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http://www.fixya.com/support/t1376265-reposting_exponential_log_equations
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math
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- If you need clarification, ask it in the comment box above.
- Better answers use proper spelling and grammar.
- Provide details, support with references or personal experience.
Tell us some more! Your answer needs to include more details to help people.You can't post answers that contain an email address.Please enter a valid email address.The email address entered is already associated to an account.Login to postPlease use English characters only.
Tip: The max point reward for answering a question is 15.
I think the problem with this equation comes when x = 0. At this point, it is undefined. Thus, it is a horizontal line y=-2, except when x=0. This is normally signified by putting an open circle when x=0.
Sorry but you are using the word EQUATION incorrectly. All you have is two numbers you want to divide one by the other.
Enter the numerator 1EE (-) 14, press the division key / then enter the 5 EE (-) 12. Press the = sign.
Note: to enter the negative sign in the exponents, use the change sign key marked (-), (+/-) or [+/-] . Do not use the regular MINUS sign.
When you use the EE key, you must not enter the 10. It is understood because EE is a shortcut for *10^.
The FX-901 calculator seems to be the equivalent of the FX 260 Solar sold in North America. It does fraction calculations, permutation, statistics but lack the Equation Solver. You cannot use it to solve equations. Casio Scientific calculators that can handle equation solving are the FX-115 and FX-991. Simplify your first equation by dividing all its terms by 3, then use elimination to carry out the solution by hand.
In equation mode, you have system of linear equations (3 unknown) you have polynomial (quadratic and cubic), and solver. Use the solver foe any type of equation (nonlinear, polynomial of order higher than 4, trigonometric, exponential, logs).
Prefixes such as milli already reflect the corresponding multiplier or divisor. One milliamp is 1/1000th of an Amp, so 37 milliamps = 0.037 of an Amp.
To keep two sides of an equation the same, you need to perform the same operation on each side. In this case, the equation is 37/1000 = 0.037. The "divide by 1000" part of this equation is included in the word "milliamp", hence 37 milliamps = 0.037 Amps.
Another way of looking at it is that a number remains unchanged if you both divide and multiply it by the same number. In this case, changing to milliamps is dividing by 1000, so you need to multiply by 1000 to keep the result unchanged.
You have actually divided by 1000 twice as a result of using x10^-3 as a multiplier when you should have used x10^3
Usually ypu know the equation then graph it. However in doing statistics, it may appear that the curve of best for a scatter plot could be could an exponential. In that case you use the STAT application to enter the x-data in List 1 and the y-data in List 2. When you finish press the f1 key to select Graph1. You should also verify by Pressing F6:set that the graph type is set to scatter, and that List 1 contains the x-data and List 2, the y-data. After you verify that all is in order press the F1: Graph 1 key. As the graph is displayed you will see several tabs with various regression models. Pan thr tabs to the right to get to F2:EXP. Select it. The equation is displayed on the next screen. At the bottom of the screen you see two tabs COPY and DRAW. Press f5:Copy to copy the equation in one of the y= functions. Press Draw and you will see the curve of best fit superposed on the scatter plot points.
The error comes from the vanishing values at the tail of the function. The exponential function does not vanish except for -infinity. Truncate the lists by removing the pairs (5,00 and (6,0) or fudge up the data (5, 1E-5) (6,1E-6) and you will get your equation of best fit.
I am including a link to a post of mine where I show how to perform an exponential regression. JUST IN CASE you need additional information on the subject.
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CC-MAIN-2016-50
| 3,950 | 19 |
http://www.cgsociety.org/index.php/CGSFeatures/CGSFeatureSpecial/emflock_2
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math
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Tue 22nd Feb 2011 | News
emFlock2 is a flocking solver for Softimage's ICE.
emFlock2 uses the three 'classic ground rules of flocking'.
1. Separation will prevent the members from crowding and colliding.
2. Alignment will make each member adapt its heading to the average heading of its visible neighbors.
3. Cohesion will make each member want to go to the average position of its visible neighbors.
• Multithreaded neighboring and flocking.
• Compounds for speed and orientation control without flipping.
• Compounds to make flocks fly along specific paths and morph into specific shapes.
• Generation of complex flight paths using nearly any desired input in any desired combination (point clouds, polygon meshes, curves, etc.).
• New neighboring compound.
• Point clouds that use emFlock2 and emFly are still just "ordinary" point clouds => they can be cached and rendered just as any other point cloud.
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CC-MAIN-2017-34
| 919 | 12 |
http://www.askphysics.com/2011/09/16/
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math
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hai I am very week in physics please help me in recovering
I think you can write an essay about a bike that you have seen only once.How? Just because of your perception
Try to understand the concepts. You may be trying to by heart the topics .Avoid that .Develop the habit of writing and studying
A rifle at a height H aimed horizontally fires a bullet to the ground. At the same time , a bullet with the same mass in dropped from the same height. Neglecting air resistance, which one hits the ground first?Explain.
Both will hit the ground simultaneously.
When a body is projected horizontally, its initial vertical velocity is zero and vertical acceleration is g, the acceleration due to gravity.
The values of a velocity and acceleration of a freely falling body are also the same.
So, both will hit the ground simultaneously
Similarities and differences in magnetic and electric field (Dilpreet posted this question)
| Both electric and magnetic field are conservative forcesBoth obey inverse square law
Both are non contact forces (Forces can be exerted without contact)
Both are attractive as well as repulsive (Like poles repel, like charges repel; unlike poles attract, unlike charges attract)
| Electric field is produced by a charge whether at rest or in motionBut magnetic field is produced only by a moving charge
The total magnetic flux through any closed surface is always zero, but the total electric flux through any closed surface is equal to the net charge enclosed by the surface multiplied by the reciprocal of absolute permitivity
Electric field lines are discontinuous as they have a starting point (+ charge) and an ending point (- charge); But magnetic field lines are continuous, they always form closed loops
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CC-MAIN-2020-40
| 1,734 | 15 |
http://www.southperry.net/showthread.php?t=70846
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math
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Pinnacle Scroll for 2H weapon for magic att 20% was the only 20% w/matk scroll that had a 10days limited.
In RED coin shop, Evolution 2H weapon for magic att 50% (+8 matk) is the only 50% +8w/matk scroll that have a 5 times purchase limit and cost 2 more coins compare to the others.
I just don't understand why is 2H weapon for matk always seems to be the one have the most restriction?
Also, what mage weapon beside Fans are 2H?
Will it be too OP for 2H magical weapons to have the same accessibility to scrolls as all other weapons?
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s3://commoncrawl/crawl-data/CC-MAIN-2016-44/segments/1476988719139.8/warc/CC-MAIN-20161020183839-00290-ip-10-171-6-4.ec2.internal.warc.gz
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CC-MAIN-2016-44
| 535 | 5 |
https://www.tapestryproject.org/how-do-you-convert-grams-to-millimeters/
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math
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How do you calculate grams into milliliters?
How many mL is 100 grams of flour?
One - 100 grams portion of white flour converted to milliliter equals to 189.27 ml. via
How do you convert grams to liters?
To convert a gram measurement to a liter measurement, divide the weight by 1,000 times the density of the ingredient or material. Thus, the weight in liters is equal to the grams divided by 1,000 times the density of the ingredient or material. via
How do you measure 1 teaspoon in grams?
How to Convert Grams to Teaspoons. To convert a gram measurement to a teaspoon measurement, divide the weight by 4.928922 times the density of the ingredient or material. Thus, the weight in teaspoons is equal to the grams divided by 4.928922 times the density of the ingredient or material. via
How many teaspoons is 4 grams?
Sliding down the label to the total carbohydrates it reads sugars “4g,” or “4 grams.” This important bit of information is your key to converting grams into teaspoons. Four grams of sugar is equal to one teaspoon. via
What weighs 1 gram exactly?
This is referring to American currency, which means it could also be stated as American paper currency weighs 1 gram. Because currency in other countries may not have the same dimensions, density of ink, or weight of paper, it cannot be generalized as all paper currency. via
How many tbsp is 100 grams?
The answer is: The change of 1 100g ( - 100 grams portion ) unit in a butter measure equals = into 7.05 tbsp ( tablespoon ) as per the equivalent measure and for the same butter type. via
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CC-MAIN-2021-43
| 1,565 | 13 |
https://www.slideshare.net/PatrickCole/physics-test-study-guide-10575566
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math
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PHYSICS STUDY GUIDEYour upcoming physics test primarily covers material from Chapters 18 and 19 in your textbook. You’ll also want to review yourfoldables, notes, in-class work, handouts, homework and quizzes. Be sure to check the class blog (website) as well. Stuff youwill need to know includes, but is not limited to: Perspective Motion (what it means) Motion Diagrams Scalar quantities vs. Vector Distance vs. Displacement Distance vs. Time graphs quantities Speed Velocity Velocity vs. Time graphs Acceleration Forces Force diagrams Balanced vs. Unbalanced Forces Stuff from our previous unit on “SPACE.” We don’t want you forgetting stuff now that we’ve moved on.1. What does it mean for an object to be in motion?2. An object is in motion if it is moving relative to a(n)______________________________3. Think about our very first physics discussion, with the ball. Why can’t you answer the question, “Is the ball moving?” with a simple “Yes,” or “No?”4. Think about your ride home from school. Imagine yourself sitting on the bus (or car), and observing the outside world. Explain why you are and are not moving on your way home from school.5. What are the four ways we discussed that an object can move? Name them, and draw the corresponding motion diagram (Hansel and Gretl dots). I. __________________________ II. __________________________ III. ___________________________ IV. ___________________________6. Describe the motion in the following diagrams. Assume that the object is moving from RIGHT TO LEFT ObjectA.
PHYSICS STUDY GUIDEB.C.7. What is SPEED? What is the equation you would use to calculate an object’s speed?8. What is the difference between AVERAGE SPEED and INSTANTANEOUS SPEED?9. What is VELOCITY? How can velocity change?10. What is ACCELERATION? How would you calculate acceleration of an object?11. Which of these two DISTANCE VS. TIME graphs shows periods of constant speed? Explain your answer.12. Describe the motion of the object in Graph A.
PHYSICS STUDY GUIDE13. Which of the two graphs on the previous page shows acceleration? How do you know?14. Using Graph A on the previous page, calculate the average speed of the object in motion from 12 s to 20 s.15. Compare Graphs A and B on the previous page. At a time of 2 seconds, which graph shows a greater velocity? How do you know?16. Identify the following VELOCITY VS. TIME graphs as CONSTANT SPEED, INCREASING SPEED or DECREASING SPEED. A._______________________ B.______________________ C._______________________17. On a DISTANCE VS. TIME graph, a flat line segment means ____________________________________. However, on a VELOCITY VS. TIME graph, a flat line segment means _____________________________.18. What is a FORCE? Explain the difference between BALANCED and UNBALANCED FORCES.19. Friction acts in a direction _____________________________________________ to the direction of an object’s motion. a. unrelated b. opposite c. equal d. perpendicular
PHYSICS STUDY GUIDE Joe pushing and cabinet moving Draw and label arrows to show which FORCES are involved in this situation. Remember to make your arrows the appropriate sizes. Indicate whether the forces are BALANCED or UNBALANCED. 20. If you increase the force on an object, its acceleration a. decreases. b. stays the same. c. also increases. d. stops. 21. The following diagrams show a skier accelerating down a hill. In these diagrams F G = Gravity. FN = Normal Force, FF = Friction and FAR = Air Resistance. THE SKIER IS NOT PUSHING HIMSELF WITH HIS POLES. In which diagram are the forces labeled correctly?A FF B FF FN FAR FG FN FG FARC D FN FN FAR FF FF FAR FG FG
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CC-MAIN-2017-51
| 3,650 | 4 |
https://lists.aswf.io/g/osl-dev/message/73
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math
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Re: Review: fix derivatives of mod()
Larry Gritz <l...@...>
On Jan 15, 2010, at 8:54 PM, Jonathan Gibbs wrote:
Well, I'm not sure what to think. The derivatives of mod(x,y) is the
Well, there's "math" and then there's "what's useful in shaders." In the strict mathematical sense, the derivative is undefined at the discontinuity (x==d), though you could speak of the limit as you approach from x<d versus x>d. My guess is going to match at least one of those! I'm not sure what we can do that's any smarter or more useful in a case like this.
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s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296944606.5/warc/CC-MAIN-20230323003026-20230323033026-00112.warc.gz
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CC-MAIN-2023-14
| 542 | 5 |
https://www.inquirymaths.com/home/number-prompts/dividing-by-a-fraction
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math
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Dividing by a fraction inquiry
Mathematical inquiry processes: Extend a pattern; find connections; generalise; reason about structure. Conceptual field of inquiry: Division of whole numbers by fractions; division and multiplication.
The prompt was designed to expose and correct a common misconception among secondary school students. When shown the prompt, they will often claim that fifty divided by a half is 25 and 20 divided by a fifth is four. Therefore, they reason, the two sides are unequal and the equation is false.
The inquiry normally starts in one of two ways. Either there will universal agreement that the prompt is not true, or one student or a few will argue that it is. (If the majority can explain why it is true, then the prompt is not appropriate for the class because it will hold no intrigue.)
In the first case, the teacher should assert the equation is true and ask each student to think why that might be before students pair up to discuss and then share their thinking. Often, at this point, students will speculate that the reason is related to 50 x 2 = 20 x 5.
The teacher, using a diagram if appropriate, should link the speculation to the fact that there are two halves in one whole, four halves in two, six halves in three and so on. Following the pattern, we have 100 halves in 50. Similarly, there are five fifths in one whole, 10 fifths in two, and 100 fifths in 20.
In the second case, the teacher will draw on the existing knowledge of one or a few students to co-construct a convincing argument.
Once the class accepts the prompt is true, the teacher might structure the inquiry by following the first line of inquiry below. Alternatively, if the class has experience in carrying out inquiries, the teacher might offer a selection of regulatory cards:
Students select a card and explain how they will continue the inquiry. They might justify their selection in one of the following ways:
Find more examples: "I will extend the the chain of equations that equal 100."
Practise a procedure: "I would like questions with which to develop fluency in dividing by a fraction."
Change the prompt: "I want to find another chain by starting with a different number or by making the equations equal to a different number (not 100)."
Explain a pattern: "I would like to explain why the prompt is true."
Lines of inquiry
1. Extending the prompt
Is it possible to find more equations of the same type (i.e. with unit fractions) that equal 100? Can we construct a chain of the equations? How do we know when the chain is complete? What is the connection with the factors of 100? Is it possible to extend the chain to the left and continue to use unit fractions? (N.B. We could extend the chain to the right using a half, a quarter, an eighth and so on, but the teacher might decide to restrict the inquiry to whole numbers in order to emphasise the connection to factors.)
2. Starting with a different number
What would happen if we were to start the chain with a different whole number? Would the chain be longer or shorter than the one we made from the prompt? What number could we start with to make the chain longer? Is it possible to make a chain of equal length?
3. Using multiples
What would happen if we were to construct a chain of equations in which the whole numbers are consecutive multiples? Would there be a connection between the fractions? Could we generalise for any chain of the same type?
4. Using a different rule
What would happen if the whole numbers were to follow a different rule? What if they doubled each time or the difference between them increased by 10 (for example, 20, 30, 50, 80)? Is there a connection between the fractions? Could we generalise for any chain of the same type?
5. Creating problems
Is it possible to find the missing fractions in these equations? Is it possible to create an equation of the same type that has no solution?
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CC-MAIN-2024-18
| 3,898 | 24 |
https://www.techspot.com/community/topics/can-anyone-explain-this.2747/
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math
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Can anyone explain this? Look closely...
Look at the circled areas... The diagonal line has been bent, bent upwards on the top one and bent downwards on the bottom one. Its very slight, thats why at first glance you don't notice it.
wow, i saw this years ago, and never able to explain it....
i finally understand...
Actually, the top and bottom pieces are exactly the same. The explaination is that there is a difference in slope between the large triangle and small triangle. When re-arranged, the difference in slope is slight, but yields enough for a whole empty unit. The graph adds to the illusion. Good job!
To think that I printed it out and cut it up with scissors...:blush:
I did that too...hehe
You probably put more effort into it than anyone else............Wasted effort maybe, but still effort:grinthumb
You ought some people just like to do this kind of challenging stuffs.
Somethings not right, I printed it out cut it out really carefully and it fits. I took lots of care
Its a mad world
Hey, never mind that, there is a way of folding a strip of paper in such a way that it has only one surface....
Flip one side over and attach the ends. You can also make a solid with one side the same way.
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s3://commoncrawl/crawl-data/CC-MAIN-2018-05/segments/1516084889660.55/warc/CC-MAIN-20180120142458-20180120162458-00699.warc.gz
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CC-MAIN-2018-05
| 1,211 | 13 |
https://math.answers.com/Q/What_is_an_object_that_is_one_kilometer_long
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math
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A Kilometer equals one thousand meters, 0.6214 miles, or 3280.8 feet. Well, if we go to the Cosmology field.. It is to be said that an Object which just hit the jupiter, if it was suppose to hit the Earth.. The length of that object would be a Kilometer Long, length of a measuring tape used by civil engineers could be 1 kilometer long.
i hope that helped
Half of two kilometers
A bridge might be one kilometer long.
..... If you have a household runway maybe
well to be exact i rlly don't know i am trying to figure out what in normal hose hold object is a kilometer long and it is due for homework i don't need to know about Jupiter!
1,000 meters is equal to 1 kilometerKilo means thousand. One thousand meters in one kilometer.
One kilometer is 39,370 inches or 3,280.8 feet.
It would depend on the velocity of the object you are measuring.
1/4 of a hour
Actually, a mile is usually not considered large. It's long. So, a mile is LONGER than a kilometer. ============ A mile is much larger than a kilometer.
One km (kilometer) equates to about 0.62 mile.
one kilometer is equal to 1000 metres
Depends on the acceleration of the object.
3.4 centimeters per year, so multiply that by one mile or one kilometer (0.6 of a mile)
One kilometer is equal to 0.621371 miles. One kilometer is equal to 1000 meters and also equal to 39370.1 inches.
1 kilometre = 1000 metres.
One kilometer is 1000 meters, so a kilometer is bigger.
A kilometer is 1,000 meters and they typical soccer field is 100 meters. Therefore, it would take ten soccer fields to be one kilometer, not 100.
1 kilometer is 0.62137 miles.
1 kilometer=0.621371192 miles
One kilometer is 1,000 meters.
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http://www.koreascience.or.kr/article/JAKO201813164519224.page
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math
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- Volume 29 Issue 2
DOI QR Code
Research Trends in Quantum Computational Algorithms for Cryptanalysis
암호해독을 위한 양자 계산 알고리즘의 최근 연구동향
- Bae, Eunok (Department of Mathematics and Research Institute for Basic Sciences, Kyung Hee University) ;
- Kim, Jeong San (Department of Applied Mathematics and Institute of Natural Sciences, Kyung Hee University) ;
- Lee, Soojoon (Department of Mathematics and Research Institute for Basic Sciences, Kyung Hee University)
- Received : 2018.02.14
- Accepted : 2018.03.12
- Published : 2018.04.25
In this paper, we mainly introduce some quantum computational algorithms that have exponential speedups over the best known classical algorithms, and summarize recent research achievements in quantum algorithms that can affect existing cryptosystems. Finally, we suggest a research direction that can improve these results more progressively.
Supported by : 한국연구재단
- J. Kim, Y. Lim, E. Bae, and D. Kim, "A research on the technique of cryptosystem security analysis using quantum computational algorithms" (in Korean), National Security Research Institute Report (Grant No. 2017-013, 2017).
- P. W. Shor, "Algorithms for quantum computation: discrete logarithms and factoring," in Proc. 35th Annual IEEE Symposium on the Foundations of Computer Science (IEEE Computer Society Press, Piscataway, NJ, USA, 1994), SIAM J. Comput. 26, 1484-1509 (1997).
- L. K. Grover, "A fast quantum mechanical algorithm for database search" in Proc. 28th Annual ACM Symposium on Theory of Computing (ACM, NY, USA, 1996), Phys. Rev. Lett. 79, 325-328 (1997).
- D. Boneh and R. Lipton, "Quantum cryptanalysis of hidden linear functions," in Proc. Crypto'95, LNCS 963, 427-437 (1995).
- A. Y. Kitaev, "Quantum measurements and the abelian stabilizer problem," arXiv:quant-ph/9511026v1 (1995).
- M. Ettinger and P. Hoyer, "A quantum observable for the graph isomorphism problem," arXiv:quant-ph/9901029v1 (1999).
- S. Hallgren, "The hidden subgroup problem and quantum computing using group representations," SIAM J. Comput. 32, 916-934 (2003). https://doi.org/10.1137/S009753970139450X
- M. Grigni, L. Schulman, M. Vazirani, and U. Vazirani, "Quantum mechanical algorithms for the non-abelian hidden subgroup problem," in Proc. 33rd Annual ACM Symposium on Theory of Computing (2001), Combinatorica 24, 137-154 (2004).
- K. Friedl, G. Ivanyos, F. Magniez, M. Santha, and P. Sen, "Hidden translation and translating coset in quantum computing," in Proc. 35th Annual ACM Symposium on Theory of Computing (2003), SIAM J. Comput. 43, 1-24 (2014).
- G. Kuperberg, "A subexponential-time quantum algorithm for the dihedral hidden subgroup problem," SIAM J. Comput. 35, 170-188 (2005). https://doi.org/10.1137/S0097539703436345
- M. Ettinger, P. Hoyer, and E. Knill, "The quantum query complexity of the hidden subgroup problem is polynomial," Inf. Process. Lett. 91, 43-48 (2004). https://doi.org/10.1016/j.ipl.2004.01.024
- D. Gavinsky, "Quantum solution to the hidden subgroup problem for poly-near-hamiltonian groups," Quantum Inf. Comput. 4, 229-235 (2004).
- Y. Inui and F. Le Gall, "Efficient quantum algorithm for the hidden subgroup problem over a class of semi-direct product groups," Quantum Inf. Comput. 7, 559-570 (2007).
- C. Moore, D. N. Rockmore, A. Russell, and L. J. Schulman, "The power of strong Fourier sampling: Quantum algorithms for affine groups and hidden shifts," in Proc. 15th Annual ACM-SIAM Symposium on Discrete Algorithms (SIAM, Philadelphia, USA, 2004), SIAM J. Comput. 37, 938-958 (2007).
- O. Regev, "A subexponential-time algorithm for the dihedral hidden subgroup problem with polynomial space," arXiv: quant-ph/0406151v1 (2004).
- D. Bacon, A. Childs, and W. van Dam, "From optimal measurement to efficient quantum algorithms for the hidden subgroup problem over semidirect product groups," in Proc. 46th Annual IEEE Symposium on the Foundations of Computer Science, 469-478 (2005).
- O. Regev, "Quantum computation and lattice problems," in Proc. 43rd Annual IEEE Symposium on the Foundations of Computer Science, 520-529 (2002).
- S. Hallgren, C. Moore, M. Rotteler, A. Russell, and P. Sen, "Limitations of quantum coset states for graph isomorphism," in Proc. 38th Annual ACM Symposium on Theory of Computing, 604-617 (2006).
- W. van Dam, S. Hallgren, and L. Ip, "Quantum algorithms for some hidden shift problems," SIAM J. Comput. 36, 763-778 (2006). https://doi.org/10.1137/S009753970343141X
- I. B. Damgard, "On the randomness of Legendre and Jacobi sequences," in Proc. Advances in Cryptology-CRYPTO 1988, 403, 163-172 (1990).
- M. Ozols, M. Roetteler, and J. Roland, "Quantum rejection sampling," in Proc. 3rd Innovations in Theoretical Computer Science Conference, 290-308 (2012).
- O. Regev, "Quantum computation and lattice problems," SIAM J. Comput. 33, 738-760 (2004). https://doi.org/10.1137/S0097539703440678
- S. Hallgren, "Polynomial-time quantum algorithm for Pell's equation and the principal ideal problem," in Proc. 34th Annual ACM Symposium on Theory of Computing (2002), J. ACM 54, 1-19 (2007).
- S. Hallgren, "Fast quantum algorithms for computing the unit group and class group of a number field," in Proc. 37th Annual ACM Symposium on Theory of Computing, 468-474 (2005).
- A. Schmidt and U. Vollmer, "Polynomial-time quantum algorithm for the computation of the unit group of a number field," in Proc. 37th Annual ACM Symposium on Theory of Computing, 475-480 (2005).
- K. Eisentrager, S. Hallgren, A. Kitaev, and F. Song, "A quanum algorithm for computing the unit group of an arbitrary degree number field," in Proc. 46th Annual ACM Symposium on Theory of Computing, 293-302 (2014).
- J. F. Biasse and F. Song, "Efficient quantum algorithms for computing class groups and solving the principal ideal problem in arbitrary degree number fields," in Proc. 27th Annual ACM-SIAM Symposium on Discrete Algorithms, (2016).
- E. Bae and S. Lee, "Quantum algorithm for continuous hidden shift problems" in preparation.
- C. Gentry and S. Halevi, "Implementing gentry's fullyhomomorphic encryption scheme," in Proc. Eurocrypt 2011, 132-150 (2011).
- V. Lyubashevsky, C. Peikert, and O. Regev, "On ideal lattices and learning with errors over rings," in Proc. Advances in cryptology-CRYPTO 2010, 6110, 1-23 (2010).
- Z. Brakerski and V. Vaikuntanathan, "Fully homomorphic encryption from ring-LWE and security for key dependent messages," in Proc. Advances in cryptology-Eurocrypt 2011, 6841, 505-524 (2011).
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| 6,522 | 43 |
https://www.utwente.nl/en/education/master/programmes/systems-control/specialisation/control-theory/
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math
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This specialisation belongs to the Master's Systems Control.
In this specialisation the main focus is laid upon the study of the core of the Systems and Control discipline: to mathematically describe a system, to study its properties and to adapt it such that it behaves in a desired way (the control problem). So you will concentrate on mathematical theory, although practical problems are always nearby, for inspiration and application. As a student, you will be able to tailor the programme to address your own individual interests and needs.
A mathematical model of a system should reflect its main features. It may be represented by difference or differential equations, but also by inequalities, algebraic equations, and logical constraints.
System models are linear or non-linear. Linear models are relatively simple and convenient. However, if non-linearities play a prominent role, then important system properties may be missed by a linear model. Take as an example an autopilot for an airplane. If it should keep the plane at a fixed heading, speed and height, then its design can be based on a linear model. If it should control take-off and landing, then non-linear models are necessary. For the design of an autopilot for the airplane, we can use a model, which is based on switching between a finite number of linear models. This yields so-called hybrid models and there are many challenging questions for this class of systems.
Simple system models can be studied analytically by working out mathematical equations. Simulation of the mathematical model, however, is necessary for the analysis of more complicated systems. Think for instance of a robot, or think of a big airplane or an economical system. However, always checks using simplified analytical models are necessary: simulations alone cannot prove the correctness of the underlying system models.
By choosing and controlling inputs or, more general, by imposing additional constraints on some of the variables, the system may be influenced so as to obtain certain desired behaviour. This is the control problem. For the case of a simple linear systems, control strategies have been well developed. However, consider for example the problem of controlling the temperature in a bulk storage room. The model is given by a partial differential equation describing the temperature distribution in the storage room. A cooling device can be switched on and off (non-linearity) to control the product temperature (which reflects a continuum of quantities to control). Recently, research has been completed to design a controller that robustly controls the switching times.
Researchers in Control Theory today develop the theories and strategies that will be used in the applications of tomorrow. But from society there is a constant pressure to invent the required theories for necessary applications today. Therefore control theory specialists also may work in larger teams working at complicated applications to provide required control solutions “on the fly”.
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s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764500250.51/warc/CC-MAIN-20230205063441-20230205093441-00114.warc.gz
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CC-MAIN-2023-06
| 3,035 | 7 |
http://www.mamzelle-deybow.com/pdf/download-epub-pdf-torrent-e-book-analysis-on-real-and-complex-126732.html
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math
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It was a universal belief among ancient civilizations that life came originally from the cosmos, and ultimately would return there after death. The shamanic journey was always to this sky-world - and it appears that it was always located in the direction of the stars of Cygnus - also known as the Northern Cross - accessed either via the Milky Way…
The next chapter is an introduction to real and complex manifolds. It contains an exposition of the theorem of Frobenius, the lemmata of Poincaré and Grothendieck with applications of Grothendieck's lemma to complex analysis, the imbedding theorem of Whitney and Thom's transversality theorem. Chapter 3 includes characterizations of linear differentiable operators, due to Peetre and Hormander. The inequalities of Garding and of Friedrichs on elliptic operators are proved and are used to prove the regularity of weak solutions of elliptic equations. The chapter ends with the approximation theorem of Malgrange-Lax and its application to the proof of the Runge theorem on open Riemann surfaces due to Behnke and Stein.
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s3://commoncrawl/crawl-data/CC-MAIN-2016-44/segments/1476988719437.30/warc/CC-MAIN-20161020183839-00336-ip-10-171-6-4.ec2.internal.warc.gz
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CC-MAIN-2016-44
| 1,397 | 8 |
https://drum.lib.umd.edu/handle/1903/5818
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math
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Mixed H2 /H∞ Optimization: A BMI Solution
MetadataShow full item record
The mixed H2 /H∞ problem arises as one means to achieve robust good performance for a controlled linear time-invariant system. The idea is to achieve H2 optimal performance subject to an H∞ bound as robustness. It has been difficult to find good algorithms for solving this problem. In this study the problem was transformed into a bilinear matrix inequality problem (BMI). Solving the BMI by the method of centers was shown to be complicated by discontinuities resulting from the unobservability of the closed-loop system. Transforming the BMI problem into a lattice of BMI subproblems makes it possible to avoid the discontinuity and solve the original problem. A robust flight control system for the F-14 is included as an example of the algorithm.
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http://fitness.stackexchange.com/questions/tagged/bodyweight-exercises+weight-loss
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math
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Physical Fitness Meta
to customize your list.
more stack exchange communities
Start here for a quick overview of the site
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Are so-called “burst workouts” effective for weight loss?
I just bought a $2.99 Kindle book called Burst Workouts which recommends very short workouts at high intensities. His main program recommends timed interval training using bodyweight exercises of 30 ...
Sep 19 '13 at 3:49
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http://mathhelpforum.com/math-topics/157554-air-resistance.html
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math
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You can get the of if you apply Newton's Second Law before the biker brakes. What are the forces on the biker before he brakes, and what is the acceleration?
Incidentally, I would think that a gradient of would imply that the angle of inclination of the hill is given by not
I don't think there's a way to get an exact numerical answer without finding , but as I've said, I think there's a way to find .
Can you see your way forward now?
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http://qs321.pair.com/~monkads/?node_id=162512
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math
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Before you get too carried away, you need to correct some business logic first. You say the following is your desired output:
1st Place: user3 (232 Points)
2nd Place: user1 (200 Points)
3rd Place: user2 (190 Points), user5 (190 Points)
4th Place: user4 (187 Points)
What you have to do is where two (or more) people tie for a place, you have to skip that number of places for the next place. Thus in the above situation you have no "4th place"; rather, user 4 is in 5th place. Of course, had three people tied for 3rd place, then the next place awarded would be 6th place.
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http://agxl.co.in/question.php?classlink=class-ii&topiclink=simplification
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math
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Free notebooks were distributed equally among children of a class. The number of notebooks each child got was one-eighth of the number of children. Had the number of children been half, each child would have got 16 notebooks. Total how many notebooks were distributed ?
In a regular week, there are 5 working days and for each day, the working hours are 8. A man gets Rs. 2.40 per hour for regular work and Rs. 3.20 per hours for overtime. If he earns Rs. 432 in 4 weeks, then how many hours does he work for ?
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https://inis.iaea.org/search/search.aspx?orig_q=author:%22Sanchez,%20N.G.%22
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math
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Results 1 - 10 of 13
Results 1 - 10 of 13. Search took: 0.019 seconds
|Sort by: date | relevance|
[en] The scattering of scalar waves from a Schwarzschild black hole is investigated for wavelengths much less than the graviational radius (r/sub s/). Explicit expressions for scattering parameters are obtained for two cases: high angular momenta and low angular momenta. In the first case we obtain the phase shifts and absorption coefficient with the JWKB method. The elastic differential cross section and the total absorption cross section are also calculated. For low angular momenta we present a method based in the DWBA (distorted wave Born approximation). With this method, the phase shifts and the absorption coefficients are obtained
[en] The cosmic microwave background power spectra are studied for different families of single field new and chaotic inflation models in the effective field theory approach to inflation. We implement a systematic expansion in 1/Ne, where Ne∼50 is the number of e-folds before the end of inflation. We study the dependence of the observables (ns, r and dns/d ln k) on the degree of the potential (2n) and confront them to the WMAP3 and large scale structure data: This shows in general that fourth degree potentials (n=2) provide the best fit to the data; the window of consistency with the WMAP3 and LSS data narrows for growing n. New inflation yields a good fit to the r and ns data in a wide range of field and parameter space. Small field inflation yields r < 0.16 while large field inflation yields r > 0.16 (for Ne=50). All members of the new inflation family predict a small but negative running -4(n+1) x 10-4 (le) dns/d ln k (le) -2 x 10-4. (The values of r, ns, dns/d ln k for arbitrary Ne follow by a simple rescaling from the Ne=50 values.) A reconstruction program is carried out suggesting quite generally that for ns consistent with the WMAP3 and LSS data and r < 0.1 the symmetry breaking scale for new inflation is |φ0|∼10 MPl while the field scale at Hubble crossing is |φc| ∼ MPl. The family of chaotic models features r (ge) 0.16 (for Ne=50) and only a restricted subset of chaotic models are consistent with the combined WMAP3 bounds on r, ns, dns/d ln k with a narrow window in field amplitude around |φc|∼15 MPl. We conclude that a measurement of r < 0.16 (for Ne=50) distinctly rules out a large class of chaotic scenarios and favors small field new inflationary models. As a general consequence, new inflation emerges more favored than chaotic inflation
[en] In TeV scale unification models, gravity propagates in 4+δ dimensions while gauge and matter fields are confined to a four dimensional brane, with gravity becoming strong at the TeV scale. For a such scenario, we study strong gravitational interactions in a effective Schwarzschild geometry. Two distinct regimes appear. For large impact parameters, the ratio ρ∼(Rs/r0)1+δ (with Rs the Schwarzschild radius and r0 the closest approach to the black hole), is small and the deflection angle χ is proportional to ρ (this is like Rutherford-type scattering). For small impact parameters, the deflection angle χ develops a logarithmic singularity and becomes infinite for ρ=ρcrit=2/(3+δ). This singularity is reflected into a strong enhancement of the backward scattering (like a glory-type effect). We suggest as distinctive signature of black hole formation in particle collisions at TeV energies, the observation of backward scattering events and their associated diffractive effects
[en] The Thomas-Fermi approach to galaxy structure determines self-consistently and non-linearly the gravitational potential of the fermionic warm dark matter (WDM) particles given their quantum distribution function f(E). This semiclassical framework accounts for the quantum nature and high number of DM particles, properly describing gravitational bounded and quantum macroscopic systems as neutron stars, white dwarfs and WDM galaxies. We express the main galaxy magnitudes as the halo radius r_h, mass M_h, velocity dispersion and phase space density in terms of the surface density which is important to confront to observations. From these expressions we derive the general equation of state for galaxies, i.e., the relation between pressure and density, and provide its analytic expression. Two regimes clearly show up: (1) Large diluted galaxies for M_h >or similar 2.3 x 10"6 M _C_i_r_c_l_e_D_o_t and effective temperatures T_0 > 0.017 K described by the classical self-gravitating WDM Boltzman gas with a space-dependent perfect gas equation of state, and (2) Compact dwarf galaxies for 1.6 x 10"6 M _C_i_r_c_l_e_D_o_t >or similar M_h >or similar M_h_,_m_i_n ≅ 3.10 x 10"4 (2 keV/m)"("1"6")"/"("5") M _C_i_r_c_l_e_D_o_t, T_0 < 0.011 K described by the quantum fermionic WDM regime with a steeper equation of state close to the degenerate state. In particular, the T_0 = 0 degenerate or extreme quantum limit yields the most compact and smallest galaxy. In the diluted regime, the halo radius r_h, the squared velocity v"2(r_h) and the temperature T_0 turn to exhibit square-root of M_h scaling laws. The normalized density profiles ρ(r)/ρ(0) and the normalized velocity profiles v"2(r)/v"2(0) are universal functions of r/r_h reflecting the WDM perfect gas behavior in this regime. These theoretical results contrasted to robust and independent sets of galaxy data remarkably reproduce the observations. For the small galaxies, 10"6 >or similar M_h ≥ M_h_,_m_i_n, the equation of state is galaxy mass dependent and the density and velocity profiles are not anymore universal, accounting to the quantum physics of the self-gravitating WDM fermions in the compact regime (near, but not at, the degenerate state). It would be extremely interesting to dispose of dwarf galaxy observations which could check these quantum effects. (orig.) 1
[en] We provide a unified formula for the quantum decay rate of heavy objects (particles) whatever they may be: topological and nontopological solitons, X particles, cosmic defects, microscopic black holes, fundamental strings, as well as the particle decays in the standard model. Extreme energy cosmic ray (EECR) top-down scenarios are based on relics from the early Universe. The key point in the top-down scenarios is the necessity to adjust the lifetime of the heavy object to the age of the Universe. This ad hoc requirement needs a very high dimensional operator to govern its decay and/or an extremely small coupling constant. The arguments produced to fine-tune the relic lifetime to the age of the Universe are critically analyzed. The natural lifetimes of such heavy objects are, however, microscopic times associated with the grand unified theory energy scale (∼10-28 sec or shorter). It is at this energy scale (by the end of inflation) that they could have been abundantly formed in the early Universe, and it seems natural that they decayed shortly after being formed. The annihilation scenario for EECRs ('wimpzillas') is also considered and its inconsistencies analyzed
[en] Recently, Warm (keV scale) Dark Matter emerged impressively over CDM (Cold Dark Matter) as the leading Dark Matter candidate. In the context of this new Dark Matter situation, which implies novelties in the astrophysical, cosmological and keV particle physics context, this 16. Paris Colloquium 2012 is devoted to the LambdaWDM Standard Model of the Universe. The topics of the colloquium are as follows: -) observational and theoretical progress on the nature of dark matter: keV scale warm dark matter, -) large and small scale structure formation in agreement with observations at large scales and small galactic scales, and -) neutrinos in astrophysics and cosmology. This document gathers the slides of the presentations.
[en] We develop the cluster expansion and the Mayer expansion for the self-gravitating thermal gas and prove the existence and stability of the thermodynamic limit N,V→∞ with N/V13 fixed. The essential (dimensionless) variable is here η=Gm2N(V13T) (which is kept fixed in the thermodynamic limit). We succeed in this way to obtain the expansion of the grand canonical partition function in powers of the fugacity. The corresponding cluster coefficients behave in the thermodynamic limit as (ηN)j-1cj, where cj are pure numbers. They are expressed as integrals associated to tree cluster diagrams. A bilinear recurrence relation for the coefficients cj is obtained from the mean field equations in the Abel's form. In this way the large j behaviour of the cj is calculated. This large j behaviour provides the position of the nearest singularity which corresponds to the critical point (collapse) of the self-gravitating gas in the grand canonical ensemble. Finally, we discuss why other attempts to define a thermodynamic limit for the self-gravitating gas fail
[en] We clarify inflaton models by considering them as effective field theories in the Ginzburg-Landau spirit. In this new approach, the precise form of the inflationary potential is constructed from the present WMAP data, and a useful scheme is prepared to confront with the forthcoming data. In this approach, the WMAP statement excluding the pure φ4 potential implies the presence of an inflaton mass term at the scale m∼1013 GeV. Chaotic, new and hybrid inflation models are studied in an unified way. In all cases the inflaton potential takes the form V(φ)=m2MPl2v(φ/MPl), where all coefficients in the polynomial v(φ) are of order one. If such potential corresponds to supersymmetry breaking, the corresponding susy breaking scale is √(mMPl)∼1016 GeV which turns to coincide with the grand unification (GUT) scale. The inflaton mass is therefore given by a seesaw formula m∼MGUT2/MPl. The observables turn to be two-valued functions: one branch corresponds to new inflation and the other to chaotic inflation, the branch point being the pure quadratic potential. For red tilted spectrum, the potential which fits the best the present data (vertical bar 1-ns vertical bar < or approx. 0.1,r < or approx. 0.1) and which best prepares the way for the forthcoming data is a trinomial polynomial with negative quadratic term (new inflation). For blue tilted spectrum, hybrid inflation turns to be the best choice. In both cases we find an analytic formula relating the inflaton mass with the ratio r of tensor to scalar perturbations and the spectral index ns of scalar perturbations: 106(m/MPl)=127√(r vertical bar 1-ns vertical bar) where the numerical coefficient is fixed by the WMAP amplitude of adiabatic perturbations. Implications for string theory are discussed
[en] Research highlights: → In Ginsburg-Landau (G-L) approach data favors new inflation over chaotic inflation. → ns and r fall inside a universal banana-shaped region in G-L new inflation. → The banana region for the observed value ns=0.964 implies 0.021< r<0.053. → Fermion condensate inflaton potential is a double well in the G-L class. - Abstract: The MCMC analysis of the CMB + LSS data in the context of the Ginsburg-Landau approach to inflation indicated that the fourth degree double-well inflaton potential in new inflation gives an excellent fit of the present CMB and LSS data. This provided a lower bound for the ratio r of the tensor to scalar fluctuations and as most probable value r ≅ 0.05, within reach of the forthcoming CMB observations. In this paper we systematically analyze the effects of arbitrarily higher order terms in the inflaton potential on the CMB observables: spectral index ns and ratio r. Furthermore, we compute in close form the inflaton potential dynamically generated when the inflaton field is a fermion condensate in the inflationary universe. This inflaton potential turns out to belong to the Ginsburg-Landau class too. The theoretical values in the (ns, r) plane for all double well inflaton potentials in the Ginsburg-Landau approach (including the potential generated by fermions) fall inside a universal banana-shaped region B. The upper border of the banana-shaped region B is given by the fourth order double-well potential and provides an upper bound for the ratio r. The lower border of B is defined by the quadratic plus an infinite barrier inflaton potential and provides a lower bound for the ratio r. For example, the current best value of the spectral index ns = 0.964, implies r is in the interval: 0.021 < r < 0.053. Interestingly enough, this range is within reach of forthcoming CMB observations.
[en] We compute the primordial scalar, vector and tensor metric perturbations arising from quantum field inflation. Quantum field inflation takes into account the nonperturbative quantum dynamics of the inflaton consistently coupled to the dynamics of the (classical) cosmological metric. For chaotic inflation, the quantum treatment avoids the unnatural requirements of an initial state with all the energy in the zero mode. For new inflation it allows a consistent treatment of the explosive particle production due to spinodal instabilities. Quantum field inflation (under conditions that are the quantum analog of slow-roll) leads, upon evolution, to the formation of a condensate starting a regime of effective classical inflation. We compute the primordial perturbations taking the dominant quantum effects into account. The results for the scalar, vector and tensor primordial perturbations are expressed in terms of the classical inflation results. For a N-component field in a O(N) symmetric model, adiabatic fluctuations dominate while isocurvature or entropy fluctuations are negligible. The results agree with the current Wilkinson Microwave Anisotropy Probe observations and predict corrections to the power spectrum in classical inflation. Such corrections are estimated to be of the order of (m2/NH2), where m is the inflaton mass and H the Hubble constant at the moment of horizon crossing. An upper estimate turns to be about 4% for the cosmologically relevant scales. This quantum field treatment of inflation provides the foundations to the classical inflation and permits to compute quantum corrections to it
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http://www.dissertation.com/abstracts/2219040
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math
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|Institution:||University of Washington|
|Full text PDF:||http://hdl.handle.net/1773/40930|
Convex optimization is more popular than ever, with extensive applications in statistics, machine learning, and engineering. Nesterov introduced optimal first-order methods for large scale convex optimization in the 1980s, and extremely fast interior point methods for small-to-medium scale convex optimization emerged in the 1990s. Today there is little reason to prefer modelling with linear programming over convex programming for computational reasons. Nonetheless, there is room to improve the already sophisticated algorithms for convex optimization. The thesis makes three primary contributions to convex optimization. First, the thesis develops new, near optimal barriers for generalized power cones. This is relevant because the performance of interior point methods depends on representing convex sets with small parameter barriers. Second, the thesis introduces an intuitive, first-order method that achieves the best theoretical convergence rate and has better performance in practice than Nesterovs method. The thesis concludes with a framework for reformulating a convex program by interchanging the objective function and a constraint function. The approach is illustrated on several examples.Advisors/Committee Members: Drusvyatskiy, Dmitriy (advisor).
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CC-MAIN-2022-33
| 1,360 | 3 |
https://www.physicsforums.com/threads/ball-spinning-around-a-rod.217235/
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math
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1. The problem statement, all variables and given/known data A 4.05 kg object is attached to a vertical rod by two strings as in Figure P6.11. The object rotates in a horizontal circle at constant speed 6.50 m/s. http://www.webassign.net/pse/p6-11.gif (a) Find the tension in the upper string. (b) Find the tension in the lower string. I have no idea where to begin. Any thoughts?
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https://socratic.org/questions/how-do-you-solve-4x-2-52-solve-using-the-square-root-property
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math
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How do you solve #4x^2=52# solve using the square root property?
Using the square root property to solve this problem involves the following steps.
Isolate the squared term. (Already isolated.)
Take the square of both sides, indicating that the solution can be positive or negative.
Solve for x.
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CC-MAIN-2024-10
| 295 | 5 |
https://findanyanswer.com/which-type-of-wave-has-the-shortest-wavelength
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math
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Which type of wave has the shortest wavelength?
Click to see full answer
Keeping this in consideration, which type of wave has the longest wavelength?
Also Know, which of the following has the smallest wavelength? Gamma Rays-have the smallest wavelengths and the most energy of any other wave in the electromagnetic spectrum.
In this way, what color has the shortest wavelength?
As the full spectrum of visible light travels through a prism, the wavelengths separate into the colors of the rainbow because each color is a different wavelength. Violet has the shortest wavelength, at around 380 nanometers, and red has the longest wavelength, at around 700 nanometers.
Which wave has highest frequency?
1 Answer. Daniel W. Gamma rays have the highest frequency.
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CC-MAIN-2021-10
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https://www.britannica.com/biography/Gregory-Adams-Kimble
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math
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Learn about this topic in these articles:
theory of learning
Recognizing this danger (and the corollary that no definition of learning is likely to be totally satisfactory) a definition proposed in 1961 by G.A. Kimble may be considered representative: Learning is a relatively permanent change in a behavioral potentiality that occurs as a result of reinforced practice. Although the definition is useful, it still leaves problems.
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| 431 | 3 |
https://brainmass.com/math/calculus-and-analysis/accounting-profit-margins-turnover-and-return-on-investme-236308
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math
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The Valve Division of Bendix, Inc, produces a small valve that is used by various companies as a component part in their production. Bendix, Inc., operates its divisions as autonomous units, giving its divisional managers great descretion in pricing and other decisions. Each division is expected to generate a mimimum required rate of return of at least 14% on its operating assets. The Valve Division has average operating assets of $700,000. The valves are sold for $5 each. Variable costs are $3 per valve, and fixed costs total $462,000 per year. The division has a capacity of 300,000 valves each year.
1.) How many valves must the Valve Division sell each year to generate the desired rate of return on its assets?
a.) What is the margin earned at this level of sales?
b.) What is the turnover at this level of sales?
2.) Assume that the Valve Division's current ROI equals the minimum required rate of 14%. In order to increase the division's ROI, the divisional manager wants to increase the selling price per valve by 4%. Market studies indicate that an increase in the selling price would cause sales to drop by 20,000 units each year. However, operating assets could be reduced by $50,000 due to decreased needs for accounts receivable and inventory. Compute the margin, turnover, and ROI if these changes are made.
3.) Refer to the original data. Assume again that the Valve Division's current ROI equals the minimum required rate of 14%. Rather than increase the selling price, the sales manager wants to reduce the selling price per valve by 4%. Market studies indicate that this would fill the plant to capacity. In order to carry the greater level of sales, however, operating assets would increase by $50,000. Compute the margin, turnover, and ROI if these changes are made.
4.) Refer to the original data. Assume that the normal volume of sales is 280,000 valves each year at a price of $5 per valve. Another division of the company is currently purchasing 20,000 valves each year from an overseas supplier, at a price of $4.25 per valve. The manager of the Valve Division has refused to meet this price, pointing out that it would result in a loss for his division:
Selling price per valve: $4.25
Cost per valve:
Fixed ($462,000/ 300,000 valves): $1.54 $4.54
Net loss per valve: $(0.29)
The manager of the Valve Division also points out that the normal $5 selling price barely allows his division to earn the required 14% rate of return. "If we take on some business at only $4.25 per unit, then our ROI is obviously going to suffer," he reasons, "and maintaining that ROI figure is the key to my future. Besides, taking on these extra units would require us to increase our operating assets by at least $50,000 due to the larger inventories and accounts receivable we would be carrying." Would you recommend that the Valve Division sel to the other division at $4.25? Show ROI computations to support your answer.© BrainMass Inc. brainmass.com June 3, 2020, 10:32 pm ad1c9bdddf
This solution explains how to calculate:
1) Expected rate of return and units produced to meet this rate.
2) Margin value based on sales
3) Turnover based on sales
4) Return on investment
These values are found multiple times during the course of four different scenarios.
