another nice property of Hausdorff spaces is that compact sets are always closed.[8] This may fail in non-Hausdorff spaces such as Sierpiński space.
The definition of a Hausdorff space says that points can be separated by neighborhoods. It turns out that this implies something which is seemingly stronger: in a Hausdorff space every pair of disjoint compact sets can also be separated by neighborhoods,[9] in other words there is a neighborhood of one set and a neighborhood of the other, such that the two neighborhoods are disjoint. This is an example of the general rule that compact sets often behave like points.
Compactness conditions together with preregularity often imply stronger separation axioms. For example, any locally compact preregular space is completely regular. Compact preregular spaces are normal, meaning that they satisfy Urysohn's lemma and the Tietze extension theorem and have partitions of unity subordinate to locally finite open covers. The Hausdorff versions of these statements are: every locally compact Hausdorff space is Tychonoff, and every compact Hausdorff space is normal Hausdorff.
The following results are some technical properties regarding maps (continuous and otherwise) to and from Hausdorff spaces.
Let f : XY be a continuous function and suppose Y is Hausdorff. Then the graph of f, $\{(x,f(x)) \mid x\in X\}$, is a closed subset of X × Y.
Let f : XY be a function and let $\operatorname{ker}(f) \triangleq \{(x,x') \mid f(x) = f(x')\}$ be its kernel regarded as a subspace of X × X.
• If f is continuous and Y is Hausdorff then ker(f) is closed.
• If f is an open surjection and ker(f) is closed then Y is Hausdorff.
• If f is a continuous, open surjection (i.e. an open quotient map) then Y is Hausdorff if and only if ker(f) is closed.
If f,g : XY are continuous maps and Y is Hausdorff then the equalizer $\mbox{eq}(f,g) = \{x \mid f(x) = g(x)\}$ is closed in X. It follows that if Y is Hausdorff and f and g agree on a dense subset of X then f = g. In other words, continuous functions into Hausdorff spaces are determined by their values on dense subsets.
Let f : XY be a closed surjection such that f−1(y) is compact for all yY. Then if X is Hausdorff so is Y.
Let f : XY be a quotient map with X a compact Hausdorff space. Then the following are equivalent
• Y is Hausdorff
• f is a closed map
• ker(f) is closed
## Preregularity versus regularity
All regular spaces are preregular, as are all Hausdorff spaces. There are many results for topological spaces that hold for both regular and Hausdorff spaces. Most of the time, these results hold for all preregular spaces; they were listed for regular and Hausdorff spaces separately because the idea of preregular spaces came later. On the other hand, those results that are truly about regularity generally don't also apply to nonregular Hausdorff spaces.
There are many situations where another condition of topological spaces (such as paracompactness or local compactness) will imply regularity if preregularity is satisfied. Such conditions often come in two versions: a regular version and a Hausdorff version. Although Hausdorff spaces aren't generally regular, a Hausdorff space that is also (say) locally compact will be regular, because any Hausdorff space is preregular. Thus from a certain point of view, it is really preregularity, rather than regularity, that matters in these situations. However, definitions are usually still phrased in terms of regularity, since this condition is better known than preregularity.
See History of the separation axioms for more on this issue.
## Variants
The terms "Hausdorff", "separated", and "preregular" can also be applied to such variants on topological spaces as uniform spaces, Cauchy spaces, and convergence spaces. The characteristic that unites the concept in all of these examples is that limits of nets and filters (when they exist) are unique (for separated spaces) or unique up to topological indistinguishability (for preregular spaces).
As it turns out, uniform spaces, and more generally Cauchy spaces, are always preregular, so the Hausdorff condition in these cases reduces to the T0 condition. These are also the spaces in which completeness makes sense, and Hausdorffness is a natural companion to completeness in these cases. Specifically, a space is complete if and only if every Cauchy net has at least one limit, while a space is Hausdorff if and only if every Cauchy net has at most one limit (since only Cauchy nets can have limits in the first place).
## Algebra of functions
The algebra of continuous (real or complex) functions on a Hausdorff space is a commutative C*-algebra, and conversely by the Banach–Stone theorem one can recover the topology of the space from the algebraic properties of its algebra of continuous functions. This leads to noncommutative geometry, where one considers noncommutative C*-algebras as representing algebras of functions on a noncommutative space.
In the Mathematics Institute of at the University of Bonn, in which Felix Hausdorff researched and lectured, there is a certain room designated the Hausdorff-Raum (German for Hausdorff Space).
## Notes
1. ^ Colin Adams and Robert Franzosa. Introduction to Topology: Pure and Applied. p. 42
2. ^ Willard, pp. 86–87.
3. ^ Willard, pp. 86–87.
4. ^ Bourbaki, p. 75.
5. ^ van Douwen, Eric K. (1993). "An anti-Hausdorff Fréchet space in which convergent sequences have unique limits". Topology and its Applications 51 (2): 147–158. DOI:10.1016/0166-8641(93)90147-6.
6. ^
7. ^ Shimrat, M. (1956). "Decomposition spaces and separation properties". Quart. J. Math. 2: 128–129.
8. ^
9. ^ Willard, p. 124.
## References
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un contenu abusif (raciste, pornographique, diffamatoire) | 3,056 | 12,119 | {"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": 3, "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-2021-04 | longest | en | 0.90637 |
https://ethereum.stackexchange.com/questions/74425/how-to-divide-and-allocate-the-amount-to-token-holders-without-loops | 1,716,389,581,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058557.6/warc/CC-MAIN-20240522132433-20240522162433-00291.warc.gz | 214,068,852 | 39,613 | # How to divide and allocate the amount to token holders without loops?
I have a smart contract that keeps record of the people holding the tokens in form of array of addresses. An owner of the smart contract, whenever, sends some ethers to the contract these ethers shall be divided among all token holders depending on how much they hold the tokens and accordingly these fund will be divided. We will allocate the funds to each holder in a form of mapping and once they want to redeem the pending reward will be sent to them.
My code is currently using a for loop to divide the received deposit among all token holders by iterating over the complete array of holders.Since using loop will eventually goes out of gas limit, so what should be the suggested approach here? How can we divide an amount among all token holders without looping/much gas?
You can use the amortisation of work pattern in order to do so. There is a detailed article that describe the similar case.
``````const uint pointMultiplier = 10e18;
struct Account {
uint balance;
uint lastDividendPoints;
}
uint totalSupply;
uint totalDividendPoints;
uint unclaimedDividends;
function dividendsOwing(address account) internal returns(uint) {
var newDividendPoints = totalDividendPoints - accounts[account].lastDividendPoints;
return (accounts[account].balance * newDividendPoints) / pointMultiplier;
}
var owing = dividendsOwing(account);
if(owing > 0) {
unclaimedDividends -= owing;
accounts[account].balance += owing;
accounts[account].lastDividendPoints = totalDividendPoints;
}
_;
}
function disburse(uint amount) {
totalDividendPoints += (amount * pointsMultiplier / totalSupply);
totalSupply += amount;
unclaimedDividends += amount;
}
``````
Basically you make users to claim their own dividends with `updateAccount` modifier and only store `totalDividendPoints` and `unclaimedDividends`
• will it keep a record that token holders that become holders late don't get deposits from earlier funds? for instance we had a fund of 10 ethers to be distributed to 5 people and all of them claimed it except two, later on we get another token holder and he claim immediately, will he get from the portion of early to people or technically he should get 0 ? Sep 1, 2019 at 12:46
• Do you have a fixed totalSupply? Sep 1, 2019 at 14:26
• Yes total supply is fixed but I want to give dividends to those who were token holders at the moment I sent funds to contract, and the fund that I will send shall be divided among all exiting token holders, and for others who come late shall not get any amount from earlier funds Sep 1, 2019 at 15:07
• then if you wanna use it like this you should keep track of every dividend distribution and how many users participated in each of those distributions. And then let users to claim them one by one Sep 4, 2019 at 13:24 | 636 | 2,825 | {"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-2024-22 | longest | en | 0.837726 |
https://scioly.org/forums/viewtopic.php?f=327&t=12172&start=30 | 1,580,170,061,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251737572.61/warc/CC-MAIN-20200127235617-20200128025617-00451.warc.gz | 646,247,732 | 12,874 | ## Density Lab B
Nba2302
Member
Posts: 67
Joined: November 7th, 2018, 2:28 pm
Division: B
State: MN
### Re: Density Lab B
Can any help me start up my studying(Im new)? What are some useful links? What is a very important thing in I need to know? where do i find density equations?
Kai0721
Member
Posts: 24
Joined: April 10th, 2018, 4:37 pm
Division: B
State: WI
### Re: Density Lab B
Can any help me start up my studying(Im new)? What are some useful links? What is a very important thing in I need to know? where do i find density equations?
I would start with the Scioly wiki page. It has a good base than just add more notes I’m also new to this event but that’s what I do for other note events.
2018-19 Events: Amazing Mechatronic , Battery Buggy , ELG, Density Lab,
Member
Posts: 224
Joined: January 22nd, 2018, 11:28 pm
Division: C
State: HI
Location: Somewhere in the Pacific Ocean
### Re: Density Lab B
Can any help me start up my studying(Im new)? What are some useful links? What is a very important thing in I need to know? where do i find density equations?
I would start with the Scioly wiki page. It has a good base than just add more notes I’m also new to this event but that’s what I do for other note events.
density is a very... simple? event. the rules are really broad and you can find everything with a quick google search. in your binder, make sure to include a periodic table (and make sure to be able to understand it), all the gas laws, basic thermodynamics conversions, and you'll be alright.
honestly for the nats exam all you really needed to know was d=M/V...
also, nasa has some pretty good practice q's
Farewell Division B! Time to get yeeted in C.
Highlands Intermediate School '16-'19
Pearl City High School '19-??
Proud DMAH Member
Killboe
Member
Posts: 205
Joined: January 29th, 2018, 7:04 am
Division: C
State: FL
### Re: Density Lab B
Can any help me start up my studying(Im new)? What are some useful links? What is a very important thing in I need to know? where do i find density equations?
I used to have this event and it was really hard to study because of the broad rules. But what I would do is start with finding some tests on it and studying the questions and then base your studying off the questions.
Stanton College Preparatory
Nba2302
Member
Posts: 67
Joined: November 7th, 2018, 2:28 pm
Division: B
State: MN
### Re: Density Lab B
Can any help me start up my studying(Im new)? What are some useful links? What is a very important thing in I need to know? where do i find density equations?
I used to have this event and it was really hard to study because of the broad rules. But what I would do is start with finding some tests on it and studying the questions and then base your studying off the questions.
Thanks!
Nba2302
Member
Posts: 67
Joined: November 7th, 2018, 2:28 pm
Division: B
State: MN
### Re: Density Lab B
Whats the best links for Density Lab for studying and notes?