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s3://commoncrawl/crawl-data/CC-MAIN-2020-34/segments/1596439739073.12/warc/CC-MAIN-20200813191256-20200813221256-00187.warc.gz
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CC-MAIN-2020-34
| 3,272 | 18 |
https://www.fixya.com/support/t1023402-ti_84_will_not_graph_multiple_functions
|
math
|
Question about Texas Instruments TI-84 Plus Calculator
I have had the TI84 for more than a year. Recently, it stopped graphing completely because somehow the PLOT 1 PLOT 2 PLOT 3 at the top of the function (Y =) screen were highlighted. Now I can get one fuction at a time....anyone know how to make the TI84 graph multiple functions simultaneously?
Already tried highlighting multiple functions (and only one equal sign will highlight at a time) and already reset Mode to SIMUL. Have also tried changing the line type at the far left of the function...no effect.
Turn off transformers. Go to apps and go all the way down until you reach transformers then hit enter. Then a menu should appear and hit uninstall that should fix it.
Posted on Sep 13, 2008
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Usually answered in minutes!
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s3://commoncrawl/crawl-data/CC-MAIN-2020-29/segments/1593655912255.54/warc/CC-MAIN-20200710210528-20200711000528-00312.warc.gz
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CC-MAIN-2020-29
| 1,334 | 17 |
http://acpt.nl/blog/forum/how-to-calculate-eps-growth-rate-58e96d
|
math
|
This growth rate is the compound annual growth rate of Diluted Normalised Earnings Per Share over the last 3 years. Although company B has higher Average EPS growth rate , it doesn't indicate that it has higher earning quality. Subtract year 1 cash flows from year 2 cash flows and then divide by year 1 cash flows. Calculate the EPS growth every year since 2002 using the following formula: =AVERAGE((B3-B2)/B2) B3 = The Current Year EPS B2= Last year's EPS. To calculate an annual percentage growth rate over one year, subtract the starting value from the final value, then divide by the starting value. Earnings per share (EPS) is the portion of a company’s profit that is allocated to each outstanding share of common stock and serves as a proxy of the company’s financial health. EPS Growth Rate, 3 Year . The compound 3-year growth rate calculated using the least squares fit over the latest two to three years' earnings per share on a running 12-month basis. The CAGR formula is the following: (current year's EPS / EPS 3 years ago) ^ (1/3) - 1 NOTE: If less than 3 years are available, a 'NA' (Not Available) code will be used. Multiply this result by 100 to get your growth rate displayed as a percentage. Earnings per share (EPS) is the portion of a company's profit allocated to each outstanding share of common stock. Company B has more "bumpy" earning than company A. Earnings Per Share (EPS) Formula. The answer is 1 or 100 percent.
Compounded EPS growth rate unable to reflect such, as it only take consideration into EPS during starting period and ending period. The EPS calculator uses the following basic formula to calculate earnings per share: EPS = (I - D) / S. Where: EPS is the earnings per share, I is the net income of a company, D is the total amount of preferred stock dividends, S is the weighted average number of common shares outstanding In this example, the growth rate is calculated by subtracting $100,000 from $200,000 and then dividing by $100,000. Growth rate will be calculated only if there is a minimum of eight trailing 4-quarter periods of positive earnings (uses a minimum of 11 quarters of data). Earnings per share serve as an indicator of a company's profitability. Calculate the growth rate from year 1 to year 2.
This will give you the EPS growth rate for 1 year period. Keep reading to learn how to calculate annual growth over multiple years!
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s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652663016853.88/warc/CC-MAIN-20220528123744-20220528153744-00364.warc.gz
|
CC-MAIN-2022-21
| 2,396 | 3 |
http://webphysics.davidson.edu/Course_Material/Py120L/Lab7/Lab7_1.html
|
math
|
If we have a system which consists of a number (i) of
masses, each with mass mi with a distance from the axis of
the moment of inertia of the system is defined as
In words, multiply each mass by the square of its
distance from the axis of rotation and sum all of these factors.
Thus we may calculate the moment of inertia of a collection of objects.
Alternatively, we can measure the moment of inertia
of an object (or a system of masses) by measuring the angular acceleration a that results from a given applied torque
rotational analog of Newton’s second law of motion then gives
I = t/a.
In this lab you will experimentally measure the
moment of inertia of an object using Equation 2, and then we will check that
measurement against the theoretical value, Equation 1.
You will use the smart pulley to measure the linear acceleration of an
object subjected to a constant applied torque.
You will use the same apparatus you used to measure centripetal force,
except that the mass and spring have been removed and the crossarm has been
replaced by a threaded rod. Four wing nuts on the threaded rod are used to clamp two
masses near its ends. The
threaded rod and its attached masses are set into rotation by wrapping a
string around the vertical shaft, running the string over the pulley,
attaching the string to a weight hanger and allowing the weight hanger to
The constant tension in the string applied tangent
to the vertical shaft gives rise to a constant torque and consequently to a
constant angular acceleration. This
acceleration will be obtained by measuring the angular velocity of rotation and
using the relation:
We can plot v as a function of t,
yielding a straight line of slope a as indicated in the equation.
(Or you can plot the angular velocity as a function of time and then
relate the linear acceleration to the angular acceleration by a=aR where R is the radius of the vertical shaft.) You will then be asked
to compare the moment of inertia obtained from the acceleration measurement, as
in Equation 2, with the moment of inertia obtained by the use of Equation 1.
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s3://commoncrawl/crawl-data/CC-MAIN-2018-05/segments/1516084891377.59/warc/CC-MAIN-20180122133636-20180122153636-00798.warc.gz
|
CC-MAIN-2018-05
| 2,090 | 33 |
https://www.teachstarter.com/us/learning-area/fractional-operations-us/
|
math
|
A wide range of colorful and engaging fractional operations teaching resources including unit and lesson plans, real-world maths investigations, worksheets, hands-on activities, PowerPoint presentations, posters and much more. Use these educational resources when teaching your students how to identify and work with fractional operations. Use these resources when learning about fractional operations.
Review important 5th-grade math standards with a student-led interactive activity that covers 12 different mathematical concepts.
Use these math mats when reviewing fraction operations with your students.
Get in all 3 Rs (reading, writing, and arithmetic) with this word problem worksheet that lets students practice adding and subtracting like fractions.
Encourage your students to work through 8 pages of 5th-grade math problems while charting their progress to measure their success.
A worksheet to practice multiplying and simplifying fractions.
A 17-slide editable PowerPoint Template to use when teaching your students how to add and subtract fractions.
A worksheet to consolidate students' understanding of adding fractions with common denominators.
Practice finding specific fractions of a set with 16 word problem task cards.
Practice how to simplify fractions before multiplying with this worksheet.
Solve word problems by dividing fractions and whole numbers with this set of 24 task cards.
A set of 3 worksheets to practice adding and subtracting fractions with like and unlike denominators.
Practice modeling and solving questions related to unit fractions and whole number division with this worksheet.
Solve a variety of word problems with this multiplying mixed numbers worksheet.
Engage students with a dominoes game while practicing how to multiply fractions.
Posters outlining the processes involved when adding, subtracting, multiplying, and dividing like and unlike fractions .
Encourage mathematical collaboration and discussion with a group of thirteen 4th-grade fractions partner activities.
Challenge your students to multiply fractions while completing 7 different tasks with this Google slides interactive activity.
Improve understanding of how to multiply mixed numbers with this set of 24 word problem task cards.
Review fraction concepts and practice mathematical constructed response questions with a set of writing about fractions task cards.
Master the ability to multiply mixed numbers with a whole-class bingo game.
Strengthen student understanding of fractional operations with this multiplying mixed numbers worksheet.
Master the ability to divide fractions with a whole-class bingo game.
Review how to multiply mixed numbers with this comparing expressions worksheet.
Improve student understanding of dividing a fraction by a fraction while working through these 3 math mazes.
Multiply mixed numbers to reveal the answer to a joke with this set of 3 riddle worksheets.
Teach your students how to divide a whole number by a fraction and divide a fraction by a whole number with this instructional slide deck and accompanying student notes.
Practice dividing fractions with this set of differentiated worksheets.
Practice multiplying mixed numbers while finding your way through this set of 3 math mazes.
Practice simplifying before multiplying fractions with this match-up activity.
Practice dividing a fraction by a fraction with this set of 24 task cards.
Practice multiplying fractions with a set of 12 task cards.
Strengthen student understanding of how to multiply mixed numbers with this match-up activity.
|
s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296945368.6/warc/CC-MAIN-20230325161021-20230325191021-00391.warc.gz
|
CC-MAIN-2023-14
| 3,552 | 33 |
http://www.math.mcgill.ca/darmon/qvnts/07-08/feigon.html
|
math
|
Brooke Feigon (Toronto)
TITLE: Averages of central L-values of Hilbert modular forms
We use the relative trace formula to obtain exact formulas for central
values of certain twisted quadratic base change L-functions averaged over
Hilbert modular forms of a fixed weight and level. We apply these formulas
to the subconvexity problem for these L-functions. This talk is based on
joint work with David Whitehouse.
|
s3://commoncrawl/crawl-data/CC-MAIN-2018-43/segments/1539583510749.37/warc/CC-MAIN-20181016093012-20181016114512-00150.warc.gz
|
CC-MAIN-2018-43
| 411 | 7 |
http://sakura-wind.deviantart.com/
|
math
|
$7 for an extra character
$5 for Complex Backgrounds (Simple BGs are free)
RULES• I'll be drawing in my style only
• SWF only, blood and gore is ok
• Max limit for characters in one picture is 3
• Send me a note with the title 'Commission' on DeviantArt to discuss about your order
• If you have any visual references, that would be great.
• If you have any questions, you can ask me.
ABOUT PAYMENT• Paypal only
• Do not send the payment until I give out the price.
|
s3://commoncrawl/crawl-data/CC-MAIN-2015-22/segments/1432207927185.70/warc/CC-MAIN-20150521113207-00236-ip-10-180-206-219.ec2.internal.warc.gz
|
CC-MAIN-2015-22
| 481 | 10 |
https://www.asknumbers.com/fahrenheit-to-celsius.aspx
|
math
|
Fahrenheit to Celsius conversion formula:
Celsius = (Fahrenheit - 32) ÷ 1.8 . To convert degrees Fahrenheit to degrees Celsius, subtract 32 from the degree Fahrenheit and divide by 1.8.
Celsius = (Fahrenheit - 32) ÷ 1.8
The 1.8 Celsius to Fahrenheit ratio means there are 1.8 Fahrenheit degrees for each Celsius degree and it is calculated by dividing the difference between boiling and freezing points for each of these temperature scales, that makes (212 °F - 32 °F) / (100 °C - 0 °C) = 1.8 .
The 32 is the difference between freezing point of Fahrenheit (32) and freezing point of Celsius (0).
How to convert Fahrenheit to Celsius?
To convert Fahrenheit to Celsius degrees, fist subtract 32, then divide the result by 1.8 ratio.
For example, to convert from 77 °F to °C, subtract 32 from Fahrenheit first and then divide by 1.8, that makes (77 °F - 32) / 1.8 = 25 °C .
What is Fahrenheit?
Fahrenheit is a temperature scale with the freezing point of water is 32 degrees and the boiling point of water is 212 degrees under standard atmospheric pressure (101.325 kPa). The symbol is "°F".
- Freezing point of water = 32 °F
- Boiling point of water = 212 °F
- Absolute zero = -459.67 °F
- Average body temperature = 98.6 °F (between 97 °F and 99 °F)
What is Celsius?
Celsius (Centigrade) is a temperature scale with the freezing point of water is 0 degree and the boiling point of water is 100 degrees under standard atmospheric pressure (101.325 kPa). The symbol is "°C".
- Freezing point of water = 0 °C
- Boiling point of water = 100 °C
- Absolute zero = -273.15 °C
- Average body temperature = 37 °C (between 36.1 °C and 37.2 °C)
|
s3://commoncrawl/crawl-data/CC-MAIN-2019-35/segments/1566027313747.38/warc/CC-MAIN-20190818083417-20190818105417-00361.warc.gz
|
CC-MAIN-2019-35
| 1,657 | 20 |
http://openstudy.com/updates/4f70e48ee4b0eb85877369ed
|
math
|
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
ok so i have this word problem and i have it half done and i got stuck
... post the problem
A triangle has an are of 77 square inches. Find the length of the base if the base is 3 inches more than the height
okay... see the area of any triangle is A = 1/2 x (height) x (base) let the height be 'k'. then base = k+3 so, (1/2 )(k) (k+3) = 77 can you solve that now??
ok good just what i got!
what have you gotten for the answer?
hmm.. now i think you got stuck?? okay listen now.. multiply both sides by 2.... then simplify the LHS. and bring every term to the left side. you 'll find you have a very familiar kind of quadratic equation.. solve for 'k' if still having problems, i ll show how to solve it
plz do show how u r gonna solve it
okay you have an answer? post what you got. i ll verify it for you!
okay there it goes....|dw:1332799036998:dw| so, i get k+14=0 and k-11=0 so k= -14 or 11 NOW... length of a side cannot be *negative*. so "-14" CANNOT be a possible solution. so k=11 only now base was (k+3)... since we assumed k is the height. so, base= 11+3 = 14
:O makes sense...
lol u know just before u posted tht i figured out how stupid i was!
lol... we 're stupid "at times".. (ans some people like me all the time!!! :D :D :p)
|
s3://commoncrawl/crawl-data/CC-MAIN-2017-51/segments/1512948515309.5/warc/CC-MAIN-20171212060515-20171212080515-00782.warc.gz
|
CC-MAIN-2017-51
| 1,931 | 14 |
https://www.natureof3laws.co.in/derive-an-expression-of-magnetic-field-at-a-point-on-the-equatorial-line-of-a-bar-magnet/
|
math
|
In this article, we will derive an expression for magnetic field at a point on the equatorial line of a bar magnet, so let’s get started…
Derivation of the magnetic field at a point on the equatorial line of a bar magnet
Consider a bar magnet NS of length and the pole strength of . Suppose the magnetic field, we wish to find at the point P, lies on the equatorial line of the bar magnet at the distance from the geometric centre of the magnet. See figure below:
Lets imagine a unit north pole placed at the point P. Then from the coulomb’s law of magnetic force, force exerted by the North pole of the bar magnet to the unit north pole placed at the point P is given as:
Similarly, the force exerted by the south pole of the magnet to the unit North pole placed at the point P is-
As the magnitude of and are equal, so their vertical components get cancelled while the horizontal components add up along PR. Hence, the magnetic field at the equatorial point P is equal to net force experienced by the unit North pole placed at point P.
Using Pythagoras theorem, the value of x can be given as . So after putting the value of x in above expression, we get-
where, , is the magnetic dipole moment of the bar magnet.
For a short bar magnet, for which , above expression can be reduced as follows:
It can be clearly seen that the direction of magnetic field at any equatorial point of a magnetic dipole (here bar magnet) is opposite to the direction of magnetic dipole moment,i.e N-pole to S-pole. So the expression of magnetic field at the point on the equatorial line is given as below:
On comparing magnetic field at axial point to the magnetic field at the equatorial point, we find that-
So from this we can say that magnetic field at axial point due to a magnetic dipole at a certain distance is twice the magnetic field due to magnetic dipole at equatorial point for same distance.
Stay tuned with Laws Of Nature for more useful and interesting content.
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s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320304947.93/warc/CC-MAIN-20220126101419-20220126131419-00667.warc.gz
|
CC-MAIN-2022-05
| 1,964 | 13 |
https://sharethefiles.com/forum/viewtopic.php?t=77934
|
math
|
A 1960s tale of betrayal and death comes back to haunt the owner of Blackwood Mansion, in this first-person, point-and-click-style "horror adventure" from Got Game Entertainment. In the role of new owner Michael Arthate, players become enthralled with the lingering mysteries of former resident James Blackwood, and his wife Catherine, whom James is thought to have killed. Whether he likes it or not, Michael seems unable to ignore the possibility that a murder took place in his own home. His imagination of the gruesome event begins to give him bad dreams. When he is awoken from a nightmare, by the sound of something scratching, Michael realizes he must take action and explore whatever secrets his old house may hold.
Code: Select all
Ü²Ü ÛÛ²²² ÛÛÛÛ²²² ÛÛÛÛÛÛ²²² ÞÛÛÛßÛÛÛÛ²²²Ý ÛÛÛÛÛÝÞÛÛÛÛ²²²² ÛÛß Ûß ÛÛÛ²²² ß²² ß Ü ÜÛÛÛÛ²²² Ü ß ÜÜÛß °Û ÛÛÛÛÛÛ²²² ß²° ßÛÜÜ ÜÜÛÛÛÛÝ ßßßßßß ÛÛÛÛÛÛ²ßßß ßßßßß Þ²²²²ÜÜ ÜÜÛÛÛÛÛÛÛß Ü²²²²²²²ÛÛÛÛÛÛ²²²²²²²Ü ßßÛÛÛÛÛ²²²²ÜÜ þÛÛÛÛÛÛÛÛÛÛÜÛÛÛÛÛÛÛÛÛÛÛÛÛÛÛÛÛÛÛÛÛÛÛÛÛ²²ÜÛÛÛÛÛÛÛ²²²²²þ ßßÛ²²²²ÛÛÛÛÛÛÛÛ²²ßßÜÛÛÛÛÛ²²²ßÛÛÛÛÛÛÛÛÛÛÛÛÛÛÛÛßß ßß²²²²Ý ß Ü ²²ÛÛÛÛ²²² ÜÜÜÜ ÞÛÛÛÛßß ßß²Ü °° ܱ²ÛÛÛÛ²²²° °°° ÜÛßß ÜÛß²Ü ÜÜ ÜÛ ° ²ÛÛÛÛÛ²²²± °°± ܲ²Ü ÜÛß ÜÛ²Ý Þ²²Ü ß²²ß ܲ²Ý ± ÛÛÛÛ²²°ßß° ܲ² Ü ßß Ü²²Ý ÜÛßÛ² Û²²² ° ÜÛÛÛ² Ü²Ü ß Ü Üß Ü ßÛß ÜÛß²Ü ÜÛÛÛ² ° ÛÛ² ܲßßß ÜÛÛÜÛÛ² ßßß²²ß ÜÛ² ² ÛÛÜ ÜÛÛÝ Þ²²Ü ÜÛ² °° ° ÛÛ²²ß Û²Ü ßÛÛß ßÛÛ²²° ±± ßÛÛ² ± ÛÛ²ßÛÛ² Û²² ÛÛ² ²± ° ÛÛ²Ý °ÞÛ²²ÜÛÛ² ÛÛ² Ü °° °ÛÛ² ° ÛÛ²ßÛÛ² ܲßß °°ÛÛßÜ ²²Ü ± ÛÜßß ° ÛÛ²ÜÛÛ² ÛÛ² ² ÜÜÜÜÜÛÛ² °ÞÛÛÜ ÜÛ²²ß ÜÜÜ Û²²Ý ÜÜÜ ± ÛÛ² °° ÛÛß ßÛ² Û۲ݰ ÛÛß ßÛ² °ßÜÛ² ÛÛ²Ý ° Û²²ßÛÛ² ²° Û²² ² ÛÛ² °° ÛÛ² ÛÛ² ÛÛÜß°° ÛÛ²°ÛÛ² ° ÛÛ² ÛÜßß ° ÛÛÜ ÜÛ² ± ÛÛ² ß ÛÛ² ±± ÛÛ² ÛÛ² ÛÛ² ° ÞÛÛ² ÛÛ² ° Û²² Û²² °°Þ²²ß ÛÛ² ° ÞÛÛÜ²ß Û ßÛÜÛ² ²² Û²²ßÛ²²ß ß²Ý Ûß ß²Ý Þ²ß ß²Ý ²ß ° ß²Ý ÛÛ²ß Û ÜÛ²ßÛß ßÛÛ Þ²ß Þ²ß Ü ßþß ßþß ßþß Ü²²Ü ßÜÛ²ß ßÛß²ÛÜ ÞÛ²Ý Ü ²Ü ß ß ß Ü Ü ßß ß Ü² Ü Þ²²Ý ßÛ²Ü ÝÞ²Ü ßÜ ß²²Ü ßÛß Üß ßÜ ß²ß Ü²²ß Üß Ü²ÝÞ Ü²Ûß ßß²²²ßßßß ß²Ü Û²ß²Ü ²Ý T E A M Þ² Ü²ß²Û Ü²ß ßßßß²²²ßß ÜÛ Ü²ß Ü²²Ü ß ßÛÛ²Ý ß²ÜÛ² R I T U E L ²ÛÜ²ß Þ²ÛÛß ß Ü²²Ü ß²Ü ÛÜ ßÛÛ²ß ßß ßß²Ü ß²²Ý Þ²²ß ܲßß ßß ß²ÛÛß Û²Ý ß² ²ß Þ²Û ²²ß ß²² Û² Scratches ²Û Û² (c) ²Û Û² Got Game ²Û Û² ²Û Û² 03/2006 :.. RELEASE.DATE ... PROTECTION ....: None ²Û Û² 50 :....... DISC(S) ... GAME.TYPE .....: Adventure ²Û ²²Ü ܲ² Û²Ý ß þÜÜ ° Ü Ü ° ÜÜþ ß Þ²Û ÜÛÛ²Ü Ü±Ü Þ²Ý ß Ü²Ý Þ²Ü ß Þ²Ý Ü±Ü Ü²ÛÛÜ ßÛ ß²Üܲ ÜÜ²ß ß²²²Ü ßÜÜ ÜÜß Ü²²²ß ß²ÜÜ ²ÜÜ²ß Ûß Ü²Ü Ü ßß²ÛÛÛ²²ÛÛÛÛ²²² ß²ÜÜÞ²² GAME ²²ÝÜÜ²ß ²²²ÛÛÛÛ²²ÛÛÛ²ßß Ü Ü±Ü ß ÜÛßßß Ü²Ü ßß ß²ÛÛ²Ý iNFOS! Þ²ÛÛ²ß ßß Ü²Ü ßßßÛÜ ß ÜÛ²Ý ß ß²² ²²ß ß Þ²ÛÜ ²² ßß ßß ²² ²² ²² Û² As a first person Horror Adventure game, you are Michael ²Û Û² Arthate, living in a haunted mansion, known as Blackwood ²Û Û² Manor. As you explore your creepy, eerie, new home, you ²Û Û² learn that it used to belong to James Blackwood who was ²Û Û² accused of murdering his wife Catherine in the early ²Û Û² 1960?s. ²Û Û² The mystery of what happened all those years ago threatens ²Û Û² to distract you from your own work. In the middle of the ²Û Û² night you awaken out of a nightmare so vivid you?d swear ²Û Û² it was real. And just then you hear the sound of something ²Û Û² scratching It may threaten your sanity, if not your life. ²Û ²²Ü ܲ² Û²Ý ß þÜÜ ° Ü Ü ° ÜÜþ ß Þ²Û ÜÛÛ²Ü Ü±Ü Þ²Ý ß Ü²Ý Þ²Ü ß Þ²Ý Ü±Ü Ü²ÛÛÜ ßÛ ß²Üܲ ÜÜ²ß ß²²²Ü ßÜÜ ÜÜß Ü²²²ß ß²ÜÜ ²ÜÜ²ß Ûß Ü²Ü Ü ßß²ÛÛÛ²²ÛÛÛÛ²²² ß²ÜÜÞ²² RiP ²²ÝÜÜ²ß ²²²ÛÛÛÛ²²ÛÛÛ²ßß Ü Ü±Ü ß ÜÛßßß Ü²Ü ßß ß²ÛÛ²Ý NOTES Þ²ÛÛ²ß ßß Ü²Ü ßßßÛÜ ß ÜÛ²Ý ß ß²² ²²ß ß Þ²ÛÜ ²² ßß ßß ²² ²² ²² Û² Ripped : Following movies : Company logo, Intro & Credits ²Û ²²Ü ܲ² Û²Ý ß þÜÜ ° Ü Ü ° ÜÜþ ß Þ²Û ÜÛÛ²Ü Ü±Ü Þ²Ý ß Ü²Ý Þ²Ü ß Þ²Ý Ü±Ü Ü²ÛÛÜ ßÛ ß²Üܲ ÜÜ²ß ß²²²Ü ßÜÜ ÜÜß Ü²²²ß ß²ÜÜ ²ÜÜ²ß Ûß Ü²Ü Ü ßß²ÛÛÛ²²ÛÛÛÛ²²² ß²ÜÜÞ²² iNSTALL ²²ÝÜÜ²ß ²²²ÛÛÛÛ²²ÛÛÛ²ßß Ü Ü±Ü ß ÜÛßßß Ü²Ü ßß ß²ÛÛ²Ý NOTES Þ²ÛÛ²ß ßß Ü²Ü ßßßÛÜ ß ÜÛ²Ý ß ß²² ²²ß ß Þ²ÛÜ ²² ßß ßß ²² ²² ²² Û² 1) Unzip ²Û Û² 2) Unace or use our installer ²Û Û² 3) Launch the setup.bat ²Û Û² 4) Play (Press escape to skip ripped movies). ²Û Û² ²Û Û² We do this for FUN. We are against any profit or ²Û Û² commercialisation of piracy. We do not spread any release, ²Û Û² others do that. We do NOT want our nfo or release listed ²Û Û² on any public place like websites, P2P netorks, ²Û Û² newsgroups, etc! It is against the original scene rules! ²Û Û² In fact, we BUY all our own games with our own hard earned ²Û Û² and worked for efforts. Which is from our own real life ²Û Û² non-scene jobs. As we love game originals. Nothing beats a ²Û Û² quality original. Support the software companies. If you ²Û Û² like this game BUY it! We did! ²Û Û² ²Û Û² We are currently looking for skilled crackers : ²Û Û² [email protected] ²Û ÛßÜ ÜßÛ Û²Ý ° ° Þ²Û Û² ß Ü²Ü ÜÜܲ²ßß RiTUEL! ßß²²ÜÜÜ Ü²Ü ß ²Û ۲ܲ²²²Ü ß ÜÜÛ²²²²²Ûßß ÜÜ ÜÜ ßßÛ²²²²²ÛÜÜ ß Ü²²²²Ü²Û Þ²²ßß ß²²Üܲ²ßß Ü²²ß Ü²ß ß²Ü ß²²Ü ßß²²Üܲ²ß ßß²²Ý ÜÛß Ü²²Ü Þ²²ß ÜÜÛ Þ²²Ý Þ²Ý ascii by mx Þ²Ý Þ²²Ý ÛÜÜ ß²²Ý ܲ²Ü ßÛÜ Ü ßß Ü²ß ÜÛß ßÜ ß²²Ü ß²Ü Ü Ü Ü²ß Ü²²ß Üß ßÛÜ ß²Ü ßß Ü ° Ü²Ü ßßßßßß ßß ß²ß ß ß²ß ßß ßßßßßß ÜÛÜ ° ß ß Ripping is a RiTUEL
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s3://commoncrawl/crawl-data/CC-MAIN-2019-39/segments/1568514572744.7/warc/CC-MAIN-20190916135948-20190916161948-00097.warc.gz
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CC-MAIN-2019-39
| 6,900 | 3 |
https://www.examyear.com/geography-hard-questions/
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math
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Applicants can download the Geography Hard Questions. Last 5 Years Papers Geography Hard Questions Solved Papers. Free downloading links of the Geography Hard Previous Papers are enclosed below.
Interested applicants can get these Geography Hard Sample Papers for free of cost. Start your preparation by the Geography Questions by downloading the Last 5 Years Question Papers.
Hard Questions on Geography
1. Endogenetic forces cause relief to ………………
(C) Get distributed
(D) Get concealed
2. Continental drift theory was not accepted because …………….
(A) The evidences cited were inadequate
(B) Explanation about force was not acceptable
(C) Jig-saw fit has gaps
(D) The evidences cited were not convincing
3. Which of the following is not true about the esker ?
(A) It is a depositional landform
(B) It is a landform of glacial action
(C) It is a long extending elevated form
(D) It is an egg shaped hill
4. Which of the following forms of transformation is not possible by wind ?
5. Which of the following landforms is predominently produced by weathering ?
(A) Abrasion platform
(B) Honeycomb and Taffoni
6. The gusty dry and warm winds of Rockies are :
7. The greenhouse gas released by livestock is :
(A) Carbon dioxide
(D) Nitrogen oxide
8. The average net radiation balance of atmosphere is ……………………. kilo langleys.
9. The cyclone formed over Arabian Sea during June 2010 was named :
10. Concept of potential evapotranspiration was proposed first in :
11. Plants of monsoon region are called :
12. Blue mud results from disintegration of rocks containing :
(A) Iron oxide and ochreous matter
(B) Silicates of potassium and glauconite
(C) Skeletons and shells
(D) Sulphide and organic matter
13. The water body with highest salinity is :
(A) Lake Van
(B) Dead Sea
(C) Lake Sambhar
(D) Great Salt Lake
14. Which one of the following is the most important element of ecosystem ?
(A) Ecological succession
(B) Energy flow
(C) Food chain
(D) Food web
15. The warm current that flows along the Peruvian Coast is called :
(B) El Nino
(C) La Nina
(D) Gulf Stream
16. Who is considered as the founder of human geography ?
17. The term possibilism is coined by :
18. Griffith Taylor promoted :
(B) Stop and go determinism
19. Miss Semple was a disciple of :
20. ‘‘Anthropogeographie’’ is authored by :
21. …………………. model appears similar to that of Von Thunen.
(A) Multiple nuclie
(D) Distance decay
22. Settlement developing in dry elevated location surrounded by marshy area is called :
(A) Wet point settlement
(B) Dry point settlement
(C) Oasis settlement
(D) Hill top settlement
23. Who introduced the concept of Rank-size rule ?
(A) R.E. Dickinson
(B) G.K. Zipf
(C) A.E. Smailes
(D) W. Christaller
24. Natural increase in population refers to :
(A) Number of births minus number of deaths
(B) Number of births minus number of deaths minus number of outmigrants
(C) Number of inmegrants minus number of outmigrants
(D) Number of births plus number of inmigrants minus number of outmigrants
25. Migration of sheperds in hilly area is termed as ………………
(B) Seasonal migration
(C) Stepwise migration
26. …………………. occupation is ‘‘Red Collar’’ job.
(C) Research scientist
27. Resources are :
(A) Natural endowments
(B) Valuable endowments
(C) Natural potentialities
(D) Exploited potentialities
28. Agglomeration economies are :
(A) Internal to firm
(B) Internal to industry
(C) External to industry
(D) Internal to region
29. The term ‘Range of goods’ refers to :
(A) Range of goods sold
(B) Spatial range of goods sold
(C) Maximum distance over which a consumer moves to purchase goods
(D) Range of customer drawn
30. Japan imports wood because she :
(A) Does not possess forests
(B) Wants to save its forests
(C) Has poor quality forests
(D) Has no accessible forests
31. Mackinder termed Eurasia as ‘‘World Island’’ because :
(A) It was surrounded by oceans
(B) It had the largest number of nations
(C) It was at the centre of the world
(D) It accounted for about 40% of land and 80% of world population
32. The worst form of Apartheid in the 20th century was in :
(D) South Africa
33. Central Asia is a part of :
(A) Islamic culture realm
(B) East Asian culture realm
(C) Meso-African culture realm
(D) European culture realm
34. Which is the smallest minority religious group in Maharashtra ?
35. Garo and Khasi tribes are found in :
(B) Arunachal Pradesh
36. ‘‘Functional’’ region is best described as an area with :
(A) Functional uniformity
(B) Functional superiority
(C) Functional nodality
(D) Common interest
37. …………………….. is the most refined technique for agglomerative regionalisation.
(B) Median and quartile
(C) Mean and standard deviation
(D) ‘Z’ score
38. India had annual plans in ………….
(A) Late 1960s
(B) Early 1970s
(C) Late 1970s
(D) Mid 1980s
39. Green Revolution aimed at :
(A) Enhancing farm production
(B) Increasing farm yield
(C) Reducing social disparities
(D) Reducing regional disparities
40. Chhattisgarh is a :
(A) Dynamic region
(B) Potential region
(C) Problem region
(D) Depressed region
41. The highest peak in India is :
(A) Mt. Everest
(B) Godwin Austin
42. Identify the pair that does not match :
(B) Red soil—Granite
(C) Clayey soil—Schist
(D) Laterites soil—Sandstone
43. The coast of Tamil Nadu has ………….. forests.
(A) Dry deciduous
(B) Moist deciduous
(C) Dry evergreen
44. Maximum proportion of area under well irrigation is in :
45. Maximum production of bauxite in India comes from :
46. For isopleth maps the locational data related to ………………………. is necessary.
47. Which of the following IRS satellite data can facilitate developing a true colour composite ?
(A) IRS 1A/B
(B) IRS 1C/D
(C) Resource sat 1
48. A ratio of mean to the standard deviation is referred as …………..
(A) coefficient of variation
(B) coefficient of correlation
(C) coefficient of regression
(D) coefficient of concentration
49. Which two of the following are not parametric tests ?
(a) Students ‘t’test
(b) Chi-square test
(c) Snedecor’s ‘F’ test
(d) Kolmogrove-Smirnov test
(A) (a) and (b)
(B) (b) and (c)
(C) (b) and (d)
(D) (a) and (c)
50. Which of the following sampling techniques is recommended to draw a sample of villages, spread in a number of Talukas, from a district ?
(A) Systematic sampling
(B) Cluster sampling
(C) Stratified sampling
(D) Random sampling
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s3://commoncrawl/crawl-data/CC-MAIN-2023-40/segments/1695233506658.2/warc/CC-MAIN-20230924155422-20230924185422-00886.warc.gz
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CC-MAIN-2023-40
| 6,472 | 172 |
https://koha.app.ist.ac.at/cgi-bin/koha/opac-detail.pl?biblionumber=371580
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math
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Current Trends in Analysis and Its Applications [electronic resource] : Proceedings of the 9th ISAAC Congress, Kraków 2013 / edited by Vladimir V. Mityushev, Michael V. Ruzhansky.
Contributor(s): Mityushev, Vladimir V [editor.] | Ruzhansky, Michael V [editor.] | SpringerLink (Online service)Material type: TextSeries: Trends in Mathematics: Publisher: Cham : Springer International Publishing : Imprint: Birkhäuser, 2015Description: XVI, 892 p. 97 illus., 48 illus. in color. online resourceContent type: text Media type: computer Carrier type: online resourceISBN: 9783319125770Subject(s): Mathematics | Functions of complex variables | Partial differential equations | Applied mathematics | Engineering mathematics | Mathematics | Partial Differential Equations | Functions of a Complex Variable | Applications of MathematicsAdditional physical formats: Printed edition:: No titleDDC classification: 515.353 LOC classification: QA370-380Online resources: Click here to access online
|Item type||Current location||Collection||Call number||Status||Date due||Barcode||Item holds|
1) complex variables and potential theory -- 2) differential equations: complex and functional analytic methods, applications -- 3) methods for applied sciences -- 4) clifford and quaternion analysis -- 5) spaces of differentiable functions of several real variables and applications -- 6) generalized functions -- 7) qualitative properties of evolution models -- 8) nonlinear infinite dimensional evolutions and control theory with applications -- 9) nonlinear pde and fixed point theory, topological and geometrical methods of analysis -- 10) didactical approaches to mathematical thinking -- 11) integral transforms and reproducing kernels -- 12) pseudo-differential operators -- 13) toeplitz operators and their applications -- 4) approximation theory and fourier analysis -- 15) differential and difference equations with applications -- 16) analytic methods in complex geometry -- 17) applications of queueing theory in modelling and performance evaluation of computer networks. .
This book is a collection of papers from the 9th International ISAAC Congress held in 2013 in Kraków, Poland. The papers are devoted to recent results in mathematics, focused on analysis and a wide range of its applications. These include up-to-date findings of the following topics: - Differential Equations: Complex and Functional Analytic Methods - Nonlinear PDE - Qualitative Properties of Evolution Models - Differential and Difference Equations - Toeplitz Operators - Wavelet Theory - Topological and Geometrical Methods of Analysis - Queueing Theory and Performance Evaluation of Computer Networks - Clifford and Quaternion Analysis - Fixed Point Theory - M-Frame Constructions - Spaces of Differentiable Functions of Several Real Variables Generalized Functions - Analytic Methods in Complex Geometry - Topological and Geometrical Methods of Analysis - Integral Transforms and Reproducing Kernels - Didactical Approaches to Mathematical Thinking Their wide applications in biomathematics, mechanics, queueing models, scattering, geomechanics etc. are presented in a concise, but comprehensible way, such that further ramifications and future directions can be immediately seen. .
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s3://commoncrawl/crawl-data/CC-MAIN-2021-10/segments/1614178374686.69/warc/CC-MAIN-20210306100836-20210306130836-00493.warc.gz
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CC-MAIN-2021-10
| 3,257 | 5 |
https://st-agnes-scilly.org/motorola-one-secret-codes/
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math
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Motorola One Secret Codes. Some code might not work on some devices. Here means pause, hold down the.
In all motorola mobile codes you can find all the android secret codes including network, gps, bluetooth, wlan. Thanks for your time guys.we will see about the secret codes for motorola mobiles. Here means pause, hold down the.
Post Your Question Or Comment.
In this app, there is complete details of the hidden codes of the mobiles to change the setting and other function of your mobile’s functions which even can’t. If your motorola cell phone is locked to a certain carrier, you can remove this lock and use your motorola with any network worldwide. 1809#*990# it opens the lg optimus 2x hidden service menu.
How Do I Change The Puk 2 Pin If I Don't Have The Original Puk2 Pin?
Let's get access to secret information about motorola one 5g ace. Thanks for your time guys.we will see about the secret codes for motorola mobiles. The description of secret codes for motorola app.
Enter The Code * # * # 2486 # * # * After Doing So, The Cqatest Menu Will Open;
Here means pause, hold down the. Display phone information, battery & history usage statistics *#*#4636#*#* 6. Click here and find out more information about secret codes.
Check Out How To Enter Hidden Mode And Use Advanced Options Of Android 9.0 Pie.
Display imei number *#06# 2. On motorola g stylus 2021: 3845#*920# it opens the lg optimus 3d hidden service.
Check Out All The Secret Codes For Your Motorola One.
Typing this secret code *#06# you can easily check your imei number. ##7764726 it opens the hidden service menu in motorola droid phones. ###119#1#, (ok) to activate efr mode ###119#0#, (ok) to deactivate efr mode code to lock keys.
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s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030334912.28/warc/CC-MAIN-20220926144455-20220926174455-00231.warc.gz
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CC-MAIN-2022-40
| 1,718 | 12 |
http://www.astm.org/Standards/C1213.htm
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math
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WITHDRAWN, NO REPLACEMENT
| ||Format||Pages||Price|| |
|1||$44.40|| ADD TO CART|
1.1 This standard is a compilation of definitions of technical terms defined in the standards developed by Committee
1.2 In the interest of common understanding and standardization, consistent word use is encouraged to help eliminate a major barrier to effective technical communication.
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CC-MAIN-2014-41
| 372 | 5 |
https://www.arxiv-vanity.com/papers/quant-ph/0508229/
|
math
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Precision characterisation of two-qubit Hamiltonians via entanglement mapping
We demonstrate a method to characterise the general Heisenberg Hamiltonian with non-uniform couplings by mapping the entanglement it generates as a function of time. Identification of the Hamiltonian in this way is possible as the coefficients of each operator control the oscillation frequencies of the entanglement function. The number of measurements required to achieve a given precision in the Hamiltonian parameters is determined and an efficient measurement strategy designed. We derive the relationship between the number of measurements, the resulting precision and the ultimate discrete error probability generated by a systematic mis-characterisation. This has important implications when implementing two-qubit gates for fault-tolerant quantum computation.
One of the key requirements for a physical system to be used for quantum information processing applications is that the system must have a controllable two-qubit coupling. This is typically realised by an interaction between a pair of two-level systems which act as qubits. It is this interaction which leads to entanglement and the ‘spooky action at a distance’ effects which give quantum computers their power. While some systems have a well defined native two-qubit interaction, this is not generally the case. In solid-state systems the interaction Hamiltonian is often a function of many control and fabrication parameters[3, 4, 5]. As such, the form of the Hamiltonian can vary from device to device and even vary within different sections of a single device. This means characterisation of some sort is critical in order to control the interaction and produce accurate gate operations for quantum computing applications.
In this paper, we show how mapping the entanglement of the system as a function of time gives a conceptually straightforward approach to determining the dynamics of the system. Specifically, we show how this method can be used to characterise a two-qubit interaction of the Heisenberg type,
where , etc. and are the Pauli operators. Many solid-state quantum computing proposals rely on this type of interaction[6, 7, 5, 8, 9, 10, 11, 12, 13, 14], as the general Heisenberg case covers a large class of quantum systems including real spin (i.e. exchange coupling) systems[6, 7, 8, 5] and pseudo spin systems such as charge based designs[13, 14]. Recent work has also shown that two-qubit gates can be designed from a Heisenberg Hamiltonian with anisotropic couplings (), as long as the components of the Hamiltonian are known accurately. In an implementation of a quantum computer consisting of nominally identical qubits, the physical interaction between any given pair of qubits is similar, so we expect the structure of the Hamiltonian to be similar across a given device. On the other hand, the size of the various couplings are a strong function of the fabrication process and will therefore vary from qubit to qubit. In these situations, not only is it important to identify the size of the relative components, but for scalable systems this characterisation must be done in an efficient manner, by which we mean that the process can be largely automated and require minimal physical modification to the original fabricated qubits.
The issue of systematic, accurate and repeatable characterisation has far reaching consequences for quantum computing, given the ongoing efforts to define an error threshold, below which arbitrary quantum computation is possible using the concepts of concatenated quantum error correction and fault-tolerance[16, 17, 18]. Recent work has put this threshold at (depending on available resources) as the probability of a discrete gate error[19, 20, 21, 22], though this is the total error probability which is a combination of environmentally induced errors, characterisation and control errors. By defining a systematic method of characterisation, we relate the number of measurements required in the initial characterisation phase to the resulting gate error rate, directly linking the required characterisation to the concatenated quantum error correction threshold.
Traditionally, characterisation has been performed using state and process tomography[23, 24, 25], where a pulse sequence is developed to realise a certain gate, assuming the basic form of the Hamiltonian is known on experimental or theoretical grounds. The effect of this gate on a complete set of input states is measured to build up the system state. This has been the method of choice for most early two-qubit experiments as the exact details of the interaction are not needed as long as the required two-qubit gate can be constructed approximately and the complete state of the system mapped. This gives extensive information about the system including the effects of decoherence or loss channels. If the gate is not ideal, then a good model is required, otherwise there is no systematic way of improving the performance of the gate or knowing whether an improvement is possible.
A method for single-qubit characterisation has been recently developed which allows the efficient determination of the terms in the system Hamiltonian and can be implemented with minimal information about the system being characterised[26, 27, 28]. Rather than assuming knowledge about the system, this method involves mapping the system evolution over time and using this to gain information about the Hamiltonian itself. While this typically requires many measurements to build up the evolution of the state of the system, it also provides detailed information about the form of the Hamiltonian. This allows any necessary gate sequence to be developed offline without the need to tomographically map every gate that may be required in a given quantum circuit.
We show how the application of an accurately characterised Hadamard gate and measurement on both qubits is sufficient to find all the couplings in the Heisenberg Hamiltonian. The result is that, using the machinery of a quantum computer architecture only, one can extract sufficient information to determine the fundamental interaction Hamiltonian and hence construct any required unitary gates. By performing a combination of single- and two-qubit characterisation, the system can be ‘boot-strapped’ from minimal knowledge of the system to provide all the required parameters for full controllability.
In contrast to spectroscopy, re-characterisation can be performed in-situ at any future time if required (e.g. to correct for long term drifts of the system parameters). Addition characterisation steps can then be performed in parallel with the quantum computer’s usual operation, whenever qubits are idle.
2 Entanglement generated by the Heisenberg Hamiltonian
Many two-qubit interactions can be described by the general Heisenberg Hamiltonian given in Eq. (1). When , this is the conventional (isotropic) Heisenberg interaction of the form , which is typical of spin based qubit coupling. If and , this is the interaction due to an Ising type coupling (), common in pseudo spin schemes. From the point of view of two-qubit gate design, for an Ising interaction it is not important which of the three terms is non-zero as the Hamiltonians , and are locally equivalent. For this analysis we consider general Hamiltonians with , and treated as parameters to be determined.
We will restrict ourselves to only consider Hamiltonians which are piecewise constant in time and we assume controllability of any single qubit terms such that they can be turned off during the two-qubit interaction, or alternatively the single-qubit terms commute with the rest of the Hamiltonian. The restrictions imposed by this assumption are discussed in section 4.
As we are interested in reducing the systematic errors introduced by imperfect characterisation (rather than random errors caused by interaction with the environment), we have assumed that the effect of decoherence is negligible within the observation time.
To begin, we analytically derive the evolution of the system described by Eq. (1) from some initial state to the state at some later time. This evolution will, in general, depend on both the initial state and the components of the Hamiltonian. To measure this evolution, the simplest method is to repeatedly initialise the system in , allow the system to evolve for a time and then measure the system. This process is then repeated for integer values of to build up the time evolution at discrete time steps separated by . The difficulty with this process is that the time evolution is both a function of the single qubit terms and two qubit terms in the Hamiltonian.
Alternatively, we can look at the entanglement generated by the interaction. By definition, if the entanglement changes with time, then a two qubit interaction must be present, since local operations alone cannot generate a change in entanglement111While a change in entanglement can be used to infer the existence of two-qubit interaction terms, it cannot be used to exclude the presence of single qubit terms within the Hamiltonian.. This leads us to the idea of using the variation in the entanglement to analyse the interaction and isolate the effect of the terms of interest in the Hamiltonian.
The entanglement of the state generated by this evolution can be quantified using the squared concurrence
where varies between , when the qubits are unentangled, to when they are maximally entangled. One method of measuring the concurrence is to measure the system in the and bases. We write the probability of measuring the qubit in the eigenstate of the operator as , where and . For example, in conventional notation this gives
In Table 1 we consider the time evolution of the entanglement given four different initial states ( to ). In each case the evolution is a simple sinusoidal function with frequency given by the combination of two of the three parameters in the Hamiltonian given in Eq. (1).
Using the set of input states to , the evolution of the system due to the Heisenberg Hamiltonian results in a significant simplification of Eqs. (5)-(7). For instance, if the systems starts in state , then for all time, whereas starting with gives . In fact, for and , and therefore these states need not be measured at all. These relations drastically reduce the number of measurements required to determine the concurrence and are true for any value of the coefficients of Eq. (1), as they lead directly from the symmetries of this Hamiltonian.
The input states considered here are either the computational states or can be reached from the computational states using a Hadamard rotation on both qubits. As the frequency of oscillation in each case is a linear combination of the coefficients , determining the frequencies for evolution from the four starting states determines all the parameters including their signs. The choice of which input states to use is largely arbitrary, depending on which frequency components are to be measured and which states can be prepared most easily. The four states discussed here are chosen purely for the fact that they can be prepared from the computational states using only Hadamard gates.
A side effect of using the Fourier transform and the squared concurrence is that it removes any sign information, hence the need for four states in general. If the sign of all the coefficients are known beforehand, or can be determined with a minimal number of measurements, then any three of these input states are sufficient for complete characterisation.
The Fourier transform of the oscillation data gives the system parameters but, in contrast to the single-qubit case, these depend on the peak positions in frequency space, rather than the peak amplitudes. While the oscillation frequencies present in the concurrence evolution are also present in the original probability evolution, the use of entanglement as a measure means the evolution is invariant under interchange of qubits and unaffected by the inclusion of single qubit terms which commute with the two-qubit interaction.
At this point an obvious question is, can we use other measures of entanglement or is concurrence somehow special? As we are only considering pure states, all bipartite entanglement measures are equivalent and so the difference comes down to implementation. In order to measure the Hamiltonian components accurately, it is important that the entanglement measure we use does not artificially introduce spurious frequencies into the evolution. This immediately rules out any entropic measure which depends on a function of the form because if varies sinusoidally, the logarithm of this function contains an infinite number of higher order harmonics. These higher order harmonics complicate the frequency analysis and prevent unambiguous discrimination of the Hamiltonian components. Most common entanglement measures are in some way related to the von Neumann entropy (i.e. ) and therefore suffer from this problem. These include the entropy of entanglement , the entanglement of formation and logarithmic negativity . Interestingly though, using the square of the negativity itself as an entanglement measure results in equivalent expressions to those obtained with the square of the concurrence.
Another consideration is how easily can the required measurements be performed experimentally, as most measures of entanglement require the complete reconstruction of the density matrix or at least a partial reconstruction. The advantage of using concurrence is that it has a closed form which requires only two measurement channels, as shown in Eqs. (5)-(7). In fact this is the minimum number of measurement channels required to characterise a Heisenberg type Hamiltonian with arbitrary coefficients.
3 Uncertainty estimation and gate errors
To illustrate our analysis procedure visually, Fig. 1 shows the evolution of the entanglement for an example Hamiltonian given entanglement measurements at each time point. Fig. 2 shows the Fourier transform of this data, showing the peaks clearly above the noise floor. From this example we see that even though the oscillations in the time domain are not well resolved, the peaks can clearly be seen above the discretisation (or ‘projection’) noise in the frequency domain.
As this characterisation process ultimately relies on accurate determination of the oscillation frequency, many of the existing techniques for frequency standards are directly applicable[35, 36]. Ultimately, there are two parameters to be chosen, the number of discrete time points, , and the number of ensemble measurements, . The minimum number of discrete time points is governed by the Nyquist criteria, giving where is the period of oscillation and is the maximum time over which the system is observed. To reduce the frequency uncertainty, should be maximised, though this will be limited by the decoherence time of the system. As we have a single frequency oscillation, the uncertainty in the frequency determination can be reduced by having large numbers of ensemble measurements on the last few time points and using this to estimate the phase of the oscillation.