UTF-8 U+6211 U+662F
Exalted Member
Posts: 1518
Joined: January 18th, 2015, 7:42 am
Division: C
State: PA
### Re: Density Lab B
Whats the best links for Density Lab for studying and notes?
Probably the best thing to do would just be to look everything in the rules up on either Wikipedia or your favorite search engine (aka DuckDuckGo)
jaah5211
Member
Posts: 55
Joined: October 22nd, 2017, 7:42 pm
Division: C
State: FL
Contact:
### Re: Density Lab B
Whats the best links for Density Lab for studying and notes?
I would study all the gas laws and maybe the phase diagrams.
Some tests use phase diagrams and ask how the density changes as its position on diagram changes.
someone1580
Member
Posts: 12
Joined: September 29th, 2018, 12:15 pm
Division: C
State: CA
### Density Lab B
It looks like this event is extremely broad and my coaches don't know much about the event. I have studied and done tests on density, buoyant force, concentrations, and gas laws. Is there anything else I should study for the event? And also how would you study for the hands-on task part of things?
MattChina
Member
Posts: 225
Joined: February 12th, 2017, 8:06 am
Division: B
State: NY
Location: somewhere over the rainbow
### Re: Density Lab B
It looks like this event is extremely broad and my coaches don't know much about the event. I have studied and done tests on density, buoyant force, concentrations, and gas laws. Is there anything else I should study for the event? And also how would you study for the hands-on task part of things?
The event is very broad and vague, I think thats pretty much what the event is about, uh as for the hands-on maybe just practice measuring things with graduated cylinders, calculating densities of non geometric solids. I dont know though, i haven't competed in this event.
2019 events: Water Quality, Battery Buggy, Elastic Launch Glider, Density Lab, Circuit Lab, Thermodynamics
R.C Murphy Co-Captain
Dank Memes Area Homeschool Team member | 1,297 | 4,931 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2020-05 | latest | en | 0.940388 |
https://ryanmaclean365.com/2020/10/14/calculate-full-months-between-two-dates-with-power-automate/?like_comment=1324&_wpnonce=43d9da6d97 | 1,624,272,963,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488269939.53/warc/CC-MAIN-20210621085922-20210621115922-00505.warc.gz | 450,587,002 | 26,784 | # Calculate Full Months between two dates with Power Automate
My friend Antti Pajunen recently asked me whether it is possible to calculate the number of months between two dates with Power Automate, akin to the DateDif function in Excel. I like a challenge, and I’ve done a lot of experimenting with dates in Power Automate so this is the solution I came up with.
## The Scenario
In this scenario Antti was working with Managed Services contracts which will always run for full months (i.e. if the Start Date and End Date were 9th January 2020 – 15th December 2020, the contract would run from 1st January 2020 – 31st December 2020) and we therefore need to be able to calculate the number of full months between those dates (in this case that would be 12 months).
## The Solution
The solution to this is a single Compose action with the following expression:
```int(first(split(string(add(div(add(div(sub(ticks(addDays(startOfMonth(addToTime(triggerBody()['date_1'], 1, 'Month')), -1)), ticks(startofmonth(triggerBody()['date']))), 864000000000),1),div(365.25,12)),0.5)),'.')))
```
As this expression is lengthy and contains quite a few different functions I’ve broken it out below:
1. Compose Start of Month for Start Date – to get the start of the month for the start date we just use the startOfMonth function.
2. Compose End of Month for End Date – as there is no endOfMonth function in Power Automate we have to get slightly more creative to calculate the end of month. For my expression I add one month to the End Date, then use the startOfMonth function to get the 1st day of that month, then subtract one day to get the end of month date.
3. Compose DiffinDays – now that we have the proper Start Date and End Date we want to work with we need to calculate the total number of days between the dates. There is no datediff function in Power Automate, so we need to convert them to their ticks representation, then subtract the start date value from the end date value, then divide the result by 864000000000 and then add 1 to get the total number of days between the values. This can be seen below:
ticks value for End Date of 2020-12-31 = 637449696000000000
ticks value for Start Date of 2020-01-01 = 637134336000000000
637449696000000000 – 637134336000000000 = 315360000000000
315360000000000 / 864000000000 = 365
365 +1 = 366
4. Compose DiffinMonths – now that we have the total number of Days between the two dates we need to calculate the total number of months. In order to do that we divide the resulting value from above by 365.25/12. We use 365.25 to account for leap years, and divide 365.25 by 12 to get an average value for months. Using the example above:
366/(365/12) = 12.0246406
5. Compose DiffinMonth Round – the final step is to round the value returned from the step above to the closest integer. In order to do this we need to add 0.5 to the value as the rounding calculation always rounds down. We then convert the float value to a string, split it on the decimal point, take the first part of the value (i.e. the integer) and convert that back to an integer value.
The result from above would then convert 12.0246406 to 12, showing that there are 12 full months between 2020-01-01 – 2020-12-31.
## Conclusion
To make this flow more efficient we combine all of the steps above into a single Compose action with the full expression:
If I’m honest, this is quite a long expression for what should be a relatively simple calculation, but in my testing it has worked consistently. I’d love to hear from others in the community if they find this useful or if they know of a more elegant solution. Drop me a comment below with your thoughts.
## 3 thoughts on “Calculate Full Months between two dates with Power Automate”
1. […] Calculate Full Months between two dates with Power Automate […]
Like
2. Thank you sooooo much for this! Excellent!
Like
3. There is a error in your code. if there is 1 day between the 2 dates, it gives a result of 1 rather than 0.
The issue is that when you are rounding, you are rounding up instead of down as you are adding 0.5 instead of taking away 0.5. So it should be sub(…) rather than add: | 1,035 | 4,177 | {"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-2021-25 | latest | en | 0.91384 |
www.pagans.eu | 1,675,678,258,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500334.35/warc/CC-MAIN-20230206082428-20230206112428-00487.warc.gz | 951,973,627 | 31,328 | The Megalithic Yard is a unit of measurement about 82.9 centimeters long, the purpose of this
article is to show that it can fluctuate slightly depending on the Earth’s latitude, and so that it could
have been used in Egypt with a value closer to 82.6 centimeters. The existence of MegYard. in
Egypt is indeed for the moment only a theory that we tell in the company of the researcher Sylvain
Tristan, because this standard was until now, only discovered and demonstrated very precisely on
the megalithic sites of Great Britain and French Brittany. Despite of the precision of the
researchers’s work in Europe, its value varies slightly depending on the sites: from 82.91 cm to
82.97 cm, does it depend on the latitude?
How the megalithic Yard was re-discovered by modern scholars?
The megalithic yard was found studying the megaliths of Great Britain and French Brittany by the
professor Alexander Thom, his discoveries were subsequently borrowed and deepened. Although
among official historians and archaeologists, this unit of measurement is open to debate and is not
recognized by a consensus ; some officials have nevertheless recognized the existence and validity
of the megalithic Yard : In a book “The big riddles of History”, written under the direction of
Jacques Marseille (Professor at the University of Paris I) and Nadège Laneyrie-Dagen (lecturer,
University of Lille III), we can read that not only the professor demonstrates that Carnac is also a
lunar observatory, but that in addition Alexander Thom remarks the use of a universal unit of
https://messagedelanuitdestemps.org/index.php/2017/04/11/le-yard-megalithique-letalon-oublie-dela-terre/
How did the elders determine this standard?
The MegYard. is bound to the polar circumference of the earth which is 40007.86 km since 366 MY
* 360 * 366 = 40007.86 Km for a MY of 82.962274 cm
It is such precise that the probability of it being a hazard tends to 0. The ancients would have been
aware of the terrestrial circumference, in this case why are we observing these slight fluctuations
from 82.91 cm to 82.97 cm? Why not find 82.96 cm which is the closest value to the multiple of the
Earth’s circumference? We think that the standard is accorded to the terrestrial circumference, but
that its determination, the fixing of the standard, must have been done locally, with a method we
will describe as esoteric, since all that is related with the invisible was stored in this category since
the Roman Church pointed at all magnetic sciences in relation to man as “Satanic” *.
The esoteric explanation of the megalithic yard of Giza
We expose here a theory that will inevitably be open to debate: the standard of the megalithic yard
would have been determined by a pendulum! These famous pendulums used by the druids, rarely
used today to connect with the occult world, we believe that the Druids controlled very well the
invisible world, the wave world containing your wireless communications Wifi , 4G, etc which can
no longer be described as satanic or new age nowadays, because science and everyday technologies
are attesting the existence of this “parallel world”. Earth’s magnetic fields, telluric energies, ether,
these are names that designate electro-magnetic energies circulating around our planet and creating
gravity. This gravitational field, formerly called the Dragon, or the Snake that surrounds Midgard in
the Nordic mythology, is thus the world of invisible energies formerly controlled and used by the
elders. The famous Tuatha de Danann was nicknamed “Snake people” or “people of the dragon” to
signal their ability to control these energies.
In this context, the pendulum is only an interface, it makes visible part of the invisible, it is a kind
of bridge between the two worlds: the one we see and the Dragon.
Technique for determining the unit of measurement
By varying the length of the string of the pendulum, we vary the oscillation frequency of the
pendulum, looking for 366 oscillations (or 183 periods) in 1/366th of day, we obtain a string of 1/2
MegYar., that is to say, an oscillation that will create a circle with a diameter of 1 megalithic Yard!
1/366th day = 86400 sec. / 366 = 263.065 sec. ; or 1 oscillation every 0.64498 seconds. So the
oscillation speed of the pendulum depends on the length of the string, but also on the gravity g. The
gravity changes very slightly according to the latitude, it is not the same in Scotland or Brittany,
even if this variation is very tiny, it necessarily involves also a tiny variation of the length of the
megalithic Yard, if it was determined with this technique, or if it was determined according to the
“Dragon”, because we repeat that the pendulum is only an interface with vibratory signals
presumably sinusoidal (the sinusoid oscillates like a snake, it is the shape of the signal of our
alternating current, see illustration below) that could have been perceived in another way by the
elders.
Knowing that gravity is the strongest at the pole, and that it is less at the equator:
g = 9.79338 at Giza (latitude 30 °); g = 9.81924 to Orkney Islands (Scotland, latitude 60 °)
With the pendulum, in Giza, there is a Megalithic Yard 82.558 cm rounded to 82.6 cm
Alésia and the Great Pyramid perfectly connected!
The distance Great Pyramid of Kheops – Pyramid of Djedefre is 8.26 Km that is to say exactly 10
Kilo MegYard.! If we extend this axis to the North with a 366 times longer distance, this axis leads
us right to the Montfault in Guillon in France, that is to say the real Alésia!
8.26 Km * 366 = 3023.16 Km that is the exact distance between the pyramid of Kheops and Alésia
(Guillon)! (Google Maps measurements, we invite you to check yourself so that you can realize the
beauty of it!)
Therefore, the construction of Alésia must be contemporaneous with Djedefre, the science and the
level of the civilization seemed to have dropped, given that the pyramid of Djedefre is in ruins and
is absolutely not comparable to those of Giza. This is confirmed by the writings of Diodorus who
tells us that Alésia was built by Heracles around 1200 BC, which corresponds in Egypt, to the
Pharaonic period and its much less beautiful constructions, more fragile and without comparison
with those of the pre-dynastic era.
Oleg de Normandie
* PS: However, the invisible world is very dangerous for the uninitiated, especially nowadays, and
that is also why the church forbade it, not only to set up their dictatorship, but also because there
were indeed more and more problems with the invisible world, these dangers are even greater
nowadays, we strongly discourage to use a pendulum.
##### 1 Comment
1. Merci pour cette article. | 1,595 | 6,681 | {"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-2023-06 | longest | en | 0.941947 |
http://new.transum.com/Podcast/ | 1,603,766,762,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107893011.54/warc/CC-MAIN-20201027023251-20201027053251-00198.warc.gz | 71,373,471 | 12,817 | # Podcast
The Transum podcast is now available as a free download. You can paste the URL below into your podcast program to subscribe.
If you like the podcast please give it a review on one of the following platforms:
A new episode is produced at the beginning of each month and contains a round up of developments on the Transum Mathematics website as well as a puzzle of the month.
## Episode Guide
71. Trains Together
How far apart will the trains be half an hour before they meet? This episode was published on Thu, 1 Oct 2020 [Running time: 11:00]
70. Six Ropes
Can the pirates win their freedom by tying the six pieces of rope into one large loop? This episode was published on Tue, 1 Sep 2020 [Running time: 6:53]
69. Bogdan and Wally
In what order should you play Ultimate Noughts and Crosses against Wally and Bogdan? This episode was published on Sat, 1 Aug 2020 [Running time: 8:18]
68. What's My Name Riddle
Work out the name from the rhyming riddles. This episode was published on Wed, 1 Jul 2020 [Running time: 8:57]
67. Mathematical Days
What is the date of the next Pythagorean Theorem Day? This episode was published on Mon, 1 Jun 2020 [Running time: 9:22]
66. Eva's Eggs and Fickle Fractions
How many eggs did Eva take to market in order to make those strange half sales? This episode was published on Fri, 1 May 2020 [Running time: 10:36]
65. Feeding Fools and Horses
How long will the horse feed last the horses after some of them left the stables? This episode was published on Wed, 1 Apr 2020 [Running time: 9:06]
64. Balancing Balloons
How many balloons must Jamie give to Ben to balance the balloon equation? This episode was published on Sun, 1 Mar 2020 [Running time: 8:40]
63. Awe-Sum, Flabbergasted and a Puzzle from Carl
Carl's puzzle is about the result of dividing the product of his parents' ages by double his own age. This episode was published on Sat, 1 Feb 2020 [Running time: 12:34]
62. Odds from evens up to 2020
Subract the sum of the odd numbers less than 2020 from the sum of the even numbers less than 2020. This episode was published on Thu, 1 Jan 2020 [Running time: 8:19]
61. Noel in Lapland
Work out the length of Noel's trip to Lapland given the details of the weather conditions. This episode was published on Sun, 1 Dec 2019 [Running time: 7:25]
60. Multiple Remainders
Find the second smallest number that satisfies the given conditions about remainders. This episode was published on Fri, 1 Nov 2019 [Running time: 12:25]
59. Coaches to Cambridge
Figure out which coach I was sitting in when I travelled by train to Cambridge last summer. This episode was published on Tue, 1 Oct 2019 [Running time: 13:42]
58. Percy Cod's Kids
How old are Percy's children given the ratio of their ages now and in ten years' time? This episode was published on Sun, 1 Sep 2019 [Running time: 9:50]
57. Carrot-Eating Critters
How many of each type of animal shared the 20 carrots? This episode was published on Thu, 1 Aug 2019 [Running time: 10:53]
56. Arrow in Square
What fraction of the square does the arrow cover? This episode was published on Mon, 1 Jul 2019 [Running time: 10:20]
55. Puzzles, Riddles and Conundrums
A puzzle (or riddle) about Percy who seems to age very quickly from 13 to 16. This episode was published on Sat, 1 Jun 2019 [Running time: 14:08]
54. Monkeys, Kittens and Dogs
A riddle about the person most likely to be able to work out the square root of 121. This episode was published on Wed, 1 May 2019 [Running time: 10:32]
53. Feed the Horses
Calculate the number of days worth of horse feed that are left for Obadiah Sloop. This episode was published on Mon, 1 Apr 2019 [Running time: 10:31]
Solve the riddle heard on the radio involving modular arithmetic. This episode was published on Fri, 1 Mar 2019 [Running time: 13:07]
51. Prime Permutations
Of all the permutations of 1 to 9 used to make nine-digit numbers, how many are prime? This episode was published on Fri, 1 Feb 2019 [Running time: 12:50]
50. 4-Digit Hotel Room Safe Code
Can you work out what number I used to lock and unlock my hotel room safe? This episode was published on Tue, 1 Jan 2019 [Running time: 9:41]
49. Choir Eye Colour
Figure out the percentage of choir members that do not have blue eyes from the clues given. This episode was published on Fri, 30 Nov 2018 [Running time: 8:20]
48. Calculator Keys at the Corners of a Rectangle
What do the four digit numbers typed into a calculator using four keys at the corners of a rectangle have in common? This episode was published on Thu, 1 Nov 2018 [Running time: 10:04]
47. Average House Numbers
Work out the house numbers from the clues about the mean, median and mode. This episode was published on Mon, 1 Oct 2018 [Running time: 8:12]
46. Clock in the Mirror
What time was the clock in the mirror really showing? This episode was published on Sat, 1 Sep 2018 [Running time: 11:00]
45. Delightfully Divisible
Find a pandigital number that is delightfully divisible. This episode was published on Mon, 30 Jul 2018 [Running time: 7:56]
44. Five integers with a product of 12
Can you find five integers that multiply together to give twelve? This episode was published on Sun, 1 Jul 2018 [Running time: 5:32]
43. Square in a Rectangle
What is the largest square that can be drawn in the corner of a 10cm by 15cm rectangle that does not cross the diagonal of the rectangle? This episode was published on Fri, 1 Jun 2018 [Running time: 8:10]
42. London Marathon
A question about the average speed required for the second half of the marathon. This episode was published on Tue, 1 May 2018 [Running time: 18:28]
41. Cutting the Lawn
Aynuk and Ayli cut half of the lawn each. How long is each side of the square lawn? This episode was published on Sun, 1 Apr 2018 [Running time: 12:56]
40. The Missing Pound's Found
A really wonderful answer to the missing pound puzzle. This episode was published on Thu, 1 Mar 2018 [Running time: 13:22]
39. Forty Five In Four Parts
Split the number 45 into four parts according to the given information. This episode was published on Thu, 1 Feb 2018 [Running time: 9:27]
38. Shrivelled Spuds
Work out the weight of the potatoes after they have been left out in the sun to dry. This episode was published on Mon, 8 Jan 2018 [Running time: 6:36]
37. Transposition Error
Work out the bank balance given information about the transposition error. This episode was published on Fri, 1 Dec 2017 [Running time: 10:04]
36. Counting Sheep
Work out the number of sheep owned by Percy and Patsy from the given clues. This episode was published on Wed, 1 Nov 2017 [Running time: 12:54]
35. Juggling with Egg Timers
Can you time exactly nine minutes using the four and seven minute egg timers? This episode was published on Sun, 1 Oct 2017 [Running time: 11:08]
34. Permutations Sum
What is sum of all the four digit numbers containing all of the digits one to four? This episode was published on Fri, 1 Sep 2017 [Running time: 12:17]
33. Odd Probability
What is the probability that two random numbers are both even if they are not both odd? This episode was published on Mon, 31 Jul 2017 [Running time: 14:24]
32. Regions in Circles
Calculate the number of regions in a circle formed by intersecting chords. This episode was published on Fri, 30 Jun 2017 [Running time: 9:32]
31. Brothers and Sisters
Can you work out the number of children in the Numlove family given the clues about brothers and sisters? This episode was published on Thu, 1 Jun 2017 [Running time: 11:58]
30. Exam Average
What mark is required in the last exam to achieve an 80% overall average? This episode was published on Fri, 30 Apr 2017 [Running time: 12:35]
29. Cake Cut
Where should the last slice of cake be cut to give two equal pieces? This episode was published on Fri, 31 Mar 2017 [Running time: 12:59]
28. Holy Sphere
Calculate the remaining volume of the sphere after a cylindrical hole has been drilled through the centre. This episode was published on Mon, 27 Feb 2017 [Running time: 11:15]
27. Jumping Flea
How many different places could the flea find itself after 8 foot-long jumps either north, south, east or west? This episode was published on Tue, 17 Jan 2017 [Running time: 10:19]
26. Last Digit
How many positive two-digit numbers are there whose square and cube both end in the same digit? This episode was published on Thu, 1 Dec 2016 [Running time: 11:47]
25. Central Station
The probability that the next train to leave will be going north is five times the probability that the next train to leave will be going south. This episode was published on Tue, 1 Nov 2016 [Running time: 6:22]
24. Canteen Queue
Is it possible to answer the question if Betsy's age is not known? This episode was published on Fri, 30 Sep 2016 [Running time: 10:14]
23. Separated Twins
Work out the combination of the safe given the clues about pairs of numbers. This episode was published on Fri, 2 Sep 2016 [Running time: 7:07]
22. Divisible By Three
A puzzle about two digit numbers that can be made from ten different digits. This episode was published on Fri, 1 Jul 2016 [Running time: 8:06]
21. Letters In Numbers
A brand new puzzle involving the letters in numbers when written as words. This episode was published on Thu, 2 Jun 2016 [Running time: 7:52]
20. Square Angled Triangle
The angles of a triangle are all square numbers. What are they? This episode was published on Tue, 3 May 2016 [Running time: 8:00]
19. Tri-Junction Puzzle
What is the probability of the three cars arriving at the road junction not being involved in an accident? This episode was published on Fri, 1 Apr 2016 [Running time: 8:17]
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14. Ant and Dec
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posted by .
What size motor [W] is required to pump 12,000 gallons of water up a water tower (with a height of 25 ft.) in 25 minutes? The density of water is 1,000 kg/cubic meter.