In the ideal case (where is large), only two measurements are necessary at all time points with the exception that measurements are taken at the final two points, giving a total number of measurements . This is in contrast to the example given in Fig. 1, where the same number of measurements are taken at each time point. The error in the phase determination on the final two points is given by the projection noise and scales as , given the uncertainty in the frequency as . The fractional uncertainty in the frequency is then given by
While this analysis is quite straight forward, for quantum computing applications it is important to link these uncertainties to typical error models to determine the probability of a gate error produced by an uncertainty in the measured system Hamiltonian. To do this we define an imperfect gate operation such that is the required gate operation followed by some error gate . Given , the effective error gate is . The effective error probability is then defined as .
If the Hamiltonian deviates from the form expected on theoretical grounds by such an amount the the error introduced by this deviation is larger than that due to characterisation uncertainties, we then use the measured Hamiltonian (rather than the theoretical one) to construct the gate. For many Hamiltonians, a two-qubit gate can be constructed using, at most, three applications of the Hamiltonian together with single qubit rotations[37, 29]. As our procedure measures the various terms in the Hamiltonian directly, it allows the construction of a pulse sequence to perform the required two-qubit gate, even when the Hamiltonian differs greatly from the theoretically expected form. Using this type of gate construction, the error rate of the gate is now governed by the characterisation uncertainties alone.
To make this more concrete, we can calculate the for two common examples of native gates, assuming they are generated from an ideal Hamiltonian (i.e. theoretical). The analysis is similar for the case of a well characterised but non-ideal Hamiltonian, though there is a cumulative effect if the two-qubit interaction is applied multiple times.
For an ideal Ising Hamiltonian (, ), the native gate is the CNOT gate, which can be constructed by applying the Ising Hamiltonian for a time combined with appropriate single-qubit rotations. Consider an example where characterisation is performed on the system, resulting in and , with the uncertainty in the peak position. We then take an imperfect gate generated by a pulse of length . This gives as the effective error probability, assuming errors in the single qubit rotations are negligible. Similarly, for an ideal isotropic Heisenberg Hamiltonian (), the native entangling gate is the square-root-of-swap (). Following the same procedure (assuming that the characterisation procedure leads to a common uncertainty in the peak positions) we obtain .
In Fig. 3, is plotted for both the Ising and Heisenberg Hamiltonians for two different values of and compared to the conservative fault-tolerant threshold of . The larger the value of , the more precise the initial estimate when . As increases, the uncertainty scales as , as expected. This allows us to calculate directly the time needed to initially characterise the system to obtain a given gate error rate. For instance, if time points are chosen, then a conservative estimate of measurements are needed to reduce the error rate to below that required to satisfy the fault-tolerant threshold, again neglecting the effects of single qubit errors. If more time points are used, the required number of measurements reduces accordingly, though this is limited by the requirement that at least two measurements are required at each time point to measure the concurrence. These estimates for the number of measurements required should be compared to the case of single qubits where to achieve a probability of error, .
4 Effect of single qubit terms
Throughout this discussion, we have assumed that the unknown Hamiltonian took the form
where is given by Eq. (1) and are single qubit terms such that . This restriction allows us to factor the evolution into separate single- and two-qubit evolution () where the single qubit evolution does not change the entanglement of the system. While this may at first appear restrictive, it actually includes several Hamiltonians of interest to solid-state quantum computing. This includes the effective exchange interaction between phosphorous donor spins in silicon and the magnetic dipolar interaction between deep donors in silicon. For both these examples, the commutation relation holds, irrespective of the value of the various coupling parameters.
A notable exception is the standard two-qubit interaction model for superconducting qubits. In this case, not only is characterisation difficult but gate design is non-trivial and requires approximate and numerical methods. In general, for a Hamiltonian of arbitrary form, the eigenstates and therefore the evolution frequencies are non-linear functions of all the system parameters.
In addition to single qubit terms which are part of the two-qubit interaction, we could also consider the effect of errors in the single qubit rotations used to prepare the input states given in Table 1. In general the system evolution is a function of six frequencies given by the sum and differences of , and and we have chosen the input states to isolated each frequency in turn. Taking an imperfect input state which is close to one of the states given in Table 1, e.g.
for some error probability , and expressing the evolution of the concurrence in a series expansion about , gives
where . The evolution now contains oscillating terms at the other five system frequencies with amplitude as well as the original evolution at a frequency given by and amplitude . As the Hamiltonian parameter estimates come from the position of the peak, the peak’s position and therefore the estimate is unaffected by small errors in the input state.
If the input state is completely unknown, the six frequency components are still present but there is now ambiguity as to which peak corresponds to which frequency. The inclusion of imperfect alignment of the measurement bases has an similar effect to imperfect state preparation, with the amplitude of the undesirable frequency components now being related to the extent of the misalignment.
We have not considered here the possibility of non-Heisenberg terms, such as or as this complicates the situation considerably, again, introducing ambiguity into the frequency spectrum. The effect of these terms is equivalent to a series of single qubit gates before and/or after the evolution[37, 29] and requires more sophisticated analysis. However, an upper bound on the size of these terms is again given by the projection noise and so scales as .
We have shown that mapping the entanglement generated by an unknown Hamiltonian provides a method of determining its structure and quantifying the various components. The Heisenberg Hamiltonian has particularly nice properties which lead to an efficient method of characterisation by mapping the time evolution of the entanglement. As this process requires finding the frequency of oscillation, the number of measurements required is typically much smaller than to precisely map the evolution of the expectation values. The required input and measurement bases can be obtained using approximate Hadmard rotations only, which relaxes some of the requirements for accurate single qubit rotations as a precursor procedure. In order to achieve precise control at, or below the fault-tolerant threshold the challenge is to be able to characterise logic gates to sufficient accuracy. Given an uncertainty in the Hamiltonian parameters and using an effective error model, we determined the probability of error due to systematic mis-characterisation and this is linked directly to the error thresholds required for fault-tolerant quantum computation. This type of characterisation procedure is of fundamental importance in experiments using two-qubit interactions, especially in the solid-state where precision control or uniformity of the Hamiltonian terms cannot be assumed a priori.
We would like to acknowledge helpful discussions with S. G. Schirmer, D. K. L. Oi and A. D. Greentree. This work was supported in part by the Australian Research Council, the US National Security Agency, the Advanced Research and Development Activity and the US Army Research Office under contract number W911NF-04-1-0290. The authors thank the von Delft group at LMU for their hospitality and, for financial support, the DFG through the SFB631. JHC and SJD acknowledge support from the Cambridge-MIT institute and LCLH was supported by the Alexander von Humboldt Foundation.
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https://www.lesswrong.com/posts/rAyKNKAahNftzoRGo/open-thread-july-16-31-2012?commentId=Jv76LxbWhKe9Ct7hT
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That example is probably just hyperbolic discounting. But CLT does say that we think differently about near/far things. In particular, we think more abstractly about distant things. That sounds like a stronger claim than yours. Try Robin Hanson's first post on the subject. Do you agree with him? with his source?
An example of hypocrisy where RH goes beyond normal CLT, but where I think it is quite fair to say that there is some connection.
His source in the first place is where I learned about construal-level theory, and I find/found it quite convincing. Hanson seems pretty accurate in his summary/analysis there, too.
In the second post: The Good Samaritan experiment seems like a stretch to apply here, but his other source is just the kind of experiment I would have thought should tell you whether CT does apply to "ideals" or not, and it appears that it does. Thanks for pointing me to these posts.
If it's worth saying, but not worth its own post, even in Discussion, it goes here.
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https://www.arxiv-vanity.com/papers/hep-ph/9808360/
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QCD analysis of inclusive B decay
M. Beneke and F. Maltoni***Permanent address: Dipartimento di Fisica dell’Università and Sez. INFN, Pisa, Italy Theory Division, CERN, CH-1211 Geneva 23
I.Z. Rothstein Department of Physics, Carnegie Mellon University,
Pittsburgh, PA 15213, U.S.A
the decay rates and -energy distributions of mesons
into the final state , where can be any one of the
-wave or -wave charmonia, at next-to-leading order in the
strong coupling. We find that a significant fraction of the
observed , and must be produced
through pairs in a colour octet state and should
therefore be accompanied by more than one light hadron. At the
same time we obtain stringent constraints on some of the
long-distance parameters for colour octet production.
PACS Nos.: 13.25.Hw, 14.40.Gx, 12.38.Bx
Exclusive decays provide us with important information on the structure of the Cabibbo-Kobayashi-Maskawa (CKM) matrix. However, the theoretical calculation of absolute branching fractions is complicated by the fact that a rather detailed knowledge of strong interaction effects is required. The theoretical situation with regard to strong interaction dynamics improves as one considers more inclusive final states. At leading order in , where is the quark mass and the strong interaction scale, the totally inclusive decay rate can be computed completely in perturbation theory. However, it is not necessary that the process be totally inclusive. A semi-inclusive decay can also be treated perturbatively in part, provided the formation of the hadron proceeds through a short-distance process. This is the case, if is a charmonium state, because the production of a charm-quark pair requires energies much larger than . The bound state dynamics of then factorizes and can be parametrized. This statement is summarized by the factorization formula
which is valid up to power corrections of order . (To this accuracy it is justified to treat the meson as a free quark.) The parameters , defined in , are sensitive to the charmonium bound state scales of order and , where is the typical charm quark velocity in the charmonium bound state. With for we consider these scales to be too small to be treated perturbatively. On the other hand the coefficient functions describe the production of a configuration at short distances and can be expanded in the strong coupling at a scale of order . We have expressed the decay rate in terms of several non-perturbative parameters . The predictive power lies in the fact that these parameters are independent of the particular charmonium production process and hence are constrained by other charmonium production processes.
Because charmonia pass as non-relativistic systems, Eq. (1) involves an expansion in and the configurations that appear in lower orders of this expansion can be usefully classified by , where , and refer to spin, orbital angular momentum and total angular momentum, respectively. In addition refers to a colour singlet or a colour octet configuration. In the present work we calculate the short-distance coefficients for
at next-to-leading order (NLO) in . As we discuss later, we believe that these terms in the velocity expansion are sufficient to reliably predict (to about 25%, barring radiative corrections in ) the decay rates into , , , and the elusive state as a function of the long-distance parameters . We find that, given the present uncertainties in the long-distance parameters, the experimentally observed branching fraction for and can easily be accounted for at NLO. However, we find it difficult to account for the observed and branching fractions simultaneously, because the expansion of the colour singlet contribution in production turns out to be untrustworthy at NLO. The NLO corrections typically enhance the decay rate by about (20-50)% in the colour octet channels and lead to bounds on the long-distance parameters, which should be useful for the phenomenology of other charmonium production processes. We also compute weights of the charmonium energy distribution, which yield additional information. The shape of the energy distribution itself, however, is difficult to predict, because it is distorted by the motion of the quark in the meson and the energy taken away in the hadronization of a state . This distortion averages out in weighted sums, as long as the weights are sufficiently smooth.
We then compare the inclusive calculation for with the sum of the measured decay rates for and . The comparison suggests a significant fraction of multi-body decays, consistent with the energy spectrum observed by CLEO . A substantial contribution from multi-body decays is also reassuring from the point of view of validity of the theoretical calculation. Factorization implies that a state hadronizes into a plus light hadrons independent of the remaining decay process up to corrections of order . If refers to a colour octet state, the conversion into charmonium requires the emission of at least one gluon. Although colour reconnections with the spectator quark in the meson must eventually occur, we expect a charmonium produced through a colour octet state to be accompanied by more than one light hadron more often than for a colour singlet state. Since we find that a large fraction of the total decay rate is from colour octet intermediate states, we also expect a large fraction of multi-body final states. This evidence also suggests to us that the energy released in the meson decay into charmonium is already large enough for an inclusive treatment to be applicable.
Inclusive production of wave charmonia has been considered in Refs. [4, 5] in the colour singlet model and at leading order (LO). In addition, the colour singlet production of the -wave state was computed in Ref. . (At LO the states and are not produced.) The colour singlet model is contained in Eq. (1) as the term where the quantum numbers of match those of the charmonium state. For -wave charmonia the colour singlet model does not coincide with the non-relativistic limit and is generally inconsistent. The authors of Ref. noted that the contribution from is leading order in for and that is leading order for . They computed the relevant short-distance coefficients to LO in . In the case of , the short-distance coefficients of states with are strongly enhanced as a consequence of the particular structure of the weak effective Lagrangian that mediates quark decay. These production channels have to be taken into account although the corresponding long-distance matrix elements are suppressed by a factor of . The relevant coefficient functions were computed in Ref. , again at LO in . Ref. adds a study of polarization effects. The only NLO calculation of charmonium production in decay is due to Bergström and Ernström , who computed the contribution of the colour singlet intermediate state to production. We repeated their calculation and comment on it later on.
The paper is organized as follows: In Section II we introduce notation and discuss the structure of important contributions to a given charmonium state. Section III provides some details on the calculation related to the handling of ultraviolet and infrared divergences at intermediate stages. Section IV contains our main results. We present expressions for the decay rates in numerical form and a comparison with existing experimental data. Analytic results for the decay rates and energy distributions are collected in two appendices for reference. Section V contains our conclusions.
The terms of interest in the effective weak Hamiltonian
contain the ‘current-current’ operators
and the QCD penguin operators . (See the review Ref. for their precise definition.) For the decays it is convenient to choose a Fierz version of the current-current operators such that the pair at the weak decay vertex is either in a colour singlet or a colour octet state. The coefficient functions are related to the usual by
and the one-loop and two-loop anomalous dimensions
The quantity is scheme-dependent and depends in particular on the treatment of . In the ‘naive dimensional regularization’ (NDR) scheme, ; in the ’t Hooft-Veltman (HV) scheme, . In the HV scheme the current-current operators, implied by the convention used in Refs. [11, 13], are not minimally subtracted. If one computes the low energy matrix elements of the weak Hamiltonian in the modified minimal subtraction () scheme, as we will do below, one has to apply an additional finite renormalization. This amounts to multiplying the coefficients by a factor of , or, equivalently, to an additional contribution to in the HV scheme. No additional renormalization is required in the NDR scheme. At NLO the strong coupling is given by
The NLO QCD corrections involve the one-loop virtual gluon correction to and the real gluon correction , where the pair is projected on one of the states in (2). The corresponding diagrams are shown in Figs. 1 and 2 respectively. The decay rate into a quarkonium can be written as the sum of partial decay rates through one of the intermediate states . At next-to-leading order the partial decay rates take the form
and . The operators are defined as in Ref. . The LO term is multiplied by if is a colour singlet state and by if is a colour octet state. We also used the fact that to high accuracy. The functions and will be given later. The LO contribution is multiplied by a correction term due to the penguin operators in (3). Likewise, we write the quarkonium energy distribution as
where . Note that to leading order in we do not distinguish the quark mass from the meson mass. To the order in the velocity expansion considered in this paper, we can also identify the momentum of the quarkonium with the momentum of the pair. (The kinematic effect of distinguishing the two is discussed in Ref. .) Hence can also be identified with , where is the quarkonium energy in the meson rest frame and the meson mass.
We now discuss which intermediate states should be taken into account for the production of a given quarkonium .
, : At leading order in the velocity expansion the spin-triplet -wave charmonium states are produced directly from a pair with the same quantum numbers, i.e. . At order relative to this colour singlet contribution, a can materialize through the colour octet states , where the subscript ‘’ implies a sum over . The suppression factor follows from the counting rules for the multipole transitions for soft gluons that convert the state into the meson . The leading order colour singlet contribution is proportional to , while the colour octet terms are proportional to . Because the weak effective Hamiltonian favours the production of colour octet pairs by a large factor
the colour octet contributions must be included, since their suppression by (for ) can easily be compensated. (The numbers serve only as order of magnitude estimates of the relative importance of the colour singlet and the colour octet contributions.) According to the velocity counting rules, there is a correction of order to the colour singlet contribution related to the derivative operator as defined in Ref. . Because it is multiplied by the small coefficient , and because we will find that the colour singlet contribution is indeed a small contribution to the total production cross section, we do not consider this additional correction in what follows. Similar derivative operators contribute to the colour octet channels. We do not take them into account, because we do not take into account other corrections of order with the large coefficient . Hence, even after including the NLO correction in , there remains an uncertainty of order in the theoretical prediction, assuming that the long-distance matrix elements were accurately known.
: The same discussion applies to the spin-singlet state. The colour singlet contribution involves . At relative order , can be produced through the colour octet states .
: At leading order in the velocity expansion, both and contribute to the production of a the spin-triplet -wave state . Because the partial production rate through the state is already multiplied by the large coefficient , it is not necessary to go to higher orders in the velocity expansion. Note that, because of the structure of the weak vertex, a pair cannot be produced in a angular momentum state at LO in .
: The same discussion as for the states applies to the spin-singlet -wave state. In this case we take into account and at NLO in . Owing to the structure of the weak vertex, a pair cannot be produced in a angular momentum state at LO in .
Iii Outline of the calculation
The Feynman diagrams shown in Figs. 1 and 2 are projected onto a colour and angular momentum state as specified in (2). The virtual corrections contain ultraviolet (UV) divergences, which can be absorbed into a renormalization of the operators in the weak effective Hamiltonian (3). The virtual corrections contain infrared (IR) divergences, which cancel against IR divergences in the real corrections. In addition, the real corrections contain IR divergences due to the emission of soft gluons from the or lines, which do not cancel with IR divergences in the virtual correction, if the pair is projected on a -wave state. These IR divergences can be factorized and absorbed into a renormalization of the non-perturbative matrix elements . In the following we provide some details on the UV and IR regularization, which are specific of the present calculation. More details on the strategy of a next-to-leading order calculation can be found in Ref. , which deals with quarkonium decay and total quarkonium production cross sections in fixed-target collisions.
iii.1 UV regularization and the treatment of
The UV divergences are regulated dimensionally and the IR divergences are regulated with a gluon mass. The UV divergences in the diagrams of Fig. 1 cancel against the UV divergences in diagrams (not shown in the figure) with the insertion of the 1-loop counterterm for . We combine the diagram with its counterterm diagram before projecting on a particular state , and before taking the 2-particle phase space integral. This has the advantages that it avoids extending the projection to dimensions and that the phase space integral can also be done in four dimensions.
The finite part of the virtual gluon correction depends on the prescription for handling in dimensions. This has to be chosen consistently with the one used to define the operators in Ref. . The prescription consists of a definition of and its anti-commutation property, together with a choice of ‘evanescent operators’. The evanescent operators are implicitly defined by specifying the order (where ) terms of the following products of Dirac matrices:
In the HV scheme, vertex diagrams are treated differently in Ref. and Ref. . As a consequence, as already mentioned above, in the HV scheme one has to multiply the coefficients defined in refs. [11, 13] by the factor , while this factor is already included in the definition of Ref. . We checked that our final result is identical in the NDR and HV schemes up to terms beyond NLO accuracy, if we use the expressions for of Sect. II including the additional factor just mentioned in the HV scheme.
The coefficient functions quoted in Sect. II refer to a Fierz version of the weak Hamiltonian different from (3) and Fierz transformations do not commute with renormalization in general. If we use the standard Fierz version rather than the singlet-octet form quoted in (3), this interchanges and in the results quoted in Appendix A.3. However, since in both schemes we used one has , either of the two Fierz versions can be used.
The NLO calculation for has already been done in Ref. in the HV scheme. We find that our result for the functions defined in (15) and given in the Appendix agrees with the result of Ref. . Nevertheless our result for the contribution of this channel to the decay rate, given by NLO terms, differs from the one given in Ref. , because the authors of Ref. used the coefficient functions of Ref. , but did not correct them (or alternatively, the low energy matrix elements) by the factor . As explained above with the conventions of Ref. this additional factor is necessary in the HV scheme to obtain a scheme-independent result.
iii.2 IR regularization and NRQCD factorization
The real and virtual corrections individually have double-logarithmic IR divergences, which we regulate by a gluon mass. However, after adding all contributions to the partonic process , the IR divergences do not cancel completely. The remaining IR divergences are associated only with emission from the and quark. This is a necessary (but not sufficient) requirement for their factorization into NRQCD matrix elements as discussed in detail in Ref. .
In addition to these IR divergences related to soft gluon emission, the last diagram in Fig. 1 exhibits the well-known Coulomb divergence, when the relative momentum of the and is set to zero. We regularize this divergence by keeping the relative momentum finite in the integrals, which would otherwise give rise to the Coulomb singularity.
In order to extract the short-distance parts (see (15)) of the partonic decay, we write
where denotes the NRQCD matrix element for a perturbative pair in the state . At NLO one has to calculate the left-hand side and to NLO.
The diagrams that contribute the correction to are shown in Fig. 3. For the first diagram (together with its complex conjugate) we obtain
where , if is a colour singlet state, if is a colour octet state and is the relative velocity of the two quarks. (The superscript ‘0’ refers to a matrix element at tree level, ‘1’ denotes a 1-loop contribution.) This renders the short-distance coefficients free of the Coulomb singularity.
The other two diagrams (called collectively ‘B’) together with their symmetry partners are UV and IR divergent. We define the NRQCD matrix elements in the scheme and denote their renormalization scale by . The IR divergence is regulated with a gluon mass to be consistent with the IR regulator used for the evaluation of the partonic process on the left-hand side of (24). The result is (compare with the Appendix of Ref. and with Ref. , where other IR regulators are used):
( denotes the gluon mass.) Note that if one breaks up the term into terms with different , one should replace
Using these results and solving for we find the IR finite short-distance coefficients for each collected in Appendix A.3.
iii.3 Difficulties with the colour singlet channels
The LO contributions to the colour singlet channels , and are proportional to the small and strongly scale dependent coefficient . One would therefore expect the NLO contribution to be particularly important for these channels. However, the strict NLO calculation leads to a negative, and therefore meaningless decay rate into these channels and to the conclusion that a reliable result can only be obtained at next-to-next-to-leading order. This problem was already identified and discussed in Ref. . (For the remainder of this section it is assumed that the reader has consulted Ref. for more details.)
Consider the three next-to-leading order terms in (15). Despite its large coefficient the -term, which comes only from a real correction, turns out to be numerically very small (see the tables in the following section). Both and (at ) are large and negative, and in particular, which comes with the larger coefficient , drives the decay rate negative.
The authors of Ref. suggested treating the decay process in a simultaneous expansion in and . This implies that one should add to the term of order all terms of order , because they also count as NLO in this rearranged expansion. On the other hand, the term (which involves ) should be neglected as being of higher order. The authors of Ref. did not actually calculate all terms of order , but estimated them by adding
to (15). This estimate can be motivated as follows: the virtual contribution to is given by the first four diagrams in Fig. 1 times the (complex conjugated) tree amplitude. All two-particle cuts to the term are given by the square of the first four diagrams in Fig. 1. Hence, ignoring the real contribution, one may argue that is close (but not equal) to the two-particle contributions to the term.
In Ref. the square of the 1-loop amplitude with a final state is computed exactly and argued to provide a better estimate than the original one of Ref. , because one leaves out only real contributions to the coefficient of , which are argued to be phase-space suppressed. However, we find that for the channel the virtual contributions alone are IR divergent. Therefore the real correction that cancels this divergence cannot be argued to be small. In our opinion this also calls into question the assumption that the real contributions are numerically small for the -wave channels. For this reason we choose to follow with a minor modification the procedure of Ref. , which adds an IR finite term by construction, since is IR finite. The minor modification is the following: the third and fourth diagrams in Fig. 1 have imaginary parts, which contribute to the real part of the square of the amplitude (and hence to the coefficient of ). The remnants of these imaginary parts after multiplying the one-loop amplitude by the complex conjugate of the tree amplitude can easily be restored from the -term (and -term in the case of ) in the results presented in Appendix A.3. If we call the restored imaginary part , then we use (28), with replaced by .
We wish to emphasize two points: first, the discussed modifications of the colour singlet channels are certainly ad hoc and should be regarded with great caution. Second, the effect on the decay rate into a particular quarkonium state is not severely affected by this uncertainty, because it is dominated by colour octet contributions, whose short-distance coefficients can be computed reliably at NLO as we shall see.
In order to gain a numerical understanding of the importance of the various terms involved in the colour singlet channels, we consider in the following three computational schemes for the decay rate: (a) the (strict) NLO calculation; (b) the NLO calculation with the term (28) added, but without the -term (‘improved’); (c) the same as (b), but with included (‘total’). For the -wave colour singlet channels, (a) and (c) yield a negative rate. They are therefore meaningless. Option (b) yields a positive result of a magnitude similar to the result of Refs. [10, 17]. It may be considered as an order-of-magnitude estimate for the colour singlet contribution, but it may well be uncertain by 100%. For the channel all three options give negative partial rates. However, since this channel mixes with and only the sum of the two is physical, a negative partial rate is not unphysical by itself.
Iv Results and Discussion
In this section we present our results for the branching fractions of decay into charmonium and moments of the quarkonium energy distributions in numerical form. The analytic expressions that enter (15) and (17) are collected in the appendices for reference.
iv.1 Branching ratios for decay into charmonium
iv.1.1 General discussion of NLO corrections
We normalize our calculation to the theoretical expression for the inclusive semileptonic decay rate
represents an excellent approximation for the 1-loop QCD correction factor. (The complete analytic result can be found in Ref. .) For any particular quarkonium state , we obtain the branching fraction in the form†††When is a -wave state, should be understood as in the following formula, so that all matrix elements have mass dimension 3. Furthermore, in the case of which refers to the state summed over , the NRQCD matrix element is chosen to be .
The overall factor is given by
where we used and as given by (16). The charm and bottom pole masses are taken to be GeV and GeV, respectively. This yields , which we use unless otherwise mentioned. The sensitivity of the charmonium production cross sections to the quark mass values will be discussed below.
We first examine the impact of the next-to-leading order correction and the dependence on the factorization scale for each intermediate state separately. We neglect the penguin contribution for this purpose. In Table 1 we show the branching fractions excluding the dimensionless normalization factor for three values of at LO and at NLO. To evaluate the LO expression we also use the Wilson coefficients at LO and 1-loop running of the strong coupling with such that both in LO and NLO. This is a large effect for the colour singlet channel, since , (in the NDR scheme) but , .
We now observe that the colour singlet contributions are, as expected, enormously scale-dependent at LO. The NRQCD matrix element is related to the radial wavefunction at the origin by up to corrections of order . Using GeV , we obtain
to be compared with the measured branching fraction .‡‡‡Note that we denote by the direct production of , excluding radiative decays into from higher-mass charmonium states. The same convention applies to all other charmonium states . The LO prediction is uncertain by a factor of about 10 for all colour singlet channels, as can be seen from Table 1. As also seen from this table, the scale uncertainty is reduced to a factor 2–3 at NLO. However, the NLO correction term renders the partial decay rates negative, as already mentioned in Section III.3.
The situation can be somewhat improved by adding the estimate (28) for the order NNLO term with the large coefficient , while treating the term as formally of higher order in a double expansion in and . The addition of (28) also reduces the factorization scale dependence further, because it contains exactly the double logarithmic correction , which is required to cancel the large scale dependence of at leading order in . In Table 2 we display the result for the partial decay rates into the colour singlet channel, which is obtained in this way (denoted ‘Impr’ in the table) and for comparison again the LO and NLO result. The improvement can and should be done only for those colour singlet channels that have non-vanishing LO contributions. The last three rows of Table 2 show the results that are obtained if we add back the term in (31) to the improved treatment. The term is sizeable and negative and therefore re-introduces a large scale dependence. The same improvement that is applied to the LO term is necessary for the term, which would require going to order . One may argue that unless this is done, it is preferable to leave the term out entirely. Therefore we shall use the ‘improved’ version (‘Impr’ in Table 2) as our default option later. While the result is certainly not accurate, we believe that this is the best we can do to the colour singlet channel without making arbitrary modifications. We note that for this gives a colour singlet contribution to the branching fraction, which is close to the lower limit in (33) and also compatible with the estimates of Refs. [10, 17]. It seems safe to conclude that colour singlet production alone is not sufficient to explain the measured branching fraction.
The partial rates in the four relevant colour octet channels are shown in the lower part of Table 1. In this case we find that the perturbative expansion is very well behaved. The NLO short-distance coefficients are larger by 20%–50% than the LO coefficients and the scale dependence is very moderate. The scale dependence is not reduced from LO to NLO. This is due to the fact that the LO coefficients depend only on the scale-insensitive , while there are sizeable coefficients of the highly scale-dependent combinations and at NLO. The numerical enhancement of the short-distance coefficients in the colour octet channels, which is evident from Table 1, is sufficient to account for the measured branching fraction, as already noted in Refs. [8, 9]. The positive NLO correction reinforces this trend. Other production processes suggest that the long-distance parameters in the colour octet channels are of the order a few times GeV. (This will be made more precise soon.) This leads to typical branching fractions of order .
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CC-MAIN-2020-45
| 28,362 | 74 |
http://allnurses.com/general-nursing-student/dosage-basic-nrsg-496482.html
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math
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I highly recommend Kaplan's "Math for Nurses" book. The book is broken down into 2 parts. The first one is "Foundation Skills" and is exactly what it says... a review of the simpler math topics. The second part is the "Applications" section and goes into detail about what nurses need to know. I have to tell you this book became my best friend and I am no longer intimidated of math thanks to it. It tells you how to do dosage calculations for oral, IV, and parental medications. It was a lifesaver. I think the best thing about it is that early on, it introduces you to 3 or 4 different math approaching strategies... if you don't like one, move on until you find one that works for you. Then, for every type of problem in the book, it shows you how to use EACH strategy to find out the answer. It rocks!
The ISBN is 978-1-4195-9955-2
I got mine at Barnes and Noble for about 15 bucks I think and it was worth every penny.
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https://zooom.ir/gre-prep/math/powers-and-roots/rationalizing/
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math
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توجیه کردنفصل: بخش ریاضی / درس: ریشه ها و قدرت ها / درس 15
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متن انگلیسی درس
Now we can talk about the very important topic of rationalizing. So throughout this module, we have seen a few fractions involving radicals. Notice that so far, no final answer has involved a radical in the denominator. And in fact I have crafted every fraction to make sure that this never happened, that it never was a result that a radical was left over in the denominator. Now why would I do that?
This is a tricky issue. Having a radical in the denominator is considered in poor taste, in mathematics. Now, why would mathematicians think this? One way to think about it is, if there are multiple fractions that need to be added or subtracted it’s helpful if all the denominators are integers so that we can find a common denominator.
A more general way to say it is. If there are multiple ways to write an answer, it’s helpful if there’s an established convention so that everyone writes the answer the same way, and that makes it much easier to compare whether two different people have found the same answer. So for whatever reason, the mathematical convention is to avoid radicals in the denominator.
Now, if a radical appears in the denominator as part of the problem, and of course if sometimes happens, we need to change the fraction to an equivalent form, so that no radicals appears in the denominator. This process of change is called rationalizing the denominator. That is to say, making the denominator a rational func, a rational number, an integer rather than an irrational square root.
So, suppose in the course of solving a problem, we found x equals 12 divided by root three. Let’s pretend that we did all the math correctly, we got this as our correct answer. This is our solution to the problem. That might be mathematically correct, but because it’s not rationalized, we won’t find it among the answers listed in this, won’t find the answer listed in that particular form.
You see all the answers on the test are rationalized, so we need to rationalize anything that we find to match it to the answer choice. If there’s a single radical in the denominator, we rationalize simply my multiplying by that radical over itself. So, we’re just gonna multiply by the square root of three, over the square root of three.
And of course, in the denominator, we get root three, twelve root three. Well, think about this. Root three is the number which, when we square it, we get three. So, if here, we multiply it by itself, so it becomes three. Then we can cancel the 12 divided by three. That becomes four.
We get four root three. That is the rationalized form, so the answer would never be written in the form 12 divided by root three. It would be written in the form four times root three. Practice rationalizing the following.
In each case you’re just gonna multiply by the radical over itself and then simplify. Pause the video and then we’ll talk about this. Okay and that first one obviously we will multiply by root five over root five and we get root five over five. In the second one, we’ll multiply by root three over root three. We’ll wind up with two root three over three.
In the third one, we’ll multiply by root 21 over root 21. We get a 21 in the denominator, that allows to cancel a factor of seven. And wind up with two root 21 over three. Those are the rationalized forms. Even if there is addition or subtraction in the numerator divided by a single root, all we have to do is multiply by the root over itself.
So, for example if we had something like four minus root six, over two root three. Really the only thing that concerns us is that root three in the denominator. So I’m gonna multiply by root three over root three. In the denominator I get a root three times a root three, which is three, and the number I distribute. That root six times root three.
Remember that root six is root two times root three. So we can multiply the two factors of root three together to get three. And then we’re just left with a root two left over. So what we wind up with, is four root three, minus three root two, over six. And that is the fully rationalized form. Things get trickier when we have addition or subtraction in the denominator.
So suppose we have to simplify something like two divided by the quantity, root five minus one. Well, hm. What we have been doing is not gonna work here. If see, if we multiply this denominator by root, by this root five, we would have to distribute across the to subtract, across the subtraction in the denominator.
The root five times five, that would also be five. But, of course, they’ll also be a one times root five. And that would be root five. And our attempt to rid the denominator of radicals would not be successful. We have to make use of a different trick, so I will say remember the difference of two squares formula from the algebra module.
P minus q times p plus q equals p squared minus q squared. If this is an unfamiliar formula I highly suggest go back to the algebra modules and watch the videos concerning this particular formula. It’s one of the most widely applicable formulas in all of mathematics. So in other words what we have here is the product of a sum and the, and the difference, as a result of squaring both terms.
If p or q, or even both of them, were radical expressions, then on the right side of the equation above, every radical expression would be squared, and there would be no radicals in the result. Very interesting. Consider any two terms, any two term expression, either addition or subtraction, in which one or more of the terms is a radical.
So we might have things like this for example. Two terms, and we’re either adding or subtracting, and one or both of the terms is a radical. If we change the addition or subtraction to its opposite, we construct what is called the conjugate expression. So that first one we have three minus root seven, the conjugate would be three plus root seven.
In the second one, we have root 13 plus two root 11. The conjugate will be root 13 minus two root 11. And so how is this helpful? Well, if we multiply any expression with radicals by its conjugate, one conjugate plays the role of a plus b, the other plays the role of a minus b. So that their product is the difference of the squares of the two terms, a squared minus b squared.
So for example if we had three root seven, three minus root seven times three plus root seven. Well essentially what we have here is an a minus b times and a plus b and of course this is gonna equal a squared minus b squared. Or in other words three squared minus root seven squared. Square each on of them we get 9 minus 7 which is 2.
So, in other words, we multiply two expressions with radicals and we get a product without radicals. That’s a really big idea. This provides a clue we need for how to rationalize a fraction with a radical expression involving addition or subtraction in the denominator. So let’s go back to that one that we couldn’t solve a moment ago.
We had something like two divided by the quantity root five minus one. Well what I’m gonna do is multiply the numerator and denominator by the conjugate of the denominator. To rationalize, we’re gonna multiply the numerator and the denominator by the conjugate of the denominator. So what we have in that denominator is root five minus one, the conjugate would be root five plus one.
So we’re gonna multiply by that conjugate, root five plus one over root five plus one. Then in that denominator, we have an a minus b and an a plus b so we’ll get an a squared minus b squared in that denominator. The numerator I’m just going to leave undistributed for a moment. And so we get a root five squared minus 14 squared.
That would be five minus one, which is four. Then I can cancel a factor of two in the numerator and denominator. And I get root five plus one over two, and that is the rationalized form. Incidentally, you don’t have to know this for the test, but that happens to be the golden ratio. So rationalize the following.
You will need to multiply by the denom, multiply the denominator by its conjugate. So I’m gonna say pause the video and try this on your own. Okay, so we have a root seven minus root three, and so what we’re gonna need to do is multiply that denominator by our root seven plus root three. So we have to multiply root seven plus root three over root seven plus root three, multiplied by the same thing over itself.
In the numerator, I’m not going to distribute just yet. In the denominator, I get the square of those terms, the difference of the squares, which just seven minus three. Seven minus three is four. And then cancel eight divided by four and we get two times root seven plus root three, and that is the fully rationalized form.
Now we can distribute it if we want to depending, doesn’t matter, the answer could be written in either of those last two forms. Rationalize the following. You will need to multiply the denominator by its conjugate. Pause the video work on this and then we’ll talk about it. Okay so we have a three plus root five in the denominator so we need to multiply by a three minus root five.
So we need to multiply both the numerator and the denominator by three minus root five. In the denominator we’re just gonna get the difference of two squares. Three squared minus root five squared. In the numerator I’m gonna have to foil. And so what I get is the product of the first, four times three is 12.
The product of the outer, four so that’s minus four, root five. The product of the inner. That’s plus six root five. And then the product of the last, which is two root five minus negative root five. Well the root five times the root five is five, two times five is ten so that’s just minus ten.
Now we’ll simplify everything in the numerator and the denominator. We wind up with two plus two with five over four. Everything is divisible by two, so we can divide it by a factor of two. And we are left with one plus root five divided by two. Here’s a practice problem.
Pause the video, and then we’ll talk about this. Okay, so we need to get x by itself, so we are, of course we are going to divide by everything in the parenthesis. We have an expression in the denominator, involving addition or subtraction with a radical.
So we’re gonna multiply by the conjugate. We have a two root five minus three. We need to multiply that by a two root five plus three. And so I’ll multiply by that. In the numerator I’ll just leave it undistributed. In the denominator, I’m gonna get two root five.
That thing squared minus three root squared. And, of course, two root pi squared as we learned from our operations of radicals a few videos ago. That’s two squared tim, times root five squared. Four times root five, four times five which is 20.
And of course three squared is nine. 20 minus nine we get 11. We can cancel a factor of 11 in the numerator, denominator and we’re left with five times the quantity two root five plus three and that is the fully rationalized form of the answer. In summary to eliminate roots from the denominator of a fraction this is called rationalizing and it’s something you always have to do.
The test answers, the answers that appear on multiple choice will always be rationalized and so we will have to rationalize to match our answers to this. If the fraction has a single root in the denominator, we rationalize simply by multiplying by that root over itself. If the denominator of the fraction contains addition or subtraction involving a radical expressions, to rationalize we need to multiply by the conjugate of the denominator over itself.
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| 12,522 | 45 |
https://projecteuclid.org/journals/kodai-mathematical-journal/volume-41/issue-1/An-isoperimetric-inequality-for-diffused-surfaces/10.2996/kmj/1521424824.short
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math
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For general varifolds in Euclidean space, we prove an isoperimetric inequality, adapt the basic theory of generalised weakly differentiable functions, and obtain several Sobolev type inequalities. We thereby intend to facilitate the use of varifold theory in the study of diffused surfaces.
"An isoperimetric inequality for diffused surfaces." Kodai Math. J. 41 (1) 70 - 85, March 2018. https://doi.org/10.2996/kmj/1521424824
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CC-MAIN-2022-05
| 425 | 2 |
https://www.slideshare.net/Bigzad/statistics-lesson-2
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math
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1.15 VariablesVariables are things that we measure, control, ormanipulate in research.Example:In studying a group of children, the weight of eachchild is a variable – it is measurable and it variesfrom child to child.Variate: Each individual measurement of a variable(e.g., each weight of a child)
Quantitative and Qualitative VariableA Quantitative Variable: whose variates can beordered by the magnitude of the characteristicsuch as weight, length, quantity and so on.e.g., number of tomatoes on a plant.A Qualitative Variable: whose variates aredifferent categories and cannot be ordered bymagnitude. (e.g., type of tree)
1.16 Observable and Hypothetical Variables.Observable Variables: Directly measurable suchas height, weight.Hypothetical Variables: Indirectly measurablesuch as inherited differences between shortdistance or long distance runners.
1.17 Functions and RelationsIf 2 variables X and Y are related that for everyspecific value x of X is associated with only onespecific value y of Y, that Y is a function of X.A domain is the set of all specific x values that Xcan assume.A range is the set of all specific y valuesassociated with the x values.
1.17 Functions and Relations• When an x value is selected, the y value is determined. Therefore, the y value ‘depends’ on x value.X is the ‘independent variable’ of the function and Y is the ‘dependent variable.’And Y is a function of X.
1.17 Functions and RelationsExample 1.29 For the relation Y = X ± 3, whatare its domain, range, and rule of association?There are two y values for every x.Domain: x values, (1, 2, 3)Range: y values. (-2 & 4, -1 & 5, 0 & 6)
1.18 Functional NotationFor Y = X2, the functional notation is y = f(x) = x2For y = f(x) = -3 + 2x + x2 , find f(0) and f(1)
1.19 Functions in StatisticsThe goal of research is to study cause and effect;to discover the factors that cause something(the effect) to occur.Example: a botanist want to know the soilcharacteristics (causes) that influence plantgrowth (effect); or an economist want to knowthe advertising factors (causes) that influencesales (effect).
1.19 Functions in StatisticsExample 1.31 In the following experiment, which isthe independent variable and which is thedependent variable?To determine the effects of water temperature onsalmon growth, you raise 2 groups of salmon (10 ineach group) under identical conditions fromhatching, except that one group is kept in 20 Cwater and the other in 24 C water. Then 200 daysafter hatching, you weigh each of the 20 salmon.
1.20 The real number line and rectangular Cartesian coordinate systemEvery number in the real number system can berepresented by a point on the real number line.
1.20 The real number line and rectangular Cartesian coordinate systemA rectangular Cartesian coordinate system (orrectangular coordinate system) is constructedby making two real number line perpendicularto each other, such that their point ofintersection (the origin) is the zero point of bothlines.Example 1.33 Plot the following points on arectangular coordinate system: A(0,0); B(-1.3);C(1,-3); D(2,1); E(-4,-2)
1.20 The real number line and rectangular Cartesian coordinate system A Rectangular Cartesian Coordinate System
1.21 Graphing FunctionsA graph is a pictorial representation of therelationship between the variables of a function.Example 1.34 Graph the function y=f(x)=4 + 2xon a rectangular coordinate system.
1.21 Graphing FunctionsQuadratic function:•Characteristics of Quadratic Functions•1. Standard form is y = ax2 + bx + c, wherea≠ 0.•2. The graph is a parabola, a u-shapedfigure.•3. The parabola will open upward ordownward.•4. A parabola that opens upwardcontains a vertex that is a minimumpoint.A parabola that opens downwardcontains a vertex that is a maximumpoint.
1.22 Sequences, Series and Summation Notation• Sequence: a function with a domain that consists of all or some part of the consecutive positive integers.• Infinite Sequence: the domain is all positive• Finite Sequence: the domain is only a part of the consecutive positive integers.• Term of the Sequence: Each number in the sequence.• f(i) = xi, for i = 1, 2, 3. the i in the xi is “subscript or an index, and xi is read “x sub I”.
1.22 Sequences, Series and Summation NotationExample 1.35 What are the terms of thissequence: f(i) = i2 – 3, for i = 2, 3, 4
1.22 Sequences, Series and Summation NotationA series is the sum of the terms of a sequence.For the infinite sequence f(i) = I + 1, for I = 1, 2,3, …, ∞, the series is the sum 2 + 3 + 4 + … + ∞.For the finite sequence f(i) = xi, for i = 1, 2, 3, theseries is x1+ x2 + x3
1.22 Sequences, Series and Summation NotationThe summation notation is a symbolicrepresentation of the series: x1+ x2 + x3 + … + xn
1.22 Sequences, Series and Summation NotationWhen it is clear that it is the entire set beingsummed, the lower and upper limits of thesummation are often omitted.Example 1.37 The height of five boys in a 3rdgrad class form the following sequence: x1 = 2.1ft, x2 = 2.0 ft, x3 = 1.9 ft, x4 = 2.0 ft, x5 = 1.8 ft.For this set of measurement, find sum.
1.23 Inequalities• THIS SIGN < means is less than.. This sign > means is greater than. In each case, the sign opens towards the larger number.• For example, 2 < 5 ("2 is less than 5"). Equivalently, 5 > 2 ("5 is greater than 2").• These are the two senses of an inequality: < and > .• the symbol ≤, "is less than or equal to;" or ≥, "is greater than or equal to."
1.23 InequalitiesExample 1.40 For the inequality 8 > 6Multiply both sides by -3Example 1.41 Solve the inequality: X + 7 > -3
Questions1.80 Using the quadratic formula to solve 4X2 = 1
QuestionsFor y = f(x) = 7x - 5, find(b) f(0)(c) f(5)1.84 Graph the linear function y = f(x) = 3- 0.5xon a rectangular coordinate system using itsslope and y intercept.
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http://omniglossiapublishing.com/notebook/graph-paper-notebook-0-5-centimeter-squares-120-pages/?post_in_lightbox=1
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math
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Art series notebook with cover by Kuindzhi, 8.5 x 11 graph paper notebook with 1/2 centimeter squares, perfect bound, ideal for graphs, math sums, composition notebook or even journal
Graph paper notebook with 120 pages with half centimeter squares in a good sized 8.5 x 11 inch format ideal for graphs, math sums, composition books and notebooks. The notebook is perfect bound so that pages will not fall out.
Squares: 1/2 cm
Numbered pages: No
Edge to Edge: No
Part of the Notebook not Ebook series with a cover depicting an artwork. This cover features a painting by the 19th century Russian artist Arkhip Kuindzhi – “After a Rain. Rainbow”. Our notebooks all have a distinctive, colorful cover. The notebook is perfect bound so that pages will not fall out and has a soft yet sturdy cover.
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s3://commoncrawl/crawl-data/CC-MAIN-2021-17/segments/1618038088471.40/warc/CC-MAIN-20210416012946-20210416042946-00066.warc.gz
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CC-MAIN-2021-17
| 799 | 6 |
https://www.coursehero.com/file/6655536/Lecture-13/
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math
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Unformatted text preview: Chapter Three
_____________________________________________________________________ Shear, Bond and Torsion 3.1 Shear Failure of RC Beams without Shear Reinforcement 3.1.1 Shear in homogeneous beams and in RC beams
• In mechanics of materials, the shear stress of homogeneous, elastic and uncracked beams can be calculated by (average shear stress)
Ib where V − shear force
Q – first moment of area ( Q = A ⋅ y ) (a) (b) (c) Fig. 3.1-1 Normal, shear and principal stresses in homogeneous uncracked
beam. (a) Flexural and shear stresses acting on elements in shear span; (b)
distribution of shear stress; (c) principal stresses on elements in shear span
103 Fig. 3.1-2 Principal compressive stress trajectories in an uncracked beam
• Crack vs flexural and shear stresses for a RC beam (g) Fig. 3.1-3 Orientation of principal stresses. (a) Geometry and loading; (b)
axial stress at section x-x; (c) shear stress distribution at section x-x; (d)
segment A; (e) segment B; (f) segment C; (g) principal compressive stress
paths 104 o Crack patterns follow the principal compressive stress paths
o Stresses in different segments Segment A (at the bottom):
v = 0 and σ = σmax Vertical flexural crack. Segment B (at the neutral axis):
v = vmax and σ = 0 about 45°−diagonal crack. Segment C (between the N.A. and the bottom):
Combination of v and σ 45°−90° diagonal crack. 3.1.2 Types of shear failure
The type of shear failure of RC beams depends mainly on various factors:
Shear span-to-depth ratio
Quantity of longitudinal reinforcement ratio ρ = As /(bd )
Geometry of the beam, etc.
• One of the most significant factors is the shear span-to-depth ratio, av / d . Shear span av is defined as the distance between points of
zero and maximum moments. Fig. 3.1-4 Shear span. (a) Geometry and loading; (b) bending moment
105 The effective shear-span/depth ratio is defined as M / Vd , where
M is the bending moment and V the corresponding shear force.
This is true for both distributed loading or concentrated loading.
• Inclined cracks must develope before complete shear failure. Inclined cracks can form by shear-web cracking or, more
commonly, by shear-flexural cracking.
• The failure mode is strongly dependent on the shear-span/depth ratio a v / d .
The shear-span/depth ratio can be divided into four general
categories. For different members falling into the same category,
the sequence of events and the nature of the failure are
approximately the same.
Category Shear span/depth ratio Category I av
d Category II 1< av
d Mode of failure
Deep beam failure
Dowel failure Category III 2.5 < av
d Shear-tension failure
(Diagonal tension failure)
Shear-bond failure Category IV 6< av
d Flexural failure 106 (1) Category I ( av / d < 1 ): Deep beam failure The structural behaviour approaches that of deep beams.
The diagonal crack occurs approximately along a line joining
the loading and support points. It forms as a result of the
splitting action of the compression force transmitted directly
from the loading point to the support.
The shear-web crack initiates frequently at about d/3 about the
bottom face, and propagates simultaneously towards the loading
and support points. When the crack has penetrated sufficiently
deeply into the concrete zone at the loading/the support points,
crashing failure of concrete occurs.
The mode of failure may be an anchorage failure at the end of
the reinforcement due to the large tensile force. 107 (2) Category II: 1 < av / d < 2.5
Shear-compression failure The diagonal crack (shear-web crack) forms independently and
not as a development of a flexural crack. Eventually, the diagonal crack penetrates into the concrete compression zone at
the loading point. Crashing failure of the concrete occurs.
Dowel failure The member can fail owing to dowel failure of the longitudinal
reinforcement at the point of the inclined crack.
(3) Category III ( 2.5 < av / d < 6 ): Diagonal tension failure 108 Flexural crack (a-b) is developed before the compressive force
is great enough to develop shear-web cracks. Then the flexural
crack (a-b) nearest the support would propagate towards
loading point, becoming an inclined crack (a-b-c) known as a
If a v / d is relative low, the diagonal crack (a-b-c) spreads but
stops at j ; the crack widens and propagates along the level of
tension reinforcement (g-h), destroying the bond between the
reinforcement and the surrounding concrete. This is the shearbond failure.