• Physics -
Convert gallons/min to kg/second.
Then multiply the kg/s pumping rate by g*(height). That will equal the rate potential energy is added to the water, in watts.
You should do the calculation.
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# What is seven eights plus one fourths plus three sixtheens in math?
Wiki User
2015-10-08 12:53:25
It is: 7/8+1/4+3/16 = 21/16
Wiki User
2015-10-08 12:53:25
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https://us.metamath.org/mpeuni/ressply1bas2.html | 1,718,458,988,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861594.22/warc/CC-MAIN-20240615124455-20240615154455-00875.warc.gz | 536,972,114 | 6,885 | Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > ressply1bas2 Structured version Visualization version GIF version
Theorem ressply1bas2 20396
Description: The base set of a restricted polynomial algebra consists of power series in the subring which are also polynomials (in the parent ring). (Contributed by Mario Carneiro, 3-Jul-2015.)
Hypotheses
Ref Expression
ressply1.s 𝑆 = (Poly1𝑅)
ressply1.h 𝐻 = (𝑅s 𝑇)
ressply1.u 𝑈 = (Poly1𝐻)
ressply1.b 𝐵 = (Base‘𝑈)
ressply1.2 (𝜑𝑇 ∈ (SubRing‘𝑅))
ressply1bas2.w 𝑊 = (PwSer1𝐻)
ressply1bas2.c 𝐶 = (Base‘𝑊)
ressply1bas2.k 𝐾 = (Base‘𝑆)
Assertion
Ref Expression
ressply1bas2 (𝜑𝐵 = (𝐶𝐾))
Proof of Theorem ressply1bas2
StepHypRef Expression
1 eqid 2824 . 2 (1o mPoly 𝑅) = (1o mPoly 𝑅)
2 ressply1.h . 2 𝐻 = (𝑅s 𝑇)
3 eqid 2824 . 2 (1o mPoly 𝐻) = (1o mPoly 𝐻)
4 ressply1.u . . 3 𝑈 = (Poly1𝐻)
5 ressply1bas2.w . . 3 𝑊 = (PwSer1𝐻)
6 ressply1.b . . 3 𝐵 = (Base‘𝑈)
74, 5, 6ply1bas 20363 . 2 𝐵 = (Base‘(1o mPoly 𝐻))
8 1on 8105 . . 3 1o ∈ On
98a1i 11 . 2 (𝜑 → 1o ∈ On)
10 ressply1.2 . 2 (𝜑𝑇 ∈ (SubRing‘𝑅))
11 eqid 2824 . 2 (1o mPwSer 𝐻) = (1o mPwSer 𝐻)
12 ressply1bas2.c . . 3 𝐶 = (Base‘𝑊)
135, 12, 11psr1bas2 20358 . 2 𝐶 = (Base‘(1o mPwSer 𝐻))
14 ressply1.s . . 3 𝑆 = (Poly1𝑅)
15 eqid 2824 . . 3 (PwSer1𝑅) = (PwSer1𝑅)
16 ressply1bas2.k . . 3 𝐾 = (Base‘𝑆)
1714, 15, 16ply1bas 20363 . 2 𝐾 = (Base‘(1o mPoly 𝑅))
181, 2, 3, 7, 9, 10, 11, 13, 17ressmplbas2 20236 1 (𝜑𝐵 = (𝐶𝐾))
Colors of variables: wff setvar class Syntax hints: → wi 4 = wceq 1538 ∈ wcel 2115 ∩ cin 3918 Oncon0 6178 ‘cfv 6343 (class class class)co 7149 1oc1o 8091 Basecbs 16483 ↾s cress 16484 SubRingcsubrg 19531 mPwSer cmps 20131 mPoly cmpl 20133 PwSer1cps1 20343 Poly1cpl1 20345 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1797 ax-4 1811 ax-5 1912 ax-6 1971 ax-7 2016 ax-8 2117 ax-9 2125 ax-10 2146 ax-11 2162 ax-12 2179 ax-ext 2796 ax-rep 5176 ax-sep 5189 ax-nul 5196 ax-pow 5253 ax-pr 5317 ax-un 7455 ax-cnex 10591 ax-resscn 10592 ax-1cn 10593 ax-icn 10594 ax-addcl 10595 ax-addrcl 10596 ax-mulcl 10597 ax-mulrcl 10598 ax-mulcom 10599 ax-addass 10600 ax-mulass 10601 ax-distr 10602 ax-i2m1 10603 ax-1ne0 10604 ax-1rid 10605 ax-rnegex 10606 ax-rrecex 10607 ax-cnre 10608 ax-pre-lttri 10609 ax-pre-lttrn 10610 ax-pre-ltadd 10611 ax-pre-mulgt0 10612 This theorem depends on definitions: df-bi 210 df-an 400 df-or 845 df-3or 1085 df-3an 1086 df-tru 1541 df-ex 1782 df-nf 1786 df-sb 2071 df-mo 2624 df-eu 2655 df-clab 2803 df-cleq 2817 df-clel 2896 df-nfc 2964 df-ne 3015 df-nel 3119 df-ral 3138 df-rex 3139 df-reu 3140 df-rmo 3141 df-rab 3142 df-v 3482 df-sbc 3759 df-csb 3867 df-dif 3922 df-un 3924 df-in 3926 df-ss 3936 df-pss 3938 df-nul 4277 df-if 4451 df-pw 4524 df-sn 4551 df-pr 4553 df-tp 4555 df-op 4557 df-uni 4825 df-int 4863 df-iun 4907 df-iin 4908 df-br 5053 df-opab 5115 df-mpt 5133 df-tr 5159 df-id 5447 df-eprel 5452 df-po 5461 df-so 5462 df-fr 5501 df-se 5502 df-we 5503 df-xp 5548 df-rel 5549 df-cnv 5550 df-co 5551 df-dm 5552 df-rn 5553 df-res 5554 df-ima 5555 df-pred 6135 df-ord 6181 df-on 6182 df-lim 6183 df-suc 6184 df-iota 6302 df-fun 6345 df-fn 6346 df-f 6347 df-f1 6348 df-fo 6349 df-f1o 6350 df-fv 6351 df-isom 6352 df-riota 7107 df-ov 7152 df-oprab 7153 df-mpo 7154 df-of 7403 df-ofr 7404 df-om 7575 df-1st 7684 df-2nd 7685 df-supp 7827 df-wrecs 7943 df-recs 8004 df-rdg 8042 df-1o 8098 df-2o 8099 df-oadd 8102 df-er 8285 df-map 8404 df-pm 8405 df-ixp 8458 df-en 8506 df-dom 8507 df-sdom 8508 df-fin 8509 df-fsupp 8831 df-oi 8971 df-card 9365 df-pnf 10675 df-mnf 10676 df-xr 10677 df-ltxr 10678 df-le 10679 df-sub 10870 df-neg 10871 df-nn 11635 df-2 11697 df-3 11698 df-4 11699 df-5 11700 df-6 11701 df-7 11702 df-8 11703 df-9 11704 df-n0 11895 df-z 11979 df-dec 12096 df-uz 12241 df-fz 12895 df-fzo 13038 df-seq 13374 df-hash 13696 df-struct 16485 df-ndx 16486 df-slot 16487 df-base 16489 df-sets 16490 df-ress 16491 df-plusg 16578 df-mulr 16579 df-sca 16581 df-vsca 16582 df-tset 16584 df-ple 16585 df-0g 16715 df-gsum 16716 df-mre 16857 df-mrc 16858 df-acs 16860 df-mgm 17852 df-sgrp 17901 df-mnd 17912 df-mhm 17956 df-submnd 17957 df-grp 18106 df-minusg 18107 df-mulg 18225 df-subg 18276 df-ghm 18356 df-cntz 18447 df-cmn 18908 df-abl 18909 df-mgp 19240 df-ur 19252 df-ring 19299 df-subrg 19533 df-psr 20136 df-mpl 20138 df-opsr 20140 df-psr1 20348 df-ply1 20350 This theorem is referenced by: ressply1bas 20397
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A065883 Remove factors of 4 from n (i.e., write n in base 4, drop final zeros, then rewrite in decimal). 15
1, 2, 3, 1, 5, 6, 7, 2, 9, 10, 11, 3, 13, 14, 15, 1, 17, 18, 19, 5, 21, 22, 23, 6, 25, 26, 27, 7, 29, 30, 31, 2, 33, 34, 35, 9, 37, 38, 39, 10, 41, 42, 43, 11, 45, 46, 47, 3, 49, 50, 51, 13, 53, 54, 55, 14, 57, 58, 59, 15, 61, 62, 63, 1, 65, 66, 67, 17, 69, 70, 71, 18, 73, 74, 75 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,2 LINKS Indranil Ghosh, Table of n, a(n) for n = 1..20000 (First 1000 terms from Harry J. Smith) FORMULA If n mod 4 = 0 then a(n) = a(n/4), otherwise a(n) = n. Multiplicative with a(p^e) = 2^(e (mod 2)) if p = 2 and a(p^e) = p^e if p is an odd prime. a(n) = n/4^A235127(n). a(n) = A214392(n) if n mod 16 != 0. - Peter Kagey, Sep 02 2015 From Robert Israel, Dec 08 2015: (Start) G.f.: x/(1-x)^2 - 3 Sum_{j>=1} x^(4^j)/(1-x^(4^j))^2. G.f. satisfies G(x) = G(x^4) + x/(1-x)^2 - 4 x^4/(1-x^4)^2. (End) Sum_{k=1..n} a(k) ~ (2/5) * n^2. - Amiram Eldar, Nov 20 2022 Dirichlet g.f.: zeta(s-1)*(4^s-4)/(4^s-1). - Amiram Eldar, Jan 04 2023 EXAMPLE a(7)=7, a(14)=14, a(28)=a(4*7)=7, a(56)=a(4*14)=14, a(112)=a(4^2*7)=7. MAPLE A065883:= n -> n/4^floor(padic:-ordp(n, 2)/2): map(A065883, [\$1..1000]); # Robert Israel, Dec 08 2015 MATHEMATICA If[Divisible[#, 4], #/4^IntegerExponent[#, 4], #]&/@Range[80] (* Harvey P. Dale, Aug 31 2013 *) PROG (PARI) baseA2B(x, a, b)= { local(d, e=0, f=1); while (x>0, d=x%b; x\=b; e+=d*f; f*=a); return(e) } { for (n=1, 1000, if (n%4, a=n, a=baseA2B(n, 10, 4); while (a%10 == 0, a\=10); a=baseA2B(a, 4, 10)); write("b065883.txt", n, " ", a) ) } \\ Harry J. Smith, Nov 03 2009 (PARI) a(n)=n/4^valuation(n, 4); \\ Joerg Arndt, Dec 09 2015 (Python) def A065883(n): return n>>((~n&n-1).bit_length()&-2) # Chai Wah Wu, Jul 09 2022 CROSSREFS Cf. A214392, A235127, A350091 (drop final 2's). Remove other factors: A000265, A038502, A132739, A244414, A242603, A004151. Sequence in context: A083346 A319652 A327938 * A214392 A071975 A350389 Adjacent sequences: A065880 A065881 A065882 * A065884 A065885 A065886 KEYWORD base,easy,nonn,mult AUTHOR Henry Bottomley, Nov 26 2001 STATUS approved
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# What is Gravity?
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posted on Aug, 2 2006 @ 07:28 PM
I feel that gravity is another form of space itself. The duality of the fabric of space, if you will. When in the force state, we experience it as gravity. Otherwise, we experience it as a unit of space. As a particle of matter migrates from one unit of space to the next, the unit is displaced/converted into gravitational force. As the force acts on the particle, it is transformed back into a spacial unit, allowing the particle to move to the next unit.
But to really figure out all the aspects of gravity would mean unification... good luck.
posted on Aug, 2 2006 @ 07:38 PM
Gravity is an electromagnetic wave and like any wave has amplitude and frequency.
...
Gravity is instantaneous and is not, as mainstream science have you believe ‘just a little faster’ than the speed of light. Any gravitation force exerted by the earth will be felt instantaneously anywhere else in the universe
John, if gravity is a wave with amplitude and frequency how can it also be instantaneous?
The frequency of a wave refers to how often the particles of the medium vibrate when a wave passes through the medium.
Here is a source.
[edit on 2-8-2006 by zoopnfunk]
posted on Aug, 2 2006 @ 07:52 PM
Gravity is not instantaneous. This has been repeatedly demonstrated.
There's a few links in this thread:
www.abovetopsecret.com...
And allow me to pimp my own thread on an implication of gravity having a fixed speed:
www.abovetopsecret.com...
posted on Aug, 3 2006 @ 02:03 PM
johnlear:
I read your post and thought, well, interesting.
My theory is based on two types of gravities and one of these could possibly be the combination of two similar separate ones (ie three total).
robertfenix:
Well, what you have said is also very close to my theory, however the semantics are somewhat different.
----------------------
It's interesting when I read how some websites explain how the earth's magnetic field is like a bar magnet. They are half right.
I think you'll find that when it's revealed, people are going to say "oh, that's how it works?"
...and wonder why it took man so long to figure it out
Cheers
JS
posted on Aug, 3 2006 @ 02:19 PM
I had a dream about gravity a long time ago when I was a kid. Unfortuately, I never learned enough mathematics to develop a formula about it.
All I can say is that gravity is a trans-dimensional force that at its smallest level behaves a lot like surface tension in water. "Blobs" of gravity float around matter and in-between dimensions, and "stick" together when they interact. It's a little hard to explain and visualize, exactly. The blobs "reach through" different dimensions and are attracted to each other, pulling the mass together along with them. They have poles, like magnets, but only one pole exists in our 3-D/4-D space, the other extends "down" (?) into another dimension.
The information was visual, and I don't know exactly what these gravity things were or why they interacted the way they did. Whoever sent the information into my brain unfortunately got the wrong guy, since I am unable to do anything with it. I assume that it/they sent it to more people than me, so maybe one of them will figure it out. But maybe that information is lost forever when I'm dead. Too bad.
posted on Aug, 3 2006 @ 03:27 PM
Originally posted by jumpspace
robertfenix:
Well, what you have said is also very close to my theory, however the semantics are somewhat different.
Semantics I can understand that, as it is a concept in my head that I am trying to express, terms are just that, like Agua and water the explain the same "thing" or concept.
What I see in reality is a ball "atom" and two arrow head leads from the ball that are attached to the ball on one end and at the other have a colored arrow head one is blue and one is red. In space or in the absence of any gravitational "pull" these arrow head leads writher around like snakes trying to point themselves to the closest gravitational "pull".
Planets are then either RED or BLUE, as well as all bodies in space.
If a strong RED body is near then the "space atom" red arrow will straighten like a tight direction rod and the atom will then be drawn to the red body in space. This is the strong force alignment. If the closest was a blue body then the blue arrow would point and then the blue mode would be the strong force alignment.
When two or more atoms connect to each other they exchange only one and one, one red for one blue and one blue for one red, but they act as a sort of conduit for the strong force of the planet (lets say gravity in the atmosphere) acting as a super highway for the matching gravitons that have been emitted by the planet core to return back to the planet core. This is the Mode Value or gravitational force of the planet.
to visualize draw two circles. one above the other seperated by three inches. like this
O
O
Then on left make a two make interchange and on the right the number of "pipes" equal to the mode value or gravitational force (lets say 5)
O
| ! ! ! ! ! !
| ! ! ! ! ! !
| ! ! ! ! ! !
| ! ! ! ! ! !
O
The right side then is the amount of Pressure/force/strong force/ bonding/ compression that each atom has on the next atom in order to exchange the same amount of RED or BLUE gravitons in the direction of the center of the planet core.
The right side gravitons are being thrown out from the core into "free air" and are absorbed by every atom and returned back to the core at the rate of the mode value or gravitational force which is determined by the mass, density and composition of the core material.
To make the atom go anti gravity then you have to switch the mode of the atom in regards to which one the atom uses for strong and weak force
O so now these two atoms can not pass through the free air
! | ! ! ! ! !
! | ! ! ! ! !
! | ! ! ! ! !
O
These two atoms then would part the other atoms above them much like a ships bow through water or a plane wing in the air, the other atoms that are not switch will seek out other not switched atoms to bond with to continue the conduit of passing the strong force gravitons back to the core.
The more you switch atmospheric atoms, the more that get drawn to your anti g machine, the more mass you can counter act against the pull of the core.
much like floating an opposite pole magnet or how the maglev train works
but in a maglev you are introducing electrons to excite the magnetic field strength, instead of attracting "graviton density" to counteract gravity.
A maglev uses an inefficient third force (electrons) intead of force counter force.
posted on Aug, 3 2006 @ 03:42 PM
APC
Abstract. Standard experimental techniques exist to determine the propagation speed of forces. When we apply these techniques to gravity, they all yield propagation speeds too great to measure, substantially faster than lightspeed. This is because gravity, in contrast to light, has no detectable aberration or propagation delay for its action, even for cases (such as binary pulsars) where sources of gravity accelerate significantly during the light time from source to target. By contrast, the finite propagation speed of light causes radiation pressure forces to have a non-radial component causing orbits to decay (the “Poynting-Robertson effect”); but gravity has no counterpart force proportional to to first order.
General relativity (GR) explains these features by suggesting that gravitation (unlike electromagnetic forces) is a pure geometric effect of curved space-time, not a force of nature that propagates. Gravitational radiation, which surely does propagate at lightspeed but is a fifth order effect in , is too small to play a role in explaining this difference in behavior between gravity and ordinary forces of nature. Problems with the causality principle also exist for GR in this connection, such as explaining how the external fields between binary black holes manage to continually update without benefit of communication with the masses hidden behind event horizons. These causality problems would be solved without any change to the mathematical formalism of GR, but only to its interpretation, if gravity is once again taken to be a propagating force of nature in flat space-time with the propagation speed indicated by observational evidence and experiments: not less than 2x1010 c.
Such a change of perspective requires no change in the assumed character of gravitational radiation or its lightspeed propagation. Although faster-than-light force propagation speeds do violate Einstein special relativity (SR), they are in accord with Lorentzian relativity, which has never been experimentally distinguished from SR—at least, not in favor of SR. Indeed, far from upsetting much of current physics, the main changes induced by this new perspective are beneficial to areas where physics has been struggling, such as explaining experimental evidence for non-locality in quantum physics, the dark matter issue in cosmology, and the possible unification of forces. Recognition of a faster-than-lightspeed propagation of gravity, as indicated by all existing experimental evidence, may be the key to taking conventional physics to the next plateau.
www.metaresearch.org...
It's a VERY interesting paper that i read some time ago and as scientist go Tom has quite the record imo ....
Get whatever you normally use, to calm yourself in times of crisis, ready as this might very well be somewhat unsettling.
Stellar
posted on Aug, 3 2006 @ 06:00 PM
1998 vs. 2002 and 2003... hmm...
If I fully understood the math we could probably duke it out
As I do not, I'll go with the more recent findings.
>
ooh! plus, 2x1010C isn't even warp 9! certainly not instantaneous. psh!
[edit on 3-8-2006 by apc]
posted on Aug, 3 2006 @ 11:52 PM
Originally posted by jumpspace
It's interesting when I read how some websites explain how the earth's magnetic field is like a bar magnet. They are half right.
I think you'll find that when it's revealed, people are going to say "oh, that's how it works?"
...and wonder why it took man so long to figure it out
Cheers
JS
May I ask then 'how does it work?'
namaste
Raphael
posted on Aug, 4 2006 @ 12:05 AM
yes these are very deep and interesting, man...
can anyone vouch for a small and sort of albert einstein way of saying a short answer, although it is big, make it smaller for better consumption.
posted on Aug, 4 2006 @ 07:27 AM
... sure.
What goes up must come down.
More Newton than Einstein though.
posted on Aug, 4 2006 @ 08:18 AM
A Simplification
Hmmm... well I tried an Einstein by giving a thought experiment - that was the real Einstein way - but let me try something simpler:
Space-time is like a mattress. When something heavy is on the mattress, the mattress sinks, and any object wants to fall towards the heavier object.
That's the extremely simple explanation - but that's an oversimplification - and because of that, it's not entirely accurate.
Why? Because in the mattress example, as stated, we're still relying on gravity to explain how something falls, even though it moves towards the surface (we're saying it rolls down from a higher to lower point).
In truth though, it's that the path that an object is taking is changing.
Perhaps a better example then would be to say,
Space-time is like the fabric of a cloth. From our perspective the fabric's material is woven in straight lines, and cutting it with scissors would give us a straight line through it. However, the fabric actually curves into a circle, and as we look back at our cut, even though the cut seemed to be straight, we realize that it turned inwards, our cut eventually led us towards the center of the cloth.
So it's not that we're "falling" inwards, or that gravity is "pulling" us down - it's that the path we're taking is actually changed. When we jump into the air, we don't come back down because of a string that pulls us back. We come back down because, slowly, the direction of "up" changes into "down". The force in our jump remains constant during the entire trip, but direction of that force is changed, until we find ourselves back where we began.
posted on Aug, 4 2006 @ 01:32 PM
Gravity is simply the distortion of spacetime.
Why matter distorts spacetime is actually unknown. But the distortion proves that the void is not really a void, because if it was, the distortion would not work.