The shear-bond failure may also occur in the Category II
members: If a v / d is relative high, the diagonal crack (a-b-c) spreads to
e, splitting the beam into two pieces along an inclined crack (ab-c-e). This is typical diagonal tension failure.
(4) Category IV ( av / d > 6 ): Flexural failure Beams with such a high value of a v / d usually fail in bending.
109 • Summary of types of failure: Fig. 3.1-5 Summary of types of failure for beams without shear reinforcement 3.1.3 Mechanisms of shear transfer
Consider a typical beam in bending and shear shown below, which is
reinforced with longitudinal steel against bending. Fig. 3.1-6 A typical beam in bending and shear
• The shear force is transmitted through the cracked beam by a combination of three mechanisms (Fig. 3.1-7):
(1) Dowel action of longitudinal reinforcement
(2) Aggregate interlock
(3) Shear stress in uncracked concrete
110 Fig. 3.1-7 Shear transfer mechanisms. (a) Dowel action; (b) aggregate
interlock; (c) shear stresses in uncracked concrete • The external shear force V is considered to be resisted by the combined action of Vcz, Va and Vd . Fig. 3.1-8 Transmitted shear forces Thus, V = Vcz + Va + Vd
111 The shear force V is carried in the approximate proportions:
Compression zone shear Vcz = 20 − 40%; Aggregate interlock
Va = 35 − 50%; Dowel action Vd = 15 − 25% .
• The shear capacity is strongly dependant upon the shear span-depth ratio a v / d , but may also be affected by:
o Tension reinforcement ratio ρ = As / (bd) ρ ↑, probably the aggregate-interlock and also dowel action ↑ Fig. 3.1-9 Effect ρ and fcu on
nominal ultimate shear stress v o Concrete strength fcu fcu ↑, possibly the shear capacities due to aggregate-interlock
and dowel action and in compression zone generally ↑; fcu ↑,
concrete tensile strength ↑, then inclined cracking load ↑.
o Aggregate type − effect on aggregate interlock capacity For lightweight concrete, the shear stress is equal to 80% of that
of normal concrete (BS 8110).
o Beam size (particularly the beam depth) − size effect: larger beams are proportionately weaker than smaller beams.
112 3.1.4 Design concrete shear stress
• The shear force carried by the concrete (incorrectly referred as) Vc = Vcz + Va + Vd
The design concrete shear stress (nominal concrete shear stress) vc = Vc
bv d (3.1) where bv = the beam width (bv = b for rectangular sections; bv=
bw for flanged sections).
• An accurate analysis for the shear strength of the concrete is not possible. The problem is solved by establishing the strength of
concrete in shear from test results.
In BS 8110: Part 1, clause 184.108.40.206, values for the design concrete
shear stress vc can be calculated by vc =
where 100 As 1/ 3 400 1/ 4
d γm (3.2) ○ γm = 1.25.
○ 100As /(bvd) should not be taken as greater than 3.
○ 400/d should not be taken less than 1.
○ For fcu > 25 N/mm2, the results may be multiplied by
( f cu / 25)1/ 3 , in which the values of fcu should not be taken as greater than 40. 113 Values of the design concrete shear stress, vc, are derived from Eq. (3.2)
and given in the following table (BS 8110: Part 1, Table 3.8): 3.1.5 Design shear stress (BS 8110: Part 1, clause 220.127.116.11)
• The design shear stress at any cross-section should be calculated
bv d (3.3) Definitions of bv can be found in Eq. (3.1).
• For a beam with web reinforcement, the shear resistance may be
regarded as being made up of the sum of the concrete resistance
and the web steel resistance: v = vc + v s (3.4) where vc is the design concrete shear stress and vs is the nominal
shear stress of web reinforcement. 114 ...
View Full Document
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https://bmt.chainalysiswhatsup.pw/physics-energy-work-equation.html
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math
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In physicswork is the energy transferred to or from an object via the application of force along a displacement. In its simplest form, it is often represented as the product of force and displacement.Wimpey laboratories vacancies
A force is said to do positive work if when applied it has a component in the direction of the displacement of the point of application. A force does negative work if it has a component opposite to the direction of the displacement at the point of application of the force. For example, when a ball is held above the ground and then dropped, the work done by the gravitational force on the ball as it falls is equal to the weight of the ball a force multiplied by the distance to the ground a displacement.
Work is a scalar quantity so it has only magnitude and no direction. Work transfers energy from one place to another, or one form to another. The SI unit of work is the joule Jthe same unit as for energy. According to Jammer, the term work was introduced in by the French mathematician Gaspard-Gustave Coriolis as "weight lifted through a height", which is based on the use of early steam engines to lift buckets of water out of flooded ore mines.
According to Rene Dugas, French engineer and historian, it is to Solomon of Caux "that we owe the term work in the sense that it is used in mechanics now". The SI unit of work is the joule Jnamed after the 19th-century English physicist James Prescott Joulewhich is defined as the work required to exert a force of one newton through a displacement of one metre.
Non-SI units of work include the newton-metre, ergthe foot-poundthe foot-poundalthe kilowatt hourthe litre-atmosphereand the horsepower-hour. Due to work having the same physical dimension as heatoccasionally measurement units typically reserved for heat or energy content, such as thermBTU and calorieare utilized as a measuring unit. The work W done by a constant force of magnitude F on a point that moves a displacement s in a straight line in the direction of the force is the product.
The work is doubled either by lifting twice the weight the same distance or by lifting the same weight twice the distance. Work is closely related to energy. The work-energy principle states that an increase in the kinetic energy of a rigid body is caused by an equal amount of positive work done on the body by the resultant force acting on that body.
Conversely, a decrease in kinetic energy is caused by an equal amount of negative work done by the resultant force. From Newton's second lawit can be shown that work on a free no fieldsrigid no internal degrees of freedom body, is equal to the change in kinetic energy KE corresponding to the linear velocity and angular velocity of that body.
The work of forces generated by a potential function is known as potential energy and the forces are said to be conservative. Therefore, work on an object that is merely displaced in a conservative force fieldwithout change in velocity or rotation, is equal to minus the change of potential energy PE of the object. These formulas show that work is the energy associated with the action of a force, so work subsequently possesses the physical dimensionsand units, of energy.
Constraint forces determine the object's displacement in the system, limiting it within a range. For example, in the case of a slope plus gravity, the object is stuck to the slope and, when attached to a taut string, it cannot move in an outwards direction to make the string any 'tauter'.
It eliminates all displacements in that direction, that is, the velocity in the direction of the constraint is limited to 0, so that the constraint forces do not perform work on the system. For a mechanical system constraint forces eliminate movement in directions that characterize the constraint. Thus the virtual work done by the forces of constraint is zero, a result which is only true if friction forces are excluded.
For example, in a pulley system like the Atwood machinethe internal forces on the rope and at the supporting pulley do no work on the system. Therefore work need only be computed for the gravitational forces acting on the bodies. Another example is the centripetal force exerted inwards by a string on a ball in uniform circular motion sideways constrains the ball to circular motion restricting its movement away from the centre of the circle.
This force does zero work because it is perpendicular to the velocity of the ball. It can change the direction of motion but never change the speed. This scalar product of force and velocity is known as instantaneous power.The concept of work in physics is much more narrowly defined than the common use of the word. Work is done on an object when an applied force moves it through a distance.
In our everyday language, work is related to expenditure of muscular effort, but this is not the case in the language of physics. A person that holds a heavy object does no physical work because the force is not moving the object through a distance. Work, according to the physics definition, is being accomplished while the heavy object is being lifted but not while the object is stationary.
Another example of the absence of work is a mass on the end of a string rotating in a horizontal circle on a frictionless surface. The centripetal force is directed toward the center of the circle and, therefore, is not moving the object through a distance; that is, the force is not in the direction of motion of the object.
However, work was done to set the mass in motion. Work is a scalar. If work is done by a varying force, the above equation cannot be used. The work performed on the object by each force is the area between the curve and the x axis. The total work done is the total area between the curve and the x axis. For example, in this case, the work done by the three successive forces is shown in Figure 1. Acting force changing with position. Kinetic energy.Cod mobile pdw setup
Kinetic energy is the energy of an object in motion. The expression for kinetic energy can be derived from the definition for work and from kinematic relationships. Consider a force applied parallel to the surface that moves an object with constant acceleration. The right side of the last equation yields the definition for kinetic energy: K. The above derivation shows that the net work is equal to the change in kinetic energy.
Potential energy. Potential energy, also referred to as stored energy, is the ability of a system to do work due to its position or internal structure. Examples are energy stored in a pile driver at the top of its path or energy stored in a coiled spring.
Potential energy is measured in units of joules. Gravitational potential energy is energy of position.When a fluid flows into a narrower channel, its speed increases.
That means its kinetic energy also increases. Where does that change in kinetic energy come from? The increased kinetic energy comes from the net work done on the fluid to push it into the channel and the work done on the fluid by the gravitational force, if the fluid changes vertical position.
Recall the work-energy theorem. There is a pressure difference when the channel narrows. This pressure difference results in a net force on the fluid: recall that pressure times area equals force.
As a result, the pressure will drop in a rapidly-moving fluidwhether or not the fluid is confined to a tube. There are a number of common examples of pressure dropping in rapidly-moving fluids. Shower curtains have a disagreeable habit of bulging into the shower stall when the shower is on. The high-velocity stream of water and air creates a region of lower pressure inside the shower, and standard atmospheric pressure on the other side.
The pressure difference results in a net force inward pushing the curtain in.
You may also have noticed that when passing a truck on the highway, your car tends to veer toward it. The reason is the same—the high velocity of the air between the car and the truck creates a region of lower pressure, and the vehicles are pushed together by greater pressure on the outside.Matris ammortizzatore di sterzo serie k con attacchi
See Figure 1. This effect was observed as far back as the mids, when it was found that trains passing in opposite directions tipped precariously toward one another. Figure 1. An overhead view of a car passing a truck on a highway. Air passing between the vehicles flows in a narrower channel and must increase its speed v 2 is greater than v1causing the pressure between them to drop P i is less than P o. Greater pressure on the outside pushes the car and truck together.
If we follow a small volume of fluid along its path, various quantities in the sum may change, but the total remains constant. In fact, each term in the equation has units of energy per unit volume. Making the same substitution into the third term in the equation, we find. Note that pressure P has units of energy per unit volume, too. To understand it better, we will look at a number of specific situations that simplify and illustrate its use and meaning.
In that case, we get.This set of 32 problems targets your ability to use equations related to work and power, to calculate the kinetic, potential and total mechanical energy, and to use the work-energy relationship in order to determine the final speed, stopping distance or final height of an object.
The more difficult problems are color-coded as blue problems. Work results when a force acts upon an object to cause a displacement or a motion or, in some instances, to hinder a motion.Bondad definicion wikipedia
Three variables are of importance in this definition - force, displacement, and the extent to which the force causes or hinders the displacement. Each of these three variables find their way into the equation for work.
That equation is:. The most complicated part of the work equation and work calculations is the meaning of the angle theta in the above equation. The angle is not just any stated angle in the problem; it is the angle between the F and the d vectors. In solving work problems, one must always be aware of this definition - theta is the angle between the force and the displacement which it causes.
If the force is in the same direction as the displacement, then the angle is 0 degrees. If the force is in the opposite direction as the displacement, then the angle is degrees. If the force is up and the displacement is to the right, then the angle is 90 degrees. This is summarized in the graphic below. Power is defined as the rate at which work is done upon an object. Like all rate quantities, power is a time-based quantity. Power is related to how fast a job is done.
Two identical jobs or tasks can be done at different rates - one slowly or and one rapidly. The work is the same in each case since they are identical jobs but the power is different.
Work, Energy, and Power
The equation for power shows the importance of time:. Special attention should be taken so as not to confuse the unit Watt, abbreviated W, with the quantity work, also abbreviated by the letter W. Combining the equations for power and work can lead to a second equation for power. If this equation is re-written as.
Thus, the equation can be re-written as. A few of the problems in this set of problems will utilize this derived equation for power. Potential energy is the stored energy of position.Samuel J. Summary 7. The work done by a force, acting over a finite path, is the integral of the infinitesimal increments of work done along the path.
The work done against a force is the negative of the work done by the force.
Definition and Mathematics of Work
The work done by a normal or frictional contact force must be determined in each particular case. The work done by the force of gravity, on an object near the surface of Earth, depends only on the weight of the object and the difference in height through which it moved.
The work done by a spring force, acting from an initial position to a final position, depends only on the spring constant and the squares of those positions. The kinetic energy of a system is the sum of the kinetic energies of all the particles in the system. Kinetic energy is relative to a frame of reference, is always positive, and is sometimes given special names for different types of motion.
This is the work-energy theorem. Alternatively, the work done, during a time interval, is the integral of the power supplied over the time interval.
Contributors and Attributions Samuel J. Work done by a force over an infinitesimal displacement. Work done by a force acting along a path from A to B. Work done by a constant force of kinetic friction. Work done going from A to B by one-dimensional spring force.Suppose that, a force is applied an object and object moves in the direction of applied force then we said work has done.
Let me explain in other words. There must be a force applied to an object and object must move in the direction of the applied force. If the motion is not in the direction of force or force is applied to an object but there is no motion then we cannot talk about work.
Now we formulize what we said above. Since force is a vector quantity both having magnitude and direction work is also a vector quantity and has same direction with applied force. We will symbolize force as F, and distance as d in formulas and exercises. If there is an angle between force and direction of motion, then we state our formula as given below.
In this case force and distance are in the same direction and angle between them is zero. Thus, cos0 is equal to 1. If the force and distance are in opposite directions then angle between them becomes degree and cos is equal to The last case shows the third situation in which force is applied perpendicularly to the distance.
Cos90 degree is zero thus, work has done is also zero. From our formula we found it kg. In other words. Look at the given examples below, we will try to clarify work with examples. Example 25 N force is applied to a box and box moves 10m. Find the work done by the force. Since the box moves in X direction, we should find the X and Y components of the applied force. Y component of the force does not responsible for the work.
Motion of the box is in X direction. So, we use the X component of the applied force. I did not mention it in the solution. If it was a different value than 1 I must write it also.Coq10 and cipro
Example Look at the given picture below. There is an apple having a force applied perpendicularly on it. However, it moves 5m in X direction. Calculate the work done by the force. Example If the box is touching to the wall and a force is applied finds the work done by the force.
Box is touching to the wall and force cannot move it. Because there is no distance we cannot talk about the work. As you can see o ur formula. If one of the variables is zero than work has done becomes zero.
Work Power Energy Exams and Solutions. Work with Examples WORK Suppose that, a force is applied an object and object moves in the direction of applied force then we said work has done. If there is an angle between force and direction of motion, then we state our formula as given below; In this case force and distance are in the same direction and angle between them is zero.
Distance If one of the variables is zero than work has done becomes zero.Power is the rate at which work is done. Mathematically, it is computed using the following equation. The standard metric unit of power is the Watt.
As is implied by the equation for power, a unit of power is equivalent to a unit of work divided by a unit of time. For historical reasons, the horsepower is occasionally used to describe the power delivered by a machine.
One horsepower is equivalent to approximately Watts. Most machines are designed and built to do work on objects. All machines are typically described by a power rating.
The power rating indicates the rate at which that machine can do work upon other objects. A car engine is an example of a machine that is given a power rating. The power rating relates to how rapidly the car can accelerate the car.
If this were the case, then a car with four times the horsepower could do the same amount of work in one-fourth the time.
The point is that for the same amount of work, power and time are inversely proportional. The power equation suggests that a more powerful engine can do the same amount of work in less time.
A person is also a machine that has a power rating. Some people are more power-full than others. That is, some people are capable of doing the same amount of work in less time or more work in the same amount of time. A common physics lab involves quickly climbing a flight of stairs and using mass, height and time information to determine a student's personal power. Despite the diagonal motion along the staircase, it is often assumed that the horizontal motion is constant and all the force from the steps is used to elevate the student upward at a constant speed.
Thus, the weight of the student is equal to the force that does the work on the student and the height of the staircase is the upward displacement. Suppose that Ben Pumpiniron elevates his kg body up the 2. If this were the case, then we could calculate Ben's power rating.
It can be assumed that Ben must apply an Newton downward force upon the stairs to elevate his body. By so doing, the stairs would push upward on Ben's body with just enough force to lift his body up the stairs. It can also be assumed that the angle between the force of the stairs on Ben and Ben's displacement is 0 degrees.
With these two approximations, Ben's power rating could be determined as shown below. This is shown below.Work and Energy : Definition of Work in Physics
This new equation for power reveals that a powerful machine is both strong big force and fast big velocity. A powerful car engine is strong and fast. A powerful piece of farm equipment is strong and fast. A powerful weightlifter is strong and fast. A powerful lineman on a football team is strong and fast.Kolumnentitel und seitenzahl word
A machine that is strong enough to apply a big force to cause a displacement in a small mount of time i. Use your understanding of work and power to answer the following questions. When finished, click the button to view the answers. Two physics students, Will N.
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http://www.indianguitartabs.com/f20/req-chords-natpukkilai-ellai-sridhar-2012-a-55500.html
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math
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Req chords: Natpukkilai ellai - Sridhar (2012)
Can friends, experts please post the chords for this?
Hear the song in E scale.
Natpukkillai Ellai... - YouTube
Natpukku Illai Ellai
Venmegam Pola Vellai
Nee Vandha Pinne Vaan Mazhai
Kaiyodu Kai Serthu
Un Tholil Thalai Saithu
Naan Pogam Idam Sorgame
Wow what a song! Rahmanish kind of intro mixed with R&B flavored beats.. Really got addicted to this tune!
Here we go:
Intro-(C#m)Ho Ho..(G#m)Ho..(C#m)Ho..(F#m)Ho Ho..(B)Ho..(C#m)Ho..(G#m)Ho..(C#m)Ho Ho..(F#m)Ho
..ven (G#m)megam pole vellai..
..nee (C#m)vantha pinbu vaanma(G#m)zhai..(B)
(E)Kaiyodu kai serthu..
..un (G#m)tholil thalai saaithu..
..naan (C#m)pogum idam sorga(B)me..
Ada (C#m)vaanam vegu thoora(G#m)me..
..nam (C#m)natpai patri pesu(F#m)me..(B)eh! eh! eh!
(C#m)Va Va..(G#m)Nam natpum rendam (F#m)thaye..(B)eh..(F#m)eh! eh! eh!
(C#m)natpe natpe..(G#m)Ada theivathukkum (C#m)theva(F#m)yi..(B)ee! ee!..(C#m)
A short melody... Really nice tune... Thanks Jimi..... Sathya
U are welcome Sathya! and thanks to Rajesh for introducing this song to the forum..Appreciated!
Originally Posted by SATHYA167
Thanks a lot, Jimi. You have made my day! Fantastic chords, completely enjoyed playing along.
Sathya, Glad you enjoyed it. I have another short melody, I will post the request.
u r welcome Rajesh!
Originally Posted by rajeshguitar
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http://axion.physics.ubc.ca/341-current/trumpet-record.html
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math
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These are the Spectra of those same three notes from the trumpet, one from the lowest notes on the trumpet(A3#) From both the record of the note itself, and from the spectrum, we A strong series of harmonics. There are wiggles of much shorter than the period of the note.
one from the mid (F5#) Again the plot would suggest a lot of higher harmonics and the spectrum bears this out. Note that the spectrum of peaks extends all the way out to the limits of hearing.
and one from the highest(A5#). This we would expect to be much "purer"-- ie fewer higher harmonics ( althought he second harmonic should be strong)
The trumpet is an instrument with very rch harmonics. This is both because of the flare of the trumpet which reduces the knee frequency for the higher harmonics, but also because of the very righ harmonic structure of the lips opening and closing, letting in the air in bursts. Again, the junk below about 100 Hz is all noise-- either from the room, or from the electronics in the computer or noise from the computer itself.
The absolute amplitude in dB is simply how far below the loudest sound that the microphone could record without clipping. It is the relative amplitudes of the notes that are important.
As explained in the notes on the Fourier transform in the course notes, the widths of the peaks arise out of the finite time that the note is recorded and the fact that the recording time is not an exact multiple of the period of the note. Both of these effect broaden out the peak.
The number of samples 16384. Since there are 44100 samples per second, this is about .4 seconds. The resolution -- the minimum frequency that can be sample is one whose period is this sample time, so the minimum frequency is about 2.5Hz. The smooth curve at low frequencies is an interpolation by the program and cannot be trusted (eg below about 50Hz)
This page and all links therefrom are copyright W. G. Unruh. They may be reproduced for non-commercial purposes but this notice must be retained in any copy. Any changes must be clearly indicated as such.
The image files were created under Mandriva Linux using audacity and xv.
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https://en.wikipedia.org/wiki/Sridhara
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math
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|This article does not cite any sources. (August 2015) (Learn how and when to remove this template message)|
Sridhar Acharya (Bengali: শ্রীধর আচার্য; c. 750 CE, India – c. ? India) was an Indian mathematician, Sanskrit pandit and philosopher. He was born in Bhurishresti (Bhurisristi or Bhurshut) village in South Radha (at present day Hughli) in the 8th Century AD. His father's name was Baladev Acharya and his mother's name was Acchoka bai. His father was a Sanskrit pandit.
He was known for two treatises: Trisatika (nitsometimes called the Patiganitasara) and the Patiganita. His major work Patiganitasara was named Trisatika because it was written in three hundred slokas. The book discusses counting of numbers, measures, natural number, multiplication, division, zero, squares, cubes, fraction, rule of three, interest-calculation, joint business or partnership and mensuration.
- He gave an exposition on zero. He wrote, "If zero is added to any number, the sum is the same number; if zero is subtracted from any number, the number remains unchanged; if zero is multiplied by any number, the product is zero".
- In the case of dividing a fraction he has found out the method of multiplying the fraction by the reciprocal of the divisor.
- He wrote on practical applications of algebra
- He separated algebra from arithmetic
- He was one of the first to give a formula for solving quadratic equations.
- Multiply both sides by 4a,
- Subtract 4ac from both sides,
- Add to both sides,
- Complete the square on the left side,
- Take square roots,
- and, divide by 2a,
Sridhara is now believed to have lived in the ninth and tenth centuries. However, there has been much dispute over his date and in different works the dates of the life of Sridhara have been placed from the seventh century to the eleventh century. The best present estimate is that he wrote around 900 AD, a date which is deduced from seeing which other pieces of mathematics he was familiar with and also seeing which later mathematicians were familiar with his work. Some historians give Bengal as the place of his birth while other historians believe that Sridhara was born in southern India.
Sridhara is known as the author of two mathematical treatises, namely the Trisatika (sometimes called the Patiganitasara ) and the Patiganita. However at least three other works have been attributed to him, namely the Bijaganita, Navasati, and Brhatpati. Information about these books was given the works of Bhaskara II (writing around 1150), Makkibhatta (writing in 1377), and Raghavabhatta (writing in 1493).
K.S. Shukla examined Sridhara's method for finding rational solutions of , , , which Sridhara gives in the Patiganita. Shukla states that the rules given there are different from those given by other Hindu mathematicians.
Sridhara was one of the first mathematicians to give a rule to solve a quadratic equation. Unfortunately, as indicated above, the original is lost and we have to rely on a quotation of Sridhara's rule from Bhaskara II:-
Multiply both sides of the equation by a known quantity equal to four times the coefficient of the square of the unknown; add to both sides a known quantity equal to the square of the coefficient of the unknown; then take the square root.
Proof of the Sridhar Acharya Formula,
Before this we have to know his famous eqn of Discriminent(D) (difference between two roots)=b^2-4ac let us consider,
- Multipling both sides by 4a,
- Subtracting from both sides,
- Then adding to both sides,
- We know that,
- Using it in the equation,
- Taking square roots,
- Hence, dividing by get
In this way, he found the proof of two roots.
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https://www.teacherspayteachers.com/Product/Composite-Functions-1582438
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math
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Composite Functions: This pack now contains three activities to help students understand the idea of composite functions.
1. Composite Functions: Putting Functions into Functions
Students start by substituting numbers into different functions and then gradually build up to inputting simple expressions, such as (x + 3) into functions. They can match their answers to the ones jumbled up at the bottom of the page.
2. Composite Functions 1
Students substitute numbers into a mixture of different composite functions.
3. Composite Functions 2
Students again are required to create composite functions, but this time they are required to multiply out brackets and work with radicals.
The answers are at the bottom of the page for each worksheet jumbled up, so that students can check their answers as they work
Core Standards: HSF.BF.A.1.C
Compose functions. For example, if T(y) is the temperature in the atmosphere as a function of height, and h(t) is the height of a weather balloon as a function of time, then T(h(t)) is the temperature at the location of the weather balloon as a function of time.
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https://en.m.wikisource.org/wiki/Posterior_Analytics_(Bouchier)/Book_II/Chapter_XV
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math
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Posterior Analytics (Bouchier)/Book II/Chapter XV
Chapter XV: How far the same Middle Term is employed for demonstrating different QuestionsEdit
- Questions for demonstration are the same when they use the same middle term. Questions may be generically the same and specifically different.
Questions for solution are the same, first from having the same middle term (as for instance all questions which can be solved by the common middle term ‘reactionary influence,’) and of these some are generically identical while possessing certain specific differences, whether of object or only of method. Take the three questions ‘What produces an Echo?’ ‘Why are objects reflected?’ ‘What causes a Rainbow?’ All these are generically one, for all involve refraction, but they differ specifically. In other cases questions differ in that the middle term of the one is subordinate to that of the other. Thus, ‘Why is the current of the Nile stronger at the end of the month?’
‘Because the end of the month is more rainy.’
‘Why then is the end of the month more rainy?’
‘Because the moon is waning.’
These two questions stand to one another in the second of the above relations.
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https://www.coursehero.com/file/5228315/Chem14A-Outline1-Review20of20Chemical2020Physical20Principles/
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math
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Unformatted text preview: • Write symbols for the elements, given their names, and vice versa. • Define a mole. • Convert between mass and moles. • Calculate the empirical formula of a compound from its mass percentage composition. • Determine the molecular formula of a compound from its empirical formula and its molar mass. • Calculate the molarity of a solute in solution, volume of solution, and mass of solute, given the other two quantities. • Balance chemical equations. • Understand and apply the concept of limiting reactant....
View Full Document
- Spring '09
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https://www.physicsforums.com/threads/direction-of-magnetic-force.714073/
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math
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1. The problem statement, all variables and given/known data Assuming the following directions of the charged particles velocity and magnetic field indicate the direction of the magnetic force exerted on the particle? 2. Relevant equations 3. The attempt at a solution The force vector is perpendicular to the B field and the right hand rule states the charge is out of the page so would the vector be down?
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| 407 | 1 |
https://analyticsindiamag.com/10-real-life-applications-of-genetic-optimization/?utm_source=rss&utm_medium=rss&utm_campaign=10-real-life-applications-of-genetic-optimization
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math
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Genetic algorithms have a variety of applications, and one of the basic applications of genetic algorithms can be the optimization of problems and solutions. We use optimization for finding the best solution to any problem. Optimization using genetic algorithms can be considered genetic optimization, and there are several benefits of performing optimization using genetic algorithms. In this article, we are going to list down 10 real-life applications of genetic optimization.
Let’s start with these interesting applications one-by-one.
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1. Traveling salesman problem (TSP)
This is one of the most common combinatorial optimization problems in real life that can be solved using genetic optimization. The main motive of this problem is to find an optimal way to be covered by the salesman, in a given map with the routes and distance between two points. If genetic algorithms are used in finding the best route structure, we don’t get the solution only once. After each iteration, we can generate offspring solutions that can inherit the qualities of parent solutions. TSP has a variety of applications like planning, logistics, and manufacturing.
2. Vehicle routing problem (VRP)
The basic vehicle routing problem (VRP) can be considered as a generalization of the TSP problem which is also a combinatorial optimization problem. In this problem we find an optimal weight of goods to be delivered or find an optimal set of delivery routes when other things like distance, weights, depot points are constrained or have any kind of restrictions. Genetic approaches are competitive with tabu search and simulated annealing algorithms in terms of solution time and quality.
3. Financial markets
In the financial market, using genetic optimization, we can solve a variety of issues because genetic optimization helps in finding an optimal set or combination of parameters that can affect the market rules and trades. For example, in the stock market, any rule is a popular tool for analysis, research, and deciding to buy or sell shares. In this example, the success of trading depends on the selection of optimal values for all parameters and combinations of parameters. Genetic algorithms can help in finding the optimal and sub-optimal combinations of parameters. Also by genetic optimization, we can find out the near-optimal value from the set of combinations.
4. Manufacturing system
One of the major applications of genetic optimization is to minimize a cost function using the optimized set of parameters. In manufacturing we can see various examples of cost function and finding an optimal set of parameters for this function can be performed by following the genetic optimization. In many cases, we can find the application of genetic optimization in product manufacturing (variation of production parameters or comparison of equipment layout). The main motive behind applying genetic optimization is to achieve an optimum production plan by taking into consideration dynamic conditions like inventories, capacity, or material quality.
5. Mechanical engineering design
In many designing procedures of mechanical components, we can also find the application of genetic optimization. We can take aircraft wing design as an example where we are required to improve the ratio of lift to drag for a complex wing. This kind of designing problem can be considered as a multidisciplinary problem, the fitness function in genetic optimization can be altered by considering some specific requirement of the design.
6. Data clustering and mining
Data clustering can be considered an unsupervised learning process where we try to segment data based on the characteristic of data points. One of the major parts of the procedure is to find out the centre point of the clusters and we know that genetic algorithms have great capability of searching for an optimal value. In data clustering and mining we can use genetic algorithms to find a data centre with an optimal error rate.
7. Image processing
There are various works and researches which show the use cases of genetic optimization in various image processing tasks. One of the major tasks related to genetic approach in image processing is image segmentation. Although these genetic optimizations can be utilized in various areas of image analysis to solve complex optimization problems. Using genetic optimization in an integrated manner with image segmentation techniques can make the whole procedure an optimization problem.
8. Neural networks
Neural networks in machine learning are one of the biggest areas where genetic algorithms have been used for optimization. One of the simplest examples of use cases of genetic optimization in neural networks is finding the best fit set of parameters for a neural network. Instead of these, we can find the use of genetic algorithms in neural network pipeline optimization, inheriting qualities of neurons, etc.
9. Wireless sensor networks
The wireless sensor network is a network that includes spatially dispersed and dedicated centres to maintain the records about the physical conditions of the environment and pass the record to a central storage system. Some notable parameters are the lifetime of the network and energy consumption for routing which plays key roles in every application. Using the genetic algorithms in WSN we can simulate the sensors and also a fitness function from GA can be used to optimize, and customize all the operational stages of WSNs.
10. Medical science
In medical science, we can find many examples of use cases of genetic optimization. The generation of a drug to diagnose any disease in the body can have the application of genetic algorithms. In various examples, we find the use of genetic optimization in predictive analysis like RNA structure prediction, operon prediction, and protein prediction, etc. also there are some use cases of genetic optimization in process alignment such as Bioinformatics Multiple Sequence Alignment, Gene expression profiling analysis, Protein folding, etc.
So these are the 10 real-life interesting applications where genetic optimization is used widely. These algorithms are part of the evolutionary algorithm family that is based on the principles of natural evaluation explained in Charles Darwin’s theory of evolution.
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http://www.rairo-ita.org/articles/ita/pdf/first/ita07037.pdf
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math
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On the power of randomization for job shop scheduling with k-units length tasks
Department of Informatics,
ETH Zurich, ETH Zentrum, 8092 Zürich, Switzerland; [email protected]
Accepted: 24 April 2008
In the job shop scheduling problem k-units-Jm, there are m machines and each machine has an integer processing time of at most k time units. Each job consists of a permutation of m tasks corresponding to all machines and thus all jobs have an identical dilation D. The contribution of this paper are the following results; (i) for jobs and every fixed k, the makespan of an optimal schedule is at most D+ o(D), which extends the result of for k=1; (ii) a randomized on-line approximation algorithm for k-units-Jm is presented. This is the on-line algorithm with the best known competitive ratio against an oblivious adversary for and k > 1; (iii) different processing times yield harder instances than identical processing times. There is no 5/3 competitive deterministic on-line algorithm for k-units-Jm, whereas the competitive ratio of the randomized on-line algorithm of (ii) still tends to 1 for .
Mathematics Subject Classification: 68W20 / 68W25
Key words: On-line algorithms / randomization / competitive ratio / scheduling
© EDP Sciences, 2008
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https://hireforstatisticsexam.com/how-to-solve-problems-involving-simpsons-paradox-in-a-statistics-exam
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math
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How to solve problems involving Simpson’s Paradox in a statistics exam? Hi Jon. I’m just starting to get over my feelings on this question so maybe you could help me out. Here’s the problem. I graduated on a position of “Comfort Management Coaching” where I taught the staff of the Com. I was supposed to train everybody for a job that evening, but for some reason I have been asked to deliver that job, so I went. As I usually do, the company’s com. was very slow but I did manage to deliver the person who could come. I honestly could have taken one job another that wasn’t too late. However, I have kept running into problems. The Com team with their own things makes trying to build teamwork, and, as such, maybe I need more testing for this question again, should I have the next one. This is where I get stuck. This is when I’ve been telling myself, that this is something I want to “fix” through testing, something that I’ve learned. This needs to be tested through testing. In my experience, when I have big problems, even big tests, as yet I should be testing a single technique, or even just a few others. I always tried to keep both methods. I really should have more testing and hopefully all the things I do that are tested are as good as they can be. Thanks for reading about this so be sure to share that comment with everyone. Just a quick question: I have 2 types of problems: I am less than ideal for training people for a job, but I am too good for this job, and also am starting to learn about what you can teach there. I would be much happier if people would know what testing is and how to get better at it. I understand testing is a lot of technology, new and fast in real life and I will be testing as long as I know it.
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I think your second description is flawed.How to solve problems involving Simpson’s Paradox in a statistics exam? In a test exam, you must fill out a PDF, a text document in a paper document of your choice, then upload this PDF to a computer in a second data-paper. On this page, you can access hundreds, thousands, hundreds of PDF documents using many different software to find all the interesting things you need. Is it normal, though, to check that you “test” your PDF with these new software software changes? Because we feel like it’s perfectly reasonable that such changes do constitute a major part of the mathematics governing the exam. But how can you test your PDF now, or be sure that it’s truly valid, when you’ve given yourself an important test paper? There are many documents that really need to be checked. That’s why your questions can’t be for those who haven’t already completed PDFs, to further educate them. What You Need: Create a PDF in Excel. Record All Dataset’s Content Check Content, Copies, Acess, Passwords, etc. Note Your Use is Just Like Making a New Paper, Written for Your Goal. Once You’ve Done Some Work You’ll Save All the PDF Documents Have a View On What You Make The PDF View page If you’d like to create the relevant paper or database in a database, but need your first section, rather than the rest of it, edit this section. If your PDF View page isn’t part of your web server — which is unusual for web users — you can look at which side of the page you’d like to open for reading. But once you’ve done that, you have to run your PDF View page into a rediculously useful terminal emulator so that the PDF link in the rediculously helpful terminal can be retrieved. If your mind isn’t thinking hard enough, there’sHow to solve problems involving Simpson’s Paradox in a statistics exam? Simple, but a painful addition! To find some solutions to such puzzles it is important to obtain a clear and fast solution to their problem. The problem is not clear or simple and the solution seems easy or the solution seems simple and useful to accomplish, or even just getting a hard problem solved with a straight answer. In this blog we are going to explain the most common problems involving Simpson’s Paradox in different ways. In this manner it’s tough to follow the logical steps of this tutorial so this is supposed to help you by the most easily prepared solutions. It would be so hard to express the key messages of the solution discussed well in our previous posts concerning the mathematical solution. The most useful solution with the most easy way of solving puzzles involves taking numbers and solving them because it is easy. The problem is the same as that of the problem YOURURL.com finding a number for solving by taking the logarithm of it. However, in our previous tutorial we pointed out this problem is very difficult because you are asking for $x-y=0$ instead of $x+y=0$.
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It could be said that we are solving the same problem wtih the same number $x$. We would have to find an alternative solution for solving the alternative problem without knowing the number for which we are going to accept the answer. This is technically a difficult problem because the solution for the alternative choice is not knowing the number because you take the logarithm and multiply it by $x$. And when you find solving the problem by taking logarithms and using a nice combination of numbers, it’s possible to follow carefully the second simple solution discussed by C. E. Jackson in his famous book “The Mathematical Theory of Simple Arguments”. Problem 1 What is the solution of this example problem 1? The simple number 4 is solved by showing, that $x=0$ and $y=
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http://cacasa2.info/advanced-econometrics-takeshi-amemiya/2015/07/monte-carlo-and-applications
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math
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Monte Carlo and Applications
Thisted (1976) compared ridge 2, modified ridge 2, ridge 3, and generalized ridge 1 by the Monte Carlo method and found the somewhat paradoxical result that ridge 2, which is minimax for the smallest subset of Л, performs best in general.
Gunst and Mason (1977) compared by the Monte Carlo method the estimators (1) least squares, (2) principal components, (3) Stein’s, and (4) ridge 2. Their conclusion was that although (3) and (4) are frequently better than (1) and (2), the improvement is not large enough to offset the advantages of (1) and (2), namely, the known distribution and the ability to select regressors.
Dempster, SchatzofF, and Wermuth (1977) compared 57 estimators, belonging to groups such as selection of regressors, principal components, Stein’s and ridge, in 160 normal linear models with factorial designs using both E(j} — — P) and E(j} — Д)’Х’Х(Д — p) as the risk function. The
winner was their version of ridge based on the empirical Bayes estimation of у defined by
The fact that their ridge beat Stein’s estimator even with respect to the risk function E(fi — — p) casts some doubt on their design of Monte
Carlo experiments, as pointed out by Efron, Morris, and Thisted in the discussion following the article.
For an application of Stein’s estimator (pulling toward the overall mean), see Efron and Morris (1975), who considered two problems, one of which is the prediction of the end-of-season batting average from the averages of the first forty at-bats. For applications of ridge estimators to the estimation of production functions, see Brown and Beattie (1975) and Vinod (1976). These authors determined у by modifications of the Hoerl and Kennard ridge trace analysis. A ridge estimator with a constant у is used by Brown and Payne (1975) in the study of election night forecasts. Aigner and Judge (1977) used generalized ridge estimators on economic data (see Section 2.2.8).
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https://homecapital.in/blog/difference-between-flat-and-reducing-interest-rate
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math
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Planning to get a loan to buy a property? You will get your loan from either a bank or a non-banking financial company. No matter from which agency or institution you get your loan sanctioned, you will be paying interest. The loan interest is calculated in two ways: flat interest rate and reducing interest rate.
How much total money you pay back (principal plus interest) depends on whether you opt for a flat interest rate or a reduced interest rate. What’s the difference? Let’s find out.
Flat interest is calculated on the full loan amount and distributed across the entire tenure. This way, you pay a fixed amount to the financial institution from which you have taken the loan. Your EMI is not subject to fluctuations in interest rates.
With a flat interest rate, you know how much you will be paying, for example, for the next 60 months, and consequently, you can plan your finances accordingly.
Contrary to a flat interest rate, in the case of reducing interest rate, your EMI is not calculated on the entire loan amount for the entire tenure, but on the reducing principal amount. A reducing interest rate is calculated on the outstanding balance or the remaining loan.
As you know that the EMI that you pay consists of interest and principal amount. It means, whenever you pay an EMI, you also pay back a part of the principal amount. So the money that you owe to your bank is constantly reducing. In reducing interest rate, your monthly payment is calculated on this reduced amount rather than the entire principal amount.
How much you pay varies every month. Your interest rate is subject to the ongoing interest rate rather than the interest rate that was in the beginning of your loan tenure.
The flat interest rate is calculated on the total principal amount and then distributed across the tenure evenly. The EMI does not take into consideration the repayment of the principal amount as the tenure progresses. The interest rate and the payable amount remain the same every month.
The flat interest rate is calculated using the following formula:
Interest payable per instalment = (original loan amount x interest rate per annum x number of years) ÷ number of instalments
That is, the original loan amount multiplied by the interest rate per annum multiplied by the number of years for which you have taken the loan, and then the entire calculation divided by the number of instalments you have agreed to pay to pay back the entire loan.
Suppose you have taken Rs. 100,000 as a loan with 5% interest that you intend to pay back in 10 years.
How many EMIs are you going to pay in 10 years? 12 months x 10 = 120 months = 120 EMIs.
This is how the flat interest rate for every month will be calculated:
(100,000 x 5% x 10) / 120 = Rs. 417 (approximately).
This is not the EMI. To calculate the EMI, to this flat interest you also need to add the principal amount.
The principal amount needs to be distributed over 120 months.
100,000 / 120 = Rs. 834 (approximately).
So, if you take a loan of Rs. 100,000 with a flat interest rate of 5% and with a tenure of 10 years, your EMI is going to be
417 + 834 = Rs. 1,251 (approximately).
As previously explained, reducing the interest rate means that the interest is calculated on the outstanding principal amount and not the whole amount. Every proceeding EMI is calculated anew based on the loan amount that is pending to be repaid. This type of interest rate is also called “diminishing rate of interest”.
Here is the formula for calculating reducing interest rate:
Interest payable per instalment = interest rate per instalment * remaining loan amount
Suppose you take a loan of Rs. 100,000 with a reducing interest rate of 5% and for a tenure of 10 years.
If you distribute 100,000 over 120 months, the principal amount for every month comes out to be 100,000 / 120 = Rs. 833 (approximately).
The interest rate for the first month will be 100,000 x 5% = Rs. 5,000.
Therefore, the EMI for the first month will be 833 + 5000 = Rs. 5,833.
For the second month, 833 is deducted from your outstanding principal amount, which will now remain 100,000 – 833 = Rs. 99,167.
The interest for the second month will be 99,167 x 5% = Rs. 4,958.35.
Hence, the EMI for the second month will be 4,958.35 + 833 = Rs. 5,791.35.
Every proceeding month, the principal amount that has already been paid is deducted before calculating interest on the amount.
Over a period of time, as the principal amount reduces, the EMI you pay is lower than the previous months. This is because your EMI is calculated not on the complete principal amount that you originally took, but the remaining amount after the number of EMIs you have already paid. In many instances, you can even reduce your tenure by adjusting your EMI (paying a little extra).
Simplicity. You always know how much you’re going to pay every month. The entire amount is equally distributed, so there are no surprises. Even someone with no financial knowledge can calculate the EMI without worrying about how much amount they’re going to have to pay next month. Also, in case the RBI increases interest rates on loans, you won’t be affected.
Every bank and non-banking financial company these days has an online calculator for calculating EMIs for flat and reducing interest rates. The calculator can give you an idea of how much EMI you will be paying for both options.
Both flat and reducing interest rates have their pros and cons. There is a reason why both options exist. Make a choice that best suits your financial plans.
Take the first step to home ownership with HomeCapital, get eligibility and in-principal sanction letter in one minute.
Click to get started.
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s3://commoncrawl/crawl-data/CC-MAIN-2024-18/segments/1712296816024.45/warc/CC-MAIN-20240412132154-20240412162154-00562.warc.gz
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CC-MAIN-2024-18
| 5,684 | 37 |
https://www.thephotoforum.com/threads/what-to-look-for-when-buying-a-used-camera.199982/
|
math
|
So I'm pretty sure my first foray into buying a DSLR will be used. Which leads me to wondering what are some signs I should look for when buying used, etc. I see a common thing to list on ebay or craigslist are how many times the shutters been clicked. I take it this is similar in regards to buying a used car? They let you know how many miles are on it. How much of a price difference should there be between a new camera and a used one?
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s3://commoncrawl/crawl-data/CC-MAIN-2018-17/segments/1524125945037.60/warc/CC-MAIN-20180421051736-20180421071736-00106.warc.gz
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CC-MAIN-2018-17
| 439 | 1 |
http://www.modelmayhem.com/mallorienoelle
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math
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Mallorie DunnModel Female Astoria, New York, US
My MM URL: http://www.modelmayhem.com/mallorienoelle
Mayhem # 385837
"Don't compromise yourself. You are all you've got." - Janis Joplin
I am an agency represented model, signed with MSA for fit modeling, and freelancing in print, showroom, and runway.
My hair is currently long bob. I have a side bang as well. It is very thick and naturally dark brown - it is not color treated in any way.
Photo Credit: Josh Sailor MM#468718
I am always interested in working with creative and talented photographers who are professional in their actions.
I am looking for paid jobs and some very limited testing. Please inquire about my rates. They are reasonable and negotiable according to the project.
Photo Credit: John Artuso MM#1943775
You may contact me through email, [email protected] and of course, here through MM.
I have four small tattoos (on my forearms) and two medium line drawing tattoos (on my back/shoulders). They can be all be covered with makeup easily, or edited out with Photoshop easily.
I am currently a freelance seamstress and tailor and also run my own clothing line, SmartGlamour (www.smartglamour.com). I graduated from Pratt Institute in 2010 receiving my Bachelors in Art and Design education. In May 2007, I received my Associates degree from FIT for fashion design.
I like to think that if you combine Audrey Hepburn, Brody Dalle, Betsey Johnson, and Gloria Steinem - you come up with someone a lot like me.
Photo Credit: Eric Walton MM#1639775
I am always willing and excited to test with my best friend (MM#635540) if you are so inclined! She is a beautiful person, inside and out.
"This above all: to thine own self be true"
Have you worked with Mallorie Dunn? Add Credits for Mallorie Dunn! >>
Photo Credit: Jito Lee MM#1165645
The Art of Fashion - House of Correia - Spring 2012 - Oct 26th, 2012.
The Art of Fashion - Keriann Correia Fall Preview - May 25th, 2011.
Fashion Keeps Fantasy Alive - Haiti Relief Charity show with Bien Abye - August 10th, 2010.
Lucullen Fashion Show - February 20th, 2010.
Viktor Luna for NY Fashion Week - February 15th, 2010.
Bombshell Clothing at Webster Hall for NY Fashion Week - February 12th, 2010.
The Big Tease 3 - Jan 23rd, 2010.
Bombshell Clothing at Webster Hall - October 16th, 2009.
Maira Houck - Brooklyn Fashion Week Show - Sept 19th, 2009.
Hillary Flowers - New York Fashion Week Show - Sept 16th, 2009.
Hillary Flowers - New York Fashion Week Show - Sept 10th, 2009.
Enigma Fashion Show - Diana Susanto, June 26th, 2009.
Enigma Fashion Show - Diana Susanto, April 11th, 2009.
Fusion Fashion Show, March 7th and 8th, 2009.
Fashion show for Sachika Shop, January 28th, 2009.
The Big Tease II, Fashion Show, January 24th, 2009.
NYU Fall FBA Fashion Show, Dec. 9th, 2008.
Confident Couture fashion show, Oct. 10th, 2008.
Ashton Warren's wedding dress collection, Fashion show June 2008.
Pratt Fashion Society Fashion show April 18th 2008.
NYU FBA Fashion show April 2008.
Frenchme Fashion show Feb 2008.
NYU FBA Fashion show fall 2007.
VerveNYC Fashion show summer 2007.
Fusion Fashion show spring 2006, 2007, and 2008.
Modeling workshop with Ford model Laura Interval summer 2004.
Newburgh Mall fashion shows from 1999-2003.
Modeling for Deb, New York and Company, New York Leather, and Sears.
http://s177.photobucket.com/albums/w240 … 010035.flv
Photo Credit: Tito Trelles MM#2480
Cosmopolitan Magazine I-Pad Application Video with DJ Pauly D - released March 25, 2011 - http://www.youtube.com/watch?v=NSl-O5yuLfE
Carlos Luna MM#1767634
Josh Sailor MM#468718
Eric Walton MM#1639775 ***
John Artuso MM#1943775
Haus of Hadz
DaVinci Pro Photography - The Event Network of NYC - MM#597774
worm carnevale MM#513555
mikey poz MM#1154329
Jito Lee MM#1165645 *****
Peter Jacobsen MM#433293
Tito Trelles MM#2480 ***
DWFK - article for Columbia MM#844953
Gary Flom - modeling for Salon Rocks MM#1030768
Jon Apostol MM##776933
Jason Groupp MM#324383
Jens Look Photography MM#451092 (not endorsing work with this photographer - PM me for details)
Fran Roberts MM#821455
Daniel Norton MM#61999 **
Hakimata Photography MM#727620
Barbarella Photography MM#1056057
C Nguyen MM#552804
Dennis Lee MM#841633
Joseph Marconi MM#903118
Ivy Photography MM#734662
Danger Photos MM#675544
Paul Tirado MM#3829
2 The Max Photography MM#446907
Christopher Shibiya MM##36267
Mike Hulsey MM#19659
fLOVE photography MM#285310
tenrocK photo MM#555381
Einat Salmon MM#671715 ****
Bulletpoint Photography MM#1087822 ***
Jason Guffy MM#80388 **
Kristian RA MM#588282 **
Pinup Photography Club run by Brandon MM#20441 *****
Bombshell Clothing - Keriann Correia **
Nick Parisse MM#249204
(* indicates number of shoots together)
Eyelashes by MUA Jessica Jade Jacob - MM#1158394
Web Tear Sheets
RealBeauty Blog - http://www.realbeauty.com/products/hair … ica-lashes
TeenVogue - http://www.teenvogue.com/style/blogs/fa … shion.html
http://confidentcouture.com/ - CC Winter 2007 - M dress and CC Fall 2009 dress #10237684
http://www.myspace.com/bombshellclothingline - Albums: Fresh Out the Wash, The Goodz, FBA Spring 08, NYC Show, and My Photos
http://2birdsonestone.com/Dresses.aspx - Custom Dresses
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s3://commoncrawl/crawl-data/CC-MAIN-2017-09/segments/1487501174154.34/warc/CC-MAIN-20170219104614-00064-ip-10-171-10-108.ec2.internal.warc.gz
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CC-MAIN-2017-09
| 5,187 | 96 |
http://www.sju.edu/int/academics/catalogs/mat-102-mathematical-explorations-ii-3-credits.htm
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math
|
MAT 102 Mathematical Explorations II (3 credits)
This is a second course for humanities majors. The course covers elementary probability, including independent and dependent events, conditional probability, binomial probability, and certain applications in a wide variety of situations. MAT 1015 is not required for MAT 102. Other topics may be covered at
Satisfies GER/GEP requirement for CPLS students only. Open to CPLS students only.