My favorite and personal theory is that it is not matter that causes gravity, but the void. This theory (that the gravity is a push, not a pull, force), explains a lot:
1) the accelerating expansion of the universe.
2) the Cassimir effect (when the two plates get together, the void between them ceases to exist).
3) the spherical shape of celestial bodies (those formed out of gases).
posted on Aug, 4 2006 @ 06:22 PM
It should be noted that "distortion", "bending", "warping", etc, are all human descriptors designed to give our brains something to relate the four-dimensional concept to. Space is not actually "bent" in the way we think of bent, but it is something our limited brains can comprehend.
posted on Aug, 4 2006 @ 09:48 PM
Perhaps(probably) I am not reading or comprehending these theories incorrectly, but many of them appear to be saying that the standard model is incorrect(or perhaps limited/incomplete?). Is this the assumption/idea you are proposing? If it is, I am not saying that the theory is wrong, I am just wondering.
(Forget this, I was misunderstanding the theory that i was asking about.)
[edit on 4-8-2006 by Liquid Swords1]
posted on Aug, 4 2006 @ 10:07 PM
posted by johnlear
Ragster,
This is what Bob Lazar had to say about gravity. Gravity is an electromagnetic wave and like any wave has amplitude and frequency. There are 2 types of gravity, Gravity A which works on an atomic scale and is the strong nuclear force . . Gravity B holds us on top of the earth, the earth and all the planets in orbit around the sun . . Gravity is instantaneous and not as mainstream science have you believe ‘just a little faster’ than the speed of light. Any gravitation force exerted by the earth will be felt instantaneously anywhere else in the universe. [Edited by Don W]
I’m sure not everyone thinks Albert Einstein knew it all, but he was the one who said nothing exceeds the speed of light, including gravity. I am unaware of anything Bob Lazar has said that is new or is accepted by the scientific community. I’ll have to go with Einstein on speed, Mr J/L. It is almost intuitive that nothing is “instantaneous” in this universe.
I prefer the standard definition, 1) electro-magnetic force which includes light; 2) strong nuclear force - keeps atoms in one piece; 3) weak nuclear force - allows for atomic decay of atoms; and 4) gravity.
posted on Aug, 4 2006 @ 10:18 PM
Gravity is an effect. Better yet a reaction to the unknown. What happens when two opposite magnets touch. An effect pushes them away. Gravity is the reaction of all the objects in our universe, and beyond. I've always felt that it is a by-product of Magentism. But I'd also love a definitive answer. Where's Hawkings?
posted on Aug, 5 2006 @ 08:53 AM
Originally posted by robertfenix
What I see in reality is a ball "atom" and two arrow head leads from the ball that are attached to the ball on one end and at the other have a colored arrow head one is blue and one is red. In space or in the absence of any gravitational "pull" these arrow head leads writher around like snakes trying to point themselves to the closest gravitational "pull".
Planets are then either RED or BLUE, as well as all bodies in space.
If a strong RED body is near then the "space atom" red arrow will straighten like a tight direction rod and the atom will then be drawn to the red body in space. This is the strong force alignment. If the closest was a blue body then the blue arrow would point and then the blue mode would be the strong force alignment.
Robert I study and compare symbols, myth, religion, iconography and science etc...looking for common denominators.
Red And Blue are significant players throughout.
Please share with me what you know of Red and Blue ... I shall apply a Cosmogony 'constant', patterns I have detected, based on the esoteric representation of numbers, colors, directions, animals etc.
namaste
Raphael
posted on Aug, 6 2006 @ 12:17 AM
red is always the dominate or strong force and blue is always the weaker return path.
there is no color to gravitons just a way to visually represent which one is which, ie when they shift then the blues become red etc, really the color is only a symbol. but when i see how gravity works in my head it is red and blue to show me the relationship.
to make anti gravity material the only thing you have to do is block and atoms ability to pass the core aligned gravitons.
in this manner an object could have up to infinite mass yet have no weight, with no measureable weight it would take very little energy to move the object.
an ionic airlifter is a great operating sample when you have only a small amount of weight you can create lift by only using static ion discharge.
by going past the netural point and attracting the opposite graviton from free air atom you increase the kinetic energy potential of the opposing forces. the craft can then hyper accelerate, the resulting "stripped" free air atoms cause a distortion and a temporary increase in specific gravity that would effect a very narrow local zone. ie someone observing close by could temporarily black out by the sudden increase in gravity at that spot. They may lose conscienceness and think strange thoughts and "lose time".
A strange paradox then happens, the faster the craft travels the slower it feels to the occupants as the opposite graviton force is built up. Possibly as it nears lightspeed or beyond the net affect may feel like instantaneous travel and for the occupants may appear to be instantaneous. Outside observers would continue on at the normal pace though.
This could spell disaster for space fairing aliens as even short hops of 40 + lightyears (80+ Lightyears round trip) could net 10,000 years or more change on their homeplanet. They could very well return to empty space or a planet vastly changed by wars for long gone only to find no civilaztion and no support to re-fuel their crafts. They may have simply died with no means to travel any further.
The problem is not travelling the great distances of space but rather what to do when you get there and what to do should you try and return to your home planet.
These travellers would be on a one way suicide mission and would become the last living entity of their species. For the odds that their kind would survive in its current civilization level are far more against them.
So an armada of space ships akin to Battlestar Gallactica would appear then to be more of the logical solution. Simply move the entire remanant of civilization from one home planet to the next. Near instantaneous travel would make several hops possible until a new home planet was found.
So the early visitors to this planet found themselves in the same situation, once here and returning to their home planet to report succeful contact with a planet with inhabitants they discovered their very own planet vastly different their new found knowledge of earth lost to possible eons of change. Travelling over 40 lightyears and having no means to send a communication signal faster then their craft could travel meant having to respond in person much like in ancient earth, when man had to the carry the message of the "new world" back to the queen. Because there were no phones in Columbus's time.
posted on Aug, 6 2006 @ 06:01 AM
We could thus measure a person's charge, his polarity and determine many aspects of his character, his political tendancies?
From the ground up everything appears to be binary?
To be confused with duality.
Here are two sites you may find interesting.
Graham Hancock is linked to this site.
www.binaryresearchinstitute.org...
and this one is supported by Dr. Milo Wolffe
www.physics-philosophy-metaphysics.com...
The above site proposes a theory that would replace Wave / Particle duality with a Wave / Wave foundation thus SpaceTime becomes SpaceMotion. 3Dimensions of Space plus Motion.
You may find your frontiers challenged.
Red being a stronger force and blue a weaker return is consistent with my findings using my 'tools'.
My tools also suggest Robert that the macrocosmic wave...the outer infinite boundary is the inverse in magnitude of the smallest / microcosmic wave known as the graviton.
So if the universe is as estimated to be 10 to the power of +35, using man as the baseline, I will suggest then that the graviton can be found using a magnification of 10 power of -35.
Other noted measurements using man as the baseline.
Size of atom 10 -5 to size of earth 10 +5
Size of proton 10 -10 to proximity to nearest star, not our sun is 10 +10
So the ancients were correct when they suggested man is the centre of his universe.
They also said the world / universe was flat
Socrates implies I should take note of patterns.
"As it is in the macrocosm so it is in the microcosm"
Or the Kabalists play around with terms like macroprosopus and microprosopus.
It means 'know thyself' is indeed a great place, a tool to help you start connecting.
Everything is connected without a doubt, and Quantum Gravity is suggesting just that, is it not?
namaste
Raphael
[edit on 6-8-2006 by Kachina]
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4.3 Conservation of Energy, the Work-Energy Principle, and Power
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Law of Conservation of Energyπ¨βπ»
The energy of a system is conserved.
When a system has no work being done on or by it, the total energy of the system is constant. This total energy can be internally converted among potential, kinetic, or thermal but never changes. This idea is the Law of Conservation of Energy and is one of the 4 major conservation laws in AP 1.
A key idea here is that this law only applies when there are no outside forces acting on the system. So a system consisting of a ball and the earth would have its energy conserved as the ball falls, but a system of a car would have its energy change as friction between the tires and the road removes kinetic energy.
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• The law of conservation of energy can be used to predict and analyze the behavior of physical systems, and to solve problems involving the transfer and transformation of energy.
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Here are some key things to remember about the work-energy principle:
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Power
Here are some key things to remember about power:
• Power is a measure of the rate at which work is done or energy is transferred.
• Power is usually measured in watts (W), or joules per second (J/s).
• Power is calculated using the formula: Power = Work / Time
• Power is a scalar quantity, meaning it has only magnitude and no direction.
• Power can be positive or negative, depending on the direction of the work or energy transfer.
• Power is a measure of how quickly a task is completed or how fast energy is transferred.
• The higher the power, the more work is being done or energy is being transferred in a given time period.
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https://www.jiskha.com/display.cgi?id=1202168579 | 1,502,999,376,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886103910.54/warc/CC-MAIN-20170817185948-20170817205948-00686.warc.gz | 927,927,651 | 4,009 | # math
posted by .
How do I take the antilog of -0.05?
• math -
Assuming the log base is 10, then
antilog (-0.05) = 10^.05 = 1/10^.05 = 1.1,2202 = 0.89125
## Similar Questions
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how would you solve for y in this problem: ln(y-1)-ln2 = x +lnx ln(y-1)-ln2 = x +lnx Solve for y... ln[(y-1)/2]]=x + lnx take the antilog of each side (y-1)/2= e^(x+lnx) solve for y.
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More Similar Questions | 631 | 2,209 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2017-34 | latest | en | 0.896018 |
https://uk.mathworks.com/matlabcentral/answers/354114-how-to-perform-knn-regression | 1,563,311,250,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195524879.8/warc/CC-MAIN-20190716201412-20190716223412-00372.warc.gz | 584,347,299 | 19,134 | ## How to perform KNN regression
### Mekala balaji (view profile)
on 26 Aug 2017
Latest activity Commented on by Image Analyst
### Image Analyst (view profile)
on 26 Aug 2017
Hi,
I have data as below:
X:
1
2
3
4
9
11
13
17
19
20
y:
0.1
0.3
0.4
0.65
1.5
2.5
2.7
2.9
3.2
3.6
I want to compute y values for at x values at 6.3, 15.5&21.5
1. for each value of x, I want to take 4 nearest points in my calculation and choose most reasonable one as the final y
2. In the second method, compare minkowski & chebychev distances and choose most reasonable one
Jan
### Jan (view profile)
on 26 Aug 2017
There is no "most reasonable" in maths. You want a valus for x=21.5, but the original data stop at x=20. There is an infinite number of possible ways to extrapolate. Without any further information like "the data follow a polynomial of order 3" it is not possible to prefer any method or even to measure the degree or reasonability. Perhaps a linear interpolation is fair, and the extrapolated values should be simply NaN.
Image Analyst
### Image Analyst (view profile)
on 26 Aug 2017
"In the second method, compare ....." sounds like a homework problem, | 342 | 1,155 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2019-30 | latest | en | 0.858135 |
http://www.nag.com/numeric/fl/manual/examples/source/f04cafe.f | 1,410,829,546,000,000,000 | text/plain | crawl-data/CC-MAIN-2014-41/segments/1410657110730.89/warc/CC-MAIN-20140914011150-00338-ip-10-196-40-205.us-west-1.compute.internal.warc.gz | 699,218,474 | 1,567 | * F04CAF Example Program Text * Mark 21 Release. NAG Copyright 2004. * .. Parameters .. INTEGER NIN, NOUT PARAMETER (NIN=5,NOUT=6) INTEGER NMAX, NRHSMX PARAMETER (NMAX=8,NRHSMX=2) INTEGER LDA, LDB PARAMETER (LDA=NMAX,LDB=NMAX) * .. Local Scalars .. DOUBLE PRECISION ERRBND, RCOND INTEGER I, IERR, IFAIL, J, N, NRHS * .. Local Arrays .. COMPLEX *16 A(LDA,NMAX), B(LDB,NRHSMX) INTEGER IPIV(NMAX) CHARACTER CLABS(1), RLABS(1) * .. External Subroutines .. EXTERNAL F04CAF, X04DBF * .. Executable Statements .. WRITE (NOUT,*) 'F04CAF Example Program Results' WRITE (NOUT,*) * Skip heading in data file READ (NIN,*) READ (NIN,*) N, NRHS IF (N.LE.NMAX .AND. NRHS.LE.NRHSMX) THEN * * Read A and B from data file * READ (NIN,*) ((A(I,J),J=1,N),I=1,N) READ (NIN,*) ((B(I,J),J=1,NRHS),I=1,N) * * Solve the equations AX = B for X * IFAIL = -1 CALL F04CAF(N,NRHS,A,LDA,IPIV,B,LDB,RCOND,ERRBND,IFAIL) * IF (IFAIL.EQ.0) THEN * * Print solution, estimate of condition number and approximate * error bound * IERR = 0 CALL X04DBF('General',' ',N,NRHS,B,LDB,'Bracketed',' ', + 'Solution','Integer',RLABS,'Integer',CLABS,80,0, + IERR) * WRITE (NOUT,*) WRITE (NOUT,*) 'Estimate of condition number' WRITE (NOUT,99999) 1.0D0/RCOND WRITE (NOUT,*) WRITE (NOUT,*) + 'Estimate of error bound for computed solutions' WRITE (NOUT,99999) ERRBND ELSE IF (IFAIL.EQ.N+1) THEN * * Matrix is numerically singular. Print estimate of * reciprocal of condition number and solution * WRITE (NOUT,*) WRITE (NOUT,*) 'Estimate of reciprocal of condition number' WRITE (NOUT,99999) RCOND * WRITE (NOUT,*) IERR = 0 CALL X04DBF('General',' ',N,NRHS,B,LDB,'Bracketed',' ', + 'Solution','Integer',RLABS,'Integer',CLABS,80,0, + IERR) * ELSE IF (IFAIL.GT.0 .AND. IFAIL.LE.N) THEN * * The upper triangualr matrix U is exactly singular. Print * details of factorization * WRITE (NOUT,*) IERR = 0 CALL X04DBF('General',' ',N,N,A,LDA,'Bracketed','F7.4', + 'Details of factorization','Integer',RLABS, + 'Integer',CLABS,80,0,IERR) * * Print pivot indices * WRITE (NOUT,*) WRITE (NOUT,*) 'Pivot indices' WRITE (NOUT,99998) (IPIV(I),I=1,N) END IF ELSE WRITE (NOUT,*) 'NMAX and/or NRHSMX too small' END IF STOP * 99999 FORMAT (8X,1P,E9.1) 99998 FORMAT ((1X,7I11)) END | 788 | 2,210 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2014-41 | latest | en | 0.570199 |
https://www.wileyindia.com/microsoft-excel-formulas-functions-for-dummies-4ed.html | 1,642,378,959,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300253.51/warc/CC-MAIN-20220117000754-20220117030754-00258.warc.gz | 1,156,254,638 | 14,806 |
# Microsoft Excel Formulas & Functions For Dummies, 4ed
ISBN: 9788126559466
408 pages
INR 499
## Description
Are you intimidated by major financial choices, like which loan to get or how to grow your savings? Don't worry—we all are! But Excel Formulas & Functions For Dummies, 4th Edition can take some of the pain out of the data organization and analysis processes. This step-by-step reference sheds light on Microsoft Excel's 150 most useful functions and offers detailed instructions on how to implement them. Additionally, each function is illustrated by helpful, real-world examples that show how they are used within a larger formula.
Introduction
Part I: Getting Started with Formulas and Functions
Chapter 1: Tapping Into Formula and Function Fundamentals
Chapter 2: Saving Time with Function Tools
Chapter 3: Saying "Array!" for Formulas and Functions
Chapter 4: Fixing Formula Boo-Boos
Part II: Doing the Math
Chapter 5: Calculating Loan Payments and Interest Rates
Chapter 6: Appreciating What You'll Get, Depreciating What You've Got
Chapter 7: Using Basic Math Functions
Part III: Solving with Statistics
Chapter 9: Throwing Statistics a Curve
Chapter 10: Using Significance Tests
Chapter 11: Rolling the Dice on Predictions and Probability
Part IV: Dancing with Data
Chapter 12: Dressing Up for Date Functions
Chapter 13: Keeping Well-Timed Functions
Chapter 14: Using Lookup, Logical and Reference Functions
Chapter 15: Digging Up the Facts
Chapter 16: Writing Home about Text Functions
Chapter 17: Playing Records with Database Functions
Part V: The Part of Tens
Chapter 18: Ten Tips for Working with Formulas
Chapter 19: Ten Functions You Really Should Know
Chapter 20: Some Really Cool Functions
Index
• Name:
• Designation:
• Name of Institute:
• Email:
• * Request from personal id will not be entertained
• Moblie:
• ISBN / Title:
• ISBN: * Please specify ISBN / Title Name clearly | 452 | 1,941 | {"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-2022-05 | latest | en | 0.759469 |
http://www.aimsedu.org/2013/09/02/penny-penning-puzzle/?replytocom=1379 | 1,502,941,335,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886102891.30/warc/CC-MAIN-20170817032523-20170817052523-00257.warc.gz | 462,355,575 | 14,132 | On the campus of
# Penny Penning Puzzle
This week’s puzzle is a good one to use early in the school year. It is fairly easy and shouldn’t frustrate students too much in their early exposure to the puzzle-solving process. To do this puzzle, students need only the student sheet depicting the nine-penny arrays and a pencil. The puzzle challenges students to draw four “pens” around the nine pennies in the array in such a way that there is an odd number of pennies in each pen. Students should begin to realize, after several trial and error attempts, that there is no way to divide nine objects into four separate sets of odd numbers since any combination of four odd numbers produces an even sum. Once this realization has been made, the puzzle seems unsolvable. This is where an important puzzle-/problem-solving skill comes into play—thinking divergently. Since there is no way to draw four separate pens (with no overlap) that each contain an odd number of pennies, one or more of the pens must overlap. Once this insight is made, it is easy to see that there are several valid solutions to the puzzle.
As in any Puzzle Corner activity, students should be encouraged to work independently and asked not to share their solutions until the appropriate sharing time at the end of the week. While you should encourage students not to give away their solutions, you may want to allow them to offer hints to their fellow students. These hints can help those students who have not yet developed their abilities to think outside the box. Your role as the teacher is to facilitate this problem-solving process—not give students the answer.
Solution
Click the arrow below to view the solution.
The Penny Penning Challenge asked students to draw four “pens” around the nine pennies in a three by three array in such a way that there were an odd number of pennies in each pen. There is no way to divide nine objects into four separate sets of odd numbers since any combination of four odd numbers produces an even sum. Therefore, to solve this puzzle students had to think divergently and realize that in any solution, one or more of the pens must overlap. When this insight has been made, there are several valid solutions to the puzzle. One of these solutions is shown below.
### 4 Responses to Penny Penning Puzzle
1. Char says:
Never mind….I went to the Wayback Machine and found it….1997, I think was the year I found it under…Thank you for posting the puzzles!!!! The Farmer puzzle is sentimental to me. A lot has changed in my teaching BUT NOT that puzzle!!
2. Char says:
I always start the beginning of the year with the Farmer, Fox, Chicken and Grain problem!!!!! I do it because it is simple…I just want to show the kids we are THINKING in my classroom….but, ugh…it is GONE!!!!! And, I cannot find a copy of it on my computer!!!!!
3. Char says:
Oh my gosh!!! Where is the full Puzzle Corner! I have been coming to this website for puzzles for gosh…at least 15 years!!!!!!! I miss the FULL Puzzle Corner!!!!!!! Is it coming back?!???? | 658 | 3,048 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.46875 | 4 | CC-MAIN-2017-34 | latest | en | 0.965138 |
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# Multiply Complex Numbers
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Tutorial
• Complex Numbers
• FOIL Review
• Multiplying Complex Numbers
## Multiply Complex Numbers
Complex Numbers
A complex number is a number in the form , containing both a real part and an imaginary part. The imaginary part is followed by i, which is the imaginary unit,
When multiplying complex numbers, we follow a similar process when multiplying binomial factors we may be familiar with when studying quadratics. The multiplication process is often referred to as FOIL, which distributes terms into the factors being multiplied. Let's take a moment to review FOIL with real numbers before looking at examples of complex number multiplication
FOIL Review
FOIL stands for First, Outside, Inside, Last, and refers to terms that are multiplied together to form individual addends to the product. Here is an example:
When multiplying two complex numbers, we will be following the same procedure, but will need to make an additional consideration when the imaginary unit is squared.
Multiplying Complex Numbers
Let's multiply the complex numbers and :
The final step here is to simplify the last term, containing the imaginary unit squared. Recall that the imaginary unit is . When this is squared, it becomes the real number .
To simplify terms, we can remove completely, but reverse the sign of its coefficient. For example, simplifies to . This is a real number that can be combined with other like terms.
To multiply complex numbers, we use the FOIL process to multiply the terms in the two complex numbers. During this process, we simplify to , which is a real number.