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s3://commoncrawl/crawl-data/CC-MAIN-2014-49/segments/1416931009179.34/warc/CC-MAIN-20141125155649-00134-ip-10-235-23-156.ec2.internal.warc.gz
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CC-MAIN-2014-49
| 437 | 3 |
https://depot.sorbonne.ae/entities/publication/5f8a7919-82e3-42bb-be45-7e582d518eb5
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math
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Properties of Unique Degree Sequences of 3-Uniform Hypergraphs
Lecture Notes in Computer Science
Discrete Geometry and Mathematical Morphology
In 2018 Deza et al. proved the NP-completeness of deciding wether there exists a 3-uniform hypergraph compatible with a given degree sequence. A well known result of Erdös and Gallai (1960) shows that the same problem related to graphs can be solved in polynomial time. So, it becomes relevant to detect classes of uniform hypergraphs that are reconstructible in polynomial time. In particular, our study concerns 3-uniform hypergraphs that are defined in the NP-completeness proof of Deza et al. Those hypergraphs are constructed starting from a non-increasing sequence s of integers and have very interesting properties. In particular, they are unique, i.e., there do not exist two non isomorphic 3-uniform hypergraphs having the same degree sequence ds . This property makes us conjecture that the reconstruction of these hypergraphs from their degree sequences can be done in polynomial time. So, we first generalize the computation of the ds degree sequences by Deza et al., and we show their uniqueness. We proceed by defining the equivalence classes of the integer sequences determining the same ds and we define a (minimal) representative. Then, we find the asymptotic growth rate of the maximal element of the representatives in terms of the length of the sequence, with the aim of generating and then reconstructing them. Finally, we show an example of a unique 3-uniform hypergraph similar to those defined by Deza et al. that does not admit a generating integer sequence s. The existence of this hypergraph makes us conjecture an extended generating algorithm for the sequences of Deza et al. to include a much wider class of unique 3-uniform hypergraphs. Further studies could also include strategies for the identification and reconstruction of those new sequences and hypergraphs.
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s3://commoncrawl/crawl-data/CC-MAIN-2024-10/segments/1707947474697.2/warc/CC-MAIN-20240228044414-20240228074414-00161.warc.gz
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CC-MAIN-2024-10
| 1,939 | 4 |
https://forums.wolfram.com/mathgroup/archive/1994/Jun/msg00029.html
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math
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Can you abort from the "Evaluate Notebook" process ?
- To: mathgroup at yoda.physics.unc.edu
- Subject: Can you abort from the "Evaluate Notebook" process ?
- From: Simon Chandler <simonc at hpcpbla.bri.hp.com>
- Date: Tue, 28 Jun 1994 13:41:47 +0100
28/6/94 Dear MathGroupers, Are there any commands that will stop further evaluation of a Notebook, where the evaluation was kicked off by the "Evaluate Notebook" item under the "Action" menu heading ? We have a particular Notebook that is used by several Mathematica illiterate users. All they do is enter a single variable name, the name of the file they want processed, and then click on the "Evaluate Notebook" command. At the end of a (very long) calculation they are presented with a result. At several points in the calculation the file's contents may be such that further calculation is silly (or impossible) so after performing the test I would like to be able to abort the calculation and issue an appropriate warning. Now, if I were writing this Notebook from scratch I would correctly 'packagize' it, and write it in a way (probably using Throw and Catch) that would allow me to easily exit from routines. However, this particular Notebook was written by another author and is rather 'linear' in style; if everything goes well the Notebook commands are executed sequentially from beginning to end, et voila - the required result. My problem arises because I have been asked by the Notebook's current users to provide the abort mechanism I discuss above - so that time is not wasted continuing with a nonsensical calculation - but without significantly changing the Notebook's contents. In particular, they don't want me to change the Cell structure, so I can't just put any required tests and subsequent conditionally dependent code into a single If command (or can I, I may be wrong. Please tell me if it _is_ possible to put an If command across different cells). Please tell me if you know how I can easily abort (non-manually) from the "Evaluate Notebook" process without significantly re-writing the Notebook. I only wish I could ! Thanks, Dr Simon Chandler Hewlett-Packard Ltd (CPB) Filton Road Stoke Gifford Bristol BS12 6QZ Tel: 0272 228109 Fax: 0272 236091 email: simonc at bri.hp.com
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s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656103669266.42/warc/CC-MAIN-20220630062154-20220630092154-00204.warc.gz
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CC-MAIN-2022-27
| 2,255 | 6 |
http://ultimateshowdowns.wikia.com/wiki/Category:Ultimate_Showdown
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math
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“ Lawrence "Larry" Whistler (born December 5, 1953) is a professional wrestler, better known...Larry Zbyszko
“The Harris Brothers were signed by the World Wrestling Federation in 1995, where they were...Blu Brothers
“ Marge Simpson is the mother of Bart and Lisa Simpson, and is the spouse of Homer Simpson...Marge Simpson
This category has the following 20 subcategories, out of 20 total.
Pages in category "Ultimate Showdown"
The following 200 pages are in this category, out of 220 total.
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s3://commoncrawl/crawl-data/CC-MAIN-2017-34/segments/1502886103910.54/warc/CC-MAIN-20170817185948-20170817205948-00473.warc.gz
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CC-MAIN-2017-34
| 501 | 6 |
https://cosmonomy.eu/eng/intro-en/deductions-12-en.htm
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math
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The revelation about
the role of matter for the complex things and the instructive
observations for the research begins in the moment, where we will
think about a Universe complete within interval of a Maximum
Period. With theoretical observations and with a sequence of reasoning, we can
think more details and observe:
1) The finite space corresponds in the
complete universe which is not present (in relative to the material world).
2) The free space is with a limit of
longest distance (of removal), same for the all things (= a finite non-Euclidean geometry space).
3) The limit in the increase
length and in the longest distance (of removal) signals a limit of a minimal curvature. Respectively, the limit of a minimal length signals a limit
for the increased curvature.
4) The finite space offers
simultaneously two limits: in removal and approach.
5) The distance in the free
space is also distance in time (because inevitably, the distance takes a time to travel).
6) The limit in the increase
of length, given that a superior speed exists in motion, is also a limit in the maximum time of an interaction. The limit of minimal length also
signals a limit for the minimal time of interaction.
7) The (empty) free space
dynamical is connected also with structural elements (with mathematical relations), because the free space identified with a constant quantity of energy, while
the structural elements with fluctuations of this constant energy.
8) The relation of finite space with the
structure of matter is direct, simultaneous for the total of matter and its behavior is wave. A permanent relation exists between visible mass with invisible wave changes, and at a part these wave phenomena include the known electromagnetic phenomena.
9) The isotropic relation of free space
with the material things, after does not exist an absolute beginning and termination in its finite length.
10) The energy of the free space is found in
balanced situation and when is disturbed, then this balance can comes back in the minimal time. (After of first calculations: minimal inertia Mmin =
0,737248 ×10-50 kg·s).
11) A fluctuation of energy is caused
when the energy is transmitted (wave and oscillation).
12) The tendency of space to transmit
energy in order to it compensates (counterbalance) the points where the energy is decreased.
13) The presence of matter and
particles are considered as moments in the change of energy of the free space, and is owed in conditions where the substitution (compensation) in
energy is prevented or becomes belatedly. (The flow impeded and delayed).
14) The structural elements
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s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323586043.75/warc/CC-MAIN-20211024142824-20211024172824-00246.warc.gz
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CC-MAIN-2021-43
| 2,615 | 38 |
https://onlinemiddleschoolmath.com/proportional-relationships/
|
math
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A proportional relationship between two quantities is a relationship in which the ratio of one quantity to the other quantity is equivalent.
Proportional Relationships in Equations
A proportional relationship can be written as an equation in the form “y=kx” where the “k” represents a rate.
Proportional Relationships in Graphs
The graph of a proportional relationship is always a straight line (linear) that goes through the origin
Proportional Relationships in Tables
A table represents a proportional relationship when there is a constant rate of change and when the input (x) equals zero and the output (y) equals zero.
Proportional vs Non-proportional
Proportional relationships are always linear however not all linear relationships are proportional. Some linear relationships are non-proportional.
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s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320302740.94/warc/CC-MAIN-20220121071203-20220121101203-00324.warc.gz
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CC-MAIN-2022-05
| 812 | 9 |
https://shakyradunn.com/slide/chemistry-a-science-for-21st-century-e9av5l
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math
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Gases Composition of air: 78% Nitrogen, 21% of Oxygen, 1% others Diatomic Gases: H2, N2 , O2, F2, Cl2 Monoatomic Gases: All the Noble gases O3 (ozone) is an allotrope of Oxygen Ionic compounds do not exist as gases at 25 oC temperature and 1 atm pressure. Boiling point of NaCl is above 1000 oC Elemental Gases and Gaseous Compounds Elements H2 (Molecular hydrogen) N2 (Molecular Nitrogen O2 (Molecular Oxygen) O3 (Ozone) F2 (Molecular Fluorine) Cl2 (Molecular Chlorine He (Helium)
Ne (Neon) Ar (Argon) Kr (Krypton) Xe (Xenon) Ra (Radon) Compounds HF (Hydrogen fluoride) HCl (Hydrogen chloride CO (Carbon monoxide CO2 (Carbon dioxide) CH4 (Methane) NH3 (Ammonia) NO (Nitric oxide) NO2 (Nitrogen dioxide) N2O (Nitrous oxide) SO2 (Sulfur dioxide) H2S (Hydrogen sulfide)
HCN (Hydrogen cyanide) Behavior of Gases: Most gases are colorless. F2, Cl2 and NO2 have color. NO2 can be visible in the polluted air (dark brown) Nobles gases are inert, they do not react with any other substances. O2 is Essential. Ozone is hazardous Compounds: H2S and HCN are poisonous gases. CO, NO2 , SO2 are toxic Physical Properties of gases 1) Gases have volume and shape of the containers 2) Gases are the most compressible of the states of the matter 3) Gases will mix evenly and completely when confined to the same container 4) Gases have much lower densities than liquids and solids Pressure of Gases Gases exert pressure Pressure is related to velocity and time
Velocity: Distance moved Elapsed time Acceleration = Change of velocity Elapsed time SI units of Velocity: m/s 2 units : m/s2 or cm/s2 Newtons second law: Force = mass x Acceleration force 1N = 1 kg m/s2 Pressure = Force/area Si units (pascal) pa 1pa = 1N/m2 Atmospheric pressure Barometer: It is an instrument to measure the atmospheric pressure Standard atmospheric pressure (1 atm) is equal to the pressure that supports 760 mm (76 cm) column of mercury at 0 oC at sea level. 1 atm = 760 mmHg or torr 1 atm = 101.325 Pa or 1.01325 x 10 5 Pa 1000 Pa = 1 kPa
1 atm = 1.01325 x 10 2 kPa Manometer Pressure-Volume Relationship-Boyles Law P (mmHg) 724 869 951 998 1230 1891 2250 V (volume) 1.50 1.33 1.22 1.18 0.94 0.61 0.58 PV 1.09 x 10 1.16 x 10 1.16 x 10 1.18 x 10 1.2 x 10 1.2 x 10 1.3 x 10 3 3 3 3 3
3 3 The data shows the inverse relationship between pressure and volume Pa 1 V Mathematical expression P = K1 x 1 V or PV = K1 Boyles Gas Law PV = K1: Pressure and Volume of gas at constant temperature and amount of gas is a constant The variation in pressure or volume can be expressed P1V1 = K1 and P2V2 = K1
Therefore P1V1 = P2V2 The Temperature Volume relationship At constant pressure, the volume of gas expands with the increase of temperature Charless law: Volume of a fixed amount of gas maintained at constant pressure is directly proportional to the absolute temperature of the gas VaT V = K2T or V = K2 T Absolute zero 0 K (kelvin scale) -273.15 0C (~ 273 0C) Variable volume and temperature equation: V1 = K2 V2 = K2 Therefore T1
T2 V 1 = V2 T1 T2 Charless Law of Temperature-pressure Relationship PaT P = K3 T P = K3 T Variable temperature/ pressure relation P1 T1 P2
T2 Relationship of Volume, Pressure, Temperature P1V1 = P2V2 Boyles Law V1 T1 V2 T1 Charless Law P1 T1
P2 T2 Charless Law Avogadros Law: Volume-Amount Relationship At the same temperature and pre ssure, e qual volumes of different gases contain the same number of atoms (monoatomic gases) or molecules (diatomic or polyatomic gases) Volume is proportional to number of moles of molecules. V an V = k4n n = number of moles K4 = Proportional constant Re lation of reactants and products in a gase ous re action 3H2 (g) + N 2(g) 3 mol es 1 mole
2NH3 (g) 2 moles At the same temperature and pres sures vol ume is proportional to number of moles Therefore: 3H2 (g) + N 2(g) 3 volumes 1 volume 2NH3 (g) 2 volumes When 2 gases react, their volumes have asimpleratio:
The ratio of molecular hydrogen and molecular nitrogen is 3 :1 If theproduct is gas, thevolume of product and reactants are in simpleratio: The ratio of ammonia to the sum of both H2 and N2 is 2 :4 or 1 :2 Relationsof gaslaws Boyles Law:At constant Temperature Vol um e decreas es Volume Vol ume increas es Higher P ress ure P = K 11 Lower P ress ure
or P = (nRT) 1 V Charless Law: At constant pressure Volume De cre ase s Lowe r Te mpe rature nRT = Cons tant V Volume V = K2 T or V = (nR) T P Volum e incre as e s Hi ghe r Te mpe rature nR/ P cons tant
Relations of Gas Laws Avogadros Law: Relation between amount and volume at constant temperature and pressure Volume decreases Volume More molecules V = K 4n Volume increases Less molecules V = (RT) n P RT/P = Constant
Ideal Gas Equation Boyles Law: V a 1/P Charless Law: V a T Avogadro: V a n V a nT P or V = R nT or P PV = nRT (R = Propor tionality constant) An ideal gas is a hypothetical gas whose pressure-volume-temperature behavior can be completely accounted for the ideal gas equation Gas Constant, R PV = nRT or R = PV
nT At 0 oC (273.5 K) and 1 atm pressure, many gases behave like ideal gas Standard Temperature and pressure (STP) : 0 oC and 1 atm. Volume of 1 mole of an ideal gas is 22.414 L (22.41L) Therefore the at STP R = 1 atm x 22.4L 1 mol x 273.5K = 0.082057 L.atm/K.mol or 0.0821 L.atm/K.mol Practice: What is the pressure (atm) of SF6 gas in a container having the volume 5.43 L at 69.5 oC Modified ideal gas equation R = PV nT R = P1V1 n1T1
R = P 2V2 n2T2 Modified equation = P1V1 = P2V2 n1T1 n2T2 n1 = n2 as the amount of gas does not change Therefore P 1V1 = P 2V2 T1 T2 Practice: At constant temperature, at 1 atm pressure the volume of helium is 0.55 L. What is the volume at 0.40 atm.? Density relation with Ideal Gas equation PV = nRT or n
= P V RT number of moles, n = m M m = mass in gram s M = Molar m ass Therefore Therefore m = MV d = M
P P RT or RT d = m V d = PM RT Calcul ate de ns ity of Carbon dioxide in g /L at 0 .9 9 0 atm and 5 5 oC Molar mass of a Gas d = PM RT
or M = dRT P 1) The density of a gas compound containing Chlorine and oxygen has 7.71g/L density at 36 oC and 2.88 atm. What is the molar mass of the gas.? 2) The density of a gas is 3.38g/L at 40oC and 1.97 atm. What is the molar mass.? Derive from both M = dRT P and n = PV RT Stoichiometry Calculate the volume of Oxygen gas (in liters) required for the combustion of 7.64 L of Acetylene at same temperature and pre ssure 2C2H 2 + 5O2 4CO2 + 2H2O
Avogadros Law: V a n at constant temperature and pressure As per e quation 2 moles of C 2H 2 = 5 moles of O2 Therefore 2L of C2H 2 = 5 L of O2 For 7.64 L of C2H2 = 7.64 L C2H2 x 5 L O2 2L C2H2 = 19.1 L of Oxyge n = Liters of O2 Stoichiometry Practice: Calculate the volume of N 2 Ge nerate d at 8 0 oC and 82 3 mmHg from the decomposition of 6 0g of NaN3 2NaN 3 (s) 2Na(s) + 3N 2(g)
Calculate the volume CO2 produced at 37 oC and 1.00 atm. when 5.60g of Glucose is decomposed C 6H12(s) + 6O2 6CO2 + 6H2O Daltons Law of Partial Pressures The total pressure of mixture of gases is the sum of the pressure of each gas in the mixture PT = P1 + P2 + P3 + . In a mixture of A and B gases the pressure of each gas can be expressed as PA = nART V Therefore PT If n = nA + nB and =
PB = nART V + PT = nRT nB RT V nBRT V V
= PT = PA + PB RT (nA + nB ) V Mole Fraction PA = nART/V Therefore PT = (nA + nB)RT/V PA =
nA PT (nA + nB = XA PA PT = nART/V (nA + nB)RT/V (mole fraction of A) The ratio of the number of moles of one to the number of moles of all gases in the mixture
Xi = ni nT Mole Fraction-Partial Pre ssure Xi = ni/n T Therefore the mole fraction is always less than 1 The relation of Partial pressure of A and B can be expressed in terms of mole fraction as XA + XB = nA nB
n A + nB PA /PT = XA or + nA + nB = 1 PA = XAPT General e quation for individual gas in a mixture: Pi = XiPT
Mole Fraction-Partial Pressure Practice 1: A mixture of gases contains 4.46 moles of Neon, 0.74 moles of Argon and 2.15 moles of Xenon. When the total pressure of the mixture is 2 atm., what is the partial pressure of each individual gases 1) calculate mole fraction of each gases with X i = ni/nT 2) Using mole fraction and total pressure, calculate the partial pressure of each gas (Pi = XiPT) Practice 2: A sample of natural gas contains 8.24 moles of methane, 0.421 mole of ethane, 0.116 mole of propane. The total pressure is 1.37 atm. Calculate partial pressure of the gases Kinetic Molecular Theory of Gases M olecular movement: Is a form of energy Energy: Capacity to do work or work done Work = Force x Distance or Energy = Force x Distance Units of Energy: J oule (J) or kilo joules (kJ ) 1 J = 1kg m 2/s2 = 1 N m 1 kJ = 1000 J Kinetic Molecular Theory of Gases 1. A gas is composed of molecules that are separated from each
other by distances far greater than their own dimensions. The molecules can be considered to be points; that is, they possess mass but have negligible volume. 2. Gas molecules are in constant motion in random directions, and they frequently collide with one another. Collisions among molecules are perfectly elastic (transfer of energy occurs during collision). 3. Gas molecules exert neither attractive nor repulsive forces on one another. 4. The average kinetic energy of the molecules is proportional to the temperature of the gas in kelvins. Any two gases at the same temperature will have the same average kinetic energy KE = mu2 m = mass and u = speed, bar means average value 5.7
Average Kinetic Ene rgy KE = 1 mu2 2 u2 = u12 + u 22 + + un2 N (u = speed, N = # of molecules) KE a T 1/2mu2 a T KE = 1 /2mu2 = CT (C = proportionality constant, T is absolute temperature) Kinetic Molecular theory: Application to the Gas laws Compressibility of Gases: In the gas phase, the molecules are separated by large distance. For this reason, gases can be compressed easily to occupy less volume Boyles Law: Molecular collisions a number density
(number density: number of molecules per unit volume) Numbe r density a 1/V Pressure a molecular collisions Therefore P a 1/V Kinetic Molecular theory: Application to the Gas laws Charless Law: KE a T Temperature results in the increase of Kinetic energy. The molecular collis ions on wall increase which in turn increases pressure KE a P PaT Kinetic Molecular theory: Application to the Gas laws Avogadros Law: P a d and P a T or P a dT d a n/V Therefore P a n/V x T
For two gases P1 a n1 T1 = V1 C n 1T1 V1 P2 a n2T2 = C n2 T2 V2 V2 If P, V, T are same n1 and n2 are e qual Kinetic Molecular theory: Application to the Gas laws
Daltons Law of Partial pressure: PT = P1 + P2 + .. If molecules do not attract or repel one another, then the pressure of one molecule is unaffected by another molecule. Consequently the total pressure is the sum of individual gas pressures Apparatus for studying molecular speed distribution 5.7 The distribution of speeds of three different gases at the same temperature The distribution of speeds for nitrogen gas molecules at three different temperatures urms =
3RT M 5.7 Gas diffusion is the gradual mixing of molecules of one gas with molecules of another by virtue of their kinetic properties. r1 r2 = M2 M1
NH4Cl NH3 17 g/mol HCl 36 g/mol 5.7 Gas effusion is the is the process by which gas under pressure escapes from one compartment of a container to another by passing through a small opening. r1 r2 = t2
t1 = M2 M1 Nickel forms a gaseous compound of the formula Ni(CO)x What is the value of x given that under the same conditions methane (CH4) effuses 3.3 times faster than the compound? r1 2 r1 = 3.3 x r2 x M1 = (3.3)2 x 16 = 174.2 M2 = r2
x = 4.1 ~ 4 58.7 + x 28 = 174.2 M1 = 16 g/mol ( ) 5.7 Deviations from Ideal Behavior 1 mole of ideal gas PV = nRT PV = 1.0 n= RT Repulsive Forces
Attractive Forces 5.8 Effect of intermolecular forces on the pressure exerted by a gas. 5.8 Van der Waals equation nonideal gas 2 an ( P + V2 ) (V nb) = nRT } }
P1/T1 = P2/T2 Charless Law: Pressure Temperature V = K4n Avogadros Law: Volume - # moles at Constant P and V PV = nRT P1V1/n1T1 = P2V2/n2T2 P1V1/T1 = P2V2/T2 Ideal Gas equation Relation of Pressure-Volume-Temperature-amount of gas Pressure-Volume-Temperature at constant n Density-Molar mass- Pressure-temperature d = PM/RT
Xi = ni/nT KE = 1/2mu2 = CT Mole Fraction Average Kinetic energy-Absolute temperature Key Equations Urms = 3RT/M Root-mean-square speed and Temperature r1/r2 = M1/M2 Grahams Law of diffusion and effusion (P + an2/V2)[V-nb] = nRT
Washington State Community and Technical College system. AS Business Administration. Central Oversight. ... May or may not utilize institution's learning management system (LMS) and other educational technologies. Follows institutional terms (typically 16, 12, 8 week) Online.
Mathematical Formulas for ship hull forms With the help of computers, use spline 'interpolation functions' to draw the "lines" using computers to "mold loft." -Prismatic or Longitudinal Coefficient: 0.55~0.80 -Waterplane Coefficient -Displacement /Length Ratio -Breadth /Length Ratio : -Draft/Length Ratio...
Condensation Polymers: Monomers with two reactive groups. Different reactive groups + different chains = different polymers. Control of Polymer Properties. Control of Polymer Properties. Property control by side group. Cross-linked Polystyrene.
Items and Itemsets An itemset is merely a set of items In LR parsing terminology an item Looks like a production with a '.' in it The '.' indicates how far the parse has gone in recognizing a string that...
Driver log archives available for inspection access with similar presentation to a paper log. Support System Capabilities. 395.15. Logs submitted within 13 days. Basic log data record content . Log data retained for 6 months. Archive in location separate from...
Mary Levinski, Sauk Rapids-Rice High School ProStart Instructor ProStart is a special program for high school students that will give you the opportunity to learn about the art of cooking and managing restaurants by training with professional chefs and getting...
This Order to bee forthwith effectually putt in Disarming & Limiting the Movements of Long Island Indians Keeping King Philip's War from Spreading to the Island, By Order of the Governor, Pt. 2 -- January 24, 1675 Execution, and a...
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https://community.trading212.com/t/cfd-minimum-quantity/8450
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I wanted to invest using the cfd options however I found the minimums of some shares very high…
For Example: Intel minimum is 20
BMW is 60
Daimler is 10
Ford is 30
These are just a few examples…
Futhermore I wanted to know how you decide upon what is the right minimum quantity.?
Thank you for your hard work:)
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https://www.greaterwrong.com/posts/5bd75cc58225bf06703754dc/smoking-lesion-steelman-iii-revenge-of-the-tickle-defense/comment/5bd75cc58225bf06703754e9
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What does the Law of Logical Causality say about CON(PA) in Sam’s probabilistic version of the troll bridge?
My intuition is that in that case, the agent would think CON(PA) would be causally downstream of itself, because the distribution of actions conditional on CON(PA) and ¬CON(PA) are different.
Can we come up with any example where the agent thinking it can control CON(PA) (or any other thing that enables accurate predictions of its actions) actually gets it into trouble?
I agree, my intuition is that LLC asserts that the troll, and even CON(PA), is downstream. And, it seems to get into trouble because it treats it as downstream.
I also suspect that Troll Bridge will end up formally outside the realm where LLC can be justified by the desire to make ratifiability imply CDT=EDT. (I’m working on another post which will go into that more.)
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http://ieeexplore.ieee.org/xpl/articleDetails.jsp?reload=true&arnumber=5457693&contentType=Conference+Publications
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Skip to Main Content
To measure the similarity between two high dimensional vector data, correlation coefficient is often used instead of Euclidean distance. For this purpose, the high dimensional vectors are mapped into hyperspherical points by normalization, and the distance between two hyperspherical data is measured as the length along geodesic on the hypersphere. Then estimations from high dimensional vector data should be resolved as minimizing appropriate energy function of the length along geodesic when high dimensional vector data are regarded as hyperspherical data. In this paper, for the first step of hyper surface fitting to hyperspehrical data, the method of curve fitting to two-dimensional spherical data by Spherical Least Squares is proposed. It is also shown that the proposed method is closely related to the curve fitting by Euclidenization of the metric.
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http://www.angelacernshiggsnightmare.com/thehumansoul-ths-2671981_reference_library/nasa_investigation.htm
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INVESTIGATION INTO THE ENTERPRISE MISSION'S PROPOSITION THAT THE 19.5 AND 33.0 DEGREE STAR ALIGNMENTS CONSTITUTE A PATTERN
Mary Anne Weaver
Copyright (c) 1999 by Delphi Technologies
For a number of years now, researcher and former NASA consultant Richard C. Hoagland of the Enterprise Mission, has observed what he has identified as a "pattern" in star positions at the time of NASA launches and/or landings. He published his discovery a few years ago as an article entitled "Kennedy's Grand NASA Plan", and has since written many more articles and performed extensive research into this topic. Later on, Mike Bara, an Aerospace design engineer with many years experience in his field, joined the research effort into this stellar alignment phenomenon and has assisted Richard Hoagland in investigating its attributes and the frequency of its occurrence.
Both Richard Hoagland and Michael Bara note that specific stars are found at particular elevations above the horizon, namely 19.5 and 33.0 degrees but also directly on the horizon and meridian, at the precise moment of a NASA launch or landing. Since 1996, Michael Bara and Richard Hoagland have done considerable research into the "star alignment" theory and posted a large amount of data. Also since that time and based on copious research into this topic, they have proposed a number of theories to account for this phenomenon.
I was intrigued by the data posted on the Enterprise Mission website and decided to investigate for myself. If there was a solid, numerical basis for the assertions made by the Enterprise Mission, it should show up in the numbers -- or not. I decided that a thorough numeric investigation could end the debate over whether such a pattern exists or not.
I have always been interested in space exploration, and have followed NASA's activities with great interest as a child and later as an adult. This interest, and my technical abilities, lead me to pursue a career in electromagnetics research and development. In 1985, I graduated from the University of Washington in Seattle, with a Bachelor's Degree in Electrical Engineering. Prior to graduation, I participated in a competitive and prestigious research and development program sponsored by GTE Labs, in which I was 1 in 50 selected nationwide. After graduation, I accepted a research position in the Antenna division at Boeing Aerospace Company in Seattle. At Boeing, I was responsible for: 3-D computer modelling, computational analysis, data analysis of radar patterns, and developing equations and analytic methods to solve for optimal antenna parameters. At Boeing, I gained experience in using rigorous experimentation and data-gathering methods to determine how accurate our "models" of the situation were, based on the radar patterns we were getting. Thus, I learned how to develop and write equations to model real life situations and how to test them; I also learned how to analyze data and real results, and make determinations based upon those analyses. These are the abilities I bring to bear on this analysis. In addition to my duties at Boeing, I've performed independent probability studies, such as a project to determine the probability of the shape and size of an object based on its radar pattern.
Since Boeing, I have worked as a computer professional, and most recently as a software design lead for a data management company. I have continued to use my mathematic and analytic skills in the computer industry, and still write programs to do computational analysis and data analysis. In order to design software of this kind, I utilize my ability with mathematics and analysis on a daily basis.
My career took a different turn in my adult years, so I never did work for NASA or help advance the space program. But, my interest in it never waned. Thus, when I learned of Richard Hoagland's research, I was intrigued and wanted to know the truth for myself. Was there really any correlation between NASA events, such as launches, and star positions? If so, what does it mean?
I don't propose to answer the "what does it mean" question in this article. I felt it important to determine first, if there a solid scientific basis for Richard Hoagland's hypothesis that these alignments form a pattern. If so, this solid foundation will provide the basis for asking and finding the answers to more questions, such as "what does it mean?"
Because of my interest in the space program, and because NASA launches and events lend themselves more easily to statistical analysis (since there are a set number of them), I decided I would limit myself to analyzing NASA events (launches, and mission activities). My task, then, was to determine numerically whether or not these NASA alignments occur more often than random chance; and if so, how much more. If there were quite a few more alignments than random chance would allow, then this could be an indication that a pattern was being followed, if all the right factors were present.
To do this, I first had to develop a correct model of the odds. For those readers who wish to view the scientific method and proofs in detail, read my technical paper. Otherwise, this summary will mention in less detail what I did and what I found.
In order to correctly model the odds, I had to come up with a simplified and restricted version of the Hoagland "star alignment model." This was for the sake of consistency, and meant to be a preliminary investigation that could catch at least some of these repetitive 19.5 and 33.0 degree star alignments.
A Brief Review of the Hoagland "Star Alignment Model"
The Hoagland "star alignment model" angle selections are based on tetrahedral geometry; specifically, a tetrahedron inscribed inside of a sphere. The circumscribed tetrahedron is descriptive of a kind of "hyperdimensional physics," which is simply a physics that takes higher (thus unseen!) spatial dimensions into account. I will not get into detail about the theory here, but rather summarize it briefly so that the reader will understand the basis of the related Hoagland/Bara star alignment work.
Hoagland's original "hyperdimensional physics" theory states that "rotation," such as that of a planet on its own axis, creates higher-dimensional dynamic forces inside a planet, which ultimately conform to a specific 3-D geometry; as a result, phenomena appear on the planets' surfaces in accordance with the geometric contact points of that geometry -- two interlaced 3-D tetrahedra inscribed inside a sphere. The lowest order touch points of these circumscribed tetrahedra (beside the poles of rotation) are at approximately 19.5 degrees, North and South latitude. Refer to Figure 1, below.
Figure 1: Two tetrahedra inscribed inside a sphere. Copyright (c) 1998 by the Enterprise Mission, used with permission.
The actual "touch points" of these tetrahedra are at 19.47 degrees N or S latitude on any particular planetary body, rounded to "19.5 degrees." This is the source of the number 19.5. Next, when one takes the Sine of the tetrahedral angle, the following number results:
Sine (19.4712) = 0.33333...
This turns out to be the vertical "height" of the 19.5 angle within a unit sphere.
Richard Hoagland and Mike Bara utilize the "shortened" version of this, which is the angle 33. However, note how many "3's" there are in the above. Since the source of the Hoagland/Bara "angle 33" and angle "3 deg 30 min" is this "repeating 3's" in the Sine of the tetrahedral angle, that will be a key thing to look for in the data.
The source of the "horizon" and "meridian" alignment emphasis is Egyptian ritual practice; specifically, Egyptian star lore. Stars, to the ancient Egyptians and Sumerians, were quite important and their position in the skies were an integral part of temple layout and design and well as ceremonial applications (sources: Star Names: Their Lore and Meaning by Richard Hinckley Allen, Astrological Origins by Cyril Fagan, and historical texts on ancient Egypt and Sumer). The horizon and meridian had important symbolic values. Stars that rose were considered to be "born," stars at the meridian had reached their "peak," and stars that were setting were considered to be "dying," or about to go into the underworld. This is common knowledge to scholars of these ancient belief systems. The horizon and meridian are, of course, the basic dividing lines for the celestial sphere as well as for terrestrial geography.
Additionally, according to Egyptian belief and mythology, the stars were actually the "abode of the gods," and in many cases, were identified with the gods themselves. The constellation of Orion and Osiris were actually identified with one another, such that the constellation was considered to be Osiris. Also, Sirius was identified with "Isis."
The Hoagland/Bara hypothesis is, then, that these star alignments are symbolic of ancient star lore and hyperdimensional physics geometry. Why such an unlikely combination? No one knows for sure at this point, but if the reader wishes to embark on further explorations of these possible connections (that have been discovered and published by Richard Hoagland, as well as others), I direct them to read the other articles on the Enterprise Mission website.
A RESTRICTED VERSION OF THE HOAGLAND/BARA STAR ALIGNMENT MODEL
Now that I have ascertained the source of the angles in question, I will define a restricted version of the above Hoagland/Bara model, which I will be using in my analysis. This model is not complete because I do not use all the "temple" locations that Richard Hoagland and Mike Bara have used in their work, nor do I use all the celestial objects they do. For example, the Hoagland/Bara temples which I do not use consist of the Mars temples -- the Viking I and II sites, and Cydonia -- as well as the Earth "temple" locations, Phoenix, JPL (Pasadena, California) and Houston. I do use Houston at one point when I am analyzing the Apollo mission events data, but not as a general rule in this analysis.
There are a total of ten "temple" sites. However, it was necessary to simplify this model for ease of numeric analysis, and to get a feel for the situation, so I only used four in this restricted model. Because this is a restricted model, I do not "catch" all the alignments (nor would I expect to) that Hoagland and Bara do; nevertheless, I did expect that this "limited" approach would at least determine whether or not the "Egyptologically-important" stars I chose to examine (in the constellations of Canis Major, Orion, and Leo -- Sirius, Mintaka, Alnilam, Alnitak, and Regulus) do appear more times than random chance would allow.
In order for a "ritual" star alignment to occur in my restricted version of the Hoagland/Bara model, the limited criteria listed below must be met.
Celestial Object must be at these Angles:
Zero degrees (either horizon)
19.5 degrees (above or below either horizon)
33.0 degrees (above or below either horizon)
Meridian (highest or lowest point a star can reach in the sky)
Other angles (such as 3 deg 30 min) which are symbolic of the numbers "33" or "19.5"
The Enterprise Mission lists many celestial bodies and stars that are used in its model, but for the purposes of this analysis, and to keep it simple, I restricted myself to specific celestial objects.
Stars used in this simplified model:
Sirius (brightest star in Canis Major)
Alnitak (Orion belt star)
Alnilam (Orion belt star)
Mintaka (Orion belt star)
Regulus (brightest star in Leo)
Locations from which stars are observed:
Planned Apollo 11 landing site
Planned Apollo 12 landing site
Planned Apollo 13/14 landing site
As previously noted, the above locations represent only four (not all ten) of the Hoagland/Bara "temple" sites, from which these alignments are observed.
Below (Figures 2 - 5) are some examples of what these configurations look like in the sky over the Earth or Moon. All of these pictures were obtained from the program RedShift, and represent the sky at the particular "temple" of interest, at the date and time I specify.
Figure 2: The above picture depicts the sky over Giza, Egypt, during the launch of Pioneer 5. Of course, this is an example of the Orion constellation, with the Orion belt star at 33 degrees altitude.
Figure 3: The above depicts the sky over the planned Apollo 12 landing site on the Moon, at the time of the launch of Apollo 15. This is an example of the constellation Canis Major, and the star Sirius, which here appears 33 degrees below the lunar horizon.
Figure 4: The sky above the Apollo 11 planned landing site on the Moon, at the time of the launch of the Mercury program's Atlas 7. An example of 19.5 degrees above the horizon, this time with the belt star Mintaka.
Figure 5: The sky above the Apollo 12 planned landing site on the Moon, once again at the time of the launch of the Mercury program's Atlas 7. This is a meridian alignment, which means that (in this case) the star Sirius has reached the highest point in the sky that it can reach. The "meridian" line is the line drawn from the Zenith through North and South. Notice that it intersects the star Sirius, indicating that Sirius has risen "to the meridian."
Though the Enterprise Mission website also lists the Sun, Comet Encke, and Mars as celestial objects used in the pattern, they were excluded from this analysis to keep the model simple.
Next, I developed a mathematical model of the odds based on how long a star stays at the elevations in question; in this case, the horizon, 19.5, 33.0, and the meridian. In the above case, the probability is expressed as the time that all stars of interest to the Hoagland/Bara ritual star alignment model stay at the altitudes of interest, at each location that Hoagland and Bara use.
NASA Programs to be Analyzed
I also limited myself to Apollo and Apollo preparation missions such as Pioneer, Ranger, Surveyor, Lunar Orbiter, Mercury, Gemini, and so on. I did this because of a time limit on how long I could spend on this project, and because all these launches have a "theme" . . . in this instance, getting ready for sending men to the Moon. Because they all had the same theme, I could therefore place them all in the same statistical grouping.
I obtained launch times (as well as landing dates and times) for Apollo, from the book "To a Rocky Moon" (Don E. Wilhelm, University of Arizona Press, Tucson, 1993). For all other spacecraft launch times, I consulted the National Space Science Data Center (NSSDC). Star positions for all the dates and times analyzed in this paper were obtained using the commercially available astronomy software program " RedShift." RedShift uses NASA measurements of star and planetary positions, in order to very accurately calculate celestial coordinates as seen from specified locations on Earth for specific dates and times, as well as for locations on other celestial bodies, such as the surface of the Moon.
I strongly encourage anyone who has an interest in this to research it themselves, as there is much more work to be done. I present the tip of the iceberg here, because I only had so much time to devote to this project; still, what I found is quite amazing.
INTRODUCTION TO PATTERN DETECTION
In order to explain how I would detect a "pattern," I will proceed to a simpler example -- a coin toss.
Suppose I enter into a coin toss game with a friend, and I want to figure out the odds of winning. If the coin is perfectly balanced, then the odds should be 1 to 1 for tossing "heads" -- in other words, I have equal chances of tossing "heads" or "tails" on any given toss.
Next, suppose I want to compute the probability of tossing "heads". Probability is given by:
Probability = Outcomes Favorable/Total Outcomes
In a coin toss, there are only two outcomes -- Heads or Tails -- and I'm just interested in computing the probability that the coin I toss will come up "heads". Then,
Probability = Outcomes Favorable/Total Outcomes
Probability(Heads) = 1/2 = 0.500.
But how can I be sure the model is right? That's straightforward, and in order to do that, I employ the Law of Large Numbers.
The Law of Large Numbers
It is a statistical fact that, the more observations I make of an event, the closer I can come to correctly defining the probability of that event's occurrence. Thus, the Law of Large Numbers states that I should draw closer and closer to the actual probability of an event with each observation I make.
So, to determine whether the formula for predicting odds is correct, I toss the coin as many times as I can in order to find the true probability. In this case, I did 100 sample "throws" with a coin-toss simulation program, and plotted the results. Refer to Fig. 6, below: Note how the graph closes in on the value 0.500, the actual probability that I computed using the "Outcomes Favorable/Total Outcomes" equation.
Fig. 6: Perfectly Weighted Coin
Number of Heads Tossed plotted against Total Tosses
To further illustrate the importance of the Law of Large Numbers, and how this Law works, refer to Fig. 7, below. It is the same as Fig. 6, except that I highlighted toss # 13.
Fig. 7: Highlight of Toss 13. Note its probability falls at 0.63 on the graph.
If I were to stop tossing coins at toss # 13, and I didn't know what the correct odds were supposed to be for a coin toss, I would incorrectly deduce that the probability of tossing heads is 0.63, or about 1.6 to 1, "for" tossing heads. Obviously, this isn't right! So, when calculating probabilities, it is important to 1) have the correct model for the odds, and 2) take enough samples to be able to make a correct estimate of the situation. Notice that, by the time I reach 80 or more samples, the probability has very closely approached what it should be -- 0.5, or 1 to 1 odds.
How can patterns or "non-random" occurrences then be separated out from a set of data? The first thing to do is determine, as I did for the "coin" above, what the "random" situation looks like. Then I can test the new situation to see if it fits the "random model."
Suppose I enter a coin-tossing game with a dishonest person who has "weighted" the coin, causing it to come up "heads" 2 times out of 3, giving 2 to 1 odds "for" tossing heads instead of 1 to 1. The odds are expressed as 2 to 1 because odds are expressed as the "number of successes to the number of failures". Probability is expressed differently; i.e. the ratio of the "number of successes"/"total events".
In order to test my theory, I have to take a few samples and see for myself, what the odds turn out to be. I already know what the true random situation is supposed to look like. In order to really test this coin and to be sure it's weighted, I should take a large number of samples. Recall how, in the "balanced coin" example, if I had only progressed to "toss 13" I would have made an incorrect assumption about the probability. I still could have estimated the trend by taking the "average" of the points, but it's far more accurate to use a large number of sample "tosses" in order to truly establish that the coin is weighted.
Again, using a coin-toss simulation program, I weighted the "coin" as described above and plotted the results in Fig. 8.
Fig. 8: Plot of Weighted Coin, weighted to produce 2 to 1 occurrence of heads
Notice how the graph closes in on the correct value for this "weighted" coin, which is 2/3 or 0.67. (Recall: Heads occur two times out of three in this example.) Note also that the data points all cluster in on a line, and that the line does not waver very far from the 0.67 value. It does not drop down to the 0.500 value, for example, but instead holds its position at 0.67. This will be important to remember later.
What can I deduce from this? First of all, that the coin is weighted, and that it's not behaving like a perfectly balanced coin would. It is very unlikely, given the behavior of this graph, that the coin would not be weighted.
For a more detailed discussion on aspects of this, such as variability of random data and so on, consult my technical paper. Basically, without getting into this in great depth, I can say that there are two factors that tell me this is probably not random: high odds against this much deviation, and the fact that the curve does not show a tendency toward convergence on the expected random value.
VERIFYING MY MODEL OF THE ODDS
Before I began analyzing NASA launch data, I had to make sure that my equations for predicting the number of these "ritual" star alignments that I should get by random chance, were correct. It's not enough to develop equations I think are accurate, i.e. that will predict the number of occurrences of 19.5 and 33.0 degree alignments. I also need to verify the "theory" by taking 100 random sample measurements. This will tell me, in effect, if I'm modelling "real life" accurately.
It's analogous to tossing a coin 100 times, in order to be certain that the probability of heads really is 1/2 or 0.5. Recall from the previous discussion, then, that the first thing I need to do is use the Law of Large Numbers to find out what the "random" situation looks like. (For details on the actual equations used to model the odds, please refer to my technical paper.)
Part of my calculation process for determining the probable occurrence of an alignment was setting "error margins" around the angles. For example, suppose I am a hypothetical conspirator who wishes to perform NASA events at the same time as specific stars are at alignment positions. Because of real world considerations, such as crew safety, device limitations, physical obstacles or unexpected delays, I can't always hit an alignment dead on. Therefore, there will be a margin of error around each angle. In order to find this hypothetical "margin" of error, I decided to analyze the data for each angle, to see what the values near 0, 19.5, 33, and the meridian would "center" around. I did not really expect to see any kind of "centering" tendencies if the data was truly random, anyway; if the data is random, then data could center around 32.7 just as easily as 33.0, for example -- and in that case, the arbitrary margins of +-0.25 degrees or +-0.5 degrees would be suitable to analyze the data. However, if someone was trying to hit the angles "dead on," the data would tend to "cluster" around 19.5, 33.0, and the horizon and meridian. Therefore, if I did see "centering," I could proceed from there on the supposition that someone was "trying" to hit these angles dead on, perhaps accepting certain limits or criteria for each angle. For a detailed explanation of this process I went through to select these "error margins," as well as a complete listing of the error margins, see my technical paper.
Why select error margins based on real data, instead of arbitrary fixed ones like +-0.5 and +-0.25?
Because the probability of an alignment event occurring within an actual observed error margin (not a "made up" fixed one) is expressed as
Time for star to traverse error margin P(alignment within error margin) = ------------------------------------------ Total rotation time
See the picture of a circle (Figure 9, below). Note how I've highlighted a small portion of that circle. That "portion" is the amount of error that I will accept around an angle; in the example below, 33 degrees.
Figure 9: Circle of 360 degrees, showing error margin around 33 degrees. Error margin is replicated as many times as it will fit into the circle of 360.
Now, if I replicate that error margin, note that it fits into the circle so many times. Recall the basic equation for probability, which is
Probability = Outcomes Favorable/Total Outcomes
In this case, "Outcomes Favorable" is that small highlighted piece of the circle; that's the error margin I'm accepting around 33 degrees. "Total Outcomes" is given by the total number of times that the error margin "fits" into a circle of 360. This describes the true probability of this event.
Of course, the above circle diagram represents the situation where I'm only looking at one occurrence of 33 degrees. In my mathematical model (see technical paper ) I also account for 33 degrees above or below the east or west horizons, which means that 33 occurs a total of four times. All this is already accounted for in my calculations; that's one reason why my equations work, and work well, in approximating the real situation. Figure 10 below illustrates this concept of "33 degrees" occurring 4 times at any given location.
Figure 10: Number of occurrences of 33 degrees, at a given observation site.
The first thing I found was that the launch data did precisely center around the Hoagland/Bara angles; i.e. 19.5, 33.0, the horizon and the meridian. This was a clear indication to me that it was entirely possible that someone had been trying to "hit" these angles dead on. I used my analysis and observations of this "centering" phenomenon to determine the appropriate error margins (see the technical paper for a listing of these error margins).
Once I had determined these "real life" error windows, in order to better approximate the real situation, I calculated from the equations I developed for star motion on the Earth and Moon, what the Probability of an alignment at Giza, Egypt or the planned Apollo 11, 12 or 13/14 lunar landing sites would be. I obtained Probability = 0.32 (or 32 hits out of 100).
Then, I took 100 sample measurements of star positions, the 100 randomly chosen dates and times spanning the years 1958 to 1978.
Just as my equations predicted, I did indeed get 32 hits out of my 100 measurements. So, now I know what the random situation looks like, and that my equations are modelling the real situation quite well. I plotted my results from my random data on the graph below (Figure 11). Notice that the data converges on the expected probability value of 0.32.
Figure 11: Graph of Random Data for the years 1958 through 1978
This convergence on the expected random value is of course due to the Law of Large Numbers, which states that if I take enough samples of an event over time, I'll figure out how often the event occurs and be able to predict it accurately. This is what science itself is based on, at least with respect to testing theories with experiments. The greater the number of experiments performed that have a specific outcome, the more likely it is that the scientist is correct in his theory.
A probability of 0.32 translates into roughly 2 to 1 odds against (odds = 1/probability - 1, to 1). This does not seem like much. But, in statistics, the important thing is -- does something beat the odds all the time?
To get an idea what this means in a "real life" situation, imagine that you really did play an unfair coin toss game with someone, where "heads" came up 2 times more often (and you bet on "tails"!). In such a case, you would be within your rights as a player in the coin-toss game, to cry "foul!" and demand that a balanced coin be used. This is because you'd know it was more likely that the coin was unbalanced, than it would that it just "coincidentally" came up heads that much of the time! A few times -- yes. A hundred times -- no.
In the previously mentioned case, where the coin comes up heads 2 times out of 3 in a set of 100 tosses, recall that the odds against that (for a perfectly balanced coin) would be 2,182 to 1. And, probably long before 100 tosses, you'd be suspecting that this coing toss game was rigged.
ANALYSIS OF NASA LAUNCH DATA
Choosing to analyze the Pioneer, Mercury, Ranger, Surveyor, Gemini, Lunar Orbiter and Apollo programs, resulted in a sample set of 82 NASA launches. Figure 12 (below) is a graph, plotting the NASA launch alignment hits, of which (in this set) there were 44, against the predicted random value of 0.32 (or 32 hits/100).
Figure 12: Graph of Actual NASA Launch Data
The odds against this much deviation from the random value are 40,192 to 1. That is approximately 20 times higher than the odds against the "rigged" coin toss!
So, this is very significant. And, while 82 launches does not in itself indicate that all of NASA follows this 19.5 and 33.0 pattern, it is enough to indicate to me that the pre-Apollo and Apollo mission launches did.
How can I tell that these results follow the pattern outlined by Richard Hoagland and Mike Bara? Because of the numbers and analysis method I selected. My analysis approach only emphasizes the frequency of star positions at specific locations. This serves as a 'filter,' because of the fact that NASA launches supposedly are tied to the position of planets, weather, and lighting conditions (the position of the Sun), NOT 'star positions.' The method of analysis that I chose emphasizes star positions and nothing else; not weather, lighting, planets, or other factors. Unless one accepts the validity of Astrology, where the positions of stars do have an effect on weather conditions or so on, there is no reason to tie star positions in with launch conditions.