Formulas to Know
Imaginary Number | 568 | 2,702 | {"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} | 4.3125 | 4 | CC-MAIN-2018-17 | latest | en | 0.878005 |
https://docs.google.com/forms/d/e/1FAIpQLScnYvLL4VEdi54dGo-8UIlUX0fexrtvykV5MVVsi6wVspMARQ/viewform?c=0&w=1&usp=send_form | 1,537,541,338,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267157203.39/warc/CC-MAIN-20180921131727-20180921152127-00233.warc.gz | 504,634,837 | 29,139 | Quiz on Special Electrical Machines
Tick the correct answer and press the submit button
Name of the student *
Roll.No *
1. What is the angle between stator direct axis and quadrature axis ?
1 point
2.Space angle, θr is measured between stator d-axis and _____
1 point
3. Reluctance motor can produce torque at ________
1 point
4.For a reluctance motor , the maximum average torque occurs when δ= __________
1 point
5. For a given reluctance motor, Rld and Rlq are ________
1 point
6. 1. For a 1.8°, 2 phase bipolar stepper motor, the stepping rate is 100 step / second. The rotational speed of the motor in r.p.m is
1 point
7.A stepper motor has a step-angle of 2.5°. Determine the resolution?
0 points
8.Both stator and rotor there are eight teeth of a 3 stack four pole stepper motor, when the excitation is changed from one stack to another to the next. The step angle is _________
1 point
9.A 3 - Φ, three stacks, variable reluctance step motor has 20 poles on each rotor and stator stack. The step angle of this motor is__________
1 point
10. A 3 stack stepper motor has 12 teeth on stator as well as on rotor. Determine the step size when each stack excited?
1 point | 301 | 1,172 | {"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-2018-39 | latest | en | 0.837948 |
https://www.teachoo.com/3922/1164/Ex-1.3--1---Let-f---1--3--4------1--2--5--g---1--2--5-/category/Composite-funcions/ | 1,695,693,523,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510130.53/warc/CC-MAIN-20230926011608-20230926041608-00208.warc.gz | 1,133,631,008 | 28,825 | Composite functions
Chapter 1 Class 12 Relation and Functions
Concept wise
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Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
### Transcript
Ex 1.3, 1 Let f: {1, 3, 4} → {1, 2, 5} and g: {1, 2, 5} → {1, 3} be given by f = {(1, 2), (3, 5), (4, 1)} and g = {(1, 3), (2, 3), (5, 1)}. Write down gof. f: {1, 3, 4} → {1, 2, 5} f = {(1, 2), (3, 5), (4, 1)} f can be denoted as g: {1, 2, 5} → {1, 3} g = {(1, 3), (2, 3), (5, 1)} g can be denoted as Finding gof So, 𝑔𝑜𝑓(1) = 3 𝑔𝑜𝑓(3) = 1 𝑔𝑜𝑓(4) = 3 ∴ 𝒈𝒐𝒇 = {(1,3), (3,1), (4,3)} | 323 | 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} | 3.78125 | 4 | CC-MAIN-2023-40 | latest | en | 0.829031 |
http://www.gnu.org/software/gnulib/manual/html_node/Checking-Integer-Overflow.html | 1,656,883,643,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104249664.70/warc/CC-MAIN-20220703195118-20220703225118-00692.warc.gz | 87,931,628 | 3,359 | Next: , Previous: , Up: Integer Properties [Contents][Index]
#### 15.7.3 Checking Integer Overflow
Signed integer arithmetic has undefined behavior on overflow in C. Although almost all modern computers use two’s complement signed arithmetic that is well-defined to wrap around, C compilers routinely optimize assuming that signed integer overflow cannot occur, which means that a C program cannot easily get at the underlying machine arithmetic. For example:
```if ((a + b < b) == (a < 0))
a += b;
else
printf ("overflow\n");
```
might not work as expected if `a` and `b` are signed, because a compiler can assume that signed overflow cannot occur and treat the entire `if` expression as if it were true. And even if `a` is unsigned, the expression might not work as expected if `b` is negative or is wider than `a`.
The following macros work around this problem by returning an overflow indication while computing the sum, difference, or product of two integers. For example, if `i` is of type `int`, `INT_ADD_OK (INT_MAX - 1, 1, &i)` sets `i` to `INT_MAX` and returns true, whereas ```INT_ADD_OK (INT_MAX, 1, &i)``` returns false.
Example usage:
```#include <intprops.h>
#include <stdio.h>
/* Compute A * B, reporting whether overflow occurred. */
void
print_product (long int a, long int b)
{
long int r;
if (INT_MULTIPLY_OK (a, b, &r))
printf ("result is %ld\n", r);
else
printf ("overflow\n");
}
```
These macros work for both signed and unsigned integers, so they can be used with integer types like `time_t` that may or may not be signed, depending on the platform.
These macros have the following restrictions:
• Their first two arguments must be integer expressions.
• Their last argument must be a non-null pointer to an integer.
• They may evaluate their arguments zero or multiple times, so the arguments should not have side effects.
• They are not necessarily constant expressions, even if all their arguments are constant expressions.
`INT_ADD_OK (a, b, r)`
Compute the sum of a and b. If it fits into `*r`, store it there and return true. Otherwise return false, possibly modifying `*r` to an unspecified value. See above for restrictions.
`INT_SUBTRACT_OK (a, b, r)`
Compute the difference between a and b. If it fits into `*r`, store it there and return true. Otherwise return false, possibly modifying `*r` to an unspecified value. See above for restrictions.
`INT_MULTIPLY_OK (a, b, r)`
Compute the product of a and b. If it fits into `*r`, store it there and return true. Otherwise return false, possibly modifying `*r` to an unspecified value. See above for restrictions.
Other macros are available if you need wrapped-around results when overflow occurs (see Wraparound Arithmetic with Integers), or if you need to check for overflow in operations other than addition, subtraction, and multiplication (see Integer Type Overflow).
Next: Wraparound Arithmetic with Integers, Previous: Integer Bounds, Up: Integer Properties [Contents][Index] | 695 | 2,986 | {"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-2022-27 | longest | en | 0.862607 |
https://encyclopedia2.thefreedictionary.com/barycentre | 1,642,801,631,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320303709.2/warc/CC-MAIN-20220121192415-20220121222415-00710.warc.gz | 283,739,351 | 11,644 | # barycentric
(redirected from barycentre)
Also found in: Dictionary.
Related to barycentre: Barycentric coordinates
## barycentric
(mathematics)
Centre of gravity, mean.
This article is provided by FOLDOC - Free Online Dictionary of Computing (foldoc.org)
References in periodicals archive ?
2) Distance of each position from the barycentre of all GPS fixes (meters), as a measure of the animal's displacement from the centre of its activity.
where [X.sub.b,j+1], [Y.sub.b,j+1]--coordinates of the (j + 1)-th barycentre with respect to the OXY, [[psi].sub.j]-- the angle between line segment connecting j-th and the (j + 1)-th barycentres and OX axis.
The photon arrival times at the X-ray device are corrected to the solar system barycentre (SSB) arrival times via the orbit data.
where [x.sub.g] is robot barycentre coordinate of x-axis, [y.sub.g] is robot barycentre coordinate of y-axis, [z.sub.g] is robot barycentre coordinate of z-axis, [[??].sub.g] is the acceleration of x-axis, and [[??].sub.g] is the acceleration of y-axis.
In general, the rounded square shape has the axis of the handle vertical through the pot body, perpendicular and pointing to the horizontal plane; and a round shape has the axis of the handle perpendicular to the generatrix of the pot body, pointing to the barycentre on the body.
4), where the barycentre of the EPN network is located.
* This refers to the International Celestial Reference System, which is the standard celestial coordinate system centered at the barycentre of the Solar System, with axes that are fixed with respect to objects in far-reaches of the cosmos.
The centre, called the Barycentre, is often within the mass of the Sun, but not always.
The initial variables have been centred, so that the barycentre of data points corresponds to the PC axis origin.
It is essentially based on the formula used to calculate the barycentre of the light signals, starting from the relative weights of the amount of photons collected by the single PMTs.
Without the searching process, we mandatorily consider that the center point is seen by not only the barycentre of the Gauss domain, but also the barycentre of the domain involving some nodes.
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Open / Close | 530 | 2,230 | {"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 | latest | en | 0.848016 |
http://www.solved-problems.com/calculus/integral/169/problem-2-7/ | 1,582,223,296,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145260.40/warc/CC-MAIN-20200220162309-20200220192309-00210.warc.gz | 233,305,756 | 13,921 | # Problem 2-7: Evaluatin Definite Integrals
Evaluate
a) $\displaystyle\int_{-1}^1 \! x^2 \, dx$
b) $\displaystyle\int_{-2}^1 \! \sin(x) \, dx$
c) $\displaystyle\int_{-1}^2 \! 2x^3+x-1 \, dx$
d) $\displaystyle\int_{-2}^2 \! |x| \, dx$
Solution
a) $\displaystyle\int_{-1}^1 \! x^2 \, dx=\frac{1}{3}x^3\Bigr|^1_{-1}=\frac{1}{3}(1)^3-\frac{1}{3}(-1)^3=\frac{2}{3}$
b) $\displaystyle\int_{-2}^1 \! \sin(x) \, dx=-\cos(x)\Bigr|^1_{-2}=-\cos(1)-(-\cos(-2))=-0.9564$
c) $\displaystyle\int_{-1}^2 \! (2x^3+x-1) \, dx=\displaystyle\int_{-1}^2 \! 2x^3 \, dx +\displaystyle\int_{-1}^2 \! x \, dx -\displaystyle\int_{-1}^2 \! 1 \, dx$
$= \frac{2}{4}x^4\Bigr|^{-1}_{2}+\frac{1}{2}x^2\Bigr|^{-1}_{2}-x\Bigr|^{-1}_{2}= (\frac{1}{2}2^4- \frac{1}{2}(-1)^4)+(\frac{1}{2}2^2-\frac{1}{2}(-1)^2)-(2-(-1))$
$= 8-\frac{1}{2}+2-\frac{1}{2}-2-1=6$
d) $\displaystyle\int_{-2}^2 \! |x| \, dx=\displaystyle\int_{-2}^0 \! |x| \, dx+\displaystyle\int_{0}^2 \! |x| \, dx=\displaystyle\int_{-2}^0 \! (-x) \, dx+\displaystyle\int_{0}^2 \! x \, dx$
$= -\frac{1}{2}{x^2}\Bigr|_{-2}^0+\frac{1}{2}{x^2}\Bigr|_{0}^2=-\frac{1}{2}{(0)^2}-\frac{1}{2}(-{(-2)^2})+\frac{1}{2}{2^2}-\frac{1}{2}{0^2}=4$
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wow,finally | 685 | 1,406 | {"found_math": true, "script_math_tex": 11, "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": 11, "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} | 4.5625 | 5 | CC-MAIN-2020-10 | latest | en | 0.457589 |
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