Patterns within Patterns
See Figure 13, below. On the left side of the figure, there are randomly arranged rectangles with mission names in them. The rectangles are oriented in different "random" directions, visually representing the concept of randomly organized launch times. Next, in the center of the Figure 13, the rectangles all assume an ordered format, which is non-random by its very nature because it is organized in a pattern. This "pattern" of rectangles represents the pattern found in the 82 launches, because in the case of the 82 launches, the missions were seen as being represented only by their launch times. When this was done, a pattern emerged in the 82 launches; therefore, the missions can be seen as being ordered in this fashion. Finally, at the far right, I decide to "magnify" one of these small "mission" rectangles in order to scrutinize it further. When I do, I find that specific events in the mission are organized in the same pattern. Thus, a "pattern" has been found within a larger pattern.
Figure 13: Illustration of the "pattern with a pattern" concept.
What if there were such a "smaller" pattern within the larger? What would this mean for the odds?
I know that the larger pattern must occur first, so that
must be expressed in a probability equation, P(large pattern). Then, the
probability of the "smaller" pattern that existing inside the Apollo program
must be written out as P(small pattern). To find out the probability of both
happening at once, I multiply them together:
P(large+small) = P(large) x P(small)
In order to test the Hoagland/Bara theory at yet another level of detail, I decided to do just that -- to see if the pattern carried itself on down through other layers of detail, into actual mission activities themselves. I analyzed the Apollo missions specifically. I was interested in finding trends, or non-random tendencies in the data within the context of the "star ritual" pattern itself. I looked for these things because non-random tendencies in the star alignment data would be extra supporting evidence of planned (with respect to star alignments) versus random launch times. Or, if I found random data, this would become apparent very quickly.
The type of events which I looked up star alignments for were activities such as docking, course corrections, landings, splashdowns, etc. Because of time constraints, I chose to focus only on the successful lunar landing missions; i.e. Apollo 11, 12, 14, 15, 16 and 17. In just these six Apollo missions, there were a total of 112 mission activities (including launches). Grouping these six missions together was appropriate, as they are all of the same "type" -- i.e., each one was a successful manned lunar landing mission, and therefore could be grouped in the same statistical category for analysis.
To do this portion of the analysis, I consulted the NSSDC again and obtained mission summaries. From these, I extracted the times of docking, engine firing, landings, etc., and looked up the corresponding star alignments that occurred at those times. If no pattern was being followed, the data should converge on a random value, just as in the "coin toss" example. Also, I felt if a pattern was followed in this data as well, it would be just as interesting, in that it indicates that the times I retrieved for mission activities from the NSSDC were selected based on a pattern!
I ask the reader to imagine the improbability of timing many maneuvers and mission activities to correspond with star alignments! If something was found, I thought it would be very unlikely.
In order to fully analyze the Apollo mission events, I decided to add "Houston" to the list of star observation sites. It is considered a major "temple" in the Hoagland/Bara model, because it was the literal "Center" of NASA's whole manned space effort, including the Apollo Program: for instance, within seconds of liftoff of every manned mission from Cape Canaveral, Houston Mission Control took over total command of all subsequent aspects of these flights. Not by accident did the phrase 'Houston' become immortalized in the language of every space enthusiast, because for over forty years Houston was also carefully designed to be the sole radio communications link between all NASA crews in space, and all the rest of us on Earth. If any location could be considered a "temple" for a NASA ritual, it would undoubtedly be "Houston."
When I did the above, the probability went up to 0.48, because I added another location where an alignment could happen. That means that, at any given time, it's more likely for me to observe an alignment. This makes sense, because the more places I choose to observe stars from, the more likely I am to see a star at 19.5 or 33.0 degrees (or on the horizon or meridian).
Much to my surprise, not only did the Apollo "mission activities" data not converge on a random value, it spectacularly conformed to the Hoagland/Bara star ritual pattern yet again! The graph below (Figure 14) illustrates this, and shows that, for 112 mission activities (including launches), the curve does not converge on a random value, but stays above it, as in the other case.
Figure 14: Graph of actual star alignments taking place during day-to-day Apollo Mission events
Graph of Alignments taking place during Apollo mission activities
In this situation, I had P(align) = 0.48, and 76 hits out of 112. This came to 100,010 to 1 odds against chance. That's how unlikely it is just for the Apollo mission activities to conform this closely to the Hoagland/Bara star ritual theory, by mere random chance! This clearly indicates that Apollo mission activities cannot be accurately modelled using random chance, because it is far too improbable to do so!
Since Houston is a "temple" in the Hoagland/Bara model, that's why this data confirms that Apollo mission activities do follow the "ritual" pattern. However, if I were to leave Houston out of it, the odds against would still be very high -- as will be seen.
The Importance of Consistency
Recall the section above, when I was speaking of "patterns within patterns." In this case, I can't multiply this Apollo mission data pattern (the "small" pattern in my previous example) times the probability of the 82 launch pattern (the "large" pattern) yet, because I used Houston in addition to Giza and the three planned Apollo landing sites in the above example. Using Houston is a different different analysis approach than I performed earlier on the 82 launches, where I used only the Apollo 11, 12, and 13/14 landing sites on the Moon, and Giza, Egypt, though it still falls within the Hoagland/Bara star alignment model.
In statistics, comparing two sets of data with two different sets of locations and/or paramenters, would be like comparing apples and oranges. Though a pattern might be found in an arrangement of both, only apples or only oranges should be considered -- not both!
The solution is, then, to examine both sets of data in the same manner. Therefore, next I will examine both sets of data using only the locations of: Giza, and the planned Apollo 11, 12 and 13/14 lunar landing sites, as I did for the 82 launches. I also use the same error tolerances for both sets of data. In this case, I use a tolerance set that gives me a Probability of alignment = 0.39 for both. (It is not 0.48 anymore, because of removing Houston from the Apollo mission data as a stellar alignment observer location. As cited above, I did this so that the Apollo mission events will be analyzed using the same locations as the ones used for the launch data.) Since Houston doesn't figure directly in any launches but those of the Apollo Program, removing it "levels the playing field" for comparison with those missions NOT controlled by Houston.
Now I will be looking at exactly the same pattern, defined in exactly the same way, for both the "large" pattern of 82 launches, and the "small" pattern I found in the Apollo lunar landing missions. Now, multiplying them together is meaningful, because the patterns will match, or not (and the numbers will indicate this one way or the other).
In the case of the Apollo mission activities data, without Houston the odds come out to be 5,152 to 1 against chance. Next, for the 82 launches, the odds are 1,489 to 1 against chance.
Note how, in both sets of data, the odds are very high "against" random chance; in the thousands, rather like the "coin toss" example shown earlier. This is important because it says that both sets of data are equally non-random within this set of tolerances; therefore, both definitely follow the same non-random tendencies.
Recall, P(large+small) = P(large)*P(small), and that Probability = 1/(odds+1). Therefore, multiplying the above results yields a likelihood for the Apollo missions conforming to the same pattern as the 82 launches, as being 7.68 million to one!
So what does this mean? It means that it is extremely unlikely that the Apollo mission activities would conform to the same pattern that the 82 launches for the "Apollo preparation" missions (in this case, the Hoagland/Bara ritual star pattern). It's one thing for the 82 launches to follow the Hoagland/Bara star ritual pattern. But, for Apollo mission activities to follow that same pattern, one that belongs to a much larger grouping (82 launches spanning decades), indicates a high degree of non-random, and organized behavior in the data. Therefore, it also indicates a high degree of planning. Consistency, by its very nature, is not random. "Random" means that the data is haphazard and follows no patterns or trends, such as was the case for the sample "coin toss". In the case of missions, sometimes the activities are performed mere minutes apart. In order for something to stay consistent, it must be ordered. If it is ordered, it is not random. That's why I express the "odds against" as "odds against chance." That's because the data is behaving in too consistent a way, for too long a time, for it to be the result of random processes.
Once again, I draw the reader's attention to the fact that, in the case of Apollo mission events, it is even less likely that these events such as spacecraft docking and manuevering should all be tied in with the position of stars in the skies of the Moon, or the skies of Earth -- because there is no "weather" to be concerned about in space. For example, how could docking and manuevering a spacecraft in the vacuum of space, be in any way dependent upon the position of stars in the skies of a planet?
More "Non-Random" Trends
Additionally, in the set of 82 launches I noticed a Sirius, Alnitak and Mintaka preference over other stars. This means that the three stars Sirius, Alnitak and Mintaka appear more often than the other stars I analyzed (in this case, Regulus and one other Orion belt star, Alnilam). I noted this and calculated the odds for it, which come to 380 to 1 against chance for the set of 82 launches (not counting Cape Canaveral data). However, the trend does not stop there.
Magically, the Sirius, Alnitak and Mintaka trend shows up in the Apollo mission activities data as well. The odds came to 17 to 1 against chance, indicating that the data favors Sirius, Alnitak and Mintaka.
Why is this significant, since the odds are lower? It is significant because the trend is the same for both the daily mission activities and the 82 launches. This ties the day-to-day mission activities in with the launch pattern in yet another way! And because of this, it makes the data yet more consistent and less random.
To express this in equations, it's necessary to multiply probabilities again:
P(trends+pattern) = P(large pattern) x P(large trend) x P(small pattern) x P(small trend)
Which yields odds of 53.5 billion to 1 against!
What do these high odds mean? For one thing, that every time I consider one more part of this picture, explaining the whole thing as being a result of "random chance" becomes more and more improbable. Also, because I'm dealing here with many samples of mission activities and launches, the message becomes all the more powerful. The high odds against express the improbability that these NASA launches I examined, plus all the Apollo lunar landing missions, all conform to random chance. These are odds against not just one mission, not just one alignment, but all those pre-Apollo launches and all the mission activities listed in the NSSDC summaries for the six Apollos I studied, behaving in the same way! The likelihood is billions to 1 against! Also, these numbers are not just saying how unlikely it is that one alignment could happen. They are expressing how unlikely it is that this much conformation to the Hoagland/Bara ritual pattern could happen by accident, in the case of the 82 launches and the Apollo lunar landing mission activities. Therefore, I must conclude that the star alignments for the mission activities and launches I studied do not happen by accident ... they must happen by design. To try and explain them via random processes results in odds of billions to one. I would not bet on the "random" side of these kind of odds ... and who would, if the odds were billions to one?
Why the Sirius, Alnitak and Mintaka Trend?
I can only speculate, but I will mention some possible connections here. There is a connection between Cape Canaveral and Egypt that goes beyond Cape Canaveral's translated "Spanish" name, which means "cape of reeds" (corresponding, perhaps, to the Egyptian "field of reeds" or the afterlife). When Mintaka is 33 degrees below the horizon at Cape Canaveral, Sirius is 33 degrees above the horizon in Giza, Egypt. Also, when Mintaka is within a degree of the meridian at Cape Canaveral, it is also at 19 degrees in Giza (depending on whether the Nadir or "Midheaven" meridian is utilized). When Alnitak is 33 degrees below the horizon at Giza, Sirius is at 19 deg 50 min below the horizon at the Cape. And so on.
I have not performed a study to be certain that Mintaka, Alnitak and Sirius have the most such alignments. So, I can only note that this is interesting. However, in light of the fact that Cape Canaveral has been and is the launch site of NASA space missions, and the fact that so many other things in this picture tie together, I would not be suprised if the Sirius, Alnitak and Mintaka alignment geometry between Cape Canaveral and Giza stood out over other possible configurations.
Another interesting fact is, the Arabs called the zodiac -- that band of sky through which the Sun and planets travel -- "Al Mintaka al Buruj," the "girdle of the Signs". (Source: Richard Hinckley Allen, "Star Names: Their Lore and Meaning," page 3.) "Mintaka" itself means "belt," and it is a star in Orion's belt. I do not know if this has significance, but I find it interesting that Mintaka is the one star in Orion's belt whose name actually means "belt," and that the entire zodiac was also named "Mintaka." Of course, "belt" is a descriptive term for the ecliptic, but the whole thing being named after "Mintaka," a star in Orion's belt, could convey a significance on Mintaka. It would be interesting to see if other mythological or statistical indicators support Mintaka's importance.
Can the Sirius/Alnitak/Mintaka trend be dismissed based on the recurring alignments at Cape Canaveral and Giza, Egypt? No, because the trend existed even after Cape Canaveral was excluded from the data, and the only remaining locations examined were Giza and lunar landing sites. This means that this improbable trend is either there by accident, which seems increasingly unlikely, or it is there by design. Naturally, since what I'm dealing with in this case is alignments pertaining to the space program, it would be consistent if the symbolism were to be interconnected in this fashion. What is more central to NASA than its Cape Canaveral launch facility?
Apollo 11 was a very special mission, historic and high profile, because it was the very first time humans successfully walked upon the surface of the Moon and returned again to Earth. Because so much preparation went into this, and the majority of the 82 launches of which Apollo is a part were part of that preparation, I thought it wise to examine other aspects of Apollo 11 for correspondences to the star ritual theory. Apollo 11 should have been laced through and through with symbolism, it being part of this pinnacle of achievement that started with earlier NASA preparatory missions.
Below, I made up a table of correspondences between Apollo 11 and the Egyptian/Masonic ritual scenario as described by Richard Hoagland.
Connection to Item "star ritual" theory ----------------------------------- --------------------------- Name of Lunar Module, the "Eagle" Eagle = Phoenix = Osiris* Date of Moon Landing, July 20 Osiris "resurrection" day Time for docking procedure 33 min Time between Moon landing and start of ceremony 33 min Mission duration 195 hours Time after ceremony starts, that Sirius reaches precise tetrahedral angle 14 min (14 = Osiris num.)** 14 = Number of ways to spin 2 Tetrahedra Apollo Program patch Orion constellation*** Astronaut identity 33rd degree Mason The name "Apollo," meaning "Sun God" Horus = Sun god Star alignments (see data page for list)
*See Ben Franklin on the "Great Seal of the United States," and
Egyptian literature on "Osiris"
**See Ancient Egyptian story of "Osiris," killed by his brother and torn into 14 pieces.
***See Ancient Egyptian literature on "Osiris," who "dwells in Orion"
If I am to calculate the improbability of the Apollo 11 "fitting the pattern" in the sense of star alignments, it also makes sense to calculate the improbability of it fitting all the rest of the picture, too, since it has so far demonstrated an amazing consistency throughout. However, the above items are very hard to quantify. There is one calculation that can be done, however.
A ceremony was conducted in the Apollo 11 lunar module on the day of the Moon landing -- July 20, 1969. Precisely at that time, the star of most importance to ancient Egyptians -- Sirius -- exactly reached the tetrahedral angle of 19.471 degrees. Now, without the context of a larger pattern, this could be dismissed . . . but not as easily in this case. So, what is so special about July 20?
I do not wish to reiterate here the research of Richard C. Hoagland and others into this date and its significance, so I will let the reader do this for themselves. A good place to start is with Richard Hoagland's first article on the subject of these star alignments. Suffice to say that, out of 365 days in a year, there is only one Osiris resurrection date, and this date also "happens" to be associated with the tetrahedron, the source of the numbers 19.5 and 33.0, which have demonstrably occurred in a non-random fashion throughout these missions. The odds are against there being as many tetrahedral numbers in the Apollo 11 "mission activities" table as there are, because the odds are against there being as many Apollo 11 alignments as there are. The numbers "19.5" and "33" refer to tetrahedral geometry, and Apollo 11's mission activities have more of these "tetrahedral" alignments than it should -- the odds are 153 to 1 against Apollo 11 having that many tetrahedral correlations. Since the odds support the "tetrahedral" correlation, therefore, it is consistent to look at the probability for the "tetrahedral Osiris" date being the very date on which this occurs.
Therefore, multiplying the probability of the trends within the 82 launches and the Apollo lunar landing missions, by 1/365 for the date July 20, yields the odds of all these events and patterns being unrelated to each other and to Apollo 11's ceremony on that one particular day:
Final Odds, Apollo 11: 19.5 trillion to 1 against chance.
This number expresses the improbability that the Apollo 11 mission could conform to the Hoagland/Bara theory in this way, conform to the rest of the pattern the way it does, and conform to the pattern which governs the 82 launches analyzed in the beginning of this paper.
Figure 15: The sky above Giza, Egypt, at the time of the launch of Gemini 5. This is an example of the constellation Leo, and the star of interest within it, Regulus. Note how the Sun is also within a degree of 33 degrees and is very close to Regulus. I did not include the Sun in my model, but I did observe that the Sun does appear at or near Hoagland/Bara alignment angles quite often. This is consistent with Egyptian mythology, since Osiris' son Horus was a Sun god. Also, each pharaoh was associated with Horus, which is another link of "kingship" with the "Sun" (see below).
Historically, Leo and the Sun have been considered "linked." The Sun is associated with kingliness, power, and authority in ancient texts on the subject; and Leo is also associated with kings, authority, and power; but more specifically, the star Regulus is associated with those same qualities. Leo and the Sun are also linked astrologically, where Leo is considered the zodiac sign over which the Sun presides; which of course, is just one more indication of this historical link. Because the Sun and Leo and so linked, and Regulus (in Leo) is linked with kingship, it is therefore valid to tie them together numerically. An error margin of +-1 degree gives me an "error window" of 2 degrees; therefore, the probability of the Sun being placed within the same degree as Regulus is, to a close approximation, 2/360, which of course gives odds of 179 to 1 against random chance.
This is just one of many correspondences which I have not had time to investigate thoroughly. However, there are so many of these unlikely "coincidental" tie-ins with the Hoagland/Bara star alignment theory that I think it would be worthwhile to investigate these "coincidental" occurrences in more depth.
THE BIG PICTURE; OR, WHAT DOES ALL THIS MEAN?
The significance of these findings is, I have shown there to be a pattern throughout 82 launches that were part of the Apollo preparation phase (and Apollo itself). Additionally, I have shown that the Apollo missions follow this same pattern on a day-to-day, mission activities level, which is even more improbable because of its consistency with the launch data. Furthermore, it is improbable that the frequency of these stellar alignments are tied to weather or lighting conditions, because of the fact that they occur for a variety of mission events, even those that do not require specific lighting or weather conditions.
The high odds express how unlikely it is for 1) the pattern in the 82 launches to occur, 2) the Apollo lunar missions being related to this pattern, and 3) the likelihood of a mission within that Apollo lunar landing grouping to conform to yet another aspect of the Hoagland/Bara star ritual theory at the same time all the rest of this is going on.
I wish to summarize now and state this briefly. I am, as a former Boeing Engineer and computer professional, only one of many who are wondering if there is any truth to this. I started this out as an impartial investigation into the Enterprise Mission data, and found some very startling and significant results. I encourage others out there who want to know more, to do the rest of this analysis . . . to analyze other missions, other launches, and so on, in order to find out if the pattern exists in other programs . . . and what it means.
Mary Anne Weaver (left) and S. Sriram, Ph.D. (right, sitting) at GTE Labs, collaborating on a research project.
Mary Anne Weaver, email address: [email protected]
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By David Cox, Andrew R. Kustin, Claudia Polini, Bernd Ulrich
Ponder a rational projective curve C of measure d over an algebraically closed box kk. There are n homogeneous types g1,...,gn of measure d in B=kk[x,y] which parameterise C in a birational, base aspect unfastened, demeanour. The authors examine the singularities of C by way of learning a Hilbert-Burch matrix f for the row vector [g1,...,gn]. within the ""General Lemma"" the authors use the generalised row beliefs of f to spot the singular issues on C, their multiplicities, the variety of branches at every one singular element, and the multiplicity of every department. enable p be a unique element at the parameterised planar curve C which corresponds to a generalised 0 of f. within the ""Triple Lemma"" the authors supply a matrix f' whose maximal minors parameterise the closure, in P2, of the blow-up at p of C in a neighbourhood of p. The authors observe the overall Lemma to f' for you to find out about the singularities of C within the first neighbourhood of p. If C has even measure d=2c and the multiplicity of C at p is the same as c, then he applies the Triple Lemma back to benefit concerning the singularities of C within the moment neighbourhood of p. think about rational aircraft curves C of even measure d=2c. The authors classify curves in line with the configuration of multiplicity c singularities on or infinitely close to C. There are 7 attainable configurations of such singularities. They classify the Hilbert-Burch matrix which corresponds to every configuration. The research of multiplicity c singularities on, or infinitely close to, a hard and fast rational airplane curve C of measure 2c is resembling the learn of the scheme of generalised zeros of the fastened balanced Hilbert-Burch matrix f for a parameterisation of C
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Secondly, the parameterization determined by ϕ is birational; that is, ϕ ∈ BHd ; thus, ϕ ∈ M Bal ∩ BHd = M . (2) To read the chart of (3), notice that the chart says for example, that there are 2 multiplicity c singularities on the curve Cc:c,c ; one of these singularities ([0 : 0 : 1]) has an infinitely near singularity of multiplicity c and the other singularity ([1 : 0 : 0]) does not have any infinitely near singularities of multiplicity c. (3) We did not include μ2 in the chart of (3) because the intersection DOBal μ2 ∩BHd , which is also called DOμ2 , is empty.
Notice that D i D ∼ = k [u k) Jac(D/k iD k) = = Jac(D i /k ei −1 D iD . i It follows that k ) = 0 ⇐⇒ Jac(D/k k) = D dim D/ Jac(D/k ⇐⇒ the roots of gcd I3 (A) are distinct and k )) = e(D/ Jac(D/k (ei − 1) and this is equal to the degree of deg gcd I3 (A) minus the number of distinct linear factors of gcd I3 (A). 34 3. 22. 19. It follows that the rings k [T dimension and this common dimension is either 0 or 1; furthermore, these two rings have the same multiplicity. The height of I3 (A) is 2 if and only if the gcd of I3 (A) is a unit.
Bal Proof. By the definition of DOBal . 21 shows how the matrices Cg and Ag are obtained from C and A . It suffices to prove the result when ϕ ∈ M Bal . We have recorded the matrices (C , A ), whose entries are linear forms from T ] and k [u u ], and which satisfy T ϕ = Q1 · · · Qμ C and C u T = A T T , for k [T μ = μ(I1 (ϕ )). The set of linearly independent forms Q1 , . . , Qμ from Bc may be extended to a basis Q1 , . . , Qc+1 for Bc . The entries of ρ(c) also form a basis for Bc ; 4. SINGULARITIES OF MULTIPLICITY EQUAL TO DEGREE DIVIDED BY TWO 45 so there is an invertible matrix υ, with entries in k , so that ρ(c) = [Q1 , .
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- Direct methods for sparse linear systems. Computational scientists often encounter problems requiring the solution ... sparse systems of linear equations. Attacking these problems efficiently requires an in-depth knowledge ... programming language, Direct Methods for Sparse Linear Systems equips readers with the working knowledge required...
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- Implementing interior point linear programming methods in the Optimization Subroutine Library. This paper discusses ... implementation of interior point (barrier) methods for linear programming within the framework...
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- objective function and constraints may be linear or nonlinear, or a mixture of both ... nonlinear functions must be smooth. Stable numerical methods are employed throughout. Features include ... basis matrix), automatic scaling of linear contraints, and automatic estimation of some or all gradients...
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- rational expectations models. We describe methods for solving general linear rational expectations models in continuous ... timing with or without exogenous variables. The methods are based on matrix eigenvalue decompositions...
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- gives a survey of interval arithmetic based methods for solving systems of equations and global ... solving linear interval equations, automatic differentiation and code list generation, interval Newton method ... results and it provides techniques to transform linear interval equations into ones which might have...
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- Regression. Estimation and inference methods for models of conditional quantiles: Linear and nonlinear parametric ... quantiles of a univariate response and several methods for handling censored survival data. Portfolio selection...
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- convex approximation concepts. The Convex Linearization method (CONLIN) exhibits many interesting features...
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CC-MAIN-2022-21
| 4,313 | 40 |
https://brainimaginginformatics.com/grey-level-index-forty-years-of-misleading-consistency/5/
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math
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4. Inconsistencies and misleading claims
A. GLI, volume fraction and stereology
First, authors keep claiming that “GLI is a well-established stereological parameter”. In fact, it has very little to do with stereology. Stereology is a science of calculation a three-dimensional value from measurements made on two-dimensional planar sections. For example, the 3-D volume of an object can be determined from the 2-D areas of its plane sections; the length of some morphological structure in 3D can be determined from counting the number of its sections in 2D, etc. The technique for determining volume fraction in 3D from area fraction on infinitely-thin sections of “spatially homogenous” samples (polished surfaces of non-transparent materials) is well known as Delesse’s principle, as it was first described by geologist A. E. Delesse in 1843. So, constantly repeating that “areal fraction is an estimate of the volume density of cell bodies” seems not only misleading, but simply wrong, because area fraction measured in microscopic sections is not equal to volume fraction. Moreover, the same authors (A. Schleicher and K. Zilles) quite clearly demonstrated with A. Wree in 1982 (the paper was described above) the difference between these two values. It is small enough only if the section thickness is between 4.6 and 5.6 μk. The exact quotation from this paper (with my underlining), section “Estimation of volume fraction by GLI measurement”, page 38, is quite clear: “a section thickness from 4.6 μk could be defined (Fig. 4a) as permitting measurement of the GLI representing an estimate of volume fractions of all grisea with a tolerable error of about ± 15%. If another section thickness is used, the measured data have to be corrected with a factor that can be ascertained from Fig. 4a.” The referenced image, taken from 1990’s paper of A. Schlicher and K. Zilles is below (I could not use A.Wree’s paper due to the poor quality of available copy).
Looking at this picture we can see quite clearly that GLI value (~35%) is close enough to the volume density for sections from 4 to 6 mkm. For 20 micron thick sections GLI value is about ~65%, which roughly gives 86% relative error (or from 78% to 390% for different structures, according to Wree et all.). Clearly, it is not a good match to unbiased values obtained from 1-mkm thick sections. In other words, by claiming that “areal fraction is an estimate of the volume density of cell bodies” the authors ignore their own results published earlier, which quite clearly demonstrate the opposite: in 20 micron-thick sections area fraction is VERY POOR estimate of volume density.
Unfortunately, this problem is not just an academic question. Correction for the section thickness might be very important, and the lack of thereof might artificially bias many results, including the increase of inter-subject variability, especially when tissue processing conditions of different brains vary. Differences in postmortem delay, time of fixation and fixative (Bodian, formalin), as, for example is described in paper “Cytoarchitectonic mapping of the human dorsal exastriate cortex” (M. Kujovic, at all, Brain Struct. Funct. 218:157-182, 2013), might have significant effect on variation of morphometric value measured in a section plane, even if the expected section thickness supposed to be the same (in this case – 20 μk ) in all measured brains.
I honestly tried to find out how the abovementioned correction of GLI is done, and was it done at all, looking in all publications of A. Schleicher and K. Zilles that followed 1982’s paper. Finally, in 2000’s paper I found direct admission: “GLI values are biased volume density estimates“. So, later claim by group, that include same authors, that GLI “profiles reflect laminar fluctuations of cells density” (S. Lorenz et all, “Two new cytoarchitectonic areas of human mid-fusiform gyrus”, Cerebral Cortex, 2015, 1-13) is quite misleading again, and contradicts earlier statement that “GLI differs from the estimation of numerical densities (number of objects per volume unit of reference space).”
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CC-MAIN-2023-50
| 4,165 | 6 |
http://forum.roulette30.com/index.php?topic=401.0
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math
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LOSING THE FIRST 2 OR 3 OR 4 SPINS SHOULD NOT HAVE ANY EFFECT IN THE ALREADY ESTABLISHED PROBABILITY OF SERIES.
It still remains in effect for the entire duration of the series. Palestis[size=78%] [/size]
You are partly right and wrong Jim.
Let me explain why,perhaps the probability for hitting 1 out of 5 instead of hitting 1 in 1 it's not the same as you rightfully claimed.This fact can be confirmed by everyone and it's been called binomial probability,there are plenty of binomial calculators on the internet free to calculate any possibility within any amount of trials,whether you call it Black/Red or Heads/Tails or Pass/Don't Pass...etc.
BUT you are mistaken non the less, by avoiding to bet the 2 first bets of the series of,let's say, 5 consecutive bets,you are not avoiding only lost bets but also bets that you could have won,this is what you are missing.
This is crucial when you calculate the totality of the possibilities,for example:
If your success rate to win only once within 5 trials is 96%,this means that you are going to 96 times out of 100,thus you would lose 4 x (5+10+20) = 4 x 35 = -140 BUT your profit won't be 96 x 5 because some wins from the total of 96 would be during the first and second trial which you are observing rather than betting.
To be exact,roughly 74 times out of the 96 winning bets would be wasted opportunities because you pass them as virtual bets,therefore 96 - 74 = 22 x 5 = 110 and 140 - 110 = -30
Because of the variance you could do better or worse than the figures above,what I showed is the average expected values.
The aftermath is that you cannot sidestep probability.
Just the facts.
When we say we will play 100 times, we don't count the times where you win in the first 2 bets. When that happens the system is abandoned. With no loss or gain. It only means lost time.
When we say 100 times we mean the 100 times where the first 2 bets were lost virtually. That means it might take 500 situations to become 100 playable situations
. It takes patience for that to happen.Betting all 5 times to take advantage of any winning opportunities
here is what will happen.
We have to place 5 bets (5+10+20+40+80)= 155 to risk.
At 96% winning rate we expect to win 5x96=480 units.
But the very small 4% losing rate will result in 4x155=620 units lost.
Playing all spins will always result in a loss. A big loss even with an impressive 96% win rate.
We are only concerned with the situation where the bets will be 0+0+5+10+20. If that's not happening we wait till it happens. Y the pressure to bet?
The risk now is 35 units while all probabilities remain the same.
The probability doesn't change in getting 1 win out of 5 trials, if we see it as (A) 5,10,20,40,80 or we see it as (B) 0,0,5,10,20. Probability has to do with number of trials. Not money amounts. The difference is that we are only looking to find a situation (B). And we abandon a situation where a win occurs in the first 2 bets. That's how being patient affects the results. We always wait for the roulette to come to OUR TERMS. We shouldn't follow her terms. That's the secret of success.
Don't forget when you were here, we never lost at Mohegan or Twin River betting the opposite after seeing 6 or more of the same EC. Remember?
Those 6 black on the board and playing red 3 times meant this: 0+0+0+0+0+0+5+10+20.
A 9 bet series where the first 6 were cost free. And I thought you liked this method very much.
And if you have unlimited bank roll you may never lose. Ever.
How many times do you think you will lose in a row, betting 3 red after seeing 6 black?
That means in order to lose, every time you see 6 black, they must become 10 black in a row.
I still haven't seen that happening 3 times in a row, never mind indefinitely.
So with a large bank roll you can never lose.
There is nothing wrong to have $15,000 bank roll and aim to win $300.
Most business have to invest a lot more than $15,000 to make less than $300. Y not in roulette?
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| 3,962 | 32 |
http://www.longwood.edu/events/calendar/?keywords=Colloquium&view=day&month=10&day=08&year=2019
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math
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Tuesday, October 08, 2019 - Tuesday, October 08, 2019
< Previous Day | Next Day >
"What is a Partial Order? Making Use of Mathematical Hierarchies" presented by Michael C. Strayer, Hampden-Sydney College. Designed for a broad audience, Longwood's Math and Computer Science Colloquium Series explores ideas, developments and careers in the fields of mathematics, computer science and mathematics education.
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CC-MAIN-2019-43
| 405 | 3 |
http://umj-old.imath.kiev.ua/article/?lang=en&article=7319
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math
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Analyticity of Higher-Order Moduli of Continuity of Real-Analytic Functions
The Perel'man's result according to which the first modulus of continuity of any real-analytic function f is a function analytic in a certain neighborhood of the origin is generalized to the case of arbitrary moduli of continuity of higher order.
English version (Springer): Ukrainian Mathematical Journal 55 (2003), no. 6, pp 905-920.
Citation Example: Dovgoshei A. A., Potemkina L. L. Analyticity of Higher-Order Moduli of Continuity of Real-Analytic Functions // Ukr. Mat. Zh. - 2003. - 55, № 6. - pp. 750-761.
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CC-MAIN-2021-04
| 591 | 4 |
https://philosophy.stackexchange.com/questions/60599/examples-for-necessary-existence-and-necessary-properties-according-to-current-m
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math
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According to the survey article by Inwagen and Sullivan on metaphysics one subject of contemporary metaphysics are questions on modality.
The authors explain „necessity de re“
as the necessary existence of an object, i.e. it is impossible that the object does not exists
or as the necessary possession of a property, i.e. it is impossible that a given object exists without having the property in question.
Inwagen and Sullivan indicate that the existence of examples for „necessity de re“ are debated. Of course properties which hold because of analyticity are necessary properties, e.g., "The circle is round" or other mathematical properties, which follow just from the definition.
My question: Which examples of necessity de re are proposed by contemporary metaphysicians, and by which arguments do they support their examples?
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CC-MAIN-2024-18
| 839 | 6 |
https://www.hindawi.com/journals/aaa/2013/237418/
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math
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A Global Curvature Pinching Result of the First Eigenvalue of the Laplacian on Riemannian Manifolds
The paper starts with a discussion involving the Sobolev constant on geodesic balls and then follows with a derivation of a lower bound for the first eigenvalue of the Laplacian on manifolds with small negative curvature. The derivation involves Moser iteration.
The Laplacian is one of the most important operator on Riemannian manifolds, and the study of its first eigenvalue is also an interesting subject in the field of geometric analysis. In general, people would like to estimate the first eigenvalue of Laplacian in terms of geometric quantities of the manifolds such as curvature, volume, diameter, and injectivity radius. In this sense, the first interesting result is that of Lichnerowicz and Obata, which proved the following result in : let be an -dimensional compact Riemannian manifold without boundary with , then the first eigenvalue of Laplacian on will satisfy that , and the inequality becomes equality if and only if .
The above result implies that the first eigenvalue of the Laplacian will have a lower bound less than if the Ricci curvature of manifolds involved has a lower bound except on a small part where the Ricci curvature satisfied that . Now a natural question arises: what is the lower bound of the first eigenvalue of Laplacian on such a manifold? In , Petersen and Sprouse gave a lower bound under the assumption that the bad part of the manifolds is small in the sense of -norm, where is a constant larger than half of the dimension of the manifold. In this paper, we are interested in the lower bound of the first eigenvalue under the global pinching of the Ricci curvature and we obtain a universal estimate of this lower bound on a certain class of manifolds.
2. A Sobolev Constant on the Geodesic Ball
The Sobolev inequality is one of the most important tools in geometric analysis, and the Sobolev constant plays an important part in the study of this field. In this section, we will obtain a general Sobolev constant only depending on the dimension of the manifold on the geodesic ball with small radius.
Definition 1. Let be a geodesic ball with radius ; we define the Sobolev constant on it to be the infimum among all the constant such that the inequality holds for all .
Definition 2. Let be a geodesic ball with radius ; we define the isoperimetric constant on it to be the supremum among all the constant such that the inequality holds for all with smooth boundary.
For any fixed point and radius , Croke proves that the equality holds , but one expects the constant to be independent on the location of the point , under some assumptions. In what follows, we will give an upper bound to independent of the point .
Let be an -dimensional Riemannian manifold, is the unit tangent bundle of , and is the canonical projective map. is the normalized geodesic from with the initial velocity . We define some notations as follows:
is the arc length from to the cut locus point along . Consider where is the standard surface measure of the unit sphere, is denoted to be the area of the unit sphere .
Definition 3. Using the Notation above, is called the visibility angle of .
If the manifold has which ensures that any minimal geodesic starting from any point in will reach the boundary before it reaches its cut locus, then the visibility angle of for any point which we denote by satisfies .
Lemma 4. Let be a closed Riemannian manifold with , then for any , the following Sobolev inequality holds on : , where and .
Proof. Croke proved the following inequality :
where , and is just the visibility angle of the domain .
As discussed above, we will have if ; then according to Croke’s inequality, we obtain . The relation between and tells us that , where is a constant only depending on the dimension .
Proposition 5. Let be a closed -dimensional Riemannian manifold with then for all , where is a constant only depending on the dimension .
Proof. Also take the inequality of Croke then the result can easily be derived from the fact that and after we integrate both sides of the inequality.
3. The First Eigenfunction and Eigenvalue
Let be a closed -dimensional Riemannian manifold; suppose that is the first eigenvalue of the Laplacian and is the first eigenfunction. In other words, they will satisfy that . By linearity, we can assume that and for the linearity. For the convenience, we call it the normalized eigenfunction. Next we will study some properties of the normalized eigenfunction and the eigenvalue.
Lemma 6. Let be a closed -dimensional Riemannian manifold with and . Then, a constant can be found such that .
Proof. One of the theorems of Yau and Schoen shows that if , where is the diameter of the manifold and is a constant depending only on .
We will now introduce some notation. Let denote the lowest eigenvalue of the Ricci curvature tensor at . For a function on , we denote . Notice that a Riemannian manifold satisfies if and only if .
The well-known Myers theorem shows that a closed manifold with would have a bounded diameter . In other words, one can deduce that if one has . We will show next a result analogous to the one in which we will use in our estimation of the eigenvalue. The proof follows identically; so it will be omitted (the reader can refer to the aforementioned article).
Lemma 7. Let be a closed -dimensional Riemannian manifold with , then for any , there exists such that if then the diameter will satisfy . In particular, there exists such that if then the diameter will satisfy . This fact, together with the volume comparison theorem, implies that , where is also a constant only dependent of .
Now, we can get a rough lower bound for the first eigenvalue.
Lemma 8. For , let as above and suppose that is a closed manifold with then there exists a constant such that .
Proof. The proof mainly belongs to Li and Yau . Let be the normalized eigenfunction of , set where . Then, we can easily get that
Denote that , and we then have by the Ricci identity on manifolds with : For the term , we have and for the term , we have Therefore, assume to be the maximum of ; then at , we have Therefore,
Denote to be the minimizing unit speed geodesic joining the maximum and minimum points of ; then integrating along , one will get:
Let ; then for any , we have .
Considering the maximum of the right hand and the upper bound of the diameter derived in Lemma 7, we can deduce that a positive constant can be found such that where is the diameter of the manifold.
Corollary 9. If the manifold one discussed satisfies all the conditions in Lemma 8 and its injectivity radius satisfies and if one let to be the normalized eigenfunction, then there exists a constant such that .
Proof. Set in the (13) from above. Then applying Lemma 6, one obtains
Proposition 10. Let be a closed -dimensional Riemannian manifold, the first eigenfunction of the Laplacian, and the corresponding eigenvalue, then holds in the sense of distribution. Moreover, if is compact with boundary, then the same conclusion holds for its Neumann boundary value problem.
Proof. From the definition, we know that holds on . Denote
According to the maximum principle of elliptic equation and the discussion about nodal set and nodal regions in , we can conclude that is a smooth manifold with dimension .
For all , integrating by parts we then have where and denote the outward normal direction with respect to the boundaries of and , respectively. Note that on and on for the definition of and . This completes the proof.
When has boundary, we can apply the same reasoning, except that the test function will require . This gives the proof.
As long as the given manifold is compact, one knows that the first normalized eigenfunction is then determined. This indicates that the first normalized eigenfunction of the Laplacian has a close relation with the geometry of the manifold. In particular, one would hope to bound the -norm of first normalized eigenvalue of Laplacian from below by the geometric quantities. In this sense, we have the following result.
Theorem 11. Let be a closed -dimensional Riemannian manifold with and . If is the normalized eigenfunction of the Laplacian, then there exists a constant such that .
Proof. We use Moser iteration to get the result. From Proposition 10, we know that holds on in the sense of distribution. Set and take the point such that .
For , denote ; is a cut-off function on , then we have by integrating by parts: However, using the identity we have therefore, using the Sobolev inequality in Lemma 4,
Putting into the inequality above and we then have by splitting the integral into three parts and using the values of on each of them: where we denote only for emphasizing the integral domain.
And putting into (25), we can derive after iteration that
Let , then
The product can be estimated as follows:
The right hand will converge to a fixed number by using the fact that and the fact is finite for some . From , we can find a positive constant such that
4. The Lower Bound of the First Eigenvalue
Using the same notation as above, we can state the following result.
Theorem 12. For , , there is an such that any closed manifold with and will satisfy that .
Proof. Assume that is the normalized eigenfunction of Laplacian on , let , direct computation shows that
Integrating both sides on , we have therefore,
if we suppose that If is the one obtained in Lemma 7, then one has:
Finally, if one chooses then as long as , and this proves the theorem.
The authors owe great thanks to the referees for their careful efforts to make the paper clearer. Research of the first author was supported by STPF of University (no. J11LA05), NSFC (no. ZR2012AM010), the Postdoctoral Fund (no. 201203030) of Shandong Province, and Postdoctoral Fund (no. 2012M521302) of China. Part of this work was done while the first author was staying at his postdoctoral mobile research station of QFNU.
S. T. Yau and R. Schoen, Lectures in Differential Geometry, Scientific Press, 1988.
P. Petersen and C. Sprouse, “Integral curvature bounds, distance estimates and applications,” Journal of Differential Geometry, vol. 50, no. 2, pp. 269–298, 1998.View at: Google Scholar | Zentralblatt MATH | MathSciNet
D. Yang, “Convergence of Riemannian manifolds with integral bounds on curvature. I,” Annales Scientifiques de l'École Normale Supérieure. Quatrième Série, vol. 25, no. 1, pp. 77–105, 1992.View at: Google Scholar | Zentralblatt MATH | MathSciNet
I. Chavel, Riemannian Geometry: A Mordern Introduction, Cambridge University Press, Cambridge, UK, 2000.View at: Publisher Site | MathSciNet
C. Sprouse, “Integral curvature bounds and bounded diameter,” Communications in Analysis and Geometry, vol. 8, no. 3, pp. 531–543, 2000.View at: Google Scholar | Zentralblatt MATH | MathSciNet
P. Li and S. T. Yau, “Estimates of eigenvalues of a compact Riemannian manifold,” in AMS Proceedings of Symposia in Pure Mathematics, pp. 205–239, American Mathematical Society, Providence, RI, USA, 1980.View at: Google Scholar | Zentralblatt MATH | MathSciNet
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CC-MAIN-2023-23
| 11,155 | 63 |
https://brainmass.com/math/calculus-and-analysis/horse-velocity-derivatives-19331
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math
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Horse Velocity Using Derivatives
This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!
The problem is in JPEG, thank you.
Quarter horses race a distance of 440 yards (a quarter mile) in a straight line. During a race the following observations where made. The top line gives the time in seconds since the race began and the bottom line gives the distance (in yards) the horse has traveled from the starting line.
1. How fast is the horse running halfway through the race?
2. Find a formula for the velocity of the horse.
3. The horse will win a bonus if the time for the race is less than 22 seconds. Decide whether you think the horse will win the bonus. Explain your reasons.
Hi, this is a great question. Start out by making your own plot of the data. I did not include a plot, because I think it is beneficial for you to do this step. If you plot the given data (distance vs. time), you will see that it is pretty much a line. The general rule is that if a distance vs. time plot is a straight positive sloping line, then the corresponding velocity is positive and constant. Therefore, the velocity in the middle can be calculated by taking the difference of two of the distances and dividing it by the corresponding time points in ...
The expert examines horse velocity using derivatives. How fast the horse is running halfway through the race is given.
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CC-MAIN-2023-23
| 1,413 | 9 |
https://projecteuclid.org/journals/journal-of-differential-geometry/volume-120/issue-3/Minimal-planes-in-asymptotically-flat-three-manifolds/10.4310/jdg/1649953568.short
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math
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In this paper, we improve a result by Chodosh and Ketover . We prove that, in an asymptotically flat $3$-manifold $M$ that contains no closed minimal surfaces, fixing $q \in M$ and $V$ a $2$-plane in $T_q M$ there is a properly embedded minimal plane $\Sigma$ in $M$ such that $q \in \Sigma$ and $T_q \Sigma = V$. We also prove that fixing three points in $M$ there is a properly embedded minimal plane passing through these three points.
"Minimal planes in asymptotically flat three-manifolds." J. Differential Geom. 120 (3) 533 - 556, March 2022. https://doi.org/10.4310/jdg/1649953568
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CC-MAIN-2022-40
| 587 | 2 |
https://www.bbc.co.uk/bitesize/guides/zyjk3k7/test
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math
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The mole is the unit for amount of substance. The number of particles in a substance can be found using the Avogadro constant. The mass of product depends upon the mass of limiting reactant.
Which of these is the best definition of a mole?
The relative atomic mass in grams of the substance
The unit for the amount of substance
The number of particles in the relative formula mass in grams of a substance
What is the definition of the Avogadro constant?
6.02 × 1023
The number of particles in one mole of a substance
What is the number of molecules in 23 g of NO2? (Mr of NO2 = 46)
1.204 × 1024
3.01 × 1023
What is the mass of 0.5 mol of HF gas? (Mr of HF = 20)
What is the amount of silver atoms in 216 g of silver, Ag? (Ar of Ag = 108)
What amount of O2 reacts with 4 mol of Al?
4Al + 3O2 → 2Al2O3
What mass of Al reacts with 96 g of oxygen? Mr of O2 = 32 and Ar of Al = 27.
What is a limiting reactant?
The reactant that is present in the smallest mass
The reactant that is left over at the end of the reaction
The reactant that is all used up in a reaction
What is the concentration of a solution formed by dissolving 2 mol of nitric acid in 4 dm3 of solution?
What is the volume of 250 cm3 of solution in dm3?
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s3://commoncrawl/crawl-data/CC-MAIN-2019-39/segments/1568514574182.31/warc/CC-MAIN-20190921022342-20190921044342-00093.warc.gz
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CC-MAIN-2019-39
| 1,219 | 22 |
https://ijnaa.semnan.ac.ir/article_6838.html
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math
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M. Abbas and T. Nazir, A new faster iteration process applied to constrained minimization and feasibility problems, Mat. Vesnik. 66 (2014), no. 2, 223–234.
A. Abkar and M. Eslamian, A fixed point theorem for generalized nonexpansive multivalued mappings, Fixed Point Theory 12 (2011), no. 2, 241–246.
R.P. Agarwal, D. Oregan and D.R. Sahu, Iterative construction of fixed points of nearly asymptotically nonexpansive mappings, J. Nonlinear Convex Anal. 8 (2007), no. 1, 61–79.
D. Ariza-Ruiz, C. Hermandez Linares, E. Llorens-Fuster and E. Moreno-Galvez, On α-nonexpansive mappings in Banach spaces, Carpath. J. Math. 32 (2016), 13–28.
K. Aoyama and F. Kohsaka, Fixed point theorem for α-nonexpansive mappings in Banach spaces, Nonlinear Anal. 74 (2011), 4387–4391.
K. Aoyama, S. Iemoto, F. Kohsaka and W. Takahashi, Fixed point and ergodic theorems for λ-hybrid mappings in Hilbert spaces, J. Nonlinear Convex Anal. 11 (2010), 335–343.
M. Bachar and M.A. Khamsi, On common approximate fixed points of monotone nonexpansive semigroups in Banach spaces, Fixed Point Theory Appl. 2015 (2015), 160.
V. Berinde, Generalized contractions and applications (Romanian), Editura Cub Press 22, Baia Mare 1997.
V. Berinde, Picard iteration converges faster than Mann iteration for a class of quasicontractive operators, Fixed Point Theory Appl. 2 (2004), 97–105.
B.A.B. Dehaish and M.A. Khamsi, Mann iteration process for monotone nonexpansive mappings, Fixed Point Theory Appl. 2015 (2015), 177.
J. Garc´ea-Falset, E. Llorens-Fuster and E. Moreno-G`a ´alvez, Fixed point theory for multivalued generalized nonexpansive mappings, Appl. Anal. Discrete Math. 6 (2012), 265–286.
K. Goebel and W.A. Kirk, Topics in Metric Fixed Point Theory, in: Cambridge Studies in Advanced Mathematics, vol. 28, Cambridge University Press, Cambridge, 1990.
K. Goebel and S. Reich, Uniform Convexity, Hyperbolic Geometry, and Nonexpansive Mappings, Monographs and Textbooks in Pure and Applied Mathematics, vol. 83, Marcel Dekker Inc., New York, 1984.
M.A. Harder, Fixed point theory and stability results for fixed point iteration procedures, PhD thesis, University of Missouri-Rolla, Missouri, 1988.
T.L. Hicks and J.R. Kubicek, On the Mann iteration process in Hilbert space, J. Math. Anal. Appl. 59 (1977), 498–504.
N. Hussain, K. Ullah and M. Arshad, Fixed point approximation for Suzuki generalized nonexpansive mappings via new iteration process, J. Nonlinear and Convex Anal. 19 (2018), no. 8, 1383–1393.
H. Iqbal, M. Abbas and S.M. Husnine, Existence and approximation of fixed points of multivalued generalized α-nonexpansive mappings in Banach spaces, Numer. Algor. 85 (2020), no. 3, 1029–1049.
S. Ishikawa, Fixed point by a new iteration method, Proc. Amer. Math. Soc. 4 (1974), no. 1, 147–150.
W.A. Kirk, A fixed point theorem for mappings which do not increase distances, Amer. Math. Month. 72 (1965), 1004–1006.
S. Maldar, F. G¨ursoy, Y. Atalan and M. Abbas, On a three-step iteration process for multivalued Reich-Suzuki type α-nonexpansive and contractive mappings, J. Appl. Math. Comput. 68 (2022), no. 2, 863–883.
W.R. Mann, Mean value methods in iteration, Proc. Am. Math. Soc. 4 (1953), 506–510.
E. Naraghirad, N.C. Wong and J.C. Yao, Approximating fixed points of α-nonexpansive mappings in uniformly convex Banach spaces and CAT(0) spaces, Fixed Point Theory Appl. 2013 (2013), 57.
M.A. Noor, New approximation schemes for general variational inequalities, J. Math. Anal. Appl. 251 (2000), 217–229.
Z. Opial, Weak convergence of the sequence of successive approximations for nonexpansive mappings, Bull. Amer. Math. Soc. 73 (1967), 591–597.
R. Pant and R. Shukla, Approximating fixed points of generalized α-nonexpansive mapping in Banach spaces, Numer. Funct. Anal. Optim. 38 (2017) 248–266.
H. Piri, B. Daraby, S. Rahrovi and M. Ghasemi, Approximating fixed points of generalized α-nonexpansive mappings in Banach spaces by new faster iteration process, Numer. Algor. 81 (2019), 1129—1148.
J. Schu, Weak and strong convergence to fixed points of asymptotically nonexpansive mappings, Bull. Aust. Math. Soc. 43 (1991), no. 1, 153–159.
N. Shahzad and H. Zegeye, On Mann and Ishikawa iteration schemes for multi-valued maps in Banach spaces, Nonlinear Anal. 71 (2009), no. 3-4, 838–844.
R. Shukla, R. Pant and M. De la Sen, Generalized α-nonexpansive mappings in Banach spaces, Fixed Point Theory and Appl. 2017 (2016), no. 1, 1–4.
S.M. Soltuz and T. Grosan, Data dependence for Ishikawa iteration when dealing with contractive like operators, Fixed Point Theory Appl. 2008 (2008), 1–7.
Y.S. Song, K. Promluang, P. Kumam and Y.J. Cho, Some convergence theorems of the Mann iteration for monotone α-nonexpansive mappings, Appl. Math. Comput. 287-288 (2016), 74–82.
T. Suzuki, Fixed point theorems and convergence theorems for some generalized nonexpansive mappings, J. Math. Anal. Appl. 340 (2008), no. 2, 1088–1095.
B.S. Thakur, D. Thakur and M. Postolache, A new iterative scheme for numerical reckoning fixed points of Suzuki’s generalized nonexpansive mappings, Appl. Math. Comput. 275 (2016), 147–155.
U.E. Udofia and D.I. Igbokwe, Convergence theorems for monotone generalized α-nonexpansive mappings in ordered banach space by a new four-step iteration process with application, Commun. Nonlinear Anal. 9 (2020), no. 2, 1 17.
U.E. Udofia and D.I. Igbokwe, A novel iterative algorithm with application to fractional differential equation, preprint.
K. Ullah, J. Ahmad and M, de la Sen, On generalized nonexpansive maps in Banach spaces, Comput. 8 (2020), no. 3, 61.
K. Ullah and M. Arshad, New iteration process and numerical reckoning fixed points in Banach spaces, U.P.B.S. Bull. Series A 79 (2017), 113–122.
K. Ullah and M. Arshad, Numerical reckoning fixed points for Suzuki generalized nonexpansive mappings via new iteration process, Filomat, 32 (2018), 187–196.
X. Weng, Fixed point iteration for local strictly pseudocontractive mapping, Proc. Amer. Math. Soc. 113 (1991), 727–731.
H.K. Xu, Inequality in Banach spaces with applications, Nonlinear Anal. 16 (1991), 1127–1138.
H. Xu, Iterative methods for the split feasibility problem in infnite-dimensional Hilbert spaces, Inverse Probl. 26 (2010), 17.
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https://studyib.net/physics/page/444/oscillations-and-waves
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math
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In this topic we look at:
- oscillations of bodies that move back and forth periodically
- progressive waves that transfer energy and information through space
- the interference effects of these progressive waves, and other phenomena
- standing waves, where waves combine to become stationary (and therefore transferring no energy at all)
You will learn about simple harmonic motion, dispersion and guitar string frequencies, and be able to answer the following questions:
- How does a force that is proportional to displacement result in an oscillation?
- Are all oscillations simple harmonic?
- What is the connection between oscillations and waves?
- Are waves really made of an infinite number of wavelets?
Another physics book classic, the mass hanging on a spring. A bit easier to analyse than a pendulum but not as easy as the mass on a spring in space. Quite obvious that the acceleration is proportional to displacement, once you realise that SHM follows.
It is easy to observe the wave motion in a string but it doesn´t tell the whole story of wave properties. A string wave can't diffract or interfere. Sometimes simple explanations cause misunderstandings. Here a string wave is polarised by a narrow slit.
We represent a sound wave by drawing lines like the coils of a slink spring but we should remember that although layers of air oscillate the individual atoms do not.
Since there is no energy transfer the pendulum wave isn't really a wave but it looks impressive. Waves motion can be represented by graphs but be careful, a graph is a graph not a picture of the wave.
A juggler may not understand the mathematical representation of phase but the are using the effect as they throw balls in the air at different times. It is important to understand the concept of phase before starting the waves section.
Wave simulations use the same mathematical equations that are used to model real waves so have the same properties. Here a ripple tank simulation will be used to show how waves reflect, refract, interfere and diffract.
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http://www.ivorcatt.co.uk/95.htm
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math
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Professor Walter H G Lewin
In his lecture, at 3 minutes into the lecture, Lewin shows the illegal current spreading out across the capacitor plate. At 11 minutes 20, the illegal sideways current has disappeared.
If any professor mentions this current, he loses his job. "Modern Physics" is dogma, not science. Lewin mentioned it, but glossed it over. He was later disgraced for other reasons.
Ivor Catt 23 December 20167
Dear Professor Walter H G Lewin,
In your MIT lecture 18 at http://videolectures.net/mit802s02_lewin_lec18/ [at 5 minutes into it] you drew lines showing electric current spreading out over the capacitor plate. That is, you mentioned that electric current flows across the capacitor plate, at right angles to the main flow. However, this was just in passing, and you did not discuss the magnetic field which must result in the horizontal direction, using Ampere's Law, or its effect on the use of Kirchhoff's First Law at the entry point into the capacitor - see Wikipedia; "At any point in an electrical circuit that does not represent a capacitor plate, the sum of currents flowing towards that point is equal to the sum of currents flowing away from that point.".
This lateral flow of electric current had been overlooked for over a century, until I pointed it out in 1978 at http://www.ivorcatt.org/icrwiworld78dec1.htm
. Please comment on the implications of this current as discussed by me in 1978. I also treat it at http://www.ivorcatt.com/411.htm
In your lecture, you drew parallel lines indicating that the electric field in the capacitor is uniform. http://www.electromagnetism.demon.co.uk/3615.htm
http://www.ivorcatt.org/icz014.htm . Please comment on my comments here.
Ivor Catt.14 May 2009 21:05
Reply within half an hour.
----- Original Message -----
Kirchhoff's current rule does not apply for a given point on the capacitor
plates. At one point in time charges can be flowing to a point and at another
point in time charge can be flowing away from that point.
I'm afraid not.
Ivor Catt. 14 May 2009 21.52
28 May 2009
Dear Professor Lewin,
. Please comment on my comments here.
Dear Professor Lewin,
I look forward to your reply to my second email, dated 28 May 2009
Ivor Catt 9 June 2009
4 January 2010. Still no reply. But on the www it says that he replies to all his emails. His Wikipedia entry says; “Lewin personally responded to hundreds of e-mail requests that he received per year from UWTV viewers”. – Ivor Catt.
https://www.youtube.com/watch?v=8ZYFYUFRblM Now, at 18 minutes, the current clearly travels sideways. More clearly than in his previous lecture. Here we see a shambolic attempt to deal with my email to him asking him about the sideways current, mentioned in 1978 and ignored, by him and by everyone else. http://www.ivorcatt.org/icrwiworld78dec1.htm Then here is his attempt to handle it, 30 or more years later.
See Heaviside here; http://www.ivorcatt.co.uk/x857.htm
“If you have got anything new …. “
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https://www.catalpatradingcompany.com/products/giant-telesphere
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math
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On your table there are three transparent tubes, each about 10.25 inches (26cm high), which have a small, stable base. In the center tube there are three large yellow balls, about 2 inches (6cm) in diameter, that are clearly visible from a distance. One ball is put in each of the tubes. The two outer tubes are covered with elegantly decorated, red/silver sleeves.
The balls migrate from the left to the right tube, then from the right to the left, and back again. At the end of the first sequence all three balls are in the center tube – as they were at the beginning.
The balls are again put into each tube, and they penetrate the closed bases at the ends of the tubes when they are stacked on top of one another. The climax is reached when the magician makes two balls disappear and immediately finds them again together with the third ball in the center tube.
You can, of course, show the sleeves empty from time to time and at the end.
You could say that Giant Telesphere is a large scale cup routine with transparent tubes. But it is visible from a distance, can be performed in front of a small audience and while surrounded, and requires absolutely no sleight of hand.
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https://opentuition.com/topic/begging-for-help-financial-instruments/
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math
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September 17, 2013 at 10:05 pm #140709perfecta1Member
Please, please,please could someone help me to understand how to do questions about financial instruments.
This is question from 2008 exam and I really don’t get it
Pingway issued a $10 million 3% convertible loan note at par on 1 April 2007 with interest payable annually in
arrears. Three years later, on 31 March 2010, the loan note is convertible into equity shares on the basis of $100 of
loan note for 25 equity shares or it may be redeemed at par in cash at the option of the loan note holder. One of the
company’s financial assistants observed that the use of a convertible loan note was preferable to a non-convertible
loan note as the latter would have required an interest rate of 8% in order to make it attractive to investors. The
assistant has also commented that the use of a convertible loan note will improve the profit as a result of lower interest
costs and, as it is likely that the loan note holders will choose the equity option, the loan note can be classified as
equity which will improve the company’s high gearing position.
The present value of $1 receivable at the end of the year, based on discount rates of 3% and 8% can be taken as:
End of year 1 0·97 0·93
2 0·94 0·86
3 0·92 0·79
show how the convertible loan note should be accounted for in Pingway’s income statement for the year ended 31 March 2008 and statement of financial position as at that date
could someone explain me please what means ‘interest rate of 8% in order to make it attractive to investors’ does it mean that the interest rate has been increased and at the end of period and 8% interest apply not 3%? Also this 3% interest is on value of loan so we will have to pay this interest every year from the amount that left to be repaid? ?
I’ve read technical article about financial instruments and itwas extremely helpful but still would like someone to go through this question with me. Thank you very much for help
September 18, 2013 at 5:52 am #140730MikeLittleKeymaster
8% is the effective rate ie the rate that would have to be paid if the loan note did not have the equity option. So 8% is the finance charge which should be reflected in the Statement of Income. That thus negates the statement from the assistant about profits being “higher because the interest amount is lower”
The instrument you have described is called a “mixed” instrument and has elements of both loan and equity. The accounting treatment is to value both elements and treat them accordingly. However, it’s extremely unlikely that anyone could value the equity element so the technique is to compute the present value of ALL payments associated with the loan (interest and capital repayment), deduct that amount from the face value of the loan and the difference will be treated as equity
So now we have Dr Cash Credit Loan Account with the present value of the loan (using 3% to calculate the annual interest payment) and Credit Other Components of Equity Account with the missing amount.
As each year goes by, calculate 8% on the outstanding amount of the loan and that is the finance charge for the Statement of Income
3% of the face value is to be paid in cash as loan interest. The difference between 8% of the brought forward obligation and the 3% interest payment is added to the obligation and carried forward into the next year
So, Dr Finance Charges Cr Cash with the 3% interest payment and
Dr Finance Charges Cr Loan Account with the difference between 8% of the brought forward obligation less 3% of the face value of the loan
September 19, 2013 at 8:35 pm #140888perfecta1Member
first of all sorry to b complete pest. I am still sitting and trying to understand the whole thing about convertible loans and Itried to focus on accounting entries first and then understand why accountant’s assumptions were incorrect.
I know how to calculate liability part and equity part in this question.
they are 8674 and 1326 respectively. Also i know why we have to charge 693290 finance cost in statement of profit and loss. but all this is in first year.
Could you tell me what will be the charge in finance costs and ammounts in SOFP in second year. I think that this would be:
CR equity 1326 (remains unchanged)
CR liabilities (9067.92 = 8674×1.08-10000×0.03)
and finance costs 1025.434 (9067.92x 0.08 +10000x 0.03)
and in 3rd year this would be
CR equity 1326
cr liability (9493*1.08-300)= 9952.44
and finance costs would be 9952.22*0.08 +300= 1096.2
I didn’t find any question prctice where it is required to calculate finance ocsts in 2nd and 3rd year hence my question.
Also is that true that after this 3 years period Following accounting entry will be required to account for the conversion of loan into shares after three years:
and this would be
DR laibility 10000
DR share option 1326
CR company shares 10000
CR share premium 1326
or this accounting is for bonds on maturity only?
September 20, 2013 at 7:11 pm #140935MikeLittleKeymaster
The second year’s finance charge would be 8% x 8,764
The third year finance charge would be 8% x (1.08% x 8,764) (I don’t have a calculator with me!)
The transfer of 1,326 seems ok to me
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http://thesis.visit-now.net/lift-of-a-flat-sleonhard-euler-an-introductionurface-in-wind/
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math
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Lift of a Flat Surface in Wind
Born in Basel, Switzerland on 15th April 1707, Leonhard Euler was arguably the brightest mathematician of all time. The Swiss mathematician and physicist is considered a pioneer in many fields of mathematics. He introduced a lot of the mathematical terminology and notation used today and he is considered the father of mathematical analysis where, for instance, he introduced the notation of a mathematical function, f(x). His contributions to the field of mathematics are in analytic and differential geometry, calculus, the calculus of variation, differential equations, series and the theory of numbers. In physics, although really all his contributions to mathematics apply to physics, he introduced both rigid body mechanics and analytical mechanics (Kline, 401-402).
If you need assistance with writing your essay, our professional essay writing service is here to help!
Born to Paul Euler and Margarete Bruckner, Leonhard was the first of six children. He grew up in Riehen but attended school in Basel. Although mathematics was not taught in his school his father had kindled his interest in the subject (Paul Euler had been friends with another great mathematician at the time, Johann Bernoulli) by giving him lessons at home. Euler entered university at the age of 13 at the University of Basel. Although his official courses of study were philosophy and law, Euler met with Johann Bernoulli who advised Euler and gave him help with his mathematical studies on Saturday afternoons (Stillwell, 188).
Euler lived and worked mainly in Russia and Germany. First he joined the faculty at St Petersburg Academy of Sciences were he worked at first in the medical department then he was quickly promoted to a senior position in the department of mathematics through the influence of his friend Daniel Bernoulli. He also helped the Russian government on many projects including serving in the Russian navy as a medical Lieutenant. After the death of Catherine I in1740 and because of the tough conditions that ensued, Euler moved to Germany at the invitation of Prussian King, Frederick II to the Prussian Academy of Science were he stayed for the next 25 years of his life. Euler gave much service to the Academy which compensated him generously. He sent most of his works to be published there, served as a representative as well as advising the Academy on its many scientific activities. It is there that he reached the peak of his career writing about 225 memoirs on almost every topic in physics and mathematics (Varadarajan, 11).
Euler returned to St. Petersburg in 1766 under the invitation of the then czarina Catherine the Great (Catherine II) to the St Petersburg Academy. During this period he lost almost all his eyesight through a series of illnesses becoming nearly totally blind by 1771. Nevertheless, his remarkable memory saw him writing about 400 memoirs during this time. It is said that he had a large slate board fixed to his desk where he wrote in large letters so that he could view dimly what was being written. He died on the 18th day of September 1783 due to cerebral hemorrhaging. It is also recorded that he was working even to his last breath; calculations of the height of flight of a hot air balloon were found on his board (Varadarajan 13).
Euler’s contribution to Mathematics and Physics was a lot. His ideas in analysis led to many advances in the field. Euler is famously known for the development of function expressions like the addition of terms, proving the power series expansion, the inverse tangent function and the number e:
∑ (xn/n!)= lim ((1/0!) + (x/1!) + (x2/2!) +…+ (xn/n!) ) =ex
The power series equation in fact helped him solve the famous 1735 Basel problem:
∑ (1/n2) = lim ((1/12) + (1/22) + (1/32) + …+(1/n2)) = π2/6
He introduced the exponential function, e, and used it plus logarithms in analytic proofs. He also defined the complex exponential function and a special case now known as the Euler’s Identity:
eiφ = cos φ + isin φ
eiπ + 1= 0 (Euler’s Identity)
In fact, De Moivre’s formula for complex functions is derived from Euler’s formula. Similarly, De Moivre is recognized for the development of calculus of variations, formulating the Euler-Lagrange equation. He was also the first to use solve problems of number theory using methods of analysis. Thus, he pioneered the theories of hyperbolic trigonometric functions, hyper geometric series, the analytic theory of continued fractions and the q-series. In fact, his work in this field led to the progress of the prime number theorem (Dunham 81).
The most prominent notation introduced by Euler is f(x) to denote the function, f that maps the variable x. In fact he is the one who introduced the notion of a function to the field of mathematics. He introduced, amongst others, the letter ∑ to mean the sum, π for the proportion regarding the perimeter of a circle up to the span or the diameter, i for the imaginary unit, √(-1) and the e (2.142…) to represent the base of the natural logarithm.
Euler also contributed to Applied Mathematics. Interestingly enough, he developed some Mathematics applications into music by which he hoped to incorporate musical theory in mathematics. This was however, not successful. This notwithstanding, Euler did solve real-world problems by applying analytical techniques. For instance, Euler incorporated the Method of Fluxions which was developed by Newton together with Leibniz’s differential calculus to develop tools that eased the application of calculus in physical problems. He is remembered for improving and furthering the numerical approximation of integrals, even coming up with the Euler approximations. More broadly, he helped to describe many applications of the constants π and e, Euler numbers, Bernoulli numbers and Venn diagrams.
The Euler-Bernoulli beam equation (one of the most fundamental equations in engineering) is just one of the contributions of the mathematician to physics. He used his analytical skills in classical mechanics and used the same methods in solving celestial problems. He determined the orbits of celestial bodies and calculated the parallax of the sun. He differed with Newton (then the authority in physics) on his corpuscular theory of light. He supported the wave theory of light proposed by Hugens.
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Euler’s contributions to graph theory are at the heart of the field of topology. He is famously known to have solved the Seven Bridges of Konigsberg problem, the solution of which is considered the first theorem of planar graph theory. He introduced the formula
It is a mathematical formula relating vertices, edges and faces of a planar graph or polyhedron. The constant in the above formula in now called the Euler characteristic.
Euler is also recognized for the use of closed curves in the provision of explanation concerning reasoning which is of syllogistic nature. Afterwards the illustration or diagrams were referred to as the Euler diagrams.
The Number Theory is perhaps the most difficult branch of mathematics. Euler used ideas in analysis linking them with the nature of prime numbers to provide evidence that the total of all the reciprocals of prime numbers diverges. He also discovered the link between the primes Riemann zeta function, what is now called the Euler product formula for the Riemann zeta function. Euler made great strides in the Lagrange four-square theorem while proving Fermat’s theorem on the sum of two squares, Fermat’s Identities and Newton’s identities. Number theory consists of several divisions which include the following: Algebraic Number Theory, Combinational Number theory, Analytic Number Theory, Transcendental number theory, Geometric number theory and lastly we have the Computational Number Theory.
For his numerous contributions to academia, Euler won numerous awards. He won the Paris Academy Prize twelve times over the course of his career. He was elected as a foreign member, in 1755, of the Royal Swedish Academy of Sciences while his image has been featured on many Russian, Swiss and German postage stamps. Above all, he was respected greatly amongst his academic peers demonstrated by a statement made by the great French mathematician, Laplace to his students to read Euler since he was the master of them all (Dunham xiii).
Though not all of the proofs of Euler are satisfactory in regard to the current standards or principles used in mathematics, the ideas created by him are of great importance. They have set a path to the current mathematical advancements.
To conclude, we can therefore say that Euler is a very significant person in the development and advancement of Mathematics. His work has contributed a lot to mathematics up to the current period.
Dunham, William. Euler: The Master of Us All. Dolciani Mathemathical Expositions Vol. 22. MAAA, 1999.
Kline, Morris. Mathematical Thoughts from Ancient to Modern Times, Vol 2. New York: Oxford University Press, 1972.
Stillwell, John. Mathematics and its History. Undergraduate Texts in Mathematics. Springer, 2002.
Varadarajan, V. S. Euler Through Time: A New Look at Old Themes. AMS Bookstore, 2006.
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http://wecareantalya.com/uqx5/p7xd.php?kmsj=5&pgjkpzj=176&wp883=advanced-trigonometry-problems-with-solutions
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math
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Advanced trigonometry problems with solutions
Advanced trigonometry problems with solutions
This book is the first of a series covering the math concepts tested on the ACT. to do the problems. concept of advanced trigonometry i problem based advanced trigonometry for ssc cgl i problem based on advanced trigonometry for ssc cgl chsl i problem based on advanced trigonometry i shortcuts Precalculus Problems Website (The development of this website was supported by a UIIP grant from the Teaching Resources Center at the University of California, Davis. Quiz.
After completing the problems you can check your work by viewing the Video Playlist of problems linked in the table. Assuming each angle given is in standard position; find the quadrant of its terminal side. This book offers a comprehensive overview of the trigonometric functions and contains a collection of 115 carefully selected introductory and advanced problems in trigonometry from world-wide renowned Olympiads and mathematical magazines, as well as original problems designed by the authors.
The study of calculus is no longer limited to those preparing for careers in mathematics and the sciences. A. This page has trigonometry rules, with sin, cos and tan triangle formulas, and problems for the exam.
A (Note: If you need a little bit of review on the more advanced trigonometry problems — like #7 — you may want to read this post. MATHEMATICS FOR ENGINEERING TRIGONOMETRY TUTORIAL 1 – TRIGONOMETRIC RATIOS, TRIGONOMETRIC TECHNIQUES AND GRAPHICAL METHODS This is the one of a series of basic tutorials in mathematics aimed at beginners or anyone wanting to refresh themselves on fundamentals. 2) 256.
This session will mainly be focused around the revision of Trigonometry JEE concepts – Trigonometric Ratios and trigonometry Problems with solutions that are very crucial to revise in the Problems Odd-numbered pages Solutions Even-numbered pages Each problem has been given a title (not used later on), to remind you what it is about if you have already looked at it, followed by a rough indication of the mathematical content in case you want to pick out questions by topic. Download print and enjoy! experiences into an advanced mathematics textbook accessible by and interesting to a relatively advanced high-school student, without being constrained by the idiosyncracies of the formal IB Further Mathematics curriculum. Sometimes we have indicated these connections.
In this post, we will see the book Solving Problems In Algebra and Trigonometry - V. Trigonometry is included on the advanced algebra and functions part of the Next Generation examination. The above question is one of our trigonometric, logarithmic, or exponential function advanced math problems.
Grade 12 trigonometry problems and questions with answers and solutions are presented. x/ D 10sin. Round measures of lengths to the nearest whole number and angles to the nearest whole degree.
Though the ancient Greeks, such as Hipparchus Introduction to Trigonometry: Trigonometric Functions, Trigonometric Angles, Inverse Trigonometry, Trigonometry Problems, Basic Trigonometry, Applications of Trigonometry, Trigonometry in the Cartesian Plane, Graphs of Trigonometric Functions, and Trigonometric Identities, examples with step by step solutions, Trigonometry Calculator TRIGONOMETRY 2 Trigonometry Self-Paced Review Module As you probably know, trigonometry is just “the measurement of trian-gles”, and that is how it got started, in connection with surveying the Mr. 1. You can answer almost every SAT trig question by using the mnemonic device for the three basic trigonometric ratios: SOH CAH […] Linear Algebra Igor Yanovsky, 2005 2 Disclaimer: This handbook is intended to assist graduate students with qualifying examination preparation.
Integers and percents, simple Algebra notes on to graph a absolute value, math problem solver, matrice calculator, ARITHEMATIC, notes on permutation and combination. 4) 1/x 6. Mathematical Circles-- A wonderful peak into Russian math training.
Basically, trigonometry is the study of triangles where we deal with the angles and sides of the triangle. The trigometric functions have a number of practical applications in real life and also help in the solutions of problems in many branches of 262 views (0) Comments · Trigonometry 144 views. Trigonometry Booster for JEE Main and Advanced has been conceptualised and produced for aspirants of various engineering entrance examinations.
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This book consists of my lectures of a freshmen-level mathematics class of-fered at Arkansas Tech University. The function f . Also, get Vedantu free study materials of textbook solutions, sample papers and board questions papers for CBSE & ICSE examinations Join Dr.
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Learning to solve right triangles provides the foundation you'll be using as you advance in trigonometry. 2-04 . Solution : First let us draw a figure for the information given in the question.
Solutions to problems. Find the lengths of all sides of the right triangle below if its area is 400. HINT : 137 = 90 + 43.
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Problems. You don’t have a scientific calculator for trigonometry A Guide to Advanced Trigonometry Before starting with Grade 12 Double and Compound Angle Identities, it is important to revise Grade 11 Trigonometry. Likewise, you will find that many topics in a calculus class require you to be able to basic trigonometry.
ted s. trigonometry facts and thinking about the unit circle (see the problems above) reveal that t D ˇ 6 and t D 5ˇ 6 are the two solutions. What are some difficult trigonometry problems? Update Cancel a U d lR M b DWeJ y ZA e B VpXY a pdfa i nLdn r ivfR e XGPk s FP D VFBB e CY v f eKr - pjp o T jCZVd o jI p qz TSAF 1 BG % hGG knwyu I Uv T xtJh ailQ T HAdXM a Z l i e rKM n j t LWgJh .
Answers are provided. The book contains about 2000 examples, problems, and exercises of which 1700 problems are for solving The IIT JEE Trigonometry problems range from the trigonometry basics to the applications of trigonometry. Learn for free about math, art, computer programming, economics, physics, chemistry, biology, medicine, finance, history, and more.
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Trigonometry in modern time is an indispensable tool in Physics, engineer-ing, computer science, biology, and in practically all the sciences. is your unrivaled introduction to this crucial subject, taught by award-winning Professor Bruce Edwards of the University of Florida. D.
About the Authors Titu Andreescu Trigonometry–Problems, exercises, etc. Trigonometry Right triangle trig: Evaluating ratios Right triangle trig: Missing sides/angles Angles and angle measure Co-terminal angles and reference angles Arc length and sector area Trig ratios of general angles Exact trig ratios of important angles The Law of Sines The Law of Cosines Graphing trig functions Translating trig functions Mathematics Standards of Learning for Virginia Public Schools – February 2009 10 Algebra II and Trigonometry The standards for this combined course in Algebra II and Trigonometry include all of the standards listed for Algebra II and Trigonometry. Je Zeager, Ph.
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Further, tan 7 9 0 in Q2. In general, to find the period of sin. Our online trigonometry trivia quizzes can be adapted to suit your requirements for taking some of the top trigonometry quizzes.
2". Assuming that the string is straight, how high above the ground is the kite? 2. Trigonometry & Calculus - powered by WebMath.
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To be more specific, trigonometry is all about a right-angled triangle. If the foot of the ladder is 6. Basic trig functions - practice problems These problems are designed to help you learn basic trigonometry ("trig") functions and how to use your calculator correctly.
Try solving these on your own (without peaking at the solutions). Check our customer testimonials to see what other parents and students are saying about our tutorials. Make grading easy with solutions to all textbook problem sets.
As in comment 1, is something that can NOT be simplified!! This Trigonometry Handbook was developed primarily through work with a number of High School and College Trigonometry classes. If you aren’t in a calculus class, you can ignore these comments. 1) Here we have the logarithmic function: 4 = log 3 81 .
When you click on the link, the problems will appear in the center column. 11. Problem 1.
Learn the concepts with our trig tutorials that show you step-by-step solutions to even the hardest trigonometry problems. Trigonometry word problems worksheet with answers is much useful to the kids who would like to practice problems on triangles in trigonometry. Solve the right triangle below.
On turning the ladder over without moving its foot, it is found that when it rests against a wall on the other side of the street it is at an angle of 15° with the ground. It also shows you how to check your answer three different ways: algebraically, graphically, and using the concept of equivalence. Only this enabled the author to squeeze about 2000 problems on plane geometry in the book of Trigonometry Basics - Problems and Solutions (WebMath) Provides solutions and answers to basic problems in trigonometry, including Right Angle Relationships, Graphing Trig Functions, Simplifying Trig Functions, and Polar Graphs.
Bx/ or cos. 2 meters. It renders complex concepts and problems of Trigonometry easy for the students and provides them as many opport Page 220 - A ladder placed at an angle of 75° just reaches the sill of a window at a height of 27 feet above the ground on one side of a street.
The problems are sorted by topic and most of them are accompanied with hints or solutions. Lecture Notes Trigonometric Identities 1 page 3 Sample Problems - Solutions 1. The text is suitable for a typical introductory Algebra & Trigonometry course, and was developed to be used flexibly.
Trigonometry is one of the important branches of mathematics and this concept is given by a Greek mathematician Hipparchus. Advanced Trigonometry Advance trigonometry covers the inverse trigonometric functions, solving equations involving trig concepts, and additional identities, including those of double and half angles Topics include: Trigonometry Problems and Questions with Solutions - Grade 12. X takes the mystery out of math by providing algebra 2 problem solver, trigonometry word problems, and an extensive library of sample math problems and solutions in video format.
two solutions if there are two, and should print "NO SOLUTION" if there are none. = . Students must practice various trigonometry problems based on trigonometric ratios and trigonometry basics so as to get acquainted with the topic.
Useful in a variety of ways in high school and college curriculums, this challenging volume will be of particular interest to teachers dealing with gifted and advanced Once P(k+1) has been proved to be true, the statement is true for all values of the variable, by Principle of Mathematical Induction. 36 offers an effective solution for trigonometry problems that you enter. Answers to the Advanced Math Problems.
1 Degree and Radian Measure 1. This course is designed for advanced students who are capable of a Download Grade 12 Trigonometry And Geometry Trigonometry Problems and Questions with Solutions Trigonometry questions with solutions and answers for grade 12. 2" Solve: 7.
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Find x and H in the right triangle below. How to solve advanced trigonometry problems. The authors are thankful to students Aparna Agarwal, Nazli Jelveh, and Trigonometry Guide currently available at www.
sinx = sin2x Find the values of x between 0<x<2π (includin 0 and 2π) So when i solved it i only got x= π/3 Howver the answer is 0, π/3, π, 5π/3, 2π I dont understand how you get all these solutions, what is the method in getting them. 0. Trigonometry for Solving Problems: This lesson offers a pair of puzzles to enforce the skills of identifying equivalent trigonometric expressons.
The topics covered in this book - Trigonometry, Vector Algebra, and Probability - are of utmost importance to engineering students. We’re sure you’ll have every angle covered ALGEBRA 2/TRIGONOMETRY Notice… A graphing calculator and a straightedge (ruler) must be available for you to use while taking this examination. A Guide to Advanced Trigonometry Before starting with Grade 12 Double and Compound Angle Identities, it is important to revise Grade 11 Trigonometry.
Answers. 2x/ has an amplitude of 10 numerical skills/pre-algebra, algebra, college algebra, geometry, and trigonometry. In quite a few probl ems you will be asked to work with trig functions, evaluate trig functions and solve trig equations.
Special attention should be given to using the general solution to solve trigonometric equations, as well as using trigonometric identities to simplify expressions. Please be aware, however, that the handbook might contain, and almost certainly contains, typos as well as incorrect or inaccurate solutions. J 7.
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Learn how to solve trigonometric equations and how to use trigonometric identities to solve various problems. Lines* 2. Let us look at the next problem on "Trigonometry word problems with solutions" Problem 3 : A string of a kite is 100 meters long and it makes an angle of 60° with horizontal.
Trigonometry A-Level Maths Revision Section on Revision Maths covers: Sine and Cosine Rule, Radians, Sin, Cos & Tan, Solving Basic Equations, Sec, Cosec & Cot, Pythagorean Identities, Compound Angle Formulae and Solving Trigonometric Equations. 5 − 2 times more space than the formulations, while still remaining complete, with no gaps whatsoever, although many of the problems are quite difficult. The following sheets list the key concepts which are taught in the specified math course.
problems in the supplemental problems (of which there are several for almost every lecture) are more challenging and less routine than would normally be found in a book of trigonometry (note there are several inexpensive problem books available for trigonometry to help supplement the text of this book if you find the problems lacking in number). Graph rational, piece-wise, power, exponential, and logarithmic functions. churcheatonschool.
Do archaeologists use trigonometry? Trigonometry is used to divide up the excavation sites properly into equal areas of work. 0 2. This exercise will familiarize the reader with the manipulation of angles, especially inverse trigonometric functions in whatever computing language is used, and will be rewarded in future more advanced applications.
The following table is a partial lists of typical equations. Addtional worksheets enhance students' abilities to appreciate and use trigonometry as a tool in problem solving. This review was originally written for my Calculus I class, but it should be accessible to anyone needing a review in some basic algebra and trig topics.
Word Problems (advanced) Surveying Problems: Problems involving finding quantities A comprehensive database of more than 40 trigonometry quizzes online, test your knowledge with trigonometry quiz questions. Trigonometry – Hard Problems Based on the illustration at right, we get the following: tan à L 90 200. Home Study Companion series provides a complete high school math experience for homeschoolers by supplementing the best existing high school math textbooks with solid teaching by an experienced teacher.
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The word “trigonometry” is derived from the Greek words trigono (τρ´ιγων o), meaning “triangle”, and metro (µǫτρω´), meaning “measure”. N. DO NOT BLINDLY APPLY powers and roots across expressions that have or signs.
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In other cases we have left them for you to discover as you learn more about mathematics. High School Math Solutions – Trigonometry Calculator, Trig Function Evaluation Limited Storage (10 problems) Placement Test Practice Problems Book II Geometry, Trigonometry, and Statistics Eric Key, University of Wisconsin-Milwaukee David Ruszkiewicz, Milwaukee Area Technical College This material is based upon work supported by the National Science Foundation under Grant No. This book has been written in a way that can be read by students.
This volume is a welcome resource for teachers seeking an undergraduate text on advanced trigonometry. Michael Algebra II with Trigonometry (book/solutions/guide) Trigonometry Advanced Placement Seventh Mathematics Describing the Real World: Precalculus and Trigonometry. On this web page "Trigonometry word problems worksheet with answers", first we are going to look at some word problems questions and then we will look answers.
The hypotenuse Free Trignometry worksheets includes visual aides, model problems, exploratory activities, practice problems, and an online component Trigonometry Worksheets (pdf) with answer keys. Moreover, when advanced concepts are employed, they are discussed in the section preceding the problems. Swamy who had an illustrious career as a renowned mathematician.
45 L° The angle 7 9 is in Q2, but tangent is defined only in Q1 and Q4. Trigonometry, at it's most basic level, is concerned with the measurement of triangles - calculations of unknown lengths and angles. Show your process to earn credit.
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ALGEBRA 2/TRIGONOMETRY Notice… A graphing calculator and a straightedge (ruler) must be available for you to use while taking this examination. org. Advanced Trigonometry Software (part 2): Continue your study of Trigonometry with the following topics: Trig Identities, Trig Proofs, Double Angle Formulas, Half Angle Formulas, Sum & Difference Formulas, Expressions & Equations, Solving Right Triangles, The Law of Cosines, The Law of Sines, Area of a Triangle, Vectors, Real World Problems, Polar Coordinates, and DeMoivre's Theorem.
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sin 2 (43°) + cos 2 (43°) = Understanding how to translate word problems into mathematical solutions is an essential skill for students to master…and easy to learn if you learn it the right way! Trigonometry word problems include problems relating to radians and degrees, circles, word problems involving trigonometric functions, and word problems involving identities. To ensure variety in the content and complexity of items within each domain, ACT Compass includes mathematics items of three general levels of cognitive complexity: basic skills, application, and analysis. Litvinenko, A.
Trigonometry is full of formulas and the students are advised to learn all the trigonometric formulas including the trigonometry basics so as to remain competitive in JEE and other engineering exams. "103 Trigonometry Problems" contains highly-selected problems and solutions used in the training and testing of the USA International Mathematical Olympiad (IMO) team. Q1 An old chestnut (general mathematics) 1 Trigonometry Booster for JEE Main and Advanced has been conceptualised and produced for aspirants of various engineering entrance examinations.
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Find the height of the kite,assuming that there is no slack in the string. Sample math problems and solutions are available for Arithmetic, Basic Algebra, Geometry, Advanced Algebra, Trigonometry and Calculus. Advanced Trigonometry Problems And Solutions - drellc.
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D 6. Quadratic equation vertex form, ways to find trig identities calculator, séries de fourier/exercices, Answers to Trigonometry Problems. Intro to Trigonometry Right Triangle Trigonometry.
advanced trigonometry problems with solutions
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https://www.jiskha.com/display.cgi?id=1269216856
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math
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posted by Andy .
A.A pair of narrow, parallel slits separated by a distance of 0.335 mm are illuminated by green laser light with wavelength = 543.0 nm. The interference pattern is observed on a screen 2.43 m from the plane of the parallel slits. Calculate the distance from the central maximum to the first bright region on either side of the central maximum.
B.What is the distance between the first and second dark bands in the interference pattern?
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https://www.osapublishing.org/oe/fulltext.cfm?uri=oe-21-10-11852&id=253364
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Several that appeared in [Opt. Express 20, 10004, (2012)] are corrected.
© 2013 OSA
In Section 3.1.1, 11th line from the bottom of p. 10010, V(x) should be written as .
In Section 3.2.3, second full paragraph on p. 10016, distributions of and omit the effect of the k 2 coefficient. It should be written that and .
In Section 5, 4th line from the bottom of p. 10025, “…appear in emission mode…” should be written as “…appear in absorption mode…”.
In Appendix A, scaling property of lognormal distributions, 1st line of p. 10029, mean parameter should be written as .
In Appendix C, p. 10032, expression for V(z) should be written as .
References and links
1. A. Ben-David and C. E. Davidson, “Probability theory for 3-layer remote sensing radiative transfer model: univariate case,” Opt. Express 20, 10004–10033 (2012), doi:. [CrossRef]
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https://rechneronline.de/chemie-rechner/half-life.php
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Original and resultung amount have the same value (e.g. Kilogramms, Milligramms, %, etc.), as well as half-life and time passed (one unit, e.g. s, min, h, d, a, etc). Please insert three values, the fourth will be calculated. You can choose which value you want to leave blank.
No responsibility is taken for the correctness of this information.
Chemical Calculators © Jumk.de Webprojects | Online Calculators | Images of Chemical Elements | Imprint & Privacy
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s3://commoncrawl/crawl-data/CC-MAIN-2023-50/segments/1700679100531.77/warc/CC-MAIN-20231204151108-20231204181108-00334.warc.gz
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https://www.astronomyclub.xyz/strange-stars/t-1.html
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0.341308 + 12.0708 e2 + 1.148889 e4 1 + 10.495346 e2 + 1.326623 e4 ' 0.614925 + 16.996055 e2 + 1.489056 e4
1 + 10.10935 e2 + 1.22184 e4 0.539409 + 2.522206 e2 + 0.178484 e4
The results of Ichimaru et al. (1987) are non-relativistic. They seem to be sufficient for most applications, because the electron gas is almost ideal at densities and temperatures, where the electrons are relativistic.
The relativistic results for the exchange energy at 0 < 1 were derived by Sha-viv & Kovetz (1972). Spurious misprints, which persisted in that and many subsequent publications, were eliminated by Stolzmann & Blocker (1996), whose result can be written as
NekBT 4n Xr3tr
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http://www.ibpsrecruitmentss.com/ibps-rrb-po-prelim-apptitute-question-paper-held-on-09-09-2017/
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math
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Recently IBPS RRB PO CONDUCTED ON 9 and 10th sept 17. Here we are providing IBPS RRB PO PRELIM APPTITUTE QUESTION PAPER HELD ON 09.09.2017 memory based. Thanks to all aspirants to share these questions.
IBPS RRB PO PRELIM APPTITUTE QUESTION PAPER HELD ON 09.09.2017:
Study the following questions before appearing coming sessions of IBPS RRB PO RELIM.
Directions (01-05): What should come in place of the question mark (?) in following number series problems?
Q01. 190, 94, 46, 22, ? , 4
(a) 12 (b) 14 (c) 10
(d) 8 (e) None of these
Q02. 5, 28, 47, 64, 77, ?
(a) 84 (b) 86 (c) 89
(d) 88 (e) None of these
Q03. 7, 4, 5, 12, 52, ?
(a) 424 (b) 428 (c) 318
(d) 440 (e) None of these
Q04. 6, 4, 5, 11, 39, ?
(a)159 (b) 169 (c) 189
(d)198 (e) None of these
Q05. 89, 88, 85, 78, 63, ?
(a) 30 (b) 34 (c) 36
(d) 32 (e) None of these
Q06. There are 3 consecutive odd numbers and 3 consecutive even numbers. The smallest even number is 9 more than largest odd number. If the square of average of all the 3 given odd number is 507 less than the square of the average of all the 3 given even number, what is the smallest odd number.
(a) 11 (b) 13 (c) 17
(d) 19 (e) 9
Q07. A can complete a task in 15 days B is 50% more efficient than A. Both A and B started working together on the task and after few days B left task and A finished the remaining of the given work.
For how many days A and B worked together.
(a) 3 (b) 5 (c) 4
(d) 6 (e) 2
Q08. A boat can travel 9.6 km downstream in 36 min. If speed of the water current is 10% of the speed of the boat in downstream. How much time will boat take to travel 19.2 km upstream.
(a) 2 hours (b) 3 hours (c) 1.25 hours
(d) 1.5 hours (e) 1 hour
Q09. A started a business with a initial investment of Rs. 1200. ‘X’ month after the start of business, B joined A with on initial investment of Rs. 1500. If total profit was 1950 at the end of year and B’s share of profit was 750. Find ‘X’
(a) 5 month (b) 6 month (c) 7 month
(d) 8 month (e) 9 month
Q10. Ratio between curved surface area and total surface area of a circular cylinder is 3 : 5. If curved surface area is 1848 cm3 then what is the height of cylinder.
(a) 28 (b) 14 (c) 17
(d) 21 (e) 7
Directions (51-55): Given below is the pie chart which shows the percentage distribution of a book xyz publishes in 5 different stores.Total books 550.
Q11. If number of female who bought the books in store E are 21 more than number of males who bought books from same store then find the number of females who bought book in store E.
(a) 75 (b) 78 (c) 71
(d) 68 (e) 73
Q12. Find the central angle for the book D.
(a) 117.5° (b) 115.2° (c) 112.8°
(d) 108.5° (e) 118.8°
Q13. If total books of another publisher ‘MNP’ is 20% more than books of ‘XYZ’ publisher then what will be total books sold by store A and B for publisher ‘MNP’. Percentage-distribution for different stores for MNP remains same as for ‘XYZ’
(a) 200 (b) 178 (c) 181
(d) 186 (e) 198
Q14. What is the ratio of total books sold by store A and C together to the total books sold by store D and E together
(a) 17 : 27 (b) 18 : 29 (c) 21 : 28
(d) 22 : 23 (e) 24 : 29
Q15. What is the difference between average of book sold by store A and E together and average books sold by store C and D together?
(a) 33 (b) 11 (c) 22
(d) 44 (e) 20
Directions (16-20): In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer
(a) if x>y (b) if x≥y
(c) if x<y (d) if x ≤y
(e) if x = y or no relationship can be established.
Q16. I. x² + 9x + 20 = 0 II. y² = 16
Q17. I. x² − 7x + 12 = 0 II. 3y² – 11y + 10 = 0
Q18. I. x² − 8x + 15 = 0 II. y² − 12y + 36 = 0
Q19. I. 2x² + 9x + 7 = 0 II. y² + 4y + 4 = 0
Q20. I. 2x² + 15x + 28 = 0 II. 2y²+ 13y + 21 = 0
Q21. Train A completely crosses train B which is 205 m long in 16 second. If they are travelling in opposite direction and sum of speed of both are 25 m/s. then find the difference (in meter) between lengths of both trains.
(a) 5 (b) 6 (c) 8
(d) 10 (e) 12
Q22. A trader mixes 14 kg rice of variety A which costs Rs. 60/kg with 18 kg of quantity of type B rice. He sells the mixture at Rs. 65/Kg and earns a profit of %. Then what was the cost price of type B rice.
(a) 30 (b) 20 (c) 40
(d) 50 (e) 45
Q23. Present age of A is 3 years less than present age of B. Ratio of B’s age 5 year ago and A’s age 4 year hence is 3 : 4 then find present age (in years) of A.
(a) 20 (b) 17 (c) 23
(d) 26 (e) 29
Q24. A bag contains 6 Red, 5 Green and 4 Yellow coloured balls. 2 balls are drawn at random after one another without replacement then what is the probability that atleat one ball is Green.
(a) (b) (c)
Q25. Cost price of B is 200 more than cost price of A. B is sold at 10% profit and A is sold at 40% loss and selling price of A and B are in the ratio 4 : 11. If A is sold at 20% loss then what will be selling price of
(a) 320 (b) 400 (c) 240
(d) 160 (e) 360
Directions (26-30): Read the following table carefully and answer the following questions.
No. of students and % of students passed out of those who appeared are given for two subjects from year 2001 to 2005 in a college XYZ.
Q26. Find the average number of students who were failed in Economics in year 2002 and year 2003 together?
(a) 1435 (b) 1565 (c) 1720
(d) 1590 (e) None of these
Q27. Number of students failed in Statistics in the year 2003 is what % of the number of students failed in Economics in the same year?
(a) 145.75% (b) 150% (c) 156.25%
(d) 158.25% (e) None of these
Q28. Find the ratio between the total number of students appeared in Economics from 2002 to 2004 together and the total number of students appeared in Statistics from year 2003 to 2005 together?
(a) 13 : 14 (b) 14 : 13 (c) 15 : 16
(d) 16 : 15 (e) None of these
Q29. Find the difference between the total number of students passed in Statistics from year 2002 and total number of students failed in Economics from year 2005.
(a) 690 (b) 385 (c) 485
(d) 550 (e) 610
Q30. Find the average number of students appeared in Economics from year 2001 to 2004 together?
(a) 3090 (b) 3015 (c) 3060
(d) 3075 (e) 3850
Direction (31-35): What approximate value should come in place of question mark (?) in the following questions? (Note: You are not expected to calculate the
Q31. ?% of (5284.89 ÷ 7.08) = 986.01 – 533. 06
(a) 42 (b) 39 (c) 74
(d) 65 (e) 60
Q32. (1041.84 + ?) ÷ 3.02 = 1816.25 ÷ 4.01
(a) 442 (b) 337 (c) 385 (d) 268 (e) 320
IBPS RRB PO REASONING QUESTION 2017.
Q33. 69.3% of 445.12 ÷ 14.06 = 623.08 ÷ ?
(a) 28 (b) 19 (c) 21
Q34. ? + 114.09 – 24.06 × 5.14 = 163.19
(a) 7 (b) 13 (c) 11
(d) 15 (e) 19
Q35. 768.16 ÷ 11.87 × 58.05 = ?
(a) 1033 (b) 1175 (c) 966
(d) 880 (e) 975
Directions (36-40): Study the following line graph carefully and answer the following questions.
Number of males and number of females are given.
They are visiting a place from Monday to Friday.
Q36. Find the ratio of the total number of males visited the place on Tuesday and Thursday together to the total number of females visited the place on Monday and Friday together?
(a) 29 : 30 (b) 30 : 29 (c) 25 : 26
(d) 26 : 25 (e) None of these
Q37. Total number of males and females together visited the place on Tuesday are what percent more/less than the total number of male and females together visited the place on Thursday ?
(a) 26 % (b) 25 % (c) 26 %
(d) 25 % (e) None of these
Q38. Find the difference between the total number of females visited the place from Monday to Wednesday and the total number of males visited the place from Wednesday to Friday?
(a) 30 (b) 60 (c) 40
(d) 50 (e) None of these
Q39. If on Saturday the number of males and number of females increased by 25% and 20% respectively as compared to that on Friday then find the total number of males and females together visited the place on Saturday?
(a) 196 (b) 306 (c) 316
(d) 206 (e) 216
Q40. Total number of males and females visited the place on Monday and Tuesday together is how much more than the total number of males and females visited the place on Thursday and Friday together?
(a) 175 (b) 125 (c) 150
(d) 160 (e)130
1.c 2.d 3.a 4.c 5.d 6.a 7.c 8.d 9.b 10.d 11.c 12.b 13.e 14.a 15.c 16.d 17.a 18.c 19.e 20.d 21.d 22.c 23.a 24.d 25.a 26.b 27.c 28.d 29.b 30.e 31.e 32.e 33.a 34.b 35.c 36.a 37.a 38.d 39.b 40.c
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https://www.oreilly.com/library/view/probability-and-statistics/9781119285427/c10.xhtml
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The probability distributions discussed in the preceding chapters will yield probabilities of the events of interest, provided that the family (or the type) of the distribution and the values of its parameters are known in advance. In practice, the family of the distribution and its associated parameters have to be estimated from data collected during the actual operation of the system under investigation.
In this chapter we investigate problems in which, from the knowledge of some characteristics of a suitably selected subset of a collection of elements, we draw inferences about the characteristics of the entire set. The collection of elements under investigation is known as the population, and its selected subset is called a sample. Methods of statistical inference help us in estimating the characteristics of the entire population based on the data collected from (or the evidence produced by) a sample. Statistical techniques are useful in both planning of the measurement activities and interpretation of the collected data.
Two aspects of the sampling process seem quite intuitive. First, as the sample size increases, the estimate generally gets closer to the “true” value, with complete correspondence being reached when the sample embraces the entire population. Second, whatever the sample size, the sample should be representative of the population. These two desirable aspects (not always satisfied) of the sampling process will ...
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https://ocw.mit.edu/courses/physics/8-422-atomic-and-optical-physics-ii-spring-2013/video-lectures/lecture-16-light-forces-part-1/
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Description: In this video, the professor discussed light forces, mechanical forces, radiation pressure force, reactive forces.
Instructor: Wolfgang Ketterle
The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free. To make a donation, or view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu.
PROFESSOR: --equal forces of light. But before we get started with it, do you have any question about the last lecture, which was on open quantum system-- Quantum Monte Carlo methods-- and in particular, the conceptual things we discussed is what happens when you do a measurement, or when you don't detect a photon, how does it change the wave function? And this was actually at the heart of realizing similar quantum systems, and simulating them using quantum Monte Carlo methods. Any question? Any points for discussion? OK.
Today is, actually conceptually, a simple lecture, because I'm not really introducing some subtleties of quantum physics. We know our Hamiltonian-- the dipole Hamiltonian-- this is the fully quantized Hamiltonian, how atoms interact with the electromagnetic field. And, of course, it's fully quantized because E perpendicular is the operator of the quantized electromagnetic field, and it's the sum of a plus a [? dega. ?] But when it comes to forces, it's actually fairly simple. We don't introduce anything else. This is our Hamiltonian. This is the energy. And the force is nothing else than the gradient of this energy.
So I decided it today-- it's rather unusual-- but I will have everything pre-written, because I think the concentration of new ideas is not so high today, so I want to go a little bit faster. But I invite everybody if you slow me down if you think I'm going too fast and if you think I'm skipping something. Also another justification is at least the first 30%, 40% of this lecture, you've already done as a homework assignment, when you looked at the classical limit of mechanical forces. You actually had already, as a preparation for this class, a homework assignment on the spontaneous scattering force and the stimulated dipole force.
PROFESSOR: Well, the only subtlety today is how do we deal with a fully quantized electromagnetic field? But what I want to show you in the next few minutes is in the end, we actually do approximations that the quantized electromagnetic field pretty much disappears from the equation. We have a classical field. And then, almost everything you have done in your homework directly applies. The only other thing we do today is the dipole moment. Well, we use, for the dipole moment of the atom, the solution of the Optical Bloch Equation. So I think now you know already everything we do in the first 30, 40 minutes.
PROFESSOR: So we get rid of all of the quantum character of the electromagnetic field in two stages. One is the following, well, we want to describe the interactions of atoms with laser fields. And that means the electromagnetic field is in a coherent state. And you know from several weeks ago, the coherent state is a superposition of a [INAUDIBLE] states, coherent superposition and such, so they're really photons inside. It's a fully quantized description of the laser beam. But now-- and this is something if you're not familiar, you should really look up-- there is an exact canonical transformation where you transform your Hamiltonian-- you do, so to speak, a basis transformation-- to another Hamiltonian. And the result is, after this unitary transformation, the coherent state is transformed to the vacuum state.
So in other words, the quantized electromagnetic field is no longer in the coherent state, it's in the vacuum state. But what appears instead is a purely classical electromagnetic field. A c number. So you can sort of say-- I'm waving my hands now-- the coherent state is the vacuum, with a displacement operator. All the quantumness of the electromagnetic field is in the vacuum state, and that's something we have to keep because this is responsible for spontaneous emission quantum fluctuations and such. But the displacement operator, which gives us an arbitrary coherent state, that can be absorbed by completely treating the electromagnetic field classically.
So therefore, we take the fully quantized electromagnetic field and we replace it now by a classical field, and simply the vacuum fluctuation. So there are no photons around anymore. There is the vacuum ready to absorb the photons, and then there's a classical electromagnetic field, which drives the atoms. And, of course, the classical electromagnetic field is not changing because of absorption emission, because it's a c number in our Hamiltonian. So this is an important conceptual step, and it's the first step in how we get rid of the quantum nature of the electromagnetic field when it interacts with atoms mechanically. OK.
So, by the way, everything I'm telling you today-- actually, in the first half of today-- can be found in Atom-Photon Interaction. Now, so this takes care actually, of most of the electromagnetic field. We come back to that in a moment. The next thing is we want to use the full quantum solution for the atomic dipole operator, and that means we want to remind ourselves of the solution of the Optical Bloch Equations.
PROFESSOR: Now, talking about the Optical Bloch Equations, I just have to remind you of what we did in class, but I have to apologize it's a little bit exercise in notation. Because in Atom-Photon Interaction they use a slightly different notation then you find in other sources. On the other hand, I think I've nicely prepared for you, and I give you the translation table. In Atom-Photon Interaction, they regard the Bloch vector, and we have introduced the components, u, v, and w, as a fictitious spin, 1/2. Therefore, you find with this kind of old-fashioned letters, Sx, Sy, Sc, in Atom-Photon Interaction.
The other thing, if you open the book, Atom-Photon Interaction is that their operators, density matrix-- the atomic density matrix-- sigma, and sigma head. One is in the rotating frame, and one is in the lab frame. So we have to go back and forth to do that, but the result is fairly trivial. So this is simply the definition. First of all, the density matrix is the full description of the atomic system, but instead of dealing with matrix elements of the density matrix, it's more convenient for two level system to use a fictitious spin or the Bloch vector. So the important equation-- the are the Optical Block Equations-- is the equation A20 in API. And it's simply the first order differential equation for the components of the Optical Bloch vector, u v, and w. Notation is as usual, the laser detuning, delta L. Omega 1 is the [INAUDIBLE] frequency. And this is the parametrization of the classical electromagnetic field which drives the system.
Now, this equation is identical to the equation we discussed in the unit on solutions of Optical Bloch Equations. We just called the [INAUDIBLE] frequency, g, delta L was called delta. And then, vectors of 2r was confusing. The components, [? r,x,y,z, ?] are two times u, v, w. And the reason for that is some people prefer that it represents spin 1/2, so they want to normalize things to 1/2. Or if you want to have unit vectors, your throw in a vector of two. But that is all, so if you're confused about a vector of two, it's exactly this substitution which has been made.
So these are the Optical Bloch Equations, and what we need now is the solution. What we need now is the solution for the dipole moment, because it is the dipole moment which is responsible for the mechanical forces. The dipole operator is, of course, the non-diagonal matrix elements, the coherencies of the density matrix. So all we want to take now from the previous unit, the expectation value of the dipole moment, which is nothing else than the trace of the statistical operator with the operator of the dipole moment.
At this point, just to guide you through, we go from the lab frame-- from the density matrix in the lab frame-- to the density matrix in the rotating frame, and this is when the laser frequency appears, is into the i omega l t. And therefore, we want to describe things in the lab frame. If you use the u and v part of the Optical Bloch vector obtained in the rotating frame, in the lab frame everything rotates at the laser frequency. If you drive an atom, if you drive a harmonic oscillator with a laser frequency, the harmonic oscillator responds at the laser frequency, not at the atomic frequency. We discussed that a while ago. So therefore, the dipole moment oscillates at the laser frequency. And their two components, u and v-- since our laser field was parametrized by definition-- epsilon 0 times cosine omega t.
We have now nicely separated, through the Optical Bloch vector the in phase part of the dipole moment, and the in quadrature phase of dipole moment. If you didn't pay attention what I said the last five minutes, that's OK, you can just start here, saying the expectation value of the dipole moment oscillates with a laser frequency. And as for any harmonic oscillator, or harmonic oscillator type system, there is one component which is in phase, and one component which is in quadrature.
Fast forward. OK. And you know from any harmonic oscillator, when it comes to the absorbed power, that it's only the quadrature component which absorbs the power. And you can immediately see that when we say-- just use what is written in black now-- the energy is nothing else than the charge times the displacement. When we divide by delta t, you'll find that the absorbed power is the electric field times the derivative of the dipole moment. Well, if you ever reach over one cycle, we are only interested in d dot, which oscillates with cosine omega t. But that means since it's d dot, d oscillates with sine omega t. And this was the in quadrature component, v.
So it is only the part, v, of the Optical Bloch vector which is responsible for exchanging energy with the electromagnetic field. And we can also, by dividing by the energy of a photon, find out what is the number of absorbed photons. But we did that already when we discussed Optical Bloch Equations.
OK, so that was our look back on the Optical Bloch Equation, the solution for the oscillating dipole moment, and now we are ready to put everything together.
So we want to know the force. The force, using Heisenberg's Equation of Motion for the momentum derivative is the expectation value of the operator, which is the gradient of the Hamiltonian. OK. So now we have to take the gradient of the Hamiltonian. This is the only expression, actually, today, I would've liked to just write it down, and build it up piece by piece. But let me step you through.
We want to take the gradient. The dipole operator of the atom does not have an r dependence, it's just x on the atom, wherever it is. But the gradient with respect to the center of mass position from the atoms comes from the operator of the electromagnetic field. So in other words, what this involves is the gradient of the operator of the fully quantized electromagnetic field. But we have already written at the electromagnetic field as an external electromagnetic field-- classical electromagnetic field-- plus a vacuum fluctuations. But the vacuum fluctuations are symmetric. There is nothing which tells vacuum fluctuations what is left and what is right. So therefore, the derivative only acts on the classical electromagnetic field.
So this is how we have now, you know, we first throw out the coherent state, and now we throw out the vacuum state. And therefore, it is only the classical part of the electromagnetic field which is responsible for the forces. Now we make one way assumption which greatly simplifies it. R is the center of mass position of the atom, and we have to take the electromagnetic field as the center of mass position. And if the atom is localized-- wave packet with a long [? dipole ?] wave-- we sort of have to involve the atomic wave function when we evaluate this operator of the electromagnetic field.
However, if the atom is very well localized-- you would actually want the atom to be localized to within an optical wave length-- then, under those circumstances, you can replace the parameter, r, which is a position of the atom, by the center of the atomic wave packet. So therefore, the kind of wave nature of the atoms-- which means the atom is smeared out-- as long as this wave packet is small, compared to the wave lengths, you can just evaluate the electromagnetic field at the center of the wave packet. This assumption requires-- since the scale of the electromagnetic field is set by the wave lengths-- this requires that the atom is localized better than an optical wave length. And that would mean that the energy of the atom, or the temperature, has to be larger than the recoil in it.
And if you're localized the, to within a wave length, your momentum spread is larger, by Heisenberg's Uncertainty Relation than h bar k, and that means your energy is larger than the recoil energy. So if you want to have a description or light forces in that limit, you may have to revise this point. But for most of laser cooling, when you start with a hot cloud and cool it down to micro-Kelvin temperatures, this is very [INAUDIBLE]. Colin?
AUDIENCE: Vacuum field doesn't contribute in sort of in the sense that the expectation value is zero. But wouldn't the rms value be non-zero? So what's the--?
PROFESSOR: Exactly. I mean, in the end, you will see that in a moment, when we talk about cooling limits, fluctuations, spontaneous emission, provides heating. It doesn't contribute if you want to find the expectation value for the force. The vacuum and the fluctuations-- I mean, that's also logical-- the vacuum has only fluctuations, it has no net force. But the fluctuation's heat. And for heating, it will be very important. But we'll come to that in a little bit later.
AUDIENCE: So what is [INAUDIBLE]?
PROFESSOR: If the light is squeezed, yes. That's a whole different thing. We really assumed here that the light is in a coherent state. Now--
AUDIENCE: The transformation that we had [INAUDIBLE].
PROFESSOR: Well, you know, you can squeeze the vacuum. And then it's squeezed. But there's so few photons, you can't really laser cool with that. So the only way how you could possibly laser cool is if you take the squeezed vacuum, and displace it. So now you have, instead of a little circle, which is a coherent state, you have an ellipse, which is displaced. I haven't done the math-- maybe that would be an excellent homework assignment for the next time I teach the course-- you could probably show that the displacement operator, which is now displacing the squeezed vacuum, again can be transformed into a c number, and the ellipse this squeezed vacuum fluctuation-- the ellipse has an even symmetry, has an 180 degree rotation symmetry. And I could imagine that it will not contribute to the force.
So in other words, if you have a displaced, squeezed vacuum, my gut feeling is nothing would change here. But what would change, aren't the fluctuations, but which would possibly change is the heating.
OK, so we within those approximations, Heisenberg's Equation of Motion for the force becomes an equation for the acceleration for the atomic wave packet. And it involves now, two terms. One is the atomic dipole moment as obtained from the Optical Bloch Equations, and the other one is nothing else than the gradient of the classical part of the electromagnetic field. And with that we are pretty much at the equations which you looked at in your homework assignment, which was a classical model for the light force. Because this part is completely classical, and the only quantum aspect which we still have is that we evaluate the dipole moment, not necessarily, it's just a driven harmonic oscillator.
When we take the Optical Bloch Equations, we will also find saturation a two-level system can be saturated, whereas a harmonic oscillator can never be saturated. So let me just repeat, so this is completely classical except for the expectation value for the dipole moment. But for no excitation, of course-- and we've discussed it many, many times-- a two-level system is nothing else for weak excitation, than a weakly excited harmonic oscillator. So then we are completely classical. But we want to discuss also saturation and that's been the Optical Bloch Equations, of course, come in very handy.
OK. Next approximation, we want to know-- approximate-- the dipole moment-- the expectation value of the dipole moment-- with the steady state solution. The steady state solution-- analytic expression-- we just plug it in, and you can have a wonderful discussion about light forces. The question is, are we allowed to do that? Now I hope you'll remember that's actually one reason why I did that when we discussed the master equation. And we discussed an [? atomic ?] [? cavity. ?] I introduced to you the adiabatic elimination of variables. We adiabatically eliminated [? of dipole ?] matrix elements.
Whenever you have something which relaxes sufficiently fast, and you're not interested in this short time scale, you can always say, I replace this quantity by its steady state value. Steady state with regard to the slowly changing parameter. And we want to do exactly the same here. So let me just translate. It's the same idea, exactly the same idea, but let me translate. If an atom moves, we have two aspects, two times [? case. ?] One is the motion of the atom. The atom may change its velocity. The atom may move from an area of strong electric field to an area of weak electric field. But this is really all that is related to the motion, to the change of momentum of a heavy object.
But now inside the atoms, you play with the, you know, Bloch vector, excited state. If the electric field is higher, you have a larger population in the excited state than in the ground state. But this usually adjusts with the damping time of gamma, with a spontaneous emission rate. And that's usually very fast. So you can make the assumption when an atom moves to a changing electric field, at every point in the electric field, it very, very quickly-- the dipole moment-- assumes the steady state solution for the local electric field. And this would now be called an adiabatic elimination of the dynamics of the atom. And we simply replace the expectation value for the dipole moment by the steady state solution of the Optical Bloch Equations.
Well, this is, of course, an approximation. And I want to take this approximation and who you what simple conclusions we can draw from that. But I also want you to know, I mean this is what I focus off in this course, to sort of think about them and get a feel where those approximations break down. So this approximation requires that the internal motion, that this, you know, two-level physics, that the internal density matrix, it has a relaxation rate of gamma. Whereas the external motion, we will actually see that very soon when we talk about molasses and damping, has a characteristic damping time which is one over the recoil energy.
The recoiled energy involves the mass. The heavier the object, the slower is the damping time. It has all the right scaling. And you will actually see in the data today, that we will naturally obtain this as the time scale over which atomic motions is changing. So, is that approximation, this hierarchy of time scales, is that usually fulfilled? It is fulfilled in almost all of the atoms you are working on. For instance, in sodium, the ratio of those two time scales is 400. But if you push it, you will find examples like, helium in the triplets statement-- metastable helium in the triplet state-- where the two time scales are comparable.
Helium is a very light, so therefore, it's faster for a light atom to change its motion by exchanging momentum with electromagnetic field. And at the same time, the transition in helium is narrower, and therefore slower. So for metastable helium, the two time scales are comparable. | if you really want to do quantitative experiments with metastable helium, you may need a more complicated theory. But, of course, if something is sort of simple in one case, the richer situation provides another opportunity for research. And I know, for instance, Hal Metcalf, a good colleague and friend of mine, he focused for a while on studying laser cooling of metastable helium, because he was interested what happens if those simple approximations don't work anymore?
Then the internal motion and the external motion becomes sort of more entangled. You have to treat them together, you cannot separate them by a hierarchy of time scales. Any questions about that?
OK, now we have done, you know, everything which is complicated has been approximated away now. And now we have a very simple result. So I'm pretty much repeating what you have done your homework, but now with expressions which have still the quantum character of the Optical Bloch Equations. So let's assume that we have an external electric field. it's parametrized like that, and at this point, by providing it like this, I leave it open whether we have a standing wave-- if you have a traveling wave, of course, we have a phase factor, kr, here-- if you have a standing wave, we don't have this phase factor. So I have parametrized the electromagnetic field in such a way that we can derive a general expression. And then we will discuss what happens in a traveling wave, and what happens in a standing wave.
I make one more assumption here, namely that the polarization is independent of position. I'm just lazy if I take derivatives of the electric field. I want to take a derivative with the amplitude of the electric field, and I want to take a derivative of the phase of the electromagnetic field. I do not want to take a derivative of the polarization. Of course, what I'm throwing out here, is all of the interesting cooling mechanisms of polarization gradient cooling, when the polarization of the light field spirals around in three dimensions. We'll talk a little bit about that later on, I just want to keep things simple here.
OK, so now if you take the gradient of the electromagnetic field, finally, we have approximated everything which complicates our life away. And we have two terms. One is the gradient x on the amplitude of the electromagnetic field, or the gradient x on the phase. And since we have taken a derivative, when we take the derivative of the phase, because of the chain rule, the cosine changes into sine. So therefore, the gradient of the electromagnetic field has a cosine in phase and an in quadrature component. And now, for the force, we have to multiply this with the dipole moment. And, of course, we will average over one cycle of the electromagnetic wave.
And, of course, that's what we discussed. There's is a u and v component of the Bloch vector, which gives rise to an in phase and in quadrature component of the dipole moment. And now, if you combine-- if you multiply-- the gradient of the electric field with the dipole moment, and average. The cosine part goes with the cosine part, the sine part goes with the sine part, and the cos terms average out. So therefore, we have the simple result that we have two contributions to the [? old ?] mechanical force of light. And the u component-- the in phase component-- of the dipole moment goes with the amplitude of the gradient. And the in quadrature component goes with the gradient of the phase.
So in other words, when it comes to light forces, the cosine component and the sine component of the dipole moment talk, actually, to two different quantities associated with the electromagnetic field. It's just in phase and in quadrature.
OK. Now I just have to tell you a few definitions and names. This component, which goes with the in phase component, is called the reactive force. Whereas the other one is called the dissipative force. The name, of course, comes that if you have a harmonic oscillator, the component which oscillates in quadrature absorbs directly energy, whereas the in phase component does not absorb energy. I'll leave it for a little bit later in our discussion how you can have a force and not exchange energy. That's a little bit a mystery when we talk about cooling with a reactive force. We will have to scrutinize what happens with energy, because at least in the most basic, the in phase component of the oscillating system does not extract energy from the electromagnetic field.
So maybe just believe that for now, and we will scrutinize it later. Any questions? Am I going too fast? Am I going too slow? Does it mean about right? OK.
So we have now obtained the reactive force, which is also called the dipole force or stimulated force, and the dissipated force, which is also called the spontaneous force. The next thing is purely nomenclature. We want to introduce appropriate vectors which point along the gradient of the phase, and which point along the gradient of the amplitude. And this is done here. So we want to say that the reactive and the dissipative force are written in a very sort of similar way, but the one off them points in the direction of a vector, alpha, the other one in the direction of a vector, beta.
The beta vector is the gradient of the phase. If the phase is kr, the gradient of the phase is simply k, the k vector of the electromagnetic wave. So that will come in handy in just a few seconds. Whereas the reactive force points into the direction of the gradient of the amplitude, but we never want to talk about electromagnetic field amplitude. For us, we always use the electric field in terms of you parametrize the Rabi frequency. The Rabi frequency is really the atomic unit of the electric field. So therefore, this vector which involves now the gradient of the amplitude, becomes a normalized gradient of the Rabi frequency. So these are the two fundamental light forces, the reactive and the dissipative force.
OK. Now we are going to have a quick look at those two forces. So therefore, I'm now discussing the two simple, but already very characteristic cases for those forces. One is we want to look at a plane travelling wave, and then we want to look at a pure standing wave. And the beauty of it is in a plane travelling wave, you've a plain wave, the amplitude of the electric field is constant everywhere, it just oscillates. And therefore, this alpha vector is zero. And in a few minutes, we look at this standing wave, and in the standing wave, the beta vector is zero. So the travelling wave and the standing wave allows us now to look at the two forces separately in two physical situations which are-- as you will see later-- highly relevant for experiments.
Well, I don't think I have to tell anybody in this room-- maybe except some people who take the course for [? Brett's ?] requirement-- that standing waves or optical lattices are common in all labs. And similar, traveling wave beams are used, for instance, for decelerating atomic beams. So we illustrate the two forces now, but we already discussing two experimentally very important geometries.
OK, plain travelling waves. I mean, I just really plug now the electromagnetic field for plain travelling wave, I plug it into the equations above. The important thing is-- and which simplifies things a lot-- is that the alpha vector is zero. The amplitude of the plain wave is-- that's what a plain wave is-- constant. And the beta vector is the gradient of the Rabi frequency is just the k vector of the light. So therefore, the dissipative or spontaneous light force was the Rabi frequency times the v component of the Optical Bloch vector times h bar k.
And this has a very simple interpretation because the steady state solution of the Optical Bloch Equation is nothing else than gamma times the excited state population. And since the excited state scatters, or emits photons at a rate, gamma, this is nothing else than the number of absorbed emitted scattered photons times h bar k. We are in steady state here, so the number of photons which are emitted into all space, into the vacuum has to be the number of photons which has been absorbed into the laser beam. So therefore, the interpretation here is very simple, you have a laser beam. Every time the atom absorbs a photon, it receives a recoil transfer, h bar k.
Now, afterwards, the photon is scattered, but the scattering is symmetric, and does not impart the force onto the atom. And this is what we discussed earlier. The quantum part of the electromagnetic field is symmetric. And here, we see sort of what it means visually. Spontaneous emission goes equally probable in opposite direction, and there is no net force, but there is heating as we discussed later.
So we can plug in the solution-- the steady state solution-- from the Optical Bloch Equation. That's our Lorentzian. We have discussed power broadening already when we discussed the Optical Bloch Equation. And, of course, I just remind you if you saturate the system with a strong laser power, what you can obtain is that half of the population is in the excited state, and half of the population is in the ground state. Under those circumstances, do you get the maximum force. And the maximum spontaneous force is the momentum per photon times the maximum rate at which a two-level atom can scatter photons.
So that's, in a nutshell, I think all you have to know about the spontaneous light force as far as the force is concerned. But we still have to discuss the heating associated with a spontaneous force, which are the force fluctuations. Questions?
Well, then let's do a similar discussion with reactive forces. Which is a little bit richer because, well, you will see. So a standing wave is parametrized here. We only have an alpha vector, not a beta vector now, because everything is in phase, there is no phase phi of r. So therefore, there is no dissipative force, the only force is a reactive force. Since the reactive force depends only on u, there is no exchange of energy. So that raises the question, what is really going on it is only the reactive part with u, how can the atom really-- How can the motion change if it cannot exchange energy?
Well, we'll come to that in different pieces. The one part I want to sort of discuss here is that you should-- I want to show you that at the level of our discussion we can already understand that the atom is actually doing a redistribution of energy and momentum. Well, if I only hold onto my standing wave, there is no energy exchange because the atom is always in phase with the standing wave-- the atomic dipole-- which is this one [INAUDIBLE] force. But I can now do what everybody does in the lab. We generate the standing wave as a superposition of two travelling waves.
And if I have a superposition of two traveling waves, there is a point in space where the phases of the traveling waves are the same. Then one phase is advanced in phase, and the other side one of the waves is lagging in phase. And for pedagogical reasons, I pick the point where the two travelling waves are 90 degree phase shifted. OK. Now the situation is that we have our electric field and what is responsible for the reactive light force is the u part of the dipole moment of the Optical Bloch vector, which oscillates in phase.
But nobody prevents me from analyzing it with respect to E1 and E2. And so if I ask now, I say, I have one laser beam. I have another laser beam. And if you want, after the laser beams have crossed, you can put in two photo-diodes, and you can not only ask, what happens to the standing wave, you can also ask, what happens to each traveling wave? Are photons absorbed? Or what happens? And now, what is obvious is if you have an harmonic oscillator, and you have an oscillating dipole moment, if the dipole moment is withing 0 and 180 degrees with a dry field, you absorb power. If it is in the two other quadrants, it emits, or delivers power.
And now you see, that we are in a situation-- which I've peak here-- where one travelling wave loses energy, and the other traveling wave gains energy. And there is a classic experiment, which was done by Bill Phillips a while ago-- he had atoms in an optical lattice, and they were sloshing. And she could really measure that when the atoms were accelerated this way, one of the laser beams gained power, and the other one lost power. So while the atoms where sloshing in momentum space, the power between the two laser beams was distributed back and forth. So this is the character of the dissipative force, that it redistributes momentum and energy between the two traveling waves.
Any questions? Boris?
AUDIENCE: But if there's a net change in the patterns of motion, the energy still has to come from somewhere. Right? Can you make that [INAUDIBLE]
PROFESSOR: This question has bothered in many different iterations. And at some point, I found a very, very easy answer to it. And this is one of your homework assignments for homework number 10. There's a very simple example where you will realize where the energy comes. What happens is, in steady state, if everything is stationary, of course, there can't be an exchange. But, if you're for instance, saying-- and I give you no part of the solution-- if you have atoms, and they are attracted by the dipole force, there's a strong laser beam and the atoms are accelerated in, and you're now asking, where does the energy come from?
Because all of what the atoms do, at least according to what I'm telling you, they redistribute photons of energy, h bar omega, into photons of energy, h bar omega. So where does the energy come from? You have to go higher in the approximation here to understand where the energy comes from. Let me just tell you one thing, this is actually something which I've encountered several times. And it can be really confusing, I've really proven to mathematically the this is false. So in this level of treatment, with these very simple concepts, we correctly obtained the optical force. But if you're now asking, where does the energy go? You have to go deeper.
And, for instance, what happens is if you take a cloud of atoms, and the cloud of atoms moves in an out of a laser beam-- just think of it as center of mass dipole oscillations of a Boson-Einstein condensate in an optical dipole trap-- what happens is when the atoms move in and out, they actually act as a phase modulator for the laser beam. The index of the refraction of your laser beam is changing. So therefore-- and we don't deal with that when he replace the electromagnetic field by a c number-- your laser beam, when the atoms move in will actually show a frequency modulation.
And you will find-- and it's a wonderful homework assignment, I'm pretty proud that we created it, and that it works out so easily-- you will find that the kinetic energy of the atoms is exactly compensated by the energy shift the photons which have passed through the atoms. But I'm giving you a very advanced answer. It's often very difficult if you have a simple picture for the force to figure out what happens to the energy. Actually, let me give you the other example.
When we have the previous example, where we have atoms and we have a laser beam which cools with a radiation pressure. This was our previous example. Where does the energy go? I have a wonderful correct-- within the assumptions-- an exact expression for the force. Where does the energy go? The atoms have a counter-propagating laser beam. They scatter photons with this rate. Every time they scatter a photon, they slow down by h bar k. They lose energy. We have a complete description what happens in the momentum picture and in forces. But where does the energy come? Or where does the energy go? The energy where the atoms have lost?
AUDIENCE: They absorb a slightly high frequency with a Doppler shift. And then slow down and then re-release a photon with slightly lower energy.
PROFESSOR: Exactly. So in that case, it actually depends which way. When the atom loses energy, the energy goes into blue-shifted photons. I'm just saying, in order to address the question, what happens to the energy, we have to bring in our physics, which we didn't even consider to address here. It is now the physics of the spontaneously emitted photons. We said for the force, spontaneous emission vacuum fluctuations can be eliminated. But if you try to understand where does the energy go, we have to go to those terms, and we even find the physics of Doppler shifts, which we didn't put in, which we didn't even assume here. But, of course, it has to be self consistent.
So therefore, you sometimes have to go to very different physics which isn't even covered by your equation to see the flip side of the coin. One side is forces, which are very simple, but the other side of the coin can be more difficult. We'll actually come back to that when we talk about cooling [INAUDIBLE].
Cooling with a stimulated force. How can a stimulated force cool, because it's all u. Well, we'll get there. And it will again be-- the energy-- will be extracted through spontaneous emission from all those side [INAUDIBLE] and all that. So we have to trace down the energy, and we successfully will do that. But energy can be subtle. Forces are simple. Other questions?
Actually, since I'm in chatty mood, when I explained to you where the energy goes in a dipole trap-- that it's a phase modulation for the laser beam-- I have been using optical dipole traps in my laboratory for many, many years, before I really figured out where the energy goes. And I'm almost 99% certain if you go to DAMOP, and ask some of the experts on cooling and trapping, and ask them to figure out this problem, most of the people will not be able to give you this answer. I've actually not found the answer in any standard textbook of laser cooling, until I eventually found it myself and posted and prepared it as a homework assignment.
So many, many people may not be able to give you the correct answer, where does the energy go when atoms slosh around in a dipole trap. Try it out at DAMOP, it may be fun. OK.
So back to the simple physics of forces. We don't need to understand the energy, because we know the forces, but we'll come back to that. So we have the reactive force. The reactive force is written here. What have we done? Well, you remember the alpha vector was the gradient of the Rabi frequency, and we have to multiply with the u component of the steady state solution of the Optical Bloch Equation. Here it is, just put together. It's a nice expression.
There are two things you should know about it. The first one is that you can actually write this force exactly as the gradient of a potential. So there is a dipole potential, and is exactly this dipole potential which is used as a trapping potential when you have dipole traps. Probably 99% of you use dipole traps in the limit where the detuning is large. And then the logarithm of 1 plus x simply becomes x, and you have the simple expression you are using. But in a way, this is remarkable. The dipole force can be derived from a potential, even if the detuning is small, and you have a hell of a lot of spontaneous scattering, which is usually not regarded as being due to conservative potential.
So this expression, that the reactive force is a gradient of a potential, even applies to a situation where gamma spontaneous scattering [INAUDIBLE] And you're not simply in-- and this is what this last term is-- in the perturbative limit of the AC Stark Effect. OK. That's number one you should know about this expression.
The second thing is if you look at this expression, the question is, what is the maximum reactive force? We talked about the maximum spontaneous, or dissipative force, which is h bar k, the momentum of a photon times gamms over 2. What is the maximum force you can get out of the reactive force? Well, of course, if you want a lot of force, you want to use a lot of laser power. Therefore, you want to use a high Rabi frequency. And now, if you want to use your laser power wisely, the question is, how should you pick the detuning? And you realize if you pick the detuning small, there is a prefector which goes to zero, but if you pick your detuning extremely large, the denominator kills you.
And well, if you analyze it and or stare at this expression for more than a second, you realize that the optimum detuning is when the detuning is on the order of the Rabi frequency. And if you plug that in, you find that under those optimum conditions, the reactive force can be written again as the momentum of a photon-- it must be, the momentum of the photon is the unit of force, the quantum of force, which has to appear-- but not the frequency is not gamma, but it is the Rabi frequency. So the picture you should sort of have is that under those situations, just assume the standing wave consists of two traveling waves, and the atom is Rabi flopping.
But it is Rabi flopping in a way that it goes up by taking a photon from one laser beam, and then it goes down by emitting the photon into the other laser beam. So during each Rabi flux cycle, it exchanges a momentum of 2 h bar k with the electromagnetic field. So therefore, the reactive force-- the stimulated force-- never saturates if you increase the Rabi frequency-- depending, of course, on your detuning, but if you choose the detuning correctly-- you can get a force which is just going further and further. Questions? Yes, Jen?
AUDIENCE: Is this the same concept as the Raman constants, except that the frequencies are just the same? Or is it just a--
PROFESSOR: It is a Raman process. Actually, everything is a Raman process here. Because in laser cooling, in everything which involves the mechanical-- the motion, the external degree freedom of the atoms-- we have an atom in one momentum state. We go-- light scattering always has to involve the excited state-- and then we go down to a different momentum state. So therefore, laser cooling is nothing else than Raman processes in the external degree of freedom of the atom. You may be used to [INAUDIBLE] Raman process more when you start in one hyperfine state, or for molecules in one vibration rotation state, and you go to another one, but, well, for vibration, and rotational states, and hyperfine states of it, Raman is used all the time. I usually used it also for the external motion, because I think it clearly brings out common features between all those different Raman processes.
So you can actually say that the reactive force is a stimulated Raman process between two momentum states, where both legs are stimulated by laser beams. In a standing wave, it would be-- the two legs-- would be stimulated by the two travelling wave components of the standing wave. Whereas the dissipative force is spontaneous Raman process, where one leg is driven by the laser, and the other one comes with spontaneous emission. Nicky?
AUDIENCE: [INAUDIBLE] when the atoms slow down, the [INAUDIBLE]
PROFESSOR: So the spontaneously emitted photon?
AUDIENCE: Yeah, I'm just thinking of actually the stimulated force. Because the atom must be stimulated. [INAUDIBLE] I'm confused how when the energy of the emitted photon must be slightly different due to the phase modulation? [INAUDIBLE] stimulated emission can be added to the frequency? [INAUDIBLE]
PROFESSOR: I mean we had the discussion after [INAUDIBLE] question, if you go up and down, and you are stimulated, you go up in a stimulated way, you go down in a stimulated way, and both laser fields which stimulate the transition have the same frequency, omega, you would actually see that there is no energy exchange. So at our current level of description we have described where the force comes from, but we don't understand yet where the energy goes, or where the energy comes from. And this is really more sophisticated, and I don't want to sort of continue the discussion.
We will encounter one situation, and this is in the [INAUDIBLE] atom picture, where we will discuss in a week or so, sisyphus cooling. And we will find out that there is symmetries in the mono-triplet. So again, we will find the missing energy in spontaneous emission. But if you don't have spontaneous emission, if you just have atoms moving in an optical lattice without spontaneous emission, it is really what I said, the phase modulation. And I would probably ask you at this point to do the homework assignment, and maybe have a discussion afterwards, because then you know exactly what I'm talking about. Does it at least-- You can take that as a preliminary answer.
So these are the two limiting cases. One is spontaneous emission, and the other one is phase modulation at a certain frequency. One can be sustained in steady state, you can spontaneously emit in sort of an atom goes to a standing wave, and this is sustainable. Whereas the phase modulation thing is actually a transient. It's an oscillation which is added in.
AUDIENCE: [INAUDIBLE] You're adding the force to the other one [INAUDIBLE]
PROFESSOR: Yeah, actually--
PROFESSOR: If that helps you, but it's the following. I would sometimes say you know, I like sort of intuitive explanations of quantum physics. Let's assume I'm an atom. I'm in [INAUDIBLE] standing wave. And you don't know it yet, but I do sisyphus cooling. By just exchanging photons, a stimulated force, I'm actually cooling. And you would say, how is it possible? Because all the photon exchanges involve the same photon, the same energy. How can I get rid of energy? And I think what really happens is the following, the atoms can lose momentum without energy, but due to Heisenberg's Uncertainty Relationship this is possible. So the atom is first taking care of its momentum-- is losing momentum-- by stimulated force, and eventually, before it's too late, it has to do some spontaneous emission where it is paying back it debts in energy to Heisenberg.
So and of course, after a certain time, everything is OK. The force was provided by exchanging photons of identical frequency. And the energy is provided by an occasional spontaneous emission event, using one of the [INAUDIBLE]. So everything is there. Everything is a perfect picture, but you have to sort of, in this description, allow a certain uncertainty that the force is the exchange of identical photons, and the energy balance is reconciled in spontaneous emission. And on any longer time scale there are enough events that the balance is perfectly matched.
But if you would take the position, no, this is not possible. You know, the atom can only-- I mean, it's sort of in diagrams. It's not that one diagram which scatters a photon has to conserve energy. You have a little bit of time. You have Heisenberg's uncertainty time to make sure that another diagram jumps in, and reconciles energy conservation. That's at least my way of looking into it, but it's a very maybe, my personal interpretation of how all these diagrams and photon scattering events work together.
OK. Let's do something simpler now. So I've explained to you the dissipative force, the reactive force, and in the next unit I want to show you simply applications of the spontaneous force. In every experiment on cold atoms, the spontaneous force is center stage. It is necessary to slow down atomic beams. It provides molasses, which is the colling of atoms to micro-Kelvin temperature, and the spontaneous force is also responsible for the Magneto-optical trap. So what I want to do here is, again, by showing you the relevant equations, how the spontaneous force, which we have just discussed, how these spontaneous force leads to those three applications. So this is one of the most experimental sections of this course, because this equation has it all in it. And I just want to show you how this equation can be applied to three different important experimental geometries.
OK. So this is our equation. It has the momentum transfer per photon. It has the maximum scatter rate, gamma over two. And then, here, it has the Lorentzian line shape. And the important thing is when we talk about molasses and beam slowing is that the detuning is the laser detuning and the Doppler detuning. So the velocity dependence enters now the spontaneous force through the detuning in the Lorentzian denominator, and it enters through the Doppler effect.
So you can pretty much say it like this, if you have a bunch of laser beam, and slow and cool your atoms, how can the lasers do the job? Well, they measure the velocity of the atoms through the Doppler shift. And if is the Doppler shift which tells the laser beams what to do, so to speak. And this is how laser cooling works. So just as an experimentalist, you should actually know what the scale is. If somebody asked you, how strong is the spontaneous scattering force? Well, a way to connect it with real time units-- with real life units-- is, what is the maximum deceleration? Well, this is 10 to the 5 G. You can ask, is 10 to the 5 G a lot? Or not? Well, for an astronaut, it would be a lot. It would. No living organism can sustain 10 to the 5 G.
But I'm really surprised when I did this calculation. When you compare it to the electric force on an ionized atom, the same force, which is provided by the spontaneous light force, would be provided by an electric field of one millivolt per centimeter. So it's not easy if you have a stainless steel chamber to avoid petch effects, which create those electric fields. And if you've a battery with 9 volt, a centimeter apart-- just a 9 volt battery-- would accelerate an ion four orders of magnitude faster than your beloved strong spontaneous light force. So in that sense, 10 to the 5 G is a lot in the macroscopic world, but if you would look at microscopic forces-- which are often electric forces-- it's absolutely tiny.
And sometimes I say, the fact that you can do 100 kilovolt per centimeter, that you can make electric forces which are seven, eight, or nine, orders of magnitude stronger, that's actually the reason why ion traps were invented before trapping of neutral particles. So some developments in the field of trapping particles-- and eventually laser cooling them-- it first started with ion traps, and then it proceeded to neutral atoms. And the reason is the ion trappers have forces at their disposal which are eight or nine orders of magnitude stronger.
OK. Optical molasses. I know some of you will hate it, because I've just explained to you that in a standing wave, we don't have any spontaneous light force, all we have is the reactive force, the stimulated force, because it is the u, the in phase oscillation of the atomic dipole operator which is responsible for everything. But now I'm just, you know, wearing another hat, and I tell you that there is a limit where the total force can be regarded with some of the two forces. So therefore, I'm pretending now that if you have near resonant light in a standing wave, that the force in the standing wave-- which you know is a purely reactive force-- can now be written as the sum of the two propagating waves. And, of course, each propagating wave does a purely dissipative force.
That is, what I'm telling you can be mathematically proven. That the stimulated force in a standing wave is equal to the sum of the two dissipative forces or each travelling wave. I just keep that in mind whenever you think you can rigorously distinguish between the dissipative force and the stimulated force. Keep in mind that a standing wave-- which has only a stimulated force, as I rigorously proved to you-- can alternatively be described as the sum of two spontaneous light forces, each of which provided by one of the travelling waves. Well, it sort of makes sense in a perturbative limit. You can just take one beam, you can take the other beam, and the combined effect of the two laser beams is higher order. So the fact that in some low intensity limit this has to be valid, is pretty clear.
Anyway, but just a warning, if you really want to use a stimulated force here, you're in big trouble. It's much, much harder to get this result out of the stimulated force. Because to get a stimulated force with velocity dependence requires you to take solutions of the Optical Bloch Equations, which are not steady state, but non-adiabatic. So just a warning. You can do it, if you want with a stimulated force, but I would strongly advise you to first use the simpler formalism. You have to go to very different approximation schemes and much more technical complexity if you want to get it out of the stimulated force.
OK. So with that assumption, you'll remember we had the dissipative force of each laser beam. Now we have two laser beams. We summed them up. Because the two laser beams come from different directions, we have a minus sign here. And for the same reason, we have plus and minus sign in the Doppler effect. So what we have is the following, each force of each laser beam has this Lorentzian envelope, shown in black, but the other force of the other laser beam has the opposite sign. And the two have opposite Doppler shifts. So if I add up the two black forces, I get the red force. And the red force is anti-symmetric with respect to velocity. So therefore, in the limit of small velocities I get a force which is minus alpha times v. And this is the force of friction.
And in a whimsical way, people called this arrangement of two traveling waves Optical Molasses because the atom literally gets stuck in this configuration like a tiny ball bearing you throw into honey. So this is optical molasses. Actually, I think some of the Europeans had to learn the word molasses for it. In the US, everybody knows what molasses is. Well, in Europe, we know what honey is, but molasses is not a standard staple. But anyway, it's optical molasses.
OK. Very simple result. And, again, it's not easy to get it out of the stimulated force. But it's possible. I will actually tell you how we get cooling out of the stimulated force later in some limit.
AUDIENCE: Is there [INAUDIBLE] why this assumption of dissipative forces is easier [INAUDIBLE] and the stimulated force will be made difficult [INAUDIBLE]?
PROFESSOR: I'm not sure if there is a simple answer why it's easier. Well, it can't be easier than that.
AUDIENCE: I mean, there has to be something wrong with our assumptions of stimulated forces so that we cannot get [INAUDIBLE]
PROFESSOR: Yeah. I think what happens is what I briefly said. To the best of my knowledge, but I'm not 100% certain. We have the simply expression because we have used the steady state solution of the Optical Bloch Equation. In other words, we have a laser beam, and it's just the power of the laser which tells us how much light scattering happens. And we use the steady state solution. And we sort of have folded that into the force for one laser beam. And then we have the same package for the other laser beam. And we have never considered that the two laser beam may have some cross talk. And this is indeed valid for all low laser power.
However, if you want to get cooling out off the stimulated force, I will show some of you later, you can no longer use the steady state solution because it would miss the effect here. You wouldn't get any cooling out of it. In order to get something which is dissipative, out of a reactive force-- reactive force is by definition not dissipative-- you need a dissipation mechanism. And the dissipation mechanism is that you're not quite steady state, there is a relaxation time, a time lag. The atom is not instantaneously in its steady state solution. It needs a little bit of time to adjust. And it is these time lag of the atom which eventually gives rise to an alpha coefficient in the stimulated force.
AUDIENCE: So then physically, you are, during cooling experiments, you can take care that you are actually not in the steady state limit?
PROFESSOR: What I'm telling you is, for the spontaneous force, we get in leading order the effect by assuming the steady state limit, but if you want to get it out of the stimulated force, as far as I know, you don't get it with the steady state limit.
PROFESSOR: If you use different approaches-- I mean this is why you want to pick your approach. Sometimes you get something already in lower order. Sometimes you get it in higher order. And I think one favorite example is you solve this problem about [INAUDIBLE] scattering and Thompson scattering, pick your Hamiltonian, d dot e, or p minus a. In one case, you have to work harder than in the other case. And I think here it's similar.
Anyway, let's just put in-- I think that's, yeah, we can do in the next few minutes. So we now want to put in the effect that the force fluctuates. And that means we have heating. Before I do that, let me just tell you what the solution is for force equals minus alpha times v. Well, it means we extract energy out of the system at a rate-- Well, energy per unit time is force times velocity. Force times velocity is now minus alpha times v square. But v square is the energy, the kinetic energy. In other words, this equation tells us that the atomic motion, the kinetic energy, is exponentially [? damned. ?] And if there were nothing else, you just need two laser beams-- optical molasses-- and you would go not just to micro-Kelvin, but to nano- and pico- Kelvin temperatures. It's an exponential decay to absolute zero.
However, we don't each reach nano- and pico- Kelvin temperatures in laser cooling because there are other processes. And what is important here is spontaneous emission. OK. So the way you treat spontaneous emission is the following, every time an atom emits spontaneously, there is a random momentum kick of h bar k. If you have n photons scattered because the momentum kicks going random direction, they only add up in a random box. You get square root N. Or if you ask, what is the average of p square due to spontaneously emission? It is the momentum of the photon squared times the number of scattering events.
So therefore, in the form of a differential equation, the heating rate, or the derivative of-- the TEMPO derivative-- of p square goes now with the number of photons per unit time, which is the scattering rate. OK. If you would stop here, and that's what many people do who explain heating in this situation, you would miss half of the heating, because what you have treated here is only the photon transfer and spontaneous emission. However, there is also fluctuation and absorption.
I mean just look at two atoms in the [INAUDIBLE]. Kind of me and another atom. And we both scatter photons. On average, in the same laser beam, we absorb in photons and get in momentum kicks. But there is Poissonian statistics how many photons I absorb, and how many photons my twin brother absorbs. So therefore, due to the randomness in absorption, or the fluctuations in absorption. There is another square root n variance in the recoil kicks which comes from the absorption process. And it so happens, for exactly that reason, that the heating-- the derivative of kinetic energy, or the difference of momentum squared due to absorption is exactly the same as in emission.
Well, if you now make different assumptions, spontaneous emission has a dipole pattern. And you can sort of factor on the order of unity, depending what the pattern of the spontaneous emission is, whether you make a 1D or 2D model of spontaneous emission that the photons can only go in one dimension, in two dimension, or three dimensions. So you have to get other prefectors, but the picture is that without going into numerical factors on the order of unity, you have fluctuations in the absorption, you have a randomness in spontaneous emission, and they both equally contribute.
So that means now the following, that we have heating rate. The heating rate we just talked about. The increase in p square. Well, if you divide by 2 times the mass, it's increasing kinetic energy. So the increase in kinetic energy is given by this expression. And it is common if you have a heating process to introduce a momentum diffusion coefficient in that way. It's just the definition. It shows you how p square increases per unit time. And now we can get the cooling limit for spontaneous emission, namely by saying, in steady state when we have-- due to photon scattering-- the same amount of heating and the same amount of cooling, the temperature will have asymptotically reached a steady state value.
The heating rate was parametrized by momentum diffusion coefficient, d over m. So this is sort of independent of the velocity of the atom. Whereas you'll remember the cooling rate was proportional to the energy because it was an exponential approach to zero energy. So therefore, when we have the heating equals the cooling, we have the energy of the atoms there, and therefore, we find that the energy of the atoms in steady state is given by, actually, the ratio of heating versus cooling. It's a simple expression. The energy or the limiting temperature for molasses is the ratio of heating versus cooling. Heating is described by momentum diffusion coefficient and alpha is described by the damping force.
I want you, actually, to keep this expression in mind. A limit in temperature in laser cooling is always-- actually, in other processes in laser cooling-- is the ratio of heating over cooling. And momentum diffusion coefficient due to heating, over a friction force due to cooling. Because we will later find polarization gradient cooling, cooling in blue molasses, we will find other cooling schemes, where we will calculate-- with the appropriate model-- the heating and the cooling. And I don't have to repeat this part. The moment I calculate the heating, the momentum diffusion, and I calculate the friction, I know what the limiting temperatures is.
Again, when I said, the energy is kt over 2, you know, of course, kinetic energy is kt over 2 times the number of degrees of freedom. I assumed 1D, here. So in everything I've said on this page, there are numerical factors which may change whether you assume one or two or three dimension.
OK. I think that's the last thing I want to tell you. And then on Wednesday we do Zeeman's slowing and Magneto-optical trapping.
We had an expression for alpha. Remember we had this Lorentz profile, for the other Lorentz profile, and alpha was just the slope. I mean, everything is known. I just didn't bother to calculate it. But it just involves derivative of Lorentzian. And now we can ask, what is the lowest temperature we can reach? Well, you can now analyze your expression for alpha for a two-level system solved by the Optical Bloch Equation. And you find that you will have the most favorable conditions for low temperature-- you get the minimum possible temperature-- in the limit of low laser power.
And your detuning-- your optimum detuning-- is half a line width away. And then you'll find this famous result, that if you cool a two-level atom, the limiting temperature is simply given by the spontaneous emission rate, or the line width of your transition, gamma. And this is the famous result for the Doppler limit. For sodium atoms, it's 250 micro-Kelvin, for rubidium and cesium, because of the heavier mass, it's lower, is 10's of micro-Kelvin, 50 or 100 micro-Kelvin. So this is a famous limit.
And the physics of it is the following, the narrower the line width is. Then you can say the better you can cool particle down to zero temperature-- and I think, Jenny, your idea of the Raman process comes now in very handy-- I can have an atom, and I can have a Raman process where I scatter photons and go to higher energy or I scatter photons and got to lower energy. And the difference between the two processes is actually, the one that whether I cool or whether I heat comes from the Doppler effect. So the more I can discriminate between through the Doppler effect, the better I can cool. But the Doppler effect-- the Doppler shift, kv-- how well I can resolve it, depends on the width of the atomic transition.
The narrower the atomic transition, the more the Doppler effect can steer the laser cooling-- the Raman process-- towards lower velocity, not towards higher velocity. So that's the natural result that the-- or an expected result-- that the natural line width appears here.
OK. I've talked a few extra minutes. Is there any question? Then I see you on Wednesday.
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